| in the coordinate plane points ( x num__1 ) and ( num__5 y ) are on line k . if line k passes through the origin and has slope num__0.2 then what are the values of x and y respectively ? <o> a ) num__4 and num__1 <o> b ) num__1 and num__5 <o> c ) num__5 and num__1 <o> d ) num__3 and num__5 <o> e ) num__5 and num__3 |
line k passes through the origin and has slope num__0.2 means that its equation is y = num__0.2 * x . thus : ( x num__1 ) = ( num__5 num__1 ) and ( num__5 y ) = ( num__51 ) - - > x = num__5 and y = num__1 answer : c <eor> c <eos> |
c |
reverse__0.2__ |
multiply__1.0__5.0__ |
| carl is facing very difficult financial times and can only pay the interest on a $ num__10000 loan he has taken . the bank charges him a quarterly compound rate of num__4.0 . what is the approximate interest he pays annually ? <o> a ) $ num__1600 <o> b ) $ num__2000 <o> c ) $ num__2150 <o> d ) $ num__2500 <o> e ) $ num__12000 |
usually you are given the annual rate of interest and it is mentioned that it is annual rate . the bank charges him a quarterly compounded annual rate of num__16.0 . here you find per quarter rate as ( num__4.0 ) % = num__4.0 i have actually never seen a question with quarter rate given but since this question did not mentionannual rate of interestand since the options did not make sense with num__4.0 annual rate of interest it is apparent that the intent was a num__4.0 quarterly rate . so the bank charges num__4.0 every quarter and compounds it in the next quarter . had it been a simple quarterly rate we would have just found num__4 * num__4.0 of num__10000 = $ num__1600 as our answer . but since the interest is compounded it will be a bit more than $ num__1600 . option ( a ) looks correct . <eor> a <eos> |
a |
percent__16.0__10000.0__ percent__16.0__10000.0__ |
percent__16.0__10000.0__ percent__16.0__10000.0__ |
| the speed at which a man can row a boat in still water is num__25 kmph . if he rows downstream where the speed of current is num__11 kmph what time will he take to cover num__80 metres ? <o> a ) num__18 seconds <o> b ) num__27 seconds <o> c ) num__26 seconds <o> d ) num__12 seconds <o> e ) num__8 seconds |
speed of the boat downstream = num__25 + num__11 = num__36 kmph = num__36 * num__0.277777777778 = num__10 m / s hence time taken to cover num__80 m = num__8.0 = num__8 seconds . answer : e <eor> e <eos> |
e |
add__25.0__11.0__ divide__80.0__10.0__ round__8.0__ |
add__25.0__11.0__ divide__80.0__10.0__ divide__80.0__10.0__ |
| if ( x ^ num__2 + num__4 x - num__11 ) / num__5 ≤ x + num__1 then x could be represented by which of the following ? <o> a ) − num__3 ≤ x ≤ num__4 <o> b ) − num__4 ≤ x ≤ num__3 <o> c ) − num__3 ≤ x ≤ num__3 <o> d ) − num__4 ≤ x ≤ − num__3 <o> e ) num__3 ≤ x ≤ num__4 |
imo a is correct answer solving through eqautions x ^ num__2 + num__4 x - num__11 < = num__5 x + num__5 ( x + num__3 ) ( x - num__4 ) < = num__0 <eor> a <eos> |
a |
add__2.0__1.0__ add__2.0__1.0__ |
add__2.0__1.0__ add__2.0__1.0__ |
| the entrance fee for a fair is $ num__5 for persons under the age of num__18 and num__20.0 more for persons older . each ride at the fair costs $ num__0.50 . if joe goes with her num__6 years old twin brothers and they each took num__3 rides in total . how much money does joe end up spending at the fair ? <o> a ) num__16 <o> b ) num__20.5 <o> c ) num__17.5 <o> d ) num__20 <o> e ) num__4.5 |
total entrance fee is ( num__2 * $ num__5 ) + ( num__1.20 * num__5 ) = $ num__16 total rides fee is ( num__0.50 * num__3 ) * num__3 = $ num__4.50 total money spent is $ num__20.50 answer is b <eor> b <eos> |
b |
reverse__0.5__ divide__6.0__5.0__ subtract__18.0__2.0__ subtract__5.0__0.5__ add__20.0__0.5__ add__20.0__0.5__ |
reverse__0.5__ divide__6.0__5.0__ subtract__18.0__2.0__ subtract__5.0__0.5__ add__20.0__0.5__ add__20.0__0.5__ |
| rs . num__825 becomes rs . num__956 in num__3 years at a certain rate of simple interest . if the rate of interest is increased by num__4.0 what amount will rs . num__825 become in num__3 years ? <o> a ) rs . num__1020.80 <o> b ) rs . num__1025 <o> c ) rs . num__1055 <o> d ) data inadequate <o> e ) none of these |
solution s . i . = rs . ( num__956 - num__825 ) = rs . num__131 rate = ( num__100 x num__0.158787878788 x num__3 ) = num__5.29292929293 % new rate = ( num__5.29292929293 + num__4 ) % = num__9.29292929293 % new s . i . = rs . ( num__825 x num__9.29292929293 x num__0.03 ) rs . num__230 . ∴ new amount = rs . ( num__825 + num__230 ) = rs . num__1055 . answer c <eor> c <eos> |
c |
percent__100.0__1055.0__ |
percent__100.0__1055.0__ |
| if q is the square of a positive integer which of the following must be equal to the square of the next positive integer ? <o> a ) √ n + num__1 <o> b ) n + num__1 <o> c ) n ^ num__2 + num__1 <o> d ) q + num__2 √ q + num__1 <o> e ) n ^ num__2 + num__2 n + num__1 |
if q is the square of a positive integer which of the following must be equal to the square of the next positive integer ? q = ( x ) ^ num__2 where x is a positive integer to calculate - ( x + num__1 ) ^ num__2 = x ^ num__2 + num__2 x + num__1 root ( q ) = x ans - q + num__2 root ( q ) + num__1 this should be d <eor> d <eos> |
d |
multiply__1.0__2.0__ |
multiply__1.0__2.0__ |
| rs . num__5600 is divided into three parts a b and c . how much a is more than c if their ratio is num__0.142857142857 : num__0.142857142857 : num__0.0714285714286 ? <o> a ) num__300 <o> b ) num__992 <o> c ) num__1120 <o> d ) num__552 <o> e ) num__312 |
num__0.142857142857 : num__0.142857142857 : num__0.0714285714286 = num__2 : num__2 : num__1 num__0.2 * num__5600 = num__1120 num__2240 - num__1120 = num__1120 answer : c <eor> c <eos> |
c |
multiply__5600.0__0.2__ multiply__2.0__1120.0__ multiply__5600.0__0.2__ |
multiply__5600.0__0.2__ multiply__2.0__1120.0__ multiply__5600.0__0.2__ |
| if a / b = num__0.75 and num__8 a + num__5 b = num__22 then find the value of a . <o> a ) num__0.5 <o> b ) num__1.5 <o> c ) num__2.5 <o> d ) num__2.0 <o> e ) num__3.5 |
( a / b ) = num__0.75 b = ( num__1.33333333333 ) a . therefore num__8 a + num__5 b = num__22 = num__8 a + num__5 * ( num__1.33333333333 ) a = num__22 num__8 a + ( num__6.66666666667 ) a = num__22 = num__44 a = num__66 = a = ( num__1.5 ) = num__1.5 answer is b . <eor> b <eos> |
b |
reverse__0.75__ subtract__8.0__1.3333__ add__22.0__44.0__ divide__66.0__44.0__ divide__66.0__44.0__ |
reverse__0.75__ divide__5.0__0.75__ add__22.0__44.0__ divide__66.0__44.0__ divide__66.0__44.0__ |
| mike took num__5 mock tests before appearing for the gmat . in each mock test he scored num__10 points more than the previous mock test . if he scored num__760 on the gmat and his average score for the mocks and the gmat was num__716.67 what was the difference in the score of his last mock and his gmat score ? <o> a ) num__20 <o> b ) num__32 <o> c ) num__40 <o> d ) num__50 <o> e ) num__60 |
one way to do this would be weighted average method . . num__1 ) let the average of num__5 mocks be x . . . so take it as a mix of num__5 quantitites of xand num__1 quantity of num__760 resulting in an average of num__716.67 . . . by alligation / weighted average . . the difference in num__760 and num__716.67 is num__0.833333333333 of difference of num__760 and x . . num__760 - num__716.67 = num__0.833333333333 * ( num__760 - x ) num__760 - x = num__43.33 * num__1.2 = num__52 . . . so x = num__708 . . the last of mock test will be num__708 + num__10 + num__10 = num__728 . . . so ans = num__760 - num__728 = num__32 b <eor> b <eos> |
b |
subtract__760.0__716.67__ reverse__0.8333__ subtract__760.0__52.0__ subtract__760.0__728.0__ subtract__760.0__728.0__ |
subtract__760.0__716.67__ reverse__0.8333__ subtract__760.0__52.0__ subtract__760.0__728.0__ subtract__760.0__728.0__ |
| john found that the average of num__15 numbers is num__40 . if num__10 is added to each number then the mean of number is ? <o> a ) num__50 <o> b ) num__45 <o> c ) num__65 <o> d ) num__78 <o> e ) num__64 |
( x + x num__1 + . . . x num__14 ) / num__15 = num__40 num__50 option a <eor> a <eos> |
a |
subtract__15.0__1.0__ add__40.0__10.0__ add__40.0__10.0__ |
subtract__15.0__1.0__ add__40.0__10.0__ add__40.0__10.0__ |
| a person is traveling at num__20 km / hr and reached his destiny in num__2.5 hr then find the distance ? <o> a ) num__53 km <o> b ) num__55 km <o> c ) num__52 km <o> d ) num__60 km <o> e ) num__50 km |
t = num__2.5 hrs = num__2.5 hrs d = t * s = num__20 * num__2.5 = num__50 km answer is e <eor> e <eos> |
e |
multiply__20.0__2.5__ round__50.0__ |
multiply__20.0__2.5__ multiply__20.0__2.5__ |
| tim has num__350 pounds of cement in num__100 num__50 and num__25 pound bags . he has an equal number of each size bag . how many bags of cement does tim have ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
let the number of each sized bag = c num__100 c + num__50 c + num__25 c = num__350 num__175 c = num__350 c = num__2 therefore tim has num__3 c = num__6 bags of cement answer : c <eor> c <eos> |
c |
divide__350.0__175.0__ multiply__2.0__3.0__ multiply__2.0__3.0__ |
divide__350.0__175.0__ multiply__2.0__3.0__ multiply__2.0__3.0__ |
| what is the least value of x so that num__2 x num__5475 is divisible by num__9 <o> a ) num__7 <o> b ) num__8 <o> c ) num__4 <o> d ) num__3 <o> e ) num__2 |
explanation : the sum of the digits of the number is divisible by num__9 . then the number is divisible by num__9 . num__2 + x + num__5 + num__4 + num__7 + num__5 = num__23 + x least value of x may be ' num__4 ' so that the total num__23 + num__4 = num__27 is divisible by num__9 . answer : option c <eor> c <eos> |
c |
subtract__9.0__5.0__ add__2.0__5.0__ add__4.0__23.0__ subtract__9.0__5.0__ |
subtract__9.0__5.0__ add__2.0__5.0__ add__4.0__23.0__ subtract__9.0__5.0__ |
| square p is inscribed in circle q . if the perimeter of p is num__40 what is the circumference of q ? <o> a ) num__11 √ num__2 π <o> b ) num__10 √ num__2 π <o> c ) num__9 √ num__2 π <o> d ) num__8 √ num__2 π <o> e ) num__7 √ num__2 π |
square forms two right angled triangles . any time we have a right angle triangle inside a circle the hypotenuse is the diameter . hypotenuse here = diagonal of the square = num__10 sqrt ( num__2 ) = diameter = > radius = num__5 sqrt ( num__2 ) circumference of the circle = num__2 pi r = num__10 pi sqrt ( num__2 ) answer is b . <eor> b <eos> |
b |
triangle_area__10.0__2.0__ |
triangle_area__10.0__2.0__ |
| a school currently maintains a fixed number of students per class . if the ratio of students per class were to be increased by num__1 num__10 fewer classes would be run for a total of num__120 students . what is the current ratio q of students per class ? <o> a ) q = num__3 <o> b ) q = num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__12 |
another way to look at the problem . . . since the total is num__120 ratio * classes = r * c = num__120 . . . . . ( i ) we are looking where ratio increases by num__1 and # of classes decreases by num__10 = ( r + num__1 ) ( c - num__10 ) = rc + c - num__10 r - num__10 = num__120 . . . . ( ii ) ( ii ) - ( i ) . . . . c = num__10 r + num__10 = num__10 ( r + num__1 ) . . . . . . . so # of classes has to be multiple of num__10 as rc = num__120 . . . . num__10 ( r + num__1 ) * r = num__120 . . . . . . . . . . . . . . . . . . . r ( r + num__1 ) = num__12 . . so num__12 is a multiple of consecutive numbers only num__3 * num__4 fits in . . . . . and r = num__3 a <eor> a <eos> |
a |
divide__120.0__10.0__ add__1.0__3.0__ multiply__1.0__3.0__ |
divide__120.0__10.0__ add__1.0__3.0__ multiply__1.0__3.0__ |
| a cheese factory sells its cheese in rectangular blocks . a normal block has a volume of num__8 cubic feet . if a small block has half the width half the depth and half the length of a normal block what is the volume of cheese in a small block in cubic feet ? <o> a ) num__1 <o> b ) num__5 <o> c ) num__4 <o> d ) num__6 <o> e ) num__2 |
volume of cube = lbh = num__8 new cube l b h are decreases of . num__5 l . num__5 b . num__5 h new volume of cube = . num__5 l * . num__5 b * . num__5 h = . num__125 * lbh = . num__125 * num__8 = num__1 answer : a <eor> a <eos> |
a |
volume_cube__5.0__ volume_cube__1.0__ |
volume_cube__5.0__ volume_cube__1.0__ |
| in a new housing development trees are to be planted along the sidewalk of a certain street . each tree takes up one square foot of sidewalk space and there are to be num__20 feet between each tree . how many trees can be planted if the road is num__148 feet long ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__16 |
let t be the number of trees . then the length required for trees on the sidewalk will be num__1 * t = t to maximize the number of trees the number of num__20 feet spaces between trees should be num__1 less than total number of trees . for example if there are num__3 trees then there should be num__2 spaces between them . so the number of num__20 feet spaces will be t - num__1 . then the length of sidewalk required for num__20 feet spaces will be num__20 * ( t - num__1 ) it is given that total length of sidewalk is num__148 feet . or num__20 ( t - num__1 ) + t = num__148 or num__20 t - num__20 + t = num__148 or num__21 t = num__168 or t = num__8 answer : - a <eor> a <eos> |
a |
square_perimeter__2.0__ square_perimeter__2.0__ |
square_perimeter__2.0__ square_perimeter__2.0__ |
| what will be the difference in simple and compound interest on num__2000 after three years at the rate of num__10 percent per annum ? <o> a ) num__160 <o> b ) num__42 <o> c ) num__62 <o> d ) num__20 <o> e ) none of these |
for num__3 years : diff . = sum × ( rate ) num__2 ( num__300 + rate ) / ( num__100 ) num__3 = num__2000 × num__10 × num__10 × num__3.1 × num__100 × num__100 = num__62 answer c <eor> c <eos> |
c |
percent__3.1__2000.0__ percent__100.0__62.0__ |
percent__3.1__2000.0__ percent__100.0__62.0__ |
| find the constant k so that : - x num__2 - ( k + num__11 ) x - num__8 = - ( x - num__2 ) ( x - num__4 ) <o> a ) num__11 <o> b ) num__12 <o> c ) num__17 <o> d ) num__14 <o> e ) num__15 |
- x num__2 - ( k + num__11 ) x - num__8 = - ( x - num__2 ) ( x - num__4 ) : given - x num__2 - ( k + num__11 ) x - num__8 = - x num__2 + num__6 x - num__8 - ( k + num__11 ) = num__6 : two polynomials are equal if their corresponding coefficients are equal . k = - num__17 : solve the above for k correct answer c <eor> c <eos> |
c |
add__2.0__4.0__ add__11.0__6.0__ add__11.0__6.0__ |
add__2.0__4.0__ add__11.0__6.0__ add__11.0__6.0__ |
| krishan and nandan jointly started a business . krishan invested six times as nandan did and invested his money for double time as compared to nandan . nandan earned rs . num__6000 . if the gain is proportional to the money invested and the time for which the money is invested then the total gain was ? <o> a ) rs . num__78000 <o> b ) rs . num__48000 <o> c ) rs . num__6000 <o> d ) rs . num__82000 <o> e ) rs . num__32000 |
num__6 : num__1 num__2 : num__1 - - - - - - num__12 : num__1 num__1 - - - - - num__6000 num__13 - - - - - ? = > rs . num__78000 answer : a <eor> a <eos> |
a |
multiply__2.0__6.0__ add__1.0__12.0__ multiply__6000.0__13.0__ multiply__6000.0__13.0__ |
multiply__2.0__6.0__ add__1.0__12.0__ multiply__6000.0__13.0__ multiply__6000.0__13.0__ |
| there are num__8 players in a chess group and each player plays each of the others once . given that each game is played by two players how many total games will be played ? <o> a ) num__10 <o> b ) num__30 <o> c ) num__28 <o> d ) num__60 <o> e ) num__90 |
num__10 players are there . two players play one game with one another . so num__8 c num__2 = num__8 * num__3.5 = num__28 so option c is correct <eor> c <eos> |
c |
subtract__10.0__8.0__ multiply__8.0__3.5__ multiply__8.0__3.5__ |
subtract__10.0__8.0__ multiply__8.0__3.5__ multiply__8.0__3.5__ |
| a pipe takes a hours to fill the tank . but because of a leakage it took num__2 times of its original time . find the time taken by the leakage to empty the tank <o> a ) num__50 min <o> b ) num__60 min <o> c ) num__90 min <o> d ) num__80 min <o> e ) num__120 min |
pipe a can do a work num__60 min . lets leakage time is x ; then num__0.0166666666667 - num__1 / x = num__0.00833333333333 x = num__120 min answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ divide__0.0167__2.0__ multiply__2.0__60.0__ round__120.0__ |
hour_to_min_conversion__ divide__0.0167__2.0__ multiply__2.0__60.0__ divide__120.0__1.0__ |
| suppose a b and c are positive integers with a < b < c such that num__1 / a + num__1 / b + num__1 / c = num__1 . what is a + b + c ? <o> a ) num__1 <o> b ) num__4 <o> c ) num__9 <o> d ) num__11 <o> e ) no such integers exist |
first note that we must have num__1 / a < num__1 so a > num__1 . since num__1 / a > num__1 / b > num__1 / c we must also have num__1 / a > num__0.333333333333 ; so a < num__3 . thus a / num__2 . now num__1 / b + num__1 / c = num__0.5 where num__2 < b < c . similar to before num__1 / b > num__0.25 so b < num__4 . thus b = num__3 . with a = num__2 and b = num__3 we have num__0.5 + num__0.333333333333 + num__1 / c = num__1 which is satisfied when c = num__6 . to conclude a + b + c = num__2 + num__3 + num__6 = num__11 . correct answer d <eor> d <eos> |
d |
subtract__3.0__1.0__ reverse__2.0__ divide__0.5__2.0__ reverse__0.25__ multiply__2.0__3.0__ multiply__1.0__11.0__ |
subtract__3.0__1.0__ reverse__2.0__ divide__0.5__2.0__ reverse__0.25__ add__2.0__4.0__ divide__11.0__1.0__ |
| the age of man is three times the sum of the ages of his two sons . four years hence his age will be double of the sum of the ages of his sons . the father â € ™ s present age is : <o> a ) num__36 years <o> b ) num__45 years <o> c ) num__50 years <o> d ) num__55 years <o> e ) num__75 years |
solution let the sum of present ages of the two sons be x years . then father ' s present age = num__3 x years . â ˆ ´ ( num__3 x + num__4 ) = num__2 ( x + num__8 ) â ‡ ” num__3 x + num__4 = num__2 x + num__16 â ‡ ” x = num__12 . hence father ' s present age = num__36 years . answer a <eor> a <eos> |
a |
multiply__2.0__4.0__ multiply__2.0__8.0__ multiply__3.0__4.0__ multiply__3.0__12.0__ multiply__3.0__12.0__ |
multiply__2.0__4.0__ multiply__2.0__8.0__ add__4.0__8.0__ multiply__3.0__12.0__ multiply__3.0__12.0__ |
| a bottle of coke contains num__200 gm in place of num__1 kg of fluid . find the actual % difference given a num__10.0 gain on initial fluid ? <o> a ) num__32.5 <o> b ) num__112.5 <o> c ) num__35.0 <o> d ) num__40.0 <o> e ) num__50 % |
fluid price of num__200 gm = num__100 + num__10 = num__110 difference = num__110 - num__20 = num__90.0 of difference = num__90 * num__1.25 = num__112.5 answer is b <eor> b <eos> |
b |
percent__10.0__200.0__ percent__100.0__112.5__ |
percent__10.0__200.0__ percent__100.0__112.5__ |
| a person can walk at a constant rate of num__8 mph and can bike at a rate of num__16 mph . if he wants to travel num__64 miles in num__8 hours using bike and walking at their constant rates how much distance would he require to walk ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__48 <o> d ) num__64 <o> e ) num__72 |
total distance = num__64 distance = speed * time walking speed = s num__1 = num__8 walking time = t num__1 bike speed = s num__2 = num__16 time traveled in bike = t num__2 d num__1 + d num__2 = num__64 s num__1 t num__1 + s num__2 t num__2 = num__64 num__8 * t num__1 + num__16 * t num__2 = num__64 t num__1 + num__2 * t num__2 = num__8 - - - - - ( num__1 ) given : t num__1 + t num__2 = num__8 - - - - - ( num__2 ) ( num__1 ) - ( num__2 ) - - > t num__2 = num__0 and t num__1 = num__8 - num__0 = num__8 walking distance = s num__1 * t num__1 = num__8 * num__8 = num__64 answer : d <eor> d <eos> |
d |
divide__16.0__8.0__ round__64.0__ |
divide__16.0__8.0__ round__64.0__ |
| num__5358 x num__51 = ? <o> a ) num__273258 <o> b ) num__273268 <o> c ) num__273348 <o> d ) num__273358 <o> e ) none of these |
explanation : num__5358 x num__51 = num__5358 x ( num__50 + num__1 ) = num__5358 x num__50 + num__5358 x num__1 = num__267900 + num__5358 = num__273258 . answer is a <eor> a <eos> |
a |
subtract__51.0__50.0__ multiply__5358.0__50.0__ multiply__5358.0__51.0__ multiply__5358.0__51.0__ |
subtract__51.0__50.0__ multiply__5358.0__50.0__ add__5358.0__267900.0__ add__5358.0__267900.0__ |
| | x + num__3 | – | num__4 - x | = | num__8 + x | how many s solutions will this equation have ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
| x | = x when x > = num__0 ( x is either positive or num__0 ) | x | = - x when x < num__0 ( note here that you can put the equal to sign here as well x < = num__0 because if x = num__0 | num__0 | = num__0 = - num__0 ( all are the same ) so the ' = ' sign can be put with x > num__0 or with x < num__0 . we usually put it with ' x > num__0 ' for consistency . a <eor> a <eos> |
a |
multiply__3.0__0.0__ |
multiply__3.0__0.0__ |
| the output of a factory was increased by num__10.0 to keep up with rising demand . to handle the holiday rush this new output was increased by num__20.0 . by approximately what percent would the output now have to be decreased in order to restore the original output ? <o> a ) num__20.0 <o> b ) num__24.0 <o> c ) num__30.0 <o> d ) num__32.0 <o> e ) num__79 % |
let initial output is o then after num__10.0 increase it will be num__1.1 o and after num__20.0 increase on this new output the latest output will be num__1.1 o * num__1.20 = num__1.32 o now we have to decrease the output by some percentage so that the new output is same as the starting output ( o ) so num__1.32 o * ( num__1 - x / num__100 ) = o = > x = num__24.24 so answer will be b <eor> b <eos> |
b |
multiply__1.1__1.2__ round_down__1.1__ round_down__24.24__ |
multiply__1.1__1.2__ round_down__1.1__ multiply__20.0__1.2__ |
| in a graduate physics course num__70 percent of the students are male and num__30 percent of the students are married . if two - sevenths of the male students are married what fraction of the male students is single ? <o> a ) num__0.285714285714 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__0.714285714286 |
let assume there are num__100 students of which num__70 are male and num__30 are females if num__30 are married then num__70 will be single . now its given that two - sevenths of the male students are married that means num__0.285714285714 of num__70 = num__20 males are married if num__30 is the total number of students who are married and out of that num__20 are males then the remaining num__10 will be females who are married . total females = num__70 married males = num__20 then single males = num__70 - num__20 = num__50 we need to find the fraction of male students who are single i . e single male students / total male student = num__0.714285714286 = num__0.714285714286 [ e ] <eor> e <eos> |
e |
percent__100.0__0.7143__ |
percent__100.0__0.7143__ |
| the cost of painting the whole surface area of a cube at the rate of num__13 paise per sq . cm is rs . num__343.98 . then the volume of the cube is <o> a ) num__8500 cm num__3 <o> b ) num__9000 cm num__3 <o> c ) num__9250 cm num__3 <o> d ) num__9261 cm num__3 <o> e ) none |
solution surface area = ( num__2646.0 ) ‹ = › num__2646 cm num__3 ‹ = › num__6 a num__2 = num__2646 ‹ = › a num__2 = num__441 ‹ = › a = num__21 . so volume = ( num__21 x num__21 x num__21 ) cm num__3 = num__9261 cm num__3 . answer d <eor> d <eos> |
d |
triangle_perimeter__13.0__2.0__6.0__ volume_cube__21.0__ volume_cube__21.0__ |
triangle_perimeter__13.0__2.0__6.0__ volume_cube__21.0__ volume_cube__21.0__ |
| a man sold num__20 articles for $ num__60 and gained num__20.0 . how many articles should he sell for $ num__50 to incur a loss num__20.0 ? <o> a ) num__25 <o> b ) num__36 <o> c ) num__40 <o> d ) num__50 <o> e ) num__48 |
production cost per article : $ num__60 * ( num__100.0 - num__20.0 ) / num__20 = $ num__2.40 required production costs for a loss of num__20.0 : $ num__50 * ( num__100.0 + num__20.0 ) = $ num__60 number of articles to be sold for $ num__60 to incur a num__20.0 loss : $ num__60 / $ num__2.40 = num__25 thus solution a is correct . <eor> a <eos> |
a |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| a man invests some money partly in num__9.0 stock at num__96 and partly in num__12.0 stock at num__120 . to obtain equal dividends from both he must invest the money in the ratio : <o> a ) num__3 : num__5 <o> b ) num__2 : num__1 <o> c ) num__16 : num__15 <o> d ) num__4 : num__5 <o> e ) none of these |
explanation : for an income of re . num__1 in num__9.0 stock at num__96 investment = rs . num__10.6666666667 = rs . num__10.6666666667 for an income re . num__1 in num__12.0 stock at num__120 investment = rs . num__10.0 = rs . num__10 . ratio of investments = ( num__10.6666666667 ) : num__10 = num__32 : num__30 = num__16 : num__15 . answer : c <eor> c <eos> |
c |
divide__96.0__9.0__ round_down__10.6667__ subtract__16.0__1.0__ multiply__1.0__16.0__ |
divide__96.0__9.0__ round_down__10.6667__ subtract__16.0__1.0__ multiply__1.0__16.0__ |
| a train num__500 m long can cross an electric pole in num__20 sec and then find the speed of the train ? <o> a ) num__95 kmph <o> b ) num__90 kmph <o> c ) num__92 kmph <o> d ) num__95 kmph <o> e ) num__98 kmph |
length = speed * time speed = l / t s = num__25.0 s = num__25 m / sec speed = num__25 * num__3.6 ( to convert m / sec in to kmph multiply by num__3.6 ) speed = num__90 kmph answer : b <eor> b <eos> |
b |
divide__500.0__20.0__ multiply__25.0__3.6__ round__90.0__ |
divide__500.0__20.0__ multiply__25.0__3.6__ multiply__25.0__3.6__ |
| the smallest number when increased by ` ` num__1 ` ` is exactly divisible by num__6 num__915 num__3545 is : <o> a ) num__631 <o> b ) num__630 <o> c ) num__359 <o> d ) num__629 <o> e ) num__600 |
lcm = num__630 num__630 - num__1 = num__629 answer : d <eor> d <eos> |
d |
subtract__630.0__1.0__ multiply__1.0__629.0__ |
subtract__630.0__1.0__ subtract__630.0__1.0__ |
| how many keystrokes are needed to type the numbers from num__1 to num__500 ? <o> a ) num__1156 <o> b ) num__1392 <o> c ) num__1480 <o> d ) num__1562 <o> e ) num__1788 |
there are num__9 one - digit numbers from num__1 to num__9 . there are num__90 two - digit numbers from num__10 to num__99 . there are num__401 three - digit numbers from num__100 to num__500 . num__9 + num__90 ( num__2 ) + num__401 ( num__3 ) = num__1392 the answer is b . <eor> b <eos> |
b |
add__1.0__9.0__ add__9.0__90.0__ subtract__500.0__99.0__ add__1.0__99.0__ add__1.0__2.0__ multiply__1.0__1392.0__ |
add__1.0__9.0__ add__9.0__90.0__ subtract__500.0__99.0__ add__1.0__99.0__ add__1.0__2.0__ multiply__1.0__1392.0__ |
| the edge of a cube is num__7 a cm . find its surface ? <o> a ) num__24 a num__8 <o> b ) num__24 a num__4 <o> c ) num__24 a num__1 <o> d ) num__24 a num__2 <o> e ) num__294 a num__2 |
num__6 a num__2 = num__6 * num__7 a * num__7 a = num__294 a num__2 answer : e <eor> e <eos> |
e |
surface_cube__7.0__ surface_cube__7.0__ |
surface_cube__7.0__ surface_cube__7.0__ |
| a shop sells chocolates it is used to sell chocolates for rs . num__2 each but there were no sales at that price . when it reduced the price all the chocolates sold out enabling the shopkeeper to realize rs num__164.90 from the chocolates alone if the new price was not less than half the original price quoted how many chocolates were sold ? <o> a ) num__1.9 <o> b ) num__1.7 <o> c ) num__1.2 <o> d ) num__1.5 <o> e ) num__1.6 |
num__16490 = num__2 × num__5 × num__17 × num__97 now now chocolate price should be greater than num__1 and less than num__2 . so num__2 x num__5 x num__17 = num__170 so total chocolates sold = num__97 and new chocolate price = rs . num__1.7 answer : b <eor> b <eos> |
b |
divide__16490.0__97.0__ divide__164.9__97.0__ divide__164.9__97.0__ |
divide__16490.0__97.0__ divide__164.9__97.0__ divide__164.9__97.0__ |
| a board num__7 ft . num__9 inches long is divided into num__3 equal parts . what is the length of each part ? <o> a ) num__31 inches <o> b ) num__32 inches <o> c ) num__33 inches <o> d ) num__34 inches <o> e ) num__35 inches |
num__7 ft num__9 in is num__84 + num__9 = num__93 inches . so num__31.0 = num__31 inches or num__2 ft . num__7 inch . answer : a <eor> a <eos> |
a |
add__9.0__84.0__ divide__93.0__3.0__ subtract__9.0__7.0__ divide__93.0__3.0__ |
add__9.0__84.0__ divide__93.0__3.0__ subtract__9.0__7.0__ divide__93.0__3.0__ |
| the diagonals of a rhombus are num__18 cm and num__22 cm . find its area ? <o> a ) num__277 <o> b ) num__266 <o> c ) num__198 <o> d ) num__288 <o> e ) num__212 |
num__0.5 * num__18 * num__22 = num__198 answer : c <eor> c <eos> |
c |
triangle_area__18.0__22.0__ triangle_area__18.0__22.0__ |
volume_rectangular_prism__18.0__22.0__0.5__ volume_rectangular_prism__18.0__22.0__0.5__ |
| if rupee one produces rupees nine over a period of num__40 years find the rate of simple interest ? <o> a ) num__22 num__0.125 % <o> b ) num__22 num__1.5 % <o> c ) num__28 num__0.5 % <o> d ) num__22 num__0.5 % <o> e ) num__32 num__0.5 % |
num__9 = ( num__1 * num__40 * r ) / num__100 r = num__22 num__0.5 % answer : d <eor> d <eos> |
d |
percent__100.0__22.0__ |
percent__100.0__22.0__ |
| the rental charge for a car is num__34 cents for the first num__0.25 mile driven and num__6 cents for every num__0.2 mile driven over the initial num__0.25 mile . if a man paid $ num__1.24 in rental charges how many miles did he drive ? <o> a ) num__2.5 <o> b ) num__3.0 <o> c ) num__3.25 <o> d ) num__3.75 <o> e ) num__4.0 |
total rent = rent for the first num__0.25 mile + rent after num__0.25 mile rent for the first num__0.25 mile = num__34 cents rent after num__0.25 mile = [ num__6 cent / ( num__0.2 mile ) ] * ( x - num__0.25 ) = num__30 ( x - num__0.25 ) therefore total rent = num__35 + num__30 ( x - num__0.25 ) num__124 = num__34 + num__30 ( x - num__0.25 ) num__90 = num__30 ( x - num__0.25 ) num__3 = x - num__0.25 x = num__3 + num__0.25 = $ num__3.25 c <eor> c <eos> |
c |
divide__6.0__0.2__ subtract__124.0__34.0__ divide__90.0__30.0__ add__0.25__3.0__ round__3.25__ |
divide__6.0__0.2__ subtract__124.0__34.0__ divide__90.0__30.0__ add__0.25__3.0__ add__0.25__3.0__ |
| num__10 women can complete a work in num__9 days and num__10 children take num__12 days to complete the work . how many days will num__6 women and num__7 children take to complete the work ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__7 <o> d ) num__8 <o> e ) num__2 |
num__1 women ' s num__1 day work = num__0.0111111111111 num__1 child ' s num__1 day work = num__0.00833333333333 ( num__6 women + num__7 children ) ' s num__1 day work = ( num__0.0666666666667 + num__0.0583333333333 ) = num__0.125 num__6 women and num__7 children will complete the work in num__8 days . d <eor> d <eos> |
d |
subtract__10.0__9.0__ add__0.0583__0.0667__ subtract__9.0__1.0__ round__8.0__ |
subtract__10.0__9.0__ add__0.0583__0.0667__ add__7.0__1.0__ add__7.0__1.0__ |
| sum of the squares of three numbers is num__351 and the sum of their products taken two at a time is num__245 . find the sum ? <o> a ) num__20 <o> b ) num__22 <o> c ) num__25 <o> d ) num__26 <o> e ) num__29 |
( a + b + c ) num__2 = a num__2 + b num__2 + c num__2 + num__2 ( ab + bc + ca ) = num__351 + num__2 * num__245 a + b + c = √ num__841 = num__29 e <eor> e <eos> |
e |
divide__841.0__29.0__ |
divide__841.0__29.0__ |
| if the population of a certain country increases at the rate of one person every num__40 seconds by how many persons does the population increase in num__1 hour ? <o> a ) num__90 <o> b ) num__120 <o> c ) num__150 <o> d ) num__180 <o> e ) num__160 |
answer = num__1.5 * num__60 = num__90 answer is a <eor> a <eos> |
a |
hour_to_min_conversion__ multiply__1.5__60.0__ round__90.0__ |
multiply__40.0__1.5__ multiply__1.5__60.0__ multiply__1.0__90.0__ |
| a train crosses a platform of num__120 m in num__15 sec same train crosses another platform of length num__180 m in num__18 sec . then find the length of the train ? <o> a ) num__288 <o> b ) num__180 <o> c ) num__288 <o> d ) num__277 <o> e ) num__265 |
length of the train be ‘ x ’ x + num__8.0 = x + num__10.0 num__6 x + num__720 = num__5 x + num__900 x = num__180 m . answer : b <eor> b <eos> |
b |
divide__120.0__15.0__ divide__180.0__18.0__ multiply__120.0__6.0__ subtract__15.0__10.0__ multiply__180.0__5.0__ round__180.0__ |
divide__120.0__15.0__ divide__180.0__18.0__ multiply__120.0__6.0__ subtract__15.0__10.0__ add__180.0__720.0__ round__180.0__ |
| how long does a train num__110 m long running at the speed of num__72 km / hr takes to cross a bridge num__132 m length ? <o> a ) num__12.8 sec <o> b ) num__12.1 sec <o> c ) num__12.2 sec <o> d ) num__12.6 sec <o> e ) num__11.1 sec |
speed = num__72 * num__0.277777777778 = num__20 m / sec total distance covered = num__110 + num__132 = num__242 m . required time = num__12.1 = num__12.1 sec . answer : b <eor> b <eos> |
b |
add__110.0__132.0__ divide__242.0__20.0__ round__12.1__ |
add__110.0__132.0__ divide__242.0__20.0__ divide__242.0__20.0__ |
| the consumption of diesel per hour of a bus varies directly as square of its speed . when the bus is travelling at num__60 kmph its consumption is num__1 litre per hour . if each litre costs $ num__60 and other expenses per hous is $ num__60 then what would be the minimum expenditure required to cover a distance of num__600 km ? <o> a ) num__120 <o> b ) num__1250 <o> c ) num__1200 <o> d ) num__1100 <o> e ) num__1150 |
num__60 kmph consumption is num__1 lt / hr so num__600 km will take num__10 hrs and the consumption is num__10 lt for entire distance . num__1 lt costs $ num__60 so num__10 lt costs $ num__600 extra expenses for num__1 hr - $ num__60 num__10 hrs - $ num__600 total expense - $ num__600 + $ num__600 = $ num__1200 answer : c <eor> c <eos> |
c |
divide__600.0__60.0__ round__1200.0__ |
divide__600.0__60.0__ divide__1200.0__1.0__ |
| a train passes a station platform in num__32 sec and a man standing on the platform in num__20 sec . if the speed of the train is num__54 km / hr . what is the length of the platform ? <o> a ) num__228 <o> b ) num__240 <o> c ) num__887 <o> d ) num__166 <o> e ) num__180 |
speed = num__54 * num__0.277777777778 = num__15 m / sec . length of the train = num__15 * num__20 = num__300 m . let the length of the platform be x m . then ( x + num__300 ) / num__32 = num__15 = > x = num__180 m . answer : e <eor> e <eos> |
e |
multiply__20.0__15.0__ round__180.0__ |
multiply__20.0__15.0__ round__180.0__ |
| two numbers are less than third number by num__30.0 and num__37.0 respectively . how much percent is the second number less than by the first <o> a ) num__8.0 <o> b ) num__9.0 <o> c ) num__10.0 <o> d ) num__11.0 <o> e ) num__12 % |
explanation : let the third number is x . then first number = ( num__100 - num__30 ) % of x = num__70.0 of x = num__7 x / num__10 second number is ( num__63 x / num__100 ) difference = num__7 x / num__10 - num__63 x / num__100 = num__7 x / num__10 so required percentage is difference is what percent of first number = > ( num__7 x / num__100 * num__1.42857142857 x * num__100 ) % = num__10.0 explanation : let the third number is x . then first number = ( num__100 - num__30 ) % of x = num__70.0 of x = num__7 x / num__10 second number is ( num__63 x / num__100 ) difference = num__7 x / num__10 - num__63 x / num__100 = num__7 x / num__10 so required percentage is difference is what percent of first number = > ( num__7 x / num__100 * num__1.42857142857 x * num__100 ) % = num__10.0 option c <eor> c <eos> |
c |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| how many numbers from num__39 to num__79 are exactly divisible by num__11 ? <o> a ) num__5 <o> b ) num__7 <o> c ) num__4 <o> d ) num__11 <o> e ) num__12 |
num__3.54545454545 = num__1 and num__7.18181818182 = num__7 = = > num__7 - num__3 = num__4 numbers answer : c <eor> c <eos> |
c |
divide__39.0__11.0__ divide__79.0__11.0__ round_down__7.1818__ round_down__3.5455__ subtract__11.0__7.0__ subtract__11.0__7.0__ |
divide__39.0__11.0__ divide__79.0__11.0__ round_down__7.1818__ round_down__3.5455__ subtract__11.0__7.0__ subtract__11.0__7.0__ |
| it is found that the volume of a cube is numerically equal to its surface area . then the measure of its edge in meters is : <o> a ) num__4 <o> b ) num__6 <o> c ) num__8 <o> d ) num__9 <o> e ) can not be determined |
if edge of cube = a volume of cube = a ^ num__3 surface area = num__6 ( a ^ num__2 ) then a ^ num__3 = num__6 ( a ^ num__2 ) a = num__6 answer : b <eor> b <eos> |
b |
multiply__2.0__3.0__ |
multiply__2.0__3.0__ |
| how long does a train num__165 meters long running at the rate of num__72 kmph take to cross a bridge num__660 meters in length ? <o> a ) num__28 <o> b ) num__41.25 <o> c ) num__55 <o> d ) num__18 <o> e ) num__12 |
t = ( num__660 + num__165 ) / num__72 * num__3.6 t = num__41.25 answer : b <eor> b <eos> |
b |
round__41.25__ |
round__41.25__ |
| a particular store purchased a stock of turtleneck sweaters and marked up its cost by num__20.0 . during the new year season it further marked up its prices by num__25.0 of the original retail price . in february the store then offered a discount of num__15.0 . what was its profit on the items sold in february ? <o> a ) num__27.5 <o> b ) num__30.0 <o> c ) num__35.0 <o> d ) num__37.5 <o> e ) num__40 % |
assume the total price = num__100 x price after num__20.0 markup = num__120 x price after num__25.0 further markup = num__1.25 * num__120 x = num__150 x price after the discount = num__0.85 * num__150 x = num__127.5 x hence total profit = num__27.5 option a <eor> a <eos> |
a |
percent__100.0__27.5__ |
percent__100.0__27.5__ |
| half of num__2 percent written as decimal is <o> a ) num__0.01 <o> b ) num__0.5 <o> c ) num__0.05 <o> d ) num__0.005 <o> e ) none of these |
explanation : it will be num__0.5 ( num__2.0 ) = num__0.5 ( num__0.02 ) = num__0.01 = num__0.01 option a <eor> a <eos> |
a |
percent__2.0__0.5__ percent__2.0__0.5__ |
percent__2.0__0.5__ percent__2.0__0.5__ |
| two trains a and b starting from two points and travelling in opposite directions reach their destinations num__9 hours and num__4 hours respectively after meeting each other . if the train a travels at num__60 kmph find the rate at which the train b runs . <o> a ) num__40 <o> b ) num__90 <o> c ) num__120 <o> d ) num__80 <o> e ) num__100 |
if two objects a and b start simultaneously from opposite points and after meeting reach their destinations in ‘ a ’ and ‘ b ’ hours respectively ( i . e . a takes ‘ a hrs ’ to travel from the meeting point to his destination and b takes ‘ b hrs ’ to travel from the meeting point to his destination ) then the ratio of their speeds is given by : sa / sb = √ ( b / a ) i . e . ratio of speeds is given by the square root of the inverse ratio of time taken . sa / sb = √ ( num__0.444444444444 ) = num__0.666666666667 this gives us that the ratio of the speed of a : speed of b as num__2 : num__3 . since speed of a is num__60 kmph speed of b must be num__60 * ( num__1.5 ) = num__90 kmph answer b <eor> b <eos> |
b |
divide__4.0__9.0__ divide__3.0__2.0__ multiply__60.0__1.5__ round__90.0__ |
divide__4.0__9.0__ divide__3.0__2.0__ multiply__60.0__1.5__ multiply__60.0__1.5__ |
| at present the ratio between the ages of amit and dhiraj is num__5 : num__4 . after num__6 years amit ’ s age will be num__26 years . what is the age of dhiraj at present ? <o> a ) num__16 <o> b ) num__77 <o> c ) num__566 <o> d ) num__197 <o> e ) num__161 |
explanation : let the present ages of amit and dhiraj be num__5 x years and num__4 x years respectively . then num__5 x + num__6 = num__26 num__5 x = num__20 x = num__4 dhiraj ’ s age = num__4 x = num__16 years answer : a <eor> a <eos> |
a |
multiply__5.0__4.0__ subtract__20.0__4.0__ subtract__20.0__4.0__ |
multiply__5.0__4.0__ subtract__20.0__4.0__ subtract__20.0__4.0__ |
| if each side of a square is increased by num__25.0 find the percentage change in its area ? <o> a ) num__65.25 <o> b ) num__56.25 <o> c ) num__65 <o> d ) num__56 <o> e ) num__25 |
let each side of the square be a then area = a x a new side = num__125 a / num__100 = num__5 a / num__4 new area = ( num__5 a x num__5 a ) / ( num__4 x num__4 ) = ( num__25 a ² / num__16 ) increased area = = ( num__25 a ² / num__16 ) - a ² increase % = [ ( num__9 a ² / num__16 ) x ( num__1 / a ² ) x num__100 ] % = num__56.25 answer : b <eor> b <eos> |
b |
percent__25.0__4.0__ percent__56.25__100.0__ |
percent__25.0__4.0__ percent__56.25__100.0__ |
| if the sum and difference of two numbers are num__20 and num__8 respectively then the difference of their square is : <o> a ) num__12 <o> b ) num__28 <o> c ) num__160 <o> d ) num__180 <o> e ) num__18 |
let the numbers be x and y . then x + y = num__20 and x - y = num__8 x num__2 - y num__2 = ( x + y ) ( x - y ) = num__20 * num__8 = num__160 . answer : c <eor> c <eos> |
c |
multiply__20.0__8.0__ multiply__20.0__8.0__ |
multiply__20.0__8.0__ multiply__20.0__8.0__ |
| a clock shows the time as num__9 a . m . if the minute hand gains num__5 minutes every hour how many minutes will the clock gain by num__5 p . m . ? <o> a ) num__30 min <o> b ) num__35 min <o> c ) num__45 min <o> d ) num__40 min <o> e ) num__55 min |
there are num__8 hours in between num__9 a . m . to num__5 p . m . num__8 * num__5 = num__40 minutes . answer : d <eor> d <eos> |
d |
multiply__5.0__8.0__ round__40.0__ |
multiply__5.0__8.0__ multiply__5.0__8.0__ |
| at a supermarket a certain item has increased from num__72 cents per pound to num__78 cents per pound . what is the % increase in the cost of item ? <o> a ) num__6.33 <o> b ) num__7.22 <o> c ) num__8.33 <o> d ) num__8.0 <o> e ) num__7 % |
percent change = ( new value - old value ) / old value x num__100 = ( ( num__78 - num__72 ) / num__72 ) x num__100 = ( num__0.0833333333333 ) * num__100 = num__8.33 there was an num__8.33 increase in the cost of item answer - c = num__8.33 <eor> c <eos> |
c |
percent__8.33__100.0__ |
percent__8.33__100.0__ |
| b completes a work in num__9 days . a alone can do it in num__10 days . if both work together the work can be completed in how many days ? <o> a ) num__3.75 days <o> b ) num__4.73 days <o> c ) num__5.75 days <o> d ) num__6.75 days <o> e ) num__7.73 days |
num__0.111111111111 + num__0.1 = num__0.211111111111 num__4.73684210526 = num__4.73 days answer : b <eor> b <eos> |
b |
add__0.1111__0.1__ round__4.73__ |
add__0.1111__0.1__ round__4.73__ |
| a vessel is filled with liquid num__3 parts of which are water and num__5 parts syrup . how much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup ? <o> a ) num__0.333333333333 <o> b ) num__0.25 <o> c ) num__0.2 <o> d ) num__0.142857142857 <o> e ) num__0.125 |
explanation : suppose the vessel initially contains num__8 litres of liquid . let x litres of this liquid be replaced with water . quantity of water in new mixture = ( num__3 - num__3 x / num__8 + x ) litres quantity of syrup in new mixture = ( num__5 - num__5 x / num__8 ) litres ( num__3 - num__3 x / num__8 + x ) = ( num__5 - num__5 x / num__8 ) = > num__5 x + num__24 = num__40 - num__5 x = > num__10 x = num__16 = > x = num__1.6 so part of the mixture replaced = ( num__1.6 x num__0.125 ) = num__0.2 . answer is c <eor> c <eos> |
c |
add__3.0__5.0__ multiply__3.0__8.0__ multiply__5.0__8.0__ subtract__40.0__24.0__ divide__8.0__5.0__ reverse__8.0__ reverse__5.0__ reverse__5.0__ |
add__3.0__5.0__ multiply__3.0__8.0__ multiply__5.0__8.0__ subtract__40.0__24.0__ divide__8.0__5.0__ reverse__8.0__ reverse__5.0__ reverse__5.0__ |
| a car is running at a speed of num__96 kmph . what distance will it cover in num__14 sec ? <o> a ) num__378 m <o> b ) num__350 m <o> c ) num__380 m <o> d ) num__200 m <o> e ) num__250 m |
speed = num__96 kmph = num__96 * num__0.277777777778 = num__27 m / s distance covered in num__14 sec = num__27 * num__14 = num__378 m answer is a <eor> a <eos> |
a |
multiply__14.0__27.0__ round__378.0__ |
multiply__14.0__27.0__ multiply__14.0__27.0__ |
| the average of num__6 observations is num__11 . a new observation is included and the new average is decreased by num__1 . the seventh observation is ? <o> a ) num__4 <o> b ) num__3 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
let seventh observation = x . then according to the question we have = > ( num__66 + x ) / num__7 = num__10 = > x = num__4 . hence the seventh observation is num__4 . answer : a <eor> a <eos> |
a |
multiply__6.0__11.0__ add__6.0__1.0__ subtract__11.0__1.0__ subtract__11.0__7.0__ subtract__11.0__7.0__ |
multiply__6.0__11.0__ add__6.0__1.0__ subtract__11.0__1.0__ subtract__11.0__7.0__ divide__4.0__1.0__ |
| if a + b = y and a - b = z then num__3 ab = can someone explain ! <o> a ) ( num__3 y ^ num__2 - num__3 z ^ num__2 ) / num__4 <o> b ) ( num__3 z ^ num__2 - num__3 y ^ num__2 ) / num__4 <o> c ) num__3 y + num__3 z / num__2 <o> d ) num__3 y - num__3 z / num__2 <o> e ) num__3 y ^ num__2 - num__3 z ^ num__1.0 |
plugging numbers is quite fast in this case : let a = num__1 b = num__2 then ; y = num__1 + num__2 = > num__3 and z = num__1 - num__2 = > - num__1 question asks num__3 ab = . . . . . . . . . as our number num__3 ab = num__3 * num__1 * num__2 = num__6 now plug the value of y and z in the answer choices . option a gives num__6 and that is the number we need . hence : ans is a . <eor> a <eos> |
a |
subtract__3.0__1.0__ multiply__3.0__2.0__ multiply__3.0__1.0__ |
subtract__3.0__1.0__ multiply__3.0__2.0__ multiply__3.0__1.0__ |
| if two numbers are in the ratio num__2 : num__3 . if num__10 is added to both of the numbers then the ratio becomes num__3 : num__4 then find the smallest number ? <o> a ) num__12 <o> b ) num__18 <o> c ) num__20 <o> d ) num__24 <o> e ) num__26 |
num__2 : num__3 num__2 x + num__10 : num__3 x + num__10 = num__3 : num__4 num__4 [ num__2 x + num__10 ] = num__3 [ num__3 x + num__10 ] num__8 x + num__40 = num__9 x + num__30 num__9 x - num__8 x = num__40 - num__30 x = num__10 then smallest number is = num__2 num__2 x = num__20 c <eor> c <eos> |
c |
multiply__2.0__4.0__ multiply__10.0__4.0__ multiply__3.0__10.0__ multiply__2.0__10.0__ multiply__2.0__10.0__ |
subtract__10.0__2.0__ multiply__10.0__4.0__ subtract__40.0__10.0__ subtract__30.0__10.0__ subtract__40.0__20.0__ |
| the average weight of num__8 person ' s increases by num__2.5 kg when a new person comes in place of one of them weighing num__55 kg . what might be the weight of the new person ? <o> a ) num__70 kg <o> b ) num__75 kg <o> c ) num__80 kg <o> d ) num__85 kg <o> e ) num__90 kg |
total weight increased = ( num__8 x num__2.5 ) kg = num__20 kg . weight of new person = ( num__55 + num__20 ) kg = num__75 kg . b ) <eor> b <eos> |
b |
multiply__8.0__2.5__ add__55.0__20.0__ add__55.0__20.0__ |
multiply__8.0__2.5__ add__55.0__20.0__ add__55.0__20.0__ |
| incomes of two companies a and b are in the ratio of num__5 : num__8 . had the income of company a been more by num__25 lakh the ratio of their incomes would have been num__5 : num__4 . what is the income of company b ? <o> a ) num__80 lakh <o> b ) num__50 lakh <o> c ) num__40 lakh <o> d ) num__60 lakh <o> e ) none of these |
let the incomes of two companies a and b be num__5 x and num__8 x respectively . from the question num__5 x + num__3.125 x = num__1.25 ⇒ num__20 x + num__100 = num__40 x ∴ x = num__5 ∴ income of company b = num__8 x = num__40 lakh answer c <eor> c <eos> |
c |
divide__25.0__8.0__ divide__5.0__4.0__ multiply__5.0__4.0__ multiply__5.0__20.0__ multiply__5.0__8.0__ multiply__5.0__8.0__ |
divide__25.0__8.0__ divide__5.0__4.0__ multiply__5.0__4.0__ multiply__5.0__20.0__ multiply__5.0__8.0__ multiply__5.0__8.0__ |
| a and b are playing mathematical puzzles . a asks b ` ` which whole numbers greater than one can divide all the nine three digit numbers num__111222 num__333444 num__555666 num__777888 and num__999 ? ' ' b immediately gave the desired answer . it was : - <o> a ) num__3 num__37 and num__119 <o> b ) num__3 num__37 and num__111 <o> c ) num__9 num__37 and num__111 <o> d ) num__3 num__9 and num__37 <o> e ) none |
explanation : each of the number can be written as a multiple of num__111 . and the factors of num__111 are num__3 and num__37 . thus the desired answer is num__3 num__37 and num__111 . answer : b <eor> b <eos> |
b |
divide__111.0__3.0__ divide__111.0__37.0__ |
divide__111.0__3.0__ divide__111.0__37.0__ |
| a num__1200 m long train crosses a tree in num__120 sec how much time will i take to pass a platform num__700 m long ? <o> a ) num__277 sec <o> b ) num__190 sec <o> c ) num__187 sec <o> d ) num__286 sec <o> e ) num__215 sec |
l = s * t s = num__10.0 s = num__10 m / sec . total length ( d ) = num__1900 m t = d / s t = num__190.0 t = num__190 sec answer : b <eor> b <eos> |
b |
divide__1200.0__120.0__ add__1200.0__700.0__ divide__1900.0__10.0__ round__190.0__ |
divide__1200.0__120.0__ add__1200.0__700.0__ divide__1900.0__10.0__ divide__1900.0__10.0__ |
| a train num__125 m long passes a man running at num__1 km / hr in the same direction in which the train is going in num__10 sec . the speed of the train is ? <o> a ) num__65 km / hr <o> b ) num__17 km / hr <o> c ) num__76 km / hr <o> d ) num__50 km / hr <o> e ) num__46 km / hr |
speed of the train relative to man = num__12.5 = num__12.5 m / sec . = num__12.5 * num__3.6 = num__45 km / hr let the speed of the train be x km / hr . then relative speed = ( x - num__1 ) km / hr . x - num__1 = num__45 = > x = num__46 km / hr . answer : e <eor> e <eos> |
e |
divide__125.0__10.0__ multiply__12.5__3.6__ add__1.0__45.0__ round__46.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ add__1.0__45.0__ divide__46.0__1.0__ |
| a work which could be finished in num__8 days was finished num__3 days earlier after num__10 more men joined . the number of men employed was ? <o> a ) num__22 <o> b ) num__30 <o> c ) num__88 <o> d ) num__71 <o> e ) num__11 |
x - - - - - - - num__8 ( x + num__10 ) - - - - num__6 x * num__8 = ( x + num__10 ) num__6 x = num__30 \ answer : b <eor> b <eos> |
b |
multiply__3.0__10.0__ round__30.0__ |
multiply__3.0__10.0__ multiply__3.0__10.0__ |
| each digit in the two - digit number g is halved to form a new two - digit number h . which of the following could be the sum of g and h <o> a ) num__153 <o> b ) num__159 <o> c ) num__137 <o> d ) num__121 <o> e ) num__89 |
let the number be x now x is halved = > x / num__2 . . the question asks what is x ? x ( number ) + x / num__2 ( half of that number ) = something = > num__3 x / num__2 = something = > x = num__2 ( something ) / num__3 so the answer choice must be divisible by num__3 eliminate c e right away check the rest b satisfies num__3 x / num__2 = num__159 = > x = num__106 verify . num__106 + num__53 = num__159 answer is b <eor> b <eos> |
b |
divide__106.0__2.0__ multiply__53.0__3.0__ |
divide__106.0__2.0__ add__53.0__106.0__ |
| the product of two numbers is num__2028 and their h . c . f . is num__13 . the number of such pairs is : <o> a ) num__9 <o> b ) num__2 <o> c ) num__7 <o> d ) num__8 <o> e ) num__1 |
let the numbers num__13 a and num__13 b . then num__13 a x num__13 b = num__2028 = > ab = num__12 . now the co - primes with product num__12 are ( num__1 num__12 ) and ( num__3 num__4 ) . [ note : two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than num__1 or equivalently if their greatest common divisor is num__1 ] so the required numbers are ( num__13 x num__1 num__13 x num__12 ) and ( num__13 x num__3 num__13 x num__4 ) . clearly there are num__2 such pairs . answer : b <eor> b <eos> |
b |
subtract__13.0__12.0__ add__1.0__3.0__ subtract__3.0__1.0__ gcd__2028.0__2.0__ |
subtract__13.0__12.0__ add__1.0__3.0__ subtract__3.0__1.0__ subtract__3.0__1.0__ |
| renu can do a piece of work in num__8 days but with the help of her friend suma she can do it in num__4 days . in what time suma can do it alone ? <o> a ) num__8 <o> b ) num__12 <o> c ) num__14 <o> d ) num__15 <o> e ) num__17 |
renu â € ™ s one day â € ™ s work = num__0.125 suma â € ™ s one day â € ™ s work = num__0.25 - num__0.125 = num__0.125 suma can do it alone in num__8 days . answer : a <eor> a <eos> |
a |
round__8.0__ |
round__8.0__ |
| the security gate at a storage facility requires a six - digit lock code . if the lock code must consist only of digits from num__1 through num__9 inclusive with no repeated digits and the first and last digits of the code must be odd how many lock codes are possible ? <o> a ) num__120 <o> b ) num__240 <o> c ) num__360 <o> d ) num__720 <o> e ) num__16800 |
x - x - x - x - x - - > there are num__5 odd digits from num__1 to num__9 inclusive thus options for the first and the last x ' s are : num__5 - x - x - x - x - num__4 . other x ' s can take following values : num__5 - num__7 - num__6 - num__5 - num__4 - num__4 - > num__5 * num__7 * num__6 * num__5 * num__4 * num__4 = num__16800 . answer : e . <eor> e <eos> |
e |
subtract__9.0__5.0__ add__1.0__5.0__ multiply__1.0__16800.0__ |
subtract__9.0__5.0__ subtract__7.0__1.0__ multiply__1.0__16800.0__ |
| if by selling an article for rs . num__60 a person loses num__0.142857142857 of money what would he have gained or lost % by selling it for rs . num__77 ? <o> a ) num__5.0 <o> b ) num__10.0 <o> c ) num__20.0 <o> d ) num__30.0 <o> e ) num__45 % |
sol . cp - = ec num__7 = num__70 num__1 - num__7 num__6 num__77 - num__70.0 profit = num__100 = num__10.0 b <eor> b <eos> |
b |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| if a light flashes every num__6 seconds how many times will it flash in ¾ of an hour ? <o> a ) num__451 times <o> b ) num__638 times <o> c ) num__838 times <o> d ) num__436 times <o> e ) num__435 times |
there are num__60 minutes in an hour . in ¾ of an hour there are ( num__60 * ¾ ) minutes = num__45 minutes . in ¾ of an hour there are ( num__60 * num__45 ) seconds = num__2700 seconds . light flashed for every num__6 seconds . in num__2700 seconds num__450.0 = num__450 times . the count start after the first flash the light will flashes num__451 times in ¾ of an hour . answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ multiply__45.0__60.0__ divide__2700.0__6.0__ round__451.0__ |
hour_to_min_conversion__ multiply__45.0__60.0__ divide__2700.0__6.0__ round__451.0__ |
| some digits were given like num__0 num__2 num__2 num__3 num__3 num__4 num__4 then how many no will be there which is greater than num__1 lac all numbers are distinct <o> a ) num__530 <o> b ) num__540 <o> c ) num__550 <o> d ) num__560 <o> e ) num__570 |
( num__6 * num__6 * num__5 * num__4 * num__3 * num__2 ) / ( num__2 ! * num__2 ! * num__2 ! ) = num__540 ways answer : b <eor> b <eos> |
b |
multiply__2.0__3.0__ add__2.0__3.0__ multiply__1.0__540.0__ |
multiply__2.0__3.0__ add__2.0__3.0__ multiply__1.0__540.0__ |
| each of the following equations has at least one solution except d . is it true ? <o> a ) - num__2 * - a = num__2 a <o> b ) num__2 * a = ( – num__2 ) * - a <o> c ) - num__4 * a = - num__3 - a <o> d ) - num__3 * a = - num__3 - a <o> e ) ( – num__2 ) - a = – num__2 - a |
if we put a = num__1 then a . num__2 b . num__2 c . - num__4 e . - num__3 d . - num__3 = - num__4 not true <eor> d <eos> |
d |
add__1.0__2.0__ add__1.0__2.0__ |
subtract__4.0__1.0__ subtract__4.0__1.0__ |
| num__10 people went to a hotel for combine dinner party num__5 of them spent rs . num__20 each on their dinner and rest spent num__4 more than the average expenditure of all the num__10 . what was the total money spent by them . <o> a ) num__1628.4 <o> b ) num__1534 <o> c ) num__1492 <o> d ) num__240 <o> e ) none of these |
solution : let average expenditure of num__10 people be x . then num__10 x = num__20 * num__5 + num__5 * ( x + num__4 ) ; or num__10 x = num__20 * num__5 + num__5 x + num__20 ; or x = num__24 ; so total money spent = num__24 * num__10 = rs . num__240 . answer : option d <eor> d <eos> |
d |
add__20.0__4.0__ multiply__10.0__24.0__ multiply__10.0__24.0__ |
add__20.0__4.0__ multiply__10.0__24.0__ multiply__10.0__24.0__ |
| if x is the product of the integers from num__1 to num__150 inclusive and num__5 ^ y is a factor of x what is the greatest possible value of y ? <o> a ) num__30 <o> b ) num__34 <o> c ) num__36 <o> d ) num__37 <o> e ) num__39 |
it basically asks for the number of num__5 s in num__150 ! num__30.0 + num__6.0 + num__1.2 = num__30 + num__6 + num__1 . hence num__37 answer : d . <eor> d <eos> |
d |
divide__150.0__5.0__ add__1.0__5.0__ divide__6.0__5.0__ multiply__1.0__37.0__ |
divide__150.0__5.0__ add__1.0__5.0__ divide__6.0__5.0__ multiply__1.0__37.0__ |
| product m is produced by mixing chemical x and chemical y in the ratio of num__5 : num__4 . chemical x is prepared by mixing two raw materials a and b in the ratio of num__1 : num__3 . chemical y is prepared by mixing raw materials b and c in the ratio of num__2 : num__1 . then the final mixture is prepared by mixing num__864 units of product m with water . if the concentration of the raw material b in the final mixture is num__50.0 how much water had been added to product m ? <o> a ) num__328 units <o> b ) num__368 units <o> c ) num__392 units <o> d ) num__616 units <o> e ) none of the above |
the final product is obtained by mixing num__864 units of product m with water . in num__864 units of product m amount of b = num__864 × num__0.712962962963 = num__616 in the final mixture concentration of b is num__50.0 . therefore the total quantity of final mixture = num__616 × num__2 = num__1232 water added = num__1232 – num__864 = num__368 the correct answer is choice b . <eor> b <eos> |
b |
multiply__2.0__616.0__ subtract__1232.0__864.0__ multiply__1.0__368.0__ |
multiply__2.0__616.0__ subtract__1232.0__864.0__ multiply__1.0__368.0__ |
| in a num__4 person race medals are awarded to the fastest num__3 runners . the first - place runner receives a gold medal the second - place runner receives a silver medal and the third - place runner receives a bronze medal . in the event of a tie the tied runners receive the same color medal . ( for example if there is a two - way tie for first - place the top two runners receive gold medals the next - fastest runner receives a silver medal and no bronze medal is awarded ) . assuming that exactly three medals are awarded and that the three medal winners stand together with their medals to form a victory circle how many z different victory circles are possible ? <o> a ) num__24 <o> b ) num__52 <o> c ) num__96 <o> d ) num__144 <o> e ) num__648 |
possible scenarios are : num__1 . gold / silver / bronze / no medal ( no ties ) - num__4 ! = num__24 ; num__2 . gold / gold / silver / no medal - num__4 ! / num__2 ! = num__12 ; num__3 . gold / silver / silver / no medal - num__4 ! / num__2 ! = num__12 ; num__4 . gold / gold / gold / no medal - num__4 ! / num__3 ! = num__4 . total z : num__24 + num__12 + num__12 + num__4 = num__52 answer : b . <eor> b <eos> |
b |
subtract__4.0__3.0__ subtract__3.0__1.0__ multiply__4.0__3.0__ multiply__1.0__52.0__ |
subtract__4.0__3.0__ subtract__3.0__1.0__ divide__24.0__2.0__ divide__52.0__1.0__ |
| x completes a work in num__20 days and y complete the same work in num__30 days . if both of them work together then the number of days required to complete the work will be ? <o> a ) num__10 days . <o> b ) num__12 days . <o> c ) num__14 days . <o> d ) num__8 days . <o> e ) num__20 days . |
if x can complete a work in x days and y can complete the same work in y days then both of them together can complete the work in x y / x + y days therefore here the required number of days = num__20 Ã — num__0.6 = num__12 days . b ) <eor> b <eos> |
b |
km_to_mile_conversion__ multiply__20.0__0.6__ round__12.0__ |
km_to_mile_conversion__ multiply__20.0__0.6__ round__12.0__ |
| a chair is bought for rs . num__300 / - and sold at rs . num__420 / - find gain or loss percentage <o> a ) num__15.0 loss <o> b ) num__40.0 gain <o> c ) num__25.0 gain <o> d ) num__30.0 gain <o> e ) num__35.0 gain |
formula = ( selling price ~ cost price ) / cost price * num__100 = ( num__450 - num__300 ) / num__300 = num__40.0 gain b <eor> b <eos> |
b |
percent__40.0__100.0__ |
percent__40.0__100.0__ |
| what is the least number to be subtracted from num__696 to make it a perfect square ? <o> a ) num__1 <o> b ) num__16 <o> c ) num__20 <o> d ) num__71 <o> e ) num__60 |
the numbers less than num__696 and are squares of certain numbers are num__676 num__625 . the least number that should be subtracted from num__696 to make it perfect square = num__696 - num__676 = num__20 . answer : c <eor> c <eos> |
c |
subtract__696.0__676.0__ subtract__696.0__676.0__ |
subtract__696.0__676.0__ subtract__696.0__676.0__ |
| if a sum of money doubles itself in num__20 years at simple interest the ratepercent per annum is <o> a ) num__12 <o> b ) num__12.5 <o> c ) num__13 <o> d ) num__5 <o> e ) num__14 |
explanation : let sum = x then simple interest = x rate = ( num__100 * x ) / ( x * num__20 ) = num__5 option d <eor> d <eos> |
d |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| in a fuel station the service costs $ num__1.15 per car every liter of fuel costs num__0.4 $ . assuming that you own num__2 sports cars and num__2 executive cars and all fuel tanks are empty . how much will it cost to fuel all cars together if a sports car tank is num__32 liters and an executive car tank is num__75.0 bigger ? <o> a ) num__37.5 $ <o> b ) num__75 $ <o> c ) num__87.5 $ <o> d ) num__94.5 $ <o> e ) num__98.4 $ |
total cars = num__4 num__1.15 * num__4 = num__4.6 - > service cost fuel cost in sports car = num__2 * num__32 * num__0.4 = num__25.6 fuel cost in executive car = num__25.6 * num__1.75 = num__44.8 total fuel cost = num__25.6 + num__44.8 = num__70.4 cost to fuel car = num__70.4 + num__4.6 = num__75 answer : b <eor> b <eos> |
b |
multiply__1.15__4.0__ multiply__1.75__25.6__ subtract__75.0__4.6__ add__70.4__4.6__ |
multiply__1.15__4.0__ multiply__1.75__25.6__ subtract__75.0__4.6__ add__70.4__4.6__ |
| company z spent num__0.25 of its revenues last year on marketing and num__0.142857142857 of the remainder on maintenance of its facilities . what fraction of last year ’ s original revenues did company z have left after its marketing and maintenance expenditures ? <o> a ) num__0.357142857143 <o> b ) num__0.5 <o> c ) num__0.607142857143 <o> d ) num__0.642857142857 <o> e ) num__0.818181818182 |
total revenues = x spent on marketing = x / num__4 remaining amount = x - x / num__4 = num__3 x / num__4 num__0.142857142857 of the remainder on maintenance of its facilities = num__3 x / num__4 * num__0.142857142857 = num__3 x / num__28 amount left = num__3 x / num__4 - num__3 x / num__28 = num__9 x / num__14 answer d <eor> d <eos> |
d |
reverse__0.25__ divide__9.0__14.0__ |
reverse__0.25__ divide__9.0__14.0__ |
| a train num__360 m long is running at a speed of num__45 km / hr . in what time will it pass a bridge num__140 m long ? <o> a ) num__40 sec <o> b ) num__42 sec <o> c ) num__45 sec <o> d ) num__48 sec <o> e ) num__58 sec |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec total distance covered = num__360 + num__140 = num__500 m required time = num__500 * num__0.08 = num__40 sec answer : a <eor> a <eos> |
a |
add__360.0__140.0__ divide__500.0__12.5__ round__40.0__ |
add__360.0__140.0__ divide__500.0__12.5__ divide__500.0__12.5__ |
| a woman complete a journey in num__20 hours . she travels first half of the journey at the rate of num__21 km / hr and second half at the rate of num__24 km / hr . find the total journey in km . <o> a ) num__334 km . <o> b ) num__216 km . <o> c ) num__314 km . <o> d ) num__448 km . <o> e ) num__544 km . |
num__0.5 x / num__21 + num__0.5 x / num__24 = num__20 - - > x / num__21 + x / num__24 = num__40 - - > x = num__448 km . d <eor> d <eos> |
d |
divide__20.0__0.5__ round__448.0__ |
divide__20.0__0.5__ round__448.0__ |
| the average height in a group of num__4 people is num__175 cm . if the average height increased when num__2 more people were added to the group which of the following can not be the heights of the two new people ? <o> a ) num__179 cm and num__172 cm <o> b ) num__181 cm and num__169 cm <o> c ) num__173 cm and num__178 cm <o> d ) num__176 cm and num__176 cm <o> e ) num__174 cm and num__177 cm |
denote x as the sum of the heights of the two new people . from the stem it follows that ( num__700 + x ) num__6 > num__175 . this reduces to x > num__350 . only the heights from b add up to num__350 cm . all other pairs add up to more than num__350 cm . answer : b <eor> b <eos> |
b |
multiply__4.0__175.0__ add__4.0__2.0__ multiply__175.0__2.0__ add__175.0__6.0__ |
multiply__4.0__175.0__ add__4.0__2.0__ multiply__175.0__2.0__ add__175.0__6.0__ |
| the sum of the present ages of two persons a and b is num__60 . if the age of a is twice that of b find the sum of their ages num__5 years hence ? <o> a ) num__70 <o> b ) num__55 <o> c ) num__60 <o> d ) num__65 <o> e ) num__80 |
a + b = num__60 a = num__2 b num__2 b + b = num__60 = = = > b = num__20 a = num__40 num__5 yrs ages num__45 and num__25 sum of their ages = num__70 answer a <eor> a <eos> |
a |
subtract__60.0__20.0__ add__5.0__40.0__ add__5.0__20.0__ add__45.0__25.0__ add__45.0__25.0__ |
subtract__60.0__20.0__ add__5.0__40.0__ add__5.0__20.0__ add__45.0__25.0__ add__45.0__25.0__ |
| find the number which when multiplied by num__15 is increased by num__196 . <o> a ) num__14 <o> b ) num__20 <o> c ) num__26 <o> d ) num__28 <o> e ) num__30 |
let the number be x . then num__15 x - x = num__196 < = > num__14 x = num__196 < = > x = num__14 . answer : a <eor> a <eos> |
a |
divide__196.0__14.0__ |
divide__196.0__14.0__ |
| a number x equals num__80.0 of the average of num__5 num__7 num__14 and a number y . if the average of x and y is num__26 the value of y is ? <o> a ) num__13 <o> b ) num__26 <o> c ) num__39 <o> d ) num__36 <o> e ) none of these |
average of num__5 num__7 num__14 and y = ( num__5 + num__7 + num__14 + y ) / num__4 therefore x = num__80.0 of ( num__5 + num__7 + num__14 + y ) / num__4 = ( num__0.8 ) x ( num__26 + y ) / num__4 = > x = ( num__26 + y ) / num__5 num__5 x - y = num__26 - - - - - ( i ) also ( x + y ) / num__2 = num__26 - - - - - ( ii ) from ( i ) and ( ii ) x = num__13 y = num__39 . answer : c <eor> c <eos> |
c |
divide__4.0__5.0__ subtract__7.0__5.0__ divide__26.0__2.0__ add__26.0__13.0__ add__26.0__13.0__ |
divide__4.0__5.0__ subtract__7.0__5.0__ divide__26.0__2.0__ add__26.0__13.0__ add__26.0__13.0__ |
| a man covers a distance on scooter . had he moved num__3 kmph faster he would have taken num__40 min less . if he had moved num__2 kmph slower he would have taken num__40 min more . the distance is . <o> a ) num__30 km <o> b ) num__40 km <o> c ) num__45 km <o> d ) num__50 km <o> e ) num__55 km |
let distance = x m usual rate = y kmph x / y – x / y + num__3 = num__0.666666666667 hr num__2 y ( y + num__3 ) = num__9 x - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ( num__1 ) x / y - num__2 – x / y = num__0.666666666667 hr y ( y - num__2 ) = num__3 x - - - - - - - - - - - ( num__2 ) divide num__1 & num__2 equations by solving we get x = num__40 km answer : b <eor> b <eos> |
b |
divide__2.0__3.0__ subtract__3.0__2.0__ round__40.0__ |
divide__2.0__3.0__ subtract__3.0__2.0__ divide__40.0__1.0__ |
| if num__3 men or num__4 women can reap a field in num__10 days how long will num__5 men and num__6 women take to reap it ? <o> a ) num__0.316666666667 <o> b ) num__0.15 <o> c ) num__0.666666666667 <o> d ) num__0.277777777778 <o> e ) num__0.25 |
explanation : num__3 men reap num__0.1 field in num__1 day num__1 man reap num__1 / ( num__3 x num__10 ) num__4 women reap num__0.1 field in num__1 day num__1 woman reap num__1 / ( num__10 x num__4 ) num__5 men and num__6 women reap ( num__5 / ( num__3 x num__10 ) + num__6 / ( num__4 x num__10 ) ) = num__0.316666666667 in num__1 day num__5 men and num__6 women will reap the field in num__0.316666666667 days answer : option a <eor> a <eos> |
a |
subtract__4.0__3.0__ multiply__1.0__0.3167__ |
subtract__4.0__3.0__ divide__0.3167__1.0__ |
| what is the greatest possible value of integer n if num__100 ! is divisible by num__15 ^ n <o> a ) num__20 <o> b ) num__21 <o> c ) num__22 <o> d ) num__23 <o> e ) num__24 |
num__15 ^ n = num__5 ^ n * num__3 ^ n highest prime factor will be the limiting factor . num__20.0 + num__4.0 = num__20 + num__4 = num__24 answer : e <eor> e <eos> |
e |
divide__15.0__5.0__ divide__100.0__5.0__ divide__20.0__5.0__ add__4.0__20.0__ add__4.0__20.0__ |
divide__15.0__5.0__ add__15.0__5.0__ divide__20.0__5.0__ add__4.0__20.0__ add__4.0__20.0__ |
| a shopkeeper gave an additional num__20 per cent concession on the reduced price after giving num__30 per cent standard concession on an article . if arun bought that article for num__1120 what was the original price ? <o> a ) num__3000 <o> b ) num__4000 <o> c ) num__2400 <o> d ) num__2000 <o> e ) none of these |
original price = num__1120 × num__100 ⁄ num__70 × num__100 ⁄ num__80 = num__2000 answer d <eor> d <eos> |
d |
subtract__100.0__30.0__ subtract__100.0__20.0__ multiply__20.0__100.0__ multiply__20.0__100.0__ |
subtract__100.0__30.0__ subtract__100.0__20.0__ multiply__20.0__100.0__ multiply__20.0__100.0__ |
| in an election between two candidates num__10.0 of votes are were declares invalid . first candidate got num__3600 votes which were num__60.0 of the total valid votes . the total number of votes enrolled in that election was : <o> a ) num__9000 <o> b ) num__98000 <o> c ) num__20000 <o> d ) num__11000 <o> e ) num__12000 |
num__100.0 - num__10.0 = num__90.0 num__54.0 - num__36.0 = num__18.0 num__18.0 - - - - - - > num__3600 ( num__18 × num__200 = num__3600 ) num__100.0 - - - - - - - > num__20000 votes ( num__100 × num__200 = num__20000 ) c <eor> c <eos> |
c |
percent__60.0__90.0__ percent__100.0__20000.0__ |
percent__60.0__90.0__ percent__100.0__20000.0__ |
| find a num__8 digit number that if multiplied by the number num__9 or any of its multiplications products ( num__18 num__27 num__36 num__45 . . ) will result in the multiplication factor repeated ( n ) number of times . <o> a ) num__12345679 <o> b ) num__42345675 <o> c ) num__22347379 <o> d ) num__52345673 <o> e ) num__21345679 |
a the number is : num__12345679 num__12345679 * num__9 = num__111111111 num__12345679 * num__18 = num__222222222 num__12345679 * num__27 = num__333333333 and so on . . . <eor> a <eos> |
a |
multiply__9.0__12345679.0__ multiply__18.0__12345679.0__ multiply__27.0__12345679.0__ divide__111111111.0__9.0__ |
multiply__9.0__12345679.0__ multiply__18.0__12345679.0__ multiply__27.0__12345679.0__ divide__111111111.0__9.0__ |
| a boat covers a certain distance downstream in num__1 hour while it comes back in num__11 â „ num__2 hours . if the speed of the stream be num__3 kmph what is the speed of the boat in still water ? <o> a ) num__31 kmph <o> b ) num__16 kmph <o> c ) num__19 kmph <o> d ) num__15 kmph <o> e ) num__14 kmph |
let the speed of the water in still water = x given that speed of the stream = num__3 kmph speed downstream = ( x + num__3 ) kmph speed upstream = ( x â ˆ ’ num__3 ) kmph he travels a certain distance downstream in num__1 hour and come back in num__11 â „ num__2 hour . i . e . distance travelled downstream in num__1 hour = distance travelled upstream in num__11 â „ num__2 hour since distance = speed à — time we have ( x + num__3 ) à — num__1 = ( x - num__3 ) num__1.5 num__2 ( x + num__3 ) = num__3 ( x - num__3 ) num__2 x + num__6 = num__3 x - num__9 x = num__6 + num__9 = num__15 kmph answer : d <eor> d <eos> |
d |
divide__3.0__2.0__ multiply__2.0__3.0__ subtract__11.0__2.0__ add__6.0__9.0__ round__15.0__ |
divide__3.0__2.0__ multiply__2.0__3.0__ subtract__11.0__2.0__ add__6.0__9.0__ add__6.0__9.0__ |
| if num__15 women or num__10 men can complete a project in num__55 days in how many days will num__5 women and num__4 men working together complete the same project ? <o> a ) num__75 <o> b ) num__8 <o> c ) num__9 <o> d ) num__85 <o> e ) none of these |
num__15 w = num__10 m now num__5 w + num__4 m = num__5 w + num__4 × num__1.5 w = num__5 w + num__6 w = = num__11 w if num__15 women can complete the project in num__55 days num__11 women can complete the same project in num__55 × num__1.36363636364 = num__75 days answer a <eor> a <eos> |
a |
divide__15.0__10.0__ subtract__10.0__4.0__ subtract__15.0__4.0__ divide__15.0__11.0__ multiply__15.0__5.0__ round__75.0__ |
divide__15.0__10.0__ subtract__10.0__4.0__ add__5.0__6.0__ divide__15.0__11.0__ multiply__15.0__5.0__ round__75.0__ |
| two pipes a and b can separately fill a cistern in num__10 and num__15 minutes respectively . a person opens both the pipes together when the cistern should have been was full he finds the waste pipe open . he then closes the waste pipe and in another num__4 minutes the cistern was full . in what time can the waste pipe empty the cistern when fill ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__7 <o> d ) num__4 <o> e ) num__1 |
num__0.1 + num__0.0666666666667 = num__0.166666666667 * num__4 = num__0.666666666667 num__1 - num__0.666666666667 = num__0.333333333333 num__0.1 + num__0.0666666666667 - num__1 / x = num__0.333333333333 x = num__8 answer : b <eor> b <eos> |
b |
add__0.1__0.0667__ divide__10.0__15.0__ multiply__10.0__0.1__ subtract__1.0__0.6667__ round__8.0__ |
add__0.1__0.0667__ divide__10.0__15.0__ multiply__10.0__0.1__ subtract__1.0__0.6667__ divide__8.0__1.0__ |
| a man is num__30 years older than his son . in two years his age will be twice the age of his son . the present age of the son is <o> a ) num__14 years <o> b ) num__28 years <o> c ) num__20 years <o> d ) num__22 years <o> e ) none |
solution let the son ' s present age be x years . then man ' s present age = ( x + num__30 ) years . then â € ¹ = â € º ( x + num__30 ) + num__2 = num__2 ( x + num__2 ) â € ¹ = â € º x + num__32 = num__2 x + num__4 x = num__28 . answer b <eor> b <eos> |
b |
add__30.0__2.0__ subtract__30.0__2.0__ subtract__30.0__2.0__ |
add__30.0__2.0__ subtract__30.0__2.0__ subtract__30.0__2.0__ |
| chris mixed num__3 pounds of raisins with num__3 pounds of nuts . if a pound of nuts costs num__3 times as much as a pound of raisins then the total cost of the raisins was what fraction of the total cost of the mixture ? <o> a ) num__0.142857142857 <o> b ) num__0.2 <o> c ) num__0.25 <o> d ) num__0.333333333333 <o> e ) num__0.428571428571 |
num__1 lbs of raisin = $ num__1 num__3 lbs of raisin = $ num__3 num__1 lbs of nuts = $ num__3 num__3 lbs of nuts = $ num__9 total value of mixture = num__9 + num__3 = num__12 fraction of the value of raisin = num__0.25 = num__0.25 ans : c <eor> c <eos> |
c |
add__3.0__9.0__ divide__3.0__12.0__ divide__3.0__12.0__ |
add__3.0__9.0__ divide__3.0__12.0__ divide__3.0__12.0__ |
| if num__2 ^ ( num__2 w ) = num__8 ^ ( w − num__2 ) what is the value of w ? <o> a ) num__3 <o> b ) num__6 <o> c ) num__9 <o> d ) num__12 <o> e ) num__15 |
num__2 ^ ( num__2 w ) = num__8 ^ ( w − num__2 ) num__2 ^ ( num__2 w ) = num__2 ^ ( num__3 * ( w − num__2 ) ) num__2 ^ ( num__2 w ) = num__2 ^ ( num__3 w - num__6 ) let ' s equate the exponents as the bases are equal . num__2 w = num__3 w - num__6 w = num__6 the answer is b . <eor> b <eos> |
b |
multiply__2.0__3.0__ multiply__2.0__3.0__ |
multiply__2.0__3.0__ multiply__2.0__3.0__ |
| a watch which gains uniformly is num__2 minutes low at noon on monday and is num__4 min . num__48 sec fast at num__2 p . m . on the following monday . when was it correct ? <o> a ) num__2 p . m . on tuesday <o> b ) num__2 p . m . on wednesday <o> c ) num__3 p . m . on thursday <o> d ) num__1 p . m . on friday <o> e ) none |
solution time from num__12 p . m . on monday to num__2 p . m . on the following monday = num__7 days num__2 hours . = num__170 hours . the watch gains = ( num__2 + num__4 x num__0.8 ) min = num__6.8 min . in num__170 hrs . now num__6.8 min are gained in num__170 hrs . num__2 min are gained in ( num__170 x num__0.147058823529 x num__2 ) hrs . watch is correct num__2 days num__2 hrs . after num__12 p . m . on monday i . e . it will be correct at num__2 p . m . on wednesday . answer b <eor> b <eos> |
b |
divide__48.0__4.0__ round__2.0__ |
divide__48.0__4.0__ round__2.0__ |
| otto and han are driving at constant speeds in opposite directions on a straight highway . at a certain time they are driving toward each other and are num__40 miles apart . one and a half hours later they are again num__40 miles apart driving away from each other . if otto drives at a speed of x miles per hour then in terms of x han drives at a speed of how many miles per hour ? <o> a ) a ) num__80 - x <o> b ) b ) num__53.33 - x <o> c ) c ) num__80 - num__2 x <o> d ) d ) num__120 - x <o> e ) e ) num__40 - x / num__2 |
let ' s say the two cars have speeds v num__1 and v num__2 . the fact that they are moving in opposite direction means that their relative speed is ( v num__1 + v num__2 ) . in other words any gap between them will be changing in size at a rate of ( v num__1 + v num__2 ) . it does n ' t matter whether they are moving toward each other or away from each other . if they are approaching each other the gap between them is decreasing at a rate of ( v num__1 + v num__2 ) . if they are moving away from each other the gap between them is increasing at a rate of ( v num__1 + v num__2 ) . either way the number for the rate of change remains the same . here the two cars approach a distance num__40 mi then move away from each other another distance of num__40 miles . that ' s a total distance of num__80 miles in num__1.5 hr which gives a rate of : r = ( num__80 mi ) / ( num__1.5 ) = num__53.33 mph that ' s the rate of change of the gap so it must equal the sum of the speeds of the two cars . one of the speeds is x and let ' s call the other y . we want y . x + y = num__53.33 y = num__53.33 - x answer = ( b ) <eor> b <eos> |
b |
multiply__40.0__2.0__ round__53.33__ |
multiply__40.0__2.0__ divide__53.33__1.0__ |
| a jeep takes num__4 hours to cover a distance of num__620 km . how much should the speed in kmph be maintained to cover the same direction in num__1.5 th of the previous time ? <o> a ) num__148 kmph <o> b ) num__152 kmph <o> c ) num__106 kmph <o> d ) num__103 kmph <o> e ) num__165 kmph |
time = num__4 distance = num__620 num__1.5 of num__4 hours = num__4 * num__1.5 = num__6 hours required speed = num__103.333333333 = num__103 kmph d ) <eor> d <eos> |
d |
multiply__4.0__1.5__ divide__620.0__6.0__ round__103.0__ |
multiply__4.0__1.5__ divide__620.0__6.0__ round__103.0__ |
| find at what time between num__8 and num__9 o ' clock will the hands of a clock be in the same straight line but not together . <o> a ) num__10 x num__1.1 <o> b ) num__10 x num__0.909090909091 <o> c ) num__11 x num__0.909090909091 <o> d ) num__12 x num__0.909090909091 <o> e ) none |
solution at num__8 o ' clock the hour is at num__8 and the minute hand is at num__12 i . e . the two hands are num__20 min . spaces apart . so the minute hand gain = ( num__30 - num__20 ) minute num__55 minutes will be gained in num__60 min . num__10 minutes spaces will be gained in ( num__1.09090909091 x num__10 ) min . = num__10 x num__0.909090909091 min . the hands will be in the same straight line but not together at num__10 x num__0.909090909091 min . past num__8 . answer b <eor> b <eos> |
b |
add__8.0__12.0__ subtract__30.0__20.0__ divide__60.0__55.0__ subtract__20.0__10.0__ |
add__8.0__12.0__ subtract__30.0__20.0__ divide__60.0__55.0__ subtract__20.0__10.0__ |
| on sunday morning pugsley and wednesday are trading pet spiders . if pugsley were to give wednesday four of his spiders wednesday would then have four times as many spiders as pugsley does . but if wednesday were to give pugsley five of her spiders pugsley would now have five fewer spiders than wednesday had before they traded . how many pet spiders does pugsley have before the trading game commences ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__10 <o> e ) num__12 |
if pugsley were to give wednesday four of his spiders wednesday would then have five times as many spiders as pugsley does : ( w + num__4 ) = num__4 ( p - num__4 ) if wednesday were to give pugsley five of her spiders pugsley would now have five fewer spiders than wednesday had before they traded : p + num__5 = w - num__5 solving gives p = num__10 and w = num__20 . answer : d . <eor> d <eos> |
d |
multiply__4.0__5.0__ round__10.0__ |
multiply__4.0__5.0__ subtract__20.0__10.0__ |
| to apply for the position of photographer at a local magazine a photographer needs to include two or three photos in an envelope accompanying the application . if the photographer has pre - selected six photos representative of her work how many choices does she have to provide the photos for the magazine ? <o> a ) num__32 <o> b ) num__35 <o> c ) num__36 <o> d ) num__40 <o> e ) num__42 |
num__6 c num__2 + num__6 c num__3 = num__15 + num__20 = num__35 the answer is b . <eor> b <eos> |
b |
divide__6.0__2.0__ add__15.0__20.0__ add__15.0__20.0__ |
divide__6.0__2.0__ add__15.0__20.0__ add__15.0__20.0__ |
| if x is an integer and y = num__3 x + num__2 which of the following can not be a divisor of y ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
just to add some more to bunuel ' s explanation if a number is a multiple of num__3 it will be in a form : n = num__3 * k where k is an integer and if a number is a multiple of num__6 it will be in a form : m = num__6 * l where l is an integer so m = num__2 * num__3 * l = ( num__2 * l ) * num__3 = num__3 * p where p is an integer . in other words the prime factorization of m must have num__3 as a factor if it ' s a multiple of num__6 but the expression ca n ' t have that ( as explained by bunuel ) so it ca n ' t be a multiple of num__6 either answer : c <eor> c <eos> |
c |
multiply__3.0__2.0__ multiply__3.0__2.0__ |
multiply__3.0__2.0__ multiply__3.0__2.0__ |
| a book store bought copies of a new book by a popular author in anticipation of robust sales . the store bought num__400 copies from their supplier each copy at wholesale price w . the store sold the first num__150 copies in the first week at num__80.0 more than w and then over the next month sold a num__100 more at num__20.0 more than w . finally to clear shelf space the store sold the remaining copies to a bargain retailer at num__40.0 less than w . what was the bookstore ’ s net percent profit or loss on the entire lot of num__400 books ? <o> a ) num__30.0 loss <o> b ) num__10.0 loss <o> c ) num__10.0 profit <o> d ) num__20.0 profit <o> e ) num__60.0 profit |
c . p . = num__400 w s . p . = num__1.8 w * num__150 + num__1.2 w * num__100 + num__0.6 w * num__150 = num__480 w since s . p . > c . p . there is net profit . profit = s . p . - c . p . = num__480 w - num__400 w = num__80 w profit % = ( profit / c . p . ) * num__100 = ( num__80 w / num__400 w ) * num__100 = num__20.0 so the correct answer is d . <eor> d <eos> |
d |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| if num__100 ! / x is not an integer which of the following could be the value of x ? <o> a ) num__5 ^ num__24 <o> b ) num__7 ^ num__16 <o> c ) num__11 ^ num__9 <o> d ) num__13 ^ num__6 <o> e ) num__17 ^ num__6 |
he answer will be a number which does n ' t completely divide num__100 ! no of num__5 ' s in num__100 ! = num__24 no of num__7 ' s in num__100 ! = num__16 no of num__11 ' s in num__100 ! = num__9 no of num__13 ' s in num__100 ! = num__7 no of num__17 ' s in num__100 ! = num__5 there are num__5 seventeens in num__100 ! however we are dividing by num__17 ^ num__6 so this can not divide num__100 ! hence our answer will be e <eor> e <eos> |
e |
subtract__16.0__5.0__ subtract__16.0__7.0__ subtract__24.0__11.0__ subtract__24.0__7.0__ subtract__11.0__5.0__ add__6.0__11.0__ |
subtract__16.0__5.0__ subtract__16.0__7.0__ subtract__24.0__11.0__ subtract__24.0__7.0__ subtract__11.0__5.0__ add__6.0__11.0__ |
| p is two years older than q who is twice as old as r . the total of the ages of p q and r is num__25 . how old is q ? <o> a ) num__7.2 years <o> b ) num__8.2 years <o> c ) num__9.8 years <o> d ) num__9.2 years <o> e ) num__9.4 years |
let age of r = x . then age of q = num__2 x age of p = num__2 + num__2 x total age of p q and r = num__25 ⇒ ( num__2 + num__2 x ) + num__2 x + x = num__25 ⇒ num__5 x = num__23 ⇒ x = num__4.6 = num__4.6 years b ' s age = num__2 x = num__2 × num__4.6 = num__9.2 years answer : d <eor> d <eos> |
d |
subtract__25.0__2.0__ divide__23.0__5.0__ multiply__2.0__4.6__ multiply__2.0__4.6__ |
subtract__25.0__2.0__ divide__23.0__5.0__ multiply__2.0__4.6__ multiply__2.0__4.6__ |
| a man can row with a speed of num__20 kmph in still water . if the stream flows at num__7 kmph then the speed in downstream is ? <o> a ) num__27 <o> b ) num__23 <o> c ) num__20 <o> d ) num__27 <o> e ) num__12 |
m = num__20 s = num__7 ds = num__20 + num__7 = num__27 answer : d <eor> d <eos> |
d |
add__20.0__7.0__ round__27.0__ |
add__20.0__7.0__ add__20.0__7.0__ |
| a theater box office sold an average ( arithmetic mean ) of num__65 tickets per staff member to a particular movie . among the daytime staff the average number sold per member was num__75 and among the evening staff the average number sold was num__60 . if there are no other employees what was the ratio of the number of daytime staff members to the number of evening staff members ? <o> a ) num__1 : num__2 <o> b ) num__1 : num__4 <o> c ) num__5 : num__11 <o> d ) num__1 : num__3 <o> e ) num__13 : num__15 |
deviation from the mean for the daytime staff = num__75 - num__65 = num__10 . deviation from the mean for the evening staff = num__65 - num__60 = num__5 . thus the ratio of the number of daytime staff members to the number of evening staff members is num__5 : num__10 = num__1 : num__2 . the answer is a . <eor> a <eos> |
a |
subtract__75.0__65.0__ subtract__65.0__60.0__ divide__10.0__5.0__ reverse__1.0__ |
subtract__75.0__65.0__ subtract__65.0__60.0__ divide__10.0__5.0__ subtract__2.0__1.0__ |
| present ages of sameer and anand are in the ratio of num__5 : num__4 respectively . four years hence the ratio of their ages will become num__11 : num__9 respectively . what is anand ' s present age in years ? <o> a ) a ) num__24 <o> b ) b ) num__89 <o> c ) c ) num__67 <o> d ) d ) num__32 <o> e ) e ) num__45 |
let the present ages of sameer and anand be num__5 x years and num__4 x years respectively . then ( num__5 x + num__1.0 x + num__4 ) = num__1.22222222222 num__9 ( num__5 x + num__4 ) = num__11 ( num__4 x + num__4 ) num__45 x + num__36 = num__44 x + num__44 num__45 x - num__44 x = num__44 - num__36 x = num__8 . anand ' s present age = num__4 x = num__32 years . answer : d <eor> d <eos> |
d |
subtract__5.0__4.0__ divide__11.0__9.0__ multiply__5.0__9.0__ multiply__4.0__9.0__ multiply__4.0__11.0__ subtract__9.0__1.0__ multiply__4.0__8.0__ multiply__4.0__8.0__ |
subtract__5.0__4.0__ divide__11.0__9.0__ multiply__5.0__9.0__ subtract__45.0__9.0__ subtract__45.0__1.0__ subtract__9.0__1.0__ subtract__36.0__4.0__ subtract__36.0__4.0__ |
| if a man can cover num__18 metres in one second how many kilometres can he cover in num__3 hours num__45 minutes ? <o> a ) num__288 <o> b ) num__162 <o> c ) num__878 <o> d ) num__168 <o> e ) num__243 |
num__18 m / s = num__12 * num__3.6 kmph num__3 hours num__45 minutes = num__3 num__0.75 hours = num__3.75 hours distance = speed * time = num__18 * num__3.6 * num__3.75 km = num__243 km . answer : e <eor> e <eos> |
e |
add__3.0__0.75__ round__243.0__ |
divide__45.0__12.0__ round__243.0__ |
| what is the ratio between perimeters of two squares one having num__3 times the diagonal then the other ? <o> a ) num__3 : num__8 <o> b ) num__3 : num__6 <o> c ) num__3 : num__7 <o> d ) num__3 : num__1 <o> e ) num__3 : num__3 |
d = num__3 d d = d a √ num__2 = num__3 d a √ num__2 = d a = num__3 d / √ num__2 a = d / √ num__2 = > num__3 : num__1 answer : d <eor> d <eos> |
d |
multiply__3.0__1.0__ |
multiply__3.0__1.0__ |
| six bells commence tolling together and toll at intervals of num__2 num__4 num__6 num__8 num__10 and num__12 seconds respectively . in num__30 minutes how many times do they toll together ? <o> a ) num__22 <o> b ) num__72 <o> c ) num__37 <o> d ) num__16 <o> e ) num__88 |
explanation : l . c . m . of num__2 num__4 num__6 num__8 num__10 num__12 is num__120 . so the bells will toll together after every num__120 seconds ( num__2 minutes ) . in num__30 minutes they will together ( num__15.0 ) + num__1 = num__16 times answer : d ) num__16 <eor> d <eos> |
d |
multiply__4.0__30.0__ divide__30.0__2.0__ multiply__2.0__8.0__ round__16.0__ |
multiply__4.0__30.0__ divide__30.0__2.0__ add__4.0__12.0__ add__4.0__12.0__ |
| a train of length num__150 metres takes num__40.5 seconds to cross a tunnel of length num__300 metres . what is the speed of the train in km / hr ? <o> a ) num__13.33 <o> b ) num__26.67 <o> c ) num__40 <o> d ) num__66.67 <o> e ) num__50 |
explanation : speed = num__150 + num__300 / num__40.5 m / sec = num__450 / num__40.5 = num__3.6 km / hr = num__40 km / hr . answer is c <eor> c <eos> |
c |
add__150.0__300.0__ round__40.0__ |
add__150.0__300.0__ round__40.0__ |
| three numbers are in the ratio num__1 : num__2 : num__3 and their h . c . f is num__3 . the numbers are <o> a ) num__12 num__24 num__30 <o> b ) num__12 num__24 num__38 <o> c ) num__12 num__24 num__362 <o> d ) num__3 num__6 num__9 <o> e ) num__12 num__24 num__321 |
explanation : let the required numbers be x num__2 x num__3 x . then their h . c . f = x . so x = num__3 \ inline \ fn _ jvn \ therefore the numbers are num__3 num__6 num__9 answer : d ) num__3 num__6 num__9 <eor> d <eos> |
d |
multiply__2.0__3.0__ add__3.0__6.0__ add__1.0__2.0__ |
multiply__2.0__3.0__ add__3.0__6.0__ add__1.0__2.0__ |
| a miniature roulette wheel is divided into num__10 equal sectors each bearing a distinct integer from num__1 to num__10 inclusive . each time the wheel is spun a ball randomly determines the winning sector by settling in that sector . if the wheel is spun two times approximately what is the probability that the product of the two winning sectors ’ integers will be even ? <o> a ) num__88.0 <o> b ) num__75.0 <o> c ) num__67.0 <o> d ) num__63.0 <o> e ) num__50 % |
the only way to have an odd product is if both integers are odd . p ( odd product ) = num__0.5 * num__0.5 = num__0.25 p ( even product ) = num__1 - num__0.25 = num__0.75 = num__75.0 the answer is b . <eor> b <eos> |
b |
subtract__1.0__0.25__ multiply__1.0__75.0__ |
subtract__1.0__0.25__ multiply__1.0__75.0__ |
| find the sum of first num__30 natural numbers <o> a ) num__470 <o> b ) num__468 <o> c ) num__465 <o> d ) num__463 <o> e ) num__487 |
explanation : sum of n natural numbers = n ( n + num__1 ) / num__2 = num__30 ( num__30 + num__1 ) / num__2 = num__30 ( num__31 ) / num__2 = num__465 option c <eor> c <eos> |
c |
add__30.0__1.0__ multiply__1.0__465.0__ |
add__30.0__1.0__ divide__465.0__1.0__ |
| john takes a trip and drives num__9 hours from town x to town z at a rate of num__60 miles per hour . on his way back john drives num__85 miles per hour and stops in town y which is midway between town x and town z . how many hours does it take scott to drive from town z to town y ? <o> a ) num__1.75 <o> b ) num__1.25 <o> c ) num__1.45 <o> d ) num__4.15 <o> e ) num__3.14 |
distance from x to z = num__9 hr * num__60 mph = num__540 miles hence distance from y to z = num__0.5 * num__540 = num__270 time = num__3.17647058824 = num__4.15 hrs imo : d <eor> d <eos> |
d |
multiply__9.0__60.0__ multiply__0.5__540.0__ divide__270.0__85.0__ round__4.15__ |
multiply__9.0__60.0__ multiply__0.5__540.0__ divide__270.0__85.0__ round__4.15__ |
| a company has num__100 employees num__40.0 of whom are employed part time . if it hires num__25 new employees num__40.0 of whom are employed part time what will be the percent increase in part time employees ? <o> a ) num__1 num__0.666666666667 % <o> b ) num__2 num__0.666666666667 % <o> c ) num__4.0 <o> d ) num__25.0 <o> e ) num__9 % |
total employee = num__100 part time empl = num__40 new part time empl = num__25 * ( num__0.4 ) = num__10 total new part time emp = num__10 + num__40 = num__50 old part time emp = num__40.0 increase = ( new - old ) * num__100 / old = num__10 * num__2.5 = num__25.0 ans - d <eor> d <eos> |
d |
percent__40.0__25.0__ percent__25.0__10.0__ percent__100.0__25.0__ |
percent__40.0__25.0__ percent__25.0__10.0__ percent__100.0__25.0__ |
| to furnish a room in a model home an interior decorator is to select num__4 chairs and num__4 tables from the collection of chairs and tables in a warehouse that are all different from each other . if there are num__5 chairs in the warehouse and if num__75 different combinations are possible how many tables are there in the warehouse ? <o> a ) a ) num__6 <o> b ) b ) num__8 <o> c ) c ) num__10 <o> d ) d ) num__15 <o> e ) e ) num__30 |
total number of ways to choose num__4 chairs from num__5 = num__5 c num__4 = num__5 total number of combination = num__75 that means we need get num__15 combinations from the selection of tables . . . . screening through the answers . . . . num__6 c num__4 = num__15 . . . . . num__15 * num__5 = num__75 answer is num__6 . . . a <eor> a <eos> |
a |
die_space__ die_space__ |
die_space__ die_space__ |
| what will come in place of the x in the following number series ? num__46080 num__3840 ? num__48 num__8 num__2 num__1 <o> a ) num__1 <o> b ) num__384 <o> c ) num__5 <o> d ) num__7 <o> e ) num__9 |
num__3840.0 = num__3840 num__384.0 = num__384 num__48.0 = num__48 num__8.0 = num__8 num__2.0 = num__2 num__1.0 = num__1 b <eor> b <eos> |
b |
multiply__48.0__8.0__ multiply__48.0__8.0__ |
multiply__48.0__8.0__ multiply__48.0__8.0__ |
| find the average of first num__20 natural numbers ? <o> a ) num__5.5 <o> b ) num__7.2 <o> c ) num__10.5 <o> d ) num__12.3 <o> e ) num__15.5 |
sum of first n natural numbers = n ( n + num__1 ) / num__2 sum of first num__20 natural numbers = num__20 * num__10.5 = num__210 average = num__10.5 = num__10.5 answer is c <eor> c <eos> |
c |
multiply__20.0__10.5__ multiply__10.5__1.0__ |
multiply__20.0__10.5__ multiply__10.5__1.0__ |
| a password of a computer used two digits where they are from num__0 and num__9 . what is the probability that the password solely consists of prime numbers and zero ? <o> a ) num__0.03125 <o> b ) num__0.0625 <o> c ) num__0.125 <o> d ) num__0.4 <o> e ) num__0.25 |
we have two digits can be used for password - num__0 num__23 num__57 . assuming that the numbers can be repeated any number of times in the password probability of selecting any one of them is num__0.5 ( out of num__10 digits from num__0 - num__9 ) so num__0.5 * num__0.5 = num__0.25 ( e ) <eor> e <eos> |
e |
union_prob__0.0__0.5__0.25__ |
union_prob__0.0__0.5__0.25__ |
| three times the first of three consecutive odd integers is num__8 more than twice the third . the third integer is : <o> a ) num__9 <o> b ) num__11 <o> c ) num__13 <o> d ) num__15 <o> e ) num__20 |
let the three integers be x x + num__2 and x + num__4 . then num__3 x = num__2 ( x + num__4 ) + num__8 x = num__16 third integer = x + num__4 = num__20 . answer : e <eor> e <eos> |
e |
divide__8.0__2.0__ multiply__8.0__2.0__ add__4.0__16.0__ add__4.0__16.0__ |
divide__8.0__2.0__ multiply__8.0__2.0__ add__4.0__16.0__ add__4.0__16.0__ |
| two trains num__111 meters and num__165 meters in length respectively are running in opposite directions one at the rate of num__60 km and the other at the rate of num__90 kmph . in what time will they be completely clear of each other from the moment they meet ? <o> a ) num__4.85 <o> b ) num__7.85 <o> c ) num__6.85 <o> d ) num__5.85 <o> e ) num__6.62 |
t = ( num__111 + num__165 ) / ( num__60 + num__90 ) * num__3.6 t = num__6.62 answer : e <eor> e <eos> |
e |
round__6.62__ |
round__6.62__ |
| a train num__500 m long is running at a speed of num__78 km / hr . if it crosses a tunnel in num__1 min then the length of the tunnel is ? <o> a ) num__277 m <o> b ) num__700 m <o> c ) num__800 m <o> d ) num__187 m <o> e ) num__1678 m |
speed = num__78 * num__0.277777777778 = num__21.6666666667 m / sec . time = num__1 min = num__60 sec . let the length of the train be x meters . then ( num__500 + x ) / num__60 = num__21.6666666667 x = num__800 m . answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ round__800.0__ |
hour_to_min_conversion__ multiply__1.0__800.0__ |
| the price of a radio was increased by num__25 percent . the new price was then increased by num__40 percent . a single increase of what percent is equivalent to these two successive increases ? <o> a ) num__80.0 <o> b ) num__75.0 <o> c ) num__65.0 <o> d ) num__50.0 <o> e ) num__45 % |
consider base price - $ num__100 num__25.0 increase = num__1.25 * num__100 = $ num__125 another num__40.0 increase on new price = num__1.4 * num__125 = $ num__175 so final price of radio - $ num__175 therefore a num__75.0 increase correct option - b <eor> b <eos> |
b |
percent__100.0__75.0__ |
percent__100.0__75.0__ |
| two trains start from p and q respectively and travel towards each other at a speed of num__50 km / hr and num__40 km / hr respectively . by the time they meet the first train has travelled num__100 km more than the second . the distance between p and q is : <o> a ) num__500 km <o> b ) num__630 km <o> c ) num__900 km <o> d ) num__660 km <o> e ) none |
sol . at the time of meeting let the distane travelled byb the second train be x km . then distance covered by the first train is ( x + num__100 ) km ∴ x / num__40 = ( x + num__100 ) / num__50 ⇔ num__50 x = num__40 x num__4000 ⇔ x = num__400 . so distance between p and q = ( x + x + num__100 ) km = num__900 km . answer c <eor> c <eos> |
c |
multiply__40.0__100.0__ round__900.0__ |
multiply__40.0__100.0__ round__900.0__ |
| the seating chart of an airplane shows num__30 rows of seats . each row has num__3 seats on each side of the center aisle and one of the seats on each side is a window saet . the view from the window seats in num__25 of the rows is obscured by the wings of the airplane . if the first person to be assigned a seat is assigned a window seat and thw window seat is assigned randomly what is the probability that the person will get a seat with an unobscured view ? <o> a ) num__0.166666666667 <o> b ) num__0.333333333333 <o> c ) num__0.666666666667 <o> d ) num__0.833333333333 <o> e ) num__0.944444444444 |
priyalr num__6 seats per row think of a boeing num__737 . we have num__30 rows therefore window num__30 seats one one side and num__30 window seats on the other totaling num__60 window seats on the whole plane . the view of the window of num__25 rows is blocked . two wings therefore num__50 window seats are blocked . total window seats = num__60 total blocked window seats = num__50 total unblocked seats = num__10 we know that a window seat was given therefore probability for not window seat is num__0.166666666667 = num__0.166666666667 ans a <eor> a <eos> |
a |
hour_to_min_conversion__ divide__30.0__3.0__ divide__10.0__60.0__ divide__10.0__60.0__ |
hour_to_min_conversion__ divide__30.0__3.0__ divide__10.0__60.0__ divide__10.0__60.0__ |
| a certain bakery sells six different - sized wedding cakes . each cake costs x dollars more than the next one below it in size and the price of the largest cake is $ num__24.50 . if the sum of the prices of the six different cakes is $ num__109.50 what is the value of x ? <o> a ) num__1.50 <o> b ) num__1.75 <o> c ) num__2.00 <o> d ) num__2.50 <o> e ) num__3.00 |
since the price of the largest cake ( num__6 th ) is $ num__24.50 then the price of the smallest cake ( num__1 st ) is $ ( num__24.50 - num__5 x ) . now the prices of the cakes are evenly spaced so the sum of the prices is ( average price ) * ( # of cakes ) = ( first + last ) / num__2 * ( # of cakes ) . so ( num__24.50 - num__5 x + num__24.50 ) / num__2 * num__6 = num__109.50 - - > x = num__2.5 . answer : d . <eor> d <eos> |
d |
subtract__6.0__1.0__ divide__5.0__2.0__ multiply__1.0__2.5__ |
subtract__6.0__1.0__ divide__5.0__2.0__ multiply__1.0__2.5__ |
| a man can row downstream at num__25 kmph and upstream at num__15 kmph . find the time to reach a destination point at a distance of num__50 km along the downstream . <o> a ) num__4 hrs . <o> b ) num__1 hrs . <o> c ) num__3 hrs . <o> d ) num__2 hrs . <o> e ) num__2 num__0.5 hrs . |
let the speed of the man in downstream = num__25 kmph . distance to be covered along the stream = num__50 kmph = > time taken = d / s = num__2.0 = num__2 hrs . answer : d <eor> d <eos> |
d |
divide__50.0__25.0__ round__2.0__ |
divide__50.0__25.0__ divide__50.0__25.0__ |
| find the number that fits somewhere into the middle of the series . some of the items involve both numbers and letters look at this series : c num__19 e num__21 __ i num__25 k num__27 . . . what number should fill the blank ? <o> a ) d num__20 <o> b ) e num__21 <o> c ) f num__22 <o> d ) g num__23 <o> e ) h num__24 |
b g num__23 in this series the letters progress by num__2 and the numbers increase by num__2 . <eor> b <eos> |
b |
subtract__21.0__19.0__ add__19.0__2.0__ |
subtract__21.0__19.0__ add__19.0__2.0__ |
| a train crosses a platform of num__120 m in num__15 sec same train crosses another platform of length num__180 m in num__18 sec . then find the length of the train ? <o> a ) num__276 m <o> b ) num__180 m <o> c ) num__286 m <o> d ) num__288 m <o> e ) num__277 m |
length of the train be ‘ x ’ x + num__8.0 = x + num__10.0 num__6 x + num__720 = num__5 x + num__900 x = num__180 m answer : b <eor> b <eos> |
b |
divide__120.0__15.0__ divide__180.0__18.0__ multiply__120.0__6.0__ subtract__15.0__10.0__ multiply__180.0__5.0__ round__180.0__ |
divide__120.0__15.0__ divide__180.0__18.0__ multiply__120.0__6.0__ subtract__15.0__10.0__ add__180.0__720.0__ round__180.0__ |
| the ratio of three numbers is num__2 : num__3 : num__5 and their sum is num__150 . the second number of the three numbers is ? <o> a ) num__24 <o> b ) num__26 <o> c ) num__27 <o> d ) num__29 <o> e ) num__45 |
num__2 : num__3 : num__5 total parts = num__10 num__10 parts - - > num__150 num__1 part - - - - > num__15 the second number of the three numbers is = num__3 * num__15 = num__45 answer : e <eor> e <eos> |
e |
multiply__2.0__5.0__ subtract__3.0__2.0__ multiply__3.0__5.0__ multiply__3.0__15.0__ multiply__3.0__15.0__ |
multiply__2.0__5.0__ subtract__3.0__2.0__ multiply__3.0__5.0__ multiply__3.0__15.0__ multiply__3.0__15.0__ |
| for any real number x the operatoris defined as : ( x ) = x ( num__2 − x ) if p + num__2 = ( p + num__2 ) then p = <o> a ) num__1 <o> b ) num__0 <o> c ) num__1 <o> d ) - num__2 <o> e ) num__3 |
( x ) = x ( num__2 − x ) ( p + num__2 ) = ( p + num__2 ) ( num__2 - p - num__2 ) = - p ( p + num__1 ) we are given that p + num__2 = ( p + num__2 ) therefore - p ( p + num__2 ) = ( p + num__2 ) or ( p + num__2 ) + p ( p + num__2 ) = num__0 ( p + num__2 ) ( p + num__1 ) = num__0 p = - num__1 p = - num__2 option d <eor> d <eos> |
d |
multiply__2.0__1.0__ |
multiply__2.0__1.0__ |
| calculate the circumference of a circular field whose radius is num__5 centimeters . <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
circumference c is given by c = num__2 π r = num__2 π * num__5 = num__10 π cm correct answer e <eor> e <eos> |
e |
multiply__5.0__2.0__ multiply__5.0__2.0__ |
multiply__5.0__2.0__ multiply__5.0__2.0__ |
| the average age of five persons is num__40 years while the average age of some other num__10 persons is num__25 years . the average age of all the num__15 persons is : <o> a ) num__27 years <o> b ) num__32 years <o> c ) num__37 years <o> d ) num__30 years <o> e ) num__35 years |
x / num__5 = = num__40 x = = num__200 after num__10 person x / num__10 = = num__25 x = = num__250 so avg btwn num__15 person is ( num__250 + num__200 ) / num__15 = = num__30 ans answer : d <eor> d <eos> |
d |
subtract__15.0__10.0__ multiply__40.0__5.0__ multiply__10.0__25.0__ subtract__40.0__10.0__ subtract__40.0__10.0__ |
subtract__15.0__10.0__ multiply__40.0__5.0__ multiply__10.0__25.0__ add__25.0__5.0__ add__25.0__5.0__ |
| a slot machine in a las vegas casino has an average profit of $ num__600 for each num__8 - hour shift for the five days sunday through thursday inclusive . if the average per - shift profit on friday and saturday is num__25.0 greater than on the other days of the week and the slot machine is in operation every hour of every day what is the total weekly profit that the casino makes from the slot machine ? <o> a ) a - num__4500 <o> b ) b - num__9000 <o> c ) c - num__13500 <o> d ) d - num__15500 <o> e ) e - num__27 |
000 |
correct answer is c - num__13500 . total profit per weekday = $ num__600 x num__3 ( shifts ) = $ num__1800 total profit per week ( apart from saturdaysunday ) = $ num__1800 x num__5 = $ num__9000 total profit per shift on weekends = $ num__600 x num__1.25 = $ num__750 total profit on weekends = $ num__750 x num__3 ( shifts ) x num__2 ( saturday sunday ) = $ num__4500 . hence the answer is - $ num__4500 + $ num__9000 = $ num__13500 . ( c ) <eor> c <eos> |
c |
c |
| what is the greatest prime factor of num__2 ^ num__10 - num__1 ? <o> a ) num__17 <o> b ) num__19 <o> c ) num__23 <o> d ) num__29 <o> e ) num__31 |
num__2 ^ num__10 - num__1 = ( num__2 ^ num__5 - num__1 ) ( num__2 ^ num__5 + num__1 ) = num__31 * num__33 the answer is e . <eor> e <eos> |
e |
divide__10.0__2.0__ add__2.0__31.0__ multiply__1.0__31.0__ |
divide__10.0__2.0__ add__2.0__31.0__ multiply__1.0__31.0__ |
| a train overtakes two persons walking along a railway track . the first person walks at num__4.5 km / hr and the other walks at num__5.4 km / hr . the train needs num__8.4 and num__8.5 seconds respectively to overtake them . what is the speed of the train if both the persons are walking in the same direction as the train ? <o> a ) num__81 km / hr <o> b ) num__88 km / hr <o> c ) num__62 km / hr <o> d ) num__46 km / hr <o> e ) num__34 km / hr |
explanation : let x is the length of the train in meter and y is its speed in kmph x / num__8.4 = ( y - num__4.5 ) ( num__0.277777777778 ) - - - ( num__1 ) x / num__8.5 = ( y - num__5.4 ) ( num__0.277777777778 ) - - - ( num__2 ) dividing num__1 by num__2 num__8.5 / num__8.4 = ( y - num__4.5 ) / ( y - num__5.4 ) = > num__8.4 y - num__8.4 × num__4.5 = num__8.5 y - num__8.5 × num__5.4 . num__1 y = num__8.5 × num__5.4 - num__8.4 × num__4.5 = > . num__1 y = num__45.9 - num__37.8 = num__8.1 = > y = num__81 km / hr answer : option a <eor> a <eos> |
a |
multiply__5.4__8.5__ multiply__4.5__8.4__ subtract__45.9__37.8__ round__81.0__ |
multiply__5.4__8.5__ multiply__4.5__8.4__ subtract__45.9__37.8__ divide__81.0__1.0__ |
| there are num__32 stations between ernakulam and chennai . how many second class tickets have to be printed so that a passenger can travel from one station to any other station ? <o> a ) num__1800 <o> b ) num__1820 <o> c ) num__1150 <o> d ) num__1122 <o> e ) num__1900 |
the total number of stations = num__34 from num__34 stations we have to choose any two stations and the direction of travel ( ernakulam to chennai is different from chennai to ernakulam ) in num__34 p num__2 ways . num__34 p num__2 = num__34 * num__33 = num__1122 answer : d <eor> d <eos> |
d |
subtract__34.0__32.0__ multiply__33.0__34.0__ round__1122.0__ |
subtract__34.0__32.0__ multiply__33.0__34.0__ multiply__33.0__34.0__ |
| a man spends num__0.2 of his salary on food num__0.1 of his salary on house rent and num__0.6 of his salary on clothes . he still has $ num__18000 left with him . find his salary . <o> a ) $ num__180000 <o> b ) $ num__280000 <o> c ) $ num__380000 <o> d ) $ num__480000 <o> e ) none |
solution : the expenditure incurred on each item is expressed as part of the total amount ( salary ) so it is an independent activity . in general for independent activities [ num__1 – ( x num__1 / y num__1 + x num__2 / y num__2 + x num__3 / y num__3 ) ] × total amount = balance amount [ num__1 – ( num__0.2 + num__0.1 + num__0.6 ) ] × total salary = $ num__18000 [ num__1 – num__0.9 ] × total salary = $ num__18000 therefore total salary = $ num__18000 × num__10 = $ num__180000 answer a <eor> a <eos> |
a |
divide__0.2__0.1__ divide__0.6__0.2__ subtract__1.0__0.1__ reverse__0.1__ divide__18000.0__0.1__ divide__18000.0__0.1__ |
divide__0.2__0.1__ divide__0.6__0.2__ subtract__1.0__0.1__ reverse__0.1__ divide__18000.0__0.1__ divide__18000.0__0.1__ |
| a person can swim in still water at num__4 km / h . if the speed of water num__2 km / h how many hours will the man take to swim back against the current for num__6 km ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__9 |
m = num__4 s = num__2 us = num__4 - num__2 = num__2 d = num__6 t = num__3.0 = num__3 answer : a <eor> a <eos> |
a |
divide__6.0__2.0__ round__3.0__ |
divide__6.0__2.0__ subtract__6.0__3.0__ |
| bradley owns b video game cartridges . if bradley ’ s total is one - third the total owned by andrew and twice the total owned by charlie how many video game cartridges do the three of them own altogether in terms of b ? <o> a ) ( num__5.33333333333 ) b <o> b ) ( num__4.25 ) b <o> c ) ( num__3.25 ) b <o> d ) ( num__4.5 ) b <o> e ) ( num__0.583333333333 ) b |
step num__1 : categorize the problem this problem is testing basic algebra and equations with an extra layer of complexity by having variables in the answer choices . step num__2 : think like the test maker what is the key detail or key details in the problem ? the key detail to immediately recognize is that the problem gives you andrew and charlie ’ s totals in relation to bradley ’ s total rather than giving you bradley ’ s total in relation to the other two ; and then the problem asks you to solve the problem in terms of b . this means that you have to relate andrew and charlie ’ s values in relation to bradley ’ s total . the test maker is attempting to force you to structure the problem in the opposite way that most people are used to structuring information . by gaining this insight it makes it easier to solve the problem . step num__3 : solve the problem b = bradley ’ s total num__3 b = andrew ’ s total ( num__0.5 ) b = charlie ’ s total add each total b + num__3 b + ( num__0.5 ) b = num__4 b + ( num__0.5 ) b = ( num__4.0 ) b + ( num__0.5 ) b = ( num__4.5 ) b therefore choose d . <eor> d <eos> |
d |
add__1.0__2.0__ reverse__2.0__ add__1.0__3.0__ add__0.5__4.0__ add__0.5__4.0__ |
add__1.0__2.0__ reverse__2.0__ add__1.0__3.0__ add__0.5__4.0__ add__0.5__4.0__ |
| if a - b = num__2 and a num__2 + b num__2 = num__25 find the value of ab . <o> a ) a ) num__10.5 <o> b ) b ) num__12 <o> c ) c ) num__15 <o> d ) d ) num__18 <o> e ) e ) num__20 |
explanation : num__2 ab = ( a num__2 + b num__2 ) - ( a - b ) num__2 = num__25 - num__4 = num__21 ab = num__10.5 answer : a <eor> a <eos> |
a |
subtract__25.0__4.0__ divide__21.0__2.0__ divide__21.0__2.0__ |
subtract__25.0__4.0__ divide__21.0__2.0__ subtract__21.0__10.5__ |
| num__3 num__7 num__15 num__31 num__63 ? <o> a ) num__125 <o> b ) num__112 <o> c ) num__153 <o> d ) num__173 <o> e ) num__127 |
e num__127 each number in the series is the preceding number multiplied by num__2 and then increased by num__1 . <eor> e <eos> |
e |
subtract__3.0__2.0__ multiply__1.0__127.0__ |
subtract__3.0__2.0__ multiply__1.0__127.0__ |
| if num__4 men can colour num__48 m long cloth in num__2 days then num__6 men can colour num__36 m long cloth in <o> a ) num__2 days <o> b ) num__1 day <o> c ) num__3 days <o> d ) num__5 days <o> e ) num__7 days |
the length of cloth painted by one man in one day = num__12.0 × num__2 = num__6 m no . of days required to paint num__36 m cloth by num__6 men = num__6.0 × num__6 = num__1 day . b ) <eor> b <eos> |
b |
divide__48.0__4.0__ round__1.0__ |
divide__48.0__4.0__ round__1.0__ |
| if two projectiles are launched at the same moment from num__1998 km apart and travel directly towards each other at num__444 km per hour and num__555 km per hour respectively how many minutes will it take for them to meet ? <o> a ) num__90 <o> b ) num__100 <o> c ) num__110 <o> d ) num__120 <o> e ) num__130 |
the projectiles travel a total of num__999 km per hour . the time to meet is num__2.0 = num__2 hours = num__120 minutes the answer is d . <eor> d <eos> |
d |
add__444.0__555.0__ divide__1998.0__999.0__ round__120.0__ |
add__444.0__555.0__ divide__1998.0__999.0__ round__120.0__ |
| one - third of rahul ' s savings in national savings certificate is equal to one - half of his savings in public provident fund . if he has rs . num__1 num__50000 as total savings how much has he saved in public provident fund ? <o> a ) rs . num__10000 <o> b ) rs . num__5896 <o> c ) rs . num__2580 <o> d ) rs . num__3697 <o> e ) rs . num__60000 |
explanation : let savings in n . s . c and p . p . f . be rs . x and rs . ( num__150000 - x ) respectively . then num__0.333333333333 x = num__0.5 ( num__150000 - x ) x / num__3 + x / num__2 = num__75000 x = num__75000 * num__1.2 = num__90000 savings in public provident fund = rs . ( num__150000 - num__90000 ) = rs . num__60000 answer : e <eor> e <eos> |
e |
divide__50000.0__150000.0__ divide__150000.0__50000.0__ reverse__0.5__ multiply__0.5__150000.0__ multiply__1.2__75000.0__ multiply__50000.0__1.2__ multiply__1.0__60000.0__ |
divide__50000.0__150000.0__ divide__150000.0__50000.0__ reverse__0.5__ multiply__0.5__150000.0__ multiply__1.2__75000.0__ multiply__50000.0__1.2__ multiply__1.0__60000.0__ |
| if the average ( arithmetic mean ) of ( num__2 a + num__16 ) and ( num__3 a - num__8 ) is num__84 what is the value of a ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__28 <o> d ) num__32 <o> e ) num__42 |
( ( num__2 a + num__16 ) + ( num__3 a - num__8 ) ) / num__2 = ( num__5 a + num__8 ) / num__2 = num__84 a = num__32 the answer is d . <eor> d <eos> |
d |
add__2.0__3.0__ multiply__2.0__16.0__ multiply__2.0__16.0__ |
add__2.0__3.0__ multiply__2.0__16.0__ multiply__2.0__16.0__ |
| consider a quarter of a circle of radius num__20 . let r be the radius of the circle inscribed in this quarter of a circle . find r . <o> a ) num__20 * ( sqr num__2 - num__1 ) <o> b ) num__8 * ( sqr num__3 - num__1 ) <o> c ) num__4 * ( sqr num__7 - num__1 ) <o> d ) num__12 * ( sqr num__7 - num__1 ) <o> e ) none of these |
i got num__20 / ( sqr num__2 + num__1 ) and just forgot to multiply by ( sqr num__2 - num__1 ) . answer a <eor> a <eos> |
a |
multiply__20.0__1.0__ |
multiply__20.0__1.0__ |
| x is the product of each integer from num__1 to num__25 inclusive and y = num__100 ^ k where k is an integer . what is the greatest value of k for which y is a factor of x ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__3 <o> d ) num__6 <o> e ) num__7 |
the number of trailing zeros in the decimal representation of n ! the factorial of a non - negative integer n can be determined with this formula : n num__5 + n num__52 + n num__53 + . . . + n num__5 k where k must be chosen such that num__5 k ≤ n x = num__1 * num__2 * num__3 . . . . * num__25 = num__25 ! no . of trailing zeros in num__25 ! = num__5.0 + num__5.0 ^ num__2 = num__6 num__100 ^ k = num__10 ^ num__2 k → k = num__3.0 = num__3 c <eor> c <eos> |
c |
add__1.0__52.0__ add__1.0__2.0__ add__1.0__5.0__ multiply__2.0__5.0__ add__1.0__2.0__ |
add__1.0__52.0__ add__1.0__2.0__ add__1.0__5.0__ multiply__2.0__5.0__ add__1.0__2.0__ |
| if n is a positive integer which of the following expressions must be even ? <o> a ) ( n − num__1 ) ( n + num__1 ) <o> b ) ( n − num__4 ) ( n + num__1 ) <o> c ) ( n − num__2 ) ( n + num__6 ) <o> d ) ( n − num__5 ) ( n + num__1 ) <o> e ) ( n − num__3 ) ( n + num__3 ) |
whether n is even or odd ( n - num__4 ) ( n + num__1 ) will have one odd factor and one even factor . the product will be even . the answer is b . <eor> b <eos> |
b |
multiply__1.0__4.0__ |
multiply__1.0__4.0__ |
| the speed at which a man can row a boat in still water is num__21 kmph . if he rows downstream where the speed of current is num__5 kmph what time will he take to cover num__90 metres ? <o> a ) num__23.46 <o> b ) num__27.46 <o> c ) num__28.46 <o> d ) num__12.46 <o> e ) num__25.46 |
speed of the boat downstream = num__21 + num__5 = num__26 kmph = num__26 * num__0.277777777778 = num__7.22 m / s hence time taken to cover num__90 m = num__90 / num__7.22 = num__12.46 seconds . answer : d <eor> d <eos> |
d |
add__21.0__5.0__ round__12.46__ |
add__21.0__5.0__ round__12.46__ |
| for every x the action [ x ] is defined : [ x ] is the greatest integer less than or equal to x . what is the value of [ num__6.5 ] x [ num__0.666666666667 ] + [ num__2 ] x num__7.2 + [ num__8.4 ] - num__6.2 ? <o> a ) num__12.6 . <o> b ) num__14.4 . <o> c ) num__15.8 . <o> d ) num__16.2 . <o> e ) num__16.4 . |
[ num__6.5 ] x [ num__0.666666666667 ] + [ num__2 ] x num__7.2 + [ num__8.4 ] - num__6.2 = num__6 * num__0 + num__2 * num__7.2 + num__8 - num__6.2 = num__0 + num__14.4 + num__1.8 num__16.2 answer d <eor> d <eos> |
d |
round_down__6.5__ round_down__0.6667__ round_down__8.4__ multiply__2.0__7.2__ divide__14.4__8.0__ add__14.4__1.8__ add__14.4__1.8__ |
round_down__6.5__ round_down__0.6667__ add__2.0__6.0__ multiply__2.0__7.2__ subtract__8.0__6.2__ add__14.4__1.8__ add__14.4__1.8__ |
| the price of a commodity ( in rupees per kilogram ) is num__100 + num__0.1 n on the nthvday of num__2007 ( n = num__12 num__100 ) and then remains constant . on the other hand the price of another commodity is num__89 + num__0.15 n on the nth day of num__2007 ( n = num__12 num__365 ) . on which date in num__2007 will the price of these twocommodities be equal ? ? ? ? <o> a ) num__11 th may <o> b ) num__12 th may <o> c ) num__20 th may <o> d ) num__22 th may <o> e ) num__24 th may |
price of both the commodity can not be same within num__100 days ( by puttng n = num__100 in both the price ) price of num__1 st commodity after num__100 days = num__100 + num__0.1 * num__100 = num__110 day on which num__2 nd commodity becomes equal to num__1 st = num__89 + num__0.15 n = num__110 n = num__140 num__140 th day of num__2007 is num__20 th may answer : c <eor> c <eos> |
c |
divide__2.0__0.1__ multiply__1.0__20.0__ |
divide__2.0__0.1__ multiply__1.0__20.0__ |
| the average age of num__15 students of a class is num__14 years . out of these the average age of num__5 students is num__14 years and that of the other num__9 students is num__16 years . tee age of the num__15 th student is ? <o> a ) num__3 years <o> b ) num__4 years <o> c ) num__5 years <o> d ) num__6 years <o> e ) num__7 years |
age of the num__15 th student = [ num__15 * num__14 - ( num__14 * num__5 + num__16 * num__9 ) ] = ( num__210 - num__214 ) = num__4 years . answer : b <eor> b <eos> |
b |
multiply__15.0__14.0__ subtract__9.0__5.0__ subtract__9.0__5.0__ |
multiply__15.0__14.0__ subtract__9.0__5.0__ subtract__9.0__5.0__ |
| a train is num__320 meter long is running at a speed of num__45 km / hour . in what time will it pass a bridge of num__140 meter length ? <o> a ) num__11 seconds <o> b ) num__38 seconds <o> c ) num__40 seconds <o> d ) num__88 seconds <o> e ) num__36.8 seconds |
speed = num__45 km / hr = num__45 * ( num__0.277777777778 ) m / sec = num__12.5 m / sec total distance = num__320 + num__140 = num__460 meter time = distance / speed = num__460 * ( num__0.08 ) = num__36.8 seconds answer : e <eor> e <eos> |
e |
add__320.0__140.0__ divide__460.0__12.5__ round__36.8__ |
add__320.0__140.0__ divide__460.0__12.5__ divide__460.0__12.5__ |
| the average age of num__15 students of a class is num__15 years . out of these the average age of num__5 students is num__14 years and that of the other num__9 students is num__16 years . tee age of the num__15 th student is ? <o> a ) num__11 years <o> b ) num__16 years <o> c ) num__33 years <o> d ) num__14 years <o> e ) num__18 years |
age of the num__15 th student = [ num__15 * num__15 - ( num__14 * num__5 + num__16 * num__9 ) ] = ( num__225 - num__214 ) = num__11 years . answer : a <eor> a <eos> |
a |
subtract__16.0__5.0__ subtract__16.0__5.0__ |
subtract__16.0__5.0__ subtract__16.0__5.0__ |
| a man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream . the ratio of the speed of the boat ( in still water ) and the stream is : <o> a ) num__3 : num__1 <o> b ) num__1 : num__3 <o> c ) num__1 : num__2 <o> d ) num__2 : num__1 <o> e ) none of these |
explanation : let speed upstream = x then speed downstream = num__2 x speed in still water = ( num__2 x + x ) / num__2 = num__3 x / num__2 speed of the stream = ( num__2 x − x ) / num__2 = x / num__2 speed in still water : speed of the stream = ( num__3 x / num__2 ) ( num__2 x / num__2 ) = num__3 : num__1 . answer : option a <eor> a <eos> |
a |
subtract__3.0__2.0__ round__3.0__ |
subtract__3.0__2.0__ add__1.0__2.0__ |
| two numbers are in the ratio of num__1 : num__2 . if num__6 be added to both their ratio changes to num__3 : num__5 . the greater number is <o> a ) num__20 <o> b ) num__24 <o> c ) num__28 <o> d ) num__32 <o> e ) num__36 |
let the ratio be x : y given x / y = num__0.5 ( x + num__6 ) / ( y + num__6 ) = num__0.6 = > x = num__12 and y = num__24 answer : b <eor> b <eos> |
b |
reverse__2.0__ divide__3.0__5.0__ multiply__2.0__6.0__ multiply__2.0__12.0__ multiply__1.0__24.0__ |
reverse__2.0__ divide__3.0__5.0__ divide__6.0__0.5__ divide__12.0__0.5__ divide__12.0__0.5__ |
| if x is num__20 percent more than y and y is num__60 percent less than z then x is what percent of z ? <o> a ) num__500.0 <o> b ) num__48.0 <o> c ) num__166.666666667 % <o> d ) num__125.0 <o> e ) num__60 % |
z = num__100 ; y = num__40 so x = num__48 x as % of z = num__0.48 * num__100 = > num__48.0 answer will be ( b ) <eor> b <eos> |
b |
subtract__60.0__20.0__ divide__48.0__100.0__ multiply__100.0__0.48__ |
subtract__60.0__20.0__ divide__48.0__100.0__ multiply__100.0__0.48__ |
| the sale price of an article including the sales tax is rs . num__616 . the rate of sales tax is num__10.0 . if the shopkeeper has made a profit of num__12.0 then the cost price of the article is : <o> a ) num__500 <o> b ) num__277 <o> c ) num__222 <o> d ) num__297 <o> e ) num__111 |
explanation : num__110.0 of s . p . = num__616 s . p . = ( num__616 * num__100 ) / num__110 = rs . num__560 c . p = ( num__110 * num__560 ) / num__112 = rs . num__500 answer : a <eor> a <eos> |
a |
percent__100.0__500.0__ |
percent__100.0__500.0__ |
| he average weight of num__8 persons increases by num__2.5 kg when a new person comes in place of one of them weighing num__75 kg . what might be the weight of the new person ? <o> a ) num__75 kg <o> b ) num__95 kg <o> c ) num__45 kg <o> d ) num__85 kg <o> e ) num__25 kg |
explanation : total weight increased = ( num__8 x num__2.5 ) kg = num__20 kg . weight of new person = ( num__75 + num__20 ) kg = num__95 kg . answer : b <eor> b <eos> |
b |
multiply__8.0__2.5__ add__75.0__20.0__ add__75.0__20.0__ |
multiply__8.0__2.5__ add__75.0__20.0__ add__75.0__20.0__ |
| the value of x is to be randomly selected from the integers from num__1 to num__11 inclusive and then substituted into the equation y = x ^ num__2 - num__4 x + num__3 . what is the probability that the value of y will be negative ? <o> a ) num__0.454545454545 <o> b ) num__0.363636363636 <o> c ) num__0.272727272727 <o> d ) num__0.181818181818 <o> e ) num__0.0909090909091 |
y will only be negative for x = num__2 . ( we can check the values from num__1 to num__11 to be certain . ) p ( y is negative ) = num__0.0909090909091 the answer is e . <eor> e <eos> |
e |
reverse__11.0__ reverse__11.0__ |
reverse__11.0__ reverse__11.0__ |
| a num__300 meter long train crosses a platform in num__42 seconds while it crosses a signal pole in num__18 seconds . what is the length of the platform ? <o> a ) num__287 m <o> b ) num__278 m <o> c ) num__400 m <o> d ) num__228 m <o> e ) num__282 m |
speed = [ num__16.6666666667 ] m / sec = num__16.6666666667 m / sec . let the length of the platform be x meters . then x + num__7.14285714286 = num__16.6666666667 num__3 ( x + num__300 ) = num__2100 è x = num__400 m . answer : c <eor> c <eos> |
c |
divide__300.0__18.0__ divide__300.0__42.0__ round__400.0__ |
divide__300.0__18.0__ divide__300.0__42.0__ round__400.0__ |
| water boils at num__212 ° f or num__100 ° c and melts at num__32 ° f or num__0 ° c . if the temperature of the particular day is num__35 ° c it is equal to <o> a ) num__40 <o> b ) num__80 <o> c ) num__95 <o> d ) num__100 <o> e ) num__120 |
let f and c denotes the temparature in fahrenheit anid celcsius respectively . then ( f - num__32 ) / ( num__212 - num__32 ) = ( c - num__0 ) / ( num__100 - num__0 ) if c = num__35 then f = num__95 . c <eor> c <eos> |
c |
round__95.0__ |
round__95.0__ |
| a cycle is bought for rs . num__900 and sold for rs . num__1080 find the gain percent ? <o> a ) num__70.0 <o> b ) num__20.0 <o> c ) num__60.0 <o> d ) num__10.0 <o> e ) num__80 % |
explanation : num__900 - - - - num__180 num__100 - - - - ? = > num__20.0 answer : b <eor> b <eos> |
b |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| the total marks obtained by a student in physics chemistry and mathematics is num__150 more than the marks obtained by him in physics . what is the average mark obtained by him in chemistry and mathematics ? <o> a ) num__48 <o> b ) num__12 <o> c ) num__75 <o> d ) num__28 <o> e ) num__91 |
let the marks obtained by the student in physics chemistry and mathematics be p c and m respectively . p + c + m = num__150 + p c + m = num__150 average mark obtained by the student in chemistry and mathematics = ( c + m ) / num__2 = num__75.0 = num__75 . option c <eor> c <eos> |
c |
divide__150.0__2.0__ divide__150.0__2.0__ |
divide__150.0__2.0__ divide__150.0__2.0__ |
| the average age of num__40 students is num__8 years . if the age of teacher is also included then their average age increases by half a year . what is the age of the teacher ? <o> a ) num__45 years <o> b ) num__48.5 years <o> c ) num__28.5 years <o> d ) num__26.5 years <o> e ) num__27.5 years |
total age of num__40 students = num__40 * num__8 = num__320 let the age of the the teacher be x then ( num__320 + x ) / num__41 = num__8 + num__0.5 = num__8 ½ . num__320 + x = num__8.5 * num__41 = num__348.5 = num__348.5 x = num__348.5 - num__320 = num__28.5 answer : c <eor> c <eos> |
c |
multiply__40.0__8.0__ add__8.0__0.5__ multiply__8.5__41.0__ subtract__348.5__320.0__ subtract__348.5__320.0__ |
multiply__40.0__8.0__ add__8.0__0.5__ multiply__8.5__41.0__ subtract__348.5__320.0__ subtract__348.5__320.0__ |
| two cars namely a and b start simultaneously from a certain place at the speed of num__40 kmph and num__55 kmph respectively . the car b reaches the destination num__2 hours earlier than a . what is the distance between the starting point and destination ? <o> a ) num__8 hours num__12 minutes <o> b ) num__6 hours num__15 minutes <o> c ) num__7 hours num__20 minutes <o> d ) num__7 hours num__12 minutes <o> e ) none |
explanation : let the time taken by car a to reach destination is t hours so the time taken by car b to reach destination is ( t – num__2 ) hours . s num__1 t num__1 = s num__2 t num__2 = > num__40 ( t ) = num__55 ( t – num__2 ) = > num__40 t = num__55 t - num__110 = > num__15 t = num__110 t = num__7 hours num__20 minutes answer – c <eor> c <eos> |
c |
multiply__55.0__2.0__ subtract__55.0__40.0__ divide__40.0__2.0__ round__7.0__ |
multiply__55.0__2.0__ subtract__55.0__40.0__ divide__40.0__2.0__ round__7.0__ |
| the length of a train and that of a platform are equal . if with a speed of num__36 k / hr the train crosses the platform in one minute then the length of the train ( in meters ) is ? <o> a ) num__299 <o> b ) num__300 <o> c ) num__299 <o> d ) num__750 <o> e ) num__261 |
speed = [ num__36 * num__0.277777777778 ] m / sec = num__10 m / sec ; time = num__1 min . = num__60 sec . let the length of the train and that of the platform be x meters . then num__2 x / num__60 = num__10 = > x = num__10 * num__30.0 = num__300 answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ divide__60.0__2.0__ multiply__10.0__30.0__ round__300.0__ |
hour_to_min_conversion__ divide__60.0__2.0__ multiply__10.0__30.0__ multiply__10.0__30.0__ |
| num__7.0 of the total quantity of rice is lost in grinding when a country has to import num__6 million tonnes but when only num__7 num__0.75 % is lost it can import num__3 million tonnes . find the quantity of rice grown in the country . <o> a ) num__200 <o> b ) num__300 <o> c ) num__400 <o> d ) num__500 <o> e ) num__600 |
let x be the total grown quantity of wheat . according to the question ( num__7.0 of x ) + num__64 num__31.0 of x ) + num__3 num__7 x num__31 x = > num__100 + num__6 = num__400 + num__3 num__1200 x = — num__400 million tonnes rice grown c <eor> c <eos> |
c |
multiply__3.0__400.0__ divide__1200.0__3.0__ |
multiply__3.0__400.0__ divide__1200.0__3.0__ |
| crazy eddie has a key chain factory . eddie managed to decrease the cost of manufacturing his key chains while keeping the same selling price and thus increased the profit from the sale of each key chain from num__30.0 of the selling price to num__50.0 of the selling price . if the manufacturing cost is now $ num__50 what was it before the decrease ? <o> a ) $ num__20 <o> b ) $ num__40 <o> c ) $ num__50 <o> d ) $ num__70 <o> e ) $ num__100 |
deargoodyear num__2013 i ' m happy to help . this is a relatively straightforward problem not very challenging . btw crazy eddiewas the actually name of an electronics chain on the east coast of the usa back in the num__1970 s . manufacturing now is $ num__50 . they now are making a num__50.0 profit so the selling price must be $ num__100 . they had this same selling price $ num__100 before they made the change and had a profit of num__30.0 so the manufacturing must have been $ num__70 . answer = ( d ) . <eor> d <eos> |
d |
subtract__100.0__30.0__ subtract__100.0__30.0__ |
subtract__100.0__30.0__ subtract__100.0__30.0__ |
| in the interior of a forest a certain number of apes equal to the square of one - fourth of the total number are playing and having great fun . the remaining four apes are on a hill and the echo of their shrieks by the adjoining hills frightens them . they came and join the apes in the forest and play with enthusiasm . what is the total number of apes ? <o> a ) num__48 <o> b ) num__8 <o> c ) num__64 <o> d ) num__80 <o> e ) num__16 or num__48 |
let total number be x no in the interior = ( x / num__4 ) ^ num__2 no outside = num__4 so : x - ( x / num__4 ) ^ num__2 = num__4 x ^ num__2 - num__16 x + num__64 = num__0 ( x - num__8 ) ^ num__2 = num__0 so x = num__8 b <eor> b <eos> |
b |
multiply__4.0__16.0__ multiply__2.0__4.0__ divide__64.0__8.0__ |
multiply__4.0__16.0__ divide__16.0__2.0__ divide__64.0__8.0__ |
| area of a square is num__0.5 hectare . the diagonal of the square is ? <o> a ) num__250 meter <o> b ) num__100 meter <o> c ) num__50 √ num__2 meter <o> d ) num__50 meter <o> e ) num__35 meter |
area = num__0.5 hectare = num__5000.0 m num__2 = num__5000 m num__2 again area = num__0.5 x ( diagonal ) num__2 so num__0.5 x ( diagonal ) num__2 = num__5000 m num__2 diagonal num__2 = num__10000 diagonal = num__100 answer : b <eor> b <eos> |
b |
square_perimeter__0.5__ multiply__5000.0__2.0__ power__10000.0__0.5__ triangle_area__2.0__100.0__ |
square_perimeter__0.5__ multiply__5000.0__2.0__ power__10000.0__0.5__ triangle_area__2.0__100.0__ |
| how much time will a train of length num__400 m moving at a speed of num__72 kmph take to cross another train of length num__300 m moving at num__36 kmph in the same direction ? <o> a ) num__40 sec <o> b ) num__50 sec <o> c ) num__60 sec <o> d ) num__70 sec <o> e ) num__80 sec |
the distance to be covered = sum of their lengths = num__400 + num__300 = num__700 m . relative speed = num__72 - num__36 = num__36 kmph = num__36 * num__0.277777777778 = num__10 mps . time required = d / s = num__70.0 = num__70 sec . answer : d <eor> d <eos> |
d |
add__400.0__300.0__ divide__700.0__10.0__ round__70.0__ |
add__400.0__300.0__ divide__700.0__10.0__ divide__700.0__10.0__ |
| what distance will be covered by a city bus moving at num__72 kmph in num__30 seconds ? <o> a ) num__200 m <o> b ) num__300 m <o> c ) num__600 m <o> d ) num__500 m <o> e ) num__400 m |
num__72 kmph = num__72 * num__0.277777777778 = num__20 mps dist = speed * time = num__20 * num__30 = num__600 m . answer c <eor> c <eos> |
c |
multiply__30.0__20.0__ round__600.0__ |
multiply__30.0__20.0__ multiply__30.0__20.0__ |
| what number has a num__5 : num__1 ratio to the number num__11 ? <o> a ) num__22 <o> b ) num__50 <o> c ) num__55 <o> d ) num__52 <o> e ) num__12 |
num__5 : num__1 = x : num__10 x = num__55 answer : c <eor> c <eos> |
c |
subtract__11.0__1.0__ multiply__5.0__11.0__ multiply__5.0__11.0__ |
subtract__11.0__1.0__ multiply__5.0__11.0__ multiply__5.0__11.0__ |
| there are two numbers . if num__50.0 of the first number is added to the second number then the second number increases to its five - fourth . find the ratio of the first number to the second number ? <o> a ) num__0.555555555556 <o> b ) num__0.714285714286 <o> c ) num__1.66666666667 <o> d ) num__0.625 <o> e ) num__2.0 |
let the two numbers be x and y . num__0.5 * x + y = num__1.25 y = > num__0.5 x = num__0.25 y = > x / y = num__2.0 answer : e <eor> e <eos> |
e |
reverse__0.5__ reverse__0.5__ |
reverse__0.5__ reverse__0.5__ |
| if m > num__0 and y is m percent of x then in terms of m x is what percent of y ? <o> a ) a ) num__100 m <o> b ) b ) num__0.01 m <o> c ) c ) num__10000 / m <o> d ) d ) num__10 / m <o> e ) e ) num__1 / m |
y = m / num__100 * x so x = num__100 / m * y so x = ( num__10000 / m ) / num__100 * y c <eor> c <eos> |
c |
percent__100.0__10000.0__ |
percent__100.0__10000.0__ |
| what is the hcf of num__0.666666666667 num__0.444444444444 and num__0.333333333333 <o> a ) num__0.155555555556 <o> b ) num__0.0444444444444 <o> c ) num__0.111111111111 <o> d ) num__0.177777777778 <o> e ) num__0.2 |
explanation : hcf of fractions = hcf of numerators / lcm of denominators = ( hcf of num__2 num__4 num__6 ) / ( lcm of num__3 num__9 num__18 ) = num__0.111111111111 = num__0.111111111111 answer : option c <eor> c <eos> |
c |
add__2.0__4.0__ divide__6.0__2.0__ add__3.0__6.0__ multiply__2.0__9.0__ reverse__9.0__ reverse__9.0__ |
add__2.0__4.0__ divide__6.0__2.0__ add__3.0__6.0__ multiply__2.0__9.0__ reverse__9.0__ reverse__9.0__ |
| a man purchased num__3 blankets @ rs . num__100 each num__5 blankets @ rs . num__150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . num__150 . find the unknown rate of two blankets ? <o> a ) num__300 <o> b ) num__350 <o> c ) num__450 <o> d ) num__470 <o> e ) num__500 |
explanation : num__10 * num__150 = num__1500 num__3 * num__100 + num__5 * num__150 = num__1050 num__1500 – num__1050 = num__450 c <eor> c <eos> |
c |
multiply__150.0__10.0__ multiply__3.0__150.0__ multiply__3.0__150.0__ |
multiply__150.0__10.0__ multiply__3.0__150.0__ multiply__3.0__150.0__ |
| from below option num__48 is divisible by which one ? <o> a ) a ) num__3 <o> b ) b ) num__5 <o> c ) c ) num__9 <o> d ) d ) num__7 <o> e ) e ) num__11 |
num__16.0 = num__16 a <eor> a <eos> |
a |
divide__48.0__16.0__ |
divide__48.0__16.0__ |
| if ( num__5 ^ num__5 ) ( num__9 ^ num__3 ) = num__3 ( num__15 ^ x ) what is the value of x ? <o> a ) num__5 <o> b ) num__9 <o> c ) num__11 <o> d ) num__13 <o> e ) num__15 |
( num__5 ^ num__5 ) ( num__9 ^ num__3 ) = num__3 ( num__15 ^ x ) = > num__5 ^ num__5 * num__3 ^ num__6 = num__3 * num__3 ^ x * num__5 ^ x = > num__5 ^ num__5 * num__3 ^ num__6 = num__3 ^ ( x + num__1 ) * num__5 ^ x value of x = num__5 answer a <eor> a <eos> |
a |
subtract__9.0__3.0__ subtract__6.0__5.0__ multiply__5.0__1.0__ |
subtract__9.0__3.0__ subtract__6.0__5.0__ multiply__5.0__1.0__ |
| a no . when divided by the sum of num__555 and num__445 gives num__2 times their difference as quotient & num__60 as remainder . find the no . is ? <o> a ) num__145646 <o> b ) num__236578 <o> c ) num__645353 <o> d ) num__456546 <o> e ) num__220060 |
( num__555 + num__445 ) * num__2 * num__110 + num__60 = num__220000 + num__60 = num__220060 e <eor> e <eos> |
e |
subtract__555.0__445.0__ add__60.0__220000.0__ add__60.0__220000.0__ |
subtract__555.0__445.0__ add__60.0__220000.0__ add__60.0__220000.0__ |
| the ratio of the two natural numbers is num__5 : num__6 . if a certain number is added to both the numbers the ratio becomes num__7 : num__8 . if the larger number exceeds the smaller number by num__10 find the number added ? <o> a ) num__17 <o> b ) num__14 <o> c ) num__10 <o> d ) num__16 <o> e ) num__20 |
let the two numbers be num__5 x and num__6 x . let the numbers added to both so that their ratio becomes num__7 : num__8 be k . ( num__5 x + k ) / ( num__6 x + k ) = num__0.875 = > num__40 x + num__8 k = num__42 x + num__7 k = > k = num__2 x . num__6 x - num__5 x = num__10 = > x = num__10 k = num__2 x = num__20 . answer : c <eor> c <eos> |
c |
divide__7.0__8.0__ multiply__5.0__8.0__ multiply__6.0__7.0__ subtract__7.0__5.0__ multiply__10.0__2.0__ multiply__5.0__2.0__ |
divide__7.0__8.0__ multiply__5.0__8.0__ multiply__6.0__7.0__ subtract__7.0__5.0__ divide__40.0__2.0__ add__8.0__2.0__ |
| if n = num__0.2 + num__0.2 ^ num__2 + num__0.2 ^ num__3 then n is multiplied by num__100 will the product be greater than num__100 ? <o> a ) less than num__100 <o> b ) greater than num__100 <o> c ) indeterminate <o> d ) equal to num__100 <o> e ) can not be determined |
we need to check if ( num__0.2 + num__0.2 ^ num__2 + num__0.2 ^ num__3 ) > num__1 if it is greater than num__1 then the product num__100 * n will be greater than num__1 else not n = num__0.2 + num__0.2 ^ num__2 + num__0.2 ^ num__3 num__5 ^ num__3 * n = num__5 ^ num__3 * num__0.2 + num__5 ^ num__3 * num__0.2 ^ num__2 + num__5 ^ num__3 * num__0.2 ^ num__3 num__125 * n = num__25 + num__5 + num__1 num__125 * n = num__31 n = num__0.248 n is less than num__1 since numerator is less than denominator . therefore num__100 n would be less than num__100 ans : a <eor> a <eos> |
a |
subtract__3.0__2.0__ reverse__0.2__ multiply__0.2__125.0__ divide__31.0__125.0__ multiply__100.0__1.0__ |
subtract__3.0__2.0__ add__2.0__3.0__ multiply__0.2__125.0__ divide__31.0__125.0__ multiply__100.0__1.0__ |
| a thief goes away with a santro car at a speed of num__22 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at num__72 kmph when will the owner over take the thief from the start ? <o> a ) num__1.66666666667 hours <o> b ) num__0.44 hours <o> c ) num__0.666666666667 hours <o> d ) num__0.333333333333 hours <o> e ) num__0.4 hours |
- - - - - - - - - - - num__22 - - - - - - - - - - - - - - - - - - - - | num__72 num__22 d = num__22 rs = num__72 â € “ num__22 = num__50 t = num__0.44 = num__0.44 hours answer : b <eor> b <eos> |
b |
subtract__72.0__22.0__ divide__22.0__50.0__ round__0.44__ |
subtract__72.0__22.0__ divide__22.0__50.0__ round__0.44__ |
| a tour group of num__25 people paid a total of $ num__1050 for entrance to a museum . if this price included a num__5.0 sales tax and all the tickets cost the same amount what was the face value of each ticket price without the sales tax ? choices <o> a ) $ num__22 <o> b ) $ num__23.94 <o> c ) $ num__40 <o> d ) $ num__25.20 <o> e ) $ num__30 |
soln : - num__42.0 = x + num__0.05 x num__42.0 = num__1.05 x x = num__40 answer : c <eor> c <eos> |
c |
divide__1050.0__25.0__ divide__42.0__1.05__ divide__42.0__1.05__ |
divide__1050.0__25.0__ divide__42.0__1.05__ divide__42.0__1.05__ |
| a train runs at the speed of num__72 kmph and crosses a num__250 metre long platform in num__26 seconds . what is the length of the train ? <o> a ) num__220 metre <o> b ) num__230 metre <o> c ) num__250 metre <o> d ) num__260 metre <o> e ) num__270 metre |
distance covered in num__26 seconds = num__26 Ã — num__72 Ã — num__0.277777777778 = num__520 meter length of the train = num__520 - num__250 = num__270 meter answer : e <eor> e <eos> |
e |
subtract__520.0__250.0__ round__270.0__ |
subtract__520.0__250.0__ subtract__520.0__250.0__ |
| sachin is younger than rahul by num__4 years . if their ages are in the respective ratio of num__7 : num__9 how old is sachin ? <o> a ) num__24.8 years <o> b ) num__28.5 years <o> c ) num__84.5 years <o> d ) num__34.5 years <o> e ) num__24.5 years |
let rahul ' s age be x years . then sachin ' s age = ( x - num__7 ) years . ( x - num__7 ) / x = num__0.777777777778 num__2 x = num__63 = > x = num__31.5 hence sachin ' s age = ( x - num__7 ) = num__24.5 years . answer : e <eor> e <eos> |
e |
divide__7.0__9.0__ subtract__9.0__7.0__ multiply__7.0__9.0__ divide__63.0__2.0__ subtract__31.5__7.0__ subtract__31.5__7.0__ |
divide__7.0__9.0__ subtract__9.0__7.0__ multiply__7.0__9.0__ divide__63.0__2.0__ subtract__31.5__7.0__ subtract__31.5__7.0__ |
| what will be the difference between simple and compound interest at num__16.0 per annum on a sum of rs . num__1000 after num__4 years ? <o> a ) num__164.19 <o> b ) num__164.12 <o> c ) num__170.6 <o> d ) num__167.1 <o> e ) num__165.11 |
s . i . = ( num__1000 * num__16 * num__4 ) / num__100 = rs . num__640 c . i . = [ num__1000 * ( num__1 + num__0.16 ) num__4 - num__1000 ] = rs . num__810.6 difference = ( num__810.6 - num__640 ) = rs . num__170.6 answer : c <eor> c <eos> |
c |
percent__16.0__1.0__ percent__100.0__170.6__ |
percent__16.0__1.0__ percent__100.0__170.6__ |
| if paint costs $ num__3.20 per quart and a quart covers num__1200 square feet how much will it cost to paint the outside of a cube num__10 feet on each edge ? <o> a ) $ num__1.60 <o> b ) $ num__16.00 <o> c ) $ num__96.00 <o> d ) $ num__108.00 <o> e ) $ num__196.00 |
total surface area = num__6 a ^ num__2 = num__6 * num__10 * num__10 = num__600 each quart covers num__20 sqr ft thus total number of quarts = num__0.5 = num__0.5 cost will be num__0.5 * num__3.2 = $ num__1.6 ans : a <eor> a <eos> |
a |
surface_cube__10.0__ multiply__10.0__2.0__ multiply__3.2__0.5__ multiply__3.2__0.5__ |
surface_cube__10.0__ multiply__10.0__2.0__ multiply__3.2__0.5__ multiply__3.2__0.5__ |
| an empty bottle weighs num__0.166666666667 th of the full bottle . when a certain percent of water was removed and the bottle was weighed the weight of the bottle turned out to be num__0.333333333333 rd of the bottle when it was full . what is the percent of water removed ? <o> a ) num__89.0 <o> b ) num__82.0 <o> c ) num__81.0 <o> d ) num__50.0 <o> e ) num__80 % |
explanation : let the weight of full bottle be num__6 kg . therefore weight of empty bottle is num__1 kg and that of water is num__5 kg . if x % of water is removed the weight of the bottle becomes num__2 kg . therefore the amount of water removed is num__4 kg . % of water removed = num__0.8 * num__100 = num__80.0 answer : e <eor> e <eos> |
e |
percent__80.0__100.0__ |
percent__80.0__100.0__ |
| a man started driving at a constant speed from the site of a blast the moment he heard the blast . he heard a second blast after a time of num__30 mins and num__20 seconds . if the second blast occurred exactly num__30 mins after the first how many meters was he from the site when he heard the second blast ? ( speed of sound = num__330 m / s ) <o> a ) num__4200 <o> b ) num__5400 <o> c ) num__6600 <o> d ) num__7800 <o> e ) num__8900 |
the distance the sound traveled to the man is num__20 * num__330 = num__6600 meters the answer is c . <eor> c <eos> |
c |
multiply__20.0__330.0__ round__6600.0__ |
multiply__20.0__330.0__ multiply__20.0__330.0__ |
| a school has num__6 maths num__5 physics and num__5 chemistry teachers each teacher can teach num__4 subjects max what is he minimum number of teachers required <o> a ) num__2 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
total subjects = num__6 + num__5 + num__5 = num__16 max subjects by num__1 teacher = num__4 so min of teachers required = num__4.0 = num__4 answer : b <eor> b <eos> |
b |
subtract__6.0__5.0__ subtract__5.0__1.0__ |
subtract__6.0__5.0__ subtract__5.0__1.0__ |
| taequan is playing a dice game where he rolls two fair six - sided dice twice . if he gets the same roll on his second try as his first he wins . if he rolled an num__8 on his first try what is the probability that he will win with his second roll ? <o> a ) num__0.0555555555556 <o> b ) num__0.0833333333333 <o> c ) num__0.111111111111 <o> d ) num__0.166666666667 <o> e ) num__0.138888888889 |
there are num__5 ways to roll an num__8 : num__2 and num__6 num__6 and num__2 num__3 and num__5 num__5 and num__3 num__4 and num__4 ( num__4 and num__4 is counted just once ) . there are num__6 * num__6 = num__36 ways to roll two six - sided dice . the dice are fair so the probability of rolling an num__8 is num__0.138888888889 . e <eor> e <eos> |
e |
subtract__8.0__2.0__ subtract__8.0__5.0__ divide__8.0__2.0__ divide__5.0__36.0__ divide__5.0__36.0__ |
subtract__8.0__2.0__ subtract__8.0__5.0__ subtract__6.0__2.0__ divide__5.0__36.0__ divide__5.0__36.0__ |
| a thief is spotted by a policeman from a distance of num__100 metres . when the policeman starts the chase the thief also starts running . if the speed of the thief be num__8 km / hr and that of the policeman num__10 km / hr how far the thief will have run before he is overtaken ? <o> a ) num__100 m <o> b ) num__200 m <o> c ) num__400 m <o> d ) num__500 m <o> e ) num__300 m |
relative speed of the policeman = ( num__10 - num__8 ) km / hr = num__2 km / hr . time taken by police man to cover num__100 m num__0.1 x num__0.5 hr = num__0.05 hr . in num__0.05 hrs the thief covers a distance of num__8 x num__0.05 km = num__0.4 km = num__400 m answer c <eor> c <eos> |
c |
subtract__10.0__8.0__ divide__10.0__100.0__ divide__0.5__10.0__ multiply__8.0__0.05__ round__400.0__ |
subtract__10.0__8.0__ divide__10.0__100.0__ divide__0.5__10.0__ subtract__0.5__0.1__ round__400.0__ |
| mangala completes a piece of work in num__18 days raju completes the same work in num__40 days . if both of them work together then the number of days required to complete the work is <o> a ) num__6 days <o> b ) num__7 days <o> c ) num__15 days <o> d ) num__9 days <o> e ) num__11 days |
if a can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days . that is the required no . of days = num__18 Ã — num__0.833333333333 = num__15 days . c <eor> c <eos> |
c |
round__15.0__ |
round__15.0__ |
| the average weight of num__8 person ' s increases by num__5 kg when a new person comes in place of one of them weighing num__35 kg . what might be the weight of the new person ? <o> a ) num__80 kg <o> b ) num__85 kg <o> c ) num__75 kg <o> d ) num__100 kg <o> e ) num__110 kg |
total weight increased = ( num__8 x num__5 ) kg = num__40 kg . weight of new person = ( num__35 + num__40 ) kg = num__75 kg . answer : c <eor> c <eos> |
c |
multiply__8.0__5.0__ add__35.0__40.0__ add__35.0__40.0__ |
add__5.0__35.0__ add__35.0__40.0__ add__35.0__40.0__ |
| a tradesman sold an article at a loss of num__20.0 . if the selling price had been increased by $ num__100 there would have been a gain of num__5.0 . what was the cost price of the article ? <o> a ) $ num__400 <o> b ) $ num__150 <o> c ) $ num__300 <o> d ) $ num__550 <o> e ) $ num__610 |
let c . p . be $ x then num__105.0 of x - num__80.0 of x = num__100 num__25.0 of x = num__100 x / num__4 = num__100 x = num__400 answer is a <eor> a <eos> |
a |
percent__5.0__80.0__ percent__100.0__400.0__ |
percent__5.0__80.0__ percent__100.0__400.0__ |
| the volumes of two cones are in the ratio num__1 : num__10 and the radii of the cones are in the ratio of num__1 : num__2 . what is the length of the wire ? <o> a ) num__2 : num__5 <o> b ) num__2 : num__7 <o> c ) num__4 : num__3 <o> d ) num__2 : num__4 <o> e ) num__2 : num__9 |
the volume of the cone = ( num__0.333333333333 ) π r num__2 h only radius ( r ) and height ( h ) are varying . hence ( num__0.333333333333 ) π may be ignored . v num__1 / v num__2 = r num__12 h num__1 / r num__22 h num__2 = > num__0.1 = ( num__1 ) num__2 h num__1 / ( num__2 ) num__2 h num__2 = > h num__1 / h num__2 = num__0.4 i . e . h num__1 : h num__2 = num__2 : num__5 answer : a <eor> a <eos> |
a |
rectangle_perimeter__1.0__10.0__ square_perimeter__0.1__ triangle_area__1.0__10.0__ multiply__1.0__2.0__ |
rectangle_perimeter__1.0__10.0__ square_perimeter__0.1__ triangle_area__1.0__10.0__ multiply__1.0__2.0__ |
| a tank is filled by num__3 pipes a b c in num__5 hours . pipe c is twice as fast as b and b is twice as fast as a . how much will pipe a alone take to fill the tank ? <o> a ) num__25 hr <o> b ) num__35 hr <o> c ) num__40 hr <o> d ) num__20 hr <o> e ) num__50 hr |
suppose pipe a alone take x hours to fill the tank then pipe b and c will take x / num__2 and x / num__4 hours respectively to fill the tank . num__1 / x + num__2 / x + num__4 / x = num__0.2 num__7 / x = num__0.2 x = num__35 hours answer is b <eor> b <eos> |
b |
subtract__5.0__3.0__ subtract__3.0__2.0__ divide__1.0__5.0__ add__3.0__4.0__ multiply__5.0__7.0__ round__35.0__ |
subtract__5.0__3.0__ subtract__3.0__2.0__ divide__1.0__5.0__ add__3.0__4.0__ divide__7.0__0.2__ divide__35.0__1.0__ |
| a can run a kilometer race in num__4 num__0.5 min while b can run same race in num__5 min . how many meters start can a give b in a kilometer race so that the race mat end in a dead heat ? <o> a ) num__200 m <o> b ) num__700 m <o> c ) num__800 m <o> d ) num__100 metre <o> e ) num__1050 m |
explanation : a can give b ( num__5 min - num__4 num__0.5 min ) = num__30 sec start . the distance covered by b in num__5 min = num__1000 m . distance covered in num__30 sec = ( num__1000 * num__30 ) / num__300 = num__100 m . a can give b num__100 m start . answer : d <eor> d <eos> |
d |
round__100.0__ |
round__100.0__ |
| a and b start walking towards each other at num__6 pm at speed of num__6 kmph and num__4 kmph . they were initially num__50 km apart . at what time do they meet ? <o> a ) num__8 pm <o> b ) num__6 pm <o> c ) num__11 pm <o> d ) num__10 pm <o> e ) num__5 pm |
time of meeting = distance / relative speed = num__12.5 + num__6 = num__5.0 = num__5 hrs after num__6 pm = num__11 pm answer is c <eor> c <eos> |
c |
divide__50.0__4.0__ add__6.0__5.0__ round__11.0__ |
divide__50.0__4.0__ add__6.0__5.0__ add__6.0__5.0__ |
| the present population of a town is num__1000 . population increase rate is num__25.0 p . a . find the population of town before num__1 years ? <o> a ) num__800 <o> b ) num__1500 <o> c ) num__1000 <o> d ) num__750 <o> e ) num__500 |
p = num__1000 r = num__25.0 required population of town = p / ( num__1 + r / num__100 ) ^ t = num__1000 / ( num__1 + num__0.25 ) = num__1000 / ( num__1.25 ) = num__800 answer is a <eor> a <eos> |
a |
percent__25.0__1.0__ percent__100.0__800.0__ |
percent__25.0__1.0__ percent__100.0__800.0__ |
| the speed of a boat in still water is num__60 kmph and the speed of the current is num__12 kmph . find the speed downstream and upstream ? <o> a ) num__87 kmph <o> b ) num__40 kmph <o> c ) num__16 kmph <o> d ) num__15 kmph <o> e ) num__48 kmph |
speed downstream = num__60 + num__12 = num__72 kmph speed upstream = num__60 - num__20 = num__48 kmph answer : e <eor> e <eos> |
e |
add__60.0__12.0__ subtract__60.0__12.0__ round__48.0__ |
add__60.0__12.0__ subtract__60.0__12.0__ subtract__60.0__12.0__ |
| a cargo ship carrying four kinds of items doohickies geegaws widgets and yamyams arrives at the port . each item weighs num__2 num__11 num__5 and num__7 pounds respectively and each item is weighed as it is unloaded . if in the middle of the unloading process the product of the individual weights of the unloaded items equals num__104350 num__400000 pounds how many widgets have been unloaded ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
we need to know the number of widgets ( which weigh num__5 pounds each ) . the number of times that num__5 divides the number is related to the number of times that num__10 divides the number . when we divide num__104350 num__400000 by num__100000 we get num__104350 num__400000 = num__1 num__043504 * num__5 ^ num__5 * num__2 ^ num__5 . num__1 num__043504 is not divisible by num__5 thus there are num__5 widgets . the answer is d . <eor> d <eos> |
d |
multiply__2.0__5.0__ subtract__11.0__10.0__ multiply__5.0__1.0__ |
multiply__2.0__5.0__ subtract__11.0__10.0__ multiply__5.0__1.0__ |
| angelina walked num__150 meters from her home to the grocery at a constant speed . she then walked num__200 meters to the gym at double the speed . she spent num__10 seconds less on her way from the grocery to the gym than on her way from home to the grocery . what was angelina ' s speed in meters per second from the grocery to the gym ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__10 <o> e ) num__12 |
let the speed be x . . . so time taken from home to grocery = num__150 / x . . the speed to gym = num__2 x . . so time taken = num__100.0 x = num__100 / x . . its given num__150 / x - num__100 / x = num__10 num__50 / x = num__10 x = num__5 m / secs . . so grocery to gym = num__2 * num__5 = num__10 m / s . . . answer : d <eor> d <eos> |
d |
divide__200.0__2.0__ subtract__150.0__100.0__ divide__10.0__2.0__ round__10.0__ |
divide__200.0__2.0__ subtract__150.0__100.0__ divide__10.0__2.0__ multiply__2.0__5.0__ |
| in a num__1000 m race usha beats shiny by num__50 m . in the same race by what time margin shiny beat mercy who runs at num__4 m / s ? <o> a ) num__100 sec . <o> b ) num__50 sec <o> c ) num__25 sec <o> d ) data not sufficient <o> e ) none of these |
speed of shiny = num__5.0 = num__5 m / s time taken by shiny to complete the race is b = num__200.0 = num__200 sec . time taken by baley to complete the race is d = num__250.0 = num__250 sec . hence d - b = num__50 sec answer : b <eor> b <eos> |
b |
divide__1000.0__5.0__ divide__1000.0__4.0__ round__50.0__ |
divide__1000.0__5.0__ divide__1000.0__4.0__ divide__200.0__4.0__ |
| insert the missing number . num__2 num__7 num__10 num__22 num__18 num__37 num__26 <o> a ) num__26 <o> b ) num__22 <o> c ) num__2 <o> d ) num__52 <o> e ) num__18 |
there are two series here num__2 num__10 num__18 num__26 . . . ( increase by num__8 ) num__7 num__22 num__37 . . . ( increase by num__15 ) hence next term is num__37 + num__15 = num__52 answer is d <eor> d <eos> |
d |
subtract__10.0__2.0__ add__7.0__8.0__ multiply__2.0__26.0__ multiply__2.0__26.0__ |
subtract__10.0__2.0__ add__7.0__8.0__ add__37.0__15.0__ add__37.0__15.0__ |
| a man buys an article for num__14.0 less than its value and sells it for num__14.0 more than its value . his gain or loss percent is : <o> a ) no profit no loss <o> b ) num__20.0 profit <o> c ) less than num__30.0 profit <o> d ) more than num__30.0 profit <o> e ) none |
let the article be worth rs . x . c . p . num__86.0 of rs . x = rs . num__86 x / num__100 s . p . = num__114.0 of rs . x = rs . num__114 x / num__100 gain = ( num__114 x / num__100 - num__86 x / num__100 ) = rs . num__7 x / num__25 gain % = num__7 x / num__25 * num__1.16279069767 x * num__100 = num__32 num__0.558139534884 % > num__30.0 answer : d <eor> d <eos> |
d |
percent__100.0__30.0__ |
percent__100.0__30.0__ |
| if x is negative and y is positive which of the following must be positive ? <o> a ) x / y <o> b ) xy <o> c ) num__2 y + x <o> d ) ( x + y ) ^ num__2 + num__1 <o> e ) num__3 x / y |
a . x / y - ve / + ve = - ve will be negative b . xy - ve * + ve = - ve will be negative c . num__2 y + x might be positive or negative . the sign will be depend on the absolute values of x and y . the expression will be positive if num__2 y > x d . ( x + y ) ^ num__2 + num__1 = ( x + y ) ^ num__2 will be positive irrespective of the absolute values of x and y will be positive always e . num__3 x / y = - ve / + ve will be negative answer d <eor> d <eos> |
d |
add__1.0__2.0__ multiply__1.0__2.0__ |
add__1.0__2.0__ multiply__1.0__2.0__ |
| a license plate in the country kerrania consists of four digits followed by two letters . the letters a b and c are used only by government vehicles while the letters d through z are used by non - government vehicles . kerrania ' s intelligence agency has recently captured a message from the country gonzalia indicating that an electronic transmitter has been installed in a kerrania government vehicle with a license plate starting with num__79 . if it takes the police num__9 minutes to inspect each vehicle what is the probability that the police will find the transmitter within three hours ? <o> a ) num__0.227848101266 <o> b ) num__0.166666666667 <o> c ) num__0.0222222222222 <o> d ) num__0.02 <o> e ) num__0.00111111111111 |
if it takes num__9 minutes to inspect one vehicle the # of vehicles that can be inspected in num__3 hours ( num__180 minutes ) = num__20.0 = num__20 . hence for calculating the probability that the police will find the transmitter within three hours the favorable cases = num__20 . now we need to figure out the total # of cases . the total # of cases = total # of such cars possible . the details given about the car is that it starts with num__79 which leaves num__2 more digits both of which can be filled by all num__10 numbers ( num__0 - num__9 ) . in addition we have num__3 letters each of which can be filled by any from the set { a b c } . hence the total # of such cars possible = num__10 * num__10 * num__3 * num__3 = num__900 so the probability that the police will find the transmitter within three hours = num__0.0222222222222 = num__0.0222222222222 . option c <eor> c <eos> |
c |
divide__180.0__9.0__ divide__20.0__2.0__ divide__20.0__900.0__ divide__20.0__900.0__ |
divide__180.0__9.0__ divide__20.0__2.0__ divide__20.0__900.0__ divide__20.0__900.0__ |
| prathik purchased num__40 shirts for rs num__3000 . he spends num__10.0 for transportation . what should be the selling price per shirt to earn profit of num__20.0 ? <o> a ) num__96 rs <o> b ) num__97 rs <o> c ) num__98 rs <o> d ) num__99 rupees <o> e ) num__100 rs |
total cp = num__3000 + num__3000 * num__0.1 = num__3300 to gain num__20.0 profit sp = ( num__100 + num__20 ) * num__33.0 = num__3960 selling price per shirt = num__99.0 = num__99 rs / shirt answer : d <eor> d <eos> |
d |
percent__99.0__100.0__ |
percent__99.0__100.0__ |
| a can do a certain job in num__16 days . b is num__60.0 more efficient than a . how many days does b alone take to do the same job ? <o> a ) num__11 num__0.25 days <o> b ) num__8 days <o> c ) num__5 days <o> d ) num__10 days <o> e ) none of them |
ratio of times taken by a and b = num__160 : num__100 = num__8 : num__5 . suppose b alone takes x days to do the job . then num__8 : num__5 : : num__16 : x = num__8 x = num__5 x num__16 = x = num__10 days . answer is d . <eor> d <eos> |
d |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| if n is a natural number then ( num__6 n ^ num__2 + num__6 n ) is always divisible by ? <o> a ) num__6 only <o> b ) num__6 and num__12 <o> c ) num__12 only <o> d ) by num__18 only <o> e ) none of these |
( num__6 n ^ num__2 + num__6 n ) = num__6 n ( n + num__1 ) which is always divisible by num__6 and num__12 both since n ( n + num__1 ) is always even . correct option : b <eor> b <eos> |
b |
multiply__6.0__2.0__ multiply__6.0__1.0__ |
multiply__6.0__2.0__ multiply__6.0__1.0__ |
| kul is num__22 years old and saras is num__33 years . find the ratio of saras ' s age to kul ' s age . <o> a ) num__3.0 <o> b ) num__1.5 <o> c ) num__2.5 <o> d ) num__1.66666666667 <o> e ) none of these |
explanation : kul is num__22 years old . saras is num__33 years old . saras ' s age : kul ' s age = num__33 : num__22 = num__3 : num__2 answer : b <eor> b <eos> |
b |
divide__33.0__22.0__ |
divide__33.0__22.0__ |
| what percent is num__240 of num__90 ? <o> a ) num__133 num__0.333333333333 % <o> b ) num__134 num__0.333333333333 % <o> c ) num__135 num__0.333333333333 % <o> d ) num__266 num__0.666666666667 % <o> e ) num__143 num__0.333333333333 % |
num__2.66666666667 = num__2.66666666667 num__2.66666666667 × num__100 = num__266.666666667 = num__266 num__0.666666666667 % d <eor> d <eos> |
d |
percent__100.0__266.0__ |
percent__100.0__266.0__ |
| mr yadav spends num__60.0 of his monthly salary on consumable items and num__50.0 of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were num__46800 how much amount per month would he have spent on clothes and transport ? <o> a ) num__4038 <o> b ) num__8076 <o> c ) num__9691.2 <o> d ) num__4845.6 <o> e ) num__3900 |
∵ amount he have spent in num__1 month on clothes transport = amount spent on saving per month ∵ amount spent on clothes and transport = num__46800 ⁄ num__12 = num__3900 answer e <eor> e <eos> |
e |
divide__46800.0__12.0__ divide__46800.0__12.0__ |
divide__46800.0__12.0__ divide__46800.0__12.0__ |
| how long does a train num__125 m long running at the speed of num__78 km / hr takes to cross a bridge num__125 m length ? <o> a ) num__12.7 sec <o> b ) num__10.1 sec <o> c ) num__11.5 sec <o> d ) num__12.1 sec <o> e ) num__11.7 sec |
speed = num__78 * num__0.277777777778 = num__21.7 m / sec total distance covered = num__125 + num__125 = num__250 m . required time = num__250 / num__21.7 ' = num__11.5 sec . answer : c <eor> c <eos> |
c |
round__11.5__ |
round__11.5__ |
| the average of seven numbers is num__18 . the average of first three numbers is num__14 and the average of last three numbers is num__23 . what is the middle number ? <o> a ) num__25 <o> b ) num__27 <o> c ) num__15 <o> d ) num__32 <o> e ) num__34 |
the total of seven numbers = num__7 x num__18 = num__126 the total of first num__3 and last num__3 numbers is = num__3 x num__14 + num__3 x num__23 = num__111 so the middle number is ( num__126 - num__111 ) = num__15 c <eor> c <eos> |
c |
multiply__18.0__7.0__ subtract__18.0__3.0__ subtract__18.0__3.0__ |
multiply__18.0__7.0__ subtract__18.0__3.0__ subtract__18.0__3.0__ |
| the compound ratio of num__5 : num__6 num__5 : num__2 and num__4 : num__5 ? <o> a ) num__1 : num__1 <o> b ) num__5 : num__3 <o> c ) num__1 : num__6 <o> d ) num__1 : num__9 <o> e ) num__1 : num__2 |
num__0.833333333333 * num__2.5 * num__0.8 = num__1.66666666667 num__5 : num__3 answer : b <eor> b <eos> |
b |
divide__5.0__6.0__ divide__5.0__2.0__ divide__2.0__2.5__ subtract__2.5__0.8333__ subtract__5.0__2.0__ multiply__2.0__2.5__ |
divide__5.0__6.0__ divide__5.0__2.0__ divide__2.0__2.5__ subtract__2.5__0.8333__ subtract__5.0__2.0__ multiply__2.0__2.5__ |
| a car runs at the speed of num__40 km per hour when not serviced and runs at num__80 kmph when serviced . after servicing the car covers a certain distance in num__8 hours . how much time will the car take to cover the same distance when not serviced ? <o> a ) num__8 hours num__12 minutes <o> b ) num__12 hours <o> c ) num__16 hours <o> d ) num__17 hours <o> e ) none |
explanation : time = num__80 * num__0.2 = num__16 hours answer â € “ c <eor> c <eos> |
c |
divide__8.0__40.0__ multiply__80.0__0.2__ round__16.0__ |
divide__8.0__40.0__ multiply__80.0__0.2__ multiply__80.0__0.2__ |
| average score for virat kohli in a series of num__10 matches is num__38.9 runs . if the average for first six matches comes out to be num__42 what is his average in the last num__4 matches of the series ? <o> a ) num__34.25 <o> b ) num__34.28 <o> c ) num__24.252 <o> d ) num__64.28 <o> e ) num__34.21 |
explanation : average runs scored by virat kohli in num__10 matches : total runs scored / num__10 — ( num__1 ) average runs scored by virat kohli in num__6 matches : ( total runs in num__6 matches ) / num__6 = > num__42 = runs / num__6 = > num__252 runs using ( num__1 ) : = > num__38.9 = runs / num__10 = > num__389 runs runs scored in rest num__4 matches : num__389 - num__252 runs = > num__137 runs average runs scored by virat kohli in num__4 matches : num__34.25 = num__34.25 runs answer : a <eor> a <eos> |
a |
subtract__10.0__4.0__ multiply__42.0__6.0__ multiply__10.0__38.9__ subtract__389.0__252.0__ divide__137.0__4.0__ multiply__34.25__1.0__ |
subtract__10.0__4.0__ multiply__42.0__6.0__ multiply__10.0__38.9__ subtract__389.0__252.0__ divide__137.0__4.0__ divide__34.25__1.0__ |
| the speed of boat in still water is num__20 km / hr . if it travels num__26 km downstream and num__14 km upstream in same time what is the speed of the stream ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__2 |
explanation : let x be speed of stream speed of boat downstream = ( num__20 + x ) speed of boat upstream = ( num__20 – x ) it is given that boat takes same time to travel num__26 km downstream and num__14 km upstream [ num__26 / ( num__20 + x ) ] = [ num__14 / ( num__20 – x ) ] num__520 – num__26 x = num__280 + num__14 x num__40 x = num__240 x = num__6 km / hr answer : a <eor> a <eos> |
a |
multiply__20.0__26.0__ multiply__20.0__14.0__ add__26.0__14.0__ subtract__520.0__280.0__ subtract__20.0__14.0__ round__6.0__ |
multiply__20.0__26.0__ multiply__20.0__14.0__ add__26.0__14.0__ subtract__520.0__280.0__ divide__240.0__40.0__ divide__240.0__40.0__ |
| set j consists of num__5 consecutive even numbers . if the smallest term in the set is - num__2 what is the range of the positive integers in set j ? <o> a ) num__0 <o> b ) num__2 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
since there are only num__5 integers another approach is the just list all num__5 . we get : - num__2 num__02 num__4 num__6 range of positive integers = num__6 - num__2 = num__4 answer : c <eor> c <eos> |
c |
add__2.0__4.0__ subtract__6.0__2.0__ |
add__2.0__4.0__ subtract__6.0__2.0__ |
| what is the measure of the acute angle between the hour and minute hands of a correctly working clock at num__4 : num__18 ? <o> a ) num__12 ◦ <o> b ) num__15 ◦ <o> c ) num__18 ◦ <o> d ) num__21 ◦ <o> e ) num__24 ◦ |
consider an axis through the center of the clock with direction the direction of the hands of the clock at num__12 : num__00 . at num__4 : num__18 the angle between the axis and the hour hand is ( num__4 + num__0.3 ) * num__360 ◦ / num__12 = num__129 ◦ the angle between the axis and the minute hand is num__0.3 * num__360 ◦ = num__108 ◦ correct answer d <eor> d <eos> |
d |
multiply__360.0__0.3__ subtract__129.0__108.0__ |
multiply__360.0__0.3__ subtract__129.0__108.0__ |
| a and b can do a piece of work in num__7 days . with the help of c they finish the work in num__5 days . c alone can do that piece of work in ? <o> a ) num__15.5 days <o> b ) num__19.5 days <o> c ) num__17.5 days <o> d ) num__16.5 days <o> e ) num__18.5 days |
c = num__0.2 – num__0.142857142857 = num__0.0571428571429 = > num__17.5 days answer : c <eor> c <eos> |
c |
subtract__0.2__0.1429__ round__17.5__ |
subtract__0.2__0.1429__ round__17.5__ |
| a train num__300 m long is running at a speed of num__99 km / hr . in what time will it pass a bridge num__195 m long ? <o> a ) num__17 <o> b ) num__18 <o> c ) num__19 <o> d ) num__20 <o> e ) num__21 |
speed = num__99 * num__0.277777777778 = num__27.5 m / sec total distance covered = num__300 + num__195 = num__495 m required time = num__495 * num__0.0363636363636 = num__18 sec answer : b <eor> b <eos> |
b |
add__300.0__195.0__ divide__495.0__27.5__ round__18.0__ |
add__300.0__195.0__ divide__495.0__27.5__ divide__495.0__27.5__ |
| of the science books in a certain supply room num__50 are on botany num__75 are on zoology num__90 are on physics . num__50 are on geology and num__110 are on chemistry . if science books are removed randomly from the supply room how many must be removed to ensure that num__80 of the books removed are on the same science ? <o> a ) num__81 <o> b ) num__59 <o> c ) num__166 <o> d ) num__285 <o> e ) num__334 |
i solve it using the tough luck technique : according to question : what is the least number of books you should pick so as to get at least num__80 books of the same science subject . num__80 books of the same science subjects is possible only for two subjects : physics = num__90 > num__80 or chemistry = num__110 > num__80 now we need to be certain that out of the books we picked there are either at least num__80 physics books or num__80 chemistry books what if we pick the first num__80 books and none of them is either physics or chemistry . possible . thus we first count all our negatives . we picked : num__50 botany books num__75 zoology books num__50 geology books now any book we pick will be either chemistry or physics . but unfortunately we ca n ' t be lucky enough to pick num__80 books and all of them will be physics right ! ! thus in order to make sure that we have num__80 books of either of these num__2 subjects we must pick num__79 * num__2 + num__1 books because we could have picked the books in following order ; num__1 st book picked : physics num__2 nd book picked : chemistry num__3 rd book picked : physics thus total = num__50 + num__75 + num__50 + num__79 * num__2 + num__1 = num__175 + num__1 + num__158 = num__334 ans : e <eor> e <eos> |
e |
subtract__80.0__79.0__ add__1.0__2.0__ multiply__2.0__79.0__ round__334.0__ |
subtract__80.0__79.0__ add__1.0__2.0__ multiply__2.0__79.0__ round__334.0__ |
| a can finish a work in num__36 days b in num__9 days and c in num__2 days b and c start the work but are forced to leave after num__3 days . the remaining work was done by a in ? <o> a ) num__15 days <o> b ) num__12 days <o> c ) num__6 days <o> d ) num__7 days <o> e ) num__8 days |
b + c num__1 day work = num__0.111111111111 + num__0.0833333333333 = num__0.194444444444 work done by b and c in num__3 days = num__0.194444444444 * num__3 = num__0.583333333333 remaining work = num__1 - num__0.583333333333 = num__0.416666666667 num__0.0277777777778 work is done by a in num__1 day num__0.416666666667 work is done by a in num__36 * num__0.416666666667 = num__15 days answer is a <eor> a <eos> |
a |
subtract__3.0__2.0__ divide__1.0__9.0__ divide__3.0__36.0__ add__0.1111__0.0833__ subtract__1.0__0.5833__ divide__1.0__36.0__ round__15.0__ |
subtract__3.0__2.0__ divide__1.0__9.0__ divide__3.0__36.0__ add__0.1111__0.0833__ subtract__1.0__0.5833__ subtract__0.1111__0.0833__ round__15.0__ |
| a and b can do a work in num__5 days and num__10 days respectively . a starts the work and b joins him after num__2 days . in how many days can they complete the remaining work ? <o> a ) num__7 days <o> b ) num__2 days <o> c ) num__5 days <o> d ) num__3 days <o> e ) num__6 days |
b num__2 days work done by a in num__2 days = num__0.4 remaining work = num__0.6 work done by both a and b in one day = num__0.2 + num__0.1 = num__0.3 remaining work = num__0.6 * num__3.33333333333 = num__2 days . <eor> b <eos> |
b |
divide__2.0__5.0__ km_to_mile_conversion__ divide__2.0__10.0__ divide__0.2__2.0__ add__0.2__0.1__ divide__2.0__0.6__ round__2.0__ |
divide__2.0__5.0__ km_to_mile_conversion__ divide__2.0__10.0__ divide__0.2__2.0__ add__0.2__0.1__ divide__2.0__0.6__ multiply__5.0__0.4__ |
| an association of mathematics teachers has num__1260 members . only num__525 of these members cast votes in the election for president of the association . what percent of the total membership voted for the winning candidate if the winning candidate received num__60 percent of the votes cast ? <o> a ) num__75.0 <o> b ) num__58.0 <o> c ) num__42.0 <o> d ) num__34.0 <o> e ) num__25 % |
total number of members = num__1260 number of members that cast votes = num__525 since winning candidate received num__60 percent of the votes cast number of votes for winning candidate = ( num__0.6 ) * num__525 = num__315 percent of total membership that voted for winning candidate = ( num__0.25 ) * num__100 = num__25.0 answer e <eor> e <eos> |
e |
percent__60.0__525.0__ percent__100.0__25.0__ |
percent__60.0__525.0__ percent__100.0__25.0__ |
| a train passes a station platform in num__32 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__180 m <o> b ) num__160 m <o> c ) num__240 m <o> d ) num__207 m <o> e ) num__202 m |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x meters . then x + num__9.375 = num__15 x + num__300 = num__480 x = num__180 m . answer : a <eor> a <eos> |
a |
multiply__20.0__15.0__ divide__300.0__32.0__ multiply__32.0__15.0__ subtract__480.0__300.0__ round__180.0__ |
multiply__20.0__15.0__ divide__300.0__32.0__ multiply__32.0__15.0__ subtract__480.0__300.0__ round__180.0__ |
| nine men went to a hotel . eight of them spent rs . num__3 each over their meals and the ninth spent rs . num__2 more than the average expenditure of all the nine . determine the total money spent by them ? <o> a ) num__29.25 <o> b ) num__29.28 <o> c ) num__79.28 <o> d ) num__29.22 <o> e ) num__29.21 |
average of num__9 = x num__9 x = num__8 * num__3 + x * num__2 x = num__3.25 total = num__9 * num__3.25 = num__29.25 answer : a <eor> a <eos> |
a |
multiply__3.25__9.0__ multiply__3.25__9.0__ |
multiply__3.25__9.0__ multiply__3.25__9.0__ |
| the current of a stream runs at the rate of num__4 kmph . a boat goes num__6 km and back to the starting point in num__2 hours then find the speed of the boat in still water ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__5 <o> e ) num__3 |
s = num__4 m = x ds = x + num__4 us = x - num__4 num__6 / ( x + num__4 ) + num__6 / ( x - num__4 ) = num__2 x = num__8 answer : c <eor> c <eos> |
c |
multiply__4.0__2.0__ round__8.0__ |
add__6.0__2.0__ add__6.0__2.0__ |
| a train covers a certain distance at a speed of num__340 kmph in num__4 hours . to cover the same distance in num__2 num__0.75 hours it must travel at a speed of <o> a ) num__495 km / hr <o> b ) num__420 km / hr <o> c ) num__450 km / hr <o> d ) num__550 km / hr <o> e ) num__650 km / hr |
explanation : distance = num__340 Ã — num__4 = num__1360 km required speed = ( num__1360 Ã — num__0.363636363636 ) = num__495 km / hr answer : option a <eor> a <eos> |
a |
multiply__340.0__4.0__ round__495.0__ |
multiply__340.0__4.0__ round__495.0__ |
| the average weight of num__8 persons increases by num__4 kg when a new person comes in place of one of them weighing num__65 kg . what might be the weight of the new person ? <o> a ) num__75 kg <o> b ) num__65 kg <o> c ) num__55 kg <o> d ) num__97 kg <o> e ) num__25 kg |
total weight increased = ( num__8 x num__4 ) kg = num__32 kg . weight of new person = ( num__65 + num__32 ) kg = num__97 kg . answer : d <eor> d <eos> |
d |
multiply__8.0__4.0__ add__65.0__32.0__ add__65.0__32.0__ |
multiply__8.0__4.0__ add__65.0__32.0__ add__65.0__32.0__ |
| which of the following statements is not correct ? a . log num__10 ( num__10 ) = num__1 b . log ( num__2 + num__3 ) = log ( num__2 x num__3 ) c . log num__10 ( num__1 ) = num__0 d . log ( num__1 + num__2 + num__3 ) = log num__1 + log num__2 + log num__3 <o> a ) log num__10 ( num__10 ) = num__1 <o> b ) log ( num__2 + num__3 ) = log ( num__2 x num__3 ) <o> c ) log num__10 ( num__1 ) = num__0 <o> d ) log ( num__1 + num__2 + num__3 ) = log num__1 + log num__2 + log num__3 <o> e ) none of these |
( a ) since loga ( a ) = num__1 so log num__10 ( num__10 ) = num__1 . ( b ) log ( num__2 + num__3 ) = log num__5 and log ( num__2 x num__3 ) = log num__6 = log num__2 + log num__3 log ( num__2 + num__3 ) log ( num__2 x num__3 ) ( c ) since loga ( num__1 ) = num__0 so log num__10 ( num__1 ) = num__0 . ( d ) log ( num__1 + num__2 + num__3 ) = log num__6 = log ( num__1 x num__2 x num__3 ) = log num__1 + log num__2 + log num__3 . so ( b ) is incorrect . answer is b . <eor> b <eos> |
b |
divide__10.0__2.0__ add__1.0__5.0__ divide__10.0__5.0__ |
add__2.0__3.0__ add__1.0__5.0__ divide__10.0__5.0__ |
| a pipe takes a hours to fill the tank . but because of a leakage it took num__7 times of its original time . find the time taken by the leakage to empty the tank <o> a ) num__50 min <o> b ) num__60 min <o> c ) num__90 min <o> d ) num__80 min <o> e ) num__70 min |
pipe a can do a work num__60 min . lets leakage time is x ; then num__0.0166666666667 - num__1 / x = num__0.00238095238095 x = num__70 min answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ divide__0.0167__7.0__ round__70.0__ |
hour_to_min_conversion__ divide__0.0167__7.0__ divide__70.0__1.0__ |
| a particular store purchased a stock of turtleneck sweaters and marked up its cost by num__20.0 . during the new year season it further marked up its prices by num__25.0 of the original retail price . in february the store then offered a discount of num__25.0 . what was its profit on the items sold in february ? <o> a ) num__12.5 <o> b ) num__30.0 <o> c ) num__35.0 <o> d ) num__37.5 <o> e ) num__40 % |
assume the total price = num__100 x price after num__20.0 markup = num__120 x price after num__25.0 further markup = num__1.25 * num__120 x = num__150 x price after the discount = num__0.75 * num__150 x = num__112.5 x hence total profit = num__12.5 option a <eor> a <eos> |
a |
percent__100.0__12.5__ |
percent__100.0__12.5__ |
| which of the following fractions terminates ? <o> a ) num__0.01875 <o> b ) num__0.00555555555556 <o> c ) num__0.0318181818182 <o> d ) num__0.075 <o> e ) num__0.02 |
for a fraction to have terminating decimal it must either have only num__2 and / or num__5 in the denominator . all other values in the denominator leads to non terminating decimal values hence num__1 . num__160 = num__2 * num__2 * num__2 * num__2 * num__2 * num__5 hence is terminating num__2 . there is a num__3 in the denominator hence it is a non terminating decimal num__3 . there is an num__11 in the denominator hence it is a non terminating decimal num__4 . there is a num__3 in the denominator hence it is a non terminating decimal num__5 . there is an num__11 in the denominator hence non terminating option a <eor> a <eos> |
a |
add__1.0__2.0__ add__1.0__3.0__ divide__3.0__160.0__ |
add__1.0__2.0__ add__1.0__3.0__ divide__3.0__160.0__ |
| a train consists of num__12 boggies each boggy num__15 metres long . the train crosses a telegraph post in num__9 seconds . due to some problem three boggies were detached . the train now crosses a telegraph post in <o> a ) num__6.75 sec <o> b ) num__12 sec <o> c ) num__15 sec <o> d ) num__20 sec <o> e ) none of these |
length of train = num__12 Ã — num__15 = num__180 m . then speed of train = num__180 â „ num__9 = num__20 m / s now length of train = num__9 Ã — num__15 = num__135 m â ˆ ´ required time = num__135 â „ num__20 = num__6.75 sec . answer a <eor> a <eos> |
a |
multiply__12.0__15.0__ divide__180.0__9.0__ multiply__15.0__9.0__ divide__135.0__20.0__ round__6.75__ |
multiply__12.0__15.0__ divide__180.0__9.0__ multiply__15.0__9.0__ divide__135.0__20.0__ divide__135.0__20.0__ |
| two trains travel in opposite directions at num__36 kmph and num__45 kmph and a man sitting in slower train passes the faster train in num__8 seconds . the length of the faster train is <o> a ) num__80 m <o> b ) num__100 m <o> c ) num__120 m <o> d ) num__180 m <o> e ) none |
solution relative speed = ( num__36 + num__45 ) km / hr = ( num__81 x num__0.277777777778 ) m / sec = ( num__22.5 ) m / sec length of the train = ( num__22.5 x num__8 ) m = num__180 m . answer d <eor> d <eos> |
d |
add__36.0__45.0__ multiply__8.0__22.5__ round__180.0__ |
add__36.0__45.0__ multiply__8.0__22.5__ round__180.0__ |
| a shopkeeper labeled the price of his articles so as to earn a profit of num__40.0 on the cost price . he then sold the articles by offering a discount of num__10.0 on the labeled price . what is the actual percent profit earned in the deal ? <o> a ) num__26.0 <o> b ) num__20.0 <o> c ) num__17.0 <o> d ) num__18.0 <o> e ) none of these |
explanation : let the cp of the article = rs . num__100 . then labeled price = rs . num__140 . sp = rs . num__140 - num__10.0 of num__140 = rs . num__140 - num__14 = rs . num__126 . gain = rs . num__126 â € “ rs . num__100 = rs . num__26 therefore gain / profit percent = num__26.0 . answer : option a <eor> a <eos> |
a |
percent__10.0__140.0__ percent__100.0__26.0__ |
percent__10.0__140.0__ percent__100.0__26.0__ |
| the ratio of two no . addition and subtraction be num__4 : num__3 . the what is the ratio of numbers ? <o> a ) num__3 : num__2 <o> b ) num__4 : num__3 <o> c ) num__5 : num__1 <o> d ) num__6 : num__5 <o> e ) num__7 : num__1 |
( x + y ) / ( x - y ) = num__1.33333333333 dividing numerator and denominator by y ( x / y ) + num__1 / ( x / y ) - num__1 = num__1.33333333333 let x / y be z z + num__1 / z - num__1 = num__1.33333333333 num__3 z + num__3 = num__4 z - num__4 z = num__7 x / y = num__7 answer e <eor> e <eos> |
e |
divide__4.0__3.0__ round_down__1.3333__ add__4.0__3.0__ add__4.0__3.0__ |
divide__4.0__3.0__ subtract__4.0__3.0__ add__4.0__3.0__ add__4.0__3.0__ |
| if a lends rs . num__3500 to b at num__10.0 per annum and b lends the same sum to c at num__11.5 per annum then the gain of b in a period of num__3 years is ? <o> a ) num__99.222 <o> b ) num__223.22 <o> c ) num__157.5 <o> d ) num__222.88 <o> e ) num__229.11 |
( num__3500 * num__1.5 * num__3 ) / num__100 = > num__157.50 answer : c <eor> c <eos> |
c |
percent__100.0__157.5__ |
percent__100.0__157.5__ |
| three workers have a productivity ratio of num__2 to num__5 to num__9 . all three workers are working on a job for num__4 hours . at the beginning of the num__5 th hour the slowest worker takes a break . the slowest worker comes back to work at the beginning of the num__9 th hour and begins working again . the job is done in ten hours . what was the ratio of the work performed by the fastest worker as compared to the slowest ? <o> a ) num__12 to num__1 <o> b ) num__6 to num__1 <o> c ) num__15 to num__2 <o> d ) num__1 to num__6 <o> e ) num__1 to num__5 |
the fastest worker who does num__9 units of job worked for all num__10 hours so he did num__9 * num__10 = num__90 units of job ; the slowest worker who does num__2 unit of job worked for only num__4 + num__2 = num__6 hours ( first num__4 hours and last num__2 hours ) so he did num__2 * num__6 = num__12 units of job ; the ratio thus is num__90 to num__12 or num__15 to num__2 . answer : c . <eor> c <eos> |
c |
multiply__2.0__5.0__ multiply__9.0__10.0__ add__2.0__4.0__ multiply__2.0__6.0__ add__5.0__10.0__ round__15.0__ |
multiply__2.0__5.0__ multiply__9.0__10.0__ add__2.0__4.0__ multiply__2.0__6.0__ add__5.0__10.0__ add__5.0__10.0__ |
| divide num__71 into num__2 parts such that num__1 part exceeds the other by num__8 . what are the num__2 no . ' s in that part ? <o> a ) num__23 <o> b ) num__28 <o> c ) num__31 <o> d ) num__36 <o> e ) num__39 |
let $ let n $ be the smaller and num__71 - n be the larger number . now since the larger number exceeds the smaller number by num__7 we can form the equation larger number – smaller number = num__7 which is equivalent to simplifying we have num__71 - num__2 n = num__7 . this gives us num__2 n = num__78 which implies that the larger number is . the smaller is num__71 - num__39 = num__31 . c <eor> c <eos> |
c |
subtract__8.0__1.0__ add__71.0__7.0__ divide__78.0__2.0__ subtract__39.0__8.0__ multiply__1.0__31.0__ |
subtract__8.0__1.0__ add__71.0__7.0__ divide__78.0__2.0__ subtract__39.0__8.0__ subtract__39.0__8.0__ |
| a light flashes every num__20 seconds how many times will it flash in ? of an hour ? <o> a ) num__550 <o> b ) num__600 <o> c ) num__650 <o> d ) num__180 <o> e ) num__750 |
num__1 flash = num__20 sec for num__1 min = num__3 flashes so for num__1 hour = num__3 * num__60 = num__180 flashes . answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ multiply__3.0__60.0__ round__180.0__ |
multiply__20.0__3.0__ multiply__3.0__60.0__ multiply__3.0__60.0__ |
| company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month num__0.666666666667 were basic and the rest were deluxe . if it takes num__1.2 as many hours to produce a deluxe stereo as it does to produce a basic stereo then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos ? <o> a ) num__0.375 <o> b ) num__0.451612903226 <o> c ) num__0.466666666667 <o> d ) num__0.485714285714 <o> e ) num__0.5 |
the easiest way for me is to plug in numbers . let the number of basic stereos produced be num__40 and number of delux stereos produced be num__20 . total of num__60 stereos . if it takes an hour to produce a basic stereo then it will take num__1.2 hours to produce a deluxe stereo . num__40 basic stereos = num__40 hours . num__20 delux stereos = num__24 hours . total hours = num__64 . then the fraction would be num__0.375 = num__0.375 . therefore answer a . <eor> a <eos> |
a |
add__40.0__20.0__ multiply__1.2__20.0__ add__40.0__24.0__ divide__24.0__64.0__ divide__24.0__64.0__ |
add__40.0__20.0__ multiply__1.2__20.0__ add__40.0__24.0__ divide__24.0__64.0__ divide__24.0__64.0__ |
| log ( a ( a ( a ) ^ num__0.25 ) ^ num__0.25 ) ^ num__0.25 here base is a . <o> a ) num__0 <o> b ) num__0.328125 <o> c ) num__1 <o> d ) num__0.349206349206 <o> e ) num__0.353846153846 |
log ( a ( a ( a ^ num__0.25 ) ^ num__0.25 ) ^ num__0.25 = > log ( a ( a ^ num__1 + num__0.25 ) ^ num__0.25 ) ^ num__0.25 = > log ( a ( a ^ num__1.25 ) ^ num__0.25 ) ^ num__0.25 = > log ( a ( a ^ num__0.3125 ) ^ num__0.25 = > log ( a ^ num__1 + num__0.3125 ) ^ num__0.25 = > log ( a ^ num__1.3125 ) ^ num__0.25 = > log ( a ^ num__0.328125 ) = > num__0.328125 log ( a ) = > num__0.328125 answer : b <eor> b <eos> |
b |
add__0.25__1.0__ multiply__0.25__1.25__ add__1.0__0.3125__ multiply__0.25__1.3125__ multiply__0.25__1.3125__ |
add__0.25__1.0__ multiply__0.25__1.25__ add__1.0__0.3125__ multiply__0.25__1.3125__ multiply__0.25__1.3125__ |
| what is the cp of rs num__100 stock at num__2 discount with num__0.2 % brokerage ? <o> a ) num__99.6 <o> b ) num__96.2 <o> c ) num__97.5 <o> d ) num__98.2 <o> e ) none of these |
explanation : use the formula cp = num__100 â € “ discount + brokerage % cp = num__100 - num__2 + num__0.2 num__98.2 thus the cp is rs num__98.2 . answer d <eor> d <eos> |
d |
percent__100.0__98.2__ |
percent__100.0__98.2__ |
| the average of the marks of num__10 students in a class is num__40 . if the marks of each student are doubled find the new average ? <o> a ) num__78 <o> b ) num__56 <o> c ) num__80 <o> d ) num__27 <o> e ) num__40 |
sum of the marks for the num__10 students = num__10 * num__40 = num__400 . the marks of each student are doubled the sum also will be doubled . the new sum = num__400 * num__2 = num__800 . so the new average = num__80.0 = num__80 . answer : c <eor> c <eos> |
c |
multiply__10.0__40.0__ multiply__2.0__400.0__ multiply__40.0__2.0__ multiply__40.0__2.0__ |
multiply__10.0__40.0__ multiply__2.0__400.0__ multiply__40.0__2.0__ multiply__40.0__2.0__ |
| a man can row upstream at num__20 kmph and downstream at num__60 kmph and then find the speed of the man in still water ? <o> a ) num__27 <o> b ) num__29 <o> c ) num__40 <o> d ) num__20 <o> e ) num__24 |
us = num__20 ds = num__60 m = ( num__20 + num__60 ) / num__2 = num__40 answer : c <eor> c <eos> |
c |
multiply__20.0__2.0__ round__40.0__ |
multiply__20.0__2.0__ round__40.0__ |
| the average mark of the students of a class in a particular exam is num__80 . if num__5 students whose average mark in that exam is num__20 are excluded the average mark of the remaining will be num__92 . find the number of students who wrote the exam . <o> a ) num__15 <o> b ) num__25 <o> c ) num__30 <o> d ) num__45 <o> e ) num__55 |
let the number of students who wrote the exam be x . total marks of students = num__80 x . total marks of ( x - num__5 ) students = num__92 ( x - num__5 ) num__80 x - ( num__5 * num__20 ) = num__92 ( x - num__5 ) num__360 = num__12 x = > x = num__30 answer : c <eor> c <eos> |
c |
subtract__92.0__80.0__ divide__360.0__12.0__ divide__360.0__12.0__ |
subtract__92.0__80.0__ divide__360.0__12.0__ divide__360.0__12.0__ |
| what should come in place of the question mark ( ? ) in the following equation ? num__28 ⁄ ? = ? ⁄ num__112 <o> a ) num__70 <o> b ) num__56 <o> c ) num__48 <o> d ) num__64 <o> e ) none of these |
num__28 ⁄ ? = ? ⁄ num__112 \ ? = √ num__28 × num__112 = num__56 answer b <eor> b <eos> |
b |
subtract__112.0__56.0__ |
subtract__112.0__56.0__ |
| find the value of num__72517 x num__9999 = m ? <o> a ) num__456578972 <o> b ) num__436567874 <o> c ) num__653658791 <o> d ) num__725097483 <o> e ) num__357889964 |
num__72517 x num__9999 = num__72517 x ( num__10000 - num__1 ) = num__72517 x num__10000 - num__72517 x num__1 = num__725170000 - num__72517 = num__725097483 d <eor> d <eos> |
d |
subtract__10000.0__9999.0__ multiply__72517.0__10000.0__ multiply__72517.0__9999.0__ multiply__72517.0__9999.0__ |
subtract__10000.0__9999.0__ multiply__72517.0__10000.0__ subtract__725170000.0__72517.0__ subtract__725170000.0__72517.0__ |
| on the coordinate plane points p and b are defined by the coordinates ( - num__10 ) and ( num__33 ) respectively and are connected to form a chord of a circle which also lies on the plane . if the area of the circle is ( num__6.25 ) π what are the coordinates of the center of the circle ? <o> a ) ( num__1.5 num__1 ) <o> b ) ( num__2 - num__5 ) <o> c ) ( num__00 ) <o> d ) ( num__1 num__1.5 ) <o> e ) ( num__22 ) |
although it took me num__3 mins to solve this question using all those equations later i thought this question can be solved easily using options . one property to keep in mind - a line passing through the centre of the circle bisects the chord ( or passes from the mid point of the chord ) . now mid point of chord here is ( - num__1 + num__3 ) / num__2 ( num__3 + num__0 ) / num__2 i . e . ( num__1 num__1.5 ) now luckily we have this in our ans . choice . so definitely this is the ans . it also indictaes that pb is the diameter of the circle . there can be a case when pb is not a diameter but in that case also the y - coordinate will remain same as it is the midpoint of the chord and we are moving up in the st . line to locate the centre of the circle . if ans choices are all distinct ( y cordinates ) only check for y cordinate and mark the ans = d <eor> d <eos> |
d |
triangle_area__1.0__3.0__ volume_cube__1.0__ |
triangle_area__1.0__3.0__ volume_cube__1.0__ |
| a does a work in num__10 days and b does the same work in num__5 days . in how many days they together will do the same work ? <o> a ) num__0.2 <o> b ) num__0.285714285714 <o> c ) num__0.3 <o> d ) num__0.272727272727 <o> e ) num__0.176470588235 |
firstly we will find num__1 day work of both a and b then by adding we can get collective days for them so a ' s num__1 day work = num__0.1 b ' s num__1 day work = num__0.0666666666667 ( a + b ) ' s num__1 day work = ( num__0.1 + num__0.2 ) = ( num__1 + num__0.2 ) = num__0.3 c <eor> c <eos> |
c |
divide__1.0__10.0__ divide__1.0__5.0__ add__0.2__0.1__ round__0.3__ |
divide__1.0__10.0__ divide__1.0__5.0__ add__0.2__0.1__ add__0.2__0.1__ |
| there is a rectangular prism made of num__1 in cubes that has been covered in tin foil . there are exactly num__128 cubes that are not touching any tin foil on any of their sides . if the width of the figure created by these num__128 cubes is twice the length and twice the height what is the measure e in inches of the width of the foil covered prism ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
if the width is w then length and height would be w / num__2 . so w * w / num__2 * w / num__2 = num__128 = > w ^ num__3 = ( num__2 ^ num__3 ) * num__64 = ( num__2 ^ num__3 ) * ( num__4 ^ num__3 ) = > w = num__2 * num__4 = num__8 in . along the width of the cuboid num__8 cubes do n ' t touch the tin foil . so the actual width will be non - touching cubes + touching cubes = num__8 + num__2 = e = num__10 ans e . <eor> e <eos> |
e |
triangle_area__1.0__128.0__ square_perimeter__1.0__ square_perimeter__2.0__ rectangle_perimeter__1.0__4.0__ rectangle_perimeter__1.0__4.0__ |
triangle_area__1.0__128.0__ square_perimeter__1.0__ power__2.0__3.0__ rectangle_perimeter__1.0__4.0__ power__10.0__1.0__ |
| the ratio between the perimeter and the width of a rectangle is num__5 : num__1 . if the area of the rectangle is num__150 sq . cm what is the length of the rectangle ? <o> a ) num__12 cm <o> b ) num__15 cm <o> c ) num__18 cm <o> d ) num__21 cm <o> e ) num__24 cm |
num__2 l + num__2 w = num__5 w l = num__3 w / num__2 w * l = num__150 num__3 w ^ num__1.0 = num__150 w ^ num__2 = num__100 w = num__10 l = num__3 ( num__10 ) / num__2 = num__15 the answer is b . <eor> b <eos> |
b |
multiply__5.0__2.0__ multiply__5.0__3.0__ multiply__5.0__3.0__ |
multiply__5.0__2.0__ multiply__5.0__3.0__ multiply__5.0__3.0__ |
| what is the diffference between the place value and face value of num__4 in the numeral num__2694 ? <o> a ) num__10 <o> b ) num__0 <o> c ) num__5 <o> d ) num__3 <o> e ) num__1 |
place value of num__4 = num__4 * num__1 = num__4 face value of num__4 = num__4 num__4 - num__4 = num__0 b <eor> b <eos> |
b |
multiply__4.0__0.0__ |
multiply__4.0__0.0__ |
| find the average of all prime numbers between num__1 and num__5 . <o> a ) num__30 <o> b ) num__3.6 <o> c ) num__5.6 <o> d ) num__6.6 <o> e ) none |
sol . there are five prime numbers between num__1 and num__5 . they are num__2 num__3 num__5 num__7 num__11 â ˆ ´ required average = [ num__2 + num__3 + num__5 + num__7 + num__2.2 ] = num__5.6 = num__5.6 answer c <eor> c <eos> |
c |
add__1.0__2.0__ add__5.0__2.0__ divide__11.0__5.0__ multiply__1.0__5.6__ |
add__1.0__2.0__ add__5.0__2.0__ divide__11.0__5.0__ multiply__1.0__5.6__ |
| peeta asks katniss ' it is num__6 am as per my watch right now . do you know what will be the time num__23 num__999 num__998 hours later ? ' katniss knows the answer . do you ? <o> a ) num__5 pm <o> b ) num__6 pm <o> c ) num__4 am <o> d ) num__3 am <o> e ) num__8 pm |
d num__3 am the time after num__24 num__000 num__000 hours will be the same as it is now . we want the time num__3 hours before that and thus you can deduct three hours from num__9 pm . so the time will be num__6 pm . <eor> d <eos> |
d |
add__6.0__3.0__ round__3.0__ |
add__6.0__3.0__ round__3.0__ |
| in how many ways a committee of num__5 members can be selected from num__4 men and num__5 ladies consisting of num__3 men and num__2 ladies ? <o> a ) num__40 <o> b ) num__152 <o> c ) num__175 <o> d ) num__200 <o> e ) num__212 |
( num__3 men out num__4 ) and ( num__2 ladies out of num__5 ) are to be chosen required number of ways = ( num__4 c num__3 * num__5 c num__2 ) = num__40 answer is a <eor> a <eos> |
a |
choose__5.0__3.0__ choose__5.0__3.0__ |
choose__5.0__3.0__ choose__5.0__3.0__ |
| in a certain lottery the probability that a number between num__12 and num__20 inclusive is drawn is num__0.2 . if the probability that a number num__12 or larger is drawn is num__0.666666666667 what is the probability that a number less than or equal to num__20 is drawn ? <o> a ) num__0.533333333333 <o> b ) num__0.166666666667 <o> c ) num__0.333333333333 <o> d ) num__0.5 <o> e ) num__0.833333333333 |
you can simply use sets concept in this question . the formula total = n ( a ) + n ( b ) - n ( a and b ) is applicable here too . set num__1 : number num__12 or larger set num__2 : number num__20 or smaller num__1 = p ( set num__1 ) + p ( set num__2 ) - p ( set num__1 and set num__2 ) ( combined probability is num__1 because every number will be either num__12 or moreor num__20 or lessor both ) num__0.666666666667 + p ( set num__2 ) - num__0.2 = num__1 p ( set num__2 ) = num__0.533333333333 answer ( a ) <eor> a <eos> |
a |
multiply__1.0__0.5333__ |
multiply__1.0__0.5333__ |
| a car is purchased on hire - purchase . the cash price is $ num__24 num__000 and the terms are a deposit of num__10.0 of the price then the balance to be paid off over num__60 equal monthly installments . interest is charged at num__12.0 p . a . what is the monthly installment ? <o> a ) $ num__503 <o> b ) $ num__504 <o> c ) $ num__555 <o> d ) $ num__576 <o> e ) $ num__587 |
explanation : cash price = $ num__24 num__000 deposit = num__10.0 Ã — $ num__24 num__000 = $ num__2400 loan amount = $ num__24000 â ˆ ’ $ num__2400 number of payments = num__60 = $ num__21600 i = p * r * t / num__100 i = num__12960 total amount = num__21600 + num__12960 = $ num__34560 regular payment = total amount / number of payments = num__576 answer : d <eor> d <eos> |
d |
percent__60.0__21600.0__ percent__24.0__2400.0__ percent__24.0__2400.0__ |
percent__60.0__21600.0__ percent__24.0__2400.0__ percent__24.0__2400.0__ |
| a dishonest dealer professes to sell goods at the cost price but uses a false weight and gains num__25.0 . find his false weight age ? <o> a ) num__600 <o> b ) num__400 <o> c ) num__500 <o> d ) num__670 <o> e ) num__800 |
e num__25 = e / ( num__1000 - e ) * num__100 num__1000 - e = num__4 e num__1000 = num__5 e = > e = num__200 num__1000 - num__200 = num__800 <eor> e <eos> |
e |
percent__100.0__800.0__ |
percent__100.0__800.0__ |
| cube of side one meter length is cut into small cubes of side num__10 cm each . how many such small cubes can be obtained ? <o> a ) num__1002 <o> b ) num__2899 <o> c ) num__1000 <o> d ) num__2807 <o> e ) num__2800 |
along one edge the number of small cubes that can be cut = num__10.0 = num__10 along each edge num__10 cubes can be cut . ( along length breadth and height ) . total number of small cubes that can be cut = num__10 * num__10 * num__10 = num__1000 answer : c <eor> c <eos> |
c |
round__1000.0__ |
round__1000.0__ |
| consider the sequence num__1 - num__23 - num__4 num__56 . . . . . what is the average of the first num__200 terms of the sequence ? <o> a ) - num__0.5 <o> b ) - num__1.5 <o> c ) num__0 <o> d ) num__0.5 <o> e ) num__1.5 |
suppose n = num__2 = > avg is ( num__1 + - num__2 ) / num__2 = - num__0.5 when n = num__4 = > avg is - num__0.5 similarly for n = num__200 avg will be - num__0.5 . answer : a <eor> a <eos> |
a |
reverse__2.0__ reverse__2.0__ |
reverse__2.0__ reverse__2.0__ |
| he average weight of num__8 persons increases by num__3.5 kg when a new person comes in place of one of them weighing num__62 kg . what might be the weight of the new person ? <o> a ) num__75 kg <o> b ) num__55 kg <o> c ) num__45 kg <o> d ) num__85 kg <o> e ) num__90 kg |
explanation : total weight increased = ( num__8 x num__3.5 ) kg = num__28 kg . weight of new person = ( num__62 + num__28 ) kg = num__90 kg . answer : e <eor> e <eos> |
e |
multiply__8.0__3.5__ add__62.0__28.0__ add__62.0__28.0__ |
multiply__8.0__3.5__ add__62.0__28.0__ add__62.0__28.0__ |
| vinoth can complete a painting work in num__20 days . prakash can do the same work in num__25 days . they start the work together but vinoth quit after num__3 days of work . how many days are required to complete the remaining painting work by prakash . <o> a ) num__15.25 days <o> b ) num__16.25 days <o> c ) num__17.25 days <o> d ) num__18.25 days <o> e ) num__19.25 days |
vinoth can complete the painting work in one day is num__0.05 prakash can complete the same work in one day is num__0.04 both of them can complete the work in num__0.05 + days = num__0.09 ( num__0.05 + num__0.04 ) they must have completed in three days = num__0.09 * num__3 = num__0.27 remaining work to be done is by prakash = num__1 - num__0.27 = num__0.73 for one work prakash can do in num__25 days for num__0.73 work he can do in num__0.73 * num__25 = num__18.25 days or num__18.25 days answer : d <eor> d <eos> |
d |
add__0.05__0.04__ multiply__3.0__0.09__ multiply__20.0__0.05__ subtract__1.0__0.27__ multiply__25.0__0.73__ round__18.25__ |
add__0.05__0.04__ multiply__3.0__0.09__ multiply__20.0__0.05__ subtract__1.0__0.27__ multiply__25.0__0.73__ multiply__25.0__0.73__ |
| a box contains nine bulbs out of which num__4 are defective . if four bulbs are chosen at random find the probability that atleast one bulb is good <o> a ) num__0.992063492063 <o> b ) num__0.968992248062 <o> c ) num__0.9765625 <o> d ) num__1.00806451613 <o> e ) num__1.03305785124 |
required probability = num__1 - num__0.00793650793651 = num__0.992063492063 . answer : a <eor> a <eos> |
a |
negate_prob__0.0079__ negate_prob__0.0079__ |
negate_prob__0.0079__ negate_prob__0.0079__ |
| a train passes a station platform in num__35 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__120 m <o> b ) num__225 m <o> c ) num__300 m <o> d ) num__360 m <o> e ) num__280 m |
speed = num__54 x num__0.277777777778 m / sec = num__15 m / sec . length of the train = ( num__15 x num__20 ) m = num__300 m . let the length of the platform be x metres . then ( x + num__300 ) / num__35 = num__15 x + num__300 = num__525 x = num__225 m . answer : option b <eor> b <eos> |
b |
subtract__35.0__20.0__ multiply__20.0__15.0__ multiply__35.0__15.0__ subtract__525.0__300.0__ round__225.0__ |
subtract__35.0__20.0__ multiply__20.0__15.0__ multiply__35.0__15.0__ subtract__525.0__300.0__ round__225.0__ |
| let n be the greatest number that will divide num__1305 num__4665 and num__6905 leaving the same remainder in each case . then sum of the digits in n is : <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
explanation : n = h . c . f . of ( num__4665 - num__1305 ) ( num__6905 - num__4665 ) and ( num__6905 - num__1305 ) = h . c . f . of num__3360 num__2240 and num__5600 = num__1120 . sum of digits in n = ( num__1 + num__1 + num__2 + num__0 ) = num__4 answer is a <eor> a <eos> |
a |
subtract__4665.0__1305.0__ subtract__6905.0__4665.0__ subtract__6905.0__1305.0__ subtract__3360.0__2240.0__ divide__2240.0__1120.0__ multiply__4.0__1.0__ |
subtract__4665.0__1305.0__ subtract__6905.0__4665.0__ subtract__6905.0__1305.0__ subtract__3360.0__2240.0__ divide__2240.0__1120.0__ multiply__4.0__1.0__ |
| a train num__120 m long crosses a platform num__120 m long in num__6 sec ; find the speed of the train ? <o> a ) num__144 kmph <o> b ) num__58 kmph <o> c ) num__54 kmph <o> d ) num__94 kmph <o> e ) num__59 kmph |
d = num__120 + num__120 = num__240 t = num__6 s = num__40.0 * num__3.6 = num__144 kmph answer : a <eor> a <eos> |
a |
divide__240.0__6.0__ multiply__40.0__3.6__ round__144.0__ |
divide__240.0__6.0__ multiply__40.0__3.6__ multiply__40.0__3.6__ |
| find the largest number of four digits which is exactly divisible by num__2718 num__1215 <o> a ) num__9700 <o> b ) num__9710 <o> c ) num__9720 <o> d ) num__9730 <o> e ) none of these |
explanation : lcm of num__27 - num__18 - num__12 - num__15 is num__540 . after dividing num__9999 by num__540 we get num__279 remainder . so answer will be num__9999 - num__279 = num__9720 answer : option c <eor> c <eos> |
c |
subtract__27.0__12.0__ subtract__9999.0__279.0__ subtract__9999.0__279.0__ |
subtract__27.0__12.0__ subtract__9999.0__279.0__ subtract__9999.0__279.0__ |
| a train num__250 meters long completely crosses a num__300 meters long bridge in num__45 seconds . what is the speed of the train is ? <o> a ) num__32 <o> b ) num__28 <o> c ) num__29 <o> d ) num__44 <o> e ) num__21 |
s = ( num__250 + num__300 ) / num__45 = num__12.2222222222 * num__3.6 = num__44 answer : d <eor> d <eos> |
d |
round__44.0__ |
round__44.0__ |
| if num__9 engines consume num__24 metric tonnes of coal when each is working num__8 hoursday bow much coal will be required for num__8 engines each running num__13 hours a day it being given that num__3 engines of former type consume as much as num__4 engines of latter type ? <o> a ) num__22 <o> b ) num__24 <o> c ) num__26 <o> d ) num__28 <o> e ) none of them |
let num__3 engines of former type consume num__1 unit in num__1 hour . then num__4 engines of latter type consume num__1 unit in num__1 hour . therefore num__1 engine of former type consumes ( num__0.333333333333 ) unit in num__1 hour . num__1 engine of latter type consumes ( num__0.25 ) unit in num__1 hour . let the required consumption of coal be x units . less engines less coal consumed ( direct proportion ) more working hours more coal consumed ( direct proportion ) less rate of consumption less coal consumed ( direct prportion ) number of engines num__9 : num__8 working hours num__8 : num__13 } : : num__24 : x rate of consumption ( num__0.333333333333 ) : ( num__0.25 ) [ num__9 x num__8 x ( num__0.333333333333 ) x x ) = ( num__8 x num__13 x ( num__0.25 ) x num__24 ) = num__24 x = num__624 = x = num__26 . hence the required consumption of coal = num__26 metric tonnes . answer is c . <eor> c <eos> |
c |
subtract__9.0__8.0__ divide__8.0__24.0__ divide__1.0__4.0__ divide__624.0__24.0__ round__26.0__ |
subtract__9.0__8.0__ divide__8.0__24.0__ divide__1.0__4.0__ divide__624.0__24.0__ round__26.0__ |
| a man walks a certain distance and rides back in num__6 num__1 ⁄ num__4 h . he can walk both ways in num__7 num__3 ⁄ num__4 h . how long it would take to ride both ways ? <o> a ) num__5 hours <o> b ) num__4 num__1 ⁄ num__2 hours <o> c ) num__4 num__3 ⁄ num__4 hours <o> d ) num__6 hours <o> e ) none of these |
we know that the relation in time taken with two different modes of transport is twalk both + tride both = num__2 ( twalk + tride ) num__31 ⁄ num__4 + tride both = num__2 × num__25 ⁄ num__4 ⇒ tride both = num__25 ⁄ num__2 - num__31 ⁄ num__4 = num__19 ⁄ num__4 = num__43 ⁄ num__4 hours answer c <eor> c <eos> |
c |
coin_space__ choose__4.0__3.0__ |
coin_space__ choose__4.0__3.0__ |
| what sum of money will produce rs . num__70 as simple interest in num__4 years at num__3 num__0.5 percent ? <o> a ) num__266 <o> b ) num__500 <o> c ) num__287 <o> d ) num__26 <o> e ) num__281 |
num__70 = ( p * num__4 * num__3.5 ) / num__100 p = num__500 answer : b <eor> b <eos> |
b |
percent__100.0__500.0__ |
percent__100.0__500.0__ |
| the hiker walking at a constant rate of num__7 miles per hour is passed by a cyclist traveling in the same direction along the same path at num__28 miles per hour . the cyclist stops to wait for the hiker num__5 minutes after passing her while the hiker continues to walk at her constant rate how many minutes must the cyclist wait until the hiker catches up ? <o> a ) num__10 <o> b ) num__15 <o> c ) num__20 <o> d ) num__25 <o> e ) num__30 |
for the num__5 minutes the cyclist continues to overtake the hiker she is going at num__21 miles per hour faster than the hiker . once the cyclist stops the hiker is going at num__7 miles per hour while the cyclist is at rest so the amount of time the hiker will take to cover the distance between them is going to be in the ratio of the relative speeds . num__3.0 * num__5 or num__15 minutes answer is ( b ) <eor> b <eos> |
b |
subtract__28.0__7.0__ divide__21.0__7.0__ multiply__5.0__3.0__ round__15.0__ |
subtract__28.0__7.0__ divide__21.0__7.0__ multiply__5.0__3.0__ multiply__5.0__3.0__ |
| tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of num__10 meters and a circumference of num__12 meters and the interior of tank b has a height of num__12 meters and a circumference of num__10 meters . the capacity of tank a is what percent of the capacity of tank b ? <o> a ) num__80.0 <o> b ) num__90.0 <o> c ) num__100.0 <o> d ) num__110.0 <o> e ) num__120 % |
the radius of tank a is num__12 / ( num__2 * pi ) . the capacity of tank a is num__10 * pi * num__144 / ( num__4 * pi ^ num__2 ) = num__360 / ( pi ) the radius of tank b is num__10 / ( num__2 * pi ) . the capacity of tank b is num__12 * pi * num__100 / ( num__4 * pi ^ num__2 ) = num__300 / ( pi ) tank a / tank b = num__1.2 = num__1.2 = num__120.0 the answer is e . <eor> e <eos> |
e |
subtract__12.0__10.0__ divide__12.0__10.0__ multiply__10.0__12.0__ round__120.0__ |
subtract__12.0__10.0__ divide__12.0__10.0__ multiply__10.0__12.0__ multiply__10.0__12.0__ |
| the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square whose area is num__1225 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is num__10 units ? <o> a ) num__140 sq . units <o> b ) num__186 sq . units <o> c ) num__176 sq . units <o> d ) num__865 sq . units <o> e ) num__143 sq . units |
given that the area of the square = num__1225 sq . units = > side of square = √ num__1225 = num__35 units the radius of the circle = side of the square = num__35 units length of the rectangle = num__0.4 * num__35 = num__14 units given that breadth = num__10 units area of the rectangle = lb = num__14 * num__10 = num__140 sq . units answer : a <eor> a <eos> |
a |
multiply__35.0__0.4__ square_perimeter__35.0__ square_perimeter__35.0__ |
multiply__35.0__0.4__ multiply__10.0__14.0__ multiply__10.0__14.0__ |
| a man buys an article for $ num__10 . and sells it for $ num__30 . find the gain percent ? <o> a ) num__25.0 <o> b ) num__50.0 <o> c ) num__20.0 <o> d ) num__150.0 <o> e ) num__300 % |
c . p . = $ num__10 s . p . = $ num__30 gain = $ num__20 gain % = num__2.0 * num__100 = num__300.0 answer is e <eor> e <eos> |
e |
percent__10.0__20.0__ percent__100.0__300.0__ |
percent__10.0__20.0__ percent__100.0__300.0__ |
| the average age of num__19 persons in a office is num__15 years . out of these the average age of num__5 of them is num__14 years and that of the other num__9 persons is num__16 years . the age of the num__15 th person is ? <o> a ) num__9 <o> b ) num__71 <o> c ) num__85 <o> d ) num__92 <o> e ) num__90 |
age of the num__15 th student = num__19 * num__15 - ( num__14 * num__5 + num__16 * num__9 ) = num__285 - num__214 = num__71 years answer is b <eor> b <eos> |
b |
multiply__19.0__15.0__ subtract__285.0__214.0__ subtract__285.0__214.0__ |
multiply__19.0__15.0__ subtract__285.0__214.0__ subtract__285.0__214.0__ |
| pavan travelled for num__11 hours . he covered the first half of the distance at num__30 kmph and remaining half of the distance at num__25 kmph . find the distance travelled by pavan . <o> a ) num__208 km <o> b ) num__637 km <o> c ) num__342 km <o> d ) num__300 kilometre <o> e ) num__543 km |
let the distance travelled be x km . total time = ( x / num__2 ) / num__30 + ( x / num__2 ) / num__25 = num__11 = > x / num__60 + x / num__50 = num__11 = > ( num__5 x + num__6 x ) / num__300 = num__11 = > x = num__300 km answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ multiply__25.0__2.0__ subtract__30.0__25.0__ subtract__11.0__5.0__ multiply__5.0__60.0__ round__300.0__ |
hour_to_min_conversion__ multiply__25.0__2.0__ subtract__30.0__25.0__ divide__30.0__5.0__ multiply__5.0__60.0__ round__300.0__ |
| there are num__3 numbers a b and c . if a : b = num__0.6 b : c = num__1.25 c : d = num__0.5 then a : d will be ? <o> a ) num__1 : num__2 <o> b ) num__3 : num__5 <o> c ) num__5 : num__7 <o> d ) num__6 : num__11 <o> e ) num__3 : num__8 |
sol . a : b = num__3 : num__5 b : c = num__5 : num__4 c : d = num__4 : num__8 ∴ a ∶ b ∶ c ∶ d = num__3 : num__5 : num__4 : num__8 thus a : d = num__3 : num__8 e <eor> e <eos> |
e |
divide__3.0__0.6__ divide__5.0__1.25__ add__3.0__5.0__ multiply__0.6__5.0__ |
divide__3.0__0.6__ divide__5.0__1.25__ add__3.0__5.0__ multiply__0.6__5.0__ |
| if num__14 men do a work in num__80 days in how many days will num__20 men do it ? <o> a ) num__18 days <o> b ) num__38 days <o> c ) num__42 days <o> d ) num__48 days <o> e ) num__56 days |
num__14 * num__80 = num__20 * x x = num__56 days answer : e <eor> e <eos> |
e |
round__56.0__ |
round__56.0__ |
| a soccer team played num__160 games and won num__65 percent of them . how many games did it win ? <o> a ) num__84 <o> b ) num__94 <o> c ) num__104 <o> d ) num__114 <o> e ) num__124 |
num__65.0 of num__160 = x num__0.65 * num__160 = x num__104 = x answer : c <eor> c <eos> |
c |
percent__65.0__160.0__ percent__65.0__160.0__ |
percent__65.0__160.0__ percent__65.0__160.0__ |
| a driver goes on a trip of num__50 kilometers the first num__25 kilometers at num__66 kilometers per hour and the remaining distance at num__33 kilometers per hour . what is the average speed of the entire trip in kilometers per hour ? <o> a ) num__38 <o> b ) num__40 <o> c ) num__44 <o> d ) num__45 <o> e ) num__48 |
the time for the first part of the trip was num__0.378787878788 hours . the time for the second part of the trip was num__0.757575757576 hours . the total time fro the trip was num__0.378787878788 + num__0.757575757576 = num__1.13636363636 = num__1.13636363636 hours . the average speed for the trip was num__50 / ( num__1.13636363636 ) = num__44 kph the answer is c . <eor> c <eos> |
c |
divide__25.0__66.0__ divide__50.0__66.0__ add__0.7576__0.3788__ round__44.0__ |
divide__25.0__66.0__ divide__50.0__66.0__ add__0.7576__0.3788__ round__44.0__ |
| it is given that num__2 ^ num__32 + num__1 is exactly divisible by a certain number . which one of the following is also divisible by the same number r ? <o> a ) a . num__2 ^ num__96 + num__1 <o> b ) b . num__2 ^ num__16 - num__1 <o> c ) c . num__2 ^ num__16 + num__1 <o> d ) d . num__7 * num__2 ^ num__33 <o> e ) e . num__2 ^ num__64 + num__1 |
a ³ + b ³ = ( a + b ) ( a ² - ab + b ² ) now let ( num__2 ^ num__32 + num__1 ) be ( a + b ) a ³ + b ³ = ( num__2 ^ num__96 + num__1 ) now as mentioned in formula above a ³ + b ³ is always divisible by ( a + b ) so any factor of r = ( a + b ) is a factor of ( a ³ + b ³ ) henca a <eor> a <eos> |
a |
multiply__2.0__1.0__ |
multiply__2.0__1.0__ |
| a cistern can be filled in num__9 hours but due to a leak at its bottom it takes num__10 hours . if the cistern is full then the time that the leak will take to make it empty will be ? <o> a ) num__20 hours <o> b ) num__19 hours <o> c ) num__90 hours <o> d ) num__80 hours <o> e ) none of these |
explanation : part filled without leak in num__1 hour = num__0.111111111111 part filled with leak in num__1 hour = num__0.1 work done by leak in num__1 hour = num__0.111111111111 − num__0.1 = num__0.0111111111111 we used subtraction as it is getting empty . so total time to empty the cistern is num__90 hours option c <eor> c <eos> |
c |
subtract__10.0__9.0__ divide__1.0__9.0__ divide__1.0__10.0__ divide__0.1111__10.0__ multiply__9.0__10.0__ round__90.0__ |
subtract__10.0__9.0__ divide__1.0__9.0__ divide__1.0__10.0__ divide__0.1111__10.0__ multiply__9.0__10.0__ round__90.0__ |
| a can give b num__100 meters start and c num__200 meters start in a kilometer race . how much start can b give c in a kilometer race ? <o> a ) num__111.12 m <o> b ) num__111.19 m <o> c ) num__111.82 m <o> d ) num__113.12 m <o> e ) num__211.12 m |
a runs num__1000 m while b runs num__900 m and c runs num__800 m . the number of meters that c runs when b runs num__1000 m = ( num__1000 * num__800 ) / num__900 = num__888.888888889 = num__888.88 m . b can give c = num__1000 - num__888.88 = num__111.12 m . answer : a <eor> a <eos> |
a |
subtract__1000.0__100.0__ subtract__900.0__100.0__ subtract__1000.0__888.88__ round__111.12__ |
subtract__1000.0__100.0__ subtract__900.0__100.0__ subtract__1000.0__888.88__ subtract__1000.0__888.88__ |
| if the perimeter of a rectangular garden is num__600 m its length when its breadth is num__95 m is ? <o> a ) num__286 m <o> b ) num__899 m <o> c ) num__200 m <o> d ) num__205 m <o> e ) num__187 m |
num__2 ( l + num__95 ) = num__600 = > l = num__205 m answer : d <eor> d <eos> |
d |
round__205.0__ |
round__205.0__ |
| how many seconds will a num__500 m long train take to cross a man walking with a speed of num__3 km / hr in the direction of the moving train if the speed of the train is num__63 km / hr ? <o> a ) num__65 sec <o> b ) num__30 sec <o> c ) num__87 sec <o> d ) num__88 sec <o> e ) num__55 sec |
speed of train relative to man = num__63 - num__3 = num__60 km / hr . = num__60 * num__0.277777777778 = num__16.6666666667 m / sec . time taken to pass the man = num__500 * num__0.06 = num__30 sec . answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__500.0__0.06__ round__30.0__ |
subtract__63.0__3.0__ multiply__500.0__0.06__ multiply__500.0__0.06__ |
| the price of a jacket is reduced by num__35.0 . during a special sale the price of the jacket is reduced another num__10.0 . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? <o> a ) num__70.9 <o> b ) num__75 <o> c ) num__48 <o> d ) num__65 <o> e ) num__67.5 |
num__1 ) let the price of jacket initially be $ num__100 . num__2 ) then it is decreased by num__35.0 therefore bringing down the price to $ num__65 . num__3 ) again it is further discounted by num__10.0 therefore bringing down the price to $ num__58.5 num__4 ) now num__58.5 has to be added byx % in order to equal the original price . num__58.5 + ( x % ) num__58.5 = num__100 . solving this eq for x we get x = num__70.9 ans is a . <eor> a <eos> |
a |
percent__100.0__70.9__ |
percent__100.0__70.9__ |
| ( num__469 + num__174 ) ( power num__2 ) - ( num__469 - num__174 ) ( power num__2 ) / ( num__469 x num__174 ) = ? <o> a ) num__2 <o> b ) num__6 <o> c ) num__4 <o> d ) num__8 <o> e ) num__10 |
formula = ( a + b ) ( power num__2 ) - ( a - b ) ( power num__2 ) / ab = num__4 ab / ab = num__4 ( where a = num__469 b = num__174 . ) answer is d . <eor> d <eos> |
d |
multiply__2.0__4.0__ |
multiply__2.0__4.0__ |
| yukou enjoys reading science - fiction . it is the new year and he has made a resolution to read a book every week for this year . he has a collection of monster books and space travel books in the ratio of num__4 : num__5 . he presently has num__20 space travel books . how many more books ( of each genre ) does he need to buy to keep his resolution without changing the ratio ? <o> a ) num__6 monster num__10 space <o> b ) num__8 monster num__10 space <o> c ) num__8 monster num__9 space <o> d ) num__4 monster num__10 space <o> e ) num__7 monster num__12 space |
there are num__52 weeks in a year . youku has num__20 space travel books so num__4 : num__5 is num__0.8 : num__1 means ( num__0.8 ) ( num__20 ) = num__16 monster books . so he has num__20 + num__16 = num__36 books . youku needs num__52 - num__36 = num__16 more books . keeping with the ratio of num__4 : num__5 and allowing for the fact that we can not have a fraction of a book the minimum ratio that satisfies these restrictions is num__8 : num__10 giving num__18 books of which num__8 are monster books and num__10 space travel books . answer : b <eor> b <eos> |
b |
divide__4.0__5.0__ subtract__5.0__4.0__ subtract__20.0__4.0__ add__20.0__16.0__ divide__8.0__0.8__ add__8.0__10.0__ multiply__1.0__8.0__ |
divide__4.0__5.0__ subtract__5.0__4.0__ subtract__20.0__4.0__ add__20.0__16.0__ divide__8.0__0.8__ add__8.0__10.0__ subtract__16.0__8.0__ |
| sripad has scored average of num__65 marks in three objects . in no subjects has he secured less than num__58 marks . he has secured more marks in maths than other two subjects . what could be his maximum score in maths ? <o> a ) num__79 <o> b ) num__28 <o> c ) num__38 <o> d ) num__27 <o> e ) num__21 |
assuming sripad has scored the least marks in subject other than science then the marks he could secure in other two are num__58 each . since the average mark of all the num__3 subject is num__65 . i . e ( num__58 + num__58 + x ) / num__3 = num__65 num__116 + x = num__195 x = num__79 marks . therefore the maximum marks he can score in maths is num__79 . answer : a <eor> a <eos> |
a |
multiply__65.0__3.0__ subtract__195.0__116.0__ subtract__195.0__116.0__ |
multiply__65.0__3.0__ subtract__195.0__116.0__ subtract__195.0__116.0__ |
| how many odd num__4 - digit positive integers t that are multiples of num__5 can be formed without using the digit num__3 ? <o> a ) num__648 <o> b ) num__729 <o> c ) num__900 <o> d ) num__1296 <o> e ) num__3240 |
i choose a . possible n ° of num__1 st digit : num__8 ( num__0 ca n ' t be the first number or else it would n ' t have num__4 digits . num__3 is exlcuded ) possible n ° of num__2 nd digit : num__9 ( num__3 is excluded ) possible n ° of num__3 rd digit : num__9 ( num__3 is excluded ) possible n ° of num__4 th digit : num__1 ( a number is a multiple of num__5 if it ends in num__5 or num__0 here we are asked for the odd numbers hence the last digit ca n ' t be num__0 ) so t = num__8 * num__9 * num__9 * num__1 = num__648 ( a ) <eor> a <eos> |
a |
subtract__4.0__3.0__ add__5.0__3.0__ subtract__5.0__3.0__ add__4.0__5.0__ multiply__1.0__648.0__ |
subtract__4.0__3.0__ add__5.0__3.0__ subtract__5.0__3.0__ add__4.0__5.0__ multiply__1.0__648.0__ |
| a cylindrical bucket of height num__36 cm and radius num__21 cm is filled with sand . the bucket is emptied on the ground and a conical heap of sand is formed the height of the heap being num__12 cm . the radius of the heap at the base is : <o> a ) num__63 cm <o> b ) num__53 cm <o> c ) num__56 cm <o> d ) num__66 cm <o> e ) none of these |
volume of the bucket = volume of the sand emptied volume of sand = π ( num__21 ) num__2 × num__36 let r be the radius of the conical heap . then num__1 ⁄ num__3 π r num__2 × num__12 = π ( num__21 ) num__2 × num__36 or r num__2 = ( num__21 ) num__2 × num__9 or r = num__21 × num__3 = num__63 answer a <eor> a <eos> |
a |
power__3.0__2.0__ multiply__21.0__3.0__ multiply__21.0__3.0__ |
power__3.0__2.0__ multiply__21.0__3.0__ multiply__21.0__3.0__ |
| louie takes out a three - month loan of $ num__1000 . the lender charges him num__10.0 interest per month compunded monthly . the terms of the loan state that louie must repay the loan in three equal monthly payments . to the nearest dollar how much does louie have to pay each month ? <o> a ) num__333 <o> b ) num__383 <o> c ) num__402 <o> d ) num__433 <o> e ) num__483 |
let the monthly payment be xx . after the num__1 st month there will be num__1000 ∗ num__1.1 − x dollars left to repay ; after the num__2 nd month there will be ( num__1000 ∗ num__1.1 − x ) ∗ num__1.1 − x = num__1210 − num__2.1 x dollars left to repay ; after the num__3 rd month there should be num__0 dollars left to repay : ( num__1210 − num__2.1 x ) ∗ num__1.1 − x = num__0 - - > num__1331 = num__3.31 x - - > x ≈ num__402 x ≈ num__402 answer : c . <eor> c <eos> |
c |
add__1.0__1.1__ add__1.0__2.0__ multiply__1.1__1210.0__ multiply__1.0__402.0__ |
add__1.0__1.1__ add__1.0__2.0__ multiply__1.1__1210.0__ multiply__1.0__402.0__ |
| num__10 x + num__2 y = - num__6 num__8 x + y = num__3 in the system of equations above what is the value of x ? <o> a ) - num__6 <o> b ) - num__4 <o> c ) - num__2 <o> d ) num__2 <o> e ) num__4 |
num__10 x + num__2 y = - num__6 can be written as num__5 x + y = - num__3 let ' s subtract the second equation from this equation . - num__3 x = - num__6 x = num__2 the answer is d . <eor> d <eos> |
d |
divide__10.0__2.0__ subtract__10.0__8.0__ |
add__2.0__3.0__ subtract__10.0__8.0__ |
| how many minutes are there in two right angles ? <o> a ) num__21 <o> b ) num__22 <o> c ) num__17 <o> d ) num__2888 <o> e ) num__2880 |
e num__2880 they are the hour - minute hour - second and minute - second . each hour the minute hand will be at a right angle twice so that ' s num__48 right angles - - num__2 for each hour of the day . each hour the second hand will go around num__60 times and have num__2 right angles for each that ' s num__120 right angles per hour or num__120 * num__24 = num__2880 <eor> e <eos> |
e |
clock_big_arm_angle__2.0__ clock_big_arm_angle__48.0__ |
clock_big_arm_angle__2.0__ clock_big_arm_angle__48.0__ |
| a cycle is bought for rs . num__900 and sold for rs . num__1150 find the gain percent ? <o> a ) num__15 <o> b ) num__20 <o> c ) num__27.77 <o> d ) num__40 <o> e ) num__12 |
num__900 - - - - num__250 num__100 - - - - ? = > num__27.77 answer : c <eor> c <eos> |
c |
percent__100.0__27.77__ |
percent__100.0__27.77__ |
| a rectangular rug with side lengths of num__2 feet and num__7 feet is placed on a square floor that has an area of num__64 square feet . if the surface of the rug does not extend beyond the area of the floor what fraction of the area of the floor is not covered by the rug ? <o> a ) a . num__0.1875 <o> b ) num__0.25 <o> c ) c . num__0.5 <o> d ) num__0.78125 <o> e ) num__0.875 |
area of the rectangular rug = num__2 * num__7 = num__14 fraction not covered by the rug = ( total area - rug area ) / total area = ( num__64 - num__14 ) / num__64 = num__0.78125 = d <eor> d <eos> |
d |
multiply__2.0__7.0__ triangle_area__2.0__0.7812__ |
multiply__2.0__7.0__ triangle_area__2.0__0.7812__ |
| the sequence x num__1 x num__2 x num__3 . . . is such that xn = num__1 / n - ( num__1 / ( n + num__1 ) ) . what is the sum of the first num__20 terms of the sequence ? <o> a ) num__2.01 <o> b ) num__0.99 <o> c ) num__0.990099009901 <o> d ) num__0.0001 <o> e ) num__0.952380952381 |
easy task and accomplish x num__1 = num__1 - num__0.5 x num__2 = num__0.5 - num__0.333333333333 x num__3 = num__0.333333333333 - num__0.25 . . . . . x num__20 = num__0.05 - num__0.047619047619 sum = x num__1 + x num__2 + x num__3 + . . . . x num__20 = num__1 - num__0.5 + num__0.5 - num__0.333333333333 + . . . . . . . num__0.05 - num__0.047619047619 = num__1 - num__0.047619047619 = num__0.952380952381 e is the answer <eor> e <eos> |
e |
reverse__2.0__ reverse__3.0__ divide__0.5__2.0__ reverse__20.0__ subtract__1.0__0.0476__ multiply__1.0__0.9524__ |
reverse__2.0__ reverse__3.0__ divide__0.5__2.0__ reverse__20.0__ subtract__1.0__0.0476__ subtract__1.0__0.0476__ |
| a larger cube has num__343 cubic inch as a volume and in the cube there are num__343 smaller cubes such that their volume is num__1 cubic inch . what is the difference between the surface areas ’ sum of the num__343 smaller cubes and the surface area of the larger cube in square inch ? <o> a ) num__54 <o> b ) num__64 <o> c ) num__81 <o> d ) num__108 <o> e ) num__1764 |
volume of larger cube = num__343 = num__7 ^ num__3 side of larger cube = num__7 volume of smaller cube = num__1 - - > side of smaller cube = num__1 surface area of larger cube = num__6 * num__7 ^ num__2 = num__294 surface area of num__27 smaller cubes = num__343 * num__6 * num__1 = num__2058 difference = num__2058 - num__294 = num__1764 answer : e <eor> e <eos> |
e |
surface_cube__1.0__ surface_cube__7.0__ volume_cube__3.0__ multiply__343.0__6.0__ multiply__294.0__6.0__ multiply__1.0__1764.0__ |
surface_cube__1.0__ surface_cube__7.0__ volume_cube__3.0__ multiply__343.0__6.0__ multiply__294.0__6.0__ multiply__1.0__1764.0__ |
| a train passes a station platform in num__36 sec and a man standing on the platform in num__20 sec . if the speed of the train is num__54 km / hr . what is the length of the platform ? <o> a ) num__287 <o> b ) num__240 <o> c ) num__772 <o> d ) num__211 <o> e ) num__288 |
speed = num__54 * num__0.277777777778 = num__15 m / sec . length of the train = num__15 * num__20 = num__300 m . let the length of the platform be x m . then ( x + num__300 ) / num__36 = num__15 = > x = num__240 m . answer : b <eor> b <eos> |
b |
multiply__20.0__15.0__ round__240.0__ |
multiply__20.0__15.0__ round__240.0__ |
| currently mangoes cost num__60 cents / pound . due to a disease affecting the mango trees it is expected that next month mangoes will cost num__200.0 more than they do currently . how much are mangoes expected to cost next month ? <o> a ) num__160 cents / pound <o> b ) num__170 cents / pound <o> c ) num__150 cents / pound <o> d ) num__190 cents / pound <o> e ) num__180 cents / pound |
if a new cost is p percent greater than the old cost then ( new cost ) = ( old cost ) + ( p / num__100 ) ( old cost ) . in this case ( new cost ) = num__60 cents / pound + ( num__2.0 ) ( num__60 cents / pound ) = num__60 cents / pound + num__120 cents / pound = num__180 cents / pound answer : e <eor> e <eos> |
e |
divide__200.0__100.0__ multiply__60.0__2.0__ add__60.0__120.0__ add__60.0__120.0__ |
divide__200.0__100.0__ multiply__60.0__2.0__ add__60.0__120.0__ add__60.0__120.0__ |
| by investing a num__6.0 stock at num__96 an income of rs . num__100 is obtained by making and investment of ? <o> a ) rs . num__1600 <o> b ) rs . num__1504 <o> c ) rs . num__1666 <o> d ) rs . num__2000 <o> e ) rs . num__5760 |
for an income of rs . num__6 investment = rs . num__96 . for an income of rs . num__100 investment = rs . ( num__16.0 ) x num__100 = rs . num__1600 answer : a <eor> a <eos> |
a |
percent__100.0__1600.0__ |
percent__100.0__1600.0__ |
| a train num__800 m long is running at a speed of num__78 km / hr . if it crosses a tunnel in num__1 min then the length of the tunnel is ? <o> a ) num__176 m <o> b ) num__786 m <o> c ) num__500 m <o> d ) num__865 m <o> e ) num__796 m |
speed = num__78 * num__0.277777777778 = num__21.6666666667 m / sec . time = num__1 min = num__60 sec . let the length of the train be x meters . then ( num__800 + x ) / num__60 = num__21.6666666667 x = num__500 m . answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ round__500.0__ |
hour_to_min_conversion__ multiply__1.0__500.0__ |
| for the positive numbers n n + num__1 n + num__2 n + num__3 and n + num__4 the mean is how much greater than the median ? <o> a ) num__0 <o> b ) num__1 <o> c ) n + l <o> d ) n + num__2 <o> e ) n + num__3 |
let ’ s first calculate the mean ( arithmetic average ) . mean = sum / quantity mean = ( n + n + num__1 + n + num__2 + n + num__3 + n + num__4 ) / num__5 mean = ( num__5 n + num__10 ) / num__5 mean = n + num__2 next we determine the median . the median is the middle value when the terms are ordered from least to greatest . the terms ordered from least to greatest are as follows : n n + num__1 n + num__2 n + num__3 n + num__4 the median is n + num__2 . finally we are asked how much greater the mean is than the median . to determine the difference we can subtract the smaller value ( the median ) from the larger value ( the mean ) and we get : n + num__2 – ( n + num__2 ) = n + num__2 – n – num__2 = num__0 the answer is a <eor> a <eos> |
a |
add__1.0__4.0__ multiply__2.0__5.0__ multiply__1.0__0.0__ |
add__1.0__4.0__ multiply__2.0__5.0__ divide__0.0__1.0__ |
| difference between the length & breadth of a rectangle is num__10 m . if its perimeter is num__206 m then its area is ? <o> a ) num__2400 m ^ num__2 <o> b ) num__1500 m ^ num__2 <o> c ) num__2520 m ^ num__2 <o> d ) num__1200 m ^ num__2 <o> e ) num__2580 m ^ num__2 |
solving the two equations we get : l = num__30 and b = num__40 . area = ( l x b ) = ( num__30 x num__40 ) m num__2 = num__1200 m ^ num__2 d <eor> d <eos> |
d |
square_perimeter__10.0__ multiply__40.0__30.0__ multiply__40.0__30.0__ |
square_perimeter__10.0__ multiply__40.0__30.0__ multiply__40.0__30.0__ |
| find the product of the local value and absolute value of num__4 in num__564823 <o> a ) num__1600 <o> b ) num__8000 <o> c ) num__16000 <o> d ) num__12000 <o> e ) num__18000 |
explanation : place value = local value face value = absolute value the place value of num__4 in num__564823 is num__4 x num__1000 = num__4000 the face value of num__4 in num__564823 is nothing but num__4 . = > num__4000 x num__4 = num__16000 answer : option c <eor> c <eos> |
c |
multiply__4.0__1000.0__ multiply__4.0__4000.0__ multiply__4.0__4000.0__ |
multiply__4.0__1000.0__ multiply__4.0__4000.0__ multiply__4.0__4000.0__ |
| a is twice as good a work man as b and together they finish the work in num__14 days . in how many days a alone can finish the work ? <o> a ) num__5 days <o> b ) num__21 days <o> c ) num__15 days <o> d ) num__25 days <o> e ) num__30 days |
explanation : wc = num__2 : num__1 num__2 x + x = num__0.0714285714286 = > x = num__0.0238095238095 num__2 x = num__0.047619047619 a can do the work in num__21 days . answer : b <eor> b <eos> |
b |
divide__1.0__14.0__ multiply__2.0__0.0238__ round__21.0__ |
divide__1.0__14.0__ multiply__2.0__0.0238__ round__21.0__ |
| a train num__240 m in length crosses a telegraph post in num__16 seconds . the speed of the train is ? <o> a ) num__40 <o> b ) num__88 <o> c ) num__77 <o> d ) num__6 <o> e ) num__23 |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec total distance covered = num__360 + num__140 = num__500 m required time = num__500 * num__0.08 = num__40 sec . answer : a <eor> a <eos> |
a |
add__360.0__140.0__ divide__500.0__12.5__ round__40.0__ |
add__360.0__140.0__ divide__500.0__12.5__ divide__500.0__12.5__ |
| a five - digit number divisible by num__3 is to be formed using numerical num__0 num__1 num__2 num__3 num__4 and num__5 without repetition . the total number w of ways this can be done is : <o> a ) num__122 <o> b ) num__210 <o> c ) num__216 <o> d ) num__217 <o> e ) num__225 |
we should determine which num__5 digits from given num__6 would form the num__5 digit number divisible by num__3 . we have six digits : num__0 num__1 num__2 num__3 num__4 num__5 . their sum = num__15 . for a number to be divisible by num__3 the sum of the digits must be divisible by num__3 . as the sum of the six given numbers is num__15 ( divisible by num__3 ) only num__5 digits good to form our num__5 digit number would be num__15 - num__0 = { num__1 num__2 num__3 num__4 num__5 } and num__15 - num__3 = { num__0 num__1 num__2 num__4 num__5 } . meaning that no other num__5 from given six will total the number divisible by num__3 . second step : we have two set of numbers : num__1 num__2 num__3 num__4 num__5 and num__0 num__1 num__2 num__4 num__5 . how many num__5 digit numbers can be formed using these two sets : num__1 num__2 num__3 num__4 num__5 - - > num__5 ! as any combination of these digits would give us num__5 digit number divisible by num__3 . num__5 ! = num__120 . num__0 num__1 num__2 num__4 num__5 - - > here we can not use num__0 as the first digit otherwise number wo n ' t be any more num__5 digit and become num__4 digit . so desired # would be total combinations num__5 ! minus combinations with num__0 as the first digit ( combination of num__4 ) num__4 ! - - > num__5 ! - num__4 ! = num__4 ! ( num__5 - num__1 ) = num__4 ! * num__4 = num__96 num__120 + num__96 = num__216 = w answer : c . <eor> c <eos> |
c |
multiply__3.0__2.0__ multiply__3.0__5.0__ add__96.0__120.0__ multiply__1.0__216.0__ |
multiply__3.0__2.0__ multiply__3.0__5.0__ add__96.0__120.0__ multiply__1.0__216.0__ |
| if f ( x ) = x ^ num__4 - num__4 x ^ num__3 - num__2 x ^ num__2 + num__5 x then f ( - num__1 ) = <o> a ) - num__4 <o> b ) - num__2 <o> c ) - num__1 <o> d ) num__1 <o> e ) num__2 |
f ( - num__1 ) = ( - num__1 ) ^ num__4 - num__4 ( - num__1 ) ^ num__3 - num__2 ( - num__1 ) ^ num__2 + num__5 ( - num__1 ) = num__1 + num__4 - num__2 - num__5 = - num__2 the answer is b . <eor> b <eos> |
b |
subtract__4.0__2.0__ |
subtract__4.0__2.0__ |
| find the value of y from ( num__12 ) ^ num__3 x num__6 ^ num__3 ÷ num__432 = y ? <o> a ) num__2134 <o> b ) num__2234 <o> c ) num__2540 <o> d ) num__2560 <o> e ) num__864 |
num__5184 e <eor> e <eos> |
e |
multiply__12.0__432.0__ divide__5184.0__6.0__ |
multiply__12.0__432.0__ divide__5184.0__6.0__ |
| a cube is painted red on all faces . it is then cut into num__27 equal smaller cubes . how many j cubes are painted on only num__2 faces ? <o> a ) num__12 <o> b ) num__8 <o> c ) num__6 <o> d ) num__10 <o> e ) num__16 |
num__1 ) draw a simple cube num__2 ) draw num__9 squares on each face of the cube ( so that it looks like a rubik ' s cube ) - this is what the cube will look like when it ' s cut into num__27 equal smaller cubes . num__3 ) remember that the outside of the cube is the part that ' s painted . . . . the mini - cubes with num__2 painted sides are all on the edge of the cube in themiddleof the edge . there are num__4 in front num__4 in back and num__4 more on thestripthat runs around the left / top / right / bottom of the cube . j = num__4 + num__4 + num__4 = num__12 . answer a <eor> a <eos> |
a |
square_perimeter__1.0__ square_perimeter__3.0__ square_perimeter__3.0__ |
square_perimeter__1.0__ square_perimeter__3.0__ square_perimeter__3.0__ |
| a certain car can travel num__20 minutes on a gallon of gasoline at num__60 miles per hour . if the car had started with a full tank and had num__8 gallons of gasoline left in its tank at the end then what percent of the tank was used to travel num__80 miles at num__60 mph ? <o> a ) num__33.0 <o> b ) num__20.0 <o> c ) num__25.0 <o> d ) num__30.0 <o> e ) num__40 % |
total time for travelling num__80 miles @ num__60 mph = num__1.33333333333 = num__1.33333333333 hour = num__80 minutes . given the car uses num__1 gallon for every num__20 minutes of driving @ num__60 mph . thus in num__80 minutes it will use = num__4 gallons . thus full tank = num__4 + num__8 = num__12 gallons - - - > num__0.333333333333 = num__33.0 of the fuel used . a is the correct answer . <eor> a <eos> |
a |
divide__80.0__60.0__ divide__80.0__20.0__ subtract__20.0__8.0__ divide__20.0__60.0__ round__33.0__ |
divide__80.0__60.0__ divide__80.0__20.0__ subtract__20.0__8.0__ subtract__1.3333__1.0__ round__33.0__ |
| two pipes a and b can fill a tank in num__10 minutes and num__20 minutes respectively . if both the taps are opened simultaneously and the tap a is closed after num__5 minutes then how much more time will it take to fill the tank by tap b ? <o> a ) num__5 min <o> b ) num__6 min <o> c ) num__10 min <o> d ) num__7 min <o> e ) num__12 min |
part filled in num__5 min . = num__5 ( num__0.1 + num__0.05 ) = num__5 * num__0.15 = num__0.75 remaining part = num__1 - num__0.75 = num__0.25 part filled by b in num__1 min . = num__0.05 num__0.05 : num__0.25 : : num__1 : x x = num__0.25 * num__1 * num__20 = num__5 min answer is a <eor> a <eos> |
a |
add__0.05__0.1__ multiply__5.0__0.15__ multiply__10.0__0.1__ divide__5.0__20.0__ round__5.0__ |
add__0.05__0.1__ multiply__5.0__0.15__ multiply__10.0__0.1__ multiply__5.0__0.05__ subtract__10.0__5.0__ |
| num__5358 x num__51 = ? <o> a ) num__273255 <o> b ) num__273257 <o> c ) num__273258 <o> d ) num__273260 <o> e ) num__273262 |
num__5358 x num__51 = num__5358 x ( num__50 + num__1 ) = num__5358 x num__50 + num__5358 x num__1 = num__267900 + num__5358 = num__273258 . c ) <eor> c <eos> |
c |
subtract__51.0__50.0__ multiply__5358.0__50.0__ multiply__5358.0__51.0__ multiply__5358.0__51.0__ |
subtract__51.0__50.0__ multiply__5358.0__50.0__ add__5358.0__267900.0__ add__5358.0__267900.0__ |
| jacob brought a scooter for a certain sum of money . he spent num__10.0 of the cost on repairs and sold the scooter for a profit of rs . num__1100 . how much did he spend on repairs if he made a profit of num__20.0 ? <o> a ) num__300 <o> b ) num__278 <o> c ) num__500 <o> d ) num__277 <o> e ) num__261 |
let the c . p . be rs . x . then num__20.0 of x = num__1100 num__0.2 * x = num__1100 = > x = num__5500 c . p . = rs . num__5500 expenditure on repairs = num__10.0 actual price = rs . ( num__100 * num__5500 ) / num__110 = rs . num__5000 expenditures on repairs = ( num__5500 - num__5000 ) = rs . num__500 . answer : c <eor> c <eos> |
c |
percent__10.0__1100.0__ percent__10.0__5000.0__ percent__10.0__5000.0__ |
percent__10.0__1100.0__ percent__10.0__5000.0__ percent__10.0__5000.0__ |
| the average salary of a person for the months of january february march and april is rs . num__8000 and that for the months february march april and may is rs . num__8600 . if his salary for the month of may is rs . num__6500 find his salary for the month of january ? <o> a ) s . num__4100 <o> b ) s . num__4270 <o> c ) s . num__4500 <o> d ) s . num__4550 <o> e ) s . num__2500 |
sum of the salaries of the person for the months of january february march and april = num__4 * num__8000 = num__32000 - - - - ( num__1 ) sum of the salaries of the person for the months of february march april and may = num__4 * num__8600 = num__34400 - - - - ( num__2 ) ( num__2 ) - ( num__1 ) i . e . may - jan = num__2400 salary of may is rs . num__6500 salary of january = rs . num__4100 answer : a <eor> a <eos> |
a |
multiply__8000.0__4.0__ multiply__8600.0__4.0__ subtract__34400.0__32000.0__ subtract__6500.0__2400.0__ subtract__6500.0__2400.0__ |
multiply__8000.0__4.0__ multiply__8600.0__4.0__ subtract__34400.0__32000.0__ subtract__6500.0__2400.0__ subtract__6500.0__2400.0__ |
| how many integers between num__100 and num__600 are there such that their unit digit is odd ? <o> a ) num__100 <o> b ) num__250 <o> c ) num__200 <o> d ) num__50 <o> e ) num__150 |
num__500 numbers between - num__100 and num__600 out of which half would be even half odd . number of odd unit digit number = num__250 . correct option is b <eor> b <eos> |
b |
subtract__600.0__100.0__ subtract__500.0__250.0__ |
subtract__600.0__100.0__ subtract__500.0__250.0__ |
| the list price of an article is rs . num__65 . a customer pays rs . num__57.33 for it . he was given two successive discounts one of them being num__10.0 . the other discount is ? <o> a ) num__8.0 <o> b ) num__7.0 <o> c ) num__10.0 <o> d ) num__12.0 <o> e ) num__2 % |
option e explanation : num__65 * ( num__0.9 ) * ( ( num__100 - x ) / num__100 ) = num__57.33 x = num__2.0 <eor> e <eos> |
e |
percent__100.0__2.0__ |
percent__100.0__2.0__ |
| when n is divided by num__15 the remainder is num__8 . what is the remainder when num__5 n is divided by num__6 ? <o> a ) num__2 <o> b ) num__6 <o> c ) num__3 <o> d ) num__4 <o> e ) num__1 |
let n = num__8 ( leaves a remainder of num__8 when divided by num__48 ) num__5 n = num__5 ( num__8 ) = num__40 which leaves a remainder of num__4 when divided by num__6 . answer d <eor> d <eos> |
d |
multiply__8.0__6.0__ multiply__8.0__5.0__ subtract__8.0__4.0__ |
multiply__8.0__6.0__ multiply__8.0__5.0__ subtract__8.0__4.0__ |
| in an express train the passengers travelling in a . c . sleeper class first class and sleeper class are in the ratio num__1 : num__2 : num__7 and rate for each class is in the ratio num__5 : num__4 : num__2 . if the total income from this train is num__54000 find the income of indian railways from a . c . sleeper class . <o> a ) num__12000 <o> b ) num__20000 <o> c ) num__22000 <o> d ) num__10000 <o> e ) none of these |
let number of passengers = x num__2 x num__7 x and rate = num__5 y num__4 y num__2 y now since income = rate × number of passengers ∴ income = num__5 xy num__8 xy num__14 xy ∴ income in ratio = num__5 : num__8 : num__14 ∴ income from a . c . sleeper class = num__1.0 + num__8 + num__14 × num__54000 = num__10000 answer d <eor> d <eos> |
d |
add__1.0__7.0__ multiply__2.0__7.0__ multiply__1.0__10000.0__ |
add__1.0__7.0__ multiply__2.0__7.0__ multiply__1.0__10000.0__ |
| a garrison of num__2000 men has provisions for num__54 days . at the end of num__15 days a reinforcement arrives and it is now found that the provisions will last only for num__20 days more . what is the reinforcement ? <o> a ) num__1900 <o> b ) num__2776 <o> c ) num__8888 <o> d ) num__1666 <o> e ) num__1878 |
num__2000 - - - - num__54 num__2000 - - - - num__39 x - - - - - num__20 x * num__20 = num__2000 * num__39 x = num__3900 num__2000 - - - - - - - num__1900 answer : a <eor> a <eos> |
a |
subtract__54.0__15.0__ subtract__3900.0__2000.0__ round__1900.0__ |
subtract__54.0__15.0__ subtract__3900.0__2000.0__ subtract__3900.0__2000.0__ |
| num__1000 men have provisions for num__20 days . if num__650 more men join them for how many days will the provisions last now ? <o> a ) num__12.9 <o> b ) num__12.5 <o> c ) num__12.6 <o> d ) num__12.2 <o> e ) num__12.1 |
num__1000 * num__20 = num__1650 * x x = num__12.1 answer : e <eor> e <eos> |
e |
add__1000.0__650.0__ round__12.1__ |
add__1000.0__650.0__ round__12.1__ |
| danny spends $ num__360 buying his favorite dolls . if he buys only small lemonhead dolls which are $ num__1 cheaper than the large lemonhead dolls he could buy num__5 more dolls than if he were to buy only large lemonhead dolls . how much does a large lemonhead doll cost ? <o> a ) $ num__5 <o> b ) $ num__6 <o> c ) $ num__7.2 <o> d ) $ num__8 <o> e ) $ num__9 |
x : price of small dolls ; y : price of big dolls . y = x + num__1 num__360 ( num__1 / x - num__1 / y ) = num__5 < = > x . y = num__72 we can make trial : num__9 x num__8 = num__72 . choose ( e ) y = num__9 . <eor> e <eos> |
e |
divide__360.0__5.0__ subtract__9.0__1.0__ multiply__1.0__9.0__ |
divide__360.0__5.0__ subtract__9.0__1.0__ add__1.0__8.0__ |
| for what range of values of ' x ' will the inequality num__15 x - num__2 / x > num__1 ? <o> a ) x > num__0.4 <o> b ) x < num__0.333333333333 <o> c ) - num__0.333333333333 < x <o> d ) - num__0.333333333333 < x and x < num__0 x > num__0.4 <o> e ) x < - num__0.333333333333 and x > num__0.4 |
in case num__1 where x > num__0 i agree . i calculated x > ( - num__0.333333333333 ) and x > num__0.4 and because x > num__0 it must be x > num__0.4 . in case num__2 where x < num__0 i do n ' t understand why you say that the range must be x > - num__0.333333333333 and x < num__0 since the values that satisfy the inequality num__15 x ^ num__2 - x - num__2 < num__0 show that x < - num__0.333333333333 and x < num__0.4 . i have the same doubt . should n ' t the answer be d ? ____________ the answer is d . <eor> d <eos> |
d |
multiply__1.0__0.3333__ |
multiply__1.0__0.3333__ |
| the radius of a wheel is num__22.4 cm . what is the distance covered by the wheel in making num__320 resolutions ? <o> a ) num__724 m <o> b ) num__704 m <o> c ) num__287 m <o> d ) num__450.6 m <o> e ) num__927 m |
in one resolution the distance covered by the wheel is its own circumference . distance covered in num__320 resolutions . = num__320 * num__2 * num__3.14285714286 * num__22.4 = num__45056 cm = num__450.6 m answer : d <eor> d <eos> |
d |
round__450.6__ |
round__450.6__ |
| the distance between delhi and mathura is num__110 kms . a starts from delhi with a speed of num__20 kmph at num__7 a . m . for mathura and b starts from mathura with a speed of num__25 kmph at num__8 p . m . from delhi . when will they meet ? <o> a ) num__12 a . m <o> b ) num__10 a . m <o> c ) num__19 a . m <o> d ) num__76 a . m <o> e ) num__37 a . m |
d = num__110 – num__20 = num__90 rs = num__20 + num__25 = num__45 t = num__2.0 = num__2 hours num__8 a . m . + num__2 = num__10 a . m . answer : b <eor> b <eos> |
b |
subtract__110.0__20.0__ add__20.0__25.0__ divide__90.0__45.0__ divide__20.0__2.0__ round__10.0__ |
subtract__110.0__20.0__ add__20.0__25.0__ divide__90.0__45.0__ add__8.0__2.0__ add__8.0__2.0__ |
| what is the probability of flipping a fair coin five times and the coin landing on heads on at least two flips ? <o> a ) num__0.625 <o> b ) num__0.875 <o> c ) num__0.6875 <o> d ) num__0.8125 <o> e ) num__0.84375 |
the number of possible outcomes is num__2 ^ num__5 = num__32 . num__0 heads : there is num__1 way to have all tails . num__1 head : there are num__5 ways to have one head . p ( num__0 or num__1 head ) = num__0.1875 = num__0.1875 p ( at least num__2 heads ) = num__1 - num__0.1875 = num__0.8125 the answer is d . <eor> d <eos> |
d |
coin_space__ vowel_space__ negate_prob__0.0__ negate_prob__0.1875__ negate_prob__0.1875__ |
coin_space__ vowel_space__ negate_prob__0.0__ negate_prob__0.1875__ negate_prob__0.1875__ |
| a runner runs the num__40 miles from marathon to athens at a constant speed . halfway through the run she injures her foot and continues to run at half her previous speed . if the second half takes her num__8 hours longer than the first half how many hours did it take the runner to run the second half ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__16 <o> d ) num__20 <o> e ) num__24 |
the runner runs the first num__20 miles at speed v and the second num__20 miles at speed v / num__2 . the time t num__2 to run the second half must be twice the time t num__1 to run the first half . t num__2 = num__2 * t num__1 = t num__1 + num__8 t num__1 = num__8 and so t num__2 = num__16 . the answer is c . <eor> c <eos> |
c |
divide__40.0__20.0__ multiply__8.0__2.0__ round__16.0__ |
divide__40.0__20.0__ multiply__8.0__2.0__ multiply__8.0__2.0__ |
| a and b can together finish a work in num__40 days . they worked together for num__10 days and then b left . after another num__12 days a finished the remaining work . in how many days a alone can finish the job ? <o> a ) num__10 <o> b ) num__25 <o> c ) num__60 <o> d ) num__16 <o> e ) num__20 |
a + b num__10 days work = num__10 * num__0.025 = num__0.25 remaining work = num__1 - num__0.25 = num__0.75 num__0.75 work is done by a in num__12 days whole work will be done by a in num__12 * num__1.33333333333 = num__16 days answer is d <eor> d <eos> |
d |
divide__10.0__40.0__ multiply__40.0__0.025__ subtract__1.0__0.25__ divide__1.0__0.75__ divide__12.0__0.75__ round__16.0__ |
multiply__10.0__0.025__ multiply__40.0__0.025__ subtract__1.0__0.25__ divide__1.0__0.75__ divide__12.0__0.75__ round__16.0__ |
| a man can do a piece of work in num__15 days but with the help of his son he can do it in num__6 days . in what time can the son do it alone ? <o> a ) num__13 <o> b ) num__9 <o> c ) num__15 <o> d ) num__8 <o> e ) num__10 |
son ' s num__1 day ' s work = ( num__0.166666666667 ) - ( num__0.0666666666667 ) = num__0.1 = num__0.1 the son alone can do the work in num__10 days answer is e <eor> e <eos> |
e |
divide__1.0__6.0__ divide__1.0__15.0__ subtract__0.1667__0.0667__ divide__1.0__0.1__ round__10.0__ |
divide__1.0__6.0__ divide__1.0__15.0__ subtract__0.1667__0.0667__ divide__1.0__0.1__ round__10.0__ |
| a jar is filled with red white and blue tokens that are equivalent except for their color . the chance of randomly selecting a red token replacing it then randomly selecting a white token is the same as the chance of randomly selecting a blue token . if the number of tokens of every color is a multiple of num__8 what is the smallest possible total number of tokens in the jar ? <o> a ) num__9 <o> b ) num__12 <o> c ) num__15 <o> d ) num__18 <o> e ) num__21 |
( red / total coins ) * ( white / total coins ) = ( blue / total coins ) í . ẹ . red * white = blue * total coins let red = num__3 a white = num__3 b blue = num__3 c total coins = num__3 ( a + b + c ) í . ẹ . num__3 a * num__3 b = num__3 c * num__3 ( a + b + c ) í . ẹ . a * b = c * ( a + b + c ) for smallest values of a b and c num__2 * num__3 = num__1 * ( num__1 + num__2 + num__3 ) í . ẹ . minimum total coins = num__3 * ( num__1 + num__2 + num__3 ) = num__15 answer : option c <eor> c <eos> |
c |
subtract__3.0__2.0__ multiply__1.0__15.0__ |
subtract__3.0__2.0__ multiply__1.0__15.0__ |
| a conical vessel whose internal radius is num__12 cm and height num__50 cm is full of liquid . the contents are emptied into a cylindrical vessel with internal radius num__10 cm . find the height to which the liquid rises in the cylindrical vessel . <o> a ) num__24 cm <o> b ) num__14 cm <o> c ) num__20 cm <o> d ) num__37 cm <o> e ) num__17 cm |
volume of the liquid in the cylindrical vessel = volume of the conical vessel = ( ( num__0.333333333333 ) * ( num__3.14285714286 ) * num__12 * num__12 * num__50 ) ) cm num__3 = ( num__22 * num__4 * num__12 * num__50 ) / num__7 cm num__3 . let the height of the liquid in the vessel be h . then ( num__3.14285714286 ) * num__10 * num__10 * h = ( num__22 * num__4 * num__12 * num__50 ) / num__7 or h = ( num__4 * num__12 * num__50 ) / num__100 = num__24 cm answer a num__24 cm <eor> a <eos> |
a |
triangle_area__50.0__4.0__ triangle_area__12.0__4.0__ triangle_area__12.0__4.0__ |
triangle_area__50.0__4.0__ triangle_area__12.0__4.0__ triangle_area__12.0__4.0__ |
| when processing flower - nectar into honey bees ' extract a considerable amount of water gets reduced . how much flower - nectar must be processed to yield num__1 kg of honey if nectar contains num__50.0 water and the honey obtained from this nectar contains num__40.0 water ? <o> a ) num__1.2 kg <o> b ) num__1.5 kg <o> c ) num__1.7 kg <o> d ) num__1.9 kg <o> e ) none of these |
explanation : flower - nectar contains num__50.0 of non - water part . in honey this non - water part constitutes num__60.0 ( num__100 - num__40 ) . therefore num__0.5 x amount of flower - nectar = num__0.60 x amount of honey = num__0.60 x num__1 kg therefore amount of flower - nectar needed = ( num__0.60 / num__0.51 ) kg = num__1.2 kgs answer : a <eor> a <eos> |
a |
add__40.0__60.0__ divide__50.0__100.0__ divide__60.0__100.0__ divide__0.6__0.5__ multiply__1.0__1.2__ |
add__40.0__60.0__ divide__50.0__100.0__ divide__60.0__100.0__ divide__0.6__0.5__ divide__1.2__1.0__ |
| a hollow iron pipe is num__21 cm long and its external diameter is num__8 cm . if the thickness of the pipe is num__1 cm and iron weighs num__8 g / cm num__3 then the weight of the pipe is : <o> a ) num__3.6 kg <o> b ) num__3.696 kg <o> c ) num__36 kg <o> d ) num__36.9 kg <o> e ) none |
solution external radius = num__4 cm internal radius = num__3 cm . volume of iron = ( num__3.14285714286 × ( num__42 ) - ( num__32 ) × num__21 ) cm num__3 = ( num__3.14285714286 × num__7 × num__1 × num__21 ) cm num__3 = num__462 cm num__3 . weight of iron = ( num__462 x num__8 ) gm = num__3696 gm = num__3.696 kg . answer b <eor> b <eos> |
b |
add__1.0__3.0__ multiply__8.0__4.0__ divide__21.0__3.0__ multiply__8.0__462.0__ multiply__1.0__3.696__ |
add__1.0__3.0__ multiply__8.0__4.0__ subtract__8.0__1.0__ multiply__8.0__462.0__ multiply__1.0__3.696__ |
| a bread merchant declares a num__20.0 discount after num__8 p . m . each bread costs rs . num__40 when it is not on sale . sam goes before num__8 p . m . to buy num__4 pieces of bread . max goes after num__8 p . m . with the same amount of money that sam had to buy bread . how many pieces did max get over sam ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
explanation : sam : num__4 pieces * num__40 = num__160 max : max had num__160 . num__160 / ( num__40 * ( num__1 - num__20.0 ) ) = num__160 / ( num__40 * num__0.80 ) = num__5.0 = num__5 pieces max - sam = num__5 - num__4 = num__1 correct option : a <eor> a <eos> |
a |
percent__20.0__4.0__ percent__20.0__5.0__ |
percent__20.0__4.0__ percent__20.0__5.0__ |
| a courtyard is num__18 meter long and num__12 meter board is to be paved with bricks of dimensions num__12 cm by num__6 cm . the total number of bricks required is : <o> a ) num__16000 <o> b ) num__18000 <o> c ) num__20000 <o> d ) num__30000 <o> e ) none of these |
explanation : number of bricks = courtyard area / num__1 brick area = ( num__1800 Ã — num__100.0 Ã — num__6 ) = num__30000 option d <eor> d <eos> |
d |
divide__1800.0__18.0__ round__30000.0__ |
divide__1800.0__18.0__ divide__30000.0__1.0__ |
| find k if num__64 / k = num__4 . <o> a ) num__16 <o> b ) num__17 <o> c ) num__18 <o> d ) num__14 <o> e ) num__13 |
since num__64 / k = num__4 and num__4.0 = num__4 then k = num__16 correct answer a <eor> a <eos> |
a |
divide__64.0__4.0__ divide__64.0__4.0__ |
divide__64.0__4.0__ divide__64.0__4.0__ |
| a train num__120 meter long passes an electric pole in num__12 seconds and another train of same length traveling in opposite direction in num__8 seconds . the speed of the second train is <o> a ) num__10 m / s <o> b ) num__20 m / s <o> c ) num__30 m / s <o> d ) num__40 m / s <o> e ) num__50 m / s |
speed of num__1 st train = num__10.0 = num__10 m / s let the speed of num__2 nd is v then num__240 / ( v + num__10 ) = num__8 num__8 v = num__160 v = num__20 m / s answer : b <eor> b <eos> |
b |
divide__120.0__12.0__ subtract__12.0__10.0__ multiply__120.0__2.0__ add__12.0__8.0__ round__20.0__ |
divide__120.0__12.0__ subtract__12.0__10.0__ multiply__120.0__2.0__ add__12.0__8.0__ add__12.0__8.0__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__378 <o> b ) num__240 <o> c ) num__772 <o> d ) num__281 <o> e ) num__213 |
speed = ( num__54 * num__0.277777777778 ) m / sec = num__15 m / sec . length of the train = ( num__15 x num__20 ) m = num__300 m . let the length of the platform be x meters . then ( x + num__300 ) / num__36 = num__15 = = > x + num__300 = num__540 = = > x = num__240 m . answer : b <eor> b <eos> |
b |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
| how many words with or without meaning can be formed by using all the letters of the word ‘ orange ’ using each letter exactly once ? <o> a ) num__700 <o> b ) num__720 <o> c ) num__750 <o> d ) num__800 <o> e ) num__820 |
the word ‘ orange ’ contains num__6 different letters . therefore required number of words = number of arrangement of num__6 letters taken all at a time = num__6 p num__6 = num__6 ! = num__6 * num__5 * num__4 * num__3 * num__2 * num__1 = num__720 answer : b <eor> b <eos> |
b |
subtract__5.0__3.0__ subtract__3.0__2.0__ multiply__1.0__720.0__ |
subtract__5.0__3.0__ subtract__3.0__2.0__ multiply__1.0__720.0__ |
| square rstu shown above is rotated in a plane about its center in a clockwise direction the minimum number of degrees necessary for r to be in the position where t is now shown . the number of degrees through which rstu is rotated is <o> a ) num__135 degree <o> b ) num__180 degree <o> c ) num__225 degree <o> d ) num__270 degree <o> e ) num__315 degree |
from the options i am assuming the positioning of t and r relative to each other to be as shown . to replace t by r focus on or . say you rotate or clockwise ( and with it the entire square ) and bring it in place of ot . how many degrees did you go ? you covered num__2 right angles i . e . num__180 degrees . answer : b <eor> b <eos> |
b |
straight_angle__ straight_angle__ |
straight_angle__ straight_angle__ |
| two taps can separately fill a cistern num__10 minutes and num__15 minutes respectively and when the waste pipe is open they can together fill it in num__18 minutes . the waste pipe can empty the full cistern in ? <o> a ) num__6 <o> b ) num__5 <o> c ) num__7 <o> d ) num__9 <o> e ) num__2 |
num__0.1 + num__0.0666666666667 - num__1 / x = num__0.0555555555556 x = num__9 answer : d <eor> d <eos> |
d |
multiply__10.0__0.1__ divide__1.0__18.0__ subtract__10.0__1.0__ round__9.0__ |
multiply__10.0__0.1__ divide__1.0__18.0__ subtract__10.0__1.0__ subtract__10.0__1.0__ |
| cereal a is num__10.0 sugar by weight whereas healthier but less delicious cereal b is num__2.0 sugar by weight . to make a delicious and healthy mixture that is num__6.0 sugar what should be the ratio of cereal a to cereal b by weight ? <o> a ) num__2 : num__9 <o> b ) num__2 : num__7 <o> c ) num__1 : num__6 <o> d ) num__1 : num__4 <o> e ) num__1 : num__1 |
ratio of a / ratio of b = ( average wt of mixture - wt of b ) / ( wt of a - average wt of mixture ) = > ratio of a / ratio of b = ( num__6 - num__2 ) / ( num__10 - num__6 ) = num__1.0 so they should be mixed in the ratio num__1 : num__1 answer - e <eor> e <eos> |
e |
reverse__1.0__ |
reverse__1.0__ |
| every day a cyclist meets a car at the station . the road is straight and both are travelling in the same direction . the cyclist travels with a speed of num__12 mph . one day the cyclist comes late by num__20 min . and meets the car num__5 miles before the station . what is the speed of the car ? <o> a ) num__40 kmph . <o> b ) num__50 kmph . <o> c ) num__60 kmph . <o> d ) num__70 kmph . <o> e ) num__80 kmph . |
if they normally meet at num__8 am then cyclist will reach at station on that day at num__8 : num__20 am . but train will reach station at num__8 am only . and he will be at at point num__5 miles before station at num__7.55 am ( i . e num__25 min before num__8 : num__20 am ) so train will take num__5 mins to travel ( to reach stn at num__8 am ) num__5 miles . so speed of train = num__60 kmph . answer : c <eor> c <eos> |
c |
subtract__20.0__12.0__ add__20.0__5.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
subtract__20.0__12.0__ add__20.0__5.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
| a merchant marks his wares num__40.0 more than the real price and allows num__20.0 discount . his profit is : <o> a ) num__20.0 <o> b ) num__18.0 <o> c ) num__16.0 <o> d ) num__12.0 <o> e ) none of these |
let the cp = num__100 rs . mark price = num__140 discount = num__20.0 selling price num__0.8 × num__140 hence profit = num__12.0 answer : d . <eor> d <eos> |
d |
percent__100.0__12.0__ |
percent__100.0__12.0__ |
| ( num__161 ) ^ num__2 - ( num__162 ) ^ num__2 = <o> a ) num__1 <o> b ) num__100 <o> c ) num__229 <o> d ) num__325 <o> e ) num__349 |
using the formula : ( a + num__1 ) ^ num__2 - a ^ num__2 = num__2 a + num__1 so answer = num__162 * num__2 + num__1 = num__324 + num__1 = num__325 = answer = d <eor> d <eos> |
d |
subtract__162.0__161.0__ multiply__2.0__162.0__ add__1.0__324.0__ add__1.0__324.0__ |
subtract__162.0__161.0__ multiply__2.0__162.0__ add__1.0__324.0__ add__1.0__324.0__ |
| a trader sold two items for rs . num__2500 each if he made a profit of num__28.0 from the first item and a loss of num__20.0 from the second item calculate his total gain or loss . <o> a ) num__2.6 <o> b ) num__12.6 <o> c ) num__3.6 <o> d ) num__7.6 <o> e ) num__5.6 |
( num__28 * num__20 ) / num__100 = num__5.6 answer : e <eor> e <eos> |
e |
percent__28.0__20.0__ percent__28.0__20.0__ |
percent__28.0__20.0__ percent__28.0__20.0__ |
| while working alone at their constant rates computer x can process num__240 files in num__4 hours and computer y can process num__240 files in num__6 hours . if all files processed by these computers are the same size how many hours would it take the two computers working at the same time at their respective constant rates to process a total of num__240 files ? <o> a ) num__2 <o> b ) num__2.2 <o> c ) num__2.4 <o> d ) num__2.6 <o> e ) num__2.8 |
both computers together process files at a rate of num__60.0 + num__40.0 = num__60 + num__40 = num__100 files per hour . the time required to process num__240 files is num__2.4 = num__2.4 hours the answer is c . <eor> c <eos> |
c |
hour_to_min_conversion__ divide__240.0__6.0__ add__40.0__60.0__ divide__240.0__100.0__ round__2.4__ |
hour_to_min_conversion__ divide__240.0__6.0__ add__40.0__60.0__ divide__240.0__100.0__ round__2.4__ |
| a sum of rs . num__1530 has been divided among a b and c such that a gets of what b gets and b gets of what c gets . b ’ s share is : <o> a ) rs . num__120 <o> b ) rs . num__160 <o> c ) rs . num__270 <o> d ) rs . num__300 <o> e ) none |
explanation let c ’ s share = rs . x then b ’ s share = rs . x / num__4 a ’ s share = rs . ( num__0.666666666667 x x / num__4 ) = rs . x / num__6 = x / num__6 + x / num__4 + x = num__1530 = > num__17 x / num__12 = num__1530 = > num__1530 x num__0.705882352941 = rs . num__1080 hence b ’ s share = rs . ( num__270.0 ) = rs . num__270 . answer c <eor> c <eos> |
c |
divide__12.0__17.0__ divide__1080.0__4.0__ divide__1080.0__4.0__ |
divide__12.0__17.0__ divide__1080.0__4.0__ divide__1080.0__4.0__ |
| the average amount with a group of seven numbers is rs . num__30 . if the newly joined member has rs . num__55 with him what was the average amount with the group before his joining the group ? <o> a ) s . num__25.6 <o> b ) s . num__25.8333333333 <o> c ) s . num__16.6 <o> d ) s . num__43.3333333333 <o> e ) s . num__25 |
total members in the group = num__7 average amount = rs . num__30 total amount with them = num__7 * num__30 = rs . num__210 one number has rs . num__55 . so the amount with remaining num__6 people = num__210 - num__55 = rs . num__155 the average amount with them = num__25.8333333333 answer : b <eor> b <eos> |
b |
multiply__30.0__7.0__ subtract__210.0__55.0__ divide__155.0__6.0__ divide__155.0__6.0__ |
multiply__30.0__7.0__ subtract__210.0__55.0__ divide__155.0__6.0__ divide__155.0__6.0__ |
| three of the sides of a rectangular prism have areas of num__91 num__39 and num__21 . what is the volume of the rectangular prism ? <o> a ) num__252 <o> b ) num__269 <o> c ) num__273 <o> d ) num__920 <o> e ) num__1911 |
one easy way to get the answer for this question is just multiply the unit ' s digits of these three numbers i . e ( num__1 * num__9 * num__1 = num__9 ) ( why the multiplication is needed ? you can see the solution suggested by bunuel in the above post ) now look out for the answer whose square gives us a num__9 in the unit place . so it is easy to pic ( c ) num__273 whose square will give us a num__9 in the unit place ( num__3 * num__3 ) and hence this is the answer . <eor> c <eos> |
c |
multiply__91.0__3.0__ |
multiply__91.0__3.0__ |
| what sum of money will produce rs . num__60 as simple interest in num__4 years at num__3 num__0.5 percent ? <o> a ) num__337 <o> b ) num__429 <o> c ) num__266 <o> d ) num__288 <o> e ) num__211 |
num__60 = ( p * num__4 * num__3.5 ) / num__100 p = num__429 answer : b <eor> b <eos> |
b |
percent__100.0__429.0__ |
percent__100.0__429.0__ |
| the sum of ages of num__5 children born at the intervals of num__3 years each is num__70 years . what is the age of the youngest child ? <o> a ) num__3 years <o> b ) num__4 years <o> c ) num__6 years <o> d ) num__7 years <o> e ) num__8 years |
let the ages of children be x ( x + num__3 ) ( x + num__6 ) ( x + num__9 ) and ( x + num__12 ) years . then x + ( x + num__3 ) + ( x + num__6 ) + ( x + num__9 ) + ( x + num__12 ) = num__70 num__5 x = num__40 x = num__8 . age of the youngest child = x = num__8 years . e ) <eor> e <eos> |
e |
add__3.0__6.0__ add__3.0__9.0__ add__5.0__3.0__ add__5.0__3.0__ |
add__3.0__6.0__ add__3.0__9.0__ add__5.0__3.0__ add__5.0__3.0__ |
| a and b go around a circular track of length num__600 m on a cycle at speeds of num__36 kmph and num__54 kmph . after how much time will they meet for the first time at the starting point ? <o> a ) num__120 sec <o> b ) num__165 sec <o> c ) num__186 sec <o> d ) num__167 sec <o> e ) num__168 sec |
time taken to meet for the first time at the starting point = lcm { length of the track / speed of a length of the track / speed of b } = lcm { num__600 / ( num__36 * num__0.277777777778 ) num__600 / ( num__54 * num__0.277777777778 ) } = lcm ( num__60 num__40 ) = num__120 sec . answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ round__120.0__ |
hour_to_min_conversion__ round__120.0__ |
| how long does a train num__110 m long traveling at num__60 kmph takes to cross a bridge of num__170 m in length ? <o> a ) num__16.5 <o> b ) num__16.0 <o> c ) num__16.4 <o> d ) num__16.8 <o> e ) num__16.1 |
d = num__110 + num__170 = num__280 m s = num__60 * num__0.277777777778 = num__16.6666666667 t = num__280 * num__0.06 = num__16.8 sec answer : d <eor> d <eos> |
d |
add__110.0__170.0__ multiply__0.06__280.0__ round__16.8__ |
add__110.0__170.0__ multiply__0.06__280.0__ multiply__0.06__280.0__ |
| what are the last two digits of num__55 * num__34 * num__86 * num__47 * num__23 * num__51 ? <o> a ) num__18 <o> b ) num__20 <o> c ) num__23 <o> d ) num__25 <o> e ) num__57 |
num__55 * num__34 * num__86 * num__47 * num__23 * num__51 = we have to focus on the last two digits only so num__55 * num__34 = num__70 * num__86 = num__20 * num__47 = num__40 num__40 * num__23 = num__20 therefore num__20 * num__51 = num__20 hence answer is b <eor> b <eos> |
b |
add__47.0__23.0__ subtract__40.0__20.0__ |
add__47.0__23.0__ subtract__40.0__20.0__ |
| a cube of side num__5 cm is painted on all its side . if it is sliced into num__1 cubic centimer cubes how many num__1 cubic centimeter cubes will have exactly one of their sides painted ? <o> a ) num__9 <o> b ) num__61 <o> c ) num__98 <o> d ) num__54 <o> e ) num__64 |
look at the image below : little cubes with exactly one painted side will be those num__3 * num__3 = num__9 which are in the center of each face . ( num__6 faces ) * ( num__9 per each ) = num__54 . answer : d . <eor> d <eos> |
d |
add__5.0__1.0__ multiply__9.0__6.0__ round__54.0__ |
add__5.0__1.0__ multiply__9.0__6.0__ multiply__1.0__54.0__ |
| a mathematics teacher tabulated the marks secured by num__35 students of num__8 th class . the average of their marks was num__72 . if the marks secured by reema was written as num__36 instead of num__86 then find the correct average marks up to two decimal places . <o> a ) num__73.41 <o> b ) num__74.31 <o> c ) num__72.43 <o> d ) num__73.43 <o> e ) can not be determined |
correct average = num__35 x num__72 + ( num__86 - num__36 ) / num__35 ≈ num__72 + num__1.43 = num__73.43 answer d <eor> d <eos> |
d |
add__72.0__1.43__ add__72.0__1.43__ |
add__72.0__1.43__ add__72.0__1.43__ |
| a grocer has a sale of rs . num__6435 rs . num__6927 rs . num__6855 rs . num__7230 and rs . num__6562 for num__5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . num__6500 ? <o> a ) rs . num__4991 <o> b ) rs . num__5991 <o> c ) rs . num__6991 <o> d ) rs . num__6001 <o> e ) none of these |
explanation : total sale for num__5 months = rs . ( num__6435 + num__6927 + num__6855 + num__7230 + num__6562 ) = rs . num__34009 . required sale = rs . [ ( num__6500 x num__6 ) - num__34009 ] = rs . ( num__39000 - num__34009 ) = rs . num__4991 . answer : a <eor> a <eos> |
a |
multiply__6500.0__6.0__ subtract__39000.0__34009.0__ subtract__39000.0__34009.0__ |
multiply__6500.0__6.0__ subtract__39000.0__34009.0__ subtract__39000.0__34009.0__ |
| if a num__7 cm cube is cut into num__1 cm cubes then what is the percentage increase in the surface area of the resulting cubes ? <o> a ) num__140.0 <o> b ) num__350.0 <o> c ) num__450.0 <o> d ) num__600.0 <o> e ) num__700 % |
the area a of the large cube is num__7 * num__7 * num__6 = num__294 square cm . the area of the num__343 small cubes is num__7 * num__7 * num__7 * num__6 = num__7 a an increase of num__600.0 . the answer is d . <eor> d <eos> |
d |
surface_cube__1.0__ surface_cube__7.0__ volume_cube__7.0__ rectangle_perimeter__294.0__6.0__ multiply__1.0__600.0__ |
surface_cube__1.0__ surface_cube__7.0__ volume_cube__7.0__ rectangle_perimeter__294.0__6.0__ multiply__1.0__600.0__ |
| out of n consecutive natural numbers four numbers are selected such that the differences of all the possible pairs are distinct . what is the least value of n for which this is possible ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
explanation : let the num__4 numbers be x num__1 x num__2 x num__3 x num__4 where x num__1 < x num__2 < x num__3 < x num__4 . let x num__1 + a num__1 = x num__2 x num__2 + a num__2 = x num__3 x num__3 + a num__3 = x num__4 . the difference of x num__1 and each of the other three of x num__2 and each of the greater two of x num__3 and x num__4 are tabulated . x num__1 x num__2 x num__3 x num__2 a num__1 x num__3 a num__1 + a num__2 a num__2 x num__4 a num__1 + a num__2 + a num__3 a num__2 + a num__3 a num__3 we require that n be minimized and these num__6 expressions have distinct values . let x num__1 = num__1 now n = num__1 + a num__1 + a num__2 + a num__3 . now a num__1 a num__2 a num__3 can be chosen as the least possible values i . e . num__12 and num__3 . however the order of chosing should be such that the six expressions ( differences ) should all be distinct . by little trial and error ( a num__1 a num__2 a num__3 ) can be ( num__1 num__3 num__2 ) or ( num__2 num__3 num__1 ) in which case n = num__7 and the four numbers can be ( num__1 num__2 num__5 num__7 ) or ( num__1 num__3 num__6 num__7 ) respectively . hence the minimum value of n is num__7 . answer : b <eor> b <eos> |
b |
coin_space__ die_space__ choose__3.0__1.0__ vowel_space__ choose__7.0__6.0__ |
coin_space__ die_space__ choose__3.0__1.0__ vowel_space__ choose__7.0__6.0__ |
| a pipe takes a hours to fill the tank . but because of a leakage it took num__5 times of its original time . find the time taken by the leakage to empty the tank <o> a ) num__50 min <o> b ) num__60 min <o> c ) num__90 min <o> d ) num__80 min <o> e ) num__75 min |
pipe a can do a work num__60 min . lets leakage time is x ; then num__0.0166666666667 - num__1 / x = num__0.00333333333333 x = num__75 min answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ divide__0.0167__5.0__ round__75.0__ |
hour_to_min_conversion__ divide__0.0167__5.0__ divide__75.0__1.0__ |
| a person is traveling at num__65 km / hr and reached his destiny in num__3 hr then find the distance ? <o> a ) num__205 km <o> b ) num__195 km <o> c ) num__295 km <o> d ) num__115 km <o> e ) num__215 km |
t = num__3 hrs d = t * s = num__65 * num__3 = num__195 km answer is b <eor> b <eos> |
b |
multiply__65.0__3.0__ round__195.0__ |
multiply__65.0__3.0__ multiply__65.0__3.0__ |
| the number of boys in the class is num__8 more than the number of girls in the class which is five times the difference between the number of girls and boys in the class . how many boys are there in the class ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__40 <o> d ) num__50 <o> e ) num__60 |
num__8 + y = num__5 ( x - y ) ; y = no of girlz & x = boys so y = num__32 & x = num__8 + y ie num__40 num__40 ans . answer : c <eor> c <eos> |
c |
add__8.0__32.0__ add__8.0__32.0__ |
add__8.0__32.0__ add__8.0__32.0__ |
| solve num__8 x – num__4 = num__4 x + num__11 <o> a ) num__5 ( num__0.333333333333 ) <o> b ) num__6 ( num__0.333333333333 ) <o> c ) num__7 ( num__0.333333333333 ) <o> d ) num__3 ( num__0.75 ) <o> e ) num__9 ( num__0.333333333333 ) |
sol . num__8 x – num__4 = num__4 x + num__11 = > num__8 x – num__4 x = num__11 + num__4 = > num__4 x = num__15 = > x = num__5.0 = num__3 ( num__0.75 ) . answer d <eor> d <eos> |
d |
add__4.0__11.0__ subtract__8.0__5.0__ divide__3.0__4.0__ subtract__8.0__5.0__ |
add__4.0__11.0__ subtract__8.0__5.0__ divide__3.0__4.0__ subtract__8.0__5.0__ |
| num__3 num__47 num__1118 num__2947 ? <o> a ) num__76 <o> b ) num__77 <o> c ) num__75 <o> d ) num__78 <o> e ) num__79 |
each number is the sum of the previous two numbers ie num__29 + num__47 = num__76 . answer : a <eor> a <eos> |
a |
add__47.0__29.0__ add__47.0__29.0__ |
add__47.0__29.0__ add__47.0__29.0__ |
| the greatest number of four digits which is divisible by num__10 num__20 num__25 and num__60 is : <o> a ) num__9000 <o> b ) num__9300 <o> c ) num__9400 <o> d ) num__9600 <o> e ) num__9700 |
greatest number of num__4 - digits is num__9999 . l . c . m . of num__10 num__20 num__25 and num__60 is num__1500 . on dividing num__9999 by num__1500 the remainder is num__999 . required number ( num__9999 - num__999 ) = num__9000 . answer : a <eor> a <eos> |
a |
multiply__25.0__60.0__ subtract__9999.0__999.0__ subtract__9999.0__999.0__ |
multiply__25.0__60.0__ subtract__9999.0__999.0__ subtract__9999.0__999.0__ |
| a company chauncy co . has an annual travel budget of $ num__59000 . the accounting department estimates that transportation expenses will increase num__5 percent in the coming year and nontransportation travel expenses will increase by num__15 percent in the coming year . last year chauncy co . spent $ num__19500 on transportation - related expenses and $ num__35000 on nontransportation travel expenses . if the annual travel budget does not increase this year and if the accounting department ’ s estimates are correct how far over the annual travel budget will expenses be this year ? <o> a ) expenses will not go over the budget . <o> b ) $ num__500 <o> c ) $ num__4225 <o> d ) $ num__1725 <o> e ) $ num__60 |
725 |
annual travel budget of $ num__59000 let transportation expenses = t = num__19500 and non - transportation expenses = n = num__35000 i . e . increased transportation expenses = num__1.05 t = num__20475 and increased non - transportation expenses = num__1.15 n = num__40250 total expense = num__20475 + num__40250 = num__60725 expense over budget = budget - expense = num__59000 - num__60725 = num__1725 answer : option d <eor> d <eos> |
d |
d |
| if each term in the sum a num__1 + a num__2 + a num__3 + . . . . . + an is either num__7 or num__77 and the sum equals num__343 which of the following could be equal to n ? <o> a ) num__38 <o> b ) num__39 <o> c ) num__40 <o> d ) num__41 <o> e ) num__42 |
since there is no num__49 in the answer choices ( num__49.0 = num__49 ) we know there is at least one num__77 . num__343 - num__77 = num__266 num__38.0 = num__38 num__38 + num__1 = num__39 . if num__39 was n ' t there i would have subtracted num__77 from num__266 and continued in a similar way . ans . b <eor> b <eos> |
b |
divide__343.0__7.0__ subtract__343.0__77.0__ divide__266.0__7.0__ add__1.0__38.0__ add__1.0__38.0__ |
divide__343.0__7.0__ subtract__343.0__77.0__ divide__266.0__7.0__ add__1.0__38.0__ add__1.0__38.0__ |
| winona can purchase a pencil and two pens for num__250 cents or two pencils and three pens for num__425 cents . if the cost of a pencil and the cost of a pen does not change compute the cost in cents of five pencils and six pens . <o> a ) num__550 cents . <o> b ) num__650 cents . <o> c ) num__750 cents . <o> d ) num__850 cents . <o> e ) num__950 cents . |
if we add one pencil and one pen an order increases in price by num__175 cents . therefore adding four pencils and four pens increases the order by num__700 cents so the cost of five pencils and six pens is num__700 + num__250 = num__950 cents . correct answer e <eor> e <eos> |
e |
subtract__425.0__250.0__ add__250.0__700.0__ add__250.0__700.0__ |
subtract__425.0__250.0__ add__250.0__700.0__ add__250.0__700.0__ |
| a sum of money deposited at c . i . amounts to rs . num__2420 in num__2 years and to rs . num__2662 in num__3 years . find the rate percent ? <o> a ) num__30.0 <o> b ) num__10.0 <o> c ) num__20.0 <o> d ) num__19.0 <o> e ) num__50 % |
num__2420 - - - num__242 num__100 - - - ? = > num__10.0 answer : b <eor> b <eos> |
b |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| two trains travelling in the same direction at num__40 and num__22 kmph completely pass off another in num__1 minute . if the length of the first train is num__125 m what is the length of the second train ? <o> a ) num__228 <o> b ) num__266 <o> c ) num__175 <o> d ) num__99 <o> e ) num__171 |
rs = num__40 – num__22 = num__18 * num__0.277777777778 = num__5 mps t = num__60 sec d = num__5 * num__60 = num__300 m num__125 - - - - - - - - num__175 m answer : c <eor> c <eos> |
c |
subtract__40.0__22.0__ hour_to_min_conversion__ multiply__5.0__60.0__ subtract__300.0__125.0__ round__175.0__ |
subtract__40.0__22.0__ hour_to_min_conversion__ multiply__5.0__60.0__ subtract__300.0__125.0__ multiply__1.0__175.0__ |
| a can do a particular work in num__6 days . b can do the same work in num__8 days . a and b signed to do it for rs . num__3200 . they completed the work in num__3 days with the help of c . how much is to be paid to c ? <o> a ) num__100 <o> b ) num__400 <o> c ) num__300 <o> d ) num__500 <o> e ) num__700 |
explanation : amount of work a can do in num__1 day = num__0.166666666667 amount of work b can do in num__1 day = num__0.125 amount of work a + b can do in num__1 day = num__0.166666666667 + num__0.125 = num__0.291666666667 amount of work a + b + c can do = num__0.333333333333 amount of work c can do in num__1 day = num__0.333333333333 - num__0.291666666667 = num__0.0416666666667 work a can do in num__1 day : work b can do in num__1 day : work c can do in num__1 day = num__0.166666666667 : num__0.125 : num__0.0416666666667 = num__4 : num__3 : num__1 amount to be paid to c = num__3200 × ( num__0.125 ) = num__400 answer : b <eor> b <eos> |
b |
divide__1.0__6.0__ divide__1.0__8.0__ add__0.125__0.1667__ divide__1.0__3.0__ divide__0.125__3.0__ add__3.0__1.0__ divide__3200.0__8.0__ round__400.0__ |
divide__1.0__6.0__ divide__1.0__8.0__ add__0.125__0.1667__ divide__1.0__3.0__ subtract__0.1667__0.125__ add__3.0__1.0__ divide__3200.0__8.0__ round__400.0__ |
| a man has rs . num__355 in the denominations of one - rupee notes num__20 - rupee notes and num__50 - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__20 <o> b ) num__15 <o> c ) num__71 <o> d ) num__40 <o> e ) num__60 |
let number of notes of each denomination be x . then x + num__20 x + num__50 x = num__355 num__71 x = num__355 x = num__5 hence total number of notes = num__3 x = num__15 . answer is b . <eor> b <eos> |
b |
divide__355.0__71.0__ subtract__20.0__5.0__ subtract__20.0__5.0__ |
divide__355.0__71.0__ subtract__20.0__5.0__ subtract__20.0__5.0__ |
| if √ num__10 = num__3.16 find the value of if √ num__2.5 <o> a ) num__1.3 <o> b ) num__1.58 <o> c ) num__2.03 <o> d ) num__2.15 <o> e ) none of these |
explanation : √ ( num__2.5 ) = √ ( num__5 × num__1.0 × num__2 ) = √ ( num__10 ) / num__2 = num__3.16 / num__2 = num__1.58 answer : b <eor> b <eos> |
b |
round_down__2.5__ divide__3.16__2.0__ divide__3.16__2.0__ |
divide__10.0__5.0__ divide__3.16__2.0__ divide__3.16__2.0__ |
| a water tank is two - fifth full . pipe a can fill a tank in num__10 minutes and pipe b can empty in num__6 minutes . if both the pipes are open how long will it take to empty or fill the tank completely ? <o> a ) num__6 min to empty <o> b ) num__7 min to full <o> c ) num__6 min to full <o> d ) num__7 min to empty <o> e ) none of these |
explanation : there are two important points to learn in this type of question first if both will open then tank will be empty first . second most important thing is if we are calculating filling of tank then we will subtract as ( filling - empting ) if we are calculating empting of thank then we will subtract as ( empting - filling ) so lets come on the question now part to emptied num__0.4 part emptied in num__1 minute = num__0.166666666667 - num__0.1 = num__0.0666666666667 = > num__0.0666666666667 : num__0.4 : : num__1 : x = > num__0.4 â ˆ — num__15 = num__6 mins answer is a <eor> a <eos> |
a |
divide__1.0__6.0__ divide__1.0__10.0__ divide__0.4__6.0__ divide__6.0__0.4__ round__6.0__ |
divide__1.0__6.0__ divide__1.0__10.0__ subtract__0.1667__0.1__ divide__6.0__0.4__ round__6.0__ |
| the l . c . m of two numbers is num__45 times their h . c . f if one of the numbers is num__125 and sum of h . c . f and l . c . m . is num__1150 the other number is : <o> a ) num__215 <o> b ) num__220 <o> c ) num__225 <o> d ) num__235 <o> e ) none of these |
let h . c . f be h and l . cm be l . then l - num__45 h and l + h = num__1150 num__45 h + h = num__1150 or h = num__25 . so l = ( num__1150 - num__25 ) - num__1125 . hence other number = ( num__25 * num__9.0 ) = num__225 . correct option : c <eor> c <eos> |
c |
gcd__125.0__1150.0__ lcm__45.0__125.0__ divide__1125.0__125.0__ lcm__45.0__25.0__ lcm__45.0__225.0__ |
gcd__125.0__1150.0__ multiply__45.0__25.0__ divide__1125.0__125.0__ multiply__9.0__25.0__ multiply__9.0__25.0__ |
| the probabilities of solving a question by num__3 students a b & c are num__1 / num__7.0 & num__0.25 respectively . the probability that the problem will be solved is ? <o> a ) num__0.5 <o> b ) num__0.75 <o> c ) num__0.6 <o> d ) num__0.428571428571 <o> e ) num__0.444444444444 |
sol . first we find the probability of not solving the problem x pto x = ( num__1 — d x - d x - d num__1 num__2 num__3 num__1 = ixix num__71 = num__4 ; num__1 num__3 required probability num__0.75 b <eor> b <eos> |
b |
coin_space__ negate_prob__0.25__ negate_prob__0.25__ |
coin_space__ negate_prob__0.25__ negate_prob__0.25__ |
| pipe a can fill a tank in num__9 hours pipe b in num__12 hours and pipe c in num__18 hours . if all the pipes are open in how many hours will the tank be filled ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
the part filled by a + b + c in num__1 hour = num__0.111111111111 + num__0.0833333333333 + num__0.0555555555556 = num__0.25 = num__0.25 all the three pipes together will fill the tank in num__4 hours . the answer is c . <eor> c <eos> |
c |
divide__1.0__9.0__ divide__1.0__12.0__ divide__1.0__18.0__ divide__1.0__0.25__ round__4.0__ |
divide__1.0__9.0__ divide__1.0__12.0__ divide__1.0__18.0__ divide__1.0__0.25__ round__4.0__ |
| a bartender removed from a bottle a certain quantity of a drink which had num__25.0 alcohol content and replaced the same quantity with another drink that had an alcohol content of num__10.0 . the alcohol content in the drink after replacement is num__20.0 what the quantity of the drink that he replaced if there was num__1 litres of the drink in the bottle <o> a ) num__0.5 <o> b ) num__0.333333333333 <o> c ) num__0.666666666667 <o> d ) num__0.6 <o> e ) num__0.625 |
num__1 litre of drink = num__1000 milliliters let x be the quantity taken out we now have num__1000 - x quantity of liquid = > there is num__0.25 ( num__1000 - x ) quantity of alcohol . - - - ( num__1 ) now we add x amount of alcohol from a num__10.0 concentrate . it means num__0.1 x alcohol is added - - - ( num__2 ) since it says final alcohol content in num__1 litre of drink is num__20.0 or num__200 milli litre from ( num__1 ) and ( num__2 ) above - ( num__1000 - x ) num__0.25 + num__0.1 x = num__200 solving for x we get num__333.33 ml or num__0.333333333333 of a litre answer : b <eor> b <eos> |
b |
percent__25.0__1.0__ percent__10.0__1.0__ percent__10.0__20.0__ percent__20.0__1000.0__ percent__0.1__333.33__ percent__0.1__333.33__ |
percent__25.0__1.0__ percent__10.0__1.0__ percent__10.0__20.0__ percent__20.0__1000.0__ percent__0.1__333.33__ percent__0.1__333.33__ |
| there are num__200 questions in a num__3 hr examination . among these questions there are num__50 mathematics problems . it is suggested that twice as much time be spent on each maths problem as for each other question . how many minutes should be spent on mathematics problems <o> a ) num__36 minutes <o> b ) num__60 minutes <o> c ) num__72 minutes <o> d ) num__100 minutes <o> e ) none of these |
explanation : let x = minutes spent on a non - math problem then num__2 x = minutes spent on a math problem x ( num__200 - num__50 ) + num__2 x ( num__50 ) = num__180 x ( num__150 ) + num__2 x ( num__50 ) = num__180 num__150 x + num__100 x = num__180 num__250 x = num__180 x = num__0.72 minutes time to spend on math problems : num__2 x ( num__50 ) = num__100 x = num__100 ( num__0.72 ) = num__72 minutes answer : c <eor> c <eos> |
c |
subtract__200.0__50.0__ divide__200.0__2.0__ add__200.0__50.0__ divide__180.0__250.0__ multiply__100.0__0.72__ round__72.0__ |
subtract__200.0__50.0__ subtract__150.0__50.0__ add__200.0__50.0__ divide__180.0__250.0__ multiply__100.0__0.72__ round__72.0__ |
| what distance ( in meters ) will be covered by a bus moving at num__54 km / hr in num__30 seconds ? <o> a ) num__390 <o> b ) num__420 <o> c ) num__450 <o> d ) num__480 <o> e ) num__510 |
num__54 km / hr = num__54 * num__0.277777777778 = num__15 m / s distance = num__15 * num__30 = num__450 meters the answer is c . <eor> c <eos> |
c |
multiply__30.0__15.0__ round__450.0__ |
multiply__30.0__15.0__ multiply__30.0__15.0__ |
| a b and c started a business with capitals of rs . num__8000 rs . num__10000 and rs . num__12000 respectively . at the end of the year the profit share of b is rs . num__1800 . the difference between the profit shares of a and c is ? <o> a ) num__720 <o> b ) num__266 <o> c ) num__155 <o> d ) num__600 <o> e ) num__441 |
ratio of investments of a b and c is num__8000 : num__10000 : num__12000 = num__4 : num__5 : num__6 and also given that profit share of b is rs . num__1800 = > num__5 parts out of num__15 parts is rs . num__1800 now required difference is num__6 - num__4 = num__2 parts required difference = num__0.4 ( num__1800 ) = rs . num__720 answer : a <eor> a <eos> |
a |
radians_to_degree__4.0__ radians_to_degree__4.0__ |
radians_to_degree__4.0__ radians_to_degree__4.0__ |
| a stick of num__30 cm is broken into two pieces what is the length of the shorter piece if the longer piece is num__6 cm longer than the other ? <o> a ) num__11 <o> b ) num__10 <o> c ) num__12 <o> d ) num__13 <o> e ) num__7 |
since num__30 is the total length . smaller is any one from answers . try back solving from answer d i . e . num__12 . num__12 + ( num__12 + num__6 ) = num__31 > num__30 hence select next smaller no . : num__12 + ( num__12 + num__6 ) = num__30 . so num__12 is the answer answer is c <eor> c <eos> |
c |
round__12.0__ |
round__12.0__ |
| num__2 |
5 num__1017 ? num__41 <o> a ) num__26 <o> b ) num__27 <o> c ) num__29 <o> d ) num__25 <o> e ) num__28 |
num__2 + num__3 = num__5 num__5 + num__5 = num__10 num__10 + num__7 = num__17 num__17 + num__11 = num__28 num__28 + num__13 = num__41 answer : e <eor> e <eos> |
e |
e |
| a train num__110 m long is running with a speed of num__60 km / hr . in what time will it pass a man who is running at num__6 km / hr in the direction opposite to that in which the train is going ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__9 <o> e ) num__2 |
speed of train relative to man = num__60 + num__6 = num__66 km / hr . = num__66 * num__0.277777777778 = num__18.3333333333 m / sec . time taken to pass the men = num__110 * num__0.0545454545455 = num__6 sec . answer : b <eor> b <eos> |
b |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ round__6.0__ |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ divide__110.0__18.3333__ |
| if a b c are integers ; a ² + b ² = num__45 and b ² + c ² = num__40 then the values of a b and c respectively are : <o> a ) num__2 num__6 num__3 <o> b ) num__3 num__2 num__6 <o> c ) num__5 num__4 num__3 <o> d ) num__4 num__53 <o> e ) none of these |
a ² + b ² = num__45 … . ( num__1 ) and b ² + c ² = num__40 subtracting we get : a ² - c ² = num__5 = > ( a + c ) ( a – c ) = num__5 . ( a + c ) = num__5 and ( a – c ) = num__1 . solving we get : a = num__3 c = num__2 . putting c = num__2 in ( ii ) we get b = num__6 . answer : b <eor> b <eos> |
b |
subtract__45.0__40.0__ subtract__3.0__1.0__ add__1.0__5.0__ add__1.0__2.0__ |
subtract__45.0__40.0__ subtract__3.0__1.0__ add__1.0__5.0__ add__1.0__2.0__ |
| in a powder mix two varieties of powders costing rs . num__15 and rs . num__20 per kg respectively so as toget a mixture worth rs . num__16.50 kg ? <o> a ) num__1 : num__3 <o> b ) num__2 : num__5 <o> c ) num__7 : num__3 <o> d ) num__7 : num__5 <o> e ) num__5 : num__8 |
by the rule of alligation : cost of num__1 kg pulses of num__1 st kindcost of num__1 kg pulses of num__2 nd kind rs . num__15 mean price rs . num__16.50 rs . num__20 num__3.50 num__1.50 required rate = num__3.50 : num__1.50 = num__7 : num__3 c <eor> c <eos> |
c |
subtract__20.0__16.5__ subtract__16.5__15.0__ multiply__2.0__3.5__ round_down__3.5__ multiply__1.0__7.0__ |
subtract__20.0__16.5__ subtract__16.5__15.0__ multiply__2.0__3.5__ round_down__3.5__ multiply__1.0__7.0__ |
| mangala completes a piece of work in num__20 days raju completes the same work in num__40 days . if both of them work together then the number of days required to complete the work is <o> a ) num__4 days <o> b ) num__6 days <o> c ) num__13.33 days <o> d ) num__10 days <o> e ) num__11 days |
if a can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days . that is the required no . of days = num__20 × num__0.666666666667 = num__13.33 days . answer : c <eor> c <eos> |
c |
round__13.33__ |
round__13.33__ |
| how much num__60.0 of num__50 is greater than num__40.0 of num__30 ? <o> a ) num__18 <o> b ) num__99 <o> c ) num__77 <o> d ) num__66 <o> e ) num__12 |
( num__0.6 ) * num__50 – ( num__0.4 ) * num__30 num__30 - num__12 = num__18 ch num__60.0 of num__50 is greater than num__40.0 of num__30 ? answer : a <eor> a <eos> |
a |
percent__40.0__30.0__ percent__60.0__30.0__ percent__60.0__30.0__ |
percent__40.0__30.0__ percent__60.0__30.0__ percent__60.0__30.0__ |
| what is the next number of the following sequence num__8 num__24 num__12 num__36 num__18 num__54 ( . . . ) <o> a ) num__30 <o> b ) num__32 <o> c ) num__34 <o> d ) num__36 <o> e ) num__38 |
the series is num__4 * num__4 = num__16 num__8.0 = num__8 num__8 * num__4 = num__32 num__16.0 = num__16 num__16 * num__4 = num__64 num__32.0 = num__32 answer : b <eor> b <eos> |
b |
subtract__12.0__8.0__ subtract__24.0__8.0__ add__8.0__24.0__ multiply__16.0__4.0__ add__8.0__24.0__ |
subtract__12.0__8.0__ subtract__24.0__8.0__ multiply__8.0__4.0__ multiply__16.0__4.0__ multiply__8.0__4.0__ |
| what is the total number of w integers between num__100 and num__200 that are divisible by num__3 ? <o> a ) num__33 <o> b ) num__32 <o> c ) num__31 <o> d ) num__30 <o> e ) num__29 |
yes there is a different way of arriving at that answer . . . . u can also use arithmetic progression to get the answer since the first term to be divisble by num__3 is num__102 . . take that as a . . the starting no and since num__198 is the last digit to be divisible by num__3 take that as n . . . since the difference is num__3 take that as d no u have to find what term is num__198 take that as nth term the formula for that is n = a + ( n - num__1 ) * d num__198 = num__102 + ( n - num__1 ) * num__3 from this u get n = num__33 <eor> a <eos> |
a |
multiply__1.0__33.0__ |
multiply__1.0__33.0__ |
| which of the following is equal to the cube of a non - integer ? <o> a ) - num__64 <o> b ) - num__1 <o> c ) num__8 <o> d ) num__9 <o> e ) num__27 |
a . - num__64 - > cube of - num__4 b . - num__1 - > cube of - num__1 c . num__8 - > cube of num__2 d . num__9 e . num__27 - > cube of num__3 all other options are cubes of integers except num__9 . num__9 will be the cube of a number slightly greater than num__2 since the cube of num__2 is num__8 . answer ( d ) <eor> d <eos> |
d |
divide__8.0__4.0__ add__1.0__8.0__ add__1.0__2.0__ add__1.0__8.0__ |
divide__8.0__4.0__ add__1.0__8.0__ subtract__4.0__1.0__ add__1.0__8.0__ |
| the speed of a train is num__90 kmph . what is the distance covered by it in num__10 minutes ? <o> a ) num__15 <o> b ) num__66 <o> c ) num__77 <o> d ) num__52 <o> e ) num__42 |
num__90 * num__0.166666666667 = num__15 kmph answer : a <eor> a <eos> |
a |
round__15.0__ |
round__15.0__ |
| if the tens digit of positive integer x is num__4 and the tens digit of positive integer y is num__4 how many possible values of the tens digit of num__2 ( x + y ) can there be ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
we only need to consider the tens and ones digits of x and y and the last two digits can be represented as num__40 + p and num__40 + q respectively . num__2 ( num__40 + p + num__40 + q ) = num__160 + num__2 ( p + q ) num__2 ( p + q ) can range from num__0 up to a maximum value of num__36 when p = q = num__9 . then the tens and ones digits of num__2 ( x + y ) can range from num__60 up to num__96 . there can be num__4 possibilities for the tens digit . the answer is c . <eor> c <eos> |
c |
choose__2.0__0.0__ |
choose__2.0__0.0__ |
| how many seconds will a num__300 m long train take to cross a man walking with a speed of num__3 km / hr in the direction of the moving train if the speed of the train is num__63 km / hr ? <o> a ) num__11 <o> b ) num__30 <o> c ) num__99 <o> d ) num__88 <o> e ) num__18 |
speed of train relative to man = num__63 - num__3 = num__60 km / hr . = num__60 * num__0.277777777778 = num__16.6666666667 m / sec . time taken to pass the man = num__300 * num__0.06 = num__18 sec . answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ divide__300.0__16.6667__ round__18.0__ |
subtract__63.0__3.0__ divide__300.0__16.6667__ divide__300.0__16.6667__ |
| there are num__10 people in a room . if each person shakes hands with exactly num__3 other people what is the total number of handshakes ? <o> a ) num__15 <o> b ) num__30 <o> c ) num__45 <o> d ) num__60 <o> e ) num__120 |
num__4 people : shrive num__555 cranky karishma and mr x we have to shake hands in this group such that each person shakes hands with num__3 people . so shrive num__555 starts : shrive num__555 - craky : num__1 handshake but num__2 hands were shaken . shrive num__555 ' s and cranky ' s shrive num__555 - karishma : num__1 handshake but num__2 hands were shaken . shrive num__555 ' s and karishma ' s shrive num__555 - mr x : num__1 handshake but num__2 hands were shaken . shrive num__555 ' s and mr x ' s now shrive num__555 has shaken hands with num__3 people . there were num__3 handshakes . but num__6 hands were shaken . now when all num__4 of us shake hands with num__3 people each person ' s hand will be shaken num__3 times . i . e . in all num__12 hands will be shaken . but they will add up to only num__6 handshakes . the other num__3 handshakes will be : cranky - karishma cranky - mr x karishma - mr x so all of us have shaken hands with exactly num__3 people . similarly when there are num__10 people each person shakes his hand num__3 times . so in all num__30 . but num__2 of these hands combined to make one handshake . so we will get only num__15 total handshakes . answer : a <eor> a <eos> |
a |
subtract__4.0__3.0__ subtract__3.0__1.0__ subtract__10.0__4.0__ add__10.0__2.0__ multiply__10.0__3.0__ add__3.0__12.0__ add__3.0__12.0__ |
subtract__4.0__3.0__ subtract__3.0__1.0__ subtract__10.0__4.0__ add__10.0__2.0__ multiply__10.0__3.0__ add__3.0__12.0__ subtract__30.0__15.0__ |
| a farmer divides his herd of x cows among his num__4 sons so that one son gets one half of the herd the second gets one - fourth the third gets one - fifth and the fourth gets num__7 cows . then x is equal to <o> a ) num__100 <o> b ) num__140 <o> c ) num__180 <o> d ) num__160 <o> e ) num__120 |
no . of cows : : x num__1 st son : x / num__2 num__2 nd son : x / num__4 num__3 rd son : x / num__5 num__4 th son : num__7 ( x ) + ( x / num__4 ) + ( x / num__5 ) + num__7 = x = > x - ( num__19 x / num__20 ) = num__7 = > ( num__20 x - num__19 x ) / num__20 = num__7 = > x = num__140 answer : b <eor> b <eos> |
b |
subtract__4.0__1.0__ add__4.0__1.0__ multiply__4.0__5.0__ multiply__7.0__20.0__ multiply__7.0__20.0__ |
subtract__4.0__1.0__ add__4.0__1.0__ add__1.0__19.0__ multiply__7.0__20.0__ divide__140.0__1.0__ |
| which of the following is closest to ( num__5 ! - num__4 ! ) / ( num__5 ! + num__4 ! ) ? <o> a ) num__0.01 <o> b ) num__0.1 <o> c ) num__0.2 <o> d ) num__0.7 <o> e ) num__10 |
( num__5 ! - num__4 ! ) / ( num__5 ! + num__4 ! ) num__4 ! ( num__5 - num__1 ) / num__4 ! ( num__5 + num__1 ) num__0.666666666667 = num__0.666666666667 = num__0.7 ( approximately ) d is the answer <eor> d <eos> |
d |
subtract__5.0__4.0__ multiply__1.0__0.7__ |
subtract__5.0__4.0__ divide__0.7__1.0__ |
| given that - num__1 ≤ v ≤ num__1 - num__2 ≤ u ≤ - num__0.5 and - num__2 ≤ z ≤ - num__0.5 and w = vz / u then which of the following is necessarily true ? <o> a ) - num__0.5 ≤ w ≤ - num__2 <o> b ) - num__4 ≤ w ≤ num__4 <o> c ) - num__4 ≤ w ≤ num__2 <o> d ) - num__2 ≤ w ≤ - num__0.5 <o> e ) none of these |
explanation : u is always negative . hence for us to have a minimum value of vz / u vz should be positive . also for the least value the numerator has to be the maximum positive value and the denominator has to be the smallest negative value . in other words vz has to be num__2 and u has to be – num__0.5 . hence the minimum value of vz / u = num__2 − num__0.5 = – num__4 . for us to get the maximum value vz has to be the smallest negative value and u has to be the highest negative value . thus vz has to be – num__2 and u has to be – num__0.5 . hence the maximum value of vz / u = num__2 / - num__0.5 = num__4 . answer : b <eor> b <eos> |
b |
divide__2.0__0.5__ multiply__1.0__4.0__ |
divide__2.0__0.5__ divide__2.0__0.5__ |
| find the next number in the sequence num__5 num__14 num__41 num__86 ? <o> a ) num__146 <o> b ) num__145 <o> c ) num__149 <o> d ) num__143 <o> e ) num__148 |
+ num__9 = > num__14 + num__27 = > num__41 + num__45 = > num__86 + num__63 = > num__149 answer is c <eor> c <eos> |
c |
subtract__14.0__5.0__ subtract__41.0__14.0__ multiply__5.0__9.0__ add__86.0__63.0__ add__86.0__63.0__ |
subtract__14.0__5.0__ subtract__41.0__14.0__ multiply__5.0__9.0__ add__86.0__63.0__ add__86.0__63.0__ |
| the diameter of the driving wheel of a bus in num__140 cm . how many revolutions per minute must the wheel make in order to keep a speed of num__33 kmph ? <o> a ) num__210 <o> b ) num__220 <o> c ) num__125 <o> d ) num__240 <o> e ) num__250 |
distance covered in num__1 min = ( num__33 * num__1000 ) / num__60 = num__550 m circumference of the wheel = ( num__2 * ( num__3.14285714286 ) * . num__70 ) = num__4.4 m no of revolution per min = num__550 / num__4.4 = num__125 answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ divide__140.0__2.0__ divide__550.0__4.4__ round__125.0__ |
hour_to_min_conversion__ divide__140.0__2.0__ divide__550.0__4.4__ divide__550.0__4.4__ |
| in an election between two candidates one got num__45.0 of the total valid votes num__20.0 of the votes were invalid . if the total number of votes was num__2000 the number of valid votes that the other candidate got was : <o> a ) num__1500 <o> b ) num__4500 <o> c ) num__3270 <o> d ) num__1100 <o> e ) num__4000 |
number of valid votes = num__80.0 of num__2000 = num__1600 . valid votes polled by other candidate = num__55.0 of num__2000 = ( num__0.55 ) x num__2000 = num__1100 answer = d <eor> d <eos> |
d |
percent__80.0__2000.0__ percent__55.0__2000.0__ percent__55.0__2000.0__ |
percent__80.0__2000.0__ percent__55.0__2000.0__ percent__55.0__2000.0__ |
| a can do a job in num__30 days and b in num__40 days . if they work on it together for num__10 days then the fraction of the work that is left is ? <o> a ) num__0.181818181818 <o> b ) num__0.230769230769 <o> c ) num__0.416666666667 <o> d ) num__0.7 <o> e ) num__0.4 |
a ' s num__1 day work = num__0.0333333333333 b ' s num__1 day work = num__0.025 a + b num__1 day work = num__0.0333333333333 + num__0.025 = num__0.0583333333333 a + b num__4 days work = num__0.0583333333333 * num__10 = num__0.583333333333 remaining work = num__1 - num__0.583333333333 = num__0.416666666667 answer is c <eor> c <eos> |
c |
divide__1.0__30.0__ divide__1.0__40.0__ add__0.025__0.0333__ divide__40.0__10.0__ subtract__1.0__0.5833__ multiply__1.0__0.4167__ |
divide__1.0__30.0__ divide__1.0__40.0__ add__0.025__0.0333__ divide__40.0__10.0__ subtract__1.0__0.5833__ multiply__1.0__0.4167__ |
| a motor cycle is moving with the speed of num__47.52 km / hr and the radius of the wheel of the motorcycle is num__21 cms . calculate the approximate no of revolutions made by the wheel in one minute . <o> a ) num__400 rpm <o> b ) num__500 rpm <o> c ) num__600 rpm <o> d ) num__700 rpm <o> e ) num__800 rpm |
given speed = num__47.52 km / h = num__47.52 * num__16.6666666667 m / s = num__792 metre / minute the radius of the wheel of the motorcycle = num__21 cm circumference of the wheels = num__2 * num__3.14285714286 * num__21 = num__132 cm = num__1.32 m number of revolutions made by the wheel in one minute = speed ( in metre / minute ) / circumference ( in m ) = num__792 / num__1.32 = num__600 so no of revolutions made by the wheel in one minute will be num__600 answer : c <eor> c <eos> |
c |
divide__792.0__1.32__ round__600.0__ |
divide__792.0__1.32__ divide__792.0__1.32__ |
| a car takes num__6 hours to cover a distance of num__469 km . how much should the speed in kmph be maintained to cover the same direction in num__1.5 th of the previous time ? <o> a ) num__48 kmph <o> b ) num__50 kmph <o> c ) num__52 kmph <o> d ) num__56 kmph <o> e ) num__60 kmph |
time = num__6 distence = num__469 num__1.5 of num__6 hours = num__6 * num__1.5 = num__9 hours required speed = num__52.1111111111 = num__52 kmph c <eor> c <eos> |
c |
multiply__6.0__1.5__ divide__469.0__9.0__ round__52.0__ |
multiply__6.0__1.5__ divide__469.0__9.0__ round__52.0__ |
| when w is divided by num__14 the reminder is num__0 . if w is three lesser than it value and when divided by num__5 its remainder is num__0 . what is the value of w ? <o> a ) num__14 <o> b ) num__28 <o> c ) num__42 <o> d ) num__45 <o> e ) num__33 |
w is divided by num__14 so that is multiple of num__14 as num__14 num__2842 . . . w - num__3 is divided by num__5 the remainder is num__0 so it is divisible by num__5 . consider from option let us take the number is num__14 it is divisible by num__14 but num__14 - num__3 is not divisible by num__5 so it is not answers so let us take num__2 nd option num__28 which is divisible by num__14 and num__28 - num__3 = num__25 is divisible by num__5 so ans is b <eor> b <eos> |
b |
subtract__5.0__3.0__ multiply__14.0__2.0__ subtract__28.0__3.0__ multiply__14.0__2.0__ |
subtract__5.0__3.0__ multiply__14.0__2.0__ subtract__28.0__3.0__ multiply__14.0__2.0__ |
| the population of a city is num__8000 . it decreases annually at the rate of num__20.0 p . a . what will be its population after num__2 years ? <o> a ) num__1000 <o> b ) num__1200 <o> c ) num__1250 <o> d ) num__3000 <o> e ) num__5120 |
formula : ( after = num__100 denominator ago = num__100 numerator ) num__8000 × num__0.8 × num__0.8 = num__5120 e <eor> e <eos> |
e |
percent__100.0__5120.0__ |
percent__100.0__5120.0__ |
| a man buys an article for rs . num__27.50 and sells it for rs num__28.60 . find his gain percent <o> a ) num__1.0 <o> b ) num__2.0 <o> c ) num__3.0 <o> d ) num__4.0 <o> e ) num__5 % |
explanation : so we have c . p . = num__27.50 s . p . = num__28.60 gain = num__28.60 - num__27.50 = rs . num__1.10 gain % = ( gaincost ∗ num__100 ) % = ( num__1 . num__1027.50 ∗ num__100 ) % = num__4.0 answer is d <eor> d <eos> |
d |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| what is the ratio between perimeters of two squares one having num__3 times the diagonal then the other ? <o> a ) num__3 : num__5 <o> b ) num__3 : num__9 <o> c ) num__3 : num__4 <o> d ) num__3 : num__1 <o> e ) num__3 : num__2 |
d = num__3 d d = d a √ num__2 = num__3 d a √ num__2 = d a = num__3 d / √ num__2 a = d / √ num__2 = > num__3 : num__1 answer : d <eor> d <eos> |
d |
multiply__3.0__1.0__ |
multiply__3.0__1.0__ |
| a part - time employee ’ s hourly wage was increased by num__40.0 . she decided to decrease the number of hours worked per week so that her total income did not change . by approximately what percent should the number of hours worked be decreased ? <o> a ) num__9.0 <o> b ) num__15.0 <o> c ) num__29.0 <o> d ) num__50.0 <o> e ) num__100 % |
let ' s plug in somenicenumbers and see what ' s needed . let ' s say the employee used to make $ num__1 / hour and worked num__100 hours / week so the total weekly income was $ num__100 / week after the num__40.0 wage increase the employee makes $ num__1.40 / hour we want the employee ' s income to remain at $ num__100 / week . so we want ( $ num__1.40 / hour ) ( new # of hours ) = $ num__100 divide both sides by num__1.40 to get : new # of hours = num__100 / num__1.40 ≈ num__71 hours so the number of hours decreases from num__100 hours to ( approximately ) num__71 hours . this represents a num__29.0 decrease ( approximately ) . answer : c <eor> c <eos> |
c |
subtract__100.0__71.0__ multiply__1.0__29.0__ |
subtract__100.0__71.0__ divide__29.0__1.0__ |
| excluding stoppages the speed of a bus is num__54 kmph and including stoppages it is num__45 kmph . for how many minutes does the bus stop per hour ? <o> a ) num__8 min <o> b ) num__5 min <o> c ) num__10 min <o> d ) num__14 min <o> e ) num__18 min |
due to stoppages it covers num__9 km less . time taken to cover num__9 km = ( num__0.166666666667 x num__60 ) min = num__10 min answer : c <eor> c <eos> |
c |
subtract__54.0__45.0__ divide__9.0__54.0__ hour_to_min_conversion__ round__10.0__ |
subtract__54.0__45.0__ divide__9.0__54.0__ hour_to_min_conversion__ round__10.0__ |
| in an election candidate smith won num__52 percent of the total vote in counties a and b . he won num__61 percent of the vote in county a . if the ratio of people who voted in county a to county b is num__3 : num__1 what percent of the vote did candidate smith win in county b ? <o> a ) num__25.0 <o> b ) num__27.0 <o> c ) num__34.0 <o> d ) num__43.0 <o> e ) num__49 % |
given voters in ratio num__3 : num__1 let a has num__300 voters & b has num__100 voters for a num__61.0 voted means num__61 * num__300 = num__183 votes combined for a & b has num__400 voters and voted num__52.0 so total votes = num__208 balance votes = num__208 - num__183 = num__25 as b has num__100 voters so num__25 votes means num__25.0 of votes required answer : a <eor> a <eos> |
a |
divide__300.0__3.0__ multiply__61.0__3.0__ add__100.0__300.0__ subtract__208.0__183.0__ multiply__1.0__25.0__ |
divide__300.0__3.0__ multiply__61.0__3.0__ add__100.0__300.0__ subtract__208.0__183.0__ multiply__1.0__25.0__ |
| how many cubes will have num__4 coloured sides and two non - coloured sides ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__3 <o> d ) num__2 <o> e ) num__1 |
only num__4 cubes situated at the corners of the cuboid will have num__4 coloured and num__2 non - coloured sides . answer b <eor> b <eos> |
b |
triangle_area__4.0__2.0__ |
triangle_area__4.0__2.0__ |
| the length of the bridge which a train num__130 meters long and travelling at num__45 km / hr can cross in num__30 seconds is ? <o> a ) num__388 <o> b ) num__277 <o> c ) num__245 <o> d ) num__288 <o> e ) num__288 |
speed = ( num__45 * num__0.277777777778 ) m / sec = ( num__12.5 ) m / sec . time = num__30 sec . let the length of bridge be x meters . then ( num__130 + x ) / num__30 = num__12.5 = = > num__2 ( num__130 + x ) = num__750 = = > x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| jill bought items costing $ num__3.45 $ num__1.99 $ num__6.59 and $ num__12.98 . she used a coupon worth $ num__2.50 . if jill had $ num__50.00 when she went into the store how much did she have when she left ? <o> a ) $ num__17.49 <o> b ) $ num__27.49 <o> c ) $ num__37.49 <o> d ) $ num__57.49 <o> e ) $ num__47.49 |
$ num__3.45 + $ num__1.99 + $ num__6.59 + $ num__12.98 = $ num__25.01 $ num__25.01 - $ num__2.50 = $ num__22.51 $ num__50.00 - $ num__22.51 = $ num__27.49 is money left correct answer b <eor> b <eos> |
b |
subtract__25.01__2.5__ subtract__50.0__22.51__ subtract__50.0__22.51__ |
subtract__25.01__2.5__ subtract__50.0__22.51__ subtract__50.0__22.51__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__21 sec . what is the length of the train ? <o> a ) num__535 m <o> b ) num__178 m <o> c ) num__186 m <o> d ) num__350 m <o> e ) num__150 m |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__21 = num__350 m answer : d <eor> d <eos> |
d |
round__350.0__ |
round__350.0__ |
| a fort had provision of food for num__150 men for num__45 days . after num__10 days num__25 men left the fort . the number of days for which the remaining food will last is : <o> a ) num__42 <o> b ) num__43 <o> c ) num__46 <o> d ) num__47 <o> e ) num__49 |
a num__42 after num__10 days : num__150 men had food for num__35 days . suppose num__125 men had food for x days . now less men more days ( indirect proportion ) num__125 : num__150 : : num__35 : x num__125 x x = num__150 x num__35 x = ( num__150 x num__35 ) / num__125 x = num__42 <eor> a <eos> |
a |
subtract__45.0__10.0__ subtract__150.0__25.0__ round__42.0__ |
subtract__45.0__10.0__ subtract__150.0__25.0__ round__42.0__ |
| a man covers a distance on scooter . had he moved num__3 kmph faster he would have taken num__40 min less . if he had moved num__2 kmph slower he would have taken num__40 min more . the distance is . <o> a ) num__30 km <o> b ) num__40 kilometre <o> c ) num__45 km <o> d ) num__50 km <o> e ) num__55 km |
let distance = x m usual rate = y kmph x / y – x / y + num__3 = num__0.666666666667 hr num__2 y ( y + num__3 ) = num__9 x - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ( num__1 ) x / y - num__2 – x / y = num__0.666666666667 hr y ( y - num__2 ) = num__3 x - - - - - - - - - - - ( num__2 ) divide num__1 & num__2 equations by solving we get x = num__40 km answer : b . <eor> b <eos> |
b |
divide__2.0__3.0__ subtract__3.0__2.0__ round__40.0__ |
divide__2.0__3.0__ subtract__3.0__2.0__ divide__40.0__1.0__ |
| after decreasing num__90.0 in the price of an article costs rs . num__320 . find the actual cost of an article ? <o> a ) num__2777 <o> b ) num__2987 <o> c ) num__3200 <o> d ) num__9977 <o> e ) num__1671 |
cp * ( num__0.1 ) = num__320 cp = num__32 * num__100 = > cp = num__3200 answer : c <eor> c <eos> |
c |
percent__100.0__3200.0__ |
percent__100.0__3200.0__ |
| a train is running at a speed of num__40 km / hr and it crosses a post in num__18 seconds . what is the length of the train ? <o> a ) num__190 metres <o> b ) num__160 metres <o> c ) num__200 metres <o> d ) num__120 metres <o> e ) num__130 metres |
explanation : speed of the train v = num__40 km / hr = num__11.1111111111 m / s = num__11.1111111111 m / s time taken to cross t = num__18 s distance covered d = vt = ( num__11.1111111111 ) × num__18 = num__200 m distance covered is equal to the length of the train = num__200 m answer : option c <eor> c <eos> |
c |
round__200.0__ |
round__200.0__ |
| in num__10 years p will be as old as q is now . twenty years ago q was twice as old as p was then . how old is p now ? <o> a ) num__30 <o> b ) num__35 <o> c ) num__40 <o> d ) num__45 <o> e ) num__50 |
q = p + num__10 q - num__20 = num__2 ( p - num__20 ) p - num__10 = num__2 p - num__40 p = num__30 the answer is a . <eor> a <eos> |
a |
twice__10.0__ divide__20.0__10.0__ twice__20.0__ triple__10.0__ triple__10.0__ |
twice__10.0__ divide__20.0__10.0__ twice__20.0__ add__10.0__20.0__ add__10.0__20.0__ |
| excluding stoppages the speed of a bus is num__72 kmph and including stoppages it is num__45 kmph . for how many minutes does the bus stop per hour ? <o> a ) num__7 min <o> b ) num__8 min <o> c ) num__9 min <o> d ) num__10 min <o> e ) num__55.8 min |
due to stoppages it covers num__9 km less . time taken to cover num__67 km = ( num__0.930555555556 ) x num__60 = num__55.8 min answer : e <eor> e <eos> |
e |
divide__67.0__72.0__ hour_to_min_conversion__ round__55.8__ |
divide__67.0__72.0__ hour_to_min_conversion__ round__55.8__ |
| bruce and bhishma are running on a circular track of length num__600 m . speed of bruce is num__30 m / s and that of bhishma is num__20 m / s . they start fro the same point at the same time in the same direction . when will they meet again for the first time ? <o> a ) num__40 <o> b ) num__45 <o> c ) num__50 <o> d ) num__60 <o> e ) num__70 |
actually arjun has to make a lead of num__600 m because when arjun will be num__600 m ahead of bhishma they will be together again as a person when completes the total length it starts retracing he same path and thus arjun and bhishma can be together again . since they make a difference of num__10 m in num__1 second . so he will create num__600 m difference in num__60 second . d <eor> d <eos> |
d |
percent__10.0__600.0__ percent__10.0__600.0__ |
percent__10.0__600.0__ percent__10.0__600.0__ |
| a snail climbing a num__20 feet high wall climbs up num__4 feet on the first day but slides down num__2 feet on the second . it climbs num__4 feet on the third day and slides down again num__2 feet on the fourth day . if this pattern continues how many days will it take the snail to reach the top of the wall ? <o> a ) num__12 <o> b ) num__16 <o> c ) num__17 <o> d ) num__20 <o> e ) num__21 |
total transaction in two days = num__4 - num__2 = num__2 feet in num__16 days it will climb num__16 feet on the num__17 th day the snail will climb num__4 feet thus reaching the top therefore total no of days required = num__17 answer : c <eor> c <eos> |
c |
subtract__20.0__4.0__ round__17.0__ |
subtract__20.0__4.0__ round__17.0__ |
| the total population of dogs in a community is estimated to be num__50.0 male and num__50.0 female . the total number of black dogs is num__20.0 greater than the total number of male black dogs . if the total number of female dogs is eight times more than the number of female black dogs what percentage of male dogs is black ? <o> a ) num__84.4 <o> b ) num__70.2 <o> c ) num__80.1 <o> d ) num__60.8 <o> e ) num__62.5 |
since we are dealing in percentage let us pick num__100 as the number of black male dogs . that means that the total number of black dogs is = num__120 ( num__20.0 more ) therefore the number of black female dogs is num__20 . the total number of female dogs is num__8 x the number of black female = num__20 * num__8 = num__160 female dogs . male dogs are num__50.0 of the dogs and because there are num__160 female dogs then there must also be num__160 male dogs therefore percentage of male dogs that are black = num__0.625 * num__100 = num__62.5 correct option is e <eor> e <eos> |
e |
add__20.0__100.0__ multiply__20.0__8.0__ divide__100.0__160.0__ multiply__0.625__100.0__ multiply__0.625__100.0__ |
add__20.0__100.0__ multiply__20.0__8.0__ divide__100.0__160.0__ multiply__0.625__100.0__ multiply__0.625__100.0__ |
| the area of a square field num__3136 sq m if the length of cost of drawing barbed wire num__3 m around the field at the rate of rs . num__1.50 per meter . two gates of num__1 m width each are to be left for entrance . what is the total cost ? <o> a ) rs . num__500 <o> b ) rs . num__250 <o> c ) rs . num__896 <o> d ) rs . num__789 <o> e ) rs . num__999 |
explanation : a ^ num__2 = num__3136 = > a = num__56 num__56 * num__4 * num__3 = num__672 – num__6 = num__666 * num__1.5 = rs . num__999 answer : e <eor> e <eos> |
e |
divide__3.0__1.5__ add__3.0__1.0__ multiply__3.0__2.0__ subtract__672.0__6.0__ multiply__1.5__666.0__ round__999.0__ |
divide__3.0__1.5__ add__3.0__1.0__ multiply__3.0__2.0__ subtract__672.0__6.0__ multiply__1.5__666.0__ multiply__1.5__666.0__ |
| in country z num__10.0 of the people do not have a university diploma but have the job of their choice and num__30.0 of the people who do not have the job of their choice have a university diploma . if num__40.0 of the people have the job of their choice what percent of the people have a university diploma ? <o> a ) num__35.0 <o> b ) num__48.0 <o> c ) num__55.0 <o> d ) num__65.0 <o> e ) num__75 % |
setting up a matrix is how i solve this one . diploma no diploma totals job of choice w / diploma job of choice w / o diploma = num__10.0 job of choice total = num__40.0 not job of choice with diploma = . num__3 x not job of choice w / o diploma = . num__7 x total not job of choice = x total with diploma total without diploma total citizen = num__100 if num__40.0 of people have their job of choice then num__60.0 of people do not have their job of choice . num__30.0 of num__60.0 = num__18.0 . we can also see that num__30.0 of the people have their job of choice and a diploma ( num__40.0 - num__10.0 = num__30.0 ) . num__30.0 + num__18.0 = num__48.0 . therefore num__48.0 of the people in country z have a diploma . ans b <eor> b <eos> |
b |
percent__10.0__30.0__ percent__30.0__60.0__ percent__100.0__48.0__ |
percent__10.0__30.0__ percent__30.0__60.0__ percent__100.0__48.0__ |
| in one alloy there is num__10.0 chromium while in another alloy it is num__6.0 . num__15 kg of the first alloy was melted together with num__35 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . <o> a ) num__7.2 <o> b ) num__9.0 <o> c ) num__9.2 <o> d ) num__8.6 <o> e ) num__8.4 % |
the amount of chromium in the new num__15 + num__35 = num__50 kg alloy is num__0.10 * num__15 + num__0.06 * num__35 = num__3.6 kg so the percentage is num__3.6 / num__50 * num__100 = num__7.2 . answer : a . <eor> a <eos> |
a |
percent__100.0__7.2__ |
percent__100.0__7.2__ |
| a pharmaceutical company received $ num__2 million in royalties on the first $ num__12 ; million in sales of generic equivalent of one of its products and then $ num__4 million in royalties on the next $ num__48 million in sales . by approximately what percent did the ratio of royalties to sales decrease from the first $ num__12 million in sales to the next $ num__48 million in sales ? <o> a ) num__20.0 <o> b ) num__25.0 <o> c ) num__30.0 <o> d ) num__50.0 <o> e ) num__55 % |
change in ratio of royalties to sales = num__0.166666666667 - num__0.0833333333333 = num__0.0833333333333 % decrease = ( num__0.0833333333333 ) / ( num__0.166666666667 ) * num__100 = num__50.0 ( approx ) answer : d ) <eor> d <eos> |
d |
divide__2.0__12.0__ reverse__12.0__ add__2.0__48.0__ add__2.0__48.0__ |
divide__2.0__12.0__ reverse__12.0__ divide__100.0__2.0__ divide__100.0__2.0__ |
| a do num__0.4 part of work in num__12 days and b do num__0.75 part of work in num__15 days . so both together in how many days can finish the work ? <o> a ) num__15 <o> b ) num__8 <o> c ) num__10 <o> d ) num__12 <o> e ) num__14 |
work of a in num__12 days = num__0.4 work of a in num__1 days = num__0.0333333333333 work of b in num__15 days = num__0.75 work of b in num__1 days = num__0.05 ( a + b ) work in num__1 days = [ num__0.0333333333333 + num__0.05 ] = num__0.0833333333333 so both together finish the work in num__12 days answer d <eor> d <eos> |
d |
divide__0.4__12.0__ divide__0.75__15.0__ divide__1.0__12.0__ round__12.0__ |
divide__0.4__12.0__ divide__0.75__15.0__ add__0.05__0.0333__ round__12.0__ |
| two pipes a and b can fill a cistern in num__37 num__0.5 minutes and num__45 minutes respectively . both pipes are opened . the cistern will be filled in just half an hour if pipe b is turned off after <o> a ) num__5 min <o> b ) num__9 min <o> c ) num__10 min <o> d ) num__15 min <o> e ) num__16 min |
explanation : solution num__1 pipe a alone can fill the cistern in num__37 num__0.5 = num__37.5 minutes . since it was open for num__30 minutes part of the cistern filled by pipe a = num__0.0266666666667 x num__30 = num__0.8 so the remaining num__0.2 part is filled by pipe b . pipe b can fill the cistern in num__45 minutes . so time required to fill num__0.2 part = num__9.0 = num__9 minutes . i . e . pipe b is turned off after num__9 minutes . answer is b <eor> b <eos> |
b |
add__37.0__0.5__ divide__1.0__37.5__ divide__30.0__37.5__ subtract__1.0__0.8__ multiply__45.0__0.2__ round__9.0__ |
add__37.0__0.5__ divide__1.0__37.5__ divide__30.0__37.5__ subtract__1.0__0.8__ multiply__45.0__0.2__ round__9.0__ |
| which of the following best approximates the value of q if num__5 ^ num__29 + num__5 ^ num__11 = num__5 ^ q ? <o> a ) num__40 <o> b ) num__30 <o> c ) num__29 <o> d ) num__27 <o> e ) num__17 |
we have : num__5 ^ num__29 + num__5 ^ num__11 = num__5 ^ q = = > because num__5 ^ num__11 > num__0 - - > num__5 ^ q must be equal or greater than num__5 ^ num__29 = = > q must be equal or greater than num__29 = = > option d and e are out immediately . divide both sides by num__5 ^ q and q > = num__29 we have : num__5 ^ ( num__29 - q ) + num__5 ^ num__2.2 ^ q = num__1 because q > = num__29 = = > num__5 ^ num__2.2 ^ q = num__0.0000 xyz that is very small we can ignore it . thus num__5 ^ ( num__29 - q ) must be approximately num__1 = = > num__29 - q = num__0 = = > q is approximately num__29 c is the answer . <eor> c <eos> |
c |
divide__11.0__5.0__ multiply__29.0__1.0__ |
divide__11.0__5.0__ multiply__29.0__1.0__ |
| which is the smallest no which divides num__2880 and gives a perfect square ? <o> a ) num__4 <o> b ) num__9 <o> c ) num__3 <o> d ) num__5 <o> e ) num__6 |
breaking num__2880 into multiples : num__2880 = num__5 * num__9 * num__64 = num__5 * num__3 ^ num__2 * num__8 ^ num__2 = num__5 * num__24 ^ num__2 num__5 being a prime number with only one multiple will divide num__2880 & result in num__9 * num__64 which is a perfect square ( num__9 * num__64 = num__24 ^ num__2 = num__576 ) answer : d <eor> d <eos> |
d |
square_perimeter__2.0__ surface_cube__2.0__ multiply__64.0__9.0__ triangle_area__2.0__5.0__ |
power__2.0__3.0__ multiply__3.0__8.0__ multiply__64.0__9.0__ triangle_area__2.0__5.0__ |
| one morning is starts to snow at a constant rate . later at num__6 : num__00 am a snow plow sets out to clear a straight street . the plow can remove a fixed volume of snow per unit time in other words its speed it inversely proportional to the depth of the snow . if the plow covered twice as much distance in the first hour as the second hour what time did it start snowing ? <o> a ) num__2 : num__22 : num__55 am <o> b ) num__5 : num__22 : num__55 am <o> c ) num__1 : num__22 : num__55 am <o> d ) num__6 : num__22 : num__55 am <o> e ) num__7 : num__22 : num__55 am |
let the depth of snow at time t to be t units . the speed of the plow at time t will be num__1 / t . define t = num__0 as the time it started snowing and t = x the time the plow started . the distance covered in the first hour is the integral from x to x + num__1 of num__1 / t dt . the antiderivative of num__1 / t is ln ( t ) so the total distance covered in the first hour is ln ( ( x + num__1 ) / x ) . by the same reasoning the distance covered in the second hour in ln ( ( x + num__2 ) / ( x + num__1 ) ) . using the fact that it the plow traveled twice as far in the first hour as the second : ln ( ( x + num__1 ) / x ) = ln ( ( x + num__2 ) / ( x + num__1 ) ) num__2 exp both sides and you have ( x + num__1 ) / x = ( ( x + num__2 ) / ( x + num__1 ) ) num__2 . solving for x you get x = ( num__25.5 - num__1 ) / num__2 which is the number of hours that elapsed between the time it started snowing and the snow plow left . it started snowing at ( num__25.5 - num__1 ) / num__2 hours before num__6 : num__00 am or num__5 : num__22 : num__55 am . correct answer b <eor> b <eos> |
b |
subtract__6.0__1.0__ round__5.0__ |
subtract__6.0__1.0__ subtract__6.0__1.0__ |
| if ( num__2 to the x ) - ( num__2 to the ( x - num__2 ) ) = num__3 ( num__2 to the num__9 ) what is the value of x ? <o> a ) num__9 <o> b ) num__11 <o> c ) num__13 <o> d ) num__15 <o> e ) num__17 |
( num__2 to the power x ) - ( num__2 to the power ( x - num__2 ) ) = num__3 ( num__2 to the power num__9 ) num__2 ^ x - num__2 ^ ( x - num__2 ) = num__3 . num__2 ^ num__9 hence x = num__11 . answer is b <eor> b <eos> |
b |
add__2.0__9.0__ add__2.0__9.0__ |
add__2.0__9.0__ add__2.0__9.0__ |
| joe went on a diet num__4 months ago when he weighed num__222 pounds . if he now weighs num__198 pounds and continues to lose at the same average monthly rate in approximately how many months will he weigh num__180 pounds ? <o> a ) num__3 <o> b ) num__3.5 <o> c ) num__3 <o> d ) num__4.5 <o> e ) num__5 |
num__222 - num__198 = num__24 pounds lost in num__4 months num__6.0 = num__6 so joe is losing weight at a rate of num__6 pounds per month . . . . in approximately how many months will he weigh num__180 pounds ? a simple approach is to just list the weights . now : num__198 lbs in num__1 month : num__192 lbs in num__2 months : num__186 lbs in num__3 months : num__180 lbs answer : c <eor> c <eos> |
c |
subtract__222.0__198.0__ divide__24.0__4.0__ subtract__198.0__6.0__ subtract__6.0__4.0__ add__180.0__6.0__ subtract__4.0__1.0__ subtract__4.0__1.0__ |
subtract__222.0__198.0__ divide__24.0__4.0__ subtract__198.0__6.0__ subtract__6.0__4.0__ subtract__192.0__6.0__ subtract__4.0__1.0__ subtract__4.0__1.0__ |
| three consecutive odd integers are in increasing order such that the sum of the last two integers is num__13 more than the first integer . find the three integers ? <o> a ) num__7 num__9 num__29 <o> b ) num__7 num__2 num__10 <o> c ) num__7 num__9 num__10 <o> d ) num__7 num__9 num__11 <o> e ) num__7 num__9 num__29 |
explanation : let the three consecutive odd integers be x x + num__2 and x + num__4 respectively . x + num__4 + x + num__2 = x + num__13 = > x = num__7 hence three consecutive odd integers are num__7 num__9 and num__11 . answer : d <eor> d <eos> |
d |
subtract__13.0__4.0__ subtract__13.0__2.0__ subtract__9.0__2.0__ |
add__2.0__7.0__ add__2.0__9.0__ subtract__9.0__2.0__ |
| what is the sum of the different positive prime factors of num__1040 ? <o> a ) num__10 <o> b ) num__14 <o> c ) num__15 <o> d ) num__18 <o> e ) num__20 |
i think answer is e : num__20 num__1040 = num__2 * num__2 * num__2 * num__2 * num__5 * num__13 sum of thedifferentpositive prime factors = num__2 + num__5 + num__13 = num__20 <eor> e <eos> |
e |
gcd__1040.0__20.0__ |
gcd__1040.0__20.0__ |
| two pipes a and b can fill a tank in num__15 minutes and num__20 minutes respectively . both the pipes are opened together but after num__4 minutes pipe a is turned off . what is the total time required to fill the tank ? <o> a ) num__12 min . num__40 sec . <o> b ) num__13 min . num__40 sec . <o> c ) num__14 min . num__30 sec . <o> d ) num__14 min . num__20 sec . <o> e ) num__14 min . num__40 sec . |
part filled in num__4 minutes = num__4 ( num__0.0666666666667 + num__0.05 ) = num__0.466666666667 . remaining part = num__1 - num__0.466666666667 = num__0.533333333333 part filled by b in num__1 minute = num__0.05 num__0.05 : num__0.533333333333 : : num__1 : x x = ( num__0.533333333333 x num__1 x num__20 ) = num__10 num__0.666666666667 min = num__10 min . num__40 sec . the tank will be full in ( num__4 min . + num__10 min . + num__40 sec . ) = num__14 min . num__40 sec . answer is e <eor> e <eos> |
e |
multiply__20.0__0.05__ subtract__1.0__0.4667__ divide__10.0__15.0__ multiply__4.0__10.0__ subtract__15.0__1.0__ round__14.0__ |
multiply__20.0__0.05__ subtract__1.0__0.4667__ divide__10.0__15.0__ multiply__4.0__10.0__ subtract__15.0__1.0__ subtract__15.0__1.0__ |
| find the value of a from ( num__15 ) ^ num__2 x num__8 ^ num__3 Ã · num__256 = a . <o> a ) num__250 <o> b ) num__420 <o> c ) num__440 <o> d ) num__650 <o> e ) num__450 |
given exp . = ( num__15 ) ^ num__2 x num__8 ^ num__3 Ã · num__256 = a = num__225 x num__512 Ã · num__256 = num__450 e <eor> e <eos> |
e |
multiply__2.0__256.0__ multiply__2.0__225.0__ multiply__2.0__225.0__ |
multiply__2.0__256.0__ multiply__2.0__225.0__ multiply__2.0__225.0__ |
| the cost of type num__1 material is rs . num__15 per kg and type num__2 material is rs . num__20 per kg . if both type num__1 and type num__2 are mixed in the ratio of num__2 : num__3 then what is the price per kg of the mixed variety of material ? <o> a ) rs . num__19 <o> b ) rs . num__16 <o> c ) rs . num__18 <o> d ) rs . num__17 <o> e ) rs . num__21 |
explanation : solution num__1 cost price ( cp ) of type num__1 material is rs . num__15 per kg cost price ( cp ) of type num__2 material is rs . num__20 per kg type num__1 and type num__2 are mixed in the ratio of num__2 : num__3 . hence cost price ( cp ) of the resultant mixture = ( num__30 + num__60 ) / num__5 = num__18.0 = num__18 = > price per kg of the mixed variety of material = rs . num__18 answer is c <eor> c <eos> |
c |
multiply__15.0__2.0__ multiply__2.0__30.0__ divide__15.0__3.0__ add__15.0__3.0__ multiply__1.0__18.0__ |
multiply__15.0__2.0__ multiply__2.0__30.0__ divide__15.0__3.0__ add__15.0__3.0__ add__15.0__3.0__ |
| sabrina is contemplating a job switch . she is thinking of leaving her job paying $ num__85000 per year to accept a sales job paying $ num__40000 per year plus num__15 percent commission for each sale made . if each of her sales is for $ num__1500 what is the least number of sales she must make per year if she is not to lose money because of the job change ? <o> a ) num__200 <o> b ) num__177 <o> c ) num__178 <o> d ) num__377 <o> e ) num__378 |
lets say she sales x items so her commission will be : x * num__1500 * num__0.15 = num__225 x salary difference is : num__85000 - num__40000 = num__45000 so commission must at least be equal to salary difference i . e . x = num__200.0 = num__200 ( approx ) . hence answer is a <eor> a <eos> |
a |
multiply__1500.0__0.15__ subtract__85000.0__40000.0__ divide__45000.0__225.0__ divide__40000.0__200.0__ |
multiply__1500.0__0.15__ subtract__85000.0__40000.0__ divide__45000.0__225.0__ divide__40000.0__200.0__ |
| the perimeter of a semi circle is num__144 cm then the radius is ? <o> a ) num__87 <o> b ) num__28 <o> c ) num__26 <o> d ) num__27 <o> e ) num__25 |
num__5.14285714286 r = num__144 = > r = num__28 answer : b <eor> b <eos> |
b |
round__28.0__ |
round__28.0__ |
| tabby is training for a triathlon . she swims at a speed of num__1 mile per hour . she runs at a speed of num__10 miles per hour . she wants to figure out her average speed for these two events . what is the correct answer for her ? <o> a ) num__8 mph <o> b ) num__5.5 mph <o> c ) num__3.5 mph <o> d ) num__4 mph <o> e ) num__0.5 mph |
( num__1 mph + num__10 mph ) / num__2 = num__5.5 mph correct option is : b <eor> b <eos> |
b |
round__5.5__ |
divide__5.5__1.0__ |
| using all the letters of the word ` ` wednesday ' ' how many different words can be formed ? <o> a ) a ) num__7 <o> b ) b ) num__9 ! <o> c ) c ) num__8 <o> d ) d ) num__7 ! <o> e ) e ) num__3 |
explanation : total number of letters = num__9 using these letters the number of num__9 letters words formed is num__8 ! . answer : option b <eor> b <eos> |
b |
choose__9.0__8.0__ |
choose__9.0__8.0__ |
| if num__4 < x < num__6 < y < num__10 then what is the greatest possible positive integer difference of x and y ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
num__4 < x < num__6 < y < num__10 ; num__4 < x y < num__10 num__4 + y < x + num__10 y - x < num__6 . positive integer difference is num__5 ( for example y = num__9.5 and x = num__4.5 ) answer : c . <eor> c <eos> |
c |
subtract__9.5__5.0__ subtract__10.0__5.0__ |
subtract__9.5__5.0__ subtract__10.0__5.0__ |
| in a certain school there are num__80 freshmen num__100 sophomores and num__220 upperclassmen drawn from three cities : a b and c . sixty percent of students are from a num__30.0 from b the rest from c and all these students from c are freshmen . half the student from b are upperclassman and the rest are split evenly between the other two grades . how many sophomores are from a ? <o> a ) num__60 <o> b ) num__70 <o> c ) num__80 <o> d ) num__90 <o> e ) num__100 |
answer = b = num__70 <eor> b <eos> |
b |
percent__100.0__70.0__ |
percent__100.0__70.0__ |
| a chair is bought for rs . num__200 / - and sold at rs . num__320 / - find gain or loss percentage <o> a ) num__75.0 loss <o> b ) num__20.0 gain <o> c ) num__60.0 gain <o> d ) num__30.0 gain <o> e ) num__35.0 gain |
formula = ( selling price ~ cost price ) / cost price * num__100 = ( num__320 - num__200 ) / num__200 = num__60.0 gain c <eor> c <eos> |
c |
percent__100.0__60.0__ |
percent__100.0__60.0__ |
| if the selling price of num__50 articles is equal to the cost price of num__45 articles then the loss or gain percent is : <o> a ) num__45.0 <o> b ) num__10.0 <o> c ) num__20.0 <o> d ) num__60.0 <o> e ) num__56 % |
c . p . of each article be re . num__1 . then c . p . of num__50 articles = rs . num__50 ; s . p . of num__50 articles = rs . num__45 . loss % = num__0.1 * num__100 = num__10.0 answer b <eor> b <eos> |
b |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| to save money arkadelphia cream cheese will reduce each dimension of its rectangular box container ( which is entirely full of cream cheese ) by num__75.0 and reduce the price it charges its consumers by num__75.0 as well . by what percentage does this increase the price - per - cubic - inch that each consumer will pay for cream cheese ? <o> a ) num__1 . no change <o> b ) num__2 . num__50.0 <o> c ) num__3 . num__100.0 <o> d ) num__4 . num__300.0 <o> e ) num__5 . num__1500 % |
l = num__20 : b = num__10 : h = num__10 of initial box and price = num__100 $ therefore price / cubic inch = num__100 / ( num__20 * num__10 * num__10 ) = num__0.05 now when dimensions are reduced by num__75.0 and price also reduced by num__75.0 l = num__5 ; b = num__2.5 ; h = num__2.5 and price = num__25 $ therefore price / cubic inch = num__25 / ( num__5 * num__2.5 * num__2.5 ) = num__0.8 percentage change = ( num__0.8 - num__0.05 ) * num__100 / num__0.05 = num__1500.0 answer is e <eor> e <eos> |
e |
reverse__20.0__ multiply__100.0__0.05__ multiply__2.5__10.0__ divide__20.0__25.0__ divide__75.0__0.05__ multiply__100.0__0.05__ |
reverse__20.0__ multiply__100.0__0.05__ multiply__2.5__10.0__ divide__20.0__25.0__ divide__75.0__0.05__ multiply__100.0__0.05__ |
| mahesh can do a piece of work in num__40 days . he works at it for num__20 days and then rajesh finished it in num__30 days . how long will y take to complete the work ? <o> a ) num__45 <o> b ) num__25 <o> c ) num__37 <o> d ) num__41 <o> e ) num__60 |
work done by mahesh in num__60 days = num__20 * num__0.025 = num__0.5 remaining work = num__1 - num__0.5 = num__0.5 num__0.5 work is done by rajesh in num__30 days whole work will be done by rajesh is num__30 * num__2 = num__60 days answer is e <eor> e <eos> |
e |
hour_to_min_conversion__ divide__20.0__40.0__ multiply__40.0__0.025__ divide__40.0__20.0__ hour_to_min_conversion__ |
hour_to_min_conversion__ multiply__20.0__0.025__ multiply__40.0__0.025__ divide__40.0__20.0__ multiply__30.0__2.0__ |
| the present ages of three persons in proportions num__4 : num__1 : num__3 . eight years ago the sum of their ages was num__56 . find their present ages ( in years ) . <o> a ) num__8 num__20 num__28 <o> b ) num__16 num__28 num__36 <o> c ) num__16 num__28 num__35 <o> d ) num__40 num__10 num__30 <o> e ) num__16 num__28 num__33 |
let their present ages be num__4 x x and num__3 x years respectively . then ( num__4 x - num__8 ) + ( x - num__8 ) + ( num__3 x - num__8 ) = num__56 num__8 x = num__80 x = num__10 . their present ages are num__4 x = num__40 years num__1 x = num__10 years and num__3 x = num__30 years respectively . answer : d <eor> d <eos> |
d |
divide__80.0__8.0__ multiply__4.0__10.0__ multiply__3.0__10.0__ multiply__4.0__10.0__ |
divide__80.0__8.0__ multiply__4.0__10.0__ subtract__40.0__10.0__ add__10.0__30.0__ |
| a dishonest dealer professes to sell his goods at cost price but still gets num__15.0 profit by using a false weight . what weight does he substitute for a kilogram ? <o> a ) num__869 num__0.142857142857 grams <o> b ) num__869 num__0.565217391304 grams <o> c ) num__869 num__2.66666666667 grams <o> d ) num__869 num__0.333333333333 grams <o> e ) num__832 num__0.0434782608696 grams |
if the cost price is rs . num__100 then to get a profit of num__15.0 the selling price should be rs . num__115 . if num__115 kg are to be sold and the dealer gives only num__100 kg to get a profit of num__15.0 . how many grams he has to give instead of one kilogram ( num__1000 gm ) . num__115 gm - - - - - - num__100 gm num__1000 gm - - - - - - ? ( num__1000 * num__100 ) / num__115 = num__869 num__0.565217391304 grams . answer : b <eor> b <eos> |
b |
percent__100.0__869.0__ |
percent__100.0__869.0__ |
| two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of num__22 kmph and num__21 kmph respectively . when they meet it is found that one train has traveled num__60 km more than the other one . the distance between the two stations is ? <o> a ) num__288 <o> b ) num__516 <o> c ) num__877 <o> d ) num__278 <o> e ) num__178 |
num__1 h - - - - - num__5 ? - - - - - - num__60 num__12 h rs = num__22 + num__21 = num__43 t = num__12 d = num__43 * num__12 = num__516 answer : b <eor> b <eos> |
b |
subtract__22.0__21.0__ divide__60.0__5.0__ add__22.0__21.0__ multiply__43.0__12.0__ round__516.0__ |
subtract__22.0__21.0__ divide__60.0__5.0__ add__22.0__21.0__ multiply__43.0__12.0__ multiply__43.0__12.0__ |
| on a num__20 mile course pat bicycled at an average rate of num__30 miles per hour for the first num__12 minutes and without a break ran the rest of the distance at an average rate of num__8 miles per hour . how many minutes did pat take to cover the entire course ? <o> a ) num__75 <o> b ) num__105 <o> c ) num__117 <o> d ) num__150 <o> e ) num__162 |
at an average rate of num__30 miles per hour in num__12 minute ( num__0.2 hours ) pat covers ( distance ) = ( time ) * ( rate ) = num__0.2 * num__30 = num__6 miles thus she should cover the remaining distance of num__20 - num__6 = num__14 miles at an average rate of num__8 miles per hour . to cover num__14 miles at an average rate of num__8 miles per hour pat needs ( time ) = ( distance ) / ( rate ) = num__1.75 = num__1.75 hours = num__105 minutes . therefore pat needs total of num__12 + num__105 = num__117 minutes to cover the entire course . answer : c . <eor> c <eos> |
c |
multiply__30.0__0.2__ subtract__20.0__6.0__ divide__14.0__8.0__ add__12.0__105.0__ round__117.0__ |
multiply__30.0__0.2__ subtract__20.0__6.0__ divide__14.0__8.0__ add__12.0__105.0__ add__12.0__105.0__ |
| to fill a tank num__25 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to num__0.2 of its present ? <o> a ) num__61.5 <o> b ) num__60.5 <o> c ) num__63.5 <o> d ) num__62.5 <o> e ) num__125 |
let capacity of num__1 bucket = x capacity of the tank = num__25 x new capacity of the bucket = num__1 x / num__5 hence number of buckets needed = num__25 x / ( num__1 x / num__5 ) = ( num__25 × num__5 ) / num__1 = num__125 answer is e . <eor> e <eos> |
e |
multiply__25.0__0.2__ divide__25.0__0.2__ round__125.0__ |
divide__1.0__0.2__ divide__25.0__0.2__ divide__25.0__0.2__ |
| murali travelled from city a to city b at a speed of num__40 kmph and from city b to city c at num__60 kmph . what is the average speed of murali from a to c given that the ratio of distances between a to b and b to c is num__2 : num__3 ? a . num__48 kmph <o> a ) num__33 <o> b ) num__50 <o> c ) num__28 <o> d ) num__27 <o> e ) num__16 |
let the distances between city a to b and b to c be num__2 x km and num__3 x km respectively . total time taken to cover from a to c = ( num__2 x ) / num__40 + ( num__3 x ) / num__60 = ( num__6 x + num__6 x ) / num__120 = num__12 x / num__120 = x / num__10 average speed = ( num__2 x + num__3 x ) / ( x / num__10 ) = num__50 kmph . answer : b <eor> b <eos> |
b |
multiply__2.0__3.0__ multiply__40.0__3.0__ subtract__60.0__48.0__ divide__60.0__6.0__ add__40.0__10.0__ round__50.0__ |
multiply__2.0__3.0__ multiply__40.0__3.0__ subtract__60.0__48.0__ divide__60.0__6.0__ add__40.0__10.0__ add__40.0__10.0__ |
| a rectangular grass field is num__75 m * num__40 m it has a path of num__2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . num__2 per sq m ? <o> a ) s . num__1350 <o> b ) s . num__1327 <o> c ) s . num__1200 <o> d ) s . num__1397 <o> e ) s . num__1927 |
area = ( l + b + num__2 d ) num__2 d = ( num__75 + num__40 + num__2.5 * num__2 ) num__2 * num__2.5 = > num__600 num__600 * num__2 = rs . num__1200 answer : c <eor> c <eos> |
c |
multiply__2.0__600.0__ triangle_area__2.0__1200.0__ |
multiply__2.0__600.0__ multiply__2.0__600.0__ |
| find the missing number : num__6 num__13 num__25 num__51 num__101 ? <o> a ) num__201 <o> b ) num__202 <o> c ) num__203 <o> d ) num__204 <o> e ) num__205 |
c num__203 num__6 x num__2 + num__1 = num__13 num__13 x num__2 - num__1 = num__25 num__25 x num__2 + num__1 = num__51 num__51 x num__2 - num__1 = num__101 num__101 x num__2 + num__1 = num__203 <eor> c <eos> |
c |
multiply__1.0__203.0__ |
multiply__1.0__203.0__ |
| there are num__6 people in the elevator . their average weight is num__150 lbs . another person enters the elevator and increases the average weight to num__151 lbs . what is the weight of the num__7 th person . <o> a ) num__157 <o> b ) num__168 <o> c ) num__189 <o> d ) num__190 <o> e ) num__200 |
solution average of num__7 people after the last one enters = num__151 . â ˆ ´ required weight = ( num__7 x num__151 ) - ( num__6 x num__150 ) = num__1057 - num__900 = num__157 . answer a <eor> a <eos> |
a |
multiply__151.0__7.0__ multiply__6.0__150.0__ add__6.0__151.0__ add__6.0__151.0__ |
multiply__151.0__7.0__ multiply__6.0__150.0__ subtract__1057.0__900.0__ subtract__1057.0__900.0__ |
| simplify : ( num__7 + num__2 ) – ( num__5 + num__3 + num__1 ) - num__1 . <o> a ) - num__1 <o> b ) – num__2 <o> c ) num__1 <o> d ) num__2 <o> e ) num__0 |
solution : ( num__7 + num__2 ) – ( num__5 + num__3 + num__1 ) - num__1 = num__9 - num__5 - num__3 + num__1 - num__1 = num__9 - num__8 + num__1 - num__1 = num__2 - num__1 = num__1 answer : ( c ) <eor> c <eos> |
c |
add__7.0__2.0__ add__7.0__1.0__ reverse__1.0__ |
add__7.0__2.0__ add__7.0__1.0__ subtract__2.0__1.0__ |
| a man has some hens and cows . if the number of heads be num__48 and the number of feet equals num__140 then the number of hens will be : <o> a ) num__26 <o> b ) num__25 <o> c ) num__23 <o> d ) num__22 <o> e ) num__20 |
let the number of hens be x and the number of cows be y . then x + y = num__48 . . . . ( i ) and num__2 x + num__4 y = num__140 x + num__2 y = num__70 . . . . ( ii ) solving ( i ) and ( ii ) we get : x = num__26 y = num__22 . the required answer = num__26 . answer is a . <eor> a <eos> |
a |
divide__140.0__2.0__ subtract__48.0__26.0__ subtract__48.0__22.0__ |
divide__140.0__2.0__ subtract__48.0__26.0__ add__4.0__22.0__ |
| the average expenditure of a labourer for num__3 months was num__85 and he fell into debt . in the next num__4 months by reducing his monthly expenses to num__60 he not only cleared off his debt but also saved num__30 . his monthly income is <o> a ) num__70.2 <o> b ) num__50.1 <o> c ) num__75.2 <o> d ) num__78.4 <o> e ) num__52.5 |
income of num__3 months = ( num__3 × num__85 ) – debt = num__255 – debt income of the man for next num__4 months = num__4 × num__60 + debt + num__30 = num__270 + debt ∴ income of num__10 months = num__525 average monthly income = num__525 ÷ num__10 = num__52.5 answer e <eor> e <eos> |
e |
multiply__3.0__85.0__ divide__30.0__3.0__ add__270.0__255.0__ divide__525.0__10.0__ divide__525.0__10.0__ |
multiply__3.0__85.0__ divide__30.0__3.0__ add__270.0__255.0__ divide__525.0__10.0__ divide__525.0__10.0__ |
| if a + b + c = num__21 what is the total number of non - negative integral solutions ? <o> a ) num__223 <o> b ) num__233 <o> c ) num__240 <o> d ) num__243 <o> e ) num__253 |
think of it is a problem of diving num__21 items to three people a b & c which may or may not receive an item . so simply use combination here as ( n - r + num__1 ) c ( r - num__1 ) i . e . ( num__21 + num__3 - num__1 ) c ( num__3 - num__1 ) = > num__23 c num__2 = = > num__253 solutions . answer : e <eor> e <eos> |
e |
subtract__3.0__1.0__ multiply__1.0__253.0__ |
subtract__3.0__1.0__ multiply__1.0__253.0__ |
| if the sum of a number and its square is num__72 what is the number ? <o> a ) num__15 <o> b ) num__26 <o> c ) num__8 <o> d ) num__91 <o> e ) none of these |
let the number be x . then x + x num__2 = num__72 ( x + num__9 ) ( x - num__8 ) = num__0 x = num__8 answer : c <eor> c <eos> |
c |
square_perimeter__2.0__ square_perimeter__2.0__ |
square_perimeter__2.0__ square_perimeter__2.0__ |
| a number when divided successively by num__4 and num__5 leaves remainder num__1 and num__4 respectively . when it is successively divided by num__5 and num__4 then the respective remainders will be ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
trying to get back the number first . num__5 x + num__4 num__4 * [ num__5 x + num__4 ] + num__1 so number is num__20 x + num__17 now divide it successively by num__5 and then num__4 ( num__20 x + num__17 ) / num__5 = num__2 remainder and num__4 x + num__3 is number left now divide it by num__4 we will get num__3 remainder answer : d <eor> d <eos> |
d |
multiply__4.0__5.0__ subtract__4.0__1.0__ subtract__4.0__1.0__ |
multiply__4.0__5.0__ add__1.0__2.0__ add__1.0__2.0__ |
| a trader sold an article on a certain price with num__20.0 profit . if he sold double of previous selling price then find its profit % <o> a ) num__140.0 <o> b ) num__120.0 <o> c ) num__100.0 <o> d ) num__90.0 <o> e ) num__80 % |
let cost price = num__100.0 selling price = num__120.0 new s . p . = num__240.0 p % = num__240 - num__100 = num__140.0 answer is a <eor> a <eos> |
a |
percent__100.0__140.0__ |
percent__100.0__140.0__ |
| a contractor undertakes to complete the construction of a tunnel num__720 meters long in num__240 days and employs num__50 men for the purpose . after num__120 days he finds that only num__240 meters of the tunnel is complete . how many more men should be employ in order to complete the work in time ? <o> a ) num__75 men <o> b ) num__80 men <o> c ) num__70 men <o> d ) num__60 men <o> e ) num__50 men |
in num__120 days only num__240 m of the tunnel is constructed by num__50 men . the remaining num__120 days num__480 m of the tunnel can be constructed by num__120 men . additional number of men required = num__120 - num__50 = num__70 men . answer : c <eor> c <eos> |
c |
subtract__720.0__240.0__ subtract__120.0__50.0__ round__70.0__ |
subtract__720.0__240.0__ subtract__120.0__50.0__ subtract__120.0__50.0__ |
| if num__9 - num__4 / x = num__7 + num__8 / x then x = <o> a ) num__15 <o> b ) num__6 <o> c ) num__1 <o> d ) num__5 <o> e ) num__12 |
we ' re given the equation num__9 - num__4 / x = num__7 + num__8 / x . we ' re asked for the value of x . the common - denominator of these num__4 numbers is x so we have to multiple num__2 of the numbers to give them that denominator . . . num__9 x - num__4 x / x = num__7 x + num__8 x / x we can then eliminate that denominator which gives us . . . . num__9 x - num__4 = num__7 x + num__8 num__12 = num__2 x both num__2 x and num__12 have a common denominator of num__2 so we divide both sides by num__2 for our final answer . . . num__6 = x b <eor> b <eos> |
b |
subtract__9.0__7.0__ add__4.0__8.0__ add__4.0__2.0__ add__4.0__2.0__ |
subtract__9.0__7.0__ add__4.0__8.0__ add__4.0__2.0__ add__4.0__2.0__ |
| two trains of equal length are running on parallel lines in the same directions at num__46 km / hr . and num__36 km / hr . the faster trains pass the slower train in num__18 seconds . the length of each train is : <o> a ) num__25 m <o> b ) num__50 m <o> c ) num__72 m <o> d ) num__80 m <o> e ) none of these |
explanation : the relative speed of train is num__46 - num__36 = num__10 km / hr = ( num__10 x num__5 ) / num__18 = num__2.77777777778 m / s num__10 × num__518 = num__259 m / s in num__18 secs the total distance traveled is num__18 x num__2.77777777778 = num__50 m . therefore the length of each train is = num__25.0 = num__25 m . answer a <eor> a <eos> |
a |
subtract__46.0__36.0__ multiply__5.0__10.0__ subtract__50.0__25.0__ |
subtract__46.0__36.0__ multiply__5.0__10.0__ subtract__50.0__25.0__ |
| find large no . from below question the difference of two numbers is num__1365 . on dividing the larger number by the smaller we get num__6 as quotient and the num__5 as remainder <o> a ) num__1637 <o> b ) num__1456 <o> c ) num__1567 <o> d ) num__1678 <o> e ) num__1635 |
let the smaller number be x . then larger number = ( x + num__1365 ) . x + num__1365 = num__6 x + num__5 num__5 x = num__1360 x = num__272 large number = num__272 + num__1365 = num__1637 a <eor> a <eos> |
a |
subtract__1365.0__5.0__ divide__1360.0__5.0__ add__1365.0__272.0__ add__1365.0__272.0__ |
subtract__1365.0__5.0__ divide__1360.0__5.0__ add__1365.0__272.0__ add__1365.0__272.0__ |
| how much interest can a person get on rs . num__8200 at num__17.5 p . a . simple interest for a period of two years and six months ? <o> a ) num__3000 <o> b ) num__4000.75 <o> c ) num__3587.5 <o> d ) num__3500 <o> e ) num__5020.75 |
i = ( num__8200 * num__2.5 * num__17.5 ) / num__100 = ( num__8200 * num__5 * num__35 ) / ( num__100 * num__2 * num__2 ) = rs . num__3587.50 answer c <eor> c <eos> |
c |
percent__100.0__3587.5__ |
percent__100.0__3587.5__ |
| the average of first five multiples of num__3 is ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__9 <o> d ) num__5 <o> e ) num__7 |
average = num__3 ( num__1 + num__2 + num__3 + num__4 + num__5 ) / num__5 = num__9.0 = num__9 . answer : b <eor> b <eos> |
b |
subtract__3.0__1.0__ add__3.0__1.0__ add__3.0__2.0__ add__4.0__5.0__ add__3.0__5.0__ |
subtract__3.0__1.0__ add__3.0__1.0__ add__3.0__2.0__ add__4.0__5.0__ add__3.0__5.0__ |
| the milk level in a rectangular box measuring num__58 feet by num__25 feet is to be lowered by num__6 inches . how many gallons of milk must be removed ? ( num__1 cu ft = num__7.5 gallons ) <o> a ) num__100 <o> b ) num__250 <o> c ) num__750 <o> d ) num__5437.5 <o> e ) num__5635.5 |
num__6 inches = num__0.5 feet ( there are num__12 inches in a foot . ) so num__58 * num__25 * num__0.5 = num__725 feet ^ num__3 of milk must be removed which equals to num__725 * num__7.5 = num__5437.5 gallons . answer : d . <eor> d <eos> |
d |
divide__6.0__0.5__ multiply__6.0__0.5__ multiply__7.5__725.0__ multiply__1.0__5437.5__ |
divide__6.0__0.5__ multiply__6.0__0.5__ multiply__7.5__725.0__ multiply__1.0__5437.5__ |
| a rectangular field is to be fenced on three sides leaving a side of num__20 feet uncovered . if the area of the field is num__390 sq . feet how many feet of fencing will be required ? <o> a ) num__34 <o> b ) num__59 <o> c ) num__68 <o> d ) num__88 <o> e ) num__92 |
given that length and area so we can find the breadth . length x breadth = area num__20 x breadth = num__390 breadth = num__19.5 feet area to be fenced = num__2 b + l = num__2 ( num__19.5 ) + num__20 = num__59 feet answer : b <eor> b <eos> |
b |
triangle_area__2.0__59.0__ |
triangle_area__2.0__59.0__ |
| find the largest number of four digits which is exactly divisible by num__2718 num__1215 <o> a ) num__9700 <o> b ) num__9710 <o> c ) num__9720 <o> d ) num__9730 <o> e ) num__9740 |
explanation : lcm of num__27 - num__18 - num__12 - num__15 is num__540 . after dividing num__9999 by num__540 we get num__279 remainder . so answer will be num__9999 - num__279 = num__9720 option c <eor> c <eos> |
c |
subtract__27.0__12.0__ subtract__9999.0__279.0__ subtract__9999.0__279.0__ |
subtract__27.0__12.0__ subtract__9999.0__279.0__ subtract__9999.0__279.0__ |
| the length of a rectangular plot is num__20 metres more than its breadth . if the cost of fencing the plot @ rs . num__26.50 per metre is rs . num__5300 what is the length of the plot in metres ? <o> a ) num__20 <o> b ) num__200 <o> c ) num__300 <o> d ) num__400 <o> e ) num__140 |
let length of plot = l meters then breadth = l - num__20 meters and perimeter = num__2 [ l + l - num__20 ] = [ num__4 l - num__40 ] meters [ num__4 l - num__40 ] * num__26.50 = num__5300 [ num__4 l - num__40 ] = num__5300 / num__26.50 = num__200 num__4 l = num__240 l = num__60.0 = num__60 meters . answer : b <eor> b <eos> |
b |
multiply__20.0__2.0__ divide__5300.0__26.5__ add__40.0__200.0__ hour_to_min_conversion__ round__200.0__ |
multiply__20.0__2.0__ divide__5300.0__26.5__ add__40.0__200.0__ add__20.0__40.0__ divide__5300.0__26.5__ |
| a bakery opened yesterday with its daily supply of num__60 dozen rolls . half of the rolls were sold by noon and num__80 percent of the remaining rolls were sold between noon and closing time . how many dozen rolls had not been sold when the bakery closed yesterday ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__6 |
answer is e . . . num__6 dozens . . . . half sold by noon - - > num__30 dozens ( num__30 * num__12 = num__360 ) remaining - - > num__30 dozens i . e num__360 eggs . . . num__80.0 sold - - > num__80 * num__3.6 = num__288 eggs remaining num__20.0 - - > num__72 eggs ( num__6 dozens ) - - > answer <eor> e <eos> |
e |
percent__60.0__6.0__ percent__80.0__360.0__ percent__20.0__360.0__ percent__20.0__30.0__ |
percent__60.0__6.0__ percent__80.0__360.0__ percent__20.0__360.0__ percent__20.0__30.0__ |
| pipe a can fill in num__20 minutes and pipe b in num__30 mins and pipe c can empty the same in num__40 mins . if all of them work together find the time taken to fill the tank <o> a ) num__17 num__0.142857142857 mins <o> b ) num__20 mins <o> c ) num__8 mins <o> d ) num__9 mins <o> e ) none of these |
a = num__20 b = num__30 c = - num__40 lcm = num__120 = > a = num__6 times b = num__4 c = - num__3 = > a + b - c = num__7 time taken to fill the tank = num__17.1428571429 mins = num__17 num__0.142857142857 mins answer : a <eor> a <eos> |
a |
divide__120.0__20.0__ divide__120.0__30.0__ divide__120.0__40.0__ add__3.0__4.0__ divide__120.0__7.0__ subtract__20.0__3.0__ subtract__17.1429__17.0__ round__17.0__ |
divide__120.0__20.0__ divide__120.0__30.0__ divide__120.0__40.0__ add__3.0__4.0__ divide__120.0__7.0__ subtract__20.0__3.0__ subtract__17.1429__17.0__ subtract__20.0__3.0__ |
| a shopkeeper wants to make a minimum net profit of num__5.0 on a pair of jeans she is selling which she bought for $ num__100 . if originally she was selling the pair at num__20.0 gain what is the largest discount she can allow ? <o> a ) num__12.5 <o> b ) num__8.0 <o> c ) num__8.3 <o> d ) num__15.0 <o> e ) num__4 % |
a if she bought the jeans at $ num__100 she is currently selling them at ( num__1.2 ) * num__100 = $ num__120 . she wants to make a profit of num__5.0 which means she needs to sell them at ( num__1.05 ) * num__100 = $ num__105 the max discount she can allow is thus ( ( num__120 - num__105 ) / num__120 ) * num__100 = num__12.5 <eor> a <eos> |
a |
percent__100.0__12.5__ |
percent__100.0__12.5__ |
| a girl goes to her school from her house at a speed of num__6 km / hr and returns at a speed of num__4 km / hr . if she takes num__10 hours in going and coming back the distance between her school and house is <o> a ) num__12 km <o> b ) num__16 km <o> c ) num__20 km <o> d ) num__24 km <o> e ) none of these |
explanation : let distance be d num__10 = d / num__4 + d / num__6 answer – d <eor> d <eos> |
d |
multiply__6.0__4.0__ |
multiply__6.0__4.0__ |
| students at a school were on average num__180 cm tall . the average female height was num__170 cm and the average male height was num__182 cms . what was the ratio of men to women ? <o> a ) num__5 : num__2 <o> b ) num__5 : num__1 <o> c ) num__4 : num__3 <o> d ) num__4 : num__1 <o> e ) num__3 : num__1 |
we ' re given a few facts to work with : num__1 ) the average height of the females is num__170 cm num__2 ) the average height of the males is num__182 cm num__3 ) the average of the group is num__180 cm we ' re asked for the ratio of men to women . w = number of women m = number of men ( num__170 w + num__182 m ) / ( w + m ) = num__180 num__170 w + num__182 m = num__180 w + num__180 m num__2 m = num__10 w m = num__5 w m / w = num__5.0 the ratio of men to women is num__5 to num__1 . b <eor> b <eos> |
b |
subtract__182.0__180.0__ add__1.0__2.0__ subtract__180.0__170.0__ add__2.0__3.0__ multiply__1.0__5.0__ |
subtract__182.0__180.0__ add__1.0__2.0__ subtract__180.0__170.0__ add__2.0__3.0__ add__2.0__3.0__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__120 m and they cross each other in num__12 sec then the speed of each train is ? <o> a ) num__10 <o> b ) num__18 <o> c ) num__63 <o> d ) num__36 <o> e ) num__72 |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__120 + num__120 ) / num__12 = > x = num__10 speed of each train = num__10 m / sec . = num__10 * num__3.6 = - num__36 km / hr . answer : d <eor> d <eos> |
d |
divide__120.0__12.0__ multiply__10.0__3.6__ round__36.0__ |
divide__120.0__12.0__ multiply__10.0__3.6__ multiply__10.0__3.6__ |
| evaluate : num__30 - num__12 ÷ num__3 × num__2 = <o> a ) num__11 <o> b ) num__22 <o> c ) num__33 <o> d ) num__44 <o> e ) num__55 |
according to order of operations num__12 ÷ num__3 × num__2 ( division and multiplication ) is done first from left to right num__12 ÷ num__3 × num__2 = num__4 × num__2 = num__8 hence num__30 - num__12 ÷ num__3 × num__2 = num__30 - num__8 = num__22 correct answer is b ) num__22 <eor> b <eos> |
b |
divide__12.0__3.0__ subtract__12.0__4.0__ subtract__30.0__8.0__ subtract__30.0__8.0__ |
divide__12.0__3.0__ subtract__12.0__4.0__ subtract__30.0__8.0__ subtract__30.0__8.0__ |
| a train running at the speed of num__72 km / hr crosses a pole in num__8 seconds . find the length of the train . <o> a ) num__150 meter <o> b ) num__145 meter <o> c ) num__160 meter <o> d ) num__135 meter <o> e ) none of these |
explanation : speed = num__72 * ( num__0.277777777778 ) m / sec = num__20 m / sec length of train ( distance ) = speed * time = num__20 * num__8 = num__160 meter option c <eor> c <eos> |
c |
multiply__8.0__20.0__ round__160.0__ |
multiply__8.0__20.0__ multiply__8.0__20.0__ |
| on the independence day bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day num__360 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ? <o> a ) num__600 <o> b ) num__620 <o> c ) num__500 <o> d ) num__520 <o> e ) num__720 |
let the number of children in the school be x . since each child gets num__2 bananas total number of bananas = num__2 x . num__2 x / ( x - num__360 ) = num__2 + num__2 ( extra ) = > num__2 x - num__720 = x = > x = num__720 . answer : e <eor> e <eos> |
e |
multiply__360.0__2.0__ multiply__360.0__2.0__ |
multiply__360.0__2.0__ multiply__360.0__2.0__ |
| the average weight of num__25 girls increases by num__1 kg when a new girl comes in place of one of them weighing num__55 kg . what might be the weight of the new girl ? <o> a ) num__85 kg <o> b ) num__90 kg <o> c ) num__83 kg <o> d ) num__80 kg <o> e ) num__82 kg |
total weight increased = num__25 x num__1 kg = num__25 kg . weight of new person = num__55 + num__25 kg = num__80 kg answer : d <eor> d <eos> |
d |
add__25.0__55.0__ add__25.0__55.0__ |
add__25.0__55.0__ add__25.0__55.0__ |
| a father said to his son ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is num__44 years now the son ' s age five years back was : <o> a ) a ) num__14 <o> b ) b ) num__17 <o> c ) c ) num__56 <o> d ) d ) num__89 <o> e ) e ) num__34 |
let the son ' s present age be x years . then ( num__44 - x ) = x num__2 x = num__44 . x = num__22 . son ' s age num__5 years back ( num__22 - num__5 ) = num__17 years . answer : b <eor> b <eos> |
b |
divide__44.0__2.0__ subtract__22.0__5.0__ subtract__22.0__5.0__ |
divide__44.0__2.0__ subtract__22.0__5.0__ subtract__22.0__5.0__ |
| the sum of ages of num__5 children born at the intervals of num__3 years each is num__50 years . what is the age of the youngest child ? <o> a ) num__4 years <o> b ) num__8 years <o> c ) num__10 years <o> d ) num__12 years <o> e ) none of these |
explanation : let the ages of children be x ( x + num__3 ) ( x + num__6 ) ( x + num__9 ) and ( x + num__12 ) years . then x + ( x + num__3 ) + ( x + num__6 ) + ( x + num__9 ) + ( x + num__12 ) = num__50 num__5 x = num__20 x = num__4 . age of the youngest child = x = num__4 years . answer : a <eor> a <eos> |
a |
add__3.0__6.0__ add__3.0__9.0__ subtract__9.0__5.0__ subtract__9.0__5.0__ |
add__3.0__6.0__ add__3.0__9.0__ subtract__9.0__5.0__ subtract__9.0__5.0__ |
| the difference between a two - digit number and the number obtained by interchanging the digit is num__36 . what is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is num__1 : num__2 ? <o> a ) num__4 <o> b ) num__8 <o> c ) num__16 <o> d ) num__18 <o> e ) none of these |
explanation : since the number is greater than the number obtained on reversing the digits so the ten ' s digit is greater than the unit ' s digit . let the ten ' s and unit ' s digits be num__2 x and x respectively . then ( num__10 * num__2 x + x ) - ( num__10 x + num__2 x ) = num__36 num__9 x = num__36 x = num__4 required difference = ( num__2 x + x ) - ( num__2 x - x ) = num__2 x = num__8 . answer is b <eor> b <eos> |
b |
subtract__10.0__1.0__ divide__36.0__9.0__ multiply__2.0__4.0__ multiply__1.0__8.0__ |
subtract__10.0__1.0__ divide__36.0__9.0__ multiply__2.0__4.0__ multiply__1.0__8.0__ |
| find the area of a parallelogram with base num__20 cm and height num__10 cm ? <o> a ) num__287 cm num__2 <o> b ) num__887 cm num__2 <o> c ) num__200 cm num__2 <o> d ) num__250 cm num__2 <o> e ) num__668 cm num__2 |
area of a parallelogram = base * height = num__20 * num__10 = num__200 cm num__2 answer : c <eor> c <eos> |
c |
multiply__20.0__10.0__ multiply__20.0__10.0__ |
multiply__20.0__10.0__ multiply__20.0__10.0__ |
| a rectangular park num__60 m long and num__40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn . if the area of the lawn is num__2109 sq . m then what is the width of the road ? <o> a ) num__2.2 m <o> b ) num__2.4 m <o> c ) num__3 m <o> d ) num__3.4 m <o> e ) num__3.6 m |
area of the park = ( num__60 x num__40 ) m num__2 = num__2400 m num__2 . area of the lawn = num__2109 m num__2 . area of the crossroads = ( num__2400 - num__2109 ) m num__2 = num__291 m num__2 . let the width of the road be x metres . then num__60 x + num__40 x - x num__2 = num__291 x num__2 - num__100 x + num__291 = num__0 ( x - num__97 ) ( x - num__3 ) = num__0 x = num__3 . answer : c <eor> c <eos> |
c |
multiply__60.0__40.0__ triangle_area__2.0__3.0__ |
multiply__60.0__40.0__ triangle_area__2.0__3.0__ |
| one night num__10 percent of the female officers on a police force were on duty . if num__200 officers were on duty that night and half of these were female officers how many female officers were on the police force ? <o> a ) num__90 <o> b ) num__180 <o> c ) num__270 <o> d ) num__500 <o> e ) num__1 |
000 |
let total number of female officers in the police force = f total number of officers on duty on that night = num__200 number of female officers on duty on that night = num__100.0 = num__100 ( num__0.1 ) * f = num__100 = > f = num__1000 answer e <eor> e <eos> |
e |
e |
| the sum an integer n and its reciprocal is equal to num__5.2 . what is the value of n ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
write equation in n as follows n + num__1 / n = num__5.2 multiply all terms by n obtain a quadratic equation and solve to obtain n = num__5 . correct answer is e ) num__5 <eor> e <eos> |
e |
round_down__5.2__ round_down__5.2__ |
round_down__5.2__ divide__5.0__1.0__ |
| num__60.0 of ram ' s marks is equal to num__20.0 of rahim ' s marks which percent is equal to num__30.0 of robert ' s marks . if robert ' s marks is num__80 then find the average marks of ram and rahim ? <o> a ) num__70 <o> b ) num__97 <o> c ) num__80 <o> d ) num__90 <o> e ) num__76 |
given num__60.0 of ram ' s marks = num__20.0 of rahim ' s marks = num__30.0 of robert ' s marks . given marks of robert = num__80 num__30.0 of num__80 = num__0.3 * num__8 = num__24 given num__60.0 of ram ' s marks = num__24 . = > ram ' s marks = ( num__24 * num__100 ) / num__60 = num__40 also num__20.0 of rahim ' s marks = num__24 = > rahim ' s marks = ( num__24 * num__100 ) / num__20 = num__120 average marks of ram and rahim = ( num__40 + num__120 ) / num__2 = num__80 . answer : c <eor> c <eos> |
c |
multiply__80.0__0.3__ add__20.0__80.0__ subtract__60.0__20.0__ add__20.0__100.0__ divide__60.0__30.0__ add__60.0__20.0__ |
multiply__80.0__0.3__ add__20.0__80.0__ subtract__60.0__20.0__ add__20.0__100.0__ divide__60.0__30.0__ add__60.0__20.0__ |
| a standard veggiematik machine can chop num__25 carrots in num__5 minutes . how many carrots can num__5 standard veggiematik machines chop in num__5 minutes ? <o> a ) num__45 <o> b ) num__75 <o> c ) num__110 <o> d ) num__125 <o> e ) num__150 |
direct relationship : - num__1 standard veggiematik machine - num__25 carrots - num__5 minutes num__1 standard veggiematik machine - num__5 carrots - num__1 minute now num__5 standard veggiematik machine - ? carrots - num__5 minutes hence = num__5 x num__5 x num__5 = num__125 carrots answer d <eor> d <eos> |
d |
multiply__25.0__5.0__ round__125.0__ |
multiply__25.0__5.0__ round__125.0__ |
| amir has num__3 desert every sunday . . . he stocked up num__10 different chocolates and num__12 diff icecreams given that he never has same combination and does n ' t num__3 chocolates nor num__3 icecreams how many sundays it requires for his stock to be over ? <o> a ) num__1045 <o> b ) num__1200 <o> c ) num__2400 <o> d ) num__7200 <o> e ) num__3600 |
he either can have num__2 chocolates and num__1 icecream or num__1 chocolate and num__2 icecream . so the solution would be . num__10 c num__2 * num__12 c num__1 + num__10 c num__1 * num__12 c num__2 = num__1200 answer : b <eor> b <eos> |
b |
subtract__12.0__10.0__ subtract__3.0__2.0__ multiply__1200.0__1.0__ |
subtract__12.0__10.0__ subtract__3.0__2.0__ multiply__1200.0__1.0__ |
| find the average of first num__27 natural numbers . <o> a ) num__14 <o> b ) num__14.2 <o> c ) num__15.2 <o> d ) num__15.8 <o> e ) num__28 |
explanation : sum of first n natural numbers = n ( n + num__1 ) / num__2 hence sum of first num__27 natural numbers = ( num__27 x num__28 ) / num__2 = num__378 therefore required average of = num__14.0 = num__14 answer : a <eor> a <eos> |
a |
add__27.0__1.0__ divide__378.0__27.0__ multiply__1.0__14.0__ |
add__27.0__1.0__ divide__378.0__27.0__ divide__14.0__1.0__ |
| a leak in the bottom of a tank can empty the full tank in num__6 hours . an inlet pipe fills water at the rate of num__4 litres a minute . when the tank is full the inlet is opened and due to the leak the tank is empty in num__8 hours . the capacity of the tank ( in litres ) is <o> a ) num__5780 litres <o> b ) num__5770 litres <o> c ) num__5760 litres <o> d ) num__5750 litres <o> e ) num__5740 litres |
explanation : work done by the inlet in num__1 hour = num__0.166666666667 − num__0.125 = num__0.0416666666667 work done by inlet in num__1 min = num__0.0416666666667 ∗ num__0.0166666666667 = num__0.000694444444444 = > volume of num__0.000694444444444 part = num__4 liters volume of whole = ( num__1440 * num__4 ) litres = num__5760 litres . option c <eor> c <eos> |
c |
divide__1.0__6.0__ divide__1.0__8.0__ divide__0.1667__4.0__ multiply__0.0417__0.0167__ multiply__4.0__1440.0__ round__5760.0__ |
divide__1.0__6.0__ divide__1.0__8.0__ divide__0.1667__4.0__ multiply__0.0417__0.0167__ multiply__4.0__1440.0__ multiply__4.0__1440.0__ |
| if the product of two numbers is num__17820 and their h . c . f . is num__12 find their l . c . m . <o> a ) num__2574 <o> b ) num__2500 <o> c ) num__1485 <o> d ) num__1574 <o> e ) none of these |
explanation : hcf * lcm = num__17820 because we know product of two numbers = product of hcf and lcm lcm = num__1485.0 = num__1485 option c <eor> c <eos> |
c |
divide__17820.0__12.0__ divide__17820.0__12.0__ |
divide__17820.0__12.0__ divide__17820.0__12.0__ |
| the length of a rectangle is increased by num__20.0 and its breadth is decreased by num__20.0 . what is the effect on its area ? <o> a ) num__1288 <o> b ) num__9600 <o> c ) num__1000 <o> d ) num__10000 <o> e ) num__2887 |
num__100 * num__100 = num__10000 num__120 * num__80 = num__9600 answer : b <eor> b <eos> |
b |
square_perimeter__20.0__ multiply__120.0__80.0__ multiply__120.0__80.0__ |
square_perimeter__20.0__ multiply__120.0__80.0__ multiply__120.0__80.0__ |
| the sum of two numbers is num__16 . the difference is num__0 . what are the two numbers ? <o> a ) num__10 - num__6 <o> b ) num__8 - num__8 <o> c ) num__9 - num__7 <o> d ) num__11 - num__5 <o> e ) num__13 - num__3 |
num__8 + num__8 = num__16 num__8 - num__8 = num__0 the answer is b <eor> b <eos> |
b |
subtract__16.0__8.0__ |
subtract__16.0__8.0__ |
| which of the following is closer to √ num__0.8 + √ num__0.05 ? <o> a ) num__0.5 <o> b ) num__0.7 <o> c ) num__0.9 <o> d ) num__1.1 <o> e ) num__1.2 |
√ num__0.8 + √ num__0.05 = approximately √ num__0.81 + √ num__0.04 = num__0.9 + num__0.2 = num__1.1 . hence the correct answer is d . <eor> d <eos> |
d |
multiply__0.8__0.05__ add__0.2__0.9__ add__0.2__0.9__ |
multiply__0.8__0.05__ add__0.2__0.9__ add__0.2__0.9__ |
| if henry were to add num__9 gallons of water to a tank that is already num__0.75 full of water the tank would be num__0.875 full . how many gallons of water would the tank hold if it were full ? <o> a ) num__25 <o> b ) num__40 <o> c ) num__64 <o> d ) num__72 <o> e ) num__96 |
num__0.875 x - num__0.75 x = num__9 galls num__0.125 * x = num__9 gallons x = num__72 gallons answer d <eor> d <eos> |
d |
subtract__0.875__0.75__ divide__9.0__0.125__ divide__9.0__0.125__ |
subtract__0.875__0.75__ divide__9.0__0.125__ divide__9.0__0.125__ |
| two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of num__14 kmph and num__21 kmph respectively . when they meet it is found that one train has traveled num__60 km more than the other one . the distance between the two stations is ? <o> a ) num__445 km <o> b ) num__415 km <o> c ) num__420 km <o> d ) num__480 km <o> e ) num__490 km |
num__1 h - - - - - num__5 ? - - - - - - num__60 num__12 h rs = num__14 + num__21 = num__35 t = num__12 d = num__35 * num__12 = num__420 answer : c <eor> c <eos> |
c |
divide__60.0__5.0__ add__14.0__21.0__ multiply__35.0__12.0__ round__420.0__ |
divide__60.0__5.0__ add__14.0__21.0__ multiply__35.0__12.0__ multiply__35.0__12.0__ |
| if the ratio of the ages of two friends a and b is in the ratio num__3 : num__5 and that of b and c is num__3 : num__5 and the sum of their ages is num__147 then how old is b ? <o> a ) num__27 years <o> b ) num__75 years <o> c ) num__45 years <o> d ) num__49 years <o> e ) num__59 years |
explanation : the ratio of the ages of a and b is num__3 : num__5 . the ratio of the ages of b and c is num__3 : num__5 . b ' s age is the common link to both these ratio . therefore if we make the numerical value of the ratio of b ' s age in both the ratios same then we can compare the ages of all num__3 in a single ratio . the can be done by getting the value of b in both ratios to be the lcm of num__3 and num__5 i . e . num__15 . the first ratio between a and b will therefore be num__9 : num__15 and the second ratio between b and c will be num__15 : num__25 . now combining the two ratios we get a : b : c = num__9 : num__15 : num__25 . let their ages be num__9 x num__15 x and num__25 x . then the sum of their ages will be num__9 x + num__15 x + num__25 x = num__49 x the question states that the sum of their ages is num__147 . i . e . num__49 x = num__147 or x = num__3 . therefore b ' s age = num__15 x = num__15 × num__3 = num__45 answer : c <eor> c <eos> |
c |
multiply__3.0__5.0__ divide__147.0__3.0__ multiply__3.0__15.0__ multiply__3.0__15.0__ |
multiply__3.0__5.0__ divide__147.0__3.0__ multiply__3.0__15.0__ multiply__3.0__15.0__ |
| how many seconds will a num__600 m long train take to cross a man walking with a speed of num__3 km / hr in the direction of the moving train if the speed of the train is num__63 km / hr ? <o> a ) num__11 sec <o> b ) num__36 sec <o> c ) num__77 sec <o> d ) num__14 sec <o> e ) num__12 |
speed of train relative to man = num__63 - num__3 = num__60 km / hr . = num__60 * num__0.277777777778 = num__16.6666666667 m / sec . time taken to pass the man = num__600 * num__0.06 = num__36 sec . answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__600.0__0.06__ round__36.0__ |
subtract__63.0__3.0__ multiply__600.0__0.06__ multiply__600.0__0.06__ |
| if num__0.7 of a pencil is green num__0.8 of the remaining is gold and the remaining num__0.5 is white what is the total length of the pencil ? <o> a ) num__5 <o> b ) num__3 <o> c ) num__1 <o> d ) num__4 <o> e ) num__2 |
green is num__0.7 gold is num__0.8 which can also be written as num__0.8 white is num__0.5 which can also be written as num__0.5 num__0.7 + num__0.8 + num__0.5 = num__2 answer is e ) num__2 <eor> e <eos> |
e |
reverse__0.5__ reverse__0.5__ |
reverse__0.5__ reverse__0.5__ |
| a man buys an article and sells it at a profit of num__20.0 . if he had bought it at num__20.0 less and sold it for rs . num__75 less he could have gained num__25.0 . what is the cost price ? <o> a ) num__388 <o> b ) num__375 <o> c ) num__288 <o> d ) num__266 <o> e ) num__269 |
cp num__1 = num__100 sp num__1 = num__120 cp num__2 = num__80 sp num__2 = num__80 * ( num__1.25 ) = num__100 num__20 - - - - - num__100 num__75 - - - - - ? = > num__375 answer : b <eor> b <eos> |
b |
percent__100.0__375.0__ |
percent__100.0__375.0__ |
| a certain store sells only black shoes and brown shoes . in a certain week the store sold x black shoes and y brown shoes . if num__0.285714285714 of all shoes sold that week were black which of the following expressions represents the value of y in terms of x ? <o> a ) x / num__3 <o> b ) num__5 x / num__2 <o> c ) num__2 x / num__3 <o> d ) num__3 x / num__2 <o> e ) num__2 x |
x / ( x + y ) = num__0.285714285714 num__7 x = num__2 x + num__2 y num__5 x = num__2 y y = num__5 x / num__2 answer b <eor> b <eos> |
b |
subtract__7.0__2.0__ subtract__7.0__2.0__ |
subtract__7.0__2.0__ subtract__7.0__2.0__ |
| the speed of a boat in upstream is num__100 kmph and the speed of the boat downstream is num__180 kmph . find the speed of the boat in still water and the speed of the stream ? <o> a ) num__10 <o> b ) num__99 <o> c ) num__88 <o> d ) num__40 <o> e ) num__23 |
speed of the boat in still water = ( num__100 + num__180 ) / num__2 = num__70 kmph . speed of the stream = ( num__180 - num__100 ) / num__2 = num__40 kmph . answer : d <eor> d <eos> |
d |
round__40.0__ |
round__40.0__ |
| at joel ’ s bookstore the current inventory is num__30.0 historical fiction . of the historical fiction books num__30.0 are new releases while num__40.0 of the other books are new releases . what fraction of all new releases are the historical fiction new releases ? <o> a ) num__0.16 <o> b ) num__0.243243243243 <o> c ) num__0.4 <o> d ) num__0.533333333333 <o> e ) num__0.666666666667 |
let there be num__100 books in all historic fiction books = num__30.0 of total = num__30 other books = num__70 new historic fiction = num__30.0 of num__30 = num__9 other new books = num__40.0 of num__70 = num__28 total new books = num__37 fraction = num__0.243243243243 ans : b <eor> b <eos> |
b |
percent__40.0__70.0__ percent__100.0__0.2432__ |
percent__40.0__70.0__ percent__100.0__0.2432__ |
| a person purchased a tv set for rs . num__16000 and a dvd player for rs . num__6250 . he sold both the items together for rs . num__35150 . what percentage of profit did he make ? <o> a ) num__22 <o> b ) num__27 <o> c ) num__58 <o> d ) num__26 <o> e ) num__11 |
the total cp = rs . num__16000 + rs . num__6250 = rs . num__22250 and sp = rs . num__35150 profit ( % ) = ( num__35150 - num__22250 ) / num__22250 * num__100 = num__58.0 . answer : c <eor> c <eos> |
c |
percent__100.0__58.0__ |
percent__100.0__58.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 seconds . what is the length of the train ? <o> a ) num__288 <o> b ) num__2667 <o> c ) num__288 <o> d ) num__150 <o> e ) num__812 |
speed = ( num__60 * num__0.277777777778 ) m / sec = ( num__16.6666666667 ) m / sec length of the train = ( speed x time ) = ( num__16.6666666667 * num__9 ) m = num__150 m . answer : d <eor> d <eos> |
d |
round__150.0__ |
round__150.0__ |
| the average marks of a class of num__30 students is num__40 and that of another class of num__50 students is num__60 . find the average marks of all the students ? <o> a ) num__50 <o> b ) num__52.5 <o> c ) num__60 <o> d ) num__62.5 <o> e ) num__65 |
sum of the marks for the class of num__30 students = num__30 * num__40 = num__1200 sum of the marks for the class of num__50 students = num__50 * num__60 = num__3000 sum of the marks for the class of num__80 students = num__1200 + num__3000 = num__4200 average marks of all the students = num__52.5 = num__52.5 answer b <eor> b <eos> |
b |
multiply__30.0__40.0__ multiply__50.0__60.0__ add__30.0__50.0__ add__1200.0__3000.0__ divide__4200.0__80.0__ divide__4200.0__80.0__ |
multiply__30.0__40.0__ multiply__50.0__60.0__ add__30.0__50.0__ add__1200.0__3000.0__ divide__4200.0__80.0__ divide__4200.0__80.0__ |
| two trains num__140 meters and num__160 meters long are running in the same direction with speeds of num__92 km / hr num__56 km / hr . in how much time will the first train cross the second <o> a ) num__37 sec <o> b ) num__25 sec <o> c ) num__30 sec <o> d ) num__28 sec <o> e ) num__32 sec |
explanation : relative speed of the trains = ( num__92 - num__56 ) km / hr = num__36 km / hr = ( num__36 Ã — num__0.277777777778 ) m / sec = num__10 m / sec . time taken by the trains to cross each other = time taken to cover ( num__140 + num__160 ) m at num__10 m / sec = ( num__30.0 ) sec = num__30 sec . answer : option c <eor> c <eos> |
c |
subtract__92.0__56.0__ round__30.0__ |
subtract__92.0__56.0__ round__30.0__ |
| if m ≠ num__1 and if nm / ( n - m ) = num__1 what is the value of n in terms of m ? <o> a ) m / ( m + num__1 ) <o> b ) ( m - num__1 ) / m <o> c ) m / ( num__1 - m ) <o> d ) ( m + num__1 ) / m <o> e ) ( num__1 - m ) / m |
this question comes with a particular ' quirk ' ( one that you probably wo n ' t see on test day ) . the gmat wo n ' t test you on the concept of undefined numbers ( re : numbers divided by num__0 ) so any time that this concept is a possibility the question writers have to add a restriction that removes the option that a num__0 could occur in the denominator . here we ' re told that b can not = num__1 which is a bit strange because that restriction does n ' t seem to impact the original equation much . in fact it impacts just one of the answer choices - so you have to ask why that restriction is even there . it ' s actually because that one answer is the correct one . final answer : c <eor> c <eos> |
c |
reverse__1.0__ |
reverse__1.0__ |
| if the l . c . m of two numbers is num__450 and their product is num__22500 find the h . c . f of the numbers . <o> a ) num__50 <o> b ) num__30 <o> c ) num__125 <o> d ) num__25 <o> e ) none of these |
h . c . f = ( product of the numbers ) / ( their l . c . m ) = num__50.0 = num__50 . answer : a <eor> a <eos> |
a |
divide__22500.0__450.0__ gcd__450.0__50.0__ |
divide__22500.0__450.0__ divide__22500.0__450.0__ |
| the diagonals of a rhombus are num__11 cm and num__20 cm . find its area ? <o> a ) num__329 <o> b ) num__288 <o> c ) num__150 <o> d ) num__238 <o> e ) num__110 |
num__0.5 * num__11 * num__20 = num__110 answer : e <eor> e <eos> |
e |
triangle_area__11.0__20.0__ triangle_area__11.0__20.0__ |
volume_rectangular_prism__11.0__20.0__0.5__ volume_rectangular_prism__11.0__20.0__0.5__ |
| a and b working together could mow a field in num__28 days and with the help of c they could have mowed it in num__21 days . how long would c take by himself ? <o> a ) num__68 days <o> b ) num__84 days <o> c ) num__90 days <o> d ) num__72 days <o> e ) num__50 days |
required answer = num__28 * num__0.75 - num__21 = num__84 days answer is b <eor> b <eos> |
b |
divide__21.0__28.0__ round__84.0__ |
divide__21.0__28.0__ round__84.0__ |
| ( num__2 ^ num__2 + num__4 ^ num__2 + num__6 ^ num__2 + . . . . . + num__22 ^ num__2 ) = ? <o> a ) num__5578 <o> b ) num__7996 <o> c ) num__9964 <o> d ) num__2024 <o> e ) num__1346 |
= ( num__1 x num__2 ) ^ num__2 + ( num__2 x num__2 ) ^ num__2 + ( num__2 x num__3 ) ^ num__3 + . . . . . . ( num__2 x num__11 ) ^ num__2 = num__2 ^ num__2 x ( num__1 ^ num__2 + num__2 ^ num__2 + num__3 ^ num__2 + . . . . . . . + num__11 ^ num__2 ) formula is = num__0.166666666667 n ( n + num__1 ) ( num__2 n + num__1 ) = ( num__4 x num__0.166666666667 x num__11 x num__12 x num__23 ) = ( num__4 x num__506 ) = num__2024 answer is d <eor> d <eos> |
d |
add__2.0__1.0__ divide__22.0__2.0__ reverse__6.0__ multiply__2.0__6.0__ add__22.0__1.0__ multiply__22.0__23.0__ multiply__4.0__506.0__ multiply__4.0__506.0__ |
add__2.0__1.0__ divide__22.0__2.0__ reverse__6.0__ add__1.0__11.0__ add__22.0__1.0__ multiply__22.0__23.0__ multiply__4.0__506.0__ multiply__4.0__506.0__ |
| a towel when bleached lost num__10.0 of its length and num__20.0 of its breadth . what is the percentage decrease in area ? <o> a ) num__28.0 <o> b ) num__30.0 <o> c ) num__44.0 <o> d ) num__54.0 <o> e ) num__64 % |
percentage change in area = ( − num__10 − num__20 + ( num__10 × num__20 ) / num__100 ) % = − num__28.0 i . e . area is decreased by num__28.0 answer : a <eor> a <eos> |
a |
percent__100.0__28.0__ |
percent__100.0__28.0__ |
| a ladder num__100 feet long is leaning against a vertical wall its lower end is num__60 foot from the bottom of the wall the side of the largest cubical box that can be placed between the wall and the ladder without disturbing the ladder is ( to the nearest foot ) . <o> a ) num__26 <o> b ) num__34 <o> c ) num__21 <o> d ) num__40 <o> e ) num__30 |
if we consider the num__2 d image then the ladder the wall and base making a triangle which height = num__80 foot now we have to calculate the largest square inside the triangle . let x be the side of square . then there are two triange of one height = x base = ( num__60 - x ) and the other of height = ( num__80 - x ) and base = x and a squre by condition ( num__60 - x ) * x / num__2 + x * ( num__80 - x ) / num__2 + x ^ num__2 = num__80 * num__30.0 x = num__34.2 answer : b <eor> b <eos> |
b |
divide__60.0__2.0__ round_down__34.2__ |
divide__60.0__2.0__ round_down__34.2__ |
| on dividing a certain number by num__5 num__7 and num__8 successively the remainders obtained are num__2 num__3 and num__4 respectively . when the order of division is reversed and the number is successively divided by num__8 num__7 and num__5 the respective remainders t will be : [ / b ] <o> a ) num__3 num__3 num__2 <o> b ) num__3 num__4 num__2 <o> c ) num__5 num__4 num__3 <o> d ) t = num__5 num__5 num__2 <o> e ) num__6 num__4 num__3 |
let the original number be x . then by the successive dividing we have followings : x = num__5 a + num__2 a = num__7 b + num__3 b = num__8 c + num__4 . so we have a = num__7 * ( num__8 c + num__4 ) + num__3 = num__7 * num__8 c + num__31 and x = num__5 * ( num__7 * num__8 c + num__31 ) + num__2 = num__5 * num__7 * num__8 c + num__157 . now by dividing x by num__8 num__7 num__5 successively we have followings : x = num__8 * ( num__5 * num__7 c + num__19 ) + num__5 num__5 * num__7 c + num__19 = num__7 * ( num__5 c + num__2 ) + num__5 num__5 c + num__2 = num__5 c + num__2 . the remainders are therefore num__5 num__5 num__2 . the answer is ( d ) . <eor> d <eos> |
d |
subtract__7.0__2.0__ |
add__2.0__3.0__ |
| salesperson a ' s compensation for any week is $ num__280 plus num__6 percent of the portion of a ' s total sales above $ num__1000 for that week . salesperson b ' s compensation for any week is num__8 percent of b ' s total sales for that week . for what amount of total weekly sales would both salespeople earn the same compensation ? <o> a ) $ num__5000 <o> b ) $ num__7000 <o> c ) $ num__9000 <o> d ) $ num__11000 <o> e ) $ num__13 |
000 |
num__280 + num__0.06 ( x - num__1000 ) = num__0.08 x num__0.02 x = num__220 x = $ num__11000 the answer is d . <eor> d <eos> |
d |
d |
| in an increasing sequence of num__10 consecutive integers the sum of the first num__5 integers is num__550 . what is the sum of the last num__5 integers in the sequence ? <o> a ) num__585 <o> b ) num__580 <o> c ) num__575 <o> d ) num__570 <o> e ) num__565 |
all num__5 integers are num__5 numbers larger than in the first sum ( eg . num__1 becomes num__6 num__2 num__7 . . . ) . num__5 * num__5 = num__25 + num__550 = num__575 c <eor> c <eos> |
c |
add__5.0__1.0__ divide__10.0__5.0__ add__5.0__2.0__ power__5.0__2.0__ add__550.0__25.0__ add__550.0__25.0__ |
add__5.0__1.0__ divide__10.0__5.0__ add__5.0__2.0__ power__5.0__2.0__ add__550.0__25.0__ add__550.0__25.0__ |
| in how many different number of ways num__4 boys and num__2 girls can sit on a bench ? <o> a ) num__700 <o> b ) num__720 <o> c ) num__740 <o> d ) num__750 <o> e ) num__760 |
npn = n ! num__6 p num__6 = num__6 × num__5 × num__4 × num__3 × num__2 × num__1 = num__720 b <eor> b <eos> |
b |
die_space__ vowel_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
die_space__ vowel_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| rs . num__6000 is lent out in two parts . one part is lent at num__7.0 p . a simple interest and the other is lent at num__8.0 p . a simple interest . the total interest at the end of one year was rs . num__450 . find the ratio of the amounts lent at the lower rate and higher rate of interest ? <o> a ) num__5 : num__1 <o> b ) num__5 : num__5 <o> c ) num__5 : num__8 <o> d ) num__5 : num__4 <o> e ) num__5 : num__2 |
let the amount lent at num__7.0 be rs . x amount lent at num__8.0 is rs . ( num__6000 - x ) total interest for one year on the two sums lent = num__0.07 x + num__0.08 ( num__6000 - x ) = num__480 - x / num__100 = > num__480 - num__0.01 x = num__450 = > x = num__3000 amount lent at num__10.0 = num__3000 required ratio = num__3000 : num__3000 = num__5 : num__5 answer : b <eor> b <eos> |
b |
percent__8.0__6000.0__ percent__100.0__5.0__ |
percent__8.0__6000.0__ percent__100.0__5.0__ |
| in a race a beats b by num__15 metres and c by num__29 metres . if b and c run over the course together b wins by num__15 metres . what is the length of the course ? <o> a ) num__225 m <o> b ) num__120 m <o> c ) num__220 m <o> d ) num__160 m <o> e ) num__180 m |
explanation : let x be the length of the course according to the question a beats b by num__15 metres and c by num__29 metres applying this if a runs x metres b runs ( x - num__15 ) and c runs ( x - num__29 ) metres . b and c together run b runs x metres and c runs ( x - num__15 ) = > if b runs num__1 metre c runs x - num__15 / x = > b runs x - num__15 m c runs ( x - num__15 / x ) * ( x - num__15 ) applying this x - num__29 = ( x - num__15 ) ( x - num__15 ) / x solving x num__2 - num__29 x = x num__2 – num__30 x + num__225 x = num__225 thus length of the course is num__225 metres . answer a <eor> a <eos> |
a |
multiply__15.0__2.0__ round__225.0__ |
multiply__15.0__2.0__ divide__225.0__1.0__ |
| one train has crossing a bridge of length num__340 m in num__42 sec and same train has crossing a another bridge of num__500 m in num__50 sec . what is the approximate speed of the train in km / hr ? <o> a ) num__60 <o> b ) num__64.8 <o> c ) num__72 <o> d ) num__76 <o> e ) num__78 |
length of train = l m speed of train = v m / s then num__340 + l = num__42 v num__500 + l = num__50 v subtract num__8 v = num__160 = > v = num__20 m / s = num__72 kmph answer : c <eor> c <eos> |
c |
subtract__50.0__42.0__ subtract__500.0__340.0__ divide__160.0__8.0__ round__72.0__ |
subtract__50.0__42.0__ subtract__500.0__340.0__ divide__160.0__8.0__ round__72.0__ |
| for every $ num__20 that a billionaire spends a millionaire spends the equivalent of num__10 cents . for every $ num__4 that a millionaire spends a yuppie spends the equivalent of $ num__1 . the ratio of money spent by a yuppie millionaire and billionaire can be expressed as <o> a ) num__1 : num__4 : num__400 <o> b ) num__1 : num__4 : num__100 <o> c ) num__20 : num__4 : num__1 <o> d ) num__1 : num__4 : num__800 <o> e ) num__400 : num__4 : num__1 |
millionaire num__10 cents = num__0.1 dollar . . therefore num__1 dollar m = num__200 dollar of b therefore num__4 dollar m = num__800 dollar of b also num__1 dollar y y : m : b = num__1 : num__4 : num__800 = num__1 : num__4 : num__800 d <eor> d <eos> |
d |
reverse__10.0__ multiply__20.0__10.0__ multiply__4.0__200.0__ reverse__1.0__ |
reverse__10.0__ multiply__20.0__10.0__ multiply__4.0__200.0__ reverse__1.0__ |
| eshan and mary each wrote two or three poem every day over a period of time eshan wrote num__43 poems while mary wrote num__61 . the number of days in this period is . . . . . <o> a ) num__19 <o> b ) num__21 <o> c ) num__20 <o> d ) num__19 <o> e ) num__18 |
suppose eshan writes num__2 poems in first num__18 days num__18 * num__2 = num__36 . num__2 poems in next num__2 days num__2 * num__2 = num__4 num__36 + num__4 = num__40 and num__3 poems in next num__1 day num__1 * num__3 = num__3 num__40 + num__3 = num__43 so ultimately he takes ( num__18 + num__2 + num__1 ) = num__21 days to write num__43 poems . now suppose marry writes num__3 poems in num__1 st num__18 days num__18 * num__3 = num__54 num__2 poems in next num__2 days num__2 * num__2 = num__4 num__54 + num__4 = num__58 num__3 poems in next num__1 day num__1 * num__3 = num__3 num__58 + num__3 = num__61 so ultimately marry takes ( num__18 + num__2 + num__1 ) = num__21 days to write num__61 poems . answer : b <eor> b <eos> |
b |
subtract__61.0__43.0__ multiply__2.0__18.0__ add__4.0__36.0__ subtract__43.0__40.0__ subtract__3.0__2.0__ subtract__61.0__40.0__ multiply__3.0__18.0__ subtract__61.0__3.0__ subtract__61.0__40.0__ |
subtract__61.0__43.0__ multiply__2.0__18.0__ add__4.0__36.0__ subtract__43.0__40.0__ subtract__3.0__2.0__ add__3.0__18.0__ multiply__3.0__18.0__ add__4.0__54.0__ multiply__1.0__21.0__ |
| income and expenditure of a man are in the ratio num__4 : num__3 . if the income of the person is rs . num__12000 then find his savings ? <o> a ) num__2600 <o> b ) num__2700 <o> c ) num__2800 <o> d ) num__2900 <o> e ) num__3000 |
let the income and the expenditure of the person be rs . num__4 x and rs . num__3 x respectively . income num__4 x = num__12000 = > x = num__3000 savings = income - expenditure = num__4 x - num__3 x = x so savings = rs . num__3000 . answer : e <eor> e <eos> |
e |
divide__12000.0__4.0__ divide__12000.0__4.0__ |
divide__12000.0__4.0__ divide__12000.0__4.0__ |
| a man buys an item at rs . num__600 and sells it at the loss of num__20 percent . then what is the selling price of that item <o> a ) rs . num__480 <o> b ) rs . num__760 <o> c ) rs . num__860 <o> d ) rs . num__960 <o> e ) none of these |
explanation : here always remember when ever x % loss it means s . p . = ( num__100 - x ) % of c . p when ever x % profit it means s . p . = ( num__100 + x ) % of c . p so here will be ( num__100 - x ) % of c . p . = num__80.0 of num__600 = num__0.8 * num__600 = num__480 option a <eor> a <eos> |
a |
percent__80.0__600.0__ percent__80.0__600.0__ |
percent__80.0__600.0__ percent__80.0__600.0__ |
| if rs . num__10 be allowed as true discount on a bill of rs . num__110 due at the end of a certain time then the discount allowed on the same sum due at the end of double the time is : <o> a ) rs . num__20 <o> b ) rs . num__21.80 <o> c ) rs . num__22 <o> d ) rs . num__22.33 <o> e ) rs . num__18.33 |
s . i . on rs . ( num__110 - num__10 ) for a certain time = rs . num__10 . s . i . on rs . num__100 for double the time = rs . num__20 . t . d . on rs . num__120 = rs . ( num__120 - num__100 ) = rs . num__20 . t . d on rs . num__110 = rs . ( num__0.166666666667 ) x num__100 = rs . num__18.33 answer : e <eor> e <eos> |
e |
percent__100.0__18.33__ |
percent__100.0__18.33__ |
| two cars car num__1 and car num__2 move towards each other from w and y respectively with respective speeds of num__20 m / s and num__15 m / s . after meeting each other car num__1 reaches y in num__10 seconds . in how many seconds does car num__2 reach w starting from y ? <o> a ) num__15.5 sec <o> b ) num__8.4 sec <o> c ) num__33.6 sec <o> d ) num__31.11 sec <o> e ) num__16.8 sec |
w - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - | - - - - - - - - - - - - - - - - - - - - - - - - - - - - y car a ( num__20 mps ) - - - - - - - - - - - - - - - - - - - - - - - - - > p < - - - - - - - - - - - - - - - car b ( num__15 mps ) let num__2 cars meet each other at point p in t seconds . car num__1 covers distance = num__20 t . car num__2 covers distance = num__15 t . so total distance wy = num__35 t . from p car num__1 reaches onto y in num__10 secs . so it covers num__15 t further . so num__15 t / num__20 = num__10 so t = num__13.3333333333 sec and total distance = ( num__35 * num__40 ) / num__3 hence car num__2 will cover total distance in ( num__35 * num__40 ) / ( num__3 * num__15 ) = num__31.11 sec approx . answer d <eor> d <eos> |
d |
add__20.0__15.0__ multiply__2.0__20.0__ add__1.0__2.0__ round__31.11__ |
add__20.0__15.0__ multiply__2.0__20.0__ divide__40.0__13.3333__ multiply__1.0__31.11__ |
| a can finish a work in num__12 days and b can do same work in half the time taken by a . then working together what part of same work they can finish in a day ? <o> a ) num__0.25 <o> b ) num__0.166666666667 <o> c ) num__0.142857142857 <o> d ) num__0.125 <o> e ) none of these |
explanation : please note in this question we need to answer part of work for a day rather than complete work . it was worth mentioning here because many do mistake at this point in hurry to solve the question so lets solve now a ' s num__1 day work = num__0.0833333333333 b ' s num__1 day work = num__0.166666666667 [ because b take half the time than a ] ( a + b ) ' s one day work = ( num__0.0833333333333 + num__0.166666666667 ) = num__0.25 so in one day num__0.25 work will be done answer : a <eor> a <eos> |
a |
divide__1.0__12.0__ add__0.1667__0.0833__ round__0.25__ |
divide__1.0__12.0__ add__0.1667__0.0833__ add__0.1667__0.0833__ |
| there are num__3 green shirts num__2 white shirts num__5 blue shirts and num__7 red shirts in the wardrobe . what is the least number of shirts that you have to take out to make sure that you will have a matching pair ? <o> a ) num__4 <o> b ) num__7 <o> c ) num__5 <o> d ) num__2 <o> e ) num__6 |
since there are num__4 colors if you take num__4 shirts you could still have num__1 shirt per color and not have a match . therefore upon taking the num__5 th shirt you will definitely have a match since the num__5 th shirt will form at least a pair with the num__1 st num__4 so num__4 + num__1 = num__5 answer is c <eor> c <eos> |
c |
subtract__7.0__3.0__ subtract__3.0__2.0__ add__3.0__2.0__ |
subtract__7.0__3.0__ subtract__3.0__2.0__ add__3.0__2.0__ |
| if num__4 a = num__16 b and num__8 b = num__13 c find a : b : c ? <o> a ) num__52 : num__13 : num__8 <o> b ) num__13 : num__7 : num__52 <o> c ) num__7 : num__13 : num__52 <o> d ) num__7 : num__13 : num__54 <o> e ) none of these |
explanation : ( num__4 a = num__16 b = = > a / b = num__4.0 ) and ( num__8 b = num__13 c = = > b / c = num__1.625 ) = = > a : b = num__16 : num__4 and b : c = num__13 : num__8 a : b : c = num__52 : num__13 : num__8 answer : option a <eor> a <eos> |
a |
divide__13.0__8.0__ multiply__4.0__13.0__ multiply__4.0__13.0__ |
divide__13.0__8.0__ multiply__4.0__13.0__ multiply__4.0__13.0__ |
| a rectangular floor is covered by a rug except for a strip num__4 meters wide along each of the four edge . if the floor is num__25 meters by num__20 meters what is the area of the rug in square meters ? <o> a ) num__256 <o> b ) num__266 <o> c ) num__204 <o> d ) num__224 <o> e ) num__324 |
a strip of num__4 meters is covering the inner rectangular rug for all num__4 sides . length of inner rug = num__25 - ( num__2 * num__4 ) breadth of inner rug = num__20 - ( num__2 * num__4 ) area of rug = num__17 * num__12 = num__204 sq . mt  nswer : c <eor> c <eos> |
c |
rectangle_perimeter__4.0__2.0__ multiply__12.0__17.0__ triangle_area__2.0__204.0__ |
rectangle_perimeter__4.0__2.0__ multiply__12.0__17.0__ multiply__12.0__17.0__ |
| the average salary per head of the entire staff of an office including the officers and clerks is rs . num__90 . the average salary of officers is rs . num__600 and that of the clerks is rs . num__84 . if the number of officers is num__2 find the number of officers in the office ? <o> a ) num__3 <o> b ) num__1 <o> c ) num__8 <o> d ) num__9 <o> e ) num__5 |
num__6 * num__22 = num__132 num__7 * num__19 = num__133 - - - - - - - - - - - - - - num__1 answer : b <eor> b <eos> |
b |
subtract__90.0__84.0__ multiply__6.0__22.0__ multiply__7.0__19.0__ subtract__133.0__132.0__ reverse__1.0__ |
subtract__90.0__84.0__ multiply__6.0__22.0__ multiply__7.0__19.0__ subtract__133.0__132.0__ subtract__2.0__1.0__ |
| in each of the following questions a number series is given with one term missing . choose the correct alternative that will continue the same pattern and fill in the blank spaces . num__2 num__7 num__14 ? num__34 num__47 <o> a ) num__31 <o> b ) num__23 <o> c ) num__36 <o> d ) num__31 <o> e ) num__33 |
b num__23 the given sequence is + num__5 + num__7 + num__9 — — ie . num__2 + num__5 = num__7 num__7 + num__7 = num__14 num__14 + num__9 = num__23 <eor> b <eos> |
b |
subtract__7.0__2.0__ add__2.0__7.0__ add__14.0__9.0__ |
subtract__7.0__2.0__ add__2.0__7.0__ add__14.0__9.0__ |
| if a store adds num__60 chairs to its current inventory the total number of chairs will be the same as three - halves the current inventory of chairs . if the manager wants to increase the current inventory by num__50.0 what will the new inventory of chairs be ? <o> a ) num__40 <o> b ) num__60 <o> c ) num__100 <o> d ) num__180 <o> e ) num__160 |
explanation : let â € ™ s say t = total current inventory of chairs . the first sentence states that num__60 + t = ( num__1.5 ) t . first solve for the current inventory : num__60 + t = ( num__1.5 ) t num__60 = ( num__1.5 ) t â ˆ ’ t num__60 = ( num__0.5 ) t num__120 = t the manager wants to increase this by num__50.0 . num__50.0 of num__120 is num__60 so the new inventory will be num__180 . answer : ( d ) . <eor> d <eos> |
d |
divide__60.0__0.5__ add__60.0__120.0__ add__60.0__120.0__ |
divide__60.0__0.5__ add__60.0__120.0__ add__60.0__120.0__ |
| if capacity of giving milk of a cow is y litre . from this num__1 litre is used to make tea . rest amount is multiply by num__2 and then divided by num__3 . calculate the result if y = num__13 litre . <o> a ) num__6 <o> b ) num__5 <o> c ) num__8 <o> d ) num__9 <o> e ) num__4 |
from num__13 litre num__1 litre used to make tea rest amount is num__12 litre . num__12 * num__2 = num__24 now num__8.0 = num__8 litre answer c <eor> c <eos> |
c |
subtract__13.0__1.0__ multiply__2.0__12.0__ divide__24.0__3.0__ multiply__1.0__8.0__ |
subtract__13.0__1.0__ multiply__2.0__12.0__ divide__24.0__3.0__ multiply__1.0__8.0__ |
| marissa keith and tony all go to disneyland many times each summer . together they have all gone a total of num__65 times . marissa and keith have gone num__55 times total while keith and tony have gone num__40 times total . how many times has keith gone just by himself ? <o> a ) num__30 <o> b ) num__20 <o> c ) num__25 <o> d ) num__40 <o> e ) num__10 |
marissa + keith + tony = num__65 m + k = num__55 k + t = num__40 plug these each into the original formula num__55 + t = num__65 t = num__10 then m + num__40 = num__65 m = num__25 so num__25 + k + num__10 = num__65 subtract num__25 and num__10 from each side to get keith ' s total k = num__30 answer : a <eor> a <eos> |
a |
subtract__65.0__55.0__ subtract__65.0__40.0__ subtract__55.0__25.0__ subtract__55.0__25.0__ |
subtract__65.0__55.0__ subtract__65.0__40.0__ subtract__55.0__25.0__ subtract__55.0__25.0__ |
| in game of num__500 points there are three participants a b and c . a game take place . a gives to b num__80 points and to c num__101 points . then how many points can b give to c <o> a ) num__15 points <o> b ) num__25 points <o> c ) num__35 points <o> d ) num__45 points <o> e ) num__55 points |
in the game of num__500 points a gives b num__80 points that means b got ( num__500 - num__80 ) = num__420 a gives b num__101 points that means b got ( num__500 - num__101 ) = num__399 so b : c = num__420 : num__399 so in the game of num__420 b gives ` ` num__21 ' ' ( num__420 - num__399 ) points to c . . so in the game of num__500 b gives to c as num__500 * num__0.05 = num__25 points . answer : b <eor> b <eos> |
b |
subtract__500.0__80.0__ subtract__500.0__101.0__ subtract__101.0__80.0__ divide__21.0__420.0__ multiply__500.0__0.05__ multiply__500.0__0.05__ |
subtract__500.0__80.0__ subtract__500.0__101.0__ subtract__101.0__80.0__ divide__21.0__420.0__ multiply__500.0__0.05__ multiply__500.0__0.05__ |
| find the ratio in which rice at rs . num__7.20 a kg be mixed with rice at rs . num__5.50 a kg to produce a mixture worth rs . num__6.30 a kg <o> a ) num__2 : num__0 <o> b ) num__2 : num__3 <o> c ) num__2 : num__1 <o> d ) num__8 : num__9 <o> e ) num__2 : num__8 |
by the rule of alligation : cost of num__1 kg rice of num__1 st kind cost of num__1 kg rice of num__2 nd kind required ratio = num__80 : num__90 = num__8 : num__9 answer : d <eor> d <eos> |
d |
add__1.0__8.0__ multiply__1.0__8.0__ |
add__1.0__8.0__ multiply__1.0__8.0__ |
| ( num__1 ^ num__3 + num__2 ^ num__3 + num__3 ^ num__3 + . . . + num__10 ^ num__3 ) = x ? <o> a ) num__360 <o> b ) num__380 <o> c ) x = num__420 <o> d ) num__440 <o> e ) num__450 |
we know that ( num__1 ^ num__3 + num__2 ^ num__3 + num__3 ^ num__3 + . . . + n ^ num__3 ) = num__1 num__2 n ( n + num__1 ) num__6 putting n = num__10 required sum = num__1 x num__10 x num__10 x num__21 = num__420.0 num__420 c <eor> c <eos> |
c |
multiply__3.0__2.0__ multiply__1.0__420.0__ |
multiply__3.0__2.0__ multiply__1.0__420.0__ |
| a train num__300 m long can cross an electric pole in num__20 sec and then find the speed of the train ? <o> a ) num__17 kmph <o> b ) num__54 kmph <o> c ) num__72 kmph <o> d ) num__18 kmph <o> e ) num__19 kmph |
length = speed * time speed = l / t s = num__15.0 s = num__15 m / sec speed = num__15 * num__3.6 ( to convert m / sec in to kmph multiply by num__3.6 ) speed = num__54 kmph answer : b <eor> b <eos> |
b |
divide__300.0__20.0__ multiply__3.6__15.0__ round__54.0__ |
divide__300.0__20.0__ multiply__3.6__15.0__ multiply__3.6__15.0__ |
| with a uniform speed a car covers a distance in num__8 hours . were the speed increased by num__4 km / hr the same distance could be covered in num__7 num__1 ⁄ num__2 hours . what is the distance covered ? <o> a ) num__640 km <o> b ) num__480 km <o> c ) num__420 km <o> d ) can not be determined <o> e ) none of these |
here d ⁄ num__7.5 - d ⁄ num__8 = num__4 ( where d is the distance in km ) ⇒ num__0.5 d = num__4 × num__8 × num__7.5 ⇒ d = num__2 × num__4 × num__8 × num__7.5 = num__480 km answer b <eor> b <eos> |
b |
subtract__8.0__7.5__ round__480.0__ |
subtract__8.0__7.5__ round__480.0__ |
| a trader mixes num__10 kg of rice at rs . num__20 per kg with num__20 kg of rice of other variety at rs . num__40 per kg and sells the mixture at rs . num__40 per kg . his profit percent is : <o> a ) no profit no loss <o> b ) num__5.0 <o> c ) num__8.0 <o> d ) num__20.0 <o> e ) none of these |
c . p . of num__30 kg rice = rs . ( num__10 x num__20 + num__20 x num__40 ) = rs . num__200 + num__800 = rs . num__1000 . s . p . of num__30 kg rice = rs . num__30 x num__40 = rs . num__1200 gain = num__0.2 x num__100.0 = num__20.0 . answer : option d <eor> d <eos> |
d |
percent__10.0__1000.0__ percent__10.0__200.0__ |
percent__10.0__1000.0__ percent__10.0__200.0__ |
| each machine of type a has num__3 steel parts and num__3 chrome parts . each machine of type b has num__6 steel parts and num__5 chrome parts . if a certain group of type a and type b machines has a total of num__60 steel parts and num__44 chrome parts how many machines are in the group <o> a ) num__12 <o> b ) num__13 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
look at the below representation of the problem : steel chrome total a num__3 num__3 num__60 > > no . of type a machines = num__10.0 = num__10 b num__6 num__5 num__44 > > no . of type b machines = num__4.0 = num__4 so the answer is num__14 i . e c . hope its clear . <eor> c <eos> |
c |
divide__60.0__6.0__ subtract__10.0__6.0__ add__4.0__10.0__ add__4.0__10.0__ |
divide__60.0__6.0__ subtract__10.0__6.0__ add__4.0__10.0__ add__4.0__10.0__ |
| when magnified num__1000 times by an electron microscope the image of a certain circular piece of tissue has a diameter of num__2 centimeter . the actual diameter of the tissue in centimeters is <o> a ) num__0.005 <o> b ) num__0.002 <o> c ) num__0.001 <o> d ) num__0.0005 <o> e ) num__0.0002 |
it is very easy if x is the diameter then the magnified length is num__1000 x . ince num__1000 x = num__2 then x = num__0.002 = num__0.002 . the answer is b <eor> b <eos> |
b |
triangle_area__2.0__0.002__ |
triangle_area__2.0__0.002__ |
| roja and pooja start moving in the opposite directions from a pole . they are moving at the speeds of num__7 km / hr and num__3 km / hr respectively . after num__4 hours what will be the distance between them ? <o> a ) num__22 km <o> b ) num__40 km <o> c ) num__65 km <o> d ) num__18 km <o> e ) num__16 km |
distance = relative speed * time = ( num__7 + num__3 ) * num__4 = num__40 km [ they are travelling in the opposite direction relative speed = sum of the speeds ] . answer : b <eor> b <eos> |
b |
round__40.0__ |
round__40.0__ |
| a cube of edge num__5 cm is cut into cubes each of edge num__1 cm . the ratio of the total surface area of one of the small cubes to that of the large cube is equal to : <o> a ) num__1 : num__25 <o> b ) num__1.225 <o> c ) num__1 : num__52 <o> d ) num__1 : num__522 <o> e ) none |
sol . required ratio = num__6 * num__1 * num__0.166666666667 * num__5 * num__5 = num__0.04 = num__1 : num__25 . answer a <eor> a <eos> |
a |
surface_cube__1.0__ volume_cube__1.0__ |
surface_cube__1.0__ multiply__25.0__0.04__ |
| from given equation find the value of x : num__2 x ² + num__9 x − num__5 <o> a ) − num__5 <o> b ) - num__4 <o> c ) - num__6 <o> d ) num__6 <o> e ) - num__4 |
that quadratic is factored as follows : num__2 x ² + num__9 x − num__5 = ( num__2 x − num__1 ) ( x + num__5 ) . lesson num__17 . now it is easy to see that the second factor will be num__0 when x = − num__5 . as for the value of x that will make num__2 x − num__1 = num__0 we must solve that little equation . ( lesson num__9 . ) we have : num__2 x = num__1 x = num__1 num__2 the solutions are : x = num__0.5 or − num__5 a <eor> a <eos> |
a |
reverse__2.0__ multiply__5.0__1.0__ |
reverse__2.0__ multiply__5.0__1.0__ |
| num__13 buckets of water fill a tank when the capacity of each bucket is num__51 litres . how many buckets will be needed to fill the same tank if the capacity of each bucket is num__17 litres ? <o> a ) num__39 <o> b ) num__28 <o> c ) num__23 <o> d ) num__12 <o> e ) num__24 |
capacity of the tank = ( num__13 × num__51 ) litre number of buckets required of capacity of each bucket is num__17 litre = num__13 × num__3.0 = num__13 × num__3 = num__39 answer is a <eor> a <eos> |
a |
divide__51.0__17.0__ multiply__13.0__3.0__ round__39.0__ |
divide__51.0__17.0__ multiply__13.0__3.0__ round__39.0__ |
| the sum of the non - prime numbers between num__50 and num__60 non - inclusive is <o> a ) num__263 <o> b ) num__293 <o> c ) num__323 <o> d ) num__353 <o> e ) num__383 |
sum of consecutive integers from num__51 to num__59 inclusive = = = = > ( a num__1 + an ) / num__2 * # of terms = ( num__51 + num__59 ) / num__2 * num__9 = num__55 * num__9 = num__495 sum of non - prime numbers b / w num__50 and num__60 non inclusive = = = > num__495 - num__112 ( i . e . num__53 + num__59 being the prime # s in the range ) = num__383 answer : e <eor> e <eos> |
e |
subtract__60.0__59.0__ subtract__60.0__51.0__ multiply__9.0__55.0__ add__2.0__51.0__ subtract__495.0__112.0__ multiply__1.0__383.0__ |
subtract__60.0__59.0__ subtract__60.0__51.0__ multiply__9.0__55.0__ add__2.0__51.0__ subtract__495.0__112.0__ multiply__1.0__383.0__ |
| the c . p of num__10 pens is equal to the s . p of num__12 pens . find his gain % or loss % ? <o> a ) num__16 num__0.25 % <o> b ) num__16 num__2.0 % <o> c ) num__16 num__0.666666666667 % <o> d ) num__18 num__0.666666666667 % <o> e ) num__16 num__25 % |
num__10 cp = num__12 sp num__12 - - - num__2 cp loss num__100 - - - ? = > num__16 num__0.666666666667 % answer : c <eor> c <eos> |
c |
percent__100.0__16.0__ |
percent__100.0__16.0__ |
| a person lent a certain sum of money at num__4.0 per annum at simple interest and in num__8 years the interest amounted to rs . num__340 less than the sum lent . what was the sum lent ? <o> a ) rs : num__567 <o> b ) rs : num__518 <o> c ) rs : num__519 <o> d ) rs : num__500 <o> e ) rs : num__520 |
p - num__340 = ( p * num__4 * num__8 ) / num__100 p = num__500 answer : d <eor> d <eos> |
d |
percent__100.0__500.0__ |
percent__100.0__500.0__ |
| how many minutes does aditya take to cover a distance of num__400 m if he runs at a speed of num__20 km / hr <o> a ) num__1 num__0.2 min <o> b ) num__2 num__0.2 min <o> c ) num__3 num__0.2 min <o> d ) num__4 num__0.2 min <o> e ) num__5 num__0.2 min |
explanation : we know that time = distance / speed speed = num__20 km / hr = num__20 ∗ num__0.277777777778 m / sec = num__5.55555555556 m / sec time = ( num__400 ∗ num__0.18 ) = num__72 sec = num__1 num__0.2 min option a <eor> a <eos> |
a |
multiply__400.0__0.18__ multiply__0.18__5.5556__ round__1.0__ |
multiply__400.0__0.18__ multiply__0.18__5.5556__ round__1.0__ |
| can anyhow help me with an easy solution for this num__3 num__4 num__49 num__5 num__3 num__64 num__7 num__6 ? <o> a ) num__121 <o> b ) num__169 <o> c ) num__144 <o> d ) num__256 <o> e ) num__225 |
num__3 num__4 num__49 num__5 num__3 num__64 num__7 num__6 ? num__3 + num__4 = num__7 and num__7 ^ num__2 = num__49 num__5 + num__3 = num__8 and num__8 ^ num__2 = num__64 num__7 + num__6 = num__13 and num__13 ^ num__2 = num__169 answer : b <eor> b <eos> |
b |
subtract__5.0__3.0__ add__3.0__5.0__ add__5.0__8.0__ power__13.0__2.0__ power__13.0__2.0__ |
subtract__5.0__3.0__ add__3.0__5.0__ add__5.0__8.0__ power__13.0__2.0__ power__13.0__2.0__ |
| there are num__4 people of different heights standing in order of increasing height . the difference is num__2 inches between the first person and the second person and also between the second person and the third person . the difference between the third person and the fourth person is num__6 inches and the average height is num__77 . how tall is the fourth person ? <o> a ) num__79 <o> b ) num__81 <o> c ) num__83 <o> d ) num__85 <o> e ) num__87 |
let x be the height of the first person . then the heights are x x + num__2 x + num__4 and x + num__10 . num__4 x + num__16 = num__4 ( num__77 ) = num__308 x = num__73 and the fourth person has a height of num__73 + num__10 = num__83 inches the answer is c . <eor> c <eos> |
c |
add__4.0__6.0__ add__6.0__10.0__ multiply__4.0__77.0__ subtract__77.0__4.0__ add__6.0__77.0__ add__6.0__77.0__ |
add__4.0__6.0__ add__6.0__10.0__ multiply__4.0__77.0__ subtract__77.0__4.0__ add__6.0__77.0__ add__6.0__77.0__ |
| a group of men decided to do a work in num__15 days but num__8 of them became absent . if the rest of the group did the work in num__18 days find the original number of men ? <o> a ) num__15 <o> b ) num__48 <o> c ) num__54 <o> d ) num__33 <o> e ) num__45 |
original number of men = num__8 * num__18 / ( num__18 - num__15 ) = num__48 answer is b <eor> b <eos> |
b |
round__48.0__ |
round__48.0__ |
| the elevator in an eleven - story office building travels at the rate of one floor per num__0.25 minute which allows time for picking up and discharging passengers . at the main floor and at the top floor the operator stops for num__1 minute . how many complete trips will an operator make during a num__6 - hour period ? <o> a ) num__88 <o> b ) num__56 <o> c ) num__42 <o> d ) num__51 <o> e ) num__64 |
complete trip = num__10 floors up and num__10 floors down = num__20 floors = num__20 * num__0.25 = num__5 minutes plus num__2 minutes = num__7 minutes . num__6 hour = num__360 minutes . in num__360 minutes operator can make num__51.4285714286 = num__51 trips . answer : d . <eor> d <eos> |
d |
multiply__0.25__20.0__ divide__10.0__5.0__ add__1.0__6.0__ divide__360.0__7.0__ round__51.0__ |
multiply__0.25__20.0__ divide__10.0__5.0__ add__1.0__6.0__ divide__360.0__7.0__ multiply__1.0__51.0__ |
| rohan spends num__40.0 of his salary on food num__20.0 on house rent num__10.0 on entertainment and num__10.0 on conveyance . if his savings at the end of a month are rs . num__1000 . then his monthly salary is <o> a ) rs . num__5000 <o> b ) rs . num__6000 <o> c ) rs . num__4000 <o> d ) rs . num__3000 <o> e ) rs . num__2000 |
sol . saving = [ num__100 - ( num__40 + num__20 + num__10 + num__10 ] % = num__20.0 . let the monthly salary be rs . x . then num__20.0 of x = num__1000 â ‡ ” num__0.2 x = num__1000 â ‡ ” x = num__1000 Ã — num__5 = num__5000 . answer a <eor> a <eos> |
a |
percent__10.0__1000.0__ percent__100.0__5000.0__ |
percent__10.0__1000.0__ percent__100.0__5000.0__ |
| a tradesman by means of his false balance defrauds to the extent of num__40.0 ? in buying goods as well as by selling the goods . what percent does he gain on his outlay ? <o> a ) num__80.0 <o> b ) num__96.0 <o> c ) num__92.0 <o> d ) num__84.0 <o> e ) num__86 % |
g % = num__40 + num__40 + ( num__40 * num__40 ) / num__100 = num__96.0 answer : b <eor> b <eos> |
b |
percent__96.0__100.0__ |
percent__96.0__100.0__ |
| present ages of x and y are in the ratio num__5 : num__6 respectively . seven years hence this ratio will become num__6 : num__7 respectively . what is x ' s present age in years ? <o> a ) num__35 years <o> b ) num__37 years <o> c ) num__16 years <o> d ) num__17 years <o> e ) num__18 years |
let the present ages of x and y be num__5 x and num__6 x years respectively . then ( num__5 x + num__7 ) / ( num__6 x + num__7 ) = num__0.857142857143 num__7 ( num__5 x + num__7 ) = num__6 ( num__6 x + num__7 ) = > x = num__7 x ' s present age = num__5 x = num__35 years . answer : a <eor> a <eos> |
a |
divide__6.0__7.0__ multiply__5.0__7.0__ multiply__5.0__7.0__ |
divide__6.0__7.0__ multiply__5.0__7.0__ multiply__5.0__7.0__ |
| what is the average of first num__21 multiples of num__4 ? <o> a ) a ) num__44 <o> b ) b ) num__77 <o> c ) c ) num__79 <o> d ) d ) num__81 <o> e ) e ) num__82 |
required average = num__7 ( num__1 + num__2 + . . . . + num__21 ) / num__21 ( num__0.190476190476 ) x ( ( num__21 x num__22 ) / num__2 ) ( because sum of first num__21 natural numbers ) = num__44 a <eor> a <eos> |
a |
divide__4.0__21.0__ add__21.0__1.0__ multiply__2.0__22.0__ multiply__1.0__44.0__ |
divide__4.0__21.0__ add__21.0__1.0__ multiply__2.0__22.0__ divide__44.0__1.0__ |
| the average age of num__10 students in a class is num__15 years . if the age of teacher is also included the average becomes num__16 years find the age of the teacher . <o> a ) num__30 <o> b ) num__26 <o> c ) num__28 <o> d ) num__29 <o> e ) num__45 |
explanation : if teacher ' s age is num__15 years there is no change in the average . but teacher has contributed num__1 year to all the students along with maintaining his age at num__16 . age of teacher = average age of all + total increase in age = num__16 + ( num__1 x num__10 ) = num__26 years answer : b <eor> b <eos> |
b |
subtract__16.0__15.0__ add__10.0__16.0__ add__10.0__16.0__ |
subtract__16.0__15.0__ add__10.0__16.0__ add__10.0__16.0__ |
| a man goes to his office from his house at a speed of num__3 km / hr and returns at a speed of num__2 km / hr . if he takes num__5 hours in going and coming what is the distance between his house and office ? <o> a ) num__8 <o> b ) num__7 <o> c ) num__4 <o> d ) num__6 <o> e ) num__09 |
if a car covers a certain distance at x kmph and an equal distance at y kmph the average speed of the whole journey = num__2 xyx + y kmphhence average speed = num__2 × num__3 × num__22 + num__3 = num__125 km / hrtotal time taken = num__5 hours ⇒ distance travelled = num__125 × num__5 = num__12 km ⇒ distance between his house and office = num__122 = num__6 km answer : d <eor> d <eos> |
d |
subtract__125.0__3.0__ multiply__3.0__2.0__ round__6.0__ |
subtract__125.0__3.0__ divide__12.0__2.0__ divide__12.0__2.0__ |
| if d is the smallest positive integer such that num__3150 multiplied by d is the square of an integer then d must be <o> a ) num__2 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__14 |
solution : this problem is testing us on the rule that when we express a perfect square by its unique prime factors every prime factor ' s exponent is an even number . let ’ s start by prime factorizing num__3150 . num__3150 = num__315 x num__10 = num__5 x num__63 x num__10 = num__5 x num__7 x num__3 x num__3 x num__5 x num__2 num__3150 = num__2 ^ num__1 x num__3 ^ num__2 x num__5 ^ num__2 x num__7 ^ num__1 ( notice that the exponents of both num__2 and num__7 are not even numbers . this tells us that num__3150 itself is not a perfect square . ) we also are given that num__3150 multiplied by d is the square of an integer . we can write this as : num__2 ^ num__1 x num__3 ^ num__2 x num__5 ^ num__2 x num__7 ^ num__1 x d = square of an integer according to our rule we need all unique prime factors ' exponents to be even numbers . thus we need one more num__2 and one more num__7 . therefore d = num__7 x num__2 = num__14 answer is e . <eor> e <eos> |
e |
rectangle_perimeter__2.0__5.0__ multiply__1.0__14.0__ |
rectangle_perimeter__2.0__5.0__ power__14.0__1.0__ |
| john takes num__6 hrs to print num__32 pages . peter takes num__5 hrs to print num__40 pages . in how many hours they will print num__110 pages together ? <o> a ) num__8 hours <o> b ) num__8 hours num__15 minutes <o> c ) num__9 hours num__15 minutes <o> d ) num__11 hours <o> e ) num__12 hours |
ram can type num__5.33333333333 pages per hour shaam can type num__8 pages per hourthey can type num__5.33333333333 + num__8 = num__13.3333333333 pages per hour so they will take num__2.75 × num__3 = num__8.25 hours or num__8 hours num__15 minutes b ) <eor> b <eos> |
b |
divide__32.0__6.0__ subtract__40.0__32.0__ add__5.3333__8.0__ divide__110.0__40.0__ divide__40.0__13.3333__ divide__110.0__13.3333__ multiply__5.0__3.0__ round__8.0__ |
divide__32.0__6.0__ subtract__40.0__32.0__ add__5.3333__8.0__ divide__110.0__40.0__ divide__40.0__13.3333__ divide__110.0__13.3333__ multiply__5.0__3.0__ add__5.0__3.0__ |
| num__70 |
78 num__8085 num__90105 num__105130 num__130130 the list consists of the times in seconds that it took each of the num__10 school children to run a distance of num__400 mts . if the standard deviation of the num__10 running times is num__22.4 rounded to the nearest tenth of a second how many of the num__10 running times are more than num__1 standard deviation below the mean of the num__10 running times . <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
how many of the num__10 running times are more than one sd below the meanmeans how many data points from given num__10 are less thanmean - num__1 sd . we are given that sd = num__22.4 so we should find mean - - > mean = num__100 - - > there is only num__1 data points below num__100 - num__22.4 = num__77.6 namely num__70 answer : a . <eor> a <eos> |
a |
a |
| the shopkeeper increased the price of a product by num__25.0 so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only num__70.0 of the required amount . what is the net difference in the expenditure on that product ? <o> a ) num__11.0 <o> b ) num__12.0 <o> c ) num__12.5 <o> d ) num__13.0 <o> e ) num__14 % |
quantity x rate = price num__1 x num__1 = num__1 num__0.7 x num__1.25 = num__0.875 decrease in price = ( num__0.125 / num__1 ) × num__100 = num__12.5 c <eor> c <eos> |
c |
multiply__1.25__0.7__ subtract__1.0__0.875__ divide__70.0__0.7__ multiply__100.0__0.125__ subtract__25.0__12.5__ |
multiply__1.25__0.7__ subtract__1.0__0.875__ divide__70.0__0.7__ multiply__100.0__0.125__ divide__12.5__1.0__ |
| the greatest number of four digits which is divisible by num__25 num__40 num__54 and num__75 is : <o> a ) num__5000 <o> b ) num__5400 <o> c ) num__5600 <o> d ) num__5800 <o> e ) num__5200 |
greatest number of num__4 - digits is num__9999 . l . c . m . of num__25 num__40 num__54 and num__75 is num__600 . on dividing num__9999 by num__5400 the remainder is num__4599 . required number ( num__9999 - num__4599 ) = num__5400 . answer : option b <eor> b <eos> |
b |
subtract__9999.0__5400.0__ subtract__9999.0__4599.0__ |
subtract__9999.0__5400.0__ subtract__9999.0__4599.0__ |
| the length of a rectangle is increased to num__2 times its original size and its width is increased to num__3 times its original size . if the area of the new rectangle is equal to num__1800 square meters what is the area of the original rectangle ? <o> a ) num__100 square meters <o> b ) num__200 square meters <o> c ) num__300 square meters <o> d ) num__400 square meters <o> e ) num__500 square meters |
if l and w be the original length and width of the rectangle and its area is given by l ? w after increase the length becomes num__2 l and the width becomes num__3 w . the area is then given by ( num__2 l ) ? ( num__3 w ) and is known . hence ( num__2 l ) ? ( num__3 w ) = num__1800 solve the above equation to find l ? w num__6 l ? w = num__1800 l ? w = num__300.0 = num__300 square meters area of original rectangle correct answer c <eor> c <eos> |
c |
multiply__2.0__3.0__ triangle_area__2.0__300.0__ |
multiply__2.0__3.0__ triangle_area__2.0__300.0__ |
| a man invests some money partly in num__12.0 stock at num__132 and partly in num__10.0 stock at num__80 . to obtain equal dividends from both he must invest the money in the ratio : <o> a ) num__3 : num__4 <o> b ) num__3 : num__5 <o> c ) num__11 : num__8 <o> d ) num__16 : num__15 <o> e ) none |
solution for an income of rs . num__1 in num__12.0 stock at num__132 investment = rs . ( num__11.0 ) = rs . num__11 . for an income of rs . num__1 in num__10.0 stock at num__80 investment = rs . ( num__8.0 ) = rs . num__8 . ∴ ratio of investments = num__11 : num__8 = num__11 : num__8 answer c <eor> c <eos> |
c |
subtract__12.0__1.0__ divide__80.0__10.0__ subtract__12.0__1.0__ |
subtract__12.0__1.0__ divide__80.0__10.0__ subtract__12.0__1.0__ |
| working alone jerry finishes cleaning half the house in a third of the time it takes nick to clean the entire house alone . jerry alone cleans the entire house in num__6 hours . how many hours will it take nick and jerry to clean the entire house if they work together ? <o> a ) num__1.5 <o> b ) num__2 <o> c ) num__2.4 <o> d ) num__3 <o> e ) num__3.6 |
answer is num__3.6 hours . jerry does the complete house in num__6 hours while nick does it in num__9 hours . num__1 / ( num__0.166666666667 + num__0.111111111111 ) = num__3.6 answer is e <eor> e <eos> |
e |
divide__1.0__6.0__ divide__1.0__9.0__ round__3.6__ |
divide__1.0__6.0__ divide__1.0__9.0__ divide__3.6__1.0__ |
| . num__008 / ? = . num__01 <o> a ) . num__8 <o> b ) . num__09 <o> c ) . num__009 <o> d ) . num__0009 <o> e ) none of them |
let . num__008 / x = . num__01 ; then x = . num__008 / . num__01 = . num__8.0 = . num__8 answer is a <eor> a <eos> |
a |
multiply__8.0__1.0__ |
divide__8.0__1.0__ |
| tanks q and b are each in the shape of a right circular cylinder . the interior of tank q has a height of num__10 meters and a circumference of num__8 meters and the interior of tank b has a height of num__8 meters and a circumference of num__10 meters . the capacity of tank q is what percent of the capacity of tank b ? <o> a ) num__75.0 <o> b ) num__80.0 <o> c ) num__100.0 <o> d ) num__120.0 <o> e ) num__125 % |
b . for q r = num__4.0 pi . its capacity = ( num__4 pi ) ^ num__2 * num__10 = num__160 pi for b r = num__10 / pi . its capacity = ( num__5 pi ) ^ num__2 * num__8 = num__200 pi q / b = num__160 pi / num__200 pi = num__0.8 <eor> b <eos> |
b |
subtract__10.0__8.0__ divide__10.0__2.0__ divide__8.0__10.0__ multiply__10.0__8.0__ |
divide__8.0__4.0__ divide__10.0__2.0__ divide__8.0__10.0__ multiply__10.0__8.0__ |
| num__10 men working num__4 hours a day can complete a work in num__18 days . how many hours a day must num__15 men work to complete the work in num__12 days ? <o> a ) num__4 hours a day <o> b ) num__5 hours a day <o> c ) num__6 hours a day <o> d ) num__7 hours a day <o> e ) num__8 hours a day |
explanation : more men less hours { indirect proportion } less days more hours { indirect proportion } [ men num__15 num__10 days num__12 num__18 ] : : num__4 : x = > x â ˆ — num__15 â ˆ — num__12 = num__10 â ˆ — num__18 â ˆ — num__4 = > x = num__10 â ˆ — num__18 â ˆ — num__0.266666666667 â ˆ — num__12 = > x = num__4 option a <eor> a <eos> |
a |
divide__4.0__15.0__ round__4.0__ |
divide__4.0__15.0__ round__4.0__ |
| if ( x + num__10 ) ^ num__2 = num__144 which of the following could be the value of num__2 x ? <o> a ) – num__26 <o> b ) – num__22 <o> c ) – num__44 <o> d ) num__66 <o> e ) num__13 |
you can also get to the answer by backsolving using answer choices starting with x = - num__22 ( x + num__10 ) ^ num__2 = num__144 ( - num__22 + num__10 ) ^ num__2 = num__144 = = = = = > ( - num__12 ) ^ num__2 = num__144 bingo clearly among the answer choices only - num__22 will result in num__144 in the equation so x must be - num__22 then num__2 x = num__2 * - num__22 = - num__44 answer : c <eor> c <eos> |
c |
add__10.0__2.0__ multiply__2.0__22.0__ multiply__2.0__22.0__ |
add__10.0__2.0__ multiply__2.0__22.0__ multiply__2.0__22.0__ |
| let f ( x y ) be defined as the remainder when ( x – y ) ! is divided by x . if x = num__36 what is the maximum value of y for which f ( x y ) = num__0 ? <o> a ) num__9 <o> b ) num__12 <o> c ) num__18 <o> d ) num__20 <o> e ) num__30 |
the question is finding y such that ( num__36 - y ) ! is a multiple of num__36 . that means we need to have num__2 ^ num__2 * num__3 ^ num__2 in ( num__36 - y ) ! num__6 ! is the smallest factorial number with num__2 ^ num__2 * num__3 ^ num__2 as a factor . num__36 - y = num__6 y = num__30 the answer is e . <eor> e <eos> |
e |
multiply__2.0__3.0__ subtract__36.0__6.0__ subtract__36.0__6.0__ |
multiply__2.0__3.0__ subtract__36.0__6.0__ subtract__36.0__6.0__ |
| what percent is num__6 gm of num__1 kg ? <o> a ) num__0.06 <o> b ) num__0.6 <o> c ) num__0.006 <o> d ) num__6.0 <o> e ) none of these |
explanation : num__1 kg = num__1000 gm num__0.006 * num__100 = num__0.6 = num__0.6 = num__0.6 answer : option b <eor> b <eos> |
b |
percent__0.6__100.0__ |
percent__0.6__100.0__ |
| the least number which when divided by num__12 num__15 num__20 and num__54 leaves in each case a remainder of num__5 is : <o> a ) num__545 <o> b ) num__488 <o> c ) num__542 <o> d ) num__548 <o> e ) num__560 |
required number = ( l . c . m . of num__12 num__15 num__20 num__54 ) + num__5 = num__540 + num__5 = num__545 . answer : a <eor> a <eos> |
a |
add__5.0__540.0__ add__5.0__540.0__ |
add__5.0__540.0__ add__5.0__540.0__ |
| a can do a piece of work in num__12 days . he worked for num__15 days and then b completed the remaining work in num__10 days . both of them together will finish it in . <o> a ) num__12 num__0.5 days <o> b ) num__12 num__0.111111111111 days <o> c ) num__12 num__4.0 days <o> d ) num__17 num__0.5 days <o> e ) num__32 num__0.5 days |
num__0.6 + num__10 / x = num__1 = > x = num__25 num__0.04 + num__0.04 = num__0.08 num__12.5 = num__12 num__0.5 days answer : a <eor> a <eos> |
a |
km_to_mile_conversion__ add__15.0__10.0__ divide__1.0__25.0__ divide__1.0__0.08__ subtract__12.5__12.0__ round__12.0__ |
km_to_mile_conversion__ add__15.0__10.0__ divide__1.0__25.0__ divide__1.0__0.08__ divide__12.5__25.0__ divide__12.0__1.0__ |
| the average age of students of a class is num__15.8 years . the average age of boys in the class is num__16.4 years and that of the girls is num__15.4 years . the ration of the number of boys to the number of girls in the class is ? <o> a ) num__2 : num__6 <o> b ) num__2 : num__3 <o> c ) num__2 : num__5 <o> d ) num__2 : num__1 <o> e ) num__2 : num__4 |
let the ratio be k : num__1 . then k * num__16.4 + num__1 * num__15.4 = ( k + num__1 ) * num__15.8 = ( num__16.4 - num__15.8 ) k = ( num__15.8 - num__15.4 ) = k = num__0.4 / num__0.6 = num__0.666666666667 required ratio = num__0.666666666667 : num__1 = num__2 : num__3 . answer : b <eor> b <eos> |
b |
subtract__16.4__15.4__ subtract__15.8__15.4__ subtract__16.4__15.8__ divide__0.4__0.6__ add__1.0__2.0__ multiply__1.0__2.0__ |
subtract__16.4__15.4__ subtract__15.8__15.4__ subtract__16.4__15.8__ divide__0.4__0.6__ add__1.0__2.0__ multiply__1.0__2.0__ |
| if num__7875 / num__5.25 = num__1500 then num__787.5 / num__52.5 is equal to ? <o> a ) num__15 <o> b ) num__19 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
answer given expression num__787.5 / num__52.5 = num__15.0 = num__7875 / ( num__525 x num__100 ) = num__15.0 num__15 correct option : a <eor> a <eos> |
a |
divide__787.5__52.5__ divide__7875.0__15.0__ divide__1500.0__15.0__ divide__7875.0__525.0__ |
divide__787.5__52.5__ divide__7875.0__15.0__ divide__1500.0__15.0__ divide__7875.0__525.0__ |
| the ratio of number of boys and girls in a school is num__2 : num__5 . if there are num__490 students in the school find the number of girls in the school ? <o> a ) num__150 <o> b ) num__350 <o> c ) num__300 <o> d ) num__370 <o> e ) num__280 |
let the number of boys and girls be num__2 x and num__5 x total students = num__490 number of girls in the school = num__5 * num__70.0 = num__350 answer is b <eor> b <eos> |
b |
multiply__5.0__70.0__ multiply__5.0__70.0__ |
multiply__5.0__70.0__ multiply__5.0__70.0__ |
| what percent is num__320 gm of num__1 kg ? <o> a ) num__25.0 <o> b ) num__40.0 <o> c ) num__10.0 <o> d ) num__8.0 <o> e ) num__32 % |
num__1 kg = num__1000 gm num__0.32 Ã — num__100 = num__32.0 = num__32.0 answer is e <eor> e <eos> |
e |
percent__100.0__32.0__ |
percent__100.0__32.0__ |
| a train moves num__100 m in num__2 sec and then find the speed travelled ? <o> a ) num__28 <o> b ) num__900 <o> c ) num__50 <o> d ) num__180 <o> e ) none |
length = speed * time speed = l / t s = num__100 m / num__2 sec s = num__50 m / sec speed = num__50 * num__3.6 ( to convert m / sec in to kmph multiply by num__3.6 ) speed = num__180 kmph . answer : d <eor> d <eos> |
d |
divide__100.0__2.0__ multiply__3.6__50.0__ round__180.0__ |
divide__100.0__2.0__ multiply__3.6__50.0__ multiply__3.6__50.0__ |
| num__7 men and num__2 boys working together can do four times as much work as a man and a boy . working capacity of man and boy is in the ratio <o> a ) num__1 : num__2 <o> b ) num__1 : num__3 <o> c ) num__2 : num__1 <o> d ) num__2 : num__3 <o> e ) none of these |
explanation : let num__1 man num__1 day work = x num__1 boy num__1 day work = y then num__7 x + num__2 y = num__4 ( x + y ) = > num__3 x = num__2 y = > x / y = num__0.666666666667 = > x : y = num__2 : num__3 option d <eor> d <eos> |
d |
subtract__7.0__4.0__ divide__2.0__3.0__ multiply__2.0__1.0__ |
add__2.0__1.0__ divide__2.0__3.0__ divide__2.0__1.0__ |
| an empty pool being filled with water at a constant rate takes num__8 hours to fill to num__0.8 of its capacity . how much more time will it take to finish filling the pool ? <o> a ) num__5 hr num__30 min <o> b ) num__5 hr num__20 min <o> c ) num__4 hr num__48 min <o> d ) num__3 hr num__12 min <o> e ) num__5 hr num__40 min |
as pool is filled to num__0.6 of its capacity then num__0.4 of its capacity is left to fill . to fill num__0.6 of the pool took num__8 hours - - > to fill num__0.4 of the pool will take num__8 / ( num__0.6 ) * num__0.4 = num__5.33333333333 hours = num__5 hours num__20 minutes ( because if t is the time needed to fill the pool then t * num__0.6 = num__8 - - > t = num__8 * num__1.66666666667 hours - - > to fill num__0.4 of the pool num__8 * num__1.66666666667 * num__0.4 = num__5.33333333333 hours will be needed ) . or plug values : take the capacity of the pool to be num__5 liters - - > num__0.6 of the pool or num__3 liters is filled in num__8 hours which gives the rate of num__0.375 liters per hour - - > remaining num__2 liters will require : time = job / rate = num__2 / ( num__0.375 ) = num__5.33333333333 hours = num__5 hours num__40 minutes . answer : e . <eor> e <eos> |
e |
round_down__5.3333__ divide__8.0__0.4__ reverse__0.6__ subtract__8.0__5.0__ divide__3.0__8.0__ divide__0.8__0.4__ multiply__8.0__5.0__ round_down__5.3333__ |
round_down__5.3333__ divide__8.0__0.4__ reverse__0.6__ subtract__8.0__5.0__ divide__3.0__8.0__ divide__0.8__0.4__ multiply__8.0__5.0__ subtract__8.0__3.0__ |
| the radii of two cones are in ratio num__2 : num__1 their volumes are equal . find the ratio of their heights . <o> a ) num__1 : num__4 <o> b ) num__1 : num__3 <o> c ) num__1 : num__2 <o> d ) num__1 : num__5 <o> e ) none of these |
explanation : let their radii be num__2 x x and their heights be h and h resp . then volume of cone = num__0.333333333333 π r num__2 h num__0.333333333333 ∗ π ∗ ( num__2 x ) num__2 ∗ h / num__0.333333333333 ∗ π ∗ x num__2 ∗ h = > hh = num__14 = > h : h = num__1 : num__4 option a <eor> a <eos> |
a |
square_perimeter__1.0__ volume_cube__1.0__ |
square_perimeter__1.0__ volume_cube__1.0__ |
| a sum of rs . num__312 was divided among num__100 boys and girls in such a way that the boy gets rs . num__3.60 and each girl rs . num__2.40 the number of girls is <o> a ) num__35 <o> b ) num__40 <o> c ) num__45 <o> d ) num__50 <o> e ) num__55 |
explanation : step ( i ) : let x be the number of boys and y be the number of girls . given total number of boys and girls = num__100 x + y = num__100 - - - - - - - - - - - - - - ( i ) step ( ii ) : a boy gets rs . num__3.60 and a girl gets rs . num__2.40 the amount given to num__100 boys and girls = rs . num__312 num__3.60 x + num__2.40 y = num__312 - - - - - - - - - - - - - - ( ii ) step ( iii ) : solving ( i ) and ( ii ) num__3.60 x + num__3.60 y = num__360 - - - - - - - - - multiply ( i ) by num__3.60 num__3.60 x + num__2.40 y = num__312 - - - - - - - - - ( ii ) num__1.20 y = num__48 y = num__48 / num__1.20 = num__40 number of girls = num__40 answer : b <eor> b <eos> |
b |
multiply__100.0__3.6__ subtract__3.6__2.4__ subtract__360.0__312.0__ divide__48.0__1.2__ divide__48.0__1.2__ |
multiply__100.0__3.6__ subtract__3.6__2.4__ subtract__360.0__312.0__ divide__48.0__1.2__ divide__48.0__1.2__ |
| a and b together can do a piece of work in num__6 days and a alone can do it in num__13 days . in how many days can b alone can do it ? <o> a ) num__12 days <o> b ) num__15 days <o> c ) num__11.1428571429 days <o> d ) num__21 days <o> e ) num__3.14285714286 days |
explanation : a and b can do work num__0.166666666667 in num__1 day a alone can do num__0.0769230769231 work in num__1 day b alone can do ( num__0.166666666667 - num__0.0769230769231 ) = num__0.0897435897436 work in num__1 day = > complete work can be done in num__11.1428571429 days by b answer : option c <eor> c <eos> |
c |
divide__1.0__13.0__ multiply__1.0__11.1429__ |
divide__1.0__13.0__ multiply__1.0__11.1429__ |
| how many bricks each measuring num__20 cm x num__10 cm x num__8 cm will be needed to build a wall num__10 m x num__8 m x num__24.5 m <o> a ) num__12250 <o> b ) num__13400 <o> c ) num__12500 <o> d ) num__14340 <o> e ) none of these |
explanation : no . of bricks = volume of the wall / volume of num__1 brick = ( num__1000 x num__800 x num__24.5 ) / ( num__20 x num__10 x num__8 ) = num__12250 answer : a <eor> a <eos> |
a |
round__12250.0__ |
divide__12250.0__1.0__ |
| in richard ' s office one half of the workers are indians quarter are from america one - fifth are from canada and the other num__100 are china . how many workers total are there in the company ? <o> a ) num__250 <o> b ) num__500 <o> c ) num__750 <o> d ) num__1000 <o> e ) num__2000 |
num__0.5 x + num__0.25 x + num__0.2 x = num__0.95 x remaining num__0.05 x = num__100 x = num__100 * num__20 = num__2000 e is the answer . <eor> e <eos> |
e |
subtract__0.25__0.2__ reverse__0.05__ divide__100.0__0.05__ divide__100.0__0.05__ |
multiply__0.25__0.2__ multiply__100.0__0.2__ multiply__100.0__20.0__ multiply__100.0__20.0__ |
| each day a man meets his wife at the train station after work and then she drives him home . she always arrives exactly on time to pick him up . one day he catches an earlier train and arrives at the station an hour early . he immediately begins walking home along the same route the wife drives . eventually his wife sees him on her way to the station and drives him the rest of the way home . when they arrive home the man notices that they arrived num__30 minutes earlier than usual . how much time did the man spend walking ? <o> a ) num__45 minutes <o> b ) num__50 minutes <o> c ) num__40 minutes <o> d ) num__55 minutes <o> e ) num__35 minutes |
as they arrived num__30 minutes earlier than usual they saved num__30 minutes on round trip from home to station ( home - station - home ) - - > num__15 minutes in each direction ( home - station ) - - > wife meets husband num__15 minutes earlier the usual meeting time - - > husband arrived an hour earlier the usual meeting time so he must have spent waking the rest of the time before their meeting which is hour - num__15 minutes = num__45 minutes . answer : a <eor> a <eos> |
a |
add__30.0__15.0__ round__45.0__ |
add__30.0__15.0__ round__45.0__ |
| how many bricks each measuring num__25 cm x num__11.25 cm x num__6 cm will be needed to build a wall of num__8 m x num__6 m x num__22.5 cm ? <o> a ) num__5600 <o> b ) num__6000 <o> c ) num__6400 <o> d ) num__7200 <o> e ) num__8600 |
number of bricks = volume of the wall / volume of num__1 brick = ( num__800 x num__600 x num__22.5 ) / ( num__25 x num__11.25 x num__6 ) = num__6400 . answer : option c <eor> c <eos> |
c |
multiply__8.0__800.0__ round__6400.0__ |
multiply__8.0__800.0__ divide__6400.0__1.0__ |
| a can do a work in num__14 days . b is num__40.0 more efficient than a . how many days does b alone take to do the same job ? <o> a ) num__9 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__8 |
ratio of times taken by a and b = num__140 : num__100 = num__7 : num__5 suppose b alone to take x days to do the work then num__7 : num__5 : : num__14 : x num__7 x = num__5 * num__14 x = num__10 days correct option is b <eor> b <eos> |
b |
subtract__140.0__40.0__ divide__140.0__14.0__ round__10.0__ |
subtract__140.0__40.0__ divide__140.0__14.0__ round__10.0__ |
| simplify : num__0.6 * num__0.6 + num__0.6 * num__0.6 <o> a ) num__0.52 <o> b ) num__0.42 <o> c ) num__0.48 <o> d ) num__0.64 <o> e ) num__0.46 |
given exp . = num__0.6 * num__0.6 + ( num__0.6 * num__6 ) = num__0.36 + num__0.1 = num__0.46 answer is e . <eor> e <eos> |
e |
divide__0.6__6.0__ add__0.36__0.1__ add__0.36__0.1__ |
divide__0.6__6.0__ add__0.36__0.1__ add__0.36__0.1__ |
| a book is bought for rs . num__600 / - and sold at a profit of num__10.0 find its selling price ? <o> a ) s . num__740 / - <o> b ) s . num__660 / - <o> c ) s . num__680 / - <o> d ) s . num__720 / - <o> e ) s . num__620 / - |
num__100.0 - - - - - - > num__600 ( num__100 * num__6 = num__600 ) num__110.0 - - - - - - > num__660 ( num__110 * num__6 = num__660 ) selling price = rs . num__660 / - option ' b ' <eor> b <eos> |
b |
percent__100.0__660.0__ |
percent__100.0__660.0__ |
| rekha took a loan of rs . num__1200 with simple interest for as many years as the rate of interest . if she paid rs . num__588 as interest at the end of the loan period what was the rate of interest ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__4 <o> d ) num__3 <o> e ) num__7 |
let rate = r % and time = r years . then ( num__1200 * r * r ) / num__100 = num__588 num__12 r num__2 = num__588 r num__2 = num__49 = > r = num__7 . answer : e <eor> e <eos> |
e |
percent__100.0__7.0__ |
percent__100.0__7.0__ |
| a cistern can be filled by a tap in num__4 hours while it can be emptied by another tap in num__9 hours . if both the taps are opened simultaneously then after how much time will the cistern get filled ? <o> a ) num__5.2 hrs <o> b ) num__2.9 hrs <o> c ) num__1.9 hrs <o> d ) num__7.2 hrs <o> e ) num__1.4 hrs |
net part filled in num__1 hour = ( num__0.25 - num__0.111111111111 ) = num__0.138888888889 the cistern will be filled in num__7.2 hrs i . e . num__7.2 hrs . answer : d <eor> d <eos> |
d |
divide__1.0__4.0__ divide__1.0__9.0__ subtract__0.25__0.1111__ round__7.2__ |
divide__1.0__4.0__ divide__1.0__9.0__ subtract__0.25__0.1111__ round__7.2__ |
| in what time will a train num__200 meters long cross an electric pole if its speed is num__128 km / hr <o> a ) num__5 seconds <o> b ) num__5.7 seconds <o> c ) num__3 seconds <o> d ) num__2.5 seconds <o> e ) none of these |
explanation : first convert speed into m / sec speed = num__128 * ( num__0.277777777778 ) = num__35 m / sec time = distance / speed = num__5.71428571429 = num__5.7 seconds option b <eor> b <eos> |
b |
divide__200.0__35.0__ round__5.7__ |
divide__200.0__35.0__ round__5.7__ |
| two trains are moving in the same direction at num__72 kmph and num__36 kmph . the faster train crosses a man in the slower train in num__27 seconds . find the length of the faster train ? <o> a ) num__270 <o> b ) num__287 <o> c ) num__269 <o> d ) num__200 <o> e ) num__271 |
relative speed = ( num__72 - num__36 ) * num__0.277777777778 = num__2 * num__5 = num__10 mps . distance covered in num__27 sec = num__27 * num__10 = num__270 m . the length of the faster train = num__270 m . answer : a <eor> a <eos> |
a |
divide__72.0__36.0__ multiply__2.0__5.0__ multiply__27.0__10.0__ round__270.0__ |
divide__72.0__36.0__ multiply__2.0__5.0__ multiply__27.0__10.0__ multiply__27.0__10.0__ |
| which of the following is closest to num__0.166666666667 + num__0.0151515151515 + num__0.0015015015015 <o> a ) num__1 / num__4.5 <o> b ) num__0.142857142857 <o> c ) num__1 / num__5.67 <o> d ) num__0.111111111111 <o> e ) num__0.125 |
num__0.166666666667 + num__0.0151515151515 + num__0.0015015015015 = num__0.16 + num__0.015 + num__0.0015 = num__0.1765 = num__1 / num__5.665 answer c <eor> c <eos> |
c |
reverse__1.0__ |
reverse__1.0__ |
| bill made a profit of num__10.0 by selling a product . if he had purchased that product for num__10.0 less and sold it at a profit of num__30.0 he would have received $ num__35 more . what was his original selling price ? <o> a ) $ num__770 <o> b ) $ num__550 <o> c ) $ num__700 <o> d ) $ num__1100 <o> e ) $ num__840 |
let the original purchase price be x so original selling price at num__10.0 profit = num__1.1 x if product is purchased at num__10.0 less of original = num__0.9 x profit of num__30.0 on this price = num__1.3 ( num__0.9 x ) he would have received $ num__35 more in second scenario = > num__1.3 ( num__0.9 x ) - num__1.1 x = num__35 = > num__0.07 x = num__35 = > x = $ num__500 original purchase price = $ num__500 hence original selling price ( at num__10.0 of profit ) = num__1.1 ( num__500 ) = $ num__550 option b <eor> b <eos> |
b |
divide__35.0__0.07__ multiply__1.1__500.0__ multiply__1.1__500.0__ |
divide__35.0__0.07__ multiply__1.1__500.0__ multiply__1.1__500.0__ |
| the side of a square is increased by num__15.0 then how much % does its area increases ? <o> a ) num__30.75 <o> b ) num__35.85 <o> c ) num__37.55 <o> d ) num__35.65 <o> e ) num__32.25 % |
a = num__100 a num__2 = num__10000 a = num__115 a num__2 = num__13225 - - - - - - - - - - - - - - - - num__10000 - - - - - - - - - num__3225 num__100 - - - - - - - ? = > num__32.25 answer : e <eor> e <eos> |
e |
power__100.0__2.0__ power__115.0__2.0__ triangle_area__32.25__2.0__ |
power__100.0__2.0__ power__115.0__2.0__ triangle_area__32.25__2.0__ |
| three cannons are firing at a target . if their individual probabilities to hit the target are num__0.4 num__0.2 and num__0.3 respectively what is the probability that none of the cannons will hit the target after one round of fire ? prob . that all the cannons will hit the target = . num__06 prob . that none of the cannons will hit = num__1 - . num__06 = . num__94 <o> a ) num__0.06 <o> b ) num__0.336 <o> c ) num__0.21 <o> d ) num__0.29 <o> e ) num__0.94 |
the probability that eachdoesn ' thit is : num__0.4 num__0.2 and num__0.3 . when we have multiple independent events we multiply the probabilities : . num__6 * . num__8 * . num__7 = num__0.336 . option : b <eor> b <eos> |
b |
add__6.0__1.0__ multiply__1.0__0.336__ |
add__6.0__1.0__ multiply__1.0__0.336__ |
| the average weight of three boys p q and r is num__54 kg while the average weight of three boys q s and t is num__60 kg . what is the average weight of p q r s and t ? <o> a ) num__338 <o> b ) num__279 <o> c ) num__279 <o> d ) num__342 <o> e ) num__211 |
explanation : total weight of ( p + q + r ) = { num__54 * num__3 } kg = num__162 kg total weight of ( q + s + t ) = ( num__60 * num__3 ) kg = num__180 kg adding both we get : p + num__2 q + s + r + t = ( num__162 + num__180 ) kg = num__342 kg so to find the average weight of p q r s & t we ought to know q ' s weight which is not given . the data is inadequate . answer : d <eor> d <eos> |
d |
multiply__54.0__3.0__ multiply__60.0__3.0__ add__162.0__180.0__ add__162.0__180.0__ |
multiply__54.0__3.0__ multiply__60.0__3.0__ add__162.0__180.0__ add__162.0__180.0__ |
| the length of a rectangular plot is num__20 mtr more than its breadth . if the cost of fencingthe plot at num__26.50 per meter is rs . num__5300 what is the length of the plot in mtr ? <o> a ) num__50 m <o> b ) num__60 m <o> c ) num__65 m <o> d ) num__75 m <o> e ) num__80 m |
let breadth = x metres . then length = ( x + num__20 ) metres . perimeter = num__5300 m = num__200 m . num__26.50 num__2 [ ( x + num__20 ) + x ] = num__200 num__2 x + num__20 = num__100 num__2 x = num__80 x = num__40 . hence length = x + num__20 = num__60 m b <eor> b <eos> |
b |
divide__5300.0__26.5__ divide__200.0__2.0__ subtract__100.0__20.0__ multiply__20.0__2.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
divide__5300.0__26.5__ divide__200.0__2.0__ subtract__100.0__20.0__ multiply__20.0__2.0__ add__20.0__40.0__ add__20.0__40.0__ |
| using all the letters of the word ` ` friday ' ' how many different words can be formed ? <o> a ) a ) num__7 <o> b ) b ) num__8 ! <o> c ) c ) num__8 <o> d ) d ) num__7 ! <o> e ) e ) num__5 ! |
explanation : total number of letters = num__5 using these letters the number of num__5 letters words formed is num__5 ! . answer : option e <eor> e <eos> |
e |
vowel_space__ vowel_space__ |
vowel_space__ vowel_space__ |
| when num__52416 is divided by num__312 the quotient is num__168 . what will be the quotient when num__522.416 is divided by num__0.0168 ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__4 <o> d ) num__3 <o> e ) num__7 |
for the num__1 st no . there are num__3 digits after decimal for the num__2 nd no . there are num__5 digits after decimal total no . of decimals = num__8 req . no . of digits = ( n - num__1 ) = ( num__8 - num__1 ) = num__7 answer : e <eor> e <eos> |
e |
subtract__3.0__1.0__ add__2.0__3.0__ add__3.0__5.0__ add__2.0__5.0__ multiply__1.0__7.0__ |
subtract__3.0__1.0__ add__2.0__3.0__ add__3.0__5.0__ subtract__8.0__1.0__ subtract__8.0__1.0__ |
| the average monthly salary of laborers and supervisors in a factory is rs . num__1250 per month ; where as the average monthly salary of num__6 supervisors is rs . num__2450 . if the average monthly salary of the laborers is rs . num__850 find the number of laborers ? <o> a ) num__14 <o> b ) num__15 <o> c ) num__16 <o> d ) num__17 <o> e ) num__18 |
num__2450 num__850 \ / num__1250 / \ num__400 num__1200 num__1 : num__3 num__1 - - > num__6 num__3 - - > ? num__18 answer : e <eor> e <eos> |
e |
subtract__1250.0__850.0__ subtract__2450.0__1250.0__ divide__1200.0__400.0__ multiply__6.0__3.0__ multiply__6.0__3.0__ |
subtract__1250.0__850.0__ subtract__2450.0__1250.0__ divide__1200.0__400.0__ multiply__6.0__3.0__ divide__18.0__1.0__ |
| a goods train runs at the speed of num__72 km / hr and crosses a num__250 m long platform in num__26 sec . what is the length of the goods train ? <o> a ) num__187 m <o> b ) num__278 m <o> c ) num__279 m <o> d ) num__270 m <o> e ) num__872 m |
speed = num__72 * num__0.277777777778 = num__20 m / sec . time = num__26 sec . let the length of the train be x meters . then ( x + num__250 ) / num__26 = num__20 x = num__270 m . answer : d <eor> d <eos> |
d |
add__250.0__20.0__ round__270.0__ |
add__250.0__20.0__ add__250.0__20.0__ |
| if the area of circle is num__616 sq cm then its circumference ? <o> a ) num__11 <o> b ) num__88 <o> c ) num__99 <o> d ) num__266 <o> e ) num__12 |
num__3.14285714286 r num__2 = num__616 = > r = num__14 num__2 * num__3.14285714286 * num__14 = num__88 answer : b <eor> b <eos> |
b |
triangle_area__88.0__2.0__ |
triangle_area__88.0__2.0__ |
| find the simple interest on rs . num__567 for num__7 months at num__9 paisa per month ? <o> a ) s . num__357.21 <o> b ) s . num__322.12 <o> c ) s . num__400 <o> d ) s . num__278.9 <o> e ) s . num__300 |
explanation : i = ( num__567 * num__7 * num__9 ) / num__100 = num__357.21 answer : option a <eor> a <eos> |
a |
percent__100.0__357.21__ |
percent__100.0__357.21__ |
| look at this series : num__97 num__97 num__93 num__93 num__89 num__89 num__85 num__85 . . . what number should come next ? <o> a ) num__75 <o> b ) num__76 <o> c ) num__78 <o> d ) num__81 <o> e ) num__80 |
explanation : in this series each number is repeated then num__4 is subtracted to arrive at the next number . answer : option d <eor> d <eos> |
d |
subtract__97.0__93.0__ subtract__85.0__4.0__ |
subtract__97.0__93.0__ subtract__85.0__4.0__ |
| joe needs to paint all the airplane hangars at the airport so he buys num__360 gallons of paint to do the job . during the first week he uses num__0.25 of all the paint . during the second week he uses num__0.333333333333 of the remaining paint . how many gallons of paint has joe used ? <o> a ) num__18 <o> b ) num__180 <o> c ) num__175 <o> d ) num__216 <o> e ) num__250 |
total paint initially = num__360 gallons paint used in the first week = ( num__0.25 ) * num__360 = num__90 gallons . remaning paint = num__270 gallons paint used in the second week = ( num__0.333333333333 ) * num__270 = num__90 gallons total paint used = num__180 gallons . option b <eor> b <eos> |
b |
multiply__360.0__0.25__ subtract__360.0__90.0__ subtract__270.0__90.0__ round__180.0__ |
multiply__360.0__0.25__ subtract__360.0__90.0__ subtract__270.0__90.0__ round__180.0__ |
| find the number that fits somewhere into the middle of the series . some of the items involve both numbers and letters look at this series : c num__99 f num__102 __ l num__108 o num__111 . . . what number should fill the blank ? <o> a ) i num__105 <o> b ) j num__106 <o> c ) k num__107 <o> d ) l num__109 <o> e ) m num__110 |
a i num__105 in this series the letters progress by num__3 and the numbers increase by num__3 . <eor> a <eos> |
a |
subtract__102.0__99.0__ add__102.0__3.0__ |
subtract__102.0__99.0__ add__102.0__3.0__ |
| the average of num__10 numbers is calculated as num__18 . it is discovered later on that while calculating the average one number namely num__36 was wrongly read as num__26 . the correct average is ? <o> a ) a ) num__16 <o> b ) b ) num__18 <o> c ) c ) num__19 <o> d ) d ) num__22 <o> e ) e ) num__24 |
explanation : num__10 * num__18 + num__36 – num__26 = num__190 = > num__19.0 = num__19 c ) <eor> c <eos> |
c |
divide__190.0__10.0__ divide__190.0__10.0__ |
divide__190.0__10.0__ divide__190.0__10.0__ |
| a coin is weighted so that the probability of heads on any flip is num__0.2 while the probability of tails is num__0.8 . if the coin is flipped num__5 times independently which of the following represents the probability that tails will appear no more than twice ? <o> a ) ( num__0.2 ) ^ num__5 + num__5 ( num__0.2 ) ^ num__4 ( num__0.8 ) + num__10 ( num__0.2 ) ^ num__3 ( num__0.8 ) ^ num__2 <o> b ) ( num__0.3 ) ^ num__5 + num__4 ( num__0.3 ) ^ num__4 ( num__0.8 ) + num__6 ( num__0.3 ) ^ num__3 ( num__0.8 ) ^ num__2 <o> c ) ( num__0.3 ) ^ num__5 + num__3 ( num__0.3 ) ^ num__4 ( num__0.8 ) + num__2 ( num__0.3 ) ^ num__3 ( num__0.8 ) ^ num__2 <o> d ) ( num__0.3 ) ^ num__5 + num__2 ( num__0.3 ) ^ num__4 ( num__0.8 ) + ( num__0.3 ) ^ num__3 ( num__0.8 ) ^ num__2 <o> e ) ( num__0.6 ) ^ num__5 + ( num__0.6 ) ^ num__4 ( num__0.4 ) + ( num__0.6 ) ^ num__3 ( num__0.4 ) ^ num__2 |
probability of head p ( h ) = num__0.2 probability of tail p ( t ) = num__0.8 tail will appear no more than twice i . e . favourable cases num__2 tails and num__3 heads probability = num__5 c num__2 * ( num__0.2 ) ^ num__3 * ( num__0.8 ) ^ num__2 num__1 tail and num__4 heads probability = num__5 c num__1 * ( num__0.2 ) ^ num__4 * ( num__0.8 ) ^ num__2 num__0 tail and num__5 heads probability = ( num__0.2 ) ^ num__5 required probability = sum of all favourable cases = ( num__0.2 ) ^ num__5 + num__5 ( num__0.2 ) ^ num__4 ( num__0.8 ) + num__10 ( num__0.2 ) ^ num__3 ( num__0.8 ) ^ num__2 answer : option a <eor> a <eos> |
a |
coin_space__ negate_prob__1.0__ negate_prob__0.8__ |
coin_space__ negate_prob__1.0__ negate_prob__0.8__ |
| if circles x and y have the same area and circle x has a circumference of num__14 π half of the radius of circle y is : <o> a ) num__16 <o> b ) num__8 <o> c ) num__4 <o> d ) num__3.5 <o> e ) num__1 |
x be radius of circle x y be radius of circle y given : pi * x ^ num__2 = pi * y ^ num__2 also num__2 * pi * x = num__14 * pi x = num__7 thus y = num__7 y / num__2 = num__3.5 ans : d <eor> d <eos> |
d |
triangle_area__3.5__2.0__ |
triangle_area__3.5__2.0__ |
| if n is an integer number and m = num__115 n then m must be divisible by which of the following ? <o> a ) num__11 <o> b ) num__15 <o> c ) num__23 <o> d ) num__25 <o> e ) num__230 |
m = num__115 n m = num__5 ∗ num__23 n so m must be divisible by num__5 & num__23 both . . . among the given options only ( c ) satisfies . <eor> c <eos> |
c |
divide__115.0__5.0__ divide__115.0__5.0__ |
divide__115.0__5.0__ divide__115.0__5.0__ |
| num__3.5 can be expressed in terms of percentage as ? <o> a ) num__0.35 <o> b ) num__3.5 <o> c ) num__35.0 <o> d ) num__350.0 <o> e ) none |
num__3.5 = num__3.5 = ( num__3.5 * num__100 ) = num__350.0 answer : d <eor> d <eos> |
d |
multiply__3.5__100.0__ multiply__3.5__100.0__ |
multiply__3.5__100.0__ multiply__3.5__100.0__ |
| the length of the bridge which a train num__130 metres long and travelling at num__45 km / hr can cross in num__30 seconds is : <o> a ) num__2878 <o> b ) num__277 <o> c ) num__245 <o> d ) num__299 <o> e ) num__271 |
speed = [ num__45 x num__0.277777777778 ] m / sec = [ num__12.5 ] m / sec time = num__30 sec let the length of bridge be x metres . then ( num__130 + x ) / num__30 = num__12.5 = > num__2 ( num__130 + x ) = num__750 = > x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| tom can hit a target num__3 times in num__6 shots marry can hit the target num__2 times in num__6 shots & mike can hit num__4 times in num__4 shots . what is the probability that atleast num__2 shots hit the target ? <o> a ) num__0.75 <o> b ) num__0.8 <o> c ) num__0.666666666667 <o> d ) num__0.714285714286 <o> e ) num__0.75 |
ashwini hits the target definitely hence required probability that atleast num__2 shots hit the target is given by karan hits tha target and raja not hit the target . or karan does not hit the target and raja hits the target . or . karan hits the target and raja hits the target = num__0.333333333333 x num__0.5 + num__0.666666666667 x num__0.5 + num__0.333333333333 x num__0.5 = num__0.666666666667 = num__0.666666666667 c <eor> c <eos> |
c |
negate_prob__0.3333__ negate_prob__0.3333__ |
negate_prob__0.3333__ negate_prob__0.3333__ |
| a took num__15 seconds to cross a rectangular field diagonally walking at the rate of num__52 m / min and b took the same time to cross the same field along its sides walking at the rate of num__68 m / min . the area of the field is ? <o> a ) num__33 <o> b ) num__60 <o> c ) num__88 <o> d ) num__27 <o> e ) num__26 |
explanation : length of the diagonal = num__52 * ( num__0.25 ) = num__13 m sum of length and breadth = num__68 * ( num__0.25 ) = num__17 m \ inline { \ color { black } \ sqrt { l ^ { num__2 } + b ^ { num__2 } } = num__13 \ ; or \ ; l + b = num__17 } area = lb = num__0.5 [ ( num__2 lb ) ] = num__0.5 [ ( l + b ) ² - ( l ² + b ² ) ] = num__0.5 [ num__17 ² - num__169 ] = num__0.5 * num__120 = num__60 sq meter answer : b ) num__60 <eor> b <eos> |
b |
multiply__52.0__0.25__ multiply__68.0__0.25__ multiply__0.25__2.0__ power__13.0__2.0__ square_perimeter__15.0__ square_perimeter__15.0__ |
multiply__52.0__0.25__ multiply__68.0__0.25__ multiply__0.25__2.0__ power__13.0__2.0__ multiply__0.5__120.0__ multiply__0.5__120.0__ |
| the difference between a two - digit number and the number obtained by interchanging the positions of its digits is num__81 . what is the difference between the two digits of that number ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
sol . let the ten ’ s digit be x and unit ’ s digit be y then ( num__10 x + y ) - ( num__10 y + x ) = num__81 ⇔ num__9 ( x - y ) = num__81 ⇔ x - y = num__9 answer e <eor> e <eos> |
e |
divide__81.0__9.0__ |
divide__81.0__9.0__ |
| in a certain lottery the probability that a number between num__12 and num__20 inclusive is drawn is num__0.166666666667 . if the probability that a number num__12 or larger is drawn is num__0.5 what is the probability that a number less than or equal to num__20 is drawn ? <o> a ) num__0.0555555555556 <o> b ) num__0.166666666667 <o> c ) num__0.666666666667 <o> d ) num__0.5 <o> e ) num__0.833333333333 |
you can simply use sets concept in this question . the formula total = n ( a ) + n ( b ) - n ( a and b ) is applicable here too . set num__1 : number num__12 or larger set num__2 : number num__20 or smaller num__1 = p ( set num__1 ) + p ( set num__2 ) - p ( set num__1 and set num__2 ) ( combined probability is num__1 because every number will be either num__12 or moreor num__20 or lessor both ) num__0.5 + p ( set num__2 ) - num__0.166666666667 = num__0.666666666667 p ( set num__2 ) = num__0.666666666667 answer ( c ) <eor> c <eos> |
c |
reverse__0.5__ add__0.1667__0.5__ add__0.1667__0.5__ |
reverse__0.5__ add__0.1667__0.5__ add__0.1667__0.5__ |
| the value of x + x ( xx ) when x = num__2 is : ( a ) num__10 ( b ) num__16 ( c ) num__18 ( d ) num__36 ( e ) num__64 <o> a ) num__10 <o> b ) num__2 <o> c ) num__8 <o> d ) num__6 <o> e ) num__4 |
x + x ( xx ) put the value of x = num__2 in the above expression we get num__2 + num__2 ( num__22 ) = num__2 + num__2 ( num__2 × num__2 ) = num__2 + num__2 ( num__4 ) = num__2 + num__8 = num__10 answer is a . <eor> a <eos> |
a |
divide__64.0__16.0__ multiply__2.0__4.0__ add__2.0__8.0__ |
divide__64.0__16.0__ multiply__2.0__4.0__ add__2.0__8.0__ |
| a certain sum of money at simple interest amounted rs . num__1240 in num__6 years at num__4.0 per annum find the sum ? <o> a ) num__998 <o> b ) num__1000 <o> c ) num__890 <o> d ) num__646 <o> e ) num__789 |
num__1240 = p [ num__1 + ( num__6 * num__4 ) / num__100 ] p = num__1000 answer : b <eor> b <eos> |
b |
percent__100.0__1000.0__ |
percent__100.0__1000.0__ |
| an aeroplane covers a certain distance at a speed of num__1500 kmph in num__2 hours . to cover the same distance in num__1 num__0.666666666667 hours it must travel at a speed of : <o> a ) num__2200 kmph <o> b ) num__1800 kmph <o> c ) num__1900 kmph <o> d ) num__2100 kmph <o> e ) none |
explanation : distance = ( num__1500 x num__2 ) = num__3000 km . speed = distance / time speed = num__3000 / ( num__1.66666666667 ) km / hr . [ we can write num__1 num__0.666666666667 hours as num__1.66666666667 hours ] required speed = num__3000 x num__0.6 km / hr = num__1800 km / hr . answer : option b <eor> b <eos> |
b |
multiply__1500.0__2.0__ add__1.0__0.6667__ km_to_mile_conversion__ multiply__0.6__3000.0__ round__1800.0__ |
multiply__1500.0__2.0__ add__1.0__0.6667__ divide__1.0__1.6667__ multiply__0.6__3000.0__ divide__1800.0__1.0__ |
| the total weight of six people of equal weights inside a car is num__12.0 of the total weight of the car ( inclusive of the six people inside ) . if three people get down from the car what will be the ratio of the weight of people inside the car to the weight of the empty car ? <o> a ) num__44 <o> b ) num__45 <o> c ) num__46 <o> d ) num__47 <o> e ) num__48 |
ans . assume total weight of the car along with people is num__100 kgs . then total weight of num__6 people is num__12.0 of num__100 = num__12 kgs . so empty car weight = num__88 kgs that is the weight of a single person = num__2.0 = num__2 kgs . if three people get down from the car then there is a six kgs reduction in the total weight and the weight of the remaining people is only num__6 kgs . the required ratio = num__6 : num__88 = num__3 : num__44 answer a <eor> a <eos> |
a |
subtract__100.0__12.0__ divide__12.0__6.0__ divide__6.0__2.0__ divide__88.0__2.0__ divide__88.0__2.0__ |
subtract__100.0__12.0__ divide__12.0__6.0__ divide__6.0__2.0__ divide__88.0__2.0__ divide__88.0__2.0__ |
| if the cost price of num__12 pens is equal to the selling price of num__7 pens the gain percent is : <o> a ) num__80.0 <o> b ) num__90.0 <o> c ) num__71.0 <o> d ) num__40.0 <o> e ) num__10 % |
explanation : let c . p . of each pen be re . num__1 . then c . p . of num__7 pens = rs . num__7 ; s . p . of num__7 pens = rs . num__12 . gain % = num__0.714285714286 * num__100 = num__71.0 answer : c <eor> c <eos> |
c |
percent__100.0__71.0__ |
percent__100.0__71.0__ |
| the sum of the two numbers is num__15 and their product is num__35 . what is the sum of the reciprocals of these numbers ? <o> a ) num__0.428571428571 <o> b ) num__0.0285714285714 <o> c ) num__4.375 <o> d ) num__0.21875 <o> e ) none of these |
let the numbers be a and b . then a + b = num__15 and ab = num__35 . a + b / ab = num__0.428571428571 ; ( num__1 / b + num__1 / a ) = num__0.428571428571 sum of reciprocals of given numbers = num__0.428571428571 . correct option : a <eor> a <eos> |
a |
divide__15.0__35.0__ divide__15.0__35.0__ |
divide__15.0__35.0__ divide__15.0__35.0__ |
| there are num__16 bees in the hive then num__7 more fly . how many bees are there in all ? <o> a ) num__7 <o> b ) num__23 <o> c ) num__12 <o> d ) num__17 <o> e ) num__25 |
num__16 + num__7 = num__23 . answer is b . <eor> b <eos> |
b |
add__16.0__7.0__ add__16.0__7.0__ |
add__16.0__7.0__ add__16.0__7.0__ |
| rs . num__2500 is divided into two parts such that if one part be put out at num__5.0 simple interest and the other at num__6.0 the yearly annual income may be rs . num__140 . how much was lent at num__5.0 ? <o> a ) rs . num__1500 <o> b ) rs . num__1300 <o> c ) rs . num__1200 <o> d ) rs . num__1000 <o> e ) rs . num__1700 |
explanation : ( x * num__5 * num__1 ) / num__100 + [ ( num__2500 - x ) * num__6 * num__1 ] / num__100 = num__140 x = num__1000 answer is d <eor> d <eos> |
d |
percent__100.0__1000.0__ |
percent__100.0__1000.0__ |
| if two dice are thrown together the probability of getting an even number on one die and an odd number on the other is ? <o> a ) num__0.125 <o> b ) num__0.5 <o> c ) num__2.0 <o> d ) num__0.111111111111 <o> e ) num__3.5 |
the number of exhaustive outcomes is num__36 . let e be the event of getting an even number on one die and an odd number on the other . let the event of getting either both even or both odd then = num__0.5 = num__0.5 p ( e ) = num__1 - num__0.5 = num__0.5 . answer : b <eor> b <eos> |
b |
negate_prob__0.5__ |
negate_prob__0.5__ |
| two taps a and b can fill a cistern in num__12 minutes and num__18 minutes respectively . they are turned on at the same time . if the tap a is turned off after num__4 minutes how long will tap b take to fill the rest of the cistern ? <o> a ) num__8 minute <o> b ) num__9 min . <o> c ) num__10 min . <o> d ) num__7 min . <o> e ) none of these |
in one min ( a + b ) fill the cistern = num__1 ⁄ num__12 + num__1 ⁄ num__18 = num__5 ⁄ num__36 th in num__4 min ( a + b ) fill the cistern = num__5 ⁄ num__36 × num__4 = num__5 ⁄ num__9 th rest part = num__1 - num__5 ⁄ num__9 = num__4 ⁄ num__9 th ∵ num__1 ⁄ num__18 th part is filled by b in one min . ∴ num__4 ⁄ num__9 th part is filled by b in num__18 × num__4 ⁄ num__9 = num__8 min answer a <eor> a <eos> |
a |
add__4.0__1.0__ add__4.0__5.0__ subtract__12.0__4.0__ round__8.0__ |
add__4.0__1.0__ add__4.0__5.0__ subtract__12.0__4.0__ subtract__12.0__4.0__ |
| if x is an integer greater than num__6 then all the following must be divisible by num__3 except <o> a ) a ) num__3 x ^ num__3 <o> b ) b ) x ( x + num__3 ) <o> c ) c ) num__9 x ^ num__2 <o> d ) d ) num__3 x ^ num__7 <o> e ) e ) x ( x + num__1 ) ( x + num__2 ) |
* any number multiplied by num__3 will be divided by num__3 for option b if we put value x = num__7 num__7 * num__10 = num__70 is not divisible by num__3 answer : b <eor> b <eos> |
b |
add__3.0__7.0__ multiply__10.0__7.0__ subtract__6.0__3.0__ |
add__3.0__7.0__ multiply__10.0__7.0__ subtract__6.0__3.0__ |
| jordan passed a point on a track at a constant speed of num__50 miles per hour . then num__15 minutes later pauline passed the same point at a constant speed of num__60 miles per hour . if both runners maintained their speeds how long after he passed the gas station did pauline catch up with jordan ? <o> a ) num__30 min <o> b ) num__45 min <o> c ) num__1 hr <o> d ) num__1 hr num__15 min <o> e ) num__1 hr num__30 min |
when pauline is at the point jordan is num__12.5 miles ahead on the ahead . ( the distance covered by him in num__15 min ) every hour pauline runs num__10 miles more than mary . how many hours will it takes him to run num__12.5 miles more ? the answer is ( num__12.5 ) / num__10 = num__1.25 = num__1 h num__15 min . answer d <eor> d <eos> |
d |
subtract__60.0__50.0__ divide__12.5__10.0__ round__1.0__ |
subtract__60.0__50.0__ divide__12.5__10.0__ round__1.0__ |
| if two girls starting from same point walking in the opposite directions with num__7 km / hr and num__3 km / hr as average speeds respectively . then the distance between them after num__12 hours is ? <o> a ) num__120 <o> b ) num__121 <o> c ) num__122 <o> d ) num__123 <o> e ) num__124 |
explanation : total distance = distance traveled by person a + distance traveled by person b = ( num__7 Ã — num__12 ) + ( num__3 Ã — num__12 ) = num__84 + num__36 = num__120 km answer : a <eor> a <eos> |
a |
multiply__7.0__12.0__ multiply__3.0__12.0__ add__36.0__84.0__ round__120.0__ |
multiply__7.0__12.0__ multiply__3.0__12.0__ add__36.0__84.0__ add__36.0__84.0__ |
| perimeter of the backwheel is num__9 feet and front wheel is num__7 feet at a certain distance . the front wheel gets num__10 revolutions more than the back wheel . what is the distance ? <o> a ) num__305 <o> b ) num__310 <o> c ) num__315 <o> d ) num__320 <o> e ) num__325 |
if distance = d number of revolutions by back wheel = d / num__9 & front wheel = d / num__7 ( d / num__7 ) - ( d / num__9 ) = num__10 num__2 d = num__630 d = num__315 answer : c <eor> c <eos> |
c |
subtract__9.0__7.0__ divide__630.0__2.0__ round__315.0__ |
subtract__9.0__7.0__ divide__630.0__2.0__ divide__630.0__2.0__ |
| angelo and isabella are both salespersons . in any given week angelo makes $ num__550 in base salary plus num__8 percent of the portion of his sales above $ num__3000 for that week . isabella makes num__10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of money ? <o> a ) num__15500 <o> b ) num__24500 <o> c ) num__25500 <o> d ) num__26500 <o> e ) num__27 |
500 |
official solution : the problem asks for the amount of weekly sales it takes for angelo and isabella to earn the same amount of money . you can write an equation that sets angelo ’ s and isabella ’ s weekly earnings equal to each other with x representing weekly sales . weekly earnings for each salesperson equal base salary plus commission . so angelo ’ s earnings are num__550 + ( num__0.08 ) ( x – num__3000 ) and isabella ’ s are num__0.10 x . set up the equation and solve : num__550 + ( num__0.08 ) ( x – num__3000 ) = num__0.10 x distribute the num__0.08 : num__550 + num__0.08 x – num__240 = num__0.10 x combine terms and subtract num__0.08 x from both sides : num__310 = num__0.02 x divide both sides by num__0.02 : num__15500 = x your answer is a . <eor> a <eos> |
a |
a |
| in a kilometer race a beats b by num__16 meters or num__10 seconds . what time does a take to complete the race ? <o> a ) num__167 sec <o> b ) num__190 sec <o> c ) num__176 sec <o> d ) num__716 sec <o> e ) num__615 sec |
time taken by b run num__1000 meters = ( num__1000 * num__10 ) / num__16 = num__625 sec . time taken by a = num__625 - num__10 = num__615 sec . answer : e <eor> e <eos> |
e |
subtract__625.0__10.0__ round__615.0__ |
subtract__625.0__10.0__ subtract__625.0__10.0__ |
| a car covers a distance of num__624 km in num__6 ½ hours . find its speed ? <o> a ) num__104 kmph <o> b ) num__776 kmph <o> c ) num__886 kmph <o> d ) num__887 kmph <o> e ) num__665 kmph |
num__104.0 = num__104 kmph answer : a <eor> a <eos> |
a |
divide__624.0__6.0__ round__104.0__ |
divide__624.0__6.0__ round__104.0__ |
| the area of one square is x ^ num__2 + num__4 x + num__4 and the area of another square is num__4 x ^ num__2 − num__12 x + num__9 . if the sum of the perimeters of both squares is num__32 what is the value of x ? <o> a ) num__0 <o> b ) num__2 <o> c ) num__2.5 <o> d ) num__4.67 <o> e ) num__3 |
spotting the pattern of equations both are in form of ( x + c ) ^ num__2 so a num__1 = ( x + num__2 ) ^ num__2 a num__2 = ( num__2 x - num__3 ) ^ num__2 l num__1 = x + num__2 l num__2 = num__2 x - num__3 p num__1 = num__4 ( x + num__2 ) p num__2 = num__4 ( num__2 x - num__3 ) p num__1 + p num__2 = num__32 num__4 ( x + num__2 ) + num__4 ( num__2 x - num__3 ) = num__32 . . . . . . . . . . . . . . > x = num__3 answer : e <eor> e <eos> |
e |
add__2.0__1.0__ add__2.0__1.0__ |
add__2.0__1.0__ add__2.0__1.0__ |
| if ( num__8 x ) / ( num__11 y / z - num__1 ) = num__3 x then which of the following is true ? <o> a ) z = - num__3 y <o> b ) z = - y <o> c ) z = - x <o> d ) z = num__3 y <o> e ) z = num__3 x |
( num__8 x ) / ( num__11 y / z - num__1 ) = num__3 x = > num__8 z / ( num__11 y - z ) = num__3 = > num__8 z = num__33 y - num__3 z = > num__11 z = num__33 y = > z = num__3 y hence ans is d <eor> d <eos> |
d |
multiply__11.0__3.0__ subtract__11.0__8.0__ |
multiply__11.0__3.0__ subtract__11.0__8.0__ |
| excluding stoppages the speed of a bus is num__5454 kmph and including stoppages it is num__4545 kmph . for how many minutes does the bus stop per hour ? <o> a ) num__12 <o> b ) num__11 <o> c ) num__10 <o> d ) num__9 <o> e ) num__8 |
explanation : speed of the bus excluding stoppages = num__54 = num__54 kmph speed of the bus including stoppages = num__45 = num__45 kmph loss in speed when including stoppages = num__54 â ˆ ’ num__45 = num__9 kmph = num__54 â ˆ ’ num__45 = num__9 kmph â ‡ ’ â ‡ ’ in num__11 hour bus covers num__99 km less due to stoppages . hence time in which the bus stops per hour = time taken to cover num__99 km = distancespeed = num__954 hour = num__16 hour = distancespeed = num__954 hour = num__16 hour = num__606 min = num__10 min answer is c <eor> c <eos> |
c |
subtract__54.0__45.0__ multiply__9.0__11.0__ divide__5454.0__9.0__ round__10.0__ |
subtract__54.0__45.0__ multiply__9.0__11.0__ divide__5454.0__9.0__ round__10.0__ |
| if nn is a positive integer and ( n + num__1 ) ( n + num__3 ) ( n + num__1 ) ( n + num__3 ) is odd then ( n + num__2 ) ( n + num__4 ) ( n + num__2 ) ( n + num__4 ) must be a multiple of which one of the following ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__8 <o> e ) num__10 |
( n + num__1 ) ( n + num__3 ) ( n + num__1 ) ( n + num__3 ) is odd only when both ( n + num__1 ) ( n + num__1 ) and ( n + num__3 ) ( n + num__3 ) are odd . this is possible only when nn is even . hence n = num__2 mn = num__2 m where mm is a positive integer . then ( n + num__2 ) ( n + num__4 ) = ( num__2 m + num__2 ) ( num__2 m + num__4 ) = num__2 ( m + num__1 ) num__2 ( m + num__2 ) = num__4 ( m + num__1 ) ( m + num__2 ) ( n + num__2 ) ( n + num__4 ) = ( num__2 m + num__2 ) ( num__2 m + num__4 ) = num__2 ( m + num__1 ) num__2 ( m + num__2 ) = num__4 ( m + num__1 ) ( m + num__2 ) = num__4 * ( product of two consecutive positive integers one which must be even ) = num__4 * ( product of two consecutive positive integers one which must be even ) = num__4 * ( an even number ) and this equals a number that is at least a multiple of num__8 = num__4 * ( an even number ) and this equals a number that is at least a multiple of num__8 hence the answer is ( d ) . <eor> d <eos> |
d |
multiply__2.0__4.0__ multiply__1.0__8.0__ |
multiply__2.0__4.0__ multiply__1.0__8.0__ |
| in how many ways can a five - letter password be chosen using the letters a b c d e f g and / or h such that at least one letter is repeated within the password ? <o> a ) num__720 <o> b ) num__864 <o> c ) num__900 <o> d ) num__936 <o> e ) num__26048 |
total number of four letter passwords = num__8 * num__8 * num__8 * num__8 * num__8 = num__32768 - - - - - - ( num__1 ) total number of passwords in which no letter repeats = num__8 c num__5 * num__5 ! = num__56 * num__120 = num__6720 - - - - - - ( num__2 ) therefore required value = ( num__1 ) - ( num__2 ) = num__32768 - num__6720 = num__26048 e <eor> e <eos> |
e |
multiply__120.0__56.0__ subtract__32768.0__6720.0__ subtract__32768.0__6720.0__ |
multiply__120.0__56.0__ subtract__32768.0__6720.0__ subtract__32768.0__6720.0__ |
| the sum an integer n and its reciprocal is equal to num__5.2 . what is the value of n ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
write equation in n as follows n + num__1 / n = num__5.2 multiply all terms by n obtain a quadratic equation and solve to obtain n = num__5 . correct answer c <eor> c <eos> |
c |
round_down__5.2__ round_down__5.2__ |
round_down__5.2__ divide__5.0__1.0__ |
| an accurate clock shows num__5 o ' clock in the morning . through how may degrees will the hour hand rotate when the clock shows num__11 o ' clock in the morning ? <o> a ) num__140 º <o> b ) num__160 º <o> c ) num__180 º <o> d ) num__200 º <o> e ) num__210 º |
angle traced by the hour hand in num__6 = ( num__360 x num__12 ) x num__6 º = num__180 º . answer : c <eor> c <eos> |
c |
clock_big_arm_angle__6.0__ straight_angle__ straight_angle__ |
clock_big_arm_angle__6.0__ straight_angle__ straight_angle__ |
| monica planned her birthday party . she prepared num__5 muffins for each of her guests and kept aside two additional muffins in case someone will want extra . after the party it turned out that one of the guests did n ' t come but every one of the guests that did come ate six muffins and num__3 muffins remained . how many guests did monica plan on ? <o> a ) num__3 . <o> b ) num__4 . <o> c ) num__5 . <o> d ) num__6 . <o> e ) num__7 . |
let x be the number of guests . number of muffins prepared = num__5 x + num__2 number of muffins eaten + number of muffins remaining = number of muffins prepared num__6 ( x - num__1 ) + num__3 = num__5 x + num__2 num__6 x - num__3 = num__5 x + num__2 x = num__5 answer : c <eor> c <eos> |
c |
subtract__5.0__3.0__ multiply__3.0__2.0__ subtract__3.0__2.0__ multiply__5.0__1.0__ |
subtract__5.0__3.0__ multiply__3.0__2.0__ subtract__3.0__2.0__ add__3.0__2.0__ |
| two pipes p and q can fill a cistern in num__10 and num__15 minutes respectively . both are opened together but at the end of num__3 minutes the first is turned off . how much longer will the cistern take to fill ? <o> a ) num__0.5 <o> b ) num__0.25 <o> c ) num__0.5 <o> d ) num__0.25 <o> e ) num__0.25 |
num__0.3 + x / num__15 = num__1 x = num__10 num__0.5 answer : a <eor> a <eos> |
a |
divide__3.0__10.0__ round__0.5__ |
divide__3.0__10.0__ divide__0.5__1.0__ |
| in a marketing survey num__60 people were asked to rank three flavors of ice cream chocolate vanilla and strawberry in order of their preference . all num__60 people responded and no two flavors were ranked equally by any of the people surveyed . if num__0.6 of the people ranked vanilla last num__0.1 of them ranked vanilla before chocolate and num__0.333333333333 of them ranked vanilla before strawberry how many people ranked vanilla first ? <o> a ) num__2 people <o> b ) num__6 <o> c ) num__14 <o> d ) num__16 <o> e ) num__24 |
first offunderstand what you are trying to find - in this case you are trying to find how many people rank vanilla first . for someone to rank vanilla first they have to rank them before chocolate or strawberry . this should be the very first hint that the correct answer can not be bigger than num__0.1 * num__60 = num__6 . so if you are time pressured on the real test at least you narrowed down the selections to two choices - not bad . but what now ? solve the easier problem . from num__60 people we are already told that num__0.6 of them rank vanilla last leaving us with num__0.4 of the people not ranking vanilla last - num__24 people . so what are those combinations that vanilla are not last ? a ) c v s b ) s v c c ) c s v ( vanilla first ) d ) s c v ( vanilla first ) what are we ultimately hunting for here ? the number of people that rank vanilla first : c + d now let ' s represent the given info into three equations num__1 ) a + b + c + d = num__24 ( num__0.4 of the people do not rank vanilla last ) num__2 ) a + c + d = num__6 ( num__0.1 rank vanilla before chocolate ) num__3 ) b + c + d = num__20 ( num__0.333333333333 rank vanilla before strawberry ) subtract equation num__1 from num__2 leaving us with b = num__24 - num__6 = num__18 sub in b into equation num__3 and solve for c + d num__18 + c + d = num__20 c + d = num__20 - num__18 = num__2 there you go ! num__2 people out of the num__60 ranked vanilla as first a <eor> a <eos> |
a |
multiply__60.0__0.1__ multiply__60.0__0.4__ add__0.6__0.4__ add__1.0__2.0__ divide__60.0__3.0__ multiply__3.0__6.0__ multiply__0.1__20.0__ |
multiply__60.0__0.1__ multiply__60.0__0.4__ add__0.6__0.4__ add__1.0__2.0__ divide__60.0__3.0__ multiply__3.0__6.0__ multiply__0.1__20.0__ |
| if ( x / y ) = ( num__1.4 ) find the value ( x ^ num__2 + y ^ num__2 ) / ( x ^ num__2 - y ^ num__2 ) <o> a ) num__3.08333333333 <o> b ) num__5.36363636364 <o> c ) num__0.662337662338 <o> d ) num__3.72727272727 <o> e ) none of them |
= ( x ^ num__2 + y ^ num__2 ) / ( x ^ num__2 - y ^ num__2 ) = ( x ^ num__2 / y ^ num__2 + num__1 ) / ( x ^ num__2 / y ^ num__2 - num__1 ) = [ ( num__1.4 ) ^ num__2 + num__1 ] / [ ( num__1.4 ) ^ num__2 - num__1 ] = [ ( num__1.96 ) + num__1 ] / [ ( num__1.96 ) - num__1 ] = num__3.08333333333 answer is a . <eor> a <eos> |
a |
round_down__1.4__ multiply__1.0__3.0833__ |
round_down__1.4__ divide__3.0833__1.0__ |
| the average marks obtained by num__120 candidates in a certain examination is num__35 . find the total marks . <o> a ) num__3800 <o> b ) num__4500 <o> c ) num__5200 <o> d ) num__3400 <o> e ) num__4200 |
following the above formula we have the total marks = num__120 * num__35 = num__4200 answer is e <eor> e <eos> |
e |
multiply__120.0__35.0__ multiply__120.0__35.0__ |
multiply__120.0__35.0__ multiply__120.0__35.0__ |
| a no . when divided by the sum of num__555 and num__445 gives num__2 times their difference as quotient & num__30 as remainder . find the no . is ? <o> a ) num__124432 <o> b ) num__145366 <o> c ) num__157768 <o> d ) num__178432 <o> e ) num__220030 |
( num__555 + num__445 ) * num__2 * num__110 + num__30 = num__220000 + num__30 = num__220030 e <eor> e <eos> |
e |
subtract__555.0__445.0__ add__30.0__220000.0__ add__30.0__220000.0__ |
subtract__555.0__445.0__ add__30.0__220000.0__ add__30.0__220000.0__ |
| if num__1.47880539499 = a + num__1 / [ b + num__1 / { c + ( d + num__1 / e ) } ] . find a * b * c * d * e ? <o> a ) num__82.5 <o> b ) num__87.5 <o> c ) num__59.3333333333 <o> d ) num__187 by num__2 <o> e ) num__98.5 |
by expanding num__1.47880539499 make form like a + num__1 / [ b + num__1 / { c + ( d + num__1 / e ) } = num__1 + ( num__0.47880539499 ) now a = num__1 = num__1 + { num__1 / ( num__2.08853118712 ) } = num__1 + { num__1 / ( num__2 + num__0.0845070422535 ) } = num__1 + { num__1 / ( num__2 + num__1 / ( num__11.8333333333 ) } now b = num__2 similarly expand c = num__11 d = num__1 e = num__4.25 finally a * b * b * c * d * e num__1 * num__2 * num__11 * num__1 * num__4.25 num__93.5 answer : d <eor> d <eos> |
d |
subtract__1.4788__1.0__ round_down__2.0885__ round_down__11.8333__ multiply__2.0__93.5__ |
subtract__1.4788__1.0__ round_down__2.0885__ round_down__11.8333__ multiply__2.0__93.5__ |
| a wire in the form of a circle of radius num__3.5 m is bent in the form of a rectangule whose length and breadth are in the ratio of num__6 : num__5 . what is the area of the rectangle ? <o> a ) num__12 cm num__2 <o> b ) num__30 cm num__2 <o> c ) num__17 cm num__2 <o> d ) num__18 cm num__2 <o> e ) num__19 cm num__2 |
the circumference of the circle is equal to the permeter of the rectangle . let l = num__6 x and b = num__5 x num__2 ( num__6 x + num__5 x ) = num__2 * num__3.14285714286 * num__3.5 = > x = num__1 therefore l = num__6 cm and b = num__5 cm area of the rectangle = num__6 * num__5 = num__30 cm num__2 answer : b <eor> b <eos> |
b |
multiply__6.0__5.0__ multiply__6.0__5.0__ |
multiply__6.0__5.0__ multiply__6.0__5.0__ |
| look at this series : num__98 num__98 num__90 num__90 num__82 num__82 . . . what number should come next ? <o> a ) num__74 <o> b ) num__15 <o> c ) num__17 <o> d ) num__19 <o> e ) num__11 |
in this series each number is repeated then num__8 is subtracted to arrive at the next number . the next number is num__74 answer : num__74 <eor> a <eos> |
a |
subtract__98.0__90.0__ subtract__82.0__8.0__ subtract__82.0__8.0__ |
subtract__98.0__90.0__ subtract__82.0__8.0__ subtract__82.0__8.0__ |
| num__12 men take num__18 days to complete a job whereas num__12 women in num__18 days can complete num__0.5 of the same job . how many days will num__10 men and num__8 women together take to complete the same job ? <o> a ) num__6 <o> b ) num__13 num__1 â „ num__2 <o> c ) num__15 num__0.428571428571 <o> d ) data inadequate <o> e ) none of these |
num__12 m à — num__18 = num__12 w à — num__18 à — num__2.0 \ w = num__0.5 m num__10 m + num__8 w = num__10 m + num__8 à — num__0.5 m = num__14 m \ num__14 men can complete the same work in num__12 à — num__1.28571428571 = num__15 num__0.428571428571 days answer c <eor> c <eos> |
c |
subtract__12.0__10.0__ add__12.0__2.0__ divide__18.0__14.0__ round__15.0__ |
subtract__12.0__10.0__ add__12.0__2.0__ divide__18.0__14.0__ round__15.0__ |
| the ages of x and y are in the ratio of num__3 : num__8 respectively and product of their ages is num__384 . compute the ratio of their ages three years hence . <o> a ) num__3 : num__5 <o> b ) num__2 : num__5 <o> c ) num__3 : num__7 <o> d ) num__3 : num__4 <o> e ) num__1 : num__4 |
as given the ages are of num__3 x num__8 x and product of their ages is num__384 so num__3 x * num__8 x = num__384 by solving we get num__24 x ^ num__2 = num__384 finally by sloving we get x ^ num__2 = num__16 x = num__4 as given three hence we add num__3 for num__3 x and num__4 x we solve it by num__3 x + num__4 : num__8 x + num__4 sub x = num__4 in above ration we get num__15 : num__35 num__3 : num__7 answer : c <eor> c <eos> |
c |
multiply__3.0__8.0__ multiply__8.0__2.0__ divide__8.0__2.0__ add__3.0__4.0__ subtract__7.0__4.0__ |
multiply__3.0__8.0__ multiply__8.0__2.0__ divide__8.0__2.0__ add__3.0__4.0__ subtract__7.0__4.0__ |
| a diagonal of a polygon is an segment between two non - adjacent vertices of the polygon . how many diagonals does a regular num__20 - sided polygon have ? <o> a ) num__875 <o> b ) num__170 <o> c ) num__1425 <o> d ) num__2025 <o> e ) num__2500 |
there ' s a direct formula for this . number of diagonals in a regular polygon = [ n * ( n - num__3 ) ] / num__2 n = number of sides of the regular polygon . here n = num__20 . plugging it in we get num__170 diagonals ! answer ( b ) . <eor> b <eos> |
b |
triangle_area__2.0__170.0__ |
triangle_area__2.0__170.0__ |
| adding num__50.0 of x to x is equivalent to multiplying x by which of the following ? <o> a ) num__12.5 <o> b ) num__1.05 <o> c ) num__1.5 <o> d ) num__1.2 <o> e ) num__1.25 |
num__150 x / num__100 = num__1.5 * x answer : c <eor> c <eos> |
c |
subtract__150.0__50.0__ divide__150.0__100.0__ divide__150.0__100.0__ |
subtract__150.0__50.0__ divide__150.0__100.0__ divide__150.0__100.0__ |
| what is the rate percent when the simple interest on rs . num__800 amount to rs . num__128 in num__4 years ? <o> a ) num__5.0 <o> b ) num__3.0 <o> c ) num__4.0 <o> d ) num__9.0 <o> e ) num__1 % |
num__128 = ( num__800 * num__4 * r ) / num__100 r = num__4.0 answer : c <eor> c <eos> |
c |
percent__4.0__100.0__ |
percent__4.0__100.0__ |
| a and b can do a piece of work in num__40 days and num__40 days respectively . they work together for num__10 days and b leaves . in how many days the whole work is completed ? <o> a ) num__30 days <o> b ) num__35 days <o> c ) num__40 days <o> d ) num__45 days <o> e ) num__50 days |
explanation : ( a + b ) ’ s num__10 days work = num__10 [ num__0.025 + num__0.025 ] = num__10 [ num__1 + num__0.025 ] = num__0.5 a complete remaining work in num__0.5 * num__40 = num__20 total work = num__10 + num__20 = num__30 days answer : option a <eor> a <eos> |
a |
multiply__40.0__0.025__ multiply__40.0__0.5__ subtract__40.0__10.0__ round__30.0__ |
multiply__40.0__0.025__ multiply__40.0__0.5__ add__10.0__20.0__ add__10.0__20.0__ |
| there are num__4 people of different heights standing in order of increasing height . the difference is num__2 inches between the first person and the second person and also between the second person and the third person . the difference between the third person and the fourth person is num__6 inches and the average height is num__79 . how tall is the fourth person ? <o> a ) num__85 <o> b ) num__87 <o> c ) num__89 <o> d ) num__91 <o> e ) num__93 |
let x be the height of the first person . then the heights are x x + num__2 x + num__4 and x + num__10 . num__4 x + num__16 = num__4 ( num__79 ) = num__316 x = num__75 and the fourth person has a height of num__75 + num__10 = num__85 inches the answer is a . <eor> a <eos> |
a |
add__4.0__6.0__ add__6.0__10.0__ multiply__4.0__79.0__ subtract__79.0__4.0__ add__6.0__79.0__ add__6.0__79.0__ |
add__4.0__6.0__ add__6.0__10.0__ multiply__4.0__79.0__ subtract__79.0__4.0__ add__6.0__79.0__ add__6.0__79.0__ |
| kamal ' s salary was decreased by num__50.0 and subsequently increased by num__30.0 . how much percent does he lose ? <o> a ) num__30.0 <o> b ) num__37.0 <o> c ) num__65.0 <o> d ) num__35.0 <o> e ) num__45 % |
explanation : solution : let original salary = rs . num__100 . new final salary = num__130.0 of ( num__50.0 of rs . num__100 ) = rs . ( num__1.3 * num__0.5 * num__100 ) = rs . num__65 . . ' . decrease = num__35.0 answer : d <eor> d <eos> |
d |
add__30.0__100.0__ divide__130.0__100.0__ divide__50.0__100.0__ multiply__50.0__1.3__ subtract__65.0__30.0__ subtract__65.0__30.0__ |
add__30.0__100.0__ divide__130.0__100.0__ divide__50.0__100.0__ multiply__50.0__1.3__ subtract__65.0__30.0__ subtract__65.0__30.0__ |
| a rectangular parking space is marked out by painting three of its sides . if the length of the unpainted side is num__99 feet and the sum of the lengths of the painted sides is num__37 feet find out the area of the parking space in square feet ? <o> a ) num__126 sq . ft <o> b ) num__64 sq . ft <o> c ) num__100 sq . ft <o> d ) num__120 sq . ft <o> e ) num__140 sq . ft |
explanation : solution num__1 length = num__9 feet breadth = num__37 − num__4.5 = num__14 feet area = num__9 × num__14 = num__126 square feet answer : option a <eor> a <eos> |
a |
triangle_area__1.0__9.0__ multiply__9.0__14.0__ multiply__1.0__126.0__ |
triangle_area__1.0__9.0__ multiply__9.0__14.0__ multiply__1.0__126.0__ |
| the h . c . f . of two numbers is num__11 and their l . c . m . is num__7700 . if one of the numbers is num__275 then the other is : <o> a ) num__279 <o> b ) num__283 <o> c ) num__308 <o> d ) num__318 <o> e ) num__342 |
other number = ( num__11 x num__7700 ) / num__275 = num__308 . answer : option c <eor> c <eos> |
c |
lcm__11.0__308.0__ |
lcm__11.0__308.0__ |
| the price of sugar is increased by num__7.0 . by how much percent should a home maker reduce her consumption of sugar to have no extra expenditure ? <o> a ) ( num__6.54205607477 ) % <o> b ) ( num__1.07 ) % <o> c ) ( num__0.934579439252 ) % <o> d ) ( num__0.07 ) % <o> e ) ( num__0.142857142857 ) % |
old price = $ num__100 ( assume ) ; old consumption = num__1 kg ( assume ) . new price = $ num__107 . we want the spendings to remain at $ num__100 . ( new consumption ) * num__107 = num__100 - - > ( new consumption ) = num__0.934579439252 kg . percent decrease = change / original * num__100 = ( num__1 - num__0.934579439252 ) / num__1 * num__100 = num__6.54205607477 % . answer : a . <eor> a <eos> |
a |
add__7.0__100.0__ divide__100.0__107.0__ multiply__1.0__6.5421__ |
add__7.0__100.0__ divide__100.0__107.0__ multiply__1.0__6.5421__ |
| num__9548 + num__7314 = num__7362 + ( ? ) <o> a ) num__8300 <o> b ) num__8400 <o> c ) num__8500 <o> d ) num__8700 <o> e ) num__9500 |
num__9548 num__16862 = num__7362 + x + num__7314 x = num__16862 - num__7362 - - - - - = num__9500 num__16862 - - - - - e ) <eor> e <eos> |
e |
add__9548.0__7314.0__ subtract__16862.0__7362.0__ subtract__16862.0__7362.0__ |
add__9548.0__7314.0__ subtract__16862.0__7362.0__ subtract__16862.0__7362.0__ |
| the current birth rate per thousand is num__32 whereas corresponding death rate is num__11 per thoudand . the net growth rate in terms of population increase in percent is given by <o> a ) num__0.0021 <o> b ) num__0.021 <o> c ) num__2.1 <o> d ) num__21.0 <o> e ) none |
solution net growth on num__1000 = ( num__32 - num__11 ) = num__21 . net growth on num__100 ‹ = › [ num__0.021 x num__100 ] % = num__2.1 . answer c <eor> c <eos> |
c |
percent__100.0__2.1__ |
percent__100.0__2.1__ |
| reena took a loan of rs . num__1200 with simple interest for as many years as the rate of interest . if she paid rs . num__432 as interest at the end of the loan period what was the rate of interest ? <o> a ) num__3.6 <o> b ) num__6 <o> c ) num__18 <o> d ) num__24 <o> e ) none of these |
let rate = r % and time = r years . then ( num__1200 x r x r ) / num__100 = num__432 num__12 r num__2 = num__432 r num__2 = num__36 r = num__6 . answer : option b <eor> b <eos> |
b |
percent__100.0__6.0__ |
percent__100.0__6.0__ |
| which of the following describes all values of x for which num__4 – x ^ num__2 > = num__0 ? <o> a ) x > = num__1 <o> b ) x < = – num__1 <o> c ) num__0 < = x < = num__1 <o> d ) x < = – num__1 or x > = num__1 <o> e ) – num__2 < = x < = num__2 |
num__4 - x ^ num__2 > = num__0 means x ^ num__2 - num__4 < = num__0 = > ( x - num__2 ) ( x + num__2 ) < = num__0 = > - num__2 < = x < = num__2 answer - e <eor> e <eos> |
e |
subtract__4.0__2.0__ |
subtract__4.0__2.0__ |
| if the cost price of num__12 pens is equal to the selling price of num__8 pens the gain percent is : <o> a ) num__25.0 <o> b ) num__33 num__0.333333333333 % <o> c ) num__50.0 <o> d ) num__66 num__0.666666666667 % <o> e ) none |
let c . p . of each pen be re . num__1 . then c . p . of num__8 pens = rs . num__8 ; s . p . of num__8 pens = rs . num__12 . gain % = num__0.5 * num__100 = num__50.0 answer : c <eor> c <eos> |
c |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| visitors to show were charged rs . num__5 each on the first day . rs . num__7.50 on the second day rs . num__2.50 on the third day and total attendance on the three days were in ratio num__2 : num__5 : num__13 respectively . the average charge per person for the whole show is ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
num__2 : num__5 : num__13 num__2 x num__5 x num__13 x num__5 num__7.5 num__2.5 num__10 x + num__37.5 x + num__32.5 x = num__80 x / num__20 x average = num__4 answer : d <eor> d <eos> |
d |
multiply__5.0__2.0__ multiply__5.0__7.5__ multiply__2.5__13.0__ multiply__2.0__10.0__ divide__10.0__2.5__ divide__10.0__2.5__ |
add__7.5__2.5__ multiply__5.0__7.5__ multiply__2.5__13.0__ multiply__2.0__10.0__ divide__10.0__2.5__ divide__10.0__2.5__ |
| the length of the bridge which a train num__130 m long and traveling at num__45 km / hr can cross in num__30 sec is ? <o> a ) num__277 <o> b ) num__299 <o> c ) num__245 <o> d ) num__200 <o> e ) num__211 |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec . time = num__30 sec let the length of bridge be x meters . then ( num__130 + x ) / num__30 = num__12.5 x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| a starts a business with a capital of rs . num__85000 . b joins in the business with rs . num__42500 after some time . for how much period does b join if the profits at the end of the year are divided in the ratio of num__3 : num__1 ? <o> a ) num__8 months <o> b ) num__7 months <o> c ) num__6 months <o> d ) num__10 months <o> e ) num__5 months |
let b joined for x months . then num__85000 × num__12 : num__42500 × x = num__3 : num__1 ⇒ num__850 × num__12 : num__425 x = num__3 : num__1 ⇒ num__850 × num__12 × num__1 = num__3 × num__425 x ⇒ num__850 × num__4 = num__425 x ⇒ x = num__8 answer is a . <eor> a <eos> |
a |
add__3.0__1.0__ subtract__12.0__4.0__ multiply__1.0__8.0__ |
add__3.0__1.0__ subtract__12.0__4.0__ multiply__1.0__8.0__ |
| if a person walks at num__14 km / hr instead of num__10 km / hr he would have walked num__20 km more . the actual distance traveled by him is ? <o> a ) num__50 <o> b ) num__99 <o> c ) num__77 <o> d ) num__61 <o> e ) num__21 |
let the actual distance traveled be x km . then x / num__10 = ( x + num__20 ) / num__14 num__4 x - num__200 = > x = num__50 km . answer : a <eor> a <eos> |
a |
subtract__14.0__10.0__ multiply__10.0__20.0__ divide__200.0__4.0__ divide__200.0__4.0__ |
subtract__14.0__10.0__ multiply__10.0__20.0__ divide__200.0__4.0__ divide__200.0__4.0__ |
| if the cost price of num__20 articles is same as the selling price of num__25 articles . find the gain or loss percentage ? <o> a ) num__20.0 loss <o> b ) num__30.0 loss <o> c ) num__40.0 gain <o> d ) num__40.0 loss <o> e ) num__50.0 loss |
explanation : num__20 cp = num__25 sp cost price cp = num__25 selling price sp = num__20 formula = ( sp - cp ) / cp * num__100 = ( num__20 - num__25 ) / num__25 * num__100 = num__20.0 loss answer : option a <eor> a <eos> |
a |
percent__20.0__100.0__ |
percent__20.0__100.0__ |
| three competing juice makers conducted a blind taste test with mall shoppers . the shoppers could choose to taste any number of the three brands of juices but had to select at least one juice that they liked . if num__100 shoppers liked brand j num__200 shoppers liked brand k num__700 shoppers liked brand l num__300 shoppers liked exactly num__2 juices and num__25 shoppers liked all three juices how many shoppers took the taste test ? <o> a ) num__1300 <o> b ) num__1000 <o> c ) num__900 <o> d ) num__700 <o> e ) num__650 |
given : atleast num__1 juice was liked by the participants . - - > neither = num__0 assume i - - > no overlap between the sets ii - - > overlap between num__2 sets iii - - > overlap between num__3 sets i + num__2 * ( ii ) + num__3 * ( iii ) = num__100 + num__200 + num__700 i + num__2 * ( num__300 ) + num__3 * ( num__25 ) = num__1000 i = num__325 total number of shoppers who took the taste test = i + ii + iii = num__325 + num__300 + num__25 = num__650 answer : e <eor> e <eos> |
e |
divide__300.0__100.0__ add__700.0__300.0__ add__300.0__25.0__ multiply__2.0__325.0__ multiply__2.0__325.0__ |
add__2.0__1.0__ add__700.0__300.0__ add__300.0__25.0__ multiply__2.0__325.0__ multiply__2.0__325.0__ |
| in what time will a train num__150 m long cross an electric pole it its speed be num__144 km / hr ? <o> a ) num__2.75 <o> b ) num__3.75 <o> c ) num__4.5 <o> d ) num__5.5 <o> e ) num__6 |
speed = num__144 * num__0.277777777778 = num__40 m / sec time taken = num__3.75 = num__3.75 sec . answer : option b <eor> b <eos> |
b |
divide__150.0__40.0__ round__3.75__ |
divide__150.0__40.0__ divide__150.0__40.0__ |
| a company pays num__15.5 dividend to its investors . if an investor buys rs . num__50 shares and gets num__25.0 on investment at what price did the investor buy the shares ? <o> a ) num__25 <o> b ) num__31 <o> c ) num__18 <o> d ) num__19 <o> e ) num__01 |
explanation : dividend on num__1 share = ( num__15.5 * num__50 ) / num__100 = rs . num__7.75 rs . num__25 is income on an investment of rs . num__100 rs . num__7.75 is income on an investment of rs . ( num__7.75 * num__100 ) / num__25 = rs . num__31 answer : b <eor> b <eos> |
b |
percent__15.5__50.0__ percent__100.0__31.0__ |
percent__15.5__50.0__ percent__100.0__31.0__ |
| the num__42 parents participating in the smithville pta have been assigned to at least num__1 of num__3 committees : festival planning classroom aid and teacher relations . num__21 parents are assigned to the festival planning committee num__18 parents are assigned to the classroom aid committee and num__19 parents are assigned to the teacher relations committee . if num__5 parents are assigned to all num__3 committees how many parents are assigned to exactly num__2 committees ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
the formula is total = a + b + c - sum of exactly two + num__2 * all three + neither num__21 + num__18 + num__19 - x - num__2 * num__5 = num__42 solving for x you get num__6 answer b <eor> b <eos> |
b |
die_space__ die_space__ |
die_space__ die_space__ |
| three workers have a productivity ratio of num__2 to num__3 to num__5 . all three workers are working on a job for num__4 hours . at the beginning of the num__5 th hour the slowest worker takes a break . the slowest worker comes back to work at the beginning of the num__9 th hour and begins working again . the job is done in ten hours . what was the ratio of the work performed by the fastest worker as compared to the slowest ? <o> a ) num__12 to num__1 <o> b ) num__6 to num__1 <o> c ) num__25 to num__6 <o> d ) num__1 to num__6 <o> e ) num__1 to num__5 |
the fastest worker who does num__6 units of job worked for all num__10 hours so he did num__5 * num__10 = num__50 units of job ; the slowest worker who does num__2 unit of job worked for only num__4 + num__2 = num__6 hours ( first num__4 hours and last num__2 hours ) so he did num__2 * num__6 = num__12 units of job ; the ratio thus is num__50 to num__12 or num__25 to num__6 . answer : c . <eor> c <eos> |
c |
multiply__2.0__3.0__ multiply__2.0__5.0__ multiply__5.0__10.0__ multiply__2.0__6.0__ divide__50.0__2.0__ round__25.0__ |
multiply__2.0__3.0__ multiply__2.0__5.0__ multiply__5.0__10.0__ multiply__2.0__6.0__ divide__50.0__2.0__ round__25.0__ |
| the tax on a commodity is diminished by num__20.0 and its consumption increased by num__15.0 . the effect on revenue is ? <o> a ) it increases by num__8.0 <o> b ) it decreases by num__8.0 <o> c ) no change in revenue <o> d ) it increases by num__10.0 <o> e ) none |
explanation : num__100 * num__100 = num__10000 num__80 * num__115 = num__9200 - - - - - - - - - - - num__10000 - - - - - - - - - - - num__800 num__100 - - - - - - - - - - - ? = > num__8.0 decrease answer is b <eor> b <eos> |
b |
subtract__100.0__20.0__ add__15.0__100.0__ multiply__80.0__115.0__ subtract__10000.0__9200.0__ divide__800.0__100.0__ divide__800.0__100.0__ |
subtract__100.0__20.0__ add__15.0__100.0__ multiply__80.0__115.0__ subtract__10000.0__9200.0__ divide__800.0__100.0__ divide__800.0__100.0__ |
| nicky and cristina are running a race . since cristina is faster than nicky she gives him a num__30 meter head start . if cristina runs at a pace of num__5 meters per second and nicky runs at a pace of only num__3 meters per second how many seconds will nicky have run before cristina catches up to him ? <o> a ) num__15 seconds <o> b ) num__18 seconds <o> c ) num__25 seconds <o> d ) num__30 seconds <o> e ) num__45 seconds |
used pluging in method say t is the time for cristina to catch up with nicky the equation will be as under : for nicky = n = num__3 * t + num__30 for cristina = c = num__5 * t @ t = num__15 n = num__75 c = num__75 right answer ans : a <eor> a <eos> |
a |
multiply__5.0__3.0__ multiply__5.0__15.0__ round__15.0__ |
multiply__5.0__3.0__ multiply__5.0__15.0__ multiply__5.0__3.0__ |
| a sum amounts to rs . num__3969 in num__2 years at the rate of num__5.0 p . a . if interest was compounded yearly then what was the principal ? <o> a ) s . num__3600 <o> b ) s . num__5000 <o> c ) s . num__4500 <o> d ) s . num__4800 <o> e ) s . num__5800 |
ci = num__3969 r = num__5 n = num__2 ci = p [ num__1 + r / num__100 ] ^ num__2 = p [ num__1 + num__0.05 ] ^ num__2 num__3969 = p [ num__1.05 ] ^ num__2 num__3969 [ num__0.952380952381 ] ^ num__2 num__3600 answer : a <eor> a <eos> |
a |
percent__5.0__1.0__ percent__100.0__3600.0__ |
percent__5.0__1.0__ percent__100.0__3600.0__ |
| a positive integer is divisible by num__3 if and only if the sum of its digits is divisible by num__3 . if the six - digit integer is divisible by num__3 and n is of the form num__1 k num__2 k num__42 where k represents a digit that occurs twice how many values could n have ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__10 |
num__1 k num__2 k num__42 taking sum of the numericals = num__1 + num__2 + num__4 + num__2 = num__9 we require the values of k such that they are also divisible by num__3 num__102042 num__132342 num__162642 num__192942 answer = num__4 = d <eor> d <eos> |
d |
add__3.0__1.0__ add__3.0__1.0__ |
add__3.0__1.0__ add__3.0__1.0__ |
| how many bricks each measuring num__50 cm x num__11.25 cm x num__6 cm will be needed to build a wall of num__8 m x num__6 m x num__22.5 cm ? <o> a ) num__6400 <o> b ) num__2400 <o> c ) num__5500 <o> d ) num__7400 <o> e ) num__3200 |
number of bricks = volume of the wall / volume of num__1 brick = ( num__800 x num__600 x num__22.5 ) / ( num__50 x num__11.25 x num__6 ) = num__3200 answer : e <eor> e <eos> |
e |
round__3200.0__ |
divide__3200.0__1.0__ |
| how much time will take for an amount of rs . num__410 to yield rs . num__81 as interest at num__4.5 per annum of simple interest ? <o> a ) num__4 years num__4 months <o> b ) num__4 years <o> c ) num__3 years <o> d ) num__9 years <o> e ) num__5 years |
time = ( num__100 * num__81 ) / ( num__410 * num__4.5 ) = num__4 years num__4 months answer : a <eor> a <eos> |
a |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| the youngest of num__4 children has siblings who are num__2 num__7 and num__11 years older than she is . if the average ( arithmetic mean ) age of the num__4 siblings is num__25 what is the age of the youngest sibling ? <o> a ) num__17 <o> b ) num__18 <o> c ) num__19 <o> d ) num__20 <o> e ) num__21 |
x + ( x + num__2 ) + ( x + num__7 ) + ( x + num__11 ) = num__100 num__4 x + num__20 = num__100 num__4 x = num__80 x = num__20 the answer is d . <eor> d <eos> |
d |
multiply__4.0__25.0__ multiply__4.0__20.0__ divide__80.0__4.0__ |
multiply__4.0__25.0__ multiply__4.0__20.0__ divide__80.0__4.0__ |
| two pipes a and b can separately fill a tank in num__20 and num__15 minutes respectively . a third pipe c can drain off num__45 liters of water per minute . if all the pipes are opened the tank can be filled in num__15 minutes . what is the capacity of the tank ? <o> a ) num__590 liters <o> b ) num__540 liters <o> c ) num__820 liters <o> d ) num__900 liters <o> e ) num__580 liters |
num__0.05 + num__0.0666666666667 - num__1 / x = num__0.0666666666667 x = num__12 num__20 * num__45 = num__900 answer : d <eor> d <eos> |
d |
multiply__20.0__0.05__ multiply__20.0__45.0__ round__900.0__ |
multiply__20.0__0.05__ multiply__20.0__45.0__ multiply__20.0__45.0__ |
| find the next term in the series num__1 num__0 - num__2 - num__3 num__8 num__95 <o> a ) num__683 <o> b ) num__684 <o> c ) num__685 <o> d ) num__686 <o> e ) num__687 |
its n ! - n ^ num__2 num__1 - num__1 ^ num__2 = num__0 num__2 ! - num__2 ^ num__2 = - num__2 num__3 ! ( num__6 ) - num__3 ^ num__2 = - num__3 num__4 ! ( num__24 ) - num__4 ^ num__2 = num__8 . . . . num__6 ! ( num__720 ) - num__6 ^ num__2 = num__684 answer : b <eor> b <eos> |
b |
multiply__2.0__3.0__ add__1.0__3.0__ multiply__3.0__8.0__ multiply__1.0__684.0__ |
subtract__8.0__2.0__ subtract__6.0__2.0__ multiply__3.0__8.0__ multiply__1.0__684.0__ |
| in an election between the two candidates the candidates who gets num__60.0 of votes polled is winned by num__280 votes majority . what is the total number of votes polled ? <o> a ) num__1200 <o> b ) num__1400 <o> c ) num__1600 <o> d ) num__1500 <o> e ) num__1540 |
note : majority ( num__20.0 ) = difference in votes polled to win ( num__60.0 ) & defeated candidates ( num__40.0 ) num__20.0 = num__60.0 - num__40.0 num__20.0 - - - - - > num__280 ( num__20 × num__14 = num__280 ) num__100.0 - - - - - > num__1400 ( num__100 × num__14 = num__1400 ) option ' b ' <eor> b <eos> |
b |
percent__100.0__1400.0__ |
percent__100.0__1400.0__ |
| a man has rs . num__496 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__90 <o> b ) num__93 <o> c ) num__96 <o> d ) num__97 <o> e ) num__99 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__496 num__16 x = num__496 x = num__31 . hence total number of notes = num__3 x = num__93 . b <eor> b <eos> |
b |
divide__496.0__16.0__ multiply__3.0__31.0__ multiply__3.0__31.0__ |
divide__496.0__16.0__ multiply__3.0__31.0__ multiply__3.0__31.0__ |
| if the difference between the length and breadth of a rectangle is num__23 m and its perimeter is num__206 m what is its area ? <o> a ) num__2800 m num__2 <o> b ) num__2740 m num__2 <o> c ) num__2520 m num__2 <o> d ) num__2200 m num__2 <o> e ) num__2500 m num__2 |
explanation : length = breadth + num__23 . therefore num__4 × num__4 × breadth + num__2 × num__23 = num__206 m ⇒ breadth = num__40 m length = num__40 + num__23 = num__63 m area = num__63 × num__40 = num__2520 m num__2 answer : option c <eor> c <eos> |
c |
multiply__40.0__63.0__ triangle_area__2.0__2520.0__ |
multiply__40.0__63.0__ triangle_area__2.0__2520.0__ |
| for a certain set the value range of its members is num__104.8 . a new set is created from the members of the old set as follows : num__12 is subtracted from a member of the old set and the result is divided by num__4 . the resulting value is a member of the new set . if this operation is done for each member of the old set what is the range of values of the members of the new set ? <o> a ) num__23.2 <o> b ) num__26.2 <o> c ) num__52.4 <o> d ) num__98.4 <o> e ) num__104.8 |
let x and z be the smallest and largest of the original set respectively . z - x = num__104.8 the smallest and largest members of the new set will be ( x - num__12 ) / num__4 and ( z - num__12 ) / num__4 . then the range is ( z - num__12 ) / num__4 - ( x - num__12 ) / num__4 = ( z - x ) / num__4 = num__104.8 / num__4 = num__26.2 the answer is b . <eor> b <eos> |
b |
divide__104.8__4.0__ divide__104.8__4.0__ |
divide__104.8__4.0__ divide__104.8__4.0__ |
| if x = num__3 and y = - num__2 what is the value of num__2 ( x - y ) ^ num__2 - xy ? <o> a ) num__152 <o> b ) num__142 <o> c ) num__121 <o> d ) num__56 <o> e ) num__111 |
x = num__3 and y = - num__2 x - y = num__3 - ( - num__2 ) = num__3 + num__2 = num__5 x * y = num__3 * - num__2 = - num__6 now we apply it in the equation num__2 ( x - y ) ^ num__2 - xy = num__2 ( num__5 ) ^ num__2 - ( - num__6 ) = = > num__2 * num__25 + num__6 = num__50 + num__6 = num__56 answer : d <eor> d <eos> |
d |
add__3.0__2.0__ multiply__3.0__2.0__ multiply__2.0__25.0__ add__6.0__50.0__ add__6.0__50.0__ |
add__3.0__2.0__ multiply__3.0__2.0__ multiply__2.0__25.0__ add__6.0__50.0__ add__6.0__50.0__ |
| a shopkeeper buys two articles for rs . num__1000 each and then sells them making num__20.0 profit on the first article and num__20.0 loss on second article . find the net profit or loss percent ? <o> a ) num__200 <o> b ) num__992 <o> c ) num__772 <o> d ) num__662 <o> e ) num__552 |
profit on first article = num__20.0 of num__1000 = num__200 . this is equal to the loss he makes on the second article . that is he makes neither profit nor loss . answer : a <eor> a <eos> |
a |
percent__20.0__1000.0__ percent__20.0__1000.0__ |
percent__20.0__1000.0__ percent__20.0__1000.0__ |
| the ratio between the number of sheep and the number of horses at the stewar farm is num__5 to num__7 . if each of horse is fed num__230 ounces of horse food per day and the farm needs a total num__12880 ounces of horse food per day . what is number sheep in the form ? ? <o> a ) num__18 <o> b ) num__28 <o> c ) num__40 <o> d ) num__56 <o> e ) num__58 |
et no of sheep and horses are num__5 k and num__7 k no of horses = num__56.0 = num__56 now num__7 k = num__56 and k = num__8 no of sheep = ( num__5 * num__8 ) = num__40 answer : c <eor> c <eos> |
c |
divide__12880.0__230.0__ divide__56.0__7.0__ multiply__5.0__8.0__ multiply__5.0__8.0__ |
divide__12880.0__230.0__ divide__56.0__7.0__ multiply__5.0__8.0__ multiply__5.0__8.0__ |
| tomy ' s age num__22 years hence will be thrice his age four years ago . find tomy ' s present age ? <o> a ) num__12 years <o> b ) num__13 years <o> c ) num__15 years <o> d ) num__17 years <o> e ) num__18 years |
let mtomy ' s present age be ' x ' years . x + num__22 = num__3 ( x - num__4 ) = > num__2 x = num__34 = > x = num__15 years . d <eor> d <eos> |
d |
add__2.0__15.0__ |
add__2.0__15.0__ |
| the maximum number of student amoung them num__1802 pens and num__1203 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : <o> a ) num__1 <o> b ) num__10 <o> c ) num__11 <o> d ) num__19 <o> e ) none of these |
solution required number of student = h . c . f of num__1802 and num__1203 = num__1 . answer a <eor> a <eos> |
a |
reverse__1.0__ |
reverse__1.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 seconds . what is the length of the train ? <o> a ) num__488 <o> b ) num__200 <o> c ) num__289 <o> d ) num__150 metre <o> e ) num__822 |
speed = ( num__60 * num__0.277777777778 ) m / sec = ( num__16.6666666667 ) m / sec length of the train = ( speed x time ) = ( num__16.6666666667 * num__9 ) m = num__150 m . answer : d <eor> d <eos> |
d |
round__150.0__ |
round__150.0__ |
| on a partly cloudy day milton decides to walk back from work . when it is sunny he walks at a speed of s miles / hr ( s is an integer ) and when it gets cloudy he increases his speed to ( s + num__1 ) miles / hr . if his average speed for the entire distance is num__2.8 miles / hr what fraction of the total distance did he cover while the sun was shining on him ? <o> a ) num__0.2 <o> b ) num__0.166666666667 <o> c ) num__0.25 <o> d ) num__0.142857142857 <o> e ) num__0.333333333333 |
if s is an integer and we know that the average speed is num__2.8 s must be = num__2 . that meanss + num__1 = num__3 . this implies that the ratio of time for s = num__2 is num__0.25 of the total time . the formula for distance / rate is d = rt . . . so the distance travelled when s = num__2 is num__2 t . the distance travelled for s + num__1 = num__3 is num__3 * num__4 t or num__12 t . therefore total distance covered while the sun was shining over him is num__0.142857142857 = num__0.142857142857 . answer : d <eor> d <eos> |
d |
round_down__2.8__ add__1.0__2.0__ reverse__0.25__ divide__3.0__0.25__ multiply__1.0__0.1429__ |
round_down__2.8__ add__1.0__2.0__ reverse__0.25__ divide__3.0__0.25__ multiply__1.0__0.1429__ |
| if num__1 kilometer is approximately num__0.6 mile which of the following best approximates the number of kilometers in num__4 miles ? <o> a ) num__2.4 <o> b ) num__4.0 <o> c ) num__3.0 <o> d ) num__2.0 <o> e ) num__6.0 |
num__1 km is approxmately equal to num__0.6 miles so num__4 km = num__4 * num__0.6 = num__2.4 miles . multiple & divide by num__10 i . e num__2.4 * num__1.0 = num__2.4 = num__2.4 answer : a <eor> a <eos> |
a |
multiply__0.6__4.0__ round__2.4__ |
multiply__0.6__4.0__ multiply__1.0__2.4__ |
| a train num__125 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is ? <o> a ) num__66 <o> b ) num__77 <o> c ) num__88 <o> d ) num__50 <o> e ) num__33 |
speed of the train relative to man = ( num__12.5 ) m / sec = ( num__12.5 ) m / sec . [ ( num__12.5 ) * ( num__3.6 ) ] km / hr = num__45 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__45 = = > x = num__50 km / hr . answer : d <eor> d <eos> |
d |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ round__50.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ multiply__5.0__10.0__ |
| the average monthly salary of num__24 employees in an organisation is rs . num__2400 . if the manager ' s salary is added then the average salary increases by rs . num__100 . what is the manager ' s monthly salary ? <o> a ) rs . num__3601 <o> b ) rs . num__3618 <o> c ) rs . num__4900 <o> d ) rs . num__3619 <o> e ) rs . num__3610 |
manager ' s monthly salary = rs . ( num__2500 * num__25 - num__2400 * num__24 ) = rs . num__4900 answer : c <eor> c <eos> |
c |
add__2400.0__100.0__ divide__2500.0__100.0__ add__2400.0__2500.0__ add__2400.0__2500.0__ |
add__2400.0__100.0__ divide__2500.0__100.0__ add__2400.0__2500.0__ add__2400.0__2500.0__ |
| two vessels p and q contain num__62.5 and num__87.5 of alcohol respectively . if num__4 litres from vessel p is mixed with num__8 litres from vessel q the ratio of alcohol and water in the resulting mixture is ? <o> a ) num__19 : num__1 <o> b ) num__19 : num__4 <o> c ) num__19 : num__8 <o> d ) num__38 : num__10 <o> e ) num__38 : num__2 |
quantity of alcohol in vessel p = num__62.5 / num__100 * num__4 = num__2.5 litres quantity of alcohol in vessel q = num__87.5 / num__100 * num__8 = num__7.0 litres quantity of alcohol in the mixture formed = num__2.5 + num__7.0 = num__9.5 = num__9.50 litres as num__12 litres of mixture is formed ratio of alcohol and water in the mixture formed = num__9.50 : num__2.50 = num__38 : num__10 . answer : d <eor> d <eos> |
d |
add__2.5__7.0__ add__4.0__8.0__ multiply__4.0__9.5__ multiply__4.0__2.5__ multiply__4.0__9.5__ |
add__2.5__7.0__ add__4.0__8.0__ multiply__4.0__9.5__ multiply__4.0__2.5__ multiply__4.0__9.5__ |
| each child has num__2 pencils and num__13 skittles . if there are num__8 children how many pencils are there in total ? <o> a ) num__16 <o> b ) num__12 <o> c ) num__18 <o> d ) num__22 <o> e ) num__08 |
num__2 * num__8 = num__16 . answer is a . <eor> a <eos> |
a |
multiply__2.0__8.0__ multiply__2.0__8.0__ |
multiply__2.0__8.0__ multiply__2.0__8.0__ |
| the product z of two prime numbers is between num__10 and num__30 . if one of the prime numbers is greater than num__2 but less than num__6 and the other prime number is greater than num__6 but less than num__24 then what is z ? <o> a ) num__21 <o> b ) num__15 <o> c ) num__14 <o> d ) num__10 <o> e ) num__6 |
the smallest possible product is num__21 which is num__3 * num__7 . all other products are too big . the answer is a . <eor> a <eos> |
a |
divide__30.0__10.0__ subtract__10.0__3.0__ subtract__24.0__3.0__ |
divide__30.0__10.0__ subtract__10.0__3.0__ multiply__3.0__7.0__ |
| mother is aged num__3 times more than her daughter rose . after num__8 years she would be two and a num__0.5 times of rose ' s age . after further num__8 years how many times would he be of rose ' s age ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
let ronit ' s present age be x years . then father ' s present age = ( x + num__3 x ) years = num__4 x years . ( num__4 x + num__8 ) = num__5 ( x + num__8 ) num__2 num__8 x + num__16 = num__5 x + num__40 num__3 x = num__24 x = num__8 . hence required ratio = ( num__4 x + num__16 ) / ( x + num__16 ) = num__2.0 = num__2 . b <eor> b <eos> |
b |
multiply__8.0__0.5__ subtract__8.0__3.0__ reverse__0.5__ divide__8.0__0.5__ multiply__8.0__5.0__ multiply__3.0__8.0__ reverse__0.5__ |
multiply__8.0__0.5__ subtract__8.0__3.0__ reverse__0.5__ divide__8.0__0.5__ multiply__8.0__5.0__ add__8.0__16.0__ reverse__0.5__ |
| the sum of three consecutive odd natural numbers each divisible by num__3 is num__78 . what is the largest among them ? <o> a ) num__21 <o> b ) num__24 <o> c ) num__29 <o> d ) num__36 <o> e ) num__45 |
let the num__3 consecutive odd numbers each divisible by num__3 be x x + num__3 x + num__6 given x + x + num__3 + x + num__6 = num__78 num__3 x = num__69 x = num__23 the numbers are num__23 num__26 num__29 therefore the largest number is num__29 answer c . <eor> c <eos> |
c |
divide__69.0__3.0__ add__3.0__23.0__ add__3.0__26.0__ add__3.0__26.0__ |
divide__69.0__3.0__ add__3.0__23.0__ add__3.0__26.0__ add__3.0__26.0__ |
| the average monthly income of p and q is rs . num__2050 . the average monthly income of q and r is rs . num__5250 and the average monthly income of p and r is rs . num__6200 . the monthly income of r is : <o> a ) num__5100 <o> b ) num__4900 <o> c ) num__9400 <o> d ) num__1500 <o> e ) num__3200 |
explanation : let p q and r represent their respective monthly incomes . then we have : p + q = ( num__2050 x num__2 ) = num__4100 . . . . ( i ) q + r = ( num__5250 x num__2 ) = num__10500 . . . . ( ii ) p + r = ( num__6200 x num__2 ) = num__12400 . . . . ( iii ) adding ( i ) ( ii ) and ( iii ) we get : num__2 ( p + q + r ) = num__27000 or p + q + r = num__13500 . . . . ( iv ) subtracting ( i ) from ( iv ) we get p = num__9400 . p ' s monthly income = rs . num__9400 . answer : c <eor> c <eos> |
c |
multiply__2050.0__2.0__ multiply__5250.0__2.0__ multiply__6200.0__2.0__ divide__27000.0__2.0__ subtract__13500.0__4100.0__ subtract__13500.0__4100.0__ |
multiply__2050.0__2.0__ multiply__5250.0__2.0__ multiply__6200.0__2.0__ divide__27000.0__2.0__ subtract__13500.0__4100.0__ subtract__13500.0__4100.0__ |
| a person crosses a num__720 m long street in num__12 minutes . what is his speed in km per hour ? <o> a ) num__3.6 <o> b ) num__4.5 <o> c ) num__4.8 <o> d ) num__5.4 <o> e ) num__5.5 |
speed = num__720 / ( num__12 x num__60 ) m / sec = num__1 m / sec . converting m / sec to km / hr = num__1 x ( num__3.6 ) km / hr = num__3.6 km / hr . answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ round__3.6__ |
divide__720.0__12.0__ divide__3.6__1.0__ |
| a person want to give his money of $ num__900 to his num__3 children a b c in the ratio num__2 : num__3 : num__4 . what is the b ' s share ? <o> a ) $ num__400 <o> b ) $ num__300 <o> c ) $ num__500 <o> d ) $ num__180 <o> e ) $ num__200 |
b ' s share = num__900 * num__0.333333333333 = $ num__300 answer is b <eor> b <eos> |
b |
reverse__3.0__ divide__900.0__3.0__ divide__900.0__3.0__ |
reverse__3.0__ divide__900.0__3.0__ divide__900.0__3.0__ |
| the length of the bridge which a train num__130 meters long and travelling at num__45 km / hr can cross in num__30 seconds is : <o> a ) num__200 m <o> b ) num__225 m <o> c ) num__245 m <o> d ) num__250 m <o> e ) num__260 m |
explanation : speed = ( num__45 * num__0.277777777778 ) m / sec = ( num__12.5 ) m / sec . time = num__30 sec . let the length of bridge be x meters . then ( num__130 + x ) / num__30 = num__12.5 = = > num__2 ( num__130 + x ) = num__750 = = > x = num__245 m . answer is c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| it takes jim two hours to wash the car and it takes julia one hour to wash the car . working together how many minutes will it take them to wash the car ? <o> a ) num__36 <o> b ) num__38 <o> c ) num__40 <o> d ) num__42 <o> e ) num__44 |
jim can wash ( num__0.5 ) of the car per hour . julia can wash num__1 car per hour . together they can wash num__0.5 + num__1 = num__1.5 cars per hour . num__1 car / ( num__1.5 ) cars per hour = num__0.666666666667 hours = num__40 minutes . the answer is c . <eor> c <eos> |
c |
add__0.5__1.0__ divide__1.0__1.5__ round__40.0__ |
add__0.5__1.0__ divide__1.0__1.5__ divide__40.0__1.0__ |
| if the cost price is num__96.0 of selling price then what is the profit percentage . <o> a ) num__7.14 <o> b ) num__7.41 <o> c ) num__4.71 <o> d ) num__4.17 <o> e ) none of them |
selling price = rs num__100 : then cost price = rs num__96 : profit = rs num__4 . profit = { ( num__0.0416666666667 ) * num__100 } % = num__4.17 answer is d . <eor> d <eos> |
d |
percent__4.17__100.0__ |
percent__4.17__100.0__ |
| after spending rs . num__5000 on rent rs . num__1500 on milk rs . num__4500 on groceries rs . num__2500 on childrens education rs . num__2000 on petrol and rs . num__2500 on miscellaneous expenses mr . kishore saved num__10.0 of his monthly salary . how much did he save in rs . ? <o> a ) num__2160 <o> b ) num__2350 <o> c ) num__2000 <o> d ) num__2300 <o> e ) none of these |
explanation : total exp = num__5000 + num__1500 + num__4500 + num__2500 + num__2000 + num__2500 = num__18000 exp in % = num__100 - num__10 = num__90.0 num__18000 = num__90.0 saving = num__10.0 = num__18000 x num__0.111111111111 = rs . num__2000 answer : c <eor> c <eos> |
c |
subtract__100.0__10.0__ divide__2000.0__18000.0__ subtract__4500.0__2500.0__ |
subtract__100.0__10.0__ divide__2000.0__18000.0__ subtract__4500.0__2500.0__ |
| the least number which should be added to num__6320 so that the sum is exactly divisible by num__4 num__6 num__7 and num__9 is : <o> a ) num__112 <o> b ) num__212 <o> c ) num__252 <o> d ) num__239 <o> e ) num__232 |
l . c . m . of num__5 num__6 num__4 and num__3 = num__252 . on dividing num__6320 by num__252 the remainder is num__20 . number to be added = ( num__252 - num__20 ) = num__232 . answer : option ' e ' <eor> e <eos> |
e |
subtract__9.0__4.0__ subtract__7.0__4.0__ multiply__4.0__5.0__ subtract__252.0__20.0__ subtract__252.0__20.0__ |
subtract__9.0__4.0__ subtract__7.0__4.0__ multiply__4.0__5.0__ subtract__252.0__20.0__ subtract__252.0__20.0__ |
| the price of commodity x increases by num__45 cents every year while the price of commodity y increases by num__20 cents every year . in num__2001 the price of commodity x was $ num__4.20 and the price of commodity y was $ num__6.30 . in which year will the price of commodity x be num__65 cents more than the price of commodity y ? <o> a ) num__2010 <o> b ) num__2011 <o> c ) num__2012 <o> d ) num__2013 <o> e ) num__2014 |
the price of commodity x increases num__25 cents each year relative to commodity y . the price difference is $ num__2.10 and commodity x needs to be num__65 cents more than commodity y . $ num__2.75 / num__25 cents = num__11 years the answer is num__2001 + num__11 years = num__2012 . the answer is c . <eor> c <eos> |
c |
subtract__45.0__20.0__ subtract__6.3__4.2__ add__2001.0__11.0__ add__2001.0__11.0__ |
subtract__45.0__20.0__ subtract__6.3__4.2__ add__2001.0__11.0__ add__2001.0__11.0__ |
| how long does a train num__120 m long running at the speed of num__70 km / hr takes to cross a bridge num__150 m length ? <o> a ) num__12.7 sec <o> b ) num__13.9 sec <o> c ) num__18.1 sec <o> d ) num__17.1 sec <o> e ) num__19.7 sec |
speed = num__70 * num__0.277777777778 = num__19.4 m / sec total distance covered = num__120 + num__150 = num__270 m . required time = num__270 / num__19.4 ' = num__13.9 sec . answer : b <eor> b <eos> |
b |
add__120.0__150.0__ round__13.9__ |
add__120.0__150.0__ round__13.9__ |
| when running a mile during a recent track meet nuria was initially credited with a final time of num__5 minutes num__39 seconds . shortly after her run officials realized that the timing mechanism malfunctioned . the stopwatch did not begin timing her until num__0.44 of a minute after she began to run . if the time was otherwise correct how long did it actually take nuria to run the mile ? <o> a ) num__5 minutes num__17.6 seconds <o> b ) num__6 minutes num__10.4 seconds <o> c ) num__5 minutes num__43.56 seconds <o> d ) num__5 minutes num__44.44 seconds <o> e ) num__5 minutes num__21.8 seconds |
one approach : the watch starts to work after nuria began his running . it means the time should be greater than credited num__5 minutes num__44 seconds . the only number is num__6 minutes num__10.4 seconds . another approach : num__0.44 close to num__30 second when added to the num__5 minutes num__44 seconds it means it passes num__6 minute . answer : b <eor> b <eos> |
b |
add__5.0__39.0__ multiply__5.0__6.0__ round__6.0__ |
add__5.0__39.0__ multiply__5.0__6.0__ round__6.0__ |
| six students are equally divided into num__3 groups then the three groups were assigned to three different topics . how many different arrangements w are possible ? <o> a ) num__30 <o> b ) num__60 <o> c ) num__90 <o> d ) num__180 <o> e ) num__540 |
num__90 is the number of ways you can assign num__3 teams formed out of num__12 people to num__3 different tasks . but now you can order the num__3 tasks in num__3 ! ways . t num__1 t num__2 t num__3 or t num__2 t num__1 t num__3 . . . . etc etc . i was confused between num__90 and num__540 but since question used the wordarrangementsdecided to go with complete arrangements w including the order of tasks . could you explain the highlighted step . . . i ' m getting num__90 = num__15 * num__3 ! suppose the students are numbered num__12 num__34 num__56 and tasks are x y and z one of the num__15 possible ways of forming teams is num__12 num__34 num__56 . these teams can be assigned to num__3 tasks in num__3 ! = num__6 ways x - - y - - z num__12 - - num__34 - - num__56 num__12 - - num__56 - - num__34 num__34 - - num__12 - - num__56 num__34 - - num__56 - - num__12 num__56 - - num__12 - - num__34 num__56 - - num__34 - - num__12 so the answer should be num__15 * num__6 = num__90 but now you can fruther decide which task you want to perform first x y or z . . = c <eor> c <eos> |
c |
subtract__3.0__1.0__ add__3.0__12.0__ subtract__90.0__34.0__ multiply__3.0__2.0__ multiply__1.0__90.0__ |
subtract__3.0__1.0__ add__3.0__12.0__ subtract__90.0__34.0__ multiply__3.0__2.0__ multiply__1.0__90.0__ |
| num__1 + num__2 + num__2 ^ num__2 = ? <o> a ) ( num__2 ^ num__3 - num__1 ) ( num__2 ^ num__3 + num__1 ) <o> b ) num__2 ^ num__6 + num__1 <o> c ) num__2 ^ num__5 - num__1 <o> d ) num__2 ^ num__5 + num__1 <o> e ) num__2 ^ num__3 - num__1 |
from num__1 + num__2 + num__2 ^ num__2 = num__1 ( num__2 ^ num__6 - num__1 ) / ( num__2 - num__1 ) = num__2 ^ num__3 - num__1 the correct answer is e . <eor> e <eos> |
e |
add__1.0__2.0__ multiply__1.0__2.0__ |
add__1.0__2.0__ divide__2.0__1.0__ |
| there are three piles of books a b and c having num__3 num__45 books respectively . these books have to be picked up and put in a shelf one by one . how many possible sequence can be formed in the shelf if a book can be picked only from top of any pile ? <o> a ) num__12 ! <o> b ) num__3 ! * num__4 ! * num__5 ! <o> c ) num__12 c num__3 * num__9 c num__4 <o> d ) num__12 ^ num__12 <o> e ) num__12 ^ num__14 |
since we should fill only from top . . . . . . . . . num__12 ! is not correct we have possibilty of filling all num__5 from c and go to a b num__0 r num__1 each from each pile or num__2 . . . . . . . . . . so num__3 ! * num__4 ! * num__5 ! and num__12 c num__3 * num__9 c num__4 are also wrong . remaining option is . . . . . . . . num__12 ^ num__12 answer : d <eor> d <eos> |
d |
vowel_space__ negate_prob__0.0__ coin_space__ choose__2.0__0.0__ choose__3.0__1.0__ |
vowel_space__ negate_prob__0.0__ coin_space__ choose__2.0__0.0__ choose__3.0__1.0__ |
| if a train travelling at a speed of num__120 kmph crosses a pole in num__6 sec then the length of train is ? <o> a ) num__281 <o> b ) num__125 <o> c ) num__200 <o> d ) num__266 <o> e ) num__121 |
d = num__120 * num__0.277777777778 * num__6 = num__200 m answer : c <eor> c <eos> |
c |
round__200.0__ |
round__200.0__ |
| num__3 women and a few men participated in a chess tournament . each player played two matches with each of the other players . if the number of matches that men played among themselves is num__78 more than those they played with the women how many more men than women participated in the tournament ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__8 <o> d ) num__9 <o> e ) num__5 |
let x be the number of men . number of matches men play among themselves is num__2 * ( xc num__2 ) number of matches men play with women is num__2 * ( num__3 x ) num__2 * ( xc num__2 ) - num__2 * ( num__3 x ) = num__78 x = num__13 difference between men and women is num__13 - num__3 = num__10 . answer a <eor> a <eos> |
a |
subtract__13.0__3.0__ subtract__13.0__3.0__ |
subtract__13.0__3.0__ subtract__13.0__3.0__ |
| what is the remainder when num__43 ^ num__87 is divided by num__5 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
the units digit of the exponents of num__3 cycle in a group of num__4 : { num__3 num__9 num__7 num__1 } num__87 has the form num__4 k + num__3 so the units digit of num__43 ^ num__87 is num__7 . the remainder when dividing by num__5 is num__2 . the answer is c . <eor> c <eos> |
c |
add__5.0__4.0__ add__3.0__4.0__ subtract__5.0__4.0__ subtract__5.0__3.0__ subtract__5.0__3.0__ |
add__5.0__4.0__ add__3.0__4.0__ subtract__5.0__4.0__ subtract__5.0__3.0__ subtract__5.0__3.0__ |
| a vessel of capacity num__90 litres is fully filled with pure milk . nine litres of milk is removed from the vessel and replaced with water . nine litres of the solution thus formed is removed and replaced with water . find the quantity of pure milk in the final milk solution ? <o> a ) num__72.8 litres <o> b ) num__72.9 litres <o> c ) num__74.9 litres <o> d ) num__72.4 litres <o> e ) num__72.7 litres |
let the initial quantity of milk in vessel be t litres . let us say y litres of the mixture is taken out and replaced by water for n times alternatively . quantity of milk finally in the vessel is then given by [ ( t - y ) / t ] n * t for the given problem t = num__90 y = num__9 and n = num__2 . hence quantity of milk finally in the vessel = [ ( num__90 - num__9 ) / num__90 ] num__2 ( num__90 ) = num__72.9 litres . answer : b <eor> b <eos> |
b |
round__72.9__ |
round__72.9__ |
| convert the num__0.361111111111 m / s into kilometers per hour ? <o> a ) num__1.7 <o> b ) num__1.5 <o> c ) num__1.3 <o> d ) num__1.1 <o> e ) num__1.2 |
num__0.361111111111 m / s = num__0.361111111111 * num__3.6 = num__1.3 = num__1.3 kmph . answer : c <eor> c <eos> |
c |
multiply__0.3611__3.6__ round__1.3__ |
multiply__0.3611__3.6__ multiply__0.3611__3.6__ |
| two letters from the word myantkar are selected at random . what is the probability that at least one of the selected letters is a consonant ? <o> a ) num__0.964285714286 <o> b ) num__0.892857142857 <o> c ) num__1.03571428571 <o> d ) num__1.14285714286 <o> e ) num__1.5 |
there are num__6 consonants and num__2 vowels in myantkar . probability that at least one of the selected letters is a consonant = num__1 - ( probability of both vowels ) probability of both vowels = num__2 c num__0.25 c num__2 = num__0.0357142857143 so answer is num__1 - num__0.0357142857143 = num__0.964285714286 answer : a <eor> a <eos> |
a |
die_space__ coin_space__ negate_prob__0.0357__ negate_prob__0.0357__ |
die_space__ coin_space__ negate_prob__0.0357__ negate_prob__0.0357__ |
| if num__6 ( a ' s capital ) = num__8 ( b ' s capital ) = num__10 ( c ' s capital ) . then the ratio of their capitals is ? <o> a ) num__20 : num__15 : num__17 <o> b ) num__20 : num__15 : num__19 <o> c ) num__20 : num__15 : num__12 <o> d ) num__20 : num__15 : num__18 <o> e ) num__20 : num__15 : num__23 |
num__6 a = num__8 b = num__10 c a : b : c = num__0.166666666667 : num__0.125 : num__0.1 = num__20 : num__15 : num__12 answer : c <eor> c <eos> |
c |
reverse__6.0__ reverse__8.0__ reverse__10.0__ subtract__20.0__8.0__ add__8.0__12.0__ |
reverse__6.0__ reverse__8.0__ reverse__10.0__ subtract__20.0__8.0__ add__8.0__12.0__ |
| at the rate of num__6.0 p . a si a sum of rs . num__2500 will earn how much interest by the end of num__5 years ? <o> a ) num__750 <o> b ) num__650 <o> c ) num__250 <o> d ) num__980 <o> e ) num__950 |
num__2500 - - - > num__5 * num__6.0 = num__30.0 ( num__2500 ) = num__750 ans a <eor> a <eos> |
a |
percent__30.0__2500.0__ percent__30.0__2500.0__ |
percent__30.0__2500.0__ percent__30.0__2500.0__ |
| if n is a prime number greater than num__5 what is the remainder when n ^ num__2 is divided by num__23 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__5 |
there are several algebraic ways to solve this question including the one under the spoiler . but the easiest way is as follows : since we can not have two correct answersjust pick a prime greater than num__5 square it and see what would be the remainder upon division of it by num__23 . n = num__5 - - > n ^ num__2 = num__49 - - > remainder upon division num__49 by num__23 is num__3 . answer : d . <eor> d <eos> |
d |
subtract__5.0__2.0__ subtract__5.0__2.0__ |
subtract__5.0__2.0__ subtract__5.0__2.0__ |
| if the compound interest on a certain sum for num__2 years in rs . num__80.80 and the simple interest rs . num__80 ; then the rate of interest per annum is <o> a ) num__2.0 <o> b ) num__1.0 <o> c ) num__3.0 <o> d ) num__4.0 <o> e ) num__5 % |
explanation : let the sum be p and rate of interest be r % per annum simple interest on rs . p for num__2 years = num__80 prt / num__100 = num__80 ( pr × num__2 ) / num__100 = num__80 pr / num__50 = num__80 pr = num__4000 - - - ( equation num__1 ) compound interest = p ( num__1 + r / num__100 ) t - p = p ( num__1 + r / num__100 ) num__2 - p = p [ ( num__1 + r / num__100 ) num__2 - num__1 ] = p [ ( num__1 + num__2 r / num__100 + r num__0.0002 ) - num__1 ] = p ( num__2 r / num__100 + r num__0.0002 ) = num__2 pr / num__100 + pr num__0.0002 = num__2 pr / num__100 + ( pr × r ) / num__10000 = [ num__2 × num__4000 ] / num__100 + ( num__4000 × r ) / num__10000 ( substituted the value of pr from equation num__1 ) = num__80 + num__0.4 r given that compound interest = rs . num__80.80 = > num__80 + num__0.4 r = num__80.80 = > num__0.4 r = num__0.80 = > r = num__0.80 / num__0.4 = num__2.0 answer : option a <eor> a <eos> |
a |
percent__2.0__50.0__ percent__80.0__1.0__ percent__2.0__100.0__ |
percent__2.0__50.0__ percent__80.0__1.0__ percent__2.0__100.0__ |
| a train moves past a post and a platform num__264 m long in num__8 seconds and num__20 seconds respectively . what is the speed of the train ? <o> a ) num__79.0 <o> b ) num__79.2 <o> c ) num__79.5 <o> d ) num__79.7 <o> e ) num__79.6 |
let x is the length of the train and v is the speed time taken to move the post = num__8 s = > x / v = num__8 = > x = num__8 v - - - ( num__1 ) time taken to cross the platform num__264 m long = num__20 s ( x + num__264 ) / v = num__20 = > x + num__264 = num__20 v - - - ( num__2 ) substituting equation num__1 in equation num__2 we get num__8 v + num__264 = num__20 v = > v = num__22.0 = num__22 m / s = num__22 × num__3.6 km / hr = num__79.2 km / hr answer is b . <eor> b <eos> |
b |
add__20.0__2.0__ multiply__3.6__22.0__ round__79.2__ |
add__20.0__2.0__ multiply__3.6__22.0__ divide__79.2__1.0__ |
| the difference of num__2 digit number & the number obtained by interchanging the digits is num__36 . what is the difference the sum and the number if the ratio between the digits of the number is num__1 : num__2 ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__11 |
let the number be xy . given xy – yx = num__36 . this means the number is greater is than the number got on reversing the digits . this shows that the ten ’ s digit x > unit digit y . also given ratio between digits is num__1 : num__2 = > x = num__2 y ( num__10 x + y ) – ( num__10 y + x ) = num__36 = > x – y = num__4 = > num__2 y – y = num__4 . hence ( x + y ) – ( x – y ) = num__3 y – y = num__2 y = num__8 b <eor> b <eos> |
b |
add__2.0__1.0__ multiply__2.0__4.0__ multiply__2.0__4.0__ |
add__2.0__1.0__ multiply__2.0__4.0__ multiply__2.0__4.0__ |
| xy = num__1 then what is ( num__4 ^ ( x + y ) ^ num__2 ) / ( num__4 ^ ( x - y ) ^ num__2 ) <o> a ) num__256 <o> b ) num__4 <o> c ) num__8 <o> d ) num__16 <o> e ) num__32 |
( x + y ) ^ num__2 - ( x - y ) ^ num__2 ( x + y + x - y ) ( x + y - x + y ) ( num__2 x ) ( num__2 y ) num__4 xy num__4 num__4 ^ num__4 = num__256 answer a <eor> a <eos> |
a |
multiply__1.0__256.0__ |
multiply__1.0__256.0__ |
| a certain galaxy is known to comprise approximately num__4 x num__10 ^ num__12 stars . of every num__50 million of these stars one is larger in mass than our sun . approximately how many stars in this galaxy are larger than the sun ? <o> a ) num__800 <o> b ) num__1250 <o> c ) num__8000 <o> d ) num__12000 <o> e ) num__80 |
000 |
num__4 * num__10 ^ num__12 num__50 mln = num__5 * num__10 ^ num__7 we divide num__10 ^ num__12 by num__10 ^ num__7 and we get ( num__10 ^ num__5 ) * num__4 = num__400000 and divide by num__5 . the result is num__80000 e <eor> e <eos> |
e |
e |
| kavi spends num__50.0 of his monthly salary on food and saves num__80.0 of the remaining amount . if his monthly salary is rs . num__19000 how much money does he save every month ? <o> a ) rs . num__2000 <o> b ) rs . num__600 <o> c ) rs . num__8000 <o> d ) rs . num__3800 <o> e ) rs . num__1200 |
explanation : kavi ' s monthly income = rs . num__19000 he spends num__50.0 on food . the total money spent on food = num__0.5 * num__19000 = rs . num__9500 now his monthly remaining income = rs . num__19000 – rs . num__9500 = rs . num__9500 out of rs . num__9500 he saves num__40.0 . amount saved = num__0.4 * num__9500 = rs . num__3800 answer : d <eor> d <eos> |
d |
percent__50.0__19000.0__ percent__50.0__80.0__ percent__80.0__0.5__ percent__40.0__9500.0__ percent__40.0__9500.0__ |
percent__50.0__19000.0__ percent__50.0__80.0__ percent__80.0__0.5__ percent__40.0__9500.0__ percent__40.0__9500.0__ |
| the area of sector of a circle whose radius is num__12 metro and whose angle at the center is num__30 ° is ? <o> a ) num__35 m num__2 <o> b ) num__36 m num__2 <o> c ) num__37 m num__2 <o> d ) num__38 m num__2 <o> e ) num__40 m num__2 |
num__0.0833333333333 * num__3.14285714286 * num__12 * num__12 = num__37 m num__2 answer : c <eor> c <eos> |
c |
triangle_area__37.0__2.0__ |
triangle_area__37.0__2.0__ |
| mary charges $ num__8.50 per hour to clean a house . if she works num__22 hours how much does she earn in one week ? <o> a ) $ num__687.00 <o> b ) $ num__112.50 <o> c ) $ num__187.00 <o> d ) $ num__487.50 <o> e ) $ num__387.50 |
multiply num__22 hours x $ num__8.50 = answer : $ num__187.00 correct answer c <eor> c <eos> |
c |
multiply__8.5__22.0__ round__187.0__ |
multiply__8.5__22.0__ round__187.0__ |
| in a class of students num__0.5 of the number of girls is equal to num__0.2 of the total number of students . what is the ratio of boys to girls in the class ? <o> a ) num__0.5 <o> b ) num__1.5 <o> c ) num__0.333333333333 <o> d ) num__0.666666666667 <o> e ) num__1.33333333333 |
( num__0.5 ) g = ( num__0.2 ) ( b + g ) num__5 g = num__2 b + num__2 g num__3 g = num__2 b b / g = num__1.5 . the answer is b . <eor> b <eos> |
b |
reverse__0.2__ reverse__0.5__ subtract__5.0__2.0__ multiply__0.5__3.0__ multiply__0.5__3.0__ |
reverse__0.2__ reverse__0.5__ subtract__5.0__2.0__ divide__3.0__2.0__ divide__3.0__2.0__ |
| there were totally num__100 men . num__84 are married . num__75 have t . v num__85 have radio num__70 have a . c . how many men have t . v radio a . c and also married ? <o> a ) num__11 <o> b ) num__12 <o> c ) num__13 <o> d ) num__14 <o> e ) num__15 |
num__100 - ( num__100 - num__84 ) - ( num__100 - num__75 ) - ( num__100 - num__85 ) - ( num__100 - num__70 ) = num__100 - num__16 - num__25 - num__15 - num__30 = num__100 - num__86 = num__14 answer : d <eor> d <eos> |
d |
subtract__100.0__84.0__ subtract__100.0__75.0__ subtract__100.0__85.0__ subtract__100.0__70.0__ add__70.0__16.0__ subtract__100.0__86.0__ subtract__100.0__86.0__ |
subtract__100.0__84.0__ subtract__100.0__75.0__ subtract__100.0__85.0__ subtract__100.0__70.0__ add__70.0__16.0__ subtract__100.0__86.0__ subtract__100.0__86.0__ |
| what will be in unit ' s place digit of num__3 to the power num__34 . <o> a ) num__1 <o> b ) num__6 <o> c ) num__3 <o> d ) num__0 <o> e ) num__9 |
as the unit place digit of num__3 to the power of num__1 is num__3 to the power of num__2 is num__9 . to the power of num__3 is num__7 to the power of num__4 is num__1 and it ' ll be repeated again in the same order . . . . so the unit place digit of num__3 to the power of num__34 is num__9 . . answer : e <eor> e <eos> |
e |
subtract__3.0__1.0__ subtract__9.0__2.0__ add__3.0__1.0__ multiply__1.0__9.0__ |
subtract__3.0__1.0__ subtract__9.0__2.0__ add__3.0__1.0__ multiply__1.0__9.0__ |
| a train running at the speed of num__56 km / hr crosses a pole in num__9 sec . what is the length of the train ? <o> a ) num__140 m <o> b ) num__786 m <o> c ) num__566 m <o> d ) num__546 m <o> e ) num__445 m |
speed = num__56 * num__0.277777777778 = num__15.5555555556 m / sec length of the train = speed * time = num__15.5555555556 * num__9 = num__140 m answer : a <eor> a <eos> |
a |
round__140.0__ |
round__140.0__ |
| if w and t are positive integers wt + w + t can not be <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
let wt + t + w = x add num__1 on both sides : wt + t + w + num__1 = x + num__1 t ( w + num__1 ) + w + num__1 = x + num__1 ( w + num__1 ) ( t + num__1 ) = x + num__1 minimum value of ( w + num__1 ) = num__2 minimum value of ( t + num__1 ) = num__2 hence x + num__1 can not be prime substitute x from the given options : num__6 + num__1 = num__7 - - > prime - - > wt + t + s can not be num__6 answer : b <eor> b <eos> |
b |
add__1.0__6.0__ multiply__1.0__6.0__ |
add__1.0__6.0__ subtract__7.0__1.0__ |
| a mobile battery in num__1 hour charges to num__20 percent . how much time ( in minute ) will it require more to charge to num__30 percent . <o> a ) num__145 <o> b ) num__90 <o> c ) num__175 <o> d ) num__160 <o> e ) num__130 |
num__1 hr = num__20 percent . thus num__15 min = num__5 percent . now to charge num__30 percent num__90 min . answer : b <eor> b <eos> |
b |
subtract__20.0__15.0__ round__90.0__ |
subtract__20.0__15.0__ round__90.0__ |
| h . c . f . of num__513 num__1134 and num__1215 is <o> a ) num__27 <o> b ) num__21 <o> c ) num__76 <o> d ) num__42 <o> e ) num__33 |
h . c . f . of num__513 num__1134 and num__1215 = num__3 × num__3 × num__3 = num__27 answer : option a <eor> a <eos> |
a |
gcd__513.0__1134.0__ gcd__513.0__1134.0__ |
gcd__513.0__1134.0__ gcd__513.0__1134.0__ |
| if num__10 men can make num__200 tables working for num__15 days at the rate of num__8 hr per day how many tables can num__5 men make working for num__10 days at the rate of num__6 hr a day ? a <o> a ) num__150 <o> b ) num__50 <o> c ) num__250 <o> d ) num__100 <o> e ) num__120 |
formula men * days * workinghour / work so m num__1 * d num__1 * t num__1 / w num__1 = m num__2 * d num__2 * t num__2 / w num__2 = num__50 answer : b <eor> b <eos> |
b |
subtract__6.0__5.0__ subtract__10.0__8.0__ multiply__10.0__5.0__ round__50.0__ |
subtract__6.0__5.0__ divide__10.0__5.0__ multiply__10.0__5.0__ multiply__10.0__5.0__ |
| a and b together can do a work in num__6 days . if a alone can do it in num__42 days . in how many days can b alone do it ? <o> a ) num__11 <o> b ) num__7 <o> c ) num__8 <o> d ) num__20 <o> e ) num__25 |
b num__7 num__0.166666666667 â € “ num__0.0238095238095 = num__0.142857142857 = > num__7 <eor> b <eos> |
b |
divide__42.0__6.0__ divide__7.0__42.0__ divide__0.1667__7.0__ divide__6.0__42.0__ round__7.0__ |
divide__42.0__6.0__ divide__7.0__42.0__ divide__0.1667__7.0__ divide__6.0__42.0__ round__7.0__ |
| a cistern can be filled by a tap in num__7 hours while it can be emptied by another tap in num__9 hours . if both the taps are opened simultaneously then after how much time will the cistern get filled ? <o> a ) num__4.5 hrs <o> b ) num__5 hrs <o> c ) num__6.5 hrs <o> d ) num__17.2 hrs <o> e ) num__31.5 hrs |
net part filled in num__1 hour = num__0.142857142857 - num__0.111111111111 = num__0.031746031746 therefore the cistern will be filled in num__31.5 hours or num__31.5 hours . answer : e <eor> e <eos> |
e |
divide__1.0__7.0__ divide__1.0__9.0__ round__31.5__ |
divide__1.0__7.0__ divide__1.0__9.0__ round__31.5__ |
| a shop sells chocolates at re . num__1 each . u can exchange num__3 wrappers for num__1 chocolate . if u have rs . num__100 how many chocolates can you totally get ? <o> a ) num__147 <o> b ) num__148 <o> c ) num__149 <o> d ) num__145 <o> e ) num__141 |
for num__100 rupees we will get num__100 chocolates then for num__3 wrappers we will get num__1 chocolate so num__33.3333333333 = = num__33 chocolates and we left with one more wraper the num__33 + num__1 = num__34 . . . num__11.3333333333 = num__11 . . . one more wrapper so num__4.0 = num__4 . . num__1.33333333333 = num__1 and one more wrapper so finally . . . num__100 + num__33 + num__11 + num__4 + num__1 = num__149 answer : c <eor> c <eos> |
c |
divide__100.0__3.0__ round_down__33.3333__ add__1.0__33.0__ divide__34.0__3.0__ round_down__11.3333__ add__1.0__3.0__ divide__4.0__3.0__ multiply__1.0__149.0__ |
divide__100.0__3.0__ round_down__33.3333__ add__1.0__33.0__ divide__34.0__3.0__ round_down__11.3333__ add__1.0__3.0__ divide__4.0__3.0__ multiply__1.0__149.0__ |
| a trader bought a car at num__20.0 discount on its original price . he sold it at a num__50.0 increase on the price he bought it . what percent of profit did he make on the original price ? <o> a ) a ) num__10.0 <o> b ) b ) num__12.0 <o> c ) c ) num__14.0 <o> d ) d ) num__20.0 <o> e ) e ) num__22 % |
original price = num__100 cp = num__80 s = num__80 * ( num__1.5 ) = num__120 num__100 - num__120 = num__20.0 d ) <eor> d <eos> |
d |
percent__20.0__100.0__ |
percent__20.0__100.0__ |
| if rupee one produces rupees nine over a period of num__25 years find the rate of simple interest ? <o> a ) num__22 num__0.142857142857 % <o> b ) num__36.0 <o> c ) num__22 num__0.125 % <o> d ) num__22 num__0.5 % <o> e ) num__22 num__0.5 % |
num__9 = ( num__1 * num__25 * r ) / num__100 r = num__36.0 answer : b <eor> b <eos> |
b |
percent__36.0__100.0__ |
percent__36.0__100.0__ |
| sripad has scored average of num__64 marks in three objects . in no subjects has he secured less than num__58 marks . he has secured more marks in maths than other two subjects . what could be his maximum score in maths ? <o> a ) num__79 <o> b ) num__28 <o> c ) num__76 <o> d ) num__27 <o> e ) num__21 |
assuming sripad has scored the least marks in subject other than science then the marks he could secure in other two are num__58 each . since the average mark of all the num__3 subject is num__64 . i . e ( num__58 + num__58 + x ) / num__3 = num__64 num__116 + x = num__192 x = num__76 marks . therefore the maximum marks he can score in maths is num__76 . answer : c <eor> c <eos> |
c |
multiply__64.0__3.0__ subtract__192.0__116.0__ subtract__192.0__116.0__ |
multiply__64.0__3.0__ subtract__192.0__116.0__ subtract__192.0__116.0__ |
| roses can be purchased individually for $ num__7.30 one dozen for $ num__36 or two dozen for $ num__50 . what is the greatest number of roses that can be purchased for $ num__680 ? <o> a ) num__156 <o> b ) num__162 <o> c ) num__316 <o> d ) num__324 <o> e ) num__325 |
buy as many $ num__50 deals as possible . we can by num__13.0 = num__13 two dozen roses thus total of num__13 * num__24 = num__312 roses . we are left with num__680 - num__650 = $ num__30 . we can buy num__30 / num__7.3 = ~ num__4 roses for that amount . total = num__312 + num__4 = num__316 . answer : c . <eor> c <eos> |
c |
multiply__13.0__24.0__ multiply__50.0__13.0__ subtract__680.0__650.0__ add__4.0__312.0__ add__4.0__312.0__ |
multiply__13.0__24.0__ multiply__50.0__13.0__ subtract__680.0__650.0__ add__4.0__312.0__ add__4.0__312.0__ |
| the average age of a class of num__42 students is num__16 yrs . if the teacher ' s age is also included the average increases by one year . find the age of the teacher <o> a ) num__45 years <o> b ) num__46 years <o> c ) num__59 years <o> d ) num__52 years <o> e ) num__54 years |
total age of students is num__42 x num__16 = num__672 years total age inclusive of teacher = num__43 x ( num__16 + num__1 ) = num__731 so teacher ' s age is num__731 - num__672 = num__59 yrs there is a shortcut for these type of problems teacher ' s age is num__16 + ( num__43 x num__1 ) = num__59 years answer : c <eor> c <eos> |
c |
multiply__42.0__16.0__ subtract__43.0__42.0__ add__16.0__43.0__ add__16.0__43.0__ |
multiply__42.0__16.0__ subtract__43.0__42.0__ add__16.0__43.0__ add__16.0__43.0__ |
| a train covers a distance in num__50 min if it runs at a speed of num__48 kmph on an average . the speed at which the train must run to reduce the time of journey to num__40 min will be . <o> a ) num__60 km / h <o> b ) num__55 km / h <o> c ) num__40 km / h <o> d ) num__70 km / h <o> e ) num__80 km / h |
time = num__0.833333333333 hr = num__0.833333333333 hr speed = num__48 mph distance = s * t = num__48 * num__0.833333333333 = num__40 km time = num__0.666666666667 hr = num__0.666666666667 hr new speed = num__40 * num__1.5 kmph = num__60 kmph answer : a . <eor> a <eos> |
a |
divide__40.0__48.0__ add__0.8333__0.6667__ hour_to_min_conversion__ hour_to_min_conversion__ |
divide__40.0__48.0__ add__0.8333__0.6667__ multiply__40.0__1.5__ multiply__40.0__1.5__ |
| a train num__650 m long is running at a speed of num__117 km / hr . in what time will it pass a bridge num__325 m long ? <o> a ) num__30 <o> b ) num__35 <o> c ) num__40 <o> d ) num__45 <o> e ) num__50 |
speed = num__117 * num__0.277777777778 = num__32.5 m / sec total distance covered = num__650 + num__325 = num__975 m required time = num__975 * num__0.0307692307692 = num__30 sec answer : a <eor> a <eos> |
a |
add__650.0__325.0__ divide__975.0__32.5__ round__30.0__ |
add__650.0__325.0__ divide__975.0__32.5__ divide__975.0__32.5__ |
| a train speeds past a pole in num__25 seconds and a platform num__150 m long in num__30 seconds . its length is ? <o> a ) num__23 <o> b ) num__22 <o> c ) num__21 <o> d ) num__30 <o> e ) num__19 |
let the length of the train be x meters and its speed be y m / sec . they x / y = num__25 = > y = x / num__25 ( x + num__150 ) / num__30 = x / num__25 x = num__30 m . answer : d <eor> d <eos> |
d |
round__30.0__ |
round__30.0__ |
| two trains are moving in opposite directions with speed of num__150 km / hr and num__90 km / hr respectively . their lengths are num__1.10 km and num__0.9 km respectively . the slower train cross the faster train in - - - seconds <o> a ) num__56 <o> b ) num__48 <o> c ) num__47 <o> d ) num__30 <o> e ) num__25 |
explanation : relative speed = num__150 + num__90 = num__240 km / hr ( since both trains are moving in opposite directions ) total distance = num__1.1 + . num__9 = num__2 km time = num__0.00833333333333 hr = num__0.00833333333333 hr = num__30.0 seconds = num__30 seconds answer : option d <eor> d <eos> |
d |
add__150.0__90.0__ add__1.1__0.9__ divide__2.0__240.0__ round__30.0__ |
add__150.0__90.0__ add__1.1__0.9__ divide__2.0__240.0__ round__30.0__ |
| lcm of num__2 / num__5.21428571429 and num__1.66666666667 is <o> a ) num__45 <o> b ) num__35 <o> c ) num__30 <o> d ) num__25 <o> e ) none |
explanation : lcm of numerators / hcf of denominators = lcm of num__2 num__3 num__5 / hcfof num__7 num__143 = num__30.0 = num__30 correct option : c <eor> c <eos> |
c |
add__2.0__3.0__ add__2.0__5.0__ round__30.0__ |
add__2.0__3.0__ add__2.0__5.0__ round__30.0__ |
| a man can do a piece of work in num__5 days but with the help of his son he can finish it in num__4 days . in what time can the son do it alone ? <o> a ) num__5 <o> b ) num__5 num__0.5 <o> c ) num__7 num__0.5 <o> d ) num__6 <o> e ) num__20 |
son ' s num__1 day work = num__0.25 - num__0.2 = num__0.05 son alone can do the work in num__20 days = num__20 days answer is e <eor> e <eos> |
e |
subtract__5.0__4.0__ divide__1.0__4.0__ divide__1.0__5.0__ divide__0.25__5.0__ multiply__5.0__4.0__ round__20.0__ |
subtract__5.0__4.0__ divide__1.0__4.0__ divide__1.0__5.0__ subtract__0.25__0.2__ multiply__5.0__4.0__ round__20.0__ |
| an error num__1.0 in excess is made while measuring the side of a square . the percentage of error in the calculated area of the square is : <o> a ) num__2.01 <o> b ) num__30.1 <o> c ) num__3.01 <o> d ) num__25.01 <o> e ) num__4.05 |
explanation : num__100 cm is read as num__101 cm . a num__1 = ( num__100 × num__100 ) cm num__2 = num__10000 and a num__2 = ( num__101 × num__101 ) cm num__2 = num__10609 ( a num__2 - a num__1 ) = num__10201 - num__10000 = num__201 = > num__0.0201 * num__100 = num__2.01 answer : a <eor> a <eos> |
a |
percent__1.0__201.0__ percent__1.0__201.0__ |
percent__1.0__201.0__ percent__1.0__201.0__ |
| if the cost price is num__81.0 of selling price then what is the profit percentage . <o> a ) num__22.45 <o> b ) num__23.45 <o> c ) num__32.45 <o> d ) num__23.54 <o> e ) num__23.55 |
selling price = rs num__100 : then cost price = rs num__81 : profit = rs num__19 . profit = { ( num__0.234567901235 ) * num__100 } % = num__23.45 answer is b . <eor> b <eos> |
b |
percent__23.45__100.0__ |
percent__23.45__100.0__ |
| alice ’ s take - home pay last year was the same each month and she saved the same fraction of her take - home pay each month . the total amount of money that she had saved at the end of the year was num__6 times the amount of that portion of her monthly take - home pay that she did not save . if all the money that she saved last year was from her take - home pay what fraction of her take - home pay did she save each month ? <o> a ) num__0.142857142857 <o> b ) num__0.166666666667 <o> c ) num__0.2 <o> d ) num__0.5 <o> e ) num__0.333333333333 |
suppose the portion of her salary that she saved is x and the remaining portion is y x + y = total take home salary given num__12 x = num__6 y or num__2 x = y total take - home salary = x + num__2 x = num__3 x hence she saved num__0.333333333333 - rd portion of her take home salary e <eor> e <eos> |
e |
divide__12.0__6.0__ divide__6.0__2.0__ reverse__3.0__ reverse__3.0__ |
divide__12.0__6.0__ divide__6.0__2.0__ reverse__3.0__ reverse__3.0__ |
| john completes a piece of work in num__320 days rose completes the same work in num__480 days . if both of them work together then the number of days required to complete the work is ? <o> a ) num__190 <o> b ) num__192 <o> c ) num__194 <o> d ) num__196 <o> e ) num__198 |
if a can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days . that is the required no . of days = num__320 × num__0.6 = num__192 days answer is b <eor> b <eos> |
b |
km_to_mile_conversion__ multiply__320.0__0.6__ round__192.0__ |
km_to_mile_conversion__ multiply__320.0__0.6__ round__192.0__ |
| two pipes a and b can separately fill a tank in num__10 and num__15 minutes respectively . a third pipe c can drain off num__45 liters of water per minute . if all the pipes are opened the tank can be filled in num__15 minutes . what is the capacity of the tank ? <o> a ) num__450 <o> b ) num__540 <o> c ) num__542 <o> d ) num__829 <o> e ) num__279 |
num__0.1 + num__0.0666666666667 - num__1 / x = num__0.0666666666667 x = num__10 num__10 * num__45 = num__450 answer : a <eor> a <eos> |
a |
multiply__10.0__0.1__ multiply__10.0__45.0__ round__450.0__ |
multiply__10.0__0.1__ multiply__10.0__45.0__ multiply__10.0__45.0__ |
| of num__30 applicants for a job num__11 had at least num__4 years ' experience num__18 had degrees and num__3 had less than num__4 years ' experience and did not have a degree . how many of the applicants had at least num__4 years ' experience and a degree ? <o> a ) num__14 <o> b ) num__13 <o> c ) num__8 <o> d ) num__7 <o> e ) num__5 |
c . num__8 num__30 - num__3 = num__27 num__27 - num__11 - num__18 = - num__8 then num__8 are in the intersection between num__4 years experience and degree . answer c <eor> c <eos> |
c |
subtract__11.0__3.0__ subtract__30.0__3.0__ subtract__11.0__3.0__ |
subtract__11.0__3.0__ subtract__30.0__3.0__ subtract__11.0__3.0__ |
| if the average ( arithmetic mean ) of num__8 consecutive odd integers is num__414 then the least of these integers is <o> a ) a ) num__407 <o> b ) b ) num__518 <o> c ) c ) num__519 <o> d ) d ) num__521 <o> e ) e ) num__525 |
a very helpful rule to know in arithmetic is the rule that in evenly spaced sets average = median . because the average will equal the median in these sets then we quickly know that the median of this set of consecutive odd integer numbers is num__414 . there are num__8 numbers in the set and in a set with an even number of terms the median is just the average of the two most median terms ( here the num__4 th and num__5 th numbers in the set ) . this means that numbers num__4 and num__5 in this set are num__413 and num__415 . because we know that number num__4 is num__413 we know that the smallest number is num__3 odd numbers below this which means that it is num__3 * num__2 = num__6 below this ( every odd number is every other number ) . therefore num__413 - num__6 = num__407 answer choice a <eor> a <eos> |
a |
subtract__8.0__5.0__ divide__8.0__4.0__ subtract__8.0__2.0__ subtract__413.0__6.0__ subtract__413.0__6.0__ |
subtract__8.0__5.0__ subtract__5.0__3.0__ subtract__8.0__2.0__ subtract__413.0__6.0__ subtract__413.0__6.0__ |
| if n = num__2 ^ num__0.15 and n ^ b = num__32 b must equal <o> a ) num__0.0375 <o> b ) num__0.6 <o> c ) num__4 <o> d ) num__33.3333333333 <o> e ) num__26.6666666667 |
num__0.15 = num__0.15 n = num__2 ^ num__0.15 n ^ b = num__2 ^ num__5 ( num__2 ^ num__0.15 ) ^ b = num__2 ^ num__5 b = num__33.3333333333 answer : d <eor> d <eos> |
d |
divide__5.0__0.15__ divide__5.0__0.15__ |
divide__5.0__0.15__ divide__5.0__0.15__ |
| the slant height of a cone is num__12 cm and radius of the base is num__4 cm find the curved surface of the cone ? <o> a ) num__87 <o> b ) num__26 <o> c ) num__48 <o> d ) num__37 <o> e ) num__20 |
π * num__12 * num__4 = num__48 answer : c <eor> c <eos> |
c |
square_perimeter__12.0__ square_perimeter__12.0__ |
multiply__12.0__4.0__ multiply__12.0__4.0__ |
| a car has runs num__10000 miles using num__5 tyres interchangably to have a equal wornout by all tyres how many miles each tyre should have run ? <o> a ) num__7000 miles . <o> b ) num__8000 miles . <o> c ) num__9000 miles . <o> d ) num__6000 miles . <o> e ) num__5000 miles . |
total distance traveled by all the tires is num__4 * num__10000 = num__40000 miles . ( since only num__4 tires can be active at a time . to make an equal run distance traveled by each tire is num__8000.0 = num__8000 miles . answer : b <eor> b <eos> |
b |
multiply__10000.0__4.0__ divide__40000.0__5.0__ round__8000.0__ |
multiply__10000.0__4.0__ divide__40000.0__5.0__ round__8000.0__ |
| a train covers a distance at a speed of num__240 kmph in num__5 hours . to cover the same distance in num__1.66666666667 hours it must travel at a speed of ? <o> a ) num__516 <o> b ) num__840 <o> c ) num__560 <o> d ) num__720 <o> e ) num__920 |
distance = num__240 * num__5 = num__1200 km required speed = num__1200 * num__0.6 = num__720 km / hr answer is d <eor> d <eos> |
d |
multiply__240.0__5.0__ km_to_mile_conversion__ multiply__1200.0__0.6__ round__720.0__ |
multiply__240.0__5.0__ km_to_mile_conversion__ multiply__1200.0__0.6__ multiply__1200.0__0.6__ |
| two pipes p and q can fill a tank in num__4 hours and num__5 hours respectively . if both pipes are opened simultaneously how much time will be taken to fill the tank ? <o> a ) num__4 hours num__20 min <o> b ) num__5 hours num__49 min <o> c ) num__3 hours num__50 min <o> d ) num__3 hours num__22 min <o> e ) num__2 hours num__13 min |
explanation : part filled by p in num__1 hour = num__0.25 part filled by q in num__1 hour = num__0.2 part filled by ( p + q ) in num__1 hour = ( num__0.25 + num__0.2 ) = ( num__0.45 ) time taken to fill the tank is ( num__2.22222222222 ) = num__2.22222222222 * num__60 = num__133 mins = num__2 hrs num__13 mins answer e <eor> e <eos> |
e |
subtract__5.0__4.0__ divide__1.0__4.0__ divide__1.0__5.0__ add__0.25__0.2__ divide__1.0__0.45__ hour_to_min_conversion__ round__2.0__ |
subtract__5.0__4.0__ divide__1.0__4.0__ divide__1.0__5.0__ add__0.25__0.2__ divide__1.0__0.45__ hour_to_min_conversion__ round__2.0__ |
| a man can row num__6 kmph in still water . when the river is running at num__1.2 kmph it takes him num__1 hour to row to a place and black . what is the total distance traveled by the man ? <o> a ) num__5.74 km <o> b ) num__5.78 km <o> c ) num__5.76 km <o> d ) num__5.79 km <o> e ) num__5.29 km |
m = num__6 s = num__1.2 ds = num__7.2 us = num__4.8 x / num__7.2 + x / num__4.8 = num__1 x = num__2.88 d = num__2.88 * num__2 = num__5.76 answer : c <eor> c <eos> |
c |
add__6.0__1.2__ subtract__6.0__1.2__ multiply__1.2__4.8__ round__5.76__ |
add__6.0__1.2__ subtract__6.0__1.2__ multiply__1.2__4.8__ multiply__1.2__4.8__ |
| what is the cp of rs num__100 stock at num__5 discount with num__0.2 % brokerage ? <o> a ) num__99.6 <o> b ) num__95.2 <o> c ) num__97.5 <o> d ) num__98.25 <o> e ) none of these |
explanation : use the formula cp = num__100 â € “ discount + brokerage % cp = num__100 - num__5 + num__0.2 num__95.2 thus the cp is rs num__95.2 . answer b <eor> b <eos> |
b |
percent__100.0__95.2__ |
percent__100.0__95.2__ |
| a train num__125 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is ? <o> a ) num__16 km / hr <o> b ) num__56 km / hr <o> c ) num__43 km / hr <o> d ) num__50 km / hr <o> e ) num__87 km / hr |
speed of the train relative to man = ( num__12.5 ) m / sec = ( num__12.5 ) m / sec . [ ( num__12.5 ) * ( num__3.6 ) ] km / hr = num__45 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__45 = = > x = num__50 km / hr . answer : d <eor> d <eos> |
d |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ round__50.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ multiply__5.0__10.0__ |
| if num__5 ^ k + num__5 ^ k = ( num__5 ^ num__9 ) ^ ( num__5 ^ num__9 ) - num__5 ^ k then k = ? <o> a ) num__5 ^ num__11 - num__1 <o> b ) num__5.5 <o> c ) num__242 <o> d ) num__3 ^ num__10 <o> e ) num__3 ^ num__11 - num__1 |
num__5 ^ k + num__5 ^ k = ( num__5 ^ num__9 ) ^ num__5 ^ num__9 - num__5 ^ k num__5 * ( num__5 ^ k ) = num__5 ^ ( num__25 * num__5 ^ num__9 ) = num__5 ^ ( num__5 ^ num__2 * num__5 ^ num__9 ) = num__5 ^ ( num__5 ^ num__11 ) num__5 ^ k + num__1 = num__5 ^ ( num__5 ^ num__11 ) so k + num__1 = num__5 ^ num__11 so k = num__5 ^ num__11 - num__1 answer is a <eor> a <eos> |
a |
add__9.0__2.0__ multiply__5.0__1.0__ |
add__9.0__2.0__ multiply__5.0__1.0__ |
| the cash realised on selling a num__14.0 stock is rs . num__107.25 brokerage being num__0.25 % is ? <o> a ) num__366 <o> b ) num__107 <o> c ) num__102 <o> d ) num__192 <o> e ) num__122 |
cash realised = rs . ( num__107.25 - num__0.25 ) = rs . num__107 . answer : b <eor> b <eos> |
b |
round_down__107.25__ round_down__107.25__ |
subtract__107.25__0.25__ subtract__107.25__0.25__ |
| an employer pays rs . num__20 for each day a worker works and for fests rs . num__3 for each day is ideal at the end of sixty days a worker gets rs . num__280 . for how many days did the worker remain ideal ? <o> a ) num__40 <o> b ) num__30 <o> c ) num__70 <o> d ) num__20 <o> e ) num__10 |
suppose a worker remained ideal for x days then he worked for num__60 - x days num__20 * ( num__60 - x ) - num__3 x = num__280 num__1200 - num__23 x = num__280 num__23 x = num__920 x = num__40 answer is a . <eor> a <eos> |
a |
hour_to_min_conversion__ multiply__20.0__60.0__ add__20.0__3.0__ subtract__1200.0__280.0__ divide__920.0__23.0__ round__40.0__ |
multiply__20.0__3.0__ multiply__20.0__60.0__ add__20.0__3.0__ subtract__1200.0__280.0__ subtract__60.0__20.0__ subtract__60.0__20.0__ |
| a car covers a distance of num__624 km in num__6 Â ½ hours . find its speed ? <o> a ) num__104 <o> b ) num__140 <o> c ) num__200 <o> d ) num__150 <o> e ) num__250 |
num__104.0 = num__104 kmph answer a <eor> a <eos> |
a |
divide__624.0__6.0__ round__104.0__ |
divide__624.0__6.0__ round__104.0__ |
| the ages of two persons differ by num__16 years . num__6 years ago the elder one was num__3 times as old as the younger one . what are their present ages of the elder person <o> a ) num__15 <o> b ) num__20 <o> c ) num__25 <o> d ) num__30 <o> e ) num__35 |
explanation : let ' s take the present age of the elder person = x and the present age of the younger person = x – num__16 ( x – num__6 ) = num__3 ( x - num__16 - num__6 ) = > x – num__6 = num__3 x – num__66 = > num__2 x = num__60 = > x = num__30.0 = num__30 answer : option d <eor> d <eos> |
d |
divide__6.0__3.0__ subtract__66.0__6.0__ divide__60.0__2.0__ divide__60.0__2.0__ |
divide__6.0__3.0__ subtract__66.0__6.0__ divide__60.0__2.0__ subtract__60.0__30.0__ |
| working alone at its constant rate machine k took num__3 hours to produce num__0.125 of the units produced last friday . then machine m started working and the two machines working simultaneously at their respective constant rates took num__6 hours to produce the rest of the units produced last friday . how many hours would it have taken machine m working alone at its constant rate to produce all of the units produced last friday ? <o> a ) num__9.6 <o> b ) num__12.4 <o> c ) num__16.2 <o> d ) num__18.3 <o> e ) num__21.5 |
machine k works at a rate of num__0.0416666666667 of the units per hour . the rate of k + m together is num__0.875 * num__0.166666666667 = num__0.145833333333 of the units per hour . the rate of machine m is num__0.145833333333 - num__0.0416666666667 = num__0.104166666667 . it would have taken machine m a total time of num__9.6 hours . the answer is a . <eor> a <eos> |
a |
divide__0.125__3.0__ add__0.125__0.0417__ divide__0.875__6.0__ round__9.6__ |
divide__0.125__3.0__ add__0.125__0.0417__ divide__0.875__6.0__ round__9.6__ |
| the number num__219 can be written as sum of the squares of num__3 different positive integers . what is the sum of these num__3 different integers ? <o> a ) num__20 <o> b ) num__21 <o> c ) num__19 <o> d ) num__17 <o> e ) num__15 |
sum of the squares of num__3 different positive integers = num__219 num__13 ^ num__2 + num__7 ^ num__2 + num__1 ^ num__2 = num__219 now sum of these num__3 different integers = num__13 + num__7 + num__1 = num__21 ans - b <eor> b <eos> |
b |
multiply__3.0__7.0__ multiply__3.0__7.0__ |
multiply__3.0__7.0__ power__21.0__1.0__ |
| a train num__100 m long is running at the speed of num__30 km / hr . find the time taken by it to pass a man standing near the railway line in seconds <o> a ) num__5 sec <o> b ) num__6 sec <o> c ) num__7 sec <o> d ) num__8 sec <o> e ) num__9 sec |
explanation : speed of the train = ( num__60 x num__0.277777777778 m / sec = num__16.6666666667 m / sec . distance moved in passing the standing man = num__100 m . required time taken = num__100 / ( num__16.6666666667 ) = ( num__100 × ( num__0.06 ) ) sec = num__6 sec answer : option b <eor> b <eos> |
b |
hour_to_min_conversion__ divide__100.0__16.6667__ round__6.0__ |
hour_to_min_conversion__ divide__100.0__16.6667__ divide__100.0__16.6667__ |
| a gardener grows cabbages in her garden that is in the shape of a square . each cabbage takes num__1 square feet of area in her garden . this year she has increased her output by num__181 cabbages as compared to last year . the shape of the area used for growing the cabbages has remained a square in both these years . how many cabbages did she produce this year ? <o> a ) num__6564 <o> b ) num__7372 <o> c ) num__8281 <o> d ) num__9331 <o> e ) can not be determined |
let the side for growing cabbages this year be x ft . thus the area is x ^ num__2 . let the side for growing cabbages last year be y ft . thus the area was y ^ num__2 . the area would have increased by num__181 sq ft as each cabbage takes num__1 sq ft space . x ^ num__2 - y ^ num__2 = num__181 ( x + y ) ( x - y ) = num__181 num__181 is a prime number and thus it will be ( num__91 + num__90 ) * ( num__91 - num__90 ) . thus x = num__91 and y = num__90 x ^ num__2 = num__91 ^ num__2 = num__8281 the answer is c . <eor> c <eos> |
c |
power__91.0__2.0__ multiply__1.0__8281.0__ |
power__91.0__2.0__ multiply__1.0__8281.0__ |
| given that num__2 x + num__15 > num__5 and num__5 x - num__13 < num__7 all values of x must be between which of the following pairs of integers ? <o> a ) - num__4 and - num__1 <o> b ) - num__5 and num__4 <o> c ) - num__4 and num__1 <o> d ) - num__2 and num__5 <o> e ) num__2 and num__5 |
num__2 x + num__15 > num__5 i . e num__2 x > num__5 - num__15 i . e . num__2 x > - num__10 i . e . x > - num__5 also num__5 x - num__13 < num__7 i . e . num__5 x < num__7 + num__13 i . e num__5 x < num__20 i . e x < num__4 i . e . - num__5 < x < num__4 answer : option b <eor> b <eos> |
b |
multiply__2.0__5.0__ multiply__2.0__10.0__ divide__20.0__5.0__ subtract__15.0__10.0__ |
subtract__15.0__5.0__ add__15.0__5.0__ divide__20.0__5.0__ subtract__15.0__10.0__ |
| difference between the length & breadth of a rectangle is num__15 m . if its perimeter is num__90 m then its area is ? ? <o> a ) num__2000 m ^ num__2 <o> b ) num__2340 m ^ num__2 <o> c ) num__450 m ^ num__2 <o> d ) num__2556 m ^ num__2 <o> e ) num__2534 m ^ num__2 |
we have : ( l - b ) = num__15 and num__2 ( l + b ) = num__90 or ( l + b ) = num__45 so num__2 l = num__60 = > l = num__30 & b = num__15 area = num__30 * num__15 = num__450 m ^ num__2 answer : c <eor> c <eos> |
c |
square_perimeter__15.0__ multiply__15.0__2.0__ triangle_area__15.0__60.0__ triangle_area__15.0__60.0__ |
square_perimeter__15.0__ multiply__15.0__2.0__ multiply__15.0__30.0__ multiply__15.0__30.0__ |
| if the average ( arithmetic mean ) of x x + num__2 and x + num__4 is num__93 what is the value of x ? <o> a ) num__95 <o> b ) num__91 <o> c ) num__102 <o> d ) num__101 <o> e ) num__85 |
am of x x + num__2 and x + num__4 = x + ( x + num__2 ) + ( x + num__4 ) / num__3 = num__3 x + num__2.0 = x + num__2 given that x + num__2 = num__93 x = num__91 answer : b <eor> b <eos> |
b |
subtract__93.0__2.0__ subtract__93.0__2.0__ |
subtract__93.0__2.0__ subtract__93.0__2.0__ |
| a rectangular courty num__3.78 metres long and num__5.25 metres wide is to be paved exactly with square tiles all of the same size . what is the largest size of the tile which could be used for the purpose ? <o> a ) num__14 cms <o> b ) num__21 cms <o> c ) num__42 cms <o> d ) none of these <o> e ) can not be determined |
solution largest size of the tile . h . c . f of num__378 cm and num__525 cm = num__21 cms . answer b <eor> b <eos> |
b |
square_perimeter__5.25__ square_perimeter__5.25__ |
square_perimeter__5.25__ square_perimeter__5.25__ |
| a man can row with a speed of num__20 kmph in still water . if the stream flows at num__5 kmph then the speed in downstream is ? <o> a ) num__22 <o> b ) num__28 <o> c ) num__25 <o> d ) num__82 <o> e ) num__34 |
m = num__20 s = num__5 ds = num__20 + num__5 = num__20 . answer : c <eor> c <eos> |
c |
add__20.0__5.0__ |
add__20.0__5.0__ |
| bruce and anne can clean their house in num__4 hours working together at their respective constant rates . if anne ’ s speed were doubled they could clean their house in num__3 hours working at their respective rates . how many q hours does it currently take anne to clean the house on her own ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__12 <o> e ) num__14 |
lets suppose anne and bruce take a and b hrs working separately so in num__1 hour they can together finish num__1 / a + num__1 / b portion of the work which equals num__0.25 ( as the work is completed in num__4 hours ) after anne doubles her rate of work the portion completed by the both is num__1 / a + num__2 / b which is equal to q = num__0.333333333333 ( as the work is completed in q = num__3 hours ) solving these num__2 equations we can find b as num__12 so d <eor> d <eos> |
d |
subtract__4.0__3.0__ divide__1.0__4.0__ subtract__3.0__1.0__ divide__1.0__3.0__ multiply__4.0__3.0__ round__12.0__ |
subtract__4.0__3.0__ divide__1.0__4.0__ subtract__3.0__1.0__ divide__1.0__3.0__ divide__3.0__0.25__ divide__3.0__0.25__ |
| an bus covers a certain distance at aspeed of num__100 kmph in num__5 hours . to cover the same distance in num__1 hr it must travel at a speed of ? <o> a ) num__560 km / h <o> b ) num__567 km / h <o> c ) num__300 km / h <o> d ) num__723 km / h <o> e ) num__720 km / h |
distance = ( num__100 x num__5 ) = num__500 km . speed = distance / time speed = num__500 / ( num__1.66666666667 ) km / hr . [ we can write num__1 hours as num__1.66666666667 hours ] required speed = num__500 x num__0.6 km / hr = num__300 km / hr . c <eor> c <eos> |
c |
multiply__100.0__5.0__ km_to_mile_conversion__ multiply__0.6__500.0__ round__300.0__ |
multiply__100.0__5.0__ divide__1.0__1.6667__ multiply__0.6__500.0__ divide__300.0__1.0__ |
| a table is bought for rs . num__1000 / - and sold at rs . num__1500 / - find gain or loss percentage <o> a ) num__25.0 gain <o> b ) num__35.0 gain <o> c ) num__45.0 loss <o> d ) num__50.0 gain <o> e ) none |
formula = ( selling price ~ cost price ) / cost price * num__100 = ( num__1500 - num__1000 ) / num__1000 = num__50.0 gain d <eor> d <eos> |
d |
percent__50.0__100.0__ |
percent__50.0__100.0__ |
| num__7 ^ num__6 n - num__1 ^ num__6 n when n is an integer > num__0 is divisible by <o> a ) num__112 <o> b ) num__127 <o> c ) num__145 <o> d ) num__187 <o> e ) num__185 |
num__127 b <eor> b <eos> |
b |
multiply__1.0__127.0__ |
multiply__1.0__127.0__ |
| the sum of three consecutive integers is num__93 . what are the integers ? <o> a ) num__20 num__2122 <o> b ) num__10 num__1112 <o> c ) num__30 num__3132 <o> d ) num__40 num__4142 <o> e ) num__30 num__31 |
32 |
first x make the first number x second x + num__1 to get the next numberwe go up one or + num__1 third x + num__2 add another num__1 ( num__2 total ) to get the third f + s + t = num__93 first ( f ) plus second ( s ) plusthird ( t ) equals num__93 ( x ) + ( x + num__1 ) + ( x + num__2 ) = num__93 replace f with x s with x + num__1 and t with x + num__2 x + x + num__1 + x + num__2 = num__93 here the parenthesis aren ′ t needed . num__3 x + num__3 = num__93 combine like terms x + x + x and num__2 + num__1 − num__3 − num__3 add num__3 to both sides num__3 x = num__90 the variable ismultiplied by num__3 num__3 num__3 divide both sides by num__3 x = num__30 our solution for x first num__30 replace x in our origional listwith num__30 second ( num__30 ) + num__1 = num__31 the numbers are num__30 num__31 and num__32 third ( num__30 ) + num__2 = num__32 correct answer e <eor> e <eos> |
e |
e |
| a rectangular plot measuring num__60 meters by num__50 meters is to be enclosed by wire fencing . if the poles of the fence are kept num__5 meters apart . how many poles will be needed ? <o> a ) num__44 m <o> b ) num__66 m <o> c ) num__26 m <o> d ) num__56 m <o> e ) num__25 m |
perimeter of the plot = num__2 ( num__60 + num__50 ) = num__220 m no of poles = num__44.0 = num__44 m answer : a <eor> a <eos> |
a |
divide__220.0__5.0__ round__44.0__ |
divide__220.0__5.0__ round__44.0__ |
| a train num__1600 m long can cross a pole in num__40 sec and then find the speed of the train ? <o> a ) num__140 <o> b ) num__142 <o> c ) num__144 <o> d ) num__146 <o> e ) num__148 |
length = speed * time speed = l / t s = num__40.0 s = num__40 m / sec speed = num__40 * num__3.6 ( to convert m / sec in to kmph multiply by num__3.6 ) speed = num__144 kmph answer : c <eor> c <eos> |
c |
multiply__40.0__3.6__ round__144.0__ |
multiply__40.0__3.6__ multiply__40.0__3.6__ |
| num__2 trains starting at the same time from num__2 stations num__200 km apart and going in opposite direction cross each other at a distance of num__110 km from one of the stations . what is the ratio of their speeds ? <o> a ) num__11 : num__9 <o> b ) num__11 : num__2 <o> c ) num__91 : num__9 <o> d ) num__11 : num__1 <o> e ) num__11 : num__5 |
in same time they cover num__110 km & num__90 km respectively so ratio of their speed = num__110 : num__90 = num__11 : num__9 answer : a <eor> a <eos> |
a |
subtract__200.0__110.0__ subtract__11.0__2.0__ round__11.0__ |
subtract__200.0__110.0__ subtract__11.0__2.0__ round__11.0__ |
| the perimeter of a triangle is num__28 cm and the inradius of the triangle is num__2.5 cm . what is the area of the triangle ? <o> a ) num__35 cm num__2 <o> b ) num__37 cm num__2 <o> c ) num__87 cm num__2 <o> d ) num__16 cm num__2 <o> e ) num__35 cm num__2 |
area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = num__2.5 * num__14.0 = num__35 cm num__2 answer : e <eor> e <eos> |
e |
triangle_area__28.0__2.5__ triangle_area__28.0__2.5__ |
multiply__2.5__14.0__ multiply__2.5__14.0__ |
| a train having a length of num__165 meter is running at a speed of num__60 kmph . in what time it will pass a man who is running at num__6 kmph in the direction opposite to that of the train <o> a ) num__9 sec <o> b ) num__8 sec <o> c ) num__6 sec <o> d ) num__4 sec <o> e ) num__2 sec |
explanation : distance = num__165 m relative speed = num__60 + num__6 = num__66 kmph ( since both the train and the man are in moving in opposite direction ) = num__66 × num__0.277777777778 mps = num__18.3333333333 mps time = distance / speed = num__165 / ( num__18.3333333333 ) = num__9 s answer : option a <eor> a <eos> |
a |
add__60.0__6.0__ divide__165.0__18.3333__ round__9.0__ |
add__60.0__6.0__ divide__165.0__18.3333__ divide__165.0__18.3333__ |
| a sum of rs . num__1360 has been divided among a b and c such that a gets of what b gets and b gets of what c gets . b ' s share is : <o> a ) num__240 <o> b ) num__120 <o> c ) num__190 <o> d ) num__200 <o> e ) num__250 |
let c ' s share = rs . x then b ' s share = rs . x / num__4 a ' s share = rs . ( num__0.666666666667 x ( x / num__4 ) ) = rs . x / num__6 therefore x / num__6 + x / num__4 + x = num__1360 = num__17 x / num__12 = num__1360 x = num__1360 x num__0.705882352941 = rs . num__960 hence b ' s share = rs . ( num__240.0 ) = rs . num__240 . answer is a . <eor> a <eos> |
a |
divide__12.0__17.0__ divide__960.0__4.0__ divide__960.0__4.0__ |
divide__12.0__17.0__ divide__960.0__4.0__ divide__960.0__4.0__ |
| a sum fetched a total simple interest of rs . num__929.20 at the rate of num__8 p . c . p . a . in num__5 years . what is the sum ? <o> a ) num__2323 <o> b ) num__1223 <o> c ) num__2563 <o> d ) num__1853 <o> e ) num__2353 |
explanation : si = rs . num__929.20 p = ? t = num__5 years r = num__8.0 p = num__100 × si / rt = num__100 × num__929.20 / num__8 × num__5 = rs . num__2323 answer : option a <eor> a <eos> |
a |
percent__100.0__2323.0__ |
percent__100.0__2323.0__ |
| joe is painting a rectangular room whose dimensions are given by a b and c meters . joe takes num__8 hours to paint a wall with dimensions a and c . he takes num__4 hours to paint a wall with dimensions b and c and num__12 hours to paint the ceiling with dimensions a and b . if joe works at a constant rate and a = num__12 then what is the volume of the room ? <o> a ) num__218 cubic meters <o> b ) num__224 cubic meters <o> c ) num__230 cubic meters <o> d ) num__288 cubic meters <o> e ) it can ’ t be determined . |
time to paint each wall as given in problem : ac = num__8 hours bc = num__4 hours ab = num__12 hours since he works at constant rate and it takes him twice as long to paint ac compared to bc ac = num__2 bc plug in num__6 for a and you find that b = num__6 since painting ab takes num__3 times as long compared to bc ab = num__3 bc plug in num__6 for a and num__3 for b and you find that c = num__4 a * b * c = num__12 * num__6 * num__4 = num__288 cubic meters answer : d <eor> d <eos> |
d |
divide__8.0__4.0__ subtract__8.0__2.0__ divide__12.0__4.0__ round__288.0__ |
divide__8.0__4.0__ subtract__8.0__2.0__ divide__12.0__4.0__ round__288.0__ |
| the time it took car p to travel num__150 miles was num__2 hours less than the time it took car r to travel the same distance . if car p ’ s average speed was num__10 miles per hour greater than that of car r what was car r ’ s average speed in miles per hour ? <o> a ) num__23 <o> b ) num__50 <o> c ) num__60 <o> d ) num__70 <o> e ) num__80 |
let speed of car r be = x then speed of car p = x + num__10 a / q ( num__150 / x ) - ( num__150 / ( x + num__10 ) ) = num__2 solving for x = num__23 miles \ hr . a <eor> a <eos> |
a |
round__23.0__ |
round__23.0__ |
| what sum of money will produce rs . num__70 as simple interest in num__3 years at num__3 num__0.5 percent ? <o> a ) num__667 <o> b ) num__500 <o> c ) num__266 <o> d ) num__288 <o> e ) num__211 |
num__70 = ( p * num__3 * num__3.5 ) / num__100 p = num__667 answer : a <eor> a <eos> |
a |
percent__100.0__667.0__ |
percent__100.0__667.0__ |
| for what values of k will the pair of equations ( num__3 x + num__4 y ) / num__2 = num__6 and kx + num__12 y = num__30 does not have a unique solution ? <o> a ) num__12 <o> b ) num__9 <o> c ) num__3 <o> d ) num__7.5 <o> e ) num__2.5 |
we have num__2 equations num__1 . ( num__3 x + num__4 y ) / num__2 = num__6 - - > num__3 x + num__4 y = num__12 - - > num__9 x + num__12 y = num__36 num__2 . kx + num__12 y = num__30 substract num__1 - num__2 we get ( num__9 - k ) x = num__6 i . e . x = num__6 / ( num__9 - k ) then by looking at options we get some value of x except for b . when we put k = num__9 x becomes num__6 / num__0 and hence answer is b <eor> b <eos> |
b |
subtract__3.0__2.0__ add__3.0__6.0__ multiply__3.0__12.0__ add__3.0__6.0__ |
subtract__3.0__2.0__ add__3.0__6.0__ add__6.0__30.0__ add__3.0__6.0__ |
| a person borrows rs . num__5000 for num__2 years at num__4.0 p . a . simple interest . he immediately lends it to another person at num__6 p . a for num__2 years . find his gain in the transaction per year . <o> a ) rs . num__112.50 <o> b ) rs . num__125 <o> c ) rs . num__150 <o> d ) rs . num__167.50 <o> e ) rs . num__175 |
gain in num__2 years = rs . [ ( num__5000 x ( num__6.25 ) x ( num__0.02 ) ) - ( ( num__5000 x num__4 x num__2 ) / num__100 ) ] = rs . ( num__625 - num__400 ) = rs . num__225 . gain in num__1 year = rs . ( num__112.5 ) = rs . num__112.50 answer : a <eor> a <eos> |
a |
percent__2.0__5000.0__ percent__0.02__5000.0__ percent__100.0__112.5__ |
percent__2.0__5000.0__ percent__0.02__5000.0__ percent__100.0__112.5__ |
| after successive discounts of num__20.0 num__10.0 and num__5.0 a certain good is sold for rs . num__6600 . find the actual price of the good . <o> a ) s . num__6000 <o> b ) s . num__9000 <o> c ) s . num__10800 <o> d ) s . num__9649 <o> e ) s . num__9980 |
let actual price was num__100 . after three successive discount this will become num__100 = = num__20.0 discount = > num__80 = = num__10.0 discount = > num__72 = = num__5.0 discount = num__68.4 now compare num__68.4 = num__6600 num__1 = num__6600 / num__68.4 num__100 = ( num__6600 * num__100 ) / num__68.4 = rs . num__9649 . answer : option d <eor> d <eos> |
d |
percent__20.0__5.0__ percent__100.0__9649.0__ |
percent__20.0__5.0__ percent__100.0__9649.0__ |
| pipes a and b can fill a cistern in num__8 and num__24 minutes respectively . they are opened an alternate minutes . find how many minutes the cistern shall be full ? <o> a ) num__22 <o> b ) num__12 <o> c ) num__88 <o> d ) num__99 <o> e ) num__27 |
num__0.125 + num__0.0416666666667 = num__0.166666666667 num__6 * num__2 = num__12 answer : b <eor> b <eos> |
b |
add__0.125__0.0417__ subtract__8.0__6.0__ divide__24.0__2.0__ round__12.0__ |
add__0.125__0.0417__ subtract__8.0__6.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
| two cars car num__1 and car num__2 move towards each other from r and y respectively with respective speeds of num__20 m / s and num__15 m / s . after meeting each other car num__1 reaches y in num__10 seconds . in how many seconds does car num__2 reach r starting from y ? <o> a ) num__15.5 sec <o> b ) num__8.4 sec <o> c ) num__33.6 sec <o> d ) num__31.11 sec <o> e ) num__16.8 sec |
r - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - | - - - - - - - - - - - - - - - - - - - - - - - - - - - - y car a ( num__20 mps ) - - - - - - - - - - - - - - - - - - - - - - - - - > p < - - - - - - - - - - - - - - - car b ( num__15 mps ) let num__2 cars meet each other at point p in t seconds . car num__1 covers distance = num__20 t . car num__2 covers distance = num__15 t . so total distance ry = num__35 t . from p car num__1 reaches onto y in num__10 secs . so it covers num__15 t further . so num__15 t / num__20 = num__10 so t = num__13.3333333333 sec and total distance = ( num__35 * num__40 ) / num__3 hence car num__2 will cover total distance in ( num__35 * num__40 ) / ( num__3 * num__15 ) = num__31.11 sec approx . answer d <eor> d <eos> |
d |
add__20.0__15.0__ multiply__2.0__20.0__ add__1.0__2.0__ round__31.11__ |
add__20.0__15.0__ multiply__2.0__20.0__ divide__40.0__13.3333__ multiply__1.0__31.11__ |
| an aeroplane covers a certain distance at a speed of num__100 kmph in num__2 hours . to cover the same distance in num__1 hours it must travel at a speed of ? <o> a ) num__100 km / hr <o> b ) num__150 km / hr <o> c ) num__200 km / hr <o> d ) num__250 km / hr <o> e ) num__300 km / hr |
distance = num__100 * num__2 = num__200 km required speed = num__200 * num__1 = num__200 km / hr answer is c <eor> c <eos> |
c |
multiply__100.0__2.0__ round__200.0__ |
multiply__100.0__2.0__ multiply__100.0__2.0__ |
| if the wheel is num__14 cm then the number of revolutions to cover a distance of num__1056 cm is ? <o> a ) num__22 <o> b ) num__29 <o> c ) num__23 <o> d ) num__12 <o> e ) num__13 |
num__2 * num__3.14285714286 * num__14 * x = num__1056 = > x = num__12 answer : d <eor> d <eos> |
d |
subtract__14.0__2.0__ round__12.0__ |
subtract__14.0__2.0__ round__12.0__ |
| a train traveling at num__100 kmph overtakes a motorbike traveling at num__64 kmph in num__12 seconds . what is the length of the train in meters ? <o> a ) num__400 meters <o> b ) num__120 meters <o> c ) num__1777 meters <o> d ) num__60 meters <o> e ) none of these |
train overtakes a bike means that we are talking about total length of the train . ( train ' s head is close to bike when it started and its tail crosses the bike when it overtakes the bike ) relative speed = num__100 - num__64 = num__36 km / h = num__36000 m / h time = num__12 seconds distance = speed * time num__36000 * num__0.00333333333333 = num__120 meters . b is the answer . <eor> b <eos> |
b |
subtract__100.0__64.0__ round__120.0__ |
subtract__100.0__64.0__ round__120.0__ |
| a can have a piece of work done in num__4 days b can work two times faster than the a c can work three times faster than a . how many days will they take to do the work together ? <o> a ) num__0.75 days <o> b ) num__0.857142857143 days <o> c ) num__0.25 days <o> d ) ca n ' t say <o> e ) none |
a ' s num__1 day work is num__0.166666666667 b ' s num__1 day work is num__0.333333333333 c ' s num__1 day work is num__0.5 ( a + b + c ) ' s num__1 day work ( num__0.166666666667 + num__0.333333333333 + num__0.5 ) is num__1.16666666667 so they can finish the work in num__0.857142857143 days answer : b <eor> b <eos> |
b |
add__0.1667__0.3333__ add__1.0__0.1667__ divide__1.0__1.1667__ multiply__1.0__0.8571__ |
add__0.1667__0.3333__ add__1.0__0.1667__ divide__1.0__1.1667__ multiply__1.0__0.8571__ |
| barbata invests $ num__2600 in the national bank at num__5.0 . how much additional money must she invest at num__8.0 so that the total annual income will be equal to num__6.0 of her entire investment ? <o> a ) num__1200 <o> b ) num__3000 <o> c ) num__1300 <o> d ) num__3600 <o> e ) num__2400 |
let the additional invested amount for num__8.0 interest be x ; equation will be ; num__2600 + num__0.05 * num__2600 + x + num__0.08 x = num__2600 + x + num__0.06 ( num__2600 + x ) num__0.05 * num__2600 + num__0.08 x = num__0.06 x + num__0.06 * num__2600 num__0.02 x = num__2600 ( num__0.06 - num__0.05 ) x = num__2600 * num__0.01 / num__0.02 = num__1300 ans : ` ` c ' ' <eor> c <eos> |
c |
subtract__0.08__0.06__ divide__0.05__5.0__ subtract__2600.0__1300.0__ |
subtract__0.08__0.06__ divide__0.05__5.0__ subtract__2600.0__1300.0__ |
| if bill can buy num__3 pairs of jeans and num__2 shirts for $ num__69 or num__2 pairs of jeans and num__3 shirts for $ num__76 how much does one shirt cost ? <o> a ) $ num__10 <o> b ) $ num__12 <o> c ) $ num__13.20 <o> d ) $ num__15 <o> e ) $ num__18 |
num__3 j + num__2 s = num__69 num__2 j + num__3 s = num__76 - - - - - - - - - - - - - - - - num__5 j + num__5 s = num__145 - - - - ( divide by num__5 ) - - - > j + s = num__29 num__3 j + num__2 s = j + num__2 ( j + s ) = j + num__58 = num__69 - - - > j = num__11 num__3 * num__11 + num__2 s = num__69 num__33 + num__2 s = num__69 num__2 s = num__36 s = num__18 answer : e <eor> e <eos> |
e |
add__3.0__2.0__ add__69.0__76.0__ divide__145.0__5.0__ multiply__2.0__29.0__ subtract__69.0__58.0__ multiply__3.0__11.0__ add__3.0__33.0__ subtract__76.0__58.0__ subtract__76.0__58.0__ |
add__3.0__2.0__ add__69.0__76.0__ divide__145.0__5.0__ multiply__2.0__29.0__ subtract__69.0__58.0__ multiply__3.0__11.0__ add__3.0__33.0__ subtract__76.0__58.0__ subtract__76.0__58.0__ |
| in a certain company the ratio of male to female employees is num__7 : num__8 . if num__3 more men were hired this ratio would increase to num__8 : num__9 . how many male employees are there in the company ? <o> a ) num__6 <o> b ) num__21 <o> c ) num__27 <o> d ) num__189 <o> e ) num__192 |
another approach is to use two variables . let m = present number of males let f = present number of females the ratio of male to female employees is num__7 : num__8 so m / f = num__0.875 cross multiply to get num__7 f = num__8 m if num__3 more men were hired this ratio would increase to num__8 : num__9 so ( m + num__3 ) / f = num__0.888888888889 cross multiply to get num__9 ( m + num__3 ) = num__8 f expand : num__9 m + num__27 = num__8 f we now have a system of two equations and two variables : num__7 f = num__8 m num__9 m + num__27 = num__8 f solve to get : m = num__189 and f = num__216 answer : d <eor> d <eos> |
d |
divide__7.0__8.0__ divide__8.0__9.0__ multiply__3.0__9.0__ multiply__7.0__27.0__ multiply__8.0__27.0__ multiply__7.0__27.0__ |
divide__7.0__8.0__ divide__8.0__9.0__ multiply__3.0__9.0__ multiply__7.0__27.0__ add__27.0__189.0__ multiply__7.0__27.0__ |
| for any number s s * is defined as the greatest positive even integer less than or equal to s . what is the value of num__5.2 – num__5.2 * ? <o> a ) num__0.2 <o> b ) num__1.2 <o> c ) num__1.8 <o> d ) num__2.2 <o> e ) num__4.0 |
since s * is defined as the greatest positive even integer less than or equal to s then num__5.2 * = num__4 ( the greatest positive even integer less than or equal to num__5.2 is num__4 ) . hence num__5.2 – num__5.2 * = num__5.2 - num__4 = num__1.2 answer : b . <eor> b <eos> |
b |
subtract__5.2__4.0__ subtract__5.2__4.0__ |
subtract__5.2__4.0__ subtract__5.2__4.0__ |
| if the volume of two cubes are in the ratio num__8 : num__1 the ratio of their edges is : <o> a ) num__2 : num__1 <o> b ) num__3 : num__2 <o> c ) num__3 : num__5 <o> d ) num__3 : num__7 <o> e ) none of these |
explanation : let the edges be a and b of two cubes then a num__3 / b num__3 = num__8.0 = > ( a / b ) num__3 = ( num__2.0 ) num__3 a / b = num__2.0 = > a : b = num__2 : num__1 option a <eor> a <eos> |
a |
multiply__1.0__2.0__ |
multiply__1.0__2.0__ |
| solve the quickfire maths brain teaser â ˆ š num__25.0 = ? <o> a ) num__30.0 <o> b ) num__40.0 <o> c ) num__50.0 <o> d ) num__19.0 <o> e ) num__29 % |
â ˆ š num__25.0 = > â ˆ š num__25 / â ˆ š num__100 = > num__0.5 = > num__0.5 = > num__50.0 c <eor> c <eos> |
c |
percent__50.0__100.0__ |
percent__50.0__100.0__ |
| num__3 candidates in an election and received num__1000 num__2000 and num__4000 votes respectively . what % of the total votes did the winning candidate got in that election ? <o> a ) num__45.0 <o> b ) num__50.0 <o> c ) num__57.14 <o> d ) num__60.0 <o> e ) num__65.12 % |
total number of votes polled = ( num__1000 + num__2000 + num__4000 ) = num__7000 so required percentage = num__0.571428571429 * num__100 = num__57.14 c <eor> c <eos> |
c |
percent__100.0__57.14__ |
percent__100.0__57.14__ |
| num__1 / ( num__3 - num__2 √ num__2 ) = ? <o> a ) num__9 + num__4 √ num__5 <o> b ) num__9 - num__4 √ num__5 <o> c ) num__3 + num__2 √ num__2 <o> d ) num__9 - num__2 √ num__5 <o> e ) num__4 + num__9 √ num__5 |
this question requires us to rationalize ( fix ) the denominator . for more on this technique seehttps : / / www . gmatprepnow . com / module / gmat . . . video / num__1044 given : num__1 / ( num__3 - num__2 √ num__2 ) multiply top and bottom by the conjugate of num__3 - num__2 √ num__2 which is num__3 + num__2 √ num__2 so num__1 / ( num__3 - num__2 √ num__2 ) = ( num__1 ) ( num__3 + num__2 √ num__2 ) / ( num__3 - num__2 √ num__2 ) ( num__3 + num__2 √ num__2 ) = ( num__3 + num__2 √ num__2 ) / ( num__1 ) = num__3 + num__2 √ num__2 = c <eor> c <eos> |
c |
multiply__1.0__3.0__ |
add__1.0__2.0__ |
| i flew my tiny seaplane to visit my mother . on the flight up i flew at num__110 mph . on the way home i flew num__72 mph . what was my average speed for the trip ? <o> a ) num__198 mph <o> b ) num__91 mph <o> c ) num__88 mph <o> d ) num__100 mph <o> e ) num__99 mph |
( num__110 mph + num__72 mph ) / num__2 = num__91 mph correct answer is : b <eor> b <eos> |
b |
round__91.0__ |
round__91.0__ |
| a retailer buys a radio for rs num__225 . his overhead expenses are rs num__20 . he sellis the radio for rs num__300 . the profit percent of the retailer is <o> a ) num__22.4 <o> b ) num__50.0 <o> c ) num__25.0 <o> d ) num__52.0 <o> e ) none of these |
explanation : cost price = ( num__225 + num__20 ) = num__245 sell price = num__300 gain = ( num__0.224489795918 ) * num__100 = num__22.4 . answer : a <eor> a <eos> |
a |
percent__100.0__22.4__ |
percent__100.0__22.4__ |
| martins made a part payment of $ num__650 toward buying a brand new washing machine which represents num__15.0 of the total cost of the washing machine how much remains to be paid ? <o> a ) $ num__1116.3 <o> b ) $ num__3583.3 <o> c ) $ num__3283.3 <o> d ) $ num__3683.3 <o> e ) $ num__2683.3 |
explanation : let ' s start with what the total price of the washing machine would be . if num__15.0 is equal to $ num__650 then num__100.0 equals $ x . we just have to multiply $ num__650 by num__6.67 to get total amount = $ num__4333.3 . out of this amount we then need to deduct the amount already paid which was $ num__650 so we have $ num__4333.3 - $ num__650 = $ num__3683.3 answer : option d <eor> d <eos> |
d |
subtract__4333.3__650.0__ subtract__4333.3__650.0__ |
subtract__4333.3__650.0__ subtract__4333.3__650.0__ |
| a teacher gave the same test to three history classes : d b and c . the average ( arithmetic mean ) scores for the three classes were num__65 num__80 and num__77 respectively . the ratio of the numbers of students in each class who took the test was num__4 to num__6 to num__5 respectively . what was the average score for the three classes combined ? <o> a ) num__74 <o> b ) num__75 <o> c ) num__76 <o> d ) num__77 <o> e ) num__78 |
ans : b ( num__75 ) lets say class d indeed has num__4 children b has num__6 children and c has num__5 children . now if the average of class d is num__65 hence total marks awarded in the class = num__65 * num__4 = num__260 similarly class b = num__80 * num__6 = num__480 class c = num__77 * num__5 = num__385 total marks provided = d + b + c = num__260 + num__480 + num__385 = num__1125 avg . marks = num__75.0 ( total no . of students ) = num__75 = b <eor> b <eos> |
b |
subtract__80.0__5.0__ multiply__65.0__4.0__ multiply__80.0__6.0__ multiply__77.0__5.0__ subtract__80.0__5.0__ |
subtract__80.0__5.0__ multiply__65.0__4.0__ multiply__80.0__6.0__ multiply__77.0__5.0__ subtract__80.0__5.0__ |
| ram and shakil run a race of num__2000 m . first ram gives shakil a start of num__200 m and beats him by num__1 minute . next ram gives shakil a start of num__6 min and is beaten by num__1000 meters . find the time in minutes in which ram and shakil can run the race seperately . <o> a ) num__12 num__18 <o> b ) num__8 num__10 <o> c ) num__18 num__20 <o> d ) num__20 num__25 <o> e ) none of these |
explanation : let speed of ram and shakil be x and y resp . case num__1 : ram gives a start of num__200 m and beats shakil by num__1 minute . num__2000 / x - num__1800 / y = - num__60 num__2000 / x = num__1800 / y - num__60 - - - - - - - - - - - - - - - - - - - - - ( num__1 ) case num__2 : ram gives shakil a start of num__6 min and is beaten by num__1000 meters distance travelled by shakil in num__6 minutes = ( num__6 * num__60 * y ) meters = num__360 y meters num__1000 / x = ( num__2000 - num__360 y ) / y - - - - - - - - - - - - - - - - - - ( num__2 ) solve equation num__1 and num__2 to obtain x and y answer is x = num__8 minutes y = num__10 minutes hence ( b ) is the corret answer . answer : b <eor> b <eos> |
b |
subtract__2000.0__200.0__ hour_to_min_conversion__ divide__2000.0__1000.0__ multiply__6.0__60.0__ add__6.0__2.0__ divide__2000.0__200.0__ round__8.0__ |
subtract__2000.0__200.0__ hour_to_min_conversion__ divide__2000.0__1000.0__ multiply__6.0__60.0__ add__6.0__2.0__ divide__2000.0__200.0__ multiply__1.0__8.0__ |
| the average temperature of monday tuesday wednesday and thursday was num__38 ∘ and that of tuesday wednesday thursday and friday was num__40 ∘ . if the temperature on monday was num__36 ∘ the temperature of friday was : <o> a ) num__44 ∘ <o> b ) num__40 ∘ <o> c ) num__38 ∘ <o> d ) num__30 ∘ <o> e ) none of these |
explanation : m + t + w + th = ( num__4 × num__38 ) = num__152 monday temperature is num__36 . so t + w + th = ( num__152 - num__36 ) = num__116 t + w + th + f = ( num__4 × num__40 ) = num__160 f = ( num__160 - num__116 ) = num__44 ∘ correct option : a <eor> a <eos> |
a |
subtract__40.0__36.0__ multiply__38.0__4.0__ subtract__152.0__36.0__ multiply__40.0__4.0__ add__40.0__4.0__ add__40.0__4.0__ |
subtract__40.0__36.0__ multiply__38.0__4.0__ subtract__152.0__36.0__ multiply__40.0__4.0__ add__40.0__4.0__ add__40.0__4.0__ |
| if num__25.0 of x is num__15 less than num__15.0 of num__1600 then x is ? <o> a ) num__872 <o> b ) num__738 <o> c ) num__900 <o> d ) num__840 <o> e ) num__83 |
num__25.0 of x = x / num__4 ; num__15.0 of num__1600 = num__0.15 * num__1600 = num__240 given that x / num__4 = num__240 - num__15 = > x / num__4 = num__225 = > x = num__900 . answer : c <eor> c <eos> |
c |
multiply__1600.0__0.15__ subtract__240.0__15.0__ multiply__225.0__4.0__ multiply__225.0__4.0__ |
multiply__1600.0__0.15__ subtract__240.0__15.0__ multiply__225.0__4.0__ multiply__225.0__4.0__ |
| three models ( k l and m ) of cars are distributed among three showrooms . the number of cars in each showrooms must be equal and each model must be represented by at least one car in every showroom . there are num__19 cars of model k num__17 cars of model l and num__15 cars of model m . what is the maximum number of cars of model k in any showroom ? <o> a ) num__17 <o> b ) num__16 <o> c ) num__15 <o> d ) num__14 <o> e ) num__13 |
the total number of cars is num__51 . so each showroom has num__17 cars ( since the number of cars in each showrooms should be equal num__17.0 = num__17 ) . moreover that the number of model k is maximum means that the numbers of model l and m should be minimum . since each model must be represented by at least one car in every showroom that minimum number should be num__1 . so maximum number of model k is num__17 - num__2 = num__15 . the answer is ( c ) <eor> c <eos> |
c |
subtract__19.0__17.0__ subtract__17.0__2.0__ |
subtract__19.0__17.0__ subtract__17.0__2.0__ |
| the circulation for magazine p in num__1961 was num__4 times the average ( arithmetic mean ) yearly circulation for magazine p for the years num__1962 - num__1970 . what is the ratio of the circulation in num__1961 to the total circulation during num__1961 - num__1970 for magazine p ? <o> a ) num__0.285714285714 <o> b ) num__0.714285714286 <o> c ) num__0.428571428571 <o> d ) num__0.571428571429 <o> e ) num__0.142857142857 |
there are num__9 years from num__1962 - num__1970 inclusive . let ' s say the average circulation every year between num__1962 - num__1970 inclusive is x . so the total circulation is num__9 x from num__1962 - num__1970 inclusive . in num__1961 the circulation is num__4 x . so total circulation for num__1961 - num__1970 is num__4 x + num__9 x = num__13 x . ratio of circulation in num__1961 to total circulation during num__1961 - num__1970 is num__4 x to num__14 x or num__0.285714285714 = num__0.285714285714 answer : a <eor> a <eos> |
a |
subtract__1970.0__1961.0__ add__4.0__9.0__ divide__4.0__14.0__ divide__4.0__14.0__ |
subtract__1970.0__1961.0__ add__4.0__9.0__ divide__4.0__14.0__ divide__4.0__14.0__ |
| find the area of a rhombus having each side equal to num__13 cm and one of whose diagonal is num__24 cm . <o> a ) num__150 sq . cm <o> b ) num__120 sq . cm <o> c ) num__130 sq . cm <o> d ) num__125 sq . cm <o> e ) num__110 sq . cm |
let abcd is a rhombus with diagonals ac and bd which intersect each other at o . ac = num__24 ⇒ ao = num__12 let bo = x and ab = num__13 cm ( given ) by pythagorean theorem c num__2 = a num__2 + b num__2 num__13 num__2 = num__12 num__2 + x num__2 num__169 = num__144 + x num__2 x num__2 = num__169 – num__144 x num__2 = num__25 x = num__5 cm bo = num__5 cm diagonal bd = num__2 x num__5 = num__10 cm . area = ½ x [ product of diagonals ] = ½ x num__24 x num__10 area = num__120 sq . cm answer is b <eor> b <eos> |
b |
power__13.0__2.0__ triangle_area__24.0__12.0__ side_by_diagonal__13.0__12.0__ multiply__2.0__5.0__ multiply__24.0__5.0__ multiply__24.0__5.0__ |
power__13.0__2.0__ triangle_area__24.0__12.0__ side_by_diagonal__13.0__12.0__ multiply__2.0__5.0__ multiply__24.0__5.0__ multiply__24.0__5.0__ |
| in the xy - coordinate system line k passes through points ( - num__7 m num__0 ) and ( num__0 num__3 m ) . which of the following is a possible equation of line k ? <o> a ) y = - num__2.33333333333 * x + num__3 m <o> b ) y = num__0.428571428571 * x - num__7 m <o> c ) y = num__2.33333333333 * x + num__3 m <o> d ) y = num__0.428571428571 * x + num__3 m <o> e ) y = - num__0.428571428571 * x - num__7 m |
the slope of the line is num__3 m / num__7 m = num__0.428571428571 . the y - intercept is num__3 m . the equation of the line is : y = num__0.428571428571 * x + num__3 m the answer is d . <eor> d <eos> |
d |
divide__3.0__7.0__ divide__3.0__7.0__ |
divide__3.0__7.0__ divide__3.0__7.0__ |
| ayesha ' s father was num__34 years of age when she was born while her mother was num__30 years old when her brother four years younger to her was born . what is the difference between the ages of her parents ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__5 <o> d ) num__1 <o> e ) num__2 |
mother ' s age when ayesha ' s brother was born = num__30 years . father ' s age when ayesha ' s brother was born = ( num__34 + num__4 ) years = num__38 years . required difference = ( num__38 - num__30 ) years = num__8 years . answer : b <eor> b <eos> |
b |
subtract__34.0__30.0__ add__34.0__4.0__ subtract__38.0__30.0__ subtract__38.0__30.0__ |
subtract__34.0__30.0__ add__34.0__4.0__ subtract__38.0__30.0__ subtract__38.0__30.0__ |
| a coin is flipped twice . what is the probability that both flips result in heads given that first flip does ? <o> a ) num__0.5 <o> b ) num__2 <o> c ) num__1 <o> d ) num__1.5 <o> e ) num__0.333333333333 |
a = first flip lands heads . b = second flips is head . p ( b | a ) = n ( a intersection b ) / n ( a ) = p ( h h ) / p [ ( h h ) ( h t ) ] = num__0.5 answer is option a <eor> a <eos> |
a |
negate_prob__0.5__ |
negate_prob__0.5__ |
| a shopkeeper sold an book offering a discount of num__5.0 and earned a profit of num__33.0 . what would have been the percentage of profit earned if no discount was offered ? <o> a ) num__140 <o> b ) num__120 <o> c ) num__130 <o> d ) num__110 <o> e ) num__150 |
let c . p . be $ num__100 . then s . p . = $ num__133 let marked price be $ x . then num__0.95 x = num__133 x = num__140.0 = $ num__140 now s . p . = $ num__140 c . p . = $ num__100 profit % = num__40.0 . a <eor> a <eos> |
a |
percent__100.0__140.0__ |
percent__100.0__140.0__ |
| a num__300 m long train crosses a platform in num__39 sec while it crosses a signal pole in num__14 sec . what is the length of the platform ? <o> a ) num__287 m <o> b ) num__350 m <o> c ) num__267 m <o> d ) num__535.71 m <o> e ) num__656 m |
speed = num__21.4285714286 = num__21.4285714286 m / sec . let the length of the platform be x meters . then ( x + num__300 ) / num__39 = num__21.4285714286 = > x = num__535.71 m . answer : d <eor> d <eos> |
d |
divide__300.0__14.0__ round__535.71__ |
divide__300.0__14.0__ round__535.71__ |
| circles x and y are concentric . if the radius of circle x is six times that of circle y what is the probability that a point selected inside circle x at random will be outside circle y ? <o> a ) num__13 <o> b ) pi / num__3 <o> c ) pi / num__2 <o> d ) num__0.972222222222 <o> e ) num__0.888888888889 |
small circle radius = a so area = pi a ^ num__2 large circle radius = num__6 a so area = num__9 . pi a ^ num__2 area of shaded region = num__36 pi a ^ num__2 - pi a ^ num__2 = num__35 pi a ^ num__2 so probability = ( num__35 pi a ^ num__2 ) / ( num__36 pi a ^ num__2 ) = num__0.972222222222 = answer = d <eor> d <eos> |
d |
square_perimeter__9.0__ triangle_area__2.0__0.9722__ |
power__6.0__2.0__ triangle_area__2.0__0.9722__ |
| how many kg of rice at rs . num__6.60 per kg . be mixed with num__56 kg of rice at rs . num__9.60 per kg to get a mixture worth rs . num__8.20 per kg <o> a ) num__56 kg <o> b ) num__52 kg <o> c ) num__44 kg <o> d ) num__49 kg <o> e ) none of these |
explanation : by the rule of alligation we have cost of num__1 kg of num__1 st kind rice cost of num__1 kg of num__2 nd kind rice num__6.6 num__9.6 price of num__1 kg of the mixture num__8.2 num__9.6 - num__8.2 = num__1.4 num__8.2 - num__6.6 = num__1.6 quantity of num__1 st kind rice : quantity of num__2 nd kind rice = num__1.4 : num__1.6 = num__7 : num__8 quantity of num__1 st kind rice : num__56 = num__7 : num__8 = > quantity of num__1 st kind rice = num__56 × num__78 = num__49 answer : option d <eor> d <eos> |
d |
subtract__9.6__8.2__ subtract__8.2__6.6__ round_down__8.2__ subtract__56.0__7.0__ subtract__56.0__7.0__ |
subtract__9.6__8.2__ subtract__8.2__6.6__ subtract__9.6__1.6__ subtract__56.0__7.0__ subtract__56.0__7.0__ |
| out of first num__20 natural numbers one number is selected at random . the probability that it is either an even number or a prime number is ? <o> a ) num__0.772727272727 <o> b ) num__0.586206896552 <o> c ) num__0.173469387755 <o> d ) num__0.85 <o> e ) num__0.62962962963 |
n ( s ) = num__20 n ( even no ) = num__10 = n ( e ) n ( prime no ) = num__8 = n ( p ) p ( e ᴜ p ) = num__0.5 + num__0.4 - num__0.05 = num__0.85 answer : d <eor> d <eos> |
d |
union_prob__0.5__0.4__0.05__ union_prob__0.5__0.4__0.05__ |
union_prob__0.5__0.4__0.05__ union_prob__0.5__0.4__0.05__ |
| when working alone painter w can paint a room in num__2 hours and working alone painter x can paint the same room in h hours . when the two painters work together and independently they can paint the room in num__0.75 of an hour . what is the value of h ? <o> a ) num__0.75 <o> b ) num__1 [ num__0.2 ] <o> c ) num__1 [ num__0.4 ] <o> d ) num__1 [ num__0.75 ] <o> e ) num__2 |
rate * time = work let painter w ' s rate be w and painter x ' s rate be x r * t = work w * num__2 = num__1 ( if the work done is same throughout the question then the work done can be taken as num__1 ) = > w = num__0.5 x * h = num__1 = > x = num__1 / h when they both work together then their rates get added up combined rate = ( w + x ) r * t = work ( w + x ) * num__0.75 = num__1 = > w + x = num__1.33333333333 = > num__0.5 + num__1 / h = num__1.33333333333 = > num__1 / h = ( num__8 - num__3 ) / num__6 = num__0.833333333333 = > h = num__1.2 = num__1 [ num__0.2 ] answer b <eor> b <eos> |
b |
divide__1.0__2.0__ divide__1.0__0.75__ add__2.0__1.0__ multiply__2.0__3.0__ subtract__1.3333__0.5__ divide__1.0__0.8333__ subtract__1.2__1.0__ round__1.0__ |
divide__1.0__2.0__ divide__1.0__0.75__ add__2.0__1.0__ multiply__2.0__3.0__ subtract__1.3333__0.5__ divide__1.0__0.8333__ subtract__1.2__1.0__ multiply__2.0__0.5__ |
| the l . c . m . of num__2 numbers is num__54 . the numbers are in the ratio num__2 : num__3 . find their sum ? <o> a ) num__36 <o> b ) num__40 <o> c ) num__45 <o> d ) num__32 <o> e ) num__56 |
let the numbers be num__2 x and num__3 x l . c . m . = num__6 x num__6 x = num__54 x = num__9 the numbers are = num__18 and num__27 required sum = num__18 + num__27 = num__45 answer is c <eor> c <eos> |
c |
multiply__2.0__3.0__ divide__54.0__6.0__ multiply__2.0__9.0__ divide__54.0__2.0__ subtract__54.0__9.0__ subtract__54.0__9.0__ |
multiply__2.0__3.0__ add__3.0__6.0__ multiply__2.0__9.0__ add__9.0__18.0__ add__18.0__27.0__ add__18.0__27.0__ |
| set a contains all the even numbers between num__22 and num__70 inclusive . set b contains all the even numbers between num__62 and num__110 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? <o> a ) num__600 <o> b ) num__800 <o> c ) num__1000 <o> d ) num__1200 <o> e ) num__1400 |
each term in set b is num__40 more than the corresponding term in set a . the difference of the sums = num__25 * num__40 = num__1000 . the answer is c . <eor> c <eos> |
c |
subtract__62.0__22.0__ multiply__40.0__25.0__ multiply__40.0__25.0__ |
subtract__62.0__22.0__ multiply__40.0__25.0__ multiply__40.0__25.0__ |
| if jake loses num__5 pounds he will weigh thrice as much as his sister . together they now weigh num__285 pounds . what is jake ' s present weight in pounds ? <o> a ) num__131 <o> b ) num__135 <o> c ) num__139 <o> d ) num__147 <o> e ) num__215 |
this problem can be solved as a simple word problem in which we must convert words to math . before we create our equations we want to define some variables . j = jake ’ s current weight in pounds s = sister ’ s current weight in pounds we are told that “ if jake loses num__8 pounds he will weigh twice as much as his sister . we put this into an equation : j – num__5 = num__3 s j = num__3 s + num__5 ( equation num__1 ) next we are told that “ together they now weigh num__275 pounds . ” we can also put this into an equation . j + s = num__285 ( equation num__2 ) to solve this equation we can substitute num__2 s + num__8 from equation num__1 for the variable j in equation num__2 : num__3 s + num__5 = num__285 - s num__4 s = num__280 s = num__70 j + num__70 = num__285 j = num__215 answer : e <eor> e <eos> |
e |
subtract__8.0__5.0__ twice__1.0__ twice__2.0__ add__5.0__275.0__ divide__280.0__4.0__ subtract__285.0__70.0__ subtract__285.0__70.0__ |
subtract__8.0__5.0__ subtract__5.0__3.0__ subtract__5.0__1.0__ add__5.0__275.0__ divide__280.0__4.0__ subtract__285.0__70.0__ subtract__285.0__70.0__ |
| what is the smallest number which when increased by num__7 is divisible by num__7 num__8 and num__24 ? <o> a ) num__154 <o> b ) num__161 <o> c ) num__168 <o> d ) num__175 <o> e ) num__182 |
lcm ( num__7 num__824 ) = num__24 x num__7 = num__168 so the least divisible number is num__168 and the number we are looking for is num__168 - num__7 = num__161 . the answer is b . <eor> b <eos> |
b |
multiply__7.0__24.0__ subtract__168.0__7.0__ subtract__168.0__7.0__ |
multiply__7.0__24.0__ subtract__168.0__7.0__ subtract__168.0__7.0__ |
| two cars start from the opposite places of a main road num__140 km apart . first car runs for num__25 km and takes a right turn and then runs num__15 km . it then turns left and then runs for another num__25 km and then takes the direction back to reach the main road . in the mean time due to minor break down the other car has run only num__35 km along the main road . what would be the distance between two cars at this point ? <o> a ) num__65 <o> b ) num__38 <o> c ) num__20 <o> d ) num__55 <o> e ) num__21 |
answer : d ) num__55 km <eor> d <eos> |
d |
round__55.0__ |
round__55.0__ |
| a fraction bears the same ratio to num__0.037037037037 as num__0.428571428571 does to num__0.555555555556 . the fraction is ? <o> a ) num__0.0322580645161 <o> b ) num__0.0285714285714 <o> c ) num__0.027027027027 <o> d ) num__0.0571428571429 <o> e ) num__0.0810810810811 |
let the fraction be x . then x : num__0.037037037037 = num__0.428571428571 : num__0.555555555556 x × num__0.555555555556 = num__0.037037037037 × num__0.428571428571 x × num__0.555555555556 = num__0.111111111111 × num__0.142857142857 x × num__0.555555555556 = num__0.015873015873 x × num__5 = num__0.142857142857 num__5 x = num__0.142857142857 = num__0.0285714285714 b ) <eor> b <eos> |
b |
multiply__0.037__0.4286__ divide__0.0159__0.5556__ divide__0.0159__0.5556__ |
multiply__0.037__0.4286__ divide__0.0159__0.5556__ divide__0.0159__0.5556__ |
| an increase of rs num__3 in the selling price of an article turn a loss of num__7.5 % into a gain of num__7.5 % . the cost price of the article is ? ? <o> a ) num__10 rs <o> b ) num__20 rs <o> c ) num__25 rs <o> d ) num__30 rs <o> e ) num__40 rs |
let cost price = x rs and selling price = y rs now condition ( num__1 ) - if we sell at y loss occurs num__7.5 % so by formula selling price = { ( num__100 - % loss ) / num__100 } * cost price so put values y = { ( num__100 - num__7.5 ) / num__100 } x y = ( num__0.925 ) * x x * num__0.91 - y = num__0 - - - - - - - - - ( num__1 ) condition num__2 - if we increase selling price by num__3 we get profit by num__7.5 % so by formula selling price = { ( num__100 + % profit ) / num__100 } * cost price put values y + num__3 = { ( num__100 + num__7.5 ) / num__100 } * x x * ( num__1.075 ) - y = num__3 - - - - - - - - ( num__2 ) now on solving equation num__1 and num__2 we get x = num__20 rs answer : b <eor> b <eos> |
b |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| a and b complete a work in num__6 days . a alone can do it in num__8 days . if both together can do the work in how many days ? <o> a ) num__3.75 days <o> b ) num__3.43 days <o> c ) num__2.99 days <o> d ) num__2.98 days <o> e ) num__2.44 days |
num__0.166666666667 + num__0.125 = num__0.291666666667 num__3.42857142857 = num__3.43 days answer : b <eor> b <eos> |
b |
add__0.1667__0.125__ round__3.43__ |
add__0.1667__0.125__ round__3.43__ |
| the addition num__742586 + num__829430 = num__1212016 is incorrect . it can be corrected by changing one digit d wherever it occurs to another digit e . find sum of d and e . <o> a ) num__1316016 <o> b ) num__1416016 <o> c ) num__1516016 <o> d ) num__1616016 <o> e ) num__1716016 |
let ' s choose num__8 = num__2 + num__6 . so consider d = num__2 and e = num__6 therefore changing every occurrence of num__2 with num__6 in the whole statement ( i . e both r . h . s and l . h . s ) we get num__746586 + num__869430 = num__1616016 answer : d <eor> d <eos> |
d |
subtract__8.0__2.0__ add__869430.0__746586.0__ add__869430.0__746586.0__ |
subtract__8.0__2.0__ add__869430.0__746586.0__ add__869430.0__746586.0__ |
| if a ( a + num__7 ) = num__18 and b ( b + num__7 ) = num__18 where a ≠ b then a + b = <o> a ) − num__48 <o> b ) − num__5 <o> c ) num__2 <o> d ) - num__7 <o> e ) num__48 |
i . e . if a = num__2 then b = - num__9 or if a = - num__9 then b = num__2 but in each case a + b = - num__9 + num__2 = - num__7 answer : option d <eor> d <eos> |
d |
add__7.0__2.0__ subtract__9.0__2.0__ |
add__7.0__2.0__ subtract__9.0__2.0__ |
| num__45 men can complete a piece of work in num__18 days . in how many days will num__27 men complete the same work ? <o> a ) num__24 <o> b ) num__77 <o> c ) num__30 <o> d ) num__25 <o> e ) num__13 |
explanation : less men means more days { indirect proportion } let the number of days be x then num__27 : num__45 : : num__18 : x [ please pay attention we have written num__27 : num__45 rather than num__45 : num__27 in indirect proportion if you get it then chain rule is clear to you : ) ] { \ color { blue } x = \ frac { num__45 \ times num__18 } { num__27 } } x = num__30 so num__30 days will be required to get work done by num__27 men . answer : c <eor> c <eos> |
c |
round__30.0__ |
round__30.0__ |
| the greatest number which on dividing num__1642 and num__1856 leaves remainders num__6 and num__4 respectively is : <o> a ) num__123 <o> b ) num__127 <o> c ) num__235 <o> d ) num__4 <o> e ) num__505 |
explanation : required number = h . c . f . of ( num__1642 - num__6 ) and ( num__1856 - num__4 ) = h . c . f . of num__1636 and num__1852 = num__4 . answer : d <eor> d <eos> |
d |
subtract__1642.0__6.0__ subtract__1856.0__4.0__ subtract__1856.0__1852.0__ |
subtract__1642.0__6.0__ subtract__1856.0__4.0__ subtract__1856.0__1852.0__ |
| a store sells brand x pens for num__5 dollars each brand y pens for $ num__4 each and brand z pens for $ num__3 each . there are a total of num__36 of these three types of pens in the store the number of brand x pens is twice sum of the number of brand y pens and brand z pens and the difference of the number of brand y pens and the number of brand z pens is num__2 . what is the largest possible total amount that three types of pens will be sold for ? <o> a ) $ num__163 <o> b ) $ num__159 <o> c ) $ num__156 <o> d ) $ num__148 <o> e ) $ num__125 |
max price = num__5 x + num__4 y + num__3 z - - - - - - - ( num__1 ) given : x + y + z = num__36 - - - - - - - ( num__2 ) given : x = num__2 ( y + z ) - - - - - - - - ( num__3 ) substitute ( num__3 ) in ( num__2 ) y + z = num__12 - - - - - ( num__4 ) given : y - z = num__2 - - - - ( num__5 ) using equations ( num__4 ) and ( num__5 ) y = num__7 and z = num__5 . substitute y and z in num__3 we get x = num__24 substitute x y and z in ( num__1 ) we get answer = num__163 i . e . answer : a <eor> a <eos> |
a |
subtract__5.0__4.0__ multiply__4.0__3.0__ add__5.0__2.0__ subtract__36.0__12.0__ multiply__1.0__163.0__ |
subtract__5.0__4.0__ multiply__4.0__3.0__ add__5.0__2.0__ subtract__36.0__12.0__ multiply__1.0__163.0__ |
| working at their respective constant rates machine a makes num__100 copies in num__8 minutes and machine b makes num__150 copies in num__10 minutes . if these machines work simultaneously at their respective rates for num__30 minutes what is the total number of copies that they will produce ? <o> a ) num__250 <o> b ) num__425 <o> c ) num__675 <o> d ) num__825 <o> e ) num__750 |
machine a can produce num__100 * num__3.75 = num__375 copies and machine b can produce num__150 * num__3.0 = num__825 copies total producing num__700 copies . d is the answer <eor> d <eos> |
d |
divide__30.0__8.0__ multiply__100.0__3.75__ divide__30.0__10.0__ round__825.0__ |
divide__30.0__8.0__ multiply__100.0__3.75__ divide__30.0__10.0__ round__825.0__ |
| how much space in cubic units is left vacant when maximum number of num__3 x num__3 x num__3 cubes are fitted in a rectangular box measuring num__6 x num__9 x num__11 ? <o> a ) num__112 <o> b ) num__111 <o> c ) num__110 <o> d ) num__109 <o> e ) num__108 |
no of cubes that can be accommodated in box = ( num__6 * num__9 * num__11 ) / ( num__3 * num__3 * num__3 ) num__6 * num__9 in numerator can be perfectly divided by num__3 * num__3 in denominator . side with length num__11 ca n ' t be perfectly divided by num__3 and hence is the limiting factor . closet multiple of num__3 less that num__11 is num__9 . so vacant area in cube = = num__6 * num__9 * ( num__11 - num__9 ) = num__6 * num__9 * num__2 = num__108 ans - e <eor> e <eos> |
e |
volume_rectangular_prism__6.0__9.0__2.0__ triangle_area__2.0__108.0__ |
volume_rectangular_prism__6.0__9.0__2.0__ volume_rectangular_prism__6.0__9.0__2.0__ |
| a fair num__2 - sided coin is flipped num__6 times . what is the probability that tails will be the result at least twice but not more than num__5 times ? <o> a ) num__0.625 <o> b ) num__0.75 <o> c ) num__0.875 <o> d ) num__0.890625 <o> e ) num__0.9375 |
total possible outcomes when coined is tossed num__6 time = num__2 ^ num__6 = num__4 * num__4 * num__4 = num__64 total possible outcomes getting num__2 or num__3 or num__4 or num__5 tails = num__6 c num__2 + num__6 c num__3 + num__6 c num__4 + num__6 c num__5 = ( num__6 * num__5 ) / num__2 + ( num__6 * num__5 * num__4 ) / ( num__3 * num__2 ) + ( num__6 * num__5 ) / num__2 + num__6 = num__15 + num__20 + num__15 + num__6 = num__56 probability of getting at least num__2 but not more than num__5 times tails = num__0.875 = num__0.875 ans = c <eor> c <eos> |
c |
subtract__6.0__2.0__ divide__6.0__2.0__ multiply__5.0__3.0__ multiply__5.0__4.0__ divide__56.0__64.0__ divide__56.0__64.0__ |
subtract__6.0__2.0__ divide__6.0__2.0__ multiply__5.0__3.0__ multiply__5.0__4.0__ divide__56.0__64.0__ divide__56.0__64.0__ |
| find the value of m from given num__72519 x num__9999 = m ? <o> a ) num__455666576 <o> b ) num__565476563 <o> c ) num__567642887 <o> d ) num__725117481 <o> e ) num__658899003 |
num__72519 x num__9999 = num__72519 x ( num__10000 - num__1 ) = num__72519 x num__10000 - num__72519 x num__1 = num__725190000 - num__72519 = num__725117481 d <eor> d <eos> |
d |
subtract__10000.0__9999.0__ multiply__72519.0__10000.0__ multiply__72519.0__9999.0__ multiply__72519.0__9999.0__ |
subtract__10000.0__9999.0__ multiply__72519.0__10000.0__ subtract__725190000.0__72519.0__ subtract__725190000.0__72519.0__ |
| a tradesman by means of his false balance defrauds to the extent of num__12.0 ? in buying goods as well as by selling the goods . what percent does he gain on his outlay ? <o> a ) num__26.4 <o> b ) num__25.44 <o> c ) num__27.5 <o> d ) num__28.54 <o> e ) num__27 |
explanation : g % = num__12 + num__12 + ( num__12 * num__12 ) / num__100 = num__25.44 answer : b <eor> b <eos> |
b |
percent__25.44__100.0__ |
percent__25.44__100.0__ |
| john was thrice as old as tom num__6 years ago . john will be num__2 times as old as tom in num__4 years . how old is tom today ? <o> a ) num__12 <o> b ) num__14 <o> c ) num__16 <o> d ) num__18 <o> e ) num__20 |
j - num__6 = num__3 ( t - num__6 ) so j = num__3 t - num__12 j + num__4 = num__2 ( t + num__4 ) ( num__3 t - num__12 ) + num__4 = num__2 t + num__8 t = num__16 the answer is c . <eor> c <eos> |
c |
divide__6.0__2.0__ multiply__6.0__2.0__ add__6.0__2.0__ multiply__2.0__8.0__ round__16.0__ |
divide__6.0__2.0__ multiply__6.0__2.0__ add__6.0__2.0__ add__4.0__12.0__ add__4.0__12.0__ |
| a train num__900 m long is running at a speed of num__78 km / hr . if it crosses a tunnel in num__1 min then the length of the tunnel is ? <o> a ) num__200 m <o> b ) num__776 m <o> c ) num__400 m <o> d ) num__187 m <o> e ) num__1678 m |
speed = num__78 * num__0.277777777778 = num__21.6666666667 m / sec . time = num__1 min = num__60 sec . let the length of the train be x meters . then ( num__900 + x ) / num__60 = num__21.6666666667 x = num__400 m . answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ round__400.0__ |
hour_to_min_conversion__ multiply__1.0__400.0__ |
| the total cost of num__100 paper plates and num__200 paper cups is $ num__8.00 at the same rates what is the total cost of num__20 of the plates and num__40 of the cups ? <o> a ) $ . num__90 <o> b ) $ num__1.00 <o> c ) $ num__1.20 <o> d ) $ num__1.50 <o> e ) $ num__1.60 |
u dont need to go through all this what u have with u is num__100 p + num__200 c = $ num__8.00 just divide the equation by num__5 and you will get what u are looking for num__20 p + num__40 c = $ num__1.60 therefore oa is e <eor> e <eos> |
e |
percent__8.0__20.0__ percent__100.0__1.6__ |
percent__8.0__20.0__ percent__100.0__1.6__ |
| rohit paid $ num__4000 for num__100 shares of stock x and $ num__2000 for num__100 shares of stock y . he later sold the same shares of both stocks gaining num__0.6 the amount he paid for stock y . if the prices he had paid for the stocks had been reversed and everything else remained the same then the net result would have been : <o> a ) a gain of num__1.4 times as much <o> b ) a gain of num__3 times as much <o> c ) a gain of num__2 times as much <o> d ) a gain of num__1.5 times as much <o> e ) a loss of num__2 times as much |
a = num__4000 ; b = num__2000 ; profit = ( num__0.6 ) * num__2000 = num__1200 after prices are reversed : a = num__2000 ; b = num__4000 ; profit = ( num__0.6 ) * num__4000 = num__2400 num__2400 - - > gain of num__2 times num__1200 answer : c <eor> c <eos> |
c |
multiply__2000.0__0.6__ multiply__4000.0__0.6__ divide__4000.0__2000.0__ divide__4000.0__2000.0__ |
multiply__2000.0__0.6__ multiply__4000.0__0.6__ divide__4000.0__2000.0__ divide__4000.0__2000.0__ |
| a thief goes away with a santro car at a speed of num__40 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at num__50 kmph when will the owner over take the thief from the start ? <o> a ) num__2 hours <o> b ) num__6 hours <o> c ) num__8 hours <o> d ) num__9 hours <o> e ) num__3 hours |
| - - - - - - - - - - - num__20 - - - - - - - - - - - - - - - - - - - - | num__50 num__40 d = num__20 rs = num__50 – num__40 = num__10 t = num__2.0 = num__2 hours num__14 . if i walk at num__3 kmph i miss answer : a <eor> a <eos> |
a |
subtract__50.0__40.0__ divide__40.0__20.0__ round__2.0__ |
subtract__50.0__40.0__ divide__40.0__20.0__ round__2.0__ |
| of the num__60 researchers in a workgroup num__40 percent will be assigned to team a and the remaining num__60 percent to team b . however num__70 percent of the researchers prefer team a and num__30 percent prefer team b . what is the lowest possible number of researchers who will not be assigned to the team they prefer ? <o> a ) num__15 <o> b ) num__18 <o> c ) num__20 <o> d ) num__25 <o> e ) num__30 |
i solved it by assuming num__100 people and then dividing my answer in two to save time on calculations . ( percentage equals number of people ) in that case num__40 will be in team a num__60 will be in team b the larger diff is num__70 want team a so diff is num__70 - num__40 = num__30 . at least num__30 people will not get their wish so for num__60 researchers the same number is num__18 . answer choice b <eor> b <eos> |
b |
percent__60.0__30.0__ percent__60.0__30.0__ |
percent__60.0__30.0__ percent__60.0__30.0__ |
| line m lies in the xy - plane . the y - intercept of line m is - num__5 and line m passes through the midpoint of the line segment whose endpoints are ( num__2 num__4 ) and ( num__6 - num__8 ) . what is the slope of line m ? <o> a ) - num__3 <o> b ) - num__1 <o> c ) - num__0.333333333333 <o> d ) num__0 <o> e ) undefined |
ans : a solution : line m goes through midpoint of ( num__2 num__4 ) and ( num__6 - num__8 ) . midpoint is ( num__4 - num__2 ) as we can see that the y axis of intercept point is ( num__0 - num__5 ) means line m is parallel to x axis slope m = - num__3 ans : a <eor> a <eos> |
a |
subtract__5.0__2.0__ subtract__5.0__2.0__ |
subtract__5.0__2.0__ subtract__5.0__2.0__ |
| a fair price shopkeeper takes num__10.0 profit on his goods . he lost num__20.0 goods during theft . his loss percent is : <o> a ) num__72.0 <o> b ) num__42.0 <o> c ) num__32.0 <o> d ) num__12.0 <o> e ) num__22 % |
explanation : suppose he has num__100 items . let c . p . of each item be re . num__1 . total cost = rs . num__100 . number of items left after theft = num__80 . s . p . of each item = rs . num__1.10 total sale = num__1.10 * num__80 = rs . num__88 hence loss % = num__0.12 * num__100 = num__12.0 answer : d <eor> d <eos> |
d |
percent__100.0__12.0__ |
percent__100.0__12.0__ |
| tom had num__63 apples . he divides all apples evenly among num__9 friends . how many apples did tom give to each of his friends ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__18 <o> d ) num__7 <o> e ) num__8 |
total number of apples = num__63 there are num__9 friends of seats on the bus . to find how many apples tom gave to each of his friends divide the total number of apples by the number of friends . we get divide num__63 by num__9 num__63 ÷ num__9 = num__7 therefore tom gives num__7 apples to each of his friends . answer is d <eor> d <eos> |
d |
divide__63.0__9.0__ divide__63.0__9.0__ |
divide__63.0__9.0__ divide__63.0__9.0__ |
| you play the following game with a friend . you share a pile of chips and you take turns removing between one and four chips from the pile . ( in particular at least one chip must be removed on each turn . ) the game ends when the last chip is removed from the pile ; the one who removes it is the loser . it is your turn and there are num__2014 chips in the pile . how many chips should you remove to guarantee that you win assuming you then make the best moves until the game is over ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) there is no way to guarantee a win even with the best play |
the key to winning this game is that if you leave your opponent with num__5 k + num__1 chips for any integer k num__0 then you win . when your opponent removes r chips you remove num__5 - r and so there are now num__5 ( k - num__1 ) + num__1 chips left and you keep using this strategy until there is only one chip left at which point your opponent takes it and loses . so you should take num__3 chips to leave num__2011 = num__5 num__402 + num__1 . correct answer c <eor> c <eos> |
c |
subtract__2014.0__3.0__ subtract__2014.0__2011.0__ |
subtract__2014.0__3.0__ subtract__2014.0__2011.0__ |
| num__36 men can complete a piece of work in num__18 days . in how many days will num__108 men complete the same work ? <o> a ) num__24 <o> b ) num__77 <o> c ) num__6 <o> d ) num__29 <o> e ) num__21 |
explanation : less men means more days { indirect proportion } let the number of days be x then num__108 : num__36 : : num__18 : x x = num__6 answer : c ) num__6 days <eor> c <eos> |
c |
divide__108.0__18.0__ round__6.0__ |
divide__108.0__18.0__ round__6.0__ |
| it was a beautiful sunny morning . the air was fresh and a mild wind was blowing against my wind screen . i was driving from bangalore to brindavan gardens . it took me num__1 hour and num__30 minutes to complete the journey . after lunch i returned to bangalore . i drove for num__90 rhinutes . how do you explain it ? <o> a ) num__75 min <o> b ) num__60 min <o> c ) num__90 min <o> d ) num__80 min <o> e ) num__88 min |
c num__90 min jansi went out for shopping . she had in her handbag approximately rs . num__15 / - in one rupee notes and num__20 p . coins . when she returned she had as many one rupee notes as she originally had and as many num__20 p . coins as she originally had one rupee notes . she actually came back with about one - third of what she had started out with . how much did she spend and exactly how much did she have with her when she started out ? there is nothing to explain here . the driving time there and back is absolutely the same because num__90 minutes and num__1 hour and num__30 minutes are one and the same thing . this problem is meant for inattentive readers who may think that there is some difference between num__90 minutes and num__1 hour num__30 minutes . <eor> c <eos> |
c |
round__90.0__ |
divide__90.0__1.0__ |
| the price of num__2 sarees and num__4 shirts is rs . num__1600 . with the same money one can buy num__1 saree and num__6 shirts . if one wants to buy num__12 shirts how much shall he have to pay ? <o> a ) rs . num__1200 <o> b ) rs . num__2400 <o> c ) rs . num__4800 <o> d ) can not be determined <o> e ) none of these |
let the price of a saree and a shirt be rs . x and rs . y respectively . then num__2 x + num__4 y = num__1600 . . . . ( i ) and x + num__6 y = num__1600 . . . . ( ii ) divide equation ( i ) by num__2 we get the below equation . = > x + num__2 y = num__800 . - - - ( iii ) now subtract ( iii ) from ( ii ) x + num__6 y = num__1600 ( - ) x + num__2 y = num__800 - - - - - - - - - - - - - - - - num__4 y = num__800 - - - - - - - - - - - - - - - - therefore y = num__200 . now apply value of y in ( iii ) = > x + num__2 x num__200 = num__800 = > x + num__400 = num__800 therefore x = num__400 solving ( i ) and ( ii ) we get x = num__400 y = num__200 . cost of num__12 shirts = rs . ( num__12 x num__200 ) = rs . num__2400 . answer : option b <eor> b <eos> |
b |
divide__1600.0__2.0__ divide__800.0__4.0__ multiply__2.0__200.0__ add__1600.0__800.0__ add__1600.0__800.0__ |
divide__1600.0__2.0__ divide__800.0__4.0__ multiply__2.0__200.0__ add__1600.0__800.0__ add__1600.0__800.0__ |
| what is the num__4 digit number in which the num__1 st digit is num__0.333333333333 of the second the num__3 rd is the sum of the num__1 st and num__2 nd and the last is three times the second ? <o> a ) num__1100 <o> b ) num__1200 <o> c ) num__1349 <o> d ) num__1400 <o> e ) num__1450 |
first digit is num__0.333333333333 second digit = > the numbers can be num__1 & num__3 num__2 & num__6 num__3 & num__9 . first + second = third = > we can eliminate num__3 & num__9 since num__3 + num__9 = num__12 . last is num__3 times the second = > we can eliminate option num__2 & num__6 since num__3 * num__6 = num__18 . hence the number is num__1349 c <eor> c <eos> |
c |
add__4.0__2.0__ add__3.0__6.0__ multiply__4.0__3.0__ multiply__3.0__6.0__ multiply__1.0__1349.0__ |
add__4.0__2.0__ add__3.0__6.0__ multiply__4.0__3.0__ multiply__3.0__6.0__ multiply__1.0__1349.0__ |
| two cars are driving toward each other . the first car is traveling at a speed of num__75 km / h which is num__25.0 slower than the second car ' s speed . if the distance between the cars is num__1050 km how many hours will it take until the two cars meet ? <o> a ) num__4 <o> b ) num__4.5 <o> c ) num__5 <o> d ) num__5.5 <o> e ) num__6 |
the speed of the first car is num__75 km / h . the speed of the second car is num__75 / num__0.75 = num__100 km / h . the two cars complete a total of num__175 km each hour . the time it takes the cars to meet is num__6.0 = num__6 hours . the answer is e . <eor> e <eos> |
e |
add__75.0__25.0__ add__75.0__100.0__ divide__1050.0__175.0__ round__6.0__ |
divide__75.0__0.75__ add__75.0__100.0__ divide__1050.0__175.0__ divide__1050.0__175.0__ |
| the dimensions of a room are num__25 feet * num__15 feet * num__12 feet . what is the cost of white washing the four walls of the room at rs . num__5 per square feet if there is one door of dimensions num__6 feet * num__3 feet and three windows of dimensions num__4 feet * num__3 feet each ? <o> a ) num__3887 <o> b ) num__2689 <o> c ) num__2678 <o> d ) num__4530 <o> e ) num__3780 |
area of the four walls = num__2 h ( l + b ) since there are doors and windows area of the walls = num__2 * num__12 ( num__15 + num__25 ) - ( num__6 * num__3 ) - num__3 ( num__4 * num__3 ) = num__906 sq . ft . total cost = num__906 * num__5 = rs . num__4530 answer : d <eor> d <eos> |
d |
multiply__5.0__906.0__ multiply__5.0__906.0__ |
multiply__5.0__906.0__ multiply__5.0__906.0__ |
| machine a produces num__100 parts twice as fast as machine b does . machine b produces num__100 parts in num__40 minutes . if each machine produces parts at a constant rate how many parts does machine a produce in num__10 minutes ? <o> a ) num__20 <o> b ) num__50 <o> c ) num__60 <o> d ) num__40 <o> e ) num__30 |
machine b produces num__100 part in num__40 minutes . machine a produces num__100 parts twice as fast as b so machine a produces num__100 parts in num__20.0 = num__20 minutes . now machine a produces num__100 parts in num__20 minutes which is num__5.0 = num__5 parts / minute . num__5 parts x a total of num__10 minutes = num__50 b <eor> b <eos> |
b |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| after num__6 games team b had an average of num__65 points per game . if it got only num__47 points in game num__7 how many more points does it need to score to get its total above num__500 ? <o> a ) num__85 <o> b ) num__74 <o> c ) num__67 <o> d ) num__63 <o> e ) num__28 |
( num__6 * num__65 ) + num__47 + x > num__500 num__390 + num__47 + x > num__500 num__437 + x > num__500 = > x > num__63 option d <eor> d <eos> |
d |
multiply__6.0__65.0__ add__47.0__390.0__ subtract__500.0__437.0__ subtract__500.0__437.0__ |
multiply__6.0__65.0__ add__47.0__390.0__ subtract__500.0__437.0__ subtract__500.0__437.0__ |
| a fruit seller had some apples . he sells num__40.0 apples and still has num__420 apples . originally he had <o> a ) num__588 apples <o> b ) num__600 apples <o> c ) num__672 apples <o> d ) num__700 apples <o> e ) none |
solution suppose originally he had x apples . then ( num__100 - num__40 ) % of x = num__420 . ‹ = › num__0.6 × x = num__420 x ‹ = › ( num__420 × num__1.66666666667 ‹ = › num__700 . answer d <eor> d <eos> |
d |
percent__100.0__700.0__ |
percent__100.0__700.0__ |
| a man can row num__30 km downstream and num__20 km upstream in num__4 hours . he can row num__45 km downstream and num__40 km upstream in num__7 hours . find the speed of man in still water ? <o> a ) num__12.9 <o> b ) num__12.6 <o> c ) num__12.4 <o> d ) num__12.5 <o> e ) num__12.1 |
let the speed of the man in still water be a kmph and let the speed of the stream be b kmph . now num__30 / ( a + b ) + num__20 / ( a - b ) = num__4 and num__45 / ( a + b ) + num__40 / ( a - b ) = num__7 solving the equation the speed of man in still water is num__12.5 kmph . answer : d <eor> d <eos> |
d |
round__12.5__ |
round__12.5__ |
| if the price of a certain computer increased num__30 percent from d dollars to num__364 dollars then num__2 d = <o> a ) num__540 <o> b ) num__570 <o> c ) num__619 <o> d ) num__649 <o> e ) num__560 |
before price increase price = d after num__30.0 price increase price = d + ( num__0.3 ) * d = num__1.3 d = num__364 ( given ) i . e . d = num__364 / num__1.3 = $ num__280 i . e . num__2 d = num__2 * num__280 = num__560 answer : option e <eor> e <eos> |
e |
divide__364.0__1.3__ multiply__2.0__280.0__ multiply__2.0__280.0__ |
divide__364.0__1.3__ multiply__2.0__280.0__ multiply__2.0__280.0__ |
| the sum of two numbers is num__50 and their product is num__375 . what will be the sum of their reciprocals ? <o> a ) num__0.025 <o> b ) num__0.106666666667 <o> c ) num__0.133333333333 <o> d ) num__9.375 <o> e ) num__12.5 |
( num__1 / a ) + ( num__1 / b ) = ( a + b ) / ab = num__0.133333333333 = num__0.133333333333 answer : c <eor> c <eos> |
c |
divide__50.0__375.0__ divide__50.0__375.0__ |
divide__50.0__375.0__ divide__50.0__375.0__ |
| find the expenditure on digging a well num__14 m deep and of num__3 m diameter at rs . num__18 per cubic meter ? <o> a ) num__2998 <o> b ) num__2799 <o> c ) num__1782 <o> d ) num__1485 <o> e ) num__2780 |
num__3.14285714286 * num__14 * num__1.5 * num__1.5 = num__99 m num__2 num__99 * num__18 = num__1782 answer : c <eor> c <eos> |
c |
divide__3.0__1.5__ multiply__18.0__99.0__ round__1782.0__ |
divide__3.0__1.5__ multiply__18.0__99.0__ multiply__18.0__99.0__ |
| a train running at num__25 km / hr takes num__18 seconds to pass a platform . next it takes num__12 seconds to pass a man standing still . find the length of the train and that of the platform . <o> a ) num__83.3 m num__41.7 m <o> b ) num__100 m num__50 m <o> c ) num__50.8 m num__45.7 m <o> d ) num__33.3333333333 m num__66.6666666667 <o> e ) none of these |
a train running at num__25 km / hr takes num__18 seconds to pass a platform . next it takes num__12 seconds to pass a man walking at num__5 km / hr in the same direction . find the length of the train and that of the platform . distance = speed * time it ' s probably best to convert to meters so we can find the length of the platform . distance = num__25000 meters / h * num__18 seconds * ( num__0.0166666666667 * num__60 ) distance = ( num__25000 * num__18 / [ num__3600 ] ) = num__125 m i ' m not sure how we take the length of the train into account ( if we do at all ) to solve this problem . for example if we were measuring by the time the very front of the train passed one end of the platform to the time the very front of the train passed the other end the result would be different than if we counted when the very front of the train passed one end of the platform to the time the back side of the train passed the end . as for the problem involving the man the man is standing still while the train is moving forward at a speed of num__25 km / h . therefore the train is moving past the man at a total speed of num__25 km / h . distance = speed * time distance = num__25 km / h * num__12 seconds distance ( length of train ) = num__83.3 m we know the length of the train is roughly num__83.3 m and the distance covered passing the platform was num__125 m . however the platform is n ' t num__125 m long - it is num__125 m minus the length of the train long ( or so i think ) which is num__125 - num__83.3 or roughly num__41.7 m . both of these answers are in a so i think that is the right answer . i am still having a bit of trouble conceptualizing why we have to subtract out the length of the train however . answer : a ) num__83.3 m num__41.7 m <eor> a <eos> |
a |
multiply__12.0__5.0__ multiply__25.0__5.0__ subtract__125.0__83.3__ subtract__125.0__41.7__ |
multiply__12.0__5.0__ multiply__25.0__5.0__ subtract__125.0__83.3__ subtract__125.0__41.7__ |
| x ÷ num__4 × num__5 + num__10 – num__12 = num__48 then x = <o> a ) num__60 <o> b ) num__40 <o> c ) num__100 <o> d ) num__28 <o> e ) num__50 |
follow reverse process num__48 + num__12 = num__60 – num__10 = num__50 ÷ num__5 = num__10 × × num__4 = num__40 option b <eor> b <eos> |
b |
multiply__5.0__12.0__ multiply__5.0__10.0__ multiply__4.0__10.0__ multiply__4.0__10.0__ |
add__12.0__48.0__ multiply__5.0__10.0__ multiply__4.0__10.0__ multiply__4.0__10.0__ |
| how many of the integers between num__25 and num__45 are even ? <o> a ) num__21 <o> b ) num__20 <o> c ) num__11 <o> d ) num__10 <o> e ) num__9 |
number start between num__25 to num__45 is num__20 numbers half of them is even . . which is num__10 answer : d <eor> d <eos> |
d |
subtract__45.0__25.0__ subtract__20.0__10.0__ |
subtract__45.0__25.0__ subtract__20.0__10.0__ |
| how many numbers from num__10 to num__1000000 are exactly divisible by num__9 ? <o> a ) num__900 <o> b ) num__8000 <o> c ) num__1100 <o> d ) num__111110 <o> e ) num__14 |
num__1.11111111111 = num__1 and num__111111.111111 = num__111111 = = > num__111111 - num__1 = num__111110 . answer : d <eor> d <eos> |
d |
divide__10.0__9.0__ round_down__1.1111__ divide__1000000.0__9.0__ round_down__111111.1111__ subtract__111111.1111__1.1111__ multiply__1.0__111110.0__ |
divide__10.0__9.0__ subtract__10.0__9.0__ divide__1000000.0__9.0__ round_down__111111.1111__ subtract__111111.1111__1.1111__ subtract__111111.1111__1.1111__ |
| the average age of a class of num__32 students is num__16 yrs . if the teacher ' s age is also included the average increases by one year . find the age of the teacher <o> a ) num__45 years <o> b ) num__46 years <o> c ) num__49 years <o> d ) num__52 years <o> e ) num__54 years |
total age of students is num__32 x num__16 = num__512 years total age inclusive of teacher = num__33 x ( num__16 + num__1 ) = num__561 so teacher ' s age is num__561 - num__512 = num__49 yrs there is a shortcut for these type of problems teacher ' s age is num__16 + ( num__33 x num__1 ) = num__49 years c <eor> c <eos> |
c |
multiply__32.0__16.0__ subtract__33.0__32.0__ add__16.0__33.0__ add__16.0__33.0__ |
multiply__32.0__16.0__ subtract__33.0__32.0__ add__16.0__33.0__ add__16.0__33.0__ |
| a fort had provision of food for num__150 men for num__45 days . after num__10 days num__25 men left the fort . find out the number of days for which the remaining food will last <o> a ) num__44 <o> b ) num__42 <o> c ) num__40 <o> d ) num__38 <o> e ) num__36 |
explanation : given that fort had provision of food for num__150 men for num__45 days hence after num__10 days the remaining food is sufficient for num__150 men for num__35 days remaining men after num__10 days = num__150 - num__25 = num__125 assume that after num__10 days the remaining food is sufficient for num__125 men for x days more men less days ( indirect proportion ) ⇒ men num__150 : num__125 } : : x : num__35 ⇒ num__150 × num__35 = num__125 x ⇒ num__6 × num__35 = num__5 x ⇒ x = num__6 × num__7 = num__42 ⇒ the remaining food is sufficient for num__125 men for num__42 days . answer : option b <eor> b <eos> |
b |
subtract__45.0__10.0__ subtract__150.0__25.0__ divide__150.0__25.0__ divide__125.0__25.0__ divide__35.0__5.0__ add__35.0__7.0__ round__42.0__ |
subtract__45.0__10.0__ subtract__150.0__25.0__ divide__150.0__25.0__ divide__125.0__25.0__ divide__35.0__5.0__ add__35.0__7.0__ round__42.0__ |
| if num__100000 microns = num__1 decimeter and num__1 num__000000 angstroms = num__1 decimeter how many angstroms equal num__1 micron ? <o> a ) num__1.0 e - num__05 <o> b ) num__0.0001 <o> c ) num__0.001 <o> d ) num__10 <o> e ) num__100 |
000 |
given that num__100000 microns = num__1 decimeter num__1 num__000000 angstroms = num__1 decimeter so num__100000 microns = num__1 num__000000 angstroms num__1 micron = num__1 num__000000 / num__100000 = num__10 answer : d <eor> d <eos> |
d |
d |
| after giving a discount of rs . num__100 the shopkeeper still gets a profit of num__30.0 if the cost price is rs . num__1000 . find the markup % ? <o> a ) . num__50 <o> b ) . num__45 <o> c ) . num__40 <o> d ) . num__48 <o> e ) . num__52 |
cost price = num__1000 s . p = num__1000 * num__1.3 = num__1300 disc = num__100 so . . . mark price = num__1300 + num__100 = num__1400 . . . . . . mark up % = num__1400 - num__1.0 = num__0.4 = . num__40 ( or ) num__40.0 answer : c <eor> c <eos> |
c |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| a football field is num__9600 square yards . if num__1200 pounds of fertilizer are spread evenly across the entire field how many pounds of fertilizer were spread over an area of the field totaling num__5600 square yards ? <o> a ) num__450 <o> b ) num__600 <o> c ) num__700 <o> d ) num__2400 <o> e ) num__3200 |
answer c ) num__9600 yards need num__1200 lbs num__1 yard will need num__0.125 = num__0.125 lbs num__3600 yards will need num__0.125 * num__5600 yards = num__700 lbs <eor> c <eos> |
c |
multiply__5600.0__0.125__ multiply__5600.0__0.125__ |
multiply__5600.0__0.125__ multiply__5600.0__0.125__ |
| when a merchant imported a certain item he paid a num__9 percent import tax on the portion of the total value of the item in excess of $ num__1000 . if the amount of the import tax that the merchant paid was $ num__85.50 what was the total value of the item ? <o> a ) $ num__1150 <o> b ) $ num__1350 <o> c ) $ num__1550 <o> d ) $ num__1750 <o> e ) $ num__1 |
950 |
let x be the value in excess of $ num__1000 . num__0.09 x = num__85.5 x = $ num__950 the total value was $ num__950 + $ num__1000 = $ num__1950 . the answer is e . <eor> e <eos> |
e |
e |
| there are num__7000 people in the particular village among them num__30.0 are educated and out of these num__28.0 are doing agriculture and out of them num__50.0 are woman ' s . then find out the number of woman ' s who are educated and doing agriculture ? <o> a ) num__294 <o> b ) num__235 <o> c ) num__350 <o> d ) num__288 <o> e ) num__580 |
total number of people in the village = num__7000 ; out of them num__30.0 educated = num__2100 ; out of them num__28.0 doing agriculture = num__588 ; out of them num__50.0 are woman ' s = num__294 . <eor> a <eos> |
a |
percent__30.0__7000.0__ percent__28.0__2100.0__ percent__50.0__588.0__ percent__50.0__588.0__ |
percent__30.0__7000.0__ percent__28.0__2100.0__ percent__50.0__588.0__ percent__50.0__588.0__ |
| what sum of money will produce rs . num__80 as simple interest in num__2 years at num__16 percent ? <o> a ) num__400 <o> b ) num__250 <o> c ) num__300 <o> d ) num__350 <o> e ) num__200 |
num__80 = ( p * num__2 * num__16 ) / num__100 p = num__250 answer : b <eor> b <eos> |
b |
percent__100.0__250.0__ |
percent__100.0__250.0__ |
| in a group of ducks and cows the total number of legs are num__24 more than twice the no . of heads . find the total no . of buffaloes . <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__16 <o> e ) num__17 |
let the number of buffaloes be x and the number of ducks be y = > num__4 x + num__2 y = num__2 ( x + y ) + num__24 = > num__2 x = num__24 = > x = num__12 b <eor> b <eos> |
b |
divide__24.0__2.0__ subtract__24.0__12.0__ |
divide__24.0__2.0__ subtract__24.0__12.0__ |
| a box has exactly num__100 balls and each ball is either red blue or white . if the box has num__5 more blue balls than white balls and thrice as many red balls as blue balls how many white balls does the box has ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
x = the number of red balls y = the number of blue balls z = the number of white balls from the first sentence we have equation # num__1 : x + y + z = num__100 . . . the box has num__5 more blue balls than white balls . . . equation # num__2 : y = num__5 + z . . . thrice as many red balls as blue balls . . . equation # num__3 : x = num__3 y solve equation # num__2 for z : z = y - num__5 now we can replace both x and z with y in equation # num__1 num__3 y + y + ( y - num__5 ) = num__100 num__5 y - num__5 = num__100 num__5 y = num__105 y = num__21 there are num__21 blue balls . this is num__5 more than the number of white balls so z = num__16 . that ' s the answer . just as a check x = num__63 and num__63 + num__21 + num__16 = num__100 . answer = num__16 ( e ) <eor> e <eos> |
e |
subtract__5.0__2.0__ add__100.0__5.0__ divide__105.0__5.0__ subtract__21.0__5.0__ multiply__3.0__21.0__ multiply__1.0__16.0__ |
subtract__5.0__2.0__ add__100.0__5.0__ divide__105.0__5.0__ subtract__21.0__5.0__ multiply__3.0__21.0__ subtract__21.0__5.0__ |
| which one of the following numbers is exactly divisible by num__11 ? <o> a ) num__11 <o> b ) num__13 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
( num__4 + num__5 + num__2 ) - ( num__1 + num__6 + num__3 ) = num__1 not divisible by num__11 . ( num__2 + num__6 + num__4 ) - ( num__4 + num__5 + num__2 ) = num__1 not divisible by num__11 . ( num__4 + num__6 + num__1 ) - ( num__2 + num__5 + num__3 ) = num__1 not divisible by num__11 . ( num__4 + num__6 + num__1 ) - ( num__2 + num__5 + num__4 ) = num__0 so num__415624 is divisible by num__11 . a <eor> a <eos> |
a |
subtract__5.0__4.0__ subtract__11.0__5.0__ add__1.0__2.0__ multiply__11.0__1.0__ |
subtract__5.0__4.0__ subtract__11.0__5.0__ add__1.0__2.0__ add__5.0__6.0__ |
| a train num__250 m long running at num__72 kmph crosses a platform in num__50 sec . what is the length of the platform ? <o> a ) num__150 m <o> b ) num__200 m <o> c ) num__250 m <o> d ) num__750 m <o> e ) num__300 m |
d = num__72 * num__0.277777777778 = num__50 = num__1000 â € “ num__250 = num__750 m answer : d <eor> d <eos> |
d |
subtract__1000.0__250.0__ round__750.0__ |
subtract__1000.0__250.0__ round__750.0__ |
| the perimeter of a square is equal to the perimeter of a rectangle of length num__15 cm and breadth num__14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) <o> a ) num__22.78 <o> b ) num__23.54 <o> c ) num__24.5 <o> d ) num__25.55 <o> e ) num__23.51 |
let the side of the square be a cm . perimeter of the rectangle = num__2 ( num__15 + num__14 ) = num__58 cm perimeter of the square = num__58 cm i . e . num__4 a = num__58 a = num__14.5 diameter of the semicircle = num__14.5 cm circumference of the semicircle = num__0.5 ( ∏ ) ( num__14.5 ) = num__0.5 ( num__3.14285714286 ) ( num__14.5 ) = num__22.78 cm to two decimal places answer : a <eor> a <eos> |
a |
rectangle_perimeter__15.0__14.0__ triangle_area__2.0__22.78__ |
rectangle_perimeter__15.0__14.0__ triangle_area__2.0__22.78__ |
| a man can row upstream at num__15 kmph and down strem num__30 kmph . he takes num__9 hours to row from lower point a to upper point b and back to a . what is the total distance traveled by him in num__9 hours ? he must return back to base point a in remaining time . <o> a ) num__80 <o> b ) num__120 <o> c ) num__140 <o> d ) num__180 <o> e ) num__200 |
upward distance traveled per hour - num__15 km distance traveled after num__6 hours = num__90 km . he remained with num__3 hours travel downward i . e . distance traveled in remaining num__3 hours downward = num__90 km so total distance traveled from a to b and back to point a = num__90 km upward + num__90 km downhill = num__180 km ( answer : d ) <eor> d <eos> |
d |
subtract__15.0__9.0__ multiply__15.0__6.0__ subtract__9.0__6.0__ multiply__30.0__6.0__ round__180.0__ |
subtract__15.0__9.0__ multiply__15.0__6.0__ subtract__9.0__6.0__ multiply__30.0__6.0__ round__180.0__ |
| calculate the circumference of a circular field whose radius is num__6 centimeters . <o> a ) num__12 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
circumference c is given by c = num__2 Ï € r = num__2 Ï € * num__6 = num__12 Ï € cm correct answer a <eor> a <eos> |
a |
multiply__6.0__2.0__ multiply__6.0__2.0__ |
multiply__6.0__2.0__ multiply__6.0__2.0__ |
| how many real roots does the equation x ^ num__4 y + num__16 xy + num__64 y = num__0 have if y < num__0 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) infinite |
x ^ num__4 y + num__16 xy + num__64 y = num__0 = > y ( x ^ num__2 + num__16 x + num__64 ) = num__0 = > y ( x + num__8 ) ^ num__2 = num__0 if y < num__0 then x = - num__6 so although there are num__2 factors they are the same x = - num__6 . the equations has num__3 distinct real root . answer d <eor> d <eos> |
d |
multiply__4.0__2.0__ add__4.0__2.0__ divide__6.0__2.0__ divide__6.0__2.0__ |
multiply__4.0__2.0__ add__4.0__2.0__ divide__6.0__2.0__ subtract__6.0__3.0__ |
| if the price of gasoline increases by num__20.0 and a driver intends to spend only num__8.0 more on gasoline by how much percent should the driver reduce the quantity of gasoline that he buys ? <o> a ) num__4.0 <o> b ) num__5.0 <o> c ) num__6.0 <o> d ) num__8.0 <o> e ) num__10 % |
let x be the amount of gasoline the driver buys originally . let y be the new amount of gasoline the driver should buy . let p be the original price per liter . ( num__1.2 * p ) y = num__1.08 ( p * x ) y = ( num__1.08 / num__1.2 ) x = num__0.9 x which is a reduction of num__10.0 . the answer is e . <eor> e <eos> |
e |
divide__1.08__1.2__ subtract__20.0__10.0__ |
divide__1.08__1.2__ subtract__20.0__10.0__ |
| the area of sector of a circle whose radius is num__12 metro and whose angle at the center is num__42 ° is ? <o> a ) num__52.7 m num__2 <o> b ) num__57.8 m num__2 <o> c ) num__52.8 metre sq <o> d ) num__72.8 m num__2 <o> e ) num__52.8 m num__2 |
num__0.116666666667 * num__3.14285714286 * num__12 * num__12 = num__52.8 m num__2 answer : c <eor> c <eos> |
c |
triangle_area__52.8__2.0__ |
triangle_area__52.8__2.0__ |
| how long will it take a sum of money invested at num__5.0 p . a . s . i . to increase its value by num__40.0 ? <o> a ) num__6 years <o> b ) num__7 years <o> c ) num__8 years <o> d ) num__9 years <o> e ) none |
sol . let the sum be x . then s . i . = num__40.0 of x = num__2 x / num__5 ; rate = num__5.0 . ∴ time = [ num__100 * num__2 x / num__5 * num__1 / x * num__5 ] = num__8 yea answer c <eor> c <eos> |
c |
percent__5.0__40.0__ percent__100.0__8.0__ |
percent__5.0__40.0__ percent__100.0__8.0__ |
| calculate the distance covered by num__150 revolutions of a wheel of radius num__20.5 cm . <o> a ) num__293.3 m <o> b ) num__197.3 m <o> c ) num__193.3 m <o> d ) num__143.3 m <o> e ) num__198.3 m |
in one resolution the distance covered by the wheel is its own circumference . distance covered in num__500 resolutions . = num__150 * num__2 * num__3.14285714286 * num__20.5 = num__19328.57 cm = num__193.3 m answer : c <eor> c <eos> |
c |
round__193.3__ |
round__193.3__ |
| given f ( x ) = num__3 x – num__5 for what value of x does num__2 * [ f ( x ) ] + num__8 = f ( num__3 x – num__6 ) <o> a ) num__0 <o> b ) num__4 <o> c ) num__6 <o> d ) num__7 <o> e ) num__13 |
answer = d = num__7 f ( x ) = num__3 x – num__5 num__2 * [ f ( x ) ] + num__8 = f ( num__3 x – num__6 ) num__2 ( num__3 x - num__5 ) + num__8 = num__3 ( num__3 x - num__6 ) - num__5 num__6 x - num__2 = num__9 x - num__23 x = num__7 <eor> d <eos> |
d |
add__5.0__2.0__ add__3.0__6.0__ add__5.0__2.0__ |
add__5.0__2.0__ add__3.0__6.0__ add__5.0__2.0__ |
| in a renowned city the average birth rate is num__9 people every two seconds and the death rate is num__3 people every two seconds . estimate the size of the population net increase that occurs in one day . <o> a ) num__32300 <o> b ) num__172800 <o> c ) num__468830 <o> d ) num__338200 <o> e ) num__259 |
200 |
every num__2 seconds num__6 persons are added ( num__9 - num__3 ) . every second num__3 persons are added . in a day num__24 hrs = num__24 * num__60 minutes = num__24 * num__60 * num__60 = num__86400 seconds . num__86400 * num__3 = num__259200 option e <eor> e <eos> |
e |
e |
| a dishonest dealer professes to sell goods at the cost price but uses a weight of num__950 grams per kg what is his percent ? <o> a ) num__2.0 <o> b ) num__25.0 <o> c ) num__5.26 <o> d ) num__29.0 <o> e ) num__45 % |
num__950 - - - num__50 num__100 - - - ? = > num__5.26 answer : c <eor> c <eos> |
c |
percent__100.0__5.26__ |
percent__100.0__5.26__ |
| by selling an article at rs . num__800 a shopkeeper makes a profit of num__25.0 . at what price should he sell the article so as to make a loss of num__25.0 ? <o> a ) num__338 <o> b ) num__480 <o> c ) num__888 <o> d ) num__266 <o> e ) num__281 |
sp = num__800 profit = num__25.0 cp = ( sp ) * [ num__100 / ( num__100 + p ) ] = num__800 * [ num__0.8 ] = num__640 loss = num__25.0 = num__25.0 of num__640 = rs . num__160 sp = cp - loss = num__640 - num__160 = rs . num__480 answer : b <eor> b <eos> |
b |
percent__25.0__640.0__ percent__100.0__480.0__ |
percent__25.0__640.0__ percent__100.0__480.0__ |
| what is the least possible value of expression e = ( x - num__1 ) ( x - num__3 ) ( x - num__4 ) ( x - num__6 ) + num__10 for real values of x ? <o> a ) num__1 <o> b ) num__10 <o> c ) num__9 <o> d ) num__0 <o> e ) num__8 |
explanation : e = ( x - num__1 ) ( x - num__6 ) ( x - num__3 ) ( x - num__4 ) + num__10 e = ( x num__2 - num__7 x + num__6 ) ( x num__2 - num__7 x + num__12 ) + num__10 let x num__2 - num__7 x + num__6 = y e = y num__2 + num__6 y + num__10 e = ( y + num__3 ) num__2 + num__1 minimum value = num__1 when y = - num__3 answer : a <eor> a <eos> |
a |
subtract__3.0__1.0__ add__1.0__6.0__ multiply__3.0__4.0__ reverse__1.0__ |
subtract__3.0__1.0__ add__1.0__6.0__ add__10.0__2.0__ subtract__3.0__2.0__ |
| there are a lot of houses such that the numbers of their doorplates are odd numbers and the first number of the doorplates is num__545 the last number of the doorplates is num__805 . how many houses are there ? <o> a ) num__125 <o> b ) num__111 <o> c ) num__101 <o> d ) num__121 <o> e ) num__131 |
so it starts from num__545 and goes like num__547 num__549 . . . . . . num__705 . and both first and last number are inclusive . since every other number is odd it ' s just num__0.5 of the numbers and since it starts with an odd and ends with an odd inclusive add one to the result . i . e . ( [ num__805 - num__545 ] [ / num__2 ] + num__1 = num__131 ans e <eor> e <eos> |
e |
reverse__0.5__ multiply__0.5__2.0__ multiply__1.0__131.0__ |
reverse__0.5__ multiply__0.5__2.0__ divide__131.0__1.0__ |
| a can run a kilometer race in num__4 num__0.5 min while b can run same race in num__5 min . how many meters start can a give b in a kilometer race so that the race mat end in a dead heat ? <o> a ) num__177 m <o> b ) num__786 m <o> c ) num__272 m <o> d ) num__100 m <o> e ) num__127 m |
a can give b ( num__5 min - num__4 num__0.5 min ) = num__30 sec start . the distance covered by b in num__5 min = num__1000 m . distance covered in num__30 sec = ( num__1000 * num__30 ) / num__300 = num__100 m . a can give b num__100 m start . answer : d <eor> d <eos> |
d |
round__100.0__ |
round__100.0__ |
| a certain company that sells only cars and trucks reported that revenues from car sales in num__1997 were down num__11 percent from num__1996 and revenues from truck sales in num__1997 were up num__7 percent from num__1996 . if total revenues from car sales and truck sales in num__1997 were up num__1 percent from num__1996 what is the ratio of revenue from car sales in num__1996 to revenue from truck sales in num__1996 ? <o> a ) num__1 : num__2 <o> b ) num__4 : num__5 <o> c ) num__1 : num__1 <o> d ) num__3 : num__2 <o> e ) num__5 : num__3 |
let c = revenue from car sales in num__1996 let t = revenue from truck sales in num__1996 equating the total revenue in num__1997 with the individual revenue from selling cars and trucks in num__1997 c ( num__0.89 ) + t ( num__1.07 ) = ( c + t ) ( num__1.01 ) dividing throughout by t ( c / t ) ( num__0.89 ) + num__1.07 = ( c / t + num__1 ) ( num__1.01 ) = > ( c / t ) ( num__0.12 ) = num__0.06 = > ( c / t ) = num__0.5 option ( a ) . <eor> a <eos> |
a |
subtract__1.01__0.89__ subtract__1.07__1.01__ divide__0.06__0.12__ reverse__1.0__ |
subtract__1.01__0.89__ subtract__1.07__1.01__ divide__0.06__0.12__ reverse__1.0__ |
| find the area of a parallelogram with base num__36 cm and height num__24 cm ? <o> a ) num__760 <o> b ) num__284 <o> c ) num__288 <o> d ) num__864 <o> e ) num__820 |
area of a parallelogram = base * height = num__36 * num__24 = num__864 cm num__2 answer : d <eor> d <eos> |
d |
multiply__36.0__24.0__ multiply__36.0__24.0__ |
multiply__36.0__24.0__ multiply__36.0__24.0__ |
| shreehari has num__500 chocolates in his self . he eats num__10.0 of the chocolates per month . how many chocolates will he have after num__2 months ? <o> a ) num__450 <o> b ) num__400 <o> c ) num__405 <o> d ) num__410 <o> e ) num__425 |
formula : ( after = num__100 denominator ago = num__100 numerator ) num__500 Ã — num__0.9 Ã — num__0.9 = num__405 c <eor> c <eos> |
c |
percent__100.0__405.0__ |
percent__100.0__405.0__ |
| a shopkeeper labeled the price of his articles so as to earn a profit of num__20.0 on the cost price . he then sold the articles by offering a discount of num__5.0 on the labeled price . what is the actual percent profit earned in the deal ? <o> a ) num__24.0 <o> b ) num__20.0 <o> c ) num__17.0 <o> d ) num__18.0 <o> e ) none of these |
explanation : let the cp of the article = rs . num__100 . then labeled price = rs . num__120 . sp = rs . num__120 - num__5.0 of num__120 = rs . num__120 - num__6 = rs . num__124 . gain = rs . num__124 â € “ rs . num__100 = rs . num__24 therefore gain / profit percent = num__24.0 . answer : option a <eor> a <eos> |
a |
percent__5.0__120.0__ percent__20.0__120.0__ percent__20.0__120.0__ |
percent__5.0__120.0__ percent__20.0__120.0__ percent__20.0__120.0__ |
| a set consist of num__2 l - num__1 element . what is the number of subsets of this set which contain at most l - num__1 elements ? <o> a ) num__2 ^ ( num__2 l - num__2 ) <o> b ) num__2 ^ ( num__2 l ) - num__2 <o> c ) num__2 ^ ( num__2 l ) - num__1 <o> d ) num__2 ^ ( num__2 l ) <o> e ) num__2 ^ ( num__2 l - num__1 ) |
i used l = num__3 so then we have num__5 ! / num__2 ! num__3 ! + num__5 ! / num__4 ! num__1 ! + num__5 ! / num__0 ! num__5 ! num__10 + num__5 + num__1 = num__16 so our target is num__16 now replace in answer choices a gives us num__2 ^ num__4 = num__16 hence a is the correct option read carefully it says at most so keep in mind that picking a small number such as num__3 will help you save time since you have to list fewer outcomes avoid num__2 since you will get num__1 arrangement ( l - num__1 ) and may be risky since num__1 is a number with certain unique properties <eor> a <eos> |
a |
add__2.0__1.0__ add__2.0__3.0__ add__1.0__3.0__ multiply__2.0__5.0__ multiply__2.0__1.0__ |
add__2.0__1.0__ add__2.0__3.0__ add__1.0__3.0__ multiply__2.0__5.0__ divide__2.0__1.0__ |
| which of these must the factor q of the product of four consecutive even integers : - num__1 ) num__48 num__2 ) num__64 num__3 ) num__96 num__4 ) num__192 num__5 ) num__80 <o> a ) num__12 only <o> b ) q = num__23 only <o> c ) q = num__1 num__23 only <o> d ) q = num__12 num__34 only <o> e ) all of them |
let the four variables be a < b < c < d . assume a worst case scenario where the a equal to a prime number ( hence odd ) . therefore a = divisible by num__1 b = a + num__1 ( divisible by num__2 ) c = a + num__3 ( divisible by num__3 ) d = a + num__4 ( divisible by num__4 ) therefore each answer choice must be divisible q by num__2 x num__3 x num__4 = num__24 only num__80 is not divisible . therefore answer = d <eor> d <eos> |
d |
divide__48.0__2.0__ divide__48.0__4.0__ |
divide__48.0__2.0__ divide__48.0__4.0__ |
| if the rectangular screen of a computer monitor is designed to have a length of x inches a perimeter of p inches and an area of a square inches which of these equations must be true ? <o> a ) x ^ num__2 + px + a = num__0 <o> b ) x ^ num__2 - px - num__2 a = num__0 <o> c ) num__2 x ^ num__2 + px + num__2 a = num__0 <o> d ) num__2 x ^ num__2 - px - num__2 a = num__0 <o> e ) num__2 x ^ num__2 - px + num__2 a = num__0 |
notice that we can discard options a and c right away . the sum of num__3 positive values can not be num__0 . now assume : length = x = num__1 inchand width = num__1 inch ; perimeter = p = num__4 inches ; area = a = num__1 square inches . plug the values of x p and a into the answer choices : only for e num__2 x ^ num__2 - px + num__2 a = num__2 - num__4 + num__2 = num__0 . answer : e . <eor> e <eos> |
e |
square_perimeter__1.0__ rectangle_perimeter__0.0__1.0__ rectangle_perimeter__0.0__1.0__ |
square_perimeter__1.0__ rectangle_perimeter__0.0__1.0__ power__2.0__1.0__ |
| a family pays $ num__800 per year for an insurance plan that pays num__70 percent of the first $ num__1000 in expenses and num__100 percent of all medical expenses thereafter . in any given year the total amount paid by the family will equal the amount paid by the plan when the family ' s medical expenses total how much ? <o> a ) $ num__1000 <o> b ) $ num__1200 <o> c ) $ num__1400 <o> d ) $ num__1800 <o> e ) $ num__2 |
200 |
assuming the medical expenses are $ num__1000 or more the family pays $ num__800 + $ num__300 = $ num__1100 . the total amount paid by insurance plan for the first $ num__1000 of expenses is $ num__700 . the insurance will pay another $ num__400 when the medical expenses are $ num__1400 . the answer is c . <eor> c <eos> |
c |
c |
| if samaira wallet have num__10 rs . note num__50 rs . note num__2000 rs . note . if num__10 rs . is num__0.6 part of his money and num__50 rs . is num__0.2 part of his money . if he has num__8000 money consist of num__2000 rs . calculate total money in samaira wallet ? <o> a ) num__18000 <o> b ) num__32000 <o> c ) num__30000 <o> d ) num__40000 <o> e ) num__35000 |
if samaira have total money is x . part of money in num__2000 rs . = ( num__1 - ( num__0.6 + num__0.2 ) ) = num__1 - num__0.8 = num__0.2 x * num__0.2 = num__8000 x = num__40000 rs . answer d <eor> d <eos> |
d |
add__0.6__0.2__ divide__8000.0__0.2__ divide__8000.0__0.2__ |
add__0.6__0.2__ divide__8000.0__0.2__ multiply__1.0__40000.0__ |
| a five - digit number divisible by num__3 is to be formed using numerical num__0 num__1 num__2 num__3 num__4 and num__5 without repetition . the total number q of ways this can be done is : <o> a ) num__122 <o> b ) num__210 <o> c ) num__216 <o> d ) num__217 <o> e ) num__225 |
we should determine which num__5 digits from given num__6 would form the num__5 digit number divisible by num__3 . we have six digits : num__0 num__1 num__2 num__3 num__4 num__5 . their sum = num__15 . for a number to be divisible by num__3 the sum of the digits must be divisible by num__3 . as the sum of the six given numbers is num__15 ( divisible by num__3 ) only num__5 digits good to form our num__5 digit number would be num__15 - num__0 = { num__1 num__2 num__3 num__4 num__5 } and num__15 - num__3 = { num__0 num__1 num__2 num__4 num__5 } . meaning that no other num__5 from given six will total the number divisible by num__3 . second step : we have two set of numbers : num__1 num__2 num__3 num__4 num__5 and num__0 num__1 num__2 num__4 num__5 . how many num__5 digit numbers can be formed using these two sets : num__1 num__2 num__3 num__4 num__5 - - > num__5 ! as any combination of these digits would give us num__5 digit number divisible by num__3 . num__5 ! = num__120 . num__0 num__1 num__2 num__4 num__5 - - > here we can not use num__0 as the first digit otherwise number wo n ' t be any more num__5 digit and become num__4 digit . so desired # would be total combinations num__5 ! minus combinations with num__0 as the first digit ( combination of num__4 ) num__4 ! - - > num__5 ! - num__4 ! = num__4 ! ( num__5 - num__1 ) = num__4 ! * num__4 = num__96 num__120 + num__96 = num__216 = q answer : c . <eor> c <eos> |
c |
multiply__3.0__2.0__ multiply__3.0__5.0__ add__96.0__120.0__ multiply__1.0__216.0__ |
multiply__3.0__2.0__ multiply__3.0__5.0__ add__96.0__120.0__ multiply__1.0__216.0__ |
| a corporation paid $ num__7 million in federal taxes on its first $ num__50 million of gross profits and then $ num__30 million in federal taxes on the next $ num__150 million in gross profits . by approximately what percent did the ratio of federal taxes to gross profits increase from the first $ num__50 million in profits to the next $ num__150 million in profits ? <o> a ) num__6.0 <o> b ) num__14.0 <o> c ) num__20.0 <o> d ) num__23.0 <o> e ) num__43 % |
initial ratio of federal taxes to gross profits : num__0.14 = num__0.14 final ratio : num__0.2 = num__0.2 thus the percentage change : ( num__0.2 - num__0.14 ) / num__0.14 * num__100 = num__0.428571428571 * num__100 = num__0.428571428571 * num__100 = just less than num__50.0 . answer : e <eor> e <eos> |
e |
divide__7.0__50.0__ divide__30.0__150.0__ subtract__150.0__50.0__ subtract__50.0__7.0__ |
divide__7.0__50.0__ divide__30.0__150.0__ subtract__150.0__50.0__ subtract__50.0__7.0__ |
| two airplanes one from atlanta to new york and the other from new york to atlanta start simultaneously . after they meet the airplanes reach their destinations after num__16 hours and num__4 hours respectively the ratio of their speeds is <o> a ) num__2 : num__3 <o> b ) num__4 : num__3 <o> c ) num__4 : num__2 <o> d ) num__1 : num__2 <o> e ) none |
solution let us name the planes as a and b . = ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š num__16 : â ˆ š num__4 = num__4 : num__2 answer c <eor> c <eos> |
c |
round__4.0__ |
round__4.0__ |
| if the roots of a quadratic equation are num__20 and - num__7 then find the equation ? <o> a ) x num__2 - x - num__15 = num__0 <o> b ) x num__2 - num__12 x - num__50 = num__0 <o> c ) x num__2 - num__13 x - num__140 = num__0 <o> d ) x num__2 - num__4 x - num__40 = num__0 <o> e ) x num__2 - num__23 x - num__150 = num__0 |
explanation : any quadratic equation is of the form x num__2 - ( sum of the roots ) x + ( product of the roots ) = num__0 - - - - ( num__1 ) where x is a real variable . as sum of the roots is num__13 and product of the roots is - num__140 the quadratic equation with roots as num__20 and - num__7 is : x num__2 - num__13 x - num__140 = num__0 . answer c <eor> c <eos> |
c |
subtract__20.0__7.0__ multiply__20.0__7.0__ multiply__1.0__2.0__ |
subtract__20.0__7.0__ multiply__20.0__7.0__ multiply__1.0__2.0__ |
| rectangle a has sides a and b and rectangle b has sides c and d . if a / c = b / d = num__0.6 what is the ratio of rectangle a ’ s area to rectangle b ’ s area ? <o> a ) num__1.66666666667 <o> b ) num__1.8 <o> c ) num__0.36 <o> d ) num__0.6 <o> e ) num__2.77777777778 |
the area of rectangle a is ab . c = num__5 a / num__3 and d = num__5 b / num__3 . the area of rectangle b is cd = num__25 ab / num__9 . the ratio of rectangle a ' s area to rectangle b ' s area is ab / ( num__25 ab / num__9 ) = num__0.36 . the answer is c . <eor> c <eos> |
c |
multiply__0.6__5.0__ divide__9.0__25.0__ divide__9.0__25.0__ |
multiply__0.6__5.0__ divide__9.0__25.0__ divide__9.0__25.0__ |
| at what rate percent of simple interest will a sum of money double itself in num__12 years ? <o> a ) num__8 num__0.125 <o> b ) num__8 num__0.333333333333 <o> c ) num__8 num__1 / num__0 <o> d ) num__8 num__0.5 <o> e ) num__8 num__1.0 |
let sum = x . then s . i . = x . rate = ( num__100 * s . i . ) / ( p * t ) = ( num__100 * x ) / ( x * num__12 ) = num__8.33333333333 = num__8 num__0.333333333333 % answer : b <eor> b <eos> |
b |
percent__100.0__8.0__ |
percent__100.0__8.0__ |
| let a be the event that a randomly selected two digit number is divisible by num__10 and let b be the event that a randomly selected two digit number is divisible by num__5 . what is p ( a and b ) ? <o> a ) num__0.02 <o> b ) num__0.0333333333333 <o> c ) num__0.0666666666667 <o> d ) num__0.1 <o> e ) num__0.2 |
p ( a and b ) = num__0.1 * num__0.2 = num__0.02 the answer is a . <eor> a <eos> |
a |
divide__0.2__10.0__ round__0.02__ |
multiply__0.2__0.1__ multiply__0.2__0.1__ |
| out of num__15 students in a class num__9 are wearing blue shirts num__3 are wearing green shirts and num__3 are wearing red shirts . four students are to be selected at random . what is the probability that at least one is wearing a green shirt ? <o> a ) num__0.459016393443 <o> b ) num__0.535211267606 <o> c ) num__0.592592592593 <o> d ) num__0.637362637363 <o> e ) num__0.673267326733 |
total possible ways to choose num__4 students out of num__15 = num__15 c num__4 = num__1365 the number of ways to choose num__4 students with no green shirts = num__12 c num__4 = num__495 p ( no green shirts ) = num__0.362637362637 = num__0.362637362637 p ( at least num__1 green shirt ) = num__1 - num__0.362637362637 = num__0.637362637363 the answer is d . <eor> d <eos> |
d |
negate_prob__0.3626__ negate_prob__0.3626__ |
negate_prob__0.3626__ negate_prob__0.3626__ |
| find the simple interest on rs . num__68000 at num__16 num__0.666666666667 % per annum for num__9 months . <o> a ) rs . num__8500 <o> b ) rs . num__8000 <o> c ) rs . num__7500 <o> d ) rs . num__7000 <o> e ) rs . num__6500 |
p = rs . num__68000 r = num__16.6666666667 % p . a and t = num__0.75 years = num__0.75 years . s . i . = ( p * r * t ) / num__100 = rs . ( num__68000 * ( num__16.6666666667 ) * ( num__0.75 ) * ( num__0.01 ) ) = rs . num__8500 answer is a . <eor> a <eos> |
a |
percent__100.0__8500.0__ |
percent__100.0__8500.0__ |
| mike took a bus from home to market that travels at num__30 kmph . while walking back at num__6 kmph halfway through he suddenly realized he was getting late and he cycled back the remaining distance in num__20 kmph . find the average speed . <o> a ) num__21.2 kmph <o> b ) num__12.1 kmph <o> c ) num__23.4 kmph <o> d ) num__20.4 kmph <o> e ) none of these |
let the distance be num__2 x ( one way ) time taken by bus = num__2 x / num__30 by walking = x / num__6 by cycling = x / num__20 hours : . average speed = total distance / total time = num__6 x / x / num__15 + x / num__6 + x / num__20 = num__6 * num__15.0 + num__10 + num__3 = num__21.2 answer : a <eor> a <eos> |
a |
divide__30.0__2.0__ subtract__30.0__20.0__ divide__30.0__10.0__ round__21.2__ |
divide__30.0__2.0__ divide__20.0__2.0__ divide__30.0__10.0__ round__21.2__ |
| if the speed of x meters per hour is equivalent to the speed of y kilometers per hour what is y in terms of x ? ( num__1 kilometer = num__1000 meters ) <o> a ) num__0.1 x <o> b ) num__0.01 x <o> c ) num__0.001 x <o> d ) num__0.0001 x <o> e ) num__0.005 x |
x meters per hour - - > - - > num__1 x meters per hour ( as there are num__3600 seconds in one hour ) ; - - > x / num__1000 = num__0.001 x kilometers per hour ( as there are num__1000 meters in one kilometer ) . answer : c . <eor> c <eos> |
c |
divide__1.0__1000.0__ divide__1.0__1000.0__ |
divide__1.0__1000.0__ divide__1.0__1000.0__ |
| a certain car increased its average speed by num__5 miles per hour in each successive num__5 - minute interval after the first interval . if in the first num__5 - minute interval its average speed was num__2 miles per hour how many miles did the car travel in the third num__5 - minute interval ? <o> a ) num__1.0 <o> b ) num__1.5 <o> c ) num__2.0 <o> d ) num__2.5 <o> e ) num__3.0 |
in the third time interval the average speed of the car was num__2 + num__5 + num__5 = num__12 miles per hour ; in num__5 minutes ( num__0.0833333333333 hour ) at that speed car would travel num__12 * num__0.0833333333333 = num__1 miles . answer : a . <eor> a <eos> |
a |
round__1.0__ |
round__1.0__ |
| the slant height of a right circular cone is num__10 m and its height is num__8 m . find the area of its curved surface . <o> a ) num__30 m num__2 <o> b ) num__40 m num__2 <o> c ) num__50 m num__2 <o> d ) num__60 m num__2 <o> e ) num__80 m num__2 |
l = num__10 m h = num__8 m . so r = l num__2 - h num__2 = ( num__10 ) num__2 - num__82 = num__6 m . curved surface area = rl = ( x num__6 x num__10 ) m num__2 = num__60 m num__2 . answer : option d <eor> d <eos> |
d |
side_by_diagonal__10.0__8.0__ multiply__10.0__6.0__ multiply__10.0__6.0__ |
side_by_diagonal__10.0__8.0__ multiply__10.0__6.0__ multiply__10.0__6.0__ |
| x and y are positive integers of v . if num__1 / x + num__1 / y < num__2 which of the following must be true ? <o> a ) x + y > num__4 <o> b ) xy > num__1 <o> c ) x / y + y / x < num__1 <o> d ) ( x - y ) ^ num__2 > num__0 <o> e ) none of the above |
answer is b : num__1 / x + num__1 / y < num__2 the maximum value of num__1 / x is num__1 because if x equals any other number greater than one it will be a fraction . the same is true with num__1 / y . so num__1 / x and num__1 / y will always be less than num__2 as long as both x and y are not both equal to one at the same time . another way of putting it is : x * y > num__1 . b <eor> b <eos> |
b |
reverse__1.0__ |
reverse__1.0__ |
| what is the sum of num__100 consecutive integers from - num__49 inclusive in a increasing order ? <o> a ) - num__29 <o> b ) num__50 <o> c ) - num__30 <o> d ) num__30 <o> e ) num__60 |
from - num__49 to - num__1 - - > num__49 nos . zero - - > num__1 number from + num__1 to + num__49 - - > num__49 nos . when we add up nos . from - num__49 to + num__49 sum will be zero . total num__99 nos will be added . num__100 th number will be num__50 . sum of these num__100 nos . = num__50 . b is the answer . <eor> b <eos> |
b |
subtract__100.0__1.0__ add__49.0__1.0__ subtract__100.0__50.0__ |
subtract__100.0__1.0__ add__49.0__1.0__ subtract__100.0__50.0__ |
| we bought num__85 hats at the store . blue hats cost $ num__6 and green hats cost $ num__7 . the total price was $ num__550 . how many green hats did we buy ? <o> a ) a ) num__36 <o> b ) b ) num__40 <o> c ) c ) num__41 <o> d ) d ) num__42 <o> e ) e ) num__44 |
let b be the number of blue hats and let g be the number of green hats . b + g = num__85 . b = num__85 - g . num__6 b + num__7 g = num__550 . num__6 ( num__85 - g ) + num__7 g = num__550 . num__510 - num__6 g + num__7 g = num__550 . g = num__550 - num__510 = num__40 . the answer is b . <eor> b <eos> |
b |
multiply__85.0__6.0__ subtract__550.0__510.0__ subtract__550.0__510.0__ |
multiply__85.0__6.0__ subtract__550.0__510.0__ subtract__550.0__510.0__ |
| a vessel is filled with liquid num__3 parts of which are water and num__5 parts of syrup . how much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup ? <o> a ) num__0.333333333333 <o> b ) num__0.25 <o> c ) num__0.2 <o> d ) num__0.142857142857 <o> e ) none |
suppose the vessel initially contains num__8 litres of liquid . let x littres of this liquid be replaced with water . quantity of water in new mixture = ( num__3 - num__3 x / num__8 + x ) litres . quantity of syrup in new mixture = ( num__5 - num__5 x / num__8 ) litres . ( num__3 - num__3 x / num__8 + x ) = ( num__5 - num__5 x / num__8 ) = num__5 x + num__24 = num__40 - num__5 x = › num__10 x = num__16 = › x = num__1.6 so part of the mixture replaced = ( num__1.6 x num__0.125 ) = num__0.2 . answer c <eor> c <eos> |
c |
add__3.0__5.0__ multiply__3.0__8.0__ multiply__5.0__8.0__ subtract__40.0__24.0__ divide__8.0__5.0__ reverse__8.0__ reverse__5.0__ reverse__5.0__ |
add__3.0__5.0__ multiply__3.0__8.0__ multiply__5.0__8.0__ subtract__40.0__24.0__ divide__8.0__5.0__ reverse__8.0__ reverse__5.0__ reverse__5.0__ |
| a fruit seller had some apples . he sells num__40.0 apples and still has num__420 apples . originally he had : <o> a ) num__700 <o> b ) num__680 <o> c ) num__720 <o> d ) num__760 <o> e ) num__660 |
apples sold = num__40.0 so remaining apples = num__60.0 num__60.0 = num__420 full ( num__100.0 ) = ( num__1.66666666667 ) x num__420 = num__700 answer : a <eor> a <eos> |
a |
percent__100.0__700.0__ |
percent__100.0__700.0__ |
| suppose you want to buy three loaves of bread that cost $ num__0.75 each and a jar of peanut butter that costs $ num__7 . a jar of jelly is $ num__2.75 but you don â € ™ t need any jelly . you have $ num__17 . how much money will you have left over ? <o> a ) $ num__7.75 <o> b ) $ num__7.50 <o> c ) $ num__3.50 <o> d ) $ num__4.50 <o> e ) $ num__5.50 |
the jelly is extra information . num__17.00 â € “ num__3 x num__0.75 â € “ num__7.00 = num__17.00 â € “ num__2.25 â € “ num__7.00 = num__7.75 . you have $ num__7.75 left . correct answer a <eor> a <eos> |
a |
multiply__0.75__3.0__ add__0.75__7.0__ add__0.75__7.0__ |
multiply__0.75__3.0__ add__0.75__7.0__ add__0.75__7.0__ |
| nobel buys an old car for $ num__1000 and spends $ num__1000 on its repairs . if he sells the scooter for $ num__2500 his gain percent is ? <o> a ) num__10.0 <o> b ) num__15.0 <o> c ) num__25.0 <o> d ) num__20.0 <o> e ) num__30 % |
c . p . = num__1000 + num__1000 = $ num__2000 s . p . = $ num__2500 gain = num__2500 - num__2000 = $ num__500 gain % = num__0.25 * num__100 = num__25.0 answer is c <eor> c <eos> |
c |
percent__25.0__100.0__ |
percent__25.0__100.0__ |
| a sum of rs . num__2665 is lent into two parts so that the interest on the first part for num__5 years at num__3.0 per annum may be equal to the interest on the second part for num__3 years at num__5.0 per annum . find the second sum ? <o> a ) rs . num__1178.55 <o> b ) rs . num__1978.25 <o> c ) rs . num__1332.5 <o> d ) rs . num__1678 <o> e ) rs . num__1675.55 |
( x * num__5 * num__3 ) / num__100 = ( ( num__2665 - x ) * num__3 * num__5 ) / num__100 num__15 x / num__100 = num__399.75 - num__15 x / num__100 num__30 x = num__39975 = > x = num__1332.5 second sum = num__2665 – num__1025 = num__1332.5 answer : c <eor> c <eos> |
c |
percent__15.0__2665.0__ percent__100.0__1332.5__ |
percent__15.0__2665.0__ percent__100.0__1332.5__ |
| what is the smallest number which when diminished by num__10 is divisible num__2 num__6 num__12 and num__24 ? <o> a ) num__35 <o> b ) num__34 <o> c ) num__20 <o> d ) num__35 <o> e ) num__342 |
required number = ( lcm of num__2 num__6 num__12 and num__24 ) + num__12 = num__24 + num__10 = num__34 option b <eor> b <eos> |
b |
add__10.0__24.0__ add__10.0__24.0__ |
add__10.0__24.0__ add__10.0__24.0__ |
| a train with a length of num__100 meters is traveling at a speed of num__72 km / hr . the train enters a tunnel num__2.3 km long . how many minutes does it take the train to pass through the tunnel from the moment the front enters to the moment the rear emerges ? <o> a ) num__2.0 <o> b ) num__2.5 <o> c ) num__3.0 <o> d ) num__3.5 <o> e ) num__4.0 |
num__72 km / hr = num__1.2 km / min the total distance is num__2.4 km . num__2.4 / num__1.2 = num__2 minutes the answer is a . <eor> a <eos> |
a |
divide__2.4__1.2__ round__2.0__ |
divide__2.4__1.2__ divide__2.4__1.2__ |
| in a certain school the ratio of boys to girls is num__5 to num__13 . if there are num__128 more girls than boys how many boys are there ? <o> a ) num__27 <o> b ) num__36 <o> c ) num__45 <o> d ) num__72 <o> e ) num__80 |
the ratio of b to g is num__5 : num__13 and the other data point is g are more than boys by num__128 . . . looking at the ratio we can say that the num__8 ( num__13 - num__5 ) extra parts caused this diff of num__128 . so num__1 part corresponds to num__16.0 = num__16 and so num__5 parts correspond to num__5 * num__16 = num__80 . e <eor> e <eos> |
e |
subtract__13.0__5.0__ divide__128.0__8.0__ multiply__5.0__16.0__ multiply__5.0__16.0__ |
subtract__13.0__5.0__ divide__128.0__8.0__ multiply__5.0__16.0__ multiply__5.0__16.0__ |
| how much time will a train of length num__200 m moving at a speed of num__72 kmph take to cross another train of length num__300 m moving at num__36 kmph in the same direction ? <o> a ) num__50 <o> b ) num__88 <o> c ) num__77 <o> d ) num__55 <o> e ) num__22 |
the distance to be covered = sum of their lengths = num__200 + num__300 = num__500 m . relative speed = num__72 - num__36 = num__36 kmph = num__36 * num__0.277777777778 = num__10 mps . time required = d / s = num__50.0 = num__50 sec . answer : a <eor> a <eos> |
a |
add__200.0__300.0__ divide__500.0__10.0__ round__50.0__ |
add__200.0__300.0__ divide__500.0__10.0__ divide__500.0__10.0__ |
| find the largest num__4 digit number which isexactly divisible by num__88 ? <o> a ) num__7890 <o> b ) num__8900 <o> c ) num__9944 <o> d ) num__9976 <o> e ) num__10000 |
largest num__4 digit number is num__9999 after doing num__9999 ÷ num__88 we get remainder num__55 hence largest num__4 digit number exactly divisible by num__88 = num__9999 - num__55 = num__9944 c <eor> c <eos> |
c |
subtract__9999.0__55.0__ subtract__9999.0__55.0__ |
subtract__9999.0__55.0__ subtract__9999.0__55.0__ |
| num__2056 x num__987 = ? <o> a ) num__1936372 <o> b ) num__2029272 <o> c ) num__1896172 <o> d ) num__1926172 <o> e ) num__1956231 |
answer : option b num__2056 x num__987 = num__2056 x ( num__1000 - num__13 ) = num__2056 x num__1000 - num__2056 x num__13 = num__2056000 - num__26728 = num__2029272 . <eor> b <eos> |
b |
subtract__1000.0__987.0__ multiply__2056.0__1000.0__ multiply__2056.0__13.0__ multiply__2056.0__987.0__ multiply__2056.0__987.0__ |
subtract__1000.0__987.0__ multiply__2056.0__1000.0__ multiply__2056.0__13.0__ subtract__2056000.0__26728.0__ subtract__2056000.0__26728.0__ |
| a watch was sold at a loss of num__46.0 . if it was sold for rs . num__140 more there would have been a gain of num__4.0 . what is the cost price ? <o> a ) num__280 <o> b ) num__288 <o> c ) num__279 <o> d ) num__277 <o> e ) num__290 |
num__54.0 num__104.0 - - - - - - - - num__50.0 - - - - num__140 num__100.0 - - - - ? = > rs . num__280 answer : a <eor> a <eos> |
a |
percent__100.0__280.0__ |
percent__100.0__280.0__ |
| what sum of money put at c . i amounts in num__2 years to rs . num__8820 and in num__3 years to rs . num__9261 ? <o> a ) num__8000 <o> b ) num__8020 <o> c ) num__2879 <o> d ) num__2686 <o> e ) num__2996 |
num__8820 - - - - num__441 num__100 - - - - ? = > num__5.0 x * num__1.05 * num__1.05 = num__8820 x * num__1.1025 = num__8820 x = num__8820 / num__1.1025 = > num__8000 answer : a <eor> a <eos> |
a |
subtract__9261.0__8820.0__ add__2.0__3.0__ divide__9261.0__8820.0__ divide__8820.0__1.1025__ divide__8820.0__1.1025__ |
subtract__9261.0__8820.0__ add__2.0__3.0__ divide__9261.0__8820.0__ divide__8820.0__1.1025__ divide__8820.0__1.1025__ |
| two taps a and b can fill a tank in num__5 hours and num__20 hours respectively . if both the taps are open then due to a leakage it took num__30 minutes more to fill the tank . if the tank is full how long will it take for the leakage alone to empty the tank ? <o> a ) num__24 hr <o> b ) num__36 hr <o> c ) num__18 hr <o> d ) num__42 hr <o> e ) num__30 hr |
part filled by a + b in num__1 hour = num__0.2 + num__0.05 = num__0.25 a and b together can fill the tank in num__4 hour work done by the leak in num__1 hour = num__0.25 - num__0.222222222222 = num__0.0277777777778 leak will empty the tank in num__36 hours . answer is b <eor> b <eos> |
b |
divide__1.0__5.0__ divide__1.0__20.0__ divide__5.0__20.0__ subtract__5.0__1.0__ subtract__0.25__0.2222__ round__36.0__ |
divide__1.0__5.0__ divide__1.0__20.0__ add__0.2__0.05__ subtract__5.0__1.0__ subtract__0.25__0.2222__ round__36.0__ |
| peter will arrange num__6 people of num__6 different heights for photograph by placing them in two rows of three so that each person in the first row is standing is standing in front of someone in the second row . the heights of the people within each row must increase from left to right and each person in the second row must be taller than the person standing in front of him or her . how many such arrangements of the num__6 people are possible ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__9 <o> d ) num__24 <o> e ) num__26 |
peter will arrange num__6 people of num__6 different heights for photograph by placing them in two rows of three so that each person in the first row is standing is standing in front of someone in the second row . person with max height is in the second row person with min height is in the first row . we need to select num__1 person in the middle of each row . . . in the middle of the first row we can put only num__2 num__3 or num__4 . in the middle of the second row we can put only num__3 num__4 num__5 . if we select { num__3 in the sec . row num__2 in the first } { num__42 } { num__52 } { num__43 } { num__53 } { num__54 } . so there are num__0 * num__1 + num__1 * num__1 + num__2 * num__1 + num__1 * num__1 + num__1 * num__1 + num__0 * num__1 = num__5 cases . . . a <eor> a <eos> |
a |
coin_space__ vowel_space__ card_space__ negate_prob__1.0__ vowel_space__ |
coin_space__ vowel_space__ card_space__ negate_prob__1.0__ vowel_space__ |
| if k > num__0 x + num__2 k = y and y + num__3 k = z what is the ratio between z - x and y - x ? <o> a ) num__3 to num__1 <o> b ) num__3 to num__2 <o> c ) num__4 to num__1 <o> d ) num__5 to num__1 <o> e ) num__5 to num__2 |
y - x = num__2 k z - y = num__3 k let ' s add the two equations together : z - x = num__5 k the ratio of z - x to y - x is num__5 : num__2 . the answer is e . <eor> e <eos> |
e |
add__2.0__3.0__ add__2.0__3.0__ |
add__2.0__3.0__ add__2.0__3.0__ |
| a man can row with a speed of num__5 kmph in still water . if the stream flows at num__5 kmph then the speed in downstream is ? <o> a ) a ) num__63 kmph <o> b ) b ) num__63 kmph <o> c ) c ) num__10 kmph <o> d ) d ) num__62 kmph <o> e ) e ) num__74 kmph |
explanation : m = num__5 s = num__5 ds = num__5 + num__5 = num__10 answer : option c <eor> c <eos> |
c |
round__10.0__ |
round__10.0__ |
| a train num__640 meters long is running with a speed of num__64 kmph . the time taken by it to cross a tunnel num__140 meters long is ? <o> a ) num__44 sec <o> b ) num__49 sec <o> c ) num__48 sec <o> d ) num__16 sec <o> e ) num__17 sec |
d = num__640 + num__140 = num__780 s = num__64 * num__0.277777777778 = num__17.7 mps t = num__780 / num__17.7 = num__44 sec answer : a <eor> a <eos> |
a |
add__640.0__140.0__ round__44.0__ |
add__640.0__140.0__ round__44.0__ |
| for every order a certain mail - order company charges a shipping fee of $ num__3 plus an additional $ num__2 if the value of the order is over $ num__50 but not over $ num__100 or an additional $ num__3 if the value of the order is over $ num__100 . how much greater are the total shipping fees for num__2 orders of $ num__75 each than the total shipping fee for num__1 order of $ num__120 ? <o> a ) $ num__1 <o> b ) $ num__2 <o> c ) $ num__3 <o> d ) $ num__4 <o> e ) $ num__5 |
total cost involved for each num__75 $ shipment : num__3 + num__2 = num__5 $ . thus for two such orders = num__10 $ total cost involved for a num__120 $ shipment : num__3 + num__3 = num__6 $ . the difference = num__4 $ . d . <eor> d <eos> |
d |
add__3.0__2.0__ multiply__2.0__5.0__ multiply__3.0__2.0__ add__3.0__1.0__ add__3.0__1.0__ |
add__3.0__2.0__ multiply__2.0__5.0__ add__1.0__5.0__ add__3.0__1.0__ add__3.0__1.0__ |
| two trains of equal length running with the speeds of num__60 and num__40 kmph take num__50 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? <o> a ) num__10 sec <o> b ) num__17 sec <o> c ) num__18 sec <o> d ) num__16 sec <o> e ) num__12 sec |
rs = num__60 - num__40 = num__20 * num__0.277777777778 = num__5.55555555556 t = num__50 d = num__50 * num__5.55555555556 = num__277.777777778 rs = num__60 + num__40 = num__100 * num__0.277777777778 t = num__277.777777778 * num__0.036 = num__10 sec answer : a <eor> a <eos> |
a |
subtract__60.0__40.0__ add__60.0__40.0__ subtract__60.0__50.0__ round__10.0__ |
subtract__60.0__40.0__ add__60.0__40.0__ subtract__60.0__50.0__ subtract__60.0__50.0__ |
| r = num__2 ^ ( n + num__1 ) then in terms of r num__9 ^ n = <o> a ) r ^ num__0.0246913580247 <o> b ) r ^ num__0.222222222222 <o> c ) r ^ num__0.5 <o> d ) r / num__3 <o> e ) r |
questions in terms of variables can be easily solved by plugging in some values : say n = num__0 r = num__2 ^ num__1 = num__2 and you need the value of num__2 ^ n which is num__2 ^ num__0 = num__1 so when you put r = num__2 in the options you should get num__1 . only option ( c ) satisfies . answer ( c ) <eor> c <eos> |
c |
reverse__2.0__ |
reverse__2.0__ |
| on a sum of money the s . i . for num__2 years is $ num__660 while the c . i . is $ num__693 the rate of interest being the same in both the cases . the rate of interest is ? <o> a ) num__10.0 <o> b ) num__32.0 <o> c ) num__72.0 <o> d ) num__14.0 <o> e ) num__82 % |
difference in c . i . and s . i for num__2 years = $ num__693 - $ num__660 = $ num__33 s . i for one year = $ num__330 s . i . on $ num__330 for num__1 year = $ num__33 rate = ( num__100 * num__33 ) / ( num__330 ) = num__10.0 the answer is a . <eor> a <eos> |
a |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| the compounded ratio of ( num__2 : num__3 ) ( num__6 : num__11 ) and ( num__11 : num__2 ) is <o> a ) num__2 : num__1 <o> b ) num__1 : num__2 <o> c ) num__36 : num__121 <o> d ) num__11 : num__24 <o> e ) none |
sol . required ratio = [ num__0.666666666667 x num__0.545454545455 x num__5.5 ] = num__2.0 = num__2 : num__1 . answer a <eor> a <eos> |
a |
divide__2.0__3.0__ divide__6.0__11.0__ divide__11.0__2.0__ subtract__3.0__2.0__ multiply__2.0__1.0__ |
divide__2.0__3.0__ divide__6.0__11.0__ divide__11.0__2.0__ subtract__3.0__2.0__ multiply__2.0__1.0__ |
| when jessica withdrew $ num__200 from her bank account her account balance decreased by num__0.4 . if she deposits an amount equal to num__0.2 of the remaining balance what will be the final balance in her bank account ? <o> a ) num__300 <o> b ) num__360 <o> c ) num__400 <o> d ) num__500 <o> e ) num__575 |
as per the question num__200 = num__2 a / num__5 thus - a which is the total amount = num__500 the amount thus left = num__300 she then deposited num__0.2 of num__300 = num__60 total amount in her account = num__360 answer b <eor> b <eos> |
b |
divide__0.4__0.2__ reverse__0.2__ divide__200.0__0.4__ subtract__500.0__200.0__ multiply__0.2__300.0__ add__300.0__60.0__ add__300.0__60.0__ |
divide__0.4__0.2__ reverse__0.2__ divide__200.0__0.4__ subtract__500.0__200.0__ divide__300.0__5.0__ add__300.0__60.0__ add__300.0__60.0__ |
| a man can ride on motorcycle at num__50 kmph upward road and num__100 kmph downward road . he takes num__12 hours to ride motorcycle uphill from lower point a to upper point b and back to a . what is the total distance traveled by him in num__12 hours ? he must return back to base point a in remaining time . <o> a ) num__1000 <o> b ) num__900 <o> c ) num__800 <o> d ) num__400 <o> e ) num__1200 |
upward distance traveled per hour - num__50 km distance traveled after num__8 hours = num__400 km . he remained with num__4 hours travel downward i . e . distance traveled in remaining num__4 hours downward = num__400 km so total distance traveled from a to b and back to point a = num__400 km upward + num__400 km downhill = num__800 km ( answer : c ) <eor> c <eos> |
c |
multiply__50.0__8.0__ subtract__12.0__8.0__ multiply__100.0__8.0__ round__800.0__ |
multiply__50.0__8.0__ subtract__12.0__8.0__ multiply__100.0__8.0__ round__800.0__ |
| a coin is tossed live times . what is the probability that there is at the least one tail ? a . num__0.96875 b . num__0.0625 <o> a ) num__0.96875 <o> b ) num__0.348314606742 <o> c ) num__1.24 <o> d ) num__1.82352941176 <o> e ) num__2.38461538462 |
let p ( t ) be the probability of getting least one tail when the coin is tossed five times . = there is not even a single tail . i . e . all the outcomes are heads . = num__0.03125 ; p ( t ) = num__1 - num__0.03125 = num__0.96875 answer : a <eor> a <eos> |
a |
negate_prob__0.9688__ negate_prob__0.0312__ |
negate_prob__0.9688__ negate_prob__0.0312__ |
| the average distance between the sun and a certain planet is approximatly num__2.1 num__10 ^ num__15 inches . which of the following is the closest to the average distence between sun and the planet in kelometers ? ( num__1 km is approx num__3.9 x num__10 ^ num__4 inches ) <o> a ) num__6.4 x ( num__10 ) ^ num__10 <o> b ) num__5.4 x ( num__10 ) ^ num__10 <o> c ) num__7.4 x ( num__10 ) ^ num__10 <o> d ) num__2.4 x ( num__10 ) ^ num__10 <o> e ) num__3.4 x ( num__10 ) ^ num__10 |
no need to solve entire problem . distance in km = num__21 * ( num__10 ) ^ num__0.384615384615 * ( num__10 ) ^ num__4 which is - num__210 * ( num__10 ) ^ num__0.358974358974 * ( num__10 ) ^ num__4 = num__5.4 x ( num__10 ) ^ num__10 answer : b <eor> b <eos> |
b |
multiply__2.1__10.0__ multiply__10.0__21.0__ multiply__1.0__5.4__ |
multiply__2.1__10.0__ multiply__10.0__21.0__ multiply__1.0__5.4__ |
| num__597 * * num__6 is divisible by both num__3 and num__11 . the non - zero digits in the hundred ’ s and ten ’ s places are respectively : <o> a ) num__3 and num__6 <o> b ) num__7 and num__9 <o> c ) num__2 and num__6 <o> d ) num__4 and num__7 <o> e ) none |
solution : let the given number be num__597 xy num__6 . then ( num__5 + num__9 + num__7 + x + y + num__6 ) = ( num__27 + x + y ) must be divisible by num__3 and ( num__6 + x + num__9 ) - ( y + num__7 + num__5 ) = ( x - y + num__3 ) must be either num__0 or divisible by num__11 . x - y + num__3 = num__0 = > y = x + num__3 num__27 + x + y ) = > ( num__27 + x + x + num__3 ) = > ( num__30 + num__2 x ) = > x = num__3 and y = num__6 . answer a <eor> a <eos> |
a |
subtract__11.0__6.0__ add__6.0__3.0__ multiply__3.0__9.0__ multiply__6.0__5.0__ divide__6.0__3.0__ subtract__6.0__3.0__ |
subtract__11.0__6.0__ add__6.0__3.0__ multiply__3.0__9.0__ add__3.0__27.0__ subtract__11.0__9.0__ subtract__6.0__3.0__ |
| andy solves problems num__78 to num__125 inclusive in a math exercise . how many problems does he solve ? <o> a ) num__53 <o> b ) num__52 <o> c ) num__51 <o> d ) num__50 <o> e ) num__48 |
num__125 - num__78 + num__1 = num__48 ' e ' is the answer <eor> e <eos> |
e |
multiply__48.0__1.0__ |
multiply__48.0__1.0__ |
| a rectangular floor that measures num__6 meters by num__8 meters is to be covered with carpet squares that each measure num__2 meters by num__2 meters . if the carpet squares cost $ num__12 apiece what is the total cost for the number of carpet squares needed to cover the floor ? <o> a ) $ num__200 <o> b ) $ num__240 <o> c ) $ num__480 <o> d ) $ num__960 <o> e ) $ num__144 |
the width of the rectangular floor ( num__6 m ) is a multiple of one side of the square ( num__2 m ) and the length of the floor ( num__8 m ) is also a multiple of the side of the square . so the number of carpets to cover the floor is ( num__3.0 ) * ( num__4.0 ) = num__12 . the total cost is num__12 * num__12 = $ num__144 the answer is therefore e . <eor> e <eos> |
e |
power__12.0__2.0__ triangle_area__2.0__144.0__ |
volume_rectangular_prism__6.0__8.0__3.0__ volume_rectangular_prism__6.0__8.0__3.0__ |
| if two numbers are in the ratio num__2 : num__3 and the numerator and denominator are added and subtracted by num__1 and num__2 respectively then the ratio becomes num__6 : num__2 then find the smallest number ? <o> a ) num__2 <o> b ) num__1 <o> c ) num__3 <o> d ) num__4 <o> e ) num__6 |
num__2 : num__3 num__2 x + num__1 : num__3 x - num__2 = num__3 : num__1 [ num__2 x + num__1 ] = num__3 [ num__3 x - num__2 ] num__2 x + num__1 = num__9 x - num__6 num__9 x - num__2 x = num__7 x = num__1 then smallest number is = num__2 x num__2 x = num__2 a <eor> a <eos> |
a |
add__3.0__6.0__ add__1.0__6.0__ multiply__2.0__1.0__ |
add__3.0__6.0__ add__1.0__6.0__ subtract__3.0__1.0__ |
| for which of the following does y ( a ) − y ( b ) = y ( a − b ) for all values of a and b ? <o> a ) y ( x ) = x ^ num__2 <o> b ) y ( x ) = x / num__2 <o> c ) y ( x ) = x + num__5 <o> d ) y ( x ) = num__2 x − num__1 <o> e ) y ( x ) = | x | |
to solve this easiest way is just put the value and see that if it equals or not . with option num__1 . y ( a ) = a ^ num__2 and y ( b ) = b ^ num__2 so l . h . s = a ^ num__2 - b ^ num__2 and r . h . s = ( a - b ) ^ num__2 = = > a ^ num__2 + b ^ num__2 - num__2 ab . so l . h . s not equal to r . h . s with option num__2 . y ( a ) = a / num__2 and y ( b ) = b / num__2 l . h . s = a / num__2 - b / num__2 = = > num__0.5 ( a - b ) r . h . s = ( a - b ) / num__2 so l . h . s = r . h . s which is the correct answer . answer : b <eor> b <eos> |
b |
reverse__2.0__ reverse__0.5__ |
reverse__2.0__ reverse__0.5__ |
| if num__102 y = num__25 then num__10 - y equals : <o> a ) - num__0.2 <o> b ) num__0.0016 <o> c ) num__0.02 <o> d ) num__0.04 <o> e ) num__0.2 |
num__102 y = num__25 ( num__10 y ) num__2 = num__52 num__10 y = num__5 num__0.1 y = num__0.2 num__10 - y = num__0.2 answer : e ) <eor> e <eos> |
e |
divide__10.0__2.0__ reverse__10.0__ reverse__5.0__ reverse__5.0__ |
divide__10.0__2.0__ reverse__10.0__ reverse__5.0__ reverse__5.0__ |
| salaries of prasanna and sathish are in the ratio num__2 : num__3 . if the salary of each is increased by rs . num__4000 the new ratio becomes num__40 : num__57 . what is sathish ' s present salary ? <o> a ) rs . num__15000 <o> b ) rs . num__18000 <o> c ) rs . num__38000 <o> d ) rs . num__43000 <o> e ) none |
sol . let the original salaries of prasanna and sathish be rs . num__2 x and rs . num__3 x respectively . then num__2 x + num__1333.33333333 x + num__4000 = num__0.701754385965 ⇔ num__57 ( num__2 x + num__4000 ) = num__40 ( num__3 x + num__4000 ) ⇔ num__6 x = num__68000 ⇔ num__3 x = num__34000 . sathish ' s present salary = ( num__3 x + num__4000 ) = ( num__34000 + num__4000 ) = rs . num__38000 . answer c <eor> c <eos> |
c |
divide__4000.0__3.0__ divide__40.0__57.0__ multiply__2.0__3.0__ divide__68000.0__2.0__ add__4000.0__34000.0__ add__4000.0__34000.0__ |
divide__4000.0__3.0__ divide__40.0__57.0__ multiply__2.0__3.0__ divide__68000.0__2.0__ add__4000.0__34000.0__ add__4000.0__34000.0__ |
| an unbiased cubic die is thrown . what is the probabiltiy of getting a multiple of num__3 or num__4 ? <o> a ) num__1 <o> b ) num__4 <o> c ) num__0.666666666667 <o> d ) num__0.25 <o> e ) num__0.5 |
total numbers in a die = num__6 p ( multiple of num__3 ) = num__0.333333333333 = num__0.333333333333 p ( multiple of num__4 ) = num__0.166666666667 p ( multiple of num__3 or num__4 ) = num__0.333333333333 + num__0.166666666667 = num__0.5 answer : e <eor> e <eos> |
e |
die_space__ negate_prob__0.5__ |
die_space__ negate_prob__0.5__ |
| the average of num__11 results is num__49 if the average of first six results is num__49 and that of the last six is num__52 . find the sixth result ? <o> a ) num__21 <o> b ) num__67 <o> c ) num__18 <o> d ) num__25 <o> e ) num__23 |
num__1 to num__11 = num__11 * num__49 = num__539 num__1 to num__6 = num__6 * num__49 = num__294 num__6 to num__11 = num__6 * num__52 = num__312 num__6 th = num__294 + num__312 â € “ num__539 = num__67 answer : b <eor> b <eos> |
b |
multiply__11.0__49.0__ multiply__49.0__6.0__ multiply__52.0__6.0__ multiply__1.0__67.0__ |
multiply__11.0__49.0__ multiply__49.0__6.0__ multiply__52.0__6.0__ multiply__1.0__67.0__ |
| if a number is chosen at random from the set { num__1 num__2 num__3 . . . . num__100 } then the probability that the chosen number is a perfect cube is <o> a ) num__0.04 <o> b ) num__0.037037037037 <o> c ) num__0.0357142857143 <o> d ) num__0.0454545454545 <o> e ) num__0.0416666666667 |
we have num__1 num__8 num__27 and num__64 as perfect cubes from num__1 to num__100 . thus the probability of picking a perfect cube is num__0.04 = num__0.04 . answer : a <eor> a <eos> |
a |
power__2.0__3.0__ power__8.0__2.0__ multiply__1.0__0.04__ |
power__2.0__3.0__ power__8.0__2.0__ multiply__1.0__0.04__ |
| the ratio of the radius of two circles is num__2 : num__3 and then the ratio of their areas is ? <o> a ) num__1 : num__8 <o> b ) num__4 : num__9 <o> c ) num__1 : num__9 <o> d ) num__1 : num__3 <o> e ) num__1 : num__2 |
r num__1 : r num__2 = num__2 : num__3 Π r num__12 : Π r num__22 r num__12 : r num__22 = num__4 : num__9 answer : b <eor> b <eos> |
b |
square_perimeter__3.0__ surface_rectangular_prism__2.0__3.0__1.0__ square_perimeter__1.0__ power__3.0__2.0__ square_perimeter__1.0__ |
square_perimeter__3.0__ surface_rectangular_prism__2.0__3.0__1.0__ square_perimeter__1.0__ power__3.0__2.0__ square_perimeter__1.0__ |
| two ants arthur and amy have discovered a picnic and are bringing crumbs back to the anthill . amy makes twice as many trips and carries one and a half times as many crumbs per trip as arthur . if arthur carries a total of c crumbs to the anthill how many crumbs will amy bring to the anthill in terms of c ? <o> a ) x / num__2 <o> b ) x <o> c ) num__3 x / num__2 <o> d ) num__2 x <o> e ) num__3 x |
lets do it by picking up numbers . let arthur carry num__2 crumbs per trip this means amy carries num__3 crumbs per trip . also let arthur make num__2 trips and so amy makes num__4 trips . thus total crumbs carried by arthur ( c ) = num__2 x num__2 = num__4 total crumbs carried by amy = num__3 x num__4 = num__12 . num__12 is num__3 times num__4 so e <eor> e <eos> |
e |
multiply__3.0__4.0__ divide__12.0__4.0__ |
multiply__3.0__4.0__ divide__12.0__4.0__ |
| raju age after num__15 years will be num__5 times his age num__5 years back what is the present age of raju <o> a ) num__15 <o> b ) num__14 <o> c ) num__10 <o> d ) num__8 <o> e ) num__9 |
explanation : clearly x + num__15 = num__5 ( x - num__5 ) < = > num__4 x = num__40 = > x = num__10 option c <eor> c <eos> |
c |
subtract__15.0__5.0__ subtract__15.0__5.0__ |
subtract__15.0__5.0__ subtract__15.0__5.0__ |
| a car is running at a speed of num__60 kmph . what distance will it cover in num__8 sec ? <o> a ) num__100 m <o> b ) num__136 m <o> c ) num__180 m <o> d ) num__200 m <o> e ) num__250 m |
speed = num__60 kmph = num__60 * num__0.277777777778 = num__17 m / s distance covered in num__8 sec = num__17 * num__8 = num__136 m answer is b <eor> b <eos> |
b |
multiply__8.0__17.0__ round__136.0__ |
multiply__8.0__17.0__ multiply__8.0__17.0__ |
| look at this series : num__32 num__31 num__33 num__32 num__34 num__33 . . . what number should come next ? <o> a ) num__38 <o> b ) num__32 <o> c ) num__35 <o> d ) num__36 <o> e ) num__31 |
here we alternatively subtracted num__1 and added num__2 then subtracted num__1 and added num__2 and so on . . the answer is option c ( num__35 ) <eor> c <eos> |
c |
subtract__32.0__31.0__ subtract__33.0__31.0__ add__33.0__2.0__ add__33.0__2.0__ |
subtract__32.0__31.0__ subtract__33.0__31.0__ add__33.0__2.0__ add__33.0__2.0__ |
| this year mbb consulting fired num__8.0 of its employees and left remaining employee salaries unchanged . sally a first - year post - mba consultant noticed that that the average ( arithmetic mean ) of employee salaries at mbb was num__10.0 more after the employee headcount reduction than before . the total salary pool allocated to employees after headcount reduction is what percent of that before the headcount reduction ? <o> a ) num__98.5 <o> b ) num__101.2 <o> c ) num__102.8 <o> d ) num__104.5 <o> e ) num__105.0 % |
num__100 employees getting num__1000 $ avg so total salary for num__100 ppl = num__100000 num__8.0 reduction in employees lead to num__92 employees and a salary increase of num__10.0 of previous avg salary thus the new avg salary is = num__10.0 ( num__1000 ) + num__1000 = num__1100 so total salary of num__92 employees is num__92 * num__1100 = num__101200 now the new salary is more than previous salary by x % . x = ( num__1.012 ) * num__100 = num__101.2 so the answer is b <eor> b <eos> |
b |
multiply__10.0__100.0__ multiply__100.0__1000.0__ subtract__100.0__8.0__ add__100.0__1000.0__ multiply__1100.0__92.0__ divide__101200.0__100000.0__ multiply__100.0__1.012__ multiply__100.0__1.012__ |
multiply__10.0__100.0__ multiply__100.0__1000.0__ subtract__100.0__8.0__ add__100.0__1000.0__ multiply__1100.0__92.0__ divide__101200.0__100000.0__ multiply__100.0__1.012__ multiply__100.0__1.012__ |
| danny drove his old car num__100 kilometers from his home to a friend . to prevent the engine from overheating he stopped his car for num__10 minutes every time he completed num__40 kilometers of continuous driving . when the car was moving danny drove it at an average speed of num__60 kilometers per hour . what was danny ' s average speed on that trip ? <o> a ) num__37.5 kilometers per hour <o> b ) num__48 kilometers per hour <o> c ) num__50 kilometers per hour <o> d ) num__75 kilometers per hour <o> e ) num__100 kilometers per hour |
time taken to reach first num__40 km is num__0.666666666667 = num__0.666666666667 hr time taken to reach next num__40 km is num__0.666666666667 = num__0.666666666667 hr time taken to reach next num__20 km is num__0.333333333333 = num__0.333333333333 hr danny stopped twice on his way at num__40 km from starting point and at num__80 km from the starting point . . . each stop was of num__10 min so two num__10 min break = num__2 * num__0.166666666667 = > num__0.333333333333 hr total time taken to reach num__100 km is ( num__0.666666666667 + num__0.666666666667 + num__0.333333333333 + num__0.333333333333 ) = num__2.0 = > num__2 hrs so average speed is num__50.0 = num__50 km / hour answer will be ( c ) num__50 <eor> c <eos> |
c |
divide__40.0__60.0__ subtract__60.0__40.0__ divide__20.0__60.0__ subtract__100.0__20.0__ divide__40.0__20.0__ divide__10.0__60.0__ divide__100.0__2.0__ round__50.0__ |
divide__40.0__60.0__ subtract__60.0__40.0__ divide__20.0__60.0__ add__60.0__20.0__ divide__40.0__20.0__ divide__10.0__60.0__ divide__100.0__2.0__ divide__100.0__2.0__ |
| if num__1400 * xy = num__1050 . then what is the value of xy = ? <o> a ) num__0.2 <o> b ) num__0.4 <o> c ) num__0.75 <o> d ) num__0.428571428571 <o> e ) num__0.571428571429 |
num__1400 x xy = num__1050 xy = num__1050 = num__0.75 num__1400 c <eor> c <eos> |
c |
divide__1050.0__1400.0__ divide__1050.0__1400.0__ |
divide__1050.0__1400.0__ divide__1050.0__1400.0__ |
| how many words with or without meaning can be formed by using all the letters of the word â € ˜ computer â € ™ using each letter exactly once ? <o> a ) num__43232 <o> b ) num__40320 <o> c ) num__43332 <o> d ) num__89000 <o> e ) num__46780 |
the word â € ˜ computer â € ™ contains num__8 different letters . therefore required number of words = number of arrangement of num__8 letters taken all at a time = num__8 p num__8 = num__8 ! = num__8 * num__7 * num__6 * num__5 * num__4 * num__3 * num__2 * num__1 = num__40320 answer : b <eor> b <eos> |
b |
subtract__7.0__4.0__ subtract__5.0__3.0__ subtract__3.0__2.0__ multiply__40320.0__1.0__ |
subtract__7.0__4.0__ subtract__5.0__3.0__ subtract__3.0__2.0__ multiply__40320.0__1.0__ |
| a squirrel runs up a cylindrical post in a perfect spiral path making one circuit for each rise of num__5 feet . how many feet does the squirrel travels if the post is num__25 feet tall and num__3 feet in circumference ? <o> a ) num__10 feet <o> b ) num__12 feet <o> c ) num__13 feet <o> d ) num__15 feet <o> e ) num__18 feet |
total circuit = num__5.0 = num__5 total feet squirrel travels = num__5 * num__3 = num__15 feet answer : d <eor> d <eos> |
d |
multiply__5.0__3.0__ multiply__5.0__3.0__ |
multiply__5.0__3.0__ multiply__5.0__3.0__ |
| num__5 women can do a work in two days . num__10 men can complete the same work in five days . what is the ratio between the capacity of a man and a woman ? <o> a ) num__1 : num__2 <o> b ) num__1 : num__5 <o> c ) num__2 : num__3 <o> d ) num__3 : num__2 <o> e ) none of these |
explanation : ( num__5 Ã — num__2 ) women can complete the work in num__1 day . â ˆ ´ num__1 woman ' s num__1 day ' s work = num__0.1 ( num__10 Ã — num__5 ) men can complete the work in num__1 day . â ˆ ´ num__1 man ' s num__1 day ' s work = num__0.02 so required ratio = num__0.1 : num__0.02 = num__1 : num__5 answer : b <eor> b <eos> |
b |
divide__10.0__5.0__ divide__1.0__10.0__ divide__0.1__5.0__ round__1.0__ |
divide__10.0__5.0__ divide__1.0__10.0__ divide__0.1__5.0__ round__1.0__ |
| the average of num__25 results is num__45 . the average of first num__12 of those is num__14 and the average of last num__12 is num__17 . what is the num__13 th result ? <o> a ) num__740 <o> b ) num__750 <o> c ) num__690 <o> d ) num__780 <o> e ) num__708 |
solution : sum of num__1 st num__12 results = num__12 * num__14 sum of last num__12 results = num__12 * num__17 num__13 th result = x ( let ) now num__12 * num__14 + num__12 * num__17 + x = num__25 * num__45 or x = num__708 . answer : option e <eor> e <eos> |
e |
subtract__14.0__13.0__ multiply__1.0__708.0__ |
subtract__14.0__13.0__ multiply__1.0__708.0__ |
| there are num__16 students in a class . on the day the test was given bob was absent . the other num__15 students took the test and their average was num__77 . the next day bob took the test and with this grade included the new average was num__78 . what was bob ' s grade on the test ? <o> a ) num__92 <o> b ) num__93 <o> c ) num__94 <o> d ) num__95 <o> e ) num__96 |
num__15 * num__77 + bob ' s grade = num__16 * num__78 alice ' s grade is num__16 * num__78 - num__15 * num__77 = num__93 . the answer is b . <eor> b <eos> |
b |
add__16.0__77.0__ add__16.0__77.0__ |
add__16.0__77.0__ add__16.0__77.0__ |
| subtracting num__6.0 of a from a is equivalent to multiplying a by how much ? <o> a ) num__0.94 <o> b ) num__9.4 <o> c ) num__0.094 <o> d ) num__94 <o> e ) none |
answer let a - num__6.0 of a = ab . ⇒ ( num__94 x a ) / num__100 = ab ∴ b = num__0.94 correct option : a <eor> a <eos> |
a |
add__6.0__94.0__ divide__94.0__100.0__ divide__94.0__100.0__ |
add__6.0__94.0__ divide__94.0__100.0__ divide__94.0__100.0__ |
| { num__3 num__5 num__8 x } for the above mentioned set mean of the set is equal to range of the set then find out possible values of x <o> a ) num__3.2 num__9.33333333333 <o> b ) num__4 num__3.2 <o> c ) num__4 num__3.2 num__9.33333333333 <o> d ) num__4 num__9.33333333333 <o> e ) num__3 num__4 num__5 |
the set can have the following forms : num__1 . { num__3 num__58 x } range = x - num__3 and mean = num__4 + x / num__4 x = num__9.33333333333 num__2 . { num__35 x num__8 } range = num__5 and mean = num__4 + x / num__4 x = num__4 num__3 . { num__3 x num__58 } range = num__5 and mean = num__4 + x / num__4 x = num__4 num__4 . { x num__3 num__58 } range = num__8 - x and mean = num__4 + x / num__4 x = num__3.2 . but as x < num__3 ( assumed here ) x = num__3.2 can not be true . hence reject this case . thus the possible values are num__4 and num__9.33333333333 . d is the correct answer . <eor> d <eos> |
d |
add__3.0__1.0__ subtract__3.0__1.0__ add__3.0__1.0__ |
add__3.0__1.0__ subtract__3.0__1.0__ add__3.0__1.0__ |
| two cyclist start from the same places in opposite directions . one is going towards north at num__18 kmph and the other is going towards south num__20 kmph . what time will they take to be num__47.5 km apart ? <o> a ) num__1 num__0.25 hours <o> b ) num__2 num__0.333333333333 hours <o> c ) num__4 hours <o> d ) num__3 num__0.75 hours <o> e ) num__6 hours |
to be ( num__18 + num__20 ) km apart they take num__1 hour to be num__47.5 km apart they take num__0.0263157894737 * num__47.5 = num__1 num__0.25 hrs answer is a <eor> a <eos> |
a |
round__1.0__ |
round__1.0__ |
| an express traveled at an average speed of num__100 km / hr stopping for num__3 min after every num__75 kn . how long did it take to reach its destination num__600 km from the starting point ? <o> a ) num__6 hrs num__21 minute <o> b ) num__6 hrs num__71 min <o> c ) num__6 hrs num__28 min <o> d ) num__6 hrs num__21 min <o> e ) num__2 hrs num__21 min |
time taken to cover num__600 km = num__6.0 = num__6 hrs . number of stoppages = num__8.0 - num__1 = num__7 total time of stoppages = num__3 * num__7 = num__21 min hence total time taken = num__6 hrs num__21 min . answer : a <eor> a <eos> |
a |
divide__600.0__100.0__ divide__600.0__75.0__ add__1.0__6.0__ multiply__3.0__7.0__ divide__600.0__100.0__ |
divide__600.0__100.0__ divide__600.0__75.0__ subtract__8.0__1.0__ multiply__3.0__7.0__ multiply__1.0__6.0__ |
| a b c d and e play a game of cards . a says to b ` ` if you give me three cards you will have as many as e has and if i give you three cards you will have as many as d has . ' ' a and b together have num__10 cards more than what d and e together have . if b has two cards more than what c has and the total number of cards be num__133 how many cards does b have ? <o> a ) num__22 <o> b ) num__23 <o> c ) num__25 <o> d ) num__35 <o> e ) num__36 |
explanation : clearly we have : b - num__3 = e . . . ( i ) b + num__3 = d . . . ( ii ) a + b = d + e + num__10 . . . ( iii ) b = c + num__2 . . . ( iv ) a + b + c + d + e = num__133 . . . ( v ) from ( i ) and ( ii ) we have : num__2 b = d + e . . . ( vi ) from ( iii ) and ( vi ) we have : a = b + num__10 . . . ( vii ) using ( iv ) ( vi ) and ( vii ) in ( v ) we get : ( b + num__10 ) + b + ( b - num__2 ) + num__2 b = num__133 num__5 b = num__125 b = num__25 . answer : c <eor> c <eos> |
c |
divide__10.0__2.0__ divide__125.0__5.0__ divide__125.0__5.0__ |
add__2.0__3.0__ divide__125.0__5.0__ divide__125.0__5.0__ |
| ( x ) + num__9088 + num__1090 - num__9156 = num__19845 . calculate the value of x <o> a ) num__11123 <o> b ) num__18153 <o> c ) num__11282 <o> d ) num__18825 <o> e ) num__18823 |
( x ) + num__9088 + num__1090 - num__9156 = num__19845 = x + num__9088 + num__1090 = num__19845 + num__9156 = x + num__10178 = num__29001 = x = num__29001 - num__10178 = num__18823 answer is e <eor> e <eos> |
e |
add__9088.0__1090.0__ add__9156.0__19845.0__ subtract__29001.0__10178.0__ subtract__29001.0__10178.0__ |
add__9088.0__1090.0__ add__9156.0__19845.0__ subtract__29001.0__10178.0__ subtract__29001.0__10178.0__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__388 <o> b ) num__240 <o> c ) num__88 <o> d ) num__66 <o> e ) num__221 |
speed = ( num__54 * num__0.277777777778 ) m / sec = num__15 m / sec . length of the train = ( num__15 x num__20 ) m = num__300 m . let the length of the platform be x meters . then ( x + num__300 ) / num__36 = num__15 = = > x + num__300 = num__540 = = > x = num__240 m . answer : b <eor> b <eos> |
b |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
| a pump can fill a tank with water in num__2 hours . because of a leak it took num__2.66666666667 hours to fill the tank . the leak can drain all the water of the tank in ? <o> a ) num__6 hours <o> b ) num__8 hours <o> c ) num__9 hours <o> d ) num__10 hours <o> e ) num__12 hours |
the pump can fill the tank in num__2 hour . but because of the leak it took num__2.66666666667 hours to fill the tank . i . e . the additional water filled by the pump in num__0.666666666667 hour is emptied by the leak in num__2.66666666667 hours . therefore the water filled by pump in num__2 bour can be emptied by the leak in ( num__2.66666666667 ) * num__3 = num__8 hours . i . e . the leak can drain all the water of the tank in num__8 hours . answer : b <eor> b <eos> |
b |
subtract__2.6667__2.0__ round__8.0__ |
subtract__2.6667__2.0__ round__8.0__ |
| what digit appears in the units place in the number obtained when num__2 ^ num__320 is multiplied out ? <o> a ) num__0 <o> b ) num__2 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
num__2 ^ num__320 = num__2 ^ num__4 k put k = num__1 last digit of num__2 ^ num__4 = num__16 last digit num__6 answer = > d ) num__6 <eor> d <eos> |
d |
add__2.0__4.0__ add__2.0__4.0__ |
add__2.0__4.0__ add__2.0__4.0__ |
| the speed at which a man can row a boat in still water is num__15 kmph . if he rows downstream where the speed of current is num__3 kmph what time will he take to cover num__60 metres ? <o> a ) num__33 <o> b ) num__88 <o> c ) num__66 <o> d ) num__12 <o> e ) num__99 |
speed of the boat downstream = num__15 + num__3 = num__18 kmph = num__18 * num__0.277777777778 = num__5 m / s hence time taken to cover num__60 m = num__12.0 = num__12 seconds . answer : d <eor> d <eos> |
d |
add__15.0__3.0__ divide__15.0__3.0__ subtract__15.0__3.0__ round__12.0__ |
add__15.0__3.0__ divide__15.0__3.0__ divide__60.0__5.0__ divide__60.0__5.0__ |
| how many of the following numbers are divisible by num__16 ? num__264 num__396 num__462 num__792 num__968 num__2178 num__5184 num__6336 <o> a ) num__2 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
by using your calculator you can calculate that the following numbers are divisible by num__16 : num__5184 num__6336 required number of number = num__2 . a <eor> a <eos> |
a |
divide__792.0__396.0__ divide__792.0__396.0__ |
divide__792.0__396.0__ divide__792.0__396.0__ |
| two pipes a and b can separately fill a cistern in num__60 minutes and num__120 minutes respectively . there is a third pipe in the bottom of the cistern to empty it . if all the three pipes are simultaneously opened then the cistern is full in num__60 minutes . in how much time the third pipe alone can empty the cistern ? <o> a ) num__90 min <o> b ) num__100 min <o> c ) num__110 min <o> d ) num__120 min <o> e ) num__130 min |
num__0.0166666666667 - ( num__0.0166666666667 + num__0.00833333333333 ) = - num__0.00833333333333 third pipe can empty in num__120 minutes answer : d <eor> d <eos> |
d |
round__120.0__ |
round__120.0__ |
| a squirrel runs up a cylindrical post in a perfect spiral path making one circuit for each rise of num__4 feet . how many feet does the squirrel travels if the post is num__16 feet tall and num__3 feet in circumference ? <o> a ) num__10 feet <o> b ) num__12 feet <o> c ) num__13 feet <o> d ) num__15 feet <o> e ) num__18 feet |
total circuit = num__4.0 = num__4 total feet squirrel travels = num__4 * num__3 = num__12 feet answer : b <eor> b <eos> |
b |
square_perimeter__3.0__ square_perimeter__3.0__ |
multiply__4.0__3.0__ multiply__4.0__3.0__ |
| in a division sum the divisor is num__10 times the quotient and num__5 times the remainder . if the remainder is num__46 what is the dividend ? <o> a ) num__4256 <o> b ) num__5336 <o> c ) num__6000 <o> d ) num__3000 <o> e ) num__4000 |
in a division sum the divisor is num__10 times the quotient and num__5 times the remainder . if the remainder is num__46 what is the dividend ? a . num__4236 b . num__4306 c . num__4336 d . num__5336 e . none of these answer : option d explanation : divisor = ( num__5 x num__46 ) = num__230 num__10 x quotient = num__230 = num__230 = num__23 num__10 dividend = ( divisor x quotient ) + remainder = ( num__230 x num__23 ) + num__46 = num__5290 + num__46 = num__5336 . answer b <eor> b <eos> |
b |
multiply__5.0__46.0__ divide__230.0__10.0__ multiply__230.0__23.0__ add__46.0__5290.0__ |
multiply__5.0__46.0__ divide__230.0__10.0__ multiply__230.0__23.0__ add__46.0__5290.0__ |
| five friends adi brian close derek and eli appeared in two aptitude tests . in the first aptitude test derek score num__50.0 less than the average score of the five people . in the second aptitude test derek score num__50.0 more than what he scored on the first aptitude test . if the score of his friend in the second aptitude test were same as their score in the first test by approximately what percentage was derek ' s score less than the average score of the num__5 people in the second aptitude ? <o> a ) num__25.0 <o> b ) num__28.0 <o> c ) num__33.0 <o> d ) num__40.0 <o> e ) num__50 % |
average score in first test be x so derby score = num__0.5 x ( first test ) derby score = num__1.5 ( num__0.5 x ) ( second test ) = . num__75 x so less than x by . num__25 answer a num__25.0 <eor> a <eos> |
a |
multiply__50.0__1.5__ multiply__50.0__0.5__ multiply__50.0__0.5__ |
multiply__50.0__1.5__ multiply__50.0__0.5__ multiply__50.0__0.5__ |
| a train num__125 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 sec . the speed of the train is ? <o> a ) num__16 km / hr <o> b ) num__50 km / hr <o> c ) num__15 km / hr <o> d ) num__17 km / hr <o> e ) num__18 km / hr |
speed of the train relative to man = num__12.5 = num__12.5 m / sec . = num__12.5 * num__3.6 = num__45 km / hr let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__45 = > x = num__50 km / hr . answer : b <eor> b <eos> |
b |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ round__50.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ multiply__5.0__10.0__ |
| a train covers a distance of num__12 km in num__10 min . if it takes num__5 sec to pass a telegraph post then the length of the train is ? <o> a ) num__298 m <o> b ) num__188 m <o> c ) num__120 m <o> d ) num__160 m <o> e ) num__189 m |
speed = ( num__1.2 * num__60 ) km / hr = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . length of the train = num__20 * num__8 = num__160 m . answer : d <eor> d <eos> |
d |
divide__12.0__10.0__ hour_to_min_conversion__ add__12.0__60.0__ subtract__20.0__12.0__ multiply__8.0__20.0__ round__160.0__ |
divide__12.0__10.0__ multiply__12.0__5.0__ multiply__1.2__60.0__ subtract__20.0__12.0__ multiply__8.0__20.0__ multiply__8.0__20.0__ |
| what will be the ratio of simple interest earned by certain amount at the same rate of interest for num__9 years and that for num__6 years ? <o> a ) num__1 : num__3 <o> b ) num__2 : num__3 <o> c ) num__1 : num__2 <o> d ) num__3 : num__2 <o> e ) num__1 : num__4 |
let the principal be p and rate of interest be r % . required ratio = ( p x r x num__9 ) / num__100 divided by ( p x r x num__6 ) / num__100 = num__9 pr / num__6 pr = num__1.5 = num__3 : num__2 . answer : d <eor> d <eos> |
d |
percent__3.0__100.0__ |
percent__3.0__100.0__ |
| if three machines working at the same rate can do num__0.75 of a job in num__30 minutes how many minutes would it take two machines working at the same rate to do num__0.6 of the job ? <o> a ) num__36 <o> b ) num__60 <o> c ) num__75 <o> d ) num__80 <o> e ) num__100 |
using the std formula m num__1 d num__1 h num__1 / w num__1 = m num__2 d num__2 h num__2 / w num__2 substituting the values we have num__3 * num__0.5 * num__1.33333333333 = num__2 * num__1.66666666667 * x ( converted num__30 min into hours = num__0.5 ) num__2 = num__3.33333333333 * x x = num__0.6 hour so num__36 minutes answer : a <eor> a <eos> |
a |
add__1.0__2.0__ divide__1.0__2.0__ divide__1.0__0.75__ divide__1.0__0.6__ divide__2.0__0.6__ round__36.0__ |
add__1.0__2.0__ divide__1.0__2.0__ divide__1.0__0.75__ divide__1.0__0.6__ divide__2.0__0.6__ divide__36.0__1.0__ |
| ravi can do a piece of work in num__30 days while prakash can do it in num__40 days . in how many days will they finish it together ? <o> a ) num__17 num__0.142857142857 days <o> b ) num__17 num__0.125 days <o> c ) num__17 num__0.111111111111 days <o> d ) num__17 num__1.0 days <o> e ) num__18 num__0.142857142857 days |
num__0.0333333333333 + num__0.025 = num__0.0583333333333 num__17.1428571429 = num__17 num__0.142857142857 days answer : a <eor> a <eos> |
a |
add__0.025__0.0333__ subtract__17.1429__17.0__ round__17.0__ |
add__0.025__0.0333__ subtract__17.1429__17.0__ round__17.0__ |
| pipes a and b can fill a cistern in num__9 and num__72 minutes respectively . they are opened an alternate minutes . find how many minutes the cistern shall be full ? <o> a ) num__13 <o> b ) num__12 <o> c ) num__16 <o> d ) num__18 <o> e ) num__19 |
: num__0.111111111111 + num__0.0138888888889 = num__0.125 num__8 * num__2 = num__16 . answer : c <eor> c <eos> |
c |
divide__9.0__72.0__ divide__72.0__9.0__ divide__2.0__0.125__ round__16.0__ |
add__0.1111__0.0139__ divide__72.0__9.0__ multiply__2.0__8.0__ multiply__2.0__8.0__ |
| look at this series : num__7 num__10 num__8 num__11 num__9 num__12 num__10 . . . what number should come next ? <o> a ) num__7 <o> b ) num__10 <o> c ) num__12 <o> d ) num__13 <o> e ) num__14 |
this is a simple alternating addition and subtraction series . in the first pattern num__3 is added ; in the second num__2 is subtracted . the answer is d . <eor> d <eos> |
d |
subtract__10.0__7.0__ subtract__10.0__8.0__ add__10.0__3.0__ |
subtract__10.0__7.0__ subtract__10.0__8.0__ add__10.0__3.0__ |
| if a and b are the two values of t that satisfy the equation t ^ num__2 â € “ num__14 t + num__24 = num__0 with a > b what is the value of a â € “ b ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
factor the left side of the equation : t ^ num__2 â € “ num__14 t + num__24 = num__0 ( t â € “ num__2 ) ( t â € “ num__12 ) = num__0 t = num__2 t = num__12 thus a = num__12 and b = num__2 . so a â € “ b = num__12 â € “ num__2 = num__10 . the answer is e . <eor> e <eos> |
e |
subtract__14.0__2.0__ subtract__24.0__14.0__ subtract__24.0__14.0__ |
subtract__14.0__2.0__ subtract__24.0__14.0__ subtract__24.0__14.0__ |
| line m lies in the xy - plane . the y - intercept of line m is - num__2 and line m passes through the midpoint of the line segment whose endpoints are ( num__2 num__4 ) and ( num__6 - num__6 ) . what is the slope of line m ? <o> a ) - num__3 <o> b ) - num__1 <o> c ) - num__0.333333333333 <o> d ) num__0 <o> e ) undefined |
ans : b solution : line m goes through midpoint of ( num__2 num__4 ) and ( num__6 - num__8 ) . midpoint is ( num__4 - num__1 ) as we can see that the y axis of intercept point is ( num__0 - num__2 ) means line m is parallel to x axis slope m = - num__1 ans : b <eor> b <eos> |
b |
multiply__2.0__4.0__ reverse__1.0__ |
multiply__2.0__4.0__ subtract__2.0__1.0__ |
| if a lends rs . num__3500 to b at num__10.0 per annum and b lends the same sum to c at num__11.5 per annum then the gain of b in a period of num__3 years is ? <o> a ) num__157.58 <o> b ) num__157.56 <o> c ) num__157.5 <o> d ) num__157.52 <o> e ) num__157.54 |
( num__3500 * num__1.5 * num__3 ) / num__100 = > num__157.50 answer : c <eor> c <eos> |
c |
percent__100.0__157.5__ |
percent__100.0__157.5__ |
| if num__20.0 of a number is equal to three - fifth of another number what is the ratio of first number to the second number ? <o> a ) num__3 : num__1 <o> b ) num__2 : num__1 <o> c ) num__1 : num__3 <o> d ) num__1 : num__2 <o> e ) num__2 : num__3 |
let num__20.0 of a = num__0.6 b . then num__20 a / num__100 = num__3 b / num__5 = > num__1 a / num__5 = num__3 b / num__5 a / b = ( num__0.6 * num__5.0 ) = num__3.0 a : b = num__3 : num__1 . answer : a <eor> a <eos> |
a |
divide__3.0__0.6__ multiply__1.0__3.0__ |
divide__3.0__0.6__ multiply__1.0__3.0__ |
| a basket has num__6 apples and num__6 oranges . three fruits are picked at random . what is the probability that at least num__2 apples are picked ? <o> a ) num__0.5 <o> b ) num__0.333333333333 <o> c ) num__0.25 <o> d ) num__0.4 <o> e ) num__0.3 |
the total possible choices are num__12 c num__3 = num__220 the ways to choose num__3 apples : num__6 c num__3 = num__20 the ways to choose num__2 apples : num__6 c num__2 * num__6 c num__1 = num__90 p ( at least num__2 apples ) = num__0.5 = num__0.5 the answer is a . <eor> a <eos> |
a |
negate_prob__0.5__ |
negate_prob__0.5__ |
| by purchasing an article at num__20.0 discount on the original price and then selling it at a price of num__25.0 above the original price a trader earns rs . num__200 as the profit . what was the original price of the article ? <o> a ) rs . num__444.44 <o> b ) rs . num__255.50 <o> c ) rs . num__100.10 <o> d ) rs . num__810 <o> e ) none of these |
explanation : let the original price of the article be rs . num__100 . hence the purchase price for the trader would be rs . num__80 and his selling price would be rs . num__125 . thus he would earn a profit of rs . num__45 ( num__125 – num__80 ) . therefore profit is rs . num__45 if the original price is rs . num__100 hence if profit is rs . num__200 then original price will be . . = num__100 x num__4.44444444444 = rs . num__444.44 answer a <eor> a <eos> |
a |
percent__100.0__444.44__ |
percent__100.0__444.44__ |
| how many paying stones each measuring num__3 * num__2 m are required to pave a rectangular court yard num__15 m long and num__6 m board ? <o> a ) num__99 <o> b ) num__18 <o> c ) num__26 <o> d ) num__15 <o> e ) num__12 |
num__15 * num__6 = num__3 * num__2 * x = > x = num__15 answer : d <eor> d <eos> |
d |
round__15.0__ |
round__15.0__ |
| the average weight of a b and c is num__45 kg . if the average weight of a and b be num__40 kg and that of b and c be num__43 kg then the weight of b is : <o> a ) num__17 kg <o> b ) num__20 kg <o> c ) num__22 kg <o> d ) num__26 kg <o> e ) num__31 kg |
let a b c represent their respective weights . then we have : a + b + c = ( num__45 x num__3 ) = num__135 . . . . ( i ) a + b = ( num__40 x num__2 ) = num__80 . . . . ( ii ) b + c = ( num__43 x num__2 ) = num__86 . . . . ( iii ) adding ( ii ) and ( iii ) we get : a + num__2 b + c = num__166 . . . . ( iv ) subtracting ( i ) from ( iv ) we get : b = num__31 . b ' s weight = num__31 kg . answer : option e <eor> e <eos> |
e |
subtract__43.0__40.0__ multiply__45.0__3.0__ subtract__45.0__43.0__ multiply__40.0__2.0__ multiply__43.0__2.0__ add__80.0__86.0__ subtract__166.0__135.0__ subtract__166.0__135.0__ |
subtract__43.0__40.0__ multiply__45.0__3.0__ subtract__45.0__43.0__ multiply__40.0__2.0__ multiply__43.0__2.0__ add__80.0__86.0__ subtract__166.0__135.0__ subtract__166.0__135.0__ |
| p alone can complete a piece of work in num__6 days . work done by q alone in one day is equal to one - third of the work done by p alone in one day . in how many days can the work be completed if p and q work together ? <o> a ) num__6 ( num__0.25 ) days <o> b ) num__6 ( num__0.75 ) days <o> c ) num__7 ( num__0.75 ) days <o> d ) num__8 ( num__0.75 ) days <o> e ) num__9 ( num__0.75 ) days |
work done by p alone in one day = num__0.111111111111 th of the total work done by q alone in one day = num__0.333333333333 ( of that done by p in one day ) = num__0.333333333333 ( num__0.111111111111 of the total ) = num__0.037037037037 of the total . work done by p and q working together in one day = num__0.111111111111 + num__0.037037037037 = num__0.148148148148 of the total they would take num__6.75 days = num__6 ( num__0.75 ) days to complete the work working together . answer : b <eor> b <eos> |
b |
multiply__0.1111__0.3333__ add__0.1111__0.037__ subtract__6.75__6.0__ round__6.0__ |
multiply__0.1111__0.3333__ add__0.1111__0.037__ subtract__6.75__6.0__ round__6.0__ |
| a runs twice as fast as b and gives b a start of num__60 m . how long should the racecourse be so that a and b might reach in the same time ? <o> a ) num__120 m . <o> b ) num__80 m . <o> c ) num__150 m . <o> d ) num__100 m . <o> e ) none of the above |
ratio of speeds of a and b is num__2 : num__1 b is num__60 m away from a but we know that a covers num__1 meter ( num__2 - num__1 ) more in every second than b the time taken for a to cover num__60 m is num__60.0 = num__60 m so the total time taken by a and b to reach = num__2 * num__60 = num__120 m answer : a <eor> a <eos> |
a |
twice__60.0__ twice__60.0__ |
twice__60.0__ twice__60.0__ |
| if num__5 is added to twice a number and this sum is multiplied by num__10 the result is the same as if the number is multiplied by num__5 and num__20 is added to the product . what is the number ? <o> a ) - num__5 <o> b ) - num__8 <o> c ) - num__2 <o> d ) num__10 <o> e ) num__11 |
let the number be x ; num__10 ( num__5 + num__2 x ) = num__5 x + num__20 ; x = - num__2 answer : c <eor> c <eos> |
c |
divide__10.0__5.0__ divide__10.0__5.0__ |
divide__10.0__5.0__ divide__10.0__5.0__ |
| laxman went num__15 kms from my house then turned left and walked num__20 kms . he then turned east and walked num__25 kms and finally turning left covered num__20 kms . how far was he from his house ? <o> a ) num__40 * num__1.414 <o> b ) num__30 * num__1.414 <o> c ) num__45 * num__1.414 <o> d ) num__20 * num__1.414 <o> e ) num__48 * num__1.414 |
if we plot a graph for the given motion then we find that total distance in a direction is num__40 km . and there is num__40 km . in perpendicular to that direction so direct distance of laxman from the initial position will be sqrt ( num__40 ^ num__2 + num__40 ^ num__2 ) = num__40 sqrt ( num__2 ) = num__40 * num__1.414 answer : a <eor> a <eos> |
a |
add__15.0__25.0__ divide__40.0__20.0__ add__15.0__25.0__ |
add__15.0__25.0__ divide__40.0__20.0__ add__15.0__25.0__ |
| a sum of money invested for a certain number of years at num__8.0 p . a . simple interest grows to rs . num__180 . the same sum of money invested for the same number of years at num__4.0 p . a . simple interest grows to rs . num__120 . for how many years was the sum invested ? <o> a ) num__25 years <o> b ) num__40 years <o> c ) num__33 years and num__4 months <o> d ) can not be determined <o> e ) none of these |
solution : from the information provided we know that principal + num__8.0 p . a . interest on principal for n years = num__180 … … . . ( num__1 ) principal + num__4.0 p . a . interest on principal for n years = num__120 … … … ( num__2 ) subtracting equation ( num__2 ) from equation ( num__1 ) we get num__4.0 p . a . interest on principal for n years = rs . num__60 . now we can substitute this value in equation ( num__2 ) i . e principal + num__60 = num__120 = principal = rs . num__60 . we know that si = pnr / num__100 where p is the principal n the number of years and r the rate percent of interest . in equation ( num__2 ) p = rs . num__60 r = num__4.0 p . a . and the simple interest = rs . num__60 . therefore num__60 = ( num__60 x n x num__4 ) / num__100 = > n = num__25.0 = num__25 years . answer a <eor> a <eos> |
a |
percent__25.0__100.0__ |
percent__25.0__100.0__ |
| a clock is set right at num__5 a . m . the clock loses num__16 minutes in num__24 hours . what will be the true time when the clock indicates num__10 p . m . on num__4 th day ? <o> a ) num__11 pm <o> b ) num__12 pm <o> c ) num__10 pm <o> d ) num__09 pm <o> e ) num__08 pm |
time from num__5 am . on a day to num__10 pm . on num__4 th day = num__89 hours . now num__23 hrs num__44 min . of this clock = num__24 hours of correct clock . num__23.7333333333 hrs of this clock = num__24 hours of correct clock num__89 hrs of this clock = ( num__24 x num__31556 x num__89 ) hrs of correct clock . = num__90 hrs of correct clock . so the correct time is num__11 p . m . answer : a <eor> a <eos> |
a |
subtract__16.0__5.0__ round__11.0__ |
subtract__16.0__5.0__ round__11.0__ |
| bill is golfing with two friends and can either buy generic golf tees that are packaged by the dozen or the higher quality aero flight tees that come by the pair . what is the minimum number of packages of aero flight tees bill must purchase to ensure that he has at least num__10 golf tees for each member of his foursome if he will buy no more than num__2 packages of the generic golf tees ? <o> a ) num__3 <o> b ) num__10 <o> c ) num__8 <o> d ) num__4 <o> e ) num__2 |
at least num__10 golf tees for each member of his threesome = total of at least num__3 * num__10 = num__30 tees . num__2 packages of the generic golf tees that are packaged by the dozen = num__2 * num__12 = num__24 tees . so bill must by at least num__6 aero tees . they come by the pair hence he must by at least num__3.0 = num__3 packages of aero flight tees . answer : a <eor> a <eos> |
a |
multiply__10.0__3.0__ add__10.0__2.0__ multiply__2.0__12.0__ multiply__2.0__3.0__ divide__6.0__2.0__ |
multiply__10.0__3.0__ add__10.0__2.0__ multiply__2.0__12.0__ multiply__2.0__3.0__ divide__6.0__2.0__ |
| an airplane flies against the wind from a to b in num__8 hours . the same airplane returns from b to a in the same direction as the wind in num__7 hours . find the ratio of the speed of the airplane ( in still air ) to the speed of the wind . <o> a ) num__5 <o> b ) num__10 <o> c ) num__15 <o> d ) num__20 <o> e ) num__25 |
let x = speed of airplane in still air y = speed of wind and d the distance between a and b . find the ratio x / y against the wind : d = num__8 ( x - y ) with the wind : d = num__7 ( x + y ) num__8 x - num__8 y = num__7 x + num__7 y hence x / y = num__15 correct answer c <eor> c <eos> |
c |
add__8.0__7.0__ round__15.0__ |
add__8.0__7.0__ add__8.0__7.0__ |
| a contractor undertakes to complete the construction of a tunnel num__720 meters long in num__240 days and employs num__60 men for the purpose . after num__120 days he finds that only num__240 meters of the tunnel is complete . how many more men should be employ in order to complete the work in time ? <o> a ) num__76 men <o> b ) num__60 men <o> c ) num__78 men <o> d ) num__87 men <o> e ) num__75 men |
in num__120 days only num__240 m of the tunnel is constructed by num__60 men . the remaining num__120 days num__480 m of the tunnel can be constructed by num__120 men . additional number of men required = num__120 - num__60 = num__60 men . answer : b <eor> b <eos> |
b |
subtract__720.0__240.0__ hour_to_min_conversion__ |
subtract__720.0__240.0__ subtract__120.0__60.0__ |
| the length of the bridge which a train num__130 metres long and travelling at num__36 km / hr can cross in num__45 seconds is : <o> a ) num__320 m <o> b ) num__225 m <o> c ) num__245 m <o> d ) num__250 m <o> e ) num__240 m |
speed = [ num__36 x num__0.277777777778 ] m / sec = num__10 m / sec time = num__45 sec let the length of bridge be x metres . then ( num__130 + x ) / num__45 = num__10 = > num__130 + x = num__450 = > x = num__320 m . answer : a <eor> a <eos> |
a |
multiply__45.0__10.0__ subtract__450.0__130.0__ round__320.0__ |
multiply__45.0__10.0__ subtract__450.0__130.0__ round__320.0__ |
| which of the following is equal to the value of num__2 ^ num__7 + num__2 ^ num__7 + num__3 ^ num__7 + num__3 ^ num__7 + num__3 ^ num__7 ? <o> a ) num__2 ^ num__8 + num__3 ^ num__8 <o> b ) num__2 ^ num__6 + num__3 ^ num__8 <o> c ) num__2 ^ num__8 + num__3 ^ num__6 <o> d ) num__2 ^ num__9 + num__3 ^ num__9 <o> e ) num__2 ^ num__5 + num__3 ^ num__5 |
num__2 ^ num__7 + num__2 ^ num__7 + num__3 ^ num__7 + num__3 ^ num__7 + num__3 ^ num__7 = num__2 ( num__2 ^ num__7 ) + num__3 ( num__3 ^ num__7 ) = num__2 ^ num__1 ( num__2 ^ num__7 ) + num__3 ^ num__1 ( num__3 ^ num__7 ) = num__2 ^ ( num__1 + num__7 ) + num__3 ^ ( num__1 + num__7 ) = num__2 ^ num__8 + num__3 ^ num__8 ans : a <eor> a <eos> |
a |
subtract__3.0__2.0__ add__7.0__1.0__ multiply__2.0__1.0__ |
subtract__3.0__2.0__ add__7.0__1.0__ multiply__2.0__1.0__ |
| fred and sam are standing num__50 miles apart and they start walking in a straight line toward each other at the same time . if fred walks at a constant speed of num__5 miles per hour and sam walks at a constant speed of num__5 miles per hour how many miles has sam walked when they meet ? <o> a ) num__5 <o> b ) num__9 <o> c ) num__25 <o> d ) num__30 <o> e ) num__45 |
relative distance = num__50 miles relative speed = num__5 + num__5 = num__10 miles per hour time taken = num__5.0 = num__5 hours distance travelled by sam = num__5 * num__5 = num__25 miles = c <eor> c <eos> |
c |
divide__50.0__5.0__ round__25.0__ |
divide__50.0__5.0__ round__25.0__ |
| on a trip a cyclist averaged num__8 miles per hour for the first num__12 miles and num__12 miles per hour for the remaining num__24 miles . if the cyclist returned immediately via the same route and took a total of num__7.5 hours for the round trip what was the average speed ( in miles per hour ) for the return trip ? <o> a ) num__9 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__13 |
the time to go num__36 miles was num__1.5 + num__2.0 = num__1.5 + num__2 = num__3.5 hours . the average speed for the return trip was num__36 miles / num__4 hours = num__9 mph . the answer is a . <eor> a <eos> |
a |
add__12.0__24.0__ divide__12.0__8.0__ divide__24.0__12.0__ add__1.5__2.0__ divide__8.0__2.0__ add__7.5__1.5__ round__9.0__ |
add__12.0__24.0__ divide__12.0__8.0__ divide__24.0__12.0__ add__1.5__2.0__ divide__8.0__2.0__ add__7.5__1.5__ add__7.5__1.5__ |
| a man can do a piece of work in num__5 days but with the help of his son he can do it in num__3 days . in what time can the son do it alone ? <o> a ) num__6.5 <o> b ) num__9.5 <o> c ) num__10.5 <o> d ) num__7.5 <o> e ) none |
son ' s one day work = ( num__0.333333333333 â ˆ ’ num__0.2 ) = num__0.133333333333 so son will do whole work in = num__7.5 days answer : c <eor> c <eos> |
c |
subtract__0.3333__0.2__ add__3.0__7.5__ |
subtract__0.3333__0.2__ add__3.0__7.5__ |
| at present the ratio between the ages of arun and deepak is num__4 : num__3 . after num__6 years arun ' s age will be num__30 years . what is the age of deepak at present ? <o> a ) num__77 years <o> b ) num__15 years <o> c ) num__66 years <o> d ) num__18 years <o> e ) num__55 years |
let the present ages of arun and deepak be num__4 x and num__3 x years respectively . then num__4 x + num__6 = num__30 = > x = num__6 deepak ' s age = num__3 x = num__18 years . answer : d <eor> d <eos> |
d |
multiply__3.0__6.0__ multiply__3.0__6.0__ |
multiply__3.0__6.0__ multiply__3.0__6.0__ |
| what least number should be added to num__1052 so that the sum is completely divisible by num__23 <o> a ) a ) num__4 <o> b ) b ) num__1 <o> c ) c ) num__6 <o> d ) d ) num__3 <o> e ) e ) num__5 |
explanation : ( num__45.7391304348 ) gives remainder num__17 num__17 + num__6 = num__23 so we need to add num__6 answer : option c <eor> c <eos> |
c |
divide__1052.0__23.0__ subtract__23.0__17.0__ subtract__23.0__17.0__ |
divide__1052.0__23.0__ subtract__23.0__17.0__ subtract__23.0__17.0__ |
| given num__2 x + y + z = num__3 and num__5 x + num__3 y + z = num__15 what is the value of x + y - z ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
( num__1 ) num__5 x + num__3 y + z = num__15 if num__2 x + y + z = num__3 then ( num__2 ) num__4 x + num__2 y + num__2 z = num__6 let ' s subtract equation ( num__2 ) for equation ( num__1 ) . x + y - z = num__9 the answer is e . <eor> e <eos> |
e |
subtract__3.0__2.0__ add__3.0__1.0__ multiply__2.0__3.0__ add__3.0__6.0__ add__3.0__6.0__ |
subtract__3.0__2.0__ add__3.0__1.0__ add__2.0__4.0__ add__3.0__6.0__ add__3.0__6.0__ |
| there is a sequence ni such in which i is a positive integer ni + num__1 = num__2 ni . if n num__1 = num__1 n num__2 = num__2 n num__3 = num__4 n num__4 = num__8 what is the scope including n num__11 ? <o> a ) num__10 ~ num__100 <o> b ) num__1000 ~ num__10000 <o> c ) num__10000 ~ num__100000 <o> d ) num__100000 ~ num__1 num__000000 <o> e ) num__1 num__000000 ~ |
- > ni = num__2 ^ ( i - num__1 ) - > num__2 ^ num__10 = num__1024 > num__1000 = num__10 ^ num__3 - > n num__11 = num__2 ^ num__10 = ( num__2 ^ num__10 ) > ( num__10 ^ num__3 ) = num__10 ^ num__3 thus the answer is b . <eor> b <eos> |
b |
add__2.0__8.0__ power__2.0__10.0__ power__10.0__3.0__ multiply__1.0__1000.0__ |
subtract__11.0__1.0__ power__2.0__10.0__ power__10.0__3.0__ power__1000.0__1.0__ |
| a and b can do a job in num__15 days and num__10 days respectively . they began the work together but a leaves after some days and b finished the remaining job in num__5 days . after how many days did a leave ? <o> a ) num__2 days <o> b ) num__3 days <o> c ) num__1 day <o> d ) num__4 days <o> e ) none of these |
a ’ s one day ’ s work = num__1 ⁄ num__15 th work . b ’ s one day ’ s work = num__1 ⁄ num__10 th work . ( a + b ) ’ s one day ’ s work = num__1 ⁄ num__15 + num__1 ⁄ num__10 = num__1 ⁄ num__6 th work . let a left after x days . ∴ ( a + b ) ’ s x days ’ work = x ⁄ num__6 th work . remaining work = num__1 - x ⁄ num__6 = ( num__6 - x ) ⁄ num__6 th work . now in num__5 days work done by b = ( num__6 - x ) ⁄ num__6 th work . ∴ in num__1 day work done by b = ( num__6 - x ) ⁄ num__30 th work . and ( num__6 - x ) ⁄ num__30 = num__1 ⁄ num__10 ∴ x = num__3 days answer b <eor> b <eos> |
b |
add__5.0__1.0__ multiply__5.0__6.0__ divide__15.0__5.0__ round__3.0__ |
add__5.0__1.0__ multiply__5.0__6.0__ divide__15.0__5.0__ subtract__6.0__3.0__ |
| a grocer is storing soap boxes in cartons that measure num__25 inches by num__42 inches by num__60 inches . if the measurement of each soap box is num__7 inches by num__12 inches by num__5 inches then what is the maximum number of soap boxes that can be placed in each carton ? <o> a ) num__210 <o> b ) num__150 <o> c ) num__280 <o> d ) num__300 <o> e ) num__420 |
however the process of dividing the volume of box by the volume of a soap seems flawed but it does work in this case due to the numbers dimensions of the box = num__25 * num__42 * num__60 dimensions of the soap = num__5 * num__12 * num__7 placing the num__7 inch side along num__42 inch side we get num__6 soaps in a line and in a similar way num__5 along num__25 and num__6 along num__60 we get = num__5 x num__6 x num__5 = num__150 so the question is why this particular arrangement in order to maximize number of soaps we need to minimize the space wasted and this is the only config where we dont waste any space so we can expect the maximum number the answer is ( b ) <eor> b <eos> |
b |
divide__42.0__7.0__ multiply__25.0__6.0__ multiply__25.0__6.0__ |
divide__42.0__7.0__ multiply__25.0__6.0__ multiply__25.0__6.0__ |
| find the value of num__72515 x num__9999 = m ? <o> a ) num__456578972 <o> b ) num__725077485 <o> c ) num__653658791 <o> d ) num__725117481 <o> e ) num__357889964 |
num__72515 x num__9999 = num__72515 x ( num__10000 - num__1 ) = num__72515 x num__10000 - num__72515 x num__1 = num__725150000 - num__72515 = num__725077485 b <eor> b <eos> |
b |
subtract__10000.0__9999.0__ multiply__72515.0__10000.0__ multiply__72515.0__9999.0__ multiply__72515.0__9999.0__ |
subtract__10000.0__9999.0__ multiply__72515.0__10000.0__ subtract__725150000.0__72515.0__ subtract__725150000.0__72515.0__ |
| if square - root of num__15 is num__3.87 . then find the value of square - root of ( num__1.66666666667 ) <o> a ) num__1.29 <o> b ) num__1.39 <o> c ) num__1.49 <o> d ) num__1.59 <o> e ) num__1.69 |
= > root num__5 / root num__3 = ( root num__5 / root num__3 ) * ( root num__3 / root num__3 ) = root num__15 / root num__9 = num__3.87 / num__3 = num__1.29 answer : a <eor> a <eos> |
a |
round_down__3.87__ divide__3.87__3.0__ divide__3.87__3.0__ |
divide__15.0__5.0__ divide__3.87__3.0__ divide__3.87__3.0__ |
| a boat covers a certain distance downstream in num__1 hour while it comes back in num__1 ½ hours . if the speed of the stream be num__3 kmph what is the speed of the boat in still water ? <o> a ) num__12 kmph <o> b ) num__13 kmph <o> c ) num__14 kmph <o> d ) num__15 kmph <o> e ) none |
sol . let the speed of the boat in still water be x kmph . then speed downstream = ( x + num__3 ) kmph speed upstream = ( x - num__3 ) kmph . ∴ ( x + num__3 ) * num__1 = ( x - num__3 ) * num__1.5 ⇔ num__2 x + num__6 = num__3 x - num__9 ⇔ x = num__15 kmph . answer d <eor> d <eos> |
d |
subtract__3.0__1.0__ multiply__3.0__2.0__ add__3.0__6.0__ add__6.0__9.0__ round__15.0__ |
subtract__3.0__1.0__ multiply__3.0__2.0__ add__3.0__6.0__ add__6.0__9.0__ multiply__1.0__15.0__ |
| what is the product of the greatest num__2 digit multiple of num__9 and the greatest num__2 digit prime number ? <o> a ) num__9312 <o> b ) num__9408 <o> c ) num__9603 <o> d ) num__9603 <o> e ) num__9 |
702 |
the greatest num__2 digit multiple of num__9 : num__99 the greatest num__2 digit prime numebr : num__97 num__97 * num__99 . num__9603 c <eor> c <eos> |
c |
c |
| joe ’ s average ( arithmetic mean ) test score across num__4 equally weighted tests was num__90 . he was allowed to drop his lowest score . after doing so his average test score improved to num__95 . what is the lowest test score that was dropped ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__55 <o> d ) num__75 <o> e ) num__80 |
the arithmetic mean of num__4 equally weighted tests was num__90 . so what we can assume is that we have num__4 test scores each num__90 . he dropped his lowest score and the avg went to num__95 . this means that the lowest score was not num__90 and other three scores had given the lowest score num__5 each to make it up to num__90 too . when the lowest score was removed the other num__3 scores got their num__5 back . so the lowest score was num__3 * num__5 = num__15 less than num__90 . so the lowest score = num__90 - num__15 = num__75 answer ( d ) <eor> d <eos> |
d |
subtract__95.0__90.0__ multiply__3.0__5.0__ subtract__90.0__15.0__ subtract__90.0__15.0__ |
subtract__95.0__90.0__ multiply__3.0__5.0__ subtract__90.0__15.0__ subtract__90.0__15.0__ |
| what is the units digit of num__28 ! + num__50 ! + num__3 ! + num__4 ! ? <o> a ) num__0 <o> b ) num__2 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
for all n greater than num__4 the units digit of n ! is num__0 . the sum of the four units digits is num__0 + num__0 + num__6 + num__4 = num__10 the units digit is num__0 . the answer is a . <eor> a <eos> |
a |
add__4.0__6.0__ multiply__28.0__0.0__ |
add__4.0__6.0__ multiply__28.0__0.0__ |
| the radius of a wheel is num__22.4 cm . what is the distance covered by the wheel in making num__2000 resolutions ? <o> a ) num__1187 m <o> b ) num__1704 m <o> c ) num__2816 m <o> d ) num__2827 m <o> e ) num__2897 m |
in one resolution the distance covered by the wheel is its own circumference . distance covered in num__2000 resolutions . = num__2000 * num__2 * num__3.14285714286 * num__22.4 = num__281600 cm = num__2816 m answer : c <eor> c <eos> |
c |
round__2816.0__ |
round__2816.0__ |
| in the county of veenapaniville there are a total of num__50 high schools of three kinds : num__25 public schools num__16 parochial schools and num__9 private independent schools . these num__50 schools are divided between three districts : a b and c . district a has num__18 high schools total . district b has num__17 high schools total and only two of those are private independent schools . if district c has an equal number of each of the three kinds of schools how many private independent schools are there in district a ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
“ in the county of veenapaniville there are a total of num__50 high schools of three kinds : num__25 public schools num__16 parochial schools and num__9 private independent schools . ” “ district a has num__18 high schools total . ” “ district b has num__17 high schools total and only two of those are private independent schools . ” “ district c has an equal number of each of the three kinds of schools . ” hmm . that ’ s not a simple number . let ’ s put that on hold for a moment . so far from the matrix it looks like we could figure out the missing number in the “ totals ” row . that number must be num__15 because num__18 + num__17 + num__15 = num__50 . now go back to that last fact : “ district c has an equal number of each of the three kinds of schools . ” now that makes sense . district c has num__15 schools so it must have num__5 of each kind : the question is asking : “ how many private independent schools are there in district a ? ” well now that ’ s just a matter of adding across the “ private independent ” row . we know num__2 + num__2 + num__5 = num__9 so the missing number must be num__2 . there ’ s still a good deal of information we can ’ t figure out from the information given in the question but we are able to answer the specific question asked here . there are num__2 private independent schools in district a so the answer = ( a ) . <eor> a <eos> |
a |
divide__50.0__25.0__ divide__50.0__25.0__ |
divide__50.0__25.0__ divide__50.0__25.0__ |
| a sum was put at simple interest at certain rate for num__3 years . had it been put at num__1.0 higher rate it would have fetched rs . num__75 more . the sum is : a . rs . num__2400 b . rs . num__2100 c . rs . num__2200 d . rs . num__2480 <o> a ) num__2500 <o> b ) num__2100 <o> c ) num__2200 <o> d ) num__2300 <o> e ) num__2400 |
num__1 percent for num__3 years = num__75 num__1 percent for num__1 year = num__25 = > num__100 percent = num__2500 answer : a <eor> a <eos> |
a |
percent__100.0__2500.0__ |
percent__100.0__2500.0__ |
| the unit digit in the product ( num__611 * num__704 * num__912 * num__261 ) is : <o> a ) num__2 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
explanation : unit digit in the given product = unit digit in ( num__1 * num__4 * num__2 * num__1 ) = num__2 answer : a <eor> a <eos> |
a |
multiply__1.0__2.0__ |
multiply__1.0__2.0__ |
| the perimeter of one square is num__48 cm and that of another is num__20 cm . find the perimeter and the diagonal of a square which is equal in area to these two combined ? <o> a ) num__13 √ num__7 <o> b ) num__13 √ num__2 <o> c ) num__13 √ num__8 <o> d ) num__13 √ num__9 <o> e ) num__12 √ num__2 |
num__4 a = num__48 num__4 a = num__20 a = num__12 a = num__5 a num__2 = num__144 a num__2 = num__25 combined area = a num__2 = num__169 = > a = num__13 d = num__13 √ num__2 answer : b <eor> b <eos> |
b |
power__12.0__2.0__ power__5.0__2.0__ triangle_perimeter__20.0__144.0__5.0__ triangle_area__2.0__13.0__ |
power__12.0__2.0__ power__5.0__2.0__ triangle_perimeter__20.0__144.0__5.0__ triangle_area__2.0__13.0__ |
| ratio of ages of three persons is num__4 : num__7 : num__9 eight years ago the sum of their ages was num__56 . find their present ages . <o> a ) num__16 num__3536 <o> b ) num__12 num__2836 <o> c ) num__16 num__2827 <o> d ) num__16 num__2836 <o> e ) none of these |
explanation : let the present ages are num__4 x num__7 x num__9 x . = > ( num__4 x - num__8 ) + ( num__7 x - num__8 ) + ( num__9 x - num__8 ) = num__56 = > num__20 x = num__80 = > x = num__4 so their present ages are : num__16 num__2836 option d <eor> d <eos> |
d |
divide__56.0__7.0__ multiply__4.0__20.0__ add__7.0__9.0__ add__7.0__9.0__ |
divide__56.0__7.0__ multiply__4.0__20.0__ add__7.0__9.0__ add__7.0__9.0__ |
| an error num__2.0 in excess is made while measuring the sideofa square . the % of error in the calculated area of the square is ? <o> a ) num__4.0 <o> b ) num__4.04 <o> c ) num__4.15 <o> d ) num__5.0 <o> e ) num__5.24 % |
num__100 cm is read as num__102 cm . a num__1 = ( num__100 x num__100 ) cm num__2 and a num__2 ( num__102 x num__102 ) cm num__2 . ( a num__2 - a num__1 ) = [ ( num__102 ) num__2 - ( num__100 ) num__2 ] = ( num__102 + num__100 ) x ( num__102 - num__100 ) = num__404 cm num__2 . percentage error = num__404 x num__100.0 = num__4.04 num__100 x num__100 b <eor> b <eos> |
b |
rectangle_perimeter__100.0__102.0__ triangle_area__2.0__4.04__ |
rectangle_perimeter__100.0__102.0__ triangle_area__2.0__4.04__ |
| which of the following describes all the values of x for which num__8 − num__4 x < num__18 x − num__18 ? <o> a ) x > num__0.714285714286 <o> b ) x < num__0.454545454545 <o> c ) num__0.454545454545 < x <o> d ) x > num__1.18181818182 <o> e ) num__1.18181818182 > x |
num__8 − num__4 x < num__18 x − num__18 i . e . num__8 + num__18 < num__18 x + num__4 x i . e . num__26 < num__22 x i . e . x > num__1.18181818182 i . e . x > num__1.18181818182 answer : option d <eor> d <eos> |
d |
add__8.0__18.0__ add__4.0__18.0__ divide__26.0__22.0__ divide__26.0__22.0__ |
add__8.0__18.0__ add__4.0__18.0__ divide__26.0__22.0__ divide__26.0__22.0__ |
| how much does a watch lose per day if its hands coincide ever num__64 minutes ? <o> a ) num__32 num__0.727272727273 <o> b ) num__32 num__0.666666666667 <o> c ) num__32 num__0.444444444444 <o> d ) num__32 num__0.470588235294 <o> e ) num__32 num__0.571428571429 |
num__55 min spaces are covered in num__60 min num__60 min . spaces are covered in ( num__1.09090909091 x num__60 ) min = num__65 + num__0.454545454545 min . loss in num__64 min = ( num__65 + num__0.454545454545 ) - num__64 = num__1.45454545455 loss in num__24 hrs ( num__1.45454545455 x num__0.015625 x num__24 x num__60 ) min = num__32 num__0.727272727273 . answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ divide__60.0__55.0__ round__32.0__ |
hour_to_min_conversion__ divide__60.0__55.0__ subtract__64.0__32.0__ |
| an athlete runs num__200 metres race in num__24 seconds . what is his speed ? <o> a ) num__23 <o> b ) num__38 <o> c ) num__37 <o> d ) num__30 <o> e ) num__28 |
speed = distancetime = num__20024 m / s = num__20024 × num__185 km / hr = num__40 × num__34 km / hr = num__10 × num__3 km / hr = num__30 km / hr answer : d <eor> d <eos> |
d |
subtract__34.0__24.0__ multiply__3.0__10.0__ round__30.0__ |
subtract__34.0__24.0__ multiply__3.0__10.0__ round__30.0__ |
| a dishonest dealer professes to sell goods at the cost price but uses a weight of num__800 grams per kg what is his percent ? <o> a ) num__65.0 <o> b ) num__67.0 <o> c ) num__45.0 <o> d ) num__78.0 <o> e ) num__25 % |
e num__800 - - - num__200 num__100 - - - ? = > num__25.0 <eor> e <eos> |
e |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| what is the diffference between the place value and face value of num__7 in the numeral num__9751 ? <o> a ) num__691 <o> b ) num__692 <o> c ) num__693 <o> d ) num__694 <o> e ) num__695 |
place value of num__7 = num__7 * num__100 = num__700 face value of num__7 = num__7 num__700 - num__7 = num__693 c <eor> c <eos> |
c |
multiply__7.0__100.0__ subtract__700.0__7.0__ subtract__700.0__7.0__ |
multiply__7.0__100.0__ subtract__700.0__7.0__ subtract__700.0__7.0__ |
| a rectangular - shaped carpet that measures x feet by y feet is priced at $ num__40 . what is the cost of the carpet in dollars per square yard ? ( num__1 square yard = num__9 square feet ) <o> a ) xy / num__360 <o> b ) num__9 xy / num__40 <o> c ) num__40 xy / num__9 <o> d ) num__360 xy <o> e ) num__360 / ( xy ) |
the area of the carpet in feet is xy . the area in square yards is xy / num__9 . the price per square yard is num__40 / ( xy / num__9 ) = num__360 / ( xy ) . the answer is e . <eor> e <eos> |
e |
multiply__40.0__9.0__ multiply__40.0__9.0__ |
multiply__40.0__9.0__ divide__360.0__1.0__ |
| a sum of money is to be distributed among a b c d in the proportion of num__5 : num__2 : num__4 : num__3 . if c gets rs . num__1000 more than d what is b ' s share ? <o> a ) rs . num__2029 <o> b ) rs . num__2028 <o> c ) rs . num__2000 <o> d ) rs . num__2022 <o> e ) rs . num__2021 |
let the shares of a b c and d be num__5 x num__2 x num__4 x and num__3 x rs . respectively . then num__4 x - num__3 x = num__1000 = > x = num__1000 . b ' s share = rs . num__2 x = num__2 * num__1000 = rs . num__2000 . answer : c <eor> c <eos> |
c |
multiply__2.0__1000.0__ multiply__2.0__1000.0__ |
multiply__2.0__1000.0__ multiply__2.0__1000.0__ |
| a woman swims downstream num__45 km and upstream num__15 km taking num__3 hours each time what is the speed of the woman in still water ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__16 <o> e ) num__18 |
num__45 - - - num__3 ds = num__15 ? - - - - num__1 num__15 - - - - num__3 us = num__5 ? - - - - num__1 m = ? m = ( num__15 + num__5 ) / num__2 = num__10 answer : a <eor> a <eos> |
a |
divide__15.0__3.0__ subtract__3.0__1.0__ subtract__15.0__5.0__ round__10.0__ |
divide__15.0__3.0__ subtract__3.0__1.0__ subtract__15.0__5.0__ subtract__15.0__5.0__ |
| num__1 num__35 num__79 . . num__50 find term of sequnce for this . <o> a ) num__1200 <o> b ) num__1345 <o> c ) num__1456 <o> d ) num__1478 <o> e ) num__2500 |
this is an arithmetic progression and we can write down a = num__1 a = num__1 d = num__2 d = num__2 n = num__50 n = num__50 . we now use the formula so that sn = num__12 n ( num__2 a + ( n − num__1 ) l ) sn = num__12 n ( num__2 a + ( n − num__1 ) l ) s num__50 = num__12 × num__50 × ( num__2 × num__1 + ( num__50 − num__1 ) × num__2 ) s num__50 = num__12 × num__50 × ( num__2 × num__1 + ( num__50 − num__1 ) × num__2 ) = num__25 × ( num__2 + num__49 × num__2 ) = num__25 × ( num__2 + num__49 × num__2 ) = num__25 × ( num__2 + num__98 ) = num__25 × ( num__2 + num__98 ) = num__2500 = num__2500 . e <eor> e <eos> |
e |
divide__50.0__2.0__ subtract__50.0__1.0__ multiply__2.0__49.0__ power__50.0__2.0__ multiply__1.0__2500.0__ |
divide__50.0__2.0__ subtract__50.0__1.0__ multiply__2.0__49.0__ power__50.0__2.0__ multiply__1.0__2500.0__ |
| audrey num__4 hours to complete a certain job . ferris can do the same job in num__3 hours . audrey and ferris decided to collaborate on the job working at their respective rates . while audrey worked continuously ferris took num__1 breaks of equal length . if the two completed the job together in num__2 hours how many minutes long was each of ferris ’ breaks ? <o> a ) num__5 <o> b ) num__30 <o> c ) num__15 <o> d ) num__20 <o> e ) num__25 |
audery and ferris collective work rate : num__0.25 + num__0.333333333333 = num__0.583333333333 collective work time = num__1.71428571429 = num__1.7 hrs job was actually done in = num__2 ( includes breaks ) breaks = actual time taken - collective work time = num__2 - num__1.7 = . num__3 hrs = num__0.5 so ferrais took num__1 breaks = . num__3.0 = . num__3 hrs = num__30 m so answer is b ) num__30 mins <eor> b <eos> |
b |
divide__1.0__4.0__ divide__1.0__3.0__ add__0.25__0.3333__ divide__1.0__2.0__ round__30.0__ |
divide__1.0__4.0__ divide__1.0__3.0__ add__0.25__0.3333__ divide__1.0__2.0__ round__30.0__ |
| if x and y are prime numbers such that x > y > num__3 then x ^ num__2 − y ^ num__2 must be divisible by which one of the following numbers ? <o> a ) num__3 <o> b ) num__7 <o> c ) num__5 <o> d ) num__9 <o> e ) num__12 |
if x = num__6 and y = num__4 then x ^ num__2 - y ^ num__2 = num__20 and num__20 is divisible only by num__5 from the options thus it must be correct . answer : c . <eor> c <eos> |
c |
multiply__3.0__2.0__ subtract__6.0__2.0__ add__3.0__2.0__ add__3.0__2.0__ |
multiply__3.0__2.0__ subtract__6.0__2.0__ add__3.0__2.0__ add__3.0__2.0__ |
| if m = n / o - num__1 < o < num__0 and num__4 < n which of the following is correct ? <o> a ) m > num__4 . <o> b ) num__0 < m < num__4 . <o> c ) - num__4 < m < num__0 . <o> d ) m < - num__4 . <o> e ) m < - num__20 . |
n is + veo is - ve eliminate ab for min . value max . numerator and minimize dinominator take n = num__4.1 m = - num__0.9 m = num__4.1 / - num__0.9 m < - num__4 ans d <eor> d <eos> |
d |
round_down__4.1__ |
divide__4.0__1.0__ |
| there are num__4 consonants num__3 vowels how many num__6 letter strings can u form with atleast one vowels in each ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
num__6 letter strings with atleast one vowel means possibilities are num__1 ) num__4 constants and num__2 vowels = num__4 c num__4 * num__3 c num__2 = num__3 num__2 ) num__3 constants and num__3 vowels = num__4 c num__3 * num__3 c num__3 = num__4 so the answer is num__3 + num__4 = num__7 answer : c <eor> c <eos> |
c |
coin_space__ choose__7.0__6.0__ |
coin_space__ choose__7.0__6.0__ |
| the radius of a cylindrical water tank is reduced by num__50.0 . however the speed by which water is filled into the tank is also decreased by num__50.0 . how much more or less time b will it take to fill the tank now ? <o> a ) num__50.0 less time <o> b ) num__50.0 more time <o> c ) num__75.0 less time <o> d ) num__75.0 more time <o> e ) num__100.0 more time |
( vc ) volume of the cylinderical vessal is directly proportional to r ^ num__2 . so if radius is num__50.0 less volume will be num__0.25 th of the original volume . ( vc / num__4 ) now if with velocity v tank can be filled in t num__1 time of volume vc so now velocity is num__50.0 less i . . e v / num__2 so time taken to fill the capacity vc / num__4 by v / num__2 velocity is t num__2 . vt num__1 = vc v / num__2 * t num__2 = vc / num__4 so t num__1 / t num__2 = num__0.5 so tank will be filled in less time . that is b = num__50.0 less . a <eor> a <eos> |
a |
reverse__0.25__ multiply__0.25__4.0__ reverse__2.0__ multiply__50.0__1.0__ |
reverse__0.25__ multiply__0.25__4.0__ reverse__2.0__ multiply__50.0__1.0__ |
| a cat chases a rat num__6 hours after the rat runs . cat takes num__4 hours to reach the rat . if the average speed of the cat is num__90 kmph what s the average speed of the rat ? <o> a ) num__32 kmph <o> b ) num__26 kmph <o> c ) num__35 kmph <o> d ) num__36 kmph <o> e ) num__32 kmph |
cat take num__10 hours and rat take num__4 hours . . . then distance chased by them is num__90 * num__4 . so speed of rat is ( num__90 * num__4 ) / num__10 = num__36 kmph . answer is d <eor> d <eos> |
d |
add__6.0__4.0__ round__36.0__ |
add__6.0__4.0__ round__36.0__ |
| how many seconds will a num__500 m long train take to cross a man walking with a speed of num__3 km / hr in the direction of the moving train if the speed of the train is num__63 km / hr ? <o> a ) num__26 sec <o> b ) num__30 sec <o> c ) num__13 sec <o> d ) num__19 sec <o> e ) num__12 sec |
speed of train relative to man = num__63 - num__3 = num__60 km / hr . = num__60 * num__0.277777777778 = num__16.6666666667 m / sec . time taken to pass the man = num__500 * num__0.06 = num__30 sec . answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__500.0__0.06__ round__30.0__ |
subtract__63.0__3.0__ multiply__500.0__0.06__ multiply__500.0__0.06__ |
| if the sum of five consecutive positive integers is a then the sum of the next three consecutive integers in terms of a is : <o> a ) num__3 a / num__5 <o> b ) num__3 a / num__5 + num__12 <o> c ) a <o> d ) num__5 a / num__3 <o> e ) num__5 a / num__3 + num__12 |
in case of consecutive integers or integers in arithmetic progression mean = median i . e . median = a / num__5 = mean = third integer first integer = a / num__5 - num__2 second integer = a / num__5 - num__1 third integer = a / num__5 fourth integer = a / num__5 + num__2 fifth integer = a / num__5 + num__2 i . e . sixth integer = a / num__5 + num__3 seventh integer = a / num__5 + num__4 eighth integer = a / num__5 + num__5 now mean of next num__3 integers = median = a / num__5 + num__4 i . e . sum of next num__3 integers = ( a / num__5 + num__4 ) * num__3 = num__3 a / num__5 + num__12 answer : option b <eor> b <eos> |
b |
add__1.0__2.0__ add__1.0__3.0__ multiply__3.0__4.0__ add__1.0__2.0__ |
add__1.0__2.0__ add__1.0__3.0__ multiply__3.0__4.0__ add__1.0__2.0__ |
| find value of x : num__121 × num__54 = x <o> a ) num__34545 <o> b ) num__46576 <o> c ) num__67887 <o> d ) num__57687 <o> e ) num__75625 |
num__121 × num__54 = num__121 × ( num__102 ) num__4 = num__121 × num__1000016 = num__7.5625 × num__10000 = num__75625 e <eor> e <eos> |
e |
multiply__7.5625__10000.0__ multiply__7.5625__10000.0__ |
multiply__7.5625__10000.0__ multiply__7.5625__10000.0__ |
| what is the smallest no . that should be added to num__45454 to make it exactly divisible by num__9 ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__5 <o> d ) num__6 <o> e ) num__9 |
if a number is divisible by num__9 the sum of its digits must be a multiple of num__9 . here num__4 + num__5 + num__4 + num__5 + num__4 = num__22 the next multiple of num__9 is num__27 . num__5 must be added to num__45454 to make it divisible by num__9 c <eor> c <eos> |
c |
subtract__9.0__4.0__ add__5.0__22.0__ subtract__9.0__4.0__ |
subtract__9.0__4.0__ add__5.0__22.0__ subtract__9.0__4.0__ |
| how many cubes of num__25 cm edge can be put in a cubical box of num__1 m edge . <o> a ) num__177 cm <o> b ) num__100 cm <o> c ) num__86 cm <o> d ) num__64 cm <o> e ) num__87 cm |
number of cubes = num__100 â ˆ — num__100 â ˆ — num__4.0 â ˆ — num__25 â ˆ — num__25 = num__64 note : num__1 m = num__100 cm answer : d <eor> d <eos> |
d |
divide__100.0__25.0__ round__64.0__ |
divide__100.0__25.0__ round__64.0__ |
| a person crosses a num__600 m long street in num__5 minutes . what is his speed in km per hour ? <o> a ) num__2 km / hr <o> b ) num__5 km / hr <o> c ) num__7.2 km / hr <o> d ) num__9 km / hr <o> e ) num__11 km / hr |
explanation : speed = ( num__120.0 * num__60 ) m / sec = num__2 m / sec . converting m / sec to km / hr = ( num__2 * num__3.6 ) km / hr = num__7.2 km / hr answer : c <eor> c <eos> |
c |
divide__600.0__5.0__ hour_to_min_conversion__ divide__120.0__60.0__ multiply__2.0__3.6__ round__7.2__ |
divide__600.0__5.0__ hour_to_min_conversion__ divide__120.0__60.0__ multiply__2.0__3.6__ multiply__2.0__3.6__ |
| if the cost price of num__17 articles is equal to the selling price of num__16 articles what is the percentage of profit or loss that the merchant makes ? <o> a ) num__20.0 loss <o> b ) num__6.25 profit <o> c ) num__33.33 loss <o> d ) num__30.33 loss <o> e ) none of these |
explanation : let cost price of num__1 article be re . num__1 . therefore cost price of num__17 articles = rs . num__17 . selling price of num__16 articles = rs . num__17 therefore selling price of num__17 articles is : - = > num__1.0625 Ã — num__17 = > num__18.06 therefore profit = selling price - cost price . = > num__18.06 â ˆ ’ num__17 = num__1.06 hence the percentage of profit = profit x num__100 / c . p . = > num__1.06 / num__17 Ã — num__100 . = > num__6.25 profit . answer : b <eor> b <eos> |
b |
percent__100.0__6.25__ |
percent__100.0__6.25__ |
| if a man can cover num__12 metres in one second how many kilometres can he cover in num__3 hours num__45 minutes ? <o> a ) num__228 <o> b ) num__162 <o> c ) num__5528 <o> d ) num__256 <o> e ) num__191 |
num__12 m / s = num__12 * num__3.6 kmph num__3 hours num__45 minutes = num__3 num__0.75 hours = num__3.75 hours distance = speed * time = num__12 * num__3.6 * num__3.75 km = num__162 km . answer : b <eor> b <eos> |
b |
add__3.0__0.75__ multiply__45.0__3.6__ round__162.0__ |
divide__45.0__12.0__ multiply__45.0__3.6__ multiply__45.0__3.6__ |
| a man buys rs . num__20 shares paying num__9.0 dividend . the man wants to have an interest of num__12.0 on his money . the market value of each share is : <o> a ) rs . num__12 <o> b ) rs . num__15 <o> c ) rs . num__18 <o> d ) rs . num__21 <o> e ) rs . num__24 |
dividend on rs . num__20 = rs . num__0.09 x num__20 = rs . num__1.8 . rs . num__12 is an income on rs . num__100 . rs . num__1.8 is an income on rs . num__8.33333333333 x num__1.8 = rs . num__15 . answer : option b <eor> b <eos> |
b |
percent__20.0__9.0__ percent__100.0__15.0__ |
percent__20.0__9.0__ percent__100.0__15.0__ |
| in the coordinate plane points ( x num__1 ) and ( num__10 y ) are on line k . if line k passes through the origin and has slope num__0.333333333333 then x + y = <o> a ) num__4.5 <o> b ) num__7 <o> c ) num__8 <o> d ) num__6.33333333333 <o> e ) num__12 |
line k passes through the origin and has slope num__0.333333333333 means that its equation is y = num__0.333333333333 * x . thus : ( x num__1 ) = ( num__3 num__1 ) and ( num__10 y ) = ( num__336.666666667 ) - - > x + y = num__3 + num__3.33333333333 = num__6.33333333333 . answer : d . <eor> d <eos> |
d |
divide__10.0__3.0__ add__3.3333__3.0__ multiply__1.0__6.3333__ |
add__0.3333__3.0__ add__3.3333__3.0__ multiply__1.0__6.3333__ |
| what will be the cost of building a fence around a square plot with area equal to num__289 sq ft if the price per foot of building the fence is rs . num__58 ? <o> a ) rs . num__3944 <o> b ) rs . num__3828 <o> c ) rs . num__4176 <o> d ) can not be determined <o> e ) none of these |
let the side of the square plot be a ft . a num__2 = num__289 = > a = num__17 length of the fence = perimeter of the plot = num__4 a = num__68 ft . cost of building the fence = num__68 * num__58 = rs . num__3944 . answer : a <eor> a <eos> |
a |
square_perimeter__17.0__ multiply__58.0__68.0__ multiply__58.0__68.0__ |
multiply__17.0__4.0__ multiply__58.0__68.0__ multiply__58.0__68.0__ |
| if the num__5 th date of a month is tuesday what date will be num__3 days after the num__3 rd friday in the month ? <o> a ) num__17 <o> b ) num__18 <o> c ) num__20 <o> d ) num__22 <o> e ) num__33 |
b num__18 num__5 th date of a month is tuesday friday will be on = num__5 + num__3 = num__8 th of a month num__1 st friday is on num__1 st of a month num__2 nd friday is on num__8 th of a month num__3 rd friday will be on num__15 th of a month num__3 days after num__15 th = num__15 + num__3 = num__18 <eor> b <eos> |
b |
add__5.0__3.0__ subtract__5.0__3.0__ multiply__5.0__3.0__ round__18.0__ |
add__5.0__3.0__ subtract__5.0__3.0__ multiply__5.0__3.0__ add__3.0__15.0__ |
| an order was placed for the supply of a carper whose length and breadth were in the ratio of num__3 : num__2 . subsequently the dimensions of the carpet were altered such that its length and breadth were in the ratio num__7 : num__3 but were was no change in its parameter . find the ratio of the areas of the carpets in both the cases . <o> a ) num__8 : num__9 <o> b ) num__8 : num__7 <o> c ) num__8 : num__1 <o> d ) num__8 : num__2 <o> e ) num__8 : num__8 |
let the length and breadth of the carpet in the first case be num__3 x units and num__2 x units respectively . let the dimensions of the carpet in the second case be num__7 y num__3 y units respectively . from the data . num__2 ( num__3 x + num__2 x ) = num__2 ( num__7 y + num__3 y ) = > num__5 x = num__10 y = > x = num__2 y required ratio of the areas of the carpet in both the cases = num__3 x * num__2 x : num__7 y : num__3 y = num__6 x num__2 : num__21 y num__2 = num__6 * ( num__2 y ) num__2 : num__21 y num__2 = num__6 * num__4 y num__2 : num__21 y num__2 = num__8 : num__7 answer : b <eor> b <eos> |
b |
vowel_space__ die_space__ choose__8.0__7.0__ |
vowel_space__ die_space__ choose__8.0__7.0__ |
| use distributive property to solve the problem below : maria bought num__12 notebooks and num__6 pens costing num__2 dollars each . how much did maria pay ? <o> a ) num__30 dollars <o> b ) num__36 dollars <o> c ) num__50 dollars <o> d ) num__60 dollars <o> e ) num__70 dollars |
solution num__2 × ( num__12 + num__6 ) = num__2 × num__12 + num__2 × num__6 = num__24 + num__12 = num__36 dollars answer b <eor> b <eos> |
b |
multiply__12.0__2.0__ add__12.0__24.0__ add__12.0__24.0__ |
multiply__12.0__2.0__ add__12.0__24.0__ add__12.0__24.0__ |
| the average age of a husband wife and their child num__3 years ago was num__27 years and that of wife and the child num__5 years ago was num__20 years . the present age of the husband is ? <o> a ) num__19 years <o> b ) num__18 years <o> c ) num__61 years <o> d ) num__14 years <o> e ) num__13 years |
sum of the present ages of husband wife and child = ( num__23 * num__2 + num__5 * num__2 ) = num__57 years . required average = num__19.0 = num__19 years . answer : a <eor> a <eos> |
a |
add__3.0__20.0__ subtract__5.0__3.0__ divide__57.0__3.0__ divide__57.0__3.0__ |
add__3.0__20.0__ subtract__5.0__3.0__ divide__57.0__3.0__ divide__57.0__3.0__ |
| together num__15 type a machines and num__7 type b machines can complete a certain job in num__4 hours . together num__8 type b machines and num__15 type c machines can complete the same job in num__11 hours . how many hours r would it take one type a machine one type b machine and one type c machine working together to complete the job ( assuming constant rates for each machine ) ? <o> a ) num__22 hours <o> b ) num__30 hours <o> c ) num__44 hours <o> d ) num__60 hours <o> e ) it can not be determined from the information above . |
say the rates of machines a b and c are a b and c respectively . together num__15 type a machines and num__7 type b machines can complete a certain job in num__4 hours - - > num__15 a + num__7 b = num__0.25 ; together num__8 type b machines and num__15 type c machines can complete the same job in num__11 hours - - > num__8 b + num__15 c = num__0.0909090909091 . sum the above : num__15 a + num__15 b + num__15 c = num__0.25 + num__0.0909090909091 = num__0.340909090909 - - > reduce by num__15 : a + b + c = num__0.0227272727273 - - > so the combined rate of the three machines is num__0.0227272727273 job / hour - - > time is reciprocal of the rate thus machines a b and c can do the job r in num__44 hours . answer : c . <eor> c <eos> |
c |
add__0.25__0.0909__ divide__0.25__11.0__ multiply__4.0__11.0__ round__44.0__ |
add__0.25__0.0909__ divide__0.25__11.0__ divide__11.0__0.25__ divide__11.0__0.25__ |
| a can do a piece of work in num__8 hours ; b and c together can do it in num__12 hours while a and c together can do it in num__16 hours . how long will b alone take to do it ? <o> a ) num__7 num__1.6 hours <o> b ) num__5 num__1.125 hours <o> c ) num__5 num__0.75 hours <o> d ) num__6 num__0.857142857143 hours <o> e ) none of the above |
a ' s num__1 hour ' s work = num__0.125 ; ( b + c ) ' s num__1 hour ' s work = num__0.0833333333333 ; ( b + c ) ' s num__1 hour ' s work = num__0.0625 ( a + b + c ) ' s num__1 hour ' s work = ( num__0.125 + num__0.0833333333333 ) = num__0.208333333333 b ' s num__1 hour ' s work = ( num__0.208333333333 - num__0.0625 ) = num__0.145833333333 b alone will take num__6.85714285714 hours to do the work . = num__6 num__0.857142857143 answer = d <eor> d <eos> |
d |
divide__1.0__8.0__ divide__1.0__12.0__ divide__1.0__16.0__ add__0.125__0.0833__ add__0.0625__0.0833__ multiply__0.125__6.8571__ round__6.0__ |
divide__1.0__8.0__ divide__1.0__12.0__ divide__1.0__16.0__ add__0.125__0.0833__ add__0.0625__0.0833__ subtract__6.8571__6.0__ subtract__12.0__6.0__ |
| an electric pump can fill a tank in num__7 hours . because of a leak in the tank it took num__14 hours to fill the tank . if the tank is full how much time will the leak take to empty it ? <o> a ) num__10 hours <o> b ) num__12 hours <o> c ) num__8 hours <o> d ) num__5 hours <o> e ) num__14 hours |
work done by the leak in num__1 hour = num__0.142857142857 - num__0.0714285714286 = num__0.0714285714286 the leak will empty the tank in num__14 hours answer is e <eor> e <eos> |
e |
divide__1.0__7.0__ divide__1.0__14.0__ round__14.0__ |
divide__1.0__7.0__ divide__1.0__14.0__ round__14.0__ |
| in the storage room of a certain bakery the ratio of sugar to flour is num__5 to num__5 and the ratio of flour to baking soda is num__10 to num__1 . if there were num__60 more pounds of baking soda in the room the ratio of flour to baking soda would be num__8 to num__1 . how many pounds of sugar are stored in the room ? <o> a ) num__600 <o> b ) num__1200 <o> c ) num__1500 <o> d ) num__1600 <o> e ) num__2400 |
sugar : flour = num__5 : num__5 = num__10 : num__10 ; flour : soda = num__10 : num__1 = num__10 : num__1 ; thus we have that sugar : flour : soda = num__10 x : num__10 x : num__1 x . also given that num__10 x / ( num__1 x + num__60 ) = num__8.0 - - > x = num__240 - - > sugar = num__10 x = num__2400 answer : e . <eor> e <eos> |
e |
multiply__10.0__240.0__ multiply__10.0__240.0__ |
multiply__10.0__240.0__ divide__2400.0__1.0__ |
| after a storm deposits num__115 billion gallons of water into the city reservoir the reservoir is num__80.0 full . if the original contents of the reservoir totaled num__245 billion gallons the reservoir was approximately what percentage full before the storm ? <o> a ) num__45.0 <o> b ) num__48.0 <o> c ) num__54.0 <o> d ) num__58.0 <o> e ) num__65 % |
when the storm deposited num__115 billion gallons volume of water in the reservoir = num__245 + num__115 = num__360 billion gallons if this is only num__80.0 of the capacity of the reservoir the total capacity of the reservoir = num__360 / num__0.8 = num__450 billion gallons therefore percentage of reservoir that was full before the storm = ( num__0.544444444444 ) * num__100 = num__54.4 option c <eor> c <eos> |
c |
add__115.0__245.0__ divide__360.0__0.8__ divide__245.0__450.0__ divide__80.0__0.8__ round_down__54.4__ |
add__115.0__245.0__ divide__360.0__0.8__ divide__245.0__450.0__ divide__80.0__0.8__ round_down__54.4__ |
| a certain company reported that the revenue on sales increased num__20.0 from num__2000 to num__2003 and increased num__40.0 from num__2000 to num__2005 . what was the approximate percent increase in revenue for this store from num__2003 to num__2005 ? <o> a ) num__50.0 <o> b ) num__40.0 <o> c ) num__35.0 <o> d ) num__32.0 <o> e ) num__17 % |
assume the revenue in num__2000 to be num__100 . then in num__2003 it would be num__120 and and in num__2005 num__140 so from num__2003 to num__2005 it increased by ( num__140 - num__120 ) / num__120 = num__0.166666666667 = num__17.0 answer : e . <eor> e <eos> |
e |
percent__100.0__17.0__ |
percent__100.0__17.0__ |
| a box contains num__100 balls numbered from num__1 to num__100 . if three balls are selected at random and with replacement from the box what is the probability k that the sum of the three numbers on the balls selected from the box will be odd ? <o> a ) num__0.25 <o> b ) num__0.375 <o> c ) num__0.5 <o> d ) num__0.625 <o> e ) num__0.75 |
the sum of the three numbers on the balls selected from the box to be odd one should select either three odd numbered balls ( odd + odd + odd = odd ) or two even numbered balls and one odd numbered ball ( even + even + odd = odd ) ; p ( ooo ) = ( num__0.5 ) ^ num__3 ; p ( eeo ) = num__3 * ( num__0.5 ) ^ num__2 * num__0.5 = num__0.375 ( you should multiply by num__3 as the scenario of two even numbered balls and one odd numbered ball can occur in num__3 different ways : eeo eoe or oee ) ; so finally k = num__0.125 + num__0.375 = num__0.5 . answer : c . <eor> c <eos> |
c |
reverse__0.5__ subtract__0.5__0.375__ reverse__2.0__ |
reverse__0.5__ subtract__0.5__0.375__ add__0.375__0.125__ |
| the average of any num__5 consecutive odd natural numbers is k . if two more such numbers just next to the previous num__5 numbers are added the new average becomes <o> a ) k + num__8 <o> b ) k + num__1 <o> c ) k + num__9 <o> d ) k + num__2 <o> e ) k + num__4 |
explanation : the num__5 consecutive odd numbers whose average is k are ( k - num__4 ) ( k - num__2 ) k ( k + num__2 ) ( k + num__4 ) again the average of ( k - num__4 ) ( k - num__2 ) k ( k + num__2 ) ( k + num__4 ) ( k + num__6 ) ( k + num__8 ) is ( k + num__2 ) answer : d <eor> d <eos> |
d |
add__2.0__4.0__ multiply__2.0__4.0__ divide__8.0__4.0__ |
add__2.0__4.0__ add__2.0__6.0__ subtract__8.0__6.0__ |
| what is the remainder when num__43717 ^ ( num__43628233 ) is divided by num__5 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
we need to find the units digit of the number . the units digit of powers of seven repeats num__7 num__9 num__3 and num__1 cyclically . since num__43628233 has the form num__4 a + num__1 the units digit is num__7 . then the remainder when dividing by num__5 is num__2 . the answer is c . <eor> c <eos> |
c |
subtract__5.0__1.0__ subtract__5.0__3.0__ subtract__5.0__3.0__ |
add__1.0__3.0__ subtract__5.0__3.0__ subtract__5.0__3.0__ |
| if num__0.25 of the time from midnight plus num__0.5 of the time from now to midnight is the present time then what is the present time ? <o> a ) num__9 : num__36 am <o> b ) num__9 : num__39 am <o> c ) num__9 : num__26 am <o> d ) num__9 : num__46 am <o> e ) num__9 : num__56 am |
midnight = num__12 am let the present time is x ( taking the total time from one midnight to other is num__24 hours ) so according to the given condition x / num__4 + ( num__24 - x ) / num__2 = x on solving we will get x = num__9.6 means num__9 hours num__0.6 * num__60 minute = num__9 : num__36 am answer : a <eor> a <eos> |
a |
divide__12.0__0.5__ reverse__0.25__ reverse__0.5__ round_down__9.6__ subtract__9.6__9.0__ multiply__0.6__60.0__ round_down__9.6__ |
divide__12.0__0.5__ reverse__0.25__ reverse__0.5__ round_down__9.6__ subtract__9.6__9.0__ multiply__0.6__60.0__ multiply__0.25__36.0__ |
| a small experimental plane has three engines one of which is redundant . that is as long as two of the engines are working the plane will stay in the air . over the course of a typical flight there is a num__0.333333333333 chance that engine one will fail . there is a num__79.0 probability that engine two will work . the third engine works only half the time . what is the probability that the plane will crash in any given flight ? <o> a ) num__0.583333333333 <o> b ) num__0.25 <o> c ) num__0.5 <o> d ) num__0.291666666667 <o> e ) num__0.708333333333 |
in probability questions the trap answer is just the multiple of the numbers in the question . i . e . if you multiply num__0.333333333333 * num__0.25 * num__0.5 = num__0.0416666666667 is trap answer the other trap answer could be num__0.666666666667 * num__0.75 * num__0.5 = num__0.25 is trap answer so lets say you have num__30 secsand you want to guess the answer then b c are ruled out because they can be traps . you best guess is a d e . so you have num__33.0 chances of being correct . a <eor> a <eos> |
a |
negate_prob__0.3333__ negate_prob__0.25__ union_prob__0.3333__0.5__0.25__ |
negate_prob__0.3333__ negate_prob__0.25__ union_prob__0.3333__0.5__0.25__ |
| a man has rs . num__312 in the denominations of one - rupee notes five - rupee notes and twenty - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__36 <o> b ) num__35 <o> c ) num__34 <o> d ) num__33 <o> e ) num__31 |
let the number of notes of each denomination be x then x + num__5 x + num__20 x = num__312 = > num__26 x = num__312 = > x = num__12.0 = num__12 the total number of notes that he has = num__3 x = num__3 × num__12 = num__36 answer is a . <eor> a <eos> |
a |
divide__312.0__26.0__ multiply__3.0__12.0__ multiply__3.0__12.0__ |
divide__312.0__26.0__ multiply__3.0__12.0__ multiply__3.0__12.0__ |
| if n is a natural number then ( num__6 n ^ num__2 + num__6 n ) is always divisible by : <o> a ) num__6 <o> b ) num__12 <o> c ) num__24 <o> d ) num__6 only <o> e ) num__6 and num__12 both |
explanation : ( num__6 n ^ num__2 + num__6 n ) = num__6 n ( n + num__1 ) which is always divisible by num__6 and num__12 both since n ( n + num__1 ) is always even . e <eor> e <eos> |
e |
multiply__6.0__2.0__ multiply__6.0__1.0__ |
multiply__6.0__2.0__ multiply__6.0__1.0__ |
| when num__0.1 percent of num__6000 is subtracted from num__0.1 of num__6000 the difference is <o> a ) num__60 <o> b ) num__600 <o> c ) num__5994 <o> d ) num__5940 <o> e ) num__594 |
( num__0.1 ) * num__6000 - ( num__0.1 ) % * num__6000 = num__600 - ( num__0.001 ) * num__6000 = num__600 - num__6 = num__594 the answer is e . <eor> e <eos> |
e |
multiply__0.1__6000.0__ multiply__6000.0__0.001__ subtract__600.0__6.0__ subtract__600.0__6.0__ |
multiply__0.1__6000.0__ multiply__6000.0__0.001__ subtract__600.0__6.0__ subtract__600.0__6.0__ |
| what is the least number which when divided by num__5 num__9 num__12 and num__18 leaves remainder num__4 in each care ? <o> a ) num__130 <o> b ) num__140 <o> c ) num__136 <o> d ) num__184 <o> e ) num__196 |
explanation : lcm of num__5 num__9 num__12 and num__18 is num__180 required number = num__180 + num__4 = num__184 answer : option d <eor> d <eos> |
d |
add__4.0__180.0__ add__4.0__180.0__ |
add__4.0__180.0__ add__4.0__180.0__ |
| find the compound interest accrued on an amount of rs . num__14800 at num__13.5 p . a at the end of two years . ( round off your answer to nearest integer ) <o> a ) rs . num__4273 <o> b ) rs . num__4613 <o> c ) rs . num__4064 <o> d ) rs . num__4266 <o> e ) none of these |
ci = num__14800 { [ num__1 + num__13.5 / num__100 ] num__2 - num__1 } = num__14800 { [ num__1 + num__0.135 ] num__2 - num__1 = num__14800 { num__2 + num__0.135 } { num__0.135 } = ( num__74 ) [ num__2 + num__0.135 ] ( num__27 ) = num__1998 [ num__2 + num__0.135 ] = num__3996 + num__269.73 = rs . num__4266 answer : d <eor> d <eos> |
d |
percent__13.5__1.0__ percent__13.5__14800.0__ percent__27.0__14800.0__ percent__13.5__1998.0__ percent__100.0__4266.0__ |
percent__13.5__1.0__ percent__13.5__14800.0__ percent__27.0__14800.0__ percent__13.5__1998.0__ percent__100.0__4266.0__ |
| out of first num__28 natural numbers one number is selected at random . the probability that it is either an even number or a prime number is - . <o> a ) num__0.5 <o> b ) num__0.842105263158 <o> c ) num__0.8 <o> d ) num__0.785714285714 <o> e ) num__0.6 |
n ( s ) = num__28 n ( even no ) = num__14 = n ( e ) n ( prime no ) = num__9 = n ( p ) p ( e ᴜ p ) = num__0.5 + num__0.321428571429 - num__0.0357142857143 = num__0.785714285714 answer : d <eor> d <eos> |
d |
union_prob__0.5__0.3214__0.0357__ union_prob__0.5__0.3214__0.0357__ |
union_prob__0.5__0.3214__0.0357__ union_prob__0.5__0.3214__0.0357__ |
| mahesh can do a piece of work in num__50 days . he works at it for num__20 days and then rajesh finished it in num__30 days . how long will y take to complete the work ? <o> a ) num__45 <o> b ) num__25 <o> c ) num__37 <o> d ) num__41 <o> e ) num__50 |
work done by mahesh in num__50 days = num__20 * num__0.02 = num__0.4 remaining work = num__1 - num__0.4 = num__0.6 num__0.6 work is done by rajesh in num__30 days whole work will be done by rajesh is num__30 * num__1.66666666667 = num__50 days answer is e <eor> e <eos> |
e |
divide__20.0__50.0__ multiply__50.0__0.02__ km_to_mile_conversion__ divide__50.0__30.0__ round__50.0__ |
multiply__20.0__0.02__ multiply__50.0__0.02__ multiply__30.0__0.02__ divide__50.0__30.0__ round__50.0__ |
| the distance between a & b is num__600 km . a person is traveling from a to b at num__70 km / hr started at num__10 am and another person is traveling from b to a at num__80 km / hr and started at same time . then at what time they meet together . <o> a ) num__2 pm <o> b ) num__4 pm <o> c ) num__1 pm <o> d ) num__3 pm <o> e ) num__5 pm |
let x hours be they will meet together distance covered by num__1 st person + distance covered by num__2 nd person = num__600 km num__70 x + num__80 x = num__600 x = num__4 hr they will meet = num__10 am + num__4 hr = num__2 pm answer is a <eor> a <eos> |
a |
round__2.0__ |
round__2.0__ |
| crazy eddie has a key chain factory . eddie managed to decrease the cost of manufacturing his key chains while keeping the same selling price and thus increased the profit from the sale of each key chain from num__25.0 of the selling price to num__50.0 of the selling price . if the manufacturing cost is now $ num__50 what was it before the decrease ? <o> a ) $ num__75 <o> b ) $ num__40 <o> c ) $ num__50 <o> d ) $ num__80 <o> e ) $ num__100 |
deargoodyear num__2013 i ' m happy to help . this is a relatively straightforward problem not very challenging . btw crazy eddiewas the actually name of an electronics chain on the east coast of the usa back in the num__1970 s . manufacturing now is $ num__50 . they now are making a num__50.0 profit so the selling price must be $ num__100 . they had this same selling price $ num__100 before they made the change and had a profit of num__25.0 so the manufacturing must have been $ num__75 . answer = ( a ) . <eor> a <eos> |
a |
add__25.0__50.0__ add__25.0__50.0__ |
add__25.0__50.0__ add__25.0__50.0__ |
| your friend collecting balls . suppose he start out with num__59 balls . you gave him another num__20 balls . how many balls do your friend have at the end ? <o> a ) num__79 <o> b ) num__97 <o> c ) num__59 <o> d ) num__20 <o> e ) num__95 |
solution friend start with num__59 balls . you gave num__20 balls : num__59 + num__20 = num__79 balls . so your friend will have num__79 balls at the end . correct answer : a <eor> a <eos> |
a |
add__59.0__20.0__ add__59.0__20.0__ |
add__59.0__20.0__ add__59.0__20.0__ |
| the speed at which a man can row a boat in still water is num__15 kmph . if he rows downstream where the speed of current is num__3 kmph what time will he take to cover num__100 metres ? <o> a ) num__22 seconds <o> b ) num__65 seconds <o> c ) num__78 seconds <o> d ) num__20 seconds <o> e ) num__21 seconds |
speed of the boat downstream = num__15 + num__3 = num__18 kmph = num__18 * num__0.277777777778 = num__5 m / s hence time taken to cover num__60 m = num__20.0 = num__20 seconds . answer : d <eor> d <eos> |
d |
add__15.0__3.0__ divide__15.0__3.0__ hour_to_min_conversion__ add__15.0__5.0__ round__20.0__ |
add__15.0__3.0__ divide__15.0__3.0__ hour_to_min_conversion__ add__15.0__5.0__ add__15.0__5.0__ |
| in a mixture of milk and water the proportion of milk by weight was num__80.0 . if in a num__180 gm mixture num__36 gms of pure milk is added what would be the percentage of milk in the mixture formed ? <o> a ) num__80.0 <o> b ) num__100.0 <o> c ) num__84.0 <o> d ) num__87.5 <o> e ) none of these |
percentage of milk in the mixture formed = [ num__0.8 ( num__180 ) + num__36 ] / ( num__180 + num__36 ) * num__100.0 = ( num__144 + num__36 ) / num__216 * num__100.0 = num__0.833333333333 * num__100.0 = num__83.33 . answer : a <eor> a <eos> |
a |
divide__80.0__0.8__ subtract__180.0__36.0__ add__180.0__36.0__ divide__180.0__216.0__ multiply__0.8333__100.0__ subtract__180.0__100.0__ |
divide__80.0__0.8__ multiply__180.0__0.8__ add__180.0__36.0__ divide__180.0__216.0__ multiply__0.8333__100.0__ multiply__0.8__100.0__ |
| calculate the circumference of a circular field whose radius is num__4 centimeters . <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
circumference c is given by c = num__2 Ï € r = num__2 Ï € * num__4 = num__8 Ï € cm correct answer d <eor> d <eos> |
d |
square_perimeter__2.0__ square_perimeter__2.0__ |
multiply__4.0__2.0__ multiply__4.0__2.0__ |
| a rectangular lawn num__55 m by num__35 m has two roads each num__4 m wide running in the middle of it . one parallel to the length and the other parallel to breadth . the cost of graveling the roads at num__75 paise per sq meter is <o> a ) num__378 <o> b ) num__278 <o> c ) num__258 <o> d ) num__287 <o> e ) num__271 |
explanation : area of cross roads = num__55 * num__4 + num__35 * num__4 - num__4 * num__4 = num__344 sq m cost of graveling = num__344 * ( num__0.75 ) = rs . num__258 answer : c ) rs . num__258 <eor> c <eos> |
c |
multiply__0.75__344.0__ multiply__0.75__344.0__ |
multiply__0.75__344.0__ multiply__0.75__344.0__ |
| a jogger running at num__9 km / hr along side a railway track is num__240 m ahead of the engine of a num__120 m long train running at num__45 km / hr in the same direction . in how much time will the train pass the jogger ? <o> a ) num__67 sec <o> b ) num__89 sec <o> c ) num__36 sec <o> d ) num__87 sec <o> e ) num__45 sec |
speed of train relative to jogger = num__45 - num__9 = num__36 km / hr . = num__36 * num__0.277777777778 = num__10 m / sec . distance to be covered = num__240 + num__120 = num__360 m . time taken = num__36.0 = num__36 sec . answer : c <eor> c <eos> |
c |
subtract__45.0__9.0__ add__240.0__120.0__ subtract__45.0__9.0__ |
subtract__45.0__9.0__ add__240.0__120.0__ subtract__45.0__9.0__ |
| the difference between the length and breadth of a rectangle is num__23 m . if its perimeter is num__206 m then its area is : <o> a ) num__2520 <o> b ) num__2420 <o> c ) num__2320 <o> d ) num__2620 <o> e ) num__2760 |
we have : ( l - b ) = num__23 and num__2 ( l + b ) = num__206 or ( l + b ) = num__103 . solving the two equations we get : l = num__63 and b = num__40 . area = ( l x b ) = ( num__63 x num__40 ) = num__2520 sq . m answer : a <eor> a <eos> |
a |
multiply__40.0__63.0__ triangle_area__2.0__2520.0__ |
multiply__40.0__63.0__ triangle_area__2.0__2520.0__ |
| a bat is bought for rs . num__400 and sold at a gain of num__40.0 find its selling price <o> a ) a ) rs . num__560 / - <o> b ) b ) rs . num__470 / - <o> c ) c ) rs . num__480 / - <o> d ) d ) rs . num__500 / - <o> e ) e ) rs . num__520 / - |
num__100.0 - - - - - - > num__400 ( num__100 * num__4 = num__400 ) num__140.0 - - - - - - > num__560 ( num__140 * num__4 = num__560 ) selling price = rs . num__560 / - answer : a <eor> a <eos> |
a |
percent__100.0__560.0__ |
percent__100.0__560.0__ |
| the angles of a triangle are in the ratio num__4 : num__3 : num__8 . find the measurement of the largest of three angles of triangle . <o> a ) num__96 ° <o> b ) num__100 ° <o> c ) num__120 ° <o> d ) num__140 ° <o> e ) num__160 ° |
if the ratio of the three angles is num__4 : num__3 : num__8 then the measures of these angles can be written as num__4 x num__3 x and num__8 x . also the sum of the three interior angles of a triangle is equal to num__180 ° . hence num__4 x + num__3 x + num__8 x = num__180 solve for x num__15 x = num__180 x = num__12 the measures of the three angles are num__4 x = num__4 x num__12 ° = num__48 num__3 x = num__3 × num__12 = num__36 ° num__8 x = num__8 × num__12 = num__96 ° a <eor> a <eos> |
a |
straight_angle__ clock_small_arm_angle__3.0__12.0__ clock_small_arm_angle__3.0__12.0__ |
straight_angle__ clock_small_arm_angle__3.0__12.0__ clock_small_arm_angle__3.0__12.0__ |
| in a camp there is a meal for num__120 men or num__200 children . if num__150 children have taken the meal how many men will be catered to with remaining meal ? <o> a ) num__41 <o> b ) num__30 <o> c ) num__35 <o> d ) num__48 <o> e ) num__49 |
b num__30 there is a meal for num__200 children . num__150 children have taken the meal . remaining meal is to be catered to num__50 children . now num__200 children num__120 men . num__50 children = ( num__0.6 ) x num__50 = num__30 men . <eor> b <eos> |
b |
subtract__150.0__120.0__ subtract__200.0__150.0__ divide__120.0__200.0__ subtract__150.0__120.0__ |
subtract__150.0__120.0__ subtract__200.0__150.0__ divide__120.0__200.0__ subtract__150.0__120.0__ |
| if m p and t are distinct positive prime numbers then ( m ^ num__3 ) ( p ) ( t ) has how many different positive factors greater than num__1 ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__12 <o> d ) num__15 <o> e ) num__27 |
umber of factors of ( num__2 ^ a ) * ( num__3 ^ b ) * ( num__5 ^ c ) . . . = ( a + num__1 ) ( b + num__1 ) ( c + num__1 ) . . . if m p and t are the distinct prime numbers then the number is already represented in its prime factorization form number of factors = ( num__3 + num__1 ) ( num__1 + num__1 ) ( num__1 + num__1 ) = num__16 out of these one factor would be num__1 . hence different positive factors greater than num__1 = num__15 correct option : d <eor> d <eos> |
d |
subtract__3.0__1.0__ add__3.0__2.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
subtract__3.0__1.0__ add__3.0__2.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
| in a certain neighborhood there are half as many beige houses as white houses and num__3 times as many white houses as brown houses . what is the ratio of the number of brown houses to the number of beige houses ? <o> a ) num__4 : num__1 <o> b ) num__3 : num__5 <o> c ) num__2 : num__3 <o> d ) num__1 : num__3 <o> e ) num__1 : num__4 |
the ratio of beige : white : brown = num__3 : num__6 : num__2 the answer is c . <eor> c <eos> |
c |
divide__6.0__3.0__ divide__6.0__3.0__ |
divide__6.0__3.0__ divide__6.0__3.0__ |
| s = { num__1 num__24 num__56 } t = { num__45 num__68 num__1210 num__714 } what is the probability that x chosen from s and y chosen from t will result x * y = odd <o> a ) num__0.333333333333 <o> b ) num__0.65625 <o> c ) num__1.81818181818 <o> d ) num__2.9 <o> e ) num__0.55 |
p : the probability that x * y is even then p = num__1 - p ( x * y is odd ) p ( x * y even ) = p ( x even ) * p ( y even ) = num__0.6 * num__0.75 = num__0.45 and p = num__1 - num__0.45 = num__0.55 option : e <eor> e <eos> |
e |
multiply__0.75__0.6__ subtract__1.0__0.45__ subtract__1.0__0.45__ |
multiply__0.75__0.6__ subtract__1.0__0.45__ subtract__1.0__0.45__ |
| a b and c have rs . num__450 between them a and c together have rs . num__200 and b and c rs . num__350 . how much does c have ? <o> a ) a ) num__100 <o> b ) b ) num__110 <o> c ) c ) num__120 <o> d ) d ) num__130 <o> e ) e ) num__140 |
a + b + c = num__450 a + c = num__200 b + c = num__350 - - - - - - - - - - - - - - a + b + num__2 c = num__550 a + b + c = num__450 - - - - - - - - - - - - - - - - c = num__100 answer : a <eor> a <eos> |
a |
add__200.0__350.0__ subtract__450.0__350.0__ subtract__450.0__350.0__ |
add__200.0__350.0__ subtract__450.0__350.0__ subtract__450.0__350.0__ |
| a can do a work in num__14 days and working together a and b can do the same work in num__10 days . in what time can b alone do the work ? <o> a ) num__35 days <o> b ) num__25 days <o> c ) num__30 days <o> d ) num__27 days <o> e ) num__23 days |
work done by b in num__1 day = num__0.1 - num__0.0714285714286 = ( num__7 - num__5 ) / num__70 = num__0.0285714285714 = num__0.0285714285714 so b alone can do the work in num__35 days . answer : a <eor> a <eos> |
a |
divide__1.0__10.0__ divide__1.0__14.0__ multiply__14.0__5.0__ subtract__0.1__0.0714__ multiply__5.0__7.0__ round__35.0__ |
divide__1.0__10.0__ divide__1.0__14.0__ divide__7.0__0.1__ subtract__0.1__0.0714__ multiply__5.0__7.0__ subtract__70.0__35.0__ |
| m is the sum of the reciprocals of the consecutive integers from num__101 to num__200 inclusive . which of the following is true ? <o> a ) num__0.333333333333 < m < num__0.5 <o> b ) num__0.2 < m < num__0.333333333333 <o> c ) num__0.5 < m < num__1 <o> d ) num__0.111111111111 < m < num__0.142857142857 <o> e ) num__0.0833333333333 < m < num__0.111111111111 |
m = num__0.00990099009901 + num__0.00980392156863 + num__0.00970873786408 + . . . . . . + num__0.005 if we replace the first num__99 terms by num__0.005 then we get a sum = num__0.5 = num__0.5 . since the actual terms are larger than num__0.005 the sum is larger than num__0.5 . if we replace the all the num__100 terms by num__0.01 we get a sum = num__1.0 = num__1 . since the actual terms are smaller than num__0.01 the sum is less than num__1 . therefore num__0.5 < m < num__1 choice c <eor> c <eos> |
c |
reverse__101.0__ reverse__200.0__ subtract__200.0__101.0__ multiply__200.0__0.5__ reverse__100.0__ subtract__101.0__100.0__ multiply__0.5__1.0__ |
reverse__101.0__ reverse__200.0__ subtract__200.0__101.0__ multiply__200.0__0.5__ reverse__100.0__ subtract__101.0__100.0__ multiply__0.5__1.0__ |
| at what rate percent on simple interest will rs . num__750 amount to rs . num__900 in num__5 years ? <o> a ) num__6.0 <o> b ) num__2.0 <o> c ) num__4.0 <o> d ) num__5.0 <o> e ) num__3 % |
num__150 = ( num__750 * num__5 * r ) / num__100 r = num__4.0 answer : c <eor> c <eos> |
c |
percent__4.0__100.0__ |
percent__4.0__100.0__ |
| the average ( arithmetic mean ) of four numbers is num__7 x + num__5 . if one of the numbers is x what is the average of the other three numbers ? <o> a ) x + num__1 <o> b ) num__3 x + num__3 <o> c ) num__9 x + num__6.66666666667 <o> d ) num__5 x + num__4 <o> e ) num__15 x + num__12 |
if the average of four numbers is num__7 x + num__5 then the sum of all four numbers if num__4 * ( num__7 x + num__5 ) = num__28 x + num__20 . if one of the numbers is x then the average of the remaining three numbers would be : ( num__28 x + num__20 - x ) / num__3 = ( num__27 x + num__20 ) / num__3 = num__9 x + num__6.66666666667 . answer is c . <eor> c <eos> |
c |
multiply__7.0__4.0__ multiply__5.0__4.0__ subtract__7.0__4.0__ add__7.0__20.0__ add__5.0__4.0__ divide__20.0__3.0__ add__5.0__4.0__ |
multiply__7.0__4.0__ multiply__5.0__4.0__ subtract__7.0__4.0__ add__7.0__20.0__ add__5.0__4.0__ divide__20.0__3.0__ add__5.0__4.0__ |
| a train running at the speed of num__90 km / hr crosses a pole in num__15 seconds . what is the length of the train ? <o> a ) num__325 <o> b ) num__350 <o> c ) num__375 <o> d ) num__400 <o> e ) num__425 |
speed = ( num__90 x ( num__0.277777777778 ) m / sec = ( num__25 ) m / sec . length of the train = ( speed x time ) . length of the train = ( ( num__25 ) x num__15 ) m = num__375 m c <eor> c <eos> |
c |
multiply__15.0__25.0__ round__375.0__ |
multiply__15.0__25.0__ round__375.0__ |
| james invested a certain sum of money in a simple interest bond whose value grew to $ num__700 at the end of num__3 years and to $ num__800 at the end of another num__2 years . what was the rate of interest in which he invested his sum ? <o> a ) num__9.09 <o> b ) num__12.5 <o> c ) num__6.67 <o> d ) num__6.25 <o> e ) num__8.33 % |
lets assume the principal amount ( initial amount invested ) to be p rate of interest to berand time as t . we need to find r now after a time of num__3 years the principal p amounts to $ num__700 and after a time of num__5 years ( question says after another num__2 years so num__3 + num__2 ) p becomes $ num__800 . formulating the above data amount ( a num__1 ) at end of num__3 years a num__1 = p ( num__1 + num__3 r / num__100 ) = num__700 amount ( a num__2 ) at end of num__5 years a num__2 = p ( num__1 + num__5 r / num__100 ) = num__800 dividing a num__2 by a num__1 we get ( num__1 + num__5 r / num__100 ) / ( num__1 + num__3 r / num__100 ) = num__1.14285714286 after cross multiplication we are left with num__11 r = num__100 which gives r = num__9.09 option : a <eor> a <eos> |
a |
percent__100.0__9.09__ |
percent__100.0__9.09__ |
| if num__0.5 of a = num__65 paise then the value of a is ? <o> a ) rs . num__130 <o> b ) rs . num__17 <o> c ) rs . num__1.70 <o> d ) rs . num__4.25 <o> e ) none |
answer ∵ num__0.5 / num__100 of a = num__0.65 ∴ a = rs . ( num__65 / num__0.5 ) = rs . num__130 correct option : a <eor> a <eos> |
a |
percent__100.0__130.0__ |
percent__100.0__130.0__ |
| if the length of an edge of cube p is twice the length of an edge of cube q what is the ratio of the volume of cube q to the volume of cube p ? <o> a ) num__0.125 <o> b ) num__0.25 <o> c ) num__0.333333333333 <o> d ) num__0.142857142857 <o> e ) num__0.111111111111 |
the length of cube q = num__1 ; the length of cube p = num__2 ; the ratio of the volume of cube q to the volume of cube p = num__1 ^ num__1.5 ^ num__3 = num__0.125 . answer : a <eor> a <eos> |
a |
multiply__1.5__2.0__ multiply__1.0__0.125__ |
multiply__1.5__2.0__ power__0.125__1.0__ |
| the area of circle o is added to its diameter . if the circumference of circle o is then subtracted from this total the result is num__12 . what is the radius of circle o ? <o> a ) – num__2 / pi <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__6 |
pi * r ^ num__2 + num__2 r - num__2 * pi * r = num__12 simplifying the equation : pi * r ( r - num__2 ) + num__2 r = num__12 without much algebraic : we can test the answers quickly then num__6 is the only possible answer that will eliminate pi from equation . answer is e <eor> e <eos> |
e |
triangle_area__2.0__6.0__ |
triangle_area__2.0__6.0__ |
| a crow leaves its nest and flies back and forth from its nest to a nearby ditch to gather worms . the distance between the nest and the ditch is num__100 meters . in one and a half hours the crow manages to bring worms to its nest num__15 times . what is the speed of the crow in kilometers per hour ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__5 <o> e ) num__7 |
the distance between the nest and the ditch is num__100 meters . num__15 times mean = a crow leaves its nest and flies back ( going and coming back ) i . e . num__2 times we get total num__30 rounds . so the distance is num__30 * num__100 = num__3000 . d = st num__3000 / num__1.5 = t i think we can take num__3000 meters as num__3 km then only we get t = num__2 . ( num__1000 meters = num__1 km ) d ) <eor> d <eos> |
d |
multiply__15.0__2.0__ multiply__100.0__30.0__ multiply__1.5__2.0__ divide__3000.0__3.0__ subtract__3.0__2.0__ divide__15.0__3.0__ |
multiply__15.0__2.0__ multiply__100.0__30.0__ multiply__1.5__2.0__ divide__3000.0__3.0__ subtract__3.0__2.0__ divide__15.0__3.0__ |
| what is the greatest positive integer x such that num__3 ^ x is a factor of num__9 ^ num__5 ? <o> a ) num__5 <o> b ) num__9 <o> c ) num__10 <o> d ) num__20 <o> e ) num__30 |
what is the greatest positive integer x such that num__3 ^ x is a factor of num__9 ^ num__5 ? num__9 ^ num__5 = ( num__3 ^ num__2 ) ^ num__5 = num__3 ^ num__10 c . num__10 <eor> c <eos> |
c |
subtract__5.0__3.0__ multiply__5.0__2.0__ multiply__5.0__2.0__ |
subtract__5.0__3.0__ multiply__5.0__2.0__ multiply__5.0__2.0__ |
| a fort had provision of food for num__150 men for num__45 days . after num__10 days num__25 men left the fort . the number of days for which the remaining food will last is : <o> a ) num__33 <o> b ) num__38 <o> c ) num__42 <o> d ) num__12 <o> e ) num__17 |
explanation : after num__10 days : num__150 men had food for num__35 days . suppose num__125 men had food for x days . now less men more days ( indirect proportion ) \ inline { \ color { blue } \ therefore } num__125 : num__150 : : num__35 : x num__125 x x = num__150 x num__35 = > x = = > x = num__42 answer : c <eor> c <eos> |
c |
subtract__45.0__10.0__ subtract__150.0__25.0__ round__42.0__ |
subtract__45.0__10.0__ subtract__150.0__25.0__ round__42.0__ |
| what is the smallest positive integer x such that num__2160 x is a perfect cube ? <o> a ) num__100 <o> b ) num__6 <o> c ) num__8 <o> d ) num__12 <o> e ) num__18 |
take out the factors of num__2160 that will come num__6 ^ num__3 * num__10 . for perfect cube you need every no . raise to the power num__3 . for num__2160 to be a perfect cube you need two num__2 and two num__5 that means num__100 a is the answer . <eor> a <eos> |
a |
power__10.0__2.0__ triangle_area__2.0__100.0__ |
power__10.0__2.0__ power__10.0__2.0__ |
| a certain computer manufacturing firm last year produced num__80 percent of its computers using parts made by its subsidiary company . if the remaining num__10800 computers were produced using parts purchased from another parts manufacturer how many computers were produced by the computer manufacturing firm altogether ? <o> a ) num__54000 <o> b ) num__64000 <o> c ) num__62000 <o> d ) num__60000 <o> e ) num__68000 |
num__80.0 parts used of subsidary company so num__20.0 parts used of other companies . now num__20.0 parts = num__10800 computers so num__1.0 parts = num__540.0 or num__100.0 parts = num__54000 computers . hence answer is ( a ) <eor> a <eos> |
a |
percent__100.0__54000.0__ |
percent__100.0__54000.0__ |
| a takes twice as much time as b or thrice as much time to finish a piece of work . working together they can finish the work in num__2 days . b can do the work alone in : <o> a ) num__5 hours <o> b ) num__6 hours <o> c ) num__7 hours <o> d ) num__8 hours <o> e ) num__4 hours |
let b takes = x hours a takes = x / num__2 num__1 / x + num__1 / ( x / num__2 ) = num__0.5 num__3 / x = num__0.5 x = num__6 hours answer : b <eor> b <eos> |
b |
divide__1.0__2.0__ add__2.0__1.0__ multiply__2.0__3.0__ round__6.0__ |
divide__1.0__2.0__ add__2.0__1.0__ divide__3.0__0.5__ divide__3.0__0.5__ |
| the average of num__10 numbers is calculated as num__16 . it is discovered later on that while calculating the average one number namely num__36 was wrongly read as num__26 . the correct average is ? <o> a ) a ) num__17 <o> b ) b ) num__18 <o> c ) c ) num__19 <o> d ) d ) num__22 <o> e ) e ) num__24 |
explanation : num__10 * num__16 + num__36 – num__26 = num__170 = > num__17.0 = num__17 a ) <eor> a <eos> |
a |
divide__170.0__10.0__ divide__170.0__10.0__ |
divide__170.0__10.0__ divide__170.0__10.0__ |
| two numbers are respectively num__20.0 and num__50.0 more than a third number . the percentage that is first of the second is ? <o> a ) num__77.0 <o> b ) num__78.0 <o> c ) num__79.0 <o> d ) num__80.0 <o> e ) num__81 % |
i ii iii num__120 num__150 num__100 num__150 - - - - - - - - - - num__120 num__100 - - - - - - - - - - - ? = > num__80.0 answer : d <eor> d <eos> |
d |
percent__80.0__100.0__ |
percent__80.0__100.0__ |
| joe ' s current weekly caloric intake is num__20000 calories . his physician has advised joe to reduce the number of calories he consumes weekly to num__14000 . by what percent does joe need to reduce his weekly caloric intake ? <o> a ) num__20.0 <o> b ) num__30.0 <o> c ) num__50.0 <o> d ) num__60.0 <o> e ) num__70 % |
a certain percentage of questions in the quant section of the gmat are just ' math questions ' - you ' ll use a formula do some calculations and you ' ll have the answer . this is one of those types of questions . you still have to write everything down and stay organized but the work involved is relatively straight - forward . here we ' re told that the original amount of calories is num__20000 and that the reduced amount of calories is num__14000 . we ' re asked what percent should the original amount of calories be reduced by ? original calories = num__20000 reduced calories = num__14000 the percent of reduced calories is relative to the original calories as shown : num__14000 / num__20000 = num__0.7 = num__70.0 . so num__14000 calories is num__70.0 of num__20000 calories ( num__100.0 ) . but the question is by what percent should the original amount of calories be reduced . num__100.0 - num__70.0 = num__30.0 . b <eor> b <eos> |
b |
percent__100.0__30.0__ |
percent__100.0__30.0__ |
| evaluate : | num__4 - num__8 ( num__3 - num__12 ) | - | num__5 - num__11 | = ? <o> a ) num__40 <o> b ) num__50 <o> c ) num__60 <o> d ) num__70 <o> e ) num__80 |
according to order of operations inner brackets first . hence | num__4 - num__8 ( num__3 - num__12 ) | - | num__5 - num__11 | = | num__4 - num__8 * ( - num__9 ) | - | num__5 - num__11 | according to order of operations multiplication within absolute value signs ( which may be considered as brackets when it comes to order of operations ) next . hence = | num__4 + num__72 | - | num__5 - num__11 | = | num__76 | - | - num__6 | = num__76 - num__6 = num__70 correct answer d ) num__70 <eor> d <eos> |
d |
add__4.0__5.0__ multiply__8.0__9.0__ add__4.0__72.0__ subtract__11.0__5.0__ subtract__76.0__6.0__ subtract__76.0__6.0__ |
add__4.0__5.0__ multiply__8.0__9.0__ add__4.0__72.0__ subtract__11.0__5.0__ subtract__76.0__6.0__ subtract__76.0__6.0__ |
| a certain sum amounts to rs . num__675 in num__3 years and rs . num__850 in num__6 years . find the rate % per annum ? <o> a ) num__10.0 <o> b ) num__8.0 <o> c ) num__11.0 <o> d ) num__9.0 <o> e ) num__7 % |
num__3 - - - num__675 num__6 - - - num__850 - - - - - - - - - - - - - - num__3 - - - num__175 n = num__1 i = num__55 r = ? p = num__675 - num__175 = num__500 num__55 = ( num__500 * num__1 * r ) / num__100 r = num__11.0 answer : c <eor> c <eos> |
c |
percent__100.0__11.0__ |
percent__100.0__11.0__ |
| if ( a – b ) is num__7 more than ( c + d ) and ( a + b ) is num__3 less than ( c – d ) then ( a – c ) is : <o> a ) num__6 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
( a – b ) – ( c + d ) = num__7 and ( c – d ) – ( a + b ) = num__3 = > ( a – c ) – ( b + d ) = num__7 and ( c – a ) – ( b + d ) = num__3 = > ( b + d ) = ( a – c ) – num__7 and ( b + d ) = ( c – a ) – num__3 = > ( a – c ) – num__7 = ( c – a ) – num__3 = > num__2 ( a – c ) = num__4 = > ( a – c ) = num__2 answer : b <eor> b <eos> |
b |
subtract__7.0__3.0__ subtract__4.0__2.0__ |
subtract__7.0__3.0__ subtract__4.0__2.0__ |
| a man buys num__12 lts of liquid which contains num__10.0 of the liquid and the rest is water . he then mixes it with num__10 lts of another mixture with num__30.0 of liquid . what is the % of water in the new mixture ? <o> a ) num__80.91 <o> b ) num__76.45 <o> c ) num__77.45 <o> d ) num__74.45 <o> e ) num__73.45 |
num__10.0 in num__12 lts is num__1.2 . so water = num__12 - num__1.2 = num__10.8 lts . num__30.0 of num__10 lts = num__3 . so water in num__2 nd mixture = num__10 - num__3 = num__7 lts . now total quantity = num__12 + num__10 = num__22 lts . total water in it will be num__10.8 + num__7 = num__17.8 lts . % of water = ( num__100 * num__17.8 ) / num__22 = num__80.91 . answer : a <eor> a <eos> |
a |
percent__12.0__10.0__ percent__10.0__30.0__ percent__100.0__80.91__ |
percent__12.0__10.0__ percent__10.0__30.0__ percent__100.0__80.91__ |
| if in a box of dimensions num__6 m * num__5 m * num__4 m smaller boxes of dimensions num__60 cm * num__50 cm * num__40 cm are kept in it then what will be the maximum number of the small boxes that can be kept in it ? <o> a ) num__500 <o> b ) num__1000 <o> c ) num__900 <o> d ) num__800 <o> e ) num__700 |
num__6 * num__5 * num__4 = num__0.6 * num__0.5 * num__0.4 * x num__1 = num__0.1 * num__0.1 * num__0.1 * x = > x = num__1000 answer : b <eor> b <eos> |
b |
km_to_mile_conversion__ subtract__6.0__5.0__ divide__6.0__60.0__ round__1000.0__ |
km_to_mile_conversion__ subtract__6.0__5.0__ divide__6.0__60.0__ round__1000.0__ |
| if x gets num__25.0 more than y and y gets num__20.0 more than z the share of z out of rs . num__740 will be : <o> a ) rs . num__300 <o> b ) rs . num__200 <o> c ) rs . num__240 <o> d ) rs . num__350 <o> e ) none of these |
z share = z y = num__1.2 z x = num__1.25 × num__1.2 z x + y + z = num__740 ( num__1.25 × num__1.2 + num__1.2 + num__1 ) z = num__74 num__3.7 z = num__740 z = num__200 answer : . b . <eor> b <eos> |
b |
divide__25.0__20.0__ round_down__1.25__ divide__74.0__20.0__ divide__740.0__3.7__ divide__740.0__3.7__ |
divide__25.0__20.0__ round_down__1.25__ divide__74.0__20.0__ divide__740.0__3.7__ divide__740.0__3.7__ |
| when num__27 is divided by positive integer n the remainder is num__4 . for how many values of n is this true ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
when num__27 is divided by n the remainder is num__4 i . e . num__4 apples left over after grouping so n must be greater than num__4 . it also means that num__23 is completely divisible by n . the factors of num__23 are num__1 and num__23 . out of these n can be num__23 . answer ( b ) . <eor> b <eos> |
b |
subtract__27.0__4.0__ reverse__1.0__ |
subtract__27.0__4.0__ reverse__1.0__ |
| two pipes can fill the cistern in num__10 hr and num__12 hr respectively while the third empty it in num__25 hr . if all pipes are opened simultaneously then the cistern will be filled in <o> a ) num__7.5 hr <o> b ) num__6.98 hr <o> c ) num__8.5 hr <o> d ) num__10 hr <o> e ) none of these |
solution : work done by all the tanks working together in num__1 hour . num__0.1 + num__0.0833333333333 − num__0.04 = num__0.142857142857 hence tank will be filled in num__6.98 hour option ( b ) <eor> b <eos> |
b |
divide__1.0__10.0__ divide__1.0__12.0__ divide__1.0__25.0__ round__6.98__ |
divide__1.0__10.0__ divide__1.0__12.0__ divide__1.0__25.0__ round__6.98__ |
| a carpenter constructed a rectangular sandbox with a capacity of num__10 cubic feet . if the carpenter were to make a similar sandbox twice as long twice as wide and twice as high as the first sandbox what would be the capacity in cubic feet of the second sandbox ? <o> a ) num__20 <o> b ) num__40 <o> c ) num__60 <o> d ) num__80 <o> e ) num__100 |
a quick note on doubling . when you double a length you have num__2 * l num__1 . when you double all lengths of a rectangle you have ( num__2 * l num__1 ) ( num__2 * l num__2 ) = a . an increase of num__2 ^ num__2 or num__4 . when you double all lengths of a rectangular prism you have ( num__2 * l num__1 ) ( num__2 * l num__2 ) ( num__2 * l num__3 ) = v . an increase of num__2 ^ num__3 or num__8 . this leads to the basic relationship : line : num__2 * original size rectangle : num__4 * original size rectangular prism : num__8 * original size answer is d <eor> d <eos> |
d |
square_perimeter__1.0__ square_perimeter__2.0__ multiply__10.0__8.0__ |
square_perimeter__1.0__ power__2.0__3.0__ multiply__10.0__8.0__ |
| salaries of ravi and sumit are in the ratio num__2 : num__3 . if the salary of each is increased by rs . num__4000 the new ratio becomes num__40 : num__57 . what is sumit ' s salary ? <o> a ) num__32000 <o> b ) num__38000 <o> c ) num__35000 <o> d ) num__33000 <o> e ) num__39000 |
let the original salaries of ravi and sumit be rs . num__2 x and rs . num__3 x respectively . then num__2 x + num__1333.33333333 x + num__4000 = num__0.701754385965 num__57 ( num__2 x + num__4000 ) = num__40 ( num__3 x + num__4000 ) num__6 x = num__68000 num__3 x = num__34000 sumit ' s present salary = ( num__3 x + num__4000 ) = rs . ( num__34000 + num__4000 ) = rs . num__38000 . answer is b . <eor> b <eos> |
b |
divide__4000.0__3.0__ divide__40.0__57.0__ multiply__2.0__3.0__ divide__68000.0__2.0__ add__4000.0__34000.0__ add__4000.0__34000.0__ |
divide__4000.0__3.0__ divide__40.0__57.0__ multiply__2.0__3.0__ divide__68000.0__2.0__ add__4000.0__34000.0__ add__4000.0__34000.0__ |
| three pipes a b and c can fill a tank from empty to full in num__30 minutes num__20 minutes and num__10 minutes respectively . when the tank is empty all the three pipes are opened . a b and c discharge chemical solutions p q and r respectively . what is the proportion of solution r in the liquid in the tank after num__3 minutes ? <o> a ) num__0.315789473684 <o> b ) num__0.545454545455 <o> c ) num__0.5 <o> d ) num__0.6 <o> e ) num__0.157894736842 |
part filled by ( a + b + c ) in num__3 minutes = num__3 ( num__0.0333333333333 + num__0.05 + num__0.1 ) = num__0.55 part filled by c in num__3 minutes = num__0.3 required ratio = num__0.3 * num__1.81818181818 = num__0.545454545455 answer : b <eor> b <eos> |
b |
divide__3.0__30.0__ divide__3.0__10.0__ multiply__1.8182__0.3__ multiply__1.8182__0.3__ |
divide__3.0__30.0__ multiply__3.0__0.1__ multiply__1.8182__0.3__ multiply__1.8182__0.3__ |
| num__9 men working num__8 hours per day dig num__30 m deep . how many extra men should be put to dig to a depth of num__50 m working num__6 hours per day ? <o> a ) num__33 <o> b ) num__66 <o> c ) num__88 <o> d ) num__11 <o> e ) num__281 |
( num__9 * num__8 ) / num__30 = ( x * num__6 ) / num__50 = > x = num__20 num__20 – num__9 = num__11 answer : d <eor> d <eos> |
d |
subtract__50.0__30.0__ subtract__20.0__9.0__ round__11.0__ |
subtract__50.0__30.0__ subtract__20.0__9.0__ round__11.0__ |
| the total cost of num__100 paper plates and num__200 paper cups is $ num__6.00 at the same rates what is the total cost of num__20 of the plates and num__40 of the cups ? <o> a ) $ . num__90 <o> b ) $ num__1.00 <o> c ) $ num__1.20 <o> d ) $ num__1.50 <o> e ) $ num__1.60 |
u dont need to go through all this what u have with u is num__100 p + num__200 c = $ num__6.00 just divide the equation by num__5 and you will get what u are looking for num__20 p + num__40 c = $ num__1.20 therefore oa is c <eor> c <eos> |
c |
percent__6.0__20.0__ percent__100.0__1.2__ |
percent__6.0__20.0__ percent__100.0__1.2__ |
| what is the least number of squares tiles required to pave the floor of a room num__15 m num__17 cm long and num__9 m num__2 cm broad ? <o> a ) num__617 <o> b ) num__746 <o> c ) num__808 <o> d ) num__812 <o> e ) num__814 |
length of largest tile = h . c . f . of num__1517 cm and num__902 cm = num__41 cm . area of each tile = ( num__41 x num__41 ) cm num__2 . required number of tiles = ( num__1517 x num__902 ) / ( num__41 x num__41 ) = num__814 . answer : option e <eor> e <eos> |
e |
round__814.0__ |
round__814.0__ |
| a train is num__400 meter long is running at a speed of num__60 km / hour . in what time will it pass a bridge of num__800 meter length ? <o> a ) num__72 seconds <o> b ) num__27 seconds <o> c ) num__40 seconds <o> d ) num__128 seconds <o> e ) num__18 seconds |
speed = num__60 km / hr = num__60 * ( num__0.277777777778 ) m / sec = num__16.6666666667 m / sec total distance = num__400 + num__800 = num__1200 meter time = distance / speed = num__1200 * ( num__0.06 ) = num__72 seconds answer : a <eor> a <eos> |
a |
add__400.0__800.0__ multiply__1200.0__0.06__ round__72.0__ |
add__400.0__800.0__ multiply__1200.0__0.06__ multiply__1200.0__0.06__ |
| the percentage profit earned by selling an article for rs . num__1920 is equal to the percentage loss incurred by selling the same article for rs . num__1280 . at what price should the article be sold to make num__25.0 profit ? <o> a ) rs . num__2000 <o> b ) rs . num__2200 <o> c ) rs . num__2400 <o> d ) rs . num__2600 <o> e ) none of these |
let c . p . be rs . x . then ( num__1920 - x ) / x * num__100 = ( x - num__1280 ) / x * num__100 num__1920 - x = x - num__1280 num__2 x = num__3200 = > x = num__1600 required s . p . = num__125.0 of rs . num__1600 = num__1.25 * num__1600 = rs . num__2000 . answer : a <eor> a <eos> |
a |
percent__100.0__2000.0__ |
percent__100.0__2000.0__ |
| look at this series : num__2 num__4 num__6 num__8 num__10 . . . what number should come next ? <o> a ) num__11 <o> b ) num__12 <o> c ) num__13 <o> d ) num__14 <o> e ) num__15 |
explanation : this is a simple addition series . each number increases by num__2 . answer : option b <eor> b <eos> |
b |
multiply__2.0__6.0__ |
multiply__2.0__6.0__ |
| the area of a square garden is a square feet and the perimeter is p feet . if a = num__2 p + num__14.25 what is the perimeter of the garden in feet ? <o> a ) num__38 <o> b ) num__36 <o> c ) num__40 <o> d ) num__56 <o> e ) num__64 |
perimeter of square = p side of square = p / num__4 area of square = ( p ^ num__2 ) / num__16 = a given that a = num__2 p + num__14.25 ( p ^ num__2 ) / num__16 = num__2 p + num__9 p ^ num__2 = num__32 p + num__228 p ^ num__2 - num__32 p - num__228 = num__0 p ^ num__2 - num__38 p + num__6 p - num__228 = num__0 p ( p - num__38 ) + num__6 ( p + num__38 ) = num__0 ( p - num__38 ) ( p + num__6 ) = num__0 p = num__38 or - num__6 discarding negative value p = num__38 answer is a <eor> a <eos> |
a |
square_perimeter__4.0__ multiply__2.0__16.0__ triangle_area__14.25__32.0__ triangle_perimeter__2.0__32.0__4.0__ triangle_perimeter__2.0__4.0__0.0__ triangle_area__2.0__38.0__ |
power__2.0__4.0__ multiply__2.0__16.0__ triangle_area__14.25__32.0__ triangle_perimeter__2.0__32.0__4.0__ triangle_perimeter__2.0__4.0__0.0__ triangle_area__2.0__38.0__ |
| in how many different ways can num__2 girls and num__2 boys form a circle such that the boys and the girls alternate ? <o> a ) num__8 <o> b ) num__1 <o> c ) num__7 <o> d ) num__5 <o> e ) num__2 |
explanation : in a circle num__2 boys can be arranged in num__2 ! ways given that the boys and the girls alternate . hence there are num__2 places for girls which can be arranged in num__2 ! ways total number of ways = num__1 ! x num__2 ! = num__1 x num__2 = num__2 . answer : option e <eor> e <eos> |
e |
coin_space__ |
coin_space__ |
| a sum amounts to rs . num__8820 in num__2 years at the rate of num__5.0 p . a . if interest was compounded yearly then what was the principal ? <o> a ) s . num__4000 <o> b ) s . num__5000 <o> c ) s . num__4500 <o> d ) s . num__4800 <o> e ) s . num__8000 |
ci = num__8820 r = num__5 n = num__2 ci = p [ num__1 + r / num__100 ] ^ num__2 = p [ num__1 + num__0.05 ] ^ num__2 num__8820 = p [ num__1.05 ] ^ num__2 num__8820 [ num__0.952380952381 ] ^ num__2 num__8000 answer : e <eor> e <eos> |
e |
percent__5.0__1.0__ percent__100.0__8000.0__ |
percent__5.0__1.0__ percent__100.0__8000.0__ |
| due to construction the speed limit along an num__10 - mile section of highway is reduced from num__55 miles per hour to num__30 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? <o> a ) a ) num__6.24 <o> b ) b ) num__8 <o> c ) c ) num__10 <o> d ) d ) num__9.1 <o> e ) e ) num__24 |
old time in minutes to cross num__10 miles stretch = num__10 * num__1.09090909091 = num__10 * num__1.09090909091 = num__10.9 new time in minutes to cross num__10 miles stretch = num__10 * num__2.0 = num__10 * num__2 = num__20 time difference = num__9.1 ans : d <eor> d <eos> |
d |
multiply__10.0__2.0__ subtract__20.0__10.9__ round__9.1__ |
multiply__10.0__2.0__ subtract__20.0__10.9__ round__9.1__ |
| felix earned num__70.0 more per month than oscar . if felix ’ s salary is decreased num__10.0 and oscar ’ s salary is increased by num__10.0 then felix would be earning what percent more than oscar ? <o> a ) num__10.0 <o> b ) num__9.09 <o> c ) num__40.0 <o> d ) num__39.09 <o> e ) it can not be determined from the information given . |
this q is all about taking the correct base for taking % s . . the very first step requires taking some numbers for each . . one should be num__100 . . . it is easy to take num__30 and num__100 here . . but the base is num__100 here so the salary can be num__100 and num__170 . . num__100 increases num__10.0 = num__110 . . num__170 decreases num__10.0 = num__153 . . so the increase is num__43 . . again the base is num__110 so answer num__0.390909090909 * num__100 = num__39.09 d without calculations answer can not be equal to or greater than num__40 as num__43 is on num__110 . . so d is the only one left <eor> d <eos> |
d |
subtract__100.0__70.0__ add__70.0__100.0__ add__10.0__100.0__ subtract__153.0__110.0__ divide__43.0__110.0__ multiply__100.0__0.3909__ subtract__70.0__30.0__ multiply__100.0__0.3909__ |
subtract__100.0__70.0__ add__70.0__100.0__ add__10.0__100.0__ subtract__153.0__110.0__ divide__43.0__110.0__ multiply__100.0__0.3909__ subtract__70.0__30.0__ multiply__100.0__0.3909__ |
| a father tells his son ` ` i was of your present age when you were born . ' ' if the father is num__36 now how old was the boy num__5 years back ? <o> a ) num__15 years <o> b ) num__13 years <o> c ) num__17 years <o> d ) num__20 years <o> e ) none of the above |
present age of father num__36 present age of son x num__36 - x = x x = num__18 five yrs back sons age num__18 - num__5 = num__13 answer : b <eor> b <eos> |
b |
subtract__18.0__5.0__ subtract__18.0__5.0__ |
subtract__18.0__5.0__ subtract__18.0__5.0__ |
| what percentage loss will a merchant incur if he marks his goods up by x % over his cost price and then offers a discount of x % on his selling price ? <o> a ) num__0.0 <o> b ) num__2 x / num__100.0 <o> c ) x ^ num__0.02 % <o> d ) x % <o> e ) num__2 x % |
answer is c . taking smart numbers : let c . p = num__100 $ and x = num__20.0 over c . p then mark up = num__120 $ after discount of x % s . p = num__80.0 of num__120 s . p = num__96 $ therefore loss is num__4 $ and hence loss in % = num__4.0 plugging numbers in options . a . num__0.0 - - - - - - - - - - - - - - - - - - - - - - - - - - - not truebecause loss is num__4.0 b . num__2 x / num__100.0 - - - - - - - - - - - - - - - - - - - - num__2 * num__0.2 = num__0.4 . hence not true c . x ^ num__0.02 % - - - - - - - - - - - - - - - - - - num__20 ^ num__0.02 = num__4.0 . true answer is c d . x % - - - - - - - - - - - - - - - - - - - - - - - - - - - num__20.0 . hence not true e . num__2 x % - - - - - - - - - - - - - - - - - - - - - - - - - - num__2 * num__20 = num__40.0 . hence not true answer is c <eor> c <eos> |
c |
percent__80.0__120.0__ percent__2.0__20.0__ percent__100.0__0.02__ |
percent__80.0__120.0__ percent__2.0__20.0__ percent__100.0__0.02__ |
| if ' a ' means ' - ' ' b ' means ' / ' ' c ' means ' + ' and ' d ' means ' x ' then find the value of the following : num__21 b num__7 c num__9 d num__10 a num__13 <o> a ) num__80 <o> b ) num__75 <o> c ) num__85 <o> d ) num__95 <o> e ) num__66 |
a num__80 explanation : num__21 b num__7 c num__9 d num__10 a num__13 = num__3.0 + num__9 x num__10 - num__13 = num__3 + num__90 - num__13 = num__80 <eor> a <eos> |
a |
divide__21.0__7.0__ multiply__9.0__10.0__ subtract__90.0__10.0__ |
subtract__10.0__7.0__ add__10.0__80.0__ subtract__90.0__10.0__ |
| b takes num__12 more days than a to finish a task . b and a start this task and a leaves the task num__12 days before the task is finished . b completes num__60.0 of the overall task . how long would b have taken to finish the task if he had worked independently ? <o> a ) num__48 days <o> b ) num__36 days <o> c ) num__28 days <o> d ) num__32 days <o> e ) num__30 days |
detailed solution let us say a and b split their share of the task and started doing their respective shares simultaneously . let ’ s say a takes a days to finish the task . therefore b takes a + num__12 days to finish the entire task . a has to finish num__40.0 of the task since b is doing the rest . so a will only take num__2 a / num__5 number of days . b only has to finish num__60.0 of the task so b will take ( num__3 ( a + num__12 ) / num__5 ) number of days . but as we know b starts working along with a and finishes num__12 days after a stops working . so ( num__3 ( a + num__12 ) / num__5 ) = ( ( num__2 a ) / num__5 + num__12 ) num__3 a + num__36 = num__2 a + num__60 a = num__24 ; b = num__36 days . answer choice b <eor> b <eos> |
b |
divide__60.0__12.0__ subtract__5.0__2.0__ multiply__12.0__3.0__ multiply__12.0__2.0__ round__36.0__ |
divide__60.0__12.0__ subtract__5.0__2.0__ multiply__12.0__3.0__ multiply__12.0__2.0__ add__12.0__24.0__ |
| if point a coordinates are ( - num__7 - num__3 ) point b coordinates are ( num__7 num__4 ) if c is along the line a - b so that ac = num__0.5 cb . what are the coordinates of point c . <o> a ) ( - num__2 - num__0.666666666667 ) <o> b ) ( num__2 - num__0.666666666667 ) <o> c ) ( num__0.666666666667 - num__2 ) <o> d ) ( num__0 num__2 ) <o> e ) ( num__2 num__0.666666666667 ) |
look at the diagram below : line segment . pngsince ac = num__0.5 cb then ac = num__0.333333333333 ab . this implies that point c is closer to a than to b which on the other hand implies that the x - coordinate of c must be less than ( - num__7 + num__7 ) / num__2 = num__0.5 . only d and a fits but as ( num__0 num__2 ) is clearly not on line segment ab then the answer must be d . answer : d . <eor> d <eos> |
d |
reverse__3.0__ reverse__0.5__ round_down__0.5__ round_down__0.5__ |
reverse__3.0__ reverse__0.5__ round_down__0.5__ divide__0.0__7.0__ |
| a seven - digit combination lock on a safe has zero exactly three times does not have the digit num__1 at all . what is the probability that exactly num__3 of its digits are odd ? <o> a ) num__0.5625 <o> b ) num__0.5 <o> c ) num__0.333333333333 <o> d ) num__0.25 <o> e ) num__0.166666666667 |
remaining number of digits after three ' num__0 ' digits = num__4 constraint - no more ' num__0 ' and no more ' num__1 ' well these four digits can be odd or even num__0.5 * num__0.5 * num__0.5 * num__0.5 = num__0.0625 ways to choose num__3 odd digits out of total num__4 digits = num__4 c num__3 = num__4 num__4 * num__0.0625 = num__0.25 d is the answer hi ! since we are talking about remaining num__4 digits . each digit can either be odd or even ( total ways to select = num__2 . desired outcome = num__1 ) so let ' s say we want to choose first num__3 digits as odd digits and last digit as even . the probability will be : - num__0.5 * num__0.5 * num__0.5 * num__0.5 = num__0.0625 but out of remaining num__4 digit any num__3 can be odd and one even as there are no constraints in the question . therefore total ways of choosing num__3 odd digits out of num__4 total digits = num__4 c num__3 so the probability = num__4 * num__0.0625 = num__0.25 answer is d <eor> d <eos> |
d |
add__1.0__3.0__ reverse__4.0__ reverse__0.5__ reverse__4.0__ |
add__1.0__3.0__ multiply__4.0__0.0625__ subtract__3.0__1.0__ multiply__4.0__0.0625__ |
| when xyz co discontinued an agency arrangement with john management held his commission of rs . num__25000 / - for one month . earlier john had availed an advance agency fees of num__8280 / - from company . but robert forgot that . after one month john asked his fees and accountant gives rs . num__18500 / - to him . what is the incentive amount given to john ? <o> a ) a ) num__1780 <o> b ) b ) num__1250 <o> c ) c ) num__10780 <o> d ) d ) num__10500 <o> e ) e ) num__8600 |
total fees = rs . num__25000 / - advance fees = num__8280 / - balance fees = num__25000 - num__8280 = num__16720 paid amount = num__18500 / - incentive amount = num__18500 - num__16720 = num__1780 / - answer is a <eor> a <eos> |
a |
subtract__25000.0__8280.0__ subtract__18500.0__16720.0__ subtract__18500.0__16720.0__ |
subtract__25000.0__8280.0__ subtract__18500.0__16720.0__ subtract__18500.0__16720.0__ |
| if a ^ num__2 + b ^ num__2 = num__144 and ab ≠ num__0 then the greatest possible value for b is between <o> a ) num__16 and num__13 <o> b ) num__12 and num__3 <o> c ) num__11 and num__4 <o> d ) num__10 and num__5 <o> e ) num__9 and num__6 |
since a ≠ num__0 greatest possible value for b will be less than num__12 but closer to it . so it lies between num__11 and num__12 answer : b <eor> b <eos> |
b |
divide__144.0__12.0__ |
divide__144.0__12.0__ |
| a is thrice as good a workman as b & therefore is able to finish a job in num__45 days less than b . in how many days they can finish the job if they work together ? <o> a ) num__10 <o> b ) num__11.25 <o> c ) num__12 <o> d ) num__13.45 <o> e ) num__14 |
a is thrice as good a workman as b = > a = num__3 b the ratio of efficiencies of a and b = num__3 : num__1 the ratio of the days taken by a and b to finish work is = num__1 : num__3 b takes num__45 days to finish the work = > num__3 parts = num__45 days a alone takes = > num__15.0 x num__1 = num__15 days a and b alone can finish the work is = ab / a + b = num__15 x num__3.0 + num__45 = num__11.25 b <eor> b <eos> |
b |
divide__45.0__3.0__ round__11.25__ |
divide__45.0__3.0__ divide__11.25__1.0__ |
| set a consists of the following unique integers : - num__2 num__17 num__3 n num__2 num__15 - num__3 and - num__27 ; which of the following could be the median of set a ? <o> a ) num__13 <o> b ) num__1 <o> c ) num__5 <o> d ) num__67 <o> e ) num__7 |
whenever a question asks you to deal with the median of a group of numbers you must put the numbers in order from least to greatest ( or at the very least group offthe numbers so that you can restrict the potential value of the median ) . here we have num__8 values ( one of the values is the variable n ) . we ' re told that the values are unique so n can not be any of the other num__7 values listed . we ' re asked which of the following answers could be the median so if we determine that one of the answer choices is a possible median then we can stop working . . . . since the group includes num__8 values the median will be the average of the num__4 th and num__5 th values . . . putting the num__7 numbers in order we have : - num__27 - num__3 - num__2 num__2 num__3 num__15 num__17 this means that the num__2 will either be the num__4 th term or the num__5 th term ( depending on the value of n ) . if . . . . n is really big then the num__4 th and num__5 th terms will be num__2 and num__3 so the median will = num__2.5 n is really small then the num__4 th and num__5 th terms will be - num__2 and num__2 so the median will = num__0 neither of those answers is among the choices though so n must be something relativelycloseto num__2 . . . . . if . . . . n = num__0 then the num__4 th and num__5 th terms will be num__0 and num__2 so the median will = num__1 . that answer is among the choices so we ' re done . b <eor> b <eos> |
b |
subtract__15.0__8.0__ subtract__7.0__3.0__ add__2.0__3.0__ divide__5.0__2.0__ subtract__3.0__2.0__ reverse__1.0__ |
subtract__15.0__8.0__ subtract__7.0__3.0__ subtract__7.0__2.0__ divide__5.0__2.0__ subtract__3.0__2.0__ subtract__2.0__1.0__ |
| find the number of different prime factors of num__6440 <o> a ) num__4 <o> b ) num__2 <o> c ) num__3 <o> d ) num__5 <o> e ) num__6 |
explanation : l . c . m of num__6440 = num__2 x num__2 x num__2 x num__5 x num__7 x num__23 num__2 num__5 num__723 number of different prime factors is num__4 . answer : option a <eor> a <eos> |
a |
add__2.0__5.0__ gcd__6440.0__4.0__ |
add__2.0__5.0__ gcd__6440.0__4.0__ |
| a train num__110 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__7 sec <o> b ) num__6 sec <o> c ) num__9 sec <o> d ) num__4 sec <o> e ) num__3 sec |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__110 * num__0.0545454545455 ] sec = num__6 sec answer : b <eor> b <eos> |
b |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ round__6.0__ |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ divide__110.0__18.3333__ |
| the total number of plums that grow during each year on a certain plum tree is equal to the number of plums that grew during the previous year less the age of the tree in years ( rounded down to the nearest integer ) . during its num__2 nd year the plum tree grew num__50 plums . if this trend continues how many plums will it grow during its num__6 th year ? <o> a ) num__36 <o> b ) num__38 <o> c ) num__40 <o> d ) num__42 <o> e ) num__48 |
num__1 st year : num__0 - num__1 ( age ) we take age = num__0 ( as the question says that we have to ( rounded down to the nearest integer ) ) num__2 ndyear : num__1 - num__2 ( age ) we take age = num__1 num__3 rd year : num__2 - num__3 ( age ) we take age = num__2 num__4 th year : num__3 - num__4 ( age ) we take age = num__3 num__5 th year : num__4 - num__5 ( age ) we take age = num__4 num__6 th year : num__5 - num__6 ( age ) we take age = num__5 thus for the num__2 nd year = num__50 num__3 rd year = num__50 - num__2 = num__48 num__4 th year = num__48 - num__3 = num__45 num__5 th year = num__45 - num__4 = num__41 . num__6 th year = num__41 - num__5 = num__36 the correct answer is a . <eor> a <eos> |
a |
add__2.0__1.0__ subtract__6.0__2.0__ add__2.0__3.0__ subtract__50.0__2.0__ subtract__50.0__5.0__ subtract__45.0__4.0__ subtract__41.0__5.0__ multiply__1.0__36.0__ |
add__2.0__1.0__ subtract__6.0__2.0__ subtract__6.0__1.0__ subtract__50.0__2.0__ subtract__50.0__5.0__ subtract__45.0__4.0__ subtract__41.0__5.0__ subtract__41.0__5.0__ |
| a thief is noticed by a policeman from a distance of num__200 m . the thief starts running and the policeman chases him . the thief and the policeman run at the rate of num__10 km and num__11 km per hour respectively . what is the distance between them after num__5 minutes ? <o> a ) num__120 m <o> b ) num__200 m <o> c ) num__170 m <o> d ) num__100 m <o> e ) num__250 m |
relative speed of the thief and policeman = num__11 - num__10 = num__1 km / hr . distance covered in num__5 minutes = num__0.0166666666667 * num__5 = num__0.0833333333333 km = num__80 m . distance between the thief and policeman = num__200 - num__80 = num__120 m a <eor> a <eos> |
a |
subtract__11.0__10.0__ subtract__200.0__80.0__ round__120.0__ |
subtract__11.0__10.0__ subtract__200.0__80.0__ subtract__200.0__80.0__ |
| ratio between two numbers is num__3 : num__4 and their sum is num__420 . find the smaller number ? <o> a ) num__23 <o> b ) num__180 <o> c ) num__287 <o> d ) num__137 <o> e ) num__192 |
explanation : num__3 x + num__4 x = num__420 x = num__60 = > num__3 x = num__180 answer : b <eor> b <eos> |
b |
multiply__3.0__60.0__ multiply__3.0__60.0__ |
multiply__3.0__60.0__ multiply__3.0__60.0__ |
| a and b together have rs . num__1210 . if num__0.266666666667 of a ' s amount is equal to num__0.4 of b ' s amount . how much amount b have . <o> a ) num__83.5 <o> b ) num__87.5 <o> c ) num__88.5 <o> d ) num__89.5 <o> e ) num__484 rupees |
explanation : in this type of question we will first try to calculate the ratio of two persons . once we get ratio then we can easily get our answer . so lets solve this . num__0.266666666667 a = num__0.4 b a = ( num__0.4 ∗ num__3.5 ) b a = num__1.5 b a / b = num__1.5 a : b = num__3 : num__2 b ' s share = num__0.4 ∗ num__1210 = num__484 option e <eor> e <eos> |
e |
round_down__3.5__ subtract__3.5__1.5__ multiply__1210.0__0.4__ multiply__1210.0__0.4__ |
round_down__3.5__ divide__3.0__1.5__ multiply__1210.0__0.4__ multiply__1210.0__0.4__ |
| a man can swim in still water at num__7.5 km / h but takes twice as long to swim upstream than downstream . the speed of the stream is ? <o> a ) num__1.7 <o> b ) num__1.4 <o> c ) num__2.5 <o> d ) num__1.5 <o> e ) num__1.1 |
m = num__7.5 s = x ds = num__7.5 + x us = num__7.5 + x num__7.5 + x = ( num__7.5 - x ) num__2 num__7.5 + x = num__15 - num__2 x num__3 x = num__7.5 x = num__2.5 answer : c <eor> c <eos> |
c |
multiply__7.5__2.0__ divide__7.5__3.0__ round__2.5__ |
multiply__7.5__2.0__ divide__7.5__3.0__ round__2.5__ |
| a man can row num__6 kmph in still water . when the river is running at num__1.2 kmph it takes him num__1 hour to row to a place and black . how far is the place ? <o> a ) num__2.89 <o> b ) num__2.88 <o> c ) num__2.82 <o> d ) num__2.87 <o> e ) num__2.84 |
m = num__6 s = num__1.2 ds = num__6 + num__1.2 = num__7.2 us = num__6 - num__1.2 = num__4.8 x / num__7.2 + x / num__4.8 = num__1 x = num__2.88 answer : b <eor> b <eos> |
b |
add__6.0__1.2__ subtract__6.0__1.2__ round__2.88__ |
add__6.0__1.2__ subtract__6.0__1.2__ divide__2.88__1.0__ |
| a boat can move upstream at num__30 kmph and downstream at num__35 kmph then the speed of the current is ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__2.5 |
us = num__30 ds = num__35 m = ( num__35 - num__30 ) / num__2 = num__2.5 answer : e <eor> e <eos> |
e |
round__2.5__ |
round__2.5__ |
| the ratio of two numbers is num__3 : num__4 and their sum is num__14 . the greater of the two numbers is ? <o> a ) num__12 <o> b ) num__14 <o> c ) num__16 <o> d ) num__8 <o> e ) num__19 |
num__3 : num__4 total parts = num__7 = num__7 parts - - > num__14 ( num__7 Ã — num__2 = num__14 ) = num__1 part - - - - > num__2 ( num__1 Ã — num__2 = num__2 ) = the greater of the two number is = num__4 = num__4 parts - - - - > num__2 ( num__2 Ã — num__4 = num__8 ) d ) <eor> d <eos> |
d |
add__3.0__4.0__ divide__14.0__7.0__ subtract__3.0__2.0__ multiply__4.0__2.0__ multiply__4.0__2.0__ |
add__3.0__4.0__ divide__14.0__7.0__ subtract__3.0__2.0__ multiply__4.0__2.0__ multiply__4.0__2.0__ |
| the average marks of a class of num__30 students is num__40 and that of another class of num__50 students is num__60 . find the average marks of all the students ? <o> a ) num__52.7 <o> b ) num__52.9 <o> c ) num__52.4 <o> d ) num__52.5 <o> e ) num__52.1 |
sum of the marks for the class of num__30 students = num__30 * num__40 = num__1200 sum of the marks for the class of num__50 students = num__50 * num__60 = num__3000 sum of the marks for the class of num__80 students = num__1200 + num__3000 = num__4200 average marks of all the students = num__52.5 = num__52.5 answer : d <eor> d <eos> |
d |
multiply__30.0__40.0__ multiply__50.0__60.0__ add__30.0__50.0__ add__1200.0__3000.0__ divide__4200.0__80.0__ divide__4200.0__80.0__ |
multiply__30.0__40.0__ multiply__50.0__60.0__ add__30.0__50.0__ add__1200.0__3000.0__ divide__4200.0__80.0__ divide__4200.0__80.0__ |
| a dishonest dealer professes to sell goods at the cost price but uses a false weight and gains num__25.0 . find his false weight age ? <o> a ) num__878 <o> b ) num__768 <o> c ) num__800 <o> d ) num__778 <o> e ) num__876 |
num__25 = e / ( num__1000 - e ) * num__100 num__1000 - e = num__4 e num__1000 = num__5 e = > e = num__200 num__1000 - num__200 = num__800 answer : c <eor> c <eos> |
c |
percent__100.0__800.0__ |
percent__100.0__800.0__ |
| a circular jogging track forms the edge of a circular lake that has a diameter of num__2 miles . johanna walked once around the track at the average speed of num__5 miles per hour . if t represents the number of hours it took johanna to walk completely around the lake which of the following is a correct statement ? <o> a ) num__0.5 < t < num__0.75 <o> b ) num__1.75 < t < num__2.0 <o> c ) num__1.0 < t < num__1.5 <o> d ) num__2.5 < t < num__3.0 <o> e ) num__3 < t < num__3.5 |
d = num__2 and r = num__1 circumference thus is num__2 pi miles we know that she walked num__5 mph knowing the formula rt = d we can deduce that t = d / r d is num__2 pi miles and r is num__5 mph t = num__2 pi / num__5 pi can be rewritten as num__3.14285714286 num__2 * num__3.14285714286 = num__6.28571428571 and multiply this by num__0.2 = num__1.25714285714 . this is greater than num__1 but less than num__1.5 therefore num__1 < t < num__1.5 . answer choice c . <eor> c <eos> |
c |
divide__1.0__5.0__ multiply__0.2__6.2857__ round__1.0__ |
divide__1.0__5.0__ multiply__0.2__6.2857__ multiply__5.0__0.2__ |
| a horse and cow were sold for rs . num__12000 each . the horse was sold at a loss of num__20.0 and the cow at a gain of num__20.0 . the entire transaction resulted in : <o> a ) no loss or gain <o> b ) loss of rs . num__1000 <o> c ) gain of rs . num__1000 <o> d ) gain of rs . num__2000 <o> e ) none of these |
explanation : in the special case of profit and loss percentages are equal and selling price is same then the transaction always results in loss . this loss percentage is given by a simple formula − ( x / num__10 ) num__2 so in this case profit % = loss % = num__20 . so x = num__20 loss percentage = − ( num__2.0 ) num__2 = − num__4 total s . p = rs . num__24000 cost price = num__24000 × num__1.04166666667 = num__25000 loss = rs . num__1000 correct option : b <eor> b <eos> |
b |
percent__20.0__10.0__ percent__4.0__25000.0__ percent__4.0__25000.0__ |
percent__20.0__10.0__ percent__4.0__25000.0__ percent__4.0__25000.0__ |
| a train passes a station platform in num__42 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__177 m <o> b ) num__176 m <o> c ) num__240 m <o> d ) num__187 m <o> e ) num__330 m |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x meters . then x + num__7.14285714286 = num__15 x + num__300 = num__630 x = num__330 m . answer : e <eor> e <eos> |
e |
multiply__20.0__15.0__ divide__300.0__42.0__ multiply__42.0__15.0__ subtract__630.0__300.0__ round__330.0__ |
multiply__20.0__15.0__ divide__300.0__42.0__ multiply__42.0__15.0__ subtract__630.0__300.0__ round__330.0__ |
| two trains are traveling on parallel tracks in the same direction . the faster train travels at num__130 miles per hour while the slower train travels at num__80 miles per hour . at num__2 o ’ clock the faster train is num__20 miles behind the slower one . how far apart are the two trains at num__5 o ' clock ? <o> a ) num__60 miles <o> b ) num__80 miles <o> c ) num__90 miles <o> d ) num__130 miles <o> e ) num__400 miles |
answer = d . num__130 miles relational speed = num__130 - num__80 = num__50 miles per hour in num__3 hours difference = num__50 * num__3 = num__150 miles fast train trailing num__20 miles so effective difference = num__150 - num__20 = num__130 miles <eor> d <eos> |
d |
subtract__130.0__80.0__ subtract__5.0__2.0__ add__130.0__20.0__ round__130.0__ |
subtract__130.0__80.0__ subtract__5.0__2.0__ multiply__3.0__50.0__ subtract__150.0__20.0__ |
| an oil cylinder was num__0.75 th full . when six bottles of oil is poured into it it is num__0.8 th full . how many bottles of oil can the full cylinder hold ? <o> a ) num__120 <o> b ) num__115 <o> c ) num__140 <o> d ) num__130 <o> e ) num__135 |
( num__0.8 - num__0.75 ) cylinder is filled by num__6 bottles num__0.8 - num__0.75 = num__0.05 cylinder is filled by num__6 bottles num__1 cylinder is filled by num__6 / num__0.05 = num__120 bottles answer : a <eor> a <eos> |
a |
subtract__0.8__0.75__ divide__6.0__0.05__ multiply__1.0__120.0__ |
subtract__0.8__0.75__ divide__6.0__0.05__ divide__6.0__0.05__ |
| how many cuboids of length num__5 m width num__6 m and height num__3 m can be farmed from a cuboid of num__18 m length num__15 m width and num__2 m height . <o> a ) num__4 <o> b ) num__6 <o> c ) num__10 <o> d ) num__8 <o> e ) num__11 |
( num__18 Ã — num__15 Ã — num__12 ) / ( num__5 Ã — num__3 Ã — num__2 ) = num__6 answer is b . <eor> b <eos> |
b |
multiply__6.0__2.0__ round__6.0__ |
multiply__6.0__2.0__ divide__18.0__3.0__ |
| what is the units digit of ( num__22 ^ num__4 ) ( num__16 ^ num__3 ) ( num__41 ^ num__8 ) ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
the units digit of num__22 ^ num__4 is the units digit of num__2 ^ num__4 which is num__6 . the units digit of num__16 ^ num__3 is the units digit of num__6 ^ num__3 which is num__6 . the units digit of num__41 ^ num__8 is the units digit of num__1 ^ num__8 which is num__1 . the units digit of num__6 * num__6 * num__1 is num__6 . the answer is e . <eor> e <eos> |
e |
divide__16.0__8.0__ subtract__22.0__16.0__ subtract__4.0__3.0__ subtract__22.0__16.0__ |
divide__16.0__8.0__ multiply__3.0__2.0__ subtract__4.0__3.0__ multiply__3.0__2.0__ |
| a bus starts from city x . the number of women in the bus is half of the number of men . in city y num__14 men leave the bus and seven women enter . now number of men and women is equal . in the beginning how many passengers entered the bus ? <o> a ) num__63 <o> b ) num__30 <o> c ) num__36 <o> d ) num__45 <o> e ) num__46 |
explanation : originally let number of women = x . then number of men = num__2 x . so in city y we have : ( num__2 x - num__14 ) = ( x + num__7 ) or x = num__21 . therefore total number of passengers in the beginning = ( x + num__2 x ) = num__3 x = num__63 . answer : a <eor> a <eos> |
a |
divide__14.0__2.0__ add__14.0__7.0__ divide__21.0__7.0__ multiply__3.0__21.0__ multiply__3.0__21.0__ |
divide__14.0__2.0__ add__14.0__7.0__ divide__21.0__7.0__ multiply__3.0__21.0__ multiply__3.0__21.0__ |
| calculate num__469157 x num__9999 = ? <o> a ) num__4586970843 <o> b ) num__4686970743 <o> c ) num__4691100843 <o> d ) num__4586870843 <o> e ) none |
answer num__469157 x num__9999 = num__469157 x ( num__10000 - num__1 ) = num__4691570000 - num__469157 = num__4691100843 . option : c <eor> c <eos> |
c |
subtract__10000.0__9999.0__ multiply__469157.0__10000.0__ multiply__469157.0__9999.0__ multiply__469157.0__9999.0__ |
subtract__10000.0__9999.0__ multiply__469157.0__10000.0__ subtract__4691570000.0__469157.0__ subtract__4691570000.0__469157.0__ |
| anmol can eat num__27 pastries in a hour . aakriti can eat num__2 pastries in num__10 minutes . divya can eat num__7 pastries in num__20 minutes . how long will it take them to eat a num__180 pastries ? <o> a ) num__2 hours . <o> b ) num__1 hours . <o> c ) num__5 hours . <o> d ) num__4 hours . <o> e ) num__3 hours . |
in one hour anmol eats num__27 pastries aakriti eats num__12 and divya eats num__21 . a total of num__60 pastries . therefore num__180 pastries would take num__180 ÷ num__60 = num__3 hours . answer e <eor> e <eos> |
e |
add__2.0__10.0__ hour_to_min_conversion__ subtract__10.0__7.0__ round__3.0__ |
add__2.0__10.0__ hour_to_min_conversion__ subtract__10.0__7.0__ round__3.0__ |
| for any non - zero a and b that satisfy | ab | = ab and | c | = - c | b - num__4 | + | ab - b | = ? <o> a ) a ) ab - num__4 <o> b ) b ) b - num__2 b + num__4 <o> c ) c ) ab + num__4 <o> d ) d ) num__2 b - ab - num__4 a <o> e ) e ) num__4 - ab |
| a | = - a suggests that ' a ' is negative | ab | = ab suggests that ' ab ' is positive ' ab ' is positive suggests that ' a ' and ' b ' have same sign i . e . either both positive or both negative but since ' a ' is negative therefore ' b ' is negative too . since b is negative so | b - num__4 | = - b + num__4 since ab is positive and b is negative so | ab - b | = ab - b i . e . | b - num__4 | + | ab - b | = - b + num__4 + ab - b = ab - num__2 b + num__4 answer : option b <eor> b <eos> |
b |
subtract__4.0__2.0__ |
subtract__4.0__2.0__ |
| there are num__6 boxes numbered num__1 num__23 num__45 and num__6 . each box is to be filled up either with a red or a green ball in such a way that at least num__1 box contains a green ball & the boxes containing green balls are consecutively numbered . the total no . of ways in which this can be done is ? <o> a ) num__18 <o> b ) num__19 <o> c ) num__21 <o> d ) num__25 <o> e ) num__27 |
if only one of the boxes has a green ball it can be any of the num__6 boxes . so this can be achieved in num__6 ways . if two of the boxes have green balls and then there are num__5 consecutive sets of num__2 boxes . num__12 num__23 num__34 num__45 num__56 . similarly if num__3 of the boxes have green balls there will be num__4 options . if num__4 boxes have green balls there will be num__3 options . if num__5 boxes have green balls then there will be num__2 options . if all num__6 boxes have green balls then there will be just num__1 options . total number of options = num__6 + num__5 + num__4 + num__3 + num__2 + num__1 = num__21 c <eor> c <eos> |
c |
subtract__6.0__1.0__ multiply__6.0__2.0__ divide__6.0__2.0__ subtract__6.0__2.0__ subtract__23.0__2.0__ multiply__1.0__21.0__ |
subtract__6.0__1.0__ multiply__6.0__2.0__ add__1.0__2.0__ add__1.0__3.0__ subtract__23.0__2.0__ multiply__1.0__21.0__ |
| a can finish a work in num__24 days b in num__9 days and c in num__12 days . b and c start the work but are forced to leave after num__3 days . when a done the work ? <o> a ) num__10 days <o> b ) num__12 days <o> c ) num__13 days <o> d ) num__9 days <o> e ) num__14 days |
b + c = = > num__0.111111111111 + num__0.0833333333333 = num__0.194444444444 b c = in num__3 days = num__0.194444444444 * num__3 = num__0.583333333333 remaining work = num__1 - num__0.583333333333 = num__0.416666666667 num__0.0416666666667 work is done by a in num__1 day num__0.416666666667 work is done a num__24 * num__0.416666666667 = num__10 days answer a <eor> a <eos> |
a |
add__0.1111__0.0833__ subtract__1.0__0.5833__ divide__1.0__24.0__ add__9.0__1.0__ round__10.0__ |
add__0.1111__0.0833__ subtract__1.0__0.5833__ divide__1.0__24.0__ add__9.0__1.0__ add__9.0__1.0__ |
| what is the area of a square with perimeter num__12 p ? <o> a ) num__16 p ^ num__2 <o> b ) num__4 p <o> c ) num__9 p ^ num__2 <o> d ) p / num__16 <o> e ) p ^ num__0.125 |
each side is num__3 p a = ( num__3 p ) ^ num__2 = num__9 p ^ num__2 answer c <eor> c <eos> |
c |
power__3.0__2.0__ triangle_area__9.0__2.0__ |
power__3.0__2.0__ power__3.0__2.0__ |
| average of first five multiples of num__3 is <o> a ) num__9 <o> b ) num__11 <o> c ) num__13 <o> d ) num__15 <o> e ) num__17 |
explanation : average = num__3 ( num__1 + num__2 + num__3 + num__4 + num__5 ) / num__5 = num__9.0 = num__9 option a <eor> a <eos> |
a |
subtract__3.0__1.0__ add__3.0__1.0__ add__3.0__2.0__ add__4.0__5.0__ multiply__1.0__9.0__ |
subtract__3.0__1.0__ add__3.0__1.0__ add__3.0__2.0__ add__4.0__5.0__ add__4.0__5.0__ |
| the average of five numbers is num__9 . the average of first two numbers is num__7 and the average of last two numbers is num__12 . what is the middle number ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__5 <o> d ) num__10 <o> e ) num__6 |
the total of five numbers = num__5 x num__9 = num__45 the total of first num__2 and last num__2 numbers is = num__2 x num__7 + num__2 x num__12 = num__38 so the middle number is ( num__45 - num__38 ) = num__7 a <eor> a <eos> |
a |
subtract__12.0__7.0__ multiply__9.0__5.0__ subtract__9.0__7.0__ subtract__45.0__7.0__ subtract__9.0__2.0__ |
subtract__12.0__7.0__ multiply__9.0__5.0__ subtract__9.0__7.0__ subtract__45.0__7.0__ subtract__9.0__2.0__ |
| a man is num__24 years older than his son . in two years his age will be twice the age of his son . the present age of this son is <o> a ) num__78 years <o> b ) num__22 years <o> c ) num__88 years <o> d ) num__66 years <o> e ) num__65 years |
let ' s son age is x then father age is x + num__24 . = > num__2 ( x + num__2 ) = ( x + num__24 + num__2 ) = > num__2 x + num__4 = x + num__26 = > x = num__22 years answer : b <eor> b <eos> |
b |
add__24.0__2.0__ subtract__24.0__2.0__ subtract__24.0__2.0__ |
add__24.0__2.0__ subtract__24.0__2.0__ subtract__24.0__2.0__ |
| what is the smallest num__6 digit number exactly divisible by num__111 ? <o> a ) num__100010 <o> b ) num__100012 <o> c ) num__100011 <o> d ) num__100013 <o> e ) num__120012 |
smallest num__6 digit number = num__100000 num__900.900900901 = num__900 remainder = num__100 . hence num__11 more should be added to num__100000 to get the smallest num__6 digit number exactly divisible by num__111 = > smallest num__6 digit number exactly divisible by num__111 = num__100000 + num__11 = num__100011 answer is c <eor> c <eos> |
c |
divide__100000.0__111.0__ round_down__900.9009__ subtract__111.0__100.0__ add__100000.0__11.0__ add__100000.0__11.0__ |
divide__100000.0__111.0__ round_down__900.9009__ subtract__111.0__100.0__ add__100000.0__11.0__ add__100000.0__11.0__ |
| the equation x / y - num__0.4 = num__0 is equivalent to each of the following except <o> a ) ( x - y ) / y = - num__0.6 <o> b ) ( num__5 x - num__3 y ) / y = num__4.2 <o> c ) ( x + num__3 y ) / y = num__3.4 <o> d ) ( x + y ) / x = num__3.5 <o> e ) x / y + num__1 = num__1.4 |
x / y - num__0.4 = num__0 num__5 x = num__2 y x = num__0.4 * y or y = num__2.5 * x plug in x or y depending on the equation . all expressions are true except for b . ( num__5 x - num__3 y ) / y = num__4.2 where y = - num__21 b . ( num__5 x - num__3 y ) / y = num__4.2 answer : b <eor> b <eos> |
b |
multiply__0.4__5.0__ reverse__0.4__ subtract__5.0__2.0__ multiply__5.0__4.2__ multiply__2.5__2.0__ |
multiply__0.4__5.0__ reverse__0.4__ subtract__5.0__2.0__ multiply__5.0__4.2__ multiply__2.5__2.0__ |
| a train travels from new york to chicago a distance of approximately num__840 miles at an average rate of num__60 miles per hour and arrives in chicago at num__7 : num__00 in evening chicago time . at what hour in the morning new york time did the train depart for chicago ? ( note : chicago time is one hour earlier than new york time ) <o> a ) num__3 : num__00 <o> b ) num__4 : num__00 <o> c ) num__5 : num__00 <o> d ) num__6 : num__00 <o> e ) num__7 : num__00 |
num__7 : num__00 in evening in chicago = num__8 : num__00 in evening in new york . so the train was in chicago num__8 : num__00 in the evening new york time . the trip took t = d / r = num__14.0 = num__14 hours . therefore the train depart from new york at num__8 : num__00 - num__14 hours = num__6 : num__00 in the morning new york time . answer : d . <eor> d <eos> |
d |
divide__840.0__60.0__ subtract__14.0__8.0__ round__6.0__ |
divide__840.0__60.0__ subtract__14.0__8.0__ subtract__14.0__8.0__ |
| a rectangular courtyard the sides of which are in the ratio of num__4 : num__3 cost rs . num__600 for paving at num__50 p per m num__2 ; find the length of the diagonal of the courtyard ? <o> a ) num__87 <o> b ) num__92 <o> c ) num__12 <o> d ) num__10 <o> e ) num__11 |
explanation : num__1 m num__2 - - - - num__0.5 ? - - - - - num__600 = > num__1200 m num__2 num__4 x * num__3 x = num__1200 = > x = num__10 answer : option d <eor> d <eos> |
d |
triangle_area__4.0__600.0__ rectangle_perimeter__4.0__1.0__ rectangle_perimeter__4.0__1.0__ |
multiply__600.0__2.0__ rectangle_perimeter__4.0__1.0__ multiply__1.0__10.0__ |
| the average age num__6 members of a committee are the same as it was num__2 years ago because an old number has been replaced by a younger number . find how much younger is the new member than the old number ? <o> a ) num__14 years <o> b ) num__17 years <o> c ) num__18 years <o> d ) num__12 years <o> e ) num__11 years |
num__6 * num__2 = num__12 answer : d <eor> d <eos> |
d |
multiply__6.0__2.0__ multiply__6.0__2.0__ |
multiply__6.0__2.0__ multiply__6.0__2.0__ |
| num__15.06 * num__0.001 = ? <o> a ) num__15060000 <o> b ) num__0.001506 <o> c ) num__0.01506 <o> d ) num__0.1506 <o> e ) none of these |
explanation : clearly after decimal num__5 digits should be there . option c <eor> c <eos> |
c |
multiply__15.06__0.001__ |
multiply__15.06__0.001__ |
| a train crosses a platform of num__120 m in num__15 sec same train crosses another platform of length num__180 m in num__18 sec . then find the length of the train ? <o> a ) num__877 m <o> b ) num__180 m <o> c ) num__786 m <o> d ) num__165 m <o> e ) num__456 m |
length of the train be ‘ x ’ x + num__8.0 = x + num__10.0 num__6 x + num__720 = num__5 x + num__900 x = num__180 m answer : b <eor> b <eos> |
b |
divide__120.0__15.0__ divide__180.0__18.0__ multiply__120.0__6.0__ subtract__15.0__10.0__ multiply__180.0__5.0__ round__180.0__ |
divide__120.0__15.0__ divide__180.0__18.0__ multiply__120.0__6.0__ subtract__15.0__10.0__ add__180.0__720.0__ round__180.0__ |
| in the xy - plane the point ( - num__2 - num__3 ) is the centre of a circle the point ( - num__2 num__2 ) lies inside the circle and the point ( num__5 - num__3 ) lies outside the circle . if the radius r of the circle r is an integer then r = <o> a ) num__6 <o> b ) num__5 <o> c ) num__4 <o> d ) num__3 <o> e ) num__2 |
can be solved without much calculations . you are given that ( - num__2 - num__3 ) is the center of the circle . point ( num__5 - num__3 ) lies inside the circle - - - > the radius is lesser than distance of ( - num__2 - num__3 ) from ( num__5 - num__3 ) - - - > lesser than num__7 units but the radius will also be greater than the distance of ( - num__2 - num__3 ) from ( - num__22 ) - - - - > greater than num__5 units . thus the radius is > num__5 but < num__7 and as it is an integer the only possible value of radius = num__6 units . a is the correct answer . <eor> a <eos> |
a |
add__2.0__5.0__ multiply__2.0__3.0__ multiply__2.0__3.0__ |
add__2.0__5.0__ multiply__2.0__3.0__ multiply__2.0__3.0__ |
| circular gears p and q start rotating at the same time at constant speeds . gear p makes num__10 revolutions per minute and gear q makes num__40 revolutions per minute . how many seconds after the gears start rotating will gear q have made exactly num__10 more revolutions than gear p ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__15 <o> d ) num__20 <o> e ) num__25 |
gear q makes num__30 more revolutions each num__60 seconds . the time to make num__10 = num__10.0 more revolutions is num__20.0 = num__20 seconds . the answer is d . <eor> d <eos> |
d |
subtract__40.0__10.0__ hour_to_min_conversion__ subtract__60.0__40.0__ round__20.0__ |
subtract__40.0__10.0__ hour_to_min_conversion__ subtract__60.0__40.0__ round__20.0__ |
| the average age of husband wife and their child num__3 years ago was num__27 years and that of wife and the child num__5 years ago was num__20 years . what is the present age of the husband ? <o> a ) num__30 <o> b ) num__40 <o> c ) num__20 <o> d ) num__50 <o> e ) num__25 |
let the present age of the husband = h present age of the wife = w present age of the child = c num__3 years ago average age of husband wife and their child = num__27 = > sum of age of husband wife and their child before num__3 years = num__3 × num__27 = num__81 = > ( h - num__3 ) + ( w - num__3 ) + ( c - num__3 ) = num__81 = > h + w + c = num__81 + num__9 = num__90 - - - equation ( num__1 ) num__5 years ago average age of wife and child = num__20 = > sum of age of wife and child before num__5 years = num__2 × num__20 = num__40 = > ( w - num__5 ) + ( c - num__5 ) = num__40 = > w + c = num__40 + num__10 = num__50 - - - equation ( num__2 ) substituting equation ( num__2 ) in equation ( num__1 ) = > h + num__50 = num__90 = > h = num__90 - num__50 = num__40 i . e . present age of the husband = num__40 answer is b . <eor> b <eos> |
b |
multiply__3.0__27.0__ divide__27.0__3.0__ add__9.0__81.0__ subtract__3.0__1.0__ multiply__20.0__2.0__ multiply__5.0__2.0__ multiply__5.0__10.0__ multiply__20.0__2.0__ |
multiply__3.0__27.0__ divide__27.0__3.0__ add__9.0__81.0__ subtract__3.0__1.0__ multiply__20.0__2.0__ add__1.0__9.0__ add__40.0__10.0__ subtract__50.0__10.0__ |
| factor : x num__4 y num__3 â € “ num__16 y num__3 <o> a ) a ) num__3 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__2 ) <o> b ) b ) y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__2 ) <o> c ) c ) num__3 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__3 ) <o> d ) d ) num__3 y num__3 ( x num__2 + num__4 ) ( x + num__3 ) ( x - num__2 ) <o> e ) e ) num__3 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__4 ) |
x num__4 y num__3 â € “ num__16 y num__3 . = y num__3 ( x num__4 â € “ num__16 ) . = y num__3 [ ( x num__2 ) num__2 - num__42 ] . = y num__3 ( x num__2 + num__4 ) ( x num__2 - num__4 ) . = y num__3 ( x num__2 + num__4 ) ( x num__2 - num__22 ) . = y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__2 ) . answer : ( b ) y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__2 ) <eor> b <eos> |
b |
gcd__3.0__42.0__ |
gcd__3.0__42.0__ |
| a can do a piece of work in num__12 days . when he had worked for num__2 days b joins him . if the complete work was finished in num__8 days . in how many days b alone can finish the work ? <o> a ) num__18 days <o> b ) num__11 days <o> c ) num__77 days <o> d ) num__188 days <o> e ) num__66 days |
num__0.666666666667 + num__6 / x = num__1 x = num__18 days answer : a <eor> a <eos> |
a |
divide__8.0__12.0__ divide__12.0__2.0__ add__12.0__6.0__ round__18.0__ |
divide__8.0__12.0__ divide__12.0__2.0__ add__12.0__6.0__ add__12.0__6.0__ |
| if x is num__20 percent more than y and y is num__70 percent less than z then x is what percent of z ? <o> a ) num__500.0 <o> b ) num__250.0 <o> c ) num__166.666666667 % <o> d ) num__36.0 <o> e ) num__60 % |
z = num__100 ; y = num__30 so x = num__36 x as % of z = num__0.36 * num__100 = > num__36.0 answer will be ( d ) <eor> d <eos> |
d |
subtract__100.0__70.0__ divide__36.0__100.0__ multiply__0.36__100.0__ |
subtract__100.0__70.0__ divide__36.0__100.0__ multiply__0.36__100.0__ |
| during a certain season a team won num__70 percent of its first num__100 games and num__50 percent of its remaining games . if the team won num__70 percent of its games for the entire season what was the total number of games that the team played ? <o> a ) num__180 <o> b ) num__170 <o> c ) num__156 <o> d ) num__150 <o> e ) num__100 |
we are first given that a team won num__70 percent of its first num__100 games . this means the team won num__0.7 x num__100 = num__70 games out of its first num__100 games . we are next given that the team won num__50 percent of its remaining games . if we use variable t to represent the total number of games in the season then we can say t – num__100 equals the number of remaining games in the season . thus we can say : num__0.5 ( t – num__100 ) = number of wins for remaining games num__0.5 t – num__50 = number of wins for remaining games lastly we are given that team won num__70 percent of all games played in the season . that is they won num__0.7 t games in the entire season . with this we can set up the equation : number of first num__100 games won + number of games won for remaining games = total number of games won in the entire season num__70 + num__0.5 t – num__50 = num__0.7 t num__20 = num__0.2 t num__200 = num__2 t num__100 = t answer is e . <eor> e <eos> |
e |
percent__50.0__200.0__ |
percent__50.0__200.0__ |
| you have been given the task of transporting num__3000 apples num__1000 miles from appleland to bananaville . your truck can carry num__1000 apples at a time . every time you travel a mile towards bananaville you must pay a tax of num__1 apple but you pay nothing when going in the other direction ( towards appleland ) . what is highest number of apples you can get to bananaville ? <o> a ) num__767 <o> b ) num__833 <o> c ) num__916 <o> d ) num__1123 <o> e ) num__907 |
step one : first you want to make num__3 trips of num__1000 apples num__333 miles . you will be left with num__2001 apples and num__667 miles to go . step two : next you want to take num__2 trips of num__1000 apples num__500 miles . you will be left with num__1000 apples and num__167 miles to go ( you have to leave an apple behind ) . step three : finally you travel the last num__167 miles with one load of num__1000 apples and are left with num__833 apples in bananaville . <eor> b <eos> |
b |
divide__3000.0__1000.0__ subtract__1000.0__333.0__ subtract__3.0__1.0__ divide__1000.0__2.0__ subtract__500.0__333.0__ subtract__1000.0__167.0__ round__833.0__ |
divide__3000.0__1000.0__ subtract__1000.0__333.0__ subtract__3.0__1.0__ divide__1000.0__2.0__ subtract__500.0__333.0__ subtract__1000.0__167.0__ round__833.0__ |
| a rectangular farm has to be fenced one long side one short side and the diagonal . if the cost of fencing is rs . num__10 per meter . the area of farm is num__1200 m num__2 and the short side is num__30 m long . how much would the job cost ? <o> a ) num__1278 <o> b ) num__1200 <o> c ) num__2887 <o> d ) num__1688 <o> e ) num__2768 |
l * num__30 = num__1200 è l = num__40 num__40 + num__30 + num__50 = num__120 num__120 * num__10 = num__1200 answer : b <eor> b <eos> |
b |
square_perimeter__10.0__ square_perimeter__30.0__ multiply__10.0__120.0__ |
square_perimeter__10.0__ square_perimeter__30.0__ multiply__10.0__120.0__ |
| in a bag there are six num__6 - sided dice ( numbered num__1 to num__6 ) three num__12 - sided dice ( numbered num__1 to num__12 ) and two num__20 - sided dice ( numbered num__1 to num__20 ) . if four of these dice are selected at random from the bag and then the four are rolled and we find the sum of numbers showing on the four dice how many different possible totals are there for this sum ? <o> a ) num__42 <o> b ) num__60 <o> c ) num__61 <o> d ) num__84 <o> e ) num__960 |
as ' complex - looking ' as this question might appear it ' s actually rather simple . pay careful attention to what the specific questions asks for - the number of different possible sums from num__4 dice . since we ' re dealing with some ' special ' dice ( some num__12 - sided and num__20 - sided dice ) we have to adjust out math accordingly but the possibilities are rather limited : num__1 ) the minimum number on any given die is num__1 num__2 ) the maximum possible sum would only occur if we took the num__4 biggest possible dice and rolled the highest possible number on each . with num__4 dice we could end up with any sum between : num__4 ( if we rolled num__1 s on all num__4 dice ) to num__64 ( if we rolled two num__20 s on the num__20 - sided dice and two num__12 s on the num__12 - sided dice ) . thus there are only num__61 possible sums . c <eor> c <eos> |
c |
subtract__6.0__4.0__ multiply__1.0__61.0__ |
subtract__6.0__4.0__ multiply__1.0__61.0__ |
| sandy is younger than molly by num__14 years . if their ages are in the respective ratio of num__7 : num__9 how old is molly ? <o> a ) num__36 <o> b ) num__45 <o> c ) num__54 <o> d ) num__63 <o> e ) num__72 |
s = m - num__14 s / m = num__0.777777777778 num__9 s = num__7 m num__9 ( m - num__14 ) = num__7 m m = num__63 the answer is d . <eor> d <eos> |
d |
divide__7.0__9.0__ multiply__7.0__9.0__ multiply__7.0__9.0__ |
divide__7.0__9.0__ multiply__7.0__9.0__ multiply__7.0__9.0__ |
| last year for every num__100 million vehicles that traveled on a certain highway num__80 vehicles were involved in accidents . if num__4 billion vehicles traveled on the highway last year how many of those vehicles were involved in accidents ? ( num__1 billion = num__1000 num__000000 ) <o> a ) num__1600 <o> b ) num__3200 <o> c ) num__800 <o> d ) num__400 <o> e ) num__200 |
to solve we will set up a proportion . we know that “ num__100 million vehicles is to num__80 accidents as num__4 billion vehicles is to x accidents ” . to express everything in terms of “ millions ” we can use num__4000 million rather than num__4 billion . creating a proportion we have : num__1.25 = num__4000 / x cross multiplying gives us : num__100 x = num__4000 * num__80 x = num__40 * num__80 = num__3200 answer : b <eor> b <eos> |
b |
multiply__4.0__1000.0__ divide__100.0__80.0__ divide__4000.0__100.0__ multiply__80.0__40.0__ multiply__80.0__40.0__ |
multiply__4.0__1000.0__ divide__100.0__80.0__ divide__4000.0__100.0__ multiply__80.0__40.0__ multiply__80.0__40.0__ |
| in track last week the boys ran sixteen laps . the girls ran four more laps . each lap is a quarter of a mile . how many miles did the girls run ? <o> a ) num__3 miles <o> b ) num__2 miles <o> c ) num__5 miles <o> d ) num__9 miles <o> e ) num__7 miles |
the girls ran num__16 + num__4 = num__20 laps . num__20 x ¼ = num__5.0 which reduces to num__5 . the girls ran num__5 miles correct answer c <eor> c <eos> |
c |
add__16.0__4.0__ divide__20.0__4.0__ divide__20.0__4.0__ |
add__16.0__4.0__ divide__20.0__4.0__ divide__20.0__4.0__ |
| two trains num__140 m and num__160 m long run at the speed of num__60 km / hr and num__40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? <o> a ) num__10.7 <o> b ) num__10.9 <o> c ) num__10.2 <o> d ) num__10.8 <o> e ) num__18.8 |
relative speed = num__60 + num__40 = num__100 km / hr . = num__100 * num__0.277777777778 = num__27.7777777778 m / sec . distance covered in crossing each other = num__140 + num__160 = num__300 m . required time = num__300 * num__0.036 = num__10.8 = num__10.8 sec . answer : d <eor> d <eos> |
d |
subtract__140.0__40.0__ add__140.0__160.0__ multiply__300.0__0.036__ round__10.8__ |
add__60.0__40.0__ add__140.0__160.0__ multiply__300.0__0.036__ multiply__300.0__0.036__ |
| the cross - section of a cannel is a trapezium in shape . if the cannel is num__9 m wide at the top and num__7 m wide at the bottom and the area of cross - section is num__560 sq m the depth of cannel is ? <o> a ) num__39 <o> b ) num__28 <o> c ) num__70 <o> d ) num__80 <o> e ) num__71 |
num__0.5 * d ( num__9 + num__7 ) = num__560 d = num__70 answer : c <eor> c <eos> |
c |
round__70.0__ |
round__70.0__ |
| according to the directions on the can of frozen orange juice concentrate num__1 can of concentrate is to be mixed with num__3 cans of water to make orange juice . how many num__12 ounces cans of the concentrate are required to prepare num__360 num__6 ounces servings of orange juice ? <o> a ) a ) num__45 <o> b ) b ) num__34 <o> c ) c ) num__50 <o> d ) d ) num__67 <o> e ) e ) num__100 |
its a . total juice rquired = num__360 * num__6 = num__2160 ounce num__12 ounce concentate makes = num__12 * num__4 = num__48 ounce juice total cans required = num__45.0 = num__45 . answer a <eor> a <eos> |
a |
multiply__360.0__6.0__ add__1.0__3.0__ multiply__12.0__4.0__ subtract__48.0__3.0__ multiply__1.0__45.0__ |
multiply__360.0__6.0__ add__1.0__3.0__ multiply__12.0__4.0__ subtract__48.0__3.0__ multiply__1.0__45.0__ |
| jack and jill are marathon runners . jack can finish a marathon ( num__42 km ) in num__7 hours and jill can run a marathon in num__4.2 hours . what is the ratio of their average running speed ? ( jack : jill ) <o> a ) num__0.933333333333 <o> b ) num__1.07142857143 <o> c ) num__1.0 <o> d ) num__1.25 <o> e ) can not be determined |
average speed of jack = distance / time = num__6.0 = num__6 average speed of jill = num__42 / ( num__4.2 ) = num__10 ratio of average speed of jack to jill = num__0.6 = num__0.6 answer c <eor> c <eos> |
c |
divide__42.0__7.0__ divide__42.0__4.2__ divide__4.2__7.0__ subtract__7.0__6.0__ |
divide__42.0__7.0__ divide__42.0__4.2__ divide__4.2__7.0__ subtract__7.0__6.0__ |
| if the sides of a square are multiplied by sqrt ( num__5 ) the area of the original square is how many times as large as the area of the resultant square ? <o> a ) num__5.0 <o> b ) num__20.0 <o> c ) num__50.0 <o> d ) num__100.0 <o> e ) num__120 % |
let x be the original length of one side . then the original area is x ^ num__2 . the new square has sides of length sqrt ( num__5 ) * x so the area is num__5 x ^ num__2 . the area of the original square is num__0.2 = num__20.0 times the area of the new square . the answer is b . <eor> b <eos> |
b |
square_perimeter__5.0__ square_perimeter__5.0__ |
square_perimeter__5.0__ volume_rectangular_prism__5.0__20.0__0.2__ |
| a sum of rs . num__12500 amounts to rs . num__16000 in num__4 years at the rate of simple interest . what is the rate of interest ? <o> a ) num__7.0 <o> b ) num__5.0 <o> c ) num__6.0 <o> d ) num__8.0 <o> e ) num__14 % |
s . i . = ( num__16000 - num__12500 ) = rs . num__3500 / - rate = ( num__100 * num__3500 ) / ( num__12500 * num__4 ) = num__7.0 answer : a <eor> a <eos> |
a |
percent__100.0__7.0__ |
percent__100.0__7.0__ |
| in a company the salary of num__40 percentage of num__400 employees is equal to num__20 percentage of x employees then what is x ? <o> a ) num__200 <o> b ) num__350 <o> c ) num__700 <o> d ) num__800 <o> e ) num__1 |
400 |
num__0.4 ( num__400 ) = num__0.2 ( x ) x = num__800 . . option d . <eor> d <eos> |
d |
d |
| a wire in the form of a circle of radius num__3.5 m is bent in the form of a rectangule whose length and breadth are in the ratio of num__6 : num__5 . what is the area of the rectangle ? <o> a ) num__60 cm num__2 <o> b ) num__30 centimeter sqaure <o> c ) num__45 cm num__2 <o> d ) num__15 cm num__2 <o> e ) none of these . |
the circumference of the circle is equal to the permeter of the rectangle . let l = num__6 x and b = num__5 x num__2 ( num__6 x + num__5 x ) = num__2 * num__3.14285714286 * num__3.5 = > x = num__1 therefore l = num__6 cm and b = num__5 cm area of the rectangle = num__6 * num__5 = num__30 cm num__2 num__14 . the area of a square is num__40 answer : b <eor> b <eos> |
b |
multiply__6.0__5.0__ square_perimeter__3.5__ rectangle_perimeter__6.0__14.0__ multiply__6.0__5.0__ |
multiply__6.0__5.0__ square_perimeter__3.5__ rectangle_perimeter__6.0__14.0__ multiply__6.0__5.0__ |
| in one hour a boat goes num__7 km / hr along the stream and num__5 km / hr against the stream . the speed of the boat in still water ( in km / hr ) is : <o> a ) num__3 km / hr <o> b ) num__4 km / hr <o> c ) num__5 km / hr <o> d ) num__6 km / hr <o> e ) num__8 km / hr |
upstream relative speed is u + v = num__7 km / hr downstream speed is u - v = num__5 where u = speed of boat in still water and v is speed of stream then adding two equations u + v + u - v = num__7 + num__5 num__2 u = num__12 finally u = num__6 . answer : d <eor> d <eos> |
d |
subtract__7.0__5.0__ add__7.0__5.0__ divide__12.0__2.0__ round__6.0__ |
subtract__7.0__5.0__ add__7.0__5.0__ divide__12.0__2.0__ divide__12.0__2.0__ |
| each of the following equations has at least one solution except <o> a ) – num__2 ^ n = ( – num__2 ) ^ - n <o> b ) num__2 ^ - n = ( – num__2 ) ^ n <o> c ) num__2 ^ n = ( – num__2 ) ^ - n <o> d ) ( – num__2 ) ^ n = – num__2 ^ n <o> e ) ( – num__2 ) ^ - n = – num__2 ^ - n |
a ) – num__2 ^ n = ( – num__2 ) ^ - n – num__2 ^ n = num__1 / ( – num__2 ) ^ n – num__2 ^ n * ( – num__2 ) ^ n = num__1 keep it . let ' s solve the other options . . ! ! b ) num__2 ^ - n = ( – num__2 ) ^ n num__0.5 ^ n = ( – num__2 ) ^ n num__1 = ( – num__2 ) ^ n * ( num__2 ^ n ) for n = num__0 l . h . s = r . h . s c ) num__2 ^ n = ( – num__2 ) ^ - n num__2 ^ n = num__1 / ( – num__2 ) ^ n ( num__2 ^ n ) * ( – num__2 ) ^ n = num__1 for n = num__0 l . h . s = r . h . s d ) ( – num__2 ) ^ n = – num__2 ^ n ( – num__2 ) ^ n / – num__2 ^ n = num__1 for n = num__1 l . h . s = r . h . s e ) ( – num__2 ) ^ - n = – num__2 ^ - n num__1 / ( – num__2 ) ^ n = num__1 / – num__2 ^ n for n = num__1 l . h . s = r . h . s answer is a <eor> a <eos> |
a |
reverse__2.0__ round_down__0.5__ reverse__0.5__ |
reverse__2.0__ round_down__0.5__ reverse__0.5__ |
| a train overtakes two person who are walking in the same direction in which the train is going at the rate of num__2 kmph and num__4 kmph and passes them completely in num__9 and num__10 seconds respectively . the length of the train is : <o> a ) num__45 m <o> b ) num__50 m <o> c ) num__54 m <o> d ) num__72 m <o> e ) none of these |
let actual speed of train = s m / sec and length of train = l m . then s - num__2 × num__0.277777777778 = l num__9 ⇒ num__9 s = l + num__5 . . . … ( i ) and s - num__4 × num__5 ⁄ num__18 = l ⁄ num__10 ⇒ num__90 s = num__9 l + num__100 . . . . . ( ii ) by ( i ) & ( ii ) we get l = num__50 m . answer b <eor> b <eos> |
b |
subtract__9.0__4.0__ multiply__2.0__9.0__ multiply__9.0__10.0__ add__10.0__90.0__ multiply__10.0__5.0__ round__50.0__ |
subtract__9.0__4.0__ multiply__2.0__9.0__ multiply__9.0__10.0__ add__10.0__90.0__ divide__100.0__2.0__ divide__100.0__2.0__ |
| a patient was given a bottle of tablets by the doctor and he was asked to take five tablets in a gap of num__5 minutes . in how much time will he be able to take all the five tablets ? <o> a ) num__1 hour . <o> b ) num__2 hour . <o> c ) num__30 min <o> d ) num__20 min <o> e ) can not be determined |
suppose he takes the first tablet at num__8 : num__00 pm . then the second will be consumed by him at num__8 : num__05 third at num__8 : num__10 fourth at num__8 : num__15 and fifth at num__8 : num__20 . time = num__20 min answer d <eor> d <eos> |
d |
add__5.0__10.0__ add__5.0__15.0__ round__20.0__ |
add__5.0__10.0__ add__5.0__15.0__ round__20.0__ |
| rohit invests a sum of money at compound interest which doubles itself in num__3 years find in how many years will it amount to four times itself ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__2 |
explanation : case ( i ) when time ( t ) = num__3 years principle p becomes num__2 p . therefore amount = num__2 p a = p ( num__1 + r / num__100 ) t num__2 p = p ( num__1 + r / num__100 ) num__3 ( num__1 + r / num__100 ) num__3 = num__2 ( num__1 + r / num__100 ) = num__3 √ num__2 - - - - - - - - ( a ) ( iii ) find t for which p becomes num__4 p when time = x and amount = num__4 p num__4 p = p ( num__1 + r / num__100 ) x ( num__1 + r / num__100 ) x = num__4 ( num__1 + r / num__100 ) = num__41 / x - - - - - - - - - ( b ) equating ( a ) and ( b ) we get num__41 / x = num__3 √ num__2 = num__7.0 num__22 / x = num__7.0 num__2 / x = num__0.333333333333 x = num__6 years answer : a <eor> a <eos> |
a |
percent__100.0__6.0__ |
percent__100.0__6.0__ |
| if num__5 men and num__2 boys working together can do four times as much work per hour as a man and a boy together . find the ratio of the work done by a man and that of a boy for a given time ? <o> a ) num__2 : num__8 <o> b ) num__2 : num__6 <o> c ) num__2 : num__2 <o> d ) num__2 : num__1 <o> e ) num__2 : num__3 |
num__5 m + num__2 b = num__4 ( num__1 m + num__1 b ) num__5 m + num__2 b = num__4 m + num__4 b num__1 m = num__2 b the required ratio of work done by a man and a boy = num__2 : num__1 answer : d <eor> d <eos> |
d |
subtract__5.0__4.0__ round__2.0__ |
subtract__5.0__4.0__ round__2.0__ |
| a wheel has a diameter of x inches and a second wheel has a diameter of y inches . the first wheel covers a distance of d feet in num__120 revolutions . how many revolutions does the second wheel make in covering d feet ? <o> a ) num__120 y / x <o> b ) num__120 - y <o> c ) num__120 - x <o> d ) num__120 x / y <o> e ) num__120 x - y |
a wheel covers num__2 π r distance in one revolution . where r = diameter / num__2 first wheel covers d feet in num__120 rev . = > d = num__120 * num__2 π * ( x / num__2 ) second wheel covers d feet in let ' s say p revolutions ; = > d = p * num__2 π * ( y / num__2 ) comparing both equations : - = > p = ( num__120 ∗ num__2 π ∗ x / num__2 ) / ( num__2 π ∗ y / num__2 ) = > num__120 x / y answer : - d <eor> d <eos> |
d |
round__120.0__ |
round__120.0__ |
| a train of length l is traveling at a constant velocity and passes a pole in t seconds . if the same train travelling at the same velocity passes a platform in num__5 t seconds then what is the length of the platform ? <o> a ) l <o> b ) num__2 l <o> c ) num__3 l <o> d ) num__4 l <o> e ) num__5 l |
the train passes a pole in t seconds so velocity v = l / t ( l + p ) / v = num__5 t ( l + p ) / ( l / t ) = num__5 t p = num__4 l the answer is d . <eor> d <eos> |
d |
round__4.0__ |
round__4.0__ |
| if the average ( arithmetic mean ) of a and b is num__210 and the average of b and c is num__160 what is the value of a − c ? <o> a ) − num__220 <o> b ) − num__100 <o> c ) num__100 <o> d ) num__135 <o> e ) it can not be determined from the information given |
question : a - c = ? ( a + b ) / num__2 = num__210 = = = > a + b = num__420 ( b + c ) / num__2 = num__160 = = = > b + c = num__320 ( a + b ) - ( b + c ) = num__420 - num__320 = = = > a + b - b - c = num__100 = = = > a - c = num__100 answer : c <eor> c <eos> |
c |
multiply__210.0__2.0__ multiply__160.0__2.0__ subtract__420.0__320.0__ subtract__420.0__320.0__ |
multiply__210.0__2.0__ multiply__160.0__2.0__ subtract__420.0__320.0__ subtract__420.0__320.0__ |
| in a division divident is num__725 divisior is num__36 and quotient is num__20 . find the remainder . <o> a ) a ) num__4 <o> b ) b ) num__3 <o> c ) c ) num__2 <o> d ) d ) num__5 <o> e ) e ) num__6 |
explanation : num__725 = num__36 x num__20 + r num__725 = num__720 + r r = num__725 - num__720 = num__5 answer : option d <eor> d <eos> |
d |
multiply__36.0__20.0__ subtract__725.0__720.0__ subtract__725.0__720.0__ |
multiply__36.0__20.0__ subtract__725.0__720.0__ subtract__725.0__720.0__ |
| a sum fetched a total simple interest of num__4020.75 at the rate of num__9.0 . p . a . in num__5 years . what is the sum ? <o> a ) num__5768 <o> b ) num__8925 <o> c ) num__2345 <o> d ) num__6474 <o> e ) num__8935 |
principal = ( num__100 x num__4020.75 ) / ( num__9 x num__5 ) = num__8935.0 = num__8935 . answer e <eor> e <eos> |
e |
percent__100.0__8935.0__ |
percent__100.0__8935.0__ |
| find the value of log y ( x num__4 ) if logx ( y num__3 ) = num__2 <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
logx ( y num__3 ) = num__2 : given x num__2 = y num__3 : rewrite in exponential form x num__4 = y num__6 : square both sides x num__4 = y num__6 : rewrite the above using the log base y logy ( x num__4 ) = logy ( y num__6 ) = num__6 correct answer c <eor> c <eos> |
c |
add__4.0__2.0__ add__4.0__2.0__ |
add__4.0__2.0__ add__4.0__2.0__ |
| num__4.2 x num__4.2 - num__1.9 x num__1.9 / num__2.3 x num__6.1 is equal to : <o> a ) num__2 <o> b ) num__1 <o> c ) num__3 <o> d ) num__4 <o> e ) num__8 |
given expression = ( a num__2 - b num__2 ) / ( a + b ) ( a - b ) = ( a num__2 - b num__2 ) / ( a num__2 - b num__2 ) = num__1 . answer is b . <eor> b <eos> |
b |
round_down__2.3__ round_down__1.9__ round_down__1.9__ |
round_down__2.3__ round_down__1.9__ reverse__1.0__ |
| in a num__1000 m race a beats b by num__200 meters or num__25 seconds . find the speed of b ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__7 <o> d ) num__5 <o> e ) num__2 |
since a beats b by num__200 m or num__25 seconds i t implies that b covers num__200 m in num__25 seconds . hence speed of b = num__8.0 = num__8 m / s . answer : a <eor> a <eos> |
a |
divide__200.0__25.0__ round__8.0__ |
divide__200.0__25.0__ divide__200.0__25.0__ |
| the kiljaro highway is marked with milestones denoting the distance to the town of kiljaro . tommy left kiljaro and drove the highway passing the num__20 km milestone at num__8 : num__30 . some time afterwards tommy got a phone call asking him to return home and he made a u - turn at the num__160 km milestone . at num__09 : a num__0 tommy passed the milestone marking num__70 km to kiljaro . the variable a represents the tens digit of the minutes in num__09 : a num__0 . assuming tommy maintained the same constant speed during the entire drive how many kilometers did tommy travel in one minute ? <o> a ) num__230 / ( num__30 + num__10 a ) <o> b ) num__230 / ( num__30 + num__60 a ) <o> c ) num__6.05263157895 a <o> d ) num__220 / ( num__30 + num__10 a ) <o> e ) num__5.5 a |
since we are dealing with the variables in the answer choices the best possible method according to me would besubstitution . substitute a with num__3 . meaning tommy would have travelled a distance of ( ( num__160 - num__20 ) + ( num__160 - num__70 ) ) in num__60 minutes . num__230 kms in num__60 minutes = = > num__3.8 km / hr . substitute a with num__3 in the answer options . option a <eor> a <eos> |
a |
hour_to_min_conversion__ add__160.0__70.0__ round__230.0__ |
hour_to_min_conversion__ add__160.0__70.0__ add__160.0__70.0__ |
| a certain sum amounts to rs . num__1600 in num__4 years and rs . num__1900 in num__7 years . find the rate % per annum ? <o> a ) num__9.33 <o> b ) num__7.33 <o> c ) num__5.33 <o> d ) num__8.33 <o> e ) num__6.33 % |
num__4 - - - num__1600 num__7 - - - num__1900 - - - - - - - - - - - - - - num__3 - - - num__300 n = num__1 i = num__100 r = ? p = num__1600 - num__400 = num__1200 num__100 = ( num__1200 * num__1 * r ) / num__100 r = num__8.33 answer : d <eor> d <eos> |
d |
percent__8.33__100.0__ |
percent__8.33__100.0__ |
| there are num__600 students in a school . the ratio of boys and girls in this school is num__3 : num__5 . find the total of girls & boys are there in this school ? <o> a ) num__265 <o> b ) num__276 <o> c ) num__375 <o> d ) num__387 <o> e ) num__390 |
in order to obtain a ratio of boys to girls equal to num__3 : num__5 the number of boys has to be written as num__3 x and the number of girls as num__5 x where x is a common factor to the number of girls and the number of boys . the total number of boys and girls is num__600 . hence num__3 x + num__5 x = num__600 solve for x num__8 x = num__600 x = num__75 number of boys num__3 x = num__3 × num__75 = num__225 number of girls num__5 x = num__5 × num__75 = num__375 c <eor> c <eos> |
c |
add__3.0__5.0__ divide__600.0__8.0__ multiply__3.0__75.0__ subtract__600.0__225.0__ subtract__600.0__225.0__ |
add__3.0__5.0__ divide__600.0__8.0__ multiply__3.0__75.0__ subtract__600.0__225.0__ subtract__600.0__225.0__ |
| if b is greater than num__1 which of the following must be negative ? <o> a ) ( num__2 - b ) ( b - num__1 ) <o> b ) ( b - num__1 ) / num__3 b <o> c ) ( num__1 - b ) ^ num__2 <o> d ) ( num__2 - b ) / ( num__1 - b ) <o> e ) ( num__1 - b ^ num__2 ) / b |
to be - ive we should remember it should not be zero apart from being positive . . . c will always be num__0 or positive . . a and d will be zero at b = num__2 . . b will always be positive . . left is e where the numerator will always be negative and denominator a positive . . ans e <eor> e <eos> |
e |
reverse__1.0__ |
subtract__2.0__1.0__ |
| for a f d are the positive integers and d | a means that “ a is divisible by d ” if d | af which of the following must be true ? <o> a ) d | a <o> b ) d | f <o> c ) d | num__2 af <o> d ) d | ( a + f ) <o> e ) d | ( a - f ) |
d / a means a is divisible by d d / af means af divisible by d . all are integers so if af is divisible by d num__1 a can be divisible by d or num__2 f can be divisble by d . so the question stem asks must true . so option a and b are could but not must . option c is num__2 af divisible by d . if ab is divisible by d then num__2 af is divisible by d . option d and e we ca n ' t predict . so option c is correct . <eor> c <eos> |
c |
multiply__1.0__2.0__ |
divide__2.0__1.0__ |
| find the value of num__72519 x num__9999 = m ? <o> a ) num__557842343 <o> b ) num__476686756 <o> c ) num__576763467 <o> d ) num__725117481 <o> e ) num__856443356 |
num__72519 x num__9999 = num__72519 x ( num__10000 - num__1 ) = num__72519 x num__10000 - num__72519 x num__1 = num__725190000 - num__72519 = num__725117481 d <eor> d <eos> |
d |
subtract__10000.0__9999.0__ multiply__72519.0__10000.0__ multiply__72519.0__9999.0__ multiply__72519.0__9999.0__ |
subtract__10000.0__9999.0__ multiply__72519.0__10000.0__ subtract__725190000.0__72519.0__ subtract__725190000.0__72519.0__ |
| each of the num__59 members in mount school class is required to sign up for a minimum of one and a maximum of three academic clubs . the three clubs to choose from are the poetry club the history club and the writing club . a total of num__22 students sign up for the poetry club num__27 students for the history club and num__28 students for the writing club . if num__6 students sign up for exactly two clubs how many students sign up for all three clubs ? <o> a ) num__2 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
each of the num__59 members in mount school class is required to sign up for a minimum of one and a maximum of three academic clubs . total = g num__1 + g num__2 + g num__3 - ( # in exactly num__2 ) - num__2 * ( # in num__3 sets ) num__59 = num__22 + num__27 + num__28 - ( num__6 ) - num__2 x so # in num__3 sets = num__6 = c <eor> c <eos> |
c |
subtract__28.0__27.0__ divide__6.0__2.0__ subtract__28.0__22.0__ |
subtract__28.0__27.0__ add__1.0__2.0__ subtract__28.0__22.0__ |
| there are num__10 stations on a railway line . the number of different journey tickets that are required by the authorities is : <o> a ) num__90 <o> b ) num__92 <o> c ) num__91 <o> d ) none of these <o> e ) can not be determined |
explanation : from a certain station there will be a ticket for each of the other num__9 stations . since there are num__10 stations on the railway line the number of different journey tickets will be num__10 × num__9 i . e num__90 . answer : a <eor> a <eos> |
a |
multiply__10.0__9.0__ multiply__10.0__9.0__ |
multiply__10.0__9.0__ multiply__10.0__9.0__ |
| laura can paint num__1 / x of a certain room in num__20 minutes . what fraction c of the same room can joseph paint in num__20 minutes if the two of them can paint the room in an hour working together at their respective rates ? <o> a ) num__1 / ( num__3 x ) <o> b ) num__3 x / ( x - num__3 ) <o> c ) ( x – num__3 ) / ( num__3 x ) <o> d ) x / ( x - num__3 ) <o> e ) ( x - num__3 ) / x |
options with variables are often done by plugging in numbers . both working together can paint the room in num__1 hr so if their individual rates were equal each would take num__2 hours alone . num__2 hours is num__120 mins so in num__20 mins each would complete c = num__6.0 = num__0.166666666667 th of the room alone . so if x = num__6 ( laura completes num__0.166666666667 th of the room in num__20 mins ) the correct option will give num__0.166666666667 . ( joseph will also paint num__0.166666666667 th of the room if their rates are same ) if you put x = num__6 in the options only option ( c ) will give num__0.166666666667 answer ( c ) <eor> c <eos> |
c |
divide__120.0__20.0__ divide__1.0__6.0__ add__1.0__2.0__ |
divide__120.0__20.0__ divide__1.0__6.0__ add__1.0__2.0__ |
| from the given equation find the value of x : x ² − num__3 x + num__2 <o> a ) num__3 <o> b ) num__2 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
( x − num__1 ) ( x − num__2 ) x = num__1 or num__2 . c <eor> c <eos> |
c |
subtract__3.0__2.0__ add__3.0__2.0__ |
subtract__3.0__2.0__ add__3.0__2.0__ |
| it takes nine minutes to load a certain video on a cellphone and fifteen seconds to load that same video on a laptop . if the two devices were connected so that they operated in concert at their respective rates how many seconds would it take them to load the video rounded to the nearest hundredth ? <o> a ) num__13.58 <o> b ) num__13.87 <o> c ) num__14.24 <o> d ) num__14.59 <o> e ) num__14.85 |
the laptop can load the video at a rate of num__0.0666666666667 of the video per second . the phone can load the video at a rate of num__1 / ( num__60 * num__9 ) = num__0.00185185185185 of the video per second . the combined rate is num__0.0666666666667 + num__0.00185185185185 = num__0.0685185185185 of the video per second . the time required to load the video is num__14.5945945946 = num__14.59 seconds . the answer is d . <eor> d <eos> |
d |
hour_to_min_conversion__ round__14.59__ |
hour_to_min_conversion__ divide__14.59__1.0__ |
| a man cycling along the road noticed that every num__15 minutes a bus overtakes him and every num__5 minutes he meets an oncoming bus . if all buses and the cyclist move at a constant speed what is the time interval between consecutive buses ? <o> a ) num__5 minutes <o> b ) num__6 minutes <o> c ) num__8 minutes <o> d ) num__9 minutes <o> e ) num__7.5 minutes |
let ' s say the distance between the buses is d . we want to determine interval = \ frac { d } { b } where b is the speed of bus . let the speed of cyclist be c . every num__15 minutes a bus overtakes cyclist : \ frac { d } { b - c } = num__15 d = num__15 b - num__15 c ; every num__5 minutes cyclist meets an oncoming bus : \ frac { d } { b + c } = num__4 d = num__4 b + num__4 c ; d = num__15 b - num__15 c = num__5 b + num__5 c - - > b = num__2 c - - > d = num__15 b - num__15 b / num__2 = num__15 b / num__2 . interval = \ frac { d } { b } = \ frac { num__7.5 b } { b } = num__7.5 answer : e ( num__7.5 minutes ) . <eor> e <eos> |
e |
divide__15.0__2.0__ round__7.5__ |
divide__15.0__2.0__ subtract__15.0__7.5__ |
| a fort had provision of food for num__150 men for num__50 days . after num__10 days num__25 men left the fort . the number of days for which the remaining food will last is : <o> a ) num__29 num__0.2 <o> b ) num__37 num__0.25 <o> c ) num__42 <o> d ) num__54 <o> e ) num__48 |
we have food for num__150 men for num__50 days . after num__10 days food left for num__150 men for num__40 days . so num__150 : num__40 now we have num__125 men and x days num__125 : num__150 : : num__40 : x x = ( num__150 * num__40 ) / num__125 = num__48 days . answer : e <eor> e <eos> |
e |
subtract__50.0__10.0__ subtract__150.0__25.0__ round__48.0__ |
subtract__50.0__10.0__ subtract__150.0__25.0__ round__48.0__ |
| a contractor undertakes to do a piece of work in num__40 days . he engages num__100 men at the begining and num__100 more after num__35 days and completes the work in stipulated time . if he had not engaged the additional men how many days behind schedule would it be finished ? <o> a ) num__2 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
explanation : \ inline \ fn _ jvn [ ( num__100 \ times num__35 ) + ( num__200 \ times num__5 ) ] men can finish the work in num__1 day \ inline \ fn _ jvn \ therefore num__4500 men can finish the work in num__1 day . num__100 men can finish it in \ inline \ fn _ jvn \ frac { num__4500 } { num__100 } = num__45 days . this is num__5 days behind schedule answer : b ) num__5 <eor> b <eos> |
b |
subtract__40.0__35.0__ add__40.0__5.0__ subtract__40.0__35.0__ |
subtract__40.0__35.0__ add__40.0__5.0__ subtract__40.0__35.0__ |
| there are n arithmetic means between num__3 and num__45 such that the sum of these arithmetic means is num__552 . find the value of n . <o> a ) num__11 <o> b ) num__15 <o> c ) num__17 <o> d ) num__21 <o> e ) num__23 |
arithmetic mean = num__552 / n = num__0.5 ( num__3 + num__45 ) = num__24 n = num__23.0 = num__23 answer : e <eor> e <eos> |
e |
divide__552.0__24.0__ divide__552.0__24.0__ |
divide__552.0__24.0__ divide__552.0__24.0__ |
| find the simple interest on $ num__6000 at num__15.0 per annum for num__9 months ? <o> a ) $ num__492 <o> b ) $ num__512 <o> c ) $ num__675 <o> d ) $ num__745 <o> e ) $ num__1020 |
p = $ num__6000 r = num__15.0 t = num__0.75 years = num__0.75 years s . i . = p * r * t / num__100 = num__6000 * num__15 * num__0.0075 = $ num__675 answer is c <eor> c <eos> |
c |
percent__100.0__675.0__ |
percent__100.0__675.0__ |
| the average ( arithmetic mean ) of four distinct positive integers is num__10 . if the average of the smaller two of these four integers is num__12 which of the following represents the maximum possible value of the largest integer ? <o> a ) num__4 <o> b ) num__16 <o> c ) num__12 <o> d ) num__8 <o> e ) num__1 |
let the distinct number be a b c and d its given a > b > c > d also a + b + c + d = num__40 and a + b = num__12 means c + d = num__28 since the question ask for the largest possible number we should choose the least value for a and b c . so d should be num__16 . if d = num__16 then c = num__12 > a b < num__12 answer : b <eor> b <eos> |
b |
subtract__40.0__12.0__ subtract__28.0__12.0__ subtract__28.0__12.0__ |
subtract__40.0__12.0__ subtract__28.0__12.0__ subtract__28.0__12.0__ |
| a is two years older than b who is twice as old as c . if the total of the ages of a b and c be num__22 the how old is b ? <o> a ) num__7 <o> b ) num__9 <o> c ) num__8 <o> d ) num__11 <o> e ) num__10 |
explanation : let c ' s age be x years . then b ' s age = num__2 x years . a ' s age = ( num__2 x + num__2 ) years . ( num__2 x + num__2 ) + num__2 x + x = num__22 ⇒ num__5 x = num__20 ⇒ x = num__4 . hence b ' s age = num__2 x = num__8 years . answer : c <eor> c <eos> |
c |
subtract__22.0__2.0__ divide__20.0__5.0__ multiply__2.0__4.0__ multiply__2.0__4.0__ |
subtract__22.0__2.0__ divide__20.0__5.0__ multiply__2.0__4.0__ multiply__2.0__4.0__ |
| village x has a population of num__72000 which is decreasing at the rate of num__1200 per year . village y has a population of num__42000 which is increasing at the rate of num__800 per year . in how many years will the population of the two villages be equal ? <o> a ) num__15 <o> b ) num__19 <o> c ) num__11 <o> d ) num__18 <o> e ) num__13 |
let the population of two villages be equal after p years then num__72000 - num__1200 p = num__42000 + num__800 p num__2000 p = num__30000 p = num__15 answer is a . <eor> a <eos> |
a |
add__1200.0__800.0__ subtract__72000.0__42000.0__ divide__30000.0__2000.0__ divide__30000.0__2000.0__ |
add__1200.0__800.0__ subtract__72000.0__42000.0__ divide__30000.0__2000.0__ divide__30000.0__2000.0__ |
| there are some num__2 wheelers and num__4 wheelers parked total number of wheels present is num__240 then how many num__4 wheelers were there <o> a ) num__20 <o> b ) num__30 <o> c ) num__40 <o> d ) num__50 <o> e ) num__60 |
num__2 x + num__4 x = num__240 ; x = num__40 ; then two whl = num__2 * num__40 = num__80 num__4 whl = num__4 * num__40 = num__160 ; ans : num__40.0 = i - e num__40 answer : c <eor> c <eos> |
c |
multiply__2.0__40.0__ multiply__2.0__80.0__ divide__160.0__4.0__ |
multiply__2.0__40.0__ multiply__2.0__80.0__ subtract__80.0__40.0__ |
| express num__50 mps in kmph ? <o> a ) num__172 <o> b ) num__160 <o> c ) num__150 <o> d ) num__180 <o> e ) num__120 |
num__25 * num__3.6 = num__180 kmph answer : d <eor> d <eos> |
d |
multiply__50.0__3.6__ round__180.0__ |
multiply__50.0__3.6__ multiply__50.0__3.6__ |
| ( num__112 x num__54 ) = ? <o> a ) num__60000 <o> b ) num__70000 <o> c ) num__80000 <o> d ) num__88000 <o> e ) none |
( num__112 x num__54 ) = num__112 x num__10 num__4 = num__112 x num__104 = num__1120000 = num__70000 num__2 num__24 num__16 b ) <eor> b <eos> |
b |
divide__1120000.0__70000.0__ divide__1120000.0__16.0__ |
divide__1120000.0__70000.0__ divide__1120000.0__16.0__ |
| num__3889 + num__14.952 - ? = num__3854.002 <o> a ) num__47.95 <o> b ) num__49.95 <o> c ) num__45.97 <o> d ) num__47.59 <o> e ) num__45.79 |
let num__3889 + num__14.952 - x = num__3854.002 . then x = ( num__3889 + num__14.952 ) - num__3854.002 = num__3903.952 - num__3854.002 = num__49.95 . answer is b <eor> b <eos> |
b |
add__3889.0__14.952__ subtract__3903.952__3854.002__ subtract__3903.952__3854.002__ |
add__3889.0__14.952__ subtract__3903.952__3854.002__ subtract__3903.952__3854.002__ |
| in may the grounds keeper at spring lake golf club built a circular green with an area of num__36 π square feet . in august he doubled the distance from the center of the green to the edge of the green . what is the total area of the renovated green ? <o> a ) num__100 pi <o> b ) num__144 pi <o> c ) num__70 <o> d ) num__80 <o> e ) num__90 |
area of circle num__36 pi sq feet = pi r ^ num__2 therefore r = num__6 now green radi doubled i . e r = num__12 area = num__144 pi b <eor> b <eos> |
b |
multiply__2.0__6.0__ square_perimeter__36.0__ square_perimeter__36.0__ |
multiply__2.0__6.0__ power__12.0__2.0__ power__12.0__2.0__ |
| in a num__800 m race around a stadium having the circumference of num__200 m the top runner meets the last runner on the num__5 th minute of the race . if the top runner runs at twice the speed of the last runner what is the time taken by the top runner to finish the race ? <o> a ) num__20 min <o> b ) num__15 min <o> c ) num__10 min <o> d ) num__5 min <o> e ) none of these |
after num__5 minutes ( before meeting ) the top runner covers num__2 rounds i . e . num__400 m and the last runner covers num__1 round i . e . num__200 m . ∴ top runner covers num__800 m race in num__10 minutes . answer c <eor> c <eos> |
c |
divide__800.0__2.0__ multiply__5.0__2.0__ round__10.0__ |
divide__800.0__2.0__ multiply__5.0__2.0__ round__10.0__ |
| if x = num__0.25 which expression has the greatest value ? <o> a ) x ^ ( num__0.333333333333 ) <o> b ) x ^ ( num__0.5 ) <o> c ) num__1 / ( num__2 x ) <o> d ) x / . num__03 <o> e ) x ^ ( - num__3 ) |
options a and b are both less than num__1 . option c : num__1 / ( num__2 ( num__0.25 ) ) = num__2.0 = num__2 option d : ( num__0.25 ) / num__0.03 = num__1 / num__0.12 = num__8.333 . . . option e : ( num__0.25 ) ^ ( - num__3 ) = num__4 ^ num__3 = num__64 the answer is e . <eor> e <eos> |
e |
divide__0.03__0.25__ add__1.0__2.0__ reverse__0.25__ add__1.0__2.0__ |
divide__0.03__0.25__ add__1.0__2.0__ reverse__0.25__ divide__3.0__1.0__ |
| if ( a – b ) is num__13 more than ( c + d ) and ( a + b ) is num__3 less than ( c – d ) then ( a – c ) is : <o> a ) num__6 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
( a – b ) – ( c + d ) = num__13 and ( c – d ) – ( a + b ) = num__3 = > ( a – c ) – ( b + d ) = num__13 and ( c – a ) – ( b + d ) = num__3 = > ( b + d ) = ( a – c ) – num__13 and ( b + d ) = ( c – a ) – num__3 = > ( a – c ) – num__13 = ( c – a ) – num__3 = > num__2 ( a – c ) = num__10 = > ( a – c ) = num__5 answer : e <eor> e <eos> |
e |
subtract__13.0__3.0__ add__3.0__2.0__ add__3.0__2.0__ |
subtract__13.0__3.0__ add__3.0__2.0__ add__3.0__2.0__ |
| if a class of num__10 students has four men how many ways can the men and women be arranged in a circle so that no two men sit next to each other ? <o> a ) num__5 ! num__4 ! <o> b ) num__5 ! num__5 ! <o> c ) num__4 ! num__3 ! <o> d ) num__10 ! <o> e ) num__10 ! / num__5 ! |
for no two men to sit together either all are in even or odd position fix one at any one position then rest three can be fixed in num__3 ! ways . . . also rest four position of women can be fixed in num__4 ! . . total ways num__4 ! num__3 ! c <eor> c <eos> |
c |
choose__4.0__3.0__ |
choose__4.0__3.0__ |
| a cask initially contains pure alcohol up to the brim . the cask can be emptied by removing exactly num__5 liters at a time . each time this is done the cask must be filled back to the brim with water . the capacity of the cask is num__15 liters . when the cask is completely emptied and filled back to the brim two times what is the ratio of alcohol to water in the cask <o> a ) num__0.8 <o> b ) num__0.6 <o> c ) num__0.666666666667 <o> d ) num__0.5 <o> e ) num__0.333333333333 |
you are given that the initial quantity of the cask is num__15 l . so for the changing of the liquids the first time you remove num__5 l of pure alcohol - - - > num__10 l of alcohol remaining and you add num__5 l of water . ratio of alcohol in the remaining mixture = num__10 / ( num__10 + num__5 ) = num__0.666666666667 = num__0.666666666667 similarly water = num__0.333333333333 . the second time you remove num__5 l of the mixture having alcohol = num__5 * num__0.666666666667 = num__3.33333333333 - - > remaining alcohol = num__10 - num__3.33333333333 = num__6.66666666667 similarly the amount of water removed = num__5 * num__0.333333333333 = num__1.66666666667 - - - > remaining water = num__5 - num__1.66666666667 = num__3.33333333333 but you also add num__5 l more of water - - > total water now = num__5 + num__3.33333333333 = num__8.33333333333 finally ratio asked = num__6.66666666667 / num__8.33333333333 = num__0.8 = num__0.8 a is the correct answer . <eor> a <eos> |
a |
subtract__15.0__5.0__ divide__10.0__15.0__ divide__5.0__15.0__ subtract__10.0__3.3333__ subtract__5.0__3.3333__ add__5.0__3.3333__ divide__6.6667__8.3333__ divide__6.6667__8.3333__ |
subtract__15.0__5.0__ divide__10.0__15.0__ divide__5.0__15.0__ subtract__10.0__3.3333__ subtract__5.0__3.3333__ add__5.0__3.3333__ divide__6.6667__8.3333__ divide__6.6667__8.3333__ |
| train p crosses a pole in num__30 seconds and train q crosses the same pole in one minute and num__15 seconds . the length of train p is three - fourths the length of train q . what is the ratio of the speed of train p to that of train q ? <o> a ) num__15 : num__8 <o> b ) num__14 : num__8 <o> c ) num__17 : num__8 <o> d ) num__13 : num__8 <o> e ) num__12 : num__8 |
a num__15 : num__8 given that train p crosses a pole in num__30 seconds and train q crosses the same pole in one minute and num__15 seconds . let the length of train p be lp and that of train q be lq given that lp = num__0.75 lq as the train p and q crosses the pole in num__30 seconds and num__75 seconds respectively = > speed of train p = vp = lp / num__30 speed of train q = vq = lq / num__75 lp = num__0.75 lq = > vp = num__0.75 lq / ( num__30 ) = lq / num__40 ratio of their speeds = vp : vq = lq / num__40 : lq / num__75 = > num__0.025 : num__0.0133333333333 = num__15 : num__8 <eor> a <eos> |
a |
divide__30.0__0.75__ divide__0.75__30.0__ round__15.0__ |
divide__30.0__0.75__ divide__0.75__30.0__ round__15.0__ |
| a man can swim in still water at num__6 km / h but takes twice as long to swim upstream than downstream . the speed of the stream is ? <o> a ) num__1.7 <o> b ) num__1.9 <o> c ) num__1.1 <o> d ) num__2 <o> e ) num__1.2 |
m = num__6 s = x ds = num__6 + x us = num__6 - x num__6 + x = ( num__6 - x ) num__2 num__6 + x = num__12 - num__2 x num__3 x = num__12 - num__6 = num__6 x = num__2 answer : d <eor> d <eos> |
d |
multiply__6.0__2.0__ divide__6.0__2.0__ round__2.0__ |
multiply__6.0__2.0__ divide__6.0__2.0__ round__2.0__ |
| convert num__300 miles into km ? <o> a ) num__360 km <o> b ) num__480 km <o> c ) num__510 km <o> d ) num__320 km <o> e ) num__500 km |
num__1 mile = num__1.6 km approximately num__300 mile = num__300 * num__1.6 = num__480 km answer is b <eor> b <eos> |
b |
mile_to_km_conversion__ multiply__300.0__1.6__ round__480.0__ |
mile_to_km_conversion__ multiply__300.0__1.6__ multiply__300.0__1.6__ |
| what number continues the sequence ? num__25 num__50 num__27 num__46 num__31 num__38 num__39 ? <o> a ) num__21 <o> b ) num__22 <o> c ) num__23 <o> d ) num__24 <o> e ) num__25 |
b num__22 there are two alternate sequences the first increases by num__2 num__4 num__8 etc and the second decreases by num__4 num__8 num__16 etc . <eor> b <eos> |
b |
divide__50.0__25.0__ subtract__50.0__46.0__ subtract__46.0__38.0__ subtract__38.0__22.0__ subtract__38.0__16.0__ |
divide__50.0__25.0__ subtract__50.0__46.0__ subtract__46.0__38.0__ subtract__38.0__22.0__ subtract__38.0__16.0__ |
| if all of the telephone extensions in a certain company must be even numbers and if each of the extensions uses all four of the digits num__1 num__2 num__3 and num__4 what is the greatest number of four - digit extensions that the company can have ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__12 <o> d ) num__16 <o> e ) num__24 |
since the phone number must be even the unit ' s digit can be either num__2 or num__4 . when the unit ' s digit is num__2 - - > number of possibilities is num__3 ! = num__6 when the unit ' s digit is num__4 - - > number of possibilities is num__3 ! = num__6 largest number of extensions = num__6 + num__6 = num__12 answer : c <eor> c <eos> |
c |
multiply__2.0__3.0__ multiply__2.0__6.0__ multiply__1.0__12.0__ |
add__2.0__4.0__ multiply__2.0__6.0__ multiply__1.0__12.0__ |
| a couple spent $ num__198 in total while dining out and paid this amount using a credit card . the $ num__198 figure included a num__20 percent tip which was paid on top of the price which already included a sales tax of num__10 percent on top of the price of the food . what was the actual price of the food before tax and tip ? <o> a ) $ num__130 <o> b ) $ num__140 <o> c ) $ num__150 <o> d ) $ num__160 <o> e ) $ num__170 |
let the price of the meal be x . after a num__10.0 sales tax addition the price is num__1.1 * x after a num__20.0 tip on this amount the total is num__1.2 * num__1.1 * x = num__1.32 x num__1.32 x = num__198 x = $ num__150 the correct answer is c . <eor> c <eos> |
c |
multiply__1.1__1.2__ divide__198.0__1.32__ divide__198.0__1.32__ |
multiply__1.1__1.2__ divide__198.0__1.32__ divide__198.0__1.32__ |
| a man rows his boat num__85 km downstream and num__45 km upstream taking num__2 num__0.5 hours each time . find the speed of the stream ? <o> a ) num__5 kmph <o> b ) num__7 kmph <o> c ) num__9 kmph <o> d ) num__8 kmph <o> e ) num__1 kmph |
speed downstream = d / t = num__85 / ( num__2 num__0.5 ) = num__34 kmph speed upstream = d / t = num__45 / ( num__2 num__0.5 ) = num__18 kmph the speed of the stream = ( num__34 - num__18 ) / num__2 = num__8 kmph answer : d <eor> d <eos> |
d |
round__8.0__ |
round__8.0__ |
| four different children have jelly beans : aaron has num__5 bianca has num__7 callie has num__8 and dante has num__19 . how many jelly beans must dante give to aaron to ensure that no child has more than num__1 fewer jelly beans than any other child ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__7 |
since bianca and callie are both within num__1 jelly bean of each other and aaron has num__5 dante must provide num__3 of his num__19 jelly beans so each child has no more than num__1 fewer jelly bean than any other child . dante + aaron = num__19 + num__5 = num__12.0 = num__12 num__19 - num__2 = num__7 so dante must provide num__3 jelly beans to aaron . answer ( e ) <eor> e <eos> |
e |
subtract__8.0__5.0__ add__5.0__7.0__ subtract__5.0__3.0__ add__5.0__2.0__ |
subtract__8.0__5.0__ add__5.0__7.0__ subtract__5.0__3.0__ add__5.0__2.0__ |
| the hypotenuse of a right triangle is num__2 centimeters more than the longer side of the triangle . the shorter side of the triangle is num__7 centimeters less than the longer side . find the length of the hypotenuse . <o> a ) num__17 <o> b ) num__18 <o> c ) num__19 <o> d ) num__20 <o> e ) num__21 |
let a and b be the two sides of the triangle such that a is longer than b . the statement ` ` the hypotenuse of a right triangle is num__2 centimeters more than the longer side of the triangle ' ' may be fomulated by h = a + num__2 or a = h - num__2 the statement ` ` the shorter side of the triangle is num__7 centimeters less than the longer side ' ' may be formulated . b = a - num__7 or b = ( h - num__2 ) - num__7 = h - num__9 we now use pythagora ' s theorem to write a third equation h num__2 = a num__2 + b num__2 substitute a by h - num__2 and b by h - num__9 in the above equation to obtain an equation in one variable only . h num__2 = ( h - num__2 ) num__2 + ( h - num__9 ) num__2 simplify and rewrite the above equation in standard form . h num__2 - num__22 h + num__85 = num__0 solve for h . h = num__5 and h = num__17 . only the solution h = num__17 gives a and b positive and it is the length of the hypotenuse of the triangle . answer a <eor> a <eos> |
a |
rectangle_perimeter__2.0__9.0__ triangle_area__2.0__17.0__ |
rectangle_perimeter__2.0__9.0__ triangle_area__2.0__17.0__ |
| a b and c are partners . a receives num__0.666666666667 of profits b and c dividing the remainder equally . a ' s income is increased by rs . num__800 when the rate to profit rises from num__5 to num__7 percent . find the capital of b ? <o> a ) num__3999 <o> b ) num__10000 <o> c ) num__2500 <o> d ) num__2772 <o> e ) num__2912 |
a : b : c = num__0.666666666667 : num__0.166666666667 : num__0.166666666667 = num__4 : num__1 : num__1 x * num__0.02 * num__0.666666666667 = num__800 b capital = num__60000 * num__0.166666666667 = num__10000 . answer : b <eor> b <eos> |
b |
subtract__5.0__4.0__ multiply__1.0__10000.0__ |
subtract__5.0__4.0__ multiply__1.0__10000.0__ |
| a train num__250 m long is running at a speed of num__27 km / hr . in what time will it pass a bridge num__200 m long ? <o> a ) num__40 <o> b ) num__45 <o> c ) num__50 <o> d ) num__55 <o> e ) num__60 |
speed = num__27 * num__0.277777777778 = num__7.5 m / sec total distance covered = num__250 + num__200 = num__450 m required time = num__450 * num__0.133333333333 = num__60 sec answer : e <eor> e <eos> |
e |
add__250.0__200.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
add__250.0__200.0__ divide__450.0__7.5__ divide__450.0__7.5__ |
| a work which could be finished in num__9 days was finished num__3 days earlier after num__10 more men joined . the number of men employed was ? <o> a ) num__24 <o> b ) num__27 <o> c ) num__20 <o> d ) num__35 <o> e ) num__25 |
c num__20 x - - - - - - - num__9 ( x + num__10 ) - - - - num__6 x * num__9 = ( x + num__10 ) num__6 x = num__20 <eor> c <eos> |
c |
subtract__9.0__3.0__ round__20.0__ |
subtract__9.0__3.0__ round__20.0__ |
| how many hours are there in num__1200 minutes ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__85 <o> d ) num__86 <o> e ) num__88 |
we know that there are num__60 minutes in num__1 hour . divide the number of minutes by the number of minutes in num__1 hour . we get divide num__1200 by num__601200 ÷ num__60 = num__20 so there are num__20 hours in num__1200 minutes . answer is b <eor> b <eos> |
b |
hour_to_min_conversion__ divide__1200.0__60.0__ round__20.0__ |
hour_to_min_conversion__ divide__1200.0__60.0__ round__20.0__ |
| a cycle is bought for rs . num__900 and sold for rs . num__1080 find the gain percent ? <o> a ) num__11 <o> b ) num__20 <o> c ) num__99 <o> d ) num__77 <o> e ) num__18 |
num__900 - - - - num__180 num__100 - - - - ? = > num__20.0 answer : b <eor> b <eos> |
b |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| when n liters of fuel was added to a tank that was already num__0.333333333333 full the tank was filled to num__2.33333333333 of its capacity . in terms of n what is the capacity of the tank in liters ? <o> a ) num__0.5 n <o> b ) num__1.33333333333 n <o> c ) num__1.5 n <o> d ) num__2.25 n <o> e ) num__2.33333333333 n |
a is the answer . to solve this problem draw diagram or use algebra . i ' m more comfortable with algebra . given - tank was already num__0.333333333333 full . - when n ltr added it became num__2.33333333333 so num__0.333333333333 + n = num__2.33333333333 n = num__2.33333333333 - num__0.333333333333 n = num__2.0 capacity of the tank is full fraction . i . e . num__1 i . e . num__1.0 so the question is num__1.0 is how much times n = num__1.0 = num__1 = xn so x = num__0.5 and total = num__0.5 n = num__0.5 n <eor> a <eos> |
a |
round_down__2.3333__ reverse__2.0__ reverse__2.0__ |
subtract__2.3333__0.3333__ reverse__2.0__ subtract__1.0__0.5__ |
| find the value of num__72518 x num__9999 = m ? <o> a ) num__345434667 <o> b ) num__246465757 <o> c ) num__465767867 <o> d ) num__725107482 <o> e ) num__645354643 |
num__72518 x num__9999 = num__72518 x ( num__10000 - num__1 ) = num__72518 x num__10000 - num__72518 x num__1 = num__725180000 - num__72518 = num__725107482 d <eor> d <eos> |
d |
subtract__10000.0__9999.0__ multiply__72518.0__10000.0__ multiply__72518.0__9999.0__ multiply__72518.0__9999.0__ |
subtract__10000.0__9999.0__ multiply__72518.0__10000.0__ subtract__725180000.0__72518.0__ subtract__725180000.0__72518.0__ |
| if the sides of a triangle are num__26 cm num__18 cm and num__10 cm what is its area ? <o> a ) num__90 <o> b ) num__110 <o> c ) num__130 <o> d ) num__140 <o> e ) num__150 |
the triangle with sides num__26 cm num__18 cm and num__10 cm is right angled where the hypotenuse is num__26 cm . area of the triangle = num__0.5 * num__18 * num__10 = num__90 cm num__2 answer : option a <eor> a <eos> |
a |
triangle_area__18.0__10.0__ square_perimeter__0.5__ triangle_area__18.0__10.0__ |
volume_rectangular_prism__18.0__10.0__0.5__ square_perimeter__0.5__ volume_rectangular_prism__18.0__10.0__0.5__ |
| a man rows his boat num__85 km downstream and num__45 km upstream taking num__2 num__0.5 hours each time . find the speed of the stream ? <o> a ) num__7 kmph <o> b ) num__6 kmph <o> c ) num__5 kmph <o> d ) num__8 kmph <o> e ) num__2 kmph |
speed downstream = d / t = num__85 / ( num__2 num__0.5 ) = num__34 kmph speed upstream = d / t = num__45 / ( num__2 num__0.5 ) = num__18 kmph the speed of the stream = ( num__34 - num__18 ) / num__2 = num__8 kmph answer : d <eor> d <eos> |
d |
round__8.0__ |
round__8.0__ |
| ramesh can finish a piece of work in num__16 days . rohan is twice as efficient as ramesh . if they work together how many days will they need to finish the same amount of work ? <o> a ) num__2.66666666667 days <o> b ) num__3.0 days <o> c ) num__5.33333333333 days <o> d ) num__3.66666666667 days <o> e ) num__2.33333333333 days |
ramesh can finish a piece of work in num__16 days . rohan is twice as efficient as ramesh . so rohan can finish a piece of work in num__8 days . days needed to finish the same amount of work = num__1 / ( num__0.125 + num__0.0625 ) = num__5.33333333333 days answer : c <eor> c <eos> |
c |
divide__1.0__8.0__ divide__1.0__16.0__ divide__5.3333__1.0__ |
divide__1.0__8.0__ divide__1.0__16.0__ divide__5.3333__1.0__ |
| a train of length num__110 meter is running at a speed of num__60 kmph . in what time it will pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__10 <o> b ) num__8 <o> c ) num__6 <o> d ) num__4 <o> e ) num__2 |
explanation : distance = num__110 m relative speed = num__60 + num__6 = num__66 kmph ( since both the train and the man are in moving in opposite direction ) = m / sec = m / sec time taken to pass the man = = num__6 s answer is c <eor> c <eos> |
c |
add__60.0__6.0__ round__6.0__ |
add__60.0__6.0__ round__6.0__ |
| a b and c are positive integers less than or equal to num__9 and multiple of num__3 . if a b and c are assembled into the three - digit number abcabc which of the following must be a factor of abcabc ? <o> a ) num__6 <o> b ) num__4 <o> c ) num__7 <o> d ) num__5 <o> e ) num__9 |
as a b and c are multiple of num__3 they can be written as a = num__3 l b = num__3 m and c = num__3 n . the sum of digits of abc is a + b + c = num__3 ( l + m + n + l + m + n ) = num__6 ( l + m + n ) . thus num__6 is always a factor . answer : a <eor> a <eos> |
a |
subtract__9.0__3.0__ subtract__9.0__3.0__ |
subtract__9.0__3.0__ subtract__9.0__3.0__ |
| y = x ^ num__2 + bx + num__128 cuts the x axis at ( h num__0 ) and ( k num__0 ) . if h and k are integers what is the least value of b ? <o> a ) - num__129 <o> b ) - num__132 <o> c ) num__32 <o> d ) num__64 <o> e ) num__128 |
as the curve cuts the x - axis at ( h num__0 ) and ( k num__0 ) . therefore h k are the roots of the quadratic equation . for the quadratic equation is in the form of ax ^ num__2 + bx + c = num__0 the product of the roots = c / a = num__128.0 = num__64 and the sum of the roots = - b / a = - b / num__1 num__128 can be expressed as product of two numbers in the following ways : num__1 * num__128 num__2 * num__64 num__4 * num__32 num__8 * num__16 the sum of the roots is maximum when the roots are num__1 and num__128 and the maximum sum is num__1 + num__128 = num__129 . the least value possible for b is therefore - num__129 . a <eor> a <eos> |
a |
divide__128.0__2.0__ divide__128.0__4.0__ multiply__2.0__4.0__ multiply__2.0__8.0__ add__128.0__1.0__ add__128.0__1.0__ |
divide__128.0__2.0__ divide__128.0__4.0__ multiply__2.0__4.0__ multiply__2.0__8.0__ add__128.0__1.0__ add__128.0__1.0__ |
| a train num__280 m long running with a speed of num__63 km / hr will pass a tree in ? <o> a ) num__22 sec <o> b ) num__16 second <o> c ) num__77 sec <o> d ) num__55 sec <o> e ) num__17 sec |
speed = num__63 * num__0.277777777778 = num__17.5 m / sec time taken = num__280 * num__0.0571428571429 = num__16 sec answer : b <eor> b <eos> |
b |
divide__280.0__17.5__ round__16.0__ |
divide__280.0__17.5__ divide__280.0__17.5__ |
| you enter a weight loss challenge game and manage to lose num__14.0 of your body weight . for the final weigh in you are forced to wear clothes that add num__2.0 to your weight . what percentage of weight loss is measured at the final weigh in ? <o> a ) num__13.0 <o> b ) num__9.22 <o> c ) num__9.0 <o> d ) num__14.0 <o> e ) num__12.28 % |
( num__100.0 - num__14.0 ) * ( num__100.0 + num__2.0 ) = num__0.86 * num__1.02 = num__12.28 the weigh in records your weight loss at num__12.28 ! the answer is e <eor> e <eos> |
e |
percent__12.28__100.0__ |
percent__12.28__100.0__ |
| find the invalid no . from the following series num__3 num__7 num__15 num__27 num__63 num__127 num__255 <o> a ) num__3 <o> b ) num__7 <o> c ) num__27 <o> d ) num__127 <o> e ) num__255 |
go on multiplying the number by num__2 and adding num__1 to it to get the next number . so num__27 is wrong . c <eor> c <eos> |
c |
subtract__3.0__2.0__ multiply__27.0__1.0__ |
subtract__3.0__2.0__ multiply__27.0__1.0__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__240 m and they cross each other in num__12 sec then the speed of each train is ? <o> a ) num__22 <o> b ) num__77 <o> c ) num__72 <o> d ) num__88 <o> e ) num__21 |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__240 + num__240 ) / num__12 = > x = num__20 speed of each train = num__20 m / sec . = num__20 * num__3.6 = - num__72 km / hr . answer : c <eor> c <eos> |
c |
divide__240.0__12.0__ multiply__3.6__20.0__ round__72.0__ |
divide__240.0__12.0__ multiply__3.6__20.0__ multiply__3.6__20.0__ |
| if taxi fares were $ num__1.00 for the first num__0.2 mile and $ num__0.40 for each num__0.2 mile there after then the taxi fare for a num__3 - mile ride was <o> a ) $ num__1.56 <o> b ) $ num__2.40 <o> c ) $ num__3.80 <o> d ) $ num__4.20 <o> e ) $ num__6.60 |
in num__3 miles initial num__0.2 mile charge is $ num__1 rest of the distance = num__3 - ( num__0.2 ) = num__2.8 rest of the distance charge = num__14 ( num__0.4 ) = $ num__5.6 ( as the charge is num__0.4 for every num__0.2 mile ) = > total charge for num__3 miles = num__1 + num__5.6 = num__6.6 answer is e . <eor> e <eos> |
e |
subtract__3.0__0.2__ divide__2.8__0.2__ multiply__0.4__14.0__ add__1.0__5.6__ multiply__1.0__6.6__ |
subtract__3.0__0.2__ divide__2.8__0.2__ multiply__0.4__14.0__ add__1.0__5.6__ add__1.0__5.6__ |
| set r consists of integers { num__3 - num__8 y num__19 - num__6 } and set b consists of integers { k - num__3 num__0 num__16 - num__5 num__9 } . number l represents the median of set r number m represents mode of set b and number z = l ^ m . if y is an integer greater than num__21 for what value of k will z be a divisor of num__26 ? <o> a ) - num__2 <o> b ) - num__1 <o> c ) num__0 <o> d ) num__1 <o> e ) num__2 |
r = { num__3 - num__8 y num__19 - num__6 } b = { k - num__3 num__016 - num__59 } y > num__21 so l = median of r = num__3 m = mode of set b z = ( num__3 ) ^ m if z is a divisor of num__26 ( num__3 ) ^ m = num__1 because num__26 does not have num__3 as a factor = > m = num__0 hence k = num__0 as m is mode and num__0 will be the most frequently occuring number in set b . answer - c <eor> c <eos> |
c |
subtract__6.0__5.0__ multiply__3.0__0.0__ |
subtract__6.0__5.0__ multiply__3.0__0.0__ |
| a shop owner sells num__30 mtr of cloth and gains sp of num__10 metres . find the gain % ? <o> a ) num__30.0 <o> b ) num__40.0 <o> c ) num__50.0 <o> d ) num__56.0 <o> e ) num__78 % |
here selling price of num__10 m cloth is obtained as profit . profit of num__10 m cloth = ( s . p . of num__30 m cloth ) – ( c . p . of num__30 m cloth ) selling price of num__20 m cloth = selling price of num__30 m of cloth let cost of each metre be rs . num__100 . therefore cost price of num__20 m cloth = rs . num__2000 and s . p . of num__20 m cloth = rs . rs . num__3000 profit % = num__10 × num__100 = num__50.0 num__20 profit of num__50.0 was made by the merchant . c <eor> c <eos> |
c |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| an aeroplane covers a certain distance at a speed of num__240 kmph in num__4 hours . to cover the same distance in num__1 num__0.666666666667 hours it must travel at a speed of : <o> a ) num__520 <o> b ) num__620 <o> c ) num__576 <o> d ) num__740 <o> e ) num__720 |
distance = ( num__240 x num__4 ) = num__960 km . speed = distance / time speed = num__960 / ( num__1.66666666667 ) km / hr . [ we can write num__1 num__0.666666666667 hours as num__1.66666666667 hours ] required speed = ( num__960 x num__0.6 ) km / hr = num__576 km / hr answer c ) num__576 km / hr <eor> c <eos> |
c |
multiply__240.0__4.0__ add__1.0__0.6667__ km_to_mile_conversion__ multiply__960.0__0.6__ round__576.0__ |
multiply__240.0__4.0__ add__1.0__0.6667__ divide__1.0__1.6667__ multiply__960.0__0.6__ divide__576.0__1.0__ |
| two pipes a and b can fill a cistern in num__37 minutes and num__45 minutes respectively . both pipes are opened . the cistern will be filled in just half an hour if the b is turned off after ? <o> a ) num__7 <o> b ) num__9 <o> c ) num__11 <o> d ) num__13 <o> e ) num__15 |
let b be turned off after x minutes . then part filled by ( a + b ) in x min . + part filled by a in ( num__30 - x ) min . = num__1 . x num__2 + num__1 + ( num__30 - x ) . num__2 = num__1 num__75 num__45 num__75 num__11 x + ( num__60 - num__2 x ) = num__1 num__225 num__75 num__11 x + num__180 - num__6 x = num__225 . x = num__9 . b ) <eor> b <eos> |
b |
add__45.0__30.0__ hour_to_min_conversion__ subtract__225.0__45.0__ divide__180.0__30.0__ subtract__11.0__2.0__ round__9.0__ |
add__45.0__30.0__ hour_to_min_conversion__ subtract__225.0__45.0__ divide__180.0__30.0__ subtract__11.0__2.0__ subtract__11.0__2.0__ |
| the lengths of the legs of a right triangle are x and y while the length of the hypotenuse is x + y - num__4 . what is the maximum radius of a circle inscribed in this triangle ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__4 <o> d ) num__22 <o> e ) can not be determined from the information given |
length of hypotenuse : x + y - num__4 = a + b . hence x + y = a + b + num__4 . sum of lengths of legs : x + y = a + b + num__2 r . therefore num__2 r = num__4 so r = num__2 . correct answer b <eor> b <eos> |
b |
subtract__4.0__2.0__ |
subtract__4.0__2.0__ |
| the ratio between the school ages of neelam and shaan is num__5 : num__6 respectively . if the ratio between the one - third age of neelam and half of shaan ' s age of num__5 : num__9 then what is the school age of shaan ? <o> a ) num__19 <o> b ) num__27 <o> c ) num__16 <o> d ) num__15 <o> e ) num__17 |
let the school ages of neelam and shaan be num__5 x and num__6 x years respectively . then ( num__0.333333333333 * num__5 x ) / ( num__0.5 * num__6 x ) = num__0.555555555556 num__15 = num__15 thus shaan ' s age can not be determined . answer : d <eor> d <eos> |
d |
divide__5.0__9.0__ add__6.0__9.0__ add__6.0__9.0__ |
divide__5.0__9.0__ add__6.0__9.0__ add__6.0__9.0__ |
| in a clothing store there are six different colored neckties ( red orange yellow green blue and indigo ) and six different colored shirts ( red orange yellow green blue and indigo ) that must be packed into boxes for gifts . if each box can only fit one necktie and one shirt what is the probability that all of the boxes will contain a necktie and a shirt of the same color <o> a ) num__0.998611111111 <o> b ) num__0.00833333333333 <o> c ) num__0.00858369098712 <o> d ) num__0.00552486187845 <o> e ) num__0.00138888888889 |
num__6 ties and num__6 shirts . . . red tie can take any of num__6 shirts . . orange can take any of the remaining num__5 shirts yellow any of remaining num__4 . . and so on till last indigo chooses the num__1 remaining . . total ways = num__6 * num__5 * num__4 * num__3 * num__2 * num__1 = num__720 out of this num__720 only num__1 way will have same colour tie and shirt . . prob = num__0.998611111111 a <eor> a <eos> |
a |
subtract__5.0__4.0__ subtract__4.0__1.0__ subtract__3.0__1.0__ multiply__1.0__0.9986__ |
subtract__5.0__4.0__ subtract__4.0__1.0__ subtract__3.0__1.0__ multiply__1.0__0.9986__ |
| the denominator of a fraction is num__1 less than twice the numerator . if the numerator and denominator are both increased by num__1 the fraction becomes num__0.6 . find the fraction ? <o> a ) num__0.714285714286 <o> b ) num__5.0 <o> c ) num__0.555555555556 . <o> d ) num__2.5 <o> e ) num__0.625 |
let the numerator and denominator of the fraction be ' n ' and ' d ' respectively . d = num__2 n - num__1 ( n + num__1 ) / ( d + num__1 ) = num__0.6 num__5 n + num__5 = num__3 d + num__3 num__5 n + num__5 = num__3 ( num__2 n - num__1 ) + num__3 = > n = num__5 d = num__2 n - num__1 = > d = num__9 hence the fraction is : num__0.555555555556 . answer : c <eor> c <eos> |
c |
add__1.0__2.0__ divide__5.0__9.0__ multiply__1.0__0.5556__ |
add__1.0__2.0__ divide__5.0__9.0__ divide__5.0__9.0__ |
| num__1 + num__2 = num__10 num__2 + num__3 = num__26 num__3 + num__4 = num__50 then num__4 + num__5 = ? <o> a ) num__80 <o> b ) num__81 <o> c ) num__82 <o> d ) num__83 <o> e ) num__84 |
num__1 + num__2 = num__10 . . . . . ( num__1 + num__2 ) ^ num__2 + num__1 = num__10 num__2 + num__3 = num__26 . . . . . . . ( num__2 + num__3 ) ^ num__2 + num__1 = num__26 num__3 + num__4 = num__50 then . . . ( num__3 + num__4 ) ^ num__2 + num__1 = num__50 num__4 + num__5 = num__82 . . . . . . . . ( num__4 + num__5 ) ^ num__2 + num__1 = num__82 answer : c <eor> c <eos> |
c |
multiply__1.0__82.0__ |
multiply__1.0__82.0__ |
| six computers each working at the same constant rate together can process a certain amount of data in num__9 days . how many additional computers each working at the same constant rate will be needed to process the same amount of data in num__6 days ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__9 <o> e ) num__12 |
explanation : if six computers require num__9 days to process the data thats a total of num__54 computer - days the product of num__6 and num__9 . if you change the number of computers or the number of days num__54 will have to remain the product whether that means num__54 days of one computer or one day with num__54 computers . in num__6 days the number of computers is : num__6 c = num__54 c = num__9 num__9 computers is num__3 more than the num__6 that it took to do the job in num__9 days so the correct choice is ( a ) . <eor> a <eos> |
a |
multiply__9.0__6.0__ subtract__9.0__6.0__ round__3.0__ |
multiply__9.0__6.0__ subtract__9.0__6.0__ subtract__9.0__6.0__ |
| a cube is divided into num__125 identical cubelets . each cut is made parallel to some surface of the cube . but before doing that the cube is painted with green on one set of opposite faces red on another set of opposite faces and blue on the third set of opposite faces . how many cubelets are painted with exactly one colour ? <o> a ) num__36 <o> b ) num__42 <o> c ) num__48 <o> d ) num__54 <o> e ) num__60 |
each side of the cube has num__5 x num__5 = num__25 cubelets . only the interior cubelets are painted one colour . on each side num__3 x num__3 = num__9 cubelets are painted one colour . since the cube has six sides the number of cubes with one colour is num__6 * num__9 = num__54 the answer is d . <eor> d <eos> |
d |
surface_cube__3.0__ surface_cube__3.0__ |
multiply__6.0__9.0__ multiply__6.0__9.0__ |
| if x / num__4 – ( x – num__3 ) / num__5 = num__1 then find the value of x . <o> a ) num__9 <o> b ) num__7 <o> c ) num__6 <o> d ) num__8 <o> e ) num__5 |
explanation : x / num__4 – ( x - num__3 ) / num__5 = num__1 ( num__5 x – num__4 ( x – num__3 ) ) / num__20 = num__1 num__5 x - num__4 x + num__12 = num__20 x = num__8 answer : a <eor> a <eos> |
a |
multiply__4.0__5.0__ multiply__4.0__3.0__ add__3.0__5.0__ add__4.0__5.0__ |
multiply__4.0__5.0__ multiply__4.0__3.0__ add__3.0__5.0__ add__4.0__5.0__ |
| you walk one mile to school every day . you leave home at the same time each day walk at a steady speed of num__3 miles per hour and arrive just as school begins . today you were distracted by the pleasant weather and walked the rst half mile at a speed of only num__2 miles per hour . at how many miles per hour must you run the last half mile in order to arrive just as school begins today ? <o> a ) num__4 miles per hour <o> b ) num__6 miles per hour <o> c ) num__8 miles per hour <o> d ) num__10 miles per hour <o> e ) num__12 miles per hour |
recall the formula s = v t where in this case s is distance ( in miles ) traveled at a constant speed of v ( miles per hour ) for a time of t hours . the rst statement indicates that it usually takes you only num__20 minutes to go to school ( solve for t in num__1 = num__3 t and change units ) today it took you num__15 minutes to cover the rst half mile ( solve for t in num__1 = num__2 = num__2 t and change units ) . you only have num__5 more minutes to arrive ( that is num__0.0833333333333 th of an hour ) and need to cover another half mile . solve for v in the equation num__1 = num__2 = ( num__1 = num__12 ) v . correct answer b <eor> b <eos> |
b |
subtract__3.0__2.0__ add__3.0__2.0__ subtract__15.0__3.0__ multiply__3.0__2.0__ |
subtract__3.0__2.0__ add__3.0__2.0__ subtract__15.0__3.0__ multiply__3.0__2.0__ |
| a train num__560 m long is running at a speed of num__45 km / hr . in what time will it pass a bridge num__140 m long ? <o> a ) num__40 <o> b ) num__56 <o> c ) num__41 <o> d ) num__42 <o> e ) num__34 |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec total distance covered = num__560 + num__140 = num__700 m required time = num__700 * num__0.08 = num__40 sec answer : option b <eor> b <eos> |
b |
add__560.0__140.0__ divide__700.0__12.5__ |
add__560.0__140.0__ divide__700.0__12.5__ |
| kim can do a work in num__3 days while david can do the same work in num__2 days . both of them finish the work together and get rs . num__150 . what is the share of kim ? <o> a ) num__22 <o> b ) num__60 <o> c ) num__88 <o> d ) num__66 <o> e ) num__12 |
kim ' s wages : david ' s wages = kim ' s num__1 day work : david ' s num__1 day work = num__0.333333333333 : num__0.5 = num__2 : num__3 kim ' s share = num__0.4 * num__150 = rs . num__60 answer : b <eor> b <eos> |
b |
subtract__3.0__2.0__ divide__1.0__3.0__ divide__1.0__2.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
subtract__3.0__2.0__ divide__1.0__3.0__ divide__1.0__2.0__ multiply__150.0__0.4__ multiply__150.0__0.4__ |
| a cyclist covers a distance of num__750 meter in num__2 minutes num__50 seconds . what is the speed in km / hr of cyclist <o> a ) num__16 km / hr <o> b ) num__17 km / hr <o> c ) num__15.84 km / hr <o> d ) num__19 km / hr <o> e ) num__20 km / hr |
explanation : speed = distance / time distance = num__750 meter time = num__2 min num__50 sec = num__170 sec speed = num__4.41176470588 = num__4.4 m / sec = > num__4.4 â ˆ — num__3.6 km / hr = num__15.84 km / hr option c <eor> c <eos> |
c |
divide__750.0__170.0__ multiply__4.4__3.6__ round__15.84__ |
divide__750.0__170.0__ multiply__4.4__3.6__ round__15.84__ |
| a goods train runs at the speed of num__72 km / hr and crosses a num__250 m long platform in num__26 sec . what is the length of the goods train ? <o> a ) num__288 <o> b ) num__277 <o> c ) num__274 <o> d ) num__270 <o> e ) num__281 |
speed = num__72 * num__0.277777777778 = num__20 m / sec . time = num__26 sec . let the length of the train be x meters . then ( x + num__250 ) / num__26 = num__20 x = num__270 m . answer : d <eor> d <eos> |
d |
add__250.0__20.0__ round__270.0__ |
add__250.0__20.0__ add__250.0__20.0__ |
| a man purchased num__1 blankets @ rs . num__100 each num__5 blankets @ rs . num__150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . num__150 . find the unknown rate of two blankets ? <o> a ) num__278 <o> b ) num__277 <o> c ) num__278 <o> d ) num__450 <o> e ) num__650 |
num__10 * num__150 = num__1500 num__1 * num__100 + num__5 * num__150 = num__850 num__1500 – num__850 = num__650 answer : e <eor> e <eos> |
e |
multiply__150.0__10.0__ subtract__1500.0__850.0__ multiply__1.0__650.0__ |
multiply__150.0__10.0__ subtract__1500.0__850.0__ multiply__1.0__650.0__ |
| a board num__7 ft . num__9 inches long is divided into num__3 equal parts . what is the length of each part ? <o> a ) num__2 ft . num__7 inches <o> b ) num__4 ft . num__7 inches <o> c ) num__1 ft . num__7 inches <o> d ) num__3 ft . num__7 inches <o> e ) none of them |
length of board = num__7 ft . num__9 inches = ( num__7 * num__12 + num__9 ) inches = num__93 inches . therefore length of each part = ( num__31.0 ) inches = num__31 inches = num__2 ft . num__7 inches answer is a . <eor> a <eos> |
a |
add__9.0__3.0__ divide__93.0__3.0__ subtract__9.0__7.0__ subtract__9.0__7.0__ |
add__9.0__3.0__ divide__93.0__3.0__ subtract__9.0__7.0__ subtract__9.0__7.0__ |
| renu can do a piece of work in num__6 days but with the help of her friend suma she can do it in num__5 days . in what time suma can do it alone ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__15 <o> e ) num__30 |
renu â € ™ s one day â € ™ s work = num__0.166666666667 suma â € ™ s one day â € ™ s work = num__0.2 - num__0.166666666667 = num__0.0333333333333 suma can do it alone in num__30 days . answer : e <eor> e <eos> |
e |
divide__0.1667__5.0__ multiply__6.0__5.0__ round__30.0__ |
subtract__0.2__0.1667__ multiply__6.0__5.0__ round__30.0__ |
| a cycle is bought for rs . num__900 and sold for rs . num__1080 find the gain percent ? <o> a ) num__20.0 <o> b ) num__30.0 <o> c ) num__15.0 <o> d ) num__10.0 <o> e ) num__35 % |
num__900 - - - - num__180 num__100 - - - - num__20.0 answer a <eor> a <eos> |
a |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| a person took some amount with some interest for num__2 years but increase the interest for num__1.0 he paid rs . num__110 / - extra then how much amount he took ? <o> a ) s . num__5500 / - <o> b ) s . num__6000 / - <o> c ) s . num__4000 / - <o> d ) s . num__7000 / - <o> e ) s . num__8000 / - |
explanation : num__2 years = rs . num__110 / - year = num__55.0 rate of interest = num__1.0 num__100.0 % × num__55.0 = rs . num__5500 / - p = rs . num__5500 / - answer : option a <eor> a <eos> |
a |
divide__110.0__2.0__ multiply__100.0__55.0__ multiply__1.0__5500.0__ |
divide__110.0__2.0__ multiply__100.0__55.0__ divide__5500.0__1.0__ |
| the cross - section of a water channel is a trapezium in shape . if the channel is num__12 meters wide at the top and num__8 meters wide at the bottom and the area of cross - section is num__700 square meters what is the depth of the channel ( in meters ) ? <o> a ) num__50 <o> b ) num__60 <o> c ) num__70 <o> d ) num__80 <o> e ) num__90 |
num__0.5 * d * ( num__12 + num__8 ) = num__700 d = num__70 the answer is c . <eor> c <eos> |
c |
round__70.0__ |
round__70.0__ |
| a salesman ’ s terms were changed from a flat commission of num__5.0 on all his sales to a fixed salary of rs . num__1000 plus num__2.5 commission on all sales exceeding rs . num__4000 . if his remuneration as per new scheme was rs . num__600 more than that by the previous schema his sales were worth ? <o> a ) num__12003 <o> b ) num__12000 <o> c ) num__12002 <o> d ) num__12022 <o> e ) num__12023 |
[ num__1000 + ( x - num__4000 ) * ( num__2.5 / num__100 ) ] - x * ( num__0.05 ) = num__600 x = num__12000 answer : b <eor> b <eos> |
b |
divide__5.0__100.0__ divide__600.0__0.05__ divide__600.0__0.05__ |
divide__5.0__100.0__ divide__600.0__0.05__ divide__600.0__0.05__ |
| find the surface area of a cuboid num__12 m long num__14 m broad and num__7 m high <o> a ) num__868 sq . cm <o> b ) num__600 sq . cm <o> c ) num__700 sq . cm <o> d ) num__900 sq . cm <o> e ) none of these |
explanation : surface area = [ num__2 ( num__12 x num__14 + num__14 x num__7 + num__12 x num__7 ) ] cm num__2 = ( num__2 x num__350 ) cm num__2 = num__700 cm num__2 . answer : c <eor> c <eos> |
c |
multiply__2.0__350.0__ triangle_area__2.0__700.0__ |
multiply__2.0__350.0__ triangle_area__2.0__700.0__ |
| the list price of an article is rs . num__65 . a customer pays rs . num__56.16 for it . he was given two successive discounts one of them being num__10.0 . the other discount is ? <o> a ) num__9.0 <o> b ) num__4.0 <o> c ) num__6.0 <o> d ) num__3.0 <o> e ) num__5 % |
num__65 * ( num__0.9 ) * ( ( num__100 - x ) / num__100 ) = num__56.16 x = num__4.0 answer : b <eor> b <eos> |
b |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| the least number which should be added to num__2497 so that the sum is exactly divisible by num__5 num__6 num__4 and num__3 is <o> a ) num__3 <o> b ) num__13 <o> c ) num__23 <o> d ) num__33 <o> e ) num__43 |
lcm of num__5 num__6 num__4 and num__3 = num__60 on dividing num__2497 by num__60 the remainder is num__37 . number to be added = num__60 - num__37 = num__23 answer : c <eor> c <eos> |
c |
subtract__60.0__37.0__ subtract__60.0__37.0__ |
subtract__60.0__37.0__ subtract__60.0__37.0__ |
| after mm students took a test there was a total of num__64.0 of correct answers . if the test contains num__50 questions what is the least number of questions that the next student have to get right to bring the total of correct answers to num__70.0 ? <o> a ) + num__20 <o> b ) + num__35 <o> c ) + num__15 <o> d ) + num__20 <o> e ) + num__45 |
denote xx as the required number of correct answers . xx must satisfy the equation num__0.64 ∗ m ∗ num__50 + x / num__50 m + num__50 = num__0.7 or num__350 m + num__350 = num__320 m + num__10 x or x = num__3 m + num__35 answer : b <eor> b <eos> |
b |
multiply__50.0__0.7__ multiply__50.0__0.7__ |
divide__350.0__10.0__ divide__350.0__10.0__ |
| the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is num__867 sq m then what is the breadth of the rectangular plot ? <o> a ) num__19 m <o> b ) num__17 m <o> c ) num__18 m <o> d ) num__14 m <o> e ) num__12 m |
let the breadth of the plot be b m . length of the plot = num__3 b m ( num__3 b ) ( b ) = num__867 num__3 b num__2 = num__867 b num__2 = num__289 = num__172 ( b > num__0 ) b = num__17 m . answer : b <eor> b <eos> |
b |
triangle_area__2.0__17.0__ |
triangle_area__2.0__17.0__ |
| there are two circles of different radii . the are of a square is num__784 sq cm and its side is twice the radius of the larger circle . the radius of the larger circle is seven - third that of the smaller circle . find the circumference of the smaller circle ? <o> a ) num__65 <o> b ) num__17 <o> c ) num__12 <o> d ) num__18 <o> e ) num__15 |
let the radii of the larger and the smaller circles be l cm and s cm respectively . let the side of the square be a cm . a num__2 = num__784 = ( num__4 ) ( num__196 ) = ( num__22 ) . ( num__142 ) a = ( num__2 ) ( num__14 ) = num__28 a = num__2 l l = a / num__2 = num__14 l = ( num__2.33333333333 ) s therefore s = ( num__0.428571428571 ) ( l ) = num__6 circumference of the smaller circle = num__2 ∏ s = num__12 ∏ cm . answer : c <eor> c <eos> |
c |
multiply__2.0__14.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
multiply__2.0__14.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
| the banker ' s discount on a bill due num__4 months hence at num__15.0 is rs . num__420 . the true discount is : <o> a ) num__400 <o> b ) num__277 <o> c ) num__268 <o> d ) num__191 <o> e ) num__123 |
t . d . = b . d . x num__100 num__100 + ( r x t ) = rs . num__420 x num__100 num__100 + num__15 x num__1 num__3 = rs . num__420 x num__100 num__105 = rs . num__400 . answer : a <eor> a <eos> |
a |
percent__100.0__400.0__ |
percent__100.0__400.0__ |
| can you replace the question mark with the number to complete the series provided the pair of numbers exhibits a similar relationship ? ? : num__4623 : : num__9 : num__647 <o> a ) ? = num__17 <o> b ) num__18 <o> c ) num__13 <o> d ) num__10 <o> e ) num__20 |
a num__17 the relationship holds for below formula : pow ( x num__3 ) - pow ( x num__2 ) - num__1 = > num__9 * num__9 * num__9 - num__9 * num__9 - num__1 = > num__729 - num__81 - num__1 = > num__647 similarly num__17 * num__17 * num__17 - num__17 * num__17 - num__1 = > num__4913 - num__289 - num__1 = > num__4623 <eor> a <eos> |
a |
subtract__3.0__2.0__ power__9.0__3.0__ power__9.0__2.0__ power__17.0__3.0__ power__17.0__2.0__ multiply__1.0__17.0__ |
subtract__3.0__2.0__ power__9.0__3.0__ power__9.0__2.0__ power__17.0__3.0__ power__17.0__2.0__ multiply__1.0__17.0__ |
| a train is num__360 meter long is running at a speed of num__45 km / hour . in what time will it pass a bridge of num__140 meter length ? <o> a ) num__87 <o> b ) num__69 <o> c ) num__40 <o> d ) num__72 <o> e ) num__21 |
speed = num__45 km / hr = num__45 * ( num__0.277777777778 ) m / sec = num__12.5 m / sec total distance = num__360 + num__140 = num__500 meter time = distance / speed = num__500 * ( num__0.08 ) = num__40 seconds answer : c <eor> c <eos> |
c |
add__360.0__140.0__ divide__500.0__12.5__ round__40.0__ |
add__360.0__140.0__ divide__500.0__12.5__ divide__500.0__12.5__ |
| the markup on a box of apples is num__10 percent of the cost . the markup is what percent of the selling price ? ( markup = selling price - cost ) <o> a ) num__9.09 <o> b ) num__10.0 <o> c ) num__12 num__0.5 % <o> d ) num__15.0 <o> e ) num__16 num__0.666666666667 % |
mp = num__0.1 cp sp = cp + num__0.1 cp = num__1.1 cp hence mp = num__0.1 / num__1.1 sp = num__1.0 sp . hence mp is num__9.09 of sp answer a <eor> a <eos> |
a |
reverse__10.0__ round_down__1.1__ multiply__1.0__9.09__ |
reverse__10.0__ round_down__1.1__ divide__9.09__1.0__ |
| the speed of a car is num__90 km in the first hour and num__60 km in the second hour . what is the average speed of the car ? <o> a ) num__22 <o> b ) num__75 <o> c ) num__44 <o> d ) num__28 <o> e ) num__12 |
s = ( num__90 + num__60 ) / num__2 = num__75 kmph answer : b <eor> b <eos> |
b |
round__75.0__ |
round__75.0__ |
| a batsman in his num__17 th innings makes a score of num__85 and their by increasing his average by num__3 . what is his average after the num__17 thinnings ? <o> a ) num__32 <o> b ) num__29 <o> c ) num__30 <o> d ) num__37 <o> e ) num__12 |
num__16 x + num__85 = num__17 ( x + num__3 ) x = num__34 + num__3 = num__37 answer : d <eor> d <eos> |
d |
add__3.0__34.0__ add__3.0__34.0__ |
add__3.0__34.0__ add__3.0__34.0__ |
| the maximum number of students among them num__1001 pens and num__910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : <o> a ) num__91 <o> b ) num__910 <o> c ) num__1001 <o> d ) num__1911 <o> e ) none of these |
explanation : required number of students = h . c . f of num__1001 and num__910 = num__91 . answer : a <eor> a <eos> |
a |
subtract__1001.0__910.0__ subtract__1001.0__910.0__ |
subtract__1001.0__910.0__ subtract__1001.0__910.0__ |
| a boy was asked to find num__0.777777777778 of a y . but he divided the y by num__0.777777777778 thus he got num__32 morethan theanswer . find the y . <o> a ) num__45 <o> b ) num__56 <o> c ) num__48 <o> d ) num__63 <o> e ) num__65 |
x / num__0.777777777778 = num__9 x / num__7 = num__32 + y actualy he should do num__7 x / num__9 = y y = ( num__9 x / num__7 ) - num__32 y = ( num__9 x - num__224 ) / num__7 ( num__9 x - num__224 ) / num__7 = num__7 x / num__9 num__81 y - num__2016 = num__49 y num__81 y - num__49 y = num__2016 num__32 y = num__2016 y = num__63 d <eor> d <eos> |
d |
multiply__32.0__7.0__ multiply__224.0__9.0__ subtract__81.0__32.0__ divide__2016.0__32.0__ divide__2016.0__32.0__ |
multiply__32.0__7.0__ multiply__224.0__9.0__ subtract__81.0__32.0__ divide__2016.0__32.0__ divide__2016.0__32.0__ |
| x y and z are consecutive numbers and x > y > z . also x + num__3 y + num__3 z = num__5 y + num__8 . what is the value of z ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
if x y and z are consecutive numbers and x > y > z then y = z + num__1 and x = z + num__2 . x + num__3 y + num__3 z = num__5 y + num__8 z + num__2 + num__3 z + num__3 + num__3 z = num__5 z + num__5 + num__8 num__2 z = num__8 z = num__4 the answer is c . <eor> c <eos> |
c |
subtract__3.0__1.0__ add__3.0__1.0__ add__3.0__1.0__ |
subtract__3.0__1.0__ add__3.0__1.0__ add__3.0__1.0__ |
| num__8 num__24 num__12 num__36 num__18 num__54 ( . . . . ) <o> a ) num__72 <o> b ) num__68 <o> c ) num__108 <o> d ) num__27 <o> e ) num__86 |
num__8 × num__3 = num__24 num__24 ÷ num__2 = num__12 num__12 × num__3 = num__36 num__36 ÷ num__2 = num__18 num__18 × num__3 = num__54 num__54 ÷ num__2 = num__27 answer is d . <eor> d <eos> |
d |
divide__24.0__8.0__ divide__24.0__12.0__ add__24.0__3.0__ add__24.0__3.0__ |
divide__24.0__8.0__ divide__24.0__12.0__ add__24.0__3.0__ add__24.0__3.0__ |
| the h . c . f . of two numbers is num__23 and the other two factors of their l . c . m . are num__9 and num__10 . the larger of the two numbers is : <o> a ) num__276 <o> b ) num__299 <o> c ) num__230 <o> d ) num__345 <o> e ) num__395 |
clearly the numbers are ( num__23 x num__9 ) and ( num__23 x num__10 ) . larger number = ( num__23 x num__10 ) = num__230 . answer : option c <eor> c <eos> |
c |
multiply__23.0__10.0__ multiply__23.0__10.0__ |
multiply__23.0__10.0__ multiply__23.0__10.0__ |
| what is the area of a square field whose diagonal of length num__20 m ? <o> a ) num__228 <o> b ) num__271 <o> c ) num__200 <o> d ) num__277 <o> e ) num__181 |
d num__1.0 = ( num__20 * num__20 ) / num__2 = num__200 answer : c <eor> c <eos> |
c |
multiply__200.0__1.0__ |
multiply__200.0__1.0__ |
| the average marks in mathematics scored by the pupils of a school at the public examination were num__39 . if four of these pupils who actually scored num__5 num__12 num__15 and num__19 marks at the examination had not been sent up the average marks for the school would have been num__44 . find the number of pupils sent up for examination from the school ? <o> a ) num__18 <o> b ) num__25 <o> c ) num__23 <o> d ) num__29 <o> e ) num__28 |
num__39 x = num__5 + num__12 + num__15 + num__19 + ( x – num__4 ) num__44 x = num__25 answer : b <eor> b <eos> |
b |
subtract__19.0__15.0__ subtract__44.0__19.0__ subtract__44.0__19.0__ |
subtract__19.0__15.0__ subtract__44.0__19.0__ subtract__44.0__19.0__ |
| if the arithmetic mean of p and q is num__10 and the arithmetic mean of q and r is num__27 what is the value of r - p ? <o> a ) num__20 <o> b ) num__10 <o> c ) num__34 <o> d ) num__40 <o> e ) num__5 |
arithmetic mean expression for p and q : ( p + q ) / num__2 = num__10 ; p + q = num__20 - - - - eq num__1 arithmetic mean expression for q and r : ( q + r ) / num__2 = num__20 ; q + r = num__54 - - - - eq num__2 subtracting eq num__1 from eq num__2 we get : r - p = num__34 hence the correct answer is c <eor> c <eos> |
c |
multiply__10.0__2.0__ multiply__27.0__2.0__ subtract__54.0__20.0__ multiply__1.0__34.0__ |
multiply__10.0__2.0__ multiply__27.0__2.0__ subtract__54.0__20.0__ divide__34.0__1.0__ |
| a sum fetched a total simple interest of rs . num__4016.25 at the rate of num__9 p . c . p . a . in num__5 years . what is the sum ? <o> a ) num__8925 <o> b ) num__8259 <o> c ) num__8529 <o> d ) num__8952 <o> e ) none |
sol . principal = rs . [ num__100 * num__4016.25 / num__9 * num__5 ] = rs . [ num__8925.0 ] = rs . num__8925 . answer a <eor> a <eos> |
a |
percent__100.0__8925.0__ |
percent__100.0__8925.0__ |
| on the coordinate plane points p and t are defined by the coordinates ( - num__10 ) and ( num__33 ) respectively and are connected to form a chord of a circle which also lies on the plane . if the area of the circle is ( num__6.25 ) π what are the coordinates of the center of the circle ? <o> a ) ( num__1.5 num__1 ) <o> b ) ( num__2 - num__5 ) <o> c ) ( num__00 ) <o> d ) ( num__1 num__1.5 ) <o> e ) ( num__22 ) |
although it took me num__3 mins to solve this question using all those equations later i thought this question can be solved easily using options . one property to keep in mind - a line passing through the centre of the circle bisects the chord ( or passes from the mid point of the chord ) . now mid point of chord here is ( - num__1 + num__3 ) / num__2 ( num__3 + num__0 ) / num__2 i . e . ( num__1 num__1.5 ) now luckily we have this in our ans . choice . so definitely this is the ans . it also indictaes that pt is the diameter of the circle . there can be a case when pt is not a diameter but in that case also the y - coordinate will remain same as it is the midpoint of the chord and we are moving up in the st . line to locate the centre of the circle . if ans choices are all distinct ( y cordinates ) only check for y cordinate and mark the ans = d <eor> d <eos> |
d |
triangle_area__1.0__3.0__ volume_cube__1.0__ |
triangle_area__1.0__3.0__ volume_cube__1.0__ |
| a and b start walking towards each other at num__4 pm at speed of num__2 kmph and num__3 kmph . they were initially num__15 km apart . at what time do they meet ? <o> a ) num__8 pm <o> b ) num__6 pm <o> c ) num__7 pm <o> d ) num__10 pm <o> e ) num__5 pm |
time of meeting = distance / relative speed = num__5.0 + num__2 = num__3.0 = num__3 hrs after num__4 pm = num__7 pm answer is c <eor> c <eos> |
c |
add__2.0__3.0__ add__4.0__3.0__ round__7.0__ |
add__2.0__3.0__ add__4.0__3.0__ add__4.0__3.0__ |
| how many seconds will a num__450 m long train take to cross a man walking with a speed of num__3 km / hr in the direction of the moving train if the speed of the train is num__63 km / hr ? <o> a ) num__27 sec <o> b ) num__30 sec <o> c ) num__86 sec <o> d ) num__16 sec <o> e ) num__18 sec |
speed of train relative to man = num__63 - num__3 = num__60 km / hr . = num__60 * num__0.277777777778 = num__16.6666666667 m / sec . time taken to pass the man = num__450 * num__0.06 = num__27 sec . answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ multiply__450.0__0.06__ round__27.0__ |
subtract__63.0__3.0__ multiply__450.0__0.06__ multiply__450.0__0.06__ |
| rani is two years older than banu who is twice as old as chitra . if the total of the age of rani banu and chitra be num__32 years then how old is banu ? <o> a ) num__7 years <o> b ) num__10 years <o> c ) num__12 years <o> d ) num__13 years <o> e ) num__14 years |
let chitra ' s age be x years . then banu ' s age = num__2 x years . rani ' s age = ( num__2 x + num__2 ) years . ( num__2 x + num__2 ) + num__2 x + x = num__32 num__5 x = num__30 x = num__6 hence banu ' s age = num__2 x = num__12 years . answer : c <eor> c <eos> |
c |
subtract__32.0__2.0__ divide__30.0__5.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
subtract__32.0__2.0__ divide__30.0__5.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
| the area of a parallelogram is num__72 sq m and its altitude is twice the corresponding base . then the length of the base is ? <o> a ) num__6 <o> b ) num__9 <o> c ) num__7 <o> d ) num__62 <o> e ) num__2 |
num__2 x * x = num__72 = > x = num__6 answer : a <eor> a <eos> |
a |
triangle_area__2.0__6.0__ |
triangle_area__2.0__6.0__ |
| the unit digit in the product ( num__891 * num__781 * num__912 * num__463 ) is : <o> a ) num__2 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
explanation : unit digit in the given product = unit digit in ( num__1 * num__1 * num__2 * num__3 ) = num__6 answer : c <eor> c <eos> |
c |
add__1.0__2.0__ multiply__2.0__3.0__ multiply__1.0__6.0__ |
add__1.0__2.0__ multiply__2.0__3.0__ multiply__1.0__6.0__ |
| if x is num__20 percent greater than num__12 then x = <o> a ) num__10.2 <o> b ) num__12.1 <o> c ) num__8.1 <o> d ) num__15.6 <o> e ) num__14.4 |
x is num__20.0 greater than num__12 means x is num__1.2 times num__12 ( in other words num__12 + num__0.2 * num__12 = num__1.2 * num__12 ) therefore x = num__1.2 * num__12 = num__14.4 answer : e <eor> e <eos> |
e |
multiply__12.0__1.2__ multiply__12.0__1.2__ |
multiply__12.0__1.2__ multiply__12.0__1.2__ |
| a train num__125 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is ? <o> a ) num__36 <o> b ) num__50 <o> c ) num__28 <o> d ) num__26 <o> e ) num__29 |
speed of the train relative to man = ( num__12.5 ) m / sec = ( num__12.5 ) m / sec . [ ( num__12.5 ) * ( num__3.6 ) ] km / hr = num__45 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__45 = = > x = num__50 km / hr answer : b <eor> b <eos> |
b |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ round__50.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ multiply__5.0__10.0__ |
| there are some people in party num__0.333333333333 rd left the party . then num__0.4 th of the remaining left the party then num__0.666666666667 rd of the remaining left the party . at last num__6 were remaining . how many people were in total ? <o> a ) num__45 <o> b ) num__27 <o> c ) num__28 <o> d ) num__26 <o> e ) num__91 |
sol : num__45 if x persons were there in total then x × ( num__1 – num__0.333333333333 ) × ( num__1 – num__0.4 ) × ( num__1 – num__0.666666666667 ) = num__6 x × num__0.666666666667 × num__0.6 × num__0.333333333333 = num__6 x = num__6 × num__5 × num__1.5 = num__45 answer : a <eor> a <eos> |
a |
add__0.3333__0.6667__ divide__0.4__0.6667__ subtract__6.0__1.0__ divide__0.6__0.4__ multiply__1.0__45.0__ |
add__0.3333__0.6667__ divide__0.4__0.6667__ subtract__6.0__1.0__ divide__0.6__0.4__ multiply__1.0__45.0__ |
| in a question on division with zero remainder a candidate took num__12 as divisor instead of num__21 . the quotient obtained by him was num__35 . the correct quotient is ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__40 <o> d ) num__43 <o> e ) num__45 |
number = ( num__35 * num__12 ) = num__420 correct quotient = num__20.0 = num__20 a ) <eor> a <eos> |
a |
multiply__12.0__35.0__ divide__420.0__21.0__ divide__420.0__21.0__ |
multiply__12.0__35.0__ divide__420.0__21.0__ divide__420.0__21.0__ |
| excluding stoppages the speed of a bus is num__70 km / hr and including stoppages it is num__40 km / hr . for how many minutes does the bus stop per hour ? <o> a ) num__118 min <o> b ) num__10 min <o> c ) num__18 min <o> d ) num__16 min <o> e ) num__25 min |
due to stoppages it covers num__30 km less . time taken to cover num__30 km = num__0.428571428571 * num__60 = num__25 min . answer : e <eor> e <eos> |
e |
subtract__70.0__40.0__ divide__30.0__70.0__ hour_to_min_conversion__ round__25.0__ |
subtract__70.0__40.0__ divide__30.0__70.0__ hour_to_min_conversion__ round__25.0__ |
| the speed of a train is num__120 kmph . what is the distance covered by it in num__10 minutes ? <o> a ) num__26 kmph <o> b ) num__20 kmph <o> c ) num__28 kmph <o> d ) num__30 kmph <o> e ) num__40 kmph |
num__120 * num__0.166666666667 = num__20 kmph answer : b <eor> b <eos> |
b |
round__20.0__ |
round__20.0__ |
| sreenivas sells a table to shiva at num__10.0 profit and shiva sells it to mahesh at num__10.0 loss . at what price did sreenivas purchase the table if mahesh paid rs . num__2178 ? <o> a ) num__2208 <o> b ) num__2201 <o> c ) num__2200 <o> d ) num__2287 <o> e ) num__2221 |
let the cost price of table for sreenivas be rs . x and given that cost price of table for mahesh = rs . num__2178 . = > ( num__90.0 ) of ( num__110.0 ) of x = rs . num__2178 . = > ( num__0.9 ) ( num__1.1 ) x = num__2178 = > x = ( num__2178 * num__100 ) / ( num__9 * num__11 ) = > x = rs . num__2200 . answer : c <eor> c <eos> |
c |
percent__10.0__90.0__ percent__10.0__110.0__ percent__100.0__2200.0__ |
percent__10.0__90.0__ percent__10.0__110.0__ percent__100.0__2200.0__ |
| after allowing a discount of num__15.0 on the marked price the selling price is $ num__6800 for an article . if it was sold at marked price there would have been a profit of num__60.0 . the cost price of the article is ? <o> a ) $ num__5400 <o> b ) $ num__5750 <o> c ) $ num__5000 <o> d ) $ num__7000 <o> e ) $ num__2000 |
c $ num__5000 given sp = $ num__6800 marked price = [ sp ( num__100 ) ] / ( num__100 - d % ) = ( num__6800 * num__100 ) / ( num__100 - num__15 ) = $ num__8000 if sp = $ num__8000 profit = num__60.0 cp = [ sp ( num__100 ) ] / ( num__100 + num__60 ) = ( num__8000 * num__100 ) / num__160 = $ num__5000 <eor> c <eos> |
c |
percent__100.0__5000.0__ |
percent__100.0__5000.0__ |
| if the cost price of num__20 articles is equal to the selling price of num__25 articles what is the % profit or loss made by the merchant ? <o> a ) num__25.0 loss <o> b ) num__25.0 profit <o> c ) num__20.0 loss <o> d ) num__20.0 profit <o> e ) num__5.0 profit |
let the cost price of num__1 article be $ num__1 . therefore cost price of num__20 articles = num__20 * num__1 = $ num__20 the selling price of num__25 articles = cost price of num__20 articles = $ num__20 . now we know the selling price of num__25 articles . let us find the cost price of num__25 articles . cost price of num__25 articles = num__25 * num__1 = $ num__25 . therefore profit made on sale of num__25 articles = selling price of num__25 articles - cost price of num__25 articles = num__20 - num__25 = - $ num__5 . as the profit is in the negative the merchant has made a loss of $ num__5 . therefore % loss = loss / cp * num__100.0 loss = - num__0.2 * num__100 = num__20.0 loss . answer : c <eor> c <eos> |
c |
percent__20.0__25.0__ percent__20.0__1.0__ percent__20.0__100.0__ |
percent__20.0__25.0__ percent__20.0__1.0__ percent__20.0__100.0__ |
| at a certain conference num__72.0 of the attendees registered at least two weeks in advance and paid their conference fee in full . if num__10.0 of the attendees who paid their conference fee in full did not register at least two weeks in advance what percent of conference attendees registered at least two weeks in advance ? <o> a ) num__18.0 <o> b ) num__62.0 <o> c ) num__79.2 <o> d ) num__80.0 <o> e ) num__82.0 % |
num__0.72 a - > at least num__2 weeks in advance and paid conference fee in full . fa - > paid the conference fee in full num__0.10 fa - > not num__2 weeks in advance paid but conference fee in full = > num__0.90 fa - > num__2 weeks in advance and paid the conference fee in full num__0.90 fa = num__0.72 a fa / a = num__0.8 * num__100 = num__80.0 answer - d <eor> d <eos> |
d |
percent__100.0__80.0__ |
percent__100.0__80.0__ |
| the effective annual rate of interest corresponding to a nominal rate of num__6.0 per annum payable half - yearly is ? <o> a ) num__6.06 <o> b ) num__6.07 <o> c ) num__6.08 <o> d ) num__6.09 <o> e ) num__6.19 % |
amount of rs . num__100 for num__1 year when compounded half - yearly = [ num__100 * ( num__1 + num__0.03 ) num__2 ] = rs . num__106.09 effective rate = ( num__106.09 - num__100 ) = num__6.09 answer : d <eor> d <eos> |
d |
percent__100.0__6.09__ |
percent__100.0__6.09__ |
| the weights of three boys are in the ratio num__4 : num__5 : num__6 . if the sum of the weights of the heaviest and the lightest boy is num__44 kg more than the weight of the third boy what is the weight of the lightest boy ? <o> a ) num__87.2 kg <o> b ) num__35.2 kg <o> c ) num__98.2 kg <o> d ) num__65.2 kg <o> e ) num__87.3 kg |
let the weights of the three boys be num__4 k num__5 k and num__6 k respectively . num__4 k + num__6 k = num__5 k + num__44 = > num__5 k = num__44 = > k = num__8.8 therefore the weight of the lightest boy = num__4 k = num__4 ( num__8.8 ) = num__35.2 kg . answer : b <eor> b <eos> |
b |
divide__44.0__5.0__ multiply__4.0__8.8__ multiply__4.0__8.8__ |
divide__44.0__5.0__ multiply__4.0__8.8__ multiply__4.0__8.8__ |
| you have three pills with you which you are required to take one after every thirty minutes . how long can the pills run for you ? <o> a ) num__45 min <o> b ) num__55 min <o> c ) num__50 min <o> d ) num__70 min <o> e ) num__60 min |
e num__60 min the pills can run for an hour . explanation : if you had ninety minutes in mind then you are wrong . you took the first pill at the zero minute . then you took the second pill after num__30 minutes and the third pill after num__30 + num__30 i . e . num__60 minutes or an hour . <eor> e <eos> |
e |
hour_to_min_conversion__ hour_to_min_conversion__ |
hour_to_min_conversion__ hour_to_min_conversion__ |
| john bought num__2 shares and sold them for $ num__84 each . if he had a profit of num__20.0 on the sale of one of the shares but a loss of num__20.0 on the sale of the other share then on the sale of both shares john had <o> a ) a profit of $ num__10 <o> b ) a profit of $ num__7 <o> c ) a loss of $ num__7 <o> d ) a loss of $ num__10 <o> e ) neither a profit nor a loss |
loss % = ( % age profit or loss / num__10 ) ^ num__2 = ( num__2.0 ) ^ num__2 = num__4.0 loss total selling price = num__84 * num__2 = $ num__168 total cost price = num__168 / ( num__0.96 ) = $ num__175 loss = num__175 - num__168 = $ num__7 answer : option c <eor> c <eos> |
c |
percent__4.0__175.0__ percent__4.0__175.0__ |
percent__4.0__175.0__ percent__4.0__175.0__ |
| a shopkeeper earns a profit of num__12.0 on selling a book at num__10.0 discount on the printed price . the ratio of the cost price and the printed price of the book is : <o> a ) num__45 : num__56 <o> b ) num__45 : num__51 <o> c ) num__47 : num__56 <o> d ) num__47 : num__51 <o> e ) none |
solution : let the cp be num__100 . hence sp = num__100 + num__12.0 of num__100 = num__112 . if the marked price be x then num__90.0 of x = num__112 = > x = ( num__112 * num__100 ) / num__90 = rs . num__124.444444444 hence required ratio = num__100 : num__124.444444444 = num__900 : num__1120 = num__45 : num__56 . answer : option a <eor> a <eos> |
a |
percent__45.0__124.4444__ percent__45.0__100.0__ |
percent__45.0__124.4444__ percent__45.0__100.0__ |
| a certain set of numbers has an average ( arithmetic mean ) of num__50 and a standard deviation of num__50.5 . if b and n two numbers in the set are both within num__2 standard deviations from the average then which of the following could be the sum of b and n ? <o> a ) - num__200 <o> b ) - num__130 <o> c ) - num__104 <o> d ) num__51 <o> e ) num__305 |
num__2 standard deviations from the average is frommean - num__2 * sdtomean + num__2 * sd thus from num__50 - num__2 * num__50.5 = num__51 to num__50 + num__2 * num__50.5 = num__151 : - num__51 < b < num__151 - num__51 < n < num__151 - num__102 < b + n < num__302 . only option d is in this range . answer : d . <eor> d <eos> |
d |
multiply__2.0__51.0__ multiply__2.0__151.0__ divide__102.0__2.0__ |
multiply__2.0__51.0__ multiply__2.0__151.0__ subtract__102.0__51.0__ |
| the batting average for num__40 innings of a cricket palyer is num__50 runs . his highest score exceeds his lowest score by num__172 runs . if these two innings are excluded the average of the remaining num__38 innings is num__48 runs . the highest score of the player is <o> a ) num__165 runs <o> b ) num__170 runs <o> c ) num__172 runs <o> d ) num__174 runs <o> e ) num__176 runs |
since scores are differ by num__172 let us consider the highest as x + num__172 lowest score would be x x + num__172 + x = num__176 num__2 x = num__4 x = num__2 so x + num__172 = num__174 answer : d <eor> d <eos> |
d |
subtract__40.0__38.0__ subtract__176.0__172.0__ add__172.0__2.0__ add__172.0__2.0__ |
subtract__40.0__38.0__ subtract__176.0__172.0__ add__172.0__2.0__ add__172.0__2.0__ |
| if the cost price is num__40.0 of selling price . then what is the profit percent . <o> a ) num__150.0 <o> b ) num__120.0 <o> c ) num__130.0 <o> d ) num__200.0 <o> e ) none of these |
explanation : let the s . p = num__100 then c . p . = num__40 profit = num__60 profit % = ( num__1.5 ) * num__100 = num__150.0 . answer : a <eor> a <eos> |
a |
percent__100.0__150.0__ |
percent__100.0__150.0__ |
| a river boat leaves silver town and travels upstream to gold town at an average speed of num__6 kilometers per hour . it returns by the same route at an average speed of num__8 kilometers per hour . what is the average speed for the round - trip in kilometers per hour ? <o> a ) num__6.8 <o> b ) num__7.1 <o> c ) num__7.2 <o> d ) num__7.5 <o> e ) num__8.0 |
pick a number which is lcm of num__8 and num__6 = num__24 . upstream time = num__4.0 = num__4 hrs downstream time = num__3.0 = num__3 hrs total time = num__7 hrs total distance = num__48 average speed = num__6.85714285714 = num__6.8 km / hr <eor> a <eos> |
a |
divide__24.0__6.0__ divide__24.0__8.0__ add__3.0__4.0__ multiply__6.0__8.0__ divide__48.0__7.0__ round__6.8__ |
divide__24.0__6.0__ divide__24.0__8.0__ add__3.0__4.0__ multiply__6.0__8.0__ divide__48.0__7.0__ round__6.8__ |
| what will be the cost of building a fence around a square plot with area equal to num__289 sq ft if the price per foot of building the fence is rs . num__58 ? <o> a ) rs . num__3944 <o> b ) rs . num__3948 <o> c ) rs . num__3942 <o> d ) rs . num__3965 <o> e ) rs . num__3929 |
let the side of the square plot be a ft . a num__2 = num__289 = > a = num__17 length of the fence = perimeter of the plot = num__4 a = num__68 ft . cost of building the fence = num__68 * num__58 = rs . num__3944 . answer : a <eor> a <eos> |
a |
square_perimeter__17.0__ multiply__58.0__68.0__ multiply__58.0__68.0__ |
multiply__17.0__4.0__ multiply__58.0__68.0__ multiply__58.0__68.0__ |
| dacid obtained num__81 num__65 num__82 num__67 and num__85 marks ( out of num__100 ) in english mathematics physics chemistry and biology . what are his average marks ? <o> a ) num__29 <o> b ) num__38 <o> c ) num__76 <o> d ) num__37 <o> e ) num__75 |
average = ( num__81 + num__65 + num__82 + num__67 + num__85 ) / num__5 = num__76 . answer : c <eor> c <eos> |
c |
subtract__81.0__5.0__ subtract__81.0__5.0__ |
subtract__81.0__5.0__ subtract__81.0__5.0__ |
| the smallest number which when diminished by num__6 is divisible num__12 num__14 num__16 num__18 and num__22 is : <o> a ) num__11099 <o> b ) num__11094 <o> c ) num__11091 <o> d ) num__10323 <o> e ) num__10483 |
required number = ( l . c . m . of num__12 num__14 num__16 num__18 num__22 ) + num__6 = num__11088 + num__6 = num__11094 answer : option b <eor> b <eos> |
b |
add__6.0__11088.0__ add__6.0__11088.0__ |
add__6.0__11088.0__ add__6.0__11088.0__ |
| num__4 num__2 - num__4 - num__8 num__32 num__26 - num__156 ? <o> a ) - num__134 <o> b ) - num__144 <o> c ) - num__154 <o> d ) - num__164 <o> e ) - num__174 |
num__4 - num__2 = num__2 num__2 * - num__2 = - num__4 - num__4 - num__4 = - num__8 - num__8 * - num__4 = num__32 num__32 - num__6 = num__26 num__26 * - num__6 = - num__156 - num__156 - num__8 = - num__164 answer : d <eor> d <eos> |
d |
add__4.0__2.0__ add__8.0__156.0__ add__8.0__156.0__ |
subtract__8.0__2.0__ add__8.0__156.0__ add__8.0__156.0__ |
| a batsman in his num__10 th innings makes a score of num__60 and thereby increases his average by num__3 . what is his average after the num__10 th innings ? he had never been ’ not out ’ . <o> a ) num__47 <o> b ) num__37 <o> c ) num__39 <o> d ) num__43 <o> e ) num__33 |
average score before num__10 th innings = num__60 - num__3 × num__10 = num__30 average score after num__10 th innings = > num__30 + num__3 = num__33 answer : e <eor> e <eos> |
e |
multiply__10.0__3.0__ add__3.0__30.0__ add__3.0__30.0__ |
multiply__10.0__3.0__ add__3.0__30.0__ add__3.0__30.0__ |
| a fruit store sells four varieties of mangoes . a b c & d . if a costs num__20.0 less than b and c costs num__25.0 less than a and d costs num__35.0 less than c what percent of b ' s cost is d ? <o> a ) num__40.0 <o> b ) num__39.0 <o> c ) num__44.0 <o> d ) num__36.0 <o> e ) num__42 % |
soln : - b = num__100 a num__20.0 less than b = num__80 c num__25.0 less than a = num__60 d num__35.0 less than c = num__39 d is what percent of b = d / b * num__100 = num__0.39 * num__100 = num__39.0 answer : b <eor> b <eos> |
b |
percent__100.0__39.0__ |
percent__100.0__39.0__ |
| a woodworker normally makes a certain number of parts in num__24 days . but he was able to increase his productivity by num__5 parts per day and so he not only finished the job in only num__22 days but also he made num__80 extra parts . how many parts does the woodworker normally makes per day and how many pieces does he make in num__24 days ? <o> a ) num__360 <o> b ) num__480 <o> c ) num__754 <o> d ) num__687 <o> e ) num__147 |
let x be the number of parts the woodworker normally makes daily . in num__24 days he makes num__24 ⋅ x pieces . his new daily production rate is x + num__5 pieces and in num__22 days he made num__22 ⋅ ( x + num__5 ) parts . this is num__80 more than num__24 ⋅ x . therefore the equation is : num__24 ⋅ x + num__80 = num__22 ( x + num__5 ) num__30 = num__2 x num__3 x = num__15 normally he makes num__15 parts a day and in num__24 days he makes num__15 ⋅ num__24 = num__360 parts . correct answer is a ) num__360 <eor> a <eos> |
a |
subtract__24.0__22.0__ subtract__5.0__2.0__ multiply__5.0__3.0__ multiply__24.0__15.0__ multiply__24.0__15.0__ |
subtract__24.0__22.0__ subtract__5.0__2.0__ multiply__5.0__3.0__ multiply__24.0__15.0__ multiply__24.0__15.0__ |
| num__6 x − num__12 = num__6 y num__4 y + num__4 x = num__12 which of the following is the number of solutions to the system of equations shown above ? <o> a ) more than three <o> b ) exactly one <o> c ) exactly two <o> d ) exactly three <o> e ) none |
num__6 x − num__12 = num__6 y = > num__6 x - num__6 y = num__12 = > x - y = num__2 - - num__1 num__4 y + num__4 x = num__12 = > x + y = num__3 - - num__2 from equation num__1 and num__2 we get num__2 x = num__5 = > x = num__2.5 y = . num__5 therefore the given system will have exactly one solution answer : b <eor> b <eos> |
b |
subtract__6.0__4.0__ divide__6.0__2.0__ subtract__6.0__1.0__ divide__5.0__2.0__ reverse__1.0__ |
subtract__6.0__4.0__ subtract__4.0__1.0__ subtract__6.0__1.0__ divide__5.0__2.0__ subtract__6.0__5.0__ |
| num__115 liters of a mixture of milk and water contains in the ratio num__3 : num__2 . how much water should now be added so that the ratio of milk and water becomes num__3 : num__4 ? <o> a ) num__12 liters <o> b ) num__32 liters <o> c ) num__46 liters <o> d ) num__50 liters <o> e ) num__34 liters |
milk = num__0.6 * num__115 = num__69 liters water = num__46 liters num__69 : ( num__46 + p ) = num__3 : num__4 num__138 + num__3 p = num__276 = > p = num__46 num__46 liters of water are to be added for the ratio become num__3 : num__4 . answer : c <eor> c <eos> |
c |
multiply__115.0__0.6__ subtract__115.0__69.0__ multiply__3.0__46.0__ multiply__2.0__138.0__ subtract__115.0__69.0__ |
multiply__115.0__0.6__ subtract__115.0__69.0__ multiply__3.0__46.0__ multiply__2.0__138.0__ subtract__115.0__69.0__ |
| let the number which when multiplied by num__12 is increased by num__198 . <o> a ) num__18 <o> b ) num__20 <o> c ) num__26 <o> d ) num__28 <o> e ) num__30 |
solution let the number be x . then num__12 x - x = num__198 ‹ = › num__11 x = num__198 x ‹ = › num__18 . answer a <eor> a <eos> |
a |
divide__198.0__11.0__ divide__198.0__11.0__ |
divide__198.0__11.0__ divide__198.0__11.0__ |
| the length of the bridge which a train num__130 metres long and travelling at num__45 km / hr can cross in num__30 seconds is : <o> a ) num__332 <o> b ) num__268 <o> c ) num__245 <o> d ) num__276 <o> e ) num__191 |
speed = [ num__45 x num__0.277777777778 ] m / sec = [ num__12.5 ] m / sec time = num__30 sec let the length of bridge be x metres . then ( num__130 + x ) / num__30 = num__12.5 = > num__2 ( num__130 + x ) = num__750 = > x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| how many seconds will a train num__100 meters long take to cross a bridge num__250 meters long if the speed of the train is num__36 kmph ? <o> a ) num__54 sec <o> b ) num__35 sec <o> c ) num__25 sec <o> d ) num__45 sec <o> e ) num__24 sec |
d = num__100 + num__250 = num__350 s = num__36 * num__0.277777777778 = num__10 mps t = num__35.0 = num__35 sec answer : b <eor> b <eos> |
b |
add__100.0__250.0__ divide__350.0__10.0__ round__35.0__ |
add__100.0__250.0__ divide__350.0__10.0__ round__35.0__ |
| in a shower num__5 cm of rain falls . the volume of water that falls on num__1.5 hectares of ground is : <o> a ) num__750 cu . m <o> b ) num__75 cu . m <o> c ) num__4500 cu . m <o> d ) num__500 cu . m <o> e ) num__600 cu . m |
hectare = num__10000 m ^ num__2 so area = ( num__1.5 * num__10000 ) m ^ num__2 = num__15000 m ^ num__2 depth = num__0.05 m = num__0.05 m volume = ( area * depth ) = num__15000 * num__0.05 = num__750 m ^ num__3 answer a <eor> a <eos> |
a |
multiply__1.5__10000.0__ multiply__0.05__15000.0__ subtract__5.0__2.0__ round__750.0__ |
multiply__1.5__10000.0__ multiply__0.05__15000.0__ multiply__1.5__2.0__ multiply__0.05__15000.0__ |
| find the area of a parallelogram with base num__30 cm and height num__12 cm ? <o> a ) num__290 cm num__2 <o> b ) num__360 cm num__2 <o> c ) num__270 cm num__2 <o> d ) num__280 cm num__2 <o> e ) num__260 cm num__2 |
area of a parallelogram = base * height = num__30 * num__12 = num__360 cm num__2 answer : b <eor> b <eos> |
b |
multiply__30.0__12.0__ multiply__30.0__12.0__ |
multiply__30.0__12.0__ multiply__30.0__12.0__ |
| if num__2994 Ã · num__14.5 = num__179 then num__29.94 Ã · num__1.45 = ? <o> a ) num__17.1 <o> b ) num__17.3 <o> c ) num__17.5 <o> d ) num__17.7 <o> e ) num__17.9 |
num__29.94 / num__1.45 = num__299.4 / num__14.5 = ( num__2994 / num__14.5 ) x num__0.1 ) [ here substitute num__179 in the place of num__2994 / num__14.5 ] = num__17.9 = num__17.9 answer is e . <eor> e <eos> |
e |
divide__29.94__299.4__ multiply__179.0__0.1__ multiply__179.0__0.1__ |
divide__29.94__299.4__ multiply__179.0__0.1__ multiply__179.0__0.1__ |
| evaluate : num__30 - num__12 * num__3 * num__2 = ? <o> a ) num__62 <o> b ) num__52 <o> c ) num__32 <o> d ) num__12 <o> e ) num__22 |
according to order of operations num__12 ? num__3 ? num__2 ( division and multiplication ) is done first from left to right num__12 * * num__2 = num__4 * num__2 = num__8 hence num__30 - num__12 * num__3 * num__2 = num__30 - num__8 = num__22 correct answer e <eor> e <eos> |
e |
divide__12.0__3.0__ subtract__12.0__4.0__ subtract__30.0__8.0__ subtract__30.0__8.0__ |
divide__12.0__3.0__ subtract__12.0__4.0__ subtract__30.0__8.0__ subtract__30.0__8.0__ |
| tough and tricky questions : number properties . what is the smallest positive integer x such that num__324 x is the cube of a positive integer ? <o> a ) num__9 <o> b ) num__12 <o> c ) num__18 <o> d ) num__24 <o> e ) num__36 |
we want to know the smallest x that will make num__324 x a cube of some number . let ' s call that number y . let ' s first figure out what we ' re working with . the prime factorization of num__324 can be visualized : . . . . . . . . . . . num__324 . . . . . . . . / . . . . . . . \ . . . . . . num__81 . . . . . . . num__4 . . . . . / . . \ . . . . . . / . . . \ . . . num__27 . . . num__3 . . . num__2 . . . . . num__2 . . . / . . \ . . num__9 . . . . num__3 / \ . num__3 num__3 so we have num__3 * num__3 * num__3 * num__3 * num__2 * num__2 that can be multiplied together to get num__324 . now we need to figure out what we need to make num__324 * x into a cube of y ( y ^ num__3 = num__324 * x ) . we have four num__3 s and two num__2 s . to arrange these numbers in identical triples ( num__2 num__33 ) we need at least one more num__2 and two num__3 ' s . each of these triples will give us the value of y ( num__2 * num__3 * num__3 = num__18 ) which multiplied by itself three times gives us num__324 * x . looking at the factors we need to complete the triples we get num__3 * num__3 * num__2 = num__18 . we know this is the smallest number possible because prime factors by definition can not be broken down any further . therefore we can go with answer choice c . if time permits we can do a sanity check . we calculated that y should be num__2 * num__3 * num__3 or num__18 . num__18 * num__18 * num__18 = num__5832 . also num__324 * num__18 = num__5832 . answer : c <eor> c <eos> |
c |
power__3.0__2.0__ triangle_perimeter__2.0__4.0__27.0__ multiply__2.0__9.0__ volume_cube__18.0__ multiply__2.0__9.0__ |
power__3.0__2.0__ triangle_perimeter__2.0__4.0__27.0__ multiply__2.0__9.0__ multiply__324.0__18.0__ multiply__2.0__9.0__ |
| bag a contains red white and blue marbles such that the red to white marble ratio is num__1 : num__3 and the white to blue marble ratio is num__2 : num__3 . bag b contains red and white marbles in the ratio of num__1 : num__4 . together the two bags contain num__26 white marbles . how many red marbles could be in bag a ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__4 <o> d ) num__2 <o> e ) num__8 |
# of red marbles in bag a can be either num__2 or num__6 . no num__2 in the choices so num__6 . d . bag a : r : w : b = num__2 : num__6 : num__9 bag b r : w = num__1 : num__4 num__6 x + num__4 y = num__26 i . e num__3 x + num__2 y = num__13 x has to be odd to make an odd sum from the eq . x = num__1 y = num__5 or x = num__3 y = num__2 so r can be num__2 x i . e num__2 or num__6 . answer d <eor> d <eos> |
d |
multiply__3.0__2.0__ add__3.0__6.0__ add__4.0__9.0__ add__1.0__4.0__ multiply__1.0__2.0__ |
add__2.0__4.0__ add__3.0__6.0__ add__4.0__9.0__ add__1.0__4.0__ multiply__1.0__2.0__ |
| the price of an article has been reduced by num__25.0 . in order to restore the original price the new price must be increased by ? <o> a ) num__33 num__0.333333333333 % <o> b ) num__9 num__0.0909090909091 % <o> c ) num__11 num__0.111111111111 % <o> d ) num__66 num__0.666666666667 % <o> e ) none |
num__100 num__75 - - - - - - - num__75 - - - - - - - num__25 num__100 - - - - - - ? = > num__33 num__0.333333333333 % answer : a <eor> a <eos> |
a |
percent__100.0__33.0__ |
percent__100.0__33.0__ |
| eesha bought num__18 sharpeners for rs . num__100 . she paid num__1 rupee more for each white sharpener than for each brown sharpener . what is the price of a white sharpener and how many white sharpener did she buy ? <o> a ) rs . num__6 num__19 <o> b ) rs . num__6 num__10 <o> c ) rs . num__6 num__16 <o> d ) rs . num__6 num__11 <o> e ) rs . num__6 num__12 |
explanation : just check the options . if she bought num__10 white sharpeners at rs . num__6 per piece she has spent rs . num__60 already . and with the remaining rs . num__40 she bought num__8 brown sharpeners at num__5.0 = rs . num__5 which is rs . num__1 less than white sharpener . sol : b <eor> b <eos> |
b |
multiply__6.0__10.0__ subtract__100.0__60.0__ subtract__18.0__10.0__ subtract__6.0__1.0__ add__1.0__5.0__ |
multiply__6.0__10.0__ subtract__100.0__60.0__ subtract__18.0__10.0__ subtract__6.0__1.0__ add__1.0__5.0__ |
| tom drives from town a to town b driving at a constant speed of num__60 miles per hour . from town b tom immediately continues to town c . the distance between a and b is twice the distance between b and c . if the average speed of the whole journey was num__36 mph then what is tom ' s speed driving from b to c in miles per hour ? <o> a ) num__12 <o> b ) num__20 <o> c ) num__24 <o> d ) num__30 <o> e ) num__36 |
let ' s assume that it takes num__4 hours to go from point a to b . then the distance between them becomes num__240 which makes distance between b and c num__120 . ( num__240 + num__120 ) / ( num__4 + x ) gives us the average speed which is num__36 . you find x = num__6 . so the question simplifies itself to num__20.0 = num__20 hence the answer is b <eor> b <eos> |
b |
multiply__60.0__4.0__ divide__120.0__6.0__ round__20.0__ |
multiply__60.0__4.0__ divide__120.0__6.0__ divide__120.0__6.0__ |
| num__66 cubic centimetres of silver is drawn into a wire num__1 mm in diameter . the length of the wire in metres will be : <o> a ) num__84 <o> b ) num__90 <o> c ) num__94 <o> d ) num__95 <o> e ) num__65 |
let the length of the wire be h . radium = num__0.5 mm = num__0.05 cm then num__3.14285714286 * num__0.05 * num__0.05 * h = num__66 h = ( num__66 * num__20 * num__20 * num__0.318181818182 ) = num__8400 cm = num__84 m answer a <eor> a <eos> |
a |
divide__1.0__0.05__ divide__1.0__3.1429__ round__84.0__ |
divide__1.0__0.05__ divide__1.0__3.1429__ multiply__1.0__84.0__ |
| the length of the bridge which a train num__130 metres long and travelling at num__45 km / hr can cross in num__30 seconds is ? <o> a ) num__776 <o> b ) num__283 <o> c ) num__245 <o> d ) num__266 <o> e ) num__288 |
speed = [ num__45 x num__0.277777777778 ] m / sec = [ num__12.5 ] m / sec time = num__30 sec let the length of bridge be x metres . then ( num__130 + x ) / num__30 = num__12.5 = > num__2 ( num__130 + x ) = num__750 = > x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| if x is num__20 percent greater than num__98 then x = <o> a ) num__68 <o> b ) num__70.4 <o> c ) num__86 <o> d ) num__105.6 <o> e ) num__117.6 |
x = num__98 * num__1.2 = num__117.6 so the answer is e . <eor> e <eos> |
e |
multiply__98.0__1.2__ multiply__98.0__1.2__ |
multiply__98.0__1.2__ multiply__98.0__1.2__ |
| two trains of length num__150 m and num__250 m are num__100 m apart . they start moving towards each other on parallel tracks at speeds num__18 kmph and num__36 kmph . in how much time will the trains cross each other ? <o> a ) num__12.4285714286 sec <o> b ) num__11.4285714286 sec <o> c ) num__33.3333333333 sec <o> d ) num__16.0 sec <o> e ) num__2.85714285714 sec |
relative speed = ( num__18 + num__36 ) * num__0.277777777778 = num__3 * num__5 = num__15 mps . the time required = d / s = ( num__100 + num__150 + num__250 ) / num__15 = num__33.3333333333 = num__33.3333333333 sec . answer : c <eor> c <eos> |
c |
subtract__18.0__3.0__ divide__100.0__3.0__ divide__100.0__3.0__ |
multiply__3.0__5.0__ divide__100.0__3.0__ divide__100.0__3.0__ |
| it is being given that ( num__232 + num__1 ) is completely divisible by a whole number . which of the following numbers is completely divisible by this number ? <o> a ) ( num__216 + num__1 ) <o> b ) ( num__216 - num__1 ) <o> c ) ( num__7 x num__223 ) <o> d ) ( num__296 + num__1 ) <o> e ) none of these |
explanation : let num__232 = x . then ( num__232 + num__1 ) = ( x + num__1 ) . let ( x + num__1 ) be completely divisible by the natural number n . then ( num__296 + num__1 ) = [ ( num__232 ) num__3 + num__1 ] = ( x num__3 + num__1 ) = ( x + num__1 ) ( x num__2 - x + num__1 ) which is completely divisible by n since ( x + num__1 ) is divisible by n . answer is d <eor> d <eos> |
d |
subtract__3.0__1.0__ multiply__1.0__296.0__ |
subtract__3.0__1.0__ multiply__1.0__296.0__ |
| a room is num__5 meters num__76 centimeters in length and num__4 meters num__32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the room . <o> a ) num__12 <o> b ) num__15 <o> c ) num__17 <o> d ) num__137 <o> e ) num__147 |
let us calculate both the length and width of the room in centimeters . length = num__5 meters and num__76 centimeters = num__576 cm width = num__4 meters and num__32 centimeters = num__432 cm as we want the least number of square tiles required it means the length of each square tile should be as large as possible . further the length of each square tile should be a factor of both the length and width of the room . hence the length of each square tile will be equal to the hcf of the length and width of the room = hcf of num__576 and num__432 = num__144 thus the number of square tiles required = ( num__576 x num__432 ) / ( num__144 x num__144 ) = num__4 x num__3 = num__12 answer : b <eor> b <eos> |
b |
divide__576.0__4.0__ divide__432.0__144.0__ multiply__4.0__3.0__ multiply__5.0__3.0__ |
divide__576.0__4.0__ divide__432.0__144.0__ multiply__4.0__3.0__ multiply__5.0__3.0__ |
| a can complete a work in num__15 days and b can do the same work in num__9 days . if a after doing num__5 days leaves the work find in how many days b will do the remaining work ? <o> a ) num__2 days <o> b ) num__4 days <o> c ) num__6 days <o> d ) num__7 days <o> e ) num__10 days |
the required answer = ( num__15 - num__5 ) * num__0.6 = num__4.0 = num__6 days answer is c <eor> c <eos> |
c |
km_to_mile_conversion__ subtract__9.0__5.0__ subtract__15.0__9.0__ round__6.0__ |
km_to_mile_conversion__ subtract__9.0__5.0__ subtract__15.0__9.0__ subtract__15.0__9.0__ |
| ( x ) + num__1789 + num__9011 - num__7819 = num__90510 . calculate the value of x <o> a ) num__87501 <o> b ) num__87599 <o> c ) num__87529 <o> d ) num__87520 <o> e ) num__87521 |
x + num__1789 + num__9011 - num__7819 = num__90510 = x + num__1789 + num__9011 = num__90510 + num__7819 = x + num__10800 = num__98329 = x = num__98329 - num__10800 = num__87529 answer is c <eor> c <eos> |
c |
add__1789.0__9011.0__ add__7819.0__90510.0__ subtract__98329.0__10800.0__ subtract__98329.0__10800.0__ |
add__1789.0__9011.0__ add__7819.0__90510.0__ subtract__98329.0__10800.0__ subtract__98329.0__10800.0__ |
| a train num__240 m long passed a pole in num__24 sec . how long will it take to pass a platform num__630 m long ? <o> a ) num__22 <o> b ) num__89 <o> c ) num__277 <o> d ) num__278 <o> e ) num__87 |
speed = num__10.0 = num__10 m / sec . required time = ( num__240 + num__630 ) / num__10 = num__87 sec . answer : e <eor> e <eos> |
e |
divide__240.0__24.0__ round__87.0__ |
divide__240.0__24.0__ round__87.0__ |
| meg and bob are among the num__5 participants in a cycling race . if each participant finishes the race and no two participants finish at the same time in how many different possible orders can the participants finish the race so that meg finishes ahead of bob ? <o> a ) num__24 <o> b ) num__30 <o> c ) num__60 <o> d ) num__90 <o> e ) num__120 |
let me redeem myself : jxxxx = > bob can be in any x . . . total num__4 ! = num__24 xjxxx = > bob can be in the last num__3 xs . . . total num__3 ! = num__6 . however since the first person can change too and it can not be bob it can only be num__3 other people . so total = num__6 * num__3 = num__18 xxjbx = > num__3 ! = num__6 total xxjxb = > num__3 ! = num__6 total so this yields total of num__12 xxxjb = > num__3 ! = num__6 ans = num__24 + num__18 + num__12 + num__6 = num__60 answer is c <eor> c <eos> |
c |
die_space__ choose__6.0__4.0__ choose__6.0__4.0__ |
die_space__ choose__6.0__4.0__ choose__6.0__4.0__ |
| in a cricket team the average of eleven players is num__28 years . out of these the average ages of three groups of three players each are num__31 years num__28 years and num__30 years respectively . if in these groups the captain and the youngest player are not included and the captain is eleven years older than the youngest player what is the age of the captain ? <o> a ) num__33 years <o> b ) num__34 years <o> c ) num__35 years <o> d ) num__26 years <o> e ) num__37 years |
explanation : let the age of youngest player be x . then age of the captain = ( x + num__11 ) . = > num__3 * num__31 + num__3 * num__28 + num__3 * num__30 + x + x + num__11 = num__11 * num__28 . = > num__93 + num__84 + num__90 + num__2 x + num__11 = num__308 so num__2 x = num__30 so x = num__15 . age of the captain = ( x + num__11 ) = num__26 years . answer : d <eor> d <eos> |
d |
subtract__31.0__28.0__ multiply__31.0__3.0__ multiply__28.0__3.0__ multiply__30.0__3.0__ subtract__30.0__28.0__ multiply__28.0__11.0__ divide__30.0__2.0__ subtract__28.0__2.0__ subtract__28.0__2.0__ |
subtract__31.0__28.0__ multiply__31.0__3.0__ multiply__28.0__3.0__ multiply__30.0__3.0__ subtract__30.0__28.0__ multiply__28.0__11.0__ divide__30.0__2.0__ add__11.0__15.0__ add__11.0__15.0__ |
| which of the following inequalities indicates the set of all values of d for which the lengths w of the three sides of a triangle can be num__34 and d ? <o> a ) num__0 < d < num__1 <o> b ) num__0 < d < num__5 <o> c ) num__0 < d < num__7 <o> d ) num__1 < d < num__5 <o> e ) num__1 < d < num__7 |
this question is a good way to apply one of the most basic relation between the num__3 sides of a triangle . in a triangle ( any triangle ) any side must be greater than the positive difference of the other two sides and less than than the sum of the other num__2 sides . let the sides of a triangle be a b c . thus | a - b | < c < a + b | b - c | < a < b + c | c - a | < b < a + c thus if the sides of the triangle are num__34 and d num__4 - num__3 < d < num__4 + num__3 - - - > num__1 < d < num__7 . thus e is the correct answer . <eor> e <eos> |
e |
subtract__3.0__2.0__ add__3.0__4.0__ reverse__1.0__ |
subtract__3.0__2.0__ add__3.0__4.0__ subtract__2.0__1.0__ |
| at what rate percent on simple interest will rs . num__750 amount to rs . num__1050 in num__5 years ? <o> a ) num__8 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
num__300 = ( num__750 * num__5 * r ) / num__100 r = num__8.0 . answer : a <eor> a <eos> |
a |
percent__8.0__100.0__ |
percent__8.0__100.0__ |
| a father said to his son ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is num__50 years now the son ' s age five years back was : <o> a ) num__15 years . <o> b ) num__20 years . <o> c ) num__25 years . <o> d ) num__30 years . <o> e ) num__35 years . |
let the son ' s present age be x years . then ( num__50 - x ) = x num__2 x = num__50 . x = num__25 . son ' s age num__5 years back ( num__25 - num__5 ) = num__20 years . answer : option b <eor> b <eos> |
b |
divide__50.0__2.0__ subtract__25.0__5.0__ subtract__25.0__5.0__ |
divide__50.0__2.0__ subtract__25.0__5.0__ subtract__25.0__5.0__ |
| a sum of money is to be distributed among a b c d in the proportion of num__6 : num__3 : num__5 : num__4 . if c gets rs . num__1000 more than d what is b ' s share ? <o> a ) num__1500 <o> b ) num__2000 <o> c ) num__2500 <o> d ) num__2800 <o> e ) num__3000 |
let the shares of a b c and d be rs . num__5 x rs . num__3 x rs . num__5 x and rs . num__4 x respectively . then num__5 x - num__4 x = num__1000 x = num__1000 . b ' s share = rs . num__3 x = rs . ( num__3 x num__1000 ) = rs . num__3000 . answer : e <eor> e <eos> |
e |
multiply__3.0__1000.0__ multiply__3.0__1000.0__ |
multiply__3.0__1000.0__ multiply__3.0__1000.0__ |
| mary jogs num__9 km at a speed of num__6 kmph . at what speed would she need to jog during the next num__1.5 hour to have an average of num__9 km per hr at the entire jogging session ? <o> a ) num__12 kmph <o> b ) num__7 kmph <o> c ) num__87 kmph <o> d ) num__74 kmph <o> e ) num__78 kmph |
let speed of the jogging be x kmph total time taken = ( num__1.5 + num__1.5 ) = num__3 hrs total distance covered = ( num__9 + num__1.5 x ) num__9 + num__1.5 x / num__3 = num__9 num__9 + num__1.5 x = num__27 num__1.5 x = num__18 x = num__18 * num__0.666666666667 = num__12 kmph answer ( a ) <eor> a <eos> |
a |
subtract__9.0__6.0__ multiply__9.0__3.0__ multiply__6.0__3.0__ divide__6.0__9.0__ add__9.0__3.0__ round__12.0__ |
subtract__9.0__6.0__ multiply__9.0__3.0__ multiply__6.0__3.0__ divide__6.0__9.0__ add__9.0__3.0__ add__9.0__3.0__ |
| the average ( arithmetic mean ) of y numbers is x . if num__15 is added to the set of numbers then the average will be x - num__5 . what is the value of y in terms of x ? <o> a ) x / num__6 - num__6 <o> b ) x / num__6 - num__5 <o> c ) x / num__7 - num__5 <o> d ) x / num__5 - num__4 <o> e ) x / num__5 - num__6 |
( a num__1 + a num__2 + . . + ay ) / y = x ( a num__1 + a num__2 + . . + ay + num__15 ) / ( y + num__1 ) = x - num__5 = > ( xy + num__15 ) / ( y + num__1 ) = x - num__5 = > xy + num__15 = yx - num__5 y + x - num__5 = > num__20 = x - num__5 y = > num__5 y = x - num__20 = > y = x / num__5 - num__4 answer - d <eor> d <eos> |
d |
add__15.0__5.0__ subtract__5.0__1.0__ multiply__5.0__1.0__ |
add__15.0__5.0__ subtract__5.0__1.0__ divide__5.0__1.0__ |
| a train num__280 m long running with a speed of num__63 km / hr will pass a tree in ? <o> a ) num__11 <o> b ) num__16 <o> c ) num__188 <o> d ) num__199 <o> e ) num__12 |
speed = num__63 * num__0.277777777778 = num__17.5 m / sec time taken = num__280 * num__0.0571428571429 = num__16 sec answer : b <eor> b <eos> |
b |
divide__280.0__17.5__ round__16.0__ |
divide__280.0__17.5__ divide__280.0__17.5__ |
| a train num__125 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 sec . the speed of the train is ? <o> a ) num__28 <o> b ) num__50 <o> c ) num__99 <o> d ) num__121 <o> e ) num__81 |
speed of the train relative to man = num__12.5 = num__12.5 m / sec . = num__12.5 * num__3.6 = num__45 km / hr let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__45 = > x = num__50 km / hr . answer : b <eor> b <eos> |
b |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ round__50.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ multiply__5.0__10.0__ |
| b @ k i num__9 d  © num__5 num__2 t $ m num__4 e j # u r num__1 a Î ´ k num__5 w num__6.0 f q num__7 h p num__8 z how many such consonants are there in the above arrangement each of which is immediately preceded by number and immediately followed by a symbol ? <o> a ) none <o> b ) one <o> c ) two <o> d ) three <o> e ) more than three |
explanation : number consonant symbol such combinations are : num__9 d  © : num__2 t $ answer is c <eor> c <eos> |
c |
coin_space__ |
coin_space__ |
| a bat is bought for rs . num__400 and sold at a gain of num__25.0 find its selling price ? <o> a ) s . num__460 / - <o> b ) s . num__480 / - <o> c ) s . num__500 / - <o> d ) s . num__520 / - <o> e ) s . num__540 / - |
num__100.0 - - - - - - > num__400 ( num__100 * num__4 = num__400 ) num__120.0 - - - - - - > num__500 ( num__125 * num__4 = num__500 ) selling price = rs . num__500 / - option ' c ' <eor> c <eos> |
c |
percent__25.0__400.0__ percent__25.0__500.0__ percent__100.0__500.0__ |
percent__25.0__400.0__ percent__25.0__500.0__ percent__100.0__500.0__ |
| the average age of father and his two sons is num__27 years . five years ago the average age of the two sons was num__12 years . if the difference between the ages of the two sons is four years what is the present age of the father ? <o> a ) num__45 <o> b ) num__46 <o> c ) num__47 <o> d ) num__48 <o> e ) num__49 |
the total present age of father and two sons is num__3 s num__27 = num__81 yrs the total present age of sons is ( num__12 + num__5 ) x num__2 = num__34 years so present age of father is num__81 – num__34 = num__47 yrs answer c <eor> c <eos> |
c |
multiply__27.0__3.0__ subtract__5.0__3.0__ subtract__81.0__34.0__ subtract__81.0__34.0__ |
multiply__27.0__3.0__ subtract__5.0__3.0__ subtract__81.0__34.0__ subtract__81.0__34.0__ |
| the sector of a circle has radius of num__21 cm and central angle num__150 o . find its perimeter ? <o> a ) num__91.5 <o> b ) num__91.4 <o> c ) num__97 <o> d ) num__91.3 <o> e ) num__91.1 |
perimeter of the sector = length of the arc + num__2 ( radius ) = ( num__0.416666666667 * num__2 * num__3.14285714286 * num__21 ) + num__2 ( num__21 ) = num__55 + num__42 = num__97 cm answer : c <eor> c <eos> |
c |
multiply__21.0__2.0__ add__42.0__55.0__ round__97.0__ |
multiply__21.0__2.0__ add__42.0__55.0__ add__42.0__55.0__ |
| the no . of gift pack bought by dexter is num__1 more than the price in rupees of each gift pack . the amount of rs . num__380 which dexter had fell short of the required amount . what is theamount by which he might have fallen short . <o> a ) num__20 <o> b ) num__40 rs <o> c ) num__50 <o> d ) num__60 <o> e ) num__90 |
let the price of gift pack be ' aa ' then number of packs bought = a + num__1 = a + num__1 hence total cost is a ( a + num__1 ) a ( a + num__1 ) it is given that num__380 < a ( a + num__1 ) num__380 < a ( a + num__1 ) if a = num__19 a = num__19 the total cost = num__19 × num__20 = num__380 = num__19 × num__20 = num__380 dexter would not have fallen short if : a = num__20 a = num__20 a ( a + num__1 ) = num__420 a ( a + num__1 ) = num__420 so he would have fallen short by rs num__40 . this is the minimum amount by which he may have fallen short . b <eor> b <eos> |
b |
add__1.0__19.0__ subtract__420.0__380.0__ multiply__1.0__40.0__ |
add__1.0__19.0__ subtract__420.0__380.0__ multiply__1.0__40.0__ |
| a is twice as good a workman as b and they took num__10 days together to do the work b alone can do it in . <o> a ) num__17 days <o> b ) num__12 days <o> c ) num__30 days <o> d ) num__25 days <o> e ) num__27 days |
wc = num__2 : num__1 num__2 x + x = num__0.1 x = num__0.0333333333333 = > num__30 days answer : c <eor> c <eos> |
c |
divide__1.0__10.0__ round__30.0__ |
divide__1.0__10.0__ round__30.0__ |
| in covering a distance of num__90 km a takes num__2 hours more than b . if a doubles his speed then he would take num__1 hour less than b . a ' s speed is : <o> a ) num__5 km / h <o> b ) num__8 km / h <o> c ) num__10 km / h <o> d ) num__15 km / h <o> e ) num__20 km / h |
let a ' s speed be x km / hr . then num__90 / x - num__45.0 x = num__3 num__45 / x = num__3 x = num__15 km / hr . answer : d <eor> d <eos> |
d |
divide__90.0__2.0__ add__2.0__1.0__ divide__45.0__3.0__ round__15.0__ |
divide__90.0__2.0__ add__2.0__1.0__ divide__45.0__3.0__ divide__45.0__3.0__ |
| if taxi fares were $ num__1.00 for the first num__0.2 mile and $ num__0.30 for each num__0.2 mile there after then the taxi fare for a num__3 - mile ride was <o> a ) $ num__1.56 <o> b ) $ num__2.40 <o> c ) $ num__3.80 <o> d ) $ num__5.20 <o> e ) $ num__2.80 |
in num__3 miles initial num__0.2 mile charge is $ num__1 rest of the distance = num__3 - ( num__0.2 ) = num__2.8 rest of the distance charge = num__14 ( num__0.3 ) = $ num__4.2 ( as the charge is num__0.3 for every num__0.2 mile ) = > total charge for num__3 miles = num__1 + num__4.2 = num__5.2 answer is d . <eor> d <eos> |
d |
subtract__3.0__0.2__ divide__2.8__0.2__ multiply__0.3__14.0__ add__1.0__4.2__ multiply__1.0__5.2__ |
subtract__3.0__0.2__ divide__2.8__0.2__ multiply__0.3__14.0__ add__1.0__4.2__ add__1.0__4.2__ |
| the average expenditure of a labourer for num__5 months was num__85 and he fell into debt . in the next num__4 months by reducing his monthly expenses to num__60 he not only cleared off his debt but also saved num__30 . his monthly income is <o> a ) num__69.5 <o> b ) num__71.2 <o> c ) num__60.4 <o> d ) num__62.1 <o> e ) num__72.1 |
income of num__6 months = ( num__5 × num__85 ) – debt = num__425 – debt income of the man for next num__4 months = num__4 × num__60 + debt + num__30 = num__270 + debt ∴ income of num__10 months = num__695 average monthly income = num__695 ÷ num__10 = num__69.5 answer a <eor> a <eos> |
a |
divide__30.0__5.0__ multiply__5.0__85.0__ add__4.0__6.0__ add__425.0__270.0__ divide__695.0__10.0__ divide__695.0__10.0__ |
divide__30.0__5.0__ multiply__5.0__85.0__ add__4.0__6.0__ add__425.0__270.0__ divide__695.0__10.0__ divide__695.0__10.0__ |
| tom divided $ num__360 among his six children for them to use for christmas gifts . his daughter kate added $ num__20 to her portion then used the money to buy num__16 gifts that each cost the same amount . what was the price of each of kate ’ s gifts ? <o> a ) $ num__5 <o> b ) $ num__10 <o> c ) $ num__15 <o> d ) $ num__20 <o> e ) $ num__25 |
$ num__360 ÷ num__6 = $ num__60 for each child $ num__60 + $ num__20 = $ num__80 kate ’ s money $ num__80 ÷ num__16 = $ num__5 cost of each gift correct answer a <eor> a <eos> |
a |
divide__360.0__6.0__ add__20.0__60.0__ divide__80.0__16.0__ divide__80.0__16.0__ |
divide__360.0__6.0__ add__20.0__60.0__ divide__80.0__16.0__ divide__80.0__16.0__ |
| if a point ( x y ) is randomly selected within the square shown in the figure above ( the vertices are on num__11 num__1 - num__1 - num__11 - num__1 - num__1 ) what are the r odds that x ^ num__2 + y ^ num__2 > num__1 ? <o> a ) num__1 . pi / num__4 <o> b ) num__2 . r = pi / num__16 <o> c ) num__3 . r = num__1 - pi / num__4 <o> d ) num__4 . num__1 - pi / num__16 <o> e ) num__5.4 - pi |
area of the square will be num__4 and the area of circle with center ( num__00 ) and radius num__1 will be pi . then succesful outcomes = num__4 - pi ( the area where x num__2 + y num__2 will be greater than num__1 ) total outcomes = num__4 therefore probability = ( num__4 - pi ) / num__4 c <eor> c <eos> |
c |
add__1.0__2.0__ |
add__1.0__2.0__ |
| sandy walked num__20 meters towards south . then sandy turned to her left and walked num__25 meters . she then turned to her left and walked num__20 meters . she then turned to her right and walked num__30 meters . what distance is she from the starting point and in which direction ? <o> a ) num__35 m east <o> b ) num__35 m north <o> c ) num__30 m west <o> d ) num__45 m west <o> e ) num__55 m east |
the net distance is num__25 + num__30 = num__55 meters to the east . the answer is e . <eor> e <eos> |
e |
add__25.0__30.0__ round__55.0__ |
add__25.0__30.0__ add__25.0__30.0__ |
| in an examination num__35.0 of the students passed and num__546 failed . how many students appeared for the examination ? <o> a ) a ) num__540 <o> b ) b ) num__400 <o> c ) c ) num__700 <o> d ) d ) num__650 <o> e ) e ) num__840 |
let the number of students appeared be x then num__65.0 of x = num__546 num__65 x / num__100 = num__546 x = num__546 * num__1.53846153846 = num__840 answer is e <eor> e <eos> |
e |
percent__100.0__840.0__ |
percent__100.0__840.0__ |
| annika hikes at a constant rate of num__12 minutes per kilometer . she has hiked num__2.75 kilometers east from the start of a hiking trail when she realizes that she has to be back at the start of the trail in num__51 minutes . if annika continues east then turns around and retraces her path to reach the start of the trail in exactly num__51 minutes for how many kilometers total did she hike east ? <o> a ) num__3.625 <o> b ) num__3 <o> c ) num__4.5 <o> d ) num__4 <o> e ) num__3.5 |
set up two r x t = d cases . num__1 . num__0.0833333333333 km / min x t = num__2.75 from which t = num__33 mins . we know total journey time now is num__51 + num__33 = num__84 . the rate is the same ie num__0.0833333333333 km / min . set up second r x t = d case . num__0.0833333333333 km / min x num__84 = num__7 km now the total journey would be halved as distance would be same in each direction . num__3.5 = num__3.5 e . <eor> e <eos> |
e |
divide__1.0__12.0__ multiply__12.0__2.75__ add__51.0__33.0__ divide__84.0__12.0__ round__3.5__ |
divide__1.0__12.0__ multiply__12.0__2.75__ add__51.0__33.0__ divide__84.0__12.0__ divide__3.5__1.0__ |
| a car going at num__10 miles per hour set out on an num__80 - mile trip at num__9 : num__00 a . m . exactly num__10 minutes later a second car left from the same place and followed the same route . how fast in miles per hour was the second car going if it caught up with the first car at num__10 : num__30 a . m . ? <o> a ) num__45 <o> b ) num__20 <o> c ) num__53 <o> d ) num__55 <o> e ) num__60 |
let car a = car that starts at num__9 am car b = car that starts at num__9 : num__10 am time for which car a travels at speed of num__10 m per hour = num__1.5 hours distance travelled by car a = num__10 * num__1.5 = num__15 miles since car b catches up car a at num__10 : num__30 time = num__80 mins = num__1.33333333333 hour speed of car b = num__15 / ( num__1.33333333333 ) = num__20 miles per hour answer b <eor> b <eos> |
b |
multiply__10.0__1.5__ subtract__30.0__10.0__ round__20.0__ |
multiply__10.0__1.5__ divide__30.0__1.5__ divide__30.0__1.5__ |
| speed of a boat in standing water is num__14 kmph and the speed of the stream is num__1.2 kmph . a man rows to a place at a distance of num__4664 km and comes back to the starting point . the total time taken by him is : <o> a ) num__784 hours <o> b ) num__794 hours <o> c ) num__780 hours <o> d ) num__684 hours <o> e ) num__884 hours |
speed downstream = ( num__12 + num__1.2 ) = num__13.2 kmph speed upstream = ( num__12 - num__1.2 ) = num__10.8 kmph total time taken = num__4664 / num__13.2 + num__4664 / num__10.8 = num__353 + num__431 = num__784 hours answer : a <eor> a <eos> |
a |
add__1.2__12.0__ subtract__12.0__1.2__ add__353.0__431.0__ round__784.0__ |
add__1.2__12.0__ subtract__12.0__1.2__ add__353.0__431.0__ add__353.0__431.0__ |
| num__70.0 of the employees in a multinational corporation have vcd players num__83 percent have microwave ovens num__80 percent have acs and num__85 percent have washing machines . at least what percentage of employees has all four gadgets ? <o> a ) ( a ) num__15.0 <o> b ) ( b ) num__5.0 <o> c ) ( c ) num__10.0 <o> d ) ( d ) num__18.0 <o> e ) ( e ) num__25 % |
so num__30.0 employees do n ' t have vcd devices . num__17.0 employees do n ' t have mo num__20.0 employees do n ' t have acs num__15.0 employees do n ' t have wm summ of employees that do n ' t have some device = num__30.0 + num__17.0 + num__20.0 + num__15.0 = num__82.0 < num__100.0 so definitely at least num__18.0 employees have num__4 devices . ans : d <eor> d <eos> |
d |
percent__100.0__18.0__ |
percent__100.0__18.0__ |
| a cistern has a leak which would empty the cistern in num__20 minutes . a tap is turned on which admits num__3 liters a minute into the cistern and it is emptied in num__24 minutes . how many liters does the cistern hold ? <o> a ) num__480 <o> b ) num__287 <o> c ) num__289 <o> d ) num__270 <o> e ) num__360 |
num__1 / x - num__0.05 = - num__0.0416666666667 x = num__120 num__120 * num__3 = num__360 answer : e <eor> e <eos> |
e |
divide__1.0__20.0__ divide__1.0__24.0__ multiply__3.0__120.0__ round__360.0__ |
divide__1.0__20.0__ divide__1.0__24.0__ multiply__3.0__120.0__ multiply__3.0__120.0__ |
| a train speeds past a pole in num__15 sec and a platform num__100 m long in num__25 sec its length is ? <o> a ) num__171 <o> b ) num__150 <o> c ) num__278 <o> d ) num__266 <o> e ) num__256 |
let the length of the train be x m and its speed be y m / sec . then x / y = num__15 = > y = x / num__15 ( x + num__100 ) / num__25 = x / num__15 = > x = num__150 m . answer : b <eor> b <eos> |
b |
round__150.0__ |
round__150.0__ |
| two trains num__140 m and num__160 m long run at the speed of num__60 km / hr and num__40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? <o> a ) num__10.9 sec <o> b ) num__14.8 sec <o> c ) num__10.6 sec <o> d ) num__10.8 sec <o> e ) num__20.8 sec |
relative speed = num__60 + num__40 = num__100 km / hr . = num__100 * num__0.277777777778 = num__27.7777777778 m / sec . distance covered in crossing each other = num__140 + num__160 = num__300 m . required time = num__300 * num__0.036 = num__10.8 = num__10.8 sec . answer : d <eor> d <eos> |
d |
subtract__140.0__40.0__ add__140.0__160.0__ multiply__300.0__0.036__ round__10.8__ |
add__60.0__40.0__ add__140.0__160.0__ multiply__300.0__0.036__ multiply__300.0__0.036__ |
| num__6 x − num__18 = num__6 y num__5 y + num__5 x = num__15 which of the following is the number of solutions to the system of equations shown above ? <o> a ) more than three <o> b ) exactly three <o> c ) exactly two <o> d ) exactly one <o> e ) none |
num__6 x − num__12 = num__6 y = > num__6 x - num__6 y = num__18 = > x - y = num__3 - - num__1 num__5 y + num__5 x = num__15 = > x + y = num__3 - - num__2 from equation num__1 and num__2 we get num__2 x = num__6 = > x = num__3 y = num__0 therefore the given system will have exactly one solution d <eor> d <eos> |
d |
subtract__18.0__6.0__ divide__18.0__6.0__ subtract__6.0__5.0__ divide__6.0__3.0__ reverse__1.0__ |
subtract__18.0__6.0__ subtract__18.0__15.0__ subtract__6.0__5.0__ subtract__5.0__3.0__ subtract__6.0__5.0__ |
| a and b can do a work in num__8 days b and c can do the same work in num__12 days . a b and c together can finish it in num__6 days . a and c together will do it in how many days ? <o> a ) num__7 <o> b ) num__6 <o> c ) num__5 <o> d ) num__8 <o> e ) num__9 |
let the total units of work be num__48 . units of work completed by a and b in one day = num__6 . units of work completed by b and c in one day = num__4 units of work completed by a b and c in one day = num__8 . from the above information we get the work completed by a and c in one day is num__6 units . therefore the number of days taken by a and c to complete the whole work = num__8.0 = num__8 days . answer : d <eor> d <eos> |
d |
multiply__8.0__6.0__ subtract__12.0__8.0__ round__8.0__ |
multiply__8.0__6.0__ subtract__12.0__8.0__ round__8.0__ |
| sides of a rectangular park are in the ratio num__3 : num__2 and its area is num__3750 sq m the cost of fencing it at num__50 ps per meter is ? <o> a ) num__125 <o> b ) num__150 <o> c ) num__200 <o> d ) num__250 <o> e ) num__300 |
num__3 x * num__2 x = num__3750 = > x = num__25 num__2 ( num__75 + num__50 ) = num__250 m num__250 * num__0.5 = rs . num__125 answer a <eor> a <eos> |
a |
triangle_area__3.0__50.0__ rectangle_perimeter__50.0__75.0__ multiply__0.5__250.0__ triangle_area__2.0__125.0__ |
multiply__3.0__25.0__ rectangle_perimeter__50.0__75.0__ multiply__0.5__250.0__ multiply__0.5__250.0__ |
| the sector of a circle has radius of num__42 cm and central angle num__135 o . find its perimeter ? <o> a ) num__91.5 cm <o> b ) num__92.2 cm <o> c ) num__28.9 cm <o> d ) num__29.2 cm <o> e ) num__183 cm |
perimeter of the sector = length of the arc + num__2 ( radius ) = ( num__0.375 * num__2 * num__3.14285714286 * num__42 ) + num__2 ( num__42 ) = num__99 + num__84 = num__183 cm answer : e <eor> e <eos> |
e |
multiply__42.0__2.0__ add__99.0__84.0__ round__183.0__ |
multiply__42.0__2.0__ add__99.0__84.0__ add__99.0__84.0__ |
| all of the citizens of a country have a three - character or four - character national identification code that is created using the num__26 letters of the alphabet and the num__10 digits from num__0 to num__9 . what is the maximum number of citizens who can be designated with these codes ? <o> a ) num__36 ^ num__7 <o> b ) num__37 ( num__36 ^ num__3 ) <o> c ) num__35 ( num__36 ^ num__4 ) <o> d ) num__35 ( num__36 ^ num__3 ) <o> e ) num__37 ( num__36 ^ num__4 ) |
the number of possible num__3 - digit codes is num__36 ^ num__3 . the number of possible num__4 - digit codes is num__36 ^ num__4 . the total sum is num__36 ^ num__3 + num__36 ^ num__4 = num__36 ^ num__3 ( num__1 + num__36 ) = num__37 ( num__36 ^ num__3 ) the answer is b . <eor> b <eos> |
b |
add__26.0__10.0__ divide__36.0__9.0__ subtract__10.0__9.0__ add__36.0__1.0__ add__36.0__1.0__ |
add__26.0__10.0__ divide__36.0__9.0__ subtract__10.0__9.0__ add__36.0__1.0__ add__36.0__1.0__ |
| find the simple interest on rs . num__64000 at num__16 num__0.666666666667 % per annum for num__9 months . <o> a ) s . num__8500 <o> b ) s . num__8000 <o> c ) s . num__7500 <o> d ) s . num__7000 <o> e ) s . num__6500 |
p = rs . num__64000 r = num__16.6666666667 % p . a and t = num__0.75 years = num__0.75 years . s . i . = ( p * r * t ) / num__100 = rs . ( num__64000 * ( num__16.6666666667 ) * ( num__0.75 ) * ( num__0.01 ) ) = rs . num__8000 answer is b . <eor> b <eos> |
b |
percent__100.0__8000.0__ |
percent__100.0__8000.0__ |
| the nth term in a certain sequence is defined for positive integer n as num__1 + ( − num__1 ) ( n ( n + num__1 ) / num__2 ) ^ num__2 . what is the total of the first num__64 terms of this sequence ? <o> a ) − num__164 <o> b ) num__0 <o> c ) num__16 <o> d ) num__32 <o> e ) num__64 |
sequence goes like this num__0 num__1 num__0 num__1 num__0 . . . . . . num__1 ( num__64 th term ) sum of num__64 terms = num__32 answer : d <eor> d <eos> |
d |
divide__64.0__2.0__ multiply__1.0__32.0__ |
divide__64.0__2.0__ multiply__1.0__32.0__ |
| if num__8 ^ z is a factor of num__256 ! which of the following could be the value of z ? <o> a ) num__32 <o> b ) num__42 <o> c ) num__30 <o> d ) num__52 <o> e ) num__36 |
num__8 ^ z is a factor of num__256 ! what values can z take . powers of num__8 in num__256 ! = [ num__32.0 ] + [ num__4.0 ] where [ x ] is the integral part of x powers of num__8 in num__256 ! = num__32 + num__4 = num__36 hence the maximum value of num__8 ^ z in num__256 ! = num__8 ^ num__36 correct option : e <eor> e <eos> |
e |
divide__256.0__8.0__ divide__32.0__8.0__ add__4.0__32.0__ add__4.0__32.0__ |
divide__256.0__8.0__ divide__32.0__8.0__ add__4.0__32.0__ add__4.0__32.0__ |
| a father said his son ` ` i was as old as you are at present at the time of your birth . ` ` if the father age is num__62 now the son age num__5 years back was <o> a ) num__14 <o> b ) num__17 <o> c ) num__11 <o> d ) num__19 <o> e ) num__26 |
let the son ' s present age be x years . then ( num__62 - x ) = x x = num__31 . son ' s age num__5 years back = ( num__31 - num__5 ) = num__26 years answer : e <eor> e <eos> |
e |
subtract__31.0__5.0__ subtract__31.0__5.0__ |
subtract__31.0__5.0__ subtract__31.0__5.0__ |
| a man gets a simple interest of rs . num__600 on a certain principal at the rate of num__5.0 p . a in two years . find the compound interest the man will get on twice the principal in two years at the same rate ? <o> a ) s . num__1000 <o> b ) s . num__1056 <o> c ) s . num__1150 <o> d ) s . num__1230 <o> e ) s . num__1256 |
let the principal be rs . p s . i at num__5.0 p . a in num__2 years on rs . p = rs . num__600 ( p ) ( num__2 ) ( num__5 ) / num__100 = num__600 p = num__6000 c . i on rs . num__2 p i . e . rs . num__12000 at num__5.0 p . a in two years = num__12000 { [ num__1 + num__0.05 ] ^ num__2 - num__1 } = num__12000 ( num__0.1025 ) = rs . num__1230 answer : d <eor> d <eos> |
d |
percent__5.0__1.0__ percent__100.0__1230.0__ |
percent__5.0__1.0__ percent__100.0__1230.0__ |
| a rower can row upstream at num__20 kph and downstream at num__34 kph . what is the speed of the rower in still water ? <o> a ) num__26 <o> b ) num__27 <o> c ) num__28 <o> d ) num__29 <o> e ) num__30 |
let v be the rower ' s speed in still water . let s be the speed of the current in the stream v - s = num__20 v + s = num__34 when we add the two equations we get : num__2 v = num__54 then v = num__27 kph . the answer is b . <eor> b <eos> |
b |
add__20.0__34.0__ divide__54.0__2.0__ round__27.0__ |
add__20.0__34.0__ divide__54.0__2.0__ subtract__54.0__27.0__ |
| the diagonals of a rhombus are num__15 cm and num__20 cm . find its area ? <o> a ) num__277 <o> b ) num__266 <o> c ) num__150 <o> d ) num__288 <o> e ) num__212 |
num__0.5 * num__15 * num__20 = num__150 answer : c <eor> c <eos> |
c |
triangle_area__15.0__20.0__ triangle_area__15.0__20.0__ |
volume_rectangular_prism__15.0__20.0__0.5__ volume_rectangular_prism__15.0__20.0__0.5__ |
| find num__12 × × num__19 <o> a ) num__238 <o> b ) num__228 <o> c ) num__208 <o> d ) num__277 <o> e ) num__101 |
mentally imagine this number as ( num__10 + num__2 ) × × num__19 = num__190 + num__38 = num__228 . answer : b <eor> b <eos> |
b |
subtract__12.0__10.0__ multiply__19.0__10.0__ multiply__19.0__2.0__ multiply__12.0__19.0__ multiply__12.0__19.0__ |
subtract__12.0__10.0__ multiply__19.0__10.0__ multiply__19.0__2.0__ add__38.0__190.0__ add__38.0__190.0__ |
| the average age of a class of num__19 students is num__20 years . the average increased by num__1 when the teacher ' s age also included . what is the age of the teacher ? <o> a ) num__39 <o> b ) num__41 <o> c ) num__40 <o> d ) num__42 <o> e ) num__43 |
total age of all students = num__19 Ã — num__20 total age of all students + age of the teacher = num__20 Ã — num__21 age of the teacher = num__20 Ã — num__21 â ˆ ’ num__20 Ã — num__19 = num__20 ( num__21 â ˆ ’ num__19 ) = num__20 Ã — num__2 = num__40 answer is c . <eor> c <eos> |
c |
add__20.0__1.0__ subtract__21.0__19.0__ add__19.0__21.0__ add__19.0__21.0__ |
add__20.0__1.0__ subtract__21.0__19.0__ add__19.0__21.0__ add__19.0__21.0__ |
| the distance between a sports store and a recreation center is approximately num__3.34 x num__10 ^ num__5 inches . which of the following is closest to the distance between the sports store and a recreation center in kilometers ? ( num__1 kilometer is approximately num__3.9 x num__10 ^ num__4 inches . ) <o> a ) num__9.33 <o> b ) num__8.98 <o> c ) num__8.56 <o> d ) num__9.21 <o> e ) num__9.03 |
we know that num__1 kilometer = num__3.9 x num__10 ^ num__4 inches . therefore we now have a ratio of ( num__1 kilometer / num__3.9 x num__10 ^ num__4 inches ) . we also know that the distance between the sports store and a recreation center is approximately num__3.34 x num__10 ^ num__5 inches . to convert inches to kilometers we need to multiply num__3.34 x num__10 ^ num__5 inches by the ratio we just found out . kilometers = num__3.34 x num__10 ^ num__5 inches x ( num__1 kilometer / num__3.9 x num__10 ^ num__4 inches ) [ note : ` ` inches ' ' cancel out leaving us with just km ] km = num__3.34 x num__10 ^ num__5 / num__3.9 x num__10 ^ num__4 km km = num__8.56 km therefore we can see that the closest answer is c . <eor> c <eos> |
c |
round__8.56__ |
divide__8.56__1.0__ |
| if num__25.0 of a class averages num__80.0 on a test num__50.0 of the class averages num__65.0 on the test and the remainder of the class averages num__90.0 on the test what is the overall class average ? <o> a ) num__60.0 <o> b ) num__65.0 <o> c ) num__70.0 <o> d ) num__75.0 <o> e ) num__80 % |
this question is a weighted average question with a series of dependent variables . the remaining portion of the class represents num__100.0 - num__25.0 - num__50.0 = num__25.0 of the class converting the portions of the class population to decimal weights we find : class average = num__0.25 x num__80 + num__0.50 x num__65 + num__0.25 x num__90 = num__75 the class average is num__75.0 final answer d ) num__75.0 <eor> d <eos> |
d |
divide__25.0__100.0__ divide__25.0__50.0__ add__25.0__50.0__ add__25.0__50.0__ |
divide__25.0__100.0__ divide__25.0__50.0__ add__25.0__50.0__ add__25.0__50.0__ |
| two numbers x and y are such that the sum of num__3.0 of x and num__4.0 of y is one - third of the sum of num__2.0 of of x and num__8.0 of y . find the ratio of x : y . <o> a ) num__3 : num__8 <o> b ) num__6 : num__7 <o> c ) num__4 : num__7 <o> d ) num__2 : num__5 <o> e ) none of these |
explanation : solution : num__3.0 of x + num__4.0 of y = num__0.333333333333 ( num__2.0 of x + num__8.0 of y ) num__3 x / num__100 + num__4 y / num__100 = num__2 x / num__300 + num__8 y / num__300 ( num__3 x + num__4 y ) num__3 = num__2 x + num__8 y num__7 x = num__4 y x / y = num__0.571428571429 answer : c <eor> c <eos> |
c |
reverse__3.0__ multiply__3.0__100.0__ add__3.0__4.0__ divide__4.0__7.0__ subtract__8.0__4.0__ |
reverse__3.0__ multiply__3.0__100.0__ add__3.0__4.0__ divide__4.0__7.0__ divide__8.0__2.0__ |
| num__20 people going to one temple . ( some time waste data ) they want rs . num__20 . for men - rs . num__3 women - rs . num__2 children - num__50 paisa . how much money spends for men children and women ? <o> a ) num__3 num__710 <o> b ) num__4 num__710 <o> c ) num__5 num__710 <o> d ) num__6 num__710 <o> e ) num__7 num__7 |
10 |
m = num__1 = > num__1 * num__3 = num__3 w = num__5 = > num__5 * num__2 = num__10 c = num__14 = > num__14 * . num__5 = num__7 num__1 + num__5 + num__14 = num__20 & num__3 + num__10 + num__7 = num__20 answer : a <eor> a <eos> |
a |
a |
| a land owner needs to fence his semicircular land leaving an opening of length of num__3 m for access . the radius of the semicircular plot is num__7 m . how long would be the fence in meters ? <o> a ) num__22 <o> b ) num__33 <o> c ) num__36 <o> d ) num__44 <o> e ) num__51 |
perimeter of the semicircle = Ï € xd / num__2 + d = Ï € x num__7 + num__14 = num__22 + num__14 = num__36 m length of the fence = num__36 - num__3 = num__33 m answer : b <eor> b <eos> |
b |
multiply__7.0__2.0__ triangle_area__3.0__22.0__ triangle_area__3.0__22.0__ |
multiply__7.0__2.0__ triangle_area__3.0__22.0__ triangle_area__3.0__22.0__ |
| a man has rs . num__560 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__45 <o> b ) num__60 <o> c ) num__75 <o> d ) num__105 <o> e ) num__120 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__560 num__16 x = num__560 x = num__35 . hence total number of notes = num__3 x = num__105 . answer = d <eor> d <eos> |
d |
divide__560.0__16.0__ multiply__35.0__3.0__ multiply__35.0__3.0__ |
divide__560.0__16.0__ multiply__35.0__3.0__ multiply__35.0__3.0__ |
| what is the smallest of six consecutive odd integers whose average ( arithmetic mean ) is x + num__2 ? <o> a ) x - num__5 <o> b ) x - num__3 <o> c ) x - num__1 <o> d ) x <o> e ) x + num__1 |
assume the integers to be n - num__5 n - num__3 n - num__1 n + num__1 n + num__3 n + num__5 am = num__6 n / num__6 = n given n = x + num__2 therefore smallest term = n - num__5 = x + num__2 - num__5 = x - num__3 correct option : b <eor> b <eos> |
b |
subtract__5.0__2.0__ subtract__3.0__2.0__ multiply__2.0__3.0__ add__2.0__1.0__ |
subtract__5.0__2.0__ subtract__3.0__2.0__ add__1.0__5.0__ add__2.0__1.0__ |
| the grade point average of one third of the classroom is num__60 ; the grade point average of the rest is num__66 . what is the grade point average of the whole class ? <o> a ) num__55 <o> b ) num__56 <o> c ) num__64 <o> d ) num__65 <o> e ) num__66 |
let n = total students in class total points for num__0.333333333333 class = num__60 n / num__3 = num__20 n total points for num__0.666666666667 class = num__66 * num__2 n / num__3 = num__44 n total points for whole class = num__20 n + num__44 n = num__64 n num__64 n total class points / n total students = num__64 grade point average for total class answer : c <eor> c <eos> |
c |
divide__60.0__3.0__ subtract__66.0__2.0__ subtract__66.0__2.0__ |
divide__60.0__3.0__ add__44.0__20.0__ add__44.0__20.0__ |
| if half of num__5 were num__3 that would one - third of num__10 be <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__1 |
explanation : half of num__5 is num__2.5 . but given as num__3 . so take num__0.5 of num__5 x = num__3 ⇒ x = num__1.2 now num__0.333333333333 ( num__10 x ) = num__0.333333333333 × num__10 × num__1.2 = num__4 . answer : b <eor> b <eos> |
b |
divide__5.0__10.0__ divide__3.0__2.5__ reverse__3.0__ divide__10.0__2.5__ divide__10.0__2.5__ |
divide__5.0__10.0__ divide__3.0__2.5__ reverse__3.0__ divide__10.0__2.5__ divide__10.0__2.5__ |
| an article is bought for rs . num__675 and sold for rs . num__900 find the gain percent ? <o> a ) num__33 num__0.142857142857 % <o> b ) num__33 num__0.666666666667 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__37 num__0.333333333333 % <o> e ) num__23 num__0.333333333333 % |
num__675 - - - - num__225 num__100 - - - - ? = > = num__33 num__0.333333333333 % answer : c <eor> c <eos> |
c |
percent__100.0__33.0__ |
percent__100.0__33.0__ |
| cream costs twice as much as skim milk . a certain brand of milk is num__0.333333333333 cream and num__0.666666666667 skim milk . what fraction of the cost of this brand is due to the cream ? <o> a ) num__0.333333333333 <o> b ) num__0.5 <o> c ) num__0.4 <o> d ) num__0.6 <o> e ) num__0.75 |
let x be the cost of the skim milk in the mixture . the cost of the cream is ( num__0.5 ) * x * num__2 = x . the total cost is x + x = num__2 x and the fraction due to the cream is x / num__2 x = num__0.5 the answer is b . <eor> b <eos> |
b |
reverse__0.5__ reverse__2.0__ |
reverse__0.5__ reverse__2.0__ |
| the number num__110 can be written as the sum of the squares of num__3 different positive integers . what is the sum of these num__3 integers ? <o> a ) num__18 <o> b ) num__16 <o> c ) num__15 <o> d ) num__14 <o> e ) num__13 |
num__7 ^ num__2 + num__5 ^ num__2 + num__6 ^ num__2 = num__49 + num__25 + num__36 = num__110 num__7 + num__5 + num__6 = num__18 hence answer is a <eor> a <eos> |
a |
multiply__3.0__2.0__ power__7.0__2.0__ power__5.0__2.0__ power__6.0__2.0__ multiply__3.0__6.0__ multiply__3.0__6.0__ |
multiply__3.0__2.0__ power__7.0__2.0__ power__5.0__2.0__ power__6.0__2.0__ multiply__3.0__6.0__ multiply__3.0__6.0__ |
| given two sets a = { num__11 num__44 num__55 } and b = { num__01 } if one number is chosen from each set at random what is the probability that the sum of both numbers is an even number <o> a ) num__0.25 <o> b ) num__0.125 <o> c ) num__0.142857142857 <o> d ) num__0.5 <o> e ) num__3 |
one way to look at it : the number from set a can be anything . the number selected from set b will determine whether the sum is odd or even . for example if a num__4 is selected from set a we need a num__0 from set b to get an even sum . if a num__5 is selected from set a we need a num__1 from set b to get an even sum . and so on . so p ( sum is even ) = p ( select any number from set aandselect the number from set b that makes the sum even ) = p ( select any number from set a ) xp ( select the number from set b that makes the sum even ) = num__1 x num__0.5 = num__0.5 = d <eor> d <eos> |
d |
divide__44.0__11.0__ divide__55.0__11.0__ subtract__1.0__0.5__ |
divide__44.0__11.0__ divide__55.0__11.0__ subtract__1.0__0.5__ |
| if the si on a certain sum of money is num__0.16 of the sum and the rate per cent equals the number yrs then the rate of interest / annum is ? <o> a ) num__4.0 <o> b ) num__5.0 <o> c ) num__6.0 <o> d ) num__7.0 <o> e ) num__8 % |
sol let the prmcipal be rs . x then the simple interest ( num__1 ) let the me ofinterestp num__3 be num__1.0 than tune m : ryears max ! max pm ‘ x : num__5 num__4.0 a <eor> a <eos> |
a |
add__1.0__3.0__ add__1.0__3.0__ |
add__1.0__3.0__ add__1.0__3.0__ |
| one side of a rectangular field is num__15 m and one of its diagonal is num__17 m . find the area of field ? <o> a ) num__120 sq . m <o> b ) num__140 sq . m <o> c ) num__125 sq . m <o> d ) num__130 sq . m <o> e ) num__110 sq . m |
other side = [ ( num__17 x num__17 ) - ( num__15 x num__15 ) ] = ( num__289 - num__225 ) = num__8 m area = num__15 x num__8 = num__120 sq . m answer : a <eor> a <eos> |
a |
side_by_diagonal__17.0__15.0__ multiply__15.0__8.0__ multiply__15.0__8.0__ |
side_by_diagonal__17.0__15.0__ multiply__15.0__8.0__ multiply__15.0__8.0__ |
| sandy gets num__3 marks for each correct sum and loses num__2 marks for each incorrect sum . sandy attempts num__30 sums and obtains num__60 marks . how many sums did sandy get correct ? <o> a ) num__18 <o> b ) num__20 <o> c ) num__22 <o> d ) num__24 <o> e ) num__26 |
let x be the correct sums and ( num__30 - x ) be the incorrect sums . num__3 x - num__2 ( num__30 - x ) = num__60 num__5 x = num__120 x = num__24 the answer is d . <eor> d <eos> |
d |
add__3.0__2.0__ multiply__2.0__60.0__ divide__120.0__5.0__ divide__120.0__5.0__ |
add__3.0__2.0__ multiply__2.0__60.0__ divide__120.0__5.0__ divide__120.0__5.0__ |
| the distance between two planets is num__4 × num__10 ^ num__6 light years . what is the distance between the two planets in parsecs ? ( num__1 parsec = num__3.26 light years ) <o> a ) num__9.3 × num__10 ^ num__5 <o> b ) num__9.9 × num__10 ^ num__6 <o> c ) num__9.3 × num__10 ^ num__7 <o> d ) num__9.9 × num__10 ^ num__7 <o> e ) num__1.2 × num__10 ^ num__6 |
num__4 * num__10 ^ num__6 light years = num__4 * num__10 ^ num__6 / num__3.26 parsecs . num__4 / num__3.26 is a little bit less than num__1 something like num__0.9 thus num__4 * num__10 ^ num__6 / num__3.26 = ~ num__1.2 * num__10 ^ num__6 . answer : e . <eor> e <eos> |
e |
round__1.2__ |
multiply__1.0__1.2__ |
| train a leaves the station traveling at num__30 miles per hour . two hours later train В leaves the same station traveling in the same direction at num__45 miles per hour . how many miles from the station was train a overtaken by train b ? <o> a ) num__100 <o> b ) num__120 <o> c ) num__140 <o> d ) num__160 <o> e ) num__180 |
after two hours train a is ahead by num__60 miles . train b can catch up at a rate of num__15 miles per hour . the time to catch up is num__4.0 = num__4 hours . in num__4 hours train a travels another num__30 * num__4 = num__120 miles for a total of num__180 miles . the answer is e . <eor> e <eos> |
e |
hour_to_min_conversion__ subtract__45.0__30.0__ divide__60.0__15.0__ multiply__30.0__4.0__ multiply__45.0__4.0__ round__180.0__ |
hour_to_min_conversion__ subtract__45.0__30.0__ divide__60.0__15.0__ multiply__30.0__4.0__ multiply__45.0__4.0__ multiply__45.0__4.0__ |
| an error num__2.0 in excess is made while measuring the side of a square . the percentage of error in the calculated area of the square is ? <o> a ) num__4.0 <o> b ) num__4.04 <o> c ) num__4.14 <o> d ) num__5.0 <o> e ) num__5.13 % |
num__100 cm is read as num__102 cm . a num__1 = ( num__100 x num__100 ) cm num__2 and a num__2 ( num__102 x num__102 ) cm num__2 . ( a num__2 - a num__1 ) = [ ( num__102 ) num__2 - ( num__100 ) num__2 ] = ( num__102 + num__100 ) x ( num__102 - num__100 ) = num__404 cm num__2 . percentage error = num__404 x num__100.0 = num__4.04 num__100 x num__100 b ) <eor> b <eos> |
b |
percent__1.0__404.0__ percent__1.0__404.0__ |
percent__1.0__404.0__ percent__1.0__404.0__ |
| how many integers n greater than and less than num__100 are there such that if the digits of n are reversed the resulting integer is n + num__9 ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
( num__10 x + y ) = ( num__10 y + x ) + num__9 = > num__9 x - num__9 y = num__9 = > x - y = num__1 and only num__8 numbers satisfy this condition and the numbers are num__2132 num__4354 num__6576 num__8798 answer : d <eor> d <eos> |
d |
subtract__10.0__9.0__ subtract__9.0__1.0__ subtract__9.0__1.0__ |
subtract__10.0__9.0__ subtract__9.0__1.0__ subtract__9.0__1.0__ |
| which of the following could be the value of x if | num__3 x – num__9 | = num__0 ? <o> a ) – num__3 <o> b ) num__0.111111111111 <o> c ) num__0.333333333333 <o> d ) - num__0.333333333333 <o> e ) num__3 |
| num__3 x – num__9 | = num__0 = > num__3 x - num__9 = num__0 or - ( num__3 x - num__9 ) = num__0 = > num__3 x = num__9 or - num__3 x = - num__9 = > x = num__3 or x = num__3 answer e <eor> e <eos> |
e |
divide__9.0__3.0__ |
divide__9.0__3.0__ |
| what least number should be added to num__1053 so that the sum is completely divisible by num__23 <o> a ) a ) num__4 <o> b ) b ) num__1 <o> c ) c ) num__2 <o> d ) d ) num__3 <o> e ) e ) num__5 |
explanation : ( num__45.7826086957 ) gives remainder num__18 num__18 + num__5 = num__23 so we need to add num__5 answer : option e <eor> e <eos> |
e |
divide__1053.0__23.0__ subtract__23.0__18.0__ subtract__23.0__18.0__ |
divide__1053.0__23.0__ subtract__23.0__18.0__ subtract__23.0__18.0__ |
| the sum of three consecutive even numbers is num__93 . find the middle number of the three ? <o> a ) num__14 <o> b ) num__35 <o> c ) num__39 <o> d ) num__31 <o> e ) num__12 |
middle value = num__31.0 = num__31 ans d <eor> d <eos> |
d |
round__31.0__ |
round__31.0__ |
| the average age of husband wife and their child num__3 years ago was num__25 years and that of wife and the child num__5 years ago was num__20 years . the present age of the husband is <o> a ) num__22 <o> b ) num__40 <o> c ) num__34 <o> d ) num__21 <o> e ) num__11 |
explanation : sum of the present ages of husband wife and child = ( num__25 x num__3 + num__3 x num__3 ) years = num__84 years . sum of the present ages of wife and child ( num__20 x num__2 + num__5 x num__2 ) years = num__50 years . husband ' s present age = ( num__84 - num__50 ) years = num__34 years . answer : c <eor> c <eos> |
c |
subtract__5.0__3.0__ multiply__25.0__2.0__ subtract__84.0__50.0__ subtract__84.0__50.0__ |
subtract__5.0__3.0__ multiply__25.0__2.0__ subtract__84.0__50.0__ subtract__84.0__50.0__ |
| the effective annual rate of interest corresponding to a nominal rate of num__14.0 per annum payable half - yearly is ? <o> a ) num__16.06 <o> b ) num__16.07 <o> c ) num__16.08 <o> d ) num__14.49 <o> e ) num__16.19 % |
amount of rs . num__100 for num__1 year when compounded half - yearly = [ num__100 * ( num__1 + num__0.07 ) num__2 ] = rs . num__14.49 effective rate = ( num__114.49 - num__100 ) = num__14.49 answer : d <eor> d <eos> |
d |
percent__100.0__14.49__ |
percent__100.0__14.49__ |
| the circumferences of two circles are num__264 meters and num__352 meters . find the difference between the areas of the larger and the smaller circles ? <o> a ) num__1877 sq m <o> b ) num__1976 sq m <o> c ) num__4312 sq m <o> d ) num__1987 sq m <o> e ) num__1678 sq m |
let the radii of the smaller and the larger circles be s m and l m respectively . num__2 ∏ s = num__264 and num__2 ∏ l = num__352 s = num__132.0 ∏ and l = num__176.0 ∏ difference between the areas = ∏ l num__2 - ∏ s num__2 = ∏ { num__1762 / ∏ num__2 - num__1322 / ∏ num__2 } = num__1762 / ∏ - num__1322 / ∏ = ( num__176 - num__132 ) ( num__176 + num__132 ) / ∏ = ( num__44 ) ( num__308 ) / ( num__3.14285714286 ) = ( num__2 ) ( num__308 ) ( num__7 ) = num__4312 sq m answer : c <eor> c <eos> |
c |
volume_rectangular_prism__2.0__7.0__308.0__ triangle_area__2.0__4312.0__ |
volume_rectangular_prism__2.0__7.0__308.0__ triangle_area__2.0__4312.0__ |
| a man can row num__5 kmph in still water . when the river is running at num__2.3 kmph it takes him num__1 hour to row to a place and black . what is the total distance traveled by the man ? <o> a ) num__2.91 <o> b ) num__3.48 <o> c ) num__2.98 <o> d ) num__3.78 <o> e ) num__4.21 |
m = num__5 s = num__2.3 ds = num__6.3 us = num__2.7 x / num__6.3 + x / num__2.7 = num__1 x = num__1.89 d = num__1.89 * num__2 = num__3.78 answer : d <eor> d <eos> |
d |
subtract__5.0__2.3__ multiply__2.0__1.89__ round__3.78__ |
subtract__5.0__2.3__ multiply__2.0__1.89__ multiply__1.0__3.78__ |
| train x crosses a stationary train y in num__60 seconds and a pole in num__25 seconds with the same speed . the length of the train x is num__300 m . what is the length of the stationary train y ? <o> a ) num__267 m <o> b ) num__420 m <o> c ) num__167 m <o> d ) num__287 m <o> e ) num__265 m |
let the length of the stationary train y be ly given that length of train x lx = num__300 m let the speed of train x be v . since the train x crosses train y and a pole in num__60 seconds and num__25 seconds respectively . = > num__300 / v = num__25 - - - > ( num__1 ) ( num__300 + ly ) / v = num__60 - - - > ( num__2 ) from ( num__1 ) v = num__12.0 = num__12 m / sec . from ( num__2 ) ( num__300 + ly ) / num__12 = num__60 = > num__300 + ly = num__60 ( num__12 ) = num__720 = > ly = num__720 - num__300 = num__420 m length of the stationary train = num__420 m answer : b <eor> b <eos> |
b |
divide__300.0__25.0__ multiply__60.0__12.0__ subtract__720.0__300.0__ round__420.0__ |
divide__300.0__25.0__ multiply__60.0__12.0__ subtract__720.0__300.0__ divide__420.0__1.0__ |
| two men a and b start from place x walking at num__4 ½ kmph and num__5 ¾ kmph respectively . how many km apart they are at the end of num__3 ½ hours if they are walking in the same direction ? <o> a ) num__4 num__0.125 km <o> b ) num__5 num__0.5 km <o> c ) num__6 num__0.125 km <o> d ) num__4 num__0.375 km <o> e ) num__3 num__0.25 km |
d num__4 num__0.375 km rs = num__5 ¾ - num__4 ½ = num__1 ¼ t = num__3 ½ h . d = num__1.25 * num__3.5 = num__4.375 = num__4 num__0.375 km <eor> d <eos> |
d |
subtract__4.0__3.0__ divide__5.0__4.0__ add__4.0__0.375__ round__4.0__ |
subtract__4.0__3.0__ divide__5.0__4.0__ multiply__1.25__3.5__ subtract__5.0__1.0__ |
| rasik walked num__20 m towards north . then he turned right and walks num__30 m . then he turns right and walks num__35 m . then he turns left and walks num__15 m . finally he turns left and walks num__15 m . in which direction and how many metres is he from the starting position ? <o> a ) num__23 <o> b ) num__38 <o> c ) num__37 <o> d ) num__45 <o> e ) num__92 |
num__45 m east answer : d <eor> d <eos> |
d |
add__30.0__15.0__ round__45.0__ |
add__30.0__15.0__ round__45.0__ |
| if num__10 ^ ( b - num__1 ) < num__0.000125 < num__10 ^ b what is the value of an integer b ? <o> a ) - num__4 <o> b ) - num__3 <o> c ) - num__2 <o> d ) num__3 <o> e ) num__4 |
- > multiply num__10 ^ num__6 - > ( num__10 ^ num__6 ) { num__10 ^ ( b - num__1 ) } < num__125 < ( num__10 ^ num__6 ) ( num__10 ^ b ) - > num__125 is bigger than num__100 - > ( num__10 ^ num__6 ) ( num__10 ^ ( b - num__1 ) ) = num__100 - > num__10 ^ ( num__6 + b - num__1 ) = num__10 ^ num__2 b + num__5 = num__2 - > b = - num__3 thus the answer is b <eor> b <eos> |
b |
divide__10.0__2.0__ add__1.0__2.0__ add__1.0__2.0__ |
subtract__6.0__1.0__ add__1.0__2.0__ add__1.0__2.0__ |
| a chair is bought for rs . num__900 / - and sold at rs . num__810 / - find the loss percentage ? <o> a ) num__10.0 loss <o> b ) num__20.0 loss <o> c ) num__30.0 loss <o> d ) num__40.0 loss <o> e ) num__50.0 loss |
formula = ( selling price ~ cost price ) / cost price * num__100 = ( num__810 ~ num__900 ) / num__900 = num__10.0 loss a ) <eor> a <eos> |
a |
divide__100.0__10.0__ |
divide__100.0__10.0__ |
| at num__1 : num__00 annie starts to bicycle along a num__84 mile road at a constant speed of num__14 miles per hour . thirty minutes earlier scott started bicycling towards annie on the same road at a constant speed of num__12 miles per hour . at what time will they meet ? <o> a ) num__2 : num__30 <o> b ) num__3 : num__00 <o> c ) num__4 : num__00 <o> d ) num__5 : num__00 <o> e ) num__6 : num__00 |
in the first num__30 minutes scott can travel num__6 miles so there are num__78 miles left . together annie and scott can complete num__26 miles . num__3.0 = num__3 so they will meet num__3 hours after num__1 : num__00 . the answer is c . <eor> c <eos> |
c |
divide__84.0__14.0__ subtract__84.0__6.0__ add__14.0__12.0__ divide__78.0__26.0__ add__1.0__3.0__ |
divide__84.0__14.0__ subtract__84.0__6.0__ add__14.0__12.0__ divide__78.0__26.0__ add__1.0__3.0__ |
| how many integers from num__20 to num__150 inclusive are divisible by num__3 but not divisible by num__7 ? <o> a ) num__37 <o> b ) num__40 <o> c ) num__42 <o> d ) num__45 <o> e ) num__55 |
we should find # of integers divisible by num__3 but not by num__3 * num__7 = num__21 . # of multiples of num__21 in the range from num__20 to num__150 inclusive is ( num__147 - num__21 ) / num__21 + num__1 = num__7 ; num__44 - num__7 = num__37 . answer : a . <eor> a <eos> |
a |
multiply__3.0__7.0__ subtract__150.0__3.0__ subtract__21.0__20.0__ subtract__44.0__7.0__ multiply__1.0__37.0__ |
multiply__3.0__7.0__ subtract__150.0__3.0__ subtract__21.0__20.0__ subtract__44.0__7.0__ multiply__1.0__37.0__ |
| the product of all the prime numbers less than num__12 is closest to which of the following powers of num__10 ? <o> a ) num__10 ^ num__7 <o> b ) num__10 ^ num__4 <o> c ) num__10 ^ num__8 <o> d ) num__10 ^ num__5 <o> e ) num__10 ^ num__9 |
product of prime numbers less than num__12 is num__2 * num__3 * num__5 * num__7 * num__11 num__2 * num__5 = num__10 . num__10 * num__11 * num__3 * num__7 = num__10 * num__10 * num__3 * num__10 ( approximately ) ~ product will be > num__10 ^ num__3 < num__10 ^ num__5 i . e . num__10 ^ num__4 b is the answer <eor> b <eos> |
b |
subtract__12.0__10.0__ divide__10.0__2.0__ subtract__12.0__5.0__ divide__12.0__3.0__ subtract__12.0__2.0__ |
subtract__12.0__10.0__ divide__10.0__2.0__ subtract__12.0__5.0__ divide__12.0__3.0__ multiply__2.0__5.0__ |
| a restaurant meal cost $ num__40 and there was no tax . if the tip was more than num__5 percent but less than num__15 percent of the cost of the meal then total amount paid must have been between : <o> a ) $ num__40 and $ num__42 <o> b ) $ num__41 and $ num__46 <o> c ) $ num__41 and num__45 <o> d ) $ num__43 and $ num__46 <o> e ) $ num__41 and $ num__47 |
let tip = t meal cost = num__40 range of tip = from num__5.0 of num__40 to num__15.0 of num__40 = num__2 to num__6 hence range of amount paid = num__40 + t = num__42 to num__46 i . e . e <eor> e <eos> |
e |
add__40.0__2.0__ add__40.0__6.0__ subtract__46.0__5.0__ |
add__40.0__2.0__ add__40.0__6.0__ subtract__46.0__5.0__ |
| the value of a num__10.5 stock in which an income of rs . num__756 is derived by investing rs . num__9000 brokerage being % is ? <o> a ) rs . num__112.75 <o> b ) rs . num__121.75 <o> c ) rs . num__124.75 <o> d ) rs . num__118.75 <o> e ) rs . num__234.75 |
for an income of rs . num__756 investment = rs . num__9000 for an income of rs . investment = = rs . num__125 for a rs . num__100 stock investment = rs . num__125 . market value of rs . num__100 stock = = rs . num__124.75 c <eor> c <eos> |
c |
percent__100.0__124.75__ |
percent__100.0__124.75__ |
| if money is invested at r percent interest compounded annually the amount of the investment will double in approximately num__70 / r years . if pat ' s parents invested $ num__5000 in a long - term bond that pays num__8 percent interest compounded annually what will be the approximate total amount of the investment num__18 years later when pat is ready for college ? <o> a ) $ num__20000 <o> b ) $ num__15000 <o> c ) $ num__12000 <o> d ) $ num__10000 <o> e ) $ num__9000 |
since investment doubles in num__70 / r years then for r = num__8 it ' ll double in num__8.75 = ~ num__9 years ( we are not asked about the exact amount so such an approximation will do ) . thus in num__18 years investment will double twice and become ( $ num__5000 * num__2 ) * num__2 = $ num__20000 ( after num__9 years investment will become $ num__5000 * num__2 = $ num__10000 and in another num__9 years it ' ll become $ num__10000 * num__2 = $ num__20000 ) . answer : a . <eor> a <eos> |
a |
divide__70.0__8.0__ divide__18.0__9.0__ multiply__5000.0__2.0__ multiply__2.0__10000.0__ |
divide__70.0__8.0__ divide__18.0__9.0__ multiply__5000.0__2.0__ multiply__2.0__10000.0__ |
| a train covers a distance of num__12 km in num__10 min . if it takes num__6 sec to pass a telegraph post then the length of the train is ? <o> a ) num__227 <o> b ) num__177 <o> c ) num__120 <o> d ) num__288 <o> e ) num__271 |
speed = ( num__1.2 * num__60 ) km / hr = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . length of the train = num__20 * num__6 = num__120 m . answer : c <eor> c <eos> |
c |
divide__12.0__10.0__ hour_to_min_conversion__ multiply__12.0__6.0__ multiply__12.0__10.0__ round__120.0__ |
divide__12.0__10.0__ multiply__10.0__6.0__ multiply__12.0__6.0__ multiply__12.0__10.0__ multiply__12.0__10.0__ |
| find the number of square tiles to cover the floor of a room measuring num__10.5 m * num__9.5 m leaving num__0.25 m space around the room . a side of square tile is given to be num__25 cms ? <o> a ) num__422 <o> b ) num__476 <o> c ) num__1440 <o> d ) num__428 <o> e ) num__413 |
floor area to be covered by tiles = num__10 * num__9 = num__90 tiles area = num__0.25 * num__0.25 = num__0.0625 no . of tiles = num__90 / num__0.0625 = num__1440 answer : c <eor> c <eos> |
c |
multiply__9.0__10.0__ divide__90.0__0.0625__ round__1440.0__ |
multiply__9.0__10.0__ divide__90.0__0.0625__ divide__90.0__0.0625__ |
| if p = num__25 * num__343 * num__0.00571428571429 how many digits are in p ? <o> a ) num__7 <o> b ) num__2 <o> c ) num__1 <o> d ) num__4 <o> e ) num__3 |
p = num__25 * num__343 * num__0.00571428571429 p = num__5 ^ num__2 * num__7 ^ num__3 * num__3 ^ num__2 / ( num__5 ^ num__2 * num__3 ^ num__2 * num__7 ) p = num__7 ^ num__2 p = num__49 ans : b <eor> b <eos> |
b |
add__2.0__5.0__ subtract__5.0__2.0__ divide__343.0__7.0__ subtract__5.0__3.0__ |
add__2.0__5.0__ subtract__5.0__2.0__ divide__343.0__7.0__ subtract__5.0__3.0__ |
| ayesha ' s father was num__38 years of age when she was born while her mother was num__36 years old when her brother four years younger to her was born . what is the difference between the ages of her parents ? <o> a ) num__8 <o> b ) num__7 <o> c ) num__6 <o> d ) num__5 <o> e ) num__1 |
mother ' s age when ayesha ' s brother was born = num__36 years . father ' s age when ayesha ' s brother was born = ( num__38 + num__4 ) = num__42 years . required difference = ( num__42 - num__36 ) = num__6 years . answer : c <eor> c <eos> |
c |
add__38.0__4.0__ subtract__42.0__36.0__ divide__36.0__6.0__ |
add__38.0__4.0__ subtract__42.0__36.0__ subtract__42.0__36.0__ |
| find the expenditure on digging a well num__14 m deep and of num__3 m diameter at rs . num__15 per cubic meter ? <o> a ) num__1979 <o> b ) num__1456 <o> c ) num__1699 <o> d ) num__1485 <o> e ) num__1689 |
num__3.14285714286 * num__14 * num__1.5 * num__1.5 = num__99 m num__2 num__99 * num__15 = num__1485 answer : d <eor> d <eos> |
d |
divide__3.0__1.5__ multiply__15.0__99.0__ round__1485.0__ |
divide__3.0__1.5__ multiply__15.0__99.0__ multiply__15.0__99.0__ |
| if a new town has num__350 residents and the population doubles every num__10 years what will be its population after num__75 years ? <o> a ) num__63357 residents <o> b ) num__9051 residents <o> c ) num__12068 residents <o> d ) num__15075 residents <o> e ) num__18102 residents |
num__350 * num__2 ^ ( num__7.5 ) = num__350 * num__2 ^ num__7.5 = num__100 * num__181.02 = num__63357 the answer is a . <eor> a <eos> |
a |
divide__75.0__10.0__ multiply__350.0__181.02__ multiply__350.0__181.02__ |
divide__75.0__10.0__ multiply__350.0__181.02__ multiply__350.0__181.02__ |
| if num__4 f ^ num__4 − num__41 f ^ num__2 + num__100 = num__0 then what is the sum of the two greatest possible values of f ? <o> a ) num__4 <o> b ) num__4.5 <o> c ) num__7 <o> d ) num__10.25 <o> e ) num__25 |
just forget about the value of f ^ num__4 for the moment and let f ^ num__2 = x . ( its better to work with power of num__2 than num__4 ) now we have equation as num__4 x ^ num__2 - num__41 x + num__100 = num__0 factoring them gives us num__4 x ^ num__2 - num__16 x - num__25 x + num__100 = num__0 . which can be solved as ( num__4 x - num__25 ) ( x - num__4 ) so x = num__4 and num__6.25 . so f ^ num__2 = num__4 and num__6.25 so f = + num__2 and minus num__2 and + num__2.5 and minus num__2.5 the two greatest values are + num__2 and + num__2.5 . so their sum num__2 + num__2.5 = num__4.5 = b <eor> b <eos> |
b |
subtract__41.0__16.0__ divide__100.0__16.0__ add__2.0__2.5__ add__2.0__2.5__ |
subtract__41.0__16.0__ divide__100.0__16.0__ add__2.0__2.5__ add__2.0__2.5__ |
| num__3 candidates in an election and received num__1136 num__7636 and num__11628 votes respectively . what % of the total votes did the winningcandidate got in that election ? <o> a ) num__45.0 <o> b ) num__54.0 <o> c ) num__57.0 <o> d ) num__60.0 <o> e ) num__65 % |
total number of votes polled = ( num__1136 + num__7636 + num__11628 ) = num__20400 so required percentage = num__0.57 * num__100 = num__57.0 c <eor> c <eos> |
c |
percent__100.0__57.0__ |
percent__100.0__57.0__ |
| if x and y are consecutive integers ( y > x ) which of the following statements could be false ? num__1 . the multiplication xy is an even number . num__2 . ( y / x ) > num__1 . num__3 . ( x + y ) is always an odd number . <o> a ) num__1 only <o> b ) num__2 only . <o> c ) num__1 and num__2 only . <o> d ) num__1 and num__3 only . <o> e ) num__1 num__2 and num__3 |
consecutive numbers means that one will be even and one will be odd . num__1 . the multiplication xy is an even number . odd * even = even . true num__2 . ( y / x ) > num__1 . if y = num__1 and x = num__0 y / x = undefined or if y = - num__1 and x = - num__2 y / x = num__0.5 < num__1 could be false num__3 . ( x + y ) is always an odd number . even + odd = odd . true b is the answer <eor> b <eos> |
b |
reverse__2.0__ reverse__0.5__ |
reverse__2.0__ reverse__0.5__ |
| of the num__100 employees at company x num__80 are full - time and num__100 have worked at company x for at least a year . there are num__20 employees at company x who aren ’ t full - time and haven ’ t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__50 <o> d ) num__80 <o> e ) num__100 |
full time employee who have not worked for at least one year = a full time employee who have worked for at least one year = b non full time employee who have worked for at least one year = c non full time employee who have not worked for at least one year = d a + b + c + d = num__150 a + b = num__80 i . e . c + d = num__70 b + c = num__100 i . e . a + d = num__50 d = num__20 i . e . c = num__70 - num__20 = num__50 i . e . b = num__100 - num__50 = num__50 i . e . a = num__80 - num__50 = num__30 b = num__20 answer : option a <eor> a <eos> |
a |
subtract__150.0__80.0__ subtract__70.0__20.0__ subtract__100.0__70.0__ subtract__100.0__80.0__ |
subtract__150.0__80.0__ subtract__70.0__20.0__ subtract__100.0__70.0__ subtract__100.0__80.0__ |
| by approximately what percent is x greater than num__0.333333333333 if ( num__0.666666666667 ) ( x ) = num__1 ? <o> a ) num__173.0 <o> b ) num__516.0 <o> c ) num__461.0 <o> d ) num__350.0 <o> e ) num__290 % |
what percent is x greater than num__0.333333333333 if ( num__0.666666666667 ) ( x ) = num__1 ? = > x = num__1.5 % change = [ ( num__1.5 - num__0.333333333333 ) / ( num__0.333333333333 ) ] * num__100 = num__350 ans d num__350.0 <eor> d <eos> |
d |
multiply__1.0__350.0__ |
multiply__1.0__350.0__ |
| by travelling at num__50 kmph a person reaches his destination on time . he covered two - third the total distance in one - third of the total time . what speed should he maintain for the remaining distance to reach his destination on time ? <o> a ) num__23 kmph <o> b ) num__24 kmph <o> c ) num__25 kmph <o> d ) num__26 kmph <o> e ) num__27 kmph |
let the time taken to reach the destination be num__3 x hours . total distance = num__50 * num__3 x = num__150 x km he covered num__0.666666666667 * num__150 x = num__100 x km in num__0.333333333333 * num__3 x = x hours so the remaining num__50 x km he has to cover in num__2 x hours . required speed = num__50 x / num__2 x = num__25 kmph . answer : c <eor> c <eos> |
c |
multiply__50.0__3.0__ subtract__150.0__50.0__ divide__50.0__150.0__ divide__100.0__50.0__ divide__50.0__2.0__ round__25.0__ |
multiply__50.0__3.0__ subtract__150.0__50.0__ divide__50.0__150.0__ divide__100.0__50.0__ divide__50.0__2.0__ divide__50.0__2.0__ |
| three times the first of three consecutive odd integers is num__4 more than twice the third . the third integer is : <o> a ) num__9 <o> b ) num__11 <o> c ) num__13 <o> d ) num__16 <o> e ) num__17 |
let the three integers be x x + num__2 and x + num__4 . then num__3 x = num__2 ( x + num__4 ) + num__4 x = num__12 third integer = x + num__4 = num__16 . answer : d <eor> d <eos> |
d |
multiply__4.0__3.0__ add__4.0__12.0__ add__4.0__12.0__ |
multiply__4.0__3.0__ add__4.0__12.0__ add__4.0__12.0__ |
| a and b take part in num__100 m race . a runs at num__5 kmph . a gives b a start of num__8 m and still beats him by num__8 seconds . the speed of b is : <o> a ) num__5.15 kmph <o> b ) num__4.14 kmph <o> c ) num__4.25 kmph <o> d ) num__4.4 kmph <o> e ) num__4.8 kmph |
in num__100 m race a ' s speed is num__5 kmph so in m / s it is num__5 * num__0.277777777778 time taken by a = d / s = num__100 * num__0.72 = num__72 sec now b would take num__72 + num__8 = num__80 seconds speed of b is = ( num__100 - num__8 ) = ( num__1.15 ) * num__3.6 = num__4.14 kmph answer : b <eor> b <eos> |
b |
multiply__100.0__0.72__ add__8.0__72.0__ multiply__5.0__0.72__ multiply__3.6__1.15__ round__4.14__ |
multiply__100.0__0.72__ add__8.0__72.0__ multiply__5.0__0.72__ multiply__3.6__1.15__ multiply__3.6__1.15__ |
| john and jacob set out together on bicycle traveling at num__15 and num__10 miles per hour respectively . after num__40 minutes john stops to fix a flat tire . if it takes john one hour to fix the flat tire and jacob continues to ride during this time how many hours will it take john to catch up to jacob assuming he resumes his ride at num__15 miles per hour ? ( consider john ' s deceleration / acceleration before / after the flat to be negligible ) <o> a ) num__1 num__0.5 <o> b ) num__3 num__0.333333333333 <o> c ) num__3 num__0.5 <o> d ) num__4 <o> e ) num__4 num__0.5 |
john ' s speed - num__15 miles / hr jacob ' s speed - num__10 miles / hr after num__40 min ( i . e num__0.666666666667 hr ) distance covered by john = num__15 x num__0.666666666667 = num__10 miles . jacob continues to ride for a total of num__1 hour and num__40 min ( until john ' s bike is repaired ) . distance covered in num__1 hour num__40 min ( i . e num__1.66666666667 hr ) = num__10 x num__1.66666666667 = num__16.7 miles . now when john starts riding back the distance between them is num__6.7 miles . jacob and john are moving in the same direction . for john to catch jacob the effective relative speed will be num__15 - num__10 = num__5 miles / hr . thus to cover num__6.7 miles at num__5 miles / hr john will take num__6.7 / num__5 = num__1.34 hours answer a <eor> a <eos> |
a |
divide__10.0__15.0__ add__1.0__0.6667__ subtract__16.7__10.0__ subtract__15.0__10.0__ divide__6.7__5.0__ reverse__1.0__ |
divide__10.0__15.0__ add__1.0__0.6667__ subtract__16.7__10.0__ subtract__15.0__10.0__ divide__6.7__5.0__ reverse__1.0__ |
| a fruit seller had some mangoes . he sells num__30.0 mangoes and still has num__280 mangoes . find the number of mangoes he had . <o> a ) num__228 mangoes <o> b ) num__400 mangoes <o> c ) num__287 mangoes <o> d ) num__408 mangoes <o> e ) num__289 mangoes |
explanation : suppose originally he had x mangoes . then ( num__100 – num__30 ) % of x = num__280 num__0.7 * x = num__280 x = ( num__280 * num__100 ) / num__70 = num__400 answer : b <eor> b <eos> |
b |
percent__100.0__400.0__ |
percent__100.0__400.0__ |
| a certain sum of money at simple interest amounted rs . num__860 in num__10 years at num__3.0 per annum find the sum ? <o> a ) num__661.54 <o> b ) num__662.54 <o> c ) num__663.54 <o> d ) num__664.54 <o> e ) num__665.54 |
num__860 = p [ num__1 + ( num__10 * num__3 ) / num__100 ] p = num__661.54 answer : a <eor> a <eos> |
a |
percent__100.0__661.54__ |
percent__100.0__661.54__ |
| a boat running downstream covers a distance of num__12 km in num__2 hours while for covering the same distance upstream it takes num__4 hours . what is the speed of the boat in still water ? <o> a ) num__4.5 km / hr <o> b ) num__6 km / hr <o> c ) num__8 km / hr <o> d ) data inadequate <o> e ) none of these |
solution rate downstream = ( num__6.0 ) kmph = num__6 kmph rate upstream = ( num__3.0 ) = num__3 kmph . ∴ speed in still water = num__0.5 ( num__6 + num__3 ) kmph = num__4.5 kmph . answer a <eor> a <eos> |
a |
divide__12.0__2.0__ divide__12.0__4.0__ divide__2.0__4.0__ add__4.0__0.5__ round__4.5__ |
add__2.0__4.0__ divide__12.0__4.0__ divide__2.0__4.0__ add__4.0__0.5__ add__4.0__0.5__ |
| a sum of money was invested in a bank at num__8.0 simple interest p . a . for num__3 years . instead had it been invested in mutual fund at num__8.5 p . a . simple interest for num__4 years the earning would have been rs . num__500 more . what is the sum invested ? <o> a ) rs . num__4500 <o> b ) rs . num__5000 <o> c ) rs . num__3500 <o> d ) rs . num__5500 <o> e ) rs . num__6500 |
let the sum be rs . x s . i from the bank = x * num__8 * num__0.03 = num__34 x / num__100 earnings in the form of interest from mutual fund = ( x * num__8.5 * num__4 ) / num__100 = num__34 x / num__100 given that num__34 x / num__100 - num__34 x / num__100 = rs . num__500 ; x = num__5000 : . the sum invested = num__5000 answer : b <eor> b <eos> |
b |
percent__100.0__5000.0__ |
percent__100.0__5000.0__ |
| in traveling from a dormitory to a certain city a student went num__0.2 of the way by foot num__0.666666666667 of the way by bus and the remaining num__6 kilometers by car . what is the distance in kilometers from the dormitory to the city ? <o> a ) num__30 <o> b ) num__45 <o> c ) num__60 <o> d ) num__90 <o> e ) num__120 |
whole trip = distance by foot + distance by bus + distance by car x = num__0.2 x + num__0.666666666667 x + num__6 x - num__0.866666666667 x = num__6 num__0.133333333333 x = num__6 = > so x = ( num__7.5 ) * num__6 = num__45 km answer b <eor> b <eos> |
b |
add__0.2__0.6667__ multiply__0.2__0.6667__ multiply__6.0__7.5__ round__45.0__ |
add__0.2__0.6667__ multiply__0.2__0.6667__ multiply__6.0__7.5__ multiply__6.0__7.5__ |
| the population of a bacteria colony doubles every day . if it was started num__8 days ago with num__3 bacteria and each bacteria lives for num__12 days how large is the colony today ? <o> a ) num__512 <o> b ) num__768 <o> c ) num__13122 <o> d ) num__2048 <o> e ) num__4096 |
num__3 ^ num__8 * ( num__2 ) = num__13122 the answer is c <eor> c <eos> |
c |
round__13122.0__ |
round__13122.0__ |
| there are num__20 poles with a constant distance between each pole . a car takes num__24 seconds to reach the num__12 th pole . how much time will it take to reach the last pole ? <o> a ) num__25.25 s <o> b ) num__17.45 s <o> c ) num__35.75 s <o> d ) num__41.45 s <o> e ) none of these |
let the distance between each pole be x m . then the distance up to num__12 th pole = num__11 xm speed = num__11 x ⁄ num__24 m / s time taken to covers the total distance of num__19 x = num__19 x × num__2.18181818182 x = num__41.45 s answer d <eor> d <eos> |
d |
divide__24.0__11.0__ round__41.45__ |
divide__24.0__11.0__ round__41.45__ |
| a train num__175 m long running at num__36 kmph crosses a platform in num__40 sec . what is the length of the platform ? <o> a ) num__271 <o> b ) num__266 <o> c ) num__225 <o> d ) num__277 <o> e ) num__232 |
length of the platform = num__36 * num__0.277777777778 * num__40 = num__400 – num__175 = num__225 answer : c <eor> c <eos> |
c |
subtract__400.0__175.0__ round__225.0__ |
subtract__400.0__175.0__ round__225.0__ |
| a certain company charges $ num__25 per package to ship packages weighing less than num__5 pounds each . for a package weighing num__5 pounds or more the company charges an initial fee of $ num__25 plus $ num__6 per pound . if the company charged $ num__36 to ship a certain package which of the following was the weight of the package in pounds ? <o> a ) num__10 <o> b ) num__6 <o> c ) num__12 <o> d ) num__18 <o> e ) num__20 |
let the weight of the package be x . then the cost is num__6 * x + num__25 = num__61 - - - > num__6 x = num__36 - - - > x = num__6 answer : b <eor> b <eos> |
b |
add__25.0__36.0__ divide__36.0__6.0__ |
add__25.0__36.0__ divide__36.0__6.0__ |
| a trader bought a car at num__15.0 discount on its original price . he sold it at a num__40.0 increase on the price he bought it . what percent of profit did he make on the original price ? <o> a ) num__119 <o> b ) num__110 <o> c ) num__112 <o> d ) num__113 <o> e ) num__115 |
original price = num__100 cp = num__85 s = num__85 * ( num__1.4 ) = num__119 num__100 - num__119 = num__19.0 answer : a <eor> a <eos> |
a |
percent__100.0__119.0__ |
percent__100.0__119.0__ |
| machine a and machine b are each used to manufacture num__330 sprockets . it takes machine a num__10 hours longer to produce num__330 sprockets than machine b . machine b produces num__10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__3 |
machine b : takes x hours to produce num__330 sprockets machine a : takes ( x + num__10 ) hours to produce num__330 sprockets machine b : in num__1 hour b makes num__330 / x sprockets machine a : in num__1 hour a makes num__330 / ( x + num__10 ) sprockets equating : num__1.1 ( num__330 / ( x + num__10 ) ) = num__330 / x num__484 / ( x + num__10 ) = num__330 / x num__363 x = num__330 x + num__3300 num__33 x = num__3300 x = num__100 a makes num__330 / ( num__110 ) = num__3 sprockets per hour answer : e <eor> e <eos> |
e |
percent__10.0__330.0__ percent__3.0__100.0__ |
percent__10.0__330.0__ percent__3.0__100.0__ |
| a deer is standing num__70 meters in from the west end of a tunnel . the deer sees a train approaching from the west at a constant speed ten times the speed the deer can run . the deer reacts by running toward the train and clears the exit when the train is num__40 meters from the tunnel . if the deer ran in the opposite direction it would barely escape out the eastern entrance just as the train came out of the eastern entrance . how long is the tunnel in meters ? <o> a ) num__160 <o> b ) num__240 <o> c ) num__520 <o> d ) num__190 <o> e ) num__720 |
let x be the length of the tunnel . when the deer runs num__70 meters west the train goes num__700 meters to a point num__40 meters from the west entrance of the tunnel . when the deer runs east the deer runs x - num__70 meters while the train goes x + num__700 + num__40 meters . x + num__700 + num__40 = num__10 ( x - num__70 ) num__9 x = num__1440 x = num__160 meters the answer is a . <eor> a <eos> |
a |
divide__700.0__70.0__ divide__1440.0__9.0__ round__160.0__ |
divide__700.0__70.0__ divide__1440.0__9.0__ round__160.0__ |
| in a certificate by mistake a candidate gave his height as num__25.0 more than actual height . in the interview panel he clarified that his height was num__5 feet num__5 nches . find the % correction made by the candidate from his stated height to his actual height ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__40 <o> d ) num__50 <o> e ) num__60 |
his height was = num__5 feet num__5 inch = num__5 + num__60 = num__65 inch . required % correction = num__65 * ( num__1.25 - num__1 ) * num__100 = num__20 b <eor> b <eos> |
b |
percent__25.0__5.0__ percent__100.0__20.0__ |
percent__25.0__5.0__ percent__100.0__20.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 seconds . find the length of the train . <o> a ) num__150 <o> b ) num__160 <o> c ) num__180 <o> d ) num__170 <o> e ) num__190 |
speed = num__60 * ( num__0.277777777778 ) m / sec = num__16.6666666667 m / sec length of train ( distance ) = speed * time ( num__16.6666666667 ) * num__9 = num__150 meter answer : option a <eor> a <eos> |
a |
round__150.0__ |
round__150.0__ |
| a num__6 ' ' cube is painted in all its faces and then it is cut down into num__1 ' ' blocks . how many num__1 ' ' blocks are there even without a single face being painted ? <o> a ) num__24 <o> b ) num__32 <o> c ) num__64 <o> d ) num__80 <o> e ) num__100 |
the unpainted blocks are the interior blocks . these blocks form a num__4 ' ' cube on the inside . the number of unpainted blocks is num__4 * num__4 * num__4 = num__64 blocks . the answer is c . <eor> c <eos> |
c |
square_perimeter__1.0__ volume_cube__4.0__ volume_cube__4.0__ |
square_perimeter__1.0__ volume_cube__4.0__ multiply__1.0__64.0__ |
| a shopkeeper purchased num__85 kg of potatoes for rs . num__595 and sold the whole lot at the rate of rs . num__8 per kg . what will be his gain percent ? <o> a ) num__18 num__0.111111111111 % <o> b ) num__18 num__3.0 % <o> c ) num__18 num__0.333333333333 % <o> d ) num__14 num__0.285714285714 % <o> e ) num__18 num__2.33333333333 % |
c . p . of num__1 kg = num__7.0 = rs . num__7 s . p . of num__1 kg = rs . num__8 gain % = num__0.142857142857 * num__100 = num__14.2857142857 = num__14 num__0.285714285714 % answer : d <eor> d <eos> |
d |
percent__100.0__14.0__ |
percent__100.0__14.0__ |
| a man spends num__0.4 of his salary on house rent num__0.3 of his salary on food and num__0.125 of his salary on conveyance . if he has $ num__1400 left with him find his expenditure on food . <o> a ) $ num__400 <o> b ) $ num__1400 <o> c ) $ num__3400 <o> d ) $ num__2400 <o> e ) $ num__4400 |
part of salary left = num__1 - ( num__0.4 + num__0.3 + num__0.125 ) = num__0.175 let the monthly salary be $ x then ( num__0.175 ) of x = num__1400 x = num__8000 expenditure on food = $ num__2400 option d <eor> d <eos> |
d |
subtract__0.3__0.125__ divide__1400.0__0.175__ multiply__0.3__8000.0__ multiply__0.3__8000.0__ |
subtract__0.3__0.125__ divide__1400.0__0.175__ multiply__0.3__8000.0__ multiply__0.3__8000.0__ |
| how many different positive integers are factors of num__20 ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__4 <o> d ) num__10 <o> e ) num__12 |
num__2 * num__10 num__4 * num__5 answer : c <eor> c <eos> |
c |
divide__20.0__2.0__ divide__20.0__4.0__ gcd__20.0__4.0__ |
divide__20.0__2.0__ divide__20.0__4.0__ gcd__20.0__4.0__ |
| in a market a dozen eggs cost as much as a pound of rice and a half - liter of kerosene costs as much as num__8 eggs . if the cost of each pound of rice is $ num__0.33 then how many q cents does a liter of kerosene cost ? [ one dollar has num__100 cents . ] <o> a ) num__0.33 <o> b ) num__0.44 <o> c ) num__0.55 <o> d ) num__44 <o> e ) num__55 |
main thing to remember is answer is asked in cents however when we calculate it comes up as num__0.44 $ just multiply by num__100 answer q = num__44 . d <eor> d <eos> |
d |
multiply__100.0__0.44__ multiply__100.0__0.44__ |
multiply__100.0__0.44__ multiply__100.0__0.44__ |
| if the cost price of num__50 articles is equal to the selling price of num__15 articles then the gain or loss percent is ? <o> a ) num__16 <o> b ) num__127 <o> c ) num__12 <o> d ) num__18 <o> e ) num__233 |
percentage of profit = num__2.33333333333 * num__100 = num__233.0 answer : e <eor> e <eos> |
e |
percent__100.0__233.0__ |
percent__100.0__233.0__ |
| chris mixed num__3 pounds of raisins with num__4 pounds of nuts . if a pound of nuts costs num__4 times as much as a pound of raisins then the total cost of the raisins was what fraction of the total cost of the mixture ? <o> a ) num__0.142857142857 <o> b ) num__0.2 <o> c ) num__0.25 <o> d ) num__0.333333333333 <o> e ) num__0.157894736842 |
num__1 lbs of raisin = $ num__1 num__3 lbs of raisin = $ num__3 num__1 lbs of nuts = $ num__4 num__4 lbs of nuts = $ num__16 total value of mixture = num__16 + num__3 = num__19 fraction of the value of raisin = num__0.157894736842 ans : e <eor> e <eos> |
e |
subtract__4.0__3.0__ add__3.0__16.0__ divide__3.0__19.0__ divide__3.0__19.0__ |
subtract__4.0__3.0__ add__3.0__16.0__ divide__3.0__19.0__ divide__3.0__19.0__ |
| a rectangular field has area equal to num__150 sq m and perimeter num__50 m . its length and breadth must be ? <o> a ) num__10 <o> b ) num__88 <o> c ) num__66 <o> d ) num__65 <o> e ) num__22 |
answer : option d explanation : lb = num__150 num__2 ( l + b ) = num__50 = > l + b = num__25 l – b = num__5 l = num__15 b = num__10 answer : a <eor> a <eos> |
a |
multiply__2.0__5.0__ multiply__2.0__5.0__ |
multiply__2.0__5.0__ multiply__2.0__5.0__ |
| two trains num__85 m and num__75 m long are running in same direction with speeds of num__62 km / hr and num__44 km / hr respectively . in what time will the first train cross the second train ? <o> a ) num__12 sec <o> b ) num__32 sec <o> c ) num__82 sec <o> d ) num__92 sec <o> e ) num__13 sec |
explanation : given that num__1 st train speed = num__62 km / hr num__2 nd train speed = num__44 km / hr hence the relative speed of two trains is num__62 - num__44 = num__18 km / hr m / s = num__5 m / s . there fore the first train crosses the second train i . e ( num__85 + num__75 ) mts of distance with a speed of num__5 m / s in time = \ frac { distance } { speed } \ frac { ( num__85 + num__75 ) } { num__5 } = num__32 sec . answer : b <eor> b <eos> |
b |
subtract__62.0__44.0__ round__32.0__ |
subtract__62.0__44.0__ divide__32.0__1.0__ |
| stacy has a num__63 page history paper due in num__3 days . how many pages per day would she have to write to finish on time ? <o> a ) num__9 <o> b ) num__8 <o> c ) num__10 <o> d ) num__8.5 <o> e ) num__21 |
num__21.0 = num__21 answer : e <eor> e <eos> |
e |
divide__63.0__3.0__ round__21.0__ |
divide__63.0__3.0__ round__21.0__ |
| a furniture dealer purchased a desk for $ num__150 and then set the selling price equal to the purchase price plus a markup that was num__25.0 of the selling price . if the dealer sold the desk at the selling price what was the amount of the dealer ' s gross profit from the purchase and the sale of the desk ? <o> a ) num__40.0 <o> b ) num__38.0 <o> c ) num__36.0 <o> d ) num__33.33 <o> e ) num__35 % |
anyway in this question there is no discount but the mark up is given as num__25.0 of the selling price . so it is not num__25.0 of $ num__150 but instead num__40.0 of selling price which is obtained by adding mark up to $ num__150 . so if selling price is s num__150 + num__25.0 of s = s s = num__200 profit = num__50 which is calculated on cost price in % terms . so num__0.333333333333 * num__100 = num__33.33 is profit . d <eor> d <eos> |
d |
percent__25.0__200.0__ percent__50.0__200.0__ percent__33.33__100.0__ |
percent__25.0__200.0__ percent__50.0__200.0__ percent__33.33__100.0__ |
| the average of runs of a cricket player of num__10 innings was num__15 . how many runs must he make in his next innings so as to increase his average of runs by num__4 ? <o> a ) num__87 <o> b ) num__16 <o> c ) num__10 <o> d ) num__76 <o> e ) num__59 |
average after num__11 innings = num__19 required number of runs = ( num__19 * num__11 ) - ( num__15 * num__10 ) = num__209 - num__150 = num__59 . answer : e <eor> e <eos> |
e |
subtract__15.0__4.0__ add__15.0__4.0__ multiply__11.0__19.0__ multiply__10.0__15.0__ subtract__209.0__150.0__ subtract__209.0__150.0__ |
subtract__15.0__4.0__ add__15.0__4.0__ multiply__11.0__19.0__ multiply__10.0__15.0__ subtract__209.0__150.0__ subtract__209.0__150.0__ |
| insert the missing number . num__2 num__7 num__8 num__17 num__16 num__27 num__22 num__37 <o> a ) num__42 <o> b ) num__52 <o> c ) num__46 <o> d ) num__28 <o> e ) num__68 |
explanation : there are two series here num__2 num__8 num__16 num__22 . . . ( increase by num__6 ) num__7 num__17 num__27 num__37 . . . ( increase by num__10 ) hence next term is num__22 + num__6 = num__28 answer : option d <eor> d <eos> |
d |
subtract__8.0__2.0__ add__2.0__8.0__ add__22.0__6.0__ add__22.0__6.0__ |
subtract__8.0__2.0__ add__2.0__8.0__ add__22.0__6.0__ add__22.0__6.0__ |
| if the diagonal of a rectangle is num__17 cm long and its perimeter is num__46 cm . find the area of the rectangle ? <o> a ) num__122 sq . cm <o> b ) num__120 sq . cm <o> c ) num__277 sq . cm <o> d ) num__118 sq . cm <o> e ) num__119 sq . cm |
let length = x and breadth = y then num__2 ( x + y ) = num__46 x + y = num__23 x ² + y ² = num__17 ² = num__289 now ( x + y ) ² = num__23 ² x ² + y ² + num__2 xy = num__529 num__289 + num__2 xy = num__529 xy = num__120 area = xy = num__120 sq . cm answer : b <eor> b <eos> |
b |
power__17.0__2.0__ triangle_area__46.0__23.0__ triangle_area__2.0__120.0__ |
power__17.0__2.0__ triangle_area__46.0__23.0__ triangle_area__2.0__120.0__ |
| ricky likes to keep a spare tyre in his car every time . on a certain day he travels num__1 num__00000 km and just to make the most of all the tyres he changes the tyres between his journey such that each tyre runs the same distance . what is the distance travelled by each tyre ? <o> a ) num__20000 <o> b ) num__35000 <o> c ) num__70000 <o> d ) num__60000 <o> e ) num__80000 |
e num__80000 . explanation : the distance travelled by each tyre : num__0.8 * num__1 num__00000 km = num__80000 km . <eor> e <eos> |
e |
round__80000.0__ |
multiply__1.0__80000.0__ |
| a man in a train notices that he can count num__41 telephone posts in one minute . if they are known to be num__50 metres apart then at what speed is the train travelling ? <o> a ) num__60 km / hr <o> b ) num__100 km / hr <o> c ) num__110 km / hr <o> d ) num__120 km / hr <o> e ) none of these |
explanation : number of gaps between num__41 poles = num__40 so total distance between num__41 poles = num__40 * num__50 = num__2000 meter = num__2 km in num__1 minute train is moving num__2 km / minute . speed in hour = num__2 * num__60 = num__120 km / hour answer : d <eor> d <eos> |
d |
multiply__50.0__40.0__ subtract__41.0__40.0__ hour_to_min_conversion__ multiply__2.0__60.0__ round__120.0__ |
multiply__50.0__40.0__ subtract__41.0__40.0__ hour_to_min_conversion__ multiply__2.0__60.0__ multiply__2.0__60.0__ |
| the standard serial numbers for a bill are num__2 letters followed by num__5 digits . how many bills are possible if letters and digits can be repeated ? <o> a ) num__26 × num__26 × num__100 num__000 <o> b ) num__24 ³ × num__10 × num__10 num__000 <o> c ) num__26 ³ × num__10 num__000 <o> d ) num__26 × num__25 × num__24 × num__10 <o> e ) num__26 x num__26 x num__10 num__000 |
official solution : ( a ) the formula for permutations of events is the product of the number of ways each event can occur . there are num__26 letters and num__10 digits . so there are num__26 × num__26 options for the two letters and num__10 × num__10 × num__10 × num__10 x num__10 for the five digits . the number of bills is num__26 × num__26 × num__10 × num__10 × num__10 × num__10 x num__10 = num__26 x num__26 x num__100 num__000 . the correct answer is choice ( a ) . <eor> a <eos> |
a |
alphabet_space__ alphabet_space__ |
alphabet_space__ alphabet_space__ |
| in may mrs lee ' s earnings were num__60 percent of the lee family ' s total income . in june mrs lee earned num__20 percent more than in may . if the rest of the family ' s income was the same both months then in june mrs lee ' s earnings were approximately what percent of the lee family ' s total income ? <o> a ) num__64.0 <o> b ) num__68.0 <o> c ) num__72.0 <o> d ) num__76.0 <o> e ) num__80 % |
say may income = num__100 l ' s income = num__60 and rest of the family = num__40 in june l ' s income = num__60 * num__1.2 = num__72 so num__1.0 + num__40 = num__64.0 answer : a <eor> a <eos> |
a |
subtract__60.0__20.0__ multiply__60.0__1.2__ round_down__1.2__ multiply__64.0__1.0__ |
subtract__60.0__20.0__ multiply__60.0__1.2__ round_down__1.2__ multiply__64.0__1.0__ |
| a trader bought a car at num__25.0 discount on its original price . he sold it at a num__40.0 increase on the price he bought it . what percent of profit did he make on the original price ? <o> a ) num__118 <o> b ) num__110 <o> c ) num__105 <o> d ) num__113 <o> e ) num__115 |
original price = num__100 cp = num__75 s = num__75 * ( num__1.4 ) = num__105 num__100 - num__105 = num__5.0 answer : c <eor> c <eos> |
c |
percent__100.0__105.0__ |
percent__100.0__105.0__ |
| what is the total number of positive integers that are less than num__1000 and that have no positive factor in common with num__1000 other than num__1 ? <o> a ) num__100 <o> b ) num__200 <o> c ) num__300 <o> d ) num__400 <o> e ) num__500 |
since num__1000 = num__2 ^ num__3 * num__5 ^ num__3 then a number can not have num__2 and / or num__5 as a factor . the odd numbers do not have num__2 as a factor and there are num__500 odd numbers from num__1 to num__1000 . we then need to eliminate the num__100 numbers that end with num__5 that is num__5 num__15 num__25 . . . num__995 . there are a total of num__500 - num__100 = num__400 such numbers between num__1 and num__1000 . the answer is d . <eor> d <eos> |
d |
add__1.0__2.0__ add__2.0__3.0__ divide__1000.0__2.0__ divide__500.0__5.0__ multiply__3.0__5.0__ subtract__1000.0__5.0__ subtract__500.0__100.0__ multiply__1.0__400.0__ |
add__1.0__2.0__ add__2.0__3.0__ divide__1000.0__2.0__ divide__500.0__5.0__ multiply__3.0__5.0__ subtract__1000.0__5.0__ subtract__500.0__100.0__ multiply__1.0__400.0__ |
| a reduction of num__10.0 in the price of rice enables a lady to obtain num__20 kgs more for rs . num__1000 find the original price per kg ? <o> a ) num__6.0 <o> b ) num__5.0 <o> c ) num__5.2 <o> d ) num__5.6 <o> e ) num__5.4 |
answer : num__1000 * ( num__0.1 ) = num__100 - - - num__20 ? - - - num__1 = > rs . num__5 num__1000 - - - num__900 ? - - - num__5 = > rs . num__5.6 . answer : d <eor> d <eos> |
d |
percent__10.0__1000.0__ percent__0.1__1000.0__ percent__100.0__5.6__ |
percent__10.0__1000.0__ percent__0.1__1000.0__ percent__100.0__5.6__ |
| set a = { num__1 num__2 num__3 num__4 num__5 num__6 q } which of the following possible values for q would cause set a to have the smallest standard deviation ? <o> a ) num__1 <o> b ) num__2.5 <o> c ) num__3 <o> d ) num__3.5 <o> e ) num__7 |
i agree . the mean of the set q = num__12 num__34 num__56 is num__3.5 . now if we add one extra number to the set in order for the standard deviation to be minimum that number must be as close as possible to the original set ( num__12 num__34 num__56 ) . therefore we have to choose the number closest to num__35 from the options we have leaving d as the best choice . answer d . <eor> d <eos> |
d |
multiply__2.0__6.0__ add__1.0__34.0__ multiply__1.0__3.5__ |
multiply__2.0__6.0__ add__1.0__34.0__ multiply__1.0__3.5__ |
| if x = the product of five distinct prime numbers how many factors does x have besides num__1 and itself ? <o> a ) num__24 <o> b ) num__30 <o> c ) num__36 <o> d ) num__42 <o> e ) num__48 |
since x has num__5 distinct prime factors x has a total of num__2 ^ num__5 = num__32 factors . besides num__1 and itself x has num__30 factors . the answer is b . <eor> b <eos> |
b |
subtract__32.0__2.0__ multiply__1.0__30.0__ |
subtract__32.0__2.0__ multiply__1.0__30.0__ |
| pipe p can drain the liquid from a tank in num__0.75 the time that it takes pipe q to drain it and in num__1.0 the time that it takes pipe r to do it . if all num__3 pipes operating simultaneously but independently are used to drain liquid from the tank then pipe q drains what portion of the liquid from the tank ? <o> a ) num__0.310344827586 <o> b ) num__0.347826086957 <o> c ) num__0.375 <o> d ) num__0.272727272727 <o> e ) num__0.75 |
suppose q can drain in num__1 hr . so rq = num__1.0 = num__1 so rp = num__1 / [ ( num__0.75 ) rq ] = num__1.33333333333 also rp = rr / ( num__1.0 ) = > num__1.33333333333 = rr / ( num__1.0 ) = > rr = num__1.33333333333 let h is the time it takes to drain by running all num__3 pipes simultaneously so combined rate = rc = num__1 / h = num__1 + num__1.33333333333 + num__1.33333333333 = num__3.66666666667 = num__1 / ( num__0.272727272727 ) thus running simultaneously pipe q will drain num__0.272727272727 of the liquid . thus answer = d . <eor> d <eos> |
d |
divide__1.0__0.75__ divide__1.0__3.6667__ multiply__1.0__0.2727__ |
divide__1.0__0.75__ divide__1.0__3.6667__ divide__1.0__3.6667__ |
| the length of the bridge which a train num__130 meters long and travelling at num__45 km / hr can cross in num__30 seconds is : <o> a ) num__2399 <o> b ) num__277 <o> c ) num__245 <o> d ) num__88 <o> e ) num__232 |
speed = ( num__45 * num__0.277777777778 ) m / sec = ( num__12.5 ) m / sec . time = num__30 sec . let the length of bridge be x meters . then ( num__130 + x ) / num__30 = num__12.5 = = > num__2 ( num__130 + x ) = num__750 = = > x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| in a factory an average of num__60 tv ' s are produced per day for the fist num__25 days of the months . a few workers fell ill for the next num__5 days reducing the daily avg for the month to num__58 sets / day . the average production per day for day last num__5 days is ? <o> a ) num__20 <o> b ) num__36 <o> c ) num__48 <o> d ) num__50 <o> e ) num__59 |
production during these num__5 days = total production in a month - production in first num__25 days . = num__30 x num__58 - num__25 x num__60 = num__240 ∴ average for last num__5 days = num__48.0 = num__48 c <eor> c <eos> |
c |
add__25.0__5.0__ divide__240.0__5.0__ divide__240.0__5.0__ |
add__25.0__5.0__ divide__240.0__5.0__ divide__240.0__5.0__ |
| what least number must be added to num__9599 so that the sum is completely divisible by num__100 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
if we divide num__9599 by num__100 remainder is num__99 num__100 - num__99 = num__1 answer : a <eor> a <eos> |
a |
subtract__100.0__99.0__ reverse__1.0__ |
subtract__100.0__99.0__ subtract__100.0__99.0__ |
| a can do a piece of work in num__8 days . b can do it in num__12 days . with the assistance of c they completed the work in num__4 days . find in how many days can c alone do it ? <o> a ) num__87 days <o> b ) num__20 days <o> c ) num__16 days <o> d ) num__24 days <o> e ) num__36 days |
c = num__0.25 - num__0.125 - num__0.0833333333333 = num__0.0416666666667 = > num__24 days answer : d <eor> d <eos> |
d |
subtract__0.125__0.0833__ round__24.0__ |
subtract__0.125__0.0833__ round__24.0__ |
| the speed at which a boy can row a boat in still water is num__60 kmph . if he rows downstream where the speed of current is num__12 kmph what time will he take to cover num__220 metres ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__11 |
speed of the boat downstream = num__60 + num__12 = num__72 kmph = num__72 * num__0.277777777778 = num__20 m / s hence time taken to cover num__220 m = num__11.0 = num__11 seconds . answer : e <eor> e <eos> |
e |
add__60.0__12.0__ divide__220.0__20.0__ round__11.0__ |
add__60.0__12.0__ divide__220.0__20.0__ divide__220.0__20.0__ |
| a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by num__50.0 while buying and by num__10.0 while selling . what is his percentage profit ? <o> a ) num__10.22 <o> b ) num__20.22 <o> c ) num__21.22 <o> d ) num__50.0 <o> e ) ca n ' t be calculated |
the owner buys num__100 kg but actually gets num__150 kg ; the owner sells num__100 kg but actually gives num__90 kg ; profit : ( num__150 - num__90 ) / num__90 * num__100 = num__50.0 answer : d . <eor> d <eos> |
d |
percent__50.0__100.0__ |
percent__50.0__100.0__ |
| a train num__360 m long is running at a speed of num__45 km / hr . in what time will it pass a bridge num__140 m long ? <o> a ) num__40 sec <o> b ) num__65 sec <o> c ) num__18 sec <o> d ) num__16 sec <o> e ) num__17 sec |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec total distance covered = num__360 + num__140 = num__500 m required time = num__500 * num__0.08 = num__40 sec answer : a <eor> a <eos> |
a |
add__360.0__140.0__ divide__500.0__12.5__ round__40.0__ |
add__360.0__140.0__ divide__500.0__12.5__ divide__500.0__12.5__ |
| after taking n tests each containing num__100 questions john had an average of num__70.0 of correct answers . how much does john need to score on the next test to make his average equal num__75.0 ? m num__13 - num__03 . <o> a ) n − num__35 <o> b ) n + num__72 <o> c ) num__2 n + num__78 <o> d ) num__2 n + num__72 <o> e ) num__2 n − num__35 |
say n = num__1 . so after num__1 test john has num__70 correct answers . in num__2 tests so in num__200 questions he needs to have num__0.75 * num__200 = num__150 correct answers so in the second test he must get num__150 - num__70 = num__80 questions correctly . now plug n = num__1 into the answer choices to see which one yields num__80 . only option d fits . answer : c . <eor> c <eos> |
c |
subtract__3.0__1.0__ multiply__100.0__2.0__ divide__75.0__100.0__ multiply__75.0__2.0__ subtract__150.0__70.0__ subtract__3.0__1.0__ |
subtract__3.0__1.0__ multiply__100.0__2.0__ divide__75.0__100.0__ multiply__75.0__2.0__ subtract__150.0__70.0__ subtract__3.0__1.0__ |
| what is the greatest power that num__3 can be raised to so that the resulting number is a factor of num__9 ! ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
the number of num__3 s in num__9 ! = num__3.0 = num__3 . . ans b <eor> b <eos> |
b |
gcd__3.0__9.0__ |
gcd__3.0__9.0__ |
| the bus fare for two persons for travelling between agra and aligarh id four - thirds the train fare between the same places for one person . the total fare paid by num__6 persons travelling by bus and num__8 persons travelling by train between the two places is rs . num__1512 . find the train fare between the two places for one person ? <o> a ) rs . num__126 <o> b ) rs . num__132 <o> c ) rs . num__120 <o> d ) rs . num__114 <o> e ) none of these |
let the train fare between the two places for one person be rs . t bus fare between the two places for two persons rs . num__1.33333333333 t = > num__3.0 ( num__1.33333333333 t ) + num__8 ( t ) = num__1512 = > num__12 t = num__1512 = > t = num__126 . answer : a <eor> a <eos> |
a |
divide__8.0__6.0__ divide__1512.0__12.0__ divide__1512.0__12.0__ |
divide__8.0__6.0__ divide__1512.0__12.0__ divide__1512.0__12.0__ |
| a can do a piece of work in num__21 days and b in num__28 days . together they started the work and b left after num__4 days . in how many days can a alone do the remaining work ? <o> a ) num__14 <o> b ) num__8 <o> c ) num__20 <o> d ) num__10 <o> e ) num__7 |
option d explanation : let a worked for x days . x / num__21 + num__0.142857142857 = num__1 = > x / num__21 = num__0.857142857143 = > x = num__18 a worked for num__18 days . so a can complete the remaining work in num__18 - num__4 = num__14 days . <eor> a <eos> |
a |
divide__4.0__28.0__ subtract__1.0__0.1429__ subtract__18.0__4.0__ round__14.0__ |
divide__4.0__28.0__ subtract__1.0__0.1429__ subtract__18.0__4.0__ subtract__28.0__14.0__ |
| if a and b are the two values of t that satisfy the equation t ^ num__2 â € “ num__7 t + num__10 = num__0 with a > b what is the value of a â € “ b ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
factor the left side of the equation : t ^ num__2 â € “ num__7 t + num__10 = num__0 ( t â € “ num__2 ) ( t â € “ num__5 ) = num__0 t = num__2 t = num__5 thus a = num__5 and b = num__2 . so a â € “ b = num__5 â € “ num__2 = num__3 . the answer is b . <eor> b <eos> |
b |
subtract__7.0__2.0__ subtract__10.0__7.0__ subtract__10.0__7.0__ |
subtract__7.0__2.0__ subtract__10.0__7.0__ subtract__10.0__7.0__ |
| a train passes a station platform in num__38 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__328 <o> b ) num__279 <o> c ) num__270 <o> d ) num__288 <o> e ) num__231 |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x meters . then x + num__7.89473684211 = num__15 x + num__300 = num__570 x = num__270 m . answer : c <eor> c <eos> |
c |
multiply__20.0__15.0__ divide__300.0__38.0__ multiply__38.0__15.0__ subtract__570.0__300.0__ round__270.0__ |
multiply__20.0__15.0__ divide__300.0__38.0__ multiply__38.0__15.0__ subtract__570.0__300.0__ round__270.0__ |
| raman mixed num__24 kg of butter at rs . num__150 per kg with num__36 kg butter at the rate of rs . num__125 per kg . at what price per kg should he sell the mixture to make a profit of num__40.0 in the transaction ? <o> a ) num__228 <o> b ) num__2767 <o> c ) num__189 <o> d ) num__266 <o> e ) num__211 |
cp per kg of mixture = [ num__24 ( num__150 ) + num__36 ( num__125 ) ] / ( num__24 + num__36 ) = rs . num__135 sp = cp [ ( num__100 + profit % ) / num__100 ] = num__135 * [ ( num__100 + num__40 ) / num__100 ] = rs . num__189 . answer : c <eor> c <eos> |
c |
percent__100.0__189.0__ |
percent__100.0__189.0__ |
| in how many q ways can num__5 people from a group of num__6 people be seated around a circular table <o> a ) num__56 <o> b ) num__80 <o> c ) num__100 <o> d ) num__120 <o> e ) num__144 |
q = num__6 c num__5 * ( num__5 - num__1 ) ! ( select num__5 out of num__6 and arrange them in circular manner ) = num__6 * num__4 ! = num__6 * num__24 = num__144 answer - e <eor> e <eos> |
e |
square_perimeter__1.0__ square_perimeter__6.0__ multiply__6.0__24.0__ multiply__6.0__24.0__ |
square_perimeter__1.0__ multiply__6.0__4.0__ multiply__6.0__24.0__ multiply__6.0__24.0__ |
| at the end of the month a certain ocean desalination plant ’ s reservoir contained num__40 million gallons of water . this amount is one fifth of the normal level . if this amount represents num__80.0 of the reservoir ’ s total capacity how many million gallons short of total capacity is the normal level ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__40 <o> d ) num__50 <o> e ) num__60 |
the q talks of total capacity normal level present level shortage etc . . so it is all about not going wrong in these terms num__40 mg = num__80.0 of total . . total = num__40 / . num__8 = num__50 mg . . normal level = num__0.2 of num__50 = num__10 mg . . shortage of normal level = num__50 - num__10 = num__40 mg . . c <eor> c <eos> |
c |
percent__80.0__50.0__ |
percent__80.0__50.0__ |
| find a sum for first num__5 prime number ' ss ? <o> a ) num__25 <o> b ) num__28 <o> c ) num__30 <o> d ) num__35 <o> e ) num__36 |
required sum = ( num__2 + num__3 + num__5 + num__7 + num__11 ) = num__28 note : num__1 is not a prime number option b <eor> b <eos> |
b |
subtract__5.0__2.0__ add__5.0__2.0__ subtract__3.0__2.0__ multiply__1.0__28.0__ |
subtract__5.0__2.0__ add__5.0__2.0__ subtract__3.0__2.0__ multiply__1.0__28.0__ |
| a man is num__29 years older than his son . in two years his age will be twice the age of his son . the present age of his son is : <o> a ) num__14 years <o> b ) num__18 years <o> c ) num__20 years <o> d ) num__27 years <o> e ) num__16 years |
let the son ' s present age be x years . then man ' s present age = ( x + num__24 ) years . ( x + num__29 ) + num__2 = num__2 ( x + num__2 ) x + num__31 = num__2 x + num__4 x = num__27 . answer : d <eor> d <eos> |
d |
add__29.0__2.0__ subtract__29.0__2.0__ subtract__29.0__2.0__ |
add__29.0__2.0__ subtract__29.0__2.0__ subtract__29.0__2.0__ |
| a can fill tank in num__9 minutes b can empty the tank in num__18 minutes . . in what time the tank be filled if both pipes work simultaneously ? <o> a ) num__14 minutes <o> b ) num__15 minutes <o> c ) num__16 minutes <o> d ) num__17 minutes <o> e ) num__18 minutes |
bigger no = num__18 ratio = num__9 : num__18 = num__1 : num__2 time taken to fill the tank = num__18 / ( num__2 - num__1 ) = num__18 minutes answer : e <eor> e <eos> |
e |
divide__18.0__9.0__ round__18.0__ |
divide__18.0__9.0__ divide__18.0__1.0__ |
| a man ' s speed with the current is num__20 kmph and speed of the current is num__1 kmph . the man ' s speed against the current will be <o> a ) num__11 kmph <o> b ) num__12 kmph <o> c ) num__18 kmph <o> d ) num__17 kmph <o> e ) none of these |
explanation : speed with current is num__20 speed of the man + it is speed of the current speed in s Ɵ ll water = num__20 - num__1 = num__19 now speed against the current will be speed of the man - speed of the current = num__19 - num__1 = num__18 kmph answer : c <eor> c <eos> |
c |
subtract__20.0__1.0__ subtract__19.0__1.0__ round__18.0__ |
subtract__20.0__1.0__ subtract__19.0__1.0__ subtract__19.0__1.0__ |
| a number is doubled and num__9 is added . if resultant is trebled it becomes num__69 . what is that number <o> a ) num__7 <o> b ) num__10 <o> c ) num__12 <o> d ) num__14 <o> e ) num__15 |
explanation : = > num__3 ( num__2 x + num__9 ) = num__69 = > num__2 x + num__9 = num__23 = > x = num__7 answer : option a <eor> a <eos> |
a |
divide__69.0__3.0__ subtract__9.0__2.0__ subtract__9.0__2.0__ |
divide__69.0__3.0__ subtract__9.0__2.0__ subtract__9.0__2.0__ |
| in a kilometer race a beats b by num__50 meters or num__10 seconds . what time does a take to complete the race ? <o> a ) num__187 sec <o> b ) num__190 sec <o> c ) num__278 sec <o> d ) num__268 sec <o> e ) num__987 sec |
time taken by b run num__1000 meters = ( num__1000 * num__10 ) / num__50 = num__200 sec . time taken by a = num__200 - num__10 = num__190 sec . answer : b <eor> b <eos> |
b |
subtract__200.0__10.0__ round__190.0__ |
subtract__200.0__10.0__ subtract__200.0__10.0__ |
| if a and b are two events such that p ( a ) = num__0.5 p ( b ) = num__0.5 and p ( a n b ) = num__0.375 find p ( not a and not b ) . <o> a ) a ) num__0.25 <o> b ) b ) num__0.375 <o> c ) c ) num__0.5 <o> d ) d ) num__0.384615384615 <o> e ) e ) num__0.75 |
p ( not a and not b ) = num__1 - ( p ( a ) + p ( b ) - p ( ab ) ) which you might find somewhere in your text . substituting in our probabilities we get : p ( not a and not b ) = num__1 - ( num__0.5 + num__0.5 - num__0.375 ) p ( not a and not b ) = num__1 - ( num__0.625 ) p ( not a and not b ) = num__0.375 . b <eor> b <eos> |
b |
subtract__1.0__0.375__ multiply__0.375__1.0__ |
subtract__1.0__0.375__ subtract__1.0__0.625__ |
| a can do a piece of work in num__12 days . he worked for num__15 days and then b completed the remaining work in num__10 days . both of them together will finish it in . <o> a ) num__12 num__0.5 days <o> b ) num__12 num__0.125 days <o> c ) num__12 num__0.5 days <o> d ) num__22 num__0.5 days <o> e ) num__12 num__3.0 days |
num__0.6 + num__10 / x = num__1 = > x = num__25 num__0.04 + num__0.04 = num__0.08 num__12.5 = num__12 num__0.5 days answer : a <eor> a <eos> |
a |
km_to_mile_conversion__ add__15.0__10.0__ divide__1.0__25.0__ divide__1.0__0.08__ subtract__12.5__12.0__ round__12.0__ |
km_to_mile_conversion__ add__15.0__10.0__ divide__1.0__25.0__ divide__1.0__0.08__ divide__12.5__25.0__ divide__12.0__1.0__ |
| a student needs num__60.0 of the marks on a test to pass the test . if the student gets num__80 marks and fails the test by num__100 marks find the maximum marks set for the test . <o> a ) num__250 <o> b ) num__300 <o> c ) num__350 <o> d ) num__400 <o> e ) num__450 |
num__60.0 = num__180 marks num__1.0 = num__3 marks num__100.0 = num__300 marks the answer is b . <eor> b <eos> |
b |
percent__100.0__300.0__ |
percent__100.0__300.0__ |
| in a flight of num__600 km an aircraft was slowed down due to bad weather . its average speed for the trip was reduced by num__200 km / hr and the time of flight increased by num__30 minutes . the duration of the flight is : <o> a ) num__1 hour <o> b ) num__2 hours <o> c ) num__3 hours <o> d ) num__4 hours <o> e ) num__5 hours |
let the duration of the flight be x hours num__600 / x - num__600 ( x + num__0.5 ) = num__200 num__600 / x - num__1200 / ( num__2 x + num__1 ) = num__200 x ( num__2 x + num__1 ) = num__3 num__2 x num__2 + x - num__3 = num__0 ( num__2 x + num__3 ) ( x - num__1 ) = num__0 x ca n ' t be negative hence x = num__1 x = num__1 hr . duration = num__1 hr answer : a <eor> a <eos> |
a |
divide__600.0__0.5__ reverse__0.5__ multiply__0.5__2.0__ divide__600.0__200.0__ round_down__0.5__ reverse__1.0__ |
divide__600.0__0.5__ reverse__0.5__ multiply__0.5__2.0__ divide__600.0__200.0__ round_down__0.5__ reverse__1.0__ |
| sonalika goes num__12 km towards north from a fixed point and then she goes num__8 km towards south from there . in the end she goes num__3 km towards east . how far and in what direction is she from her starting point ? <o> a ) num__5 km north east <o> b ) num__4 km north east <o> c ) num__5 km north west <o> d ) num__6 km north east <o> e ) num__6 km north west |
she is currently at num__5 km north east from the starting point answer : a <eor> a <eos> |
a |
subtract__8.0__3.0__ subtract__8.0__3.0__ |
subtract__8.0__3.0__ subtract__8.0__3.0__ |
| a train speeds past a pole in num__15 sec and a platform num__150 m long in num__25 sec its length is ? <o> a ) num__50 m <o> b ) num__150 m <o> c ) num__225 m <o> d ) num__300 m <o> e ) none of these |
let the length of the train be x m and its speed be y m / sec . then x / y = num__15 = > y = x / num__15 ( x + num__150 ) / num__25 = x / num__15 = > x = num__225 m . answer : c <eor> c <eos> |
c |
round__225.0__ |
round__225.0__ |
| carol spends num__0.25 of her savings on a stereo and num__0.2 less than she spent on the stereo for a television . what fraction of her savings did she spend on the stereo and television ? <o> a ) num__0.25 <o> b ) num__0.285714285714 <o> c ) num__0.416666666667 <o> d ) num__0.45 <o> e ) num__0.583333333333 |
total savings = s amount spent on stereo = ( num__0.25 ) s amount spent on television = ( num__1 - num__0.2 ) ( num__0.25 ) s = ( num__0.8 ) * ( num__0.25 ) * s = ( num__0.2 ) s ( stereo + tv ) / total savings = s ( num__0.25 + num__0.2 ) / s = num__0.45 answer : d <eor> d <eos> |
d |
divide__0.2__0.25__ add__0.25__0.2__ add__0.25__0.2__ |
divide__0.2__0.25__ add__0.25__0.2__ add__0.25__0.2__ |
| the jogging track in a sports complex is num__561 m in circumference . deepak and his wife start from the same point and walk in opposite directions at num__4.5 km / hr and num__3.75 km / hr respectively . they will meet for the first time in ? <o> a ) num__5.29 min <o> b ) num__5.28 min <o> c ) num__4.08 min <o> d ) num__9.28 min <o> e ) num__5.988 min |
clearly the two will meet when they are num__561 m apart . to be ( num__4.5 + num__3.75 ) = num__8.25 km apart they take num__1 hour . to be num__561 m apart they take ( num__0.121212121212 * num__0.561 ) hrs = ( num__0.068 * num__60 ) min = num__4.08 min . answer : c <eor> c <eos> |
c |
add__4.5__3.75__ reverse__8.25__ multiply__0.1212__0.561__ multiply__60.0__0.068__ multiply__1.0__4.08__ |
add__4.5__3.75__ reverse__8.25__ multiply__0.1212__0.561__ multiply__60.0__0.068__ multiply__1.0__4.08__ |
| if num__6 ! / num__3 ^ x is an integer what is the greatest possible value of x ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
num__6 - num__2 * num__3 num__3 - num__1 * num__3 hence max of num__3 ^ num__2 is allowed . imo a . <eor> a <eos> |
a |
divide__6.0__3.0__ subtract__3.0__2.0__ divide__6.0__3.0__ |
divide__6.0__3.0__ subtract__3.0__2.0__ subtract__3.0__1.0__ |
| what is the remainder when num__9614241 is divided by num__99 <o> a ) num__50 <o> b ) num__65 <o> c ) num__45 <o> d ) num__54 <o> e ) num__60 |
num__97113.5454545 num__1068249.0 * num__11 num__97113.5454545 = > rem = num__6 we have cancelled it by num__9 reqd remainder = num__9 * num__6 = num__54 answer : d <eor> d <eos> |
d |
divide__9614241.0__99.0__ divide__1068249.0__97113.5455__ divide__9614241.0__1068249.0__ multiply__6.0__9.0__ multiply__6.0__9.0__ |
divide__9614241.0__99.0__ divide__1068249.0__97113.5455__ divide__9614241.0__1068249.0__ multiply__6.0__9.0__ multiply__6.0__9.0__ |
| alfred buys an old scooter for $ num__4700 and spends $ num__800 on its repairs . if he sells the scooter for $ num__6400 his gain percent is ? <o> a ) num__5.45 <o> b ) num__16.36 <o> c ) num__7.0 <o> d ) num__8.12 <o> e ) num__10 % |
c . p . = num__4700 + num__800 = $ num__5500 s . p . = $ num__6400 gain = num__6400 - num__5500 = $ num__900 gain % = num__0.163636363636 * num__100 = num__16.36 answer is b <eor> b <eos> |
b |
percent__16.36__100.0__ |
percent__16.36__100.0__ |
| n a series of numbers the next number is formed by adding num__1 to the sum of the previous numbers and the num__10 th number is num__1280 . then what is the first number in the series ? ( series will be like this x x + num__1 ( x + ( x + num__1 ) ) + num__1 . . . . . . . ) <o> a ) num__2 <o> b ) num__4 <o> c ) num__5 <o> d ) num__7 <o> e ) num__9 |
answer : option b sol . the given series is x x + num__1 num__2 x + num__2 num__4 x + num__4 . . . . . . if you observe the pattern here the coefficient of x + num__1 is in the powers of num__2 . so num__4 th term has a power of num__2 num__5 th term has a power of num__3 . . . num__10 th term has a power of num__8 . so tenth term would be num__2828 ( x + num__1 ) = num__256 ( x + num__1 ) . given num__256 ( x + num__1 ) = num__1280 x = num__4 . answer : b <eor> b <eos> |
b |
add__1.0__4.0__ add__1.0__2.0__ subtract__10.0__2.0__ divide__1280.0__5.0__ add__1.0__3.0__ |
add__1.0__4.0__ add__1.0__2.0__ add__3.0__5.0__ divide__1280.0__5.0__ add__1.0__3.0__ |
| alice deposited rs num__2 million in bank at num__4.0 simple interest per year . after num__10 year how many rs she will get from bank ? <o> a ) num__75000 <o> b ) num__90000 <o> c ) num__10000 <o> d ) num__80000 <o> e ) num__60000 |
i = ( num__200000 * num__10 * num__4 ) / num__100 i = num__80000 answer : d <eor> d <eos> |
d |
percent__100.0__80000.0__ |
percent__100.0__80000.0__ |
| by selling num__33 metres of cloth one gains the selling price of num__11 metres . find the gain percent . <o> a ) num__96.0 <o> b ) num__48.0 <o> c ) num__50.0 <o> d ) num__36.0 <o> e ) num__45 % |
( sp of num__33 m ) - ( cp of num__33 m ) = gain = sp of num__11 m sp of num__22 m = cp of num__33 m let cp of each metre be re . num__1 then cp of num__22 m = rs . num__22 sp of num__22 m = rs . num__33 . gain % = [ ( num__0.5 ) * num__100 ] % = num__50.0 ans : c <eor> c <eos> |
c |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| there are many numbers that can be expressed as the sum of three squares in three different ways . can you find out the smallest of such number ? <o> a ) num__55 <o> b ) num__54 <o> c ) num__49 <o> d ) num__50 <o> e ) num__51 |
b num__54 num__7 ^ num__2 + num__2 ^ num__2 + num__1 ^ num__1 num__6 ^ num__2 + num__3 ^ num__2 + num__3 ^ num__2 num__2 ^ num__2 + num__5 ^ num__2 + num__5 ^ num__2 <eor> b <eos> |
b |
subtract__7.0__1.0__ add__1.0__2.0__ add__2.0__3.0__ multiply__1.0__54.0__ |
subtract__7.0__1.0__ add__1.0__2.0__ add__2.0__3.0__ multiply__1.0__54.0__ |
| the average temperature for monday tuesday wednesday and thursday was num__48 degrees and for tuesday wednesday thursday and friday was num__46 degrees . if the temperature on monday was num__42 degrees . find the temperature on friday ? <o> a ) num__22 <o> b ) num__679 <o> c ) num__62 <o> d ) num__34 <o> e ) num__12 |
m + tu + w + th = num__4 * num__48 = num__192 tu + w + th + f = num__4 * num__46 = num__184 m = num__42 tu + w + th = num__192 - num__42 = num__150 f = num__184 – num__150 = num__34 . answer : d <eor> d <eos> |
d |
subtract__46.0__42.0__ multiply__48.0__4.0__ multiply__46.0__4.0__ subtract__192.0__42.0__ subtract__184.0__150.0__ subtract__184.0__150.0__ |
subtract__46.0__42.0__ multiply__48.0__4.0__ multiply__46.0__4.0__ subtract__192.0__42.0__ subtract__184.0__150.0__ subtract__184.0__150.0__ |
| working alone a can complete a certain kind of job in num__10 hours . a and d working together at their respective rates can complete one of these jobs in num__5 hours . in how many hours can d working alone complete one of these jobs ? <o> a ) num__9 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__13 |
let total time taken by d to complete the job = d total time taken by a to complete the job = num__10 work done by a in an hour num__1 / a = num__0.1 working together a and d can complete the job in num__5 hours num__1 / a + num__1 / d = num__0.2 = > num__1 / d = num__0.2 - num__1 / a = num__0.2 - num__0.1 = num__0.1 = > d = num__10 hours answer b <eor> b <eos> |
b |
divide__1.0__10.0__ divide__1.0__5.0__ round__10.0__ |
divide__1.0__10.0__ divide__1.0__5.0__ divide__10.0__1.0__ |
| susan tim and kim need to be seated in num__3 identical chairs in straight line so that susan is seated always left to tim . how many such arrangements are possible ? <o> a ) a ) num__6 <o> b ) b ) num__120 <o> c ) c ) num__80 <o> d ) d ) num__240 <o> e ) e ) num__60 |
total number of arrangements = num__3 ! = num__6 in exactly half susan will be to the left of tim which gives us num__3 arrangements option ( a ) <eor> a <eos> |
a |
die_space__ die_space__ |
die_space__ die_space__ |
| simplify num__682 * num__9 <o> a ) num__2459 <o> b ) num__4236 <o> c ) num__6895 <o> d ) num__6138 <o> e ) num__1478 |
explanation : although it is a simple question but the trick is to save time in solving this . rather than multiplying it we can do as follows : num__682 * ( num__10 - num__1 ) = num__6820 - num__682 = num__6138 answer : option d <eor> d <eos> |
d |
subtract__10.0__9.0__ multiply__682.0__10.0__ multiply__682.0__9.0__ multiply__682.0__9.0__ |
subtract__10.0__9.0__ multiply__682.0__10.0__ multiply__682.0__9.0__ multiply__682.0__9.0__ |
| two cyclist start from the same places in opposite directions . one is going towards north at num__18 kmph and the other is going towards south num__20 kmph . what time will they take to be num__76 km apart ? <o> a ) num__1 num__0.25 hours <o> b ) num__2 num__0.333333333333 hours <o> c ) num__4 hours <o> d ) num__3 num__0.75 hours <o> e ) num__2 hours |
to be ( num__18 + num__20 ) km apart they take num__1 hour to be num__76 km apart they take num__0.0263157894737 * num__76 = num__2 hrs answer is e <eor> e <eos> |
e |
subtract__20.0__18.0__ round__2.0__ |
subtract__20.0__18.0__ round__2.0__ |
| calculate the l . c . m of num__0.444444444444 num__0.714285714286 num__0.692307692308 num__0.466666666667 is : <o> a ) num__1230 <o> b ) num__1290 <o> c ) num__1260 <o> d ) num__1240 <o> e ) num__2260 |
required l . c . m = l . c . m . of num__4 num__5 num__9 num__7 / h . c . f . of num__9 num__7 num__13 num__15 = num__1260.0 = num__1260 answer is c <eor> c <eos> |
c |
add__4.0__5.0__ add__4.0__9.0__ gcd__0.4444__1260.0__ |
add__4.0__5.0__ add__4.0__9.0__ gcd__0.4444__1260.0__ |
| in a garden num__26 trees are planted at equal distances along a yard num__700 metres long one tree being at each end of the yard . what is the distance between two consecutive trees ? <o> a ) num__10 <o> b ) num__28 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
num__26 trees have num__25 gaps between them . length of each gap = num__28.0 = num__28 i . e . distance between two consecutive trees = num__28 answer is b . <eor> b <eos> |
b |
divide__700.0__25.0__ round__28.0__ |
divide__700.0__25.0__ round__28.0__ |
| in a tree num__0.4 of the birds are robins while the rest are bluejays . if num__0.333333333333 of the robins are female and num__0.666666666667 of the bluejays are female what fraction of the birds in the tree are male ? <o> a ) num__0.6 <o> b ) num__0.266666666667 <o> c ) num__0.466666666667 <o> d ) num__0.533333333333 <o> e ) num__0.566666666667 |
the fraction of birds that are male robins is ( num__0.666666666667 ) ( num__0.4 ) = num__0.266666666667 . the fraction of birds that are male bluejays is ( num__0.333333333333 ) ( num__0.6 ) = num__0.2 . the total fraction of male birds is num__0.266666666667 + num__0.2 = num__0.466666666667 . the answer is c . <eor> c <eos> |
c |
multiply__0.4__0.6667__ divide__0.4__0.6667__ multiply__0.3333__0.6__ subtract__0.6667__0.2__ subtract__0.6667__0.2__ |
multiply__0.4__0.6667__ add__0.3333__0.2667__ multiply__0.3333__0.6__ add__0.2667__0.2__ add__0.2667__0.2__ |
| the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is num__42 kmph find the speed of the stream ? <o> a ) num__19 kmph <o> b ) num__18 kmph <o> c ) num__14 kmph <o> d ) num__13 kmph <o> e ) num__17 kmph |
the ratio of the times taken is num__2 : num__1 . the ratio of the speed of the boat in still water to the speed of the stream = ( num__2 + num__1 ) / ( num__2 - num__1 ) = num__3.0 = num__3 : num__1 speed of the stream = num__14.0 = num__14 kmph . answer : c <eor> c <eos> |
c |
add__1.0__2.0__ divide__42.0__3.0__ round__14.0__ |
add__1.0__2.0__ divide__42.0__3.0__ divide__42.0__3.0__ |
| two trains are traveling on a collision course . if train a is traveling at a speed of num__350 mph and train b is traveling num__28.0 slower how much time will it take the trains to collide if the initial distance between the two is num__2408 miles ? <o> a ) two hours and num__30 minutes . <o> b ) one hour and num__10 minutes . <o> c ) two hours and num__25 minutes . <o> d ) three hours and num__15 minutes . <o> e ) four hours and num__0 minutes . |
answer is e . four hours and num__0 minutes . since they are racing towards each other the speed is accumulative ie . num__350 mph + num__350 * num__0.72 = num__602 ( num__0.72 because the second train is num__28.0 slower ) so time before collision is total distance / total speed = num__4.0 = num__4 hours and num__0 mins <eor> e <eos> |
e |
divide__2408.0__602.0__ round__4.0__ |
divide__2408.0__602.0__ divide__2408.0__602.0__ |
| what is the least number which when divided by num__5 num__6 num__9 and num__18 leaves remainder num__4 in each care ? <o> a ) num__30 <o> b ) num__44 <o> c ) num__36 <o> d ) num__56 <o> e ) num__94 |
explanation : lcm of num__5 num__6 num__9 and num__18 is num__90 required number = num__90 + num__4 = num__94 answer : option e <eor> e <eos> |
e |
multiply__5.0__18.0__ add__4.0__90.0__ add__4.0__90.0__ |
multiply__5.0__18.0__ add__4.0__90.0__ add__4.0__90.0__ |
| if it takes num__30 construction workers num__70 days to build a skyscraper how long would it take num__50 workers ? <o> a ) num__52 days <o> b ) num__47 days <o> c ) num__42 days <o> d ) num__37 days <o> e ) num__32 days |
num__30 * num__70 = num__50 * x x = num__42 answer : c <eor> c <eos> |
c |
round__42.0__ |
round__42.0__ |
| sides of a rectangular park are in the ratio num__3 : num__2 and its area is num__5766 sq m the cost of fencing it at num__50 ps per meter is ? <o> a ) num__287 <o> b ) num__1287 <o> c ) num__125 <o> d ) num__155 <o> e ) num__271 |
num__3 x * num__2 x = num__5766 = > x = num__31 num__2 ( num__93 + num__62 ) = num__310 m num__310 * num__0.5 = rs . num__155 answer : d <eor> d <eos> |
d |
multiply__3.0__31.0__ multiply__2.0__31.0__ rectangle_perimeter__93.0__62.0__ multiply__0.5__310.0__ triangle_area__2.0__155.0__ |
multiply__3.0__31.0__ multiply__2.0__31.0__ rectangle_perimeter__93.0__62.0__ multiply__0.5__310.0__ multiply__0.5__310.0__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__420 m and they cross each other in num__12 sec then the speed of each train is ? <o> a ) num__11 <o> b ) num__26 <o> c ) num__126 <o> d ) num__99 <o> e ) num__27 |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__420 + num__420 ) / num__12 = > x = num__35 speed of each train = num__10 m / sec . = num__35 * num__3.6 = num__126 km / hr . answer : c <eor> c <eos> |
c |
divide__420.0__12.0__ subtract__12.0__2.0__ multiply__35.0__3.6__ round__126.0__ |
divide__420.0__12.0__ subtract__12.0__2.0__ multiply__35.0__3.6__ multiply__35.0__3.6__ |
| a number when divided by a divisor leaves a remainder of num__24 . when twice the original number is divided by the same divisor the remainder is num__11 . what is the value of the divisor ? <o> a ) num__13 <o> b ) num__59 <o> c ) num__35 <o> d ) num__37 <o> e ) num__38 |
let the original number be ' a ' let the divisor be ' d ' let the quotient of the division of aa by dd be ' x ' therefore we can write the relation as a / d = xa / d = x and the remainder is num__24 . i . e . a = dx + num__24 when twice the original number is divided by d num__2 a is divided by d . we know that a = dx + num__24 . therefore num__2 a = num__2 dx + num__48 the problem states that ( num__2 dx + num__48 ) / dleaves a remainder of num__11 . num__2 dx is perfectly divisible by dd and will therefore not leave a remainder . the remainder of num__11 was obtained by dividing num__48 by d . when num__48 is divided by num__37 the remainder that one will obtain is num__11 . hence the divisor is num__37 . option ( d ) <eor> d <eos> |
d |
multiply__24.0__2.0__ subtract__48.0__11.0__ subtract__48.0__11.0__ |
multiply__24.0__2.0__ subtract__48.0__11.0__ subtract__48.0__11.0__ |
| a train num__605 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__51 <o> b ) num__64 <o> c ) num__11 <o> d ) num__22 <o> e ) num__33 |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__605 * num__0.0545454545455 ] sec = num__33 sec answer : e <eor> e <eos> |
e |
add__60.0__6.0__ round__33.0__ |
add__60.0__6.0__ round__33.0__ |
| the product of two numbers is num__4107 . if the h . c . f . of these numbers is num__37 then the greater number is : <o> a ) num__101 <o> b ) num__107 <o> c ) num__111 <o> d ) num__185 <o> e ) num__196 |
let the numbers be num__37 a and num__37 b . then num__37 a x num__37 b = num__4107 ab = num__3 . now co - primes with product num__3 are ( num__1 num__3 ) . so the required numbers are ( num__37 x num__1 num__37 x num__3 ) i . e . ( num__37 num__111 ) . greater number = num__111 . answer : option c <eor> c <eos> |
c |
divide__4107.0__37.0__ divide__4107.0__37.0__ |
divide__4107.0__37.0__ divide__4107.0__37.0__ |
| a man takes num__3 hours num__45 minutes to row a boat num__20 km downstream of a river and num__2 hours num__30 minutes to cover a distance of num__3 km upstream . find the speed of the current . <o> a ) num__1 km / hr <o> b ) num__2 km / hr <o> c ) num__3 km / hr <o> d ) num__4 km / hr <o> e ) none of these |
explanation : first of all we know that speed of current = num__0.5 ( speed downstream - speed upstream ) [ important ] so we need to calculate speed downstream and speed upstream first . speed = distance / time [ important ] speed upstream = ( num__6.66666666667 num__0.75 ) km / hr = num__5.33333333333 km / hr speed downstream = ( num__1.5 num__0.5 ) km / hr = num__1.2 km / hr so speed of current = num__0.5 ( num__5.33333333333 − num__1.2 ) = num__2 km / hr option b <eor> b <eos> |
b |
divide__20.0__3.0__ divide__3.0__2.0__ round__2.0__ |
divide__20.0__3.0__ divide__3.0__2.0__ divide__3.0__1.5__ |
| a sells a bicycle to b at a profit of num__60.0 and b sells it to c at a loss of num__40.0 . find the resultant profit or loss . <o> a ) - num__4.0 <o> b ) num__5.0 <o> c ) - num__5.0 <o> d ) num__6.0 <o> e ) - num__7 % |
the resultant profit or loss = num__60 - num__40 - ( num__60 * num__40 ) / num__100 = - num__4.0 loss = num__4.0 answer is a <eor> a <eos> |
a |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| for integers a and b num__16 a = num__32 ^ b . which of the following correctly expresses a in terms of b ? <o> a ) a = num__2 ^ b <o> b ) a = num__4 ^ b <o> c ) a = num__2 ^ ( num__5 b − num__4 ) <o> d ) a = num__4 ^ ( num__5 b − num__4 ) <o> e ) a = num__2 ^ ( num__5 b ) |
i tried this one with an example although i am not quite sure if this is correct . i would appreciate if there would be a common rule for this . i tried it out with b = num__2 and only c correctly expresses the following : num__16 a = num__32 ^ b b = num__2 num__16 a = num__1024 a = num__64 choice c - - > a = num__2 ^ num__6 = num__64 however this is a quite time consuming process . is there something faster ? answer c <eor> c <eos> |
c |
divide__32.0__16.0__ multiply__32.0__2.0__ divide__32.0__16.0__ |
divide__32.0__16.0__ multiply__32.0__2.0__ divide__32.0__16.0__ |
| a man has rs . num__412 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__45 <o> b ) num__60 <o> c ) num__77.25 <o> d ) num__90 <o> e ) num__105 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__412 num__16 x = num__412 x = num__25.75 . hence total number of notes = num__3 x = num__77.25 . answer = c <eor> c <eos> |
c |
divide__412.0__16.0__ multiply__25.75__3.0__ multiply__25.75__3.0__ |
divide__412.0__16.0__ multiply__25.75__3.0__ multiply__25.75__3.0__ |
| the average ( arithmetic mean ) of num__8 numbers is z and one of the numbers is num__14 . if num__14 is replaced with num__28 then what is the new average ( arithmetic mean ) in terms of z ? <o> a ) z + num__0.5 <o> b ) z + num__1.75 <o> c ) z + num__2 <o> d ) z + num__4 <o> e ) num__2 z + num__1 |
( num__8 z − num__14 + num__28 ) / num__8 or ( num__8 z + num__14 ) / num__8 or ( num__4 z + num__7 ) / num__4 or z + num__1.75 so correct answer will be ( b ) z + num__1.75 <eor> b <eos> |
b |
divide__28.0__4.0__ divide__14.0__8.0__ divide__14.0__8.0__ |
divide__28.0__4.0__ divide__14.0__8.0__ divide__14.0__8.0__ |
| a plane flying north at num__500 kmph passes over a city at num__12 noon . a plane flying east at the same attitude passes over the same city at num__12.30 pm . the plane is flying east at num__400 kmph . to the nearest hundred km how far apart are the two planes at num__2 pm ? <o> a ) num__600 km <o> b ) num__1000 km <o> c ) num__1100 km <o> d ) num__1200 km <o> e ) num__1300 km |
the plane flying north would have traveled for num__2 hours at num__2 pm . it travels at num__500 kmph . hence the plane flying north would have covered num__1000 kms from the city at num__2 pm . the plane flying east would have traveled for num__1.5 hours at num__2 pm . it travels at num__400 kmph . hence the plane flying east would have covered num__1.5 * num__400 = num__600 kms from the city at num__2 pm . the two planes are flying at right angles to each other . the shortest distance between the two planes is the length of the hypotenuse of the right triangle whose legs are the distances covered by the two planes . hence distance = num__1000 square + num__200 square = num__1166 km the nearest num__100 km distance between the two planes = num__1200 kms . answer d <eor> d <eos> |
d |
multiply__500.0__2.0__ multiply__400.0__1.5__ divide__400.0__2.0__ subtract__500.0__400.0__ multiply__12.0__100.0__ round__1200.0__ |
multiply__500.0__2.0__ multiply__400.0__1.5__ divide__400.0__2.0__ subtract__500.0__400.0__ multiply__12.0__100.0__ multiply__12.0__100.0__ |
| john started driving on a highway at a constant speed of r miles per hour at num__13 : num__00 . then num__2 hours later tom started driving on the same highway at a constant speed of num__2 r miles per hour at num__15 : num__00 . if both drivers maintained their speed how many miles had john driven on the highway when tom caught up with john ? <o> a ) num__2 r <o> b ) num__3 r <o> c ) num__4 r <o> d ) num__5 r <o> e ) num__6 r |
in num__2 hours john drove num__2 r miles . tom can catch john at a rate of num__2 r - r = r miles per hour . the time it takes tom to catch john is num__2 r / r = num__2 hours . in num__2 hours john can drive another num__2 r miles for a total of num__4 r miles . the answer is c . <eor> c <eos> |
c |
round__4.0__ |
round__4.0__ |
| if n = num__5 p where p is a prime number greater than num__1 how many different positive even divisors does n have including n ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__32 <o> d ) num__4 <o> e ) num__3 |
n = num__5 p and that p is a prime number greater than num__1 . let ' s test p = num__2 ; so n = num__10 the question now asks how many different positive even divisors does num__10 have including num__10 ? num__10 : num__110 num__52 how many of these divisors are even ? num__2 num__10 … . . num__2 even divisors . answer : b <eor> b <eos> |
b |
coin_space__ card_space__ coin_space__ |
coin_space__ card_space__ coin_space__ |
| which of the following numbers is between num__2 ⁄ num__3 and num__1 ⁄ num__2 ? <o> a ) . num__45 <o> b ) . num__35 <o> c ) . num__29 <o> d ) . num__22 <o> e ) . num__55 |
is n ' t it e ? num__0.666666666667 = . num__66 num__0.5 = . num__50 the only answer between that is e . num__55 . <eor> e <eos> |
e |
divide__2.0__3.0__ reverse__2.0__ multiply__1.0__55.0__ |
divide__2.0__3.0__ reverse__2.0__ multiply__1.0__55.0__ |
| what is the dividend . divisor num__17 the quotient is num__9 and the remainder is num__5 ? <o> a ) num__130 <o> b ) num__134 <o> c ) num__148 <o> d ) num__158 <o> e ) num__160 |
d = d * q + r d = num__17 * num__9 + num__5 d = num__153 + num__5 d = num__158 <eor> d <eos> |
d |
multiply__17.0__9.0__ add__5.0__153.0__ add__5.0__153.0__ |
multiply__17.0__9.0__ add__5.0__153.0__ add__5.0__153.0__ |
| bruce and anne can clean their house in num__4 hours working together at their respective constant rates . if anne ’ s speed were doubled they could clean their house in num__3 hours working at their respective rates . how many w hours does it currently take anne to clean the house on her own ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__12 <o> e ) num__14 |
lets suppose anne and bruce take a and b hrs working separately so in num__1 hour they can together finish num__1 / a + num__1 / b portion of the work which equals num__0.25 ( as the work is completed in num__4 hours ) after anne doubles her rate of work the portion completed by the both is num__1 / a + num__2 / b which is w equal to num__0.333333333333 ( as the work is completed in w = num__3 hours ) solving these num__2 equations we can find b as num__12 so d <eor> d <eos> |
d |
subtract__4.0__3.0__ divide__1.0__4.0__ subtract__3.0__1.0__ divide__1.0__3.0__ multiply__4.0__3.0__ round__12.0__ |
subtract__4.0__3.0__ divide__1.0__4.0__ subtract__3.0__1.0__ divide__1.0__3.0__ divide__3.0__0.25__ divide__3.0__0.25__ |
| find ( num__7 x + num__4 y ) / ( x - num__2 y ) if x / num__2 y = num__2.5 ? <o> a ) num__13 <o> b ) num__25 <o> c ) num__26 <o> d ) num__27 <o> e ) num__29 |
x / num__2 y = num__1.5 = > x = num__10 y / num__2 = num__5 y = > ( num__7 x + num__4 y ) / ( x - num__2 y ) = ( ( num__7 * ( num__5 y ) ) + num__4 y ) / ( num__5 y - num__2 y ) = > num__39 y / num__3 y = num__13 answer : a <eor> a <eos> |
a |
subtract__4.0__2.5__ multiply__4.0__2.5__ subtract__7.0__2.0__ subtract__7.0__4.0__ add__3.0__10.0__ add__3.0__10.0__ |
subtract__4.0__2.5__ multiply__4.0__2.5__ subtract__7.0__2.0__ subtract__7.0__4.0__ add__3.0__10.0__ add__3.0__10.0__ |
| a policeman noticed a criminal from a distance of num__180 km . the criminal starts running and the policeman chases him . the criminal and the policeman run at the rate of num__8 km and num__9 km per hour respectively . what is the distance between them after num__5 minutes ? <o> a ) num__96.7 m <o> b ) num__120 m <o> c ) num__130 m <o> d ) num__150 m <o> e ) none of these |
explanation : solution : relative speed = ( num__9 - num__8 ) = num__1 km / hr . distance covered in num__3 minutes = ( num__1 * num__0.0833333333333 ) km = num__0.0833333333333 km = num__83.3 m . . ' . distance between the criminal and policeman = ( num__180 - num__83.3 ) m = num__96.7 m . answer : a <eor> a <eos> |
a |
subtract__9.0__8.0__ subtract__8.0__5.0__ subtract__180.0__83.3__ round__96.7__ |
subtract__9.0__8.0__ subtract__8.0__5.0__ subtract__180.0__83.3__ subtract__180.0__83.3__ |
| mr . wayne bungee jumps from the top of a building straight to the ground at a constant speed . num__3 seconds after he starts plummeting he passes the num__20 th floor . num__15 seconds after he starts plummeting he passes the num__5 th floor . each floor is num__3.6 meters high . what is mr . wayne ' s speed in meters per second ? <o> a ) num__3 <o> b ) num__3.75 <o> c ) num__4.5 <o> d ) num__4.25 <o> e ) num__5 |
imo : c num__15 floors * num__3.6 meter / floor = num__54 meters time = num__12 sec rate * num__12 = num__54 rate = num__4.5 = num__4.5 answer c <eor> c <eos> |
c |
multiply__15.0__3.6__ subtract__15.0__3.0__ divide__54.0__12.0__ round__4.5__ |
multiply__15.0__3.6__ subtract__15.0__3.0__ divide__54.0__12.0__ divide__54.0__12.0__ |
| the speed of a boat in still water is num__50 kmph and the speed of the current is num__20 kmph . find the speed downstream and upstream ? <o> a ) num__40 num__68 kmph <o> b ) num__40 num__30 kmph <o> c ) num__70 num__30 kmph <o> d ) num__40 num__60 kmph <o> e ) num__20 num__60 kmph |
speed downstream = num__50 + num__20 = num__70 kmph speed upstream = num__50 - num__20 = num__30 kmph answer : c <eor> c <eos> |
c |
add__50.0__20.0__ subtract__50.0__20.0__ round__70.0__ |
add__50.0__20.0__ subtract__50.0__20.0__ add__50.0__20.0__ |
| a man rows his boat num__85 km downstream and num__45 km upstream taking num__2 num__0.5 hours each time . find the speed of the stream ? <o> a ) num__76 kmph <o> b ) num__6 kmph <o> c ) num__14 kmph <o> d ) num__8 kmph <o> e ) num__4 kmph |
speed downstream = d / t = num__85 / ( num__2 num__0.5 ) = num__34 kmph speed upstream = d / t = num__45 / ( num__2 num__0.5 ) = num__18 kmph the speed of the stream = ( num__34 - num__18 ) / num__2 = num__8 kmph answer : d <eor> d <eos> |
d |
round__8.0__ |
round__8.0__ |
| a num__300 m long train crosses a platform in num__39 sec while it crosses a signal pole in num__18 sec . what is the length of the platform ? <o> a ) num__389 m <o> b ) num__350 m <o> c ) num__289 m <o> d ) num__299 m <o> e ) num__219 m |
speed = num__16.6666666667 = num__16.6666666667 m / sec . let the length of the platform be x meters . then ( x + num__300 ) / num__39 = num__16.6666666667 num__3 x + num__900 = num__1950 = > x = num__350 m . answer : b <eor> b <eos> |
b |
divide__300.0__18.0__ multiply__300.0__3.0__ round__350.0__ |
divide__300.0__18.0__ multiply__300.0__3.0__ round__350.0__ |
| ratio between rahul and deepak is num__4 : num__3 after num__6 years rahul age will be num__26 years . what is deepak present age . <o> a ) num__14 <o> b ) num__15 <o> c ) num__20 <o> d ) num__22 <o> e ) num__23 |
explanation : present age is num__4 x and num__3 x = > num__4 x + num__6 = num__26 = > x = num__5 so deepak age is = num__3 ( num__5 ) = num__15 answer : option b <eor> b <eos> |
b |
multiply__3.0__5.0__ multiply__3.0__5.0__ |
multiply__3.0__5.0__ multiply__3.0__5.0__ |
| a train leaves mumabai at num__9 am at a speed of num__40 kmph . after one hour another train leaves mumbai in the same direction as that of the first train at a speed of num__60 kmph . when and at what distance from mumbai do the two trains meet ? <o> a ) num__187 <o> b ) num__279 <o> c ) num__120 <o> d ) num__278 <o> e ) num__379 |
when the second train leaves mumbai the first train covers num__40 * num__1 = num__40 km so the distance between first train and second train is num__40 km at num__10.00 am time taken by the trains to meet = distance / relative speed = num__40 / ( num__60 - num__40 ) = num__2 hours so the two trains meet at num__12 a . m . the two trains meet num__2 * num__60 = num__120 km away from mumbai . answer : c <eor> c <eos> |
c |
add__9.0__1.0__ add__2.0__10.0__ multiply__60.0__2.0__ round__120.0__ |
add__9.0__1.0__ add__2.0__10.0__ multiply__60.0__2.0__ multiply__60.0__2.0__ |
| on a man ' s tombstone it is said that one sixth of his life was spent in childhood and one twelfth as a teenager . one seventh of his life passed between the time he became an adult and the time he married ; five years later his son was born . alas the son died four years before he did . he lived to be twice as old as his son did . how old did the man live to be ? <o> a ) num__82 <o> b ) num__83 <o> c ) num__84 <o> d ) num__85 <o> e ) num__86 |
let the age of man is x so that of son = x / num__2 ( given ) now according to the given condition ( x / num__6 ) + ( x / num__12 ) + ( x / num__7 ) + num__5 + ( x / num__2 ) + num__4 = x or x = num__84 answer : c <eor> c <eos> |
c |
multiply__2.0__6.0__ subtract__7.0__2.0__ subtract__6.0__2.0__ multiply__7.0__12.0__ multiply__7.0__12.0__ |
multiply__2.0__6.0__ subtract__7.0__2.0__ subtract__6.0__2.0__ multiply__7.0__12.0__ multiply__7.0__12.0__ |
| if the annual interest on a principal is num__16.0 how many years before the amount is double ? <o> a ) num__6.25 <o> b ) num__20 <o> c ) num__17.2 <o> d ) num__15.2 <o> e ) num__15.2 |
p = ( p * num__16 * r ) / num__100 r = num__6.25 answer : a <eor> a <eos> |
a |
percent__6.25__100.0__ |
percent__6.25__100.0__ |
| josh spends a total of $ num__7.50 buying n items in the convenience store . if each of the items is either a num__5 cents single bubblegum or a num__50 cents bubblegum pack then n may be which of the following ? <o> a ) num__99 <o> b ) num__100 <o> c ) num__101 <o> d ) num__109 <o> e ) num__141 |
let x be num__0.05 $ single bubblegum and y be the total num__0.5 $ packs - - - > num__0.05 x + num__0.5 y = num__7.5 - - - > x + num__10 y = num__150 . . . ( num__1 ) also x + y = n . . . ( num__2 ) solving num__1 and num__2 we get n = num__150 - num__9 y - - - > only value satisfied is when y = num__1 - - - > n = num__150 - num__9 = num__141 . e is the correct answer . other options are not equal to multiples of num__9 removed from num__150 . <eor> e <eos> |
e |
divide__5.0__0.5__ divide__7.5__0.05__ reverse__0.5__ subtract__10.0__1.0__ subtract__150.0__9.0__ multiply__1.0__141.0__ |
divide__5.0__0.5__ divide__7.5__0.05__ reverse__0.5__ subtract__10.0__1.0__ subtract__150.0__9.0__ subtract__150.0__9.0__ |
| the area of circle o is added to its diameter . if twice the circumference of circle o is then subtracted from this total the result is num__8 . what is the diameter of circle o ? <o> a ) num__4 / π <o> b ) num__4 <o> c ) num__8 <o> d ) num__10 <o> e ) num__6 |
area + diameter - num__2 * circumference = num__8 pi * r ^ num__2 + num__2 r = num__8 + num__2 * num__2 pi * r isolate r and get r ( pi * r + num__2 ) = num__8 + num__4 pi * r r = ( num__8 + num__4 pi * r ) / ( pi * r + num__2 ) = > num__4 ( pi * r + num__2 ) / ( pi * r + num__2 ) r = num__4 diameter = num__2 r = num__8 c <eor> c <eos> |
c |
square_perimeter__2.0__ |
multiply__2.0__4.0__ |
| a reduction of num__20.0 in the price of oil enables a house wife to obtain num__5 kgs more for rs . num__800 what is the reduced price for kg ? <o> a ) num__30 <o> b ) num__31 <o> c ) num__32 <o> d ) num__33 <o> e ) num__34 |
num__800 * ( num__0.2 ) = num__160 - - - - num__5 ? - - - - num__1 = > rs . num__32 answer : c <eor> c <eos> |
c |
percent__20.0__800.0__ percent__20.0__5.0__ percent__20.0__160.0__ percent__20.0__160.0__ |
percent__20.0__800.0__ percent__20.0__5.0__ percent__20.0__160.0__ percent__20.0__160.0__ |
| the angle between the minute hand and the hour hand of a clock when the time is num__4.20 is : <o> a ) num__0 <o> b ) num__5 <o> c ) num__12 <o> d ) num__8 <o> e ) num__10 |
angle num__4.33333333333 hrs = num__30.0 * num__4.33333333333 = num__130 deg angle hand in num__20 min num__6.0 * num__20 = num__120 deg req = num__130 - num__120 = num__10 deg answer e <eor> e <eos> |
e |
multiply__6.0__20.0__ subtract__130.0__120.0__ round__10.0__ |
multiply__6.0__20.0__ subtract__130.0__120.0__ subtract__130.0__120.0__ |
| in a mixture of num__45 litres the ratio of milk to water is num__4 : num__1 . additional num__18 litres of water is added to the mixture . find the ratio of milk to water in the resulting mixture . <o> a ) num__2.0 <o> b ) num__1.33333333333 <o> c ) num__0.666666666667 <o> d ) num__0.75 <o> e ) num__1.5 |
given that milk / water = num__4 x / x and num__4 x + x = num__45 - - > x = num__9 . thus milk = num__4 x = num__36 liters and water = x = num__9 liters . new ratio = num__36 / ( num__9 + num__18 ) = num__1.33333333333 = num__1.33333333333 . answer : b . <eor> b <eos> |
b |
subtract__45.0__9.0__ multiply__1.0__1.3333__ |
subtract__45.0__9.0__ divide__1.3333__1.0__ |
| without any stoppages a person travels a certain distance at an average speed of num__40 km / hr and with stoppages he covers the same distance at an average of num__20 km / hr . how many minutes per hour does he stop ? <o> a ) num__15 minute <o> b ) num__20 minutes <o> c ) num__30 minutes <o> d ) num__45 minutes <o> e ) none of these |
explanation : in one hour the distance covered at actual speed = num__40 km and with stoppages it covers only num__20 km so to travel num__20 km at original speed i . e . num__20 = num__40 * t so t = num__0.5 hour = num__30 minutes answer – c <eor> c <eos> |
c |
divide__20.0__40.0__ round__30.0__ |
divide__20.0__40.0__ round__30.0__ |
| divide rs . num__720 among a b and c so that a receives num__0.333333333333 as much as b and c together and b receives num__0.666666666667 as a and c together . a ' s share is ? <o> a ) s . num__800 <o> b ) s . num__400 <o> c ) s . num__600 <o> d ) s . num__180 <o> e ) s . num__900 |
a + b + c = num__720 a = num__0.333333333333 ( b + c ) ; b = num__0.666666666667 ( a + c ) a / ( b + c ) = num__0.333333333333 a = num__0.25 * num__720 = > num__180 answer : d <eor> d <eos> |
d |
multiply__720.0__0.25__ multiply__720.0__0.25__ |
multiply__720.0__0.25__ multiply__720.0__0.25__ |
| in a game of billiards a can give b num__20 points in num__60 and he can give c num__40 points in num__60 . how many points can b give c in a game of num__100 ? <o> a ) num__18 <o> b ) num__27 <o> c ) num__50 <o> d ) num__21 <o> e ) num__17 |
a scores num__80 while b score num__40 and c scores num__20 . the number of points that c scores when b scores num__100 = ( num__100 * num__20 ) / num__40 = num__50 . in a game of num__100 points b gives ( num__100 - num__50 ) = num__50 points to c . answer : c <eor> c <eos> |
c |
add__20.0__60.0__ subtract__100.0__50.0__ |
subtract__100.0__20.0__ subtract__100.0__50.0__ |
| the perimeter of a triangle is num__40 cm and the inradius of the triangle is num__2.5 cm . what is the area of the triangle <o> a ) a ) num__72 <o> b ) b ) num__828 <o> c ) c ) num__50 <o> d ) d ) num__34 <o> e ) e ) num__35 |
explanation : area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = num__2.5 * num__20.0 = num__50 cm num__2 answer : option c <eor> c <eos> |
c |
triangle_area__40.0__2.5__ triangle_area__40.0__2.5__ |
multiply__2.5__20.0__ multiply__2.5__20.0__ |
| the length q of a rectangle is decreased by num__15.0 and its width is increased by num__40.0 . does the area of the rectangle decrease or increase and by what percent ? <o> a ) decreases by num__19.0 <o> b ) decreases by num__25.0 <o> c ) increases by num__6.0 <o> d ) increases by num__19.0 <o> e ) increases by num__25 % |
let the length q of the rectangle be num__100 x and width be num__100 y . area = num__100 x * num__100 y = num__10000 xy now after the change length = num__85 x and width = num__140 y . area = num__11900 xy % change = ( num__11900 xy - num__10000 xy ) / ( num__10000 xy ) = num__19.0 increase . hence d . <eor> d <eos> |
d |
percent__100.0__19.0__ |
percent__100.0__19.0__ |
| which of the following number should be added to num__11158 to make it exactly divisible by num__77 ? <o> a ) num__6 <o> b ) num__5 <o> c ) num__8 <o> d ) num__7 <o> e ) num__9 |
ondividing num__11158 by num__77 the remainder is num__70 thus number to be added = ( num__77 - num__70 ) = num__7 answer : d <eor> d <eos> |
d |
subtract__77.0__70.0__ subtract__77.0__70.0__ |
subtract__77.0__70.0__ subtract__77.0__70.0__ |
| if p is a positive integer and p ^ num__2 is divisible by num__12 then the largest positive integer that must divide p ^ num__3 is <o> a ) num__2 ^ num__3 <o> b ) num__2 ^ num__6 <o> c ) num__3 ^ num__3 <o> d ) num__6 ^ num__3 <o> e ) num__12 ^ num__2 |
since p is an integer so p can not have a num__2 and sqrt num__3 ( because squaring this will give us a num__2 ^ num__2 and num__3 ( making the product as num__12 and making p ^ num__2 as a multiple of num__12 ) ) p ^ num__2 is divisible by num__12 ( num__12 = num__2 * num__2 * num__3 ) so p should have at least one num__2 and one num__3 so that p ^ num__2 has a num__2 ^ num__2 and two num__3 so p will have a num__2 and a num__3 . or p will be a multiple of num__6 so largest possible integer than should divide p ^ num__3 is num__6 ^ num__3 so answer will be d <eor> d <eos> |
d |
multiply__2.0__3.0__ multiply__2.0__3.0__ |
multiply__2.0__3.0__ multiply__2.0__3.0__ |
| num__40 num__45 num__50 num__55 num__65 num__75 num__75 num__100 num__100 num__100 . the list above shows the scores of num__10 schoolchildren on a certain test . if the standard deviation of the num__10 scores is num__22.2 rounded to the nearest tenth how many of the scores are more than num__1 standard deviation below the mean of the num__10 scores ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
the average of { num__40 num__45 num__50 num__55 num__65 num__75 num__75 num__100 num__100 num__100 } is num__70.5 num__1 standard deviation below the mean is num__70.5 - num__22.2 = num__48.3 . hence there are two scores ( num__40 and num__45 ) more than num__1 standard deviation below the mean . answer b . <eor> b <eos> |
b |
subtract__70.5__22.2__ divide__100.0__50.0__ |
subtract__70.5__22.2__ divide__100.0__50.0__ |
| each of the products produced yesterday was checked by worker x or worker y . num__0.5 of the products checked by worker x are defective and num__0.8 of the products checked by worker y are defective . if the total defective rate of all the products checked by worker x and worker y is num__0.74 what fraction of the products was checked by worker y ? <o> a ) num__0.5 <o> b ) num__0.666666666667 <o> c ) num__0.6 <o> d ) num__0.75 <o> e ) num__0.8 |
x : num__0.5 is num__0.24 - points from num__0.74 . y : num__0.8 is num__0.06 - points from num__0.74 . therefore the ratio of products checked by y : x is num__4 : num__1 . thus worker y checked num__0.8 of the products . the answer is e . <eor> e <eos> |
e |
subtract__0.74__0.5__ subtract__0.8__0.74__ divide__0.24__0.06__ multiply__0.8__1.0__ |
subtract__0.74__0.5__ subtract__0.8__0.74__ divide__0.24__0.06__ multiply__0.8__1.0__ |
| a father said to his son ` ` i was as old as you are at present at the time of your birth . ' ' if the father ' s age is num__42 years now the son ' s age five years back was : <o> a ) num__19 years <o> b ) num__14 years <o> c ) num__33 years <o> d ) num__16 years <o> e ) num__39 years |
let the son ' s present age be x years . then ( num__42 - x ) = x num__2 x = num__42 = > x = num__21 son ' s age num__5 years back = ( num__21 - num__5 ) = num__16 years . answer : d <eor> d <eos> |
d |
divide__42.0__2.0__ subtract__21.0__5.0__ subtract__21.0__5.0__ |
divide__42.0__2.0__ subtract__21.0__5.0__ subtract__21.0__5.0__ |
| in the circular region shown above sections a and b represent num__0.375 and num__0.454545454545 respectively of the area of the circular region . section c represents what fractional part of the area of the circular region ? <o> a ) num__0.170454545455 <o> b ) num__0.181818181818 <o> c ) num__0.227272727273 <o> d ) num__0.511363636364 <o> e ) num__0.829545454545 |
a b and c represent complete area of circle and c fraction be x . num__0.375 + num__0.454545454545 + x = num__1 . = > x = num__1 - ( num__0.375 + num__0.454545454545 ) x = num__0.170454545455 . answer : option a is correct answer . . <eor> a <eos> |
a |
multiply__1.0__0.1705__ |
multiply__1.0__0.1705__ |
| num__10 men working num__6 hrs per day manufacture num__200 shoes in num__18 days . how many hours a day must num__15 men work to manufacture the same num__200 shoes in num__12 days ? <o> a ) num__6 <o> b ) num__10 <o> c ) num__12 <o> d ) num__15 <o> e ) num__18 |
let the required number of hours be x . more men less hours per day ( indirect proportion ) less days more working hours per day ( indirect proportion ) men num__15 : num__10 } : : num__6 : x days num__12 : num__18 . ' . num__15 * num__12 * x = num__10 * num__18 * num__6 = x = num__10 * num__18 * num__6 / ( num__15 * num__12 ) = x = num__6 answer : a <eor> a <eos> |
a |
round__6.0__ |
round__6.0__ |
| which of the following inequalities is equivalent to − num__1 < x < num__5 ? <o> a ) | x - num__1 | < num__7 <o> b ) | x + num__2 | < num__3 <o> c ) | x + num__3 | < num__5 <o> d ) | x - num__2 | < num__3 <o> e ) none of the above |
we know that | x | < a means - a < x < a where sum of lower limit of x ( i . e - a ) and the upper limit of x ( i . e a ) is num__0 given is - num__1 < x < num__5 let ' s say by adding y to this inequality we will get into the above format - num__1 + y < x + y < num__5 + y now to move this into the mod format we need to have ( - num__1 + y ) + ( num__5 + y ) = num__0 = > y = - num__2 thus - num__3 < x - num__2 < num__3 = > | x - num__2 | < num__3 . hence answer is d <eor> d <eos> |
d |
add__1.0__2.0__ multiply__1.0__2.0__ |
add__1.0__2.0__ subtract__5.0__3.0__ |
| a part - time employee ’ s hourly wage was increased by num__10.0 . she decided to decrease the number of hours worked per week so that her total income did not change . by approximately what percent should the number of hours worked be decreased ? <o> a ) num__9.0 <o> b ) num__15.0 <o> c ) num__25.0 <o> d ) num__50.0 <o> e ) num__100 % |
let ' s say the employee used to make $ num__1 / hour and worked num__100 hours / week so the total weekly income was $ num__100 / week after the num__10.0 wage increase the employee makes $ num__1.10 / hour we want the employee ' s income to remain at $ num__100 / week . so we want ( $ num__1.10 / hour ) ( new # of hours ) = $ num__100 divide both sides by num__1.10 to get : new # of hours = num__100 / num__1.10 ≈ num__90 hours so the number of hours decreases from num__100 hours to ( approximately ) num__90 hours . this represents a num__10.0 decrease ( approximately ) . answer choice a is the closest answer . <eor> a <eos> |
a |
subtract__100.0__10.0__ subtract__10.0__1.0__ |
subtract__100.0__10.0__ divide__90.0__10.0__ |
| the c . i . on a certain sum for num__2 years rs . num__41 and the simple interest is rs . num__40 . what is the rate percent ? <o> a ) num__8.0 <o> b ) num__5.0 <o> c ) num__6.0 <o> d ) num__4.0 <o> e ) num__2 % |
si = num__20 + num__20 ci = num__20 + num__21 num__20 - - - - num__1 num__100 - - - - ? = > num__5.0 answer : b <eor> b <eos> |
b |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| what is the sum of two consecutive even numbers the difference of whose squares is num__84 ? <o> a ) num__34 <o> b ) num__38 <o> c ) num__42 <o> d ) num__46 <o> e ) num__48 |
solution let the numbers be x and x + num__2 . then ( x + num__2 ) num__2 – x num__2 = num__84 ‹ = › num__4 x + num__4 = num__84 ‹ = › num__4 x = num__80 ‹ = › x = num__20 . required sum = x + ( x + num__2 ) = num__2 x + num__2 = num__42 . answer c <eor> c <eos> |
c |
subtract__84.0__4.0__ divide__80.0__4.0__ divide__84.0__2.0__ divide__84.0__2.0__ |
subtract__84.0__4.0__ divide__80.0__4.0__ divide__84.0__2.0__ divide__84.0__2.0__ |
| in what time will a train num__140 meters long cross an electric pole if its speed is num__210 km / hr <o> a ) num__5 seconds <o> b ) num__4.5 seconds <o> c ) num__3 seconds <o> d ) num__2.4 seconds <o> e ) none of these |
explanation : first convert speed into m / sec speed = num__210 * ( num__0.277777777778 ) = num__58 m / sec time = distance / speed = num__2.41379310345 = num__2.4 seconds option d <eor> d <eos> |
d |
divide__140.0__58.0__ round__2.4__ |
divide__140.0__58.0__ round__2.4__ |
| there is a num__100 pound watermelon laying out in the sun . num__99 percent of the watermelon ' s weight is water . after laying out for a few hours num__98 percent of the watermelon ' s weight is water . how much water evaporated ? <o> a ) a ) num__20 <o> b ) b ) num__30 <o> c ) c ) num__40 <o> d ) d ) num__50 <o> e ) e ) num__10 |
in the beginning it is num__99 pounds water and num__1 pound other stuff . at the end the num__1 pound other stuff is num__2 percent so the total weight is num__50 pounds . num__50 pounds - num__1 pound other stuff = num__49 pounds water . so num__99 pounds - num__49 pounds = num__50 pounds water lost . answer is d <eor> d <eos> |
d |
percent__98.0__50.0__ percent__100.0__50.0__ |
percent__98.0__50.0__ percent__100.0__50.0__ |
| what is the dividend . divisor num__19 the quotient is num__9 and the remainder is num__5 ? <o> a ) a ) num__130 <o> b ) b ) num__134 <o> c ) c ) num__148 <o> d ) d ) num__176 <o> e ) e ) num__160 |
d = d * q + r d = num__19 * num__9 + num__5 d = num__171 + num__5 d = num__176 answer d <eor> d <eos> |
d |
multiply__19.0__9.0__ add__5.0__171.0__ add__5.0__171.0__ |
multiply__19.0__9.0__ add__5.0__171.0__ add__5.0__171.0__ |
| steve traveled the first num__2 hours of his journey at num__40 mph and the last num__3 hours of his journey at num__80 mph . what is his average speed of travel for the entire journey ? <o> a ) num__61 mph <o> b ) num__62 mph <o> c ) num__63 mph <o> d ) num__64 mph <o> e ) num__65 mph |
( num__2 * num__40 + num__3 * num__80 ) / num__5 = num__64 mph answer : d <eor> d <eos> |
d |
add__2.0__3.0__ round__64.0__ |
add__2.0__3.0__ round__64.0__ |
| x y a and b are positive integers of h . when x is divided by y the remainder is num__6 . when a is divided by b the remainder is num__9 . which of the following is not a possible value for y + b ? <o> a ) num__24 <o> b ) num__21 <o> c ) num__20 <o> d ) num__17 <o> e ) num__15 |
x y a and b are positive integers of h . when x is divided by y the remainder is num__6 . when a is divided by b the remainder is num__9 . answer : e . <eor> e <eos> |
e |
add__6.0__9.0__ |
add__6.0__9.0__ |
| a sum of money amounts to rs . num__6690 after num__3 years and to rs . num__10035 after num__6 years on compound interest . find the sum . <o> a ) num__333 <o> b ) num__4460 <o> c ) num__37687 <o> d ) num__26698 <o> e ) num__2911 |
explanation : let the sum be rs . p . then p ( num__1 + r / num__100 ) ^ num__3 = num__6690 … ( i ) and p ( num__1 + r / num__100 ) ^ num__6 = num__10035 … ( ii ) on dividing we get ( num__1 + r / num__100 ) ^ num__3 = num__1.49850523169 = num__1.5 . substituting this value in ( i ) we get : p * ( num__1.5 ) = num__6690 or p = ( num__6690 * num__0.666666666667 ) = num__4460 hence the sum is rs . num__4460 . answer : b ) num__4460 <eor> b <eos> |
b |
percent__100.0__4460.0__ |
percent__100.0__4460.0__ |
| a searchlight on top of the watchtower makes num__4 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least num__12 seconds ? <o> a ) num__0.5 <o> b ) num__0.666666666667 <o> c ) num__0.333333333333 <o> d ) num__0.25 <o> e ) num__0.2 |
the searchlight completes one revolution every num__15 seconds . the probability that the man ' s area will be lit up is num__0.8 = num__0.8 . the probability that he will stay in the dark is num__1 - num__0.8 = num__0.2 the answer is e . <eor> e <eos> |
e |
divide__12.0__15.0__ subtract__1.0__0.8__ round__0.2__ |
divide__12.0__15.0__ subtract__1.0__0.8__ subtract__1.0__0.8__ |
| john and steve are speed walkers in a race . john is num__16 meters behind steve when he begins his final push . john blazes to the finish at a pace of num__4.2 m / s while steve maintains a blistering num__3.7 m / s speed . if john finishes the race num__2 meters ahead of steve how long was john ’ s final push ? <o> a ) num__13 seconds <o> b ) num__17 seconds <o> c ) num__36 seconds <o> d ) num__34 seconds <o> e ) num__51 seconds |
let t be the time that john spent for his final push . thus per the question num__4.2 t = num__3.7 t + num__16 + num__2 - - - > num__0.5 t = num__18 - - - > t = num__36 seconds . c is the correct answer . <eor> c <eos> |
c |
subtract__4.2__3.7__ add__16.0__2.0__ multiply__2.0__18.0__ round__36.0__ |
subtract__4.2__3.7__ add__16.0__2.0__ multiply__2.0__18.0__ round__36.0__ |
| how many words with or without meaning can be formed using all letters of the word good using each letter exactly once ? <o> a ) num__18 <o> b ) num__20 <o> c ) num__22 <o> d ) num__23 <o> e ) num__24 |
the word good has exactly num__4 letters which are all different . therefore the number of words that can be formed = number of permutations of num__4 letters taken all at a time . = p ( num__4 num__4 ) = num__4 ! = num__4 x num__3 x num__2 × num__1 = num__24 answer : e <eor> e <eos> |
e |
subtract__3.0__2.0__ multiply__24.0__1.0__ |
subtract__3.0__2.0__ multiply__24.0__1.0__ |
| at what time between num__7 and num__8 o ' clock will the hands of a clock be in the same straight line but not together ? <o> a ) num__5 min past num__7 <o> b ) num__5 x num__0.545454545455 min past num__7 <o> c ) num__5 x num__0.227272727273 min past num__7 <o> d ) num__5 x num__0.454545454545 min past num__7 <o> e ) num__0.227272727273 min past num__7 |
explanation : when the hands of the clock are in the same straight line but not together they are num__30 minute spaces apart . at num__7 o ' clock they are num__25 min . spaces apart . minute hand will have to gain only num__5 min . spaces . num__55 min . spaces are gained in num__60 min . num__5 min . spaces are gained in num__1.09090909091 x num__5 min = num__5 x num__0.454545454545 min . required time = num__5 x num__0.454545454545 min . past num__7 . answer is d <eor> d <eos> |
d |
subtract__30.0__25.0__ add__25.0__30.0__ add__5.0__55.0__ divide__60.0__55.0__ divide__25.0__55.0__ divide__25.0__5.0__ |
subtract__30.0__25.0__ add__25.0__30.0__ add__5.0__55.0__ divide__60.0__55.0__ divide__25.0__55.0__ divide__25.0__5.0__ |
| a man rows num__750 m in num__675 seconds against the stream and returns in num__7 and half minutes . his rowing speed in s Ɵ ll water is <o> a ) num__4 kmph <o> b ) num__5 kmph <o> c ) num__6 kmph <o> d ) num__7 kmph <o> e ) none of these |
explanation : rate upstream = ( num__1.11111111111 ) = num__1.11111111111 m / sec rate downstream ( num__1.66666666667 ) m / sec = num__1.66666666667 m / sec rate in still water = ( num__0.5 ) * [ ( num__1.11111111111 ) + ( num__1.66666666667 ) ] m / sec . = num__1.38888888889 m / sec = ( num__1.38888888889 ) * ( num__3.6 ) kmph = num__5 kmph answer : b <eor> b <eos> |
b |
divide__750.0__675.0__ multiply__1.3889__3.6__ round__5.0__ |
divide__750.0__675.0__ multiply__1.3889__3.6__ multiply__1.3889__3.6__ |
| four staff members at a certain company worked on a project . the amounts of time that the four staff members worked on the project were in the ratio num__2 to num__3 to num__5 to num__6 . if one of the four staff members worked on the project for num__60 hours which of the following can not be the total number of hours that the four staff members worked on the project ? <o> a ) num__480 <o> b ) num__160 <o> c ) num__320 <o> d ) num__192 <o> e ) num__240 |
four members worked in ration num__2 : num__3 : num__5 : num__6 hence as everyone mentioned individual work could be taken as num__2 x num__3 x num__5 x and num__6 x . also this gives us total work as num__16 x . but we are told that one of these individual works is num__60 hrs . hence possible scenarios if ( num__1 ) num__2 x = num__60 = > num__16 x = num__480 ( num__2 ) num__3 x = num__60 = > num__16 x = num__320 ( num__3 ) num__5 x = num__60 = > num__16 x = num__192 ( num__4 ) num__6 x = num__60 = > num__16 x = num__160 hence answer is e <eor> e <eos> |
e |
subtract__3.0__2.0__ add__3.0__1.0__ divide__480.0__3.0__ multiply__60.0__4.0__ |
subtract__3.0__2.0__ add__3.0__1.0__ divide__480.0__3.0__ multiply__60.0__4.0__ |
| what is the least number which should be added to num__2597 so that the sum is exactly divisible by num__5 num__6 num__4 and num__3 ? <o> a ) num__17 <o> b ) num__23 <o> c ) num__26 <o> d ) num__35 <o> e ) num__43 |
l . c . m . of num__5 num__6 num__4 and num__3 = num__60 . when dividing num__2597 by num__60 the remainder is num__17 . the number to be added = num__60 - num__17 = num__43 . the answer is e . <eor> e <eos> |
e |
subtract__60.0__17.0__ subtract__60.0__17.0__ |
subtract__60.0__17.0__ subtract__60.0__17.0__ |
| the age of a person is thrice the total ages of his num__2 daughters . num__0.5 decades hence his age will be twice of the total ages of his daughters . then what is the father ’ s current age ? [ num__0.5 decades = num__5 years ] <o> a ) num__45 years <o> b ) num__55 years <o> c ) num__42 years <o> d ) num__46 years <o> e ) num__52 years |
a num__45 years let total of current ages of the num__2 daughters is a years . then father ’ s current age = num__3 a years . ( num__3 a + num__5 ) = num__2 ( a + num__10 ) num__3 a + num__5 = num__2 a + num__20 a = num__15 therefore father ’ s current age = num__45 years . <eor> a <eos> |
a |
subtract__5.0__2.0__ multiply__2.0__5.0__ multiply__2.0__10.0__ multiply__5.0__3.0__ multiply__3.0__15.0__ |
subtract__5.0__2.0__ multiply__2.0__5.0__ multiply__2.0__10.0__ add__5.0__10.0__ multiply__3.0__15.0__ |
| rerty traveled a distance of num__30 km covering the first num__10 km in x minutes the next num__10 km in y minutes and the last num__10 km in z minutes . if he totally took num__3 y minutes to cover the whole distance then which of the following can not be true ? assume x y and z are different . <o> a ) z = num__3 x <o> b ) x = num__3 z <o> c ) y = num__2 x <o> d ) x = num__2 y <o> e ) y = num__3 x |
rerty travelled for x y and for z min . total time : x + y + z which is equal to : num__3 y equating both sides we get x + y + z = num__3 y = > x + z = num__2 y . . . . . . . . eqn num__1 looking out at options d says x = num__2 y using it in eqn num__1 num__2 y + z = num__2 y = > z = num__0 mins which i guess is not possible . the ans isd <eor> d <eos> |
d |
subtract__3.0__2.0__ subtract__3.0__1.0__ |
subtract__3.0__2.0__ subtract__3.0__1.0__ |
| a man can row num__10 kmph in still water . when the river is running at num__1.2 kmph it takes him num__1 hour to row to a place and back . what is the total distance traveled by the man ? <o> a ) num__6.24 km <o> b ) num__6 km <o> c ) num__9.86 km <o> d ) num__5.66 km <o> e ) num__10 km |
m = num__10 s = num__1.2 ds = num__11.2 us = num__8.8 x / num__11.2 + x / num__8.8 = num__1 x = num__4.93 d = num__4.93 * num__2 = num__9.86 answer : c <eor> c <eos> |
c |
add__10.0__1.2__ subtract__10.0__1.2__ multiply__2.0__4.93__ round__9.86__ |
add__10.0__1.2__ subtract__10.0__1.2__ multiply__2.0__4.93__ multiply__1.0__9.86__ |
| find the number of zeroes in num__58 ! ( num__58 factorial ) <o> a ) num__12 <o> b ) num__13 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
no of zeroes is num__11.6 = num__11 num__2.2 = num__2 num__11 + num__2 = num__13 answer : b <eor> b <eos> |
b |
add__2.0__11.0__ add__2.0__11.0__ |
add__2.0__11.0__ add__2.0__11.0__ |
| cost of an item is rs num__12.60 & profit is num__10.0 over selling price what is the selling price <o> a ) num__12.86 <o> b ) num__13.86 <o> c ) num__14.86 <o> d ) num__15.86 <o> e ) num__16.86 |
profit = ( sp - cp / cp ) * num__100 cp = num__12.60 ( sp - num__12.60 ) / num__12.60 * num__100 = num__10 sp = num__13.86 answer : b <eor> b <eos> |
b |
percent__100.0__13.86__ |
percent__100.0__13.86__ |
| before being simplified the instructions for computing income tax in country r were to add num__2 percent of one ' s annual income to the average ( arithmetic mean ) of num__100 units of country r ' s currency and num__1 percent of one ' s annual income . which of the following represents the simplified formula for computing the income tax in country r ' s currency for a person in that country whose annual income is a ? <o> a ) num__50 + a / num__200 <o> b ) num__50 + num__3 a / num__100 <o> c ) num__50 + a / num__40 <o> d ) num__100 + a / num__50 <o> e ) num__100 + num__3 a / num__100 |
the simplified formula for computing the income tax in country r ' s currency for a person in that country annual income is a : num__50 + a / num__40 answer : c <eor> c <eos> |
c |
divide__100.0__2.0__ divide__100.0__2.0__ |
divide__100.0__2.0__ divide__100.0__2.0__ |
| v q and r are positive integers . if v q and r are assembled into the six - digit number vqrvqr which one of the following must be a factor of vqrvqr ? <o> a ) num__23 <o> b ) num__19 <o> c ) num__17 <o> d ) num__7 <o> e ) none of the above |
one short way - vqrvqr = num__1000 vqr + vqr = ( num__1000 + num__1 ) vqr = num__1001 vqr therefore any factor of num__1001 is a factor of vqrvqr num__7 is a factor of num__1001 so d <eor> d <eos> |
d |
add__1000.0__1.0__ multiply__1.0__7.0__ |
add__1000.0__1.0__ multiply__1.0__7.0__ |
| there is num__60.0 increase in an amount in num__6 years at s . i . what will be the c . i . of rs . num__12000 after num__3 years at the same rate ? <o> a ) num__2387 <o> b ) num__2978 <o> c ) num__3972 <o> d ) num__2671 <o> e ) num__1871 |
explanation : let p = rs . num__100 . then s . i . rs . num__60 and t = num__6 years . r = ( num__100 * num__60 ) / ( num__100 * num__6 ) = num__10.0 p . a . now p = rs . num__12000 t = num__3 years and r = num__10.0 p . a . c . i . = [ num__12000 * { ( num__1 + num__0.1 ) num__3 - num__1 } ] = num__12000 * num__0.331 = rs . num__3972 answer : c <eor> c <eos> |
c |
percent__1.0__10.0__ percent__100.0__3972.0__ |
percent__1.0__10.0__ percent__100.0__3972.0__ |
| the number num__86 can be written as the sum of the squares of num__3 different positive integers . what is the sum of these num__3 integers ? <o> a ) num__17 <o> b ) num__16 <o> c ) num__15 <o> d ) num__14 <o> e ) num__13 |
num__7 ^ num__2 + num__6 ^ num__2 + num__1 ^ num__2 = num__49 + num__36 + num__1 = num__86 num__7 + num__6 + num__1 = num__14 hence answer is d <eor> d <eos> |
d |
multiply__3.0__2.0__ power__7.0__2.0__ power__6.0__2.0__ rectangle_perimeter__1.0__6.0__ rectangle_perimeter__1.0__6.0__ |
multiply__3.0__2.0__ power__7.0__2.0__ power__6.0__2.0__ rectangle_perimeter__1.0__6.0__ power__14.0__1.0__ |
| a leak in the bottom of a tank can empty the full tank in num__6 hours . an inlet pipe fills water at the rate of num__4 liters per minute . when the tank is full in inlet is opened and due to the leak the tank is empties in num__8 hours . the capacity of the tank is ? <o> a ) num__5729 <o> b ) num__5760 <o> c ) num__2889 <o> d ) num__2870 <o> e ) num__2799 |
num__1 / x - num__0.166666666667 = - num__0.125 x = num__24 hrs num__24 * num__60 * num__4 = num__5760 . answer : b <eor> b <eos> |
b |
divide__1.0__6.0__ divide__1.0__8.0__ multiply__6.0__4.0__ hour_to_min_conversion__ round__5760.0__ |
divide__1.0__6.0__ divide__1.0__8.0__ multiply__6.0__4.0__ hour_to_min_conversion__ divide__5760.0__1.0__ |
| when the integer n is divided by num__8 the remainder is num__5 . which of the following is not an even number ? <o> a ) n + num__3 <o> b ) n / num__2 + num__3.5 <o> c ) n – num__3 <o> d ) num__3 n + num__1 <o> e ) num__5 n + num__2 |
given number = n = num__8 p + num__5 - - - > the number n is odd . thus a . n + num__3 n + num__3 = odd + odd = even . thus not correct . b . n / num__2 + num__3.5 as n is odd thus n / num__2 = abc . num__5 ( where c = even ) - - - > abc . num__5 + num__3.5 = abc + num__3 + num__0.5 + num__0.5 = even + odd + num__1 = odd + num__1 = even . thus not correct . c . n – num__3 n - num__3 = odd - odd = even . thus not correct . d . num__3 n + num__1 as n = odd num__3 n = odd and num__3 n + num__1 = odd + odd = even . thus not correct . e . num__5 n + num__2 as n = odd num__5 n = odd - - - > num__5 n + num__2 = odd + even = odd . thus correct . answer : e <eor> e <eos> |
e |
subtract__8.0__5.0__ subtract__5.0__3.0__ reverse__2.0__ multiply__0.5__2.0__ subtract__8.0__3.0__ |
subtract__8.0__5.0__ subtract__5.0__3.0__ reverse__2.0__ subtract__3.0__2.0__ subtract__8.0__3.0__ |
| the sum of first eight prime numbers is ? <o> a ) num__12 <o> b ) num__23 <o> c ) num__24 <o> d ) num__77 <o> e ) num__30 |
required sum = ( num__2 + num__3 + num__5 + num__7 + num__11 + num__13 + num__17 + num__19 ) = num__77 . note : num__1 is not a prime number . definition : a prime number ( or a prime ) is a natural number that has exactly two distinct natural number divisors : num__1 and itself . d <eor> d <eos> |
d |
add__2.0__3.0__ add__2.0__5.0__ add__2.0__11.0__ add__2.0__17.0__ multiply__7.0__11.0__ subtract__3.0__2.0__ multiply__1.0__77.0__ |
add__2.0__3.0__ add__2.0__5.0__ add__2.0__11.0__ add__2.0__17.0__ multiply__7.0__11.0__ subtract__3.0__2.0__ multiply__1.0__77.0__ |
| it takes joey the postman num__1 hours to run a num__2 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is num__5 mile / hour what is the speed with which joey returns ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__13 <o> d ) num__14 <o> e ) num__15 |
let his speed for one half of the journey be num__2 miles an hour let the other half be x miles an hour now avg speed = num__5 mile an hour num__2 * num__2 * x / num__2 + x = num__5 num__4 x = num__5 x + num__10 = > x = num__10 a <eor> a <eos> |
a |
subtract__5.0__1.0__ multiply__2.0__5.0__ round__10.0__ |
subtract__5.0__1.0__ multiply__2.0__5.0__ multiply__1.0__10.0__ |
| a train num__200 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__11 <o> b ) num__10 <o> c ) num__5 <o> d ) num__9 <o> e ) num__8 |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__200 * num__0.0545454545455 ] sec = num__11 sec answer : a <eor> a <eos> |
a |
add__60.0__6.0__ divide__66.0__6.0__ round__11.0__ |
add__60.0__6.0__ divide__66.0__6.0__ divide__66.0__6.0__ |
| if | num__4 x + num__10 | = num__46 what is the sum of all the possible values of x ? <o> a ) num__2 <o> b ) - num__2 <o> c ) num__4 <o> d ) - num__5 <o> e ) num__6 |
there will be two cases num__4 x + num__10 = num__46 or num__4 x + num__10 = - num__46 = > x = num__9 or x = - num__14 sum of both the values will be - num__14 + num__9 = - num__5 answer is d <eor> d <eos> |
d |
add__4.0__10.0__ subtract__9.0__4.0__ subtract__10.0__5.0__ |
add__4.0__10.0__ subtract__9.0__4.0__ subtract__10.0__5.0__ |
| the ratio between the present ages of a and b is num__6 : num__3 respectively . the ratio between a ' s age num__4 years ago and b ' s age num__4 years hence is num__1 : num__1 . what is the ratio between a ' s age num__4 years hence and b ' s age num__4 years ago ? <o> a ) num__5 : num__4 <o> b ) num__3 : num__0 <o> c ) num__5 : num__1 <o> d ) num__3 : num__2 <o> e ) num__3 : num__7 |
let the present ages of a and b be num__6 x and num__3 x years respectively . then ( num__6 x - num__4 ) / ( num__3 x + num__4 ) = num__1.0 num__3 x = num__8 = > x = num__2.67 required ratio = ( num__5 x + num__4 ) : ( num__3 x - num__4 ) = num__20 : num__4 = num__5 : num__1 . answer : c <eor> c <eos> |
c |
subtract__6.0__1.0__ multiply__4.0__5.0__ subtract__6.0__1.0__ |
subtract__6.0__1.0__ multiply__4.0__5.0__ subtract__6.0__1.0__ |
| how many odd factors does num__210 have ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__8 |
start with the prime factorization : num__210 = num__2 * num__3 * num__5 * num__7 for odd factors we put aside the factor of two and look at the other prime factors . set of exponents = { num__1 num__1 num__1 } plus num__1 to each = { num__2 num__2 num__2 } product = num__2 * num__2 * num__2 = num__8 therefore there are num__8 odd factors of num__210 . in case you are curious they are { num__1 num__3 num__5 num__7 num__15 num__21 num__35 and num__105 } answer : e . <eor> e <eos> |
e |
add__2.0__3.0__ add__2.0__5.0__ subtract__3.0__2.0__ add__1.0__7.0__ multiply__3.0__5.0__ multiply__3.0__7.0__ multiply__5.0__7.0__ divide__210.0__2.0__ add__1.0__7.0__ |
add__2.0__3.0__ add__2.0__5.0__ subtract__3.0__2.0__ add__1.0__7.0__ multiply__3.0__5.0__ multiply__3.0__7.0__ multiply__5.0__7.0__ multiply__3.0__35.0__ add__1.0__7.0__ |
| the difference between a two - digit number and the number obtained by interchanging the positions of its digits is num__72 . what is the difference between the two digits of that number ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
sol . let the ten ’ s digit be x and unit ’ s digit be y then ( num__10 x + y ) - ( num__10 y + x ) = num__72 ⇔ num__9 ( x - y ) = num__72 ⇔ x - y = num__8 answer d <eor> d <eos> |
d |
divide__72.0__9.0__ divide__72.0__9.0__ |
divide__72.0__9.0__ divide__72.0__9.0__ |
| in traveling from a dormitory to a certain city a student went num__0.2 of the way by foot num__0.666666666667 of the way by bus and the remaining num__12 kilometers by car . what is the distance in kilometers from the dormitory to the city ? <o> a ) num__30 <o> b ) num__45 <o> c ) num__60 <o> d ) num__64 <o> e ) num__120 |
i believe there is a better way to do it . basically one of the options should satisfy the given criteria . num__60 did num__0.2 * num__60 = num__12 num__0.666666666667 * num__60 = num__40 so total distance num__52 + remaining num__12 = num__64 answer d <eor> d <eos> |
d |
hour_to_min_conversion__ add__12.0__40.0__ add__12.0__52.0__ round__64.0__ |
hour_to_min_conversion__ add__12.0__40.0__ add__12.0__52.0__ add__12.0__52.0__ |
| mangala completes a piece of work in num__12 days raju completes the same work in num__36 days . if both of them work together then the number of days required to complete the work is <o> a ) num__6 days <o> b ) num__7 days <o> c ) num__8 days <o> d ) num__9 days <o> e ) num__11 days |
if a can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days . that is the required no . of days = num__12 Ã — num__0.75 = num__9 days . d <eor> d <eos> |
d |
multiply__12.0__0.75__ round__9.0__ |
multiply__12.0__0.75__ round__9.0__ |
| two sets of num__6 consecutive positive integers have exactly one integer in common . the sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set ? <o> a ) num__14 <o> b ) num__42 <o> c ) num__28 <o> d ) num__12 <o> e ) it can not be determined from the information given . |
a = ( num__12 num__34 num__56 ) sum of this = num__21 b = ( num__6 num__78 num__910 num__1112 ) sum of this = num__63 the differenct between num__63 - num__21 = num__42 hence num__42 is the answer i . e . b <eor> b <eos> |
b |
subtract__63.0__21.0__ subtract__63.0__21.0__ |
subtract__63.0__21.0__ subtract__63.0__21.0__ |
| if num__12 men can reap num__120 acres of land in num__16 days how many acres of land can num__36 men reap in num__32 days ? <o> a ) num__269 <o> b ) num__512 <o> c ) num__369 <o> d ) num__720 <o> e ) num__450 |
num__12 men num__120 acres num__16 days num__36 men ? num__32 days num__120 * num__3.0 * num__2.0 num__120 * num__3 * num__2 num__120 * num__6 = num__720 answer : d <eor> d <eos> |
d |
divide__36.0__12.0__ divide__32.0__16.0__ divide__12.0__2.0__ multiply__120.0__6.0__ round__720.0__ |
divide__36.0__12.0__ divide__32.0__16.0__ multiply__2.0__3.0__ multiply__120.0__6.0__ multiply__120.0__6.0__ |
| - num__69 * num__39 + num__450 = ? <o> a ) num__2736 <o> b ) num__2309 <o> c ) - num__2801 <o> d ) - num__2241 <o> e ) none of these |
= > - num__69 * ( num__40 - num__1 ) + num__450 ; = > - ( num__69 * num__40 ) + num__69 + num__450 ; = > - num__2760 + num__519 = - num__2241 . correct option : d <eor> d <eos> |
d |
subtract__40.0__39.0__ multiply__69.0__40.0__ add__69.0__450.0__ subtract__2760.0__519.0__ multiply__1.0__2241.0__ |
subtract__40.0__39.0__ multiply__69.0__40.0__ add__69.0__450.0__ subtract__2760.0__519.0__ multiply__1.0__2241.0__ |
| a train of num__25 carriages each of num__60 meters length when an engine also of num__60 meters length is running at a speed of num__60 kmph . in what time will the train cross a bridge num__2.5 km long ? <o> a ) num__4 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__9 |
d = num__25 * num__60 + num__2500 = num__4000 m t = num__66.6666666667 * num__3.6 = num__240 sec = num__4 mins answer : a <eor> a <eos> |
a |
divide__4000.0__60.0__ divide__240.0__60.0__ round__4.0__ |
divide__4000.0__60.0__ divide__240.0__60.0__ round__4.0__ |
| everyone shakes hands with everyone else in a room . total number of handshakes is num__55 . number of persons = ? <o> a ) num__14 <o> b ) num__12 <o> c ) num__11 <o> d ) num__15 <o> e ) num__16 |
in a room of n people the number of possible handshakes is c ( n num__2 ) or n ( n - num__1 ) / num__2 so n ( n - num__1 ) / num__2 = num__55 or n ( n - num__1 ) = num__110 or n = num__11 answer is ( c ) <eor> c <eos> |
c |
multiply__55.0__2.0__ multiply__1.0__11.0__ |
multiply__55.0__2.0__ divide__11.0__1.0__ |
| in x game of billiards x can give y num__10 points in num__60 and he can give z num__30 points in num__60 . how many points can y give z in x game of num__100 ? <o> a ) num__30 <o> b ) num__20 <o> c ) num__25 <o> d ) num__40 <o> e ) num__50 |
x scores num__60 while y score num__50 and z scores num__30 . the number of points that z scores when y scores num__100 = ( num__100 * num__30 ) / num__50 = num__60 . in x game of num__100 points y gives ( num__100 - num__60 ) = num__40 points to c . d <eor> d <eos> |
d |
subtract__60.0__10.0__ add__10.0__30.0__ add__10.0__30.0__ |
subtract__60.0__10.0__ subtract__100.0__60.0__ subtract__100.0__60.0__ |
| num__63 + num__5 * num__12 / ( num__60.0 ) = ? <o> a ) num__22 <o> b ) num__77 <o> c ) num__29 <o> d ) num__64 <o> e ) num__21 |
num__63 + num__5 * num__12 / ( num__60.0 ) = num__63 + num__5 * num__12 / ( num__60 ) = num__63 + ( num__5 * num__12 ) / num__60 = num__63 + num__1 = num__64 . answer : d <eor> d <eos> |
d |
add__63.0__1.0__ add__63.0__1.0__ |
add__63.0__1.0__ add__63.0__1.0__ |
| when n is divided by num__5 the remainder is num__3 . what is the remainder when n ^ num__2 is divided by num__5 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
n = num__5 x + num__3 for some integer x n ^ num__2 = ( num__5 x + num__3 ) ^ num__2 = num__5 y + num__9 for some integer y when we divide this by num__5 the remainder is num__4 . the answer is e . <eor> e <eos> |
e |
subtract__9.0__5.0__ subtract__9.0__5.0__ |
subtract__9.0__5.0__ subtract__9.0__5.0__ |
| a watch was sold at a loss of num__10.0 . if it was sold for rs . num__210 more there would have been a gain of num__4.0 . what is the cost price ? <o> a ) s : num__1000 <o> b ) s : num__1500 <o> c ) s : num__1278 <o> d ) s : num__1028 <o> e ) s : num__1027 |
num__90.0 num__104.0 - - - - - - - - num__14.0 - - - - num__210 num__100.0 - - - - ? = > rs : num__1500 answer : b <eor> b <eos> |
b |
percent__100.0__1500.0__ |
percent__100.0__1500.0__ |
| i gain num__70 paise on rs . num__70 . my gain percent is : <o> a ) num__0.1 <o> b ) num__1.0 <o> c ) num__7.0 <o> d ) num__10.0 <o> e ) noneof these |
solution num__11.1 gain % = ( num__0.70 / num__70 x num__100 ) % = num__1.0 . answer b <eor> b <eos> |
b |
percent__1.0__100.0__ |
percent__1.0__100.0__ |
| evaluate : num__986 x num__237 + num__986 x num__863 <o> a ) num__956000 <o> b ) num__996000 <o> c ) num__986000 <o> d ) num__976000 <o> e ) none of them |
= num__986 x num__137 + num__986 x num__863 = num__986 x ( num__137 + num__863 ) = num__986 x num__1000 = num__986000 . answer is c . <eor> c <eos> |
c |
add__863.0__137.0__ multiply__986.0__1000.0__ multiply__986.0__1000.0__ |
add__863.0__137.0__ multiply__986.0__1000.0__ multiply__986.0__1000.0__ |
| mahesh marks an article num__15.0 above the cost price of rs . num__540 . what must be his discount percentage if he sells it at rs . num__462 ? <o> a ) num__25.6 <o> b ) num__21.0 <o> c ) num__20.0 <o> d ) num__19.0 <o> e ) none of these |
cp = rs . num__540 mp = num__540 + num__15.0 of num__540 = rs . num__621 sp = rs . num__462 discount = num__621 - num__462 = num__159 discount % = num__0.256038647343 * num__100 = num__25.6 answer : a <eor> a <eos> |
a |
percent__100.0__25.6__ |
percent__100.0__25.6__ |
| find the last digit of num__1 ! + num__2 ! + num__3 ! + num__4 ! + . . . . . . . . . . . . . . . num__100 ! <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
num__5 ! + num__6 ! . . . . = always unit digit o . num__1 ! + num__2 ! + num__3 ! + num__4 ! = num__33 so num__33 + * num__0 = * num__3 so the unit digit will be num__3 answer : d <eor> d <eos> |
d |
add__1.0__4.0__ add__1.0__5.0__ add__1.0__2.0__ |
add__1.0__4.0__ add__1.0__5.0__ add__1.0__2.0__ |
| one fourth of one third of two fifth of a number is num__20 . what will be num__40.0 of that number <o> a ) num__240 <o> b ) num__150 <o> c ) num__180 <o> d ) num__200 <o> e ) num__250 |
explanation : ( num__0.25 ) * ( num__0.333333333333 ) * ( num__0.4 ) * x = num__20 then x = num__20 * num__30 = num__600 num__40.0 of num__600 = num__240 answer : option a <eor> a <eos> |
a |
percent__40.0__600.0__ percent__40.0__600.0__ |
percent__40.0__600.0__ percent__40.0__600.0__ |
| the area of a square garden is a square feet and the perimeter is p feet . if a = p + num__5 what is the perimeter of the garden in feet ? <o> a ) num__28 <o> b ) num__20 <o> c ) num__40 <o> d ) num__56 <o> e ) num__64 |
perimeter of square = p side of square = p / num__4 area of square = ( p ^ num__2 ) / num__16 = a given that a = p + num__5 ( p ^ num__2 ) / num__16 = p + num__5 p ^ num__2 = num__16 p + num__80 p ^ num__2 - num__16 p - num__80 = num__0 p ^ num__2 - num__20 p + num__4 p - num__80 = num__0 p ( p - num__20 ) + num__4 ( p + num__20 ) = num__0 ( p - num__20 ) ( p + num__4 ) = num__0 p = num__20 o r - num__4 discarding negative value p = num__20 answer is b <eor> b <eos> |
b |
square_perimeter__4.0__ multiply__5.0__16.0__ square_perimeter__5.0__ square_perimeter__5.0__ |
power__2.0__4.0__ multiply__5.0__16.0__ square_perimeter__5.0__ square_perimeter__5.0__ |
| cereal a is num__10.0 sugar by weight whereas healthier but less delicious cereal b is num__2.0 sugar by weight . to make a delicious and healthy mixture that is num__5.0 sugar what should be the ratio of cereal a to cereal b by weight ? <o> a ) num__3 : num__5 <o> b ) num__2 : num__7 <o> c ) num__1 : num__6 <o> d ) num__1 : num__4 <o> e ) num__1 : num__3 |
( num__0.1 ) a + ( num__0.02 ) b = ( num__0.05 ) ( a + b ) num__5 a = num__3 b = > a / b = num__0.6 answer is a . <eor> a <eos> |
a |
reverse__10.0__ divide__0.1__5.0__ divide__0.1__2.0__ subtract__5.0__2.0__ divide__3.0__5.0__ subtract__5.0__2.0__ |
reverse__10.0__ divide__0.1__5.0__ divide__0.1__2.0__ subtract__5.0__2.0__ divide__3.0__5.0__ subtract__5.0__2.0__ |
| the average amount with a group of seven numbers is rs . num__30 . if the newly joined member has rs . num__60 with him what was the average amount with the group before his joining the group ? <o> a ) s . num__25.6 <o> b ) s . num__25 <o> c ) s . num__16.6 <o> d ) s . num__26 <o> e ) s . num__25.6 |
total members in the group = num__7 average amount = rs . num__30 total amount with them = num__7 * num__30 = rs . num__210 one number has rs . num__60 . so the amount with remaining num__6 people = num__210 - num__60 = rs . num__150 the average amount with them = num__25.0 = rs . num__25 answer : b <eor> b <eos> |
b |
multiply__30.0__7.0__ subtract__210.0__60.0__ divide__150.0__6.0__ divide__150.0__6.0__ |
multiply__30.0__7.0__ subtract__210.0__60.0__ divide__150.0__6.0__ divide__150.0__6.0__ |
| a goods train runs at the speed of num__72 kmph and crosses a num__250 m long platform in num__26 seconds . what is the length of the goods train ? <o> a ) num__270 . <o> b ) num__480 <o> c ) num__920 <o> d ) num__280 <o> e ) num__290 |
speed = num__72 x num__0.277777777778 m / sec = num__20 m / sec . time = num__26 sec . let the length of the train be x metres . then x + num__9.61538461538 = num__20 x + num__250 = num__520 x = num__270 . answer : a <eor> a <eos> |
a |
divide__250.0__26.0__ multiply__26.0__20.0__ add__250.0__20.0__ round__270.0__ |
divide__250.0__26.0__ multiply__26.0__20.0__ add__250.0__20.0__ add__250.0__20.0__ |
| the difference in compound interest earned on a deposit ( compounded annually ) in year num__1 and year num__2 is $ num__40 . had the interest rate been three times its present value the difference w would have been how much ? <o> a ) num__13.3333333333 <o> b ) num__40 <o> c ) num__120 <o> d ) num__360 <o> e ) num__420 |
case num__1 : deposit = $ x ; rate of increase = r . interest yearned in num__1 year = xr . deposit in num__1 year = x + xr . interest yearned in num__2 year = ( x + xr ) r . the difference w = ( x + xr ) r - xr = xr ^ num__2 = num__40 . case num__2 : deposit = $ x ; rate of increase = num__3 r . interest yearned in num__1 year = x ( num__3 r ) . deposit in num__1 year = x + num__3 xr . interest yearned in num__2 year = ( x + num__3 xr ) num__3 r . the difference = ( x + num__3 xr ) num__3 r - num__3 xr = num__9 xr ^ num__2 . since from case num__1 we know that xr ^ num__2 = num__40 then num__9 xr ^ num__2 = num__9 * num__40 = num__360 . answer : d . <eor> d <eos> |
d |
add__1.0__2.0__ multiply__40.0__9.0__ multiply__1.0__360.0__ |
add__1.0__2.0__ multiply__40.0__9.0__ multiply__1.0__360.0__ |
| how many minutes does aditya take to cover a distance of num__400 m if he runs at a speed of num__20 km / hr ? <o> a ) num__1 min <o> b ) num__1 ( num__0.2 ) min <o> c ) num__2 min <o> d ) num__3 min <o> e ) num__4 min |
aditya â € ™ s speed = num__20 km / hr = { num__20 * num__0.277777777778 } min = num__5.55555555556 m / s time taken to cover num__400 m = { num__400 * num__0.18 } sec = num__72 sec = num__1 ( num__0.2 ) min answer b <eor> b <eos> |
b |
multiply__400.0__0.18__ multiply__0.18__5.5556__ round__1.0__ |
multiply__400.0__0.18__ multiply__0.18__5.5556__ multiply__0.18__5.5556__ |
| what is the difference between the largest number and the least number written with the digits num__9 num__3 num__5 num__7 ? <o> a ) num__6084 <o> b ) num__3788 <o> c ) num__2077 <o> d ) num__2721 <o> e ) num__6174 |
explanation : num__3579 num__9753 - - - - - - - - - - - - num__6174 answer : e <eor> e <eos> |
e |
subtract__9753.0__3579.0__ subtract__9753.0__3579.0__ |
subtract__9753.0__3579.0__ subtract__9753.0__3579.0__ |
| the average of nine numbers is num__27 . if one number is excluded the average becomes num__25 . the excluded number is <o> a ) num__25 <o> b ) num__27 <o> c ) num__30 <o> d ) num__35 <o> e ) num__43 |
sol . therefore excluded number = ( num__27 × num__9 ) - ( num__25 × num__8 ) = num__243 – num__200 = num__43 . answer e <eor> e <eos> |
e |
multiply__27.0__9.0__ multiply__25.0__8.0__ subtract__243.0__200.0__ subtract__243.0__200.0__ |
multiply__27.0__9.0__ multiply__25.0__8.0__ subtract__243.0__200.0__ subtract__243.0__200.0__ |
| a grocer stacked oranges in a pile . the bottom layer was rectangular with num__4 rows of num__6 oranges each . in the second layer from the bottom each orange rested on num__4 oranges from the bottom layer and in the third layer each orange rested on num__4 oranges from the second layer . which of the following is the maximum number of oranges that could have been in the third layer ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
bottom layer = num__6 x num__4 = num__24 iind layer = ( num__6 - num__1 ) x ( num__4 - num__1 ) = num__15 iiird layer = ( num__5 - num__1 ) x ( num__3 - num__1 ) = num__8 answer = num__8 = d <eor> d <eos> |
d |
multiply__4.0__6.0__ add__4.0__1.0__ subtract__4.0__1.0__ add__3.0__5.0__ multiply__1.0__8.0__ |
multiply__4.0__6.0__ subtract__6.0__1.0__ subtract__4.0__1.0__ add__3.0__5.0__ multiply__1.0__8.0__ |
| a rectangular tank measuring num__5 m × num__4.5 m × num__2.1 m is dug in the centre of the field measuring num__13.5 m × num__2.5 . the earth dug out is spread evenly over the remaining portion of a field . how much is the level of the field raised ? <o> a ) num__4.0 m <o> b ) num__4.1 m <o> c ) num__4.2 m <o> d ) num__4.3 m <o> e ) none of these |
area of the field = num__13.5 × num__2.5 = num__33.75 m num__2 area covered by the rectangular tank = num__5 × num__4.5 = num__22.50 m num__2 area of the field on which the earth dug out is to be spread = num__33.75 – num__22.50 = num__11.25 m num__2 let the required height be h . then num__11.25 × h = num__5 × num__4.5 × num__2.1 or h = num__4.2 m answer c <eor> c <eos> |
c |
multiply__13.5__2.5__ divide__5.0__2.5__ multiply__5.0__4.5__ multiply__4.5__2.5__ multiply__2.1__2.0__ round__4.2__ |
multiply__13.5__2.5__ divide__5.0__2.5__ multiply__5.0__4.5__ multiply__4.5__2.5__ multiply__2.1__2.0__ round__4.2__ |
| a sum of money deposited at c . i . amounts to rs . num__2420 in num__2 years and to rs . num__2662 in num__3 years . find the rate percent ? <o> a ) num__17.0 <o> b ) num__10.0 <o> c ) num__20.0 <o> d ) num__18.0 <o> e ) num__40 % |
num__2420 - - - num__242 num__100 - - - ? = > num__10.0 answer : b <eor> b <eos> |
b |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| a father said to his son ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is num__38 years now the son ' s age five years back was : <o> a ) num__8 years <o> b ) num__14 years <o> c ) num__4 years <o> d ) num__2 years <o> e ) num__8 years |
explanation : let the son ' s present age be x years . then ( num__38 - x ) = x num__2 x = num__38 . x = num__19 . son ' s age num__5 years back ( num__19 - num__5 ) = num__14 years . answer : b <eor> b <eos> |
b |
divide__38.0__2.0__ subtract__19.0__5.0__ subtract__19.0__5.0__ |
divide__38.0__2.0__ subtract__19.0__5.0__ subtract__19.0__5.0__ |
| a shopkeeper buys mangoes at the rate of num__5 a rupee and sells them at num__4 a rupee . find his net profit or loss percent ? <o> a ) num__33 num__0.125 % <o> b ) num__33 num__2.33333333333 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__25.0 <o> e ) num__34 num__0.333333333333 % |
the total number of mangoes bought by the shopkeeper be num__20 . if he buys num__5 a rupee his cp = num__4 he selling at num__4 a rupee his sp = num__5 profit = sp - cp = num__5 - num__4 = num__1 profit percent = num__0.25 * num__100 = num__25.0 answer : d <eor> d <eos> |
d |
percent__5.0__20.0__ percent__25.0__100.0__ |
percent__5.0__20.0__ percent__25.0__100.0__ |
| if n is a positive integer such that n ! / ( n - num__2 ) ! = num__342 find n . <o> a ) num__13 <o> b ) num__17 <o> c ) num__15 <o> d ) num__11 <o> e ) num__19 |
let expand n as follows n ! = n * ( n - num__1 ) * ( n - num__2 ) ! n ! / ( n - num__2 ) ! = n * ( n - num__1 ) * ( n - num__2 ) ! = n ( n - num__1 ) = num__342 solve the equation n ( n - num__1 ) = num__342 and select the positive integer solution which is n = num__19 correct answer e <eor> e <eos> |
e |
multiply__1.0__19.0__ |
multiply__1.0__19.0__ |
| the average of num__13 result is num__20 . average of the first num__8 of them is num__20 and that of the last num__8 is num__40 . find the num__8 th result ? <o> a ) num__35 <o> b ) num__37 <o> c ) num__46 <o> d ) num__90 <o> e ) num__100 |
sum of all the num__13 results = num__13 * num__20 = num__260 sum of the first num__7 of them = num__8 * num__20 = num__160 sum of the last num__7 of them = num__8 * num__40 = num__320 so the num__8 th number = num__260 + num__160 - num__320 = num__100 . e <eor> e <eos> |
e |
multiply__13.0__20.0__ subtract__20.0__13.0__ multiply__20.0__8.0__ multiply__8.0__40.0__ subtract__260.0__160.0__ subtract__260.0__160.0__ |
multiply__13.0__20.0__ subtract__20.0__13.0__ multiply__20.0__8.0__ multiply__8.0__40.0__ subtract__260.0__160.0__ subtract__260.0__160.0__ |
| at a round table num__7 knights are to be seated around a circular table . two seating arrangement are considered different only when the position of the people are different relative to each other . what is the total number of possible seating arrangement for the group ? <o> a ) a . num__5 <o> b ) b . num__10 <o> c ) c . num__24 <o> d ) d . num__32 <o> e ) e . num__720 |
for num__5 people seating arrangement around a circular table = ( num__7 - num__1 ) ! = num__6 ! = num__6 * num__5 * num__4 * num__3 * num__2 * num__1 = num__720 oa is e <eor> e <eos> |
e |
vowel_space__ die_space__ coin_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
vowel_space__ die_space__ coin_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| a train consists of num__12 boggies each boggy num__15 metres long . the train crosses a telegraph post in num__9 seconds . due to some problem one boggies were detached . the train now crosses a telegraph post in <o> a ) num__18 sec <o> b ) num__12 sec <o> c ) num__15 sec <o> d ) num__8.25 sec <o> e ) none of these |
length of train = num__12 Ã — num__15 = num__180 m . then speed of train = num__180 â „ num__9 = num__20 m / s now length of train = num__11 Ã — num__15 = num__165 m â ˆ ´ required time = num__165 â „ num__20 = num__8.25 sec . answer d <eor> d <eos> |
d |
multiply__12.0__15.0__ divide__180.0__9.0__ subtract__20.0__9.0__ multiply__15.0__11.0__ divide__165.0__20.0__ round__8.25__ |
multiply__12.0__15.0__ divide__180.0__9.0__ subtract__20.0__9.0__ multiply__15.0__11.0__ divide__165.0__20.0__ divide__165.0__20.0__ |
| sam taken a loan rs . num__15000 / - from co - operative society with an interest @ num__11.5 per month . at the same time he deposited rs . num__10000 / - as fixed deposit with an interest @ num__9.5 per month . after one week sam asked the manager to calculate the interest to be paid . what is the interest amount for num__7 days ? <o> a ) num__165 <o> b ) num__220 <o> c ) num__310 <o> d ) num__185 <o> e ) num__181 |
loan amount : rs . num__15000 / - @ num__11.5 interest per month = num__15000 / - * num__11.5 = rs . num__1725 interest for one day = num__57.5 = num__57.50 interest for num__7 days is = num__57.50 * num__7 = num__403 fd amount is = rs . num__10000 / - @ num__9.5 interest per month = num__10000 * num__9.5 = num__950 / - interest for num__7 days = num__31.6666666667 * num__7 = num__222 interest amount to be paid by sam = num__403 - num__222 = num__181 / - for num__7 days answer is e <eor> e <eos> |
e |
subtract__403.0__222.0__ subtract__403.0__222.0__ |
subtract__403.0__222.0__ subtract__403.0__222.0__ |
| danny and steve are running towards each other each one from his own house . danny can reach steve ' s house in num__33 minutes of running which is half the time it takes steve to reach danny ' s house . if the two started to run at the same time how much time longer will it take steve to reach the halfway point between their houses than danny to reach the halfway point between their houses ? <o> a ) num__16.5 minutes <o> b ) num__33 minutes <o> c ) num__49.5 minutes <o> d ) num__66 minutes <o> e ) num__99 minutes |
danny ' s time is num__33 minutes . the time to reach halfway is num__16.5 minutes . steve ' s time is num__66 minutes . the time to reach halfway is num__33 minutes . the time difference is num__33 - num__16.5 = num__16.5 minutes the answer is a . <eor> a <eos> |
a |
round__16.5__ |
subtract__33.0__16.5__ |
| the average expenditure of a labourer for num__8 months was num__85 and he fell into debt . in the next num__4 months by reducing his monthly expenses to num__60 he not only cleared off his debt but also saved num__30 . his monthly income is <o> a ) num__80 <o> b ) num__85 <o> c ) num__75 <o> d ) num__95 <o> e ) num__100 |
income of num__8 months = ( num__8 × num__85 ) – debt = num__680 – debt income of the man for next num__4 months = num__4 × num__60 + debt + num__30 = num__270 + debt ∴ income of num__10 months = num__950 average monthly income = num__950 ÷ num__10 = num__95 answer d <eor> d <eos> |
d |
multiply__8.0__85.0__ add__270.0__680.0__ add__85.0__10.0__ add__85.0__10.0__ |
multiply__8.0__85.0__ add__270.0__680.0__ add__85.0__10.0__ add__85.0__10.0__ |
| if the length of the longest chord of a certain circle is num__10 what is the radius of that certain circle ? <o> a ) num__2.5 <o> b ) num__5 <o> c ) num__10 <o> d ) num__15 <o> e ) num__20 |
longest chord of a circle is the diameter of the circle diameter = num__2 * radius if diameter of the circle is given as num__10 = num__2 * num__5 so radius of the circle = num__5 correct answer - b <eor> b <eos> |
b |
triangle_area__2.0__5.0__ |
triangle_area__2.0__5.0__ |
| in num__1986 the book value of a certain car was num__0.666666666667 original price and in num__1988 its book value was num__0.5 the original purchase price . by what percent did the book value for this car decrease from num__1986 to num__1988 ? <o> a ) num__16 num__0.666666666667 % <o> b ) num__25.0 <o> c ) num__33 num__0.333333333333 <o> d ) num__50.0 <o> e ) num__75 % |
in num__1986 - num__200 in num__1988 - num__150 decrease of num__50 over num__200 between num__86 and num__88 so num__25.0 answer : b <eor> b <eos> |
b |
subtract__200.0__150.0__ multiply__0.5__50.0__ multiply__0.5__50.0__ |
subtract__200.0__150.0__ multiply__0.5__50.0__ subtract__50.0__25.0__ |
| in a class there are num__20 boys whose average age is decreased by num__2 months when one boy aged num__20 years replaced by a new boy . the age of the new boy is ? <o> a ) num__14 years num__8 months <o> b ) num__15 years <o> c ) num__16 years num__8 months <o> d ) num__17 years num__10 months <o> e ) num__17 years |
total decrease = ( num__20 x num__2 ) months = num__3 years num__4 months age of the new boy = num__20 years - num__3 years num__4 months . = num__16 years num__8 months . answer : c <eor> c <eos> |
c |
subtract__20.0__4.0__ multiply__2.0__4.0__ subtract__20.0__4.0__ |
subtract__20.0__4.0__ multiply__2.0__4.0__ subtract__20.0__4.0__ |
| at a certain college num__40 percent of the total number of students are freshmen . if num__50 percent of the fresh - men are enrolled in the school of liberal arts and of these num__50 percent are psychology majors what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal arts ? <o> a ) num__10.0 <o> b ) num__12.0 <o> c ) num__14.0 <o> d ) num__16.0 <o> e ) num__18 % |
let ' s say there is a total of num__100 students at this college . num__40 percent of the total number of students are freshmen . # of freshmen = num__40.0 of num__100 = num__40 num__50 percent of the fresh - men are enrolled in the school of liberal arts . . . number of liberal arts freshmen = num__50.0 of num__40 = num__20 . . . and of these num__50 percent are psychology majors . . . number of liberal arts freshmen who are psychology majors = num__50.0 of num__20 = num__10 what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal arts ? num__0.1 = num__10.0 answer : a <eor> a <eos> |
a |
subtract__50.0__40.0__ reverse__10.0__ reverse__0.1__ |
subtract__50.0__40.0__ reverse__10.0__ subtract__50.0__40.0__ |
| a ( num__7 w ^ num__3 ) is the ( x y ) coordinate of point located on the parabola y = x ^ num__2 + num__15 . what is the value of w ? <o> a ) num__3 . <o> b ) num__4 . <o> c ) num__5 . <o> d ) num__6 . <o> e ) num__9 . |
y = x ^ num__2 + num__15 w ^ num__3 = num__7 ^ num__2 + num__15 w ^ num__3 = num__64 w = num__4 answer b <eor> b <eos> |
b |
subtract__7.0__3.0__ subtract__7.0__3.0__ |
subtract__7.0__3.0__ subtract__7.0__3.0__ |
| find the principal which yields a simple interest of rs . num__20 and compound interest of rs . num__24 in two years at the same percent rate per annum ? <o> a ) s . num__50 <o> b ) s . num__48 <o> c ) s . num__42 <o> d ) s . num__20 <o> e ) s . num__60 |
explanation : si in num__2 years = rs . num__20 si in num__1 year = rs . num__10 ci in num__2 years = rs . num__24.0 rate per annum = [ ( ci – si ) / ( si in num__1 year ) ] * num__100 = [ ( num__24 – num__20 ) / num__20 ] * num__100 = num__20.0 p . a . let the principal be rs . x time = t = num__2 years % rate = num__20.0 p . a . si = ( prt / num__100 ) num__20 = ( x * num__20 * num__2 ) / num__100 x = rs . num__50 answer : a <eor> a <eos> |
a |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| a man has rs . num__512 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__96 <o> b ) num__70 <o> c ) num__50 <o> d ) num__80 <o> e ) num__60 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__512 num__16 x = num__512 x = num__32 . hence total number of notes = num__3 x = num__96 . answer is a . <eor> a <eos> |
a |
divide__512.0__16.0__ multiply__32.0__3.0__ multiply__32.0__3.0__ |
divide__512.0__16.0__ multiply__32.0__3.0__ multiply__32.0__3.0__ |
| a man is num__25 years older than his son . in two years his age will be twice the age of his son . the present age of the son is ? <o> a ) num__11 <o> b ) num__23 <o> c ) num__27 <o> d ) num__22 <o> e ) num__91 |
let the son ' s present age be x years . then man ' s present age = ( x + num__25 ) years . ( x + num__25 ) + num__2 = num__2 ( x + num__2 ) x + num__27 = num__2 x + num__4 = > x = num__23 . answer : b <eor> b <eos> |
b |
add__25.0__2.0__ subtract__25.0__2.0__ subtract__25.0__2.0__ |
add__25.0__2.0__ subtract__25.0__2.0__ subtract__25.0__2.0__ |
| if it is num__5 : num__22 in the evening on a certain day what time in the morning was it exactly num__2 num__880695 minutes earlier ? ( assume standard time in one location . ) <o> a ) num__5 : num__30 <o> b ) num__5 : num__32 <o> c ) num__5 : num__47 <o> d ) num__5 : num__49 <o> e ) num__5 : num__54 |
num__5 : num__22 minus num__2 num__880695 must end with num__7 the only answer choice which ends with num__7 is c . answer : c . <eor> c <eos> |
c |
add__5.0__2.0__ round__5.0__ |
add__5.0__2.0__ round__5.0__ |
| the sum of all the integers k such that – num__23 < k < num__24 is <o> a ) num__0 <o> b ) - num__43 <o> c ) - num__25 <o> d ) - num__49 <o> e ) - num__51 |
- num__22 - - - - - - - - - - - - - - - - - - num__0 - - - - - - - - - - - - - - - - - num__23 values upto + num__23 cancels outwe are left with only - num__22 - num__21 sum of which is - num__43 . hence option d . b <eor> b <eos> |
b |
add__21.0__22.0__ add__21.0__22.0__ |
add__21.0__22.0__ add__21.0__22.0__ |
| if b does not equal zero and ab = b / num__2 what is the value of a ? <o> a ) a ) num__0.125 <o> b ) b ) num__0.25 <o> c ) c ) num__0.333333333333 <o> d ) d ) num__0.5 <o> e ) of the above |
explanation : to solve for a divide both sides of the equation by b : ab = b / num__2 ( ab ) / b = ( b / num__2 ) / b a = ( b / num__2 ) * num__1 / b a = num__0.5 answer : ( d ) . <eor> d <eos> |
d |
reverse__2.0__ reverse__2.0__ |
reverse__2.0__ reverse__2.0__ |
| find large number from below question the difference of two no . ' s is num__1365 . on dividing the larger no . by the smaller we get num__6 as quotient and the num__15 as remainder ? <o> a ) num__1456 <o> b ) num__1563 <o> c ) num__1546 <o> d ) num__1643 <o> e ) num__1635 |
let the smaller number be x . then larger number = ( x + num__1365 ) . x + num__1365 = num__6 x + num__15 num__5 x = num__1350 x = num__270 large number = num__270 + num__1365 = num__1635 e <eor> e <eos> |
e |
subtract__1365.0__15.0__ divide__1350.0__5.0__ add__1365.0__270.0__ add__1365.0__270.0__ |
subtract__1365.0__15.0__ divide__1350.0__5.0__ add__1365.0__270.0__ add__1365.0__270.0__ |
| excluding stoppages the speed of a bus is num__40 kmph and including stoppages it is num__30 kmph . for how many minutes does the bus stop per hour ? <o> a ) num__10 min <o> b ) num__12 min <o> c ) num__18 min <o> d ) num__15 min <o> e ) num__20 min |
due to stoppages it covers num__10 km less . time taken to cover num__10 km = ( num__0.25 x num__60 ) min = num__15 min answer : d <eor> d <eos> |
d |
subtract__40.0__30.0__ divide__10.0__40.0__ hour_to_min_conversion__ multiply__0.25__60.0__ round__15.0__ |
subtract__40.0__30.0__ divide__10.0__40.0__ hour_to_min_conversion__ multiply__0.25__60.0__ round__15.0__ |
| if the difference between compound interest ( interest compounded yearly ) and simple interest on a sum for num__2 years at num__10.0 p . a . is rs . num__150 then sum is <o> a ) rs . num__12000 <o> b ) rs . num__15000 <o> c ) rs . num__13000 <o> d ) rs . num__10000 <o> e ) rs . num__14 |
000 |
compund interest = p [ num__1 + r / num__100 ] ^ t - p ci = p [ num__0.21 ] simple interest = ptr / num__100 si = p [ num__0.2 ] difference p [ num__0.21 ] - p [ num__0.2 ] = num__150 p = num__15000 answer : b <eor> b <eos> |
b |
b |
| jessica can text num__85 words per minute while maurice can text num__25 words per minute . an abridged english translation of tolstoy ’ s war and peace contains num__410400 words . working together how many days will it take jessica and maurice to text the entire abridged translation of war and peace ? <o> a ) num__0.3 <o> b ) num__1.3 <o> c ) num__3 <o> d ) num__2.6 <o> e ) num__11.3 |
since the answer choices are reasonably spaced out we can approximate and save time combined words per day = num__110 * num__60 * num__24 num__2.59090909091 will be closest to only num__2.6 so answer should be d . <eor> d <eos> |
d |
add__85.0__25.0__ hour_to_min_conversion__ round__2.6__ |
add__85.0__25.0__ hour_to_min_conversion__ round__2.6__ |
| if the sides of a cube are in the ratio num__5 : num__3 . what is the ratio of their diagonals ? <o> a ) num__5 : num__9 <o> b ) num__9 : num__3 <o> c ) num__5 : num__4 <o> d ) num__3 : num__5 <o> e ) num__5 : num__3 |
explanation : diagonal of a cube = a √ num__3 where a is side a num__1 : a num__2 = num__5 : num__3 d num__1 : d num__2 = num__5 : num__3 where √ num__3 cancelled both side answer : e <eor> e <eos> |
e |
multiply__5.0__1.0__ |
multiply__5.0__1.0__ |
| if f ( num__1 num__3 ) = num__10 and f ( num__1 num__4 ) = num__17 what is the value of f ( num__1 num__5 ) ? <o> a ) num__107 <o> b ) num__127 <o> c ) num__147 <o> d ) none of these <o> e ) num__26 |
solution : the function f ( a b ) = a num__3 + b num__2 f ( num__1 num__3 ) therefore is = num__10 and f ( num__1 num__4 ) = num__13 + num__62 = num__17 . therefore f ( num__1 num__5 ) = num__1 + num__25 = num__26 answer e <eor> e <eos> |
e |
subtract__3.0__1.0__ add__3.0__10.0__ add__1.0__25.0__ add__1.0__25.0__ |
subtract__3.0__1.0__ add__3.0__10.0__ add__1.0__25.0__ add__1.0__25.0__ |
| a and b can do a piece of work in num__18 days ; band c can do it in num__24 days a and c can do it in num__36 days . in how many days will a band c finish it together ? <o> a ) num__14 <o> b ) num__16 <o> c ) num__19 <o> d ) num__18 <o> e ) num__20 |
( a + b ) ' s num__1 day ' s work = ( num__0.0555555555556 ) ( b + c ) ' s num__1 day ' s work = ( num__0.0416666666667 ) and ( a + c ) ' s num__1 day ' s work = ( num__0.0277777777778 ) adding we get : num__2 ( a + b + c ) ' s num__1 day ' s work = ¬ ( num__0.0555555555556 + num__0.0416666666667 + num__0.0277777777778 ) = num__0.125 = num__0.125 ( a + b + c ) ' s num__1 day ' s work = num__0.0625 thus a band c together can finish the work in num__16 days . answer b num__16 <eor> b <eos> |
b |
divide__1.0__18.0__ divide__1.0__24.0__ divide__1.0__36.0__ divide__36.0__18.0__ divide__0.125__2.0__ subtract__18.0__2.0__ round__16.0__ |
divide__1.0__18.0__ divide__1.0__24.0__ divide__1.0__36.0__ divide__36.0__18.0__ divide__0.125__2.0__ subtract__18.0__2.0__ round__16.0__ |
| as a treat for her two crying children a mother runs to the freezer in which she has two cherry ice pops four orange ice pops and four lemon - lime ice pops . if she chooses two at random to bring outside to the children but realizes as she runs out the door that she can not bring them different flavors without one invariably being jealous of the other and getting even more upset what is the probability that she has to return to the freezer to make sure that they each receive the same flavor ? <o> a ) num__0.111111111111 <o> b ) num__0.166666666667 <o> c ) num__0.277777777778 <o> d ) num__0.0888888888889 <o> e ) num__0.833333333333 |
probability of not getting the same flavor - > favorable - > cherry - orange [ num__2 c num__1 * num__4 c num__1 [ or simply num__2 * num__4 ] or cherry - lemon [ num__2 * num__4 ] or orange - lemon [ num__4 * num__4 ] prob = ( num__2 * num__4 + num__2 * num__4 + num__4 * num__4 ) / num__9 c num__2 = num__0.0888888888889 = num__0.0888888888889 answer - > d <eor> d <eos> |
d |
multiply__1.0__0.0889__ |
multiply__1.0__0.0889__ |
| alex takes a loan of $ num__6000 to buy a used truck at the rate of num__8.0 simple interest . calculate the annual interest to be paid for the loan amount . <o> a ) num__680 <o> b ) num__700 <o> c ) num__720 <o> d ) num__480 <o> e ) num__420 |
from the details given in the problem principle = p = $ num__6000 and r = num__8.0 or num__0.08 expressed as a decimal . as the annual interest is to be calculated the time period t = num__1 . plugging these values in the simple interest formula i = p x t x r = num__6000 x num__1 x num__0.08 = num__480.00 annual interest to be paid = $ num__480 answer : d <eor> d <eos> |
d |
percent__8.0__6000.0__ percent__8.0__6000.0__ |
percent__8.0__6000.0__ percent__8.0__6000.0__ |
| if the diameter of circle r is num__60.0 of the diameter of circle s the area of circle r is what percent of the area of circle s ? <o> a ) num__50.0 <o> b ) num__46.0 <o> c ) num__36.0 <o> d ) num__26.0 <o> e ) num__38 % |
let diameter of circle r dr = num__60 and diameter of circle s ds = num__100 radius of circle r rr = num__30 radius of circle s rs = num__50 area of circle r / area of circle s = ( pi * rr ^ num__2 ) / ( pi * rs ^ num__2 ) = ( num__0.6 ) ^ num__2 = ( num__0.6 ) ^ num__2 = num__36.0 answer : c <eor> c <eos> |
c |
multiply__60.0__0.6__ multiply__60.0__0.6__ |
multiply__60.0__0.6__ multiply__60.0__0.6__ |
| find the number of square tiles to cover the floor of a room measuring num__6.5 m * num__8.5 m leaving num__0.25 m space around the room . a side of square tile is given to be num__25 cms ? <o> a ) num__768 <o> b ) num__476 <o> c ) num__429 <o> d ) num__428 <o> e ) num__413 |
floor area to be covered by tiles = num__6 * num__8 = num__48 tiles area = num__0.25 * num__0.25 = num__0.0625 no . of tiles = num__48 / num__0.0625 = num__768 answer : a <eor> a <eos> |
a |
multiply__6.0__8.0__ divide__48.0__0.0625__ round__768.0__ |
multiply__6.0__8.0__ divide__48.0__0.0625__ divide__48.0__0.0625__ |
| k = num__2 ^ n + num__6 where n is an integer greater than num__1 . if k is divisible by num__9 which of the following must be divisible by num__9 ? <o> a ) num__2 ^ n - num__8 <o> b ) num__2 ^ n - num__2 <o> c ) num__2 ^ n - num__3 <o> d ) num__2 ^ n + num__4 <o> e ) num__2 ^ n + num__5 |
given : k m and n are integers if k is a divisor of both n and m then k is a divisor of n + m ( and n – m and m – n ) we ' re told that num__9 is a divisor of num__2 ^ n + num__6 we also know that num__9 is a divisor of num__9 . so applying the abov erule num__9 is a divisor of num__2 ^ n + num__6 + num__9 and num__9 is a divisor of num__2 ^ n + num__6 - num__9 = num__2 ^ n - num__3 c <eor> c <eos> |
c |
add__2.0__1.0__ multiply__2.0__1.0__ |
add__2.0__1.0__ subtract__3.0__1.0__ |
| a train speeds past a pole in num__15 seconds and a platfrom num__100 m long in num__25 seconds . its length is : <o> a ) num__100 m <o> b ) num__125 m <o> c ) num__130 m <o> d ) num__150 m <o> e ) none |
sol . let the length of the train be x metres and its speed be y m / sec . they x / y = num__15 ⇒ y = x / num__15 ∴ x + num__4.0 = x / num__15 ⇔ x = num__150 m . answer d <eor> d <eos> |
d |
divide__100.0__25.0__ round__150.0__ |
divide__100.0__25.0__ round__150.0__ |
| raju rahim rajesh together can do a work in num__5 days . raju alone can do the work in num__10 days and rahim alone can do the same work in num__20 days . find in what time rajesh alone can do that work ? <o> a ) num__10 days <o> b ) num__15 days <o> c ) num__20 days <o> d ) num__30 days <o> e ) num__18 days |
the required answer is = num__5 * num__10 * num__2.0 * num__20 - num__5 ( num__10 + num__20 ) = num__20.0 = num__20 days answer is c <eor> c <eos> |
c |
divide__10.0__5.0__ round__20.0__ |
divide__10.0__5.0__ multiply__10.0__2.0__ |
| a certain list consists of num__21 different numbers . if n is in the list and n is num__4 times the average ( arithmetic mean ) of the other num__20 numbers in the list then n is what fraction e of the sum of the num__21 numbers in the list ? <o> a ) num__0.05 <o> b ) num__0.166666666667 <o> c ) num__0.2 <o> d ) num__0.190476190476 <o> e ) num__0.238095238095 |
this is how i used to calculate which i think works pretty well : if you let the average of the num__20 other numbers equal a can you write this equation for sum of the list ( s ) n + num__20 a = s the question tells us that n = num__4 a plug this back into the first equation and you get that the sum is num__24 a num__4 a + num__20 a = num__24 a therefore fraction e of n to the total would be num__4 a / num__24 a or num__0.166666666667 answer b <eor> b <eos> |
b |
add__4.0__20.0__ divide__4.0__24.0__ divide__4.0__24.0__ |
add__4.0__20.0__ divide__4.0__24.0__ divide__4.0__24.0__ |
| what is num__15 percent of num__68 ? <o> a ) num__7.6 <o> b ) num__8.7 <o> c ) num__9.1 <o> d ) num__10.2 <o> e ) num__11.6 |
( num__0.15 ) * num__68 = num__10.2 the answer is d . <eor> d <eos> |
d |
percent__15.0__68.0__ percent__15.0__68.0__ |
percent__15.0__68.0__ percent__15.0__68.0__ |
| what is num__92.0 of num__0.6 ? <o> a ) num__0.552 <o> b ) num__69.0 <o> c ) num__0.6845 <o> d ) num__0.6859 <o> e ) num__0.69 |
num__92.0 * ( num__0.6 ) = num__0.92 * num__0.6 = num__0.552 answer : a <eor> a <eos> |
a |
percent__92.0__0.6__ percent__92.0__0.6__ |
percent__92.0__0.6__ percent__92.0__0.6__ |
| the area of a rectangular plot is num__460 square metres . if the length is num__15.0 more than the breadth what is the breadth of the plot ? <o> a ) num__14 metres <o> b ) num__20 metres <o> c ) num__18 metres <o> d ) num__16 metres <o> e ) num__17 metres |
explanation : length = num__115.0 of breadth . length × breadth = num__460 m num__2 ⇒ num__115.0 of breadth × breadth = num__460 m num__2 ⇒ num__1.15 × breadth × breadth = num__460 m num__2 ⇒ breadth × breadth = num__400 m num__2 ⇒ breadth = num__20 m answer : option b <eor> b <eos> |
b |
triangle_area__2.0__20.0__ |
triangle_area__2.0__20.0__ |
| a man opens a shop . he buys water bottles for num__50 cents each fruit for num__25 cents each and snacks for a dollar each . he wants to start a bundle of selling a bottle a water num__3 snacks and num__2 fruits for d dollars . every num__5 th bundle sold will be sold for num__2 dollars and a complimentary snack at the owner ' s expense . what should he charge at least to not lose any money ? <o> a ) num__4.6 <o> b ) num__3.0 <o> c ) num__4.1 <o> d ) num__3.2 <o> e ) num__2.9 |
w = num__0.50 f = num__0.25 s = num__1.00 b = w + num__2 f + num__3 s = num__4.00 they pay he pays num__4 net loss num__0 fifth bundle = num__2 they pay he pays num__5 net loss num__3 dollars num__0.6 = num__60 cents . charging num__60 more cents on each bundle will be safe . answer : a <eor> a <eos> |
a |
reverse__2.0__ divide__0.5__2.0__ subtract__3.0__2.0__ reverse__0.25__ round_down__0.5__ divide__3.0__5.0__ add__4.0__0.6__ |
reverse__2.0__ divide__0.5__2.0__ subtract__3.0__2.0__ add__3.0__1.0__ round_down__0.5__ divide__3.0__5.0__ add__4.0__0.6__ |
| a bat is bought for rs . num__400 and sold at a gain of num__10.0 find its selling price ? <o> a ) s . num__460 / - <o> b ) s . num__480 / - <o> c ) s . num__500 / - <o> d ) s . num__520 / - <o> e ) s . num__440 / - |
num__100.0 - - - - - - > num__400 ( num__100 * num__4 = num__400 ) num__110.0 - - - - - - > num__440 ( num__120 * num__4 = num__440 ) selling price = rs . num__440 / - option ' e ' <eor> e <eos> |
e |
percent__100.0__440.0__ |
percent__100.0__440.0__ |
| two water pumps working simultaneously at their respective constant rates took exactly num__5 hours to fill a swimming pool . if the constant rate of one pump was num__1.5 times the constant rate of the other how many hours would it have taken the faster pump to fill the pool if it had worked alone at its constant rate ? <o> a ) num__1.66666666667 <o> b ) num__5.33333333333 <o> c ) num__5.66666666667 <o> d ) num__7.33333333333 <o> e ) num__8.33333333333 |
let the rate of the slow pump be x . then the rate of the fast pump is num__1.5 x . ( x + num__1.5 x ) * num__5 = num__2.5 x * num__5 = num__12.5 x which is the capacity of the whole pool . working alone : num__1.5 x * time = num__12.5 x time = num__12.5 / num__1.5 = num__8.33333333333 = num__8.33333333333 hours the answer is e . <eor> e <eos> |
e |
multiply__5.0__2.5__ divide__12.5__1.5__ divide__12.5__1.5__ |
multiply__5.0__2.5__ divide__12.5__1.5__ divide__12.5__1.5__ |
| the population of bacteria culture doubles every num__2 minutes . approximately how many minutes will it take for the population to grow from num__1000 to num__500000 bacteria ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__18 <o> e ) num__20 |
num__2000 ^ num__12000 ^ num__22000 ^ num__32000 ^ num__4 . . . . . . . num__2000 ^ num__9 population increases in this sequence taking num__9 * num__2 = num__18 answer : d <eor> d <eos> |
d |
multiply__2.0__1000.0__ multiply__2.0__9.0__ round__18.0__ |
multiply__2.0__1000.0__ multiply__2.0__9.0__ multiply__2.0__9.0__ |
| the sides of a square region measured to the nearest centimeter are num__7 centimeters long . the least possible value of the actual area of the square region is <o> a ) num__42.25 sq cm <o> b ) num__45.25 sq cm <o> c ) num__43.25 sq cm <o> d ) num__50.25 sq cm <o> e ) num__52.25 sq cm |
though there might be some technicalities concerning the termnearest ( as num__6.5 is equidistant from both num__6 and num__7 ) the answer still should be a : num__6.5 ^ num__2 = num__42.25 . answer : a <eor> a <eos> |
a |
power__6.5__2.0__ power__6.5__2.0__ |
power__6.5__2.0__ power__6.5__2.0__ |
| a department store perceived a run on a lower priced garment and raised the price by num__20.0 to keep stock available until another shipment arrived . customers fumed and the department store manager allowed them to purchase the garments at num__80.0 of the new price to save face . what was the difference in the new retail price and the price the customers were allowed to pay instead ? <o> a ) num__4.0 <o> b ) num__5.0 <o> c ) num__6.0 <o> d ) num__7.0 <o> e ) num__8.0 % |
quantity x rate = price num__1 x num__1 = num__1 num__0.8 x num__1.20 = num__0.960 decrease in price = ( num__0.040 / num__1 ) Ã — num__100 = num__4.0 answer = option a <eor> a <eos> |
a |
multiply__0.8__1.2__ subtract__1.0__0.96__ add__20.0__80.0__ divide__80.0__20.0__ divide__80.0__20.0__ |
multiply__0.8__1.2__ divide__0.8__20.0__ divide__80.0__0.8__ divide__80.0__20.0__ divide__80.0__20.0__ |
| the least number which should be added to num__4298 so that the sum is exactly divisible by num__5 num__6 num__4 and num__3 is ? <o> a ) num__12 <o> b ) num__22 <o> c ) num__32 <o> d ) num__42 <o> e ) num__52 |
l . c . m . of num__5 num__6 num__4 and num__3 = num__60 . on dividing num__4298 by num__60 the remainder is num__38 . number to be added = ( num__60 - num__38 ) = num__22 b ) <eor> b <eos> |
b |
subtract__60.0__38.0__ subtract__60.0__38.0__ |
subtract__60.0__38.0__ subtract__60.0__38.0__ |
| a cistern is normally filled in num__6 hours but takes two hours longer to fill because of a leak in its bottom . if the cistern is full the leak will empty it in ? <o> a ) num__33 <o> b ) num__88 <o> c ) num__24 <o> d ) num__99 <o> e ) num__11 |
num__0.166666666667 - num__1 / x = num__0.125 x = num__24 answer : c <eor> c <eos> |
c |
round__24.0__ |
divide__24.0__1.0__ |
| find the num__37.5 of num__976 ? <o> a ) num__368 <o> b ) num__655 <o> c ) num__366 <o> d ) num__437 <o> e ) num__545 |
num__37.5 of num__976 = num__37.5 / num__100 * num__976 = num__0.375 * num__976 = num__0.375 * num__976 = num__3 * num__122 = num__366 . answer : c <eor> c <eos> |
c |
percent__37.5__976.0__ percent__37.5__976.0__ |
percent__37.5__976.0__ percent__37.5__976.0__ |
| it takes a boat num__3 hours to travel down a river from point a to point b and num__5 hours to travel up the river from b to a . how long would it take the same boat to go from a to b in still water ? <o> a ) num__9 hours and num__45 minutes <o> b ) num__3 hours and num__45 minutes <o> c ) num__1 hours and num__45 minutes <o> d ) num__6 hours and num__45 minutes <o> e ) num__8 hours and num__45 minutes |
let : s be the speed of the boat in still water r be the rate of the water current and d the distance between a and b . d = num__3 ( s + r ) : boat traveling down river d = num__5 ( s - r ) : boat traveling up river num__3 ( s + r ) = num__5 ( s - r ) r = s / num__4 : solve above equation for r d = num__3 ( s + s / num__4 ) : substitute r by s / num__4 in equation b d / s = num__3.75 hours = let : s be the speed of the boat in still water r be the rate of the water current and d the distance between a and b . d = num__3 ( s + r ) : boat traveling down river d = num__5 ( s - r ) : boat traveling up river num__3 ( s + r ) = num__5 ( s - r ) r = s / num__4 : solve above equation for r d = num__3 ( s + s / num__4 ) : substitute r by s / num__4 in equation b d / s = num__3.75 hours = num__3 hours and num__45 minutes . correct answer b <eor> b <eos> |
b |
round__3.0__ |
round__3.0__ |
| two numbers n and num__16 have lcm = num__48 and gcf = num__8 . find n . <o> a ) num__35 <o> b ) num__56 <o> c ) num__76 <o> d ) num__87 <o> e ) num__24 |
the product of two integers is equal to the product of their lcm and gcf . hence . num__16 * n = num__48 * num__8 n = num__48 * num__0.5 = num__24 correct answer e <eor> e <eos> |
e |
divide__8.0__16.0__ add__16.0__8.0__ round__24.0__ |
divide__8.0__16.0__ multiply__48.0__0.5__ multiply__48.0__0.5__ |
| a goods train leaves a station at a certain time and at a fixed speed . after ^ hours an express train leaves the same station and moves in the same direction at a uniform speed of num__90 kmph . this train catches up the goods train in num__4 hours . find the speed of the goods train . <o> a ) num__9 kmph <o> b ) num__52 kmph <o> c ) num__13 kmph <o> d ) num__36 kmph <o> e ) num__45 kmph |
let the speed of the goods train be x kmph . distance covered by goods train in num__10 hours = distance covered by express train in num__4 hours num__10 x = num__4 x num__90 or x = num__36 . so speed of goods train = num__36 kmph . ans : d <eor> d <eos> |
d |
round__36.0__ |
round__36.0__ |
| what is the thousandths digit in the decimal equivalent of num__0.0114 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__3 <o> d ) num__5 <o> e ) num__6 |
num__0.0114 = num__57 / ( num__5 * num__10 ^ num__3 ) = ( num__11.4 ) * num__10 ^ - num__3 = num__11.4 * num__10 ^ - num__3 = . num__0114 thousandths digit = num__1 answer b <eor> b <eos> |
b |
divide__57.0__5.0__ multiply__10.0__11.4__ reverse__1.0__ |
divide__57.0__5.0__ multiply__10.0__11.4__ reverse__1.0__ |
| a clock is set right at num__8 a . m . the clock gains num__10 minutes in num__24 hours will be the true time when the clock indicates num__1 p . m . on the following day ? <o> a ) num__48 min . past num__12 <o> b ) num__48 min . past num__17 <o> c ) num__48 min . past num__13 <o> d ) num__48 min . past num__11 <o> e ) num__48 min . past num__16 |
time from num__8 a . m . on a day to num__1 p . m . on the following day = num__29 hours . num__24 hours num__10 min . of this clock = num__24 hours of the correct clock . hrs of this clock = num__24 hours of the correct clock . num__29 hours of this clock = hrs of the correct clock = num__28 hrs num__48 min of the correct clock . therefore the correct time is num__28 hrs num__48 min . after num__8 a . m . this is num__48 min . past num__12 . answer : a <eor> a <eos> |
a |
subtract__29.0__1.0__ round__48.0__ |
subtract__29.0__1.0__ round__48.0__ |
| the average marks of num__20 students in a class is num__100 . but a student mark is wrongly noted as num__50 instead of num__10 then find the correct average marks ? <o> a ) num__78 <o> b ) num__82 <o> c ) num__98 <o> d ) num__91 <o> e ) num__85 |
correct avg marks = num__100 + ( num__10 - num__50 ) / num__20 avg = num__100 - num__2 = num__98 answer is c <eor> c <eos> |
c |
divide__20.0__10.0__ subtract__100.0__2.0__ subtract__100.0__2.0__ |
divide__20.0__10.0__ subtract__100.0__2.0__ subtract__100.0__2.0__ |
| a num__6.0 stock yields num__8.0 . the market value of the stock is : <o> a ) rs . num__48 <o> b ) rs . num__75 <o> c ) rs . num__96 <o> d ) rs . num__133.33 <o> e ) rs . num__143.33 |
let the face value of the stock is rs . num__100 it yields rs . num__6 market value of the stock = ( num__0.75 ) * num__100 = rs . num__75 answer : b <eor> b <eos> |
b |
percent__75.0__100.0__ |
percent__75.0__100.0__ |
| what percent is num__70 of num__280 ? <o> a ) num__25.0 <o> b ) num__30.0 <o> c ) num__35.0 <o> d ) num__40.0 <o> e ) num__42 % |
num__0.25 = num__0.25 num__0.25 × num__100 = num__25.0 a ) <eor> a <eos> |
a |
percent__25.0__100.0__ |
percent__25.0__100.0__ |
| the angle between the minute hand and the hour hand of a clock when the time is num__9.30 is <o> a ) num__35 ° <o> b ) num__65 ° <o> c ) num__45 ° <o> d ) num__105 ° <o> e ) num__95 ° |
angle between hands of a clock when the minute hand is behind the hour hand the angle between the two hands at m minutes past h ' o clock = num__30 ( h − m / num__5 ) + m / num__2 degree when the minute hand is ahead of the hour hand the angle between the two hands at m minutes past h ' o clock = num__30 ( m / num__5 − h ) − m / num__2 degree here h = num__9 m = num__30 and minute hand is behind the hour hand . hence the angle = num__30 ( h − m / num__5 ) + m / num__2 = num__30 ( num__9 − num__6.0 ) + num__15.0 = num__30 ( num__9 − num__6 ) + num__15 = num__30 × num__3 + num__15 = num__105 ° answer is d . <eor> d <eos> |
d |
divide__30.0__5.0__ add__6.0__9.0__ subtract__5.0__2.0__ round__105.0__ |
divide__30.0__5.0__ add__6.0__9.0__ divide__6.0__2.0__ round__105.0__ |
| an art gallery owner is hanging paintings for a new show . of the three paintings she has to choose from she can only hang two on the main wall of the gallery . assuming that she hangs as many as possible on that wall in how many ways can she arrange the paintings ? <o> a ) num__6 <o> b ) num__30 <o> c ) num__64 <o> d ) num__120 <o> e ) num__216 |
arrangment means permutations questions . . num__3 p num__2 = num__3 ! / num__2 ! = num__6 . imo option a is correct answer . . <eor> a <eos> |
a |
coin_space__ die_space__ die_space__ |
coin_space__ die_space__ die_space__ |
| a man swims downstream num__30 km and upstream num__18 km taking num__3 hours each time what is the speed of the man in still water ? <o> a ) num__1 <o> b ) num__8 <o> c ) num__9 <o> d ) num__7 <o> e ) num__5 |
num__30 - - - num__3 ds = num__10 ? - - - - num__1 num__18 - - - - num__3 us = num__6 ? - - - - num__1 m = ? m = ( num__10 + num__6 ) / num__2 = num__8 answer : b <eor> b <eos> |
b |
divide__30.0__3.0__ divide__18.0__3.0__ subtract__3.0__1.0__ subtract__18.0__10.0__ round__8.0__ |
divide__30.0__3.0__ divide__18.0__3.0__ subtract__3.0__1.0__ subtract__18.0__10.0__ subtract__18.0__10.0__ |
| water consists of hydrogen and oxygen and the approximate ratio by mass of hydrogen to oxygen is num__2 : num__16 . approximately how many grams of oxygen are there in num__126 grams of water ? <o> a ) num__16 <o> b ) num__72 <o> c ) num__112 <o> d ) num__128 <o> e ) num__142 |
solution : we are given that the ratio of hydrogen to oxygen in water by mass is num__2 : num__16 . using our ratio multiplier we can re - write this as num__2 x : num__16 x . we can now use these expressions to determine how much oxygen is in num__126 grams of water . num__2 x + num__16 x = num__126 num__18 x = num__126 x = num__7 since x is num__7 we know that there are num__16 x num__7 = num__112 grams of oxygen in num__126 grams of water . answer c . <eor> c <eos> |
c |
add__2.0__16.0__ divide__126.0__18.0__ multiply__16.0__7.0__ multiply__16.0__7.0__ |
add__2.0__16.0__ divide__126.0__18.0__ multiply__16.0__7.0__ multiply__16.0__7.0__ |
| if p and q are positive integers related by the equation p = num__10 q - num__5 what is the remainder when p is divided by num__5 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
this question asks what is . . . ( the answer ) so we know that the answer will be definite . as such we can test values to quickly get the solution . we ' re told that p and q are positive integers and p = num__10 q - num__5 . we ' re asked for the remainder when p is divided by num__5 . if . . . . q = num__1 p = num__5 num__1.0 = num__1 remainder num__0 final answer : a <eor> a <eos> |
a |
multiply__10.0__0.0__ |
multiply__10.0__0.0__ |
| if the selling price of num__60 articles is equal to the cost price of num__30 articles then the loss or gain percent is : <o> a ) num__55.0 <o> b ) num__35.0 <o> c ) num__66.0 <o> d ) num__45.0 <o> e ) num__50 % |
let c . p . of each article be re . num__1 . then c . p . of num__60 articles = rs . num__60 ; s . p . of num__60 articles = rs . num__30 . loss % = num__0.5 * num__100 = num__50.0 answer : e <eor> e <eos> |
e |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| a b and c can do a work in num__9 num__30 and num__45 days respectively . if they work together in how many days will they complete the work ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__11 |
one days ' s work of a b and c = num__0.111111111111 + num__0.0333333333333 + num__0.0222222222222 = ( num__10 + num__3 + num__2 ) / num__90 = num__0.166666666667 a b and c together can do the work in num__6 days . answer : b <eor> b <eos> |
b |
divide__30.0__10.0__ multiply__9.0__10.0__ subtract__9.0__3.0__ round__6.0__ |
divide__30.0__10.0__ multiply__9.0__10.0__ subtract__9.0__3.0__ round__6.0__ |
| the principal that amounts to rs . num__4913 in num__3 years at num__6 num__0.25 % per annum c . i . compounded annually is ? <o> a ) rs . num__3096 <o> b ) rs . num__4076 <o> c ) rs . num__4085 <o> d ) rs . num__4096 <o> e ) rs . num__5096 |
principal = [ num__4913 / ( num__1 + num__25 / ( num__4 * num__100 ) ) num__3 ] = num__4913 * num__0.941176470588 * num__0.941176470588 * num__0.941176470588 = rs . num__4096 . answer : d <eor> d <eos> |
d |
percent__100.0__4096.0__ |
percent__100.0__4096.0__ |
| due to reduction of num__25.0 in price of oranges a customer can purchase num__4 oranges more for rs . num__16 . what is original price of an orange ? <o> a ) num__1.39 <o> b ) num__1.33 <o> c ) num__1.31 <o> d ) num__1.32 <o> e ) num__1.38 |
explanation : recall it is based on inverse proportion or product constancy concept . reducion in price increase in amount num__25.0 \ inline \ left ( \ frac { num__1 } { num__4 } \ right ) \ inline \ left ( \ frac { num__1 } { num__3 } \ right ) num__33.33 = num__4 oranges it means original number of oranges = num__4 x num__3 = num__12 \ therefore original price of oranges = \ inline \ frac { num__16 } { num__12 } = rs . num__1.33 answer : b ) rs num__1.33 <eor> b <eos> |
b |
subtract__4.0__1.0__ multiply__4.0__3.0__ multiply__1.0__1.33__ |
subtract__4.0__1.0__ multiply__4.0__3.0__ multiply__1.0__1.33__ |
| the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is num__42 kmph find the speed of the stream ? <o> a ) num__15 kmph <o> b ) num__17 kmph <o> c ) num__14 kmph <o> d ) num__18 kmph <o> e ) num__19 kmph |
the ratio of the times taken is num__2 : num__1 . the ratio of the speed of the boat in still water to the speed of the stream = ( num__2 + num__1 ) / ( num__2 - num__1 ) = num__3.0 = num__3 : num__1 speed of the stream = num__14.0 = num__14 kmph . answer : c <eor> c <eos> |
c |
add__1.0__2.0__ divide__42.0__3.0__ round__14.0__ |
add__1.0__2.0__ divide__42.0__3.0__ divide__42.0__3.0__ |
| if paint costs $ num__3.20 per quart and a quart covers num__60 square feet how much will it cost to paint the outside of a cube num__10 feet on each edge ? <o> a ) $ num__1.60 <o> b ) $ num__32.00 <o> c ) $ num__96.00 <o> d ) $ num__108.00 <o> e ) $ num__196.00 |
total surface area = num__6 a ^ num__2 = num__6 * num__10 * num__10 = num__600 each quart covers num__20 sqr ft thus total number of quarts = num__10.0 = num__10 cost will be num__10 * num__3.2 = $ num__32 ans : b <eor> b <eos> |
b |
surface_cube__10.0__ multiply__10.0__2.0__ multiply__3.2__10.0__ multiply__3.2__10.0__ |
multiply__60.0__10.0__ multiply__10.0__2.0__ multiply__3.2__10.0__ multiply__3.2__10.0__ |
| sandy gets num__3 marks for each correct sum and loses num__2 marks for each incorrect sum . sandy attempts num__30 sums and obtains num__55 marks . how many sums did sandy get correct ? <o> a ) num__19 <o> b ) num__21 <o> c ) num__23 <o> d ) num__25 <o> e ) num__27 |
let x be the correct sums and ( num__30 - x ) be the incorrect sums . num__3 x - num__2 ( num__30 - x ) = num__55 num__5 x = num__115 x = num__23 the answer is c . <eor> c <eos> |
c |
add__3.0__2.0__ divide__115.0__5.0__ divide__115.0__5.0__ |
add__3.0__2.0__ divide__115.0__5.0__ divide__115.0__5.0__ |
| a certain telescope increases the visual range at a particular location from num__90 kilometers to num__150 kilometers . by what percent is the visual range increased by using the telescope ? <o> a ) num__30.0 <o> b ) num__33 num__0.5 % <o> c ) num__40.0 <o> d ) num__60.0 <o> e ) num__66 num__0.666666666667 % |
original visual range = num__90 km new visual range = num__150 km percent increase in the visual range by using the telescope = ( num__150 - num__90 ) / num__90 * num__100.0 = num__0.666666666667 * num__100.0 = num__66.67 answer e <eor> e <eos> |
e |
divide__100.0__150.0__ multiply__0.6667__100.0__ round_down__66.67__ |
divide__100.0__150.0__ multiply__0.6667__100.0__ round_down__66.67__ |
| if p : q = num__3 : num__4 and q : r = num__5 : num__6 then p : q : r is : <o> a ) num__3 : num__4 : num__6 <o> b ) num__3 : num__5 : num__6 <o> c ) num__15 : num__20 : num__24 <o> d ) num__5 : num__6 : num__3 <o> e ) none |
p : q = num__3 : num__4 q : r = num__5 : num__6 p : q : r = num__3 × num__5 : num__5 × num__4 : num__6 × num__4 ⇒ num__15 : num__20 : num__24 answer : c . <eor> c <eos> |
c |
multiply__3.0__5.0__ multiply__4.0__5.0__ multiply__4.0__6.0__ multiply__3.0__5.0__ |
multiply__3.0__5.0__ multiply__4.0__5.0__ multiply__4.0__6.0__ multiply__3.0__5.0__ |
| at the end of the first quarter the share price of a certain mutual fund was num__30 percent higher than it was at the beginning of the year . at the end of the second quarter the share price was num__75 percent higher than it was at the beginning of the year . what was the percent increase in the share price from the end of the first quarter to the end of the second quarter ? <o> a ) num__20.0 <o> b ) num__25.0 <o> c ) num__30.0 <o> d ) num__34.6 <o> e ) num__40 % |
another method is to use the formula for num__2 successive percentage changes : total = a + b + ab / num__100 num__75 = num__30 + b + num__30 b / num__100 b = num__34.6 answer ( d ) <eor> d <eos> |
d |
percent__34.6__100.0__ |
percent__34.6__100.0__ |
| a tank is num__25 m long num__12 m wide and num__6 m deep . the cost of plastering its walls and bottom at num__45 paise per sq m is <o> a ) rs . num__234.80 <o> b ) rs . num__334.80 <o> c ) rs . num__434.80 <o> d ) rs . num__534.80 <o> e ) none of these |
explanation : area to be plastered = [ num__2 ( l + b ) à — h ] + ( l à — b ) = [ num__2 ( num__25 + num__12 ) à — num__6 ] + ( num__25 à — num__12 ) = num__744 sq m cost of plastering = num__744 à — ( num__0.45 ) = rs . num__334.80 answer : b <eor> b <eos> |
b |
divide__12.0__6.0__ multiply__744.0__0.45__ round__334.8__ |
divide__12.0__6.0__ multiply__744.0__0.45__ round__334.8__ |
| the total of company c ' s assets in num__1994 was num__400.0 greater than the total in num__1993 which in turn was num__400.0 greater than the total in num__1992 . if the total of company c ' s assets in in num__1992 was n dollars which one of the following represents company c ' s assets in num__1994 : <o> a ) num__7 n <o> b ) num__8 n <o> c ) num__9 n <o> d ) num__12 n <o> e ) num__25 n |
let ' s use actual numbers . starting in num__1992 let ' s say company c had $ num__100 in assets . in num__1993 the total assets were num__400.0 greater which means num__1992 plus num__400.0 of num__1992 : $ num__100 + num__4 x $ num__100 = $ num__100 + $ num__400 = $ num__500 in num__1994 the total assets were num__400.0 greater than they were in num__1993 which means num__1993 plus num__400.0 of num__1993 : $ num__500 + num__4 x $ num__500 = $ num__500 + $ num__2000 = $ num__2500 this is num__25 times the num__1992 number so the correct answer is num__25 n . e <eor> e <eos> |
e |
divide__400.0__100.0__ add__400.0__100.0__ multiply__500.0__4.0__ add__2000.0__500.0__ divide__100.0__4.0__ divide__100.0__4.0__ |
divide__400.0__100.0__ add__400.0__100.0__ multiply__500.0__4.0__ add__2000.0__500.0__ divide__100.0__4.0__ divide__100.0__4.0__ |
| num__30 square stone slabs of equal size were needed to cover a floor area of num__120 sq . m . find the length of each stone slab ? <o> a ) num__120 cm <o> b ) num__200 cm <o> c ) num__88 cm <o> d ) num__666 cm <o> e ) num__776 cm |
area of each slab = num__4.0 m num__2 = num__4 m num__2 length of each slab √ num__4 = num__2 m = num__200 cm b <eor> b <eos> |
b |
triangle_area__200.0__2.0__ |
triangle_area__200.0__2.0__ |
| for what value of x between − num__5 and num__5 inclusive is the value of x ^ num__2 − num__10 x + num__16 the greatest ? <o> a ) − num__5 <o> b ) − num__2 <o> c ) num__0 <o> d ) num__2 <o> e ) num__4 |
we can see from the statement that two terms containing x x ^ num__2 will always be positive and - num__10 x will be positive if x is - ive . . so the equation will have greatest value if x is - ive and lower the value of x greater is the equation . so - num__5 will give the greatest value . . ans a <eor> a <eos> |
a |
subtract__10.0__5.0__ |
subtract__10.0__5.0__ |
| num__1397 x num__1397 <o> a ) num__1951609 <o> b ) num__1981709 <o> c ) num__18362619 <o> d ) num__2031719 <o> e ) none of these |
num__1397 x num__1397 = ( num__1397 ) num__2 = ( num__1400 - num__3 ) num__2 = ( num__1400 ) num__2 + ( num__3 ) num__2 - ( num__2 x num__1400 x num__3 ) = num__1960000 + num__9 - num__8400 = num__1960009 - num__8400 = num__1951609 . answer : option a <eor> a <eos> |
a |
subtract__1400.0__1397.0__ add__1960000.0__9.0__ subtract__1960009.0__8400.0__ subtract__1960009.0__8400.0__ |
subtract__1400.0__1397.0__ add__1960000.0__9.0__ subtract__1960009.0__8400.0__ subtract__1960009.0__8400.0__ |
| how many num__9 ' s are there preceded by num__5 but not followed by num__1 ? num__5 num__9 num__3 num__2 num__1 num__7 num__4 num__2 num__5 num__9 num__7 num__4 num__6 num__1 num__3 num__2 num__8 num__7 num__4 num__1 num__3 num__8 num__3 num__5 num__9 num__1 num__7 num__4 num__3 num__9 num__5 num__8 num__2 num__0 num__1 num__8 num__7 num__4 num__6 num__3 <o> a ) num__4 <o> b ) num__2 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
num__5 num__9 num__3 num__5 num__9 num__7 only at these places num__2 is preceded by num__5 but not followed by num__1 answer : b <eor> b <eos> |
b |
subtract__9.0__7.0__ |
subtract__9.0__7.0__ |
| every year taylor goes to the same carnival and he attempts to shoot a basketball into a small hoop hoping that he will win a ten - foot tall teddy bear . if the probability that taylor does not make a basket is num__0.333333333333 what is the probability that taylor makes exactly one basket in three tries ? <o> a ) num__0.111111111111 <o> b ) num__0.222222222222 <o> c ) num__0.4 <o> d ) num__0.6 <o> e ) num__0.428571428571 |
probability of basket = num__0.666666666667 probability of no basket = num__0.333333333333 required probability = num__3 * num__0.666666666667 * num__0.333333333333 * num__0.333333333333 = num__0.222222222222 b is the answer . <eor> b <eos> |
b |
multiply__0.3333__0.6667__ multiply__0.3333__0.6667__ |
multiply__0.3333__0.6667__ multiply__0.3333__0.6667__ |
| num__100 people are standing in a circle . the person standing at number num__1 is having a sword . he kills the person next to him with the sword and then gives the sword to the third person . this process is carried out till there is just one person left . can you find out who survives ? <o> a ) num__93 <o> b ) num__73 <o> c ) num__90 <o> d ) num__75 <o> e ) num__87 |
b the person on num__73 rd number will survive . till the number is the power of num__2 the last person to survive will be the one who started it . but since the number here is not the power of num__2 we will take the greatest power of num__2 that is less than the number which is num__64 . num__100 - num__64 = num__36 num__36 people are killed as num__2 num__4 num__6 . . . num__72 . thus the sword will now be given to the num__73 rd person . now he is the first person to start in the remaining num__64 people . thus he will be the one to survive . <eor> b <eos> |
b |
square_perimeter__1.0__ surface_cube__1.0__ multiply__2.0__36.0__ multiply__1.0__73.0__ |
square_perimeter__1.0__ surface_cube__1.0__ multiply__2.0__36.0__ multiply__1.0__73.0__ |
| anmol can eat num__27 pastries in a hour . aakriti can eat num__2 pastries in num__10 minutes . divya can eat num__7 pastries in num__20 minutes . how long will it take them to eat a num__420 pastries ? <o> a ) num__2 hours . <o> b ) num__1 hours . <o> c ) num__5 hours . <o> d ) num__4 hours . <o> e ) num__7 hours . |
in one hour anmol eats num__27 pastries aakriti eats num__12 and divya eats num__21 . a total of num__60 pastries . therefore num__420 pastries would take num__420 ÷ num__60 = num__7 hours . answer e <eor> e <eos> |
e |
add__2.0__10.0__ divide__420.0__20.0__ hour_to_min_conversion__ round__7.0__ |
add__2.0__10.0__ divide__420.0__20.0__ hour_to_min_conversion__ round__7.0__ |
| total rs . num__700 are divided among num__3 persons ( a b and c ) . a gets num__0.5 of b and b gets num__0.5 of c . how much c have ? <o> a ) num__300 <o> b ) num__400 <o> c ) num__500 <o> d ) num__600 <o> e ) num__700 |
a gets num__100 rs . b gets num__200 rs . c gets num__400 rs . a = b / num__2 b = c / num__2 a + b + c = num__700 ( b / num__2 ) + b + ( num__2 b ) = num__700 b = num__200 a = b / num__2 = num__100.0 = num__100 c = num__2 b = num__2 * num__200 = num__400 answer : b <eor> b <eos> |
b |
divide__100.0__0.5__ divide__200.0__0.5__ reverse__0.5__ divide__200.0__0.5__ |
divide__100.0__0.5__ divide__200.0__0.5__ reverse__0.5__ divide__200.0__0.5__ |
| how much water must be added to num__60 litres of milk at num__1 num__1 ⁄ num__2 litres for num__20 so as to have a mixture worth num__10 num__2 ⁄ num__3 a litre ? <o> a ) num__10 litres <o> b ) num__12 litres <o> c ) num__15 litres <o> d ) num__18 litres <o> e ) none of these |
c . p . of num__1 litre of milk = ( num__20 × num__2 ⁄ num__3 ) = num__40 ⁄ num__3 ∴ ratio of water and milk = num__8 ⁄ num__3 : num__32 ⁄ num__3 = num__8 : num__32 = num__1 : num__4 ∴ quantity of water to be added to num__60 litres of milk = ( num__1 ⁄ num__4 × num__60 ) litres = num__15 litres . answer c <eor> c <eos> |
c |
subtract__60.0__20.0__ subtract__10.0__2.0__ subtract__40.0__8.0__ add__1.0__3.0__ divide__60.0__4.0__ divide__60.0__4.0__ |
subtract__60.0__20.0__ subtract__10.0__2.0__ subtract__40.0__8.0__ add__1.0__3.0__ divide__60.0__4.0__ divide__60.0__4.0__ |
| what is the least value of k . so that num__123 k num__578 is divisible by num__11 . <o> a ) num__8 <o> b ) num__7 <o> c ) num__5 <o> d ) num__6 <o> e ) num__4 |
explanation : the difference of the sum of the digits at odd place and the sum of the digits at even place of the number is either num__0 or a number divisible by num__11 . ( num__1 + num__3 + num__5 + num__8 ) - ( num__2 + k + num__7 ) = num__17 - num__9 - k = num__8 - k therefore if k = num__8 then the value become zero . k = num__8 is the least value so that num__123 k num__578 is divisible by num__11 . answer : option a <eor> a <eos> |
a |
subtract__11.0__3.0__ subtract__3.0__1.0__ add__2.0__5.0__ subtract__11.0__2.0__ subtract__11.0__3.0__ |
subtract__11.0__3.0__ subtract__3.0__1.0__ add__2.0__5.0__ subtract__11.0__2.0__ subtract__11.0__3.0__ |
| what is the greatest positive integer x such that num__6 ^ x is a factor of num__216 ^ num__10 ? <o> a ) num__5 <o> b ) num__9 <o> c ) num__10 <o> d ) num__20 <o> e ) num__30 |
num__216 ^ num__10 = ( num__6 ^ num__3 ) ^ num__10 = num__6 ^ num__30 answer : e <eor> e <eos> |
e |
multiply__10.0__3.0__ multiply__10.0__3.0__ |
multiply__10.0__3.0__ multiply__10.0__3.0__ |
| brindavan express leave chennai central station every day at num__07.50 am and goes to bangalore city railway station . this train is very popular among the travelers . on num__25 th july num__2012 number of passengers traveling by i class and ii class was in the ratio num__1 : num__4 . the fare for this travel is in the ratio num__3 : num__1 . the total fare collected was rs . num__224000 / . ( rs . two lakhs twenty four thousand only ) . what was the fare collected from i class passengers on that day ? <o> a ) rs . num__32000 <o> b ) rs . num__96000 <o> c ) rs . num__128000 <o> d ) rs . num__5 num__00000 <o> e ) none of these |
solution : let the number of passenger traveling by first class be x . then number of passenger traveling by second class will be num__4 x . but the fare is in the ratio num__3 : num__1 in other words if num__3 y fare is collected per i class passenger y would be collected per ii class passenger . fares of i class passengers : fares of ii class passengers = x * num__3 y : num__4 x * y = num__3 : num__4 the above ratio can be interpreted as follows . if total fare is num__3 + num__4 = num__7 then i class passengers should pay rs . num__3 similarly we can calculate the fare of i class passengers when total was num__224000 total fare i class fare num__7 num__3 num__224000 ? = num__224000 * ( num__0.428571428571 ) = rs . num__96000 . answer : option b <eor> b <eos> |
b |
round_down__7.5__ divide__3.0__7.0__ multiply__1.0__96000.0__ |
add__4.0__3.0__ divide__3.0__7.0__ multiply__1.0__96000.0__ |
| there is a square with sides of num__13 . what is the area of the biggest circle that can be cut out of this square ? <o> a ) num__132.73 <o> b ) num__231.92 <o> c ) num__530.93 <o> d ) num__113.1 <o> e ) num__204.33 |
area of a circle = a = Ï € r ^ num__2 square is num__13 wide so circle ' s diameter would be num__13 and radius would be num__6.5 a = Ï € num__6.5 ^ num__2 which is approximately num__132.73 answer is a <eor> a <eos> |
a |
triangle_area__2.0__132.73__ |
triangle_area__2.0__132.73__ |
| a cycle is bought for rs . num__600 and sold for rs . num__1080 find the gain percent ? <o> a ) num__22 <o> b ) num__20 <o> c ) num__90 <o> d ) num__80 <o> e ) num__11 |
num__600 - - - - num__180 num__100 - - - - ? = > num__80.0 answer : d <eor> d <eos> |
d |
percent__100.0__80.0__ |
percent__100.0__80.0__ |
| ratio between rahul and deepak is num__4 : num__3 after num__10 years rahul age will be num__26 years . what is deepak present age . <o> a ) num__12 <o> b ) num__15 <o> c ) num__20 <o> d ) num__22 <o> e ) num__24 |
explanation : present age is num__4 x and num__3 x = > num__4 x + num__10 = num__26 = > x = num__16 so deepak age is = num__3 ( num__4 ) = num__12 option a <eor> a <eos> |
a |
subtract__26.0__10.0__ multiply__4.0__3.0__ multiply__4.0__3.0__ |
subtract__26.0__10.0__ multiply__4.0__3.0__ multiply__4.0__3.0__ |
| running at their respective constant rate machine x takes num__2 days longer to produce w widgets than machines y . at these rates if the two machines together produce num__5 w / num__4 widgets in num__3 days how many days would it take machine x alone to produce num__4 w widgets . <o> a ) num__4 <o> b ) num__6 <o> c ) num__8 <o> d ) num__10 <o> e ) num__24 |
i am getting num__12 . e . hope havent done any calculation errors . . approach . . let y = no . of days taken by y to do w widgets . then x will take y + num__2 days . num__1 / ( y + num__2 ) + num__1 / y = num__0.416666666667 ( num__0.416666666667 is because ( num__1.25 ) w widgets are done in num__3 days . so x widgets will be done in num__2.4 days or num__0.416666666667 th of a widget in a day ) solving we have y = num__4 = > x takes num__6 days to doing x widgets . so he will take num__24 days to doing num__4 w widgets . answer : e <eor> e <eos> |
e |
multiply__4.0__3.0__ subtract__5.0__4.0__ divide__5.0__12.0__ divide__5.0__4.0__ divide__3.0__1.25__ add__2.0__4.0__ multiply__2.0__12.0__ multiply__2.0__12.0__ |
multiply__4.0__3.0__ subtract__5.0__4.0__ divide__5.0__12.0__ divide__5.0__4.0__ divide__3.0__1.25__ add__2.0__4.0__ multiply__2.0__12.0__ divide__24.0__1.0__ |
| num__8 men can dig a pit in num__20 days . if a work half as much again as a boy then num__4 men and num__9 boys can dig a similiar pit . find the days for num__15 days can dig ? <o> a ) num__20 <o> b ) num__14 <o> c ) num__16 <o> d ) num__10 <o> e ) num__21 |
num__1 man = num__1.5 boys . ( num__4 men + num__9 boys ) = num__15 boys . num__8 men = ( num__1.5 ) * num__8 = num__12 boys . now num__12 boys can dig the bit in num__20 days . num__15 boys can dig = num__20 * num__0.8 = num__16 days . the option c is answer <eor> c <eos> |
c |
subtract__9.0__8.0__ add__8.0__4.0__ divide__12.0__15.0__ subtract__20.0__4.0__ round__16.0__ |
subtract__9.0__8.0__ add__8.0__4.0__ divide__12.0__15.0__ multiply__20.0__0.8__ multiply__20.0__0.8__ |
| the side of a square is increased by num__30.0 then how much % does its area increases ? <o> a ) num__30.0 <o> b ) num__36.25 <o> c ) num__69.0 <o> d ) num__52.25 <o> e ) num__56.55 % |
a = num__100 a num__2 = num__10000 a = num__130 a num__2 = num__16900 - - - - - - - - - - - - - - - - num__10000 - - - - - - - - - num__6900 num__100 - - - - - - - ? = > num__69.0 answer : c <eor> c <eos> |
c |
power__100.0__2.0__ power__130.0__2.0__ triangle_area__2.0__69.0__ |
power__100.0__2.0__ power__130.0__2.0__ triangle_area__2.0__69.0__ |
| in a room with num__7 people num__4 people have exactly num__1 friend in the room and num__3 people have exactly num__2 friends in the room ( assuming that friendship is a mutual relationship i . e . if jane is paul ' s friend paul is jane ' s friend ) . if two individuals are selected from the room at random what is the probability that those two individuals are not friends ? <o> a ) num__0.238095238095 <o> b ) num__0.428571428571 <o> c ) num__0.571428571429 <o> d ) num__0.714285714286 <o> e ) num__0.761904761905 |
( num__0.571428571429 ) ( num__0.833333333333 ) + ( num__0.428571428571 ) ( num__0.666666666667 ) if you choose one of the num__4 with one other friend then you have a num__0.833333333333 chance of not picking their friend num__2 nd . if you choose one of the num__3 with num__2 friends you have a num__0.666666666667 chance of not picking one of their friends second . add them up . num__0.47619047619 + num__0.285714285714 num__0.761904761905 = num__0.761904761905 e . num__0.761904761905 <eor> e <eos> |
e |
divide__4.0__7.0__ subtract__1.0__0.5714__ divide__2.0__3.0__ divide__2.0__7.0__ add__0.2857__0.4762__ multiply__1.0__0.7619__ |
divide__4.0__7.0__ subtract__1.0__0.5714__ divide__2.0__3.0__ divide__2.0__7.0__ add__0.2857__0.4762__ add__0.2857__0.4762__ |
| if f ( a ) = num__3 a + num__6 what is the value of ( f ( num__2 a ) + f ( a ) ) ? <o> a ) num__2 a - num__4 <o> b ) num__8 a - num__7 <o> c ) num__9 a + num__5 <o> d ) num__8 a + num__12 <o> e ) num__3 a + num__5 |
f ( a ) = num__3 a + num__6 f ( num__2 a ) = num__5 a + num__6 ( f ( num__2 a ) + f ( a ) ) = num__5 a + num__6 + num__3 a + num__6 = num__8 a + num__12 answer is d <eor> d <eos> |
d |
add__3.0__2.0__ add__3.0__5.0__ multiply__6.0__2.0__ add__3.0__5.0__ |
add__3.0__2.0__ add__3.0__5.0__ multiply__6.0__2.0__ add__3.0__5.0__ |
| sum of num__53 odd numbers is ? <o> a ) num__2709 <o> b ) num__2809 <o> c ) num__2909 <o> d ) num__3009 <o> e ) num__3109 |
sum of num__1 st n odd no . s = num__1 + num__3 + num__5 + num__7 + . . . = n ^ num__2 so sum of num__1 st num__53 odd numbers = num__53 ^ num__2 = num__2809 answer : b <eor> b <eos> |
b |
subtract__3.0__1.0__ multiply__1.0__2809.0__ |
subtract__3.0__1.0__ multiply__1.0__2809.0__ |
| given that num__2 x + num__11 > num__5 and num__5 x - num__13 < num__7 all values of x must be between which of the following pairs of integers ? <o> a ) - num__4 and - num__1 <o> b ) - num__1 and num__4 <o> c ) - num__4 and num__1 <o> d ) - num__3 and num__4 <o> e ) num__2 and num__5 |
num__2 x + num__11 > num__5 i . e num__2 x > num__5 - num__11 i . e . num__2 x > - num__6 i . e . x > - num__3 also num__5 x - num__13 < num__7 i . e . num__5 x < num__7 + num__13 i . e num__5 x < num__20 i . e x < num__4 i . e . - num__3 < x < num__4 answer : option d <eor> d <eos> |
d |
subtract__11.0__5.0__ subtract__5.0__2.0__ add__13.0__7.0__ subtract__11.0__7.0__ subtract__5.0__2.0__ |
subtract__11.0__5.0__ subtract__5.0__2.0__ add__13.0__7.0__ subtract__11.0__7.0__ subtract__5.0__2.0__ |
| on dividing num__136 by a number the quotient is num__9 and the remainder is num__1 . find the divisor ? <o> a ) a ) num__12 <o> b ) b ) num__15 <o> c ) c ) num__16 <o> d ) d ) num__17 <o> e ) e ) num__18 |
d = ( d - r ) / q = ( num__136 - num__1 ) / num__9 = num__15.0 = num__15 b ) <eor> b <eos> |
b |
multiply__1.0__15.0__ |
divide__15.0__1.0__ |
| the ratio of number of boys and girls in a school is num__3 : num__2 . if there are num__100 students in the school find the number of girls in the school ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__35 <o> d ) num__40 <o> e ) num__50 |
let the number of boys and girls be num__3 x and num__2 x total students = num__100 number of girls in the school = num__2 * num__20.0 = num__40 answer is d <eor> d <eos> |
d |
multiply__2.0__20.0__ multiply__2.0__20.0__ |
multiply__2.0__20.0__ multiply__2.0__20.0__ |
| a train speeds past a pole in num__15 seconds and a platform num__150 meters long in num__25 seconds . what is the length of the train ( in meters ) ? <o> a ) num__205 <o> b ) num__215 <o> c ) num__225 <o> d ) num__235 <o> e ) num__245 |
let the length of the train be x meters . the speed of the train is x / num__15 . then x + num__150 = num__25 * ( x / num__15 ) num__10 x = num__2250 x = num__225 meters the answer is c . <eor> c <eos> |
c |
divide__150.0__15.0__ multiply__15.0__150.0__ divide__2250.0__10.0__ round__225.0__ |
divide__150.0__15.0__ multiply__15.0__150.0__ divide__2250.0__10.0__ divide__2250.0__10.0__ |
| - num__64 x num__29 + num__165 = ? <o> a ) num__2436 <o> b ) num__2801 <o> c ) - num__2801 <o> d ) - num__1601 <o> e ) none of them |
given exp . = - num__64 x ( num__30 - num__1 ) + num__165 = - ( num__64 x num__30 ) + num__64 + num__165 = - num__1830 + num__229 = - num__1601 answer is d <eor> d <eos> |
d |
subtract__30.0__29.0__ add__64.0__165.0__ subtract__1830.0__229.0__ multiply__1.0__1601.0__ |
subtract__30.0__29.0__ add__64.0__165.0__ subtract__1830.0__229.0__ subtract__1830.0__229.0__ |
| three friends are buying a gift for a friend . declan contributes num__4 dollars more than num__0.25 the cost of the gift ed contributes num__6 dollar less than num__0.333333333333 the cost of the gift and frank contributes the remaining num__22 dollars . what is the cost of the gift ? <o> a ) num__48 <o> b ) num__54 <o> c ) num__60 <o> d ) num__66 <o> e ) num__72 |
declan = d ed = e frank = f t = total d + e + f = t ( t / num__4 + num__4 ) + ( t / num__3 - num__6 ) + num__22 = t t = num__20 + ( num__7 t / num__12 ) num__12 t = num__20 ( num__12 ) + num__7 t num__5 t = num__5 * num__4 ( num__12 ) t = num__48 the correct answer is a . <eor> a <eos> |
a |
add__4.0__3.0__ multiply__4.0__3.0__ multiply__0.25__20.0__ multiply__4.0__12.0__ multiply__4.0__12.0__ |
add__4.0__3.0__ multiply__4.0__3.0__ multiply__0.25__20.0__ multiply__4.0__12.0__ multiply__4.0__12.0__ |
| find the greatest number which leaves the same remainder when it divides num__25 num__57 and num__105 . <o> a ) num__23 <o> b ) num__28 <o> c ) num__18 <o> d ) num__16 <o> e ) num__12 |
explanation : num__105 - num__57 = num__48 num__57 - num__25 = num__32 num__105 - num__25 = num__80 the h . c . f of num__32 num__48 and num__80 is num__16 . answer : d <eor> d <eos> |
d |
subtract__105.0__57.0__ subtract__57.0__25.0__ subtract__105.0__25.0__ subtract__48.0__32.0__ subtract__32.0__16.0__ |
subtract__105.0__57.0__ subtract__57.0__25.0__ subtract__105.0__25.0__ subtract__48.0__32.0__ subtract__32.0__16.0__ |
| a small rectangular park has a perimeter of num__560 feet and a diagonal measurement of num__600 feet . what is its area in square feet ? <o> a ) num__172800 <o> b ) num__19600 <o> c ) num__20000 <o> d ) num__20400 <o> e ) num__20 |
800 |
you can avoid a lot of work in this problem by recognizing that with the info provided the diagonal forms a triangle inside the rectangle with sides that have a num__3 : num__4 : num__5 ratio . diagonal = num__200 num__2 x + num__2 y = num__560 or x + y = num__280 a ^ num__2 + b ^ num__2 = c ^ num__2 for each the sides of the triangle using the ratio num__3 : num__4 : num__5 for sides and knowing c = num__600 you can deduce the following a = num__360 b = num__480 num__360 x num__480 = num__172800 a is the answer . <eor> a <eos> |
a |
a |
| if the remainder is num__3 when positive integer n is divided by num__18 what is the remainder when n is divided by num__6 ? <o> a ) num__3 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
assume x is quotient here n = num__18 x + num__3 - - - - - - - - - - ( num__1 ) and n = num__6 x + ? we can also write first term as n = ( num__18 x + num__6 ) = num__6 ( num__3 x + num__1 ) + num__3 ie num__6 ( num__3 x + num__1 ) + num__3 ie the first term is perfectly divisible by num__6 . so the remainder left is num__3 . so answer ( a ) is right choice . <eor> a <eos> |
a |
multiply__3.0__1.0__ |
subtract__6.0__3.0__ |
| if s and t are positive integers such that s / t = num__56.18 which of the following could be the remainder when s is divided by t ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__8 <o> d ) num__20 <o> e ) num__90 |
the remainder is num__0.18 or num__0.18 . you can go one step further and say that x / y = num__0.18 where x and y are whole numbers . plug in all the answer choices for x and see which one makes y a whole number . one thing that jumps out at me is that a b c and d are all even . e on the other hand is odd . why ? maybe i ' ll start plugging in here first . num__90 / y = num__0.18 num__18 y = num__9000 y = num__500 answer : e . <eor> e <eos> |
e |
divide__9000.0__18.0__ multiply__0.18__500.0__ |
divide__9000.0__18.0__ multiply__0.18__500.0__ |
| if three numbers in the ratio num__3 : num__2 : num__5 be such that the sum of their squares is num__1862 the middle number will be <o> a ) num__10 <o> b ) num__14 <o> c ) num__18 <o> d ) num__22 <o> e ) num__24 |
explanation : let the numbers be num__3 x num__2 x and num__5 x . then num__9 x + num__4 x + num__25 x = num__1862 ⇒ num__38 x = num__1862 ⇒ x = num__49 ⇒ x = num__7 . middle number = num__2 x = num__14 option b <eor> b <eos> |
b |
subtract__9.0__5.0__ divide__1862.0__38.0__ add__3.0__4.0__ multiply__2.0__7.0__ multiply__2.0__7.0__ |
subtract__9.0__5.0__ divide__1862.0__38.0__ add__3.0__4.0__ add__5.0__9.0__ add__5.0__9.0__ |
| in measuring the sides of a rectangle one side is taken num__5.0 in excess and the other num__4.0 in deficit . find the error percent in the error percent in the area calculated from these measurements . <o> a ) num__0.8 <o> b ) num__0.2 <o> c ) num__0.4 <o> d ) num__0.9 <o> e ) num__0.6 % |
let x and y be the sides of the rectangle . then correct area = xy . calculated area = ( num__1.05 * x ) * ( num__0.96 * y ) = num__1.008 * xy . error in measurement = num__1.008 * xy ) - xy = num__0.008 * xy . error % = [ num__0.008 * xy * num__1 / xy * num__100 ] % = num__0.8 = num__0.8 . answer is a <eor> a <eos> |
a |
percent__100.0__0.8__ |
percent__100.0__0.8__ |
| there num__3 kinds of books in the library physics chemistry and biology . ratio of physics to chemistry is num__3 to num__2 ; ratio of chemistry to biology is num__4 to num__3 and the total of the books is more than num__3000 . which one of following can be the total e of the book ? <o> a ) num__3003 <o> b ) num__3027 <o> c ) num__3024 <o> d ) num__3021 <o> e ) num__3018 |
first you have to find the common ratio for all num__3 books . you have : p : c : b num__3 : num__2 - - > multiply by num__2 ( gives you row num__3 ) num__4 : num__6 num__6 : num__4 : num__3 hence : p : c : b : t ( total ) num__6 : num__4 : num__3 : num__13 - - - - > this means the total number e must be a multiple of num__13 . answer a is correct since num__299 is divisible by num__13 hence is num__2990 and so is num__3003 ( num__2990 + num__13 ) . <eor> a <eos> |
a |
multiply__3.0__2.0__ add__3.0__3000.0__ add__3.0__3000.0__ |
add__2.0__4.0__ add__3.0__3000.0__ add__3.0__3000.0__ |
| from the given equation find the value of x : x ² − num__3 x + num__2 <o> a ) num__1 <o> b ) num__3 <o> c ) num__2 <o> d ) num__5 <o> e ) num__6 |
( x − num__1 ) ( x − num__2 ) x = num__1 or num__2 . c <eor> c <eos> |
c |
subtract__3.0__2.0__ subtract__3.0__1.0__ |
subtract__3.0__2.0__ subtract__3.0__1.0__ |
| sam taken a loan rs . num__15000 / - from co - operative society with an interest @ num__11.5 per month . at the same time he deposited rs . num__10000 / - as fixed deposit with an interest @ num__9.5 per month . after one week sam asked the manager to calculate the interest to be paid . what is the interest amount for num__7 days ? <o> a ) a ) num__165 <o> b ) b ) num__220 <o> c ) c ) num__310 <o> d ) d ) num__185 <o> e ) e ) num__181 |
loan amount : rs . num__15000 / - @ num__11.5 interest per month = num__15000 / - * num__11.5 = rs . num__1725 interest for one day = num__57.5 = num__57.50 interest for num__7 days is = num__57.50 * num__7 = num__403 fd amount is = rs . num__10000 / - @ num__9.5 interest per month = num__10000 * num__9.5 = num__950 / - interest for num__7 days = num__31.6666666667 * num__7 = num__222 interest amount to be paid by sam = num__403 - num__222 = num__181 / - for num__7 days answer is e <eor> e <eos> |
e |
subtract__403.0__222.0__ subtract__403.0__222.0__ |
subtract__403.0__222.0__ subtract__403.0__222.0__ |
| calculate the time it will take for a train that is num__200 meter long to pass a bridge of num__180 meter length if the speed of the train is num__65 km / hour ? <o> a ) num__20 seconds <o> b ) num__4 seconds <o> c ) num__31.04 seconds <o> d ) num__11.04 seconds <o> e ) num__21.04 seconds |
speed = num__65 km / hr = num__65 * ( num__0.277777777778 ) m / sec = num__18.06 m / sec total distance = num__200 + num__180 = num__380 meter time = distance / speed = num__380 * ( num__0.0553709856035 ) = num__21.04 seconds answer : e <eor> e <eos> |
e |
add__200.0__180.0__ round__21.04__ |
add__200.0__180.0__ round__21.04__ |
| train e and f num__455 miles apart are traveling toward each other at constant rates and in the same time zone . if train e left at num__4 pm traveling at a speed of num__60 miles per hour and train f left at num__5 : num__45 pm and traveling at num__45 miles per hour then at what time would they pass each other ? <o> a ) num__9.03 pm <o> b ) num__9.05 pm <o> c ) num__9.07 pm <o> d ) num__9.08 pm <o> e ) num__9.10 pm |
first since e has a headstart then in that num__1 hr num__45 min or num__1.75 hrs he travels num__105 miles then remaining distance to be traveled will be num__455 - num__105 = num__350 miles now using relative rates ( num__105 ) ( t ) = num__350 this gives num__3.33333333333 hours now num__5.45 pm + num__3.33333333333 hours gives us num__9.05 pm hence answer is b <eor> b <eos> |
b |
subtract__5.0__4.0__ add__60.0__45.0__ subtract__455.0__105.0__ divide__350.0__105.0__ round__9.05__ |
subtract__5.0__4.0__ add__60.0__45.0__ subtract__455.0__105.0__ divide__350.0__105.0__ round__9.05__ |
| an athlete runs num__200 metres race in num__24 seconds . his speed is <o> a ) num__20 km / hr <o> b ) num__24 km / hr <o> c ) num__28.5 km / hr <o> d ) num__30 km / hr <o> e ) none |
sol . speed = num__8.33333333333 m / sec = num__8.33333333333 m / sec = ( num__8.33333333333 x num__3.6 ) km / hr = num__30 km / hr . answer d <eor> d <eos> |
d |
divide__200.0__24.0__ round__30.0__ |
divide__200.0__24.0__ round__30.0__ |
| if the ratio of the change in the book ' s price to its new price is num__1 : num__5 what is the % price increase ? <o> a ) num__25.0 <o> b ) num__50.0 <o> c ) num__75.0 <o> d ) num__90.0 <o> e ) num__100 % |
solution : - given ratio of change in book price / new price = num__0.2 = > ( new price − old price ) / new price = num__0.2 = > old price / new price = num__0.8 - - - - equation num__1.0 increase is always with respect in the initial or old price . = > % increase = ( ( new price - old price ) / old price ) * num__100 - - - equation num__2 substituting num__1 in num__2 we have % increase = ( ( new − ( num__0.8 ) new ) / ( ( num__0.8 ) new ) ) ∗ num__100 = num__25.0 answer : a <eor> a <eos> |
a |
reverse__5.0__ subtract__1.0__0.2__ divide__5.0__0.2__ multiply__1.0__25.0__ |
reverse__5.0__ subtract__1.0__0.2__ divide__5.0__0.2__ multiply__1.0__25.0__ |
| if num__4 x + y = num__34 num__2 x - y = num__20 for integers of x and y y ^ num__2 = ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__0 <o> d ) num__10 <o> e ) num__3 |
num__4 x + y = num__34 num__2 x - y = num__20 num__6 x = num__54 x = num__9 y = - num__2 y ^ num__2 = num__4 answer is b <eor> b <eos> |
b |
add__4.0__2.0__ add__34.0__20.0__ divide__54.0__6.0__ subtract__6.0__2.0__ |
add__4.0__2.0__ add__34.0__20.0__ divide__54.0__6.0__ subtract__6.0__2.0__ |
| the average of first num__10 odd numbers is ? <o> a ) num__11 <o> b ) average = num__10 <o> c ) num__18 <o> d ) num__12 <o> e ) num__19 |
explanation : sum of num__10 odd no . = num__100 average = num__10.0 = num__10 answer : b <eor> b <eos> |
b |
divide__100.0__10.0__ |
divide__100.0__10.0__ |
| what will be the cost of gardening num__1 meter boundary around a rectangular plot having perimeter of num__340 meters at the rate of rs . num__10 per square meter ? <o> a ) num__22887 <o> b ) num__3440 <o> c ) num__2777 <o> d ) num__2689 <o> e ) num__2771 |
in this question we are having perimeter . we know perimeter = num__2 ( l + b ) right so num__2 ( l + b ) = num__340 as we have to make num__1 meter boundary around this so area of boundary = ( ( l + num__2 ) + ( b + num__2 ) - lb ) = num__2 ( l + b ) + num__4 = num__340 + num__4 = num__344 so required cost will be = num__344 * num__10 = num__3440 answer : b <eor> b <eos> |
b |
square_perimeter__1.0__ multiply__10.0__344.0__ multiply__1.0__3440.0__ |
square_perimeter__1.0__ multiply__10.0__344.0__ multiply__1.0__3440.0__ |
| a man buys s cycle for $ num__1400 and sells it at a loss of num__15.0 . what is the selling price of the cycle ? <o> a ) $ num__950 <o> b ) $ num__1006 <o> c ) $ num__1190 <o> d ) $ num__1245 <o> e ) $ num__1521 |
s . p . = num__85.0 of the $ num__1400 = num__0.85 * num__1400 = $ num__1190 answer is c <eor> c <eos> |
c |
percent__85.0__1400.0__ percent__85.0__1400.0__ |
percent__85.0__1400.0__ percent__85.0__1400.0__ |
| let the number which when multiplied by num__15 is increased by num__280 . <o> a ) num__14 <o> b ) num__20 <o> c ) num__26 <o> d ) num__28 <o> e ) num__30 |
solution let the number be x . then num__15 x - x = num__280 ‹ = › num__14 x = num__280 x ‹ = › num__20 . answer b <eor> b <eos> |
b |
divide__280.0__14.0__ divide__280.0__14.0__ |
divide__280.0__14.0__ divide__280.0__14.0__ |
| a dishonest dealer professes to sell goods at the cost price but uses a weight of num__700 grams per kg what is his percent ? <o> a ) num__11 <o> b ) num__25 <o> c ) num__77 <o> d ) num__42 <o> e ) num__11 |
num__700 - - - num__300 num__100 - - - ? = > num__42.0 answer : d <eor> d <eos> |
d |
percent__42.0__100.0__ |
percent__42.0__100.0__ |
| the length of a rectangle is halved while its breadth is tripled . wa tis the % changein area ? <o> a ) num__34.0 <o> b ) num__45.0 <o> c ) num__50.0 <o> d ) num__60.0 <o> e ) num__67 % |
let original length = x and original breadth = y . original area = xy . new length = x . num__2 new breadth = num__3 y . new area = x x num__3 y = num__3 xy . num__2 num__2 increase % = num__1 xy x num__1 x num__100.0 = num__50.0 . num__2 xy c <eor> c <eos> |
c |
triangle_area__1.0__100.0__ triangle_area__1.0__100.0__ |
triangle_area__1.0__100.0__ triangle_area__1.0__100.0__ |
| the ratio of money with ram and gopal is num__7 : num__17 and that with gopal and krishan is num__7 : num__17 . if ram has rs . num__490 krishan has ? <o> a ) rs . num__2890 <o> b ) rs . num__2330 <o> c ) rs . num__1190 <o> d ) rs . num__1620 <o> e ) rs . num__2680 |
ram : gopal = num__7 : num__17 = num__49 : num__119 gopal : krishan = num__7 : num__17 = num__119 : num__289 ram : gopal : krishan = num__49 : num__119 : num__289 ram : krishan = num__49 : num__289 thus num__49 : num__289 = num__490 : n & there n = num__289 x num__10.0 = rs . num__2890 answer : a <eor> a <eos> |
a |
multiply__7.0__17.0__ subtract__17.0__7.0__ multiply__289.0__10.0__ multiply__289.0__10.0__ |
multiply__7.0__17.0__ subtract__17.0__7.0__ multiply__289.0__10.0__ multiply__289.0__10.0__ |
| in what time a sum of money double itself at num__5.0 per annum simple interest ? <o> a ) num__33 num__0.125 % <o> b ) num__33 num__0.333333333333 % <o> c ) num__33 num__2.33333333333 % <o> d ) num__32 num__0.333333333333 % <o> e ) num__20 % |
p = ( p * num__5 * r ) / num__100 r = num__20.0 answer : e <eor> e <eos> |
e |
percent__20.0__100.0__ |
percent__20.0__100.0__ |
| if ram and gohul can do a job in num__10 days and num__15 days independently how many days would they take to complete the same job working simultaneously ? <o> a ) num__10 <o> b ) num__5 <o> c ) num__8 <o> d ) num__6 <o> e ) num__13 |
if total work is x . ram working rate = x / num__10 per day . working rate of gohul = x / num__15 per day . rate of work = ( x / num__10 ) + ( x + num__15 ) = num__30 x / num__5 x = num__6 days the answer is option d <eor> d <eos> |
d |
subtract__15.0__10.0__ divide__30.0__5.0__ round__6.0__ |
subtract__15.0__10.0__ divide__30.0__5.0__ divide__30.0__5.0__ |
| num__12 num__36 ? num__324 num__972 <o> a ) num__108 <o> b ) num__125 <o> c ) num__215 <o> d ) num__312 <o> e ) num__368 |
series pattern every next element = num__3 x previous element ∴ missing term = num__3 x num__36 = num__108 answer : a <eor> a <eos> |
a |
divide__36.0__12.0__ multiply__36.0__3.0__ multiply__36.0__3.0__ |
divide__36.0__12.0__ multiply__36.0__3.0__ multiply__36.0__3.0__ |
| to apply for the position of photographer at a local magazine a photographer needs to include three or four photos in an envelope accompanying the application . if the photographer has pre - selected six photos representative of her work how many choices does she have to provide the photos for the magazine ? <o> a ) num__35 <o> b ) num__36 <o> c ) num__40 <o> d ) num__42 <o> e ) num__45 |
num__6 c num__3 + num__6 c num__4 = num__20 + num__15 = num__35 the answer is a . <eor> a <eos> |
a |
add__15.0__20.0__ add__15.0__20.0__ |
add__15.0__20.0__ add__15.0__20.0__ |
| a can run a kilometer race in num__4 num__0.5 min while b can run same race in num__5 min . how many meters start can a give b in a kilometer race so that the race mat end in a dead heat ? <o> a ) num__200 m <o> b ) num__700 m <o> c ) num__800 m <o> d ) num__100 metre <o> e ) num__1300 m |
explanation : a can give b ( num__5 min - num__4 num__0.5 min ) = num__30 sec start . the distance covered by b in num__5 min = num__1000 m . distance covered in num__30 sec = ( num__1000 * num__30 ) / num__300 = num__100 m . a can give b num__100 m start . answer : d <eor> d <eos> |
d |
round__100.0__ |
round__100.0__ |
| if each participant of a chess tournament plays exactly one game with each of the remaining participants then num__253 games will be played during the tournament . what is the number of participants ? <o> a ) num__21 <o> b ) num__22 <o> c ) num__23 <o> d ) num__24 <o> e ) num__25 |
let n be the number of participants . the number of games is nc num__2 = n * ( n - num__1 ) / num__2 = num__253 n * ( n - num__1 ) = num__506 = num__23 * num__22 ( trial and error ) the answer is c . <eor> c <eos> |
c |
multiply__253.0__2.0__ subtract__23.0__1.0__ add__1.0__22.0__ |
multiply__253.0__2.0__ subtract__23.0__1.0__ divide__23.0__1.0__ |
| num__1 + num__1 + num__1 + num__2 × num__1 ^ num__2 + num__2 × num__1 ^ num__1 + num__2 × num__1 ^ num__4 + num__2 × num__1 ^ num__5 + num__2 × num__1 ^ num__6 + num__2 × num__1 ^ num__7 = <o> a ) num__1 ^ num__7 <o> b ) num__1 ^ num__8 <o> c ) num__1 ^ num__14 <o> d ) num__1 ^ num__28 <o> e ) num__3 ^ num__30 |
we have the sum of num__9 terms . now if all terms were equal to the largest term num__2 * num__1 ^ num__7 we would have : sum = num__9 * ( num__2 * num__1 ^ num__7 ) = num__2 * num__1 ^ num__9 = ~ num__1 ^ num__10 so the actual sum is less than num__1 ^ num__10 and more than num__1 ^ num__7 ( option a ) as the last term is already more than that . so the answer is clearly b . answer : b . <eor> b <eos> |
b |
add__2.0__7.0__ add__1.0__9.0__ reverse__1.0__ |
add__2.0__7.0__ multiply__2.0__5.0__ reverse__1.0__ |
| on a test students receive num__6 points for each correct answer and are penalized by losing num__2 points for each incorrect answer . there are num__6 questions on the test and each question has num__4 answer options a b c and d . it is known that num__5 of the questions have option b as the correct answer and one question has option c as the correct answer . if a student marks b for the first num__3 questions and c for the last num__3 questions what is the minimum possible score that student can receive ? <o> a ) - num__4 <o> b ) - num__1 <o> c ) num__0 <o> d ) num__1 <o> e ) num__4 |
for the minimum possible score let us take the worst case scenario suppose he gets all of the last num__3 qs wrong whose correct answer options are b and one of the first num__3 qs wrong whose correct option is c in that case he will get only num__2 out of the first num__3 qs right . therefore minimum possible score = ( num__6 * num__2 ) - ( num__4 * num__2 ) = num__4 answer e <eor> e <eos> |
e |
choose__4.0__3.0__ |
choose__4.0__3.0__ |
| the length of the bridge which a train num__130 m long and traveling at num__45 km / hr can cross in num__30 sec is ? <o> a ) num__188 <o> b ) num__166 <o> c ) num__245 <o> d ) num__777 <o> e ) num__765 |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec . time = num__30 sec let the length of bridge be x meters . then ( num__130 + x ) / num__30 = num__12.5 x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| a man covered a certain distance at some speed . if he had moved num__3 kmph faster he would have taken num__40 minutes less . if he had moved num__2 kmph slower he would have taken num__40 minutes more . what is the distance in km ? <o> a ) num__36 <o> b ) num__38 <o> c ) num__40 <o> d ) num__42 <o> e ) num__45 |
explanation : speed = num__2 v num__1 v num__2 / v num__1 − v num__2 = num__2 × num__3 × num__23 − num__2 = num__12 km / h = hr distance = vt num__1 ( num__1 + v / v num__1 ) = num__12 × num__0.666666666667 ( num__1 + num__4.0 ) = num__40 km answer is c <eor> c <eos> |
c |
subtract__3.0__2.0__ divide__2.0__3.0__ add__3.0__1.0__ round__40.0__ |
subtract__3.0__2.0__ divide__2.0__3.0__ add__3.0__1.0__ divide__40.0__1.0__ |
| the purchase price of an article is $ num__48 . in order to include num__30.0 of cost for overhead and to provide $ num__12 of net profit the markup should be <o> a ) num__15.0 <o> b ) num__25.0 <o> c ) num__35.0 <o> d ) num__40.0 <o> e ) num__55 % |
cost price of article = num__48 $ % of overhead cost = num__30 net profit = num__12 $ we need to calculate % markup net profit as % of cost price = ( num__0.25 ) * num__100 = num__25.0 total markup should be = num__25 + num__30 = num__55.0 answer e <eor> e <eos> |
e |
percent__100.0__55.0__ |
percent__100.0__55.0__ |
| a particular store purchased a stock of turtleneck sweaters and marked up its cost by num__20.0 . during the new year season it further marked up its prices by num__25.0 of the original retail price . in february the store then offered a discount of num__10.0 . what was its profit on the items sold in february ? <o> a ) num__27.5 <o> b ) num__30.0 <o> c ) num__35.0 <o> d ) num__37.5 <o> e ) num__40 % |
assume the total price = num__100 x price after num__20.0 markup = num__120 x price after num__25.0 further markup = num__1.25 * num__120 x = num__150 x price after the discount = num__0.9 * num__150 x = num__135 x hence total profit = num__35.0 option c <eor> c <eos> |
c |
percent__35.0__100.0__ |
percent__35.0__100.0__ |
| a and b walk around a circular track . a and b walk at a speed of num__2 rounds per hour and num__3 rounds per hour respectively . if they start at num__8 a . m . from the same point in opposite directions how many times shall they cross each other before num__9.30 a . m . ? <o> a ) num__9.39 am <o> b ) num__9.80 am <o> c ) num__9.30 am <o> d ) num__9.32 am <o> e ) num__5.30 am |
relative speed = speed of a + speed of b ( ∴ they walk in opposite directions ) = num__2 + num__3 = num__5 rounds per hour = > they cross each other num__5 times in num__1 hour and num__2 times in num__0.5 hour time duration from num__8 am to num__9.30 am = num__1.5 hour hence they cross each other num__7 times before num__9.30 am answer : c <eor> c <eos> |
c |
add__2.0__3.0__ subtract__3.0__2.0__ divide__1.0__2.0__ subtract__2.0__0.5__ add__2.0__5.0__ round__9.3__ |
add__2.0__3.0__ subtract__3.0__2.0__ divide__1.0__2.0__ add__0.5__1.0__ add__2.0__5.0__ round__9.3__ |
| if q < num__0 which of the following must be true ? <o> a ) – num__100 < q < num__100 <o> b ) q ≤ – num__1 <o> c ) q ≤ num__1 <o> d ) q ^ num__2 ≥ num__0.25 <o> e ) – num__100 < q < num__0 |
q is a negative number . a . – num__100 < q < num__100 q can be a negative number less than - num__100 b . q ≤ – num__1 q can be a fraction in which case it would be greater than - num__1 c . q ≤ num__1 must be true since num__1 is positive and q is negative d . q ^ num__2 ≥ num__0.25 not true if m is a negative fraction like - num__0.2 e . – num__100 < q < num__0 q can be a negative number less than - num__100 = c <eor> c <eos> |
c |
reverse__1.0__ |
power__1.0__0.25__ |
| the greatest possible length which can be used to measure exactly the length num__10 m num__10 m num__80 cm num__10 m num__90 cm is <o> a ) num__15 cm <o> b ) num__25 cm <o> c ) num__35 cm <o> d ) num__10 cm <o> e ) num__43 cm |
explanation : required length = h . c . f of num__1000 cm num__1080 cm and num__1090 c = num__10 cm . answer : d <eor> d <eos> |
d |
add__80.0__1000.0__ add__10.0__1080.0__ round__10.0__ |
add__80.0__1000.0__ add__10.0__1080.0__ round__10.0__ |
| in an elaction between two candidates num__10.0 of votes are were declares invalid . first candidate got num__1800 votes which were num__60.0 of the total valid votes . the total number of votes enrolled in that election was ? <o> a ) num__9000 votes <o> b ) num__10000 votes <o> c ) num__11000 votes <o> d ) num__12000 votes <o> e ) num__13000 votes |
num__100.0 - num__10.0 = num__90.0 num__54.0 - num__36.0 = num__18.0 num__18.0 - - - - - - > num__1800 ( num__18 × num__100 = num__1800 ) num__100.0 - - - - - - - > num__10000 votes ( num__100 × num__100 = num__10000 ) b ) <eor> b <eos> |
b |
percent__60.0__90.0__ percent__100.0__10000.0__ |
percent__60.0__90.0__ percent__100.0__10000.0__ |
| a fair sided dice labeled num__1 to num__6 is tossed num__2 times . what is the probability the sum of the num__2 throws is num__5 ? <o> a ) num__0.166666666667 <o> b ) num__0.0324074074074 <o> c ) num__0.111111111111 <o> d ) num__0.0416666666667 <o> e ) num__0.0509259259259 |
the total possible number of combination if the dice is thrown num__2 times is num__36 . there are num__4 possible outcomes that the sum of the number is num__5 which is ( num__2 + num__3 ) ( num__3 + num__2 ) ( num__1 + num__4 ) and ( num__4 + num__1 ) answer : c <eor> c <eos> |
c |
subtract__6.0__2.0__ add__1.0__2.0__ divide__4.0__36.0__ |
subtract__6.0__2.0__ add__1.0__2.0__ divide__4.0__36.0__ |
| a car traveled num__75.0 of the way from town a to town b at an average speed of num__75 mph . the car traveled at an average speed of v mph for the remaining part of the trip . the average speed for the entire trip was num__50 mph . what is v in mph ? <o> a ) num__30 <o> b ) num__35 <o> c ) num__25 <o> d ) num__40 <o> e ) num__45 |
assume total distance = num__100 miles time taken for num__75 miles = num__1.0 = num__1 hour time taken for the rest of the num__25 miles = num__25 / v hours . average speed = num__50 therefore the total time needed = num__2 hours . num__2 = num__1 + num__25 / v hence v = num__25 mph answer : c <eor> c <eos> |
c |
subtract__75.0__50.0__ divide__50.0__25.0__ subtract__75.0__50.0__ |
subtract__75.0__50.0__ divide__50.0__25.0__ divide__50.0__2.0__ |
| a plane was originally flying at an altitude of x feet when it ascended num__2000 feet and then descended num__5000 feet . if the plane ' s altitude after these two changes was num__0.2 its original altitude then the solution of which of the following equations gives the plane ' s original altitude in feet ? <o> a ) x + num__2000 = num__0.333333333333 * ( x - num__3000 ) <o> b ) num__0.333333333333 * ( x - num__3000 ) = x <o> c ) x + num__3000 = num__0.333333333333 * x <o> d ) x - num__7000 = num__0.333333333333 * x <o> e ) x - num__3000 = num__0.2 * x |
plane ' s original altitude = x plane ' s new altitude after ascending num__2000 ft = x + num__2000 plane ' s new altitude after descending num__5000 ft from previous altitude = x + num__2000 - num__5000 = x - num__3000 so after two changes plane is at num__0.2 its original altitude = > x - num__3000 = x / num__5 answer ( e ) <eor> e <eos> |
e |
subtract__5000.0__2000.0__ reverse__0.2__ subtract__5000.0__2000.0__ |
subtract__5000.0__2000.0__ reverse__0.2__ subtract__5000.0__2000.0__ |
| a train num__400 m long can cross an electric pole in num__20 sec and then find the speed of the train ? <o> a ) num__987 <o> b ) num__271 <o> c ) num__72 <o> d ) num__27 <o> e ) num__28 |
length = speed * time speed = l / t s = num__20.0 s = num__20 m / sec speed = num__20 * num__3.6 ( to convert m / sec in to kmph multiply by num__3.6 ) speed = num__72 kmph answer : c <eor> c <eos> |
c |
multiply__20.0__3.6__ round__72.0__ |
multiply__20.0__3.6__ multiply__20.0__3.6__ |
| an inspector rejects num__0.08 of the meters as defective . how many will be examine to project ? <o> a ) num__1000 <o> b ) num__1500 <o> c ) num__2000 <o> d ) num__2300 <o> e ) num__2500 |
let the number of meters to be examined be x . then num__0.08 of x = num__2 [ ( num__0.08 ) * ( num__0.01 ) * x ] = num__2 x = [ ( num__2 * num__100 * num__100 ) / num__8 ] = num__2500 . answer e num__2500 <eor> e <eos> |
e |
percent__100.0__2500.0__ |
percent__100.0__2500.0__ |
| in a division divident is num__686 divisior is num__36 and quotient is num__19 . find the remainder . <o> a ) a ) num__4 <o> b ) b ) num__3 <o> c ) c ) num__2 <o> d ) d ) num__5 <o> e ) e ) num__6 |
explanation : num__686 = num__36 x num__19 + r num__686 = num__684 + r r = num__686 - num__684 = num__2 answer : option c <eor> c <eos> |
c |
multiply__36.0__19.0__ subtract__686.0__684.0__ subtract__686.0__684.0__ |
multiply__36.0__19.0__ subtract__686.0__684.0__ subtract__686.0__684.0__ |
| when x + p is divided by p the quotient is n and there is no remainder . which of the following must be the value of x ? <o> a ) p ( n - num__1 ) <o> b ) n ( p − num__1 ) <o> c ) p ( n + num__1 ) + num__2 n <o> d ) p ( n - num__1 ) + num__2 n <o> e ) n ( p + num__1 ) ( p − num__1 ) |
given in question ( x + p ) / p = n = > x + p = np = > x = np - p now we are good to check the options x = p ( n - num__1 ) answer a <eor> a <eos> |
a |
reverse__1.0__ |
reverse__1.0__ |
| john buys num__100 shares of par value rs . num__10 each of a company which pays an annual dividend of num__12.0 at such a price that he gets num__10.0 on his investment . find the market value of a share . <o> a ) num__12 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__8 |
face value of each share = rs . num__10 total dividend received by john = num__100 à — num__10 à — num__0.12 = rs . num__120 let market value of num__100 shares = rs . x x à — num__0.1 = num__120 x = num__1200 ie market value of num__100 shares = rs . num__1200 hence market value of each share = rs . num__12 answer is a <eor> a <eos> |
a |
percent__100.0__12.0__ |
percent__100.0__12.0__ |
| calculate the time it will take for a full tank of water to become completely empty due to a leak given that the tank could be filled in num__10 hours but due to the leak in its bottom it takes num__11 hours to be filled ? <o> a ) num__90 hours <o> b ) num__80 hours <o> c ) num__50 hours <o> d ) num__120 hours <o> e ) num__110 hours |
part filled without leak in num__1 hour = num__0.1 part filled with leak in num__1 hour = num__0.0909090909091 work done by leak in num__1 hour = num__0.1 â ˆ ’ num__0.0909090909091 = num__110 hours answer : e <eor> e <eos> |
e |
subtract__11.0__10.0__ divide__1.0__10.0__ divide__1.0__11.0__ multiply__10.0__11.0__ round__110.0__ |
subtract__11.0__10.0__ divide__1.0__10.0__ divide__1.0__11.0__ multiply__10.0__11.0__ round__110.0__ |
| the cost price of an article is rs . num__100 . to gain num__50.0 after allowing a num__50.0 discount the market price of the article is : <o> a ) rs . num__200 <o> b ) rs . num__400 <o> c ) rs . num__350 <o> d ) rs . num__300 <o> e ) rs . num__380 |
discount id always on mp . . . . so let m . p be x . . . . . nd s . p is num__100 + ( num__50.0 of num__100 ) = num__150 so ( num__0.5 ) * x = num__150 on solving x = num__300 answer : d <eor> d <eos> |
d |
percent__100.0__300.0__ |
percent__100.0__300.0__ |
| the product of the squares of two positive integers is num__400 . how many pairs of positive integers satisfy this condition ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
first break down num__200 into num__20 * num__20 and further into the prime factors num__2 * num__2 * num__5 * num__2 * num__2 * num__5 . now we are looking for all the possible pairs ( num__2 numbers ) of squares whose product results in num__400 . num__1 st : num__2 ^ num__2 * num__10 ^ num__2 ( i . e . the first two num__2 ' s and two times num__2 * num__5 = num__10 ) num__2 nd : num__4 ^ num__2 * num__5 ^ num__2 ( i . e . two times num__2 * num__2 = num__4 = num__4 ^ num__2 and num__5 ^ num__2 ) . num__3 rd : num__1 ^ num__2 * num__20 ^ num__2 ( i . e . two times num__2 * num__2 * num__5 and num__1 ^ num__2 = num__1 ) answer d . <eor> d <eos> |
d |
triangle_area__1.0__20.0__ square_perimeter__1.0__ side_by_diagonal__5.0__4.0__ multiply__1.0__3.0__ |
multiply__2.0__5.0__ square_perimeter__1.0__ side_by_diagonal__5.0__4.0__ multiply__1.0__3.0__ |
| the average of num__11 numbers is num__50 . if the average of first num__6 numbers is num__49 and that of last num__6 is num__52 . find the num__6 th number . <o> a ) num__54 <o> b ) num__55 <o> c ) num__56 <o> d ) num__57 <o> e ) num__58 |
avg = num__50 so total sum of num__11 numbers = num__550 let num__6 th number is x so according to the given condition sum of first num__5 numbers = num__49 * num__6 - x = num__294 - x sum of last num__5 numbers = num__52 * num__6 - x = num__312 - x so num__6 th number x = num__550 - ( num__294 - x ) - ( num__312 - x ) = num__2 x - num__56 so x = num__2 x - num__56 implies x = num__56 answer : c <eor> c <eos> |
c |
multiply__11.0__50.0__ subtract__11.0__6.0__ multiply__6.0__49.0__ multiply__6.0__52.0__ subtract__52.0__50.0__ add__50.0__6.0__ add__50.0__6.0__ |
multiply__11.0__50.0__ subtract__11.0__6.0__ multiply__6.0__49.0__ multiply__6.0__52.0__ subtract__52.0__50.0__ add__50.0__6.0__ add__50.0__6.0__ |
| the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square whose area is num__1225 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is num__10 units ? <o> a ) num__140 sq . units <o> b ) num__786 sq . units <o> c ) num__167 sq . units <o> d ) num__178 sq . units <o> e ) num__176 sq . units |
given that the area of the square = num__1225 sq . units = > side of square = √ num__1225 = num__35 units the radius of the circle = side of the square = num__35 units length of the rectangle = num__0.4 * num__35 = num__14 units given that breadth = num__10 units area of the rectangle = lb = num__14 * num__10 = num__140 sq . units answer : a <eor> a <eos> |
a |
multiply__35.0__0.4__ square_perimeter__35.0__ square_perimeter__35.0__ |
multiply__35.0__0.4__ multiply__10.0__14.0__ multiply__10.0__14.0__ |
| a circular logo is enlarged to fit the lid of a jar . the new diameter is num__60 per cent larger than the original . by what percentage has the area of the logo increased ? <o> a ) num__50 <o> b ) num__80 <o> c ) num__100 <o> d ) num__143.36 <o> e ) num__250 |
let old diameter be num__4 so radius is num__2 old area = num__4 π new diameter is num__6.24 so radius is num__3.12 new area = num__9.7344 π increase in area is num__5.7344 π % increase in area = num__5.7344 / num__4 * num__100 so % increase is num__143.36 answer will be ( d ) <eor> d <eos> |
d |
power__3.12__2.0__ triangle_area__2.0__143.36__ |
power__3.12__2.0__ triangle_area__2.0__143.36__ |
| of the total amount that jill spent on a shopping trip excluding taxes she spent num__60 percent on clothing num__20 percent on food and num__20 percent on other items . if jill paid a num__10 percent tax on the clothing no tax on the food and an num__10 percent tax on all other items then the total tax that she paid was what percent of the total amount that she spent excluding taxes ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__30 <o> d ) num__25 <o> e ) num__15 |
assume she has $ num__200 to spend . tax clothing = num__60.0 = $ num__120 = $ num__12.00 food = num__20.0 = $ num__40 = $ num__0.00 items = num__20.0 = $ num__40 = $ num__8.00 total tax = $ num__20.0 of total amount = num__0.1 * num__100 = num__10.0 answer a <eor> a <eos> |
a |
multiply__20.0__10.0__ divide__120.0__10.0__ subtract__60.0__20.0__ subtract__20.0__12.0__ reverse__10.0__ add__60.0__40.0__ reverse__0.1__ |
multiply__20.0__10.0__ divide__120.0__10.0__ subtract__60.0__20.0__ subtract__20.0__12.0__ reverse__10.0__ add__60.0__40.0__ multiply__100.0__0.1__ |
| there are two examinations rooms a and b . if num__10 students are sent from a to b then the number of students in each room is the same . if num__20 candidates are sent from b to a then the number of students in a is double the number of students in b . the number of students in room a is : <o> a ) num__20 <o> b ) num__80 <o> c ) num__100 <o> d ) num__415 <o> e ) num__150 |
let the number of students in rooms a and b be x and y respectively . then x - num__10 = y + num__10 x - y = num__20 . . . . ( i ) and x + num__20 = num__2 ( y - num__20 ) x - num__2 y = - num__60 . . . . ( ii ) solving ( i ) and ( ii ) we get : x = num__100 y = num__80 . the required answer a = num__100 . anawer is a <eor> a <eos> |
a |
divide__20.0__10.0__ add__20.0__60.0__ multiply__10.0__2.0__ |
divide__20.0__10.0__ add__20.0__60.0__ subtract__100.0__80.0__ |
| he time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is num__42 kmph find the speed of the stream ? <o> a ) num__35 kmph <o> b ) num__28 kmph <o> c ) num__14 kmph <o> d ) num__88 kmph <o> e ) num__12 kmph |
the ratio of the times taken is num__2 : num__1 . the ratio of the speed of the boat in still water to the speed of the stream = ( num__2 + num__1 ) / ( num__2 - num__1 ) = num__3.0 = num__3 : num__1 speed of the stream = num__14.0 = num__14 kmph . answer : c <eor> c <eos> |
c |
add__1.0__2.0__ divide__42.0__3.0__ round__14.0__ |
add__1.0__2.0__ divide__42.0__3.0__ divide__42.0__3.0__ |
| find the missing figures : ? % of num__40 = num__2.125 <o> a ) num__5.31 <o> b ) num__6.51 <o> c ) num__8.71 <o> d ) num__7.71 <o> e ) num__4.51 |
( i ) let x % of num__40 = num__2.125 . then ( x / num__100 ) * num__40 = num__2.125 x = ( num__2.125 * num__2.5 ) = num__5.31 answer is a . <eor> a <eos> |
a |
percent__100.0__5.31__ |
percent__100.0__5.31__ |
| the current of a stream at num__1 kmph . a motor boat goes num__35 km upstream and back to the starting point in num__12 hours . the speed of the motor boat in still water is ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__4 |
s = num__1 m = x ds = x + num__1 us = x - num__1 num__35 / ( x + num__1 ) + num__35 / ( x - num__1 ) = num__12 x = num__6 answer : a <eor> a <eos> |
a |
round__6.0__ |
subtract__12.0__6.0__ |
| a certain sum is invested at simple interest at num__15.0 p . a . for two years instead of investing at num__12.0 p . a . for the same time period . therefore the interest received is more by rs . num__840 . find the sum ? <o> a ) rs . num__7000 <o> b ) rs . num__9000 <o> c ) rs . num__14000 <o> d ) rs . num__17000 <o> e ) rs . num__27000 |
let the sum be rs . x . ( x * num__15 * num__2 ) / num__100 - ( x * num__12 * num__2 ) / num__100 = num__840 = > num__30 x / num__100 - num__24 x / num__100 = num__840 = > num__6 x / num__100 = num__840 = > x = num__14000 . answer : c <eor> c <eos> |
c |
percent__100.0__14000.0__ |
percent__100.0__14000.0__ |
| a completes a work in num__12 days and b complete the same work in num__24 days . if both of them work together then the number of days required to complete the work will be ? <o> a ) num__8 days <o> b ) num__10 days <o> c ) num__12 days <o> d ) num__14 days <o> e ) none |
if a can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days therefore here the required number of days = num__12 × num__0.666666666667 = num__8 days . a ) <eor> a <eos> |
a |
round__8.0__ |
round__8.0__ |
| a train covers a distance of num__12 km in num__10 min . if it takes num__1 sec to pass a telegraph post then the length of the train is ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__30 <o> d ) num__15 <o> e ) num__45 |
speed = ( num__1.2 * num__60 ) km / hr = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . length of the train = num__20 * num__1 = num__20 m . answer : option b <eor> b <eos> |
b |
divide__12.0__10.0__ hour_to_min_conversion__ add__12.0__60.0__ round__20.0__ |
divide__12.0__10.0__ hour_to_min_conversion__ multiply__1.2__60.0__ multiply__1.0__20.0__ |
| the length of a rectangle is increased by num__25.0 and its breadth is decreased by num__20.0 . what is the effect on its area ? <o> a ) num__10000 <o> b ) num__297 <o> c ) num__9279 <o> d ) num__2767 <o> e ) num__2676 |
num__100 * num__100 = num__10000 num__125 * num__80 = num__10000 answer : a <eor> a <eos> |
a |
square_perimeter__25.0__ surface_rectangular_prism__25.0__20.0__100.0__ square_perimeter__20.0__ multiply__80.0__125.0__ |
square_perimeter__25.0__ surface_rectangular_prism__25.0__20.0__100.0__ square_perimeter__20.0__ multiply__80.0__125.0__ |
| the simple interest on a certain sum of money for num__2 l / num__2 years at num__12.0 per annum is rs . num__40 less tban the simple interest on the same sum for num__3 ½ years at num__10.0 per annum . find the sum . <o> a ) rs . num__700 . <o> b ) rs . num__200 . <o> c ) rs . num__600 . <o> d ) rs . num__800 . <o> e ) rs . num__400 . |
let the sum be rs . x then ( ( x * num__10 * num__7 ) / ( num__100 * num__2 ) ) – ( ( x * num__12 * num__5 ) / ( num__100 * num__2 ) ) = num__40 ( num__7 x / num__20 ) - ( num__3 x / num__10 ) = num__40 x = ( num__40 * num__20 ) = num__800 . hence the sum is rs . num__800 . answer is d . <eor> d <eos> |
d |
subtract__10.0__3.0__ add__2.0__3.0__ multiply__2.0__10.0__ multiply__40.0__20.0__ multiply__40.0__20.0__ |
subtract__10.0__3.0__ subtract__12.0__7.0__ multiply__2.0__10.0__ multiply__40.0__20.0__ multiply__40.0__20.0__ |
| the percentage profit earned by selling an article for $ num__1920 is equal to the percentage loss incurred by selling the same article for $ num__1280 . at what price should the article be sold to make num__25.0 profit ? <o> a ) $ num__3000 <o> b ) $ num__1000 <o> c ) $ num__2000 <o> d ) $ num__5000 <o> e ) $ num__6000 |
c $ num__2000 let c . p . be $ x . then ( num__1920 - x ) / x * num__100 = ( x - num__1280 ) / x * num__100 num__1920 - x = x - num__1280 num__2 x = num__3200 = > x = num__1600 required s . p . = num__125.0 of $ num__1600 = num__1.25 * num__1600 = $ num__2000 . <eor> c <eos> |
c |
percent__100.0__2000.0__ |
percent__100.0__2000.0__ |
| there is num__60.0 increase in an amount in num__6 years at si . what will be the ci of rs . num__15000 after num__3 years at the same rate ? <o> a ) num__2372 <o> b ) num__2572 <o> c ) num__4542 <o> d ) num__4965 <o> e ) num__3972 |
let p = rs . num__100 . then s . i . rs . num__60 and t = num__6 years . r = num__100 x num__60 = num__10.0 p . a . num__100 x num__6 now p = rs . num__15000 . t = num__3 years and r = num__10.0 p . a . c . i . = rs . num__15000 x num__1 + num__10 num__3 - num__1 num__100 = rs . num__15000 x num__331 num__1000 = num__4965 . d <eor> d <eos> |
d |
percent__100.0__4965.0__ |
percent__100.0__4965.0__ |
| henry earns $ num__120 a week from his job . his income increased and now makes $ num__180 a week . what is the percent increase ? <o> a ) num__70.0 <o> b ) num__60.0 <o> c ) num__50.0 <o> d ) num__40.0 <o> e ) num__30 % |
increase = ( num__0.5 ) * num__100 = ( num__0.5 ) * num__100 = num__50.0 . c <eor> c <eos> |
c |
percent__50.0__100.0__ |
percent__50.0__100.0__ |
| there are num__5 blue pillow covers num__3 white pillow covers and num__9 green pillow covers in the wardrobe . what is the least number of pillow covers that you have to take out to make sure that you will have a matching pair ? <o> a ) num__4 <o> b ) num__3 <o> c ) num__12 <o> d ) num__6 <o> e ) num__9 |
since there are num__3 colors if you take num__3 pillow covers you could still have num__1 pillow cover per color and not have a match . therefore upon taking the num__4 th pillow cover you will definitely have a match since the num__4 th pillow cover will form at least a pair with the num__1 st num__3 so num__3 + num__1 = num__4 answer is a <eor> a <eos> |
a |
subtract__5.0__1.0__ subtract__5.0__1.0__ |
add__3.0__1.0__ add__3.0__1.0__ |
| x y a and b are positive integers of s . when x is divided by y the remainder is num__6 . when a is divided by b the remainder is num__9 . which of the following is not a possible value for y + b ? <o> a ) num__24 <o> b ) num__21 <o> c ) num__20 <o> d ) num__17 <o> e ) num__15 |
x y a and b are positive integers of s . when x is divided by y the remainder is num__6 . when a is divided by b the remainder is num__9 . answer : e . <eor> e <eos> |
e |
add__6.0__9.0__ |
add__6.0__9.0__ |
| for which of the following does f ( x ) − f ( y ) = f ( x − y ) for all values of x and y ? <o> a ) f ( a ) = a ^ num__2 <o> b ) f ( a ) = a / num__2 <o> c ) f ( a ) = a + num__5 <o> d ) f ( a ) = num__2 a − num__1 <o> e ) f ( a ) = | a | |
to solve this easiest way is just put the value and see that if it equals or not . with option num__1 . f ( x ) = x ^ num__2 and f ( y ) = y ^ num__2 so l . h . s = x ^ num__2 - y ^ num__2 and r . h . s = ( x - y ) ^ num__2 = = > x ^ num__2 + y ^ num__2 - num__2 xy . so l . h . s not equal to r . h . s with option num__2 . f ( x ) = x / num__2 and f ( y ) = y / num__2 l . h . s = x / num__2 - y / num__2 = = > num__0.5 ( x - y ) r . h . s = ( x - y ) / num__2 so l . h . s = r . h . s which is the correct answer . b <eor> b <eos> |
b |
reverse__2.0__ reverse__0.5__ |
reverse__2.0__ reverse__0.5__ |
| a man a woman and a boy can together complete a piece of work in num__4 days . if a man alone can do it in num__6 days and a boy alone in num__18 days how long will a woman take to complete the work ? <o> a ) num__36 days <o> b ) num__32 days <o> c ) num__34 days <o> d ) num__42 days <o> e ) num__49 days |
explanation : ( num__1 man + num__1 woman + num__1 boy ) ’ s num__1 day ’ s work = num__0.25 num__1 man ’ s num__1 day work = num__0.166666666667 num__1 boy ’ s num__1 day ’ s work = num__0.0555555555556 ( num__1 man + num__1 boy ) ‘ s num__1 day ’ s work = num__0.166666666667 + num__0.0555555555556 = num__0.222222222222 therefore num__1 woman ’ s num__1 day ’ s work = num__0.25 – num__0.222222222222 = num__0.0277777777778 therefore the woman alone can finish the work in num__36 days . answer : option a <eor> a <eos> |
a |
divide__1.0__4.0__ divide__1.0__6.0__ divide__1.0__18.0__ divide__4.0__18.0__ subtract__0.25__0.2222__ round__36.0__ |
divide__1.0__4.0__ divide__1.0__6.0__ divide__1.0__18.0__ divide__4.0__18.0__ subtract__0.25__0.2222__ round__36.0__ |
| a train is num__410 meter long is running at a speed of num__45 km / hour . in what time will it pass a bridge of num__140 meter length <o> a ) num__20 seconds <o> b ) num__44 seconds <o> c ) num__40 seconds <o> d ) num__50 seconds <o> e ) none of these |
explanation : speed = num__45 km / hr = num__45 * ( num__0.277777777778 ) m / sec = num__12.5 m / sec total distance = num__410 + num__140 = num__550 meter time = distance / speed = num__550 ∗ num__0.08 = num__44 seconds option b <eor> b <eos> |
b |
add__410.0__140.0__ divide__550.0__12.5__ round__44.0__ |
add__410.0__140.0__ divide__550.0__12.5__ divide__550.0__12.5__ |
| the ratio between the length and the breadth of a rectangular park is num__3 : num__2 . if a man cycling along the boundary of the park at the speed of num__12 km / hr completes one round in num__8 minutes then what is the area of the park ( in sq . m ) ? <o> a ) num__153500 <o> b ) num__153650 <o> c ) num__153600 <o> d ) num__153700 <o> e ) num__153750 |
let length = num__3 x km breadth = num__2 x km distance travelled by the man at the speed of num__12 km / hr in num__8 minutes = num__2 ( num__3 x + num__2 x ) = num__10 x therefore num__12 × num__860 = num__10 xx = num__425 km = num__160 m area = num__3 x × num__2 x = num__6 x num__2 = num__6 × num__1602 = num__153600 m num__2 answer is c . <eor> c <eos> |
c |
add__2.0__8.0__ multiply__3.0__2.0__ round__153600.0__ |
add__2.0__8.0__ divide__12.0__2.0__ round__153600.0__ |
| the c . p of num__17 books is equal to the s . p of num__20 books . find his gain % or loss % ? <o> a ) num__16 num__0.666666666667 % loss <o> b ) num__16 num__0.25 % loss <o> c ) num__16 num__1.0 % loss <o> d ) num__36 num__0.666666666667 % loss <o> e ) num__15.0 loss |
num__17 cp = num__20 sp num__20 - - - num__3 cp loss num__100 - - - ? = > num__15.0 loss answer : e <eor> e <eos> |
e |
percent__100.0__15.0__ |
percent__100.0__15.0__ |
| a person deposits num__0.0625 of his income as provident fund and num__0.0666666666667 of the remaining as insurance premium . if he spends num__0.714285714286 of the balance on domestic needs and deposits an amount of rs . num__50 in the bank his total income would be <o> a ) num__150 <o> b ) num__200 <o> c ) num__250 <o> d ) num__300 <o> e ) num__350 |
( num__1 - ( num__0.0625 ) ) ( num__1 - ( num__0.0666666666667 ) ) ( num__1 - ( num__0.714285714286 ) ) of income = num__50 hence income = num__200 answer : b <eor> b <eos> |
b |
multiply__1.0__200.0__ |
multiply__1.0__200.0__ |
| factor : num__5 x num__4 y num__3 â € “ num__80 y num__3 <o> a ) a ) num__5 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__2 ) <o> b ) b ) num__2 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__2 ) <o> c ) c ) num__3 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__3 ) <o> d ) d ) num__3 y num__3 ( x num__2 + num__4 ) ( x + num__3 ) ( x - num__2 ) <o> e ) e ) num__3 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__4 ) |
num__5 x num__4 y num__3 â € “ num__80 y num__3 . = num__5 y num__3 ( x num__4 â € “ num__16 ) . = num__5 y num__3 [ ( x num__2 ) num__2 - num__42 ] . = num__5 y num__3 ( x num__2 + num__4 ) ( x num__2 - num__4 ) . = num__5 y num__3 ( x num__2 + num__4 ) ( x num__2 - num__22 ) . = num__5 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__2 ) . answer : ( a ) num__5 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__2 ) <eor> a <eos> |
a |
divide__80.0__5.0__ subtract__5.0__3.0__ gcd__5.0__80.0__ |
divide__80.0__5.0__ subtract__5.0__3.0__ add__3.0__2.0__ |
| the average earning of amechanic for the first num__4 days of a week is rs . num__32 and for the last four days is rs . num__22 . if heearns rs . num__20 on the fourth day his average earning forthe whole week is ? <o> a ) s . num__10 <o> b ) s . num__20 <o> c ) s . num__32 <o> d ) s . num__40 <o> e ) s . num__28 |
total earning for the week = sum of earning of first four days + sum of earning of last four days - earning of num__4 th day = num__4 x num__32 + num__4 x num__22 - num__20 = rs . num__196 â ˆ ´ average earning = num__28.0 = rs . num__28 e <eor> e <eos> |
e |
subtract__32.0__4.0__ subtract__32.0__4.0__ |
subtract__32.0__4.0__ subtract__32.0__4.0__ |
| the speed of a boat in still water in num__37 km / hr and the rate of current is num__13 km / hr . the distance travelled downstream in num__10 minutes is : <o> a ) num__10.44 km <o> b ) num__10.6 km <o> c ) num__11.4 km <o> d ) num__11.22 km <o> e ) num__8.33 km |
explanation : speed downstream = ( num__37 + num__13 ) = num__50 kmph time = num__24 minutes = num__0.166666666667 hour = num__0.166666666667 hour distance travelled = time × speed = ( num__0.166666666667 ) × num__50 = num__8.33 km answer : option e <eor> e <eos> |
e |
add__37.0__13.0__ subtract__37.0__13.0__ round__8.33__ |
add__37.0__13.0__ subtract__37.0__13.0__ round__8.33__ |
| a salesperson received a commission of num__3 percent of the sale price for each of the first num__100 machines that she sold and num__4 percent of the sale price for each machine that she sold after the first num__100 . if the sale price of each machine was $ num__10000 and the salesperson received a $ num__40000 commission how many machines did she sell ? <o> a ) num__90 <o> b ) num__103 <o> c ) num__105 <o> d ) num__115 <o> e ) num__125 |
first num__100 machines = num__3.0 commission = num__0.03 * num__100 * num__10000 = num__30000 commission from sale of next machines = num__40000 - num__30000 = num__10000 so num__25 more machines . . total = num__125 machines imo e . . . <eor> e <eos> |
e |
percent__100.0__125.0__ |
percent__100.0__125.0__ |
| how many seconds will a num__800 meter long train moving with a speed of num__63 km / hr take to cross a man walking with a speed of num__3 km / hr in the direction of the train ? <o> a ) num__48 <o> b ) num__36 <o> c ) num__26 <o> d ) num__11 <o> e ) num__18 |
explanation : here distance d = num__800 mts speed s = num__63 - num__3 = num__60 kmph = num__60 x num__0.277777777778 m / s time t = = num__48 sec . answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ round__48.0__ |
subtract__63.0__3.0__ round__48.0__ |
| for any real number x the operatoris defined as : ( x ) = x ( num__4 − x ) if p + num__1 = ( p + num__1 ) then p = <o> a ) - num__2 <o> b ) num__0 <o> c ) num__1 <o> d ) num__2 <o> e ) num__3 |
( x ) = x ( num__4 − x ) ( p + num__1 ) = ( p + num__1 ) ( num__4 - p - num__1 ) = ( num__3 - p ) ( p + num__1 ) we are given that p + num__1 = ( p + num__1 ) therefore ( num__3 - p ) ( p + num__1 ) = ( p + num__1 ) or ( p + num__1 ) + ( p - num__3 ) ( p + num__1 ) = num__0 ( p + num__1 ) ( p - num__3 ) = num__0 p = - num__1 p = num__3 option e <eor> e <eos> |
e |
subtract__4.0__1.0__ subtract__4.0__1.0__ |
subtract__4.0__1.0__ subtract__4.0__1.0__ |
| the number of arrangements that can be made with the letters of the word meadows so that the vowels occupy the num__1 st and last places ? <o> a ) a ) num__340 <o> b ) b ) num__360 <o> c ) c ) num__256 <o> d ) d ) num__196 <o> e ) e ) num__400 |
explanation : the word meadows has num__7 letters of which num__3 are vowels . v - - - - - v as the vowels have to occupy num__1 st and last places they can be arranged in the num__3 c num__2 i . e . num__3 ways . while the consonants and vowel can be arranged among themselves in the remaining num__5 places in num__5 ! i . e . num__120 ways . hence the total ways are num__120 * num__3 = num__360 . b <eor> b <eos> |
b |
coin_space__ vowel_space__ choose__5.0__2.0__ choose__5.0__2.0__ |
coin_space__ vowel_space__ choose__5.0__2.0__ choose__5.0__2.0__ |
| in an election between two candidates one got num__55.0 of the total valid votes num__20.0 of the votes were invalid . if the total number of votes was num__7500 the number of valid votes that the other candidate got was <o> a ) num__548 <o> b ) num__9674 <o> c ) num__2397 <o> d ) num__2700 <o> e ) num__3000 |
explanation : number of valid votes = num__80.0 of num__7500 = num__6000 . valid votes polled by other candidate = num__45.0 of num__6000 = ( num__0.45 * num__6000 ) = num__2700 answer : d <eor> d <eos> |
d |
percent__80.0__7500.0__ percent__45.0__6000.0__ percent__45.0__6000.0__ |
percent__80.0__7500.0__ percent__45.0__6000.0__ percent__45.0__6000.0__ |
| the tens digit of a two - digit number is two more than its unit digit . the two - digit number is num__7 times the sum of the digits . find the units digits ? <o> a ) num__7 <o> b ) num__2 <o> c ) num__4 <o> d ) num__5 <o> e ) num__1 |
explanation : let the two - digit number be num__10 a + b a = b + num__2 - - - ( num__1 ) num__10 a + b = num__7 ( a + b ) = > a = num__2 b substituting a = num__2 b in equation ( num__1 ) we get num__2 b = b + num__2 = > b = num__2 hence the units digit is : num__2 . answer : b <eor> b <eos> |
b |
multiply__1.0__2.0__ |
multiply__1.0__2.0__ |
| a can give b num__300 meters start and c num__600 meters start in a kilometer race . how much start can b give c in a kilometer race ? <o> a ) num__111.12 <o> b ) num__111.67 <o> c ) num__428.57 <o> d ) num__111.11 <o> e ) num__101.12 |
a runs num__1000 m while b runs num__700 m and c runs num__400 m . the number of meters that c runs when b runs num__1000 m = ( num__1000 * num__400 ) / num__700 = num__571.43 m . b can give c = num__1000 - num__571.43 = num__428.57 m . answer : c <eor> c <eos> |
c |
subtract__1000.0__300.0__ subtract__1000.0__600.0__ subtract__1000.0__571.43__ round__428.57__ |
subtract__1000.0__300.0__ subtract__1000.0__600.0__ subtract__1000.0__571.43__ subtract__1000.0__571.43__ |
| if n = num__2.0254 and n * is the decimal obtained by rounding n to the nearest hundredth what is the value of n * – n ? <o> a ) - num__0.0053 <o> b ) - num__0.0003 <o> c ) num__0.0007 <o> d ) num__0.0046 <o> e ) num__0.0153 |
n * = num__2.03 n * - n = num__2.03 - num__2.0254 num__0.0046 answer : d <eor> d <eos> |
d |
subtract__2.03__2.0254__ subtract__2.03__2.0254__ |
subtract__2.03__2.0254__ subtract__2.03__2.0254__ |
| a collection of num__20 coins each with a face value of either num__10 cents or num__35 cents has a total face value of $ num__2.5 . how many of the coins have a face value of num__35 cents ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__7 |
num__0.10 ( x ) + num__0.35 ( num__20 - x ) = num__2.5 num__0.10 x + num__7 - num__0.35 x = num__2.5 so x = num__18 num__20 - x = num__2 ans is a <eor> a <eos> |
a |
coin_space__ coin_space__ |
coin_space__ coin_space__ |
| a retailer buys a radio for rs num__225 . his overhead expenses are rs num__15 . he sellis the radio for rs num__350 . the profit percent of the retailer is <o> a ) num__10.0 <o> b ) num__45.8 <o> c ) num__25.0 <o> d ) num__52.0 <o> e ) none of these |
explanation : cost price = ( num__225 + num__15 ) = num__240 sell price = num__350 gain = ( num__0.458333333333 ) * num__100 = num__45.8 . answer : b <eor> b <eos> |
b |
percent__100.0__45.8__ |
percent__100.0__45.8__ |
| how much water must be added to a num__30 quarts of a num__75.0 acid solution to reduce it to a num__15.0 solution of acid ? <o> a ) num__120 qts <o> b ) num__220 qts <o> c ) num__320 qts <o> d ) num__420 qts <o> e ) num__520 qts |
i ’ m adding water so we have water + water = water starting off with num__30 quarts and adding x quarts we should end up with ( num__30 + x ) quarts num__30 + x = ( num__30 + x ) the problem describes the solution in terms of acid we have set it up in terms of water . so we have to change the percentages and put them in the problem . . num__25 ( num__30 ) + x = . num__85 ( num__30 + x ) again we multiply both sides of the equation by num__100 to get rid of the decimal point . num__25 ( num__30 ) + num__100 x = num__85 ( num__30 + x ) num__750 + num__100 x = num__2550 + num__85 x num__15 x = num__1800 x = num__120 qts correct answer a <eor> a <eos> |
a |
add__75.0__25.0__ multiply__30.0__25.0__ multiply__30.0__85.0__ subtract__2550.0__750.0__ divide__1800.0__15.0__ divide__1800.0__15.0__ |
add__75.0__25.0__ multiply__30.0__25.0__ multiply__30.0__85.0__ subtract__2550.0__750.0__ divide__1800.0__15.0__ divide__1800.0__15.0__ |
| if a and b are positive integers such that a / b = num__2.14 which of the following must be a divisor of a ? <o> a ) num__15 <o> b ) num__12 <o> c ) num__25 <o> d ) num__102 <o> e ) num__107 |
num__1 ) a and b are positive integers num__2 ) a / b = num__2.14 a = num__214 b = num__100 num__2.14 = num__2.14 these values are not the smallest possible values though ( since they ' re both even we can divide both by num__2 ) . . . a = num__107 b = num__50 num__2.14 = num__2.14 there is no other way to reduce this fraction so a must be a multiple of num__107 and b must be an equivalent multiple of num__50 . at this point though the value of b is irrelevant to the question . we ' re asked for what must divide into a . . . . since a is a multiple of num__107 we have to ' factor - down ' num__107 . this gives us only num__107 apart from num__1 . so this integer must be a factor of a . you ' ll find the match in the answer choices . answer : e <eor> e <eos> |
e |
round_down__2.14__ divide__214.0__2.14__ divide__214.0__2.0__ divide__100.0__2.0__ multiply__2.14__50.0__ |
round_down__2.14__ divide__214.0__2.14__ divide__214.0__2.0__ divide__100.0__2.0__ divide__107.0__1.0__ |
| num__3639 + num__11.95 - x = num__3054 . find the value of x . <o> a ) num__407.09 <o> b ) num__479.75 <o> c ) num__523.93 <o> d ) num__596.95 <o> e ) none of these |
explanation : let num__3639 + num__11.95 – x = num__3054 then x = ( num__3639 + num__11.95 ) – num__3054 = num__3650.95 – num__3054 = num__596.95 answer : d <eor> d <eos> |
d |
add__3639.0__11.95__ subtract__3650.95__3054.0__ subtract__3650.95__3054.0__ |
add__3639.0__11.95__ subtract__3650.95__3054.0__ subtract__3650.95__3054.0__ |
| the sume of the present ages of a father and his son is num__60 years . six years ago father ’ s ago was five times the age of the son . after num__6 years son ’ s age will be : <o> a ) num__12 years <o> b ) num__14 years <o> c ) num__18 years <o> d ) num__20 years <o> e ) none of these |
solution let the present ages of son and father be x and ( num__60 - x ) years recepectively . then ( num__60 - x ) - num__6 = num__5 ( x - num__6 ) ⇔ num__54 - x = num__5 x - num__30 ⇔ num__6 x = num__84 ⇔ x = num__14 . ∴ son ' s age after num__6 years = ( x + num__6 ) = num__20 years . answer d <eor> d <eos> |
d |
subtract__60.0__6.0__ multiply__6.0__5.0__ add__54.0__30.0__ divide__84.0__6.0__ add__6.0__14.0__ add__6.0__14.0__ |
subtract__60.0__6.0__ multiply__6.0__5.0__ add__54.0__30.0__ divide__84.0__6.0__ add__6.0__14.0__ add__6.0__14.0__ |
| on the first day of her vacation louisa traveled num__160 miles . on the second day traveling at the same average speed she traveled num__280 miles . if the num__160 - mile trip took num__3 hours less than the num__280 - mile trip what was the average speed in miles per hour ? <o> a ) num__40 <o> b ) num__45 <o> c ) num__46 <o> d ) num__48 <o> e ) num__50 |
( time ) * ( rate ) = ( distance ) - - > ( rate ) = ( distance ) / ( time ) - - > given : ( rate ) = num__160 / t = num__280 / ( t + num__3 ) - - > num__4 / t = num__7 / ( t + num__3 ) - - > num__4 t + num__12 = num__7 t - - - - > num__3 t = num__12 . t = num__4 - - - - > ( rate ) = num__40.0 = num__40 answer : a <eor> a <eos> |
a |
add__3.0__4.0__ multiply__3.0__4.0__ divide__160.0__4.0__ round__40.0__ |
add__3.0__4.0__ multiply__3.0__4.0__ divide__160.0__4.0__ divide__160.0__4.0__ |
| a traveler changes num__150 pounds into rupees at the rate of rs . num__7000 for num__100 pounds . he spends rs . num__9060 and changes the remaining amount back to pounds at the rate of num__100 pounds to rs . num__7200 . how many pounds will he get ? <o> a ) num__20 pounds <o> b ) num__77 pounds <o> c ) num__66 pounds <o> d ) num__55 pounds <o> e ) num__99 pounds |
explanation : amount of rupees for exchange of num__150 pounds = ( num__1.5 ) * ( num__7000.0 ) = rs . num__10500 out of these rs . num__10500 traveler spent rs . num__9060 remaining balance in rupees = num__10500 – num__9060 = rs . num__1440 now these rs . num__1440 are exchanged back to pounds with the rate of num__100 pounds for rs . num__7200 amount in pounds = ( num__1440 * num__100 ) / num__7200 = num__20 pounds answer : a <eor> a <eos> |
a |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| how long does a train num__110 m long running at the speed of num__72 km / hr takes to cross a bridge num__140 m length ? <o> a ) num__82.1 sec . <o> b ) num__12.5 sec <o> c ) num__19.1 sec . <o> d ) num__17.1 sec . <o> e ) num__42.1 sec . |
speed = num__72 * num__0.277777777778 = num__20 m / sec total distance covered = num__110 + num__140 = num__250 m . required time = num__12.5 = num__12.5 sec . answer : b <eor> b <eos> |
b |
add__110.0__140.0__ divide__250.0__20.0__ round__12.5__ |
add__110.0__140.0__ divide__250.0__20.0__ divide__250.0__20.0__ |
| if in a given number num__5 num__8 num__9 num__4 num__3 num__2 num__7 num__6 num__1 num__4 we interchange the first and the second digits the third and the fourth the fifth and the sixth and so on then counting from the right end which digit will be sixth ? <o> a ) num__3 <o> b ) num__2 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
after interchange num__8 num__5 num__4 num__9 num__2 num__3 num__6 num__7 num__1 num__4 now count from right sixth digit be num__2 answer : b <eor> b <eos> |
b |
round__2.0__ |
round__2.0__ |
| from mumbai railway station two trains a and b start simultaneously from stations num__400 miles apart and travel the same route toward each other on adjacent parallel tracks . if train a and train b travel at a constant rate of num__40 miles per hour and num__40 miles per hour respectively how many miles will train a have traveled when the trains pass each other to the nearest mile ? <o> a ) num__112 <o> b ) num__133 <o> c ) num__150 <o> d ) num__167 <o> e ) num__180 |
since we know the distance ( num__400 ) and the combined rate ( num__80 ) we plug it into the formula : distance = rate * time num__400 = num__80 * time we can solve for the time they will meet cause we added the rate of train a and train b together . so the time will be num__5.0 from dividing num__80 on both sides to isolate time in the equation above . time will be num__20 hours so now you can plug that in for train a ’ s distance . distance = rate * time distance = num__40 * num__20 distance = num__180 according to answer choice e . <eor> e <eos> |
e |
divide__400.0__80.0__ round__180.0__ |
divide__400.0__80.0__ round__180.0__ |
| find out the square of a number which when doubled exceeds its one eighth by num__15 ? <o> a ) num__16 <o> b ) num__25 <o> c ) num__19 <o> d ) num__64 <o> e ) num__17 |
let the number be p then the square will be p ^ num__2 according to question : num__2 p = ( p / num__8 ) + num__15 = > num__16 p = p + num__120 = > p = num__8 p ^ num__2 = num__8 ^ num__2 = num__64 . answer : d <eor> d <eos> |
d |
square_perimeter__2.0__ multiply__2.0__8.0__ multiply__15.0__8.0__ square_perimeter__16.0__ square_perimeter__16.0__ |
square_perimeter__2.0__ multiply__2.0__8.0__ multiply__15.0__8.0__ power__8.0__2.0__ power__8.0__2.0__ |
| find the principle on a certain sum of money at num__5.0 per annum for num__2 num__0.4 years if the amount being rs . num__1120 ? <o> a ) num__1000 <o> b ) num__2777 <o> c ) num__9999 <o> d ) num__2777 <o> e ) num__2661 |
num__1120 = p [ num__1 + ( num__5 * num__2.4 ) / num__100 ] p = num__1000 answer : a <eor> a <eos> |
a |
add__2.0__0.4__ multiply__1.0__1000.0__ |
add__2.0__0.4__ multiply__1.0__1000.0__ |
| a baseball card decreased in value num__20.0 in its first year and num__10.0 in its second year . what was the total percent decrease of the card ' s value over the two years ? <o> a ) num__28.0 <o> b ) num__30.0 <o> c ) num__32.0 <o> d ) num__36.0 <o> e ) num__72 % |
let the initial value of baseball card = num__100 after first year value of baseball card = ( num__1 - num__0.2 ) * num__100 = num__80 after second year value of baseball card = ( num__1 - num__0.1 ) * num__80 = num__72 total percent decrease of the card ' s value over the two years = ( num__100 - num__72 ) / num__100 * num__100.0 = num__28.0 answer a <eor> a <eos> |
a |
percent__20.0__1.0__ percent__10.0__1.0__ percent__100.0__28.0__ |
percent__20.0__1.0__ percent__10.0__1.0__ percent__100.0__28.0__ |
| num__1 num__8 num__27 num__64 . . . <o> a ) num__25 <o> b ) num__48 <o> c ) num__59 <o> d ) num__63 <o> e ) num__125 |
explanation : numbers are num__1 ^ num__3 = num__1 num__2 ^ num__3 = num__8 num__3 ^ num__3 = num__27 num__4 ^ num__3 = num__17 num__5 ^ num__3 = num__125 answer : e <eor> e <eos> |
e |
subtract__3.0__1.0__ add__1.0__3.0__ add__1.0__4.0__ power__5.0__3.0__ multiply__1.0__125.0__ |
subtract__3.0__1.0__ add__1.0__3.0__ add__1.0__4.0__ power__5.0__3.0__ power__5.0__3.0__ |
| a block of wood has dimensions num__10 cm x num__10 cm x num__90 cm . the block is painted red and then cut evenly at the num__45 cm mark parallel to the sides to form two rectangular solids of equal volume . what percentage of the surface area of each of the new solids is not painted red ? <o> a ) num__5.0 <o> b ) num__10.0 <o> c ) num__15.0 <o> d ) num__20.0 <o> e ) num__25 % |
the area of each half is num__100 + num__4 ( num__450 ) + num__100 = num__2000 the area that is not painted is num__100 . the fraction that is not painted is num__0.05 = num__0.05 = num__5.0 the answer is a . <eor> a <eos> |
a |
triangle_area__10.0__90.0__ multiply__0.05__100.0__ multiply__0.05__100.0__ |
triangle_area__10.0__90.0__ multiply__0.05__100.0__ multiply__0.05__100.0__ |
| ashok secured average of num__78 marks in num__6 subjects . if the average of marks in num__5 subjects is num__74 how many marks did he secure in the num__6 th subject ? <o> a ) num__38 <o> b ) num__98 <o> c ) num__99 <o> d ) num__17 <o> e ) num__80 |
explanation : number of subjects = num__6 average of marks in num__6 subjects = num__78 therefore total marks in num__6 subjects = num__78 * num__6 = num__468 now no . of subjects = num__5 total marks in num__5 subjects = num__74 * num__5 = num__370 therefore marks in num__6 th subject = num__468 – num__370 = num__98 answer : b <eor> b <eos> |
b |
multiply__78.0__6.0__ multiply__5.0__74.0__ subtract__468.0__370.0__ subtract__468.0__370.0__ |
multiply__78.0__6.0__ multiply__5.0__74.0__ subtract__468.0__370.0__ subtract__468.0__370.0__ |
| the cost of num__20 packets of sugar each weighing num__900 grams is rs . num__28 . what will be the cost of num__27 packets if each packet weighs num__1 kg ? <o> a ) rs num__42 <o> b ) rs num__56 <o> c ) rs num__58.50 <o> d ) rs num__64.75 <o> e ) none of these |
explanation : let the required cost be rs . x . then more packets more cost ( direct proportion ) more weight more cost ( direct proportion ) packets num__20 : num__27 weight num__900 : num__1000 : : num__28 : x ( num__20 x num__900 x x ) = ( num__27 x num__1000 x num__28 ) x = ( num__27 x num__1000 x num__28 ) / num__20 x num__900 = num__42 answer a <eor> a <eos> |
a |
multiply__1.0__42.0__ |
divide__42.0__1.0__ |
| num__4 num__5 num__7 num__11 num__19 ( . . . ) <o> a ) num__22 <o> b ) num__35 <o> c ) num__27 <o> d ) num__32 <o> e ) num__25 |
explanation : num__4 num__4 × num__2 - num__3 = num__5 num__5 × num__2 - num__3 = num__7 num__7 × num__2 - num__3 = num__11 num__11 × num__2 - num__3 = num__19 num__19 × num__2 - num__3 = num__35 answer : option b <eor> b <eos> |
b |
subtract__7.0__5.0__ subtract__5.0__2.0__ multiply__5.0__7.0__ multiply__5.0__7.0__ |
subtract__7.0__5.0__ subtract__5.0__2.0__ multiply__5.0__7.0__ multiply__5.0__7.0__ |
| num__31 of the scientists that attended a certain workshop were wolf prize laureates and num__12 of these num__31 were also nobel prize laureates . of the scientists that attended that workshop and had not received the wolf prize the number of scientists that had received the nobel prize was num__3 greater than the number of scientists that had not received the nobel prize . if num__50 of the scientists attended that workshop how many of them were nobel prize laureates ? <o> a ) a ) num__11 <o> b ) b ) num__23 <o> c ) c ) num__24 <o> d ) d ) num__29 <o> e ) d ) num__36 |
lets solve by creating equation . . w = num__31 . . total = num__50 . . not w = num__50 - num__31 = num__19 . . now let people who were neither be x so out of num__19 who won nobel = x + num__3 . . so x + x + num__3 = num__19 or x = num__8 . . so who won nobel but not wolf = x + num__3 = num__11 . . but people who won both w and n = num__12 . . so total who won n = num__11 + num__12 = num__23 . . b <eor> b <eos> |
b |
subtract__31.0__12.0__ add__3.0__8.0__ subtract__31.0__8.0__ round__23.0__ |
subtract__31.0__12.0__ add__3.0__8.0__ subtract__31.0__8.0__ subtract__31.0__8.0__ |
| what is the cost of leveling the field in the form of parallelogram at the rate of rs . num__7.0 sq . metre whose base & perpendicular distance from the other side being num__84 m & num__24 m respectively ? <o> a ) s . num__2400 <o> b ) s . num__2016 <o> c ) s . num__1400 <o> d ) s . num__3480 <o> e ) s . num__2000 |
area of the parallelogram = length of the base * perpendicular height = num__84 * num__24 = num__2016 m . total cost of levelling = rs . num__2016 b <eor> b <eos> |
b |
multiply__84.0__24.0__ round__2016.0__ |
multiply__84.0__24.0__ multiply__84.0__24.0__ |
| the respective ages of father and his son are num__40 and num__16 years . in how many years will the father be twice as old as his son ? <o> a ) num__19 years <o> b ) num__8 years <o> c ) num__10 years <o> d ) num__15 years <o> e ) num__12 years |
suppose x years later the father will be twice as old as his son . x + num__40 = num__2 ( x + num__16 ) x = num__40 - num__32 = num__8 years answer : b <eor> b <eos> |
b |
multiply__16.0__2.0__ subtract__40.0__32.0__ subtract__40.0__32.0__ |
multiply__16.0__2.0__ subtract__40.0__32.0__ subtract__40.0__32.0__ |
| sonia ' s father was num__38 years of age when she was born while her mother was num__36 years old when her brother four years younger to her was born . what is the difference between the ages of her parents ? <o> a ) num__8 years <o> b ) num__6 years <o> c ) num__4 years <o> d ) num__2 years <o> e ) none |
explanation : mother ' s age when sonia ' s brother was born = num__36 years . father ' s age when sonia ' s brother was born = ( num__38 + num__4 ) years = num__42 years . required difference = ( num__42 - num__36 ) years = num__6 years . answer : b <eor> b <eos> |
b |
add__38.0__4.0__ subtract__42.0__36.0__ divide__36.0__6.0__ |
add__38.0__4.0__ subtract__42.0__36.0__ subtract__42.0__36.0__ |
| the average age of num__40 students in a class is num__15 years . if the age of teacher is also included the average becomes num__16 years find the age of the teacher . <o> a ) num__22 <o> b ) num__27 <o> c ) num__28 <o> d ) num__26 <o> e ) num__56 |
explanation : if teacher ' s age is num__15 years there is no change in the average . but teacher has contributed num__1 year to all the students along with maintaining his age at num__16 . age of teacher = average age of all + total increase in age = num__16 + ( num__1 x num__40 ) = num__56 years answer : e <eor> e <eos> |
e |
subtract__16.0__15.0__ add__40.0__16.0__ add__40.0__16.0__ |
subtract__16.0__15.0__ add__40.0__16.0__ add__40.0__16.0__ |
| two taps can separately fill a cistern num__10 minutes and num__15 minutes respectively and when the waste pipe is open they can together fill it in num__15 minutes . the waste pipe can empty the full cistern in ? <o> a ) a ) num__7 <o> b ) b ) num__6 <o> c ) c ) num__8 <o> d ) d ) num__10 <o> e ) e ) num__5 |
num__0.1 + num__0.0666666666667 - num__1 / x = num__0.0666666666667 x = num__10 answer : d <eor> d <eos> |
d |
multiply__10.0__0.1__ round__10.0__ |
multiply__10.0__0.1__ divide__10.0__1.0__ |
| num__3 ^ ( num__2 x / num__3 ) = num__1 ^ ( num__5 x ) what is the value of x ? <o> a ) - num__1 <o> b ) num__0.666666666667 <o> c ) num__0 <o> d ) num__0.5 <o> e ) num__0.75 |
num__3 ^ ( num__2 x / num__3 ) = num__1 num__2 x / num__3 = num__0 ; x = num__0 = ans c <eor> c <eos> |
c |
multiply__3.0__0.0__ |
divide__0.0__3.0__ |
| you have a bucket of jelly beans . some are red some are blue and some green . with your eyes closed pick out num__2 of a like color . how many do you have to grab to be sure you have num__2 of the same ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
if you grab num__3 it might be all different . so take one extra to make sure you have two of the same . so take atleast num__4 answer : b <eor> b <eos> |
b |
choose__4.0__3.0__ |
choose__4.0__3.0__ |
| in a series of football matches the probability that team p wins a match against team q is num__0.25 and the probability that the match ends in a draw is num__0.333333333333 . if the two teams play five matches what is the probability that all five matches end in a draw ? <o> a ) num__0.00411522633745 <o> b ) num__0.0123456790123 <o> c ) num__0.037037037037 <o> d ) num__0.111111111111 <o> e ) num__0.333333333333 |
p ( num__5 draws ) = ( num__0.333333333333 ) ^ num__5 = num__0.00411522633745 the answer is a . <eor> a <eos> |
a |
power__0.3333__5.0__ power__0.3333__5.0__ |
power__0.3333__5.0__ power__0.3333__5.0__ |
| ayesha ’ s father was num__38 years of age when she was born while her mother was num__32 years old when her brother four years younger to her was born . what is the difference between the ages of her parents ? <o> a ) num__2 years <o> b ) num__4 years <o> c ) num__6 years <o> d ) num__10 years <o> e ) none |
explanation mother ’ s age when ayesha ’ s brother was born = num__32 years . father ’ s age when ayesha ’ s brother was born = ( num__38 + num__4 ) years = num__42 years . required difference = ( num__42 – num__32 ) years = num__10 years . answer d <eor> d <eos> |
d |
add__38.0__4.0__ subtract__42.0__32.0__ subtract__42.0__32.0__ |
add__38.0__4.0__ subtract__42.0__32.0__ subtract__42.0__32.0__ |
| how many seconds will a num__500 meter long train take to cross a man walking with a speed of num__3 km / hr in the direction of the moving train if the speed of the train is num__63 km / hr ? <o> a ) num__767 meters <o> b ) num__426 meters <o> c ) num__500 meters <o> d ) num__165 meters <o> e ) num__156 meters |
let length of tunnel is x meter distance = num__800 + x meter time = num__1 minute = num__60 seconds speed = num__78 km / hr = num__78 * num__0.277777777778 m / s = num__21.6666666667 m / s distance = speed * time num__800 + x = ( num__21.6666666667 ) * num__60 num__800 + x = num__20 * num__65 = num__1300 x = num__1300 - num__800 = num__500 meters answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ divide__60.0__3.0__ add__500.0__800.0__ round__500.0__ |
subtract__63.0__3.0__ divide__60.0__3.0__ add__500.0__800.0__ divide__500.0__1.0__ |
| a particular library has num__75 books in a special collection all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if by the end of the month num__80 percent of books that were loaned out are returned and there are num__65 books in the special collection at that time how many books of the special collection were loaned out during that month ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__35 <o> d ) num__40 <o> e ) num__50 |
there are num__10 books less ( num__75 - num__65 ) which represents num__20.0 of the loaned books ( num__100 - num__80 ) so total loaned out books = num__50 answer e <eor> e <eos> |
e |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| a batsman scored num__120 runs which included num__6 boundaries and num__4 sixes . what % of his total score did he make by running between the wickets <o> a ) num__40.0 <o> b ) num__60.0 <o> c ) num__65.0 <o> d ) num__70.0 <o> e ) num__75 % |
number of runs made by running = num__110 - ( num__6 x num__4 + num__4 x num__6 ) = num__120 - ( num__48 ) = num__72 now we need to calculate num__60 is what percent of num__120 . = > num__0.6 * num__100 = num__60.0 b <eor> b <eos> |
b |
subtract__120.0__48.0__ divide__72.0__120.0__ divide__60.0__0.6__ subtract__120.0__60.0__ |
subtract__120.0__48.0__ divide__72.0__120.0__ divide__60.0__0.6__ subtract__120.0__60.0__ |
| find the cost of fencing around a circular field of diameter num__28 m at the rate of rs . num__1.50 a meter ? <o> a ) a ) rs . num__150 <o> b ) b ) rs . num__132 <o> c ) c ) rs . num__100 <o> d ) d ) rs . num__125 <o> e ) e ) rs . num__225 |
num__2 * num__3.14285714286 * num__14 = num__88 num__88 * num__1 num__0.5 = rs . num__132 answer : b <eor> b <eos> |
b |
divide__28.0__2.0__ subtract__1.5__1.0__ multiply__1.5__88.0__ round__132.0__ |
divide__28.0__2.0__ subtract__1.5__1.0__ multiply__1.5__88.0__ multiply__1.5__88.0__ |
| n is the greatest number which divides num__1305 num__4665 and num__6905 and gives the same remainder in each case . what is the sum of the digits in n ? <o> a ) num__4 <o> b ) num__3 <o> c ) num__6 <o> d ) num__5 <o> e ) num__2 |
num__6905 - num__1305 = num__5600 num__6905 - num__4665 = num__2240 num__4665 - num__1305 = num__3360 hence the greatest number which divides num__1305 num__4665 and num__6905 and gives the same remainder n = hcf of num__5600 num__2240 num__3360 = num__1120 sum of digits in n = sum of digits in num__1120 = num__1 + num__1 + num__2 + num__0 = num__4 answer : option a <eor> a <eos> |
a |
subtract__6905.0__1305.0__ subtract__6905.0__4665.0__ subtract__4665.0__1305.0__ subtract__3360.0__2240.0__ divide__2240.0__1120.0__ multiply__4.0__1.0__ |
subtract__6905.0__1305.0__ subtract__6905.0__4665.0__ subtract__4665.0__1305.0__ subtract__3360.0__2240.0__ divide__2240.0__1120.0__ multiply__4.0__1.0__ |
| operation # is defined as adding a randomly selected two digit multiple of num__12 to a randomly selected two digit prime number and reducing the result by half . if operation # is repeated num__10 times what is the probability that it will yield at least two integers ? <o> a ) num__0.0 <o> b ) num__10.0 <o> c ) num__20.0 <o> d ) num__30.0 <o> e ) num__40 % |
any multiple of num__12 is even . any two - digit prime number is odd . ( even + odd ) / num__2 is not an integer . thus # does not yield an integer at all . therefore p = num__0 . answer : a . <eor> a <eos> |
a |
subtract__12.0__10.0__ multiply__12.0__0.0__ |
subtract__12.0__10.0__ divide__0.0__12.0__ |
| john had $ num__100 to buy drinks and sandwiches for his birhtday party . he bought num__5 small boxes of drinks at $ num__4 each box and num__8 boxes of sandwiches at $ num__6 each box . how much money was left after the shopping ? <o> a ) $ num__62 <o> b ) $ num__92 <o> c ) $ num__32 <o> d ) $ num__22 <o> e ) $ num__82 |
money spent on drinks num__5 ? num__4 = $ num__20 money spent on sandwiches num__8 ? num__6 = $ num__48 total money spent num__20 + num__48 = $ num__68 money left after shopping num__100 - num__68 = $ num__32 correct answer c <eor> c <eos> |
c |
divide__100.0__5.0__ multiply__8.0__6.0__ add__48.0__20.0__ subtract__100.0__68.0__ subtract__100.0__68.0__ |
divide__100.0__5.0__ multiply__8.0__6.0__ add__48.0__20.0__ subtract__100.0__68.0__ subtract__100.0__68.0__ |
| mr . depak has $ num__429774 in his bank account . what is the least amount of money ( in whole number of dollars ) that he must add to his account if he wants to split this money evenly among her six best friends ? options : <o> a ) $ num__1 <o> b ) $ num__2 <o> c ) $ num__3 <o> d ) $ num__4 <o> e ) $ num__6 |
to find the least amount deepak should add to his saving account to split the money evenly among his num__6 best friends he needs to make the total divisible by num__6 simply add the individual digits of the total = num__4 + num__2 + num__9 + num__7 + num__7 + num__4 = num__33 if you add num__3 the number is divisible by num__6 ( num__33 + num__3 ) correct option : c <eor> c <eos> |
c |
subtract__6.0__4.0__ subtract__9.0__2.0__ divide__6.0__2.0__ divide__6.0__2.0__ |
subtract__6.0__4.0__ subtract__9.0__2.0__ divide__6.0__2.0__ divide__6.0__2.0__ |
| in a rectangular coordinate system points o ( num__22 ) p ( num__26 ) and q ( num__102 ) represent the sites of three proposed housing developments . if a fire station can be built at any point in the coordinate system at which point would it be equidistant from all three developments ? <o> a ) ( num__37 ) <o> b ) ( num__53 ) <o> c ) ( num__34 ) <o> d ) ( num__64 ) <o> e ) ( num__55 ) |
all points equidistant from o and q lie on the line x = num__6 so the fire station should lie on this line . all points equidistant from o and p lie on the line y = num__4 so the fire station should lie on this line . these two intersect at ( num__64 ) and that will be the point equidistant from all num__3 points . the answer is d . <eor> d <eos> |
d |
volume_cube__4.0__ volume_cube__4.0__ |
volume_cube__4.0__ volume_cube__4.0__ |
| num__8 friends went to a hotel and decided to pay the bill amount equally . but num__7 of them could pay rs . num__80 each as a result num__8 th has to pay rs . num__70 extra than his share . find the amount paid by him . <o> a ) num__140 <o> b ) num__121 <o> c ) num__110 <o> d ) num__160 <o> e ) none |
explanation : average amount paid by num__7 persons = rs . num__80 increase in average due to rs . num__70 paid extra by the num__8 th men = rs . num__10.0 = rs . num__10 therefore average expenditure of num__8 friends = rs . num__80 + rs . num__10 = rs . num__90 therefore amount paid by the num__11 th men = rs . num__90 + rs . num__70 = rs . num__160 correct option : d <eor> d <eos> |
d |
divide__80.0__8.0__ add__80.0__10.0__ add__70.0__90.0__ add__70.0__90.0__ |
divide__80.0__8.0__ add__80.0__10.0__ add__70.0__90.0__ add__70.0__90.0__ |
| two identical circles intersect such that their centers and the points at which they intersect form a square of side num__1 cm . what is the area of the region where the two circles intersect ? <o> a ) pi / num__2 - num__1 <o> b ) pi / num__4 - num__1 <o> c ) num__1 - pi / num__4 <o> d ) pi / num__8 + num__1 <o> e ) num__4 - pi |
drawing a diagram we see that the radius of the circle equals the side of the square . the area of intersection is : pi * r ^ num__0.5 - ( r ^ num__2 - pi * r ^ num__0.5 ) = pi * r ^ num__1.0 - r ^ num__2 = pi / num__2 - num__1 the answer is a . <eor> a <eos> |
a |
divide__1.0__0.5__ round__2.0__ |
divide__1.0__0.5__ divide__1.0__0.5__ |
| the population of a city increases by num__8.0 per year but due to migration it decrease by num__1.0 per years . what will be the percentage increase in population in num__3 years ? <o> a ) num__9.0 <o> b ) num__9.27 <o> c ) num__22.5 <o> d ) num__12.0 <o> e ) none of these |
actual increase in population = num__7.0 let earlier population = num__100 then the population after num__3 years = num__100 ( num__1 + num__0.07 ) ^ num__3 = num__122.5043 ∴ required percentage = num__22.5 answer : c <eor> c <eos> |
c |
subtract__8.0__1.0__ divide__7.0__100.0__ multiply__1.0__22.5__ |
subtract__8.0__1.0__ divide__7.0__100.0__ multiply__1.0__22.5__ |
| there is a cycle race going on in a circular track and num__0.2 th of the total in front a person and num__0.833333333333 th of the total behind that person gives up the total number of participants . total how many participants are there ? <o> a ) num__30 <o> b ) num__31 <o> c ) num__32 <o> d ) num__33 <o> e ) num__34 |
assume there are x participants in the race . participants in front of a person wil b x - num__1 and that behind him wil b x - num__1 . so num__0.2 ( x - num__1 ) + num__0.833333333333 ( x - num__1 ) = x we get x = num__31 answer : b <eor> b <eos> |
b |
multiply__1.0__31.0__ |
multiply__1.0__31.0__ |
| a train of length num__250 m crosses a bridge of length num__150 m in num__32 seconds . what is the speed of train ? <o> a ) num__33 <o> b ) num__27 <o> c ) num__25 <o> d ) num__22 <o> e ) num__45 |
sol : ( length of train + length of bridge ) = speed of train x time ( num__250 + num__150 ) = num__32 x speed speed = num__12.5 = num__12.5 m / s = num__45 km / h answer = e <eor> e <eos> |
e |
round__45.0__ |
round__45.0__ |
| in a room filled with num__7 people num__3 people have exactly num__1 sibling in the room and num__4 people have exactly num__2 siblings in the room . if two individuals are selected from the room at random what is the probability that those two individuals are not siblings ? <o> a ) num__0.238095238095 <o> b ) num__0.428571428571 <o> c ) num__0.571428571429 <o> d ) num__0.714285714286 <o> e ) num__0.619047619048 |
there are suppose a b c d e f g members in the room num__4 people who have exactly one sibling . . . . a b c d . . . . ( a is bs ∘ sssibl ∈ g ∘ ssand ∘ ssviceversa ) ∘ ss ( c ∘ ssis ∘ ssds ∘ sssibl ∈ g ∘ ssand ∘ ssviceversa ) ∘ ss ( c ∘ ssis ∘ ssdssibl ∈ gandviceversa ) ( cisds sibling and viceversa ) ( c is ds sibling and viceversa ) . . . now remaning efg are num__4 people who have exactly num__2 siblings . . . . ( e has f and g as his / her sibling and so on . . ) there are now num__3 different set of siblings ( a and b ) ( c and d ) ; ( efg ) now first selecting num__2 people out of num__7 is num__7 c num__2 = num__21 first sibling pair - - - - ( a and b ) - - selecting num__2 people - - num__2 c num__2 = num__1 second sibling pair ( c and d ) - - selecting num__2 people - - num__2 c num__2 = num__1 third sibling pair ( e f g ) - - selecting num__2 out of num__4 - - num__4 c num__2 = num__6 total = num__1 + num__1 + num__6 = num__8 but a / c to formula p ( success ) - num__1 - p ( fail ) here p ( failure ) is selecting num__2 people who are siblings = num__0.380952380952 ( num__21 is num__7 c num__2 ) = num__1 - num__0.380952380952 = num__0.619047619048 ans e <eor> e <eos> |
e |
multiply__7.0__3.0__ subtract__7.0__1.0__ add__7.0__1.0__ divide__8.0__21.0__ subtract__1.0__0.381__ subtract__1.0__0.381__ |
multiply__7.0__3.0__ subtract__7.0__1.0__ add__7.0__1.0__ divide__8.0__21.0__ subtract__1.0__0.381__ subtract__1.0__0.381__ |
| a sum of money amounts to rs . num__9800 after num__5 years and rs . num__12005 after num__8 years at the same rate of simple interest . the rate of interest per annum is <o> a ) num__6.0 <o> b ) num__5.0 <o> c ) num__12.0 <o> d ) num__10.0 <o> e ) num__8 % |
s . i . for num__3 years = rs . ( num__12005 - num__9800 ) = rs . num__2205 . s . i . for num__5 years = rs . num__735.0 x num__5 = rs . num__3675 principal = rs . ( num__9800 - num__3675 ) = rs . num__6125 . hence rate = ( num__100 x num__3675 ) / ( num__6125 x num__5 ) % = num__12.0 answer : c <eor> c <eos> |
c |
percent__100.0__12.0__ |
percent__100.0__12.0__ |
| a sum amount to rs . num__1344 in two years at simple interest . what will be the compound interest on the same sum at the same rate of interest for the same period ? <o> a ) num__10.28 <o> b ) num__10.35 <o> c ) num__10.25 <o> d ) num__14.25 <o> e ) num__11.25 % |
num__100 num__5 - - - - i num__5 num__0.25 - - - ii - - - - - - - - - - - num__10.25 answer : c <eor> c <eos> |
c |
percent__10.25__100.0__ |
percent__10.25__100.0__ |
| a and b complete a job in num__6 days . a alone can do the job in num__24 days . if b works alone how many days will it take to complete the job ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
the rate of a and b is num__0.166666666667 a ' s rate is num__0.0416666666667 b ' s rate is num__0.166666666667 - num__0.0416666666667 = num__0.125 b can complete the job in num__8 days . the answer is a . <eor> a <eos> |
a |
subtract__0.1667__0.0417__ round__8.0__ |
subtract__0.1667__0.0417__ round__8.0__ |
| for num__1 rs num__1 p interest wat will be for num__2000 rs ? <o> a ) num__45 rs <o> b ) num__50 rs <o> c ) num__20 rs <o> d ) num__30 rs <o> e ) num__48 rs |
for num__1 rs num__1 p interest for num__2000 rs x x = num__2000.0 * num__1 p = = > num__2000 paise to express in rs num__20.0 = num__20 rs answer : c <eor> c <eos> |
c |
percent__1.0__2000.0__ percent__1.0__2000.0__ |
percent__1.0__2000.0__ percent__1.0__2000.0__ |
| if the difference of two numbers is num__3 and the difference of their square is num__39 then the larger number is : <o> a ) num__8 <o> b ) num__9 <o> c ) num__12 <o> d ) num__14 <o> e ) num__13 |
let the numbers be x and y then x num__2 - y num__2 = num__39 and x - y = num__3 we get x + y = num__13 solving x - y = num__3 x + y = num__13 x = num__8 y = num__5 larger number num__8 answer a num__8 <eor> a <eos> |
a |
square_perimeter__2.0__ square_perimeter__2.0__ |
square_perimeter__2.0__ square_perimeter__2.0__ |
| num__10 + num__19 <o> a ) num__29 <o> b ) num__13 <o> c ) num__28 <o> d ) num__6 <o> e ) num__2 |
a <eor> a <eos> |
a |
add__10.0__19.0__ |
add__10.0__19.0__ |
| two numbers are respectively num__20.0 and num__50.0 more than a third number . the ratio of the two numbers is : <o> a ) num__2 : num__5 <o> b ) num__3 : num__5 <o> c ) num__4 : num__5 <o> d ) num__6 : num__7 <o> e ) num__7 : num__8 |
explanation : let the third number be x . then first number = num__120.0 of x = num__120 x / num__100 = num__6 x / num__5 second number = num__150.0 of x = num__150 x / num__100 = num__3 x / num__2 ratio of first two numbers = ( num__6 x / num__5 - num__3 x / num__2 ) = num__12 x : num__15 x = num__4 : num__5 . answer is c <eor> c <eos> |
c |
subtract__120.0__20.0__ divide__120.0__20.0__ divide__100.0__20.0__ add__50.0__100.0__ divide__150.0__50.0__ divide__100.0__50.0__ multiply__2.0__6.0__ subtract__20.0__5.0__ divide__20.0__5.0__ divide__20.0__5.0__ |
subtract__120.0__20.0__ divide__120.0__20.0__ divide__100.0__20.0__ add__50.0__100.0__ divide__150.0__50.0__ divide__100.0__50.0__ multiply__2.0__6.0__ subtract__20.0__5.0__ divide__20.0__5.0__ divide__20.0__5.0__ |
| the ratio between the present ages of a and b is num__5 : num__3 respectively . the ratio between a ' s age num__4 years ago and b ' s age num__4 years hence is num__1 : num__1 . what is the ratio between a ' s age num__4 years hence and b ' s age num__4 years ago ? <o> a ) num__1 : num__3 <o> b ) num__2 : num__1 <o> c ) num__3 : num__1 <o> d ) num__4 : num__1 <o> e ) none of these |
let the present ages of a and b be num__5 x and num__3 x years respectively . then ( num__5 x - num__4 ) / ( num__3 x + num__4 ) = num__1.0 num__2 x = num__8 = > x = num__4 required ratio = ( num__5 x + num__4 ) : ( num__3 x - num__4 ) = num__24 : num__8 = num__3 : num__1 answer : c <eor> c <eos> |
c |
subtract__5.0__3.0__ add__5.0__3.0__ multiply__3.0__8.0__ subtract__5.0__2.0__ |
subtract__5.0__3.0__ add__5.0__3.0__ multiply__3.0__8.0__ subtract__5.0__2.0__ |
| the population of a bacteria colony doubles every day . if it was started num__8 days ago with num__4 bacteria and each bacteria lives for num__12 days how large is the colony today ? <o> a ) num__512 <o> b ) num__768 <o> c ) num__1024 <o> d ) num__131072 <o> e ) num__409600 |
num__4 ^ num__8 ( num__2 ) = num__131072 the answer is d . <eor> d <eos> |
d |
divide__8.0__4.0__ round__131072.0__ |
divide__8.0__4.0__ round__131072.0__ |
| an article is bought for rs . num__675 and sold for rs . num__900 find the gain percent ? <o> a ) num__33 num__0.142857142857 % <o> b ) num__33 num__0.0 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__37 num__0.333333333333 % <o> e ) num__32 num__0.333333333333 % |
num__675 - - - - num__225 num__100 - - - - ? = > num__33 num__0.333333333333 % answer : c <eor> c <eos> |
c |
percent__100.0__33.0__ |
percent__100.0__33.0__ |
| the average age of a group of num__10 students is num__14 years . if num__5 more students join the group the average age rises by num__1 year . the average age of the new students is : <o> a ) num__22 <o> b ) num__38 <o> c ) num__11 <o> d ) num__17 <o> e ) num__91 |
explanation : total age of the num__10 students = num__10 × num__14 = num__140 total age of num__15 students including the newly joined num__5 students = num__15 × num__15 = num__225 total age of the new students = num__225 − num__140 = num__85 average age = num__17.0 = num__17 years answer : d <eor> d <eos> |
d |
multiply__10.0__14.0__ add__10.0__5.0__ subtract__225.0__140.0__ divide__85.0__5.0__ multiply__1.0__17.0__ |
multiply__10.0__14.0__ add__10.0__5.0__ subtract__225.0__140.0__ divide__85.0__5.0__ multiply__1.0__17.0__ |
| a big container is num__30.0 full with water . if num__18 liters of water is added the container becomes num__0.75 full . what is the capacity of the big container ? <o> a ) num__36 liters <o> b ) num__40 liters <o> c ) num__45 liters <o> d ) num__54 liters <o> e ) num__60 liters |
a big container is num__30.0 full with water and after num__18 liters of water is added the container becomes num__75.0 full . hence these num__18 liters account for num__45.0 of the container which means that the capacity of it is num__18 / num__0.45 = num__40 liters . or : if the capacity of the container is x liters then : num__0.3 x + num__18 = num__0.75 x - - > x = num__40 liters . answer : b . <eor> b <eos> |
b |
subtract__75.0__30.0__ divide__30.0__0.75__ subtract__0.75__0.45__ divide__30.0__0.75__ |
subtract__75.0__30.0__ divide__30.0__0.75__ subtract__0.75__0.45__ divide__30.0__0.75__ |
| the average of first num__10 odd numbers is ? <o> a ) num__44 <o> b ) num__10 <o> c ) num__99 <o> d ) num__77 <o> e ) num__62 |
sum of num__10 odd no . = num__100 average = num__10.0 = num__10 answer : b <eor> b <eos> |
b |
divide__100.0__10.0__ |
divide__100.0__10.0__ |
| there are num__24 students in chad â € ™ s class . he brought num__26 cookies to pass out for his birthday . how many cookies will each student get ? will there be any cookies left over ? <o> a ) num__1 - num__9 <o> b ) num__1 - num__7 <o> c ) num__2 - num__8 <o> d ) num__1 - num__2 <o> e ) num__3 - num__10 |
num__1.08333333333 = num__1 r num__2 chad will give each student num__1 cookie and there will be num__2 cookies left over correct answer d <eor> d <eos> |
d |
divide__26.0__24.0__ subtract__26.0__24.0__ round__1.0__ |
divide__26.0__24.0__ subtract__26.0__24.0__ round__1.0__ |
| if a @ b = a * b ^ ( num__0.5 ) then num__4 @ num__9 = ? self made <o> a ) num__12 <o> b ) num__6 <o> c ) num__3 <o> d ) num__8 <o> e ) num__4 |
a @ b = a * b ^ ( num__0.5 ) num__4 @ num__9 = num__4 * num__9 ^ ( num__0.5 ) = num__4 * num__3 = num__12 correct option : a <eor> a <eos> |
a |
multiply__4.0__3.0__ multiply__4.0__3.0__ |
multiply__4.0__3.0__ multiply__4.0__3.0__ |
| if n is the product of the integers from num__1 to num__20 inclusive what is the greatest integer k for which num__2 k num__2 k is a factor of n ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__15 <o> d ) num__18 <o> e ) num__20 |
there are num__10 numbers divisible by num__2 there are num__5 numbers divisible by num__4 there are num__2 numbers divisible by num__8 there is num__1 number divisible by num__16 . hence the total number of num__2 ’ s in num__20 ! are num__10 + num__5 + num__2 + num__1 = num__18 answer : d <eor> d <eos> |
d |
divide__20.0__2.0__ divide__10.0__2.0__ divide__20.0__5.0__ multiply__2.0__4.0__ subtract__20.0__4.0__ subtract__20.0__2.0__ multiply__1.0__18.0__ |
divide__20.0__2.0__ divide__10.0__2.0__ divide__20.0__5.0__ multiply__2.0__4.0__ subtract__20.0__4.0__ add__2.0__16.0__ add__2.0__16.0__ |
| a batch of widgets costs p + num__15 dollars for a company to produce and each batch sells for p ( num__11 – p ) dollars . for which of the following values of p does the company make a profit ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
to make profit selling price should be greater than cost price p ( num__11 - p ) - p - num__15 > num__0 = > p ^ num__2 - num__10 * p + num__15 < num__0 hence p = num__8 a <eor> a <eos> |
a |
subtract__10.0__2.0__ subtract__10.0__2.0__ |
subtract__10.0__2.0__ subtract__10.0__2.0__ |
| a certain automobile company ’ s best - selling model is the speedster . the speedster like all of their other models comes in coupe and convertible styles . num__0.75 of the current inventory is speedsters of which num__0.6 are convertibles . if there are num__30 vehicles that are not speedsters how many speedster convertibles are there ? <o> a ) num__30 <o> b ) num__54 <o> c ) num__60 <o> d ) num__70 <o> e ) num__90 |
total vehicle = num__0.75 of speedster + num__0.25 of others . speedster convertibles = num__0.75 total vehicle * num__0.6 given : num__0.25 constitutes num__30 vehicles . hence num__0.75 constitutes num__90 speedster convertibls = num__90 * num__0.6 = num__54 b <eor> b <eos> |
b |
multiply__0.6__90.0__ round__54.0__ |
multiply__0.6__90.0__ multiply__0.6__90.0__ |
| a train passes a man standing on the platform . if the train is num__150 meters long and its speed is num__72 kmph how much time it took in doing so ? <o> a ) num__6 Â ½ sec <o> b ) num__6 Â ½ sec <o> c ) num__9 Â ½ sec <o> d ) num__8 Â ½ sec <o> e ) num__7 Â ½ sec |
e num__7 Â ½ sec s = num__72 * num__0.277777777778 = num__20 mps t = num__7.5 = num__7 Â ½ sec <eor> e <eos> |
e |
divide__150.0__20.0__ round__7.0__ |
divide__150.0__20.0__ round__7.0__ |
| the average score of a cricketer in num__2 matches is num__30 and in other num__3 matches is num__40 . then find the average score in all the num__5 matches ? <o> a ) num__25 <o> b ) num__27 <o> c ) num__30 <o> d ) num__35 <o> e ) num__36 |
average in num__5 matches = ( num__2 * num__30 + num__3 * num__40 ) / num__2 + num__3 = num__60 + num__24.0 = num__36.0 = num__36 answer is e <eor> e <eos> |
e |
multiply__2.0__30.0__ subtract__60.0__24.0__ subtract__60.0__24.0__ |
multiply__2.0__30.0__ subtract__60.0__24.0__ subtract__60.0__24.0__ |
| find missing number - num__5 num__1 - num__11 num__23 num__29 num__17 - num__20 - num__14 ? <o> a ) - num__26 <o> b ) - num__27 <o> c ) - num__28 <o> d ) - num__29 <o> e ) - num__30 |
( - num__5 ) + num__6 = num__1 num__23 + num__6 = num__29 ( - num__20 ) + num__6 = ( - num__14 ) num__1 - num__12 = - num__11 num__29 - num__12 = num__17 ( - num__14 ) - num__12 = ( - num__26 ) ans is - num__26 answer : a <eor> a <eos> |
a |
add__5.0__1.0__ add__1.0__11.0__ add__20.0__6.0__ multiply__1.0__26.0__ |
add__5.0__1.0__ add__1.0__11.0__ add__20.0__6.0__ add__20.0__6.0__ |
| a vantakes num__6 hours to cover a distance of num__540 km . how much should the speed in kmph be maintained to cover the same direction in num__1.5 th of the previous time ? <o> a ) num__60 kmph <o> b ) num__61 kmph <o> c ) num__62 kmph <o> d ) num__64 kmph <o> e ) num__66 kmph |
time = num__6 distence = num__540 num__1.5 of num__6 hours = num__6 * num__1.5 = num__9 hours required speed = num__60.0 = num__60 kmph a ) <eor> a <eos> |
a |
multiply__6.0__1.5__ hour_to_min_conversion__ hour_to_min_conversion__ |
multiply__6.0__1.5__ hour_to_min_conversion__ hour_to_min_conversion__ |
| if a / b = num__1.25 then ( num__4 a + num__3 b ) / ( num__4 a - num__3 b ) = ? <o> a ) num__7 <o> b ) num__6 <o> c ) num__3 <o> d ) num__5 <o> e ) num__4 |
answer dividing numerator as well as denominator by b we get given exp . = ( num__4 a + num__3 b ) / ( num__4 a - num__3 b ) = ( num__4 a / b + num__3 ) / ( num__4 a / b - num__3 ) since a / b = num__1.25 this implies that = [ ( num__4 * num__5 ) / num__4 + num__3 ] / [ ( num__4 * num__5 ) / num__4 - num__3 ) ] = ( num__5 + num__3 ) / ( num__5 - num__3 ) = num__4 option : e <eor> e <eos> |
e |
multiply__1.25__4.0__ divide__5.0__1.25__ |
multiply__1.25__4.0__ divide__5.0__1.25__ |
| ( num__0.756 x num__0.75 ) terms of rate percent is equivalent to ? <o> a ) num__18.9 <o> b ) num__37.8 <o> c ) num__56.7 <o> d ) num__75.0 <o> e ) none |
answer ( num__0.756 x num__0.75 ) = ( num__0.756 ) x ( num__0.75 ) x num__100.0 = num__56.7 correct option : c <eor> c <eos> |
c |
percent__100.0__56.7__ |
percent__100.0__56.7__ |
| the parameter of a square is equal to the perimeter of a rectangle of length num__16 cm and breadth num__14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) <o> a ) num__23.53 cm <o> b ) num__23.52 cm <o> c ) num__23.51 cm <o> d ) num__23.37 cm <o> e ) num__23.57 cm |
explanation : let the side of the square be a cm . parameter of the rectangle = num__2 ( num__16 + num__14 ) = num__60 cm parameter of the square = num__60 cm i . e . num__4 a = num__60 a = num__15 diameter of the semicircle = num__15 cm circimference of the semicircle = num__0.5 ( ∏ ) ( num__15 ) = num__0.5 ( num__3.14285714286 ) ( num__15 ) = num__23.5714285714 = num__23.57 cm to two decimal places answer : option e <eor> e <eos> |
e |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
| how many days are there in x weeks x days ? <o> a ) num__8 x <o> b ) num__14 x <o> c ) num__7 <o> d ) num__7 x <o> e ) num__8 |
x weeks x days = ( num__7 x + x ) days = num__8 x days . answer is a . <eor> a <eos> |
a |
round__8.0__ |
round__8.0__ |
| the area of a square is equal to five times the area of a rectangle of dimensions num__32 cm * num__10 cm . what is the perimeter of the square ? <o> a ) num__289 cm <o> b ) num__160 cm <o> c ) num__829 cm <o> d ) num__288 cm <o> e ) num__289 cm |
area of the square = s * s = num__5 ( num__32 * num__10 ) = num__1600 = > s = num__40 = num__40 cm perimeter of the square = num__4 * num__40 = num__160 cm . answer : b <eor> b <eos> |
b |
volume_rectangular_prism__32.0__10.0__5.0__ square_perimeter__10.0__ square_perimeter__40.0__ square_perimeter__40.0__ |
volume_rectangular_prism__32.0__10.0__5.0__ square_perimeter__10.0__ multiply__32.0__5.0__ multiply__32.0__5.0__ |
| the average weight of num__10 men is increased by num__1 ½ kg when one of the men who weighs num__58 kg is replaced by a new man . what is the weight of the new man ? <o> a ) num__80 kg <o> b ) num__73 kg <o> c ) num__70 kg <o> d ) num__75 kg <o> e ) num__85 kg |
since the average has increased by num__1.5 kg the weight of the man who stepped in must be equal to num__58 + num__10 x num__1.5 num__58 + num__15 = num__73 kg ans : ' b ' <eor> b <eos> |
b |
multiply__10.0__1.5__ add__58.0__15.0__ multiply__1.0__73.0__ |
multiply__10.0__1.5__ add__58.0__15.0__ add__58.0__15.0__ |
| a cycle is bought for rs . num__900 and sold for rs . num__1160 find the gain percent ? <o> a ) num__11 <o> b ) num__29 <o> c ) num__99 <o> d ) num__77 <o> e ) num__18 |
num__900 - - - - num__260 num__100 - - - - ? = > num__29.0 answer : b <eor> b <eos> |
b |
percent__29.0__100.0__ |
percent__29.0__100.0__ |
| the ratio of male to female in a class is num__2 : num__7 . the career preference of the students in the class are to be represented in a circle graph . if the area of the graph allocated to each career preference is to be proportional to the number of students who have that career preference how many degrees of the circle should be used to represent a career that is preferred by one third of the males and two - third of the females in the class ? <o> a ) a ) num__160 degree <o> b ) b ) num__168 degree <o> c ) c ) num__191 degree <o> d ) d ) num__192 degree <o> e ) e ) num__213 degree |
here is my approach = > males = > num__2 x and females = num__7 x = > total = num__9 x now num__9 x = > num__360 therefore num__16 x / num__3 = > num__213 degree . p . s = > num__16 x / num__3 is nothing but total number of students with the given preference answer e <eor> e <eos> |
e |
triangle_area__2.0__213.0__ |
triangle_area__2.0__213.0__ |
| two trains num__100 meters and num__120 meters long are running in the same direction with speeds of num__72 km / hr num__54 km / hr in how much time will the first train cross the second <o> a ) num__36 sec <o> b ) num__40 sec <o> c ) num__44 sec <o> d ) num__46 sec <o> e ) num__48 sec |
explanation : relative speed of the trains = ( num__72 - num__54 ) km / hr = num__18 km / hr = ( num__18 × num__0.277777777778 ) m / sec = num__5 m / sec . time taken by the trains to cross each other = time taken to cover ( num__100 + num__120 ) m at num__5 m / sec = ( num__44.0 ) sec = num__44 sec . answer : option c <eor> c <eos> |
c |
subtract__72.0__54.0__ round__44.0__ |
subtract__72.0__54.0__ round__44.0__ |
| what is the remainder when num__135 ^ num__77 is divided by num__7 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
( num__15 ^ num__77 * num__9 ^ num__77 ) / num__7 ; ( num__15 ^ num__77 ) / num__7 * ( num__9 ^ num__77 ) / num__7 ; num__1 * num__2 ^ num__11.0 ; ( ( num__2 ^ num__3 ) ^ num__25 * num__2 ^ num__2 ) / num__7 ; num__1 * num__4 = num__4 answer : e <eor> e <eos> |
e |
divide__135.0__15.0__ subtract__9.0__7.0__ divide__77.0__7.0__ add__1.0__2.0__ subtract__7.0__3.0__ subtract__7.0__3.0__ |
divide__135.0__15.0__ subtract__9.0__7.0__ divide__77.0__7.0__ add__1.0__2.0__ subtract__7.0__3.0__ divide__4.0__1.0__ |
| walking num__0.857142857143 th of his usual speed a man is num__12 minutes too late . what is the usual time taken by him to cover that distance ? <o> a ) num__1 hour num__10 minutes <o> b ) num__1 hour num__82 minutes <o> c ) num__2 hour num__12 minutes <o> d ) num__1 hour num__12 minutes <o> e ) num__1 hour num__62 minutes |
new speed = num__0.857142857143 of usual speed speed and time are inversely proportional . hence new time = num__1.16666666667 of usual time hence num__1.16666666667 of usual time - usual time = num__12 minutes = > num__0.166666666667 of usual time = num__12 minutes = > usual time = num__12 x num__6 = num__72 minutes = num__1 hour num__12 minutes answer : d <eor> d <eos> |
d |
multiply__12.0__6.0__ multiply__0.8571__1.1667__ round__1.0__ |
multiply__12.0__6.0__ subtract__1.1667__0.1667__ subtract__1.1667__0.1667__ |
| the speed of a boat in upstream is num__60 kmph and the speed of the boat downstream is num__80 kmph . find the speed of the boat in still water and the speed of the stream ? <o> a ) num__76 kmph <o> b ) num__10 kmph <o> c ) num__29 kmph <o> d ) num__25 kmph <o> e ) num__16 kmph |
speed of the boat in still water = ( num__60 + num__80 ) / num__2 = num__70 kmph . speed of the stream = ( num__80 - num__60 ) / num__2 = num__10 kmph . answer : b <eor> b <eos> |
b |
subtract__80.0__70.0__ round__10.0__ |
subtract__80.0__70.0__ subtract__80.0__70.0__ |
| a man can swim in still water at num__3 km / h but takes twice as long to swim upstream than downstream . the speed of the stream is ? <o> a ) num__1 <o> b ) num__4.2 <o> c ) num__5.3 <o> d ) num__1.5 <o> e ) num__5.2 |
m = num__3 s = x ds = num__3 + x us = num__3 - x num__3 + x = ( num__3 - x ) num__2 num__3 + x = num__6 - num__2 x num__3 x = num__3 x = num__1 answer : a <eor> a <eos> |
a |
multiply__3.0__2.0__ subtract__3.0__2.0__ round__1.0__ |
multiply__3.0__2.0__ subtract__3.0__2.0__ subtract__3.0__2.0__ |
| find the lowest common multiple of num__24 num__30 and num__40 . <o> a ) num__360 <o> b ) num__420 <o> c ) num__120 <o> d ) num__320 <o> e ) num__280 |
lcm = num__2 * num__2 * num__2 * num__3 * num__5 = num__120 . answer is c <eor> c <eos> |
c |
add__2.0__3.0__ multiply__24.0__5.0__ multiply__24.0__5.0__ |
add__2.0__3.0__ multiply__24.0__5.0__ multiply__24.0__5.0__ |
| a man can row num__3 kmph in still water . when the river is running at num__1.2 kmph it takes him num__1 hour to row to a place and black . what is the total distance traveled by the man ? <o> a ) num__2.9 <o> b ) num__2.8 <o> c ) num__2.4 <o> d ) num__2.8 <o> e ) num__2.2 |
m = num__3 s = num__1.2 ds = num__3.6 us = num__1.8 x / num__3.6 + x / num__1.8 = num__1 x = num__1.2 d = num__1.2 * num__2 = num__2.4 answer : c <eor> c <eos> |
c |
multiply__3.0__1.2__ subtract__3.0__1.2__ subtract__3.0__1.0__ multiply__1.2__2.0__ round__2.4__ |
multiply__3.0__1.2__ subtract__3.0__1.2__ divide__3.6__1.8__ multiply__1.2__2.0__ multiply__1.2__2.0__ |
| pipe a and pipe b fill water into a tank of capacity num__2000 litres at a rate of num__200 l / min and num__50 l / min . pipe c drains at a rate of num__25 l / min . pipe a is open for num__1 min and closed then pipe b is open for num__2 min and closed . further the pipe c is opened and drained for another num__2 min . this process is repeated until the tank is filled . how long will it take to fill the tank ? <o> a ) num__14 min <o> b ) num__18 min <o> c ) num__25 min <o> d ) num__32 min <o> e ) num__40 min |
tank capacity : num__2000 l num__1 st - num__200 l / min for num__1 min volume filled : num__200 l num__2 nd - num__100 l / min for num__2 min volume filled : num__100 l num__3 rd ( water draining ) : num__25 l / min * num__2 : num__50 l total : ( num__200 + num__100 ) - num__50 = num__250 l filled for num__1 cycle number of num__250 in num__2000 l tank : num__8.0 = num__8 time taken to fill : num__8 * total time = num__8 * num__5 = num__40 ( option e ) <eor> e <eos> |
e |
divide__200.0__2.0__ add__1.0__2.0__ add__200.0__50.0__ divide__2000.0__250.0__ add__2.0__3.0__ divide__2000.0__50.0__ round__40.0__ |
divide__200.0__2.0__ add__1.0__2.0__ add__200.0__50.0__ divide__2000.0__250.0__ add__2.0__3.0__ divide__2000.0__50.0__ divide__2000.0__50.0__ |
| num__2 pipes p num__1 and p num__2 together can fill a tank in num__6 hrs and p num__3 alone can fill the tank in num__10 hrs if they work together what is the time taken to fill the tank ? <o> a ) num__1.75 hrs <o> b ) num__2.25 hrs <o> c ) num__3.25 hrs <o> d ) num__3.75 hrs <o> e ) num__4.0 hrs |
explanation : ( p num__1 + p num__2 ) can fill num__0.166666666667 tank in num__1 hr p num__3 can fill num__0.1 tank in num__1 hr ( p num__1 + p num__2 + p num__3 ) can fill tank ( num__0.166666666667 + num__0.1 ) = num__0.266666666667 = > num__3 pipes can fill the tank in num__3.75 hrs answer : option d <eor> d <eos> |
d |
divide__1.0__6.0__ divide__1.0__10.0__ add__0.1__0.1667__ round__3.75__ |
divide__1.0__6.0__ divide__1.0__10.0__ add__0.1__0.1667__ round__3.75__ |
| the total of the ages of jayant prem and saransh is num__75 years . ten years ago the ratio of their ages was num__2 : num__3 : num__4 . what is the present age of jayant ? <o> a ) num__20 years <o> b ) num__32 years <o> c ) num__34 years <o> d ) num__38 years <o> e ) none |
solution let the ages of jayant prem and saransh num__10 years ago be num__2 x num__3 x and num__4 x years respectively . then ( num__2 x + num__10 ) + ( num__3 x + num__10 ) + ( num__4 x + num__10 ) = num__75 . ‹ = › num__9 x = num__45 x = num__5 . ∴ jayant ' s present age = num__2 x + num__10 = num__20 years . answer a <eor> a <eos> |
a |
add__2.0__3.0__ multiply__2.0__10.0__ multiply__2.0__10.0__ |
add__2.0__3.0__ multiply__2.0__10.0__ multiply__2.0__10.0__ |
| two trains are moving at num__50 kmph and num__70 kmph in opposite directions . their lengths are num__150 m and num__100 m respectively . the time they will take to pass each other completely is ? <o> a ) num__7 num__0.142857142857 sec <o> b ) num__7 num__3.5 sec <o> c ) num__7 num__0.125 sec <o> d ) num__7 num__0.5 sec <o> e ) num__7 num__1.5 sec |
num__70 + num__50 = num__120 * num__0.277777777778 = num__33.3333333333 mps d = num__150 + num__100 = num__250 m t = num__250 * num__0.03 = num__7.5 = num__7 num__0.5 sec answer : d <eor> d <eos> |
d |
add__50.0__70.0__ add__150.0__100.0__ multiply__0.03__250.0__ divide__50.0__100.0__ round__7.0__ |
add__50.0__70.0__ add__150.0__100.0__ multiply__0.03__250.0__ divide__50.0__100.0__ round__7.0__ |
| the length of the bridge which a train num__130 meters long and travelling at num__45 km / hr can cross in num__30 seconds is <o> a ) num__288 <o> b ) num__278 <o> c ) num__245 <o> d ) num__397 <o> e ) num__252 |
speed = ( num__45 * num__0.277777777778 ) m / sec = ( num__12.5 ) m / sec . time = num__30 sec . let the length of bridge be x meters . then ( num__130 + x ) / num__30 = num__12.5 = = > num__2 ( num__130 + x ) = num__750 = = > x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| if | z | / w = num__2 which of the following must be true ? <o> a ) z = - num__2 w <o> b ) z = num__2 w <o> c ) z ^ num__2 = num__4 w ^ num__2 <o> d ) z ^ num__2 = num__4 w ^ num__3 <o> e ) z ^ num__3 = num__8 w ^ num__3 |
| z | / w = num__2 . | z | = num__2 w . then z = num__2 w or - num__2 w and so z ^ num__2 = num__4 w ^ num__2 . the answer is c . <eor> c <eos> |
c |
subtract__4.0__2.0__ |
subtract__4.0__2.0__ |
| the speed of a train is num__90 kmph . what is the distance covered by it in num__10 minutes ? <o> a ) num__15 kmph <o> b ) num__17 kmph <o> c ) num__75 kmph <o> d ) num__18 kmph <o> e ) num__87 kmph |
num__90 * num__0.166666666667 = num__15 kmph answer : a <eor> a <eos> |
a |
round__15.0__ |
round__15.0__ |
| following an increase in prices the price of a candy box was num__10 pounds and the price of a can of soda was num__6 pounds . if the price of a candy box was raised by num__25.0 and the price of a can of soda was raised by num__50.0 . what was the price of a box of candy plus a can of soda before prices were raised ? <o> a ) num__11 . <o> b ) num__12 <o> c ) num__13 <o> d ) num__14 <o> e ) num__14.5 |
let the candy and soda before the increase be c and s respectively so increased price of can = c ( num__1 + num__0.25 ) = num__10 - - - - - - - - - - - > c = num__8 increase price of soda : s ( num__1 + num__0.5 ) = num__6 - - - - - - - - - - - > s = num__4 hence total = num__8 + num__4 = num__12 . . hence answer : b <eor> b <eos> |
b |
divide__25.0__50.0__ reverse__0.25__ divide__6.0__0.5__ divide__6.0__0.5__ |
divide__25.0__50.0__ subtract__10.0__6.0__ add__4.0__8.0__ add__4.0__8.0__ |
| bob bikes to school every day at a steady rate of d miles per hour . on a particular day bob had a flat tire exactly halfway to school . he immediately started walking to school at a steady pace of y miles per hour . he arrived at school exactly t hours after leaving his home . how many miles is it from the school to bob ' s home ? <o> a ) ( d + y ) / t <o> b ) num__2 ( d + t ) / dy <o> c ) num__2 dyt / ( d + y ) <o> d ) num__2 ( d + y + t ) / dy <o> e ) x ( y + t ) + y ( x + t ) |
if we choose for d ( distance ) : num__10 miles for d num__10 and for y num__5 . t would be num__90 minutes or num__15 hours . if i try this for answer choice c it fits . <eor> c <eos> |
c |
add__10.0__5.0__ divide__10.0__5.0__ |
add__10.0__5.0__ divide__10.0__5.0__ |
| excluding stoppages the speed of a bus is num__54 kmph and including stoppages it is num__45 kmph . for how many minutes does the bus stop per hour ? <o> a ) num__12 <o> b ) num__10 <o> c ) num__11 <o> d ) num__9 <o> e ) num__8 |
explanation : speed of the bus excluding stoppages = num__54 kmph speed of the bus including stoppages = num__45 kmph loss in speed when including stoppages = num__54 - num__45 = num__9 kmph = > in num__1 hour bus covers num__9 km less due to stoppages hence time that the bus stop per hour = time taken to cover num__9 km = distance / speed = num__0.166666666667 hour = num__0.166666666667 hour = num__10.0 min = num__10 min answer : c <eor> c <eos> |
c |
subtract__54.0__45.0__ divide__9.0__54.0__ add__1.0__9.0__ add__1.0__10.0__ |
subtract__54.0__45.0__ divide__9.0__54.0__ add__1.0__9.0__ add__1.0__10.0__ |
| a train covers a distance of num__12 km in num__10 min . if it takes num__4 sec to pass a telegraph post then the length of the train is ? <o> a ) num__20 <o> b ) num__110 <o> c ) num__120 <o> d ) num__80 <o> e ) num__60 |
speed = ( num__1.2 * num__60 ) km / hr = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . length of the train = num__20 * num__4 = num__80 m . answer : option d <eor> d <eos> |
d |
divide__12.0__10.0__ hour_to_min_conversion__ add__12.0__60.0__ multiply__4.0__20.0__ round__80.0__ |
divide__12.0__10.0__ hour_to_min_conversion__ multiply__1.2__60.0__ multiply__4.0__20.0__ multiply__4.0__20.0__ |
| a factory makes num__134341 jam packets which need to be placed in boxes and each box can contain at most num__9 packets . how many packets will be in the last unfilled box ? <o> a ) num__1 packet <o> b ) num__2 packets <o> c ) num__3 packets <o> d ) num__4 packets <o> e ) num__5 packets |
in order to divide the sum in num__9 parts the amount must be divisible by num__9 divisibility rule of num__9 : the sum of the digits must be divisible by num__9 sum of digits of num__134341 = num__16 and num__9 is divisible by num__9 . hence we need to add num__2 to this number for it to be divisible by num__9 correct option : b <eor> b <eos> |
b |
gcd__16.0__2.0__ |
gcd__16.0__2.0__ |
| a person incurs a loss of num__5.0 be selling a watch for rs . num__1140 . at what price should the watch be sold to earn num__5.0 profit <o> a ) rs . num__1200 <o> b ) rs . num__1230 <o> c ) rs . num__1260 <o> d ) rs . num__1290 <o> e ) none of these |
explanation : let the new s . p . be x then . ( num__100 - loss % ) : ( num__1 st s . p . ) = ( num__100 + gain % ) : ( num__2 nd s . p . ) = > ( num__0.0833333333333 = num__105 / x ) = > x = num__1260 option c <eor> c <eos> |
c |
percent__100.0__1260.0__ |
percent__100.0__1260.0__ |
| if two numbers are in the ratio num__4 : num__3 . if num__10 is added to both of the numbers then the ratio becomes num__5 : num__4 then find the biggest number . <o> a ) num__60 <o> b ) num__80 <o> c ) num__70 <o> d ) num__40 <o> e ) num__100 |
num__4 : num__3 num__4 x + num__20 : num__3 x + num__20 = num__5 : num__4 num__4 [ num__4 x + num__20 ] = num__5 [ num__3 x + num__20 ] num__16 x + num__80 = num__15 x + num__100 num__16 x - num__15 x = num__100 - num__80 x = num__20 then biggest number is = num__4 x num__4 x = num__80 b <eor> b <eos> |
b |
multiply__4.0__5.0__ subtract__20.0__4.0__ multiply__4.0__20.0__ multiply__3.0__5.0__ multiply__5.0__20.0__ multiply__4.0__20.0__ |
multiply__4.0__5.0__ subtract__20.0__4.0__ multiply__4.0__20.0__ add__10.0__5.0__ add__80.0__20.0__ subtract__100.0__20.0__ |
| a man swims downstream num__72 km and upstream num__45 km taking num__9 hours each time ; what is the speed of the current ? <o> a ) num__1.6 <o> b ) num__1.5 <o> c ) num__1.2 <o> d ) num__1.9 <o> e ) num__1.3 |
num__72 - - - num__9 ds = num__8 ? - - - - num__1 num__45 - - - - num__9 us = num__5 ? - - - - num__1 s = ? s = ( num__8 - num__5 ) / num__2 = num__1.5 answer : b <eor> b <eos> |
b |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ round__1.5__ |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ divide__1.5__1.0__ |
| find the number of square tiles to cover the floor of a room measuring num__4 m * num__9 m leaving num__0.25 m space around the room . a side of square tile is given to be num__50 cms ? <o> a ) num__187 <o> b ) num__476 <o> c ) num__119 <o> d ) num__208 <o> e ) num__276 |
area of the room = num__3.5 * num__8.5 = num__29.75 area of the tile = num__0.5 * num__0.5 = num__0.25 no . of tiles = num__29.75 / num__0.25 = num__119 answer : c <eor> c <eos> |
c |
multiply__3.5__8.5__ subtract__4.0__3.5__ multiply__4.0__29.75__ round__119.0__ |
multiply__3.5__8.5__ subtract__4.0__3.5__ multiply__4.0__29.75__ multiply__4.0__29.75__ |
| the ratio between the length and the breadth of a rectangular park is num__3 : num__2 . if a man cycling along the boundary of the park at the speed of num__13 km / hr completes one round in num__8 minutes then the area of the park ( in sq . m ) is <o> a ) num__345600 <o> b ) num__153400 <o> c ) num__153600 <o> d ) num__154000 <o> e ) num__154200 |
perimeter = distance covered in num__8 min . = num__18000 x num__8 m = num__2400 m . num__60 let length = num__3 x metres and breadth = num__2 x metres . then num__2 ( num__3 x + num__2 x ) = num__2400 or x = num__240 . length = num__720 m and breadth = num__480 m . area = ( num__720 x num__480 ) m num__2 = num__345600 m num__2 . a <eor> a <eos> |
a |
hour_to_min_conversion__ multiply__3.0__240.0__ multiply__2.0__240.0__ multiply__720.0__480.0__ round__345600.0__ |
hour_to_min_conversion__ multiply__3.0__240.0__ multiply__2.0__240.0__ multiply__720.0__480.0__ round__345600.0__ |
| if n = num__3 x num__4 x p where p is a prime number greater than num__3 how many different positive non - prime divisors does n have excluding num__1 and n ? <o> a ) six <o> b ) seven <o> c ) eight <o> d ) nine <o> e ) ten |
n = xa ∗ yb where x and y are prime then number of factors of n are given by the formula ( a + num__1 ) ( b + num__1 ) { notice i have simply increased the powers of x and y by one and then multiplied them . also make sure x and y are distinct primes ) so in your question n = num__3 x num__4 x p where p is prime . i can n = num__31 ∗ num__22 ∗ p ^ num__1 or number of factors = ( num__1 + num__1 ) * ( num__2 + num__1 ) * ( num__1 + num__1 ) = num__12 . but notice we need to find non prime and other than num__1 and n so we have three prime numbers ( num__23 p ) num__1 and n . excluding these we have num__12 - num__5 = num__7 such factors . hence answer is b <eor> b <eos> |
b |
subtract__3.0__1.0__ multiply__3.0__4.0__ add__1.0__22.0__ add__3.0__2.0__ add__3.0__4.0__ add__3.0__4.0__ |
subtract__3.0__1.0__ multiply__3.0__4.0__ add__1.0__22.0__ add__3.0__2.0__ add__3.0__4.0__ add__3.0__4.0__ |
| a train num__540 meters long is running with a speed of num__54 kmph . the time taken by it to cross a tunnel num__180 meters long is ? <o> a ) num__66 sec <o> b ) num__46 sec <o> c ) num__48 sec <o> d ) num__65 sec <o> e ) num__64 sec |
d = num__540 + num__180 = num__720 s = num__54 * num__0.277777777778 = num__15 mps t = num__48.0 = num__48 sec answer : c <eor> c <eos> |
c |
add__540.0__180.0__ divide__720.0__15.0__ round__48.0__ |
add__540.0__180.0__ divide__720.0__15.0__ round__48.0__ |
| a doctor prescribed num__18 cubic centimeters of a certain drug to a patient whose body weight was num__120 pounds . if the typical dosage is num__2 cubic centimeters per num__10 pounds of the body weight by what percent was the prescribed dosage lesser than the typical dosage ? <o> a ) num__8.0 <o> b ) - num__37.5 <o> c ) num__11.0 <o> d ) num__12.5 <o> e ) num__14.8 % |
typical dosage per num__10 pound of the body weight = num__2 c . c typical dosage per num__120 pound of the body weight = num__2 * ( num__12.0 ) = num__2 * num__12 = num__24 c . c dosage prescribed by doctor for num__120 pound patient = num__18 c . c % prescribed dosage greater than the typical dosage = ( num__18 - num__1.5 ) * num__100.0 = ( - num__0.375 ) * num__100.0 = - num__37.5 answer b <eor> b <eos> |
b |
divide__120.0__10.0__ multiply__2.0__12.0__ divide__18.0__12.0__ multiply__0.375__100.0__ multiply__0.375__100.0__ |
divide__120.0__10.0__ multiply__2.0__12.0__ divide__18.0__12.0__ multiply__0.375__100.0__ multiply__0.375__100.0__ |
| ganesh ram and sohan together can do a work in num__16 days . if ganesh and ram together can do a the same work in num__24 days then how long will take sohan to do the same work ? <o> a ) num__48 <o> b ) num__58 <o> c ) num__68 <o> d ) num__78 <o> e ) num__100 |
solution : work done by three of them together in num__1 day = num__0.0625 . . . . . . . . . . . . ( num__1 ) work done by ganesh and ram together in num__1 day = num__0.0416666666667 . . . . . ( num__2 ) so work done by sohan in one day = ( num__1 ) - ( num__2 ) = ( num__0.0625 ) - ( num__0.0416666666667 ) = num__0.0208333333333 = > total days taken by sohan to complete the work alone = num__48 days . answer a <eor> a <eos> |
a |
divide__1.0__16.0__ divide__1.0__24.0__ subtract__0.0625__0.0417__ multiply__24.0__2.0__ round__48.0__ |
divide__1.0__16.0__ divide__1.0__24.0__ subtract__0.0625__0.0417__ multiply__24.0__2.0__ round__48.0__ |
| five boys picked up num__30 mangoes . in how many ways can they divide them if all mangoes be identical ? <o> a ) a ) num__4 ^ num__30 <o> b ) b ) num__2 ^ num__30 <o> c ) c ) num__1 ^ num__30 <o> d ) d ) num__3 ^ num__30 <o> e ) e ) num__5 ^ num__30 |
each mango can be given to any one of the four people or in other words . . num__1 mango can be divided into num__5 ways . . . so all num__30 can be divided in num__5 ^ num__30 ways ans : e <eor> e <eos> |
e |
vowel_space__ vowel_space__ |
vowel_space__ vowel_space__ |
| the prices of tea and coffee per kg were the same in june . in july the price of coffee shot up by num__20.0 and that of tea dropped by num__20.0 . if in july a mixture containing equal quantities of tea and coffee costs num__40 / kg . how much did a kg of coffee cost in june ? <o> a ) num__40 <o> b ) num__60 <o> c ) num__80 <o> d ) num__100 <o> e ) num__120 |
let the price of tea and coffee be x per kg in june . price of tea in july = num__1.2 x price of coffee in july = num__0.8 x . in july the price of num__0.5 kg ( num__400 gm ) of tea and num__0.5 kg ( num__400 gm ) of coffee ( equal quantities ) = num__40 num__1.2 x ( num__0.5 ) + num__0.8 x ( num__0.5 ) = num__40 = > x = num__40 thus proved . . . option a . <eor> a <eos> |
a |
divide__20.0__40.0__ divide__20.0__0.5__ |
divide__20.0__40.0__ divide__20.0__0.5__ |
| a farmer has an apple orchard consisting of fuji and gala apple trees . due to high winds this year num__10.0 of his trees cross pollinated . the number of his trees that are pure fuji plus the cross - pollinated ones totals num__136 while num__0.75 of all his trees are pure fuji . how many of his trees are pure gala ? <o> a ) num__24 <o> b ) num__33 <o> c ) num__55 <o> d ) num__77 <o> e ) num__88 |
let f = pure fuji g = pure gala and c - cross pollinated . c = num__10.0 of x where x is total trees . c = . num__1 x also num__3 x / num__4 = f and c + f = num__136 = > . num__1 x + num__0.75 x = num__136 = > x = num__160 num__160 - num__136 = pure gala = num__24 . a <eor> a <eos> |
a |
add__1.0__3.0__ subtract__160.0__136.0__ multiply__1.0__24.0__ |
add__1.0__3.0__ subtract__160.0__136.0__ subtract__160.0__136.0__ |
| simplify : num__100 x num__100 - num__50 x num__50 <o> a ) num__7612 <o> b ) num__7614 <o> c ) num__7500 <o> d ) num__7615 <o> e ) none of them |
( num__100 ) ^ num__2 - ( num__50 ) ^ num__2 = ( num__100 + num__50 ) ( num__100 - num__50 ) = num__150 x num__50 = num__7500 . answer is c . <eor> c <eos> |
c |
divide__100.0__50.0__ add__100.0__50.0__ multiply__50.0__150.0__ multiply__50.0__150.0__ |
divide__100.0__50.0__ add__100.0__50.0__ multiply__50.0__150.0__ multiply__50.0__150.0__ |
| bob wants to run a mile in the same time as his sister . if bob ’ s time for a mile is currently num__10 minutes and his sister ’ s time is currently num__8 minutes num__30 seconds by what percent does bob need to improve his time in order run a mile in the same time as his sister ? <o> a ) num__3.0 <o> b ) num__15.0 <o> c ) num__8.0 <o> d ) num__10.0 <o> e ) num__12 % |
bob ' s time = num__600 secs . his sis ' time = num__510 secs . percent increase needed = ( num__640 - num__0.85 ) * num__100 = num__0.15 * num__100 = num__15.0 . ans ( b ) . <eor> b <eos> |
b |
divide__510.0__600.0__ multiply__100.0__0.15__ round__15.0__ |
divide__510.0__600.0__ multiply__100.0__0.15__ subtract__30.0__15.0__ |
| an unbiased cubic die is thrown . what is the probabiltiy of getting a multiple of num__3 or num__4 ? <o> a ) num__0.0833333333333 <o> b ) num__0.5 <o> c ) num__0.75 <o> d ) num__0.166666666667 <o> e ) num__0.111111111111 |
s = { num__12 num__34 num__56 } p ( e ) = n ( e ) / n ( s ) n ( s ) = total numbers in a die = num__6 let e be the event of getting the multiple of num__3 then e = { num__36 } p ( mutliple of num__3 ) = num__0.333333333333 = num__0.333333333333 let e be the event of getting the multiple of num__4 then e = { num__4 } p ( multiple of num__4 ) = num__0.166666666667 p ( multiple of num__3 or num__4 ) = num__0.333333333333 + num__0.166666666667 = num__0.5 ans b <eor> b <eos> |
b |
die_space__ negate_prob__0.5__ |
die_space__ negate_prob__0.5__ |
| rectangular floors x and y have equal area . if floor x is num__18 feet by num__18 feet and floor y is num__9 feet wide what is the length of floor y in feet ? <o> a ) num__13 num__0.5 <o> b ) num__18 <o> c ) num__18 num__0.75 <o> d ) num__21 <o> e ) num__36 |
the area of a rectangle is : area = length x width we are given that floor x is num__12 feet by num__18 feet and that floor y is num__9 feet wide . so we can say : length of x = num__18 width of x = num__18 width of y = num__9 length of y = n we also can say : area of floor x = area of floor y ( length of x ) ( width of x ) = ( length of y ) ( width of y ) ( num__18 ) ( num__18 ) = num__9 n ( num__18 ) ( num__2 ) = n num__36 = n answer e . <eor> e <eos> |
e |
square_perimeter__9.0__ square_perimeter__9.0__ |
square_perimeter__9.0__ square_perimeter__9.0__ |
| find the smallest number in gp whose sum is num__38 and product is num__1728 <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
let x y z be the numbers in geometric progression ? y ^ num__2 = xz x + y + z = num__38 xyz = num__1728 xyz = xzy = y ^ num__2 y = y ^ num__3 = num__1728 y = num__12 y ^ num__2 = xz = num__144 z = num__144 / x x + y + z = x + num__12 + num__144 / x = num__38 x ^ num__2 + num__12 x + num__144 = num__38 x x ^ num__2 - num__26 x + num__144 = num__0 ( x - num__18 ) ( x - num__8 ) = num__0 x = num__818 if x = num__8 z = num__38 - num__8 - num__12 = num__18 the numbers are num__812 num__18 their sum is num__38 their product is num__1728 the smallest number is num__8 answer : d <eor> d <eos> |
d |
divide__1728.0__12.0__ subtract__38.0__12.0__ divide__144.0__18.0__ divide__144.0__18.0__ |
divide__1728.0__12.0__ subtract__38.0__12.0__ divide__144.0__18.0__ divide__144.0__18.0__ |
| if ywusq is num__25 - num__23 - num__21 - num__19 - num__17 then mkigf is <o> a ) num__9 - num__8 - num__7 - num__6 - num__5 <o> b ) num__1 - num__2 - num__3 - num__5 - num__7 <o> c ) num__7 - num__8 - num__4 - num__5 - num__3 <o> d ) num__13 - num__11 - num__9 - num__7 - num__6 <o> e ) none |
in alphabet has num__26 words coresponding as a position num__1 e position num__5 z position num__26 similarly as given in question y = num__25 ; w = num__23 ; u = num__21 ; s = num__19 ; q = num__17 for m = num__13 ; k = num__11 ; i = num__9 and so on answer : d <eor> d <eos> |
d |
subtract__26.0__25.0__ subtract__26.0__21.0__ subtract__26.0__17.0__ multiply__1.0__13.0__ |
subtract__26.0__25.0__ subtract__26.0__21.0__ subtract__26.0__17.0__ multiply__1.0__13.0__ |
| a train moves fast a telegraph post and a bridge num__264 m long in num__8 sec and num__20 sec respectively . what is the speed of the train ? <o> a ) num__79.5 <o> b ) num__79.0 <o> c ) num__79.4 <o> d ) num__79.2 <o> e ) num__79.3 |
let the length of the train be x m and its speed be y m / sec . then x / y = num__8 = > x = num__8 y ( x + num__264 ) / num__20 = y y = num__22 speed = num__22 m / sec = num__22 * num__3.6 = num__79.2 km / hr . answer : d <eor> d <eos> |
d |
multiply__3.6__22.0__ round__79.2__ |
multiply__3.6__22.0__ multiply__3.6__22.0__ |
| alex and brenda both stand at point x . alex begins to walk away from brenda in a straight line at a rate of num__6 miles per hour . one hour later brenda begins to ride a bicycle in a straight line in the opposite direction at a rate of r miles per hour . if r > num__8 which of the following represents the amount of time in terms of r that alex will have been walking when brenda has covered twice as much distance as alex ? <o> a ) r - num__4 <o> b ) r / ( r + num__4 ) <o> c ) r / ( r - num__12 ) <o> d ) num__8 / ( r - num__8 ) <o> e ) num__2 r - num__4 |
let t be the time thatalexwill have been walking when brenda has covered twice as much distance as alex . in t hours alex will cover num__4 t miles ; since brenda begins her journey num__1 hour later than alex then total time for her will be t - num__1 hours and the distance covered in that time will be r ( t - num__1 ) ; we want the distance covered by brenda to be twice as much as that of alex : num__2 * num__6 t = r ( t - num__1 ) - - > num__12 t = rt - r - - > t = r / ( r - num__12 ) . answer : c . <eor> c <eos> |
c |
subtract__6.0__4.0__ multiply__6.0__2.0__ round__12.0__ |
subtract__6.0__4.0__ multiply__6.0__2.0__ multiply__6.0__2.0__ |
| a train covers a distance of num__90 km in num__20 min . if it takes num__9 sec to pass a telegraph post then the length of the train is ? <o> a ) num__675 <o> b ) num__680 <o> c ) num__685 <o> d ) num__690 <o> e ) num__695 |
speed = ( num__4.5 * num__60 ) km / hr = ( num__270 * num__0.277777777778 ) m / sec = num__75 m / sec . length of the train = num__75 * num__9 = num__675 m . answer : option a <eor> a <eos> |
a |
divide__90.0__20.0__ hour_to_min_conversion__ multiply__4.5__60.0__ multiply__9.0__75.0__ round__675.0__ |
divide__90.0__20.0__ hour_to_min_conversion__ multiply__4.5__60.0__ multiply__9.0__75.0__ multiply__9.0__75.0__ |
| two trains of length num__100 m and num__200 m are num__100 m apart . they start moving towards each other on parallel tracks at speeds num__54 kmph and num__72 kmph . after how much time will the trains meet ? <o> a ) num__2.22222222222 sec <o> b ) num__10.0 sec <o> c ) num__2.85714285714 sec <o> d ) num__4.44444444444 sec <o> e ) num__20.0 sec |
they are moving in opposite directions relative speed is equal to the sum of their speeds . relative speed = ( num__54 + num__72 ) * num__0.277777777778 = num__7 * num__5 = num__35 mps . the time required = d / s = num__2.85714285714 = num__2.85714285714 sec . answer : c <eor> c <eos> |
c |
multiply__5.0__7.0__ divide__100.0__35.0__ divide__100.0__35.0__ |
multiply__5.0__7.0__ divide__100.0__35.0__ divide__100.0__35.0__ |
| how much time will a train of length num__200 m moving at a speed of num__72 kmph take to cross another train of length num__300 m moving at num__36 kmph in the same direction ? <o> a ) num__50 sec <o> b ) num__77 <o> c ) num__55 <o> d ) num__44 <o> e ) num__11 |
answer : option a explanation : the distance to be covered = sum of their lengths = num__200 + num__300 = num__500 m . relative speed = num__72 - num__36 = num__36 kmph = num__36 * num__0.277777777778 = num__10 mps . time required = d / s = num__50.0 = num__50 sec . answer : a <eor> a <eos> |
a |
add__200.0__300.0__ divide__500.0__10.0__ round__50.0__ |
add__200.0__300.0__ divide__500.0__10.0__ divide__500.0__10.0__ |
| the average monthly income of p and q is rs . num__5050 . the average monthly income of q and r is rs . num__6250 and the average monthly income of p and r is rs . num__5200 . the monthly income of p is : <o> a ) num__3500 <o> b ) num__4000 <o> c ) num__4050 <o> d ) num__5000 <o> e ) num__6000 |
explanation : let p q and r represent their respective monthly incomes . then we have : p + q = ( num__5050 x num__2 ) = num__10100 . . . . ( i ) q + r = ( num__6250 x num__2 ) = num__12500 . . . . ( ii ) p + r = ( num__5200 x num__2 ) = num__10400 . . . . ( iii ) adding ( i ) ( ii ) and ( iii ) we get : num__2 ( p + q + r ) = num__33000 or p + q + r = num__16500 . . . . ( iv ) subtracting ( ii ) from ( iv ) we get p = num__4000 . p ' s monthly income = rs . num__4000 . answer : b <eor> b <eos> |
b |
multiply__5050.0__2.0__ multiply__6250.0__2.0__ multiply__5200.0__2.0__ divide__33000.0__2.0__ subtract__16500.0__12500.0__ subtract__16500.0__12500.0__ |
multiply__5050.0__2.0__ multiply__6250.0__2.0__ multiply__5200.0__2.0__ divide__33000.0__2.0__ subtract__16500.0__12500.0__ subtract__16500.0__12500.0__ |
| a batsman makes a score of num__50 runs in the num__17 th inning and thus increases his averages by num__2 . what is his average after num__17 th inning ? <o> a ) num__39 <o> b ) num__18 <o> c ) num__42 <o> d ) num__40.5 <o> e ) num__41.5 |
let the average after num__16 th inning = x then total run after num__16 th inning = num__16 x then total run after num__17 th inning = num__16 x + num__50 then average run after num__17 th inning = ( num__16 x + num__50 ) / num__17 ( num__16 x + num__50 ) / num__17 = x + num__2 = > num__16 x + num__50 = num__17 x + num__34 = > x = num__16 x = num__16 ; average after num__17 th inning = num__16 + num__2 = num__18 answer : b <eor> b <eos> |
b |
subtract__50.0__16.0__ add__2.0__16.0__ add__2.0__16.0__ |
subtract__50.0__16.0__ add__2.0__16.0__ add__2.0__16.0__ |
| the average age of a b and c is num__25 years . if the average age of a and c is num__29 years what is the age of b in years ? <o> a ) num__17 <o> b ) num__35 <o> c ) num__20 <o> d ) num__32 <o> e ) num__21 |
age of b = age of ( a + b + c ) â € “ age of ( a + c ) = num__25 Ã — num__3 â € “ num__29 Ã — num__2 = num__75 â € “ num__58 = num__17 years a <eor> a <eos> |
a |
multiply__25.0__3.0__ multiply__29.0__2.0__ subtract__75.0__58.0__ subtract__75.0__58.0__ |
multiply__25.0__3.0__ multiply__29.0__2.0__ subtract__75.0__58.0__ subtract__75.0__58.0__ |
| the population of a town is num__8000 . it decreases annually at the rate of num__10.0 p . a . what was its population num__2 years ago ? <o> a ) num__9876 <o> b ) num__8000 <o> c ) num__8500 <o> d ) num__9500 <o> e ) num__10000 |
formula : ( after = num__100 denominator ago = num__100 numerator ) num__8000 Ã — num__1.11111111111 Ã — num__1.11111111111 = num__9876 a ) <eor> a <eos> |
a |
percent__100.0__9876.0__ |
percent__100.0__9876.0__ |
| the shopkeeper increased the price of a product by num__20.0 so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only num__70.0 of the required amount . what is the net difference in the expenditure on that product ? <o> a ) a ) num__12.5 <o> b ) b ) num__13.0 <o> c ) c ) num__13.15 <o> d ) d ) num__12.0 <o> e ) e ) num__15 % |
quantity x rate = price num__1 x num__1 = num__1 num__0.7 x num__1.20 = num__0.84 decrease in price = ( num__0.12 / num__1 ) × num__100 = num__12.0 d ) <eor> d <eos> |
d |
multiply__1.2__0.7__ divide__70.0__0.7__ multiply__100.0__0.12__ multiply__1.0__12.0__ |
multiply__1.2__0.7__ divide__70.0__0.7__ multiply__100.0__0.12__ divide__12.0__1.0__ |
| a bank issued credit card numbers and the corresponding pin ( personal identification number ) . both are num__3 - digit numbers up to num__996 . pinaki was the last to get the credit card and so he had the last possible credit card number . he was afraid of forgetting his pin . he wrote down the number num__123 in his diary to remember his pin . he also wrote out the way to calculate num__123 : ` ` multiply the card number by pin . divide the product by num__997 . the remainder is num__123 ' ' . once prafull saw his diary in which pinaki wrote this number num__123 . prafull did a lot of purchasing as he now knows pinaki ' s pin . what is pinaki ' s pin ? <o> a ) num__874 <o> b ) num__875 <o> c ) num__876 <o> d ) num__877 <o> e ) none |
explanation : let the pin is x . according to the question the card number = num__996 and remainder = num__123 . thus ( num__996 × x ) / num__997 = num__123 . = > x = num__874 . answer : a <eor> a <eos> |
a |
subtract__997.0__123.0__ subtract__997.0__123.0__ |
subtract__997.0__123.0__ subtract__997.0__123.0__ |
| in a group of dogs and peacocks the number of legs are num__18 less than four times the number of heads how many peacocks are there in that group ? <o> a ) num__9 <o> b ) num__8 <o> c ) num__7 <o> d ) num__6 <o> e ) num__5 |
let the number of dogs be ' x ' and the number of peacocks by ' y ' . then number of legs in the group = num__4 x + num__2 y . number of heads in the group = x + y so num__4 x + num__2 y = num__4 ( x + y ) – num__18 ⇒ num__2 y = num__18 ⇒ y = num__9 number of peacocks in that group = num__9 . answer : a <eor> a <eos> |
a |
divide__18.0__2.0__ subtract__18.0__9.0__ |
divide__18.0__2.0__ subtract__18.0__9.0__ |
| convert the following unit : num__8 hectares in m  ² <o> a ) num__60000 m  ² <o> b ) num__70000 m  ² <o> c ) num__80000 m  ² <o> d ) num__85000 m  ² <o> e ) num__90000 m  ² |
num__8 hectares in m  ² num__1 hectare = num__10000 m  ² therefore num__8 hectares = num__8 à — num__10000 m  ² = num__80000 m  ² answer : option c <eor> c <eos> |
c |
multiply__8.0__10000.0__ multiply__8.0__10000.0__ |
multiply__8.0__10000.0__ multiply__8.0__10000.0__ |
| a man can hit a target once in num__4 shots . if he fires num__4 shots in succession what is the probability that he will hit his target ? <o> a ) num__0.575255102041 <o> b ) num__0.71556122449 <o> c ) num__0.68359375 <o> d ) num__0.461146496815 <o> e ) num__1.12611012433 |
the man will hit the target if he hits it once or twice or thrice or all four times in the four shots that he takes . so the only possibility when the man will not hit the target is when he fails to hit the target in even one of the four shots that he takes . the event of not hitting the target even once is the complement of the event of hitting the target at least once . the probability that he will not hit the target in any given shot = num__1 – num__0.25 = ¾ therefore the probability that he will not hit the target in any of the four shots = num__0.75 * num__0.75 * num__0.75 * num__0.75 = num__0.31640625 the probability that he will hit the target at least in one of the four shots = num__1 - num__0.31640625 = num__0.68359375 ans : c <eor> c <eos> |
c |
negate_prob__0.25__ negate_prob__0.3164__ negate_prob__0.3164__ |
negate_prob__0.25__ negate_prob__0.3164__ negate_prob__0.3164__ |
| a man buy a book in rs num__50 & sale it rs num__50 . what is the rate of profit ? ? ? <o> a ) num__10.0 <o> b ) num__0.0 <o> c ) num__30.0 <o> d ) num__25.0 <o> e ) num__28 % |
cp = num__50 sp = num__50 profit = num__50 - num__50 = num__0.0 = num__0.0 * num__100 = num__0.0 answer : b <eor> b <eos> |
b |
percent__50.0__0.0__ |
percent__50.0__0.0__ |
| tough and tricky questions : combinations . eight contestants representing four different countries advance to the finals of a fencing championship . assuming all competitors have an equal chance of winning how many possibilities are there with respect to how a first - place and second - place medal can be awarded ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__12 <o> d ) num__16 <o> e ) num__24 |
eight contestants representing four different countries advance to the finals of a fencing championship . assuming all competitors have an equal chance of winning how many possibilities are there with respect to how a first - place and second - place medal can be awarded ? we have num__2 slots to be filled using num__8 contestants : num__8 options for slot num__1 * num__3 option for slot num__2 = num__8 * num__3 = num__24 ans . e ) num__24 <eor> e <eos> |
e |
add__1.0__2.0__ multiply__8.0__3.0__ multiply__8.0__3.0__ |
add__1.0__2.0__ multiply__8.0__3.0__ multiply__8.0__3.0__ |
| a man buys a cycle for rs . num__1400 and sells it at a loss of num__15.0 . what is the selling price of the cycle ? <o> a ) rs num__970 <o> b ) rs num__1075 <o> c ) rs num__1190 <o> d ) rs num__1210 <o> e ) rs num__1810 |
s . p . = num__85.0 of rs . num__1400 = rs . num__0.85 x num__1400 = rs . num__1190 answer : c <eor> c <eos> |
c |
percent__85.0__1400.0__ percent__85.0__1400.0__ |
percent__85.0__1400.0__ percent__85.0__1400.0__ |
| an bus covers a certain distance at a speed of num__280 kmph in num__5 hours . to cover the samedistance in num__1 hr it must travel at a speed of ? <o> a ) num__600 km / hr <o> b ) num__720 km / hr <o> c ) num__730 km / hr <o> d ) num__840 km / hr <o> e ) num__760 km / hr |
distance = ( num__280 x num__5 ) = num__1400 km . speed = distance / time speed = num__1400 / ( num__1.66666666667 ) km / hr . [ we can write num__1 hours as num__1.66666666667 hours ] required speed = num__1400 x num__3 km / hr = num__840 km / hr . d <eor> d <eos> |
d |
multiply__280.0__5.0__ multiply__280.0__3.0__ round__840.0__ |
multiply__280.0__5.0__ multiply__280.0__3.0__ divide__840.0__1.0__ |
| ten different letters of alphabet are given words with num__5 letters are formed from these given letters . then the number of words which have at least one letter repeated is : <o> a ) num__69760 <o> b ) num__30240 <o> c ) num__99748 <o> d ) num__42386 <o> e ) none |
solution : number of words which have at least one letter replaced = total number of words - total number of words in which no letter is repeated . = > num__105 – num__16 p num__5 . = > num__100000 − num__30240 = num__69760 . answer : option a <eor> a <eos> |
a |
subtract__100000.0__30240.0__ subtract__100000.0__30240.0__ |
subtract__100000.0__30240.0__ subtract__100000.0__30240.0__ |
| in a survey of political preferences num__78.0 of those asked were in favour of at least one of the proposals : i ii and iii . num__50.0 of those asked favoured proposal i num__30.0 favoured proposal ii and num__20.0 favoured proposal iii . if num__5.0 of those asked favoured all three of the proposals what w percentage of those asked favoured more than one of the num__3 proposals . <o> a ) num__10 <o> b ) num__12 <o> c ) num__17 <o> d ) num__22 <o> e ) num__30 |
bunuel my answer for exactly num__2 people was num__17 and this was my approach : num__100.0 = ( a + b + c ) - ( anb + anc + bnc ) - num__5.0 + num__22.0 which leads me to w = num__100.0 = ( num__50 + num__30 + num__20 ) - ( at least num__2 people ) - num__5.0 + num__22.0 . c <eor> c <eos> |
c |
percent__100.0__17.0__ |
percent__100.0__17.0__ |
| a shopkeeper loses num__15.0 if an article is sold for rs . num__187 . what should be the selling price of the article to gain num__20.0 ? <o> a ) s . num__247 <o> b ) s . num__248 <o> c ) s . num__264 <o> d ) s . num__329 <o> e ) s . num__412 |
given that sp = rs . num__187 and loss = num__15.0 cp = [ num__100 ( sp ) ] / ( num__100 - l % ) = ( num__100 * num__187 ) / num__85 = num__20 * num__6 = rs . num__220 . to get num__20.0 profit new sp = [ ( num__100 + p % ) cp ] / num__100 = ( num__220 * num__120 ) / num__100 = rs . num__264 answer : c <eor> c <eos> |
c |
percent__100.0__264.0__ |
percent__100.0__264.0__ |
| the average price of three items of furniture is rs . num__15000 . if their prices are in the ratio num__3 : num__5 : num__7 the price of the cheapest item is <o> a ) num__4888 <o> b ) num__2090 <o> c ) num__9028 <o> d ) num__9000 <o> e ) num__9202 |
explanation : let their prices be num__3 x num__5 x and num__7 x . then num__3 x + num__5 x + num__7 x = ( num__15000 * num__3 ) or x = num__3000 . cost of cheapest item = num__3 x = rs . num__9000 . answer : d <eor> d <eos> |
d |
divide__15000.0__5.0__ multiply__3.0__3000.0__ multiply__3.0__3000.0__ |
divide__15000.0__5.0__ multiply__3.0__3000.0__ multiply__3.0__3000.0__ |
| rose is two years older than bruce who is twice as old as chris . if the total of the age of rose b and chris be num__27 years then how old is bruce ? <o> a ) num__8 years <o> b ) num__10 years <o> c ) num__12 years <o> d ) num__14 years <o> e ) num__16 years |
let chris ' s age be x years . then bruce ' s age = num__2 x years . rose ' s age = ( num__2 x + num__2 ) years . ( num__2 x + num__2 ) + num__2 x + x = num__27 num__5 x = num__25 x = num__5 . hence bruce ' s age = num__2 x = num__10 years . b <eor> b <eos> |
b |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
| what is the value of ( log num__2 num__3 ) ( log num__3 num__4 ) ( log num__4 num__5 ) . . . ( log num__63 num__64 ) ? <o> a ) num__0.166666666667 <o> b ) num__2 <o> c ) num__2.5 <o> d ) num__6 <o> e ) num__32 |
using the identity ( loga b ) ( logb c ) = loga c repeatedly we obtain ( log num__2 num__3 ) ( log num__3 num__4 ) ( log num__4 num__5 ) . . . ( log num__63 num__64 ) = log num__2 num__64 = num__6 correct answer d <eor> d <eos> |
d |
multiply__2.0__3.0__ multiply__2.0__3.0__ |
multiply__2.0__3.0__ multiply__2.0__3.0__ |
| a father said to his son ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is num__68 years now the son ' s age five years back was : <o> a ) num__14 <o> b ) num__17 <o> c ) num__19 <o> d ) num__29 <o> e ) num__24 |
let the son ' s present age be x years . then ( num__68 - x ) = x num__2 x = num__68 . x = num__34 . son ' s age num__5 years back ( num__34 - num__5 ) = num__29 years . answer : d <eor> d <eos> |
d |
divide__68.0__2.0__ subtract__34.0__5.0__ subtract__34.0__5.0__ |
divide__68.0__2.0__ subtract__34.0__5.0__ subtract__34.0__5.0__ |
| each of x alarm tolls at regular intervals . all of them tolls together twelve times a day . no two alarm at equal intervals of time . if each alarm tolls after a whole number of minutes what is the maximum possible value of x ? <o> a ) num__14 <o> b ) num__16 <o> c ) num__18 <o> d ) num__20 <o> e ) num__22 |
explanation : the alarm tolls together twelves times a day . therefore they toll together once every num__2 hours i . e in num__120 minutes . since no two alarms toll at equal intervals of time the total number of distinct factors of num__120 including num__1 and num__120 itself = num__23 × num__3 × num__5 . the number of factors = ( num__3 + num__1 ) × num__2 × num__2 = num__16 . hence the maximum value of x is num__16 . answer : b <eor> b <eos> |
b |
add__1.0__2.0__ add__2.0__3.0__ multiply__1.0__16.0__ |
add__1.0__2.0__ add__2.0__3.0__ multiply__1.0__16.0__ |
| the winning relay team in a high school sports competition clocked num__48 minutes for a distance of num__13.2 km . its runners a b c and d maintained speeds of num__16 kmph num__17 kmph num__18 kmph and num__19 kmph respectively . what is the ratio of the time taken by b to than taken by d ? <o> a ) num__5 : num__16 <o> b ) num__19 : num__17 <o> c ) num__9 : num__8 <o> d ) num__8 : num__9 <o> e ) none of these |
explanation : since it is a relay race all the runners ran the same distance . hence for a same distance ( ratio of times ) = num__1 / ( ratio of speeds ) . hence ratio of times taken by b t & d = num__19 : num__17 answer : b <eor> b <eos> |
b |
subtract__17.0__16.0__ round__19.0__ |
subtract__17.0__16.0__ divide__19.0__1.0__ |
| a b and c are partners . a receives num__0.666666666667 of profits b and c dividing the remainder equally . a ' s income is increased by rs . num__200 when the rate to profit rises from num__5 to num__7 percent . find the capital of b ? <o> a ) rs . num__2450 <o> b ) rs . num__3600 <o> c ) rs . num__2500 <o> d ) rs . num__3100 <o> e ) rs . num__2100 |
a : b : c = num__0.666666666667 : num__0.166666666667 : num__0.166666666667 = num__4 : num__1 : num__1 x * num__0.02 * num__0.666666666667 = num__200 b capital = num__15000 * num__0.166666666667 = num__2500 answer : c <eor> c <eos> |
c |
subtract__5.0__4.0__ divide__4.0__200.0__ multiply__1.0__2500.0__ |
subtract__5.0__4.0__ divide__4.0__200.0__ multiply__1.0__2500.0__ |
| a town ' s oldest inhabitant is x years older than the sum of the ages of the lee triplets . if the oldest inhabitants is now j years old how old will one of the triplets f be in num__20 years ? j - x - num__13.3333333333 is my answers . j = x + l + l + l is the initial situation after num__20 years j + num__20 = x + l + l + l + num__60 . . . num__20 years for each triplet so num__60 years totally . j - x - num__13.3333333333 = l is my answer . what wrong am i doing ? since the age asked is after num__20 years i also consider adding num__20 years to j . <o> a ) ( j - num__50 ) / num__3 <o> b ) num__3 ( j + num__20 ) / x <o> c ) ( j + x - num__50 ) / num__3 <o> d ) ( j - x + num__60 ) / num__3 <o> e ) ( j + x - num__20 ) / num__3 |
here it goes : oldest inhabitant = sum of age of triplets + x j = num__3 l + x so l = ( j - x ) / num__3 after num__20 years = l + num__20 f = ( j - x ) / num__3 + num__20 = ( j - x + num__60 ) / num__3 = d <eor> d <eos> |
d |
divide__60.0__20.0__ multiply__20.0__3.0__ |
divide__60.0__20.0__ multiply__20.0__3.0__ |
| a num__3 ' ' cube is painted in all its faces and then it is cut down into num__1 ' ' blocks . how many num__1 ' ' blocks are there even without a single face being painted ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
the unpainted blocks are the interior blocks . this block forms a num__1 ' ' cube on the inside . the answer is b . <eor> b <eos> |
b |
volume_cube__1.0__ |
volume_cube__1.0__ |
| the volume of a sphere with radius r is ( num__1.33333333333 ) * pi * r ^ num__3 and the surface area is num__4 * pi * r ^ num__3 . if a sperical balloon has a volume of num__36 pi cubic centimeters what is hte surface area of the balloon in square centimeters ? <o> a ) a . num__40 <o> b ) b . num__100 <o> c ) c . num__400 <o> d ) d . num__1000 <o> e ) e . num__10 |
000 |
the surface area is num__4 . pi . r ^ num__2 ( its area remember not volume ) as num__1.33333333333 . pi . r ^ num__3 = num__36 pi r = num__3 so area = num__4 . pi . r ^ num__2 = num__36 . pi = num__36 x num__3.14 = num__100 ( approx ) b <eor> b <eos> |
b |
b |
| a dress on sale in a shop is marked at $ d . during the discount sale its price is reduced by num__15.0 . staff are allowed a further num__10.0 reduction on the discounted price . if a staff member buys the dress what will she have to pay in terms of d ? <o> a ) a ) num__0.75 d <o> b ) b ) num__0.76 d <o> c ) c ) num__0.765 d <o> d ) d ) num__0.775 d <o> e ) e ) num__0.805 d |
successive doscounts = - a - b + ab / num__100 putting the values total discount = - num__15 - num__10 + num__1.5 = - num__25 + num__1.5 = num__23.5 discount hence the final price of the dress = d - num__0.235 d = num__0.765 d correct option : c <eor> c <eos> |
c |
percent__15.0__10.0__ percent__100.0__0.765__ |
percent__15.0__10.0__ percent__100.0__0.765__ |
| what is the value of num__4 ^ num__4 + num__4 ^ num__6 ? <o> a ) num__4 ^ num__12 <o> b ) num__4 ^ num__35 <o> c ) num__17 ( num__4 ^ num__5 ) <o> d ) num__8 ^ num__12 <o> e ) num__17 ( num__4 ^ num__4 ) |
num__4 ^ num__4 + num__4 ^ num__6 = num__4 ^ num__4 ( num__1 + num__4 ^ num__2 ) = num__4 ^ num__4 * num__17 answer e <eor> e <eos> |
e |
subtract__6.0__4.0__ multiply__1.0__17.0__ |
subtract__6.0__4.0__ multiply__1.0__17.0__ |
| num__15 men take num__21 days of num__8 hrs . each to do a piece of work . how many days of num__5 hrs . each would it take for num__21 women if num__3 women do as much work as num__2 men ? <o> a ) num__30 <o> b ) num__20 <o> c ) num__16 <o> d ) num__26 <o> e ) num__36 |
let num__1 man does num__1 unit / hr of work num__15 m in num__21 days of num__8 hrs will do ( num__15 * num__21 * num__8 ) units num__3 w = num__2 m num__1 w = ( num__0.666666666667 ) units / hr num__21 w with num__6 hrs a day will take ( num__15 * num__21 * num__8 ) / ( num__21 * num__5 * ( num__0.666666666667 ) ) days = > num__36 days answer : e <eor> e <eos> |
e |
subtract__3.0__2.0__ divide__2.0__3.0__ subtract__21.0__15.0__ add__15.0__21.0__ round__36.0__ |
subtract__3.0__2.0__ divide__2.0__3.0__ multiply__3.0__2.0__ add__15.0__21.0__ divide__36.0__1.0__ |
| a certain collage has total of num__200 seniors each majoring in exactly one of six subjects . a minimum of num__20 seniors major in each six subjects . if three quarter of the seniors major in one of four subjects what is the greatest possible number of seniors majoring in one of the other two subjects ? <o> a ) num__100 <o> b ) num__80 <o> c ) num__25 <o> d ) num__60 <o> e ) num__50 |
answer c num__200 = num__150 + num__20 + x = > x = num__25 = num__0.75 * num__200 in num__4 subjects + num__20 min for num__5 th subject + x = num__200 <eor> c <eos> |
c |
divide__150.0__200.0__ divide__20.0__4.0__ add__20.0__5.0__ |
divide__150.0__200.0__ divide__20.0__4.0__ add__20.0__5.0__ |
| look at this series : num__1.5 num__2.3 num__3.1 num__3.9 . . . what number should come next ? <o> a ) num__4.7 <o> b ) num__4.5 <o> c ) num__6.7 <o> d ) num__8.9 <o> e ) num__9.2 |
in this simple addition series each number increases by num__0.8 . answer a <eor> a <eos> |
a |
subtract__2.3__1.5__ add__3.9__0.8__ |
subtract__2.3__1.5__ add__3.9__0.8__ |
| a train num__132 m long pass a telegraph pole in num__6 seconds . find the speed of the train <o> a ) num__66 km / hr <o> b ) num__68.4 km / hr <o> c ) num__72 km / hr <o> d ) num__79.2 km / hr <o> e ) none |
sol . speed = [ num__22.0 ] m / sec = [ num__22 * num__3.6 ] km / hr = num__79.2 km / hr . answer d <eor> d <eos> |
d |
divide__132.0__6.0__ multiply__3.6__22.0__ round__79.2__ |
divide__132.0__6.0__ multiply__3.6__22.0__ multiply__3.6__22.0__ |
| if a boat goes num__7 km upstream in num__42 minutes and the speed of the stream is num__3 kmph then the speed of the boat in still water is : <o> a ) num__2.5 km / hr <o> b ) num__4.2 km / hr <o> c ) num__5 km / hr <o> d ) num__10.5 km / hr <o> e ) none of these |
solution speed of stream = num__0.5 ( num__13 - num__8 ) km / hr = num__2.5 kmph . answer a <eor> a <eos> |
a |
subtract__3.0__0.5__ round__2.5__ |
subtract__3.0__0.5__ subtract__3.0__0.5__ |
| a man walks at a speed of num__3 km / hr and runs at a speed of num__7 km / hr . how much time will the man require to cover a distance of num__10 num__0.5 km if he completes half of the distance i . e . ( num__5 num__0.25 ) km on foot and the other half by running ? <o> a ) num__2 num__0.142857142857 hours <o> b ) num__2 num__0.5 hours <o> c ) num__2 num__3.5 hours <o> d ) num__4 num__0.5 hours <o> e ) num__2 num__0.2 hours |
required time = ( num__5 num__0.25 ) / num__3 + ( num__5 num__0.25 ) / num__7 = num__2 num__0.5 hours . answer : b <eor> b <eos> |
b |
subtract__7.0__5.0__ round__2.0__ |
divide__10.0__5.0__ divide__10.0__5.0__ |
| given that a “ num__12 - inch pizza ” means circular pizza with a diameter of num__12 inches changing from an num__4 - inch pizza to a num__12 - inch pizza gives you approximately what percent increase in the total amount of pizza ? <o> a ) num__330 <o> b ) num__500 <o> c ) num__670 <o> d ) num__800 <o> e ) num__125 |
area of num__4 inch pizza = num__2 * num__2 * pi = num__4 pi area of num__12 inch pizza = num__6 * num__6 * pi = num__36 pi num__8.0 * num__100 = num__800.0 increase hence answer is d <eor> d <eos> |
d |
divide__12.0__2.0__ subtract__12.0__4.0__ multiply__100.0__8.0__ multiply__100.0__8.0__ |
divide__12.0__2.0__ multiply__4.0__2.0__ multiply__100.0__8.0__ multiply__100.0__8.0__ |
| a train num__125 m long passes a man running at num__3 km / hr in the same direction in which the train is going in num__10 sec . the speed of the train is ? <o> a ) num__65 km / hr <o> b ) num__17 km / hr <o> c ) num__76 km / hr <o> d ) num__50 km / hr <o> e ) num__48 km / hr |
speed of the train relative to man = num__12.5 = num__12.5 m / sec . = num__12.5 * num__3.6 = num__45 km / hr let the speed of the train be x km / hr . then relative speed = ( x - num__3 ) km / hr . x - num__3 = num__45 = > x = num__48 km / hr . answer : e <eor> e <eos> |
e |
divide__125.0__10.0__ multiply__12.5__3.6__ add__3.0__45.0__ round__48.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ add__3.0__45.0__ round__48.0__ |
| let a b c be distinct digits . consider a two - digit number ‘ ab ’ and a three - digit number ‘ ccb ’ both defined under the usual decimal number system if ( ab ) num__2 = ccb > num__300 then the value of b is <o> a ) num__1 <o> b ) num__0 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
explanation : given ( ab ) num__2 = ccb . [ only num__1 num__5 and num__6 on squaring will result with same digits in units place . ] taking a = num__2 b = num__1 = > ( num__21 ) ^ num__2 = num__441 > num__300 . hence the required value of b = num__1 answer : a <eor> a <eos> |
a |
add__1.0__5.0__ reverse__1.0__ |
add__1.0__5.0__ reverse__1.0__ |
| can you find the missing number in the sequence given below ? num__10 num__17 num__24 num__11 num__18 ? num__12 num__19 num__26 num__13 num__20 num__27 <o> a ) num__25 <o> b ) num__23 <o> c ) num__20 <o> d ) num__18 <o> e ) num__19 |
let ' s break the given series as below : num__10 num__17 num__24 num__11 num__18 ? num__12 num__19 num__26 num__13 num__20 num__27 now read the number from left hand side from top to bottom as : so the number that will replace ' ? ' is num__25 answer : a <eor> a <eos> |
a |
add__12.0__13.0__ add__12.0__13.0__ |
add__12.0__13.0__ add__12.0__13.0__ |
| how many seconds will a num__500 metre long train take to cross a man walking with a speed of num__3 km / hr in the direction of the moving train if the speed of the train is num__63 km / hr ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__40 <o> d ) num__45 <o> e ) num__35 |
speed of the train relative to man = ( num__63 - num__3 ) km / hr = num__60 km / hr = ( num__60 x ( num__0.277777777778 ) ) m / sec = num__16.6666666667 m / sec . therefore time taken to pass the man = ( num__500 x ( num__0.06 ) ) sec = num__30 sec . answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__500.0__0.06__ round__30.0__ |
subtract__63.0__3.0__ multiply__500.0__0.06__ subtract__60.0__30.0__ |
| the average monthly income of p and q is rs . num__5050 . the average monthly income of q and r is rs . num__6250 and the average monthly income of p and r is rs . num__5200 . the monthly income of p is : <o> a ) num__3500 <o> b ) num__4000 <o> c ) num__4050 <o> d ) num__5000 <o> e ) num__5050 |
let p q and r represent their respective monthly incomes . then we have : p + q = ( num__5050 x num__2 ) = num__10100 . . . . ( i ) q + r = ( num__6250 x num__2 ) = num__12500 . . . . ( ii ) p + r = ( num__5200 x num__2 ) = num__10400 . . . . ( iii ) adding ( i ) ( ii ) and ( iii ) we get : num__2 ( p + q + r ) = num__33000 or p + q + r = num__16500 . . . . ( iv ) subtracting ( ii ) from ( iv ) we get p = num__4000 . p ' s monthly income = rs . num__4000 . answer : option b <eor> b <eos> |
b |
multiply__5050.0__2.0__ multiply__6250.0__2.0__ multiply__5200.0__2.0__ divide__33000.0__2.0__ subtract__16500.0__12500.0__ subtract__16500.0__12500.0__ |
multiply__5050.0__2.0__ multiply__6250.0__2.0__ multiply__5200.0__2.0__ divide__33000.0__2.0__ subtract__16500.0__12500.0__ subtract__16500.0__12500.0__ |
| a jar of num__220 marbles is divided equally among a group of marble - players today . if num__2 people joined the group in the future each person would receive num__1 marble less . how many people are there in the group today ? <o> a ) num__20 <o> b ) num__21 <o> c ) num__22 <o> d ) num__23 <o> e ) num__24 |
num__220 = num__20 * num__11 = num__22 * num__10 there are num__20 people in the group today . the answer is a . <eor> a <eos> |
a |
divide__220.0__20.0__ add__2.0__20.0__ divide__220.0__22.0__ divide__220.0__11.0__ |
divide__220.0__20.0__ multiply__2.0__11.0__ divide__220.0__22.0__ multiply__2.0__10.0__ |
| how long does a train num__110 meters long running at the speed of num__72 km / hour take to cross a bridge num__132 meters in length ? <o> a ) num__22.1 seconds <o> b ) num__12.1 seconds <o> c ) num__17.1 seconds <o> d ) num__12.9 seconds <o> e ) num__11.1 seconds |
speed = num__72 km / hour = num__72 * ( num__0.277777777778 ) m / sec = num__20 m / sec total distance to be covered = num__110 + num__132 = num__142 meters time = distance / speed = num__12.1 = num__12.1 seconds answer : b <eor> b <eos> |
b |
round__12.1__ |
round__12.1__ |
| the time it took car p to travel num__800 miles was num__2 hours less than the time it took car r to travel the same distance . if car p ’ s average speed was num__10 miles per hour greater than that of car r what was car r ’ s average speed in miles per hour ? <o> a ) num__43 <o> b ) num__58 <o> c ) num__60 <o> d ) num__70 <o> e ) num__80 |
let speed of car r be = x then speed of car p = x + num__10 a / q ( num__800 / x ) - ( num__800 / ( x + num__10 ) ) = num__2 solving for x = num__58 miles \ hr . b <eor> b <eos> |
b |
round__58.0__ |
round__58.0__ |
| by selling num__150 mangoes a fruit - seller gains the selling price of num__30 mangoes . find the gain percent ? <o> a ) num__15.0 <o> b ) num__25.0 <o> c ) num__115.0 <o> d ) num__75.0 <o> e ) num__55 % |
explanation : sp = cp + g num__150 sp = num__150 cp + num__30 sp num__120 sp = num__150 cp num__120 - - - num__30 cp num__100 - - - ? = > num__25.0 answer : b <eor> b <eos> |
b |
percent__25.0__100.0__ |
percent__25.0__100.0__ |
| how long will it take a sum of money invested at num__5.0 p . a . s . i . to increase its value by num__100.0 ? <o> a ) num__15 years <o> b ) num__16 years <o> c ) num__25 years <o> d ) num__20 years <o> e ) num__18 years |
sol . let the sum be x . then s . i . = num__100.0 of x = x ; rate = num__5.0 . â ˆ ´ time = [ num__100 * num__2 x / num__5 * num__1 / x * num__5 ] = num__20 years . answer : d <eor> d <eos> |
d |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| a worker ' s daily wage is increased by num__50.0 and the new wage is $ num__42 per day . what was the worker ' s daily wage before the increase ? <o> a ) $ num__22 <o> b ) $ num__24 <o> c ) $ num__28 <o> d ) $ num__32 <o> e ) $ num__36 |
let x be the daily wage before the increase . num__1.5 x = $ num__42 x = $ num__28 the answer is c . <eor> c <eos> |
c |
divide__42.0__1.5__ round__28.0__ |
divide__42.0__1.5__ round__28.0__ |
| in what time will a railway train num__60 m long moving at the rate of num__36 kmph pass a telegraph post on its way ? <o> a ) num__7 sec <o> b ) num__8 sec <o> c ) num__2 sec <o> d ) num__6 sec <o> e ) num__9 sec |
t = num__1.66666666667 * num__3.6 = num__6 sec answer : d <eor> d <eos> |
d |
divide__60.0__36.0__ round__6.0__ |
divide__60.0__36.0__ round__6.0__ |
| what is the different between the place value and face value of num__6 in the numeral num__5468 ? <o> a ) num__45 <o> b ) num__54 <o> c ) num__60 <o> d ) num__36 <o> e ) num__44 |
place value of num__6 = num__6 * num__10 = num__60 face value of num__6 = num__6 num__60 - num__6 = num__54 answer is b <eor> b <eos> |
b |
multiply__6.0__10.0__ subtract__60.0__6.0__ subtract__60.0__6.0__ |
multiply__6.0__10.0__ subtract__60.0__6.0__ subtract__60.0__6.0__ |
| the parameter of a square is equal to the perimeter of a rectangle of length num__16 cm and breadth num__14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) ? <o> a ) num__23.52 cm <o> b ) num__23.87 cm <o> c ) num__23.37 cm <o> d ) num__22.57 cm <o> e ) num__23.57 cm |
let the side of the square be a cm . parameter of the rectangle = num__2 ( num__16 + num__14 ) = num__60 cm parameter of the square = num__60 cm i . e . num__4 a = num__60 a = num__15 diameter of the semicircle = num__15 cm circimference of the semicircle = num__0.5 ( ∏ ) ( num__15 ) = num__0.5 ( num__3.14285714286 ) ( num__15 ) = num__23.5714285714 = num__23.57 cm to two decimal places answer : e <eor> e <eos> |
e |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
| what is the difference between the place value of num__2 in the numeral num__7229 ? <o> a ) num__150 <o> b ) num__160 <o> c ) num__180 <o> d ) num__190 <o> e ) num__200 |
option ' c ' num__200 - num__20 = num__180 <eor> c <eos> |
c |
subtract__200.0__20.0__ subtract__200.0__20.0__ |
subtract__200.0__20.0__ subtract__200.0__20.0__ |
| a certain list consists of num__21 different numbers . if n is in the list and n is num__8 times the average ( arithmetic mean ) of the other num__20 numbers in the list then n is what fraction of the sum of the num__21 numbers in the list ? <o> a ) num__0.05 <o> b ) num__0.166666666667 <o> c ) num__0.285714285714 <o> d ) num__0.190476190476 <o> e ) num__0.238095238095 |
series : a num__1 a num__2 . . . . a num__20 n sum of a num__1 + a num__2 + . . . + a num__20 = num__20 * x ( x = average ) so n = num__8 * x hence a num__1 + a num__2 + . . + a num__20 + n = num__28 x so the fraction asked = num__8 x / num__28 x = num__0.285714285714 c <eor> c <eos> |
c |
subtract__21.0__20.0__ add__8.0__20.0__ divide__8.0__28.0__ divide__8.0__28.0__ |
subtract__21.0__20.0__ add__8.0__20.0__ divide__8.0__28.0__ divide__8.0__28.0__ |
| express a speed of num__56 kmph in meters per second ? <o> a ) num__10.2 mps <o> b ) num__05 mps <o> c ) num__09.1 mps <o> d ) num__12 mps <o> e ) num__15.5 mps |
e num__15.5 mps num__56 * num__0.277777777778 = num__15.5 mps <eor> e <eos> |
e |
round__15.5__ |
round__15.5__ |
| in an examination num__20.0 of total students failed in hindi num__70.0 failed in english and num__10.0 in both . the percentage of these who passed in both the subjects is : <o> a ) num__10.0 <o> b ) num__20.0 <o> c ) num__30.0 <o> d ) num__40.0 <o> e ) num__50 % |
pass percentage = num__100 - ( num__20 + num__70 - num__10 ) = num__100 - num__80 = num__20 answer : b <eor> b <eos> |
b |
add__70.0__10.0__ subtract__100.0__80.0__ |
add__70.0__10.0__ subtract__100.0__80.0__ |
| a can do a job in num__9 days and b can do it in num__27 days . a and b working together will finish twice the amount of work in - - - - - - - days ? <o> a ) num__22 days <o> b ) num__18 days <o> c ) num__22 num__3.0 days <o> d ) num__27 days <o> e ) num__9 days |
num__0.111111111111 + num__0.037037037037 = num__0.111111111111 = num__0.111111111111 num__9.0 = num__9 * num__2 = num__18 days answer : b <eor> b <eos> |
b |
multiply__9.0__2.0__ round__18.0__ |
multiply__9.0__2.0__ multiply__9.0__2.0__ |
| if the number num__992 num__13224 x is divisible by num__11 what must be the value of x ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__0 |
multiplication rule of num__11 : ( sum of digits at odd places - sum of digits at even places ) should be divisible by num__11 given number : num__992 num__13224 x sum of digits at odd places = num__9 + num__2 + num__3 + num__2 + x = num__16 + x ( i ) sum of digits at even places = num__9 + num__1 + num__2 + num__4 = num__16 ( ii ) ( i ) - ( ii ) = num__16 + x - num__16 = x - num__0 hence x should be = num__0 to make this a multiple of num__11 ( num__0 ) option e <eor> e <eos> |
e |
subtract__11.0__9.0__ subtract__3.0__2.0__ add__1.0__3.0__ multiply__992.0__0.0__ |
subtract__11.0__9.0__ subtract__3.0__2.0__ add__1.0__3.0__ multiply__992.0__0.0__ |
| the speed at which a man can row a boat in still water is num__15 kmph . if he rows downstream where the speed of current is num__3 kmph what time will he take to cover num__60 metres ? <o> a ) num__11 <o> b ) num__77 <o> c ) num__16 <o> d ) num__12 <o> e ) num__10 |
speed of the boat downstream = num__15 + num__3 = num__18 kmph = num__18 * num__0.277777777778 = num__5 m / s hence time taken to cover num__60 m = num__12.0 = num__12 seconds . answer : d <eor> d <eos> |
d |
add__15.0__3.0__ divide__15.0__3.0__ subtract__15.0__3.0__ round__12.0__ |
add__15.0__3.0__ divide__15.0__3.0__ divide__60.0__5.0__ divide__60.0__5.0__ |
| two goods trains each num__500 m long are running in opposite directions on parallel tracks . their speeds are num__45 km / hr and num__30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? <o> a ) num__22 <o> b ) num__28 <o> c ) num__48 <o> d ) num__277 <o> e ) num__12 |
relative speed = num__45 + num__30 = num__75 km / hr . num__75 * num__0.277777777778 = num__20.8333333333 m / sec . distance covered = num__500 + num__500 = num__1000 m . required time = num__1000 * num__0.048 = num__48 sec . answer : c <eor> c <eos> |
c |
add__45.0__30.0__ multiply__1000.0__0.048__ round__48.0__ |
add__45.0__30.0__ multiply__1000.0__0.048__ multiply__1000.0__0.048__ |
| if the length of a rectangle is halved and its breadth is tripled what is the percentage change in its area ? <o> a ) num__25.0 increase <o> b ) num__25.0 decrease <o> c ) num__50.0 increase <o> d ) num__50.0 decrease <o> e ) none |
explanation : length is halved . i . e . length is decreased by num__50.0 num__50.0 breadth is tripled i . e . breadth is increased by num__200.0 num__200.0 change in area = ( − num__50 + num__200 − num__50 × num__2.0 ) % = num__50.0 i . e . area is increased by num__50.0 answer : option d <eor> d <eos> |
d |
square_perimeter__50.0__ triangle_area__50.0__2.0__ |
square_perimeter__50.0__ triangle_area__50.0__2.0__ |
| frank is num__12 years younger then john . in num__5 years john will be twice as old as frank . how old will frank be in four years ? <o> a ) num__8 <o> b ) num__11 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
frank is num__15 years younger then john - - > f + num__12 = j ; in num__5 years john will be twice as old as frank ( in num__5 year john will be j + num__5 years old and frank will be f + num__5 years old ) - - > j + num__5 = num__2 * ( f + num__5 ) - - > ( f + num__12 ) + num__5 = num__2 * ( f + num__5 ) - - > f = num__7 ; in num__4 years frank will be num__7 + num__4 = num__11 years old . answer : b . <eor> b <eos> |
b |
triple__5.0__ subtract__12.0__5.0__ twice__2.0__ add__4.0__7.0__ add__4.0__7.0__ |
triple__5.0__ subtract__12.0__5.0__ twice__2.0__ add__4.0__7.0__ add__4.0__7.0__ |
| how many such num__3 ' s are there in the following number sequence which are immediately preceded by an odd number and immediately followed by an even number ? num__5 num__3 num__8 num__9 num__4 num__3 num__7 num__2 num__3 num__8 num__1 num__3 num__8 num__4 num__2 num__3 num__5 num__7 num__3 num__4 num__2 num__3 num__6 <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 three <o> d ) num__4 <o> e ) more than four |
a is followed by b means : a comes first b comes next a preceded by b means : b comes first a comes next the sets which satisfy the condition is : num__5 num__3 num__8 num__1 num__3 num__8 num__7 num__3 num__4 answer : c <eor> c <eos> |
c |
multiply__3.0__1.0__ |
multiply__3.0__1.0__ |
| three times the first of three consecutive odd integers is num__3 more than twice the third . the third integer is : <o> a ) num__9 <o> b ) num__11 <o> c ) num__13 <o> d ) num__15 <o> e ) num__17 |
let the three integers be x x + num__2 and x + num__4 . then num__3 x = num__2 ( x + num__4 ) + num__3 x = num__11 third integer = x + num__4 = num__15 . answer : d <eor> d <eos> |
d |
add__11.0__4.0__ add__11.0__4.0__ |
add__11.0__4.0__ add__11.0__4.0__ |
| by investing in num__1623.0 stock at num__64 one earns rs . num__1700 . the investment made is <o> a ) s . num__9600 <o> b ) s . num__7500 <o> c ) s . num__5640 <o> d ) s . num__6528 <o> e ) s . num__6760 |
explanation : market value = rs . num__64 face value is not given and hence take it as rs . num__100 num__16 num__0.666666666667 % of the face value = num__16.6666666667 ie to earn num__16.6666666667 investment = rs . num__64 hence to earn rs . num__1700 investment needed = num__64 × num__3 × num__34.0 = num__6528 answer : option d <eor> d <eos> |
d |
percent__100.0__6528.0__ |
percent__100.0__6528.0__ |
| robert ate num__12 chocolates nickel ate num__3 chocolates . how many more chocolates did robert ate than nickel ? <o> a ) a ) num__4 <o> b ) b ) num__7 <o> c ) c ) num__9 <o> d ) d ) num__5 <o> e ) e ) num__2 |
num__12 - num__3 = num__9 . answer is c <eor> c <eos> |
c |
subtract__12.0__3.0__ subtract__12.0__3.0__ |
subtract__12.0__3.0__ subtract__12.0__3.0__ |
| a producer of tea blends two varieties of tea from two tea gardens one costing rs num__18 per kg and another rs num__20 per kg in the ratio num__5 : num__3 . if he sells the blended variety at rs num__25 per kg then his gain percent is <o> a ) num__12.0 <o> b ) num__33.0 <o> c ) num__14.0 <o> d ) num__15.0 <o> e ) num__16 % |
explanation : suppose he bought num__5 kg and num__3 kg of tea . cost price = rs . ( num__5 x num__18 + num__3 x num__20 ) = rs . num__150 . selling price = rs . ( num__8 x num__25 ) = rs . num__200 . profit = num__200 - num__150 = num__50 so profit % = ( num__0.333333333333 ) * num__100 = num__33.0 option b <eor> b <eos> |
b |
percent__25.0__200.0__ percent__50.0__200.0__ percent__33.0__100.0__ |
percent__25.0__200.0__ percent__50.0__200.0__ percent__33.0__100.0__ |
| a large tank can filled by a and b in num__60 minutes and num__40 minutes respectively . how many minutes will it take to fill the tanker from empty state if b is used for half the time and a and b fill it together for the other half ? <o> a ) num__10 min <o> b ) num__15 min <o> c ) num__30 min <o> d ) num__25 min <o> e ) num__42 min |
part filled by a + b in num__1 minute = num__0.0166666666667 + num__0.025 = num__0.0416666666667 suppose the tank is filled in x minutes then x / num__2 ( num__0.0416666666667 + num__0.025 ) = num__1 x / num__2 * num__0.0666666666667 = num__1 x = num__30 min answer is c <eor> c <eos> |
c |
divide__1.0__60.0__ divide__1.0__40.0__ add__0.025__0.0167__ add__0.025__0.0417__ divide__60.0__2.0__ round__30.0__ |
divide__1.0__60.0__ divide__1.0__40.0__ add__0.025__0.0167__ add__0.025__0.0417__ divide__60.0__2.0__ divide__60.0__2.0__ |
| if a person walks at num__10 km / hr instead of num__8 km / hr he would have walked num__14 km more . the actual distance traveled by him is ? <o> a ) num__50 <o> b ) num__40 <o> c ) num__56 <o> d ) num__16 <o> e ) num__20 |
let the actual distance traveled be x km . then x / num__8 = ( x + num__14 ) / num__10 num__2 x - num__112 = > x = num__56 km . answer : c <eor> c <eos> |
c |
subtract__10.0__8.0__ multiply__8.0__14.0__ divide__112.0__2.0__ divide__112.0__2.0__ |
subtract__10.0__8.0__ multiply__8.0__14.0__ divide__112.0__2.0__ divide__112.0__2.0__ |
| the average temperature for tuesday wednesday and thursday was num__52 ° c . the average temperature for wednesday thursday and friday was num__54 ° c . if the temperature on friday be num__53 ° c what was the temperature on tuesday ? <o> a ) num__39 ° c <o> b ) num__44 ° c <o> c ) num__37 ° c <o> d ) num__47 ° c <o> e ) none of these |
explanation : t + w + t = num__52 × num__3 = num__156 ° c w + t + f = num__54 × num__3 = num__162 ° c also temperature on friday = num__53 ° c temperature on tuesday = num__156 + num__53 - num__162 = num__47 ° c answer : option d <eor> d <eos> |
d |
multiply__52.0__3.0__ multiply__54.0__3.0__ round__47.0__ |
multiply__52.0__3.0__ multiply__54.0__3.0__ round__47.0__ |
| let n ~ be defined for all positive integers n as the remainder when ( n - num__1 ) ! is divided by n . what is the value of num__34 ~ ? <o> a ) num__6 <o> b ) num__1 <o> c ) num__2 <o> d ) num__0 <o> e ) num__31 |
n ~ = ( n - num__1 ) ! so num__34 ~ = ( num__34 - num__1 ) ! = num__33 ! when num__33 ! / num__34 we have num__16 * num__2 inside num__33 ! hence num__34 gets cancelled and we get remainder as num__0 d <eor> d <eos> |
d |
subtract__34.0__1.0__ multiply__1.0__0.0__ |
subtract__34.0__1.0__ multiply__1.0__0.0__ |
| in a weight - lifting competition the total weight of joe ' s two lifts was num__450 pounds . if twice the weight of his first lift was num__300 pounds more than the weight of his second lift what was the weight in pounds of his first lift ? <o> a ) num__250 <o> b ) num__275 <o> c ) num__325 <o> d ) num__350 <o> e ) num__400 |
this problem is a general word translation . we first define variables and then set up equations . we can define the following variables : f = the weight of the first lift s = the weight of the second lift we are given that the total weight of joe ' s two lifts was num__450 pounds . we sum the two variables to obtain : f + s = num__450 we are also given that twice the weight of his first lift was num__300 pounds more than the weight of his second lift . we express this as : num__2 f = num__300 + s num__2 f – num__300 = s we can now plug in ( num__2 f – num__300 ) for s into the first equation so we have : f + num__2 f – num__300 = num__450 num__3 f = num__750 f = num__250 answer is a . <eor> a <eos> |
a |
add__450.0__300.0__ divide__750.0__3.0__ divide__750.0__3.0__ |
add__450.0__300.0__ divide__750.0__3.0__ divide__750.0__3.0__ |
| the ratio by weight measured in pounds of books to clothes to electronics in a suitcase initially stands at num__5 : num__4 : num__2 . someone removes num__9 pounds of clothing from the suitcase thereby doubling the ratio of books to clothes . how many pounds do the electronics in the suitcase weigh ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__11 |
the weights of the items in the suitcase are num__5 k num__4 k and num__2 k . if removing num__9 pounds of clothes doubles the ratio of books to clothes then num__9 pounds represents half the weight of the clothes . num__2 k = num__9 pounds and then k = num__4.5 pounds . the electronics weigh num__2 ( num__4.5 ) = num__9 pounds . the answer is c . <eor> c <eos> |
c |
divide__9.0__2.0__ add__5.0__4.0__ |
divide__9.0__2.0__ add__5.0__4.0__ |
| a rectangular field is to be fenced on three sides leaving a side of num__20 feet uncovered . if the area of the field is num__10 sq . feet how many feet of fencing will be required ? <o> a ) num__34 <o> b ) num__40 <o> c ) num__68 <o> d ) num__21 <o> e ) num__78 |
we have : l = num__20 ft and lb = num__10 sq . ft . so b = num__0.5 ft . length of fencing = ( l + num__2 b ) = ( num__20 + num__1 ) ft = num__21 ft . answer : d <eor> d <eos> |
d |
square_perimeter__0.5__ multiply__0.5__2.0__ rectangle_perimeter__10.0__0.5__ rectangle_perimeter__10.0__0.5__ |
square_perimeter__0.5__ multiply__0.5__2.0__ rectangle_perimeter__10.0__0.5__ rectangle_perimeter__10.0__0.5__ |
| the output of a factory is increased by num__10.0 to keep up with rising demand . to handle the holiday rush this new output is increased by num__20.0 . by approximately what percent would the output of the factory now have to be decreased in order to restore the original output ? <o> a ) num__20.0 <o> b ) num__24.0 <o> c ) num__30.0 <o> d ) num__32.0 <o> e ) num__79 % |
let the output of a factory is num__100 first increased by num__10.0 = num__110 due to holiday rush it increase by num__20.0 = num__110 + num__22 = num__132 num__132 - ( x / num__100 * num__132 ) = num__100 num__24.24 answer : b <eor> b <eos> |
b |
add__10.0__100.0__ add__110.0__22.0__ round_down__24.24__ |
add__10.0__100.0__ add__110.0__22.0__ round_down__24.24__ |
| a boat with still water speed of a is traveling upstream where the streams speed is b . after covering certain distance s the boat turns back and tries to come to the starting place . but the boat stops at num__0.75 th of the intended distance due to a technical fault . what is the ratio of the time taken for the return journey to that of the upstream journey . <o> a ) num__3 ( a - b ) / num__4 ( a + b ) <o> b ) num__4 ( a - b ) / num__3 ( a + b ) <o> c ) num__3 ( a + b ) / num__4 ( a - b ) <o> d ) num__1.33333333333 <o> e ) num__0.5 |
speed of boat upstream = a - b speed of boat downstream = a + b for upstream journey time taken = distance / speed = s / ( a - b ) during return journey the boat had covered on num__0.75 th of s : therefore for return journey time taken = distance / speed = ( num__3 s / num__4 ) / ( a + b ) ratio of time taken for return journey to the upstream journey = ( num__3 s / num__4 ) / ( a + b ) / ( s / ( a - b ) ) = num__3 ( a - b ) / num__4 ( a + b ) answer : a <eor> a <eos> |
a |
divide__3.0__0.75__ round__3.0__ |
divide__3.0__0.75__ round__3.0__ |
| two taps can separately fill a cistern num__10 minutes and num__15 minutes respectively and when the waste pipe is open they can together fill it in num__18 minutes . the waste pipe can empty the full cistern in ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__9 <o> e ) num__1 |
num__0.1 + num__0.0666666666667 - num__1 / x = num__0.0555555555556 x = num__9 answer : d <eor> d <eos> |
d |
multiply__10.0__0.1__ divide__1.0__18.0__ subtract__10.0__1.0__ round__9.0__ |
multiply__10.0__0.1__ divide__1.0__18.0__ subtract__10.0__1.0__ subtract__10.0__1.0__ |
| if p and q are the roots of the equation x num__2 - bx + c = num__0 what is the equation if the roots are ( pq + p + q ) and ( pq - p - q ) ? <o> a ) x num__2 - num__2 cx + ( c num__2 - b num__2 ) = num__0 <o> b ) x num__2 - num__2 bx + ( b num__2 + c num__2 ) = num__0 <o> c ) bcx num__2 - num__2 ( b + c ) x + c num__2 = num__0 <o> d ) x num__2 + num__2 bx - ( c num__2 - b num__2 ) = num__0 <o> e ) none |
explanatory answer in the given quadratic equation x num__2 - bx + c = num__0 the sum of the roots p + q = b - - - ( num__1 ) and the product of the roots pq = c - - - ( num__2 ) we have to formulate a quadratic equation whose roots are ( pq + p + q ) and ( pq - p - q ) . the sum of the two roots = pq + p + q + pq - p - q = num__2 pq but from eqn ( num__2 ) we know that pq = c therefore the sum of the roots = num__2 c the product of the roots = ( pq + p + q ) ( pq - p - q ) = ( pq ) num__2 - ( p + q ) num__2 from equation ( num__1 ) and ( num__2 ) we know that pq = c and p + q = b therefore the product of the roots = c num__2 - b num__2 we know the sum of the roots and the product of the roots . therefore the quadratic equation is x num__2 - ( sum of the roots ) x + product of the roots = num__0 = > x num__2 - num__2 cx + c num__2 - b num__2 = num__0 answer a <eor> a <eos> |
a |
multiply__2.0__1.0__ |
multiply__2.0__1.0__ |
| if a and b are positive integers and x = num__9 ^ a and y = num__5 ^ b which of the following is a possible units digit of xy ? <o> a ) num__1 <o> b ) num__4 <o> c ) num__5 <o> d ) num__7 <o> e ) num__8 |
the units digit of num__9 ^ a is either num__1 or num__9 . the units digit of num__5 ^ b is num__5 . num__1 * num__5 = num__5 and num__9 * num__5 = num__45 . the units digit of xy is num__5 . the answer is c . <eor> c <eos> |
c |
multiply__9.0__5.0__ multiply__5.0__1.0__ |
multiply__9.0__5.0__ multiply__5.0__1.0__ |
| if n is a prime number greater than num__3 what is the remainder when n ^ num__2 is divided by num__11 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__5 |
there are several algebraic ways to solve this question including the one under the spoiler . but the easiest way is as follows : since we can not have two correct answersjust pick a prime greater than num__3 square it and see what would be the remainder upon division of it by num__11 . n = num__5 - - > n ^ num__2 = num__25 - - > remainder upon division num__25 by num__11 is num__3 . answer : d . <eor> d <eos> |
d |
add__3.0__2.0__ subtract__5.0__2.0__ |
add__3.0__2.0__ subtract__5.0__2.0__ |
| a train num__130 metres long travels at a speed of num__45 km / hr crosses a bridge in num__30 seconds . what is the length of the bridge ? <o> a ) num__210 <o> b ) num__225 <o> c ) num__235 <o> d ) num__245 <o> e ) num__260 |
speed = num__45 km / hr = num__45 × num__0.277777777778 m / s = num__12.5 m / s distance travelled = ( speed x time ) distance travelled = num__12.5 × num__30 = num__375 m length of the bridge = num__375 - num__130 = num__245 m answer : d <eor> d <eos> |
d |
multiply__30.0__12.5__ subtract__375.0__130.0__ round__245.0__ |
multiply__30.0__12.5__ subtract__375.0__130.0__ subtract__375.0__130.0__ |
| mike is num__50.0 less efficient than alice . how much time will they working together take to complete a job which alice alone could have done in num__15 days ? <o> a ) num__5 days <o> b ) num__10 days <o> c ) num__9 days <o> d ) num__18 days <o> e ) num__15 days |
ratio of times taken by alice and mike = num__100 : num__150 = num__2 : num__3 . suppose mike takes x days to do the work . then num__2 : num__3 : : num__15 : x â ‡ ’ x = ( num__15 Ã — num__3 ) / num__2 â ‡ ’ x = num__45 â „ num__2 alice â € ™ s num__1 day â € ™ s work = num__1 â „ num__15 ; mike â € ™ s num__1 days work = num__2 â „ num__45 ( alice + mike ) â € ™ s num__1 day â € ™ s work = ( num__1 â „ num__15 + num__2 â „ num__45 ) = num__1 â „ num__9 â ˆ ´ alice and mike together can complete the job in num__9 days . answer c <eor> c <eos> |
c |
add__50.0__100.0__ divide__100.0__50.0__ divide__150.0__50.0__ multiply__15.0__3.0__ subtract__3.0__2.0__ round__9.0__ |
add__50.0__100.0__ divide__100.0__50.0__ divide__150.0__50.0__ multiply__15.0__3.0__ subtract__3.0__2.0__ divide__9.0__1.0__ |
| a company organized a recruiting process for num__3 vacant positions of assistant manager for its product launches . the company ' s efforts yielded num__11 eligible candidates . how many sets of num__3 candidates can the company choose to fill the num__3 assistant manager positions ? <o> a ) num__2060 <o> b ) num__1320 <o> c ) num__165 <o> d ) num__315 <o> e ) num__220 |
num__11 * num__10 * num__3.0 * num__2 * num__1 = num__165 c <eor> c <eos> |
c |
subtract__3.0__2.0__ multiply__1.0__165.0__ |
subtract__3.0__2.0__ multiply__1.0__165.0__ |
| fresh grapes contain num__90.0 by weight while dried grapes contain num__20.0 water by weight . what is the weight of dry grapes available from num__40 kg of fresh grapes ? <o> a ) num__5 kg <o> b ) num__2.4 kg <o> c ) num__2.5 kg <o> d ) num__10 kg <o> e ) none of these |
the weight of non - water in num__20 kg of fresh grapes ( which is num__100 - num__90 = num__10.0 of whole weight ) will be the same as the weight of non - water in x kg of dried grapes ( which is num__100 - num__20 = num__80.0 of whole weight ) so num__40 â ˆ — num__0.1 = x â ˆ — num__0.8 - - > x = num__5 answer : a <eor> a <eos> |
a |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| a can do half the work in one day where as b can do it full . b can also do half the work of c in one day . ratio in their efficiency will be ? <o> a ) num__1 : num__2 : num__6 <o> b ) num__1 : num__2 : num__9 <o> c ) num__1 : num__2 : num__5 <o> d ) num__1 : num__2 : num__4 <o> e ) num__1 : num__3 : num__4 |
wc of a : b = num__1 : num__2 b : c = num__1 : num__2 - - - - - - - - - - - - - - - - - - - - - a : b : c = num__1 : num__2 : num__4 answer : d <eor> d <eos> |
d |
round__1.0__ |
subtract__2.0__1.0__ |
| in order to obtain an income of rs . num__650 from num__10.0 stock at rs . num__96 one must make an investment of <o> a ) rs . num__3100 <o> b ) rs . num__6500 <o> c ) rs . num__6240 <o> d ) rs . num__9600 <o> e ) rs . num__9800 |
explanation : market value = rs . num__96 . required income = rs . num__650 . here face value is not given . take face value as rs . num__100 if it is not given in the question to obtain rs . num__10 ( ie num__10.0 of the face value num__100 ) investment = rs . num__96 to obtain rs . num__650 investment = num__9.6 × num__650 = num__6240 answer : option c <eor> c <eos> |
c |
percent__10.0__96.0__ percent__100.0__6240.0__ |
percent__10.0__96.0__ percent__100.0__6240.0__ |
| the product of the ages of syam and sunil is num__240 . if twice the age of sunil is more than syam ' s age by num__4 years what is sunil ' s age ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
we can solve the quadratic equation by factorization as x ( x − num__2 ) = num__120 ⇒ x num__2 − num__2 x − num__120 = num__0 ⇒ ( x − num__12 ) ( x + num__10 ) = num__0 ⇒ x = num__12 or − num__10 since x is age and can not be negative x = num__12 answer is e . <eor> e <eos> |
e |
divide__240.0__2.0__ divide__120.0__12.0__ divide__120.0__10.0__ |
divide__240.0__2.0__ divide__120.0__12.0__ add__2.0__10.0__ |
| the edge of a cube is num__6 a cm . find its surface ? <o> a ) num__116 a num__2 cm num__2 <o> b ) num__126 a num__2 cm num__2 <o> c ) num__256 a num__2 cm num__2 <o> d ) num__150 a num__2 cm num__2 <o> e ) num__216 a num__2 cm num__2 |
num__6 a num__2 = num__6 * num__6 a * num__6 a = num__216 a num__2 answer : e <eor> e <eos> |
e |
volume_cube__6.0__ volume_cube__6.0__ |
volume_cube__6.0__ volume_cube__6.0__ |
| if x > num__0 x / num__5 + x / num__25 is what percent of x ? <o> a ) num__6.0 <o> b ) num__24.0 <o> c ) num__37 num__0.5 % <o> d ) num__60.0 <o> e ) num__75 % |
just plug and chug . since the question asks for percents pick num__100 . ( but any number will do . ) num__20.0 + num__4.0 = num__20 + num__4 = num__24 num__24 is num__24.0 of num__100 = b <eor> b <eos> |
b |
subtract__25.0__5.0__ divide__20.0__5.0__ add__20.0__4.0__ add__20.0__4.0__ |
subtract__25.0__5.0__ divide__20.0__5.0__ add__20.0__4.0__ add__20.0__4.0__ |
| right now the ratio between the ages of sandy and molly is num__4 : num__3 . after num__6 years sandy ’ s age will be num__34 years . what is molly ' s age right now ? <o> a ) num__12 <o> b ) num__15 <o> c ) num__18 <o> d ) num__21 <o> e ) num__24 |
now sandy is num__34 - num__6 = num__28 molly ' s age is ( num__0.75 ) * num__28 = num__21 the answer is d . <eor> d <eos> |
d |
subtract__34.0__6.0__ divide__3.0__4.0__ multiply__0.75__28.0__ multiply__0.75__28.0__ |
subtract__34.0__6.0__ divide__3.0__4.0__ multiply__0.75__28.0__ multiply__0.75__28.0__ |
| what is the length of the longest pole which can be kept in a room num__12 m long num__5 m broad and num__3 m high ? <o> a ) num__7 <o> b ) num__9 <o> c ) num__11 <o> d ) num__13.34 <o> e ) none |
explanation : d num__2 = num__122 + num__52 + num__32 = num__13.34 d ) <eor> d <eos> |
d |
subtract__5.0__3.0__ round__13.34__ |
subtract__5.0__3.0__ round__13.34__ |
| find the roots of the quadratic equation : x num__2 + num__2 x - num__15 = num__0 ? <o> a ) - num__5 num__3 <o> b ) num__3 num__5 <o> c ) - num__3 num__5 <o> d ) - num__3 - num__5 <o> e ) num__5 num__2 |
explanation : x num__2 + num__5 x - num__3 x - num__15 = num__0 x ( x + num__5 ) - num__3 ( x + num__5 ) = num__0 ( x - num__3 ) ( x + num__5 ) = num__0 = > x = num__3 or x = - num__5 . answer is a <eor> a <eos> |
a |
divide__15.0__5.0__ add__2.0__3.0__ |
subtract__5.0__2.0__ add__2.0__3.0__ |
| a man has rs . num__560 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__33 <o> b ) num__105 <o> c ) num__37 <o> d ) num__90 <o> e ) num__28 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__560 ⇒ num__16 x = num__560 ∴ x = num__35 . hence total number of notes = num__3 x = num__105 answer : b <eor> b <eos> |
b |
divide__560.0__16.0__ multiply__35.0__3.0__ multiply__35.0__3.0__ |
divide__560.0__16.0__ multiply__35.0__3.0__ multiply__35.0__3.0__ |
| pipe a can fill a tank in num__16 minutes and pipe b cam empty it in num__24 minutes . if both the pipes are opened together after how many minutes should pipe b be closed so that the tank is filled in num__30 minutes ? <o> a ) num__76 <o> b ) num__87 <o> c ) num__76 <o> d ) num__56 <o> e ) num__21 |
let the pipe b be closed after x minutes . num__1.875 - x / num__24 = num__1 = > x / num__24 = num__1.875 - num__1 = num__0.875 = > x = num__0.875 * num__24 = num__21 . answer : e <eor> e <eos> |
e |
divide__30.0__16.0__ subtract__1.875__1.0__ multiply__24.0__0.875__ round__21.0__ |
divide__30.0__16.0__ subtract__1.875__1.0__ multiply__24.0__0.875__ multiply__24.0__0.875__ |
| how many multiples of num__3 are there between num__81 and num__358 ? <o> a ) num__94 <o> b ) num__95 <o> c ) num__93 <o> d ) num__97 <o> e ) num__98 |
num__3 * num__27 = num__81 num__3 * num__119 = num__357 total multiples of num__3 = ( num__119 - num__27 ) + num__1 = num__93 answer is c . <eor> c <eos> |
c |
divide__81.0__3.0__ multiply__3.0__119.0__ subtract__358.0__357.0__ multiply__1.0__93.0__ |
divide__81.0__3.0__ multiply__3.0__119.0__ subtract__358.0__357.0__ multiply__1.0__93.0__ |
| a man can row num__6 kmph in still water . when the river is running at num__1.2 kmph it takes him num__1 hour to row to a place and black . how far is the place ? <o> a ) num__2.89 <o> b ) num__2.88 <o> c ) num__2.86 <o> d ) num__2.82 <o> e ) num__2.81 |
m = num__6 s = num__1.2 ds = num__6 + num__1.2 = num__7.2 us = num__6 - num__1.2 = num__4.8 x / num__7.2 + x / num__4.8 = num__1 x = num__2.88 . answer : b <eor> b <eos> |
b |
add__6.0__1.2__ subtract__6.0__1.2__ round__2.88__ |
add__6.0__1.2__ subtract__6.0__1.2__ divide__2.88__1.0__ |
| a rectangular field is to be fenced on three sides leaving a side of num__30 feet uncovered . if the area of the field is num__600 sq . ft how many feet of fencing will be required ? <o> a ) num__70 feet <o> b ) num__52 feet <o> c ) num__32 feet <o> d ) num__12 feet <o> e ) num__55 feet |
explanation : we are given with length and area so we can find the breadth . as length * breadth = area = > num__30 * breadth = num__600 = > breadth = num__20 feet area to be fenced = num__2 b + l = num__2 * num__20 + num__30 = num__70 feet answer : option a <eor> a <eos> |
a |
triangle_area__2.0__70.0__ |
triangle_area__2.0__70.0__ |
| x and y are integers . x + y < num__11 and x > num__6 . what is the smallest possible value of x - y ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__4 <o> d ) - num__2 <o> e ) - num__4 |
because x & y are integers the two equations can be simply written as : x + y < = num__10 - - - eq . num__1 x > = num__7 - - - - eq . num__2 for smallest value of x - y x has to be the smallest whereas y has to be the largest . from the above eq . num__2 smallest value of x = num__7 plugging it in eq . num__1 : num__7 + y < = num__10 or y < = num__3 therefore largest value of y = num__3 & smallest values of x = num__7 x - y = num__7 - num__3 = num__4 answer is c <eor> c <eos> |
c |
subtract__11.0__10.0__ add__6.0__1.0__ divide__6.0__2.0__ subtract__11.0__7.0__ subtract__11.0__7.0__ |
subtract__11.0__10.0__ add__6.0__1.0__ add__1.0__2.0__ subtract__11.0__7.0__ subtract__11.0__7.0__ |
| the product of two numbers is num__120 and the sum of their squares is num__289 . find the sum ? <o> a ) num__23 <o> b ) num__20 <o> c ) num__15 <o> d ) num__26 <o> e ) num__28 |
nos a and b ab = num__120 and a num__2 + b num__2 = num__289 ( a + b ) num__2 = num__529 a + b = root ( num__529 ) = num__23 answer a <eor> a <eos> |
a |
divide__529.0__23.0__ |
divide__529.0__23.0__ |
| in a certain state gasoline stations compute the price per gallon p in dollars charged at the pump by adding a num__7 percent sales tax to the dealer ' s price per gallon d in dollars and then adding a gasoline tax of $ num__0.18 per gallon . which of the following gives the dealer ' s price per gallon d in terms of the price per gallon p charged at the pump ? <o> a ) d = p / num__1.07 - num__0.18 <o> b ) d = p - num__0.25 <o> c ) d = ( p - num__0.18 ) / num__1.07 <o> d ) d = p / num__1.25 <o> e ) d = ( p - num__0.07 ) / num__1.18 |
let dealers price ( d ) be num__1 . so adding num__7.0 to dealers price is d + num__7.0 of d . i . e . num__1 + num__7.0 of num__1 which is num__1 + num__0.07 . then add num__0.18 to the value . now num__1.07 + num__0.18 . this is now num__1.25 . you have the gasoline stations price ( p ) as num__1.25 dollars . now sub num__1.25 in the options to know which option gave you d = num__1 . d must equal num__1 because you earlier picked num__1 as the value of d in the question . ps : always remember to start from e upwards . answer : c <eor> c <eos> |
c |
add__1.0__0.07__ add__0.18__1.07__ multiply__0.18__1.0__ |
add__1.0__0.07__ add__0.18__1.07__ multiply__0.18__1.0__ |
| each of the three people individually can complete a certain job in num__3 num__5 and num__6 hours respectively . what is the lowest fraction of the job that can be done in num__1 hour by num__2 of the people working together at their respective rates ? <o> a ) num__0.266666666667 <o> b ) num__0.233333333333 <o> c ) num__0.366666666667 <o> d ) num__0.388888888889 <o> e ) num__0.277777777778 |
the two slowest people work at rates of num__0.2 and num__0.166666666667 of the job per hour . the sum of these rates is num__0.2 + num__0.166666666667 = num__0.366666666667 of the job per hour . the answer is c . <eor> c <eos> |
c |
divide__1.0__5.0__ divide__1.0__6.0__ add__0.1667__0.2__ multiply__1.0__0.3667__ |
divide__1.0__5.0__ divide__1.0__6.0__ add__0.1667__0.2__ add__0.1667__0.2__ |
| machine – a produces num__40.0 of the total output and machine - b produces num__60.0 of the total output . an average of nine units out of a thousand goods manufactured by machine - a and one unit of num__300 units produced by machine - b prove to be defective . what is the probability that a unit chosen at random from the total daily output of the factory is defective ? <o> a ) a . num__0.56 <o> b ) b . num__0.056 <o> c ) c . num__0.0056 <o> d ) d . num__0.00056 <o> e ) e . num__0.000056 |
let total production be num__10000 units . . . . a produces num__4000 units and num__36 units are defective b produces num__6000 units and num__20 units are defective so out of total num__10000 units num__56 units are defective . . . so the required probability = num__0.0056 = > num__0.0056 answer will be ( c ) <eor> c <eos> |
c |
subtract__10000.0__4000.0__ subtract__60.0__40.0__ add__36.0__20.0__ divide__56.0__10000.0__ divide__56.0__10000.0__ |
subtract__10000.0__4000.0__ subtract__60.0__40.0__ add__36.0__20.0__ divide__56.0__10000.0__ divide__56.0__10000.0__ |
| find large number from below question the difference of two numbers is num__1515 . on dividing the larger number by the smaller we get num__16 as quotient and the num__15 as remainder <o> a ) num__1209 <o> b ) num__1615 <o> c ) num__1245 <o> d ) num__1300 <o> e ) num__1635 |
let the smaller number be x . then larger number = ( x + num__1515 ) . x + num__1515 = num__16 x + num__15 num__15 x = num__1500 x = num__100 large number = num__100 + num__1515 = num__1615 answer : b <eor> b <eos> |
b |
subtract__1515.0__15.0__ divide__1500.0__15.0__ add__1515.0__100.0__ add__1515.0__100.0__ |
subtract__1515.0__15.0__ divide__1500.0__15.0__ add__1515.0__100.0__ add__1515.0__100.0__ |
| suppose num__8 monkeys take num__8 minutes to eat num__8 bananas . how many monkeys would it take to eat num__48 bananas in num__48 minutes <o> a ) num__5 monkeys <o> b ) num__6 monkeys <o> c ) num__7 monkeys <o> d ) num__8 monkeys <o> e ) num__9 monkeys |
num__8 monkeys take num__8 minutes to eat num__8 bananas means each monkey takes num__8 min for num__1 banana in num__48 mins one monkey can eat num__6 bananas . . . . which means that num__8 monkeys can eat num__48 bananas in num__48 minutes so answer is num__8 monkeys answer : d <eor> d <eos> |
d |
divide__48.0__8.0__ round__8.0__ |
divide__48.0__8.0__ round__8.0__ |
| how many odd num__4 - digit positive integers e that are multiples of num__5 can be formed without using the digit num__3 ? <o> a ) num__648 <o> b ) num__729 <o> c ) num__900 <o> d ) num__1296 <o> e ) num__3240 |
i choose a . possible n ° of num__1 st digit : num__8 ( num__0 ca n ' t be the first number or else it would n ' t have num__4 digits . num__3 is exlcuded ) possible n ° of num__2 nd digit : num__9 ( num__3 is excluded ) possible n ° of num__3 rd digit : num__9 ( num__3 is excluded ) possible n ° of num__4 th digit : num__1 ( a number is a multiple of num__5 if it ends in num__5 or num__0 here we are asked for the odd numbers hence the last digit ca n ' t be num__0 ) so e = num__8 * num__9 * num__9 * num__1 = num__648 ( a ) <eor> a <eos> |
a |
subtract__4.0__3.0__ add__5.0__3.0__ subtract__5.0__3.0__ add__4.0__5.0__ multiply__1.0__648.0__ |
subtract__4.0__3.0__ add__5.0__3.0__ subtract__5.0__3.0__ add__4.0__5.0__ multiply__1.0__648.0__ |
| the average age of a husband and his wife was num__23 years at the time of their marriage . after five years they have a one - year old child . the average age of the family now is <o> a ) num__12 years <o> b ) num__27 years <o> c ) num__19 years <o> d ) num__18 years <o> e ) num__37 years |
sum of the present ages of husband wife and child = ( num__23 * num__2 + num__5 * num__2 ) + num__1 = num__57 years . required average = ( num__19.0 ) = num__19 years . answer : c <eor> c <eos> |
c |
multiply__1.0__19.0__ |
multiply__1.0__19.0__ |
| if p ( a ) = num__0.6 and p ( b ) = num__0.4 find p ( a n b ) if a and b are independent events . <o> a ) num__0.24 <o> b ) num__0.12 <o> c ) num__0.32 <o> d ) num__0.153846153846 <o> e ) num__0.176470588235 |
p ( a n b ) = p ( a ) . p ( b ) p ( a n b ) = num__0.6 . num__0.4 p ( a n b ) = num__0.24 . a <eor> a <eos> |
a |
multiply__0.6__0.4__ multiply__0.6__0.4__ |
multiply__0.6__0.4__ multiply__0.6__0.4__ |
| suppose you work for a manufacturing plant that pays you $ num__12.50 an hour plus $ num__0.16 for each widget you make . how many widgets must you produce in a num__40 hour week to earn $ num__580 ( before payroll deductions ) ? <o> a ) num__400 <o> b ) num__500 <o> c ) num__600 <o> d ) num__700 <o> e ) num__800 |
total pay = num__40 * $ num__12.50 + $ num__0.16 * x = num__580 x = num__80 / num__0.16 = num__500 the answer is b . <eor> b <eos> |
b |
multiply__12.5__40.0__ round__500.0__ |
multiply__12.5__40.0__ multiply__12.5__40.0__ |
| by the how much is two - fifth of num__200 greater than three - fifths of num__125 ? <o> a ) num__15 <o> b ) num__3 <o> c ) num__5 <o> d ) num__30 <o> e ) none of these |
reqd no . = num__2 ⁄ num__5 × num__200 - num__3 ⁄ num__5 × num__125 = num__80 - num__75 = num__5 answer c <eor> c <eos> |
c |
subtract__5.0__2.0__ subtract__200.0__125.0__ add__2.0__3.0__ |
subtract__5.0__2.0__ subtract__200.0__125.0__ subtract__80.0__75.0__ |
| the respective heights and volumes of a hemisphere and a right circular cylinder are equal then the ratio of their radii is : <o> a ) num__1.225 <o> b ) num__2.225 <o> c ) num__3.225 <o> d ) num__1.825 <o> e ) num__0.225 |
height of hemisphere = radius of hemisphere = height of right circular cylinder = h and radius of right circular cylinder = r then ( num__0.666666666667 ) * pi * h ^ num__3 = pi * ( r ^ num__2 ) * h h / r = root ( num__1.5 ) = root num__1.5 = num__1.225 answer : a <eor> a <eos> |
a |
triangle_area__2.0__1.225__ |
triangle_area__2.0__1.225__ |
| the average of seven numbers is num__18 . the average of first three numbers is num__14 and the average of last three numbers is num__16 . what is the middle number ? <o> a ) num__27 <o> b ) num__29 <o> c ) num__36 <o> d ) num__34 <o> e ) num__35 |
the total of seven numbers = num__7 x num__18 = num__126 the total of first num__3 and last num__3 numbers is = num__3 x num__14 + num__3 x num__16 = num__90 so the middle number is ( num__126 - num__90 ) = num__36 c <eor> c <eos> |
c |
multiply__18.0__7.0__ subtract__126.0__90.0__ subtract__126.0__90.0__ |
multiply__18.0__7.0__ subtract__126.0__90.0__ subtract__126.0__90.0__ |
| on a wedding catering service an experienced chef can prepare a service for a wedding in num__12 hours while an novice chef would finish the preparations in num__24 hours . if the catering service employs the same number of novice and experienced chefs then how many chefs would it take to prepare a wedding service in num__1 hour and num__36 minutes ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__6 <o> e ) num__10 |
experienced chefs work = num__1 wedding / num__12 hours novice chefs work = num__1 wedding / num__24 hours since we do n ' t know the number of experienced or novice chefs but know that there is an equal number each let the number of chefs for each group equalx num__1 hr and num__36 mins = num__1.6 an hour x / num__12 + x / num__24 = num__1 wedding / ( num__1.6 ) x / num__12 + x / num__24 = num__0.625 x = num__5 so there are num__5 novice chefs and num__5 experienced chefs . in total there are num__10 . the answer is e . <eor> e <eos> |
e |
mile_to_km_conversion__ divide__1.0__1.6__ round__10.0__ |
mile_to_km_conversion__ divide__1.0__1.6__ divide__10.0__1.0__ |
| if num__11 persons meet at a reunion and each person shakes hands exactly once with each of the others what is the total number of handshakes ? <o> a ) num__11 × num__10 × num__9 × num__8 × num__7 × num__6 × num__5 × num__4 × num__3 × num__2 × num__1 <o> b ) num__10 × num__9 × num__8 × num__7 × num__6 × num__5 × num__4 × num__3 × num__2 × num__1 <o> c ) num__11 × num__10 <o> d ) num__55 <o> e ) num__45 |
total number of handshakes = c num__112 = num__11 ∗ num__5.0 = num__55 answer : d <eor> d <eos> |
d |
multiply__11.0__5.0__ multiply__11.0__5.0__ |
multiply__11.0__5.0__ multiply__11.0__5.0__ |
| a reduction of num__22.0 in the price of salt enables a lady to obtain num__10 kgs more for rs . num__100 find the original price per kg ? <o> a ) s . num__2.82 <o> b ) s . num__2.4 <o> c ) s . num__2.5 <o> d ) s . num__2.2 <o> e ) s . num__2.1 |
num__100 * ( num__0.22 ) = num__22 - - - num__10 ? - - - num__1 = > rs . num__2.2 num__100 - - - num__78 ? - - - num__2.2 = > rs . num__2.82 answer : a <eor> a <eos> |
a |
percent__22.0__10.0__ percent__100.0__2.82__ |
percent__22.0__10.0__ percent__100.0__2.82__ |
| source : knewton a cyclist ' s speed varies depending on the terrain between num__6.0 miles per hour and num__13.0 miles per hour inclusive . what is the maximum distance in miles that the cyclist could travel in num__7 hours ? <o> a ) num__42 <o> b ) num__56 <o> c ) num__70 <o> d ) num__91 <o> e ) num__140 |
we are told that : generallya cyclist ' s speed varies depending on the terrain between num__6.0 miles per hour and num__13.0 miles per hour inclusive . is it possible the cyclist to travel with maximum speed for some time ? why not if there is right terrain for that . so if there is long enough terrain for the maximum speed of num__13 mph then the maximum distance in miles that the cyclist could travel in num__7 hours would be num__7 * num__13 = num__91 miles . answer : d . <eor> d <eos> |
d |
multiply__13.0__7.0__ round__91.0__ |
multiply__13.0__7.0__ multiply__13.0__7.0__ |
| there are num__400 female managers in a certain company . find the total number of female employees in the company if num__0.4 of all the employees are managers and num__0.4 of all male employees are managers . <o> a ) num__800 <o> b ) num__900 <o> c ) num__1000 <o> d ) num__1100 <o> e ) none of these |
as per question stem num__0.4 m ( portion of men employees who are managers ) + num__400 ( portion of female employees who are managers ) = num__0.4 t ( portion of total number of employees who are managers ) thus we get that num__0.4 m + num__400 = num__0.4 t or num__0.4 ( t - m ) = num__400 from here we get that t - m = num__1000 that would be total number of female employees and the answer ( c ) <eor> c <eos> |
c |
divide__400.0__0.4__ divide__400.0__0.4__ |
divide__400.0__0.4__ divide__400.0__0.4__ |
| two trains travelling in the same direction at num__50 and num__40 kmph completely pass off another in num__1 num__0.5 minute . if the length of the first train is num__125 m what is the length of the second train ? <o> a ) num__300 m <o> b ) num__287 m <o> c ) num__125 m <o> d ) num__250 m <o> e ) num__167 m |
rs = num__50 â € “ num__40 = num__10 * num__0.277777777778 = num__2.77777777778 mps t = num__90 sec d = num__2.77777777778 * num__90 = num__250 m num__125 - - - - - - - - num__125 m answer : c <eor> c <eos> |
c |
subtract__50.0__40.0__ add__50.0__40.0__ divide__125.0__0.5__ round__125.0__ |
subtract__50.0__40.0__ add__50.0__40.0__ divide__125.0__0.5__ multiply__1.0__125.0__ |
| the function f is defined by subtracting num__25 from the square of a number and the function s is defined as the square root of one - half of a number . if s ( f ( x ) ) = num__10 then which of the following is a possible value of x ? <o> a ) - num__15 <o> b ) - num__5 <o> c ) num__0 <o> d ) num__5 <o> e ) num__25 |
f ( x ) = x ^ num__2 - num__25 s ( x ) = sqrt ( x / num__2 ) not sqrt ( x ) / num__2 because the question clearly says its square root of ( half of the number ) . s ( f ( x ) ) = num__10 s ( x ^ num__2 - num__25 ) = num__10 sqrt ( ( x ^ num__2 - num__25 ) / num__2 ) = num__10 = > ( x ^ num__2 - num__25 ) / num__2 = num__100 = > x ^ num__2 = num__225 = > x = num__15 or - num__15 answer is a . <eor> a <eos> |
a |
subtract__25.0__10.0__ subtract__25.0__10.0__ |
subtract__25.0__10.0__ subtract__25.0__10.0__ |
| a train num__150 m long is running with a apecd of num__68 kmph . in what time will it pass a man who is running at num__8 kmph in the same direction in which the train is going ? <o> a ) num__6 sec . <o> b ) num__7 sec . <o> c ) num__9 sec . <o> d ) num__11 sec . <o> e ) none |
solution speed of the train relative to man = ( num__68 - num__8 ) = num__60 kmph = num__60 x num__0.277777777778 = num__16.6666666667 m / sec . time taken by it to cover num__150 m at ( num__16.6666666667 ) m / sec = ( num__112 - num__50 ) kmph = ( num__150 x num__0.06 ) sec = num__9 sec . answer c <eor> c <eos> |
c |
hour_to_min_conversion__ divide__150.0__16.6667__ round__9.0__ |
subtract__68.0__8.0__ divide__150.0__16.6667__ divide__150.0__16.6667__ |
| if num__30.0 of the num__880 students at a certain college are enrolled in biology classes how many students at the college are not enrolled in a biology class ? <o> a ) num__110 <o> b ) num__616 <o> c ) num__550 <o> d ) num__430 <o> e ) num__880 |
students enrolled in biology are num__30.0 and therefore not enrolled are num__70.0 . so of num__880 is num__880 * . num__7 = num__616 answer is b num__616 <eor> b <eos> |
b |
percent__70.0__880.0__ percent__70.0__880.0__ |
percent__70.0__880.0__ percent__70.0__880.0__ |
| a rectangular tiled patio is composed of num__96 square tiles . the rectangular patio will be rearranged so that there will be num__2 fewer columns of tiles and num__4 more rows of tiles . after the change in layout the patio will still have num__96 tiles and it will still be rectangular . how many rows are in the tile patio before the change in layout ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__8 <o> d ) num__13 <o> e ) num__28 |
suppose there are c columns and there are r rows original situation so number of tiles = c * r = num__96 also . reach column has r tiles and each row has c tiles new situation number of tiles in each column is r - num__2 and number of tiles in each row is c + num__4 so number of rows = r - num__2 and number of columns is c + num__4 so number of tiles = ( r - num__2 ) * ( c + num__4 ) = num__96 comparing both of them we get c * r = ( r - num__2 ) * ( c + num__4 ) = > num__4 r - num__2 c = num__8 c = num__2 r - num__4 putting it in c * r = num__96 ( num__2 r - num__4 ) * r = num__96 num__2 r ^ num__2 - num__4 r - num__96 = num__0 r can not be negative so r = num__8 and c = num__12 so answer will be c <eor> c <eos> |
c |
square_perimeter__2.0__ rectangle_perimeter__2.0__4.0__ square_perimeter__2.0__ |
multiply__2.0__4.0__ rectangle_perimeter__2.0__4.0__ multiply__2.0__4.0__ |
| if m is the least common multiple of num__90196 and num__300 which of the following is not a factor of m ? <o> a ) num__600 <o> b ) num__700 <o> c ) num__900 <o> d ) num__2100 <o> e ) num__4 |
900 |
first calculate the lcm of the given numbers num__90 = num__2 * num__3 * num__3 * num__5 num__196 = num__2 * num__2 * num__7 * num__7 num__300 = num__2 * num__2 * num__3 * num__5 * num__5 lcm = num__2 * num__2 * num__3 * num__3 * num__5 * num__5 * num__7 * num__7 this is the number m . now check each number whether a factor of m . answer a is num__600 . <eor> a <eos> |
a |
a |
| a sum of rs . num__66000 is divided into three parts such that the simple interests accrued on them for six two and eleven years respectively may be equal . find the amount deposited for num__11 years . <o> a ) num__6500 <o> b ) num__2000 <o> c ) num__4500 <o> d ) num__3000 <o> e ) num__6000 |
let the amounts be x y z in ascending order of value . as the interest rate and interest accrued are same for num__2 years num__6 years and num__11 years i . e . num__2 x = num__6 y = num__11 z = k . l . c . m . of num__2 num__611 = num__66 so x : y : z : = num__33000 : num__11000 : num__6000 the amount deposited for num__11 years = num__6000 answer : e <eor> e <eos> |
e |
multiply__11.0__6.0__ divide__66000.0__2.0__ divide__66000.0__6.0__ divide__66000.0__11.0__ divide__66000.0__11.0__ |
multiply__11.0__6.0__ divide__66000.0__2.0__ divide__66000.0__6.0__ divide__66000.0__11.0__ divide__66000.0__11.0__ |
| the c . p of num__10 pens is equal to the s . p of num__12 pens . find his gain % or loss % ? <o> a ) num__16 num__0.285714285714 % <o> b ) num__16 num__0.333333333333 % <o> c ) num__16 num__0.666666666667 % <o> d ) num__16 num__3.0 % <o> e ) num__16 num__0.333333333333 % |
num__10 cp = num__12 sp num__12 - - - num__2 cp loss num__100 - - - ? = > num__16 num__0.666666666667 % answer : c <eor> c <eos> |
c |
percent__100.0__16.0__ |
percent__100.0__16.0__ |
| which of the following is equal to the value of num__2 ^ num__4 + num__2 ^ num__4 + num__3 ^ num__4 + num__3 ^ num__4 + num__3 ^ num__4 ? <o> a ) num__2 ^ num__5 + num__3 ^ num__10 <o> b ) num__2 ^ num__5 + num__3 ^ num__9 <o> c ) num__2 ^ num__5 + num__3 ^ num__8 <o> d ) num__2 ^ num__5 + num__3 ^ num__5 <o> e ) num__2 ^ num__5 + num__3 ^ num__7 |
num__2 ^ num__4 + num__2 ^ num__4 + num__3 ^ num__4 + num__3 ^ num__4 + num__3 ^ num__4 = num__2 ( num__2 ^ num__4 ) + num__3 ( num__3 ^ num__4 ) = num__2 ^ num__1 ( num__2 ^ num__4 ) + num__3 ^ num__1 ( num__3 ^ num__4 ) = num__2 ^ ( num__1 + num__4 ) + num__3 ^ ( num__1 + num__4 ) = num__2 ^ num__5 + num__3 ^ num__5 ans : d <eor> d <eos> |
d |
subtract__4.0__3.0__ add__2.0__3.0__ multiply__2.0__1.0__ |
subtract__4.0__3.0__ add__2.0__3.0__ multiply__2.0__1.0__ |
| in store a there are num__10 pairs of pants for every num__30 store b has . the price ratio between the pants in store b and the pants in store a is num__3 : num__4 . if all the pants were sold in both places until the stock ran out what is the ratio between the total amount stores a earned to the total amount store b earned ? <o> a ) num__3 : num__16 . <o> b ) num__2 : num__3 . <o> c ) num__4 : num__9 <o> d ) num__3 : num__4 . <o> e ) num__2 : num__5 . |
num__1 st statement : ratio of pants store a : store b num__10 x : num__30 x x : num__3 x price : num__4 y : num__3 y total revenue num__4 xy : num__9 xy num__4 : num__9 answer : c <eor> c <eos> |
c |
subtract__4.0__3.0__ subtract__10.0__1.0__ add__3.0__1.0__ |
subtract__4.0__3.0__ subtract__10.0__1.0__ add__3.0__1.0__ |
| a man is num__24 years older than his son . in two years his age will be twice the age of his son . the present age of his son is : <o> a ) num__14 years <o> b ) num__18 years <o> c ) num__20 years <o> d ) num__22 years <o> e ) num__24 years |
let the son ' s present age be x years . then man ' s present age = ( x + num__24 ) years . ( x + num__24 ) + num__2 = num__2 ( x + num__2 ) x + num__26 = num__2 x + num__4 x = num__22 . answer : option d <eor> d <eos> |
d |
add__24.0__2.0__ subtract__24.0__2.0__ subtract__24.0__2.0__ |
add__24.0__2.0__ subtract__24.0__2.0__ subtract__24.0__2.0__ |
| kurt a french painter has num__9 jars of paint : num__4 jars of yellow paint num__2 jars of red paint and num__3 jars of brown paint . kurt pours the contents of num__3 jars of paint into a new container to make a new color which he will name according to the following conditions : the paint will be namedbrun yif it contains num__2 jars of brown paint and no yellow . the paint will be namedbrun xif the paint contains num__3 jars of brown paint . the paint will be namedjaune xif the paint contains at least num__2 jars of yellow . the paint will be namedjaune yif the paint contains exactly num__1 jar of yellow . what is the probability m that the new color will be one of thejaunecolors ? <o> a ) num__0.119047619048 <o> b ) num__0.880952380952 <o> c ) num__0.047619047619 <o> d ) num__0.444444444444 <o> e ) num__0.555555555556 |
i get b . jaune y = ( num__4 choose num__1 ) * ( num__5 choose num__2 ) = num__4 * num__10 = num__40 jaune x = ( num__4 choose num__2 ) * ( num__5 choose num__1 ) + ( num__4 choose num__3 ) = num__6 * num__5 + num__4 = num__34 total combinations = num__9 choose num__3 = num__84 probability of jaune mm = ( num__40 + num__34 ) / num__84 = num__0.880952380952 . b <eor> b <eos> |
b |
subtract__9.0__4.0__ add__9.0__1.0__ multiply__4.0__10.0__ subtract__9.0__3.0__ subtract__40.0__6.0__ multiply__1.0__0.881__ |
add__4.0__1.0__ add__9.0__1.0__ multiply__4.0__10.0__ add__4.0__2.0__ subtract__40.0__6.0__ multiply__1.0__0.881__ |
| tough and tricky questions : word problems . if bill ' s salary increased by num__16 percent he would earn $ num__812 per month . if instead his salary were to increase by only num__10 percent how much money t would he earn per month ? <o> a ) $ num__700 <o> b ) $ num__754 <o> c ) $ num__770 <o> d ) $ num__782 <o> e ) $ num__893 |
official solution : ( c ) if bill ’ s salary increased by num__16.0 he would earn $ num__812 . algebraically this can be written as : $ num__812 = num__1.16 s where s is his current salary . then s = $ num__812 / num__1.16 = $ num__700 . now that we know his current salary is $ num__700 we can calculate what his salary would be if it were increased by num__10.0 . we know that num__10.0 of $ num__700 is $ num__70 so his salary would be : t = $ num__700 + $ num__70 = $ num__770 . the correct answer is choice ( c ) . <eor> c <eos> |
c |
divide__812.0__1.16__ divide__700.0__10.0__ add__70.0__700.0__ add__70.0__700.0__ |
divide__812.0__1.16__ divide__700.0__10.0__ add__70.0__700.0__ add__70.0__700.0__ |
| when a positive integer a is divided by num__7 and num__3 the remainders obtained are num__1 and num__2 respectively . when the positive integer b is divided by num__7 and num__3 the remainders obtained are num__1 and num__2 respectively . which of the following is a factor of ( a - b ) ? <o> a ) num__10 <o> b ) num__14 <o> c ) num__21 <o> d ) num__28 <o> e ) num__36 |
a = num__7 k + num__1 = num__3 j + num__2 b = num__7 m + num__1 = num__3 n + num__2 a - b = num__7 ( k - m ) = num__3 ( j - n ) a - b is a multiple of both num__7 and num__3 so it is a multiple of num__21 . the answer is c . <eor> c <eos> |
c |
multiply__7.0__3.0__ multiply__7.0__3.0__ |
multiply__7.0__3.0__ multiply__7.0__3.0__ |
| a goods train leaves a station at a certain time and at a fixed speed . after ^ hours an express train leaves the same station and moves in the same direction at a uniform speed of num__90 kmph . this train catches up the goods train in num__4 hours . find the speed of the goods train . <o> a ) num__36 kmph <o> b ) num__16 kmph <o> c ) num__26 kmph <o> d ) num__46 kmph <o> e ) num__50 kmph |
. let the speed of the goods train be x kmph . distance covered by goods train in num__10 hours = distance covered by express train in num__4 hours num__10 x = num__4 x num__90 or x = num__36 . so speed of goods train = num__36 kmph answer a <eor> a <eos> |
a |
round__36.0__ |
round__36.0__ |
| of all the homes on gotham street num__0.666666666667 are termite - ridden and num__0.4 of these are collapsing . what fraction of the homes are termite - ridden but not collapsing ? <o> a ) a ) num__0.133333333333 <o> b ) b ) num__0.2 <o> c ) c ) num__0.8 <o> d ) d ) num__0.4 <o> e ) e ) num__0.2 |
let total homes be num__15 termite ridden = num__0.666666666667 ( num__15 ) = num__10 termite ridden and collapsing = num__0.4 ( num__10 ) = num__4 thus homes that are termite ridden but not collapsing = num__10 - num__4 = num__6 thus required ratio = num__0.4 = num__0.4 answer d <eor> d <eos> |
d |
multiply__0.4__10.0__ multiply__0.4__15.0__ divide__4.0__10.0__ |
multiply__0.4__10.0__ subtract__10.0__4.0__ divide__4.0__10.0__ |
| if the total income of a state increases by num__50.0 and the population of the state increases by num__20.0 what is the percent change in per capita income for the state ? <o> a ) a net increase of num__40.0 <o> b ) a net increase of num__30.0 <o> c ) a net increase of num__25.0 <o> d ) a net increase of num__20.0 <o> e ) a net decrease of num__20 % |
we ' re told that the total income of a state increases by num__50.0 and the total population increases by num__20.0 to start : total income = $ num__100 total population = num__100 people $ num__1.0 = $ num__1 per person after the increase : new total income = $ num__150 new total population = $ num__120 $ num__1.25 = $ num__1.25 per person the question asks for the percent increase in per capita income which means that we need to use the percentage change formula : percent change = ( new - old ) / old new = num__1.25 old = num__1 ( num__1.25 - num__1 ) / num__1 = ( num__0.25 ) / num__1 = num__0.25 = num__25.0 increase c <eor> c <eos> |
c |
percent__20.0__1.25__ percent__100.0__25.0__ |
percent__20.0__1.25__ percent__100.0__25.0__ |
| in a box of num__10 pens a total of num__3 are defective . if a customer buys num__2 pens selected at random from the box what is the probability that neither pen will be defective ? <o> a ) num__0.545454545455 <o> b ) num__0.466666666667 <o> c ) num__0.470588235294 <o> d ) num__0.409090909091 <o> e ) num__0.434782608696 |
p ( neither pen is defective ) = num__0.7 * num__0.666666666667 = num__0.466666666667 the answer is b . <eor> b <eos> |
b |
divide__2.0__3.0__ multiply__0.7__0.6667__ multiply__0.7__0.6667__ |
divide__2.0__3.0__ multiply__0.7__0.6667__ multiply__0.7__0.6667__ |
| if a young child was num__15 months old one year ago how old was the child in months x months ago ? <o> a ) x − num__27 <o> b ) x − num__12 <o> c ) num__15 − x <o> d ) num__24 − x <o> e ) num__27 − x |
the child ' s age today is num__15 months + num__12 months = num__27 months x months ago the child ' s age was ( num__27 - x ) months old . the answer is e . <eor> e <eos> |
e |
add__15.0__12.0__ add__15.0__12.0__ |
add__15.0__12.0__ add__15.0__12.0__ |
| how many positive factors do num__180 and num__90 have in common ? <o> a ) num__6 <o> b ) num__12 <o> c ) num__16 <o> d ) num__18 <o> e ) num__24 |
the number of common factors will be same as number of factors of the highest common factor ( hcf ) hcf of num__180 and num__90 is num__90 number of factors of num__90 = num__12 answer : b <eor> b <eos> |
b |
gcd__180.0__12.0__ |
gcd__180.0__12.0__ |
| a car is purchased on hire - purchase . the cash price is $ num__22 num__000 and the terms are a deposit of num__10.0 of the price then the balance to be paid off over num__60 equal monthly installments . interest is charged at num__12.0 p . a . what is the monthly installment ? <o> a ) $ num__503 <o> b ) $ num__504 <o> c ) $ num__515 <o> d ) $ num__528 <o> e ) $ num__537 |
explanation : cash price = $ num__22 num__000 deposit = num__10.0 Ã — $ num__22 num__000 = $ num__2200 loan amount = $ num__22000 â ˆ ’ $ num__2200 number of payments = num__60 = $ num__19800 i = p * r * t / num__100 i = num__11880 total amount = num__19800 + num__11880 = $ num__31680 regular payment = total amount / number of payments = num__528 answer : d <eor> d <eos> |
d |
percent__60.0__19800.0__ percent__100.0__528.0__ |
percent__60.0__19800.0__ percent__100.0__528.0__ |
| a motor cyclist participant of a race says ` ` we drove with the speed of num__10 miles an hour one way but while returning because of less traffic we drove on the same route with num__15 miles per hour . ' ' what was their average speed in the whole journey ? <o> a ) num__10 miles / hour <o> b ) num__12 miles / hour <o> c ) num__11 miles / hour <o> d ) num__13 miles / hour <o> e ) num__14 miles / hour |
num__2 xy / x + y = num__2 * num__6.0 = num__12 miles / hour answer : b <eor> b <eos> |
b |
add__10.0__2.0__ round__12.0__ |
add__10.0__2.0__ add__10.0__2.0__ |
| yesterday it took robert num__4 hours to drive from city a to city b . today it took robert num__2.5 hours to drive back from city В to city a along the same route . if he had saved num__15 minutes in both trips the speed for the round trip would be num__70 miles per hour . what is the distance between city a and city b ? <o> a ) num__90 <o> b ) num__120 <o> c ) num__150 <o> d ) num__210 <o> e ) num__300 |
num__2 d / num__70 = num__6 ( because time = num__4 + num__2.5 - num__0.5 hrs ) = > d = num__210 answer - d <eor> d <eos> |
d |
add__4.0__2.0__ subtract__2.5__2.0__ round__210.0__ |
add__4.0__2.0__ subtract__2.5__2.0__ round__210.0__ |
| a certain company that sells only cars and trucks reported that revenues from car sales in num__1997 were down num__11 percent from num__1996 and revenues from truck sales were up num__7 percent from num__1996 . if total revenues from car sales and truck sales in num__1997 were up num__1 percent from num__1996 what is the ratio of revenue from car sales in num__1996 to revenue from truck sales in num__1996 ? <o> a ) num__1 : num__2 <o> b ) num__4 : num__5 <o> c ) num__1 : num__1 <o> d ) num__3 : num__2 <o> e ) num__5 : num__3 |
this is a weighted average question . average of - num__11.0 and + num__7.0 is + num__1.0 . using w num__1 / w num__2 = ( a num__2 - avg ) / ( avg - a num__1 ) we get w num__1 / w num__2 = ( num__7 - num__1 ) / ( num__1 - ( - num__11 ) ) = num__0.5 revenue from car : revenue from trucks = num__1 : num__2 answer : a <eor> a <eos> |
a |
reverse__2.0__ reverse__1.0__ |
subtract__11.0__7.0__ subtract__3.0__1.0__ reverse__1.0__ |
| a sum amounts to rs . num__5292 in num__2 years at the rate of num__5.0 p . a . if interest was compounded yearly then what was the principal ? <o> a ) s . num__4000 <o> b ) s . num__5000 <o> c ) s . num__4500 <o> d ) s . num__4800 <o> e ) s . num__5800 |
ci = num__5292 r = num__5 n = num__2 ci = p [ num__1 + r / num__100 ] ^ num__2 = p [ num__1 + num__0.05 ] ^ num__2 num__5292 = p [ num__1.05 ] ^ num__2 num__5292 [ num__0.952380952381 ] ^ num__2 num__4800 answer : d <eor> d <eos> |
d |
percent__5.0__1.0__ percent__100.0__4800.0__ |
percent__5.0__1.0__ percent__100.0__4800.0__ |
| how many litres of pure acid are there in num__8 litres of a num__20.0 solution ? <o> a ) num__1.4 <o> b ) num__1.5 <o> c ) num__1.6 <o> d ) num__2.4 <o> e ) none |
solution quantity of pure acid = num__20.0 of num__8 litres = ( num__0.2 × num__8 ) litres = num__1.6 litres . answer c <eor> c <eos> |
c |
percent__8.0__20.0__ percent__8.0__20.0__ |
percent__8.0__20.0__ percent__8.0__20.0__ |
| { - num__10 - num__6 - num__5 - num__4 - num__2.5 - num__1 num__0 num__2.5 num__4 num__5 num__7 num__10 } a number is to be selected at random from the set above . what is the probability that the number selected will be a solution of the equation ( x - num__5 ) ( x + num__10 ) ( num__2 x - num__5 ) = num__0 ? <o> a ) num__0.0833333333333 <o> b ) num__0.166666666667 <o> c ) num__0.25 <o> d ) num__0.333333333333 <o> e ) num__0.5 |
roots of the equation ( x - num__5 ) ( x + num__10 ) ( num__2 x - num__5 ) = num__0 are x = num__5 x = - num__10 andx = num__2.5 = num__2.5 . so three solutions present in our set of num__12 distinct numbers therefore p = num__0.25 = num__0.25 . answer : c . <eor> c <eos> |
c |
add__10.0__2.0__ reverse__4.0__ reverse__4.0__ |
add__10.0__2.0__ reverse__4.0__ reverse__4.0__ |
| anitha runs her own baking company . this morning anitha worker baked num__810 plum pies . they also baked some more after lunch . in total they baked num__888 plum pies . how many plum pies did anitha workers bake after lunch ? <o> a ) num__28 <o> b ) num__56 <o> c ) num__78 <o> d ) num__13 <o> e ) num__34 |
num__888 - num__810 = num__78 . answer is c . <eor> c <eos> |
c |
subtract__888.0__810.0__ subtract__888.0__810.0__ |
subtract__888.0__810.0__ subtract__888.0__810.0__ |
| if | num__7 x + num__2 | = num__16 then find the product of the values of x ? <o> a ) - num__5.14 <o> b ) num__6.19 <o> c ) - num__7.18 <o> d ) - num__8.62 <o> e ) num__5.69 |
| num__7 x + num__2 | = num__16 num__7 x + num__2 = num__16 or num__7 x + num__2 = - num__16 num__7 x = num__14 or num__7 x = - num__18 x = num__2 or x = - num__2.57 product = - num__2.57 * num__2 = - num__5.14 answer is a <eor> a <eos> |
a |
multiply__7.0__2.0__ add__2.0__16.0__ multiply__2.0__2.57__ multiply__2.0__2.57__ |
multiply__7.0__2.0__ add__2.0__16.0__ multiply__2.0__2.57__ multiply__2.0__2.57__ |
| there are num__16 teams in a soccer league and each team plays each of the others once . given that each game is played by two teams how many total games will be played ? . <o> a ) num__256 <o> b ) num__230 <o> c ) num__196 <o> d ) num__169 <o> e ) num__120 |
let ' s call the teams . . . abcde fghij klmno p team a plays each of the other num__15 teams so that ' s num__15 games . team b already played team a so it plays num__14 other games . team c already played teams a and b so it plays num__13 other games . team d already played teams a b and c so it plays num__12 other games . etc . the sum of all of these games is . . . num__15 + num__14 + num__13 . . . . . + num__3 + num__2 + num__1 = num__120 final answer : e <eor> e <eos> |
e |
subtract__16.0__13.0__ subtract__16.0__14.0__ subtract__16.0__15.0__ multiply__1.0__120.0__ |
subtract__16.0__13.0__ subtract__16.0__14.0__ subtract__16.0__15.0__ multiply__1.0__120.0__ |
| the divisor is num__21 the quotient is num__14 and the remainder is num__7 . what is the dividend ? <o> a ) num__201 <o> b ) num__394 <o> c ) num__302 <o> d ) num__301 <o> e ) num__294 |
d = d * q + r d = num__21 * num__14 + num__7 d = num__294 + num__7 d = num__301 <eor> d <eos> |
d |
multiply__21.0__14.0__ add__7.0__294.0__ add__7.0__294.0__ |
multiply__21.0__14.0__ add__7.0__294.0__ add__7.0__294.0__ |
| anne and beth will participate in a sack race ( in a sack race people hop to reach the finish line ) . in the time that anne takes num__3 hops beth takes num__4 hops but the distance covered by anne in num__4 hops is equal to distance covered by beth in num__5 hops . what is the ratio of anne ’ s speed : beth ’ s speed ? <o> a ) num__3 : num__5 <o> b ) num__12 : num__20 <o> c ) num__15 : num__16 <o> d ) num__1 : num__1 <o> e ) num__5 : num__3 |
the distance covered by anne in num__4 hops is equal to distance covered by beth in num__5 hops - - > num__4 * ( distance of ann ' s num__1 hop ) = num__5 * ( distance of beth ' s num__1 hop ) . the time that anne takes num__3 hops beth takes num__4 hops - - > time for ann to take num__12 hops - - > num__3 * ( time for ann ' s num__1 hop ) = num__4 * ( time for beth ' s num__1 hop ) . divide one by another : num__1.33333333333 * ( distance of ann ' s num__1 hop ) / ( time for ann ' s num__1 hop ) = num__1.25 * ( distance of beth ' s num__1 hop ) / ( time for beth ' s num__1 hop ) ; num__1.33333333333 * ( anne ' s speed ) = num__1.25 * ( beth ' s speed ) ; num__1.33333333333 * ( anne ' s speed ) / ( beth ' s speed ) = num__0.9375 . answer : c . <eor> c <eos> |
c |
subtract__4.0__3.0__ multiply__3.0__4.0__ divide__4.0__3.0__ divide__5.0__4.0__ divide__1.25__1.3333__ multiply__3.0__5.0__ |
subtract__4.0__3.0__ multiply__3.0__4.0__ divide__4.0__3.0__ divide__5.0__4.0__ divide__1.25__1.3333__ multiply__3.0__5.0__ |
| compound interest earned on a sum for the second and the third years are rs . num__1200 and rs . num__1440 respectively . find the rate of interest ? <o> a ) num__80.0 p . a <o> b ) num__50.0 p . a <o> c ) num__20.0 p . a <o> d ) num__27.0 p . a <o> e ) num__23.0 p . a |
rs . num__1440 - num__1200 = rs . num__240 is the interest on rs . num__1200 for one year . rate of interest = ( num__100 * num__240 ) / ( num__100 * num__1 ) = num__20.0 p . a answer : c <eor> c <eos> |
c |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| what is the unit digit in num__7 ^ num__105 ? <o> a ) num__1 <o> b ) num__5 <o> c ) num__7 <o> d ) num__9 <o> e ) none of them |
unit digit in num__7 ^ num__105 = unit digit in [ ( num__7 ^ num__4 ) ^ num__26 x num__7 ] but unit digit in ( num__7 ^ num__4 ) ^ num__26 = num__1 therefore unit digit in num__7 ^ num__105 = ( num__1 x num__7 ) = num__7 answer is c <eor> c <eos> |
c |
multiply__7.0__1.0__ |
multiply__7.0__1.0__ |
| if x dollars is invested at num__10 percent for one year and y dollars is invested at num__8 percent for one year the annual income from the num__10 percent investment will exceed the annual income from the num__8 percent investment by $ num__83 . if $ num__2000 is the total amount invested how much is invested at num__8 percent ? <o> a ) $ num__550 <o> b ) $ num__650 <o> c ) $ num__750 <o> d ) $ num__850 <o> e ) $ num__950 |
num__0.1 x = num__0.08 ( num__2000 - x ) + num__83 num__0.18 x = num__243 x = num__1350 then the amount invested at num__8.0 is $ num__2000 - $ num__1350 = $ num__650 the answer is b . <eor> b <eos> |
b |
reverse__10.0__ add__0.1__0.08__ divide__243.0__0.18__ subtract__2000.0__1350.0__ subtract__2000.0__1350.0__ |
reverse__10.0__ add__0.1__0.08__ divide__243.0__0.18__ subtract__2000.0__1350.0__ subtract__2000.0__1350.0__ |
| an equilateral triangle is inscribed in a circle as shown above . what is the area of the shaded region if the area of the circle is num__2 ? <o> a ) num__2 - num__3 √ num__3 <o> b ) num__2 - num__3 √ num__3 / ∏ <o> c ) num__2 - num__3 √ num__0.75 <o> d ) num__2 - num__3 √ num__3 / ( num__2 ∏ ) <o> e ) num__2 ∏ - num__3 √ num__3 / ( num__2 ∏ ) |
num__1 ) since we are subtracting from overall so first term should be num__2 . . . e out num__2 ) the area of triangle that is to be subtracted should be in terms of π . . . a and c are out . . also a gives you answer in negative num__3 ) in b ans would be around num__0.3 and in d ans will be around num__1.2 . . . clearly the shaded and unshaded are close to num__0.5 each . . d is the answer <eor> d <eos> |
d |
square_perimeter__0.3__ square_perimeter__0.5__ |
square_perimeter__0.3__ square_perimeter__0.5__ |
| let x be the smallest positive integer such that num__10 and num__14 are factors of num__250 + x so the value of x is : <o> a ) num__0 <o> b ) num__10 <o> c ) num__20 <o> d ) num__30 <o> e ) num__40 |
you can calculate the lcm ( num__1014 ) = num__70 . after that you should check which of the possible x values will result in a number divisible by num__70 . the only possibility for this is num__30 ( because num__250 + num__30 = num__280 = num__70 * num__4 ) answer d <eor> d <eos> |
d |
add__250.0__30.0__ subtract__14.0__10.0__ subtract__280.0__250.0__ |
add__250.0__30.0__ subtract__14.0__10.0__ subtract__280.0__250.0__ |
| two friends p & q started a business investing amounts in the ratio of num__5 : num__6 . r joined them after six months investing an amount equal to that of q ’ s amount . at the end of the year num__20.0 profit was earned which was equal to num__98000 . what was the amount invested by r ? <o> a ) num__2 num__10000 <o> b ) num__1 num__05000 <o> c ) num__1 num__75000 <o> d ) data inadequate <o> e ) none of these |
ratio for amount invested by p q & r = num__5 x × num__12 : num__6 x × num__12 : num__6 x × num__6 = num__60 x : num__72 x : num__36 x = num__5 x : num__6 x : num__3 x profit = num__98000 = num__20.0 of t where t = total amount t = num__490000 amount received by r = num__3 x / num__3 x + num__6 x + num__5 x ( num__490000 ) = num__105000 answer b <eor> b <eos> |
b |
percent__5.0__60.0__ percent__5.0__20.0__ |
percent__5.0__60.0__ percent__5.0__20.0__ |
| in how many h ways can a four - letter password be chosen using the letters a b c d e and / or f such that at least one letter is repeated within the password ? <o> a ) num__720 <o> b ) num__864 <o> c ) num__900 <o> d ) num__936 <o> e ) num__1 |
296 |
total number of four letter passwords = num__6 * num__6 * num__6 * num__6 = num__1296 - - - - - - ( num__1 ) total number of passwords in which no letter repeats = num__6 c num__4 * num__4 ! = num__15 * num__24 = num__360 - - - - - - ( num__2 ) therefore required value h = ( num__1 ) - ( num__2 ) = num__1296 - num__360 = num__936 . d <eor> d <eos> |
d |
d |
| line m lies in the xy - plane . the y - intercept of line m is - num__2 and line m passes through the midpoint of the line segment whose endpoints are ( num__2 num__8 ) and ( num__144 ) . what is the slope of line m ? <o> a ) - num__2 <o> b ) - num__1 <o> c ) num__0 <o> d ) num__1 <o> e ) num__2 |
the midpoint of ( num__28 ) and ( num__144 ) is ( num__86 ) . the slope of a line through ( num__0 - num__2 ) and ( num__86 ) is ( num__6 - ( - num__2 ) ) / ( num__8 - num__0 ) = num__1.0 = num__1 the answer is d . <eor> d <eos> |
d |
subtract__8.0__2.0__ reverse__1.0__ |
subtract__8.0__2.0__ reverse__1.0__ |
| the average age of a group of num__5 members is num__20 years . two years later a new member joins the group . the average age of the group becomes num__21 years . what is the age of the new member ? <o> a ) num__20 years <o> b ) num__21 years <o> c ) num__16 years <o> d ) num__23 years <o> e ) num__24 years |
the average age of a group of num__5 members is num__20 years - - > the sum of the ages is num__5 * num__20 = num__100 ; two years later the sum of the ages of these num__5 members would be num__100 + num__5 * num__2 = num__110 ; now say the age of the new member is x years so the sum of the ages of a new num__6 member group is num__110 + x . since given that the average age of this group of num__6 members is num__21 years then : num__21 * num__6 = num__110 + x - - > x = num__16 . answer : c . <eor> c <eos> |
c |
multiply__5.0__20.0__ subtract__21.0__5.0__ subtract__21.0__5.0__ |
multiply__5.0__20.0__ subtract__21.0__5.0__ subtract__21.0__5.0__ |
| a certain family has num__3 sons : richard is num__6 years older than david and david is num__8 years older than scott . if in num__8 years richard will be twice as old as scott then how old was david num__3 years ago ? <o> a ) num__8 <o> b ) num__11 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
let ' s say age of richard isr age of david isd age of scott iss now richard is num__6 years older than david i . e . r = d + num__6 david is num__8 years older than scott i . e . d = s + num__8 if in num__8 years richard will be twice as old as scott i . e . r + num__8 = num__2 x ( s + num__8 ) i . e . r + num__8 = num__2 s + num__16 i . e . r = num__2 s + num__8 but r = d + num__6 = ( s + num__8 ) + num__6 = s + num__14 therefore num__2 s + num__8 = s + num__14 i . e . s = num__6 i . e . r = num__20 i . e . d = num__14 now how old was david num__3 years ago ? i . e . d - num__3 = num__14 - num__3 = num__11 years answer : option b <eor> b <eos> |
b |
divide__6.0__3.0__ twice__8.0__ add__6.0__8.0__ add__6.0__14.0__ add__3.0__8.0__ add__3.0__8.0__ |
subtract__8.0__6.0__ twice__8.0__ add__6.0__8.0__ add__6.0__14.0__ add__3.0__8.0__ add__3.0__8.0__ |
| margaret is num__7 years more than twice the age of his son . the age of son is num__12 . find the age of mother and find the difference between their ages <o> a ) num__19 yrs <o> b ) num__20 yrs <o> c ) num__30 yrs <o> d ) num__40 yrs <o> e ) num__50 yrs |
let age of son x = num__12 margaret is num__7 years more than twice the age of his son y = num__7 + num__2 ( x ) = num__7 + num__2 ( num__12 ) = num__7 + num__24 = num__31 yrs difference = num__31 - num__12 = num__19 yrs answer : a <eor> a <eos> |
a |
multiply__12.0__2.0__ add__7.0__24.0__ add__7.0__12.0__ add__7.0__12.0__ |
multiply__12.0__2.0__ add__7.0__24.0__ add__7.0__12.0__ add__7.0__12.0__ |
| barbata invests $ num__2800 in the national bank at num__5.0 . how much additional money must she invest at num__8.0 so that the total annual income will be equal to num__6.0 of her entire investment ? <o> a ) num__1200 <o> b ) num__3000 <o> c ) num__1000 <o> d ) num__1400 <o> e ) num__2400 |
let the additional invested amount for num__8.0 interest be x ; equation will be ; num__2800 + num__0.05 * num__2800 + x + num__0.08 x = num__2800 + x + num__0.06 ( num__2800 + x ) num__0.05 * num__2800 + num__0.08 x = num__0.06 x + num__0.06 * num__2800 num__0.02 x = num__2800 ( num__0.06 - num__0.05 ) x = num__2800 * num__0.01 / num__0.02 = num__1400 ans : ` ` d ' ' <eor> d <eos> |
d |
subtract__0.08__0.06__ divide__0.05__5.0__ subtract__2800.0__1400.0__ |
subtract__0.08__0.06__ divide__0.05__5.0__ subtract__2800.0__1400.0__ |
| in a class of num__40 students num__2 students did not borrow any books from the library num__12 students each borrowed num__1 book num__11 students each borrowed num__2 books and the rest borrowed at least num__3 books . if the average number of books per student was num__2 what is the maximum number of books any single student could have borrowed ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
the class borrowed a total of num__40 * num__2 = num__80 books . the num__25 students who borrowed num__0 num__1 or num__2 books borrowed a total of num__12 + num__11 * num__2 = num__34 . to maximize the number of books borrowed by num__1 student let ' s assume that num__14 students borrowed num__3 books and num__1 student borrowed the rest . num__80 - num__34 - num__3 * num__14 = num__4 the maximum number of books borrowed by any student is num__4 . the answer is b . <eor> b <eos> |
b |
multiply__40.0__2.0__ add__2.0__12.0__ divide__12.0__3.0__ divide__12.0__3.0__ |
multiply__40.0__2.0__ add__2.0__12.0__ add__1.0__3.0__ add__1.0__3.0__ |
| a taxi leaves point a num__3 hours after a bus left the same spot . the bus is traveling num__30 mph slower than the taxi . find the speed of the taxi if it overtakes the bus in three hours . <o> a ) num__60 <o> b ) num__72 <o> c ) num__48 <o> d ) num__36 <o> e ) num__64 |
let the speed of bus be v - num__30 speed of taxi be v the bus travelled a total of num__6 hrs and taxi a total of num__3 hrs . hence num__6 * ( v - num__30 ) = num__3 v num__6 v - num__180 = num__3 v num__3 v = num__180 v = num__60 mph a <eor> a <eos> |
a |
multiply__30.0__6.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
multiply__30.0__6.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
| on sunday bill ran num__4 more miles than he ran on saturday . julia did not run on saturday but she ran twice the number of miles on sunday that bill ran on sunday . if bill and julia ran a total of num__32 miles on saturday and sunday how many miles did bill run on sunday ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
let bill run x on saturday so he will run x + num__4 on sunday . . julia will run num__2 * ( x + num__4 ) on sunday . . totai = x + x + num__4 + num__2 x + num__8 = num__32 . . num__4 x + num__12 = num__32 . . x = num__5 . . ans = x + num__4 = num__5 + num__4 = num__9 answer e <eor> e <eos> |
e |
multiply__4.0__2.0__ add__4.0__8.0__ add__4.0__5.0__ round__9.0__ |
multiply__4.0__2.0__ add__4.0__8.0__ add__4.0__5.0__ add__4.0__5.0__ |
| if a > num__1 and b = num__2 ^ ( a − num__1 ) then num__4 ^ a = <o> a ) num__16 b ^ num__2 <o> b ) num__4 b ^ num__2 <o> c ) b ^ num__2 <o> d ) b ^ num__0.5 <o> e ) b ^ num__0.125 |
if a > num__1 and b = num__2 ^ ( a − num__1 ) then num__4 ^ a given a > num__1 so let ' s assume a = num__2 b = num__2 ^ ( a - num__1 ) = num__2 ^ ( num__2 - num__1 ) = num__2 so b = num__2 hence num__4 ^ a = num__4 ^ num__2 = num__16 only num__1 ans . choice can satisfy this : a ) num__16 b ^ num__2 - - > clearly > num__16 b ) num__4 b ^ num__2 - - > num__4 * num__2 ^ num__2 = num__16 ( we can stop after this as there can be only num__1 right answer ) c ) b ^ num__2 - - > clearly < num__16 d ) b ^ num__0.5 - - > clearly < num__16 e ) b ^ num__0.125 - - > clearly < num__16 ans . b ) num__4 b ^ num__2 <eor> b <eos> |
b |
reverse__2.0__ divide__2.0__16.0__ multiply__1.0__4.0__ |
reverse__2.0__ divide__2.0__16.0__ multiply__1.0__4.0__ |
| a man can row a boat at num__20 kmph in still water . if the speed of the stream is num__8 kmph what is the time taken to row a distance of num__60 km downstream ? <o> a ) num__1.66666666667 <o> b ) num__1.03448275862 <o> c ) num__2.14285714286 <o> d ) num__2.30769230769 <o> e ) num__0.365853658537 |
speed downstream = num__20 + num__8 = num__28 kmph . time required to cover num__60 km downstream = d / s = num__2.14285714286 = num__2.14285714286 hours . answer : c <eor> c <eos> |
c |
add__20.0__8.0__ divide__60.0__28.0__ divide__60.0__28.0__ |
add__20.0__8.0__ divide__60.0__28.0__ divide__60.0__28.0__ |
| a person spends num__0.333333333333 rd of the money with him on clothes num__0.2 th of the remaining on food and num__0.25 th of the remaining on travel . now he is left with rs num__500 . how much did he have with him in the beginning ? <o> a ) s num__200 <o> b ) s num__1250 <o> c ) s num__300 <o> d ) s num__450 <o> e ) s num__550 |
suppose the amount in the beginning was rs ’ x ’ money spent on clothes = rs num__1 x / num__3 balance = rs num__2 x / num__3 money spent on food = num__0.2 of num__2 x / num__3 = rs num__2 x / num__15 balance = num__2 x / num__3 - num__2 x / num__15 = rs num__8 x / num__15 money spent on travel = num__0.25 of num__8 x / num__15 = rs num__2 x / num__15 = num__8 x / num__15 - num__2 x / num__15 = num__6 x / num__15 = rs num__2 x / num__5 therefore num__2 x / num__5 = num__500 = num__1250 answer : b <eor> b <eos> |
b |
subtract__3.0__1.0__ divide__3.0__0.2__ divide__2.0__0.25__ multiply__2.0__3.0__ reverse__0.2__ multiply__1.0__1250.0__ |
subtract__3.0__1.0__ divide__3.0__0.2__ divide__2.0__0.25__ subtract__8.0__2.0__ reverse__0.2__ divide__1250.0__1.0__ |
| if a store owner increases a product ’ s price by num__20 percent and then increases this price by another num__15 percent what percent of the original price is the total price increase ? <o> a ) num__20.0 <o> b ) num__35.0 <o> c ) num__38.0 <o> d ) num__65.0 <o> e ) num__135 % |
let the initial price be num__100 num__1 st increase = num__20.0 the price now is num__120 num__2 nd increase of num__15.0 on num__120 = num__18 final price = num__138 total increase = num__38.0 answer : c <eor> c <eos> |
c |
add__20.0__100.0__ subtract__20.0__2.0__ add__18.0__120.0__ add__20.0__18.0__ add__20.0__18.0__ |
add__20.0__100.0__ subtract__20.0__2.0__ add__18.0__120.0__ add__20.0__18.0__ add__20.0__18.0__ |
| today is peter ' s birthday . one year from today he will be twice as old as he was num__9 years ago . how old is peter today ? <o> a ) num__25 years <o> b ) num__19 years <o> c ) num__17 years <o> d ) num__23 years <o> e ) num__21 years |
let peter ' s age = x x + num__1 = num__2 ( x - num__9 ) x = num__19 answer : b <eor> b <eos> |
b |
round__19.0__ |
round__19.0__ |
| the marks obtained by vijay and amith are in the ratio num__4 : num__4 and those obtained by amith and abhishek in the ratio of num__3 : num__2 . the marks obtained by vijay and abhishek are in the ratio of ? <o> a ) num__3 : num__2 <o> b ) num__6 : num__1 <o> c ) num__6 : num__5 <o> d ) num__6 : num__2 <o> e ) num__6 : num__3 |
num__4 : num__4 num__3 : num__2 - - - - - - - num__12 : num__12 : num__8 num__12 : num__8 num__3 : num__2 answer : a <eor> a <eos> |
a |
multiply__4.0__3.0__ multiply__4.0__2.0__ divide__12.0__4.0__ |
multiply__4.0__3.0__ subtract__12.0__4.0__ divide__12.0__4.0__ |
| a work as fast as b . if b can complete a work in num__24 days independently the number of days in which a and b can together finish the work in ? <o> a ) num__2 days <o> b ) num__3 days <o> c ) num__8 days <o> d ) num__5 days <o> e ) num__6 days |
ratio of rates of working of a and b = num__2 : num__1 ratio of times taken = num__1 : num__2 a ' s num__1 day work = num__0.0833333333333 b ' s num__1 day work = num__0.0416666666667 a + b num__1 day work = num__0.0833333333333 + num__0.0416666666667 = num__0.125 = num__0.125 a and b can finish the work in num__8 days answer is c <eor> c <eos> |
c |
divide__2.0__24.0__ divide__1.0__24.0__ add__0.0417__0.0833__ divide__1.0__0.125__ round__8.0__ |
divide__2.0__24.0__ divide__1.0__24.0__ add__0.0417__0.0833__ divide__1.0__0.125__ round__8.0__ |
| an investor bought num__200 shares of stock in abcd company in num__1990 . by num__1992 the investment was worth only num__0.666666666667 of its original value . by num__1995 the num__200 shares were worth only num__0.2 of their value in num__1990 . by what percent did the value of the investment drop from num__1992 to num__1995 ? <o> a ) num__16 num__0.666666666667 % <o> b ) num__70.0 <o> c ) num__33 num__0.333333333333 % <o> d ) num__50.0 <o> e ) num__66 num__0.666666666667 % |
let price of each share in num__1990 = x . total cost in num__1990 = num__200 x now price in num__1992 = num__0.666666666667 * num__200 x = num__133.333333333 * x now price in num__1995 = num__0.2 * num__200 x = num__40 x % change in num__1995 from num__1992 = [ ( num__133.333333333 * x - num__40 x ) / ( num__133.333333333 * x ) ] * num__100 = num__70.0 answer b <eor> b <eos> |
b |
percent__100.0__70.0__ |
percent__100.0__70.0__ |
| the parameter of a square is equal to the perimeter of a rectangle of length num__16 cm and breadth num__14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) <o> a ) num__23.58 cm <o> b ) num__23.17 cm <o> c ) num__28.17 cm <o> d ) num__23.77 cm <o> e ) num__23.57 cm |
let the side of the square be a cm . parameter of the rectangle = num__2 ( num__16 + num__14 ) = num__60 cm parameter of the square = num__60 cm i . e . num__4 a = num__60 a = num__15 diameter of the semicircle = num__15 cm circimference of the semicircle = num__0.5 ( ∏ ) ( num__15 ) = num__0.5 ( num__3.14285714286 ) ( num__15 ) = num__23.5714285714 = num__23.57 cm to two decimal places answer : e <eor> e <eos> |
e |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
| a train when moves at an average speed of num__40 kmph reaches its destination on time . when its average speed becomes num__35 kmph then it reaches its destination num__15 minutes late . find the length of the journey <o> a ) num__40 km <o> b ) num__50 km <o> c ) num__60 km <o> d ) num__70 km <o> e ) num__80 km |
difference between timings = num__15 min = num__0.25 hr . let the length of journey be x km . then x / num__35 - x / num__40 = num__0.25 num__8 x - num__7 x = num__70 x = num__70 km . answer : d <eor> d <eos> |
d |
subtract__15.0__8.0__ round__70.0__ |
subtract__15.0__8.0__ round__70.0__ |
| in a rectangular coordinate system what is the area of a triangle whose vertices have the coordinates ( num__2 num__0 ) ( num__7 num__4 ) and ( num__7 - num__4 ) ? <o> a ) num__18 <o> b ) num__19 <o> c ) num__20 <o> d ) num__21 <o> e ) num__22 |
the triangle is symmetric about the x - axis . the part above the x - axis forms a triangle with a base of num__5 and a height of num__4 . the area of this part is ( num__0.5 ) ( num__5 ) ( num__4 ) . we can double this to find the area of the whole triangle . the total area is ( num__2 ) ( num__0.5 ) ( num__5 ) ( num__4 ) = num__20 . the answer is c . <eor> c <eos> |
c |
square_perimeter__5.0__ square_perimeter__5.0__ |
square_perimeter__5.0__ square_perimeter__5.0__ |
| the area of square abcd above is num__22 . the shaded region is the intersection of the square and a circular region centered at c . what is the area of the shaded region ? <o> a ) num__18 − num__2.25 ∗ π <o> b ) num__22 − num__5.5 ∗ π <o> c ) num__9 + num__2.25 ∗ π <o> d ) num__9 + num__4.5 ∗ π <o> e ) num__4.5 − num__2.25 ∗ π |
= square - quarter circle = num__22 - pi * num__5.5 = num__22 - pi * num__5.5 answer : b <eor> b <eos> |
b |
square_perimeter__5.5__ |
square_perimeter__5.5__ |
| if num__9 is added to twice a number and this sum is multiplied by num__3 the result is the same as if the number is multiplied by num__4 and num__13 is added to the product . what is the number ? <o> a ) - num__7 <o> b ) - num__8 <o> c ) - num__9 <o> d ) - num__10 <o> e ) - num__11 |
let the number be x ; num__3 ( num__9 + num__2 x ) = num__4 x + num__13 ; x = - num__7 answer : a <eor> a <eos> |
a |
subtract__9.0__2.0__ subtract__9.0__2.0__ |
subtract__9.0__2.0__ subtract__9.0__2.0__ |
| a man can row with a speed of num__14 kmph in still water . if the stream flows at num__8 kmph then the speed in downstream is ? <o> a ) num__22 <o> b ) num__28 <o> c ) num__20 <o> d ) num__82 <o> e ) num__34 |
m = num__14 s = num__8 ds = num__14 + num__8 = num__22 . answer : a <eor> a <eos> |
a |
add__14.0__8.0__ round__22.0__ |
add__14.0__8.0__ add__14.0__8.0__ |
| a shopkeeper sold an article offering a discount of num__4.0 and earned a profit of num__32.0 . what would have been the percentage of profit earned if no discount was offered ? <o> a ) num__24.5 <o> b ) num__28.5 <o> c ) num__30.5 <o> d ) num__32.5 <o> e ) num__37.5 |
let c . p . be rs . num__100 . then s . p . = rs . num__132 let marked price be rs . x . then num__0.96 x = num__132 x = num__137.5 = rs . num__137.5 now s . p . = rs . num__137.5 c . p . = rs . num__100 profit % = num__37.5 . answer : e <eor> e <eos> |
e |
percent__100.0__37.5__ |
percent__100.0__37.5__ |
| the denominator of a fraction is num__1 less than twice the numerator . if the numerator and denominator are both increased by num__1 the fraction becomes num__0.6 . find the fraction ? <o> a ) num__0.333333333333 <o> b ) num__0.666666666667 <o> c ) num__0.555555555556 <o> d ) num__0.714285714286 <o> e ) num__0.636363636364 |
explanation : let the numerator and denominator of the fraction be ' n ' and ' d ' respectively . d = num__2 n - num__1 ( n + num__1 ) / ( d + num__1 ) = num__0.6 num__5 n + num__5 = num__3 d + num__3 num__5 n + num__5 = num__3 ( num__2 n - num__1 ) + num__3 = > n = num__5 d = num__2 n - num__1 = > d = num__9 hence the fraction is : num__0.555555555556 c <eor> c <eos> |
c |
add__1.0__2.0__ divide__5.0__9.0__ multiply__1.0__0.5556__ |
add__1.0__2.0__ divide__5.0__9.0__ divide__5.0__9.0__ |
| how many different subsets of the set { num__0 num__1 num__2 } do not contain num__0 ? <o> a ) a . num__6 <o> b ) b . num__7 <o> c ) c . num__1 <o> d ) d . num__4 <o> e ) e . num__2 |
number of subset since we have num__2 digits other than num__0 we can take any numbers from the set of num__2 to make a subset . also it is a matter of selection and not arrangement . so we will consider combinations . num__2 c num__1 + num__2 c num__2 = num__3 and one set is the null set having no elements in it so num__3 + num__1 = num__4 . answer d . <eor> d <eos> |
d |
choose__2.0__0.0__ choose__2.0__0.0__ |
choose__2.0__0.0__ choose__2.0__0.0__ |
| a b and c completed a piece of work costing rs . num__1800 . a worked for num__6 days b for num__4 days and c for num__9 days . if their daily wages are in the ratio num__5 : num__6 : num__4 how much amount will be received by a ? <o> a ) rs . num__800 <o> b ) rs . num__600 <o> c ) rs . num__900 <o> d ) rs . num__750 <o> e ) rs . num__850 |
explanation : the new ratio = wages ratio × no . of days a = num__5 × num__6 = num__30 b = num__6 × num__4 = num__24 c = num__4 × num__9 = num__36 therefore the ratio of the amount received = num__30 : num__24 : num__36 = num__5 : num__4 : num__6 total ratio = num__15 num__1 unit of ratio = num__120.0 = rs . num__120 therefore amount received by a = num__5 units = num__5 × num__120 = rs . num__600 . answer : option b <eor> b <eos> |
b |
multiply__6.0__5.0__ multiply__6.0__4.0__ add__6.0__30.0__ add__6.0__9.0__ subtract__6.0__5.0__ divide__1800.0__15.0__ multiply__5.0__120.0__ multiply__5.0__120.0__ |
multiply__6.0__5.0__ multiply__6.0__4.0__ add__6.0__30.0__ add__6.0__9.0__ subtract__6.0__5.0__ divide__1800.0__15.0__ multiply__5.0__120.0__ multiply__5.0__120.0__ |
| which answer is closest to the √ num__17 ? <o> a ) a num__4.1 <o> b ) b num__4.2 <o> c ) c num__4.4 <o> d ) d num__4.5 <o> e ) e num__4.6 |
try filling the numbers into the y ' s . y x y = num__4.1 x num__4.1 = num__16.81 num__4.2 x num__4.2 = num__17.64 answer : a <eor> a <eos> |
a |
divide__16.81__4.1__ |
divide__16.81__4.1__ |
| find the value of num__72518 x num__9999 = m ? <o> a ) num__456578972 <o> b ) num__436567874 <o> c ) num__653658791 <o> d ) num__725117481 <o> e ) num__725107482 |
num__72518 x num__9999 = num__72518 x ( num__10000 - num__1 ) = num__72518 x num__10000 - num__72518 x num__1 = num__725180000 - num__72518 = num__725107482 e <eor> e <eos> |
e |
subtract__10000.0__9999.0__ multiply__72518.0__10000.0__ multiply__72518.0__9999.0__ multiply__72518.0__9999.0__ |
subtract__10000.0__9999.0__ multiply__72518.0__10000.0__ subtract__725180000.0__72518.0__ subtract__725180000.0__72518.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 sec . what is the length of the train ? <o> a ) num__118 <o> b ) num__150 <o> c ) num__277 <o> d ) num__258 <o> e ) num__191 |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__9 = num__150 m answer : b <eor> b <eos> |
b |
round__150.0__ |
round__150.0__ |
| a man buys num__58 pens at marked price of num__46 pens from a whole seller . if he sells these pens giving a discount of num__1.0 what is the profit percent ? <o> a ) num__7.6 <o> b ) num__7.7 <o> c ) num__24.82 <o> d ) num__13.6 <o> e ) num__27.82 % |
explanation : let marked price be re . num__1 each c . p . of num__58 pens = rs . num__46 s . p . of num__58 pens = num__99.0 of rs . num__58 = rs . num__57.42 profit % = ( profit / c . p . ) x num__100 profit % = ( num__11.42 / num__46 ) x num__100 = num__24.82 answer : c <eor> c <eos> |
c |
percent__58.0__99.0__ percent__100.0__24.82__ |
percent__58.0__99.0__ percent__100.0__24.82__ |
| a b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = num__2 : num__3 and b : c = num__2 : num__5 . if the total runs scored by all of them are num__100 the runs scored by c are ? <o> a ) num__59 <o> b ) num__58 <o> c ) num__60 <o> d ) num__61 <o> e ) num__12 |
a : b = num__2 : num__3 b : c = num__2 : num__5 a : b : c = num__4 : num__6 : num__15 num__0.6 * num__100 = num__60 answer : c <eor> c <eos> |
c |
multiply__2.0__3.0__ multiply__3.0__5.0__ divide__3.0__5.0__ multiply__100.0__0.6__ multiply__100.0__0.6__ |
multiply__2.0__3.0__ multiply__3.0__5.0__ divide__3.0__5.0__ multiply__100.0__0.6__ multiply__100.0__0.6__ |
| if num__0.75 : x : : num__5 : num__11 then x is equal to : <o> a ) num__1.12 <o> b ) num__1.65 <o> c ) num__1.2 <o> d ) num__1.3 <o> e ) none of these |
explanation : ( x * num__5 ) = ( num__0.75 * num__11 ) x = num__8.25 / num__5 = num__1.65 answer : b <eor> b <eos> |
b |
multiply__0.75__11.0__ divide__8.25__5.0__ divide__8.25__5.0__ |
multiply__0.75__11.0__ divide__8.25__5.0__ divide__8.25__5.0__ |
| the size of a flat - screen television is given as the length of the screen ’ s diagonal . how many square inches greater is the screen of a square num__28 - inch flat - screen television than a square num__25 - inch flat - screen television ? <o> a ) num__79.5 <o> b ) num__89.3 <o> c ) num__85.5 <o> d ) num__75.4 <o> e ) num__72.5 |
if we take a square with side length x and draw a diagonal we get two isosceles right triangles . if we focus on one such right triangle we see that the legs have length x . square num__28 - inch flat - screen television the diagonal ( hypotenuse ) = num__28 so we can apply the pythagorean theorem to get x ² + x ² = num__28 ² simplify : num__2 x ² = num__28 ² divide both sides by num__2 to get : x ² = num__28 ² / num__2 since the area of the square = x ² we can see that the area of this square is num__28 ² / num__2 square num__25 - inch flat - screen television the diagonal ( hypotenuse ) = num__25 so we can apply the pythagorean theorem to get x ² + x ² = num__25 ² simplify : num__2 x ² = num__25 ² divide both sides by num__2 to get : x ² = num__25 ² / num__2 since the area of the square = x ² we can see that the area of this square is num__25 ² / num__2 difference in areas = num__28 ² / num__2 - num__25 ² / num__2 = ( num__28 ² - num__25 ² ) / num__2 = ( num__784 - num__625 ) / num__2 = num__79.5 = num__79.5 a <eor> a <eos> |
a |
power__28.0__2.0__ power__25.0__2.0__ triangle_area__79.5__2.0__ |
power__28.0__2.0__ power__25.0__2.0__ triangle_area__79.5__2.0__ |
| if the c . i . on a sum for num__2 years at num__12 num__0.5 % per annum is rs . num__510 the s . i . on the same sum at the same rate for the same period of time is ? <o> a ) rs . num__400 <o> b ) rs . num__450 <o> c ) rs . num__460 <o> d ) rs . num__480 <o> e ) rs . num__580 |
let the sum be rs . p . then [ p ( num__1 + num__25 / ( num__2 * num__100 ) ) num__2 - p ] = num__510 p [ ( num__1.125 ) num__2 - num__1 ] = num__510 . sum = rs . num__1920 so s . i . = ( num__1920 * num__25 * num__2 ) / ( num__2 * num__100 ) = rs . num__480 answer : d <eor> d <eos> |
d |
percent__25.0__1920.0__ percent__100.0__480.0__ |
percent__25.0__1920.0__ percent__100.0__480.0__ |
| ayush was born two years after his father ' s marriage . his mother is five years younger than his father but num__22 years older than ayush who is num__10 years old . at what age did the father get married ? <o> a ) num__22 years <o> b ) num__23 years <o> c ) num__24 years <o> d ) num__25 years <o> e ) num__26 years |
explanation : ayush ' s present age = num__10 years . his mother ' s present age = ( num__10 + num__22 ) years = num__32 years . ayush ' s father ' s present age = ( num__32 + num__5 ) years = num__37 years . ayush ' s father ' s age at the time of ayush ' s birth = ( num__37 - num__10 ) years = num__27 years . therefore ayush ' s father ' s age at the time of marriage = ( num__27 - num__2 ) years = num__25 years . answer : d ) num__25 year <eor> d <eos> |
d |
add__22.0__10.0__ add__32.0__5.0__ add__22.0__5.0__ divide__10.0__5.0__ subtract__27.0__2.0__ subtract__27.0__2.0__ |
add__22.0__10.0__ add__32.0__5.0__ add__22.0__5.0__ divide__10.0__5.0__ subtract__27.0__2.0__ subtract__27.0__2.0__ |
| there is num__6 letter d f g h r t . then how many word of num__2 letter can be form through these letters ? <o> a ) num__720 <o> b ) num__360 <o> c ) num__48 <o> d ) num__30 <o> e ) num__24 |
it will comes like num__6 p num__2 num__6 ! / num__4 ! = ( num__1 * num__2 * num__3 * num__4 * num__5 * num__6 ) / num__1 * num__2 * num__3 * num__4 = num__30 answer d <eor> d <eos> |
d |
subtract__6.0__2.0__ divide__6.0__2.0__ subtract__6.0__1.0__ multiply__6.0__5.0__ multiply__6.0__5.0__ |
subtract__6.0__2.0__ divide__6.0__2.0__ subtract__6.0__1.0__ multiply__6.0__5.0__ multiply__6.0__5.0__ |
| what is num__5 + num__7 <o> a ) num__2 <o> b ) num__4 <o> c ) num__10 <o> d ) num__12 <o> e ) num__11 |
d <eor> d <eos> |
d |
add__5.0__7.0__ |
add__5.0__7.0__ |
| if two dice are thrown together the probability of getting an even number on one die and an odd number on the other is ? <o> a ) num__0.166666666667 <o> b ) num__0.5 <o> c ) num__0.111111111111 <o> d ) num__0.2 <o> e ) num__1.0 |
the number of exhaustive outcomes is num__36 . let e be the event of getting an even number on one die and an odd number on the other . let the event of getting either both even or both odd then = num__0.5 = num__0.5 p ( e ) = num__1 - num__0.5 = num__0.5 . answer : b <eor> b <eos> |
b |
negate_prob__0.5__ |
negate_prob__0.5__ |
| if num__233 / num__0.233 = num__23.3 / x what is the value of x <o> a ) num__233 <o> b ) num__23.3 <o> c ) num__0.233 <o> d ) num__0.0233 <o> e ) none of these |
explanation : num__233 / num__0.233 = num__23.3 / x ⇒ num__1000.0 = num__23.3 / x ⇒ num__1000 = num__23.3 / x ⇒ x = num__23.3 / num__1000 = num__0.0233 . answer : option d <eor> d <eos> |
d |
divide__233.0__0.233__ divide__23.3__1000.0__ divide__23.3__1000.0__ |
divide__233.0__0.233__ divide__23.3__1000.0__ divide__23.3__1000.0__ |
| latha took a loan of rs . num__2000 with simple interest for as many years as the rate of interest . if she paid rs . num__180 as interest at the end of the loan period what was the rate of interest ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__7 <o> d ) num__3 <o> e ) num__9 |
let rate = r % and time = r years . then ( num__2000 * r * r ) / num__100 = num__180 num__20 r num__2 = num__180 r num__2 = num__9 = > r = num__3 answer : d <eor> d <eos> |
d |
percent__3.0__100.0__ |
percent__3.0__100.0__ |
| a four years nsc certificate was purchased for rs . num__500 with rs . num__1000 being the maturity value . find the rate of s . i . <o> a ) num__24.0 <o> b ) num__22.0 <o> c ) num__16.0 <o> d ) num__25.0 <o> e ) num__32 % |
explanation : p = rs . num__500 si = rs . num__500 t = num__4 r = ? r = ( num__100 × si ) / pt = ( num__100 × num__500 ) / ( num__500 × num__4 ) = num__25.0 = num__25.0 answer : option d <eor> d <eos> |
d |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| the length of a bridge in meters which a train num__90 - meters long and traveling at num__45 km / hr can cross in num__30 seconds is ? <o> a ) num__145 <o> b ) num__215 <o> c ) num__265 <o> d ) num__285 <o> e ) num__375 |
num__45 km / h = num__45000 m / num__3600 s = num__12.5 m / s in num__30 seconds the train can go num__30 ( num__12.5 ) = num__375 meters let x be the length of the bridge . x + num__90 = num__375 meters x = num__285 meters the answer is d . <eor> d <eos> |
d |
divide__45000.0__3600.0__ multiply__30.0__12.5__ subtract__375.0__90.0__ round__285.0__ |
divide__45000.0__3600.0__ multiply__30.0__12.5__ subtract__375.0__90.0__ round__285.0__ |
| if num__20 typists can type num__60 letters in num__20 minutes then how many letters will num__30 typists working at the same rate complete in num__1 hour ? <o> a ) num__240 <o> b ) num__200 <o> c ) num__270 <o> d ) num__300 <o> e ) num__310 |
no . of letters typing by num__20 typists in num__20 minutes = num__60 no . of letters typing by num__20 typists in num__60 minutes = num__60 * num__3 = num__180 no . of letters typing by num__30 typists in num__60 minutes = num__9.0 * num__30 = num__270 answer : c <eor> c <eos> |
c |
divide__60.0__20.0__ multiply__60.0__3.0__ divide__180.0__20.0__ multiply__30.0__9.0__ round__270.0__ |
divide__60.0__20.0__ multiply__60.0__3.0__ divide__180.0__20.0__ multiply__30.0__9.0__ multiply__30.0__9.0__ |
| num__1 / num__0.08 is equal to <o> a ) num__25.5 <o> b ) num__2.5 <o> c ) num__12.5 <o> d ) . num__25 <o> e ) none of these |
explanation : num__1 / num__0.08 = ( num__1 * num__100 ) / num__8 = num__12.5 = num__12.5 option c <eor> c <eos> |
c |
multiply__0.08__100.0__ reverse__0.08__ reverse__0.08__ |
multiply__0.08__100.0__ reverse__0.08__ reverse__0.08__ |
| t = { num__2 num__3 num__4 num__5 } b = { num__4 num__5 num__6 num__7 num__8 } two integers will be randomly selected from the sets above one integer from set t and one integer from set b . what is the probability that the sum of the two integers will equal num__9 ? <o> a ) num__0.15 <o> b ) num__0.20 <o> c ) num__0.25 <o> d ) num__0.30 <o> e ) num__0.33 |
the total number of pairs t b possible is num__4 * num__5 = num__20 . out of these num__20 pairs only num__4 sum up to num__9 : ( num__2 num__7 ) ; ( num__3 num__6 ) ( num__4 num__5 ) and ( num__5 num__4 ) . the probability thus is num__0.2 = num__0.2 . answer : b . <eor> b <eos> |
b |
multiply__4.0__5.0__ reverse__5.0__ reverse__5.0__ |
multiply__4.0__5.0__ reverse__5.0__ reverse__5.0__ |
| on an order of num__5 dozen boxes of a consumer product a retailer receives an extra dozen free . this is equivalent to allowing him a discount of ? <o> a ) num__16 num__0.333333333333 % <o> b ) num__16 num__2.66666666667 % <o> c ) num__16 num__0.666666666667 % <o> d ) num__16 num__0.25 % <o> e ) num__16 num__0.333333333333 % |
clearly the retailer gets num__1 dozen out of num__6 dozens free . equivalent discount = num__0.166666666667 * num__100 = num__16 num__0.666666666667 % . answer : c <eor> c <eos> |
c |
add__5.0__1.0__ reverse__6.0__ multiply__1.0__16.0__ |
add__5.0__1.0__ reverse__6.0__ multiply__1.0__16.0__ |
| let abcd be a convex quadrilateral with ba = bc and da = dc . let e and f be the midpoints of bc and cd respectively and let bf and de intersect at g . if the area of cegf is num__50 what is the area of abgd ? <o> a ) num__100 <o> b ) num__200 <o> c ) num__300 <o> d ) num__400 <o> e ) num__500 |
let [ xy z ] denote the area of a polygon xy z . observe that bf and de are two medians of num__4 bcd so g is the centroid of num__4 bcd . since the medians of a triangle divide the triangle into six parts of equal area we have [ cgf ] = [ cge ] = num__0.5 [ cegf ] = num__25 : it follows that [ cbd ] = num__6 num__25 = num__150 and [ bgd ] = num__2 num__25 = num__50 . furthermore since ba = bc da = dc and bd = bd we have that num__4 abd = num__4 cbd so [ abd ] = [ cbd ] = num__150 . finally [ abgd ] = [ abd ] + [ bgd ] = num__150 + num__50 = num__200 correct answer b <eor> b <eos> |
b |
multiply__50.0__0.5__ multiply__6.0__25.0__ reverse__0.5__ multiply__50.0__4.0__ multiply__50.0__4.0__ |
multiply__50.0__0.5__ multiply__6.0__25.0__ reverse__0.5__ add__50.0__150.0__ add__50.0__150.0__ |
| two friends decide to get together ; so they start riding bikes towards each other . they plan to meet halfway . each is riding at num__6 mph . they live num__36 miles apart . one of them has a pet carrier pigeon and it starts flying the instant the friends start traveling . the pigeon flies back and forth at num__15 mph between the num__2 friends until the friends meet . how many miles does the pigeon travel ? <o> a ) num__54 <o> b ) num__66 <o> c ) num__80 <o> d ) num__45 <o> e ) num__96 |
d num__45 it takes num__3 hours for the friends to meet ; so the pigeon flies for num__3 hours at num__18 mph = num__45 miles <eor> d <eos> |
d |
divide__6.0__2.0__ multiply__6.0__3.0__ round__45.0__ |
divide__6.0__2.0__ multiply__6.0__3.0__ round__45.0__ |
| in how many ways can num__8 different gifts be divided among four children such that each child receives exactly two gifts ? <o> a ) num__16 ^ num__4 <o> b ) ( num__4 ! ) ^ num__4 <o> c ) num__8 ! / ( num__2 ! ) ^ num__4 <o> d ) num__8 ! / num__4 ! <o> e ) num__4 ^ num__16 |
total num__8 different gifts and num__4 children . thus any one child gets num__8 c num__2 gifts then the other child gets num__6 c num__2 gifts ( num__8 total - num__2 already given ) then the third one gets num__4 c num__2 gifts and the last child gets num__2 c num__2 gifts . since order in which each child gets the gift is not imp thus ans : num__8 c num__2 * num__6 c num__2 * num__4 c num__2 * num__2 c num__2 = num__8 ! / ( num__2 ! ) ^ num__4 ans : c . <eor> c <eos> |
c |
divide__8.0__4.0__ subtract__8.0__2.0__ multiply__2.0__4.0__ |
divide__8.0__4.0__ subtract__8.0__2.0__ multiply__2.0__4.0__ |
| of num__10 applicants for a job num__3 had at least num__5 years of prior work experience num__4 had advanced degrees and num__2 had at least num__5 years of prior work experience and advanced degrees . how many of the applicants had neither num__5 years of prior work experience nor advanced degrees ? <o> a ) num__3 <o> b ) num__9 <o> c ) num__5 <o> d ) num__4 <o> e ) num__8 |
had advanced degrees and num__5 years work experience : num__2 only had num__5 years work experience : num__3 - num__2 = num__1 only had advanced degrees : num__4 - num__2 = num__2 num__10 - num__2 - num__1 - num__2 = num__5 answer : c <eor> c <eos> |
c |
vowel_space__ |
vowel_space__ |
| the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is num__867 sq m then what is the breadth of the rectangular plot ? <o> a ) num__16 m <o> b ) num__17 m <o> c ) num__87 m <o> d ) num__27 m <o> e ) num__34 m |
let the breadth of the plot be b m . length of the plot = num__3 b m ( num__3 b ) ( b ) = num__867 num__3 b num__2 = num__867 b num__2 = num__289 = num__172 ( b > num__0 ) b = num__17 m . answer : b <eor> b <eos> |
b |
triangle_area__2.0__17.0__ |
triangle_area__2.0__17.0__ |
| if an average hard drive had a capacity of num__0.3 tb in num__2000 and average hard drive capacities double every num__5 years what will be the average hard drive capacity in num__2050 ? <o> a ) num__3072 <o> b ) num__512 <o> c ) num__768 <o> d ) num__1024 <o> e ) num__1536 |
num__0.3 * num__2 ^ num__10 = num__0.3 * num__1024 = num__307.2 the answer is a . <eor> a <eos> |
a |
multiply__5.0__2.0__ multiply__0.3__1024.0__ multiply__307.2__10.0__ |
multiply__5.0__2.0__ multiply__0.3__1024.0__ multiply__307.2__10.0__ |
| if num__42.24 = k ( num__14 + m / num__50 ) where k and m are positive integers and m < num__50 then what is the value of k + m ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
num__42.24 = num__14 k + km / num__50 . . . we can rewrite the number as follows : num__42 + num__0.24 = num__14 k + km / num__50 . . . . . . . . since k is integer then num__42 = num__14 k . . . . . . . . . . k = num__3 num__0.24 = km / num__50 . . . . . . num__0.24 = num__3 m / num__50 . . . . . . m = num__4 k + m = num__3 + num__4 = num__7 answer : b <eor> b <eos> |
b |
round_down__42.24__ subtract__42.24__42.0__ divide__42.0__14.0__ add__3.0__4.0__ subtract__14.0__7.0__ |
round_down__42.24__ subtract__42.24__42.0__ divide__42.0__14.0__ add__3.0__4.0__ add__3.0__4.0__ |
| a batsman scored num__120 runs which included num__5 boundaries and num__8 sixes . what % of his total score did he make by running between the wickets <o> a ) num__43.33 <o> b ) num__50.0 <o> c ) num__65.0 <o> d ) num__70.0 <o> e ) num__75 % |
number of runs made by running = num__120 - ( num__5 x num__4 + num__8 x num__6 ) = num__120 - ( num__68 ) = num__52 now we need to calculate num__60 is what percent of num__120 . = > num__0.433333333333 * num__100 = num__43.33 a <eor> a <eos> |
a |
subtract__120.0__68.0__ add__8.0__52.0__ divide__52.0__120.0__ multiply__100.0__0.4333__ multiply__100.0__0.4333__ |
subtract__120.0__68.0__ add__8.0__52.0__ divide__52.0__120.0__ multiply__100.0__0.4333__ multiply__100.0__0.4333__ |
| if num__11 men working num__6 hours a day can do a work in num__88 days . then num__6 men working num__8 hours a day can do it in how many days ? <o> a ) num__78 days . <o> b ) num__121 days . <o> c ) num__49 days . <o> d ) num__48 days . <o> e ) num__67 days . |
b num__121 days . from the above formula i . e ( m num__1 * t num__1 / w num__1 ) = ( m num__2 * t num__2 / w num__2 ) so ( num__11 * num__6 * num__88.0 ) = ( num__6 * num__8 * d / num__1 ) on solving d = num__121 days . <eor> b <eos> |
b |
subtract__8.0__6.0__ round__121.0__ |
subtract__8.0__6.0__ divide__121.0__1.0__ |
| the g . c . d . of num__1.08 num__0.36 and num__0.4 is : <o> a ) num__0.04 <o> b ) num__0.9 <o> c ) num__0.18 <o> d ) num__0.108 <o> e ) num__0.118 |
given numbers are num__1.08 num__0.36 and num__0.40 . h . c . f . of num__108 num__36 and num__40 is num__18 h . c . f . of given numbers = num__0.04 . answer : option a <eor> a <eos> |
a |
subtract__0.4__0.36__ subtract__0.4__0.36__ |
subtract__0.4__0.36__ subtract__0.4__0.36__ |
| since num__2001 the standard serial numbers on a new york state license plate are num__3 letters followed by num__4 digits . how many different license plates are possible if letters and digits can be repeated ? <o> a ) num__26 × num__3 × num__10 × num__4 <o> b ) num__26 × num__25 × num__24 × num__10 × num__9 × num__8 × num__7 <o> c ) num__26 ³ × num__9 × num__9 × num__9 × num__9 <o> d ) num__26 × num__25 × num__24 × num__10 num__000 <o> e ) num__26 ³ × num__10 num__000 |
solution : num__3 repeated letters can be chosen as num__26 * num__26 * num__26 num__4 repeated digits can be chosen as num__10 * num__10 * num__10 * num__10 answer e <eor> e <eos> |
e |
alphabet_space__ alphabet_space__ |
alphabet_space__ alphabet_space__ |
| there are some fruits in box num__0.333333333333 rd eaten by children . then num__0.4 th of the remaining by men then num__0.666666666667 rd of the remaining by women . at last num__4 were remaining . how many fruits were in total ? <o> a ) num__45 <o> b ) num__30 <o> c ) num__28 <o> d ) num__26 <o> e ) num__91 |
sol : if x fruits were there in total then x × ( num__1 – num__0.333333333333 ) × ( num__1 – num__0.4 ) × ( num__1 – num__0.666666666667 ) = num__4 x × num__0.666666666667 × num__0.6 × num__0.333333333333 = num__30 x = num__30 answer : b <eor> b <eos> |
b |
add__0.3333__0.6667__ divide__0.4__0.6667__ multiply__1.0__30.0__ |
add__0.3333__0.6667__ divide__0.4__0.6667__ multiply__1.0__30.0__ |
| the average age of num__15 students of a class is num__15 years . out of these the average age of num__3 students is num__14 years and that of the other num__11 students is num__16 years . the age of the num__15 th student is <o> a ) num__9 years <o> b ) num__11 years <o> c ) num__14 years <o> d ) num__7 years <o> e ) num__25 years |
solution age of the num__15 th student = [ num__15 x num__15 - ( num__14 x num__3 + num__16 x num__11 ) ] = ( num__225 - num__218 ) = num__7 years . answer d <eor> d <eos> |
d |
subtract__225.0__218.0__ subtract__14.0__7.0__ |
subtract__225.0__218.0__ subtract__14.0__7.0__ |
| a company chauncy co . has an annual travel budget of $ num__51000 . the accounting department estimates that transportation expenses will increase num__5 percent in the coming year and nontransportation travel expenses will increase by num__15 percent in the coming year . last year chauncy co . spent $ num__19500 on transportation - related expenses and $ num__35000 on nontransportation travel expenses . if the annual travel budget does not increase this year and if the accounting department ’ s estimates are correct how far over the annual travel budget will expenses be this year ? <o> a ) expenses will not go over the budget . <o> b ) $ num__500 <o> c ) $ num__4225 <o> d ) $ num__9725 <o> e ) $ num__60 |
725 |
annual travel budget of $ num__51000 let transportation expenses = t = num__19500 and non - transportation expenses = n = num__35000 i . e . increased transportation expenses = num__1.05 t = num__20475 and increased non - transportation expenses = num__1.15 n = num__40250 total expense = num__20475 + num__40250 = num__60725 expense over budget = budget - expense = num__51000 - num__60725 = num__9725 answer : option d <eor> d <eos> |
d |
d |
| on a certain test bill scored num__20 more points than john but half as many points as sue . if the sum of the scores of the three students was num__140 points how many points did bill receive ? <o> a ) num__30 <o> b ) num__40 <o> c ) num__50 <o> d ) num__80 <o> e ) num__100 |
consider john ' s points as x so bill ' s points are x + num__20 so sue ' s points are num__2 x + num__40 total of all three students is num__4 x + num__60 but this total is given as num__140 x = num__20 so bill ' s points = num__20 + num__20 = num__40 correct answer option b <eor> b <eos> |
b |
multiply__20.0__2.0__ add__20.0__40.0__ multiply__20.0__2.0__ |
multiply__20.0__2.0__ add__20.0__40.0__ multiply__20.0__2.0__ |
| a sporting goods store ordered an equal number of white and yellow tennis balls . the tennis ball company delivered num__30 extra white balls making the ratio of white balls to yellow balls num__6 : num__5 . how many tennis balls did the store originally order ? <o> a ) num__120 <o> b ) num__150 <o> c ) num__180 <o> d ) num__300 <o> e ) num__330 |
( x + num__30 ) / ( x ) = num__1.2 cross multiply : num__5 x + num__150 = num__6 x x = num__150 and since we originally ordered num__2 x balls the answer is num__300 . d <eor> d <eos> |
d |
divide__6.0__5.0__ multiply__30.0__5.0__ multiply__2.0__150.0__ multiply__2.0__150.0__ |
divide__6.0__5.0__ multiply__30.0__5.0__ multiply__2.0__150.0__ multiply__2.0__150.0__ |
| can you find a seven digit number which describes itself . the first digit is the number of zeros in the number . the second digit is the number of ones in the number etc . for example in the number num__21200 there are num__2 zeros num__1 one num__2 twos num__0 threes and num__0 fours . <o> a ) num__4211000 <o> b ) num__1211000 <o> c ) num__5211000 <o> d ) num__2211000 <o> e ) num__3211000 |
e num__3211000 <eor> e <eos> |
e |
round__3211000.0__ |
round__3211000.0__ |
| a car travels a distance of num__160 miles in num__2 hours and num__40 minutes what is the speed of the car in miles per hour ? <o> a ) num__54 <o> b ) num__60 <o> c ) num__84 <o> d ) num__116 <o> e ) num__120 |
given time = num__2 hours num__40 mins num__40 mins can be written as num__0.666666666667 hours = num__0.666666666667 hrs total time = num__2 + ( num__0.666666666667 ) = num__2.66666666667 hrs therefore speed = num__160 / ( num__0.266666666667 ) = num__60 option b <eor> b <eos> |
b |
add__2.0__0.6667__ hour_to_min_conversion__ hour_to_min_conversion__ |
add__2.0__0.6667__ hour_to_min_conversion__ hour_to_min_conversion__ |
| a katiland ' s train rails across an open track at num__250 kilometers per hour . a regular passenger train travels at num__68.0 of the katiland ' s train speed . if the two trains start moving from the same station at the same time how much time longer will it take the passenger train than the katiland ' s to travel num__850 kilometers ? <o> a ) num__1 hour and num__24 minutes . <o> b ) num__1 hour and num__36 minutes . <o> c ) num__2 hours and num__24 minutes . <o> d ) num__2 hours and num__36 minutes . <o> e ) num__5 hours . |
difference in time = time taken by passenger train - time taken by katiland ' s train num__850 / ( num__250 * num__68 ) * num__100 - num__3.4 num__850 ( num__0.4 * num__68 - num__0.004 ) num__850 * num__32 / ( num__250 * num__68 ) num__1.6 hrs or num__1 hr and num__36 mins b is the answer <eor> b <eos> |
b |
divide__850.0__250.0__ divide__100.0__250.0__ divide__0.4__100.0__ subtract__100.0__68.0__ mile_to_km_conversion__ multiply__250.0__0.004__ subtract__68.0__32.0__ round__1.0__ |
divide__850.0__250.0__ divide__100.0__250.0__ divide__0.4__100.0__ subtract__100.0__68.0__ mile_to_km_conversion__ multiply__250.0__0.004__ subtract__68.0__32.0__ multiply__250.0__0.004__ |
| the list price of an article is rs . num__67 . a customer pays rs . num__56.16 for it . he was given two successive discounts one of them being num__10.0 . the other discount is ? <o> a ) num__3.86 <o> b ) num__4.86 <o> c ) num__5.86 <o> d ) num__6.86 <o> e ) num__7.86 % |
num__67 * ( num__0.9 ) * ( ( num__100 - x ) / num__100 ) = num__56.16 x = num__6.86 answer : d <eor> d <eos> |
d |
percent__6.86__100.0__ |
percent__6.86__100.0__ |
| the weights of one liter vegetable ghee packet of two brands ‘ a ’ and ‘ b ’ are num__900 gm and num__700 gm respectively . if they are mixed in the ratio of num__3 : num__2 by volumes to form a mixture of num__4 liters what is the weight ( in kg ) of the mixture ? <o> a ) num__3.84 <o> b ) num__1.75 <o> c ) num__3.28 <o> d ) num__2.72 <o> e ) none of these |
here ' s how i did it . my notes from reading the problem were : num__1 l a = num__900 gm num__1 l b = num__700 gm we are mixing five parts ( num__3 parts a plus num__2 parts b num__5 parts total ) to get num__4 l so num__5 x = num__4 - - - > x = num__0.8 . each part is num__0.8 of a liter . so if we have num__3 parts a we have num__900 * num__3 * ( num__0.8 ) = num__2160 if we have num__2 parts b we have num__700 * num__2 * ( num__0.8 ) = num__1120 num__2160 + num__1120 = num__3280 solving for units gives us num__3.28 so the answer is c <eor> c <eos> |
c |
subtract__3.0__2.0__ add__3.0__2.0__ divide__4.0__5.0__ add__1120.0__2160.0__ multiply__1.0__3.28__ |
subtract__3.0__2.0__ add__3.0__2.0__ divide__4.0__5.0__ add__1120.0__2160.0__ multiply__1.0__3.28__ |
| how many factors of num__990 are odd numbers greater than num__1 ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
prime factors of num__330 are num__2 ^ num__13 ^ num__25 ^ num__1 and num__11 ^ num__1 total divisors = ( power if a prime factor + num__1 ) total no . of odd factors ( num__3 num__511 ) = ( num__2 + num__1 ) ( num__1 + num__1 ) ( num__1 + num__1 ) = num__12 since we need odd divisors other than num__1 = > num__12 - num__1 = num__11 odd divisors a is the answer <eor> a <eos> |
a |
subtract__13.0__2.0__ divide__990.0__330.0__ add__1.0__11.0__ gcd__990.0__3.0__ |
subtract__13.0__2.0__ add__1.0__2.0__ add__1.0__11.0__ add__1.0__2.0__ |
| the ratio between the perimeter and the width of a rectangle is num__5 : num__1 . if the area of the rectangle is num__294 sq . cm what is the width of the rectangle ? <o> a ) num__10 cm <o> b ) num__12 cm <o> c ) num__14 cm <o> d ) num__16 cm <o> e ) num__18 cm |
num__2 l + num__2 w = num__5 w l = num__3 w / num__2 w * l = num__294 num__3 w ^ num__1.0 = num__294 w ^ num__2 = num__196 w = num__14 the answer is c . <eor> c <eos> |
c |
rectangle_perimeter__5.0__2.0__ rectangle_perimeter__5.0__2.0__ |
rectangle_perimeter__5.0__2.0__ power__14.0__1.0__ |
| if num__15 men working num__9 hours a day can reap a field in num__16 days in how many days will num__18 men reap the field working num__8 hours a day ? <o> a ) num__13 <o> b ) num__14 <o> c ) num__15 <o> d ) num__16 <o> e ) num__17 |
since the number of days is to be found out we compare each item with the number of days . more men less days ( indirect ) less working hours more days ( indirect ) men num__18 : num__15 } . . working hours num__8 : num__9 } . . num__16 : x num__18 x num__8 x x = num__15 x num__9 x num__16 or x = num__15 required number of days = num__15 answer : c <eor> c <eos> |
c |
round__15.0__ |
round__15.0__ |
| machine a can make num__350 widgets in num__1 hour and machine b can make num__250 widgets in num__1 hour . if both machines work together how much time will it take them to make a total of num__800 widgets ? <o> a ) num__1 hour and num__20 minutes <o> b ) num__1 hour and num__24 minutes <o> c ) num__1 hour and num__30 minutes <o> d ) num__1 hour and num__36 minutes <o> e ) num__1 hour and num__40 minutes |
( num__350 + num__250 ) t = num__800 num__600 t = num__800 t = num__1.33333333333 t = num__1 hour and num__20 minutes answer : a <eor> a <eos> |
a |
add__350.0__250.0__ divide__800.0__600.0__ round__1.0__ |
add__350.0__250.0__ divide__800.0__600.0__ round__1.0__ |
| roy is now num__6 years older than julia and half of that amount older than kelly . if in num__2 years roy will be twice as old as julia then in num__2 years what would be roy ’ s age multiplied by kelly ’ s age ? <o> a ) num__84 <o> b ) num__96 <o> c ) num__100 <o> d ) num__108 <o> e ) num__120 |
r = j + num__6 = k + num__3 r + num__2 = num__2 ( j + num__2 ) ( j + num__6 ) + num__2 = num__2 j + num__4 j = num__4 r = num__10 k = num__7 in num__2 years ( r + num__2 ) ( k + num__2 ) = num__12 * num__9 = num__108 the answer is d . <eor> d <eos> |
d |
divide__6.0__2.0__ subtract__6.0__2.0__ add__6.0__4.0__ add__3.0__4.0__ multiply__6.0__2.0__ add__6.0__3.0__ multiply__9.0__12.0__ multiply__9.0__12.0__ |
divide__6.0__2.0__ subtract__6.0__2.0__ add__6.0__4.0__ add__3.0__4.0__ multiply__6.0__2.0__ add__6.0__3.0__ multiply__9.0__12.0__ multiply__9.0__12.0__ |
| the ratio of the length and the width of a rectangle is num__4 : num__3 and the area of the rectangle is num__4800 sq cm . what is the ratio of the width and the area of the rectangle ? <o> a ) num__1 : num__72 <o> b ) num__1 : num__76 <o> c ) num__1 : num__80 <o> d ) num__1 : num__84 <o> e ) num__1 : num__88 |
let the length and the width be num__4 x and num__3 x respectively . area = ( num__4 x ) ( num__3 x ) = num__4800 num__12 x ^ num__2 = num__4800 x ^ num__2 = num__400 x = num__20 the ratio of the width and the area is num__3 x : num__12 x ^ num__2 = num__1 : num__4 x = num__1 : num__80 the answer is c . <eor> c <eos> |
c |
square_perimeter__3.0__ square_perimeter__20.0__ volume_cube__1.0__ |
square_perimeter__3.0__ square_perimeter__20.0__ power__1.0__4.0__ |
| a sum of rs . num__150 has two types of coins rs . num__1 and rs . num__2 . if total number of coins is num__100 then number of rs . num__2 coins is <o> a ) num__50 <o> b ) num__44 <o> c ) num__65 <o> d ) num__40 <o> e ) none of the above |
x = num__1 rupee coin y = num__2 rupee coin x + num__2 y = num__150 x + y = num__100 answer : a <eor> a <eos> |
a |
subtract__150.0__100.0__ |
subtract__150.0__100.0__ |
| the sum and the product of two numbers are num__12 and num__20 respectively the difference of the number is ? <o> a ) num__1 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__2 |
explanation : x + y = num__12 xy = num__20 ( x - y ) num__2 = ( x + y ) num__2 - num__4 xy ( x - y ) num__2 = num__144 - num__80 = > ( x - y ) = num__8 answer : d <eor> d <eos> |
d |
multiply__20.0__4.0__ subtract__12.0__4.0__ subtract__12.0__4.0__ |
multiply__20.0__4.0__ subtract__12.0__4.0__ subtract__12.0__4.0__ |
| find the odd man out num__1 num__2 num__5 num__14 num__41 num__125 <o> a ) num__2 <o> b ) num__14 <o> c ) num__124 <o> d ) num__41 <o> e ) num__125 |
num__1 * num__3 - num__1 = num__2 num__2 * num__3 - num__1 = num__5 num__5 * num__3 - num__1 = num__14 num__14 * num__3 - num__1 = num__41 num__41 * num__3 - num__1 = num__122 answer : e <eor> e <eos> |
e |
add__1.0__2.0__ subtract__125.0__3.0__ multiply__1.0__125.0__ |
subtract__5.0__2.0__ subtract__125.0__3.0__ multiply__1.0__125.0__ |
| the average age of students of a class is num__15.8 years . the average age of boys in the class is num__16.6 years and that of the girls is num__15.4 years . the ration of the number of boys to the number of girls in the class is : <o> a ) num__1 : num__2 <o> b ) num__2 : num__3 <o> c ) num__2 : num__4 <o> d ) num__2 : num__1 <o> e ) num__2 : num__9 |
let the ratio be k : num__1 . then k * num__16.6 + num__1 * num__15.4 = ( k + num__1 ) * num__15.8 = ( num__16.6 - num__15.8 ) k = ( num__15.8 - num__15.4 ) = k = num__0.4 / num__0.6 = num__0.5 required ratio = num__1.0 : num__1 = num__1 : num__2 . answer : a <eor> a <eos> |
a |
subtract__15.8__15.4__ subtract__1.0__0.4__ reverse__0.5__ reverse__1.0__ |
subtract__15.8__15.4__ subtract__1.0__0.4__ reverse__0.5__ reverse__1.0__ |
| if an integer n is to be selected at random from num__1 to num__100 inclusive what is probability n ( n + num__1 ) will be divisible by num__5 ? <o> a ) num__0.2 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__0.4 |
for n ( n + num__1 ) to be a multiple of num__5 either n or n + num__1 has to be a multiple of num__5 . thus n must be of the form num__5 k or num__5 k - num__1 . the probability is num__0.4 . the answer is e . <eor> e <eos> |
e |
multiply__1.0__0.4__ |
multiply__1.0__0.4__ |
| the diameter of the driving wheel of a bus is num__140 cm . how many revolution per minute must the wheel make in order to keep a speed of num__66 kmph ? <o> a ) num__129 <o> b ) num__250 <o> c ) num__228 <o> d ) num__119 <o> e ) num__112 |
circumference = no . of revolutions * distance covered distance to be covered in num__1 min . = ( num__66 x num__1000 ) / num__60 m = num__1100 m . circumference of the wheel = num__2 x ( num__3.14285714286 ) x num__0.70 m = num__4.4 m . number of revolutions per min . = ( num__1100 / num__4.4 ) = num__250 . answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ divide__1100.0__4.4__ round__250.0__ |
hour_to_min_conversion__ divide__1100.0__4.4__ divide__1100.0__4.4__ |
| how many seconds will a num__420 metre long train take to cross a man running with a speed of num__6 km / hr in the direction of the moving train if the speed of the train is num__30 km / hr ? <o> a ) num__25 <o> b ) num__63 <o> c ) num__40 <o> d ) num__45 <o> e ) num__60 |
explanation : speed of train relatively to man = ( num__30 - num__6 ) km / hr = num__24 km / hr = ( num__24 x num__0.277777777778 ) m / sec = num__6.66 m / sec time taken to pass the man = ( num__420 / num__6.66 ) sec = num__63 sec . answer : b <eor> b <eos> |
b |
subtract__30.0__6.0__ round__63.0__ |
subtract__30.0__6.0__ round__63.0__ |
| which of the following is equal to the average ( arithmetic mean ) of ( x + num__2 ) ^ num__2 and ( x - num__4 ) ^ num__2 ? <o> a ) x ^ num__2 <o> b ) x ^ num__2 + num__2 <o> c ) x ^ num__2 - num__2 x + num__10 <o> d ) x ^ num__2 + num__2 x + num__10 <o> e ) x ^ num__2 + num__4 x + num__5 |
avg = [ ( x + num__2 ) ^ num__2 + ( x - num__4 ) ^ num__2 ] / num__2 expanding and simplifying ( x ^ num__2 + num__4 x + num__4 + x ^ num__2 - num__8 x + num__16 ) / num__2 = x ^ num__2 - num__2 x + num__10 answer c . <eor> c <eos> |
c |
multiply__2.0__4.0__ multiply__2.0__8.0__ add__2.0__8.0__ subtract__4.0__2.0__ |
multiply__2.0__4.0__ multiply__2.0__8.0__ add__2.0__8.0__ subtract__4.0__2.0__ |
| num__2 x + y = num__16 | y | < = num__16 for how many ordered pairs ( x y ) that are solutions of the system above are x and y both integers ? <o> a ) num__12 <o> b ) num__13 <o> c ) num__16 <o> d ) num__17 <o> e ) num__19 |
hi - the questions says | y | < = num__16 so the values of y will range from num__0 to num__16 . so the values of y will be num__0 num__12 num__34 num__56 num__78 num__910 num__1112 num__1314 num__1516 . so num__17 ( d ) is the answer . remember sign does n ' t matter for y as y will always be positive . d <eor> d <eos> |
d |
divide__34.0__2.0__ divide__34.0__2.0__ |
divide__34.0__2.0__ subtract__34.0__17.0__ |
| two pipes a and b together can fill a cistern in num__4 hours . had they been opened separately then b would have taken num__6 hours more than a to fill cistern . how much time will be taken by a to fill the cistern separately ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__6 <o> d ) num__78 <o> e ) num__9 |
let the cistern be filled by pipe a alone in x hours . then pipe b will fill it in ( x + num__6 ) hours . num__1 / x + num__1 / ( x + num__6 ) = num__0.25 x num__2 - num__2 x - num__24 = num__0 ( x - num__6 ) ( x + num__4 ) = num__0 = > x = num__6 . answer : c <eor> c <eos> |
c |
divide__1.0__4.0__ subtract__6.0__4.0__ multiply__4.0__6.0__ round__6.0__ |
divide__1.0__4.0__ subtract__6.0__4.0__ divide__6.0__0.25__ add__4.0__2.0__ |
| in the coordinate plane a slope of the line k is num__3 times the y - intercept of the line k . what is the x - intercept of the line k ? <o> a ) - num__4 <o> b ) num__4 <o> c ) - num__0.25 <o> d ) - num__0.333333333333 <o> e ) num__2 |
as y = num__3 mx + m from num__0 = num__3 mx + m we get x = - num__0.333333333333 . hence the correct answer choice is d . <eor> d <eos> |
d |
reverse__3.0__ reverse__3.0__ |
reverse__3.0__ reverse__3.0__ |
| in a sequence of num__41 numbers each term except for the first one is num__7 less than the previous term . if the greatest term in the sequence is num__281 what is the smallest term in the sequence ? <o> a ) num__1 <o> b ) - num__2 <o> c ) num__0 <o> d ) num__8 <o> e ) num__6 |
which term is the greatest ? the first or the last ? it is given to you that every term is num__7 less than the previous term . hence as you go on your terms keep becoming smaller and smaller . the first term is the greatest term . an = num__281 + ( num__41 - num__1 ) * ( - num__7 ) an = num__281 - num__280 = num__1 a <eor> a <eos> |
a |
subtract__281.0__1.0__ reverse__1.0__ |
subtract__281.0__1.0__ subtract__281.0__280.0__ |
| it takes joey the postman num__1 hours to run a num__5 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is num__8 mile / hour what is the speed with which joey returns ? <o> a ) num__11 <o> b ) num__12 <o> c ) num__13 <o> d ) num__14 <o> e ) num__20 |
let his speed for one half of the journey be num__5 miles an hour let the other half be x miles an hour now avg speed = num__8 mile an hour num__2 * num__5 * x / num__5 + x = num__8 num__10 x = num__8 x + num__40 = > num__2 x = num__40 = > x = num__20 e <eor> e <eos> |
e |
multiply__5.0__2.0__ multiply__5.0__8.0__ multiply__2.0__10.0__ round__20.0__ |
multiply__5.0__2.0__ multiply__5.0__8.0__ multiply__2.0__10.0__ multiply__1.0__20.0__ |
| three years ago the average age of a family of seven members was num__22 years . a boy have been born the average age of the family is the same today . what is the age of the boy ? <o> a ) a ) num__1 <o> b ) b ) num__2 <o> c ) c ) num__3 <o> d ) d ) num__4 <o> e ) e ) num__5 |
num__7 * num__25 = num__175 num__8 * num__22 = num__176 - - - - - - - - - - - - - - num__1 answer : a <eor> a <eos> |
a |
multiply__7.0__25.0__ multiply__22.0__8.0__ subtract__8.0__7.0__ reverse__1.0__ |
multiply__7.0__25.0__ multiply__22.0__8.0__ subtract__8.0__7.0__ subtract__8.0__7.0__ |
| of the num__200 students at university xyz majoring in one or more of the engineering disciplines num__120 are majoring in electrical and num__150 are majoring in mechanical . if at least num__30 of the students are not majoring in either electrical or mechanical then the number of students majoring in both electrical and mechanical could be any number from <o> a ) num__30 to num__70 <o> b ) num__50 to num__80 <o> c ) num__110 to num__130 <o> d ) num__130 to num__150 <o> e ) num__150 to num__170 |
if there are num__120 students majoring in electrical then there must be num__80 person not majoring electrical at all since both of the student not majoring electrical and mechanical is at least num__30 so the number of the student who are not majoring electrical but majoring mechanical will be at least num__50 . if there are num__150 students majoring in mechanical there must be num__50 students who are not majoring mechanical at all since the number of the student who are not majoring electrical but majoring mechanical will be at least num__50 hence the number of students both majoring mechanical and electrical will be at least num__130 so there must be at least num__130 students who major in both see the answers option the only possible answer is d ! <eor> d <eos> |
d |
subtract__200.0__120.0__ subtract__200.0__150.0__ add__80.0__50.0__ add__80.0__50.0__ |
subtract__200.0__120.0__ subtract__200.0__150.0__ add__80.0__50.0__ add__80.0__50.0__ |
| in a certain sequence each term except for the first term is one less than twice the previous term . if the first term is num__1 then the num__3 rd term is which of the following ? <o> a ) − num__1.5 <o> b ) − num__1 <o> c ) num__1 . <o> d ) num__0.5 <o> e ) num__2 |
first = num__1 second = num__2 * num__1 - num__1 = num__1 . second = num__2 * num__1 - num__1 = num__1 . answer : option c <eor> c <eos> |
c |
subtract__3.0__1.0__ reverse__1.0__ |
subtract__3.0__1.0__ subtract__3.0__2.0__ |
| if num__13 men do a work in num__80 days in how many days will num__20 men do it ? <o> a ) num__18 days <o> b ) num__38 days <o> c ) num__42 days <o> d ) num__48 days <o> e ) num__52 days |
num__13 * num__80 = num__20 * x x = num__52 days answer : e <eor> e <eos> |
e |
round__52.0__ |
round__52.0__ |
| if a person walks at num__15 km / hr instead of num__10 km / hr he would have walked num__20 km more . the actual distance traveled by him is : <o> a ) num__50 km <o> b ) num__56 km <o> c ) num__60 km <o> d ) num__40 km <o> e ) num__80 km |
let the actual distance travelled be x km . x / num__10 = ( x + num__20 ) / num__15 num__15 x = num__10 x + num__200 num__5 x = num__200 x = num__40 km . answer : d <eor> d <eos> |
d |
multiply__10.0__20.0__ subtract__15.0__10.0__ divide__200.0__5.0__ divide__200.0__5.0__ |
multiply__10.0__20.0__ subtract__15.0__10.0__ divide__200.0__5.0__ divide__200.0__5.0__ |
| every digit of a number written in binary is either num__0 or num__1 . to translate a number from binary multiply the nth digit ( reading from right to left ) by num__2 ^ ( n - num__1 ) what is the largest prime number ( written in binary ) that is a factor of both num__10010000 and num__100100000 ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__1001 <o> d ) num__1011 <o> e ) num__10001 |
binary divison can provide a quick answer if you are comfortable with it . as option e is the biggest binary number we try with it first : num__99910.0899101 = num__10000 num__999100.899101 = num__100000 so answer is option is c <eor> c <eos> |
c |
divide__10010000.0__10000.0__ |
divide__10010000.0__10000.0__ |
| look at this series : num__53 num__53 num__39 num__39 num__25 num__25 . . . what number should come next ? <o> a ) a ) num__12 <o> b ) b ) num__11 <o> c ) c ) num__27 <o> d ) d ) num__53 <o> e ) e ) num__86 |
in this series each number is repeated then num__14 is subtracted to arrive at the next number . answer : b <eor> b <eos> |
b |
subtract__53.0__39.0__ subtract__25.0__14.0__ |
subtract__53.0__39.0__ subtract__25.0__14.0__ |
| b and c work together can complete work in num__8 days a and b together can complete same work in num__12 days and a & c together complete in num__16 days in total how many days a b and c complete the same work together ? <o> a ) num__3 num__0.692307692308 <o> b ) num__7 num__0.384615384615 <o> c ) num__7 num__0.416666666667 <o> d ) num__3 num__0.416666666667 <o> e ) num__4 num__0.416666666667 |
explanation : ( a + b + c ) ’ s num__2 day work = ( num__0.125 + num__0.0833333333333 + num__0.0625 ) = num__0.270833333333 on day work = num__0.135416666667 so they complete work in num__7.38461538462 = num__7 num__0.384615384615 days answer : option b <eor> b <eos> |
b |
divide__16.0__8.0__ divide__2.0__16.0__ divide__0.125__2.0__ divide__0.2708__2.0__ subtract__7.3846__7.0__ round__7.0__ |
divide__16.0__8.0__ divide__2.0__16.0__ divide__0.125__2.0__ divide__0.2708__2.0__ subtract__7.3846__7.0__ round__7.0__ |
| two - third of a positive number and num__0.0740740740741 of its reciprocal are equal . find the positive number . <o> a ) num__0.36 <o> b ) num__3.5 <o> c ) num__0.333333333333 <o> d ) num__5.76 <o> e ) num__4.96551724138 |
explanation : let the positive number be x . then num__0.666666666667 x = num__0.0740740740741 * num__1 / x x num__2 = num__0.0740740740741 * num__1.5 = num__0.111111111111 x = √ num__0.111111111111 = num__0.333333333333 . answer : c <eor> c <eos> |
c |
multiply__0.0741__1.5__ subtract__1.0__0.6667__ subtract__1.0__0.6667__ |
multiply__0.0741__1.5__ divide__0.6667__2.0__ divide__0.1111__0.3333__ |
| if a @ b = a * b ^ ( num__0.5 ) then num__5 @ num__9 = ? self made <o> a ) num__12 <o> b ) num__16 <o> c ) num__13 <o> d ) num__15 <o> e ) num__14 |
a @ b = a * b ^ ( num__0.5 ) num__5 @ num__9 = num__5 * num__9 ^ ( num__0.5 ) = num__5 * num__3 = num__15 correct option : d <eor> d <eos> |
d |
multiply__5.0__3.0__ multiply__5.0__3.0__ |
multiply__5.0__3.0__ multiply__5.0__3.0__ |
| two trains are running at num__40 km / hr and num__20 km / hr respectively in the same direction . fast train completely passes a man sitting in the slower train in num__5 sec . what is the length of the fast train ? <o> a ) num__35 <o> b ) num__24 <o> c ) num__27 <o> d ) num__27 num__0.777777777778 <o> e ) num__67 |
relative speed = ( num__40 - num__20 ) = num__20 km / hr . = num__20 * num__0.277777777778 = num__5.55555555556 m / sec . length of faster train = num__5.55555555556 * num__5 = num__27.7777777778 = num__27 num__0.777777777778 m . answer : option d <eor> d <eos> |
d |
subtract__27.7778__27.0__ round__27.0__ |
subtract__27.7778__27.0__ subtract__27.7778__0.7778__ |
| large medium and small ships are used to bring water . num__4 large ships carry as much water as num__7 small ships ; num__3 medium ships carry as much water as num__2 large ships and num__1 small ship . if num__15 large num__7 medium and num__14 small ships each made num__36 journeys to bring a certain quantity of water then in how many journeys will num__12 large num__14 medium and num__21 small ships bring the same quantity of water ? <o> a ) num__29 journeys <o> b ) num__30 <o> c ) num__20 <o> d ) num__50 <o> e ) num__55 |
explanation : here large medium and small ships are denoted by the symbol l m and s . now according to the question num__4 l = num__7 s . - - - - - - ( i ) num__3 m = num__2 l + s . - - - - - - ( ii ) using above equations the ratios of the capacity of the large medium and small ships are : - num__7 : num__6 : num__4 . let the number of journeys required be x . since quantity of water remains the same so : - = ( ( num__15 × num__7 + num__7 × num__6 + num__14 × num__4 ) num__36 ) = x ( num__12 × num__7 + num__14 × num__6 + num__21 × num__4 ) . = > ( ( num__15 × num__7 + num__7 × num__6 + num__14 × num__4 ) num__36 ) / ( num__12 × num__7 + num__14 × num__6 + num__21 × num__4 ) = x = > x = num__29.0 = > x = num__29 . answer : a <eor> a <eos> |
a |
add__4.0__2.0__ add__15.0__14.0__ multiply__1.0__29.0__ |
add__4.0__2.0__ add__15.0__14.0__ add__15.0__14.0__ |
| num__30 square stone slabs of equal size were needed to cover a floor area of num__50.7 sq . m . find the length of each stone slab ? <o> a ) num__130 cm <o> b ) num__767 cm <o> c ) num__88 cm <o> d ) num__666 cm <o> e ) num__776 cm |
area of each slab = num__50.7 / num__30 m num__2 = num__1.69 m num__2 length of each slab √ num__1.69 = num__1.3 m = num__130 cm <eor> a <eos> |
a |
triangle_area__2.0__130.0__ |
triangle_area__2.0__130.0__ |
| six students wrote science exam . their average marks are num__70 . five students got num__65 num__75 num__55 num__72 and num__69 marks respectively . therefore what is the marks of the sixth student ? <o> a ) num__84 <o> b ) num__68 <o> c ) num__85 <o> d ) num__75 <o> e ) num__42 |
explanation total marks of num__5 students = ( num__65 + num__75 + num__55 + num__72 + num__69 ) = num__336 required marks = [ ( num__70 x num__6 ) – num__336 ] = ( num__420 – num__336 ) = num__84 answer a <eor> a <eos> |
a |
subtract__70.0__65.0__ subtract__75.0__69.0__ multiply__70.0__6.0__ divide__420.0__5.0__ divide__420.0__5.0__ |
subtract__70.0__65.0__ subtract__75.0__69.0__ multiply__70.0__6.0__ divide__420.0__5.0__ divide__420.0__5.0__ |
| average weight of num__10 people increased by num__1.5 kg when one person of num__45 kg is replaced by a new man . then weight of the new man is <o> a ) num__50 <o> b ) num__55 <o> c ) num__60 <o> d ) num__65 <o> e ) num__70 |
explanation : total weight increased is num__1.5 * num__10 = num__15 . so weight of new person is num__45 + num__15 = num__60 option c <eor> c <eos> |
c |
multiply__10.0__1.5__ add__45.0__15.0__ add__45.0__15.0__ |
multiply__10.0__1.5__ add__45.0__15.0__ add__45.0__15.0__ |
| num__1 num__6 num__13 num__22 num__33 ? <o> a ) num__35 <o> b ) num__46 <o> c ) num__48 <o> d ) num__49 <o> e ) num__38 |
the pattern is + num__5 + num__7 + num__9 + num__11 . . . . answer : b . <eor> b <eos> |
b |
subtract__6.0__1.0__ add__1.0__6.0__ subtract__22.0__13.0__ add__6.0__5.0__ add__13.0__33.0__ |
subtract__6.0__1.0__ add__1.0__6.0__ subtract__22.0__13.0__ add__6.0__5.0__ add__13.0__33.0__ |
| jerry an electrician worked num__9 months out of the year . what percent of the year did he work ? ( round answer to the nearest hundredth ) what percent num__12 is num__9 ? num__12 months = num__1 year <o> a ) num__58.33 <o> b ) num__68.33 <o> c ) num__78.33 <o> d ) num__88.33 <o> e ) num__75.00 % |
num__1 . multiply the opposites num__9 x num__100 = num__900 num__100 = num__0.75 divide by the remaining number num__75.0 correct answer e <eor> e <eos> |
e |
percent__75.0__100.0__ |
percent__75.0__100.0__ |
| the twice the square of a natural number is increased by num__8 the number is equal to num__458 more than the number . <o> a ) num__3 <o> b ) num__6 <o> c ) num__8 <o> d ) num__12 <o> e ) num__15 |
explanation : let the number be x . then num__2 x num__2 + num__8 = num__458 = > num__2 x num__2 = num__450 = > x num__2 - num__225 = > x = num__15 answer : option e <eor> e <eos> |
e |
subtract__458.0__8.0__ divide__450.0__2.0__ divide__225.0__15.0__ |
subtract__458.0__8.0__ divide__450.0__2.0__ divide__225.0__15.0__ |
| in a college the ratio of the numbers of boys to the girls is num__8 : num__5 . if there are num__190 girls the total number of students in the college is ? <o> a ) num__562 <o> b ) num__356 <o> c ) num__452 <o> d ) num__494 <o> e ) num__512 |
let the number of boys and girls be num__8 x and num__5 x then num__5 x = num__190 x = num__38 total number of students = num__13 x = num__13 * num__38 = num__494 answer is d <eor> d <eos> |
d |
divide__190.0__5.0__ add__8.0__5.0__ multiply__38.0__13.0__ multiply__38.0__13.0__ |
divide__190.0__5.0__ add__8.0__5.0__ multiply__38.0__13.0__ multiply__38.0__13.0__ |
| a sum was put at simple interest at a certain rate for num__9 years had it been put at num__5.0 higher rate it would have fetched num__1350 more . find the sum . <o> a ) num__3000 <o> b ) num__3100 <o> c ) num__3200 <o> d ) num__3300 <o> e ) num__3400 |
difference in s . i . = p × t / num__100 ( r num__1 − r num__2 ) ⇒ num__1350 = p × num__9 x num__0.05 ( ∵ r num__1 - r num__2 = num__2 ) ⇒ p = num__1350 × num__11.1111111111 x num__5 = num__3000 answer a <eor> a <eos> |
a |
percent__5.0__1.0__ percent__100.0__3000.0__ |
percent__5.0__1.0__ percent__100.0__3000.0__ |
| the remainder when w = num__1 + num__3 + num__3 ^ num__2 + num__3 ^ num__3 + . . . . . . . . . . + num__3 ^ num__200 is divided num__13 . <o> a ) num__12 <o> b ) num__7 <o> c ) num__0 <o> d ) num__5 <o> e ) num__3 |
w = num__1 + num__3 + num__3 ^ num__2 + num__3 ^ num__3 + . . . . . . . . . . . . . . . . . . + num__3 ^ num__200 is a geometric progression having common ratio as ' num__3 ' and number of terms as ' num__201 ' . since sum to n terms in gp = a ( r ^ n - num__1 ) / ( r - num__1 ) where a = first term and r = common ration hence num__1 * ( num__3 ^ num__201 - num__1 ) / ( num__3 - num__1 ) rem of ( num__3 ^ num__201 - num__1 ) / num__2 divided by num__13 num__3 ^ num__201 - num__0.0384615384615 wkt num__3 ^ num__3 = num__27 = num__26 + num__1 { ( num__26 + num__1 ) ^ num__67 - num__1 } / num__26 { num__1 - num__1 } / num__26 = > num__0 . c <eor> c <eos> |
c |
add__1.0__200.0__ multiply__2.0__13.0__ divide__201.0__3.0__ round_down__0.0385__ round_down__0.0385__ |
add__1.0__200.0__ multiply__2.0__13.0__ divide__201.0__3.0__ round_down__0.0385__ multiply__1.0__0.0__ |
| two trains are moving in opposite directions at num__60 km / hr and num__90 km / hr . their lengths are num__1.10 km and num__0.9 km respectively . the time taken by the slower train to cross the faster train in seconds is ? <o> a ) num__76 sec <o> b ) num__88 sec <o> c ) num__48 sec <o> d ) num__44 sec <o> e ) num__33 sec |
relative speed = num__60 + num__90 = num__150 km / hr . = num__150 * num__0.277777777778 = num__41.6666666667 m / sec . distance covered = num__1.10 + num__0.9 = num__2 km = num__2000 m . required time = num__2000 * num__0.024 = num__48 sec . answer : c <eor> c <eos> |
c |
add__60.0__90.0__ add__1.1__0.9__ multiply__2000.0__0.024__ round__48.0__ |
add__60.0__90.0__ add__1.1__0.9__ multiply__2000.0__0.024__ multiply__2000.0__0.024__ |
| currently apples cost num__75 cents / pound . due to a disease affecting the apple trees it is expected that next month apples will cost num__150.0 more than they do currently . how much are apples expected to cost next month ? <o> a ) num__195 cents / pound <o> b ) num__192 cents / pound <o> c ) num__187.50 cents / pound <o> d ) num__187 cents / pound <o> e ) num__190 cents / pound |
if a new cost is p percent greater than the old cost then ( new cost ) = ( old cost ) + ( p / num__100 ) ( old cost ) . in this case ( new cost ) = num__75 cents / pound + ( num__1.5 ) ( num__75 cents / pound ) = num__75 cents / pound + num__112.50 cents / pound = num__187.50 cents / pound answer : c <eor> c <eos> |
c |
divide__150.0__100.0__ multiply__75.0__1.5__ add__75.0__112.5__ add__75.0__112.5__ |
divide__150.0__100.0__ multiply__75.0__1.5__ add__75.0__112.5__ add__75.0__112.5__ |
| a bus starts from city x . the number of women in the bus is half of the number of men . in city y num__12 men leave the bus and six women enter . now number of men and women is equal . in the beginning how many passengers entered the bus ? <o> a ) num__15 <o> b ) num__30 <o> c ) num__36 <o> d ) num__45 <o> e ) num__54 |
explanation : originally let number of women = x . then number of men = num__2 x . so in city y we have : ( num__2 x - num__12 ) = ( x + num__6 ) or x = num__18 . therefore total number of passengers in the beginning = ( x + num__2 x ) = num__3 x = num__54 . answer : e <eor> e <eos> |
e |
divide__12.0__2.0__ add__12.0__6.0__ divide__6.0__2.0__ multiply__3.0__18.0__ multiply__3.0__18.0__ |
divide__12.0__2.0__ add__12.0__6.0__ divide__6.0__2.0__ multiply__3.0__18.0__ multiply__3.0__18.0__ |
| a and b can finish a work num__30 days if they work together . they worked together for num__20 days and then b left . a finished the remaining work in another num__20 days . in how many days a alone can finish the work ? <o> a ) num__60 <o> b ) num__50 <o> c ) num__40 <o> d ) num__30 <o> e ) num__20 |
explanation : amount of work done by a and b in num__1 day = num__0.0333333333333 amount of work done by a and b in num__20 days = num__20 × ( num__0.0333333333333 ) = num__0.666666666667 = num__0.666666666667 remaining work – num__1 – num__0.666666666667 = num__0.333333333333 a completes num__0.333333333333 work in num__20 days amount of work a can do in num__1 day = ( num__0.333333333333 ) / num__20 = num__0.0166666666667 = > a can complete the work in num__60 days answer : option a <eor> a <eos> |
a |
divide__1.0__30.0__ divide__20.0__30.0__ subtract__1.0__0.6667__ divide__0.3333__20.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
divide__1.0__30.0__ divide__20.0__30.0__ subtract__1.0__0.6667__ divide__0.3333__20.0__ hour_to_min_conversion__ divide__60.0__1.0__ |
| a father said to his son ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is num__46 years now the son ' s age five years back was <o> a ) num__16 years <o> b ) num__17 years <o> c ) num__18 years <o> d ) num__19 years <o> e ) num__20 years |
let the son ' s present age be x years . then ( num__46 - x ) = x num__2 x = num__46 . x = num__23 . son ' s age num__5 years back ( num__23 - num__5 ) = num__18 years . c ) <eor> c <eos> |
c |
divide__46.0__2.0__ subtract__23.0__5.0__ subtract__23.0__5.0__ |
divide__46.0__2.0__ subtract__23.0__5.0__ subtract__23.0__5.0__ |
| for any integer n greater than num__1 n * denotes the product of all the integers from num__1 to n inclusive . how many multiples of num__4 are there between num__4 * and num__5 * inclusive ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__20 <o> d ) num__24 <o> e ) num__25 |
here num__4 * = num__4 ! and num__5 * = num__5 ! question is asking no . of terms in an ap whose first term is num__24 ( num__4 ! ) last term is num__120 ( num__5 ! ) and common difference is num__4 . last term or nth term = first term + ( no . of tems - num__1 ) ( common difference ) tn = a + ( n - num__1 ) d num__120 = num__24 + ( n - num__1 ) num__4 num__24.0 = n - num__1 n = num__25 . . . . . e <eor> e <eos> |
e |
multiply__5.0__24.0__ add__1.0__24.0__ add__1.0__24.0__ |
multiply__5.0__24.0__ add__1.0__24.0__ add__1.0__24.0__ |
| two buses and a van were employed to transport a class of students on a field trip . num__0.6 of the class boarded the first bus . num__0.5 of the remaining students boarded the second bus and the rest of the students boarded the van . when the second bus broke down num__0.5 of the students on the second bus boarded the first bus . what fraction of the class was on board the first bus ? <o> a ) a ) num__0.5 <o> b ) b ) num__0.7 <o> c ) c ) num__0.733333333333 <o> d ) d ) num__0.766666666667 <o> e ) e ) num__0.8 |
this is a pure ratio question ( we are n ' t given any numbers for anything ) so you can just choose any starting number you like and use it to solve the problem . the best number to pick is almost always the product of all the denominators of your fractions so here we could start with num__30 students . then we have : • num__0.6 of these or num__18 students board the first bus • there are num__12 students left . num__0.5 of these or num__6 students board the second bus • this bus breaks down and num__0.5 of the num__6 students or num__3 students board the first bus • the first bus now contains num__21 out of the original num__30 students so the answer is num__0.7 = num__0.7 answer b <eor> b <eos> |
b |
multiply__0.6__30.0__ subtract__30.0__18.0__ multiply__0.5__12.0__ multiply__0.5__6.0__ add__3.0__18.0__ divide__21.0__30.0__ round__0.7__ |
multiply__0.6__30.0__ subtract__30.0__18.0__ multiply__0.5__12.0__ multiply__0.5__6.0__ add__3.0__18.0__ divide__21.0__30.0__ round__0.7__ |
| the average of num__11 numbers is num__30 . if the average of first six numbers is num__17.5 and that of last six is num__42.5 then what is the sixth number ? <o> a ) num__30 <o> b ) num__36 <o> c ) num__45 <o> d ) num__47 <o> e ) num__43 |
given : average of num__11 numbers = num__30 step num__1 : calculate total of num__11 numbers by multiplying it by average value num__30 = num__11 x num__30 = num__330 step num__2 : calculate total of first six members by multiplying it by average value num__17.5 = num__17.5 x num__6 = num__105 step num__3 : calculate total of last six members by multiplying it by average value num__42.5 = num__42.5 x num__6 = num__255 therefore we can find sixth number by adding value of first six and last six numbers and subtracting it from the total value of num__11 numbers . sixth number = ( num__105 + num__255 ) - num__330 = num__30 answer is a <eor> a <eos> |
a |
multiply__11.0__30.0__ multiply__17.5__6.0__ add__1.0__2.0__ multiply__42.5__6.0__ multiply__30.0__1.0__ |
multiply__11.0__30.0__ multiply__17.5__6.0__ add__1.0__2.0__ multiply__42.5__6.0__ multiply__30.0__1.0__ |
| for a finite sequence of non zero numbers the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative . what is the number of variations in sign for the sequence num__1 - num__3 num__2 num__5 - num__4 - num__7 - num__6 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
given sequence : { num__1 - num__3 num__2 num__5 - num__4 - num__6 } the questions basically asks : how manypairs of consecutive termsare there in the sequence such that the product of these consecutive terms is negative . num__1 * ( - num__3 ) = - num__3 = negative ; - num__3 * num__2 = - num__6 = negative ; num__2 * num__5 = num__10 = positive ; num__5 * ( - num__4 ) = - num__20 = negative ; ( - num__4 ) * ( - num__6 ) = num__24 = positive . so there are num__5 pairs of consecutive terms of the sequence for which the product is negative . answer : e . <eor> e <eos> |
e |
add__3.0__7.0__ multiply__2.0__10.0__ multiply__4.0__6.0__ multiply__1.0__5.0__ |
multiply__2.0__5.0__ multiply__2.0__10.0__ multiply__4.0__6.0__ multiply__1.0__5.0__ |
| sonika deposited rs . num__8000 which amounted to rs . num__9200 after num__3 years at simple interest . had the interest been num__2.0 more . she would get how much ? <o> a ) num__9680 <o> b ) num__2338 <o> c ) num__1278 <o> d ) num__1098 <o> e ) num__1279 |
( num__8000 * num__3 * num__2 ) / num__100 = num__480 num__9200 - - - - - - - - num__9680 answer : a <eor> a <eos> |
a |
percent__100.0__9680.0__ |
percent__100.0__9680.0__ |
| a bar over a sequence of digits in a decimal indicates that the sequence repeats indefinitely . what is the value of ( num__10 ^ num__4 - num__10 ^ num__2 ) ( num__0.0014 ) ? <o> a ) num__13.86 <o> b ) num__14.86 <o> c ) num__15.86 <o> d ) num__12.86 <o> e ) num__11.86 |
you get num__10 ^ num__2 ( num__100 - num__1 ) ( . num__0014 ) we know num__0.0014 = . num__0014 num__10 ^ num__2 ( num__99 ) * ( num__0.0014 ) num__10 ^ num__2 = num__100 and num__100 ^ num__2 = num__10000 cancel out the num__100 with the num__10 ^ num__2 left with num__99 ( num__0.14 ) . . num__99 * num__14 = num__13.86 answer : a <eor> a <eos> |
a |
power__10.0__2.0__ add__10.0__4.0__ subtract__100.0__1.0__ power__10.0__4.0__ multiply__0.0014__100.0__ multiply__99.0__0.14__ multiply__1.0__13.86__ |
power__10.0__2.0__ add__10.0__4.0__ subtract__100.0__1.0__ power__10.0__4.0__ multiply__0.0014__100.0__ multiply__99.0__0.14__ multiply__1.0__13.86__ |
| two pipes a and b can fill a tank in num__20 and num__30 minutes respectively . if both the pipes are used together then how long will it take to fill the tank ? <o> a ) num__15 minutes . <o> b ) num__17 minutes . <o> c ) num__12 minutes . <o> d ) num__10 minutes . <o> e ) num__11 minutes . |
part filled by a in num__1 min = num__0.05 part filled by b in num__1 min = num__0.0333333333333 . part filled by ( a + b ) in num__1 min = num__0.05 + num__0.0333333333333 = num__0.0833333333333 . both pipes can fill the tank in num__12 minutes . answer is d <eor> d <eos> |
d |
divide__1.0__20.0__ divide__1.0__30.0__ add__0.05__0.0333__ subtract__30.0__20.0__ |
divide__1.0__20.0__ divide__1.0__30.0__ add__0.05__0.0333__ subtract__30.0__20.0__ |
| a certain company reported that the revenue on sales increased num__40.0 from num__2000 to num__2003 and increased num__60.0 from num__2000 to num__2005 . what was the approximate percent increase in revenue for this store from num__2003 to num__2005 ? <o> a ) num__50.0 <o> b ) num__40.0 <o> c ) num__14.0 <o> d ) num__32.0 <o> e ) num__29 % |
assume the revenue in num__2000 to be num__100 . then in num__2003 it would be num__140 and and in num__2005 num__160 so from num__2003 to num__2005 it increased by ( num__160 - num__140 ) / num__140 = num__0.142857142857 = num__0.142857142857 = ~ num__14.0 . answer : c <eor> c <eos> |
c |
percent__100.0__14.0__ |
percent__100.0__14.0__ |
| radius of a circle is num__5 cm if we draw a rectangle of maximum size what is the area of rectangle ? <o> a ) num__30 <o> b ) num__40 <o> c ) num__50 <o> d ) num__60 <o> e ) num__70 |
area of rectangle of max size inside circle will be a square with diagonal equal to num__10 cms . then side of square = root ( num__50.0 ) = root num__50 area of square = root num__50 * root num__50 = num__50 sq cms . answer : c <eor> c <eos> |
c |
multiply__5.0__10.0__ multiply__5.0__10.0__ |
multiply__5.0__10.0__ multiply__5.0__10.0__ |
| together andrea and brian weigh p pounds ; brian weighs num__14 pounds more than andrea . brian and andrea ' s dog cubby weighs p / num__4 pounds more than andrea . in terms of p what is cubby ' s weight in pounds ? <o> a ) p / num__2 - num__10 <o> b ) num__3 p / num__4 - num__7 <o> c ) num__3 p / num__2 - num__5 <o> d ) num__5 p / num__4 - num__10 <o> e ) num__5 p - num__5 |
together andrea and brian weigh p pounds - - > a + b = p . brian weighs num__10 pounds more than andrea - - > b = a + num__14 - - > a + ( a + num__14 ) = p - - > a = ( p - num__14 ) / num__2 = p / num__2 - num__7 . brian and andrea ' s dog cubby weighs p / num__4 pounds more than andrea - - > c = a + p / num__4 = ( p / num__2 - num__7 ) + p / num__4 = num__3 p / num__4 - num__7 . answer : b . <eor> b <eos> |
b |
subtract__14.0__4.0__ divide__14.0__2.0__ subtract__7.0__4.0__ subtract__7.0__4.0__ |
subtract__14.0__4.0__ divide__14.0__2.0__ subtract__7.0__4.0__ subtract__7.0__4.0__ |
| num__45.0 of the employees of a company are men . num__60.0 of the men in the company speak french and num__40.0 of the employees of the company speak french . what is % of the women in the company who do not speak french ? <o> a ) num__4.0 <o> b ) num__10.0 <o> c ) num__76.36 <o> d ) num__90.16 <o> e ) num__20 % |
no of employees = num__100 ( say ) men = num__45 women = num__55 men speaking french = num__0.6 * num__45 = num__27 employees speaking french = num__0.55 * num__100 = num__55 therefore women speaking french = num__55 - num__42 = num__13 and women not speaking french = num__55 - num__13 = num__42.0 of women not speaking french = num__0.763636363636 * num__100 = num__76.36 answer c <eor> c <eos> |
c |
percent__45.0__60.0__ percent__100.0__76.36__ |
percent__45.0__60.0__ percent__100.0__76.36__ |
| x = num__3 num__1 num__5 y = num__27 num__7 num__81 then what is the relation between x and y <o> a ) num__3 * x ) ^ num__1 <o> b ) num__3 * x ) ^ num__2 <o> c ) num__3 * x ) ^ num__4 <o> d ) num__3 * x ) ^ num__3 <o> e ) num__3 * x ) ^ num__5 |
y = ( num__3 * x ) ^ num__2 num__81 = ( num__3 * num__3 ) ^ num__2 num__81 = num__81 answer : b <eor> b <eos> |
b |
subtract__3.0__1.0__ multiply__3.0__1.0__ |
subtract__3.0__1.0__ multiply__3.0__1.0__ |
| on sunday morning pugsley and wednesday are trading pet spiders . if pugsley were to give wednesday two of his spiders wednesday would then have three times as many spiders as pugsley does . but if wednesday were to give pugsley one of her spiders pugsley would now have two fewer spiders than wednesday had before they traded . how many pet spiders does pugsley have before the trading game commences ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
if pugsley were to give wednesday two of his spiders wednesday would then have three times as many spiders as pugsley does : ( w + num__2 ) = num__3 ( p - num__2 ) if wednesday were to give pugsley one of her spiders pugsley would now have three fewer spiders than wednesday had before they traded : p + num__1 = w - num__3 solving gives p = num__6 and w = num__10 . answer : c . <eor> c <eos> |
c |
subtract__3.0__2.0__ multiply__2.0__3.0__ round__6.0__ |
subtract__3.0__2.0__ multiply__2.0__3.0__ round__6.0__ |
| the simple interest on a certain sum of money for num__2 l / num__2 years at num__12.0 per year is rs . num__40 less than the simple interest on the same sum for num__3 ½ years at num__10.0 per year . find the sum . <o> a ) num__800 <o> b ) num__700 <o> c ) num__600 <o> d ) num__500 <o> e ) none of them |
let the sum be rs . x then ( ( x * num__10 * num__7 ) / ( num__100 * num__2 ) ) – ( ( x * num__12 * num__5 ) / ( num__100 * num__2 ) ) = num__40 = ( num__7 x / num__20 ) - ( num__3 x / num__10 ) = num__40 x = ( num__40 * num__20 ) = num__800 . hence the sum is rs . num__800 . answer is a . <eor> a <eos> |
a |
subtract__10.0__3.0__ add__2.0__3.0__ multiply__2.0__10.0__ multiply__40.0__20.0__ multiply__40.0__20.0__ |
subtract__10.0__3.0__ subtract__12.0__7.0__ multiply__2.0__10.0__ multiply__40.0__20.0__ multiply__40.0__20.0__ |
| half a number plus num__5 is num__13 . what is the number ? <o> a ) num__8 <o> b ) num__16 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
let x be the number . always replace ` ` is ' ' with an equal sign ( num__0.5 ) x + num__5 = num__13 ( num__0.5 ) x = num__13 - num__5 ( num__0.5 ) x = num__8 x = num__16 correct answer is b <eor> b <eos> |
b |
subtract__13.0__5.0__ divide__8.0__0.5__ divide__8.0__0.5__ |
subtract__13.0__5.0__ divide__8.0__0.5__ divide__8.0__0.5__ |
| if num__9 gallons of gasoline are added to a tank that is already filled to num__0.75 of its capacity the tank is then filled to num__0.9 of its capacity . how many gallons does the tank hold ? <o> a ) num__20 <o> b ) num__24 <o> c ) num__36 <o> d ) num__40 <o> e ) num__54 |
let the capacity of the tank = c ( num__0.75 ) c + num__9 = ( num__0.9 ) c = > ( num__0.9 ) c - ( num__0.75 ) c = num__9 = > ( num__0.15 ) c = num__9 = > c = ( num__9 * num__20 ) / num__3 = num__60 number of gallons of gasoline that the tank currently holds = num__0.75 * c + num__9 = num__45 + num__9 = num__54 answer e <eor> e <eos> |
e |
subtract__0.9__0.75__ multiply__0.15__20.0__ divide__9.0__0.15__ multiply__0.75__60.0__ add__9.0__45.0__ add__9.0__45.0__ |
subtract__0.9__0.75__ multiply__0.15__20.0__ divide__9.0__0.15__ multiply__0.75__60.0__ add__9.0__45.0__ add__9.0__45.0__ |
| robert left from a pvt company . management hold his salary rs . num__15000 / - for one month . earlier robert earned a performance incentive rs . num__7280 / - from company . but robert forgot that . after one month robert asked his salary and accountant gives rs . num__18500 / - to him . what is the bonus amount given to robert ? <o> a ) a ) num__9500 <o> b ) b ) num__12500 <o> c ) c ) num__10780 <o> d ) d ) num__10500 <o> e ) e ) num__8600 |
total salary = rs . num__15000 / - incentive earned earlier = num__7280 / - balance salary = num__15000 - num__7280 = num__7720 paid amount = num__18500 / - bonus = num__18500 - num__7720 = num__10780 / - answer is c <eor> c <eos> |
c |
subtract__15000.0__7280.0__ subtract__18500.0__7720.0__ subtract__18500.0__7720.0__ |
subtract__15000.0__7280.0__ subtract__18500.0__7720.0__ subtract__18500.0__7720.0__ |
| souju ' s age is num__118.0 of what it was num__10 years ago but num__86 num__0.666666666667 % of what it will be after num__10 years . what is her present age ? <o> a ) num__56 years <o> b ) num__65 years <o> c ) num__60 years <o> d ) num__66 years <o> e ) num__75 years |
let the age before num__10 years = x . then num__118 x / num__100 = x + num__10 â ‡ ’ num__118 x = num__100 x + num__1000 â ‡ ’ x = num__55.5555555556 = num__55 present age = x + num__10 = num__55 + num__10 = num__65 answer : b <eor> b <eos> |
b |
multiply__10.0__100.0__ round_down__55.5556__ add__10.0__55.0__ add__10.0__55.0__ |
multiply__10.0__100.0__ round_down__55.5556__ add__10.0__55.0__ add__10.0__55.0__ |
| pipe a can fill a tank in num__5 hours pipe b in num__10 hours and pipe c in num__30 hours . if all the pipes are open in how many hours will the tank be filled ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__3 <o> d ) num__6 <o> e ) num__8 |
part filled by a + b + c in num__1 hour = num__0.2 + num__0.1 + num__0.0333333333333 = num__0.333333333333 all the three pipes together will fill the tank in num__3 hours . answer : c <eor> c <eos> |
c |
divide__1.0__5.0__ divide__1.0__10.0__ divide__1.0__30.0__ divide__10.0__30.0__ divide__30.0__10.0__ round__3.0__ |
divide__1.0__5.0__ divide__1.0__10.0__ divide__1.0__30.0__ divide__10.0__30.0__ divide__30.0__10.0__ round__3.0__ |
| a cashier mentally reversed the digits of one customer ' s correct amount of change and thus gave the customer an incorrect amount of change . if the cash register contained num__18 cents more than it should have as a result of this error which of the following could have been the correct amount of change in cents ? <o> a ) num__14 <o> b ) num__45 <o> c ) num__54 <o> d ) num__65 <o> e ) num__42 |
just check the answers and reverse the numbers until you get num__18 . num__42 - num__24 = num__18 answer e <eor> e <eos> |
e |
subtract__42.0__18.0__ add__18.0__24.0__ |
subtract__42.0__18.0__ add__18.0__24.0__ |
| if num__0 < a < b and k = ( num__2 a + num__3 b ) / b which of the following must be true ? <o> a ) k < num__2 <o> b ) k < num__5 <o> c ) k < num__9 <o> d ) k > num__9 <o> e ) k > num__11 |
here ' s another approach : k = ( num__2 a + num__3 b ) / b = num__2 a / b + num__3 b / b = num__2 ( a / b ) + num__3 since num__0 < a < b we know that a / b is less than num__1 which means that num__2 ( a / b ) is some number less than num__2 . so we get k = ( some number less than num__2 ) + num__3 from here we can see that k must be less than num__5 answer : b <eor> b <eos> |
b |
subtract__3.0__2.0__ add__2.0__3.0__ add__2.0__3.0__ |
subtract__3.0__2.0__ add__2.0__3.0__ add__2.0__3.0__ |
| a wire in the form of a circle of radius num__3.5 m is bent in the form of a rectangule whose length and breadth are in the ratio of num__6 : num__5 . what is the area of the rectangle ? <o> a ) num__65 cm num__2 <o> b ) num__30 cm num__2 <o> c ) num__16 cm num__2 <o> d ) num__17 cm num__2 <o> e ) num__15 cm num__2 |
the circumference of the circle is equal to the permeter of the rectangle . let l = num__6 x and b = num__5 x num__2 ( num__6 x + num__5 x ) = num__2 * num__3.14285714286 * num__3.5 = > x = num__1 therefore l = num__6 cm and b = num__5 cm area of the rectangle = num__6 * num__5 = num__30 cm num__2 answer : b <eor> b <eos> |
b |
multiply__6.0__5.0__ multiply__6.0__5.0__ |
multiply__6.0__5.0__ multiply__6.0__5.0__ |
| two trains num__140 m and num__160 m long run at the speed of num__60 km / hr and num__40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? <o> a ) num__10.2 <o> b ) num__10.5 <o> c ) num__10.9 <o> d ) num__10.8 <o> e ) num__10.1 |
relative speed = num__60 + num__40 = num__100 km / hr . = num__100 * num__0.277777777778 = num__27.7777777778 m / sec . distance covered in crossing each other = num__140 + num__160 = num__300 m . required time = num__300 * num__0.036 = num__10.8 = num__10.8 sec . answer : d <eor> d <eos> |
d |
subtract__140.0__40.0__ add__140.0__160.0__ multiply__300.0__0.036__ round__10.8__ |
add__60.0__40.0__ add__140.0__160.0__ multiply__300.0__0.036__ multiply__300.0__0.036__ |
| claire can paint the living room in num__7 hours and barry can paint the living room in num__5 hours . how many hours will it take for both of them working together to paint the living room ? <o> a ) num__2 <o> b ) num__2 num__0.916666666667 <o> c ) num__3 num__0.454545454545 <o> d ) num__4 num__0.5 <o> e ) num__5 |
work hrs = ab / ( a + b ) = num__2.91666666667 = num__2 num__0.916666666667 answer is b <eor> b <eos> |
b |
subtract__7.0__5.0__ subtract__2.9167__2.0__ round__2.0__ |
subtract__7.0__5.0__ subtract__2.9167__2.0__ round__2.0__ |
| two persons start running simultaneously around a circular track of length num__600 m from the same point at speeds of num__15 km / hr and num__25 km / hr . when will they meet for the first time any where on the track if they are moving in opposite directions ? <o> a ) num__11 <o> b ) num__54 <o> c ) num__28 <o> d ) num__27 <o> e ) num__12 |
time taken to meet for the first time anywhere on the track = length of the track / relative speed = num__600 / ( num__15 + num__25 ) num__0.277777777778 = num__600 * num__0.45 * num__5 = num__54 seconds . answer : b <eor> b <eos> |
b |
round__54.0__ |
round__54.0__ |
| the area of a triangle will be when a = num__3 m b = num__4 m c = num__5 m a b c being lengths of respective sides ? <o> a ) num__3 <o> b ) num__6 <o> c ) num__4 <o> d ) num__9 <o> e ) num__1 |
s = ( num__3 + num__4 + num__5 ) / num__2 = num__6 answer : b <eor> b <eos> |
b |
triangle_area__3.0__4.0__ triangle_area__3.0__4.0__ |
triangle_area__3.0__4.0__ triangle_area__3.0__4.0__ |
| three pipes a b and c can fill a tank from empty to full in num__30 minutes num__20 minutes and num__10 minutes respectively . when the tank is empty all the three pipes are opened . a b and c discharge chemical solutions p q and r respectively . what is the proportion of solution r in the liquid in the tank after num__3 minutes ? <o> a ) num__0.454545454545 <o> b ) num__0.545454545455 <o> c ) num__0.636363636364 <o> d ) num__0.727272727273 <o> e ) num__0.818181818182 |
part filled by ( a + b + c ) in num__3 minutes = num__3 ( num__0.0333333333333 + num__0.05 + num__0.1 ) = num__0.55 part filled by c in num__3 minutes = num__0.3 required ratio = num__0.3 * num__1.81818181818 = num__0.545454545455 answer : b <eor> b <eos> |
b |
divide__3.0__30.0__ divide__3.0__10.0__ multiply__1.8182__0.3__ multiply__1.8182__0.3__ |
divide__3.0__30.0__ multiply__3.0__0.1__ multiply__1.8182__0.3__ multiply__1.8182__0.3__ |
| a certain family has num__3 sons : richard is num__6 years older than david and david is num__8 years older than scott . if in num__8 years richard will be twice as old as scott then how old was david num__2 years ago ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
let ' s say age of richard isr age of david isd age of scott iss now richard is num__6 years older than david i . e . r = d + num__6 david is num__8 years older than scott i . e . d = s + num__8 if in num__8 years richard will be twice as old as scott i . e . r + num__8 = num__2 x ( s + num__8 ) i . e . r + num__8 = num__2 s + num__16 i . e . r = num__2 s + num__8 but r = d + num__6 = ( s + num__8 ) + num__6 = s + num__14 therefore num__2 s + num__8 = s + num__14 i . e . s = num__6 i . e . r = num__20 i . e . d = num__14 now how old was david num__2 years ago ? i . e . d - num__2 = num__14 - num__2 = num__12 years answer : option c <eor> c <eos> |
c |
twice__8.0__ add__6.0__8.0__ add__6.0__14.0__ twice__6.0__ twice__6.0__ |
twice__8.0__ add__6.0__8.0__ add__6.0__14.0__ subtract__14.0__2.0__ subtract__14.0__2.0__ |
| look carefully for the pattern and then choose which pair of numbers comes next . num__28 num__25 num__5 num__21 num__18 num__5 num__14 <o> a ) num__11 num__5 <o> b ) num__10 num__7 <o> c ) num__11 num__8 <o> d ) num__5 num__10 <o> e ) num__10 num__5 |
explanation : this is an alternating subtraction series with the interpolation of a random number num__5 as every third number . in the subtraction series num__3 is subtracted then num__4 then num__3 and so on . answer : option a <eor> a <eos> |
a |
subtract__28.0__25.0__ subtract__25.0__21.0__ subtract__25.0__14.0__ |
subtract__28.0__25.0__ subtract__25.0__21.0__ subtract__25.0__14.0__ |
| a fruit drink is made of orange watermelon and grape juice where num__25 percent of the drink is orange juice and num__40 percent is watermelon juice . if the drink is made with num__105 ounces of grape juice how many ounces is the drink total ? <o> a ) num__220 <o> b ) num__250 <o> c ) num__280 <o> d ) num__300 <o> e ) num__340 |
let the total number of ounces in the drink be x . % of orange = num__25.0 % of watermelon = num__40.0 % of grape = num__100.0 - num__65.0 = num__35.0 num__0.35 x = num__105 x = num__300 therefore there are a total of num__300 ounces in the drink . the answer is d . <eor> d <eos> |
d |
add__25.0__40.0__ subtract__100.0__65.0__ divide__35.0__100.0__ divide__105.0__0.35__ divide__105.0__0.35__ |
subtract__105.0__40.0__ subtract__100.0__65.0__ divide__35.0__100.0__ divide__105.0__0.35__ divide__105.0__0.35__ |
| in an election candidate a got num__75.0 of the total valid votes . if num__15.0 of the total votes were declared invalid and the total numbers of votes is num__560000 find the number of valid vote polled in favor of candidate . <o> a ) num__330000 <o> b ) num__340000 <o> c ) num__347000 <o> d ) num__356000 <o> e ) num__357000 |
total number of invalid votes = num__15.0 of num__560000 = num__0.15 × num__560000 = num__84000.0 = num__84000 total number of valid votes num__560000 – num__84000 = num__476000 percentage of votes polled in favour of candidate a = num__75.0 therefore the number of valid votes polled in favour of candidate a = num__75.0 of num__476000 = num__0.75 × num__476000 = num__357000.0 = num__357000 e ) <eor> e <eos> |
e |
percent__15.0__560000.0__ percent__75.0__476000.0__ percent__75.0__476000.0__ |
percent__15.0__560000.0__ percent__75.0__476000.0__ percent__75.0__476000.0__ |
| a sum of money triples itself in twelve years at simple interest . find the rate of interest ? <o> a ) num__16 num__0.285714285714 % <o> b ) num__16 num__2.66666666667 % <o> c ) num__17 num__0.666666666667 % <o> d ) num__11 num__0.666666666667 % <o> e ) num__16 num__0.666666666667 % |
let the pricipal be rs . x then amount = num__3 x ( where r = rate of interest ) = > interest = num__3 x - x = rs . num__2 x r = ( num__100 * num__2 x ) / ( x * num__12 ) = num__16.6666666667 % = num__16 num__0.666666666667 % . answer : e <eor> e <eos> |
e |
percent__100.0__16.0__ |
percent__100.0__16.0__ |
| in how many different ways can the letters of the word ' leading ' be arranged in such a way that the vowels always come together ? <o> a ) num__720 <o> b ) num__850 <o> c ) num__200 <o> d ) num__365 <o> e ) num__750 |
the word ' leading ' has num__7 different letters . when the vowels eai are always together they can be supposed to form one letter . then we have to arrange the letters lndg ( eai ) . now num__5 ( num__4 + num__1 = num__5 ) letters can be arranged in num__5 ! = num__120 ways . the vowels ( eai ) can be arranged among themselves in num__3 ! = num__6 ways . required number of ways = ( num__120 x num__6 ) = num__720 . answer a <eor> a <eos> |
a |
vowel_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
vowel_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| if a boat is moving in upstream with velocity of num__14 km / hr and goes downstream with a velocity of num__40 km / hr then what is the speed of the stream ? <o> a ) num__13 <o> b ) num__44 <o> c ) num__55 <o> d ) num__6 <o> e ) num__77 |
num__13 km / hr answer : a <eor> a <eos> |
a |
round__13.0__ |
round__13.0__ |
| a driver would have reduced the time it took to drive from home to the store by num__0.25 if the average speed had been increased by num__12 miles per hour . what was the actual average speed in miles per hour when the driver drove from home to the store ? <o> a ) num__24 <o> b ) num__28 <o> c ) num__32 <o> d ) num__36 <o> e ) num__40 |
since the distance remains the same ( we ' re just changing the rate and time ) any increase in rate or time is met with a decrease in the other term . decreasing the time by num__0.25 would give us : d = ( r ) ( t ) = ( num__3 t / num__4 ) ( x * r ) x = num__1.33333333333 since ( num__3 t / num__4 ) ( num__4 r / num__3 ) = ( r ) ( t ) = d num__4 r / num__3 = r + num__12 r / num__3 = num__12 r = num__36 the answer is d . <eor> d <eos> |
d |
multiply__0.25__12.0__ reverse__0.25__ divide__4.0__3.0__ multiply__12.0__3.0__ multiply__12.0__3.0__ |
multiply__0.25__12.0__ reverse__0.25__ divide__4.0__3.0__ multiply__12.0__3.0__ multiply__12.0__3.0__ |
| a man has rs . num__544 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__80 <o> b ) num__90 <o> c ) num__95 <o> d ) num__98 <o> e ) num__102 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__544 num__16 x = num__544 x = num__34 . hence total number of notes = num__3 x = num__102 e <eor> e <eos> |
e |
divide__544.0__16.0__ multiply__34.0__3.0__ multiply__34.0__3.0__ |
divide__544.0__16.0__ multiply__34.0__3.0__ multiply__34.0__3.0__ |
| two trains are running at num__40 km / hr and num__20 km / hr respectively in the same direction . fast train completely passes a man sitting in the slower train in num__5 sec . what is the length of the fast train ? <o> a ) num__27 num__0.875 <o> b ) num__27 num__1.16666666667 <o> c ) num__27 num__1.75 <o> d ) num__27 num__7 by num__9 <o> e ) num__27 num__7.0 |
relative speed = ( num__40 - num__20 ) = num__20 km / hr . = num__20 * num__0.277777777778 = num__5.55555555556 m / sec . length of faster train = num__5.55555555556 * num__5 = num__27.7777777778 = num__27 num__0.777777777778 m . answer : d <eor> d <eos> |
d |
subtract__27.7778__27.0__ round__27.0__ |
subtract__27.7778__27.0__ subtract__27.7778__0.7778__ |
| num__20 throws of a die produces following results score - - number of occurrences - - - num__1 - - - - - - - - - - - - - - - - - - - num__4 - - - num__2 - - - - - - - - - - - - - - - - - - - num__3 - - - num__3 - - - - - - - - - - - - - - - - - - - num__5 - - - num__4 - - - - - - - - - - - - - - - - - - - num__2 - - - num__5 - - - - - - - - - - - - - - - - - - - num__2 - - - num__6 - - - - - - - - - - - - - - - - - - - num__4 what is the probability m that one more throw to this series will increase the mean score ? <o> a ) num__0.166666666667 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__0.833333333333 |
i get the mean as num__3.35 - - > num__3.35 . thus in order to raise this mean we would need a num__45 or num__6 . thus probability mis num__0.5 . c <eor> c <eos> |
c |
reverse__2.0__ reverse__2.0__ |
reverse__2.0__ subtract__1.0__0.5__ |
| the average weight of num__8 persons increases by num__2.5 kg when a new person comes in place of one of them weighing num__65 kg . what might be the weight of the new person ? <o> a ) num__75 kg <o> b ) num__65 kg <o> c ) num__55 kg <o> d ) num__85 kg <o> e ) num__25 kg |
total weight increased = ( num__8 x num__2.5 ) kg = num__20 kg . weight of new person = ( num__65 + num__20 ) kg = num__85 kg . answer : d <eor> d <eos> |
d |
multiply__8.0__2.5__ add__65.0__20.0__ add__65.0__20.0__ |
multiply__8.0__2.5__ add__65.0__20.0__ add__65.0__20.0__ |
| two pipes a and b can fill a tank in num__40 and num__120 minutes respectively . if both the pipes are used together then how long will it take to fill the tank ? <o> a ) num__30 <o> b ) num__35 <o> c ) num__25 <o> d ) num__82 <o> e ) num__22 |
part filled by a in num__1 min . = num__0.025 part filled by b in num__1 min . = num__0.00833333333333 part filled by ( a + b ) in num__1 min . = num__0.025 + num__0.00833333333333 = num__0.0333333333333 . both the pipes can fill the tank in num__30 minutes . answer : a <eor> a <eos> |
a |
divide__1.0__40.0__ divide__1.0__120.0__ add__0.025__0.0083__ round__30.0__ |
divide__1.0__40.0__ divide__1.0__120.0__ add__0.025__0.0083__ round__30.0__ |
| num__100 oranges were bought at the rate of rs . num__350 and sold at the rate of rs . num__48 per dozen . what is the percentage of profit or loss ? <o> a ) num__19 num__0.285714285714 % <o> b ) num__17 num__0.285714285714 % <o> c ) num__14 num__0.285714285714 % <o> d ) num__15 num__0.285714285714 % <o> e ) num__5 num__0.882352941176 % |
num__100 * num__4.0 * num__350 - num__1 ) num__100 = num__0.142857142857 * num__100 = num__14 num__0.285714285714 % i . e . profit percentage = num__14 num__0.285714285714 % answer : c <eor> c <eos> |
c |
percent__4.0__350.0__ percent__100.0__14.0__ |
percent__4.0__350.0__ percent__100.0__14.0__ |
| a waitress ' s income consists of her salary and tips . during one week her tips were num__1.75 of her salary . what fraction of her income for the week came from tips ? <o> a ) num__0.111111111111 <o> b ) num__0.166666666667 <o> c ) num__0.666666666667 <o> d ) num__0.444444444444 <o> e ) num__0.636363636364 |
her tips were num__1.75 of her salary . let ' s say her salary = $ num__4 this mean her tips = ( num__1.75 ) ( $ num__4 ) = $ num__7 so her total income = $ num__4 + $ num__7 = $ num__11 what fraction of her income for the week came from tips $ num__7 / $ num__11 = num__0.636363636364 = e <eor> e <eos> |
e |
multiply__1.75__4.0__ add__4.0__7.0__ divide__7.0__11.0__ divide__7.0__11.0__ |
multiply__1.75__4.0__ add__4.0__7.0__ divide__7.0__11.0__ divide__7.0__11.0__ |
| the length of a room is num__5.5 m and width is num__3.75 m . find the cost of paving the floor by slabs at the rate of rs . num__800 per sq . metre . <o> a ) rs . num__15000 <o> b ) rs . num__15550 <o> c ) rs . num__15600 <o> d ) rs . num__16500 <o> e ) none of these |
solution area of the floor = ( num__5.5 × num__3.75 ) m num__2 = num__20.625 m num__2 ∴ cost of paving = rs . ( num__800 × num__20.625 ) = num__16500 . answer d <eor> d <eos> |
d |
multiply__5.5__3.75__ multiply__800.0__20.625__ round__16500.0__ |
multiply__5.5__3.75__ multiply__800.0__20.625__ round__16500.0__ |
| what is theleast number should be added to num__1056 so the sum of the number is completely divisible by num__23 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__5 <o> d ) num__6 <o> e ) num__8 |
( num__45.9130434783 ) gives remainder num__21 num__21 + num__2 = num__23 so we need to add num__2 b <eor> b <eos> |
b |
divide__1056.0__23.0__ subtract__23.0__21.0__ subtract__23.0__21.0__ |
divide__1056.0__23.0__ subtract__23.0__21.0__ subtract__23.0__21.0__ |
| the unit digit in the product ( num__635 * num__767 * num__984 * num__489 ) is : <o> a ) num__0 <o> b ) num__8 <o> c ) num__3 <o> d ) num__2 <o> e ) num__1 |
explanation : unit digit in the given product = unit digit in ( num__5 * num__7 * num__4 * num__9 ) = num__0 answer : a <eor> a <eos> |
a |
add__4.0__5.0__ multiply__635.0__0.0__ |
add__4.0__5.0__ multiply__635.0__0.0__ |
| if b does not equal zero and ab = b / num__8 what is the value of a ? <o> a ) a ) num__0.125 <o> b ) b ) num__0.25 <o> c ) c ) num__0.333333333333 <o> d ) d ) num__0.5 <o> e ) of the above |
explanation : to solve for a divide both sides of the equation by b : ab = b / num__8 ( ab ) / b = ( b / num__8 ) / b a = ( b / num__8 ) * num__1 / b a = num__0.125 answer : ( a ) . <eor> a <eos> |
a |
reverse__8.0__ reverse__8.0__ |
reverse__8.0__ reverse__8.0__ |
| a tradesman by means of his false balance defrauds to the extent of num__20.0 ? in buying goods as well as by selling the goods . what percent does he gain on his outlay ? <o> a ) num__11 <o> b ) num__22 <o> c ) num__44 <o> d ) num__55 <o> e ) num__79 |
g % = num__20 + num__20 + ( num__20 * num__20 ) / num__100 = num__44.0 answer : c <eor> c <eos> |
c |
percent__100.0__44.0__ |
percent__100.0__44.0__ |
| the average weight of num__8 person ' s increases by num__3.5 kg when a new person comes in place of one of them weighing num__65 kg . what might be the weight of the new person ? <o> a ) kg <o> b ) num__70 kg <o> c ) num__80 kg <o> d ) num__90 kg <o> e ) num__93 kg |
total weight increased = ( num__8 x num__3.5 ) kg = num__28 kg . weight of new person = ( num__65 + num__28 ) kg = num__93 kg . e ) <eor> e <eos> |
e |
multiply__8.0__3.5__ add__65.0__28.0__ add__65.0__28.0__ |
multiply__8.0__3.5__ add__65.0__28.0__ add__65.0__28.0__ |
| mary ’ s annual income is $ num__15000 and john ’ s annual income is $ num__18000 . by how much must mary ’ s annual income increase so that it constitutes num__70.0 of mary and john ’ s combined income ? <o> a ) $ num__3000 <o> b ) $ num__4000 <o> c ) $ num__7000 <o> d ) $ num__11000 <o> e ) $ num__27 |
000 |
let mary ' s income increase by x then the equation will be num__15000 + x = ( num__0.7 ) * ( num__15000 + x + num__18000 ) num__15000 + x = ( num__0.7 ) * ( num__33000 + x ) num__300000 + num__20 x = num__14 x + num__462000 num__6 x = num__162000 x = num__27000 so answer will be e <eor> e <eos> |
e |
e |
| in how many different ways can the letters of the word ' optical ' be arranged so that the vowels always come together ? <o> a ) num__610 <o> b ) num__720 <o> c ) num__825 <o> d ) num__920 <o> e ) none of these |
explanation : the word ' optical ' has num__7 letters . it has the vowels ' o ' ' i ' ' a ' in it and these num__3 vowels should always come together . hence these three vowels can be grouped and considered as a single letter . that is ptcl ( oia ) . hence we can assume total letters as num__5 . and all these letters are different . number of ways to arrange these letters = num__5 ! = [ num__5 x num__4 x num__3 x num__2 x num__1 ] = num__120 all the num__3 vowels ( oia ) are different number of ways to arrange these vowels among themselves = num__3 ! = [ num__3 x num__2 x num__1 ] = num__6 hence required number of ways = num__120 x num__6 = num__720 . answer : option b <eor> b <eos> |
b |
vowel_space__ coin_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
vowel_space__ coin_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| a rectangular plot measuring num__15 meters by num__50 meters is to be enclosed by wire fencing . if the poles of the fence are kept num__5 meters apart . how many poles will be needed ? <o> a ) num__46 m <o> b ) num__26 m <o> c ) num__26 m <o> d ) num__56 m <o> e ) num__25 m |
perimeter of the plot = num__2 ( num__15 + num__50 ) = num__130 m no of poles = num__26.0 = num__26 m answer : b <eor> b <eos> |
b |
divide__130.0__5.0__ round__26.0__ |
divide__130.0__5.0__ round__26.0__ |
| how much time will a train of length num__200 m moving at a speed of num__72 kmph take to cross another train of length num__300 m moving at num__36 kmph in the same direction ? <o> a ) num__50 <o> b ) num__99 <o> c ) num__88 <o> d ) num__77 <o> e ) num__51 |
the distance to be covered = sum of their lengths = num__200 + num__300 = num__500 m . relative speed = num__72 - num__36 = num__36 kmph = num__36 * num__0.277777777778 = num__10 mps . time required = d / s = num__50.0 = num__50 sec . answer : a <eor> a <eos> |
a |
add__200.0__300.0__ divide__500.0__10.0__ round__50.0__ |
add__200.0__300.0__ divide__500.0__10.0__ divide__500.0__10.0__ |
| in the manufacture of a certain product num__9 percent of the units produced are defective and num__4 percent of the defective units are shipped for sale . what percent of the units produced are defective units that are shipped for sale ? <o> a ) num__0.125 <o> b ) num__0.36 <o> c ) num__0.8 <o> d ) num__1.25 <o> e ) num__2.0 % |
percent of defective produced = num__9.0 percent of the defective units that are shipped for sale = num__4.0 percent of units produced are defective units that are shipped for sale = ( num__0.04 ) * ( num__0.09 ) * num__100.0 = ( num__0.0036 ) * num__100.0 = ( num__0.36 ) % = . num__36.0 answer b <eor> b <eos> |
b |
percent__9.0__0.04__ percent__9.0__4.0__ percent__9.0__4.0__ |
percent__9.0__0.04__ percent__9.0__4.0__ percent__9.0__4.0__ |
| if janice was num__21 years old z years ago and lisa will be num__19 years old in p years what was the average ( arithmetic mean ) of their ages num__8 years ago ? <o> a ) ( z + p ) / num__2 <o> b ) ( z - p + num__24 ) / num__4 <o> c ) ( z - p + num__14 ) / num__4 <o> d ) ( z + p + num__34 ) / num__2 <o> e ) ( z - p + num__24 ) / num__2 |
today j = z + num__21 and l = num__19 - p num__8 years ago j = z + num__13 and l = num__11 - p the average of their ages was ( z - p + num__24 ) / num__2 the answer is e . <eor> e <eos> |
e |
subtract__21.0__8.0__ subtract__19.0__8.0__ add__11.0__13.0__ subtract__21.0__19.0__ add__11.0__13.0__ |
subtract__21.0__8.0__ subtract__19.0__8.0__ add__11.0__13.0__ subtract__21.0__19.0__ add__11.0__13.0__ |
| find all positive integers n such that n ^ num__2 + num__1 is divisible by n + num__1 . <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
there is only one such positive integer : n = num__1 . in fact n ^ num__2 + num__1 = n ( n + num__1 ) - ( n - num__1 ) ; thus if n + num__1 | n ^ num__2 + num__1 then n + num__1 | n - num__1 which for positive integer n is possible only if n - num__1 = num__0 hence if n = num__1 . answer is num__1 <eor> a <eos> |
a |
reverse__1.0__ |
subtract__2.0__1.0__ |
| num__5 men are equal to as many women as are equal to num__7 boys . all of them earn rs . num__90 only . men ’ s wages are ? <o> a ) num__8 rs . <o> b ) num__7 rs . <o> c ) num__6 rs . <o> d ) num__4 rs . <o> e ) num__5 rs . |
num__5 m = xw = num__7 b num__5 m + xw + num__7 b - - - - - num__90 rs . num__5 m + num__5 m + num__5 m - - - - - num__90 rs . num__15 m - - - - - - num__90 rs . = > num__1 m = num__6 rs . answer : c <eor> c <eos> |
c |
add__5.0__1.0__ add__5.0__1.0__ |
add__5.0__1.0__ add__5.0__1.0__ |
| in a village there are four tribes located at different distances from each other . tribe r is num__60 miles away from tribe p ; tribe s num__40 miles away from tribe b and tribe r is num__10 miles nearer to the tribe s than it is to tribe q . how far is the tribe s located from the tribe p ? <o> a ) num__23 km <o> b ) num__24 km <o> c ) num__25 km <o> d ) num__26 km <o> e ) num__27 |
solution : num__25 distance between tribe q and tribe s = num__40 km = > distance between tribe q and tribe r + distance between tribe r and tribe s = num__40 km = > distance between tribe q and tribe r + ( distance between tribe q and tribe r - num__10 ) = num__40 km = > num__2 * distance between tribe q and tribe r = num__50 km = > distance between tribe q and tribe r = num__25 km answer c <eor> c <eos> |
c |
subtract__60.0__10.0__ round__25.0__ |
subtract__60.0__10.0__ subtract__50.0__25.0__ |
| a boat can travel with a speed of num__16 km / hr in still water . if the rate of stream is num__5 km / hr then find the time taken by the boat to cover distance of num__168 km downstream . <o> a ) num__4 hours <o> b ) num__5 hours <o> c ) num__6 hours <o> d ) num__7 hours <o> e ) num__8 hours |
explanation : it is very important to check if the boat speed given is in still water or with water or against water . because if we neglect it we will not reach on right answer . i just mentioned here because mostly mistakes in this chapter are of this kind only . lets see the question now . speed downstream = ( num__16 + num__5 ) = num__21 kmph time = distance / speed = num__8.0 = num__8 hours option e <eor> e <eos> |
e |
add__16.0__5.0__ divide__168.0__21.0__ round__8.0__ |
add__16.0__5.0__ divide__168.0__21.0__ divide__168.0__21.0__ |
| look at this series : u num__32 v num__29 __ x num__23 y num__20 . . . what number should fill the blank ? <o> a ) m num__43 <o> b ) d num__14 <o> c ) w num__26 <o> d ) f num__24 <o> e ) k num__81 |
c w num__26 in this series the letters progress by num__1 ; the numbers decrease by num__3 . <eor> c <eos> |
c |
subtract__32.0__29.0__ subtract__29.0__3.0__ |
subtract__32.0__29.0__ subtract__29.0__3.0__ |
| if the length of the longest chord of a certain circle is num__22 what is the radius of that certain circle ? <o> a ) num__11 <o> b ) num__5 <o> c ) num__10 <o> d ) num__15 <o> e ) num__20 |
longest chord of a circle is the diameter of the circle diameter = num__2 * radius if diameter of the circle is given as num__22 = num__2 * num__11 so radius of the circle = num__11 correct answer - a <eor> a <eos> |
a |
triangle_area__2.0__11.0__ |
triangle_area__2.0__11.0__ |
| mary peter and lucy were picking chestnuts . mary picked three times as much as much chestnuts than peter . lucy picked num__4 kg more than peter . together the three of them picked num__29 kg of chestnuts . how many kilograms of chestnuts did mary peter and lucy pick respectively ? <o> a ) num__9 num__5 and num__15 <o> b ) num__5 num__15 and num__9 <o> c ) num__15 num__5 and num__9 <o> d ) num__9 num__15 and num__5 <o> e ) num__5 num__9 and num__15 |
m = num__3 p l = p + num__4 m + p + l = num__29 num__3 p + p + ( p + num__4 ) = num__29 p = num__5 m = num__15 l = num__9 therefore mary peter and lucy picked num__15 num__5 and num__9 kg respectively . the answer is c . <eor> c <eos> |
c |
multiply__3.0__5.0__ add__4.0__5.0__ multiply__3.0__5.0__ |
multiply__3.0__5.0__ add__4.0__5.0__ multiply__3.0__5.0__ |
| sum of the squares of three numbers is num__414 and the sum of their products taken two at a time is num__131 . find the sum ? <o> a ) num__20 <o> b ) num__22 <o> c ) num__25 <o> d ) num__26 <o> e ) num__29 |
( a + b + c ) num__2 = a num__2 + b num__2 + c num__2 + num__2 ( ab + bc + ca ) = num__414 + num__2 * num__131 a + b + c = √ num__676 = num__26 d <eor> d <eos> |
d |
divide__676.0__26.0__ |
divide__676.0__26.0__ |
| john is three times as old as sam . if john will be twice as old as sam in six years how old was sam four years ago ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__16 |
j = num__3 s after num__6 years j + num__6 = num__2 ( s + num__6 ) j = num__2 s + num__6 num__2 s + num__6 = num__3 s s = num__6 four years ago s = num__6 - num__4 = num__2 a is the answer <eor> a <eos> |
a |
divide__6.0__3.0__ subtract__6.0__2.0__ subtract__4.0__2.0__ |
divide__6.0__3.0__ subtract__6.0__2.0__ subtract__4.0__2.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__15 sec . what is the length of the train ? <o> a ) num__120 m <o> b ) num__190 m <o> c ) num__115 m <o> d ) num__110 m <o> e ) num__250 m |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__15 = num__250 m answer : e <eor> e <eos> |
e |
round__250.0__ |
round__250.0__ |
| in num__2008 the profits of company m were num__5 percent of revenues . in num__2009 the revenues of company n fell by num__20 percent but profits were num__15 percent of revenues . the profits in num__2009 were what percent of the profits in num__2008 ? <o> a ) num__80.0 <o> b ) num__105.0 <o> c ) num__124.0 <o> d ) num__240.0 <o> e ) num__138 % |
the profit num__0 f num__2009 in terms of num__2008 = num__0.8 * num__3.0 * num__100 = num__240.0 d <eor> d <eos> |
d |
percent__20.0__15.0__ percent__100.0__240.0__ |
percent__20.0__15.0__ percent__100.0__240.0__ |
| the speed of a car is num__85 km in the first hour and num__45 km in the second hour . what is the average speed of the car ? <o> a ) num__72 kmph <o> b ) num__65 kmph <o> c ) num__30 kmph <o> d ) num__80 kmph <o> e ) num__82 kmph |
s = ( num__85 + num__45 ) / num__2 = num__65 kmph answer : b <eor> b <eos> |
b |
round__65.0__ |
round__65.0__ |
| if x is a prime number and x - num__1 is the median of the set { x - num__1 num__3 x + num__3 num__2 x - num__4 } then what is the average ( arithmetic mean ) of the set ? <o> a ) num__2 <o> b ) num__1.66666666667 <o> c ) num__3 <o> d ) num__3.33333333333 <o> e ) num__4.66666666667 |
if x is a prime number and x - num__1 is the median of the set { x - num__1 num__3 x + num__3 num__2 x - num__4 } then what is the average ( arithmetic mean ) of the set ? a . num__2 b . num__1.66666666667 c . num__3 d . num__3.33333333333 e . num__4.66666666667 solution : x - num__1 is the median of the set implies if we arrange the set in ascending order the set would be [ num__2 x - num__4 x - num__1 num__3 x + num__3 ] . this means that : num__2 x - num__4 < x - num__2 < num__3 x + num__3 solving inequality gives x > - num__2 and x < num__3 . since x is prime number and the only prime number < num__3 is num__2 so put x = num__2 . hence set becomes : [ num__0 num__1 num__9 ] and avg = num__3.33333333333 . d <eor> d <eos> |
d |
add__3.0__1.6667__ multiply__1.0__3.3333__ |
add__3.0__1.6667__ multiply__1.0__3.3333__ |
| the h . c . f . of two numbers is num__12 and their l . c . m . is num__600 . if one of the number is num__60 find the other ? <o> a ) num__100 <o> b ) num__240 <o> c ) num__120 <o> d ) num__200 <o> e ) num__150 |
other number = num__12 * num__10.0 = num__120 answer is c <eor> c <eos> |
c |
divide__600.0__60.0__ multiply__12.0__10.0__ lcm__12.0__120.0__ |
divide__600.0__60.0__ multiply__12.0__10.0__ multiply__12.0__10.0__ |
| the average weight of a group of boys is num__30 kg . after a boy of weight num__37 kg joins the group the average weight of the group goes up by num__1 kg . find the number of boys in the group originally ? <o> a ) num__4 <o> b ) num__8 <o> c ) num__5 <o> d ) num__2 <o> e ) num__6 |
let the number off boys in the group originally be x . total weight of the boys = num__30 x after the boy weighing num__37 kg joins the group total weight of boys = num__30 x + num__37 so num__30 x + num__37 + num__31 ( x + num__1 ) = > x = num__6 . answer : e <eor> e <eos> |
e |
add__30.0__1.0__ subtract__37.0__31.0__ subtract__37.0__31.0__ |
add__30.0__1.0__ subtract__37.0__31.0__ subtract__37.0__31.0__ |
| buses a and b start from a common bus stop x . bus a begins to travel in a straight line away from bus b at a constant rate of num__60 miles per hour . one hour later bus b begins to travel in a straight line in the exact opposite direction at a constant rate of num__80 miles per hour . if both buses travel indefinitely what is the positive difference in minutes between the amount of time it takes bus b to cover the exact distance that bus a has covered and the amount of time it takes bus b to cover twice the distance that bus a has covered ? <o> a ) num__36 <o> b ) num__72 <o> c ) num__132 <o> d ) num__144 <o> e ) num__180 |
num__1 st part : - in num__1 hr bus a covers num__30 miles . relative speed of bus abus b is ( num__80 - num__30 ) = num__50 mph . so time required for bus b to cover the exact distance as a is num__50 * t = num__30 t = num__0.6 = num__36 min num__2 nd part num__80 * t = num__2 d - b has to cover twice the distance num__30 * ( t + num__1 ) = d - a traveled num__1 hr more and has to travel only only d so d / num__30 - num__2 d / num__80 = num__1 d = num__120 t = num__3 hrs = num__180 min question asks for + ve difference between part num__1 and part num__2 in minutes = num__180 - num__36 = num__180 min e <eor> e <eos> |
e |
subtract__80.0__30.0__ km_to_mile_conversion__ multiply__60.0__0.6__ divide__60.0__30.0__ multiply__60.0__2.0__ add__1.0__2.0__ multiply__60.0__3.0__ round__180.0__ |
subtract__80.0__30.0__ divide__30.0__50.0__ multiply__60.0__0.6__ divide__60.0__30.0__ multiply__60.0__2.0__ add__1.0__2.0__ multiply__60.0__3.0__ multiply__60.0__3.0__ |
| the sum of two numbers is num__40 and their product is num__350 . what will be the sum of their reciprocals ? <o> a ) num__0.114285714286 <o> b ) num__0.106666666667 <o> c ) num__18.75 <o> d ) num__9.375 <o> e ) num__12.5 |
( num__1 / a ) + ( num__1 / b ) = ( a + b ) / ab = num__0.114285714286 = num__0.114285714286 answer : a <eor> a <eos> |
a |
divide__40.0__350.0__ divide__40.0__350.0__ |
divide__40.0__350.0__ divide__40.0__350.0__ |
| rs . num__800 amounts to rs . num__920 in num__3 years at simple interest . if the interest is increased by num__3.0 it would amount to how much ? <o> a ) num__299 <o> b ) num__277 <o> c ) num__266 <o> d ) num__992 <o> e ) num__188 |
( num__800 * num__3 * num__3 ) / num__100 = num__72 num__920 + num__72 = num__992 answer : d <eor> d <eos> |
d |
percent__100.0__992.0__ |
percent__100.0__992.0__ |
| x men can do a work in num__120 days . if there were num__20 men less the work would have taken num__60 days more . what is the value of x ? <o> a ) num__60 <o> b ) num__77 <o> c ) num__26 <o> d ) num__66 <o> e ) num__281 |
we have m num__1 d num__1 = m num__2 d num__2 num__120 x = ( x - num__20 ) num__180 = > num__2 x = ( x - num__20 ) num__3 = > num__2 x = num__3 x - num__60 = > x = num__60 answer : a <eor> a <eos> |
a |
divide__120.0__60.0__ add__120.0__60.0__ divide__60.0__20.0__ subtract__120.0__60.0__ |
divide__120.0__60.0__ add__120.0__60.0__ divide__60.0__20.0__ subtract__120.0__60.0__ |
| what number has a num__5 : num__1 ratio to the number num__10 ? <o> a ) num__36 <o> b ) num__50 <o> c ) num__76 <o> d ) num__56 <o> e ) num__62 |
num__5 : num__1 = x : num__10 x = num__50 answer : b <eor> b <eos> |
b |
multiply__5.0__10.0__ multiply__5.0__10.0__ |
multiply__5.0__10.0__ multiply__5.0__10.0__ |
| how many seconds will a num__500 m long train take to cross a man walking with a speed of num__3 km / hr in the direction of the moving train if the speed of the train is num__63 km / hr ? <o> a ) num__11 <o> b ) num__30 <o> c ) num__99 <o> d ) num__88 <o> e ) num__61 |
speed of train relative to man = num__63 - num__3 = num__60 km / hr . = num__60 * num__0.277777777778 = num__16.6666666667 m / sec . time taken to pass the man = num__500 * num__0.06 = num__30 sec . answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__500.0__0.06__ round__30.0__ |
subtract__63.0__3.0__ multiply__500.0__0.06__ multiply__500.0__0.06__ |
| if the wheel is num__14 cm then the number of revolutions to cover a distance of num__1056 cm is ? <o> a ) num__15 <o> b ) num__10 <o> c ) num__14 <o> d ) num__12 <o> e ) num__11 |
num__2 * num__3.14285714286 * num__14 * x = num__1056 = > x = num__12 answer : d <eor> d <eos> |
d |
subtract__14.0__2.0__ round__12.0__ |
subtract__14.0__2.0__ round__12.0__ |
| a train num__650 m long is running at a speed of num__81 km / hr . in what time will it pass a bridge num__340 m long ? <o> a ) num__42 <o> b ) num__43 <o> c ) num__44 <o> d ) num__45 <o> e ) num__46 |
speed = num__81 * num__0.277777777778 = num__22.5 m / sec total distance covered = num__650 + num__340 = num__990 m required time = num__990 * num__0.0444444444444 = num__44 sec answer : c <eor> c <eos> |
c |
add__650.0__340.0__ divide__990.0__22.5__ round__44.0__ |
add__650.0__340.0__ divide__990.0__22.5__ divide__990.0__22.5__ |
| on a map num__1 inch represents num__28 miles . how many x xxinches would be necessary to represent a distance of num__383.6 miles ? <o> a ) num__5.2 <o> b ) num__7.4 <o> c ) num__13.7 <o> d ) num__21.2 <o> e ) num__28.7 |
x inches necessary to represent a distance of num__383.6 miles = num__383.6 / num__28 = num__13.7 answer c <eor> c <eos> |
c |
divide__383.6__28.0__ round__13.7__ |
divide__383.6__28.0__ divide__383.6__28.0__ |
| a b and c start swimming in a pool simultaneously from the same end . to complete num__12 laps a takes num__10 minutes b takes num__8 minutes and c takes num__6 minutes . what is the ratio of speeds a : b : c ? <o> a ) num__3 : num__4 : num__5 <o> b ) num__10 : num__15 : num__20 <o> c ) num__5 : num__4 : num__3 <o> d ) num__4 : num__6 : num__5 <o> e ) num__12 : num__15 : num__25 |
the slowest one is a then comes b and the fastest one is c . so the the ratio of speeds a : b : c must be in ascending order . eliminate options c d and e . a ' s speed ( distance ) / ( time ) = num__1.2 = num__1.2 lap per minute = num__10 laps in num__10 minute ; b ' s speed ( distance ) / ( time ) = num__1.5 = num__1.5 lap per minute = num__15 laps in num__10 minutes ; c ' s speed ( distance ) / ( time ) = num__2.0 = num__2 lap per minute = num__20 laps in num__10 minutes / therefore the the ratio of speeds a : b : c is num__12 : num__15 : num__20 . answer : b . <eor> b <eos> |
b |
divide__12.0__10.0__ divide__12.0__8.0__ multiply__10.0__1.5__ subtract__12.0__10.0__ add__12.0__8.0__ round__10.0__ |
divide__12.0__10.0__ divide__12.0__8.0__ multiply__10.0__1.5__ divide__12.0__6.0__ add__12.0__8.0__ divide__12.0__1.2__ |
| a woman complete a journey in num__15 hours . she travels first half of the journey at the rate of num__21 km / hr and second half at the rate of num__24 km / hr . find the total journey in km . <o> a ) num__336 km . <o> b ) num__216 km . <o> c ) num__314 km . <o> d ) num__224 km . <o> e ) num__544 km . |
num__0.5 x / num__21 + num__0.5 x / num__24 = num__15 - - > x / num__21 + x / num__24 = num__30 - - > x = num__336 km . a <eor> a <eos> |
a |
divide__15.0__0.5__ round__336.0__ |
divide__15.0__0.5__ round__336.0__ |
| p can do a work in the same time in which q and r together can do it . if p and q work together the work can be completed in num__10 days . r alone needs num__50 days to complete the same work . then q alone can do it in <o> a ) num__30 days <o> b ) num__25 days <o> c ) num__20 days <o> d ) num__15 days <o> e ) num__10 days |
explanation : work done by p and q in num__1 day = num__0.1 work done by r in num__1 day = num__0.02 work done by p q and r in num__1 day = num__0.1 + num__0.02 = num__0.12 but work done by p in num__1 day = work done by q and r in num__1 day . hence the above equation can be written as work done by p in num__1 day × num__2 = num__0.12 = > work done by p in num__1 day = num__0.06 = > work done by q and r in num__1 day = num__0.06 hence work done by q in num__1 day = num__0.06 – num__0.02 = num__0.04 = num__0.04 so q alone can do the work in num__25 days answer : option b <eor> b <eos> |
b |
divide__1.0__10.0__ divide__1.0__50.0__ add__0.1__0.02__ divide__0.12__2.0__ divide__2.0__50.0__ divide__50.0__2.0__ round__25.0__ |
divide__1.0__10.0__ divide__1.0__50.0__ add__0.1__0.02__ divide__0.12__2.0__ divide__2.0__50.0__ divide__50.0__2.0__ round__25.0__ |
| raju age after num__55 years will be num__5 times his age num__5 years back what is the present age of raju <o> a ) num__28 <o> b ) num__17 <o> c ) num__10 <o> d ) num__77 <o> e ) num__20 |
clearly x + num__55 = num__5 ( x - num__5 ) < = > num__4 x = num__80 = > x = num__20 answer : e <eor> e <eos> |
e |
multiply__5.0__4.0__ multiply__5.0__4.0__ |
multiply__5.0__4.0__ multiply__5.0__4.0__ |
| two trains each num__100 m long moving in opposite directions cross other in num__20 sec . if one is moving twice as fast the other then the speed of the faster train is ? <o> a ) num__22 <o> b ) num__98 <o> c ) num__60 <o> d ) num__88 <o> e ) num__24 |
let the speed of the slower train be x m / sec . then speed of the train = num__2 x m / sec . relative speed = ( x + num__2 x ) = num__3 x m / sec . ( num__100 + num__100 ) / num__20 = num__3 x = > x = num__3.33333333333 . so speed of the faster train = num__6.66666666667 = num__6.66666666667 * num__3.6 = num__24 km / hr . answer : e <eor> e <eos> |
e |
divide__20.0__3.0__ round__24.0__ |
divide__20.0__3.0__ round__24.0__ |
| what is the average of num__120 num__130 num__140 num__510 num__520 num__530 num__1115 num__1120 and num__1125 ? <o> a ) num__419 <o> b ) num__551 <o> c ) num__601 <o> d ) num__590 <o> e ) num__721 |
add num__120 num__130 num__140 num__510 num__520 num__530 num__1115 num__1120 and num__1125 grouping numbers together may quicken the addition sum = num__5310 num__590.0 = num__590 . d <eor> d <eos> |
d |
subtract__1120.0__530.0__ subtract__1120.0__530.0__ |
subtract__1120.0__530.0__ subtract__1120.0__530.0__ |
| the sum of first n consecutive odd integers is n ^ num__2 . what is the sum of all odd integers between num__9 and num__39 inclusive . <o> a ) num__351 <o> b ) num__384 <o> c ) num__410 <o> d ) num__424 <o> e ) num__450 |
we ' re dealing with a sequence of consecutive odd integers : num__9 to num__39 inclusive . we ' re asked for the sum of this group . num__1 ) start with the sum of the smallest and the biggest : num__9 + num__39 = num__48 num__2 ) now look at the ' next smallest ' and the ' next biggest ' : num__11 + num__37 = num__48 now we have proof that there is no middle term . we have num__8 bunches of num__48 num__8 ( num__48 ) = num__384 b <eor> b <eos> |
b |
add__9.0__39.0__ add__2.0__9.0__ subtract__39.0__2.0__ subtract__9.0__1.0__ multiply__8.0__48.0__ multiply__384.0__1.0__ |
add__9.0__39.0__ add__2.0__9.0__ subtract__39.0__2.0__ subtract__9.0__1.0__ multiply__8.0__48.0__ multiply__384.0__1.0__ |
| sakshi can do a piece of work in num__20 days . tanya is num__25.0 more efficient than sakshi . the number of days taken by tanya to do the same piece of work is : <o> a ) num__15 <o> b ) num__16 <o> c ) num__18 <o> d ) num__25 <o> e ) none of these |
ratio of times taken by sakshi and tanya = num__125 : num__100 = num__5 : num__4 suppose tanya takes x days to do the work ; num__5 : num__4 : : num__20 : x x = ( num__4 x num__4.0 ) x = num__16 days . correct option : b <eor> b <eos> |
b |
subtract__125.0__25.0__ subtract__25.0__20.0__ divide__20.0__5.0__ subtract__20.0__4.0__ round__16.0__ |
subtract__125.0__25.0__ subtract__25.0__20.0__ divide__20.0__5.0__ subtract__20.0__4.0__ round__16.0__ |
| a man can row a boat at num__20 kmph in still water . if the speed of the stream is num__5 kmph what is the time taken to row a distance of num__60 km downstream ? <o> a ) num__1.66666666667 hours <o> b ) num__2.84615384615 hours <o> c ) num__6.15384615385 hours <o> d ) num__2.4 hours <o> e ) num__0.697674418605 hours |
speed downstream = num__20 + num__5 = num__25 kmph . time required to cover num__60 km downstream = d / s = num__2.4 = num__2.4 hours . answer : d <eor> d <eos> |
d |
add__20.0__5.0__ divide__60.0__25.0__ round__2.4__ |
add__20.0__5.0__ divide__60.0__25.0__ divide__60.0__25.0__ |
| a tin of oil was num__0.8 full . when num__6 bottles of oil were taken out and four bottles of oil were poured into it it was ¾ full . how many bottles of oil can the tin contain ? <o> a ) num__20 <o> b ) num__40 <o> c ) num__60 <o> d ) num__80 <o> e ) num__90 |
suppose x bottles can fill the tin completely then num__0.8 x - num__0.75 x = num__6 - num__4 x / num__20 = num__2 x = num__40 therefore required no of bottles = num__40 answer b <eor> b <eos> |
b |
subtract__6.0__4.0__ multiply__2.0__20.0__ multiply__2.0__20.0__ |
subtract__6.0__4.0__ multiply__2.0__20.0__ multiply__2.0__20.0__ |
| a can complete the job in num__2 hours and b can complete the same job in num__4 hours . a works for num__1 hour and then b joins and both complete the job . what fraction of the job did b complete <o> a ) num__0.166666666667 <o> b ) num__0.3 <o> c ) num__0.5 <o> d ) num__0.833333333333 <o> e ) num__0.888888888889 |
a = num__0.166666666667 <eor> a <eos> |
a |
multiply__1.0__0.1667__ |
multiply__1.0__0.1667__ |
| num__3 pumps working num__8 hours a day can empty a tank in num__2 days . how many hours a day must num__6 pumps work to empty the tank in num__1 day ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__13 |
num__3 pumps take num__16 hrs total ( num__8 hrs a day ) if num__1 pump will be working then it will need num__16 * num__3 = num__48 hrs num__1 pump need num__48 hrs if i contribute num__6 pumps then num__8.0 = num__8 hrs . answer : a <eor> a <eos> |
a |
multiply__8.0__2.0__ multiply__3.0__16.0__ round__8.0__ |
multiply__8.0__2.0__ multiply__3.0__16.0__ round__8.0__ |
| the h . c . f . of two numbers is num__23 and the other two factors of their l . c . m . are num__13 and num__15 . the larger of the two numbers is : <o> a ) num__276 <o> b ) num__299 <o> c ) num__322 <o> d ) num__345 <o> e ) num__395 |
clearly the numbers are ( num__23 x num__13 ) and ( num__23 x num__15 ) . larger number = ( num__23 x num__15 ) = num__345 . answer : option d <eor> d <eos> |
d |
multiply__23.0__15.0__ multiply__23.0__15.0__ |
multiply__23.0__15.0__ multiply__23.0__15.0__ |
| if x * y = xy – num__2 ( x + y ) for all integers x and y then num__4 * ( – num__3 ) = <o> a ) – num__14 <o> b ) – num__11 <o> c ) – num__4 <o> d ) num__4 <o> e ) num__16 |
num__4 * ( - num__3 ) = num__4 * ( - num__3 ) - num__2 ( num__4 + ( - num__3 ) ) = - num__12 - num__2 = - num__14 option ( a ) <eor> a <eos> |
a |
multiply__4.0__3.0__ add__2.0__12.0__ add__2.0__12.0__ |
multiply__4.0__3.0__ add__2.0__12.0__ add__2.0__12.0__ |
| one - fourth of the workers in palabras bookstore have read the latest book by j . saramago and num__0.625 of the workers have read the latest book by h . kureishi . the number of workers that have read neither book is one less than the number of the workers that have read the latest saramago book and have not read the latest kureishi book . if there are num__72 workers in the palabras bookstore how many of them have read both books ? <o> a ) num__13 <o> b ) num__12 <o> c ) num__9 <o> d ) num__8 <o> e ) num__4 |
there are total num__72 workers . one - fourth of the workers in palabras bookstore have read the latest book by j . saramago so num__18 have read saramago . num__0.625 of the workers have read the latest book by h . kureishi . so ( num__0.625 ) * num__72 = num__45 have read kureishi the number of workers that have read neither book is one less than the number of the workers that have read the latest saramago book and have not read the latest kureishi book if b workers have read both books num__18 - b have read saramago but not kureishi . so ( num__18 - b - num__1 ) have read neither . total = n ( a ) + n ( b ) - both + neither num__72 = num__18 + num__45 - b + ( num__18 - b - num__1 ) b = num__4 answer ( e ) <eor> e <eos> |
e |
multiply__0.625__72.0__ divide__72.0__18.0__ divide__72.0__18.0__ |
multiply__0.625__72.0__ divide__72.0__18.0__ multiply__1.0__4.0__ |
| the amount of an investment will double in approximately num__70 / p years where p is the percent interest compounded annually . if thelma invests $ num__40000 in a long - term cd that pays num__5 percent interest compounded annually what will be the approximate total value of the investment when thelma is ready to retire num__42 years later ? <o> a ) $ num__280000 <o> b ) $ num__320000 <o> c ) $ num__360000 <o> d ) $ num__450000 <o> e ) $ num__540 |
000 |
the amount of an investment will double in approximately num__70 / p years where p is the percent interest compounded annually . if thelma invests $ num__40000 in a long - term cd that pays num__5 percent interest compounded annually what will be the approximate total value of the investment when thelma is ready to retire num__42 years later ? the investment gets doubled in num__70 / p years . therefore the investment gets doubled in num__14.0 = every num__14 years . after num__42 years the investment will get doubled num__3.0 = num__3 times . so the amount invested will get doubled thrice . so num__40000 * num__2 = num__80000 num__80000 * num__2 = num__160000 num__160000 * num__2 = num__320000 hence the answer is b . <eor> b <eos> |
b |
b |
| rectangular floors x and y have equal area . if floor x is num__10 feet by num__18 feet and floor y is num__9 feet wide what is the length of floor y in feet ? <o> a ) num__13 num__0.5 <o> b ) num__20 <o> c ) num__18 num__0.75 <o> d ) num__21 <o> e ) num__24 |
the area of a rectangle is : area = length x width we are given that floor x is num__10 feet by num__18 feet and that floor y is num__9 feet wide . so we can say : length of x = num__10 width of x = num__18 width of y = num__9 length of y = n we also can say : area of floor x = area of floor y ( length of x ) ( width of x ) = ( length of y ) ( width of y ) ( num__10 ) ( num__18 ) = num__9 n ( num__10 ) ( num__2 ) = n num__20 = n answer b . <eor> b <eos> |
b |
multiply__10.0__2.0__ multiply__10.0__2.0__ |
multiply__10.0__2.0__ multiply__10.0__2.0__ |
| the figure above shows the dimensions of a semicircular cross section of a one - way tunnel . the single traffic lane is num__12 feet wide and is equidistant from the sides of the tunnel . if vehicles must clear the top of the tunnel by at least ½ foot when they are inside the traffic lane what should be the limit l on the height of vehicles that are allowed to use the tunnel ? <o> a ) num__5 ½ ft <o> b ) num__7 ½ ft <o> c ) num__8 ½ ft <o> d ) num__9 ½ ft <o> e ) num__10 ft |
let ' s label the midpoint of the circle o . since the base of the semi - circle is num__20 we know that the diameter is num__20 and accordingly the radius is num__10 . we also know that the traffic lane is num__12 feet long and there ' s an equal amount of space on either side so the traffic lane extends num__6 feet on either side of o . let ' s call the leftmost point on the base of the traffic lane a . so the distance oa is num__6 . now draw a line straight up from a to the top of the tunnel . let ' s label the point at which the line intersects the circle b . the answer to the question will therefore be the height ab - . num__5 feet ( we need to leave . num__5 feet of clearance ) . here ' s the key to solving the question : if we draw a line from o to b that line is a radius of the circle and therefore has length num__10 . we now have right triangle oab ( the right angle is at point a ) with leg oa = num__6 and hypotenuse ob = num__10 . we can now solve for leg ab = num__8 ( either by applying the pythagorean theorum or by applying the num__0.75 / num__5 special right triangle ratio ) . finally : ab = num__8 so the correct answer l is num__8 - . num__5 = num__7.5 . . . choose ( b ) ! <eor> b <eos> |
b |
subtract__20.0__12.0__ divide__6.0__8.0__ multiply__0.75__10.0__ subtract__12.0__5.0__ |
subtract__20.0__12.0__ divide__6.0__8.0__ multiply__0.75__10.0__ subtract__12.0__5.0__ |
| x and y are integers x is even and negative y is odd and positive . which of the following could be false ? num__1 . ( y + x ) is an odd number . num__2 . y ^ ( x + y ) is an integer . num__3 . x ^ y is a positive number . <o> a ) num__2 only . <o> b ) num__3 only . <o> c ) num__1 only . <o> d ) num__2 and num__3 only . <o> e ) num__1 num__2 and num__3 . |
statement i is true statement ii is not true statement iii is not true answer : c <eor> c <eos> |
c |
reverse__1.0__ |
reverse__1.0__ |
| a soccer store typically sells replica jerseys at a discount of num__30 percent to num__50 percent off list price . during the annual summer sale everything in the store is an additional num__40 percent off the original list price . if a replica jersey ' s list price is $ num__80 approximately what percent of the list price is the lowest possible sale price ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__30 <o> d ) num__40 <o> e ) num__50 |
let the list price be num__2 x for min sale price the first discount given should be num__50.0 num__2 x becomes x here now during summer sale additional num__20.0 off is given ie sale price becomes num__0.8 x it is given lise price is $ num__80 = > num__2 x = num__80 = > x = num__20 and num__0.8 x = num__32 so lowest sale price is num__32 which is num__40.0 of num__80 hence a is the answer <eor> a <eos> |
a |
divide__80.0__40.0__ subtract__50.0__30.0__ divide__40.0__50.0__ add__30.0__2.0__ subtract__50.0__30.0__ |
divide__80.0__40.0__ subtract__50.0__30.0__ divide__40.0__50.0__ add__30.0__2.0__ subtract__50.0__30.0__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__80 m and they cross each other in num__12 sec then the speed of each train is ? <o> a ) num__24 <o> b ) num__35 <o> c ) num__36 <o> d ) num__37 <o> e ) num__38 |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__80 + num__80 ) / num__12 = > x = num__6.66 speed of each train = num__10 m / sec . = num__6.66 * num__3.6 = num__24 km / hr . answer : option a <eor> a <eos> |
a |
subtract__12.0__2.0__ multiply__12.0__2.0__ round__24.0__ |
subtract__12.0__2.0__ multiply__12.0__2.0__ multiply__12.0__2.0__ |
| how many multiples of num__4 are there between num__4 and num__64 ? <o> a ) num__12 <o> b ) num__64 <o> c ) num__14 <o> d ) num__16 <o> e ) num__60 |
it should be mentioned whether num__4 and num__64 are inclusive . if num__4 and num__64 are inclusive then the answer is ( num__64 - num__4 ) / num__4 + num__1 = num__16 . if num__4 and num__64 are not inclusive then the answer is ( num__60 - num__8 ) / num__4 + num__1 = num__14 . since oa is c then we have not inclusive case . <eor> c <eos> |
c |
divide__64.0__4.0__ subtract__64.0__4.0__ multiply__1.0__14.0__ |
divide__64.0__4.0__ subtract__64.0__4.0__ divide__14.0__1.0__ |
| the simple form of the ratio num__0.666666666667 : num__0.4 is ? <o> a ) num__5 : num__8 <o> b ) num__5 : num__5 <o> c ) num__5 : num__9 <o> d ) num__5 : num__3 <o> e ) num__5 : num__1 |
num__0.666666666667 : num__0.4 = num__10 : num__6 = num__5 : num__3 answer : d <eor> d <eos> |
d |
subtract__10.0__5.0__ |
subtract__10.0__5.0__ |
| if a b and c together can finish a piece of work in num__4 days . a alone in num__12 days and b in num__9 days then c alone can do it in ? <o> a ) num__11 <o> b ) num__17 <o> c ) num__18 <o> d ) num__19 <o> e ) num__16 |
c = num__0.25 - num__0.0833333333333 – num__0.111111111111 = num__0.0555555555556 = > num__18 days ' answer : c <eor> c <eos> |
c |
round__18.0__ |
round__18.0__ |
| a father said to his son ` ` i was as old as you are at present at the time of your birth . ' ' if the father ' s age is num__38 years now the son ' s age six years back was ? <o> a ) num__13 yr <o> b ) num__15 yr <o> c ) num__20 yr <o> d ) num__25 yr <o> e ) num__30 yr |
let the son ' s present age be x years then num__38 - x = x x = num__19 son ' s age num__6 years back = num__19 - num__6 = num__13 years answer is a <eor> a <eos> |
a |
subtract__19.0__6.0__ subtract__19.0__6.0__ |
subtract__19.0__6.0__ subtract__19.0__6.0__ |
| a num__400 meter long train crosses a man standing on the platform in num__10 sec . what is the speed of the train ? <o> a ) num__144 <o> b ) num__108 <o> c ) num__278 <o> d ) num__126 <o> e ) num__112 |
s = num__40.0 * num__3.6 = num__144 kmph answer : a <eor> a <eos> |
a |
divide__400.0__10.0__ multiply__40.0__3.6__ round__144.0__ |
divide__400.0__10.0__ multiply__40.0__3.6__ multiply__40.0__3.6__ |
| a cheese factory sells its cheese in rectangular blocks . a normal block has a volume of num__5 cubic feet . if a large block has three times the width twice the depth and half the length of a normal block what is the volume of cheese in a large block in cubic feet ? <o> a ) num__10 <o> b ) num__30 <o> c ) num__15 <o> d ) num__20 <o> e ) num__25 |
volume of cube = lbh = num__5 new cube l b h are increases of num__3 b num__2 h and a decrease of . num__5 l new volume of cube = . num__5 l * num__3 b * num__2 h = num__3 * lbh = num__3 * num__5 = num__15 answer : c <eor> c <eos> |
c |
multiply__5.0__3.0__ multiply__5.0__3.0__ |
multiply__5.0__3.0__ multiply__5.0__3.0__ |
| carol spends num__0.25 of her savings on a stereo and num__0.5 less than she spent on the stereo for a television . what fraction of her savings did she spend on the stereo and television ? <o> a ) num__0.25 <o> b ) num__0.285714285714 <o> c ) num__0.416666666667 <o> d ) num__0.375 <o> e ) num__0.583333333333 |
total savings = s amount spent on stereo = ( num__0.25 ) s amount spent on television = ( num__1 - num__0.5 ) ( num__0.25 ) s = ( num__0.5 ) * ( num__0.25 ) * s = ( num__0.125 ) s ( stereo + tv ) / total savings = s ( num__0.25 + num__0.125 ) / s = num__0.375 answer : d <eor> d <eos> |
d |
multiply__0.25__0.5__ add__0.25__0.125__ add__0.25__0.125__ |
multiply__0.25__0.5__ add__0.25__0.125__ add__0.25__0.125__ |
| a computer is programmed to multiply consecutive even integers num__2 * num__4 * num__6 * num__8 * … * n until the product is divisible by num__1551 what is the value of n ? <o> a ) num__22 <o> b ) num__38 <o> c ) num__62 <o> d ) num__94 <o> e ) num__672 |
factorise num__1551 . . num__3 * num__11 * num__47 . . so n has to be a multiple of largest prime number num__61 . . so n = num__2 * num__47 = num__94 . . ans : d <eor> d <eos> |
d |
divide__6.0__2.0__ add__8.0__3.0__ multiply__2.0__47.0__ multiply__2.0__47.0__ |
divide__6.0__2.0__ add__8.0__3.0__ multiply__2.0__47.0__ multiply__2.0__47.0__ |
| if a person having rs . num__2000 and he want to distribute this to his five children in the manner that each son having rs . num__30 more than the younger one what will be the share of youngest child ? <o> a ) rs . num__175 <o> b ) rs . num__325 <o> c ) rs . num__340 <o> d ) rs . num__260 <o> e ) rs . num__230 |
explanation : assume first child ( the youngest ) get = rs . x according to the question ; each son having rs . num__30 more than the younger one second child will get = rs . x + num__30 third child will get = rs . x + num__30 + num__30 = x + num__60 forth child will get = rs . x + num__30 + num__30 + num__30 = x + num__90 fifth child will get = rs . x + num__30 + num__30 + num__30 + num__30 = x + num__120 total amount they got = rs . num__2000 x + ( x + num__30 ) + ( x + num__60 ) + ( x + num__90 ) + ( x + num__120 ) = num__2000 num__5 x + num__300 = num__2000 num__5 x = num__1700 x = rs . num__340 so the youngest child will get rs . num__340 . answer is c <eor> c <eos> |
c |
add__30.0__60.0__ add__30.0__90.0__ multiply__5.0__60.0__ subtract__2000.0__300.0__ divide__1700.0__5.0__ divide__1700.0__5.0__ |
add__30.0__60.0__ add__30.0__90.0__ multiply__5.0__60.0__ subtract__2000.0__300.0__ divide__1700.0__5.0__ divide__1700.0__5.0__ |
| it takes joey the postman num__1 hours to run a num__7 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is num__8 mile / hour what is the speed with which joey returns ? <o> a ) num__11 <o> b ) num__12 <o> c ) num__13 <o> d ) num__8 <o> e ) num__15 |
let his speed for one half of the journey be num__7 miles an hour let the other half be x miles an hour now avg speed = num__8 mile an hour num__2 * num__7 * x / num__7 + x = num__8 num__14 x = num__8 x + num__56 = > x = num__8 d <eor> d <eos> |
d |
multiply__7.0__2.0__ multiply__7.0__8.0__ round__8.0__ |
multiply__7.0__2.0__ multiply__7.0__8.0__ add__1.0__7.0__ |
| each machine of type a has num__2 steel parts and num__2 chrome parts . each machine of type b has num__6 steel parts and num__5 chrome parts . if a certain group of type a and type b machines has a total of num__20 steel parts and num__22 chrome parts how many machines are in the group <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__6 <o> e ) num__7 |
look at the below representation of the problem : steel chrome total a num__2 num__2 num__20 > > no . of type a machines = num__5.0 = num__5 b num__6 num__5 num__22 > > no . of type b machines = num__2.0 = num__2 so the answer is num__7 i . e e . hope its clear . <eor> e <eos> |
e |
add__2.0__5.0__ add__2.0__5.0__ |
add__2.0__5.0__ add__2.0__5.0__ |
| the length of a rectangle is halved while its breadth is tripled . watis the % change in area ? <o> a ) num__40.0 <o> b ) num__30.0 <o> c ) num__50.0 <o> d ) num__60.0 <o> e ) num__70 % |
let original length = x and original breadth = y . original area = xy . new length = x . num__2 new breadth = num__3 y . new area = x x num__3 y = num__3 xy . num__2 num__2 increase % = num__1 xy x num__1 x num__100.0 = num__50.0 . num__2 xy c <eor> c <eos> |
c |
triangle_area__1.0__100.0__ triangle_area__1.0__100.0__ |
triangle_area__1.0__100.0__ triangle_area__1.0__100.0__ |
| find the sum and number of divisors of num__544 excluding num__1 and num__544 . <o> a ) num__589 <o> b ) num__599 <o> c ) num__618 <o> d ) num__659 <o> e ) num__648 |
if n = p ^ a * q ^ b * r ^ c * . . . total no . of factors of n = ( a + num__1 ) * ( b + num__1 ) * ( c + num__1 ) * . . . & sum of all of the factors of n = ( num__1 + p + p ^ num__2 + . . . + p ^ a ) * ( num__1 + q + . . + q ^ b ) * ( num__1 + r + r ^ num__2 + . . . + r ^ c ) num__544 = num__2 ^ num__5 * num__17 number of divisors of num__544 excluding num__1 and num__544 = ( num__5 + num__1 ) * ( num__1 + num__1 ) - num__2 = num__10 sum of divisors of num__544 excluding num__1 and num__544 = ( num__1 + num__2 + num__2 ^ num__2 + num__2 ^ num__3 + num__2 ^ num__4 + num__2 ^ num__5 ) * ( num__1 + num__17 ) - num__1 - num__544 = num__589 answer : a <eor> a <eos> |
a |
multiply__2.0__5.0__ add__1.0__2.0__ add__1.0__3.0__ multiply__1.0__589.0__ |
multiply__2.0__5.0__ add__1.0__2.0__ add__1.0__3.0__ multiply__1.0__589.0__ |
| in what time will a railway train num__56 m long moving at the rate of num__39 kmph pass a telegraph post on its way ? <o> a ) num__3 sec <o> b ) num__4 sec <o> c ) num__5 sec <o> d ) num__6 sec <o> e ) num__7 sec |
t = num__1.4358974359 * num__3.6 = num__5 sec answer : c <eor> c <eos> |
c |
divide__56.0__39.0__ round__5.0__ |
divide__56.0__39.0__ round__5.0__ |
| in covering a distance of num__40 km a takes num__2 hours more than b . if a doubles his speed then he would take num__1 hour less than b . a ' s speed is : <o> a ) num__5 km / h <o> b ) num__6.7 km / h <o> c ) num__10 km / h <o> d ) num__15 km / h <o> e ) num__20 km / h |
let a ' s speed be x km / hr . then num__40 / x - num__20.0 x = num__3 num__20 / x = num__3 x = num__6.7 km / hr . answer : b <eor> b <eos> |
b |
divide__40.0__2.0__ add__2.0__1.0__ round__6.7__ |
divide__40.0__2.0__ add__2.0__1.0__ divide__6.7__1.0__ |
| a car travelling with num__0.714285714286 km of its actual speed covers num__42 km in num__1 hr num__40 min num__48 sec find the actual speed of the car ? <o> a ) num__78 kmph <o> b ) num__85 kmph <o> c ) num__35 kmph <o> d ) num__24 kmph <o> e ) num__74 kmph |
time taken = num__1 hr num__40 min num__48 sec = num__1.68 hrs let the actual speed be x kmph then num__0.714285714286 x * num__1.68 = num__42 x = ( num__42 * num__7 * num__75 ) / num__5 * num__126 = num__35 kmph answer ( c ) <eor> c <eos> |
c |
multiply__75.0__1.68__ subtract__42.0__7.0__ round__35.0__ |
multiply__75.0__1.68__ multiply__5.0__7.0__ divide__35.0__1.0__ |
| what was the principal if at num__12.0 per annum the interest after num__3 years amounted to rs . num__5888 less than the sum lent ? <o> a ) num__9000 <o> b ) num__3000 <o> c ) num__5000 <o> d ) num__1300 <o> e ) num__9200 |
p - num__2336 = ( p * num__12 * num__3 ) / num__100 p = num__9200 answer : e <eor> e <eos> |
e |
percent__100.0__9200.0__ |
percent__100.0__9200.0__ |
| for one toss of a certain coin the probability that the outcome is heads is num__0.4 . if this coin is tossed num__5 times which of the following is the probability that the outcome will be heads at least num__4 times ? <o> a ) ( num__0.4 ) ^ num__5 <o> b ) num__2 ( num__0.4 ) ^ num__4 <o> c ) num__3 ( num__0.4 ) ^ num__4 <o> d ) num__4 ( num__0.4 ) ^ num__4 ( num__0.6 ) + ( num__0.4 ) ^ num__5 <o> e ) num__5 ( num__0.4 ) ^ num__4 ( num__0.6 ) + ( num__0.4 ) ^ num__5 |
p ( num__5 heads ) = ( num__0.4 ) ^ num__5 there are num__5 ways to get num__4 heads : hhhht hhhth hhthh hthhh thhhh p ( num__4 heads ) = num__5 * ( num__0.4 ) ^ num__4 ( num__0.6 ) the answer is e . <eor> e <eos> |
e |
negate_prob__0.4__ vowel_space__ |
negate_prob__0.4__ vowel_space__ |
| for each color copy print shop x charges $ num__1.25 and print shop y charges $ num__2.75 . how much greater is the charge for num__60 color copies at print shop y than at print shop x ? <o> a ) $ num__80 <o> b ) $ num__90 <o> c ) $ num__100 <o> d ) $ num__110 <o> e ) $ num__120 |
the difference in the two prices is $ num__2.75 - $ num__1.25 = $ num__1.50 for each color copy . each color copy will cost an extra $ num__1.50 at print shop y . num__60 * $ num__1.50 = $ num__90 the answer is b . <eor> b <eos> |
b |
subtract__2.75__1.25__ multiply__60.0__1.5__ multiply__60.0__1.5__ |
subtract__2.75__1.25__ multiply__60.0__1.5__ multiply__60.0__1.5__ |
| calculate the circumference of a circular field whose radius is num__8 centimeters . <o> a ) num__18 π cm <o> b ) num__1 π cm <o> c ) num__8 π cm <o> d ) num__19 π cm <o> e ) num__28 π cm |
circumference c is given by c = num__2 π r = num__2 π * num__9 = num__18 π cm answer : a <eor> a <eos> |
a |
multiply__9.0__2.0__ multiply__9.0__2.0__ |
multiply__9.0__2.0__ multiply__9.0__2.0__ |
| the average monthly income of p and q is rs . num__5050 . the average monthly income of q and r is num__6250 and the average monthly income of p and r is rs . num__5200 . the monthly income of qis ? <o> a ) a ) rs . num__4078 <o> b ) b ) rs . num__6100 <o> c ) c ) rs . num__4029 <o> d ) d ) rs . num__4027 <o> e ) e ) rs . num__4020 |
let p q and r represent their respective monthly incomes . then we have : p + q = ( num__5050 * num__2 ) = num__10100 - - - ( i ) q + r = ( num__6250 * num__2 ) = num__12500 - - - ( ii ) p + r = ( num__5200 * num__2 ) = num__10400 - - - ( iii ) adding ( i ) ( ii ) and ( iii ) we get : num__2 ( p + q + r ) = num__33000 = p + q + r = num__16500 - - - ( iv ) subtracting ( iii ) from ( iv ) we get p = num__6100 . p ' s monthly income = rs . num__6100 . answer : b <eor> b <eos> |
b |
multiply__5050.0__2.0__ multiply__6250.0__2.0__ multiply__5200.0__2.0__ divide__33000.0__2.0__ subtract__16500.0__10400.0__ subtract__16500.0__10400.0__ |
multiply__5050.0__2.0__ multiply__6250.0__2.0__ multiply__5200.0__2.0__ divide__33000.0__2.0__ subtract__16500.0__10400.0__ subtract__16500.0__10400.0__ |
| excluding stoppages the speed of a train is num__45 kmph and including stoppages it is num__32 kmph . of how many minutes does the train stop per hour ? <o> a ) e num__982 <o> b ) num__27 <o> c ) num__17.3 <o> d ) num__121 <o> e ) num__28 |
explanation : t = num__0.288888888889 * num__60 = num__17.3 answer : option c <eor> c <eos> |
c |
hour_to_min_conversion__ round__17.3__ |
hour_to_min_conversion__ round__17.3__ |
| the sum of five numbers is num__400 . the average of the first two numbers is num__50 and the third number is num__60 . find the average of the two numbers ? <o> a ) num__70 <o> b ) num__65 <o> c ) num__85 <o> d ) num__80 <o> e ) num__90 |
let the five numbers be p q r s and t . = > p + q + r + s + t = num__400 ( p + q ) / num__2 = num__50 and r = num__60 p + q = num__100 and r = num__60 p + q + r = num__160 s + t = num__400 - ( p + q + r ) = num__240 average of the last two numbers = ( s + t ) / num__2 = num__80 answer : d <eor> d <eos> |
d |
multiply__50.0__2.0__ add__60.0__100.0__ subtract__400.0__160.0__ divide__160.0__2.0__ divide__160.0__2.0__ |
multiply__50.0__2.0__ add__60.0__100.0__ subtract__400.0__160.0__ divide__160.0__2.0__ divide__160.0__2.0__ |
| a man buys a cycle for rs . num__2300 and sells it at a loss of num__30.0 . what is the selling price of the cycle ? <o> a ) s . num__1690 <o> b ) s . num__1610 <o> c ) s . num__1890 <o> d ) s . num__1602 <o> e ) s . num__1092 |
s . p . = num__70.0 of rs . num__2300 = rs . num__70 x num__23.0 = rs . num__1610 answer : option b <eor> b <eos> |
b |
percent__70.0__2300.0__ percent__70.0__2300.0__ |
percent__70.0__2300.0__ percent__70.0__2300.0__ |
| if log x / num__2 = log y / num__3 = log z / num__5 then yz in terms of x is <o> a ) x <o> b ) x num__2 <o> c ) x num__3 <o> d ) x num__4 <o> e ) none of these |
logx / num__2 = logy / num__3 = logz / num__5 = k ( say ) . = > logx = num__2 k logy = num__3 k log z = num__5 k = > logyz = num__3 k + num__5 k = num__8 k ; logx num__4 = num__8 k thus logyz = logx num__4 = > yz = x num__4 answer : d <eor> d <eos> |
d |
add__3.0__5.0__ divide__8.0__2.0__ divide__8.0__2.0__ |
add__3.0__5.0__ divide__8.0__2.0__ divide__8.0__2.0__ |
| a cask full of milk ' e ' liters is removed and water is mixed . this process is repeated and now the ratio in milk to water is num__16 : num__9 . then what is the amount of milk in liter . <o> a ) e + num__9 <o> b ) num__9 e <o> c ) num__5 e <o> d ) num__8 e <o> e ) cant be determined |
after ' n ' operations the amount of pure liquid remaining is = [ x { ( num__1 - e / x ) ^ n } ] . . . where x is original amount of liquid and e is amount of liquid removed . . this is standard formula in rs aggarwal . . . now in our case the operations is performed twice . . . let the original milk be x liters . . e liters removed twice n = num__2 milk remaining ( m ) = [ x { ( num__1 - e / x ) ^ num__2 } ] . . . . water is ( w ) = x - [ x { ( num__1 - e / x ) ^ num__2 } ] m / w = num__1.77777777778 take ( num__1 - e / x ) ^ num__2 = z now solving we get z = num__0.64 that is ( num__1 - e / x ) ^ num__2 = ( num__0.8 ) ^ num__2 num__1 - e / x = num__0.8 x = num__5 e answer : c <eor> c <eos> |
c |
divide__16.0__9.0__ multiply__1.0__5.0__ |
divide__16.0__9.0__ divide__5.0__1.0__ |
| the value of log num__5 ( num__0.008 ) is : <o> a ) num__3 <o> b ) - num__3 <o> c ) num__0.333333333333 <o> d ) - num__0.333333333333 <o> e ) none of these |
log num__5 ( num__0.008 ) = log num__5 ( num__0.2 ) ^ num__3 = num__3 log num__5 ( num__0.2 ) = num__3 ( log num__5 num__1 - log num__5 num__5 ) = num__3 ( num__0 - num__1 ) = - num__3 answer : b <eor> b <eos> |
b |
reverse__5.0__ multiply__5.0__0.2__ round_down__0.008__ multiply__1.0__3.0__ |
reverse__5.0__ multiply__5.0__0.2__ round_down__0.008__ multiply__1.0__3.0__ |
| num__4 |
25 num__49121 <o> a ) num__149 <o> b ) num__169 <o> c ) num__189 <o> d ) num__209 <o> e ) num__219 |
num__13 ^ num__2 = num__169 because follow sequence of square of the prime numbers answer : b <eor> b <eos> |
b |
b |
| a certain barrel which is a right circular cylinder is filled to capacity with num__60 gallons of oil . the first barrel is poured into a second barrel also a right circular cylinder which is empty . the second barrel is twice as tall as the first barrel and has twice the diameter of the first barrel . if all of the oil in the first barrel is poured into the second barrel how much empty capacity in gallons is left in the second barrel ? <o> a ) there is no empty capacity <o> b ) num__100 gallons <o> c ) num__300 gallons <o> d ) num__420 gallons <o> e ) num__840 gallons |
radius of first cylinder = r diameter = num__2 r height = h radius of second cylinder = num__2 r diamter = num__2 d and height = num__2 h volume of first cylinder = pie ( r ^ num__2 ) * h = num__60 volume of second cylinder = pie ( num__2 r ^ num__2 ) num__2 h put the value of pie ( r ^ num__2 ) * h = num__60 in the second cylinder volume = pie ( r ^ num__2 ) * num__4 * num__2 = num__60 * num__8 = num__480 gallons empty capacity = num__420 gallons answer d <eor> d <eos> |
d |
square_perimeter__2.0__ multiply__60.0__8.0__ triangle_area__2.0__420.0__ |
multiply__2.0__4.0__ multiply__60.0__8.0__ triangle_area__2.0__420.0__ |
| the length of a bridge in meters which a train num__80 - meters long and traveling at num__45 km / hr can cross in num__30 seconds is ? <o> a ) num__215 <o> b ) num__235 <o> c ) num__255 <o> d ) num__275 <o> e ) num__295 |
num__45 km / h = num__45000 m / num__3600 s = num__12.5 m / s in num__30 seconds the train can go num__30 ( num__12.5 ) = num__375 meters let x be the length of the bridge . x + num__80 = num__375 meters x = num__295 meters the answer is e . <eor> e <eos> |
e |
multiply__80.0__45.0__ divide__45000.0__3600.0__ multiply__30.0__12.5__ subtract__375.0__80.0__ round__295.0__ |
multiply__80.0__45.0__ divide__45000.0__3600.0__ multiply__30.0__12.5__ subtract__375.0__80.0__ round__295.0__ |
| there are num__6 people in the elevator . their average weight is num__156 lbs . another person enters the elevator and increases the average weight to num__151 lbs . what is the weight of the num__7 th person . <o> a ) num__121 <o> b ) num__168 <o> c ) num__189 <o> d ) num__190 <o> e ) num__200 |
solution average of num__7 people after the last one enters = num__151 . â ˆ ´ required weight = ( num__7 x num__151 ) - ( num__6 x num__156 ) = num__1057 - num__936 = num__121 . answer a <eor> a <eos> |
a |
multiply__151.0__7.0__ multiply__6.0__156.0__ subtract__1057.0__936.0__ subtract__1057.0__936.0__ |
multiply__151.0__7.0__ multiply__6.0__156.0__ subtract__1057.0__936.0__ subtract__1057.0__936.0__ |
| water boils at num__212 ° f or num__100 ° c and ice melts at num__32 ° f or num__0 ° c . if the temperature of a pot of water is num__40 ° c what is the temperature of the pot of water in ° f ? <o> a ) num__85 ° f <o> b ) num__92 ° f <o> c ) num__96 ° f <o> d ) num__99 ° f <o> e ) num__104 ° f |
let f and c denote the temperature in fahrenheit and celsius respectively . ( f - num__32 ) / ( num__212 - num__32 ) = ( c - num__0 ) / ( num__100 - num__0 ) f = num__9 c / num__5 + num__32 f = num__9 ( num__40 ) / num__5 + num__32 = num__104 ° f the answer is e . <eor> e <eos> |
e |
round__104.0__ |
round__104.0__ |
| marla starts running around a circular track at the same time nick starts walking around the same circular track . marla completes num__10 laps around the track per hour and nick completes num__5 laps around the track per hour . how many minutes after marla and nick begin moving will marla have completed num__4 more laps around the track than nick ? <o> a ) num__5 <o> b ) num__48 <o> c ) num__12 <o> d ) num__15 <o> e ) num__20 |
maria ' s rate - num__10 laps per hour - - > num__0.166666666667 laps / min nick ' s rate - num__5 laps per hour - - > num__0.0833333333333 laps / min lets set equations : num__0.166666666667 * t = num__4 ( since maria had to run num__4 laps before nick would start ) num__0.0833333333333 * t = num__0 ( hick has just started and has n ' t run any lap yet ) ( num__0.166666666667 - num__0.0833333333333 ) * t = num__4 - num__0 ( since nick was chasing maria ) t = num__48 min needed maria to run num__4 laps answer : b <eor> b <eos> |
b |
round__48.0__ |
round__48.0__ |
| when x dividedby num__288 the remainder is num__47 . find the remainder when the same x is divided by num__24 ? <o> a ) num__20 <o> b ) num__23 <o> c ) num__25 <o> d ) num__27 <o> e ) num__29 |
num__23 option b <eor> b <eos> |
b |
subtract__47.0__24.0__ subtract__47.0__24.0__ |
subtract__47.0__24.0__ subtract__47.0__24.0__ |
| a car covers a distance of num__636 km in num__6 ½ hours . find its speed ? <o> a ) num__104 kmph <o> b ) num__176 kmph <o> c ) num__298 kmph <o> d ) num__106 kmph <o> e ) num__268 kmph |
num__106.0 = num__106 kmph answer : d <eor> d <eos> |
d |
divide__636.0__6.0__ round__106.0__ |
divide__636.0__6.0__ round__106.0__ |
| a cistern can be filled by num__3 pipes in num__30 num__40 and num__60 min and emptied by an escape pipe in half an hour the num__3 taps are turned on at noon but the escape pipe is at the same time accidentally left open and not closed for a quarter of an hour at what time will the cistern be full ? <o> a ) num__60 min <o> b ) num__90 min <o> c ) num__120 min <o> d ) num__150 min <o> e ) num__170 min |
explanation : let the total time be x min x / num__30 + x / num__40 + x / num__60 – x / num__15 = num__1 = > x = num__120 min answer : option c <eor> c <eos> |
c |
multiply__3.0__40.0__ round__120.0__ |
multiply__3.0__40.0__ divide__120.0__1.0__ |
| rahul travels from a to b a distance of num__200 miles in num__5 hours . he returns to a in num__4 hours . find his average speed ? <o> a ) num__52.2 mph <o> b ) num__61.9 mph <o> c ) num__44.4 mph <o> d ) num__35.7 mph <o> e ) num__65.6 mph |
speed from a to b = num__40.0 = num__40 mph speed from b to a = num__50.0 = num__50 mph average speed = num__2 * num__40 * num__0.555555555556 = num__44.4 mph answer is c <eor> c <eos> |
c |
divide__200.0__5.0__ divide__200.0__4.0__ round__44.4__ |
divide__200.0__5.0__ divide__200.0__4.0__ round__44.4__ |
| two people measure each other ' s height the height of the taller person is h and the height of the other person is l . if the difference in their heights is equal to the average height what is the value of h / l <o> a ) num__0.333333333333 . <o> b ) num__0.5 . <o> c ) num__3 . <o> d ) num__2 . <o> e ) num__6 . |
difference = average h - l = ( h + l ) / num__2 solving for h / l gives num__3 . a quick check h be num__9 and l be num__3 num__9 - num__3 = ( num__9 + num__3 ) / num__2 c <eor> c <eos> |
c |
divide__9.0__3.0__ |
divide__9.0__3.0__ |
| the function f is defined for all the positive integers h by the following rule : f ( h ) is the number of positive integers each of which is less than h and has no positive factor in common with h other than num__1 . if p is a prime number then f ( p ) ? <o> a ) p - num__1 <o> b ) p - num__2 <o> c ) ( p + num__1 ) / num__2 <o> d ) ( p - num__1 ) / num__2 <o> e ) num__2 |
the moment you put a prime number in the function f ( h ) notice that all the numbers lesser than h have no divisor clashing with divisor of h since h is prime ! ! . for instance f ( num__7 ) = { num__6 num__5 num__4 num__3 num__2 num__1 } thus for f ( p ) number of integers falling under this set will be p - num__1 answer : - a <eor> a <eos> |
a |
subtract__7.0__1.0__ subtract__6.0__1.0__ subtract__5.0__1.0__ subtract__4.0__1.0__ subtract__3.0__1.0__ reverse__1.0__ |
subtract__7.0__1.0__ subtract__6.0__1.0__ subtract__5.0__1.0__ subtract__4.0__1.0__ subtract__3.0__1.0__ subtract__2.0__1.0__ |
| in num__24 minutes the hour hand of a clock moves through an angle of : <o> a ) num__60 ° <o> b ) num__24 ° <o> c ) num__12 ° <o> d ) num__5 ° <o> e ) none |
num__12 hour = num__360 ˚ num__1 hr . = num__30.0 = num__30 ˚ num__60 min = num__30 ˚ num__1 min num__0.5 = . num__5 ˚ num__24 min . = num__0.5 × num__24 = num__12 ˚ answer : c <eor> c <eos> |
c |
divide__360.0__12.0__ hour_to_min_conversion__ divide__12.0__24.0__ divide__60.0__12.0__ round__12.0__ |
divide__360.0__12.0__ hour_to_min_conversion__ divide__12.0__24.0__ divide__60.0__12.0__ round__12.0__ |
| a copying machine ` ` p ' ' can copy num__135 pages in num__9 minutes p and q together can copy num__300 pages in num__12 minutes . in what time q can copy num__80 pages ? <o> a ) num__6 minute <o> b ) num__7 minute <o> c ) num__9 minute <o> d ) num__8 minute <o> e ) num__10 minute |
p copying num__135 pages in num__9 minutes . so in num__1 minute p copying = num__15.0 = num__15 pages p and q both copying num__300 pages in num__12 minute . so in num__12 minute p coying the num__12 * num__15 = num__180 pages . so q copying alone = num__300 - num__180 = num__120 pages . according to question q takes num__12 minute for copying the pages . so for num__1 minute num__10.0 = num__10 pages . because for num__10 pages takes num__1 minute so num__80 pages it ' s takes num__8 minute so ans is num__8 minute . answer : d <eor> d <eos> |
d |
divide__135.0__9.0__ multiply__12.0__15.0__ subtract__135.0__15.0__ add__9.0__1.0__ subtract__9.0__1.0__ round__8.0__ |
divide__135.0__9.0__ multiply__12.0__15.0__ subtract__135.0__15.0__ add__9.0__1.0__ subtract__9.0__1.0__ subtract__9.0__1.0__ |
| a man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream . the ratio of the speed of the boat ( in still water ) and the stream is : <o> a ) num__1 : num__2 <o> b ) num__2 : num__5 <o> c ) num__6 ' ' num__2 <o> d ) num__3 : num__1 <o> e ) num__4 : num__7 |
explanation : let speed upstream = x hen speed downstream = num__2 x speed in still water = num__2 x + x / num__2 speed of the stream = num__2 x - x / num__2 speed in still water : speed of the stream = num__3 x / num__2 : x / num__2 = num__3 : num__1 answer : d <eor> d <eos> |
d |
subtract__3.0__2.0__ round__3.0__ |
subtract__3.0__2.0__ add__1.0__2.0__ |
| a man can row upstream at num__25 kmph and downstream at num__35 kmph and then find the speed of the man in still water ? <o> a ) num__39 <o> b ) num__27 <o> c ) num__30 <o> d ) num__98 <o> e ) num__287 |
us = num__25 ds = num__35 m = ( num__35 + num__25 ) / num__2 = num__30 answer : c <eor> c <eos> |
c |
round__30.0__ |
round__30.0__ |
| if x ^ num__2 > x ^ num__7 > x ^ num__5 which of the following could be the value of x ? <o> a ) - num__9 <o> b ) - num__2 <o> c ) - num__0.625 <o> d ) num__5 <o> e ) num__9 |
x ^ num__2 > x ^ num__7 > x ^ num__5 num__1 ) here we have an even power of a variable ( x ) that is greater than both a larger odd power of the variable x and a smaller odd power of the variable x . this can be possible only if the base is negative ; therefore the variable x must be a negative number . num__2 ) now if the variable x is negative the higher power results in a smaller number if x < - num__1 and a higher power results in a larger number if num__0 > x > - num__1 . with this information we know that x is a negative number between num__0 and - num__1 ; therefore c ( - num__0.625 ) is the only option . <eor> c <eos> |
c |
multiply__1.0__0.625__ |
multiply__1.0__0.625__ |
| num__5 num__916 num__2848 __ <o> a ) num__79 <o> b ) num__80 <o> c ) num__81 <o> d ) num__82 <o> e ) num__83 |
( num__9 + num__5 ) + num__2 = num__16 ( num__16 + num__9 ) + num__3 = num__28 ( num__28 + num__16 ) + num__4 = num__48 ( num__48 + num__28 ) + num__5 = num__81 answer : c <eor> c <eos> |
c |
subtract__5.0__2.0__ subtract__9.0__5.0__ multiply__3.0__16.0__ power__3.0__4.0__ power__3.0__4.0__ |
subtract__5.0__2.0__ subtract__9.0__5.0__ multiply__3.0__16.0__ power__3.0__4.0__ power__3.0__4.0__ |
| vikas bought a grinder with num__15.0 discount on the labelled price . he sold the grinder for rs . num__2880 with num__20.0 profit on the labelled price . at what price did he buy the grinder ? <o> a ) rs num__2000 <o> b ) rs num__2040 <o> c ) rs num__2240 <o> d ) rs num__2400 <o> e ) none of these |
explanation : let the labelled price be rs . x . then num__120.0 of x = num__2880 therefore x = ( num__24000.8333333 ) = num__2400 . c . p = num__85.0 of rs . num__2400 = rs ( num__0.85 x num__2400 ) = rs . num__2040 . answer : b <eor> b <eos> |
b |
percent__85.0__2400.0__ percent__85.0__2400.0__ |
percent__85.0__2400.0__ percent__85.0__2400.0__ |
| ram professes to sell his goods at the cost price but he made use of num__900 grms instead of a kg what is the gain percent ? <o> a ) num__11 num__0.125 % <o> b ) num__11 num__0.888888888889 % <o> c ) num__11 num__0.111111111111 % <o> d ) num__19 num__0.111111111111 % <o> e ) num__11 num__1.0 % |
num__900 - - - num__100 num__100 - - - ? = > num__11 num__0.111111111111 % answer : c <eor> c <eos> |
c |
percent__100.0__11.0__ |
percent__100.0__11.0__ |
| if ( num__64 ) num__2 - ( num__36 ) num__2 = num__10 x then x = ? <o> a ) num__200 <o> b ) num__220 <o> c ) num__210 <o> d ) num__280 <o> e ) none of these |
explanation : a num__2 − b num__2 = ( a − b ) ( a + b ) ( num__64 ) num__2 - ( num__36 ) num__2 = ( num__64 - num__36 ) ( num__64 + num__36 ) = num__28 × num__100 given that ( num__64 ) num__2 - ( num__36 ) num__2 = num__10 x = > num__28 × num__100 = num__10 x = > x = num__280 . answer : option d <eor> d <eos> |
d |
subtract__64.0__36.0__ add__64.0__36.0__ multiply__10.0__28.0__ multiply__10.0__28.0__ |
subtract__64.0__36.0__ add__64.0__36.0__ multiply__10.0__28.0__ multiply__10.0__28.0__ |
| there is rain at a rate of num__12 centimeters per hour all over new jersey . somewhere downtown in new jersey a group of students are waiting for the rain to stop . if the rain filled a tank the with a base area of num__250 square centimeters and a depth of num__6 centimeters how long did the students wait for the rain to stop ? <o> a ) num__8 hours <o> b ) num__4 hours num__25 minutes <o> c ) num__3 hours num__15 minutes <o> d ) num__6 hours <o> e ) num__2 hours |
answer is : e num__4 hours and num__30 mins the volume of the tank is irrelevant and only height matters since rain fell all over the city . thus it takes only . num__2.0 = num__2 hours of rain to fill the tank . <eor> e <eos> |
e |
divide__12.0__6.0__ round__2.0__ |
divide__12.0__6.0__ round__2.0__ |
| two vessels p and q contain num__62.5 and num__87.5 of alcohol respectively . if num__8 litres from vessel p is mixed with num__4 litres from vessel q the ratio of alcohol and water in the resulting mixture is ? <o> a ) num__17 : num__7 <o> b ) num__18 : num__8 <o> c ) num__19 : num__9 <o> d ) num__19 : num__5 <o> e ) num__19 : num__4 |
quantity of alcohol in vessel p = num__62.5 / num__100 * num__8 = num__5 litres quantity of alcohol in vessel q = num__87.5 / num__100 * num__4 = num__3.5 litres quantity of alcohol in the mixture formed = num__5 + num__3.5 = num__8.5 = num__8.5 litres as num__12 litres of mixture is formed ratio of alcohol and water in the mixture formed = num__8.5 : num__3.5 = num__17 : num__7 . answer : a <eor> a <eos> |
a |
add__5.0__3.5__ add__8.0__4.0__ add__5.0__12.0__ subtract__12.0__5.0__ add__5.0__12.0__ |
add__5.0__3.5__ add__8.0__4.0__ add__5.0__12.0__ subtract__12.0__5.0__ add__5.0__12.0__ |
| the difference between simple and compound interest on rs . num__1300 for one year at num__10.0 per annum reckoned half - yearly is ? <o> a ) num__8.0 <o> b ) num__3.25 <o> c ) num__9.15 <o> d ) num__3.13 <o> e ) num__2.0 |
s . i . = ( num__1300 * num__10 * num__1 ) / num__100 = rs . num__130 c . i . = [ num__1300 * ( num__1 + num__0.05 ) num__2 - num__1300 ] = rs . num__133.25 difference = ( num__133.25 - num__130 ) = rs . num__3.25 answer : b <eor> b <eos> |
b |
percent__10.0__1300.0__ percent__100.0__3.25__ |
percent__10.0__1300.0__ percent__100.0__3.25__ |
| a car averages num__45 mph for the first num__4 hours of a trip and averages num__75 mph for each additional hour . the average speed for the entire trip was num__65 mph . how many hours long is the trip ? <o> a ) num__12 <o> b ) num__11 <o> c ) num__10 <o> d ) num__9 <o> e ) num__8 |
let the time for which car averages num__75 mph = t num__65 * ( t + num__4 ) = num__45 * num__4 + num__75 t = > num__10 t = num__80 = > t = num__8 total duration of the trip = num__8 + num__4 = num__12 answer a <eor> a <eos> |
a |
subtract__75.0__65.0__ divide__80.0__10.0__ add__4.0__8.0__ add__4.0__8.0__ |
subtract__75.0__65.0__ divide__80.0__10.0__ add__4.0__8.0__ add__4.0__8.0__ |
| num__39 persons can repair a road in num__12 days working num__5 hours a day . in how many days will num__30 persons working num__6 hours a day complete the work ? <o> a ) num__10 <o> b ) num__16 <o> c ) num__13 <o> d ) num__18 <o> e ) num__19 |
c num__13 let the required number of days be x . less persons more days ( indirect proportion ) more working hours per day less days ( indirect proportion ) persons num__30 : num__39 : : num__12 : x working hours / day num__6 : num__5 num__30 x num__6 x x = num__39 x num__5 x num__12 x = ( num__39 x num__5 x num__12 ) / ( num__30 x num__6 ) x = num__13 <eor> c <eos> |
c |
round__13.0__ |
round__13.0__ |
| pam and stanley packed several boxes with reams of paper . while both packed pam packed num__20.0 of the boxes . after pam stopped stanley packed the same number of boxes that he had packed while working with pam . what is the ratio of the number of boxes pam packed to the number of boxes stanley packed ? <o> a ) num__1 to num__4 <o> b ) num__1 to num__8 <o> c ) num__3 to num__5 <o> d ) num__3 to num__4 <o> e ) num__3 to num__2 |
correct answer : b solution : b . we know that when pam and stanley were both working the ratio was num__1 boxes by pam to num__4 boxes by stanley . we also know that stanley continued working after pam stopped . he packed as many boxes alone as he had packed whlie working with pam effectively doubling his number of boxes . thus the ratio of pam ' s boxes to stanley ' s boxes is num__1 to num__8 . answer b is correct . <eor> b <eos> |
b |
reverse__1.0__ |
reverse__1.0__ |
| a clothing manufacturer has determined that she can sell num__100 suits a week at a selling price of rs . num__200 each . for each rise of rs . num__4 in the selling price she will sell num__2 less suits a week . if she sells the suits for rs . x each how many rupees a week will she receive from the sales of the suits ? <o> a ) x num__1.0 <o> b ) num__200 − x / num__2 <o> c ) num__50 x - ( x ^ num__2 ) / num__4 ) <o> d ) num__150 x - ( x ^ num__2 ) / num__4 ) <o> e ) num__200 x - ( x ^ num__2 ) / num__2 ) |
explanatory answer assume that she sells the suits at rs . num__204 ; then she will sell two suits lesser or num__98 suits . substitute x = num__204 in each of the answer choices and check where the answer is num__98 . for choice ( num__2 ) the answer is num__98 and this is the quickest way to find the answer . answer b <eor> b <eos> |
b |
add__200.0__4.0__ subtract__100.0__2.0__ multiply__100.0__2.0__ |
add__200.0__4.0__ subtract__100.0__2.0__ multiply__100.0__2.0__ |
| which of the following is equal to num__1 ( num__0.166666666667 ) % ? <o> a ) num__0.012 / num__100 <o> b ) num__0.12 / num__100 <o> c ) num__1.1 / num__100 <o> d ) num__0.12 <o> e ) num__1.2 |
this notation may be confusing for some since it looks like we ' re multiplying num__1 and num__0.166666666667 how about adding a space : which of the following is equal to ( num__1 num__0.166666666667 ) % ( num__1 num__0.166666666667 ) % = num__1.1 = num__1.1 / num__100 answer : c <eor> c <eos> |
c |
multiply__1.0__1.1__ |
divide__1.1__1.0__ |
| a rectangular window is twice as long as it is wide . if its perimeter is num__18 feet then what are its dimensions in feet ? <o> a ) num__1.5 by num__3.5 <o> b ) num__1.66666666667 by num__3.33333333333 <o> c ) num__2 by num__4 <o> d ) num__3 by num__6 <o> e ) num__3.33333333333 by num__6.66666666667 |
let x be the width of the window . then the length is num__2 x . x + num__2 x + x + num__2 x = num__18 num__6 x = num__18 x = num__3.0 = num__3 the width is num__3 and the length is num__6 . the answer is d . <eor> d <eos> |
d |
triangle_area__2.0__3.0__ |
triangle_area__2.0__3.0__ |
| if difference between compound interest and simple interest on a sum at num__10.0 p . a . for num__2 years is rs . num__36 then sum is <o> a ) s . num__5000 <o> b ) s . num__5100 <o> c ) s . num__5800 <o> d ) s . num__6000 <o> e ) s . num__3600 |
p ( r / num__100 ) ^ num__2 = c . i - s . i p ( num__0.1 ) ^ num__2 = num__36 num__3600 answer : e <eor> e <eos> |
e |
percent__100.0__3600.0__ |
percent__100.0__3600.0__ |
| if a train travelling at a speed of num__90 kmph crosses a pole in num__5 sec then the length of train is ? <o> a ) num__338 <o> b ) num__125 <o> c ) num__726 <o> d ) num__268 <o> e ) num__171 |
d = num__90 * num__0.277777777778 * num__5 = num__125 m answer : b <eor> b <eos> |
b |
round__125.0__ |
round__125.0__ |
| the salaries of a and b together amount to $ num__500 . a spends num__90.0 of his salary and b num__80.0 of his . if now their savings are the same what is a ' s salary ? <o> a ) $ num__50 <o> b ) $ num__150 <o> c ) $ num__69 <o> d ) $ num__34 <o> e ) $ num__52 |
let a ' s salary is x b ' s salary = num__500 - x ( num__100 - num__90 ) % of x = ( num__100 - num__80 ) % of ( num__500 - x ) x = $ num__34 ( approximately ) answer is d <eor> d <eos> |
d |
percent__100.0__34.0__ |
percent__100.0__34.0__ |
| a gardener changed the size of his rectangle shaped garden by increasing it ' s length by num__40.0 & decreasing is ' s width by num__20.0 . find area of new garden . <o> a ) num__1.04 <o> b ) num__1.12 <o> c ) num__1.24 <o> d ) num__1.4 <o> e ) num__1.5 |
a num__1 = l * b a num__2 = ( l * num__1.4 ) * ( b * num__0.8 ) = num__1.12 * lb so area of garden is by num__1.12 times of old area . answer : b <eor> b <eos> |
b |
multiply__0.8__1.4__ multiply__1.0__1.12__ |
multiply__0.8__1.4__ multiply__1.0__1.12__ |
| d q and r are positive integers . if d q and r are assembled into the six - digit number dqrdqr which one of the following must be a factor of dqrdqr ? <o> a ) num__23 <o> b ) num__19 <o> c ) num__17 <o> d ) num__7 <o> e ) none of the above |
one short way - dqrdqr = num__1000 dqr + dqr = ( num__1000 + num__1 ) dqr = num__1001 dqr therefore any factor of num__1001 is a factor of dqrdqr num__7 is a factor of num__1001 so d <eor> d <eos> |
d |
add__1000.0__1.0__ multiply__1.0__7.0__ |
add__1000.0__1.0__ multiply__1.0__7.0__ |
| sheila works num__8 hours per day on monday wednesday and friday and num__6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ num__288 per week . how much does she earn in dollars per hour ? <o> a ) num__11 <o> b ) num__10 <o> c ) num__9 <o> d ) num__8 <o> e ) num__7 |
let sheila earn x dollars per hour so on monday wednesday and friday she earns num__8 x each and on tuesday and thursday she earns num__6 x each in total over the week she should earn num__3 ( num__8 x ) + num__2 ( num__6 x ) = num__36 x she earns $ num__288 per week num__36 x = num__288 x = num__8 correct option : d <eor> d <eos> |
d |
subtract__8.0__6.0__ divide__288.0__8.0__ round__8.0__ |
subtract__8.0__6.0__ divide__288.0__8.0__ add__6.0__2.0__ |
| a license plate in the country kerrania consists of four digits followed by two letters . the letters a b and c are used only by government vehicles while the letters d through z are used by non - government vehicles . kerrania ' s intelligence agency has recently captured a message from the country gonzalia indicating that an electronic transmitter has been installed in a kerrania government vehicle with a license plate starting with num__79 . if it takes the police num__20 minutes to inspect each vehicle what is the probability that the police will find the transmitter within three hours ? <o> a ) num__0.227848101266 <o> b ) num__0.166666666667 <o> c ) num__0.04 <o> d ) num__0.02 <o> e ) num__0.01 |
so there are num__900 cars that they have to search . . each takes num__20 mins total of num__18000 mins . . have to find in num__180 mins . . prob num__0.01 = num__0.01 <eor> e <eos> |
e |
multiply__20.0__900.0__ divide__180.0__18000.0__ round__0.01__ |
multiply__20.0__900.0__ divide__180.0__18000.0__ round__0.01__ |
| ♠ n denotes the number obtained when n is rounded to the nearest tenth . for example ♠ num__4.31 = num__4.3 ♠ num__3.32 - ♠ num__2.42 = <o> a ) num__1.05 <o> b ) num__0.9 <o> c ) - num__1.05 <o> d ) - num__1.0 <o> e ) - num__0.1 |
♠ num__3.32 - ♠ num__2.42 = num__3.3 - num__2.4 = num__0.9 answer : b <eor> b <eos> |
b |
subtract__3.32__2.42__ subtract__3.32__2.42__ |
subtract__3.32__2.42__ subtract__3.32__2.42__ |
| in a certain deck of cards each card has a positive integer written on it in a multiplication game a child draws a card and multiplies the integer on the card with the next large integer . if the each possible product is between num__35 and num__200 then the least and greatest integer on the card would be <o> a ) num__3 and num__15 <o> b ) num__3 and num__20 <o> c ) num__4 and num__13 <o> d ) num__4 and num__14 <o> e ) num__5 and num__14 |
given : num__35 < x ( x + num__1 ) < num__200 . now it ' s better to test the answer choices here rather than to solve : if x = num__5 then x ( x + num__1 ) = num__30 > num__35 - - > so the least value is num__5 . test for the largest value : if x = num__14 then x ( x + num__1 ) = num__14 * num__15 = num__210 > num__200 answer : e . <eor> e <eos> |
e |
subtract__35.0__5.0__ add__1.0__14.0__ multiply__14.0__15.0__ subtract__35.0__30.0__ |
subtract__35.0__5.0__ add__1.0__14.0__ multiply__14.0__15.0__ subtract__35.0__30.0__ |
| a car gets num__33 miles to the gallon . if it is modified to use a solar panel it will use only num__75 percent as much fuel as it does now . if the fuel tank holds num__16 gallons how many more miles will the car be able to travel per full tank of fuel after it has been modified ? <o> a ) num__168 <o> b ) num__172 <o> c ) num__176 <o> d ) num__180 <o> e ) num__184 |
originally the distance the car could go on a full tank was num__16 * num__33 = num__528 miles . after it has been modified the car can go num__33 / num__0.75 = num__44 miles per gallon . on a full tank the car can go num__16 * num__44 = num__704 miles thus num__176 miles more . the answer is c . <eor> c <eos> |
c |
multiply__33.0__16.0__ divide__33.0__0.75__ multiply__16.0__44.0__ subtract__704.0__528.0__ round__176.0__ |
multiply__33.0__16.0__ divide__33.0__0.75__ multiply__16.0__44.0__ subtract__704.0__528.0__ round__176.0__ |
| a train num__275 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__15 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__275 * num__0.0545454545455 ] sec = num__15 sec answer : a <eor> a <eos> |
a |
add__60.0__6.0__ divide__275.0__18.3333__ round__15.0__ |
add__60.0__6.0__ divide__275.0__18.3333__ divide__275.0__18.3333__ |
| in an arithmetic progression the difference between the any two consecutive terms is a constant . what is the arithmetic mean of all of the terms from the first to the num__27 th in an arithmetic progression if the sum of the num__12 th and num__16 th terms of the sequence is num__98 ? <o> a ) num__47 <o> b ) num__49 <o> c ) num__51 <o> d ) num__53 <o> e ) num__55 |
let x be the difference between any two consecutive terms . the mean of a sequence like this is the middle term thus the num__14 th term in the sequence . then the mean of the num__13 th and num__15 th term is also equal to the overall mean because the num__13 th term is ( num__14 th term - x ) and the num__15 th term is ( num__14 th term + x ) . similarly the mean of the num__12 th and num__16 th term is also equal to the mean . etc . . . thus the mean is num__49.0 = num__49 the answer is b . <eor> b <eos> |
b |
subtract__27.0__14.0__ subtract__27.0__12.0__ subtract__98.0__49.0__ |
subtract__27.0__14.0__ subtract__27.0__12.0__ subtract__98.0__49.0__ |
| in num__120 m race a covers the distance in num__36 seconds and b in num__45 seconds . in this race a beats b by : <o> a ) num__24 m <o> b ) num__25 m <o> c ) num__22.5 m <o> d ) num__9 m <o> e ) num__12 m |
distance covered by b in num__9 sec . = num__2.66666666667 x num__9 m = num__24 m . a beats b by num__24 metres . answer : option a <eor> a <eos> |
a |
subtract__45.0__36.0__ divide__120.0__45.0__ round__24.0__ |
subtract__45.0__36.0__ divide__120.0__45.0__ round__24.0__ |
| if x = - num__0.875 and y = - num__0.5 what is the value of the expression - num__2 x – y ^ num__2 ? <o> a ) - num__1.5 <o> b ) - num__1 <o> c ) num__1 <o> d ) num__1.5 <o> e ) num__1.75 |
- num__2 x - y ^ num__2 - num__2 x - num__0.875 - ( num__0.5 ) ^ num__2 num__1.75 - num__0.25 num__1.5 d correct <eor> d <eos> |
d |
divide__0.875__0.5__ divide__0.5__2.0__ subtract__2.0__0.5__ subtract__2.0__0.5__ |
divide__0.875__0.5__ subtract__2.0__1.75__ subtract__2.0__0.5__ subtract__2.0__0.5__ |
| the surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are num__8 cm each . the radius of the sphere is <o> a ) num__3 cm <o> b ) num__4 cm <o> c ) num__6 cm <o> d ) num__8 cm <o> e ) none |
solution num__4 Î r num__2 = num__2 Î num__4 x num__8 â ‡ ’ r num__2 = ( num__4 x num__4.0 ) â ‡ ’ num__16 â ‡ ’ r = num__4 cm . answer b <eor> b <eos> |
b |
square_perimeter__4.0__ triangle_area__2.0__4.0__ |
square_perimeter__4.0__ triangle_area__2.0__4.0__ |
| the average ( arithmetic mean ) of eight numbers is num__49.1 . if the sum of half of these numbers is num__158.4 what is the average of the other half ? <o> a ) num__12.8 <o> b ) num__24.2 <o> c ) num__58.6 <o> d ) num__72.1 <o> e ) num__96.8 |
arithmetic mean = sum / total numbers sum = num__49.1 * num__8 = num__392.8 sum of half of these numbers is num__158.4 . so num__4 numbers sum is num__158.4 . rest num__4 numbers sum = num__392.8 - num__158.4 = num__234.4 arithmetic mean of the num__4 nos = num__234.4 / num__4 = num__58.6 hence c is the answer . <eor> c <eos> |
c |
multiply__49.1__8.0__ subtract__392.8__158.4__ divide__234.4__4.0__ divide__234.4__4.0__ |
multiply__49.1__8.0__ subtract__392.8__158.4__ divide__234.4__4.0__ divide__234.4__4.0__ |
| a number when divided by num__6 leaves a remainder num__3 . when the square of the same number is divided by num__6 the remainder is <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
sol . let x = num__6 q + num__3 . then x num__2 = ( num__6 q + num__3 ) num__2 = num__36 q num__2 + num__36 q + num__9 = num__6 ( num__6 q num__2 + num__6 q + num__1 ) + num__3 . so when x num__2 is divided by num__6 remainder = num__3 . answer d <eor> d <eos> |
d |
power__6.0__2.0__ triangle_area__6.0__3.0__ triangle_area__6.0__1.0__ |
power__6.0__2.0__ triangle_area__6.0__3.0__ triangle_area__6.0__1.0__ |
| find the odd man out num__515 num__4560 num__90180 <o> a ) num__60 <o> b ) num__180 <o> c ) num__90 <o> d ) num__5 <o> e ) num__15 |
num__5 * num__3 = num__15 num__15 * num__3 = num__45 num__45 * num__2 = num__90 num__90 * num__2 = num__180 num__180 * num__2 = num__360 answer : a <eor> a <eos> |
a |
multiply__3.0__5.0__ multiply__3.0__15.0__ subtract__5.0__3.0__ multiply__2.0__45.0__ multiply__2.0__90.0__ multiply__2.0__180.0__ add__45.0__15.0__ |
multiply__3.0__5.0__ multiply__3.0__15.0__ subtract__5.0__3.0__ multiply__2.0__45.0__ multiply__2.0__90.0__ multiply__2.0__180.0__ add__45.0__15.0__ |
| sonika deposited rs . num__7000 which amounted to rs . num__9200 after num__3 years at simple interest . had the interest been num__2.0 more . she would get how much ? <o> a ) num__9680 <o> b ) num__2277 <o> c ) num__9620 <o> d ) num__2774 <o> e ) num__1212 |
( num__7000 * num__3 * num__2 ) / num__100 = num__420 num__9200 - - - - - - - - num__9620 answer : c <eor> c <eos> |
c |
percent__100.0__9620.0__ |
percent__100.0__9620.0__ |
| an error num__2.0 in excess is made while measuring the side of a square . what is the percentage of error in the calculated area of the square ? <o> a ) num__4.04 <o> b ) num__2.02 <o> c ) num__4.0 <o> d ) num__2.0 <o> e ) num__2.40 % |
explanation : error = num__2.0 while measuring the side of a square . let correct value of the side of the square = num__100 then measured value = num__100 + num__2 = num__102 ( ∵ num__2 is num__2.0 of num__100 ) correct area of the square = num__100 × num__100 = num__10000 calculated area of the square = num__102 × num__102 = num__10404 error = num__10404 − num__10000 = num__404 percentage error = error / actual value × num__100 = num__0.0404 × num__100 = num__4.04 answer : option a <eor> a <eos> |
a |
percent__0.0404__10000.0__ percent__4.04__100.0__ |
percent__0.0404__10000.0__ percent__4.04__100.0__ |
| a man goes from a to b at a speed of num__20 kmph and comes back to a at a speed of num__30 kmph . find his average speed for the entire journey ? <o> a ) num__76 kmph <o> b ) num__24 kmph <o> c ) num__54 kmph <o> d ) num__56 kmph <o> e ) num__76 kmph |
distance from a and b be ' d ' average speed = total distance / total time average speed = ( num__2 d ) / [ ( d / num__20 ) + ( d / num__30 ) ] = ( num__2 d ) / [ num__5 d / num__60 ) = > num__24 kmph . answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ round__24.0__ |
hour_to_min_conversion__ round__24.0__ |
| two trains t num__1 and t num__2 start simultaneously from two stations x and y respectively towards each other . if they are num__60 km apart both num__3 and num__6 hours after start then find the distance between the two stations . <o> a ) num__240 km <o> b ) num__200 km <o> c ) num__220 km <o> d ) num__180 km <o> e ) num__210 km |
in first num__3 hours t num__1 travels r km and t num__2 travels s km . after num__6 hours they traveled r + s + num__60 + num__60 num__2 ( r + s ) = r + s + num__120 r + s = num__120 hence distance between xy is r + s + num__60 = num__120 + num__60 = num__200 answer : b <eor> b <eos> |
b |
multiply__2.0__60.0__ round__200.0__ |
multiply__2.0__60.0__ round__200.0__ |
| a boat can travel with a speed of num__25 km / hr in still water . if the speed of the stream is num__5 km / hr find the time taken by the boat to go num__90 km downstream . <o> a ) num__1 hr <o> b ) num__2 hrs <o> c ) num__3 hrs <o> d ) num__4 hrs <o> e ) num__5 hrs |
speed downstream = ( num__25 + num__5 ) km / hr = num__30 km / hr . time taken to travel num__90 km downstream = num__3.0 hrs = num__3 hrs . answer : c <eor> c <eos> |
c |
add__25.0__5.0__ divide__90.0__30.0__ round__3.0__ |
add__25.0__5.0__ divide__90.0__30.0__ divide__90.0__30.0__ |
| an incredible punch is composed of mango orange juice and milk . how many pints of orange juice are required to make num__9 gallons of punch containing twice as much mango as orange juice and three times as much orange juice as milk ? ( num__1 gallon = num__8 pints ) <o> a ) num__13 pints <o> b ) num__7.2 pints <o> c ) num__5 pints <o> d ) num__12 pints <o> e ) num__6.5 pints |
num__1 gallon = num__8 pints num__9 gallons = num__72 mango = num__2 orange orange = num__3 milk mango + orange + milk = num__72 num__2 orange + orange + num__0.333333333333 orange = num__72 num__10 orange = num__72 orange = num__7.2 pints answer b <eor> b <eos> |
b |
multiply__9.0__8.0__ add__1.0__2.0__ reverse__3.0__ add__9.0__1.0__ divide__72.0__10.0__ multiply__1.0__7.2__ |
multiply__9.0__8.0__ add__1.0__2.0__ reverse__3.0__ add__9.0__1.0__ divide__72.0__10.0__ multiply__1.0__7.2__ |
| if a man can cover num__12 metres in num__0 ne second how many kilometres can he cover in num__3 hours num__45 minutes ? <o> a ) num__132 km <o> b ) num__167 km <o> c ) num__143 km <o> d ) num__162 kilometres <o> e ) num__245 km |
d num__12 m / s = num__12 * num__3.6 kmph num__3 hours num__45 minutes = num__3 num__0.75 hours = num__3.75 hours distance = speed * time = num__12 * num__3.6 * num__3.75 km = num__162 km . <eor> d <eos> |
d |
add__3.0__0.75__ multiply__45.0__3.6__ round__162.0__ |
divide__45.0__12.0__ multiply__45.0__3.6__ multiply__45.0__3.6__ |
| ajay can ride num__50 km in num__1 hour . in how many hours he can ride num__900 km ? <o> a ) num__10 hrs <o> b ) num__15 hrs <o> c ) num__20 hrs <o> d ) num__25 hrs <o> e ) num__18 hrs |
num__1 hour he ride num__50 km he ride num__900 km in = num__18.0 * num__1 = num__18 hours answer is e <eor> e <eos> |
e |
divide__900.0__50.0__ round__18.0__ |
divide__900.0__50.0__ multiply__1.0__18.0__ |
| a window num__2 inches wide is placed on the back of a rectangular painted area with dimensions num__8 inches by num__12 inches . what is the area of the window in square inches ? <o> a ) num__44 <o> b ) num__96 <o> c ) num__128 <o> d ) num__144 <o> e ) num__168 |
this question is an example of a ' punch out ' question - we have to find the area of everything then ' punch out ' the part that we do n ' t want . we ' re told that a window num__2 inches wide is placed around a rectangular painted area with dimensions num__8 inches by num__12 inches . we ' re asked for the area of the window in square inches . area of a rectangle = ( length ) ( width ) so the area of the window is . . . ( num__8 ) ( num__12 ) = num__96 the window ' adds ' num__2 inches to the top bottom left and right ' sides ' of the painted area so the area of everything is . . . ( num__8 + num__2 + num__2 ) ( num__12 + num__2 + num__2 ) = ( num__12 ) ( num__16 ) = num__196 when we ' punch out ' the area of the painted area we ' ll be left with the area of the window : num__192 - num__96 = num__96 final answer : b <eor> b <eos> |
b |
multiply__8.0__12.0__ multiply__2.0__8.0__ rectangle_perimeter__2.0__96.0__ multiply__2.0__96.0__ triangle_area__2.0__96.0__ |
multiply__8.0__12.0__ multiply__2.0__8.0__ rectangle_perimeter__2.0__96.0__ multiply__2.0__96.0__ triangle_area__2.0__96.0__ |
| two trains of equal length are running on parallel lines in the same direction at num__46 km / hr and num__36 km / hr . the faster train passes the slower train in num__36 seconds . the length of each train is : <o> a ) num__50 m <o> b ) num__72 m <o> c ) num__80 m <o> d ) num__82 m <o> e ) num__84 m |
let the length of each train be x metres . then distance covered = num__2 x metres . relative speed = ( num__46 - num__36 ) km / hr ( num__10 x ( num__0.277777777778 ) ) m / s = num__2.77777777778 m / s num__2 x / num__36 = num__2.77777777778 num__2 x = num__100 x = num__50 answer : a <eor> a <eos> |
a |
subtract__46.0__36.0__ divide__10.0__36.0__ divide__100.0__2.0__ divide__100.0__2.0__ |
subtract__46.0__36.0__ divide__10.0__36.0__ divide__100.0__2.0__ divide__100.0__2.0__ |
| if janice was num__24 years old z years ago and lisa will be num__22 years old in p years what was the average ( arithmetic mean ) of their ages num__6 years ago ? <o> a ) ( z + p ) / num__2 <o> b ) ( z - p + num__34 ) / num__4 <o> c ) ( z - p + num__24 ) / num__4 <o> d ) ( z + p + num__44 ) / num__2 <o> e ) ( z - p + num__34 ) / num__2 |
today j = z + num__24 and l = num__22 - p num__6 years ago j = z + num__18 and l = num__16 - p the average of their ages was ( z - p + num__34 ) / num__2 the answer is e . <eor> e <eos> |
e |
subtract__24.0__6.0__ subtract__22.0__6.0__ add__16.0__18.0__ subtract__24.0__22.0__ add__16.0__18.0__ |
subtract__24.0__6.0__ subtract__22.0__6.0__ add__16.0__18.0__ subtract__24.0__22.0__ add__16.0__18.0__ |
| a train num__225 m long passes a man running at num__10 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is : <o> a ) num__81 <o> b ) num__90 <o> c ) num__91 <o> d ) num__85 <o> e ) num__96 |
speed of the train relative to man = num__22.5 m / sec = num__22.5 m / sec . = num__22.5 x num__3.6 km / hr = num__81 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__10 = num__81 = num__91 km / hr . answer : c <eor> c <eos> |
c |
divide__225.0__10.0__ multiply__22.5__3.6__ add__10.0__81.0__ round__91.0__ |
divide__225.0__10.0__ multiply__22.5__3.6__ add__10.0__81.0__ round__91.0__ |
| if janice was num__22 years old z years ago and lisa will be num__20 years old in p years what was the average ( arithmetic mean ) of their ages num__7 years ago ? <o> a ) ( z + p ) / num__2 <o> b ) ( z - p + num__28 ) / num__4 <o> c ) ( z - p + num__18 ) / num__4 <o> d ) ( z + p + num__38 ) / num__2 <o> e ) ( z - p + num__28 ) / num__2 |
today j = z + num__22 and l = num__20 - p num__7 years ago j = z + num__15 and l = num__13 - p the average of their ages was ( z - p + num__28 ) / num__2 the answer is e . <eor> e <eos> |
e |
subtract__22.0__7.0__ subtract__20.0__7.0__ add__13.0__15.0__ subtract__22.0__20.0__ add__13.0__15.0__ |
subtract__22.0__7.0__ subtract__20.0__7.0__ add__13.0__15.0__ subtract__22.0__20.0__ add__13.0__15.0__ |
| it takes john exactly num__30 minutes to rake a lawn and it takes his son todd exactly num__60 minutes to rake the same lawn . if john and todd decide to rake the lawn together and both work at the same rate that they did previously how many minutes will it take them to rake the lawn <o> a ) num__16 <o> b ) num__45 <o> c ) num__36 <o> d ) num__20 <o> e ) num__90 |
num__1 ) poe - john takes num__30 min so if he takes help of someone else it has to be less than num__30 min . . only a and b are left . . if both do the work in num__30 mis each the combined time will be num__15 mins so num__16 is slightly less when the other person does in num__60 mins . . ans num__20 d <eor> d <eos> |
d |
add__1.0__15.0__ round__20.0__ |
add__1.0__15.0__ round__20.0__ |
| the price of a certain product increased by the same percent from num__1960 to num__1970 as from num__1970 to num__1980 . if its price of $ num__2.00 in num__1970 was num__150 percent of its price in num__1960 what was its price in num__1980 ? <o> a ) a ) $ num__1.80 <o> b ) b ) $ num__2.00 <o> c ) c ) $ num__2.40 <o> d ) d ) $ num__2.70 <o> e ) e ) $ num__3.00 |
the price in num__1970 was num__150 percent of its price in num__1960 means that the percent increase was num__50.0 from num__1960 to num__1970 ( and from num__1970 to num__1980 ) . therefore the price in num__1980 = $ num__2.0 * num__1.5 = $ num__3.00 . answer : e . <eor> e <eos> |
e |
multiply__2.0__1.5__ multiply__2.0__1.5__ |
multiply__2.0__1.5__ multiply__2.0__1.5__ |
| machine a and machine g are each used to manufacture num__660 sprockets . it takes machine a num__10 hours longer to produce num__660 sprockets than machine g . machine g produces num__10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ? <o> a ) num__6 <o> b ) num__6.6 <o> c ) num__60 <o> d ) num__100 <o> e ) num__110 |
[ reveal ] spoiler : timeg : num__660 / x timea : [ num__660 / x + num__10 ] num__660 / x = [ num__660 / x + num__10 ] * num__1.1 num__660 / x = num__66 * num__11 / x + num__10 num__660 x + num__10 = num__66 * num__11 * x num__660 x + num__6600 = num__66 * num__11 * x x = num__100 plug in back to timea num__6.6 + num__10 = > num__6.0 = num__6 <eor> a <eos> |
a |
percent__10.0__660.0__ percent__10.0__66.0__ percent__100.0__6.0__ |
percent__10.0__660.0__ percent__10.0__66.0__ percent__100.0__6.0__ |
| eleven bags are bought for rs . num__1000 and sold at num__10 for rs . num__1200 . what is the gain or loss in percentage ? <o> a ) num__10.0 <o> b ) num__22.0 <o> c ) num__25.0 <o> d ) num__20.0 <o> e ) none of these |
explanation : as selling price is rs . num__1200 for num__10 bags each bag is sold at rs . num__120 . hence the profit is rs . num__100 after selling num__10 bags plus the selling price of num__1 bag = num__100 + num__120 = rs . num__220 . % profit = num__0.22 * num__100 = num__22.0 answer b <eor> b <eos> |
b |
percent__10.0__1200.0__ percent__10.0__1000.0__ percent__10.0__220.0__ percent__10.0__220.0__ |
percent__10.0__1200.0__ percent__10.0__1000.0__ percent__10.0__220.0__ percent__10.0__220.0__ |
| i . x + num__2 y + num__3 z = num__2 ii . x + y - z = num__0 iii . num__2 x + num__2 y - z = num__1 what is the value of y in the system above ? <o> a ) - num__2 <o> b ) - num__1 <o> c ) num__0 <o> d ) num__1 <o> e ) num__2 |
num__1 . from the given system of equations i can make equation ii . to be x + y = z num__2 . now put that in iii . : num__2 x + num__2 y - ( x + y ) = num__1 = num__2 x + num__2 y - x - y = num__1 = x + y = num__1 that gives usz = num__1 ! num__3 . put z = num__1 in i . and solve : x + num__2 y + num__3 = num__2 ; calculate minus num__3 = x + num__2 y = - num__1 ; calculate minus x = num__2 y = - num__1 - x ; num__4 . now put num__2 y = - num__1 - x in iii . and solve : num__2 x + ( - num__1 - x ) - num__1 = num__1 = x - num__2 = num__1 ; calculate plus num__2 that gives usx = num__3 num__5 . now solve for y with x = num__3 and z = num__1 : x + y = num__1 = num__3 + y = num__1 ; calculate minus num__3 that gives y = - num__2 and hence answer ( a ) - num__2 <eor> a <eos> |
a |
add__3.0__1.0__ add__2.0__3.0__ multiply__2.0__1.0__ |
add__3.0__1.0__ add__2.0__3.0__ subtract__3.0__1.0__ |
| steve gets on the elevator at the num__11 th floor of a building and rides up at a rate of num__47 floors per minute . at the same time joyce gets on an elevator on the num__51 st floor of the same building and rides down at a rate of num__53 floors per minute . if they continue traveling at these rates at which floor will their paths cross ? <o> a ) num__19 <o> b ) num__28 <o> c ) num__30 <o> d ) num__33 <o> e ) num__44 |
steve gets on the elevator at the num__11 th floor of a building and rides up at a rate of num__47 floors per minute . at the same time joyce gets on an elevator on the num__51 st floor of the same building and rides down at a rate of num__53 floors per minute . if they continue traveling at these rates at which floor will their paths cross ? num__40 floors / num__100 floors per minute = num__0.4 minutes num__11 + num__19.0 = num__33 num__51 - num__21.0 = num__33 answer : d <eor> d <eos> |
d |
percent__33.0__100.0__ |
percent__33.0__100.0__ |
| a tap can fill the tank in num__15 minutes and another can empty it in num__8 minutes . if the tank is already half full and both the taps are opened together the tank will be : <o> a ) filled in num__12 min <o> b ) emptied in num__12 min <o> c ) filled in num__8 min <o> d ) will be empty in num__8 min <o> e ) emptied in num__10 min |
work done by both taps = num__0.0666666666667 + ( - num__0.125 ) = - num__0.0583333333333 so tank will be emptied becoz of minus sign . so time taken = num__17.1428571429 given the tank is already half filled so another half will be emptied in num__17.1428571429 * num__0.5 nearly num__8 minutes answer : d <eor> d <eos> |
d |
subtract__0.125__0.0667__ round__8.0__ |
subtract__0.125__0.0667__ round__8.0__ |
| after m students took a test the average score was num__56.0 . if the test has num__50 questions what is the least number of questions that the next student has to get right to bring the average score up to num__60.0 ? <o> a ) m + num__30 <o> b ) num__0.56 m + num__30 <o> c ) m + num__28 <o> d ) num__2 m + num__28 <o> e ) num__2 m + num__30 |
the total number of correct answers for m students is ( num__0.56 ) * num__50 * m = num__28 * m for an average of num__75.0 : ( total correct answers ) / ( m + num__1 ) = num__0.6 * num__50 = num__30 let x be the number of correct answers for the next student . ( x + num__28 m ) / m + num__1 = num__30 x + num__28 m = num__30 m + num__30 x = num__2 m + num__30 the answer is e . <eor> e <eos> |
e |
multiply__50.0__0.56__ multiply__50.0__0.6__ divide__56.0__28.0__ divide__56.0__28.0__ |
multiply__50.0__0.56__ multiply__50.0__0.6__ divide__56.0__28.0__ divide__56.0__28.0__ |
| find the area of the sector whose are is making an angle of num__90 ° at the center of a circle of radius num__3.2 cm . <o> a ) num__56.72 sq . cm <o> b ) num__56.22 sq . cm <o> c ) num__56.82 sq . cm <o> d ) num__56.42 sq . cm <o> e ) num__56.32 sq . cm |
area of the sector = num__0.25 × π r ( power ) num__2 = num__0.25 × num__3.14285714286 × num__3.2 × num__3.2 = ( num__11 × num__10.24 ) / num__2 = num__112.64 / num__2 = num__56.32 sq . cm answer is e . <eor> e <eos> |
e |
multiply__11.0__10.24__ divide__112.64__2.0__ round__56.32__ |
multiply__11.0__10.24__ divide__112.64__2.0__ divide__112.64__2.0__ |
| if p = q - r - num__1 < r < num__0 and q = num__0 which of the following is correct ? <o> a ) p < - num__1 . <o> b ) num__0 < p < num__2 <o> c ) p > num__0 <o> d ) num__0 < p < num__4 <o> e ) p < - num__20 |
q is num__0 & r is - ve for min . value max . numerator and minimize denominator take q = num__0 & r = - num__0.9 p = num__0 - ( - num__0.9 ) p > num__0 ans c <eor> c <eos> |
c |
round_down__0.9__ |
round_down__0.9__ |
| approximately what percentage of the world ’ s forested area is represented by finland given that finland has num__53.42 million hectares of forested land of the world ’ s num__8.076 billion hectares of forested land . <o> a ) num__0.0066 <o> b ) num__0.066 <o> c ) num__0.66 percent <o> d ) num__6.6 <o> e ) num__66 % |
approximation is a strategy that helps us arrive at less than an exact number and the inclusion in this problem of the word “ approximately ” is an obvious clue . first num__8.076 billion is num__8076 million . next num__8076 million rounds to num__8000 million and num__53.42 million rounds to num__53 million . dividing num__53 million by num__8000 million we arrive at num__0.0066 ( num__53 m / num__8000 m ) . we convert this decimal figure to a percentage by multiplying by num__100 ( or moving the decimal point two places to the right ) and adding a percent sign in order to obtain our answer of num__0.66 . note that the shortcut method involves comparing num__53 million to num__1.0 of num__8000 million or num__80 million . since num__53 million is approximately two - thirds of num__80 million then the answer is some two - thirds of num__1.0 or num__0.66 . answer : c . <eor> c <eos> |
c |
round_down__53.42__ divide__53.42__8076.0__ multiply__100.0__0.0066__ divide__8000.0__100.0__ multiply__1.0__0.66__ |
round_down__53.42__ divide__53.42__8076.0__ multiply__100.0__0.0066__ divide__8000.0__100.0__ divide__0.66__1.0__ |
| kim can walk num__4 kilometers in one hour . how long does it take kim to walk num__18 kilometers ? <o> a ) num__7 hours and num__30 minutes . <o> b ) num__8 hours and num__30 minutes . <o> c ) num__9 hours and num__30 minutes . <o> d ) num__3 hours and num__30 minutes . <o> e ) num__4 hrs and num__30 min |
the time it takes kim to walk num__18 kilometers is equal to num__4.5 = num__4.5 hours = num__4 hours + num__0.5 * num__60 minutes = num__4 hours and num__30 minutes . correct answer e <eor> e <eos> |
e |
divide__18.0__4.0__ subtract__4.5__4.0__ hour_to_min_conversion__ multiply__0.5__60.0__ round__4.0__ |
divide__18.0__4.0__ subtract__4.5__4.0__ hour_to_min_conversion__ multiply__0.5__60.0__ round__4.0__ |
| if y = num__2 + num__2 k and y ≠ num__0 y ≠ num__0 then num__2 / y + num__1 / y + num__1 / y + num__1 / y = ? <o> a ) num__1 / ( num__8 + num__8 k ) <o> b ) num__2.5 ( num__1 + k ) <o> c ) num__1 / ( num__8 + k ) <o> d ) num__4 / ( num__8 + k ) <o> e ) num__4 / ( num__1 + k ) |
num__2 / y + num__1 / y + num__1 / y + num__1 / y = num__5 / y = num__5 / ( num__2 + num__2 k ) = num__2.5 ( num__1 + k ) answer : b <eor> b <eos> |
b |
divide__5.0__2.0__ multiply__1.0__2.5__ |
divide__5.0__2.0__ divide__2.5__1.0__ |
| two trains are running at num__40 km / hr and num__20 km / hr respectively in the same direction . fast train completely passes a man sitting in the slower train in num__5 sec . what is the length of the fast train ? <o> a ) num__27 num__0.777777777778 <o> b ) num__27 num__1.16666666667 <o> c ) num__27 num__3.5 <o> d ) num__27 num__0.777777777778 <o> e ) num__27 num__7.0 |
relative speed = ( num__40 - num__20 ) = num__20 km / hr . = num__20 * num__0.277777777778 = num__5.55555555556 m / sec . length of faster train = num__5.55555555556 * num__5 = num__27.7777777778 = num__27 num__0.777777777778 m . answer : d <eor> d <eos> |
d |
subtract__27.7778__27.0__ round__27.0__ |
subtract__27.7778__27.0__ subtract__27.7778__0.7778__ |
| p q r enter into a partnership . p initially invests num__25 lakh and adds another num__10 lakh after one year . q initially invests num__35 lakh and withdraws num__10 lakh after num__2 years . r ' s investment is rs num__30 lakh . in what ratio should the profit be divided at the end of num__3 years ? <o> a ) num__19 : num__18 : num__19 <o> b ) num__18 : num__19 : num__19 <o> c ) num__19 : num__17 : num__18 <o> d ) num__19 : num__19 : num__18 <o> e ) num__17 : num__18 : num__19 |
p : q : r = ( num__25 × num__1 + num__35 × num__2 ) : ( num__35 × num__2 + num__25 × num__1 ) : ( num__30 × num__3 ) = num__95 : num__95 : num__90 = num__19 : num__19 : num__18 answer is d . <eor> d <eos> |
d |
subtract__3.0__2.0__ multiply__30.0__3.0__ subtract__19.0__1.0__ multiply__19.0__1.0__ |
subtract__3.0__2.0__ multiply__30.0__3.0__ subtract__19.0__1.0__ add__1.0__18.0__ |
| the length of a train and that of a platform are equal . if with a speed of num__90 k / hr the train crosses the platform in one minute then the length of the train ( in meters ) is ? <o> a ) num__752 <o> b ) num__799 <o> c ) num__719 <o> d ) num__750 <o> e ) num__712 |
speed = [ num__90 * num__0.277777777778 ] m / sec = num__25 m / sec ; time = num__1 min . = num__60 sec . let the length of the train and that of the platform be x meters . then num__2 x / num__60 = num__25 è x = num__25 * num__30.0 = num__750 answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ subtract__90.0__60.0__ multiply__25.0__30.0__ round__750.0__ |
hour_to_min_conversion__ divide__60.0__2.0__ multiply__25.0__30.0__ divide__750.0__1.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__6 seconds . what is the length of the train ? <o> a ) num__100 <o> b ) num__130 <o> c ) num__150 <o> d ) num__170 <o> e ) num__160 |
speed = ( num__60 x num__0.277777777778 ) m / sec = ( num__16.6666666667 ) m / sec . length of the train = ( speed x time ) . length of the train = ( num__16.6666666667 x num__6 ) m = num__100 m . answer is a . <eor> a <eos> |
a |
round__100.0__ |
round__100.0__ |
| if after successive discounts of num__20.0 and num__6.25 % on the marked price of an article a trader gets num__20.0 profit on the cost price then by what percent is the marked price above the cost price ? <o> a ) num__50.0 <o> b ) num__60.0 <o> c ) num__70.0 <o> d ) num__80.0 <o> e ) num__90 % |
let the market price be rs num__100 . then after first discount s . p becomes num__80 and after another discount it becomes rs . num__75 . so s . p = num__75 . now with this s . p he has made a profit of num__20.0 and hence the c . p = ( num__0.625 ) * num__100 = num__62.5 so percentage change with respect to market price is = ( num__37.5 / num__62.5 ) * num__100 giving num__60.0 answer : b <eor> b <eos> |
b |
percent__75.0__80.0__ percent__100.0__60.0__ |
percent__75.0__80.0__ percent__100.0__60.0__ |
| num__0.444444444444 × num__2.0 + num__0.407407407407 – num__0.222222222222 = ? <o> a ) num__27 ( num__0.208333333333 ) <o> b ) num__13 ( num__0.285714285714 ) <o> c ) num__24 ( num__0.185185185185 ) <o> d ) num__17 ( num__0.142857142857 ) <o> e ) none of these |
explanation : = num__0.444444444444 × num__2.0 + num__0.407407407407 – num__0.222222222222 = num__0.444444444444 × num__2.0 + num__0.407407407407 – num__0.222222222222 = num__12 * num__54 + num__11 - num__0.222222222222 = num__24.1851851852 = num__24 ( num__0.185185185185 ) answer : option c <eor> c <eos> |
c |
round_down__24.1852__ subtract__0.4074__0.2222__ round_down__24.1852__ |
multiply__2.0__12.0__ subtract__0.4074__0.2222__ multiply__2.0__12.0__ |
| when num__30 is divided by the positive integer k the remainder is num__6 for how many different values of k is this true ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__8 |
when num__30 is divided by k remainder is num__6 i . e . there are num__6 balls leftover after grouping . so k must be greater than num__6 . it also means that num__30 is completely divisible by k . factors of num__24 are num__1 num__2 num__3 num__4 num__6 num__8 num__12 num__24 out of these k can be num__4 num__6 num__8 num__12 num__24 . answer ( c ) <eor> c <eos> |
c |
coin_space__ choose__3.0__1.0__ vowel_space__ |
coin_space__ choose__3.0__1.0__ vowel_space__ |
| what least value should be replaced by * in num__223 * num__431 so the number become divisible by num__9 <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
explanation : trick : number is divisible by num__9 if sum of all digits is divisible by num__9 so ( num__2 + num__2 + num__3 + * + num__4 + num__3 + num__1 ) = num__15 + * should be divisible by num__9 num__15 + num__3 will be divisible by num__9 so that least number is num__3 . answer : option a <eor> a <eos> |
a |
subtract__3.0__2.0__ divide__9.0__3.0__ |
subtract__3.0__2.0__ add__1.0__2.0__ |
| what is the sum of the integers from - num__194 to num__195 inclusive ? <o> a ) num__0 <o> b ) num__5 <o> c ) num__195 <o> d ) num__875 <o> e ) num__965 |
sum / n = average . sum = ( average ) ( n ) average = a + b / num__2 = num__194 + num__97.5 = num__0.5 number of items ( n ) = b - a + num__1 = num__195 - ( - num__194 ) + num__1 = num__195 + num__195 = num__390 . sum = average * n = num__0.5 * num__390 = num__195 . answer is c <eor> c <eos> |
c |
divide__195.0__2.0__ reverse__2.0__ subtract__195.0__194.0__ divide__195.0__0.5__ add__194.0__1.0__ |
divide__195.0__2.0__ reverse__2.0__ subtract__195.0__194.0__ divide__195.0__0.5__ add__194.0__1.0__ |
| what is the tens ' digit r of the sum of the first num__40 terms of num__1 num__11 num__111 num__1111 num__11111 num__111111 . . . ? <o> a ) r = num__2 <o> b ) r = num__3 <o> c ) r = num__4 <o> d ) num__8 <o> e ) num__9 |
all of the first num__40 terms have num__1 at the units place . except the first term the tens digit of all the remaining num__39 terms is num__1 . so now if you do the addition num__1 num__1 num__1 num__1 . . . . . . . . . num__1 num__1 num__0 num__1 num__3 num__0 answer is b <eor> b <eos> |
b |
subtract__40.0__1.0__ reverse__111111.0__ multiply__1.0__3.0__ |
subtract__40.0__1.0__ reverse__111111.0__ multiply__1.0__3.0__ |
| a person can swim in still water at num__4 km / h . if the speed of water num__2 km / h how many hours will the man take to swim back against the current for num__6 km ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__2 <o> d ) num__5 <o> e ) num__7 |
m = num__4 s = num__2 us = num__4 - num__2 = num__2 d = num__6 t = num__3.0 = num__3 answer : a <eor> a <eos> |
a |
divide__6.0__2.0__ round__3.0__ |
divide__6.0__2.0__ subtract__6.0__3.0__ |
| eight people are planning to share equally the cost of a rental car . if one person withdraws from the arrangement and the others share equally the entire cost of the car then the share of each of the remaining persons increased by : <o> a ) num__1 by num__7 <o> b ) num__0.125 <o> c ) num__0.111111111111 <o> d ) num__0.875 <o> e ) num__0.2 |
original share of num__1 person = num__0.125 new share of num__1 person = num__0.142857142857 increase = ( num__0.142857142857 - num__0.125 ) = num__0.0178571428571 required fraction = ( num__0.0178571428571 ) / ( num__0.125 ) = ( num__0.0178571428571 ) * num__8 = num__0.142857142857 answer : a <eor> a <eos> |
a |
multiply__0.125__0.1429__ reverse__0.125__ reverse__1.0__ |
multiply__0.125__0.1429__ reverse__0.125__ reverse__1.0__ |
| a solid yellow stripe is to be painted in the middle of a certain highway . if num__1 gallon of paint covers an area of r square feet of highway how many gallons of paint will be needed to paint a stripe of t inches wide on a stretch of highway m miles long ? ( num__1 mile = num__5280 feet and num__1 foot = num__12 inches ) <o> a ) ( num__5280 mt ) / num__12 r <o> b ) ( num__5280 pt ) / num__12 m <o> c ) ( num__5280 pmt ) / num__12 <o> d ) ( num__5280 ) ( num__12 m ) / pt <o> e ) ( num__5280 ) ( num__12 p ) / mt |
given that : num__1 gallon of paint covers an area ofrsquare feet . question : how many gallonsof paint will be needed . . . in any case you will have : ( total area in square feet ) / ( gallons per feet ) = ( total area in square feet ) / r so r must be in the denominator : eliminate all but a and d . now lets see where should be t : ( area in square feet ) = ( width in feet ) * ( length in feet ) - - > width = tinchesas num__1 feet = num__12 inchesthent inches = t / num__12 feet so ( area in square feet ) = ( t / num__12 ) * ( length in feet ) so t must be in the nominator : only a is left . answer : a . <eor> a <eos> |
a |
round__5280.0__ |
multiply__1.0__5280.0__ |
| the ratio of men to women in the snyder community choir is num__4 to num__7 . the ratio of men to women in the leigh community choir is num__5 to num__6 . if the two choirs merged the ratio of men to women in the combined choir would be num__22 to num__27 . if snyder has num__4 more men and num__6 more women than leigh how many women are in the snyder choir ? <o> a ) num__20 <o> b ) num__24 <o> c ) num__28 <o> d ) num__32 <o> e ) num__34 |
this question has a number of patterns in it that we can take advantage of . we can also test the answers ( but we can actually eliminate most of the answer choices rather quickly . we ' re told that the ratio of the men to women in the snyder choir is num__4 to num__7 so the number of men must be a multiple of num__4 and the number of women must be a multiple of num__4 the question asks how many women are in the snyder choir so the answer must be a multiple of num__7 . we can eliminate answersa b d and e ( since they ' re not multiples of num__7 answer c <eor> c <eos> |
c |
multiply__4.0__7.0__ |
multiply__4.0__7.0__ |
| if ‘ + ’ means ‘ × ’ ‘ - ‘ means ‘ + ’ ‘ × ’ means ‘ ÷ ’ and ‘ ÷ ’ means ‘ - ‘ then find the value of ; num__6 – num__9 + num__8 × num__3 ÷ num__25 = <o> a ) num__6 <o> b ) num__10 <o> c ) num__12 <o> d ) num__5 <o> e ) num__8 |
d num__5 <eor> d <eos> |
d |
subtract__8.0__3.0__ subtract__8.0__3.0__ |
subtract__8.0__3.0__ subtract__8.0__3.0__ |
| the length of the rectangular field is double its width . inside the field there is square shaped pond num__8 m long . if the area of the pond is num__0.125 of the area of the field . what is the length of the field ? <o> a ) num__22 <o> b ) num__32 <o> c ) num__77 <o> d ) num__66 <o> e ) num__8 |
a / num__8 = num__8 * num__8 = > a = num__8 * num__8 * num__8 x * num__2 x = num__8 * num__8 * num__8 x = num__16 = > num__2 x = num__32 nswer : b <eor> b <eos> |
b |
multiply__8.0__2.0__ square_perimeter__8.0__ square_perimeter__8.0__ |
multiply__8.0__2.0__ multiply__16.0__2.0__ multiply__16.0__2.0__ |
| in store a there are num__10 pairs of pants for every num__50 store b has . the price ratio between the pants in store b and the pants in store a is num__3 : num__4 . if all the pants were sold in both places until the stock ran out what is the ratio between the total amount stores a earned to the total amount store b earned ? <o> a ) num__3 : num__16 . <o> b ) num__2 : num__3 . <o> c ) num__4 : num__15 . <o> d ) num__3 : num__4 . <o> e ) num__2 : num__5 . |
num__1 st statement : ratio of pants store a : store b num__10 x : num__50 x x : num__5 x price : num__4 y : num__3 y total revenue num__4 xy : num__15 xy num__4 : num__15 answer : c <eor> c <eos> |
c |
subtract__4.0__3.0__ divide__50.0__10.0__ add__10.0__5.0__ add__3.0__1.0__ |
subtract__4.0__3.0__ divide__50.0__10.0__ add__10.0__5.0__ add__3.0__1.0__ |
| find the value of x in num__5 ( x + num__4 ) = num__25 <o> a ) num__1 <o> b ) num__1.5 <o> c ) num__2 <o> d ) num__2.5 <o> e ) num__3 |
num__5 ( x + num__4 ) = num__25 dividing both sides by num__5 gives ( x + num__4 ) = num__5 subtracing both sides with num__4 gives x = num__1 answer : a <eor> a <eos> |
a |
subtract__5.0__4.0__ reverse__1.0__ |
subtract__5.0__4.0__ reverse__1.0__ |
| when positive integer d is divided by positive integer b the result is num__4.35 . which of the following could be the reminder when d is divided by b ? <o> a ) num__13 <o> b ) num__14 <o> c ) num__15 <o> d ) num__16 <o> e ) num__17 |
the remainder will be obtained from the decimal part when d is divided by b i . e . num__0.35 num__0.35 = num__0.35 = num__0.35 so possible remainders are num__714 num__2128 . only option b - num__14 satisfies this ps : for b - num__14 d = num__174 and b = num__40 <eor> b <eos> |
b |
divide__174.0__4.35__ multiply__40.0__0.35__ |
divide__174.0__4.35__ multiply__40.0__0.35__ |
| jancy invested an amount of rs . num__9000 at the rate of num__10.0 p . a simple interest and another amount at the rate of num__15.0 p . a . simple interest . the total interest earned at the end of one year on the total amount invested became num__14.0 p . a . find the total amount invested ? <o> a ) num__15800 <o> b ) num__12800 <o> c ) num__11800 <o> d ) num__16800 <o> e ) num__17800 |
let the second amount be rs . x . then ( num__9000 * num__10 * num__1 ) / num__100 + ( x * num__15 * num__1 ) / num__100 = [ ( num__12000 + x ) * num__14 * num__1 ] / num__100 num__9000 + num__15 x = num__16800 + num__14 x x = num__7800 total investment = num__10000 + num__7800 = rs . num__17800 . answer : e <eor> e <eos> |
e |
percent__100.0__17800.0__ |
percent__100.0__17800.0__ |
| num__1600 men have provisions for num__28 days in the temple . if after num__4 days num__400 men leave the temple how long will the food last now ? <o> a ) num__76 days <o> b ) num__98 days <o> c ) num__32 days <o> d ) num__45 days <o> e ) num__24 days |
num__1600 - - - - num__28 days num__1600 - - - - num__24 num__1200 - - - - ? num__1600 * num__24 = num__1200 * x x = num__32 days answer : c <eor> c <eos> |
c |
subtract__28.0__4.0__ subtract__1600.0__400.0__ add__28.0__4.0__ round__32.0__ |
subtract__28.0__4.0__ subtract__1600.0__400.0__ add__28.0__4.0__ round__32.0__ |
| calculate the largest num__3 digit number which is exactly divisible by num__89 ? <o> a ) num__911 <o> b ) num__969 <o> c ) num__979 <o> d ) num__977 <o> e ) num__971 |
largest num__4 digit number is num__999 after doing num__999 ÷ num__89 we get remainder num__20 hence largest num__3 digit number exactly divisible by num__89 = num__999 - num__20 = num__979 c <eor> c <eos> |
c |
subtract__999.0__20.0__ subtract__999.0__20.0__ |
subtract__999.0__20.0__ subtract__999.0__20.0__ |
| robert is travelling on his cycle and has calculated to reach point a at num__2 p . m . if he travels at num__10 km / hr ; he will reach there at num__12 noon if he travels at num__15 km / hr . at what speed must he travel to reach a at num__1 p . m . ? <o> a ) num__17 kmph <o> b ) num__19 kmph <o> c ) num__15 kmph <o> d ) num__12 kmph <o> e ) num__16 kmph |
d num__12 kmph let the distance traveled be x km . then x / num__10 - x / num__15 = num__2 num__3 x - num__2 x = num__60 = > x = num__60 km . time taken to travel num__60 km at num__10 km / hr = num__6.0 = num__6 hrs . so robert started num__6 hours before num__2 . p . m . i . e . at num__8 a . m . required speed = num__12.0 = num__12 kmph . <eor> d <eos> |
d |
add__2.0__1.0__ hour_to_min_conversion__ multiply__2.0__3.0__ add__2.0__6.0__ round__12.0__ |
subtract__15.0__12.0__ hour_to_min_conversion__ divide__12.0__2.0__ subtract__10.0__2.0__ divide__12.0__1.0__ |
| find the constant k so that : - x num__2 - ( k + num__10 ) x - num__8 = - ( x - num__2 ) ( x - num__4 ) <o> a ) num__11 <o> b ) num__12 <o> c ) num__16 <o> d ) num__14 <o> e ) num__15 |
- x num__2 - ( k + num__10 ) x - num__8 = - ( x - num__2 ) ( x - num__4 ) : given - x num__2 - ( k + num__10 ) x - num__8 = - x num__2 + num__6 x - num__8 - ( k + num__10 ) = num__6 : two polynomials are equal if their corresponding coefficients are equal . k = - num__16 : solve the above for k correct answer c <eor> c <eos> |
c |
add__2.0__4.0__ multiply__2.0__8.0__ multiply__2.0__8.0__ |
add__2.0__4.0__ add__10.0__6.0__ add__10.0__6.0__ |
| multiply : ( x – num__4 ) ( x + num__5 ) <o> a ) ( a ) x num__2 + num__5 x - num__20 <o> b ) ( b ) x num__2 - num__4 x - num__20 <o> c ) ( c ) x num__2 - x - num__20 <o> d ) ( d ) x num__2 + x - num__20 <o> e ) ( e ) x num__2 + x + num__20 |
( x – num__4 ) ( x + num__5 ) . = x ( x + num__5 ) - num__4 ( x + num__5 ) . = x num__2 + num__5 x – num__4 x – num__20 . = x num__2 + x - num__20 the answer is ( d ) <eor> d <eos> |
d |
multiply__4.0__5.0__ subtract__4.0__2.0__ |
multiply__4.0__5.0__ subtract__4.0__2.0__ |
| a computer store originally bought num__1000 modems at a total cost of d dollars . if each modem was sold for num__25 percent more than its original cost what was the individual price in terms of d of each modem sold ? <o> a ) d / num__800 <o> b ) num__5 d / num__1000 <o> c ) num__125 d <o> d ) d / num__1000 + num__25 <o> e ) num__125 / ( num__1000 d ) |
since you ' re choosing to approach this algebraically i ' m going to add some labeling to your work . . . . num__1000 purchased for a total of d dollars . . . . . each modem costs d / num__1000 dollars . . . . each modem is then sold for num__25.0 above the cost : ( num__1.25 ) ( d / num__1000 ) = num__1.25 d / num__1000 num__1.25 d / num__1000 = num__125 d / num__100000 num__125 / num__1000 = num__0.125 so . . . . num__125 / num__100000 = num__0.00125 so the sell price of each modem is d / num__800 final answer : a <eor> a <eos> |
a |
divide__125.0__1000.0__ divide__1.25__1000.0__ divide__1000.0__1.25__ divide__1000.0__1.25__ |
divide__125.0__1000.0__ divide__1.25__1000.0__ divide__1000.0__1.25__ divide__1000.0__1.25__ |
| a standard veggiematik machine can chop num__36 carrots in num__4 minutes . how many carrots can num__6 standard veggiematik machines chop in num__4 minutes ? <o> a ) num__36 <o> b ) num__54 <o> c ) num__108 <o> d ) num__216 <o> e ) num__324 |
direct relationship : - num__1 standard veggiematik machine - num__36 carrots - num__4 minutes num__1 standard veggiematik machine - num__9 carrots - num__1 minute now num__6 standard veggiematik machine - ? carrots - num__4 minutes hence = num__9 x num__6 x num__4 = num__216 carrots answer d <eor> d <eos> |
d |
divide__36.0__4.0__ multiply__36.0__6.0__ round__216.0__ |
divide__36.0__4.0__ multiply__36.0__6.0__ round__216.0__ |
| in a num__100 metre race a can allow b to start the race either num__10 metres ahead of the starting point or num__2 seconds before a so that both reach the finishing line at the same time . what is the speed of b in m / sec and how long will a take to run num__100 m ? <o> a ) num__6 m / sec ; num__18 sec <o> b ) num__5 m / s ; num__18 sec <o> c ) num__4 m / s ; num__15 sec <o> d ) num__3 m / s ; num__12 secs <o> e ) num__2 m / s ; num__6 secs |
a can allow b to start the race num__10 metres ahead so the time a need to cover num__100 meters equals to the time b needs to cover num__90 meters - - > num__100 / a = num__90 / b where a and b are the rates of a and b respectively . a can allow b to start the race num__2 seconds before a so the time a need to cover num__100 meters is num__2 seconds less than the time b needs to cover num__100 meters - - > num__100 / a + num__2 = num__100 / b . subtract one from another : num__2 = num__100 / b - num__90 / b - - > b = num__5 . answer : b . or directly : b needs num__2 seconds more to cover num__100 meters than to cover num__90 meters - - > num__2 = num__100 / b - num__90 / b - - > b = num__5 . <eor> b <eos> |
b |
subtract__100.0__10.0__ divide__10.0__2.0__ round__5.0__ |
subtract__100.0__10.0__ divide__10.0__2.0__ divide__10.0__2.0__ |
| in a num__4000 meter race around a circular stadium having a circumference of num__1000 meters the fastest runner and the slowest runner reach the same point at the end of the num__5 th minute for the first time after the start of the race . all the runners have the same starting point and each runner maintains a uniform speed throughout the race . if the fastest runner runs at twice the speed of the slowest runner what is the time taken by the fastest runner to finish the race ? <o> a ) num__20 min <o> b ) num__15 min <o> c ) num__10 min <o> d ) num__5 min <o> e ) none of these |
explanation : the ratios of the speed of the fastest and the slowest is num__2 : num__1 . hence they should meet only at one point at the circumference i . e the starting point ( as the difference in the ratios in reduced form is num__1 ) . for the two of them to meet for the first time the faster shuld have completed one complete round over the slower one . since the two of them meet for the first time after num__5 min the faster one should have completed num__2 rounds ( i . e num__2000 m ) and the slower one should have completed num__1 round ( num__1000 in this time ) . thus the faster one would complete the race ( i . e num__4000 m ) in num__10 min . answer : c <eor> c <eos> |
c |
divide__4000.0__2.0__ multiply__5.0__2.0__ round__10.0__ |
divide__4000.0__2.0__ multiply__5.0__2.0__ round__10.0__ |
| a sun is divided among x y and z in such a way that for each rupee x gets y gets num__45 paisa and z gets num__30 paisa . if the share of y is rs . num__54 what is the total amount ? <o> a ) num__166 <o> b ) num__105 <o> c ) num__178 <o> d ) num__177 <o> e ) num__210 |
x : y : z = num__100 : num__45 : num__30 num__20 : num__9 : num__6 num__9 - - - num__54 num__35 - - - ? = > num__210 answer : e <eor> e <eos> |
e |
subtract__54.0__45.0__ divide__54.0__9.0__ multiply__35.0__6.0__ multiply__35.0__6.0__ |
subtract__54.0__45.0__ divide__54.0__9.0__ multiply__35.0__6.0__ multiply__35.0__6.0__ |
| two pipes a and b can fill a cistern in num__15 and num__30 minutes respectively and a third pipe c can empty it in num__40 minutes . how long will it take to fill the cistern if all the three are opened at the same time ? <o> a ) num__17 num__0.25 min <o> b ) num__28 num__0.142857142857 min <o> c ) num__17 num__0.142857142857 min <o> d ) num__13 num__0.428571428571 min <o> e ) num__17 num__0.714285714286 min |
num__0.0666666666667 + num__0.0333333333333 - num__0.025 = num__0.075 num__13.3333333333 = num__13 num__0.428571428571 answer : d <eor> d <eos> |
d |
round__13.0__ |
round__13.0__ |
| a can do a job in num__18 days and b can do it in num__30 days . in how many days will they finish it together ? <o> a ) num__0.111111111111 <o> b ) num__0.0888888888889 <o> c ) num__0.285714285714 <o> d ) num__2.0 <o> e ) num__2.5 |
explanation : num__0.0555555555556 + num__0.0333333333333 = num__0.0888888888889 = num__0.0888888888889 so both together will finish the work in num__0.0888888888889 days . answer : b <eor> b <eos> |
b |
add__0.0556__0.0333__ add__0.0556__0.0333__ |
add__0.0556__0.0333__ add__0.0556__0.0333__ |
| a specialized type of sand consists of num__40.0 mineral x by volume and num__60.0 mineral y by volume . if mineral x weighs num__4 grams per cubic centimeter and mineral y weighs num__3 grams per cubic centimeter how many grams does a cubic meter of specialized sand combination weigh ? ( num__1 meter = num__100 centimeters ) <o> a ) num__5 num__500000 <o> b ) num__2 num__400000 <o> c ) num__55000 <o> d ) num__28000 <o> e ) num__280 |
let the volume be num__1 m ^ num__3 = num__1 m * num__1 m * num__1 m = num__100 cm * num__100 cm * num__100 cm = num__1 num__000000 cm ^ num__3 by volume num__40.0 is x = num__400000 cm ^ num__3 num__60.0 is y = num__600000 cm ^ num__3 by weight in num__1 cm ^ num__3 x is num__4 gms in num__400000 cm ^ num__3 x = num__4 * num__400000 = num__1 num__600000 grams in num__1 cm ^ num__3 y is num__3 gms in num__600000 cm ^ num__3 y = num__3 * num__600000 = num__1 num__800000 gms total gms in num__1 m ^ num__3 = num__1 num__600000 + num__1 num__800000 = num__2 num__400000 answer : b <eor> b <eos> |
b |
triangle_area__3.0__400000.0__ triangle_area__4.0__400000.0__ triangle_area__4.0__1.0__ triangle_area__4.0__1.0__ |
triangle_area__3.0__400000.0__ triangle_area__4.0__400000.0__ triangle_area__4.0__1.0__ power__2.0__1.0__ |
| two negative numbers are multiplied to give a product of num__48 . if the lesser number is num__10 less than thrice the greater number what is the greater number ? <o> a ) - num__6 <o> b ) - num__8 <o> c ) - num__5 <o> d ) - num__9 <o> e ) - num__10 |
test the options . the options give you the greater number . ( a ) - num__6 triple of - num__6 is - num__18 and num__10 less is - num__8 . - num__8 * - num__6 = num__48 ( correct ) correct answer ( a ) <eor> a <eos> |
a |
divide__48.0__6.0__ divide__48.0__8.0__ |
subtract__18.0__10.0__ divide__48.0__8.0__ |
| a number is said to be prime saturated if the product of all the different positive prime factors of e is less than the square root of e . what is the greatest two digit prime saturated integer ? <o> a ) num__99 <o> b ) num__98 <o> c ) num__97 <o> d ) num__96 <o> e ) num__95 |
e = num__96 = num__3 * num__32 = num__3 * num__2 ^ num__5 answer is d . <eor> d <eos> |
d |
divide__96.0__3.0__ add__2.0__3.0__ multiply__32.0__3.0__ |
divide__96.0__3.0__ add__2.0__3.0__ multiply__32.0__3.0__ |
| a can finish a work in num__18 days and b can do the same work in num__15 days . b worked for num__10 days and left the job . in how many days a alone can finish the remaining work ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__3 <o> d ) num__2 <o> e ) num__1 |
b ' s num__10 day ' s work = num__0.666666666667 = num__0.666666666667 remaining work = ( num__1 - ( num__0.666666666667 ) ) = num__0.333333333333 now num__0.0555555555556 work is done by a in num__1 day . therefore num__0.333333333333 work is done by a in num__18 * ( num__0.333333333333 ) = num__6 days . answer b <eor> b <eos> |
b |
divide__10.0__15.0__ subtract__1.0__0.6667__ divide__1.0__18.0__ round__6.0__ |
divide__10.0__15.0__ subtract__1.0__0.6667__ divide__1.0__18.0__ round__6.0__ |
| john got a monthly raise of $ num__237 . if he gets paid every other week write an integer describing how the raise will affect his paycheck . <o> a ) num__327 <o> b ) num__723 <o> c ) num__732 <o> d ) num__231 <o> e ) num__237 |
let the num__1 st paycheck be x ( integer ) . john got a monthly raise of $ num__237 . so after completing the num__1 st month he will get $ ( x + num__237 ) . similarly after completing the num__2 nd month she will get $ ( x + num__237 ) + $ num__237 . = $ ( x + num__237 + num__237 ) = $ ( x + num__474 ) so in this way end of every month his salary will increase by ( e ) $ num__237 . <eor> e <eos> |
e |
multiply__237.0__2.0__ multiply__237.0__1.0__ |
multiply__237.0__2.0__ multiply__237.0__1.0__ |
| num__1 num__9 num__17 num__33 num__49 num__73 ? <o> a ) num__89 <o> b ) num__69 <o> c ) num__97 <o> d ) num__88 <o> e ) num__60 |
c num__97 the pattern is + num__8 + num__8 + num__16 + num__24 . . . <eor> c <eos> |
c |
subtract__9.0__1.0__ subtract__17.0__1.0__ subtract__33.0__9.0__ multiply__1.0__97.0__ |
subtract__9.0__1.0__ subtract__17.0__1.0__ add__8.0__16.0__ add__73.0__24.0__ |
| a train num__700 m long is running at a speed of num__78 km / hr . if it crosses a tunnel in num__1 min then the length of the tunnel is ? <o> a ) num__277 m <o> b ) num__700 m <o> c ) num__600 m <o> d ) num__187 m <o> e ) num__1678 m |
speed = num__78 * num__0.277777777778 = num__21.6666666667 m / sec . time = num__1 min = num__60 sec . let the length of the train be x meters . then ( num__700 + x ) / num__60 = num__21.6666666667 x = num__600 m . answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ round__600.0__ |
hour_to_min_conversion__ multiply__1.0__600.0__ |
| p is num__30.0 more efficient than q . p can complete a work in num__23 days . if p and q work together how much time will it take to complete the same work ? <o> a ) num__7 days <o> b ) num__8 days <o> c ) num__13 <o> d ) num__5 days <o> e ) num__2 days |
explanation : work done by p in num__1 day = num__0.0434782608696 let work done by q in num__1 day = q q × ( num__1.3 ) = num__0.0434782608696 = > q = num__100 / ( num__23 × num__130 ) = num__10 / ( num__23 × num__13 ) work done by p and q in num__1 day = num__0.0434782608696 + num__10 / ( num__23 × num__13 ) = num__23 / ( num__23 × num__13 ) = num__0.0769230769231 = > p and q together can do the work in num__13 days answer : c <eor> c <eos> |
c |
divide__1.0__23.0__ add__30.0__100.0__ subtract__23.0__10.0__ divide__1.0__13.0__ round__13.0__ |
divide__1.0__23.0__ add__30.0__100.0__ divide__130.0__10.0__ divide__1.0__13.0__ divide__130.0__10.0__ |
| find the length of the wire required to go num__10 times round a square field containing num__53824 m num__2 . <o> a ) num__15840 <o> b ) num__9280 <o> c ) num__2667 <o> d ) num__8766 <o> e ) num__66711 |
a num__2 = num__53824 = > a = num__232 num__4 a = num__928 num__928 * num__10 = num__9280 answer : b <eor> b <eos> |
b |
square_perimeter__232.0__ multiply__10.0__928.0__ multiply__10.0__928.0__ |
multiply__4.0__232.0__ multiply__10.0__928.0__ multiply__10.0__928.0__ |
| in each term of a sequence num__7 is added to get the next term . if the first term is num__3 what is the thirty - seventh term ? <o> a ) num__222 <o> b ) num__284 <o> c ) num__252 <o> d ) num__263 <o> e ) num__255 |
num__1 rst term + num__36 terms = num__3 + num__7 + num__7 + num__7 + num__7 + num__7 + num__7 + num__7 + . . . + num__7 ( num__36 times ) num__3 + ( num__7 x num__36 ) = num__3 + num__252 = num__255 answer e <eor> e <eos> |
e |
multiply__7.0__36.0__ add__3.0__252.0__ add__3.0__252.0__ |
multiply__7.0__36.0__ add__3.0__252.0__ add__3.0__252.0__ |
| sandy is younger than molly by num__12 years . if the ratio of their ages is num__7 : num__9 how old is sandy ? <o> a ) num__42 <o> b ) num__49 <o> c ) num__56 <o> d ) num__63 <o> e ) num__70 |
let sandy ' s age be num__7 x and let molly ' s age be num__9 x . num__9 x - num__7 x = num__12 x = num__6 sandy is num__42 years old . the answer is a . <eor> a <eos> |
a |
multiply__7.0__6.0__ multiply__7.0__6.0__ |
multiply__7.0__6.0__ multiply__7.0__6.0__ |
| in how many different number of ways a committee of num__2 person of can be selected from num__2 boys and num__2 girls such that at least num__1 girl is included in the committe <o> a ) num__3 <o> b ) num__4 <o> c ) num__2 <o> d ) num__5 <o> e ) num__9 |
num__1 g num__1 b num__2 g = ( num__2 c num__1 * num__2 c num__1 ) + num__2 c num__2 = ( num__2 * num__2 ) + num__1 = num__5 total num__2 m num__2 w num__4 c num__2 = num__6 num__2 c num__2 = num__1 at least one girl = total - with out girl at least one girl = num__6 - num__1 = num__5 d ) <eor> d <eos> |
d |
vowel_space__ die_space__ vowel_space__ |
vowel_space__ die_space__ vowel_space__ |
| if n is a natural number then num__6 n ^ num__2 + num__6 n is always divisible by ? <o> a ) num__6 only <o> b ) num__6 and num__12 <o> c ) num__12 only <o> d ) num__18 only <o> e ) num__20 only |
num__6 n ^ num__2 + num__6 n = num__6 n ( n + num__1 ) which is always divisible by num__6 and num__12 both since n ( n + num__1 ) is always even . answer is b <eor> b <eos> |
b |
multiply__6.0__2.0__ multiply__6.0__1.0__ |
multiply__6.0__2.0__ multiply__6.0__1.0__ |
| a board num__5 ft . num__2 inches long is divided into num__3 equal parts . what is the length of each part ? <o> a ) num__5 ft . num__7 inches <o> b ) num__3 ft . num__7 inches <o> c ) num__4 ft . num__7 inches <o> d ) num__1 ft . num__8.7 inches <o> e ) num__1 ft . num__7 inches |
length of board = num__5 ft . num__2 inches = ( num__5 * num__12 + num__2 ) inches = num__62 inches . therefore length of each part = ( num__20.6666666667 ) inches = num__20.7 inches = num__1 ft . num__8.7 inches answer is d . <eor> d <eos> |
d |
divide__62.0__3.0__ subtract__3.0__2.0__ subtract__20.7__12.0__ reverse__1.0__ |
divide__62.0__3.0__ subtract__3.0__2.0__ subtract__20.7__12.0__ reverse__1.0__ |
| a train is moving at a speed of num__50 km / hr and its length is num__500 m . find the time taken by it to pass a man standing near the railway line ? <o> a ) num__30 sec <o> b ) num__45 sec <o> c ) num__36 sec <o> d ) num__29 sec <o> e ) num__52 sec |
speed of the train = num__50 * num__0.277777777778 = num__13.8888888889 m / sec distance moved in passing the standing man = num__500 m required time taken = num__500 / ( num__13.8888888889 ) = num__500 * num__0.072 = num__36 sec answer is c <eor> c <eos> |
c |
multiply__500.0__0.072__ round__36.0__ |
multiply__500.0__0.072__ multiply__500.0__0.072__ |
| william is mixing up a salad dressing . regardless of the number of servings the recipe requires that num__0.625 of the finished dressing mix be coconut oil num__0.25 vinegar and the remainder an even mixture of salt pepper and sugar . if william accidentally doubles the vinegar and forgets the sugar altogether what proportion of the botched dressing will be coconut oil ? <o> a ) num__0.51724137931 <o> b ) num__0.625 <o> c ) num__0.3125 <o> d ) num__0.5 <o> e ) num__0.481481481481 |
coconut oil = num__0.625 = num__0.625 - - > num__15 parts out of num__24 ; vinegar = num__0.25 = num__0.25 - - > num__6 parts out of num__24 ; salt + pepper + sugar = num__1 - ( num__0.625 + num__0.25 ) = num__0.125 so each = num__0.0416666666667 - - > num__1 part out of num__24 each ; if vinegar = num__12 ( instead of num__6 ) and sugar = num__0 ( instead of num__1 ) then total = num__15 + num__12 + num__1 + num__1 + num__0 = num__29 parts out of which num__15 parts are coconut oil - - > proportion = num__0.51724137931 . answer : a . <eor> a <eos> |
a |
divide__15.0__0.625__ multiply__0.25__24.0__ reverse__24.0__ round_down__0.625__ divide__15.0__29.0__ multiply__1.0__0.5172__ |
divide__15.0__0.625__ multiply__0.25__24.0__ reverse__24.0__ round_down__0.625__ divide__15.0__29.0__ multiply__1.0__0.5172__ |
| if rs . num__450 amount to rs . num__540 in num__4 years what will it amount to in num__6 years at the same rate % per annum ? <o> a ) num__589 <o> b ) num__580 <o> c ) num__585 <o> d ) num__582 <o> e ) num__523 |
num__90 = ( num__450 * num__4 * r ) / num__100 r = num__5.0 i = ( num__450 * num__6 * num__5 ) / num__100 = num__135 num__450 + num__135 = num__585 answer : c <eor> c <eos> |
c |
percent__100.0__585.0__ |
percent__100.0__585.0__ |
| a train num__105 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is : <o> a ) num__22 <o> b ) num__50 <o> c ) num__42.8 <o> d ) num__288 <o> e ) num__12 |
speed of the train relative to man = ( num__10.5 ) m / sec = ( num__10.5 ) m / sec . [ ( num__10.5 ) * ( num__3.6 ) ] km / hr = num__37.8 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__37.8 = = > x = num__42.8 km / hr . answer : c <eor> c <eos> |
c |
divide__105.0__10.0__ multiply__10.5__3.6__ add__5.0__37.8__ round__42.8__ |
divide__105.0__10.0__ multiply__10.5__3.6__ add__5.0__37.8__ round__42.8__ |
| if the sides of a triangle are num__26 cm num__24 cm and num__10 cm what is its area ? <o> a ) num__120 cm num__2 <o> b ) num__765 cm num__2 <o> c ) num__216 cm num__2 <o> d ) num__197 cm num__2 <o> e ) num__275 cm num__2 |
the triangle with sides num__26 cm num__24 cm and num__10 cm is right angled where the hypotenuse is num__26 cm . area of the triangle = num__0.5 * num__24 * num__10 = num__120 cm num__2 answer : a <eor> a <eos> |
a |
triangle_area__24.0__10.0__ square_perimeter__0.5__ triangle_area__24.0__10.0__ |
volume_rectangular_prism__24.0__10.0__0.5__ square_perimeter__0.5__ volume_rectangular_prism__24.0__10.0__0.5__ |
| in a kilometer race a beats b by num__80 meters or num__10 seconds . what time does a take to complete the race ? <o> a ) num__180 sec <o> b ) num__190 sec <o> c ) num__115 sec <o> d ) num__490 sec <o> e ) num__390 sec |
time taken by b run num__1000 meters = ( num__1000 * num__10 ) / num__80 = num__125 sec . time taken by a = num__125 - num__10 = num__115 sec . answer : c <eor> c <eos> |
c |
subtract__125.0__10.0__ round__115.0__ |
subtract__125.0__10.0__ subtract__125.0__10.0__ |
| how many numbers from num__10 to num__1000 are exactly divisible by num__9 ? <o> a ) num__100 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__14 |
num__1.11111111111 = num__1 and num__111.111111111 = num__111 = = > num__111 - num__1 = num__100 . therefore num__100 answer : a <eor> a <eos> |
a |
divide__10.0__9.0__ round_down__1.1111__ divide__1000.0__9.0__ round_down__111.1111__ divide__1000.0__10.0__ divide__1000.0__10.0__ |
divide__10.0__9.0__ subtract__10.0__9.0__ divide__1000.0__9.0__ round_down__111.1111__ divide__1000.0__10.0__ divide__1000.0__10.0__ |
| in a certain sequence the first term is num__4 and each successive term is num__1 more than the reciprocal of the term that immediately precedes it . what is the fifth term in this sequence ? <o> a ) num__1.625 <o> b ) num__1.64285714286 <o> c ) num__1.6 <o> d ) num__0.625 <o> e ) num__0.615384615385 |
let five terms in the sequence be a b c d e a = num__4 b = num__1 + num__0.25 = num__1.25 c = num__1 + num__0.8 = num__1.8 d = num__1 + num__0.555555555556 = num__1.55555555556 e = num__1 + num__0.642857142857 = num__1.64285714286 hence answer should be b . <eor> b <eos> |
b |
reverse__4.0__ add__1.0__0.25__ reverse__1.25__ add__1.0__0.8__ reverse__1.8__ add__1.0__0.5556__ add__1.0__0.6429__ multiply__1.0__1.6429__ |
reverse__4.0__ add__1.0__0.25__ reverse__1.25__ add__1.0__0.8__ reverse__1.8__ add__1.0__0.5556__ add__1.0__0.6429__ add__1.0__0.6429__ |
| toby is four years younger than debby . thrice the sum of the ages of toby and debby equals their mother ’ s age . if the age of the mother is num__54 find the ages of toby and debby ? <o> a ) num__10 and num__14 <o> b ) num__5 and num__9 <o> c ) num__5 and num__10 <o> d ) num__6 and num__10 <o> e ) num__7 and num__11 |
let the age of debby be x and toby be x - num__4 num__3 ( x + x - num__4 ) = num__54 x = num__11 the ages of toby and debby are num__7 and num__11 . answer : e <eor> e <eos> |
e |
subtract__11.0__4.0__ subtract__11.0__4.0__ |
subtract__11.0__4.0__ subtract__11.0__4.0__ |
| what is the lowest positive integer that is divisible by each of the integers num__1 through num__11 inclusive ? <o> a ) num__420 <o> b ) num__840 <o> c ) num__1260 <o> d ) num__2520 <o> e ) num__27 |
720 |
we have to find the lcm of num__1 num__2 num__3 num__2 ^ num__2 num__5 num__2 * num__3 num__7 num__2 ^ num__3 num__3 ^ num__2 num__2 * num__5 and num__11 . the lcm is num__1 * num__2 ^ num__3 * num__3 ^ num__2 * num__5 * num__7 * num__11 = num__27720 the answer is e . <eor> e <eos> |
e |
e |
| a tap can fill a tank in num__10 hours . after half the tank is filled two more similar taps are opened . what is the total time taken to fill the tank completely ? <o> a ) num__1 hr num__20 min <o> b ) num__4 hr <o> c ) num__3 hr <o> d ) num__6 hr num__40 min <o> e ) num__2 hr num__50 min |
explanation : num__1 tap can fill the tank in num__10 hours . therefore num__1 tap can fill half tank in num__5 hours . num__3 taps can fill the tank in num__3.33333333333 hour . therefore num__3 taps can fill half tank in num__1.66666666667 hour = num__1 hour num__40 minutes . total time taken = num__6 hour num__40 minutes answer : option d <eor> d <eos> |
d |
divide__10.0__3.0__ subtract__5.0__3.3333__ add__1.0__5.0__ round__6.0__ |
divide__10.0__3.0__ subtract__5.0__3.3333__ add__1.0__5.0__ round__6.0__ |
| a train num__240 m long passed a pole in num__24 sec . how long will it take to pass a platform num__650 m long ? <o> a ) num__22 sec <o> b ) num__89 sec <o> c ) num__77 sec <o> d ) num__99 sec <o> e ) num__55 sec |
speed = num__10.0 = num__10 m / sec . required time = ( num__240 + num__650 ) / num__10 = num__89 sec . answer : b <eor> b <eos> |
b |
divide__240.0__24.0__ round__89.0__ |
divide__240.0__24.0__ round__89.0__ |
| john buys a bike for rs . num__4700 and spends rs . num__800 on its repairs . if he sells the scooter for rs . num__5800 what is his gain percent ? <o> a ) num__5 num__0.555555555556 % <o> b ) num__5 num__0.454545454545 % <o> c ) num__6.0 <o> d ) num__6 num__0.545454545455 % <o> e ) num__6 num__0.538461538462 % |
cost price ( c . p . ) = rs . ( num__4700 + num__800 ) = rs . num__5500 . selling price ( s . p . ) = rs . num__5800 . gain = ( s . p . ) - ( c . p . ) = rs . ( num__5800 - num__5500 ) = rs . num__300 . gain % = num__300 x num__100.0 = num__5 num__0.454545454545 num__5500.0 answer b <eor> b <eos> |
b |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| if the speed of a man is num__51 km per hour then what is the distance traveled by him in num__30 seconds ? <o> a ) num__275 m <o> b ) num__360 m <o> c ) num__375 m <o> d ) num__425 m <o> e ) num__440 m |
the distance traveled in num__30 sec = num__51 * ( num__0.277777777778 ) * num__30 = num__425 m answer : d <eor> d <eos> |
d |
round__425.0__ |
round__425.0__ |
| a work as fast as b . if b can complete a work in num__16 days independently the number of days in which a and b can together finish the work in ? <o> a ) num__2 days <o> b ) num__3 days <o> c ) num__4 days <o> d ) num__5 num__0.333333333333 days <o> e ) num__6 days |
ratio of rates of working of a and b = num__2 : num__1 ratio of times taken = num__1 : num__2 a ' s num__1 day work = num__0.125 b ' s num__1 day work = num__0.0625 a + b num__1 day work = num__0.125 + num__0.0625 = num__0.1875 = > num__5.33333333333 = num__5 num__0.333333333333 a and b can finish the work in num__5 num__0.333333333333 days answer is d <eor> d <eos> |
d |
divide__2.0__16.0__ divide__0.125__2.0__ add__0.125__0.0625__ divide__1.0__0.1875__ multiply__0.0625__5.3333__ round__5.0__ |
divide__2.0__16.0__ divide__0.125__2.0__ add__0.125__0.0625__ divide__1.0__0.1875__ multiply__0.0625__5.3333__ round__5.0__ |
| working alone at its constant rate machine a produces x boxes in num__5 minutes and working alone at its constant rate machine b produces num__2 x boxes in num__10 minutes . how many minutes does it take machines a and b working simultaneously at their respective constant rates to produce num__3 x boxes ? <o> a ) num__3 minutes <o> b ) num__7.5 minutes <o> c ) num__5 minutes <o> d ) num__6 minutes <o> e ) num__12 minutes |
rate = work / time given rate of machine a = num__2 x / num__10 min machine b produces num__2 x boxes in num__10 min hence machine b produces num__2 x boxes in num__10 min . rate of machine b = num__2 x / num__10 we need tofind the combined time that machines a and b working simultaneouslytakeat their respective constant rates let ' s first find the combined rate of machine a and b rate of machine a = num__2 x / num__10 min + rate of machine b = num__2 x / num__10 = num__4 x / num__10 now combine time = combine work needs to be done / combine rate = num__3 x / num__4 x * num__10 = num__7.5 min ans : b <eor> b <eos> |
b |
round__7.5__ |
round__7.5__ |
| mudit ' s age num__22 years hence will be thrice his age four years ago . find mudit ' s present age ? <o> a ) num__12 <o> b ) num__17 <o> c ) num__27 <o> d ) num__18 <o> e ) num__15 |
explanation : let mudit ' s present age be ' m ' years . m + num__22 = num__3 ( m - num__4 ) = > num__2 m = num__34 = > m = num__17 years . answer : b <eor> b <eos> |
b |
divide__34.0__2.0__ divide__34.0__2.0__ |
divide__34.0__2.0__ subtract__34.0__17.0__ |
| a man sells two articles for rs . num__3600 each and he gains num__30.0 on the first and loses num__30.0 on the next . find his total gain or loss ? <o> a ) num__9.0 loss <o> b ) num__2.0 loss <o> c ) num__7.0 loss <o> d ) num__8.0 loss <o> e ) num__1.0 loss |
explanation : ( num__30 * num__30 ) / num__100 = num__9.0 loss answer : a <eor> a <eos> |
a |
percent__9.0__100.0__ |
percent__9.0__100.0__ |
| the simple interest accrued on an amount rs . num__10000 at the end of two years is same as the compound interest on rs . num__8000 at the end of two years . the rate of interest is same in both the cases . what is the rate of interest ? <o> a ) num__80.0 p . a <o> b ) num__100.0 p . a . <o> c ) num__40.0 p . a <o> d ) num__44.0 p . a <o> e ) num__50.0 p . a |
given that rs . num__10000 is invested in s . i for two years and rs . num__8000 in c . i for two years = > c . i - s . i = > num__9000 { [ num__1 + r / num__100 ] num__2 - num__1 } = ( num__10000 ) num__2 r / num__100 = > num__9 { num__1 + num__2 r / num__100 + r num__2 / ( num__100 ) num__2 - num__1 } = r / num__5 = > r = num__0.0 of num__40.0 since r ! = num__0.0 r = num__40.0 answer : c <eor> c <eos> |
c |
percent__1.0__10000.0__ percent__100.0__40.0__ |
percent__1.0__10000.0__ percent__100.0__40.0__ |
| at a local appliance manufacturing facility the workers received a num__30.0 hourly pay raise due to extraordinary performance . if one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged by approximately what percent would he reduce the number of hours that he worked ? <o> a ) num__83.0 <o> b ) num__80.0 <o> c ) num__23.0 <o> d ) num__17.0 <o> e ) num__12 % |
let ' s say he works usually num__10 hours and earns num__100 per hour . num__10 * num__100 = num__1000 num__10 * num__130 = num__1300 ( this are the new earnings after the raise ) to figure out how much he needs to work with the new salary in order to earn the original num__1000 : num__7.69230769231 = num__7.67 so he can reduce his work by num__2.33 hours . which is > num__23.0 . answer c <eor> c <eos> |
c |
percent__100.0__23.0__ |
percent__100.0__23.0__ |
| in what ratio must a grocer mix two varieties of pulses costing rs . num__15 and rs . num__20 per kg respectively to obtain a mixture worth rs . num__16.50 per kg ? <o> a ) num__1 : num__2 <o> b ) num__2 : num__1 <o> c ) num__3 : num__7 <o> d ) num__7 : num__3 <o> e ) none of these |
explanation : by the rule of alligation we have cp of num__1 kg of num__1 st variety pulse cp of num__1 kg of num__2 nd variety pulse num__15 num__20 mean price num__16.5 num__20 - num__16.5 = num__3.5 num__16.5 - num__15 = num__1.5 required ratio = num__3.5 : num__1.5 = num__35 : num__15 = num__7 : num__3 answer : option d <eor> d <eos> |
d |
subtract__20.0__16.5__ subtract__16.5__15.0__ add__15.0__20.0__ multiply__2.0__3.5__ round_down__3.5__ multiply__1.0__7.0__ |
subtract__20.0__16.5__ subtract__16.5__15.0__ add__15.0__20.0__ multiply__2.0__3.5__ round_down__3.5__ multiply__1.0__7.0__ |
| in how many ways can the word leading arranged if vowels always come together ? <o> a ) num__120 <o> b ) num__720 <o> c ) num__60 <o> d ) num__8220 <o> e ) num__100 |
leading = has num__3 vowels ( eai ) and num__4 consonant ( ldng ) total letters = num__7 for consonant = num__5 ! ( that is num__4 + num__1 letters . one comes from the vowels ) = num__5 * num__4 * num__3 * num__2 * num__1 = num__120 ways for vowels = num__3 ! = num__3 * num__2 * num__1 = num__6 total number of ways = num__120 * num__6 = num__720 ways option b <eor> b <eos> |
b |
vowel_space__ coin_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
vowel_space__ coin_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| find the value of ( num__875 num__0.259176863181 ) × num__899 <o> a ) num__786600 <o> b ) num__786658 <o> c ) num__786679 <o> d ) num__786858 <o> e ) num__786890 |
( num__875 num__0.259176863181 ) × num__899 ( num__786625 + num__233 ) / num__899 × num__899 num__875.259176863 × num__899 num__786858 d ) <eor> d <eos> |
d |
multiply__875.0__899.0__ add__875.0__0.2592__ add__786625.0__233.0__ add__786625.0__233.0__ |
multiply__875.0__899.0__ add__875.0__0.2592__ add__786625.0__233.0__ add__786625.0__233.0__ |
| kim finds a num__2 - meter tree branch and marks it off in thirds and fifths . she then breaks the branch along all the markings and removes one piece of every distinct length . what fraction of the original branch remains ? <o> a ) num__0.6 <o> b ) num__0.4 <o> c ) num__1.4 <o> d ) num__0.5 <o> e ) num__0.533333333333 |
num__3 pieces of num__0.2 length and two piece each of num__0.0666666666667 and num__0.133333333333 lengths . removing one piece each from pieces of each kind of lengths the all that will remain will be num__2 pieces of num__0.2 i . e num__0.4 num__1 piece of num__0.0666666666667 and num__1 piece of num__0.133333333333 which gives us num__0.4 + num__0.0666666666667 + num__0.133333333333 - - - - - > num__0.6 answer is a <eor> a <eos> |
a |
divide__0.2__3.0__ subtract__0.2__0.0667__ multiply__2.0__0.2__ subtract__3.0__2.0__ km_to_mile_conversion__ km_to_mile_conversion__ |
divide__0.2__3.0__ subtract__0.2__0.0667__ multiply__2.0__0.2__ subtract__3.0__2.0__ subtract__1.0__0.4__ subtract__1.0__0.4__ |
| the speed of a boat in upstream is num__60 kmph and the speed of the boat downstream is num__80 kmph . find the speed of the boat in still water and the speed of the stream ? <o> a ) num__10 kmph <o> b ) num__11 kmph <o> c ) num__16 kmph <o> d ) num__18 kmph <o> e ) num__19 kmph |
speed of the boat in still water = ( num__60 + num__80 ) / num__2 = num__70 kmph . speed of the stream = ( num__80 - num__60 ) / num__2 = num__10 kmph . answer : a <eor> a <eos> |
a |
subtract__80.0__70.0__ round__10.0__ |
subtract__80.0__70.0__ subtract__80.0__70.0__ |
| a room num__5 m num__55 cm long and num__3 m num__74 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor ? <o> a ) num__222 <o> b ) num__267 <o> c ) num__176 <o> d ) num__287 <o> e ) num__261 |
explanation : area of the room = ( num__544 * num__374 ) size of largest square tile = h . c . f of num__544 & num__374 = num__34 cm area of num__1 tile = ( num__34 x num__34 ) number of tiles required = = [ ( num__544 x num__374 ) / ( num__34 x num__34 ) ] = num__176 answer : c ) num__176 <eor> c <eos> |
c |
round__176.0__ |
divide__176.0__1.0__ |
| a car moves at num__80 km / hr . what is the speed of the car in meters per second ? <o> a ) num__2029 msec <o> b ) num__2229 msec <o> c ) num__2429 msec <o> d ) num__2629 msec <o> e ) none of these |
explana Ɵ on : speed = ( num__80 ∗ num__518 ) m / sec = num__2009 m / sec = num__2229 msec answer : b <eor> b <eos> |
b |
round__2229.0__ |
round__2229.0__ |
| a leak in the bottom of a tank can empty the full tank in num__6 hours . an inlet pipe fills water at the rate of num__4.5 liters per minute . when the tank is full in inlet is opened and due to the leak the tank is empties in num__8 hours . the capacity of the tank is ? <o> a ) num__5729 <o> b ) num__5760 <o> c ) num__2889 <o> d ) num__6480 <o> e ) num__2799 |
num__1 / x - num__0.166666666667 = - num__0.125 x = num__24 hrs num__24 * num__60 * num__4.5 = num__6480 . answer : d <eor> d <eos> |
d |
divide__1.0__6.0__ divide__1.0__8.0__ hour_to_min_conversion__ round__6480.0__ |
divide__1.0__6.0__ divide__1.0__8.0__ hour_to_min_conversion__ divide__6480.0__1.0__ |
| a certain characteristic in a large population has a distribution that is symmetric about the mean m . if num__68 percent of the distribution lies within one standard deviation d of the mean what percent e of the distribution is less than m + d ? <o> a ) num__16.0 <o> b ) num__32.0 <o> c ) num__48.0 <o> d ) num__84.0 <o> e ) num__92 % |
d the prompt says that num__68.0 of the population lies between m - d and m + d . thus num__32.0 of the population is less than m - d or greater than m + d . since the population is symmetric half of this num__32.0 is less than m - d and half is greater than m + d . thus e = ( num__68 + num__16 ) % or ( num__100 - num__16 ) % of the population is less than m + d . d <eor> d <eos> |
d |
add__68.0__32.0__ add__68.0__16.0__ |
add__68.0__32.0__ add__68.0__16.0__ |
| if p and r are integers and p ^ num__2 = num__14 r then r must be divisible by which of the following ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__5 <o> d ) num__7 <o> e ) num__14 |
ans : d solution : for p to be an int num__14 r must be whole square of a number . num__14 r = num__7 * num__2 * r to make it whole square we need num__7 so r can must be divisible by num__7 y where y is itself a whole square . so d is the ans <eor> d <eos> |
d |
divide__14.0__2.0__ divide__14.0__2.0__ |
divide__14.0__2.0__ divide__14.0__2.0__ |
| a train covers a certain distance at a speed of num__320 kmph in num__4 hours . to cover the same distance in num__2 hours it must travel at a speed of <o> a ) num__850 km / hr <o> b ) num__800 km / hr <o> c ) num__740 km / hr <o> d ) num__640 km / hr <o> e ) num__600 km / hr |
explanation : distance = num__320 × num__4 = num__1280 km required speed = ( num__640.0 ) = num__640 km / hr answer : option d <eor> d <eos> |
d |
multiply__320.0__4.0__ multiply__320.0__2.0__ round__640.0__ |
multiply__320.0__4.0__ divide__1280.0__2.0__ divide__1280.0__2.0__ |
| the cross - section of a cannel is a trapezium in shape . if the cannel is num__11 m wide at the top and num__9 m wide at the bottom and the area of cross - section is num__960 sq m the depth of cannel is ? <o> a ) num__69 <o> b ) num__92 <o> c ) num__94 <o> d ) num__49 <o> e ) num__96 |
num__0.5 * d ( num__11 + num__9 ) = num__960 d = num__96 answer : e <eor> e <eos> |
e |
round__96.0__ |
round__96.0__ |
| by selling num__45 lemons for rs num__40 a man loses num__20.0 . how many should he sell for rs num__24 to gain num__20.0 in the transaction ? <o> a ) num__16 <o> b ) num__18 <o> c ) num__20 <o> d ) num__22 <o> e ) num__25 |
explanation : let s . p . of num__45 lemons be rs . x . then num__80 : num__40 = num__120 : x or x = num__40 x num__1.5 = num__60 for rs . num__60 lemons sold = num__45 for rs . num__24 lemons sold = num__0.75 x num__24 = num__18 . answer is b <eor> b <eos> |
b |
percent__45.0__40.0__ percent__45.0__40.0__ |
percent__45.0__40.0__ percent__45.0__40.0__ |
| if rs . num__1440 / - are divided among a b and c so that a receives num__0.333333333333 rd as much as b and b receives num__0.25 th as much as c . the amount b received is ? <o> a ) num__250 <o> b ) num__270 <o> c ) num__310 <o> d ) num__330 <o> e ) num__350 |
a : b : c = num__1 : num__3 : num__12 total parts = num__16 b ' s share is = num__3 parts num__16 - - - - - > num__1440 num__1 - - - - - > num__90 num__3 - - - - - > num__270 ( b ' s share is num__270 ) b ) <eor> b <eos> |
b |
divide__3.0__0.25__ divide__1440.0__16.0__ multiply__3.0__90.0__ multiply__1.0__270.0__ |
divide__3.0__0.25__ divide__1440.0__16.0__ multiply__3.0__90.0__ multiply__1.0__270.0__ |
| a football player scores num__5 goals in his fifth match thus increasing his average goals score by num__0.2 . the total number of goals in his num__5 matches would be <o> a ) num__14 <o> b ) num__16 <o> c ) num__18 <o> d ) num__10 <o> e ) num__21 |
while this question can be solved with a rather straight - forward algebra approach ( as the other posters have noted ) it can also be solved by testing the answers . one of those numbers must be the total number of goals . . . from a tactical standpoint it ' s best to test either answer b or answer d so if the answer is not correct then you would have a gauge for whether you should gohigherorlowerwith your next test . here i ' ll start with answer e = num__21 goals if . . . . total goals = num__21 goals num__5 th game = num__5 goals num__1 st num__4 games = num__16 goals avg . for num__1 st num__4 games = num__4.0 = num__4 goal / game avg . for all num__5 games = num__4.2 = num__4.2 goals / game this is an exact match for what we ' re told in the prompt so answer e must be the answer . <eor> e <eos> |
e |
multiply__5.0__0.2__ subtract__5.0__1.0__ subtract__21.0__5.0__ add__0.2__4.0__ add__5.0__16.0__ |
multiply__5.0__0.2__ subtract__5.0__1.0__ subtract__21.0__5.0__ divide__21.0__5.0__ divide__4.2__0.2__ |
| if p ( a ) = num__0.571428571429 and p ( b ) = num__0.4 find p ( a n b ) if a and b are independent events . <o> a ) num__0.28 <o> b ) num__0.12 <o> c ) num__0.228571428571 <o> d ) num__0.153846153846 <o> e ) num__0.176470588235 |
p ( a n b ) = p ( a ) . p ( b ) p ( a n b ) = num__0.571428571429 . num__0.4 p ( a n b ) = num__0.228571428571 . c <eor> c <eos> |
c |
multiply__0.5714__0.4__ multiply__0.5714__0.4__ |
multiply__0.5714__0.4__ multiply__0.5714__0.4__ |
| an amount at compound interest sums to rs . num__17640 / - in num__2 years and to rs . num__20286 / - in num__3 years at the same rate of interest . find the rate percentage ? <o> a ) num__5.0 <o> b ) num__7.0 <o> c ) num__9.0 <o> d ) num__15.0 <o> e ) num__12 % |
explanation : the difference of two successive amounts must be the simple interest in num__1 year on the lower amount of money . s . i = num__20286 / - - num__17640 / - = rs . num__2646 / - rate of interest = ( num__0.15 ) × ( num__100.0 ) = > num__15.0 answer : option d <eor> d <eos> |
d |
subtract__3.0__2.0__ subtract__20286.0__17640.0__ divide__2646.0__17640.0__ multiply__100.0__0.15__ multiply__1.0__15.0__ |
subtract__3.0__2.0__ subtract__20286.0__17640.0__ divide__2646.0__17640.0__ multiply__100.0__0.15__ divide__15.0__1.0__ |
| if the given two numbers are respectively num__6.0 and num__30.0 of a third number then what percentage is the first of the second ? <o> a ) num__20.0 <o> b ) num__25.0 <o> c ) num__18.0 <o> d ) num__30.0 <o> e ) none of these |
here l = num__6 and m = num__30 therefore first number = l / m x num__100.0 of second number = num__0.2 x num__100.0 of second number = num__20.0 of second number answer : a <eor> a <eos> |
a |
percent__20.0__100.0__ |
percent__20.0__100.0__ |
| a man is riding a bike with front and back wheel circumference of num__40 inches and num__70 inches respectively . if the man rides the bike on a straight road without slippage how many inches will the man have travelled when the front wheel has made num__15 revolutions more than the back wheel ? <o> a ) num__3388 <o> b ) num__2882 <o> c ) num__1400 <o> d ) num__277 <o> e ) num__2282 |
explanation : given the ratio of the circumference of front wheel is num__40 inches and back wheel is num__70 inches distance covered = circumference of the wheel à — no . of revolutions made by the wheel if n is the number of revolutions made by back wheel the number of revolutions made by front wheel is n + num__15 distance covered by both the wheels is the same num__40 * ( n + num__3 ) = num__70 n n = num__20 front wheel : back wheel circumference num__40 : num__70 revolutions num__35 : num__20 distance covered \ small \ rightarrow num__40 à — num__35 = num__70 à — num__20 = num__1400 inches . answer : c <eor> c <eos> |
c |
add__15.0__20.0__ multiply__40.0__35.0__ multiply__40.0__35.0__ |
add__15.0__20.0__ multiply__40.0__35.0__ multiply__40.0__35.0__ |
| the incomes of two persons a and b are in the ratio num__3 : num__4 . if each saves rs . num__100 per month the ratio of their expenditures is num__1 : num__2 . find their incomes ? <o> a ) num__150 num__208 <o> b ) num__150 num__206 <o> c ) num__150 num__200 <o> d ) num__150 num__202 <o> e ) num__150 num__201 |
the incomes of a and b be num__3 p and num__4 p . expenditures = income - savings ( num__3 p - num__100 ) and ( num__4 p - num__100 ) the ratio of their expenditure = num__1 : num__2 ( num__3 p - num__100 ) : ( num__4 p - num__100 ) = num__1 : num__2 num__2 p = num__100 = > p = num__50 their incomes = num__150 num__200 answer : c <eor> c <eos> |
c |
divide__100.0__2.0__ multiply__3.0__50.0__ multiply__4.0__50.0__ multiply__3.0__50.0__ |
divide__100.0__2.0__ multiply__3.0__50.0__ multiply__4.0__50.0__ subtract__200.0__50.0__ |
| what is the product of the greatest num__2 digit multiple of num__15 and the greatest num__2 digit prime number ? <o> a ) num__9312 <o> b ) num__9408 <o> c ) num__9506 <o> d ) num__9603 <o> e ) num__8730 |
the greatest num__2 digit multiple of num__15 : num__90 the greatest num__2 digit prime numebr : num__97 num__97 * num__90 . num__8730 e <eor> e <eos> |
e |
multiply__97.0__90.0__ multiply__97.0__90.0__ |
multiply__97.0__90.0__ multiply__97.0__90.0__ |
| in the science city kolkata the rate of the ticket is increased by num__50.0 to increased the revenue but simultaneously num__20.0 of the visitor decreased . what is percentage change in the revenue . if it is known that the science city collects one revenue only from the visitors and it has no other financial supports : <o> a ) + num__20.0 <o> b ) - num__25.0 <o> c ) + num__305 <o> d ) - num__30.0 <o> e ) can not determined |
solution : let the initial revenue be num__100 . num__100 - - - - - num__50.0 ↑ ( ticket up ) - - - - - > num__150 - - - - - num__20.0 ↓ ( visitors down ) - - - - - > num__120 . there is num__20.0 increase in the revenue . answer : option a <eor> a <eos> |
a |
percent__20.0__100.0__ |
percent__20.0__100.0__ |
| a person can row at num__9 kmph and still water . he takes num__6 num__0.5 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is num__1 kmph ? <o> a ) num__60 km <o> b ) num__87 km <o> c ) num__89 km <o> d ) num__29 km <o> e ) num__20 km |
let the distance between a and b be x km . total time = x / ( num__9 + num__1 ) + x / ( num__9 - num__1 ) = num__6.5 = > x / num__10 + x / num__8 = num__6.5 = > ( num__4 x + num__5 x ) / num__40 = num__6.5 = > x = num__29 km . answer : d <eor> d <eos> |
d |
add__6.0__0.5__ add__9.0__1.0__ subtract__9.0__1.0__ multiply__0.5__8.0__ subtract__9.0__4.0__ multiply__4.0__10.0__ round__29.0__ |
add__6.0__0.5__ add__9.0__1.0__ subtract__9.0__1.0__ subtract__10.0__6.0__ subtract__9.0__4.0__ multiply__4.0__10.0__ divide__29.0__1.0__ |
| two twins sisters sita and geeta were standing back to back and suddenly they started running in opposite directions for num__10 km each . then they turned left and ran for another num__7.5 km . what is the distance ( in kilometers ) between the the two twins when they stop ? <o> a ) num__21 <o> b ) num__23 <o> c ) num__25 <o> d ) num__27 <o> e ) num__30 |
the distance between them is the hypotenuse of a right angle triangle with sides num__15 km and num__20 km . the hypotenuse = sqrt ( num__15 ^ num__2 + num__20 ^ num__2 ) = num__25 the answer is c . <eor> c <eos> |
c |
divide__20.0__10.0__ add__10.0__15.0__ round__25.0__ |
divide__20.0__10.0__ add__10.0__15.0__ add__10.0__15.0__ |
| a rectangular courtyard num__3.78 m lang and num__5.25 m broad is to be paved exactly with square tiles all of the same size . the minimum number of such tiles is : <o> a ) num__425 <o> b ) num__430 <o> c ) num__440 <o> d ) num__450 <o> e ) num__460 |
l = num__378 cm and b = num__525 cm maximum length of a square tile = hcf of ( num__378525 ) = num__21 cm number of tiles = ( num__378 × num__525 ) / ( num__21 × num__21 ) = ( num__18 × num__25 ) = num__450 answer is d . <eor> d <eos> |
d |
square_perimeter__5.25__ multiply__18.0__25.0__ multiply__18.0__25.0__ |
square_perimeter__5.25__ multiply__18.0__25.0__ multiply__18.0__25.0__ |
| each of the three people individually can complete a certain job in num__3 num__4 and num__6 hours respectively . what is the lowest fraction of the job that can be done in num__1 hour by num__2 of the people working together at their respective rates ? <o> a ) num__0.333333333333 <o> b ) num__0.45 <o> c ) num__0.533333333333 <o> d ) num__0.416666666667 <o> e ) num__0.666666666667 |
the two slowest people work at rates of num__0.25 and num__0.166666666667 of the job per hour . the sum of these rates is num__0.25 + num__0.166666666667 = num__0.416666666667 of the job per hour . the answer is d . <eor> d <eos> |
d |
divide__1.0__4.0__ divide__1.0__6.0__ add__0.25__0.1667__ multiply__1.0__0.4167__ |
divide__1.0__4.0__ divide__1.0__6.0__ add__0.25__0.1667__ add__0.25__0.1667__ |
| the average weight of num__8 person ' s increases by num__2.5 kg when a new person comes in place of one of them weighing num__65 kg . what might be the weight of the new person ? <o> a ) num__65 kg <o> b ) num__70 kg <o> c ) num__75 kg <o> d ) num__80 kg <o> e ) num__85 kg |
total weight increased = ( num__8 x num__2.5 ) kg = num__20 kg . weight of new person = ( num__65 + num__20 ) kg = num__85 kg . answer : e <eor> e <eos> |
e |
multiply__8.0__2.5__ add__65.0__20.0__ add__65.0__20.0__ |
multiply__8.0__2.5__ add__65.0__20.0__ add__65.0__20.0__ |
| a and b started a partnership business investing capital in the ratio of num__3 : num__5 . c joined in the partnership after six months with an amount equal to that of b . at the end of one year the profit should be distributed among a b and c in - - - proportion . <o> a ) num__6 : num__10 : num__5 <o> b ) num__5 : num__10 : num__6 <o> c ) num__10 : num__5 : num__6 <o> d ) num__10 : num__6 : num__5 <o> e ) num__5 : num__6 : num__10 |
initial investment capital ratio of a and b = num__3 : num__5 let initial capital of a and b be num__3 x and num__5 x respectively . amount that c invested after num__6 months = num__5 x ( since it is equal to b ' s investment ) ratio in which profit should be distributed after num__1 year = num__3 x × num__12 : num__5 x × num__12 : num__5 x × num__6 = num__3 × num__12 : num__5 × num__12 : num__5 × num__6 = num__3 × num__2 : num__5 × num__2 : num__5 = num__6 : num__10 : num__5 answer is a . <eor> a <eos> |
a |
subtract__6.0__5.0__ subtract__3.0__1.0__ multiply__5.0__2.0__ multiply__3.0__2.0__ |
subtract__6.0__5.0__ subtract__3.0__1.0__ multiply__5.0__2.0__ multiply__3.0__2.0__ |
| a bullet train covers a distance in num__50 min if it runs at a speed of num__48 kmph on an average . the speed at which the bullet train must run to reduce the time of journey to num__40 min will be . <o> a ) num__80 kmph <o> b ) num__40 kmph <o> c ) num__60 kmph <o> d ) num__90 kmph <o> e ) num__30 kmph |
c num__60 kmph time = num__0.833333333333 hr = num__0.833333333333 hr speed = num__48 mph distance = s * t = num__48 * num__0.833333333333 = num__40 km time = num__0.666666666667 hr = num__0.666666666667 hr new speed = num__40 * num__1.5 kmph = num__60 kmph <eor> c <eos> |
c |
hour_to_min_conversion__ divide__50.0__60.0__ divide__40.0__60.0__ add__0.8333__0.6667__ hour_to_min_conversion__ |
hour_to_min_conversion__ divide__50.0__60.0__ divide__40.0__60.0__ add__0.8333__0.6667__ multiply__40.0__1.5__ |
| four extra - large sandwiches of exactly the same size were ordered for m students where m > num__9 . three of the sandwiches were evenly divided among the students . since num__4 students did not want any of the fourth sandwich it was evenly divided among the remaining students . if carol ate one piece from each of the four sandwiches the amount of sandwich that she ate would be what fraction of a whole extra - large sandwich ? <o> a ) ( num__4 m - num__27 ) / [ m ( m - num__3 ) ] <o> b ) ( num__4 m - num__27 ) / [ m ( m - num__6 ) ] <o> c ) ( num__4 m - num__27 ) / [ m ( m - num__9 ) ] <o> d ) ( num__4 m - num__27 ) / [ m ( m - num__8 ) ] <o> e ) ( num__4 m - num__20 ) / [ m ( m - num__10 ) ] |
three of the sandwiches were evenly divided among the students i . e . amount of sandwiches with each students = num__3 / m amount of num__4 th sandwich with remaining ( m - num__9 ) students = num__1 / ( m - num__9 ) the amount of sandwich that carol ate = num__3 / m + num__1 / ( m - num__9 ) = ( num__3 m - num__27 + m ) / [ m ( m - num__9 ) ] = ( num__4 m - num__27 ) / [ m ( m - num__9 ) ] answer : c <eor> c <eos> |
c |
subtract__4.0__3.0__ multiply__9.0__3.0__ multiply__4.0__1.0__ |
subtract__4.0__3.0__ multiply__9.0__3.0__ divide__4.0__1.0__ |
| if a light flashes every num__6 seconds how many times will it flash in num__0.25 of an hour ? <o> a ) num__51 <o> b ) num__151 <o> c ) num__251 <o> d ) num__351 <o> e ) num__451 |
in num__0.25 of an hour there are num__15 * num__60 = num__900 seconds the number of num__6 - second intervals = num__150.0 = num__150 after the first flash there will be num__150 more flashes for a total of num__151 . the answer is b . <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__15.0__60.0__ divide__900.0__6.0__ round__151.0__ |
hour_to_min_conversion__ multiply__15.0__60.0__ divide__900.0__6.0__ round__151.0__ |
| a person covered one - fourth of the total distance at num__17 kmph and remaining distance at num__24 kmph . what is the average speed for the total distance ? <o> a ) num__21 ( num__0.125 ) <o> b ) num__21 ( num__0.76 ) <o> c ) num__21 ( num__0.32 ) <o> d ) num__29 ( num__0.04 ) <o> e ) num__21 ( num__0.16 ) |
let the total distance be x km total time taken = ( x / num__4 ) / num__17 + ( num__3 x / num__4 ) / num__24 = num__25 x / num__544 average speed = x / ( num__25 x / num__544 ) = = num__21 ( num__0.76 ) kmph . answer : b <eor> b <eos> |
b |
add__17.0__4.0__ add__17.0__4.0__ |
add__17.0__4.0__ add__17.0__4.0__ |
| dhanumjay invested an amount of rs . num__17400 for two years . find the rate of compound interest that will fetch him an amount of rs . num__1783.50 at the end of two years ? <o> a ) num__8.0 p . a . <o> b ) num__5.0 p . a . <o> c ) num__6.0 p . a . <o> d ) num__4.0 p . a . <o> e ) num__3.0 p . a |
explanation : let the rate of interest be r % p . a . num__17400 [ num__1 + r / num__100 ] num__2 = num__17400 + num__1783.50 [ num__1 + r / num__100 ] num__2 = ( num__17400 + num__1783.50 ) / num__17400 = num__1 + num__0.1025 = num__1 + num__0.1025 = num__1.1025 = [ num__1.05 ] num__2 [ num__1 + r / num__100 ] = num__1.05 r / num__100 = num__0.05 ∴ r = num__5 answer : option b <eor> b <eos> |
b |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| a father said to his son ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is num__38 years now the son ' s age five years back was : <o> a ) num__14 years <o> b ) num__19 years <o> c ) num__33 years <o> d ) num__38 years <o> e ) num__39 years |
let the son ' s present age be x years . then ( num__38 - x ) = x num__2 x = num__38 . x = num__19 . son ' s age num__5 years back ( num__19 - num__5 ) = num__14 years . answer : a <eor> a <eos> |
a |
divide__38.0__2.0__ subtract__19.0__5.0__ subtract__19.0__5.0__ |
divide__38.0__2.0__ subtract__19.0__5.0__ subtract__19.0__5.0__ |
| how many anagrams can we make with the word ' cacao ' ? <o> a ) num__5 ! / ( num__3 ! ) num__2 <o> b ) num__5 ! / ( num__2 ! ) num__2 num__2 ! <o> c ) num__5 ! / ( num__3 ! ) num__2 num__3 ! <o> d ) num__4 ! / ( num__2 ! ) num__2 <o> e ) num__5 ! / ( num__2 ! ) num__2 |
n items of which p are alike of one kind q alike of the other r alike of another kind and the remaining are distinct can be arranged in a row in n ! / p ! q ! r ! ways . the letter pattern ' cacao ' consists of num__5 letters of which there are num__2 c ' s num__2 a ' s and num__1 o . number of arrangements = num__5 ! / ( num__2 ! ) num__2 answer : e <eor> e <eos> |
e |
multiply__1.0__5.0__ |
divide__5.0__1.0__ |
| if num__9 is a factor of num__2 k then which of the following may not be an integer ? <o> a ) num__6 k / num__54 + num__2 k / num__3 <o> b ) ( num__4 k - num__18 ) / num__9 <o> c ) ( num__2 k + num__27 ) / num__9 <o> d ) ( num__81 - num__4 k ^ num__2 ) / num__81 <o> e ) ( num__2 k - num__3 ) / num__3 |
i got the right answer a after plugging in num__18 for k . while going through kaplan ' s explanation i could n ' t understand the part about choice a . ` ` num__6 k / num__54 + num__2 k / num__3 = num__0.5 * num__2 k / num__9 + num__2 k / num__3 . since num__9 is a factor of num__2 k num__3 is also factor of num__2 k . so num__2 k / num__3 is an integer as is num__2 k / num__9 . but num__0.5 = num__0.5 so if num__2 k / num__9 is not even the expression num__0.5 * num__2 k / num__9 will not be even . ' ' but is n ' t num__2 k / num__9 always even ? num__9 is a factor of num__2 k which means that we could have num__2 ( num__9 ) num__2 ( num__2 ) ( num__9 ) num__2 ( num__3 ) ( num__9 ) and so forth . the num__9 in the denominator cancels out the num__9 in the numerator . so we are left with num__2 * something . so num__2 k / num__9 is even . a <eor> a <eos> |
a |
multiply__9.0__2.0__ multiply__9.0__6.0__ subtract__9.0__6.0__ reverse__2.0__ subtract__9.0__3.0__ |
multiply__9.0__2.0__ multiply__9.0__6.0__ divide__6.0__2.0__ reverse__2.0__ multiply__2.0__3.0__ |
| if the selling price of num__50 books is equal to the cost price of num__40 books then the loss or gain percent is : <o> a ) num__10 <o> b ) num__30 <o> c ) num__20 <o> d ) num__40 <o> e ) num__50 |
c num__20.0 let c . p . of each books be $ num__1 . then c . p . of num__50 books = $ num__50 ; s . p . of num__50 books = $ num__40 . loss % = num__0.2 * num__100 = num__20.0 <eor> c <eos> |
c |
percent__50.0__40.0__ percent__1.0__20.0__ percent__50.0__40.0__ |
percent__50.0__40.0__ percent__1.0__20.0__ percent__50.0__40.0__ |
| at a certain resort each of the num__39 food service employees is trained to work in a minimum of num__1 restaurant and a maximum of num__3 restaurants . the num__3 restaurants are the family buffet the dining room and the snack bar . exactly num__19 employees are trained to work in the family buffet num__18 are trained to work in the dining room and num__12 are trained to work in the snack bar . if num__2 employees are trained to work in exactly num__2 restaurants how many employees are trained to work in all num__3 restaurants ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
num__39 = num__19 + num__18 + num__12 - num__2 - num__2 x num__2 x = num__19 + num__18 + num__12 - num__2 - num__39 = num__45 - num__37 = num__8 x = num__4 c <eor> c <eos> |
c |
subtract__39.0__2.0__ subtract__45.0__37.0__ add__1.0__3.0__ round__4.0__ |
subtract__39.0__2.0__ subtract__45.0__37.0__ add__1.0__3.0__ add__1.0__3.0__ |
| the area of a circle of radius num__5 is numerically what percent of its circumference ? <o> a ) num__200 <o> b ) num__225 <o> c ) num__240 <o> d ) num__250 <o> e ) none |
solution required % = [ Π x ( num__5 ) num__1.0 Π x num__5 x num__100 ] % ‹ = › num__250.0 . answer d <eor> d <eos> |
d |
triangle_area__5.0__100.0__ triangle_area__5.0__100.0__ |
triangle_area__5.0__100.0__ triangle_area__5.0__100.0__ |
| the length of the bridge which a train num__130 meters long and travelling at num__45 km / hr can cross in num__30 seconds is ? <o> a ) num__297 m <o> b ) num__178 m <o> c ) num__245 m <o> d ) num__278 m <o> e ) num__297 m |
speed = ( num__45 * num__0.277777777778 ) m / sec = ( num__12.5 ) m / sec . time = num__30 sec . let the length of bridge be x meters . then ( num__130 + x ) / num__30 = num__12.5 = = > num__2 ( num__130 + x ) = num__750 = = > x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| how many paying stones each measuring num__2 m * num__2 m are required to pave a rectangular court yard num__30 m long and num__18 m board ? <o> a ) num__99 <o> b ) num__18 <o> c ) num__135 <o> d ) num__17 <o> e ) num__12 |
num__30 * num__18 = num__2 * num__2 * x = > x = num__135 answer : c <eor> c <eos> |
c |
round__135.0__ |
round__135.0__ |
| company kw is being sold and both company a and company b were considering the purchase . the price of company kw is num__30.0 more than company a has in assets and this same price is also num__100.0 more than company b has in assets . if companies a and b were to merge and combine their assets the price of company kw would be approximately what percent of these combined assets ? <o> a ) num__66.0 <o> b ) num__79.0 <o> c ) num__86.0 <o> d ) num__116.0 <o> e ) num__150 % |
let the price of company a ' s assets be num__100 price of assets of kw is num__30.0 more than company a ' s assets which is num__130 price of assets of kw is num__100.0 more than company b ' s assets which means price of company b ' s assets is half the price of kw = num__65 a + b = num__165 kw = num__130 kw / ( a + b ) * num__100 = num__0.787878787879 * num__100 = num__78.78 or num__79.0 b <eor> b <eos> |
b |
percent__100.0__79.0__ |
percent__100.0__79.0__ |
| find the distance covered by a man walking for num__72 min at a speed of num__10 km / hr ? <o> a ) num__12 km <o> b ) num__3 km <o> c ) num__4 km <o> d ) num__5 km <o> e ) num__6 km |
distance = num__10 * num__1.2 = num__12 km answer is a <eor> a <eos> |
a |
multiply__10.0__1.2__ round__12.0__ |
multiply__10.0__1.2__ multiply__10.0__1.2__ |
| the sum of the heights of two high - rises is x feet . if the first high rise is num__37 feet taller than the second how tall will the second high rise be ? <o> a ) ( x - num__37 ) <o> b ) num__2 x − ( num__37 + z ) <o> c ) ( x − num__37 ) / num__2 + z <o> d ) x / num__2 - num__37 + z <o> e ) ( num__2 x − num__37 ) / z |
i will note h num__1 the height of high - rise num__1 and h num__2 the height of high - rise num__2 . so : h num__1 + h num__2 = x and h num__1 = h num__2 + num__37 = > h num__2 + h num__2 + num__37 = x = > num__2 h num__2 = x - num__37 = ? h num__2 = ( x - num__37 ) correct answer a <eor> a <eos> |
a |
multiply__37.0__1.0__ |
multiply__37.0__1.0__ |
| cistern may be filled by a tap in num__4 hours while it can be emptied by another tap in num__9 hours . if both the taps are opened simultaneously then after how much time cistern will get filled ? <o> a ) num__8 hrs <o> b ) num__8.5 hrs <o> c ) num__7 hrs <o> d ) num__7.2 hrs <o> e ) num__9 hrs |
filled in num__1 hr = num__0.25 empties in num__1 hr = num__0.111111111111 net filled in num__1 hr = num__0.25 - num__0.111111111111 = = > filled in num__7.2 = = > num__7.2 hrs answer d <eor> d <eos> |
d |
divide__1.0__4.0__ divide__1.0__9.0__ round__7.2__ |
divide__1.0__4.0__ divide__1.0__9.0__ round__7.2__ |
| a man is num__24 years older than his son . in two years his age will be twice the age of his son . what is the present age of his son ? <o> a ) num__23 years <o> b ) num__22 years <o> c ) num__21 years <o> d ) num__20 years <o> e ) num__19 years |
let present age of the son = x years then present age the man = ( x + num__24 ) years given that in num__2 years man ' s age will be twice the age of his son ⇒ ( x + num__24 ) + num__2 = num__2 ( x + num__2 ) ⇒ x = num__22 answer : b <eor> b <eos> |
b |
subtract__24.0__2.0__ subtract__24.0__2.0__ |
subtract__24.0__2.0__ subtract__24.0__2.0__ |
| excluding stoppages the speed of a train is num__45 kmph and including stoppages it is num__32 kmph . of how many minutes does the train stop per hour ? <o> a ) num__17 <o> b ) num__83 <o> c ) num__12 <o> d ) num__83 <o> e ) num__28 |
explanation : t = num__0.288888888889 * num__60 = num__17 answer : option a <eor> a <eos> |
a |
hour_to_min_conversion__ round__17.0__ |
hour_to_min_conversion__ round__17.0__ |
| if num__9 / ( num__1 + num__4 / x ) = num__1 then x = <o> a ) num__3 <o> b ) num__0.5 <o> c ) num__0.333333333333 <o> d ) - num__0.333333333333 <o> e ) - num__3 |
the expression num__9 / ( num__1 + num__4 / x ) = num__1 should have been equal to something . if num__9 / ( num__1 + num__4 / x ) = num__1 = num__1 = > num__9 x / ( x + num__4 ) = num__1 = > num__9 x = x + num__4 = > num__8 x = num__4 = > x = num__0.5 correct option : b <eor> b <eos> |
b |
subtract__9.0__1.0__ divide__4.0__8.0__ subtract__1.0__0.5__ |
subtract__9.0__1.0__ divide__4.0__8.0__ divide__4.0__8.0__ |
| a rectangular lawn of dimensions num__80 m * num__60 m has two roads each num__10 m wide running in the middle of the lawn one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . num__3 per sq m ? <o> a ) rs . num__3600 <o> b ) rs . num__3700 <o> c ) rs . num__3800 <o> d ) rs . num__3900 <o> e ) rs . num__4900 |
area = ( l + b – d ) d ( num__80 + num__60 – num__10 ) num__10 = > num__1300 m num__2 num__1300 * num__3 = rs . num__3900 answer : d <eor> d <eos> |
d |
multiply__3.0__1300.0__ multiply__3.0__1300.0__ |
multiply__3.0__1300.0__ multiply__3.0__1300.0__ |
| in a can there is a mixture of milk and water in the ratio num__5 : num__3 . if the can is filled with an additional num__8 liters of milk the can would be full and the ratio of milk and water would become num__2 : num__1 . find the capacity of the can ? <o> a ) num__56 <o> b ) num__60 <o> c ) num__64 <o> d ) num__68 <o> e ) num__72 |
let c be the capacity of the can . ( num__0.625 ) * ( c - num__8 ) + num__8 = ( num__0.666666666667 ) * c num__15 c - num__120 + num__192 = num__16 c c = num__72 the answer is e . <eor> e <eos> |
e |
divide__5.0__8.0__ divide__2.0__3.0__ multiply__5.0__3.0__ multiply__8.0__15.0__ divide__120.0__0.625__ multiply__8.0__2.0__ subtract__192.0__120.0__ multiply__1.0__72.0__ |
divide__5.0__8.0__ divide__2.0__3.0__ multiply__5.0__3.0__ multiply__8.0__15.0__ divide__120.0__0.625__ multiply__8.0__2.0__ subtract__192.0__120.0__ multiply__1.0__72.0__ |
| if x ^ num__2 = y ^ num__3 where x and y are non negative and non zero integers what is the greatest possible value of x * y from options below ? <o> a ) num__343 <o> b ) num__625 <o> c ) num__243 <o> d ) num__921 <o> e ) num__741 |
c . x = num__27 y = num__9 num__27 ^ num__2 = num__729 num__9 ^ num__3 = num__729 x * y = num__27 * num__9 = num__243 my answer is ( c ) num__243 <eor> c <eos> |
c |
divide__27.0__3.0__ multiply__9.0__27.0__ multiply__9.0__27.0__ |
divide__27.0__3.0__ multiply__9.0__27.0__ multiply__9.0__27.0__ |
| the ratio between the speeds of two trains is num__7 : num__9 . if the second train runs num__540 kms in num__3 hours then the speed of the first train is ? <o> a ) num__140 km / hr <o> b ) num__130 km / hr <o> c ) num__110 km / hr <o> d ) num__120 km / hr <o> e ) none of these |
explanation : let the speeds of two trains be num__7 x and num__9 x km / hr . num__9 / x = num__180.0 = > x = num__20 km / hr so speed of first train is num__20 * num__7 = num__140 km / hr option a <eor> a <eos> |
a |
divide__540.0__3.0__ divide__180.0__9.0__ multiply__7.0__20.0__ round__140.0__ |
divide__540.0__3.0__ divide__180.0__9.0__ multiply__7.0__20.0__ multiply__7.0__20.0__ |
| vehicle x is num__22 miles ahead of vehicle y which is traveling the same direction along the same route as vehicle x . if vehicle x is traveling at an average speed of num__36 miles per hour and vehicle y is traveling at an average speed of num__45 miles per hour how long will it take vehicle y to overtake and drive num__23 miles ahead of vehicle x ? <o> a ) num__5 hours <o> b ) num__7 hours num__20 minutes <o> c ) num__4 hours <o> d ) num__3 hours num__45 minutes <o> e ) num__2 hours num__30 minutes |
relative speed = num__45 - num__36 = num__9 miles per hour dist required = num__22 + num__23 = num__45 miles time taken to overtake = num__5.0 = num__5 hours . a is the answer . <eor> a <eos> |
a |
subtract__45.0__36.0__ divide__45.0__9.0__ round__5.0__ |
subtract__45.0__36.0__ divide__45.0__9.0__ round__5.0__ |
| currently guavas cost num__30 cents / pound . due to a disease affecting the guava trees it is expected that next month guavas will cost num__50.0 more than they do currently . how much are guavas expected to cost next month ? <o> a ) num__50 cents / pound <o> b ) num__45 cents / pound <o> c ) num__55 cents / pound <o> d ) num__60 cents / pound <o> e ) num__65 cents / pound |
if a new cost is p percent greater than the old cost then ( new cost ) = ( old cost ) + ( p / num__100 ) ( old cost ) . in this case ( new cost ) = num__30 cents / pound + ( num__0.5 ) ( num__30 cents / pound ) = num__30 cents / pound + num__15 cents / pound = num__45 cents / pound answer : b <eor> b <eos> |
b |
divide__50.0__100.0__ multiply__30.0__0.5__ add__30.0__15.0__ add__30.0__15.0__ |
divide__50.0__100.0__ multiply__30.0__0.5__ add__30.0__15.0__ add__30.0__15.0__ |
| num__8.008 / num__4.004 <o> a ) num__2 <o> b ) num__0.02 <o> c ) num__0.2 <o> d ) num__20 <o> e ) num__2 |
answer is num__2 move the decimal forward three places for both numerator and denominator or just multiply both by a thousand . the result is num__2.0 = num__2 answer a <eor> a <eos> |
a |
divide__8.008__4.004__ divide__8.008__4.004__ |
divide__8.008__4.004__ divide__8.008__4.004__ |
| if the sides of a cube are in the ratio num__4 : num__3 . what is the ratio of their diagonals ? <o> a ) num__4 : num__3 <o> b ) num__4 : num__0 <o> c ) num__4 : num__5 <o> d ) num__4 : num__7 <o> e ) num__4 : num__2 |
explanation : a num__1 : a num__2 = num__4 : num__3 d num__1 : d num__2 = num__4 : num__3 answer : option a <eor> a <eos> |
a |
triangle_area__4.0__1.0__ square_perimeter__1.0__ |
triangle_area__4.0__1.0__ square_perimeter__1.0__ |
| find the roots of quadratic equation : num__3 x num__2 - num__7 x - num__6 = num__0 ? <o> a ) - num__6 num__3 <o> b ) num__3 - num__0.666666666667 <o> c ) - num__5 num__2 <o> d ) - num__9 num__2 <o> e ) none of these |
num__3 x num__2 - num__9 x + num__2 x - num__6 = num__0 num__3 x ( x - num__3 ) + num__2 ( x - num__3 ) = num__0 ( x - num__3 ) ( num__3 x + num__2 ) = num__0 = > x = num__3 - num__0.666666666667 answer : b <eor> b <eos> |
b |
add__3.0__6.0__ divide__2.0__3.0__ subtract__6.0__3.0__ |
add__3.0__6.0__ divide__2.0__3.0__ subtract__6.0__3.0__ |
| in a ground there are men and dogs . in all there are num__20 heads and num__72 feet . how many men and how many horses are in the ground ? <o> a ) num__8 men and num__14 dogs . <o> b ) num__7 men and num__14 dogs . <o> c ) num__6 men and num__14 digs . <o> d ) num__5 men and num__14 dogs . <o> e ) num__4 men and num__14 dogs . |
num__6 men and num__14 dogs . let m = men and d = dogs . we can come up with num__2 equations . m + d = num__20 num__2 m + num__4 d = num__72 solving the num__2 equations will give num__6 men and num__14 dogs . answer c <eor> c <eos> |
c |
subtract__20.0__6.0__ subtract__6.0__2.0__ subtract__20.0__14.0__ |
subtract__20.0__6.0__ subtract__6.0__2.0__ add__2.0__4.0__ |
| a num__28 cm long wire is to be cut into two pieces so that one piece will be num__0.4 th of the other how many centimeters will the shorter piece be ? <o> a ) num__12 <o> b ) num__8 <o> c ) num__88 <o> d ) num__77 <o> e ) num__14 |
num__1 : num__0.4 = num__5 : num__2 num__0.285714285714 * num__28 = num__8 answer : b <eor> b <eos> |
b |
multiply__0.4__5.0__ round__8.0__ |
multiply__0.4__5.0__ round__8.0__ |
| speed of a boat in standing water is num__9 kmph and the speed of the stream is num__1 . num__5 kmph . a man rows to a place at a distance of num__105 km and comes back to the starting point . the total time taken by him is <o> a ) num__16 hours <o> b ) num__18 hours <o> c ) num__20 hours <o> d ) num__24 hours <o> e ) none |
solution speed upsteram = num__7.5 kmph . speed downsteram = num__10.5 kmph . total time taken = [ num__10.5 / num__7.5 + num__105 / num__10.5 ] hours = num__24 hours . answer d <eor> d <eos> |
d |
round__24.0__ |
divide__24.0__1.0__ |
| if m = | | n – num__3 | – num__2 | for how many values of n is m = num__6 ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__2 |
m = | | n – num__3 | – num__2 | can be num__4 only and only when n - num__3 = + / - num__8 . so there are num__2 values of n answer : e <eor> e <eos> |
e |
subtract__6.0__2.0__ add__2.0__6.0__ round__2.0__ |
subtract__6.0__2.0__ add__2.0__6.0__ divide__6.0__3.0__ |
| pascal has num__96 miles remaining to complete his cycling trip . if he reduced his current speed by num__4 miles per hour the remainder of the trip would take him num__16 hours longer than it would if he increased his speed by num__50.0 . what is his current speed w ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__12 <o> e ) num__16 |
let the current speed be x miles per hour . time taken if speed is num__50.0 faster ( i . e . num__3 x / num__2 = num__1.5 x ) = num__96 / num__1.5 x time taken if speed is reduced by num__4 miles / hr ( i . e . ( x - num__4 ) ) = num__96 / ( x - num__4 ) as per question num__96 / ( x - num__4 ) - num__96 / num__1.5 x = num__16 solving this w we get x = num__8 . b . <eor> b <eos> |
b |
divide__3.0__2.0__ multiply__4.0__2.0__ round__8.0__ |
divide__3.0__2.0__ divide__16.0__2.0__ divide__16.0__2.0__ |
| a merchant has num__100 lbs of sugar part of which he sells at num__7.0 profit and the rest at num__15.0 profit . he gains num__10.0 on the whole . find how much is sold at num__7.0 profit ? <o> a ) num__70 lbs <o> b ) num__40 lbs <o> c ) num__30 lbs <o> d ) num__50 lbs <o> e ) num__63 lbs |
these types ofweighted averagequestions can be solved in a variety of ways so you can choose whichever method you find easiest / fastest . here ' s another variation on the weighted average formula : a = # of pounds sold at num__7.0 profit b = # of pounds sold at num__15.0 profit a + b = num__100 pounds ( . num__07 a + . num__15 b ) / ( a + b ) = . num__10 . num__07 a + . num__15 b = . num__1 a + . num__1 b . num__05 b = . num__03 a num__5 b = num__3 a num__1.66666666667 = a / b so for every num__5 pounds of a we have num__3 pounds of b . with num__100 pounds total we have num__63 pounds of a and num__37 pounds of b . option e <eor> e <eos> |
e |
percent__100.0__63.0__ |
percent__100.0__63.0__ |
| the difference between two integers is num__5 . their product is num__500 . find the numbers . <o> a ) num__15 num__20 <o> b ) num__20 num__25 <o> c ) num__30 num__25 <o> d ) num__21 num__26 <o> e ) num__25 num__27 |
explanation : let the integers be x and ( x + num__5 ) . then x ( x + num__5 ) = num__500 x num__2 + num__5 x - num__500 = num__0 ( x + num__25 ) ( x - num__20 ) = num__0 x = num__20 so the numbers are num__20 and num__25 . answer is b <eor> b <eos> |
b |
divide__500.0__25.0__ divide__500.0__25.0__ |
subtract__25.0__5.0__ subtract__25.0__5.0__ |
| source : knewton a cyclist ' s speed varies depending on the terrain between num__6.0 miles per hour and num__12.0 miles per hour inclusive . what is the maximum distance in miles that the cyclist could travel in num__7 hours ? <o> a ) num__42 <o> b ) num__84 <o> c ) num__70 <o> d ) num__98 <o> e ) num__140 |
we are told that : generallya cyclist ' s speed varies depending on the terrain between num__6.0 miles per hour and num__12.0 miles per hour inclusive . is it possible the cyclist to travel with maximum speed for some time ? why not if there is right terrain for that . so if there is long enough terrain for the maximum speed of num__12 mph then the maximum distance in miles that the cyclist could travel in num__7 hours would be num__7 * num__12 = num__84 miles . answer : b . <eor> b <eos> |
b |
multiply__12.0__7.0__ round__84.0__ |
multiply__12.0__7.0__ multiply__12.0__7.0__ |
| a can do a piece of work in num__15 days . a does the work for num__5 days only and leaves the job . b does the remaining work in num__10 days . in how many days b alone can do the work ? <o> a ) num__5 days <o> b ) num__15 days <o> c ) num__12 days <o> d ) num__9 days <o> e ) num__10 days |
explanation : a ’ s num__5 day work = num__5 * num__0.0666666666667 = num__0.333333333333 remaining work = num__1 - num__0.333333333333 = num__0.666666666667 b completes num__0.666666666667 work in num__6 days b alone can do in x days num__0.666666666667 * x = num__10 x = num__15 days answer : option b <eor> b <eos> |
b |
divide__5.0__15.0__ divide__10.0__15.0__ add__5.0__1.0__ round__15.0__ |
divide__5.0__15.0__ subtract__1.0__0.3333__ add__5.0__1.0__ round__15.0__ |
| the average of first ten prime numbers which are odd is ? <o> a ) num__15.9 <o> b ) num__15.5 <o> c ) num__15.2 <o> d ) num__15.8 <o> e ) num__13.2 |
sum of first num__10 prime no . which are odd = num__158 average = num__15.8 = num__15.8 answer : d <eor> d <eos> |
d |
divide__158.0__10.0__ divide__158.0__10.0__ |
divide__158.0__10.0__ divide__158.0__10.0__ |
| if we have num__8 people in a meeting in how many ways they can sit around a table ? <o> a ) num__10 * num__9 ! <o> b ) num__9 * num__8 ! <o> c ) num__8 * num__7 ! <o> d ) num__7 * num__6 ! <o> e ) num__6 * num__5 ! |
if there are ` ` n ' ' people sitting around a table there are ` ` ( n - num__1 ) ! ' ' possible arrangements : here n = num__8 ( n - num__1 ) ! = num__7 ! = num__7 * num__6 ! ans : ' ' d ' ' <eor> d <eos> |
d |
die_space__ choose__7.0__6.0__ |
die_space__ choose__7.0__6.0__ |
| the area of a square is equal to four times the area of a rectangle of dimensions num__52 cm * num__13 cm . what is the perimeter of the square ? <o> a ) num__200 <o> b ) num__208 <o> c ) num__210 <o> d ) num__220 <o> e ) num__228 |
area of the square = s * s = num__4 ( num__52 * num__13 ) = > s = num__52 cm perimeter of the square = num__4 * num__52 = num__208 cm . answer : option b <eor> b <eos> |
b |
square_perimeter__52.0__ square_perimeter__52.0__ |
multiply__52.0__4.0__ multiply__52.0__4.0__ |
| two employees x and y are paid a total of rs . num__650 per week by their employer . if x is paid num__120 percent of the sum paid to y how much is y paid per week ? <o> a ) s . num__200.45 <o> b ) s . num__250.45 <o> c ) s . num__295.45 <o> d ) s . num__300.45 <o> e ) s . num__310.45 |
let the amount paid to x per week = x and the amount paid to y per week = y then x + y = num__650 but x = num__120.0 of y = num__120 y / num__100 = num__12 y / num__10 ∴ num__12 y / num__10 + y = num__650 ⇒ y [ num__1.2 + num__1 ] = num__650 ⇒ num__22 y / num__10 = num__650 ⇒ num__22 y = num__6500 ⇒ y = num__295.454545455 = rs . num__295.45 c ) <eor> c <eos> |
c |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__12.0__10.0__ multiply__650.0__10.0__ divide__6500.0__22.0__ multiply__1.0__295.45__ |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__12.0__10.0__ multiply__650.0__10.0__ divide__6500.0__22.0__ divide__295.45__1.0__ |
| if s and t are positive integers such that s / t = num__4.12 which of the following could be the remainder when s is divided by t ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__8 <o> d ) num__20 <o> e ) num__30 |
the remainder is num__0.12 or num__0.12 . you can go one step further and say that x / y = num__0.12 where x and y are whole numbers . plug in all the answer choices for x and see which one makes y a whole number . one thing that jumps out at me is that a b c and d are all even . e on the other hand is odd . why ? maybe i ' ll start plugging in here first . num__30 / y = num__0.12 num__12 y = num__3000 y = num__250 answer : e . <eor> e <eos> |
e |
divide__3000.0__12.0__ multiply__250.0__0.12__ |
divide__3000.0__12.0__ multiply__250.0__0.12__ |
| a table is bought for rs . num__400 / - and sold at rs . num__500 / - find gain or loss percentage <o> a ) num__25.0 loss <o> b ) num__25.0 gain <o> c ) num__15.0 gain <o> d ) num__15.0 loss <o> e ) none |
formula = ( selling price ~ cost price ) / cost price * num__100 = ( num__500 - num__400 ) / num__400 = num__25.0 gain b <eor> b <eos> |
b |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| a semicircle has a radius of num__11 . what is the approximate perimeter of the semicircle ? <o> a ) num__45 <o> b ) num__51 <o> c ) num__57 <o> d ) num__63 <o> e ) num__69 |
the perimeter of a circle is num__2 * pi * r . the perimeter of a semicircle is num__2 * pi * r / num__2 + num__2 r = pi * r + num__2 r the perimeter is pi * num__11 + num__2 * num__11 which is about num__57 . the answer is c . <eor> c <eos> |
c |
triangle_area__57.0__2.0__ |
triangle_area__57.0__2.0__ |
| the cubic f ( x ) = x num__3 - num__7 x - num__6 has three distinct roots a b and c . compute num__1 / a + num__1 / b + num__1 / c . <o> a ) - num__0.833333333333 <o> b ) - num__1.0 <o> c ) - num__1.16666666667 <o> d ) - num__1.33333333333 <o> e ) - num__1.5 |
num__1 / a + num__1 / b + num__1 / c = ab + bc + ca / abc = - num__7 / - num__6 = - num__1.16666666667 correct answer c <eor> c <eos> |
c |
divide__7.0__6.0__ divide__7.0__6.0__ |
divide__7.0__6.0__ divide__7.0__6.0__ |
| if x > = y and y > num__1 what is the maximum value of log ( base x ) - ( x / y ) + log ( base y ) x ^ num__2 / y ? <o> a ) - num__0.5 <o> b ) num__0 <o> c ) num__0.5 <o> d ) num__1 <o> e ) num__2 |
log - ( x / y ) + log ( x ^ num__2 ) / y = > - log x / y + num__2 log ( x / y ) x y x y by formula ( i ) - log x / y = log y / x ( ii ) log y / x = log y - log x x x x x x log y - log x + num__2 { log x - log y } x x y y log y - num__1 + num__2 log x - num__2 = = > log y + num__2 log x - num__3 x y x y by sub x & y as num__2 by question condition - - - > ans num__1 + num__2 - num__3 = num__0 answer : b <eor> b <eos> |
b |
add__1.0__2.0__ multiply__1.0__0.0__ |
add__1.0__2.0__ divide__0.0__1.0__ |
| cereal a is num__9.0 sugar by weight whereas healthier but less delicious cereal b is num__2.0 sugar by weight . to make a delicious and healthy mixture that is num__4.0 sugar what should be the ratio of cereal a to cereal b by weight ? <o> a ) num__2 : num__9 <o> b ) num__2 : num__5 <o> c ) num__1 : num__6 <o> d ) num__1 : num__4 <o> e ) num__1 : num__3 |
( num__0.09 ) a + ( num__0.02 ) b = ( num__0.04 ) ( a + b ) num__5 a = num__2 b = > a / b = num__0.4 answer is b . <eor> b <eos> |
b |
multiply__2.0__0.02__ subtract__9.0__4.0__ divide__2.0__5.0__ subtract__4.0__2.0__ |
multiply__2.0__0.02__ subtract__9.0__4.0__ divide__2.0__5.0__ divide__4.0__2.0__ |
| a man is num__40 years older than his son . in six years his age will be twice the age of his son . the present age of this son is <o> a ) num__31 years <o> b ) num__32 years <o> c ) num__33 years <o> d ) num__34 years <o> e ) num__35 years |
explanation : let ' s son age is x then father age is x + num__40 . = > num__2 ( x + num__6 ) = ( x + num__40 + num__6 ) = > num__2 x + num__12 = x + num__46 = > x = num__34 years answer : option d <eor> d <eos> |
d |
multiply__2.0__6.0__ add__40.0__6.0__ subtract__40.0__6.0__ subtract__40.0__6.0__ |
multiply__2.0__6.0__ add__40.0__6.0__ subtract__40.0__6.0__ subtract__40.0__6.0__ |
| a vendor buys num__10 t - shirts at an average price of $ num__14 per t - shirt . he then buys num__15 more t - shirts at an average price of $ num__11 per t - shirt . what is the average price s per t - shirt that the vendor paid for these purchases ? <o> a ) $ num__12.20 <o> b ) $ num__12.50 <o> c ) $ num__12.55 <o> d ) $ num__12.70 <o> e ) $ num__13.00 |
correct answer : a explanation : the relevant formula for this problem is average s = ( sum ) / ( number of terms ) . another way to look at the formula is sum = average x number of terms . for the first purchase the vendor ' s sum ( total cost ) was $ num__140 since num__14 x num__10 = num__140 . for the second purchase the vendor ' s cost was $ num__165 since num__11 x num__15 = num__165 . the grand sum is then $ num__140 + $ num__165 which equals $ num__305 . the total number of shirts purchased was num__25 so to get the average price per shirt we divide num__305 by num__25 which equals $ num__12.20 . as a result the correct answer is a . note : a relative understanding of weighted average offers a shortcut to this problem . because the true average of num__11 and num__14 is num__12.5 but the vendor sells more shirts at the lower price than at the higher price the weighted average must be less than $ num__12.50 ; only answer choice a is a possibility . <eor> a <eos> |
a |
multiply__10.0__14.0__ multiply__15.0__11.0__ add__165.0__140.0__ add__10.0__15.0__ divide__305.0__25.0__ divide__305.0__25.0__ |
multiply__10.0__14.0__ multiply__15.0__11.0__ add__165.0__140.0__ add__10.0__15.0__ divide__305.0__25.0__ divide__305.0__25.0__ |
| a boat travels num__80 km downstream in num__8 hours and num__48 km upstream in num__12 hours . find the speed of the boat in still water and the speed of the water current . <o> a ) num__7 km / hr num__3 km / hr <o> b ) num__5 km / hr num__4 km / hr <o> c ) num__1 km / hr num__6 km / hr <o> d ) num__2 km / hr num__1 km / hr <o> e ) num__4 km / hr num__1 km / hr |
downstream speed = num__80 km / num__8 hrs = num__10 kmph upstream speed = num__48 km / num__12 hrs = num__4 kmph speed of boat = avg of downstream and upstream speeds speed of boat = ( num__10 + num__4 ) / num__2 kmph = num__7 kmph . current speed = half of the difference of downstream and upstream speeds currend speed = ( num__10 - num__4 ) / num__2 kmph = num__3 kmph ans is num__7 km / hr num__3 km / hr answer : a <eor> a <eos> |
a |
divide__80.0__8.0__ divide__48.0__12.0__ divide__8.0__4.0__ divide__12.0__4.0__ round__7.0__ |
divide__80.0__8.0__ divide__48.0__12.0__ divide__8.0__4.0__ divide__12.0__4.0__ add__3.0__4.0__ |
| ten years ago a mother was four times older than her daughter . after ten years the mother will be twice older than daughter . the present age of the daughter is <o> a ) num__22 <o> b ) num__21 <o> c ) num__24 <o> d ) num__20 <o> e ) num__19 |
let the mother ’ s present age be ’ x ’ let the daughter ’ s age be ’ y ’ ; num__10 years ago ; ( x - num__10 ) = num__4 ( y - num__10 ) or x - num__4 y = - num__30 . . ( num__1 ) num__10 years later ; ( x + num__10 ) = num__2 ( y + num__10 ) or x - num__2 y = num__10 . . ( num__2 ) solving ( num__1 ) & ( num__2 ) y = num__20 answer : d <eor> d <eos> |
d |
multiply__2.0__10.0__ multiply__1.0__20.0__ |
subtract__30.0__10.0__ subtract__30.0__10.0__ |
| the jogging track in a sports complex is num__1000 meters in circumference . deepak and his wife start from the same point and walk in opposite directions at num__20 km / hr and num__13 km / hr respectively . they will meet for the first time in ? <o> a ) num__50 min <o> b ) num__33 min <o> c ) num__35 min <o> d ) num__25 min <o> e ) num__20 min |
clearly the two will meet when they are num__1000 m apart to be num__20 + num__13 = num__33 km apart they take num__1 hour to be num__1000 m apart they take num__33 * num__1.0 = num__33 min . answer is b <eor> b <eos> |
b |
add__20.0__13.0__ add__20.0__13.0__ |
add__20.0__13.0__ add__20.0__13.0__ |
| from the below series find the remainder ? num__1201 × num__1203 × num__1205 × num__1207 is divided by num__6 ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__8 |
if you do n ' t know the above rule this problem is really calculation intensive . but by applying the above rule when num__1201 num__1201 num__1203 num__1204 divided by num__6 leaves remainders num__1 num__3 num__5 num__1 . the product of these remainders = num__15 . when num__15 is divided by num__6 remainder is num__3 a <eor> a <eos> |
a |
subtract__1205.0__1204.0__ subtract__1207.0__1204.0__ subtract__6.0__1.0__ multiply__3.0__5.0__ subtract__1207.0__1204.0__ |
subtract__1205.0__1204.0__ subtract__1207.0__1204.0__ subtract__6.0__1.0__ multiply__3.0__5.0__ subtract__1207.0__1204.0__ |
| if num__0 < a < b and k = ( num__2 a + num__12 b ) / b which of the following must be true ? <o> a ) k < num__2 <o> b ) k < num__7 <o> c ) k < num__9 <o> d ) k > num__9 <o> e ) k > num__11 |
please follow posting guidelines ( link in my signature ) especially writing the correct topic title . also do mention the source of the question if you select the tag : source - other please specify . as for your question it is very simple and straightforward with number plug in . assume a = num__1 and b = num__2 as a < b you get k = ( num__2 + num__24 ) / num__2 = num__13 . this eliminates all but option e making it the correct answer . <eor> e <eos> |
e |
multiply__2.0__12.0__ add__12.0__1.0__ subtract__12.0__1.0__ |
multiply__2.0__12.0__ add__12.0__1.0__ subtract__12.0__1.0__ |
| a plant manager must assign num__12 new workers to one of five shifts . she needs a first second and third shift and two alternate shifts . each of the shifts will receive num__2 new workers . how many different ways can she assign the new workers ? <o> a ) num__23760 <o> b ) num__47520 <o> c ) num__33000 <o> d ) num__48600 <o> e ) num__54000 |
whatever : my take selecting team of num__2 out of num__12 to assign to the shifts = num__12 c num__2 = num__66 ways . now num__2 out of num__12 means total of num__6 group possible . so putting them in shifts = counting methode : first second third alt alt = num__6 * num__5 * num__4 * num__3 * num__2 * num__1 = num__720 here alt and alt are the same : so num__360.0 = num__360 ways . total ways of selecting = ( selecting num__2 out of num__12 ) * arranging those teams in shifts = num__66 * num__360 = num__23760 ans : a <eor> a <eos> |
a |
divide__12.0__2.0__ subtract__6.0__2.0__ divide__12.0__4.0__ subtract__3.0__2.0__ divide__720.0__2.0__ multiply__360.0__66.0__ round__23760.0__ |
divide__12.0__2.0__ subtract__6.0__2.0__ divide__12.0__4.0__ subtract__3.0__2.0__ divide__720.0__2.0__ multiply__360.0__66.0__ multiply__360.0__66.0__ |
| in the coordinate plane line a has a slope of - num__1 and an x - intercept of num__1 . line b has a slope of num__5 and a y - intercept of - num__5 . if the two lines intersect at the point ( a b ) what is the sum a + b ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
the equation of line a is y = - x + num__1 the equation of line b is y = num__5 x - num__5 num__5 x - num__5 = - x + num__1 x = num__1 y = num__0 the point of intersection is ( num__10 ) and then a + b = num__1 . the answer is b . <eor> b <eos> |
b |
reverse__1.0__ |
reverse__1.0__ |
| how many trailing zeroes does num__53 ! + num__54 ! have ? <o> a ) num__12 <o> b ) num__13 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
num__53 ! + num__54 ! = num__53 ! + num__54 * num__53 ! = num__53 ! ( num__1 + num__54 ) = num__53 ! * num__55 number of trailing num__0 s in num__53 ! = number of num__5 s in the expansion of num__53 ! = num__10 + num__2 = num__12 there is num__1 more num__5 in num__55 . hence total number of trailing num__0 s = num__12 + num__1 = num__13 answer ( b ) <eor> b <eos> |
b |
subtract__54.0__53.0__ add__54.0__1.0__ divide__10.0__5.0__ add__2.0__10.0__ add__1.0__12.0__ add__1.0__12.0__ |
subtract__54.0__53.0__ add__54.0__1.0__ divide__10.0__5.0__ add__2.0__10.0__ add__1.0__12.0__ add__1.0__12.0__ |
| a person took some amount with some interest for num__5 years but increase the interest for num__1.0 he paid rs . num__250 / - extra then how much amount he took ? <o> a ) rs . num__5500 / - <o> b ) rs . num__6000 / - <o> c ) rs . num__4000 / - <o> d ) rs . num__5000 / - <o> e ) none of these |
explanation : num__5 years = rs . num__250 / - year = num__50.0 rate of interest = num__1.0 num__100.0 % × num__50.0 = rs . num__5000 / - p = rs . num__5000 / - answer : option d <eor> d <eos> |
d |
divide__250.0__5.0__ multiply__100.0__50.0__ multiply__1.0__5000.0__ |
divide__250.0__5.0__ multiply__100.0__50.0__ divide__5000.0__1.0__ |
| one millisecond is num__0.001 of a second . the costs for a single run of a computer program are $ num__1.07 for operating - system overhead $ num__0.019 per millisecond of computer time and $ num__4.35 for the mounting of a data tape . what is the total of these three costs for num__1 run of a program that requires num__1.5 seconds of computer time ? <o> a ) $ num__7.15 <o> b ) $ num__8.87 <o> c ) $ num__33.92 <o> d ) $ num__35.57 <o> e ) $ num__39.62 |
operating system overhead for num__1 run = num__1.07 $ mounting of data tape = num__4.35 $ cost per num__1 millisecond of computer time = . num__019 $ total cost of num__1 run of a program that requires num__1.5 seconds of computer time = num__1.07 + ( . num__019 * num__1500 ) + num__4.35 = num__1.07 + num__28.5 + num__4.35 = num__33.92 $ answer c <eor> c <eos> |
c |
multiply__0.019__1500.0__ multiply__1.0__33.92__ |
multiply__0.019__1500.0__ multiply__1.0__33.92__ |
| two trains one from p to q and the other from q to p start simultaneously . after they meet the trains reach their destinations after num__9 hours and num__36 hours respectively . the ratio of their speeds is <o> a ) num__4 : num__1 <o> b ) num__4 : num__2 <o> c ) num__4 : num__5 <o> d ) num__6 : num__3 <o> e ) num__3 : num__6 |
ratio of their speeds = speed of first train : speed of second train = √ num__36 − − √ num__9 = num__6 : num__3 answer is d . <eor> d <eos> |
d |
subtract__9.0__6.0__ round__6.0__ |
subtract__9.0__6.0__ round__6.0__ |
| if num__1 + num__2 + num__3 + . . . + n = n ( n + num__1 ) then num__3 ( num__1 + num__3 + num__5 + . . . . + num__69 ) = ? <o> a ) num__3675 <o> b ) num__3575 <o> c ) num__3475 <o> d ) num__3375 <o> e ) num__3275 |
explanation : to solve this use the formula of ap sn = ( n / num__2 ) ( a + l ) . . . . . . . . . . . . . . . . ( num__1 ) to find n use = > tn = a + ( n - num__1 ) d = > num__69 = num__1 + ( n - num__1 ) num__2 = > n = num__35 use value of n in ( num__1 ) then sn = ( num__17.5 ) ( num__1 + num__69 ) = num__1225 ans : - num__3 ( sn ) = num__3675 answer : a <eor> a <eos> |
a |
divide__35.0__2.0__ multiply__3.0__1225.0__ multiply__1.0__3675.0__ |
divide__35.0__2.0__ multiply__3.0__1225.0__ divide__3675.0__1.0__ |
| the length of a train and that of a platform are equal . if with a speed of num__108 k / hr the train crosses the platform in one minute then the length of the train ( in meters ) is ? <o> a ) num__227 <o> b ) num__299 <o> c ) num__276 <o> d ) num__750 <o> e ) num__900 |
speed = [ num__108 * num__0.277777777778 ] m / sec = num__30 m / sec ; time = num__1 min . = num__60 sec . let the length of the train and that of the platform be x meters . then num__2 x / num__60 = num__30 è x = num__30 * num__30.0 = num__900 answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ divide__60.0__30.0__ round__900.0__ |
hour_to_min_conversion__ divide__60.0__30.0__ divide__900.0__1.0__ |
| given that e and f are events such that p ( e ) = num__0.16 p ( f ) = num__0.4 and p ( e n f ) = num__0.4 find p ( e | f ) and p ( f | e ) <o> a ) num__0.5 <o> b ) num__0.142857142857 <o> c ) num__0.25 <o> d ) num__0.333333333333 <o> e ) num__0.2 |
here e and f are events p ( e | f ) = p ( enf ) / p ( f ) = num__0.4 / num__0.4 = num__1 p ( f | e ) = p ( enf ) / p ( e ) = num__0.4 / num__0.16 = num__0.25 . c <eor> c <eos> |
c |
multiply__0.25__1.0__ |
divide__0.25__1.0__ |
| in a covering a certain distance the speeds of a and b are in the ratio of num__3 : num__4 . a takes num__30 minutes more than b to reach the destination . the time taken by a to reach the destination is ? <o> a ) num__2 hrs <o> b ) num__3 hrs <o> c ) num__5 hrs <o> d ) num__6 hrs <o> e ) num__4 hrs |
a ratio of speeds = num__3 : num__4 ratio of times taken = num__4 : num__3 suppose a takes num__4 x hrs and b takes num__3 x hrs to reach the destination . then num__4 x - num__3 x = num__0.5 = > x = num__0.5 time taken by a = num__4 x hrs = num__4 * num__0.5 = num__2 hrs . <eor> a <eos> |
a |
multiply__4.0__0.5__ round__2.0__ |
multiply__4.0__0.5__ multiply__4.0__0.5__ |
| what is the average of first num__21 multiples of num__6 ? <o> a ) a ) num__66 <o> b ) b ) num__77 <o> c ) c ) num__79 <o> d ) d ) num__81 <o> e ) e ) num__82 |
required average = num__6 ( num__1 + num__2 + . . . . + num__21 ) / num__21 ( num__0.285714285714 ) x ( ( num__21 x num__22 ) / num__2 ) ( because sum of first num__21 natural numbers ) = num__66 a <eor> a <eos> |
a |
divide__6.0__21.0__ add__21.0__1.0__ multiply__1.0__66.0__ |
divide__6.0__21.0__ add__21.0__1.0__ divide__66.0__1.0__ |
| machine a and machine b are each used to manufacture num__550 sprockets . it takes machine a num__10 hours longer to produce num__660 sprockets than machine b . machine b produces num__10.0 more sprockets per hour than machine a . how many sprockets per hour does machineaproduce ? <o> a ) num__5.5 <o> b ) num__6.6 <o> c ) num__60 <o> d ) num__100 <o> e ) num__110 |
time taken by b = t time taken by a = t + num__10 qty produced by a = q qty produced by b = num__1.1 q for b : t ( num__1.1 q ) = num__550 qt = num__500 for a : ( t + num__10 ) ( q ) = num__550 qt + num__10 q = num__550 num__500 + num__10 q = num__550 q = num__5 so a can produce num__5 / hour . then b can produce = num__5 ( num__1.1 ) = num__5.5 / hour . a <eor> a <eos> |
a |
percent__1.1__500.0__ percent__1.1__500.0__ |
percent__1.1__500.0__ percent__1.1__500.0__ |
| which of the following numbers is two more than the square of an odd integer ? <o> a ) num__14173 <o> b ) num__14361 <o> c ) num__15131 <o> d ) num__15737 <o> e ) num__15 |
981 |
if a square has an odd unit ' s digit then its ten ' s digit must be even . only num__15131 - num__2 = num__15129 satisfies this condition . the answer is c . <eor> c <eos> |
c |
c |
| how many positive integers s less than num__20 are either a multiple of num__2 an odd multiple of num__9 or the sum of a positive multiple of num__2 and a positive multiple of num__9 ? <o> a ) num__19 <o> b ) num__18 <o> c ) num__17 <o> d ) num__16 <o> e ) num__15 |
we ' re asked to deal with the positive integers less than num__20 . there are only num__19 numbers in that group ( num__1 to num__19 inclusive ) . we ' re asked to find all of the numbers that fit one ( or more ) of the given descriptions . looking at the answer choices we have every value form num__15 to num__19 inclusive so most ( if not all ) of the numbers from num__1 to num__19 fit one ( or more ) of the descriptions . how long would it take you to find the ones that do n ' t fit . . . . . ? the first several should be pretty easy to find - the prompt wants us to focus on multiples of num__2 and num__9 ( and sums of those multiples ) . so what odd numbers are less than num__9 ? num__1 num__3 num__5 num__7 none of these values fit the given descriptions . that ' s num__4 out of num__19 that we know for sure do not fit . num__19 - num__4 = num__15 = e <eor> e <eos> |
e |
subtract__20.0__19.0__ add__2.0__1.0__ subtract__20.0__15.0__ add__2.0__5.0__ divide__20.0__5.0__ subtract__20.0__5.0__ |
subtract__20.0__19.0__ add__2.0__1.0__ subtract__20.0__15.0__ subtract__9.0__2.0__ subtract__9.0__5.0__ subtract__20.0__5.0__ |
| ayesha ' s father was num__38 years of age when she was born while her mother was num__36 years old when her brother six years younger to her was born . what is the difference between the ages of her parents ? <o> a ) num__2 years <o> b ) num__4 years <o> c ) num__6 years <o> d ) num__7 years <o> e ) num__8 years |
mother ' s age when ayesha ' s brother was born = num__36 years . father ' s age when ayesha ' s brother was born = ( num__38 + num__6 ) years = num__44 years . required difference = ( num__44 - num__36 ) years = num__8 years . answer : option e <eor> e <eos> |
e |
add__38.0__6.0__ subtract__44.0__36.0__ subtract__44.0__36.0__ |
add__38.0__6.0__ subtract__44.0__36.0__ subtract__44.0__36.0__ |
| the difference between a number and its two - fifth is num__510 . what is num__12.0 of that number ? <o> a ) num__58 <o> b ) num__110 <o> c ) num__76 <o> d ) num__102 <o> e ) num__86 |
let the number be x . then x - num__0.4 x = num__510 x = ( num__510 * num__5 ) / num__3 = num__850 num__12.0 of num__850 = num__102 . answer : d <eor> d <eos> |
d |
divide__510.0__5.0__ divide__510.0__5.0__ |
divide__510.0__5.0__ divide__510.0__5.0__ |
| a farming field can be ploughed by num__6 tractors in num__4 days . when num__6 tractors work together each of them ploughs num__120 hectares a day . if two of the tractors were moved to another field then the remaining num__4 tractors could plough the same field in num__5 days . how many hectares a day would one tractor plough then ? <o> a ) num__144 <o> b ) num__121 <o> c ) num__256 <o> d ) num__320 <o> e ) num__169 |
if each of num__6 tractors ploughed num__120 hectares a day and they finished the work in num__4 days then the whole field is : num__120 ⋅ num__6 ⋅ num__4 = num__720 ⋅ num__4 = num__2880 hectares . let ' s suppose that each of the four tractors ploughed x hectares a day . therefore in num__5 days they ploughed num__5 ⋅ num__4 ⋅ x = num__20 ⋅ x hectares which equals the area of the whole field num__2880 hectares . so we get num__20 x = num__2880 x = num__288020 = num__144 . hence each of the four tractors would plough num__144 hectares a day . so answer is a <eor> a <eos> |
a |
multiply__6.0__120.0__ multiply__4.0__720.0__ multiply__4.0__5.0__ divide__2880.0__20.0__ round__144.0__ |
multiply__6.0__120.0__ multiply__4.0__720.0__ multiply__4.0__5.0__ divide__2880.0__20.0__ round__144.0__ |
| subash can copy num__50 pages in num__10 hrs . subash and prakash together can copy num__300 pages in num__40 hours . in how much time prakash can copy num__10 pages . <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__4 <o> e ) num__16 |
subhas ' s num__1 hr copy page = num__5.0 = num__5 page ( subhas + prakash ) ' s num__1 hr copy page = num__7.5 = num__7.5 page from above prakash ' s num__1 hr copy page = num__2.5 page so time taken in num__30 page ' s copy = ( num__10 / num__2.5 ) = num__4 hrs answer : d <eor> d <eos> |
d |
divide__50.0__10.0__ divide__300.0__40.0__ subtract__10.0__7.5__ divide__300.0__10.0__ divide__10.0__2.5__ divide__10.0__2.5__ |
divide__50.0__10.0__ divide__300.0__40.0__ subtract__10.0__7.5__ divide__300.0__10.0__ divide__10.0__2.5__ divide__10.0__2.5__ |
| what is the absolute value of twice the difference of the roots of the equation w = num__5 y ^ num__2 − num__20 y + num__15 = num__0 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
w = num__5 y ^ num__2 − num__20 y + num__15 = num__0 - - > y ^ num__2 − num__4 y + num__3 = num__0 - - > y = num__1 or y = num__3 . the difference is num__2 twice the difference is num__4 . answer : e . <eor> e <eos> |
e |
divide__20.0__5.0__ subtract__5.0__2.0__ subtract__5.0__4.0__ subtract__5.0__1.0__ |
divide__20.0__5.0__ subtract__5.0__2.0__ subtract__5.0__4.0__ subtract__5.0__1.0__ |
| the product z of two prime numbers is between num__15 and num__36 . if one of the prime numbers is greater than num__2 but less than num__6 and the other prime number is greater than num__8 but less than num__24 then what is z ? <o> a ) num__35 <o> b ) num__33 <o> c ) num__28 <o> d ) num__21 <o> e ) num__15 |
the smallest possible product is num__33 which is num__3 * num__11 . all other products are too big . the answer is b . <eor> b <eos> |
b |
subtract__36.0__33.0__ add__8.0__3.0__ subtract__36.0__3.0__ |
subtract__36.0__33.0__ add__8.0__3.0__ multiply__11.0__3.0__ |
| laura can paint num__1 / x of a certain room in num__20 minutes . what fraction z of the same room can joseph paint in num__20 minutes if the two of them can paint the room in an hour working together at their respective rates ? <o> a ) num__1 / ( num__3 x ) <o> b ) num__3 x / ( x - num__3 ) <o> c ) ( x – num__3 ) / ( num__3 x ) <o> d ) x / ( x - num__3 ) <o> e ) ( x - num__3 ) / x |
options with variables are often done by plugging in numbers . both working together can paint the room in num__1 hr so if their individual rates were equal each would take num__2 hours alone . num__2 hours is num__120 mins so in num__20 mins each would complete z = num__6.0 = num__0.166666666667 th of the room alone . so if x = num__6 ( laura completes num__0.166666666667 th of the room in num__20 mins ) the correct option will give num__0.166666666667 . ( joseph will also paint num__0.166666666667 th of the room if their rates are same ) if you put x = num__6 in the options only option ( c ) will give num__0.166666666667 answer ( c ) <eor> c <eos> |
c |
divide__120.0__20.0__ divide__1.0__6.0__ add__1.0__2.0__ |
divide__120.0__20.0__ divide__1.0__6.0__ add__1.0__2.0__ |
| the malibu country club needs to drain its pool for refinishing . the hose they use to drain it can remove num__60 cubic feet of water per minute . if the pool is num__80 feet wide by num__150 feet long by num__10 feet deep and is currently at num__100.0 capacity how long will it take to drain the pool ? <o> a ) num__1400 <o> b ) num__1600 <o> c ) num__1800 <o> d ) num__1500 <o> e ) num__2000 |
volume of pool = num__80 * num__150 * num__10 cu . ft num__100.0 full = num__60 * num__100 * num__10 * num__1 cu . ft water is available to drain . draining capacity = num__60 cu . ft / min therefore time taken = num__80 * num__150 * num__10 * num__0.0166666666667 min = num__2000 min e <eor> e <eos> |
e |
percent__100.0__2000.0__ |
percent__100.0__2000.0__ |
| if one - third of one - fourth of a number is num__25 then three - tenth of that number is : a . num__35 <o> a ) num__23 <o> b ) num__90 <o> c ) num__26 <o> d ) num__54 <o> e ) num__01 |
explanation : the number is num__0.333333333333 of num__0.25 is = num__15 then num__0.333333333333 Ã — num__0.25 = num__25 number is num__300 then num__300 Ã — num__0.3 = num__90 answer : b <eor> b <eos> |
b |
multiply__0.3__300.0__ multiply__0.3__300.0__ |
multiply__0.3__300.0__ multiply__0.3__300.0__ |
| num__13 sheeps and num__9 pigs were bought for rs . num__1291.85 . if the average price of a sheep be rs . num__74 . what is the average price of a pig . <o> a ) rs . num__26.65 <o> b ) rs . num__36.55 <o> c ) rs . num__35.65 <o> d ) rs . num__36.65 <o> e ) rs . num__33.65 |
average price of a sheep = rs . num__74 : . total price of num__13 sheeps = ( num__74 * num__13 ) = rs . num__962 but total price of num__13 sheeps and num__9 pigs = rs . num__1291.85 total price of num__9 pigs = rs . ( num__1291.85 - num__962 ) = rs . num__329.85 hence average price of a pig = ( num__329.85 / num__9 ) = rs . num__36.65 answer is d . <eor> d <eos> |
d |
multiply__13.0__74.0__ subtract__1291.85__962.0__ divide__329.85__9.0__ divide__329.85__9.0__ |
multiply__13.0__74.0__ subtract__1291.85__962.0__ divide__329.85__9.0__ divide__329.85__9.0__ |
| what percent is num__36 paisa ' s of num__12 rupees ? <o> a ) num__3.0 <o> b ) num__4.0 <o> c ) num__5.0 <o> d ) num__6.0 <o> e ) num__7 % |
num__12 rupees = num__1200 paisa ' s num__0.03 × num__100 = num__0.25 num__4.0 = num__3.0 a <eor> a <eos> |
a |
percent__0.25__1200.0__ percent__0.25__1200.0__ |
percent__0.25__1200.0__ percent__0.25__1200.0__ |
| without stoppages a train travels certain distance with an average speed of num__400 km / h and with stoppages it covers the same distance with an average speed of num__360 km / h . how many minutes per hour the train stops ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
due to stoppages it covers num__40 km less . time taken to cover num__40 km = num__40 â „ num__400 h = num__1 â „ num__10 h = num__1 â „ num__10 Ã — num__60 min = num__6 min answer d <eor> d <eos> |
d |
subtract__400.0__360.0__ divide__400.0__40.0__ hour_to_min_conversion__ divide__360.0__60.0__ round__6.0__ |
subtract__400.0__360.0__ divide__400.0__40.0__ hour_to_min_conversion__ divide__360.0__60.0__ round__6.0__ |
| earl can stuff advertising circulars into envelopes at the rate of num__36 envelopes per minutes and ellen requires a minutes and half to stuff the same number of envelops . working together how long will it take earl and ellen to stuff num__240 envelopes <o> a ) num__6 minutes <o> b ) num__5 minutes <o> c ) num__7 minutes <o> d ) num__3 minutes <o> e ) num__4 minutes |
earl takes num__1 min . for num__36 envelopes . ellen takes num__1.5 mins for the same . so ellen can stuff ( ( num__36 ) / ( num__1.5 ) ) in num__1 min . i . e . num__24 envlpes a min . so both of them when work together can stuff num__36 + num__24 = num__60 envelopes in num__1 min . for num__240 envelopes they will take num__4.0 mins . i . e . num__4 mins . answer : e <eor> e <eos> |
e |
divide__36.0__1.5__ hour_to_min_conversion__ divide__240.0__60.0__ round__4.0__ |
divide__36.0__1.5__ add__36.0__24.0__ divide__240.0__60.0__ divide__240.0__60.0__ |
| a man buys num__25 lts of liquid which contains num__20.0 of the liquid and the rest is water . he then mixes it with num__25 lts of another mixture with num__30.0 of liquid . what is the % of water in the new mixture ? <o> a ) num__55 <o> b ) num__82 <o> c ) num__73 <o> d ) num__75 <o> e ) num__85 |
num__20.0 in num__25 lts is num__5 . so water = num__25 - num__5 = num__20 lts . num__30.0 of num__25 lts = num__7.5 . so water in num__2 nd mixture = num__25 - num__7.5 = num__17.5 lts . now total quantity = num__25 + num__25 = num__50 lts . total water in it will be num__20 + num__17.5 = num__37.5 lts . % of water = ( num__100 * num__37.5 ) / num__50 = num__75 . answer : d <eor> d <eos> |
d |
percent__25.0__20.0__ percent__25.0__30.0__ percent__100.0__75.0__ |
percent__25.0__20.0__ percent__25.0__30.0__ percent__100.0__75.0__ |
| num__1300 boys and num__700 girls are examined in a test ; num__52.0 of the boys and num__43.0 of the girls pass . the percentage of the total who failed is ? <o> a ) a ) num__63.3 <o> b ) b ) num__52.4 <o> c ) c ) num__81.2 <o> d ) d ) num__51.15 <o> e ) e ) num__69.1 % |
total number of students = num__1300 + num__700 = num__2000 number of students passed = ( num__52.0 of num__1300 + num__43.0 of num__700 ) = num__676 + num__301 = num__977 number of failures = num__1023 * num__0.05 = num__51.15 answer is d <eor> d <eos> |
d |
add__1300.0__700.0__ add__676.0__301.0__ subtract__2000.0__977.0__ multiply__0.05__1023.0__ multiply__0.05__1023.0__ |
add__1300.0__700.0__ add__676.0__301.0__ subtract__2000.0__977.0__ multiply__0.05__1023.0__ multiply__0.05__1023.0__ |
| he population of a city is num__160000 . if it increases at the rate of num__6.0 per annum then what will be its population num__2 years hence ? <o> a ) num__1797787 <o> b ) num__1797723 <o> c ) num__179776 <o> d ) num__179728 <o> e ) num__179718 |
population after n years = p [ num__1 + ( r / num__100 ) ] num__2 population after num__2 years = num__160000 * [ num__1 + num__0.06 ] num__2 = ( num__160000 x num__1.06 x num__1.06 ) = num__179776 answer : c <eor> c <eos> |
c |
percent__6.0__1.0__ percent__100.0__179776.0__ |
percent__6.0__1.0__ percent__100.0__179776.0__ |
| rewrite ( num__10 ^ num__38 ) – num__85 as a base num__10 integer what is the sum of the digits in that integer ? <o> a ) num__354 <o> b ) num__357 <o> c ) num__330 <o> d ) num__370 <o> e ) num__360 |
we know that ( num__10 ^ num__38 ) is ending num__00 so ( num__10 ^ num__38 ) – num__85 = num__9 . . . . num__9915 total number of digits in ( num__10 ^ num__38 ) – num__85 is num__38 or num__36 digits of num__9 and two digits num__1 and num__5 . answer choice is num__36 * num__9 + num__6 = num__330 answer is c num__330 <eor> c <eos> |
c |
subtract__10.0__9.0__ add__1.0__5.0__ multiply__1.0__330.0__ |
subtract__10.0__9.0__ add__1.0__5.0__ multiply__1.0__330.0__ |
| jc has to visit at least num__2 european cities on his vacation trip . if he can visit only london paris rome or madrid how many different itineraries defined as the sequence of visited cities can jc create ? <o> a ) num__12 <o> b ) num__36 <o> c ) num__48 <o> d ) num__60 <o> e ) num__72 |
the question asks the different itineraries and we need to count the order . for example visiting city a first and visiting city b later is different from visiting city b first and visiting city a later . to visit num__2 cities from num__4 cities there are num__4 Ã — num__3 = num__124 Ã — num__3 = num__12 different itineraries to visit num__3 cities from num__4 cities there are num__4 Ã — num__3 Ã — num__2 = num__244 Ã — num__3 Ã — num__2 = num__24 different itineraries to visit num__4 cities from num__4 cities there are num__4 Ã — num__3 Ã — num__2 Ã — num__1 = num__244 Ã — num__3 Ã — num__2 Ã — num__1 = num__24 different itineraries the total different itineraries are num__12 + num__24 + num__24 = num__6012 + num__24 + num__24 = num__60 different itineraries . answer : d <eor> d <eos> |
d |
multiply__3.0__4.0__ multiply__2.0__12.0__ subtract__3.0__2.0__ multiply__1.0__60.0__ |
multiply__3.0__4.0__ multiply__2.0__12.0__ subtract__3.0__2.0__ multiply__1.0__60.0__ |
| if two of the four expressions a + b b + num__7 b a - b and num__7 a - b are chosen at random what is the probability that their product will be of the form of a ^ num__2 - ( xb ) ^ num__2 where x is an integer ? <o> a ) num__0.5 <o> b ) num__0.142857142857 <o> c ) num__0.166666666667 <o> d ) num__0.125 <o> e ) num__0.2 |
total number of expressions if num__2 are multiplied = num__4 c num__2 = num__4 ! / num__2 ! * num__2 ! = num__6 now we have to find the expressions that are in form of x ^ num__2 - ( by ) ^ num__2 one is ( a + b ) ( a - b ) = a ^ num__2 - ( num__1 b ) ^ num__2 if you see any other combination we are always going to have one term of ab since num__7 a and num__7 b are there so there can be only one type of combination like this . therefore probablity is num__0.166666666667 answer : c <eor> c <eos> |
c |
add__2.0__4.0__ subtract__7.0__6.0__ reverse__6.0__ reverse__6.0__ |
add__2.0__4.0__ subtract__7.0__6.0__ reverse__6.0__ reverse__6.0__ |
| a small experimental plane has three engines one of which is redundant . that is as long as two of the engines are working the plane will stay in the air . over the course of a typical flight there is a num__0.333333333333 chance that engine one will fail . there is a num__65.0 probability that engine two will work . the third engine works only half the time . what is the probability that the plane will crash in any given flight ? <o> a ) num__0.583333333333 <o> b ) num__0.25 <o> c ) num__0.5 <o> d ) num__0.291666666667 <o> e ) num__0.708333333333 |
in probability questions the trap answer is just the multiple of the numbers in the question . i . e . if you multiply num__0.333333333333 * num__0.25 * num__0.5 = num__0.0416666666667 is trap answer the other trap answer could be num__0.666666666667 * num__0.75 * num__0.5 = num__0.25 is trap answer so lets say you have num__30 secsand you want to guess the answer then b c are ruled out because they can be traps . you best guess is a d e . so you have num__33.0 chances of being correct . e <eor> e <eos> |
e |
negate_prob__0.3333__ negate_prob__0.25__ union_prob__0.25__0.5__0.0417__ |
negate_prob__0.3333__ negate_prob__0.25__ union_prob__0.25__0.5__0.0417__ |
| in a certain pond num__90 fish were caught tagged and returned to the pond . a few days later num__90 fish were caught again of which num__2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond what ` s the approximate number of fish in the pond ? <o> a ) num__400 <o> b ) num__625 <o> c ) num__1250 <o> d ) num__2500 <o> e ) num__4091 |
the percent of tagged fish in the second catch is num__0.0222222222222 * num__100 = num__2.22 . we are told that num__2.22 approximates the percent of tagged fish in the pond . since there are num__90 tagged fish then we have num__0.022 x = num__90 - - > x = num__4091 . answer : e . <eor> e <eos> |
e |
percent__100.0__4091.0__ |
percent__100.0__4091.0__ |
| how many seconds will a num__500 meter long train take to cross a man walking with a speed of num__3 km / hr in the direction of the moving train if the speed of the train is num__63 km / hr ? <o> a ) num__338 <o> b ) num__277 <o> c ) num__500 <o> d ) num__127 <o> e ) num__181 |
let length of tunnel is x meter distance = num__800 + x meter time = num__1 minute = num__60 seconds speed = num__78 km / hr = num__78 * num__0.277777777778 m / s = num__21.6666666667 m / s distance = speed * time num__800 + x = ( num__21.6666666667 ) * num__60 num__800 + x = num__20 * num__65 = num__1300 x = num__1300 - num__800 = num__500 meters answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ divide__60.0__3.0__ add__500.0__800.0__ round__500.0__ |
subtract__63.0__3.0__ divide__60.0__3.0__ add__500.0__800.0__ divide__500.0__1.0__ |
| if the number num__653 ab is divisible by num__90 then ( a + b ) = ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
num__90 = num__10 x num__9 clearly num__653 ab is divisible by num__10 so b = num__0 now num__653 a num__0 is divisible by num__9 . so ( num__6 + num__5 + num__3 + a + num__0 ) = ( num__14 + a ) is divisible by num__9 . so a = num__4 . hence ( a + b ) = ( num__4 + num__0 ) = num__4 . answer : c <eor> c <eos> |
c |
divide__90.0__10.0__ subtract__9.0__6.0__ add__5.0__9.0__ subtract__9.0__5.0__ subtract__9.0__5.0__ |
divide__90.0__10.0__ subtract__9.0__6.0__ add__5.0__9.0__ subtract__9.0__5.0__ subtract__9.0__5.0__ |
| for all positive integers x f ( x ) = f ( x + num__1 ) . if f ( num__3 ) = num__15 then what is f ( num__7 ) ? <o> a ) num__8 <o> b ) num__12 <o> c ) num__15 <o> d ) num__21 <o> e ) num__35 |
here we go - - - f ( x ) = f ( x + num__1 ) - - - ( num__1 ) f ( num__3 ) = num__15 - - - - ( num__2 ) now f ( num__7 ) = f ( num__6 + num__1 ) - - - - - > f ( num__6 ) = f ( num__5 + num__1 ) - - - - - - - > f ( num__5 ) = f ( num__4 + num__1 ) - - - - - - > f ( num__4 ) = f ( num__3 + num__1 ) - - - > f ( num__3 ) = num__15 now backtrack f ( num__4 ) = num__15 - - - - - > f ( num__5 ) = num__15 - - - - - - > f ( num__6 ) = num__15 - - - - > f ( num__7 ) = num__15 though i picked option c i am still not num__100.0 sure of the answer ( do n ' t know why ) my bets on c . <eor> c <eos> |
c |
subtract__3.0__1.0__ multiply__3.0__2.0__ add__3.0__2.0__ add__1.0__3.0__ multiply__1.0__15.0__ |
subtract__3.0__1.0__ subtract__7.0__1.0__ add__3.0__2.0__ add__1.0__3.0__ multiply__1.0__15.0__ |
| in the new budget the price of milk rose by num__20.0 . by how much percent must a person reduce his consumption so that his expenditure on it does not increase ? <o> a ) num__7.5 <o> b ) num__9.1 <o> c ) num__10.9 <o> d ) num__16.67 <o> e ) num__15 % |
reduce in consumption = r / ( num__100 + r ) * num__100.0 = num__0.166666666667 * num__100 = num__16.67 answer is d <eor> d <eos> |
d |
multiply__0.1667__100.0__ multiply__0.1667__100.0__ |
multiply__0.1667__100.0__ multiply__0.1667__100.0__ |
| a car is running at a speed of num__90 kmph . what distance will it cover in num__15 second ? <o> a ) num__375 m <o> b ) num__395 m <o> c ) num__373 m <o> d ) num__372 m <o> e ) num__371 m |
explanation : given : speed = num__108 kmph = ( num__90 x ( num__0.277777777778 ) ) m / sec = num__25 m / sec distance covered in num__15 second = ( num__25 x num__15 ) m = num__375 m . answer : a <eor> a <eos> |
a |
multiply__15.0__25.0__ round__375.0__ |
multiply__15.0__25.0__ round__375.0__ |
| a motor car starts with the speed of num__70 km / hr with its speed increasing every two hours by num__10 kmph . in how many hours will it cover num__345 kms <o> a ) num__4 hrs <o> b ) num__5 hrs <o> c ) num__6 hrs <o> d ) num__4 ½ hrs <o> e ) num__5 hrs |
explanation : distance covered in first num__2 hours = num__70 × num__2 = num__140 km distance covered in next num__2 hours = num__80 × num__2 = num__160 km remaining distance = num__345 - num__140 - num__160 = num__45 km speed of the fifth hour = num__90 km / hr time taken to cover num__45 km = num__0.5 = ½ hr total time taken = num__2 + num__2 + ½ = num__4 ½ hrs answer : option d <eor> d <eos> |
d |
multiply__70.0__2.0__ add__70.0__10.0__ multiply__2.0__80.0__ add__10.0__80.0__ divide__70.0__140.0__ divide__2.0__0.5__ round__4.0__ |
multiply__70.0__2.0__ add__70.0__10.0__ multiply__2.0__80.0__ add__10.0__80.0__ divide__70.0__140.0__ divide__2.0__0.5__ divide__2.0__0.5__ |
| in a apartment num__30.0 of the people speak english num__20.0 speak hindi and num__10.0 speak both . if a people is selected at random what is the probability that he has speak english or hindi ? <o> a ) num__0.333333333333 <o> b ) num__0.285714285714 <o> c ) num__0.4 <o> d ) num__0.428571428571 <o> e ) num__0.272727272727 |
p ( e ) = num__0.3 = num__0.3 p ( h ) = num__0.2 = num__0.2 and p ( e ∩ h ) = num__0.1 = num__0.1 p ( e or h ) = p ( e u h ) = p ( e ) + p ( h ) - p ( e ∩ h ) = ( num__0.3 ) + ( num__0.2 ) - ( num__0.1 ) = num__0.4 = num__0.4 c <eor> c <eos> |
c |
union_prob__0.3__0.2__0.1__ union_prob__0.3__0.2__0.1__ |
union_prob__0.3__0.2__0.1__ union_prob__0.3__0.2__0.1__ |
| if a fraction is multiplied by itself and then divided by the reciprocal of the same fraction the result is num__18 num__0.962962962963 . find the fraction . <o> a ) num__0.296296296296 <o> b ) num__1 num__0.333333333333 <o> c ) num__2 num__0.666666666667 <o> d ) num__1.0 <o> e ) none of these |
answer if the required fraction be p according to the question ( p x p ) / ( num__1 / p ) = num__67.6296296296 ⇒ p num__3 = num__18.962962963 ∴ p = num__2.66666666667 = num__2 num__0.666666666667 correct option : c <eor> c <eos> |
c |
add__18.0__0.963__ round_down__2.6667__ divide__2.0__3.0__ round_down__2.6667__ |
add__18.0__0.963__ round_down__2.6667__ divide__2.0__3.0__ divide__2.0__1.0__ |
| if num__6 spiders make num__3 webs in num__7 days then how many days are needed for num__1 spider to make num__1 web ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
explanation : let num__1 spider make num__1 web in x days . more spiders less days ( indirect proportion ) more webs more days ( direct proportion ) hence we can write as ( spiders ) num__6 : num__1 ( webs ) num__1 : num__7 } : : x : num__3 â ‡ ’ num__6 Ã — num__1 Ã — num__7 = num__1 Ã — num__3 Ã — x â ‡ ’ x = num__14 answer : option c <eor> c <eos> |
c |
round__14.0__ |
round__14.0__ |
| num__3 years ago the average age of a family of num__5 members was num__20 years . a baby having been born the average age of the family is the same today . the present age of the baby is <o> a ) num__1 year . <o> b ) num__1.5 years . <o> c ) num__2 years . <o> d ) num__3 years . <o> e ) num__5 years . |
we ' re told that num__3 years ago the average age of a family of num__5 members was num__20 years . this means that the sum of their ages was ( num__5 ) ( num__20 ) = num__100 years . a baby was born ( meaning that there are now num__6 family members ) . we ' re told that the present average age of the family members is the same . this means that the sum of the ages is ( num__6 ) ( num__20 ) = num__120 years . we ' re asked for the present age of the baby . since each of the original num__5 family members has aged num__3 years since the initial average was calculated the sum of their present ages is . . . num__100 + ( num__5 ) ( num__3 ) = num__115 . num__120 - num__115 = num__5 e <eor> e <eos> |
e |
multiply__5.0__20.0__ add__20.0__100.0__ subtract__120.0__5.0__ divide__100.0__20.0__ |
multiply__5.0__20.0__ add__20.0__100.0__ subtract__120.0__5.0__ subtract__120.0__115.0__ |
| the salary of a worker is first increased by num__30.0 and afterwards reduced by num__30.0 . what is net change in his salary ? <o> a ) num__910.0 increase <o> b ) num__10.0 increase <o> c ) num__9.0 decrease <o> d ) num__12.0 decrease <o> e ) num__15.0 increase |
( num__30 * num__30 ) / num__100 = num__9.0 decrease answer c <eor> c <eos> |
c |
percent__9.0__100.0__ |
percent__9.0__100.0__ |
| set a consists of integers - num__9 num__8 num__3 num__10 and j ; set b consists of integers - num__2 num__5 num__0 num__7 - num__6 and q . if r is the median of set a and w is the mode of set b and r ^ w is a factor of num__34 what is the value of q if j is negative ? <o> a ) - num__2 <o> b ) q = num__0 <o> c ) num__1 <o> d ) num__2 <o> e ) num__5 |
since j is negative therefore median of the set will be num__3 . which is an odd number . also we know that odd ^ n will always be odd . now since num__34 is even therefore num__3 ^ w will have only one factor with num__34 for w = num__0 . mode of the set is a number which occurs maximum number of time in the set . now since q is the mode of the set . therefore q = num__0 . hence answer is b <eor> b <eos> |
b |
multiply__9.0__0.0__ |
multiply__9.0__0.0__ |
| to asphalt num__1 km road num__30 men spent num__12 days working num__8 hours per day . how many days num__20 men will spend to asphalt a road of num__2 km working num__14 hours a day ? <o> a ) num__24 <o> b ) num__23 <o> c ) num__24 <o> d ) num__20.57 <o> e ) num__22 |
man - hours required to asphalt num__1 km road = num__30 * num__12 * num__8 = num__2880 man - hours required to asphalt num__2 km road = num__2880 * num__2 = num__5760 man - hours available per day = num__20 * num__14 = num__280 therefore number of days = num__20.5714285714 = num__20.57 days ans = d <eor> d <eos> |
d |
multiply__2.0__2880.0__ multiply__20.0__14.0__ divide__5760.0__280.0__ round__20.5714__ round__20.5714__ |
multiply__2.0__2880.0__ multiply__20.0__14.0__ divide__5760.0__280.0__ round__20.5714__ multiply__1.0__20.57__ |
| the average age of a committee of num__10 members is num__25 years . a member aged num__55 years retired and his place was taken by another member aged num__35 years . the average age of present committee is ; <o> a ) num__23 years <o> b ) num__38 years <o> c ) num__36 years <o> d ) num__35 years <o> e ) num__37 years |
total age of the committee = num__25 * num__10 = num__250 total age when a member is retired and a new one was joined = num__250 - num__55 + num__35 = num__230 average age of present committee = num__23.0 = num__23 . answer : a <eor> a <eos> |
a |
multiply__10.0__25.0__ divide__230.0__10.0__ divide__230.0__10.0__ |
multiply__10.0__25.0__ divide__230.0__10.0__ divide__230.0__10.0__ |
| a board num__6 ft . num__9 inches long is divided into num__3 equal parts . what is the length of each part ? <o> a ) num__5 ft . num__7 inches <o> b ) num__3 ft . num__7 inches <o> c ) num__2 ft . num__3 inches <o> d ) num__2 ft . num__7 inches <o> e ) num__1 ft . num__7 inches |
length of board = num__6 ft . num__9 inches = ( num__6 * num__12 + num__9 ) inches = num__81 inches . therefore length of each part = ( num__27.0 ) inches = num__27 inches = num__2 ft . num__3 inches answer is c . <eor> c <eos> |
c |
add__9.0__3.0__ multiply__9.0__3.0__ divide__6.0__3.0__ divide__6.0__3.0__ |
add__9.0__3.0__ multiply__9.0__3.0__ divide__6.0__3.0__ divide__6.0__3.0__ |
| a company pays num__18.5 dividend to its investors . if an investor buys rs . num__50 shares and gets num__25.0 on investment at what price did the investor buy the shares ? <o> a ) num__25 <o> b ) num__66 <o> c ) num__18 <o> d ) num__37 <o> e ) num__01 |
explanation : dividend on num__1 share = ( num__18.5 * num__50 ) / num__100 = rs . num__9.25 rs . num__25 is income on an investment of rs . num__100 rs . num__9.25 is income on an investment of rs . ( num__9.25 * num__100 ) / num__25 = rs . num__37 answer : d <eor> d <eos> |
d |
percent__18.5__50.0__ percent__100.0__37.0__ |
percent__18.5__50.0__ percent__100.0__37.0__ |
| a man saves num__30.0 of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by num__30.0 he is only able to save rs . num__300 per month . what is his monthly salary ? <o> a ) rs . num__4000 <o> b ) rs . num__3750 <o> c ) rs . num__5000 <o> d ) rs . num__5500 <o> e ) rs . num__6500 |
income = rs . num__100 expenditure = rs . num__70 savings = rs . num__30 present expenditure num__70 + num__70 * ( num__0.3 ) = rs . num__91 present savings = num__100 – num__91 = rs . num__8 if savings is rs . num__8 salary = rs . num__100 if savings is rs . num__300 salary = num__12.5 * num__300 = num__3750 answer : b <eor> b <eos> |
b |
subtract__100.0__30.0__ divide__30.0__100.0__ divide__100.0__8.0__ multiply__300.0__12.5__ multiply__300.0__12.5__ |
subtract__100.0__30.0__ divide__30.0__100.0__ divide__100.0__8.0__ multiply__300.0__12.5__ multiply__300.0__12.5__ |
| what least number should be subtracted from num__3381 so that the remainder when divided by num__9 num__11 and num__17 will leave in each case the same remainder num__8 ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__12 <o> e ) num__15 |
the lcm of num__9 num__11 and num__17 is num__1683 . the next multiple is num__2 * num__1683 = num__3366 . num__3366 + { remainder } = num__3366 + num__8 = num__3374 which is num__7 less than num__3381 . answer : b . <eor> b <eos> |
b |
subtract__11.0__9.0__ multiply__2.0__1683.0__ add__8.0__3366.0__ subtract__3381.0__3374.0__ subtract__3381.0__3374.0__ |
subtract__11.0__9.0__ multiply__2.0__1683.0__ add__8.0__3366.0__ subtract__3381.0__3374.0__ subtract__3381.0__3374.0__ |
| company x provides bottled water to its employees in num__10 liter bottles $ num__2 each . at least how many num__10 l bottles must company x buy monthly so that new contract with $ num__50 fixed monthly rate and $ num__1 for num__50 l bottle each paid off ? ( assume that no other costs apply ) <o> a ) num__28 <o> b ) num__29 <o> c ) num__30 <o> d ) num__31 <o> e ) num__32 |
let the no . of num__10 liter bottles be x so the no . of num__50 l bottles will be x / num__5 ( to equate the vol . ) since the total cost will be equal num__2 x = num__50 + x / num__5 so x = num__27.77 or num__28 . answer is ( a ) . <eor> a <eos> |
a |
divide__10.0__2.0__ multiply__1.0__28.0__ |
divide__10.0__2.0__ divide__28.0__1.0__ |
| x completes a work in num__120 days and y complete the same work in num__30 days . if both of them work together then the number of days required to complete the work will be ? <o> a ) num__24 days . <o> b ) num__22 days . <o> c ) num__34 days . <o> d ) num__40 days . <o> e ) num__45 days . |
if x can complete a work in x days and y can complete the same work in y days then both of them together can complete the work in x y / x + y days therefore here the required number of days = num__120 Ã — num__0.2 = num__24 days . a ) <eor> a <eos> |
a |
multiply__120.0__0.2__ round__24.0__ |
multiply__120.0__0.2__ round__24.0__ |
| the lcm of two numbers is num__2310 and hcf is num__30 . if one of the numbers is num__210 . then what is the other number ? <o> a ) num__715 <o> b ) num__825 <o> c ) num__330 <o> d ) num__582 <o> e ) num__465 |
first number * second number = lcm * hcf other number = num__2310 * num__0.142857142857 = num__11 * num__30 = num__330 answer : c <eor> c <eos> |
c |
divide__30.0__210.0__ divide__2310.0__210.0__ multiply__30.0__11.0__ round__330.0__ |
divide__30.0__210.0__ divide__2310.0__210.0__ multiply__30.0__11.0__ multiply__30.0__11.0__ |
| mixture contains alcohol and water in the ratio num__4 : num__3 . if num__8 liters of water is added to the mixture the ratio becomes num__4 : num__5 . find the quantity of alcohol in the given mixture . <o> a ) num__10 <o> b ) num__99 <o> c ) num__27 <o> d ) num__4 <o> e ) num__29 |
let the quantity of alcohol and water be num__4 x litres and num__3 x litres respectively num__32 x = num__4 ( num__3 x + num__5 ) num__20 x = num__20 x = num__1 quantity of alcohol = ( num__4 x num__1 ) litres = num__4 litres . answer : d <eor> d <eos> |
d |
multiply__4.0__8.0__ multiply__4.0__5.0__ subtract__4.0__3.0__ multiply__4.0__1.0__ |
multiply__4.0__8.0__ multiply__4.0__5.0__ subtract__4.0__3.0__ add__3.0__1.0__ |
| a is num__20 percent more efficient than b . if the two person can complete a piece of work in num__60 days . in how many days . a working alone can complete the work <o> a ) num__110 <o> b ) num__277 <o> c ) num__287 <o> d ) num__279 <o> e ) num__2781 |
as a is num__20.0 more efficient than b if b ' s per day work is num__100 units then a ' s num__120 . both persons together completes ( num__100 + num__120 ) units = num__220 units a day . they took num__60 days to complete the work . so total work = num__60 x num__220 if a alone set to complete the work he takes = num__60 × num__220120 = num__11060 × num__220120 = num__110 days answer : a <eor> a <eos> |
a |
add__20.0__100.0__ add__100.0__120.0__ round__110.0__ |
add__20.0__100.0__ add__100.0__120.0__ round__110.0__ |
| num__65 + num__5 * num__12 / ( num__60.0 ) = ? <o> a ) num__22 <o> b ) num__77 <o> c ) num__29 <o> d ) num__66 <o> e ) num__21 |
num__65 + num__5 * num__12 / ( num__60.0 ) = num__65 + num__5 * num__12 / ( num__60 ) = num__65 + ( num__5 * num__12 ) / num__60 = num__65 + num__1 = num__66 . answer : d <eor> d <eos> |
d |
add__65.0__1.0__ add__65.0__1.0__ |
add__65.0__1.0__ add__65.0__1.0__ |
| mother is aged num__3 times more than her daughter rose . after num__8 years she would be two and a num__0.5 times of rose ' s age . after further num__8 years how many times would he be of rose ' s age ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__5 <o> e ) num__6 |
let ronit ' s present age be x years . then father ' s present age = ( x + num__3 x ) years = num__4 x years . ( num__4 x + num__8 ) = num__5 ( x + num__8 ) num__2 num__8 x + num__16 = num__5 x + num__40 num__3 x = num__24 x = num__8 . hence required ratio = ( num__4 x + num__16 ) / ( x + num__16 ) = num__2.0 = num__2 . b <eor> b <eos> |
b |
multiply__8.0__0.5__ subtract__8.0__3.0__ reverse__0.5__ divide__8.0__0.5__ multiply__8.0__5.0__ multiply__3.0__8.0__ reverse__0.5__ |
multiply__8.0__0.5__ subtract__8.0__3.0__ reverse__0.5__ divide__8.0__0.5__ multiply__8.0__5.0__ add__8.0__16.0__ reverse__0.5__ |
| if ( num__1 – num__1.25 ) n = num__3 then n = <o> a ) − num__400 <o> b ) − num__140 <o> c ) − num__4 <o> d ) num__4 <o> e ) - num__12 |
( num__1 – num__1.25 ) n = num__3 simplify to get : - num__0.25 n = num__3 rewrite as ( - num__0.25 ) n = num__3 multiply both sides by - num__4 to get : n = - num__12 answer : e <eor> e <eos> |
e |
subtract__1.25__1.0__ reverse__0.25__ multiply__3.0__4.0__ multiply__1.0__12.0__ |
subtract__1.25__1.0__ reverse__0.25__ multiply__3.0__4.0__ multiply__1.0__12.0__ |
| if a and b are positive integers which of the following can not be odd ? <o> a ) ( num__2 + num__4 a ) / ( num__4 + num__4 b ) <o> b ) ( num__4 a ) / ( b ) <o> c ) a / b <o> d ) ( num__4 + a ) / ( num__2 + num__4 b ) <o> e ) ( num__4 + a ) / ( num__1 + num__4 b ) |
a . ( num__2 + num__4 a ) / ( num__4 + num__4 b ) - under any condition this ca n ' t be odd . answer b . ( num__4 a ) / ( b ) = num__4.1 / num__4 = num__1 = odd c . a / b = num__1.0 = odd d . ( num__4 + a ) / ( num__2 + num__4 b ) = ( num__4 + num__2 ) / ( num__2 + num__4.1 ) = num__1 = odd e . ( num__4 + a ) / ( num__1 + num__4 b ) = ( num__4 + num__1 ) / ( num__1 + num__4.1 ) = num__1 = odd answer : a <eor> a <eos> |
a |
multiply__2.0__1.0__ |
divide__2.0__1.0__ |
| mary rolls a die num__5 times . what is the probability that she gets a one on the last one of the five rolls ? <o> a ) num__0.1 <o> b ) num__0.999871399177 <o> c ) none <o> d ) num__1 <o> e ) num__1.00794551646 |
the die has num__6 sides and it was rolle num__5 times then num__6 * num__6 * num__6 * num__6 * num__6 = num__7776 the number of ways to get num__1 is num__1 then num__0.000128600823045 the probability would be num__1 - num__0.000128600823045 = num__0.999871399177 answer b <eor> b <eos> |
b |
die_space__ negate_prob__0.0001__ negate_prob__0.0001__ |
die_space__ negate_prob__0.0001__ negate_prob__0.0001__ |
| sakshi can do a piece of work in num__20 days . tanya is num__25.0 more efficient than sakshi . the number of days taken by tanya to do the same piece of work is ? <o> a ) num__15 <o> b ) num__16 <o> c ) num__18 <o> d ) num__25 <o> e ) num__20 |
explanation : ratio of times taken by sakshi and tanys = num__125 : num__100 = num__5 : num__4 suppose tanya takes x days to do the work . num__5 : num__4 : : num__20 : x = > x = num__16 days . hence tanya takes num__16 days to complete the work . answer is b <eor> b <eos> |
b |
subtract__125.0__25.0__ subtract__25.0__20.0__ divide__20.0__5.0__ subtract__20.0__4.0__ round__16.0__ |
subtract__125.0__25.0__ subtract__25.0__20.0__ divide__20.0__5.0__ subtract__20.0__4.0__ round__16.0__ |
| three grades of milk are num__1 percent num__4 percent and num__3 percent fat by volume . if x gallons of the num__1 percent grade y gallons of the num__2 percent grade and z gallons of the num__3 percent grade are mixed to give x + y + z gallons of a num__1.5 percent grade what is x in terms of y and z ? <o> a ) y + num__3 z <o> b ) ( y + z ) / num__4 <o> c ) num__2 y + num__3 z <o> d ) num__3 y + z <o> e ) num__4 y + num__3 z |
soln : the resulting equation is = > ( . num__01 x + . num__04 y + . num__03 z ) / ( x + y + z ) = num__1.5 / num__100 = > x + num__4 y + num__3 z = num__1.5 x + num__1.5 y + num__1.5 z taking x to one side and y and z to other side we get = > x = num__4 y + num__3 z ans is e <eor> e <eos> |
e |
multiply__1.0__4.0__ |
add__1.0__3.0__ |
| peter bought an item at num__20.0 discount on its original price . he sold it with num__40.0 increase on the price he bought it . the sale price is by what percent more than the original price ? <o> a ) num__7.2 <o> b ) num__7.5 <o> c ) num__10 <o> d ) num__12 <o> e ) none |
sol . let the original price be rs . num__100 . then c . p . = rs . num__80 . s . p . = num__140.0 of rs . num__80 = rs . [ num__112 - num__100.0 ] = num__12.0 . answer d <eor> d <eos> |
d |
percent__80.0__140.0__ percent__100.0__12.0__ |
percent__80.0__140.0__ percent__100.0__12.0__ |
| if p and q are positive integers how many integers are larger than pq and smaller than p ( q + num__5 ) ? <o> a ) num__3 <o> b ) p + num__2 <o> c ) p – num__2 <o> d ) num__2 p – num__1 <o> e ) num__5 p - num__1 |
the number of integers between x and y where x > y is ( x - y ) - num__1 . for example the number of integers between num__1 and num__5 is ( num__5 - num__1 ) - num__1 = num__3 : num__2 num__3 and num__4 . thus the number of integers between pq and p ( q + num__5 ) = pq + num__5 p is ( pq + num__5 p - pq ) - num__1 = num__5 p - num__1 . answer : e . <eor> e <eos> |
e |
subtract__5.0__3.0__ subtract__5.0__1.0__ multiply__5.0__1.0__ |
subtract__5.0__3.0__ subtract__5.0__1.0__ add__1.0__4.0__ |
| a b and c have rs . num__500 between them a and c together have rs . num__200 and b and c rs . num__350 . how much does c have ? <o> a ) num__50 <o> b ) num__49 <o> c ) num__87 <o> d ) num__76 <o> e ) num__66 |
a + b + c = num__500 a + c = num__200 b + c = num__350 - - - - - - - - - - - - - - a + b + num__2 c = num__550 a + b + c = num__500 - - - - - - - - - - - - - - - - c = num__50 answer : a <eor> a <eos> |
a |
add__200.0__350.0__ subtract__550.0__500.0__ subtract__550.0__500.0__ |
add__200.0__350.0__ subtract__550.0__500.0__ subtract__550.0__500.0__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__388 <o> b ) num__378 <o> c ) num__240 <o> d ) num__388 <o> e ) num__771 |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x meters . then x + num__8.33333333333 = num__15 x + num__300 = num__540 x = num__240 m . answer : c <eor> c <eos> |
c |
multiply__20.0__15.0__ divide__300.0__36.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ divide__300.0__36.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
| the average age of applicants for a new job is num__32 with a standard deviation of num__8 . the hiring manager is only willing to accept applicants whose age is within one standard deviation of the average age . assuming that all applicants ' ages are integers and that the endpoints of the range are included what is the maximum number of different ages of the applicants ? <o> a ) num__8 <o> b ) num__77 <o> c ) num__7 <o> d ) num__18 <o> e ) num__34 |
minimum age = average - num__1 standard deviation = num__32 - num__8 = num__24 maximum age = average + num__1 standard deviation = num__32 + num__8 = num__40 maximum number of different ages of the applicants = num__30 - num__24 + num__1 = num__7 answer c <eor> c <eos> |
c |
subtract__32.0__8.0__ add__32.0__8.0__ subtract__8.0__1.0__ subtract__8.0__1.0__ |
subtract__32.0__8.0__ add__32.0__8.0__ subtract__8.0__1.0__ subtract__8.0__1.0__ |
| peter has $ num__642986 in his savings account . what is the least amount of money ( in whole number of dollars ) that he must add to his account if he wants to split this money evenly among his nine children ? <o> a ) $ num__642986 <o> b ) $ num__642987 <o> c ) $ num__642988 <o> d ) $ num__642989 <o> e ) $ num__642 |
990 |
to find the least amount the man should add to his saving account to split the money evenly among his num__9 children he needs to make the total divisible by num__9 simply add the individual digits of the total = num__6 + num__4 + num__2 + num__9 + num__8 + num__6 = num__35 if you add num__1 the number is divisible by num__9 ( num__35 + num__1 ) correct option : b <eor> b <eos> |
b |
b |
| the ratio of two natural numbers is num__5 : num__6 . if a certain number is added to both the numbers the ratio becomes num__7 : num__8 . if the larger number exceeds the smaller number by num__10 find the number added ? <o> a ) num__27 <o> b ) num__298 <o> c ) num__20 <o> d ) num__28 <o> e ) num__12 |
let the two numbers be num__5 x and num__6 x . let the number added to both so that their ratio becomes num__7 : num__8 be k . ( num__5 x + k ) / ( num__6 x + k ) = num__0.875 num__42 x = num__7 k = > k = num__2 x . num__6 x - num__5 x = num__10 = > x = num__10 k = num__2 x = num__20 . answer : c <eor> c <eos> |
c |
divide__7.0__8.0__ multiply__6.0__7.0__ subtract__7.0__5.0__ multiply__10.0__2.0__ multiply__10.0__2.0__ |
divide__7.0__8.0__ multiply__6.0__7.0__ subtract__7.0__5.0__ multiply__10.0__2.0__ multiply__10.0__2.0__ |
| the average age of the mother and her six children is num__12 years which is reduced by num__5 years if the age of the mother is excluded . how old is the mother ? <o> a ) num__40 years <o> b ) num__42 years <o> c ) num__50 years <o> d ) num__52 years <o> e ) none |
solution age of the mother = ( num__12 x num__7 - num__7 x num__6 ) = num__42 years . answer b <eor> b <eos> |
b |
subtract__12.0__5.0__ multiply__6.0__7.0__ multiply__6.0__7.0__ |
subtract__12.0__5.0__ multiply__6.0__7.0__ multiply__6.0__7.0__ |
| an aeroplane covers a certain distance at a speed of num__240 kmph in num__5 hours . to cover the same distance in num__1 num__0.666666666667 hours it must travel at a speed of : <o> a ) num__300 km / hour <o> b ) num__360 km / hour <o> c ) num__600 km / hour <o> d ) num__720 km / hour <o> e ) none of these |
distance = ( num__240 * num__5 ) km = num__1200 km . required speed = ( num__1200 * num__0.6 ) km / hr = num__720 kilometer / hour . correct option : d <eor> d <eos> |
d |
multiply__240.0__5.0__ km_to_mile_conversion__ multiply__1200.0__0.6__ round__720.0__ |
multiply__240.0__5.0__ km_to_mile_conversion__ multiply__1200.0__0.6__ multiply__1.0__720.0__ |
| carina has num__85 ounces of coffee divided into num__5 - and num__10 - ounce packages . if she has num__2 more num__5 - ounce packages than num__10 - ounce packages how many num__10 - ounce packages does she have ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
lets say num__5 and num__10 ounce packages be x and y respectively . given that num__5 x + num__10 y = num__85 and x = y + num__2 . what is the value of y . substituting the x in first equation num__5 y + num__10 + num__10 y = num__85 - > y = num__5.0 . = num__5 a <eor> a <eos> |
a |
subtract__10.0__5.0__ |
subtract__10.0__5.0__ |
| sundar invested rs . num__4455 in rs . num__10 shares quoted at rs . num__8.25 . if the rate of dividend be num__12.0 his annual income <o> a ) num__600 <o> b ) num__648 <o> c ) num__500 <o> d ) num__700 <o> e ) num__750 |
no of shares = num__4455 / num__8.25 = num__540 face value = num__540 * num__10 = num__5400 income = num__0.12 * num__5400 = num__648 answer b <eor> b <eos> |
b |
percent__12.0__5400.0__ percent__12.0__5400.0__ |
percent__12.0__5400.0__ percent__12.0__5400.0__ |
| in a family each daughter has the same number of brothers as she has sisters and each son has twice as many sisters as he has brothers . how many sons are there in the family ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
explanation : let d and s represent the number of daughters and sons respectively . then we have : d - num__1 = s and num__2 ( s - num__1 ) = d . solving these two equations we get : d = num__4 s = num__3 . answer : b <eor> b <eos> |
b |
twice__1.0__ twice__2.0__ triple__1.0__ triple__1.0__ |
twice__1.0__ twice__2.0__ subtract__4.0__1.0__ subtract__4.0__1.0__ |
| [ ( num__5.5 ÷ num__0.5 × num__5.5 ) ] / [ ( num__5.5 × num__0.5 ÷ num__5.5 ) ] = ? <o> a ) num__111 <o> b ) num__101 <o> c ) num__91 <o> d ) num__81 <o> e ) num__121 |
explanation : [ ( num__5.5 ÷ num__0.5 × num__5.5 ) ] / [ ( num__5.5 × num__0.5 ÷ num__5.5 ) ] = [ num__5.5 × num__2.0 × num__5.5 ] / [ num__5.5 × num__0.5 × num__0.181818181818 ] = num__60.5 / num__0.5 = num__60.5 x num__2.0 = num__121 answer : option e <eor> e <eos> |
e |
reverse__0.5__ reverse__5.5__ multiply__2.0__60.5__ multiply__2.0__60.5__ |
reverse__0.5__ reverse__5.5__ divide__60.5__0.5__ divide__60.5__0.5__ |
| a motorcyclist started riding at highway marker a drove num__120 miles to highway marker b and then without pausing continued to highway marker c where she stopped . the average speed of the motorcyclist over the course of the entire trip was num__25 miles per hour . if the ride from marker a to marker b lasted num__3 times as many hours as the rest of the ride and the distance from marker b to marker c was half of the distance from marker a to marker b what was the average speed in miles per hour of the motorcyclist while driving from marker b to marker c ? <o> a ) num__40 <o> b ) num__45 <o> c ) num__108 <o> d ) num__55 <o> e ) num__60 |
a - b = num__120 miles b - c = num__60 miles avg speed = num__25 miles time taken for a - b num__3 t and b - c be t avg speed = ( num__120 + num__60 ) / total time num__25 = num__45.0 t t = num__108 b - c = num__108 mph answer c <eor> c <eos> |
c |
hour_to_min_conversion__ round__108.0__ |
hour_to_min_conversion__ round__108.0__ |
| b completes a work in num__8 days . a alone can do it in num__10 days . if both work together the work can be completed in how many days ? <o> a ) num__3.75 days <o> b ) num__4.44 days <o> c ) num__5 days <o> d ) num__6.44 days <o> e ) num__7 days |
num__0.125 + num__0.1 = num__0.225 num__4.44444444444 = num__4.44 days answer : b <eor> b <eos> |
b |
add__0.125__0.1__ round__4.44__ |
add__0.125__0.1__ round__4.44__ |
| the sum of ages of num__4 children born num__2 years different each is num__40 yrs . what is the age of the elder child ? <o> a ) num__13 <o> b ) num__9 <o> c ) num__16 <o> d ) num__17 <o> e ) num__18 |
let the ages of children be x ( x + num__2 ) ( x + num__4 ) ( x + num__6 ) years . then x + ( x + num__2 ) + ( x + num__4 ) + ( x + num__6 ) = num__40 num__4 x + num__12 = num__40 = > num__4 x = num__28 x = num__7 x + num__6 = num__7 + num__6 = num__13 answer : a <eor> a <eos> |
a |
add__4.0__2.0__ multiply__2.0__6.0__ subtract__40.0__12.0__ divide__28.0__4.0__ add__6.0__7.0__ add__6.0__7.0__ |
add__4.0__2.0__ multiply__2.0__6.0__ subtract__40.0__12.0__ divide__28.0__4.0__ add__6.0__7.0__ add__6.0__7.0__ |
| lisa and robert have taken the same number of photos on their school trip . lisa has taken num__3 times as many photos as claire and robert has taken num__28 more photos than claire . how many photos has claire taken ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__12 <o> e ) num__14 |
l = r l = num__3 c r = c + num__28 num__3 c = c + num__28 c = num__14 the answer is e . <eor> e <eos> |
e |
subtract__28.0__14.0__ |
subtract__28.0__14.0__ |
| if the wheel is num__42 cm then the number of revolutions to cover a distance of num__1056 cm is ? <o> a ) num__18 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__19 |
num__2 * num__3.14285714286 * num__42 * x = num__1056 = > x = num__4 answer : b <eor> b <eos> |
b |
round__4.0__ |
round__4.0__ |
| the present worth of rs . num__1404 due in two equal half - yearly installments at num__8.0 per annum simple interest is : <o> a ) rs . num__1325 <o> b ) rs . num__1300 <o> c ) rs . num__1350 <o> d ) rs . num__1500 <o> e ) rs . num__1800 |
solution required sum = p . w . of rs . num__702 due num__6 months hence + p . w . of rs . num__702 due num__1 year hence = rs . [ ( num__100 x num__7.02 + num__8 x num__0.5 ) + ( num__100 x num__7.02 + num__8 x num__1 ) ] = rs . ( num__675 + num__650 ) = rs . num__1325 . answer a <eor> a <eos> |
a |
percent__1.0__702.0__ percent__100.0__1325.0__ |
percent__1.0__702.0__ percent__100.0__1325.0__ |
| if a : b = num__2 : num__3 and b : c = num__5 : num__7 then find a : b : c . <o> a ) num__10 : num__15 : num__27 <o> b ) num__10 : num__15 : num__21 <o> c ) num__10 : num__15 : num__22 <o> d ) num__10 : num__15 : num__12 <o> e ) num__10 : num__15 : num__28 |
a : b : c = ( num__2 x num__5 ) : ( num__3 x num__5 ) : ( num__3 x num__7 ) = ( num__10 ) : ( num__15 ) : ( num__21 ) answer : a <eor> a <eos> |
a |
multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__3.0__7.0__ multiply__2.0__5.0__ |
multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__3.0__7.0__ multiply__2.0__5.0__ |
| jaydeep purchased num__25 kg of rice at the rate of num__16.50 per kg and num__35 kg of rice at the rate of num__24.50 per kg . he mixed the two and sold the mixture . approximately at what price per kg did he sell the mixture to make num__25 per cent profit ? <o> a ) num__26.5 <o> b ) num__27.5 <o> c ) num__28.5 <o> d ) num__30.0 <o> e ) num__29.0 |
cp = num__25 × num__16.50 + num__35 × num__24.50 = num__1270 sp = num__1270 × num__125 ⁄ num__100 = num__1587.50 price per kg = num__1587.50 ⁄ num__60 ≈ num__26.50 answer a <eor> a <eos> |
a |
percent__100.0__26.5__ |
percent__100.0__26.5__ |
| a car covers a distance of num__624 km in num__2 num__0.4 hours . find its speed ? <o> a ) num__104 kmph <o> b ) num__194 kmph <o> c ) num__109 kmph <o> d ) num__260 kmph <o> e ) num__271 kmph |
num__312.0 num__0.4 = num__260 kmph answer : d <eor> d <eos> |
d |
divide__624.0__2.0__ round__260.0__ |
divide__624.0__2.0__ round__260.0__ |
| a man traveled a total distance of num__900 km . he traveled one - third of the whole trip by plane and the distance traveled by train is two - thirds of the distance traveled by bus . if he traveled by train plane and bus how many kilometers did he travel by bus ? <o> a ) num__240 <o> b ) num__280 <o> c ) num__320 <o> d ) num__360 <o> e ) num__400 |
total distance traveled = num__900 km . distance traveled by plane = num__300 km . distance traveled by bus = x distance traveled by train = num__2 x / num__3 x + num__2 x / num__3 + num__300 = num__900 num__5 x / num__3 = num__600 x = num__360 km the answer is d . <eor> d <eos> |
d |
divide__900.0__300.0__ add__2.0__3.0__ subtract__900.0__300.0__ round__360.0__ |
divide__900.0__300.0__ add__2.0__3.0__ subtract__900.0__300.0__ round__360.0__ |
| the difference between a two - digit number and the number obtained by interchanging the digits is num__36 . what is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is num__1 : num__2 ? <o> a ) num__4 <o> b ) num__8 <o> c ) num__10 <o> d ) num__9 <o> e ) num__11 |
since the number is greater than the number obtained on reversing the digits so the ten ' s digit is greater than the unit ' s digit . let ten ' s and unit ' s digits be num__2 x and x respectively . then ( num__10 x num__2 x + x ) - ( num__10 x + num__2 x ) = num__36 num__9 x = num__36 x = num__4 . required difference = ( num__2 x + x ) - ( num__2 x - x ) = num__2 x = num__8 . answer : b <eor> b <eos> |
b |
subtract__10.0__1.0__ divide__36.0__9.0__ multiply__2.0__4.0__ multiply__1.0__8.0__ |
subtract__10.0__1.0__ divide__36.0__9.0__ subtract__9.0__1.0__ subtract__9.0__1.0__ |
| if w is divisible by num__10 and num__4 w must be a multiple of which of the following ? <o> a ) num__8 <o> b ) num__12 <o> c ) num__20 <o> d ) num__24 <o> e ) num__36 |
if w is divisible by num__2 * num__2 and num__2 * num__5 then w is divisible by num__2 * num__2 * num__5 = num__20 . the answer is c . <eor> c <eos> |
c |
divide__10.0__2.0__ multiply__10.0__2.0__ multiply__10.0__2.0__ |
divide__10.0__2.0__ multiply__10.0__2.0__ multiply__10.0__2.0__ |
| what amount does kiran get if he invests rs . num__18000 at num__15.0 p . a . simple interest for four years ? <o> a ) num__3888 <o> b ) num__26989 <o> c ) num__26700 <o> d ) num__28800 <o> e ) num__2879 |
simple interest = ( num__18000 * num__4 * num__15 ) / num__100 = rs . num__10800 amount = p + i = num__18000 + num__10800 = rs . num__28800 answer : d <eor> d <eos> |
d |
percent__100.0__28800.0__ |
percent__100.0__28800.0__ |
| a cistern has a leak which would empty the cistern in num__20 minutes . a tap is turned on which admits num__3 liters a minute into the cistern and it is emptied in num__24 minutes . how many liters does the cistern hold ? <o> a ) num__360 <o> b ) num__487 <o> c ) num__481 <o> d ) num__729 <o> e ) num__268 |
num__1 / x - num__0.05 = - num__0.0416666666667 x = num__120 num__120 * num__3 = num__360 answer : a <eor> a <eos> |
a |
divide__1.0__20.0__ divide__1.0__24.0__ multiply__3.0__120.0__ round__360.0__ |
divide__1.0__20.0__ divide__1.0__24.0__ multiply__3.0__120.0__ multiply__3.0__120.0__ |
| a parking garage rents parking spaces for $ num__10 per week or $ num__24 per month . how much does a person save in a year by renting by the month rather than by the week ? <o> a ) $ num__140 <o> b ) $ num__160 <o> c ) $ num__220 <o> d ) $ num__232 <o> e ) $ num__260 |
num__10 $ per week ! an year has num__52 weeks . annual charges per year = num__52 * num__10 = num__520 $ num__30 $ per month ! an year has num__12 months . annual charges per year = num__12 * num__24 = num__288 $ num__520 - num__288 = num__232 ans d <eor> d <eos> |
d |
multiply__10.0__52.0__ multiply__24.0__12.0__ subtract__520.0__288.0__ subtract__520.0__288.0__ |
multiply__10.0__52.0__ multiply__24.0__12.0__ subtract__520.0__288.0__ subtract__520.0__288.0__ |
| a train crosses a platform of num__150 m in num__15 sec same train crosses another platform of length num__250 m in num__20 sec . then find the length of the train ? <o> a ) num__150 m <o> b ) num__120 m <o> c ) num__186 m <o> d ) num__167 m <o> e ) num__189 m |
length of the train be ‘ x ’ x + num__10.0 = x + num__12.5 num__4 x + num__600 = num__3 x + num__750 x = num__150 m answer : a <eor> a <eos> |
a |
divide__150.0__15.0__ divide__250.0__20.0__ multiply__150.0__4.0__ add__150.0__600.0__ round__150.0__ |
divide__150.0__15.0__ divide__250.0__20.0__ multiply__150.0__4.0__ add__150.0__600.0__ round__150.0__ |
| the first photo shoot takes num__3 minutes long and then the following shoots are taken at a rate of num__25 seconds / shoot as the model is already at the scene . what is the maximum number of photo shoots taken under num__10 minuntes ? <o> a ) num__13 <o> b ) num__14 <o> c ) num__15 <o> d ) num__16 <o> e ) num__17 |
a must be an integer as it is the number shoots at a rate of num__25 sec / shoot num__3 * num__60 + num__25 a = num__10 * num__60 num__25 a = num__420 a = num__16 the total number of shoots - - > num__1 + num__16 = num__17 and num__18 th shoot will be taken at num__605 seconds which is above num__10 minutes answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ add__1.0__16.0__ add__1.0__17.0__ round__17.0__ |
hour_to_min_conversion__ add__1.0__16.0__ add__1.0__17.0__ add__1.0__16.0__ |
| what is the length of the longest pole which can be kept in a room num__12 m long num__4 m broad and num__2 m high ? <o> a ) num__7 <o> b ) num__12.8 <o> c ) num__11 <o> d ) num__13 <o> e ) none |
explanation : d num__2 = num__122 + num__42 + num__22 = num__12.8 b ) <eor> b <eos> |
b |
round__12.8__ |
round__12.8__ |
| mangala completes a piece of work in num__10 days raju completes the same work in num__40 days . if both of them work together then the number of days required to complete the work is ? <o> a ) num__8 days <o> b ) num__12 days <o> c ) num__14 days <o> d ) num__16 days <o> e ) num__18 days |
if a can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days . that is the required no . of days = num__10 × num__0.8 = num__8 days a ) <eor> a <eos> |
a |
multiply__10.0__0.8__ round__8.0__ |
multiply__10.0__0.8__ round__8.0__ |
| raghu working alone can paint a room in num__6 hours . peter and john working independently can paint the same room in num__3 hours and num__2 hours respectively . tom starts painting the room and works on his own for one hour . he is then joined by peter and they work together for an hour . finally john joins them and the three of them work together to finish the room each one working at his respective rate . what fraction of the whole job was done by peter ? <o> a ) num__0.444444444444 <o> b ) num__0.166666666667 <o> c ) num__0.333333333333 <o> d ) num__0.388888888889 <o> e ) num__0.111111111111 |
let ' s use smart numbers here - - > work = num__18 rate * time = work tom : num__3 x num__6 = num__18 peter : num__6 x num__3 = num__18 john : num__9 x num__2 = num__18 before john joined tom and peter : tom worked num__2 hours - > num__2 * num__3 = num__6 and peter num__1 * num__6 = num__6 gives us num__12 . so we are left with num__18 - num__12 = num__6 for all three of them - - > ( num__3 + num__6 + num__9 ) * t = num__6 thus t = num__0.333333333333 this means that peter worked num__2 + num__0.333333333333 hours = num__6 + num__2 = num__8 - - > num__0.444444444444 = num__0.444444444444 at least this approach helps me . . . do n ' t like dealind with fractions when you ' re tired . answer : a <eor> a <eos> |
a |
multiply__6.0__3.0__ add__6.0__3.0__ subtract__3.0__2.0__ multiply__6.0__2.0__ divide__6.0__18.0__ add__6.0__2.0__ divide__8.0__18.0__ multiply__1.0__0.4444__ |
multiply__6.0__3.0__ add__6.0__3.0__ subtract__3.0__2.0__ multiply__6.0__2.0__ divide__6.0__18.0__ add__6.0__2.0__ divide__8.0__18.0__ multiply__1.0__0.4444__ |
| what is the measure of the radius of the circle inscribed in a triangle whose sides measure num__4 num__11 and num__12 units ? <o> a ) num__1.6 units <o> b ) num__6 units <o> c ) num__3 units <o> d ) num__5 units <o> e ) num__12 units |
sides are num__4 num__11 and num__12 . . . thus it is right angle triangle since num__12 ^ num__2 = num__4 ^ num__2 + num__11 ^ num__2 therefore area = num__0.5 * num__11 * num__4 = num__22 we have to find in - radius therefore area of triangle = s * r . . . . where s = semi - perimeter and r = in - radius now s = semi - perimeter = num__12 + num__11 + num__2.0 = num__135 thus num__22 = num__135 * r and hence r = in - radius = num__1.6 option a <eor> a <eos> |
a |
triangle_area__4.0__11.0__ triangle_area__2.0__1.6__ |
multiply__11.0__2.0__ volume_rectangular_prism__0.5__2.0__1.6__ |
| the parameter of a square is equal to the perimeter of a rectangle of length num__16 cm and breadth num__14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) <o> a ) num__23.56 <o> b ) num__23.54 <o> c ) num__23.51 <o> d ) num__23.52 <o> e ) num__23.57 |
let the side of the square be a cm . parameter of the rectangle = num__2 ( num__16 + num__14 ) = num__60 cm parameter of the square = num__60 cm i . e . num__4 a = num__60 a = num__15 diameter of the semicircle = num__15 cm circimference of the semicircle = num__0.5 ( ∏ ) ( num__15 ) = num__0.5 ( num__3.14285714286 ) ( num__15 ) = num__23.5714285714 = num__23.57 cm to two decimal places answer : e <eor> e <eos> |
e |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
| a garrison of num__2000 men has provisions for num__54 days . at the end of num__21 days a reinforcement arrives and it is now found that the provisions will last only for num__20 days more . what is the reinforcement ? <o> a ) num__1977 <o> b ) num__1893 <o> c ) num__1979 <o> d ) num__1900 <o> e ) num__1300 |
num__2000 - - - - num__54 num__2000 - - - - num__33 x - - - - - num__20 x * num__20 = num__2000 * num__33 x = num__3300 num__2000 - - - - - - - num__1300 answer : e <eor> e <eos> |
e |
subtract__54.0__21.0__ subtract__3300.0__2000.0__ round__1300.0__ |
subtract__54.0__21.0__ subtract__3300.0__2000.0__ subtract__3300.0__2000.0__ |
| carol is three times alice ’ s age but only twice as old as betty . alice is twenty years younger than carol . how old is betty ? <o> a ) num__26 <o> b ) num__29 <o> c ) num__32 <o> d ) num__28 <o> e ) num__15 |
beyond the obvious algebra approach to this question it can also be solved by testing the answers . we ' re told num__3 facts about the relative ages of num__3 people : num__1 ) carol ' s age is num__3 times alice ' s age num__2 ) carol ' s age is num__2 times betty ' s age num__3 ) carol is num__20 years older than alice we ' re asked for betty ' s age . from the answer choices and the information provided carol ca n ' t be that old ( the difference of num__20 years = num__3 times ; that relationship can only occur when the numbers are relatively small ) . since carol is twice betty ' s age betty is clearly younger than carol so we ' ll test a smaller answer first . if . . . . betty = num__15 then carol = num__2 ( num__15 ) = num__30 then alice = num__30 - num__20 = num__10 and num__10.0 = num__10 all of these values mesh perfectly with the facts and with one another so betty must be num__15 e <eor> e <eos> |
e |
subtract__3.0__1.0__ multiply__2.0__15.0__ divide__20.0__2.0__ multiply__1.0__15.0__ |
subtract__3.0__1.0__ multiply__2.0__15.0__ subtract__30.0__20.0__ subtract__30.0__15.0__ |
| a pharmaceutical company received $ num__3 million in royalties on the first $ num__20 million in sales of and then $ num__9 million in royalties on the next $ num__108 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ num__20 million in sales to the next $ num__108 million in sales ? <o> a ) num__8.0 <o> b ) num__15.0 <o> c ) num__45.0 <o> d ) num__52.0 <o> e ) num__56 % |
first ratio = num__0.15 = num__15.0 second ratio = ( num__3 + num__12 ) / ( num__9 + num__108 ) = num__0.128205128205 = num__128.0 decrease = num__22.0 num__22.0 / num__15.0 = num__15.0 approximately . answer : b <eor> b <eos> |
b |
divide__3.0__20.0__ add__3.0__9.0__ add__20.0__108.0__ add__3.0__12.0__ |
divide__3.0__20.0__ add__3.0__9.0__ add__20.0__108.0__ add__3.0__12.0__ |
| joe drives num__600 miles at num__60 miles per hour and then he drives the next num__120 miles at num__40 miles per hour . what is his average speed for the entire trip in miles per hour ? <o> a ) num__42 <o> b ) num__55 <o> c ) num__50 <o> d ) num__54 <o> e ) num__56 |
t num__1 = num__10.0 = num__10 hours t num__2 = num__3.0 = num__3 hours t = t num__1 + t num__2 = num__13 hours avg speed = total distance / t = num__55.3846153846 = num__55 mph = b <eor> b <eos> |
b |
divide__600.0__60.0__ divide__120.0__60.0__ divide__120.0__40.0__ add__3.0__10.0__ round__55.0__ |
divide__600.0__60.0__ divide__120.0__60.0__ divide__120.0__40.0__ add__3.0__10.0__ divide__55.0__1.0__ |
| two pipes a and b can fill a cistern in num__12 and num__20 minutes respectively and a third pipe c can empty it in num__30 minutes . how long will it take to fill the cistern if all the three are opened at the same time ? <o> a ) num__15 min <o> b ) num__13 min <o> c ) num__12 min <o> d ) num__10 min <o> e ) num__17 min |
num__0.0833333333333 + num__0.05 - num__0.0333333333333 = num__0.1 num__10.0 = num__10 answer : d <eor> d <eos> |
d |
subtract__0.0833__0.05__ subtract__30.0__20.0__ round__10.0__ |
subtract__0.0833__0.05__ subtract__30.0__20.0__ subtract__20.0__10.0__ |
| what number should replace the question mark ? num__24 num__30 ? num__60 num__84 num__114 <o> a ) num__45 <o> b ) num__44 <o> c ) num__48 <o> d ) num__42 <o> e ) num__49 |
d num__42 the sequence progresses + num__6 + num__12 + num__18 + num__24 + num__30 . <eor> d <eos> |
d |
subtract__30.0__24.0__ subtract__42.0__30.0__ subtract__24.0__6.0__ add__24.0__18.0__ |
subtract__30.0__24.0__ subtract__42.0__30.0__ add__6.0__12.0__ add__24.0__18.0__ |
| num__4 pipes can fill a reservoir in num__15 num__20 num__30 and num__60 hrs the first was opened at num__6 am and num__2 nd at num__7 am and thirds at num__8 am and num__4 th at num__9 am when will the reservoir be full ? <o> a ) num__6 hrs <o> b ) num__7 hrs <o> c ) num__8 hrs <o> d ) num__9 hrs <o> e ) num__10 hrs |
explanation : let time be t hrs after num__6 am t / num__15 + ( t - num__1 ) / num__30 + ( t - num__2 ) / num__30 + ( t - num__3 ) / num__60 = num__1 t = num__7 hrs answer : option b <eor> b <eos> |
b |
subtract__7.0__6.0__ subtract__4.0__1.0__ add__4.0__3.0__ |
subtract__7.0__6.0__ subtract__4.0__1.0__ add__4.0__3.0__ |
| a man took loan from a bank at the rate of num__8.0 p . a . simple interest . after num__4 years he had to pay $ num__250 interest only for the period . the principal amount borrowed by him was : <o> a ) $ num__881.25 <o> b ) $ num__781.25 <o> c ) $ num__761.35 <o> d ) $ num__661.25 <o> e ) $ num__681.55 |
principal ( p ) = ? time ( t ) = num__4 years simple interest ( si ) = $ num__250 r = num__8.0 p = num__100 × si / rt = num__100 × num__31.25 × num__4 = $ num__781.25 answer : b <eor> b <eos> |
b |
percent__100.0__781.25__ |
percent__100.0__781.25__ |
| a train is approaching a tunnel which is ab in length . ther is a cat inside the tunnel which is num__0.375 distance from the starting point of the tunnel . when the train whistes the cat starts to run . the train catches the cat exactly at the entry point of the tunnel . if the cat runs towards the exit the train catches tha cat exactly at the exit point . the speed of the train is greater than the speed of the cat in what order ? <o> a ) num__1 : num__4 <o> b ) num__4 : num__1 <o> c ) num__1 : num__3 <o> d ) num__3 : num__1 <o> e ) num__1 : num__2 |
d / v num__1 = ( num__0.375 ) x / v num__2 . . . ( num__1 ) ( d + x ) / v num__1 = ( num__0.625 ) x / v num__2 = > d / v num__1 = ( num__0.625 ) x / v num__2 - x / v num__1 . . . ( num__2 ) from equns . ( num__1 ) and ( num__2 ) ( num__0.375 ) x / v num__2 = ( num__0.625 ) x / v num__2 - x / v num__1 = > num__3 / ( num__8 v num__2 ) = num__5 ( num__8 v num__2 ) - num__1 / v num__1 = > num__1 / v num__1 = ( num__0.625 - num__0.375 ) / v num__2 = > num__4 v num__2 = v num__1 answer : b <eor> b <eos> |
b |
subtract__1.0__0.375__ add__1.0__2.0__ divide__3.0__0.375__ add__2.0__3.0__ add__1.0__3.0__ round__4.0__ |
subtract__1.0__0.375__ add__1.0__2.0__ divide__3.0__0.375__ add__2.0__3.0__ add__1.0__3.0__ add__1.0__3.0__ |
| if m and n are positive integers that have remainders of num__2 and num__3 respectively when divided by num__5 which of the following could not be a possible value of m + n ? <o> a ) num__80 <o> b ) num__52 <o> c ) num__35 <o> d ) num__25 <o> e ) num__10 |
m = num__5 p + num__2 n = num__5 q + num__3 we need m + n = num__5 p + num__2 + num__5 q + num__3 = num__5 ( p + q ) + num__5 pick numbers for pq since it is an addition of pq and the answer to this expression should be an integer ( because all the numbers being added are integers ) we just need to choose values so that we get integer multiples of num__5 so p + q = num__0 ; m + n = num__5 p + q = num__1 ; m + n = num__10 p + q = num__2 ; m + n = num__15 and so on so basically you get something like - num__5 num__10 num__1520 num__2530 . . . . . all are multiples of num__5 except num__52 so answer b . <eor> b <eos> |
b |
subtract__3.0__2.0__ multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__1.0__52.0__ |
subtract__3.0__2.0__ multiply__2.0__5.0__ add__5.0__10.0__ multiply__1.0__52.0__ |
| samantha is driving from city a to city b . the distance between the two cities is num__88 miles . she is driving at a constant speed of num__22 miles per hour and has been travelling for half an hour after starting from city a . what percent of the total distance she is yet to cover ? <o> a ) num__87 <o> b ) num__87.5 <o> c ) num__88 <o> d ) num__90 <o> e ) num__97.5 |
she is driving at the rate of num__22 miles per hour for half an hour . means the distance she has covered so far is = speed x time = num__22 miles per hour x num__0.5 hour = num__11 miles . therefore the distance she is yet to cover is num__88 - num__11 = num__77 miles . we want to know num__77 miles is what percent of the total distance ( num__88 miles ) . let it be x % of the total distance . this means x % of num__88 = num__77 solving we get x = num__87.5 ans b <eor> b <eos> |
b |
multiply__22.0__0.5__ subtract__88.0__11.0__ subtract__88.0__0.5__ round__87.5__ |
multiply__22.0__0.5__ subtract__88.0__11.0__ subtract__88.0__0.5__ subtract__88.0__0.5__ |
| laura took out a charge account at the general store and agreed to pay num__8.0 simple annual interest . if she charges $ num__35 on her account in january how much will she owe a year later assuming she does not make any additional charges or payments ? <o> a ) $ num__2.10 <o> b ) $ num__37.10 <o> c ) $ num__37.16 <o> d ) $ num__37.8 <o> e ) $ num__38.80 |
principal that is amount taken by laura at year beginning = num__35 $ rate of interest = num__8.0 interest = ( num__0.08 ) * num__35 = num__2.8 $ total amount that laura owes a year later = num__35 + num__2.8 = num__37.8 $ answer d <eor> d <eos> |
d |
multiply__35.0__0.08__ add__35.0__2.8__ add__35.0__2.8__ |
multiply__35.0__0.08__ add__35.0__2.8__ add__35.0__2.8__ |
| what is the sum of all num__3 digit numbers that leave a remainder of ' num__2 ' when divided by num__3 ? <o> a ) num__897 <o> b ) num__164850 <o> c ) num__164749 <o> d ) num__149700 <o> e ) num__156 |
720 |
explanatory answer step num__1 : identify the series the smallest num__3 digit number that will leave a remainder of num__2 when divided by num__3 is num__101 . the next couple of numbers that will leave a remainder of num__2 when divided by num__3 are num__104 and num__107 . the largest num__3 digit number that will leave a remainder of num__2 when divided by num__3 is num__998 . it is evident that any number in the sequence will be a num__3 digit positive integer of the form ( num__3 n + num__2 ) . so the given numbers are in an arithmetic sequence with the first term being num__101 and the last term being num__998 and the common difference being num__3 . step num__2 : compute the sum sum of an arithmetic progression ( ap ) = [ first term + last term / num__2 ] n we know the first term : num__101 we know the last term : num__998 . the only unknown is the number of terms n . in an a . p . the nth term an = a num__1 + ( n - num__1 ) * d in this case therefore num__998 = num__101 + ( n - num__1 ) * num__3 or num__897 = ( n - num__1 ) * num__3 ( n - num__1 ) = num__299 or n = num__300 . sum of the ap will therefore be [ num__101 + num__499.0 ] ∗ num__300 = num__164850 choice b is the correct answer . <eor> b <eos> |
b |
b |
| insert the missing number . num__8 num__7 num__11 num__12 num__14 num__17 num__17 ( . . . . ) <o> a ) num__27 <o> b ) num__20 <o> c ) num__22 <o> d ) num__24 <o> e ) num__26 |
there are two series ( num__8 num__11 num__14 num__17 ) and ( num__7 num__12 num__17 num__22 ) increasing by num__3 and num__5 respectively . answer : c <eor> c <eos> |
c |
add__8.0__14.0__ subtract__11.0__8.0__ subtract__8.0__3.0__ add__8.0__14.0__ |
add__8.0__14.0__ subtract__11.0__8.0__ subtract__8.0__3.0__ add__8.0__14.0__ |
| car e and car y traveled the same num__80 - mile route . if car e took num__2 hours and car y traveled at an average speed that was num__50 percent faster than the average speed of car e how many hours did it take car y to travel the route ? <o> a ) num__0.666666666667 <o> b ) num__1 <o> c ) num__1.33333333333 <o> d ) num__1.6 <o> e ) num__3 |
the speed of car e is ( distance ) / ( time ) = num__40.0 = num__40 miles per hour . the speed of car y = num__1.5 * num__40 = num__60 miles per hour - - > ( time ) = ( distance ) / ( speed ) = num__1.33333333333 = num__1.33333333333 hours . answer : c . or : to cover the same distance at num__1.5 as fast rate num__0.666666666667 as much time is needed - - > ( time ) * num__0.666666666667 = num__2 * num__0.666666666667 = num__1.33333333333 hours . answer : c . <eor> c <eos> |
c |
divide__80.0__2.0__ hour_to_min_conversion__ divide__80.0__60.0__ subtract__2.0__1.3333__ divide__80.0__60.0__ |
divide__80.0__2.0__ multiply__1.5__40.0__ divide__80.0__60.0__ subtract__2.0__1.3333__ divide__80.0__60.0__ |
| what is the product of all the prime factors of num__22 ? <o> a ) num__18 <o> b ) num__9 <o> c ) num__22 <o> d ) num__6 <o> e ) num__3 |
num__22 factors are num__2 num__111 . now again prime numbers are only two i . e num__2 and num__11 this time we would have num__2 * num__11 = num__22 answer : c <eor> c <eos> |
c |
divide__22.0__2.0__ lcm__22.0__2.0__ |
divide__22.0__2.0__ multiply__2.0__11.0__ |
| in a pair of fractions fraction a is twice the fraction b and the product of two fractions is num__0.08 . what is the value of fraction a ? <o> a ) num__0.222222222222 <o> b ) num__1.0 <o> c ) num__0.4 <o> d ) num__2.0 <o> e ) num__0.285714285714 |
explanation : a = num__2 b = > b = num__0.5 a so ab = num__0.08 num__0.5 a num__2 = num__0.08 a num__2 = num__0.16 a = num__0.4 answer : c <eor> c <eos> |
c |
reverse__2.0__ divide__0.08__0.5__ divide__0.16__0.4__ |
reverse__2.0__ divide__0.08__0.5__ divide__0.16__0.4__ |
| given positive even integer y which of the following can not be evenly divisible by y ? <o> a ) y + num__2 <o> b ) y + num__8 <o> c ) num__2 y + num__4 <o> d ) num__3 y − num__1 <o> e ) num__2 y - num__2 |
this can be easily solved by substituting values : assume y to be num__2 a . y + num__2 : num__4 is evenly divisible by num__2 b . y + num__8 : num__10 is evenly divisible by num__2 c . num__2 y + num__4 : num__8 is evenly divisible by num__2 d . num__3 y − num__1 : num__5 is not evenly divisible by num__2 e . num__2 y + num__2 : num__2 is evenly divisible by num__2 option d <eor> d <eos> |
d |
multiply__2.0__4.0__ add__2.0__8.0__ subtract__3.0__2.0__ add__1.0__4.0__ add__1.0__2.0__ |
multiply__2.0__4.0__ add__2.0__8.0__ subtract__3.0__2.0__ add__1.0__4.0__ add__1.0__2.0__ |
| the probability of sam passing the exam is num__0.25 . the probability of sam passing the exam and michael passing the driving test is num__0.333333333333 . what is the probability of michael passing his driving test ? <o> a ) num__0.0416666666667 . <o> b ) num__0.5 . <o> c ) num__0.333333333333 . <o> d ) num__0.666666666667 . <o> e ) num__0.4 |
num__0.333333333333 * num__1 / m = num__0.166666666667 num__1 / m = num__0.5 answer b <eor> b <eos> |
b |
add__0.3333__0.1667__ divide__0.25__0.5__ |
add__0.3333__0.1667__ divide__0.25__0.5__ |
| a train running at a speed of num__36 kmph crosses an electric pole in num__12 seconds . in how much time will it cross a num__340 m long platform ? <o> a ) num__37 min <o> b ) num__46 min <o> c ) num__47 min <o> d ) num__67 min <o> e ) num__45 min |
b num__46 min let the length of the train be x m . when a train crosses an electric pole the distance covered is its own length . so x = num__12 * num__36 * num__0.277777777778 m = num__120 m . time taken to cross the platform = ( num__120 + num__340 ) / num__36 * num__0.277777777778 = num__46 min . <eor> b <eos> |
b |
round__46.0__ |
round__46.0__ |
| a cyclist covers a distance of num__750 meter in num__2 minutes num__30 seconds . what is the speed in km / hr of cyclist <o> a ) num__16 km / hr <o> b ) num__17 km / hr <o> c ) num__18 km / hr <o> d ) num__19 km / hr <o> e ) num__20 km / hr |
explanation : speed = distance / time distance = num__750 meter time = num__2 min num__30 sec = num__150 sec speed = num__5.0 = num__5 m / sec = > num__5 ∗ num__3.6 km / hr = num__18 km / hr option c <eor> c <eos> |
c |
divide__750.0__150.0__ multiply__5.0__3.6__ round__18.0__ |
divide__750.0__150.0__ multiply__5.0__3.6__ round__18.0__ |
| a sum of money is to be distributed among a b c d in the proportion of num__5 : num__1 : num__5 : num__3 . if c gets rs . num__1000 more than d what is b ' s share ? <o> a ) rs . num__500 <o> b ) rs . num__1000 <o> c ) rs . num__1500 <o> d ) rs . num__2000 <o> e ) none |
sol . let the shares of a b c and d be rs . num__5 x rs . x rs . num__5 x and rs . num__3 x respectively . then num__5 x - num__3 x = num__1000 ⇔ num__2 x = num__1000 ⇔ x = num__500 . ∴ b ' s share = rs . x = rs . num__500 . answer a <eor> a <eos> |
a |
subtract__5.0__3.0__ divide__1000.0__2.0__ multiply__1.0__500.0__ |
subtract__5.0__3.0__ divide__1000.0__2.0__ subtract__1000.0__500.0__ |
| a tradesman by means of his false balance defrauds to the extent of num__20.0 ? in buying goods as well as by selling the goods . what percent does he gain on his outlay ? a . num__20.0 b . num__45.0 <o> a ) num__77 <o> b ) num__66 <o> c ) num__44 <o> d ) num__99 <o> e ) num__11 |
g % = num__20 + num__20 + ( num__20 * num__20 ) / num__100 = num__44.0 answer : c <eor> c <eos> |
c |
percent__100.0__44.0__ |
percent__100.0__44.0__ |
| a man walking at a constant rate of num__12 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of num__24 miles per hour . the woman stops to wait for the man num__10 minutes after passing him while the man continues to walk at his constant rate . how many minutes must the woman wait until the man catches up ? <o> a ) num__13 <o> b ) num__10 <o> c ) num__15 <o> d ) num__11 <o> e ) num__9 |
when the woman passes the man they are aligned ( m and w ) . they are moving in the same direction . after num__5 minutes the woman ( w ) will be ahead the man ( m ) : m - - - - - - m - - - - - - - - - - - - - - - w w in the num__5 minutes after passing the man the woman walks the distance mw = ww which is num__10 * num__0.4 = num__4 miles and the man walks the distance mm which is num__10 * num__0.2 = num__2 mile . the difference of num__4 - num__2 = num__2 miles ( mw ) will be covered by the man in ( num__2 ) / num__9 = num__0.222222222222 of an hour which is ~ num__13 minutes . answer a . <eor> a <eos> |
a |
multiply__10.0__0.4__ subtract__12.0__10.0__ add__4.0__5.0__ divide__2.0__9.0__ add__4.0__9.0__ round__13.0__ |
multiply__10.0__0.4__ subtract__12.0__10.0__ add__4.0__5.0__ divide__2.0__9.0__ add__4.0__9.0__ round__13.0__ |
| the sum of first five prime numbers is ? <o> a ) num__12 <o> b ) num__28 <o> c ) num__56 <o> d ) num__24 <o> e ) num__22 |
required sum = ( num__2 + num__3 + num__5 + num__7 + num__11 ) = num__28 . note : num__1 is not a prime number . definition : a prime number ( or a prime ) is a natural number that has exactly two distinct natural number divisors : num__1 and itself . b <eor> b <eos> |
b |
add__2.0__3.0__ add__2.0__5.0__ subtract__3.0__2.0__ multiply__1.0__28.0__ |
add__2.0__3.0__ add__2.0__5.0__ subtract__3.0__2.0__ multiply__1.0__28.0__ |
| if num__8 ^ x = num__2 ^ num__9 what is x ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
num__2 ^ num__3 x = num__2 ^ num__9 num__3 x = num__9 therefore x = num__3 answer b <eor> b <eos> |
b |
divide__9.0__3.0__ |
divide__9.0__3.0__ |
| in what ratio must rice of rs . num__16 per kg be mixed with rice of rs . num__24 per kg so that cost of mixture is rs . num__18 per kg ? <o> a ) num__4 : num__1 <o> b ) num__2 : num__1 <o> c ) num__3 : num__1 <o> d ) num__4 : num__3 <o> e ) num__4 : num__5 |
( num__18 - num__24 ) / ( num__16 - num__18 ) = num__3.0 = num__3 : num__1 answer : c <eor> c <eos> |
c |
multiply__3.0__1.0__ |
divide__3.0__1.0__ |
| a man can swim in still water at num__4.5 km / h but takes twice as long to swim upstream than downstream . the speed of the stream is ? <o> a ) num__1.7 <o> b ) num__1.9 <o> c ) num__1.1 <o> d ) num__1.5 <o> e ) num__1.2 |
m = num__4.5 s = x ds = num__4.5 + x us = num__4.5 + x num__4.5 + x = ( num__4.5 - x ) num__2 num__4.5 + x = num__9 - num__2 x num__3 x = num__4.5 x = num__1.5 answer : d <eor> d <eos> |
d |
multiply__4.5__2.0__ divide__4.5__3.0__ round__1.5__ |
multiply__4.5__2.0__ subtract__4.5__3.0__ subtract__4.5__3.0__ |
| a man swims downstream num__72 km and upstream num__45 km taking num__9 hours each time ; what is the speed of the current ? <o> a ) num__1.6 <o> b ) num__1.5 <o> c ) num__1.3 <o> d ) num__1.1 <o> e ) num__1.2 |
num__72 - - - num__9 ds = num__8 ? - - - - num__1 num__45 - - - - num__9 us = num__5 ? - - - - num__1 s = ? s = ( num__8 - num__5 ) / num__2 = num__1.5 answer : b <eor> b <eos> |
b |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ round__1.5__ |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ divide__1.5__1.0__ |
| a certain computer manufacturing firm last year produced num__82 percent of its computers using parts made by its subsidiary company . if the remaining num__10818 computers were produced using parts purchased from another parts manufacturer how many computers were produced by the computer manufacturing firm altogether ? <o> a ) num__60000 <o> b ) num__60100 <o> c ) num__70000 <o> d ) num__75000 <o> e ) num__65000 |
num__82.0 parts used of subsidary company so num__18.0 parts used of other companies . now num__18.0 parts = num__10818 computers so num__1.0 parts = num__601.0 or num__100.0 parts = num__60100 computers . hence answer is ( b ) <eor> b <eos> |
b |
percent__100.0__60100.0__ |
percent__100.0__60100.0__ |
| the sum of present age of abe and the age before num__7 years is num__35 . find the present age of abe . what will be his age after num__7 years ? <o> a ) num__25 <o> b ) num__26 <o> c ) num__27 <o> d ) num__28 <o> e ) num__29 |
present age = x before num__7 yrs y = x - num__7 after num__7 yrs z = x + num__7 by the qn x + ( x - num__7 ) = num__35 num__2 x - num__7 = num__35 num__2 x = num__35 + num__7 x = num__21.0 x = num__21 z = x + num__7 = num__21 + num__7 = num__28 answer : d <eor> d <eos> |
d |
add__7.0__21.0__ add__7.0__21.0__ |
add__7.0__21.0__ add__7.0__21.0__ |
| a computer system uses alphanumeric case sensitive characters for its passwords . when the system was created it required users to create passwords having num__5 characters in length . this year it added the option of creating passwords having num__6 characters in length . which of the following gives the expression for the total number of passwords the new computer system can accept ? assume there are num__62 unique alphanumeric case sensitive characters . <o> a ) num__63 ^ num__4 <o> b ) num__62 ^ num__5 <o> c ) num__62 ( num__62 ^ num__4 ) <o> d ) num__63 ( num__62 ^ num__5 ) <o> e ) num__63 ( num__62 ^ num__6 ) |
total number of passwords = number of num__5 character password + number of num__6 character password = num__62 ^ num__5 + num__62 ^ num__6 ( since there is no limitation on repetition each character can be chosen in num__62 ways ) = num__62 ^ num__5 ( num__1 + num__62 ) = num__62 ^ num__5 * num__63 answer d <eor> d <eos> |
d |
subtract__6.0__5.0__ add__62.0__1.0__ add__62.0__1.0__ |
subtract__6.0__5.0__ add__62.0__1.0__ add__62.0__1.0__ |
| a vessel of capacity num__90 litres is fully filled with pure milk . nine litres of milk is removed from the vessel and replaced with water . nine litres of the solution thus formed is removed and replaced with water . find the quantity of pure milk in the final milk solution ? <o> a ) num__77.9 litres <o> b ) num__72.9 litres <o> c ) num__72.8 litres <o> d ) num__62.9 litres <o> e ) num__72.5 litres |
let the initial quantity of milk in vessel be t litres . let us say y litres of the mixture is taken out and replaced by water for n times alternatively . quantity of milk finally in the vessel is then given by [ ( t - y ) / t ] n * t for the given problem t = num__90 y = num__9 and n = num__2 . hence quantity of milk finally in the vessel = [ ( num__90 - num__9 ) / num__90 ] num__2 ( num__90 ) = num__72.9 litres . answer : b <eor> b <eos> |
b |
round__72.9__ |
round__72.9__ |
| in one hour a boat goes num__11 km / hr along the stream and num__5 km / hr against the stream . the speed of the boat in still water ( in km / hr ) is : <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__8 <o> e ) num__9 |
speed in still water = ( num__11 + num__5 ) / num__2 kmph = num__8 kmph . answer : d <eor> d <eos> |
d |
round__8.0__ |
round__8.0__ |
| a tank contains num__7500 gallons of a solution that is num__6 percent sodium chloride by volume . if num__2000 gallons of water evaporate from the tank the remaining solution will be approximately what percent sodium chloride ? <o> a ) num__5.18 <o> b ) num__6.18 <o> c ) num__7.18 <o> d ) num__8.18 <o> e ) . num__7 % |
we start with num__7500 gallons of a solution that is num__6.0 sodium chloride by volume . this means that there are num__0.06 x num__7500 = num__450 gallons of sodium chloride . when num__2000 gallons of water evaporate we are left with num__5500 gallons of solution . from here we can determine what percent of the num__5500 gallon solution is sodium chloride . ( sodium chloride / total solution ) x num__100 = ? ( num__450 / num__5500 ) x num__100 = ? num__0.081 x num__100 = ? = num__8.18 answer is d . <eor> d <eos> |
d |
percent__6.0__7500.0__ percent__100.0__8.18__ |
percent__6.0__7500.0__ percent__100.0__8.18__ |
| if num__20 men can build a water fountain num__56 metres long in num__7 days what length of a similar water fountain can be built by num__35 men in num__3 days ? <o> a ) num__40 m <o> b ) num__42 m <o> c ) num__47 m <o> d ) num__49 m <o> e ) num__50 m |
explanation : let the required length be x metres more men more length built ( direct proportion ) less days less length built ( direct proportion ) men num__20 : num__35 days num__7 : num__3 : : num__56 : x therefore ( num__20 x num__7 x x ) = ( num__35 x num__3 x num__56 ) x = ( num__35 x num__3 x num__56 ) / num__140 = num__42 hence the required length is num__42 m . answer : b <eor> b <eos> |
b |
multiply__20.0__7.0__ add__7.0__35.0__ round__42.0__ |
multiply__20.0__7.0__ add__7.0__35.0__ round__42.0__ |
| everyone shakes hands with everyone else in a room . total number of handshakes is num__78 . number of persons = ? <o> a ) num__13 <o> b ) num__12 <o> c ) num__11 <o> d ) num__15 <o> e ) num__16 |
in a room of n people the number of possible handshakes is c ( n num__2 ) or n ( n - num__1 ) / num__2 so n ( n - num__1 ) / num__2 = num__78 or n ( n - num__1 ) = num__156 or n = num__13 answer is ( a ) <eor> a <eos> |
a |
multiply__78.0__2.0__ multiply__1.0__13.0__ |
multiply__78.0__2.0__ divide__13.0__1.0__ |
| a charitable association sold an average of num__66 raffle tickets per member . among the female members the average was num__70 raffle tickets . the male to female ratio of the association is num__1 : num__2 . what was the average number r of tickets sold by the male members of the association <o> a ) num__50 <o> b ) num__56 <o> c ) num__58 <o> d ) num__62 <o> e ) num__66 |
given that total average r sold is num__66 male / female = num__0.5 and female average is num__70 . average of male members isx . ( num__70 * f + x * m ) / ( m + f ) = num__66 - > solving this equation after substituting num__2 m = f x = num__58 . ans c . <eor> c <eos> |
c |
reverse__2.0__ multiply__1.0__58.0__ |
reverse__2.0__ multiply__1.0__58.0__ |
| in the new budget the price of rise rose by num__5.0 . by how much percent must a person reduce his consumption so that his expenditure on it does not increase ? <o> a ) num__7.5 <o> b ) num__9.1 <o> c ) num__4.76 <o> d ) num__12.6 <o> e ) num__15 % |
reduce in consumption = r / ( num__100 + r ) * num__100.0 = num__0.047619047619 * num__100 = num__4.76 answer is c <eor> c <eos> |
c |
multiply__0.0476__100.0__ multiply__0.0476__100.0__ |
multiply__0.0476__100.0__ multiply__0.0476__100.0__ |
| how many seconds are there in a year ? <o> a ) num__31536000 <o> b ) num__21536000 <o> c ) num__11536000 <o> d ) none <o> e ) can not be determined |
solution : = = mathematical answer = = mathematical answer = number of seconds in a minute * number of minute in an hour * number hour in a day * number of day in year num__60 * num__60 * num__24 * num__365 = num__31536000 answer a <eor> a <eos> |
a |
hour_to_min_conversion__ round__31536000.0__ |
hour_to_min_conversion__ round__31536000.0__ |
| what will come in place of the x in the following number series ? num__46080 num__3840 num__384 num__48 num__8 num__2 x <o> a ) num__1 <o> b ) num__2 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
num__3840.0 = num__3840 num__384.0 = num__384 num__48.0 = num__48 num__8.0 = num__8 num__2.0 = num__2 num__1.0 = num__1 a <eor> a <eos> |
a |
reverse__1.0__ |
reverse__1.0__ |
| the difference between the simple interest received from two different sources on rs . num__1500 for num__3 years is rs . num__13.50 . the difference between their rates of interest is ? <o> a ) num__0.65 <o> b ) num__3.3 <o> c ) num__0.3 <o> d ) num__0.4 <o> e ) num__0.6 |
( num__1500 * r num__1 * num__3 ) / num__100 - ( num__1500 * r num__2 * num__3 ) / num__100 = num__13.50 num__4500 ( r num__1 - r num__2 ) = num__1350 r num__1 - r num__2 = num__0.3 . answer : c <eor> c <eos> |
c |
percent__100.0__0.3__ |
percent__100.0__0.3__ |
| when a number is added to another number the total becomes num__33 num__1 ⁄ num__3 per cent of the second number . what is the ratio between the first and the second number ? <o> a ) num__3 : num__7 <o> b ) num__7 : num__4 <o> c ) num__7 : num__3 <o> d ) data inadequate <o> e ) none of these |
let the first and second numbers be x and y respectively . then x + y = num__10 ⁄ num__3 y or x = num__7 ⁄ num__3 y ∴ x : y = num__7 : num__3 answer c <eor> c <eos> |
c |
subtract__10.0__3.0__ multiply__1.0__7.0__ |
subtract__10.0__3.0__ multiply__1.0__7.0__ |
| the area of playground is num__5900 sq . meters . what will be the cost of covering it with grass sheet num__1 cm deep if cost of grass sheet is $ num__2.80 per cubic meter . <o> a ) a ) $ num__144 <o> b ) b ) $ num__150.50 <o> c ) c ) $ num__165 <o> d ) d ) $ num__158.60 <o> e ) e ) $ num__160.70 |
total volume * unit cost = total cost or num__5900 * num__0.01 * num__2.8 = total cost = num__165 = c <eor> c <eos> |
c |
round__165.0__ |
multiply__1.0__165.0__ |
| the length of a rectangle i s two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square whose area is num__1225 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is num__10 units ? <o> a ) num__140 sq . units <o> b ) num__149 <o> c ) num__148 <o> d ) num__17 <o> e ) num__143 |
given that the area of the square = num__1225 sq . units = > side of square = √ num__1225 = num__35 units the radius of the circle = side of the square = num__35 units length of the rectangle = num__0.4 * num__35 = num__14 units given that breadth = num__10 units area of the rectangle = lb = num__14 * num__10 = num__140 sq . units answer : option a <eor> a <eos> |
a |
multiply__35.0__0.4__ square_perimeter__35.0__ square_perimeter__35.0__ |
multiply__35.0__0.4__ multiply__10.0__14.0__ multiply__10.0__14.0__ |
| little texas drilling company has three wells each producing oil at a constant rate . well a produces one barrel every two minutes . well b produces one barrel every three minutes . well c produces one barrel every four minutes . how many hours does it take little texas drilling company to produce num__265 barrels of oil ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__180 |
well a produces num__0.5 a barrel per minute . wells b and c produce num__0.333333333333 + num__0.25 = num__0.583333333333 ( a little more than half ) a barrel per minute . so all three wells combined produce a little more than num__1 barrel per minute . so for num__265 barrels they will take a bit less than num__265 mins which will be more than num__4 hrs but less than num__5 hrs . answer ( d ) <eor> d <eos> |
d |
add__0.25__0.3333__ divide__1.0__0.25__ add__4.0__1.0__ round__5.0__ |
add__0.25__0.3333__ divide__1.0__0.25__ add__4.0__1.0__ add__4.0__1.0__ |
| the cost price of a particular amount of articles is same as the number of articles . selling price of the articles comes out to be num__20 . over the whole transaction num__25.0 is gained . what is x ? <o> a ) num__16 <o> b ) num__77 <o> c ) num__15 <o> d ) num__18 <o> e ) num__161 |
explanation : according to the question let n be the number of articles . = > cost price of n articles = n = > selling price of n articles = num__20 = > profit % = num__25.0 = > profit % = ( sp - cp ) / cp x num__100 putting the values in the formula ( num__20 - n ) / n x num__100 = num__25 = > num__5 n / num__4 = num__20 = > n = num__16 answer : a <eor> a <eos> |
a |
percent__20.0__25.0__ percent__100.0__16.0__ |
percent__20.0__25.0__ percent__100.0__16.0__ |
| a company pays project contractors a rate of a dollars for the first hour and b dollars for each additional hour after the first where a > b . in a given month a contractor worked on two different projects that lasted num__3 and num__5 hours respectively . the company has the option to pay for each project individually or for all the projects at the end of the month . which arrangement would be cheaper for the company and how much would the company save ? <o> a ) per month with savings of $ ( num__2 a + num__2 b ) <o> b ) per month with savings of $ ( num__2 a - num__2 b ) <o> c ) the two options would cost an equal amount . <o> d ) per project with savings of $ ( num__2 a + num__2 b ) <o> e ) per project with savings of $ ( num__2 a - num__2 b ) |
per project company will pay as follows : for num__3 hours work = a + num__2 b for num__5 hours work = a + num__4 b total = num__2 a + num__6 b per month company will pay for num__8 hours work = a + num__5 b total per contract - total per month num__3 a + num__5 b - ( a + num__7 b ) num__2 a - num__2 b since a > b amount num__3 a + num__5 b ( per contract amount ) > a + num__7 b ( per project amount ) by num__2 a - num__2 b . hence per month payment will be cheaper by num__2 a - num__2 b . oa b <eor> b <eos> |
b |
subtract__5.0__3.0__ multiply__3.0__2.0__ add__3.0__5.0__ add__3.0__4.0__ subtract__5.0__3.0__ |
subtract__5.0__3.0__ add__2.0__4.0__ add__3.0__5.0__ add__3.0__4.0__ subtract__5.0__3.0__ |
| num__3 x < num__2 y < num__0 which of the following is the greatest ? <o> a ) num__2 x + num__3 y <o> b ) - ( num__3 x + y ) <o> c ) num__2 x - y <o> d ) x + y <o> e ) num__0 |
plugging in - num__1 for both x and y we get a . num__2 x + num__3 y = - num__5 b . - ( num__3 x + y ) = num__4 c . num__2 x - y = num__1 d . x + y = - num__2 e . num__0 = num__0 answer : b <eor> b <eos> |
b |
subtract__3.0__2.0__ add__3.0__2.0__ add__3.0__1.0__ multiply__3.0__1.0__ |
subtract__3.0__2.0__ add__3.0__2.0__ add__3.0__1.0__ add__2.0__1.0__ |
| in the parking lot there are num__44 vehicles num__24 of them are buses and the rest are cars . the color of num__24 vehicles is red of which num__18 are buses . how many cars can be found in the parking lot which are not colored red ? <o> a ) num__11 <o> b ) num__12 <o> c ) num__13 <o> d ) num__14 <o> e ) num__15 |
the number of cars is num__44 - num__24 = num__20 . the number of red cars is num__24 - num__18 = num__6 . the number of cars which are not red is num__20 - num__6 = num__14 . the answer is d . <eor> d <eos> |
d |
subtract__44.0__24.0__ subtract__24.0__18.0__ subtract__20.0__6.0__ subtract__20.0__6.0__ |
subtract__44.0__24.0__ subtract__24.0__18.0__ subtract__20.0__6.0__ subtract__20.0__6.0__ |
| if a certain sample of data has a mean of num__23.0 and a standard deviation of num__3.0 which of the following values is more than num__2.5 standard deviations from the mean ? <o> a ) num__12.0 <o> b ) num__13.5 <o> c ) num__15.0 <o> d ) num__23.5 <o> e ) num__26.5 |
value ismore than num__2.5 sdfrom the mean means that the distance between the mean and the value must be more than num__2.5 * sd = num__7.5 . so the value must be either less than num__23 - num__7.5 = num__15.5 or more than num__23 + num__7.5 = num__30.5 . answer : c . <eor> c <eos> |
c |
multiply__3.0__2.5__ subtract__23.0__7.5__ add__23.0__7.5__ round_down__15.5__ |
multiply__3.0__2.5__ subtract__23.0__7.5__ add__23.0__7.5__ subtract__30.5__15.5__ |
| series problem like num__4 num__12 x num__44 num__46 num__132 num__134 begin of the skype highlighting num__44 num__46 num__132 num__134 end of the skype highlighting . find x ? <o> a ) num__14 <o> b ) num__15 <o> c ) num__16 <o> d ) num__17 <o> e ) num__18 |
num__134 - num__132 = num__2 num__46 - num__44 = num__2 x - num__12 = num__2 x = num__14 answer : a <eor> a <eos> |
a |
subtract__46.0__44.0__ add__12.0__2.0__ add__12.0__2.0__ |
subtract__46.0__44.0__ add__12.0__2.0__ add__12.0__2.0__ |
| how many no . ' s with num__3 different digits num__2 num__3 num__5 num__6 num__7 and num__9 which are divisible by num__5 ? <o> a ) num__18 <o> b ) num__20 <o> c ) num__25 <o> d ) num__28 <o> e ) num__31 |
since each desired number is divisible by num__5 so we must have num__5 at the unit place . so there is num__1 way of doing it . the tens place can now be filled by any of the remaining num__5 digits ( num__2 num__3 num__6 num__7 num__9 ) . so there are num__5 ways of filling the tens place . the hundreds place can now be filled by any of the remaining num__4 digits . so there are num__4 ways of filling it . required number of numbers = ( num__1 x num__5 x num__4 ) = num__20 b <eor> b <eos> |
b |
subtract__3.0__2.0__ add__3.0__1.0__ multiply__5.0__4.0__ multiply__5.0__4.0__ |
subtract__3.0__2.0__ add__3.0__1.0__ multiply__5.0__4.0__ multiply__5.0__4.0__ |
| a plane flies num__420 miles with the wind and num__350 miles against the wind in the same length of time . if the speed of the wind is num__23 mph what is the speed of the plain in still air ? <o> a ) num__153 mph <o> b ) num__253 mph <o> c ) num__353 mph <o> d ) num__453 mph <o> e ) num__553 mph |
the speed of the plane in still air = x miles / hour the speed of the wind is num__23 mph speed with the wind = ( x + num__23 ) mph speed against the wind = ( x – num__23 ) mph time = distance / speed according to the problem num__420 / ( x + num__23 ) = num__350 / ( x – num__23 ) num__420 ( x – num__23 ) = num__350 ( x + num__23 ) num__420 x – num__9660 = num__350 x + num__805 num__420 x – num__350 x = num__8050 + num__9660 num__70 x = num__17710 x = num__253.0 x = num__253 therefore the speed of the plane in still air = num__253 mph . correct answer b ) num__253 mph <eor> b <eos> |
b |
multiply__420.0__23.0__ multiply__350.0__23.0__ subtract__420.0__350.0__ add__8050.0__9660.0__ divide__17710.0__70.0__ round__253.0__ |
multiply__420.0__23.0__ multiply__350.0__23.0__ subtract__420.0__350.0__ add__8050.0__9660.0__ divide__17710.0__70.0__ divide__17710.0__70.0__ |
| num__32 num__0.2 num__16015 num__0.5 num__3028 num__0.7 ? <o> a ) num__30 <o> b ) num__40 <o> c ) num__50 <o> d ) num__60 <o> e ) num__70 |
( num__32 num__0.2 num__160 ) ( num__15 num__0.5 num__30 ) ( num__28 num__0.7 ? ) num__32 / num__0.2 = num__160 num__15 / num__0.5 = num__30 num__28 / num__0.7 = num__40 answer : b <eor> b <eos> |
b |
divide__32.0__0.2__ divide__15.0__0.5__ divide__28.0__0.7__ divide__28.0__0.7__ |
divide__32.0__0.2__ divide__15.0__0.5__ divide__28.0__0.7__ divide__28.0__0.7__ |
| at pat ' s pet shop num__25 cups of bird seed are used every num__5 days to feed num__20 parakeets . how many cups of bird seed would be required to feed num__15 parakeets for num__7 days ? <o> a ) num__35.25 <o> b ) num__15 <o> c ) num__64 <o> d ) num__104 <o> e ) num__26 num__0.25 |
let ' s go step - by - step . num__25 cups are used over a num__5 day period which means num__5 cups a day . this feeds num__20 parakeets which means each parakeet needs num__0.25 of a cup every day . for num__15 parakeets we need num__15 * num__0.25 cups = num__3 num__0.75 cups a day . over num__7 days we need num__26 num__0.25 cups . choice e . <eor> e <eos> |
e |
divide__5.0__20.0__ divide__15.0__5.0__ divide__15.0__20.0__ round__26.0__ |
divide__5.0__20.0__ divide__15.0__5.0__ multiply__0.25__3.0__ round__26.0__ |
| what is the greatest possible common divisor of two different positive integers which are less than num__144 ? can someone explain why the answer is num__71 if we assume that the integers are num__143 and num__142 ? <o> a ) num__143 <o> b ) num__142 <o> c ) num__72 <o> d ) num__71 <o> e ) num__12 |
first of all what is the greatest common divisor of num__143 and num__142 ? it is num__1 . you are looking for the common divisor . num__142 and num__143 will have no common divisor except num__1 . think : num__2 and num__3 have gcd ( greatest common divisor ) of num__1 num__2 and num__4 have gcd of num__2 . num__3 and num__4 have gcd ( greatest common divisor ) of num__1 so if you were to select num__2 numbers less than num__5 with the greatest gcd you need to select num__2 and num__4 not num__3 and num__4 . now think : num__143 = num__11 * num__13 the greatest possible divisor it will have with another number less than num__144 will be either num__11 or num__13 . let ' s move on . num__142 = num__2 * num__71 the greatest possible divisor it can have with another number less than num__144 can be num__71 ( say if the other selected integer is num__71 ) do you think another number less than num__144 could have a gcd of greater than num__71 ? no because when you split a number into two factors one of them will be at least num__2 . if it is greater than num__2 the other factor will obviously be less than num__71 . <eor> d <eos> |
d |
subtract__144.0__143.0__ subtract__144.0__142.0__ add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ divide__143.0__11.0__ multiply__71.0__1.0__ |
subtract__144.0__143.0__ subtract__144.0__142.0__ add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ divide__143.0__11.0__ multiply__71.0__1.0__ |
| a number y is chosen at random from the numbers num__1 - num__3 - num__1 num__0 num__1 num__2 num__3 . what is the probability that | x | < num__3 ? <o> a ) num__0.428571428571 <o> b ) num__0.25 <o> c ) num__0.2 <o> d ) num__0.142857142857 <o> e ) num__0.4 |
| x | | x | can take num__7 values . to get | x | < num__2 | x | < num__2 ( i . e . − num__2 < x < + num__2 − num__2 < x ⇒ p ( | x | < num__2 ) = favourable casestotal casesp ( | x | < num__2 ) = favourable casestotal cases = num__0.428571428571 a <eor> a <eos> |
a |
divide__3.0__7.0__ multiply__1.0__0.4286__ |
divide__3.0__7.0__ multiply__1.0__0.4286__ |
| the length of a rectangular field is num__1.4 its width . if the perimeter of the field is num__288 meters what is the width of the field ? <o> a ) num__50 <o> b ) num__60 <o> c ) num__70 <o> d ) num__80 <o> e ) num__90 |
let l be the length and w be the width . l = ( num__1.4 ) w perimeter : num__2 l + num__2 w = num__288 num__2 ( num__1.4 ) w + num__2 w = num__288 solve the above equation to find : w = num__60 m and l = num__84 m . correct answer b ) num__60 <eor> b <eos> |
b |
multiply__1.4__60.0__ triangle_area__2.0__60.0__ |
multiply__1.4__60.0__ triangle_area__2.0__60.0__ |
| a clock shows the time as num__11 a . m . if the minute hand gains num__5 minutes every hour how many minutes will the clock gain by num__6 p . m . ? <o> a ) num__45 minutes <o> b ) num__55 minutes <o> c ) num__35 minutes <o> d ) num__25 minutes <o> e ) num__40 minutes |
there are num__7 hours in between num__11 a . m . to num__6 p . m . num__7 * num__5 = num__35 minutes . answer : c <eor> c <eos> |
c |
multiply__5.0__7.0__ round__35.0__ |
multiply__5.0__7.0__ multiply__5.0__7.0__ |
| | x + num__3 | – | num__4 - x | = | num__8 + x | how many x solutions will this equation have ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
x = | x | = x when x > = num__0 ( x is either positive or num__0 ) | x | = - x when x < num__0 ( note here that you can put the equal to sign here as well x < = num__0 because if x = num__0 | num__0 | = num__0 = - num__0 ( all are the same ) so the ' = ' sign can be put with x > num__0 or with x < num__0 . we usually put it with ' x > num__0 ' for consistency . a <eor> a <eos> |
a |
multiply__3.0__0.0__ |
multiply__3.0__0.0__ |
| num__90 num__180 num__12 num__50 num__100 num__200 ? num__2 num__50 num__4 num__25 num__2 num__6 num__30 num__3 <o> a ) num__130 <o> b ) num__150 <o> c ) num__180 <o> d ) num__100 <o> e ) num__220 |
num__30 * num__3 = num__90 num__6 * num__30 = num__180 num__6 * num__2 = num__12 num__25 * num__2 = num__50 num__4 * num__25 = num__100 num__50 * num__4 = num__200 num__2 * num__50 = num__100 ans is num__100 answer : d <eor> d <eos> |
d |
multiply__50.0__2.0__ |
multiply__50.0__2.0__ |
| what will be the product of ( num__50 - num__1 ) * ( num__50 - num__2 ) * ( num__50 - num__3 ) * . . . . . . . . . . . . . . * ( num__50 - num__79 ) * ( num__50 - num__80 ) ? <o> a ) less than - num__100000 <o> b ) - num__2500 <o> c ) num__0 <o> d ) num__2500 <o> e ) more than num__100 |
000 |
one of the terms is ( num__50 - num__50 ) so the product is num__0 . the answer is c . <eor> c <eos> |
c |
c |
| a man goes from a to b at a speed of num__14 kmph and comes back to a at a speed of num__10 kmph . find his average speed for the entire journey ? <o> a ) num__11.6 kmph <o> b ) num__12.9 kmph <o> c ) num__12.8 kmph <o> d ) num__11.9 kmph <o> e ) num__12.6 kmph |
distance from a and b be ' d ' average speed = total distance / total time average speed = ( num__2 d ) / [ ( d / num__14 ) + ( d / num__10 ] = ( num__2 d ) / [ num__12 d / num__70 ) = > num__11.6 kmph . answer : a <eor> a <eos> |
a |
subtract__14.0__2.0__ round__11.6__ |
add__10.0__2.0__ round__11.6__ |
| in a certain city num__40.0 of the registered voters are democrats and the rest are republicans . in a mayoral race if num__75.0 of the registered voters who are democrats and num__20.0 of the registered voters who are republicans are expected to vote for candidate a what % of the registered voters are expected to vote for candidate a ? <o> a ) a ) num__50.0 <o> b ) b ) num__53.0 <o> c ) c ) num__54.0 <o> d ) d ) num__55.0 <o> e ) e ) num__57 % |
let total voters = num__100 so d = num__60 vote for the mayor num__60 * num__75.0 = num__45 ( num__60 * num__0.75 ) r = num__40 vote for the mayor num__40 * num__20.0 = ( num__40 * num__0.2 ) num__45 + num__8 = num__53 num__0.53 * num__100 = num__55.0 d <eor> d <eos> |
d |
percent__75.0__60.0__ percent__40.0__20.0__ percent__100.0__55.0__ |
percent__75.0__60.0__ percent__40.0__20.0__ percent__100.0__55.0__ |
| if n divided by num__7 has a remainder of num__1 what is the remainder when num__3 times n is divided by num__7 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__5 <o> e ) num__6 |
as per question = > n = num__7 p + num__1 for some integer p hence num__3 n = > num__21 q + num__3 = > remainder = > num__6 for some integer q alternatively = > n = num__2 > num__3 n = > num__3 = > num__3 divided by num__7 will leave a remainder num__3 hence c <eor> c <eos> |
c |
multiply__7.0__3.0__ subtract__7.0__1.0__ subtract__3.0__1.0__ multiply__1.0__3.0__ |
multiply__7.0__3.0__ subtract__7.0__1.0__ subtract__3.0__1.0__ add__1.0__2.0__ |
| a train crosses a platform of num__120 m in num__15 sec same train crosses another platform of length num__180 m in num__18 sec . then find the length of the train ? <o> a ) num__298 m <o> b ) num__180 m <o> c ) num__287 m <o> d ) num__297 m <o> e ) num__261 m |
length of the train be ‘ x ’ x + num__8.0 = x + num__10.0 num__6 x + num__720 = num__5 x + num__900 x = num__180 m answer : b <eor> b <eos> |
b |
divide__120.0__15.0__ divide__180.0__18.0__ multiply__120.0__6.0__ subtract__15.0__10.0__ multiply__180.0__5.0__ round__180.0__ |
divide__120.0__15.0__ divide__180.0__18.0__ multiply__120.0__6.0__ subtract__15.0__10.0__ add__180.0__720.0__ round__180.0__ |
| in a particular state num__62.0 of the counties received some rain on monday and num__54.0 of the counties received some rain on tuesday . no rain fell either day in num__28.0 of the counties in the state . what percent of the counties received some rain on monday and tuesday ? <o> a ) num__25.0 <o> b ) num__44.0 <o> c ) num__56.0 <o> d ) num__62.0 <o> e ) num__70 % |
num__62 + num__54 + num__28 = num__144.0 the number is num__44.0 above num__100.0 because num__44.0 of the counties were counted twice . the answer is b . <eor> b <eos> |
b |
percent__100.0__44.0__ |
percent__100.0__44.0__ |
| in a game of billiards a can give b num__20 points in num__60 and he can give c num__30 points in num__60 . how many points can b give c in a game of num__100 ? <o> a ) num__25 <o> b ) num__15 <o> c ) num__65 <o> d ) num__82 <o> e ) num__95 |
explanation : a scores num__60 while b score num__40 and c scores num__30 . the number of points that c scores when b scores num__100 = ( num__100 * num__30 ) / num__40 = num__25 * num__3 = num__75 . in a game of num__100 points b gives ( num__100 - num__75 ) = num__25 points to c . answer a <eor> a <eos> |
a |
subtract__60.0__20.0__ divide__60.0__20.0__ subtract__100.0__25.0__ subtract__100.0__75.0__ |
subtract__60.0__20.0__ divide__60.0__20.0__ subtract__100.0__25.0__ subtract__100.0__75.0__ |
| today my car meter reads as num__63736 kms . i notes that this is a palindrome . how many minimum kms i need to travel so my car meter find another palindrom . <o> a ) num__100 kms <o> b ) num__115 kms <o> c ) num__110 kms <o> d ) num__210 kms <o> e ) num__120 kms |
a num__100 kms num__63736 + num__100 = num__63636 a palindrome <eor> a <eos> |
a |
subtract__63736.0__100.0__ round__100.0__ |
subtract__63736.0__100.0__ round__100.0__ |
| the sum of three numbers is num__98 . the ratio of the first to the second is num__0.666666666667 and the ratio of the second to the third is num__0.625 . the second number is : <o> a ) num__15 <o> b ) num__20 <o> c ) num__30 <o> d ) num__32 <o> e ) num__33 |
let the three numbers be x y and z . sum of the numbers is num__98 . x + y + z = num__98 … … … … … … ( i ) the ratio of the first to the second is num__0.666666666667 . x / y = num__0.666666666667 . x = num__0.666666666667 ? y . x = num__2 y / num__3 . the ratio of the second to the third is num__0.625 . y / z = num__0.625 . z / y = num__1.6 . z = num__1.6 ? y . z = num__8 y / num__5 . put the value of x = num__2 y / num__3 and z = num__8 y / num__5 in ( i ) . num__2 y / num__3 + y + num__8 y / num__5 = num__98 num__49 y / num__15 = num__98 . num__49 y = num__98 ? num__15 . num__49 y = num__1470 . y = num__30.0 . y = num__30 . therefore the second number is num__30 correct answer c <eor> c <eos> |
c |
reverse__0.625__ multiply__0.625__8.0__ divide__98.0__2.0__ multiply__3.0__5.0__ multiply__98.0__15.0__ multiply__2.0__15.0__ multiply__2.0__15.0__ |
reverse__0.625__ add__3.0__2.0__ divide__98.0__2.0__ multiply__3.0__5.0__ multiply__98.0__15.0__ divide__1470.0__49.0__ divide__1470.0__49.0__ |
| roy is now num__8 years older than julia and half of that amount older than kelly . if in num__2 years roy will be twice as old as julia then in num__2 years what would be roy ’ s age multiplied by kelly ’ s age ? <o> a ) num__160 <o> b ) num__168 <o> c ) num__176 <o> d ) num__184 <o> e ) num__192 |
r = j + num__8 = k + num__4 r + num__2 = num__2 ( j + num__2 ) ( j + num__8 ) + num__2 = num__2 j + num__4 j = num__6 r = num__14 k = num__10 in num__2 years ( r + num__2 ) ( k + num__2 ) = num__16 * num__12 = num__192 the answer is e . <eor> e <eos> |
e |
divide__8.0__2.0__ subtract__8.0__2.0__ add__8.0__6.0__ add__8.0__2.0__ multiply__8.0__2.0__ add__8.0__4.0__ multiply__12.0__16.0__ multiply__12.0__16.0__ |
divide__8.0__2.0__ add__2.0__4.0__ add__8.0__6.0__ add__8.0__2.0__ multiply__8.0__2.0__ add__8.0__4.0__ multiply__12.0__16.0__ multiply__12.0__16.0__ |
| ravi invested certain amount for two rates of simple interests at num__6.0 p . a . and num__7.0 p . a . what is the ratio of ravi ' s investments if the interests from those investments are equal ? <o> a ) num__7 : num__9 <o> b ) num__7 : num__5 <o> c ) num__7 : num__2 <o> d ) num__7 : num__6 <o> e ) num__7 : num__1 |
let x be the investment of ravi in num__6.0 and y be in num__7.0 x ( num__6 ) ( n ) / num__100 = y ( num__7 ) ( n ) / num__100 = > x / y = num__1.16666666667 x : y = num__7 : num__6 answer : d <eor> d <eos> |
d |
percent__7.0__100.0__ |
percent__7.0__100.0__ |
| a can do a work in num__15 days and b in num__20 days . if they work on it together for num__4 days then the fraction of the work that is left is ? <o> a ) num__8 <o> b ) num__0.727272727273 <o> c ) num__0.615384615385 <o> d ) num__0.533333333333 <o> e ) none |
a ' s num__1 day ' s work = num__1 ; num__15 b ' s num__1 day ' s work = num__1 ; num__20 ( a + b ) ' s num__1 day ' s work = ( num__1 + num__1 ) = num__7 . num__15 num__20 num__60 ( a + b ) ' s num__4 day ' s work = ( num__7 x num__4 ) = num__7 . num__60 num__15 therefore remaining work = ( num__1 - num__7 ) = num__8 . num__15 num__15 option d <eor> d <eos> |
d |
hour_to_min_conversion__ subtract__15.0__7.0__ divide__8.0__15.0__ |
hour_to_min_conversion__ subtract__15.0__7.0__ divide__8.0__15.0__ |
| a man walked diagonally across a square lot . approximately what was the percent saved by not walking along the edges ? <o> a ) num__30 <o> b ) num__39 <o> c ) num__87 <o> d ) num__27 <o> e ) num__91 |
let the side of the square ( abcd ) be x meters . then ab + bc = num__2 x metres . ac = = ( num__1.41 x ) m . saving on num__2 x metres = ( num__0.59 x ) m . saving % = = num__30.0 ( approx ) answer : a <eor> a <eos> |
a |
triangle_area__2.0__30.0__ |
triangle_area__2.0__30.0__ |
| convert num__150 inches into centimeter ? <o> a ) num__312 cm <o> b ) num__381 cm <o> c ) num__350 cm <o> d ) num__310 cm <o> e ) num__354 cm |
num__1 inch = num__2.54 cm num__150 inches = num__110 * num__2.54 = num__381 cm answer is b <eor> b <eos> |
b |
multiply__150.0__2.54__ round__381.0__ |
multiply__150.0__2.54__ multiply__150.0__2.54__ |
| in how many different ways can the letters of the word ' optical ' be arranged so that the vowels always come together ? <o> a ) num__800 <o> b ) num__880 <o> c ) num__720 <o> d ) num__1500 <o> e ) num__1110 |
the word ' optical ' has num__7 letters . it has the vowels ' o ' ' i ' ' a ' in it and these num__3 vowels should always come together . hence these three vowels can be grouped and considered as a single letter . that is ptcl ( oia ) . hence we can assume total letters as num__5 and all these letters are different . number of ways to arrange these letters = num__5 ! = num__5 × num__4 × num__3 × num__2 × num__1 = num__120 all the num__3 vowels ( oia ) are different number of ways to arrange these vowels among themselves = num__3 ! = num__3 × num__2 × num__1 = num__6 hence required number of ways = num__120 × num__6 = num__720 c <eor> c <eos> |
c |
vowel_space__ coin_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
vowel_space__ coin_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| a chair is bought for rs . num__800 / - and sold at a loss of num__10.0 find its selling price <o> a ) rs . num__720 / - <o> b ) rs . num__600 / - <o> c ) rs . num__630 / - <o> d ) rs . num__820 / - <o> e ) rs . num__970 |
explanation : num__100.0 - - - - - - > num__800 ( num__100 * num__8 = num__800 ) num__90.0 - - - - - - > num__720 ( num__90 * num__8 = num__720 ) selling price = rs . num__720 / - answer : option a <eor> a <eos> |
a |
percent__90.0__800.0__ percent__90.0__800.0__ |
percent__90.0__800.0__ percent__90.0__800.0__ |
| a person purchased a tv set for rs . num__1000 and a dvd player for rs . num__650 . he sold both the items together for rs . num__2500 . what percentage of profit did he make ? <o> a ) num__51.51 <o> b ) num__96.96 <o> c ) num__10.11 <o> d ) num__15.12 <o> e ) num__23.32 % |
the total cp = rs . num__1000 + rs . num__650 = rs . num__1650 and sp = rs . num__2500 profit ( % ) = ( num__2500 - num__1650 ) / num__1650 * num__100 = num__51.51 answer : a <eor> a <eos> |
a |
percent__100.0__51.51__ |
percent__100.0__51.51__ |
| on a ferry there are num__38 cars and num__12 trucks . the cars have an average mass of num__950 kg and the trucks have an average mass of num__3100 kg . what is the average mass of all num__50 vehicles on the ferry ? <o> a ) num__1442 <o> b ) num__1448 <o> c ) num__1454 <o> d ) num__1460 <o> e ) num__1466 |
the total mass of the vehicles is num__38 * num__950 + num__12 * num__3100 = num__73300 the average mass is num__1466.0 = num__1466 the answer is e . <eor> e <eos> |
e |
divide__73300.0__50.0__ divide__73300.0__50.0__ |
divide__73300.0__50.0__ divide__73300.0__50.0__ |
| in a party every person shakes hands with every other person . if there were a total of num__105 handshakes in the party then what is the number of persons present in the party ? <o> a ) num__15 <o> b ) num__16 <o> c ) num__17 <o> d ) num__18 <o> e ) num__19 |
explanation : let the number of persons be n â ˆ ´ total handshakes = nc num__2 = num__105 n ( n - num__1 ) / num__2 = num__105 â ˆ ´ n = num__15 answer : a <eor> a <eos> |
a |
multiply__15.0__1.0__ |
divide__15.0__1.0__ |
| a collection of books went on sale and num__0.666666666667 of them were sold for $ num__3.50 each . if none of the num__40 remaining books were sold what was the total amount received for the books that were sold ? <o> a ) $ num__180 <o> b ) $ num__250 <o> c ) $ num__260 <o> d ) $ num__280 <o> e ) $ num__300 |
if num__40 books constitute num__0.333333333333 rd of the total then num__0.666666666667 rd of the total = num__80 books amount received for sold books = num__80 * num__3.5 = $ num__280 answer : d <eor> d <eos> |
d |
multiply__3.5__80.0__ multiply__3.5__80.0__ |
multiply__3.5__80.0__ multiply__3.5__80.0__ |
| when positive integer n is divided by num__5 the remainder is num__1 . when n is divided by num__7 the remainder is num__3 . what is the smallest positive integer k such that k + n is a multiple of num__39 ? <o> a ) num__3 <o> b ) num__8 <o> c ) num__12 <o> d ) num__32 <o> e ) num__35 |
n = num__5 p + num__1 = num__611 num__1621 num__2631 n = num__7 q + num__3 = num__3 num__1017 num__2431 = > n = num__39 m + num__31 to get this we need to take lcm of co - efficients of p and q and first common number in series . so we need to add num__8 more to make it num__39 m + num__39 answer - b <eor> b <eos> |
b |
add__5.0__3.0__ add__5.0__3.0__ |
add__5.0__3.0__ add__5.0__3.0__ |
| in order to fence a square manish fixed num__48 poles . if the distance between two poles is num__2 metres then what will be the area of the square so formed ? <o> a ) num__133 cm num__2 <o> b ) num__276 cm num__2 <o> c ) num__2500 cm num__2 <o> d ) num__576 cm num__2 <o> e ) none of these |
let the side of the square be x m . ∴ perimeter of the square = num__48 × num__2 = num__4 x ∴ x = num__24 m ∴ area = ( num__24 ) num__2 = num__576 m num__2 answer d <eor> d <eos> |
d |
divide__48.0__2.0__ round__576.0__ |
divide__48.0__2.0__ round__576.0__ |
| while flying over the pacific an airplane makes a num__25 ° turn to the right to avoid a storm . if as a result the airplane is traveling in a direction num__7 ° east of north in what direction was it originally flying ? <o> a ) ( a ) num__30 ° west of north <o> b ) ( b ) num__30 ° east of north <o> c ) ( c ) num__18 ° west of north <o> d ) ( d ) num__18 ° east of north <o> e ) ( e ) num__5 ° west of north |
after a turn of num__25 ° you are num__7 ° north east with a num__18 ° turn you would be perfectly pointing at north . so you were before the turn num__18 ° to the other side ( west ) . answer c ) <eor> c <eos> |
c |
subtract__25.0__7.0__ round__18.0__ |
subtract__25.0__7.0__ round__18.0__ |
| how many of the positive factors of num__45 num__16 and how many common factors are there in numbers ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
factors of num__45 - num__1 num__3 num__9 num__15 and num__45 factors of num__16 - num__1 num__2 num__4 num__8 and num__16 comparing both we have three common factors of num__4516 - num__1 answer ( a ) <eor> a <eos> |
a |
divide__45.0__3.0__ subtract__3.0__1.0__ add__1.0__3.0__ divide__16.0__2.0__ subtract__16.0__15.0__ |
subtract__16.0__1.0__ subtract__3.0__1.0__ add__1.0__3.0__ subtract__9.0__1.0__ subtract__16.0__15.0__ |
| tea worth rs . num__126 per kg and rs . num__135 per kg are mixed with a third variety in the ratio num__1 : num__1 : num__2 . if the mixture is worth rs . num__153 per kg the price of the third variety per kg will be : <o> a ) rs . num__169 <o> b ) rs . num__169.50 <o> c ) rs . num__175.50 <o> d ) rs . num__180 <o> e ) rs . num__180.50 |
since first and second varieties are mixed in equal proportions . so their average price = rs . num__126 + num__135 = rs . num__130.50 num__2 so the mixture is formed by mixing two varieties one at rs . num__130.50 per kg and the other at say rs . x per kg in the ratio num__2 : num__2 i . e . num__1 : num__1 . we have to find x . by the rule of alligation we have : cost of num__1 kg of num__1 st kindcost of num__1 kg tea of num__2 nd kind rs . num__130.50 mean price rs . num__153 rs . x ( x - num__153 ) num__22.50 x - num__153 = num__1 num__22.50 x - num__153 = num__22.50 x = num__175.50 option c <eor> c <eos> |
c |
subtract__153.0__130.5__ add__153.0__22.5__ multiply__1.0__175.5__ |
subtract__153.0__130.5__ add__153.0__22.5__ add__153.0__22.5__ |
| two identical machines have the ability to produce both nuts and bolts . however it takes num__3 second to produce a bolt but num__1 seconds to produce a nut . what is the fastest the two machines working together can produce num__1000 nuts and num__1000 bolts ? <o> a ) num__1250 seconds <o> b ) num__1500 seconds <o> c ) num__1750 seconds <o> d ) num__2000 seconds <o> e ) num__3000 seconds |
to minimize the amount of time we should make both machines work all the time while producing num__1000 nuts and num__1000 bolts . two machines to produce num__1000 bolts will need ( num__1000 * num__3 ) / num__2 = num__1500 seconds . two machines to produce num__1000 nuts will need ( num__1000 * num__1 ) / num__2 = num__500 seconds . total = num__1500 + num__500 = num__2000 . answer : d <eor> d <eos> |
d |
subtract__3.0__1.0__ divide__1000.0__2.0__ multiply__1000.0__2.0__ round__2000.0__ |
subtract__3.0__1.0__ divide__1000.0__2.0__ multiply__1000.0__2.0__ multiply__1.0__2000.0__ |
| a train takes num__6 hours to cover a distance of num__540 km . how much should the speed in kmph be maintained to cover the same direction in num__1.5 th of the previous time ? <o> a ) num__50 kmph <o> b ) num__54 kmph <o> c ) num__60 kmph <o> d ) num__59 kmph <o> e ) num__60 kmph |
time = num__6 distance = num__540 num__1.5 of num__6 hours = num__6 * num__1.5 = num__9 hours required speed = num__60.0 = num__60 kmph c ) <eor> c <eos> |
c |
multiply__6.0__1.5__ hour_to_min_conversion__ hour_to_min_conversion__ |
multiply__6.0__1.5__ hour_to_min_conversion__ hour_to_min_conversion__ |
| how many seconds will a num__400 metre long train take to cross a man running with a speed of num__6 km / hr in the direction of the moving train if the speed of the train is num__46 km / hr ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__40 <o> d ) num__45 <o> e ) num__36 |
explanation : speed of train relatively to man = ( num__46 - num__6 ) km / hr = num__40 km / hr = ( num__40 x num__0.277777777778 ) m / sec = num__11.11 m / sec time taken to pass the man = ( num__400 / num__11.11 ) sec = num__36 sec . answer : e <eor> e <eos> |
e |
subtract__46.0__6.0__ round__36.0__ |
subtract__46.0__6.0__ round__36.0__ |
| a pair of prime numbers that can be expressed in the form k { p ( p + num__6 ) } is de fi ned as a pair of “ sexy primes . ” a “ sexy triplet ” is a group of three primes that can be expressed in the form { p ( p + num__6 ) ( p + num__12 ) } . all of the following prime numbers are the middle term of a sexy triplet except <o> a ) num__11 <o> b ) num__13 <o> c ) num__17 <o> d ) num__19 <o> e ) num__23 |
definetly a sitter . plug in the values for middle term and calculate . we can see clearly that k for num__19 num__19 + num__6 gives num__25 which is not a prime therefore correct answer d <eor> d <eos> |
d |
add__6.0__19.0__ subtract__25.0__6.0__ |
add__6.0__19.0__ subtract__25.0__6.0__ |
| the average of runs of a cricket player of num__10 innings was num__32 . how many runs must he make in his next innings so as to increase his average of runs by num__4 ? <o> a ) num__98 <o> b ) num__76 <o> c ) num__23 <o> d ) num__16 <o> e ) num__12 |
average = total runs / no . of innings = num__32 so total = average x no . of innings = num__32 x num__10 = num__320 . now increase in avg = num__4 runs . so new avg = num__32 + num__4 = num__36 runs total runs = new avg x new no . of innings = num__36 x num__11 = num__396 runs made in the num__11 th inning = num__396 - num__320 = num__76 answer b <eor> b <eos> |
b |
multiply__10.0__32.0__ add__32.0__4.0__ multiply__11.0__36.0__ subtract__396.0__320.0__ subtract__396.0__320.0__ |
multiply__10.0__32.0__ add__32.0__4.0__ multiply__11.0__36.0__ subtract__396.0__320.0__ subtract__396.0__320.0__ |
| if the integer n has exactly three positive divisors including num__1 and n how many positive divisors does n ^ num__3 have ? <o> a ) num__7 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
take the example of num__4 . . . it has num__3 positive divisors ( num__1 num__24 ) now take the example of num__64 . . . it has only num__7 divisors . . so a is the ans <eor> a <eos> |
a |
add__1.0__3.0__ add__3.0__4.0__ multiply__1.0__7.0__ |
add__1.0__3.0__ add__3.0__4.0__ multiply__1.0__7.0__ |
| two employees m and n are paid a total of $ num__605 per week by their employer . if m is paid num__120 percent of the salary paid to n how much is n paid per week ? <o> a ) $ num__245 <o> b ) $ num__255 <o> c ) $ num__265 <o> d ) $ num__275 <o> e ) $ num__285 |
num__1.2 n + n = num__605 num__2.2 n = num__605 n = num__275 the answer is d . <eor> d <eos> |
d |
divide__605.0__2.2__ divide__605.0__2.2__ |
divide__605.0__2.2__ divide__605.0__2.2__ |
| a company has num__15 managers and num__75 associates . the num__15 managers have an average salary of $ num__150000 . the num__75 associates have an average salary of $ num__30000 . what is the average salary for the company ? <o> a ) $ num__35000 <o> b ) $ num__45000 <o> c ) $ num__50000 <o> d ) $ num__65000 <o> e ) $ num__75 |
000 |
another method is to get ratios say num__30000 = a and we know the # of people are in num__1 : num__5 ratio average = ( num__5 a * num__1 + a * num__5 ) / num__6 = num__10 a / num__6 = num__50000 answer is c . $ num__50000 <eor> c <eos> |
c |
c |
| together num__15 type a machines and num__7 type b machines can complete a certain job in num__4 hours . together num__8 type b machines and num__15 type c machines can complete the same job in num__11 hours . how many q hours would it take one type a machine one type b machine and one type c machine working together to complete the job ( assuming constant rates for each machine ) ? <o> a ) num__22 hours <o> b ) num__30 hours <o> c ) num__44 hours <o> d ) num__60 hours <o> e ) it can not be determined from the information above . |
say the rates of machines a b and c are a b and c respectively . together num__15 type a machines and num__7 type b machines can complete a certain job in num__4 hours - - > num__15 a + num__7 b = num__0.25 ; together num__8 type b machines and num__15 type c machines can complete the same job in num__11 hours - - > num__8 b + num__15 c = num__0.0909090909091 . sum the above : num__15 a + num__15 b + num__15 c = num__0.25 + num__0.0909090909091 = num__0.340909090909 - - > reduce by num__15 : a + b + c = num__0.0227272727273 - - > so the combined rate of the three machines is num__0.0227272727273 job / hour - - > time is reciprocal of the rate thus machines a b and c can do the job q in num__44 hours . answer : c . <eor> c <eos> |
c |
add__0.25__0.0909__ divide__0.25__11.0__ multiply__4.0__11.0__ round__44.0__ |
add__0.25__0.0909__ divide__0.25__11.0__ divide__11.0__0.25__ divide__11.0__0.25__ |
| a man can swim in still water at num__9 km / h but takes twice as long to swim upstream than downstream . the speed of the stream is ? <o> a ) num__3 <o> b ) num__7.5 <o> c ) num__2.25 <o> d ) num__1.5 <o> e ) num__4 |
m = num__9 s = x ds = num__9 + x us = num__9 - x num__9 + x = ( num__9 - x ) num__2 num__9 + x = num__18 - num__2 x num__3 x = num__9 x = num__3 answer : a <eor> a <eos> |
a |
multiply__9.0__2.0__ round__3.0__ |
multiply__9.0__2.0__ round__3.0__ |
| two friends run the new york city marathon one friend finishes the marathon in num__400 minutes while the second friend finishes the marathon in num__625 minutes . what is the ratio of their speed ? <o> a ) num__10 : num__12 <o> b ) num__1 : num__2 <o> c ) num__5 : num__10 <o> d ) num__20 : num__25 <o> e ) none |
solution let us name the friends as a and b . = ( friend a ' s speed ) : ( friend b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š num__400 : â ˆ š num__625 = num__20 : num__25 answer d <eor> d <eos> |
d |
round__20.0__ |
round__20.0__ |
| the probability that a speaks truth is num__0.6 and that of b speaking truth is num__0.571428571429 . what is the probability that they agree in stating the same fact ? <o> a ) num__0.514285714286 <o> b ) num__0.473684210526 <o> c ) num__0.5 <o> d ) num__1.125 <o> e ) num__0.58064516129 |
if both agree stating the same fact either both of them speak truth of both speak false . probability = num__0.6 * num__0.571428571429 + num__0.4 * num__0.428571428571 = num__0.342857142857 + num__0.171428571429 = num__0.514285714286 answer : a <eor> a <eos> |
a |
negate_prob__0.6__ negate_prob__0.5714__ union_prob__0.6__0.3429__0.4286__ union_prob__0.6__0.4286__0.5143__ |
negate_prob__0.6__ negate_prob__0.5714__ union_prob__0.6__0.3429__0.4286__ union_prob__0.6__0.4286__0.5143__ |
| num__1391 x num__1391 = ? <o> a ) a ) num__1951609 <o> b ) b ) num__1951601 <o> c ) c ) num__1951602 <o> d ) d ) num__1951603 <o> e ) e ) num__1934881 |
num__1391 x num__1391 = ( num__1391 ) num__2 = ( num__1400 - num__9 ) num__2 = ( num__1400 ) num__2 + ( num__9 ) num__2 - ( num__2 x num__1400 x num__9 ) = num__1960000 + num__81 - num__25200 = num__1960081 - num__25200 = num__1934881 . answer : e <eor> e <eos> |
e |
subtract__1400.0__1391.0__ add__1960000.0__81.0__ subtract__1960081.0__25200.0__ subtract__1960081.0__25200.0__ |
subtract__1400.0__1391.0__ add__1960000.0__81.0__ subtract__1960081.0__25200.0__ subtract__1960081.0__25200.0__ |
| a type t machine can complete a job in num__5 hours and a type b machine can complete the job in num__7 hours . how many hours will it take num__2 type t machines and num__3 type b machines working together and independently to complete the job ? <o> a ) num__0.2 <o> b ) num__0.828571428571 <o> c ) num__0.833333333333 <o> d ) num__1.20689655172 <o> e ) num__2.91666666667 |
now d should be the answer . t need num__5 hours to complete and b needs num__7 hours to compete so num__2 t + num__3 b will complete num__0.4 + num__0.428571428571 or num__0.828571428571 portion of the job in num__1 hour so the whole job will take num__1.20689655172 hours . . . . = d <eor> d <eos> |
d |
divide__2.0__5.0__ divide__3.0__7.0__ add__0.4__0.4286__ subtract__3.0__2.0__ divide__1.0__0.8286__ divide__1.0__0.8286__ |
divide__2.0__5.0__ divide__3.0__7.0__ add__0.4__0.4286__ subtract__3.0__2.0__ divide__1.0__0.8286__ divide__1.0__0.8286__ |
| integers num__3 n + num__2 and num__8 n + num__1 are divisible by an integer p . if p is not equal to num__1 then p equals to ? <o> a ) num__2 <o> b ) num__5 <o> c ) num__7 <o> d ) num__11 <o> e ) num__13 |
if p divides num__3 n + num__2 then p divides num__24 n + num__16 . if p divides num__8 n + num__1 then p divides num__24 n + num__3 . then p must equal num__13 . the answer is e . <eor> e <eos> |
e |
multiply__3.0__8.0__ multiply__2.0__8.0__ subtract__16.0__3.0__ multiply__1.0__13.0__ |
multiply__3.0__8.0__ multiply__2.0__8.0__ subtract__16.0__3.0__ multiply__1.0__13.0__ |
| a circular rim a having a diameter of num__45 inches is rotating at the rate of x inches / min . another circular rim b with a diameter of num__30 inches is rotating at the rate of y inches / min . what is the value of y in terms of x if both the rims reach their starting positions at the same time after every rotation . <o> a ) num__3 x / num__2 <o> b ) num__4 x / num__5 <o> c ) num__7 x / num__5 <o> d ) num__5 x / num__7 <o> e ) num__3 x / num__4 |
t = s num__1 / v num__1 = s num__2 / v num__2 or num__45 / x = num__30 / y or y = num__45 x / num__30 = num__3 x / num__2 ( answer a ) <eor> a <eos> |
a |
multiply__1.0__3.0__ |
multiply__1.0__3.0__ |
| a number a is squared and then multiplied by negative num__1 . the result of this operation is equal to num__1 times the sum of num__2 times a and num__1 . what is one possible value of a ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
- num__1 * a ^ num__2 = num__1 ( num__2 a + num__1 ) a = - num__1 or - num__1 a = - num__1 = b answer : a <eor> a <eos> |
a |
reverse__1.0__ |
subtract__2.0__1.0__ |
| revenues were recorded for store a and store b over a period of three months . in the first month store a ' s revenues were $ num__12000 higher than store b ' s revenues . in the second month store a ' s revenues were $ num__8000 higher than store b ' s revenues . if store a ' s average ( arithmetic mean ) monthly revenue for the three months was $ num__2000 greater than store b ' s average monthly revenue then store b ' s revenue in the third month was how much greater than store a ' s revenue ? <o> a ) $ num__14000 <o> b ) $ num__15000 <o> c ) $ num__42000 <o> d ) $ num__46000 <o> e ) $ num__50 |
000 |
answer : cit may be tempting to come up with a lot of variables one each for each month ' s revenue for each company . however focus on the differences . in the first month the difference was + num__12 in favor of a . ( note that we can drop the thousands since every number in the question is in terms of thousands . ) in the second the difference was + num__8 in favor of a . the average was + num__2 in favor of a . with these numbers use the average formula to find the third month ( t ) : ( num__12 + num__8 + t ) / num__3 = num__2 num__20 + t = num__6 t = - num__14 since positive numbers indicate a difference in favor of a negative numbers are in favor of b . - num__14 represents a $ num__14000 advantage in favor of store b . choice ( a ) is correct . <eor> a <eos> |
a |
a |
| if num__36 men can do a piece of work in num__25 hours in how many hours will num__10 men do it ? <o> a ) num__76 hours <o> b ) num__66 hours <o> c ) num__57 hours <o> d ) num__90 hours <o> e ) num__18 hours |
explanation : let the required numbers of hours be x . less men more hours ( indirect proportion ) therefore num__10 : num__36 : : num__25 : x = ( num__10 x x ) = ( num__36 x num__25 ) = x = ( num__36 x num__25 ) / num__10 = num__90 . hence num__10 men can do it in num__90 hours . answer : d <eor> d <eos> |
d |
round__90.0__ |
round__90.0__ |
| a boat can move upstream at num__15 kmph and downstream at num__75 kmph then the speed of the current is ? <o> a ) num__30 <o> b ) num__20 <o> c ) num__45 <o> d ) num__10 <o> e ) num__20 |
us = num__15 ds = num__75 m = ( num__75 - num__15 ) / num__2 = num__30 answer : a <eor> a <eos> |
a |
multiply__15.0__2.0__ round__30.0__ |
multiply__15.0__2.0__ round__30.0__ |
| on a map num__1.5 inches represent num__24 miles . how many miles approximately is the distance if you measured num__49 centimeters assuming that num__1 - inch is num__2.54 centimeters ? <o> a ) num__174.2 <o> b ) num__212 <o> c ) num__288.1 <o> d ) num__296 <o> e ) num__308 |
num__1.5 inch = num__2.54 * num__1.5 cm . so num__2.54 * num__1.5 represents num__24 miles . so for num__49 cm . : num__49 / ( num__2.54 * num__1.5 ) = x / num__24 - - - > x = num__24 * num__49 / ( num__3.81 ) = num__308 answer will be e . <eor> e <eos> |
e |
multiply__1.5__2.54__ round__308.0__ |
multiply__1.5__2.54__ multiply__1.0__308.0__ |
| solve the number series by identifying the last number . num__1 num__4 num__5 num__6 num__7 num__9 num__11 ? <o> a ) num__100 <o> b ) num__120 <o> c ) num__98 <o> d ) num__89 <o> e ) num__131 |
a num__100 series of number that do not contain albhabet ' t ' <eor> a <eos> |
a |
multiply__1.0__100.0__ |
multiply__1.0__100.0__ |
| if a ^ num__2 + b ^ num__2 = num__15 and ab = num__25 what is the value of the expression ( a - b ) ^ num__2 + ( a + b ) ^ num__2 ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__30 <o> d ) num__60 <o> e ) num__50 |
( a - b ) ^ num__2 = a ^ num__2 + b ^ num__2 - num__2 ab = num__15 - num__25 = - num__10 ( a + b ) ^ num__2 = a ^ num__2 + b ^ num__2 + num__2 ab = num__15 + num__25 = num__40 so ( a + b ) ^ num__2 + ( a - b ) ^ num__2 = num__40 - num__10 = num__50 e <eor> e <eos> |
e |
subtract__25.0__15.0__ add__15.0__25.0__ multiply__2.0__25.0__ multiply__2.0__25.0__ |
subtract__25.0__15.0__ add__15.0__25.0__ add__40.0__10.0__ add__40.0__10.0__ |
| a student traveled num__25 percent of the distance of the trip alone continued another num__20 miles with a friend and then finished the last half of the trip alone . how many miles long was the trip ? <o> a ) num__60 <o> b ) num__80 <o> c ) num__100 <o> d ) num__120 <o> e ) num__150 |
let x be the total length of the trip . num__0.25 x + num__20 miles + num__0.5 x = x num__20 miles = num__0.25 x x = num__80 miles the answer is b . <eor> b <eos> |
b |
divide__20.0__0.25__ round__80.0__ |
divide__20.0__0.25__ round__80.0__ |
| m = { - num__6 - num__5 - num__4 - num__3 - num__2 - num__1 } t = { - num__4 - num__3 - num__2 - num__1 num__0 num__1 num__2 num__3 num__4 num__5 } if an integer is to be randomly selected from set m above and an integer is to be randomly selected from set t above what is the probability that the product of the two integers will be negative ? <o> a ) num__0 <o> b ) num__0.333333333333 <o> c ) num__0.4 <o> d ) num__0.5 <o> e ) num__0.6 |
we will have a negative product only if num__1 num__2 num__3 num__4 or num__5 are selected from set t . p ( negative product ) = num__0.5 = num__0.5 the answer is d . <eor> d <eos> |
d |
reverse__2.0__ reverse__2.0__ |
reverse__2.0__ reverse__2.0__ |
| eight coins are tossed simultaneously . in how many of the outcomes will the third coin turn up a head ? <o> a ) num__2 ^ num__9 <o> b ) num__2 ^ num__10 <o> c ) num__3 * num__2 ^ num__8 <o> d ) num__2 ^ num__7 <o> e ) num__3 * num__2 ^ num__10 |
fix the third coin as h . the remaining num__7 coins have num__2 ^ num__7 outcomes . ans : d <eor> d <eos> |
d |
coin_space__ coin_space__ |
coin_space__ coin_space__ |
| if a and b are positive even integers and the least common multiple of a and b is expressed as a * b / e which of the following statements could be false ? <o> a ) n is a factor of both a and b <o> b ) ( a * b ) / n < ab <o> c ) ab is multiple of num__2 . <o> d ) ( a * b ) / n is a multiple of num__2 . <o> e ) e is a multiple of num__4 . |
if we divide by e and get a multiple ofaand ofb then e must be a common factor . answer a must be true . if a and b are even integers dividing by a factor ( also an integer ) results in something smaller than the productab . answer b must be true . if eitheraorbis even thenabmust be even . answer c must be true . if both are even the smallest either number can be is num__2 . thus the smallest common factor must be num__2 . answer d must be true . ifbis num__6 andais num__2 nwould be num__6 which is not num__4 . e can be false . <eor> e <eos> |
e |
subtract__6.0__2.0__ subtract__6.0__2.0__ |
subtract__6.0__2.0__ subtract__6.0__2.0__ |
| a person goes to his office at num__0.333333333333 rd of the speed at which he returns from his office . if the avg speed during the whole trip is num__27 m / h . what is the speedof the person while he was going to his office ? <o> a ) num__8 km / h <o> b ) num__10 km / h <o> c ) num__12 km / h <o> d ) num__13 km / h <o> e ) num__18 km / h |
u = k v = num__3 k \ inline \ therefore \ frac { num__2 uv } { u + v } \ : \ : \ rightarrow \ frac { num__2 \ times k \ times num__3 k } { ( k + num__3 k ) } = num__27 \ inline \ rightarrow num__1.5 k = num__27 \ inline \ rightarrow k = num__18 km / h e <eor> e <eos> |
e |
divide__3.0__2.0__ divide__27.0__1.5__ round__18.0__ |
divide__3.0__2.0__ divide__27.0__1.5__ divide__27.0__1.5__ |
| a man ' s regular pay is $ num__3 per hour up to num__40 hours . overtime is twice the payment for regular time . if he was paid $ num__198 how many hours overtime did he work ? <o> a ) num__8 <o> b ) num__5 <o> c ) num__9 <o> d ) num__6 <o> e ) num__13 |
at $ num__3 per hour up to num__40 hours regular pay = $ num__3 x num__40 = $ num__120 if total pay = $ num__168 overtime pay = $ num__198 - $ num__120 = $ num__78 overtime rate ( twice regular ) = num__2 x $ num__3 = $ num__6 per hour = > number of overtime hours = $ num__78 / $ num__6 = num__13 ans is e <eor> e <eos> |
e |
multiply__3.0__40.0__ subtract__198.0__120.0__ multiply__3.0__2.0__ divide__78.0__6.0__ round__13.0__ |
multiply__3.0__40.0__ subtract__198.0__120.0__ multiply__3.0__2.0__ divide__78.0__6.0__ divide__78.0__6.0__ |
| in a certain store the profit is num__320.0 of the cost . if the cost increases by num__25.0 but the selling price remains constant approximately what percentage of the selling price is the profit <o> a ) num__70.0 <o> b ) num__80.0 <o> c ) num__90.0 <o> d ) num__100.0 <o> e ) none of above |
explanation : let c . p . = rs . num__100 . then profit = rs . num__320 s . p . = rs . num__420 . new c . p . = num__125.0 of rs . num__100 = rs . num__125 new s . p . = rs . num__420 . profit = rs . ( num__420 - num__125 ) = rs . num__295 required percentage = ( num__0.702380952381 ) * num__100 = num__70.0 ( approx ) answer : a <eor> a <eos> |
a |
percent__70.0__100.0__ |
percent__70.0__100.0__ |
| a set of numbers has the property that for any number x in the set x + num__1 is also in the set . if - num__1 is in the set which of the following must also be in the set num__1 . num__2 num__2 . num__0 num__3 . - num__2 <o> a ) num__1 only <o> b ) num__2 only <o> c ) num__1 and num__2 only <o> d ) num__2 and num__3 only <o> e ) num__12 and num__3 |
if x is in the set than x + num__1 is also in the set if - num__1 in the set than - num__1 + num__1 = num__0 is also in the set since + num__1 is present num__1 + num__1 = num__2 is also present in the set . therefore ans c <eor> c <eos> |
c |
reverse__1.0__ |
subtract__2.0__1.0__ |
| num__4 num__6 num__12 num__14 num__28 num__30 ? <o> a ) num__20 <o> b ) num__40 <o> c ) no = num__60 <o> d ) num__80 <o> e ) num__100 |
c num__60 the given sequence is a combination of two series num__4 num__12 num__28 . . . . and num__6 num__14 num__30 . . . . the pattern is + num__8 + num__16 + num__32 . so the missing number = ( num__28 + num__32 ) = num__60 <eor> c <eos> |
c |
subtract__12.0__4.0__ add__4.0__12.0__ add__4.0__28.0__ add__28.0__32.0__ |
subtract__12.0__4.0__ add__4.0__12.0__ add__4.0__28.0__ add__28.0__32.0__ |
| find the middle one when the sum of num__3 consecutive even numbers is num__36 ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__16 <o> e ) num__17 |
num__3 consecutive numbers can be a - num__1 a a + num__1 so sum of numbers = num__3 a = num__36 . hence a = num__12 . b <eor> b <eos> |
b |
divide__36.0__3.0__ round__12.0__ |
divide__36.0__3.0__ round__12.0__ |
| a man walking at the rate of num__5 km / hr crosses a bridge in num__15 minutes . the length of the bridge ( in metres ) is <o> a ) num__600 <o> b ) num__750 <o> c ) num__1000 <o> d ) num__1250 <o> e ) num__1500 |
explanation : speed = ( num__5 x num__0.277777777778 ) m / sec = num__1.38888888889 m / sec . distance covered in num__15 minutes = ( num__1.38888888889 x num__15 x num__60 ) m = num__1250 m . answer is d <eor> d <eos> |
d |
hour_to_min_conversion__ round__1250.0__ |
hour_to_min_conversion__ round__1250.0__ |
| difference between the compound interest and the simple interest accrued on an amount of rs . num__18000 in two year is rs . num__405 . what was the rate of interest ? <o> a ) num__14.0 <o> b ) num__16.0 <o> c ) num__12.5 <o> d ) num__15.0 <o> e ) num__15.5 % |
c . i - s . i = p ( r / num__100 ) ^ num__2 num__405 = num__18000 ( r ^ num__0.0002 ) ( num__405 * num__10000 ) / num__18000 = r ^ num__2 r ^ num__2 = num__225 ans r = num__15.0 answer : d <eor> d <eos> |
d |
percent__100.0__15.0__ |
percent__100.0__15.0__ |
| a man can row num__6 kmph in still water . when the river is running at num__1.2 kmph it takes him num__1 hour to row to a place and black . what is the total distance traveled by the man ? <o> a ) num__5.78 <o> b ) num__5.86 <o> c ) num__5.76 <o> d ) num__8.76 <o> e ) num__5.46 |
m = num__6 s = num__1.2 ds = num__7.2 us = num__4.8 x / num__7.2 + x / num__4.8 = num__1 x = num__2.88 d = num__2.88 * num__2 = num__5.76 answer : c <eor> c <eos> |
c |
add__6.0__1.2__ subtract__6.0__1.2__ multiply__1.2__4.8__ round__5.76__ |
add__6.0__1.2__ subtract__6.0__1.2__ multiply__1.2__4.8__ multiply__1.2__4.8__ |
| num__6 litres of water are poured into an aquarium of dimensions num__50 cm length num__30 cm breadth and num__40 cm height . how high ( in cm ) will the water rise ? ( num__1 litre = num__1000 cm ³ ) <o> a ) num__4 <o> b ) num__8 <o> c ) num__10 <o> d ) num__20 <o> e ) num__40 |
lxbxh = num__12000 h = num__120.0 * num__30 = num__4 cm ' a ' is the answer . <eor> a <eos> |
a |
divide__120.0__30.0__ round__4.0__ |
divide__120.0__30.0__ multiply__1.0__4.0__ |
| the difference of two numbers is num__1365 . on dividing the larger number by the smaller we get num__6 as quotient and the num__15 as remainder . what is the smaller number ? <o> a ) num__120 <o> b ) num__180 <o> c ) num__270 <o> d ) num__260 <o> e ) num__230 |
let the smaller number be x . then larger number = ( x + num__1365 ) . x + num__1365 = num__6 x + num__15 num__5 x = num__1350 x = num__270 smaller number = num__270 . answer : c <eor> c <eos> |
c |
subtract__1365.0__15.0__ divide__1350.0__5.0__ divide__1350.0__5.0__ |
subtract__1365.0__15.0__ divide__1350.0__5.0__ divide__1350.0__5.0__ |
| two taps can separately fill a cistern num__7 minutes and num__14 minutes respectively and when the waste pipe is open they can together fill it in num__8 minutes . the waste pipe can empty the full cistern in ? <o> a ) num__6.2 <o> b ) num__8.2 <o> c ) num__11.2 <o> d ) num__9.2 <o> e ) num__7.2 |
num__0.142857142857 + num__0.0714285714286 - num__1 / x = num__0.125 x = num__11.2 answer : c <eor> c <eos> |
c |
subtract__8.0__7.0__ divide__1.0__8.0__ round__11.2__ |
subtract__8.0__7.0__ divide__1.0__8.0__ divide__11.2__1.0__ |
| in a num__500 m race the ratio of the speeds of two contestants a and b is num__3 : num__4 . a has a start of num__170 m . then a wins by : <o> a ) num__60 m <o> b ) num__20 m <o> c ) num__43 m <o> d ) num__20 m <o> e ) num__23 m |
to reach the winning post a will have to cover a distance of ( num__500 - num__170 ) m i . e . num__330 m . while a covers num__3 m b covers num__4 m . while a covers num__330 m b covers num__4 x num__110.0 m = num__440 m . thus when a reaches the winning post b covers num__440 m and therefore remains num__60 m behind . a wins by num__60 m . answer : a <eor> a <eos> |
a |
subtract__500.0__170.0__ divide__330.0__3.0__ multiply__4.0__110.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
subtract__500.0__170.0__ divide__330.0__3.0__ multiply__4.0__110.0__ subtract__500.0__440.0__ subtract__500.0__440.0__ |
| remainder of num__2 ^ num__1.0 ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
( num__2 ^ ( num__1.02173913043 ) ) / num__47 = num__2 ^ num__0.0212765957447 = num__2 answer : a <eor> a <eos> |
a |
reverse__47.0__ multiply__2.0__1.0__ |
reverse__47.0__ divide__2.0__1.0__ |
| bruno and sacha are running in the same direction around a stadium . sacha runs at a constant speed of num__6 meters per second and bruno runs at a constant speed of num__5 meters per second . at a certain point sacha overtakes bruno . if five minute afterward sacha stops and waits for bruno to reach him then how many seconds does he have to wait ? <o> a ) num__12 <o> b ) num__24 <o> c ) num__36 <o> d ) num__60 <o> e ) num__72 |
the difference of the speed is num__1 m per second so in five minute sacha will be num__300 m ahead of bruno . . bruno will cover this in num__60.0 = num__60 secs . . d <eor> d <eos> |
d |
subtract__6.0__5.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
subtract__6.0__5.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
| the unit ’ s digit in the product ( num__3127 ) ^ num__173 is : <o> a ) num__1 <o> b ) num__3 <o> c ) num__7 <o> d ) num__9 <o> e ) num__11 |
unit digit in ( num__3127 ) ^ num__173 = unit digit in num__7 ^ num__173 . now num__7 ^ num__4 gives unit digit num__1 . therefore num__7 ^ num__173 = ( num__74 ) ^ num__43 * num__7 ^ num__1 . thus num__7 ^ num__173 gives unit digit num__7 . answer : c <eor> c <eos> |
c |
multiply__1.0__7.0__ |
multiply__1.0__7.0__ |
| a boatman goes num__2 km against the current of the stream in num__2 hour and goes num__1 km along the current in num__20 minutes . how long will it take to go num__5 km in stationary water ? <o> a ) num__4 hr <o> b ) num__25 min <o> c ) num__45 min <o> d ) num__6 hr <o> e ) num__2 hr num__30 min |
explanation : speed upstream = num__1.0 = num__1 kn / hr speed downstream = num__1 / ( num__0.333333333333 ) = num__3 km / hr speed in still water = num__0.5 ( num__3 + num__1 ) = num__2 km / hr time taken to travel num__5 km in still water = num__2.5 = num__2 num__0.5 hours = num__2 hour num__30 minutes answer : e <eor> e <eos> |
e |
add__2.0__1.0__ divide__1.0__2.0__ add__2.0__0.5__ round__2.0__ |
add__2.0__1.0__ divide__1.0__2.0__ add__2.0__0.5__ divide__2.0__1.0__ |
| linda spent num__0.75 of her savings on furniture and the rest on a tv . if the tv cost her $ num__200 what were her original savings ? <o> a ) $ num__900 <o> b ) $ num__300 <o> c ) $ num__600 <o> d ) $ num__700 <o> e ) $ num__800 |
if linda spent num__0.75 of her savings on furnitute the rest num__1.0 - num__0.75 = num__0.25 on a tv but the tv cost her $ num__200 . so num__0.25 of her savings is $ num__200 . so her original savings are num__4 times $ num__200 = $ num__800 correct answer e <eor> e <eos> |
e |
subtract__1.0__0.75__ reverse__0.25__ multiply__200.0__4.0__ multiply__200.0__4.0__ |
subtract__1.0__0.75__ reverse__0.25__ multiply__200.0__4.0__ multiply__200.0__4.0__ |
| the average weight of num__8 people increases by num__2.5 kg when a new person comes in place of one of them weighing num__45 kg . what is the weight of the new person ? <o> a ) num__75 kg <o> b ) num__85 kg <o> c ) num__95 kg <o> d ) num__65 kg <o> e ) num__55 kg |
the total weight increase = ( num__8 x num__2.5 ) kg = num__20 kg weight of new person = ( num__45 + num__20 ) kg = num__65 kg the answer is d . <eor> d <eos> |
d |
multiply__8.0__2.5__ add__45.0__20.0__ add__45.0__20.0__ |
multiply__8.0__2.5__ add__45.0__20.0__ add__45.0__20.0__ |
| as part of a game five people each must secretly choose an integer between num__1 and num__5 inclusive . what is the approximate likelihood that all five people will choose different numbers ? <o> a ) num__4.0 <o> b ) num__12.0 <o> c ) num__16.0 <o> d ) num__20.0 <o> e ) num__25 % |
ans : a ) num__1 st person has option no ' s - ( num__1 num__23 num__45 ) - there fore probability of getting a no = num__5 c num__0.2 c num__1 = num__1 num__2 nd person has option no ' s any four he has to choose a no from four no ' s - there fore probability of getting a no = num__4 c num__0.2 c num__1 = num__0.8 num__3 rd person has option no ' s any three he has to choose a no from three no ' s - there fore probability of getting a no = num__3 c num__0.2 c num__1 = num__0.6 num__4 th person has option no ' s any two - there fore probability of getting a no = num__2 c num__0.2 c num__1 = num__0.4 num__4 th person has option no ' s only one - there fore probability of getting a no = num__1 c num__0.2 c num__1 = num__0.2 = num__1 * num__0.8 * num__0.6 * num__0.4 * num__0.2 = num__4.0 <eor> a <eos> |
a |
coin_space__ negate_prob__0.2__ negate_prob__0.6__ choose__4.0__3.0__ |
coin_space__ negate_prob__0.2__ negate_prob__0.6__ choose__4.0__3.0__ |
| the market price of an article was num__40.0 more than its cost price . i was going to sell it at market price to a customer but he showed me some defects in the article due to which i gave him a discount of num__28.57 . next day he came again and showed me some more defects hence i gave him another discount that was equal to num__12.5 of the cost price . what was the approximate loss to me ? <o> a ) loss of num__10.0 <o> b ) loss of num__12.5 <o> c ) loss of num__15.0 <o> d ) loss of num__25.0 <o> e ) none of these |
let the cost price be rs num__100 then market price is rs num__140 now the first discount is of num__28.57 ( approx . ) = num__0.285714285714 th of market price . hence its selling price = num__140 x num__0.714285714286 = rs num__100 now since you are selling at cost price any further discount will be equal to loss percentage . answer : b <eor> b <eos> |
b |
percent__12.5__100.0__ |
percent__12.5__100.0__ |
| which of the following is equal to num__1 ( num__0.5 ) % ? <o> a ) num__0.012 / num__100 <o> b ) num__0.12 / num__100 <o> c ) num__1.5 / num__100 <o> d ) num__0.12 <o> e ) num__1.2 |
this notation may be confusing for some since it looks like we ' re multiplying num__1 and num__0.5 how about adding a space : which of the following is equal to ( num__1 num__0.5 ) % ( num__1 num__0.5 ) % = num__1.5 = num__1.5 / num__100 answer : c <eor> c <eos> |
c |
add__1.0__0.5__ add__1.0__0.5__ |
add__1.0__0.5__ divide__1.5__1.0__ |
| a man took loan from a bank at the rate of num__12.0 p . a . simple interest . after num__3 years he had to pay rs . num__2520 interest only for the period . the principal amount borrowed by him was : <o> a ) s . num__2000 <o> b ) s . num__1000 <o> c ) s . num__5000 <o> d ) s . num__8000 <o> e ) s . num__7 |
000 |
principal = rs . ( num__100 x num__2520 ) / ( num__12 x num__3 ) = rs . num__7000 . answer : option e <eor> e <eos> |
e |
e |
| the average of a set of five distinct integers is num__370 . if each number is less than num__2000 and the median of the set is the greatest possible value what is the sum of the two smallest numbers ? <o> a ) - num__4494 <o> b ) - num__3997 <o> c ) - num__3494 <o> d ) - num__4144 <o> e ) the answer can not be determined from the information given |
sum of the set = num__370 * num__5 = num__1850 each number is less than num__2000 and median of the set is the greatest possible value . therefore last num__3 numbers can be num__19991998 and num__1997 . their sum = num__5994 . therefore sum of two smallest numbers = num__1850 - num__5994 = - num__4144 answer = d <eor> d <eos> |
d |
multiply__370.0__5.0__ subtract__2000.0__3.0__ subtract__5994.0__1850.0__ subtract__5994.0__1850.0__ |
multiply__370.0__5.0__ subtract__2000.0__3.0__ subtract__5994.0__1850.0__ subtract__5994.0__1850.0__ |
| a crate measures num__7 feet by num__8 feet by num__12 feet on the inside . a stone pillar in the shape of a right circular cylinder must fit into the crate for shipping so that it rests upright when the crate sits on at least one of its six sides . what is the radius in feet of the pillar with the largest volume that could still fit in the crate ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__7 <o> e ) num__12 |
we can find the radius of all the three cases of cylinders . the only crux to find the answer faster is that : voulme is pi * r ^ num__2 * h . the volume is a function of r ^ num__2 . so r has to be the highest to find the largest volume . so r = num__7 for the surface num__8 * num__12 face . volume = num__343 pi answer d <eor> d <eos> |
d |
volume_cube__7.0__ triangle_area__7.0__2.0__ |
volume_cube__7.0__ triangle_area__7.0__2.0__ |
| a car dealer has only red and yellow cars in his show room . he currently has num__20.0 more yellow cars then red . if he has num__20 red cars how many yellow cars does he have ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__35 <o> d ) num__40 <o> e ) num__45 |
since we know the dealer only has yellow and red cars in their showroom we can conclude that red and yellow cars make up num__100.0 of the cars . since we know the difference between the amount of cars is num__20.0 we can assume that the red cars make up num__40.0 of the cars . num__20 / x = num__0.4 using cross multiplication we can find that num__100 x num__20 = num__2000 and num__50.0 = num__50 so now we know that there are num__50 cars total . we can subtract the number of red cars from the total to find the number of yellow cars . num__50 - num__20 = num__30 there are a total of num__30 yellow cars . the answer is b <eor> b <eos> |
b |
percent__100.0__30.0__ |
percent__100.0__30.0__ |
| if num__4 men working num__10 hours a day earn rs . num__1000 per week then num__9 men working num__6 hours a day will earn how much per week ? <o> a ) rs num__840 <o> b ) rs num__1350 <o> c ) rs num__1620 <o> d ) rs num__1680 <o> e ) none of these |
explanation : ( men num__4 : num__9 ) : ( hrs / day num__10 : num__6 ) : : num__1000 : x hence num__4 * num__10 * x = num__9 * num__6 * num__1000 or x = num__9 * num__6 * num__250.0 * num__10 = num__1350 answer : b <eor> b <eos> |
b |
divide__1000.0__4.0__ round__1350.0__ |
divide__1000.0__4.0__ round__1350.0__ |
| working together num__7 identical pumps can empty a pool in num__6 hours . how many hours will it take num__6 pumps to empty the same pool ? <o> a ) num__6 <o> b ) num__5 <o> c ) num__4 <o> d ) num__8 <o> e ) num__7 |
the answer is e num__7 pumps take num__6 hours . num__1 pump will take num__6 * num__7 hours . hence num__6 pumps will take ( num__6 * num__7 ) / num__6 = num__7 hours <eor> e <eos> |
e |
subtract__7.0__6.0__ round__7.0__ |
subtract__7.0__6.0__ divide__7.0__1.0__ |
| mr . ram is on tour and he has rs num__360 for his expenses . if he exceeds his tour by num__4 days he must cut down daily expenses by rs num__3 . the number of days of mr . ram ' s tour programme is <o> a ) num__28 days <o> b ) num__24 days <o> c ) num__22 days <o> d ) num__20 days <o> e ) none of these |
solution : let ram under takes a tour of x days . then expenses for each day = num__360 / x num__360 / x + num__4 = num__360 / x − num__3 x = num__20 x = num__20 and − num__24 − num__24 hence x = num__20 days . option ( d ) <eor> d <eos> |
d |
add__4.0__20.0__ round__20.0__ |
add__4.0__20.0__ round__20.0__ |
| two pipes a and b can fill a tank in num__6 hours and num__4 hours respectively . if they are opened on alternate hours and if pipe a is opened first in how many hours the tank shall be full ? <o> a ) num__4 hrs <o> b ) num__5 hrs <o> c ) num__7 hrs <o> d ) num__9 hrs <o> e ) none |
solution a ' s work in num__1 hour = num__0.166666666667 . b ' s work in num__1 hour = num__0.25 . ( a + b ) ' s num__2 hour ' s work when opened alternately = ( num__0.166666666667 + num__0.25 ) = num__0.416666666667 . ( a + b ) ' s num__4 hour ' s work when opened alternately = num__0.833333333333 = num__0.833333333333 . remaining part = ( num__1 - num__0.833333333333 ) = num__0.166666666667 . therefore total tank to fill the tank ( num__4 + num__1 ) hrs = num__5 hrs . answer b <eor> b <eos> |
b |
divide__1.0__6.0__ divide__1.0__4.0__ subtract__6.0__4.0__ add__0.25__0.1667__ subtract__1.0__0.1667__ subtract__6.0__1.0__ round__5.0__ |
divide__1.0__6.0__ divide__1.0__4.0__ subtract__6.0__4.0__ add__0.25__0.1667__ subtract__1.0__0.1667__ subtract__6.0__1.0__ subtract__6.0__1.0__ |
| carl bought num__16 gumballs lewis bought num__12 gumballs and carey bought x gumballs . the average ( arithmetic mean ) number of gumballs the three bought is between num__19 and num__25 inclusive . what is the difference between the greatest number and the smallest number of gumballs carey could have bought ? <o> a ) num__20 <o> b ) num__16 <o> c ) num__22 <o> d ) num__18 <o> e ) num__24 |
smallest gumballs = ( num__19 - num__16 ) + ( num__19 - num__12 ) + num__19 = num__29 largest gumballs = ( num__25 - num__16 ) + ( num__25 - num__12 ) + num__25 = num__47 difference = num__47 - num__29 = num__18 d <eor> d <eos> |
d |
subtract__47.0__29.0__ subtract__47.0__29.0__ |
subtract__47.0__29.0__ subtract__47.0__29.0__ |
| the sum of three numbers is num__98 . if the ratio of the first to second is num__2 : num__3 and that of the second to the third is num__5 : num__8 then the second number is : <o> a ) num__10 <o> b ) num__50 <o> c ) num__20 <o> d ) num__30 <o> e ) num__40 |
let the three parts be a b c . then a : b = num__2 : num__3 and b : c = num__5 : num__8 = ( num__5 x ( num__0.6 ) ) : ( num__8 x ( num__0.6 ) ) = num__3 : num__4.8 a : b : c = num__2 : num__3 : num__4.8 = num__10 : num__15 : num__24 b = ( num__98 x num__0.30612244898 ) = num__30 . answer is d . <eor> d <eos> |
d |
divide__3.0__5.0__ multiply__8.0__0.6__ multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__3.0__8.0__ multiply__2.0__15.0__ multiply__2.0__15.0__ |
divide__3.0__5.0__ multiply__8.0__0.6__ multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__3.0__8.0__ multiply__2.0__15.0__ multiply__2.0__15.0__ |
| a woman took some money for borrowed for num__3 years the total will be rs . num__4000 and num__5 years it will be rs . num__5000 / - . then how much amount she borrowed ? <o> a ) rs . num__2500 / - <o> b ) rs . num__2550 / - <o> c ) rs . num__2590 / - <o> d ) rs . num__2600 / - <o> e ) rs . num__2650 / - |
num__3 years - - - - - - - - > rs . num__4000 / - num__5 years - - - - - - - - > rs . num__5000 / - ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - num__2 years - - - - - - - - - > rs . num__1000 / - num__1 year - - - - - - - - - - > rs . num__500 / - num__3 years * rs . num__500 / - = rs . num__1500 / - p = rs . num__4000 / - - rs . num__1500 / - = rs . num__2500 / - p = rs . num__2500 / - a <eor> a <eos> |
a |
subtract__5.0__3.0__ subtract__5000.0__4000.0__ subtract__3.0__2.0__ divide__1000.0__2.0__ multiply__3.0__500.0__ subtract__4000.0__1500.0__ subtract__4000.0__1500.0__ |
subtract__5.0__3.0__ subtract__5000.0__4000.0__ subtract__3.0__2.0__ divide__1000.0__2.0__ multiply__3.0__500.0__ subtract__4000.0__1500.0__ subtract__4000.0__1500.0__ |
| if we have num__5 people in a meeting in how many ways they can sit around a table ? <o> a ) num__10 * num__9 ! <o> b ) num__9 * num__8 ! <o> c ) num__8 * num__7 ! <o> d ) num__7 * num__6 ! <o> e ) num__4 * num__3 ! |
if there arenpeople sitting around a table there are ( n - num__1 ) ! possible arrangements : here n = num__5 ( n - num__1 ) ! = num__4 ! = num__4 * num__3 ! ans : e <eor> e <eos> |
e |
choose__4.0__3.0__ |
choose__4.0__3.0__ |
| how many seconds will a num__500 meter long train take to cross a man walking with a speed of num__3 km / hr in the direction of the moving train if the speed of the train is num__63 km / hr ? <o> a ) num__878 meters <o> b ) num__154 meters <o> c ) num__500 meters <o> d ) num__184 meters <o> e ) num__157 meters |
let length of tunnel is x meter distance = num__800 + x meter time = num__1 minute = num__60 seconds speed = num__78 km / hr = num__78 * num__0.277777777778 m / s = num__21.6666666667 m / s distance = speed * time num__800 + x = ( num__21.6666666667 ) * num__60 num__800 + x = num__20 * num__65 = num__1300 x = num__1300 - num__800 = num__500 meters answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ divide__60.0__3.0__ add__500.0__800.0__ round__500.0__ |
subtract__63.0__3.0__ divide__60.0__3.0__ add__500.0__800.0__ divide__500.0__1.0__ |
| the present ages of lewis and brown are in the ratio num__1 : num__2 . three years from now the ages will be in the ratio num__3 : num__5 . find the present age of brown . <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
l : b = num__1 : num__2 & ( l + num__3 ) : ( b + num__3 ) = num__3 : num__5 then num__2 l = b & num__5 l + num__15 = num__3 b + num__9 by solving above we get b = num__12 answer : e <eor> e <eos> |
e |
multiply__3.0__5.0__ add__3.0__9.0__ multiply__1.0__12.0__ |
multiply__3.0__5.0__ add__3.0__9.0__ add__3.0__9.0__ |
| a crew can row a certain course up the stream in num__84 minutes ; they can row the same course down stream in num__6 minutes less than they can row it in still water . how long would they take to row down with the stream <o> a ) num__45 or num__23 minutes <o> b ) num__63 or num__12 minutes <o> c ) num__78 minutes <o> d ) num__19 minutes <o> e ) num__25 minutes |
a method has been discussed here to nullify the calculations though . all i can input here is speed of a crew in still water = num__0.5 ( speed upstream + speed downstream ) approximating this for time t = num__0.5 ( num__84 + ( t - num__6 ) ) giving t = num__78 c . <eor> c <eos> |
c |
subtract__84.0__6.0__ round__78.0__ |
subtract__84.0__6.0__ subtract__84.0__6.0__ |
| what is the minimum possible range in scores of the three test - takers ? three people each took num__5 tests . if the ranges of their scores in the num__5 practice tests were num__17 num__25 and num__36 . <o> a ) num__36 <o> b ) num__17 <o> c ) num__25 <o> d ) num__40 <o> e ) num__35 |
i simply looked at the num__3 different possible scores for each individual test : num__17 num__3625 we have to find the minimum range : num__36 - num__17 = num__9 num__36 - num__25 = num__11 num__25 - num__17 = num__8 the find the minimum range you have to make the set of the num__5 scores as small as possible . which means that num__4 of the num__5 scores of each individual person is zero . num__8 * num__5 = num__40 answer : d <eor> d <eos> |
d |
subtract__36.0__25.0__ add__5.0__3.0__ divide__36.0__9.0__ multiply__5.0__8.0__ multiply__5.0__8.0__ |
subtract__36.0__25.0__ subtract__17.0__9.0__ subtract__9.0__5.0__ multiply__5.0__8.0__ multiply__5.0__8.0__ |
| in a maths test students were asked to find num__0.3125 of a certain number . one of the students by mistake found num__0.833333333333 th of that number and his answer was num__50 more than the correct answer . find the number . <o> a ) num__96 <o> b ) num__280 <o> c ) num__384 <o> d ) num__400 <o> e ) num__500 |
explanation : let the number be x . num__5 * x / num__6 = num__5 * x / num__16 + num__50 num__25 * x / num__48 = num__50 x = num__96 answer a <eor> a <eos> |
a |
divide__5.0__0.3125__ multiply__6.0__16.0__ multiply__6.0__16.0__ |
divide__5.0__0.3125__ multiply__6.0__16.0__ multiply__6.0__16.0__ |
| man is num__20 years older than his son . in two years his age will be twice the age of his son . the present age of his son is : <o> a ) num__14 years <o> b ) num__26 years <o> c ) num__17 years <o> d ) num__18 years <o> e ) num__22 years |
let the son ' s present age be x years . then man ' s present age = ( x + num__20 ) years . ( x + num__20 ) + num__2 = num__2 ( x + num__2 ) x + num__22 = num__2 x + num__4 x = num__18 . answer : d <eor> d <eos> |
d |
add__20.0__2.0__ subtract__20.0__2.0__ subtract__20.0__2.0__ |
add__20.0__2.0__ subtract__20.0__2.0__ subtract__20.0__2.0__ |
| from the word ' laparoscopy ' how many independent meaningful words can be made without changing the order of the letters and using each letter only once ? <o> a ) num__8 <o> b ) num__7 <o> c ) num__6 <o> d ) num__2 <o> e ) num__3 |
explanation : the words are lap and copy . answer : d ) num__2 <eor> d <eos> |
d |
coin_space__ coin_space__ |
coin_space__ coin_space__ |
| when positive integer n is divided by num__4 the remainder is num__2 . when n is divided by num__7 the remainder is num__5 . how many values less than num__100 can n take ? <o> a ) num__0 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
a quick approac to this q is . . the equation we can form is . . num__3 x + num__2 = num__7 y + num__5 . . num__3 x - num__3 = num__7 y . . . num__3 ( x - num__1 ) = num__7 y . . . so ( x - num__1 ) has to be a multiple of num__7 as y then will take values of multiple of num__3 . . here we can see x can be num__1 num__815 num__2229 so num__5 values till num__100 is reached as ( num__29 - num__1 ) * num__3 = num__84 and next multiple of num__7 will be num__84 + num__21 > num__100 . . ans num__3 . . c <eor> c <eos> |
c |
subtract__7.0__4.0__ subtract__4.0__3.0__ multiply__7.0__3.0__ subtract__4.0__1.0__ |
subtract__7.0__4.0__ subtract__4.0__3.0__ multiply__7.0__3.0__ subtract__4.0__1.0__ |
| a train num__360 m long runs with a speed of num__45 km / hr . what time will it take to pass a platform of num__340 m long ? <o> a ) num__38 sec <o> b ) num__35 sec <o> c ) num__44 sec <o> d ) num__40 sec <o> e ) num__56 sec |
explanation : speed = num__45 km / hr = num__45 × ( num__0.277777777778 ) m / s = num__12.5 = num__12.5 = num__12.5 m / s total distance = length of the train + length of the platform = num__360 + num__340 = num__700 meter time taken to cross the platform = num__700 / ( num__12.5 ) = num__700 × num__0.08 = num__56 seconds answer : option e <eor> e <eos> |
e |
add__360.0__340.0__ divide__700.0__12.5__ round__56.0__ |
add__360.0__340.0__ divide__700.0__12.5__ divide__700.0__12.5__ |
| find the value of y from ( num__12 ) ^ num__3 x num__6 ^ num__4 ÷ num__432 = y ? <o> a ) num__2234 <o> b ) num__4350 <o> c ) num__4560 <o> d ) num__5132 <o> e ) y = num__5184 |
given exp . = ( num__12 ) num__3 x num__64 = ( num__12 ) num__3 x num__64 = ( num__12 ) num__2 x num__62 = ( num__72 ) num__2 = num__5184 num__432 num__12 x num__62 e <eor> e <eos> |
e |
divide__12.0__6.0__ subtract__64.0__2.0__ multiply__12.0__6.0__ multiply__12.0__432.0__ multiply__12.0__432.0__ |
divide__12.0__6.0__ subtract__64.0__2.0__ multiply__12.0__6.0__ multiply__12.0__432.0__ multiply__12.0__432.0__ |
| numbers from num__1 to num__10 ^ num__2012 are given there are ' m ' no of num__1 ' s and ' n ' no of num__2 ' s . then what is ( m - n ) . <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
let us consider num__1 - num__100 here no of num__1 ' s m = num__21 and no . of num__2 ' s n = num__20 there by m - n = num__1 similary from num__1 to ( num__10 ^ num__2012 - num__1 ) no . of num__1 ' s = no . of num__2 ' s but in last term num__10 ^ num__2012 a one is added there by the ans is num__1 . answer : b <eor> b <eos> |
b |
multiply__10.0__2.0__ reverse__1.0__ |
subtract__21.0__1.0__ subtract__2.0__1.0__ |
| if z num__2 + num__4 y num__2 = num__4 zy then z : y is : <o> a ) num__2 : num__1 <o> b ) num__1 : num__2 <o> c ) num__1 : num__1 <o> d ) num__1 : num__4 <o> e ) num__1 : num__5 |
as the coefficient of z is num__1 and coefficient of y is num__4 answer : d <eor> d <eos> |
d |
reverse__1.0__ |
reverse__1.0__ |
| a sells a bicycle to b at a profit of num__50.0 and b sells it to c at a loss of num__40.0 . find the resultant profit or loss . <o> a ) - num__4.0 <o> b ) num__5.0 <o> c ) - num__5.0 <o> d ) num__6.0 <o> e ) - num__10 % |
the resultant profit or loss = num__50 - num__40 - ( num__50 * num__40 ) / num__100 = - num__10.0 loss = num__10.0 answer is e <eor> e <eos> |
e |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| the sum of three consecutive even numbers is num__42 . find the middle number of the three ? <o> a ) num__14 <o> b ) num__86 <o> c ) num__16 <o> d ) num__28 <o> e ) num__23 |
three consecutive even numbers ( num__2 p - num__2 ) num__2 p ( num__2 p + num__2 ) . ( num__2 p - num__2 ) + num__2 p + ( num__2 p + num__2 ) = num__42 num__6 p = num__42 = > p = num__7 . the middle number is : num__2 p = num__14 . answer : a <eor> a <eos> |
a |
divide__42.0__6.0__ multiply__2.0__7.0__ round__14.0__ |
divide__42.0__6.0__ multiply__2.0__7.0__ round__14.0__ |
| arabica coffee costs $ num__0.5 per ounce while robusta coffee costs $ num__0.3 per ounce . if the blend of arabica and robusta costs $ num__0.33 per ounce what is the share of arabica in this blend ? <o> a ) num__15.0 <o> b ) num__24.0 <o> c ) num__30.0 <o> d ) num__33.0 <o> e ) num__40 % |
a = amount of arabica coffee num__1 - a = amount of robusta coffee . because if you subtract a from the num__1 ounce the remaining amount is robusta therefore : . num__5 a + . num__3 ( num__1 - a ) = . num__33 . num__5 a + . num__3 - . num__3 a = . num__33 a = . num__1.5 therefore : . num__1.5 / num__1 ounce = num__15.0 . therefore the answer should be a <eor> a <eos> |
a |
add__0.5__1.0__ multiply__3.0__5.0__ multiply__1.0__15.0__ |
add__0.5__1.0__ multiply__3.0__5.0__ divide__15.0__1.0__ |
| a shopkeeper sold an article offering a discount of num__4.0 and earned a profit of num__44.0 . what would have been the percentage of profit earned if no discount was offered ? <o> a ) num__24 <o> b ) num__28 <o> c ) num__30 <o> d ) num__32 <o> e ) num__50 |
let c . p . be rs . num__100 . then s . p . = rs . num__144 let marked price be rs . x . then num__0.96 x = num__144 x = num__150.0 = rs . num__150 now s . p . = rs . num__150 c . p . = rs . num__100 profit % = num__50.0 . answer : e <eor> e <eos> |
e |
percent__50.0__100.0__ |
percent__50.0__100.0__ |
| what is the least number of squares tiles required to pave the floor of a room num__15 m num__17 cm long and num__9 m num__2 cm broad <o> a ) num__814 <o> b ) num__256 <o> c ) num__3258 <o> d ) num__896 <o> e ) num__745 |
explanation : length of largest tile = h . c . f . of num__1517 cm and num__902 cm = num__41 cm . area of each tile = ( num__41 x num__41 ) cm ^ num__2 . required number of tiles = ( num__1517 x num__22.0 * num__41 ) = num__814 . answer : a <eor> a <eos> |
a |
divide__902.0__41.0__ round__814.0__ |
divide__902.0__41.0__ round__814.0__ |
| jerry ’ s average ( arithmetic mean ) score on the first num__3 of num__4 tests is num__81 . if jerry wants to raise his average by num__2 points what score must he earn on the fourth test ? <o> a ) num__87 <o> b ) num__89 <o> c ) num__90 <o> d ) num__93 <o> e ) num__95 |
total score on num__3 tests = num__81 * num__3 = num__243 jerry wants the average to be = num__83 hence total score on num__4 tests should be = num__83 * num__4 = num__332 score required on the fourth test = num__332 - num__243 = num__89 option b <eor> b <eos> |
b |
multiply__3.0__81.0__ add__81.0__2.0__ multiply__4.0__83.0__ subtract__332.0__243.0__ subtract__332.0__243.0__ |
multiply__3.0__81.0__ add__81.0__2.0__ multiply__4.0__83.0__ subtract__332.0__243.0__ subtract__332.0__243.0__ |
| what is remainder of the division ( num__1625 * num__1627 * num__1629 ) / num__12 ? <o> a ) num__3 <o> b ) num__2 <o> c ) num__1 <o> d ) num__0 <o> e ) num__4 |
remainder will be number / num__100 here as the divisor is two digit number = num__12 . hence checking for the last two digits = num__5 * num__7 * num__9 = num__15 thus remainder = num__3 . answer : a <eor> a <eos> |
a |
subtract__12.0__5.0__ subtract__12.0__9.0__ subtract__12.0__9.0__ |
subtract__12.0__5.0__ divide__15.0__5.0__ divide__9.0__3.0__ |
| a certain quantity of num__40.0 concentration solution is replaced with num__25.0 concentration solution such that the concentration of the combined amount is num__35.0 . what ' s the ratio of the amount of solution that was replaced to the amount of solution that was not replaced ? <o> a ) num__1 : num__3 <o> b ) num__1 : num__2 <o> c ) num__2 : num__3 <o> d ) num__2 : num__1 <o> e ) num__3 : num__1 |
from the wording in the prompt we ' re essentially mixing two liquids : liquid x is num__40.0 concentrate liquid y is num__25.0 concentrate we ' re supposed to mix some of each and end up with a solution that is num__35.0 concentrate . the prompt does n ' t give us specific amounts of liquids to work with and the question itself asks for the ratio of the two liquids so we could even test values if doing it that way made it easier for you . for now though here ' s the algebra : x = # of ounces of liquid x y = # of ounces of liquid y ( . num__4 x + . num__25 y ) / ( x + y ) = . num__35 . num__4 x + . num__25 y = . num__35 x + . num__35 y . num__05 x = . num__1 y num__5 x = num__10 y from here you have to be careful to answer the question that is asked . num__5 x = num__10 y y / x = num__0.5 = num__0.5 answer : b <eor> b <eos> |
b |
percent__25.0__4.0__ percent__40.0__25.0__ percent__5.0__10.0__ percent__25.0__4.0__ |
percent__25.0__4.0__ percent__40.0__25.0__ percent__5.0__10.0__ percent__25.0__4.0__ |
| tough and tricky questions : exponents . if num__6 ^ ( num__3 x + num__4 ) * num__2 ^ ( num__2 y + num__1 ) = num__36 ^ x * num__16 ^ y then x + y = <o> a ) - num__3.5 <o> b ) - num__0.142857142857 <o> c ) num__0.428571428571 <o> d ) num__3.5 <o> e ) - num__0.428571428571 |
here is my solution . num__6 ^ ( num__3 x + num__4 ) * num__2 ^ ( num__2 y + num__1 ) = num__36 ^ x * num__16 ^ y here rhs num__36 ^ x * num__16 ^ y = num__6 ^ ( num__2 x ) * num__2 ^ ( num__4 y ) equating powers on both sides - - > num__3 x + num__4 = num__2 x thus x = - num__4 and num__2 y + num__1 = num__4 y giving y = num__0.5 so x + y = - num__3.5 option : a <eor> a <eos> |
a |
reverse__2.0__ add__3.0__0.5__ add__3.0__0.5__ |
reverse__2.0__ add__3.0__0.5__ add__3.0__0.5__ |
| marcela karla isabel and viviana all got their math tests back . the average of marcela karla and isabel ' s grades was an num__85.0 . the average of karla isabel and viviana ' s grades was a num__78.0 . if viviana got a num__75.0 what did marcela get ? <o> a ) num__90.0 <o> b ) num__92.0 <o> c ) num__96.0 <o> d ) num__88.0 <o> e ) num__98 % |
solution : average grade for marcela karla and isabel = num__85.0 total grade points = num__3 * num__85 = num__255 average grade for karla isabel and viviana = num__78.0 total grade points = num__3 * num__78 = num__234 viviana got a num__75.0 . now ( marcela + karla + isabel ) - ( karla + isabel + viviana ) = num__255 - num__234 marcela - viviana = num__21 marcela ' s score = viviana ' s + num__21 = num__96.0 answer : option c <eor> c <eos> |
c |
subtract__78.0__75.0__ multiply__85.0__3.0__ multiply__78.0__3.0__ subtract__255.0__234.0__ add__75.0__21.0__ add__75.0__21.0__ |
subtract__78.0__75.0__ multiply__85.0__3.0__ multiply__78.0__3.0__ subtract__255.0__234.0__ add__75.0__21.0__ add__75.0__21.0__ |
| if num__6 parallel lines in a plane is intersected by a family of another num__8 parallel lines how many parallelograms are there in the network thus formed ? <o> a ) num__80 <o> b ) num__63 <o> c ) num__420 <o> d ) num__160 <o> e ) num__1260 |
parallelogram can formed by num__2 horizontal and num__2 vertical lines for horizontal num__6 c num__2 for vertical num__8 c num__2 total parallelogram is num__6 c num__2 * num__8 c num__2 = num__15 * num__28 = num__420 answer : c <eor> c <eos> |
c |
subtract__8.0__6.0__ multiply__15.0__28.0__ round__420.0__ |
subtract__8.0__6.0__ multiply__15.0__28.0__ multiply__15.0__28.0__ |
| here are num__6 periods in each working day of a school . in how many ways can one organize num__5 subjects such that each subject is allowed at least one period ? <o> a ) num__1023 <o> b ) num__6000 <o> c ) num__1980 <o> d ) num__1800 <o> e ) num__1450 |
num__5 subjects can be arranged in num__6 periods in num__6 p num__5 ways . any of the num__5 subjects can be organized in the remaining period ( num__5 c num__1 ways ) . two subjects are alike in each of the arrangement . so we need to divide by num__2 ! to avoid overcounting . total number of arrangements = num__6 p num__5 × num__5 c num__0.5 ! = num__1800 option d <eor> d <eos> |
d |
subtract__6.0__5.0__ divide__1.0__2.0__ round__1800.0__ |
subtract__6.0__5.0__ divide__1.0__2.0__ round__1800.0__ |
| the avg . age of a group of num__28 students is num__20 years . if num__4 more students join the group the avg age increases by num__1 year . the avg age of the new student is ? <o> a ) num__22 years <o> b ) num__23 years <o> c ) num__24 years <o> d ) num__25 years <o> e ) num__28 years |
total age of num__28 students = num__28 * num__20 = num__560 if total age of num__4 students = x then ( num__560 + x ) / ( num__28 + num__4 ) = ( num__20 + num__1 ) x = num__112 so average age of new students = num__28.0 = num__28 years answer : e <eor> e <eos> |
e |
multiply__28.0__20.0__ multiply__28.0__4.0__ multiply__28.0__1.0__ |
multiply__28.0__20.0__ multiply__28.0__4.0__ multiply__28.0__1.0__ |
| there are num__15 teams in the hockey league and each team faces all the other teams num__10 times each . how many games are played in the season ? <o> a ) num__650 <o> b ) num__750 <o> c ) num__850 <o> d ) num__950 <o> e ) num__1050 |
the number of ways to choose two teams is num__15 c num__2 = num__15 * num__7.0 = num__105 the total number of games in the season is num__10 * num__105 = num__1050 . the answer is e . <eor> e <eos> |
e |
multiply__15.0__7.0__ multiply__10.0__105.0__ multiply__10.0__105.0__ |
multiply__15.0__7.0__ multiply__10.0__105.0__ multiply__10.0__105.0__ |
| the ratio of the amount of the oil bill for the month of february to the amount of the oil bill for the month of january was num__3 : num__2 . if the oil bill for february had been $ num__10 more the corresponding ratio would have been num__5 : num__3 . how much was the oil bill for january ? <o> a ) $ num__60 <o> b ) $ num__80 <o> c ) $ num__100 <o> d ) $ num__120 <o> e ) $ num__140 |
num__3 : num__2 = num__9 : num__6 and num__5 : num__3 = num__10 : num__6 . an increase in $ num__10 increases the ratio by num__1 : num__6 . therefore january ' s bill was num__6 ( $ num__10 ) = $ num__60 . the answer is a . <eor> a <eos> |
a |
multiply__3.0__2.0__ subtract__3.0__2.0__ multiply__10.0__6.0__ multiply__10.0__6.0__ |
multiply__3.0__2.0__ subtract__3.0__2.0__ multiply__10.0__6.0__ multiply__10.0__6.0__ |
| the speed at which a man can row a boat in still water is num__16 kmph . if he rows downstream where the speed of current is num__4 kmph what time will he take to cover num__100 metres ? <o> a ) num__23 <o> b ) num__27 <o> c ) num__28 <o> d ) num__12 <o> e ) num__18 |
speed of the boat downstream = num__16 + num__4 = num__20 kmph = num__20 * num__0.277777777778 = num__5.55 m / s hence time taken to cover num__100 m = num__100 / num__5.55 = num__18 seconds . answer : e <eor> e <eos> |
e |
add__16.0__4.0__ round__18.0__ |
add__16.0__4.0__ round__18.0__ |
| the difference between c . i . and s . i . on an amount of rs . num__15000 for num__2 years is rs . num__96 . what is the rate of interest per annum ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
[ num__15000 * ( num__1 + r / num__100 ) num__2 - num__15000 ] - ( num__15000 * r * num__2 ) / num__100 = num__96 num__15000 [ ( num__1 + r / num__100 ) num__2 - num__1 - num__2 r / num__100 ] = num__96 num__15000 [ ( num__100 + r ) num__2 - num__10000 - num__200 r ] / num__10000 = num__96 r num__2 = ( num__96 * num__2 ) / num__3 = num__64 = > r = num__8 rate = num__8.0 answer : a <eor> a <eos> |
a |
percent__2.0__10000.0__ percent__100.0__8.0__ |
percent__2.0__10000.0__ percent__100.0__8.0__ |
| in how many ways can five different rings be worn on four fingers of one hand ? <o> a ) num__10 <o> b ) num__5 <o> c ) num__15 <o> d ) num__16 <o> e ) num__18 |
required number of ways = ways of selecting num__4 objects out of num__5 given objects = num__5 c num__4 = num__5 * num__4 * num__3 * num__0.0833333333333 = num__5 answer b <eor> b <eos> |
b |
vowel_space__ vowel_space__ |
vowel_space__ vowel_space__ |
| a company has num__3 directors and num__3 managers . how many different committees with num__5 people can be chosen having at least num__1 director ? <o> a ) num__500 <o> b ) num__720 <o> c ) num__4500 <o> d ) num__6 <o> e ) num__55 |
atleast one means exactly one and more than one . different committees with num__5 people can be chosen having at least num__1 director is ( num__3 c num__2 * num__3 c num__3 ) + ( num__3 c num__3 * num__3 c num__2 ) = num__6 ; hence d . <eor> d <eos> |
d |
coin_space__ die_space__ die_space__ |
coin_space__ die_space__ die_space__ |
| a box contains num__5 pairs of shoes ( num__10 shoes in total ) . if two shoes are selected at random what it is the probability that they are matching shoes ? <o> a ) num__0.00526315789474 <o> b ) num__0.05 <o> c ) num__0.111111111111 <o> d ) num__0.1 <o> e ) num__0.111111111111 |
the problem with your solution is that we do n ' t choose num__1 shoe from num__10 but rather choose the needed one after we just took one and need the second to be the pair of it . so the probability would simply be : num__1.0 * num__0.111111111111 ( as after taking one at random there are num__9 shoes left and only one is the pair of the first one ) = num__0.111111111111 answer : c . <eor> c <eos> |
c |
subtract__10.0__1.0__ reverse__9.0__ |
subtract__10.0__1.0__ reverse__9.0__ |
| a and b take part in num__100 m race . a runs at num__5 kmph . a gives b a start of num__8 m and still beats him by num__8 seconds . the speed of b is : <o> a ) num__1.33 kmph <o> b ) num__4.14 kmph <o> c ) num__2.33 kmph <o> d ) num__2.88 kmph <o> e ) num__4.44 kmph |
explanation : a ' s speed = \ fn _ jvn { \ color { black } \ left ( num__5 \ times \ frac { num__5 } { num__18 } \ right ) m / sec } = \ fn _ jvn { \ color { black } \ frac { num__25 } { num__18 } m / sec } time taken by a to cover num__100 m = \ fn _ jvn { \ color { black } ( num__100 \ times \ frac { num__18 } { num__15 } ) sec } = num__72 sec \ fn _ jvn { \ color { black } \ therefore } time taken by b to cover num__92 m = ( num__72 + num__8 ) = num__80 sec . \ fn _ jvn { \ color { black } \ therefore } b ' s speed = \ fn _ jvn { \ color { black } \ left ( \ frac { num__92 } { num__80 } \ times \ frac { num__18 } { num__5 } \ right ) kmph } = num__4.14 kmph answer : b <eor> b <eos> |
b |
subtract__100.0__8.0__ add__8.0__72.0__ round__4.14__ |
subtract__100.0__8.0__ add__8.0__72.0__ round__4.14__ |
| find the number of square tiles to cover the floor of a room measuring num__4 m * num__9 m leaving num__0.25 m space around the room . a side of square tile is given to be num__25 cms ? <o> a ) num__197 <o> b ) num__476 <o> c ) num__169 <o> d ) num__260 <o> e ) num__256 |
num__3 num__0.5 * num__8 num__0.5 = num__0.25 * num__0.25 * x = > x = num__476 answer : b <eor> b <eos> |
b |
divide__4.0__0.5__ round__476.0__ |
divide__4.0__0.5__ round__476.0__ |
| the diagonals of a rhombus are num__12 cm and num__15 cm . find its area ? <o> a ) num__30 <o> b ) num__45 <o> c ) num__90 <o> d ) num__120 <o> e ) num__180 |
area = num__0.5 x num__12 x num__15 = num__90 cm  ² answer : c <eor> c <eos> |
c |
triangle_area__12.0__15.0__ triangle_area__12.0__15.0__ |
triangle_area__12.0__15.0__ triangle_area__12.0__15.0__ |
| which of the following inequalities is equivalent to num__10 + num__2 x > num__100 + num__5 x ? <o> a ) x < - num__30 <o> b ) x > - num__30 <o> c ) x > - num__20 <o> d ) x < - num__10 <o> e ) x > - num__11 |
you can work with the inequality in the same way as you do when you have the equal to sign . you can take terms to the other side by flipping their sign you can multiply / divide the inequality by the same term on both sides of the equation etc . the only important thing to note is the following : when you multiply / divide by a negative number the sign of the inequality flips . to illustrate num__10 + num__2 x > num__100 + num__5 x - num__90 > num__5 x - num__2 x ( correct ) - num__90 > num__3 x ( correct ) - num__30 > num__3 x ( correct ) the correct option is a <eor> a <eos> |
a |
subtract__100.0__10.0__ subtract__5.0__2.0__ multiply__10.0__3.0__ multiply__10.0__3.0__ |
subtract__100.0__10.0__ subtract__5.0__2.0__ divide__90.0__3.0__ divide__90.0__3.0__ |
| in what ratio p : q should the mixture p of milk and water in the ratio of num__6 : num__1 be mixed with another mixture q of milk and water in the ratio num__3 : num__4 so that the resultant mixture contains equal quantities of milk and water ? <o> a ) num__1 : num__3 <o> b ) num__1 : num__4 <o> c ) num__1 : num__5 <o> d ) num__2 : num__5 <o> e ) num__3 : num__4 |
( num__0.857142857143 ) * p + ( num__0.428571428571 ) * q = ( num__0.142857142857 ) * p + ( num__0.571428571429 ) * q num__5 p = q p / q = num__0.2 the answer is c . <eor> c <eos> |
c |
subtract__1.0__0.8571__ subtract__1.0__0.4286__ subtract__6.0__1.0__ reverse__5.0__ reverse__1.0__ |
divide__0.4286__3.0__ subtract__1.0__0.4286__ add__1.0__4.0__ reverse__5.0__ reverse__1.0__ |
| if k = num__2 n - num__1 where n is an integer what is the remainder of k ^ num__0.25 ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) can not be determined from the information given . |
let us assume that k = num__4 ( n ) ( n - num__1 ) this is divisible by num__8 . so k + num__1 when divided by num__8 will give reminder num__1 . for example consider n = num__2 we have num__4 * num__2 * num__1 + num__1 = num__9 when we divide this by num__8 we get reminder num__1 . answer : a <eor> a <eos> |
a |
reverse__0.25__ divide__2.0__0.25__ add__1.0__8.0__ reverse__1.0__ |
reverse__0.25__ multiply__2.0__4.0__ add__1.0__8.0__ subtract__2.0__1.0__ |
| guna has num__8 flavors of ice cream in him parlor . how many options are there for thilak to pick a one - flavor two - flavor three - flavor four - flavor five - flavor six - flavor seven - flavor or eight - flavor order ? <o> a ) num__223 <o> b ) num__395 <o> c ) num__448 <o> d ) num__774 <o> e ) num__835 |
num__8 c num__1 + num__8 c num__2 + num__8 c num__3 + num__8 c num__4 + num__8 c num__5 + num__8 c num__6 + num__8 c num__7 + num__8 c num__8 = num__395 . answer : b <eor> b <eos> |
b |
add__1.0__2.0__ divide__8.0__2.0__ subtract__8.0__3.0__ subtract__8.0__2.0__ subtract__8.0__1.0__ multiply__1.0__395.0__ |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ add__1.0__5.0__ add__1.0__6.0__ multiply__1.0__395.0__ |
| of the employees in a company num__40.0 are female who have a master degree . if num__40.0 of the female employees do not have a master degree what percent of the employees in the company are female ? <o> a ) num__66.67 <o> b ) num__67.67 <o> c ) num__68.67 <o> d ) num__65.67 <o> e ) num__64.67 % |
let e be total no . of employees and f be total no of female employees . question asked what is f / e x num__100.0 ? no . of female employees with masters = num__0.4 x e ( from the question ) no . of female employees without masters = num__0.4 x f ( from the question ) therefore no . of female employees with masters = f - num__0.4 f = num__0.6 f the num__2 expressions equal each other therefore num__0.6 f = num__0.4 e ; f / e = num__0.4 / num__0.6 = num__66.67 ans : a <eor> a <eos> |
a |
percent__100.0__66.67__ |
percent__100.0__66.67__ |
| a house wife saved $ num__3.50 in buying an item on sale . if she spent $ num__32.50 for the item approximately how much percent she saved in the transaction ? <o> a ) num__8.0 <o> b ) num__9.0 <o> c ) num__10.0 <o> d ) num__11.0 <o> e ) num__12 % |
actual price = num__32.50 + num__3.50 = $ num__36 saving = num__3.50 / num__36 * num__100 = num__10.0 approximately answer is c <eor> c <eos> |
c |
add__3.5__32.5__ divide__100.0__10.0__ |
add__3.5__32.5__ divide__100.0__10.0__ |
| the average age of num__15 students of a class is num__15 years . out of these the average age of num__8 students is num__14 years and that of the other num__6 students is num__16 years . the age of the num__15 th student is <o> a ) num__9 years <o> b ) num__11 years <o> c ) num__17 years <o> d ) num__21 years <o> e ) num__25 years |
solution age of the num__15 th student = [ num__15 x num__15 - ( num__14 x num__8 + num__16 x num__6 ) ] = ( num__225 - num__208 ) = num__17 years . answer c <eor> c <eos> |
c |
subtract__225.0__208.0__ subtract__225.0__208.0__ |
subtract__225.0__208.0__ subtract__225.0__208.0__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__338 <o> b ) num__240 <o> c ) num__287 <o> d ) num__267 <o> e ) num__191 |
speed = ( num__54 * num__0.277777777778 ) m / sec = num__15 m / sec . length of the train = ( num__15 x num__20 ) m = num__300 m . let the length of the platform be x meters . then ( x + num__300 ) / num__36 = num__15 = = > x + num__300 = num__540 = = > x = num__240 m . answer : b <eor> b <eos> |
b |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
| when num__200 is divided by positive integer x the remainder is num__3 . what is the remainder when num__297 is divided by x ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
f num__200 / x leaves a reminder num__3 then ( num__200 - num__3 ) i . e . num__197 is divisible by x so ( num__200 + num__197 ) / x leaves a reminder rem ( num__200 / x ) + rem ( num__197 / x ) = > num__3 + num__0 = num__3 answer : b <eor> b <eos> |
b |
subtract__200.0__3.0__ subtract__200.0__197.0__ |
subtract__200.0__3.0__ subtract__200.0__197.0__ |
| if x + y = num__2 and x ^ num__2 - xy - num__10 - num__2 y ^ num__2 = num__0 what does x − num__2 y equal ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__5 <o> e ) num__10 |
if x + y = num__2 and x ^ num__2 - xy - num__10 - num__2 y ^ num__2 = num__0 what does x − num__2 y equal ? a . num__0 b . num__1 c . num__2 d . num__5 e . num__10 since x + y = num__2 we have y = num__2 - x . so num__0 = x ^ num__2 - xy - num__10 - num__2 y ^ num__2 = x ^ num__2 - x ( num__2 - x ) - num__10 - num__2 ( num__2 - x ) ^ num__2 = x ^ num__2 - num__2 x + x ^ num__2 - num__10 - num__8 + num__8 x - num__2 x ^ num__2 = x = num__6 x - num__18 . - - - > x = num__3 - - - > y = - num__1 - - - > x - num__2 y = num__5 the answer is ( d ) . <eor> d <eos> |
d |
divide__10.0__2.0__ subtract__10.0__2.0__ add__1.0__5.0__ add__10.0__8.0__ add__2.0__1.0__ add__2.0__3.0__ |
divide__10.0__2.0__ subtract__10.0__2.0__ add__1.0__5.0__ add__10.0__8.0__ add__2.0__1.0__ add__2.0__3.0__ |
| in a num__1000 m race a beats b by num__10 m and b beats c by num__100 m . in the same race by how many meters does a beat c ? <o> a ) a ) num__109 m <o> b ) b ) num__829 m <o> c ) c ) num__822 m <o> d ) d ) num__929 m <o> e ) e ) num__132 mj |
by the time a covers num__1000 m b covers ( num__1000 - num__10 ) = num__990 m . by the time b covers num__1000 m c covers ( num__1000 - num__100 ) = num__900 m . so the ratio of speeds of a and c = num__1.0101010101 * num__1.11111111111 = num__1.12233445567 so by the time a covers num__1000 m c covers num__891 m . so in num__1000 m race a beats c by num__1000 - num__891 = num__109 m . answer : a <eor> a <eos> |
a |
subtract__1000.0__10.0__ subtract__1000.0__100.0__ divide__1000.0__990.0__ divide__1000.0__900.0__ multiply__1.1111__1.0101__ subtract__1000.0__891.0__ round__109.0__ |
subtract__1000.0__10.0__ subtract__1000.0__100.0__ divide__1000.0__990.0__ divide__1000.0__900.0__ multiply__1.1111__1.0101__ subtract__1000.0__891.0__ subtract__1000.0__891.0__ |
| a person crosses a num__600 m long street in num__3 minnutes . what is his speed in km per hour ? <o> a ) num__5.8 km / hr <o> b ) num__7.2 km / hr <o> c ) num__9 km / hr <o> d ) num__12 km / hr <o> e ) num__3 km / hr |
speed = num__200.0 * num__60 = num__3.33 m / sec = num__3.33 * num__3.6 = num__12 km / hr answer is d <eor> d <eos> |
d |
divide__600.0__3.0__ hour_to_min_conversion__ round__12.0__ |
divide__600.0__3.0__ hour_to_min_conversion__ round__12.0__ |
| the average of num__9 observations was num__8 that of the num__1 st of num__5 being num__10 and that of the last num__5 being num__8 . what was the num__5 th observation ? <o> a ) num__18 <o> b ) num__12 <o> c ) num__15 <o> d ) num__17 <o> e ) num__19 |
explanation : num__1 to num__9 = num__8 * num__9 = num__72 num__1 to num__5 = num__5 * num__10 = num__50 num__5 to num__9 = num__5 * num__8 = num__40 num__5 th = num__50 + num__40 = num__90 – num__72 = num__18 option a <eor> a <eos> |
a |
multiply__9.0__8.0__ multiply__5.0__10.0__ multiply__8.0__5.0__ multiply__9.0__10.0__ add__8.0__10.0__ add__8.0__10.0__ |
multiply__9.0__8.0__ multiply__5.0__10.0__ multiply__8.0__5.0__ multiply__9.0__10.0__ add__8.0__10.0__ add__8.0__10.0__ |
| how many seconds will a num__1200 m long train take to cross a man walking with a speed of num__3 km / hr in the direction of the moving train if the speed of the train is num__63 km / hr ? <o> a ) num__26 sec <o> b ) num__65 sec <o> c ) num__55 sec <o> d ) num__19 sec <o> e ) num__72 sec |
speed of train relative to man = num__63 - num__3 = num__60 km / hr . = num__60 * num__0.277777777778 = num__16.6666666667 m / sec . time taken to pass the man = num__1200 * num__0.06 = num__72 sec . answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ multiply__1200.0__0.06__ round__72.0__ |
subtract__63.0__3.0__ multiply__1200.0__0.06__ multiply__1200.0__0.06__ |
| two buses start at the same time one from p to q and the other from q to p . if both buses reach after num__4 hours and num__16 hours at q and p respectively after they cross each other what would be the ratio of speeds of the bus starting from p and that of the one starting from point q ? <o> a ) num__2 : num__1 <o> b ) num__1 : num__2 <o> c ) num__2 : num__2 <o> d ) num__1 : num__4 <o> e ) num__5 : num__2 |
hint : sp / sq = √ tq / √ tp sp and sq are speeds of two the buses at points p and q respectively . tp = num__18 hrs and tq = num__4 hrs sp / sq = √ num__16 / √ num__4 therefore ratio of speeds sp / sq = num__2.0 = num__2.0 one bus travels at a speed twice of the other bus . answer is a <eor> a <eos> |
a |
subtract__18.0__16.0__ round__2.0__ |
subtract__18.0__16.0__ divide__4.0__2.0__ |
| the average age of a husband and his wife was num__20 years at the time of their marriage . after five years they have a one - year old child . the average age of the family now is : <o> a ) num__11 <o> b ) num__18 <o> c ) num__19 <o> d ) num__287 <o> e ) num__17 |
explanation : sum of the present ages of husband wife and child = ( num__20 * num__2 + num__5 * num__2 ) + num__1 = num__51 years . required average = ( num__17.0 ) = num__17 years . answer : e ) num__17 <eor> e <eos> |
e |
multiply__1.0__17.0__ |
multiply__1.0__17.0__ |
| we run a business that rents out canoes and kayaks . a canoe rental costs $ num__11 per day and a kayak rental costs $ num__16 dollars per day . one day our business rents out num__4 canoes for every num__3 kayaks and receives a total of $ num__460 in revenue . how many more canoes than kayaks were rented out ? <o> a ) num__8 <o> b ) num__7 <o> c ) num__6 <o> d ) num__5 <o> e ) num__4 |
let x be the number of canoes . then num__3 x / num__4 is the number of kayaks . num__11 x + ( num__3 x / num__4 ) * num__16 = num__460 num__11 x + num__12 x = num__460 num__23 x = num__460 x = num__20 ( canoes ) num__3 x / num__4 = num__15 ( kayaks ) there were num__20 - num__15 = num__5 more canoes rented out . the answer is d . <eor> d <eos> |
d |
subtract__16.0__4.0__ add__11.0__12.0__ add__16.0__4.0__ add__11.0__4.0__ subtract__16.0__11.0__ subtract__16.0__11.0__ |
subtract__16.0__4.0__ add__11.0__12.0__ add__16.0__4.0__ add__11.0__4.0__ subtract__16.0__11.0__ subtract__16.0__11.0__ |
| a num__14.0 stock yielding num__8.0 is quoted at ? <o> a ) num__165 <o> b ) num__170 <o> c ) num__175 <o> d ) num__180 <o> e ) num__185 |
assume that face value = rs . num__100 as it is not given to earn rs . num__8 money invested = rs . num__100 to earn rs . num__14 money invested = num__100 × num__1.75 = rs . num__175 ie market value of the stock = rs . num__175 answer is c . <eor> c <eos> |
c |
percent__100.0__175.0__ |
percent__100.0__175.0__ |
| in how many different ways can the letters of the word ` ` aim ' ' be rearrangement ? <o> a ) num__4 <o> b ) num__2 <o> c ) num__7 <o> d ) num__9 <o> e ) num__6 |
option ' e ' the total number of arrangements is num__3 p num__3 = num__3 ! = num__6 <eor> e <eos> |
e |
die_space__ die_space__ |
die_space__ die_space__ |
| the value of x is one quarter of z . the sum of x y and z is equal to num__26 . if the value of y is twice the value of z what is the largest factor of the sum of y and z ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__8 <o> d ) num__12 <o> e ) num__24 is answer |
while most test takers would treat this as a ' system ' algebra question and solve it that way there are some interesting number property patterns in this prompt that will actually allow you to use the answer choices ( and some pattern - matching ) to get to the solution . to start i ' m going to deal with the last piece of information first : y is twice z the prompt asks for the biggest factor of the sum of y and z . since y = num__2 z the sum of y and z is ( num__2 z ) + z = num__3 z . this means that the biggest factor is a multiple of num__3 ( and likely one of the larger multiples of num__3 in the answer choices ) . the second piece of information tells us . . . x + y + z = num__26 by extension . . . x + ( y + z ) = num__26 x + ( larger multiple of num__3 ) = num__26 from the answer choices there are really only num__2 options that make any sense : answers d and e . . . . with answer d : if . . . . y + z = num__12 then x = num__14 but . . . . the first piece of information tells us that x = num__0.25 of z so x has to be relatively small . . . . in this example it ' s larger than z so d is not the answer . at this point the answer would have to be e but here ' s the proof . . . if . . . y + z = num__24 and y = num__2 z . . . . y = num__16 z = num__8 x is num__0.25 of z so x = ( num__0.25 ) ( num__8 ) = num__2 x + y + z = num__26 num__2 + num__16 + num__8 = num__26 all of the numbers ' fit ' what we were told so this must be the answer . e <eor> e <eos> |
e |
subtract__26.0__12.0__ divide__3.0__12.0__ subtract__26.0__2.0__ add__2.0__14.0__ divide__2.0__0.25__ subtract__26.0__2.0__ |
subtract__26.0__12.0__ divide__3.0__12.0__ subtract__26.0__2.0__ add__2.0__14.0__ subtract__24.0__16.0__ subtract__26.0__2.0__ |
| a merchant mixes three varieties of rice costing rs . num__20 / kg rs . num__24 / kg and rs . num__30 / kg and sells the mixture at a profit of num__20.0 at rs . num__30 / kg . how many kgs of the second variety will be in the mixture if num__2 kgs of the third variety is there in the mixture ? <o> a ) num__1 kgs <o> b ) num__3 kgs <o> c ) num__5 kgs <o> d ) num__6 kgs <o> e ) none of these |
explanation : if the selling price of mixture is rs . num__30 / kg and the merchant makes a profit of num__20.0 then the cost price of the mixture = num__30 / num__1.2 = rs . num__25 / kg . we need to find out the ratio in which the three varieties are mixed to obtain a mixture costing rs . num__25 / kg . let variety a cost rs . num__20 / kg variety b cost rs . num__24 / kg and variety c cost rs . num__30 / kg . the mean desired price falls between b and c . step num__1 : find out the ratio qa : qc using alligation rule . qa / qc = ( num__30 − num__25 ) / ( num__25 − num__20 ) = num__1.0 . step num__2 : find out the ratio qb : qc using alligation rule . qb / qc = num__30 − num__25 / ( num__25 − num__24 ) = num__5.0 . step num__3 : qc is found by adding the value of qc in step num__1 and step num__2 = num__1 + num__1 = num__2 . therefore the required ratio = num__1 : num__5 : num__2 . if there are num__2 kgs of the third variety in the mixture then there will be num__5 kgs of the second variety in the mixture . answer : c <eor> c <eos> |
c |
percent__20.0__25.0__ percent__20.0__25.0__ |
percent__20.0__25.0__ percent__20.0__25.0__ |
| find the area of a parallelogram with base num__24 cm and height num__12 cm ? <o> a ) num__198 cm num__2 <o> b ) num__288 cm num__2 <o> c ) num__279 cm num__2 <o> d ) num__128 cm num__2 <o> e ) num__297 cm num__2 |
area of a parallelogram = base * height = num__24 * num__12 = num__288 cm num__2 answer : b <eor> b <eos> |
b |
multiply__24.0__12.0__ multiply__24.0__12.0__ |
multiply__24.0__12.0__ multiply__24.0__12.0__ |
| arjun started a business with rs . num__20000 and is joined afterwards by anoop with rs . num__40 num__000 . after how many months did anoop join if the profits at the end of the year are divided equally ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
suppose anoop joined after num__3 months . then num__20000 * num__12 = num__40000 * ( num__12 – x ) = > = > x = num__6 . answer : d <eor> d <eos> |
d |
subtract__12.0__6.0__ |
subtract__12.0__6.0__ |
| the difference between the simple interest received from two different sources on rs . num__1400 for num__3 years is rs . num__13.50 . the difference between their rates of interest is ? <o> a ) a ) num__2.4 <o> b ) b ) num__2.6 <o> c ) c ) num__0.2 <o> d ) d ) num__2.0 <o> e ) e ) num__1.3 % |
( num__1400 * r num__1 * num__3 ) / num__100 - ( num__1400 * r num__2 * num__3 ) / num__100 = num__13.50 num__4200 ( r num__1 - r num__2 ) = num__1350 r num__1 - r num__2 = num__0.2 answer : c <eor> c <eos> |
c |
percent__100.0__0.2__ |
percent__100.0__0.2__ |
| a num__1200 m long train crosses a tree in num__120 sec how much time will i take to pass a platform num__600 m long ? <o> a ) num__176 sec <o> b ) num__180 sec <o> c ) num__178 sec <o> d ) num__267 sec <o> e ) num__276 sec |
l = s * t s = num__10.0 s = num__10 m / sec . total length ( d ) = num__1800 m t = d / s t = num__180.0 t = num__180 sec answer : b <eor> b <eos> |
b |
divide__1200.0__120.0__ add__1200.0__600.0__ divide__1800.0__10.0__ round__180.0__ |
divide__1200.0__120.0__ add__1200.0__600.0__ divide__1800.0__10.0__ divide__1800.0__10.0__ |
| the average of the marks of num__10 students in a class is num__80 . if the marks of each student are doubled find the new average ? <o> a ) num__80 <o> b ) num__120 <o> c ) num__160 <o> d ) num__270 <o> e ) num__110 |
sum of the marks for the num__10 students = num__10 * num__80 = num__800 . the marks of each student are doubled the sum also will be doubled . the new sum = num__800 * num__2 = num__1600 . so the new average = num__160.0 = num__160 . answer : c <eor> c <eos> |
c |
multiply__10.0__80.0__ multiply__800.0__2.0__ multiply__80.0__2.0__ multiply__80.0__2.0__ |
multiply__10.0__80.0__ multiply__800.0__2.0__ multiply__80.0__2.0__ multiply__80.0__2.0__ |
| water tax is increased by num__20.0 but its consumption is decreased by num__20.0 . then the increase or decrease in the expenditure of the money is : <o> a ) no change <o> b ) num__5.0 decrease <o> c ) num__4.0 increase <o> d ) num__4.0 decrease <o> e ) none |
explanation : let tax = rs . num__100 and consumption = num__100 units original expenditure = rs . ( num__100 × num__100 ) = rs . num__10000 ( num__100 × num__100 ) = rs . num__10000 new expenditure = rs . ( num__120 × num__80 ) = rs . num__9600 ( num__120 × num__80 ) = rs . num__9600 decrease in expenditure = ( num__0.04 × num__100 ) % = num__4.0 correct option : d <eor> d <eos> |
d |
add__20.0__100.0__ subtract__100.0__20.0__ multiply__80.0__120.0__ divide__80.0__20.0__ divide__80.0__20.0__ |
add__20.0__100.0__ subtract__100.0__20.0__ multiply__80.0__120.0__ divide__80.0__20.0__ divide__80.0__20.0__ |
| a shopkeeper buys mangoes at the rate of num__5 a rupee and sells them at num__3 a rupee . find his net profit or loss percent ? <o> a ) num__66 num__0.666666666667 % <o> b ) num__33 num__2.33333333333 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__33 num__0.625 % <o> e ) num__34 num__0.333333333333 % |
the total number of mangoes bought by the shopkeeper be num__15 . if he buys num__5 a rupee his cp = num__3 he selling at num__3 a rupee his sp = num__5 profit = sp - cp = num__5 - num__3 = num__2 profit percent = num__0.666666666667 * num__100 = num__66 num__0.666666666667 % answer : a <eor> a <eos> |
a |
percent__100.0__66.0__ |
percent__100.0__66.0__ |
| for real number x int ( x ) denotes integer part of x . int ( x ) is the largest integer less than or equal to x . int ( num__12 ) = num__1 int ( - num__24 ) = - num__3 . find the value of int ( num__0.5 ) + int ( num__0.5 + num__100 ) + int ( num__0.5 + num__0.02 ) + . . . . + int ( num__0.5 + num__0.99 ) <o> a ) num__150 <o> b ) num__98 <o> c ) num__278 <o> d ) num__277 <o> e ) num__1811 |
int ( num__0.5 ) = num__0 int ( num__0.5 + num__100 ) = num__100 into ( num__0.5 + num__0.02 ) = num__0 . . . . . . int ( num__0.5 + num__0.5 ) = num__1 int ( num__0.5 + num__0.51 ) = num__1 . . . . . . . int ( num__0.5 + num__0.99 ) = num__1 so num__100 + num__1 + num__1 + . . . . . num__50 times = num__150 answer : a <eor> a <eos> |
a |
round_down__0.5__ reverse__0.02__ divide__3.0__0.02__ multiply__1.0__150.0__ |
round_down__0.5__ reverse__0.02__ add__100.0__50.0__ add__100.0__50.0__ |
| when fair coin is tossed num__9 times find the probability of getting head at least once . <o> a ) num__0.001953125 <o> b ) num__0.998046875 <o> c ) num__0.017578125 <o> d ) num__9.86274509804 <o> e ) num__0.986274509804 |
num__1 - num__0.5 ^ num__9 = num__1 - num__0.001953125 = num__0.998046875 answer : b <eor> b <eos> |
b |
negate_prob__0.002__ negate_prob__0.002__ |
negate_prob__0.002__ negate_prob__0.002__ |
| num__0.35 represents what percent of num__0.0007 ? <o> a ) num__0.05 <o> b ) num__0.5 <o> c ) num__5.0 <o> d ) num__500.0 <o> e ) num__50000 % |
one more method num__0.35 represents what percent of num__0.0007 ? adjusting the decimal num__3500 represents what percent of num__7 ? divide by num__7 num__500 represents what percent of num__1 ? answer = num__500 * num__100 = num__50000.0 = e <eor> e <eos> |
e |
percent__100.0__50000.0__ |
percent__100.0__50000.0__ |
| in what time will a train num__100 meters long cross an electric pole if its speed is num__360 km / hr <o> a ) num__1 second <o> b ) num__4.5 seconds <o> c ) num__3 seconds <o> d ) num__2.5 seconds <o> e ) none of these |
explanation : first convert speed into m / sec speed = num__360 * ( num__0.277777777778 ) = num__100 m / sec time = distance / speed = num__1.0 = num__1 second answer : a <eor> a <eos> |
a |
divide__100.0__360.0__ round__1.0__ |
divide__100.0__360.0__ round__1.0__ |
| two employees a and b are paid a total of rs . num__550 per week by their employer . if a is paid num__120 percent of the sum paid to b how much is b paid per week ? <o> a ) rs . num__220 <o> b ) rs . num__250 <o> c ) rs . num__260 <o> d ) rs . num__270 <o> e ) rs . num__280 |
let the amount paid to a per week = x and the amount paid to b per week = y then x + y = num__550 but x = num__120.0 of y = num__120 y / num__100 = num__12 y / num__10 ∴ num__12 y / num__10 + y = num__550 ⇒ y [ num__1.2 + num__1 ] = num__550 ⇒ num__22 y / num__10 = num__550 ⇒ num__22 y = num__5500 ⇒ y = num__250.0 = num__250.0 = rs . num__250 b <eor> b <eos> |
b |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__10.0__12.0__ multiply__550.0__10.0__ divide__5500.0__22.0__ multiply__1.0__250.0__ |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__10.0__12.0__ multiply__550.0__10.0__ divide__5500.0__22.0__ divide__250.0__1.0__ |
| machine a takes num__100 hours to complete a certain job and starts that job at num__7 am . after two hour of working alone machine a is joined by machine b and together they complete the job at num__12 pm . how long would it have taken machine b to complete the jobif it had worked alone for the entire job ? <o> a ) num__100.11 <o> b ) num__10.98 <o> c ) num__18.07 <o> d ) num__45.01 <o> e ) num__35.09 |
let us assume total job = num__100 units a finishes num__100 units in num__100 hrs ( given ) hence a ( working rate ) = num__1 units / hr now given that a works for num__2 hr ( so num__2 units done ) then a and b finish total work in num__15 hours . hence a and b finish num__98 units in num__15 hours . of these num__1 x num__15 = num__15 units were done by a . hence b did num__83 units in num__15 hours . hence b ( working rate ) = num__5.53333333333 units / hr hence b takes num__100 x num__0.180722891566 = num__18.07 hours to complete the job . answer c . <eor> c <eos> |
c |
subtract__100.0__2.0__ subtract__98.0__15.0__ divide__83.0__15.0__ divide__1.0__5.5333__ multiply__100.0__0.1807__ round__18.07__ |
subtract__100.0__2.0__ subtract__98.0__15.0__ divide__83.0__15.0__ divide__1.0__5.5333__ multiply__100.0__0.1807__ divide__18.07__1.0__ |
| if num__0 < l < y l is an odd number and y is a prime number which of the following can be the value of l + y ? <o> a ) num__11 . <o> b ) num__13 . <o> c ) num__17 . <o> d ) num__10 . <o> e ) num__7 . |
if l is odd implies l ( odd ) + y ( odd / even prime ) = ( even / odd ) if y is even the num__2 is the even prime therefore y = num__2 l = num__1 since num__0 < l < y therefore l + y = num__3 . but num__3 is not an option which implies y is odd now l ( odd ) + y ( odd ) = even num__10 is the only option that is even = d <eor> d <eos> |
d |
add__1.0__2.0__ multiply__1.0__10.0__ |
add__1.0__2.0__ divide__10.0__1.0__ |
| a sells a bicycle to b at a profit of num__60.0 and b sells it to c at a loss of num__60.0 . find the resultant profit or loss . <o> a ) - num__4.0 <o> b ) num__0.0 <o> c ) - num__5.0 <o> d ) num__6.0 <o> e ) - num__36 % |
the resultant profit or loss = num__60 - num__60 - ( num__60 * num__60 ) / num__100 = - num__36.0 loss = num__36.0 answer is e <eor> e <eos> |
e |
percent__100.0__36.0__ |
percent__100.0__36.0__ |
| an amount of rs . num__2000 is to be distributed amongst p q r and s such that “ p ” gets twice as that of “ q ” and “ s ” gets four times as that of “ r ” . if “ q ” and “ r ” are to receive equal amount what is the difference between the amounts received by s and p ? <o> a ) num__2378 <o> b ) num__268 <o> c ) num__269 <o> d ) num__500 <o> e ) num__6971 |
explanation : we have p = num__2 q & s = num__4 r further q = r & p + q + r + s = num__2000 thus we get num__2 q + q + q + num__4 q = num__2000 num__8 q = num__2000 or q = rs . num__250 thus r = rs . num__250 p = num__500 & s = rs . num__1000 hence the required difference = ( s – p ) = ( num__1000 – num__500 ) = rs . num__500 answer : d <eor> d <eos> |
d |
multiply__2.0__4.0__ divide__2000.0__8.0__ divide__2000.0__4.0__ divide__2000.0__2.0__ divide__2000.0__4.0__ |
multiply__2.0__4.0__ divide__2000.0__8.0__ divide__2000.0__4.0__ divide__2000.0__2.0__ divide__2000.0__4.0__ |
| pascal has num__96 miles remaining to complete his cycling trip . if he reduced his current speed by num__4 miles per hour the remainder of the trip would take him num__16 hours longer than it would if he increased his speed by num__50.0 . what is his current speed z ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__12 <o> e ) num__16 |
let the current speed be x miles per hour . time taken if speed is num__50.0 faster ( i . e . num__3 x / num__2 = num__1.5 x ) = num__96 / num__1.5 x time taken if speed is reduced by num__4 miles / hr ( i . e . ( x - num__4 ) ) = num__96 / ( x - num__4 ) as per question num__96 / ( x - num__4 ) - num__96 / num__1.5 x = num__16 solving this z we get x = num__8 . b . <eor> b <eos> |
b |
divide__3.0__2.0__ multiply__4.0__2.0__ round__8.0__ |
divide__3.0__2.0__ divide__16.0__2.0__ divide__16.0__2.0__ |
| a num__180 meter long train crosses a man standing on the platform in num__6 sec . what is the speed of the train ? <o> a ) num__48 kmph <o> b ) num__52 kmph <o> c ) num__108 kmph <o> d ) num__100 kmph <o> e ) num__98 kmph |
s = num__30.0 * num__3.6 = num__108 kmph answer : c <eor> c <eos> |
c |
divide__180.0__6.0__ multiply__3.6__30.0__ round__108.0__ |
divide__180.0__6.0__ multiply__3.6__30.0__ multiply__3.6__30.0__ |
| a searchlight on top of the watchtower makes num__2 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least num__5 seconds ? <o> a ) num__0.833333333333 <o> b ) num__0.416666666667 <o> c ) num__0.583333333333 <o> d ) num__0.916666666667 <o> e ) num__0.733333333333 |
the searchlight completes one revolution every num__30 seconds . the probability that the man ' s area will be lit up is num__0.166666666667 = num__0.166666666667 . the probability that he will stay in the dark is num__1 - num__0.166666666667 = num__0.833333333333 the answer is a . <eor> a <eos> |
a |
divide__5.0__30.0__ subtract__1.0__0.1667__ subtract__1.0__0.1667__ |
divide__5.0__30.0__ subtract__1.0__0.1667__ subtract__1.0__0.1667__ |
| if t and y are different prime numbers less than ten what must be true about the sum of t and y ? <o> a ) the sum is even <o> b ) the sum is odd <o> c ) the sum is less than ten <o> d ) the sum is greater than ten <o> e ) the sum is less than num__13 |
this has to be solved by checking for each option : prime numbers less that num__10 = num__2 num__3 num__5 num__7 a . the sum is even may or may not be true . t = num__3 y = num__5 - > true . t = num__2 y = num__3 - true b . the sum is odd may or may not be true . t = num__3 y = num__5 - > false . t = num__2 y = num__3 - true c . the sum is less than ten may or may not be true t = num__5 y = num__7 sum = num__12 t = num__2 y = num__3 . sum = num__5 d . the sum is greater than ten may or may not be true t = num__5 y = num__7 sum = num__12 t = num__2 y = num__3 . sum = num__5 e . the sum is less than num__13 this will always be true . maximum sum = num__7 + num__5 = num__12 . this is always less than num__13 . correct option : e <eor> e <eos> |
e |
add__2.0__3.0__ add__2.0__5.0__ add__2.0__10.0__ add__3.0__10.0__ add__3.0__10.0__ |
add__2.0__3.0__ add__2.0__5.0__ add__2.0__10.0__ add__3.0__10.0__ add__3.0__10.0__ |
| the height of the wall is num__4 times its width and length of the wall is num__3 times its height . if the volume of the wall be num__10368 cu . m . its width is <o> a ) num__4 m <o> b ) num__5 m <o> c ) num__6 m <o> d ) num__7 m <o> e ) num__8 m |
explanation : let width = x then height = num__4 x and length = num__12 x num__12 x à — num__4 x à — x = num__10368 x = num__6 answer : c <eor> c <eos> |
c |
square_perimeter__3.0__ triangle_area__4.0__3.0__ triangle_area__4.0__3.0__ |
square_perimeter__3.0__ triangle_area__4.0__3.0__ triangle_area__4.0__3.0__ |
| a military garrison in a wild frontier has num__500 men who have provisions to last num__26 weeks . at the end of num__6 weeks they are joined by another num__300 men . how long will the provisions last ? <o> a ) num__13 num__0.333333333333 <o> b ) num__12 num__0.333333333333 <o> c ) num__11 num__1.0 <o> d ) num__12 num__0.5 <o> e ) num__12 num__0.75 |
total provisions = num__500 * num__26 * num__7 = num__91000 provisions used for in num__6 weeks = num__6 * num__7 * num__500 = num__21000 remaining provisions = num__70000 remaining provisions need to be split between num__800 ( num__500 + num__300 ) people . number of provisions per person = num__87.5 we need to find the answer in terms of weeks = ( num__87.5 ) / num__7 = num__12.5 answer : d <eor> d <eos> |
d |
subtract__91000.0__21000.0__ add__500.0__300.0__ divide__70000.0__800.0__ divide__87.5__7.0__ round_down__12.5__ |
subtract__91000.0__21000.0__ add__500.0__300.0__ divide__70000.0__800.0__ divide__87.5__7.0__ round_down__12.5__ |
| a watch was sold at a loss of num__10.0 . if it was sold for rs . num__140 more there would have been a gain of num__4.0 . what is the cost price ? <o> a ) num__1000 <o> b ) num__3783 <o> c ) num__1837 <o> d ) num__3799 <o> e ) num__3786 |
num__90.0 num__104.0 - - - - - - - - num__14.0 - - - - num__140 num__100.0 - - - - ? = > rs . num__1000 answer : a <eor> a <eos> |
a |
percent__10.0__140.0__ percent__100.0__1000.0__ |
percent__10.0__140.0__ percent__100.0__1000.0__ |
| two trains one from p to q and the other from q to p start simultaneously . after they meet the trains reach their destinations after num__9 hours and num__25 hours respectively . the ratio of their speeds is <o> a ) num__4 : num__1 <o> b ) num__4 : num__2 <o> c ) num__3 : num__5 <o> d ) num__5 : num__3 <o> e ) num__4 : num__6 |
ratio of their speeds = speed of first train : speed of second train = √ num__25 − − √ num__9 = num__5 : num__3 answer is d . <eor> d <eos> |
d |
round__5.0__ |
round__5.0__ |
| a sum was put at simple interest at a certain rate for num__6 years had it been put at num__4.0 higher rate it would have fetched num__144 more . find the sum . <o> a ) num__500 <o> b ) num__600 <o> c ) num__700 <o> d ) num__800 <o> e ) num__900 |
difference in s . i . = p × t / num__100 ( r num__1 − r num__2 ) ⇒ num__144 = p × num__6 x num__0.04 ( ∵ r num__1 - r num__2 = num__2 ) ⇒ p = num__144 × num__16.6666666667 × num__4 = num__600 answer b <eor> b <eos> |
b |
percent__4.0__1.0__ percent__100.0__600.0__ |
percent__4.0__1.0__ percent__100.0__600.0__ |
| avg cost of num__5 pens and num__4 pencils isrs . num__36 . the avg cost of num__5 pens and num__9 pencils is rs . num__56 . find the total cost of num__25 pens and num__25 pencils ? <o> a ) num__80 <o> b ) num__90 <o> c ) num__100 <o> d ) num__110 <o> e ) num__120 |
avg cost of num__5 pens and num__4 pencils is rs . num__36 ( num__5 pen + num__4 pencils ) num__9 = num__36 therefore num__1 pen cost is num__4.0 = num__4 so num__5 pen cost is num__5 x num__4 = num__20 and num__4 pencils cost is num__4 x num__4 = num__16 again avg cost of num__5 pens and num__9 pencils is rs . num__56 ( num__5 pen + num__9 pencils ) num__14 = num__56 therefore num__1 pen cost is num__4.0 = num__4 so num__5 pen cost is num__5 x num__4 = num__20 and num__9 pencils cost is num__9 x num__4 = num__36 so num__25 pen cost is ( num__25 / a ) = num__100 and num__25 pencils cost is ( num__25 x num__4 ) = num__100 answer : c <eor> c <eos> |
c |
subtract__5.0__4.0__ multiply__5.0__4.0__ subtract__36.0__20.0__ add__5.0__9.0__ multiply__5.0__20.0__ multiply__5.0__20.0__ |
subtract__5.0__4.0__ multiply__5.0__4.0__ subtract__36.0__20.0__ add__5.0__9.0__ multiply__5.0__20.0__ divide__100.0__1.0__ |
| the average weight of a group of boys is num__30 kg . after a boy of weight num__35 kg joins the group the average weight of the group goes up by num__1 kg . find the number of boys in the group originally ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__9 |
let the number off boys in the group originally be x . total weight of the boys = num__30 x after the boy weighing num__35 kg joins the group total weight of boys = num__30 x + num__35 so num__30 x + num__35 + num__31 ( x + num__1 ) = > x = num__4 . answer : a <eor> a <eos> |
a |
add__30.0__1.0__ subtract__35.0__31.0__ subtract__35.0__31.0__ |
add__30.0__1.0__ subtract__35.0__31.0__ subtract__35.0__31.0__ |
| calculate the value of num__5.11 x num__10 ^ num__8 ? <o> a ) num__511000000 <o> b ) num__5110000 <o> c ) num__511000 <o> d ) num__51100000 <o> e ) num__51100 |
num__5.11 x num__10 ^ num__6 = num__5.11 x num__100000000 = num__511000000 . answer = a <eor> a <eos> |
a |
multiply__5.11__100000000.0__ multiply__5.11__100000000.0__ |
multiply__5.11__100000000.0__ multiply__5.11__100000000.0__ |
| a hollow iron pipe is num__21 cm long and its external diameter is num__8 cm . if the thickness of the pipe is num__1 cm and iron weighs num__8 g / cm ^ num__3 then the weight of the pipe is <o> a ) num__4.996 kg <o> b ) num__5.696 kg <o> c ) num__3.796 kg <o> d ) num__3.696 kg <o> e ) num__3.996 kg |
external radius = num__4 cm internal radius = num__3 cm . volume of iron = [ num__3.14285714286 x ( num__4 ^ num__2 x num__3 ^ num__2 ) num__21 ] cm ^ num__3 = num__462 cm ^ num__3 weight of iron = ( num__462 x num__8 ) gm = num__3696 gm = num__3.696 kg answer : d <eor> d <eos> |
d |
add__1.0__3.0__ divide__8.0__4.0__ multiply__8.0__462.0__ multiply__1.0__3.696__ |
add__1.0__3.0__ divide__8.0__4.0__ multiply__8.0__462.0__ multiply__1.0__3.696__ |
| mario and nina each have a bag of marbles each of which contains num__10 blue marbles num__10 red marbles an num__10 white marbles . if mario and nina each select one marble from their respective bags what is the probability that either mario or nina select a red marble ? <o> a ) num__0.75 <o> b ) num__0.222222222222 <o> c ) num__0.25 <o> d ) num__0.125 <o> e ) num__0.0625 |
mario and nina each select one marble from their respective bags . probability that either mario or nina select a red marble = probability that mario selects a red marble + probability that nina selects a red marble probability that either mario or nina select a red marble = ( num__0.333333333333 ) * ( num__0.333333333333 ) + ( num__0.333333333333 ) * ( num__0.333333333333 ) = num__2 * ( num__0.111111111111 ) probability that either mario or nina select a red marble = num__0.222222222222 answer would be b . <eor> b <eos> |
b |
coin_space__ union_prob__0.1111__0.3333__0.2222__ |
coin_space__ union_prob__0.1111__0.3333__0.2222__ |
| the bankers discount and the true discount of a sum at num__20.0 per annum simple interest for the same time are rs . num__200 and rs . num__150 respectively . what is the sum and the time ? <o> a ) sum = rs . num__600 and time = num__1.6 years <o> b ) sum = rs . num__500 and time = num__1.6 years <o> c ) sum = rs . num__400 and time = num__1.6 years <o> d ) sum = rs . num__600 and time = num__3 years <o> e ) none of these |
explanation : bd = rs . num__200 td = rs . num__150 r = num__20.0 f = bd × td / ( bd – td ) = num__200 × num__150 / ( num__200 – num__150 ) = num__200 × num__3.0 = rs . num__600 bd = simple interest on the face value of the bill for unexpired time = ftr / num__100 ⇒ num__200 = ( num__600 × t × num__20 ) / num__100 ⇒ num__200 = num__6 × t × num__20 ⇒ num__20 = num__12 × t ⇒ t = num__1.66666666667 = num__1.6 years answer : option a <eor> a <eos> |
a |
percent__3.0__200.0__ percent__6.0__200.0__ percent__100.0__600.0__ |
percent__3.0__200.0__ percent__6.0__200.0__ percent__100.0__600.0__ |
| the average weight of num__8 persons increases by num__3.5 kg when a new person comes in place of one of them weighing num__65 kg . what might be the weight of the new person ? <o> a ) num__75 kg <o> b ) num__65 kg <o> c ) num__55 kg <o> d ) num__85 kg <o> e ) num__93 kg |
total weight increased = ( num__8 x num__3.5 ) kg = num__28 kg . weight of new person = ( num__65 + num__28 ) kg = num__93 kg . answer : e <eor> e <eos> |
e |
multiply__8.0__3.5__ add__65.0__28.0__ add__65.0__28.0__ |
multiply__8.0__3.5__ add__65.0__28.0__ add__65.0__28.0__ |
| d and e are two points respectively on sides ab and ac of triangle abc such that de is parallel to bc . if the ratio of area of triangle ade to that of the trapezium decb is num__144 : num__25 and de = num__13 cm then find the length of bc . <o> a ) num__5 <o> b ) num__13 <o> c ) num__14 <o> d ) num__11 <o> e ) num__15 |
abc and ade are similar triangles . so ( side of abc / side of ade ) ^ num__2 = num__0.147928994083 side of abc / side of ade = num__0.384615384615 so the length of bc = num__5 answer - a <eor> a <eos> |
a |
triangle_area__2.0__5.0__ |
triangle_area__2.0__5.0__ |
| the banker ' s gain of a certain sum due num__2 years hence at num__10.0 per annum is rs . num__24 . what is the present worth ? <o> a ) rs . num__600 <o> b ) rs . num__500 <o> c ) rs . num__400 <o> d ) rs . num__300 <o> e ) none of these |
explanation : t = num__2 years r = num__10.0 td = ( bg × num__100 ) / tr = ( num__24 × num__100 ) / ( num__2 × num__10 ) = num__12 × num__10 = rs . num__120 td = ( pw × tr ) / num__100 ⇒ num__120 = ( pw × num__2 × num__10 ) / num__100 ⇒ num__1200 = pw × num__2 pw = num__600.0 = rs . num__600 answer : option a <eor> a <eos> |
a |
percent__100.0__600.0__ |
percent__100.0__600.0__ |
| two pipes a and b can separately fill a tank in num__12 and num__15 minutes respectively . a third pipe c can drain off num__45 liters of water per minute . if all the pipes are opened the tank can be filled in num__15 minutes . what is the capacity of the tank ? <o> a ) num__33 <o> b ) num__540 <o> c ) num__287 <o> d ) num__278 <o> e ) num__191 |
num__0.0833333333333 + num__0.0666666666667 - num__1 / x = num__0.0666666666667 x = num__12 num__12 * num__45 = num__540 answer : b <eor> b <eos> |
b |
multiply__12.0__45.0__ round__540.0__ |
multiply__12.0__45.0__ multiply__12.0__45.0__ |
| john and amanda stand at opposite ends of a straight road and start running towards each other at the same moment . their rates are randomly selected in advance so that john runs at a constant rate of num__2 num__3 num__4 or num__5 miles per hour and amanda runs at a constant rate of num__4 num__5 num__6 or num__7 miles per hour . what is the probability that john has traveled farther than amanda by the time they meet ? <o> a ) num__0.0625 <o> b ) num__0.3125 <o> c ) num__0.375 <o> d ) num__0.5 <o> e ) num__0.25 |
john will run farther if he runs at num__5 mph and amanda runs at num__4 mph . p ( john runs farther ) = num__0.25 * num__0.25 = num__0.0625 the answer is a . <eor> a <eos> |
a |
divide__0.25__4.0__ divide__0.25__4.0__ |
divide__0.25__4.0__ divide__0.25__4.0__ |
| oranges are bought at num__7 for rs . num__6 . at what rate per hundred must they be sold to gain num__33.0 ? <o> a ) num__57 <o> b ) num__68 <o> c ) num__95 <o> d ) num__114 <o> e ) num__125 |
num__7 oranges bought for rs num__6 num__1 orange cost = num__0.857142857143 therefore for num__100 oranges = num__100 * ( num__0.857142857143 ) gain = num__33.0 num__85.7142857143 + ( num__0.33 ) * num__85.7142857143 = rate per hundred he want to sell ans : num__114 answer : d <eor> d <eos> |
d |
percent__33.0__1.0__ percent__100.0__114.0__ |
percent__33.0__1.0__ percent__100.0__114.0__ |
| if the ratio of a to b is num__11 to num__3 and the ratio of b to c is num__1 to num__5 what is the ratio of a to c ? <o> a ) num__0.266666666667 <o> b ) num__0.733333333333 <o> c ) num__0.4 <o> d ) num__0.8 <o> e ) num__1.16666666667 |
a : b = num__11 : num__3 - - num__1 b : c = num__1 : num__5 = > b : c = num__3 : num__15 - - num__2 from num__1 and num__2 we get a : c = num__11 : num__15 answer b <eor> b <eos> |
b |
multiply__3.0__5.0__ subtract__3.0__1.0__ divide__11.0__15.0__ |
multiply__3.0__5.0__ subtract__3.0__1.0__ divide__11.0__15.0__ |
| the face value of a share is rs num__100 . it is sold at a discount of rs num__5 . how many such shares can be bought by investing rs num__38000 ? <o> a ) num__360 <o> b ) num__380 <o> c ) num__420 <o> d ) num__400 <o> e ) num__480 |
face value of a share = rs . num__100 discount = num__5.0 so share is bought at rs . num__95 investing amount = rs . num__38000 so num__380 shares are bought without discount with discount num__400 shares can be bought answer : d <eor> d <eos> |
d |
percent__100.0__400.0__ |
percent__100.0__400.0__ |
| each term of a certain sequence is num__4 less than the previous term . the first term of this sequence is num__19 . if the sum of the first n terms of the sequence is n what is the value of positive integer n ? <o> a ) num__1 <o> b ) num__13 <o> c ) num__15 <o> d ) num__19 <o> e ) num__47 |
we know that it is a decreasing sequence i . e . the terms keep decreasing till num__0 and then negative terms start . what does this imply ? if the sum of the first n terms of the sequence is n since number of terms will definitely be positive we are looking for a positive sum . num__19 + num__16 + num__13 + num__10 + num__7 + num__4 + num__1 - num__2 - num__5 - num__7 - num__10 . . . . and so on note that the first num__7 terms are positive and all others negative . every negative term has greater absolute value than the corresponding positive terms i . e . - num__2 absolute value is greater than num__1 absolute value - num__5 absolute value is greater than num__4 absolute value and so on . . . since we have num__7 positive terms we must have less than num__7 negative terms to get the sum as positive . if we have num__6 negative terms we will have a total of num__13 terms . of the given options only num__13 is possible and hence it must be the answer . answer ( c ) <eor> c <eos> |
c |
add__4.0__1.0__ add__4.0__2.0__ subtract__19.0__4.0__ |
add__4.0__1.0__ add__4.0__2.0__ subtract__19.0__4.0__ |
| you have a bag of num__9 letters : num__3 xs num__3 ys and num__3 zs . you are given a box divided into num__3 rows and num__3 columns for a total of num__9 areas . how many q different ways can you place one letter into each area such that there are no rows or columns with num__2 or more of the same letter ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__9 <o> d ) num__12 <o> e ) num__18 |
consider one particular arrangement of the first row : xyz then we can construct only two boxes with so that no rows or columns have num__2 or more of the same letter : xyz yzx zxy and : xyz zxy yzx now the first row itself can be arranged in num__3 ! = num__6 ways ( since there are three distinct letter ) hence the total number of boxes possible q is num__2 * num__6 = num__12 . answer : d . <eor> d <eos> |
d |
subtract__9.0__3.0__ add__9.0__3.0__ add__9.0__3.0__ |
multiply__3.0__2.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
| how many seconds will a num__500 m long train take to cros a man walking with a speed of num__3 kmph in the direction of the moving train if the speed of the train is num__63 kmph <o> a ) num__25 <o> b ) num__30 <o> c ) num__40 <o> d ) num__45 <o> e ) num__50 |
time = distance ( here length of the train ) / relative speed ( num__63 - num__3 ) thus time = num__8.33333333333 * num__0.277777777778 = num__500 * num__0.3 * num__5 = num__30 seconds answer : b <eor> b <eos> |
b |
round__30.0__ |
round__30.0__ |
| if num__5 √ num__5 x num__5 ^ num__3 + num__5 ^ - num__1.5 = num__5 ^ ( a + num__2 ) then the value of a is : <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
num__5 ^ num__1 x num__5 ^ num__0.5 x num__5 ^ num__3 + num__5 ^ - num__1.5 = num__5 ^ ( a + num__2 ) num__5 ^ num__1.5 x num__5 ^ num__3 + num__5 ^ - num__1.5 num__5 ^ num__4.5 + num__5 ^ - num__1.5 num__5 ^ num__3.0 num__5 ^ num__3 = num__5 ^ ( a + num__2 ) now num__3 = a + num__2 a = num__1 answer : b <eor> b <eos> |
b |
round_down__1.5__ reverse__2.0__ subtract__5.0__0.5__ round_down__1.5__ |
subtract__3.0__2.0__ subtract__1.5__1.0__ subtract__5.0__0.5__ subtract__3.0__2.0__ |
| the diagonals of a rhombus are num__10 cm and num__12 cm . find its area ? <o> a ) num__158 <o> b ) num__60 <o> c ) num__150 <o> d ) num__123 <o> e ) num__117 |
num__0.5 * num__10 * num__12 = num__60 answer : b <eor> b <eos> |
b |
triangle_area__10.0__12.0__ triangle_area__10.0__12.0__ |
volume_rectangular_prism__10.0__12.0__0.5__ volume_rectangular_prism__10.0__12.0__0.5__ |
| what is the least number should be added to num__1100 so the sum of the number is completely divisible by num__23 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__5 <o> d ) num__4 <o> e ) num__8 |
( num__47.8260869565 ) gives remainder num__19 num__19 + num__4 = num__23 so we need to add num__4 answer : d <eor> d <eos> |
d |
divide__1100.0__23.0__ subtract__23.0__19.0__ subtract__23.0__19.0__ |
divide__1100.0__23.0__ subtract__23.0__19.0__ subtract__23.0__19.0__ |
| a train num__125 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is : <o> a ) num__22 <o> b ) num__50 <o> c ) num__99 <o> d ) num__288 <o> e ) num__12 |
speed of the train relative to man = ( num__12.5 ) m / sec = ( num__12.5 ) m / sec . [ ( num__12.5 ) * ( num__3.6 ) ] km / hr = num__45 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__45 = = > x = num__50 km / hr . answer : b <eor> b <eos> |
b |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ round__50.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ multiply__5.0__10.0__ |
| a circular mat with diameter num__18 inches is placed on a square tabletop each of whose sides is num__24 inches long . which of the following is closest to the fraction of the tabletop covered by the mat ? <o> a ) num__0.416666666667 <o> b ) num__0.6 <o> c ) num__0.5 <o> d ) num__0.375 <o> e ) num__0.833333333333 |
so we are looking for the area of the cloth over the area of the table area of the cloth = ( pi ) ( r ) ^ num__2 which is about ( num__3 ) ( num__9 ) ( num__9 ) area of the table = ( num__24 ) ( num__24 ) so the quick way to estimate is looking at the fraction like this : ( num__0.125 ) ( num__3.375 ) i hope this is easy to follow so with some simplification i get ( num__0.125 ) ( num__3 ) = ( num__0.375 ) answer is d <eor> d <eos> |
d |
power__3.0__2.0__ volume_rectangular_prism__0.125__3.0__9.0__ multiply__0.125__3.0__ multiply__0.125__3.0__ |
power__3.0__2.0__ volume_rectangular_prism__0.125__3.0__9.0__ multiply__0.125__3.0__ multiply__0.125__3.0__ |
| if a b and c together can finish a piece of work in num__4 days . a alone in num__12 days and b in num__24 days then c alone can do it in ? <o> a ) num__1 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__6 |
c = num__0.25 - num__0.0833333333333 – num__0.0416666666667 = num__0.125 = > num__8 days ' answer : c <eor> c <eos> |
c |
add__0.0417__0.0833__ subtract__12.0__4.0__ round__8.0__ |
add__0.0417__0.0833__ subtract__12.0__4.0__ subtract__12.0__4.0__ |
| if the price of an article went up by num__20.0 then by what percent should it be brought down to bring it back to its original price ? <o> a ) num__16 num__0.666666666667 % <o> b ) num__16 num__0.333333333333 % <o> c ) num__16 num__2.66666666667 % <o> d ) num__17 num__0.666666666667 % <o> e ) num__16 num__2.0 % |
let the price of the article be rs . num__100 . num__20.0 of num__100 = num__20 . new price = num__100 + num__20 = rs . num__120 required percentage = ( num__120 - num__100 ) / num__120 * num__100 = num__0.166666666667 * num__100 = num__16.6666666667 = num__16 num__0.666666666667 % . answer : a <eor> a <eos> |
a |
percent__100.0__16.0__ |
percent__100.0__16.0__ |
| both robert and alice leave from the same location at num__6 : num__00 a . m . driving in the same direction but in separate cars . robert drives num__30 miles per hour while alice drives num__40 miles per hour . after num__6 hours alice ’ s car stops . at what time will robert ’ s car reach alice ’ s car ? <o> a ) num__2 p . m . <o> b ) num__3 p . m . <o> c ) num__4 p . m . <o> d ) num__8 p . m . <o> e ) num__9 p . m . |
num__6 : num__00 am so num__6 hours later is num__12 : num__00 pm in six hours robert will have driven num__6 * num__30 = num__180 miles in six hours alive will have driven num__6 * num__40 = num__240 miles so robert needs num__240 - num__180 = num__60 miles do catch alice up . so at num__30 mph he will need num__2 hours num__12 : num__00 pm + num__2 hours = num__2 : num__00 pm ans : a <eor> a <eos> |
a |
multiply__6.0__30.0__ multiply__6.0__40.0__ hour_to_min_conversion__ divide__12.0__6.0__ round__2.0__ |
multiply__6.0__30.0__ multiply__6.0__40.0__ subtract__240.0__180.0__ divide__12.0__6.0__ round__2.0__ |
| a merchant sold an article at a loss of num__12.0 . if the selling price had been increased by $ num__50 the merchant would have made a profit of num__8.0 . what is the cost price of the article ? <o> a ) $ num__550 <o> b ) $ num__500 <o> c ) $ num__510 <o> d ) $ num__565 <o> e ) $ num__520 |
let c . p . be $ x then num__108.0 of x - num__88.0 of x = num__100 num__20.0 of x = num__100 x / num__5 = num__100 x = num__500 answer is b <eor> b <eos> |
b |
percent__100.0__500.0__ |
percent__100.0__500.0__ |
| a glass was filled with num__25 ounces of water and num__0.04 ounce of the water evaporated each day during a num__10 - day period . what percent of the original amount of water evaporated during this period ? <o> a ) num__0.004 <o> b ) num__0.04 <o> c ) num__0.4 <o> d ) num__1.6 <o> e ) num__20 % |
in num__10 days num__10 * num__0.04 = num__0.4 ounces of water evaporated which is num__0.4 / num__25 â ˆ — num__100 = num__1.6 of the original amount of water . answer : d . <eor> d <eos> |
d |
percent__100.0__1.6__ |
percent__100.0__1.6__ |
| how long does a train num__110 m long running at the speed of num__36 km / hr takes to cross a bridge num__132 m length ? <o> a ) num__12.9 sec <o> b ) num__12.1 sec <o> c ) num__24.2 sec <o> d ) num__16.8 sec <o> e ) num__14.9 sec |
speed = num__36 * num__0.277777777778 = num__10 m / sec total distance covered = num__110 + num__132 = num__242 m . required time = num__24.2 = num__24.2 sec . answer : c <eor> c <eos> |
c |
add__110.0__132.0__ divide__242.0__10.0__ round__24.2__ |
add__110.0__132.0__ divide__242.0__10.0__ divide__242.0__10.0__ |
| the wheel of a motorcycle num__70 cm in diameter makes num__40 revolutions in every num__10 sec . what is the speed of motorcycle n km / hr ? <o> a ) num__31.65 <o> b ) num__31.68 <o> c ) num__31.62 <o> d ) num__31.61 <o> e ) num__31.67 |
explanation : in this type of question we will first calculate the distance covered in given time . distance covered will be : number of revolutions x circumference so we will be having distance and time from which we can calculate the speed . radius of wheel = num__35.0 = num__35 cm distance covered in num__40 revolutions will be = num__88 m distance covered in num__1 sec = answer : b ) num__31.68 kmph <eor> b <eos> |
b |
round__31.68__ |
round__31.68__ |
| a garrison of num__500 persons had provisions for num__27 days . after num__3 days a reinforcement of num__300 persons arrived . for how many more days will the remaining food last now ? <o> a ) num__12 days <o> b ) num__16 days <o> c ) num__14 days <o> d ) num__15 days <o> e ) none of these |
explanation : given that fort had provision for num__500 persons for num__27 days hence after num__3 days the remaining food is sufficient for num__500 persons for num__24 days remaining persons after num__3 days = num__500 + num__300 = num__800 assume that after num__10 days the remaining food is sufficient for num__800 persons for x days more men less days ( indirect proportion ) ⇒ men num__500 : num__800 } : : x : num__24 ⇒ num__500 × num__24 = num__800 x ⇒ num__5 × num__24 = num__8 x ⇒ x = num__5 × num__3 = num__15 . answer : option d <eor> d <eos> |
d |
subtract__27.0__3.0__ add__500.0__300.0__ add__3.0__5.0__ multiply__3.0__5.0__ round__15.0__ |
subtract__27.0__3.0__ add__500.0__300.0__ add__3.0__5.0__ add__5.0__10.0__ add__5.0__10.0__ |
| a highway is to be divided into num__3 lanes . for this purpose two yellow stripes are painted so that stripes divide the highway into num__3 lanes . if num__2 gallons of paint cover an area of num__3 p square feet of highway how many gallons of paint are needed to paint two stripes x inches wide on a stretch of highway m miles long ? ( num__1 mile = num__5280 feet and num__1 feet = num__12 inches ) <o> a ) num__440 p / mx <o> b ) num__880 mx / p <o> c ) num__880 mx / num__3 p <o> d ) num__1760 mx / num__3 p <o> e ) num__1760 p / num__3 mx |
num__1 square foot needs num__0.666666666667 p gallons of paint . the width of each stripe is x / num__12 feet . the length of each stripe is num__5280 m feet . the area of each stripe is ( num__5280 m ) ( x / num__12 ) = num__440 mx square feet . for two stripes the total area is num__880 mx square feet . the number of gallons needed is ( num__880 mx ) * ( num__0.666666666667 p ) = num__1760 mx / num__3 p gallons . the answer is d . <eor> d <eos> |
d |
divide__2.0__3.0__ divide__5280.0__12.0__ multiply__2.0__440.0__ multiply__2.0__880.0__ multiply__2.0__880.0__ |
divide__2.0__3.0__ divide__5280.0__12.0__ multiply__2.0__440.0__ multiply__2.0__880.0__ multiply__2.0__880.0__ |
| a train num__330 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__18 <o> e ) num__9 |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__330 * num__0.0545454545455 ] sec = num__18 sec answer : d <eor> d <eos> |
d |
add__60.0__6.0__ divide__330.0__18.3333__ round__18.0__ |
add__60.0__6.0__ divide__330.0__18.3333__ divide__330.0__18.3333__ |
| what sum of money will produce rs . num__70 as simple interest in num__4 years at num__3 num__0.5 percent ? <o> a ) num__288 <o> b ) num__500 <o> c ) num__299 <o> d ) num__2766 <o> e ) num__2912 |
num__70 = ( p * num__4 * num__3.5 ) / num__100 p = num__500 . answer : b <eor> b <eos> |
b |
percent__100.0__500.0__ |
percent__100.0__500.0__ |
| some articles were bought at num__6 articles for rs . num__5 and sold at num__5 articles for rs . num__6 . gain percent is : <o> a ) num__30.0 <o> b ) num__33.3333333333 % <o> c ) num__35.0 <o> d ) num__44.0 <o> e ) none of these |
explanation : c . p . of num__6 articles = rs . num__5 = > c . p . of num__1 article = rs . num__0.833333333333 s . p . of num__5 articles = rs . num__6 = > s . p . of num__1 article = rs . num__1.2 profit earned per article = num__1.2 - num__0.833333333333 = num__0.366666666667 % profit = num__0.366666666667 ∗ num__1.2 ∗ num__100.0 = num__44.0 answer : d <eor> d <eos> |
d |
percent__100.0__44.0__ |
percent__100.0__44.0__ |
| in a flight of num__600 km an aircraft was slowed down due to bad weather . its average speed for the trip was reduced by num__200 km / hr and the time of flight increased by num__30 minutes . the duration of the flight is : <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__1 |
flight = = > x hrs num__600 / x - num__600 / x + num__0.5 = num__200 x ( num__2 x + num__1 ) = num__3 num__2 x num__2 + x - num__3 = num__0 x = num__1 answer e <eor> e <eos> |
e |
reverse__0.5__ multiply__0.5__2.0__ divide__600.0__200.0__ round_down__0.5__ reverse__1.0__ |
reverse__0.5__ multiply__0.5__2.0__ divide__600.0__200.0__ round_down__0.5__ reverse__1.0__ |
| at present the ratio between the ages of arun and deepak is num__4 : num__3 . after num__6 years arun ' s age will be num__38 years . what is the age of deepak at present ? <o> a ) num__77 years <o> b ) num__15 years <o> c ) num__66 years <o> d ) num__24 years <o> e ) num__55 years |
let the present ages of arun and deepak be num__4 x and num__3 x years respectively . then num__4 x + num__6 = num__38 = > x = num__8 deepak ' s age = num__3 x = num__24 years . answer : d <eor> d <eos> |
d |
multiply__4.0__6.0__ multiply__4.0__6.0__ |
multiply__4.0__6.0__ multiply__4.0__6.0__ |
| a can do a piece of work in num__20 days and b can do it in num__30 days and c can do it num__10 days . they started the work together and a leaves after num__2 days and c leaves after num__4 days from the beginning . how long will work lost ? <o> a ) num__13 <o> b ) num__16 <o> c ) num__25 <o> d ) num__14 <o> e ) num__15 |
num__0.1 + x / num__30 + num__0.4 = num__1 x = num__15.0 = num__15 answer : e <eor> e <eos> |
e |
divide__2.0__20.0__ divide__4.0__10.0__ multiply__10.0__0.1__ divide__30.0__2.0__ round__15.0__ |
divide__2.0__20.0__ divide__4.0__10.0__ multiply__10.0__0.1__ divide__30.0__2.0__ divide__30.0__2.0__ |
| two trains running in opposite directions cross a man standing on the platform in num__20 seconds and num__18 seconds respectively and they cross each other in num__19 seconds . the ratio of their speeds is : <o> a ) num__1 : num__1 <o> b ) num__3 : num__2 <o> c ) num__3 : num__8 <o> d ) num__3 : num__25 <o> e ) num__3 : num__4 |
let the speeds of the two trains be x m / sec and y m / sec respectively . then length of the first train = num__20 x meters and length of the second train = num__18 y meters . ( num__20 x + num__15 y ) / ( x + y ) = num__19 = = > num__20 x + num__18 y = num__19 x + num__19 y = = > x = y = = > x / y = num__1.0 answer : option a <eor> a <eos> |
a |
subtract__20.0__19.0__ round__1.0__ |
subtract__20.0__19.0__ round__1.0__ |
| find the value of num__72519 x num__9999 = m ? <o> a ) num__34545481 <o> b ) num__45461743 <o> c ) num__24117456 <o> d ) num__725117481 <o> e ) num__45541177 |
num__72519 x num__9999 = num__72519 x ( num__10000 - num__1 ) = num__72519 x num__10000 - num__72519 x num__1 = num__725190000 - num__72519 = num__725117481 d <eor> d <eos> |
d |
subtract__10000.0__9999.0__ multiply__72519.0__10000.0__ multiply__72519.0__9999.0__ multiply__72519.0__9999.0__ |
subtract__10000.0__9999.0__ multiply__72519.0__10000.0__ subtract__725190000.0__72519.0__ subtract__725190000.0__72519.0__ |
| a man invested rs . num__14400 in rs . num__100 shares of a company at num__20.0 premium . if the company declares num__5.0 dividend at the end of the year then how much does he get ? <o> a ) num__500 <o> b ) num__600 <o> c ) num__650 <o> d ) num__700 <o> e ) num__750 |
no of shares = num__120.0 = num__120 face value = num__100 * num__120 = num__12000 annual income = num__0.05 * num__12000 = num__600 answer b <eor> b <eos> |
b |
percent__5.0__12000.0__ percent__100.0__600.0__ |
percent__5.0__12000.0__ percent__100.0__600.0__ |
| a trader sells num__75 meters of cloth for rs . num__4950 at the profit of rs . num__15 per metre of cloth . what is the cost price of one metre of cloth ? <o> a ) num__51 <o> b ) num__88 <o> c ) num__90 <o> d ) num__42 <o> e ) num__22 |
sp of num__1 m of cloth = num__66.0 = rs . num__66 cp of num__1 m of cloth = sp of num__1 m of cloth - profit on num__1 m of cloth = rs . num__66 - rs . num__15 = rs . num__51 answer : a <eor> a <eos> |
a |
divide__4950.0__75.0__ subtract__66.0__15.0__ round__51.0__ |
divide__4950.0__75.0__ subtract__66.0__15.0__ subtract__66.0__15.0__ |
| what is the sum of the different positive prime factors of num__380 ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__15 <o> d ) num__17 <o> e ) num__26 |
prime factorization of num__380 = num__38 * num__10 = num__2 * num__19 * num__10 = num__2 * num__19 * num__2 * num__5 = num__2 ^ num__2 * num__5 * num__19 sum of the different positive prime factors of num__380 = num__2 + num__5 + num__19 = num__26 answer e <eor> e <eos> |
e |
divide__380.0__38.0__ gcd__38.0__10.0__ divide__38.0__2.0__ divide__10.0__2.0__ lcm__2.0__26.0__ |
divide__380.0__38.0__ gcd__38.0__10.0__ divide__38.0__2.0__ divide__10.0__2.0__ lcm__2.0__26.0__ |
| if the population of a certain country increases at the rate of num__1 person every num__60 seconds by how many persons does the population increase in num__2 hours ? <o> a ) num__120 <o> b ) num__80 <o> c ) num__100 <o> d ) num__150 <o> e ) num__140 |
answer = num__1 * num__120 ( num__2 hour = num__120 minutes ) = num__120 answer = a <eor> a <eos> |
a |
multiply__60.0__2.0__ round__120.0__ |
multiply__60.0__2.0__ multiply__1.0__120.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 sec . what is the length of the train ? <o> a ) num__535 m <o> b ) num__178 m <o> c ) num__186 m <o> d ) num__168 m <o> e ) num__150 m |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__9 = num__150 m answer : e <eor> e <eos> |
e |
round__150.0__ |
round__150.0__ |
| factor | percent of respondents | user - friendly | num__56.0 | fast response time | num__48.0 | bargain prices | num__42.0 | the table gives three factors to be considered when choosing an internet service provider and the percent of the num__1100 respondents to a survey who cited that factor as important . if num__30 percent of the respondents cited both “ user - friendly ” and “ fast response time ” what is the maximum possible number of respondents who cited “ bargain prices ” but neither “ user - friendly ” nor “ fast response time ? ” <o> a ) num__286 <o> b ) num__336 <o> c ) num__360 <o> d ) num__384 <o> e ) num__420 |
the way i looked at is is as follows : userfriendly ( uf ) + fastresponse ( fr ) = num__30.0 uf leftover = num__56 - num__30 = num__26.0 fr leftover = num__48 - num__30 = num__18.0 sum these = num__74.0 ( ie . num__74.0 were either uf + fr uf fr ) num__26.0 leftover is the maximum number for bargain price num__0.26 * num__1100 = num__26 * num__11 = num__286 ( a ) . <eor> a <eos> |
a |
percent__26.0__1100.0__ percent__26.0__1100.0__ |
percent__26.0__1100.0__ percent__26.0__1100.0__ |
| a type of extra - large suv averages num__12.2 miles per gallon ( mpg ) on the highway but only num__7.6 mpg in the city . what is the maximum distance in miles that this suv could be driven on num__25 gallons of gasoline ? <o> a ) num__190 <o> b ) num__284.6 <o> c ) num__300 <o> d ) num__305 <o> e ) num__312 |
to maximize the distance that suv could be driven on num__25 gallons of gasoline we need consider only highway driving . max distance = num__25 * num__12.2 = num__305 answer d <eor> d <eos> |
d |
multiply__12.2__25.0__ multiply__12.2__25.0__ |
multiply__12.2__25.0__ multiply__12.2__25.0__ |
| square rstu shown above is rotated in a plane about its center in a clockwise direction the minimum number of degrees necessary for t to be in the position where r is now shown . the number of degrees through which rstu is rotated is <o> a ) num__135 degree <o> b ) num__180 degree <o> c ) num__225 degree <o> d ) num__270 degree <o> e ) num__315 degree |
from the options i am assuming the positioning of r and t relative to each other to be as shown . to replace r by t focus on ot . say you rotate ot clockwise ( and with it the entire square ) and bring it in place of or . how many degrees did you go ? you covered num__2 right angles i . e . num__180 degrees . answer : b <eor> b <eos> |
b |
straight_angle__ straight_angle__ |
straight_angle__ straight_angle__ |
| a man swims downstream num__35 km and upstream num__20 km taking num__5 hours each time what is the speed of the man in still water ? <o> a ) num__6.5 <o> b ) num__5.5 <o> c ) num__5.2 <o> d ) num__4.6 <o> e ) num__9.7 |
num__35 - - - num__5 ds = num__7 ? - - - - num__1 num__20 - - - - num__5 us = num__4 ? - - - - num__1 m = ? m = ( num__7 + num__4 ) / num__2 = num__5.5 answer : b <eor> b <eos> |
b |
divide__35.0__5.0__ divide__20.0__5.0__ subtract__7.0__5.0__ round__5.5__ |
divide__35.0__5.0__ divide__20.0__5.0__ subtract__7.0__5.0__ divide__5.5__1.0__ |
| an investment compounds annually at an interest rate of num__33.34 what is the smallest investment period by which time the investment will more than triple in value ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__5 <o> e ) num__4 |
assume initial amount is x annual interest is num__34.1 so after num__1 year the amount will become x * ( num__100 + num__33.34 ) / num__100 = > x * num__1.33333333333 now we need to find n for x * ( num__1.33333333333 ) ^ n = num__3 x or in other words n = num__4 e <eor> e <eos> |
e |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| in the x - y plane there are num__4 points ( num__00 ) ( num__04 ) ( num__94 ) and ( num__90 ) . if these num__4 points makes a rectangle what is the probability that x + y < num__4 ? <o> a ) num__0.4 <o> b ) num__0.6 <o> c ) num__0.428571428571 <o> d ) num__0.571428571429 <o> e ) num__0.222222222222 |
the line y = - x + num__4 intersects the rectangle and these three points of intersection ( num__00 ) ( num__04 ) and ( num__40 ) form a triangle . the points below the line y = - x + num__4 satisfy x + y < num__4 . the area of this triangle is ( num__0.5 ) ( num__4 ) ( num__4 ) = num__8 the area of the rectangle is num__36 . p ( x + y < num__4 ) = num__0.222222222222 = num__0.222222222222 the answer is e . <eor> e <eos> |
e |
divide__4.0__0.5__ subtract__40.0__4.0__ divide__8.0__36.0__ divide__8.0__36.0__ |
divide__4.0__0.5__ subtract__40.0__4.0__ divide__8.0__36.0__ divide__8.0__36.0__ |
| if ( m - num__8 ) is a factor of m ^ num__2 - sm - num__24 then s = <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__11 <o> e ) num__16 |
( m - num__8 ) ( m - a ) = m ^ num__2 - sm - num__24 a = - num__3 s = num__8 + a = num__5 = d = b <eor> b <eos> |
b |
divide__24.0__8.0__ subtract__8.0__3.0__ subtract__8.0__3.0__ |
divide__24.0__8.0__ subtract__8.0__3.0__ subtract__8.0__3.0__ |
| the radius of a wheel is num__22.4 cm . what is the distance covered by the wheel in making num__500 resolutions ? <o> a ) num__187 m <o> b ) num__704 m <o> c ) num__179 m <o> d ) num__127 m <o> e ) num__297 m |
in one resolution the distance covered by the wheel is its own circumference . distance covered in num__500 resolutions . = num__500 * num__2 * num__3.14285714286 * num__22.4 = num__70400 cm = num__704 m answer : b <eor> b <eos> |
b |
round__704.0__ |
round__704.0__ |
| by selling num__150 mangoes a fruit - seller gains the selling price of num__30 mangoes . find the gain percent ? <o> a ) num__28.0 <o> b ) num__25.0 <o> c ) num__95.0 <o> d ) num__75.0 <o> e ) num__45 % |
sp = cp + g num__150 sp = num__150 cp + num__30 sp num__120 sp = num__150 cp num__120 - - - num__30 cp num__100 - - - ? = > num__25.0 answer : b <eor> b <eos> |
b |
percent__25.0__100.0__ |
percent__25.0__100.0__ |
| a sales staff is composed of a sales manager and two sales people all of whom earn commission as a percentage of sales . each sales person earns num__5.0 commission on sales . in a given week the sales staff earned a total of $ num__1500 in commissions on $ num__5000 worth of sales . what commission rate did the sales manager earn during that week ? <o> a ) num__20.0 <o> b ) num__30.0 <o> c ) num__35.0 <o> d ) num__40.0 <o> e ) num__45 % |
a for me . let managers comminsion rate be m . m * num__5000 + num__2 * num__0.05 * num__5000 = num__1500 num__5000 * m = num__1000 m = num__0.2 = num__20.0 <eor> a <eos> |
a |
percent__2.0__1000.0__ percent__2.0__1000.0__ |
percent__2.0__1000.0__ percent__2.0__1000.0__ |
| a certain lab experiments with white and brown mice only . in one experiment num__0.4 of the mice are brown . if there are num__12 white mice in the experiment how many mice in total are in the experiment ? <o> a ) num__20 <o> b ) num__12 <o> c ) num__15 <o> d ) num__18 <o> e ) num__16 |
let total number of mice = m number of brown mice = num__0.4 m number of white mice = num__0.6 m = num__12 = > m = num__20 answer a <eor> a <eos> |
a |
divide__12.0__0.6__ divide__12.0__0.6__ |
divide__12.0__0.6__ divide__12.0__0.6__ |
| the greatest number of four digits which is divisible by num__15 num__25 num__40 and num__75 is : <o> a ) num__9000 <o> b ) num__9400 <o> c ) num__9600 <o> d ) num__9800 <o> e ) none |
explanation greatest number of num__4 - digits is num__9999 . l . c . m . of num__15 num__25 num__40 and num__75 is num__600 . on dividing num__9999 by num__600 the remainder is num__399 . required number ( num__9999 – num__399 ) = num__9600 . answer c <eor> c <eos> |
c |
multiply__15.0__40.0__ subtract__9999.0__399.0__ subtract__9999.0__399.0__ |
multiply__15.0__40.0__ subtract__9999.0__399.0__ subtract__9999.0__399.0__ |
| an article is bought for rs . num__600 and sold for rs . num__500 find the loss percent ? <o> a ) num__16 num__1.33333333333 % <o> b ) num__33.3333333333 % <o> c ) num__16.0 <o> d ) num__16 num__0.666666666667 % <o> e ) num__18 % |
explanation : num__600 - - - - num__100 num__100 - - - - ? = > num__16 num__0.666666666667 % answer is d <eor> d <eos> |
d |
percent__16.0__100.0__ |
percent__16.0__100.0__ |
| two pipes can fill a tank in num__20 and num__24 minutes respectively and a waste pipe can empty num__3 gallons per minute . all the three pipes working together can fill the tank in num__15 minutes . the capacity of the tank is ? <o> a ) num__887 gallons <o> b ) num__176 gallons <o> c ) num__120 gallons <o> d ) num__289 gallons <o> e ) num__976 gallons |
work done by the waste pipe in num__1 minute = num__0.0666666666667 - ( num__0.05 + num__0.0416666666667 ) = - num__0.025 volume of num__0.025 part = num__3 gallons volume of whole = num__3 * num__40 = num__120 gallons . answer : c <eor> c <eos> |
c |
divide__1.0__15.0__ divide__1.0__20.0__ divide__1.0__24.0__ subtract__0.0667__0.0417__ divide__1.0__0.025__ divide__3.0__0.025__ round__120.0__ |
divide__1.0__15.0__ divide__1.0__20.0__ divide__1.0__24.0__ subtract__0.0667__0.0417__ divide__1.0__0.025__ multiply__3.0__40.0__ multiply__3.0__40.0__ |
| the food in a camp lasts for num__20 men for num__50 days . if twenty more men join how many days will the food last ? <o> a ) num__22 days <o> b ) num__25 days <o> c ) num__28 days <o> d ) num__16 days <o> e ) num__27 days |
one man can consume the same food in num__20 * num__50 = num__1000 days . num__20 more men join the total number of men = num__40 the number of days the food will last = num__25.0 = num__25 days . answer : b <eor> b <eos> |
b |
multiply__20.0__50.0__ divide__1000.0__40.0__ round__25.0__ |
multiply__20.0__50.0__ divide__1000.0__40.0__ round__25.0__ |
| what sum of money will produce rs . num__70 as simple interest in num__4 years at num__3 num__0.5 percent ? <o> a ) rs . num__525 <o> b ) rs . num__500 <o> c ) rs . num__550 <o> d ) rs . num__555 <o> e ) rs . num__625 |
explanation : num__70 = ( p * num__4 * num__3.5 ) / num__100 p = num__500 answer is b <eor> b <eos> |
b |
percent__100.0__500.0__ |
percent__100.0__500.0__ |
| a train num__110 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__3 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__110 * num__0.0545454545455 ] sec = num__6 sec answer : b <eor> b <eos> |
b |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ round__6.0__ |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ divide__110.0__18.3333__ |
| express num__25 mps in kmph ? <o> a ) num__16 <o> b ) num__76 <o> c ) num__90 <o> d ) num__87 <o> e ) num__12 |
num__25 * num__3.6 = num__90 kmph answer : c <eor> c <eos> |
c |
multiply__25.0__3.6__ round__90.0__ |
multiply__25.0__3.6__ multiply__25.0__3.6__ |
| a woman begins bicycling with a pace of num__10 kmph and she increases her pace every two hours by num__2 kmph . she rests from bicycling num__2 hours after her pace reaches her maximum speed of num__14 kmph . she does a cool - down for the final num__8 km at a pace of num__2 kmph . how far in km did she bicycle ? <o> a ) num__96 km <o> b ) num__100 km <o> c ) num__106 km <o> d ) num__116 km <o> e ) num__126 km |
distance covered in first two hours = num__10 Ã — num__2 = num__20 km distance covered in next two hours = num__12 Ã — num__2 = num__24 km distance covered in next two hours = num__14 Ã — num__2 = num__48 km distance covered in first six hours num__20 + num__24 + num__48 = num__92 km remaining distance = num__8 km num__92 + num__8 = num__100 km answer b <eor> b <eos> |
b |
multiply__10.0__2.0__ add__10.0__2.0__ add__10.0__14.0__ multiply__2.0__24.0__ add__8.0__92.0__ round__100.0__ |
multiply__10.0__2.0__ add__10.0__2.0__ add__10.0__14.0__ multiply__2.0__24.0__ add__8.0__92.0__ add__8.0__92.0__ |
| how many multiples of num__2 are there between num__1 and num__58 exclusive ? <o> a ) num__21 <o> b ) num__22 <o> c ) num__24 <o> d ) num__26 <o> e ) num__28 |
num__28 multiples of num__2 between num__1 and num__58 exclusive . from num__2 * num__1 upto num__2 * num__28 ( num__12 num__34 . . . num__28 ) . hence num__28 multiples ! correct option is e <eor> e <eos> |
e |
multiply__1.0__28.0__ |
multiply__1.0__28.0__ |
| in a series of five football matches between germany and argentina the probability of germany winning a match is num__0.333333333333 and the probability that the match ends in a draw is num__0.25 . if a win gets the team num__2 points a draw num__1 point and a loss num__0 points what is the probability that germany will end up in num__5 draws in the series ? <o> a ) num__0.0416666666667 <o> b ) num__0.144675925926 <o> c ) num__0.04 <o> d ) num__0.0009765625 <o> e ) num__0.2 |
since the probability of a draw is num__0.25 then the probability of num__5 draws in num__5 matches is ( num__0.25 ) ^ num__5 = num__0.0009765625 . answer : d . <eor> d <eos> |
d |
power__0.25__5.0__ power__0.25__5.0__ |
power__0.25__5.0__ power__0.25__5.0__ |
| evaluate : | num__4 - num__8 ( num__3 - num__12 ) | - | num__5 - num__11 | = <o> a ) num__40 <o> b ) num__50 <o> c ) num__60 <o> d ) num__70 <o> e ) num__80 |
according to order of operations inner brackets first . hence | num__4 - num__8 ( num__3 - num__12 ) | - | num__5 - num__11 | = | num__4 - num__8 * ( - num__9 ) | - | num__5 - num__11 | according to order of operations multiplication within absolute value signs ( which may be considered as brackets when it comes to order of operations ) next . hence = | num__4 + num__72 | - | num__5 - num__11 | = | num__76 | - | - num__6 | = num__76 - num__6 = num__70 correct answer is d ) num__70 <eor> d <eos> |
d |
add__4.0__5.0__ multiply__8.0__9.0__ add__4.0__72.0__ subtract__11.0__5.0__ subtract__76.0__6.0__ subtract__76.0__6.0__ |
add__4.0__5.0__ multiply__8.0__9.0__ add__4.0__72.0__ subtract__11.0__5.0__ subtract__76.0__6.0__ subtract__76.0__6.0__ |
| a team of eight entered for a shooting competition . the best marks man scored num__82 points . if he had scored num__92 points the average scores for . the team would have been num__85 . how many points altogether did the team score ? <o> a ) num__662 <o> b ) num__672 <o> c ) num__652 <o> d ) num__642 <o> e ) num__670 |
num__8 * num__85 = num__680 – num__10 = num__670 answer : e <eor> e <eos> |
e |
multiply__85.0__8.0__ subtract__92.0__82.0__ subtract__680.0__10.0__ subtract__680.0__10.0__ |
multiply__85.0__8.0__ subtract__92.0__82.0__ subtract__680.0__10.0__ subtract__680.0__10.0__ |
| roger can read a book in k minutes . what part of the book can he read in num__4 minutes ? ( k > num__8 ) <o> a ) num__8 + k <o> b ) num__8 / k <o> c ) k / num__8 <o> d ) ( k + num__8 ) / k <o> e ) ( k - num__8 ) / k |
let ' s sayk = num__24 that is it takes num__24 minutes to read the entire book . so in num__8 minutes roger can read num__0.333333333333 of the book so we ' re looking for the answer choice that yields an output of num__0.333333333333 whenk = num__24 a ) num__8 + num__24 = num__32 eliminate b ) num__0.333333333333 = num__0.333333333333 keep c ) num__3.0 = num__3 eliminate d ) ( num__24 + num__8 ) / num__24 = num__1.33333333333 = num__1.33333333333 eliminate e ) ( num__24 - num__8 ) / num__24 = num__0.666666666667 = num__0.666666666667 eliminate answer : a <eor> a <eos> |
a |
divide__8.0__24.0__ multiply__4.0__8.0__ divide__24.0__8.0__ divide__4.0__3.0__ round__8.0__ |
divide__8.0__24.0__ add__8.0__24.0__ divide__24.0__8.0__ divide__4.0__3.0__ divide__32.0__4.0__ |
| a person goes to a bank and quotes x rs and y paise on a cheque . the cashier misreads it and gives y rs and x paise . the man comes out and donates num__5 paise to a begger . now the man has exactly double the amount he has quoted on the cheque . . <o> a ) num__31 rs num__63 paise <o> b ) num__30 rs num__63 paise <o> c ) num__31 rs num__33 paise <o> d ) num__31 rs num__3 paise <o> e ) num__33 rs num__63 paise |
he gets num__100 y + x paise instead of num__100 x + y paise ( num__100 y + x ) - ( num__100 x + y ) - num__5 = ( num__100 x + y ) num__99 y - num__99 x - num__5 = num__100 x + y num__98 y = num__199 x + num__5 y = num__2 x + ( num__3 x + num__5 ) / num__98 only an integer value of x = num__31 will make y an integer with the constraints that both x & y < num__100 substituing y = num__2 * num__31 + num__1 = num__63 ans : num__31 rs num__63 paise answer : a <eor> a <eos> |
a |
add__99.0__100.0__ subtract__100.0__98.0__ subtract__5.0__2.0__ subtract__99.0__98.0__ multiply__1.0__31.0__ |
add__99.0__100.0__ subtract__100.0__98.0__ subtract__5.0__2.0__ subtract__99.0__98.0__ multiply__1.0__31.0__ |
| num__1 = num__52 = num__253 = num__2534 = num__2545 = ? <o> a ) num__1 <o> b ) num__255 <o> c ) num__345 <o> d ) num__445 <o> e ) num__235 |
num__1 = num__52 = num__253 = num__2534 = num__2545 = ? num__5 = num__1 check the first eqn . answer : a <eor> a <eos> |
a |
reverse__1.0__ |
reverse__1.0__ |
| if num__70.0 of a number is equal to two - third of another number what is the ratio of first number to the second number ? <o> a ) num__5 : num__7 <o> b ) num__18 : num__23 <o> c ) num__17 : num__19 <o> d ) num__20 : num__21 <o> e ) num__2 : num__3 |
let num__70.0 of a = num__0.666666666667 b . then num__70 a / num__100 = num__2 b / num__3 = > num__7 a / num__10 = num__2 b / num__3 a / b = ( num__0.666666666667 * num__1.42857142857 ) = num__0.952380952381 a : b = num__20 : num__21 . answer : d <eor> d <eos> |
d |
divide__70.0__7.0__ divide__100.0__70.0__ multiply__1.4286__0.6667__ multiply__2.0__10.0__ multiply__3.0__7.0__ multiply__2.0__10.0__ |
divide__70.0__7.0__ divide__100.0__70.0__ multiply__1.4286__0.6667__ multiply__2.0__10.0__ multiply__3.0__7.0__ multiply__2.0__10.0__ |
| with both valves open the pool will be filled with water in num__48 minutes . the first valve alone would fill the pool in num__2 hours . if the second valve emits num__50 cubic meters of water more than the first every minute then what is the capacity t of the pool ? <o> a ) num__9000 cubic meters <o> b ) num__10500 cubic meters <o> c ) num__11750 cubic meters <o> d ) num__12000 cubic meters <o> e ) num__12500 cubic meters |
d . num__12000 cubic meters . if both hte valves fill the pool in num__48 minutes and valve num__1 only fills in num__120 minutes then valve num__2 alone will fill the pool in ( num__48 * num__120 ) / ( num__120 - num__48 ) = num__80 minutes . now if valve num__1 admits x cubic meter of water per minute then the capacity of pool will be num__120 x and also num__80 ( x + num__50 ) . or num__120 x = num__80 ( x + num__50 ) . or x = num__100 . hence the capacity of pool = num__120 x = num__12000 cubic meters . <eor> d <eos> |
d |
multiply__2.0__50.0__ round__12000.0__ |
multiply__2.0__50.0__ divide__12000.0__1.0__ |
| what is the present worth of rs . num__121 due in num__2 years at num__5.0 simple interest per annum <o> a ) num__110 <o> b ) num__120 <o> c ) num__130 <o> d ) num__140 <o> e ) none of these |
explanation : let the present worth be rs . x then s . i . = rs . ( num__121 - x ) = ( x * num__5 * num__0.02 ) = num__121 - x = num__10 x = num__12100 - num__100 x = num__110 x = num__12100 x = num__110 answer : a <eor> a <eos> |
a |
percent__100.0__110.0__ |
percent__100.0__110.0__ |
| if | x - num__12 | = num__100 what is the sum of all the possible values of x ? <o> a ) - num__12 <o> b ) - num__22 <o> c ) num__24 <o> d ) num__36 <o> e ) num__42 |
there will be two cases x - num__12 = num__100 or x - num__12 = - num__100 = > x = num__112 or x = - num__88 sum of both the values will be - num__88 + num__112 = num__24 answer is c <eor> c <eos> |
c |
add__12.0__100.0__ subtract__100.0__12.0__ subtract__112.0__88.0__ subtract__112.0__88.0__ |
add__12.0__100.0__ subtract__100.0__12.0__ subtract__112.0__88.0__ subtract__112.0__88.0__ |
| there are two positive numbers in the ratio num__5 : num__8 . if the larger number exceeds the smaller by num__45 then find the smaller number ? <o> a ) num__25 <o> b ) num__66 <o> c ) num__77 <o> d ) num__88 <o> e ) num__75 |
let the two positive numbers be num__5 x and num__8 x respectively . num__8 x - num__5 x = num__45 num__3 x = num__45 = > x = num__15 = > smaller number = num__5 x = num__75 . answer : e <eor> e <eos> |
e |
subtract__8.0__5.0__ multiply__5.0__3.0__ multiply__5.0__15.0__ multiply__5.0__15.0__ |
subtract__8.0__5.0__ multiply__5.0__3.0__ multiply__5.0__15.0__ multiply__5.0__15.0__ |
| a can be divided by num__11 with no remainder . which of the following expressions could be divided by num__11 leaving a remainder of num__1 ? <o> a ) a - num__20 . <o> b ) a - num__12 . <o> c ) a - num__9 . <o> d ) a - num__10 . <o> e ) a - num__13 . |
we could use the following logic : since a is a multiple of num__11 we could very well say a = num__11 k ( where k is num__12 . . . n ) . the condition here should now be num__11 k + num__1 = < operation > such that it gives us an integer num__11 k + num__1 = a - num__20 = > a = num__11 k + num__21 not divisible by num__11 num__11 k + num__1 = a - num__12 = > a = num__11 k + num__31 not divisible by num__11 num__11 k + num__1 = a - num__9 = > a = num__11 k + num__10 not divisible by num__11 num__11 k + num__1 = a - num__10 = > a = num__11 k + num__11 divisible by num__11 num__11 k + num__1 = a - num__13 = > a = num__11 k + num__14 not divisible by num__11 answer therefore is a - num__10 ( d ) <eor> d <eos> |
d |
add__11.0__1.0__ add__1.0__20.0__ add__11.0__20.0__ subtract__20.0__11.0__ subtract__11.0__1.0__ add__1.0__12.0__ add__1.0__13.0__ subtract__11.0__1.0__ |
add__11.0__1.0__ add__1.0__20.0__ add__11.0__20.0__ subtract__20.0__11.0__ subtract__11.0__1.0__ add__1.0__12.0__ add__1.0__13.0__ subtract__11.0__1.0__ |
| the winning relay team in a high school sports competition clocked num__48 minutes for a distance of num__13.2 km . its runners a b c and d maintained speeds of num__17 kmph num__18 kmph num__19 kmph and num__20 kmph respectively . what is the ratio of the time taken by b to than taken by d ? <o> a ) num__5 : num__16 <o> b ) num__5 : num__17 <o> c ) num__9 : num__8 <o> d ) num__10 : num__9 <o> e ) none of these |
explanation : since it is a relay race all the runners ran the same distance . hence for a same distance ( ratio of times ) = num__1 / ( ratio of speeds ) . hence ratio of times taken by b t & d = num__20 : num__18 = num__10 : num__9 . answer : d <eor> d <eos> |
d |
subtract__18.0__17.0__ subtract__19.0__10.0__ round__10.0__ |
subtract__18.0__17.0__ subtract__19.0__10.0__ divide__10.0__1.0__ |
| a train of length num__560 metres takes num__90 seconds to cross a tunnel of length num__340 metres . what is the speed of the train in km / hr ? <o> a ) num__36 km / hr . <o> b ) num__46 km / hr . <o> c ) num__30 km / hr . <o> d ) num__48 km / hr . <o> e ) none |
sol . speed = [ num__560 + num__3.77777777778 ] m / sec = [ num__10.0 * num__3.6 ] km / hr = num__36 km / hr . answer a <eor> a <eos> |
a |
divide__340.0__90.0__ multiply__3.6__10.0__ round__36.0__ |
divide__340.0__90.0__ multiply__3.6__10.0__ multiply__3.6__10.0__ |
| in the coordinate plane points ( x num__5 ) and ( num__9 y ) are on line k . if line k passes through the origin and has slope num__0.555555555556 then x - y = <o> a ) num__4 <o> b ) num__6 <o> c ) num__5 <o> d ) num__8 <o> e ) num__7 |
line k passes through the origin and has slope num__0.25 means that its equation is y = num__0.555555555556 * x . thus : ( x num__5 ) = ( num__9 num__5 ) and ( num__9 y ) = ( num__95 ) - - > x - y = num__9 - num__5 = num__4 . answer : a <eor> a <eos> |
a |
reverse__0.25__ reverse__0.25__ |
subtract__9.0__5.0__ subtract__9.0__5.0__ |
| a bus can hold num__120 passengers . if there are num__12 rows of seats on the bus how many seats are in each row ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__45 <o> d ) num__8 <o> e ) num__9 |
no . of seats = num__10.0 = num__10 there are num__10 seats in each row answer : a <eor> a <eos> |
a |
divide__120.0__12.0__ divide__120.0__12.0__ |
divide__120.0__12.0__ divide__120.0__12.0__ |
| a typist can type num__120 words in num__100 seconds . at that rate how many seconds would it take her to type num__258 words ? <o> a ) num__205 <o> b ) num__210 <o> c ) num__215 <o> d ) num__220 <o> e ) num__225 |
num__1.2 = num__258 / n num__120 * n = num__258 * num__100 num__120 n = num__25800 n = num__215 answer : c <eor> c <eos> |
c |
percent__100.0__215.0__ |
percent__100.0__215.0__ |
| if a lends rs . num__3150 to b at num__8.0 per annum and b lends the same sum to c at num__12.5 per annum then the gain of b in a period of num__2 years is ? <o> a ) num__280 <o> b ) num__295 <o> c ) num__283.5 <o> d ) num__245 <o> e ) num__200 |
( num__3150 * num__4.5 * num__2 ) / num__100 = > num__283.5 answer : c <eor> c <eos> |
c |
percent__100.0__283.5__ |
percent__100.0__283.5__ |
| on a certain road num__10.0 of the motorists exceed the posted speed limit and receive speeding tickets but num__20.0 of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on the road exceed the posted speed limit ? <o> a ) num__10.5 <o> b ) num__12.5 <o> c ) num__15.0 <o> d ) num__22.0 <o> e ) num__30 % |
say there are num__100 motorists . { # of motorists who exceed speed & receive tickets } + { # of motorists who exceed speed & do n ' t receive tickets } = { total # of motorist who exceed speed } ; given : { # of motorists who exceed speed & receive tickets } = num__10 ; also if { total # of motorist who exceed speed } = x then num__0.2 x = { # of motorists who exceed speed & do n ' t receive tickets } ; num__10 + num__0.2 x = x - - > x = num__12.5 answer : b . <eor> b <eos> |
b |
percent__12.5__100.0__ |
percent__12.5__100.0__ |
| at the opening of a trading day at a certain stock exchange the price per share of stock m was $ num__25 . if the price per share of stock m was $ num__28 at the closing of the day what was the percent increase in the price per share of stock m for that day ? <o> a ) num__1.4 <o> b ) num__5.9 <o> c ) num__12.0 <o> d ) num__12.5 <o> e ) num__23.6 % |
opening = num__25 closing = num__28 rise in price = num__3 so percent increase = num__0.12 * num__100 = num__12 answer : c <eor> c <eos> |
c |
subtract__28.0__25.0__ divide__3.0__25.0__ multiply__100.0__0.12__ round__12.0__ |
subtract__28.0__25.0__ divide__3.0__25.0__ multiply__100.0__0.12__ multiply__100.0__0.12__ |
| what is the sum of the greatest common factor and the lowest common multiple of num__16 and num__36 ? <o> a ) num__124 <o> b ) num__152 <o> c ) num__148 <o> d ) num__123 <o> e ) num__412 |
prime factorization of num__16 = num__2 x num__2 x num__2 x num__2 prime factorization of num__36 = num__2 x num__2 x num__3 x num__3 gcf = num__4 lcm = num__144 sum = num__148 option : c <eor> c <eos> |
c |
multiply__36.0__4.0__ add__144.0__4.0__ add__144.0__4.0__ |
multiply__36.0__4.0__ add__144.0__4.0__ add__144.0__4.0__ |
| john and mary were each paid x dollars in advance to do a certain job together . john worked on the job for num__10 hours and mary worked num__4 hours less than john . if mary gave john y dollars of her payment so that they would have received the same hourly wage what was the dollar amount in terms of y that john was paid in advance ? <o> a ) num__4 y <o> b ) num__5 y <o> c ) num__6 y <o> d ) num__8 y <o> e ) num__9 y |
let $ x be the advance that both receive = num__4 x amount earned per hour by john and mary = x / num__10 and x / num__6 mary gives $ y to john to make the wages earned equal hence john wage per hr = ( x + y ) num__10 which is now equal to mary ' s wage ( x - y ) / num__6 solve ( x + y ) num__10 = ( x - y ) / num__6 num__6 x + num__6 y = num__10 x - num__10 y num__4 x = num__16 y x = num__4 y ans . a <eor> a <eos> |
a |
subtract__10.0__4.0__ add__10.0__6.0__ subtract__10.0__6.0__ |
subtract__10.0__4.0__ add__10.0__6.0__ subtract__10.0__6.0__ |
| for one toss of a certain coin the probability that the outcome is heads is num__0.6 . if this coin is tossed num__5 times which of the following is the probability that the outcome will be heads at least num__4 times ? <o> a ) ( num__0.6 ) ^ num__5 <o> b ) num__2 ( num__0.6 ) ^ num__4 <o> c ) num__3 ( num__0.6 ) ^ num__4 <o> d ) num__4 ( num__0.6 ) ^ num__4 ( num__0.4 ) + ( num__0.6 ) ^ num__5 <o> e ) num__5 ( num__0.6 ) ^ num__4 ( num__0.4 ) + ( num__0.6 ) ^ num__5 |
p ( h ) = num__0.6 p ( h ) = num__0.6 so p ( t ) = num__0.4 p ( t ) = num__0.4 . we want to determine the probability of at least num__4 heads in num__5 tries . at least num__4 heads means num__4 or num__5 . let ' s calculate each one : num__5 heads : p ( h = num__5 ) = num__0.6 ^ num__5 ; num__4 h and num__1 t : p ( h = num__4 ) = num__5 ! / num__4 ! ∗ num__0.6 ^ num__4 ∗ num__0.4 = num__5 ∗ num__0.6 ^ num__4 ∗ num__0.4 multiplying by num__5 as num__4 h and num__1 t may occur in num__5 different ways : hhhht hhhth hhthh hthhh thhhh so p ( h ≥ num__4 ) = num__0.6 ^ num__5 + num__5 ∗ num__0.6 ^ num__4 ∗ num__0.4 answer : e . <eor> e <eos> |
e |
negate_prob__0.6__ vowel_space__ |
negate_prob__0.6__ vowel_space__ |
| three numbers are in the ratio num__5 : num__4 : num__3 and their average is num__360 . the largest number is : <o> a ) num__30 <o> b ) num__450 <o> c ) num__27 <o> d ) num__21 <o> e ) num__22 |
explanation : let the numbers be num__5 x num__4 x and num__3 x then ( num__5 x + num__4 x + num__3 x ) / num__3 = num__360 = > num__12 x = num__360 * num__3 = > x = num__90 largest number num__5 x = num__5 * num__90 = num__450 answer : b <eor> b <eos> |
b |
multiply__4.0__3.0__ divide__360.0__4.0__ multiply__5.0__90.0__ multiply__5.0__90.0__ |
multiply__4.0__3.0__ divide__360.0__4.0__ multiply__5.0__90.0__ multiply__5.0__90.0__ |
| a radio station has to choose num__6 days of the seven in a week to broadcast a certain program and that set will repeat each week . the program can be broadcast equally on any of the seven weekdays — - weekdays vs . weekends don ’ t matter at all — - nor does it matter whether the days the program airs are adjacent or not . absolutely any num__6 of the seven weekdays can be chosen . how many different num__6 - day combinations of the seven weekdays can be constructed ? <o> a ) num__7 <o> b ) num__15 <o> c ) num__21 <o> d ) num__35 <o> e ) num__56 |
pick num__6 days out of num__7 . . num__7 c num__6 = num__7 answer : a <eor> a <eos> |
a |
choose__7.0__6.0__ |
choose__7.0__6.0__ |
| a seller sells his apple at cp but uses a weight of num__800 gm instead of kg weight . what is his profit % ? <o> a ) num__20.0 <o> b ) num__23.0 <o> c ) num__25.0 <o> d ) num__30.0 <o> e ) num__32 % |
if a trader professes to sell his goods at cost price but uses false weights then gain % = [ error ( true value − error ) × num__100 ] % so here profit percentage = [ num__200 ( num__1000 − num__200 ) × num__100 ] % = [ num__200800 × num__100 ] % = num__25.0 c <eor> c <eos> |
c |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| m and b are integers . the expression ( m + num__1 ) ( b + num__1 ) is even . what can be said about m and b ? <o> a ) they are both even numbers . <o> b ) at least one of them is even . <o> c ) at least one of them is odd . <o> d ) they are both odd . <o> e ) nothing can be said surly on m and b |
odd x odd = odd odd x even = even even x even = even to fulfill condition either ( m + num__1 ) or ( b + num__1 ) needs to be even so either m or b needs to be odd or at least one of them is odd . at least one of them is odd = c <eor> c <eos> |
c |
reverse__1.0__ |
reverse__1.0__ |
| a shirt goes on sale for num__80.0 of its original price . one week later the sale price is marked down num__10.0 . the final price is what percent of the original price ? <o> a ) num__72.0 <o> b ) num__70.0 <o> c ) num__52.0 <o> d ) num__50.0 <o> e ) num__28 % |
just assume original price is num__100 . sale price = num__80 then it is marked down by num__10.0 = num__80 - num__8 = num__72 . hence it is num__72.0 of the original price . hence answer is a <eor> a <eos> |
a |
percent__80.0__10.0__ percent__100.0__72.0__ |
percent__80.0__10.0__ percent__100.0__72.0__ |
| sakshi can do a piece of work in num__12 days . tanya is num__20.0 more efficient than sakshi . the number of days taken by tanya to do the same piece of work is ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__16 <o> e ) num__18 |
ratio of times taken by sakshi and tanys = num__120 : num__100 = num__6 : num__5 suppose tanya takes x days to do the work . num__6 : num__5 : : num__12 : x = > x = num__10 days . hence tanya takes num__10 days to complete the work . answer : a <eor> a <eos> |
a |
subtract__120.0__20.0__ divide__120.0__20.0__ divide__100.0__20.0__ divide__120.0__12.0__ round__10.0__ |
subtract__120.0__20.0__ divide__120.0__20.0__ divide__100.0__20.0__ divide__120.0__12.0__ round__10.0__ |
| the lcm and hcf of two numbers are num__100000 and num__10000 respectively . find the larger of the two numbers if their sum is num__110000 . <o> a ) num__100000 <o> b ) num__46 <o> c ) num__577 <o> d ) num__577 <o> e ) num__767 |
there are num__2 approaches in solving this . methode num__1 . hcf * lcm = the actual number . num__100000 * num__10000 = num__1000000000 so the answer which we are looking for has to be a factor of num__1000000000 . so among the options shortlist the answers by eliminating those numbers which is not divisible by num__1000000000 . and then take the highest number as the answer as the question asks abt the highest number . answer is a <eor> a <eos> |
a |
multiply__100000.0__10000.0__ round__100000.0__ |
multiply__100000.0__10000.0__ round__100000.0__ |
| in an university college the ratio of the number of boys and girls is num__5 : num__9 . if the percentage increase in the number of boys and girls be num__20.0 and num__40.0 respectively . what will be the new ratio ? <o> a ) num__21 : num__10 <o> b ) num__12 : num__17 <o> c ) num__15 : num__23 <o> d ) num__10 : num__21 <o> e ) none of these |
explanation : solution : originally let the number of boys and girls in the university college be num__5 x and num__9 x respectively . their increased number is ( num__20.0 of num__5 x ) : ( num__40.0 of num__9 x ) = > ( num__120 * num__5 x / num__100 ) : ( num__140 * num__9 x / num__100 ) = num__60 x : num__126 x = > x : y = num__10 : num__21 . answer : d <eor> d <eos> |
d |
multiply__5.0__20.0__ add__20.0__120.0__ add__20.0__40.0__ subtract__20.0__10.0__ |
multiply__5.0__20.0__ add__20.0__120.0__ add__20.0__40.0__ divide__100.0__10.0__ |
| john was thrice as old as tom num__6 years ago . john will be num__1.5 times as old as tom in num__6 years . how old is tom today ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__12 <o> e ) num__14 |
j - num__6 = num__3 ( t - num__6 ) so j = num__3 t - num__12 j + num__6 = num__1.5 * ( t + num__6 ) num__2 j + num__12 = num__3 t + num__18 num__2 ( num__3 t - num__12 ) + num__12 = num__3 t + num__18 num__3 t = num__30 t = num__10 the answer is c . <eor> c <eos> |
c |
divide__6.0__3.0__ multiply__6.0__3.0__ add__12.0__18.0__ subtract__12.0__2.0__ subtract__12.0__2.0__ |
divide__6.0__3.0__ multiply__6.0__3.0__ add__12.0__18.0__ subtract__12.0__2.0__ subtract__12.0__2.0__ |
| a sum of money amounts to rs num__9800 after num__5 years and rs num__12005 after num__8 years at the same rate of simple interest . the rate of interest per annum is <o> a ) num__9.0 <o> b ) num__10.0 <o> c ) num__11.0 <o> d ) num__12.0 <o> e ) num__13 % |
explanation : we can get si of num__3 years = num__12005 - num__9800 = num__2205 si for num__5 years = ( num__735.0 ) * num__5 = num__3675 [ so that we can get principal amount after deducting si ] principal = num__12005 - num__3675 = num__6125 so rate = ( num__100 * num__3675 ) / ( num__6125 * num__5 ) = num__12.0 option d <eor> d <eos> |
d |
percent__100.0__12.0__ |
percent__100.0__12.0__ |
| the sum of the first num__30 positive even integers is num__930 . what is the sum of the first num__30 odd integers ? <o> a ) num__930 <o> b ) num__900 <o> c ) num__960 <o> d ) num__975 <o> e ) num__980 |
sum of first n even numbers = n ( n + num__1 ) = num__930 sum of first n odd numbers = n ^ num__2 = num__30 * num__30 = num__900 ( here n = num__30 ) answer : b <eor> b <eos> |
b |
subtract__930.0__30.0__ subtract__930.0__30.0__ |
subtract__930.0__30.0__ multiply__1.0__900.0__ |
| if a ( a + num__7 ) = num__60 and b ( b + num__7 ) = num__60 where a ≠ b then a + b = <o> a ) − num__7 <o> b ) − num__2 <o> c ) num__2 <o> d ) num__46 <o> e ) num__48 |
i . e . if a = num__5 then b = - num__12 or if a = - num__12 then b = num__5 but in each case a + b = - num__12 + num__5 = - num__7 answer : option a <eor> a <eos> |
a |
add__7.0__5.0__ subtract__12.0__5.0__ |
add__7.0__5.0__ subtract__12.0__5.0__ |
| find the remainder when the polynomial x num__4 - num__3 x num__2 + num__7 x - num__10 is divided by ( x - num__2 ) . <o> a ) num__8 <o> b ) - num__20 <o> c ) num__18 <o> d ) num__0 <o> e ) none |
explanatory answer by remainder theorem : if a polynomial in one variable ' x ' is divided by ( x - a ) where ' a ' is any real number then the remainder is the value of the polynomial at x = a . therefore remainder = ( num__2 ) num__4 - num__3 ( num__2 ) num__2 + num__7 ( num__2 ) - num__10 = num__8 answer a <eor> a <eos> |
a |
multiply__4.0__2.0__ multiply__4.0__2.0__ |
subtract__10.0__2.0__ subtract__10.0__2.0__ |
| what two - digit number is less than the sum of the square of its digits by num__15 and exceeds their doubled product by num__5 ? <o> a ) num__95 <o> b ) num__99 <o> c ) num__26 <o> d ) num__73 <o> e ) none of the above |
let the digits be x and y . the number would be num__10 x + y . we are given that num__2 xy + num__5 = num__10 x + y = x ^ num__2 y ^ num__2 - num__15 thus num__2 xy + num__5 = x ^ num__2 + y ^ num__2 - num__15 x ^ num__2 + y ^ num__2 - num__2 xy = num__16 ( x - y ) ^ num__2 = num__16 ( x - y ) = num__4 or - num__4 substituting the values of ( x - y ) in the equation num__2 xy + num__5 = num__10 x + y x comes out to be num__1 or num__9 . . . thus the two numbers can be num__15 or num__73 thus the answer is d <eor> d <eos> |
d |
subtract__15.0__5.0__ divide__10.0__5.0__ subtract__5.0__4.0__ add__5.0__4.0__ multiply__1.0__73.0__ |
subtract__15.0__5.0__ divide__10.0__5.0__ subtract__5.0__4.0__ add__5.0__4.0__ multiply__1.0__73.0__ |
| the wages of num__24 men and num__16 women amount to num__11600 per day . half the number of men and num__37 women has same money . the daily wages paid to each man is <o> a ) num__223 <o> b ) num__350 <o> c ) num__31887 <o> d ) num__2797 <o> e ) num__911 |
num__24 m + num__16 w = num__11600 num__12 m + num__37 w = num__11600 solving we get num__12 m = num__21 w substituting in the first equation we get num__42 w + num__16 w = num__11600 ⇒ ⇒ w = num__200 m = num__350 answer : b <eor> b <eos> |
b |
subtract__37.0__16.0__ round__350.0__ |
subtract__37.0__16.0__ round__350.0__ |
| num__1397 x num__1397 = ? <o> a ) num__1951000 <o> b ) num__1951500 <o> c ) num__1951509 <o> d ) num__1951511 <o> e ) num__1951609 |
num__1397 x num__1397 = ( num__1397 ) ^ num__2 = ( num__1400 - num__3 ) ^ num__2 = ( num__1400 ) num__2 + ( num__3 ) num__2 - ( num__2 x num__1400 x num__3 ) = num__1960000 + num__9 - num__8400 = num__1960009 - num__8400 = num__1951609 . e ) <eor> e <eos> |
e |
subtract__1400.0__1397.0__ add__1960000.0__9.0__ subtract__1960009.0__8400.0__ subtract__1960009.0__8400.0__ |
subtract__1400.0__1397.0__ add__1960000.0__9.0__ subtract__1960009.0__8400.0__ subtract__1960009.0__8400.0__ |
| if n is an integer from num__1 to num__96 ( inclusive ) what is the probability for n * ( n + num__1 ) * ( n + num__2 ) being divisible by num__8 ? <o> a ) num__25.0 <o> b ) num__50.0 <o> c ) num__62.5 <o> d ) num__72.5 <o> e ) num__75 % |
( a ) num__25.0 to be divisible by num__8 the number needs to have num__3 num__2 s in it . only a multiple of num__4 can provide that . number of numbers divisible by num__4 = num__24.0 = num__24 . so p ( num__8 ) = num__0.25 = num__25.0 . <eor> a <eos> |
a |
add__1.0__2.0__ add__1.0__3.0__ divide__96.0__4.0__ reverse__4.0__ add__1.0__24.0__ |
add__1.0__2.0__ add__1.0__3.0__ divide__96.0__4.0__ reverse__4.0__ add__1.0__24.0__ |
| num__36 workers can reap a field in num__6 days . if the work is to be completed in num__4 days the extra workers required are <o> a ) num__10 <o> b ) num__25 <o> c ) num__18 <o> d ) num__15 <o> e ) num__16 |
explanation : m num__1 d num__1 = m num__2 d num__2 num__36 ( num__6 ) = x ( num__4 ) = num__54 extra required workers are num__54 - num__36 = num__18 answer : option c <eor> c <eos> |
c |
subtract__6.0__4.0__ divide__36.0__2.0__ round__18.0__ |
subtract__6.0__4.0__ subtract__54.0__36.0__ subtract__36.0__18.0__ |
| ten years ago p was half of q ' s age . if the ratio of their present ages is num__3 : num__4 what will be the total of their present ages ? <o> a ) num__45 <o> b ) num__40 <o> c ) num__35 <o> d ) num__30 <o> e ) num__25 |
let present age of p and q be num__3 x and num__4 x respectively . ten years ago p was half of q ' s age ⇒ ( num__3 x − num__10 ) = num__1 ( num__4 x − num__10 ) / num__2 ⇒ num__6 x − num__20 = num__4 x − num__10 ⇒ num__2 x = num__10 ⇒ x = num__5 total of their present ages num__3 x + num__4 x = num__7 x = num__7 * num__5 = num__35 answer : c <eor> c <eos> |
c |
subtract__4.0__3.0__ subtract__3.0__1.0__ multiply__3.0__2.0__ multiply__2.0__10.0__ add__3.0__2.0__ add__3.0__4.0__ multiply__5.0__7.0__ multiply__1.0__35.0__ |
subtract__4.0__3.0__ subtract__3.0__1.0__ multiply__3.0__2.0__ multiply__2.0__10.0__ add__3.0__2.0__ add__3.0__4.0__ multiply__5.0__7.0__ multiply__1.0__35.0__ |
| a rectangular wall is covered entirely with two kinds of decorative tiles : regular and jumbo . num__0.333333333333 of the tiles are jumbo tiles which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles . if regular tiles cover num__60 square feet of the wall and no tiles overlap what is the area of the entire wall ? <o> a ) num__160 <o> b ) num__240 <o> c ) num__330 <o> d ) num__440 <o> e ) num__560 |
the number of jumbo tiles = x . the number of regular tiles = num__2 x . assume the ratio of the dimensions of a regular tile is a : a - - > area = a ^ num__2 . the dimensions of a jumbo tile is num__3 a : num__3 a - - > area = num__9 a ^ num__2 . the area of regular tiles = num__2 x * a ^ num__2 = num__60 . the area of jumbo tiles = x * num__9 a ^ num__2 = num__4.5 ( num__2 x * a ^ num__2 ) = num__4.5 * num__60 = num__270 . total area = num__60 + num__270 = num__330 . answer : c . <eor> c <eos> |
c |
power__3.0__2.0__ multiply__60.0__4.5__ triangle_area__2.0__330.0__ |
power__3.0__2.0__ multiply__60.0__4.5__ triangle_area__2.0__330.0__ |
| n * is defined as num__0.5 ( n + num__10 ) for any number n . what is n if ( ( n * ) * ) * = num__15 ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__40 <o> d ) num__50 <o> e ) num__60 |
n * = num__0.5 ( n + num__10 ) ( n * ) * = num__0.5 ( num__0.5 n + num__5 + num__10 ) = num__0.25 n + num__7.5 ( ( n * ) * ) * = num__0.5 ( num__0.25 n + num__7.5 + num__10 ) = num__0.125 n + num__8.75 we know that ( ( n * ) * ) * = num__15 num__0.125 n + num__8.75 = num__15 num__0.125 n = num__6.25 n = num__50.0 = num__50 answer : d <eor> d <eos> |
d |
multiply__0.5__10.0__ multiply__0.5__15.0__ multiply__0.5__0.25__ subtract__15.0__8.75__ multiply__10.0__5.0__ multiply__10.0__5.0__ |
multiply__0.5__10.0__ multiply__0.5__15.0__ multiply__0.5__0.25__ subtract__15.0__8.75__ multiply__10.0__5.0__ multiply__10.0__5.0__ |
| in an election between two candidates num__10.0 of votes are were declares invalid . first candidate got num__1800 votes which were num__60.0 of the total valid votes . the total number of votes enrolled in that election was : <o> a ) num__9000 <o> b ) num__9900 <o> c ) num__10000 <o> d ) num__11000 <o> e ) num__12000 |
num__100.0 - num__10.0 = num__90.0 num__54.0 - num__36.0 = num__18.0 num__18.0 - - - - - - > num__1800 ( num__18 × num__100 = num__1800 ) num__100.0 - - - - - - - > num__10000 votes ( num__100 × num__100 = num__10000 ) c <eor> c <eos> |
c |
percent__60.0__90.0__ percent__100.0__10000.0__ |
percent__60.0__90.0__ percent__100.0__10000.0__ |
| a man drive his car to the office at num__20 miles / hr . after reaching office he realize that its a new year holiday so he went back home at a speed of num__30 miles / hr . discounting the time spent in the stoppage what was his average speed of his journey ? <o> a ) num__21 <o> b ) num__20 <o> c ) num__25 <o> d ) num__24 <o> e ) num__29 |
d num__24 miles / hr ( not num__25 miles / hr which might be guessed by many ) d : distance traveled t num__1 : time of going to office t num__2 : time of returning back y : average speed d = num__20 * t num__1 t num__1 = d / num__20 d = num__30 * t num__2 t num__2 = d / num__30 num__2 d = y ( t num__2 + t num__1 ) num__2 d = y ( d / num__20 + d / num__30 ) num__2 d = y ( num__3 d / num__60 + num__2 d / num__60 ) num__2 d = y ( num__5 d / num__60 ) y = num__120 d / num__5 d = > y = num__24 <eor> d <eos> |
d |
subtract__25.0__24.0__ add__1.0__2.0__ hour_to_min_conversion__ subtract__30.0__25.0__ multiply__2.0__60.0__ round__24.0__ |
subtract__25.0__24.0__ add__1.0__2.0__ multiply__20.0__3.0__ add__2.0__3.0__ multiply__2.0__60.0__ divide__120.0__5.0__ |
| pat is now num__22 years older than dan . if in num__6 years pat will be twice as old as dan how old will pat be in num__15 years ? <o> a ) num__47 <o> b ) num__49 <o> c ) num__51 <o> d ) num__53 <o> e ) num__55 |
p = d + num__22 so d = p - num__22 . p + num__6 = num__2 ( d + num__6 ) . p + num__6 = num__2 ( p - num__22 + num__6 ) . p + num__6 = num__2 p - num__32 . p = num__38 . in num__15 years pat will be num__53 years old . the answer is d . <eor> d <eos> |
d |
add__6.0__32.0__ add__15.0__38.0__ add__15.0__38.0__ |
add__6.0__32.0__ add__15.0__38.0__ add__15.0__38.0__ |
| the side of a rhombus is num__26 m and length of one of its diagonals is num__20 m . the area of the rhombus is ? <o> a ) num__299 <o> b ) num__278 <o> c ) num__278 <o> d ) num__480 <o> e ) num__281 |
num__262 – num__102 = num__242 d num__1 = num__20 d num__2 = num__48 num__0.5 * num__20 * num__48 = num__480 answer : d <eor> d <eos> |
d |
triangle_perimeter__26.0__20.0__2.0__ triangle_area__20.0__48.0__ triangle_area__20.0__48.0__ |
triangle_perimeter__26.0__20.0__2.0__ volume_rectangular_prism__20.0__0.5__48.0__ multiply__1.0__480.0__ |
| calculate the average of first num__14 even numbers is ? <o> a ) num__11 <o> b ) num__15 <o> c ) num__16 <o> d ) num__14 <o> e ) num__12 |
explanation : sum of num__14 even numbers = num__14 * num__15 = num__210 average = num__15.0 = num__15 answer : option b <eor> b <eos> |
b |
multiply__14.0__15.0__ divide__210.0__14.0__ |
multiply__14.0__15.0__ divide__210.0__14.0__ |
| if num__10 ^ ( z - num__1 ) < num__0.000125 < num__10 ^ z what is the value of an integer z ? <o> a ) - num__4 <o> b ) - num__3 <o> c ) - num__2 <o> d ) num__3 <o> e ) num__4 |
- > multiply num__10 ^ num__6 - > ( num__10 ^ num__6 ) { num__10 ^ ( z - num__1 ) } < num__125 < ( num__10 ^ num__6 ) ( num__10 ^ z ) - > num__125 is bigger than num__100 - > ( num__10 ^ num__6 ) ( num__10 ^ ( z - num__1 ) ) = num__100 - > num__10 ^ ( num__6 + z - num__1 ) = num__10 ^ num__2 z + num__5 = num__2 - > z = - num__3 thus the answer is b <eor> b <eos> |
b |
divide__10.0__2.0__ add__1.0__2.0__ add__1.0__2.0__ |
subtract__6.0__1.0__ add__1.0__2.0__ add__1.0__2.0__ |
| without any stoppage a person travels a certain distance at an average speed of num__42 km / h and with stoppages he covers the same distance at an average speed of num__28 km / h . how many minutes per hour does he stop ? <o> a ) num__30 minutes <o> b ) num__28 minutes <o> c ) num__11 minutes <o> d ) num__17 minutes <o> e ) num__20 minutes |
let the total distance to be covered is num__84 kms . time taken to cover the distance without stoppage = num__2.0 hrs = num__2 hrs time taken to cover the distance with stoppage = num__3.0 = num__3 hrs . thus he takes num__60 minutes to cover the same distance with stoppage . therefore in num__1 hour he stops for num__20 minutes . answer : e <eor> e <eos> |
e |
divide__84.0__42.0__ divide__84.0__28.0__ hour_to_min_conversion__ subtract__3.0__2.0__ divide__60.0__3.0__ round__20.0__ |
divide__84.0__42.0__ divide__84.0__28.0__ hour_to_min_conversion__ subtract__3.0__2.0__ divide__60.0__3.0__ round__20.0__ |
| what is the smallest positive integer that can be multiplied by num__1008 to make it a perfect square ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__11 |
num__1008 = num__2 ^ num__4 x num__3 ^ num__2 x num__7 . therefore the smallest integer to multiplied to num__1008 to make it a perfect square is num__7 . answer d . <eor> d <eos> |
d |
triangle_area__2.0__7.0__ |
triangle_area__2.0__7.0__ |
| of the final grades received by the students in a certain math course num__0.2 are a ' s num__0.25 are b ' s num__0.5 are c ' s and the remaining num__25 grades are d ' s . what is the number of students in the course ? <o> a ) num__80 <o> b ) num__110 <o> c ) num__160 <o> d ) num__500 <o> e ) num__400 |
we start by creating a variable for the total number of students in the math course . we can say : t = total number of students in the math course next we can use variable t in an equation that we translate from the given information . we are given that of the final grades received by the students in a certain math course num__0.2 are a ' s num__0.25 are b ' s num__0.5 are c ' s and the remaining num__25 grades are d ' s . since this represents all the grades in the class it represents all the students in the class . thus we know : # a ’ s + # b ’ s + # c ’ s + # d ’ s = total number of students in the class num__0.2 ( t ) + ¼ ( t ) + ½ ( t ) + num__25 = t we can multiply the entire equation by num__20 to cancel out the denominators of the fractions and we have : num__4 t + num__5 t + num__10 t + num__500 = num__20 t num__19 t + num__500 = num__20 t num__500 = t there are a total of num__500 students in the math class . answer is d . <eor> d <eos> |
d |
reverse__0.25__ reverse__0.2__ multiply__0.5__20.0__ multiply__25.0__20.0__ multiply__25.0__20.0__ |
reverse__0.25__ reverse__0.2__ multiply__0.5__20.0__ multiply__25.0__20.0__ multiply__25.0__20.0__ |
| there are num__300 female managers in a certain company . find the total number of female employees in the company if num__0.4 of all the employees are managers and num__0.4 of all male employees are managers . <o> a ) num__650 <o> b ) num__700 <o> c ) num__750 <o> d ) num__800 <o> e ) none of these |
as per question stem num__0.4 m ( portion of men employees who are managers ) + num__200 ( portion of female employees who are managers ) = num__0.4 t ( portion of total number of employees who are managers ) thus we get that num__0.4 m + num__300 = num__0.4 t or num__0.4 ( t - m ) = num__300 from here we get that t - m = num__750 that would be total number of female employees and the answer ( c ) <eor> c <eos> |
c |
divide__300.0__0.4__ divide__300.0__0.4__ |
divide__300.0__0.4__ divide__300.0__0.4__ |
| a car is running at a speed of num__90 kmph . what distance will it cover in num__15 second ? <o> a ) num__100 m <o> b ) num__255 m <o> c ) num__375 m <o> d ) can not be determined <o> e ) none of these |
explanation : given : speed = num__108 kmph = ( num__90 x ( num__0.277777777778 ) ) m / sec = num__25 m / sec distance covered in num__15 second = ( num__25 x num__15 ) m = num__375 m . answer : c <eor> c <eos> |
c |
multiply__15.0__25.0__ round__375.0__ |
multiply__15.0__25.0__ round__375.0__ |
| two years ago claire put $ num__2000 into a savings account . at the end of the first year her account had accrued $ num__100 in interest bringing her total balance to $ num__2100 . the next year her account balance increased by num__10.0 . at the end of the two years by what percent has claire ' s account balance increased from her initial deposit of $ num__2000 ? <o> a ) num__10.0 <o> b ) num__12.0 <o> c ) num__15.5 <o> d ) num__17.0 <o> e ) num__20 % |
investment num__2000 dollars num__1 st year total gained = num__100 total amount end of first year = num__2100 second year account increased by num__10.0 = num__2100 * num__0.1 = num__210 therefore total amount by second year end = num__2310 so total percentage increase in money = ( num__2310 - num__2000 ) * num__0.05 = num__15.5 correct answer c = num__15.5 <eor> c <eos> |
c |
percent__10.0__1.0__ percent__10.0__2100.0__ percent__100.0__15.5__ |
percent__10.0__1.0__ percent__10.0__2100.0__ percent__100.0__15.5__ |
| the sum of two numbers is num__120 . if the greater number is four times the less what are the numbers ? <o> a ) num__80 <o> b ) num__84 <o> c ) num__85 <o> d ) num__96 <o> e ) none |
let x = the less number num__4 x = the greater number the sum of two numbers is num__120 x + num__4 x = num__120 num__5 x = num__120 x = num__24 num__4 x = num__96 option d <eor> d <eos> |
d |
divide__120.0__5.0__ subtract__120.0__24.0__ subtract__120.0__24.0__ |
divide__120.0__5.0__ subtract__120.0__24.0__ subtract__120.0__24.0__ |
| a man can row num__6 kmph in still water . when the river is running at num__1.2 kmph it takes him num__1 hour to row to a place and black . what is the total distance traveled by the man ? <o> a ) num__5.78 <o> b ) num__5.79 <o> c ) num__5.76 <o> d ) num__5.74 <o> e ) num__5.721 |
m = num__6 s = num__1.2 ds = num__7.2 us = num__4.8 x / num__7.2 + x / num__4.8 = num__1 x = num__2.88 d = num__2.88 * num__2 = num__5.76 . answer : c <eor> c <eos> |
c |
add__6.0__1.2__ subtract__6.0__1.2__ multiply__1.2__4.8__ round__5.76__ |
add__6.0__1.2__ subtract__6.0__1.2__ multiply__1.2__4.8__ multiply__1.2__4.8__ |
| what percent is num__120 of num__90 ? <o> a ) num__133 num__0.333333333333 % <o> b ) num__134 num__0.333333333333 % <o> c ) num__135 num__0.333333333333 % <o> d ) num__140 num__0.333333333333 % <o> e ) num__143 num__0.333333333333 % |
num__1.33333333333 = num__1.33333333333 num__1.33333333333 × num__100 = num__133.333333333 = num__133 num__0.333333333333 % a <eor> a <eos> |
a |
percent__100.0__133.0__ |
percent__100.0__133.0__ |
| it takes avery num__2.5 hours to build a brick wall while tom can do it in num__5 hours . if the two start working together and after an hour avery leaves how much time will it take tom to complete the wall on his own ? <o> a ) num__15 minutes . <o> b ) num__30 minutes . <o> c ) num__1 hour and num__30 minutes . <o> d ) num__1 hour and num__40 minutes <o> e ) num__2 hours |
avery ' s efficiency is num__100 / num__2.5 = num__40.0 tom ' s = num__20.0 = num__20.0 they worked together for num__1 hour and finished num__60.0 of the job remaining = num__40.0 tom will complete num__20.0 in num__60 minutes num__40.0 in num__120 minutes time taken by tom to finish the remaining on his own = num__120 minutes answer : e <eor> e <eos> |
e |
divide__100.0__2.5__ divide__100.0__5.0__ hour_to_min_conversion__ add__100.0__20.0__ divide__5.0__2.5__ |
divide__100.0__2.5__ divide__100.0__5.0__ hour_to_min_conversion__ add__100.0__20.0__ divide__5.0__2.5__ |
| the dimensions of a room are num__25 feet * num__15 feet * num__12 feet . what is the cost of white washing the four walls of the room at rs . num__7 per square feet if there is one door of dimensions num__6 feet * num__3 feet and three windows of dimensions num__4 feet * num__3 feet each ? <o> a ) num__4000 <o> b ) num__345 <o> c ) num__5673 <o> d ) num__6342 <o> e ) num__4566 |
area of the four walls = num__2 h ( l + b ) since there are doors and windows area of the walls = num__2 * num__12 ( num__15 + num__25 ) - ( num__6 * num__3 ) - num__3 ( num__4 * num__3 ) = num__906 sq . ft . total cost = num__906 * num__7 = rs . num__6342 answer : option d <eor> d <eos> |
d |
multiply__7.0__906.0__ multiply__7.0__906.0__ |
multiply__7.0__906.0__ multiply__7.0__906.0__ |
| calculate the distance covered by num__420 revolutions of a wheel of radius num__23.1 cm . <o> a ) num__609.84 m <o> b ) num__409.84 m <o> c ) num__509.84 m <o> d ) num__109.84 m <o> e ) num__709.84 m |
in one resolution the distance covered by the wheel is its own circumference . distance covered in num__500 resolutions . = num__420 * num__2 * num__3.14285714286 * num__23.1 = num__60984 cm = num__609.84 m answer : a <eor> a <eos> |
a |
round__609.84__ |
round__609.84__ |
| sawyer is mixing up a salad dressing . regardless of the number of servings the recipe requires that num__0.625 of the finished dressing mix be peanut oil num__0.25 vinegar and the remainder an even mixture of salt pepper and sugar . if sawyer accidentally doubles the vinegar and forgets the sugar altogether what proportion of the botched dressing will be peanut oil ? <o> a ) num__0.51724137931 <o> b ) num__0.625 <o> c ) num__0.3125 <o> d ) num__0.5 <o> e ) num__0.481481481481 |
peanut oil = num__0.625 = num__0.625 - - > num__15 parts out of num__24 ; vinegar = num__0.25 = num__0.25 - - > num__6 parts out of num__24 ; salt + pepper + sugar = num__1 - ( num__0.625 + num__0.25 ) = num__0.125 so each = num__0.0416666666667 - - > num__1 part out of num__24 each ; if vinegar = num__12 ( instead of num__6 ) and sugar = num__0 ( instead of num__1 ) then total = num__15 + num__12 + num__1 + num__1 + num__0 = num__29 parts out of which num__15 parts are peanut oil - - > proportion = num__0.51724137931 . answer : a . <eor> a <eos> |
a |
divide__15.0__0.625__ multiply__0.25__24.0__ reverse__24.0__ round_down__0.625__ divide__15.0__29.0__ multiply__1.0__0.5172__ |
divide__15.0__0.625__ multiply__0.25__24.0__ reverse__24.0__ round_down__0.625__ divide__15.0__29.0__ multiply__1.0__0.5172__ |
| the ratio of the area of a square to that of the square drawn on its diagonal is ? <o> a ) num__1 : num__2 <o> b ) num__2 : num__4 <o> c ) num__3 : num__5 <o> d ) num__4 : num__6 <o> e ) num__5 : num__7 |
explanation : a ^ num__2 : ( a √ num__2 ) ^ num__2 a ^ num__2 : num__2 a ^ num__2 = num__1 : num__2 answer : a <eor> a <eos> |
a |
volume_cube__1.0__ |
power__1.0__2.0__ |
| a box contains nine bulbs out of which num__4 are defective . if four bulbs are chosen at random find the probability that atleast one bulb is good . <o> a ) num__0.952380952381 <o> b ) num__0.992063492063 <o> c ) num__0.00793650793651 <o> d ) num__0.0449438202247 <o> e ) num__0.0471204188482 |
explanation : required probability = num__1 - num__0.00793650793651 = num__0.992063492063 answer b <eor> b <eos> |
b |
negate_prob__0.0079__ negate_prob__0.0079__ |
negate_prob__0.0079__ negate_prob__0.0079__ |
| if p ( x ) = ax ^ num__4 + bx ^ num__3 + cx ^ num__2 + dx + e has roots at x = num__1 num__2 num__3 num__4 and p ( num__0 ) = num__48 what is p ( num__5 ) ? <o> a ) num__48 <o> b ) num__24 <o> c ) num__0 <o> d ) num__50 <o> e ) num__40 |
roots are num__12 num__34 so we can write this equation as p ( x ) = a ( x - num__1 ) ( x - num__2 ) ( x - num__3 ) ( x - num__4 ) . . . . . eq num__1 put x = num__0 here we p ( num__0 ) = num__48 = num__24 a then a = num__2 now put a in eq num__1 and get the answer . . . p ( num__5 ) = a . num__4.3 . num__2.1 = num__48 answer : a <eor> a <eos> |
a |
multiply__4.0__3.0__ multiply__2.0__12.0__ multiply__4.0__12.0__ |
multiply__4.0__3.0__ multiply__2.0__12.0__ multiply__4.0__12.0__ |
| if | x - num__9 | = num__3 x then x = ? <o> a ) num__1.5 <o> b ) num__3.1 <o> c ) num__4.5 <o> d ) - num__2.25 <o> e ) - num__4.5 |
| x - num__9 | = num__3 x . . . ( given ) x ^ num__2 - num__18 x + num__81 = num__9 x ^ num__2 num__8 * x ^ num__2 + num__18 * x - num__81 = num__0 . . . . ( by solving above eq . we get ) x = - num__4.5 or num__2.25 = = = > ans - e <eor> e <eos> |
e |
multiply__9.0__2.0__ divide__9.0__2.0__ divide__4.5__2.0__ subtract__9.0__4.5__ |
multiply__9.0__2.0__ divide__9.0__2.0__ divide__4.5__2.0__ subtract__9.0__4.5__ |
| if two positive numbers are in the ratio num__0.125 : num__0.142857142857 then by what percent is the second number more than the first ? <o> a ) num__70.0 <o> b ) num__14.28 <o> c ) num__60.0 <o> d ) num__50.0 <o> e ) num__65 % |
given ratio = num__0.125 : num__0.333333333333 = num__3 : num__8 let first number be num__7 x and the second number be num__8 x . the second number is more than first number by num__1 x . required percentage = num__1 x / num__7 x * num__100 = num__14.28 . answer : b <eor> b <eos> |
b |
reverse__0.125__ multiply__0.125__8.0__ multiply__1.0__14.28__ |
reverse__0.125__ multiply__0.125__8.0__ multiply__1.0__14.28__ |
| a warehouse is labeling computer inventory with num__3 - digit codes . each code is to consist of four digits between the numbers num__0 and num__8 . the manager is debating whether to allow any num__3 - digit codes to have leading zero ' s before a non - zero number is used . if not allowed then numbers like num__0025 cannot be used . the manager wants to measure the magnitude of the number of num__3 - digit code possibilities that are lost if he disallows the use of leading zero ' s . how many such codes would be lost ? <o> a ) num__981 <o> b ) num__1000 <o> c ) num__1011 <o> d ) num__1101 <o> e ) num__200 |
the number of codes possible if leading zero is allowed = num__10 * num__10 * num__10 = num__10 ^ num__3 the number of codes possible if leading zero is not allowed = num__8 * num__10 * num__10 = num__9 * num__10 ^ num__2 the difference = num__10 ^ num__3 - num__8 * num__10 ^ num__2 = num__10 ^ num__2 ( num__10 - num__8 ) = num__2 * num__10 ^ num__2 . answer : e . similar questions to practice : hope it helps . <eor> e <eos> |
e |
subtract__10.0__8.0__ multiply__8.0__25.0__ |
subtract__10.0__8.0__ multiply__8.0__25.0__ |
| find two integers neither of which ends in a zero and whose product is exactly num__1 num__00000 <o> a ) num__64 and num__15625 <o> b ) num__32 and num__3125 <o> c ) num__64 and num__15620 <o> d ) num__64 and num__15635 <o> e ) num__64 and num__16 |
625 |
num__1 num__00000 = num__10 ^ num__5 = num__10 x num__10 x num__10 x num__10 x num__10 = ( num__2 x num__5 ) x ( num__2 x num__5 ) x ( num__2 x num__5 ) x ( num__2 x num__5 ) x ( num__2 x num__5 ) = ( num__2 ^ num__5 ) x ( num__5 ^ num__5 ) = num__32 x num__3125 so the numbers are num__32 and num__3125 answer : b <eor> b <eos> |
b |
b |
| how many points ( x y ) lie on the line segment between ( num__22 num__12 num__0.666666666667 ) and ( num__16 num__17 num__0.666666666667 ) such that x and y are both integers ? <o> a ) num__1 <o> b ) num__5 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
slope = ( num__17 num__0.666666666667 - num__12 num__0.666666666667 ) / ( num__16 - num__22 ) = - num__0.833333333333 y = mx + b = > num__12 num__0.666666666667 = - num__22 * num__0.833333333333 + b = > b = num__31 y = - num__5 x / num__6 + num__31 only integer values work and the only multiples of num__6 between num__16 and num__22 for x values is num__18 thus num__1 point . a <eor> a <eos> |
a |
subtract__22.0__17.0__ subtract__22.0__16.0__ add__12.0__6.0__ subtract__17.0__16.0__ reverse__1.0__ |
subtract__22.0__17.0__ subtract__22.0__16.0__ add__12.0__6.0__ subtract__17.0__16.0__ reverse__1.0__ |
| on sunday bill ran num__4 more miles than he ran on saturday . julia did not run on saturday but she ran twice the number of miles on sunday that bill ran on sunday . if bill and julia ran a total of num__20 miles on saturday and sunday how many miles did bill run on sunday ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
let bill run x on saturday so he will run x + num__4 on sunday . . julia will run num__2 * ( x + num__4 ) on sunday . . totai = x + x + num__4 + num__2 x + num__8 = num__20 . . num__4 x + num__12 = num__20 . . x = num__2 . . ans = x + num__4 = num__2 + num__4 = num__6 answer b <eor> b <eos> |
b |
multiply__4.0__2.0__ add__4.0__8.0__ add__4.0__2.0__ round__6.0__ |
multiply__4.0__2.0__ add__4.0__8.0__ add__4.0__2.0__ add__4.0__2.0__ |
| a dishonest dealer professes to sell his goods at cost price but still gets num__20.0 profit by using a false weight . what weight does he substitute for a kilogram ? <o> a ) num__833 num__0.142857142857 <o> b ) num__833 num__0.333333333333 <o> c ) num__833 num__0.5 <o> d ) num__833 num__0.111111111111 <o> e ) num__833 num__0.2 |
if the cost price is rs . num__100 then to get a profit of num__20.0 the selling price should be rs . num__120 . if num__120 kg are to be sold and the dealer gives only num__100 kg to get a profit of num__20.0 . how many grams he has to give instead of one kilogram ( num__1000 gm ) . num__120 gm - - - - - - num__100 gm num__1000 gm - - - - - - ? ( num__1000 * num__100 ) / num__120 = num__833.333333333 = num__833 num__0.333333333333 grams . answer : b <eor> b <eos> |
b |
percent__100.0__833.0__ |
percent__100.0__833.0__ |
| the price of num__3 pens and num__6 pencils is rs . num__1500 . with the same money one can buy num__1 pen and num__12 pencils . if one wants to buy num__23 pencils how much shall he have to pay ? <o> a ) num__2400 <o> b ) num__2300 <o> c ) num__2200 <o> d ) num__2100 <o> e ) num__2000 |
let the price of a pen and a pencil be rs . x and rs . y respectively . then num__3 x + num__6 y = num__1500 . . . . ( i ) and x + num__12 y = num__1500 . . . . ( ii ) divide equation ( i ) by num__3 we get the below equation . = x + num__2 y = num__500 . - - - ( iii ) now subtract ( iii ) from ( ii ) x + num__12 y = num__1500 ( - ) x + num__2 y = num__500 - - - - - - - - - - - - - - - - num__10 y = num__1000 - - - - - - - - - - - - - - - - y = num__100 cost of num__23 pencils = num__23 * num__100 = num__2300 answer : b <eor> b <eos> |
b |
subtract__3.0__1.0__ divide__1500.0__3.0__ subtract__12.0__2.0__ subtract__1500.0__500.0__ divide__1000.0__10.0__ multiply__23.0__100.0__ multiply__1.0__2300.0__ |
subtract__3.0__1.0__ divide__1500.0__3.0__ subtract__12.0__2.0__ subtract__1500.0__500.0__ divide__1000.0__10.0__ multiply__23.0__100.0__ multiply__1.0__2300.0__ |
| num__0.40 represents what percent of num__0.008 ? <o> a ) num__0.05 <o> b ) num__0.5 <o> c ) num__5.0 <o> d ) num__500.0 <o> e ) num__5000 % |
one more method num__0.40 represents what percent of num__0.008 ? adjusting the decimal num__400 represents what percent of num__8 ? divide by num__8 num__50 represents what percent of num__1 ? answer = num__50 * num__100 = num__5000.0 = e <eor> e <eos> |
e |
percent__100.0__5000.0__ |
percent__100.0__5000.0__ |
| a num__10 liter mixture of milk and water contains num__30 percent water . two liters of this mixture is taken away . how many liters of water should now be added so that the amount of milk in the mixture is double that of water ? <o> a ) num__3.3 <o> b ) num__3.4 <o> c ) num__0.4 <o> d ) num__4.9 <o> e ) num__3.87 |
explanation : two liters were taken away so we have only num__8 liters of mixture . amount of milk in num__8 liters of mixture = num__8 × num__70.0 = num__5.6 liters amount of water in num__8 lit of mix = num__8 - num__5.6 = num__2.4 liters . half of milk i . e half of num__5.6 = num__2.8 liters . we need ( num__2.8 - num__2.4 ) liters water more = num__0.4 lit answer : c <eor> c <eos> |
c |
subtract__8.0__5.6__ subtract__2.8__2.4__ subtract__2.8__2.4__ |
subtract__8.0__5.6__ subtract__2.8__2.4__ subtract__2.8__2.4__ |
| the circumference of the front wheel of a cart is num__30 ft long and that of the back wheel is num__33 ft long . what is the distance traveled by the cart when the front wheel has done five more revolutions than the rear wheel ? <o> a ) num__20 ft <o> b ) num__25 ft <o> c ) num__750 ft <o> d ) num__900 ft <o> e ) num__1650 ft |
point to note : both the wheels would have traveled the same distance . now consider no . of revolutions made by back wheel as x which implies that the number of revolutions made by the front wheel is ( x + num__5 ) . equating the distance traveled by front wheel to back wheel : ( x + num__5 ) * num__30 = x * num__33 . ( formula for calculating the distance traveled by each wheel is : # of revolutions * circumference . ) solving this eqn . gives x = num__50 . sub x = num__50 either in ( x + num__5 ) * num__30 or in x * num__33 to get the distance which is num__1650 . so the correct choice is e . <eor> e <eos> |
e |
multiply__33.0__50.0__ round__1650.0__ |
multiply__33.0__50.0__ multiply__33.0__50.0__ |
| num__60 percent of movie theatres in town x have num__2 screens or less . num__20.0 of those theatres sell an average of more than $ num__300 worth of popcorn per showing . num__56 percent of all the movie theatres in town x sell $ num__300 or less of popcorn per showing . what percent of all the stores on the street have num__4 or more screens and sell an average of more than $ num__300 worth of popcorn per day ? <o> a ) num__12 <o> b ) num__18 <o> c ) num__32 <o> d ) num__40 <o> e ) num__44 |
lets take numbers here . assume that the total number of movie theaters in the town = num__100 then number of movie theaters with num__3 screens or less = num__60 = > number of movie theaters with num__4 screens or more = num__40 movie theaters with num__3 screens or less selling popcorn at more than $ num__300 = num__20.0 of num__60 = num__12 number of movie theaters selling popcorn at $ num__300 or less = num__56 = > number of movie theaters selling popcorn at more than $ num__300 = num__100 - num__56 = num__44 of these num__44 theaters num__12 are those with num__3 screens or less therefore num__40 ( num__44 - num__12 ) must be those with four screens or more d is the answer <eor> d <eos> |
d |
divide__60.0__20.0__ subtract__60.0__20.0__ multiply__4.0__3.0__ subtract__56.0__12.0__ subtract__60.0__20.0__ |
divide__60.0__20.0__ subtract__60.0__20.0__ multiply__4.0__3.0__ subtract__56.0__12.0__ subtract__60.0__20.0__ |
| in the xy - plane the points ( c d ) ( c - d ) and ( - c - d ) are three vertices of a certain square . if c < num__0 and d > num__0 which of the following points r is in the same quadrant as the fourth vertex of the square ? <o> a ) ( - num__5 - num__3 ) <o> b ) ( - num__5 num__3 ) <o> c ) ( num__5 - num__3 ) <o> d ) ( num__3 - num__5 ) <o> e ) ( num__3 num__5 ) |
the question : in the xy - plane the points ( c d ) ( c - d ) and ( - c - d ) are three vertices of a certain square . if c < num__0 and d > num__0 which of the following points r is in the same quadrant as the fourth vertex of the square ? i marked the tricky part in red . it seems c is anegativenumber and d is a positive number . this means vertex # num__1 = ( c d ) is in qii ( that is negative x and positive y ) vertex # num__2 = ( c - d ) is in qiii ( that is both xy negative ) vertex # num__3 = ( - c - d ) is in qiv ( that is y is negative but x is positive ) that means the last vertex should be in the first quadrant - - - the only first quadrant point is ( num__5 num__3 ) answer = e . <eor> e <eos> |
e |
add__1.0__2.0__ add__2.0__3.0__ add__1.0__2.0__ |
add__1.0__2.0__ add__2.0__3.0__ subtract__5.0__2.0__ |
| solve ( num__0.76 × num__0.76 × num__0.76 − num__0.008 ) / ( num__0.76 × num__0.76 + num__0.76 × num__0.2 + num__0.04 ) <o> a ) num__0.56 <o> b ) num__0.58 <o> c ) num__0.6 <o> d ) num__0.57 <o> e ) num__0.59 |
num__0.56 option ' a ' <eor> a <eos> |
a |
subtract__0.76__0.2__ subtract__0.76__0.2__ |
subtract__0.76__0.2__ subtract__0.76__0.2__ |
| the length of a rectangle is four times its width . if the area is num__100 m num__2 what is the length of the rectangle ? <o> a ) num__20 <o> b ) num__40 <o> c ) num__50 <o> d ) num__80 <o> e ) num__60 |
let l be the length and w be the width of the rectangle . hence l = num__4 w we now use the area to write num__100 = l ? w substitute l by num__4 w in the equation above num__100 = num__4 w ? w = num__4 w num__2 solve for w and find l num__4 w num__2 = num__100 w num__2 = num__25 w = num__5 and l = num__4 w = num__20 m correct answer a <eor> a <eos> |
a |
square_perimeter__5.0__ square_perimeter__5.0__ |
square_perimeter__5.0__ square_perimeter__5.0__ |
| how many integral divisors does the number num__120 have ? <o> a ) num__14 <o> b ) num__16 <o> c ) num__12 <o> d ) num__20 <o> e ) none of these |
explanatory answer express the number in terms of its prime factors num__120 = num__2 ( num__3 ) * num__3 * num__5 . the three prime factors are num__2 num__3 and num__5 . the powers of these prime factors are num__3 num__1 and num__1 respectively . find the number of factors as follows to find the number of factors / integral divisors that num__120 has increment the powers of each of the prime factors by num__1 and then multiply them . number of factors = ( num__3 + num__1 ) * ( num__1 + num__1 ) * ( num__1 + num__1 ) = num__4 * num__2 * num__2 = num__16 . choice b <eor> b <eos> |
b |
add__2.0__3.0__ subtract__3.0__2.0__ add__1.0__3.0__ power__2.0__4.0__ multiply__1.0__16.0__ |
add__2.0__3.0__ subtract__3.0__2.0__ add__1.0__3.0__ power__2.0__4.0__ multiply__1.0__16.0__ |
| num__30 quintals is what percent of num__2 metric tonnes ? <o> a ) num__15.0 <o> b ) num__1.5 <o> c ) num__150.0 <o> d ) num__30.0 <o> e ) none |
answer required percent = { num__30 / ( num__2 x num__10 ) } x num__100.0 = num__150.0 correct option : c <eor> c <eos> |
c |
percent__100.0__150.0__ |
percent__100.0__150.0__ |
| q is a set of nine distinct prime numbers . if the sum of the integers in q is even and the number x is a member of q then what is the least value that x can be ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__7 |
num__2 + num__3 + num__5 + num__7 + num__11 + num__13 + num__17 + num__23 + num__29 = num__100 ( sum is even ) least number = num__2 a <eor> a <eos> |
a |
add__2.0__3.0__ add__2.0__5.0__ add__2.0__11.0__ subtract__5.0__3.0__ |
add__2.0__3.0__ add__2.0__5.0__ add__2.0__11.0__ subtract__5.0__3.0__ |
| a number x is chosen at random from the set of positive integers less than num__12 . what is the probability that ( num__9 / x ) > x ? <o> a ) num__0.0909090909091 <o> b ) num__0.222222222222 <o> c ) num__0.181818181818 <o> d ) num__0.666666666667 <o> e ) num__0.777777777778 |
number x has to be chosen from numbers num__1 - num__11 ( num__9 / x ) > x = > num__9 > x ^ num__2 = > x ^ num__2 - num__9 < num__0 x can have num__2 values only num__1 num__2 therefore probability = num__0.181818181818 answer c <eor> c <eos> |
c |
subtract__12.0__1.0__ subtract__11.0__9.0__ divide__2.0__11.0__ multiply__1.0__0.1818__ |
subtract__12.0__1.0__ subtract__11.0__9.0__ divide__2.0__11.0__ divide__2.0__11.0__ |
| the least number which should be added to num__28523 so that the sum is exactly divisible by num__3 num__5 num__7 and num__8 is <o> a ) num__40 <o> b ) num__37 <o> c ) num__35 <o> d ) num__45 <o> e ) num__48 |
lcm of num__3 num__57 and num__8 = num__840 num__33.955952381 = num__33 remainder = num__803 least no which should be added = num__840 - num__803 = num__37 answer b <eor> b <eos> |
b |
divide__28523.0__840.0__ round_down__33.956__ subtract__840.0__803.0__ subtract__840.0__803.0__ |
divide__28523.0__840.0__ round_down__33.956__ subtract__840.0__803.0__ subtract__840.0__803.0__ |
| what is the sum of all even numbers from num__1 to num__601 ? <o> a ) num__122821 <o> b ) num__281228 <o> c ) num__90300 <o> d ) num__122850 <o> e ) num__128111 |
explanation : num__300.0 = num__300 num__300 * num__301 = num__90300 answer : c <eor> c <eos> |
c |
add__1.0__300.0__ multiply__300.0__301.0__ multiply__1.0__90300.0__ |
add__1.0__300.0__ multiply__300.0__301.0__ multiply__1.0__90300.0__ |
| the length of a rectangular plot is num__20 metres more than its breadth . if the cost of fencing the plot @ rs . num__26.50 per metre is rs . num__5300 what is the length of the plot in metres ? <o> a ) num__333 <o> b ) num__200 <o> c ) num__288 <o> d ) num__276 <o> e ) num__1999 |
let length of plot = l meters then breadth = l - num__20 meters and perimeter = num__2 [ l + l - num__20 ] = [ num__4 l - num__40 ] meters [ num__4 l - num__40 ] * num__26.50 = num__5300 [ num__4 l - num__40 ] = num__5300 / num__26.50 = num__200 num__4 l = num__240 l = num__60.0 = num__60 meters . answer : b <eor> b <eos> |
b |
multiply__20.0__2.0__ divide__5300.0__26.5__ add__40.0__200.0__ hour_to_min_conversion__ round__200.0__ |
multiply__20.0__2.0__ divide__5300.0__26.5__ add__40.0__200.0__ add__20.0__40.0__ divide__5300.0__26.5__ |
| the ratio of the number of females to males at a party was num__1 : num__2 but when num__7 females and num__7 males left the ratio became num__1 : num__3 . how many people were at the party originally ? <o> a ) num__28 <o> b ) num__32 <o> c ) num__36 <o> d ) num__42 <o> e ) num__50 |
the total number of people are x females + num__2 x males . num__3 * ( x - num__7 ) = num__2 x - num__7 x = num__14 there were num__3 x = num__42 people at the party originally . the answer is d . <eor> d <eos> |
d |
multiply__2.0__7.0__ multiply__3.0__14.0__ multiply__1.0__42.0__ |
multiply__2.0__7.0__ multiply__3.0__14.0__ multiply__1.0__42.0__ |
| the total cost of a vacation was divided among num__3 people . if the total cost of the vacation had been divided equally among num__5 people the cost per person would have been $ num__50 less . what was the total cost cost of the vacation ? <o> a ) $ num__200 <o> b ) $ num__375 <o> c ) $ num__400 <o> d ) $ num__500 <o> e ) $ num__600 |
c for cost . p price per person . c = num__3 * p c = num__5 * p - num__250 substituting the value of p from the first equation onto the second we get p = num__125 . plugging in the value of p in the first equation we get c = num__375 . which leads us to answer choice b <eor> b <eos> |
b |
multiply__5.0__50.0__ multiply__3.0__125.0__ multiply__3.0__125.0__ |
multiply__5.0__50.0__ multiply__3.0__125.0__ multiply__3.0__125.0__ |
| in a camp there is a meal for num__100 men or num__200 children . if num__150 children have taken the meal how many men will be catered to with the remaining meal ? <o> a ) num__25 men <o> b ) num__32 men <o> c ) num__34 men <o> d ) num__30 men <o> e ) none of these |
explanation : there is meal for num__200 children . num__150 children have taken the meal . remaining meal is to be catered to num__50 children . now num__200 children = num__100 men so num__50 children = num__0.5 * num__50 = num__25 men answer : a <eor> a <eos> |
a |
subtract__200.0__150.0__ divide__100.0__200.0__ multiply__0.5__50.0__ multiply__0.5__50.0__ |
subtract__200.0__150.0__ divide__100.0__200.0__ multiply__0.5__50.0__ multiply__0.5__50.0__ |
| when you multiply a number by num__4 and then subtract num__8 the result is the same as if you first subtracted num__4 from the same number and then multiplied by num__12 . what is the number ? <o> a ) a ) num__5 <o> b ) b ) num__13 <o> c ) c ) num__19 <o> d ) d ) num__31 <o> e ) e ) num__10 |
let x equal the number and then change the words into an equation and solve for x : num__4 x – num__8 = num__12 ( x – num__4 ) num__4 x – num__8 = num__12 x – num__48 num__40 = num__8 x num__5 = x correct answer a ) num__5 <eor> a <eos> |
a |
multiply__4.0__12.0__ subtract__48.0__8.0__ divide__40.0__8.0__ divide__40.0__8.0__ |
multiply__4.0__12.0__ subtract__48.0__8.0__ divide__40.0__8.0__ divide__40.0__8.0__ |
| find four consecutive even integers so that the sum of the first two added to twice the sum of the last two is equal to num__742 . <o> a ) num__120 num__122 num__124 num__126 <o> b ) num__120 num__122 num__124 num__128 <o> c ) num__120 num__121 num__122 num__123 <o> d ) num__123 num__125 num__127 num__129 <o> e ) none of these |
let x x + num__2 x + num__4 and x + num__6 be the four integers . the sum of the first two x + ( x + num__2 ) twice the sum of the last two is written as num__2 ( ( x + num__4 ) + ( x + num__6 ) ) = num__4 x + num__20 sum of the first two added to twice the sum of the last two is equal to num__742 is written as x + ( x + num__2 ) + num__4 x + num__20 = num__742 solve for x and find all four numbers x = num__120 x + num__2 = num__122 x + num__4 = num__124 x + num__6 = num__126 answer a <eor> a <eos> |
a |
add__2.0__4.0__ multiply__6.0__20.0__ add__2.0__120.0__ add__2.0__122.0__ add__2.0__124.0__ multiply__6.0__20.0__ |
add__2.0__4.0__ multiply__6.0__20.0__ add__2.0__120.0__ add__2.0__122.0__ add__2.0__124.0__ multiply__6.0__20.0__ |
| if num__3 / p = num__8 & num__3 / q = num__18 then p - q = ? <o> a ) num__0.208333333333 <o> b ) num__0.25 <o> c ) num__0.291666666667 <o> d ) num__0.333333333333 <o> e ) num__0.375 |
p = num__0.375 q = num__0.166666666667 = > q = num__0.166666666667 therefore p - q = ( num__0.375 ) - ( num__0.166666666667 ) = num__0.208333333333 answer : a <eor> a <eos> |
a |
divide__3.0__8.0__ divide__3.0__18.0__ subtract__0.375__0.1667__ subtract__0.375__0.1667__ |
divide__3.0__8.0__ divide__3.0__18.0__ subtract__0.375__0.1667__ subtract__0.375__0.1667__ |
| a group of three girls and three boys have tickets for six adjacent seats in one row of a theater . if the three boys will not sit in three adjacent seats how many possible different seating arrangements are there for these num__6 theatre - goers ? <o> a ) num__6 ! – num__2 ! num__3 ! num__2 ! <o> b ) num__6 ! – num__6 ! num__3 ! <o> c ) num__7 × num__2 ! num__3 ! num__2 ! <o> d ) num__6 ! – num__4 ! num__3 ! <o> e ) num__2 ! num__3 ! num__2 ! |
there are num__3 boys and num__3 girls we want to calculate the seating arrangements if three boys do not sit together like mmm . let ' s calculate the # of arrangements when they sit together and subtract from total # of arrangements of these num__6 persons without restriction . thus we ' ll get the # of arrangements asked in the question . num__1 . total # of arrangements of num__6 is num__6 ! . num__2 . # of arrangements when num__3 boys are seated together like mmm ; among themselves these num__3 boys can sit in num__3 ! # of ways now consider these num__3 boys as one unit like this { mmm } . we ' ll have total of num__4 units : { mmm } { w } { w } { w } the # of arrangements of these num__4 units is num__4 ! . hence total # of arrangements when num__3 men sit together is : num__3 ! num__4 ! . # of arrangements when num__3 boys do not sit together would be : num__6 ! - num__3 ! num__4 ! . answer : d <eor> d <eos> |
d |
coin_space__ die_space__ |
coin_space__ die_space__ |
| a family consists of grandparents parents and three grand children . the average age of the grandparents is num__67 years that of the parents is num__38 years and that of the grandchildren is num__6 years . what is the average age of the family ? <o> a ) num__28 num__4 â „ num__7 years <o> b ) num__31 num__5 â „ num__7 years <o> c ) num__32 num__4 â „ num__7 years <o> d ) num__27 num__1 â „ num__2 years <o> e ) none of these |
required average = ( num__67 Ã — num__2 + num__38 Ã — num__2 + num__6 Ã — num__1.5 + num__2 + num__3 ) = num__32.5714285714 = num__32 num__4 â „ num__7 years answer c <eor> c <eos> |
c |
divide__6.0__2.0__ round_down__32.5714__ divide__6.0__1.5__ add__3.0__4.0__ round_down__32.5714__ |
divide__6.0__2.0__ round_down__32.5714__ divide__6.0__1.5__ add__3.0__4.0__ round_down__32.5714__ |
| find the simple interest on rs . num__500 for num__9 months at num__6 paisa per month ? <o> a ) num__277 <o> b ) num__270 <o> c ) num__288 <o> d ) num__266 <o> e ) num__121 |
i = ( num__500 * num__9 * num__6 ) / num__100 = num__270 answer : b <eor> b <eos> |
b |
percent__100.0__270.0__ |
percent__100.0__270.0__ |
| a is twice as good a work man as b and together they finish a piece of work in num__36 days . the number of days taken by b alone to finish the work is : <o> a ) num__100 days . <o> b ) num__108 days . <o> c ) num__98 days . <o> d ) num__47 days . <o> e ) num__103 days . |
solution ( a ’ s num__1 day ’ s work ) : ( b ’ s num__1 day ’ s work ) = num__2 : num__1 . ( a + b ) ' s num__1 day ’ s work = num__0.0277777777778 divide num__0.0277777777778 in the ratio num__2 : num__1 . ∴ b ’ s num__1 day ’ s work = ( num__0.0277777777778 x num__0.333333333333 ) = num__0.00925925925926 hence b alone can finish the work in num__108 days . answer b <eor> b <eos> |
b |
divide__1.0__36.0__ multiply__0.0278__0.3333__ round__108.0__ |
divide__1.0__36.0__ multiply__0.0278__0.3333__ round__108.0__ |
| when positive integer a is divided by positive integer b the remainder is num__5 . if a / b = num__58.25 what is the value of b ? <o> a ) num__18 <o> b ) num__19 <o> c ) num__15 <o> d ) num__20 <o> e ) num__13 |
when positive integer a is divided by positive integer b the remainder is num__5 - - > a = qb + num__5 ; a / b = num__58.25 - - > a = num__58 b + num__0.25 b ( so q above equals to num__58 ) ; num__0.25 b = num__6 - - > b = num__20 . answer : d . <eor> d <eos> |
d |
round_down__58.25__ subtract__58.25__58.0__ divide__5.0__0.25__ divide__5.0__0.25__ |
round_down__58.25__ subtract__58.25__58.0__ divide__5.0__0.25__ divide__5.0__0.25__ |
| how many numbers between num__11 and num__90 are divisible by num__7 ? <o> a ) num__11 <o> b ) num__13 <o> c ) num__15 <o> d ) num__12 <o> e ) num__10 |
the required numbers are num__14 num__21 num__28 num__35 . . . . num__77 num__84 . this is an a . p . with a = num__14 and d = ( num__21 - num__14 ) = num__7 . let it contain n terms . then tn = num__84 = > a + ( n - num__1 ) d = num__84 = > num__14 + ( n - num__1 ) x num__7 = num__84 or n = num__11 . required number of terms = num__11 answer is a . <eor> a <eos> |
a |
add__7.0__14.0__ add__7.0__21.0__ add__7.0__28.0__ multiply__11.0__7.0__ add__7.0__77.0__ multiply__11.0__1.0__ |
add__7.0__14.0__ add__7.0__21.0__ add__7.0__28.0__ multiply__11.0__7.0__ add__7.0__77.0__ multiply__11.0__1.0__ |
| the cash realised on selling a num__14.0 stock is rs . num__108.25 brokerage being num__0.25 % is ? <o> a ) num__366 <o> b ) num__106 <o> c ) num__108 <o> d ) num__192 <o> e ) num__122 |
cash realised = rs . ( num__108.25 - num__0.25 ) = rs . num__108 . answer : c <eor> c <eos> |
c |
round_down__108.25__ round_down__108.25__ |
subtract__108.25__0.25__ subtract__108.25__0.25__ |
| num__4 ^ x + num__4 * num__4 ^ - x = num__4 what is the value of x ? <o> a ) num__0.5 <o> b ) - num__0.5 <o> c ) - num__1 <o> d ) num__0.5 <o> e ) num__1 |
let num__4 ^ x = t t + num__4 / t = num__4 t ^ num__2 - num__4 t + num__4 = num__0 ( t - num__2 ) ^ num__2 = num__0 ( t - num__2 ) = num__0 t = num__2 num__4 ^ x = num__4 ^ num__0.5 x = num__0.5 ans : a <eor> a <eos> |
a |
reverse__2.0__ reverse__2.0__ |
reverse__2.0__ reverse__2.0__ |
| a box contains num__4 black and num__5 red balls and another box contains num__5 black and num__4 red balls . one ball is to drawn from either of two boxes . what is the probability of drawing a black ball ? <o> a ) num__0.111111111111 <o> b ) num__0.222222222222 <o> c ) num__0.444444444444 <o> d ) num__0.5 <o> e ) num__0.666666666667 |
explanation : probability of choosing first box is num__0.5 and second box is num__0.5 . ∴ probability of choosing one balck ball from the first box is num__4 c num__0.111111111111 c num__1 x num__0.5 = num__0.444444444444 x num__0.5 = num__0.222222222222 probability of choosing one black ball from the second box is = num__0.5 x num__5 c num__0.111111111111 c num__1 = num__0.277777777778 ∴ p ( e ) = num__0.222222222222 + num__0.277777777778 = num__4 + num__0.277777777778 = num__0.5 = num__0.5 answer : option d <eor> d <eos> |
d |
union_prob__0.5__0.2222__0.4444__ negate_prob__0.5__ |
union_prob__0.5__0.2222__0.4444__ negate_prob__0.5__ |
| in a group of ducks and cows the total number of legs are num__26 more than twice the no . of heads . find the total no . of buffaloes . <o> a ) num__10 <o> b ) num__12 <o> c ) num__13 <o> d ) num__15 <o> e ) num__16 |
let the number of buffaloes be x and the number of ducks be y = > num__4 x + num__2 y = num__2 ( x + y ) + num__26 = > num__2 x = num__26 = > x = num__13 c <eor> c <eos> |
c |
divide__26.0__2.0__ divide__26.0__2.0__ |
divide__26.0__2.0__ divide__26.0__2.0__ |
| in a camp there is a meal for num__120 men or num__200 children . if num__125 children have taken the meal how many men will be catered to with remaining meal ? <o> a ) num__30 <o> b ) num__20 <o> c ) num__40 <o> d ) num__45 <o> e ) num__63 |
there is a meal for num__200 children . num__125 children have taken the meal . remaining meal is to be catered to num__75 children . now num__200 children num__120 men . num__75 children = ( num__0.6 ) x num__75 = num__45 men . d <eor> d <eos> |
d |
subtract__200.0__125.0__ divide__120.0__200.0__ subtract__120.0__75.0__ subtract__120.0__75.0__ |
subtract__200.0__125.0__ divide__120.0__200.0__ subtract__120.0__75.0__ subtract__120.0__75.0__ |
| the sum of the cube of three numbers is num__138 while the sum of their products taken two at a time is num__131 . their sum is : <o> a ) num__20 <o> b ) num__21 <o> c ) num__22 <o> d ) num__24 <o> e ) num__25 |
let the numbers be a b and c . then a num__2 + b num__2 + c num__2 = num__138 and ( ab + bc + ca ) = num__131 . ( a + b + c ) num__2 = a num__2 + b num__2 + c num__2 + num__2 ( ab + bc + ca ) = num__138 + num__2 x num__131 = num__400 . ( a + b + c ) = num__400 = num__20 a <eor> a <eos> |
a |
divide__400.0__20.0__ |
divide__400.0__20.0__ |
| the mean of num__50 observations was num__36 . it was found later that an observation num__48 was wrongly taken as num__23 . the corrected new mean is : <o> a ) num__36.0 <o> b ) num__36.5 <o> c ) num__36.2 <o> d ) num__36.1 <o> e ) num__36.8 |
explanation : correct sum = ( num__36 * num__50 + num__48 - num__23 ) = num__1825 . correct mean = = num__36.5 = num__36.5 answer : b ) num__36.5 <eor> b <eos> |
b |
divide__1825.0__50.0__ divide__1825.0__50.0__ |
divide__1825.0__50.0__ divide__1825.0__50.0__ |
| the average amount with a group of seven numbers is rs . num__20 . if the newly joined member has rs . num__26 with him what was the average amount with the group before his joining the group ? <o> a ) s . num__17 <o> b ) s . num__12 <o> c ) s . num__15 <o> d ) s . num__29 <o> e ) s . num__19 |
total members in the group = num__7 average amount = rs . num__20 total amount with them = num__7 * num__20 = rs . num__140 one number has rs . num__26 . so the amount with remaining num__6 people = num__140 - num__26 = rs . num__114 the average amount with them = num__19.0 = rs . num__19 . answer : e <eor> e <eos> |
e |
multiply__20.0__7.0__ subtract__26.0__20.0__ subtract__140.0__26.0__ subtract__26.0__7.0__ subtract__26.0__7.0__ |
multiply__20.0__7.0__ subtract__26.0__20.0__ subtract__140.0__26.0__ subtract__26.0__7.0__ subtract__26.0__7.0__ |
| find the third proportion to num__12 num__36 ? <o> a ) num__106 <o> b ) num__107 <o> c ) num__108 <o> d ) num__109 <o> e ) num__110 |
let x be num__3 rd pro . then num__12 : num__36 = num__36 : x so x = num__108 answer : c <eor> c <eos> |
c |
divide__36.0__12.0__ multiply__36.0__3.0__ multiply__36.0__3.0__ |
divide__36.0__12.0__ multiply__36.0__3.0__ multiply__36.0__3.0__ |
| if num__1 + num__9 + num__11 = num__1 then what is the value of num__12 + num__11 + num__9 = ? <o> a ) num__11 <o> b ) num__15 <o> c ) num__17 <o> d ) num__12 <o> e ) num__10 |
e num__10 equation num__1 + num__9 + num__11 = num__1 can be derived from one ( o ) + nine ( n ) + eleven ( e ) = one = > num__1 similarly for equation num__12 + num__11 + num__9 twelve ( t ) + eleven ( e ) + nine ( n ) = > ten ( num__10 ) <eor> e <eos> |
e |
add__1.0__9.0__ add__1.0__9.0__ |
add__1.0__9.0__ add__1.0__9.0__ |
| if pq = p ^ num__4 + q ^ num__2 – num__2 pq for what value of q is pq equal to p ^ num__4 for all values of p ? <o> a ) - num__2 <o> b ) - num__1 <o> c ) num__2 <o> d ) num__1 <o> e ) num__0 |
c p num__4 + q num__2 â € “ num__2 pq = p num__4 = > q num__2 - num__2 pq = num__0 = > q ( q - num__2 p ) = num__0 = > q = num__0 or q = num__2 p im not following the logic you used here . . how did you replace pq with num__0 could someone help explain it in more detai e <eor> e <eos> |
e |
multiply__4.0__0.0__ |
multiply__4.0__0.0__ |
| a completes a work in num__10 days and b complete the same work in num__12 days . if both of them work together then the number of days required to complete the work will be <o> a ) num__6.8 days <o> b ) num__9 days <o> c ) num__10 days <o> d ) num__12 days <o> e ) num__13 days |
if a can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days therefore here the required number of days = num__10 × num__0.545454545455 = num__6.8 days . a <eor> a <eos> |
a |
round__6.8__ |
round__6.8__ |
| if x is num__30 percent greater than num__88 then x = <o> a ) num__68 <o> b ) num__70.4 <o> c ) num__86 <o> d ) num__114.4 <o> e ) num__108 |
x = num__88 * num__1.3 = num__114.4 so the answer is d . <eor> d <eos> |
d |
multiply__88.0__1.3__ multiply__88.0__1.3__ |
multiply__88.0__1.3__ multiply__88.0__1.3__ |
| the cost price of a radio is rs . num__1500 and it was sold for rs . num__1230 find the loss % ? <o> a ) num__18 <o> b ) num__99 <o> c ) num__77 <o> d ) num__66 <o> e ) num__41 |
num__1500 - - - - num__270 num__100 - - - - ? = > num__18.0 answer : a <eor> a <eos> |
a |
percent__100.0__18.0__ |
percent__100.0__18.0__ |
| three people each took num__5 tests . if the ranges of their scores in the num__5 practice tests were num__17 num__28 and num__35 num__45 what is the minimum possible range in scores of the three test - takers ? <o> a ) a num__17 <o> b ) b ) num__28 <o> c ) c ) num__35 <o> d ) d ) num__45 <o> e ) e ) num__80 |
it is finding the minimum range between all their scores if all test taker scores are between num__0 and maximum range we will have : a - num__0 and num__17 b - num__0 and num__28 c - num__0 and num__35 d - num__0 and num__45 therefore the minimum range is num__45 it cant be any lower however you play with the numbers . d <eor> d <eos> |
d |
add__17.0__28.0__ |
add__17.0__28.0__ |
| the g . c . d . of num__1.08 num__0.31 and num__0.9 is : <o> a ) num__0.01 <o> b ) num__0.9 <o> c ) num__0.18 <o> d ) num__0.108 <o> e ) none |
explanation given numbers are num__1.08 num__0.31 and num__0.90 . h . c . f . of num__108 num__31 and num__90 is num__1 h . c . f . of given numbers = num__0.01 . answer a <eor> a <eos> |
a |
divide__1.08__108.0__ divide__1.08__108.0__ |
divide__1.08__108.0__ divide__1.08__108.0__ |
| during a two - week period the price of an ounce of silver increased by num__20 percent by the end of the first week and then decreased by num__10 percent of this new price by the end of the second week . if the price of silver was x dollars per ounce at the beginning of the two - week period what was the price in dollars per ounce by the end of the period ? <o> a ) num__1.02 x <o> b ) num__1.04 x <o> c ) num__1.06 x <o> d ) num__1.08 x <o> e ) num__1.1 x |
the price at the end is num__0.9 ( num__1.2 x ) = num__1.08 x the answer is d . <eor> d <eos> |
d |
multiply__0.9__1.2__ multiply__0.9__1.2__ |
multiply__0.9__1.2__ multiply__0.9__1.2__ |
| the sector of a circle has radius of num__28 cm and central angle num__135 o . find its perimeter ? <o> a ) num__91.5 <o> b ) num__91.4 <o> c ) num__91.7 <o> d ) num__91.3 <o> e ) num__122 |
perimeter of the sector = length of the arc + num__2 ( radius ) = ( num__0.375 * num__2 * num__3.14285714286 * num__28 ) + num__2 ( num__28 ) = num__66 + num__56 = num__122 cm answer : e <eor> e <eos> |
e |
multiply__28.0__2.0__ add__66.0__56.0__ round__122.0__ |
multiply__28.0__2.0__ add__66.0__56.0__ add__66.0__56.0__ |
| adam spent num__0.166666666667 of his lifetime in adolescence . his facial hair started growing after num__0.0833333333333 more of his life . he successfully married his girlfriend diana after num__0.142857142857 more of his life . their son was born after num__5 years from then . the son lived just num__0.5 of what adam lived . adam dies four years after his son died . <o> a ) num__84 <o> b ) num__90 <o> c ) num__81 <o> d ) num__75 <o> e ) num__66 |
a let us assume that adam lived for p years . p / num__6 + p / num__12 + p / num__7 + num__5 + p / num__2 + num__4 = x = > p = num__84 . therefore adam lived for num__84 years . <eor> a <eos> |
a |
divide__6.0__0.5__ subtract__12.0__5.0__ reverse__0.5__ divide__2.0__0.5__ multiply__7.0__12.0__ multiply__7.0__12.0__ |
divide__6.0__0.5__ subtract__12.0__5.0__ reverse__0.5__ divide__2.0__0.5__ multiply__7.0__12.0__ multiply__7.0__12.0__ |
| if two pipes function simultaneously the reservoir will be filled in num__12 hours one pipe fills the reservoir num__10 hours faster than the other . how many hours it takes the second pipe to fill the reservoir ? <o> a ) num__15 hours <o> b ) num__18 hours <o> c ) num__30 hours <o> d ) num__20 hours <o> e ) num__22 hours |
c num__30 hours <eor> c <eos> |
c |
round__30.0__ |
round__30.0__ |
| the ratio between the length and the breadth of a rectangular park is num__1 : num__4 . if a man cycling along the boundary of the park at the speed of num__12 km / hr completes one round in num__8 min then the area of the park ( in sq . m ) is ? <o> a ) num__124545 m <o> b ) num__134561 m <o> c ) num__156787 m <o> d ) num__15450 m <o> e ) num__102400 m |
perimeter = distance covered in num__8 min . = num__12000 x num__8 m = num__1600 m . num__60 let length = num__1 x metres and breadth = num__4 x metres . then num__2 ( num__1 x + num__4 x ) = num__1600 or x = num__160 . length = num__160 m and breadth = num__640 m . area = ( num__160 x num__640 ) m num__2 = num__102400 m e <eor> e <eos> |
e |
triangle_area__1.0__4.0__ surface_rectangular_prism__4.0__12.0__2.0__ square_perimeter__160.0__ multiply__160.0__640.0__ multiply__1.0__102400.0__ |
triangle_area__1.0__4.0__ surface_rectangular_prism__4.0__12.0__2.0__ square_perimeter__160.0__ multiply__160.0__640.0__ multiply__1.0__102400.0__ |
| baby isabel plays with blocks . each block is num__1.5 inches tall . she has a collection of num__20 blocks . if she could stack all of the blocks up one on top of the other how many inches tall would her tower of blocks be ? <o> a ) num__67 ' ' <o> b ) num__30 ' ' <o> c ) num__106 ' ' <o> d ) num__120 ' ' <o> e ) num__151.7 ' ' |
num__1.5 ' ' * num__20 = num__30 ' ' . answer is b . <eor> b <eos> |
b |
multiply__1.5__20.0__ multiply__1.5__20.0__ |
multiply__1.5__20.0__ multiply__1.5__20.0__ |
| a trail mix company keeps costs down by employing the peanuts : cashews : almonds ratio of num__10 : num__4 : num__1 in each bag of up to num__75 total nuts . what is the maximum percentage by which the company could decrease its number of peanuts per bag and still have peanuts constitute more than half the total amount of nuts ? <o> a ) num__40.0 <o> b ) num__48.0 <o> c ) num__49.0 <o> d ) num__50.0 <o> e ) num__58 % |
peanuts = num__50 cashews = num__20 almonds = num__5 we want to remove as many peanuts as possible while still having peanuts represent more than half of the mixture . . . the number of cashews and almonds will stay the same though so we have num__20 + num__5 = num__25 of those non - peanuts in total . if we had num__25 peanuts and num__25 non - peanuts then that would be num__50.0 exactly . we want more than num__50.0 though so we need to add in num__1 more peanut . this gives us . . . peanuts = num__26 cashews = num__20 almonds = num__5 the question asked for the decrease in the number of peanuts as a percentage . we started with num__50 peanuts and removed num__24 = num__0.48 = num__48.0 ; answer : b <eor> b <eos> |
b |
add__4.0__1.0__ subtract__75.0__50.0__ add__1.0__25.0__ add__4.0__20.0__ divide__24.0__50.0__ multiply__1.0__48.0__ |
add__4.0__1.0__ subtract__75.0__50.0__ add__1.0__25.0__ add__4.0__20.0__ divide__24.0__50.0__ multiply__1.0__48.0__ |
| in the list num__3 num__3 num__4 num__4 num__5 num__5 num__5 num__5 num__7 num__11 num__21 what fraction of the data is less than the mode ? <o> a ) num__0.222222222222 <o> b ) num__0.333333333333 <o> c ) num__0.363636363636 <o> d ) num__0.666666666667 <o> e ) num__0.777777777778 |
mode : the mode of any set is the term which has the highest frequency ( occurrence ) highest frequent term in the set is num__5 ( with frequency num__4 ) hence mode = num__5 two terms ( num__3 num__3 num__4 num__4 ) out of a total of num__11 terms are less than mode of the set . fraction of set that are less than mode of set = num__0.363636363636 answer : option c <eor> c <eos> |
c |
divide__4.0__11.0__ divide__4.0__11.0__ |
divide__4.0__11.0__ divide__4.0__11.0__ |
| a car takes num__6 hours to cover a distance of num__540 km . how much should the speed in kmph be maintained to cover the same direction in num__1.5 th of the previous time ? <o> a ) num__60 kmph <o> b ) num__70 kmph <o> c ) num__80 kmph <o> d ) num__85 kmph <o> e ) num__90 kmph |
time = num__6 distence = num__540 num__1.5 of num__6 hours = num__6 * num__1.5 = num__9 hours required speed = num__60.0 = num__60 kmph a <eor> a <eos> |
a |
multiply__6.0__1.5__ hour_to_min_conversion__ hour_to_min_conversion__ |
multiply__6.0__1.5__ hour_to_min_conversion__ hour_to_min_conversion__ |
| a cycle is bought for rs . num__900 and sold for rs . num__1080 find the gain percent ? <o> a ) num__11 <o> b ) num__20 <o> c ) num__28 <o> d ) num__26 <o> e ) num__21 |
num__900 - - - - num__180 num__100 - - - - ? = > num__20.0 answer : b <eor> b <eos> |
b |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| a grocer buys apples for num__100 ¢ per pound . if num__20.0 of the apples goes bad and he still wants to make a num__10.0 profit over his purchase price what should be the sales price ? <o> a ) num__136 ¢ <o> b ) num__130 ¢ <o> c ) num__140 ¢ <o> d ) num__137.5 ¢ <o> e ) num__142 ¢ |
for making calculation simple let us consider he buys num__10 pounds @ num__100 / pound = num__1000 ¢ in total . num__20.0 apples lost . . . means num__8 pounds left . further he wants num__10.0 profit on num__1000 ¢ means he wants to earn num__1100 ¢ sales price should be num__137.5 = ¢ num__137.5 ans d <eor> d <eos> |
d |
percent__100.0__137.5__ |
percent__100.0__137.5__ |
| of the three numbers the first is twice as second and three times the third . the average of the three numbers is num__88 and the three numbers in order are ? <o> a ) num__116 num__5836 <o> b ) num__98 num__4933 <o> c ) num__126 num__6336 <o> d ) num__144 num__7248 <o> e ) none of these |
explanation : solution : let a = x b = x / num__2 c = x / num__3 . = x + x / num__2 + x / num__1.0 = num__88 by solving we will get x = num__144 . hence a = num__144 b = num__72.0 = num__72 c = num__48.0 = num__48 answer : d <eor> d <eos> |
d |
subtract__3.0__2.0__ divide__144.0__2.0__ divide__144.0__3.0__ multiply__1.0__144.0__ |
subtract__3.0__2.0__ divide__144.0__2.0__ divide__144.0__3.0__ divide__144.0__1.0__ |
| if two numbers are in the ratio num__4 : num__3 . if num__20 is added to both of the numbers then the ratio becomes num__5 : num__4 then find the smallest number ? <o> a ) num__60 <o> b ) num__30 <o> c ) num__40 <o> d ) num__20 <o> e ) num__50 |
num__4 : num__3 num__4 x + num__20 : num__3 x + num__20 = num__5 : num__4 num__4 [ num__4 x + num__20 ] = num__5 [ num__3 x + num__20 ] num__16 x + num__80 = num__15 x + num__100 num__16 x - num__15 x = num__100 - num__80 x = num__20 then smallest number is = num__2 num__3 x = num__60 a <eor> a <eos> |
a |
subtract__20.0__4.0__ multiply__4.0__20.0__ multiply__3.0__5.0__ multiply__20.0__5.0__ subtract__5.0__3.0__ multiply__4.0__15.0__ multiply__4.0__15.0__ |
subtract__20.0__4.0__ multiply__4.0__20.0__ subtract__20.0__5.0__ add__20.0__80.0__ subtract__5.0__3.0__ subtract__80.0__20.0__ subtract__80.0__20.0__ |
| if num__8 cats can kill num__8 rats in num__8 minutes how long will it take num__100 cats to kill num__100 rats ? <o> a ) num__6 minutes <o> b ) num__7 minutes <o> c ) num__8 minutes <o> d ) num__9 minutes <o> e ) num__10 minutes |
it will take num__8 minutes for num__100 cats to kill num__100 rats . num__1 cat can kill num__1 rat in num__8 minutes so num__100 cats can kill num__100 rats in num__8 minutes answer c <eor> c <eos> |
c |
round__8.0__ |
round__8.0__ |
| a cistern is normally filled in num__2 hrs but takes num__2 hrs longer to fill because of a leak on its bottom if cistern is full how much time citern would empty ? <o> a ) num__4 hours <o> b ) num__20 hours <o> c ) num__30 hours <o> d ) num__40 hours <o> e ) num__50 hours |
if leakage / hour = num__1 / x then num__0.5 - num__1 / x = num__0.25 solving num__1 / x = num__0.25 so in num__4 hours full cistern will be empty . answer : a <eor> a <eos> |
a |
divide__1.0__2.0__ divide__0.5__2.0__ divide__2.0__0.5__ round__4.0__ |
divide__1.0__2.0__ divide__0.5__2.0__ divide__2.0__0.5__ divide__2.0__0.5__ |
| last year the price per share of stock x increased by k percent and the earnings per share of stock x increased by m percent where k is greater than m . by what percent did the ratio of price per share to earnings per share increase in terms of k and m ? <o> a ) k m % <o> b ) ( k – m ) % <o> c ) [ num__100 ( k – m ) ] / ( num__100 + k ) % <o> d ) [ num__100 ( k – m ) ] / ( num__100 + m ) % <o> e ) [ num__100 ( k – m ) ] / ( num__100 + k + m ) % |
price per share ( pps ) = x earning per share ( eps ) = y increased pps = x ( num__1 + k / num__100 ) increased eps = y ( num__1 + m / num__100 ) % increase in the ratios ( pps / eps ) = { [ x ( num__100 + k ) ] / [ y ( num__100 + m ) ] - ( x / y ) } / ( x / y ) on simplifying % increase = [ num__100 ( k - m ) ] / ( num__100 + m ) % answer : d <eor> d <eos> |
d |
multiply__1.0__100.0__ |
divide__100.0__1.0__ |
| a certain protective additive increases from num__50 days to num__60 days the time between required maintenance checks on an industrial vehicle . by what percent is the time between maintenance checks increased by using the additive ? <o> a ) num__25.0 <o> b ) num__33 num__0.333333333333 % <o> c ) num__50.0 <o> d ) num__66 num__0.666666666667 % <o> e ) num__20 % |
general formula for percent increase or decrease ( percent change ) : percent = change / original ∗ num__100 so the time between maintenance checks increased by num__60 − num__1.0 ∗ num__100 = num__20 answer : e . <eor> e <eos> |
e |
multiply__20.0__1.0__ |
divide__20.0__1.0__ |
| the relationship between quantities m and n is expressed by the equation num__30 m = num__5 ( n - num__42 ) . if the difference between the two chosen values of n is num__30 what is the difference in the corresponding values of m ? <o> a ) num__5 <o> b ) num__3.81818181818 <o> c ) num__5.45454545455 <o> d ) num__13.6363636364 <o> e ) num__14 |
another simple way num__30 m = num__5 ( n - num__42 ) assume n = num__42 to make one side equal to num__0 then num__30 m = num__0 and m = num__0 repeat it by assuming n as num__42 + num__30 i . e num__72 num__30 m - num__5 ( num__72 - num__42 ) num__30 m = num__150 m = num__5.0 diff = num__5 - num__0 = num__5 answer : a <eor> a <eos> |
a |
add__30.0__42.0__ multiply__30.0__5.0__ divide__150.0__30.0__ |
add__30.0__42.0__ multiply__30.0__5.0__ divide__150.0__30.0__ |
| two trains started at the same time one from a to b and the other from b to a . if they arrived at b and a respectively num__16 hours and num__25 hours after they passed each other the ratio of the speeds of the two trains was <o> a ) num__2 : num__1 <o> b ) num__3 : num__2 <o> c ) num__4 : num__3 <o> d ) num__5 : num__4 <o> e ) num__1 : num__2 |
explanation : note : if two trains ( or bodies ) start at the same time from points a and b towards each other and after crossing they take a and b sec in reaching b and a respectively then : ( a ' s speed ) : ( b ' s speed ) = ( b : a ) therefore ratio of the speeds of two trains = = num__5 : num__4 . answer : d <eor> d <eos> |
d |
round__5.0__ |
round__5.0__ |
| if | num__20 x - num__10 | = num__150 then find the product of the values of x ? <o> a ) - num__45 <o> b ) num__56 <o> c ) - num__62 <o> d ) num__35 <o> e ) - num__30 |
| num__20 x - num__10 | = num__150 num__20 x - num__10 = num__150 or num__20 x - num__10 = - num__150 num__20 x = num__160 or num__20 x = - num__140 x = num__8 or x = - num__7 product = - num__7 * num__8 = - num__56 answer is b <eor> b <eos> |
b |
add__10.0__150.0__ subtract__150.0__10.0__ divide__160.0__20.0__ divide__140.0__20.0__ multiply__7.0__8.0__ multiply__7.0__8.0__ |
add__10.0__150.0__ subtract__150.0__10.0__ divide__160.0__20.0__ divide__140.0__20.0__ multiply__7.0__8.0__ multiply__7.0__8.0__ |
| jack has two dice one has six equally probable sides labeled num__1 num__2 num__3 num__4 num__5 num__6 and the other has seven equally probable sides labeled num__1 num__2 num__3 num__4 num__5 num__6 num__7 . if jack rolls both dice what is the probability that both of the numbers will be even ? <o> a ) num__0.214285714286 <o> b ) num__0.285714285714 <o> c ) num__0.333333333333 <o> d ) num__0.5 <o> e ) num__0.571428571429 |
probability that the number on first die is even = num__0.5 [ because num__3 out of num__6 faces are even ] probability that the number on second die is even = num__0.428571428571 [ because num__3 out of num__7 faces are even ] probability that both dice result in odd numbers = ( num__0.5 ) * ( num__0.428571428571 ) = num__0.214285714286 answer : option a <eor> a <eos> |
a |
reverse__2.0__ divide__3.0__7.0__ multiply__0.5__0.4286__ multiply__1.0__0.2143__ |
reverse__2.0__ divide__3.0__7.0__ multiply__0.5__0.4286__ multiply__1.0__0.2143__ |
| five people are planning to share equally the cost of a rental car . if one person withdraws from the arrangement and the others share equally the entire cost of the car then the share of each of the remaining persons increased by : <o> a ) num__0.875 <o> b ) num__0.125 <o> c ) num__0.25 <o> d ) num__0.142857142857 <o> e ) num__0.111111111111 |
original share of num__1 person = num__0.2 new share of num__1 person = num__0.25 increase = num__0.25 - num__0.2 = num__0.05 required fraction = ( num__0.05 ) / ( num__0.2 ) = num__0.25 answer is c . <eor> c <eos> |
c |
subtract__0.25__0.2__ multiply__0.25__1.0__ |
subtract__0.25__0.2__ divide__0.25__1.0__ |
| a can have a piece of work done in num__8 days b can work three times faster than the a c can work five times faster than a . how many days will they take to do the work together ? <o> a ) num__3 days <o> b ) num__0.888888888889 days <o> c ) num__4 days <o> d ) ca n ' t say <o> e ) none |
a ' s num__1 day work is num__0.125 b ' s num__1 day work is num__0.375 c ' s num__1 day work is num__0.625 ( a + b + c ) ' s num__1 day work ( num__0.125 + num__0.375 + num__0.625 ) is num__1.125 so they can finish the work in num__0.888888888889 days answer : b <eor> b <eos> |
b |
divide__1.0__8.0__ subtract__1.0__0.375__ add__0.125__1.0__ divide__1.0__1.125__ multiply__1.0__0.8889__ |
divide__1.0__8.0__ subtract__1.0__0.375__ add__0.125__1.0__ divide__1.0__1.125__ multiply__1.0__0.8889__ |
| calculate the sum of first num__28 natural numbers . <o> a ) num__426 <o> b ) num__406 <o> c ) num__401 <o> d ) num__409 <o> e ) num__405 |
solution we know that ( num__1 + num__2 + num__3 + . . . . . + num__28 ) = n ( n + num__1 ) / num__2 therefore ( num__1 + num__2 + num__3 + . . . . + num__28 ) = ( num__28 × num__14.5 ) = num__406 . answer b <eor> b <eos> |
b |
add__1.0__2.0__ multiply__28.0__14.5__ multiply__28.0__14.5__ |
add__1.0__2.0__ multiply__28.0__14.5__ divide__406.0__1.0__ |
| the average age of num__15 students of a class is num__15 years . out of these the average age of num__5 students is num__14 years and that of the other num__9 students is num__16 years . the age of the num__15 th student is <o> a ) num__9 years <o> b ) num__11 years <o> c ) num__14 years <o> d ) num__21 years <o> e ) num__25 years |
solution age of the num__15 th student = [ num__15 x num__15 - ( num__14 x num__5 + num__16 x num__9 ) ] = ( num__225 - num__214 ) = num__11 years . answer b <eor> b <eos> |
b |
subtract__16.0__5.0__ subtract__16.0__5.0__ |
subtract__16.0__5.0__ subtract__16.0__5.0__ |
| the average of seven numbers is num__26 . the average of first three numbers is num__14 and the average of last three numbers is num__19 . what is the middle number ? <o> a ) num__25 <o> b ) num__83 <o> c ) num__29 <o> d ) num__32 <o> e ) num__34 |
the total of seven numbers = num__7 x num__26 = num__182 the total of first num__3 and last num__3 numbers is = num__3 x num__14 + num__3 x num__19 = num__99 so the middle number is ( num__182 - num__99 ) = num__83 b <eor> b <eos> |
b |
subtract__26.0__19.0__ multiply__26.0__7.0__ subtract__182.0__99.0__ subtract__182.0__99.0__ |
subtract__26.0__19.0__ multiply__26.0__7.0__ subtract__182.0__99.0__ subtract__182.0__99.0__ |
| a trained covered x km at num__40 kmph and another num__2 x km at num__20 kmph . find the average speed of the train in covering the entire num__3 x km . <o> a ) num__22 <o> b ) num__99 <o> c ) num__24 <o> d ) num__66 <o> e ) num__887 |
total time taken = x / num__40 + num__2 x / num__20 hours = num__5 x / num__40 = x / num__8 hours average speed = num__3 x / ( x / num__8 ) = num__24 kmph answer : c <eor> c <eos> |
c |
add__2.0__3.0__ divide__40.0__5.0__ multiply__3.0__8.0__ multiply__3.0__8.0__ |
add__2.0__3.0__ divide__40.0__5.0__ multiply__3.0__8.0__ multiply__3.0__8.0__ |
| the average age of a group of n people is num__15 years old . one more person aged num__37 joins the group and the new average is num__17 years old . what is the value of n ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__16 <o> e ) num__18 |
num__15 n + num__37 = num__17 ( n + num__1 ) num__2 n = num__20 n = num__10 the answer is a . <eor> a <eos> |
a |
subtract__17.0__15.0__ subtract__37.0__17.0__ divide__20.0__2.0__ multiply__1.0__10.0__ |
subtract__17.0__15.0__ subtract__37.0__17.0__ divide__20.0__2.0__ multiply__1.0__10.0__ |
| the sum of ages of num__5 children born at the intervals of num__2 years each is num__55 years . what is the age of the youngest child ? <o> a ) num__4 <o> b ) num__7 <o> c ) num__6 <o> d ) num__9 <o> e ) num__8 |
let the ages of children be x ( x + num__2 ) ( x + num__4 ) ( x + num__6 ) and ( x + num__8 ) years . then x + ( x + num__2 ) + ( x + num__4 ) + ( x + num__6 ) + ( x + num__8 ) = num__50 num__5 x = num__55 - num__20 = num__35 x = num__7 . age of the youngest child = x = num__7 years . answer : b <eor> b <eos> |
b |
add__2.0__4.0__ multiply__2.0__4.0__ subtract__55.0__5.0__ multiply__5.0__4.0__ subtract__55.0__20.0__ add__5.0__2.0__ add__5.0__2.0__ |
add__2.0__4.0__ add__2.0__6.0__ subtract__55.0__5.0__ multiply__5.0__4.0__ subtract__55.0__20.0__ add__5.0__2.0__ add__5.0__2.0__ |
| a train num__400 m long can cross an electric pole in num__20 sec and then find the speed of the train ? <o> a ) num__35 km / hr <o> b ) num__48 km / hr <o> c ) num__57 km / hr <o> d ) num__72 km / hr <o> e ) num__83 km / hr |
length = speed * time speed = l / t s = num__20.0 s = num__20 m / sec speed = num__20 * num__3.6 ( to convert m / sec in to km / hr multiply by num__3.6 ) speed = num__72 km / hr answer : d <eor> d <eos> |
d |
multiply__20.0__3.6__ round__72.0__ |
multiply__20.0__3.6__ multiply__20.0__3.6__ |
| a electricity company charges $ num__3 for the first num__500 units consumption plus $ num__0.05 for each additional unit . what would this company charge for a consumption of num__800 units ? <o> a ) $ num__10 <o> b ) $ num__12 <o> c ) $ num__15 <o> d ) $ num__16 <o> e ) $ num__18 |
first num__500 units harges = $ num__3 rest of the units = num__800 - ( num__500 ) = num__300 charge for the rest of the units = num__300 * num__05 = num__15 total charge = num__3 + num__15 = num__18 answer is e <eor> e <eos> |
e |
subtract__800.0__500.0__ multiply__3.0__5.0__ add__3.0__15.0__ add__3.0__15.0__ |
subtract__800.0__500.0__ multiply__3.0__5.0__ add__3.0__15.0__ add__3.0__15.0__ |
| if num__638521 is to be divisible by num__5 what is the least whole number that should be added to ? <o> a ) num__8 <o> b ) num__7 <o> c ) num__5 <o> d ) num__4 <o> e ) num__3 |
a number is divisible by num__5 if the last digit is either num__0 or num__5 . here num__638521 = num__1 ( last digit is neither num__0 or num__5 ) . the next multiple of num__5 i . e to make the last digit num__0 or num__5 add num__4 num__4 must be added to num__638521 to make it divisible by num__5 d <eor> d <eos> |
d |
reverse__638521.0__ subtract__5.0__1.0__ subtract__5.0__1.0__ |
reverse__638521.0__ subtract__5.0__1.0__ subtract__5.0__1.0__ |
| a b and c can do a piece of work in num__68 and num__12 days respectively . in how many days would all of them complete the same job working together ? <o> a ) num__2 num__0.75 days . <o> b ) num__2 num__0.666666666667 days . <o> c ) num__2 num__1.5 days . <o> d ) num__3 num__0.5 days . <o> e ) num__3 num__0.75 days . |
explanation : in this type of questions we first get the finishing of work in num__1 day for a b c then we will add them to get the result as : part finished by a in num__1 day = num__0.166666666667 part finished by b in num__1 day = num__0.125 part finished by c in num__1 day = num__0.0833333333333 part finished by ( a + b + c ) in num__1 day = num__0.375 = num__0.375 remaining work = num__1 - num__0.45 = num__0.55 . number days to finish the work by a + b + c = num__2 num__0.666666666667 days . option b <eor> b <eos> |
b |
divide__1.0__12.0__ subtract__1.0__0.45__ round__2.0__ |
divide__1.0__12.0__ subtract__1.0__0.45__ round__2.0__ |
| the food in a camp lasts for num__10 men for num__20 days . if num__30 more men join how many days will the food last ? <o> a ) num__8 days <o> b ) num__4 days <o> c ) num__5 days <o> d ) num__10 days <o> e ) num__17 days |
one man can consume the same food in num__10 * num__20 = num__200 days . num__30 more men join the total number of men = num__40 the number of days the food will last = num__5.0 = num__5 days . answer : c <eor> c <eos> |
c |
multiply__10.0__20.0__ add__10.0__30.0__ divide__200.0__40.0__ round__5.0__ |
multiply__10.0__20.0__ add__10.0__30.0__ divide__200.0__40.0__ round__5.0__ |
| a sequence consists of num__16 consecutive even integers written in increasing order . the sum of the first num__8 of these even integers is num__456 . what is the sum of the last num__8 of the even integers ? <o> a ) num__580 <o> b ) num__584 <o> c ) num__588 <o> d ) num__592 <o> e ) num__596 |
the sum of the first num__8 numbers is : x + ( x + num__2 ) + . . . + ( x + num__14 ) = num__456 the sum of the next num__8 numbers is : ( x + num__16 ) + ( x + num__2 + num__16 ) + . . . + ( x + num__14 + num__16 ) = num__456 + num__8 ( num__16 ) = num__456 + num__128 = num__584 the answer is b . <eor> b <eos> |
b |
divide__16.0__8.0__ subtract__16.0__2.0__ multiply__16.0__8.0__ add__456.0__128.0__ add__456.0__128.0__ |
divide__16.0__8.0__ subtract__16.0__2.0__ multiply__16.0__8.0__ add__456.0__128.0__ add__456.0__128.0__ |
| the perimeter of one square is num__48 cm and that of another is num__20 cm . find the perimeter and the diagonal of a square which is equal in area to these two combined ? <o> a ) num__13 √ num__6 <o> b ) num__13 √ num__2 <o> c ) num__13 √ num__0 <o> d ) num__13 √ num__2 <o> e ) num__13 √ num__1 |
num__4 a = num__48 num__4 a = num__20 a = num__12 a = num__5 a num__2 = num__144 a num__2 = num__25 combined area = a num__2 = num__169 = > a = num__13 d = num__13 √ num__2 answer : b <eor> b <eos> |
b |
power__12.0__2.0__ power__5.0__2.0__ triangle_perimeter__20.0__144.0__5.0__ triangle_area__2.0__13.0__ |
power__12.0__2.0__ power__5.0__2.0__ triangle_perimeter__20.0__144.0__5.0__ triangle_area__2.0__13.0__ |
| in a recent election ms . robinson received num__8000 voters cast by independent voters that is voters not registered with a specific political party . she also received num__40 percent of the votes cast by those voters registered with a political party . if n is the total number of votes cast in the election and num__40 percent of the votes cast were cast by independent voters which of the following represents the number of votes that mrs . robbins received ? <o> a ) num__0.06 n + num__3200 <o> b ) num__0.1 n + num__7200 <o> c ) num__0.24 n + num__8000 <o> d ) num__0.1 n + num__8000 <o> e ) num__0.06 n + num__8000 |
num__40.0 of n are independent voters hence num__60.0 of n are not independent voters . from this group she received num__40.0 votes so num__0.4 * num__0.6 * n plus num__8000 votes from independents : total = num__0.4 * num__0.6 * n + num__8000 = num__0.24 * n + num__8000 . answer : c . <eor> c <eos> |
c |
multiply__0.4__0.6__ multiply__0.4__0.6__ |
multiply__0.4__0.6__ multiply__0.4__0.6__ |
| by selling an article at rs . num__800 a shopkeeper makes a profit of num__25.0 . at what price should he sell the article so as to make a loss of num__40.0 ? <o> a ) num__228 <o> b ) num__384 <o> c ) num__267 <o> d ) num__288 <o> e ) num__276 |
sp = num__800 profit = num__25.0 cp = ( sp ) * [ num__100 / ( num__100 + p ) ] = num__800 * [ num__0.8 ] = num__640 loss = num__40.0 = num__40.0 of num__640 = rs . num__256 sp = cp - loss = num__640 - num__256 = rs . num__384 answer : b <eor> b <eos> |
b |
percent__40.0__640.0__ percent__100.0__384.0__ |
percent__40.0__640.0__ percent__100.0__384.0__ |
| twelve years ago p was half of q ' s age . if the ratio of their present ages is num__3 : num__4 what will be the total of their present ages ? <o> a ) num__33 <o> b ) num__37 <o> c ) num__29 <o> d ) num__31 <o> e ) num__42 |
e let present age of p and q be num__3 x num__3 x and num__4 x num__4 x respectively . twelve years ago p was half of q ' s age â ‡ ’ ( num__3 x â ˆ ’ num__12 ) = num__0.5 ( num__4 x â ˆ ’ num__12 ) â ‡ ’ num__6 x â ˆ ’ num__24 = num__4 x â ˆ ’ num__12 â ‡ ’ num__2 x = num__12 â ‡ ’ x = num__6 total of their present ages = num__3 x + num__4 x = num__7 x = num__7 Ã — num__6 = num__42 <eor> e <eos> |
e |
multiply__3.0__4.0__ divide__3.0__0.5__ multiply__4.0__6.0__ reverse__0.5__ add__3.0__4.0__ multiply__6.0__7.0__ multiply__6.0__7.0__ |
multiply__3.0__4.0__ divide__3.0__0.5__ multiply__4.0__6.0__ reverse__0.5__ add__3.0__4.0__ multiply__6.0__7.0__ multiply__6.0__7.0__ |
| ( num__6 ) num__6.5 × ( num__36 ) num__4.5 ÷ ( num__216 ) num__4.5 = ( num__6 ) ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
explanation : ( num__6 ) num__6.5 × ( num__36 ) num__4.5 ÷ ( num__216 ) num__4.5 = ( num__6 ) num__6.5 × [ ( num__6 ) num__2 ] num__4.5 ÷ [ ( num__6 ) num__3 ] num__4.5 = ( num__6 ) num__6.5 × ( num__6 ) num__9 ÷ ( num__6 ) num__13.5 = ( num__6 ) ( num__6.5 + num__9 - num__13.5 ) = ( num__6 ) num__2 answer : option b <eor> b <eos> |
b |
subtract__6.5__4.5__ divide__6.0__2.0__ add__6.0__3.0__ multiply__4.5__3.0__ divide__6.0__3.0__ |
subtract__6.5__4.5__ divide__6.0__2.0__ add__6.0__3.0__ add__4.5__9.0__ subtract__6.5__4.5__ |
| a train num__200 m long is running at a speed of num__68 kmph . how long does it take to pass a man who is running at num__8 kmph in the same direction as the train ? <o> a ) num__5 sec <o> b ) num__9 sec <o> c ) num__12 sec <o> d ) num__15 sec <o> e ) num__18 sec |
speed of the train relative to man = ( num__68 - num__8 ) kmph = ( num__60 * num__0.277777777778 ) m / sec = ( num__16.6666666667 ) m / sec time taken by the train to cross the man = time taken by it to cover num__200 m at num__16.6666666667 m / sec = num__200 * num__0.06 sec = num__12 sec answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ divide__200.0__16.6667__ round__12.0__ |
subtract__68.0__8.0__ divide__200.0__16.6667__ divide__200.0__16.6667__ |
| what inequality represents the condition num__1 < x < num__7 ? <o> a ) | x | < num__3 <o> b ) | x + num__5 | < num__4 <o> c ) | x - num__1 | < num__9 <o> d ) | - num__4 + x | < num__3 <o> e ) | num__3 + x | < num__5 |
num__1 < x < num__7 represents segment with length : num__7 - num__1 = num__6 center : ( num__7 + num__1 ) / num__2 = num__4 for any | x - a | < b a is a center and b is a half of length . so | x - num__4 | < num__3.0 represents our inequality ( answer d ) <eor> d <eos> |
d |
subtract__7.0__1.0__ subtract__6.0__2.0__ add__1.0__2.0__ add__1.0__3.0__ |
subtract__7.0__1.0__ subtract__6.0__2.0__ add__1.0__2.0__ add__1.0__3.0__ |
| free notebooks were distributed equally among children of a class . the number of notebooks each child got was one - eighth of the number of children . had the number of children been half each child would have got num__16 notebooks . total how many notebooks were distributed ? <o> a ) num__502 <o> b ) num__512 <o> c ) num__523 <o> d ) num__521 <o> e ) num__500 |
let n be the total number of children . more children less notebooks ( indirect proportion ) if the number children = n number of books each child will get = n / num__8 if the number children = n / num__2 number of books each child will get = num__16 hence we can write as children n : n / num__2 } : : num__16 : n / num__8 ⇒ n × n / num__8 = n / num__2 × num__16 ⇒ n num__8 = num__8.0 = num__8 ⇒ n = num__8 × num__8 = num__64 then total number of notebooks those were distributed = n × n / num__8 = ( num__64 × num__64 ) / num__8 = num__64 × num__8 = num__512 answer is b . <eor> b <eos> |
b |
divide__16.0__8.0__ multiply__8.0__64.0__ multiply__8.0__64.0__ |
divide__16.0__8.0__ multiply__8.0__64.0__ multiply__8.0__64.0__ |
| the total marks obtained by a student in physics chemistry and mathematics is num__150 more than the marks obtained by him in physics . what is the average mark obtained by him in chemistry and mathematics ? <o> a ) num__75 <o> b ) num__28 <o> c ) num__279 <o> d ) num__271 <o> e ) num__27 |
let the marks obtained by the student in physics chemistry and mathematics be p c and m respectively . p + c + m = num__150 + p c + m = num__150 average mark obtained by the student in chemistry and mathematics = ( c + m ) / num__2 = num__75.0 = num__75 . answer : a <eor> a <eos> |
a |
divide__150.0__2.0__ divide__150.0__2.0__ |
divide__150.0__2.0__ divide__150.0__2.0__ |
| a sum of money doubles itself in num__7 years . in how many years it becomes four fold ? <o> a ) num__35 years <o> b ) num__28 years <o> c ) num__14 years <o> d ) num__21 years <o> e ) none of these |
double in seven years . so num__4 times will be in = num__7 × ( num__4 − num__1 ) = num__21 years . answer : d . <eor> d <eos> |
d |
multiply__1.0__21.0__ |
multiply__1.0__21.0__ |
| a car drove from town a to town b without stopping . the car traveled the first num__50 miles of its journey at an average speed of num__25 miles per hour . what was the car ’ s average speed in miles per hour for the remaining num__110 miles if the car ’ s average speed for the entire trip was num__40 miles per hour ? <o> a ) num__28 <o> b ) num__48 <o> c ) num__50 <o> d ) num__55 <o> e ) num__70 |
kaplanofficial solution : to solve this problem you must remember that average speed means total distance divided by total time over an entire journey . this should not be confused with the average of the speeds . the total distance in this problem is num__160 miles but we will need to express the total time in a more complex way . for the first num__50 miles the car traveled at num__25 miles per hour . therefore we can say that the time this took was num__2.0 because distance divided by rate equals time . for the last num__110 miles we are trying to solve for the rate so we can call this speed r . thus the time for the final num__110 miles is num__110 / r . if we put all of this together knowing that the average speed over the entire journey is num__40 miles per hour we get the equation num__160 / ( num__2.0 + num__110 / r ) = num__40 . we can now solve for r and reach our answer . num__160 / ( num__2.0 + num__110 / r ) = num__40 r = num__55 num__55 is choice ( d ) and that ’ s our correct answer . <eor> d <eos> |
d |
add__50.0__110.0__ divide__50.0__25.0__ divide__110.0__2.0__ round__55.0__ |
add__50.0__110.0__ divide__50.0__25.0__ divide__110.0__2.0__ divide__110.0__2.0__ |
| a train sets off at num__2 p . m . at the speed of num__70 kmph . another train starts at num__2 : num__30 p . m . in the same direction at the rate of num__85 kmph . at what time the trains will meet ? <o> a ) num__10.18 p . m <o> b ) num__10.29 p . m <o> c ) num__10.30 p . m <o> d ) num__10.38 p . m <o> e ) num__9.30 p . m |
d = num__70 * num__1 ½ = num__105 km rs = num__85 – num__70 = num__15 t = num__7.0 = num__7 h num__2.30 + num__7 h = num__9.30 p . m . answer : e <eor> e <eos> |
e |
divide__30.0__2.0__ divide__105.0__15.0__ add__7.0__2.3__ round__9.3__ |
divide__30.0__2.0__ divide__105.0__15.0__ add__7.0__2.3__ add__7.0__2.3__ |
| for each num__6 - month period during a light bulb ' s life span the odds of it not burning out from over - use are half what they were in the previous num__6 - month period . if the odds of a light bulb burning out during the first num__6 - month period following its purchase are num__0.125 what are the odds of it burning out during the period from num__6 months to num__1 year following its purchase ? <o> a ) num__0.185185185185 <o> b ) num__0.222222222222 <o> c ) num__0.5 <o> d ) num__0.444444444444 <o> e ) num__0.666666666667 |
p ( of not burning out in a six mnth period ) = num__0.5 of p ( of not burning out in prev num__6 mnth period ) p ( of burning out in num__1 st num__6 mnth ) = num__0.125 - - - > p ( of not burning out in num__1 st num__6 mnth ) = num__1 - num__0.125 = num__0.875 - - - - > p ( of not burning out in a six mnth period ) = num__0.5 * num__0.875 = num__0.444444444444 - - - > p ( of burning out in a six mnth period ) = num__1 - num__0.444444444444 = num__0.571428571429 now p ( of burning out in num__2 nd six mnth period ) = p ( of not burning out in num__1 st six mnth ) * p ( of burning out in a six mnth ) = num__0.875 * num__0.571428571429 = num__0.5 ans c <eor> c <eos> |
c |
subtract__1.0__0.125__ divide__0.5__0.875__ reverse__0.5__ reverse__2.0__ |
subtract__1.0__0.125__ divide__0.5__0.875__ reverse__0.5__ subtract__1.0__0.5__ |
| a person can row at num__10 kmph in still water . if the velocity of the current is num__2 kmph and it takes him num__20 hour to row to a place and come back how far is the place ? <o> a ) num__24 km <o> b ) num__30 km <o> c ) num__96 km <o> d ) num__12 km <o> e ) num__15 km |
speed of down stream = num__10 + num__2 = num__12 kmph speed of upstream = num__10 - num__2 = num__8 kmph let the required distance be xkm x / num__12 + x / num__8 = num__20 num__2 x + num__3 x = num__480 x = num__96 km answer is c <eor> c <eos> |
c |
add__10.0__2.0__ subtract__10.0__2.0__ multiply__8.0__12.0__ round__96.0__ |
add__10.0__2.0__ subtract__10.0__2.0__ multiply__8.0__12.0__ round__96.0__ |
| a b and c have rs . num__500 between them a and c together have rs . num__200 and b and c rs . num__350 . how much does c have ? <o> a ) num__50 <o> b ) num__99 <o> c ) num__88 <o> d ) num__77 <o> e ) num__52 |
a + b + c = num__500 a + c = num__200 b + c = num__350 - - - - - - - - - - - - - - a + b + num__2 c = num__550 a + b + c = num__500 - - - - - - - - - - - - - - - - c = num__50 answer : a <eor> a <eos> |
a |
add__200.0__350.0__ subtract__550.0__500.0__ subtract__550.0__500.0__ |
add__200.0__350.0__ subtract__550.0__500.0__ subtract__550.0__500.0__ |
| sachin was twice as old as ajay num__10 years back . how old is ajay today if sachin will be num__40 years old num__10 years hence <o> a ) num__76 years <o> b ) num__88 years <o> c ) num__97 years <o> d ) num__20 years <o> e ) num__55 years |
sachin ' s age today = num__30 years . sachin ' s age num__10 years back = num__20 years . ajay ' s age num__10 years back = num__10 years . ajay ' s age today = num__20 years answer : d <eor> d <eos> |
d |
subtract__40.0__10.0__ subtract__30.0__10.0__ round__20.0__ |
subtract__40.0__10.0__ subtract__30.0__10.0__ round__20.0__ |
| a man buys a cycle for rs . num__1400 and sells it at a loss of num__15.0 . what is the selling price of the cycle ? <o> a ) rs . num__1090 <o> b ) rs . num__1160 <o> c ) rs . num__1190 <o> d ) rs . num__1202 <o> e ) rs . num__1256 |
s . p . = num__85.0 of rs . num__1400 = rs . num__0.85 x num__1400 = rs . num__1190 answer : c <eor> c <eos> |
c |
percent__85.0__1400.0__ percent__85.0__1400.0__ |
percent__85.0__1400.0__ percent__85.0__1400.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 sec . what is the length of the train ? <o> a ) num__100 <o> b ) num__110 <o> c ) num__120 <o> d ) num__130 <o> e ) num__150 |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__9 = num__150 m answer : option e <eor> e <eos> |
e |
round__150.0__ |
round__150.0__ |
| two brothers take the same route to school on their bicycles one gets to school in num__25 minutes and the second one gets to school in num__36 minutes . the ratio of their speeds is <o> a ) num__4 : num__5 <o> b ) num__1 : num__2 <o> c ) num__6 : num__7 <o> d ) num__5 : num__6 <o> e ) none |
solution let us name the brothers as a and b . = ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š num__25 : â ˆ š num__36 = num__5 : num__6 answer d <eor> d <eos> |
d |
round__5.0__ |
round__5.0__ |
| a number whose fifth part increased by num__5 is equal to its fourth part diminished by num__5 is <o> a ) num__160 <o> b ) num__180 <o> c ) num__200 <o> d ) num__220 <o> e ) none of these |
x / num__5 + num__5 = x / num__4 - num__5 = > x / num__5 - x / num__4 = num__10 x / num__20 = num__10 = > x = num__200 answer : c . <eor> c <eos> |
c |
multiply__5.0__4.0__ multiply__20.0__10.0__ multiply__20.0__10.0__ |
multiply__5.0__4.0__ multiply__20.0__10.0__ multiply__20.0__10.0__ |
| a train is num__460 meter long is running at a speed of num__45 km / hour . in what time will it pass a bridge of num__140 meter length ? <o> a ) num__76 seconds <o> b ) num__18 seconds <o> c ) num__40 seconds <o> d ) num__48 seconds <o> e ) num__45 seconds |
speed = num__45 km / hr = num__45 * ( num__0.277777777778 ) m / sec = num__12.5 m / sec total distance = num__460 + num__140 = num__600 meter time = distance / speed = num__600 * ( num__0.08 ) = num__48 seconds answer : d <eor> d <eos> |
d |
add__460.0__140.0__ divide__600.0__12.5__ round__48.0__ |
add__460.0__140.0__ divide__600.0__12.5__ divide__600.0__12.5__ |
| find the distance covered by a man walking for num__30 min at a speed of num__5 km / hr ? <o> a ) num__1 km <o> b ) num__3 km <o> c ) num__4 km <o> d ) num__2.5 km <o> e ) num__6 km |
distance = num__5 * num__0.5 = num__2.5 km answer is d <eor> d <eos> |
d |
multiply__5.0__0.5__ round__2.5__ |
multiply__5.0__0.5__ multiply__5.0__0.5__ |
| num__90.0 of the population of a village is num__45000 . the total population of the village is ? <o> a ) num__26799 <o> b ) num__24000 <o> c ) num__50000 <o> d ) num__29973 <o> e ) num__12312 |
x * ( num__0.9 ) = num__45000 x = num__500 * num__100 x = num__50000 answer : c <eor> c <eos> |
c |
divide__45000.0__90.0__ divide__90.0__0.9__ divide__45000.0__0.9__ divide__45000.0__0.9__ |
divide__45000.0__90.0__ divide__90.0__0.9__ multiply__100.0__500.0__ multiply__100.0__500.0__ |
| find out the wrong number in the given sequence of numbers . num__1 num__2 num__6 num__15 num__31 num__56 num__94 <o> a ) num__2 <o> b ) num__6 <o> c ) num__15 <o> d ) num__31 <o> e ) num__94 |
num__1 ( result ) + ( num__1 * num__1 ) = num__2 . num__2 ( result ) + ( num__2 * num__2 ) = num__6 . num__6 ( result ) + ( num__3 * num__3 ) = num__15 . num__15 ( result ) + ( num__4 * num__4 ) = num__31 . num__31 ( result ) + ( num__5 * num__5 ) = num__56 . num__56 ( result ) + ( num__6 * num__6 ) = num__92 . now we are getting num__92 not num__94 . . so num__94 is the wrong number of the given . answer : e <eor> e <eos> |
e |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ subtract__94.0__2.0__ multiply__1.0__94.0__ |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ subtract__94.0__2.0__ multiply__1.0__94.0__ |
| if f ( x num__2 – num__1 ) = x num__4 – num__7 x num__2 + k num__1 and f ( x num__3 – num__2 ) = x num__6 – num__9 x num__3 + k num__2 then the value of ( k num__2 – k num__1 ) is _____ <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) none of the above |
explanatory answer given data f ( x num__2 – num__1 ) = x num__4 – num__7 x num__2 + k num__1 f ( x num__3 – num__2 ) = x num__6 – num__9 x num__3 + k num__2 approach and solution when x num__2 = num__1 f ( x num__2 – num__1 ) = f ( num__1 - num__1 ) = f ( num__0 ) = ( num__1 ) num__2 - num__7 ( num__1 ) + k num__1 f ( num__0 ) = - num__6 + k num__1 . . . . . . . . . . ( num__1 ) essentially we have replaced all x num__2 with num__1 . when x num__3 = num__2 f ( x num__3 – num__2 ) = f ( num__2 - num__2 ) = f ( num__0 ) = ( num__2 ) num__2 - num__9 ( num__2 ) + k num__2 f ( num__0 ) = - num__14 + k num__2 . . . . . . . . . . ( num__2 ) essentially we have replaced all x num__3 with num__2 . equating f ( num__0 ) in equations ( num__1 ) and ( num__2 ) ( - num__6 + k num__1 ) = ( - num__14 + k num__2 ) or k num__2 - k num__1 = num__8 the correct answer is choice c . <eor> c <eos> |
c |
multiply__2.0__7.0__ multiply__2.0__4.0__ multiply__2.0__4.0__ |
multiply__2.0__7.0__ add__2.0__6.0__ add__2.0__6.0__ |
| the probability is num__0.5 that a certain coin turns up heads on any given toss . if the coin is tossed three times what is the probability that the coin turns up tails on at least one of the tosses ? <o> a ) num__0.875 <o> b ) num__0.625 <o> c ) num__0.375 <o> d ) num__0.75 <o> e ) num__0.4375 |
p ( num__3 heads ) = num__0.5 * num__0.5 * num__0.5 = num__0.125 . p ( at least one tail ) = num__1 - num__0.125 = num__0.875 . the answer is a . <eor> a <eos> |
a |
negate_prob__0.125__ negate_prob__0.125__ |
negate_prob__0.125__ negate_prob__0.125__ |
| look at this series : num__3 num__4 num__7 num__8 num__11 num__12 . . . what number should come next ? <o> a ) num__11 <o> b ) num__14 <o> c ) num__15 <o> d ) num__17 <o> e ) num__19 |
c num__15 this alternating addition series begins with num__3 ; then num__1 is added to give num__4 ; then num__3 is added to give num__7 ; then num__1 is added and so on . <eor> c <eos> |
c |
add__3.0__12.0__ subtract__4.0__3.0__ add__3.0__12.0__ |
add__3.0__12.0__ subtract__4.0__3.0__ add__3.0__12.0__ |
| if y > num__0 ( num__10 y ) / num__20 + ( num__3 y ) / num__10 is what percent of y ? <o> a ) num__40.0 <o> b ) num__50.0 <o> c ) num__60.0 <o> d ) num__70.0 <o> e ) num__80 % |
can be reduced to y / num__2 + num__3 y / num__10 = num__4 y / num__5 = num__80.0 e <eor> e <eos> |
e |
divide__20.0__10.0__ divide__10.0__2.0__ multiply__20.0__4.0__ multiply__20.0__4.0__ |
divide__20.0__10.0__ divide__10.0__2.0__ multiply__20.0__4.0__ multiply__20.0__4.0__ |
| the average weight of num__16 pupils in a class is num__50.25 kg and that of the remaining num__8 pupils is num__45.15 kg . find the average weights of all the pupils in the class . <o> a ) a ) num__48.55 <o> b ) b ) num__49 <o> c ) c ) num__51 <o> d ) d ) num__61 <o> e ) e ) num__62 |
required average = ( num__50.25 x num__16 + num__45.15 x num__8 ) / ( num__16 + num__8 ) = ( num__804 + num__361.20 ) / num__24 = num__1165.20 / num__24 = num__48.55 kg answer is a <eor> a <eos> |
a |
multiply__16.0__50.25__ multiply__8.0__45.15__ add__16.0__8.0__ add__804.0__361.2__ divide__1165.2__24.0__ divide__1165.2__24.0__ |
multiply__16.0__50.25__ multiply__8.0__45.15__ add__16.0__8.0__ add__804.0__361.2__ divide__1165.2__24.0__ divide__1165.2__24.0__ |
| the speed of a boat in still water is num__60 kmph and the speed of the current is num__10 kmph . find the speed downstream and upstream ? <o> a ) num__40 num__68 kmph <o> b ) num__70 num__50 kmph <o> c ) num__90 num__60 kmph <o> d ) num__40 num__60 kmph <o> e ) num__20 num__60 kmph |
speed downstream = num__60 + num__10 = num__70 kmph speed upstream = num__60 - num__10 = num__50 kmph answer : b <eor> b <eos> |
b |
add__60.0__10.0__ subtract__60.0__10.0__ round__70.0__ |
add__60.0__10.0__ subtract__60.0__10.0__ add__60.0__10.0__ |
| the total of the ages of amar akbar and anthony is num__80 years . what was the total of their ages three years ago ? <o> a ) num__71 years <o> b ) num__72 years <o> c ) num__74 years <o> d ) num__77 years <o> e ) num__75 years |
explanation : required sum = ( num__80 - num__3 x num__3 ) years = ( num__80 - num__9 ) years = num__71 years . answer is a <eor> a <eos> |
a |
subtract__80.0__9.0__ subtract__80.0__9.0__ |
subtract__80.0__9.0__ subtract__80.0__9.0__ |
| two roads xy and yz of num__15 metres and num__20 metres length respectively are perpendicular to each other . what is the distance between x & z by the shortest route ? <o> a ) num__35 metres <o> b ) num__30 metres <o> c ) num__24 metres <o> d ) num__25 metres <o> e ) none of these |
xyz is a right - angled triangle xz = √ num__15 ( num__2 ) + num__20 ( num__2 ) = √ = num__625 = num__25 m answer d <eor> d <eos> |
d |
round__25.0__ |
round__25.0__ |
| the sum of two numbers is num__25 and their difference is num__15 . find their product . <o> a ) num__104 <o> b ) num__100 <o> c ) num__114 <o> d ) num__325 <o> e ) none |
sol . let the numbers be x and y . then x + y = num__25 x - y = num__15 num__2 x = num__40 = > x = num__20 so y = num__5 xy = num__20 * num__5 = num__100 answer b <eor> b <eos> |
b |
add__25.0__15.0__ divide__40.0__2.0__ subtract__25.0__20.0__ multiply__5.0__20.0__ multiply__5.0__20.0__ |
add__25.0__15.0__ divide__40.0__2.0__ subtract__25.0__20.0__ multiply__5.0__20.0__ multiply__5.0__20.0__ |
| for any positive integer n n > num__1 thelengthof n is the number of positive primes ( not necessary distinct ) whose product is n . for ex the length of num__50 is num__5 since num__50 = num__2 x num__5 x num__5 . what is the greatest possible length of a positive integer less than num__1000 . <o> a ) num__10 <o> b ) num__9 <o> c ) num__8 <o> d ) num__7 <o> e ) num__6 |
you are missing something in your post : for any positive integer n n > num__1 thelengthof n is the number of positive primes ( not distinct ) whose product is n . for example the length of num__50 is num__3 since num__50 = ( num__2 ) ( num__5 ) ( num__5 ) the lenght of num__1000 = ( num__2 ) ( num__5 ) ( num__2 ) ( num__5 ) ( num__2 ) ( num__5 ) = num__6 but we need n < num__1000 using num__2 as the base = ( num__2 ) ( num__2 ) ( num__2 ) ( num__2 ) ( num__2 ) ( num__2 ) ( num__2 ) ( num__2 ) ( num__2 ) = num__7 the length of num__512 . the answer is ( d ) <eor> d <eos> |
d |
add__1.0__2.0__ add__1.0__5.0__ add__1.0__6.0__ add__1.0__6.0__ |
add__1.0__2.0__ add__1.0__5.0__ add__1.0__6.0__ add__1.0__6.0__ |
| a is the average ( arithmetic mean ) of the first num__7 positive multiples of num__4 and b is the median of the first num__3 positive multiples of positive integer n . if the value of a ^ num__2 – b ^ num__2 is zero what is the value of n ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
if a ^ num__2 - b ^ num__2 = num__0 then let ' s assume that a = b . a must equal the num__4 th positive multiple of num__4 thus a = num__16 which also equals b . b is the second positive multiple of n thus n = num__8.0 = num__8 . the answer is d . <eor> d <eos> |
d |
multiply__4.0__2.0__ multiply__4.0__2.0__ |
multiply__4.0__2.0__ subtract__16.0__8.0__ |
| if x is to be chosen at random from the set { num__56 num__78 } and y is to be chosen at random from the set { num__7 num__89 } what ` s the probability that xy will be even ? <o> a ) num__0.166666666667 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.833333333333 <o> e ) num__0.666666666667 |
the product to be even either one or both must be even so : p ( x = even or y = even ) = p ( x = even ) + p ( y = even ) - p ( x = even and y = even ) = num__0.5 + num__0.333333333333 - num__0.5 * num__0.333333333333 = num__0.666666666667 ; or : p ( xy = even ) = num__1 - p ( xy = odd ) = num__1 - num__0.5 * num__0.666666666667 = num__0.666666666667 so basically we find the probability of the product to be odd ( which happens when both x and y are odd ) and subtract it from num__1 . answer : e . <eor> e <eos> |
e |
negate_prob__0.3333__ negate_prob__0.3333__ |
negate_prob__0.3333__ negate_prob__0.3333__ |
| if num__7 spiders make num__7 webs in num__7 days then num__1 spider will make num__1 web in how many days ? <o> a ) num__6 <o> b ) num__1 <o> c ) num__3 <o> d ) num__9 <o> e ) num__7 |
let the required number days be x . less spiders more days ( indirect proportion ) less webs less days ( direct proportion ) spiders num__1 : num__7 : : num__7 : x webs num__7 : num__1 : . num__1 * num__7 * x = num__7 * num__1 * num__7 x = num__7 . answer is e . <eor> e <eos> |
e |
round__7.0__ |
round__7.0__ |
| if num__7 spiders make num__7 webs in num__7 days then num__1 spider will make num__1 web in how many days ? <o> a ) num__8 <o> b ) num__6 <o> c ) num__7 <o> d ) num__5 <o> e ) num__1 |
explanation : let the required number days be x . less spiders more days ( indirect proportion ) less webs less days ( direct proportion ) { \ color { blue } therefore } num__1 x num__7 x x = num__7 x num__1 x num__7 = > x = num__7 answer : c <eor> c <eos> |
c |
round__7.0__ |
round__7.0__ |
| in town x num__64 percent of the population are employed and num__48 percent of the population are employed males . what percent of the employed people in town x are females ? <o> a ) num__16.0 <o> b ) num__25.0 <o> c ) num__32.0 <o> d ) num__40.0 <o> e ) num__52 % |
let population be num__100 total employed people = num__64.0 of the population = num__64 employed males = num__48.0 of the population = num__48 employed females = num__16.0 of the population = num__16 employed females % in terms of employed people = ( num__0.25 ) * num__100 = num__25.0 answer is b . <eor> b <eos> |
b |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| two trains of equal length are running on parallel lines in the same direction at num__46 km / hr and num__36 km / hr . the faster train passes the slower train in num__36 seconds . the length of each train is ? <o> a ) num__40 meter <o> b ) num__45 meter <o> c ) num__50 meter <o> d ) num__55 meter <o> e ) none of these |
explanation : let the length of each train is x meter distance will be x + x = num__2 x relative speed = num__46 - num__36 = num__10 km / hr = num__10 * ( num__0.277777777778 ) = num__2.77777777778 m / sec distance = speed * time num__2 x = num__2.77777777778 ∗ num__36 num__2 x = num__100 = > x = num__50 so length of both the trains are num__50 meters option c <eor> c <eos> |
c |
subtract__46.0__36.0__ divide__10.0__36.0__ divide__100.0__2.0__ divide__100.0__2.0__ |
subtract__46.0__36.0__ divide__10.0__36.0__ divide__100.0__2.0__ divide__100.0__2.0__ |
| the cost of five chairs and three tables is rs . num__3110 . cost of one chair is rs . num__210 less than cost one table . what is the cost of two tables and two chairs ? <o> a ) rs . num__1660 / - <o> b ) rs . num__1860 / - <o> c ) rs . num__2600 / - <o> d ) can not be determined <o> e ) none of these |
explanation : let ‘ x ’ be the cost of one chair then cost of num__1 table = x + num__210 given num__5 x + num__3 ( x + num__210 ) = rs . num__3110 num__8 x = num__3110 – num__630 = num__2480 cost of one chair ( x ) = num__310.0 = num__310 cost of one table = num__310 + num__210 = rs . num__520 : . cost of num__2 tables + num__2 chairs = num__2 x num__520 + num__2 x num__310 = rs . num__1660 answer : option a <eor> a <eos> |
a |
add__3.0__5.0__ multiply__210.0__3.0__ subtract__3110.0__630.0__ divide__2480.0__8.0__ add__210.0__310.0__ subtract__3.0__1.0__ multiply__1.0__1660.0__ |
add__3.0__5.0__ multiply__210.0__3.0__ subtract__3110.0__630.0__ divide__2480.0__8.0__ add__210.0__310.0__ subtract__3.0__1.0__ multiply__1.0__1660.0__ |
| a man is num__22 years older than his son . in two years his age will be twice the age of his son . the present age of his son is <o> a ) num__20 years <o> b ) num__21 years <o> c ) num__22 years <o> d ) num__24 years <o> e ) num__26 years |
explanation : let the son ' s present age be x years . then man ' s present age = ( x + num__22 ) years = > ( x + num__22 ) + num__2 = num__2 ( x + num__2 ) = > x + num__24 = num__2 x + num__4 so x = num__20 answer : option a <eor> a <eos> |
a |
add__22.0__2.0__ subtract__22.0__2.0__ subtract__22.0__2.0__ |
add__22.0__2.0__ subtract__22.0__2.0__ subtract__22.0__2.0__ |
| one hour after yolanda started walking from x to y a distance of num__60 miles bob started walking along the same road from y to x . if yolanda â s walking rate was num__5 miles per hour and bob â s was num__6 miles per hour how many miles had bob walked when they met ? <o> a ) num__40 <o> b ) num__25 <o> c ) num__28 <o> d ) num__30 <o> e ) num__30.5 |
let t be the number of hours that bob had walked when he met yolanda . then when they met bob had walked num__4 t miles and yolanda had walked num__5 ( t + num__1 ) miles . these distances must sum to num__60 miles so num__6 t + num__5 ( t + num__1 ) = num__60 which may be solved for t as follows num__6 t + num__5 ( t + num__1 ) = num__60 num__6 t + num__5 t + num__5 = num__60 num__11 t = num__55 t = num__5 ( hours ) therefore bob had walked num__6 t = num__6 ( num__5 ) = num__30 miles when they met . the best answer is d . <eor> d <eos> |
d |
subtract__5.0__4.0__ add__5.0__6.0__ subtract__60.0__5.0__ multiply__5.0__6.0__ round__30.0__ |
subtract__5.0__4.0__ add__5.0__6.0__ subtract__60.0__5.0__ multiply__5.0__6.0__ round__30.0__ |
| two trains leave the same train station at num__6 : num__00 am and num__6 : num__45 am and they travel at num__100 kph and num__125 kph respectively . how many kilometers from the train station will the two trains be together ? <o> a ) num__300 <o> b ) num__325 <o> c ) num__350 <o> d ) num__375 <o> e ) num__400 |
at num__6 : num__45 the first train is num__75 km ahead . the second train gains num__25 km each hour . the time it takes the second train to catch the first train is num__3.0 = num__3 hours . in this time the second train travels num__3 * num__125 = num__375 km . the answer is d . <eor> d <eos> |
d |
subtract__100.0__75.0__ divide__75.0__25.0__ multiply__125.0__3.0__ round__375.0__ |
subtract__100.0__75.0__ divide__75.0__25.0__ multiply__125.0__3.0__ multiply__125.0__3.0__ |
| the sector of a circle has radius of num__21 cm and central angle num__135 o . find its perimeter ? <o> a ) num__91.5 cm <o> b ) num__91.6 cm <o> c ) num__91.2 cm <o> d ) num__91.3 cm <o> e ) num__91.9 cm |
perimeter of the sector = length of the arc + num__2 ( radius ) = ( num__0.375 * num__2 * num__3.14285714286 * num__21 ) + num__2 ( num__21 ) = num__49.5 + num__42 = num__91.5 cm answer : a <eor> a <eos> |
a |
multiply__21.0__2.0__ add__49.5__42.0__ round__91.5__ |
multiply__21.0__2.0__ add__49.5__42.0__ add__49.5__42.0__ |
| what amount does kiran get if he invests rs . num__18000 at num__15.0 p . a . simple interest for five years ? <o> a ) num__31500 <o> b ) num__27773 <o> c ) num__29989 <o> d ) num__28800 <o> e ) num__29883 |
simple interest = ( num__18000 * num__5 * num__15 ) / num__100 = rs . num__13500 amount = p + i = num__18000 + num__13500 = rs . num__31500 answer : a <eor> a <eos> |
a |
percent__100.0__31500.0__ |
percent__100.0__31500.0__ |
| if num__2 ab - c = num__2 a ( b - num__2 c ) which of the following must be true ? <o> a ) a = num__0 and c = num__0 <o> b ) a = num__0.5 and b = num__2 <o> c ) b = num__1 and c = num__0 <o> d ) a = num__1 or b = num__0 <o> e ) a = num__0.25 or c = num__0 |
num__2 ab - c = num__2 a ( b - num__2 c ) num__2 ab - c = num__2 ab - num__4 ac c = num__4 ac num__4 ac - c = num__0 c ( num__4 a - num__1 ) = num__0 either c = num__0 ; or a = num__0.25 e is the answer <eor> e <eos> |
e |
reverse__4.0__ reverse__4.0__ |
reverse__4.0__ reverse__4.0__ |
| eric nick and archi make contributions to the society of nature protection in the ratio of num__5 : num__3 : num__2.5 . if altogether they contribute num__4914 nis how much more money does nick contribute than archi ? <o> a ) num__128 nis <o> b ) num__212 nis <o> c ) num__234 nis <o> d ) num__245 nis <o> e ) num__288 nis |
e : n : a num__5 : num__3 : num__2.5 so total = num__5 x + num__3 x + num__2.5 x = num__10.5 x = num__4914 so x = num__468 nick contribution is num__0.5 more than archi so num__0.5 x = num__0.5 * num__468 = num__234 so c . num__234 nis is the correct answer <eor> c <eos> |
c |
divide__4914.0__10.5__ subtract__3.0__2.5__ multiply__0.5__468.0__ multiply__0.5__468.0__ |
divide__4914.0__10.5__ subtract__3.0__2.5__ multiply__0.5__468.0__ multiply__0.5__468.0__ |
| the sum of ages of num__5 children born num__3 years different each is num__50 yrs . what is the age of the elder child ? <o> a ) num__7 <o> b ) num__9 <o> c ) num__16 <o> d ) num__18 <o> e ) num__21 |
let the ages of children be x ( x + num__3 ) ( x + num__6 ) ( x + num__9 ) and ( x + num__12 ) years . then x + ( x + num__3 ) + ( x + num__6 ) + ( x + num__9 ) + ( x + num__12 ) = num__50 num__5 x = num__20 x = num__4 . x + num__12 = num__4 + num__12 = num__16 c <eor> c <eos> |
c |
add__3.0__6.0__ add__3.0__9.0__ subtract__9.0__5.0__ add__4.0__12.0__ add__4.0__12.0__ |
add__3.0__6.0__ add__3.0__9.0__ subtract__9.0__5.0__ add__4.0__12.0__ add__4.0__12.0__ |
| a man can row downstream at num__18 kmph and upstream at num__12 kmph . find the speed of the man in still water and the speed of stream respectively ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__4 <o> e ) num__9 |
let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = num__18 - - - ( num__1 ) and x - y = num__12 - - - ( num__2 ) from ( num__1 ) & ( num__2 ) num__2 x = num__30 = > x = num__15 y = num__3 . answer : a <eor> a <eos> |
a |
add__18.0__12.0__ divide__30.0__2.0__ subtract__18.0__15.0__ round__3.0__ |
add__18.0__12.0__ divide__30.0__2.0__ subtract__18.0__15.0__ subtract__18.0__15.0__ |
| if r = num__2 ^ num__3 * num__5 ^ num__2 * num__7 and s = num__2 ^ num__3 * num__3 ^ num__2 * num__5 which of the following is equal to the greatest common divisor of r and s ? <o> a ) num__2 * num__5 <o> b ) num__2 ^ num__3 * num__5 <o> c ) num__2 ^ num__3 * num__5 ^ num__2 <o> d ) num__2 * num__3 * num__5 * num__7 <o> e ) num__2 ^ num__3 * num__3 ^ num__2 * num__5 ^ num__2 * num__7 |
gcd = product of prime factors raised to the least power = num__2 ^ num__3 * num__5 the answer is b . <eor> b <eos> |
b |
subtract__5.0__3.0__ |
subtract__5.0__3.0__ |
| what is the square root of num__36 divided by num__6 ? <o> a ) num__9 <o> b ) num__36 <o> c ) num__122 <o> d ) num__6 <o> e ) num__1 |
square root is a number times itself square root of num__36 = num__6 num__1.0 = num__1 ( e ) num__1 <eor> e <eos> |
e |
volume_cube__1.0__ |
volume_cube__1.0__ |
| in num__10 years a will be twice as old as b was num__10 years ago . if a is now num__9 years older than b the present age of b is : <o> a ) num__19 years <o> b ) num__29 years <o> c ) num__39 years <o> d ) num__49 years <o> e ) num__59 years |
solution let b ' s present age = x years . then a ' s present age = ( x + num__9 ) years . ∴ ( x + num__9 ) + num__10 = num__2 ( x - num__10 ) ⇔ x + num__19 = num__2 x - num__20 ⇔ x = num__39 . answer c <eor> c <eos> |
c |
add__10.0__9.0__ multiply__10.0__2.0__ add__19.0__20.0__ add__19.0__20.0__ |
add__10.0__9.0__ multiply__10.0__2.0__ add__19.0__20.0__ add__19.0__20.0__ |
| i flew my tiny seaplane to visit my mother . on the flight up i flew at num__140 mph . on the way home i flew num__88 mph . what was my average speed for the trip ? <o> a ) num__114 mph <o> b ) num__110 mph <o> c ) num__88 mph <o> d ) num__100 mph <o> e ) num__99 mph |
( num__140 mph + num__88 mph ) / num__2 = num__114 mph correct answer is : a <eor> a <eos> |
a |
round__114.0__ |
round__114.0__ |
| pradeep has to obtain num__25.0 of the total marks to pass . he got num__185 marks and failed by num__25 marks . the maximum marks are <o> a ) num__840 <o> b ) num__600 <o> c ) num__800 <o> d ) num__1000 <o> e ) num__900 |
explanation : let their maximum marks be x . then num__25.0 of x = num__185 + num__25 = > num__0.25 x = num__210 x = ( num__8404.0 ) x = num__840 . answer : a <eor> a <eos> |
a |
add__25.0__185.0__ divide__210.0__0.25__ divide__210.0__0.25__ |
add__25.0__185.0__ divide__210.0__0.25__ divide__210.0__0.25__ |
| a and b are two stations num__390 km apart . a train starts from a at num__10 a . m . and travels towards b at num__65 kmph . another train starts from b at num__11 a . m . and travels towards a at num__35 kmph . at what time do they meet ? <o> a ) num__2.15 pm <o> b ) num__2.15 am <o> c ) num__3.15 pm <o> d ) num__3.15 am <o> e ) none of them |
suppose they meet x hours after num__10 a . m . then ( distance moved by first in x hrs ) + [ distance moved by second in ( x - num__1 ) hrs ] = num__390 . num__65 x + num__35 ( x - num__1 ) = num__390 = > num__100 x = num__425 = > x = num__4.25 so they meet num__4 hrs . num__15 min . after num__10 a . m i . e . at num__2.15 p . m . answer is a . <eor> a <eos> |
a |
subtract__11.0__10.0__ add__65.0__35.0__ add__390.0__35.0__ divide__425.0__100.0__ add__11.0__4.0__ round__2.15__ |
subtract__11.0__10.0__ add__65.0__35.0__ add__390.0__35.0__ divide__425.0__100.0__ add__11.0__4.0__ round__2.15__ |
| num__2 no ' s are respectively num__40.0 & num__60.0 more than num__3 rdnumber . find the ration of two numbers ? <o> a ) num__4 : num__5 <o> b ) num__5 : num__7 <o> c ) num__7 : num__8 <o> d ) num__7 : num__9 <o> e ) num__7 : num__11 |
step num__1 : let the third number is a then first number is num__140.0 of a = num__140 x a / num__100 = num__7 a / num__5 and second number is num__160.0 of b = num__160 x b / num__100 = num__8 b / num__5 . step num__2 : now ratio of first and second number is num__7 a / num__5 : num__8 b / num__5 = num__35 a : num__40 b = num__7 : num__8 . c <eor> c <eos> |
c |
subtract__3.0__2.0__ add__40.0__60.0__ add__2.0__3.0__ add__60.0__100.0__ divide__40.0__5.0__ subtract__40.0__5.0__ add__2.0__5.0__ |
subtract__3.0__2.0__ add__40.0__60.0__ add__2.0__3.0__ add__60.0__100.0__ divide__40.0__5.0__ subtract__40.0__5.0__ divide__35.0__5.0__ |
| a five - digit number divisible by num__3 is to be formed using numerical num__0 num__1 num__2 num__3 num__4 and num__5 without repetition . the total number of ways this can be done is : <o> a ) num__122 <o> b ) num__210 <o> c ) num__216 <o> d ) num__217 <o> e ) num__225 |
first step : we should determine which num__5 digits from given num__6 would form the num__5 digit number divisible by num__3 . we have six digits : num__0 num__1 num__2 num__3 num__4 num__5 . their sum = num__15 . for a number to be divisible by num__3 the sum of the digits must be divisible by num__3 . as the sum of the six given numbers is num__15 ( divisible by num__3 ) only num__5 digits good to form our num__5 digit number would be num__15 - num__0 = { num__1 num__2 num__3 num__4 num__5 } and num__15 - num__3 = { num__0 num__1 num__2 num__4 num__5 } . meaning that no other num__5 from given six will total the number divisible by num__3 . second step : we have two set of numbers : num__1 num__2 num__3 num__4 num__5 and num__0 num__1 num__2 num__4 num__5 . how many num__5 digit numbers can be formed using these two sets : num__1 num__2 num__3 num__4 num__5 - - > num__5 ! as any combination of these digits would give us num__5 digit number divisible by num__3 . num__5 ! = num__120 . num__0 num__1 num__2 num__4 num__5 - - > here we can not use num__0 as the first digit otherwise number wo n ' t be any more num__5 digit and become num__4 digit . so desired # would be total combinations num__5 ! minus combinations with num__0 as the first digit ( combination of num__4 ) num__4 ! - - > num__5 ! - num__4 ! = num__4 ! ( num__5 - num__1 ) = num__4 ! * num__4 = num__96 num__120 + num__96 = num__216 answer : c . <eor> c <eos> |
c |
multiply__3.0__2.0__ multiply__3.0__5.0__ add__96.0__120.0__ multiply__1.0__216.0__ |
multiply__3.0__2.0__ multiply__3.0__5.0__ add__96.0__120.0__ multiply__1.0__216.0__ |
| a single discount equivalent to the discount series of num__10.0 num__10.0 and num__5.0 is ? <o> a ) num__23.05 <o> b ) num__31.95 <o> c ) num__21.65 <o> d ) num__23.45 <o> e ) num__23.15 |
num__100 * ( num__0.9 ) * ( num__0.9 ) * ( num__0.95 ) = num__76.95 num__100 - num__76.95 = num__23.05 answer : a <eor> a <eos> |
a |
percent__23.05__100.0__ |
percent__23.05__100.0__ |
| the average age of the district level hockey team of eleven is num__22 years . the average age gets increased by num__1 year when the coach age is also included . what is the age of the coach ? <o> a ) num__22 yrs <o> b ) num__34 yrs <o> c ) num__27 yrs <o> d ) num__28 yrs <o> e ) num__19 yrs |
total age of players in team = num__22 x num__11 = num__242 when coach is included total members = num__12 avg age increases by one becomes num__23 total age when coach is included = num__23 x num__12 = num__276 age of coach = num__276 - num__242 = num__34 yrs . answer : b <eor> b <eos> |
b |
multiply__22.0__11.0__ add__1.0__11.0__ add__22.0__1.0__ multiply__12.0__23.0__ add__22.0__12.0__ add__22.0__12.0__ |
multiply__22.0__11.0__ add__1.0__11.0__ add__22.0__1.0__ multiply__12.0__23.0__ subtract__276.0__242.0__ subtract__276.0__242.0__ |
| at what rate percent on simple interest will a sum of money double itself in num__30 years ? <o> a ) num__3 num__0.111111111111 % <o> b ) num__3 num__2.33333333333 % <o> c ) num__7 num__0.333333333333 % <o> d ) num__5 num__0.333333333333 % <o> e ) num__1 num__0.333333333333 % |
p = ( p * num__30 * r ) / num__100 r = num__3 num__0.333333333333 % answer : a <eor> a <eos> |
a |
percent__3.0__100.0__ |
percent__3.0__100.0__ |
| how many three digit numbers can having only two consecutive digits identical is <o> a ) num__153 <o> b ) num__162 <o> c ) num__168 <o> d ) num__163 <o> e ) none of these |
when num__0 is the repeated digit like num__100 num__200 . . . . num__9 in number when num__0 occurs only once like num__110 num__220 . . . . . num__9 in number when num__0 does not occur like num__112211 . . . . . num__2 × ( num__8 × num__9 ) = num__144 in number . hence total = num__9 + num__9 + num__144 = num__162 . answer b <eor> b <eos> |
b |
divide__200.0__100.0__ round__162.0__ |
divide__200.0__100.0__ round__162.0__ |
| if m ^ num__2 - num__8 m + num__1 = num__0 then m ^ num__3 + m ^ ( - num__3 ) = ? <o> a ) num__487 <o> b ) num__488 <o> c ) num__489 <o> d ) num__457 <o> e ) num__495 |
here from equation num__1 we will get m ^ num__2 - num__8 m + num__1 = num__0 m ^ num__2 = num__8 m - num__1 m = ( num__8 m - num__1 ) / m . . . . . . . . . ( num__1 ) m = num__8 - num__1 / m so m + num__1 / m = num__8 now in equation num__2 m ^ num__3 + num__1 / m ^ num__3 ( m + num__1 / m ) ( m ^ num__2 - m * num__1 / m + num__1 / m ^ num__2 ) ( m + num__1 / m ) ( m ^ num__2 - num__1 + num__1 / m ^ num__2 ) ( m + num__1 / m ) ( m ^ num__2 + num__1 / m ^ num__2 - num__1 ) ( m + num__1 / m ) { ( m + num__1 / m ) ^ num__2 - num__2 m * num__1 / m - num__1 } { since a ^ num__2 + b ^ num__2 = ( a + b ) ^ num__2 - num__2 ab } ( m + num__1 / m ) { ( m + num__1 / m ) ^ num__2 - num__2 - num__1 } now putting the value of m from equation num__1 num__8 ( num__8 ^ num__2 - num__2 - num__1 ) num__8 ( num__64 - num__3 ) = num__488 answer : b <eor> b <eos> |
b |
multiply__1.0__488.0__ |
multiply__1.0__488.0__ |
| the ratio of the radius of two circles is num__2 : num__7 and then the ratio of their areas is ? <o> a ) num__1 : num__7 <o> b ) num__2 : num__9 <o> c ) num__1 : num__9 <o> d ) num__4 : num__49 <o> e ) num__3 : num__4 |
r num__1 : r num__2 = num__2 : num__7 Î r num__1 ^ num__2 : Î r num__2 ^ num__2 r num__1 ^ num__2 : r num__2 ^ num__2 = num__4 : num__49 answer : d <eor> d <eos> |
d |
square_perimeter__1.0__ power__7.0__2.0__ square_perimeter__1.0__ |
square_perimeter__1.0__ power__7.0__2.0__ power__4.0__1.0__ |
| a girl was asked to multiply a certain number by num__43 . she multiplied it by num__34 and got his answer less than the correct one by num__1215 . find the number to be multiplied . <o> a ) num__130 <o> b ) num__132 <o> c ) num__135 <o> d ) num__136 <o> e ) num__138 |
let the required number be x . then num__43 x – num__34 x = num__1215 or num__9 x = num__1215 or x = num__135 . required number = num__135 answer : c <eor> c <eos> |
c |
subtract__43.0__34.0__ divide__1215.0__9.0__ divide__1215.0__9.0__ |
subtract__43.0__34.0__ divide__1215.0__9.0__ divide__1215.0__9.0__ |
| num__3889 + num__16.952 - ? = num__3854.002 <o> a ) num__51.95 <o> b ) num__49.75 <o> c ) num__45.97 <o> d ) num__47.59 <o> e ) num__45.79 |
let num__3889 + num__16.952 - x = num__3854.002 . then x = ( num__3889 + num__16.952 ) - num__3854.002 = num__3905.952 - num__3854.002 = num__51.95 . answer is a <eor> a <eos> |
a |
add__3889.0__16.952__ subtract__3905.952__3854.002__ subtract__3905.952__3854.002__ |
add__3889.0__16.952__ subtract__3905.952__3854.002__ subtract__3905.952__3854.002__ |
| a train crosses a platform of num__120 m in num__15 sec same train crosses another platform of length num__180 m in num__18 sec . then find the length of the train ? <o> a ) num__299 <o> b ) num__180 <o> c ) num__288 <o> d ) num__127 <o> e ) num__122 |
length of the train be ‘ x ’ x + num__8.0 = x + num__10.0 num__6 x + num__720 = num__5 x + num__900 x = num__180 m answer : b <eor> b <eos> |
b |
divide__120.0__15.0__ divide__180.0__18.0__ multiply__120.0__6.0__ subtract__15.0__10.0__ multiply__180.0__5.0__ round__180.0__ |
divide__120.0__15.0__ divide__180.0__18.0__ multiply__120.0__6.0__ subtract__15.0__10.0__ add__180.0__720.0__ round__180.0__ |
| num__8597 - ? = num__7429 - num__4358 <o> a ) num__5426 <o> b ) num__5706 <o> c ) num__5526 <o> d ) num__5476 <o> e ) none of these |
explanation : num__7429 let num__8597 - x = num__3071 - num__4358 then x = num__8597 - num__3071 - - - - = num__5526 num__3071 - - - - answer : c <eor> c <eos> |
c |
subtract__7429.0__4358.0__ subtract__8597.0__3071.0__ subtract__8597.0__3071.0__ |
subtract__7429.0__4358.0__ subtract__8597.0__3071.0__ subtract__8597.0__3071.0__ |
| rajan got married num__8 years ago . his present age is num__1.2 times his age at the time of his marriage . rajan ' s sister was num__10 years younger to him at the time of his marriage . the age of rajan ' s sister is <o> a ) num__32 years <o> b ) num__36 years <o> c ) num__38 years <o> d ) num__40 years <o> e ) none |
solution let rajan ' s present age be x years . then his age at the time of marriage = ( x - num__8 ) years . therefore x = num__1.2 ( x - num__6 ) num__5 x = num__6 x - num__48 x = num__48 . rajan ' s sister ' s age at the time of his marriage = ( x - num__8 ) - num__10 = ( x - num__18 ) = num__30 rajan ' s sister ' s present age = ( num__30 + num__8 ) = num__38 years . answer c <eor> c <eos> |
c |
divide__6.0__1.2__ multiply__8.0__6.0__ add__8.0__10.0__ multiply__5.0__6.0__ add__8.0__30.0__ add__8.0__30.0__ |
divide__6.0__1.2__ multiply__8.0__6.0__ add__8.0__10.0__ subtract__48.0__18.0__ add__8.0__30.0__ add__8.0__30.0__ |
| which of the following is closer to ( num__20 ! + num__19 ! ) / ( num__20 ! - num__19 ! ) ? <o> a ) num__11 <o> b ) num__13 <o> c ) num__1 <o> d ) num__3 <o> e ) num__7 |
notice that num__20 ! = ( num__20 ) ( num__19 ! ) ( num__20 ! + num__19 ! ) * ( num__20 ! - num__19 ! ) = [ num__19 ! ( num__20 + num__1 ) ] * [ num__19 ! ( num__20 - num__1 ) ] = num__1.10526315789 ≈ num__1 answer : c <eor> c <eos> |
c |
subtract__20.0__19.0__ reverse__1.0__ |
subtract__20.0__19.0__ subtract__20.0__19.0__ |
| let x be the smallest positive integer such that num__28 and num__42 are factors of num__678 - x so the value of x is : <o> a ) num__28 <o> b ) num__48 <o> c ) num__6 <o> d ) num__62 <o> e ) num__34 |
you can simply check which of the possible values for x will make num__678 - x divisible by num__28 and num__42 simultaneously . the only possible value is x = num__6 ( because num__678 - num__6 = num__672 = num__28 * num__24 = num__42 * num__16 ) . answer c <eor> c <eos> |
c |
subtract__678.0__6.0__ divide__672.0__28.0__ divide__672.0__42.0__ subtract__678.0__672.0__ |
subtract__678.0__6.0__ divide__672.0__28.0__ divide__672.0__42.0__ subtract__678.0__672.0__ |
| if | num__8 y − num__3 | = | num__2 y + num__7 | which of the following could be a value of y ? <o> a ) num__1 <o> b ) num__1.66666666667 <o> c ) num__0.4 <o> d ) - num__0.666666666667 <o> e ) - num__1 |
num__8 y − num__3 = num__2 y + num__7 or num__8 y − num__3 = - num__2 y - num__7 num__6 y = num__10 or num__10 y = - num__4 y = num__1.66666666667 or y = - num__0.4 answer : b <eor> b <eos> |
b |
subtract__8.0__2.0__ add__8.0__2.0__ divide__8.0__2.0__ divide__10.0__6.0__ divide__4.0__10.0__ divide__10.0__6.0__ |
subtract__8.0__2.0__ add__8.0__2.0__ subtract__7.0__3.0__ divide__10.0__6.0__ divide__4.0__10.0__ divide__10.0__6.0__ |
| the least number of complete years in which a sum of money put out at num__20.0 compound interest will be more than doubled is : <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
explanation : p ( num__1 + num__0.2 ) n > num__2 p = > ( num__1.2 ) n > num__2 . now ( num__1.2 x num__1.2 x num__1.2 x num__1.2 ) > num__2 . so n = num__4 years . answer is b <eor> b <eos> |
b |
add__1.0__0.2__ multiply__20.0__0.2__ multiply__20.0__0.2__ |
add__1.0__0.2__ multiply__20.0__0.2__ multiply__20.0__0.2__ |
| a train num__710 m long is running at a speed of num__78 km / hr . if it crosses a tunnel in num__1 min then the length of the tunnel is ? <o> a ) num__2898 <o> b ) num__277 <o> c ) num__500 <o> d ) num__297 <o> e ) num__590 |
speed = num__78 * num__0.277777777778 = num__21.6666666667 m / sec . time = num__1 min = num__60 sec . let the length of the train be x meters . then ( num__710 + x ) / num__60 = num__21.6666666667 x = num__590 m . answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ round__590.0__ |
hour_to_min_conversion__ multiply__1.0__590.0__ |
| jane makes toy bears . when she works with an assistant she makes num__53 percent more bears per week and works num__10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ? <o> a ) num__20.0 <o> b ) num__70.0 <o> c ) num__100.0 <o> d ) num__180.0 <o> e ) num__200 % |
let ' s assume just jane num__40 bears per num__40 / hrs a week so that is num__1 bear / hr . with an assistant she makes num__61.2 bears per num__36 hours a week or num__1.7 bears / hr ( [ num__40 bears * num__1.53 ] / [ num__40 hrs * . num__90 ] ) . [ ( num__1.7 - num__1 ) / num__1 ] * num__100.0 = num__70.0 answer : b <eor> b <eos> |
b |
divide__61.2__36.0__ divide__61.2__40.0__ add__10.0__90.0__ round__70.0__ |
divide__61.2__36.0__ divide__61.2__40.0__ add__10.0__90.0__ divide__70.0__1.0__ |
| two sets of num__7 consecutive positive integers have exactly one integer in common . the sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set ? <o> a ) num__24 <o> b ) num__17 <o> c ) num__28 <o> d ) num__42 <o> e ) it can not be determined from the information given . |
a = ( num__1 num__23 num__45 num__67 ) sum of this = num__28 b = ( num__7 num__89 num__1011 num__1213 ) sum of this = num__70 the differenct between num__70 - num__28 = num__42 hence num__42 is the answer i . e . d <eor> d <eos> |
d |
subtract__70.0__28.0__ multiply__1.0__42.0__ |
subtract__70.0__28.0__ subtract__70.0__28.0__ |
| the length of rectangle is thrice its breadth and its perimeter is num__88 m find the area of the rectangle ? <o> a ) num__432 sq m <o> b ) num__363 sq m <o> c ) num__452 sq m <o> d ) num__428 sq m <o> e ) num__528 sq m |
num__2 ( num__3 x + x ) = num__88 l = num__33 b = num__11 lb = num__33 * num__11 = num__363 answer : b <eor> b <eos> |
b |
multiply__33.0__11.0__ multiply__33.0__11.0__ |
multiply__33.0__11.0__ multiply__33.0__11.0__ |
| if num__5 and num__8 are factors of num__60 n what is the minimum value of n ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__7 <o> d ) num__14 <o> e ) num__56 |
num__60 n / num__7 * num__8 should be integer = > num__2 * num__2 * num__3 * num__5 * n / num__5 * num__2 * num__2 * num__2 = num__3 * n / num__2 must be an integer for this to be true n must multiple of num__2 thus min of n = num__2 hence a <eor> a <eos> |
a |
subtract__7.0__5.0__ subtract__5.0__2.0__ subtract__5.0__3.0__ |
subtract__7.0__5.0__ subtract__5.0__2.0__ subtract__5.0__3.0__ |
| find the average of first num__40 natural numbers . <o> a ) num__20.5 <o> b ) num__18 <o> c ) num__19.5 <o> d ) num__19 <o> e ) num__20 |
sum of first n natural numbers = n ( n + num__1 ) / num__2 so sum of first num__40 natural numbers = num__40 * num__20.5 = num__820 required average = num__20.5 = num__20.5 answer : a <eor> a <eos> |
a |
multiply__40.0__20.5__ multiply__1.0__20.5__ |
multiply__40.0__20.5__ multiply__1.0__20.5__ |
| a person want to give his money of $ num__2700 to his num__3 children a b c in the ratio num__5 : num__6 : num__9 . what is the a + c share ? <o> a ) $ num__1890 <o> b ) $ num__2002 <o> c ) $ num__3500 <o> d ) $ num__1500 <o> e ) $ num__5640 |
a ' s share = num__2700 * num__0.25 = $ num__675 c ' s share = num__2700 * num__0.45 = $ num__1215 a + d = $ num__1890 answer is a <eor> a <eos> |
a |
multiply__2700.0__0.25__ multiply__2700.0__0.45__ add__675.0__1215.0__ add__675.0__1215.0__ |
multiply__2700.0__0.25__ multiply__2700.0__0.45__ add__675.0__1215.0__ add__675.0__1215.0__ |
| in what time will a train num__110 m long cross an electric pole it its speed be num__144 km / hr ? <o> a ) num__2.75 sec <o> b ) num__4.25 sec <o> c ) num__5 sec <o> d ) num__12.5 sec <o> e ) num__6 sec |
speed = num__144 * num__0.277777777778 = num__40 m / sec time taken = num__2.75 = num__2.75 sec . answer : a <eor> a <eos> |
a |
divide__110.0__40.0__ round__2.75__ |
divide__110.0__40.0__ divide__110.0__40.0__ |
| ( num__3 * num__10 ^ num__2 ) * ( num__4 * num__10 ^ - num__2 ) = ? <o> a ) num__14 <o> b ) num__12 <o> c ) num__1200 <o> d ) num__1.2 <o> e ) num__14.11 |
num__3 * num__10 ^ num__2 = num__300 num__4 * num__10 ^ - num__2 = num__0.04 ( num__3 * num__10 ^ num__2 ) * ( num__4 * num__10 ^ - num__2 ) = num__300 * num__0.04 = num__12.00 the answer is option b <eor> b <eos> |
b |
multiply__3.0__4.0__ multiply__3.0__4.0__ |
multiply__3.0__4.0__ multiply__3.0__4.0__ |
| the price of commodity x increases by num__40 paise every year while the price of commodity y increases by num__15 paise every year . if in num__1988 the price of commodity x was num__4.20 and that of y was num__6.30 in which year will commodity x cost num__40 paise more than commodity y ? <o> a ) num__1997 <o> b ) num__1998 <o> c ) num__1999 <o> d ) num__2000 <o> e ) none of these |
suppose in ‘ n ’ years the price of comodity x will be more by num__40 paise than that of commodity y . ∴ num__420 + num__40 n – num__630 – num__15 n = num__40 or num__25 n – num__210 = num__40 or num__25 n = num__250 or n = num__250 ⁄ num__25 = num__10 years . answer b <eor> b <eos> |
b |
subtract__40.0__15.0__ subtract__630.0__420.0__ add__40.0__210.0__ subtract__25.0__15.0__ add__1988.0__10.0__ |
subtract__40.0__15.0__ subtract__630.0__420.0__ add__40.0__210.0__ subtract__25.0__15.0__ add__1988.0__10.0__ |
| how long does a train num__110 m long running at the speed of num__72 km / hr takes to cross a bridge num__132 m length ? <o> a ) num__13.1 sec <o> b ) num__12.1 second <o> c ) num__22.1 sec <o> d ) num__32.1 sec <o> e ) num__13.2 sec |
speed = num__72 * num__0.277777777778 = num__20 m / sec total distance covered = num__110 + num__132 = num__242 m . required time = num__12.1 = num__12.1 sec . answer : b <eor> b <eos> |
b |
add__110.0__132.0__ divide__242.0__20.0__ round__12.1__ |
add__110.0__132.0__ divide__242.0__20.0__ divide__242.0__20.0__ |
| the parameter of a square is equal to the perimeter of a rectangle of length num__16 cm and breadth num__14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ? <o> a ) num__23.56 <o> b ) num__23.59 <o> c ) num__23.76 <o> d ) num__23.12 <o> e ) num__23.57 |
let the side of the square be a cm . parameter of the rectangle = num__2 ( num__16 + num__14 ) = num__60 cm parameter of the square = num__60 cm i . e . num__4 a = num__60 a = num__15 diameter of the semicircle = num__15 cm circimference of the semicircle = num__0.5 ( ∏ ) ( num__15 ) = num__0.5 ( num__3.14285714286 ) ( num__15 ) = num__23.5714285714 = num__23.57 cm to two decimal places answer : e <eor> e <eos> |
e |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
| a rectangular region has a fence along three sides and a wall along the fourth side . the fenced side opposite the wall is twice the length of each of the other two fenced sides . if the area of the rectangular region is num__200 square feet what is the total length of the fence in feet ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__30 <o> d ) num__40 <o> e ) num__80 |
two sides each = x the third = num__2 x and the wall length is thus num__2 x too x * num__2 x = num__2 x ^ num__2 = num__200 ie x ^ num__2 = num__100 ie x = num__10 l = num__20 w = num__10 total lenght of fence = num__2 * num__10 + num__20 = num__40 my answer is d <eor> d <eos> |
d |
multiply__2.0__10.0__ square_perimeter__10.0__ square_perimeter__10.0__ |
multiply__2.0__10.0__ multiply__2.0__20.0__ multiply__2.0__20.0__ |
| the price of rice is reduced by num__2.0 . how many kilograms of rice can now be bought for the money which was sufficient to buy num__49 kg of rice earlier ? <o> a ) num__48 kg <o> b ) num__49 kg <o> c ) num__50 kg <o> d ) num__51 kg <o> e ) none |
sol . let the original price be rs . num__100 per kg . money required to buy num__49 kg of rice = rs . ( num__100 x num__49 ) = rs . num__4900 . new price = rs . num__98 per kg . ∴ quantity of rice bought = [ num__50.0 ] kg = num__50 kg . answer c <eor> c <eos> |
c |
percent__2.0__4900.0__ percent__100.0__50.0__ |
percent__2.0__4900.0__ percent__100.0__50.0__ |
| two pipes a and b can fill a tank in num__12 and num__24 minutes respectively . if both the pipes are used together how long will it take to fill the tank ? <o> a ) num__9 min <o> b ) num__8 min <o> c ) num__4 min <o> d ) num__6 min <o> e ) none of these |
explanation : tank can be filled by pipe a in num__12 minutes and pipe b in num__24 minutes . = > part filled by a in num__1 minute = num__0.0833333333333 = > part filled by b in num__1 minute = num__0.0416666666667 = > part filled by both in num__1 minute = num__0.0833333333333 + num__0.0416666666667 = num__0.125 = num__0.125 time taken to fill the entire tank = num__8 minutes answer b <eor> b <eos> |
b |
divide__1.0__12.0__ divide__1.0__24.0__ add__0.0417__0.0833__ divide__1.0__0.125__ round__8.0__ |
divide__1.0__12.0__ divide__1.0__24.0__ add__0.0417__0.0833__ divide__1.0__0.125__ round__8.0__ |
| a train sets off at num__2 : num__00 pm at the speed of num__70 km / h . another train starts at num__3 : num__00 pm in the same direction at the rate of num__80 km / h . at what time will the second train catch the first train ? <o> a ) num__10 : num__00 <o> b ) num__10 : num__30 <o> c ) num__11 : num__00 <o> d ) num__11 : num__30 <o> e ) num__12 : num__00 |
in one hour the first train travels num__70 km . the second train catches the first train at a rate of num__80 km / h - num__70 km / h = num__10 km / h . the second train will catch the first train in num__7.0 = num__7 hours so at num__10 : num__00 pm . the answer is a . <eor> a <eos> |
a |
subtract__80.0__70.0__ divide__70.0__10.0__ round__10.0__ |
subtract__80.0__70.0__ divide__70.0__10.0__ divide__70.0__7.0__ |
| the h . c . f . of two numbers is num__12 and their difference is num__12 . the numbers are <o> a ) num__66 num__78 <o> b ) num__70 num__82 <o> c ) num__94 num__106 <o> d ) num__84 num__96 <o> e ) none |
solution out of the given numbers the two with h . c . f . num__12 and difference num__12 are num__84 and num__96 answer d <eor> d <eos> |
d |
add__12.0__84.0__ lcm__12.0__84.0__ |
add__12.0__84.0__ lcm__12.0__84.0__ |
| a positive number x is multiplied by num__4 and this product is then divided by num__3 . if the positive square root of the result of these two operations equals x what is the value of x ? <o> a ) num__2.25 <o> b ) num__1.5 <o> c ) num__1.33333333333 <o> d ) num__0.333333333333 <o> e ) num__0.5 |
we need to produce an equation from the information given in the problem stem . we are first given that x is multiplied by num__2 and then the product is divided by num__3 . this gives us : num__2 x / num__3 next we are given that the positive square root of the result ( which is num__2 x / num__3 ) is equal to x . this gives us √ ( num__2 x / num__3 ) = x num__2 x / num__3 = x ^ num__2 num__2 x = num__3 x ^ num__2 num__3 x ^ num__2 – num__2 x = num__0 x ( num__3 x – num__2 ) = num__0 x = num__0 or num__3 x – num__2 = num__0 num__3 x = num__2 x = num__0.333333333333 because x is positive x = num__0.333333333333 . the answer is d . <eor> d <eos> |
d |
reverse__3.0__ reverse__3.0__ |
reverse__3.0__ reverse__3.0__ |
| a ( num__5 w ^ num__2 ) is the ( x y ) coordinate of point located on the parabola y = x ^ num__2 + num__11 . what is the value of w ? <o> a ) num__3 . <o> b ) num__4 . <o> c ) num__5 . <o> d ) num__6 . <o> e ) num__9 . |
y = x ^ num__2 + num__11 w ^ num__2 = num__5 ^ num__2 + num__11 w ^ num__2 = num__36 w = num__6 answer d <eor> d <eos> |
d |
subtract__11.0__5.0__ subtract__11.0__5.0__ |
subtract__11.0__5.0__ subtract__11.0__5.0__ |
| in a vessel there are num__10 litres of alcohol . an operation is defined as taking out five litres of what is present in the vessel and adding num__10 litres of pure water to it . what is the ratio of alcohol to water after two operations ? <o> a ) num__1 : num__0 <o> b ) num__1 : num__5 <o> c ) num__1 : num__1 <o> d ) num__1 : num__2 <o> e ) num__1 : num__9 |
final concentration = initial concentration ( num__1 − replacement quantityfinal volume ) ( num__1 − replacement quantityfinal volume ) final concentration = = num__1 × ( num__1 − num__1015 ) = num__13 = num__1 × ( num__1 − num__1015 ) = num__13 final concentration = num__13 × ( num__1 − num__1020 ) = num__1613 × ( num__1 − num__1020 ) = num__16 so ratio of alcohol : water = num__1 : num__5 answer : b <eor> b <eos> |
b |
subtract__1020.0__1015.0__ reverse__1.0__ |
subtract__1020.0__1015.0__ reverse__1.0__ |
| if a b and c together can finish a piece of work in num__2 days . a alone in num__12 days and b in num__18 days then c alone can do it in ? <o> a ) num__5 days <o> b ) num__4 days <o> c ) num__3 days <o> d ) num__2 days <o> e ) num__7 days |
c = num__0.5 - num__0.0833333333333 – num__0.0555555555556 = num__0.333333333333 = > num__3 days answer : c <eor> c <eos> |
c |
round__3.0__ |
round__3.0__ |
| six students - num__3 boys and num__3 girls - are to sit side by side for a makeup exam . how many ways could they arrange themselves given that no two boys and no two girls can sit next to one another ? <o> a ) a ) num__12 <o> b ) b ) num__36 <o> c ) c ) num__72 <o> d ) d ) num__240 <o> e ) e ) num__720 |
in order not to have two boys or two girls next to one another we must make an alternate sitting arrangement . ( b - g - b - g - b - g ) or ( g - b - g - b - g - b ) . total ways = num__2 c num__1 * num__3 ! * num__3 ! = num__2 * num__6 * num__6 = num__72 ans - c <eor> c <eos> |
c |
subtract__3.0__2.0__ multiply__3.0__2.0__ multiply__72.0__1.0__ |
subtract__3.0__2.0__ multiply__3.0__2.0__ multiply__72.0__1.0__ |
| a dishonest dealer professes to sell his goods at cost price but still gets num__20.0 profit by using a false weight . what weight does he substitute for a kilogram ? <o> a ) num__833 num__0.333333333333 grams <o> b ) num__850 num__0.333333333333 grams <o> c ) num__825 num__0.333333333333 grams <o> d ) num__810 num__0.333333333333 grams <o> e ) num__870 num__0.333333333333 grams |
a num__833 num__0.333333333333 grams if the cost price is rs . num__100 then to get a profit of num__20.0 the selling price should be rs . num__120 . if num__120 kg are to be sold and the dealer gives only num__100 kg to get a profit of num__20.0 . how many grams he has to give instead of one kilogram ( num__1000 gm ) . num__120 gm - - - - - - num__100 gm num__1000 gm - - - - - - ? ( num__1000 * num__100 ) / num__120 = num__833.333333333 = num__833 num__0.333333333333 grams . <eor> a <eos> |
a |
percent__100.0__833.0__ |
percent__100.0__833.0__ |
| a man divides $ num__8400 among num__5 sons num__4 daughters and num__1 nephews . if each daughter receives four times as much as each nephews and each son receives five times as much as each nephews how much does each daughter receive ? <o> a ) a ) $ num__200 <o> b ) b ) $ num__1000 <o> c ) c ) $ num__800 <o> d ) d ) $ num__1200 <o> e ) e ) $ num__400 |
let the share of each nephews be $ x . then share of each daughter = $ num__4 x share of each son = $ num__5 x . so num__5 * num__5 x + num__4 * num__4 x + num__1 * x = num__8400 num__25 x + num__16 x + x = num__8600 num__42 x = num__8400 x = num__200 . daughter receives four times of nephew so num__4 * num__200 = num__800 . so each daughter receives $ num__800 . answer is option c ) $ num__800 . <eor> c <eos> |
c |
divide__8400.0__42.0__ multiply__4.0__200.0__ multiply__4.0__200.0__ |
divide__8400.0__42.0__ multiply__4.0__200.0__ multiply__4.0__200.0__ |
| the vertices of a rectangle in the standard ( x y ) coordinate place are ( num__00 ) ( num__06 ) ( num__100 ) and ( num__106 ) . if a line through ( num__11 ) partitions the interior of this rectangle into num__2 regions that have equal areas what is the slope of this line ? <o> a ) num__0.5 <o> b ) num__0.333333333333 <o> c ) num__0.25 <o> d ) num__0.2 <o> e ) num__0.166666666667 |
we should know that a rectangle can be divided into two equal areas by a straight line only when the straight line passes through the center of the rectangle . the center of the rectangle is the point ( num__53 ) . the slope of line passing through points ( num__11 ) and ( num__53 ) = ( num__3 - num__1 ) / ( num__5 - num__1 ) = num__0.5 the answer is a . <eor> a <eos> |
a |
divide__106.0__2.0__ divide__6.0__2.0__ subtract__3.0__2.0__ subtract__6.0__1.0__ reverse__2.0__ reverse__2.0__ |
divide__106.0__2.0__ divide__6.0__2.0__ subtract__3.0__2.0__ subtract__6.0__1.0__ reverse__2.0__ reverse__2.0__ |
| if x ^ num__6 + y ^ num__6 = num__500 then the greatest possible value of x is between <o> a ) num__0 and num__3 <o> b ) num__3 and num__6 <o> c ) num__6 and num__9 <o> d ) num__9 and num__12 <o> e ) num__12 and num__15 |
for the greatest possible value of x ^ num__6 we must minimize the value of y ^ num__6 i . e . lets say y ^ num__6 = num__0 then we need to find a number x such that x ^ num__6 < num__500 . num__2 ^ num__6 = num__64 and num__3 ^ num__6 = num__729 so we can say that the maximum possible value of x can be a little less than num__3 hence answer = between num__0 and num__3 hence a <eor> a <eos> |
a |
divide__6.0__2.0__ multiply__6.0__0.0__ |
divide__6.0__2.0__ multiply__6.0__0.0__ |
| a cistern which could be filled in num__9 hours takes one hour more to be filled owing to a leak in its bottom . if the cistern is full in what time will the leak empty it ? <o> a ) num__90 <o> b ) num__25 <o> c ) num__40 <o> d ) num__23 <o> e ) num__60 |
num__0.111111111111 - num__1 / x = num__0.1 = > num__90 hrs answer a <eor> a <eos> |
a |
divide__9.0__0.1__ round__90.0__ |
divide__9.0__0.1__ divide__9.0__0.1__ |
| a reduction of num__25.0 in the price of oil enables a house wife to obtain num__5 kgs more for rs . num__1100 what is the reduced price for kg ? <o> a ) s . num__40 <o> b ) s . num__46 <o> c ) s . num__55 <o> d ) s . num__41 <o> e ) s . num__42 |
num__800 * ( num__0.25 ) = num__275 - - - - num__5 ? - - - - num__1 = > rs . num__55 answer : c <eor> c <eos> |
c |
percent__25.0__1100.0__ percent__5.0__1100.0__ percent__5.0__1100.0__ |
percent__25.0__1100.0__ percent__5.0__1100.0__ percent__5.0__1100.0__ |
| a rectangular field is to be fenced on three sides leaving a side of num__30 feet uncovered . if the area of the field is num__810 sq . feet how many feet of fencing will be required ? <o> a ) num__34 <o> b ) num__40 <o> c ) num__68 <o> d ) num__84 <o> e ) none |
explanation we have : l = num__30 ft and lb = num__810 sq . ft . so b = num__27 ft . length of fencing = ( l + num__2 b ) = ( num__30 + num__54 ) ft = num__84 ft . answer d <eor> d <eos> |
d |
multiply__2.0__27.0__ triangle_area__2.0__84.0__ |
multiply__2.0__27.0__ triangle_area__2.0__84.0__ |
| a train num__130 m long passes a man running at num__6 kmph in the direction opposite to that of the train in num__6 seconds . the speed of the train is <o> a ) num__54 kmph <o> b ) num__60 kmph <o> c ) num__66 kmph <o> d ) num__72 kmph <o> e ) num__82 kmph |
speed of train relative to man : num__21.6666666667 * num__3.6 km / hr = num__78 km / hr let speed of train = x therefore x + num__6 = num__78 x = num__78 - num__6 x = num__72 km / hr answer : d <eor> d <eos> |
d |
divide__130.0__6.0__ subtract__78.0__6.0__ round__72.0__ |
divide__130.0__6.0__ subtract__78.0__6.0__ subtract__78.0__6.0__ |
| the table below shows how many coaches work with each of the major sports teams at kristensen school . although no single coach works with all three teams num__3 coaches work with both the track and tennis teams num__2 coaches work with both the track and baseball teams and num__3 coach works with both the tennis and baseball teams . how many different coaches work with these three teams ? sports no of coaches track num__8 tennis num__5 baseball num__4 <o> a ) num__6 <o> b ) num__9 <o> c ) num__11 <o> d ) num__12 <o> e ) num__17 |
x = num__8 + num__5 + num__4 - ( num__3 + num__2 + num__3 ) = num__9 answer is b <eor> b <eos> |
b |
choose__9.0__8.0__ |
choose__9.0__8.0__ |
| a sum of rs . num__2600 is lent out in num__2 parts si at num__10.0 for num__5 yrs is equal to si on num__2 nd part at num__9.0 rate for num__6 yrs . find the ratio of parts . <o> a ) num__21 : num__23 <o> b ) num__27 : num__25 <o> c ) num__23 : num__27 <o> d ) num__23 : num__24 <o> e ) num__23 : num__26 |
solution : given si num__1 = si num__2 p num__1 : p num__2 = num__1 / r num__1 t num__1 : num__1 / r num__2 t num__2 = num__1 / ( num__10 * num__5 ) : num__1 / ( num__9 : num__6 ) num__27 : num__25 b <eor> b <eos> |
b |
subtract__10.0__9.0__ subtract__27.0__2.0__ add__2.0__25.0__ |
subtract__10.0__9.0__ subtract__27.0__2.0__ multiply__1.0__27.0__ |
| a train running at a speed of num__36 kmph crosses an electric pole in num__12 seconds . in how much time will it cross a num__360 m long platform ? <o> a ) num__37 min <o> b ) num__55 min <o> c ) num__48 min <o> d ) num__67 min <o> e ) num__45 min |
c num__48 min let the length of the train be x m . when a train crosses an electric pole the distance covered is its own length . so x = num__12 * num__36 * num__0.277777777778 m = num__120 m . time taken to cross the platform = ( num__120 + num__360 ) / num__36 * num__0.277777777778 = num__48 min . <eor> c <eos> |
c |
add__36.0__12.0__ round__48.0__ |
add__36.0__12.0__ add__36.0__12.0__ |
| how long will it take a sum of money invested at num__6.0 p . a . s . i . to increase its value by num__60.0 ? <o> a ) num__11 years <o> b ) num__12 years <o> c ) num__10 years <o> d ) num__15 years <o> e ) num__8 years |
sol . let the sum be x . then s . i . = num__60.0 of x = num__3 x / num__5 ; rate = num__6.0 . â ˆ ´ time = [ num__100 * num__3 x / num__5 * num__1 / x * num__6 ] = num__10 years answer c <eor> c <eos> |
c |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| if a - b = num__3 and a num__2 + b num__2 = num__29 find the value of ab . <o> a ) num__10 <o> b ) num__12 <o> c ) num__15 <o> d ) num__18 <o> e ) num__20 |
explanation : num__2 ab = ( a num__2 + b num__2 ) - ( a - b ) num__2 = num__29 - num__9 = num__20 ab = num__10 . answer : a <eor> a <eos> |
a |
subtract__29.0__9.0__ divide__20.0__2.0__ divide__20.0__2.0__ |
subtract__29.0__9.0__ divide__20.0__2.0__ subtract__20.0__10.0__ |
| ( x ) + num__4671 + num__6514 - num__7687 = num__19190 . calculate the value of x <o> a ) num__15615 <o> b ) num__15692 <o> c ) num__15687 <o> d ) num__15112 <o> e ) num__15690 |
x + num__4671 + num__6514 - num__7687 = num__19190 = x + num__4671 + num__6514 = num__19190 + num__7687 = x + num__11185 = num__26877 = x = num__26877 - num__11185 = num__15692 answer is b <eor> b <eos> |
b |
add__4671.0__6514.0__ add__7687.0__19190.0__ subtract__26877.0__11185.0__ subtract__26877.0__11185.0__ |
add__4671.0__6514.0__ add__7687.0__19190.0__ subtract__26877.0__11185.0__ subtract__26877.0__11185.0__ |
| a father said to his son ` ` i was as old as you are at present at the time of your birth . ' ' if the father ' s age is num__50 years now the son ' s age num__10 years back was ? <o> a ) num__12 yr <o> b ) num__15 yr <o> c ) num__14 yr <o> d ) num__10 yr <o> e ) num__20 yr |
let the son ' s present age be x years then num__50 - x = x x = num__25 son ' s age num__10 years back = num__25 - num__10 = num__15 years answer is b <eor> b <eos> |
b |
subtract__25.0__10.0__ subtract__25.0__10.0__ |
subtract__25.0__10.0__ subtract__25.0__10.0__ |
| the volume of a cube is num__1728 cc . find its surface ? <o> a ) num__864 <o> b ) num__288 <o> c ) num__299 <o> d ) num__267 <o> e ) num__289 |
a num__3 = num__1728 = > a = num__12 num__6 a num__2 = num__6 * num__12 * num__12 = num__864 answer : a <eor> a <eos> |
a |
square_perimeter__3.0__ surface_cube__12.0__ surface_cube__12.0__ |
square_perimeter__3.0__ surface_cube__12.0__ surface_cube__12.0__ |
| deepak can read x pages in num__10 minutes . at this rate how long will it take him to read y pages ? <o> a ) ( num__12 / x ) * y <o> b ) ( num__5 / x ) * y <o> c ) ( num__10 / x ) * y <o> d ) ( num__10 x ) * y <o> e ) ( num__13 / x ) * y |
x page in num__10 mins i . e . num__1 page in num__10 / x mins i . e . y pages in ( num__10 / x ) * y answer : option c <eor> c <eos> |
c |
multiply__10.0__1.0__ |
multiply__10.0__1.0__ |
| the speed of a railway engine is num__84 km per hour when no compartment is attached and the reduction in speed is directly proportional to the square root of the number of compartments attached . if the speed of the train carried by this engine is num__24 km per hour when num__9 compartments are attached the maximum number of compartments that can be carried by the engine is : <o> a ) num__19 <o> b ) num__18 <o> c ) num__16 <o> d ) num__17 <o> e ) num__14 |
the reduction in speed is directly proportional to the square root of the number of compartments attached doesreductionmean amount subtracted ? or percentage decrease ? there are at least two interpretations and the wording does not provide a clear interpretation between them . evidently what the question intends is the subtraction interpretation . what is subtracted from the speed is directly proportional to the square root of the number of compartments attached . in other words if s = speed and n = number of compartments then s = num__84 - k * sqrt ( n ) wherekis a constant of the proportionality . in general if a is directly proportional to b we can write a = k * b and solve for k . if n = num__9 then s = num__24 num__24 = num__84 - k * sqrt ( num__9 ) = num__84 - num__3 k k = num__20 now we need to know : what value of n makes s go to zero ? num__0 = num__84 - num__20 * sqrt ( n ) num__20 * sqrt ( n ) = num__84 sqrt ( n ) = num__4.2 n = num__4.2 ^ num__2 > num__17 with num__18 compartments the train does not budge . therefore it would budge if there were one fewer cars . thus num__17 is the maximum number of cars the engine can pull and still move . d <eor> d <eos> |
d |
divide__84.0__20.0__ subtract__20.0__3.0__ multiply__9.0__2.0__ round__17.0__ |
divide__84.0__20.0__ subtract__20.0__3.0__ multiply__9.0__2.0__ subtract__20.0__3.0__ |
| the area of a square field is num__4802 m ( power ) num__2 the length of its diagonal is : <o> a ) num__98 m <o> b ) num__102 m <o> c ) num__95 m <o> d ) num__105 m <o> e ) num__96 m |
let the diagonal be d metres then num__0.5 d ( power ) num__2 = num__4802 d num__2 = num__9604 d = √ num__9604 d = num__98 m answer is a . <eor> a <eos> |
a |
multiply__4802.0__2.0__ power__9604.0__0.5__ triangle_area__2.0__98.0__ |
multiply__4802.0__2.0__ power__9604.0__0.5__ triangle_area__2.0__98.0__ |
| a tank is filled by three pipes with uniform flow . the first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone . the second pipe fills the tank num__5 hours faster than the first pipe and num__4 hours slower than the third pipe . find the time required by the first pipe to fill the tank ? <o> a ) num__10 hours <o> b ) num__15 hours <o> c ) num__17 hours <o> d ) num__18 hours <o> e ) num__20 hours |
explanation : suppose first pipe alone takes x hours to fill the tank . then second and third pipes will take ( x - num__5 ) and ( x - num__9 ) hours respectively to fill the tank . as per question we get num__1 / x + num__1 / x − num__5 = num__1 / x − num__9 = > x − num__5 + x / x ( x − num__5 ) = num__1 / x − num__9 = > ( num__2 x − num__5 ) ( x − num__9 ) = x ( x − num__5 ) = > x num__2 − num__18 x + num__45 = num__0 after solving this euation we get ( x - num__15 ) ( x + num__3 ) = num__0 as value can not be negative so x = num__15 option b <eor> b <eos> |
b |
add__5.0__4.0__ subtract__5.0__4.0__ multiply__2.0__9.0__ multiply__5.0__9.0__ subtract__5.0__2.0__ round__15.0__ |
add__5.0__4.0__ subtract__5.0__4.0__ multiply__2.0__9.0__ multiply__5.0__9.0__ subtract__5.0__2.0__ divide__45.0__3.0__ |
| x starts a business with rs . num__54000 . y joins in the business after num__9 months with rs . num__27000 . what will be the ratio in which they should share the profit at the end of the year ? <o> a ) num__1 : num__2 <o> b ) num__2 : num__1 <o> c ) num__8 : num__3 <o> d ) num__3 : num__5 <o> e ) num__8 : num__1 |
explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = num__54000 * num__12 : num__27000 * num__3 = num__54 * num__12 : num__27 * num__3 = num__2 * num__4 : num__1 = num__8 : num__1 . answer : option e <eor> e <eos> |
e |
subtract__12.0__9.0__ multiply__9.0__3.0__ divide__54000.0__27000.0__ divide__12.0__3.0__ subtract__3.0__2.0__ subtract__9.0__1.0__ subtract__9.0__1.0__ |
subtract__12.0__9.0__ multiply__9.0__3.0__ divide__54000.0__27000.0__ divide__12.0__3.0__ subtract__3.0__2.0__ multiply__2.0__4.0__ multiply__1.0__8.0__ |
| a circular rim num__28 inches in diameter rotates the same number of inches per second as a circular rim num__49 inches in diameter . if the smaller rim makes x revolutions per second how many revolutions per minute does the larger rim makes in terms of x ? <o> a ) num__48 pi / x <o> b ) num__75 x <o> c ) num__34.2 x <o> d ) num__24 x <o> e ) x / num__75 |
c = ( pi ) d c ( small ) : ( pi ) * num__28 c ( large ) : ( pi ) * num__49 lets say the time horizon is num__60 seconds so during that time the smaller rim covers a distance of ( pi ) * num__28 * num__60 = ( pi ) * ( num__1680 ) inches ( pi ) * ( num__1680 ) = ( pi ) * ( num__49 ) ( x ) pi * ( num__34.2 ) = pi * ( x ) num__34.2 = x answer : c . num__34.2 x <eor> c <eos> |
c |
hour_to_min_conversion__ multiply__28.0__60.0__ round__34.2__ |
hour_to_min_conversion__ multiply__28.0__60.0__ round__34.2__ |
| three business people who wish to invest in a new company . each person is willing to pay one third of the total investment . . after careful calculation they realize that each of them would pay $ num__7800 less if they could find two more equal investors . how much is the total investment in the new business . <o> a ) a ) $ num__58500 <o> b ) b ) $ num__54000 <o> c ) c ) $ num__21000 <o> d ) d ) $ num__5400 <o> e ) e ) $ num__3 |
600 |
initially each invest in x . hence total investment is num__3 x . total investment is also num__5 ( x - num__7800 ) . num__3 x = num__5 ( x - num__7800 ) x = num__5 * num__3900.0 = num__19500 num__3 x = num__58500 and the answer is a . <eor> a <eos> |
a |
a |
| find the largest number which divides num__62 num__132237 to leave the same reminder <o> a ) num__30 <o> b ) num__32 <o> c ) num__35 <o> d ) num__45 <o> e ) num__55 |
explanation : trick is hcf of ( num__237 - num__132 ) ( num__132 - num__62 ) ( num__237 - num__62 ) = hcf of ( num__70 num__105175 ) = num__35 option c <eor> c <eos> |
c |
subtract__132.0__62.0__ subtract__70.0__35.0__ |
subtract__132.0__62.0__ subtract__70.0__35.0__ |
| laura took out a charge account at the general store and agreed to pay num__9.0 simple annual interest . if she charges $ num__35 on her account in january how much will she owe a year later assuming she does not make any additional charges or payments ? <o> a ) $ num__2.10 <o> b ) $ num__37.10 <o> c ) $ num__37.16 <o> d ) $ num__38.10 <o> e ) $ num__38.15 |
principal that is amount taken by laura at year beginning = num__35 $ rate of interest = num__9.0 interest = ( num__0.09 ) * num__35 = num__3.15 $ total amount that laura owes a year later = num__35 + num__3.15 = num__38.15 $ answer e <eor> e <eos> |
e |
multiply__35.0__0.09__ add__35.0__3.15__ add__35.0__3.15__ |
multiply__35.0__0.09__ add__35.0__3.15__ add__35.0__3.15__ |
| a number of num__47 marbles is to be divided and contain with boxes . if each box is to contain num__3 num__4 or num__5 marbles what is the largest possible number of boxes ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__15 <o> d ) num__16 <o> e ) num__17 |
to maximize # of boxes we should minimize marbles per box : num__14 * num__3 + num__1 * num__5 = num__47 - - > num__14 + num__1 = num__15 . answer : c . <eor> c <eos> |
c |
subtract__4.0__3.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
subtract__4.0__3.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
| in the rectangular coordinate system points ( num__2 num__0 ) and ( – num__2 num__0 ) both lie on circle c . what is the maximum possible value of the radius of c ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__8 <o> d ) num__16 <o> e ) none of the above |
the answer is a it takes num__3 distinct points to define a circle . only num__2 are given here . the two points essentially identify a single chord of the circle c . since no other information is provided however the radius of the circle can essentially be anything . all this information tell us is that the radius isgreater num__2 <eor> a <eos> |
a |
side_by_diagonal__2.0__0.0__ |
side_by_diagonal__2.0__0.0__ |
| in an office totally there are num__6400 employees and num__65.0 of the total employees are males . num__25.0 of the males in the office are at - least num__50 years old . find the number of males aged below num__50 years ? <o> a ) num__3127 <o> b ) num__3128 <o> c ) num__3120 <o> d ) num__3122 <o> e ) num__3124 |
number of male employees = num__6400 * num__0.65 = num__4160 required number of male employees who are less than num__50 years old = num__4160 * ( num__100 - num__25 ) % = num__4160 * num__0.75 = num__3120 . answer : c <eor> c <eos> |
c |
multiply__6400.0__0.65__ divide__65.0__0.65__ multiply__4160.0__0.75__ multiply__4160.0__0.75__ |
multiply__6400.0__0.65__ divide__65.0__0.65__ multiply__4160.0__0.75__ multiply__4160.0__0.75__ |
| the population of a town is num__10000 . it increases annually at the rate of num__15.0 p . a . what will be its population after num__2 years ? <o> a ) num__13300 <o> b ) num__13350 <o> c ) num__13225 <o> d ) num__13500 <o> e ) num__13600 |
formula : ( after = num__100 denominator ago = num__100 numerator ) num__10000 Ã — num__1.15 Ã — num__1.15 = num__13225 c <eor> c <eos> |
c |
percent__100.0__13225.0__ |
percent__100.0__13225.0__ |
| ifaequals the sum of the even integers from num__2 to num__100 inclusive andbequals the sum of the odd integers from num__1 to num__99 inclusive what is the value of a - b ? <o> a ) num__1 <o> b ) num__50 <o> c ) num__19 <o> d ) num__20 <o> e ) num__21 |
this is a solution from beatthegmat : even numbers : ( num__100 - num__2 ) / num__2 + num__1 = num__50 even integers . ( num__100 + num__2 ) / num__2 = num__51 is the average of the even set . sum = avg * ( # of elements ) = num__51 * num__50 = num__2550 = a odd numbers : ( num__99 - num__1 ) / num__2 + num__1 = num__50 odd integers . ( num__99 + num__1 ) / num__2 = num__50 is the average of the odd set . sum = avg * ( # of elements ) = num__50 * num__50 = num__2500 = b a - b = num__2550 - num__2500 = num__50 . ( b ) <eor> b <eos> |
b |
divide__100.0__2.0__ add__1.0__50.0__ multiply__50.0__51.0__ subtract__2550.0__50.0__ divide__100.0__2.0__ |
divide__100.0__2.0__ add__1.0__50.0__ multiply__50.0__51.0__ subtract__2550.0__50.0__ divide__100.0__2.0__ |
| maximun number of identical pieces ( of same size ) of a cake by making only num__3 cuts ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__5 |
since cake is a num__3 d object it will have length breadth height . . . first cut across length now we will get num__2 pieces then cut across breadth we will get num__4 parts in total now . . . now cut across height finally we will get num__8 parts . . . answer : c <eor> c <eos> |
c |
multiply__2.0__4.0__ multiply__2.0__4.0__ |
multiply__2.0__4.0__ multiply__2.0__4.0__ |
| the ratio of two numbers is num__4 : num__5 and their h . c . f . is num__4 . their l . c . m . is <o> a ) num__48 <o> b ) num__22 <o> c ) num__56 <o> d ) num__80 <o> e ) num__67 |
explanation : let the numbers be num__4 x and num__5 x . then their h . c . f . = x . so x = num__4 . so the numbers num__16 and num__20 . l . c . m . of num__16 and num__20 = num__80 . option d <eor> d <eos> |
d |
multiply__4.0__5.0__ multiply__4.0__20.0__ multiply__4.0__20.0__ |
multiply__4.0__5.0__ multiply__4.0__20.0__ multiply__4.0__20.0__ |
| if ajay drives at num__0.8 th of his usual speed to his office he is num__6 minutes late . what is his usual time to reach his office ? <o> a ) num__36 <o> b ) num__24 <o> c ) num__30 <o> d ) num__18 <o> e ) num__19 |
explanation : let t be his usual time to reach his office and v be his usual speed . v = d / t … … … . ( d is the distance ajay travels while going to his office ) vt = d at v num__1 = num__4 v / num__5 ; t num__1 = t + num__6 num__4 v / num__5 = d / ( t + num__6 ) num__4 v / num__5 * ( t + num__6 ) = d num__4 v / num__5 * ( t + num__6 ) = vt on solving we get t = num__24 minutes answer b <eor> b <eos> |
b |
subtract__6.0__1.0__ multiply__6.0__4.0__ round__24.0__ |
add__1.0__4.0__ multiply__6.0__4.0__ multiply__6.0__4.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 seconds . find the length of the train ? <o> a ) num__150 meter <o> b ) num__299 meter <o> c ) num__666 meter <o> d ) num__155 meter <o> e ) num__144 meter |
speed = num__60 * ( num__0.277777777778 ) m / sec = num__16.6666666667 m / sec length of train ( distance ) = speed * time ( num__16.6666666667 ) * num__9 = num__150 meter answer : a <eor> a <eos> |
a |
round__150.0__ |
round__150.0__ |
| in a num__500 m race the ratio of the speeds of two contestants a and b is num__3 : num__4 . a has a start of num__155 m . then a wins by : <o> a ) num__60 m <o> b ) num__20 m <o> c ) num__40 m <o> d ) num__20 m <o> e ) num__23 m |
to reach the winning post a will have to cover a distance of ( num__500 - num__155 ) m i . e . num__345 m . while a covers num__3 m b covers num__4 m . while a covers num__345 m b covers num__4 x num__115.0 m = num__460 m . thus when a reaches the winning post b covers num__460 m and therefore remains num__40 m behind . a wins by num__40 m . answer : c <eor> c <eos> |
c |
subtract__500.0__155.0__ divide__345.0__3.0__ multiply__4.0__115.0__ subtract__500.0__460.0__ round__40.0__ |
subtract__500.0__155.0__ divide__345.0__3.0__ multiply__4.0__115.0__ subtract__500.0__460.0__ subtract__500.0__460.0__ |
| a train passes a station platform in num__35 sec and a man standing on the platform in num__20 sec . if the speed of the train is num__54 km / hr . what is the length of the platform ? <o> a ) num__228 <o> b ) num__240 <o> c ) num__225 <o> d ) num__166 <o> e ) num__1811 |
speed = num__54 * num__0.277777777778 = num__15 m / sec . length of the train = num__15 * num__20 = num__300 m . let the length of the platform be x m . then ( x + num__300 ) / num__35 = num__15 = > x = num__225 m . answer : c <eor> c <eos> |
c |
subtract__35.0__20.0__ multiply__20.0__15.0__ round__225.0__ |
subtract__35.0__20.0__ multiply__20.0__15.0__ round__225.0__ |
| sam purchased num__20 dozens of toys at the rate of rs . num__375 per dozen . he sold each one of them at the rate of rs . num__33 . what was his percentage profit ? <o> a ) num__5.9 <o> b ) num__5.6 <o> c ) num__5.6 <o> d ) num__9.6 <o> e ) num__1.6 % |
c . p . of num__1 toy = num__31.25 = rs . num__31.25 s . p of num__1 toy = rs . num__33 profit = num__1.75 / num__31.25 * num__100 = num__5.6 = num__5.6 answer : c <eor> c <eos> |
c |
percent__100.0__5.6__ |
percent__100.0__5.6__ |
| city a and city b are num__140 miles apart . train c departs city a heading towards city b at num__4 : num__00 and travels at num__40 miles per hour . train d departs city b heading towards city a at num__4 : num__35 and travels at num__20 miles per hour . the trains travel on parallel tracks . at what time do the two trains meet ? <o> a ) num__5 : num__00 <o> b ) num__5 : num__30 <o> c ) num__6 : num__00 <o> d ) num__6 : num__35 <o> e ) num__7 : num__00 |
train c has traveled num__20 mi in the half hour before train d has started its journey . num__140 - num__20 = num__120 num__40 + num__20 = num__60 mph num__120 mi / num__60 mph = num__2 hrs num__4 : num__35 pm + num__2 hrs = num__6 : num__35 pm answer : d . num__6 : num__35 <eor> d <eos> |
d |
subtract__140.0__20.0__ hour_to_min_conversion__ divide__40.0__20.0__ add__4.0__2.0__ round__6.0__ |
subtract__140.0__20.0__ add__40.0__20.0__ divide__40.0__20.0__ add__4.0__2.0__ add__4.0__2.0__ |
| two full tanks one shaped like the cylinder and the other like a cone contain liquid fuel the cylindrical tank held num__500 lts more then the conical tank after num__200 lts of fuel is pumped out from each tank the cylindrical tank now contains twice the amount of fuel in the canonical tank how many lts of fuel did the cylindrical tank have when it was full ? <o> a ) num__1200 <o> b ) num__279 <o> c ) num__2879 <o> d ) num__27976 <o> e ) num__27711 |
let the cylindrical tank capacity x + num__500 then the conical tank capacity = x after num__200 lts pumped out then remaining fuel with the tanks = x + num__300 x - num__200 given that first term is doubt the second . x + num__300 x − num__200 = num__21 x + num__300 x − num__200 = num__21 solving we get x = num__700 cylindrical tank capacity = num__1200 lts answer : a <eor> a <eos> |
a |
square_perimeter__300.0__ square_perimeter__300.0__ |
square_perimeter__300.0__ square_perimeter__300.0__ |
| a mixture of certain quantity of milk with num__16 liters of water is worth num__90 p per liter . if pure milk be worth rs num__1.08 per liter how much milk is there in the mixture ? <o> a ) num__100 <o> b ) num__90 <o> c ) num__80 <o> d ) num__75 <o> e ) num__70 |
the mean value is num__90 p and price of water is num__0 p by the alligation rule : quantity of milk / quantity of water = ( num__90 - num__0 ) / ( num__108 - num__90 ) = num__5.0 = num__5.0 proportion milk is num__5.0 there in the mixture therefore num__5 * num__16 = num__80 liters milk is there in the mixture . ans - c <eor> c <eos> |
c |
multiply__16.0__5.0__ multiply__16.0__5.0__ |
multiply__16.0__5.0__ multiply__16.0__5.0__ |
| tough and tricky questions : arithmetic . ( num__56 ^ num__2 + num__56 ^ num__2 ) / num__28 ^ num__2 = <o> a ) num__4 <o> b ) num__18 <o> c ) num__29 <o> d ) num__8 <o> e ) num__116 |
ans is num__8 my approach was : ( num__56 ^ num__2 + num__56 ^ num__2 ) / num__28 ^ num__2 = num__56 ( num__56 + num__56 ) / num__28 * num__28 = num__56 * num__4.0 * num__28 = num__2 * num__4 = num__8 d <eor> d <eos> |
d |
divide__8.0__2.0__ multiply__2.0__4.0__ |
divide__8.0__2.0__ multiply__2.0__4.0__ |
| a sum of money invested at compound interest to rs . num__800 in num__3 years and to rs num__840 in num__4 years . the rate on interest per annum is . <o> a ) num__4.0 <o> b ) num__5.0 <o> c ) num__6.0 <o> d ) num__7.0 <o> e ) num__8 % |
s . i . on rs num__800 for num__1 year = num__40 rate = ( num__100 * num__40 ) / ( num__800 * num__1 ) = num__5.0 answer : option b <eor> b <eos> |
b |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| a corporation ten times its annual bonus to num__25 of its employees . what percent of the employees ’ new bonus is the increase ? <o> a ) num__25.0 <o> b ) num__44.0 <o> c ) num__38.0 <o> d ) num__9.0 <o> e ) num__90 % |
let the annual bonus be x . a corporation triples its annual bonus . so new bonus = num__10 x . increase = num__10 x - x = num__9 x the increase is what percent of the employees ’ new bonus = ( num__9 x / num__10 x ) * num__100 = num__90.0 hence e . <eor> e <eos> |
e |
multiply__9.0__10.0__ multiply__9.0__10.0__ |
multiply__9.0__10.0__ multiply__9.0__10.0__ |
| in a tennis match if the first serve is not legal that means a fault a second serve is allowed . normally num__20 to num__40 serves are required in a set of tennis match . at the end of a set set summary stats reveal that first serves percentage was num__63.0 . if m out of n were first serves then m + n is equal to <o> a ) num__1.43 n <o> b ) num__1.53 n <o> c ) num__1.63 n <o> d ) num__2.63 n <o> e ) num__3.63 n |
m = n * ( num__0.63 ) = num__63 n / num__100 so m + n = num__63 n / num__100 + n = num__163 n / num__100 = num__1.63 n answer : c <eor> c <eos> |
c |
divide__63.0__0.63__ add__63.0__100.0__ divide__163.0__100.0__ divide__163.0__100.0__ |
divide__63.0__0.63__ add__63.0__100.0__ divide__163.0__100.0__ divide__163.0__100.0__ |
| in an arithmetic progression the difference between the any two consecutive terms is a constant . what is the arithmetic mean of all of the terms from the first to the num__25 th in an arithmetic progression if the sum of the num__11 th and num__15 th terms of the sequence is num__96 ? <o> a ) num__48 <o> b ) num__63 <o> c ) num__55 <o> d ) num__96 <o> e ) num__108 |
let x be the difference between any two consecutive terms . the mean of a sequence like this is the middle term thus the num__13 th term in the sequence . then the mean of the num__12 th and num__14 th term is also equal to the overall mean because the num__12 th term is ( num__13 th term - x ) and the num__14 th term is ( num__13 th term + x ) . similarly the mean of the num__11 th and num__15 th term is also equal to the mean . etc . . . thus the mean is num__48.0 = num__48 the answer is a . <eor> a <eos> |
a |
subtract__25.0__13.0__ subtract__25.0__11.0__ subtract__96.0__48.0__ |
subtract__25.0__13.0__ subtract__25.0__11.0__ subtract__96.0__48.0__ |
| a train running at the speed of num__90 km / hr crosses a pole in num__10 seconds . find the length of the train . <o> a ) num__150 <o> b ) num__180 <o> c ) num__250 <o> d ) num__200 <o> e ) num__225 |
speed = num__90 * ( num__0.277777777778 ) m / sec = num__25 m / sec length of train ( distance ) = speed * time num__25 * num__10 = num__250 meter answer : c <eor> c <eos> |
c |
multiply__10.0__25.0__ round__250.0__ |
multiply__10.0__25.0__ multiply__10.0__25.0__ |
| exactly two sides of a certain num__10 - sided die are red . what is the probability that kumar rolls the die num__2 times and the die lands with a red side up for the first time on the second roll ? <o> a ) num__0.16 <o> b ) num__0.2 <o> c ) num__0.3 <o> d ) num__0.32 <o> e ) num__0.56 |
total no . of sides = num__10 sides that are red = num__2 probability that the die lands with red side up = num__0.2 therefore probability that the die does not land with red side up = num__1 - num__0.2 = num__0.8 probability that kumar rolls the die num__2 times and the die lands with a red side up for the first time on the second roll = ( num__1 st roll - non red face ) x ( num__2 nd roll - red face ) = num__0.8 x num__0.2 = num__0.16 = num__0.16 = num__0.128 a <eor> a <eos> |
a |
divide__2.0__10.0__ subtract__1.0__0.2__ multiply__0.2__0.8__ multiply__0.8__0.16__ multiply__1.0__0.16__ |
divide__2.0__10.0__ subtract__1.0__0.2__ multiply__0.2__0.8__ multiply__0.8__0.16__ multiply__1.0__0.16__ |
| the average salary per head of the entire staff of an college including the lecturer and professors is rs . num__90 . the average salary of professors is rs . num__600 and that of the lecturer is rs . num__84 . if the number of professors is num__2 find the number of professors in the office ? <o> a ) num__900 <o> b ) num__1000 <o> c ) num__1020 <o> d ) num__1200 <o> e ) num__1240 |
num__1 - > num__12 num__85 - > num__1020 c <eor> c <eos> |
c |
add__84.0__1.0__ multiply__12.0__85.0__ multiply__1.0__1020.0__ |
add__84.0__1.0__ multiply__12.0__85.0__ multiply__1.0__1020.0__ |
| a lawn is in the form of a rectangle having its sides in the ratio num__2 : num__3 . the area of the lawn is ( num__0.166666666667 ) hectares . find breadth of the lawn . <o> a ) num__15 m <o> b ) num__20 m <o> c ) num__25 m <o> d ) num__35 m <o> e ) num__50 m |
let length = num__2 x metres and breadth = num__3 x metre . now area = ( num__0.166666666667 ) x num__1000 m num__2 = num__1666.66666667 m num__2 so num__2 x * num__3 x = num__1666.66666667 < = > x num__2 = num__277.777777778 < = > x = num__16.6666666667 therefore breadth = num__3 x = num__3 ( num__16.6666666667 ) m = num__50 m . answer e num__50 m <eor> e <eos> |
e |
triangle_area__2.0__50.0__ |
triangle_area__2.0__50.0__ |
| in how many years will a sum of money doubles itself at num__5.0 per annum simple interest <o> a ) num__25 years <o> b ) num__28 years <o> c ) num__10 years <o> d ) num__20 years <o> e ) num__21 years |
let the initial sum be x then amount = num__2 x ( because sum will be double ) amount = principal + s . i num__2 x = x + s . i s . i = x s . i = ( p * r * t ) / num__100 x = ( x * num__5 * t ) / num__100 therefore t = num__20 years answer : d <eor> d <eos> |
d |
percent__20.0__100.0__ |
percent__20.0__100.0__ |
| a person starts walking at a speed of num__5 km / hr through half the distance rest of the distance he covers with aspeed num__4 km / hr . total time of travel is num__18 hours . what is the maximum distance he can cover ? <o> a ) num__20 km <o> b ) num__40 km <o> c ) num__60 km <o> d ) num__80 km <o> e ) num__90 km |
t = d / s so num__18 = x / num__2 * num__0.2 + x / num__2 * num__0.25 ( because half distance with num__5 km / ph and remaining half with num__4 km / hr ) num__18 = x ( num__0.225 ) x = num__80 km answer : d <eor> d <eos> |
d |
divide__18.0__0.225__ round__80.0__ |
divide__18.0__0.225__ divide__18.0__0.225__ |
| a vessel of capacity num__2 litre has num__25.0 of alcohol and another vessel of capacity num__6 litre had num__30.0 alcohol . the total liquid of num__8 litre was poured out in a vessel of capacity num__10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? <o> a ) num__23.0 . <o> b ) num__71.0 . <o> c ) num__49.0 . <o> d ) num__29.0 . <o> e ) num__51.0 . |
num__25.0 of num__2 litres = num__0.5 litres num__30.0 of num__6 litres = num__1.8 litres therefore total quantity of alcohol is num__2.3 litres . this mixture is in a num__10 litre vessel . hence the concentration of alcohol in this num__10 litre vessel is num__23.0 answer : a <eor> a <eos> |
a |
reverse__2.0__ add__0.5__1.8__ subtract__25.0__2.0__ subtract__25.0__2.0__ |
reverse__2.0__ add__0.5__1.8__ subtract__25.0__2.0__ subtract__25.0__2.0__ |
| a car starts from x and moves towards y . at the same time another car starts from y and moves towards x . both travel at a constant speed and meet after num__4 hours . during that time the faster car traveled num__0.333333333333 of the distance xy . how long would it take the slower car to travel the distance xy ? <o> a ) num__6 hours <o> b ) num__8 hours <o> c ) num__10 hours <o> d ) num__12 hours <o> e ) num__15 hours |
if one car travelled num__0.333333333333 then the other car must have travelled only num__0.666666666667 as they are meeting after num__4 hours in a certain point . so : ( num__0.666666666667 ) xy = num__4 * y - - > it took the car num__4 hours to travel num__0.666666666667 of the distance at a constant speed y . so if we solve this last equation : num__2 xy = num__3 * num__4 * y = num__6 * y - - > it will take this car num__6 hours in total to reach its final destination . answer : a <eor> a <eos> |
a |
add__4.0__2.0__ round__6.0__ |
multiply__2.0__3.0__ multiply__2.0__3.0__ |
| if num__0.45 : x : : num__4 : num__2 then x is equal to <o> a ) num__0.225 <o> b ) num__0.228 <o> c ) num__0.254 <o> d ) num__0.256 <o> e ) none |
sol . ( x × num__4 ) = ( num__0.45 × num__2 ) ⇒ x = num__0.9 / num__4 = num__0.225 . answer a <eor> a <eos> |
a |
multiply__0.45__2.0__ divide__0.45__2.0__ divide__0.45__2.0__ |
multiply__0.45__2.0__ divide__0.45__2.0__ divide__0.45__2.0__ |
| a num__1200 m long train crosses a tree in num__120 sec how much time will i take to pass a platform num__800 m long ? <o> a ) num__200 sec <o> b ) num__190 sec <o> c ) num__167 sec <o> d ) num__197 sec <o> e ) num__179 sec |
l = s * t s = num__10.0 s = num__10 m / sec . total length ( d ) = num__2000 m t = d / s t = num__200.0 t = num__200 sec answer : a <eor> a <eos> |
a |
divide__1200.0__120.0__ add__1200.0__800.0__ divide__2000.0__10.0__ round__200.0__ |
divide__1200.0__120.0__ add__1200.0__800.0__ divide__2000.0__10.0__ divide__2000.0__10.0__ |
| the present population of a town is num__3888 . population increase rate is num__20.0 p . a . find the population of town before num__2 years ? <o> a ) num__2500 <o> b ) num__2100 <o> c ) num__3500 <o> d ) num__3600 <o> e ) num__2700 |
p = num__3888 r = num__20.0 required population of town = p / ( num__1 + r / num__100 ) ^ t = num__3888 / ( num__1 + num__0.2 ) ^ num__2 = num__3888 / ( num__1.2 ) ^ num__2 = num__2700 ( approximately ) answer is e <eor> e <eos> |
e |
percent__20.0__1.0__ percent__100.0__2700.0__ |
percent__20.0__1.0__ percent__100.0__2700.0__ |
| a and b together can do a piece of work in num__8 days . if a alone can do the same work in num__12 days then b alone can do the same work in ? <o> a ) num__25 days <o> b ) num__24 <o> c ) num__22 days <o> d ) num__27 days <o> e ) num__21 days |
b num__24 days b = num__0.125 – num__0.5 = num__0.0416666666667 = > num__24 days <eor> b <eos> |
b |
divide__12.0__24.0__ divide__0.5__12.0__ round__24.0__ |
divide__12.0__24.0__ divide__0.5__12.0__ round__24.0__ |
| a shopkeeper buys mangoes at the rate of num__4 a rupee and sells them at num__3 a rupee . find his net profit or loss percent ? <o> a ) num__33 num__0.142857142857 % <o> b ) num__33 num__3.0 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__39 num__0.333333333333 % <o> e ) num__23 num__0.333333333333 % |
the total number of mangoes bought by the shopkeeper be num__12 . if he buys num__4 a rupee his cp = num__3 he selling at num__3 a rupee his sp = num__4 profit = sp - cp = num__4 - num__3 = num__1 profit percent = num__0.333333333333 * num__100 = num__33 num__0.333333333333 % answer : c <eor> c <eos> |
c |
percent__33.0__100.0__ |
percent__33.0__100.0__ |
| if price of t . v set is reduced by num__18.0 then its sale increases by num__72.0 find net effect on sale value <o> a ) num__41 <o> b ) num__45 <o> c ) num__46 <o> d ) num__47 <o> e ) num__48 |
- a + b + ( ( - a ) ( b ) / num__100 ) = - num__18 + num__72 + ( - num__18 * num__72 ) / num__100 = - num__18 + num__72 - num__13 = num__41 answer : a <eor> a <eos> |
a |
percent__41.0__100.0__ |
percent__41.0__100.0__ |
| if g h and i are each equal to num__1 or num__0 which of the following options will not be a correct value for e given the equation e = g / num__5 + h / num__5 ^ num__2 + i / num__5 ^ num__3 . <o> a ) num__0.08 <o> b ) num__0.2 <o> c ) num__0.04 <o> d ) num__0.24 <o> e ) num__0.048 |
e = g / num__5 + h / num__5 ^ num__2 + i / num__5 ^ num__3 = g / num__5 + h / num__25 + i / num__125 = ( num__25 g + num__5 h + i ) / num__125 depending on whether g h and i take num__0 or num__1 : substitute num__0 or num__1 for the different values of g h and i . answer a <eor> a <eos> |
a |
multiply__5.0__25.0__ divide__2.0__25.0__ |
multiply__5.0__25.0__ divide__2.0__25.0__ |
| if the speed of x meters per second is equivalent to the speed of y meters per hour what is y in terms of x ? ( num__1 kilometer = num__1000 meters ) <o> a ) num__15 x / num__18 <o> b ) num__3600 x <o> c ) num__18 x / num__5 <o> d ) num__60 x <o> e ) num__3600000 x |
x = y / num__3600 y = num__3600 x answer : b <eor> b <eos> |
b |
round__3600.0__ |
divide__3600.0__1.0__ |
| what least number must be added to num__1056 so that the sum is completely divisible by num__23 ? <o> a ) num__2 <o> b ) num__1 <o> c ) num__3 <o> d ) num__23 <o> e ) num__9 |
num__45.9130434783 = num__45.91 that is num__23 * num__45 = num__1035 now num__1056 - num__1035 = num__21 to get completely divided by num__23 add num__2 to num__1056 ans - a <eor> a <eos> |
a |
divide__1056.0__23.0__ round_down__45.91__ multiply__23.0__45.0__ subtract__1056.0__1035.0__ subtract__23.0__21.0__ subtract__23.0__21.0__ |
divide__1056.0__23.0__ round_down__45.91__ multiply__23.0__45.0__ subtract__1056.0__1035.0__ subtract__23.0__21.0__ subtract__23.0__21.0__ |
| a car dealership has num__40 cars on the lot num__20.0 of which are silver . if the dealership receives a new shipment of num__80 cars num__50.0 of which are not silver what percentage of total number of cars are silver ? <o> a ) num__30.0 <o> b ) num__35.0 <o> c ) num__40.0 <o> d ) num__45.0 <o> e ) num__50 % |
the number of silver cars is num__0.2 * num__40 + num__0.5 * num__80 = num__48 the percentage of cars which are silver is num__0.4 = num__40.0 the answer is c . <eor> c <eos> |
c |
percent__80.0__0.5__ percent__80.0__50.0__ |
percent__80.0__0.5__ percent__80.0__50.0__ |
| using all the letters of the word ` ` live ' ' how many words can be formed which begin with l and end with e ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__2 |
there are four letters in the given word . consider num__4 blanks . . . . the first blank and last blank must be filled with l and e all the remaining two blanks can be filled with the remaining num__2 letters in num__2 ! ways . the number of words = num__2 ! = num__2 . answer : e <eor> e <eos> |
e |
coin_space__ coin_space__ |
coin_space__ coin_space__ |
| h . c . f of num__136 num__144 and a third number is num__8 and their l . c . m is num__2 ^ num__4 * num__3 ^ num__2 * num__17 * num__7 . the third number is : <o> a ) num__154 <o> b ) num__248 <o> c ) num__360 <o> d ) num__424 <o> e ) num__168 |
num__136 = num__2 ^ num__3 * num__17 ; num__144 = num__2 ^ num__4 * num__3 ^ num__2 h . c . f = num__2 ^ num__3 = num__8 since h . c . f is the product of lowest powers of common factors so the third number must have ( num__2 ^ num__3 ) as its factor . since l . c . m is the product of highest powers of common prime factors so the third number must have num__3 and num__7 as its factors . third number = num__2 ^ num__3 * num__3 * num__7 = num__168 answer : e <eor> e <eos> |
e |
lcm__8.0__168.0__ |
lcm__8.0__168.0__ |
| in a school of num__800 students num__44.0 wear blue shirts num__28.0 wear red shirts num__10.0 wear green shirts and the remaining students wear other colors . how many students wear other colors ( not blue not red not green ) ? <o> a ) num__120 <o> b ) num__132 <o> c ) num__144 <o> d ) num__156 <o> e ) num__168 |
num__44 + num__28 + num__10 = num__82.0 num__100 – num__82 = num__18.0 num__800 * num__0.18 = num__144 the answer is c . <eor> c <eos> |
c |
percent__18.0__800.0__ percent__100.0__144.0__ |
percent__18.0__800.0__ percent__100.0__144.0__ |
| if num__10 gallons of grape juice are added to num__30 gallons of a mixture which contains num__10 percent grape juice then what percent of the resulting mixture is grape juice ? <o> a ) num__14.0 <o> b ) num__25.0 <o> c ) num__28.0 <o> d ) num__32.5 <o> e ) num__50 % |
official solution : if we start with num__40 gallons of a mixture that is num__10.0 grape juice then we have : num__30 × num__0.10 = num__3 gallons of grape juice . num__30 × num__0.90 = num__27 gallons of other components . if we add num__10 gallons of grape juice we will end up with num__13 gallons of grape juice and num__36 gallons of other components and we will have a total of num__40 gallons of the mixture . so num__0.325 of the new mixture is grape juice . now we convert this to a percent : percent grape juice = num__325.0 . the correct answer is choice ( d ) <eor> d <eos> |
d |
add__10.0__30.0__ reverse__10.0__ divide__30.0__10.0__ subtract__30.0__3.0__ add__10.0__3.0__ multiply__40.0__0.9__ divide__13.0__40.0__ divide__325.0__10.0__ |
add__10.0__30.0__ reverse__10.0__ divide__30.0__10.0__ subtract__30.0__3.0__ add__10.0__3.0__ multiply__40.0__0.9__ divide__13.0__40.0__ divide__325.0__10.0__ |
| the average of num__10 numbers is num__40.2 . later it is found that two numbers have been wrongly copied . the first is num__17 greater than the actual number and the second number added is num__13 instead of num__31 . find the correct average . <o> a ) num__40.3 <o> b ) num__40.4 <o> c ) num__40.6 <o> d ) num__40.8 <o> e ) none of the above |
sum of num__10 numbers = num__402 corrected sum of num__10 numbers = num__402 – num__13 + num__31 – num__17 = num__403 hence new average = num__403 ⁄ num__10 = num__40.3 answer a <eor> a <eos> |
a |
multiply__10.0__40.2__ multiply__13.0__31.0__ divide__403.0__10.0__ divide__403.0__10.0__ |
multiply__10.0__40.2__ multiply__13.0__31.0__ divide__403.0__10.0__ divide__403.0__10.0__ |
| a man gains num__40.0 by selling an article for a certain price . if he sells it at double the price the percentage of profit will be . <o> a ) num__130.0 <o> b ) num__180.0 <o> c ) num__150.0 <o> d ) num__160.0 <o> e ) num__170 % |
explanation : let the c . p . = x then s . p . = ( num__1.4 ) x = num__7 x / num__5 new s . p . = num__2 ( num__7 x / num__5 ) = num__14 x / num__5 profit = num__14 x / num__5 - x = num__9 x / num__5 profit % = ( profit / c . p . ) * num__100 = > ( num__9 x / num__5 ) * ( num__1 / x ) * num__100 = num__180.0 option b <eor> b <eos> |
b |
percent__40.0__5.0__ percent__100.0__180.0__ |
percent__40.0__5.0__ percent__100.0__180.0__ |
| on a certain farm the ratio of horses to cows is num__6 : num__1 . if the farm were to sell num__15 horses and buy num__15 cows the ratio of horses to cows would then be num__3 : num__1 . after the transaction how many more horses than cows would the farm own ? <o> a ) num__40 <o> b ) num__50 <o> c ) num__60 <o> d ) num__70 <o> e ) num__80 |
originally there were num__6 k horses and k cows . ( num__6 k - num__15 ) = num__3 ( k + num__15 ) num__6 k - num__3 k = num__45 + num__15 num__3 k = num__60 k = num__20 the difference between horses and cows is ( num__6 k - num__15 ) - ( k + num__15 ) = num__5 k - num__30 = num__70 the answer is d . <eor> d <eos> |
d |
multiply__15.0__3.0__ add__15.0__45.0__ divide__60.0__3.0__ subtract__6.0__1.0__ multiply__6.0__5.0__ multiply__1.0__70.0__ |
multiply__15.0__3.0__ add__15.0__45.0__ divide__60.0__3.0__ subtract__6.0__1.0__ subtract__45.0__15.0__ multiply__1.0__70.0__ |
| a train num__200 m long is running with a speed of num__60 km / hr . in what time will it pass a man who is running at num__10 km / hr in the direction opposite to that in which the train is going ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__10 <o> d ) num__9 <o> e ) num__5 |
speed of train relative to man = num__60 + num__10 = num__70 km / hr . = num__70 * num__0.277777777778 = num__19.4444444444 m / sec . time taken to pass the men = num__200 * num__0.0514285714286 = num__10 sec . answer : c <eor> c <eos> |
c |
add__60.0__10.0__ round__10.0__ |
add__60.0__10.0__ round__10.0__ |
| excluding stoppages the speed of a train is num__45 kmph and including stoppages it is num__36 kmph . of how many minutes does the train stop per hour ? <o> a ) num__19 <o> b ) num__17 <o> c ) num__12 <o> d ) num__15 <o> e ) num__18 |
t = num__0.2 * num__60 = num__12 . answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ multiply__60.0__0.2__ round__12.0__ |
hour_to_min_conversion__ multiply__60.0__0.2__ multiply__60.0__0.2__ |
| a num__120 metres long train running at the speed of num__120 kmph crosses another train running in opposite direction at the speed of num__80 kmph in num__9 seconds . what is the length of the other train ? <o> a ) num__220 meter <o> b ) num__225 meter <o> c ) num__380 meter <o> d ) num__435 meter <o> e ) none of these |
explanation : as trains are running in opposite directions so their relative speed will get added so relative speed = num__120 + num__80 = num__200 kmph = num__200 * ( num__0.277777777778 ) = num__55.5555555556 m / sec let the length of other train is x meter then x + num__13.3333333333 = num__55.5555555556 = > x + num__120 = num__500 = > x = num__380 so the length of the train is num__380 meters option c <eor> c <eos> |
c |
add__120.0__80.0__ divide__120.0__9.0__ subtract__500.0__120.0__ round__380.0__ |
add__120.0__80.0__ divide__120.0__9.0__ subtract__500.0__120.0__ round__380.0__ |
| the city club has exactly num__5 new members at the end of its first week . every subsequent week each of the previous week ' s new members ( and only these members ) brings exactly p new members into the club . if y is the number of new members brought into the club during the twelfth week which of the following could be y ? <o> a ) num__5 ^ num__0.0833333333333 <o> b ) num__3 ^ num__11 * num__5 ^ num__11 <o> c ) num__3 ^ num__12 * num__5 ^ num__12 <o> d ) num__3 ^ num__11 * num__5 ^ num__12 <o> e ) num__60 ^ num__12 |
at the end of the first week there are num__5 new members ; at the end of the second week there are num__5 p new members ( since each num__5 new members from the previous week brings p new members ) ; at the end of the third week there are num__5 p ^ num__2 new members ( since each num__5 p new members from the previous week brings p new members ) ; . . . at the end of the twelfth week there are num__5 p ^ num__11 new members ( since each num__5 p ^ num__10 new members from the previous week brings p new members ) . we are given that num__5 p ^ num__11 = y . out of the answers only d yields integer value for p : num__5 p ^ num__11 = num__3 ^ num__11 * num__5 ^ num__12 - - > p = num__3 * num__5 = num__15 . answer : d . <eor> d <eos> |
d |
multiply__5.0__2.0__ subtract__5.0__2.0__ add__2.0__10.0__ multiply__5.0__3.0__ subtract__5.0__2.0__ |
multiply__5.0__2.0__ subtract__5.0__2.0__ add__2.0__10.0__ multiply__5.0__3.0__ subtract__5.0__2.0__ |
| if num__10 spiders make num__5 webs in num__5 days then how many days are needed for num__1 spider to make num__1 web ? <o> a ) num__10 <o> b ) num__9 <o> c ) num__5 <o> d ) num__15 <o> e ) num__20 |
explanation : let num__1 spider make num__1 web in x days . more spiders less days ( indirect proportion ) more webs more days ( direct proportion ) hence we can write as ( spiders ) num__10 : num__1 ( webs ) num__1 : num__5 } : : x : num__5 â ‡ ’ num__10 Ã — num__1 Ã — num__5 = num__1 Ã — num__5 Ã — x â ‡ ’ x = num__10 answer : option a <eor> a <eos> |
a |
round__10.0__ |
round__10.0__ |
| the cost of cultivating a square field at the rate of rs . num__135 per hectare is rs . num__1215 . the cost of putting a fence around it at the rate of num__75 paise per meter would be : <o> a ) s . num__360 <o> b ) s . num__810 <o> c ) rs . num__900 <o> d ) s . num__1800 <o> e ) s . num__1900 |
area = total cost / rate = ( num__9.0 ) hectares = ( num__9 * num__10000 ) sq . m . therefore side of the square = Ö num__90000 = num__300 m . perimeter of the field = ( num__300 * num__4 ) m = num__1200 m cost of fencing = rs . ( num__1200 * num__0.75 ) = rs . num__900 answer : c <eor> c <eos> |
c |
percent__75.0__1200.0__ percent__75.0__1200.0__ |
percent__75.0__1200.0__ percent__75.0__1200.0__ |
| a department of ten people - four men and six women - needs to send a team of five to a conference . if they want to make sure that there are no more than three members of the team from any one gender how many distinct groups are possible to send ? <o> a ) num__100 <o> b ) num__120 <o> c ) num__150 <o> d ) num__180 <o> e ) num__200 |
they can make a team of num__3 men and num__2 women . the number of ways to do this is num__4 c num__3 * num__6 c num__2 = num__4 * num__15 = num__60 they can make a team of num__2 men and num__3 women . the number of ways to do this is num__4 c num__2 * num__6 c num__3 = num__6 * num__20 = num__120 the total number of distinct groups is num__180 . the answer is d . <eor> d <eos> |
d |
multiply__2.0__3.0__ multiply__4.0__15.0__ divide__60.0__3.0__ multiply__2.0__60.0__ multiply__3.0__60.0__ multiply__3.0__60.0__ |
multiply__2.0__3.0__ multiply__4.0__15.0__ divide__60.0__3.0__ multiply__2.0__60.0__ multiply__3.0__60.0__ multiply__3.0__60.0__ |
| after decreasing num__24.0 in the price of an article costs rs . num__912 . find the actual cost of an article ? <o> a ) num__1400 <o> b ) num__1300 <o> c ) num__1200 <o> d ) num__1100 <o> e ) num__1000 |
explanation : cp * ( num__0.76 ) = num__912 cp = num__12 * num__100 = > cp = num__1200 answer is c <eor> c <eos> |
c |
percent__100.0__1200.0__ |
percent__100.0__1200.0__ |
| on dividing a number by num__999 the quotient is num__366 and the remainder is num__103 . the number is : <o> a ) num__364724 <o> b ) num__365387 <o> c ) num__365737 <o> d ) num__366757 <o> e ) num__366767 |
required number = num__999 * num__366 + num__103 = ( num__1000 - num__1 ) * num__366 + num__103 = num__366000 - num__366 + num__103 = num__365737 . answer : c <eor> c <eos> |
c |
subtract__1000.0__999.0__ multiply__366.0__1000.0__ multiply__1.0__365737.0__ |
subtract__1000.0__999.0__ multiply__366.0__1000.0__ multiply__1.0__365737.0__ |
| two trains are moving at num__75 kmph and num__70 kmph in opposite directions . their lengths are num__150 m and num__100 m respectively . the time they will take to pass each other completely is ? <o> a ) num__6 num__0.206896551724 sec <o> b ) num__7 num__0.142857142857 sec <o> c ) num__7 num__1.0 sec <o> d ) num__8 num__0.5 sec <o> e ) num__7 num__0.0526315789474 sec |
num__70 + num__75 = num__145 * num__0.277777777778 = num__40.2777777778 mps d = num__150 + num__100 = num__250 m t = num__250 * num__0.0248275862069 = num__6.20689655172 = num__6 num__0.206896551724 sec answer : a <eor> a <eos> |
a |
add__75.0__70.0__ add__150.0__100.0__ divide__250.0__40.2778__ subtract__6.2069__6.0__ round__6.0__ |
add__75.0__70.0__ add__150.0__100.0__ divide__250.0__40.2778__ subtract__6.2069__6.0__ round__6.0__ |
| the distance traveled by earth in one year approximately num__6590000000000 miles . the distance traveled by the earth by in num__100 years ? <o> a ) num__10 < num__16 > <o> b ) num__10 < num__10 > <o> c ) num__10 < num__13 > <o> d ) num__10 < num__15 > <o> e ) num__10 < num__18 > |
the distance traveled by earth in one year = num__65 num__900000 num__000000 miles the distance traveled by earth in num__100 years = num__65 num__900000 num__000000 * num__100 years = num__6590 num__000000 num__000000 miles = num__659 * num__10 < num__13 > ( exponent ) answer : c <eor> c <eos> |
c |
divide__100.0__6.59e+12__ divide__6590.0__659.0__ round__10.0__ |
divide__100.0__6.59e+12__ divide__6590.0__659.0__ round__10.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__7 sec . what is the length of the train ? <o> a ) num__112.21 <o> b ) num__27.21 <o> c ) num__117.66 <o> d ) num__277 <o> e ) num__116.66 |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__7 = num__116.66 answer : e <eor> e <eos> |
e |
round__116.66__ |
round__116.66__ |
| the ratio of three numbers is num__3 : num__4 : num__7 and their product is num__18144 . the numbers are <o> a ) num__9 num__1221 <o> b ) num__15 num__0025 <o> c ) num__18 num__2442 <o> d ) num__24 num__1617 <o> e ) num__30 num__15 |
17 |
let the numbers be num__3 x num__4 x and num__7 x num__3 x x num__4 x x num__7 x = num__18144 x num__3 = num__216 x = num__6 the numbers are num__18 num__2442 answer c <eor> c <eos> |
c |
c |
| num__8 hens weigh equal to as many duck as are equal to num__7 cats . all of them weigh num__264 kg only . hen ' s weighs are ? <o> a ) num__6 <o> b ) num__5 <o> c ) num__2 <o> d ) num__7 <o> e ) num__11 |
num__8 h = xd = num__7 c num__8 h + xd + num__7 c - - - - - num__264 kg . num__8 h + num__8 h + num__8 h - - - - - num__264 kg . num__24 h - - - - - - num__264 kgs . = > num__1 h = num__11 kg answer : e <eor> e <eos> |
e |
subtract__8.0__7.0__ divide__264.0__24.0__ divide__264.0__24.0__ |
subtract__8.0__7.0__ divide__264.0__24.0__ divide__264.0__24.0__ |
| the length of the bridge which a train num__150 metres long and travelling at num__54 km / hr can cross in num__30 seconds is ? <o> a ) num__340 m . <o> b ) num__350 m . <o> c ) num__320 m . <o> d ) num__330 m . <o> e ) num__360 m . |
speed = [ num__54 x num__0.277777777778 ] m / sec = [ num__15 ] m / sec time = num__30 sec let the length of bridge be x metres . then ( num__130 + x ) / num__30 = num__15 = > ( num__130 + x ) / num__30 = num__320 = > x = num__320 m . answer : c <eor> c <eos> |
c |
round__320.0__ |
round__320.0__ |
| during a certain week a seal ate num__65.0 of the first num__80 smelt it came across and num__30.0 of the remaining smelt it came across . if the seal ate num__40.0 of the smelt it came across during the entire week how many smelt did it eat ? <o> a ) num__32 <o> b ) num__40 <o> c ) num__55 <o> d ) num__64 <o> e ) num__112 |
total smelt = x . then num__0.65 * num__80 + num__0.3 ( x - num__80 ) = num__0.4 * x - - > x = num__280 - - > num__0.4 * x = num__112 . answer : e . <eor> e <eos> |
e |
percent__40.0__280.0__ percent__40.0__280.0__ |
percent__40.0__280.0__ percent__40.0__280.0__ |
| what is the sum of all the prime numbers greater than num__30 but less than num__50 ? <o> a ) num__126 <o> b ) num__199 <o> c ) num__198 <o> d ) num__188 <o> e ) num__122 |
required sum = ( num__31 + num__37 + num__41 + num__43 + num__47 ) = num__199 note : num__1 is not a prime number answer b <eor> b <eos> |
b |
subtract__31.0__30.0__ multiply__1.0__199.0__ |
subtract__31.0__30.0__ multiply__1.0__199.0__ |
| find the invalid no . from the following series num__18 num__24 num__29 num__35 num__40 num__42 <o> a ) num__13 <o> b ) num__18 <o> c ) num__40 <o> d ) num__37 <o> e ) num__42 |
the differences between two successive terms from the beginning are num__6 num__56 num__5 num__6 num__5 . so num__42 is wrong . answer : e <eor> e <eos> |
e |
subtract__24.0__18.0__ subtract__29.0__24.0__ add__18.0__24.0__ |
subtract__24.0__18.0__ subtract__29.0__24.0__ add__18.0__24.0__ |
| a truck covers a distance of num__296 km at a certain speed in num__8 hours . how much time would a car take at an average speed which is num__18 kmph more than that of the speed of the truck to cover a distance which is num__6.5 km more than that travelled by the truck ? <o> a ) num__6 hours <o> b ) num__5 hours <o> c ) num__5.5 hours <o> d ) num__8 hours <o> e ) none |
explanation : speed of the truck = distance / time = num__37.0 = num__37 kmph now speed of car = ( speed of truck + num__18 ) kmph = ( num__37 + num__18 ) = num__55 kmph distance travelled by car = num__296 + num__6.5 = num__302.5 km time taken by car = distance / speed = num__302.5 / num__55 = num__5.5 hours . answer – c <eor> c <eos> |
c |
divide__296.0__8.0__ add__18.0__37.0__ add__296.0__6.5__ divide__302.5__55.0__ round__5.5__ |
divide__296.0__8.0__ add__18.0__37.0__ add__296.0__6.5__ divide__302.5__55.0__ divide__302.5__55.0__ |
| how many zeroes are there at the end of the number n if n = num__90 ! + num__180 ! ? <o> a ) num__15 <o> b ) num__18 <o> c ) num__21 <o> d ) num__24 <o> e ) num__27 |
the number of zeroes at the end of num__90 ! will be less than the number of zeroes at the end of num__180 ! hence it is sufficient to calculate the number of zeroes at the end of num__90 ! the number of zeroes = [ num__18.0 ] + [ num__3.6 ] + [ num__0.72 ] = num__18 + num__3 + num__0 = num__21 the answer is c . <eor> c <eos> |
c |
round_down__3.6__ round_down__0.72__ add__3.0__18.0__ add__3.0__18.0__ |
round_down__3.6__ round_down__0.72__ add__3.0__18.0__ add__3.0__18.0__ |
| two trains of equal lengths take num__5 sec and num__15 sec respectively to cross a telegraph post . if the length of each train be num__120 m in what time will they cross other travelling in opposite direction ? <o> a ) num__22 <o> b ) num__12 <o> c ) num__7.5 <o> d ) num__99 <o> e ) num__21 |
speed of the first train = num__24.0 = num__24 m / sec . speed of the second train = num__8.0 = num__8 m / sec . relative speed = num__24 + num__8 = num__32 m / sec . required time = ( num__120 + num__120 ) / num__32 = num__7.5 sec . answer : c <eor> c <eos> |
c |
divide__120.0__5.0__ divide__120.0__15.0__ add__24.0__8.0__ round__7.5__ |
divide__120.0__5.0__ divide__120.0__15.0__ add__24.0__8.0__ round__7.5__ |
| in the floor of a particular kitchen owned by an abstract artist each row of tiles to the right of the first row contains two fewer tiles than the row directly to its left . if there are nine rows in all and a total of num__540 tiles in the floor how many tiles does the leftmost row contain ? <o> a ) num__52 <o> b ) num__56 <o> c ) num__60 <o> d ) num__64 <o> e ) num__68 |
this question can be solved in a variety of ways : with algebra by testing the answers and by using a great number property shortcut involving consecutive integers . we ' re given a few facts to work with : num__1 ) there are num__9 rows of tiles . num__2 ) when going from ' left to right ' each row contains two fewer tiles than the one next to it . num__3 ) there are a total of num__540 tiles we ' re asked how many tiles the left - most most row holds ( meaning the one with the most tiles ) . to start num__540 is divisible by num__9 so we can figure out the average number of tiles per row . that is num__60.0 = num__60 . since we ' re dealing with a set of num__9 consecutive integers that differ by num__2 each we know that the ' num__5 th row ' will have num__60 tiles ( the average ) . then we just have to ' add num__2 s ' until we get to the first row . . . num__60 + num__2 + num__2 + num__2 + num__2 = num__68 . final answer : e <eor> e <eos> |
e |
add__1.0__2.0__ divide__540.0__9.0__ add__2.0__3.0__ multiply__1.0__68.0__ |
add__1.0__2.0__ divide__540.0__9.0__ add__2.0__3.0__ multiply__1.0__68.0__ |
| the area of sector of a circle whose radius is num__12 metro and whose angle at the center is num__36 Â ° is ? <o> a ) num__52.6 <o> b ) num__45.3 <o> c ) num__52.8 <o> d ) num__52.1 <o> e ) num__52.2 |
num__0.1 * num__3.14285714286 * num__12 * num__12 = num__45.3 m num__2 answer : b <eor> b <eos> |
b |
triangle_area__45.3__2.0__ |
triangle_area__45.3__2.0__ |
| kim can do a work in num__3 days while david can do the same work in num__2 days . both of them finish the work together and get rs . num__150 . what is the share of kim ? <o> a ) rs . num__30 <o> b ) rs . num__60 <o> c ) rs . num__70 <o> d ) rs . num__75 <o> e ) none of these |
kim ' s wages : david ' s wages = kim ' s num__1 day ' s work : david ' s num__1 day ' s work = num__0.333333333333 : num__0.5 = num__2 : num__3 kim ' s share = rs . ( num__0.4 * num__150 ) = rs . num__60 correct option : b <eor> b <eos> |
b |
subtract__3.0__2.0__ divide__1.0__3.0__ divide__1.0__2.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
subtract__3.0__2.0__ divide__1.0__3.0__ divide__1.0__2.0__ multiply__150.0__0.4__ multiply__150.0__0.4__ |
| the list price of an article is rs . num__65 . a customer pays rs . num__56.16 for it . he was given two successive discounts one of them being num__10.0 . the other discount is ? <o> a ) num__7.0 <o> b ) num__4 percent <o> c ) num__5.0 <o> d ) num__2.0 <o> e ) num__1 % |
num__65 * ( num__0.9 ) * ( ( num__100 - x ) / num__100 ) = num__56.16 x = num__4.0 answer : b <eor> b <eos> |
b |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| it is given that num__2 ^ num__32 + num__1 is exactly divisible by a certain number . which one of the following is also divisible by the same number x ? <o> a ) a . num__2 ^ num__96 + num__1 <o> b ) b . num__2 ^ num__16 - num__1 <o> c ) c . num__2 ^ num__16 + num__1 <o> d ) d . num__7 * num__2 ^ num__33 <o> e ) e . num__2 ^ num__64 + num__1 |
a ³ + b ³ = ( a + b ) ( a ² - ab + b ² ) now let ( num__2 ^ num__32 + num__1 ) be ( a + b ) a ³ + b ³ = ( num__2 ^ num__96 + num__1 ) now as mentioned in formula above a ³ + b ³ is always divisible by ( a + b ) so any factor of x = ( a + b ) is a factor of ( a ³ + b ³ ) henca a <eor> a <eos> |
a |
multiply__2.0__1.0__ |
multiply__2.0__1.0__ |
| an uneducated retailer marks all its goods at num__50.0 above the cost price and thinking that he will still make num__25.0 profit offers a discount of num__25.0 on the marked price . what is the actual profit on the sales ? <o> a ) num__12.5 <o> b ) num__10.5 <o> c ) num__9.5 <o> d ) num__14.5 <o> e ) none of them |
let c . p = rs num__100 . then marked price = rs num__100 s . p = num__75.0 of rs num__150 = rs num__112.50 hence gain % = num__12.5 answer is a . <eor> a <eos> |
a |
percent__75.0__150.0__ percent__50.0__25.0__ percent__50.0__25.0__ |
percent__75.0__150.0__ percent__50.0__25.0__ percent__50.0__25.0__ |
| a b and c start swimming in a pool simultaneously from the same end . to complete num__10 laps a takes num__10 minutes b takes num__6 minutes and c takes num__6 minutes . what is the ratio of speeds a : b : c ? <o> a ) num__3 : num__5 : num__5 <o> b ) num__12 : num__15 : num__20 <o> c ) num__5 : num__4 : num__3 <o> d ) num__4 : num__6 : num__5 <o> e ) num__12 : num__15 : num__25 |
the slowest one is a then comes b and the fastest one is c . so the the ratio of speeds a : b : c must be in ascending order . eliminate options c d and e . a ' s speed ( distance ) / ( time ) = num__1.0 = num__1 lap per minute = num__3 laps in num__3 minute ; b ' s speed ( distance ) / ( time ) = num__1.66666666667 = num__1.66666666667 lap per minute = num__5 laps in num__3 minutes ; c ' s speed ( distance ) / ( time ) = num__1.66666666667 = num__1.66666666667 lap per minute = num__5 laps in num__3 minutes / therefore the the ratio of speeds a : b : c is num__3 : num__5 : num__5 . answer : a . <eor> a <eos> |
a |
divide__10.0__6.0__ subtract__6.0__1.0__ round__3.0__ |
divide__10.0__6.0__ subtract__6.0__1.0__ divide__3.0__1.0__ |
| niharika covers a distance in num__30 min if she drives at a speed of num__60 km / h on an average . find the speed at which she must drive at to reduce the time of the journey by num__30.0 . <o> a ) num__84.71 kmph . <o> b ) num__85.74 kmph . <o> c ) num__83.71 kmph . <o> d ) num__85.71 kmph . <o> e ) num__84.71 kmph . |
given speed = num__60 kmph . it means niharika covered num__60 km in num__60 min ( num__1 hour ) . so in num__30 min he will cover num__30 km . num__30.0 time reduced = num__30 - num__30.0 of num__30 = num__21 min . thus niharika needs to cover num__30 km in num__21 min ( num__0.35 hour ) . speed * time = distance . speed * num__0.35 = num__30 speed = num__85.71 kmph . answer : option d <eor> d <eos> |
d |
divide__21.0__60.0__ multiply__1.0__85.71__ |
divide__21.0__60.0__ multiply__1.0__85.71__ |
| in an election contested by two parties party d secured num__12.0 of the total votes more than party r . if party r got num__132000 votes by how many votes did it lose the election ? <o> a ) num__24000 <o> b ) num__36000 <o> c ) num__168000 <o> d ) num__240000 <o> e ) num__300 |
000 |
d + r = num__100 - - - - ( num__1 ) d - r = num__12 - - - - ( num__2 ) solving ( num__1 ) and ( num__2 ) d = num__56.0 and r = num__44.0 num__44.0 of total = num__132000 total = num__300000 d = num__168000 difference = num__168000 - num__132000 = num__36000 answer : b <eor> b <eos> |
b |
b |
| at a reception one - third of the guests departed at a certain time . later two - fifths of the guests departed . even later two - thirds of the remaining guests departed . if six people were left how many were originally present at the party ? <o> a ) num__28 <o> b ) num__37 <o> c ) num__20 <o> d ) num__45 <o> e ) num__27 |
explanation : let original members be x first one third guest departed i . e x / num__3 remaining guests = x – ( x / num__3 ) = num__2 x / num__3 now from the remaining ( num__2 x / num__3 ) two - fifths departed = num__0.4 ( num__2 x / num__3 ) = num__4 x / num__15 i . e . now remaining guests will be ( num__2 x / num__3 – num__4 x / num__15 ) = num__2 x / num__5 now from remaining ( num__2 x / num__5 ) two - thirds departed = num__0.666666666667 ( num__2 x / num__5 ) = num__4 x / num__15 now remaining guests = ( num__2 x / num__5 – num__4 x / num__15 ) = num__2 x / num__15 given num__2 x / num__15 = num__6 ⇒ x = num__45 answer : d <eor> d <eos> |
d |
add__2.0__3.0__ divide__2.0__3.0__ multiply__2.0__3.0__ multiply__3.0__15.0__ multiply__3.0__15.0__ |
divide__2.0__0.4__ divide__2.0__3.0__ multiply__2.0__3.0__ multiply__3.0__15.0__ multiply__3.0__15.0__ |
| a dishonest dealer professes to sell goods at the cost price but uses a false weight and gains num__25.0 . find his false weight age ? <o> a ) num__337 <o> b ) num__2399 <o> c ) num__800 <o> d ) num__287 <o> e ) num__221 |
explanation : num__25 = e / ( num__1000 - e ) * num__100 num__1000 - e = num__4 e num__1000 = num__5 e = > e = num__200 num__1000 - num__200 = num__800 answer : c <eor> c <eos> |
c |
percent__100.0__800.0__ |
percent__100.0__800.0__ |
| pipe a can fill a tank in num__8 minutes and pipe b cam empty it in num__24 minutes . if both the pipes are opened together after how many minutes should pipe b be closed so that the tank is filled in num__30 minutes ? <o> a ) num__18 <o> b ) num__27 <o> c ) num__98 <o> d ) num__27 <o> e ) num__66 |
let the pipe b be closed after x minutes . num__3.75 - x / num__24 = num__1 = > x / num__24 = num__3.75 - num__1 = num__2.75 = > x = num__2.75 * num__24 = num__66 . answer : e <eor> e <eos> |
e |
divide__30.0__8.0__ subtract__3.75__1.0__ multiply__24.0__2.75__ round__66.0__ |
divide__30.0__8.0__ subtract__3.75__1.0__ multiply__24.0__2.75__ multiply__24.0__2.75__ |
| how many words can be formed by using all letters of the word ' around ' ? <o> a ) num__712 <o> b ) num__720 <o> c ) num__757 <o> d ) num__736 <o> e ) num__742 |
the word around contains num__6 different letters required number of words = num__6 p num__6 = num__6 ! = num__6 * num__5 * num__4 * num__3 * num__2 * num__1 = num__120 answer is b <eor> b <eos> |
b |
die_space__ vowel_space__ coin_space__ choose__6.0__3.0__ |
die_space__ vowel_space__ coin_space__ choose__6.0__3.0__ |
| a tank of num__100 litres is filled with kerosene . ram replaces num__10.0 of kerosene with water and repeats this procedure num__4 times . what is the amount of kerosene left in the tank ? <o> a ) num__65.60 litres <o> b ) num__65.71 litres <o> c ) num__25.61 litres <o> d ) num__65.61 litres <o> e ) num__65.31 litres |
explanation : as ram replaces num__10.0 of kerosene with water the amount of kerosene left is num__90.0 . as the replacement is done four times after num__1 st replacement kerosene left is : num__0.9 * num__100 = num__90 litres after num__2 nd replacement kerosene left is : num__0.9 * num__90 = num__81 litres after num__3 rd replacement kerosene left is : num__0.9 * num__81 = num__72.9 litres after num__4 th replacement kerosene left is : num__0.9 * num__72.9 = num__65.61 litres answer : d <eor> d <eos> |
d |
subtract__100.0__10.0__ divide__90.0__100.0__ multiply__0.9__90.0__ subtract__4.0__1.0__ multiply__81.0__0.9__ multiply__72.9__0.9__ round__65.61__ |
subtract__100.0__10.0__ divide__90.0__100.0__ multiply__0.9__90.0__ subtract__4.0__1.0__ multiply__81.0__0.9__ multiply__72.9__0.9__ multiply__72.9__0.9__ |
| a gets num__3 times as much money as b gets b gets only rs . num__25 more then what c gets . the three gets rs . num__675 in all . find the share of b ? <o> a ) num__130 <o> b ) num__120 <o> c ) num__218 <o> d ) num__140 <o> e ) num__145 |
a + b + c = num__675 a = num__3 b num__3 b + b + b - num__25 = num__675 num__5 b = num__700 b = num__140 answer : d <eor> d <eos> |
d |
add__25.0__675.0__ divide__700.0__5.0__ divide__700.0__5.0__ |
add__25.0__675.0__ divide__700.0__5.0__ divide__700.0__5.0__ |
| if a : b : c = num__3 : num__4 : num__7 then the ratio ( a + b + c ) : c is equal to <o> a ) num__2 : num__1 <o> b ) num__14 : num__3 <o> c ) num__7 : num__2 <o> d ) num__1 : num__2 <o> e ) none |
solution : ( a + b + c ) = num__3 + num__4 + num__7 = num__14 and c = num__7 then ( a + b + c ) : c = num__2 : num__1 . answer : option a <eor> a <eos> |
a |
divide__14.0__7.0__ subtract__3.0__2.0__ subtract__3.0__1.0__ |
divide__14.0__7.0__ subtract__3.0__2.0__ subtract__3.0__1.0__ |
| a driver goes on a trip of num__50 kilometers the first num__25 kilometers at num__60 kilometers per hour and the remaining distance at num__30 kilometers per hour . what is the average speed of the entire trip in kilometers per hour ? <o> a ) num__35 <o> b ) num__36 <o> c ) num__40 <o> d ) num__42 <o> e ) num__45 |
the time for the first part of the trip was num__0.416666666667 = num__0.416666666667 hours . the time for the second part of the trip was num__0.833333333333 = num__0.833333333333 hours . the total time for the trip was num__0.416666666667 + num__0.833333333333 = num__1.25 = num__1.25 hours . the average speed for the trip was num__50 / ( num__1.25 ) = num__40 kph the answer is c . <eor> c <eos> |
c |
divide__25.0__60.0__ divide__50.0__60.0__ add__0.4167__0.8333__ divide__50.0__1.25__ round__40.0__ |
divide__25.0__60.0__ divide__50.0__60.0__ add__0.4167__0.8333__ divide__50.0__1.25__ divide__50.0__1.25__ |
| the number num__110 can be written as sum of the squares of num__3 different positive integers . what is the sum of these num__3 different integers ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__18 <o> d ) num__14 <o> e ) num__17 |
sum of the squares of num__3 different positive integers = num__110 num__5 ^ num__2 + num__6 ^ num__2 + num__7 ^ num__2 = num__110 now sum of these num__3 different integers = num__5 + num__6 + num__7 = num__18 ans - c <eor> c <eos> |
c |
multiply__3.0__2.0__ multiply__3.0__6.0__ multiply__3.0__6.0__ |
multiply__3.0__2.0__ multiply__3.0__6.0__ multiply__3.0__6.0__ |
| a full bottle contains num__40.0 oil num__20.0 vinegar and num__40.0 water . the bottle is poured into a larger bottle four times as big as original . the remaining space in the larger bottle is then filled with water . if there was num__8 ml of oil in the original bottle how much of water is in the final mixture ? <o> a ) num__66 <o> b ) num__67 <o> c ) num__68 <o> d ) num__69 <o> e ) num__70 |
let the total volume of original bottle be x then question says num__40.0 of x is oil . also question mentions that there is num__8 ml of oil in original bottle . set up equation - - > num__40.0 of x = num__8 num__40 * x / num__100 = num__8 x = num__20 . hence oil in original container is = num__8 ml ( num__40.0 of total volume ) vinegar in original container is = num__4 ml ( num__20.0 of total volume ) water in original container is = num__8 ml . ( num__40.0 of total volume ) now we take a container which has num__4 times volume of original container . so volume of new container = num__4 * num__20 = num__80 in this new container we pour oil vinegar and water from original container . volume occupied in new container by oil vinegar and water from original container = num__8 + num__4 + num__8 = num__20 ml hence volume left in new container = num__80 - num__20 = num__60 ml as per question remaining space in the larger bottle is then filled with water . i . e num__60 ml is filled with water . hence total water in new container = num__60 ml + num__8 ml ( this one poured from old container ) = num__68 ml answer : c <eor> c <eos> |
c |
multiply__20.0__4.0__ add__40.0__20.0__ add__8.0__60.0__ add__8.0__60.0__ |
multiply__20.0__4.0__ add__40.0__20.0__ add__8.0__60.0__ add__8.0__60.0__ |
| a train num__360 m long is running at a speed of num__45 km / hr . in what time will it pass a bridge num__190 m long ? <o> a ) num__44 <o> b ) num__99 <o> c ) num__88 <o> d ) num__77 <o> e ) num__21 |
: speed = num__45 * num__0.277777777778 = num__12.5 m / sec total distance covered = num__360 + num__190 = num__550 m required time = num__550 * num__0.08 = num__44 sec answer : a <eor> a <eos> |
a |
add__360.0__190.0__ divide__550.0__12.5__ round__44.0__ |
add__360.0__190.0__ divide__550.0__12.5__ divide__550.0__12.5__ |
| the sum of ages of num__5 children born num__3 years different each is num__50 years . what is the age of the elder child ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__16 <o> e ) num__17 |
let the ages of children be x ( x + num__3 ) ( x + num__6 ) ( x + num__9 ) and ( x + num__12 ) years . then x + ( x + num__3 ) + ( x + num__6 ) + ( x + num__9 ) + ( x + num__12 ) = num__50 num__5 x = num__20 x = num__4 . x + num__12 = num__4 + num__12 = num__16 d <eor> d <eos> |
d |
add__3.0__6.0__ add__3.0__9.0__ subtract__9.0__5.0__ add__4.0__12.0__ add__4.0__12.0__ |
add__3.0__6.0__ add__3.0__9.0__ subtract__9.0__5.0__ add__4.0__12.0__ add__4.0__12.0__ |
| a man is walking at a speed of num__10 km per hour . after every kilometre he takes rest for num__3 minutes . how much time will be take to cover a distance of num__3 kilometres ? <o> a ) num__48 min . <o> b ) num__50 min . <o> c ) num__45 min . <o> d ) num__55 min . <o> e ) num__24 min . |
rest time = number of rest à — time for each rest = num__2 à — num__3 = num__6 minutes total time to cover num__3 km = ( num__3 â „ num__10 à — num__60 ) minutes + num__6 minutes = num__24 minutes answer e <eor> e <eos> |
e |
multiply__3.0__2.0__ hour_to_min_conversion__ round__24.0__ |
multiply__3.0__2.0__ hour_to_min_conversion__ round__24.0__ |
| when i was married num__10 years ago my wife is the num__6 th member of the family . today my father died and a baby born to me . the average age of my family during my marriage is same as today . what is the age of father when he died ? <o> a ) num__40 yrs <o> b ) num__50 yrs <o> c ) num__60 yrs <o> d ) num__70 yrs <o> e ) num__80 yrs |
let average age num__10 yrs ago was x yrs total age num__10 yrs before was num__6 x yr at present father died now members remains = num__5 now total age after num__10 yrs is ( no . of member ) * ( x + num__10 ) yrs i . e . total age is num__5 ( x + num__10 ) and after baby born total age = num__5 ( x + num__10 ) + num__0 because baby age is num__0 yrs as per ques . both time age is same i . e . num__6 x = num__5 x + num__50 + num__0 after solving . . x = num__50 i . e . father ' s age before num__10 yrs was num__50 yrs now father died after num__10 yrs hence age at dead time = num__50 + num__10 = num__60 yrs . answer : c <eor> c <eos> |
c |
multiply__10.0__5.0__ multiply__10.0__6.0__ multiply__10.0__6.0__ |
multiply__10.0__5.0__ multiply__10.0__6.0__ multiply__10.0__6.0__ |
| at a company the average wage of male employees is $ num__4 per hour and the average wage of female employees is $ num__8 per hour . if the average wage of all employees is $ num__7 per hour what is the ratio of the number of male employees to the number of female employees ? <o> a ) num__1 to num__3 <o> b ) num__2 to num__3 <o> c ) num__3 to num__2 <o> d ) num__5 to num__3 <o> e ) num__2 to num__1 |
average hourly wage of male employees = num__4 $ average hourly wage of female employees = num__8 $ average hourly wage of all employees = num__7 $ let number of male employees = m number of female employees = f num__7 = ( num__4 m + num__8 f ) / m + f = > num__7 m + num__7 f = num__4 m + num__8 f = > num__3 m = num__1 f = > m / f = num__0.333333333333 answer a <eor> a <eos> |
a |
subtract__7.0__4.0__ subtract__4.0__3.0__ reverse__3.0__ reverse__1.0__ |
subtract__7.0__4.0__ subtract__4.0__3.0__ reverse__3.0__ reverse__1.0__ |
| if num__20.0 of x is num__15 less than num__15.0 of num__1500 then x is ? <o> a ) num__872 <o> b ) num__738 <o> c ) num__837 <o> d ) num__840 <o> e ) num__1050 |
num__20.0 of x = x / num__5 ; num__15.0 of num__1500 = num__0.15 * num__1500 = num__225 given that x / num__5 = num__225 - num__15 = > x / num__5 = num__210 = > x = num__1050 . answer : e <eor> e <eos> |
e |
subtract__20.0__15.0__ multiply__1500.0__0.15__ subtract__225.0__15.0__ multiply__5.0__210.0__ multiply__5.0__210.0__ |
subtract__20.0__15.0__ multiply__1500.0__0.15__ subtract__225.0__15.0__ multiply__5.0__210.0__ multiply__5.0__210.0__ |
| a contractor is engaged for num__30 days on the condition that he receives rs . num__25 for each dayhe works & is fined rs . num__7.50 for each day is absent . he gets rs . num__425 in all . for how many dayswas he absent ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__15 <o> e ) num__17 |
num__30 * num__25 = num__750 num__425 - - - - - - - - - - - num__325 num__25 + num__7.50 = num__32.5 num__325 / num__32.5 = num__10 b <eor> b <eos> |
b |
multiply__30.0__25.0__ subtract__750.0__425.0__ add__25.0__7.5__ divide__325.0__32.5__ round__10.0__ |
multiply__30.0__25.0__ subtract__750.0__425.0__ add__25.0__7.5__ divide__325.0__32.5__ divide__325.0__32.5__ |
| an e - commerce website marks his wares num__30.0 more than the real price and allows num__10.0 discount . their profit is : <o> a ) num__17.0 <o> b ) num__19.0 <o> c ) num__21.0 <o> d ) num__23.0 <o> e ) none of these |
let the cp = num__100 rs . mark price = num__130 discount = num__10.0 selling price num__0.9 Ã — num__130 hence profit = num__17.0 answer : a . <eor> a <eos> |
a |
percent__17.0__100.0__ |
percent__17.0__100.0__ |
| there is food for num__760 men for num__22 days . how many more men should join after two days so that the same food may last for num__19 days more ? <o> a ) num__38 <o> b ) num__40 <o> c ) num__83 <o> d ) num__87 <o> e ) num__81 |
num__760 - - - - num__22 num__760 - - - - num__20 x - - - - - num__19 x * num__19 = num__760 * num__20 x = num__800 num__760 - - - - - - - num__40 answer : b <eor> b <eos> |
b |
divide__760.0__19.0__ round__40.0__ |
subtract__800.0__760.0__ subtract__800.0__760.0__ |
| workers at a campaign office have num__1000 fliers to send out . if they send out num__0.2 of them in the morning and num__0.25 of the remaining ones out during the afternoon how many are left for the next day ? <o> a ) num__300 <o> b ) num__800 <o> c ) num__1100 <o> d ) num__600 <o> e ) num__1900 |
( num__0.2 ) * num__1000 = num__200 remaining = num__1000 - num__200 = num__800 ( num__0.25 ) of remaining = ( num__0.25 ) * num__800 = num__200 remaining now = num__800 - num__200 = num__600 answer : option d <eor> d <eos> |
d |
multiply__1000.0__0.2__ subtract__1000.0__200.0__ subtract__800.0__200.0__ round__600.0__ |
multiply__1000.0__0.2__ subtract__1000.0__200.0__ subtract__800.0__200.0__ subtract__800.0__200.0__ |
| if the average ( arithmetic mean ) of a and b is num__20 and c – a = num__30 what is the average of b and c ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__40 <o> d ) num__35 <o> e ) num__45 |
a + b / num__2 = num__20 = > a + b = num__40 a = c - num__30 . . . sub this value c - num__30 + b = num__40 = > c + b = num__70 = > c + b / num__2 = num__35 answer : d <eor> d <eos> |
d |
multiply__20.0__2.0__ add__30.0__40.0__ divide__70.0__2.0__ divide__70.0__2.0__ |
multiply__20.0__2.0__ add__30.0__40.0__ divide__70.0__2.0__ divide__70.0__2.0__ |
| how many members are there between num__1 to num__100 which are divisible by num__9 and the sum of two digits is num__9 ? <o> a ) num__13 <o> b ) num__7 <o> c ) num__11 <o> d ) num__9 <o> e ) num__10 |
there are num__11 numbers divisible by num__9 . but in number num__99 num__9 + num__9 is not equal to num__9 . so the ans is num__10 numbers answer : e <eor> e <eos> |
e |
subtract__100.0__1.0__ add__1.0__9.0__ add__1.0__9.0__ |
subtract__100.0__1.0__ add__1.0__9.0__ add__1.0__9.0__ |
| solution a is num__20.0 salt and solution b is num__70.0 salt . if you have num__30 ounces of solution a and num__60 ounces of solution b in what ratio could you mix solution a with solution b to produce num__50 ounces of a num__50.0 salt solution ? <o> a ) num__6 : num__4 <o> b ) num__6 : num__14 <o> c ) num__2 : num__3 <o> d ) num__4 : num__6 <o> e ) num__3 : num__7 |
forget the volumes for the time being . you have to mix num__20.0 and num__80.0 solutions to get num__50.0 . this is very straight forward since num__50 is int he middle of num__20 and num__80 so we need both solutions in equal quantities . if this does n ' t strike use w num__1 / w num__2 = ( a num__2 - aavg ) / ( aavg - a num__1 ) w num__1 / w num__2 = ( num__70 - num__50 ) / ( num__50 - num__20 ) = num__0.666666666667 so the volume of the two solutions will be equal . answer has to be num__2 : num__3 c <eor> c <eos> |
c |
add__20.0__60.0__ divide__60.0__30.0__ divide__20.0__30.0__ divide__60.0__20.0__ divide__60.0__30.0__ |
add__20.0__60.0__ divide__60.0__30.0__ divide__20.0__30.0__ divide__60.0__20.0__ divide__60.0__30.0__ |
| the sum of the present ages of two persons a and b is num__60 . if the age of a is twice that of b find the sum of their ages num__5 years hence ? <o> a ) num__50 <o> b ) num__60 <o> c ) num__70 <o> d ) num__80 <o> e ) num__90 |
a + b = num__60 a = num__2 b num__2 b + b = num__60 = > b = num__20 then a = num__40 . num__5 years their ages will be num__45 and num__25 . sum of their ages = num__45 + num__25 = num__70 . answer : c <eor> c <eos> |
c |
subtract__60.0__20.0__ add__5.0__40.0__ add__5.0__20.0__ add__45.0__25.0__ add__45.0__25.0__ |
subtract__60.0__20.0__ add__5.0__40.0__ add__5.0__20.0__ add__45.0__25.0__ add__45.0__25.0__ |
| the average age of an adult class is num__50 years . num__12 new students with an avg age of num__32 years join the class . therefore decreasing the average by num__4 years . find what was the original average age of the class ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__42 <o> d ) num__20 <o> e ) num__22 |
let original strength = y then num__50 y + num__12 x num__32 = ( y + num__12 ) x num__46 â ‡ ’ num__50 y + num__384 = num__46 y + num__552 â ‡ ’ num__4 y = num__168 â ˆ ´ y = num__42 c <eor> c <eos> |
c |
subtract__50.0__4.0__ multiply__12.0__32.0__ multiply__12.0__46.0__ subtract__552.0__384.0__ subtract__46.0__4.0__ subtract__46.0__4.0__ |
subtract__50.0__4.0__ multiply__12.0__32.0__ multiply__12.0__46.0__ subtract__552.0__384.0__ subtract__46.0__4.0__ subtract__46.0__4.0__ |
| a man distributes num__0.375 of his money to his wife and num__0.4 to his son . he still has rs . num__3375 left with him . how much did his wife get ? <o> a ) rs . num__3625 / - <o> b ) rs . num__4625 / - <o> c ) rs . num__5625 / - <o> d ) rs . num__6625 / - <o> e ) rs . num__7625 / - |
let total money = m . so m * ( num__1 - num__0.375 - num__0.4 ) = num__3375 ; = > m = num__3375 / ( num__1 - num__0.375 - num__0.4 ) ; = > m = num__15000 so his wife got = num__15000 * ( num__0.375 ) = num__5625 . so his wife got rs . num__5625 / - answer : c <eor> c <eos> |
c |
multiply__0.375__15000.0__ multiply__0.375__15000.0__ |
multiply__0.375__15000.0__ multiply__0.375__15000.0__ |
| in a certain animal shelter the ratio of the number of dogs to the number of cats is num__15 to num__7 . if num__16 additional cats were to be taken in by the shelter the ratio of the number of dogs to the number of cats would be num__15 to num__11 . how many dogs are in the shelter ? <o> a ) num__15 <o> b ) num__25 <o> c ) num__30 <o> d ) num__45 <o> e ) num__60 |
this ratio question can be solved in a couple of different ways . here ' s an algebraic approach . . . we ' re told that the ratio of the number of dogs to the number of cats is num__15 : num__7 . we ' re then told that num__16 more cats are added to this group and the ratio becomes num__15 : num__11 . we ' re asked for the number of dogs . algebraically since the number of dogs is a multiple of num__15 and the number of cats is a multiple of num__7 we can write this initial relationship as . . . num__15 x / num__7 x when we add the num__12 cats and factor in the ' ending ratio ' we have an equation . . . . num__15 x / ( num__7 x + num__16 ) = num__1.36363636364 here we have num__1 variable and num__1 equation so we can solve for x . . . . ( num__15 x ) ( num__11 ) = ( num__7 x + num__16 ) ( num__15 ) ( x ) ( num__11 ) = ( num__7 x + num__16 ) ( num__1 ) num__11 x = num__7 x + num__16 num__4 x = num__16 x = num__4 with this x we can figure out the initial number of dogs and cats . . . initial dogs = num__15 x = num__15 ( num__4 ) = num__60 final answer : e <eor> e <eos> |
e |
divide__15.0__11.0__ round_down__1.3636__ subtract__15.0__11.0__ multiply__15.0__4.0__ multiply__15.0__4.0__ |
divide__15.0__11.0__ round_down__1.3636__ subtract__15.0__11.0__ multiply__15.0__4.0__ divide__60.0__1.0__ |
| num__24 buckets of water fill a tank when the capacity of each bucket is num__13.5 litres . how many buckets will be required to fill the same tank if the capacity of each bucket is num__9 litres ? <o> a ) num__30 <o> b ) num__36 <o> c ) num__60 <o> d ) data inadequate <o> e ) none of these |
capacity of the tank = num__24 Ã — num__13.5 = num__324 litres when the capacity of each bucket = num__9 litres then the required no . of buckets = num__324 â „ num__9 = num__36 answer b <eor> b <eos> |
b |
multiply__24.0__13.5__ divide__324.0__9.0__ round__36.0__ |
multiply__24.0__13.5__ divide__324.0__9.0__ round__36.0__ |
| if a man reduces the selling price of a fan from rs . num__420 to rs . num__402 his loss increases by num__3.0 . the cost price of the fan is : <o> a ) rs . num__600 <o> b ) rs . num__650 <o> c ) rs . num__475 <o> d ) rs . num__525 <o> e ) none of these |
explanation : solution : let c . p . be rs . x . then num__3.0 of x = ( num__420 - num__402 ) = num__18 = > num__3 x / num__100 = num__18 = > x = rs . num__600 . answer : a <eor> a <eos> |
a |
percent__100.0__600.0__ |
percent__100.0__600.0__ |
| bruno and sacha are running in the same direction around a stadium . sacha runs at a constant speed of num__6 meters per second and bruno runs at a constant speed of num__5 meters per second . at a certain point sacha overtakes bruno . if four minute afterward sacha stops and waits for bruno to reach him then how many seconds does he have to wait ? <o> a ) num__12 <o> b ) num__24 <o> c ) num__36 <o> d ) num__48 <o> e ) num__72 |
the difference of the speed is num__1 m per second so in four minute sacha will be num__240 m ahead of bruno . . bruno will cover this in num__48.0 = num__48 secs . . d <eor> d <eos> |
d |
subtract__6.0__5.0__ divide__240.0__5.0__ round__48.0__ |
subtract__6.0__5.0__ divide__240.0__5.0__ round__48.0__ |
| to save money arkadelphia cream cheese will reduce each dimension of its rectangular box container ( which is entirely full of cream cheese ) by num__10.0 and reduce the price it charges its consumers by num__10.0 as well . by what percentage does this increase the price - per - cubic - inch that each consumer will pay for cream cheese ? <o> a ) num__1 . num__23.5 <o> b ) num__2 . num__50.0 <o> c ) num__3 . num__100.0 <o> d ) num__4 . num__300.0 <o> e ) num__5 . num__400 % |
take smart numbers let l = num__20 : b = num__10 : h = num__10 of initial box and price = num__50 $ therefore price / cubic inch = num__50 / ( num__20 * num__10 * num__10 ) = num__0.025 now when dimensions are reduced by num__10.0 and price also reduced by num__10.0 l = num__18 ; b = num__9 ; h = num__9 and price = num__45 $ therefore price / cubic inch = num__45 / ( num__18 * num__9 * num__9 ) = num__0.030 percentage change = ( num__0.030 - num__0.025 ) * num__100 / num__0.025 = num__23.5 answer is a <eor> a <eos> |
a |
subtract__10.0__9.0__ |
subtract__10.0__9.0__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__120 metres and they cross each other in num__12 seconds then the speed of each train ( in km / hr ) is : <o> a ) num__12 <o> b ) num__18 <o> c ) num__36 <o> d ) num__72 <o> e ) num__38 |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__120 + num__120 ) / num__12 = > num__2 x = num__20 = > x = num__10 . therefore speed of each train = num__10 m / sec = ( num__10 x ( num__3.6 ) ) km / hr = num__36 km / hr . answer : c <eor> c <eos> |
c |
divide__120.0__12.0__ multiply__10.0__3.6__ round__36.0__ |
divide__120.0__12.0__ multiply__10.0__3.6__ round__36.0__ |
| the cost to rent a small bus for a trip is x dollars which is to be shared equally among the people taking the trip . if num__10 people take the trip rather than num__20 how many more dollars in terms of x will it cost per person ? <o> a ) x / num__6 <o> b ) x / num__16 <o> c ) x / num__20 <o> d ) num__3 x / num__40 <o> e ) num__3 x / num__80 |
just plugging in values x = num__200 cost per person if we consider num__10 = num__20 cost per person if we consider num__20 = num__10 difference between costs is num__10 dollars . plugging in the value of x into the answer choices let ' s see which one yields a result of num__10 . and that is c . <eor> c <eos> |
c |
multiply__10.0__20.0__ divide__200.0__10.0__ |
multiply__10.0__20.0__ divide__200.0__10.0__ |
| a parking garage rents parking spaces for $ num__12 per week or $ num__30 per month . how much does a person save in a year by renting by the month rather than by the week ? <o> a ) $ num__140 <o> b ) $ num__160 <o> c ) $ num__220 <o> d ) $ num__240 <o> e ) $ num__264 |
num__12 $ per week ! an year has num__52 weeks . annual charges per year = num__52 * num__12 = num__624 $ num__30 $ per month ! an year has num__12 months . annual charges per year = num__12 * num__30 = num__360 $ num__624 - num__360 = num__264 ans e <eor> e <eos> |
e |
multiply__12.0__52.0__ multiply__12.0__30.0__ subtract__624.0__360.0__ subtract__624.0__360.0__ |
multiply__12.0__52.0__ multiply__12.0__30.0__ subtract__624.0__360.0__ subtract__624.0__360.0__ |
| simplify : num__5005 - num__5000 + num__10 <o> a ) num__4505 <o> b ) num__4509 <o> c ) num__4501 <o> d ) num__4500 <o> e ) none of them |
num__5005 - num__5000 + num__10 = num__5005 - ( num__500.0 ) = num__5005 - num__500 = num__4505 answer is a <eor> a <eos> |
a |
divide__5000.0__10.0__ subtract__5005.0__500.0__ subtract__5005.0__500.0__ |
divide__5000.0__10.0__ subtract__5005.0__500.0__ subtract__5005.0__500.0__ |
| amy ’ s retirement portfolio contains only stocks and bonds . at the beginning of num__2016 her portfolio had an allocation of num__60.0 stocks and num__40.0 bonds . over the course of num__2016 the total value of her portfolio increased by num__8.0 with the value of her stock holdings increasing by num__10.0 . by what percent did the value of her bond holdings increase ? <o> a ) num__4.0 <o> b ) num__5.0 <o> c ) num__6.0 <o> d ) num__7.0 <o> e ) num__7.5 % |
thus total value of bond at the end of num__2016 is num__420 ( ie num__1080 - num__660 ) so per unit value of bond at the end of num__2016 is num__10.5 = num__10.50 hence percentage increase is num__0.50 / num__10 ∗ num__100 = num__5.0 answer will be ( b ) <eor> b <eos> |
b |
percent__5.0__100.0__ |
percent__5.0__100.0__ |
| a train num__390 metres long is moving at a speed of num__25 kmph . it will cross a man coming from the opposite direction at a speed of num__2 km per hour in : <o> a ) num__30 sec <o> b ) num__32 sec <o> c ) num__36 sec <o> d ) num__38 sec <o> e ) num__52 sec |
relative speed = ( num__25 + num__2 ) km / hr = num__27 km / hr = ( num__27 × num__0.277777777778 ) m / sec = num__7.5 m / sec . time taken by the train to pass the man = ( num__390 × num__0.133333333333 ) sec = num__52 sec answer : e <eor> e <eos> |
e |
add__25.0__2.0__ divide__390.0__7.5__ round__52.0__ |
add__25.0__2.0__ divide__390.0__7.5__ divide__390.0__7.5__ |
| how many bricks each measuring num__25 cm * num__11.25 cm * num__6 cm will be needed to build a wall num__8 m * num__6 m * num__22.5 m <o> a ) num__6100 <o> b ) num__6200 <o> c ) num__6300 <o> d ) num__6400 <o> e ) none of these |
explanation : to solve this type of question simply divide the volume of wall with the volume of brick to get the numbers of required bricks so lets solve this number of bricks = volume of wall / volume of num__1 brick = num__800 ∗ num__600 ∗ num__22.5 / num__25 ∗ num__11.25 ∗ num__6 = num__6400 option d <eor> d <eos> |
d |
multiply__8.0__800.0__ round__6400.0__ |
multiply__8.0__800.0__ divide__6400.0__1.0__ |
| in what time will a railway train num__80 m long moving at the rate of num__36 kmph pass a telegraph post on its way ? <o> a ) num__8 sec <o> b ) num__1 sec <o> c ) num__9 sec <o> d ) num__6 sec <o> e ) num__2 sec |
t = num__2.22222222222 * num__3.6 = num__8 sec answer : a <eor> a <eos> |
a |
divide__80.0__36.0__ round__8.0__ |
divide__80.0__36.0__ round__8.0__ |
| in an electric circuit two resistors with resistances x and y are connected in parallel . if r is the combined resistance of these two resistors then the reciprocal of r is equal to the sum of the reciprocals of x and y . what is r if x is num__4 ohms and y is num__5 ohms ? <o> a ) num__0.777777777778 <o> b ) num__2.22222222222 <o> c ) num__0.35 <o> d ) num__0.45 <o> e ) num__0.55 |
num__1 / r = num__1 / x + num__1 / y num__1 / r = num__0.25 + num__0.2 = num__0.45 r = num__2.22222222222 the answer is b . <eor> b <eos> |
b |
subtract__5.0__4.0__ reverse__4.0__ reverse__5.0__ add__0.25__0.2__ reverse__0.45__ reverse__0.45__ |
subtract__5.0__4.0__ reverse__4.0__ reverse__5.0__ add__0.25__0.2__ reverse__0.45__ reverse__0.45__ |
| what is the maximum value of vx - yz . if the value of v x y z have to be chosen from the set a where a ( - num__3 - num__2 - num__1 num__01 num__23 ) <o> a ) num__15 <o> b ) num__66 <o> c ) num__25 <o> d ) num__88 <o> e ) num__17 |
explanation : to maximize the value of vx - yz we make yz negative and vx as maximum as possible using given value . vx − yz = ( − num__3 ) num__2 − ( − num__3 × num__2 ) vx − yz = ( − num__3 ) num__2 − ( − num__3 × num__2 ) = num__15 answer : a <eor> a <eos> |
a |
multiply__1.0__15.0__ |
multiply__1.0__15.0__ |
| excluding stoppages the speed of a train is num__35 kmph and including stoppages it is num__22 kmph . of how many minutes does the train stop per hour ? <o> a ) num__82 <o> b ) num__17 <o> c ) num__22 <o> d ) num__82 <o> e ) num__18 |
explanation : t = num__0.371428571429 * num__60 = num__22 answer : option c <eor> c <eos> |
c |
hour_to_min_conversion__ round__22.0__ |
hour_to_min_conversion__ round__22.0__ |
| set a consists of num__19 elements . the average of set a is l . if a new element is added to the set and the average grows by k what is the value of the new element ? <o> a ) a ) l ( num__1 + k / num__5 ) <o> b ) b ) l * ( k / num__100 ) - num__20 l <o> c ) c ) num__20 l ( num__1 + k / num__100 ) <o> d ) d ) num__20 ( num__1 + k / num__100 ) - num__19 l <o> e ) e ) l * ( k / num__5 ) - num__19 |
let ' s assume the value of new element to bea . so converting the word problem into equation ( num__19 * l + a ) / ( num__19 + num__1 ) = l + k = = > after solving the equation we will get a ( value of newly added element ) = l + num__20 k but according to the answer options problem writer wanted to conveyk % percent rise in average value . if we consider this case than resulting equation will be ( num__19 * l + a ) / ( num__19 + num__1 ) = l + ( k / num__100 ) * l = = > num__19 l + a = num__20 [ l + kl / num__100 ] = = > a = num__20 l + kl / num__5 - num__19 l = = > a = l + kl / num__5 which is equivalent to option [ a ] <eor> a <eos> |
a |
add__19.0__1.0__ divide__100.0__20.0__ reverse__1.0__ |
add__19.0__1.0__ divide__100.0__20.0__ reverse__1.0__ |
| the weight of every type a widget is the same the weight of every type b widget is the same and the weight of every type c widget is the same . if the weight of num__7 type a widgets is equal to the weight of num__2 type b widgets and the weight of num__4 type b widgets is equal to the weight of num__7 type c widgets . what is the ratio of the total weight of num__1 type a widget and num__1 type b widget to the total weight of num__1 type b widget and num__1 type c widget ? <o> a ) num__3 : num__5 <o> b ) num__5 : num__7 <o> c ) num__7 : num__9 <o> d ) num__9 : num__11 <o> e ) num__11 : num__13 |
num__4 b = num__7 c and so b = num__7 c / num__4 num__7 a = num__2 b and so a = num__2 b / num__7 = c / num__2 a + b = c / num__2 + num__7 c / num__4 = num__9 c / num__4 b + c = num__7 c / num__4 + c = num__11 c / num__4 the ratio of a + b : b + c = num__9 : num__11 the answer is d . <eor> d <eos> |
d |
add__7.0__2.0__ add__7.0__4.0__ add__7.0__2.0__ |
add__7.0__2.0__ add__7.0__4.0__ add__7.0__2.0__ |
| the product of two positive integers is num__675 and their difference is num__2 . what is the bigger number ? <o> a ) num__21 <o> b ) num__23 <o> c ) num__25 <o> d ) num__27 <o> e ) num__29 |
let ' s use trial and error to find the two numbers . num__25 * num__23 = num__575 ( too low ) num__27 * num__25 = num__675 the answer is d . <eor> d <eos> |
d |
subtract__25.0__2.0__ multiply__25.0__23.0__ divide__675.0__25.0__ divide__675.0__25.0__ |
subtract__25.0__2.0__ multiply__25.0__23.0__ divide__675.0__25.0__ divide__675.0__25.0__ |
| a and b are two partially filled buckets of water . if num__5 liters are transferred from a to b then a would contain one - third of the amount of water in b . alternatively if num__4 liters are transferred from b to a b would contain one - half of the amount of water in a . bucket a contains how many liters of water ? <o> a ) num__11 <o> b ) num__10.4 <o> c ) num__17 <o> d ) num__21 <o> e ) num__23 |
let a contains a b contains b liters so ( a - num__5 ) / ( b + num__5 ) = num__0.333333333333 . . . . . . . ( num__1 ) again ( b - num__4 ) / ( a + num__4 ) = num__0.5 . . . . . . . . . . ( num__2 ) from ( num__1 ) ( num__2 ) we find a = num__10.4 ans : ( b ) <eor> b <eos> |
b |
subtract__5.0__4.0__ reverse__0.5__ multiply__1.0__10.4__ |
subtract__5.0__4.0__ reverse__0.5__ divide__10.4__1.0__ |
| average of num__10 matches is num__32 how many runs one should should score to increase his average by num__5 runs . <o> a ) a ) num__70 <o> b ) b ) num__76 <o> c ) c ) num__78 <o> d ) d ) num__80 <o> e ) e ) num__87 |
explanation : average after num__11 innings should be num__37 so required score = ( num__11 * num__37 ) - ( num__10 * num__32 ) = num__407 - num__320 = num__87 answer : option e <eor> e <eos> |
e |
add__32.0__5.0__ multiply__37.0__11.0__ multiply__10.0__32.0__ subtract__407.0__320.0__ subtract__407.0__320.0__ |
add__32.0__5.0__ multiply__37.0__11.0__ multiply__10.0__32.0__ subtract__407.0__320.0__ subtract__407.0__320.0__ |
| in a race of num__200 m a can beat b by num__21 m and c by num__15 m . in a race of num__425 m c will beat b by ? <o> a ) num__30 m <o> b ) num__75 m <o> c ) num__85 m <o> d ) num__25 m <o> e ) num__15 m |
explanation : a : b = num__100 : num__79 a : c = num__100 : num__85 = > b : c = num__79 : num__85 when c covers num__85 m b covers num__79 m when c covers num__425 m b covers = > num__395 m so c beats b by ( num__425 - num__395 ) = num__30 m . answer : a <eor> a <eos> |
a |
subtract__100.0__21.0__ subtract__100.0__15.0__ subtract__425.0__395.0__ round__30.0__ |
subtract__100.0__21.0__ subtract__100.0__15.0__ subtract__425.0__395.0__ subtract__425.0__395.0__ |
| positive integer a gives the remainder of num__12 when divided by another positive integer b . if a / b = num__47.24 what is the value of b ? <o> a ) num__96 <o> b ) num__75 <o> c ) num__48 <o> d ) num__25 <o> e ) num__50 |
. num__24 of b = remainder . num__24 of b = num__12 b = ( num__12 * num__100 ) / num__24 = num__50 . e <eor> e <eos> |
e |
subtract__100.0__50.0__ |
subtract__100.0__50.0__ |
| at a certain supplier a machine of type a costs $ num__22000 and a machine of type b costs $ num__50000 . each machine can be purchased by making a num__20 percent down payment and repaying the remainder of the cost and the finance charges over a period of time . if the finance charges are equal to num__40 percent of the remainder of the cost how much less would num__2 machines of type a cost than num__1 machine of type b under this arrangement ? <o> a ) $ num__7920 <o> b ) $ num__11200 <o> c ) $ num__12000 <o> d ) $ num__12800 <o> e ) $ num__13 |
200 |
total cost of num__2 machines of type a = num__20.0 of ( cost of num__2 machine a ) + remainder + num__40.0 remainder = num__20.0 of num__44000 + ( num__44000 - num__20.0 of num__44000 ) + num__40.0 of ( num__44000 - num__20.0 of num__44000 ) = num__58080 total cost of num__1 machine of type b = num__20.0 of ( cost of num__1 machine b ) + remainder + num__40.0 remainder = num__20.0 of num__50000 + ( num__50000 - num__20.0 of num__50000 ) + num__40.0 of ( num__50000 - num__20.0 of num__50000 ) = num__66000 diff = num__66000 - num__58080 = num__7900 hence a . <eor> a <eos> |
a |
a |
| the average age of father and his two sons is num__24 years . five years ago the average age of the two sons was num__15 years . if the difference between the ages of the two sons is four years what is the present age of the father ? <o> a ) num__45 <o> b ) num__32 <o> c ) num__47 <o> d ) num__48 <o> e ) num__49 |
the total present age of father and two sons is num__3 x num__24 = num__72 yrs the total present age of sons is ( num__15 + num__5 ) x num__2 = num__40 years so present age of father is num__72 â € “ num__40 = num__32 yrs answer b <eor> b <eos> |
b |
multiply__24.0__3.0__ divide__15.0__3.0__ subtract__5.0__3.0__ subtract__72.0__40.0__ subtract__72.0__40.0__ |
multiply__24.0__3.0__ divide__15.0__3.0__ subtract__5.0__3.0__ subtract__72.0__40.0__ subtract__72.0__40.0__ |
| what is the cost of leveling the field in the form of parallelogram at the rate of rs . num__3.0 sq . metre whose base & perpendicular distance from the other side being num__64 m & num__74 m respectively ? <o> a ) s . num__6000 <o> b ) s . num__6300 <o> c ) s . num__4736 <o> d ) s . num__6480 <o> e ) s . num__7000 |
area of the parallelogram = length of the base * perpendicular height = num__64 * num__74 = num__4736 m . total cost of levelling = rs . num__4736 c <eor> c <eos> |
c |
multiply__64.0__74.0__ round__4736.0__ |
multiply__64.0__74.0__ multiply__64.0__74.0__ |
| due to construction the speed limit along an num__9 - mile section of highway is reduced from num__55 miles per hour to num__20 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? <o> a ) a ) num__5.61 <o> b ) b ) num__8 <o> c ) c ) num__10 <o> d ) d ) num__17.19 <o> e ) e ) num__24 |
old time in minutes to cross num__9 miles stretch = num__9 * num__1.09090909091 = num__9 * num__1.09090909091 = num__9.81 new time in minutes to cross num__9 miles stretch = num__9 * num__3.0 = num__9 * num__3.0 = num__27 time difference = num__17.19 ans : d <eor> d <eos> |
d |
multiply__9.0__3.0__ subtract__27.0__9.81__ round__17.19__ |
multiply__9.0__3.0__ subtract__27.0__9.81__ round__17.19__ |
| find the odd man out num__2 num__5 num__11 num__23 num__48 num__95 <o> a ) num__5 <o> b ) num__11 <o> c ) num__23 <o> d ) num__48 <o> e ) num__28 |
num__2 * num__2 + num__1 = num__5 num__5 * num__2 + num__1 = num__11 num__11 * num__2 + num__1 = num__23 num__23 * num__2 + num__1 = num__47 num__47 * num__2 + num__1 = num__95 answer : d <eor> d <eos> |
d |
subtract__48.0__1.0__ multiply__48.0__1.0__ |
subtract__48.0__1.0__ multiply__48.0__1.0__ |
| rahim and his uncle differ in their ages by num__30 years . after num__7 years if the sum of their ages is num__66 what will be the age of the uncle ? <o> a ) num__41 years <o> b ) num__36 <o> c ) num__48 <o> d ) num__33 <o> e ) num__44 |
a num__41 let uncle ’ s present age = x rahim ' s present age = y y – x = num__30 . . . ( i ) after num__7 year ( x + num__7 ) + ( y + num__7 ) = num__66 x + y + num__14 = num__66 x + y = num__52 . . . ( ii ) combining ( i ) and ( ii ) we get ( x + y = num__52 ) + ( x – y = num__30 ) num__2 x = num__82 x = num__41 uncle ' s age is num__41 <eor> a <eos> |
a |
subtract__66.0__14.0__ divide__14.0__7.0__ add__30.0__52.0__ subtract__82.0__41.0__ |
subtract__66.0__14.0__ divide__14.0__7.0__ add__30.0__52.0__ subtract__82.0__41.0__ |
| find the value of x if the mean of the series num__245 num__564458 num__12536 x is num__250 <o> a ) num__50 <o> b ) num__72 <o> c ) num__36 <o> d ) num__45 <o> e ) num__112 |
mean = num__245 + num__564 + num__458 + num__125 + num__36 + x / num__6 = num__250 num__1428 + x = num__1500 x = num__72 answer is b <eor> b <eos> |
b |
multiply__250.0__6.0__ subtract__1500.0__1428.0__ subtract__1500.0__1428.0__ |
multiply__250.0__6.0__ subtract__1500.0__1428.0__ subtract__1500.0__1428.0__ |
| if num__6 is one solution of the equation x ^ num__2 + num__3 x + k = num__10 where k is a constant what is the other solution ? <o> a ) - num__7 <o> b ) - num__4 <o> c ) - num__3 <o> d ) - num__9 <o> e ) num__6 |
the phrase “ num__6 is one solution of the equation ” means that one value of x is num__6 . thus we first must plug num__6 for x into the given equation to determine the value of k . so we have num__6 ^ num__2 + ( num__3 ) ( num__6 ) + k = num__10 num__36 + num__18 + k = num__10 num__54 + k = num__10 k = - num__44 next we plug - num__44 into the given equation for k and then solve for x . x ^ num__2 + num__3 x – num__44 = num__10 x ^ num__2 + num__3 x – num__54 = num__0 ( x + num__9 ) ( x - num__6 ) = num__0 x = - num__9 or x = num__6 thus - num__9 is the other solution . answer d . <eor> d <eos> |
d |
multiply__6.0__3.0__ multiply__3.0__18.0__ subtract__54.0__10.0__ add__6.0__3.0__ add__6.0__3.0__ |
multiply__6.0__3.0__ add__36.0__18.0__ subtract__54.0__10.0__ add__6.0__3.0__ add__6.0__3.0__ |
| the volume of a cube is num__3375 cc . find its surface . <o> a ) num__864 <o> b ) num__556 <o> c ) num__255 <o> d ) num__287 <o> e ) num__1350 |
a num__3 = num__3375 = > a = num__15 num__6 a num__2 = num__6 * num__15 * num__15 = num__1350 answer : e <eor> e <eos> |
e |
surface_cube__15.0__ surface_cube__15.0__ |
surface_cube__15.0__ surface_cube__15.0__ |
| a certain company retirement plan has arule of num__70 provision that allows an employee to retire when the employee ' s age plus years of employment with the company total at least num__70 . in what year could a female employee hired in num__1990 on her num__32 nd birthday first be eligible to retire under this provision ? <o> a ) num__2003 <o> b ) num__2009 <o> c ) num__2005 <o> d ) num__2006 <o> e ) num__2007 |
she must gain at least num__70 points now she has num__31 and every year gives her two more points : one for age and one for additional year of employment so num__32 + num__2 * ( # of years ) = num__70 - - > ( # of years ) = num__19 - - > num__1990 + num__19 = num__2009 . answer : b . <eor> b <eos> |
b |
add__1990.0__19.0__ add__1990.0__19.0__ |
add__1990.0__19.0__ add__1990.0__19.0__ |
| two trains start from p and q respectively and travel towards each other at a speed of num__50 km / hr and num__40 km / hr respectively . by the time they meet the first train has traveled num__100 km more than the second . the distance between p and q is ? <o> a ) num__237 <o> b ) num__278 <o> c ) num__277 <o> d ) num__900 <o> e ) num__261 |
at the time of meeting let the distance traveled by the second train be x km . then distance covered by the first train is ( x + num__100 ) km . x / num__40 = ( x + num__100 ) / num__50 num__50 x = num__40 x + num__4000 = > x = num__400 so distance between p and q = ( x + x + num__100 ) km = num__900 km . answer : d <eor> d <eos> |
d |
multiply__40.0__100.0__ round__900.0__ |
multiply__40.0__100.0__ round__900.0__ |
| in a pvt company num__15 staffs are working . salary for num__10 staffs @ num__15000 / - and num__5 staffs @ num__9000 / - per month . owner deducted amount of num__15 leaves from every one . then how much amount will be given in total for one month ? <o> a ) num__195000 <o> b ) num__202000 <o> c ) num__192000 <o> d ) num__188000 <o> e ) num__188500 |
salary for num__10 staffs @ num__15000 / - : num__1 num__50000 / - salary for num__5 staffs @ num__9000 / - : num__45000 / - total salary : num__195000 / - salary cutting num__1 day for num__10 staffs : num__5000 / - salary cutting num__1 days for num__5 staffs : num__1500 / - total salary for one month : num__195000 / - - num__6500 / = answer is e <eor> e <eos> |
e |
multiply__5.0__9000.0__ divide__50000.0__10.0__ divide__15000.0__10.0__ add__5000.0__1500.0__ subtract__195000.0__6500.0__ |
multiply__5.0__9000.0__ divide__50000.0__10.0__ divide__15000.0__10.0__ add__5000.0__1500.0__ subtract__195000.0__6500.0__ |
| what is the length of the longest pole which can be kept in a room num__10 m long num__4 m broad and num__3 m high ? <o> a ) num__7 <o> b ) num__9 <o> c ) num__11.18 <o> d ) num__13 <o> e ) none |
explanation : d num__2 = num__102 + num__42 + num__32 = num__11.18 c ) <eor> c <eos> |
c |
subtract__42.0__10.0__ round__11.18__ |
subtract__42.0__10.0__ round__11.18__ |
| how long does a train num__110 m long running at the speed of num__72 km / hr takes to cross a bridge num__175 m length ? <o> a ) num__12.9 sec <o> b ) num__12.1 sec <o> c ) num__17.9 sec <o> d ) num__16.8 sec <o> e ) num__14.25 sec |
speed = num__72 * num__0.277777777778 = num__20 m / sec total distance covered = num__110 + num__175 = num__285 m . required time = num__14.25 = num__14.25 sec . answer : e <eor> e <eos> |
e |
add__110.0__175.0__ divide__285.0__20.0__ round__14.25__ |
add__110.0__175.0__ divide__285.0__20.0__ divide__285.0__20.0__ |
| pipe a can fill a tank in num__4 hours . due to a leak at the bottom it takes num__8 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? <o> a ) num__13 <o> b ) num__17 <o> c ) num__18 <o> d ) num__8 <o> e ) num__12 |
let the leak can empty the full tank in x hours num__0.25 - num__1 / x = num__0.125 = > num__1 / x = num__0.25 - num__0.125 = ( num__2 - num__1 ) / num__8 = num__0.125 = > x = num__8 . answer : d <eor> d <eos> |
d |
multiply__4.0__0.25__ divide__1.0__8.0__ divide__8.0__4.0__ round__8.0__ |
multiply__4.0__0.25__ divide__1.0__8.0__ divide__8.0__4.0__ divide__8.0__1.0__ |
| what is the total number of integers between num__200 and num__500 that are divisible by num__25 ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__14 <o> d ) num__15 <o> e ) num__18 |
num__125 num__150 num__175 . . . num__450475 this is an equally spaced list ; you can use the formula : n = ( largest - smallest ) / ( ' space ' ) + num__1 = ( num__475 - num__125 ) / ( num__25 ) + num__1 = num__14.0 + num__1 = num__14 + num__1 = num__15 answer is d <eor> d <eos> |
d |
add__25.0__125.0__ subtract__200.0__25.0__ subtract__500.0__25.0__ add__1.0__14.0__ multiply__1.0__15.0__ |
add__25.0__125.0__ subtract__200.0__25.0__ subtract__500.0__25.0__ add__1.0__14.0__ add__1.0__14.0__ |
| if rs . num__400 amount to rs . num__540 in num__4 years what will it amount to in num__6 years at the same rate % per annum ? <o> a ) s . num__575 <o> b ) s . num__595 <o> c ) s . num__590 <o> d ) s . num__610 <o> e ) s . num__585 |
num__80 = ( num__400 * num__4 * r ) / num__100 r = num__8.75 i = ( num__400 * num__6 * num__8.75 ) / num__100 = num__210 num__400 + num__210 = num__610 answer : d <eor> d <eos> |
d |
percent__100.0__610.0__ |
percent__100.0__610.0__ |
| a car runs num__10000 miles using num__5 tyres interchangeably . to have equal worn out by all tyres how many miles each tyre should have run ? <o> a ) num__5000 <o> b ) num__8000 <o> c ) num__4000 <o> d ) num__3000 <o> e ) num__7000 |
for num__5 tyres running interchangeably num__10000 miles the car runs for num__1 tyre = num__2000.0 = num__2000 as you all know i suppose the car has num__4 tyre to have equal worn out by num__4 tyres we need = num__2000 * num__4 = num__8000 answer : b <eor> b <eos> |
b |
divide__10000.0__5.0__ subtract__5.0__1.0__ subtract__10000.0__2000.0__ round__8000.0__ |
divide__10000.0__5.0__ subtract__5.0__1.0__ multiply__2000.0__4.0__ multiply__2000.0__4.0__ |
| if num__1 = num__5 num__2 = num__10 num__3 = num__15 num__4 = num__20 num__5 = num__25 num__6 = num__30 then num__25 = ? hint : its a logic riddle not a mathematical riddle <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
e num__5 as stated num__5 = num__25 = > num__25 = num__5 answer is e <eor> e <eos> |
e |
multiply__1.0__5.0__ |
multiply__1.0__5.0__ |
| three bells commence tolling together and toll at intervals of num__2 num__4 num__6 seconds respectively . in num__10 minutes how many times do they toll together ? <o> a ) num__51 <o> b ) num__52 <o> c ) num__53 <o> d ) num__54 <o> e ) num__55 |
lcm of num__2 num__4 num__6 is num__12 . so after each num__12 seconds they would toll together . hence in num__10 minutes they would toll num__10 * num__60 seconds / num__12 seconds = num__50 times but then the question says they commence tolling together . so they basically also toll at thebeginning ( num__0 second ) . so total tolls together = num__50 + num__1 = num__51 answer : a <eor> a <eos> |
a |
multiply__2.0__6.0__ hour_to_min_conversion__ subtract__60.0__10.0__ add__1.0__50.0__ round__51.0__ |
multiply__2.0__6.0__ multiply__6.0__10.0__ subtract__60.0__10.0__ add__1.0__50.0__ add__1.0__50.0__ |
| swetha takes a trip and drives num__8 hours from town a to town c at a rate of num__40 miles per hour . on her way back swetha drives num__30 miles per hour and stops in town y which is midway between town a and town c . how many hours does it take swetha to drive from town c to town b ? <o> a ) num__1.01 <o> b ) num__1.25 <o> c ) num__1.53 <o> d ) num__5.33 <o> e ) num__2.0 |
distance from a to c = num__8 hr * num__40 mph = num__320 miles hence distance from b to c = num__0.5 * num__320 = num__160 time = num__5.33333333333 = num__5.33 hrs imo : d <eor> d <eos> |
d |
multiply__8.0__40.0__ multiply__0.5__320.0__ divide__160.0__30.0__ round__5.3333__ round__5.3333__ |
multiply__8.0__40.0__ multiply__0.5__320.0__ divide__160.0__30.0__ round__5.3333__ round__5.3333__ |
| in each term of a sequence num__12 is added to get the next term . if the first term is num__5 what is the forty - fifth term ? <o> a ) num__533 <o> b ) num__595 <o> c ) num__513 <o> d ) num__562 <o> e ) num__531 |
num__1 rst term + num__44 terms = num__5 + num__12 + num__12 + num__12 + num__12 + num__12 + num__12 + num__12 + . . . + num__12 ( num__44 times ) num__5 + ( num__12 x num__44 ) = num__5 + num__528 = num__533 answer a <eor> a <eos> |
a |
multiply__12.0__44.0__ add__5.0__528.0__ add__5.0__528.0__ |
multiply__12.0__44.0__ add__5.0__528.0__ add__5.0__528.0__ |
| a two digit number is num__18 less than the sum of the squares of its digits . how many such numbers are there ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
suppose the digit num__10 x + y . so num__10 x + y = x ^ num__2 + y ^ num__2 - num__18 = > ( x - num__6 ) ( x - num__4 ) = ( y - num__7 ) ( - y - num__6 ) so conddition satisfied for x = num__64 and y = num__7 so no of digit num__2 . num__6747 . answer : b <eor> b <eos> |
b |
subtract__6.0__2.0__ subtract__4.0__2.0__ |
subtract__6.0__2.0__ subtract__4.0__2.0__ |
| the average age of a class of num__32 students is num__16 yrs . if the teacher ' s age is also included the average increases by one year . find the age of the teacher <o> a ) num__30 years <o> b ) num__31 years <o> c ) num__32 years <o> d ) num__34 years <o> e ) num__49 years |
total age of students is num__32 x num__16 = num__512 years total age inclusive of teacher = num__33 x ( num__16 + num__1 ) = num__561 so teacher ' s age is num__561 - num__512 = num__49 yrs there is a shortcut for these type of problems teacher ' s age is num__16 + ( num__33 x num__1 ) = num__49 years e <eor> e <eos> |
e |
multiply__32.0__16.0__ subtract__33.0__32.0__ add__16.0__33.0__ add__16.0__33.0__ |
multiply__32.0__16.0__ subtract__33.0__32.0__ add__16.0__33.0__ add__16.0__33.0__ |
| city a and city b are num__140 miles apart . train c departs city a heading towards city b at num__4 : num__00 and travels at num__40 miles per hour . train d departs city b heading towards city a at num__5 : num__30 and travels at num__20 miles per hour . the trains travel on parallel tracks . at what time do the two trains meet ? <o> a ) num__5 : num__00 <o> b ) num__7 : num__30 <o> c ) num__6 : num__00 <o> d ) num__6 : num__30 <o> e ) num__7 : num__00 |
train c has traveled num__20 mi in the half hour before train d has started its journey . num__140 - num__20 = num__120 num__40 + num__20 = num__60 mph num__120 mi / num__60 mph = num__2 hrs num__5 : num__30 pm + num__2 hrs = num__7 : num__30 pm answer : b . num__7 : num__30 <eor> b <eos> |
b |
subtract__140.0__20.0__ hour_to_min_conversion__ divide__40.0__20.0__ divide__140.0__20.0__ round__7.0__ |
subtract__140.0__20.0__ add__40.0__20.0__ divide__40.0__20.0__ divide__140.0__20.0__ divide__140.0__20.0__ |
| a and b can do a piece of work in num__6 days b and c in num__8 days c and a in num__12 days . how long will c take to do it ? <o> a ) num__48 days <o> b ) num__55 days <o> c ) num__24 days <o> d ) num__33 days <o> e ) num__40 days |
num__2 c = num__0.125 + num__0.0833333333333 – num__0.166666666667 = num__0.0416666666667 c = num__0.0208333333333 = > num__48 days answer : a <eor> a <eos> |
a |
subtract__8.0__6.0__ divide__2.0__12.0__ subtract__0.125__0.0833__ divide__0.125__6.0__ multiply__6.0__8.0__ round__48.0__ |
subtract__8.0__6.0__ divide__2.0__12.0__ subtract__0.125__0.0833__ divide__0.125__6.0__ multiply__6.0__8.0__ round__48.0__ |
| if x ^ num__2 + num__9 / x ^ num__2 = num__10 what is the value of x - num__3 / x <o> a ) num__36 <o> b ) num__25 <o> c ) num__2 <o> d ) num__5 <o> e ) num__3 |
to find : x - num__3 / x . let it be t . = > x - num__3 / x = t = > ( x ^ num__2 + num__9 / x ^ num__2 ) - num__2 * x * num__3 / x = t ^ num__2 ( squaring both sides ) . = > ( num__10 ) - num__2 * num__3 = num__4 = > t ^ num__2 = num__4 . thus t = num__2 or t = - num__2 . answer c <eor> c <eos> |
c |
subtract__4.0__2.0__ |
subtract__4.0__2.0__ |
| four people need to cross a dark river at night . * they have only one torch and the river is too risky to cross without the torch . * if all people cross simultaneously then torch light wont be sufficient . * speed of each person of crossing the river is different . cross time for each person is num__1 min num__2 minutes num__7 minutes and num__10 minutes . what is the shortest time needed for all four of them to cross the river ? <o> a ) num__16 minutes <o> b ) num__17 minutes <o> c ) num__18 minutes <o> d ) num__19 minutes <o> e ) num__20 minutes |
solution : num__17 min the initial solution most people will think of is to use the fastest person as an usher to guide everyone across . how long would that take ? num__10 + num__1 + num__7 + num__1 + num__2 = num__21 minutes . is that it ? no . that would make this question too simple even as a warm up question . let ' s brainstorm a little further . to reduce the amount of time we should find a way for num__10 and num__7 to go together . if they cross together then we need one of them to come back to get the others . that would not be ideal . how do we get around that ? maybe we can have num__1 waiting on the other side to bring the torch back . ahaa we are getting closer . the fastest way to get num__1 across and be back is to use num__2 to usher num__1 across . so let ' s put all this together . num__1 and num__2 go cross num__2 comes back num__7 and num__10 go across num__1 comes back num__1 and num__2 go across ( done ) total time = num__2 + num__2 + num__10 + num__1 + num__2 = num__17 minutes answer b <eor> b <eos> |
b |
add__7.0__10.0__ round__17.0__ |
add__7.0__10.0__ add__7.0__10.0__ |
| a water tank is one - fifth full . pipe a can fill a tank in num__10 minutes and pipe b can empty it in num__6 minutes . if both the pipes are open how long will it take to empty or fill the tank completely ? <o> a ) num__6 min . to empty <o> b ) num__9 min . to empty <o> c ) num__5 min . to empty <o> d ) num__4 min . to empty <o> e ) num__3 min . to empty |
explanation : clearly pipe b is faster than pipe a and so the tank will be emptied . part to be emptied = num__0.2 part emptied by ( a + b ) in num__1 minute = ( num__0.166666666667 - num__0.1 ) = num__0.0666666666667 so the tank will be emptied in num__3 min answer : e <eor> e <eos> |
e |
divide__1.0__6.0__ divide__1.0__10.0__ subtract__0.1667__0.1__ round__3.0__ |
divide__1.0__6.0__ divide__1.0__10.0__ subtract__0.1667__0.1__ subtract__6.0__3.0__ |
| if x + y + z = num__1 . then xy + yz + zx is <o> a ) num__0.5 <o> b ) num__0.333333333333 <o> c ) num__0.25 <o> d ) num__0.2 <o> e ) num__0.166666666667 |
num__0.5 + num__0.5 + num__0 = num__1 xy = num__0.5 * num__0.5 = num__0.25 answer : c <eor> c <eos> |
c |
round_down__0.5__ multiply__1.0__0.25__ |
round_down__0.5__ multiply__1.0__0.25__ |
| a circular rim num__14 inches in diameter rotates the same number of inches per second as a circular rim num__35 inches in diameter . if the smaller rim makes x revolutions per second how many revolutions per minute does the larger rim makes in terms of x ? <o> a ) num__48 pi / x <o> b ) num__75 x <o> c ) num__48 x <o> d ) num__24 x <o> e ) x / num__75 |
the larger rim must circulate for the same number of inches the smaller rim does . c = ( pi ) d c ( small ) : ( pi ) * num__14 c ( large ) : ( pi ) * num__35 lets say the time horizon is num__60 seconds so during that time the smaller rim covers a distance of ( pi ) * num__14 * num__60 = ( pi ) * ( num__840 ) inches ( pi ) * ( num__840 ) = ( pi ) * ( num__35 ) ( x ) pi * ( num__24 ) = pi * ( x ) num__24 = x answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ multiply__14.0__60.0__ divide__840.0__35.0__ round__24.0__ |
hour_to_min_conversion__ multiply__14.0__60.0__ divide__840.0__35.0__ round__24.0__ |
| a veterinarian surveys num__26 of his patrons . he discovers that num__14 have dogs num__10 have cats and num__5 have fish . four have dogs and cats num__3 have dogs and fish and one has a cat and fish . if no one has all three kinds of pets how many patrons have none of these pets ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
no . of patron have dogs = num__14 no . of patron have cats = num__10 no . of patron have fish = num__5 no . of patron have dogs and cats = num__4 no . of patron have dogs and fish = num__3 no . of patron have cats and fish = num__1 no . of patron have all three pets = num__0 no . of patron have only dogs = num__7 no . of patron have only cats = num__5 no . of patron have only dogs = num__1 so total no . of patron have pets = num__7 + num__5 + num__1 + num__4 + num__3 + num__1 = num__21 so no . of patron do n ' t have any pets = num__26 - num__21 = num__5 num__5 patrons do n ' t have any these pets answer : c <eor> c <eos> |
c |
subtract__14.0__10.0__ subtract__5.0__4.0__ subtract__10.0__3.0__ subtract__26.0__5.0__ subtract__26.0__21.0__ |
subtract__14.0__10.0__ subtract__5.0__4.0__ subtract__10.0__3.0__ subtract__26.0__5.0__ subtract__26.0__21.0__ |
| at a meeting of the num__9 joint chiefs of staff the chief of naval operations does not want to sit next to the chief of the national guard bureau . how many ways can the num__9 chiefs of staff be seated around a circular table ? <o> a ) num__120 <o> b ) num__480 <o> c ) num__960 <o> d ) num__2520 <o> e ) num__6050 |
bunuel i ' m also a little confused with the number of arrangements of n distinct objects in a circle . why is it given by ( n - num__1 ) ! . in theveritasanswer they say : answer e ( num__6050 ) should be the number of ways to arrange all num__7 without the seating restriction given . is this incorrect ? e <eor> e <eos> |
e |
multiply__1.0__6050.0__ |
multiply__1.0__6050.0__ |
| find the odd man out . num__1 num__9 num__16 num__51 num__121 num__169 num__225 <o> a ) num__169 <o> b ) num__51 <o> c ) num__16 <o> d ) num__1 <o> e ) num__121 |
explanation : each of the given numbers except num__51 is a perfect square answer : option b <eor> b <eos> |
b |
multiply__1.0__51.0__ |
multiply__1.0__51.0__ |
| how long does a train num__100 m long traveling at num__60 kmph takes to cross another train of num__170 m in length has a speed of num__40 kmph ? <o> a ) num__1.42 sec <o> b ) num__16.8 sec <o> c ) num__5 sec <o> d ) num__1.68 sec <o> e ) num__3.6 sec |
d = num__100 m s = ( num__60 + num__40 ) * num__0.277777777778 = num__83.3333333333 t = num__100 * num__0.036 = num__3.6 sec answer : e <eor> e <eos> |
e |
multiply__100.0__0.036__ round__3.6__ |
multiply__100.0__0.036__ multiply__100.0__0.036__ |
| the captain of a cricket team of num__11 members is num__26 years old and the wicket keeper is num__3 years older . if the ages of these two are excluded the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? <o> a ) num__23 years <o> b ) num__24 years <o> c ) num__25 years <o> d ) num__26 years <o> e ) none of these |
explanation : let the average age of the whole team by x years . num__11 x - ( num__26 + num__29 ) = num__9 ( x - num__1 ) num__11 x - num__9 x = num__46 num__2 x = num__46 x = num__23 . so average age of the team is num__23 years . answer is a <eor> a <eos> |
a |
add__26.0__3.0__ subtract__11.0__9.0__ subtract__26.0__3.0__ subtract__26.0__3.0__ |
add__26.0__3.0__ subtract__11.0__9.0__ subtract__26.0__3.0__ subtract__26.0__3.0__ |
| the area of a triangle is with base num__4 m and height num__5 m ? <o> a ) num__11 <o> b ) num__10 <o> c ) num__88 <o> d ) num__26 <o> e ) num__32 |
num__0.5 * num__4 * num__5 = num__10 m num__2 answer : b <eor> b <eos> |
b |
triangle_area__4.0__5.0__ square_perimeter__0.5__ triangle_area__4.0__5.0__ |
volume_rectangular_prism__4.0__5.0__0.5__ multiply__4.0__0.5__ multiply__5.0__2.0__ |
| the ratio of number of boys and girls in a school is num__2 : num__5 . if there are num__420 students in the school find the number of girls in the school ? <o> a ) num__150 <o> b ) num__250 <o> c ) num__300 <o> d ) num__370 <o> e ) num__280 |
let the number of boys and girls be num__2 x and num__5 x total students = num__420 number of girls in the school = num__5 * num__60.0 = num__300 answer is c <eor> c <eos> |
c |
multiply__5.0__60.0__ multiply__5.0__60.0__ |
multiply__5.0__60.0__ multiply__5.0__60.0__ |
| if x is to be chosen at random from the set { num__1 num__2 num__3 num__4 } and y is to be chosen at random from the set { num__6 num__7 } what is the probability that xy will be even ? <o> a ) num__0.166666666667 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__0.833333333333 |
probably the best way to solve would be to use num__1 - p ( opposite event ) = num__1 - p ( odd ) = num__1 - p ( odd ) * p ( odd ) = num__1 - num__0.5 * num__0.666666666667 = num__0.666666666667 = num__0.166666666667 answer : a <eor> a <eos> |
a |
union_prob__0.5__0.6667__1.0__ union_prob__0.5__0.6667__1.0__ |
union_prob__0.5__0.6667__1.0__ union_prob__0.5__0.6667__1.0__ |
| jake wants to invite num__9 friends to his num__8 th birthday party . each invitation requires num__2 stamps to mail costing num__12 cents each . how much will jake have to spend on stamps to send invitations to his num__9 friends ? <o> a ) $ num__1.08 <o> b ) $ num__1.41 <o> c ) $ num__1.74 <o> d ) $ num__1.98 <o> e ) $ num__2.16 |
num__2 stamps multiplied by num__12 cents = num__24 cents per invitation . num__24 cents multiplied by num__9 invitations = num__216 cents = $ num__2.16 hence answer : e . <eor> e <eos> |
e |
multiply__2.0__12.0__ multiply__9.0__24.0__ round__2.16__ |
multiply__2.0__12.0__ multiply__9.0__24.0__ round__2.16__ |
| a cycle is bought for rs . num__900 and sold for rs . num__1080 find the gain percent ? <o> a ) num__28.0 <o> b ) num__20.0 <o> c ) num__22.0 <o> d ) num__16.0 <o> e ) num__27 % |
num__900 - - - - num__180 num__100 - - - - ? = > num__20.0 answer : b <eor> b <eos> |
b |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| find the angle between the hour and the minute hand of a clock when the time is num__3.25 . <o> a ) num__47.5 <o> b ) num__47.9 <o> c ) num__47.1 <o> d ) num__47.3 <o> e ) num__47.6 |
formula : θ = ∣ ∣ ∣ num__30 h − num__112 m ∣ ∣ ∣ θ = | num__30 h − num__112 m | angle = num__5.5 × num__25 – num__30 × num__3 = num__47.5 = num__47.5 answer : a <eor> a <eos> |
a |
round__47.5__ |
round__47.5__ |
| the cost price of num__21 articles is equal to selling price of num__18 articles . find gain or loss percentage . <o> a ) num__15.0 % <o> b ) num__3.33333333333 % <o> c ) num__13.3333333333 % <o> d ) num__6.66666666667 % <o> e ) num__16.6666666667 % |
cost price of each article be rs num__1 cost price of num__18 articles = rs num__18 selling price of num__18 articles = rs num__21 . gain % = [ ( num__0.166666666667 ) * num__100 ] % = num__16.6666666667 % answer is e . <eor> e <eos> |
e |
percent__100.0__16.6667__ |
percent__100.0__16.6667__ |
| x can finish a work in num__18 days . y can finish the same work in num__15 days . yworked for num__10 days and left the job . how many days does x alone need to finish the remaining work ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__9 |
work done by x in num__1 day = num__0.0555555555556 work done by y in num__1 day = num__0.0666666666667 work done by y in num__10 days = num__0.666666666667 = num__0.666666666667 remaining work = num__1 – num__0.666666666667 = num__0.333333333333 number of days in which x can finish the remaining work = ( num__0.333333333333 ) / ( num__0.0555555555556 ) = num__6 c <eor> c <eos> |
c |
divide__1.0__18.0__ divide__1.0__15.0__ divide__10.0__15.0__ subtract__1.0__0.6667__ round__6.0__ |
divide__1.0__18.0__ divide__1.0__15.0__ divide__10.0__15.0__ subtract__1.0__0.6667__ divide__6.0__1.0__ |
| a box contains num__10 tablets of medicine a and num__16 tablets of medicine b . what is the least number of tablets that should be taken from the box to ensure that at least two tablets of each kind are among the extracted . <o> a ) num__12 <o> b ) num__15 <o> c ) num__17 <o> d ) num__18 <o> e ) num__21 |
the worst case scenario will be if we remove all num__16 tablets of medicine b first . the next num__2 tablets we remove have to be of medicine a so to guarantee that at least two tablets of each kind will be taken we should remove minimum of num__16 + num__2 = num__18 tablets . answer : d . <eor> d <eos> |
d |
add__16.0__2.0__ add__16.0__2.0__ |
add__16.0__2.0__ add__16.0__2.0__ |
| the number num__189 is equal to the sum of the cubes of two integers . what is the product of those integers ? <o> a ) num__8 <o> b ) num__15 <o> c ) num__20 <o> d ) num__27 <o> e ) num__39 |
num__4 ^ num__3 + num__5 ^ num__3 = num__189 the number is num__4 * num__5 = num__20 c <eor> c <eos> |
c |
multiply__4.0__5.0__ multiply__4.0__5.0__ |
multiply__4.0__5.0__ multiply__4.0__5.0__ |
| two trains are traveling on parallel tracks in the same direction . the faster train travels at num__130 miles per hour while the slower train travels at num__100 miles per hour . at num__2 o ’ clock the faster train is num__25 miles behind the slower one . how far apart are the two trains at num__5 o ' clock ? <o> a ) num__60 miles <o> b ) num__80 miles <o> c ) num__90 miles <o> d ) num__65 miles <o> e ) num__400 miles |
answer = d . num__65 miles relational speed = num__130 - num__100 = num__30 miles per hour in num__3 hours difference = num__30 * num__3 = num__90 miles fast train trailing num__25 miles so effective difference = num__90 - num__25 = num__65 miles <eor> d <eos> |
d |
divide__130.0__2.0__ subtract__130.0__100.0__ subtract__5.0__2.0__ add__25.0__65.0__ round__65.0__ |
divide__130.0__2.0__ subtract__130.0__100.0__ subtract__5.0__2.0__ multiply__3.0__30.0__ subtract__130.0__65.0__ |
| a shipment of num__8 tv sets contains num__6 black and white sets and num__2 color sets . if num__2 tv sets are to be chosen at random from this shipment what is the probability that at least num__1 of the num__2 sets chosen will be a black and white set ? <o> a ) num__0.142857142857 <o> b ) num__0.25 <o> c ) num__0.357142857143 <o> d ) num__0.392857142857 <o> e ) num__0.964285714286 |
num__0.25 * num__0.142857142857 = num__0.0357142857143 num__1 - ( num__0.0357142857143 ) = num__0.964285714286 = num__0.964285714286 answer e <eor> e <eos> |
e |
negate_prob__0.0357__ negate_prob__0.0357__ |
negate_prob__0.0357__ negate_prob__0.0357__ |
| two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of num__16 kmph and num__21 kmph respectively . when they meet it is found that one train has traveled num__60 km more than the other one . the distance between the two stations is ? <o> a ) num__457 km <o> b ) num__444 km <o> c ) num__547 km <o> d ) num__645 km <o> e ) num__453 km |
num__1 h - - - - - num__5 ? - - - - - - num__60 num__12 h rs = num__16 + num__21 = num__37 t = num__12 d = num__37 * num__12 = num__444 answer : b <eor> b <eos> |
b |
subtract__21.0__16.0__ divide__60.0__5.0__ add__16.0__21.0__ multiply__12.0__37.0__ round__444.0__ |
subtract__21.0__16.0__ divide__60.0__5.0__ add__16.0__21.0__ multiply__12.0__37.0__ multiply__12.0__37.0__ |
| num__0.002 x num__0.6 = ? <o> a ) num__0.0012 <o> b ) num__0.001 <o> c ) num__0.01 <o> d ) num__0.1 <o> e ) none of these |
explanation : num__2 x num__6 = num__12 . sum of decimal places = num__4 num__0.002 x num__0.6 = num__0.0012 answer - a <eor> a <eos> |
a |
multiply__2.0__6.0__ subtract__6.0__2.0__ multiply__0.002__0.6__ multiply__0.002__0.6__ |
multiply__2.0__6.0__ subtract__6.0__2.0__ multiply__0.002__0.6__ multiply__0.002__0.6__ |
| kate and her boyfriend ben completed the painting of their new apartment together in num__6 days . if they were to work separately how long will it take each of them to complete it if ben can complete the painting num__5 days earlier than kate . <o> a ) num__6 and num__11 <o> b ) num__5 and num__10 <o> c ) num__10 and num__15 <o> d ) num__8 and num__13 <o> e ) num__6 and num__10 |
work = ( a ) ( b ) / ( a + b ) where a and b are the individual times of each entity . here we ' re told that ( working together ) the two workers would complete a job in num__12 days . this means that ( individually ) each of them would take more than num__10 days to do the job . answers e a and c are illogical since the individual times must both be greater than num__10 days . so we can test the values for answers b and d . using the values for answers c and d . . . answer c : ( num__10 ) ( num__15 ) / ( num__10 + num__15 ) = num__6.0 = num__6 this is a match final answer : c <eor> c <eos> |
c |
add__5.0__10.0__ round__10.0__ |
add__5.0__10.0__ round__10.0__ |
| a man walks at a speed of num__3 km / hr and runs at a speed of num__6 km / hr . how much time will the man require to cover a distance of num__12 km if he completes half of the distance i . e . ( num__6 ) km on foot and the other half by running ? <o> a ) num__2.6 <o> b ) num__2.2 <o> c ) num__2.5 <o> d ) num__2 <o> e ) num__3 |
required time = ( num__6 ) / num__3 + ( num__6 ) / num__6 = num__3 hours . answer : e <eor> e <eos> |
e |
round__3.0__ |
round__3.0__ |
| josh spends a total of $ num__5.00 buying n items in the convenience store . if each of the items is either a num__5 cents single bubblegum or a num__50 cents bubblegum pack then n may be which of the following ? <o> a ) num__91 <o> b ) num__100 <o> c ) num__101 <o> d ) num__109 <o> e ) num__140 |
let x be num__0.05 $ single bubblegum and y be the total num__0.5 $ packs - - - > num__0.05 x + num__0.5 y = num__5.0 - - - > x + num__10 y = num__100 . . . ( num__1 ) also x + y = n . . . ( num__2 ) solving num__1 and num__2 we get n = num__110 - num__9 y - - - > only value satisfied is when y = num__1 - - - > n = num__100 - num__9 = num__91 . a is the correct answer . other options are not equal to multiples of num__9 removed from num__100 . <eor> a <eos> |
a |
divide__5.0__0.5__ divide__5.0__0.05__ reverse__0.5__ add__100.0__10.0__ subtract__10.0__1.0__ subtract__100.0__9.0__ multiply__1.0__91.0__ |
divide__5.0__0.5__ divide__5.0__0.05__ reverse__0.5__ add__100.0__10.0__ subtract__10.0__1.0__ subtract__100.0__9.0__ subtract__100.0__9.0__ |
| the average salary of a person for the months of january february march and april is rs . num__8000 and that for the months february march april and may is rs . num__8500 . if his salary for the month of may is rs . num__6500 find his salary for the month of january ? <o> a ) rs . num__4580 <o> b ) rs . num__4570 <o> c ) rs . num__4500 <o> d ) rs . num__4550 <o> e ) rs . num__2500 |
sum of the salaries of the person for the months of january february march and april = num__4 * num__8000 = num__32000 - - - - ( num__1 ) sum of the salaries of the person for the months of february march april and may = num__4 * num__8500 = num__34000 - - - - ( num__2 ) ( num__2 ) - ( num__1 ) i . e . may - jan = num__2000 salary of may is rs . num__6500 salary of january = rs . num__4500 answer : c <eor> c <eos> |
c |
multiply__8000.0__4.0__ multiply__8500.0__4.0__ divide__8000.0__4.0__ subtract__6500.0__2000.0__ subtract__6500.0__2000.0__ |
multiply__8000.0__4.0__ multiply__8500.0__4.0__ subtract__8500.0__6500.0__ subtract__6500.0__2000.0__ subtract__6500.0__2000.0__ |
| two trains are moving in opposite directions at num__60 km / hr and num__90 km / hr . their lengths are num__1.10 km and num__0.9 km respectively . the time taken by the slower train to cross the faster train in seconds is ? <o> a ) num__18 sec <o> b ) num__17 sec <o> c ) num__48 sec <o> d ) num__19 sec <o> e ) num__12 sec |
relative speed = num__60 + num__90 = num__150 km / hr . = num__150 * num__0.277777777778 = num__41.6666666667 m / sec . distance covered = num__1.10 + num__0.9 = num__2 km = num__2000 m . required time = num__2000 * num__0.024 = num__48 sec . answer : c <eor> c <eos> |
c |
add__60.0__90.0__ add__1.1__0.9__ multiply__2000.0__0.024__ round__48.0__ |
add__60.0__90.0__ add__1.1__0.9__ multiply__2000.0__0.024__ multiply__2000.0__0.024__ |
| david bought num__4 toys with the average cost of $ num__8 . if david also buys the fifth toy with the price of $ num__10 what is the average ( arithmetic mean ) price of those num__5 toys ? <o> a ) $ num__5 <o> b ) $ num__6 <o> c ) $ num__8.4 <o> d ) $ num__10 <o> e ) $ num__11 |
num__4 toys - > total cost = num__4 * avg cost = num__4 * num__8 = num__32 fifth toy cost = num__10 total cost for num__5 toys = num__32 + num__10 = num__42 am = num__8.4 = num__8.4 hence c <eor> c <eos> |
c |
multiply__4.0__8.0__ add__10.0__32.0__ divide__42.0__5.0__ divide__42.0__5.0__ |
multiply__4.0__8.0__ add__10.0__32.0__ divide__42.0__5.0__ divide__42.0__5.0__ |
| wanda baked num__168 cookies packaged them in boxes of num__12 and sold each box for $ num__1.20 . how many boxes of num__12 can be packaged if there are num__168 cookies ? what was the total amount of money that wanda made for all of the boxes sold ? <o> a ) $ num__26.80 <o> b ) $ num__36.80 <o> c ) $ num__46.80 <o> d ) $ num__76.80 <o> e ) $ num__16.80 |
divide num__168 ÷ num__12 = num__14 multiply $ num__1.20 x num__14 = $ num__16.80 correct answer e <eor> e <eos> |
e |
divide__168.0__12.0__ multiply__1.2__14.0__ multiply__1.2__14.0__ |
divide__168.0__12.0__ multiply__1.2__14.0__ multiply__1.2__14.0__ |
| a work can be finished in num__16 days by twenty women . the same work can be finished in fifteen days by sixteen men . the ratio between the capacity of a man and a woman is <o> a ) num__4 : num__3 <o> b ) num__3 : num__4 <o> c ) num__2 : num__3 <o> d ) num__3 : num__2 <o> e ) num__4 : num__5 |
work done by num__20 women in num__1 day = num__0.0625 work done by num__1 woman in num__1 day = num__1 / ( num__16 × num__20 ) work done by num__16 men in num__1 day = num__0.0666666666667 work done by num__1 man in num__1 day = num__1 / ( num__15 × num__16 ) ratio of the capacity of a man and woman = num__1 / ( num__15 × num__16 ) : num__1 / ( num__16 × num__20 ) = num__0.0666666666667 : num__0.05 = num__0.333333333333 : num__0.25 = num__4 : num__3 answer is a . <eor> a <eos> |
a |
divide__1.0__16.0__ subtract__16.0__1.0__ divide__1.0__20.0__ multiply__16.0__0.25__ subtract__4.0__1.0__ round__4.0__ |
divide__1.0__16.0__ subtract__16.0__1.0__ divide__1.0__20.0__ divide__1.0__0.25__ subtract__4.0__1.0__ divide__16.0__4.0__ |
| of the five numbers in a sequence the first term is num__20000 and each of the following terms is num__20.0 of the previous term . what is the value range of the five numbers ? <o> a ) num__9375 <o> b ) num__9750 <o> c ) num__9975 <o> d ) num__9984 <o> e ) num__19 |
968 |
num__1 st number = num__20000 num__2 nd number = num__20.0 of num__20000 = num__4000 num__3 rd number = num__20.0 of num__4000 = num__800 num__4 th number = num__20.0 of num__800 = num__160 num__5 th number = num__20.0 of num__160 = num__32 range = num__20000 - num__32 = num__19968 e is the answer <eor> e <eos> |
e |
e |
| jane makes toy bears . when she works with an assistant she makes num__90 percent more bears per week and works num__10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ? <o> a ) num__20.0 <o> b ) num__80.0 <o> c ) num__100.0 <o> d ) num__120.0 <o> e ) num__200 % |
we can use fractional equivalents here to solve the problem num__80.0 = num__0.8 ; this means that in num__1 st case if she prepares num__5 bears in num__2 nd case she prepares num__9 bears num__10.0 = num__0.1 ; this means that in num__1 st case if she needs num__10 hours in num__2 nd case she needs num__9 hours now we come to productivity based on above fractional values the productivity in num__1 st case is num__0.5 bears / hour and in the num__2 nd case it is num__1 bear / hour hence the productivity is double with the assistant i . e . the increase in productivity is num__120.0 d <eor> d <eos> |
d |
subtract__90.0__10.0__ divide__10.0__5.0__ divide__90.0__10.0__ divide__1.0__10.0__ divide__1.0__2.0__ round__120.0__ |
subtract__90.0__10.0__ divide__10.0__5.0__ divide__90.0__10.0__ divide__1.0__10.0__ divide__1.0__2.0__ divide__120.0__1.0__ |
| if r = num__199999 and s = num__991999 which of the following is the units digit of r ^ num__4 + s ^ num__3 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__8 <o> e ) num__9 |
the exponents of num__9 cycle between num__9 ( odd exponents ) and num__1 ( even exponents ) . then the sum of r ^ num__4 + s ^ num__3 will have the units digit of num__1 + num__9 = num__10 as a units digit . the answer is a . <eor> a <eos> |
a |
subtract__4.0__3.0__ add__9.0__1.0__ reverse__199999.0__ |
subtract__4.0__3.0__ add__9.0__1.0__ reverse__199999.0__ |
| if a boat goes num__4 km upstream in num__12 minutes and the speed of the stream is num__8 kmph then the speed of the boat in still water is ? <o> a ) num__28 <o> b ) num__29 <o> c ) num__30 <o> d ) num__31 <o> e ) num__32 |
rate upsteram = ( num__0.333333333333 * num__60 ) kmph = num__20 kmph . speed of the stream = num__8 kmph let speed in still water be xkm / hr . then speed upstream = ( x - num__8 ) km / hr . x - num__8 = num__20 = = > x = num__28 km / hr answer ( a ) <eor> a <eos> |
a |
divide__4.0__12.0__ hour_to_min_conversion__ add__12.0__8.0__ add__8.0__20.0__ round__28.0__ |
divide__4.0__12.0__ hour_to_min_conversion__ add__12.0__8.0__ add__8.0__20.0__ round__28.0__ |
| a candidate got num__35.0 of the votes polled and he lost to his rival by num__2460 votes . how many votes were cast ? <o> a ) num__7500 <o> b ) num__3388 <o> c ) num__2665 <o> d ) num__8200 <o> e ) num__2661 |
num__35.0 - - - - - - - - - - - l num__65.0 - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - num__30.0 - - - - - - - - - - num__2460 num__100.0 - - - - - - - - - ? = > num__8200 answer : d <eor> d <eos> |
d |
percent__100.0__8200.0__ |
percent__100.0__8200.0__ |
| a woman put $ num__580 into a savings account for one year . the rate of interest on the account was num__6 ½ % . how much was the interest for the year in dollars and cents ? ( round to the nearest cent ) <o> a ) $ num__37.70 <o> b ) $ num__47.70 <o> c ) $ num__57.70 <o> d ) $ num__67.70 <o> e ) $ num__77.70 |
we have num__6 num__0.5 / num__100 = x / num__580 multiply the opposites : num__6 ½ x num__580 = num__3770 we take num__37.70 / num__3770 . num__00 num__100 $ num__37.70 correct answer a <eor> a <eos> |
a |
percent__37.7__100.0__ |
percent__37.7__100.0__ |
| if x / y = num__3 and ( num__2 a - x ) / ( num__5 b - y ) = num__3 then the value of a / b is ? <o> a ) num__7.5 <o> b ) - num__2 <o> c ) num__1 <o> d ) num__2 <o> e ) num__3 |
x = num__3 y num__2 a - num__3 y = num__3 ( num__5 b - y ) num__2 a - num__3 y = num__15 b - num__3 y num__2 a = num__15 b a / b = num__7.5 answer : a <eor> a <eos> |
a |
multiply__3.0__5.0__ divide__15.0__2.0__ divide__15.0__2.0__ |
multiply__3.0__5.0__ divide__15.0__2.0__ divide__15.0__2.0__ |
| there are num__18 students in a class . on the day the test was given alice was absent . the other num__17 students took the test and their average was num__78 . the next day alice took the test and with this grade included the new average was num__79 . what was alice ' s grade on the test ? <o> a ) num__96 <o> b ) num__95 <o> c ) num__94 <o> d ) num__93 <o> e ) num__92 |
num__17 * num__78 + alice ' s grade = num__18 * num__79 alice ' s grade is num__18 * num__79 - num__17 * num__78 = num__96 . the answer is a . <eor> a <eos> |
a |
add__18.0__78.0__ add__18.0__78.0__ |
add__18.0__78.0__ add__18.0__78.0__ |
| the difference of two numbers is num__1097 . on dividing the larger number by the smaller we get num__10 as quotient and the num__17 as remainder . what is the smaller number ? <o> a ) num__230 <o> b ) num__120 <o> c ) num__95 <o> d ) num__310 <o> e ) num__320 |
solution : let the smaller number be x . then larger number = ( x + num__1097 ) x + num__1097 = num__10 x + num__17 num__9 x = num__1080 x = num__120 answer b <eor> b <eos> |
b |
subtract__1097.0__17.0__ divide__1080.0__9.0__ divide__1080.0__9.0__ |
subtract__1097.0__17.0__ divide__1080.0__9.0__ divide__1080.0__9.0__ |
| if num__0.3 of a number is equal to num__0.06 of another number the ratio of the numbers i <o> a ) num__3 : num__15 <o> b ) num__1 : num__2 <o> c ) num__3 : num__20 <o> d ) num__20 : num__3 <o> e ) none |
sol . num__0.3 a = num__0.06 b â ‡ ” a / b = num__0.06 / num__0.30 = num__0.2 = num__0.2 . â ˆ ´ a : b = num__3 : num__15 . answer a <eor> a <eos> |
a |
divide__0.06__0.3__ divide__3.0__0.2__ multiply__0.2__15.0__ |
divide__0.06__0.3__ divide__3.0__0.2__ multiply__0.2__15.0__ |
| y can do a work in num__0.666666666667 the time it takes x . z can do the same work in ¾ the time it takes y . when all three are typing at the same time what fraction of their total work does y do <o> a ) num__0.333333333333 <o> b ) num__0.307692307692 <o> c ) num__0.391304347826 <o> d ) num__0.275862068966 <o> e ) num__0.310344827586 |
given the ratio of the time of y and x is num__0.666666666667 : num__1 therefore the ratio of speed of the x and y becomes num__2 : num__3 . similarly the ratio of the speed of the z and y given as num__4 : num__3 therefore the speed of the y and z becomes num__3 : num__4 . now comparing the ratio of the x y and z it becomes num__2 : num__3 : : num__3 : num__4 hence the y is doing num__0.333333333333 hence it is doing num__0.333333333333 work . answer : a <eor> a <eos> |
a |
add__1.0__2.0__ add__1.0__3.0__ reverse__3.0__ reverse__3.0__ |
add__1.0__2.0__ add__1.0__3.0__ reverse__3.0__ reverse__3.0__ |
| an amusement park charges $ num__8 for an adults ' ticket and $ num__6 for a children ' s ticket . on a certain day a total of num__150 tickets were sold for a total cost of $ num__1020 . how many more children ' s tickets were sold than adults ' tickets ? <o> a ) num__30 <o> b ) num__50 <o> c ) num__80 <o> d ) num__90 <o> e ) num__120 |
childcount = ? adultcount = ? totalcount = num__150 childadmission = $ num__6 adultadmission = $ num__8 totaladmission = $ num__1020 a ) totalcount = adultcount + childcount b ) totaladmission = ( childadmission * childcount ) + ( adultadmission * adultcount ) solve for a ) for childcount = totalcount - adultcount substitute into b ) : totaladmission = ( childadmission * ( totalcount - adultcount ) ) + ( adultadmission * adultcount ) = > adultcount = ( totaladmission - ( childadmission * totalcount ) ) / ( adultadmission - childadmission ) so adultcount = num__60 and childcount = num__90 so the difference is num__30 more children then adults a . num__30 <eor> a <eos> |
a |
subtract__150.0__60.0__ subtract__90.0__60.0__ subtract__90.0__60.0__ |
subtract__150.0__60.0__ subtract__90.0__60.0__ subtract__90.0__60.0__ |
| num__40 percent of the members of a study group are women and num__20 percent of those women are lawyers . if one member of the study group is to be selected at random what is the probability that the member selected is a woman lawyer ? <o> a ) num__0.2 <o> b ) num__0.08 <o> c ) num__0.16 <o> d ) num__0.14 <o> e ) num__0.15 |
say there are num__100 people in that group then there would be num__0.4 * num__0.20 * num__100 = num__8 women lawyers which means that the probability that the member selected is a woman lawyer is favorable / total = num__0.08 . answer : b <eor> b <eos> |
b |
percent__40.0__20.0__ percent__40.0__0.2__ percent__40.0__0.2__ |
percent__40.0__20.0__ percent__40.0__0.2__ percent__40.0__0.2__ |
| the c . p of num__15 books is equal to the s . p of num__18 books . find his gain % or loss % ? <o> a ) num__16 num__0.666666666667 % <o> b ) num__16 num__0.285714285714 % <o> c ) num__16 num__0.0 % <o> d ) num__18 num__0.666666666667 % <o> e ) num__26 num__0.666666666667 % |
num__15 cp = num__18 sp num__18 - - - num__3 cp loss num__100 - - - ? = > num__16 num__0.666666666667 % loss answer : a <eor> a <eos> |
a |
percent__100.0__16.0__ |
percent__100.0__16.0__ |
| a train with a length of num__100 meters is traveling at a speed of num__72 km / hr . the train enters a tunnel num__1.7 km long . how many minutes does it take the train to pass through the tunnel from the moment the front enters to the moment the rear emerges ? <o> a ) num__0.9 <o> b ) num__1.2 <o> c ) num__1.5 <o> d ) num__1.8 <o> e ) num__2.1 |
num__72 km / hr = num__1.2 km / min the total distance is num__1.8 km . num__1.8 / num__1.2 = num__1.5 minutes the answer is c . <eor> c <eos> |
c |
divide__1.8__1.2__ round__1.5__ |
divide__1.8__1.2__ divide__1.8__1.2__ |
| if in a race of num__130 m a covers the distance in num__20 seconds and b in num__25 seconds then a beats b by : <o> a ) num__20 m <o> b ) num__26 m <o> c ) num__11 m <o> d ) num__10 m <o> e ) num__15 m |
explanation : the difference in the timing of a and b is num__5 seconds . hence a beats b by num__5 seconds . the distance covered by b in num__5 seconds = ( num__130 * num__5 ) / num__25 = num__26 m hence a beats b by num__26 m . answer b <eor> b <eos> |
b |
subtract__25.0__20.0__ divide__130.0__5.0__ round__26.0__ |
subtract__25.0__20.0__ divide__130.0__5.0__ divide__130.0__5.0__ |
| a father said to his son ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is num__48 years now the son ' s age five years back was : <o> a ) num__14 <o> b ) num__17 <o> c ) num__19 <o> d ) num__22 <o> e ) num__24 |
let the son ' s present age be x years . then ( num__48 - x ) = x num__2 x = num__48 . x = num__24 . son ' s age num__5 years back ( num__24 - num__5 ) = num__19 years . answer : c <eor> c <eos> |
c |
divide__48.0__2.0__ subtract__24.0__5.0__ subtract__24.0__5.0__ |
divide__48.0__2.0__ subtract__24.0__5.0__ subtract__24.0__5.0__ |
| a can give b num__100 meters start and c num__200 meters start in a kilometer race . how much start can b give c in a kilometer race ? <o> a ) num__111.12 m <o> b ) num__119.12 m <o> c ) num__111.13 m <o> d ) num__121.12 m <o> e ) num__111.52 m |
a runs num__1000 m while b runs num__900 m and c runs num__800 m . the number of meters that c runs when b runs num__1000 m = ( num__1000 * num__800 ) / num__900 = num__888.888888889 = num__888.88 m . b can give c = num__1000 - num__888.88 = num__111.12 m . answer : a <eor> a <eos> |
a |
subtract__1000.0__100.0__ subtract__900.0__100.0__ subtract__1000.0__888.88__ round__111.12__ |
subtract__1000.0__100.0__ subtract__900.0__100.0__ subtract__1000.0__888.88__ subtract__1000.0__888.88__ |
| if each participant of a chess tournament plays exactly one game with each of the remaining participants then num__105 games will be played during the tournament . find the number of participants . <o> a ) num__15 <o> b ) num__16 <o> c ) num__17 <o> d ) num__18 <o> e ) num__19 |
let p be the number of participants . pc num__2 = num__105 ( p ) ( p - num__1 ) = num__210 = num__15 * num__14 p = num__15 the answer is a . <eor> a <eos> |
a |
multiply__105.0__2.0__ subtract__15.0__1.0__ add__1.0__14.0__ |
multiply__105.0__2.0__ subtract__15.0__1.0__ add__1.0__14.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__12 sec . what is the length of the train ? <o> a ) num__287 <o> b ) num__699 <o> c ) num__677 <o> d ) num__168 <o> e ) num__200 |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__12 = num__200 m answer : e <eor> e <eos> |
e |
round__200.0__ |
round__200.0__ |
| compound interest earned on a sum for the second and the third years are rs . num__1200 and rs . num__1440 respectively . find the rate of interest ? <o> a ) num__70.0 p . a <o> b ) num__50.0 p . a <o> c ) num__20.0 p . a <o> d ) num__10.0 p . a <o> e ) num__90.0 p . a |
explanation : rs . num__1440 - num__1200 = rs . num__240 is the interest on rs . num__1200 for one year . rate of interest = ( num__100 * num__240 ) / ( num__100 * num__1 ) = num__20.0 p . a answer : c <eor> c <eos> |
c |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| john has $ num__1200 at the beginning of his trip after spending money he still has exactly $ num__800 less than he spent on the trip . how much money does john still have ? <o> a ) $ num__200 <o> b ) $ num__400 <o> c ) $ num__600 <o> d ) $ num__800 <o> e ) $ num__1 |
200 |
suppose total money spent = x not spend ( money he still has ) = x - num__800 x + x - num__800 = num__1200 x = num__1000 money not spend = num__1000 - num__800 = num__200 answer : a <eor> a <eos> |
a |
a |
| a train passes a station platform in num__36 sec and a man standing on the platform in num__20 sec . if the speed of the train is num__54 km / hr . what is the length of the platform ? <o> a ) num__258 m <o> b ) num__240 m <o> c ) num__278 m <o> d ) num__289 m <o> e ) num__268 m |
speed = num__54 * num__0.277777777778 = num__15 m / sec . length of the train = num__15 * num__20 = num__300 m . let the length of the platform be x m . then ( x + num__300 ) / num__36 = num__15 = > x = num__240 m . answer : b <eor> b <eos> |
b |
multiply__20.0__15.0__ round__240.0__ |
multiply__20.0__15.0__ round__240.0__ |
| an accurate clock shows num__8 o ’ clock in the morning . through how many degrees will the hour hand rotate when the clock shows num__4 o ’ clock in the afternoon ? <o> a ) num__60 ° <o> b ) num__90 ° <o> c ) num__180 ° <o> d ) num__320 ° <o> e ) num__240 ° |
sol . angle traced by the hour hand in num__8 hours = [ num__30.0 * num__8 ] ° = num__240 ° answer e <eor> e <eos> |
e |
clock_big_arm_angle__4.0__ clock_big_arm_angle__4.0__ |
clock_big_arm_angle__4.0__ clock_big_arm_angle__4.0__ |
| num__10 litres of water are poured into an aquarium of dimensions num__50 cm length num__10 cm breadth and num__40 cm height . how high ( in cm ) will the water rise ? ( num__1 litre = num__1000 cm ³ ) <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__20 <o> e ) num__40 |
lxbxh = num__10000 h = num__200.0 * num__10 = num__20 cm ' d ' is the answer . <eor> d <eos> |
d |
multiply__10.0__1000.0__ divide__10000.0__50.0__ divide__1000.0__50.0__ round__20.0__ |
multiply__10.0__1000.0__ divide__10000.0__50.0__ divide__1000.0__50.0__ round__20.0__ |
| anna and carol buy cds and tapes at a music store that sells each of its cds for a certain price and each of its tapes for a certain price . anna spends twice as much as carol spends buying three times as many cds and the same number of tapes . if carol spends $ num__40.00 on four cds and five tapes how much does one tape cost ? <o> a ) $ num__4.00 <o> b ) $ num__6.25 <o> c ) $ num__12.00 <o> d ) $ num__25.00 <o> e ) $ num__100.00 |
cd = > c type = > t carol : num__4 c + num__5 t = num__40 anna : num__12 c + num__5 t = num__80 num__12 c - num__4 c = num__40 num__8 c = num__40 c = num__5 num__4 c + num__5 t = num__40 num__20 + num__5 t = num__40 num__5 t = num__20 t = num__4 answer is a <eor> a <eos> |
a |
divide__40.0__5.0__ multiply__4.0__5.0__ subtract__12.0__8.0__ |
subtract__12.0__4.0__ add__12.0__8.0__ subtract__12.0__8.0__ |
| a lends rs . num__2500 to b and a certain to c at the same time at num__7.0 p . a . simple interest . if after num__4 years a altogether receives rs . num__1120 as interest from b and c then the sum lent to c is ? <o> a ) num__1522 <o> b ) num__1500 <o> c ) num__1789 <o> d ) num__2689 <o> e ) num__2689 |
let the sum lent to c be rs . x . then ( num__2500 * num__7 * num__4 ) / num__100 + ( x * num__7 * num__4 ) / num__100 = num__1120 num__0.28 x = ( num__1120 - num__700 ) = > x = num__1500 answer : b <eor> b <eos> |
b |
percent__4.0__2500.0__ percent__7.0__4.0__ percent__100.0__1500.0__ |
percent__4.0__2500.0__ percent__7.0__4.0__ percent__100.0__1500.0__ |
| there are num__100 men in town . out of which num__85.0 were married num__70.0 have a phone num__75.0 own a car num__80.0 own a house . what is the maximum number of people who are married own a phone own a car and own a house ? <o> a ) num__20 <o> b ) num__15 <o> c ) num__10 <o> d ) num__5 <o> e ) none of these |
% of married = num__85.0 % of phone = num__70.0 % of car = num__75.0 % of house = num__80 . therefore % of not married = num__15.0 % of not having phone = num__30.0 % of not owning car = num__25.0 % of not having house = num__20.0 so % of not having any of these things = num__15 + num__25 + num__30 + num__20 = num__90.0 so % of people having all = num__100 - num__90 = num__10.0 thus people = num__10 . answer - option c <eor> c <eos> |
c |
percent__80.0__25.0__ percent__100.0__10.0__ |
percent__80.0__25.0__ percent__100.0__10.0__ |
| to num__15 lts of water containing num__20.0 alcohol we add num__5 lts of pure water . what is % alcohol . <o> a ) num__25.0 <o> b ) num__35.0 <o> c ) num__15.0 <o> d ) num__18.0 <o> e ) num__19 % |
since num__20.0 alcohol in num__15 lit = ( num__15 * num__20 ) / num__100 = num__3 lit since total water = ( num__15 - num__3 ) + num__5 = num__17 total mix = num__3 + num__17 = num__20.0 of alcohol = ( num__0.15 ) * num__100 = num__15.0 answer : c <eor> c <eos> |
c |
percent__15.0__20.0__ percent__5.0__3.0__ percent__15.0__100.0__ |
percent__15.0__20.0__ percent__5.0__3.0__ percent__15.0__100.0__ |
| a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by num__12.0 while buying and by num__30.0 while selling . what is his percentage profit ? <o> a ) num__10.22 <o> b ) num__20.22 <o> c ) num__21.22 <o> d ) num__60.0 <o> e ) ca n ' t be calculated |
the owner buys num__100 kg but actually gets num__112 kg ; the owner sells num__100 kg but actually gives num__70 kg ; profit : ( num__112 - num__70 ) / num__70 * num__100 = ~ num__60.0 answer : d . <eor> d <eos> |
d |
percent__60.0__100.0__ |
percent__60.0__100.0__ |
| in a class of num__35 students num__26 play football and play num__20 long tennis if num__17 play above many play neither ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__12 <o> e ) num__14 |
num__26 + num__20 - num__17 = num__29 num__35 - num__29 = num__6 play neither answer is a <eor> a <eos> |
a |
die_space__ die_space__ |
die_space__ die_space__ |
| a and b are two circles . the radius of a is thrice as large as the diameter of b . what is the ratio between the areas of the circles ? <o> a ) num__1 : num__18 . <o> b ) num__1 : num__20 . <o> c ) num__1 : num__4 . <o> d ) num__1 : num__36 . <o> e ) num__1 : num__16 . |
given : the radius of a is thrice as large as the diameter of b . = > r ( a ) = num__3 * d ( b ) = num__3 * num__2 * r ( b ) = num__6 r ( b ) . the radius are in ratio of num__1 : num__6 thus the area will be in the ratio of square of radius . num__1 : num__36 . hence d . <eor> d <eos> |
d |
multiply__2.0__3.0__ power__6.0__2.0__ volume_cube__1.0__ |
multiply__2.0__3.0__ volume_rectangular_prism__2.0__3.0__6.0__ volume_cube__1.0__ |
| the sum of num__3 hours num__45 minutes and num__4 hours num__55 minutes is approximately what percent of a day ? <o> a ) num__33.0 <o> b ) num__34.0 <o> c ) num__35.0 <o> d ) num__36.0 <o> e ) num__37 % |
since the question is asking for an approximate percentage num__3 : num__45 + num__4 : num__55 ~ num__9 hours % of day = num__9 * num__4.16666666667 ~ num__9 * num__4.0 = num__36.0 answer is d . <eor> d <eos> |
d |
subtract__45.0__9.0__ round__36.0__ |
multiply__4.0__9.0__ multiply__4.0__9.0__ |
| if it is num__6 : num__17 in the evening on a certain day what time in the morning was it exactly num__2 num__880717 minutes earlier ? ( assume standard time in one location . ) <o> a ) num__6 : num__22 <o> b ) num__6 : num__24 <o> c ) num__6 : num__27 <o> d ) num__6 : num__20 <o> e ) num__6 : num__32 |
num__6 : num__17 minus num__2 num__880717 in any way must end with num__0 the only answer choice which ends with num__0 is d . answer : d . <eor> d <eos> |
d |
divide__6.0__880717.0__ round__6.0__ |
divide__6.0__880717.0__ round__6.0__ |
| three numbers are in the ratio num__1 : num__2 : num__3 and their h . c . f is num__12 . the numbers are : <o> a ) num__4 num__8 num__12 <o> b ) num__5 num__10 num__15 <o> c ) num__10 num__20 num__30 <o> d ) num__12 num__24 num__36 <o> e ) num__12 num__24 num__39 |
let the required numbers be x num__2 x and num__3 x . then their h . c . f = x . so x = num__12 . the numbers are num__12 num__24 num__36 . answer : d <eor> d <eos> |
d |
multiply__2.0__12.0__ multiply__3.0__12.0__ multiply__1.0__12.0__ |
multiply__2.0__12.0__ multiply__3.0__12.0__ multiply__1.0__12.0__ |
| the combined area of the two black squares is equal to num__1000 square units . a side of the larger black square is num__8 units longer than a side of the smaller black square . what is the combined area of the two white rectangles in square units ? <o> a ) num__928 <o> b ) num__936 <o> c ) num__948 <o> d ) num__968 <o> e ) num__972 |
the length of a smaller square - x ; the length of a larger square - x + num__8 ; the area of entire square - x ^ num__2 + ( x + num__8 ) ^ num__2 = num__1000 - - > num__2 x ^ num__2 + num__16 x = num__936 ; the combined area of the two white rectangles - x ( x + num__8 ) + x ( x + num__8 ) = num__2 x ^ num__2 + num__16 x - - > look up : num__2 x ^ num__2 + num__16 x = num__936 . answer : b . <eor> b <eos> |
b |
multiply__8.0__2.0__ triangle_area__2.0__936.0__ |
multiply__8.0__2.0__ triangle_area__2.0__936.0__ |
| if x is to be chosen at random from the set { num__1 num__2 num__3 num__4 } and y is to be chosen at random from the set { num__5 num__6 num__7 } what is the probability that xy will be even ? <o> a ) num__0.166666666667 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__2 by num__3 <o> e ) num__0.833333333333 |
in order to make even nos . by multiplication we should have even * odd odd * even or even * even total even nos . possible by multiplying nos . from the num__2 sets : ( num__1 * num__6 ) ; num__2 * any of the three from set b ; num__3 * num__6 ; & num__4 * any of the three from set b num__1 + num__3 + num__1 + num__3 = num__8 total possibilities = num__4 * num__3 = num__12 p ( event ) = num__0.666666666667 or num__0.666666666667 ans d <eor> d <eos> |
d |
choose__3.0__1.0__ coin_space__ |
choose__3.0__1.0__ coin_space__ |
| num__100 kg of an alloy a is mixed with num__200 kg of alloy b . if alloy a has lead and tin in the ratio num__5 : num__3 and alloy b has tin and copper in the ratio num__2 : num__3 then the amount of tin in the new alloy is ? <o> a ) num__100.6 kg <o> b ) num__120.3 kg <o> c ) num__117.5 kg <o> d ) num__110.8 kg <o> e ) num__114 kg |
quantity of tin in num__100 kg of a = num__100 * num__0.375 = num__37.5 kg quantity of tin in num__200 kg of b = num__200 * num__0.4 = num__80 kg quantity of tin in the new alloy = num__37.5 + num__80 = num__117.5 kg answer is c <eor> c <eos> |
c |
multiply__100.0__0.375__ divide__2.0__5.0__ multiply__200.0__0.4__ add__37.5__80.0__ add__37.5__80.0__ |
multiply__100.0__0.375__ divide__2.0__5.0__ multiply__200.0__0.4__ add__37.5__80.0__ add__37.5__80.0__ |
| susan drove an average speed of num__15 miles per hour for the first num__40 miles of a tripthen at a average speed of num__60 miles / hr for the remaining num__20 miles of the trip if she made no stops during the trip what was susan ' s avg speed in miles / hr for the entire trip <o> a ) num__35 <o> b ) num__20 <o> c ) num__45 <o> d ) num__50 <o> e ) num__55 |
avg . speed = total distance / total time total distance = num__60 miles total time = num__2.66666666667 + num__0.333333333333 = num__3 avg . speed = num__20 . answer - b <eor> b <eos> |
b |
divide__40.0__15.0__ divide__20.0__60.0__ divide__60.0__20.0__ round__20.0__ |
divide__40.0__15.0__ divide__20.0__60.0__ divide__60.0__20.0__ subtract__40.0__20.0__ |
| a man is num__35 years older than his son . in two years his age will be twice the age of his son . the present age of this son is <o> a ) num__78 years <o> b ) num__22 years <o> c ) num__33 years <o> d ) num__66 years <o> e ) num__65 years |
let ' s son age is x then father age is x + num__35 . = > num__2 ( x + num__2 ) = ( x + num__35 + num__2 ) = > num__2 x + num__4 = x + num__37 = > x = num__33 years answer : c <eor> c <eos> |
c |
add__35.0__2.0__ subtract__35.0__2.0__ subtract__35.0__2.0__ |
add__35.0__2.0__ subtract__35.0__2.0__ subtract__35.0__2.0__ |
| a merchant sells an item at a num__20.0 discount but still makes a gross profit of num__20 percent of the cost . what percent t of the cost would the gross profit on the item have been if it had been sold without the discount ? <o> a ) num__20.0 <o> b ) num__40.0 <o> c ) num__50.0 <o> d ) num__60.0 <o> e ) num__75 % |
let the market price of the product is mp . let the original cost price of the product is cp . selling price ( discounted price ) = num__100.0 of mp - num__20.0 mp = num__80.0 of mp . - - - - - - - - - - - - - - - - ( num__1 ) profit made by selling at discounted price = num__20.0 of cp - - - - - - - - - - - - - - ( num__2 ) apply the formula : profit t = selling price - original cost price = > num__20.0 of cp = num__80.0 of mp - num__100.0 cp = > mp = num__120 cp / num__80 = num__1.5 ( cp ) now if product is sold without any discount then profit = selling price ( without discount ) - original cost price = market price - original cost price = mp - cp = num__1.5 cp - cp = num__0.5 cp = num__50.0 of cp thus answer should bec . <eor> c <eos> |
c |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| if x is num__11 percent greater than num__80 then x = <o> a ) num__88.8 <o> b ) num__91.0 <o> c ) num__88.0 <o> d ) num__70.9 <o> e ) num__71.2 |
num__11.0 of num__80 = ( num__80 * num__0.11 ) = num__8.8 num__11.0 greater than num__80 = num__80 + num__8.8 = num__88.8 answer is clearly a . <eor> a <eos> |
a |
multiply__80.0__0.11__ add__80.0__8.8__ add__80.0__8.8__ |
multiply__80.0__0.11__ add__80.0__8.8__ add__80.0__8.8__ |
| if a sample of data has mean of num__24 and sd of num__3 which of the following values is more than num__2.5 sds from the mean ? <o> a ) a . num__19 <o> b ) b . num__18.5 <o> c ) c . num__15 <o> d ) d . num__23.5 <o> e ) e . num__26.5 |
value ismore than num__2.5 sdfrom the mean means that the distance between the mean and the value must be more than num__2.5 sd = num__7.5 . so the value either < num__24 - num__7.5 = num__16.5 or > num__24 + num__7.5 = num__31.5 . answer : c . <eor> c <eos> |
c |
multiply__3.0__2.5__ subtract__24.0__7.5__ add__24.0__7.5__ subtract__31.5__16.5__ |
multiply__3.0__2.5__ subtract__24.0__7.5__ add__24.0__7.5__ subtract__31.5__16.5__ |
| a fruit seller has oranges apples and guavas in the ratio num__2 : num__5 : num__8 . the number of apples is more than the number of oranges by a number that is a multiple of both num__6 and num__11 . what is the minimum number of fruits in his shop ? <o> a ) num__240 <o> b ) num__360 <o> c ) num__120 <o> d ) num__90 <o> e ) num__80 |
detailed solution let the number of fruits be num__2 k num__5 k and num__8 k given num__5 k – num__2 k = multiple of num__6 and num__8 lcm of num__6 and num__8 is num__24 let ’ s say num__5 k – num__2 k = num__24 n num__3 k = num__24 n for k to be a natural number and have minimum value n should be equal to num__1 num__3 k = num__24 or k = num__8 hence the number of fruits = num__2 k + num__5 k + num__8 k = num__15 * num__8 = num__120 correct answer : c <eor> c <eos> |
c |
subtract__5.0__2.0__ subtract__6.0__5.0__ multiply__5.0__3.0__ multiply__5.0__24.0__ multiply__5.0__24.0__ |
subtract__5.0__2.0__ subtract__6.0__5.0__ multiply__5.0__3.0__ multiply__5.0__24.0__ multiply__5.0__24.0__ |
| find the area of the quadrilateral of one of its diagonals is num__40 cm and its off sets num__9 cm and num__6 cm ? <o> a ) num__189 cm num__2 <o> b ) num__300 cm num__2 <o> c ) num__127 cm num__2 <o> d ) num__177 cm num__2 <o> e ) num__187 cm num__2 |
num__0.5 * num__40 ( num__9 + num__6 ) = num__300 cm num__2 answer : b <eor> b <eos> |
b |
square_perimeter__0.5__ triangle_area__2.0__300.0__ |
square_perimeter__0.5__ volume_rectangular_prism__0.5__2.0__300.0__ |
| tom and jerry are running on the same road towards each other . if tom is running at a speed of num__2.5 meters per second and jerry is running num__36.0 slower how much time t will it take them to meet if the initial distance between the two is num__50 meters and tom started running num__20 seconds before jerry did ? <o> a ) num__2 minute and num__12 seconds . <o> b ) two minutes . <o> c ) num__44 seconds . <o> d ) num__20 seconds . <o> e ) num__12 seconds . |
tom is running alone for num__20 seconds . so he will cover a distance of num__20 * num__2.5 = num__50 m in num__20 seconds = t basically jerry has not started yet and tom has covered the distance alone and met jerry on the other side d is the answer <eor> d <eos> |
d |
round__20.0__ |
round__20.0__ |
| find the number of factors of num__300 excluding num__1 and itself . <o> a ) num__22 <o> b ) num__28 <o> c ) num__16 <o> d ) num__88 <o> e ) num__26 |
explanation : num__300 = num__4.75 = num__4 . num__25.3 = num__22 . num__52.31 number of factors of num__300 = ( num__2 + num__1 ) ( num__2 + num__1 ) ( num__1 + num__1 ) = num__18 . there are num__16 factors of num__300 excluding num__1 and itself . answer : c <eor> c <eos> |
c |
gcd__300.0__4.75__ gcd__300.0__22.0__ subtract__22.0__4.0__ subtract__18.0__2.0__ multiply__1.0__16.0__ |
gcd__300.0__4.75__ gcd__300.0__22.0__ subtract__22.0__4.0__ subtract__18.0__2.0__ multiply__1.0__16.0__ |
| at a certain restaurant the average ( arithmetic mean ) number of customers served for the past x days was num__60 . if the restaurant serves num__120 customers today raising the average to num__70 customers per day what is the value of x ? <o> a ) num__2 <o> b ) num__5 <o> c ) num__9 <o> d ) num__15 <o> e ) num__30 |
withoutusing the formula we can see that today the restaurant served num__50 customers above the average . the total amount above the average must equal total amount below the average . this additional num__50 customers must offset the “ deficit ” below the average of num__70 created on the x days the restaurant served only num__60 customers per day . num__5.0 = num__5 days . choice ( a ) . withthe formula we can set up the following : num__70 = ( num__60 x + num__120 ) / ( x + num__1 ) num__70 x + num__70 = num__60 x + num__120 num__10 x = num__50 x = num__5 answer choice ( b ) <eor> b <eos> |
b |
subtract__120.0__70.0__ subtract__60.0__50.0__ multiply__1.0__5.0__ |
subtract__120.0__70.0__ divide__50.0__5.0__ divide__5.0__1.0__ |
| of all the homes on gotham street num__0.333333333333 are termite - ridden and num__0.4 of these are collapsing . what fraction of the homes are termite - ridden but not collapsing ? <o> a ) num__0.133333333333 <o> b ) num__0.2 <o> c ) num__0.8 <o> d ) num__0.4 <o> e ) num__0.2 |
the fraction of homes which are termite - infested but not collapsing is num__0.6 * num__0.333333333333 = num__0.2 the answer is e . <eor> e <eos> |
e |
multiply__0.3333__0.6__ multiply__0.3333__0.6__ |
multiply__0.3333__0.6__ multiply__0.3333__0.6__ |
| num__16 men take num__21 days of num__8 hours each to do a piece of work . how many days of num__6 hours each would num__21 women take to do the same . if num__3 women do as much work as num__2 men ? <o> a ) num__32 <o> b ) num__87 <o> c ) num__30 <o> d ) num__99 <o> e ) num__77 |
num__3 w = num__2 m num__16 m - - - - - - num__21 * num__8 hours num__21 w - - - - - - x * num__6 hours num__14 m - - - - - - x * num__6 num__16 * num__21 * num__8 = num__14 * x * num__6 x = num__32 answer : a <eor> a <eos> |
a |
subtract__16.0__2.0__ multiply__16.0__2.0__ round__32.0__ |
subtract__16.0__2.0__ multiply__16.0__2.0__ multiply__16.0__2.0__ |
| if four fair dice are thrown simultaneously what is the probability of getting at least one pair ? <o> a ) num__0.166666666667 <o> b ) num__0.277777777778 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__0.722222222222 |
num__1 st roll = any number = num__1.0 num__2 nd roll = not a match to the first = num__0.833333333333 num__3 rd roll = not a match to the num__1 st or num__2 nd = num__0.666666666667 num__4 th roll = not a match to the num__1 st or num__2 nd or num__3 rd = num__0.5 ( num__1.0 ) ( num__0.833333333333 ) ( num__0.666666666667 ) ( num__0.5 ) = ( num__1 ) ( num__0.833333333333 ) ( num__0.666666666667 ) ( num__0.5 ) = num__0.277777777778 num__0.277777777778 is the probability of rolling num__0 matching numbers so . . . num__1 - num__0.277777777778 = num__0.722222222222 = the probability of rolling at least one matching pair of numbers num__0.722222222222 = num__0.722222222222 answer : e <eor> e <eos> |
e |
coin_space__ union_prob__0.8333__0.6667__1.0__ negate_prob__1.0__ negate_prob__0.2778__ negate_prob__0.2778__ |
coin_space__ union_prob__0.8333__0.6667__1.0__ negate_prob__1.0__ negate_prob__0.2778__ negate_prob__0.2778__ |
| the average age num__10 members of a committee are the same as it was num__2 years ago because an old number has been replaced by a younger number . find how much younger is the new member than the old number ? <o> a ) num__14 years <o> b ) num__17 years <o> c ) num__20 years <o> d ) num__12 years <o> e ) num__11 years |
num__10 * num__2 = num__20 answer : c <eor> c <eos> |
c |
multiply__10.0__2.0__ multiply__10.0__2.0__ |
multiply__10.0__2.0__ multiply__10.0__2.0__ |
| the difference of a larger number and a smaller number is num__6 . the sum of the larger number and twice the smaller is num__15 . what is the larger number ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__11 |
let x be the larger number and y be the smaller number . x - y = num__6 x + num__2 ( y ) = num__15 solve by substitution : y = x - num__6 x + num__2 ( x - num__6 ) = num__15 x + num__2 x - num__12 = num__15 num__3 x = num__27 x = num__9 the larger number is num__9 so answer c is correct . <eor> c <eos> |
c |
multiply__6.0__2.0__ divide__6.0__2.0__ add__15.0__12.0__ add__6.0__3.0__ add__6.0__3.0__ |
multiply__6.0__2.0__ subtract__15.0__12.0__ add__15.0__12.0__ add__6.0__3.0__ add__6.0__3.0__ |
| two trains running in opposite directions cross a man standing on the platform in num__27 seconds and num__17 seconds respectively and they cross each other in num__23 seconds . the ratio of their speeds is : <o> a ) num__0.333333333333 <o> b ) num__1.5 <o> c ) num__0.5 <o> d ) num__1.5 <o> e ) num__0.333333333333 |
let the speeds of the two trains be x m / sec and y m / sec respectively . then length of the first train = num__27 x meters and length of the second train = num__17 y meters . ( num__27 x + num__17 y ) / ( x + y ) = num__23 = = > num__27 x + num__17 y = num__23 x + num__23 y = = > num__4 x = num__6 y = = > x / y = num__1.5 . answer : b <eor> b <eos> |
b |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ round__1.5__ |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ divide__6.0__4.0__ |
| find the area of the square field if a train num__800 metres long passes the field with a speed of num__120 kmph in one minute . <o> a ) num__1.44 sq . km <o> b ) num__4 sq . km <o> c ) num__2 sq . km <o> d ) num__2.64 sq . km <o> e ) none of these |
explanation : num__120 km / hr = num__120 * num__0.277777777778 = num__33.33 m / s v = d / t ; num__33.33 = d / num__60 d = num__2000 m hence in one minute the train travels num__2000 m . but as the train is num__800 m long and it passes the field the length of the field is num__2000 – num__800 = num__1200 m . area = num__1200 * num__1200 = num__1.44 sq . km answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ subtract__2000.0__800.0__ round__1.44__ |
hour_to_min_conversion__ subtract__2000.0__800.0__ round__1.44__ |
| the average of ten numbers is num__210 . the average of first five numbers is num__40 and the average of last four numbers is num__200 . what is the middle number ? <o> a ) a ) num__200 <o> b ) b ) num__2000 <o> c ) c ) num__800 <o> d ) d ) num__2100 <o> e ) e ) num__1100 |
the total of ten numbers = num__10 x num__210 = num__2100 the total of first num__5 and last num__4 numbers is = num__5 x num__40 + num__4 x num__200 = num__1000 so the middle number is ( num__2100 - num__1000 ) = num__1100 e <eor> e <eos> |
e |
subtract__210.0__200.0__ multiply__210.0__10.0__ divide__200.0__40.0__ divide__40.0__10.0__ multiply__200.0__5.0__ subtract__2100.0__1000.0__ subtract__2100.0__1000.0__ |
subtract__210.0__200.0__ multiply__210.0__10.0__ divide__200.0__40.0__ divide__40.0__10.0__ multiply__200.0__5.0__ subtract__2100.0__1000.0__ subtract__2100.0__1000.0__ |
| worker a takes num__8 hours to do a job . worker b takes num__12 hours to do the same job . how long it take both a & b working together but independently to do the same job ? <o> a ) num__40 days <o> b ) num__4.8 days <o> c ) num__4.33333333333 days <o> d ) num__3.33333333333 days <o> e ) num__6.66666666667 days |
a ' s one hour work = num__0.125 . b ' s one hour work = num__0.0833333333333 . ( a + b ) ' s one hour work = num__0.125 + num__0.0833333333333 = num__0.208333333333 . both a & b can finish the work in num__4.8 days b <eor> b <eos> |
b |
add__0.125__0.0833__ round__4.8__ |
add__0.125__0.0833__ round__4.8__ |
| find the compound interest accrued on an amount of rs . num__14800 at num__13.5 p . a at the end of two years . ( round off your answer to nearest integer ) <o> a ) num__4269 <o> b ) num__4267 <o> c ) num__4260 <o> d ) num__4266 <o> e ) num__4262 |
ci = num__14800 { [ num__1 + num__13.5 / num__100 ] num__2 - num__1 } = num__14800 { [ num__1 + num__0.135 ] num__2 - num__1 = num__14800 { num__2 + num__0.135 } { num__0.135 } = ( num__74 ) [ num__2 + num__0.135 ] ( num__27 ) = num__1998 [ num__2 + num__0.135 ] = num__3996 + num__269.73 = rs . num__4266 answer : d <eor> d <eos> |
d |
percent__13.5__1.0__ percent__13.5__14800.0__ percent__27.0__14800.0__ percent__13.5__1998.0__ percent__100.0__4266.0__ |
percent__13.5__1.0__ percent__13.5__14800.0__ percent__27.0__14800.0__ percent__13.5__1998.0__ percent__100.0__4266.0__ |
| a train num__550 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__30 <o> b ) num__60 <o> c ) num__25 <o> d ) num__40 <o> e ) num__10 |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__550 * num__0.0545454545455 ] sec = num__30 sec answer : a <eor> a <eos> |
a |
add__60.0__6.0__ round__30.0__ |
add__60.0__6.0__ round__30.0__ |
| the floor of a rectangular room is num__18 m long and num__12 m wide . the room is surrounded by a veranda of width num__2 m on all its sides . the area of the veranda is : <o> a ) num__124 m num__2 <o> b ) num__126 m num__2 <o> c ) num__136 m num__2 <o> d ) num__158 m num__2 <o> e ) none of these |
area of the outer rectangle = num__22 Ã — num__16 = num__352 m num__2 area of the inner rectangle = num__18 Ã — num__12 = num__216 m num__2 required area = ( num__352 â € “ num__216 ) = num__136 m num__2 answer c <eor> c <eos> |
c |
multiply__16.0__22.0__ multiply__18.0__12.0__ triangle_area__2.0__136.0__ |
multiply__16.0__22.0__ multiply__18.0__12.0__ triangle_area__2.0__136.0__ |
| a no . when divided by the sum of num__555 and num__445 gives num__2 times their difference as quotient & num__10 as remainder . find the no . is ? <o> a ) num__220010 <o> b ) num__145778 <o> c ) num__220110 <o> d ) num__235467 <o> e ) num__220001 |
( num__555 + num__445 ) * num__2 * num__110 + num__10 = num__220000 + num__10 = num__220010 a <eor> a <eos> |
a |
subtract__555.0__445.0__ add__10.0__220000.0__ add__10.0__220000.0__ |
subtract__555.0__445.0__ add__10.0__220000.0__ add__10.0__220000.0__ |
| a car traveled num__281 miles in num__4 hours num__41 minutes . what was the average speed of the car in miles per hour ? <o> a ) num__30 miles / hour <o> b ) num__90 miles / hour <o> c ) num__60 miles / hour <o> d ) num__80 miles / hour <o> e ) num__70 miles / hour |
we first convert the time of num__4 hours num__41 minutes in minutes num__4 hours num__41 minutes = num__4 ? num__60 + num__41 = num__281 minutes average speed s is given by distance / time . hence s = num__281 miles / num__281 minutes = num__1 mile / minute = num__60 miles / hour correct answer c <eor> c <eos> |
c |
hour_to_min_conversion__ hour_to_min_conversion__ |
hour_to_min_conversion__ divide__60.0__1.0__ |
| num__4 / x + num__3 x = num__3 ( x + num__8 ) <o> a ) - num__9 <o> b ) - num__0.333333333333 <o> c ) num__0.166666666667 <o> d ) num__0.111111111111 <o> e ) num__9 |
we can solve - expand the right side multiply by x on both sides and then subtract away the num__3 x ^ num__2 terms : ( num__4 / x ) + num__3 x = num__3 ( x + num__8 ) ( num__4 / x ) + num__3 x = num__3 x + num__24 num__4 + num__3 x ^ num__2 = num__3 x ^ num__2 + num__24 x num__4 = num__24 x num__0.166666666667 = x and to confirm you can plug that answer back into the original equation to see that it makes the left and right sides equal . c <eor> c <eos> |
c |
divide__8.0__4.0__ multiply__3.0__8.0__ divide__4.0__24.0__ divide__4.0__24.0__ |
divide__8.0__4.0__ multiply__3.0__8.0__ divide__4.0__24.0__ divide__4.0__24.0__ |
| xy = num__1 then what is ( num__3 ^ ( x + y ) ^ num__2 ) / ( num__3 ^ ( x - y ) ^ num__2 ) <o> a ) num__2 <o> b ) num__4 <o> c ) num__81 <o> d ) num__16 <o> e ) num__32 |
( x + y ) ^ num__2 - ( x - y ) ^ num__2 ( x + y + x - y ) ( x + y - x + y ) ( num__2 x ) ( num__2 y ) num__4 xy num__4 num__3 ^ num__4 = num__81 answer c <eor> c <eos> |
c |
add__1.0__3.0__ multiply__1.0__81.0__ |
add__1.0__3.0__ multiply__1.0__81.0__ |
| among the two clocks clock a gains num__20 seconds per minute . if clock a and b are set at num__2 num__0 ' clock when clock a is at num__7 : num__20 what does clock b show ? <o> a ) num__5 : num__30 <o> b ) num__6 : num__00 <o> c ) num__5 : num__45 <o> d ) num__6 : num__20 <o> e ) num__3 : num__30 |
clock a gains num__20 seconds per minute num__1200 seconds per hour or num__20 minutes per hour . the two clocks show num__2 : num__00 at num__2 num__0 ' clock at num__3 : num__00 - clock b is at num__3 : num__00 clock a is at num__3 : num__20 ( num__1 hour + gains num__20 minutes ) at num__4 : num__00 - clock b is at num__4 : num__00 clock a is at num__4 : num__40 ( num__2 hours + gains num__40 minutes ) in num__4 hours the clock a gains num__4 * num__20 = num__80 minutes or num__1 hour num__20 minutes if clock a is at num__7 : num__20 the clock b is at num__6 : num__00 answer is b <eor> b <eos> |
b |
subtract__3.0__2.0__ subtract__7.0__3.0__ multiply__20.0__2.0__ multiply__20.0__4.0__ multiply__2.0__3.0__ round__6.0__ |
subtract__3.0__2.0__ subtract__7.0__3.0__ multiply__20.0__2.0__ multiply__20.0__4.0__ multiply__2.0__3.0__ multiply__2.0__3.0__ |
| the value of ( ( x – y ) ³ + ( y - z ) ³ + ( z – x ) ³ ) / ( num__12 ( x – y ) ( y – z ) ( z – x ) ) is equal to : <o> a ) num__0 <o> b ) num__0.0833333333333 <o> c ) num__1 <o> d ) num__0.25 <o> e ) num__0.333333333333 |
since ( x – y ) + ( y – z ) + ( z – x ) = num__0 so ( x – y ) ³ + ( y – z ) ³ + ( z – x ) ³ = num__3 ( x – y ) ( y – z ) ( z – x ) . ( num__3 ( x – y ) ( y – z ) ( z – x ) ) / ( num__12 ( x – y ) ( y – z ) ( z – x ) ) = num__0.25 . answer : d <eor> d <eos> |
d |
divide__3.0__12.0__ divide__3.0__12.0__ |
divide__3.0__12.0__ divide__3.0__12.0__ |
| if sharon ' s weekly salary increased by num__10 percent she would earn $ num__330 per week . if instead her weekly salary were to increase by num__8 percent how much would she earn per week ? <o> a ) rs . num__424 <o> b ) rs . num__524 <o> c ) rs . num__324 <o> d ) rs . num__350 <o> e ) rs . num__354 |
soln : - ( num__3.0 ) num__108 = num__324 in this case long division does not take much time . ( num__3.0 ) num__108 = num__324 answer : c <eor> c <eos> |
c |
multiply__3.0__108.0__ multiply__3.0__108.0__ |
multiply__3.0__108.0__ multiply__3.0__108.0__ |
| a num__1200 m long train crosses a tree in num__120 sec how much time will i take to pass a platform num__700 m long ? <o> a ) num__180 sec <o> b ) num__190 sec <o> c ) num__170 sec <o> d ) num__175 sec <o> e ) num__115 sec |
answer : option b l = s * t s = num__10.0 s = num__10 m / sec . total length ( d ) = num__1900 m t = d / s t = num__190.0 t = num__190 sec <eor> b <eos> |
b |
divide__1200.0__120.0__ add__1200.0__700.0__ divide__1900.0__10.0__ round__190.0__ |
divide__1200.0__120.0__ add__1200.0__700.0__ divide__1900.0__10.0__ divide__1900.0__10.0__ |
| if p ^ num__2 – num__6 p + num__8 = q and p is a positive integer between num__1 and num__10 inclusive what is the probability that q < num__0 ? <o> a ) num__0.2 <o> b ) num__0.1 <o> c ) num__0.4 <o> d ) num__0.6 <o> e ) num__0.3 |
( p - num__2 ) ( p - num__4 ) = q p = num__3 for this to be true so num__0.1 ans : b <eor> b <eos> |
b |
subtract__6.0__2.0__ add__2.0__1.0__ reverse__10.0__ reverse__10.0__ |
subtract__6.0__2.0__ subtract__4.0__1.0__ reverse__10.0__ reverse__10.0__ |
| if a coin is tossed twice what is the probability that it will land either heads both times or tails both times ? <o> a ) a ) num__0.125 <o> b ) b ) num__0.166666666667 <o> c ) c ) num__0.25 <o> d ) d ) num__0.5 <o> e ) e ) num__1 |
prob to get num__2 heads = num__0.5 * num__0.5 = num__0.25 prob to get num__2 tails = num__0.5 * num__0.5 = num__0.25 so prob to get either num__2 heads or num__2 tails = num__0.25 + num__0.25 = num__0.5 answer will be d . <eor> d <eos> |
d |
coin_space__ negate_prob__0.5__ |
coin_space__ negate_prob__0.5__ |
| two pipes a and b can separately fill a tank in num__2 minutes and num__15 minutes respectively . both the pipes are opened together but num__4 minutes after the start the pipe a is turned off . how much time will it take to fill the tank ? <o> a ) num__5 <o> b ) num__2 <o> c ) num__10 <o> d ) num__12 <o> e ) num__15 |
num__0.333333333333 + x / num__15 = num__1 x = num__10 answer c <eor> c <eos> |
c |
round__10.0__ |
divide__10.0__1.0__ |
| a retailer bought a shirt at wholesale and marked it up num__80.0 to its initial price of $ num__36 . by how many more dollars does he need to increase the price to achieve a num__100.0 markup ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
let x be the wholesale price . then num__1.8 x = num__36 and x = num__36 / num__1.8 = num__20 . to achieve a num__100.0 markup the price needs to be $ num__40 . the retailer needs to increase the price by $ num__4 more . the answer is b . <eor> b <eos> |
b |
divide__36.0__1.8__ divide__80.0__20.0__ divide__80.0__20.0__ |
divide__36.0__1.8__ divide__80.0__20.0__ divide__80.0__20.0__ |
| the area of a rectangular field is equal to num__300 square meters . its perimeter is equal to num__70 meters . find the length of this rectangle . <o> a ) l = num__60 <o> b ) l = num__20 <o> c ) l = num__40 <o> d ) l = num__30 <o> e ) l = num__50 |
l * w = num__300 : area l is the length and w is the width . num__2 l + num__2 w = num__70 : perimeter l = num__35 - w : solve for l ( num__35 - w ) * w = num__300 : substitute in the area equation w = num__15 and l = num__20 : solve for w and find l using l = num__35 - w . correct answer b <eor> b <eos> |
b |
triangle_area__2.0__20.0__ |
triangle_area__2.0__20.0__ |
| a and b together have rs . num__1210 . if of a ' s amount is equal to of b ' s amount how much amount does b have ? <o> a ) num__568 <o> b ) num__484 <o> c ) num__456 <o> d ) num__258 <o> e ) num__125 |
num__0.8 a = num__0.4 b a = ( num__0.4 * num__3.75 ) b a = num__1.5 b a / b = num__1.5 b ' s share = rs ( num__1210 * num__0.4 ) = num__484 <eor> b <eos> |
b |
multiply__3.75__0.4__ multiply__1210.0__0.4__ multiply__1210.0__0.4__ |
multiply__3.75__0.4__ multiply__1210.0__0.4__ multiply__1210.0__0.4__ |
| the average of ten numbers is num__7 . if each number is multiplied by num__10 then what is the average of the new set of numbers ? <o> a ) num__7 <o> b ) num__17 <o> c ) num__70 <o> d ) num__100 <o> e ) num__107 |
the sum of the ten numbers is num__7 * num__10 = num__70 if each number is multiplied by num__10 the new sum is num__10 * ( num__70 ) . the average is then num__10 * num__7 = num__70 the answer is c . <eor> c <eos> |
c |
multiply__7.0__10.0__ multiply__7.0__10.0__ |
multiply__7.0__10.0__ multiply__7.0__10.0__ |
| if f ( x ) = num__3 x ^ num__4 - num__4 x ^ num__3 - num__3 x ^ num__2 + num__6 x then f ( - num__1 ) = <o> a ) - num__2 <o> b ) - num__1 <o> c ) num__0 <o> d ) num__1 <o> e ) num__2 |
f ( - num__1 ) = num__3 ( - num__1 ) ^ num__4 - num__4 ( - num__1 ) ^ num__3 - num__3 ( - num__1 ) ^ num__2 + num__6 ( - num__1 ) = num__3 + num__4 - num__3 - num__6 = - num__2 the answer is a . <eor> a <eos> |
a |
subtract__3.0__1.0__ |
subtract__3.0__1.0__ |
| at their respective rates pump a b and c can fulfill an empty tank or pump - out the full tank in num__2 num__3 and num__6 hours . if a and b are used to pump - out water from the half - full tank while c is used to fill water into the tank in how many hours the tank will be empty ? <o> a ) num__0.666666666667 <o> b ) num__1 <o> c ) num__0.75 <o> d ) num__1.5 <o> e ) num__2 |
rate of a = num__0.5 tank / hour ; rate of b = num__0.333333333333 tank / hour ; rate of c = num__0.166666666667 tank / hour . combined rate when a and b are used to pump - out water while c is used to fill water into the tank is num__0.5 + num__0.333333333333 - num__0.166666666667 = num__0.666666666667 tank hour . so to empty the full tank num__1.5 hours ( reciprocal of rate ) are needed . to empty the half - full tank half of that time would be needed : num__0.5 * num__1.5 = num__0.75 hours . answer : c . <eor> c <eos> |
c |
divide__3.0__6.0__ divide__2.0__6.0__ divide__0.5__3.0__ divide__2.0__3.0__ subtract__2.0__0.5__ multiply__0.5__1.5__ round__0.75__ |
divide__3.0__6.0__ divide__2.0__6.0__ divide__0.5__3.0__ divide__2.0__3.0__ subtract__2.0__0.5__ multiply__0.5__1.5__ multiply__0.5__1.5__ |
| a die is drawed find the probability that an odd number is draw ? <o> a ) num__0.5 <o> b ) num__0.5 <o> c ) num__1.5 <o> d ) num__0.2 <o> e ) num__0.4 |
let us first write the sample space s of the experiment . s = { num__12 num__34 num__56 } let e be the event ` ` an odd number is obtained ' ' and write it down . e = { num__1 num__35 } p ( e ) = n ( e ) / n ( s ) = num__0.5 = num__0.5 a <eor> a <eos> |
a |
negate_prob__0.5__ |
negate_prob__0.5__ |
| a bucket full of nuts was discovered by the crow living in the basement . the crow eats a fifth of the total number of nuts in num__4 hours . how many hours in total will it take the crow to finish a quarter of the nuts ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
in one hour the crow eats num__0.05 of the nuts . ( num__0.25 ) / ( num__0.05 ) = num__5 hours the answer is b . <eor> b <eos> |
b |
reverse__4.0__ divide__0.25__0.05__ divide__0.25__0.05__ |
reverse__4.0__ divide__0.25__0.05__ divide__0.25__0.05__ |
| if the simple interest on a sum of money for num__2 years at num__5.0 per annum is rs . num__56 what is the compound interest on the same sum at the rate and for the same time ? <o> a ) s . num__57.40 <o> b ) s . num__57.22 <o> c ) s . num__51.219 <o> d ) s . num__56.18 <o> e ) s . num__53.11 |
explanation : sum = ( num__56 * num__100 ) / ( num__2 * num__5 ) = rs . num__560 amount = [ num__560 * ( num__1 + num__0.05 ) num__2 ] = rs . num__617.4 c . i . = ( num__617.4 - num__560 ) = rs . num__57.4 . answer : a <eor> a <eos> |
a |
percent__5.0__1.0__ percent__100.0__57.4__ |
percent__5.0__1.0__ percent__100.0__57.4__ |
| a rectangular pig farm has a fence along three sides and a wall along the fourth side . the fenced side opposite the wall is twice the length of each of the other two fenced sides . if the area of the rectangular region is num__1250 square feet what is the total length of the fence in feet ? <o> a ) num__25 <o> b ) num__50 <o> c ) num__100 <o> d ) num__150 <o> e ) num__200 |
two sides each = x the third = num__2 x and the wall length is thus num__2 x too x * num__2 x = num__2 x ^ num__2 = num__1250 ie x ^ num__2 = num__625 ie x = num__25 l = num__50 w = num__25 total length of fence = num__2 * num__25 + num__50 = num__100 my answer is c <eor> d <eos> |
d |
multiply__2.0__25.0__ square_perimeter__25.0__ rectangle_perimeter__50.0__25.0__ |
multiply__2.0__25.0__ multiply__2.0__50.0__ rectangle_perimeter__50.0__25.0__ |
| if a : b = num__1 : num__3 and b : c = num__2 : num__4 find a : b : c ? <o> a ) num__1 : num__6 : num__3 <o> b ) num__1 : num__3 : num__6 <o> c ) num__6 : num__1 : num__3 <o> d ) num__6 : num__3 : num__1 <o> e ) num__3 : num__6 : num__1 |
a : b = num__1 : num__3 b : c = num__2 : num__4 num__1 : num__3 num__2 : num__4 ( a = a × b b = b × b and c = b × c ) a : b : c = num__1 : num__3 : num__6 b <eor> b <eos> |
b |
multiply__3.0__2.0__ reverse__1.0__ |
multiply__3.0__2.0__ reverse__1.0__ |
| in a bag containing num__3 balls a white ball was placed & then num__1 ball was taken out atrandom . what is the probability that the extracted ball would turn on to be white if all possible hypothesis concerning the color of theballs that initiallyin the bag were equally possible ? <o> a ) num__3 <o> b ) num__0.625 <o> c ) num__27 <o> d ) num__54 <o> e ) num__81 |
since all possible hypothesis regarding the colour of the balls are equally likely therefore these could be num__3 white balls initially in the bag . ∴ required probability = num__0.25 [ num__1 + num__0.75 + num__0.5 + num__0.25 ] = num__0.25 [ ( num__4 + num__3 + num__2 + num__1 ) / num__4 ] = num__0.625 b <eor> b <eos> |
b |
negate_prob__0.25__ union_prob__1.0__0.25__0.75__ coin_space__ union_prob__1.0__0.25__0.625__ |
negate_prob__0.25__ union_prob__1.0__0.25__0.75__ coin_space__ union_prob__1.0__0.25__0.625__ |
| what ratio must a shopkeepermix peas and soybean of rs . num__16 and rs . num__25 / kg as to obtain a mixture of rs . num__20 ? <o> a ) num__10 : num__7 <o> b ) num__9 : num__8 <o> c ) num__5 : num__4 <o> d ) num__13 : num__11 <o> e ) num__14 : num__8 |
correct option : ( c ) use rule of alligation to determine the ratio the required ratio of soybean and peas = num__5 : num__4 <eor> c <eos> |
c |
subtract__25.0__20.0__ subtract__20.0__16.0__ subtract__25.0__20.0__ |
subtract__25.0__20.0__ subtract__20.0__16.0__ subtract__25.0__20.0__ |
| in num__2008 the profits of company n were num__10 percent of revenues . in num__2009 the revenues of company n fell by num__20 percent but profits were num__9 percent of revenues . the profits in num__2009 were what percent of the profits in num__2008 ? <o> a ) num__80.0 <o> b ) num__72.0 <o> c ) num__120.0 <o> d ) num__124.2 <o> e ) num__138 % |
x = profits r = revenue x / r = num__01 x = num__10 r = num__100 num__2009 : r = num__80 x / num__80 = num__009 = num__0.09 x = num__80 * num__0.09 x = num__7.2 num__7.2 / num__10 = num__0.72 = num__72.0 answer b <eor> b <eos> |
b |
percent__9.0__1.0__ percent__9.0__80.0__ percent__10.0__7.2__ percent__100.0__72.0__ |
percent__9.0__1.0__ percent__9.0__80.0__ percent__10.0__7.2__ percent__100.0__72.0__ |
| a grocer has a sale of rs . num__1500 rs . num__1500 rs . num__2000 rs . num__2600 for num__4 consecutive months . how much sale must he have in the fifth month so that he gets an average sale of rs . num__2000 ? <o> a ) rs . num__4980 <o> b ) rs . num__3400 <o> c ) rs . num__4000 <o> d ) rs . num__2400 <o> e ) rs . num__5000 |
total sale for num__4 months = rs . ( num__1500 + num__1500 + num__2000 + num__2600 ) = rs . num__7600 required sale = rs . [ ( num__2000 x num__5 ) - num__7600 ] = rs . ( num__10000 - num__7600 ) = rs . num__2400 . d <eor> d <eos> |
d |
multiply__2000.0__5.0__ subtract__10000.0__7600.0__ subtract__10000.0__7600.0__ |
multiply__2000.0__5.0__ subtract__10000.0__7600.0__ subtract__10000.0__7600.0__ |
| a man is num__18 years older than his son . in two years his age will be twice the age of his son . the present age of this son is <o> a ) num__21 years <o> b ) num__22 years <o> c ) num__23 years <o> d ) num__12 years <o> e ) num__16 years |
explanation : let ' s son age is x then father age is x + num__18 . = > num__2 ( x + num__2 ) = ( x + num__18 + num__2 ) = > num__2 x + num__4 = x + num__20 = > x = num__16 years option e <eor> e <eos> |
e |
add__18.0__2.0__ subtract__18.0__2.0__ subtract__18.0__2.0__ |
add__18.0__2.0__ subtract__18.0__2.0__ subtract__18.0__2.0__ |
| a water tank is two - fifth full . pipe a can fill a tank in num__10 minutes and pipe b can empty in num__6 minutes . if both the pipes are open how long will it take to empty or fill the tank completely ? <o> a ) num__7 min to empty <o> b ) num__5 min to full <o> c ) num__4 min to full <o> d ) num__2 min to empty <o> e ) num__6 min to empty |
num__0.166666666667 - num__0.1 = = num__0.0666666666667 num__0.0666666666667 : num__0.4 : : num__1 : x num__0.4 * num__15 = num__6 min answer e <eor> e <eos> |
e |
subtract__0.1667__0.1__ multiply__10.0__0.1__ divide__6.0__0.4__ round__6.0__ |
subtract__0.1667__0.1__ multiply__10.0__0.1__ divide__6.0__0.4__ multiply__0.4__15.0__ |
| the distance between two cities a and b is num__870 km . a train starts from a at num__8 a . m . and travel towards b at num__60 km / hr . another train starts from b at num__9 a . m and travels towards a at num__75 km / hr . at what time do they meet ? <o> a ) num__09 pm <o> b ) num__07 pm <o> c ) num__03 pm <o> d ) num__05 am <o> e ) num__03 am |
explanation : suppose they meet x hrs after num__8 a . m then [ distance moved by first in x hrs ] + [ distance moved by second in ( x - num__1 ) hrs ] = num__870 . therefore num__60 x + num__75 ( x - num__1 ) = num__870 . = > x = num__7 . so they meet at ( num__8 + num__7 ) i . e num__15 = > num__3 pm answer : c ) <eor> c <eos> |
c |
subtract__9.0__8.0__ subtract__8.0__1.0__ add__8.0__7.0__ round__3.0__ |
subtract__9.0__8.0__ subtract__8.0__1.0__ add__8.0__7.0__ round__3.0__ |
| the number num__311311311311311311311 is : <o> a ) divisible by num__3 but not by num__11 <o> b ) divisible by num__11 but not by num__3 <o> c ) divisible by both num__3 and num__11 <o> d ) neither divisible by num__3 nor by num__11 . <o> e ) none of thses |
sum of digits = num__35 and so it is not divisible by num__3 . ( sum of digits at odd places ) - ( sum of digits at even places ) = num__19 - num__16 = num__3 not divisible by num__11 . so the given number is neither divided by num__3 nor by num__11 . answer : d <eor> d <eos> |
d |
subtract__19.0__3.0__ subtract__19.0__16.0__ |
subtract__19.0__3.0__ subtract__19.0__16.0__ |
| rs . num__4500 amounts to rs . num__5544 in two years at compound interest compounded annually . if the rate of the interest for the first year is num__12.0 find the rate of interest for the second year ? <o> a ) num__10.0 <o> b ) num__16.0 <o> c ) num__70.0 <o> d ) num__80.0 <o> e ) num__20 % |
let the rate of interest during the second year be r % . given num__4500 * { ( num__100 + num__12 ) / num__100 } * { ( num__100 + r ) / num__100 } = num__5544 r = num__10.0 answer : a <eor> a <eos> |
a |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| a box contains nine bulbs out of which num__4 are defective . if four bulbs are chosen at random find the probability that atleast one bulb is good ? <o> a ) num__0.748502994012 <o> b ) num__1.0593220339 <o> c ) num__0.992063492063 <o> d ) num__0.9765625 <o> e ) num__0.968992248062 |
required probability = num__1 - num__0.00793650793651 = num__0.992063492063 answer : c <eor> c <eos> |
c |
negate_prob__0.0079__ negate_prob__0.0079__ |
negate_prob__0.0079__ negate_prob__0.0079__ |
| the average marks scored by ganesh in english science mathematics and history is less than num__16 from that scored by him in english history geography and mathematics . what is the difference of marks in science and geography scored by him ? <o> a ) num__40 <o> b ) num__50 <o> c ) num__64 <o> d ) data inadequate <o> e ) none of these |
e + h + g + m / num__4 â ˆ ’ e + s + m + h / num__4 = num__16 â ‡ ’ g - s = num__64 answer c <eor> c <eos> |
c |
multiply__16.0__4.0__ multiply__16.0__4.0__ |
multiply__16.0__4.0__ multiply__16.0__4.0__ |
| a house wife saved $ num__2.50 in buying an item on sale . if she spent $ num__25 for the item approximately how much percent she saved in the transaction ? <o> a ) num__8.0 <o> b ) num__9.0 <o> c ) num__10.0 <o> d ) num__11.0 <o> e ) num__12 % |
actual price = num__25 + num__2.50 = $ num__27.50 saving = num__2.50 / num__27.50 * num__100 = num__9.09090909091 = num__9.0 approximately answer is b <eor> b <eos> |
b |
add__2.5__25.0__ round_down__9.0909__ round_down__9.0909__ |
add__2.5__25.0__ round_down__9.0909__ round_down__9.0909__ |
| there are num__1500 jelly beans divided between two jars jar x and jar y . if there are num__200 fewer jelly beans in jar x than three times the number of beans in jar y how many beans are in jar x ? <o> a ) num__1050 <o> b ) num__1075 <o> c ) num__1100 <o> d ) num__1125 <o> e ) num__1150 |
x + y = num__1500 so y = num__1500 - x x = num__3 y - num__200 x = num__3 ( num__1500 - x ) - num__200 num__4 x = num__4300 x = num__1075 the answer is b . <eor> b <eos> |
b |
divide__4300.0__4.0__ divide__4300.0__4.0__ |
divide__4300.0__4.0__ divide__4300.0__4.0__ |
| if the simple interest on a certain amount in at num__4.0 rate num__5 years amounted to rs . num__2000 less than the principal . what was the principal ? <o> a ) num__1500 <o> b ) num__2500 <o> c ) num__2507 <o> d ) num__3200 <o> e ) num__11500 |
p - num__2000 = ( p * num__5 * num__4 ) / num__100 p = num__2500 answer : b <eor> b <eos> |
b |
percent__5.0__2000.0__ percent__100.0__2500.0__ |
percent__5.0__2000.0__ percent__100.0__2500.0__ |
| a train num__360 m long runs with a speed of num__45 km / hr . what time will it take to pass a platform of num__130 m long ? <o> a ) num__39.2 sec <o> b ) num__35 sec <o> c ) num__44 sec <o> d ) num__40 sec <o> e ) none of these |
speed = num__45 km / hr = num__45 Ã — ( num__0.277777777778 ) m / s = num__12.5 = num__12.5 = num__12.5 m / s total distance = length of the train + length of the platform = num__360 + num__130 = num__490 meter time taken to cross the platform = num__490 / ( num__12.5 ) = num__490 Ã — num__0.08 = num__39.2 seconds answer : a <eor> a <eos> |
a |
add__360.0__130.0__ divide__490.0__12.5__ round__39.2__ |
add__360.0__130.0__ divide__490.0__12.5__ divide__490.0__12.5__ |
| the length and breadth of a room are num__8 m and num__6 m respectively . a cat runs along all the four walls and finally along a diagonal order to catch a rat . how much total distance is covered by the cat ? <o> a ) num__10 <o> b ) num__14 <o> c ) num__38 <o> d ) num__48 <o> e ) num__47 |
diagonal = num__10 num__2 ( l + b ) + d = num__38 answer : c <eor> c <eos> |
c |
subtract__8.0__6.0__ round__38.0__ |
subtract__8.0__6.0__ round__38.0__ |
| at a meeting of num__5 chiefs of staff the chief of naval operations does not want to sit next to the chief of the national guard bureau . how many ways can the chiefs of staff be seated around a circular table ? <o> a ) num__9 <o> b ) num__12 <o> c ) num__15 <o> d ) num__18 <o> e ) num__21 |
num__5 people can be arranged around a table in num__4 ! ways . consider the two chiefs who will not sit together as a single unit . the number of ways to arrange num__4 units around a table is num__3 ! we need to multiply this by num__2 as the two people can be switched in each arrangement . the total number of ways to arrange the chiefs is num__4 ! - num__2 * num__3 ! = num__2 * num__3 ! = num__12 the answer is b . <eor> b <eos> |
b |
subtract__5.0__3.0__ multiply__3.0__4.0__ multiply__3.0__4.0__ |
subtract__5.0__3.0__ multiply__3.0__4.0__ multiply__3.0__4.0__ |
| if num__11.25 m of a uniform iron rod weight num__42.75 kg what will be the weight of num__6 m of the same rod ? <o> a ) num__22.8 kg <o> b ) num__25.6 kg <o> c ) num__28 kg <o> d ) num__26.5 kg <o> e ) none of these |
solution let the required weight be x kg . then less length less weight ∴ num__11.25 : num__6 : : num__42.75 : x ⇔ num__11.25 × x = num__6 × num__42.75 ⇔ x = ⇔ x = ( num__6 x num__42.75 ) / num__11.25 = num__22.8 . answer a <eor> a <eos> |
a |
round__22.8__ |
round__22.8__ |
| if a and b are positive integers of y such that a / b = num__2.86 which of the following must be a divisor of a ? <o> a ) num__10 <o> b ) b . num__13 <o> c ) num__18 <o> d ) num__26 <o> e ) num__50 |
the prompt gives us a couple of facts to work with : num__1 ) a and b are positive integers num__2 ) a / b = num__2.86 we can use these facts to figure out possible values of a and b . the prompt asks us for what must be a divisor of a . since we ' re dealing with a fraction a and b could be an infinite number of different integers so we have to make both as small as possible ; in doing so we ' ll be able to find the divisors that always divide in ( and eliminate the divisors that only sometimes divide in ) . the simplest place to start is with . . . a = num__286 b = num__100 num__2.86 = num__2.86 these values are not the smallest possible values though ( since they ' re both even we can divide both by num__2 ) . . . a = num__143 b = num__50 num__2.86 = num__2.86 there is no other way to reduce this fraction so a must be a multiple of num__143 and b must be an equivalent multiple of num__50 . at this point though the value of b is irrelevant to the question . we ' re asked for what must divide into a . . . . since a is a multiple of num__143 we have to ' factor - down ' num__143 . this gives us ( num__11 ) ( num__13 ) . so both of those integers must be factors of a . you ' ll find the match in the answer choices . final answer : b <eor> b <eos> |
b |
round_down__2.86__ divide__286.0__2.86__ divide__286.0__2.0__ divide__100.0__2.0__ add__2.0__11.0__ multiply__1.0__13.0__ |
round_down__2.86__ divide__286.0__2.86__ divide__286.0__2.0__ divide__100.0__2.0__ divide__143.0__11.0__ divide__13.0__1.0__ |
| the cost of num__2 chairs and num__3 tables is rs . num__1300 . the cost of num__3 chairs and num__2 tables is rs . num__1200 . the cost of each table is more than that of each chair by ? <o> a ) num__377 <o> b ) num__268 <o> c ) num__297 <o> d ) num__272 <o> e ) num__100 |
num__2 c + num__3 t = num__1300 - - - ( num__1 ) num__3 c + num__3 t = num__1200 - - - ( num__2 ) subtracting num__2 nd from num__1 st we get - c + t = num__100 = > t - c = num__100 answer : e <eor> e <eos> |
e |
subtract__3.0__2.0__ subtract__1300.0__1200.0__ subtract__1300.0__1200.0__ |
subtract__3.0__2.0__ subtract__1300.0__1200.0__ subtract__1300.0__1200.0__ |
| a goods train runs at the speed of num__72 kmph and crosses a num__250 m long platform in num__26 seconds . what is the length of the goods train ? <o> a ) num__210 m <o> b ) num__230 m <o> c ) num__240 m <o> d ) num__260 m <o> e ) num__270 m |
speed = num__72 x ( num__0.277777777778 ) m / sec = num__20 m / sec time = num__26 sec . let the length of the train be x metres . ( x + num__250 ) / num__26 = num__20 x + num__250 = num__520 x = num__270 . answer : e <eor> e <eos> |
e |
multiply__26.0__20.0__ add__250.0__20.0__ round__270.0__ |
multiply__26.0__20.0__ add__250.0__20.0__ add__250.0__20.0__ |
| a man walks at a speed of num__6 km / hr and runs at a speed of num__7 km / hr . how much time will the man require to cover a distance of num__10 num__0.5 km if he completes half of the distance i . e . ( num__5 num__0.25 ) km on foot and the other half by running ? <o> a ) num__1 num__2.0 hours <o> b ) num__1 num__0.625 hours <o> c ) num__2 num__0.166666666667 hours <o> d ) num__2 num__1.0 hours <o> e ) num__2 num__0.5 hours |
required time = ( num__5 num__0.25 ) / num__6 + ( num__5 num__0.25 ) / num__7 = num__1 num__0.625 hours . answer : b <eor> b <eos> |
b |
subtract__6.0__5.0__ round__1.0__ |
subtract__6.0__5.0__ round__1.0__ |
| a sucrose solution contains num__10 grams of sucrose per num__100 cubic centimeters of solution . if num__60 cubic centimeters of the solution were poured into an empty container how many grams of sucrose would be in the container ? <o> a ) num__4.00 <o> b ) num__6.00 <o> c ) num__5.50 <o> d ) num__6.50 <o> e ) num__6.75 |
we are given that a sucrose solution contains num__10 grams of sucrose per num__100 cubic centimeters of solution . since we are dealing with a solution we know that the grams of sucrose is proportional to the number of cubic centimeters of solution . thus to determine how many grams of sucrose would be in the container when we have num__60 cubic centimeters of solution we can set up a proportion . we can say : “ num__10 grams of sucrose is to num__100 cubic centimeters of solution as x grams of sucrose is to num__60 cubic centimeters of solution . ” let ’ s now set up the proportion and solve for x . num__0.1 = x / num__60 when we cross multiply we obtain : ( num__10 ) ( num__60 ) = num__100 x num__600 = num__100 x num__6.00 = x there are num__6.00 grams of sucrose in the solution in the container . the answer is b . <eor> b <eos> |
b |
divide__10.0__100.0__ multiply__10.0__60.0__ divide__60.0__10.0__ round__6.0__ |
divide__10.0__100.0__ divide__60.0__0.1__ divide__60.0__10.0__ divide__60.0__10.0__ |
| the product of the squares of two positive integers is num__9 . how many pairs of positive integers satisfy this condition ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
ans : b - num__1 pairs ( x ˆ num__2 ) ( y ˆ num__2 ) = num__9 [ square root both sides ] xy = num__3 num__3 = num__1 x num__3 num__3 x num__1 cancel the repeats this leaves us with exactly num__1 options . hence b <eor> b <eos> |
b |
volume_cube__1.0__ |
volume_cube__1.0__ |
| num__4 num__7 num__13 num__25 num__49 ( . . . ) <o> a ) num__22 <o> b ) num__35 <o> c ) num__97 <o> d ) num__32 <o> e ) num__25 |
explanation : num__4 num__4 × num__2 - num__1 = num__7 num__7 × num__2 - num__1 = num__13 num__13 × num__2 - num__1 = num__25 num__25 × num__2 - num__1 = num__49 num__49 × num__2 - num__1 = num__97 answer : option c <eor> c <eos> |
c |
multiply__1.0__97.0__ |
multiply__1.0__97.0__ |
| what is the greatest possible length which can be used to measure exactly the lengths num__8 m num__4 m num__20 cm and num__12 m num__20 cm ? <o> a ) num__20 cm <o> b ) num__23 cm <o> c ) num__25 cm <o> d ) num__10 cm <o> e ) num__28 cm |
required length = hcf of num__800 cm num__420 cm num__1220 cm = num__20 cm answer is a <eor> a <eos> |
a |
add__800.0__420.0__ round__20.0__ |
add__800.0__420.0__ round__20.0__ |
| the price of a jacket is reduced by num__25.0 . during a special sale the price of the jacket is reduced another num__20.0 . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? <o> a ) num__66.67 <o> b ) num__35 <o> c ) num__48 <o> d ) num__65 <o> e ) num__67.5 |
num__1 ) let the price of jacket initially be $ num__100 . num__2 ) then it is decreased by num__25.0 therefore bringing down the price to $ num__75 . num__3 ) again it is further discounted by num__20.0 therefore bringing down the price to $ num__60 . num__4 ) now num__60 has to be added byx % in order to equal the original price . num__60 + ( x % ) num__60 = num__100 . solving this eq for x we get x = num__66.67 ans is a . <eor> a <eos> |
a |
percent__100.0__66.67__ |
percent__100.0__66.67__ |
| walking at ¾ of his usual speed a man is late by num__2 ½ hr . the usual time is <o> a ) num__7 num__0.111111111111 <o> b ) num__7 num__0.5 <o> c ) num__7 num__0.2 <o> d ) num__7 num__0.125 <o> e ) num__7 num__0.142857142857 |
usual speed = s usual time = t distance = d new speed is ¾ s new time is num__1.33333333333 t num__1.33333333333 t – t = num__2.5 t = num__7.5 = num__7 num__0.5 aswer : b <eor> b <eos> |
b |
subtract__2.5__2.0__ round__7.0__ |
subtract__2.5__2.0__ round__7.0__ |
| united telephone charges a base rate of $ num__6.00 for service plus an additional charge of $ num__0.25 per minute . atlantic call charges a base rate of $ num__12.00 for service plus an additional charge of $ num__0.20 per minute . for what number of minutes would the bills for each telephone company be the same ? <o> a ) num__20 minutes <o> b ) num__110 minutes <o> c ) num__120 minutes <o> d ) num__140 minutes <o> e ) num__160 minutes |
lets take number of minutesx . given that num__6 + num__0.25 x = num__12 + num__0.2 x - > num__0.05 x = num__2 - > x = num__120 minutes ans c <eor> c <eos> |
c |
subtract__0.25__0.2__ divide__12.0__6.0__ divide__6.0__0.05__ divide__6.0__0.05__ |
subtract__0.25__0.2__ divide__12.0__6.0__ divide__6.0__0.05__ divide__6.0__0.05__ |
| how long does a train num__150 m long traveling at num__36 kmph takes to cross a signal post ? <o> a ) num__8 sec <o> b ) num__15 sec <o> c ) num__20 sec <o> d ) num__10 sec <o> e ) num__22 sec |
d = num__150 m s = num__36 * num__0.277777777778 = num__10 t = num__150 * num__0.1 = num__15 sec answer : b <eor> b <eos> |
b |
multiply__150.0__0.1__ round__15.0__ |
multiply__150.0__0.1__ multiply__150.0__0.1__ |
| a boat running downstream covers a distance of num__20 km in num__2 hours while for covering the same distance upstream it takes num__5 hours . what is the speed of the boat in still water ? <o> a ) num__4 km / hr <o> b ) num__6 km / hr <o> c ) num__7 km / hr <o> d ) data inadequate <o> e ) none of these |
solution rate downstream = ( num__10.0 ) kmph = num__10 kmph rate upstream = ( num__4.0 ) = num__4 kmph . ∴ speed in still water = num__0.5 ( num__10 + num__4 ) kmph = num__7 kmph . answer c <eor> c <eos> |
c |
divide__20.0__2.0__ divide__20.0__5.0__ divide__2.0__4.0__ add__2.0__5.0__ round__7.0__ |
divide__20.0__2.0__ divide__20.0__5.0__ divide__2.0__4.0__ add__2.0__5.0__ add__2.0__5.0__ |
| tough and tricky questions : arithmetic . ( num__14 ^ num__2 + num__14 ^ num__2 ) / num__7 ^ num__2 = <o> a ) num__4 <o> b ) num__58 <o> c ) num__29 <o> d ) num__8 <o> e ) num__116 |
ans is num__8 my approach was : ( num__14 ^ num__2 + num__14 ^ num__2 ) / num__29 ^ num__2 = num__14 ( num__14 + num__14 ) / num__7 * num__7 = num__14 * num__4.0 * num__7 = num__2 * num__4 = num__8 d <eor> d <eos> |
d |
divide__8.0__2.0__ multiply__2.0__4.0__ |
divide__8.0__2.0__ multiply__2.0__4.0__ |
| find out difference between si and ci if p = num__1000000 r = num__4.0 ant t = num__3 yr ? <o> a ) num__4864 <o> b ) num__4854 <o> c ) num__4844 <o> d ) num__4868 <o> e ) num__4866 |
p = num__1000000 r = num__4.0 annual t = num__3 yr si = p * r * t / num__100 = num__1000000 * num__4 * num__0.03 = num__120000 ci = p [ ( num__1 + r / num__100 ) ^ t - num__1 ] = num__1000000 [ ( num__1 + num__0.04 ) ^ num__3 - num__1 ] = num__124864 difference = ci - si = num__124864 - num__120000 = num__4864 answer : a <eor> a <eos> |
a |
divide__3.0__100.0__ subtract__4.0__3.0__ divide__4.0__100.0__ subtract__124864.0__120000.0__ multiply__1.0__4864.0__ |
divide__3.0__100.0__ subtract__4.0__3.0__ divide__4.0__100.0__ subtract__124864.0__120000.0__ multiply__1.0__4864.0__ |
| the length of rectangle is thrice its breadth and its perimeter is num__96 m find the area of the rectangle ? <o> a ) num__432 <o> b ) num__787 <o> c ) num__288 <o> d ) num__212 <o> e ) num__233 |
num__2 ( num__3 x + x ) = num__96 l = num__36 b = num__12 lb = num__36 * num__12 = num__432 answer : a <eor> a <eos> |
a |
square_perimeter__3.0__ multiply__36.0__12.0__ triangle_area__2.0__432.0__ |
square_perimeter__3.0__ multiply__36.0__12.0__ multiply__36.0__12.0__ |
| find the curved surface area if the radius of a cone is num__21 m and slant height is num__15 m ? <o> a ) num__660 <o> b ) num__770 <o> c ) num__880 <o> d ) num__900 <o> e ) num__990 |
cone curved surface area = Ï € rl num__3.14285714286 Ã — num__21 Ã — num__15 = num__66 Ã — num__15 = num__990 m ( power num__2 ) answer is e . <eor> e <eos> |
e |
multiply__15.0__66.0__ multiply__15.0__66.0__ |
multiply__15.0__66.0__ multiply__15.0__66.0__ |
| a train is num__360 meter long is running at a speed of num__54 km / hour . in what time will it pass a bridge of num__140 meter length ? <o> a ) num__27 seconds <o> b ) num__33 seconds <o> c ) num__40 seconds <o> d ) num__11 seconds <o> e ) num__12 seconds |
speed = num__54 km / hr = num__54 * ( num__0.277777777778 ) m / sec = num__15 m / sec total distance = num__360 + num__140 = num__500 meter time = distance / speed = num__500 * ( num__0.0666666666667 ) = num__33 seconds answer : b <eor> b <eos> |
b |
add__360.0__140.0__ round__33.0__ |
add__360.0__140.0__ round__33.0__ |
| find the value of ( num__70 + num__0.28 ) × num__100 <o> a ) num__7028 <o> b ) num__4028 <o> c ) num__3128 <o> d ) num__3256 <o> e ) num__5264 |
( num__7000 + num__28 ) / num__100 * num__100 = num__7028 answer : a <eor> a <eos> |
a |
multiply__70.0__100.0__ multiply__0.28__100.0__ add__7000.0__28.0__ add__7000.0__28.0__ |
multiply__70.0__100.0__ multiply__0.28__100.0__ add__7000.0__28.0__ add__7000.0__28.0__ |
| a circular well with a diameter of num__2 meters is dug to a depth of num__14 meters . what is the volume of the earth dug out . <o> a ) num__48 m num__3 <o> b ) num__94 m num__3 <o> c ) num__44 m num__3 <o> d ) num__04 m num__3 <o> e ) num__14 m num__3 |
volume = π r num__2 hvolume = ( num__227 ∗ num__1 ∗ num__1 ∗ num__14 ) m num__3 = num__44 m num__3 answer : c <eor> c <eos> |
c |
triangle_area__2.0__44.0__ |
triangle_area__2.0__44.0__ |
| what is the smallest five digit number that is divisible by num__15 num__32 num__45 and num__54 ? <o> a ) num__11260 <o> b ) num__11860 <o> c ) num__12360 <o> d ) num__12960 <o> e ) num__13560 |
num__15 = num__3 * num__5 num__32 = num__2 ^ num__5 num__45 = num__3 ^ num__2 * num__5 num__54 = num__2 * num__3 ^ num__3 lcm = num__2 ^ num__5 * num__3 ^ num__3 * num__5 = num__4320 the smallest five - digit number that is a multiple of num__4320 is num__3 * num__4320 = num__12960 the answer is d . <eor> d <eos> |
d |
divide__45.0__15.0__ divide__15.0__3.0__ subtract__5.0__3.0__ multiply__4320.0__3.0__ multiply__4320.0__3.0__ |
divide__45.0__15.0__ divide__15.0__3.0__ subtract__5.0__3.0__ multiply__4320.0__3.0__ multiply__4320.0__3.0__ |
| horse started to chase a dog as it relived stable two hours ago . and horse started to ran with average speed of num__22 kmph horse crossed num__10 meters road and two ponds with depth num__3 meters and it crossed num__2 small streets with num__200 meters length . after travelling num__6 hours num__2 hours after the sunset it got dog . compute the speed of dog ? <o> a ) num__16.5 kmph <o> b ) num__16.6 kmph <o> c ) num__16.4 kmph <o> d ) num__16.9 kmph <o> e ) num__16.2 kmph |
in an average speed of num__22 kmph horse got dog in num__6 hours so distance = speed * time distance = num__22 * num__6 = num__132 km dog takes num__8 hours to travel num__132 km . so the speed if dog is speed = distance / time speed = num__16.5 = num__16.5 kmph answer : num__16.5 kmph answer a <eor> a <eos> |
a |
multiply__22.0__6.0__ subtract__10.0__2.0__ divide__132.0__8.0__ round__16.5__ |
multiply__22.0__6.0__ subtract__10.0__2.0__ divide__132.0__8.0__ divide__132.0__8.0__ |
| a = num__4 ^ num__15 - num__625 ^ num__3 and a / x is an integer where x is a positive integer greater than num__1 such that it does not have a factor p such that num__1 < p < x then how many different values for x are possible ? <o> a ) none <o> b ) one <o> c ) two <o> d ) three <o> e ) four |
this is a tricky worded question and i think the answer is should be d not c . . . here is my reason : the stem says that x is a positive integer such that has no factor grater than num__2 and less than x itself . the stem wants to say that x is a prime number . because any prime number has no factor grater than num__1 and itself . on the other hand the stem says that x could get how many different number not must get different number ( this is very important issue ) as our friends say if we simplify numerator more we can obtain : num__5 ^ num__12 ( num__5 ^ num__3 - num__1 ) = num__5 ^ num__12 ( num__124 ) = num__5 ^ num__12 ( num__31 * num__2 * num__2 ) divided by x and we are told that this fraction is an integer . so x could be ( not must be ) num__5 num__31 or num__2 ! ! ! so x could get num__2 different values and answer is c . . . . <eor> c <eos> |
c |
subtract__3.0__1.0__ add__4.0__1.0__ multiply__4.0__3.0__ divide__124.0__4.0__ subtract__4.0__2.0__ |
subtract__3.0__1.0__ add__4.0__1.0__ multiply__4.0__3.0__ divide__124.0__4.0__ subtract__4.0__2.0__ |
| the average distance between the sun and a certain planet is approximatly num__2.6 num__10 ^ num__9 inches . which of the following is the closest to the average distence between sun and the planet in kelometers ? ( num__1 km is approx num__3.9 x num__10 ^ num__4 inches ) <o> a ) num__7 x ( num__10 ) ^ num__5 <o> b ) num__7 x ( num__10 ) ^ num__7 <o> c ) num__6.7 x ( num__10 ) ^ num__4 <o> d ) num__7 x ( num__10 ) ^ num__8 <o> e ) num__7 x ( num__10 ) ^ num__12 |
no need to solve entire problem . distance in km = num__26 * ( num__10 ) ^ num__0.230769230769 * ( num__10 ) ^ num__4 which is - num__260 * ( num__10 ) ^ num__0.205128205128 * ( num__10 ) ^ num__4 = num__6.7 x ( num__10 ) ^ num__4 answer : c <eor> c <eos> |
c |
multiply__2.6__10.0__ multiply__10.0__26.0__ multiply__1.0__6.7__ |
multiply__2.6__10.0__ multiply__10.0__26.0__ multiply__1.0__6.7__ |
| a is faster than b . a and b each walk num__24 km . the sum of their speeds is num__7 km / hr and the sum of times taken by them is num__14 hours . then a ' s speed is equal to ? <o> a ) num__5 <o> b ) num__3 <o> c ) num__4 <o> d ) num__7 <o> e ) num__9 |
let a ' s speed = x km / hr . then b ' s speed = ( num__7 - x ) km / hr . so num__24 / x + num__24 / ( num__7 - x ) = num__14 x num__2 - num__98 x + num__168 = num__0 ( x - num__3 ) ( x - num__4 ) = num__0 = > x = num__3 or num__4 . since a is faster than b so a ' s speed = num__4 km / hr and b ' s speed = num__3 km / hr . answer : b <eor> b <eos> |
b |
divide__14.0__7.0__ multiply__7.0__14.0__ multiply__24.0__7.0__ subtract__7.0__3.0__ subtract__7.0__4.0__ |
divide__14.0__7.0__ multiply__7.0__14.0__ multiply__24.0__7.0__ subtract__7.0__3.0__ subtract__7.0__4.0__ |
| a train running at the speed of num__52 km / hr crosses a pole in num__9 sec . what is the length of the train ? <o> a ) num__130 m <o> b ) num__786 m <o> c ) num__566 m <o> d ) num__546 m <o> e ) num__440 m |
speed = num__52 * num__0.277777777778 = num__14.4444444444 m / sec length of the train = speed * time = num__14.4444444444 * num__9 = num__130 m answer : a <eor> a <eos> |
a |
round__130.0__ |
round__130.0__ |
| a brick measures num__20 cm * num__10 cm * num__7.5 cm how many bricks will be required for a wall num__25 m * num__2 m * num__0.75 m ? <o> a ) num__29798 <o> b ) num__27908 <o> c ) num__78902 <o> d ) num__25000 <o> e ) num__27991 |
num__25 * num__2 * num__0.75 = num__0.2 * num__0.1 * num__7.5 / num__100 * x num__25 = num__0.01 * x = > x = num__25000 answer : d <eor> d <eos> |
d |
divide__2.0__10.0__ divide__2.0__20.0__ divide__20.0__0.2__ divide__0.2__20.0__ round__25000.0__ |
divide__2.0__10.0__ divide__2.0__20.0__ divide__20.0__0.2__ divide__0.2__20.0__ round__25000.0__ |
| in x game of billiards x can give y num__20 points in num__60 and he can give z num__50 points in num__60 . how many points can y give z in x game of num__100 ? <o> a ) num__30 <o> b ) num__20 <o> c ) num__25 <o> d ) num__40 <o> e ) num__75 |
x scores num__60 while y score num__40 and z scores num__10 . the number of points that z scores when y scores num__100 = ( num__100 * num__50 ) / num__40 = num__25 . in x game of num__100 points y gives ( num__100 - num__25 ) = num__75 points to c . e <eor> e <eos> |
e |
subtract__60.0__20.0__ subtract__60.0__50.0__ add__50.0__25.0__ add__50.0__25.0__ |
subtract__60.0__20.0__ subtract__60.0__50.0__ subtract__100.0__25.0__ subtract__100.0__25.0__ |
| num__20 num__19 num__17 ? num__10 num__5 <o> a ) num__15 <o> b ) num__14 <o> c ) num__13 <o> d ) num__12 <o> e ) num__11 |
the pattern is - num__1 - num__2 - num__3 . . . answer : b . <eor> b <eos> |
b |
subtract__20.0__19.0__ divide__20.0__10.0__ subtract__20.0__17.0__ subtract__19.0__5.0__ |
subtract__20.0__19.0__ subtract__19.0__17.0__ subtract__20.0__17.0__ subtract__19.0__5.0__ |
| two pipes a and b can separately fill a tank in num__2 minutes and num__15 minutes respectively . both the pipes are opened together but num__4 minutes after the start the pipe a is turned off . how much time will it take to fill the tank ? <o> a ) num__15 <o> b ) num__10 <o> c ) num__81 <o> d ) num__71 <o> e ) num__12 |
num__0.333333333333 + x / num__15 = num__1 x = num__10 answer : b <eor> b <eos> |
b |
round__10.0__ |
divide__10.0__1.0__ |
| the average age of num__25 students in a class is num__12 years . if teacher ' s age is also included then average increases num__1 year then find the teacher ' s age ? <o> a ) num__40 <o> b ) num__35 <o> c ) num__29 <o> d ) num__38 <o> e ) num__42 |
total age of num__25 students = num__25 * num__12 = num__300 total age of num__26 persons = num__26 * num__13 = num__338 age of teacher = num__338 - num__300 = num__38 years answer is d <eor> d <eos> |
d |
multiply__25.0__12.0__ add__25.0__1.0__ subtract__25.0__12.0__ multiply__13.0__26.0__ add__25.0__13.0__ add__25.0__13.0__ |
multiply__25.0__12.0__ add__25.0__1.0__ subtract__25.0__12.0__ multiply__13.0__26.0__ subtract__338.0__300.0__ subtract__338.0__300.0__ |
| in a certain production lot num__40 percent of the toys are red and the remaining toys are green . half of toys are small and half are large . if num__10 percent of the toys are red and small and num__40 toys are green and large . how many of the toys are red and large ? <o> a ) num__40 <o> b ) num__60 <o> c ) num__70 <o> d ) num__80 <o> e ) num__90 |
num__40.0 toys are red . num__50.0 toys are small - num__50.0 are large red and small - num__10.0 red and large must be num__30.0 ( since there are total num__40.0 red toys ) but num__50.0 toys are large . if num__30.0 are red and large the rest num__20.0 large toys must be green . so large and green toys are num__20.0 . these num__20.0 large and green toys are actually num__40 so total number of toys must be num__200 . since red and large toys are num__30.0 they must be num__60 . answer : b <eor> b <eos> |
b |
percent__40.0__50.0__ percent__30.0__200.0__ percent__30.0__200.0__ |
percent__40.0__50.0__ percent__30.0__200.0__ percent__30.0__200.0__ |
| a clock is started at noon . by num__10 minutes past num__2 the hour hand has turned through how many degrees ? <o> a ) num__55 Â ° <o> b ) num__60 Â ° <o> c ) num__65 Â ° <o> d ) num__70 Â ° <o> e ) num__75 Â ° |
the angle traced by the hour hand in num__12 hrs is num__360 Â ° the angle traced by the hour hand in num__1 hour is num__30 Â ° the angle traced by the hour hand in num__10 minutes is num__5 Â ° the angle traced by the hour hand in num__2 hours and num__10 minutes is num__2 * num__30 Â ° + num__5 Â ° = num__65 Â ° the answer is c . <eor> c <eos> |
c |
add__10.0__2.0__ divide__360.0__12.0__ divide__10.0__2.0__ round__65.0__ |
add__10.0__2.0__ divide__360.0__12.0__ divide__10.0__2.0__ round__65.0__ |
| if g is the smallest positive integer such that num__3150 multiplied by g is the square of an integer then g must be <o> a ) num__2 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__14 |
solution : this problem is testing us on the rule that when we express a perfect square by its unique prime factors every prime factor ' s exponent is an even number . let ’ s start by prime factorizing num__3150 . num__3150 = num__315 x num__10 = num__5 x num__63 x num__10 = num__5 x num__7 x num__3 x num__3 x num__5 x num__2 num__3150 = num__2 ^ num__1 x num__3 ^ num__2 x num__5 ^ num__2 x num__7 ^ num__1 ( notice that the exponents of both num__2 and num__7 are not even numbers . this tells us that num__3150 itself is not a perfect square . ) we also are given that num__3150 multiplied by g is the square of an integer . we can write this as : num__2 ^ num__1 x num__3 ^ num__2 x num__5 ^ num__2 x num__7 ^ num__1 x g = square of an integer according to our rule we need all unique prime factors ' exponents to be even numbers . thus we need one more num__2 and one more num__7 . therefore g = num__7 x num__2 = num__14 answer is e . <eor> e <eos> |
e |
rectangle_perimeter__2.0__5.0__ multiply__1.0__14.0__ |
rectangle_perimeter__2.0__5.0__ power__14.0__1.0__ |
| a lent rs . num__5000 to b for num__2 years and rs . num__3000 to c for num__4 years on simple interest at the same rate of interest and received rs . num__2200 in all from both of them as interest . the rate of interest per annum is ? <o> a ) num__11 <o> b ) num__66 <o> c ) num__88 <o> d ) num__10 <o> e ) num__27 |
let the rate be r % p . a . then ( num__5000 * r * num__2 ) / num__100 + ( num__3000 * r * num__4 ) / num__100 = num__2200 num__100 r + num__120 r = num__2200 r = num__10.0 answer : d <eor> d <eos> |
d |
percent__2.0__5000.0__ percent__4.0__3000.0__ percent__100.0__10.0__ |
percent__2.0__5000.0__ percent__4.0__3000.0__ percent__100.0__10.0__ |
| how many kg of pure salt must be added to num__100 kg of num__10.0 solution of salt and water to increase it to a num__30.0 solution ? <o> a ) a ) num__6.7 <o> b ) b ) num__1.3 <o> c ) c ) num__9.6 <o> d ) d ) num__12.5 <o> e ) e ) num__28.6 |
amount salt in num__100 kg solution = num__10 * num__1.0 = num__10 kg let x kg of pure salt be added then ( num__10 + x ) / ( num__100 + x ) = num__0.3 num__100 + num__10 x = num__300 + num__3 x num__7 x = num__200 x = num__28.6 answer is e <eor> e <eos> |
e |
divide__30.0__100.0__ multiply__10.0__30.0__ multiply__10.0__0.3__ subtract__10.0__3.0__ subtract__300.0__100.0__ multiply__1.0__28.6__ |
divide__30.0__100.0__ multiply__10.0__30.0__ multiply__10.0__0.3__ subtract__10.0__3.0__ subtract__300.0__100.0__ multiply__1.0__28.6__ |
| a can do a piece of work in num__15 days and b in num__20 days . they began the work together but num__5 days before the completion of the work a leaves . the work was completed in ? <o> a ) num__8 days <o> b ) num__10 days <o> c ) num__15 days <o> d ) num__11 num__0.428571428571 days <o> e ) num__16 days |
explanation : ( x – num__5 ) / num__15 + x / num__20 = num__1 x = num__11 num__0.428571428571 days answer : d <eor> d <eos> |
d |
round__11.0__ |
divide__11.0__1.0__ |
| a watch was sold at a loss of num__10.0 . if it was sold for rs . num__140 more there would have been a gain of num__4.0 . what is the cost price ? <o> a ) rs : num__1000 <o> b ) rs : num__1067 <o> c ) rs : num__1278 <o> d ) rs : num__1028 <o> e ) rs : num__1027 |
num__90.0 num__104.0 - - - - - - - - num__14.0 - - - - num__140 num__100.0 - - - - ? = > rs : num__1000 answer : a <eor> a <eos> |
a |
percent__10.0__140.0__ percent__100.0__1000.0__ |
percent__10.0__140.0__ percent__100.0__1000.0__ |
| evaluate : num__13 + sqrt ( - num__4 + num__5 Ã — num__3 Ã · num__3 ) = ? <o> a ) num__12 <o> b ) num__13 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
according to order of operations inner brackets first where num__5 x num__3 Ã · num__3 is first calculated since it has a multiplication and a division . num__5 Ã — num__3 Ã · num__3 = num__15 Ã · num__3 = num__5 hence num__13 + sqrt ( - num__4 + num__5 Ã — num__3 Ã · num__3 ) = num__13 + sqrt ( - num__4 + num__5 ) = num__13 + sqrt ( num__1 ) = num__13 + num__1 = num__14 correct answer c ) num__14 <eor> c <eos> |
c |
multiply__5.0__3.0__ subtract__4.0__3.0__ add__13.0__1.0__ add__13.0__1.0__ |
multiply__5.0__3.0__ subtract__4.0__3.0__ add__13.0__1.0__ add__13.0__1.0__ |
| if a tire rotates at num__400 revolutions per minute when the car is traveling num__120 km / h what is the circumference of the tire ? <o> a ) num__2 <o> b ) num__1 <o> c ) num__4 <o> d ) num__3 <o> e ) num__5 |
num__400 rev / minute = num__400 * num__60 rev / num__60 minutes = num__24000 rev / hour num__24000 * c = num__120000 m : c is the circumference c = num__5 meters correct answer e <eor> e <eos> |
e |
hour_to_min_conversion__ multiply__400.0__60.0__ divide__120000.0__24000.0__ round__5.0__ |
hour_to_min_conversion__ multiply__400.0__60.0__ divide__120000.0__24000.0__ divide__120000.0__24000.0__ |
| how many num__5 - digit positive integers exist the sum of whose digits are odd ? <o> a ) num__36000 <o> b ) num__38000 <o> c ) num__45000 <o> d ) num__90000 <o> e ) none |
solution : there are num__9 � num__104 = num__90000 num__5 - digit positive integers . out of these num__90000 positive integers the sum of the digits of half of the numbers will add up to an odd number and the remaining half will add up to an even number . hence there are num__45000.0 = num__45000 num__5 - digit positive integers whose sum add up to an odd number . answer : option c <eor> c <eos> |
c |
subtract__90000.0__45000.0__ |
subtract__90000.0__45000.0__ |
| machines a and b produce bottles at their respective constant rates . machine a produces k bottles in num__3 hours and machine b produces k bottles in num__5 hours . how many hours does it take machines a and b working simultaneously to produce num__16 k bottles ? <o> a ) num__24 <o> b ) num__26 <o> c ) num__28 <o> d ) num__30 <o> e ) num__32 |
a ' s rate = k / num__3 b ' s rate = k / num__5 k / num__3 + k / num__5 = num__8 k / num__15 num__16 k / ( num__8 k / num__15 ) = num__30 hours the answer is d . <eor> d <eos> |
d |
add__3.0__5.0__ multiply__3.0__5.0__ round__30.0__ |
add__3.0__5.0__ multiply__3.0__5.0__ round__30.0__ |
| a certain hall contains two cuckoo clocks . if the first clock chirps num__20 times per hour and the second clock chirps num__9 times per hour and both clocks chirp for the first time at num__2 : num__03 pm at what time will the first clock have chirped three times as many times as the second clock can chirp per hour ? <o> a ) num__2 : num__42 pm <o> b ) num__2 : num__50 pm <o> c ) num__3 : num__07 pm <o> d ) num__3 : num__21 pm <o> e ) num__3 : num__30 pm |
the question is asking when the first clock will have chirped num__27 times ( num__3 * num__9 chirps ) . the first clock chirps once every num__3 minutes . if the first chirp is at num__2 : num__03 then the num__27 th chirp is at num__81 minutes after num__2 : num__00 . the answer is d . <eor> d <eos> |
d |
multiply__9.0__3.0__ multiply__3.0__27.0__ round__3.0__ |
multiply__9.0__3.0__ multiply__3.0__27.0__ round__3.0__ |
| a jogger running at num__9 km / hr along side a railway track is num__200 m ahead of the engine of a num__120 m long train running at num__45 km / hr in the same direction . in how much time will the train pass the jogger ? <o> a ) num__32 <o> b ) num__20 <o> c ) num__36 <o> d ) num__88 <o> e ) num__34 |
speed of train relative to jogger = num__45 - num__9 = num__36 km / hr . = num__36 * num__0.277777777778 = num__10 m / sec . distance to be covered = num__200 + num__120 = num__320 m . time taken = num__32.0 = num__32 sec . answer : option a <eor> a <eos> |
a |
subtract__45.0__9.0__ add__200.0__120.0__ divide__320.0__10.0__ divide__320.0__10.0__ |
subtract__45.0__9.0__ add__200.0__120.0__ divide__320.0__10.0__ divide__320.0__10.0__ |
| if sn is the sum of the first n terms of a certain sequence and if sn = n ( n + num__6 ) for all positive integers n what is the third term of the sequence ? <o> a ) num__10 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
tn = nth term t num__1 = num__1 st term t num__2 = num__2 nd term and so on . . . sn = sum of first n terms of the sequence sn = num__1 st term + num__2 nd term + num__3 rd term + . . . + nth term sn = t num__1 + t num__2 + t num__3 + . . . . tn you are given here thatsn is the sum of first n terms . . . so you have sn = t num__1 + t num__2 + t num__3 + . . . . tn = n ( n + num__1 ) so s num__1 = t num__1 s num__2 = t num__1 + t num__2 s num__3 = t num__1 + t num__2 + t num__3 and so on s num__1 = t num__1 = num__1 * ( num__1 + num__1 ) = num__2 s num__2 = t num__1 + t num__2 = num__2 + t num__2 = num__2 * ( num__2 + num__1 ) = num__6 so t num__2 = num__4 s num__3 = t num__1 + t num__2 + t num__3 = num__2 + num__4 + t num__3 = num__3 * ( num__3 + num__1 ) = num__12 so t num__3 = num__10 ( this is what we wanted ) the third term is num__10 . a <eor> a <eos> |
a |
divide__6.0__2.0__ subtract__6.0__2.0__ multiply__6.0__2.0__ add__6.0__4.0__ add__6.0__4.0__ |
add__1.0__2.0__ add__1.0__3.0__ multiply__6.0__2.0__ add__6.0__4.0__ add__6.0__4.0__ |
| a train moves fast a telegraph post and a bridge num__264 m long in num__8 sec and num__20 sec respectively . what is the speed of the train ? <o> a ) num__45 <o> b ) num__46 <o> c ) num__47 <o> d ) num__79.2 <o> e ) num__76 |
let the length of the train be x m and its speed be y m / sec . then x / y = num__8 = > x = num__8 y ( x + num__264 ) / num__20 = y y = num__22 speed = num__22 m / sec = num__22 * num__3.6 = num__79.2 km / hr . answer : option d <eor> d <eos> |
d |
multiply__3.6__22.0__ round__79.2__ |
multiply__3.6__22.0__ multiply__3.6__22.0__ |
| a train is num__360 meter long is running at a speed of num__48 km / hour . in what time will it pass a bridge of num__140 meter length . <o> a ) num__27.5 seconds <o> b ) num__37.5 seconds <o> c ) num__47.5 seconds <o> d ) num__57.5 seconds <o> e ) none of these |
explanation : speed = num__48 km / hr = num__48 * ( num__0.277777777778 ) m / sec = num__13.3333333333 m / sec total distance = num__360 + num__140 = num__500 meter time = distance / speed = num__12.5 / num__3 = num__37.5 seconds answer : b <eor> b <eos> |
b |
add__360.0__140.0__ multiply__12.5__3.0__ round__37.5__ |
add__360.0__140.0__ multiply__12.5__3.0__ multiply__12.5__3.0__ |
| num__20 litres of water are poured into an aquarium of dimensions num__50 cm length num__20 cm breadth and num__40 cm height . how high ( in cm ) will the water rise ? ( num__1 litre = num__1000 cm ³ ) <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__20 <o> e ) num__40 |
lxbxh = num__12000 h = num__400.0 * num__20 = num__20 cm ' d ' is the answer . <eor> d <eos> |
d |
round__20.0__ |
round__20.0__ |
| a cistern num__8 m long and num__4 m wide contains water up to a depth of num__1 m num__25 cm . the total area of the wet surface is : <o> a ) num__62 m num__2 <o> b ) num__50 m num__2 <o> c ) num__53.5 m num__2 <o> d ) num__55 m num__2 <o> e ) num__57 m num__2 |
area of the wet surface = [ num__2 ( lb + bh + lh ) - lb ] = num__2 ( bh + lh ) + lb = [ num__2 ( num__4 x num__1.25 + num__8 x num__1.25 ) + num__8 x num__4 ] m num__2 = num__62 m num__2 . answer : option a <eor> a <eos> |
a |
divide__8.0__4.0__ round__62.0__ |
divide__8.0__4.0__ round__62.0__ |
| arrange the words given below in a meaningful sequence . num__1 . poverty num__2 . population num__3 . death num__4 . unemployment num__5 . disease <o> a ) num__2 num__4 num__1 num__5 num__8 <o> b ) num__2 num__4 num__1 num__5 num__7 <o> c ) num__2 num__4 num__1 num__5 num__3 <o> d ) num__2 num__4 num__1 num__5 num__9 <o> e ) num__2 num__4 num__7 num__5 num__3 |
answer : c ) num__2 num__4 num__1 num__5 num__3 explanation : the correct order is : population unemployment poverty disease death num__2 num__4 num__1 num__5 num__3 answer : c <eor> c <eos> |
c |
multiply__1.0__2.0__ |
multiply__1.0__2.0__ |
| what is the least number which when doubled will be exactly divisible by num__12 num__14 num__18 and num__22 ? <o> a ) num__1286 <o> b ) num__1436 <o> c ) num__1216 <o> d ) num__1386 <o> e ) num__1298 |
lcm of num__12 num__14 num__18 and num__22 = num__2772 hence the least number which will be exactly divisible by num__12 num__14 num__18 and num__22 = num__2772 num__2772 ÷ num__2 = num__1386 num__1386 is the number which when doubled we get num__2772 answer : option d hence num__1386 is the least number which when doubled will be exactly divisible by num__12 num__14 num__18 and num__22 . <eor> d <eos> |
d |
subtract__14.0__12.0__ divide__2772.0__2.0__ divide__2772.0__2.0__ |
subtract__14.0__12.0__ divide__2772.0__2.0__ divide__2772.0__2.0__ |
| a space shuttle orbits the earth at about num__9 kilometers per second . this speed is equal to how many kilometers per hour ? <o> a ) num__480 <o> b ) num__2880 <o> c ) num__4800 <o> d ) num__28800 <o> e ) num__32 |
400 |
seconds in num__1 hours : num__60 s in num__1 min num__60 min in num__1 hr num__60 * num__60 = num__3600 sec in num__1 hr num__9 * num__3600 = num__32400 answer : e <eor> e <eos> |
e |
e |
| average age of num__18 men is decreased by num__1 year when one man whose age is num__49 years is replaced by a new man . find age of the new man <o> a ) num__31 <o> b ) num__77 <o> c ) num__88 <o> d ) num__29 <o> e ) num__22 |
explanation : if the replaced persons age is same with the existing average there is no change in the average . but by replacement overall decrease in the age is num__18 × num__1 = num__18 years . this is the change bought by the new man . age of new man = age of man replaced - total decrease in age = num__49 - ( num__1 × num__18 ) = num__31 years \ answer : a <eor> a <eos> |
a |
subtract__49.0__18.0__ multiply__1.0__31.0__ |
subtract__49.0__18.0__ subtract__49.0__18.0__ |
| if a : b = num__2 : num__5 b : c = num__3 : num__7 and c : d = num__1 : num__3 find a : b : c : d ? <o> a ) num__6 : num__15 : num__25 : num__115 <o> b ) num__6 : num__15 : num__35 : num__115 <o> c ) num__6 : num__15 : num__35 : num__105 <o> d ) num__6 : num__15 : num__35 : num__125 <o> e ) none of these |
explanation : a : b = num__2 : num__5 b : c = num__3 : num__7 c : d = num__1 : num__3 num__2 : num__5 num__3 : num__7 ( a = num__2 × num__3 = num__6 b = num__5 × num__3 = num__15 and c = num__5 × num__7 = num__35 ) ( a = a × b b = b × b and c = b × c ) a : b : c = num__6 : num__15 : num__35 a : b : c = num__6 : num__15 : num__35 and c : d = num__1 : num__3 ( note : first a b c multiplication with c means num__1 and last c means num__35 multiplication with d means num__3 a : b : c : d = num__6 : num__15 : num__35 : num__105 answer : option c <eor> c <eos> |
c |
multiply__2.0__3.0__ multiply__5.0__3.0__ multiply__5.0__7.0__ multiply__3.0__35.0__ multiply__2.0__3.0__ |
multiply__2.0__3.0__ multiply__5.0__3.0__ multiply__5.0__7.0__ multiply__3.0__35.0__ multiply__2.0__3.0__ |
| how many num__5 digit number contain number num__3 ? <o> a ) num__31512 <o> b ) num__32512 <o> c ) num__33512 <o> d ) num__34512 <o> e ) num__37512 |
total num__5 digit no . = num__9 * num__10 * num__10 * num__10 * num__10 = num__90000 not containing num__3 = num__8 * num__9 * num__9 * num__9 * num__9 = num__52488 total num__5 digit number contain num__3 = num__90000 - num__52488 = num__37512 answer : e <eor> e <eos> |
e |
add__5.0__3.0__ subtract__90000.0__52488.0__ subtract__90000.0__52488.0__ |
add__5.0__3.0__ subtract__90000.0__52488.0__ subtract__90000.0__52488.0__ |
| which of the following is equal to num__1 ( num__0.142857142857 ) % ? <o> a ) num__0.012 / num__100 <o> b ) num__0.12 / num__100 <o> c ) num__1.14 / num__100 <o> d ) num__0.12 <o> e ) num__1.2 |
this notation may be confusing for some since it looks like we ' re multiplying num__1 and num__0.142857142857 how about adding a space : which of the following is equal to ( num__1 num__0.142857142857 ) % ( num__1 num__0.142857142857 ) % = num__1.14 = num__1.14 / num__100 answer : c <eor> c <eos> |
c |
multiply__1.0__1.14__ |
divide__1.14__1.0__ |
| incomes of two companies a and b are in the ratio of num__5 : num__8 . had the income of company a been more by num__30 lakh the ratio of their incomes would have been num__5 : num__4 . what is the income of company b ? <o> a ) num__80 lakh <o> b ) num__50 lakh <o> c ) num__48 lakh <o> d ) num__60 lakh <o> e ) none of these |
let the incomes of two companies a and b be num__5 x and num__8 x respectively . from the question num__5 x + num__3.75 x = num__1.25 ⇒ num__20 x + num__120 = num__40 x ∴ x = num__6 ∴ income of company b = num__8 x = num__48 lakh answer c <eor> c <eos> |
c |
divide__30.0__8.0__ divide__5.0__4.0__ multiply__5.0__4.0__ multiply__30.0__4.0__ multiply__5.0__8.0__ divide__30.0__5.0__ multiply__8.0__6.0__ multiply__8.0__6.0__ |
divide__30.0__8.0__ divide__5.0__4.0__ multiply__5.0__4.0__ multiply__30.0__4.0__ multiply__5.0__8.0__ divide__30.0__5.0__ add__8.0__40.0__ add__8.0__40.0__ |
| how many positive integers less than num__5000 are there in which the sum of the digits equals num__5 ? <o> a ) num__64 <o> b ) num__62 <o> c ) num__63 <o> d ) num__61 <o> e ) num__56 |
basically the question asks how many num__4 digit numbers ( including those in the form num__0 xxx num__00 xx and num__000 x ) have digits which add up to num__5 . think about the question this way : we know that there is a total of num__5 to be spread among the num__4 digits we just have to determine the number of ways it can be spread . let x represent a sum of num__1 and | represent a seperator between two digits . as a result we will have num__5 x ' s ( digits add up to the num__5 ) and num__3 | ' s ( num__3 digit seperators ) . so for example : xx | x | x | x = num__2111 | | xxx | xx = num__0032 etc . there are num__8 c num__3 ways to determine where to place the separators . hence the answer is num__8 c num__3 = num__56 . e <eor> e <eos> |
e |
subtract__5.0__4.0__ subtract__4.0__1.0__ add__5.0__3.0__ multiply__1.0__56.0__ |
subtract__5.0__4.0__ subtract__4.0__1.0__ add__5.0__3.0__ multiply__1.0__56.0__ |
| in each series look for the degree and direction of change between the numbers . in other words do the numbers increase or decrease and by how much ? look at this series : num__95 num__10 num__75 num__20 num__55 . . . what number should come next ? <o> a ) num__35 <o> b ) num__40 <o> c ) num__30 <o> d ) num__25 <o> e ) num__90 |
c num__30 this is an alternating addition and subtraction series . in the first pattern num__20 is subtracted from each number to arrive at the next . in the second num__10 is added to each number to arrive at the next . <eor> c <eos> |
c |
add__10.0__20.0__ add__10.0__20.0__ |
add__10.0__20.0__ add__10.0__20.0__ |
| ben bought num__2 q steaks for w dollars . jerome buys r steaks for a num__50.0 discount how much will the steaks cost him in cents ? <o> a ) num__50 rw / q . <o> b ) num__50 qr / w . <o> c ) num__25 rq / w . <o> d ) num__25 rw / q . <o> e ) rw / ( num__4 q ) . |
ben bought num__2 q steaks for w dollars so num__1 steak = w / num__2 q jerome buys r steaks for a num__50.0 discount : r * ( w / num__4 q ) in cents the answer will be : r * ( num__100 w / num__4 q ) = num__25 rw / q = d <eor> d <eos> |
d |
percent__2.0__50.0__ percent__25.0__100.0__ |
percent__2.0__50.0__ percent__25.0__100.0__ |
| calculate the profit loss percentage if a trader bought num__7 balloons per rupee and sells the balloons at num__6 per rupee . <o> a ) num__13.7 <o> b ) num__12.7 <o> c ) num__15.3 <o> d ) num__16.7 <o> e ) num__13.7 % |
let he total number of oranges bought by the shopkeeper be num__12 . if he buys num__7 a rupee his cp = num__6 he selling at num__6 a rupee his sp = num__7 profit = sp - cp = num__7 - num__6 = num__1 profit percent = num__0.166666666667 * num__100 = num__16.7 answer : d <eor> d <eos> |
d |
percent__100.0__16.7__ |
percent__100.0__16.7__ |
| the food in a camp lasts for num__30 men for num__40 days . if ten more men join how many days will the food last ? <o> a ) num__22 days <o> b ) num__30 days <o> c ) num__87 days <o> d ) num__16 days <o> e ) num__17 days |
one man can consume the same food in num__30 * num__40 = num__1200 days . num__10 more men join the total number of men = num__40 the number of days the food will last = num__30.0 = num__30 days . answer : b <eor> b <eos> |
b |
multiply__30.0__40.0__ subtract__40.0__30.0__ round__30.0__ |
multiply__30.0__40.0__ subtract__40.0__30.0__ round__30.0__ |
| how much num__60.0 of num__50 is greater than num__40.0 of num__30 ? <o> a ) num__18 <o> b ) num__28 <o> c ) num__29 <o> d ) num__11 <o> e ) num__12 |
( num__0.6 ) * num__50 – ( num__0.4 ) * num__30 num__30 - num__12 = num__18 answer : a <eor> a <eos> |
a |
percent__40.0__30.0__ percent__60.0__30.0__ percent__60.0__30.0__ |
percent__40.0__30.0__ percent__60.0__30.0__ percent__60.0__30.0__ |
| a list of measurements in increasing order is num__4 num__5 num__6 num__8 num__10 and x . if the median of these measurements is num__0.666666666667 times their arithmetic mean what is the value of x ? <o> a ) num__24 <o> b ) num__26 <o> c ) num__28 <o> d ) num__30 <o> e ) num__32 |
the median is ( num__6 + num__8 ) / num__2 = num__7 the mean is ( num__4 + num__5 + num__6 + num__8 + num__10 + x ) / num__6 = ( num__33 + x ) / num__6 ( num__33 + x ) / num__6 * num__0.666666666667 = num__7 x = num__63 - num__33 = num__30 the answer is d . <eor> d <eos> |
d |
subtract__6.0__4.0__ add__5.0__2.0__ multiply__5.0__6.0__ multiply__5.0__6.0__ |
subtract__6.0__4.0__ add__5.0__2.0__ multiply__5.0__6.0__ multiply__5.0__6.0__ |
| tough and tricky questions : absolute values . if | a | = num__0.2 and | b | = num__0.4 which of the following can not be the result of a + b ? <o> a ) num__0.6 <o> b ) - num__0.6 <o> c ) - num__0.2 <o> d ) num__0.666666666667 <o> e ) num__0.2 |
| a | = num__0.2 absolute value of ' a ' can have two values = num__0.2 and - ( num__0.2 ) | b | = num__0.4 absolute value of ' b ' can have two values = num__0.4 and - ( num__0.4 ) now different combinations of a + b are as follows : a + b = ( num__0.2 ) + ( num__0.4 ) = num__0.6 - a - b = - ( num__0.2 ) - ( num__0.4 ) = - num__0.6 a - b = ( num__0.2 ) - ( num__0.4 ) = - ( num__0.2 ) - a + b = - ( num__0.2 ) + ( num__0.4 ) = num__0.2 cross verifying with the given options left over option is d . <eor> d <eos> |
d |
add__0.2__0.4__ divide__0.4__0.6__ |
add__0.2__0.4__ divide__0.4__0.6__ |
| a can do a half of certain work in num__70 days and b one third of the same in num__35 days . they together will do the whole work in . <o> a ) num__77 days <o> b ) num__55 days <o> c ) num__44 days <o> d ) num__60 days <o> e ) num__33 days |
a = num__140 days b = num__105 days num__0.00714285714286 + num__0.00952380952381 = num__0.0166666666667 = num__0.0166666666667 = > num__60 days answer : d <eor> d <eos> |
d |
add__70.0__35.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
add__70.0__35.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
| x does a work in num__15 days . y does the same work in num__30 days . in how many days they together will do the same work ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__20 <o> d ) num__30 <o> e ) num__15 |
x ' s num__1 day ' s work = num__0.0666666666667 y ' s num__1 day ' s work = num__0.0333333333333 ( x + y ) ' s num__1 day ' s work = ( num__0.0666666666667 + num__0.0333333333333 ) = num__0.1 both together will finish the work in num__10 days . correct option is a <eor> a <eos> |
a |
divide__1.0__15.0__ divide__1.0__30.0__ add__0.0667__0.0333__ divide__1.0__0.1__ round__10.0__ |
divide__1.0__15.0__ divide__1.0__30.0__ add__0.0667__0.0333__ divide__1.0__0.1__ round__10.0__ |
| eight people are planning to share equally the cost of a rental car . if one person withdraws from the arrangement and the others share equally the entire cost of the car then the share of each of the remaining persons increased by : <o> a ) num__0.875 <o> b ) num__0.125 <o> c ) num__0.333333333333 <o> d ) num__0.142857142857 <o> e ) num__0.111111111111 |
original share of num__1 person = num__0.125 new share of num__1 person = num__0.142857142857 increase = num__0.142857142857 - num__0.125 = num__0.0178571428571 required fraction = ( num__0.0178571428571 ) / ( num__0.125 ) = ( num__0.0178571428571 ) * ( num__8.0 ) = num__0.142857142857 answer is d . <eor> d <eos> |
d |
multiply__0.125__0.1429__ reverse__0.125__ add__0.125__0.0179__ |
multiply__0.125__0.1429__ reverse__0.125__ divide__0.1429__1.0__ |
| a car traveling at a certain constant speed takes num__2 seconds longer to travel num__1 kilometer than it would take to travel num__1 kilometer at num__60 kilometers per hour . at what speed in kilometers per hour is the car traveling ? <o> a ) num__59 <o> b ) num__57 <o> c ) num__59.5 <o> d ) num__58 <o> e ) num__55.5 |
d num__75 * t = num__1 km = > t = num__0.0166666666667 km / h v * ( t + num__0.000555555555556 ) = num__1 v ( num__0.0166666666667 + num__0.000555555555556 ) = num__1 = > v = num__58 km / h <eor> d <eos> |
d |
divide__1.0__60.0__ subtract__60.0__2.0__ round__58.0__ |
divide__1.0__60.0__ subtract__60.0__2.0__ multiply__1.0__58.0__ |
| a and b complete a work in num__7 days . a alone can do it in num__14 days . if both together can do the work in how many days ? <o> a ) num__4.6 days <o> b ) num__4.78 days <o> c ) num__5.65 days <o> d ) num__3.77 days <o> e ) num__5.75 days |
num__0.142857142857 + num__0.0714285714286 = num__0.214285714286 num__4.66666666667 = num__4.6 days answer : a <eor> a <eos> |
a |
add__0.0714__0.1429__ round__4.6__ |
add__0.0714__0.1429__ round__4.6__ |
| if a * b * c = ( √ ( a + num__2 ) ( b + num__3 ) ) / ( c + num__1 ) find the value of num__6 * num__15 * num__5 . <o> a ) num__2 <o> b ) num__5 <o> c ) num__11 <o> d ) num__3 <o> e ) num__4 |
num__6 * num__15 * num__5 = ( √ ( num__6 + num__2 ) ( num__15 + num__3 ) ) / ( num__5 + num__1 ) = ( √ num__8 * num__18 ) / num__6 = ( √ num__144 ) / num__6 = num__2.0 = num__2 answer is a <eor> a <eos> |
a |
add__2.0__6.0__ multiply__3.0__6.0__ multiply__8.0__18.0__ multiply__2.0__1.0__ |
add__2.0__6.0__ multiply__3.0__6.0__ multiply__8.0__18.0__ multiply__2.0__1.0__ |
| the watch store down the street from wendy ' s house has num__2 leather watches num__1 gold watch and num__1 silver watch . how many watches does the store have in all ? <o> a ) num__10 <o> b ) num__19 <o> c ) num__04 <o> d ) num__12 <o> e ) num__07 |
add the numbers of watches . num__2 + num__1 + num__1 = num__4 . answer is c . <eor> c <eos> |
c |
multiply__1.0__4.0__ |
multiply__1.0__4.0__ |
| a man bought a bike at num__40.0 discount on its original price . he sold it at a num__80.0 increase on the price he bought it . what percent of profit did he make on the original price ? <o> a ) num__2.0 <o> b ) num__8.0 <o> c ) num__6.0 <o> d ) num__4.0 <o> e ) num__10 % |
original price = num__100 cp = num__60 s = num__60 * ( num__1.8 ) = num__108 num__100 - num__108 = num__8.0 answer : b <eor> b <eos> |
b |
percent__8.0__100.0__ |
percent__8.0__100.0__ |
| two isosceles triangles have equal vertical angles and their areas are in the ratio num__4 : num__9 . find the ratio of their corresponding heights . <o> a ) num__0.8 <o> b ) num__1.25 <o> c ) num__1.5 <o> d ) num__0.714285714286 <o> e ) num__0.666666666667 |
we are basically given that the triangles are similar . in two similar triangles the ratio of their areas is the square of the ratio of their sides and also the square of the ratio of their corresponding heights . therefore area / area = height ^ num__2 / height ^ num__2 = num__0.444444444444 - - > height / height = num__0.666666666667 . answer : e . <eor> e <eos> |
e |
triangle_area__2.0__0.6667__ |
triangle_area__2.0__0.6667__ |
| if the graph of y = f ( x ) is transformed into the graph of num__2 y - num__6 = - num__4 f ( x - num__3 ) point ( a b ) on the graph of y = f ( x ) becomes point ( a b ) on the graph of num__2 y - num__6 = - num__4 f ( x - num__3 ) where a and b are given by <o> a ) a = a - num__3 b = b <o> b ) a = a - num__3 b = b num__2 <o> c ) a = a + num__3 b = - num__2 b <o> d ) a = a + num__3 b = - num__2 b + num__3 <o> e ) none |
solution we first solve num__2 y - num__6 = - num__4 f ( x - num__3 ) for y . y = - num__2 f ( x - num__3 ) + num__3 the graph of y = - num__2 f ( x - num__3 ) + num__3 is that of y = f ( x ) shifted num__3 units to the right stretched vertically by a factor of num__2 reflected on the x axis and shifted up by num__3 units . a point of y = f ( x ) will undergo the same transforamtions . hence point ( a b ) on the graph of y = f ( x ) becomes ( a + num__3 b ) on the graph of y = f ( x - num__3 ) : shifted num__3 units to the right becomes ( a + num__3 num__2 b ) on the graph of y = num__2 f ( x - num__3 ) : stretched vertically by num__2 becomes ( a + num__3 - num__2 b ) on the graph of y = - num__2 f ( x - num__3 ) : reflected on x axis becomes ( a + num__3 - num__2 b + num__3 ) on the graph of y = - num__2 f ( x - num__3 ) + num__3 : shifted up num__3 units answer d <eor> d <eos> |
d |
divide__6.0__2.0__ |
subtract__6.0__3.0__ |
| a num__1200 m long train crosses a tree in num__120 sec how much time will i take to pass a platform num__1200 m long ? <o> a ) num__200 sec <o> b ) num__240 sec <o> c ) num__167 sec <o> d ) num__197 sec <o> e ) num__179 sec |
l = s * t s = num__10.0 s = num__10 m / sec . total length ( d ) = num__2400 m t = d / s t = num__240.0 t = num__240 sec answer : b <eor> b <eos> |
b |
divide__1200.0__120.0__ divide__2400.0__10.0__ round__240.0__ |
divide__1200.0__120.0__ divide__2400.0__10.0__ divide__2400.0__10.0__ |
| if x is a sum of all even integers on the interval num__13 . . . num__61 and y is their number what is the gcd ( x y ) ? <o> a ) num__1 <o> b ) num__13 <o> c ) num__26 <o> d ) num__24 <o> e ) num__1014 |
x = num__14 + num__16 + . . . + num__60 = ( largest + smallest ) / num__2 * ( # of terms ) = ( num__14 + num__60 ) / num__2 * num__24 = num__37 * num__24 . gcd of num__24 and num__37 * num__22 is num__24 . answer : d . <eor> d <eos> |
d |
subtract__16.0__14.0__ add__13.0__24.0__ subtract__24.0__2.0__ subtract__61.0__37.0__ |
subtract__16.0__14.0__ add__13.0__24.0__ subtract__24.0__2.0__ add__2.0__22.0__ |
| during a trip on an expressway jon drove a total of x miles . his average speed on a certain num__5 - mile section of the expressway was num__30 miles per hour and his average speed for the remainder of the trip was num__60 miles per hour . his travel time for the x - mile trip was what percent greater than it would have been if he had traveled at a constant rate of num__60 miles per hour for the entire trip ? <o> a ) num__8.5 <o> b ) num__50.0 <o> c ) x / num__12.0 <o> d ) num__60 / x % <o> e ) num__500 / x % |
so jon drove ( x - num__5 ) miles at num__60 mph and num__5 miles at num__30 mph : ( x - num__5 ) / num__60 + ( num__0.166666666667 ) = ( x + num__5 ) / num__60 - current time to drive the whole distance if the entire trip @ num__60 mph than time = x / num__60 so to get percent [ ( x + num__5 ) / num__60 ] / [ x / num__60 ] x num__100 = ( x + num__5 ) / x * num__100 = num__100 + num__500 / x so increase of num__500 / x % [ / u ] = e <eor> e <eos> |
e |
divide__5.0__30.0__ multiply__5.0__100.0__ multiply__5.0__100.0__ |
divide__5.0__30.0__ multiply__5.0__100.0__ multiply__5.0__100.0__ |
| tickets numbered num__1 to num__10 are mixed up and then a ticket is drawn at random . what is the probability that the ticket drawn bears a number which is a multiple of num__2 ? <o> a ) num__0.333333333333 <o> b ) num__0.4 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__0.25 |
here s = { num__1 num__23 . . . . . . num__910 } e = event getting a multiple of num__3 = { num__2 num__46 num__810 } probability = num__0.5 = num__0.5 correct option is c <eor> c <eos> |
c |
add__1.0__2.0__ multiply__2.0__23.0__ reverse__2.0__ reverse__2.0__ |
add__1.0__2.0__ multiply__2.0__23.0__ reverse__2.0__ reverse__2.0__ |
| what is the dividend . divisor num__17 the quotient is num__9 and the remainder is num__5 ? <o> a ) num__150 <o> b ) num__154 <o> c ) num__158 <o> d ) num__160 <o> e ) num__164 |
d = d * q + r d = num__17 * num__9 + num__5 d = num__153 + num__5 d = num__158 c ) <eor> c <eos> |
c |
multiply__17.0__9.0__ add__5.0__153.0__ add__5.0__153.0__ |
multiply__17.0__9.0__ add__5.0__153.0__ add__5.0__153.0__ |
| in the quadrilateral pqrs d = num__10 cm h num__1 = num__5 cm and h num__2 = num__2.5 cm . find the area <o> a ) num__21 <o> b ) num__22 <o> c ) num__37.5 <o> d ) num__24 <o> e ) num__25 |
area of quad . = num__0.5 * any diagonal * ( sum of num__2 perpendiculars which is drawn on that diagona ) so num__0.5 * num__10 * ( num__5 + num__2.5 ) = num__37.5 answer : c <eor> c <eos> |
c |
divide__1.0__2.0__ round__37.5__ |
divide__1.0__2.0__ multiply__1.0__37.5__ |
| if ( c - a ) / ( c - b ) = num__1 then ( num__5 b - num__4 a ) / ( c - a ) = <o> a ) num__0.5 <o> b ) num__1 <o> c ) num__1.5 <o> d ) num__2 <o> e ) num__2.5 |
let ' s say c = num__3 b = num__1 a = num__1 so that our num__1 st expression holds true . now ibsert those numbers in the second expression and we ' ll get num__0.5 answer a ( hopefully ) ) ) <eor> a <eos> |
a |
subtract__4.0__1.0__ subtract__1.0__0.5__ |
subtract__4.0__1.0__ subtract__1.0__0.5__ |
| if n denotes a number to the left of num__0 on the number line such that the square of n is less than num__0.111111111111 then the reciprocal of n must be <o> a ) less than - num__3 <o> b ) between - num__1 and - num__0.333333333333 <o> c ) between - num__0.333333333333 and num__0 <o> d ) between num__0 and num__0.333333333333 <o> e ) greater than num__3 |
n ^ num__2 < num__0.111111111111 abs ( n ) < num__0.333333333333 since n < num__0 then num__0 > n > - num__0.333333333333 num__1 / n < - num__3 the answer is a . <eor> a <eos> |
a |
add__1.0__2.0__ add__1.0__2.0__ |
divide__0.3333__0.1111__ divide__3.0__1.0__ |
| a man engaged a servant on the condition that he would pay him rs . num__900 and a uniform after num__1 year service . he served only for num__9 months and receiveduniform and rs . num__650 find the price of the uniform ? <o> a ) rs . num__90 <o> b ) rs . num__100 <o> c ) rs . num__130 <o> d ) rs . num__170 <o> e ) rs . num__190 |
num__0.75 = num__0.75 * num__900 = num__675 num__650 - - - - - - - - - - - - - num__25 num__0.25 - - - - - - - - num__25 num__1 - - - - - - - - - ? = > rs . num__100 b <eor> b <eos> |
b |
multiply__900.0__0.75__ subtract__675.0__650.0__ subtract__1.0__0.75__ divide__900.0__9.0__ divide__900.0__9.0__ |
multiply__900.0__0.75__ subtract__675.0__650.0__ subtract__1.0__0.75__ divide__900.0__9.0__ multiply__1.0__100.0__ |
| a completes num__80.0 of a work in num__20 days . then b also joins and a and b together finish the remaining work in num__3 days . how long does it need for b if he alone completes the work ? <o> a ) num__37 ½ days <o> b ) num__22 days <o> c ) num__31 days <o> d ) num__20 days <o> e ) num__24 days |
explanation : work done by a in num__20 days = num__0.8 = num__0.8 = num__0.8 work done by a in num__1 day = ( num__0.8 ) / num__20 = num__0.04 = num__0.04 - - - ( num__1 ) work done by a and b in num__3 days = num__0.2 = num__0.2 ( because remaining num__20.0 is done in num__3 days by a and b ) work done by a and b in num__1 day = num__0.0666666666667 - - - ( num__2 ) work done by b in num__1 day = num__0.0666666666667 – num__0.04 = num__0.0266666666667 = > b can complete the work in num__37.5 days = num__37 ½ days answer : option a <eor> a <eos> |
a |
divide__0.8__20.0__ subtract__1.0__0.8__ divide__0.2__3.0__ subtract__3.0__1.0__ subtract__0.0667__0.04__ round__37.0__ |
divide__0.8__20.0__ subtract__1.0__0.8__ divide__0.2__3.0__ subtract__3.0__1.0__ subtract__0.0667__0.04__ divide__37.0__1.0__ |
| a certain factory produces buttons and buckles at a uniform weight . if the total weight of num__2 buttons and num__2 buckles is one third of num__11 buckles and num__3 buttons then the weight of num__3 buttons and num__3 buckles is how many times that of num__5 buckles and num__6 buttons ? <o> a ) num__0.466666666667 <o> b ) num__0.444444444444 <o> c ) num__0.545454545455 <o> d ) num__0.555555555556 <o> e ) num__0.533333333333 |
let x be the weight of a button and let y be the weight of a buckle . num__2 x + num__2 y = ( num__0.333333333333 ) ( num__3 x + num__11 y ) num__3 x = num__5 y x = num__5 y / num__3 num__3 x + num__3 y = a ( num__6 x + num__5 y ) num__8 y = a ( num__15 y ) a = num__0.533333333333 the answer is e . <eor> e <eos> |
e |
reverse__3.0__ add__2.0__6.0__ multiply__3.0__5.0__ divide__8.0__15.0__ divide__8.0__15.0__ |
reverse__3.0__ add__2.0__6.0__ multiply__3.0__5.0__ divide__8.0__15.0__ divide__8.0__15.0__ |
| in what ratio mental a at rs . num__68 per kg be mixed with another metal at rs . num__96 per kg so that cost of alloy ( mixture ) is rs . num__82 per kg ? <o> a ) num__5 : num__8 <o> b ) num__4 : num__7 <o> c ) num__3 : num__7 <o> d ) num__1 : num__1 <o> e ) num__9 : num__8 |
( num__96 - num__82 ) / ( num__82 - num__68 ) = num__1.0 = num__1.0 answer : d <eor> d <eos> |
d |
reverse__1.0__ |
reverse__1.0__ |
| a boat can travel with a speed of num__16 km / hr in still water . if the rate of stream is num__5 km / hr then find the time taken by the boat to cover distance of num__63 km downstream . <o> a ) num__3 hours <o> b ) num__5 hours <o> c ) num__6 hours <o> d ) num__7 hours <o> e ) num__8 hours |
explanation : it is very important to check if the boat speed given is in still water or with water or against water . because if we neglect it we will not reach on right answer . i just mentioned here because mostly mistakes in this chapter are of this kind only . lets see the question now . speed downstream = ( num__16 + num__5 ) = num__21 kmph time = distance / speed = num__3.0 = num__3 hours option a <eor> a <eos> |
a |
add__16.0__5.0__ divide__63.0__21.0__ round__3.0__ |
add__16.0__5.0__ divide__63.0__21.0__ divide__63.0__21.0__ |
| a sum of rs . num__1460 is divided into num__13 students so that each boy gets rs . num__120 while each girl gets rs . num__100 . find the number of boys and number of girls <o> a ) num__8 b num__5 g <o> b ) num__5 b num__8 g <o> c ) num__4 b num__9 g <o> d ) num__9 b num__4 g <o> e ) num__9 b num__5 g |
num__120 b + num__100 g = num__1460 - - - - - i b + g = num__13 - - - - - - - ii from i num__100 b + num__20 b + num__100 g = num__1460 num__100 ( b + g ) + num__20 b = num__1460 num__100 ( num__13 ) + num__20 b = num__1460 by simplifing we get b = num__8 ; g = num__5 answer : a <eor> a <eos> |
a |
subtract__120.0__100.0__ subtract__13.0__8.0__ subtract__13.0__5.0__ |
subtract__120.0__100.0__ subtract__13.0__8.0__ subtract__13.0__5.0__ |
| a man can swim in still water at num__4.5 km / h but takes twice as long to swim upstream than downstream . the speed of the stream is ? <o> a ) num__2 : num__0 <o> b ) num__3 : num__4 <o> c ) num__4 : num__9 <o> d ) num__1.5 <o> e ) num__1 : num__5 |
m = num__4.5 s = x ds = num__4.5 + x us = num__4.5 + x num__4.5 + x = ( num__4.5 - x ) num__2 num__4.5 + x = num__9 - num__2 x num__3 x = num__4.5 x = num__1.5 answer : d <eor> d <eos> |
d |
multiply__4.5__2.0__ divide__4.5__3.0__ round__1.5__ |
multiply__4.5__2.0__ subtract__4.5__3.0__ subtract__4.5__3.0__ |
| a train covers a distance of num__12 km in num__10 min . if it takes num__11 sec to pass a telegraph post then the length of the train is ? <o> a ) num__298 <o> b ) num__220 <o> c ) num__120 <o> d ) num__776 <o> e ) num__991 |
speed = ( num__1.2 * num__60 ) km / hr = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . length of the train = num__20 * num__11 = num__220 m . answer : b <eor> b <eos> |
b |
divide__12.0__10.0__ hour_to_min_conversion__ add__12.0__60.0__ multiply__11.0__20.0__ round__220.0__ |
divide__12.0__10.0__ hour_to_min_conversion__ multiply__1.2__60.0__ multiply__11.0__20.0__ multiply__11.0__20.0__ |
| one fourth of a solution that was num__10.0 sugar by weight was replaced by a second solution resulting in a solution that was num__16 percent sugar by weight . the second solution was what percent sugar by weight ? <o> a ) num__34.0 <o> b ) num__24.0 <o> c ) num__22.0 <o> d ) num__18.0 <o> e ) num__8.5 % |
et num__100 be the total volume . total salt = num__10 salt taken out = num__2.5 = num__2.5 to make solution num__16.0 total salt = num__16 salt added = num__16 - num__7.5 = num__8.5 solution had = num__8.5 / num__25 * num__100 = num__34.0 sugar answer : a <eor> a <eos> |
a |
percent__34.0__100.0__ |
percent__34.0__100.0__ |
| a man sells two articles for rs . num__3600 each and he gains num__42.0 on the first and loses num__42.0 on the next . find his total gain or loss ? <o> a ) num__9.0 loss <o> b ) num__400 <o> c ) num__4000 <o> d ) num__17.64 loss <o> e ) num__8.0 loss |
( num__42 * num__42 ) / num__100 = num__17.64 loss answer : d <eor> d <eos> |
d |
percent__17.64__100.0__ |
percent__17.64__100.0__ |
| robert spent $ num__30 in buying raw materials $ num__105 in buying machinery and num__10.0 of the total amount he had as cash with him . what was the total amount ? <o> a ) a ) $ num__150 <o> b ) b ) $ num__210 <o> c ) c ) $ num__250 <o> d ) d ) $ num__160 <o> e ) e ) $ num__200 |
let the total amount be x then ( num__100 - num__10 ) % of x = num__30 + num__105 num__90.0 of x = num__135 num__90 x / num__100 = num__135 x = $ num__150 answer is a <eor> a <eos> |
a |
percent__100.0__150.0__ |
percent__100.0__150.0__ |
| ab + cd = aaa where ab and cd are two - digit numbers and aaa is a three digit number ; a b c and d are distinct positive integers . in the addition problem above what is the value of c ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__7 <o> d ) num__9 <o> e ) can not be determined |
ab and cd are two digit integers their sum can give us only one three digit integer of a kind of aaa it ' s num__111 . so a = num__1 . num__1 b + cd = num__111 now c can not be less than num__9 because no to digit integer with first digit num__1 ( mean that it ' s < num__20 ) can be added to two digit integer less than num__90 to have the sum num__111 ( if cd < num__90 meaning c < num__9 cd + num__1 b < num__111 ) - - > c = num__9 answer : d . <eor> d <eos> |
d |
multiply__1.0__9.0__ |
multiply__1.0__9.0__ |
| a certain company reported that the revenue on sales increased num__40.0 from num__2000 to num__2003 and increased num__90.0 from num__2000 to num__2005 . what was the approximate percent increase in revenue for this store from num__2003 to num__2005 ? <o> a ) num__50.0 <o> b ) num__36.0 <o> c ) num__30.0 <o> d ) num__32.0 <o> e ) num__29 % |
assume the revenue in num__2000 to be num__100 . then in num__2003 it would be num__140 and and in num__2005 num__190 so from num__2003 to num__2005 it increased by ( num__190 - num__140 ) / num__140 = num__0.357142857143 = ~ num__29.0 . answer : b <eor> b <eos> |
b |
percent__40.0__90.0__ |
percent__40.0__90.0__ |
| a can do a work in num__6 days b can do a work in num__8 days and c can do it in num__12 days . b left work after num__6 days . for how many number of days should a and c should work together to complete the remaining work ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
b work num__0.125 * num__6 = num__0.75 remaining work = num__1 - num__0.75 = num__0.25 a and c work together = num__0.166666666667 + num__0.0833333333333 = num__0.25 = num__0.25 take reciprocial num__4 * remaining work = num__4 * num__0.25 = num__1 answer : a <eor> a <eos> |
a |
divide__6.0__8.0__ multiply__8.0__0.125__ subtract__1.0__0.75__ divide__0.125__0.75__ subtract__0.25__0.1667__ subtract__12.0__8.0__ round__1.0__ |
multiply__6.0__0.125__ multiply__8.0__0.125__ subtract__1.0__0.75__ divide__0.125__0.75__ subtract__0.25__0.1667__ subtract__12.0__8.0__ multiply__8.0__0.125__ |
| what is the greatest value of n such that num__4 ^ n is a factor of num__23 ! ? <o> a ) num__8 <o> b ) num__4 <o> c ) num__7 <o> d ) num__5 <o> e ) num__9 |
pretty simple really . if m = num__9 then num__4 m = num__36 which is num__18 x num__2 both of which are included in num__23 ! since num__9 is the largest number here its the answer . answer is e <eor> e <eos> |
e |
multiply__4.0__9.0__ gcd__4.0__18.0__ gcd__9.0__18.0__ |
multiply__4.0__9.0__ gcd__4.0__18.0__ gcd__9.0__18.0__ |
| pipe a can fill a tank in num__6 hours pipe b in num__12 hours and pipe c in num__36 hours . if all the pipes are open in how many hours will the tank be filled ? <o> a ) num__3.2 <o> b ) num__3 <o> c ) num__3.3 <o> d ) num__3.5 <o> e ) num__3.6 |
num__0.166666666667 + num__0.0833333333333 + num__0.0277777777778 = num__0.277777777778 = num__1 / num__3.6 . so num__3.6 hrs answer : e <eor> e <eos> |
e |
divide__6.0__36.0__ divide__0.1667__6.0__ round__3.6__ |
divide__6.0__36.0__ divide__0.1667__6.0__ divide__3.6__1.0__ |
| a tank is filled by three pipes with uniform flow . the first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone . the second pipe fills the tank num__5 hours faster than the first pipe and num__4 hours slower than the third pipe . time required by the first pipe to fill the tank is <o> a ) num__30 hours <o> b ) num__15 hours <o> c ) num__10 hours <o> d ) num__6 hours <o> e ) num__7 hours |
explanation : suppose the first pipe alone can fill the tank in x hours . then second pipe alone can fill the tank in ( x - num__5 ) hours third pipe alone can fill the tank in ( x â ˆ ’ num__5 ) - num__4 = ( x â ˆ ’ num__9 ) hours . part filled by first pipe and second pipe together in num__1 hr = part filled by third pipe in num__1 hr we can not take the value ( x â ˆ ’ num__3 ) because ( x â ˆ ’ num__9 ) becomes negative which is not possible because the third pipe can fill the tank in ( x â ˆ ’ num__9 ) hours . hence x = num__15 answer is b <eor> b <eos> |
b |
add__5.0__4.0__ subtract__5.0__4.0__ subtract__4.0__1.0__ multiply__5.0__3.0__ round__15.0__ |
add__5.0__4.0__ subtract__5.0__4.0__ subtract__4.0__1.0__ multiply__5.0__3.0__ round__15.0__ |
| the average of num__10 numbers is calculated as num__16 . it is discovered later on that while calculating the average the number num__55 was incorrectly read as num__25 and this incorrect number was used in the calculation . what is the correct average ? <o> a ) num__17 <o> b ) num__19 <o> c ) num__25 <o> d ) num__31 <o> e ) num__46 |
the total sum of the numbers should be increased by num__30 . then the average will increase by num__3.0 = num__3 . the correct average is num__19 . the answer is b . <eor> b <eos> |
b |
subtract__55.0__25.0__ divide__30.0__10.0__ add__16.0__3.0__ add__16.0__3.0__ |
subtract__55.0__25.0__ divide__30.0__10.0__ add__16.0__3.0__ add__16.0__3.0__ |
| the factorial expression num__10 ! / num__5 ! is divisible by which of the following integers ? <o> a ) num__5 <o> b ) num__31 <o> c ) num__21 <o> d ) num__11 <o> e ) num__19 |
num__10 ! / num__5 ! = > num__10 x num__9 x num__8 x num__7 x num__6 ( a ) num__5 can divide num__10 ( b ) num__31 none of the multiplicands present ( c ) num__21 none of the multiplicands present ( d ) num__11 none of the multiplicands present ( e ) num__19 none of the multiplicands present hence answer will be ( a ) <eor> a <eos> |
a |
subtract__31.0__10.0__ add__5.0__6.0__ add__10.0__9.0__ subtract__10.0__5.0__ |
subtract__31.0__10.0__ add__5.0__6.0__ add__10.0__9.0__ subtract__10.0__5.0__ |
| if from a group of num__5 people an old member is replaced by a new one the average age is same as it was num__3 years ago . what is the difference between the ages of the old member and the new one ? <o> a ) num__12 <o> b ) num__15 <o> c ) num__11 <o> d ) num__18 <o> e ) num__16 |
explanation : the present average age is ( x + num__3 ) when the old member is present and it equals to x when an old member is replaced by a new one . the difference in the ages of the old member and the new one is equal to the difference in the total age before and after the replacement = num__15 years . answer : b <eor> b <eos> |
b |
multiply__5.0__3.0__ multiply__5.0__3.0__ |
multiply__5.0__3.0__ multiply__5.0__3.0__ |
| a total of num__324 coins of num__20 paise and num__25 paise make a sum of rs . num__71 . the number of num__25 - paise coins is <o> a ) num__120 <o> b ) num__124 <o> c ) num__144 <o> d ) num__200 <o> e ) num__220 |
explanation : let the number of num__20 - paise coins be x . then number of num__25 - paise coins = ( num__324 - x ) . therefore num__0.20 x x + num__0.25 ( num__324 - x ) = num__71 num__20 x + num__25 ( num__324 - x ) = num__7100 num__5 x = num__1000 x = num__200 . hence number of num__25 - paise coins = ( num__324 - x ) - num__124 . answer : b <eor> b <eos> |
b |
reverse__0.2__ multiply__0.2__1000.0__ subtract__324.0__200.0__ subtract__324.0__200.0__ |
subtract__25.0__20.0__ multiply__0.2__1000.0__ subtract__324.0__200.0__ subtract__324.0__200.0__ |
| there are num__600 students in a school . the ratio of boys and girls in this school is num__3 : num__5 . find the total of girls & boys are there in this school ? <o> a ) num__225 <o> b ) num__257 <o> c ) num__375 <o> d ) num__380 <o> e ) num__390 |
in order to obtain a ratio of boys to girls equal to num__3 : num__5 the number of boys has to be written as num__3 x and the number of girls as num__5 x where x is a common factor to the number of girls and the number of boys . the total number of boys and girls is num__600 . hence num__3 x + num__5 x = num__600 solve for x num__8 x = num__600 x = num__75 number of boys num__3 x = num__3 × num__75 = num__225 number of girls num__5 x = num__5 × num__75 = num__375 c <eor> c <eos> |
c |
add__3.0__5.0__ divide__600.0__8.0__ multiply__3.0__75.0__ subtract__600.0__225.0__ subtract__600.0__225.0__ |
add__3.0__5.0__ divide__600.0__8.0__ multiply__3.0__75.0__ subtract__600.0__225.0__ subtract__600.0__225.0__ |
| if a person walks at num__15 km / hr instead of num__9 km / hr he would have walked num__20 km more . the time traveled by him is ? <o> a ) num__8 hours <o> b ) num__1.8 hours <o> c ) num__3.33333333333 hours <o> d ) num__6 hours <o> e ) num__3.66666666667 hours |
let the actual distance traveled be x km . then x / num__9 = ( x + num__20 ) / num__15 num__5 x - num__3 x = num__60 = > x = num__30 km . travel time is = num__3.33333333333 hours = num__3.33333333333 hours answer : c <eor> c <eos> |
c |
subtract__20.0__15.0__ divide__15.0__5.0__ multiply__20.0__3.0__ divide__30.0__9.0__ divide__30.0__9.0__ |
subtract__20.0__15.0__ divide__15.0__5.0__ multiply__20.0__3.0__ divide__30.0__9.0__ divide__30.0__9.0__ |
| the value of ( ( x – y ) ³ + ( y - z ) ³ + ( z – x ) ³ ) / ( num__21 ( x – y ) ( y – z ) ( z – x ) ) is equal to : <o> a ) num__0 <o> b ) num__0.0833333333333 <o> c ) num__0.142857142857 <o> d ) num__0.25 <o> e ) num__0.333333333333 |
since ( x – y ) + ( y – z ) + ( z – x ) = num__0 so ( x – y ) ³ + ( y – z ) ³ + ( z – x ) ³ = num__3 ( x – y ) ( y – z ) ( z – x ) . ( num__3 ( x – y ) ( y – z ) ( z – x ) ) / ( num__21 ( x – y ) ( y – z ) ( z – x ) ) = num__0.142857142857 . answer : c <eor> c <eos> |
c |
divide__3.0__21.0__ divide__3.0__21.0__ |
divide__3.0__21.0__ divide__3.0__21.0__ |
| ratio between rahul and deepak is num__4 : num__3 after num__6 years rahul age will be num__22 years . what is deepak present age ? <o> a ) num__22 <o> b ) num__15 <o> c ) num__77 <o> d ) num__266 <o> e ) num__12 |
present age is num__4 x and num__3 x = > num__4 x + num__6 = num__22 = > x = num__4 so deepak age is = num__3 ( num__4 ) = num__12 answer : e <eor> e <eos> |
e |
multiply__4.0__3.0__ multiply__4.0__3.0__ |
multiply__4.0__3.0__ multiply__4.0__3.0__ |
| if a b c d e and f are integers and ( ab + cdef ) < num__0 then what is the maximum number a of integers that can be negative ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
minimuum should be num__1 maximum should be num__4 : num__1 out of a or b to make the multiplication negative num__3 out of c d e or f to make the multiplication negative . negative + negative < num__0 answer : c maximum will be num__5 . . you dont require both the multiplicatin to be negative for entire equation to be negative . . . any one a or b can be negative to make ab negative and it can still be more ( away from num__0 ) than the multiplication of num__4 other - ve numbers . . . actually by writing minimum required as num__1 out of num__6 you are actually meaning num__5 out of num__6 also possible as you will see a = num__5 or num__1 will give you same equation . . ans d <eor> d <eos> |
d |
subtract__4.0__1.0__ add__1.0__4.0__ add__1.0__5.0__ add__1.0__4.0__ |
subtract__4.0__1.0__ add__1.0__4.0__ add__1.0__5.0__ add__1.0__4.0__ |
| bill has a small deck of num__10 playing cards made up of only num__2 suits of num__5 cards each . each of the num__5 cards within a suit has a different value from num__1 to num__5 ; thus there are num__2 cards in the deck that have the same value . bill likes to play a game in which he shuffles the deck turns over num__4 cards and looks for pairs of cards that have the same value . what is the chance that bill finds at least one pair of cards that have the same value ? <o> a ) num__0.466666666667 <o> b ) num__0.5625 <o> c ) num__0.611111111111 <o> d ) num__0.619047619048 <o> e ) num__0.64 |
p ( no pairs ) = num__0.888888888889 * num__0.75 * num__0.571428571429 = num__0.380952380952 p ( at least one pair ) = num__1 - num__0.380952380952 = num__0.619047619048 the answer is d . <eor> d <eos> |
d |
negate_prob__0.381__ negate_prob__0.381__ |
negate_prob__0.381__ negate_prob__0.381__ |
| ( num__935421 x num__625 ) = ? <o> a ) num__583038125 <o> b ) num__583538125 <o> c ) num__583738125 <o> d ) num__584038125 <o> e ) num__584638125 |
num__935421 x num__625 = num__935421 x num__54 = num__935421 x num__10 num__4 num__2 = num__935421 x num__104 = num__9354210000 num__24 num__16 = num__584638125 e ) <eor> e <eos> |
e |
multiply__935421.0__625.0__ multiply__935421.0__625.0__ |
multiply__935421.0__625.0__ multiply__935421.0__625.0__ |
| how many seconds will it take for a car that is traveling at a constant rate of num__120 miles per hour to travel a distance of num__77 yards ? ( num__1 mile = num__1160 yards ) <o> a ) num__1.25 <o> b ) num__1.31 <o> c ) num__1.58 <o> d ) num__1.87 <o> e ) num__2.3 |
speed = num__120 miles / hr = num__58.67 yard / s distance = num__77 yards time = distance / speed = num__77 / num__58.67 = num__1.31 sec ans - b <eor> b <eos> |
b |
round__1.31__ |
divide__1.31__1.0__ |
| mixture contains alcohol and water in the ratio num__4 : num__3 . if num__7 liters of water is added to the mixture the ratio becomes num__4 : num__5 . find the quantity of alcohol in the given mixture . <o> a ) num__10 <o> b ) num__99 <o> c ) num__5 <o> d ) num__22 <o> e ) num__29 |
let the quantity of alcohol and water be num__4 x litres and num__3 x litres respectively num__28 x = num__4 ( num__3 x + num__5 ) num__16 x = num__20 x = num__1.25 quantity of alcohol = ( num__4 x num__1.25 ) litres = num__5 litres . answer : c <eor> c <eos> |
c |
multiply__4.0__7.0__ multiply__4.0__5.0__ divide__5.0__4.0__ multiply__4.0__1.25__ |
multiply__4.0__7.0__ add__4.0__16.0__ divide__5.0__4.0__ multiply__4.0__1.25__ |
| aman can row num__1 km upstream in num__20 min and back to his starting point in num__12 min . how long would he take to row num__1 km in still water ? . <o> a ) num__12 mins <o> b ) num__13 mins <o> c ) num__14 mins <o> d ) num__15 mins <o> e ) num__16 mins |
if his speed is x mtr / min in still water and w mtr / minis speed of water then x - w = num__50.0 = num__50 x + w = num__83.3333333333 = num__83.3333333333 adding num__2 x = num__50 + num__83.3333333333 = num__133.333333333 x = num__66.6666666667 mtrs / min time taken to row num__1 kms in still water = num__1000 / ( num__66.6666666667 ) = num__15 mins . answer : d <eor> d <eos> |
d |
add__83.3333__50.0__ divide__133.3333__2.0__ multiply__20.0__50.0__ divide__1000.0__66.6667__ multiply__1.0__15.0__ |
add__83.3333__50.0__ divide__133.3333__2.0__ multiply__20.0__50.0__ divide__1000.0__66.6667__ divide__1000.0__66.6667__ |
| the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square whose area is num__1600 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is num__10 units ? <o> a ) num__140 <o> b ) num__150 <o> c ) num__160 <o> d ) num__170 <o> e ) num__180 |
given that the area of the square = num__1600 sq . units = > side of square = √ num__1600 = num__40 units the radius of the circle = side of the square = num__40 units length of the rectangle = num__0.4 * num__40 = num__16 units given that breadth = num__10 units area of the rectangle = lb = num__16 * num__10 = num__160 sq . units answer : option c <eor> c <eos> |
c |
square_perimeter__10.0__ multiply__0.4__40.0__ square_perimeter__40.0__ square_perimeter__40.0__ |
square_perimeter__10.0__ multiply__0.4__40.0__ multiply__10.0__16.0__ multiply__10.0__16.0__ |
| for the set { num__2 num__2 num__3 num__3 num__4 num__4 x } which of the following values of x will most increase the standard deviation ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) ( num__2 ) ^ num__2 |
standard deviation step num__1 as pointed out by others is to find out the mean = num__3 step num__2 for each number : subtract the mean and square the result = ( num__1 - num__3 ) ^ num__2 = ( - num__2 ) ^ num__2 ( num__2 - num__3 ) ^ num__2 = ( - num__1 ) ^ num__2 ( num__3 - num__3 ) ^ num__2 = ( num__0 ) ^ num__2 ( num__4 - num__3 ) ^ num__2 = ( num__1 ) ^ num__2 ( num__5 - num__3 ) ^ num__2 = ( num__2 ) ^ num__2 clearly ( num__1 - num__3 ) ^ num__2 = ( - num__2 ) ^ num__2 will give you the greatest value among all the other options . hence e <eor> e <eos> |
e |
subtract__3.0__2.0__ add__2.0__3.0__ multiply__2.0__1.0__ |
subtract__3.0__2.0__ add__2.0__3.0__ subtract__3.0__1.0__ |
| a salesman sold twice as much pears in the afternoon than in the morning . if he sold num__360 kilograms of pears that day how many kilograms did he sell in the morning and how many in the afternoon ? <o> a ) num__240 <o> b ) num__377 <o> c ) num__269 <o> d ) num__277 <o> e ) num__768 |
let xx be the number of kilograms he sold in the morning . then in the afternoon he sold num__2 x num__2 x kilograms . so the total is x + num__2 x = num__3 xx + num__2 x = num__3 x . this must be equal to num__360 . num__3 x = num__3603 x = num__360 x = num__3603 x = num__3603 x = num__120 x = num__120 therefore the salesman sold num__120 kg in the morning and num__2 ⋅ num__120 = num__2402 ⋅ num__120 = num__240 kg in the afternoon . answer : a <eor> a <eos> |
a |
divide__360.0__3.0__ subtract__360.0__120.0__ round__240.0__ |
divide__360.0__3.0__ subtract__360.0__120.0__ round__240.0__ |
| a certain number of persons can finish a piece of work in num__100 days . if there were num__10 persons less it would take num__10 more days finish the work . how many persons were there originally ? <o> a ) num__90 <o> b ) num__100 <o> c ) num__110 <o> d ) num__120 <o> e ) none of these |
explanation : assume that x persons can finish a piece of work in num__100 days also it is given that ( x - num__10 ) persons can finish a piece of work in num__110 days ( ∵ num__100 + num__10 = num__110 ) more persons less days ( indirect proportion ) hence we can write as personsx : ( x − num__10 ) } : : num__110 : num__100 ⇒ num__100 x = num__110 ( x − num__10 ) ⇒ num__100 x = num__110 x − num__1100 ⇒ num__10 x = num__1100 ⇒ x = num__110.0 = num__110 . answer : option c <eor> c <eos> |
c |
add__100.0__10.0__ multiply__10.0__110.0__ round__110.0__ |
add__100.0__10.0__ multiply__10.0__110.0__ add__100.0__10.0__ |
| if x ^ num__2 is divisible by num__240 what is the least possible value of integer x ? <o> a ) num__120 <o> b ) num__30 <o> c ) num__90 <o> d ) num__60 <o> e ) num__12 |
num__240 can be written as ( num__2 ^ num__4 ) * num__3 * num__5 . for x ^ num__2 to be divisible by num__240 it should contain at least num__2 ^ num__4 and num__3 and num__5 in its factors . we can leave out option e because num__12 doesnt have num__5 as one of its factor . now if we check for option b num__30 can be written as num__2 * num__3 * num__5 hence num__30 ^ num__2 will have num__2 as the maximum power of num__2 so we can leave out this option too . option d is the right answer if we follow the same method as we followed for other two previous options . num__60 = ( num__2 ^ num__2 ) * num__3 * num__5 ; num__60 ^ num__2 = ( num__2 ^ num__4 ) * ( num__3 ^ num__2 ) * ( num__5 ^ num__2 ) . so it shows that num__60 ^ num__2 is divisible by num__240 and hence the answer . answer : d <eor> d <eos> |
d |
add__2.0__3.0__ multiply__3.0__4.0__ multiply__2.0__30.0__ multiply__2.0__30.0__ |
add__2.0__3.0__ multiply__3.0__4.0__ multiply__2.0__30.0__ multiply__2.0__30.0__ |
| if $ num__10000 is invested at x percent simple annual interest for n years which of the following represents the total amount of interest in dollars that will be earned by this investment in the n years ? <o> a ) num__10000 ( x ^ n ) <o> b ) num__10000 ( x / num__100 ) ^ n <o> c ) num__10000 n ( x / num__100 ) <o> d ) num__10000 ( num__1 + x / num__100 ) ^ n <o> e ) num__10000 n ( num__1 + x / num__100 ) |
the reasoning is from the following : the formula for simple rate of interest is f = p ( num__1 + rt ) where f = future value p = present value r = rate t = time we can substitute x for r n for t and num__10000 in for p f = num__10000 ( num__1 + xn ) this formula will tell you the principal ( original amount invested ) + interest but we only want interest so the ` ` num__1 + ' ' portion of the formula in unnecessary . - - > f = num__10000 ( xn ) now we need to get the x into terms of a percentage so we divide x by num__100 - - > f = num__10000 ( n * x / num__100 ) if we move the n outside the brackets we get - - > f = num__10000 n ( x / num__100 ) answer : c <eor> c <eos> |
c |
percent__1.0__10000.0__ percent__100.0__10000.0__ |
percent__1.0__10000.0__ percent__100.0__10000.0__ |
| a b c d and e are num__5 consecutive points on a straight line . if bc = num__3 cd de = num__8 ab = num__5 and ac = num__11 what is the length of ae ? <o> a ) num__15 <o> b ) num__17 <o> c ) num__19 <o> d ) num__21 <o> e ) num__23 |
ac = num__11 and ab = num__5 so bc = num__6 . bc = num__3 cd so cd = num__2 . the length of ae is ab + bc + cd + de = num__5 + num__6 + num__2 + num__8 = num__21 the answer is d . <eor> d <eos> |
d |
subtract__11.0__5.0__ subtract__5.0__3.0__ round__21.0__ |
subtract__11.0__5.0__ subtract__5.0__3.0__ round__21.0__ |
| a sum of money is to be distributed among a b c d in the proportion of num__5 : num__2 : num__4 : num__3 . if c gets $ num__500 more than d what is d ' s share ? <o> a ) $ num__1000 <o> b ) $ num__1200 <o> c ) $ num__1500 <o> d ) $ num__1800 <o> e ) $ num__2000 |
let the shares of a b c and d be num__5 x num__2 x num__4 x and num__3 x respectively . then num__4 x - num__3 x = num__500 x = $ num__500 d ' s share = num__3 x = num__3 * $ num__500 = $ num__1500 the answer is c . <eor> c <eos> |
c |
multiply__3.0__500.0__ multiply__3.0__500.0__ |
multiply__3.0__500.0__ multiply__3.0__500.0__ |
| the difference between a two - digit number and the number obtained by interchanging the positions of its digits is num__36 . what is the difference between the two digits of that number ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__9 <o> d ) can not be determined <o> e ) none of these |
let the ten ' s digit be x and unit ' s digit be y . then ( num__10 x + y ) - ( num__10 y + x ) = num__36 num__9 ( x - y ) = num__36 x - y = num__4 . answer : b <eor> b <eos> |
b |
divide__36.0__9.0__ divide__36.0__9.0__ |
divide__36.0__9.0__ divide__36.0__9.0__ |
| how many of the positive factors of num__15 num__45 and how many common factors are there in numbers ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
factors of num__15 - num__1 num__3 num__5 and num__15 factors of num__45 - num__1 num__3 num__9 num__15 and num__45 comparing both we have three common factors of num__4516 - num__3 answer c <eor> c <eos> |
c |
divide__45.0__15.0__ divide__15.0__3.0__ divide__45.0__5.0__ gcd__15.0__3.0__ |
divide__45.0__15.0__ divide__15.0__3.0__ divide__45.0__5.0__ gcd__15.0__3.0__ |
| what number has a num__5 : num__1 ratio to the number num__10 ? <o> a ) num__22 <o> b ) num__50 <o> c ) num__99 <o> d ) num__72 <o> e ) num__12 |
num__5 : num__1 = x : num__10 x = num__50 answer : b <eor> b <eos> |
b |
multiply__5.0__10.0__ multiply__5.0__10.0__ |
multiply__5.0__10.0__ multiply__5.0__10.0__ |
| three business people who wish to invest in a new company . each person is willing to pay one third of the total investment . . after careful calculation they realize that each of them would pay $ num__7600 less if they could find two more equal investors . how much is the total investment in the new business . <o> a ) a ) $ num__64000 <o> b ) b ) $ num__54000 <o> c ) c ) $ num__21000 <o> d ) d ) $ num__5400 <o> e ) e ) $ num__57 |
000 |
initially each invest in x . hence total investment is num__3 x . total investment is also num__5 ( x - num__7600 ) . num__3 x = num__5 ( x - num__7600 ) x = num__5 * num__3800.0 = num__19000 num__3 x = num__57000 and the answer is e . <eor> e <eos> |
e |
e |
| the chennai express of num__200 m runs at a speed of num__62 km / hr and a person runs on the platform at a speed of num__20 km / hr in the direction opposite to that of train . find the time taken by the train to cross the running person ? <o> a ) num__8.77 sec <o> b ) num__9.77 sec <o> c ) num__12.77 sec <o> d ) num__13.00 sec <o> e ) none of these |
explanation : given : length of train = num__200 m speed of train = num__62 km / hr speed of person = num__20 km / hr num__1 ) convert km / hr into m / s - num__62 km / hr = num__62 x num__0.277777777778 = num__17.22 m / s - num__20 km / hr = num__20 x num__0.277777777778 = num__5.55 m / s as the train and the running person move in opposite directions their speed values are added to find the relative speed . relative speed ( speed of train relative to man ) = num__17.22 + num__5.55 = num__22.77 m / s we know speed = distance / time therefore time taken by the train to cross the running person = time taken by the train to cover num__200 m at a relative of num__22.77 m / s = num__200 / num__22.78 = num__8.77 sec answer is a <eor> a <eos> |
a |
add__5.55__17.22__ round__8.77__ |
add__5.55__17.22__ divide__8.77__1.0__ |
| in a num__1500 m race usha beats shiny by num__50 m . in the same race by what time margin shiny beat mercy who runs at num__4 m / s ? <o> a ) num__100 sec . <o> b ) num__50 sec <o> c ) num__75 sec <o> d ) data not sufficient <o> e ) none of these |
speed of shiny = num__5.0 = num__5 m / s time taken by shiny to complete the race is b = num__300.0 = num__300 sec . time taken by baley to complete the race is d = num__375.0 = num__275 sec . hence d - b = num__75 sec answer : c <eor> c <eos> |
c |
divide__1500.0__5.0__ divide__1500.0__4.0__ divide__300.0__4.0__ round__75.0__ |
divide__1500.0__5.0__ divide__1500.0__4.0__ divide__300.0__4.0__ divide__300.0__4.0__ |
| the total number of plums that grew during each year on a certain plum tree was equal to the number of plums that grew during the previous year less the age of the tree in years ( rounded to the nearest lower integer . ) during its num__3 rd year this plum tree grew num__50 plums . how many plums did it grow during its num__6 th year ? <o> a ) a ) num__41 <o> b ) b ) num__38 <o> c ) c ) num__35 <o> d ) d ) num__29 <o> e ) e ) num__28 |
the answer shud be num__38 . ( num__50 - num__3 - num__4 - num__5 = num__38 ) yes if you go the num__50 - num__3 way then it will be num__38 . if instead you go the num__50 - num__4 way ( which they have suggested ) the answer will be num__35 . as i said there is ambiguity here . answer is c <eor> c <eos> |
c |
subtract__38.0__3.0__ subtract__38.0__3.0__ |
subtract__38.0__3.0__ subtract__38.0__3.0__ |
| cole drove from home to work at an average speed of num__75 kmh . he then returned home at an average speed of num__105 kmh . if the round trip took a total of num__4 hours how many minutes did it take cole to drive to work ? <o> a ) num__66 <o> b ) num__70 <o> c ) num__95 <o> d ) num__112 <o> e ) num__140 |
first round distance travelled ( say ) = d speed = num__75 k / h time taken t num__2 = d / num__75 hr second round distance traveled = d ( same distance ) speed = num__105 k / h time taken t num__2 = d / num__105 hr total time taken = num__4 hrs therefore num__4 = d / num__75 + d / num__105 lcm of num__75 and num__105 = num__525 num__4 = d / num__75 + d / num__105 = > num__4 = num__7 d / num__525 + num__5 d / num__525 = > d = num__175.0 km therefore t num__1 = d / num__75 = > t num__1 = num__525 / ( num__3 x num__75 ) = > t num__1 = ( num__7 x num__60 ) / num__3 - - in minutes = > t num__1 = num__140 minutes . e <eor> e <eos> |
e |
divide__525.0__75.0__ subtract__7.0__2.0__ subtract__5.0__4.0__ subtract__4.0__1.0__ hour_to_min_conversion__ round__140.0__ |
divide__525.0__75.0__ subtract__7.0__2.0__ subtract__5.0__4.0__ subtract__4.0__1.0__ hour_to_min_conversion__ divide__140.0__1.0__ |
| a person purchased a tv set for rs . num__16000 and a dvd player for rs . num__1500 . he sold both the items together for rs . num__31150 . what percentage of profit did he make ? <o> a ) num__78.0 <o> b ) num__35.0 <o> c ) num__40.0 <o> d ) num__76.0 <o> e ) none of these |
the total cp = rs . num__16000 + rs . num__1500 = rs . num__17500 and sp = rs . num__31150 profit ( % ) = ( num__31150 - num__17500 ) / num__17500 * num__100 = num__78.0 answer : a <eor> a <eos> |
a |
percent__100.0__78.0__ |
percent__100.0__78.0__ |
| midway through his round a golfer hits a magnificent num__210 yard drive which brings his average length per drive for the round up to now from num__156 to num__162 yards . how far would he have had to hit the drive to bring his average length of drive up from num__156 to num__165 yards ? <o> a ) num__243 yards <o> b ) num__239 yards <o> c ) num__189 yards <o> d ) num__245 yards <o> e ) num__237 yards |
e num__237 yards eight holes average num__156 = num__1 num__248 yards nine holes average num__162 = num__1 num__458 yards ( + num__210 ) nine holes average num__165 = num__1 num__485 yards ( + num__237 ) . <eor> e <eos> |
e |
add__210.0__248.0__ add__237.0__248.0__ multiply__1.0__237.0__ |
add__210.0__248.0__ add__237.0__248.0__ multiply__1.0__237.0__ |
| a car takes num__6 hours to cover a distance of num__270 km . how much should the speed in kmph be maintained to cover the same direction in num__1.5 th of the previous time ? <o> a ) num__30 kmph <o> b ) num__60 kmph <o> c ) num__65 kmph <o> d ) num__70 kmph <o> e ) none |
time = num__6 distence = num__270 num__1.5 of num__6 hours = num__6 * num__1.5 = num__9 hours required speed = num__30.0 = num__30 kmph a ) <eor> a <eos> |
a |
multiply__6.0__1.5__ divide__270.0__9.0__ round__30.0__ |
multiply__6.0__1.5__ divide__270.0__9.0__ round__30.0__ |
| the distance between delhi and mathura is num__110 kms . a starts from delhi with a speed of num__20 kmph at num__7 a . m . for mathura and b starts from mathura with a speed of num__25 kmph at num__8 p . m . from delhi . when will they meet ? <o> a ) num__40.00 a . m . <o> b ) num__10.00 a . m . <o> c ) num__12.00 a . m . <o> d ) num__18.00 a . m . <o> e ) num__19.00 a . m . |
explanation : d = num__110 – num__20 = num__90 rs = num__20 + num__25 = num__45 t = num__2.0 = num__2 hours num__8 a . m . + num__2 = num__10 a . m . answer : option b <eor> b <eos> |
b |
subtract__110.0__20.0__ add__20.0__25.0__ divide__90.0__45.0__ divide__20.0__2.0__ round__10.0__ |
subtract__110.0__20.0__ add__20.0__25.0__ divide__90.0__45.0__ add__8.0__2.0__ add__8.0__2.0__ |
| the speed of a boat in upstream is num__60 kmph and the speed of the boat downstream is num__80 kmph . find the speed of the boat in still water and the speed of the stream ? <o> a ) num__10 kmph <o> b ) num__17 kmph <o> c ) num__19 kmph <o> d ) num__17 kmph <o> e ) num__29 kmph |
speed of the boat in still water = ( num__60 + num__80 ) / num__2 = num__70 kmph . speed of the stream = ( num__80 - num__60 ) / num__2 = num__10 kmph . answer : a <eor> a <eos> |
a |
subtract__80.0__70.0__ round__10.0__ |
subtract__80.0__70.0__ subtract__80.0__70.0__ |
| a train num__400 m long can cross an electric pole in num__20 sec and then find the speed of the train ? <o> a ) num__76 kmph <o> b ) num__54 kmph <o> c ) num__72 kmph <o> d ) num__34 kmph <o> e ) num__91 kmph |
length = speed * time speed = l / t s = num__20.0 s = num__20 m / sec speed = num__20 * num__3.6 ( to convert m / sec in to kmph multiply by num__3.6 ) speed = num__72 kmph answer : c <eor> c <eos> |
c |
multiply__20.0__3.6__ round__72.0__ |
multiply__20.0__3.6__ multiply__20.0__3.6__ |
| if num__4 men working num__10 hours a day earn rs . num__1200 per week then num__9 men working num__6 hours a day will earn how much per week ? <o> a ) rs num__840 <o> b ) rs num__1320 <o> c ) rs num__1620 <o> d ) rs num__1680 <o> e ) none of these |
explanation : ( men num__4 : num__9 ) : ( hrs / day num__10 : num__6 ) : : num__1200 : x hence num__4 * num__10 * x = num__9 * num__6 * num__1200 or x = num__9 * num__6 * num__300.0 * num__10 = num__1620 answer : c <eor> c <eos> |
c |
divide__1200.0__4.0__ round__1620.0__ |
divide__1200.0__4.0__ round__1620.0__ |
| tanya prepared num__4 different letters to be sent to num__4 different addresses . for each letter she prepared an envelope with its correct address . if the num__4 letters are to be put in num__4 envelopes at random what is the probability that only num__1 letter will be put into the envelope with its correct address ? <o> a ) num__0.0416666666667 <o> b ) num__0.125 <o> c ) num__0.25 <o> d ) num__0.333333333333 <o> e ) num__0.375 |
possible sequences are ciii icii and so on . . . c = correct i = incorrect total possible sequences are num__4 p num__1 = num__4 ways . lets take the probability of having one particular sequence ciii . p = num__0.25 * num__0.666666666667 * num__0.5 * num__1 = num__0.0833333333333 . so the probability of having any of the sequence is num__4 * num__0.0833333333333 = num__0.333333333333 ans : d <eor> d <eos> |
d |
union_prob__0.25__0.5__0.6667__ negate_prob__0.6667__ negate_prob__0.6667__ |
union_prob__0.25__0.5__0.6667__ negate_prob__0.6667__ negate_prob__0.6667__ |
| if x = num__1 - num__4 t and y = num__2 t - num__2 then for what value of t does x = y ? <o> a ) num__2.5 <o> b ) num__0.5 <o> c ) num__0.666666666667 <o> d ) num__0.4 <o> e ) num__0 |
we are given x = num__1 – num__4 t and y = num__2 t – num__2 and we need to determine the value for t when x = y . we should notice that both x and y are already in terms of t . thus we can substitute num__1 – num__4 t for x and num__2 t – num__2 for y in the equation x = y . this gives us : num__1 – num__4 t = num__2 t – num__2 num__3 = num__6 t num__0.5 = t the answer is b . <eor> b <eos> |
b |
add__1.0__2.0__ add__4.0__2.0__ reverse__2.0__ reverse__2.0__ |
add__1.0__2.0__ add__4.0__2.0__ reverse__2.0__ reverse__2.0__ |
| a batsman scored num__120 runs whichincluded num__3 boundaries and num__8 sixes . what % of his total score did he make by running between the wickets <o> a ) num__40.0 <o> b ) num__50.0 <o> c ) num__60.0 <o> d ) num__70.0 <o> e ) num__80 % |
number of runs made by running = num__110 - ( num__3 x num__4 + num__8 x num__6 ) = num__120 - ( num__60 ) = num__60 now we need to calculate num__60 is what percent of num__120 . = > num__0.5 * num__100 = num__50.0 b <eor> b <eos> |
b |
divide__3.0__6.0__ multiply__0.5__100.0__ multiply__0.5__100.0__ |
divide__3.0__6.0__ multiply__0.5__100.0__ multiply__0.5__100.0__ |
| the cost price of num__20 articles is equal to the selling price of num__25 articles . the loss percent in the transaction is <o> a ) num__5 <o> b ) num__20 <o> c ) num__25 <o> d ) num__30 <o> e ) none of these |
let c . p . of num__1 article = num__1 then c . p . of num__25 articles = num__25 and s . p . of num__25 articles = num__20 ∴ loss % = num__25 − num__1.0 × num__100 = num__25.0 answer c <eor> c <eos> |
c |
percent__25.0__100.0__ |
percent__25.0__100.0__ |
| num__2664 ÷ num__12 ÷ num__6 = ? <o> a ) num__43 <o> b ) num__41 <o> c ) num__37 <o> d ) num__33 <o> e ) num__21 |
explanation : num__2664 ÷ num__12 = num__222 num__222 ÷ num__6 = num__37 . answer : option c <eor> c <eos> |
c |
divide__2664.0__12.0__ divide__222.0__6.0__ divide__222.0__6.0__ |
divide__2664.0__12.0__ divide__222.0__6.0__ divide__222.0__6.0__ |
| point ( g h ) is on the circle represented by g ^ num__2 + h ^ num__2 = num__10 and g h are integers . how many such points are possible ? <o> a ) num__0 <o> b ) num__2 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
g ^ num__2 + h ^ num__2 = num__10 and g h are integers means that num__10 is the sum of two perfect squares . num__10 is the sum of only one pair of perfect squares num__1 and num__9 . so there can be num__8 such points num__4 in each quadrant : ( num__1 num__3 ) ; ( num__1 - num__3 ) ; ( - num__1 num__3 ) ; ( - num__1 - num__3 ) ; ( num__3 num__1 ) ; ( num__3 - num__1 ) ; ( - num__3 num__1 ) ; ( - num__3 - num__1 ) . answer : e . <eor> e <eos> |
e |
subtract__10.0__1.0__ subtract__10.0__2.0__ divide__8.0__2.0__ add__2.0__1.0__ multiply__2.0__4.0__ |
subtract__10.0__1.0__ subtract__10.0__2.0__ divide__8.0__2.0__ add__2.0__1.0__ subtract__10.0__2.0__ |
| if janice was num__22 years old z years ago and lisa will be num__20 years old in p years what was the average ( arithmetic mean ) of their ages num__8 years ago ? <o> a ) ( z + p ) / num__2 <o> b ) ( z - p + num__26 ) / num__4 <o> c ) ( z - p + num__16 ) / num__4 <o> d ) ( z + p + num__36 ) / num__2 <o> e ) ( z - p + num__26 ) / num__2 |
today j = z + num__22 and l = num__20 - p num__8 years ago j = z + num__14 and l = num__12 - p the average of their ages was ( z - p + num__26 ) / num__2 the answer is e . <eor> e <eos> |
e |
subtract__22.0__8.0__ subtract__20.0__8.0__ add__12.0__14.0__ subtract__22.0__20.0__ add__12.0__14.0__ |
subtract__22.0__8.0__ subtract__20.0__8.0__ add__12.0__14.0__ subtract__22.0__20.0__ add__12.0__14.0__ |
| how many integers are divisible by num__2 between num__10 ! and num__10 ! + num__20 inclusive ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__21 <o> d ) num__9 <o> e ) num__10 |
c = num__21 num__10 ! is divisible by num__2 there are num__20 numbers between num__10 ! and num__10 ! + num__20 that are divisible by num__1 . hence num__21 <eor> c <eos> |
c |
subtract__21.0__20.0__ add__20.0__1.0__ |
subtract__21.0__20.0__ add__20.0__1.0__ |
| y and f represent two distinct digits . if the number num__45 y num__89 f is divisible by num__36 what is the value of ( y + f ) ? <o> a ) a ) num__9 <o> b ) b ) num__3 <o> c ) c ) num__12 <o> d ) d ) num__6 <o> e ) e ) num__10 |
a no . divisible by num__36 means it is div by num__49 . to be div by num__4 last num__2 digits to be multiple of num__4 so f could be either num__2 or num__6 only similarly for a number to be div by num__9 its sum must be multiple of num__9 first . filtering we find y could be either num__1 num__47 only . to be divisible by num__9 only num__1 works . ( if we choose f = num__6 then y have to be num__6 but as per condition yf are distinct ) so y = num__1 f = num__2 ans : - num__3 . b <eor> b <eos> |
b |
subtract__49.0__45.0__ add__2.0__4.0__ subtract__45.0__36.0__ add__45.0__2.0__ add__1.0__2.0__ add__1.0__2.0__ |
subtract__49.0__45.0__ add__2.0__4.0__ subtract__45.0__36.0__ subtract__49.0__2.0__ subtract__9.0__6.0__ subtract__6.0__3.0__ |
| a river num__2 m deep and num__45 m wide is flowing at the rate of num__3 kmph the amount of water that runs into the sea per minute is ? <o> a ) num__4500 m num__3 <o> b ) num__27000 m num__3 <o> c ) num__3000 m num__3 <o> d ) num__2700 m num__3 <o> e ) num__3700 m num__3 |
( num__3000 * num__2 * num__5 ) / num__60 = num__4500 m num__3 answer : a <eor> a <eos> |
a |
add__2.0__3.0__ hour_to_min_conversion__ round__4500.0__ |
add__2.0__3.0__ hour_to_min_conversion__ round__4500.0__ |
| what is the num__4 digit number in which the num__1 st digit is num__0.333333333333 of the second the num__3 rd is the sum of the num__1 st and num__2 nd and the last is three times the num__2 nd ? <o> a ) num__965 <o> b ) num__1245 <o> c ) num__1349 <o> d ) num__1368 <o> e ) num__1578 |
first digit is num__0.333333333333 second digit = > the numbers can be num__1 & num__3 num__2 & num__6 num__3 & num__9 . first + second = third = > we can eliminate num__3 & num__9 since num__3 + num__9 = num__12 . last is num__3 times the second = > we can eliminate option num__2 & num__6 since num__3 * num__6 = num__18 . hence the number is num__1349 c <eor> c <eos> |
c |
add__4.0__2.0__ add__3.0__6.0__ multiply__4.0__3.0__ multiply__3.0__6.0__ multiply__1.0__1349.0__ |
add__4.0__2.0__ add__3.0__6.0__ multiply__4.0__3.0__ multiply__3.0__6.0__ multiply__1.0__1349.0__ |
| an association of mathematics teachers has num__1260 members . only num__525 of these members cast votes in the election for president of the association . what percent of the total membership voted for the winning candidate if the winning candidate received num__72 percent of the votes cast ? <o> a ) num__75.0 <o> b ) num__58.0 <o> c ) num__30.0 <o> d ) num__34.0 <o> e ) num__25 % |
total number of members = num__1260 number of members that cast votes = num__525 since winning candidate received num__72 percent of the votes cast number of votes for winning candidate = ( num__0.72 ) * num__525 = num__378 percent of total membership that voted for winning candidate = ( num__0.3 ) * num__100 = num__30.0 answer c <eor> c <eos> |
c |
percent__72.0__525.0__ percent__100.0__30.0__ |
percent__72.0__525.0__ percent__100.0__30.0__ |
| a financier claims to be lending money at simple interest but he includes the interest every six months for calculating the principal . if he is charging an interest of num__10.0 the effective rate of interest becomes <o> a ) num__10.25 <o> b ) num__10.0 <o> c ) num__9.25 <o> d ) num__9.0 <o> e ) none of these |
explanation : let the sum is num__100 . as financier includes interest every six months . then we will calculate si for num__6 months then again for six months as below : si for first six months = ( num__100 * num__10 * num__1 ) / ( num__100 * num__2 ) = rs . num__5 important : now sum will become num__100 + num__5 = num__105 si for last six months = ( num__105 * num__10 * num__1 ) / ( num__100 * num__2 ) = rs . num__5.25 so amount at the end of year will be ( num__100 + num__5 + num__5.25 ) = num__110.25 effective rate = num__110.25 - num__100 = num__10.25 option a <eor> a <eos> |
a |
percent__5.0__105.0__ percent__100.0__10.25__ |
percent__5.0__105.0__ percent__100.0__10.25__ |
| the cash realised on selling a num__14.0 stock is rs . num__104.25 brokerage being num__0.25 % is <o> a ) num__123 <o> b ) num__106 <o> c ) num__100 <o> d ) num__156 <o> e ) num__104 |
explanation : cash realised = rs . ( num__104.25 - num__0.25 ) = rs . num__104 . answer : e <eor> e <eos> |
e |
round_down__104.25__ round_down__104.25__ |
subtract__104.25__0.25__ subtract__104.25__0.25__ |
| the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is num__42 kmph find the speed of the stream ? <o> a ) num__22 kmph <o> b ) num__77 kmph <o> c ) num__14 kmph <o> d ) num__17 kmph <o> e ) num__18 kmph |
the ratio of the times taken is num__2 : num__1 . the ratio of the speed of the boat in still water to the speed of the stream = ( num__2 + num__1 ) / ( num__2 - num__1 ) = num__3.0 = num__3 : num__1 speed of the stream = num__14.0 = num__14 kmph . answer : c <eor> c <eos> |
c |
add__1.0__2.0__ divide__42.0__3.0__ round__14.0__ |
add__1.0__2.0__ divide__42.0__3.0__ divide__42.0__3.0__ |
| simplify : num__5005 - num__5000 + num__10 <o> a ) num__4505 <o> b ) num__4509 <o> c ) num__4501 <o> d ) num__4508 <o> e ) none of them |
num__5005 - num__5000 + num__10 = num__5005 - ( num__500.0 ) = num__5005 - num__500 = num__4505 . answer is a <eor> a <eos> |
a |
divide__5000.0__10.0__ subtract__5005.0__500.0__ subtract__5005.0__500.0__ |
divide__5000.0__10.0__ subtract__5005.0__500.0__ subtract__5005.0__500.0__ |
| x can finish a work in num__18 days . y can finish the same work in num__15 days . yworked for num__10 days and left the job . how many days does x alone need to finish the remaining work ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
work done by x in num__1 day = num__0.0555555555556 work done by y in num__1 day = num__0.0666666666667 work done by y in num__10 days = num__0.666666666667 = num__0.666666666667 remaining work = num__1 – num__0.666666666667 = num__0.333333333333 number of days in which x can finish the remaining work = ( num__0.333333333333 ) / ( num__0.0555555555556 ) = num__6 c <eor> c <eos> |
c |
divide__1.0__18.0__ divide__1.0__15.0__ divide__10.0__15.0__ subtract__1.0__0.6667__ round__6.0__ |
divide__1.0__18.0__ divide__1.0__15.0__ divide__10.0__15.0__ subtract__1.0__0.6667__ divide__6.0__1.0__ |
| company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month num__0.666666666667 were basic and the rest were deluxe . if it takes num__2.2 as many hours to produce a deluxe stereo as it does to produce a basic stereo then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos ? <o> a ) num__0.52380952381 <o> b ) num__0.451612903226 <o> c ) num__0.466666666667 <o> d ) num__0.485714285714 <o> e ) num__0.5 |
# of basic stereos was num__0.666666666667 of total and # of deluxe stereos was num__0.333333333333 of total let ' s assume total = num__15 then basic = num__10 and deluxe = num__5 . now if time needed to produce one deluxe stereo is num__1 unit than time needed to produce one basic stereo would be num__2.2 units . total time for basic would be num__10 * num__1 = num__10 and total time for deluxe would be num__5 * num__2.2 = num__11 - - > total time for both of them would be num__10 + num__11 = num__21 - - > deluxe / total = num__0.52380952381 . answer : a . <eor> a <eos> |
a |
subtract__15.0__10.0__ add__0.6667__0.3333__ multiply__2.2__5.0__ add__10.0__11.0__ divide__11.0__21.0__ multiply__1.0__0.5238__ |
subtract__15.0__10.0__ add__0.6667__0.3333__ multiply__2.2__5.0__ add__10.0__11.0__ divide__11.0__21.0__ multiply__1.0__0.5238__ |
| if finn was num__19 months old one year ago how old was he in months x months ago ? <o> a ) x − num__30 <o> b ) x − num__12 <o> c ) num__18 − x <o> d ) num__24 − x <o> e ) num__31 − x |
age today = num__19 months + num__12 months ( num__1 year ) = num__31 months x months ago - today ' s age - x = num__31 - x ans - e <eor> e <eos> |
e |
add__19.0__12.0__ add__19.0__12.0__ |
add__19.0__12.0__ add__19.0__12.0__ |
| fox jeans regularly sell for $ num__15 a pair and pony jeans regularly sell for $ num__18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ num__5 is saved by purchasing num__5 pairs of jeans : num__3 pairs of fox jeans and num__2 pairs of pony jeans . if the sum of the two discounts rates is num__18 percent what is the discount rate on pony jeans ? <o> a ) num__9.0 <o> b ) num__10.0 <o> c ) num__34.4 <o> d ) num__12.0 <o> e ) num__15 % |
you know that fox jeans costs $ num__15 and pony jeans costs $ num__18 you also know that num__3 pairs of fox jeans and num__2 pairs of pony jeans were purchased . so num__3 ( num__15 ) = num__45 - fox num__2 ( num__18 ) = num__36 - pony the total discount discount is $ num__5 and you are asked to find the percent discount of pony jeans so num__45 ( num__18 - x ) / num__100 + num__36 ( x ) / num__100 = num__5 or num__45 * num__18 - num__45 * x + num__36 * x = num__5 * num__100 or num__9 x = - num__5 * num__100 + num__45 * num__18 x = num__34.4444444444 = num__34.4 c <eor> c <eos> |
c |
percent__34.4__100.0__ |
percent__34.4__100.0__ |
| the age of man is three times the sum of the ages of his two sons . seven years hence his age will be double of the sum of the ages of his sons . the father â € ™ s present age is : <o> a ) num__40 years <o> b ) num__45 years <o> c ) num__50 years <o> d ) num__55 years <o> e ) num__63 years |
solution let the sum of present ages of the two sons be x years . then father ' s present age = num__3 x years . â ˆ ´ ( num__3 x + num__7 ) = num__2 ( x + num__14 ) â ‡ ” num__3 x + num__7 = num__2 x + num__28 â ‡ ” x = num__21 . hence father ' s present age = num__63 years . answer e <eor> e <eos> |
e |
multiply__2.0__7.0__ multiply__2.0__14.0__ multiply__3.0__7.0__ multiply__3.0__21.0__ multiply__3.0__21.0__ |
multiply__2.0__7.0__ multiply__2.0__14.0__ add__7.0__14.0__ multiply__3.0__21.0__ multiply__3.0__21.0__ |
| what is the smallest positive integer x such that num__5000 x is a perfect cube ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__8 <o> d ) num__25 <o> e ) num__18 |
take out the factors of num__5000 that will come num__10 ^ num__3 * num__5 . for perfect cube you need every no . raise to the power num__3 . for num__5000 x to be a perfect cube you need two num__5 that means num__25 . d is the answer . <eor> d <eos> |
d |
triangle_area__10.0__5.0__ triangle_area__10.0__5.0__ |
triangle_area__10.0__5.0__ triangle_area__10.0__5.0__ |
| the cost price of num__13 articles is equal to the selling price of num__11 articles . find the profit percent ? <o> a ) num__18 num__0.133333333333 <o> b ) num__18 num__0.181818181818 <o> c ) num__18 num__0.142857142857 <o> d ) num__18 num__0.125 <o> e ) num__18 num__0.166666666667 |
num__13 cp = num__11 sp num__11 - - - num__2 cp num__100 - - - ? = > num__18 num__0.181818181818 % . answer : b <eor> b <eos> |
b |
percent__100.0__18.0__ |
percent__100.0__18.0__ |
| the total number of digits used in numbering the pages of a book having num__346 pages is <o> a ) num__732 <o> b ) num__990 <o> c ) num__1098 <o> d ) num__930 <o> e ) num__1405 |
total number of digits = ( no . of digits in num__1 - digit page nos . + no . of digits in num__2 - digit page nos . + no . of digits in num__3 - digit page nos . ) = ( num__1 x num__9 + num__2 x num__90 + num__3 x num__247 ) = ( num__9 + num__180 + num__741 ) = num__930 . answer : d <eor> d <eos> |
d |
add__1.0__2.0__ multiply__2.0__90.0__ multiply__3.0__247.0__ multiply__1.0__930.0__ |
add__1.0__2.0__ multiply__2.0__90.0__ multiply__3.0__247.0__ multiply__1.0__930.0__ |
| the length of a rectangular plot is num__20 metres more than its breadth . if the cost of fencing the plot @ rs . num__26.50 per metre is rs . num__5300 what is the length of the plot in metres ? <o> a ) num__55 <o> b ) num__60 <o> c ) num__65 <o> d ) num__70 <o> e ) num__75 |
length of the fence = num__530026.50 = num__200 m ⇒ num__2 ( length + breadth ) = num__200 m ⇒ num__2 ( breadth + num__20 + breadth ) = num__200 m ( ∵ length = breadth + num__20 ) ⇒ breadth + num__20 + breadth = num__100 m ⇒ breadth = num__40 m length = num__40 + num__20 = num__60 m answer is b . <eor> b <eos> |
b |
divide__5300.0__26.5__ divide__200.0__2.0__ multiply__20.0__2.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
divide__5300.0__26.5__ divide__200.0__2.0__ multiply__20.0__2.0__ add__20.0__40.0__ add__20.0__40.0__ |
| how many days are there in x weeks x days <o> a ) num__14 x <o> b ) num__8 x <o> c ) num__7 x ^ num__2 <o> d ) num__7 <o> e ) num__6 |
explanation : x weeks x days = ( num__7 × x ) + x = num__7 x + x = num__8 x days answer : option b <eor> b <eos> |
b |
round__8.0__ |
round__8.0__ |
| the positive integer b is divisible by num__5 . what is the remainder when dividing num__3 ( num__6 + b ) ( num__5 + b ) by num__5 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
when we multiply any of these options : num__3 * num__11 * num__10 ( b = num__5 ) or num__3 * num__16 * num__15 ( b = num__10 ) we end up with numbers ending in num__0 which are exactly divisible by num__5 leaving no remainder . hence answer is a . <eor> a <eos> |
a |
add__5.0__6.0__ add__5.0__11.0__ multiply__5.0__3.0__ multiply__5.0__0.0__ |
add__5.0__6.0__ add__5.0__11.0__ multiply__5.0__3.0__ multiply__5.0__0.0__ |
| the salary of a worker is first increased by num__30.0 and afterwards reduced by num__30.0 . what is net change in his salary ? <o> a ) num__3.0 decrease <o> b ) num__5.0 decrease <o> c ) num__9.0 decrease <o> d ) num__8.0 decrease <o> e ) num__6.0 decrease |
( num__30 * num__30 ) / num__100 = num__9.0 decrease answer : c <eor> c <eos> |
c |
percent__9.0__100.0__ |
percent__9.0__100.0__ |
| the ratio of three numbers is num__1 : num__2 : num__4 and the sum of their squares is num__4725 . the sum of the numbers is ? <o> a ) a ) num__10 <o> b ) b ) num__12 <o> c ) c ) num__15 <o> d ) d ) num__14 <o> e ) e ) num__9 |
let the numbers be x num__2 x num__4 x then x ^ num__2 + num__4 x ^ num__2 + num__16 x ^ num__2 = num__4725 num__21 x ^ num__2 = num__4725 x ^ num__2 = num__225 x = num__15 answer is c <eor> c <eos> |
c |
divide__4725.0__21.0__ subtract__16.0__1.0__ multiply__1.0__15.0__ |
divide__4725.0__21.0__ subtract__16.0__1.0__ multiply__1.0__15.0__ |
| tough and tricky questions : word problems . mike sarah and david decided to club together to buy a present . each of them gave equal amount of money . luckily sarah negotiated a num__20.0 discount for the present so that each of them paid num__4 dollars less . how much did they pay for a present ? <o> a ) num__20 <o> b ) num__36 <o> c ) num__48 <o> d ) num__60 <o> e ) num__72 |
answer c . we know that sarah negotiated a discount of num__20.0 so each of them paid $ num__4 less . since there are three people the num__20.0 of the original price amounts to $ num__12 . num__5 times num__12 $ is num__60 $ so the original price before sarah negotiated the discount had been $ num__60 . they paid $ num__12 less than the base price so they spent $ num__48 . <eor> c <eos> |
c |
divide__20.0__4.0__ multiply__5.0__12.0__ multiply__4.0__12.0__ multiply__4.0__12.0__ |
divide__20.0__4.0__ multiply__5.0__12.0__ multiply__4.0__12.0__ multiply__4.0__12.0__ |
| the area of a triangle is with base num__5.5 m and height num__6 m ? <o> a ) num__11 m num__2 <o> b ) num__16.5 m num__2 <o> c ) num__18.5 m num__2 <o> d ) num__19.5 m num__2 <o> e ) num__12 m num__2 |
num__0.5 * num__5.5 * num__6 = num__16.5 m num__2 answer : b <eor> b <eos> |
b |
triangle_area__5.5__6.0__ square_perimeter__0.5__ triangle_area__5.5__6.0__ |
volume_rectangular_prism__5.5__6.0__0.5__ square_perimeter__0.5__ volume_rectangular_prism__5.5__6.0__0.5__ |
| two trains of length num__100 meters and num__200 meters are num__840 meters apart . they are moving towards each other on parallel tracks at speeds of num__54 km / h and num__72 km / h . after how many seconds will the trains meet ? <o> a ) num__20 <o> b ) num__22 <o> c ) num__24 <o> d ) num__26 <o> e ) num__28 |
the speeds are num__15.0 = num__15 m / s and num__20.0 = num__20 m / s the relative speed is num__35 m / s . time = num__24.0 = num__24 seconds the answer is c . <eor> c <eos> |
c |
add__20.0__15.0__ divide__840.0__35.0__ round__24.0__ |
add__20.0__15.0__ divide__840.0__35.0__ divide__840.0__35.0__ |
| how many three - element subsets of { num__12 num__34 } are there that do not contain the pair of elements num__2 and num__4 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
required subsets are = { num__1 num__23 } { num__1 num__34 } = num__2 answer : b <eor> b <eos> |
b |
multiply__2.0__1.0__ |
multiply__2.0__1.0__ |
| if a ( a + num__6 ) = num__27 and b ( b + num__6 ) = num__27 where a ≠ b then a + b = <o> a ) − num__48 <o> b ) − num__6 <o> c ) num__2 <o> d ) num__46 <o> e ) num__48 |
a ( a + num__6 ) = num__27 and b ( b + num__6 ) = num__27 = > a b must be integers and if a is - num__9 or num__3 b will be num__3 and - num__9 respectively = > a + b = - num__6 ans : b <eor> b <eos> |
b |
divide__27.0__9.0__ subtract__9.0__3.0__ |
subtract__9.0__6.0__ subtract__9.0__3.0__ |
| a train is num__360 meter long is running at a speed of num__45 km / hour . in what time will it pass a bridge of num__140 meter length ? <o> a ) num__65 seconds <o> b ) num__46 seconds <o> c ) num__40 seconds <o> d ) num__97 seconds <o> e ) num__26 seconds |
speed = num__45 km / hr = num__45 * ( num__0.277777777778 ) m / sec = num__12.5 m / sec total distance = num__360 + num__140 = num__500 meter time = distance / speed = num__500 * ( num__0.08 ) = num__40 seconds answer : c <eor> c <eos> |
c |
add__360.0__140.0__ divide__500.0__12.5__ round__40.0__ |
add__360.0__140.0__ divide__500.0__12.5__ divide__500.0__12.5__ |
| a cistern has a leak which would empty the cistern in num__20 minutes . a tap is turned on which admits num__1 liters a minute into the cistern and it is emptied in num__24 minutes . how many liters does the cistern hold ? <o> a ) num__480 <o> b ) num__120 <o> c ) num__289 <o> d ) num__270 <o> e ) num__927 |
num__1 / x - num__0.05 = - num__0.0416666666667 x = num__120 num__120 * num__1 = num__120 answer : b <eor> b <eos> |
b |
divide__1.0__20.0__ divide__1.0__24.0__ round__120.0__ |
divide__1.0__20.0__ divide__1.0__24.0__ multiply__1.0__120.0__ |
| a palindrome is a word or a number that reads the same forward and backward . for example num__2442 and num__111 are palindromes . if num__5 - digit palindromes are formed using one or more of the digits num__1 num__2 num__3 num__4 and num__5 how many palindromes are possible ? <o> a ) num__25 <o> b ) num__75 <o> c ) num__125 <o> d ) num__225 <o> e ) num__625 |
there are num__5 choices for each of the first three digits . the number of possible palindromes is num__5 ^ num__3 = num__125 . the answer is c . <eor> c <eos> |
c |
multiply__1.0__125.0__ |
multiply__1.0__125.0__ |
| if ( num__20 ) ² is subtracted from the square of a number the answer so obtained is num__4321 . what is the number ? <o> a ) num__68 <o> b ) num__69 <o> c ) num__70 <o> d ) num__71 <o> e ) num__72 |
x ^ num__2 = num__4321 + num__400 = num__4721 num__4761 = num__69 * num__69 x = num__69 answer : b <eor> b <eos> |
b |
power__20.0__2.0__ triangle_area__2.0__69.0__ |
power__20.0__2.0__ triangle_area__2.0__69.0__ |
| in a certain large company the ratio of college graduates with a graduate degree to non - college graduates is num__1 : num__8 and ratio of college graduates without a graduate degree to non - college graduates is num__2 : num__3 . if one picks a random college graduate at this large company what is the probability t this college graduate has a graduate degree ? <o> a ) num__0.0909090909091 <o> b ) num__0.0833333333333 <o> c ) num__0.0769230769231 <o> d ) num__0.157894736842 <o> e ) num__0.0697674418605 |
in believe the answer is d . please see below for explanation . num__0 ) we are told the following ratios cgd - college graduate with degree ncg - non college graduate cgn - college graduate no degree cgd ncg cgn num__1 num__8 num__3 num__2 in order to make cgd and cgn comparable we need to find the least common multiple of num__8 and num__3 and that is num__24 multiplying the first ratio by num__3 and the second ratio by num__8 we get cgd ncg cgn num__3 num__24 num__16 if one picks a random college graduate at this large company what is the probability this college graduate has a graduate degree ? nr of cgd = num__3 nr of cg = num__3 + num__16 = num__19 probability t of cgd / ( cg ) - > num__0.157894736842 answer d <eor> d <eos> |
d |
multiply__8.0__3.0__ multiply__8.0__2.0__ add__3.0__16.0__ divide__3.0__19.0__ multiply__1.0__0.1579__ |
multiply__8.0__3.0__ subtract__24.0__8.0__ add__3.0__16.0__ divide__3.0__19.0__ divide__3.0__19.0__ |
| the greatest common factor of positive integers m and n is num__14 . what is the greatest common factor of ( num__2 m ^ num__2 num__2 n ^ num__2 ) ? <o> a ) num__14 <o> b ) num__28 <o> c ) num__196 <o> d ) num__392 <o> e ) num__784 |
m = num__14 j and n = num__14 k where the greatest common factor of j and k is num__1 . num__2 m ^ num__2 = num__2 * num__14 * num__14 * j ^ num__2 and num__2 n ^ num__2 = num__2 * num__14 * num__14 * k ^ num__2 the greatest common factor of num__2 m ^ num__2 and num__2 n ^ num__2 is num__2 * num__14 * num__14 = num__392 the answer is d . <eor> d <eos> |
d |
lcm__14.0__392.0__ |
multiply__392.0__1.0__ |
| if xy > num__0 num__1 / x + num__1 / y = num__15 and num__1 / xy = num__5 then ( x + y ) / num__5 = ? <o> a ) num__0.04 <o> b ) num__0.166666666667 <o> c ) num__0.6 <o> d ) num__5 <o> e ) num__6 |
( num__1 / x + num__1 / y ) = num__15 canbe solved as { ( x + y ) / xy } = num__5 . substituting for num__1 / xy = num__5 we get x + y = num__3.0 = = > ( x + y ) / num__5 = num__15 / ( num__5 * num__5 ) = num__0.6 . c <eor> c <eos> |
c |
divide__15.0__5.0__ divide__3.0__5.0__ multiply__1.0__0.6__ |
divide__15.0__5.0__ divide__3.0__5.0__ multiply__1.0__0.6__ |
| a train passes a man standing on the platform . if the train is num__160 meters long and its speed is num__72 kmph how much time it took in doing so ? <o> a ) num__5 sec <o> b ) num__6 sec <o> c ) num__8 sec <o> d ) num__9 sec <o> e ) num__7 sec |
d = num__160 s = num__72 * num__0.277777777778 = num__20 mps t = num__8.0 = num__8 sec answer : c <eor> c <eos> |
c |
divide__160.0__20.0__ round__8.0__ |
divide__160.0__20.0__ round__8.0__ |
| length of a rectangular plot is num__12 mtr more than its breadth . if the cost of fencing the plot at num__26.50 per meter is rs . num__5300 what is the length of the plot in mtr ? <o> a ) num__56 m <o> b ) num__60 m <o> c ) num__80 m <o> d ) num__82 m <o> e ) num__84 m |
let breadth = x metres . then length = ( x + num__12 ) metres . perimeter = num__5300 m = num__200 m . num__26.50 num__2 [ ( x + num__12 ) + x ] = num__200 num__2 x + num__12 = num__100 num__2 x = num__88 x = num__44 hence length = x + num__12 = num__56 m a <eor> a <eos> |
a |
divide__5300.0__26.5__ divide__200.0__2.0__ subtract__100.0__12.0__ divide__88.0__2.0__ add__12.0__44.0__ round__56.0__ |
divide__5300.0__26.5__ divide__200.0__2.0__ subtract__100.0__12.0__ divide__88.0__2.0__ add__12.0__44.0__ add__12.0__44.0__ |
| on a partly cloudy day derek decides to walk back from work . when it is sunny he walks at a speed of s miles / hr ( s is an integer ) and when it gets cloudy he increases his speed to ( s + num__1 ) miles / hr . if his average speed for the entire distance is num__2.8 miles / hr what fraction r of the total distance did he cover while the sun was shining on him ? <o> a ) num__0.25 <o> b ) num__0.8 <o> c ) num__0.2 <o> d ) num__0.166666666667 <o> e ) num__0.142857142857 |
if s is an integer and we know that the average speed is num__2.8 s must be = num__2 . that meanss + num__1 = num__3 . this implies that the ratio of time for s = num__2 is num__0.25 of the total time . the formula for distance / rate is d = rt . . . so the distance travelled when s = num__2 is num__2 t . the distance travelled for s + num__1 = num__3 is num__3 * num__4 t or num__12 t . therefore total distance covered while the sun was shining over him is r = num__0.142857142857 = num__0.142857142857 . answer : e <eor> e <eos> |
e |
round_down__2.8__ add__1.0__2.0__ reverse__0.25__ divide__3.0__0.25__ multiply__1.0__0.1429__ |
round_down__2.8__ add__1.0__2.0__ reverse__0.25__ divide__3.0__0.25__ multiply__1.0__0.1429__ |
| num__4 num__2549 num__121169 num__289361 <o> a ) num__149 <o> b ) num__169 <o> c ) num__189 <o> d ) num__529 <o> e ) num__219 |
num__23 ^ num__2 = num__529 because follow sequence of square of the prime numbers answer : d <eor> d <eos> |
d |
power__23.0__2.0__ power__23.0__2.0__ |
power__23.0__2.0__ power__23.0__2.0__ |
| the average runs scored by a batsman in num__20 matches is num__40 . in the next num__10 matches the batsman scored an average of num__13 runs . find his average in all the num__30 matches ? <o> a ) num__31 <o> b ) num__66 <o> c ) num__55 <o> d ) num__77 <o> e ) num__81 |
total score of the batsman in num__20 matches = num__800 . total score of the batsman in the next num__10 matches = num__130 . total score of the batsman in the num__30 matches = num__930 . average score of the batsman = num__31.0 = num__31 . answer : a <eor> a <eos> |
a |
multiply__20.0__40.0__ multiply__10.0__13.0__ add__800.0__130.0__ divide__930.0__30.0__ divide__930.0__30.0__ |
multiply__20.0__40.0__ multiply__10.0__13.0__ add__800.0__130.0__ divide__930.0__30.0__ divide__930.0__30.0__ |
| two men and three women can repair a bridge in num__10 days while three men and two women can do same work in num__8 days . if two men and one woman are used to finish this work in how many days they will complete it ? <o> a ) num__9.7 days <o> b ) num__12.5 days <o> c ) num__10 days <o> d ) num__11.3 days <o> e ) num__12.7 days |
work done by num__2 men and num__3 women in num__1 day = num__0.1 work done by num__3 men and num__2 women in num__1 day = num__0.125 let num__1 men does m work in num__1 day and num__1 women does w work in num__1 day . the above equations can be written as num__2 m + num__3 w = num__0.1 - - - ( num__1 ) num__3 m + num__2 w = num__0.125 - - - ( num__2 ) solving equation ( num__1 ) and ( num__2 ) we get m = num__0.07 and w = num__0.01 num__2 men and num__1 woman together can finish the work in = num__0.07 + num__0.01 = num__0.08 = num__12.5 days b <eor> b <eos> |
b |
subtract__10.0__8.0__ subtract__3.0__2.0__ divide__1.0__10.0__ divide__1.0__8.0__ divide__0.1__10.0__ multiply__8.0__0.01__ divide__0.125__0.01__ round__12.5__ |
subtract__10.0__8.0__ subtract__3.0__2.0__ divide__1.0__10.0__ divide__1.0__8.0__ divide__0.1__10.0__ add__0.07__0.01__ divide__0.125__0.01__ round__12.5__ |
| a tank is filled by three pipes with uniform flow . the first two pipes operating simultaneously fill the tank in the same during which the tank is filled by the third pipe alone . the second pipe fills the tank num__5 hours faster than the first pipe and num__4 hours slower than the third pipe . the time required by the first pipe is ? <o> a ) num__16 <o> b ) num__13 <o> c ) num__15 <o> d ) num__18 <o> e ) num__12 |
suppose first pipe alone takes x hours to fill the tank . then second and third pipes will take ( x - num__5 ) and ( x - num__9 ) hours respectively to fill the tank . num__1 / x + num__1 / ( x - num__5 ) = num__1 / ( x - num__9 ) ( num__2 x - num__5 ) ( x - num__9 ) = x ( x - num__5 ) x num__2 - num__18 x + num__45 = num__0 ( x - num__15 ) ( x - num__3 ) = num__0 = > x = num__15 answer : c <eor> c <eos> |
c |
add__5.0__4.0__ subtract__5.0__4.0__ multiply__2.0__9.0__ multiply__5.0__9.0__ subtract__5.0__2.0__ round__15.0__ |
add__5.0__4.0__ subtract__5.0__4.0__ multiply__2.0__9.0__ multiply__5.0__9.0__ subtract__5.0__2.0__ divide__45.0__3.0__ |
| a number increased by num__10.0 gives num__660 . the number is ? <o> a ) num__200 <o> b ) num__300 <o> c ) num__500 <o> d ) num__600 <o> e ) num__400 |
formula = total = num__100.0 increase = ` ` + ' ' decrease = ` ` - ' ' a number means = num__100.0 that same number increased by num__10.0 = num__110.0 num__110.0 - - - - - - - > num__660 ( num__110 Ã — num__6 = num__660 ) num__100.0 - - - - - - - > num__600 ( num__100 Ã — num__6 = num__600 ) option ' d ' <eor> d <eos> |
d |
percent__100.0__600.0__ |
percent__100.0__600.0__ |
| a driver covers a certain distance by car driving at num__60 km / hr and returns back to the starting point riding on a scooter at num__10 km / hr . what was the average speed for the whole journey ? <o> a ) num__14.2 km / h <o> b ) num__17.1 km / h <o> c ) num__19.3 km / h <o> d ) num__21.7 km / h <o> e ) num__23.5 km / h |
time num__1 = d / num__60 time num__2 = d / num__10 total time = d / num__60 + d / num__10 = num__7 d / num__60 average speed = total distance / total time = num__2 d / ( num__7 d / num__60 ) = num__17.1 km / h the answer is b . <eor> b <eos> |
b |
multiply__1.0__17.1__ |
divide__17.1__1.0__ |
| walking at num__0.75 of her normal speed a worker is num__12 minutes later than usual in reaching her office . the usual time ( in minutes ) taken by her to cover the distance between her home and her office is <o> a ) num__36 <o> b ) num__48 <o> c ) num__60 <o> d ) num__62 <o> e ) num__66 |
let v be her normal speed and let t be her normal time . d = ( num__0.75 ) v * ( t + num__12 ) since the distance is the same we can equate this to a regular day which is d = v * t v * t = ( num__0.75 ) v * ( t + num__12 ) t / num__4 = num__9 t = num__36 the answer is a . <eor> a <eos> |
a |
multiply__0.75__12.0__ multiply__4.0__9.0__ round__36.0__ |
multiply__0.75__12.0__ multiply__4.0__9.0__ multiply__4.0__9.0__ |
| the ratio of a to b to c is num__2 to num__3 to num__4 and a b c are positive integers . if the average ( arithmetic mean ) of the three numbers is num__27 what is the value of a ? <o> a ) num__18 <o> b ) num__20 <o> c ) num__21 <o> d ) num__22 <o> e ) num__24 |
let a = num__2 k b = num__3 k and c = num__4 k . a + b + c = num__2 k + num__3 k + num__4 k = num__9 k since the average is num__27 : num__9 k / num__3 = num__27 num__3 k = num__27 k = num__9 then a = num__18 . the answer is a . <eor> a <eos> |
a |
divide__27.0__3.0__ multiply__2.0__9.0__ multiply__2.0__9.0__ |
divide__27.0__3.0__ multiply__2.0__9.0__ multiply__2.0__9.0__ |
| a train speeds past a pole in num__15 seconds and a platform num__100 m long in num__25 seconds . its length is ? <o> a ) num__188 m <o> b ) num__876 m <o> c ) num__251 m <o> d ) num__150 m <o> e ) num__145 m |
let the length of the train be x meters and its speed be y m / sec . they x / y = num__15 = > y = x / num__15 x + num__4.0 = x / num__15 x = num__150 m . answer : d <eor> d <eos> |
d |
divide__100.0__25.0__ round__150.0__ |
divide__100.0__25.0__ round__150.0__ |
| if num__76 is divided into four parts proportional to num__75 num__34 then the smallest part is <o> a ) num__17 <o> b ) num__15 <o> c ) num__16 <o> d ) num__19 <o> e ) num__12 |
given ratio = num__7 : num__5 : num__3 : num__4 sum of ratio trems = num__19 smallest part = ( num__76 x num__0.157894736842 ) = num__12 answer e num__12 <eor> e <eos> |
e |
subtract__7.0__3.0__ divide__76.0__4.0__ divide__3.0__19.0__ multiply__3.0__4.0__ multiply__3.0__4.0__ |
subtract__7.0__3.0__ divide__76.0__4.0__ divide__3.0__19.0__ multiply__3.0__4.0__ multiply__3.0__4.0__ |
| the positive integers p and q leave remainders of num__2 and num__3 respectively when divided by num__6 . p > q . what is the remainder when p – q is divided by num__6 ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__3 <o> d ) num__2 <o> e ) num__1 |
let ' s test out some values of p and q that satisfy the given information . p leaves are remainder of num__2 when divided by num__6 so p could equal num__8 q leaves are remainder of num__3 when divided by num__6 so q could equal num__3 what is the remainder when p – q is divided by num__6 ? so p - q = num__8 - num__3 = num__5 and when we divide num__5 by num__6 we get num__0 with remainder num__5 answer : a <eor> a <eos> |
a |
add__2.0__6.0__ add__2.0__3.0__ add__2.0__3.0__ |
add__2.0__6.0__ subtract__8.0__3.0__ subtract__8.0__3.0__ |
| simplify : num__0.2 * num__0.4 + num__0.6 * num__0.8 <o> a ) num__0.52 <o> b ) num__0.42 <o> c ) num__0.48 <o> d ) num__0.64 <o> e ) num__0.56 |
given exp . = num__0.2 * num__0.4 + ( num__0.6 * num__0.8 ) = num__0.08 + num__0.48 = num__0.56 answer is e . <eor> e <eos> |
e |
multiply__0.2__0.4__ add__0.4__0.08__ add__0.48__0.08__ add__0.48__0.08__ |
multiply__0.2__0.4__ add__0.4__0.08__ add__0.48__0.08__ add__0.48__0.08__ |
| the current birth rate per certain number of people is num__32 whereas corresponding death rate is num__11 per same number of people . if the net growth rate in terms of population increase is num__2.1 percent find number of persons . ( initally ) <o> a ) num__1000 <o> b ) num__10000 <o> c ) num__100 <o> d ) num__2000 <o> e ) num__20000 |
sol . net growth on x = ( num__32 - num__11 ) = num__21 . net growth on num__100 = ( num__21 / x à — num__100 ) % = num__2.1 . then x = num__1000 answer : a <eor> a <eos> |
a |
percent__100.0__1000.0__ |
percent__100.0__1000.0__ |
| a and b together can do a piece of work in num__10 days . b alone can finish it in num__20 days . in how many days can a alone finish the work ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__21 <o> d ) num__22 <o> e ) num__23 |
time taken by a to finish the work = xy / ( y - x ) = num__10 x num__20 / ( num__20 - num__10 ) = num__20.0 = num__20 days answer : option b <eor> b <eos> |
b |
round__20.0__ |
round__20.0__ |
| num__5358 x num__52 = ? <o> a ) num__273762 <o> b ) num__278616 <o> c ) num__273298 <o> d ) num__273258 <o> e ) num__277382 |
b num__5358 x num__52 = num__5358 x ( num__50 + num__2 ) = num__5358 x num__50 + num__5358 x num__2 = num__267900 + num__10716 = num__278616 . <eor> b <eos> |
b |
subtract__52.0__50.0__ multiply__5358.0__50.0__ multiply__5358.0__2.0__ multiply__5358.0__52.0__ multiply__5358.0__52.0__ |
subtract__52.0__50.0__ multiply__5358.0__50.0__ multiply__5358.0__2.0__ add__10716.0__267900.0__ add__10716.0__267900.0__ |
| a can do a piece of work in num__12 days which b can do in num__12 days . b worked at it for num__9 days . a can finish the remaining work in ? <o> a ) num__3 days <o> b ) num__5 days <o> c ) num__6 days <o> d ) num__7 days <o> e ) num__11 days |
b ' s num__9 day ' s work = num__9 x ( num__0.0833333333333 ) = num__0.75 remaining work = ( num__1 - num__0.75 ) = num__0.25 num__0.25 work is done by a in = num__12 x ( num__0.25 ) = num__3 days . answer : a <eor> a <eos> |
a |
divide__9.0__12.0__ subtract__1.0__0.75__ subtract__12.0__9.0__ round__3.0__ |
divide__9.0__12.0__ subtract__1.0__0.75__ subtract__12.0__9.0__ subtract__12.0__9.0__ |
| a reduction of num__25.0 in the price of oil enables a house wife to obtain num__5 kgs more for rs . num__700 what is the reduced price for kg ? <o> a ) num__35 <o> b ) num__60 <o> c ) num__70 <o> d ) num__80 <o> e ) num__20 |
a num__700 * ( num__0.25 ) = num__175 - - - - num__5 ? - - - - num__1 = > rs . num__35 <eor> a <eos> |
a |
percent__25.0__700.0__ percent__5.0__700.0__ percent__5.0__700.0__ |
percent__25.0__700.0__ percent__5.0__700.0__ percent__5.0__700.0__ |
| the average age of num__20 students in a class is num__15 years . if the age of teacher is also included the average becomes num__16 years find the age of the teacher . <o> a ) num__35 <o> b ) num__38 <o> c ) num__45 <o> d ) num__36 <o> e ) num__60 |
explanation : if teacher ' s age is num__15 years there is no change in the average . but teacher has contributed num__1 year to all the students along with maintaining his age at num__16 . age of teacher = average age of all + total increase in age = num__16 + ( num__1 x num__20 ) = num__36 years answer : d <eor> d <eos> |
d |
subtract__16.0__15.0__ add__20.0__16.0__ add__20.0__16.0__ |
subtract__16.0__15.0__ add__20.0__16.0__ add__20.0__16.0__ |
| a b and c run around a circular track starting from the same point simultaneously and in the same direction at speeds of num__4 kmph num__6 kmph and num__8 kmph respectively . if the length of the track is num__400 meters when will a b and c meet at the starting point for the first time after they started the race ? <o> a ) num__1 minute <o> b ) num__5 minutes <o> c ) num__12 minutes <o> d ) num__20 minutes <o> e ) num__24 minutes |
num__4 kmph num__6 kmph and num__8 kmph is equal to num__66.6666666667 mtrs / min num__100.0 mtrs / min and num__133.333333333 mtrs / min or num__66.6666666667 num__100 num__133.333333333 mtrs / min respectively . seeing this we can infer that the answer should be at least divisible by num__3 . ( minutes in options are integers . so if the answer is not divisible by num__3 we will have num__66.6666666667 * answer = distance traveled by a in fractions whereas num__100 * answer = distance traveled by a as integer ) so a b and d are out . at num__12 min a b and c will travel num__800 num__1200 and num__1600 mtrs respectively . length of the circle is given as num__400 mtrs . as num__800 num__1200 and num__1600 are divisible by num__400 we can say that a b and c will be at starting point after num__12 min . hence c will be the answer . <eor> c <eos> |
c |
divide__400.0__6.0__ divide__400.0__4.0__ divide__400.0__133.3333__ add__4.0__8.0__ multiply__8.0__100.0__ add__400.0__800.0__ multiply__4.0__400.0__ round__12.0__ |
divide__400.0__6.0__ divide__400.0__4.0__ divide__400.0__133.3333__ multiply__4.0__3.0__ multiply__8.0__100.0__ multiply__400.0__3.0__ multiply__4.0__400.0__ multiply__4.0__3.0__ |
| the area of a sector of a circle of radius num__5 cm formed by an arc of length num__3.5 cm is ? <o> a ) num__8.78 <o> b ) num__8.67 <o> c ) num__8.75 <o> d ) num__8.98 <o> e ) num__8.28 |
( num__5 * num__3.5 ) / num__2 = num__8.75 answer : c <eor> c <eos> |
c |
triangle_area__5.0__3.5__ triangle_area__5.0__3.5__ |
triangle_area__5.0__3.5__ triangle_area__5.0__3.5__ |
| a man traveled a total distance of num__1800 km . he traveled one - third of the whole trip by plane and the distance traveled by train is three - fifth of the distance traveled by bus . if he traveled by train plane and bus then find the distance traveled by bus ? <o> a ) num__450 km <o> b ) num__850 km <o> c ) num__1200 km <o> d ) num__750 km <o> e ) none of these |
total distance traveled = num__1800 km . distance traveled by plane = num__600 km . distance traveled by bus = x distance traveled by train = num__3 x / num__5 = > x + num__3 x / num__5 + num__600 = num__1800 = > num__8 x / num__5 = num__1200 = > x = num__750 km . answer : d <eor> d <eos> |
d |
divide__1800.0__600.0__ add__3.0__5.0__ subtract__1800.0__600.0__ round__750.0__ |
divide__1800.0__600.0__ add__3.0__5.0__ subtract__1800.0__600.0__ round__750.0__ |
| which is greater num__0.666666666667 or num__0.75 ? <o> a ) num__0.666666666667 <o> b ) num__0.75 <o> c ) both are equal <o> d ) can not be determined <o> e ) both are equal and equal to num__1 |
to know which is greater num__0.666666666667 or num__0.75 = multiply both by num__4 * num__3 = num__12 ( product of denominator ) = num__0.666666666667 * num__12 and num__0.75 * num__12 respectively = num__8 and num__9 respectively which means num__0.75 is greater . ans : b <eor> b <eos> |
b |
multiply__0.75__4.0__ multiply__3.0__4.0__ subtract__12.0__4.0__ multiply__0.75__12.0__ divide__3.0__4.0__ |
multiply__0.75__4.0__ multiply__3.0__4.0__ subtract__12.0__4.0__ multiply__0.75__12.0__ divide__3.0__4.0__ |
| a money lender lent a total of $ num__1700 to two of his customers . he charged at the rate of num__5.0 p . a . to one of them and num__6.0 p . a . to the other . if he earned an average of num__5.67 on $ num__1700 how much did he lend to each of the two customers ? <o> a ) num__700 ; num__1100 <o> b ) num__1139 ; num__561 <o> c ) num__1000 ; num__800 <o> d ) num__1200 ; num__800 <o> e ) none of the above |
the method given above is the one given in the book . however the method i used was ( num__1 * x * num__5 ) / num__100 + ( num__1700 - x ) * num__0.06 = num__1700 * num__5.67 / num__100 simplifying we get x = num__561 b <eor> b <eos> |
b |
subtract__6.0__5.0__ divide__6.0__100.0__ subtract__1700.0__561.0__ |
subtract__6.0__5.0__ divide__6.0__100.0__ subtract__1700.0__561.0__ |
| two trains are running at num__40 km / hr and num__20 km / hr respectively in the same direction . fast train completely passes a man sitting in the slower train in num__5 sec . what is the length of the fast train ? <o> a ) num__27 num__0.875 m <o> b ) num__77 num__0.777777777778 m <o> c ) num__27 num__0.888888888889 m <o> d ) num__27 num__0.777777777778 m <o> e ) num__22 num__0.777777777778 m |
relative speed = ( num__40 - num__20 ) = num__20 km / hr . = num__20 * num__0.277777777778 = num__5.55555555556 m / sec . length of faster train = num__5.55555555556 * num__5 = num__27.7777777778 = num__27 num__0.777777777778 m . answer : d <eor> d <eos> |
d |
subtract__27.7778__27.0__ round__27.0__ |
subtract__27.7778__27.0__ subtract__27.7778__0.7778__ |
| a class average mark in an exam is num__70 . the average of students who scored below num__60 is num__50 . the average of students who scored num__60 or more is num__75 . if the total number of students in this class is num__20 how many students scored below num__60 ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__38 <o> d ) num__42 <o> e ) num__33 |
let n the number of students who scored below num__60 and n the number of students who scored num__60 or more . xi the grades below num__60 and yi the grades num__60 or above . [ sum ( xi ) + sum ( yi ) ] / num__20 = num__70 : class average sum ( xi ) / n = num__50 : average for less that num__60 sum ( yi ) / n = num__75 : average for num__60 or more num__50 n + num__75 n = num__1400 : combine the above equations n + n = num__20 : total number of students n = num__4 and n = num__16 : solve the above system correct answer b <eor> b <eos> |
b |
multiply__70.0__20.0__ subtract__20.0__4.0__ subtract__70.0__50.0__ |
multiply__70.0__20.0__ subtract__20.0__4.0__ add__4.0__16.0__ |
| it is currently num__9 : num__53 pm . what time was it in the morning exactly num__149061 minutes ago ? <o> a ) num__9 : num__12 <o> b ) num__9 : num__18 <o> c ) num__9 : num__26 <o> d ) num__9 : num__32 <o> e ) num__9 : num__38 |
divide by num__60 to convert num__149061 minutes to hours : num__2484.35 = num__2484 r num__21 . that is num__2484 hours and num__21 minutes . all of the answers are in the num__9 am hour before : num__53 thus we can assume num__2484 previous must be num__9 : num__53 am . num__21 minutes before that is num__9 : num__32 am . d <eor> d <eos> |
d |
hour_to_min_conversion__ divide__149061.0__60.0__ subtract__53.0__21.0__ round__9.0__ |
hour_to_min_conversion__ divide__149061.0__60.0__ subtract__53.0__21.0__ round__9.0__ |
| a photograph is copied onto a sheet of paper num__8.5 inches by num__10 inches . a num__1.5 inch margin is left all around . what area in square inches does the picture cover ? <o> a ) num__76 <o> b ) num__65 <o> c ) num__59.5 <o> d ) num__49 <o> e ) num__38.5 |
area covered by photograph = ( num__8.5 - num__3 ) * ( num__10 - num__3 ) = num__5.5 * num__7 = num__38.5 answer e <eor> e <eos> |
e |
multiply__5.5__7.0__ multiply__5.5__7.0__ |
multiply__5.5__7.0__ multiply__5.5__7.0__ |
| the ratio between the speeds of two trains is num__7 : num__8 . if the second train runs num__400 kms in num__4 hours then the speed of the first train is ? <o> a ) num__83.5 km / hr <o> b ) num__84.5 km / hr <o> c ) num__86.5 km / hr <o> d ) num__87.5 km / hr <o> e ) none of these |
explanation : let the speeds of two trains be num__7 x and num__8 x km / hr . num__8 / x = num__100.0 = > x = num__12.5 km / hr so speed of first train is num__12.5 * num__7 = num__87.5 km / hr option d <eor> d <eos> |
d |
divide__400.0__4.0__ divide__100.0__8.0__ multiply__7.0__12.5__ round__87.5__ |
divide__400.0__4.0__ divide__100.0__8.0__ multiply__7.0__12.5__ multiply__7.0__12.5__ |
| how many terminating zeroes q does num__200 ! have ? <o> a ) num__40 <o> b ) num__48 <o> c ) num__49 <o> d ) num__55 <o> e ) num__64 |
you have num__40 multiples of num__5 num__8 of num__25 and num__1 of num__125 . this will give num__49 zeros . c <eor> c <eos> |
c |
divide__200.0__40.0__ divide__40.0__5.0__ divide__200.0__8.0__ multiply__5.0__25.0__ multiply__1.0__49.0__ |
divide__200.0__40.0__ divide__40.0__5.0__ divide__200.0__8.0__ multiply__5.0__25.0__ multiply__1.0__49.0__ |
| evaluate ( ( num__5.68 ) num__2 – ( num__4.32 ) num__2 ) / num__5.68 - num__4.32 <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
given expression = a num__2 - b num__2 / a - b = ( a + b ) ( a - b ) / a - b = ( a + b ) ( num__5.68 ) num__2 – ( num__4.32 ) num__2 / num__5.68 - num__4.32 = ( num__5.68 + num__4.32 ) = num__10 answer : c <eor> c <eos> |
c |
add__5.68__4.32__ add__5.68__4.32__ |
add__5.68__4.32__ add__5.68__4.32__ |
| a man has some hens and cows . if the number of heads be num__48 and the number of feet equals num__140 then the number of hens will be : <o> a ) num__22 <o> b ) num__23 <o> c ) num__24 <o> d ) num__26 <o> e ) num__25 |
explanation : let the number of hens be x and the number of cows be y . then x + y = num__48 . . . . ( i ) and num__2 x + num__4 y = num__140 x + num__2 y = num__70 . . . . ( ii ) solving ( i ) and ( ii ) we get : x = num__26 y = num__22 . the required answer = num__26 . answer is d <eor> d <eos> |
d |
divide__140.0__2.0__ subtract__48.0__26.0__ subtract__48.0__22.0__ |
divide__140.0__2.0__ subtract__48.0__26.0__ add__4.0__22.0__ |
| skier sarah completes a straight num__300 - meter downhill run in t seconds and at an average speed of ( x + num__10 ) meters per second . she then rides a chairlift back up the mountain the same distance at an average speed of ( x - num__8 ) meters per second . if the ride up the mountain took num__135 seconds longer than her run down the mountain what was her average speed in meters per second during her downhill run ? <o> a ) num__10 <o> b ) num__15 <o> c ) num__20 <o> d ) num__25 <o> e ) num__30 |
backsolving is the preferred approach for this one num__300 / ( x - num__8 ) - num__300 / ( x + num__10 ) = num__135 we are asked for the value of x + num__10 so we can start with c = x + num__10 = num__20 so x = num__10 num__150.0 - num__15.0 = num__135 hence answer is c <eor> c <eos> |
c |
divide__300.0__20.0__ divide__300.0__15.0__ |
divide__300.0__20.0__ divide__300.0__15.0__ |
| john is measuring a banner for the school play and needs to know how many inches are in num__5 yards and num__2 feet . can you help him out ? <o> a ) num__235 inches . <o> b ) num__204 inches . <o> c ) num__220 inches . <o> d ) num__206 inches . <o> e ) num__208 inches . |
b num__204 inches . num__5 yards is the same as num__15 feet ( num__3 feet = num__1 yard ) . num__15 feet + num__2 feet = num__17 feet . there are num__12 inches in each foot . so num__17 feet x num__12 = num__204 inches . <eor> b <eos> |
b |
subtract__5.0__2.0__ subtract__3.0__2.0__ add__2.0__15.0__ divide__204.0__17.0__ multiply__1.0__204.0__ |
subtract__5.0__2.0__ subtract__3.0__2.0__ add__2.0__15.0__ divide__204.0__17.0__ multiply__1.0__204.0__ |
| a train num__150 m long running at num__72 kmph crosses a platform in num__25 sec . what is the length of the platform ? <o> a ) num__443 m <o> b ) num__354 m <o> c ) num__450 m <o> d ) num__350 m <o> e ) num__250 m |
d num__350 d = num__72 * num__0.277777777778 = num__25 = num__500 – num__150 = num__350 <eor> d <eos> |
d |
add__150.0__350.0__ round__350.0__ |
add__150.0__350.0__ round__350.0__ |
| the diagonals of a rhombus are num__13 cm and num__20 cm . find its area ? <o> a ) num__329 <o> b ) num__288 <o> c ) num__150 <o> d ) num__130 <o> e ) num__31 |
num__0.5 * num__13 * num__20 = num__130 answer : d <eor> d <eos> |
d |
triangle_area__13.0__20.0__ triangle_area__13.0__20.0__ |
volume_rectangular_prism__13.0__20.0__0.5__ volume_rectangular_prism__13.0__20.0__0.5__ |
| find the area of trapezium whose parallel sides are num__10 cm and num__18 cm long and the distance between them is num__10 cm . <o> a ) num__140 cm num__2 <o> b ) num__185 cm num__2 <o> c ) num__185 cm num__2 <o> d ) num__185 cm num__2 <o> e ) num__195 cm num__2 |
explanation : area of a trapezium = num__0.5 ( sum of parallel sides ) * ( perpendicular distance between them ) = num__0.5 ( num__10 + num__18 ) * ( num__10 ) = num__140 cm num__2 answer : option a <eor> a <eos> |
a |
round__140.0__ |
round__140.0__ |
| on a ferry there are num__32 cars and num__8 trucks . the cars have an average mass of num__900 kg and the trucks have an average mass of num__3000 kg . what is the average mass of all num__40 vehicles on the ferry ? <o> a ) num__1200 <o> b ) num__1240 <o> c ) num__1280 <o> d ) num__1320 <o> e ) num__1360 |
the total mass of the vehicles is num__32 * num__900 + num__8 * num__3000 = num__52800 the average mass is num__1320.0 = num__1320 the answer is d . <eor> d <eos> |
d |
divide__52800.0__40.0__ divide__52800.0__40.0__ |
divide__52800.0__40.0__ divide__52800.0__40.0__ |
| if the sides of a triangle are num__26 cm num__24 cm and num__12 cm what is its area ? <o> a ) num__120 cm num__2 <o> b ) num__112 cm num__2 <o> c ) num__144 cm num__2 <o> d ) num__666 cm num__2 <o> e ) num__886 cm num__2 |
the triangle with sides num__26 cm num__24 cm and num__10 cm is right angled where the hypotenuse is num__26 cm . area of the triangle = num__0.5 * num__24 * num__12 = num__144 cm num__2 answer : c <eor> c <eos> |
c |
side_by_diagonal__26.0__24.0__ triangle_area__24.0__12.0__ square_perimeter__0.5__ triangle_area__24.0__12.0__ |
side_by_diagonal__26.0__24.0__ volume_rectangular_prism__24.0__12.0__0.5__ square_perimeter__0.5__ volume_rectangular_prism__24.0__12.0__0.5__ |
| which one of the following numbers is a solution of the equation ( x ( num__2 sq ) - num__2 ) ( x ( num__2 sq ) + num__6 ) ( x ( third num__3 ) - num__8 ) = num__0 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
one checks each of the ve given possible answers . since num__2 - num__3 third - - num__8 = num__0 the answer is num__2 . correct answer b <eor> b <eos> |
b |
divide__6.0__3.0__ |
subtract__8.0__6.0__ |
| three partners a b c in a business invested money such that num__3 ( a â € ™ s capital ) = num__2 ( b â € ™ s capital ) = num__5 ( c â € ™ s capital ) then the ratio of their capitals is <o> a ) num__63 : num__45 : num__34 <o> b ) num__63 : num__54 : num__34 <o> c ) num__36 : num__54 : num__28 <o> d ) num__63 : num__45 : num__35 <o> e ) num__10 : num__15 : num__6 |
explanation : let num__3 ( a â € ™ s capital ) = num__2 ( b â € ™ s capital ) = num__5 ( c â € ™ s capital ) = rs . x then a â € ™ s capital = rs x / num__3 b â € ™ s capital = rs . x / num__2 and c â € ™ s capital = rs . x / num__5 . a : b : c = x / num__3 : x / num__2 : x / num__5 num__10 : num__15 : num__6 answer : option e <eor> e <eos> |
e |
multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__3.0__2.0__ multiply__2.0__5.0__ |
multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__3.0__2.0__ multiply__2.0__5.0__ |
| rs . num__850 becomes rs . num__956 in num__3 years at a certain rate of simple interest . if the rate of interest is increased by num__4.0 what amount will rs . num__850 become in num__3 years ? <o> a ) rs . num__1020.80 <o> b ) rs . num__1025 <o> c ) rs . num__1058 <o> d ) data inadequate <o> e ) none of these |
solution s . i . = rs . ( num__956 - num__850 ) = rs . num__106 rate = ( num__100 x num__0.124705882353 x num__3 ) = num__4.1568627451 % new rate = ( num__4.1568627451 + num__4 ) % = num__8.1568627451 % new s . i . = rs . ( num__850 x num__8.1568627451 x num__0.03 ) rs . num__208 . ∴ new amount = rs . ( num__850 + num__208 ) = rs . num__1058 . answer c <eor> c <eos> |
c |
percent__100.0__1058.0__ |
percent__100.0__1058.0__ |
| eight years ago p was half of q ' s age . if the ratio of their present ages is num__3 : num__4 what will be the total of their present ages ? <o> a ) num__33 <o> b ) num__67 <o> c ) num__28 <o> d ) num__31 <o> e ) num__35 |
let present age of p and q be num__3 x num__3 x and num__4 x num__4 x respectively . eight years ago p was half of q ' s age â ‡ ’ ( num__3 x â ˆ ’ num__8 ) = num__0.5 ( num__4 x â ˆ ’ num__8 ) â ‡ ’ num__6 x â ˆ ’ num__16 = num__4 x â ˆ ’ num__8 â ‡ ’ num__2 x = num__8 â ‡ ’ x = num__4 total of their present ages = num__3 x + num__4 x = num__7 x = num__7 Ã — num__4 = num__28 c <eor> c <eos> |
c |
divide__4.0__8.0__ divide__3.0__0.5__ divide__8.0__0.5__ reverse__0.5__ add__3.0__4.0__ multiply__4.0__7.0__ multiply__4.0__7.0__ |
divide__4.0__8.0__ divide__3.0__0.5__ divide__8.0__0.5__ reverse__0.5__ add__3.0__4.0__ multiply__4.0__7.0__ multiply__4.0__7.0__ |
| the probability is num__0.6 that an “ unfair ” coin will turn up tails on any given toss . if the coin is tossed num__3 times what is the probability that at least one of the tosses will turn up tails ? <o> a ) num__0.064 <o> b ) num__0.36 <o> c ) num__0.64 <o> d ) num__0.784 <o> e ) num__0.936 |
probability that at least one of the tosses will turn up tails = num__1 - probability that all will be heads = num__1 - ( num__0.4 * num__0.4 * num__0.4 ) = num__1 - num__0.064 = num__0.936 e <eor> e <eos> |
e |
negate_prob__0.6__ negate_prob__0.064__ negate_prob__0.064__ |
negate_prob__0.6__ negate_prob__0.064__ negate_prob__0.064__ |
| if x is a positive even number then each of the following is even odd except <o> a ) ( x + num__3 ) ( x + num__5 ) <o> b ) x ^ num__2 + num__4 <o> c ) x ^ num__2 + num__6 x + num__9 <o> d ) num__3 x ^ num__2 + num__4 <o> e ) num__5 ( x + num__3 ) |
since it is given tht x is even number so any integer multiplied with x will also be even . . so we should concentrate only on other terms . . lets see the choices . . a . ( x + num__3 ) ( x + num__5 ) we have two terms with x and each is added with a odd number . . each bracket becomes odd and odd * odd = odd b . x ^ num__2 + num__5 here we are adding an odd number to even . . so e + o = o c . x ^ num__2 + num__6 x + num__9 here we are again adding an odd number to even . . so e + e + o = o d . num__3 x ^ num__2 + num__4 here we are adding an even number to even . . so e + e = e . . so tjis is our answer e . num__5 ( x + num__3 ) again o * o = o b <eor> b <eos> |
b |
subtract__5.0__3.0__ multiply__2.0__3.0__ power__3.0__2.0__ subtract__6.0__2.0__ subtract__4.0__2.0__ |
subtract__5.0__3.0__ multiply__2.0__3.0__ power__3.0__2.0__ subtract__6.0__2.0__ subtract__4.0__2.0__ |
| barbata invests $ num__2400 in the national bank at num__5.0 . how much additional money must she invest at num__10.0 so that the total annual income will be equal to num__6.0 of her entire investment ? <o> a ) num__120 <o> b ) num__600 <o> c ) num__1000 <o> d ) num__360 <o> e ) num__240 |
let the additional invested amount for num__10.0 interest be x ; equation will be ; num__2400 + num__0.05 * num__2400 + x + num__0.10 x = num__2400 + x + num__0.06 ( num__2400 + x ) num__0.05 * num__2400 + num__0.10 x = num__0.06 x + num__0.06 * num__2400 num__0.04 x = num__2400 ( num__0.06 - num__0.05 ) x = num__2400 * num__0.01 / num__0.04 = num__600 ans : b <eor> b <eos> |
b |
reverse__10.0__ subtract__0.1__0.06__ divide__0.05__5.0__ divide__6.0__0.01__ divide__6.0__0.01__ |
reverse__10.0__ subtract__0.1__0.06__ divide__0.05__5.0__ divide__6.0__0.01__ divide__6.0__0.01__ |
| num__54 men working num__8 hours per day dig num__30 m deep . how many extra men should be put to dig to a depth of num__50 m working num__6 hours per day ? <o> a ) a ) num__57 <o> b ) b ) num__77 <o> c ) c ) num__66 <o> d ) d ) num__55 <o> e ) e ) num__67 |
( num__54 * num__8 ) / num__30 = ( x * num__6 ) / num__50 = > x = num__120 num__120 – num__54 = num__66 answer : c <eor> c <eos> |
c |
subtract__120.0__54.0__ round__66.0__ |
subtract__120.0__54.0__ round__66.0__ |
| how many bricks each measuring num__25 cm x num__11.25 cm x num__6 cm will be needed to build a wall of num__8 m x num__6 m x num__22.5 cm ? <o> a ) num__3488 <o> b ) num__37799 <o> c ) num__6400 <o> d ) num__2777 <o> e ) num__2911 |
explanation : number of bricks \ frac { num__800 \ times num__600 \ times num__22.5 } { num__25 \ times num__11.25 \ times num__6 } = num__6400 answer : c ) num__6400 <eor> c <eos> |
c |
multiply__8.0__800.0__ round__6400.0__ |
multiply__8.0__800.0__ round__6400.0__ |
| there are num__18 stations between hyderabad and bangalore . how many second class tickets have to be printed so that a passenger can travel from any station to any other station ? <o> a ) num__156 <o> b ) num__167 <o> c ) num__157 <o> d ) num__352 <o> e ) num__380 |
the total number of stations = num__20 from num__20 stations we have to choose any two stations and the direction of travel ( i . e . hyderabad to bangalore is different from bangalore to hyderabad ) in ² ⁰ p ₂ ways . ² ⁰ p ₂ = num__20 * num__19 = num__380 . answer : e <eor> e <eos> |
e |
multiply__19.0__20.0__ round__380.0__ |
multiply__19.0__20.0__ multiply__19.0__20.0__ |
| find the area of trapezium whose parallel sides are num__20 cm and num__18 cm long and the distance between them is num__15 cm . <o> a ) num__335 cm num__2 <o> b ) num__885 cm num__2 <o> c ) num__285 cm num__2 <o> d ) num__825 cm num__2 <o> e ) num__725 cm num__2 |
area of a trapezium = num__0.5 ( sum of parallel sides ) * ( perpendicular distance between them ) = num__0.5 ( num__20 + num__18 ) * ( num__15 ) = num__285 cm num__2 answer : c <eor> c <eos> |
c |
subtract__20.0__18.0__ round__285.0__ |
subtract__20.0__18.0__ round__285.0__ |
| car x began traveling at an average speed of num__35 miles per hour . after num__48 minutes car y began traveling at an average speed of num__39 miles per hour . when both cars had traveled the same distance both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ? <o> a ) num__105 <o> b ) num__140 <o> c ) num__175 <o> d ) num__210 <o> e ) num__245 |
in num__48 minutes car x travels num__28 miles . car y gains num__4 miles each hour so it takes num__7 hours to catch car x . in num__7 hours car x travels num__245 miles . the answer is e . <eor> e <eos> |
e |
subtract__39.0__35.0__ subtract__35.0__28.0__ multiply__35.0__7.0__ round__245.0__ |
subtract__39.0__35.0__ subtract__35.0__28.0__ multiply__35.0__7.0__ round__245.0__ |
| if the cost price of num__12 items is equal to the selling price of num__16 items the loss percent is <o> a ) num__20.0 <o> b ) num__25.0 <o> c ) num__30.0 <o> d ) num__35.0 <o> e ) none of these |
explanation : let the cost price of num__1 item = re . num__1 cost price of num__16 items = num__16 selling price of num__16 items = num__12 loss = num__16 - num__12 = rs num__4 loss % = ( num__0.25 ) * num__100 = num__25.0 option b <eor> b <eos> |
b |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| avinash covers a distance of num__8 km in num__50 minutes . if he covers num__3 km distance in num__0.4 th of time then what speed should he maintain to cover the remaining distance in the remaining time ? <o> a ) num__15 kmph <o> b ) num__17 kmph <o> c ) num__21 kmph <o> d ) num__23 kmph <o> e ) num__27 kmph |
total distance = num__8 km total time = num__50 km time taken to cover the distence of num__3 km = num__50 * num__0.4 = num__20 min = num__0.333333333333 hours remaining distance = num__8 - num__3 = num__5 km required speed = num__5.0 / num__3 = num__15 kmph option ' a ' <eor> a <eos> |
a |
divide__8.0__0.4__ subtract__8.0__3.0__ multiply__3.0__5.0__ round__15.0__ |
divide__8.0__0.4__ subtract__8.0__3.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
| the population of a town is num__45000 ; num__0.555555555556 th of them are males and the rest females num__40.0 of the males are married . what is the percentage of married females ? <o> a ) num__28.0 <o> b ) num__50.0 <o> c ) num__32.0 <o> d ) num__34.0 <o> e ) num__39 % |
male = num__45000 * num__0.555555555556 = num__25000 female = num__45000 * num__0.444444444444 = num__20000 married male = num__25000 * num__0.4 = num__10000 married female = num__10000 num__20000 - - - - - - - - - - - - num__10000 num__100 - - - - - - - - - - - - - ? = > num__50.0 answer : b <eor> b <eos> |
b |
percent__40.0__25000.0__ percent__0.4__25000.0__ percent__100.0__50.0__ |
percent__40.0__25000.0__ percent__0.4__25000.0__ percent__100.0__50.0__ |
| two pipes a and b can fill a cistern in num__12 and num__15 minutes respectively . both are opened together but after num__4 minutes a is turned off . after how much more time will the cistern be filled ? <o> a ) num__6 <o> b ) num__8 num__1.0 <o> c ) num__8 num__0.25 <o> d ) num__8 num__0.5 <o> e ) num__8 num__0.428571428571 |
num__0.333333333333 + ( num__4 + x ) / num__15 = num__1 x = num__6 answer : a <eor> a <eos> |
a |
divide__4.0__12.0__ round__6.0__ |
divide__4.0__12.0__ divide__6.0__1.0__ |
| at a particular graduation party with num__300 guests num__70.0 of the guests brought gifts and num__40.0 of the female guests brought gifts . if num__36 males did not bring gifts to the party how many females did bring gifts ? <o> a ) num__18 <o> b ) num__36 <o> c ) num__42 <o> d ) num__68 <o> e ) num__70 |
the correct method total = num__300 . . num__70.0 of num__300 = num__210 got gifts . . num__90 did not get gift out of which num__48 are males so remaining num__90 - num__36 = num__54 are females . . but num__40.0 females brought gift so num__60.0 did not get it . . so num__60.0 = num__54 num__100.0 = num__54 * num__1.66666666667 = num__90 . . ans num__40.0 of num__90 = num__36 b <eor> b <eos> |
b |
percent__70.0__300.0__ percent__40.0__90.0__ |
percent__70.0__300.0__ percent__40.0__90.0__ |
| a train num__280 m long running with a speed of num__63 km / hr will pass a tree in ? <o> a ) num__18 sec <o> b ) num__16 sec <o> c ) num__76 sec <o> d ) num__15 sec <o> e ) num__17 sec |
speed = num__63 * num__0.277777777778 = num__17.5 m / sec time taken = num__280 * num__0.0571428571429 = num__16 sec answer : b <eor> b <eos> |
b |
divide__280.0__17.5__ round__16.0__ |
divide__280.0__17.5__ divide__280.0__17.5__ |
| calculate the circumference of a circular field whose radius is num__2 centimeters . <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
circumference c is given by c = num__2 Ï € r = num__2 Ï € * num__2 = num__4 Ï € cm correct answer b <eor> b <eos> |
b |
triangle_area__2.0__4.0__ |
triangle_area__2.0__4.0__ |
| two trains each num__100 m long are moving in opposite directions . they cross each other in num__8 seconds . if one is moving twice as fast the other the speed of the faster train is <o> a ) num__75 km / hr <o> b ) num__60 km / hr <o> c ) num__35 km / hr <o> d ) num__70 km / hr <o> e ) num__80 km / hr |
total distance covered = num__100 + num__100 = num__200 m time = num__8 s let speed of slower train is v . then the speed of the faster train is num__2 v ( since one is moving twice as fast the other ) relative speed = v + num__2 v = num__3 v num__3 v = num__25.0 m / s = num__25 m / s = > v = num__8.33333333333 m / s speed of the faster train = num__2 v = num__16.6666666667 m / s = ( num__16.6666666667 ) × ( num__3.6 ) km / hr = num__5 × num__12.0 = num__5 × num__12 = num__60 km / hr answer : b <eor> b <eos> |
b |
divide__200.0__100.0__ divide__200.0__8.0__ divide__25.0__3.0__ subtract__25.0__8.3333__ subtract__8.0__3.0__ divide__100.0__8.3333__ hour_to_min_conversion__ hour_to_min_conversion__ |
divide__200.0__100.0__ divide__200.0__8.0__ divide__25.0__3.0__ subtract__25.0__8.3333__ add__2.0__3.0__ divide__100.0__8.3333__ hour_to_min_conversion__ hour_to_min_conversion__ |
| a car crosses a num__600 m long bridge in num__5 minutes . whatis the speed of car in km per hour ? <o> a ) num__1 m / sec <o> b ) num__7.2 km / hr <o> c ) num__3 m / sec <o> d ) num__4 m / sec <o> e ) num__5 m / sec |
speed = num__600 m / sec . num__5 x num__60 = num__2 m / sec . converting m / sec to km / hr ( see important formulas section ) = num__2 x num__18 km / hr num__5 = num__7.2 km / hr b <eor> b <eos> |
b |
hour_to_min_conversion__ round__7.2__ |
hour_to_min_conversion__ round__7.2__ |
| how long does a train num__110 m long running at the speed of num__72 km / hr takes to cross a bridge num__132 m length ? <o> a ) num__82.1 sec . <o> b ) num__12.1 sec <o> c ) num__19.1 sec . <o> d ) num__17.1 sec . <o> e ) num__42.1 sec . |
speed = num__72 * num__0.277777777778 = num__20 m / sec total distance covered = num__110 + num__132 = num__242 m . required time = num__12.1 = num__12.1 sec . answer : b <eor> b <eos> |
b |
add__110.0__132.0__ divide__242.0__20.0__ round__12.1__ |
add__110.0__132.0__ divide__242.0__20.0__ divide__242.0__20.0__ |
| when the positive integer x is divided by num__9 the remainder is num__5 . what is the remainder when num__4 x is divided by num__9 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__4 <o> e ) num__6 |
i tried plugging in numbers x = num__9 q + num__5 x = num__14 num__4 x = num__56 num__4 x / num__9 = num__9 * num__6 + num__2 remainder is num__2 . answer is c <eor> c <eos> |
c |
add__9.0__5.0__ multiply__4.0__14.0__ subtract__6.0__4.0__ subtract__4.0__2.0__ |
add__9.0__5.0__ multiply__4.0__14.0__ subtract__6.0__4.0__ divide__4.0__2.0__ |
| in how many ways can num__16 different gifts be divided among four children such that each child receives exactly four gifts ? <o> a ) num__16 ^ num__4 <o> b ) ( num__4 ! ) ^ num__4 <o> c ) num__16 ! / ( num__4 ! ) ^ num__4 <o> d ) num__16 ! / num__4 ! <o> e ) num__4 ^ num__16 |
total num__16 different gifts and num__4 children . thus any one child gets num__16 c num__4 gifts then the other child gets num__12 c num__4 gifts ( num__16 total - num__4 already given ) then the third one gets num__8 c num__4 gifts and the last child gets num__4 c num__4 gifts . since order in which each child gets the gift is not imp thus ans : num__16 c num__4 * num__12 c num__4 * num__8 c num__4 * num__4 c num__4 = num__16 ! / ( num__4 ! ) ^ num__4 ans : c . <eor> c <eos> |
c |
subtract__16.0__4.0__ subtract__12.0__4.0__ add__12.0__4.0__ |
subtract__16.0__4.0__ subtract__12.0__4.0__ add__12.0__4.0__ |
| consider below info to be a table : amount of bacteria time - amount num__1 pm - num__10 gms num__4 pm - x gms num__7 pm - num__19.6 gms data for a certain biology experiment are given in the table above . if the amount of bacteria present increased by the same factor during each of the two num__3 - hour periods shown how many grams of bacteria were present at num__4 pm ? <o> a ) num__12.0 <o> b ) num__13.0 <o> c ) num__12.2 <o> d ) num__12.3 <o> e ) num__12.4 |
let say the factor is x the at num__7 pm quantity = num__10 ( i . e . quantity @ num__1 pm ) * x ^ num__2 = num__19.6 = > x ^ num__2 = num__1.96 or x = num__1.3 = > @ num__4 pm quantity = num__10 x = num__10 * num__1.3 = num__13 . answer b <eor> b <eos> |
b |
subtract__3.0__1.0__ divide__19.6__10.0__ add__10.0__3.0__ multiply__1.0__13.0__ |
subtract__3.0__1.0__ divide__19.6__10.0__ multiply__10.0__1.3__ multiply__1.0__13.0__ |
| the area of a square field num__3136 sq m if the length of cost of drawing barbed wire num__3 m around the field at the rate of rs . num__1.50 per meter . two gates of num__1 m width each are to be left for entrance . what is the total cost ? <o> a ) num__288 <o> b ) num__200 <o> c ) num__999 <o> d ) num__277 <o> e ) num__271 |
a num__2 = num__3136 = > a = num__56 num__56 * num__4 * num__3 = num__672 – num__6 = num__666 * num__1.5 = num__999 answer : c <eor> c <eos> |
c |
divide__3.0__1.5__ add__3.0__1.0__ multiply__3.0__2.0__ subtract__672.0__6.0__ multiply__1.5__666.0__ round__999.0__ |
divide__3.0__1.5__ add__3.0__1.0__ multiply__3.0__2.0__ subtract__672.0__6.0__ multiply__1.5__666.0__ multiply__1.5__666.0__ |
| find the numbers which are in the ratio num__3 : num__2 : num__2 such that the sum of the first and the second added to the difference of the third and the second is num__20 ? <o> a ) num__12 num__88 <o> b ) num__4 num__422 <o> c ) num__9 num__332 <o> d ) num__9 num__612 <o> e ) num__9 num__2 |
23 |
let the numbers be a b and c . a : b : c = num__3 : num__2 : num__2 given ( a + b ) + ( c - b ) = num__20 = > a + c = num__20 = > num__3 x + num__2 x = num__20 = > x = num__4 a b c are num__3 x num__2 x num__2 x a b c are num__12 num__8 num__8 . answer : a <eor> a <eos> |
a |
a |
| there is a church tower num__150 feet tall and another catholic tower at a distance of num__350 feet from it which is num__200 feet tall . there is one each bird sitting on top of both the towers . they fly at a constant speed and time to reach a grain in b / w the towers at the same time . at what distance from the church is the grain ? <o> a ) num__170 meters <o> b ) num__160 meters <o> c ) num__150 meters <o> d ) num__140 meters <o> e ) num__130 meters |
if x ft is the distance of grain from num__200 ft tall tower then num__150 ^ num__2 + ( num__350 - x ) ^ num__2 = num__200 ^ num__2 + x ^ num__2 then x = num__150 meters answer : c <eor> c <eos> |
c |
round__150.0__ |
subtract__350.0__200.0__ |
| how many num__3 - digit numbers are completely divisible num__6 ? <o> a ) num__149 <o> b ) num__150 <o> c ) num__151 <o> d ) num__166 <o> e ) num__172 |
explanation : num__3 - digit number divisible by num__6 are : num__102 num__108 num__114 . . . num__996 this is an a . p . in which a = num__102 d = num__6 and l = num__996 let the number of terms be n . then tn = num__996 . a + ( n - num__1 ) d = num__996 num__102 + ( n - num__1 ) x num__6 = num__996 num__6 x ( n - num__1 ) = num__894 ( n - num__1 ) = num__149 n = num__150 number of terms = num__150 . answer is b <eor> b <eos> |
b |
add__6.0__102.0__ add__6.0__108.0__ subtract__996.0__102.0__ divide__894.0__6.0__ add__1.0__149.0__ add__1.0__149.0__ |
add__6.0__102.0__ add__6.0__108.0__ subtract__996.0__102.0__ divide__894.0__6.0__ add__1.0__149.0__ add__1.0__149.0__ |
| in each series look for the degree and direction of change between the numbers . in other words do the numbers increase or decrease and by how much look at this series : num__2 num__1 ( num__0.5 ) ( num__0.25 ) . . . what number should come next ? <o> a ) num__0.125 <o> b ) num__0.25 <o> c ) num__0.4 <o> d ) num__0.285714285714 <o> e ) num__1.0 |
a num__0.125 this is a simple division series ; each number is one - half of the previous number . in other terms to say the number is divided by num__2 successively to get the next result . num__2.0 = num__2 num__1.0 = num__1 num__0.5 = num__0.5 ( num__0.5 ) / num__2 = num__0.25 ( num__0.25 ) / num__2 = num__0.125 and so on . <eor> a <eos> |
a |
multiply__0.5__0.25__ multiply__1.0__0.125__ |
divide__0.25__2.0__ divide__0.25__2.0__ |
| a train running at num__0.333333333333 of its own speed reached a place in num__9 hours . how much time could be saved if the train would have run at its own speed ? <o> a ) num__8 hrs <o> b ) num__10 hrs <o> c ) num__12 hrs <o> d ) num__15 hrs <o> e ) num__6 hrs |
time taken if run its own speed = num__0.333333333333 * num__9 = num__3 hrs time saved = num__9 - num__3 = num__6 hrs answer : e <eor> e <eos> |
e |
subtract__9.0__3.0__ round__6.0__ |
subtract__9.0__3.0__ subtract__9.0__3.0__ |
| the ratio of two numbers is num__1 : num__3 and the sum of their cubes is num__945 . the difference of number is ? <o> a ) num__5.23 <o> b ) num__1.23 <o> c ) num__3.93 <o> d ) num__3.23 <o> e ) num__4.23 |
num__1 x num__3 x num__1 x cube + num__27 x cube = num__945 num__28 x cube = num__945 x cube = num__33.75 = > x = num__3.23 answer : d <eor> d <eos> |
d |
add__1.0__27.0__ divide__945.0__28.0__ multiply__1.0__3.23__ |
add__1.0__27.0__ divide__945.0__28.0__ multiply__1.0__3.23__ |
| a man buys rs . num__20 shares paying num__9.0 dividend . the man wants to have an interest of num__12.0 on his money . the market value of each share is : <o> a ) num__12 <o> b ) num__15 <o> c ) num__18 <o> d ) num__20 <o> e ) num__22 |
explanation : dividend on rs . num__20 = rs . ( num__0.09 ) x num__20 = rs . num__1.8 . rs . num__12 is an income on rs . num__100 . rs . num__1.8 is an income on rs . [ ( num__8.33333333333 ) x ( num__1.8 ) ] = rs . num__15 . answer : b <eor> b <eos> |
b |
percent__20.0__9.0__ percent__100.0__15.0__ |
percent__20.0__9.0__ percent__100.0__15.0__ |
| in a shower num__5 cm of rain falls . the volume of water that falls on num__1.5 hectares of ground is : <o> a ) num__75 cu . m <o> b ) num__750 cu . m <o> c ) num__7500 cu . m <o> d ) num__75000 cu . m <o> e ) num__7.50 cu . m |
num__1 hectare = num__10000 sq mtr the volume of water that falls on num__1.5 hectares of ground = num__0.05 * num__1.5 * num__10000 = num__75 * num__10 cu mtr = num__750 cub mtr . answer : b <eor> b <eos> |
b |
multiply__10.0__75.0__ round__750.0__ |
multiply__10.0__75.0__ multiply__10.0__75.0__ |
| the effective annual rate of interest corresponding to a nominal rate of num__6.0 per annum payable half - yearly is : <o> a ) num__6.06 <o> b ) num__6.07 <o> c ) num__6.08 <o> d ) num__6.09 <o> e ) num__6.10 % |
explanation : amount of rs . num__100 for num__1 year = rs . [ num__100 x ( num__1 + num__0.03 ) num__2 ] = rs . num__106.09 when compounded half - yearly effective rate = ( num__106.09 - num__100 ) % = num__6.09 answer is d <eor> d <eos> |
d |
percent__100.0__6.09__ |
percent__100.0__6.09__ |
| the compound interest earned on a sum for the second and the third years are $ num__1400 and $ num__1540 respectively . what is the rate of interest ? <o> a ) num__2.0 <o> b ) num__4.0 <o> c ) num__6.0 <o> d ) num__8.0 <o> e ) num__10 % |
num__1540 - num__1400 = num__140 is the rate of interest on $ num__1400 for one year . the rate of interest = ( num__100 * num__140 ) / ( num__1400 ) = num__10.0 the answer is e . <eor> e <eos> |
e |
percent__10.0__100.0__ |
percent__10.0__100.0__ |
| p can do a work in the same time in which q and r together can do it . if p and q work together the work can be completed in num__10 days . r alone needs num__30 days to complete the same work . then q alone can do it in <o> a ) num__10 <o> b ) num__22 <o> c ) num__25 <o> d ) num__27 <o> e ) num__30 |
work done by p and q in num__1 day = num__0.1 work done by r in num__1 day = num__0.0333333333333 work done by p q and r in num__1 day = num__0.1 + num__0.0333333333333 = num__0.133333333333 but work done by p in num__1 day = work done by q and r in num__1 day . hence the above equation can be written as work done by p in num__1 day à — num__2 = num__0.133333333333 = > work done by p in num__1 day = num__0.0666666666667 = > work done by q and r in num__1 day = num__0.0666666666667 hence work done by q in num__1 day = num__0.0666666666667 â € “ num__0.0333333333333 = num__0.0333333333333 = num__0.0333333333333 so q alone can do the work in num__30 days answer is e . <eor> e <eos> |
e |
divide__1.0__10.0__ divide__1.0__30.0__ add__0.1__0.0333__ divide__2.0__30.0__ round__30.0__ |
divide__1.0__10.0__ divide__1.0__30.0__ add__0.1__0.0333__ divide__2.0__30.0__ round__30.0__ |
| num__53 num__53 num__40 num__40 num__27 num__27 ? <o> a ) num__14 <o> b ) num__12 <o> c ) num__16 <o> d ) num__18 <o> e ) num__20 |
first each number is repeated then num__13 is subtracted to arrive at the next number . answer : a . <eor> a <eos> |
a |
subtract__53.0__40.0__ subtract__27.0__13.0__ |
subtract__53.0__40.0__ subtract__27.0__13.0__ |
| the average height of num__35 boys in a class was calculated as num__180 cm . it has later found that the height of one of the boys in the class was wrongly written as num__166 cm whereas his actual height was num__106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ) . <o> a ) num__178.27 <o> b ) num__178.29 <o> c ) num__178.21 <o> d ) num__178.19 <o> e ) num__178.55 |
calculated average height of num__35 boys = num__180 cm . wrong total height of num__35 boys = num__180 * num__35 cm . this was as a result of an actual height of num__106 cm being wrongly written as num__166 cm . correct total height of num__35 boys = num__180 * num__35 cm - num__166 cm + num__106 cm = num__180 * num__35 cm - num__166 cm + num__106 cm / num__35 = num__180 cm - num__1.71428571429 cm = num__180 cm - num__1.71 cm = num__178.29 cm . answer : b <eor> b <eos> |
b |
subtract__180.0__1.71__ subtract__180.0__1.71__ |
subtract__180.0__1.71__ subtract__180.0__1.71__ |
| if it takes a hrs for x to repair num__1 machine and b hrs for y to repair num__1 machine then how many hrs will they take to repair num__750 machines if they work together ? <o> a ) num__750 × ab / ( a + b ) <o> b ) num__750 ( a + b ) / ab <o> c ) num__750 ab <o> d ) num__750 / ( a + b ) <o> e ) none of these |
explanation : x repair machine in a hrs he work num__1 / a for num__1 hr y repairs machine in b hrs he works num__1 / b for num__1 hr ( x + y ) work per num__1 hr = num__1 / a + num__1 / b = ( a + b ) / ab time = ab / ( a + b ) for num__750 machines time = num__750 × ab / ( a + b ) answer : option a <eor> a <eos> |
a |
round__750.0__ |
divide__750.0__1.0__ |
| in a box of num__11 pens a total of num__3 are defective . if a customer buys num__2 pens selected at random from the box what is the probability that neither pen will be defective ? <o> a ) num__0.454545454545 <o> b ) num__0.484848484848 <o> c ) num__0.509090909091 <o> d ) num__0.493506493506 <o> e ) num__0.511363636364 |
# defective pens = num__3 # good pens = num__8 probability of the num__1 st pen being good = num__0.727272727273 probability of the num__2 nd pen being good = num__0.7 total probability = num__0.727272727273 * num__0.7 = num__0.509090909091 the answer is c . <eor> c <eos> |
c |
subtract__11.0__3.0__ subtract__3.0__2.0__ divide__8.0__11.0__ multiply__0.7273__0.7__ multiply__1.0__0.5091__ |
subtract__11.0__3.0__ subtract__3.0__2.0__ divide__8.0__11.0__ multiply__0.7273__0.7__ multiply__1.0__0.5091__ |
| after allowing a discount of num__15.0 on the marked price the selling price is rs . num__6800 for an article . if it was sold at marked price there would have been a profit of num__60.0 . the cost price of the article is ? <o> a ) num__2338 <o> b ) num__2298 <o> c ) num__5000 <o> d ) num__2871 <o> e ) num__1171 |
explanation : given sp = rs . num__6800 marked price = [ sp ( num__100 ) ] / ( num__100 - d % ) = ( num__6800 * num__100 ) / ( num__100 - num__15 ) = rs . num__8000 if sp = rs . num__8000 profit = num__60.0 cp = [ sp ( num__100 ) ] / ( num__100 + num__60 ) = ( num__8000 * num__100 ) / num__160 = rs . num__5000 answer : c <eor> c <eos> |
c |
percent__100.0__5000.0__ |
percent__100.0__5000.0__ |
| a tradesman by means of his false balance defrauds to the extent of num__30.0 ? in buying goods as well as by selling the goods . what percent does he gain on his outlay ? <o> a ) num__60.0 <o> b ) num__70.0 <o> c ) num__69.0 <o> d ) num__71.0 <o> e ) num__68 % |
g % = num__30 + num__30 + ( num__30 * num__30 ) / num__100 = num__69.0 answer : c <eor> c <eos> |
c |
percent__100.0__69.0__ |
percent__100.0__69.0__ |
| if a earns num__33.0 % more than b how much percent does b earn less then a ? <o> a ) num__42.0 <o> b ) num__14.0 <o> c ) num__25.0 <o> d ) num__15.0 <o> e ) num__35 % |
required percentage = [ ( ( num__33.3333333333 ) * num__100 ) / [ num__100 + ( num__33.3333333333 ) ] ] % = [ ( num__0.25 ) * num__100 ] % = num__25.0 answer is c . <eor> c <eos> |
c |
multiply__0.25__100.0__ multiply__0.25__100.0__ |
multiply__0.25__100.0__ multiply__0.25__100.0__ |
| the sum of the present ages of a son and his father is num__60 years . six years ago father ' s age was five times the age of the son . after num__6 years what will be son ' s age ? <o> a ) num__100 <o> b ) num__40 <o> c ) num__80 <o> d ) num__20 <o> e ) num__10 |
let the present age of the son = x then present age of the father = ( num__60 − x ) six years ago father ' s age was num__5 times the age of the son ( num__60 - x ) - num__6 = num__5 ( x - num__6 ) num__84 = num__6 x x = num__14.0 = num__14 son ' s age after num__6 years = x + num__6 = num__14 + num__6 = num__20 answer is d . <eor> d <eos> |
d |
divide__84.0__6.0__ add__6.0__14.0__ add__6.0__14.0__ |
divide__84.0__6.0__ add__6.0__14.0__ add__6.0__14.0__ |
| by the how much is two - fifth of num__150 greater than three - fifths of num__25 ? <o> a ) num__15 <o> b ) num__3 <o> c ) num__45 <o> d ) num__30 <o> e ) none of these |
reqd no . = num__2 â „ num__5 Ã — num__150 - num__3 â „ num__5 Ã — num__25 = num__60 - num__15 = num__45 answer c <eor> c <eos> |
c |
subtract__5.0__2.0__ multiply__3.0__5.0__ multiply__3.0__15.0__ multiply__3.0__15.0__ |
subtract__5.0__2.0__ multiply__3.0__5.0__ subtract__60.0__15.0__ subtract__60.0__15.0__ |
| what is the difference between the place value and face value of num__3 in the numeral num__1375 ? <o> a ) num__287 <o> b ) num__290 <o> c ) num__297 <o> d ) num__390 <o> e ) num__398 |
place value of num__3 = num__3 * num__100 = num__300 face value of num__3 = num__3 num__300 - num__3 = num__297 c <eor> c <eos> |
c |
multiply__3.0__100.0__ subtract__300.0__3.0__ subtract__300.0__3.0__ |
multiply__3.0__100.0__ subtract__300.0__3.0__ subtract__300.0__3.0__ |
| jerry an electrician worked num__6 months out of the year . what percent of the year did he work ? ( round answer to the nearest hundredth ) what percent num__12 is num__6 ? num__12 months = num__1 year <o> a ) num__58.33 <o> b ) num__68.33 <o> c ) num__78.33 <o> d ) num__50.0 <o> e ) num__98.33 % |
num__1 . multiply the opposites num__6 x num__100 = num__600 num__100 = num__0.5 num__50.0 ( rounded to hundredth ) correct answer d <eor> d <eos> |
d |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| a textile manufacturing firm employees num__70 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the num__70 looms is rs num__5 num__00000 and the monthly manufacturing expenses is rs num__1 num__50000 . assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over the number of looms . monthly establishment charges are rs num__75000 . if one loom breaks down and remains idle for one month the decrease in profit is : <o> a ) num__13000 <o> b ) num__7000 <o> c ) num__10000 <o> d ) num__5000 <o> e ) none of these |
explanation : profit = num__5 num__00000 â ˆ ’ ( num__1 num__50000 + num__75000 ) = rs . num__2 num__75000 . since such loom contributes equally to sales and manufacturing expenses . but the monthly charges are fixed at rs num__75000 . if one loan breaks down sales and expenses will decrease . new profit : - = > num__500000 Ã — ( num__0.985714285714 ) â ˆ ’ num__150000 Ã — ( num__0.985714285714 ) â ˆ ’ num__75000 . = > rs num__2 num__70000 . decrease in profit = > num__2 num__75000 â ˆ ’ num__2 num__70000 = > rs . num__5000 . answer : d <eor> d <eos> |
d |
multiply__75000.0__2.0__ subtract__75000.0__70000.0__ multiply__1.0__5000.0__ |
multiply__75000.0__2.0__ subtract__75000.0__70000.0__ subtract__75000.0__70000.0__ |
| jeya invested an amount of rs . num__9500 at the rate of num__13.0 p . a simple interest and another amount at the rate of num__15.0 p . a . simple interest . the total interest earned at the end of one year on the total amount invested became num__14.0 p . a . find the total amount invested ? <o> a ) num__15550 <o> b ) num__18450 <o> c ) num__13450 <o> d ) num__14450 <o> e ) num__14650 |
let the second amount be rs . x . then ( num__9500 * num__13 * num__1 ) / num__100 + ( x * num__15 * num__1 ) / num__100 = [ ( num__12000 + x ) * num__14 * num__1 ] / num__100 num__12350 + num__15 x = num__16800 + num__14 x x = num__4450 total investment = num__10000 + num__4450 = rs . num__14450 . answer : d <eor> d <eos> |
d |
percent__100.0__14450.0__ |
percent__100.0__14450.0__ |
| in a classroom num__12 students brought apples and num__8 students brought bananas . if exactly num__10 students brought only one of the two types of fruits how many students brought both types of fruits ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__12 <o> e ) num__14 |
say x students brought both fruits . ( num__12 - x ) + ( num__8 - x ) = num__10 - - > x = num__5 . answer : a . <eor> a <eos> |
a |
vowel_space__ vowel_space__ |
vowel_space__ vowel_space__ |
| find the value of num__72519 x num__9999 = m ? <o> a ) num__434539873 <o> b ) num__355797990 <o> c ) num__435453490 <o> d ) m = num__725117481 <o> e ) num__873430134 |
num__72519 x num__9999 = num__72519 x ( num__10000 - num__1 ) = num__72519 x num__10000 - num__72519 x num__1 = num__725190000 - num__72519 = num__725117481 d <eor> d <eos> |
d |
subtract__10000.0__9999.0__ multiply__72519.0__10000.0__ multiply__72519.0__9999.0__ multiply__72519.0__9999.0__ |
subtract__10000.0__9999.0__ multiply__72519.0__10000.0__ subtract__725190000.0__72519.0__ subtract__725190000.0__72519.0__ |
| at a certain university num__69.0 of the professors are women and num__70.0 of the professors are tenured . if num__90.0 of the professors are women tenured or both then what percent of the men are tenured ? <o> a ) num__25 <o> b ) num__37.5 <o> c ) num__50 <o> d ) num__52.5 <o> e ) num__75 |
answer is num__75.0 total women = num__69.0 total men = num__40.0 total tenured = num__70.0 ( both men and women ) therefore women tenured + women professors + men tenured = num__90.0 men tenured = num__21.0 but question wants to know the percent of men that are tenured num__21.0 / num__40.0 = num__52.5 d <eor> d <eos> |
d |
percent__70.0__75.0__ percent__70.0__75.0__ |
percent__70.0__75.0__ percent__70.0__75.0__ |
| two trains of equal lengths take num__12 sec and num__15 sec respectively to cross a telegraph post . if the length of each train be num__120 m in what time will they cross other travelling in opposite direction ? <o> a ) num__22 <o> b ) num__13 <o> c ) num__77 <o> d ) num__99 <o> e ) num__21 |
speed of the first train = num__10.0 = num__10 m / sec . speed of the second train = num__8.0 = num__8 m / sec . relative speed = num__10 + num__8 = num__18 m / sec . required time = ( num__120 + num__120 ) / num__18 = num__13 sec . answer : b <eor> b <eos> |
b |
divide__120.0__12.0__ divide__120.0__15.0__ add__8.0__10.0__ round__13.0__ |
divide__120.0__12.0__ divide__120.0__15.0__ add__8.0__10.0__ round__13.0__ |
| it ' s valentine ' s day and alan has promised to cook his girlfriend alana her favorite meal : fish fingers and custard . alana will be arriving in one hour exactly . alan only has a microwave . the fish fingers take num__13 minutes to defrost in the microwave and num__6 minutes to cook in the microwave . the custard must be prepared after the fish fingers and takes num__24 minutes to cook in the microwave and then num__6 minutes to cool sufficiently so as not to burn someone ' s tongue . it will take alan num__1 minute to plate the food . however alan is on his xbox and really wants to play one more game of his football video game before alana arrives . what is the maximum length of each half of the match that alan can play to ensure the meal is ready for alana ' s arrival ? <o> a ) num__5 min <o> b ) num__6 minutes <o> c ) num__7 minutes <o> d ) num__8 minutes <o> e ) num__10 minutes |
alana will arrive in num__60 minutes . the total meal preparation time is num__13 ( fish finger defrost ) + num__6 ( fish finger cook ) + num__24 ( custard cook ) + num__6 ( custard cool ) + num__1 ( plate time ) = num__50 minutes . alan therefore has num__10 minutes ( num__60 - num__50 = num__10 ) to play his football video game . as football matches have two halves alan have select num__5 minute halves ( num__5.0 = num__5 ) answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ divide__60.0__6.0__ subtract__6.0__1.0__ round__5.0__ |
hour_to_min_conversion__ subtract__60.0__50.0__ subtract__6.0__1.0__ subtract__6.0__1.0__ |
| if x - y = num__2 x - num__2 z x - num__2 y = num__4 z and x + y + z = num__21 what is the value of y / z ? <o> a ) - num__4.5 . <o> b ) - num__2 . <o> c ) - num__1.7 . <o> d ) - num__0.667 . <o> e ) num__2.5 . |
x - y = num__2 x - num__2 z y = - x + num__2 z - - - - - - - - - - num__1 x - num__2 y = num__4 z x - num__4 z = num__2 y - - - - - - - - - num__2 adding equation num__1 from equation num__2 - num__2 z = num__3 y y / z = - num__0.667 d is the answer <eor> d <eos> |
d |
add__2.0__1.0__ multiply__1.0__0.667__ |
add__2.0__1.0__ divide__0.667__1.0__ |
| in an exam amar scored num__64 percent bhavan scored num__36 percent and chetan num__44 percent . the maximum score awarded in the exam is num__400 . find the average mark scored by all the three boys ? <o> a ) num__384 <o> b ) num__192 <o> c ) num__207 <o> d ) num__269 <o> e ) num__208 |
average mark scored by all the three boys = [ num__0.64 ( num__400 ) + num__0.36 ( num__400 ) + num__0.44 ( num__400 ) ] / num__3 = num__192 answer : b <eor> b <eos> |
b |
multiply__64.0__3.0__ multiply__64.0__3.0__ |
multiply__64.0__3.0__ multiply__64.0__3.0__ |
| a man buys an song for num__10.0 less than its value and sells it for num__10.0 more than its value . his gain or loss percent is : <o> a ) < num__25 <o> b ) < num__10 <o> c ) > num__20 <o> d ) > num__30 <o> e ) < num__30 |
c > num__20 let the song be worth $ x . c . p . num__90.0 of $ x = $ num__9 x / num__10 s . p . = num__110.0 of $ x = $ num__11 x / num__10 gain = ( num__11 x / num__10 - num__9 x / num__10 ) = $ x / num__5 gain % = x / num__5 * num__1.11111111111 x * num__100 = num__22 num__0.222222222222 % > num__20.0 <eor> c <eos> |
c |
percent__10.0__90.0__ percent__10.0__110.0__ percent__20.0__110.0__ percent__1.1111__20.0__ percent__100.0__20.0__ |
percent__10.0__90.0__ percent__10.0__110.0__ percent__20.0__110.0__ percent__1.1111__20.0__ percent__100.0__20.0__ |
| an empty bucket being filled with paint at a constant rate takes num__6 minutes to be filled to num__0.7 of its capacity . how much more time b will it take to fill the bucket to full capacity ? <o> a ) num__0.388888888889 <o> b ) num__0.5 <o> c ) num__2 <o> d ) num__2.57142857143 <o> e ) num__3.6 |
solution - work and time are directly proportional . w num__1 / w num__2 = t num__1 / t num__2 num__0.7 work in num__6 mins num__1 work in t mins ( num__0.7 ) / num__1 = num__6 / t - > t = num__8.57142857143 mins . remaining minutes to fill the tank b = num__8.57142857143 - num__6 = num__2.57142857143 mins . ans d . <eor> d <eos> |
d |
divide__6.0__0.7__ subtract__8.5714__6.0__ multiply__1.0__2.5714__ |
divide__6.0__0.7__ subtract__8.5714__6.0__ divide__2.5714__1.0__ |
| a is two years older than b who is twice as old as c . if the total of the ages of a b and c be num__27 then how old is b ? <o> a ) num__17 years <o> b ) num__19 years <o> c ) num__29 years <o> d ) num__10 years <o> e ) num__12 years |
let c ' s age be x years . then b ' s age = num__2 x years . a ' s age = ( num__2 x + num__2 ) years . ( num__2 x + num__2 ) + num__2 x + x = num__27 num__5 x = num__25 = > x = num__5 hence b ' s age = num__2 x = num__10 years . answer : d <eor> d <eos> |
d |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
| if n = num__2 ^ num__0.20 and n ^ b = num__16 b must equal <o> a ) num__0.0375 <o> b ) num__0.6 <o> c ) num__20 <o> d ) num__1.66666666667 <o> e ) num__26.6666666667 |
num__0.2 = num__0.2 n = num__2 ^ num__0.2 n ^ b = num__2 ^ num__4 ( num__2 ^ num__0.2 ) ^ b = num__2 ^ num__4 b = num__20 answer : c <eor> c <eos> |
c |
add__16.0__4.0__ add__16.0__4.0__ |
add__16.0__4.0__ add__16.0__4.0__ |
| a cistern is normally filled in num__8 hours but takes two hours longer to fill because of a leak in its bottom . if the cistern is full the leak will empty it in ? <o> a ) num__27 <o> b ) num__29 <o> c ) num__40 <o> d ) num__89 <o> e ) num__22 |
num__0.125 - num__1 / x = num__0.1 x = num__40 answer : c <eor> c <eos> |
c |
multiply__8.0__0.125__ round__40.0__ |
multiply__8.0__0.125__ divide__40.0__1.0__ |
| cole drove from home to work at an average speed of num__30 kmh . he then returned home at an average speed of num__90 kmh . if the round trip took a total of num__2 hours how many minutes did it take cole to drive to work ? <o> a ) num__66 <o> b ) num__70 <o> c ) num__72 <o> d ) num__75 <o> e ) num__90 |
let the distance one way be x time from home to work = x / num__30 time from work to home = x / num__90 total time = num__2 hrs ( x / num__30 ) + ( x / num__90 ) = num__2 solving for x we get x = num__45 time from home to work in minutes = ( num__45 ) * num__2.0 = num__90 minutes ans = e <eor> e <eos> |
e |
divide__90.0__2.0__ round__90.0__ |
divide__90.0__2.0__ multiply__2.0__45.0__ |
| two trains of length num__180 m and num__280 m are running towards each other on parallel lines at num__42 kmph and num__30 kmph respectively . in what time will they be clear of each other from the moment they meet ? <o> a ) num__28 <o> b ) num__266 <o> c ) num__990 <o> d ) num__20 <o> e ) num__23 |
relative speed = ( num__42 + num__30 ) * num__0.277777777778 = num__4 * num__5 = num__20 mps . distance covered in passing each other = num__180 + num__280 = num__460 m . the time required = d / s = num__23.0 = num__23 sec . answer : e <eor> e <eos> |
e |
multiply__4.0__5.0__ add__180.0__280.0__ divide__460.0__20.0__ round__23.0__ |
multiply__4.0__5.0__ add__180.0__280.0__ divide__460.0__20.0__ divide__460.0__20.0__ |
| a trader bought a car at num__30.0 discount on its original price . he sold it at a num__40.0 increase on the price he bought it . what percent of profit did he make on the original price ? <o> a ) num__118 <o> b ) num__110 <o> c ) num__112 <o> d ) num__113 <o> e ) num__98 |
original price = num__100 cp = num__70 s = num__70 * ( num__1.4 ) = num__98 num__100 - num__112 = num__2.0 answer : e <eor> e <eos> |
e |
percent__98.0__100.0__ |
percent__98.0__100.0__ |
| a team of five entered for a shooting competition . the best marks man scored num__85 points . if he had scored num__92 points the average scores for . the team would have been num__84 . how many points altogether did the team score ? <o> a ) num__288 <o> b ) num__413 <o> c ) num__168 <o> d ) num__127 <o> e ) num__664 |
explanation : num__5 * num__84 = num__420 - num__7 = num__413 answer : b <eor> b <eos> |
b |
multiply__84.0__5.0__ subtract__92.0__85.0__ subtract__420.0__7.0__ subtract__420.0__7.0__ |
multiply__84.0__5.0__ subtract__92.0__85.0__ subtract__420.0__7.0__ subtract__420.0__7.0__ |
| the weight of a glass of jar is num__40.0 of the weight of the jar filled with coffee beans . after some of the beans have been removed the weight of the jar and the remaining beans is num__60.0 of the original total weight . what fraction part of the beans remain in the jar ? <o> a ) num__0.2 <o> b ) num__0.333333333333 <o> c ) num__0.4 <o> d ) num__0.5 <o> e ) num__0.666666666667 |
let weight of jar filled with beans = num__100 g weight of jar = num__40 g weight of coffee beans = num__60 g weight of jar and remaining beans = num__60 g weight of remaining beans = num__20 g fraction remaining = num__0.333333333333 = num__0.333333333333 answer is b . <eor> b <eos> |
b |
percent__100.0__0.3333__ |
percent__100.0__0.3333__ |
| the speed of a boat in still water is num__15 km / hr and the rate of current is num__3 km / hr the distance travelled downstream in num__12 minutes is : <o> a ) num__2 km <o> b ) num__2.6 km <o> c ) num__3.6 km <o> d ) num__4 km <o> e ) none |
sol . speed downstream = ( num__15 + num__3 ) kmph = num__18 kmph . distance travelled = [ num__18 * num__0.2 ] km = num__3.6 km . answer c <eor> c <eos> |
c |
add__15.0__3.0__ divide__3.0__15.0__ multiply__18.0__0.2__ round__3.6__ |
add__15.0__3.0__ divide__3.0__15.0__ multiply__18.0__0.2__ multiply__18.0__0.2__ |
| in a num__1000 m race a beats b by num__60 m and b beats c by num__100 m . in the same race by how many meters does a beat c ? <o> a ) num__154 m <o> b ) num__176 m <o> c ) num__268 m <o> d ) num__129 m <o> e ) num__218 m |
by the time a covers num__1000 m b covers ( num__1000 - num__60 ) = num__940 m . by the time b covers num__1000 m c covers ( num__1000 - num__100 ) = num__900 m . so the ratio of speeds of a and c = num__1.06382978723 * num__1.11111111111 = num__1.18203309693 so by the time a covers num__1000 m c covers num__846 m . so in num__1000 m race a beats c by num__1000 - num__846 = num__154 m . answer : a <eor> a <eos> |
a |
subtract__1000.0__60.0__ subtract__1000.0__100.0__ divide__1000.0__940.0__ divide__1000.0__900.0__ multiply__1.1111__1.0638__ subtract__1000.0__846.0__ round__154.0__ |
subtract__1000.0__60.0__ subtract__1000.0__100.0__ divide__1000.0__940.0__ divide__1000.0__900.0__ multiply__1.1111__1.0638__ subtract__1000.0__846.0__ subtract__1000.0__846.0__ |
| the captain of a cricket team of num__11 members is num__26 years old and the wicket keeper is num__3 years older . if the ages of these two are excluded the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? <o> a ) num__20 years <o> b ) num__21 years <o> c ) num__22 years <o> d ) num__23 years <o> e ) num__24 years |
explanation : let the average age of the whole team by x years . num__11 x – ( num__26 + num__29 ) = num__9 ( x - num__1 ) num__11 x – num__9 x = num__46 num__2 x = num__46 x = num__23 . so average age of the team is num__23 years . answer d <eor> d <eos> |
d |
add__26.0__3.0__ subtract__11.0__9.0__ subtract__26.0__3.0__ subtract__26.0__3.0__ |
add__26.0__3.0__ subtract__11.0__9.0__ subtract__26.0__3.0__ subtract__26.0__3.0__ |
| if ( a + b ) = num__5 ( b + c ) = num__9 and ( c + d ) = num__3 what is the value of ( a + d ) ? <o> a ) num__16 . <o> b ) num__8 . <o> c ) num__7 . <o> d ) num__2 . <o> e ) - num__1 . |
given a + b = num__5 b + c = num__9 c + d = num__3 ( a + b ) - ( b + c ) + ( c + d ) = ( a + d ) = > num__5 - num__9 + num__3 = - num__1 . option e . . . <eor> e <eos> |
e |
reverse__1.0__ |
reverse__1.0__ |
| a club has exactly num__5 new members at the end of its first week . every subsequent week each of the previous week ' s new members ( and only these members ) brings exactly a new members into the club . if b is the number of new members brought into the club during the twelfth week which of the following could be b ? <o> a ) num__5 ^ num__0.0833333333333 <o> b ) num__3 ^ num__11 * num__5 ^ num__11 <o> c ) num__3 ^ num__12 * num__5 ^ num__12 <o> d ) num__3 ^ num__11 * num__5 ^ num__12 <o> e ) num__60 ^ num__12 |
at the end of the first week there are num__5 new members ; at the end of the second week there are num__5 a new members ( since each num__5 new members from the previous week brings a new members ) ; at the end of the third week there are num__5 a ^ num__2 new members ( since each num__5 a new members from the previous week brings a new members ) ; . . . at the end of the twelfth week there are num__5 a ^ num__11 new members ( since each num__5 a ^ num__10 new members from the previous week brings a new members ) . we are given that num__5 a ^ num__11 = b . out of the answers only d yields integer value for a : num__5 a ^ num__11 = num__3 ^ num__11 * num__5 ^ num__12 - - > a = num__3 * num__5 = num__15 . answer : d . <eor> d <eos> |
d |
multiply__5.0__2.0__ subtract__5.0__2.0__ add__2.0__10.0__ multiply__5.0__3.0__ subtract__5.0__2.0__ |
multiply__5.0__2.0__ subtract__5.0__2.0__ add__2.0__10.0__ multiply__5.0__3.0__ subtract__5.0__2.0__ |
| rita was running on a num__500 metre track at a rate of num__100 metres per minute . one minute after rita started liz started running from the same starting point on the track at num__150 metres per minute . what is the shortest time that liz could run on the track in order to meet up with rita ? <o> a ) num__4 minutes <o> b ) num__3 minutes <o> c ) num__2 minutes <o> d ) num__1.6 minutes <o> e ) num__1 minute |
we do n ' t need to assume here the direction . we need to find which will take less time so there are num__2 cases num__1 ) both moving in same direction . . . then time would be num__2 minutes num__2 ) both moving in opposite direction then time would be num__1.6 minutes as we need the shortest time it would be the second case answer : d <eor> d <eos> |
d |
mile_to_km_conversion__ mile_to_km_conversion__ |
mile_to_km_conversion__ mile_to_km_conversion__ |
| if num__20 liters of chemical x are added to num__80 liters of a mixture that is num__25.0 chemical x and num__75.0 chemical y then what percentage of the resulting mixture is chemical x ? <o> a ) num__28.0 <o> b ) num__32.0 <o> c ) num__36.0 <o> d ) num__40.0 <o> e ) num__44 % |
the amount of chemical x in the solution is num__20 + num__0.25 ( num__80 ) = num__40 liters . num__40 liters / num__100 liters = num__40.0 the answer is d . <eor> d <eos> |
d |
divide__20.0__80.0__ add__20.0__80.0__ subtract__80.0__40.0__ |
divide__20.0__80.0__ add__20.0__80.0__ subtract__80.0__40.0__ |
| two trains start from a & b and travel towards each other at speed of num__50 kmph and num__60 kmph resp . at the time of the meeting the second train has traveled num__100 km more than the first . the distance between them . <o> a ) num__800 km <o> b ) num__900 km <o> c ) num__1000 km <o> d ) num__1100 km <o> e ) num__1200 km |
let the distance traveled by the first train be x km then distance covered by the second train is x + num__100 km x / num__50 = x + num__1.66666666667 x = num__500 so the distance between a & b is x + x + num__100 = num__1100 km answer is d . <eor> d <eos> |
d |
divide__100.0__60.0__ round__1100.0__ |
divide__100.0__60.0__ round__1100.0__ |
| john is going with num__10 friends on a trip to sidney for spring break . airfare and hotel costs a total of $ num__12100.00 for the group of num__11 friends . how much does each person have to pay for their hotel and airfare ? <o> a ) $ num__1010 <o> b ) $ num__1100 <o> c ) $ num__1110 <o> d ) $ num__1101 <o> e ) $ num__1200 |
answer = b the total cost of the trip ( $ num__12100.00 ) divided by num__11 equals $ num__1100.00 . <eor> b <eos> |
b |
divide__12100.0__11.0__ divide__12100.0__11.0__ |
divide__12100.0__11.0__ divide__12100.0__11.0__ |
| num__6000 - num__5000 ÷ num__20.00 = ? <o> a ) num__900 <o> b ) num__500 <o> c ) num__5750 <o> d ) num__5000 <o> e ) none |
answer given expression = num__6000 - num__5000 ÷ num__20.00 = num__6000 - num__250 = num__5750 correct option : c <eor> c <eos> |
c |
divide__5000.0__20.0__ subtract__6000.0__250.0__ subtract__6000.0__250.0__ |
divide__5000.0__20.0__ subtract__6000.0__250.0__ subtract__6000.0__250.0__ |
| if x = num__1 + √ num__2 then what is the value of x num__4 - num__4 x num__3 + num__4 x num__2 + num__3 ? <o> a ) - num__1 <o> b ) num__0 <o> c ) num__4 <o> d ) num__2 <o> e ) num__3 |
answer x = num__1 + √ num__2 ∴ x num__4 - num__4 x num__3 + num__4 x num__2 + num__5 = x num__2 ( x num__2 - num__4 x + num__4 ) + num__3 = x num__2 ( x - num__2 ) num__2 + num__3 = ( num__1 + √ num__2 ) num__2 ( num__1 + √ num__2 - num__2 ) num__2 + num__3 = ( √ num__2 + num__1 ) num__2 ( √ num__2 - num__1 ) num__2 + num__3 = [ ( √ num__2 ) num__2 - ( num__1 ) num__2 ] num__2 + num__3 = ( num__2 - num__1 ) num__2 = num__1 + num__3 = num__4 correct option : c <eor> c <eos> |
c |
add__1.0__4.0__ multiply__1.0__4.0__ |
add__1.0__4.0__ add__1.0__3.0__ |
| if $ num__30000 interest is invested in x percent simple annual interest for num__5 years which of the following represents the total amount of interest in dollars that will be earned by this investment in the n years ? <o> a ) num__2000 x <o> b ) num__3000 x <o> c ) num__4500 x <o> d ) num__1200 x <o> e ) num__1500 x |
num__30000 * x / num__100 * num__5 = num__1500 x answer : e <eor> e <eos> |
e |
percent__5.0__30000.0__ percent__5.0__30000.0__ |
percent__5.0__30000.0__ percent__5.0__30000.0__ |
| in what time a sum of money double itself at num__3.0 per annum simple interest ? <o> a ) num__33 num__0.142857142857 % <o> b ) num__33 num__0.333333333333 % <o> c ) num__33 num__0.5 % <o> d ) num__33 num__2.33333333333 % <o> e ) num__33 num__1.66666666667 % |
p = ( p * num__3 * r ) / num__100 r = num__33 num__0.333333333333 % answer : b <eor> b <eos> |
b |
percent__33.0__100.0__ |
percent__33.0__100.0__ |
| if x / y = num__3 / z then num__4 x ^ num__2 = <o> a ) y / z <o> b ) xy <o> c ) y ² / z ² <o> d ) num__36 y ² / z ² <o> e ) num__15 y ² / z ² |
this question is most easily solved by isolating y in the equation and substituting into the expression num__4 x ² : x / y = num__3 / z x = num__3 y / z if we substitute num__3 y / z into the expression for x we get : num__4 ( num__3 y / z ) ² = num__4 ( num__9 y ² / z ² ) = num__36 y ² / z ² . the correct answer is choice ( d ) . <eor> d <eos> |
d |
multiply__4.0__9.0__ multiply__4.0__9.0__ |
multiply__4.0__9.0__ multiply__4.0__9.0__ |
| if the sum of the first n positive integers is s what is the sum of the first n positive integers divisible by num__3 in terms of s ? <o> a ) s / num__2 <o> b ) s <o> c ) num__2 s <o> d ) num__3 s <o> e ) num__4 s |
answer is d given sum of num__1 num__2 num__3 . . . . . n = s sum of num__3 num__6 num__9 . . . . num__3 n = num__3 [ sum ( num__1 num__2 num__3 . . . . n ) ] = num__3 s <eor> d <eos> |
d |
subtract__3.0__1.0__ multiply__3.0__2.0__ add__3.0__6.0__ multiply__3.0__1.0__ |
subtract__3.0__1.0__ multiply__3.0__2.0__ add__3.0__6.0__ multiply__3.0__1.0__ |
| an equal number of desks and bookcases are to be placed along a library wall that is num__15 meters long . each desk is num__2 meters long and each bookshelf is num__1.5 meters long . if the maximum possible number of desks and bookcases are to be placed along the wall then the space along the wall that is left over will be how many meters k long ? <o> a ) num__0.5 <o> b ) num__1 <o> c ) num__1.5 <o> d ) num__2 <o> e ) num__3 |
let x be the number of desks and bookcases that are placed along the library wall . num__2 x + num__1.5 x < num__15 num__3.5 x < num__15 since x is a non negative integer the largest number x can be is num__4 . when x is num__4 the desks and bookcases take up num__3.5 * num__4 = num__14 m = k leaving num__1 m of empty space . thus i believe the answer is b ) num__1 <eor> b <eos> |
b |
add__2.0__1.5__ multiply__3.5__4.0__ round_down__1.5__ round_down__1.5__ |
add__2.0__1.5__ multiply__3.5__4.0__ round_down__1.5__ round_down__1.5__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__21 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__225 <o> b ) num__240 <o> c ) num__288 <o> d ) num__267 <o> e ) num__211 |
speed = ( num__54 * num__0.277777777778 ) m / sec = num__15 m / sec . length of the train = ( num__15 x num__21 ) m = num__315 m . let the length of the platform be x meters . then ( x + num__315 ) / num__36 = num__15 = = > x + num__315 = num__540 = = > x = num__225 m . answer : a <eor> a <eos> |
a |
subtract__36.0__21.0__ multiply__21.0__15.0__ multiply__36.0__15.0__ subtract__540.0__315.0__ round__225.0__ |
subtract__36.0__21.0__ multiply__21.0__15.0__ multiply__36.0__15.0__ subtract__540.0__315.0__ round__225.0__ |
| a b c are three consecutive positive integers ( a > b > c ) . what is the value of the expression num__2 a + b + num__2 c ? <o> a ) num__6 a + num__7 . <o> b ) num__5 a + num__1 . <o> c ) num__5 a - num__5 . <o> d ) num__6 a - num__5 . <o> e ) num__6 a - num__7 . |
b = a - num__1 c = a - num__2 putting these values in num__2 a + b + num__2 c we get num__5 a - num__5 c is the answer <eor> c <eos> |
c |
multiply__1.0__5.0__ |
multiply__1.0__5.0__ |
| the average weight of num__7 persons increases by num__1.5 kg . if a person weighing num__65 kg is replaced by a new person what could be the weight of the new person ? <o> a ) num__76 kg <o> b ) num__77 kg <o> c ) num__75.5 kg <o> d ) data inadequate <o> e ) none of these |
total weight increases = num__7 × num__1.5 = num__10.5 kg so the weight of new person = num__65 + num__10.5 = num__75.5 kg answer c <eor> c <eos> |
c |
multiply__7.0__1.5__ add__65.0__10.5__ add__65.0__10.5__ |
multiply__7.0__1.5__ add__65.0__10.5__ add__65.0__10.5__ |
| if in a kindergarten class there are five children num__3 feet num__4 inches tall ( indicated by notation num__3 ' num__4 ' ' ) two children num__4 ' num__3 ' ' tall and thirteen children num__2 ' num__5 ' ' tall which of the following is the median height of the children ? <o> a ) num__2 ' num__5 ' ' <o> b ) num__2 ' num__6 ' ' <o> c ) num__3 ' num__3 ' ' <o> d ) num__3 ' num__4 ' ' <o> e ) num__4 ' num__3 |
total number of children in class = num__20 so median must be between num__10 th and num__11 th child when they are arranged in ascending order . when arranged in ascending order num__5 th and num__6 th child will have the height of num__3 ' num__4 ' ' so required median height = ( num__2 ' num__5 ' ' + num__2 ' num__5 ' ' ) / num__2 = num__2 ' num__5 ' ' ans : option a <eor> a <eos> |
a |
multiply__4.0__5.0__ multiply__2.0__5.0__ multiply__3.0__2.0__ subtract__4.0__2.0__ |
multiply__4.0__5.0__ divide__20.0__2.0__ add__4.0__2.0__ divide__4.0__2.0__ |
| when jessica withdrew $ num__200 from her bank account her account balance decreased by num__0.4 . if she deposits an amount equal to num__0.5 of the remaining balance what will be the final balance in her bank account ? <o> a ) num__300 <o> b ) num__375 <o> c ) num__450 <o> d ) num__500 <o> e ) num__575 |
as per the question num__200 = num__2 a / num__5 thus - a which is the total amount = num__500 the amount thus left = num__300 she then deposited num__0.5 of num__300 = num__150 total amount in her account = num__450 answer c <eor> c <eos> |
c |
reverse__0.5__ divide__2.0__0.4__ divide__200.0__0.4__ subtract__500.0__200.0__ multiply__0.5__300.0__ add__300.0__150.0__ add__300.0__150.0__ |
reverse__0.5__ divide__2.0__0.4__ divide__200.0__0.4__ subtract__500.0__200.0__ divide__300.0__2.0__ add__300.0__150.0__ add__300.0__150.0__ |
| two cogged wheels of which one has num__32 cogs and other num__54 cogs work into each other . if the latter turns num__80 times in three quarters of a minute how often does the other turn in num__8 seconds ? <o> a ) num__48 <o> b ) num__24 <o> c ) num__38 <o> d ) num__36 <o> e ) num__35 |
explanation : less cogs more turns and less time less turns number of turns required = num__80 Ã — Ã — = num__24 times answer is b <eor> b <eos> |
b |
subtract__32.0__8.0__ round__24.0__ |
subtract__32.0__8.0__ round__24.0__ |
| a store sells a certain product at a fixed price per unit . at the product ' s current price a units cost a total of exactly $ num__300 . if the price were lowered by $ num__5 from its current value then a + num__2 n units would cost exactly $ num__300 ; if the price were raised by $ num__5 then a – n units would cost exactly $ num__300 . what is the value of a ? <o> a ) num__10 <o> b ) num__15 <o> c ) num__20 <o> d ) num__25 <o> e ) num__30 |
it got too complicated when i used algebra . using plugging in it was quite fast . price quantity total value p a pa = num__300 p - num__5 a + num__2 n ( p - num__5 ) ( a + num__2 n ) = num__300 p + num__5 a - n ( p + num__5 ) ( a - n ) = num__300 solving three equations for three unknowns . tough ! ! plugging in i always start with c . c was the answer here so saved calculation ! putting values in above equations : price quantity total value num__15 num__20 num__300 num__10 num__20 + num__2 n num__300 - > num__10 ( num__20 + num__2 n ) = num__300 - > num__200 + num__20 n = num__300 - > num__20 n = num__100 - > n = num__5 num__20 num__15 num__300 so a = num__20 satisfies all equations ! ! <eor> c <eos> |
c |
divide__300.0__15.0__ multiply__5.0__2.0__ multiply__10.0__20.0__ subtract__300.0__200.0__ divide__300.0__15.0__ |
add__5.0__15.0__ subtract__15.0__5.0__ multiply__10.0__20.0__ subtract__300.0__200.0__ add__5.0__15.0__ |
| the average weight of num__46 students in a class is num__52 kg . num__5 of them whose average weight is num__48 kg leave the class and other num__5 students whose average weight is num__54 kg join the class . what is the new average weight ( in kg ) of the class ? <o> a ) num__51 num__1 ⁄ num__23 <o> b ) num__52 num__0.652173913043 <o> c ) num__52 num__15 ⁄ num__3 <o> d ) num__43.42 <o> e ) none of these |
total weight of num__46 students = num__46 × num__52 = num__2392 kg total weight of num__5 students who leave = num__5 × num__48 = num__240 kg total weight of num__5 students who join = num__5 × num__54 = num__270 kg therefore new total weight of num__46 students = num__2392 – num__240 + num__270 = num__2422 ⇒ new average weight = num__2422 ⁄ num__46 = num__52 num__0.652173913043 kg answer b <eor> b <eos> |
b |
multiply__46.0__52.0__ multiply__5.0__48.0__ multiply__5.0__54.0__ divide__2392.0__46.0__ |
multiply__46.0__52.0__ multiply__5.0__48.0__ multiply__5.0__54.0__ divide__2392.0__46.0__ |
| look at this series : num__7 num__10 num__8 num__11 num__9 num__12 . . . what number should come next ? <o> a ) num__7 <o> b ) num__10 <o> c ) num__12 <o> d ) num__13 <o> e ) num__14 |
explanation : this is a simple alternating addition and subtraction series . in the first pattern num__3 is added ; in the second num__2 is subtracted . answer : option b <eor> b <eos> |
b |
subtract__10.0__7.0__ subtract__10.0__8.0__ add__7.0__3.0__ |
subtract__10.0__7.0__ subtract__10.0__8.0__ add__7.0__3.0__ |
| there are num__48 students in a class . find the numbers of ways in which a committee of num__2 students is to be formed ? <o> a ) num__1128 <o> b ) num__1978 <o> c ) num__2546 <o> d ) num__3121 <o> e ) num__2400 |
required number of ways = num__48 c num__2 = num__48 * num__23.5 = num__1128 answer is a <eor> a <eos> |
a |
multiply__48.0__23.5__ multiply__48.0__23.5__ |
multiply__48.0__23.5__ multiply__48.0__23.5__ |
| in a certain state gasoline stations compute the price per gallon p in dollars charged at the pump by adding a num__3 percent sales tax to the dealer ' s price per gallon d in dollars and then adding a gasoline tax of $ num__0.18 per gallon . which of the following gives the dealer ' s price per gallon d in terms of the price per gallon p charged at the pump ? <o> a ) d = ( p - num__0.03 ) / num__1.18 <o> b ) d = p / num__1.21 <o> c ) d = ( p - num__0.18 ) / num__1.03 <o> d ) d = p - num__0.20 <o> e ) d = p / num__1.03 - num__0.18 |
let dealers price ( d ) be num__1 . so adding num__3.0 to dealers price is d + num__3.0 of d . i . e . num__1 + num__3.0 of num__1 which is num__1 + num__0.03 . then add num__0.18 to the value . now num__1.03 + num__0.18 . this is now num__1.21 . you have the gasoline stations price ( p ) as num__1.21 dollars . now sub num__1.21 in the options to know which option gave you d = num__1 . d must equal num__1 because you earlier picked num__1 as the value of d in the question . ps : always remember to start from e upwards . answer : c <eor> c <eos> |
c |
add__1.0__0.03__ add__0.18__1.03__ multiply__0.18__1.0__ |
add__1.0__0.03__ add__0.18__1.03__ multiply__0.18__1.0__ |
| a man sold num__10 eggs for num__5 rupees and gained num__20.0 . how many eggs did he buy for num__5 rupees ? <o> a ) num__10 eggs <o> b ) num__12 eggs <o> c ) num__14 eggs <o> d ) num__16 eggs <o> e ) none of these |
s . p . for num__1 egg = num__5 ⁄ num__10 = num__1 ⁄ num__2 ∴ c . p . for num__1 egg = num__100 / ( num__100 + num__20 ) × num__12 = num__512 ⇒ he bought num__12 eggs for num__5 rupees . answer b <eor> b <eos> |
b |
percent__5.0__20.0__ percent__10.0__20.0__ percent__100.0__12.0__ |
percent__5.0__20.0__ percent__10.0__20.0__ percent__100.0__12.0__ |
| ( √ num__27 + √ num__243 ) / √ num__75 = ? <o> a ) num__2 √ num__2 <o> b ) num__2 √ num__3 <o> c ) num__3 √ num__2 <o> d ) num__3 √ num__3 <o> e ) num__2.4 |
( √ num__27 + √ num__243 ) / √ num__75 = ( num__3 √ num__3 + num__9 √ num__3 ) / num__5 √ num__3 = num__12 √ num__0.6 √ num__3 = num__2.4 . hence the correct answer is e <eor> e <eos> |
e |
divide__27.0__3.0__ add__3.0__9.0__ divide__3.0__5.0__ subtract__3.0__0.6__ subtract__3.0__0.6__ |
divide__27.0__3.0__ add__3.0__9.0__ divide__3.0__5.0__ divide__12.0__5.0__ divide__12.0__5.0__ |
| two sets of num__4 consecutive positive even integers have exactly one integer in common . the sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set ? <o> a ) num__14 <o> b ) num__7 <o> c ) num__18 <o> d ) num__24 <o> e ) it can not be determined from the information given . |
a = ( num__24 num__68 ) sum of this = num__20 b = ( num__810 num__1214 ) sum of this = num__44 the differenct between num__44 - num__20 = num__24 hence num__24 is the answer i . e . d <eor> d <eos> |
d |
subtract__24.0__4.0__ subtract__68.0__24.0__ add__4.0__20.0__ |
subtract__24.0__4.0__ subtract__68.0__24.0__ subtract__68.0__44.0__ |
| how many num__0.5 s are there in num__37 num__0.5 ? <o> a ) num__75 <o> b ) num__150 <o> c ) num__300 <o> d ) num__600 <o> e ) num__700 |
required number = ( num__37.5 ) / ( num__0.5 ) = ( num__37.5 x num__2.0 ) = num__75 answer : a <eor> a <eos> |
a |
add__0.5__37.0__ reverse__0.5__ divide__37.5__0.5__ divide__37.5__0.5__ |
add__0.5__37.0__ reverse__0.5__ divide__37.5__0.5__ divide__37.5__0.5__ |
| find the odd number : num__15 num__1117 num__2329 <o> a ) num__1 <o> b ) num__3 <o> c ) num__4 <o> d ) num__6 <o> e ) num__7 |
explanation : all the given numbers are prime numbers . num__1 is not a prime number because it is divisible by only one factor that is num__1 . answer : a <eor> a <eos> |
a |
reverse__1.0__ |
reverse__1.0__ |
| the sum of all the integers k such that – num__25 < k < num__24 is <o> a ) num__0 <o> b ) - num__2 <o> c ) - num__25 <o> d ) - num__49 <o> e ) - num__47 |
- num__24 - - - - - - - - - - - - - - - - - - num__0 - - - - - - - - - - - - - - - - - num__23 values upto + num__23 cancels outwe are left with only - num__24 - num__23 sum of which is - num__47 . hence option d . e <eor> e <eos> |
e |
add__24.0__23.0__ add__24.0__23.0__ |
add__24.0__23.0__ add__24.0__23.0__ |
| crazy eddie has a key chain factory . eddie managed to decrease the cost of manufacturing his key chains while keeping the same selling price and thus increased the profit from the sale of each key chain from num__50.0 of the selling price to num__50.0 of the selling price . if the manufacturing cost is now $ num__50 what was it before the decrease ? <o> a ) $ num__20 <o> b ) $ num__40 <o> c ) $ num__50 <o> d ) $ num__80 <o> e ) $ num__100 |
deargoodyear num__2013 i ' m happy to help . this is a relatively straightforward problem not very challenging . btw crazy eddiewas the actually name of an electronics chain on the east coast of the usa back in the num__1970 s . manufacturing now is $ num__50 . they now are making a num__50.0 profit so the selling price must be $ num__100 . they had this same selling price $ num__100 before they made the change and had a profit of num__50.0 so the manufacturing must have been $ num__50 . answer = ( c ) . <eor> c <eos> |
c |
subtract__100.0__50.0__ |
subtract__100.0__50.0__ |
| in the biology lab of ` ` jefferson ' ' high school there are num__0.036 * num__10 ^ num__5 germs equally divided among num__45000 * num__10 ^ ( - num__3 ) petri dishes . how many germs live happily in a single dish ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__30 <o> d ) num__60 <o> e ) num__80 |
num__0.036 * num__10 ^ num__5 can be written as num__3600 num__45000 * num__10 ^ ( - num__3 ) can be written as num__45 required = num__80.0 = num__80 answer : e <eor> e <eos> |
e |
divide__3600.0__45.0__ divide__3600.0__45.0__ |
divide__3600.0__45.0__ divide__3600.0__45.0__ |
| a cylinder is sliced by a plane to form the solid shown . the base edge of the solid is a circle of radius num__3 . the top edge is an ellipse . the highest point on the ellipse is num__6 units above the base . the lowest point on the ellipse is num__2 units above the base . what is the volume in cubic units of the solid ? <o> a ) num__24 π <o> b ) num__30 π <o> c ) num__36 π <o> d ) num__42 π <o> e ) num__48 π |
the plane parallel to the base passing through the center of the ellipse will cross the ellipse along its minor axis . if we cut the solid along this plane and flip the top piece that is sliced off num__180 degrees around the minor axis the solid becomes a cylinder with the base of area num__9 and height ( num__2 + num__6 ) / num__2 = num__4 . the volume is therefore num__36 π . correct answer c <eor> c <eos> |
c |
triangle_area__3.0__6.0__ square_perimeter__9.0__ square_perimeter__9.0__ |
triangle_area__3.0__6.0__ square_perimeter__9.0__ square_perimeter__9.0__ |
| the average salary of all the workers in a workshop is rs . num__8000 . the average salary of num__7 technicians is rs . num__20000 and the average salary of the rest is rs . num__6000 . the total number of workers in the workshop is <o> a ) num__20 <o> b ) num__49 <o> c ) num__22 <o> d ) num__23 <o> e ) none |
sol . let the total number of workers be x . then num__8000 x = ( num__20000 × num__7 ) + num__6000 ( x – num__7 ) ‹ = › num__2000 x = num__98000 ‹ = › x = num__49 . answer b <eor> b <eos> |
b |
subtract__8000.0__6000.0__ divide__98000.0__2000.0__ divide__98000.0__2000.0__ |
subtract__8000.0__6000.0__ divide__98000.0__2000.0__ divide__98000.0__2000.0__ |
| mr . gangadhar mr . ramesh and mr . shridhar together earned num__19800 . the ratio of earnings between mr . gangadhar and mr . ramesh is num__2 : num__1 while that between mr . ramesh and mr . shridhar is num__3 : num__2 . how much did mr . ramesh earn ? <o> a ) num__3600 <o> b ) num__5400 <o> c ) num__1800 <o> d ) num__6300 <o> e ) none of these |
ratio of investment = num__6 : num__3 : num__2 ∴ share of mr . ramesh = num__3 ⁄ num__11 × num__19800 = num__5400 answer b <eor> b <eos> |
b |
multiply__2.0__3.0__ multiply__1.0__5400.0__ |
multiply__2.0__3.0__ multiply__1.0__5400.0__ |
| when w is divided by num__13 the reminder is num__0 . if w is one more than it value and when divided by num__10 its remainder is num__0 . what is the value of w ? <o> a ) num__39 <o> b ) num__26 <o> c ) num__14 <o> d ) num__13 <o> e ) num__52 |
w is divided by num__13 so that is multiple of num__14 as num__13 num__2639 . . . w + num__1 is divided by num__10 the remainder is num__0 so it is divisible by num__10 . consider from option let us take the number is num__39 it is divisible by num__13 but num__39 + num__1 is divisible by num__10 so ans is a <eor> a <eos> |
a |
subtract__14.0__13.0__ multiply__1.0__39.0__ |
subtract__14.0__13.0__ multiply__1.0__39.0__ |
| a dishonest dealer claims to sell a product at its cost price . he uses a counterfeit weight which is num__20.0 less than the real weight . further greed overtook him and he added num__40.0 impurities to the product . find the net profit percentage of the dealer ? <o> a ) num__94.0 <o> b ) num__87.5 <o> c ) num__55.0 <o> d ) num__56.25 <o> e ) num__36 % |
the dealer uses weight which is num__20.0 less than the real weight . or ( num__1 - num__0.2 ) or num__0.8 of real weight . it means that he is selling $ num__4 worth of product for $ num__5 . the dealer then further added num__50.0 impurities to the product . it means that he is selling $ num__5 worth of product for $ num__7.5 . so his profit is $ num__7.5 - $ num__4 = $ num__3.5 and his profit percent is ( num__3.5 / num__4 ) * num__100 = num__87.5 answer : - b <eor> b <eos> |
b |
subtract__1.0__0.2__ multiply__20.0__0.2__ reverse__0.2__ divide__40.0__0.8__ subtract__7.5__4.0__ multiply__20.0__5.0__ multiply__1.0__87.5__ |
subtract__1.0__0.2__ multiply__20.0__0.2__ reverse__0.2__ divide__40.0__0.8__ subtract__7.5__4.0__ multiply__20.0__5.0__ multiply__1.0__87.5__ |
| if a exceeds b by num__40.0 b is less than c by num__20.0 then a : c is : <o> a ) num__28 : num__25 <o> b ) num__26 : num__25 <o> c ) num__3 : num__2 <o> d ) num__3 : num__1 <o> e ) num__28 : num__27 |
a = b + num__40.0 of b a = num__1.4 * b now a / b = num__1.4 b = c - num__20.0 of c b = num__0.8 * c b / c = num__0.8 so a / c = ( a / b ) * ( b / c ) = ( num__14 * num__8 ) / ( num__10 * num__10 ) = num__28 : num__25 . answer - option a <eor> a <eos> |
a |
divide__14.0__1.4__ add__20.0__8.0__ divide__20.0__0.8__ add__20.0__8.0__ |
divide__14.0__1.4__ add__20.0__8.0__ divide__20.0__0.8__ add__20.0__8.0__ |
| if f ( x ) = num__4 * ( x ^ num__4 - num__1 ) / ( x ^ num__2 ) what is f ( num__1 / x ) in terms of f ( x ) ? <o> a ) f ( x ) <o> b ) - f ( x ) <o> c ) - num__4 * f ( x ) <o> d ) - num__1 / f ( x ) <o> e ) num__2 * f ( x ) |
num__4 * f ( num__1 / x ) = num__4 * ( ( num__1 / x ) ^ num__4 - num__1 ) / ( ( num__1 / x ) ^ num__2 ) = num__4 * ( ( num__1 / x ^ num__4 ) - num__1 ) / ( num__1 / x ^ num__2 ) = num__4 * ( ( num__1 - x ^ num__4 ) / ( x ^ num__4 ) ) / ( num__1 / x ^ num__2 ) = num__4 * ( num__1 - x ^ num__4 ) / ( x ^ num__2 ) = - num__4 * ( ( x ^ num__4 ) - num__1 ) / ( x ^ num__2 ) = - num__4 f ( x ) answer is c . <eor> c <eos> |
c |
multiply__4.0__1.0__ |
multiply__4.0__1.0__ |
| a cheetah can at num__65 mph but only for a maximum of num__6 minutes before it exhausts itself . how many miles can the cheetah run before tiring out ? round to the nearest hundredths place . <o> a ) num__5 miles <o> b ) num__5.50 miles <o> c ) num__6 miles <o> d ) num__6.50 miles <o> e ) num__7 miles |
num__65 mph x ( num__0.1 ) = num__6.5 miles answer d <eor> d <eos> |
d |
multiply__65.0__0.1__ round__6.5__ |
multiply__65.0__0.1__ round__6.5__ |
| a man took a loan from a bank at the rate of num__8.0 p . a . simple interest . after num__3 years he had to pay rs . num__5400 interest only for the period . the principal amount borrowed by him was : <o> a ) rs . num__2000 <o> b ) rs . num__10500 <o> c ) rs . num__15500 <o> d ) rs . num__22500 <o> e ) none |
solution principal = rs . ( num__100 x num__675.0 x num__3 ) = rs . num__22500 . answer d <eor> d <eos> |
d |
percent__100.0__22500.0__ |
percent__100.0__22500.0__ |
| six bells commence tolling together and toll at intervals of num__12 num__34 num__56 seconds respectively . in num__30 minutes how many times do they toll together ? <o> a ) num__44 <o> b ) num__31 <o> c ) num__15 <o> d ) num__16 <o> e ) num__17 |
l . c . m of num__12 num__34 num__56 is num__180 . i . e after each num__1 min they will toll together . so in num__30 min they will toll num__30 times . as they have initially tolled once the answer will be num__30 + num__1 = num__31 . answer : b <eor> b <eos> |
b |
add__30.0__1.0__ round__31.0__ |
add__30.0__1.0__ add__30.0__1.0__ |
| the ratio between the speeds of two trains is num__7 : num__8 . if the second train runs num__400 km in num__4 hours then the speed of the first train is : a . b . c . d . answer : option d explanation : <o> a ) num__70 km / hr <o> b ) num__75 km / hr <o> c ) num__87.5 km / hr <o> d ) num__84 km / hr <o> e ) num__75.6 km / hr |
let the speed of two trains be num__7 x and num__8 x km / hr . then num__8 x = num__100.0 = num__100 x = num__12.5 = num__12.5 num__8 speed of first train = ( num__7 x num__12.5 ) km / hr = num__87.5 km / hr . answer : c <eor> c <eos> |
c |
divide__400.0__4.0__ divide__100.0__8.0__ multiply__7.0__12.5__ round__87.5__ |
divide__400.0__4.0__ divide__100.0__8.0__ multiply__7.0__12.5__ round__87.5__ |
| if the average of num__10 consecutive integers is num__23.5 then the num__10 th integer is : - <o> a ) num__15 <o> b ) num__20 <o> c ) num__23 <o> d ) num__28 <o> e ) num__25 |
the average falls between the num__5 th and num__6 th integers integer num__5 = num__23 integer num__6 = num__24 . counting up to the tenth integer we get num__28 . answer : d <eor> d <eos> |
d |
round_down__23.5__ add__5.0__23.0__ add__5.0__23.0__ |
round_down__23.5__ add__5.0__23.0__ add__5.0__23.0__ |
| how many num__4 ' s are there preceded by num__7 but not followed by num__3 ? num__5 num__9 num__3 num__2 num__1 num__7 num__4 num__2 num__6 num__9 num__7 num__4 num__6 num__1 num__3 num__2 num__8 num__7 num__4 num__1 num__3 num__8 num__3 num__2 num__5 num__6 num__7 num__4 num__3 num__9 num__5 num__8 num__2 num__0 num__1 num__8 num__7 num__4 num__6 num__3 <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
num__7 num__4 num__2 num__7 num__4 num__6 num__7 num__4 num__1 num__7 num__4 num__6 only at these placees num__4 is preceded by num__7 but not followed by num__3 answer : a <eor> a <eos> |
a |
multiply__4.0__1.0__ |
multiply__4.0__1.0__ |
| alex has num__4 pens worth of { num__22 num__25 num__30 num__40 } what is the total mean and median of the worth of pens ? <o> a ) num__3.42 <o> b ) num__6.16 <o> c ) num__8.32 <o> d ) num__2.0 <o> e ) num__1.75 |
this is a good question to understand the difference between mean and median . mean : average of all the numbers . ( sum of all the elements divided by the number of elements ) median : arrange the elements of the set in increasing order . if the number of terms is odd the middle term is the median . if the number of terms is even the average of middle two terms is the median coming to this question mean = ( num__22 + num__25 + num__30 + num__40 ) / num__4 = num__29.25 median = ( num__25 + num__30 ) / num__2 = num__27.5 total = num__1.75 option e <eor> e <eos> |
e |
subtract__29.25__27.5__ subtract__29.25__27.5__ |
subtract__29.25__27.5__ subtract__29.25__27.5__ |
| oy started cycling along the boundaries of a square field from corner point a . after half an hour he reached the corner point c diagonally opposite to a . if his speed was num__8 km / hr the area of the filed in square km is ? <o> a ) num__64 <o> b ) num__4 <o> c ) num__16 <o> d ) num__9 <o> e ) num__5 |
explanation : distance covered by roy in num__0.5 hr = num__4 kmtherefore side of the square = num__2.0 = num__2 kmhence area = num__2 × num__2 = num__4 square km answer : b <eor> b <eos> |
b |
multiply__8.0__0.5__ square_perimeter__0.5__ multiply__8.0__0.5__ |
multiply__8.0__0.5__ square_perimeter__0.5__ multiply__8.0__0.5__ |
| two pots are in side - by - side . one pot which is num__20 inches tall casts a shadow that is num__10 inches long . the other pot is num__40 inches tall . compute in inches the length of the shadow that the taller pot casts . <o> a ) num__25 <o> b ) num__30 <o> c ) num__10 <o> d ) num__15 <o> e ) num__20 |
the ratio of shadow to height is constant so if x is the length of the shadow then num__2.0 = num__40 / x and x = num__20 . correct answer e <eor> e <eos> |
e |
divide__20.0__10.0__ multiply__10.0__2.0__ |
divide__20.0__10.0__ divide__40.0__2.0__ |
| water boils at num__212 ° f or num__100 ° c and ice melts at num__32 ° f or num__0 ° c . if the temperature of a pot of water is num__45 ° c what is the temperature of the pot of water in ° f ? <o> a ) num__92 ° f <o> b ) num__97 ° f <o> c ) num__104 ° f <o> d ) num__113 ° f <o> e ) num__118 ° f |
let f and c denote the temperature in fahrenheit and celsius respectively . ( f - num__32 ) / ( num__212 - num__32 ) = ( c - num__0 ) / ( num__100 - num__0 ) f = num__9 c / num__5 + num__32 f = num__9 ( num__45 ) / num__5 + num__32 = num__113 ° f the answer is d . <eor> d <eos> |
d |
divide__45.0__9.0__ round__113.0__ |
divide__45.0__9.0__ round__113.0__ |
| ayush was born two years after his father ' s marriage . his mother is five years younger than his father but num__20 years older than ayush who is num__10 years old . at what age did the father get married ? <o> a ) num__23 years <o> b ) num__25 years <o> c ) num__33 years <o> d ) num__35 years <o> e ) num__37 years |
explanation : ayush ' s present age = num__10 years . his mother ' s present age = ( num__10 + num__20 ) years = num__30 years . ayush ' s father ' s present age = ( num__30 + num__5 ) years = num__35 years . ayush ' s father ' s age at the time of ayush ' s birth = ( num__35 - num__10 ) years = num__25 years . therefore ayush ' s father ' s age at the time of marriage = ( num__25 - num__2 ) years = num__23 years . answer : a ) num__23 year <eor> a <eos> |
a |
add__20.0__10.0__ add__5.0__30.0__ add__20.0__5.0__ divide__20.0__10.0__ subtract__25.0__2.0__ subtract__25.0__2.0__ |
add__20.0__10.0__ add__5.0__30.0__ add__20.0__5.0__ divide__20.0__10.0__ subtract__25.0__2.0__ subtract__25.0__2.0__ |
| a man saves num__25.0 of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by num__10.0 he is only able to save rs . num__175 per month . what is his monthly salary ? <o> a ) rs . num__1000 <o> b ) rs . num__2000 <o> c ) rs . num__1500 <o> d ) rs . num__3000 <o> e ) rs . num__3100 |
income = rs . num__100 expenditure = rs . num__75 savings = rs . num__25 present expenditure num__75 + num__75 * ( num__0.1 ) = rs . num__82.5 present savings = num__100 – num__82.50 = rs . num__17.50 if savings is rs . num__17.50 salary = rs . num__100 if savings is rs . num__175 salary = num__100 / num__17.5 * num__175 = num__1000 answer : a <eor> a <eos> |
a |
subtract__175.0__100.0__ reverse__10.0__ divide__175.0__10.0__ multiply__10.0__100.0__ multiply__10.0__100.0__ |
subtract__175.0__100.0__ reverse__10.0__ divide__175.0__10.0__ multiply__10.0__100.0__ multiply__10.0__100.0__ |
| all the milk in container a which was filled to its brim was poured into two containers b and c . the quantity of milk in container b was num__62.5 less than the capacity of container a . if num__158 liters was now transferred from c to b then both the containers would have equal quantities of milk . what was the initial quantity of milk in container a ? <o> a ) num__1264 <o> b ) num__1723 <o> c ) num__1129 <o> d ) num__2613 <o> e ) num__1372 |
a b has num__62.5 or ( num__0.625 ) of the milk in a . therefore let the quantity of milk in container a ( initially ) be num__8 k . quantity of milk in b = num__8 k - num__5 k = num__3 k . quantity of milk in container c = num__8 k - num__3 k = num__5 k container : a b c quantity of milk : num__8 k num__3 k num__5 k it is given that if num__158 liters was transferred from container c to container b then both the containers would have equal quantities of milk . num__5 k - num__158 = num__3 k + num__158 = > num__2 k = num__316 = > k = num__158 the initial quantity of milk in a = num__8 k = num__8 * num__158 = num__1264 liters . <eor> a <eos> |
a |
multiply__0.625__8.0__ subtract__8.0__5.0__ subtract__5.0__3.0__ multiply__158.0__2.0__ multiply__158.0__8.0__ multiply__158.0__8.0__ |
multiply__0.625__8.0__ subtract__8.0__5.0__ subtract__5.0__3.0__ multiply__158.0__2.0__ multiply__158.0__8.0__ multiply__158.0__8.0__ |
| in the formula y = num__1 / ( num__2 x + num__2 ) ^ num__3 if x is halved then y is multiplied by <o> a ) num__27 <o> b ) num__3.375 <o> c ) num__0.296296296296 <o> d ) num__8 <o> e ) num__0.037037037037 |
say x = num__2 = > y num__1 = num__0.00925925925926 when x = num__1 ; y num__2 = num__0.03125 y num__2 = num__3.375 * y num__1 . answer : b <eor> b <eos> |
b |
multiply__1.0__3.375__ |
multiply__1.0__3.375__ |
| if w is a positive integer and w / num__15 = num__8.2 what is the remainder when w is divided by num__15 ? <o> a ) a . num__1 <o> b ) b . num__2 <o> c ) c . num__3 <o> d ) d . num__4 <o> e ) e . num__8 |
if w were a multiple of num__15 the quotient w / num__15 would be an integer . the fact that it ' s a decimal tells us that num__15 goes into w eight whole times and some decimal part of a time . this decimal part num__0.2 is the remainder divided by the divisor . let r be the remainder . r / num__15 = num__0.2 = num__0.2 r = ( num__15 ) * ( num__0.2 ) = num__3 answer = ( c ) <eor> c <eos> |
c |
multiply__15.0__0.2__ multiply__15.0__0.2__ |
multiply__15.0__0.2__ multiply__15.0__0.2__ |
| how many positive integers less than num__50 are multiples of num__3 but not multiples of num__5 ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__8 <o> d ) num__10 <o> e ) num__12 |
imo the answer is c ( num__8 numbers ) the lcm of num__3 and num__5 is num__15 . if x < num__50 and x is divisible by num__3 not by num__5 - - > x is not divisible by num__15 . from num__1 - - > num__50 we have num__3 numbers which is divisible by num__15 : num__15 num__30 num__45 . from num__1 - - > num__50 we have ( num__45 - num__3 ) / num__3 + num__1 = num__15 numbers divisible by num__3 . therefore our answer is num__15 - num__3 = num__12 numbers . e <eor> e <eos> |
e |
add__3.0__5.0__ multiply__3.0__5.0__ subtract__50.0__5.0__ subtract__15.0__3.0__ multiply__1.0__12.0__ |
add__3.0__5.0__ multiply__3.0__5.0__ subtract__50.0__5.0__ subtract__15.0__3.0__ divide__12.0__1.0__ |
| find the surface area of a num__10 cm x num__4 cm x num__2 cm brick . <o> a ) num__136 cu cm <o> b ) num__124 cu cm <o> c ) num__164 cu cm <o> d ) num__180 cu cm <o> e ) none |
sol . surface area = [ num__2 ( num__10 x num__4 + num__4 x num__2 + num__10 x num__2 ) ] = ( num__2 x num__68 ) = num__136 cu cm answer a <eor> a <eos> |
a |
triangle_area__4.0__68.0__ triangle_area__4.0__68.0__ |
triangle_area__4.0__68.0__ triangle_area__4.0__68.0__ |
| a series of boxes are placed on a conveyor belt at one end and unloaded at the other . the boxes are each num__0.5 meters on each side and are placed immediately adjacent to one another as they are loaded . if the conveyor belt is num__30 meters long and moves the boxes at a speed of num__0.25 m / s how long does it take for num__100 boxes to be loaded and unloaded ? <o> a ) num__4 min num__30 sec <o> b ) num__4 min num__50 sec <o> c ) num__5 min num__20 sec <o> d ) num__5 min num__40 sec <o> e ) num__6 min num__10 sec |
c num__5 min num__20 sec d = num__100 * num__0.5 m + num__30 m = num__80 m t = num__80 m / num__0.25 m / s = num__320 s = num__5 min num__20 sec <eor> c <eos> |
c |
divide__100.0__5.0__ subtract__100.0__20.0__ divide__80.0__0.25__ round__5.0__ |
divide__100.0__5.0__ divide__20.0__0.25__ divide__80.0__0.25__ multiply__0.25__20.0__ |
| the average earning of a mechanic for the first - four days of a week is rs . num__18 and for the last four days is rs . num__22 . if he earns rs . num__6 on the fourth day his average earning for the whole week is ? <o> a ) rs . num__18.95 <o> b ) rs num__16 <o> c ) rs . num__22 <o> d ) rs . num__25.71 <o> e ) none of these |
answer total earning for the week = sum of earning of first four days + sum of earning of last four days - earning of num__4 th day = num__4 x num__18 + num__4 x num__22 - num__6 = rs . num__154 â ˆ ´ average earning = num__22.0 = rs . num__22 correct option : c <eor> c <eos> |
c |
subtract__22.0__18.0__ add__18.0__4.0__ |
subtract__22.0__18.0__ add__18.0__4.0__ |
| the average age of students of a class is num__15.8 years . the average age of boys in the class is num__16.4 years and that of the girls is num__15.7 years . the ration of the number of boys to the number of girls in the class is ? <o> a ) num__1 : num__6 <o> b ) num__2 : num__3 <o> c ) num__2 : num__5 <o> d ) num__2 : num__1 <o> e ) num__2 : num__4 |
let the ratio be k : num__1 . then k * num__16.4 + num__1 * num__15.7 = ( k + num__1 ) * num__15.8 = ( num__16.4 - num__15.8 ) k = ( num__15.8 - num__15.7 ) = k = num__0.1 / num__0.6 = num__0.166666666667 required ratio = num__0.166666666667 : num__1 = num__1 : num__6 . answer : a <eor> a <eos> |
a |
subtract__15.8__15.7__ subtract__16.4__15.8__ divide__0.1__0.6__ divide__0.6__0.1__ reverse__1.0__ |
subtract__15.8__15.7__ subtract__16.4__15.8__ divide__0.1__0.6__ divide__0.6__0.1__ reverse__1.0__ |
| in a certain egg - processing plant every egg must be inspected and is either accepted for processing or rejected . for every num__96 eggs accepted for processing num__4 eggs are rejected . if on a particular day num__12 additional eggs were accepted but the overall number of eggs inspected remained the same the ratio of those accepted to those rejected would be num__99 to num__1 . how many e eggs does the plant process per day ? <o> a ) num__100 <o> b ) num__300 <o> c ) num__400 <o> d ) num__3000 <o> e ) num__4 |
000 |
straight pluggin in for me . as usual i started with c and got the answer . lets ' back calculate and see what we get let us consider eggs processed each day to be num__400 so initial ratio of eggs processed and rejected is num__96 : num__4 or num__24 : num__1 so out of num__400 eggs there will be num__384 eggs processed and num__16 rejected . now if the no . of eggs inspected remain and num__12 more eggs get accepted that means there e = num__384 + num__12 = num__396 eggs accepted and num__4 rejected . . . and the ratio will be num__99 : num__1 bingo . . . this is what the questions says . . . . its always a good idea to start with c . <eor> c <eos> |
c |
c |
| what will be the cost of building a fence around a square plot with area equal to num__289 sq ft if the price per foot of building the fence is rs . num__58 ? <o> a ) rs . num__3944 <o> b ) rs . num__3988 <o> c ) rs . num__3928 <o> d ) rs . num__3928 <o> e ) rs . num__3943 |
let the side of the square plot be a ft . a num__2 = num__289 = > a = num__17 length of the fence = perimeter of the plot = num__4 a = num__68 ft . cost of building the fence = num__68 * num__58 = rs . num__3944 . answer : a <eor> a <eos> |
a |
square_perimeter__17.0__ multiply__58.0__68.0__ multiply__58.0__68.0__ |
multiply__17.0__4.0__ multiply__58.0__68.0__ multiply__58.0__68.0__ |
| a goods train runs at the speed of num__72 km / hr and crosses a num__300 m long platform in num__26 sec . what is the length of the goods train ? <o> a ) num__278 <o> b ) num__166 <o> c ) num__151 <o> d ) num__220 <o> e ) num__109 |
speed = num__72 * num__0.277777777778 = num__20 m / sec . time = num__26 sec . let the length of the train be x meters . then ( x + num__300 ) / num__26 = num__20 x = num__220 m . answer : d <eor> d <eos> |
d |
round__220.0__ |
round__220.0__ |
| an athlete takes num__10 seconds to run num__100 m . what is his avg . speed in miles per hour ? <o> a ) num__22.37 <o> b ) num__26.66 <o> c ) num__24.35 <o> d ) num__36.0 <o> e ) num__42.44 |
his average speed is num__10 m / s . which is num__36 km / hr . but num__36 km = num__22.37 miles . the average speed of the athlete is num__22.37 mph answer : a <eor> a <eos> |
a |
round__22.37__ |
round__22.37__ |
| there are num__7 dozen mangoes in a box . if there are num__29 such boxes how many mangoes are there in all the boxes together ? <o> a ) num__516 <o> b ) num__2436 <o> c ) num__6192 <o> d ) num__628 <o> e ) none |
number of mangoes = num__7 dozens = num__7 × num__12 = num__84 ∴ number of mangoes in num__29 boxes = num__29 × num__84 = num__2436 answer b <eor> b <eos> |
b |
multiply__7.0__12.0__ multiply__29.0__84.0__ multiply__29.0__84.0__ |
multiply__7.0__12.0__ multiply__29.0__84.0__ multiply__29.0__84.0__ |
| if ( num__5 ^ num__13 ) ( num__9 ^ num__7 ) = num__3 ( num__15 ^ x ) what is the value of x ? <o> a ) num__7 <o> b ) num__9 <o> c ) num__11 <o> d ) num__13 <o> e ) num__15 |
( num__5 ^ num__13 ) ( num__9 ^ num__7 ) = num__3 ( num__15 ^ x ) = > num__5 ^ num__13 * num__3 ^ num__14 = num__3 * num__3 ^ x * num__5 ^ x = > num__5 ^ num__13 * num__3 ^ num__14 = num__3 ^ ( x + num__1 ) * num__5 ^ x value of x = num__13 answer d <eor> d <eos> |
d |
add__5.0__9.0__ subtract__15.0__14.0__ multiply__13.0__1.0__ |
add__5.0__9.0__ subtract__15.0__14.0__ multiply__13.0__1.0__ |
| a light flashes every num__15 seconds how many times will it flash in ? of an hour ? <o> a ) num__550 <o> b ) num__600 <o> c ) num__240 <o> d ) num__700 <o> e ) num__750 |
num__1 flash = num__15 sec for num__1 min = num__4 flashes so for num__1 hour = num__4 * num__60 = num__240 flashes . answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ multiply__4.0__60.0__ round__240.0__ |
multiply__15.0__4.0__ multiply__4.0__60.0__ multiply__4.0__60.0__ |
| solve for x and check : num__6 x = num__54 <o> a ) num__12 <o> b ) num__15 <o> c ) num__5 <o> d ) num__9 <o> e ) none of these |
solution : dividing each side by num__6 we obtain ( num__6 x / num__6 ) = ( num__9.0 ) therefore : x = num__9 check : num__6 x = num__54 ( num__6 * num__9 ) = num__54 num__54 = num__54 answer : d <eor> d <eos> |
d |
divide__54.0__6.0__ divide__54.0__6.0__ |
divide__54.0__6.0__ divide__54.0__6.0__ |
| which is odd one num__3 num__5 num__7 num__12 num__17 num__19 <o> a ) num__19 <o> b ) num__17 <o> c ) num__5 <o> d ) num__12 <o> e ) num__14 |
each of the numbers is a prime number except num__12 . answer : option d <eor> d <eos> |
d |
add__5.0__7.0__ |
add__5.0__7.0__ |
| how much time will it take for an amount of num__500 to yield num__100 as interest at num__5.0 per annum of simple interest ? <o> a ) num__3 year <o> b ) num__4 year <o> c ) num__6 year <o> d ) num__5 year <o> e ) num__7 year |
time = ( num__100 x num__100 ) / ( num__500 x num__5 ) years = num__4 years . answer b <eor> b <eos> |
b |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| find the average of first num__40 natural numbers ? <o> a ) num__12.3 <o> b ) num__15.4 <o> c ) num__63.2 <o> d ) num__45.2 <o> e ) num__20.5 |
sum of first n natural numbers = n ( n + num__1 ) / num__2 ; so sum of num__40 natural numbers = ( num__40 * num__41 ) / num__2 = num__820 . therefore the required average = ( num__20.5 ) = num__20.5 . ans : e <eor> e <eos> |
e |
add__40.0__1.0__ divide__41.0__2.0__ multiply__20.5__1.0__ |
add__40.0__1.0__ divide__41.0__2.0__ multiply__20.5__1.0__ |
| which of the following is not a multiple of num__3 ! + num__4 ? <o> a ) num__5 ! + num__20 <o> b ) num__4 ! + num__1 <o> c ) num__5 ! - num__60 <o> d ) num__4 ! + num__36 <o> e ) num__5 ! + num__20 |
the factorial num__3 ! means num__3 * num__2 therefore num__3 ! + num__4 = ( num__3 * num__2 ) + num__4 which gives num__6 + num__4 = num__10 all the options except option b ) num__4 ! + num__1 are multiples of num__3 ! + num__4 the correct option is b ) <eor> b <eos> |
b |
multiply__3.0__2.0__ add__4.0__6.0__ subtract__3.0__2.0__ add__3.0__1.0__ |
multiply__3.0__2.0__ add__4.0__6.0__ subtract__3.0__2.0__ add__3.0__1.0__ |
| a tradesman by means of his false balance defrauds to the extent of num__22.0 ? in buying goods as well as by selling the goods . what percent does he gain on his outlay ? <o> a ) num__56.25 <o> b ) num__49.56 <o> c ) num__84.46 <o> d ) num__24.5 <o> e ) num__45.35 % |
g % = num__25 + num__25 + ( num__25 * num__25 ) / num__100 = num__56.25 answer : a <eor> a <eos> |
a |
percent__56.25__100.0__ |
percent__56.25__100.0__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__225 m and they cross each other in num__18 sec then the speed of each train is ? <o> a ) num__22 <o> b ) num__77 <o> c ) num__36 <o> d ) num__88 <o> e ) num__45 |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__225 + num__225 ) / num__18 = > x = num__12.5 speed of each train = num__12.5 m / sec . = num__12.5 * num__3.6 = - num__45 km / hr . answer : e <eor> e <eos> |
e |
divide__225.0__18.0__ multiply__12.5__3.6__ round__45.0__ |
divide__225.0__18.0__ multiply__12.5__3.6__ multiply__12.5__3.6__ |
| divide $ num__1000 among a b in the ratio num__1 : num__3 . how many $ that a get ? <o> a ) $ num__50 <o> b ) $ num__500 <o> c ) $ num__150 <o> d ) $ num__250 <o> e ) $ num__600 |
sum of ratio terms = num__1 + num__3 = num__4 a = num__1000 * num__0.25 = $ num__250 answer is d <eor> d <eos> |
d |
add__1.0__3.0__ reverse__4.0__ multiply__1000.0__0.25__ multiply__1000.0__0.25__ |
add__1.0__3.0__ reverse__4.0__ multiply__1000.0__0.25__ multiply__1000.0__0.25__ |
| elena ’ s bread recipe calls for num__3 ounces of butter for each num__4 cups of flour used . she needs to make num__4 times the original recipe . if num__12 ounces of butter is used then how many cups of flour are needed ? <o> a ) num__1 <o> b ) num__4 <o> c ) num__9 <o> d ) num__13 <o> e ) num__16 |
number of cups flour needed for num__3 ounces of butter = num__4 number of cups flour needed for num__1 ounce of butter = num__1.33333333333 number of cups flour needed for num__12 ounces of butter = num__1.33333333333 * num__12 = num__16 answer e <eor> e <eos> |
e |
subtract__4.0__3.0__ divide__4.0__3.0__ add__4.0__12.0__ add__4.0__12.0__ |
subtract__4.0__3.0__ divide__4.0__3.0__ add__4.0__12.0__ multiply__1.0__16.0__ |
| average temperature of first num__4 days of a week is num__6 ° c and that of the last num__4 days is num__40.3 ° c . if the average temperature of the week be num__39.1 ° c the temperature on num__4 th day is ? <o> a ) num__36.7 ° c <o> b ) num__38.6 ° c <o> c ) num__39.8 ° c <o> d ) num__41.9 ° c <o> e ) num__51.9 ° c |
let temperature on num__4 th day be x ° c therefore num__4 x num__38.6 + num__4 x num__40.3 - x = num__7 x num__39.1 = > x = num__41.9 therefore temperature on num__4 th day = num__41.9 ° c . answer : d <eor> d <eos> |
d |
round__41.9__ |
round__41.9__ |
| if the price of an item is decreased by num__40.0 and then increased by num__40.0 the net effect on the price of the item is <o> a ) a decrease of num__99.0 <o> b ) no change <o> c ) a decrease of num__16.0 <o> d ) a increase of num__1.0 <o> e ) none |
initially assume num__100 rupees num__40.0 discount in num__100 gives price of num__60 rupees then num__40.0 raise in num__60 is only num__24 rupees . therefore total price = num__84 rupees . hence num__16.0 is the loss answer : c <eor> c <eos> |
c |
subtract__100.0__40.0__ add__24.0__60.0__ subtract__40.0__24.0__ subtract__40.0__24.0__ |
subtract__100.0__40.0__ add__24.0__60.0__ subtract__40.0__24.0__ subtract__40.0__24.0__ |
| what is the rate percent when the simple interest on rs . num__800 amount to rs . num__160 in num__4 years ? <o> a ) num__5.0 <o> b ) num__8.0 <o> c ) num__9.0 <o> d ) num__4.0 <o> e ) num__1 % |
num__160 = ( num__180 * num__4 * r ) / num__100 r = num__5.0 answer : a <eor> a <eos> |
a |
percent__5.0__100.0__ |
percent__5.0__100.0__ |
| a num__18.0 stock yielding num__12.0 is quoted at : <o> a ) s . num__83.33 <o> b ) s . num__110 <o> c ) s . num__112 <o> d ) s . num__120 <o> e ) s . num__150 |
income of rs num__12 on investment of rs num__100 income of rs num__18 on investment of ? = ( num__18 * num__100 ) / num__12 = num__150 answer : e <eor> e <eos> |
e |
percent__100.0__150.0__ |
percent__100.0__150.0__ |
| the area of a square is num__4096 sq cm . find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is num__24 cm less than the side of the square . <o> a ) num__5 : num__12 <o> b ) num__5 : num__13 <o> c ) num__5 : num__167 <o> d ) num__5 : num__19 <o> e ) num__5 : num__16 |
let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a num__2 = num__4096 = num__212 a = ( num__212 ) num__0.5 = num__26 = num__64 l = num__2 a and b = a - num__24 b : l = a - num__24 : num__2 a = num__40 : num__128 = num__5 : num__16 answer : option e <eor> e <eos> |
e |
power__4096.0__0.5__ rectangle_perimeter__24.0__40.0__ rectangle_perimeter__0.5__2.0__ triangle_area__0.5__64.0__ rectangle_perimeter__0.5__2.0__ |
power__4096.0__0.5__ rectangle_perimeter__24.0__40.0__ rectangle_perimeter__0.5__2.0__ triangle_area__0.5__64.0__ rectangle_perimeter__0.5__2.0__ |
| at an art exhibit the entrance costs is $ num__1.15 per person and every print sold costs $ num__5 . assuming that you bought prints that came in a bag that could hold num__12 and another that holds num__50.0 more . how much money do you end up spending all together at the exhibition ? <o> a ) num__180 <o> b ) num__181.15 <o> c ) num__1.15 <o> d ) num__81.15 <o> e ) num__200.15 |
entrance fee is $ num__1.15 . total prints bought is num__12 + ( num__1.50 * num__12 ) = num__36 total prints cost num__36 * $ num__5 = $ num__180 . total spent at exhibit is $ num__1.15 + $ num__180 = $ num__181.15 answer is b . <eor> b <eos> |
b |
multiply__5.0__36.0__ add__1.15__180.0__ add__1.15__180.0__ |
multiply__5.0__36.0__ add__1.15__180.0__ add__1.15__180.0__ |
| a num__450 m long train crosses a platform in num__39 sec while it crosses a signal pole in num__18 sec . what is the length of the platform ? <o> a ) num__600 <o> b ) num__525 <o> c ) num__360 <o> d ) num__370 <o> e ) num__380 |
speed = num__25.0 = num__25 m / sec . let the length of the platform be x meters . then ( x + num__450 ) / num__39 = num__25 = > x = num__975 m . l = num__975 - num__450 = num__525 answer : option b <eor> b <eos> |
b |
divide__450.0__18.0__ multiply__39.0__25.0__ subtract__975.0__450.0__ round__525.0__ |
divide__450.0__18.0__ multiply__39.0__25.0__ subtract__975.0__450.0__ subtract__975.0__450.0__ |
| look at this series : num__7 num__10 num__8 num__11 num__9 num__12 num__10 num__13 num__11 . . . what number should come next ? <o> a ) num__12 <o> b ) num__13 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
this is a simple alternating addition and subtraction series . in the first pattern num__3 is added ; in the second num__2 is subtracted . the answer is c . <eor> c <eos> |
c |
subtract__10.0__7.0__ subtract__10.0__8.0__ multiply__7.0__2.0__ |
subtract__10.0__7.0__ subtract__10.0__8.0__ multiply__7.0__2.0__ |
| how many num__1 ' s are there preceded by num__2 but not followed by num__0 ? num__5 num__9 num__3 num__2 num__1 num__7 num__4 num__2 num__6 num__9 num__7 num__4 num__2 num__1 num__3 num__2 num__8 num__7 num__0 num__1 num__3 num__8 num__3 num__2 num__5 num__6 num__7 num__4 num__3 num__9 num__5 num__8 num__2 num__2 num__1 num__0 num__7 num__4 num__6 num__3 <o> a ) num__4 <o> b ) num__2 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
num__2 num__1 num__7 num__2 num__1 num__3 only at these places num__2 is preceded by num__2 but not followed by num__0 answer : b <eor> b <eos> |
b |
multiply__1.0__2.0__ |
multiply__1.0__2.0__ |
| pipe a can fill a tank in num__6 hours . due to a leak at the bottom it takes num__9 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? <o> a ) num__13 <o> b ) num__17 <o> c ) num__18 <o> d ) num__19 <o> e ) num__12 |
let the leak can empty the full tank in x hours num__0.166666666667 - num__1 / x = num__0.111111111111 = > num__1 / x = num__0.166666666667 - num__0.111111111111 = ( num__3 - num__2 ) / num__18 = num__0.0555555555556 = > x = num__18 . answer : c <eor> c <eos> |
c |
divide__1.0__9.0__ subtract__9.0__6.0__ divide__6.0__3.0__ multiply__6.0__3.0__ divide__1.0__18.0__ round__18.0__ |
divide__1.0__9.0__ subtract__9.0__6.0__ divide__6.0__3.0__ multiply__6.0__3.0__ divide__1.0__18.0__ divide__18.0__1.0__ |
| in the olympic track represented above num__8 runners are going to compete to reach the finish line in the shortest amount of time . between tom in line num__1 and jack in line num__2 there ' s a distance of num__1.2 meters the same as in the subsequent lanes . if each runner is to race onto their assigned line ( and not in the middle of the lane ) and tom is to run on line num__1 and bob on line num__8 . bob will have to start - position himself approximately how many meters in front of tom for the two to run the same distance before reaching the finishing line ? <o> a ) num__37 <o> b ) num__51 <o> c ) num__53 <o> d ) num__64 <o> e ) num__85 |
first perimeter : num__402 meters second perimeter : num__455 for the two of them to run num__402 meters the second athlete will have to start num__53 meters ahead . answer : c <eor> c <eos> |
c |
subtract__455.0__402.0__ round__53.0__ |
subtract__455.0__402.0__ round__53.0__ |
| the average of first five multiples of num__4 is ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__9 <o> d ) num__5 <o> e ) num__12 |
average = num__4 ( num__1 + num__2 + num__3 + num__4 + num__5 ) / num__5 = num__12.0 = num__12 . answer : e <eor> e <eos> |
e |
subtract__4.0__1.0__ add__4.0__1.0__ multiply__4.0__3.0__ multiply__4.0__3.0__ |
add__1.0__2.0__ add__4.0__1.0__ multiply__4.0__3.0__ divide__12.0__1.0__ |
| a man can row a boat three quarters of a kilometre in num__11.25 minutes . what is the speed of the boat in still water ? <o> a ) num__4 kmph <o> b ) num__8 kmph <o> c ) num__6 kmph <o> d ) num__5 kmph <o> e ) num__12 kmph |
explanation : simple division . no stream current mentioned . trick question . speed = distance / time . speed = num__0.75 km / ( num__11.25 / num__60 ) hr therefore speed = num__4 km / hr answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ round__4.0__ |
hour_to_min_conversion__ round__4.0__ |
| yes bank offers an interest of num__5.0 per annum compounded annually on all its deposits . if $ num__10000 is deposited what will be the ratio of the interest earned in the num__4 th year to the interest earned in the num__5 th year ? <o> a ) num__1 : num__5 <o> b ) num__625 : num__3125 <o> c ) num__100 : num__105 <o> d ) num__100 ^ num__4 : num__105 ^ num__4 <o> e ) num__725 : num__3225 |
hi bunuel here is my approach : is this correct ? yes bank offers an interest of num__5.0 per annum compounded annually on all its deposits . interest earned in num__4 year = num__10000 ( num__1 + num__0.05 ) ^ num__4 interest earned in num__5 year = num__10000 ( num__1 + num__0.05 ) ^ num__5 ratio = { num__10000 ( num__1.05 ) ^ num__4 } / { num__10000 ( num__1.05 ^ num__5 ) } = > num__1.05 ^ num__4 / num__1.05 ^ num__5 = > num__1 / num__1.05 multiplied by num__100 in both numerator and denominator gives num__100 : num__105 hence ans : c <eor> c <eos> |
c |
percent__5.0__1.0__ percent__1.0__10000.0__ percent__1.05__10000.0__ percent__1.0__10000.0__ |
percent__5.0__1.0__ percent__1.0__10000.0__ percent__1.05__10000.0__ percent__1.0__10000.0__ |
| ayesha ' s father was num__38 years of age when she was born while her mother was num__36 years old when her brother four years younger to her was born . what is the difference between the ages of her parents ? <o> a ) num__2 years <o> b ) num__4 years <o> c ) num__6 years <o> d ) num__7 years <o> e ) num__8 years |
mother ' s age when ayesha ' s brother was born = num__36 years . father ' s age when ayesha ' s brother was born = ( num__38 + num__4 ) years = num__42 years . required difference = ( num__42 - num__36 ) years = num__6 years . answer : option c <eor> c <eos> |
c |
add__38.0__4.0__ subtract__42.0__36.0__ divide__36.0__6.0__ |
add__38.0__4.0__ subtract__42.0__36.0__ subtract__42.0__36.0__ |
| a small rectangular park has a perimeter of num__560 feet and a diagonal measurement of num__300 feet . what is its area in square feet ? <o> a ) num__43200 <o> b ) num__19600 <o> c ) num__20000 <o> d ) num__20400 <o> e ) num__20 |
800 |
you can avoid a lot of work in this problem by recognizing that with the info provided the diagonal forms a triangle inside the rectangle with sides that have a num__3 : num__4 : num__5 ratio . diagonal = num__200 num__2 x + num__2 y = num__560 or x + y = num__280 a ^ num__2 + b ^ num__2 = c ^ num__2 for each the sides of the triangle using the ratio num__3 : num__4 : num__5 for sides and knowing c = num__300 you can deduce the following a = num__180 b = num__240 num__180 x num__240 = num__43200 a is the answer . <eor> a <eos> |
a |
a |
| right triangle abc is the base of the prism in the figure above . if ab = ac = √ num__2 and the height of the prism is num__3 what is the volume of the prism ? <o> a ) num__1 <o> b ) num__1.5 <o> c ) num__2 <o> d ) num__3 <o> e ) num__6 |
volume of prism = area of base * height = num__0.5 * ( square root of num__2 ) * ( square root of num__2 ) * num__3 = num__3 answer : d <eor> d <eos> |
d |
triangle_area__2.0__3.0__ |
volume_rectangular_prism__2.0__3.0__0.5__ |
| john want to buy a $ num__100 trouser at the store but he think it ’ s too expensive . finally it goes on sale for $ num__30 . what is the percent decrease ? <o> a ) num__20.0 <o> b ) num__30.0 <o> c ) num__40.0 <o> d ) num__70.0 <o> e ) num__80 % |
the is always the difference between our starting and ending points . in this case it ’ s num__100 – num__30 = num__70 . the “ original ” is our starting point ; in this case it ’ s num__100 . ( num__0.7 ) * num__100 = ( num__0.7 ) * num__100 = num__70.0 . d <eor> d <eos> |
d |
subtract__100.0__30.0__ divide__70.0__100.0__ subtract__100.0__30.0__ |
subtract__100.0__30.0__ divide__70.0__100.0__ multiply__100.0__0.7__ |
| the cost price of a radio is rs . num__1500 and it was sold for rs . num__1230 find the loss % ? <o> a ) num__18 <o> b ) num__19 <o> c ) num__81 <o> d ) num__16 <o> e ) num__13 |
num__1500 - - - - num__270 num__100 - - - - ? = > num__18.0 answer : a <eor> a <eos> |
a |
percent__100.0__18.0__ |
percent__100.0__18.0__ |
| when num__28 is divided by the positive integer a the remainder is num__1 . what is the sum of all the possible values of a for which this is true ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__9 <o> d ) num__30 <o> e ) num__39 |
the only possible values of the form num__28 = ap + num__1 are num__39 or num__27 the sum = num__3 + num__9 + num__27 = num__39 . e is the correct answer . we dont need to look for values > num__28 as all these values will leave a remainder of num__28 and not num__1 . <eor> e <eos> |
e |
subtract__28.0__1.0__ divide__27.0__3.0__ multiply__1.0__39.0__ |
subtract__28.0__1.0__ divide__27.0__3.0__ multiply__1.0__39.0__ |
| aamir saves num__32.0 of his monthly salary . if he spends rs . num__27200 then find his savings ? <o> a ) num__12877 <o> b ) num__12817 <o> c ) num__12171 <o> d ) num__12800 <o> e ) num__128722 |
let the monthly salary of aamir be rs . x . num__68.0 of x = num__27200 = > x = ( num__27200 * num__100 ) / num__68 = num__40000 his savings = num__0.32 * num__40000 = num__12800 . answer : d <eor> d <eos> |
d |
percent__32.0__40000.0__ percent__32.0__40000.0__ |
percent__32.0__40000.0__ percent__32.0__40000.0__ |
| if an object travels at num__85 feet per second how many feet does it travel in num__30 minutes ? <o> a ) num__3000 <o> b ) num__1500 <o> c ) num__1800 <o> d ) num__9000 <o> e ) num__2500 |
if an object travels at num__85 feet per second it covers num__85 x num__60 feet in one minute and num__5 x num__60 x num__30 feet in num__30 minutes . answer = num__9000 answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ round__9000.0__ |
hour_to_min_conversion__ round__9000.0__ |
| num__100 liters of a mixture contains milk and water in the ratio num__4 : num__1 . if num__10 liters of this mixture be replaced by num__10 liters of milk the ratio of milk to water in the new mixture would be ? <o> a ) num__7 : num__5 <o> b ) num__9 : num__2 <o> c ) num__6 : num__1 <o> d ) num__5 : num__4 <o> e ) num__11 : num__4 |
quantity of milk in num__100 liters if mix = num__100 * num__0.8 = num__80 liters quantity of milk in num__110 liters of new mix = num__80 + num__10 = num__90 liters quantity of water in it = num__110 - num__90 = num__20 liters ratio of milk and water in new mix = num__90 : num__20 = num__9 : num__2 answer is b <eor> b <eos> |
b |
multiply__100.0__0.8__ add__100.0__10.0__ subtract__100.0__10.0__ subtract__100.0__80.0__ subtract__10.0__1.0__ divide__20.0__10.0__ multiply__1.0__9.0__ |
multiply__100.0__0.8__ add__100.0__10.0__ subtract__100.0__10.0__ subtract__100.0__80.0__ subtract__10.0__1.0__ divide__20.0__10.0__ multiply__1.0__9.0__ |
| - num__74 x num__29 + num__265 = ? <o> a ) num__2436 <o> b ) - num__1881 <o> c ) - num__2801 <o> d ) - num__2071 <o> e ) none of them |
given exp . = - num__74 x ( num__30 - num__1 ) + num__265 = - ( num__74 x num__30 ) + num__74 + num__265 = - num__2220 + num__339 = - num__1881 answer is b <eor> b <eos> |
b |
subtract__30.0__29.0__ multiply__74.0__30.0__ add__74.0__265.0__ subtract__2220.0__339.0__ multiply__1.0__1881.0__ |
subtract__30.0__29.0__ multiply__74.0__30.0__ add__74.0__265.0__ subtract__2220.0__339.0__ subtract__2220.0__339.0__ |
| find the odd man out num__3 num__5 num__7 num__11 num__14 num__17 num__19 num__23 <o> a ) num__3 <o> b ) num__5 <o> c ) num__11 <o> d ) num__14 <o> e ) num__23 |
as num__14 is only odd number . num__14 is not prime answer : d <eor> d <eos> |
d |
add__3.0__11.0__ |
add__3.0__11.0__ |
| every monday marina eats one croissant and every tuesday she eats two croissants . on each subsequent day of the week she eats a number of croissants equal to the sum of the croissants eaten on the two previous days with the exception that if she eats more than four croissants on any particular day the next day she will eat only one croissant . at the end of the week ( which runs from monday through sunday ) the cycle resets and marina goes back to eating one croissant on monday two on tuesday and so forth . if a particular month begins on a thursday how many croissants will marina eat on the num__27 th of that month ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__5 <o> e ) num__6 |
she eat as follow mon - num__1 tue - num__2 wed - num__3 thr - num__5 fri - num__1 ( since she had more than num__4 previous day ) sat - num__6 sunday - num__1 ( since she had more than num__4 previous day ) so num__27 th day of month she will have num__2 . answer is b <eor> b <eos> |
b |
add__1.0__2.0__ add__2.0__3.0__ add__1.0__3.0__ add__1.0__5.0__ multiply__1.0__2.0__ |
add__1.0__2.0__ add__2.0__3.0__ subtract__5.0__1.0__ add__1.0__5.0__ subtract__3.0__1.0__ |
| a and b together do a work in num__20 days . b and c together in num__15 days and c and a in num__12 days . then b alone can finish same work in how many days ? <o> a ) num__50 <o> b ) num__60 <o> c ) num__45 <o> d ) num__35 <o> e ) num__48 |
( a + b ) work in num__1 day = num__0.05 ( b + c ) work in num__1 days = num__0.0666666666667 . ( c + a ) work in num__1 days = num__0.0833333333333 ( num__1 ) adding = num__2 [ a + b + c ] in num__1 day work = [ num__0.05 + num__0.0666666666667 + num__0.0833333333333 ] = num__0.2 ( a + b + c ) work in num__1 day = num__0.1 b work in num__1 days = [ a + b + c ] work in num__1 days - work of ( a + c ) in num__1 days = [ num__0.1 - num__0.0833333333333 ] = num__0.0166666666667 b alone finish work in num__60 days answer b <eor> b <eos> |
b |
divide__1.0__20.0__ divide__1.0__15.0__ divide__1.0__12.0__ divide__2.0__20.0__ divide__0.2__12.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
divide__1.0__20.0__ divide__1.0__15.0__ divide__1.0__12.0__ divide__2.0__20.0__ subtract__0.1__0.0833__ hour_to_min_conversion__ hour_to_min_conversion__ |
| three brothers live in a farm . they agreed to buy new seeds : adam and ben would go and charlie stayed to protect fields . ben bought num__75 sacks of wheat in the market whereas adam bought num__45 sacks . at home they split the sacks equally . charlie had paid num__1400 dollars for the wheat . how much dollars did ben and adam get of the sum considering equal split of the sacks ? <o> a ) b $ num__1150 - a $ num__180 <o> b ) b $ num__1400 - a $ num__160 <o> c ) b $ num__1225 - a $ num__175 <o> d ) b $ num__1327 - a $ num__199 <o> e ) b $ num__1876 - a $ num__250 |
every farmer ' s part is num__0.333333333333 ( num__45 + num__75 ) = num__40 sacks . charlie paid $ num__1400 for num__40 sacks then num__1 sack costs $ num__35.0 = $ num__35 / sack . adam got $ num__35 * ( num__45 - num__40 ) = num__35 * num__5 = $ num__175 . ben got $ num__35 * ( num__75 - num__40 ) = num__35 * num__35 = $ num__1225 . answer : ben $ num__1225 adam $ num__175 <eor> c <eos> |
c |
subtract__75.0__40.0__ subtract__45.0__40.0__ multiply__35.0__5.0__ subtract__1400.0__175.0__ subtract__1400.0__175.0__ |
subtract__75.0__40.0__ subtract__45.0__40.0__ multiply__35.0__5.0__ subtract__1400.0__175.0__ subtract__1400.0__175.0__ |
| if lcm of two number is num__693 hcf of two numbers is num__11 and one number is num__99 then find other <o> a ) num__34 <o> b ) num__77 <o> c ) num__12 <o> d ) num__45 <o> e ) num__67 |
explanation : for any this type of question remember product of two numbers = product of their hcf and lcm so other number = num__693 × num__0.111111111111 = num__77 option b <eor> b <eos> |
b |
divide__11.0__99.0__ round__77.0__ |
divide__11.0__99.0__ round__77.0__ |
| a man bought num__20 shares paying num__9.0 dividend . the man wants to have an interest of num__12.0 on his money . the market value of each share is <o> a ) num__12 <o> b ) num__14 <o> c ) num__15 <o> d ) num__16 <o> e ) num__17 |
dividend on num__20 = ( num__0.09 * num__20 ) = num__1.8 num__12 is an income on num__100 num__1.8 is an income on ( num__8.33333333333 * num__1.8 ) = num__15 option c <eor> c <eos> |
c |
percent__20.0__9.0__ percent__100.0__15.0__ |
percent__20.0__9.0__ percent__100.0__15.0__ |
| if a and b are integers and ( num__2 ^ a ) ( num__3 ^ b ) is a factor of num__900 ^ num__40 what is the largest possible value of ab ? <o> a ) num__2 a <o> b ) num__5 a <o> c ) num__20 a <o> d ) num__40 a <o> e ) num__80 a |
( num__2 ^ a ) ( num__3 ^ b ) is a factor of num__900 ^ num__40 we need to find the largest possible value of ab . we know that num__900 = num__2 ^ num__2 * num__5 ^ num__2 * num__3 ^ num__2 therefore num__900 ^ num__40 will have num__2 powers of num__3 in it . hence in ( num__2 ^ a ) ( num__3 ^ b ) b has to num__2 therefore value of ab = num__2 a correct option : a <eor> a <eos> |
a |
add__2.0__3.0__ gcd__2.0__900.0__ |
add__2.0__3.0__ gcd__2.0__900.0__ |
| aarti can do a piece of work in num__9 days . in how many days will she complete three time of work of same type ? <o> a ) num__6 days <o> b ) num__18 days <o> c ) num__21 days <o> d ) num__27 days <o> e ) num__13 days |
we have the important relation more work more time ( days ) a piece of work can be done in num__9 days . three times of work of same type can be done in num__9 x num__3 = num__27 days answer d <eor> d <eos> |
d |
multiply__9.0__3.0__ round__27.0__ |
multiply__9.0__3.0__ round__27.0__ |
| num__50 men do a work in num__100 days . how many men are needed to finish the work in num__20 days ? <o> a ) num__50 <o> b ) num__100 <o> c ) num__250 <o> d ) num__300 <o> e ) num__400 |
men required to finish the work in num__2 days = num__50 * num__5.0 = num__250 answer is c <eor> c <eos> |
c |
divide__100.0__50.0__ divide__100.0__20.0__ multiply__50.0__5.0__ round__250.0__ |
divide__100.0__50.0__ divide__100.0__20.0__ multiply__50.0__5.0__ multiply__50.0__5.0__ |
| pipe a can fill a tank in num__9 hours . due to a leak at the bottom it takes num__12 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? <o> a ) num__36 <o> b ) num__88 <o> c ) num__18 <o> d ) num__26 <o> e ) num__12 |
let the leak can empty the full tank in x hours num__0.111111111111 - num__1 / x = num__0.0833333333333 = > num__1 / x = num__0.111111111111 - num__0.0833333333333 = num__0.0833333333333 = > x = num__36 . answer : a <eor> a <eos> |
a |
divide__1.0__12.0__ round__36.0__ |
divide__1.0__12.0__ divide__36.0__1.0__ |
| what is the units digit of ( num__52 ^ num__5 ) ( num__37 ^ num__3 ) ( num__71 ^ num__9 ) ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
the units digit of num__52 ^ num__5 is the units digit of num__2 ^ num__5 which is num__2 . the units digit of num__37 ^ num__3 is the units digit of num__7 ^ num__3 which is num__3 . the units digit of num__71 ^ num__9 is the units digit of num__1 ^ num__9 which is num__1 . the units digit of num__2 * num__3 * num__1 is num__6 . the answer is c . <eor> c <eos> |
c |
subtract__5.0__3.0__ add__5.0__2.0__ subtract__3.0__2.0__ add__5.0__1.0__ add__5.0__1.0__ |
subtract__5.0__3.0__ add__5.0__2.0__ subtract__3.0__2.0__ multiply__3.0__2.0__ multiply__3.0__2.0__ |
| the average amount with a group of seven numbers is rs . num__20 . if the newly joined member has rs . num__20 with him what was the average amount with the group before his joining the group ? <o> a ) s . num__17 <o> b ) s . num__12 <o> c ) s . num__15 <o> d ) s . num__29 <o> e ) s . num__20 |
total members in the group = num__7 average amount = rs . num__20 total amount with them = num__7 * num__20 = rs . num__140 one number has rs . num__20 . so the amount with remaining num__6 people = num__140 - num__20 = rs . num__20 the average amount with them = num__20.0 = rs . num__20 . answer : e <eor> e <eos> |
e |
multiply__20.0__7.0__ divide__140.0__7.0__ |
multiply__20.0__7.0__ divide__140.0__7.0__ |
| if num__1 tic equals num__4 tacs and num__5 tacs equal num__8 tocs what is the ratio of one tic to one toc ? <o> a ) num__7.5 <o> b ) num__1.2 <o> c ) num__0.833333333333 <o> d ) num__6.4 <o> e ) num__0.0666666666667 |
tic = num__4 * tac and num__5 * tac = num__8 * toc ; num__5 * tic = num__20 * tac and num__20 * tac = num__32 * toc - - > num__5 * tic = num__32 * toc - - > tic / toc = num__6.4 . answer : d . <eor> d <eos> |
d |
multiply__4.0__5.0__ multiply__4.0__8.0__ divide__32.0__5.0__ multiply__1.0__6.4__ |
multiply__4.0__5.0__ multiply__4.0__8.0__ divide__32.0__5.0__ multiply__1.0__6.4__ |
| two pipes a and b can separately fill a tank in num__2 minutes and num__15 minutes respectively . both the pipes are opened together but num__4 minutes after the start the pipe a is turned off . how much time will it take to fill the tank ? <o> a ) num__22 <o> b ) num__10 minutes <o> c ) num__88 <o> d ) num__566 <o> e ) num__222 |
num__0.333333333333 + x / num__15 = num__1 x = num__10 . answer : b <eor> b <eos> |
b |
round__10.0__ |
divide__10.0__1.0__ |
| a ratio between two numbers is num__7 : num__9 and their l . c . m . is num__189 . the second number is <o> a ) num__35 <o> b ) num__25 <o> c ) num__20 <o> d ) num__27 <o> e ) none |
sol . let the required numbers be num__7 x and num__9 x . then their l . c . m . is num__63 x . ∴ num__63 x = num__189 ⇔ x = num__3 . hence the second number is num__27 . answer d <eor> d <eos> |
d |
multiply__7.0__9.0__ divide__189.0__63.0__ multiply__9.0__3.0__ multiply__9.0__3.0__ |
multiply__7.0__9.0__ divide__189.0__63.0__ multiply__9.0__3.0__ multiply__9.0__3.0__ |
| a train num__125 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is ? <o> a ) num__11 <o> b ) num__50 <o> c ) num__88 <o> d ) num__65 <o> e ) num__22 |
speed of the train relative to man = ( num__12.5 ) m / sec = ( num__12.5 ) m / sec . [ ( num__12.5 ) * ( num__3.6 ) ] km / hr = num__45 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__45 = = > x = num__50 km / hr . answer : b <eor> b <eos> |
b |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ round__50.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ multiply__5.0__10.0__ |
| num__1 . num__12.91 is the first sunday . which is the fourth tuesday of december num__91 ? <o> a ) num__20 . num__12.91 <o> b ) num__22 . num__12.91 <o> c ) num__24 . num__12.91 <o> d ) num__25 . num__12.91 <o> e ) none of these |
explanation : given that num__1 . num__12.91 is the first sunday hence we can assume that num__3 . num__12.91 is the first tuesday if we add num__7 days to num__3 . num__12.91 we will get second tuesday if we add num__14 days to num__3 . num__12.91 we will get third tuesday if we add num__21 days to num__3 . num__12.91 we will get fourth tuesday = > fourth tuesday = ( num__3 . num__12.91 + num__21 days ) = num__24 . num__12.91 . answer : option c <eor> c <eos> |
c |
multiply__3.0__7.0__ add__3.0__21.0__ round__24.0__ |
add__7.0__14.0__ add__3.0__21.0__ add__3.0__21.0__ |
| a rectangular parking space is marked out by painting three of its sides . if the length of the unpainted side is num__9 feet and the sum of the lengths of the painted sides is num__37 feet find out the area of the parking space in square feet ? <o> a ) num__124 sq . ft . <o> b ) num__120 sq . ft . <o> c ) num__126 sq . ft . <o> d ) num__129 sq . ft . <o> e ) num__132 sq . ft . |
let l = num__9 ft . then l + num__2 b = num__37 = > num__2 b = num__37 - l = num__37 - num__9 = num__28 = > b = num__14.0 = num__14 ft area = lb = num__9 × num__14 = num__126 sq . ft . answer is c . <eor> c <eos> |
c |
multiply__9.0__14.0__ multiply__9.0__14.0__ |
multiply__9.0__14.0__ multiply__9.0__14.0__ |
| working at their respective constant rates paul abdul and adam alone can finish a certain work in num__3 num__4 and num__5 hours respectively . if all three work together to finish the work what fraction q of the work will be done by adam ? <o> a ) num__0.25 <o> b ) num__0.255319148936 <o> c ) num__0.333333333333 <o> d ) num__0.416666666667 <o> e ) num__0.425531914894 |
let the total work be num__60 units . pual does num__20.0 = num__20 units of work per hr . abdul does num__15 units per hr and adam does num__12 units per hr . if all work together they do ( num__20 + num__15 + num__12 ) units per hr = num__47 units per hr . so the time taken to finish the work = num__1.27659574468 hrs . adam will do num__1.27659574468 * num__12 units of work in num__1.27659574468 hr . fraction of work adam does = work done by adam / total work q > ( num__1.27659574468 * num__12 ) / num__60 = num__0.255319148936 answer b <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__4.0__5.0__ multiply__3.0__5.0__ multiply__3.0__4.0__ divide__60.0__47.0__ divide__1.2766__5.0__ divide__1.2766__5.0__ |
hour_to_min_conversion__ multiply__4.0__5.0__ multiply__3.0__5.0__ multiply__3.0__4.0__ divide__60.0__47.0__ divide__1.2766__5.0__ divide__1.2766__5.0__ |
| the average weight of num__10 oarsmen in a boat is increased by num__1.8 kg when one of the crew who weighs num__53 kg is replaced by a new man . find the weight of the new man . <o> a ) num__69 <o> b ) num__72 <o> c ) num__75 <o> d ) num__71 <o> e ) none of them |
total weight increased = ( num__1.8 x num__10 ) kg = num__18 kg . : . weight of the new man = ( num__53 + num__18 ) kg = num__71 kg . answer is d <eor> d <eos> |
d |
multiply__10.0__1.8__ add__53.0__18.0__ add__53.0__18.0__ |
multiply__10.0__1.8__ add__53.0__18.0__ add__53.0__18.0__ |
| if the tens digit x and the units digit y of a positive integer n are reversed the resulting integer is num__45 more than n . what is y in terms of x ? <o> a ) x + num__5 <o> b ) x + num__3 <o> c ) x + num__4 <o> d ) x - num__3 <o> e ) x - num__4 |
original digits = xy i . e . number = num__10 x + y after reversing the digits : digits = yx i . e . number = num__10 y + x num__10 y + x is num__45 more than num__10 x + y num__10 x + y + num__45 = num__10 y + x num__10 x - x + num__45 = num__10 y - y num__9 x + num__45 = num__9 y x + num__5 = y or y = x + num__5 answer : a <eor> a <eos> |
a |
divide__45.0__9.0__ divide__45.0__9.0__ |
divide__45.0__9.0__ subtract__10.0__5.0__ |
| a man has rs . num__630 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__45 <o> b ) num__60 <o> c ) num__75 <o> d ) num__90 <o> e ) num__118.125 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__630 num__16 x = num__630 x = num__39.375 . hence total number of notes = num__3 x = num__118.125 . answer = e <eor> e <eos> |
e |
divide__630.0__16.0__ multiply__39.375__3.0__ multiply__39.375__3.0__ |
divide__630.0__16.0__ multiply__39.375__3.0__ multiply__39.375__3.0__ |
| a shopkeeper buys mangoes at the rate of num__6 a rupee and sells them at num__5 a rupee . find his net profit or loss percent ? <o> a ) num__33 num__0.125 % <o> b ) num__33 num__2.33333333333 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__33 num__0.625 % <o> e ) num__20 % |
the total number of mangoes bought by the shopkeeper be num__30 . if he buys num__6 a rupee his cp = num__5 he selling at num__5 a rupee his sp = num__6 profit = sp - cp = num__6 - num__5 = num__1 profit percent = num__0.2 * num__100 = num__20.0 answer : e <eor> e <eos> |
e |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| the list price of an article is rs . num__66 . a customer pays rs . num__56.16 for it . he was given two successive discounts one of them being num__10.0 . the other discount is ? <o> a ) num__3.45 <o> b ) num__4.45 <o> c ) num__5.45 <o> d ) num__6.45 <o> e ) num__7.45 % |
num__66 * ( num__0.9 ) * ( ( num__100 - x ) / num__100 ) = num__56.16 x = num__5.45 answer : c <eor> c <eos> |
c |
percent__5.45__100.0__ |
percent__5.45__100.0__ |
| a shopkeeper buys two articles for rs . num__1000 each and then sells them making num__30.0 profit on the first article and num__30.0 loss on second article . find the net profit or loss percent ? <o> a ) num__200 <o> b ) num__768 <o> c ) num__276 <o> d ) num__300 <o> e ) num__279 |
profit on first article = num__30.0 of num__1000 = num__300 . this is equal to the loss he makes on the second article . that is he makes neither profit nor loss . answer : d <eor> d <eos> |
d |
percent__30.0__1000.0__ percent__30.0__1000.0__ |
percent__30.0__1000.0__ percent__30.0__1000.0__ |
| excluding the stoppages the speed of a bus is num__12 km / hr and including the stoppages the speed of the bus is num__6 km / hr . for how many minutes does the bus stop per hour ? <o> a ) num__15 min <o> b ) num__30 min <o> c ) num__12 min <o> d ) num__20 min <o> e ) num__18 min |
speed of the bus without stoppage = num__12 km / hr speed of the bus with stoppage = num__6 km / hr difference in speed = num__6 km / hr so the time taken in the stoppages = time taken to cover num__6 km = ( num__0.5 ) hr = num__0.5 hr = num__30 min answer : b <eor> b <eos> |
b |
divide__6.0__12.0__ round__30.0__ |
divide__6.0__12.0__ round__30.0__ |
| a square is of side num__1 km . a man travels first num__2 sides with the num__30 mph and third side with num__60 kmph . what is the speed that man has to travel the fourth side if the average speed is num__60 kmph <o> a ) num__100 kmph <o> b ) num__110 kmph <o> c ) num__120 kmph <o> d ) num__130 kmph <o> e ) num__140 kmph |
num__1 mile = num__1.6 km total time taken to travel num__3 sides = ( num__0.0208333333333 ) + ( num__0.0208333333333 ) + ( num__0.0166666666667 ) = num__0.0583333333333 hr time taken to travel num__4 sides at avg speed = num__0.0666666666667 hr so reqred speed for fourth side = num__1 / { ( num__0.0666666666667 ) - ( num__0.0583333333333 ) } = num__120 kmph answer : c <eor> c <eos> |
c |
mile_to_km_conversion__ add__1.0__2.0__ divide__1.0__60.0__ add__1.0__3.0__ divide__2.0__30.0__ multiply__2.0__60.0__ round__120.0__ |
mile_to_km_conversion__ add__1.0__2.0__ divide__1.0__60.0__ add__1.0__3.0__ divide__2.0__30.0__ multiply__2.0__60.0__ divide__120.0__1.0__ |
| if a store adds num__100 chairs to its current inventory the total number of chairs will be the same as three - halves the current inventory of chairs . if the manager wants to increase the current inventory by num__10.0 what will the new inventory of chairs be ? <o> a ) num__40 <o> b ) num__60 <o> c ) num__100 <o> d ) num__140 <o> e ) num__220 |
explanation : let â € ™ s say t = total current inventory of chairs . the first sentence states that num__100 + t = ( num__1.5 ) t . first solve for the current inventory : num__100 + t = ( num__1.5 ) t num__100 = ( num__1.5 ) t â ˆ ’ t num__100 = ( num__0.5 ) t num__200 = t the manager wants to increase this by num__10.0 . num__10.0 of num__200 is num__20 so the new inventory will be num__220 answer : e <eor> e <eos> |
e |
divide__100.0__0.5__ divide__10.0__0.5__ add__200.0__20.0__ add__200.0__20.0__ |
divide__100.0__0.5__ divide__10.0__0.5__ add__200.0__20.0__ add__200.0__20.0__ |
| a can do a work in num__24 days and b can do it in num__12 days . in how many days a and b can do the work ? <o> a ) num__20 days <o> b ) num__10 days <o> c ) num__6 days <o> d ) num__5 days <o> e ) num__8 days |
explanation : a ' s num__1 day ' s work = num__0.0416666666667 b ' s num__1 day ' s work = num__0.0833333333333 they work together = num__0.0416666666667 + num__0.0833333333333 = num__0.125 = num__8.0 = num__8 days answer : option e <eor> e <eos> |
e |
divide__1.0__24.0__ divide__1.0__12.0__ add__0.0833__0.0417__ divide__1.0__0.125__ round__8.0__ |
divide__1.0__24.0__ divide__1.0__12.0__ add__0.0833__0.0417__ divide__1.0__0.125__ round__8.0__ |
| how many prime numbers between num__1 and num__100 are factors of num__143 ? <o> a ) num__6 <o> b ) num__5 <o> c ) num__4 <o> d ) num__2 <o> e ) num__3 |
factor of num__143 = num__11 * num__13 - - - num__2 prime numbers d <eor> d <eos> |
d |
divide__143.0__11.0__ subtract__13.0__11.0__ multiply__1.0__2.0__ |
divide__143.0__11.0__ subtract__13.0__11.0__ multiply__1.0__2.0__ |
| solve below question y : num__6 y - num__27 + num__3 y = num__4 + num__9 - y <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
num__9 y + y = num__13 + num__27 num__10 y = num__40 = > y = num__4 b <eor> b <eos> |
b |
add__4.0__9.0__ add__6.0__4.0__ add__27.0__13.0__ divide__40.0__10.0__ |
add__4.0__9.0__ add__6.0__4.0__ add__27.0__13.0__ divide__40.0__10.0__ |
| what is the area of a circle having the same perimeter as that of a square whose area is num__121 . <o> a ) num__306 <o> b ) num__307 <o> c ) num__308 <o> d ) num__309 <o> e ) num__310 |
side of the square = sqrt ( num__12 ) = num__11 perimeter of the square = num__44 num__2 * pie * r = num__44 r = num__22.0 * pie area of the circle = pie * r ^ num__2 = num__308 answer : c <eor> c <eos> |
c |
square_perimeter__11.0__ multiply__2.0__11.0__ triangle_area__2.0__308.0__ |
square_perimeter__11.0__ multiply__2.0__11.0__ triangle_area__2.0__308.0__ |
| a train passes a man standing on the platform . if the train is num__170 meters long and its speed is num__72 kmph how much time it took in doing so ? <o> a ) num__8 num__0.166666666667 sec <o> b ) num__8 num__0.111111111111 sec <o> c ) num__8 num__1.0 sec <o> d ) num__8 num__0.5 sec <o> e ) num__2 num__0.5 sec |
d = num__170 s = num__72 * num__0.277777777778 = num__20 mps t = num__8.5 = num__8 num__0.5 sec answer : d <eor> d <eos> |
d |
divide__170.0__20.0__ subtract__8.5__8.0__ round__8.0__ |
divide__170.0__20.0__ subtract__8.5__8.0__ round__8.0__ |
| erik ' s mother gave him $ num__86 to go to the store . erik bought num__3 loaves of bread and num__3 cartons of orange juice . each loaf of bread cost $ num__3 and each carton of orange juice cost $ num__6 . how much money does erik have left ? <o> a ) $ num__24 <o> b ) $ num__27 <o> c ) $ num__59 <o> d ) $ num__75 <o> e ) $ num__35 |
step num__1 : find the cost of the loaves of bread . num__3 × $ num__3 = $ num__9 step num__2 : find the cost of the orange juice . num__3 × $ num__6 = $ num__18 step num__3 : find the total cost of the groceries . $ num__9 + $ num__18 = $ num__27 step num__4 : find the amount of money left . $ num__86 – $ num__27 = $ num__59 erik has $ num__59 left . answer is c . <eor> c <eos> |
c |
add__3.0__6.0__ subtract__3.0__1.0__ multiply__3.0__6.0__ multiply__3.0__9.0__ add__3.0__1.0__ subtract__86.0__27.0__ subtract__86.0__27.0__ |
add__3.0__6.0__ subtract__3.0__1.0__ multiply__3.0__6.0__ add__9.0__18.0__ add__3.0__1.0__ subtract__86.0__27.0__ subtract__86.0__27.0__ |
| a can lay railway track between two given stations in num__16 days and b can do the same job in num__12 days . with help of c they did the job in num__4 days only . then c alone can do the job in : <o> a ) num__9 num__0.2 days <o> b ) num__9 num__0.4 days <o> c ) num__9 num__0.6 days <o> d ) num__10 <o> e ) num__15 |
explanation : ( a + b + c ) ' s num__1 day ' s work = num__0.25 a ' s num__1 day ' s work = num__0.0625 b ' s num__1 day ' s work = num__0.0833333333333 c ' s num__1 day ' s work = num__0.25 - ( num__0.0625 + num__0.0833333333333 ) = num__0.104166666667 so c alone can do the work in num__9.6 = num__9 num__0.6 days answer is c <eor> c <eos> |
c |
divide__4.0__16.0__ divide__0.25__4.0__ divide__1.0__12.0__ km_to_mile_conversion__ round__9.0__ |
divide__4.0__16.0__ divide__0.25__4.0__ divide__1.0__12.0__ subtract__9.6__9.0__ subtract__9.6__0.6__ |
| if x and y are integers such that | y + num__3 | ≤ num__3 and num__2 y – num__3 x + num__6 = num__0 what is the least possible value e of the product xy ? <o> a ) - num__12 <o> b ) - num__3 <o> c ) num__0 <o> d ) num__2 <o> e ) none of the above |
how to deal with inequalities involving absolute values ? first example shows us the so callednumber case in this case we have | y + num__3 | ≤ num__3 which is generalized | something | ≤ some number . first we solve as if there were no absolute value brackets : y + num__3 ≤ num__3 y ≤ num__0 so y is num__0 or negative second scenario - remove the absolute value brackets . put a negative sign around the other side of the inequality andflip the sign : y + num__3 > = - num__3 y > = - num__6 therefore we have a possible range for y : - num__6 = < y < = num__0 ok so far so good we ' re half way through . what about x ? here ' s the formula : num__2 y – num__3 x + num__6 = num__0 rewrite it as num__2 y + num__6 = num__3 x . you can say that num__2 y + num__6 is a multiple of num__3 ( = num__3 x ) . so all values which must be integer must also satisfy this constraint . i ' m just saying that so it ' s easier to evaluate all the possible numbers ( - num__6 - num__3 num__0 ) . if you plug in y = num__0 x will be num__2 and xy = num__0 as the lowest possible value e . hence answer choice c is the one to go . <eor> c <eos> |
c |
multiply__3.0__0.0__ |
multiply__3.0__0.0__ |
| look at this series num__8 num__22 num__8 num__28 num__8 . . . what number should come next ? <o> a ) num__34 <o> b ) num__8 <o> c ) num__12 <o> d ) num__44 <o> e ) num__30 |
simple addition series with a random number num__8 - interpolated as every other number . in the series num__6 is added to each number except num__8 to arrive at the next number . answer a <eor> a <eos> |
a |
subtract__28.0__22.0__ add__28.0__6.0__ |
subtract__28.0__22.0__ add__28.0__6.0__ |
| num__2 + num__2 + num__2 ² + num__2 ³ . . . + num__2 ^ num__11 <o> a ) num__2 ^ num__9 <o> b ) num__2 ^ num__10 <o> c ) num__2 ^ num__16 <o> d ) num__2 ^ num__12 <o> e ) num__2 ^ num__37 |
num__2 + num__2 = num__2 ^ num__2 num__2 ^ num__2 + num__2 ^ num__2 = ( num__2 ^ num__2 ) * ( num__1 + num__1 ) = num__2 ^ num__3 num__2 ^ num__3 + num__2 ^ num__3 = ( num__2 ^ num__3 ) * ( num__1 + num__1 ) = num__2 ^ num__4 so you can notice the pattern . . . in the end you will have num__2 ^ num__11 + num__2 ^ num__11 which will give you num__2 ^ num__12 answer d <eor> d <eos> |
d |
add__2.0__1.0__ add__1.0__3.0__ add__11.0__1.0__ multiply__2.0__1.0__ |
add__2.0__1.0__ add__1.0__3.0__ add__11.0__1.0__ multiply__2.0__1.0__ |
| one fourth of one third of two fifth of a number is num__30 . what will be num__40.0 of that number <o> a ) a ) num__360 <o> b ) b ) num__150 <o> c ) c ) num__180 <o> d ) d ) num__200 <o> e ) e ) num__220 |
explanation : ( num__0.25 ) * ( num__0.333333333333 ) * ( num__0.4 ) * x = num__30 then x = num__30 * num__30 = num__900 num__40.0 of num__900 = num__360 answer : option a <eor> a <eos> |
a |
percent__40.0__900.0__ percent__40.0__900.0__ |
percent__40.0__900.0__ percent__40.0__900.0__ |
| num__30 square stone slabs of equal size were needed to cover a floor area of num__67.5 sq . m . find the length of each stone slab ? <o> a ) num__150 cm <o> b ) num__767 cm <o> c ) num__88 cm <o> d ) num__666 cm <o> e ) num__776 cm |
area of each slab = num__67.5 / num__30 m num__2 = num__2.25 m num__2 length of each slab √ num__2.25 = num__1.5 m = num__150 cm <eor> a <eos> |
a |
triangle_area__2.0__150.0__ |
triangle_area__2.0__150.0__ |
| if a = num__105 and a ^ num__3 = num__21 * num__25 * num__15 * b what is the value of b ? <o> a ) num__35 <o> b ) num__42 <o> c ) num__45 <o> d ) num__147 <o> e ) num__54 |
first step will be to break down all the numbers into their prime factors . num__105 = num__3 * num__5 * num__7 num__21 = num__7 * num__3 num__25 = num__5 * num__5 num__15 = num__3 * num__5 so ( num__105 ) ^ num__3 = num__3 * num__7 * num__5 * num__5 * num__3 * num__5 * b therefore ( num__3 * num__5 * num__7 ) ^ num__3 = num__3 ^ num__2 * num__5 ^ num__3 * num__7 * b therefore b = num__3 ^ num__3 * num__5 ^ num__3 * num__7 ^ num__1.0 ^ num__2 * num__5 ^ num__3 * num__7 b = num__3 * num__7 ^ num__2 = num__3 * num__49 = num__147 correct answer d . <eor> d <eos> |
d |
divide__105.0__21.0__ divide__105.0__15.0__ subtract__5.0__3.0__ subtract__3.0__2.0__ multiply__3.0__49.0__ multiply__3.0__49.0__ |
divide__105.0__21.0__ divide__105.0__15.0__ subtract__5.0__3.0__ subtract__3.0__2.0__ multiply__3.0__49.0__ multiply__3.0__49.0__ |
| a train having a length of num__240 m passes a post in num__24 seconds . how long will it take to pass a platform having a length of num__640 m ? <o> a ) num__120 sec <o> b ) num__99 sec <o> c ) num__89 sec <o> d ) num__80 sec <o> e ) num__88 sec |
explanation : v = num__10.0 ( where v is the speed of the train ) = num__10 m / s t = ( num__240 + num__640 ) / num__10 = num__88 seconds answer : option e <eor> e <eos> |
e |
divide__240.0__24.0__ round__88.0__ |
divide__240.0__24.0__ round__88.0__ |
| p q and r can do a work in num__5 num__7 and num__10 days respectively . they completed the work and got rs . num__310 . what is the share of q and r respectively ? <o> a ) num__100 and num__80 <o> b ) num__100 and num__90 <o> c ) num__100 and num__110 <o> d ) num__70 and num__100 <o> e ) num__100 and num__70 |
the ratio of their working rates = num__0.2 : num__0.142857142857 : num__0.1 = num__14 : num__10 : num__7 . since they work together the share of q = num__0.322580645161 * num__310 = rs . num__100 the share of r = num__0.225806451613 * num__310 = rs . num__70 the share of q and r respectively is rs . num__100 and rs . num__70 answer : e <eor> e <eos> |
e |
divide__10.0__0.1__ multiply__5.0__14.0__ round__100.0__ |
divide__10.0__0.1__ multiply__5.0__14.0__ round__100.0__ |
| two trains travelling in the same direction at num__40 and num__22 kmph completely pass off another in num__1 minute . if the length of the first train is num__125 m what is the length of the second train ? <o> a ) num__156 m <o> b ) num__189 m <o> c ) num__175 m <o> d ) num__134 m <o> e ) num__972 m |
rs = num__40 – num__22 = num__18 * num__0.277777777778 = num__5 mps t = num__60 sec d = num__5 * num__60 = num__300 m num__125 - - - - - - - - num__175 m answer : c <eor> c <eos> |
c |
subtract__40.0__22.0__ hour_to_min_conversion__ multiply__5.0__60.0__ subtract__300.0__125.0__ round__175.0__ |
subtract__40.0__22.0__ hour_to_min_conversion__ multiply__5.0__60.0__ subtract__300.0__125.0__ multiply__1.0__175.0__ |
| the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square whose area is num__1225 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is num__10 units ? <o> a ) num__140 <o> b ) num__156 <o> c ) num__175 <o> d ) num__214 <o> e ) none of these |
explanation : given that the area of the square = num__1225 sq . units = > side of square = â ˆ š num__1225 = num__35 units the radius of the circle = side of the square = num__35 units length of the rectangle = num__0.4 * num__35 = num__14 units given that breadth = num__10 units area of the rectangle = lb = num__14 * num__10 = num__140 sq . units answer is a <eor> a <eos> |
a |
multiply__35.0__0.4__ square_perimeter__35.0__ square_perimeter__35.0__ |
multiply__35.0__0.4__ multiply__10.0__14.0__ multiply__10.0__14.0__ |
| jennifer works num__20 days a month at d dollars per day for m months out of the year . which of the following represents her yearly pay ? <o> a ) m / ( num__20 d ) <o> b ) num__20 d <o> c ) num__10 md / num__6 <o> d ) num__20 d / m <o> e ) num__20 md |
total income of jennifer in a year = d * num__20 * m dollars . = num__20 dm answer e <eor> e <eos> |
e |
round__20.0__ |
round__20.0__ |
| a swimming pool num__9 m wide and num__12 m long is num__1 m deep on the shallow side and num__4 m deep on the deeper side . its volume is : <o> a ) num__260 <o> b ) num__262 <o> c ) num__270 <o> d ) num__272 <o> e ) none of these |
explanation : volume will be length * breadth * height but in this case two heights are given so we will take average volume = ( num__12 ∗ num__9 ∗ ( num__1 + num__2.0 ) ) m num__312 ∗ num__9 ∗ num__2.5 m num__3 = num__270 m num__3 option c <eor> c <eos> |
c |
subtract__12.0__9.0__ round__270.0__ |
add__1.0__2.0__ multiply__1.0__270.0__ |
| what is the smallest prime number ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
explanation : smallest prime number is num__2 . num__0 and num__1 are neither prime numbers nor composite numbers . answer : option c <eor> c <eos> |
c |
multiply__1.0__2.0__ |
multiply__1.0__2.0__ |
| what will come in place of the x in the following number series ? num__6 num__12 num__21 num__33 x <o> a ) num__33 <o> b ) num__35 <o> c ) num__39 <o> d ) num__48 <o> e ) num__42 |
the pattern is + num__6 + num__9 + num__12 + num__15 â € ¦ â € ¦ â € ¦ . . so the missing term is = num__33 + num__15 = num__48 answer : d <eor> d <eos> |
d |
subtract__21.0__12.0__ add__6.0__9.0__ add__33.0__15.0__ add__33.0__15.0__ |
subtract__21.0__12.0__ add__6.0__9.0__ add__33.0__15.0__ add__33.0__15.0__ |
| before being simplified the instructions for computing income tax in country r were to add num__4 percent of one ' s annual income to the average ( arithmetic mean ) of num__100 units of country r ' s currency and num__1 percent of one ' s annual income . which of the following represents the simplified formula for computing the income tax in country r ' s currency for a person in that country whose annual income is i ? <o> a ) num__50 + i / num__200 <o> b ) num__50 + num__3 i / num__100 <o> c ) num__50 + i / num__40 <o> d ) num__100 + i / num__50 <o> e ) num__50 + num__9 i / num__200 |
income of a person is i num__4 percent of ones annual income = num__4 i / num__100 num__100 units of country r ' s currency = num__100 num__1 percent of one ' s annual income = i / num__100 average of num__100 units and num__1 percent annual income = num__50 + i / num__200 sum of average and num__4 percent = num__4 i / num__100 + num__50 + i / num__200 = num__50 + num__9 i / num__200 so answer is e . <eor> e <eos> |
e |
multiply__4.0__50.0__ subtract__100.0__50.0__ |
multiply__4.0__50.0__ divide__200.0__4.0__ |
| a train running at a speed of num__36 kmph crosses an electric pole in num__12 seconds . in how much time will it cross a num__350 m long platform ? <o> a ) num__33 <o> b ) num__66 <o> c ) num__88 <o> d ) num__47 <o> e ) num__88 |
let the length of the train be x m . when a train crosses an electric pole the distance covered is its own length . so x = num__12 * num__36 * num__0.277777777778 m = num__120 m . time taken to cross the platform = ( num__120 + num__350 ) / num__36 * num__0.277777777778 = num__47 min . answer : d <eor> d <eos> |
d |
round__47.0__ |
round__47.0__ |
| during a trip on an expressway don drove a total of x miles . his average speed on a certain num__5 - mile section of the expressway was num__30 miles per hour and his average speed for the remainder of the trip was num__60 miles per hour . his travel time for the x - mile trip was what percent greater than it would have been if he had traveled at a constant rate of num__60 miles per hour for the entire trip ? <o> a ) num__8.5 <o> b ) num__50.0 <o> c ) x / num__12.0 <o> d ) num__60 / x % <o> e ) num__500 / x % |
total dist = x miles time take to clear num__5 mile section at num__30 miles / hr = num__0.166666666667 = num__0.166666666667 hr time taken to clear x - num__5 miles section at num__60 miles / hr = x - num__0.0833333333333 hr total time = num__0.166666666667 + x - num__0.0833333333333 = x + num__0.0833333333333 hr total time to clear x miles if he traveled at constant rate of num__60 miles / hr = x / num__60 hr extra time = num__0.0833333333333 = num__0.0833333333333 hr percentage greater = ( num__0.0833333333333 ) / ( x / num__60 ) * num__100.0 = num__500 / x % answer : e <eor> e <eos> |
e |
divide__5.0__30.0__ divide__5.0__60.0__ multiply__5.0__100.0__ multiply__5.0__100.0__ |
divide__5.0__30.0__ divide__5.0__60.0__ multiply__5.0__100.0__ multiply__5.0__100.0__ |
| when positive integer k is divided by num__6 the remainder is num__3 . which of the following can not be an even integer ? <o> a ) k + num__1 <o> b ) k - num__11 <o> c ) num__4 k + num__2 <o> d ) ( k - num__3 ) / num__3 + num__2 <o> e ) k / num__3 |
first because k / num__6 leaves a remainder of num__3 we know k must be odd . ( k = num__6 x + num__3 = num__3 ( num__2 x + num__1 ) = odd * odd = odd ) knowing k is odd we can examine the answer choices : a . k + num__1 - - > odd + num__1 - - > must be even b . k - num__11 - - > odd - odd - - > must be even c . num__4 k + num__1 - - > num__4 * odd + num__1 = even + num__1 - - > must be odd d . ( k - num__3 ) / num__3 + num__2 - - > ( num__6 x + num__3 - num__3 ) / num__3 + num__2 = num__2 x + num__2 - - > must be even e . k / num__3 - - > ( num__6 x + num__3 ) / num__3 = num__2 x + num__1 = must be odd answer : c <eor> c <eos> |
c |
divide__6.0__3.0__ subtract__3.0__2.0__ subtract__6.0__2.0__ subtract__6.0__2.0__ |
divide__6.0__3.0__ subtract__3.0__2.0__ subtract__6.0__2.0__ subtract__6.0__2.0__ |
| what is the smallest integer that is multiple of num__3 num__59 <o> a ) a ) num__70 <o> b ) b ) num__45 <o> c ) c ) num__200 <o> d ) d ) num__280 <o> e ) e ) num__140 |
it is the lcm of num__3 num__5 and num__9 which is num__45 . the answer is b . <eor> b <eos> |
b |
multiply__9.0__5.0__ multiply__9.0__5.0__ |
multiply__9.0__5.0__ multiply__9.0__5.0__ |
| f ( x ) is a function such that f ( x ) + num__3 f ( num__8 - x ) = x for all real numbers x . find the value of f ( num__2 ) <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
f ( x ) + num__3 f ( num__8 - x ) = f ( num__2 ) + num__3 f ( num__6 ) = num__2 : x = num__2 above f ( num__6 ) + num__3 f ( num__2 ) = num__6 : x = num__6 above f ( num__6 ) = num__6 - num__3 f ( num__2 ) : solve equation c for f ( num__6 ) f ( num__2 ) + num__3 ( num__6 - num__3 f ( num__2 ) ) = num__2 : substitute f ( num__2 ) = num__2 : solve above equation correct answer c <eor> c <eos> |
c |
multiply__3.0__2.0__ subtract__8.0__6.0__ |
subtract__8.0__2.0__ subtract__8.0__6.0__ |
| a swimming pool num__9 m wide and num__12 m long is num__1 m deep on the shallow side and num__4 m deep on the deeper side . its volume is ? <o> a ) num__870 m num__3 <o> b ) num__277 m num__3 <o> c ) num__270 m num__3 <o> d ) num__220 m num__3 <o> e ) num__170 m num__3 |
volume will be length * breadth * height but in this case two heights are given so we will take average volume = ( num__12 ∗ num__9 ∗ ( num__1 + num__42 ) ) m num__312 ∗ num__9 ∗ num__2.5 m num__3 = num__270 m num__3 answer : c <eor> c <eos> |
c |
subtract__12.0__9.0__ subtract__312.0__42.0__ round__270.0__ |
subtract__12.0__9.0__ subtract__312.0__42.0__ multiply__1.0__270.0__ |
| a little old lady lives in a house with only her children and the children each have pets . she has twice as many sons as daughters . each son has a dog and each daughter has a cat . if the total number of pets and children living with her are y then how many dogs live in the house in terms of y <o> a ) y / num__8 <o> b ) y / num__2 <o> c ) num__2 * y <o> d ) num__3 * y <o> e ) y / num__3 |
since y is the total of all children and pets y = boys + girls + dogs + cats since each boy has a dog then boys = dogs . since each girl has a cat then girls = cats thus y = num__2 * boys + num__2 * girls since there are twice as many boys as girls girls = boys / num__2 . thus y = num__2 * boys + num__2 * ( boys / num__2 ) which simplifies to y = num__2 * boys + num__1 * boy or y = num__3 * boys since boys = dogs . . . y = num__3 * dogs which converts to dogs = y / num__3 answer : e <eor> e <eos> |
e |
add__1.0__2.0__ add__1.0__2.0__ |
add__1.0__2.0__ add__1.0__2.0__ |
| a bus starts from city x . the number of women in the bus is half of the number of men . in city y num__20 men leave the bus and ten women enter . now number of men and women is equal . in the beginning how many passengers entered the bus ? <o> a ) num__15 <o> b ) num__90 <o> c ) num__36 <o> d ) num__45 <o> e ) num__46 |
explanation : originally let number of women = x . then number of men = num__2 x . so in city y we have : ( num__2 x - num__20 ) = ( x + num__10 ) or x = num__30 . therefore total number of passengers in the beginning = ( x + num__2 x ) = num__3 x = num__90 . answer : b <eor> b <eos> |
b |
divide__20.0__2.0__ add__20.0__10.0__ divide__30.0__10.0__ multiply__3.0__30.0__ multiply__3.0__30.0__ |
divide__20.0__2.0__ add__20.0__10.0__ divide__30.0__10.0__ multiply__3.0__30.0__ multiply__3.0__30.0__ |
| the average height of num__35 boys in a class was calculated as num__183 cm . it has later found that the height of one of the boys in the class was wrongly written as num__166 cm whereas his actual height was num__106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ) . ? <o> a ) num__178.27 cm <o> b ) num__181.29 cm <o> c ) num__978.29 cm <o> d ) num__178.89 cm <o> e ) num__176.29 cm |
calculated average height of num__35 boys = num__183 cm . wrong total height of num__35 boys = num__183 * num__35 cm . this was as a result of an actual height of num__106 cm being wrongly written as num__166 cm . correct total height of num__35 boys = num__183 cm - ( num__166 cm - num__106 cm ) / num__35 = num__183 cm - num__1.71428571429 cm = num__183 cm - num__1.71 cm = num__181.29 cm . answer : b <eor> b <eos> |
b |
subtract__183.0__1.71__ subtract__183.0__1.71__ |
subtract__183.0__1.71__ subtract__183.0__1.71__ |
| the banker ' s discount on rs . num__1650 due a certain time hence is rs . num__165 . find the true discount and the banker ' s gain . <o> a ) num__12 <o> b ) num__15 <o> c ) num__18 <o> d ) num__21 <o> e ) none |
solution sum = b . d x t . d / b . d - t . d = b . d x t . d / b . g t . d / b . g = sum / b . d = num__10.0 = num__10.0 if b . d is rs . num__11 t . d = rs . num__10 . if b . d is rs . num__165 t . d = rs . ( num__0.909090909091 x num__165 ) = rs . num__150 . and b . g = rs ( num__165 - num__150 ) = rs . num__15 . answer b <eor> b <eos> |
b |
percent__10.0__150.0__ percent__10.0__150.0__ |
percent__10.0__150.0__ percent__10.0__150.0__ |
| an article is bought for rs . num__695 and sold for rs . num__900 find the gain percent ? <o> a ) num__30 num__0.333333333333 % <o> b ) num__33 num__0.333333333333 % <o> c ) num__23 num__0.333333333333 % <o> d ) num__35 num__0.333333333333 % <o> e ) num__29 num__0.496402877698 % |
e num__29 num__0.496402877698 num__695.0 - - - - num__205 num__100 - - - - ? = > num__29 num__0.496402877698 % <eor> e <eos> |
e |
percent__100.0__29.0__ |
percent__100.0__29.0__ |
| a b worker are doing a job . worker a takes num__8 hours to do a job . b takes num__10 hours to do a job . how long should it take both a and b working together to do same job . <o> a ) num__4 num__0.444444444444 <o> b ) num__3 num__0.333333333333 <o> c ) num__3 num__0.222222222222 <o> d ) num__5 num__0.222222222222 <o> e ) num__6 num__0.444444444444 |
in this type of questions first we need to calculate num__1 hours work then their collective work as a ' s num__1 hour work is num__0.125 b ' s num__1 hour work is num__0.1 ( a + b ) ' s num__1 hour work = num__0.125 + num__0.1 = num__0.225 so both will finish the work in num__4.44444444444 hours = num__4 num__0.444444444444 answer a <eor> a <eos> |
a |
divide__1.0__8.0__ divide__1.0__10.0__ add__0.125__0.1__ divide__1.0__0.225__ multiply__0.1__4.4444__ round__4.0__ |
divide__1.0__8.0__ divide__1.0__10.0__ add__0.125__0.1__ divide__1.0__0.225__ multiply__0.1__4.4444__ round__4.0__ |
| a shopkeeper sold an article at $ num__100 with num__40.0 profit . then find its cost price ? <o> a ) $ num__120 <o> b ) $ num__100 <o> c ) $ num__91 <o> d ) $ num__71 <o> e ) $ num__69 |
cost price = selling price * num__100 / ( num__100 + profit ) c . p . = num__100 * num__0.714285714286 = $ num__71 ( approximately ) answer is d <eor> d <eos> |
d |
percent__100.0__71.0__ |
percent__100.0__71.0__ |
| an investment of d dollars at k percent simple annual interest yields $ num__200 over a num__2 year period . in terms of d what dollar amount invested at the same rate will yield $ num__2400 over a num__3 year period ? <o> a ) ( num__2 d ) / num__3 <o> b ) ( num__3 d ) / num__4 <o> c ) ( num__4 d ) / num__3 <o> d ) ( num__3 d ) / num__2 <o> e ) num__8 d |
num__200 dollars in num__2 years means num__100 dollars in num__1 year . to get num__2400 dollars ' it will take num__24 years . to get num__2400 in num__3 years we need num__8.0 = num__8 times money . answer is e . <eor> e <eos> |
e |
percent__1.0__2400.0__ percent__100.0__8.0__ |
percent__1.0__2400.0__ percent__100.0__8.0__ |
| if n = num__16 × num__10 ^ ( - p ) and − num__4 < p < num__4 how many different integer values of p will make n a perfect square ? <o> a ) num__0 <o> b ) num__2 <o> c ) num__3 <o> d ) num__5 <o> e ) num__7 |
i think the answer should be c . thoose are p - values that satisfy given restriction : - num__2 num__0 num__2 ( note a fraction can be also a perfect square ) - - > num__16 * num__100 num__16 * num__1 num__0.16 <eor> c <eos> |
c |
divide__16.0__100.0__ subtract__4.0__1.0__ |
divide__16.0__100.0__ subtract__4.0__1.0__ |
| how many numbers between num__100 and num__672 are divisible by num__2 num__3 and num__7 together ? <o> a ) num__112 <o> b ) num__77 <o> c ) num__267 <o> d ) num__14 <o> e ) num__99 |
explanation : as the division is by num__2 num__3 num__7 together the numbers are to be divisible by : num__2 * num__3 * num__7 = num__42 the limits are num__100 and num__672 the first number divisible is num__42 * num__3 = num__126 to find out the last number divisible by num__42 within num__672 : num__16.0 = num__14 hence num__42 * num__16 = num__672 is the last number divisible by num__42 within num__672 hence total numbers divisible by num__2 num__3 num__7 together are ( num__16 â € “ num__2 ) = num__14 answer : d <eor> d <eos> |
d |
multiply__3.0__42.0__ divide__672.0__42.0__ multiply__2.0__7.0__ multiply__2.0__7.0__ |
multiply__3.0__42.0__ divide__672.0__42.0__ multiply__2.0__7.0__ multiply__2.0__7.0__ |
| two trains of length num__120 m and num__280 m are running towards each other on parallel lines at num__42 kmph and num__30 kmph respectively . in what time will they be clear of each other from the moment they meet ? <o> a ) num__28 <o> b ) num__266 <o> c ) num__990 <o> d ) num__20 <o> e ) num__11 |
relative speed = ( num__42 + num__30 ) * num__0.277777777778 = num__4 * num__5 = num__20 mps . distance covered in passing each other = num__120 + num__280 = num__400 m . the time required = d / s = num__20.0 = num__20 sec . answer : d <eor> d <eos> |
d |
divide__120.0__30.0__ multiply__4.0__5.0__ add__120.0__280.0__ round__20.0__ |
divide__120.0__30.0__ multiply__4.0__5.0__ add__120.0__280.0__ multiply__4.0__5.0__ |
| in how many different ways can the letters of the word ' optical ' be arranged so that the vowels always come together ? <o> a ) num__120 <o> b ) num__720 <o> c ) num__4320 <o> d ) num__2160 <o> e ) none |
the word ' optical ' contains num__7 different letters . when the vowels oia are always together they can be supposed to form one letter . then we have to arrange the letters ptcl ( oia ) . now num__5 letters can be arranged in num__5 ! = num__120 ways . the vowels ( oia ) can be arranged among themselves in num__3 ! = num__6 ways . required number of ways = ( num__120 x num__6 ) = num__720 . option b <eor> b <eos> |
b |
vowel_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
vowel_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| the g . c . d . of num__1.08 num__0.36 and num__0.15 is ? <o> a ) num__0.03 <o> b ) num__0.9 <o> c ) num__0.03 <o> d ) num__0.108 <o> e ) none of these |
given numbers are num__1.08 num__0.36 and num__0.15 . h . c . f of num__108 num__36 and num__15 is num__3 h . c . f of given numbers = num__0.03 . correct options : c <eor> c <eos> |
c |
divide__1.08__0.36__ divide__1.08__36.0__ divide__1.08__36.0__ |
divide__1.08__0.36__ divide__1.08__36.0__ divide__1.08__36.0__ |
| which of the following is equal to num__10 ^ - ( - num__3 ) ^ num__1 ? <o> a ) num__1 / ( num__10 ^ num__5 ) <o> b ) num__1 / ( num__10 ^ num__6 ) <o> c ) num__1 / ( num__10 ^ num__3 ) <o> d ) num__1 / ( num__10 ^ num__8 ) <o> e ) num__1 / ( num__10 ^ num__9 ) |
num__10 ^ - ( - num__3 ) ^ num__1 = > num__10 ^ - ( - num__3 ^ num__1 ) = > num__10 ^ - ( num__3 ) = num__1 / ( num__10 ^ num__3 ) answer : c <eor> c <eos> |
c |
reverse__1.0__ |
reverse__1.0__ |
| a work can be completed by num__12 boys in num__24 days and num__12 girls in num__12 days . in how many days would the num__6 boys and num__6 girls working together complete the work ? <o> a ) num__12 <o> b ) num__18 <o> c ) num__16 <o> d ) num__20 <o> e ) num__24 |
in num__1 day num__12 boys does num__0.0416666666667 of the total work . = > in num__1 day num__1 boy does num__1 / ( num__24 * num__12 ) of the total work = > in num__1 days num__6 boys do num__6 / ( num__24 * num__12 ) = num__0.0208333333333 of the total work in num__1 day num__12 girls do num__0.0833333333333 of the total work . = > in num__1 day num__1 girl does num__1 / ( num__12 * num__12 ) of the total work = > in num__1 day num__6 girls do num__6 / ( num__12 * num__12 ) = num__0.0416666666667 of the total work in num__1 day num__6 boys and num__6 girls do num__0.0208333333333 + num__0.0416666666667 of the total work = > in num__1 days num__6 boys and num__6 girls do num__0.0625 = num__0.0625 of the total work so num__6 boys and num__6 girls complete all work in num__16 days . hence the answer is c <eor> c <eos> |
c |
divide__1.0__24.0__ divide__1.0__12.0__ add__0.0417__0.0208__ divide__1.0__0.0625__ round__16.0__ |
divide__1.0__24.0__ divide__1.0__12.0__ add__0.0417__0.0208__ divide__1.0__0.0625__ divide__1.0__0.0625__ |
| find the greatest number of five digits to which if num__7143 is added the final number becomes exactly divisible by num__18 num__24 num__30 num__32 and num__36 . <o> a ) num__22391 <o> b ) num__81111 <o> c ) num__27999 <o> d ) num__27711 <o> e ) num__99417 |
lcm of num__18 num__24 num__30 num__32 and num__36 is num__1440 . num__99999 + num__7143 = num__107142 . dividing num__107142 by num__1440 the remainder is num__582 . ∴ required number = num__99999 – num__582 = num__99417 alternate method : cross check with options . answer : e <eor> e <eos> |
e |
add__7143.0__99999.0__ subtract__99999.0__582.0__ subtract__99999.0__582.0__ |
add__7143.0__99999.0__ subtract__99999.0__582.0__ subtract__99999.0__582.0__ |
| in the xy - coordinate system what is the slope of the line that goes through the point ( num__4 num__4 ) and is equidistant from the two points p = ( num__0 num__2 ) and q = ( num__12 num__8 ) ? <o> a ) num__0.1 <o> b ) num__0.3 <o> c ) num__0.5 <o> d ) num__0.7 <o> e ) num__0.9 |
first get the middle coordinate between ( num__02 ) and ( num__128 ) . . . x = num__0 + ( num__12 - num__0 ) / num__2 = num__6 y = num__2 + ( num__8 - num__2 ) / num__2 = num__5 second get the slope of ( num__65 ) and ( num__44 ) . m = num__5 - num__0.666666666667 - num__4 = num__0.5 = num__0.5 answer : c <eor> c <eos> |
c |
add__4.0__2.0__ divide__4.0__6.0__ reverse__2.0__ reverse__2.0__ |
add__4.0__2.0__ divide__4.0__6.0__ reverse__2.0__ reverse__2.0__ |
| num__10 positive integers are arranged in ascending order . the range of first seven is num__25 and last num__7 is num__30 . what will be the difference in minimum range of the set in following num__2 cases num__1 ) when num__10 integers are distinct . num__2 ) when the integers can be repeated in the set ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__7 <o> d ) num__25 <o> e ) num__30 |
let ' s assume the set starts from num__1 . case num__1 ) when num__10 integers are distinct . num__1 st position = num__1 num__4 th position = num__4 num__7 th position = num__26 num__10 th position = num__34 minimum range of the set = num__33 case num__2 ) when the integers can be repeated in the set . num__1 st position = num__1 num__4 th position = num__1 num__7 th position = num__26 num__10 th position = num__31 minimum range of set = num__30 difference in minimum range = num__33 - num__30 = num__3 answer : a <eor> a <eos> |
a |
alphabet_space__ choose__3.0__2.0__ |
alphabet_space__ choose__3.0__2.0__ |
| a man can swim in still water at num__4.5 km / h but takes twice as long to swim upstream than downstream . the speed of the stream is ? <o> a ) num__1.6 <o> b ) num__1.3 <o> c ) num__1.4 <o> d ) num__1.5 <o> e ) num__1.1 |
m = num__4.5 s = x ds = num__4.5 + x us = num__4.5 + x num__4.5 + x = ( num__4.5 - x ) num__2 num__4.5 + x = num__9 - num__2 x num__3 x = num__4.5 x = num__1.5 answer : d <eor> d <eos> |
d |
multiply__4.5__2.0__ divide__4.5__3.0__ round__1.5__ |
multiply__4.5__2.0__ subtract__4.5__3.0__ subtract__4.5__3.0__ |
| john began driving from home on a trip averaging num__30 miles per hour . how many miles per hour must carla drive on average to catch up to him in exactly num__3 hours if she leaves num__30 minutes after john ? <o> a ) num__35 <o> b ) num__55 <o> c ) num__39 <o> d ) num__40 <o> e ) num__60 |
carla starts num__30 minutes later and it takes num__3 hr for carla to meet john so john total time travelled = num__3 hr + num__30 minutes john distance = num__30 * ( num__3 num__0.5 ) = num__105 so carla need to travle num__105 to meet john in num__3 hrs speed of carla = num__35.0 = num__35 miles per hour answer is a <eor> a <eos> |
a |
divide__105.0__3.0__ round__35.0__ |
divide__105.0__3.0__ round__35.0__ |
| a restaurant sells salmon pizzas for $ num__13 each and pepperoni pizzas for $ num__9 each . on every salmon pizza the restaurant makes a profit of $ num__3 while on every pepperoni pizza it makes a profit of $ num__1 . if on a given day the restaurant ' s sales amounted to $ num__468 which of the following can not be the profit made on that day ? <o> a ) num__108 <o> b ) num__94 <o> c ) num__88 <o> d ) num__80 <o> e ) num__66 |
$ num__88 is the only option that does not fit into a combination of salmon and pepperoni pizza sales profits given the total sales of the day . answer : c <eor> c <eos> |
c |
multiply__1.0__88.0__ |
multiply__1.0__88.0__ |
| percentage of profit earned by selling a book for $ num__3360 is equal to the percentage loss incurred by selling the same book for $ num__2640 . what price should the book be sold to make num__20.0 profit ? <o> a ) $ num__3600 <o> b ) $ num__4000 <o> c ) $ num__3500 <o> d ) $ num__3700 <o> e ) $ num__4000 |
let c . p . be rs . x . then ( num__3360 - x ) x num__100 = ( x - num__2640 ) x num__100 num__3360 - x = x - num__2640 num__2 x = num__6000 x = num__3000 required s . p . = num__120.0 of num__3000 = num__1.20 x num__3000 = $ num__3600 answer : a <eor> a <eos> |
a |
percent__2.0__6000.0__ percent__100.0__3600.0__ |
percent__2.0__6000.0__ percent__100.0__3600.0__ |
| at a certain resort each of the num__39 food service employees is trained to work in a minimum of num__1 restaurant and a maximum of num__3 restaurants . the num__3 restaurants are the family buffet the dining room and the snack bar . exactly num__21 employees are trained to work in the family buffet num__18 are trained to work in the dining room and num__12 are trained to work in the snack bar . if num__4 employees are trained to work in exactly num__2 restaurants how many employees are trained to work in all num__3 restaurants ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
num__39 = num__21 + num__18 + num__12 - num__4 - num__2 x num__2 x = num__21 + num__18 + num__12 - num__4 - num__39 = num__47 - num__39 = num__8 x = num__4 c <eor> c <eos> |
c |
subtract__12.0__4.0__ round__4.0__ |
subtract__12.0__4.0__ add__1.0__3.0__ |
| a sum of rs . num__2665 is lent into two parts so that the interest on the first part for num__8 years at num__3.0 per annum may be equal to the interest on the second part for num__3 years at num__5.0 per annum . find the second sum ? <o> a ) rs . num__450 <o> b ) rs . num__1640 <o> c ) rs . num__523 <o> d ) rs . num__458 <o> e ) rs . num__796 |
explanation : ( x * num__8 * num__3 ) / num__100 = ( ( num__2665 - x ) * num__3 * num__5 ) / num__100 num__24 x / num__100 = num__399.75 - num__15 x / num__100 num__39 x = num__39975 = > x = num__1025 second sum = num__2665 – num__1025 = rs . num__1640 answer : b <eor> b <eos> |
b |
percent__100.0__1640.0__ |
percent__100.0__1640.0__ |
| what is the greatest prime factor of num__4 ^ num__17 - num__2 ^ num__26 ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__5 <o> d ) num__127 <o> e ) num__11 |
i ' m happy to help with this . we know num__4 = num__2 ^ num__2 so num__4 ^ num__17 = ( num__2 ^ num__2 ) ^ num__17 = num__2 ^ ( num__2 * num__17 ) = num__2 ^ num__34 that takes advantage of a law of exponents that says ( a ^ n ) ^ m = a ^ ( n * m ) so num__4 ^ num__17 - num__2 ^ num__26 = num__2 ^ num__34 - num__2 ^ num__26 = num__2 ^ ( num__26 + num__8 ) - num__2 ^ num__26 = ( num__2 ^ num__26 ) * ( num__2 * num__8 ) - num__2 ^ num__26 = ( num__2 ^ num__8 - num__1 ) * ( num__2 ^ num__26 ) = ( num__128 - num__1 ) * ( num__2 ^ num__26 ) = num__127 * ( num__2 ^ num__26 ) the prime factors of num__63 are num__127 so the largest prime factor is num__127 answer choice d . here ' s a blog you may find helpful . http : / / magoosh . com / gmat / num__2012 / gmat - math - factors / does all that make sense ? please let me know if you have any further questions . mike wow . i am floored by how great of an explanation you provided . posts like that make me really think that doing thousands of practice problems with good explanations beats out reading books on math every day of the week . d <eor> d <eos> |
d |
multiply__17.0__2.0__ multiply__4.0__2.0__ subtract__128.0__1.0__ subtract__128.0__1.0__ |
multiply__17.0__2.0__ multiply__4.0__2.0__ subtract__128.0__1.0__ subtract__128.0__1.0__ |
| calculate the average of first num__9 even numbers is ? <o> a ) num__10 <o> b ) num__13 <o> c ) num__12 <o> d ) num__15 <o> e ) num__17 |
explanation : sum of num__10 even numbers = num__9 * num__10 = num__90 average = num__10.0 = num__10 answer : option a <eor> a <eos> |
a |
multiply__9.0__10.0__ divide__90.0__9.0__ |
multiply__9.0__10.0__ divide__90.0__9.0__ |
| there are num__7 fictions and num__6 non - fictions . how many cases are there such that num__2 fictions and num__2 non - fictions are selected from them ? <o> a ) num__90 <o> b ) num__120 <o> c ) num__315 <o> d ) num__180 <o> e ) num__200 |
number of ways of selecting num__2 fiction books = num__7 c num__2 number of ways of selecting num__2 non fiction books = num__6 c num__2 num__7 c num__2 * num__6 c num__2 = num__21 * num__15 = num__315 answer : c <eor> c <eos> |
c |
subtract__21.0__6.0__ multiply__15.0__21.0__ multiply__15.0__21.0__ |
subtract__21.0__6.0__ multiply__15.0__21.0__ multiply__15.0__21.0__ |
| if num__5 a = num__6 b and ab ≠ num__0 what is the ratio of a / num__6 to b / num__5 ? <o> a ) num__1.44 <o> b ) num__0.833333333333 <o> c ) num__1 <o> d ) num__0.2 <o> e ) num__0.694444444444 |
a nice fast approach is the first find a pair of numbers that satisfy the given equation : num__5 a = num__6 b here ' s one pair : a = num__6 and b = num__5 what is the ratio of a / num__6 to b / num__5 ? in other words what is the value of ( a / num__6 ) / ( b / num__5 ) ? plug in values to get : ( a / num__6 ) / ( b / num__5 ) = ( num__1.0 ) / ( num__1.0 ) = num__1.0 = num__1 c <eor> c <eos> |
c |
subtract__6.0__5.0__ reverse__1.0__ |
subtract__6.0__5.0__ reverse__1.0__ |
| in what time will a train num__140 m long cross an electric pole it its speed be num__190 km / hr ? <o> a ) num__2.6 sec <o> b ) num__2.9 sec <o> c ) num__2.7 sec <o> d ) num__8.7 sec <o> e ) num__8.5 sec |
speed = num__190 * num__0.277777777778 = num__53 m / sec time taken = num__2.64150943396 = num__2.6 sec . answer : a <eor> a <eos> |
a |
divide__140.0__53.0__ round__2.6__ |
divide__140.0__53.0__ round__2.6__ |
| simplify : num__200 x num__200 - num__150 x num__150 <o> a ) num__761200 <o> b ) num__761400 <o> c ) num__761800 <o> d ) num__17500 <o> e ) none of them |
( num__200 ) ^ num__2 - ( num__150 ) ^ num__2 = ( num__200 + num__150 ) ( num__200 - num__150 ) = num__350 x num__50 = num__17500 . answer is d <eor> d <eos> |
d |
add__200.0__150.0__ subtract__200.0__150.0__ multiply__50.0__350.0__ multiply__50.0__350.0__ |
add__200.0__150.0__ subtract__200.0__150.0__ multiply__50.0__350.0__ multiply__50.0__350.0__ |
| you need to unlock a secret code using following clues can you ? here you have the clues : clue - num__1 : num__0 num__7 num__9 ( one of the numbers is correct and is placed in its correct position ) clue - num__2 : num__0 num__3 num__2 ( nothing is correct ) clue - num__3 : num__1 num__0 num__8 ( two numbers are correct but not placed at its correct position . ) clue - num__4 : num__9 num__2 num__6 ( one number is correct but not placed at its correct position . ) clue - num__5 : num__6 num__7 num__8 ( one number is correct but not placed at its correct position . ) <o> a ) num__819 <o> b ) num__918 <o> c ) num__198 <o> d ) num__189 <o> e ) num__891 |
num__0 num__7 num__9 : num__9 is placed correctly * num__0 num__3 num__2 : none of the numbers is in the code * num__1 num__0 num__8 : num__1 and num__8 are correct number of code but placed at wrong position * num__9 num__2 num__6 : number num__9 is there but placed at wrong position * num__6 num__7 num__8 : number num__8 is there but placed at wrong position correct answer is a ) num__819 <eor> a <eos> |
a |
round__819.0__ |
multiply__1.0__819.0__ |
| on a map num__1 inch represents num__28 miles . how many v inches would be necessary to represent a distance of num__383.6 miles ? <o> a ) num__5.2 <o> b ) num__7.4 <o> c ) num__13.7 <o> d ) num__21.2 <o> e ) num__28.7 |
v inches necessary to represent a distance of num__383.6 miles = num__383.6 / num__28 = num__13.7 answer c <eor> c <eos> |
c |
divide__383.6__28.0__ round__13.7__ |
divide__383.6__28.0__ divide__383.6__28.0__ |
| in a particular state num__60.0 of the counties received some rain on monday and num__55.0 of the counties received some rain on tuesday . no rain fell either day in num__25.0 of the counties in the state . what percent of the counties received some rain on monday and tuesday ? <o> a ) num__12.5 <o> b ) num__40.0 <o> c ) num__50.0 <o> d ) num__60.0 <o> e ) num__67.5 % |
num__60 + num__55 + num__25 = num__140.0 the number is num__40.0 above num__100.0 because num__40.0 of the counties were counted twice . the answer is b . <eor> b <eos> |
b |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| a washing machine is available for $ num__39000 cash or $ num__17000 as down payment followed by five equal monthly instalments of $ num__4800 each . the simple rate of interest per annum under the instalment plan would be <o> a ) num__18.0 <o> b ) num__19.0 <o> c ) num__21.2 <o> d ) num__21.81 <o> e ) num__22.07 % |
simple interest si = ( p * r * t ) / num__100 where p is the principal amount r is the rate of interest and t is time in years the way i see it : washing machine on down payment of num__17000 customer is not paying any interest . it is the remaining sum which will be paid for num__5 months that will bear an interest . therefore the principal amount for which interest is being charged is num__39000 - num__17000 = num__22000 for this num__22000 a total sum of num__5 * num__4800 = num__24000 was pain . ( time is five months so t = num__0.416666666667 as t is in years . ) thus si = num__2000 or num__2000 = ( p * r * t ) / num__100 num__2000 = ( num__22000 * r * num__5 ) / ( num__100 * num__12 ) r = ( num__2000 * num__12 * num__100 ) / num__22000 * num__5 r = num__21.81 therefore answer is d <eor> d <eos> |
d |
percent__21.81__100.0__ |
percent__21.81__100.0__ |
| certain stocks in january were num__30.0 greater than they were in february and num__20.0 lesser than they were in march . what was the percentage decrease in the stocks from february to march ? <o> a ) num__44.0 <o> b ) num__30.0 <o> c ) num__40.0 <o> d ) num__47.5 <o> e ) num__50 % |
let stocks value in feb = num__100 = > in jan = num__70 and march = num__70 * num__0.8 = num__56 thus percent decrease in stocks from feb to march = num__100 - num__56 = num__44.0 hence answer is a <eor> a <eos> |
a |
subtract__100.0__30.0__ multiply__0.8__70.0__ subtract__100.0__56.0__ subtract__100.0__56.0__ |
subtract__100.0__30.0__ multiply__0.8__70.0__ subtract__100.0__56.0__ subtract__100.0__56.0__ |
| a solid yellow stripe is to be painted in the middle of a certain highway . if num__1 gallon of paint covers an area of e square feet of highway how many gallons of paint will be needed to paint a stripe of t inches wide on a stretch of highway m miles long ? ( num__1 mile = num__5280 feet and num__1 foot = num__12 inches ) <o> a ) ( num__5280 mt ) / num__12 e <o> b ) ( num__5280 pt ) / num__12 m <o> c ) ( num__5280 pmt ) / num__12 <o> d ) ( num__5280 ) ( num__12 m ) / pt <o> e ) ( num__5280 ) ( num__12 p ) / mt |
given that : num__1 gallon of paint covers an area ofesquare feet . question : how many gallonsof paint will be needed . . . in any case you will have : ( total area in square feet ) / ( gallons per feet ) = ( total area in square feet ) / e so e must be in the denominator : eliminate all but a and d . now lets see where should be t : ( area in square feet ) = ( width in feet ) * ( length in feet ) - - > width = tinchesas num__1 feet = num__12 inchesthent inches = t / num__12 feet so ( area in square feet ) = ( t / num__12 ) * ( length in feet ) so t must be in the nominator : only a is left . answer : a . <eor> a <eos> |
a |
round__5280.0__ |
multiply__1.0__5280.0__ |
| in a certain book club the members are voting between two books – a mystery novel and a biography – to decide which book the group will read next . there are num__4 times as many votes for the mystery novel as there are votes for the biography . the ratio of biography votes to the total number of votes is <o> a ) num__1 to num__2 <o> b ) num__1 to num__3 <o> c ) num__1 to num__4 <o> d ) num__1 to num__5 <o> e ) num__3 to num__4 |
imo : answer d num__1 to num__5 <eor> d <eos> |
d |
add__4.0__1.0__ reverse__1.0__ |
add__4.0__1.0__ reverse__1.0__ |
| a cement block balances evenly on the scales with num__3 quarters of a pound and num__3 quarters of a block . what is the weight of the whole block ? <o> a ) num__1 pounds <o> b ) num__2 pounds <o> c ) num__3 pounds <o> d ) num__4 pounds <o> e ) num__5 pounds |
x = num__0.75 + x * num__0.75 x / num__4 = num__0.75 x = num__3 pounds answer : c <eor> c <eos> |
c |
divide__3.0__0.75__ multiply__0.75__4.0__ |
divide__3.0__0.75__ multiply__0.75__4.0__ |
| hcf and lcm two numbers are num__12 and num__396 respectively . if one of the numbers is num__24 then the other number is ? <o> a ) num__36 <o> b ) num__198 <o> c ) num__132 <o> d ) num__264 <o> e ) num__364 |
num__12 * num__396 = num__24 * x x = num__198 answer : b <eor> b <eos> |
b |
round__198.0__ |
round__198.0__ |
| if a man walks at a rate of num__10 kmph he misses a train by num__14 minutes . however if he walks at the rate of num__12 kmph he reaches the station num__10 minutes before the arrival of the train . find the distance covered by him to reach the station . <o> a ) num__6 km <o> b ) num__15 <o> c ) num__12 <o> d ) num__4 <o> e ) num__8 |
let the required distance x km difference in the times taken at two speeds = num__12 min = num__0.2 hr ( x / num__10 ) - ( x / num__12 ) = num__0.1 x = num__6 the required distance is num__6 km answer is a <eor> a <eos> |
a |
round__6.0__ |
subtract__12.0__6.0__ |
| in a class of num__10 students the teacher writes a number on the board . the first student tells that number is divisible by num__1 the second student tells that the number is divisible by num__2 the third by num__3 the forth by num__4 and so on till num__10 . if the statement of one of the students is wrong then what is the least number possible . <o> a ) num__320 <o> b ) num__340 <o> c ) num__360 <o> d ) num__380 <o> e ) num__400 |
statement of num__7 th student was wrong . no is not divisible by num__7 . lcm of num__23 num__45 num__68 num__910 = num__360 answer : c <eor> c <eos> |
c |
subtract__10.0__3.0__ add__45.0__23.0__ multiply__1.0__360.0__ |
subtract__10.0__3.0__ add__45.0__23.0__ multiply__1.0__360.0__ |
| a shopkeeper purchased num__90 kg of potatoes for rs . num__540 and sold the whole lot at the rate of rs . num__8 per kg . what will be his gain percent ? <o> a ) num__38 num__0.111111111111 % <o> b ) num__38 num__3.0 % <o> c ) num__38 num__0.333333333333 % <o> d ) num__32 num__0.333333333333 % <o> e ) num__33 num__0.333333333333 % |
c . p . of num__1 kg = num__6.0 = rs . num__6 s . p . of num__1 kg = rs . num__8 gain % = num__0.333333333333 * num__100 = num__33.3333333333 = num__33 num__0.333333333333 % answer : e <eor> e <eos> |
e |
percent__100.0__33.0__ |
percent__100.0__33.0__ |
| six years ago the ratio of the ages of kunal and sagar was num__6 : num__5 . four years hence the ratio of their ages will be num__11 : num__10 . what is sagar ' s age at present ? <o> a ) num__16 years <o> b ) num__18 years <o> c ) num__20 years <o> d ) num__22 years <o> e ) num__24 years |
explanation : let the ages of kunal and sagar num__6 years ago be num__6 x and num__5 x years respectively . then ( num__6 x + num__6 ) + num__4 = num__11 ( num__5 x + num__6 ) + num__4 = num__10 ⇒ num__10 ( num__6 x + num__10 ) = num__11 ( num__5 x + num__10 ) ⇒ num__5 x = num__10 ⇒ x = num__2 . sagar ' s present age = ( num__5 x + num__6 ) = num__16 years . answer : a <eor> a <eos> |
a |
subtract__10.0__6.0__ subtract__6.0__4.0__ add__6.0__10.0__ add__6.0__10.0__ |
subtract__10.0__6.0__ subtract__6.0__4.0__ add__6.0__10.0__ add__6.0__10.0__ |
| two employees x and y are paid a total of rs . num__616 per week by their employer . if x is paid num__120 percent of the sum paid to y how much is y paid per week ? <o> a ) s . num__250 <o> b ) s . num__280 <o> c ) s . num__290 <o> d ) s . num__299 <o> e ) s . num__300 |
let the amount paid to x per week = x and the amount paid to y per week = y then x + y = num__616 but x = num__120.0 of y = num__120 y / num__100 = num__12 y / num__10 ∴ num__12 y / num__10 + y = num__616 ⇒ y [ num__1.2 + num__1 ] = num__616 ⇒ num__22 y / num__10 = num__616 ⇒ num__22 y = num__6160 ⇒ y = num__280.0 = num__280.0 = rs . num__280 b <eor> b <eos> |
b |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__12.0__10.0__ multiply__616.0__10.0__ divide__6160.0__22.0__ multiply__1.0__280.0__ |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__12.0__10.0__ multiply__616.0__10.0__ divide__6160.0__22.0__ divide__6160.0__22.0__ |
| a sum of money is to be distributed among a b c d in the proportion of num__5 : num__2 : num__4 : num__3 . if c gets rs . num__1000 more than d what is b ' s share ? <o> a ) rs . num__500 <o> b ) rs . num__1500 <o> c ) rs . num__2000 <o> d ) rs . num__2500 <o> e ) none of these |
let the shares of a b c and d be rs . num__5 x rs . num__2 x rs . num__4 x and rs . num__3 x respectively . then num__4 x - num__3 x = num__1000 x = num__1000 . b ' s share = rs . num__2 x = rs . ( num__2 x num__1000 ) = rs . num__2000 . answer : option c <eor> c <eos> |
c |
multiply__2.0__1000.0__ multiply__2.0__1000.0__ |
multiply__2.0__1000.0__ multiply__2.0__1000.0__ |
| in what time will a train num__100 m long cross an electric pole it its speed be num__144 km / hr ? <o> a ) num__2.5 <o> b ) num__7.3 <o> c ) num__2.3 <o> d ) num__1.4 <o> e ) num__1.1 |
speed = num__144 * num__0.277777777778 = num__40 m / sec time taken = num__2.5 = num__2.5 sec . answer : a <eor> a <eos> |
a |
divide__100.0__40.0__ round__2.5__ |
divide__100.0__40.0__ divide__100.0__40.0__ |
| a cylindrical tank of diameter num__35 cm is full of water . if num__11 litres of water is drawn off the water level in the tank will drop by : <o> a ) num__11 num__0.428571428571 cm <o> b ) num__11 num__0.285714285714 cm <o> c ) num__11 num__0.142857142857 cm <o> d ) num__11 cm <o> e ) none of these |
explanation : let the drop in the water level be h cm then volume of cylinder = π r num__2 h = > num__3.14285714286 ∗ num__17.5 ∗ num__17.5 ∗ h = num__11000 = > h = num__11000 ∗ num__7 ∗ num__0.181818181818 ∗ num__35 ∗ num__35 cm = num__11.4285714286 cm = num__11 num__0.428571428571 cm option a <eor> a <eos> |
a |
divide__35.0__2.0__ divide__2.0__11.0__ subtract__11.4286__11.0__ round__11.0__ |
divide__35.0__2.0__ divide__2.0__11.0__ subtract__11.4286__11.0__ round__11.0__ |
| if the area of a circle is num__144 pi square feet find its circumference . <o> a ) num__65 pi feet <o> b ) num__24 pi feet <o> c ) num__42 pi feet <o> d ) num__18 pi feet <o> e ) num__64 pi feet |
the area is given by pi * r * r . hence pi * r * r = num__144 pi r * r = num__81 ; hence r = num__12 feet the circumference is given by num__2 * pi * r = num__2 * pi * num__12 = num__24 pi feet correct answer b <eor> b <eos> |
b |
surface_cube__2.0__ surface_cube__2.0__ |
multiply__2.0__12.0__ multiply__2.0__12.0__ |
| a light flashes every num__6 seconds how many times will it flash in ? of an hour ? <o> a ) num__550 <o> b ) num__600 <o> c ) num__650 <o> d ) num__700 <o> e ) num__750 |
num__1 flash = num__6 sec for num__1 min = num__10 flashes so for num__1 hour = num__10 * num__60 = num__600 flashes . answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__10.0__60.0__ round__600.0__ |
multiply__6.0__10.0__ multiply__10.0__60.0__ multiply__10.0__60.0__ |
| suppose a game is played between two players a and b . on each turn of the game exactly one of a or b gets a point . suppose a is better than b and has a probability of num__0.666666666667 of getting a point on each turn of the game . the first person to get two points ahead in the game is the winner . what is the probability that a wins the game ? <o> a ) num__0.555555555556 <o> b ) num__0.571428571429 <o> c ) num__0.666666666667 <o> d ) num__0.8 <o> e ) num__0.888888888889 |
let w num__0 be the event that player a wins the game . the goal is to compute p ( w num__0 ) . let w num__1 be the event that player a wins the game given that she is ahead by one point . let w - num__1 be the event player a wins the game given that she is behind by one point . then the probability that player a wins the game is p ( w num__0 ) = p ( a is ahead by num__1 point ) p ( w num__1 ) + p ( a is behind by num__1 point ) p ( w - num__1 ) = ( num__0.666666666667 ) p ( w num__1 ) + ( num__0.333333333333 ) p ( w - num__1 ) : ( num__1 ) now p ( w num__1 ) = p ( player a gets a point on the next turn ) + p ( player b gets a point on the next turn ) p ( w num__0 ) = ( num__0.666666666667 ) + ( num__0.333333333333 ) p ( w num__0 ) ; and p ( w - num__1 ) = p ( player a gets a point on the next turn ) p ( w num__0 ) = ( num__0.666666666667 ) p ( w num__0 ) : substituting these equations into ( num__1 ) we have p ( w num__0 ) = ( num__0.666666666667 ) [ ( num__0.666666666667 ) + ( num__0.333333333333 ) p ( w num__0 ) ] + ( num__0.333333333333 ) ( num__0.666666666667 ) p ( w num__0 ) = num__0.444444444444 + ( num__0.444444444444 ) p ( w num__0 ) : therefore p ( w num__0 ) = ( num__0.444444444444 ) ( num__1.8 ) = num__0.8 correct answer d <eor> d <eos> |
d |
round_down__0.6667__ subtract__1.0__0.6667__ subtract__1.8__1.0__ multiply__1.0__0.8__ |
round_down__0.6667__ subtract__1.0__0.6667__ subtract__1.8__1.0__ subtract__1.8__1.0__ |
| if the ratio of apples to bananas is num__5 to num__2 and the ratio of bananas to cucumbers is num__1 to num__3 what is the ratio of apples to cucumbers ? <o> a ) num__2 : num__3 <o> b ) num__5 : num__3 <o> c ) num__5 : num__6 <o> d ) num__1 : num__4 <o> e ) num__3 : num__4 |
the ratio of bananas to cucumbers is num__1 to num__3 which equals num__2 to num__6 . the ratio of apples to bananas to cucumbers is num__5 to num__2 to num__6 . the ratio of apples to cucumbers is num__5 to num__6 . the answer is c . <eor> c <eos> |
c |
add__5.0__1.0__ multiply__5.0__1.0__ |
add__5.0__1.0__ multiply__5.0__1.0__ |
| a man ' s speed with the current is num__20 kmph and speed of the current is num__3 kmph . the man ' s speed against the current will be <o> a ) num__11 kmph <o> b ) num__12 kmph <o> c ) num__14 kmph <o> d ) num__17 kmph <o> e ) none of these |
explanation : speed with current is num__20 speed of the man + it is speed of the current speed in s Ɵ ll water = num__20 - num__3 = num__17 now speed against the current will be speed of the man - speed of the current = num__17 - num__3 = num__14 kmph answer : c <eor> c <eos> |
c |
subtract__20.0__3.0__ subtract__17.0__3.0__ round__14.0__ |
subtract__20.0__3.0__ subtract__17.0__3.0__ subtract__17.0__3.0__ |
| a certain sum amounts to rs . num__1725 in num__3 years and rs . num__1875 in num__5 years . find the rate % per annum ? <o> a ) num__2 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
num__3 - - - num__1725 num__5 - - - num__1875 - - - - - - - - - - - - - - num__2 - - - num__150 n = num__1 i = num__75 r = ? p = num__1725 - num__225 = num__1500 num__75 = ( num__1500 * num__1 * r ) / num__100 r = num__5.0 . answer : b <eor> b <eos> |
b |
percent__5.0__100.0__ |
percent__5.0__100.0__ |
| jaclyn buys $ num__20 num__000 worth of debentures in a company . she earns num__9.5 p . a . simple interest paid to her quarterly ( that is every num__3 months ) . if the agreed period of the debenture was num__18 months : calculate the amount of interest jaclyn will earn for each quarter <o> a ) num__475 <o> b ) num__234 <o> c ) num__289 <o> d ) num__345 <o> e ) none of these |
explanation : i = ( p x r x t ) / num__100 = num__30000 * num__9.5 / num__100 * ( num__1.5 ) ^ num__0.166666666667 = num__475 answer : a <eor> a <eos> |
a |
percent__100.0__475.0__ |
percent__100.0__475.0__ |
| a train num__280 m long is running at num__70 kmph . in how much time will it pass a platform num__210 m long ? <o> a ) num__1.5 seconds <o> b ) num__9.56 seconds <o> c ) num__28.56 seconds <o> d ) num__30.56 seconds <o> e ) num__3.5 seconds |
distance travelled = num__280 + num__210 m = num__490 m speed = num__70 * num__0.625 = num__51.25 m time = num__490 * num__0.019512195122 = num__9.56 seconds answer : b . <eor> b <eos> |
b |
add__280.0__210.0__ round__9.56__ |
add__280.0__210.0__ round__9.56__ |
| a trained covered x km at num__40 kmph and another num__2 x km at num__20 kmph . find the average speed of the train in covering the entire num__3 x km . <o> a ) num__12 kmph <o> b ) num__88 kmph <o> c ) num__24 kmph <o> d ) num__17 kmph <o> e ) num__19 kmph |
total time taken = x / num__40 + num__2 x / num__20 hours = num__5 x / num__40 = x / num__8 hours average speed = num__3 x / ( x / num__8 ) = num__24 kmph answer : c <eor> c <eos> |
c |
add__2.0__3.0__ divide__40.0__5.0__ multiply__3.0__8.0__ multiply__3.0__8.0__ |
add__2.0__3.0__ divide__40.0__5.0__ multiply__3.0__8.0__ multiply__3.0__8.0__ |
| how many seconds will a num__500 meter long train take to cross a man walking with a speed of num__3 km / hr in the direction of the moving train if the speed of the train is num__63 km / hr ? <o> a ) num__299 meters <o> b ) num__278 meters <o> c ) num__500 meters <o> d ) num__276 meters <o> e ) num__299 meters |
let length of tunnel is x meter distance = num__800 + x meter time = num__1 minute = num__60 seconds speed = num__78 km / hr = num__78 * num__0.277777777778 m / s = num__21.6666666667 m / s distance = speed * time num__800 + x = ( num__21.6666666667 ) * num__60 num__800 + x = num__20 * num__65 = num__1300 x = num__1300 - num__800 = num__500 meters answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ divide__60.0__3.0__ add__500.0__800.0__ round__500.0__ |
subtract__63.0__3.0__ divide__60.0__3.0__ add__500.0__800.0__ divide__500.0__1.0__ |
| if p / num__6 = r / num__2 and r = num__0.75 s what is the ratio of p to r to s ? <o> a ) num__2 : num__3 : num__4 <o> b ) num__8 : num__12 : num__9 <o> c ) num__9 : num__6 : num__8 <o> d ) num__6 : num__4 : num__3 <o> e ) num__18 : num__6 : num__8 |
p / num__6 = r / num__2 r = num__0.75 s what is p : r : s ? we get p / r = num__3.0 and r / s = num__0.75 so num__2.0 * num__1.5 and num__1.0 * num__0.75 num__3.0 and num__0.75 e <eor> e <eos> |
e |
divide__6.0__2.0__ multiply__2.0__0.75__ round_down__1.5__ multiply__6.0__3.0__ |
divide__6.0__2.0__ multiply__2.0__0.75__ round_down__1.5__ multiply__6.0__3.0__ |
| a man buys num__20 lts of liquid which contains num__30.0 of the liquid and the rest is water . he then mixes it with num__20 lts of another mixture with num__20.0 of liquid . what is the % of water in the new mixture ? <o> a ) num__59 <o> b ) num__95 <o> c ) num__80 <o> d ) num__75 <o> e ) num__65 |
num__30.0 in num__20 lts is num__6 . so water = num__20 - num__6 = num__14 lts . num__20.0 of num__20 lts = num__4 . so water in num__2 nd mixture = num__20 - num__4 = num__16 lts . now total quantity = num__20 + num__20 = num__40 lts . total water in it will be num__14 + num__16 = num__30 lts . % of water = ( num__100 * num__30 ) / num__40 = num__75 . answer : d <eor> d <eos> |
d |
percent__20.0__30.0__ percent__75.0__100.0__ |
percent__20.0__30.0__ percent__75.0__100.0__ |
| the base of a right triangle is num__8 and hypotenuse is num__10 . its area is ? <o> a ) num__12 <o> b ) num__80 <o> c ) num__59 <o> d ) num__24 <o> e ) num__25 |
explanation : h num__2 = ( num__10 ) num__2 - ( num__8 ) num__2 - ( num__6 ) num__2 - > h = num__6 num__0.5 * num__8 * num__6 = num__24 answer is d <eor> d <eos> |
d |
side_by_diagonal__10.0__8.0__ surface_cube__2.0__ surface_cube__2.0__ |
side_by_diagonal__10.0__8.0__ volume_rectangular_prism__8.0__0.5__6.0__ volume_rectangular_prism__8.0__0.5__6.0__ |
| two trains num__140 m and num__160 m long run at the speed of num__60 km / hr and num__40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? <o> a ) num__20.8 sec <o> b ) num__10.7 sec <o> c ) num__10.3 sec <o> d ) num__10.8 sec <o> e ) num__10.2 sec |
relative speed = num__60 + num__40 = num__100 km / hr . = num__100 * num__0.277777777778 = num__27.7777777778 m / sec . distance covered in crossing each other = num__140 + num__160 = num__300 m . required time = num__300 * num__0.036 = num__10.8 = num__10.8 sec . answer : d <eor> d <eos> |
d |
subtract__140.0__40.0__ add__140.0__160.0__ multiply__300.0__0.036__ round__10.8__ |
add__60.0__40.0__ add__140.0__160.0__ multiply__300.0__0.036__ multiply__300.0__0.036__ |
| a train speeds past a pole in num__20 seconds and passes through a tunnel num__500 m long in num__40 seconds . its length is : <o> a ) num__200 m <o> b ) num__400 m <o> c ) num__500 m <o> d ) num__700 m <o> e ) none of the above |
sol . let the length of the train be x metres and its speed be y m / sec . then x / y = num__20 ⇒ y = x / num__20 ∴ ( x + num__500 ) / num__40 = x / num__20 ⇔ x = num__500 m . answer c <eor> c <eos> |
c |
round__500.0__ |
round__500.0__ |
| in how many ways can num__16 different gifts be divided among four children such that each child receives exactly three gifts ? <o> a ) num__16 ^ num__4 <o> b ) ( num__4 ! ) ^ num__4 <o> c ) num__16 ! / ( num__3 ! ) ^ num__4 <o> d ) num__16 ! / num__4 ! <o> e ) num__4 ^ num__16 |
total num__16 different gifts and num__4 children . thus any one child gets num__16 c num__3 gifts then the other child gets num__13 c num__3 gifts ( num__16 total - num__3 already given ) then the third one gets num__10 c num__3 gifts and the last child gets num__7 c num__3 gifts . since order in which each child gets the gift is not imp thus ans : num__16 c num__3 * num__13 c num__3 * num__10 c num__3 * num__7 c num__3 = num__16 ! / ( num__3 ! ) ^ num__4 ans : c . <eor> c <eos> |
c |
subtract__16.0__3.0__ subtract__13.0__3.0__ add__3.0__4.0__ add__3.0__13.0__ |
subtract__16.0__3.0__ subtract__13.0__3.0__ subtract__10.0__3.0__ add__3.0__13.0__ |
| a can give b num__100 meters start and c num__170 meters start in a kilometer race . how much start can b give c in a kilometer race ? <o> a ) num__11.77 meters <o> b ) num__55.77 meters <o> c ) num__77.77 meters <o> d ) num__113.77 meters <o> e ) none of these |
explanation : a runs num__1000 meters while b runs num__900 meters and c runs num__830 meters . therefore b runs num__900 meters while c runs num__830 meters . so the number of meters that c runs when b runs num__1000 meters = ( num__1000 x num__830 ) / num__900 = num__922.22 meters thus b can give c ( num__1000 - num__922.22 ) = num__77.77 meters start answer : c <eor> c <eos> |
c |
subtract__1000.0__100.0__ subtract__1000.0__170.0__ round__77.77__ |
subtract__1000.0__100.0__ subtract__1000.0__170.0__ round__77.77__ |
| a card marked num__1 and num__2 on its two faces . two colors white and black are used to paint the two faces of the card . if the card is painted with the different colors in how many ways can the card be painted ? <o> a ) num__3 <o> b ) num__6 <o> c ) num__2 <o> d ) num__12 <o> e ) num__27 |
if the base of the card is white then in order the card to be painted with the different colors the top must be white . another side can be painted in black . but we can have the base painted in either of the two colors thus the total number of ways to paint the card is num__2 * num__1 = num__2 . answer : c . <eor> c <eos> |
c |
coin_space__ |
coin_space__ |
| if the sample interest on a sum of money num__20.0 per annum for num__2 years is $ num__400 find the compound interest on the same sum for the same period at the same rate ? <o> a ) $ num__460 <o> b ) $ num__510 <o> c ) $ num__440 <o> d ) $ num__500 <o> e ) $ num__550 |
rate = num__20.0 time = num__2 years s . i . = $ num__400 principal = num__100 * num__20.0 * num__2 = $ num__1000 amount = num__1000 ( num__1 + num__0.2 ) ^ num__2 = $ num__1440 c . i . = num__1440 - num__1000 = $ num__440 answer is c <eor> c <eos> |
c |
percent__20.0__1.0__ percent__100.0__440.0__ |
percent__20.0__1.0__ percent__100.0__440.0__ |
| a train covers a distance of num__12 km in num__10 min . if it takes num__7 sec to pass a telegraph post then the length of the train is ? <o> a ) num__298 <o> b ) num__288 <o> c ) num__140 <o> d ) num__776 <o> e ) num__991 |
speed = ( num__1.2 * num__60 ) km / hr = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . length of the train = num__20 * num__7 = num__140 m . answer : c <eor> c <eos> |
c |
divide__12.0__10.0__ hour_to_min_conversion__ add__12.0__60.0__ multiply__7.0__20.0__ round__140.0__ |
divide__12.0__10.0__ hour_to_min_conversion__ multiply__1.2__60.0__ multiply__7.0__20.0__ multiply__7.0__20.0__ |
| let y = num__2 m + x ^ num__2 and m = num__5 x + num__2 . if m ^ num__2 = num__49 then which of the following can be a value of num__2 y + num__3 m ? <o> a ) num__48 <o> b ) num__56 <o> c ) num__61 <o> d ) num__51 <o> e ) num__71 |
m ^ num__2 = num__49 = > m = num__7 num__7 = num__5 x + num__2 = > x = num__1 y = num__2 * num__7 + num__1 ^ num__2 = num__15 num__2 y + num__3 m = num__2 * num__15 + num__3 * num__7 = num__51 answer : d <eor> d <eos> |
d |
add__2.0__5.0__ subtract__3.0__2.0__ multiply__5.0__3.0__ add__2.0__49.0__ add__2.0__49.0__ |
add__2.0__5.0__ subtract__3.0__2.0__ multiply__5.0__3.0__ add__2.0__49.0__ add__2.0__49.0__ |
| a trained covered x km at num__40 kmph and another num__2 x km at num__20 kmph . find the average speed of the train in covering the entire num__5 x km . <o> a ) num__20 kmph <o> b ) num__25 kmph <o> c ) num__30 kmph <o> d ) num__35 kmph <o> e ) num__40 kmph |
total time taken = x / num__40 + num__2 x / num__20 hours = num__5 x / num__40 = x / num__8 hours average speed = num__5 x / ( x / num__8 ) = num__40 kmph answer : e <eor> e <eos> |
e |
divide__40.0__5.0__ multiply__2.0__20.0__ |
divide__40.0__5.0__ multiply__2.0__20.0__ |
| a man whose speed is num__4.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at num__1.5 kmph find his average speed for the total journey ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__32 |
m = num__45 s = num__1.5 ds = num__6 us = num__3 as = ( num__2 * num__6 * num__3 ) / num__9 = num__4 answer : b <eor> b <eos> |
b |
add__4.5__1.5__ subtract__4.5__1.5__ divide__3.0__1.5__ multiply__4.5__2.0__ round_down__4.5__ round_down__4.5__ |
add__4.5__1.5__ divide__4.5__1.5__ divide__3.0__1.5__ multiply__4.5__2.0__ divide__6.0__1.5__ divide__6.0__1.5__ |
| a cycle is bought for rs . num__900 and sold for rs . num__1140 find the gain percent ? <o> a ) num__11 <o> b ) num__27 <o> c ) num__99 <o> d ) num__77 <o> e ) num__18 |
num__900 - - - - num__240 num__100 - - - - ? = > num__27.0 answer : b <eor> b <eos> |
b |
percent__100.0__27.0__ |
percent__100.0__27.0__ |
| in one year the population of a village increased by num__30.0 and in the next year it decreased by num__30.0 . if at the end of num__2 nd year the population was num__13650 what was it in the beginning ? <o> a ) num__7787 <o> b ) num__8000 <o> c ) num__15000 <o> d ) num__1277 <o> e ) num__2081 |
x * num__1.3 * num__0.7 = num__13650 x * num__0.91 = num__13650 x = num__13650 / num__0.91 = > num__15000 answer : c <eor> c <eos> |
c |
subtract__2.0__1.3__ multiply__0.7__1.3__ divide__13650.0__0.91__ divide__13650.0__0.91__ |
subtract__2.0__1.3__ multiply__0.7__1.3__ divide__13650.0__0.91__ divide__13650.0__0.91__ |
| the equation x = num__2 y ^ num__2 + num__5 y - num__19 describes a parabola in the xy coordinate plane . if line l with slope of num__3 intersects the parabola in the upper - left quadrant at x = - num__5 the equation for l is <o> a ) num__3 x + y + num__15 = num__0 <o> b ) y - num__3 x - num__11 = num__0 <o> c ) - num__3 x + y - num__19 = num__0 <o> d ) - num__2 x - y - num__7 = num__0 <o> e ) - num__3 x + y + num__13.5 = num__0 |
the question is made to look difficult though it is pretty simple if you focus on just the line and use process of elimination . ( remember that gmat does not focus on parabolas so basically the question should be quite do - able even if someone does n ' t know how to handle parabolas . ) we need equation of l . its slope must be num__3 . slope in option a and d is not num__3 so we are left with b c and e the line has a point ( - num__5 y ) on it where y is positive ( since the point lies in upper left quadrant ) . in options b and e if you put x = - num__5 you get - ve value for y co - ordinate . so ignore them . answer must be ( c ) <eor> c <eos> |
c |
subtract__5.0__2.0__ |
subtract__5.0__2.0__ |
| a man buys num__50 pens at marked price of num__46 pens from a whole seller . if he sells these pens giving a discount of num__1.0 what is the profit percent ? <o> a ) num__7.6 <o> b ) num__7.7 <o> c ) num__7.32 <o> d ) num__3.6 <o> e ) num__7.8 % |
explanation : let marked price be re . num__1 each c . p . of num__50 pens = rs . num__46 s . p . of num__50 pens = num__99.0 of rs . num__50 = rs . num__49.50 profit % = ( profit / c . p . ) x num__100 profit % = ( num__3.50 / num__46 ) x num__100 = num__7.6 answer : a <eor> a <eos> |
a |
percent__50.0__99.0__ percent__100.0__7.6__ |
percent__50.0__99.0__ percent__100.0__7.6__ |
| if a tap could fill entire tank in num__20 hrs due to leakage then in how much time tank can be emptied by leakage if tap can fill entire tank in num__12 hrs without leakage <o> a ) num__30 hrs <o> b ) num__24 hrs <o> c ) num__36 hrs <o> d ) num__48 hrs <o> e ) num__52 hrs |
time take to fill [ withleakage ] = num__20 hrs so workdone in num__1 hr = num__0.05 time tkae to fill [ without leakage ] = num__12 hrs so workdone in num__1 hr = num__0.0833333333333 if u subtract both u ' ll get time taken by leakage to empty . . . num__0.05 - num__0.0833333333333 = num__0.0333333333333 = num__0.0333333333333 so num__30 hrs answer : a <eor> a <eos> |
a |
reverse__20.0__ reverse__12.0__ subtract__0.0833__0.05__ multiply__1.0__30.0__ |
reverse__20.0__ reverse__12.0__ subtract__0.0833__0.05__ multiply__1.0__30.0__ |
| what is the tenth digit of ( num__5 ! * num__5 ! - num__5 ! * num__3 ! ) / num__5 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__0 <o> d ) num__6 <o> e ) num__7 |
( num__5 ! * num__5 ! + num__5 ! * num__3 ! ) / num__5 = num__5 ! ( num__5 ! + num__3 ! ) / num__5 = num__120 ( num__120 + num__6 ) / num__5 = num__3024 units digit of the above product will be equal to num__2 answer b <eor> b <eos> |
b |
subtract__5.0__3.0__ subtract__5.0__3.0__ |
divide__6.0__3.0__ divide__6.0__3.0__ |
| a man can row at num__5 kmph in still water . if the velocity of the current is num__1 kmph and it takes him num__1 hour to row to a place and come back . how far is that place . <o> a ) num__4 km <o> b ) num__1.4 km <o> c ) num__2.4 km <o> d ) num__3.4 km <o> e ) num__4.4 km |
explanation : let the distance is x km rate downstream = num__5 + num__1 = num__6 kmph rate upstream = num__5 - num__1 = num__4 kmph then x / num__6 + x / num__4 = num__1 [ because distance / speed = time ] = > num__2 x + num__3 x = num__12 = > x = num__2.4 = num__2.4 km answer is c <eor> c <eos> |
c |
add__5.0__1.0__ subtract__5.0__1.0__ subtract__6.0__4.0__ subtract__5.0__2.0__ multiply__2.0__6.0__ divide__12.0__5.0__ round__2.4__ |
add__5.0__1.0__ subtract__5.0__1.0__ subtract__6.0__4.0__ subtract__5.0__2.0__ multiply__2.0__6.0__ divide__12.0__5.0__ divide__2.4__1.0__ |
| if the quantity num__3 ^ num__4 + num__3 ^ num__6 + num__3 ^ num__8 is written as ( a + b ) ( a – b ) in which both a and b are integers which of the following could be the value of b ? <o> a ) num__57 <o> b ) num__26 <o> c ) num__52 <o> d ) num__81 <o> e ) num__25 |
( a + b ) ( a - b ) = a ^ num__2 - b ^ num__2 num__3 ^ num__4 + num__3 ^ num__6 + num__3 ^ num__8 = num__3 ^ num__4 ( num__1 + num__3 ^ num__2 + num__3 ^ num__4 ) if the above expression is solved even then every term of the expression will remain a multiple of num__3 ^ num__4 which is out of parenthesis hence b must be a multiple of num__3 ^ num__4 i . e . num__81 answer : option d <eor> d <eos> |
d |
divide__6.0__3.0__ subtract__3.0__2.0__ multiply__1.0__81.0__ |
subtract__6.0__4.0__ subtract__3.0__2.0__ multiply__1.0__81.0__ |
| balls of equal size are arranged in rows to form an equilateral triangle . the top most row consists of one ball the num__2 nd row of two balls and so on . if num__19 balls are added then all the balls can be arranged in the shape of square and each of the sides of the square contain num__8 balls less than the each side of the triangle did . how many balls made up the triangle ? <o> a ) num__465 <o> b ) num__2209 <o> c ) num__2878 <o> d ) num__1210 <o> e ) num__1560 |
as expected this question boils down to num__2 equation consider total number of balls in triangle = t and number of balls in last row = x . num__1 + num__2 + num__3 + . . . + x = t x ( x + num__1 ) / num__2 = t - - - - ( a ) as mentioned in the question side of a square will be ( x - num__8 ) and total number of balls in square will be ( t + num__19 ) ( x - num__8 ) ^ num__2 = t + num__19 - - - - - ( b ) now the hardest part of the question will be to solve these num__2 equations and this looks like time consuming but the easy way will be plug and play . also we ' ve to find a value of t ( from num__5 optiosn given below ) which can make a square of a a number . one we know this it will be a cake walk . we can see that option a fits this criteria in eq ( b ) . add - num__465 + num__19 = num__484 = num__22 ^ num__2 = ( x - num__8 ) ^ num__2 hence x = num__30 cross check by putting in eq ( a ) = x ( x + num__1 ) / num__2 = t = > num__30 * num__15.5 = num__465 hence answer is a . <eor> a <eos> |
a |
add__2.0__1.0__ add__2.0__3.0__ add__19.0__465.0__ add__19.0__3.0__ add__8.0__22.0__ divide__465.0__30.0__ multiply__15.5__30.0__ |
add__2.0__1.0__ add__2.0__3.0__ add__19.0__465.0__ add__19.0__3.0__ add__8.0__22.0__ divide__465.0__30.0__ multiply__15.5__30.0__ |
| the average of a couple was num__28 yrs when they were married num__5 yrs ago . the avg age of the couple and a child who was born during the interval is num__23 yrs now . how old is the child now ? <o> a ) num__2 yrs <o> b ) num__4 yrs <o> c ) num__3 yrs <o> d ) num__1 yrs <o> e ) none of these |
( a + b - num__5 - num__5 ) / num__2 = num__28 ; a + b = num__66 ; a + b + c / num__3 = num__23 ; solving both c = num__3 answer : c <eor> c <eos> |
c |
subtract__5.0__2.0__ subtract__5.0__2.0__ |
subtract__5.0__2.0__ subtract__5.0__2.0__ |
| if num__8 men can reap num__80 hectares in num__24 days then how many hectares can num__36 men reap in num__30 days ? <o> a ) num__127 <o> b ) num__237 <o> c ) num__287 <o> d ) num__450 <o> e ) num__281 |
explanation : let the required no of hectares be x . then more men more hectares ( direct proportion ) more days more hectares ( direct proportion ) men \ : num__8 : num__36 \ \ days \ : num__24 : num__30 \ end { matrix } \ right \ } : num__80 : x \ inline \ fn _ jvn \ therefore \ inline \ fn _ jvn num__8 \ times num__24 \ times x = num__36 \ times num__30 \ times num__80 \ inline \ fn _ jvn \ leftrightarrow \ inline \ fn _ jvn x = \ frac { num__36 \ times num__30 \ times num__80 } { num__8 \ times num__24 } \ inline \ fn _ jvn \ leftrightarrow x = num__450 answer : d <eor> d <eos> |
d |
round__450.0__ |
round__450.0__ |
| what proximate value should come in place of the question mark ( ? ) in the following question ? num__6.695 Ã — num__1084 + num__2568.34 â € “ num__1708.34 = ? <o> a ) num__6000 <o> b ) num__12000 <o> c ) num__10000 <o> d ) num__8096 <o> e ) num__9 |
000 |
? â ‰ ˆ num__6.7 Ã — num__1080 + num__2560 â € “ num__1700 â ‰ ˆ num__7236 + num__860 â ‰ ˆ num__8096 answer d <eor> d <eos> |
d |
d |
| the average of first num__6 natural numbers is ? <o> a ) num__5.2 <o> b ) num__3.5 <o> c ) num__5.3 <o> d ) num__5.9 <o> e ) num__5.1 |
sum of num__6 natural no . = num__21.0 = num__21 average = num__3.5 = num__3.5 answer : b <eor> b <eos> |
b |
divide__21.0__6.0__ divide__21.0__6.0__ |
divide__21.0__6.0__ divide__21.0__6.0__ |
| twenty four men can do a work in num__48 days . how many men are required to complete the work in num__12 days ? <o> a ) num__85 <o> b ) num__46 <o> c ) num__50 <o> d ) num__96 <o> e ) num__60 |
d num__96 we have m num__1 d num__1 = m num__2 d num__2 so num__24 * num__48 = m num__2 * num__12 = > m num__2 = num__96 . <eor> d <eos> |
d |
divide__96.0__48.0__ divide__48.0__2.0__ round__96.0__ |
divide__96.0__48.0__ multiply__12.0__2.0__ multiply__48.0__2.0__ |
| a train num__360 m long is running at a speed of num__45 kmph . in what time will it pass a bridge num__140 m long ? <o> a ) num__40 sec <o> b ) num__42 sec <o> c ) num__45 sec <o> d ) num__48 sec <o> e ) num__49 sec |
speed of the train num__45 * num__0.277777777778 = num__12.5 m / sec time = ( num__360 + num__140 ) / num__12.5 = num__500 * num__0.08 = num__40 sec answer : a <eor> a <eos> |
a |
add__360.0__140.0__ divide__500.0__12.5__ round__40.0__ |
add__360.0__140.0__ divide__500.0__12.5__ divide__500.0__12.5__ |
| kumar saves num__32.0 of his monthly salary . if he spends rs . num__27200 then find his savings ? <o> a ) num__15400 <o> b ) num__14300 <o> c ) num__13200 <o> d ) num__12800 <o> e ) num__9500 |
answer : option d let the monthly salary of kumar be rs . x . num__68.0 of x = num__27200 = > x = ( num__27200 * num__100 ) / num__68 = num__40000 his savings = num__0.32 * num__40000 = num__12800 . <eor> d <eos> |
d |
percent__32.0__40000.0__ percent__32.0__40000.0__ |
percent__32.0__40000.0__ percent__32.0__40000.0__ |
| when a whole no . n is divided by num__4 we will get num__3 as remainder . what will be the remainder if num__2 n is divided by num__4 ? <o> a ) num__1 <o> b ) num__4 <o> c ) num__2 <o> d ) num__6 <o> e ) num__7 |
let n ÷ num__4 = p remainder = num__3 = > n = num__4 p + num__3 num__2 n = num__2 ( num__4 p + num__3 ) = num__8 p + num__6 = num__8 p + num__4 + num__2 = num__4 ( num__2 p + num__1 ) + num__2 hence if num__2 n is divided by num__4 we will get num__2 as remainder . c <eor> c <eos> |
c |
multiply__4.0__2.0__ add__4.0__2.0__ subtract__4.0__3.0__ subtract__4.0__2.0__ |
multiply__4.0__2.0__ add__4.0__2.0__ subtract__4.0__3.0__ subtract__4.0__2.0__ |
| to mail a package the rate is num__10 cents for the first pound and num__5 cents for each additional pound . two packages weighing num__4 pounds and num__6 pounds respectively can be mailed seperately or combined as one package . which method is cheaper and how much money is saved ? <o> a ) combined with a saving of num__15 cents <o> b ) combined with a saving of num__10 cents <o> c ) combined with a saving of num__5 cents <o> d ) separately with a saving of num__5 cents <o> e ) separately with a saving of num__15 cents |
num__4 pounds cost = num__10 + num__3 * num__5 = num__25 num__6 pounds cost = num__10 + num__5 * num__5 = num__35 total = num__60 num__10 pounds cost = num__10 + num__9 * num__5 = num__55 answer : c <eor> c <eos> |
c |
add__10.0__25.0__ multiply__10.0__6.0__ add__5.0__4.0__ subtract__60.0__5.0__ subtract__10.0__5.0__ |
add__10.0__25.0__ multiply__10.0__6.0__ add__5.0__4.0__ subtract__60.0__5.0__ subtract__10.0__5.0__ |
| the probability of sam passing the exam is num__0.2 . the probability of sam passing the exam and michael passing the driving test is num__0.166666666667 . what is the probability of michael passing his driving test ? <o> a ) num__0.833333333333 . <o> b ) num__0.5 . <o> c ) num__0.333333333333 . <o> d ) num__0.666666666667 . <o> e ) num__0.4 |
num__0.2 * num__1 / m = num__0.166666666667 num__1 / m = num__0.833333333333 answer a <eor> a <eos> |
a |
subtract__1.0__0.1667__ subtract__1.0__0.1667__ |
subtract__1.0__0.1667__ multiply__1.0__0.8333__ |
| if the compound interest on an amount of rs . num__29000 in two years is rs . num__9352.5 what is the rate of interest ? <o> a ) num__11 <o> b ) num__9 <o> c ) num__15 <o> d ) num__18 <o> e ) num__25 |
let interest rate be x . ( num__2 x + ( x ^ num__2 ) / num__100 ) * ( num__0.01 ) * num__29000 = num__9352.5 x = num__15 answer : c <eor> c <eos> |
c |
percent__100.0__15.0__ |
percent__100.0__15.0__ |
| my watch gains num__5 min every hour . how many degrees the second hand moves in every min ? <o> a ) num__300 <o> b ) num__350 <o> c ) num__390 <o> d ) num__400 <o> e ) num__450 |
gain num__5 min every num__1 hr degrees moves = = > num__390 answer c <eor> c <eos> |
c |
round__390.0__ |
round__390.0__ |
| carol is three times alice ’ s age but only twice as old as betty . alice is twenty - four years younger than carol . how old is betty ? <o> a ) num__26 <o> b ) num__19 <o> c ) num__22 <o> d ) num__18 <o> e ) num__24 |
beyond the obvious algebra approach to this question it can also be solved by testing the answers . we ' re told num__3 facts about the relative ages of num__3 people : num__1 ) carol ' s age is num__3 times alice ' s age num__2 ) carol ' s age is num__2 times betty ' s age num__3 ) carol is num__24 years older than alice we ' re asked for betty ' s age . from the answer choices and the information provided carol ca n ' t be that old ( the difference of num__12 years = num__3 times ; that relationship can only occur when the numbers are relatively small ) . since carol is twice betty ' s age betty is clearly younger than carol so we ' ll test a smaller answer first . if . . . . betty = num__18 then carol = num__2 ( num__18 ) = num__36 then alice = num__36 - num__24 = num__12 and num__12.0 = num__12 all of these values mesh perfectly with the facts and with one another so betty must be num__18 d <eor> d <eos> |
d |
subtract__3.0__1.0__ divide__24.0__2.0__ multiply__2.0__18.0__ multiply__1.0__18.0__ |
subtract__3.0__1.0__ divide__24.0__2.0__ multiply__2.0__18.0__ subtract__36.0__18.0__ |
| if the sides of a triangle are num__9 cm num__16 cm and num__19 cm what is its area ? <o> a ) num__69 <o> b ) num__72 <o> c ) num__75 <o> d ) num__89 <o> e ) num__93 |
the triangle with sides num__9 cm num__16 cm and num__19 cm is right angled where the hypotenuse is num__19 cm . area of the triangle = num__0.5 * num__9 * num__16 = num__72 cm num__2 answer : option b <eor> b <eos> |
b |
triangle_area__9.0__16.0__ square_perimeter__0.5__ triangle_area__9.0__16.0__ |
volume_rectangular_prism__9.0__16.0__0.5__ square_perimeter__0.5__ volume_rectangular_prism__9.0__16.0__0.5__ |
| three candidates contested an election and received num__5000 num__22000 and num__34800 votes respectively . what percentage of the total votes did the winning candidate get ? <o> a ) num__56.31 <o> b ) num__45.0 <o> c ) num__75.2 <o> d ) num__86.0 <o> e ) num__95.15 % |
explanation : total number of votes polled = ( num__5000 + num__22000 + num__34800 ) = num__61800 . required percentage = ( num__0.563106796117 ) x num__100 = num__56.31 . answer : a <eor> a <eos> |
a |
divide__34800.0__61800.0__ multiply__100.0__0.5631__ multiply__100.0__0.5631__ |
divide__34800.0__61800.0__ multiply__100.0__0.5631__ multiply__100.0__0.5631__ |
| by selling an article at rs . num__500 a profit of num__25.0 is made . find its cost price ? <o> a ) num__400 <o> b ) num__267 <o> c ) num__287 <o> d ) num__480 <o> e ) num__811 |
sp = num__500 cp = ( sp ) * [ num__100 / ( num__100 + p ) ] = num__500 * [ num__100 / ( num__100 + num__25 ) ] = num__500 * [ num__0.8 ] = rs . num__400 answer : a <eor> a <eos> |
a |
percent__100.0__400.0__ |
percent__100.0__400.0__ |
| a group of college friends plan to rent a house together for x dollars per month which is to be shared equally among them . if four rather than six people rent the house together how many more dollars in terms of x will each person have to pay for rent per month ? <o> a ) x / num__4 <o> b ) x / num__6 <o> c ) x / num__12 <o> d ) x / num__10 <o> e ) x / num__24 |
if num__6 people share the rent each person would pay x / num__6 . if num__4 people share the rent each person will pay x / num__4 . the extra amount each person pays is x / num__4 - x / num__6 = num__3 x / num__12 - num__2 x / num__12 = x / num__12 the answer is c . <eor> c <eos> |
c |
multiply__3.0__4.0__ divide__12.0__6.0__ multiply__2.0__6.0__ |
multiply__3.0__4.0__ divide__12.0__6.0__ multiply__2.0__6.0__ |
| the average monthly salary of num__8 workers and one supervisor in a factory was num__430 . @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss num__430 . @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss num__430 . whenthesupervisor whosesalarywas num__430 . when the supervisor whose salary was num__870 per month retired a new person was appointed and then the average salary of num__9 people was $ $ num__430 per month . the salary of the new supervisor is : <o> a ) num__870 <o> b ) num__600 <o> c ) num__287 <o> d ) num__771 <o> e ) num__191 |
explanation : total salary of num__8 workers and supervisor together = num__9 Ã — num__430 = num__3870 now total salary of num__8 workers = num__3870 â ˆ ’ num__870 = num__3000 total salary of num__9 workers including the new supervisor = num__9 Ã — num__430 = num__3870 salary of the new supervisor = num__3870 â ˆ ’ num__3000 = num__870 answer : a <eor> a <eos> |
a |
multiply__430.0__9.0__ subtract__3870.0__870.0__ subtract__3870.0__3000.0__ |
multiply__430.0__9.0__ subtract__3870.0__870.0__ subtract__3870.0__3000.0__ |
| a rectangular park num__60 m long and num__40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn . if the area of the lawn is num__2109 sq . m then what is the width of the road ? <o> a ) num__2.91 m <o> b ) num__3 m <o> c ) num__5.82 m <o> d ) num__4.32 m <o> e ) none of these |
area of the park = ( num__60 x num__40 ) m num__2 = num__2400 m num__2 . area of the lawn = num__2109 m num__2 . area of the crossroads = ( num__2400 - num__2109 ) m num__2 = num__291 m num__2 . let the width of the road be x metres . then num__60 x + num__40 x - x num__2 = num__291 x num__2 - num__100 x + num__291 = num__0 ( x - num__97 ) ( x - num__3 ) = num__0 x = num__3 . answer : option b <eor> b <eos> |
b |
multiply__60.0__40.0__ triangle_area__2.0__3.0__ |
multiply__60.0__40.0__ triangle_area__2.0__3.0__ |
| lionel left his house and walked towards walt ' s house num__48 miles away . two hours later walt left his house and ran towards lionel ' s house . if lionel ' s speed was num__3 miles per hour and walt ' s num__4 miles per hour how many miles had lionel walked when he met walt ? <o> a ) num__12 <o> b ) num__16 <o> c ) num__20 <o> d ) num__24 <o> e ) num__28 |
in the first num__2 hours lionel at the rate of num__3 miles per hour covered distance = rate * time = num__3 * num__2 = num__6 miles . so the distance between him and walt was num__48 - num__6 = num__42 miles when walt left his house . now their combined rate to cover this distance was num__3 + num__4 = num__7 miles per hour hence they will meet ( they will cover that distance ) in time = distance / rate = num__6.0 = num__6 hours . total time that lionel was walking is num__2 + num__6 = num__8 hours which means that he covered in that time interval distance = rate * time = num__3 * num__8 = num__24 miles . answer : d . <eor> d <eos> |
d |
multiply__3.0__2.0__ subtract__48.0__6.0__ add__3.0__4.0__ divide__48.0__6.0__ divide__48.0__2.0__ round__24.0__ |
multiply__3.0__2.0__ subtract__48.0__6.0__ add__3.0__4.0__ divide__48.0__6.0__ divide__48.0__2.0__ divide__48.0__2.0__ |
| the average weight of num__5 person ' s increases by num__4 kg when a new person comes in place of one of them weighing num__50 kg . what is the weight of the new person ? <o> a ) num__70 <o> b ) num__86.5 <o> c ) num__80 <o> d ) num__88.5 <o> e ) num__75 |
total increase in weight = num__5 × num__4 = num__20 if x is the weight of the new person total increase in weight = x − num__50 = > num__20 = x - num__50 = > x = num__20 + num__50 = num__70 answer : a <eor> a <eos> |
a |
multiply__5.0__4.0__ add__50.0__20.0__ add__50.0__20.0__ |
multiply__5.0__4.0__ add__50.0__20.0__ add__50.0__20.0__ |
| what is the units digit of num__63 ^ num__3 * num__17 ^ num__3 * num__49 ^ num__3 ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__9 |
the units digit of num__63 ^ num__3 is the units digit of num__3 * num__3 * num__3 = num__27 which is num__7 . the units digit of num__17 ^ num__3 is the units digit of num__7 * num__7 * num__7 = num__343 which is num__3 . the units digit of num__49 ^ num__3 is the units digit of num__9 * num__9 * num__9 = num__729 which is num__9 . the units digit of num__7 * num__3 * num__9 = num__189 is num__9 . the answer is e . <eor> e <eos> |
e |
multiply__49.0__7.0__ divide__63.0__7.0__ multiply__63.0__3.0__ divide__63.0__7.0__ |
multiply__49.0__7.0__ divide__63.0__7.0__ multiply__63.0__3.0__ divide__63.0__7.0__ |
| what is the present worth of rs . num__176 due in num__2 years at num__5.0 simple interest per annum ? <o> a ) num__288 <o> b ) num__160 <o> c ) num__120 <o> d ) num__277 <o> e ) num__6123 |
let the present worth be rs . x . then s . i . = ( num__176 - x ) ( x * num__5 * num__2 ) / num__100 = ( num__176 - x ) num__10 x = num__17600 - num__100 x num__110 x = num__17600 = > x = num__160 . answer : b <eor> b <eos> |
b |
percent__100.0__160.0__ |
percent__100.0__160.0__ |
| if log num__27 = num__1.431 then the value of log num__9 is : num__0.934 b . num__0.945 c . num__0.954 d . num__0.958 <o> a ) num__0.934 <o> b ) num__0.945 <o> c ) num__0.954 <o> d ) num__0.958 <o> e ) num__0.964 |
log num__27 = num__1.431 log ( num__3 ^ num__3 ) = num__1.431 num__3 log num__3 = num__1.431 log num__3 = num__0.477 log num__9 = log ( num__3 ^ num__2 ) = num__2 log num__3 = ( num__2 x num__0.477 ) = num__0.954 . answer is c . <eor> c <eos> |
c |
divide__27.0__9.0__ subtract__1.431__0.954__ divide__0.954__0.477__ subtract__1.431__0.477__ |
divide__27.0__9.0__ subtract__1.431__0.954__ divide__0.954__0.477__ subtract__1.431__0.477__ |
| there are num__28 girls in hostel whose average age is decreased by num__2 months when one girl num__18 yrs is replaced by a new boy . find the age of new girl . <o> a ) num__1 year <o> b ) num__1 year num__2 months <o> c ) num__4 years num__8 months <o> d ) num__3 years <o> e ) none |
total decrease = num__28 * num__2 = num__56 months = num__4 years num__8 months c <eor> c <eos> |
c |
multiply__28.0__2.0__ multiply__2.0__4.0__ divide__8.0__2.0__ |
multiply__28.0__2.0__ multiply__2.0__4.0__ divide__8.0__2.0__ |
| a tap can fill a tank in num__2 hours . after half the tank is filled two more similar taps are opened . what is the total time taken to fill the tank completely ? <o> a ) num__1 hr num__40 min <o> b ) num__4 hr <o> c ) num__3 hr <o> d ) num__1 hr num__20 min <o> e ) num__2 hr num__50 min |
explanation : num__1 tap can fill the tank in num__2 hours . therefore num__1 tap can fill half tank in num__1 hours . num__3 taps can fill the tank in num__0.666666666667 hour . therefore num__3 taps can fill half tank in num__0.333333333333 hour = num__20 minutes . total time taken = num__1 hour num__20 minutes answer : option d <eor> d <eos> |
d |
add__2.0__1.0__ divide__2.0__3.0__ divide__1.0__3.0__ round__1.0__ |
add__2.0__1.0__ divide__2.0__3.0__ divide__1.0__3.0__ round__1.0__ |
| given that e and f are events such that p ( e ) = num__8 p ( f ) = num__2 and p ( e n f ) = num__8 find p ( e | f ) and p ( f | e ) <o> a ) num__5 <o> b ) num__4 <o> c ) num__12 <o> d ) num__36 <o> e ) num__98 |
here e and f are events p ( e | f ) = p ( enf ) / p ( f ) = num__1.0 = num__1 p ( f | e ) = p ( enf ) / p ( e ) = num__4.0 = num__4 . b ) <eor> b <eos> |
b |
divide__8.0__2.0__ divide__8.0__2.0__ |
divide__8.0__2.0__ divide__8.0__2.0__ |
| two good train each num__900 m long are running in opposite directions on parallel tracks . their speeds are num__45 km / hr and num__30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one . <o> a ) num__12.2 sec <o> b ) num__24.9 sec <o> c ) num__86.4 sec <o> d ) num__60.2 sec <o> e ) none |
sol . relative speed = ( num__45 + num__30 ) km / hr = ( num__75 x num__0.277777777778 ) m / sec = ( num__20.8333333333 ) m / sec . distance covered = ( num__900 + num__900 ) m = num__1000 m . required time = ( num__1800 x num__0.048 ) sec = num__86.4 sec . answer c <eor> c <eos> |
c |
add__45.0__30.0__ multiply__0.048__1800.0__ round__86.4__ |
add__45.0__30.0__ multiply__0.048__1800.0__ round__86.4__ |
| a train num__360 m long is running at a speed of num__45 km / hr . in what time will it pass a bridge num__140 m long ? <o> a ) num__40 <o> b ) num__99 <o> c ) num__88 <o> d ) num__66 <o> e ) num__21 |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec total distance covered = num__360 + num__140 = num__500 m required time = num__500 * num__0.08 = num__40 sec . answer : a <eor> a <eos> |
a |
add__360.0__140.0__ divide__500.0__12.5__ round__40.0__ |
add__360.0__140.0__ divide__500.0__12.5__ divide__500.0__12.5__ |
| a man has rs . num__416 in the denominations of one - rupee notes five - rupee notes and ten rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__45 <o> b ) num__60 <o> c ) num__78 <o> d ) num__90 <o> e ) num__95 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__416 num__16 x = num__416 x = num__26 . hence total number of notes = num__3 x = num__78 answer : option c <eor> c <eos> |
c |
divide__416.0__16.0__ multiply__3.0__26.0__ multiply__3.0__26.0__ |
add__10.0__16.0__ multiply__3.0__26.0__ multiply__3.0__26.0__ |
| pipe a can fill a tank in num__3 hours . due to a leak at the bottom it takes num__9 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? <o> a ) num__4.5 <o> b ) num__17 <o> c ) num__18 <o> d ) num__19 <o> e ) num__12 |
let the leak can empty the full tank in x hours num__0.333333333333 - num__1 / x = num__0.111111111111 = > num__1 / x = num__0.333333333333 - num__0.111111111111 = ( num__3 - num__1 ) / num__9 = num__0.222222222222 = > x = num__4.5 = num__4.5 . answer : a <eor> a <eos> |
a |
divide__3.0__9.0__ divide__1.0__9.0__ subtract__0.3333__0.1111__ round__4.5__ |
divide__3.0__9.0__ divide__1.0__9.0__ subtract__0.3333__0.1111__ subtract__9.0__4.5__ |
| if the product num__4864 x num__9 p num__2 is divisible by num__12 then the value of p is : <o> a ) num__1 <o> b ) num__5 <o> c ) num__8 <o> d ) num__6 <o> e ) num__4 |
explanation : clearly num__4864 is divisible by num__4 . so num__9 p num__2 must be divisible by num__3 . so ( num__9 + p + num__2 ) must be divisible by num__3 . p = num__1 . answer a <eor> a <eos> |
a |
subtract__12.0__9.0__ subtract__3.0__2.0__ reverse__1.0__ |
subtract__12.0__9.0__ subtract__3.0__2.0__ reverse__1.0__ |
| a man has some hens and cows . if the number of heads be num__48 and the number of feet equals num__140 then the number of hens will be <o> a ) num__26 <o> b ) num__27 <o> c ) num__28 <o> d ) num__29 <o> e ) num__30 |
let number of hens = h and number of cows = c number of heads = num__48 = > h + c = num__48 - - - ( equation num__1 ) number of feet = num__140 = > num__2 h + num__4 c = num__140 = > h + num__2 c = num__70 - - - ( equation num__2 ) ( equation num__2 ) - ( equation num__1 ) gives num__2 c - c = num__70 - num__48 = > c = num__22 substituting the value of c in equation num__1 we get h + num__22 = num__48 = > h = num__48 - num__22 = num__26 i . e . number of hens = num__26 answer is a . <eor> a <eos> |
a |
divide__140.0__2.0__ subtract__70.0__48.0__ subtract__48.0__22.0__ subtract__48.0__22.0__ |
divide__140.0__2.0__ subtract__70.0__48.0__ subtract__48.0__22.0__ subtract__48.0__22.0__ |
| if the true discount on s sum due num__2 years hence at num__14.0 per annum be rs . num__182 the sum due is : <o> a ) s . num__832 <o> b ) s . num__968 <o> c ) s . num__1960 <o> d ) s . num__2400 <o> e ) s . num__2800 |
td = pw * r * t / num__100 so num__182 = pw * num__14 * num__0.02 so pw = num__650 sum = pw + td . . sum = num__650 + num__182 = num__832 answer : a <eor> a <eos> |
a |
percent__100.0__832.0__ |
percent__100.0__832.0__ |
| a train covers a distance of num__12 km in num__10 min . if it takes num__6 sec to pass a telegraph post then the length of the train is ? <o> a ) num__288 <o> b ) num__277 <o> c ) num__120 <o> d ) num__285 <o> e ) num__121 |
speed = ( num__1.2 * num__60 ) km / hr = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . length of the train = num__20 * num__6 = num__120 m . answer : c <eor> c <eos> |
c |
divide__12.0__10.0__ hour_to_min_conversion__ multiply__12.0__6.0__ multiply__12.0__10.0__ round__120.0__ |
divide__12.0__10.0__ multiply__10.0__6.0__ multiply__12.0__6.0__ multiply__12.0__10.0__ multiply__12.0__10.0__ |
| the average mark of num__20 students in a class is num__70 years . the average age of num__10 students is num__65 . what is the total mark of remaining num__10 students ? <o> a ) num__850 <o> b ) num__800 <o> c ) num__750 <o> d ) num__700 <o> e ) num__680 |
sum of the ages of num__10 students = ( num__20 * num__70 ) - ( num__10 * num__65 ) = num__1400 - num__650 = num__750 required total marks = num__750 answer : c <eor> c <eos> |
c |
multiply__20.0__70.0__ multiply__10.0__65.0__ subtract__1400.0__650.0__ subtract__1400.0__650.0__ |
multiply__20.0__70.0__ multiply__10.0__65.0__ subtract__1400.0__650.0__ subtract__1400.0__650.0__ |
| a coin has two sides . one side has the number num__1 on it and the other side has the number num__2 on it . if the coin is flipped three times what is the probability that the sum of the numbers on the landing side of the coin will be greater than num__6 ? <o> a ) num__0.375 <o> b ) num__0.0625 <o> c ) num__0.125 <o> d ) num__0.5 <o> e ) num__0 |
one approach to solve the problem is to list the different possibilities for a toss of coin three times . because there are two outcomes and the coin is tossed three times the table will have num__2 * num__2 * num__2 or num__8 rows . next add the resulting rows together to find the sum ( the fourth column in the table below ) . toss num__1 | toss num__2 | toss num__3 | sum num__1 - - - - - - - - - - num__1 - - - - - - - - num__1 - - - - - - num__3 num__1 - - - - - - - - - - num__1 - - - - - - - - num__2 - - - - - - num__4 num__1 - - - - - - - - - - num__2 - - - - - - - - num__1 - - - - - - num__4 num__1 - - - - - - - - - - num__2 - - - - - - - - num__2 - - - - - - num__5 num__2 - - - - - - - - - - num__1 - - - - - - - - num__1 - - - - - - num__4 num__2 - - - - - - - - - - num__1 - - - - - - - - num__2 - - - - - - num__5 num__2 - - - - - - - - - - num__2 - - - - - - - - num__1 - - - - - - num__5 num__2 - - - - - - - - - - num__2 - - - - - - - - num__2 - - - - - - num__6 from the table we see that there are num__0 situations where the sum of the tosses will be greater than num__6 . and there are num__8 possible combinations resulting in a probability of num__0.0 or a probability of num__0 . so the correct answer is e . <eor> e <eos> |
e |
vowel_space__ negate_prob__1.0__ negate_prob__1.0__ |
vowel_space__ negate_prob__1.0__ negate_prob__1.0__ |
| if num__4 and num__8 are factors of num__60 n what is the minimum value of n ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__8 <o> d ) num__14 <o> e ) num__56 |
num__60 n / num__4 * num__8 should be integer = > num__2 * num__2 * num__3 * num__5 * n / num__2 * num__2 * num__2 * num__2 * num__2 = num__3 * num__5 * n / num__8 must be an integer for this to be true n must multiple of num__8 thus min of n = num__8 hence c <eor> c <eos> |
c |
divide__8.0__4.0__ subtract__8.0__3.0__ lcm__4.0__8.0__ |
divide__8.0__4.0__ subtract__8.0__3.0__ multiply__4.0__2.0__ |
| in how many ways can be num__3 boys and num__3 girls sit around circular table so that no two boys sit next to each other ? <o> a ) ( num__5 ! ) ^ num__2 <o> b ) ( num__6 ! ) ^ num__2 <o> c ) num__3 ! num__2 ! <o> d ) num__11 ! <o> e ) ( num__5 ! ) ^ num__2 * num__6 ! |
first fix one boy and place other num__2 in alt seats so total ways is num__3 ! now place each girl between a pair of boys . . . total ways of seating arrangement of girls num__3 ! total is num__2 ! * num__3 ! ans c <eor> c <eos> |
c |
coin_space__ choose__3.0__2.0__ |
coin_space__ choose__3.0__2.0__ |
| if ( x + yi ) / i = ( num__7 + num__9 i ) where x and y are real what is the value of ( x + yi ) ( x - yi ) ? <o> a ) num__120 <o> b ) num__130 <o> c ) num__150 <o> d ) num__170 <o> e ) num__200 |
( x + yi ) / i = ( num__7 + num__9 i ) ( x + yi ) = i ( num__7 + num__9 i ) = - num__9 + num__7 i ( x + yi ) ( x - yi ) = ( - num__9 + num__7 i ) ( - num__9 - num__7 i ) = num__81 + num__49 = num__130 correct answer is b ) num__130 <eor> b <eos> |
b |
add__49.0__81.0__ add__49.0__81.0__ |
add__49.0__81.0__ add__49.0__81.0__ |
| a num__600 meter long train crosses a signal post in num__40 seconds . how long will it take to cross a num__3 kilometer long bridge at the same speed ? <o> a ) num__4 min <o> b ) num__6 min <o> c ) num__8 min <o> d ) num__2 min <o> e ) num__9 min |
s = num__15.0 = num__15 mps s = num__240.0 = num__240 sec = num__4 min answer : a <eor> a <eos> |
a |
divide__600.0__40.0__ round__4.0__ |
divide__600.0__40.0__ round__4.0__ |
| the difference of the cubes of two consecutive even integers is divisible by which of the following integers ? <o> a ) num__3 <o> b ) num__6 <o> c ) num__4 <o> d ) num__5 <o> e ) num__2 |
solution : let take num__2 consecutive even numbers num__2 and num__4 . = > ( num__4 * num__4 * num__4 ) - ( num__2 * num__2 * num__2 ) = num__64 - num__8 = num__56 which is divisible by num__4 . answer c <eor> c <eos> |
c |
multiply__2.0__4.0__ subtract__64.0__8.0__ divide__8.0__2.0__ |
multiply__2.0__4.0__ subtract__64.0__8.0__ subtract__8.0__4.0__ |
| if a two - digit positive integer has its digits reversed the resulting integer differs from the original by num__54 . by how much do the two digits differ ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
given that ( num__10 a + b ) - ( num__10 b + a ) = num__54 - - > num__9 a - num__9 b = num__54 - - > a - b = num__6 . answer : d . <eor> d <eos> |
d |
divide__54.0__9.0__ divide__54.0__9.0__ |
divide__54.0__9.0__ divide__54.0__9.0__ |
| if x + y = num__280 x - y = num__200 for integers of x and y y = ? <o> a ) num__200 <o> b ) num__240 <o> c ) num__50 <o> d ) num__115 <o> e ) num__150 |
x + y = num__280 x - y = num__200 num__2 x = num__80 x = num__40 y = num__240 answer is b <eor> b <eos> |
b |
subtract__280.0__200.0__ divide__80.0__2.0__ subtract__280.0__40.0__ subtract__280.0__40.0__ |
subtract__280.0__200.0__ divide__80.0__2.0__ subtract__280.0__40.0__ subtract__280.0__40.0__ |
| how many multiples of num__7 are there between num__14 and num__343 exclusive ? <o> a ) num__48 <o> b ) num__47 <o> c ) num__46 <o> d ) num__45 <o> e ) num__44 |
num__7 * num__2 = num__14 num__7 * num__49 = num__343 total multiples = ( num__49 - num__2 ) + num__1 = num__48 exclude num__7 num__343 = num__48 - num__2 = num__46 answer is ( c ) <eor> c <eos> |
c |
divide__14.0__7.0__ divide__343.0__7.0__ subtract__49.0__1.0__ subtract__48.0__2.0__ multiply__1.0__46.0__ |
divide__14.0__7.0__ divide__343.0__7.0__ subtract__49.0__1.0__ subtract__48.0__2.0__ multiply__1.0__46.0__ |
| a luxury liner queen marry ii is transporting several cats as well as the crew ( sailors a cook and captain ) to a nearby port . altogether these passengers have num__15 heads and num__42 legs . how many cats does the ship host ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
sa ' s + co + ca + cats = num__15 . sa ' s + num__1 + num__1 + cats = num__15 or sa ' s + cats = num__13 . sa ' s ( num__2 ) + num__2 + num__2 + cats * num__4 = num__42 sa ' s * num__2 + cats * num__4 = num__38 or sa ' s + cats * num__2 = num__19 or num__13 - cats + cat * num__2 = num__19 then cats = num__6 b <eor> b <eos> |
b |
subtract__15.0__13.0__ subtract__42.0__4.0__ add__15.0__4.0__ add__2.0__4.0__ multiply__1.0__6.0__ |
subtract__15.0__13.0__ subtract__42.0__4.0__ add__15.0__4.0__ add__2.0__4.0__ multiply__1.0__6.0__ |
| what will come in place of the x in the following number series ? num__6 num__12 num__21 x num__48 <o> a ) num__33 <o> b ) num__35 <o> c ) num__39 <o> d ) num__41 <o> e ) num__42 |
( a ) the pattern is + num__6 + num__9 + num__12 + num__15 … … … . . so the missing term is = num__21 + num__12 = num__33 <eor> a <eos> |
a |
subtract__21.0__12.0__ add__6.0__9.0__ add__12.0__21.0__ add__12.0__21.0__ |
subtract__21.0__12.0__ add__6.0__9.0__ add__12.0__21.0__ add__12.0__21.0__ |
| sunil invested $ num__1800 fir num__2 years and $ num__1400 for num__3 years at the same rate of simple interest . if the total interest from these investments is $ num__810 what is the rate of interest ? <o> a ) num__12 num__0.75 % <o> b ) num__25 num__0.333333333333 % <o> c ) num__10 num__0.25 % <o> d ) num__8.0 <o> e ) num__13 num__0.5 % |
( num__1800 x num__2 xr ) / num__100 + ( num__1400 x num__3 xr ) / num__100 = num__810 num__78 r = num__810 r = num__10 num__0.25 % answer is c <eor> c <eos> |
c |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| to furnish a room in a model home an interior decorator is to select num__2 chairs and num__2 tables from the collection of chairs and tables in a warehouse that are all different from each other . if there are num__5 chairs in the warehouse and if num__150 different combinations are possible how many tables are there in the warehouse ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__15 <o> e ) num__30 |
total number of ways to choose num__2 chairs from num__5 = num__5 c num__2 = num__10 total number of combination = num__150 that means we need get num__15 combinations from the selection of tables . . . . screening through the answers . . . . num__6 c num__2 = num__15 . . . . . num__15 * num__10 = num__150 answer is num__6 . . . a <eor> a <eos> |
a |
die_space__ die_space__ |
die_space__ die_space__ |
| two kinds of vodka are mixed in the ratio num__1 : num__2 and num__2 : num__1 and they are sold fetching the profit num__10.0 and num__10.0 respectively . if the vodkas are mixed in equal ratio and the individual profit percent on them are increased by num__1.33333333333 and num__1.66666666667 times respectively then the mixture will fetch the profit of <o> a ) num__10.0 <o> b ) num__20.0 <o> c ) num__21.0 <o> d ) num__23.0 <o> e ) can not be determined |
answer : a . <eor> a <eos> |
a |
multiply__1.0__10.0__ |
multiply__1.0__10.0__ |
| a cistern is two - third full of water . pipe a can fill the remaining part in num__12 minutes and pipe b in num__8 minutes . once the cistern is emptied how much time will they take to fill it together completely ? <o> a ) num__12 minutes <o> b ) num__12.5 min <o> c ) num__14.4 min <o> d ) num__10.2 min <o> e ) num__14.66 min |
as pipe a takes num__12 min . to fill remaining one - third it takes num__36 min . to fill completely . similarly pipe b takes num__24 min . to fill completely so total time taken by both together is reciprocal of : ( num__0.0277777777778 ) + ( num__0.0416666666667 ) = num__0.0694444444444 ans : num__14.4 min . answer : c <eor> c <eos> |
c |
subtract__36.0__12.0__ round__14.4__ |
subtract__36.0__12.0__ round__14.4__ |
| in a bag containing num__3 balls a white ball was placed and then num__1 ball was taken out at random . what is the probability that the extracted ball would turn on to be white if all possible hypothesis concerning the color of theballs that initiallyin the bag were equally possible ? <o> a ) num__0.666666666667 <o> b ) num__0.625 <o> c ) num__0.6 <o> d ) num__0.571428571429 <o> e ) num__0.444444444444 |
since all possible hypothesis regarding the colour of the balls are equally likely therefore these could be num__3 white balls initially in the bag . ∴ required probability = num__0.25 [ num__1 + num__0.75 + num__0.5 + num__0.25 ] = num__0.25 [ ( num__4 + num__3 + num__2 + num__1 ) / num__4 ] = num__0.625 b <eor> b <eos> |
b |
negate_prob__0.25__ union_prob__1.0__0.25__0.75__ coin_space__ union_prob__1.0__0.25__0.625__ |
negate_prob__0.25__ union_prob__1.0__0.25__0.75__ coin_space__ union_prob__1.0__0.25__0.625__ |
| the function f is defined by subtracting num__25 from the square of a number and the function a is defined as the square root of one - half of a number . if a ( f ( x ) ) = num__10 then which of the following is a possible value of x ? <o> a ) - num__15 <o> b ) - num__5 <o> c ) num__0 <o> d ) num__5 <o> e ) num__25 |
f ( x ) = x ^ num__2 - num__25 a ( x ) = sqrt ( x / num__2 ) not sqrt ( x ) / num__2 because the question clearly says its square root of ( half of the number ) . a ( f ( x ) ) = num__10 a ( x ^ num__2 - num__25 ) = num__10 sqrt ( ( x ^ num__2 - num__25 ) / num__2 ) = num__10 = > ( x ^ num__2 - num__25 ) / num__2 = num__100 = > x ^ num__2 = num__225 = > x = num__15 or - num__15 answer is a . <eor> a <eos> |
a |
subtract__25.0__10.0__ subtract__25.0__10.0__ |
subtract__25.0__10.0__ subtract__25.0__10.0__ |
| the triplicate ratio of num__1 : num__2 is ? <o> a ) num__1 : num__64 <o> b ) num__1 : num__8 <o> c ) num__1 : num__10 <o> d ) num__3 : num__8 <o> e ) num__5 : num__64 |
num__1 ^ num__3 : num__2 ^ num__3 = num__1 : num__8 answer : b <eor> b <eos> |
b |
add__1.0__2.0__ reverse__1.0__ |
add__1.0__2.0__ reverse__1.0__ |
| in a num__1000 m race a beats b by num__200 meters or num__25 seconds . find the speed of b ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__7 <o> d ) num__6 <o> e ) num__5 |
since a beats b by num__200 m or num__25 seconds i t implies that b covers num__200 m in num__25 seconds . hence speed of b = num__8.0 = num__8 m / s . answer : a <eor> a <eos> |
a |
divide__200.0__25.0__ round__8.0__ |
divide__200.0__25.0__ divide__200.0__25.0__ |
| if each digit in the set of a = { num__01 num__23 num__45 num__67 num__89 } is exactly used once in how many ways can the digits be arranged such that the num__2 and num__5 are not adjacent ? <o> a ) num__10 ! / num__2 ! <o> b ) num__10 ! - num__9 ! <o> c ) num__9 ! <o> d ) num__10 ! <o> e ) num__8 * num__9 ! |
the answer should be the ( total arrangements ) - ( arrangements when num__2 and num__5 are adjacent ) . the number of ways to arrange num__10 units is num__10 ! if we think of num__2 and num__5 as one single unit the number of arrangements is num__9 ! note that num__2 and num__5 can have two arrangements so we multiply by num__2 to get num__2 * num__9 ! the total number of valid arrangements is num__10 ! - num__2 * num__9 ! = num__10 * num__9 ! - num__2 * num__9 ! = num__8 * num__9 ! the answer is e . <eor> e <eos> |
e |
multiply__2.0__5.0__ divide__45.0__5.0__ subtract__9.0__1.0__ multiply__1.0__8.0__ |
multiply__2.0__5.0__ subtract__10.0__1.0__ subtract__9.0__1.0__ multiply__1.0__8.0__ |
| if re . num__1 amounts to rs . num__9 over a period of num__10 years . what is the rate of simple interest ? <o> a ) num__90.0 <o> b ) num__30.0 <o> c ) num__50.0 <o> d ) num__80.0 <o> e ) num__70 % |
num__8 = ( num__1 * num__10 * r ) / num__100 r = num__80.0 answer : d <eor> d <eos> |
d |
percent__100.0__80.0__ |
percent__100.0__80.0__ |
| if a and b are the roots of the equation x num__2 - num__7 x + num__7 = num__0 then the value of a num__2 + b num__2 is : <o> a ) num__35 <o> b ) num__24 <o> c ) num__17 <o> d ) num__6 <o> e ) num__5 |
sol . ( b ) the sum of roots = a + b = num__7 product of roots = ab = num__8 now a num__2 + b num__2 = ( a + b ) num__2 - num__2 ab = num__49 - num__14 = num__35 answer a <eor> a <eos> |
a |
multiply__2.0__7.0__ subtract__49.0__14.0__ subtract__49.0__14.0__ |
multiply__2.0__7.0__ subtract__49.0__14.0__ subtract__49.0__14.0__ |
| if the simple interest on a certain sum of money for num__4 years is one – fifth of the sum then the rate of interest per annum is <o> a ) num__4.0 <o> b ) num__7.0 <o> c ) num__6.0 <o> d ) num__5.0 <o> e ) num__3 % |
explanation : let the principal ( p ) be x then simple interest ( si ) = x / num__5 time ( t ) = num__4 years rate of interest per annum ( r ) = ( num__100 × si ) / pt = ( num__100 × ( x / num__5 ) / ( x × num__4 ) = num__5.0 = num__5.0 answer : option d <eor> d <eos> |
d |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| a train num__180 m long passed a pole in num__10 sec . how long will it take to pass a platform num__800 m long ? <o> a ) num__80 <o> b ) num__89 <o> c ) num__54 <o> d ) num__67 <o> e ) num__56 |
speed = num__18.0 = num__18 m / sec . required time = ( num__180 + num__800 ) / num__18 = num__54 sec . answer : option c <eor> c <eos> |
c |
divide__180.0__10.0__ round__54.0__ |
divide__180.0__10.0__ round__54.0__ |
| if two numbers are in the ratio num__2 : num__3 . if num__17 is added to both of the numbers then the ratio becomes num__3 : num__4 then find the smallest number ? <o> a ) a ) num__10 <o> b ) b ) num__20 <o> c ) c ) num__34 <o> d ) d ) num__30 <o> e ) e ) num__35 |
num__2 : num__3 num__2 x + num__17 : num__3 x + num__17 = num__3 : num__4 num__4 [ num__2 x + num__17 ] = num__3 [ num__3 x + num__17 ] num__8 x + num__68 = num__9 x + num__51 num__9 x - num__8 x = num__68 - num__51 x = num__17 then smallest number is = num__2 num__2 x = num__34 correct option c <eor> c <eos> |
c |
multiply__2.0__4.0__ multiply__17.0__4.0__ subtract__17.0__8.0__ multiply__3.0__17.0__ multiply__2.0__17.0__ multiply__2.0__17.0__ |
multiply__2.0__4.0__ multiply__17.0__4.0__ subtract__17.0__8.0__ subtract__68.0__17.0__ subtract__51.0__17.0__ subtract__51.0__17.0__ |
| the area of a square field is num__900 km num__2 . how long will it take for a horse to run around at the speed of num__12 km / h ? <o> a ) num__12 h <o> b ) num__10 h <o> c ) num__8 h <o> d ) num__6 h <o> e ) none of these |
explanation area of field = num__900 km num__2 . then each side of field = √ num__900 = num__30 km distance covered by the horse = perimeter of square field = num__30 × num__4 = num__120 km ∴ time taken by horse = distances / peed = num__10.0 = num__10 h answer b <eor> b <eos> |
b |
square_perimeter__30.0__ triangle_area__2.0__10.0__ |
square_perimeter__30.0__ triangle_area__2.0__10.0__ |
| the sum of the ages of num__5 children born at the intervals of num__3 years each is num__80 years . what is the age of the youngest child ? <o> a ) num__2 years <o> b ) num__4 years <o> c ) num__6 years <o> d ) num__8 years <o> e ) num__10 years |
let the ages of the children be x ( x + num__3 ) ( x + num__6 ) ( x + num__9 ) and ( x + num__12 ) years . then x + ( x + num__3 ) + ( x + num__6 ) + ( x + num__9 ) + ( x + num__12 ) = num__80 num__5 x = num__50 = > x = num__10 . age of youngest child = x = num__10 years . answer : e <eor> e <eos> |
e |
add__3.0__6.0__ add__3.0__9.0__ divide__50.0__5.0__ divide__50.0__5.0__ |
add__3.0__6.0__ add__3.0__9.0__ divide__50.0__5.0__ divide__50.0__5.0__ |
| the ratio between the number of sheep and the number of horses at the stewart farm is num__4 to num__7 if each horse is fed num__230 ounces of horse food per day and the farm needs a total num__12880 ounces of horse food per day what is the number of sheep in the farm ? <o> a ) num__18 <o> b ) num__28 <o> c ) num__32 <o> d ) num__56 <o> e ) num__60 |
actual number of horses = num__56.0 ( hold it since it is a harder calculation ) ratio multiplier = num__1288 / ( num__23 * num__7 ) actual number of sheep = num__1288 / ( num__23 * num__7 ) * num__4 . first divide num__1288 by num__7 to get : num__184 * num__0.173913043478 use the last digit to figure out that num__23 will go num__8 times into num__184 . confirm . num__8 * num__4 = num__32 answer ( c ) <eor> c <eos> |
c |
divide__12880.0__230.0__ divide__1288.0__56.0__ divide__1288.0__7.0__ divide__4.0__23.0__ divide__184.0__23.0__ multiply__4.0__8.0__ multiply__4.0__8.0__ |
divide__12880.0__230.0__ divide__1288.0__56.0__ divide__1288.0__7.0__ divide__4.0__23.0__ divide__184.0__23.0__ multiply__4.0__8.0__ multiply__4.0__8.0__ |
| a cistern can be filled by a tap in num__4 hours while it can be emptied by another tap in num__9 hours . if both the taps are opened simultaneously then after how much time cistern will get filled ? <o> a ) num__7 hours <o> b ) num__7.1 hours <o> c ) num__7.2 hours <o> d ) num__7.3 hours <o> e ) num__7.4 hours |
explanation : when we have question like one is filling the tank and other is empting it then we subtraction as filled in num__1 hour = num__0.25 empties in num__1 hour = num__0.111111111111 net filled in num__1 hour = num__0.25 - num__0.111111111111 = num__0.138888888889 so cistern will be filled in num__7.2 hours i . e . num__7.2 hours answer is c <eor> c <eos> |
c |
divide__1.0__4.0__ divide__1.0__9.0__ subtract__0.25__0.1111__ round__7.2__ |
divide__1.0__4.0__ divide__1.0__9.0__ subtract__0.25__0.1111__ round__7.2__ |
| an aeroplane covers a certain distance of num__480 kmph in num__4 hours . to cover the same distance in num__4 num__0.333333333333 hours it must travel at a speed of <o> a ) num__440 <o> b ) num__540 <o> c ) num__443 <o> d ) num__740 <o> e ) num__250 |
speed of aeroplane = num__480 kmph distance travelled in num__4 hours = num__480 * num__4 = num__1920 km speed of aeroplane to acver num__1920 km in num__4.33333333333 = num__1920 * num__0.230769230769 = num__443 km answer c . <eor> c <eos> |
c |
multiply__480.0__4.0__ add__4.0__0.3333__ round__443.0__ |
multiply__480.0__4.0__ add__4.0__0.3333__ round__443.0__ |
| the list price of an article is rs . num__65 . a customer pays rs . num__56.16 for it . he was given two successive discounts one of them being num__10.0 . the other discount is ? <o> a ) num__6.0 <o> b ) num__3.0 <o> c ) num__8.0 <o> d ) num__1.0 <o> e ) num__4 % |
num__65 * ( num__0.9 ) * ( ( num__100 - x ) / num__100 ) = num__56.16 x = num__4.0 answer : e <eor> e <eos> |
e |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| two trains are running at num__40 km / hr and num__20 km / hr respectively in the same direction . fast train completely passes a man sitting in the slower train in num__5 sec . what is the length of the fast train ? <o> a ) num__27 num__1.4 <o> b ) num__27 num__0.777777777778 <o> c ) num__27 num__3.5 <o> d ) num__27 num__0.777777777778 <o> e ) num__24 num__3.5 |
relative speed = ( num__40 - num__20 ) = num__20 km / hr . = num__20 * num__0.277777777778 = num__5.55555555556 m / sec . length of faster train = num__5.55555555556 * num__5 = num__27.7777777778 = num__27 num__0.777777777778 m . answer : d <eor> d <eos> |
d |
subtract__27.7778__27.0__ round__27.0__ |
subtract__27.7778__27.0__ subtract__27.7778__0.7778__ |
| both robert and alice leave from the same location at num__7 : num__00 a . m . driving in the same direction but in separate cars . robert drives num__30 miles per hour while alice drives num__50 miles per hour . after num__6 hours alice ’ s car stops . at what time will robert ’ s car reach alice ’ s car ? <o> a ) num__1 p . m . <o> b ) num__3 p . m . <o> c ) num__4 p . m . <o> d ) num__7 p . m . <o> e ) num__9 p . m . |
num__7 : num__00 am so num__6 hours later is num__1 : num__00 pm in six hours robert will have driven num__6 * num__30 = num__180 miles in six hours alive will have driven num__6 * num__50 = num__300 miles so robert needs num__300 - num__180 = num__120 miles do catch alice up . so at num__30 mph he will need num__2 hours num__1 : num__00 pm + num__6 hours = num__7 : num__00 pm ans : d <eor> d <eos> |
d |
subtract__7.0__6.0__ multiply__30.0__6.0__ multiply__50.0__6.0__ subtract__300.0__180.0__ round__7.0__ |
subtract__7.0__6.0__ multiply__30.0__6.0__ multiply__50.0__6.0__ subtract__300.0__180.0__ add__6.0__1.0__ |
| if the diameter of circle r is num__40.0 of the diameter of circle s the area of circle r is what percent of the area of circle s ? <o> a ) num__16.0 <o> b ) num__18.0 <o> c ) num__20.0 <o> d ) num__22.0 <o> e ) num__24 % |
let diameter of circle r dr = num__40 and diameter of circle s ds = num__100 radius of circle r rr = num__20 radius of circle s rs = num__50 area of circle r / area of circle s = ( pi * rr ^ num__2 ) / ( pi * rs ^ num__2 ) = ( num__0.4 ) ^ num__2 = ( num__0.4 ) ^ num__2 = num__16.0 answer : a <eor> a <eos> |
a |
multiply__40.0__0.4__ multiply__40.0__0.4__ |
multiply__40.0__0.4__ multiply__40.0__0.4__ |
| look at this series : num__53 num__53 num__40 num__40 num__27 num__27 . . . what number should come next ? <o> a ) num__14 <o> b ) num__15 <o> c ) num__17 <o> d ) num__19 <o> e ) num__11 |
a num__14 in this series each number is repeated then num__13 is subtracted to arrive at the next number . <eor> a <eos> |
a |
subtract__53.0__40.0__ subtract__27.0__13.0__ |
subtract__53.0__40.0__ subtract__27.0__13.0__ |
| each student is given a five - character identification code . the first two characters are selected from the numbers num__0 to num__9 inclusive and the last three characters are selected from the num__26 letters of the alphabet . if characters may be repeated and the same characters used in a different order constitute a different code how many different identification codes can be generated following these rules ? <o> a ) num__1 num__491000 <o> b ) num__1 num__527200 <o> c ) num__1 num__638400 <o> d ) num__1 num__757600 <o> e ) num__1 num__841 |
800 |
the number of possible codes is num__10 * num__10 * num__26 * num__26 * num__26 = num__1 num__757600 . the answer is d . <eor> d <eos> |
d |
d |
| find s . p when cp = rs num__80.40 loss = num__5.0 <o> a ) rs num__66.34 <o> b ) rs num__69.34 <o> c ) rs num__58.34 <o> d ) rs num__68.34 <o> e ) rs num__60.34 |
cp = rs num__80.40 loss = num__5.0 sol : sp = num__85.0 of rs num__80.40 = rs { ( num__0.85 ) * num__80.40 } = rs num__68.34 . answer is d . <eor> d <eos> |
d |
percent__80.4__85.0__ percent__80.4__85.0__ |
percent__80.4__85.0__ percent__80.4__85.0__ |
| jacob is num__18 years old . he is num__2 times as old as his brother . how old will jacob be when he is twice as old ? <o> a ) num__36 <o> b ) num__38 <o> c ) num__33 <o> d ) num__32 <o> e ) num__38 |
j = num__18 ; j = num__2 b ; b = num__9.0 = num__9 ; twice as old so b = num__9 ( now ) + ( num__9 ) = num__18 ; jacob is num__18 + num__18 = num__36 answer : a <eor> a <eos> |
a |
divide__18.0__2.0__ multiply__18.0__2.0__ multiply__18.0__2.0__ |
divide__18.0__2.0__ multiply__18.0__2.0__ multiply__18.0__2.0__ |
| the total price of a basic computer and printer are $ num__2500 . if the same printer had been purchased with an enhanced computer whose price was $ num__500 more than the price of the basic computer then the price of the printer would have been num__0.25 of that total . what was the price of the basic computer ? <o> a ) num__1500 <o> b ) num__1600 <o> c ) num__1750 <o> d ) num__1900 <o> e ) num__2000 |
let the price of basic computer be c and the price of the printer be p : c + p = $ num__2500 . the price of the enhanced computer will be c + num__500 and total price for that computer and the printer will be num__2500 + num__500 = $ num__3000 . now we are told that the price of the printer is num__0.25 of that new total price : p = num__0.25 * $ num__3000 = $ num__750 . plug this value in the first equation : c + num__750 = $ num__2500 - - > c = $ num__1750 answer : c . <eor> c <eos> |
c |
add__2500.0__500.0__ multiply__0.25__3000.0__ subtract__2500.0__750.0__ subtract__2500.0__750.0__ |
add__2500.0__500.0__ multiply__0.25__3000.0__ subtract__2500.0__750.0__ subtract__2500.0__750.0__ |
| in num__1998 the profits of company n were num__10 percent of revenues . in num__1999 the revenues of company n fell by num__30 percent but profits were num__14 percent of revenues . the profits in num__1999 were what percent of the profits in num__1998 ? <o> a ) num__80.0 <o> b ) num__105.0 <o> c ) num__120.0 <o> d ) num__124.2 <o> e ) num__98 % |
num__0 |
098 r = x / num__100 * num__0.1 r answer e <eor> e <eos> |
e |
e |
| a certain store sells small medium and large toy trucks in each of the colors red blue green and yellow . the store has an equal number of trucks of each possible color - size combination . if paul wants a medium red truck and his mother will randomly select one the trucks in the store what is the probability that the truck she selects will have at least one of the two features paul wants ? <o> a ) num__0.25 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.583333333333 <o> e ) num__0.666666666667 |
no . of colors = num__4 no . of sizes = num__3 total outcomes = num__12 remaining outcomes = num__6 hence probability that truck is neither red nor of medium size is num__0.047619047619 / num__12 . hence probability that the truck she selects will have at least one of the two features paul wants will be num__1 − num__0.0495867768595 − num__0.5 or num__0.5 answer : c <eor> c <eos> |
c |
die_space__ negate_prob__0.5__ |
die_space__ negate_prob__0.5__ |
| the ratio by volume of bleach to detergent to water in a certain solution is num__2 : num__40 : num__100 . the solution will be altered so that the ratio of bleach ( b ) to detergent is tripled while the ratio of detergent to water is halved . if the altered solution will contain num__300 liters of water how many liters of detergent will it contain ? <o> a ) num__40 <o> b ) num__60 <o> c ) num__50 <o> d ) num__30 <o> e ) num__70 |
b : d : w = num__2 : num__40 : num__100 bnew / dnew = ( num__0.333333333333 ) * ( num__0.05 ) = ( num__0.0166666666667 ) dnew / wnew = ( num__0.5 ) * ( num__0.4 ) = ( num__0.2 ) wnew = num__300 dnew = wnew / num__5 = num__60.0 = num__60 so answer will be b b <eor> b <eos> |
b |
multiply__0.05__0.3333__ triangle_perimeter__0.05__0.3333__0.0167__ square_perimeter__0.05__ rectangle_perimeter__2.0__0.5__ triangle_area__300.0__0.4__ triangle_area__2.0__60.0__ |
multiply__0.05__0.3333__ triangle_perimeter__0.05__0.3333__0.0167__ multiply__0.5__0.4__ multiply__100.0__0.05__ multiply__300.0__0.2__ multiply__300.0__0.2__ |
| how long will it take a sum of money invested at num__5.0 p . a . s . i . to increase its value by num__50.0 ? <o> a ) num__10 years . <o> b ) num__4 years . <o> c ) num__8 years . <o> d ) num__12 years . <o> e ) num__14 years . |
sol . let the sum be x . then s . i . = num__50.0 of x = x / num__2 ; rate = num__5.0 . â ˆ ´ time = [ num__100 * x / num__2 * num__1 / x * num__5 ] = num__10 years . answer a <eor> a <eos> |
a |
percent__50.0__2.0__ percent__100.0__10.0__ |
percent__50.0__2.0__ percent__100.0__10.0__ |
| find the difference between the average of first n even numbers and the average of all the even numbers up to n ( n is an even number ) . <o> a ) n <o> b ) ( n - num__1 ) / num__2 <o> c ) n / num__2 <o> d ) can not be determined <o> e ) none of these |
explanation : ( n + num__1 ) - ( ( n / num__2 ) + num__1 ) answer : c <eor> c <eos> |
c |
multiply__1.0__2.0__ |
divide__2.0__1.0__ |
| which of the following is closest to num__10 ^ num__180 – num__10 ^ num__20 ? <o> a ) num__10 ^ num__210 <o> b ) num__10 ^ num__180 <o> c ) num__10 ^ num__150 <o> d ) num__10 ^ num__90 <o> e ) num__10 ^ num__6 |
num__10 ^ num__180 – num__10 ^ num__20 num__10 ^ num__20 * ( num__10 ^ num__160 – num__1 ) as we know num__10 ^ num__2 - num__1 means num__100 - num__1 and we get num__99 which is approximately num__100 . hence ( num__10 ^ num__160 – num__1 ) would remain as num__10 ^ num__160 . and num__10 ^ num__20 * num__10 ^ num__160 = num__10 ^ num__180 . answer is b . <eor> b <eos> |
b |
subtract__180.0__20.0__ divide__20.0__10.0__ power__10.0__2.0__ subtract__100.0__1.0__ multiply__10.0__1.0__ |
subtract__180.0__20.0__ divide__20.0__10.0__ power__10.0__2.0__ subtract__100.0__1.0__ multiply__10.0__1.0__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__10 seconds . if the speed of the train is num__72 km / hr what is the length of the platform ? <o> a ) num__338 <o> b ) num__240 <o> c ) num__520 <o> d ) num__267 <o> e ) num__191 |
speed = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . length of the train = ( num__20 x num__10 ) m = num__200 m . let the length of the platform be x meters . then ( x + num__200 ) / num__36 = num__20 = = > x + num__200 = num__720 = = > x = num__520 m . answer : c <eor> c <eos> |
c |
divide__10.0__36.0__ multiply__10.0__20.0__ multiply__36.0__20.0__ subtract__720.0__200.0__ round__520.0__ |
divide__10.0__36.0__ multiply__10.0__20.0__ multiply__36.0__20.0__ subtract__720.0__200.0__ round__520.0__ |
| a grocer stacked oranges in a pile . the bottom layer was rectangular with num__3 rows of num__7 oranges each . in the second layer from the bottom each orange rested on num__4 oranges from the bottom layer and in the third layer each orange rested on num__4 oranges from the second layer . which of the following is the maximum number of oranges that could have been in the third layer ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__3 <o> d ) num__2 <o> e ) num__1 |
bottom layer = num__7 x num__3 = num__21 iind layer = ( num__7 - num__1 ) x ( num__3 - num__1 ) = num__12 iiird layer = ( num__6 - num__1 ) x ( num__2 - num__1 ) = num__5 answer = num__5 = a <eor> a <eos> |
a |
multiply__3.0__7.0__ subtract__4.0__3.0__ multiply__3.0__4.0__ subtract__7.0__1.0__ subtract__3.0__1.0__ add__3.0__2.0__ add__3.0__2.0__ |
multiply__3.0__7.0__ subtract__4.0__3.0__ multiply__3.0__4.0__ subtract__7.0__1.0__ subtract__3.0__1.0__ subtract__7.0__2.0__ subtract__7.0__2.0__ |
| best friends sprite and icey have are playing a game of marbles . they invite num__2 of their friends to play with them . there are num__36 marbles in the bag . if all num__4 people are playing how many marbles does each person get ? <o> a ) num__3 <o> b ) num__9 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
answer = b num__9.0 = num__9 answer = b <eor> b <eos> |
b |
divide__36.0__4.0__ divide__36.0__4.0__ |
divide__36.0__4.0__ divide__36.0__4.0__ |
| if num__5 ^ r is a factor of num__30 ! which of the following is the maximum possible value of r ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
num__5 ^ r is a factor of num__30 ! we need the maximum value of r or in other words we need the maximum power of num__5 in num__30 ! powers of num__5 in num__30 ! = [ num__6.0 ] + [ num__1.2 ] = num__7 where [ x ] is the integral part of x other way is to calculate the number of num__5 ' s in num__30 ! in the multiplication upto num__30 we will encounter : num__5 num__10 num__15 num__2025 num__30 each of num__5 num__10 num__15 num__20 num__30 contain one power of num__5 num__25 contains num__2 powers of num__5 . hence total powers of num__5 in num__30 ! = num__5 + num__2 = num__7 correct option : d <eor> d <eos> |
d |
divide__30.0__5.0__ divide__6.0__5.0__ add__5.0__10.0__ add__5.0__15.0__ add__5.0__20.0__ divide__30.0__15.0__ add__5.0__2.0__ |
divide__30.0__5.0__ divide__6.0__5.0__ add__5.0__10.0__ add__5.0__15.0__ add__5.0__20.0__ divide__30.0__15.0__ add__5.0__2.0__ |
| in a num__1000 m race a beats b by num__50 m and b beats c by num__100 m . in the same race by how many meters does a beat c ? <o> a ) num__145 <o> b ) num__266 <o> c ) num__276 <o> d ) num__027 <o> e ) num__266 |
by the time a covers num__1000 m b covers ( num__1000 - num__50 ) = num__950 m . by the time b covers num__1000 m c covers ( num__1000 - num__100 ) = num__900 m . so the ratio of speeds of a and c = num__1.05263157895 * num__1.11111111111 = num__1.16959064327 so by the time a covers num__1000 m c covers num__855 m . so in num__1000 m race a beats c by num__1000 - num__855 = num__145 m . answer : a <eor> a <eos> |
a |
subtract__1000.0__50.0__ subtract__1000.0__100.0__ divide__1000.0__950.0__ divide__1000.0__900.0__ subtract__1000.0__855.0__ round__145.0__ |
subtract__1000.0__50.0__ subtract__1000.0__100.0__ divide__1000.0__950.0__ divide__1000.0__900.0__ subtract__1000.0__855.0__ subtract__1000.0__855.0__ |
| what amount does an investor receive if the investor invests $ num__3000 at num__10.0 p . a . compound interest for two years compounding done annually ? <o> a ) $ num__3420 <o> b ) $ num__3630 <o> c ) $ num__3870 <o> d ) $ num__4040 <o> e ) $ num__4220 |
a = ( num__1 + r / num__100 ) ^ n * p ( num__1.1 ) ^ num__2 * num__5000 = num__1.21 * num__5000 = num__3630 the answer is b . <eor> b <eos> |
b |
percent__100.0__3630.0__ |
percent__100.0__3630.0__ |
| a train running at a speed of num__60 kmph crosses a pole in num__6 seconds . what is the length of the train ? <o> a ) num__120 m <o> b ) num__180 m <o> c ) num__100 m <o> d ) num__150 m <o> e ) num__160 m |
num__60 kmph = num__16.6666666667 m / sec num__16.6666666667 * num__6 = num__100 m answer : c <eor> c <eos> |
c |
round__100.0__ |
round__100.0__ |
| p is able to do a piece of work in num__10 days and q can do the same work in num__12 days . if they can work together for num__5 days what is the fraction of work left ? <o> a ) num__0.583333333333 <o> b ) num__0.416666666667 <o> c ) num__0.0833333333333 <o> d ) num__0.25 <o> e ) num__0.5 |
explanation : amount of work p can do in num__1 day = num__0.1 amount of work q can do in num__1 day = num__0.0833333333333 amount of work p and q can do in num__1 day = num__0.1 + num__0.0833333333333 = num__0.183333333333 amount of work p and q can together do in num__5 days = num__5 × ( num__0.183333333333 ) = num__0.916666666667 fraction of work left = num__1 – num__0.916666666667 = num__0.0833333333333 answer : option c <eor> c <eos> |
c |
divide__1.0__10.0__ divide__1.0__12.0__ add__0.1__0.0833__ subtract__1.0__0.0833__ divide__1.0__12.0__ |
divide__1.0__10.0__ divide__1.0__12.0__ add__0.1__0.0833__ subtract__1.0__0.0833__ divide__1.0__12.0__ |
| two equilateral triangles of side num__12 cm are placed one on top of another such that a num__6 pointed star is formed . if the six verticals lie on a circle what is the area of the circle not enclosed by the star ? <o> a ) num__68 <o> b ) num__83 <o> c ) num__57 <o> d ) num__61 <o> e ) num__62 |
area of star = √ num__0.75 * ( num__12 ) ^ num__2 + num__3 * √ num__0.75 * ( num__4.0 ) ^ num__2 = num__48 √ num__3 if r be radius of circle then r = num__6 / cos num__30 = > r = num__12 / √ num__3 = > r ^ num__2 = num__48 area of circle = pi * num__48 area of the circle not enclosed by the star = num__48 * pi - num__48 √ num__3 = num__48 ( num__3.14 - num__1.732 ) = num__68 ( approax ) answer : a <eor> a <eos> |
a |
square_perimeter__0.75__ square_perimeter__12.0__ rectangle_perimeter__12.0__3.0__ rectangle_perimeter__4.0__30.0__ triangle_area__2.0__68.0__ |
square_perimeter__0.75__ multiply__12.0__4.0__ rectangle_perimeter__12.0__3.0__ rectangle_perimeter__4.0__30.0__ triangle_area__2.0__68.0__ |
| a b and c can do a piece of work in num__20 num__30 and num__60 days respectively . in how many days can a do the work if he is assisted by b and c on every third day ? <o> a ) num__12 days <o> b ) num__15 days <o> c ) num__18 days <o> d ) num__20 days <o> e ) num__22 days |
a ’ s two day ’ s work = num__0.1 = num__0.1 ( a + b + c ) ’ s one day ’ s work = num__0.05 + num__0.0333333333333 + num__0.0166666666667 = num__0.1 = num__0.1 work done in num__3 days = ( num__0.1 + num__0.1 ) = num__0.2 now num__0.2 work is done in num__3 days therefore whole work will be done in ( num__3 x num__5 ) = num__15 days . answer : c <eor> c <eos> |
c |
subtract__0.05__0.0333__ multiply__30.0__0.1__ subtract__20.0__5.0__ add__3.0__15.0__ |
subtract__0.05__0.0333__ multiply__30.0__0.1__ subtract__20.0__5.0__ add__3.0__15.0__ |
| the sum of ages of num__5 children born at the intervals of num__3 years each is num__50 years . what is the age of the youngest child ? <o> a ) num__4 years <o> b ) num__5 years <o> c ) num__6 years <o> d ) num__7 years <o> e ) num__8 years |
let the ages of children be x ( x + num__3 ) ( x + num__6 ) ( x + num__9 ) and ( x + num__12 ) years . then x + ( x + num__3 ) + ( x + num__6 ) + ( x + num__9 ) + ( x + num__12 ) = num__50 num__5 x = num__20 x = num__4 . age of the youngest child = x = num__4 years . answer : a <eor> a <eos> |
a |
add__3.0__6.0__ add__3.0__9.0__ subtract__9.0__5.0__ subtract__9.0__5.0__ |
add__3.0__6.0__ add__3.0__9.0__ subtract__9.0__5.0__ subtract__9.0__5.0__ |
| a tradesman is marketing his goods num__20.0 above the cost price of the goods . he gives num__10.0 discount on cash payment find his gain percent . <o> a ) num__12.0 <o> b ) num__8.0 <o> c ) num__15.0 <o> d ) num__18.0 <o> e ) none of these |
let the c . p . of the goods be num__100 ⇒ marked price of the goods = num__120 discount = num__10.0 ⇒ s . p . is num__90.0 of num__120 = num__108 ∴ gain = ( num__108 – num__100 ) i . e . num__8.0 . answer b <eor> b <eos> |
b |
percent__90.0__120.0__ percent__100.0__8.0__ |
percent__90.0__120.0__ percent__100.0__8.0__ |
| a teacher gave the same test to three history classes : a b and c . the average ( arithmetic mean ) scores for the three classes were num__65 num__90 and num__77 respectively . the ratio of the numbers of students in each class who took the test was num__4 to num__6 to num__5 respectively . what was the average score for the three classes combined ? <o> a ) num__74 <o> b ) num__75 <o> c ) num__76 <o> d ) num__77 <o> e ) num__79 |
ratio is num__4 : num__6 : num__5 numbers are num__4 x num__6 x num__5 x total scores of each class is ( num__65 * num__4 x + num__6 x * num__90 + num__77 * num__5 x ) = num__260 x + num__540 x + num__385 x = num__1185 x total number of students = num__15 x average = num__1185 x / num__15 x = num__79 e is the answer <eor> e <eos> |
e |
multiply__65.0__4.0__ multiply__90.0__6.0__ multiply__77.0__5.0__ divide__90.0__6.0__ divide__1185.0__15.0__ divide__1185.0__15.0__ |
multiply__65.0__4.0__ multiply__90.0__6.0__ multiply__77.0__5.0__ divide__90.0__6.0__ divide__1185.0__15.0__ divide__1185.0__15.0__ |
| the average weight of num__10 students decreases by num__6 kg when one of them weighing num__120 kg is replaced by a new student . the weight of the student is <o> a ) num__62 kg <o> b ) num__60 kg <o> c ) num__70 kg <o> d ) num__72 kg <o> e ) none of these |
explanation : let the weight of student be x kg . given difference in average weight = num__6 kg = > ( num__120 - x ) / num__10 = num__6 = > x = num__60 answer : b <eor> b <eos> |
b |
multiply__10.0__6.0__ multiply__10.0__6.0__ |
multiply__10.0__6.0__ subtract__120.0__60.0__ |
| the price of a jacket is reduced by num__10.0 . during a special sale the price of the jacket is reduced another num__30.0 . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? <o> a ) num__32.5 <o> b ) num__58.7 <o> c ) num__48 <o> d ) num__65 <o> e ) num__67.5 |
num__1 ) let the price of jacket initially be $ num__100 . num__2 ) then it is decreased by num__10.0 therefore bringing down the price to $ num__90 . num__3 ) again it is further discounted by num__30.0 therefore bringing down the price to $ num__63 . num__4 ) now num__67.5 has to be added byx % in order to equal the original price . num__63 + ( x % ) num__63 = num__100 . solving this eq for x we get x = num__58.7 ans is b <eor> b <eos> |
b |
percent__10.0__30.0__ percent__100.0__58.7__ |
percent__10.0__30.0__ percent__100.0__58.7__ |
| an author received $ num__0.80 in royalties for each of the first num__100000 copies of her book sold and $ num__0.80 in royalties for each additional copy sold . if she received a total of $ num__260000 in royalties how many copies of her book were sold ? <o> a ) num__130000 <o> b ) num__300000 <o> c ) num__325000 <o> d ) num__400000 <o> e ) num__420 |
000 |
total royalties for first num__100.000 books = . num__8 * num__100000 = num__80000 total royalties for the rest of the books = num__260000 - num__80000 = num__180000 remaining books = num__180000 / num__0.8 = num__225000 total books = num__225000 + num__100000 = num__325000 answer c <eor> c <eos> |
c |
c |
| in a bread recipe the ratio of water to flour is num__13 to num__20 . there is num__168 g more flour than water . how much flour does the recipe call for ? <o> a ) num__400 g <o> b ) num__440 g <o> c ) num__480 g <o> d ) num__510 g <o> e ) num__525 g |
let x equal the amount of water and y equal the amount of flour . we know that x = num__0.65 y . we also know that x = y - num__168 . so we can simplify : y - num__168 = num__0.65 y so to isolate y we get : num__0.35 y = num__168 and to solve for y : y = num__480 answer : c <eor> c <eos> |
c |
divide__13.0__20.0__ divide__168.0__0.35__ divide__168.0__0.35__ |
divide__13.0__20.0__ divide__168.0__0.35__ divide__168.0__0.35__ |
| five persons - a b c d and e are being compared in weight and height . the second heaviest person . d is the shortest . a is the num__2 nd tallest and shorter than e the heaviest person is the third tallest person . there is only one person shorter than b who is lighter than e and a respectively . what is the position of a in height and weight respectively ? <o> a ) num__2 nd num__4 th <o> b ) num__4 th num__3 rd <o> c ) num__2 nd num__1 st <o> d ) num__5 th num__3 rd <o> e ) num__2 nd num__5 th |
explanation : a is the second tallest person . there is only one person shorter than b . so b is the fourth tallest person . d is the shortest and second heaviest . a is shorter than e . so the third tallest person is c . who is the heaviest person . b is lighter than e and a respectively . so the final arrangement is as follows . a is the second tallest and the fourth heaviest person . answer is a <eor> a <eos> |
a |
round__2.0__ |
round__2.0__ |
| evaluate combination num__100 c num__98 = num__100 ! / ( num__98 ) ! ( num__2 ) ! <o> a ) num__4950 <o> b ) num__1510 <o> c ) num__4170 <o> d ) num__3170 <o> e ) none of these |
explanation : ncr = n ! / ( r ) ! ( n − r ) ! num__100 c num__98 = num__100 ! / ( num__98 ) ! ( num__2 ) ! = num__100 ∗ num__99 ∗ num__98 ! / ( num__98 ) ! ( num__2 ) ! = num__100 ∗ num__49.5 ∗ num__1 = num__100 ∗ num__49.5 ∗ num__1 = num__4950 option a <eor> a <eos> |
a |
divide__99.0__2.0__ subtract__100.0__99.0__ multiply__100.0__49.5__ multiply__100.0__49.5__ |
divide__99.0__2.0__ subtract__100.0__99.0__ multiply__100.0__49.5__ divide__4950.0__1.0__ |
| if ( num__1 – num__1.25 ) n = num__2 then n = <o> a ) − num__400 <o> b ) − num__140 <o> c ) − num__8 <o> d ) num__4 <o> e ) num__400 |
( num__1 – num__1.25 ) n = num__2 simplify to get : - num__0.25 n = num__2 rewrite as ( - num__0.25 ) n = num__2 multiply both sides by - num__4 to get : n = - num__8 answer : c <eor> c <eos> |
c |
subtract__1.25__1.0__ reverse__0.25__ multiply__2.0__4.0__ multiply__1.0__8.0__ |
subtract__1.25__1.0__ reverse__0.25__ multiply__2.0__4.0__ multiply__1.0__8.0__ |
| john traveled from point a to b in num__4 merter per sec and from b to a in num__6 meter per sec . what is his average speed ? <o> a ) num__4.6 m / s <o> b ) num__4.8 m / s <o> c ) num__5.2 m / s <o> d ) num__5.4 m / s <o> e ) num__5.8 m / s |
since the distance traveled is same we can apply direct formula = num__2 xs num__1 xs num__2 / ( s num__1 + s num__2 ) num__2 x num__4 x num__0.6 = num__4.8 . ' b ' is the answer . <eor> b <eos> |
b |
subtract__6.0__4.0__ km_to_mile_conversion__ round__4.8__ |
subtract__6.0__4.0__ km_to_mile_conversion__ divide__4.8__1.0__ |
| in an election only two candidates contested . a candidate secured num__70.0 of the valid votes and won by a majority of num__172 votes . find the total number of valid votes ? <o> a ) num__430 <o> b ) num__287 <o> c ) num__267 <o> d ) num__262 <o> e ) num__927 |
let the total number of valid votes be x . num__70.0 of x = num__0.7 * x = num__7 x / num__10 number of votes secured by the other candidate = x - num__7 x / num__100 = num__3 x / num__10 given num__7 x / num__10 - num__3 x / num__10 = num__172 = > num__4 x / num__10 = num__172 = > num__4 x = num__1720 = > x = num__430 . answer : a <eor> a <eos> |
a |
percent__100.0__430.0__ |
percent__100.0__430.0__ |
| a man can row upstream at num__8 kmph and downstream at num__13 kmph . the speed of the stream is <o> a ) num__2.5 km / hr <o> b ) num__4.2 km / hr <o> c ) num__5 km / hr <o> d ) num__10.5 km / hr <o> e ) none |
solution speed of stream = num__0.5 ( num__13 - num__8 ) kmph = num__0.5 x num__5 = num__2.5 = num__2.5 answer a <eor> a <eos> |
a |
subtract__13.0__8.0__ multiply__0.5__5.0__ round__2.5__ |
subtract__13.0__8.0__ multiply__0.5__5.0__ subtract__5.0__2.5__ |
| in a mixture of num__60 liters the ratio of milk and water is num__2 : num__1 . what amount of water must be added to make the ratio of milk and water as num__1 : num__2 ? <o> a ) num__65 liters <o> b ) num__68 liters <o> c ) num__60 liters <o> d ) num__26 liters <o> e ) num__27 liters |
num__2 : num__1 - - - num__60 num__1 : num__2 num__2 : num__4 - - - - - - - - - - - - - - - num__3 num__3 - - - - - - - - - - - num__60 num__3 - - - - - - - - - - - ? = > num__60 liters answer : c <eor> c <eos> |
c |
add__2.0__1.0__ multiply__60.0__1.0__ |
subtract__4.0__1.0__ multiply__60.0__1.0__ |
| pavan travelled for num__11 hours . he covered the first half of the distance at num__30 kmph and remaining half of the distance at num__25 kmph . find the distance travelled by pavan . <o> a ) num__240 km <o> b ) num__225 km <o> c ) num__452 km <o> d ) num__300 km <o> e ) num__290 km |
d num__300 km let the distance travelled be x km . total time = ( x / num__2 ) / num__30 + ( x / num__2 ) / num__25 = num__11 = > x / num__60 + x / num__50 = num__11 = > ( num__5 x + num__6 x ) / num__300 = num__11 = > x = num__300 km <eor> d <eos> |
d |
hour_to_min_conversion__ multiply__25.0__2.0__ subtract__30.0__25.0__ subtract__11.0__5.0__ round__300.0__ |
hour_to_min_conversion__ multiply__25.0__2.0__ divide__300.0__60.0__ divide__30.0__5.0__ round__300.0__ |
| the average of num__11 numbers is num__10.7 . if the average of first six is num__10.5 and that of the last six is num__11.4 the sixth number is ? <o> a ) num__9 <o> b ) num__9.2 <o> c ) num__10 <o> d ) num__13.7 <o> e ) num__12 |
explanation : num__1 to num__11 = num__11 * num__10.7 = num__117.7 num__1 to num__6 = num__6 * num__10.5 = num__63 num__6 to num__11 = num__6 * num__11.4 = num__68.4 num__63 + num__68.4 = num__131.4 – num__117.7 = num__13.7 num__6 th number = num__13.7 d ) <eor> d <eos> |
d |
multiply__11.0__10.7__ multiply__10.5__6.0__ multiply__11.4__6.0__ add__68.4__63.0__ subtract__131.4__117.7__ multiply__1.0__13.7__ |
multiply__11.0__10.7__ multiply__10.5__6.0__ multiply__11.4__6.0__ add__68.4__63.0__ subtract__131.4__117.7__ multiply__1.0__13.7__ |
| a and b can do a piece of work in num__12 days and num__16 days respectively . both work for num__3 days and then a goes away . find how long will b take to complete the remaining work ? <o> a ) num__15 days <o> b ) num__12 days <o> c ) num__10 days <o> d ) num__9 days <o> e ) num__8 days |
explanation : num__0.25 + ( num__3 + x ) / num__16 = num__1 x = num__9 days answer : d <eor> d <eos> |
d |
divide__3.0__12.0__ subtract__12.0__3.0__ round__9.0__ |
divide__3.0__12.0__ subtract__12.0__3.0__ divide__9.0__1.0__ |
| a number when divided by d leaves a remainder of num__8 and when divided by num__3 d leaves a remainder of num__21 . what is the remainder left when twice the number is divided by num__3 d ? <o> a ) num__13 <o> b ) num__12 <o> c ) num__3 <o> d ) num__2 <o> e ) num__42 |
making eqn as per the question a = dx + num__8 . . . . . . num__1 a = num__3 dy + num__21 . . . . num__2 multiplying num__3 in eqn num__1 we get eqn num__3 num__3 a = num__3 dx + num__24 . . . . num__3 a = num__3 dy + num__21 . . . . . . num__2 now subtract num__2 from num__3 we get . . num__2 a = num__3 d ( x - y ) + num__3 so here ' s the remainder we got is num__3 answer : c <eor> c <eos> |
c |
subtract__3.0__1.0__ multiply__8.0__3.0__ multiply__3.0__1.0__ |
subtract__3.0__1.0__ add__3.0__21.0__ add__1.0__2.0__ |
| a batsman makes a score of num__90 runs in the num__17 th inning and thus increases his averages by num__3 . what is his average after num__17 th inning ? <o> a ) num__39 <o> b ) num__35 <o> c ) num__42 <o> d ) num__40.5 <o> e ) num__42 |
let the average after num__16 th inning = x then total run after num__16 th inning = num__16 x then total run after num__17 th inning = num__16 x + num__90 then average run after num__17 th inning = ( num__16 x + num__90 ) / num__17 ( num__16 x + num__90 ) / num__17 = x + num__3 = > num__16 x + num__90 = num__17 x + num__51 = > x = num__39 x = num__39 ; average after num__17 th inning = num__39 + num__3 = num__42 answer : e <eor> e <eos> |
e |
multiply__17.0__3.0__ subtract__90.0__51.0__ add__3.0__39.0__ add__3.0__39.0__ |
multiply__17.0__3.0__ subtract__90.0__51.0__ add__3.0__39.0__ add__3.0__39.0__ |
| on a certain road num__10.0 of the motorists exceed the posted speed limit and receive speeding tickets but num__60.0 of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on that road exceed the posted speed limit ? <o> a ) num__10.5 <o> b ) num__12.5 <o> c ) num__15.0 <o> d ) num__25.0 <o> e ) num__30 % |
suppose there are x motorists . num__10.0 of them exceeded the speed limit and received the ticket i . e . x / num__10 . again suppose total no . of motorists who exceeded the speed limit are y . num__60.0 of y exceeded the speed limit but did n ' t received the ticket i . e . num__3 y / num__5 . it means num__2 y / num__5 received the ticket . hence num__2 y / num__5 = x / num__10 or y / x = num__0.25 or y / x * num__100 = num__0.25 * num__100 = num__25.0 d <eor> d <eos> |
d |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| a shopping cart contains only apples oranges and pears . if there are four times as many oranges as apples and five times as many pears as oranges then the apples are equivalent to what fraction of the pears ? <o> a ) num__0.125 <o> b ) num__0.111111111111 <o> c ) num__0.0833333333333 <o> d ) num__0.0625 <o> e ) num__0.05 |
o = num__4 a p = num__5 o = num__20 a a = p / num__20 the answer is e . <eor> e <eos> |
e |
multiply__4.0__5.0__ reverse__20.0__ |
multiply__4.0__5.0__ reverse__20.0__ |
| how many num__3 - letter words can be formed using the letters of the english alphabet that contain num__1 different vowels and num__2 different consonants ? <o> a ) num__4 ! * num__5 c num__2 * num__21 c num__2 <o> b ) num__4 ! * num__5 c num__3 * num__21 c num__2 <o> c ) num__3 ! * num__5 c num__1 * num__21 c num__2 <o> d ) num__5 ! * num__5 c num__2 * num__21 c num__2 <o> e ) num__6 ! * num__5 c num__2 * num__21 c num__2 |
combination for num__1 vowels out of num__5 = num__5 c num__1 ways combination for num__2 consonants of num__21 = num__21 c num__2 ways additionally the arrangement can be = num__3 ! total arrangements = num__3 ! * num__5 c num__1 * num__21 c num__2 c <eor> c <eos> |
c |
vowel_space__ choose__3.0__2.0__ |
vowel_space__ choose__3.0__2.0__ |
| if ( x + num__4 ) num__2 − − − − − − − √ = num__3 ( x + num__4 ) num__2 = num__3 which of the following could be the value of x − num__4 x − num__4 ? <o> a ) - num__11 <o> b ) - num__7 <o> c ) - num__4 <o> d ) - num__3 <o> e ) num__5 |
x + num__4 ) num__2 = num__9 ( x + num__4 ) num__2 = num__9 ( x − num__4 + num__8 ) num__2 = num__9 ( x − num__4 + num__8 ) num__2 = num__9 let x - num__4 = a ( a + num__8 ) num__2 = num__9 ( a + num__8 ) num__2 = num__9 now sub answer choices in a a ) ( − num__11 + num__8 ) num__2 = ( − num__3 ) num__2 = num__9 ( − num__11 + num__8 ) num__2 = ( − num__3 ) num__2 = num__9 a is your answer <eor> a <eos> |
a |
multiply__4.0__2.0__ add__2.0__9.0__ add__2.0__9.0__ |
multiply__4.0__2.0__ add__2.0__9.0__ add__2.0__9.0__ |
| what is the smallest number should be added to num__5377 so that the sum is completely divisible by num__7 ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__6 <o> d ) num__2 <o> e ) num__1 |
solution : divide num__5377 with num__7 we get remainder as num__1 . so add num__6 to the given number so that it will divisible by num__7 . answer c <eor> c <eos> |
c |
subtract__7.0__1.0__ subtract__7.0__1.0__ |
subtract__7.0__1.0__ subtract__7.0__1.0__ |
| if each of its sides of a rectangle is increased by num__18.0 what is the percentage increase in the area of the rectangle ? <o> a ) num__39.24 <o> b ) num__38.24 <o> c ) num__36.24 <o> d ) num__37.24 <o> e ) num__38.24 % |
assume original length = num__100 m by num__50 m original area = num__100 * num__50 = num__5000 num__18.0 increase on each side l = num__100 + num__18 = num__118 w = num__50 + num__9 = num__59 new area = num__118 * num__59 = num__6962 increase in area = num__6962 - num__5000 = num__1962.0 increase = num__39.24 answer a <eor> a <eos> |
a |
percent__18.0__50.0__ percent__50.0__118.0__ percent__100.0__39.24__ |
percent__18.0__50.0__ percent__50.0__118.0__ percent__100.0__39.24__ |
| if num__4 : num__6 : : x : num__36 then find the value of x <o> a ) num__24 <o> b ) num__22 <o> c ) num__28 <o> d ) num__30 <o> e ) num__18 |
explanation : treat num__4 : num__6 as num__0.666666666667 and x : num__36 as x / num__36 treat : : as = so we get num__0.666666666667 = x / num__36 = > num__6 x = num__144 = > x = num__24 option a <eor> a <eos> |
a |
divide__4.0__6.0__ multiply__4.0__36.0__ multiply__4.0__6.0__ multiply__4.0__6.0__ |
divide__4.0__6.0__ multiply__4.0__36.0__ divide__144.0__6.0__ divide__144.0__6.0__ |
| one - third of rahul ' s savings in national savings certificate is equal to one - half of his savings in public provident fund . if he has rs . num__1 num__50000 as total savings how much has he saved in public provident fund ? <o> a ) rs . num__30000 <o> b ) rs . num__50000 <o> c ) rs . num__60000 <o> d ) rs . num__90000 <o> e ) rs . num__70 |
000 |
n / num__3 = p / num__2 n + p = num__150000 num__60000 answer : c <eor> c <eos> |
c |
c |
| a is thrice as good as workman as b and therefore is able to finish a job in num__60 days less than b . working together they can do it in ? <o> a ) num__22 num__0.5 <o> b ) num__22 num__0.333333333333 <o> c ) num__23 <o> d ) num__23 num__0.5 <o> e ) num__23 num__0.333333333333 |
ratio of times taken by a and b = num__1 : num__3 . the time difference is ( num__3 - num__1 ) num__2 days while b take num__3 days and a takes num__1 day . if difference of time is num__2 days b takes num__3 days . if difference of time is num__60 days b takes num__3 x num__60 = num__90 days . num__2 so a takes num__30 days to do the work . a ' s num__1 day ' s work = num__1 num__30 b ' s num__1 day ' s work = num__1 num__90 ( a + b ) ' s num__1 day ' s work = num__1 + num__1 = num__4 = num__2 num__30 num__90 num__90 num__45 a and b together can do the work in num__45 = num__22 num__1 days . num__2 num__2 answer is a <eor> a <eos> |
a |
subtract__3.0__1.0__ divide__60.0__2.0__ add__1.0__3.0__ divide__90.0__2.0__ round__22.0__ |
subtract__3.0__1.0__ subtract__90.0__60.0__ add__1.0__3.0__ divide__90.0__2.0__ round__22.0__ |
| ten circular coins are thrown on the street so that no two of them overlap or touch and no three of them have a tangent line in common . what is the total number of lines which are tangent to two of the coins ? <o> a ) num__60 <o> b ) num__100 <o> c ) num__180 <o> d ) num__240 <o> e ) num__360 |
there are n ( n - num__1 ) = num__2 different ways of choosing num__2 objects from a given collection of n objects . thus there are num__10 * num__4.5 = num__45 different ways of choosing num__2 of the num__10 coins . for each selection of two coins one can form num__4 tangent lines as shown at the right . therefore the answer is num__4 * num__45 = num__180 . correct answer c <eor> c <eos> |
c |
straight_angle__ straight_angle__ |
straight_angle__ straight_angle__ |
| the diameter of a wheel of cycle is num__21 cm . it moves slowly along a road . how far will it go in num__500 revolutions ? <o> a ) num__220 <o> b ) num__330 <o> c ) num__360 <o> d ) num__390 <o> e ) num__410 |
in revolution distance that wheel covers = circumference of wheel diameter of wheel = num__21 cm therefore circumference of wheel = π d = num__3.14285714286 × num__21 = num__66 cm so in num__1 revolution distance covered = num__66 cm in num__500 revolution distance covered = num__66 × num__500 cm = num__33000 cm = num__330.0 m = num__330 m answer b <eor> b <eos> |
b |
multiply__500.0__66.0__ round__330.0__ |
multiply__500.0__66.0__ round__330.0__ |
| a metallic sheet is of rectangular shape with dimensions num__40 m x num__30 m . from each of its corners a square is cut off so as to make an open box . if the length of the square is num__8 m the volume of the box ( in m cube ) is : <o> a ) num__3220 m cube <o> b ) num__4140 m cube <o> c ) num__2688 m cube <o> d ) num__4000 m cube <o> e ) none of these |
explanation : l = ( num__40 - num__16 ) m = num__24 m [ because num__8 + num__8 = num__16 ] b = ( num__30 - num__16 ) m = num__14 m h = num__8 m . volume of the box = ( num__24 x num__14 x num__8 ) m cube = num__2688 m cube . option c <eor> c <eos> |
c |
volume_rectangular_prism__8.0__14.0__24.0__ volume_rectangular_prism__8.0__14.0__24.0__ |
volume_rectangular_prism__8.0__14.0__24.0__ volume_rectangular_prism__8.0__14.0__24.0__ |
| a work can be finished in num__13 days by ten women . the same work can be finished in fifteen days by ten men . the ratio between the capacity of a man and a woman is <o> a ) num__14 : num__13 <o> b ) num__13 : num__15 <o> c ) num__12 : num__13 <o> d ) num__13 : num__12 <o> e ) num__14 : num__15 |
work done by num__10 women in num__1 day = num__0.0769230769231 work done by num__1 woman in num__1 day = num__1 / ( num__13 Ã — num__10 ) work done by num__16 men in num__1 day = num__0.0666666666667 work done by num__1 man in num__1 day = num__1 / ( num__15 Ã — num__10 ) ratio of the capacity of a man and woman = num__1 / ( num__15 Ã — num__10 ) : num__1 / ( num__13 Ã — num__10 ) = num__0.0666666666667 : num__0.0769230769231 = num__0.0666666666667 : num__0.0769230769231 = num__13 : num__15 answer is b . <eor> b <eos> |
b |
divide__1.0__13.0__ subtract__16.0__1.0__ round__13.0__ |
divide__1.0__13.0__ subtract__16.0__1.0__ divide__13.0__1.0__ |
| a num__3.0 stock produces num__10.0 . the market value of the stock is : <o> a ) rs . num__15 <o> b ) rs . num__25 <o> c ) rs . num__20 <o> d ) rs . num__30 <o> e ) rs . num__40 |
let the face value of the product is rs . num__100 it stock rs . num__3 market value of the stock = ( num__0.3 ) * num__100 = rs . num__30 answer : d <eor> d <eos> |
d |
percent__3.0__10.0__ percent__100.0__30.0__ |
percent__3.0__10.0__ percent__100.0__30.0__ |
| there are num__20 poles with a constant distance between each pole . a car takes num__20 second to reach the num__12 th pole . how much will it take to reach the last pole . <o> a ) num__34.4543 <o> b ) num__34.5455 <o> c ) num__34.45128 <o> d ) num__34.51288 <o> e ) num__34.41222 |
assuming the car starts at the first pole . to reach the num__12 th pole the car need to travel num__11 poles ( the first pole does n ' t count as the car is already there ) . num__11 poles num__20 seconds num__1 pole ( num__1.81818181818 ) seconds to reach the last ( num__20 th ) pole the car needs to travel num__19 poles . num__19 pole num__19 x ( num__1.81818181818 ) seconds = num__34.5455 seconds answer : b <eor> b <eos> |
b |
subtract__12.0__11.0__ divide__20.0__11.0__ subtract__20.0__1.0__ multiply__1.0__34.5455__ |
subtract__12.0__11.0__ divide__20.0__11.0__ subtract__20.0__1.0__ multiply__1.0__34.5455__ |
| murali travelled from city a to city b at a speed of num__40 kmph and from city b to city c at num__60 kmph . what is the average speed of murali from a to c given that the ratio of distances between a to b and b to c is num__10 : num__5 ? <o> a ) num__16 <o> b ) num__45 <o> c ) num__277 <o> d ) num__92 <o> e ) num__11 |
let the distances between city a to b and b to c be num__10 x km and num__5 x km respectively . total time taken to cover from a to c = ( num__10 x ) / num__40 + ( num__5 x ) / num__60 = ( num__30 x + num__10 x ) / num__120 = num__40 x / num__120 = x / num__3 average speed = ( num__10 x + num__5 x ) / ( x / num__3 ) = num__45 kmph . answer : b <eor> b <eos> |
b |
subtract__40.0__10.0__ divide__120.0__40.0__ add__40.0__5.0__ round__45.0__ |
subtract__40.0__10.0__ divide__120.0__40.0__ add__40.0__5.0__ add__40.0__5.0__ |
| a sells a bicycle to b at a profit of num__70.0 and b sells it to c at a loss of num__40.0 . find the resultant profit or loss . <o> a ) - num__4.0 <o> b ) num__2.0 <o> c ) - num__5.0 <o> d ) num__6.0 <o> e ) - num__7 % |
the resultant profit or loss = num__70 - num__40 - ( num__70 * num__40 ) / num__100 = + num__2.0 profit = num__2.0 answer is b <eor> b <eos> |
b |
percent__2.0__100.0__ |
percent__2.0__100.0__ |
| an empty pool being filled with water at a constant rate takes num__5 hours to fill to num__0.6 of its capacity . how much more time will it take to finish filling the pool ? <o> a ) num__5 hr num__30 min <o> b ) num__5 hr num__20 min <o> c ) num__4 hr num__48 min <o> d ) num__3 hr num__20 min <o> e ) num__2 hr num__40 min |
( num__0.6 ) of a pool / num__5 hours = num__0.12 ( the rate ) ( num__3 pools / num__25 hours ) = ( num__0.4 * pool ) / x hours cross multiply num__3 x = ( num__0.4 ) num__25 x = num__3 num__0.333333333333 num__0.333333333333 of an hour = num__20 minutes d <eor> d <eos> |
d |
divide__0.6__5.0__ multiply__5.0__0.6__ divide__3.0__0.12__ reverse__3.0__ subtract__25.0__5.0__ multiply__5.0__0.6__ |
divide__0.6__5.0__ multiply__5.0__0.6__ divide__3.0__0.12__ reverse__3.0__ subtract__25.0__5.0__ multiply__5.0__0.6__ |
| if num__5 x + y = num__15 num__5 y + z = num__25 and num__2 z + x = num__2 what is the value of x ? <o> a ) num__2 <o> b ) num__7 <o> c ) num__9 <o> d ) num__12 <o> e ) none of these |
explanation : num__5 x + y = num__15 – - - - - - - - i * num__5 num__5 y + z = num__25 – - - - - - - ii num__2 z + x = num__2 - - - - - - - - iii subtract ( i ) and ( ii ) num__25 x + num__5 y + num__0 = num__75 - num__0 + num__5 y + z = num__25 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - num__25 x + num__0 – z = num__50 … . . . . . . . . . . iv multiply ( iv ) * num__2 we get num__50 x – num__2 z = num__100 adding ( iii ) and ( iv ) we get num__50 x – num__2 z = num__100 + x + num__2 z = num__2 - - - - - - - - - - - - - - - - - - - - - - - - num__51 x = num__102 x = num__2 the value of x = num__2 answer a <eor> a <eos> |
a |
multiply__5.0__15.0__ multiply__25.0__2.0__ add__25.0__75.0__ add__2.0__100.0__ divide__100.0__50.0__ |
multiply__5.0__15.0__ multiply__25.0__2.0__ add__25.0__75.0__ add__2.0__100.0__ subtract__102.0__100.0__ |
| num__7 ^ num__8 is equal to which of the following ? <o> a ) num__5 num__764787 <o> b ) num__5 num__764788 <o> c ) num__5 num__764799 <o> d ) num__5 num__764800 <o> e ) num__5 num__764 |
801 |
units digit spotting can be fun cyclicity of num__7 to the power of anything num__7 ^ num__1 = num__7 num__7 ^ num__2 = num__9 num__7 ^ num__3 = num__3 num__7 ^ num__4 = num__1 num__7 ^ num__5 = num__7 - - - pattern formed - - so the cycle of this is num__4 . the fifth power will start the cycle again . therefore num__7 to the power of num__8 will give a units digit of num__1 . answer e <eor> e <eos> |
e |
e |
| if a company allocates num__15 percent of its budget to advertising num__10 percent to capital improvements and num__55 percent to salaries what fraction of its budget remains for other allocations ? <o> a ) num__0.8 <o> b ) num__0.6 <o> c ) num__0.3 <o> d ) num__0.2 <o> e ) num__0.1 |
num__100.0 - ( num__15.0 + num__10.0 + num__55.0 ) = num__20.0 remains for other allocations - - > num__20.0 = num__0.2 = num__0.2 . answer : d . <eor> d <eos> |
d |
percent__100.0__0.2__ |
percent__100.0__0.2__ |
| a can finish a work in num__8 days and b can do same work in half the time taken by a . then working together what part of same work they can finish in a day ? <o> a ) num__0.6 <o> b ) num__0.5 <o> c ) num__0.428571428571 <o> d ) num__0.375 <o> e ) none of these |
explanation : please note in this question we need to answer part of work for a day rather than complete work . it was worth mentioning here because many do mistake at this point in hurry to solve the question so lets solve now a ' s num__1 day work = num__0.125 b ' s num__1 day work = num__0.25 [ because b take half the time than a ] ( a + b ) ' s one day work = ( num__0.125 + num__0.25 ) = num__0.375 so in one day num__0.375 work will be done answer : d <eor> d <eos> |
d |
divide__1.0__8.0__ add__0.125__0.25__ add__0.125__0.25__ |
divide__1.0__8.0__ add__0.125__0.25__ add__0.125__0.25__ |
| if each side of the square is increased by num__25.0 find the percentage change in its area ? <o> a ) num__89.45 <o> b ) num__56.25 <o> c ) num__89.98 <o> d ) num__54.48 <o> e ) num__45.0 % |
let each side of the square be a . then area = a ^ num__2 new side = num__125 a / num__100 = num__5 a / num__4 new area = ( num__5 a / num__4 ) ^ num__2 = num__25 a ^ num__0.125 increase in area = ( num__25 a ^ num__0.125 - a ^ num__2 ) = num__9 a ^ num__0.125 increased % = ( num__9 a ^ num__0.125 * num__1 / a ^ num__2 * num__100 ) = num__56.25 answer is ( b ) <eor> b <eos> |
b |
percent__25.0__4.0__ percent__100.0__56.25__ |
percent__25.0__4.0__ percent__100.0__56.25__ |
| a circular mat with diameter num__16 inches is placed on a square tabletop each of whose sides is num__24 inches long . which of the following is closest to the fraction of the tabletop covered by the mat ? <o> a ) num__0.416666666667 <o> b ) num__0.4 <o> c ) num__0.333333333333 <o> d ) num__0.75 <o> e ) num__0.833333333333 |
so we are looking for the area of the cloth over the area of the table area of the cloth = ( pi ) ( r ) ^ num__2 which is about ( num__3 ) ( num__8 ) ( num__8 ) area of the table = ( num__24 ) ( num__24 ) so the quick way to estimate is looking at the fraction like this : ( num__0.125 ) ( num__2.66666666667 ) i hope this is easy to follow so with some simplification i get ( num__0.125 ) ( num__2.66666666667 ) = ( num__0.333333333333 ) answer is c <eor> c <eos> |
c |
square_perimeter__2.0__ multiply__0.125__2.6667__ multiply__0.125__2.6667__ |
power__2.0__3.0__ multiply__0.125__2.6667__ multiply__0.125__2.6667__ |
| a and b can do a work in num__12 days b and c in num__15 days c and a in num__20 days . if a b and c work together they will complete the work in : <o> a ) num__5 days <o> b ) num__7 num__0.833333333333 days <o> c ) num__10 days <o> d ) num__15 num__0.666666666667 days <o> e ) none of these |
( a + b ) ' s num__1 day ' s work = num__0.0833333333333 ( b + c ) ' s num__1 day ' s work = num__0.0666666666667 ; ( a + c ) ' s num__1 day ' s work = num__0.05 adding we get : num__2 ( a + b + c ) ' s num__1 day ' s work = ( num__0.0833333333333 + num__0.0666666666667 + num__0.05 ) = num__0.2 = num__0.2 ( a + b + c ) ' s num__1 day ' s work = num__0.1 so a b and c together can complete the work in num__10 days . correct option : c <eor> c <eos> |
c |
divide__1.0__12.0__ divide__1.0__15.0__ divide__1.0__20.0__ divide__2.0__20.0__ subtract__12.0__2.0__ round__10.0__ |
divide__1.0__12.0__ divide__1.0__15.0__ divide__1.0__20.0__ divide__2.0__20.0__ subtract__12.0__2.0__ round__10.0__ |
| a student scored an average of num__85 marks in num__3 subjects : physics chemistry and mathematics . if the average marks in physics and mathematics is num__90 and that in physics and chemistry is num__70 what are the marks in physics ? <o> a ) num__86 <o> b ) num__16 <o> c ) num__76 <o> d ) num__65 <o> e ) num__26 |
given m + p + c = num__85 * num__3 = num__255 - - - ( num__1 ) m + p = num__90 * num__2 = num__180 - - - ( num__2 ) p + c = num__70 * num__2 = num__140 - - - ( num__3 ) where m p and c are marks obtained by the student in mathematics physics and chemistry . p = ( num__2 ) + ( num__3 ) - ( num__1 ) = num__180 + num__140 - num__255 = num__65 answer : d <eor> d <eos> |
d |
multiply__85.0__3.0__ subtract__3.0__1.0__ multiply__90.0__2.0__ multiply__70.0__2.0__ multiply__1.0__65.0__ |
multiply__85.0__3.0__ subtract__3.0__1.0__ multiply__90.0__2.0__ multiply__70.0__2.0__ multiply__1.0__65.0__ |
| the distance from city a to city b is num__30 miles . while driving from city a to city b bob drives at a constant speed of num__40 miles per hour . alice leaves city a num__30 minutes after bob . what is the minimum constant speed in miles per hour that alice must exceed in order to arrive in city b before bob ? <o> a ) num__45 <o> b ) num__88 <o> c ) num__120 <o> d ) num__152 <o> e ) num__154 |
the time it takes bob to drive to city b is num__0.75 = num__0.75 hours . alice needs to take less than num__0.25 hours for the trip . alice needs to exceed a constant speed of num__30 / num__0.25 = num__120 miles per hour . the answer is c . <eor> c <eos> |
c |
divide__30.0__40.0__ divide__30.0__0.25__ round__120.0__ |
divide__30.0__40.0__ divide__30.0__0.25__ divide__30.0__0.25__ |
| sreenivas sells a table to shiva at num__10.0 profit and shiva sells it to mahesh at num__10.0 loss . at what price did sreenivas purchase the table if mahesh paid rs . num__2178 ? <o> a ) rs . num__2100 <o> b ) rs . num__2150 <o> c ) rs . num__2200 <o> d ) rs . num__2250 <o> e ) none of these |
let the cost price of table for sreenivas be rs . x and given that cost price of table for mahesh = rs . num__2178 . = > ( num__90.0 ) of ( num__110.0 ) of x = rs . num__2178 . = > ( num__0.9 ) ( num__1.1 ) x = num__2178 = > x = ( num__2178 * num__100 ) / ( num__9 * num__11 ) = > x = rs . num__2200 answer : c <eor> c <eos> |
c |
percent__10.0__90.0__ percent__10.0__110.0__ percent__100.0__2200.0__ |
percent__10.0__90.0__ percent__10.0__110.0__ percent__100.0__2200.0__ |
| if num__1 / a ^ num__2 + a ^ num__2 represents the diameter of circle o and num__1 / a + a = num__4 which of the following best approximates the circumference of circle o ? <o> a ) num__28 <o> b ) num__22 <o> c ) num__20 <o> d ) num__16 <o> e ) num__12 |
given that ( num__1 / a ) + a = num__3 square both sides of the equation : we get [ ( num__1 / a ) ^ num__2 + a ^ num__2 + num__2 * ( num__1 / a ) * a ] = num__9 = > ( num__1 / a ) ^ num__2 + a ^ num__2 + num__2 = num__9 = > ( num__1 / a ) ^ num__2 + a ^ num__2 = num__7 - - - - - - - - - - - - - - - - - ( num__1 ) diameter d = ( num__1 / a ) ^ num__2 + a ^ num__2 = num__7 ( from ( num__1 ) ) so radius = d / num__2 = num__3.5 circumference = num__2 * pi * r = num__2 * ( num__3.14285714286 ) * ( num__3.5 ) = num__20 so the answer should be c <eor> c <eos> |
c |
add__1.0__2.0__ add__4.0__3.0__ divide__7.0__2.0__ multiply__1.0__20.0__ |
add__1.0__2.0__ add__4.0__3.0__ divide__7.0__2.0__ multiply__1.0__20.0__ |
| the difference between simple and compound interest on rs . num__1500 for one year at num__10.0 per annum reckoned half - yearly is ? <o> a ) num__8.0 <o> b ) num__3.0 <o> c ) num__9.5 <o> d ) num__3.75 <o> e ) num__2.15 |
s . i . = ( num__1500 * num__10 * num__1 ) / num__100 = rs . num__150 c . i . = [ num__1500 * ( num__1 + num__0.05 ) num__2 - num__1500 ] = rs . num__153.75 difference = ( num__153.75 - num__150 ) = rs . num__3.75 answer : d <eor> d <eos> |
d |
percent__10.0__1500.0__ percent__100.0__3.75__ |
percent__10.0__1500.0__ percent__100.0__3.75__ |
| if - num__3 x + num__2 y = num__28 and num__3 x + num__6 y = num__84 what is the product of x and y ? <o> a ) num__264 . <o> b ) num__428 <o> c ) num__0 <o> d ) num__462 <o> e ) num__642 |
given - num__3 x + num__2 y = num__28 - - - eq num__1 num__3 x + num__6 y = num__84 - - eq num__2 sum both eqns we get num__8 y = num__112 = > y = num__14 sum num__2 y in eq num__1 = > - num__3 x - num__28 = num__28 . = > x = num__0 now xy = num__0 * num__14 = num__0 . option c is correct answer . <eor> c <eos> |
c |
subtract__3.0__2.0__ add__2.0__6.0__ add__28.0__84.0__ divide__28.0__2.0__ multiply__3.0__0.0__ |
subtract__3.0__2.0__ add__2.0__6.0__ add__28.0__84.0__ add__6.0__8.0__ multiply__3.0__0.0__ |
| num__0.03 x num__0.5 = ? <o> a ) num__0.0001 <o> b ) num__0.001 <o> c ) num__0.015 <o> d ) num__0.1 <o> e ) none of these |
explanation : num__3 x num__5 = num__15 . sum of decimal places = num__3 num__0.02 x num__0.5 = num__0.015 answer - c <eor> c <eos> |
c |
multiply__3.0__5.0__ multiply__0.03__0.5__ multiply__0.03__0.5__ |
multiply__3.0__5.0__ multiply__0.03__0.5__ subtract__0.03__0.015__ |
| a train num__100 meters long takes num__6 seconds to cross a man walking at num__5 kmph in the direction opposite to that of the train . find the speed of the train . <o> a ) num__45 kmph <o> b ) num__50 kmph <o> c ) num__55 kmph <o> d ) num__60 kmph <o> e ) num__70 kmph |
explanation : let the speed of the train be x kmph . speed of the train relative to man = ( x + num__5 ) kmph = ( x + num__5 ) × num__0.277777777778 m / sec . therefore num__100 / ( ( x + num__5 ) × num__0.277777777778 ) = num__6 < = > num__30 ( x + num__5 ) = num__1800 < = > x = num__55 speed of the train is num__55 kmph . answer : option c <eor> c <eos> |
c |
multiply__6.0__5.0__ round__55.0__ |
multiply__6.0__5.0__ round__55.0__ |
| a present value of a machine is $ num__2000 . its value depletiation rate is num__20.0 per annum then find the machine value before num__2 years ? <o> a ) $ num__3125 <o> b ) $ num__2945 <o> c ) $ num__3012 <o> d ) $ num__2545 <o> e ) $ num__3150 |
p = $ num__2000 r = num__20.0 t = num__2 years machine value before num__2 years = p / [ ( num__1 - r / num__100 ) ^ t ] = num__2000 * num__1.25 * num__1.25 = $ num__3125 answer is a <eor> a <eos> |
a |
percent__100.0__3125.0__ |
percent__100.0__3125.0__ |
| amit and ian paint a wall in alternating shifts . first amit paints alone then ian paints alone then amit paints alone etc . during each of his shifts amit paints num__0.5 of the remaining unpainted area of the wall while ian paints num__0.333333333333 of the remaining unpainted area of the wall during each of his shifts . if amit goes first what fraction of the wall ' s area will remain unpainted after amit has completed his num__2 th shift ? <o> a ) num__0.037037037037 <o> b ) num__0.0185185185185 <o> c ) num__0.0123456790123 <o> d ) num__0.166666666667 <o> e ) num__0.00462962962963 |
fraction of area unpainted after first shift of amit = num__1 - ( num__0.5 ) = num__0.5 fraction of area unpainted after first shift of ian = ( num__0.5 ) - ( num__0.333333333333 ) ( num__0.5 ) = ( num__0.666666666667 ) ( num__0.5 ) fraction of area unpainted after second shift of amit = ( num__0.5 ) ( num__0.666666666667 ) ( num__0.5 ) = num__0.166666666667 answer : option d <eor> d <eos> |
d |
multiply__0.5__2.0__ triangle_area__0.5__0.6667__ surface_cube__0.1667__ |
multiply__0.5__2.0__ triangle_area__0.5__0.6667__ surface_cube__0.1667__ |
| if n is a positive integer and n ^ num__2 is divisible by num__134 then the largest positive integer that must divide n is <o> a ) num__6 <o> b ) num__12 <o> c ) num__24 <o> d ) num__36 <o> e ) num__48 |
the question asks aboutthe largest positive integer that must divide n not could divide n . since the least value of n for which n ^ num__2 is a multiple of num__72 is num__12 then the largest positive integer that must divide n is num__12 . complete solution of this question is given above . please ask if anything remains unclear . i spent a few hours on this one alone and i ' m still not clear . i chose num__12 at first but then changed to num__48 . i ' m not a native speaker so here is how i interpreted this question : the largest positive integer that must divide n = the largest positive factor of n . since n is a variable ( i . e . n is moving ) so is its largest factor . please correct if i ' m wrong here . i know that if n = num__12 n ^ num__2 = num__144 = num__2 * num__72 ( satisfy the condition ) . when n = num__12 the largest factor of n is n itself which is num__12 . check : num__12 is the largest positive number that must divide num__12 - - > true however if n = num__48 n ^ num__2 = num__48 * num__48 = num__32 * num__72 ( satisfy the condition too ) . when n = num__48 the largest factor of n is n itself which is num__48 . check : num__48 is the largest positive number that must divide num__48 - - > true so i also notice that the keyword ismust notcould . the question is why is num__48 notmust divide num__48 but instead onlycould divide num__48 ? i ' m not clear right here . why is num__12 must divide num__12 ? what ' s the difference between them ? only restriction we have on positive integer n is that n ^ num__2 is divisible by num__72 . the least value of n for which n ^ num__2 is divisible by num__72 is num__12 thus nmustbe divisible by num__12 ( n is in any case divisible by num__12 ) . for all other values of n for which n ^ num__2 is divisible by num__72 n will still be divisible by num__12 . this means that n is always divisible by num__12 if n ^ num__2 is divisible by num__72 . now ask yourself : if n = num__24 is n divisible by num__48 ? no . so n is not always divisible by num__48 . c <eor> c <eos> |
c |
multiply__2.0__72.0__ multiply__2.0__12.0__ multiply__2.0__12.0__ |
multiply__2.0__72.0__ multiply__2.0__12.0__ multiply__2.0__12.0__ |
| an bus covers a certain distance at a speed of num__250 kmph in num__5 hours . to cover the samedistance in num__1 hr it must travel at a speed of ? <o> a ) num__600 km / hr <o> b ) num__720 km / hr <o> c ) num__730 km / hr <o> d ) num__750 km / hr <o> e ) num__760 km / hr |
distance = ( num__250 x num__5 ) = num__1250 km . speed = distance / time speed = num__1250 / ( num__1.66666666667 ) km / hr . [ we can write num__1 hours as num__1.66666666667 hours ] required speed = num__1250 x num__3 km / hr = num__750 km / hr . d <eor> d <eos> |
d |
multiply__250.0__5.0__ multiply__250.0__3.0__ round__750.0__ |
multiply__250.0__5.0__ multiply__250.0__3.0__ divide__750.0__1.0__ |
| if two numbers are in the ratio num__2 : num__3 . if num__10 is added to both of the numbers then the ratio becomes num__3 : num__4 then find the smallest number ? <o> a ) num__18 <o> b ) num__20 <o> c ) num__22 <o> d ) num__25 <o> e ) num__26 |
num__2 : num__3 num__2 x + num__10 : num__3 x + num__10 = num__3 : num__4 num__4 [ num__2 x + num__10 ] = num__3 [ num__3 x + num__10 ] num__8 x + num__40 = num__9 x + num__30 num__9 x - num__8 x = num__40 - num__30 x = num__10 then smallest number is = num__2 num__2 x = num__20 a <eor> a <eos> |
a |
multiply__2.0__4.0__ multiply__10.0__4.0__ multiply__3.0__10.0__ multiply__2.0__10.0__ multiply__2.0__9.0__ |
subtract__10.0__2.0__ multiply__10.0__4.0__ subtract__40.0__10.0__ subtract__30.0__10.0__ add__10.0__8.0__ |
| simplify : num__5793405 x num__9999 <o> a ) num__57928256595 <o> b ) num__59278256595 <o> c ) num__57928262559 <o> d ) num__59722422582 <o> e ) none of them |
num__5793405 x num__9999 = num__5793405 ( num__10000 - num__1 ) = num__57934050000 - num__5793405 = num__57928256595 . answer is a <eor> a <eos> |
a |
subtract__10000.0__9999.0__ multiply__5793405.0__10000.0__ multiply__5793405.0__9999.0__ multiply__5793405.0__9999.0__ |
subtract__10000.0__9999.0__ multiply__5793405.0__10000.0__ subtract__57934050000.0__5793405.0__ subtract__57934050000.0__5793405.0__ |
| a train sets off at num__2 p . m . at the speed of num__70 kmph . another train starts at num__1 : num__30 p . m . in the same direction at the rate of num__85 kmph . at what time the trains will meet ? <o> a ) num__10.18 p . m <o> b ) num__10.29 p . m <o> c ) num__8.30 p . m <o> d ) num__10.38 p . m <o> e ) num__10.32 p . m |
d = num__70 * num__1 ½ = num__105 km rs = num__85 – num__70 = num__15 t = num__7.0 = num__7 h num__1.30 + num__7 h = num__8.30 p . m . answer : c <eor> c <eos> |
c |
divide__30.0__2.0__ divide__105.0__15.0__ add__7.0__1.3__ round__8.3__ |
divide__30.0__2.0__ divide__105.0__15.0__ add__7.0__1.3__ multiply__1.0__8.3__ |
| the function f is defined for all positive integers u by the following rule . f ( u ) is the number of positive integers each of which is less than u and has no positive factor in common with u other than num__1 . if p is any prime number then f ( p ) = <o> a ) ( p + num__1 ) / num__2 <o> b ) p - num__2 <o> c ) p - num__1 <o> d ) ( p - num__1 ) / num__2 <o> e ) num__2 |
if not the wording the question would n ' t be as tough as it is now . the gmat often hides some simple concept in complicated way of delivering it . this question for instance basically asks : how many positive integers are less than given prime number p which have no common factor with p except num__1 . well as p is a prime all positive numbers less than p have no common factors with p ( except common factor num__1 ) . so there would be p - num__1 such numbers ( as we are looking number of integers less than p ) . for example : if p = num__5 how many numbers are less than num__5 having no common factors with num__5 : num__1 num__2 num__3 num__4 - - > num__5 - num__1 = num__4 . answer : c . <eor> c <eos> |
c |
add__1.0__2.0__ add__1.0__3.0__ reverse__1.0__ |
subtract__5.0__2.0__ subtract__5.0__1.0__ subtract__2.0__1.0__ |
| a rectangular paper when folded into two congruent parts had a perimeter of num__34 cm foer each part folded along one set of sides and the same is num__38 cm . when folded along the other set of sides . what is the area of the paper ? <o> a ) num__140 <o> b ) num__788 <o> c ) num__277 <o> d ) num__288 <o> e ) num__167 |
explanation : when folded along the breadth we have num__2 ( l / num__2 + b ) = num__34 or l + num__2 b = num__34 . . . . . . . . . . . ( num__1 ) when folded along the length we have num__2 ( l + b / num__2 ) = num__38 or num__2 l + b = num__38 . . . . . ( num__2 ) from num__1 & num__2 we get l = num__14 and b = num__10 area of the paper = num__14 * num__10 = num__140 sq cm answer : a ) num__140 <eor> a <eos> |
a |
multiply__10.0__14.0__ multiply__1.0__140.0__ |
multiply__10.0__14.0__ multiply__1.0__140.0__ |
| if jan spent num__40.0 of her monthly income on housing and num__20.0 less than she spent on housing on a car payment what percent of her monthly income did she have left after making both her housing payment and her car payment ? <o> a ) num__20.0 <o> b ) num__28.0 <o> c ) num__42.0 <o> d ) num__50.0 <o> e ) num__60 % |
lets say the total income of jan is num__100 then she spends num__40 on housing num__20.0 less than she spent on housing = num__40 − num__0.2 ∗ num__40 = num__3240 − num__0.2 ∗ num__40 = num__32 total expenditure = num__40 + num__32 = num__72 remaining = num__100 - num__72 = num__28 hence the answer is b num__28.0 <eor> b <eos> |
b |
percent__100.0__28.0__ |
percent__100.0__28.0__ |
| two trains started from the same point . at num__8 : num__00 am the first train traveled east at the rate of num__80 mph . at num__9 : num__00 am the second train traveled west at the rate of num__100 mph . at what time were they num__530 miles apart ? <o> a ) num__11 : num__30 am <o> b ) num__10 : num__30 am <o> c ) num__8 : num__30 am <o> d ) num__7 : num__30 am <o> e ) num__9 : num__30 am |
when the first train has traveled for t hours the second train will have traveled ( t - num__1 ) hours since it started num__1 hour late . hence if d num__1 and d num__2 are the distances traveled by the two trains then d num__1 = num__80 t and d num__2 = num__100 ( t - num__1 ) since the trains are traveling in opposite direction the total distance d between the two trains is given by d = d num__1 + d num__2 = num__180 t - num__100 for d to be num__530 miles we need to have num__180 t - num__100 = num__530 solve for t t = num__3 hours num__30 minutes . num__8 am + num__3 : num__30 = num__11 : num__30 am answer a <eor> a <eos> |
a |
subtract__9.0__8.0__ add__80.0__100.0__ add__1.0__2.0__ add__8.0__3.0__ round__11.0__ |
subtract__9.0__8.0__ add__80.0__100.0__ add__1.0__2.0__ add__8.0__3.0__ add__8.0__3.0__ |
| the average age of a husband and a wife is num__23 years when they were married six years ago but now the average age of the husband wife and child is num__20 years ( the child was born during the interval ) . what is the present age of the child ? <o> a ) num__7 years <o> b ) num__9 years <o> c ) num__6 years <o> d ) num__4 years <o> e ) num__2 years |
num__29 * num__2 = num__58 num__20 * num__3 = num__60 - - - - - - - - - - - num__2 years answer : e <eor> e <eos> |
e |
multiply__2.0__29.0__ subtract__23.0__20.0__ multiply__20.0__3.0__ divide__58.0__29.0__ |
multiply__2.0__29.0__ subtract__23.0__20.0__ multiply__20.0__3.0__ subtract__60.0__58.0__ |
| an article cost is rs num__912 after decreasing num__24.0 . so find the actual cost of the article ? <o> a ) num__1200 <o> b ) num__1500 <o> c ) num__1300 <o> d ) num__1000 <o> e ) num__1700 |
cp * ( num__0.76 ) = > num__912 cp = num__12 * num__100 = num__1200 answer a <eor> a <eos> |
a |
percent__100.0__1200.0__ |
percent__100.0__1200.0__ |
| six machines at a certain factory operate at the same constant rate . if five of these machines operating simultaneously take num__42 hours to fill a certain production order how many fewer hours does it take all six machines operating simultaneously to fill the same production order ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__7 <o> d ) num__9 <o> e ) num__12 |
the total work is num__5 * num__42 = num__210 machine hours the time required for six machines is num__35.0 = num__35 hours thus num__7 fewer hours . the answer is c . <eor> c <eos> |
c |
multiply__42.0__5.0__ subtract__42.0__35.0__ subtract__42.0__35.0__ |
multiply__42.0__5.0__ subtract__42.0__35.0__ subtract__42.0__35.0__ |
| num__3 pumps working num__8 hours a day can empty a tank in num__2 days . how many hours a day must num__12 pumps work to empty the tank in num__1 day ? <o> a ) num__4 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__13 |
num__3 pumps take num__16 hrs total ( num__8 hrs a day ) if num__1 pump will be working then it will need num__16 * num__3 = num__48 hrs num__1 pump need num__48 hrs if i contribute num__12 pumps then num__4.0 = num__4 hrs . answer : a <eor> a <eos> |
a |
multiply__8.0__2.0__ multiply__3.0__16.0__ add__3.0__1.0__ round__4.0__ |
multiply__8.0__2.0__ multiply__3.0__16.0__ add__3.0__1.0__ round__4.0__ |
| a stationary engine has enough fuel to move num__12 hours when its tank is num__0.8 full how much hours will it run when the tank is num__0.333333333333 full <o> a ) num__3 hours <o> b ) num__4 hours <o> c ) num__5 hours <o> d ) num__6 hours <o> e ) num__7 hours |
when engine is ( num__0.8 ) full it moves for num__12 hr so when tank is full it will move = num__12 / ( num__0.8 ) = num__15 hr so when it is num__0.333333333333 full it will move = num__15 * ( num__0.333333333333 ) = num__5 hours answer : c <eor> c <eos> |
c |
divide__12.0__0.8__ round__5.0__ |
divide__12.0__0.8__ round__5.0__ |
| if each edge of cube increased by num__50.0 the percentage increase in <o> a ) num__100 <o> b ) num__114 <o> c ) num__115 <o> d ) num__125 <o> e ) num__105 |
num__100 × ( num__150 ) / num__100 × ( num__150 ) / num__100 = num__225 = > num__125.0 answer is d . <eor> d <eos> |
d |
percent__100.0__125.0__ |
percent__100.0__125.0__ |
| what is the simplified result of following the steps below in order ? ( num__1 ) add num__5 y to num__2 z ( num__2 ) multiply the sum by num__3 ( num__3 ) subtract z + y from the product <o> a ) num__5 z + num__14 y <o> b ) num__5 x + num__16 y <o> c ) num__5 x + num__5 y <o> d ) num__6 x + num__4 y <o> e ) num__3 x + num__12 y |
num__3 ( num__5 y + num__2 z ) - z - y = num__14 y + num__5 z ' a ' is the answer <eor> a <eos> |
a |
multiply__1.0__5.0__ |
add__2.0__3.0__ |
| a train num__360 m long is running at a speed of num__45 km / hr . in what time will it pass a bridge num__140 m long ? <o> a ) num__40 sec <o> b ) num__42 sec <o> c ) num__45 sec <o> d ) num__48 sec <o> e ) num__50 sec |
explanation : formula for converting from km / hr to m / s : x km / hr = ( x x num__0.625 ) m / s . therefore speed = ( num__45 x num__0.277777777778 ) m / sec = num__12.5 m / sec . total distance to be covered = ( num__360 + num__140 ) m = num__500 m . formula for finding time = ( distance / speed ) required time = ( num__500 x num__0.08 ) sec = num__40 sec . answer is a <eor> a <eos> |
a |
add__360.0__140.0__ divide__500.0__12.5__ round__40.0__ |
add__360.0__140.0__ divide__500.0__12.5__ divide__500.0__12.5__ |
| two trains of length num__100 m and num__200 m are num__100 m apart . they start moving towards each other on parallel tracks at speeds num__108 kmph and num__72 kmph . in how much time will the trains cross each other ? <o> a ) num__8 <o> b ) num__7 <o> c ) num__9 <o> d ) num__2 <o> e ) num__1 |
relative speed = ( num__108 + num__72 ) * num__0.277777777778 = num__50 mps . the time required = d / s = ( num__100 + num__100 + num__200 ) / num__50 = num__8.0 = num__8 sec . answer : a <eor> a <eos> |
a |
subtract__108.0__100.0__ round__8.0__ |
subtract__108.0__100.0__ round__8.0__ |
| what is the probability of getting a number less than num__4 when a die is rolled ? <o> a ) num__0.333333333333 <o> b ) num__0.25 <o> c ) num__0.2 <o> d ) num__0.166666666667 <o> e ) num__0.5 |
total number of outcomes possible when a die is rolled = num__6 ( â ˆ µ any one face out of the num__6 faces ) n ( s ) = num__6 e = getting a no less than num__4 = { num__1 num__23 } hence n ( e ) = num__3 p ( e ) = n ( e ) / n ( s ) = num__0.5 answer e <eor> e <eos> |
e |
die_space__ negate_prob__0.5__ |
die_space__ negate_prob__0.5__ |
| a train num__110 m long is running with a speed of num__60 km / hr . in what time will it pass a man who is running at num__6 km / hr in the direction opposite to that in which the train is going ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
speed of train relative to man = num__60 + num__6 = num__66 km / hr . = num__66 * num__0.277777777778 = num__18.3333333333 m / sec . time taken to pass the men = num__110 * num__0.0545454545455 = num__6 sec . answer : option b <eor> b <eos> |
b |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ round__6.0__ |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ divide__110.0__18.3333__ |
| what is the units digit of num__18 ! + num__2 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
any number above num__4 ! such as num__5 ! num__6 ! etc . . . are always multiples of num__10 so their units digit is num__0 . the units digit of num__18 ! + num__2 is num__2 . the answer is b . <eor> b <eos> |
b |
add__2.0__4.0__ multiply__2.0__5.0__ subtract__4.0__2.0__ |
add__2.0__4.0__ add__4.0__6.0__ subtract__4.0__2.0__ |
| a and b can finish a work in num__14 days while a alone can do the same work in num__24 days . in how many days b alone will complete the work ? <o> a ) num__33 num__0.333333333333 days <o> b ) num__35 num__0.666666666667 days <o> c ) num__33 num__0.666666666667 days <o> d ) num__31 num__0.666666666667 days <o> e ) num__32 num__0.666666666667 days |
b = num__0.0714285714286 – num__0.0416666666667 = num__0.0297619047619 = > num__33.6 days = > num__33 num__0.666666666667 days answer : c <eor> c <eos> |
c |
round__33.0__ |
round__33.0__ |
| how many words with or without meaning can be formed using all letters of the word sharp using each letter exactly once ? <o> a ) num__90 <o> b ) num__100 <o> c ) num__110 <o> d ) num__120 <o> e ) num__130 |
the word sharp has exactly num__5 letters which are all different . therefore the number of words that can be formed = number of permutations of num__5 letters taken all at a time . = p ( num__5 num__5 ) = num__5 ! = num__5 x num__4 x num__3 x num__2 × num__1 = num__120 answer : d <eor> d <eos> |
d |
subtract__5.0__3.0__ subtract__3.0__2.0__ multiply__1.0__120.0__ |
subtract__5.0__3.0__ subtract__3.0__2.0__ multiply__1.0__120.0__ |
| in a college the ratio of the numbers of boys to the girls is num__8 : num__5 . if there are num__210 girls the total number of students in the college is ? <o> a ) num__562 <o> b ) num__356 <o> c ) num__452 <o> d ) num__416 <o> e ) num__546 |
let the number of boys and girls be num__8 x and num__5 x then num__5 x = num__210 x = num__42 total number of students = num__13 x = num__13 * num__42 = num__546 answer is e <eor> e <eos> |
e |
divide__210.0__5.0__ add__8.0__5.0__ multiply__42.0__13.0__ multiply__42.0__13.0__ |
divide__210.0__5.0__ add__8.0__5.0__ multiply__42.0__13.0__ multiply__42.0__13.0__ |
| if the average of num__6 consecutive number is num__27 then the largest of the number is ? <o> a ) num__27.5 <o> b ) num__28.5 <o> c ) num__29 <o> d ) num__29.5 <o> e ) num__30 |
x + x + num__1 + x + num__2 + x + num__3 + x + num__4 + x + num__0.833333333333 = num__27 num__6 x + num__15 = num__162 x = num__24.5 x + num__5 = num__29.5 answer : d <eor> d <eos> |
d |
divide__6.0__2.0__ subtract__6.0__2.0__ multiply__6.0__27.0__ subtract__6.0__1.0__ add__24.5__5.0__ add__24.5__5.0__ |
add__1.0__2.0__ add__1.0__3.0__ multiply__6.0__27.0__ add__1.0__4.0__ add__24.5__5.0__ add__24.5__5.0__ |
| for any a and b that satisfy | a – b | = b – a and a > num__0 then | a + num__5 | + | - b | + | b – a | + | ab | = <o> a ) a - b <o> b ) num__2 a - num__2 b <o> c ) num__2 a - num__2 b - num__5 <o> d ) ab + num__2 b + num__5 <o> e ) a - b - num__5 |
observation - num__1 : | a – b | = b – a which is possible only when signs of a and b are same since given a > num__0 so we figure out that a and b are both positive observation - num__2 : | a – b | must be non - negative and so should be the value of b - a which is possible only when absolute value of b is greater than or equal to absolute value of a now you may choose the values of a and b based on above observations e . g . b = num__5 and a = num__1 and check the value of given functions and options | a + num__5 | + | - b | + | b – a | + | ab | = | num__1 + num__5 | + | - num__5 | + | num__5 – num__1 | + | num__1 * num__5 | = num__20 ab + num__2 b + num__5 = num__1 * num__5 + num__10 + num__5 = num__20 all other options are not equal to num__20 answer : d <eor> d <eos> |
d |
multiply__5.0__2.0__ multiply__1.0__2.0__ |
multiply__5.0__2.0__ multiply__1.0__2.0__ |
| in a can there is a mixture of milk and water in the ratio num__1 : num__5 . if it is filled with an additional num__2 litres of milk the can would be full and ratio of milk and water would become num__2 : num__5 . find the capacity of the can ? <o> a ) num__14 <o> b ) num__44 <o> c ) num__48 <o> d ) num__50 <o> e ) num__56 |
let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = num__0.166666666667 ( t - num__2 ) after adding milk quantity of milk in the mixture = num__0.285714285714 t . num__2 t / num__7 - num__2 = num__0.166666666667 ( t - num__2 ) num__5 t = num__84 - num__14 = > t = num__14 . answer : a <eor> a <eos> |
a |
add__5.0__2.0__ multiply__2.0__7.0__ multiply__1.0__14.0__ |
add__5.0__2.0__ multiply__2.0__7.0__ divide__14.0__1.0__ |
| gold is num__19 times as heavy as water and copper num__9 times as heavy as water . the ratio in which these two metals be mixed so that the mixture is num__15 times as heavy as water is : <o> a ) num__1 : num__2 <o> b ) num__2 : num__3 <o> c ) num__3 : num__2 <o> d ) num__19 : num__135 <o> e ) none of these |
explanation : this question can be solved using weighted average formula . if two quantities of weights m n have concentrations x y are mixed then final concentration = mx + ny / m + n take num__1 unit of gold and x units of copper . num__1 × num__19 + x × num__9.0 + x = num__15 ⇒ num__19 + num__9 x = num__15 + num__15 x ⇒ x = num__0.666666666667 so they are to be mixed in the ratio num__1 : x = num__1 : num__0.666666666667 or num__3 : num__2 correct option : c <eor> c <eos> |
c |
subtract__3.0__1.0__ divide__9.0__3.0__ |
subtract__3.0__1.0__ divide__9.0__3.0__ |
| a began a business with rs . num__85000 . he was joined afterwards by b with ks . num__42500 . for how much period does b join if the profits at the end of the year are divided in the ratio of num__3 : num__1 ? <o> a ) num__4 months <o> b ) num__5 months <o> c ) num__6 months <o> d ) num__8 months <o> e ) num__9 months |
suppose b joined for x months . then ( num__85000 * num__12 ) / ( num__42500 * x ) = num__3 . or x = ( num__85000 * num__12 ) / ( num__42500 * num__3 ) = num__8 . so b joined for num__8 months . answer d <eor> d <eos> |
d |
multiply__1.0__8.0__ |
multiply__1.0__8.0__ |
| the length of the bridge which a train num__160 metres long and travelling at num__45 km / hr can cross in num__30 seconds is : <o> a ) num__200 m <o> b ) num__215 m <o> c ) num__245 m <o> d ) num__250 m <o> e ) num__270 m |
let the length of the bridge : l the equation now is l + num__160 / num__12.5 m / s ( num__45 km / hr or num__12.5 m / s ) = num__30 solving l = num__215 m answer : b <eor> b <eos> |
b |
round__215.0__ |
round__215.0__ |
| what is the thousandths digit in the decimal equivalent of num__0.0086 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__3 <o> d ) num__5 <o> e ) num__8 |
num__0.0086 = num__43 / ( num__5 * num__10 ^ num__3 ) = ( num__8.6 ) * num__10 ^ - num__3 = num__8.6 * num__10 ^ - num__3 = . num__0086 thousandths digit = num__8 answer e <eor> e <eos> |
e |
divide__43.0__5.0__ multiply__10.0__8.6__ round_down__8.6__ round_down__8.6__ |
divide__43.0__5.0__ multiply__10.0__8.6__ round_down__8.6__ round_down__8.6__ |
| eesha bought two varities of rice costing num__50 rs per kg and num__60 rs per kg and mixed them in some ratio . then she sold that mixture at num__70 rs per kg making a profit of num__20.0 what was the ratio of the micxture ? <o> a ) num__1 : num__9 <o> b ) num__1 by num__5 <o> c ) num__1 : num__7 <o> d ) num__1 : num__3 <o> e ) num__1 : num__1 |
selling price of the mixture = num__70 and profit = num__20.0 cost price of the mixture = num__70 × num__100120 = num__70 × num__5670 × num__100120 = num__70 × num__56 by applying alligation rule : so ratio = num__60 − num__1753 : num__1753 − num__5060 − num__1753 : num__1753 − num__50 = num__1 : num__5 answer : b <eor> b <eos> |
b |
percent__20.0__5.0__ |
percent__20.0__5.0__ |
| a train covers a distance of num__12 km in num__10 min . if it takes num__6 sec to pass a telegraph post then the length of the train is ? <o> a ) num__999 <o> b ) num__279 <o> c ) num__120 metre <o> d ) num__99 <o> e ) num__21 |
speed = ( num__1.2 * num__60 ) km / hr = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . length of the train = num__20 * num__6 = num__120 m . answer : c <eor> c <eos> |
c |
divide__12.0__10.0__ hour_to_min_conversion__ multiply__12.0__6.0__ multiply__12.0__10.0__ round__120.0__ |
divide__12.0__10.0__ multiply__10.0__6.0__ multiply__12.0__6.0__ multiply__12.0__10.0__ multiply__12.0__10.0__ |
| in a pair of fractions fractions a is twice the fraction b and the product of two fractions is num__0.08 . what is the value of fraction a ? <o> a ) num__0.2 <o> b ) num__0.04 <o> c ) num__0.4 <o> d ) num__0.6 <o> e ) data inadequate |
a = num__2 b ab = num__0.08 num__2 b * b = num__0.08 b ^ num__2 = num__0.04 b = num__0.2 a = num__0.4 answer : c <eor> c <eos> |
c |
divide__0.08__2.0__ divide__0.08__0.2__ divide__0.08__0.2__ |
divide__0.08__2.0__ multiply__0.2__2.0__ multiply__0.2__2.0__ |
| a boat running downstream covers a distance of num__22 km in num__4 hours while for covering the same distance upstream it takes num__5 hours . what is the speed of the boat in still water ? <o> a ) num__5 kmph <o> b ) num__4 kmph <o> c ) num__4.75 kmph <o> d ) num__4.95 kmph <o> e ) num__4.65 kmph |
speed downstream = num__5.5 = num__5.5 kmph speed upstream = num__4.4 = num__4.4 kmph speed of the boat in still water = num__5.5 + num__4.4 / num__2 = num__4.95 kmph answer : option d <eor> d <eos> |
d |
divide__22.0__4.0__ divide__22.0__5.0__ round__4.95__ |
divide__22.0__4.0__ divide__22.0__5.0__ round__4.95__ |
| what is the least number of squares tiles required to pave the floor of a room num__15 m num__17 cm long and num__9 m num__2 cm broad ? <o> a ) num__814 tiles <o> b ) num__800 <o> c ) num__100 <o> d ) num__200 <o> e ) num__456 |
length of largest tile = h . c . f . of num__1517 cm and num__902 cm = num__41 cm . area of each tile = ( num__41 x num__41 ) cm num__2 . required number of tiles = ( num__1517 x num__902 ) / num__41 x num__41 = num__814 . answer : a <eor> a <eos> |
a |
round__814.0__ |
round__814.0__ |
| the sum of two consecutive integers is num__31 . find the numbers . <o> a ) num__17 num__18 <o> b ) num__7 num__8 <o> c ) num__15 num__16 <o> d ) num__1 num__2 <o> e ) num__8 num__9 |
n + ( n + num__1 ) = num__31 num__2 n + num__1 = num__31 num__2 n = num__30 n = num__15 answer : c <eor> c <eos> |
c |
subtract__31.0__1.0__ divide__30.0__2.0__ round__15.0__ |
subtract__31.0__1.0__ divide__30.0__2.0__ round__15.0__ |
| there are two examinations rooms a and b . if num__10 students are sent from a to b then the number of students in each room is the same . if num__20 candidates are sent from b to a then the number of students in a is double the number of students in b . the number of students in room a is : <o> a ) num__20 <o> b ) num__80 <o> c ) num__100 students <o> d ) num__200 <o> e ) num__120 |
let the number of students in rooms a and b be x and y respectively . then x - num__10 = y + num__10 x - y = num__20 . . . . ( i ) and x + num__20 = num__2 ( y - num__20 ) x - num__2 y = - num__60 . . . . ( ii ) solving ( i ) and ( ii ) we get : x = num__100 y = num__80 . the required answer a = num__100 . answer : c <eor> c <eos> |
c |
divide__20.0__10.0__ add__20.0__60.0__ add__20.0__80.0__ |
divide__20.0__10.0__ add__20.0__60.0__ add__20.0__80.0__ |
| the original price of a camera was displayed as a whole dollar amount . after adding sales tax of num__8 percent the final price was also a whole dollar amount . which of the following could be the final price of the camera ? <o> a ) $ num__115 <o> b ) $ num__101 <o> c ) $ num__100 <o> d ) $ num__108 <o> e ) $ num__109 |
final price = ( num__1 + num__0.08 ) * original price = num__1.08 * original price from options given only num__108 is divisible by num__1.08 as it is stated op is whole dollar amount . hence d <eor> d <eos> |
d |
add__1.0__0.08__ multiply__1.0__108.0__ |
add__1.0__0.08__ multiply__1.0__108.0__ |
| in a barrel of juice there is num__30 liters ; in a barrel of beer there are num__50 liters . if the price ratio between barrels of juice to a barrel of beer is num__3 : num__4 what is the price ratio between one liter of juice and one liter of beer ? <o> a ) num__3 : num__2 . <o> b ) num__2 : num__1 . <o> c ) num__3 : num__1 . <o> d ) num__4 : num__3 . <o> e ) num__15 : num__4 |
price of num__30 l juice = num__3 x num__1 l = num__3 x / num__30 price of num__50 l beer = num__4 x num__1 l = num__4 x / num__50 ratio of num__1 l price = num__3 x / num__7.5 x / num__50 = num__15 : num__4 e is the answer <eor> e <eos> |
e |
subtract__4.0__3.0__ divide__30.0__4.0__ subtract__30.0__15.0__ |
subtract__4.0__3.0__ divide__30.0__4.0__ divide__15.0__1.0__ |
| a tour group of num__25 people paid a total of $ num__630 for entrance to a museum . if this price included a num__5.0 sales tax and all the tickets cost the same amount what was the face value of each ticket price without the sales tax ? choices <o> a ) $ num__22 <o> b ) $ num__23.94 <o> c ) $ num__24 <o> d ) $ num__25.20 <o> e ) $ num__30 |
soln : - num__25.2 = x + num__0.05 x num__25.2 = num__1.05 x x = ( num__630 * num__100 ) / ( num__25 * num__105 ) = num__63 * num__10 * num__0.190476190476 * num__5 = num__24 answer : c <eor> c <eos> |
c |
divide__630.0__25.0__ divide__5.0__0.05__ add__5.0__100.0__ divide__630.0__63.0__ divide__25.2__1.05__ divide__25.2__1.05__ |
divide__630.0__25.0__ divide__5.0__0.05__ add__5.0__100.0__ divide__630.0__63.0__ divide__25.2__1.05__ divide__25.2__1.05__ |
| after getting num__2 successive discounts a shirt with a list price of rs num__150 is available at rs num__105 . if the second discount is num__12.55 find the first discount <o> a ) num__38.0 <o> b ) num__15.0 <o> c ) num__20.0 <o> d ) num__12.0 <o> e ) num__36 % |
let the first discount be x % then num__87.5 of ( num__100 - x ) % of num__150 = num__105 num__87.5 / num__100 * ( num__100 - x ) / num__100 * num__450 = num__150 = > num__105 = > num__100 - x = ( num__105 * num__100 * num__100 ) / ( num__150 * num__87.5 ) = num__80 x = ( num__100 - num__80 ) = num__20 first discount = num__20.0 ans : c <eor> c <eos> |
c |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| a father can do a certain job in x hours . his son takes twice as long to do the job . working together they can do the job in num__6 hours . how many hours does the father take to do the job ? <o> a ) num__9 <o> b ) num__18 <o> c ) num__12 <o> d ) num__20 <o> e ) num__16 |
explanatory answers the father completes the job in x hours . so the son will take num__2 x hours to complete the same job . in an hour the father will complete num__1 / x of the total job . in an hour the son will complete num__0.5 x of the total job . so if the father and son work together in an hour they will complete num__1 / x + num__0.5 x of the job . i . e . in an hour they will complete num__1.5 x of the job . the question states that they complete the entire task in num__6 hours if they work together . i . e . they complete num__0.166666666667 < sup > th < / sup > of the task in an hour . equating the two information we get num__1.5 x = num__0.166666666667 cross multiplying and solving for x we get num__2 x = num__18 or x = num__9 . the father takes num__9 hours to complete the job . answer a <eor> a <eos> |
a |
divide__1.0__2.0__ add__0.5__1.0__ divide__1.0__6.0__ multiply__6.0__1.5__ round__9.0__ |
divide__1.0__2.0__ add__0.5__1.0__ divide__1.0__6.0__ divide__18.0__2.0__ divide__9.0__1.0__ |
| num__12 persons can complete the work in num__18 days . after working for num__6 days num__4 more persons added to complete the work fast . in how many more days they will complete the work ? <o> a ) num__10 days <o> b ) num__9 days <o> c ) num__8 days <o> d ) num__7 days <o> e ) num__6 days |
total work num__12 * num__18 = num__216 units after num__6 days work finished num__6 * num__12 = num__72 units remaining work num__216 - num__72 = num__144 units remaing days = num__144 ( num__12 + num__4 ) = num__9 days answer : b <eor> b <eos> |
b |
multiply__12.0__18.0__ multiply__12.0__6.0__ subtract__216.0__72.0__ round__9.0__ |
multiply__12.0__18.0__ multiply__12.0__6.0__ subtract__216.0__72.0__ subtract__18.0__9.0__ |
| if num__0 < a < b and k = ( num__2 a + num__6 b ) / b which of the following must be true ? <o> a ) k < num__2 <o> b ) k < num__7 <o> c ) k < num__8 <o> d ) k > num__9 <o> e ) k > num__11 |
here ' s another approach : k = ( num__2 a + num__6 b ) / b = num__2 a / b + num__6 b / b = num__2 ( a / b ) + num__6 since num__0 < a < b we know that a / b is less than num__1 which means that num__2 ( a / b ) is some number less than num__2 . so we get k = ( some number less than num__2 ) + num__6 from here we can see that k must be less than num__8 answer : c <eor> c <eos> |
c |
add__2.0__6.0__ add__2.0__6.0__ |
add__2.0__6.0__ add__2.0__6.0__ |
| a man buys an article and sells it at a profit of num__20.0 . if he had bought it at num__20.0 less and sold it for rs . num__78 less he could have gained num__25.0 . what is the cost price ? <o> a ) s . num__370 <o> b ) s . num__375 <o> c ) s . num__375 <o> d ) s . num__390 <o> e ) s . num__300 |
cp num__1 = num__100 sp num__1 = num__120 cp num__2 = num__80 sp num__2 = num__80 * ( num__1.25 ) = num__100 num__20 - - - - - num__100 num__78 - - - - - ? = > num__390 answer : d <eor> d <eos> |
d |
percent__100.0__390.0__ |
percent__100.0__390.0__ |
| a car moves at the speed of num__80 km / hr . what is the speed of the car in metres per second ? <o> a ) num__8 m / sec <o> b ) num__20 num__0.111111111111 m / sec <o> c ) num__22 num__2 / num__0 m / sec <o> d ) can not be determined <o> e ) none of these |
speed = ( num__80 * num__0.277777777778 ) m / sec = num__22.2222222222 m / sec = num__22 num__0.222222222222 m / sec . correct option : c <eor> c <eos> |
c |
subtract__22.2222__22.0__ round__22.0__ |
subtract__22.2222__22.0__ round__22.0__ |
| p can do a work in num__60 days . q can do the same work in num__9 days and r can do the same in num__12 days . q and r start the work and leave after num__3 days . p finishes the remaining work in - - - days . <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__15 <o> e ) num__25 |
work done by p in num__1 day = num__0.0166666666667 work done by q in num__1 day = num__0.111111111111 work done by r in num__1 day = num__0.0833333333333 work done by q and r in num__1 day = num__0.111111111111 + num__0.0833333333333 = num__0.194444444444 work done by q and r in num__3 days = num__3 Ã — num__0.194444444444 = num__0.583333333333 remaining work = num__1 â € “ num__0.583333333333 = num__0.416666666667 number of days in which p can finish the remaining work = ( num__0.416666666667 ) / ( num__0.0166666666667 ) = num__25 option e <eor> e <eos> |
e |
divide__1.0__60.0__ divide__1.0__9.0__ divide__1.0__12.0__ add__0.1111__0.0833__ subtract__1.0__0.5833__ round__25.0__ |
divide__1.0__60.0__ divide__1.0__9.0__ divide__1.0__12.0__ add__0.1111__0.0833__ subtract__1.0__0.5833__ divide__25.0__1.0__ |
| if the price of num__25 toys is num__500 then what will the price of num__10 toys ? <o> a ) num__144 <o> b ) num__361 <o> c ) num__117 <o> d ) num__287 <o> e ) num__200 |
one toy price = num__20.0 = num__20 num__10 toy price = num__10 * num__20 = num__200 answer : e <eor> e <eos> |
e |
divide__500.0__25.0__ multiply__10.0__20.0__ multiply__10.0__20.0__ |
divide__500.0__25.0__ multiply__10.0__20.0__ multiply__10.0__20.0__ |
| the average of five integers is num__71 and none of these integers is greater than num__100 . if the average of three of the integers is num__76 what is the least possible value of one of the other two integers ? <o> a ) num__5 <o> b ) num__15 <o> c ) num__20 <o> d ) num__21 <o> e ) num__27 |
when it comes to averages we know thataverage value = ( sum of n values ) / n we can rewrite this into a useful formula : sum of n values = ( average value ) ( n ) the average of five integers is num__71 so the sum of all num__5 integers = ( num__71 ) ( num__5 ) = num__355 the average of three of the integers is num__65 so the sum of the num__3 integers = ( num__76 ) ( num__3 ) = num__228 so the sum of the num__2 remaining integers = num__355 - num__228 = num__127 if the sum of the num__2 remaining integers = num__127 and we want to minimize one value we must maximize the other value . num__100 is the maximum value so let num__1 integer = num__100 which means the other must equal num__27 answer : e <eor> e <eos> |
e |
subtract__76.0__71.0__ multiply__71.0__5.0__ multiply__76.0__3.0__ subtract__5.0__3.0__ subtract__355.0__228.0__ subtract__3.0__2.0__ subtract__127.0__100.0__ multiply__1.0__27.0__ |
subtract__76.0__71.0__ multiply__71.0__5.0__ multiply__76.0__3.0__ subtract__5.0__3.0__ subtract__355.0__228.0__ subtract__3.0__2.0__ subtract__127.0__100.0__ divide__27.0__1.0__ |
| in a flight of num__600 km an aircraft was slowed down due to bad weather . its average speed for the trip was reduced by num__200 km / hr and the time of flight increased by num__30 minutes . the duration of the flight is ? <o> a ) num__1 hr <o> b ) num__3 hr <o> c ) num__2 hr <o> d ) num__6 hr <o> e ) num__8 hr |
a num__1 hr let the duration of the flight be x hours . then num__600 / x - num__600 / ( x + num__0.5 ) = num__200 x ( num__2 x + num__1 ) = num__3 num__2 x num__2 + x - num__3 = num__0 ( num__2 x + num__3 ) ( x - num__1 ) = num__0 x = num__1 hr <eor> a <eos> |
a |
reverse__0.5__ divide__600.0__200.0__ round_down__0.5__ reverse__1.0__ |
reverse__0.5__ divide__600.0__200.0__ round_down__0.5__ reverse__1.0__ |
| one pump drains one - half of a pond in num__8 hours and then a second pump starts draining the pond . the two pumps working together finish emptying the pond in one - half hour . how long would it take the second pump to drain the pond if it had to do the job alone ? <o> a ) num__1 hour <o> b ) num__1.1 hour <o> c ) num__3 hours <o> d ) num__5 hours <o> e ) num__6 hours |
first pump drains num__0.5 of the tank in num__8 hours so num__16 hours it will take to drain the full tank . let num__2 nd pump drains the full tank in a hours so both together can drain ( num__0.0625 + num__1 / a ) part in num__1 hour son in num__0.5 hour they drain num__0.5 * ( num__0.0625 + num__1 / a ) part of the tank given that in num__0.5 hour they drain num__0.5 of the tank hence we can say num__0.5 * ( num__0.0625 + num__1 / a ) = num__0.5 solving u get a = num__1.06666666667 = num__1.1 hence answer is b <eor> b <eos> |
b |
divide__8.0__0.5__ divide__16.0__8.0__ divide__0.5__8.0__ multiply__0.5__2.0__ round__1.1__ |
divide__8.0__0.5__ divide__16.0__8.0__ divide__0.5__8.0__ multiply__0.5__2.0__ divide__1.1__1.0__ |
| at what rate of compound interest per annum will a sum of $ num__1200 becomes num__1348.32 in num__2 years ? <o> a ) num__5.0 <o> b ) num__6.0 <o> c ) num__7.0 <o> d ) num__8.0 <o> e ) num__10 % |
let the rate be r % p . a . num__1200 * ( num__1 + r / num__100 ) ^ num__2 = num__1348.32 ( num__1 + r / num__100 ) ^ num__2 = num__1.1236 = num__1.1236 ( num__1 + r / num__100 ) ^ num__2 = ( num__1.06 ) ^ num__2 num__1 + r / num__100 = num__1.06 r = num__6.0 answer is b <eor> b <eos> |
b |
percent__100.0__6.0__ |
percent__100.0__6.0__ |
| if two numbers are in the ratio num__5 : num__2 . if num__20 is subtracted from both of the numbers then the ratio becomes num__4 : num__2 then find the smallest number ? <o> a ) num__30 <o> b ) num__20 <o> c ) num__40 <o> d ) num__50 <o> e ) num__12 |
num__5 : num__2 num__5 x - num__20 : num__3 x - num__20 = num__4 : num__2 num__2 [ num__5 x - num__20 ] = num__4 [ num__3 x - num__20 ] num__10 x - num__40 = num__12 x - num__80 num__12 x - num__10 x = num__80 - num__40 x = num__20 then smallest number is = num__2 x num__2 x = num__40 c <eor> c <eos> |
c |
subtract__5.0__2.0__ multiply__5.0__2.0__ multiply__2.0__20.0__ add__2.0__10.0__ multiply__2.0__40.0__ multiply__2.0__20.0__ |
subtract__5.0__2.0__ multiply__5.0__2.0__ multiply__2.0__20.0__ add__2.0__10.0__ multiply__2.0__40.0__ subtract__80.0__40.0__ |
| a train num__150 m long passes a man running at num__6 kmph in the direction opposite to that of the train in num__6 seconds . the speed of the train is <o> a ) num__54 kmph <o> b ) num__60 kmph <o> c ) num__66 kmph <o> d ) num__72 kmph <o> e ) num__84 kmph |
speed of train relative to man : num__25.0 * num__3.6 km / hr = num__90 km / hr let speed of train = x therefore x + num__6 = num__90 x = num__90 - num__6 x = num__84 km / hr answer : e <eor> e <eos> |
e |
divide__150.0__6.0__ multiply__25.0__3.6__ subtract__90.0__6.0__ round__84.0__ |
divide__150.0__6.0__ multiply__25.0__3.6__ subtract__90.0__6.0__ subtract__90.0__6.0__ |
| of num__800 surveyed students num__20.0 of those who read book a also read book b and num__25.0 of those who read book b also read book a . if each student read at least one of the books what is the difference between the number of students who read only book a and the number of students who read only book b ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__30 <o> d ) num__55 <o> e ) num__100 |
say the number of students who read book a is a and the number of students who read book b is b . given that num__20.0 of those who read book a also read book b and num__25.0 of those who read book b also read book a so the number of students who read both books is num__0.2 a = num__0.25 b - - > a = num__1.25 b . since each student read at least one of the books then { total } = { a } + { b } - { both } - - > num__800 = num__1.25 b + b - num__0.25 b - - > b = num__400 a = num__1.25 b = num__500 and { both } = num__0.25 b = num__100 . the number of students who read only book a is { a } - { both } = num__500 - num__100 = num__400 ; the number of students who read only book b is { b } - { both } = num__400 - num__100 = num__300 ; the difference is num__400 - num__300 = num__100 . answer : e . <eor> e <eos> |
e |
percent__20.0__500.0__ percent__20.0__500.0__ |
percent__20.0__500.0__ percent__20.0__500.0__ |
| what is the ratio w of the surface area of a cube to the surface area of a rectangular solid identical to the cube in all ways except that its length has been doubled ? <o> a ) num__0.25 <o> b ) num__0.375 <o> c ) num__0.5 <o> d ) num__0.6 <o> e ) num__2 |
let x be the side of the cube . therefore x ^ num__2 * num__6 = surface area . the rectangle is the same other than length is num__2 x . the width and height are the same as the cube . num__2 * w * h + num__2 * w * l + num__2 * h * l = num__2 x ^ num__2 + num__4 x ^ num__2 + num__4 x ^ num__2 = num__10 x ^ num__2 . num__6 x ^ num__0.2 x ^ num__2 = num__0.6 = w . answer = d <eor> d <eos> |
d |
triangle_area__6.0__0.2__ triangle_area__2.0__0.6__ |
triangle_area__6.0__0.2__ triangle_area__2.0__0.6__ |
| a fruit seller had some apples . he sells num__40.0 apples and still he has num__420 . originally he had how many apples ? <o> a ) num__400 <o> b ) num__500 <o> c ) num__700 <o> d ) num__450 <o> e ) num__650 |
suppose originally he had x apples then ( num__100 - num__40 ) % of x = num__420 num__60 x / num__100 = num__420 x = num__700 answer is c <eor> c <eos> |
c |
percent__100.0__700.0__ |
percent__100.0__700.0__ |
| if two girls starting from same point walking in the opposite directions with num__5 km / hr and num__10 km / hr as average speeds respectively . then the distance between them after num__5 hours is ? <o> a ) num__45 <o> b ) num__55 <o> c ) num__65 <o> d ) num__75 <o> e ) num__85 |
explanation : total distance = distance traveled by person a + distance traveled by person b = ( num__5 Ã — num__5 ) + ( num__10 Ã — num__5 ) = num__25 + num__50 = num__75 km answer : d <eor> d <eos> |
d |
multiply__5.0__10.0__ add__25.0__50.0__ round__75.0__ |
multiply__5.0__10.0__ add__25.0__50.0__ add__25.0__50.0__ |
| mary sold boxes of butter cookies . ann sold num__5 times as much as she did . num__18 boxes of cookies were sold that day how many boxes did mary sell ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__10 <o> e ) num__18 |
# of boxes of cookies mary sold = x ann sold num__5 times more = num__5 x x + num__5 x = num__18 num__6 x = num__18 x = num__3.0 = num__3 answer : a <eor> a <eos> |
a |
divide__18.0__6.0__ divide__18.0__6.0__ |
divide__18.0__6.0__ divide__18.0__6.0__ |
| what will come in place of the x in the following number series ? num__46080 num__3840 num__384 num__48 ? num__2 num__1 <o> a ) num__1 <o> b ) num__3 <o> c ) num__5 <o> d ) num__8 <o> e ) num__9 |
num__3840.0 = num__3840 num__384.0 = num__384 num__48.0 = num__48 num__8.0 = num__8 num__2.0 = num__2 num__1.0 = num__1 d <eor> d <eos> |
d |
divide__384.0__48.0__ divide__384.0__48.0__ |
divide__384.0__48.0__ divide__384.0__48.0__ |
| in a jar there are num__3 golf balls and num__2 ping pong balls . what is the probability of drawing at least one golf ball when drawing two consecutive balls randomly ? <o> a ) num__0.25 <o> b ) num__0.5 <o> c ) num__0.6 <o> d ) num__0.9 <o> e ) num__0.8 |
p ( at least one gold ball ) = num__1 - p ( no golf ball so num__2 ping pong balls ) = num__1 - num__0.4 * num__0.25 = num__0.9 . answer : d . <eor> d <eos> |
d |
subtract__3.0__2.0__ round__0.9__ |
subtract__3.0__2.0__ round__0.9__ |
| a train num__360 m long runs with a speed of num__45 km / hr . what time will it take to pass a platform of num__150 m long ? <o> a ) num__38 sec <o> b ) num__35 sec <o> c ) num__44 sec <o> d ) num__40.8 sec <o> e ) none of these |
speed = num__45 km / hr = num__45 Ã — ( num__0.277777777778 ) m / s = num__12.5 = num__12.5 = num__12.5 m / s total distance = length of the train + length of the platform = num__360 + num__150 = num__510 meter time taken to cross the platform = num__510 / ( num__12.5 ) = num__510 Ã — num__0.08 = num__40.8 seconds answer : d <eor> d <eos> |
d |
add__360.0__150.0__ multiply__0.08__510.0__ round__40.8__ |
add__360.0__150.0__ divide__510.0__12.5__ divide__510.0__12.5__ |
| if x + y = num__3 and x – y = num__3 then x ^ num__2 - y ^ num__2 = <o> a ) - num__4 <o> b ) num__4 <o> c ) num__10 <o> d ) num__9 <o> e ) num__40 |
the fastest approach has already been shown . here ' s one more option . given : x + y = num__3 x – y = num__3 add the two equations to get : num__2 x = num__6 which means x = num__3 if x = num__3 we can plug that value into either equation to conclude that y = num__0 if x = num__3 and y = num__0 then x ² - y ² = num__9 answer : d <eor> d <eos> |
d |
multiply__3.0__2.0__ add__3.0__6.0__ add__3.0__6.0__ |
multiply__3.0__2.0__ add__3.0__6.0__ add__3.0__6.0__ |
| the average weight of num__8 person ' s increases by num__2.5 kg when a new person comes in place of one of them weighing num__65 kg . what might be the weight of the new person ? <o> a ) num__76 kg <o> b ) num__76.5 kg <o> c ) num__85 kg <o> d ) data inadequate <o> e ) none of these |
total weight increased = ( num__8 x num__2.5 ) kg = num__20 kg . weight of new person = ( num__65 + num__20 ) kg = num__85 kg . answer : option c <eor> c <eos> |
c |
multiply__8.0__2.5__ add__65.0__20.0__ add__65.0__20.0__ |
multiply__8.0__2.5__ add__65.0__20.0__ add__65.0__20.0__ |
| a certain country had a total annual expenditure of $ num__2.4 x num__10 ^ num__11 last year . if the population of the country was num__240 million last year what was the per capita expenditure ? <o> a ) $ num__500 <o> b ) $ num__1000 <o> c ) $ num__2000 <o> d ) $ num__3000 <o> e ) $ num__5 |
000 |
total expenditure / population = per capita expenditure hence ( num__24 x num__10 ^ num__11 ) / num__240 num__000 num__000 = ( num__24 x num__10 ^ num__11 ) / ( num__24 x num__10 ^ num__8 ) = num__1 x num__10 ^ ( num__11 - num__8 ) = num__1 x num__10 ^ num__3 = num__1000 . answer is b . <eor> b <eos> |
b |
b |
| a cycle is bought for rs . num__900 and sold for rs . num__1260 find the gain percent ? <o> a ) num__40.0 <o> b ) num__20.0 <o> c ) num__80.0 <o> d ) num__30.0 <o> e ) num__24 % |
num__900 - - - - num__360 num__100 - - - - ? = > num__40.0 answer : a <eor> a <eos> |
a |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| for positive integers m and n which of the following can be written as n ^ num__2 ? <o> a ) m + num__2 <o> b ) m + num__3 <o> c ) m + num__1 <o> d ) m + num__4 <o> e ) m - num__2 |
plug values : if m = num__8 then m + num__1 = num__9 the question asks which of the following can be written as y ^ num__2 . if m = num__8 then m + num__1 can be written as num__3 ^ num__2 . answer : d <eor> d <eos> |
d |
add__8.0__1.0__ add__2.0__1.0__ divide__8.0__2.0__ |
add__8.0__1.0__ add__2.0__1.0__ add__1.0__3.0__ |
| a and b can together finish a work in num__40 days . they worked together for num__10 days and then b left . after another num__6 days a finished the remaining work . in how many days a alone can finish the job ? <o> a ) num__10 <o> b ) num__25 <o> c ) num__60 <o> d ) num__8 <o> e ) num__20 |
a + b num__10 days work = num__10 * num__0.025 = num__0.25 remaining work = num__1 - num__0.25 = num__0.75 num__0.75 work is done by a in num__6 days whole work will be done by a in num__6 * num__1.33333333333 = num__8 days answer is d <eor> d <eos> |
d |
divide__10.0__40.0__ multiply__40.0__0.025__ subtract__1.0__0.25__ divide__1.0__0.75__ divide__6.0__0.75__ round__8.0__ |
multiply__10.0__0.025__ multiply__40.0__0.025__ subtract__1.0__0.25__ divide__1.0__0.75__ divide__6.0__0.75__ round__8.0__ |
| a train having a length of num__275 meter is running at a speed of num__60 kmph . in what time it will pass a man who is running at num__6 kmph in the direction opposite to that of the train <o> a ) num__10 sec <o> b ) num__8 sec <o> c ) num__15 sec <o> d ) num__4 sec <o> e ) num__2 sec |
explanation : distance = num__275 m relative speed = num__60 + num__6 = num__66 kmph ( since both the train and the man are in moving in opposite direction ) = num__66 × num__0.277777777778 mps = num__18.3333333333 mps time = distance / speed = num__275 / ( num__18.3333333333 ) = num__15 s answer : option c <eor> c <eos> |
c |
add__60.0__6.0__ divide__275.0__18.3333__ round__15.0__ |
add__60.0__6.0__ divide__275.0__18.3333__ divide__275.0__18.3333__ |
| solve for x : num__2 x – y = ( num__0.75 ) x + num__6 . <o> a ) ( y + num__6 ) / num__5 <o> b ) num__4 ( y + num__6 ) / num__5 <o> c ) ( y + num__6 ) <o> d ) num__4 ( y - num__6 ) / num__5 <o> e ) num__1.25 ( y - num__6 ) |
solution : num__2 x – y = ( num__0.75 ) x + num__6 . or num__2 x - ( num__0.75 ) x = y + num__6 . or ( num__8 x - num__3 x ) / num__4 = y + num__6 . or num__5 x / num__4 = y + num__6 . or num__5 x = num__4 ( y + num__6 ) . or num__5 x = num__4 y + num__24 . or x = ( num__4 y + num__24 ) / num__5 . therefore x = num__4 ( y + num__6 ) / num__5 . correct answer b <eor> b <eos> |
b |
add__2.0__6.0__ divide__6.0__2.0__ subtract__6.0__2.0__ add__2.0__3.0__ multiply__6.0__4.0__ subtract__6.0__2.0__ |
add__2.0__6.0__ divide__6.0__2.0__ subtract__6.0__2.0__ add__2.0__3.0__ multiply__6.0__4.0__ subtract__6.0__2.0__ |
| the average of num__9 numbers is num__7 and the average of num__7 other numbers is num__6 . what is the average of all num__16 numbers ? <o> a ) num__6 num__0.5625 <o> b ) num__8 <o> c ) num__7 num__7 ⁄ num__8 <o> d ) num__7 num__1 ⁄ num__2 <o> e ) num__7 num__1 ⁄ num__4 |
the average of num__9 numbers is num__7 . . . . the sum of those numbers is ( num__9 ) ( num__7 ) = num__63 . the average of num__7 other numbers is num__6 . . . . the sum of those numbers is ( num__7 ) ( num__6 ) = num__42 the sum of all num__16 numbers is num__63 + num__42 = num__105 . the average of those numbers is num__6.5625 = num__6 num__0.5625 . final answer : [ reveal ] spoiler : a <eor> a <eos> |
a |
multiply__9.0__7.0__ multiply__7.0__6.0__ add__42.0__63.0__ divide__105.0__16.0__ divide__9.0__16.0__ round_down__6.5625__ |
multiply__9.0__7.0__ multiply__7.0__6.0__ add__42.0__63.0__ divide__105.0__16.0__ divide__9.0__16.0__ round_down__6.5625__ |
| num__3 x ^ num__2 - x - num__4 = <o> a ) ( num__3 x - num__4 ) ( x - num__2 ) <o> b ) ( num__3 x - num__4 ) ( x + num__2 ) <o> c ) ( num__3 x - num__4 ) ( x - num__3 ) <o> d ) ( num__3 x - num__2 ) ( x + num__1 ) <o> e ) none of above |
num__3 x ^ num__2 + num__2 x - num__8 = num__3 x ^ num__2 + num__3 x - num__4 x - num__4 = num__3 x ( x + num__1 ) - num__4 ( x + num__1 ) = ( num__3 x - num__4 ) ( x + num__1 ) option d <eor> d <eos> |
d |
multiply__2.0__4.0__ subtract__3.0__2.0__ multiply__3.0__1.0__ |
multiply__2.0__4.0__ subtract__3.0__2.0__ add__2.0__1.0__ |
| in a garden there are num__10 rows and num__12 columns of mango trees . the distance between the two trees is num__2 metres and a distance of five metre is left from all sides of the boundary of the garden . what is the length of the garden ? <o> a ) num__32 <o> b ) num__24 <o> c ) num__26 <o> d ) num__28 <o> e ) num__30 |
between the num__12 mango trees there are num__11 gaps and each gap has num__2 meter length also num__5 meter is left from all sides of the boundary of the garden . hence length of the garden = ( num__11 Ã — num__2 ) + num__5 + num__5 = num__32 meter answer is a . <eor> a <eos> |
a |
divide__10.0__2.0__ round__32.0__ |
divide__10.0__2.0__ round__32.0__ |
| what number is that to which if its half be added and from the sum num__20 be subtracted the remainder will be a fourth of the number itself ? <o> a ) num__5.33333333333 <o> b ) num__8.0 <o> c ) num__4.0 <o> d ) num__16.0 <o> e ) none |
let x be put for the number required . then by the conditions proposed x + x / num__2 - num__20 = x / num__4 and reducing the equation x = num__16 . proof num__16 + num__8.0 - num__20 = num__4.0 . answer c <eor> c <eos> |
c |
subtract__20.0__4.0__ divide__16.0__2.0__ subtract__20.0__16.0__ |
subtract__20.0__4.0__ divide__16.0__2.0__ subtract__20.0__16.0__ |
| how many of the positive factors of num__33 are not factors of num__45 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
factors of num__33 - num__1 num__3 num__11 num__33 factors of num__45 - num__1 num__3 num__5 num__9 num__15 num__45 comparing both we have three factors of num__33 which are not factors of num__45 - num__11 num__33 the answer is c <eor> c <eos> |
c |
gcd__33.0__45.0__ divide__33.0__3.0__ divide__45.0__5.0__ divide__45.0__3.0__ subtract__3.0__1.0__ |
gcd__33.0__45.0__ divide__33.0__3.0__ divide__45.0__5.0__ divide__45.0__3.0__ subtract__3.0__1.0__ |
| a box contains ten bulbs out of which num__5 are defective . if five bulbs are chosen at random find the probability that atleast one bulb is good ? <o> a ) num__0.996031746032 <o> b ) num__0.995884773663 <o> c ) num__0.996138996139 <o> d ) num__0.996240601504 <o> e ) num__0.995967741935 |
required probability = num__1 - num__5 c num__0.5 c num__5 = num__1 - num__0.00396825396825 = num__0.996031746032 answer : a <eor> a <eos> |
a |
negate_prob__0.004__ negate_prob__0.004__ |
negate_prob__0.004__ negate_prob__0.004__ |
| in how many ways can num__5 people a b c d e be seated in a row such that c and d are not seated next to each other as well as a and b are not seated next to each other ? <o> a ) num__384 <o> b ) num__396 <o> c ) num__576 <o> d ) num__24 <o> e ) num__696 |
number of total arrangements = num__5 ! restriction num__1 = abcd not next to each other - - > let say ab and cd are considered as one unit respectively restriction num__2 = ab is not the same as ba + cd is not the same as dc - - > the number will increase by num__2 * num__2 total number of arrangements - number out of restrictions = result num__5 ! - ( num__4 ! * num__2 * num__2 ) = num__120 - ( num__24 * num__2 * num__2 ) = num__24 answer d <eor> d <eos> |
d |
coin_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
coin_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
| a train num__280 m long running with a speed of num__63 km / hr will pass a tree in ? <o> a ) num__15 sec <o> b ) num__16 sec <o> c ) num__18 sec <o> d ) num__20 sec <o> e ) num__25 sec |
speed = num__63 * num__0.277777777778 = num__17.5 m / sec time taken = num__280 * num__0.0571428571429 = num__16 sec answer : b <eor> b <eos> |
b |
divide__280.0__17.5__ round__16.0__ |
divide__280.0__17.5__ divide__280.0__17.5__ |
| the cost of carpeting a room num__18 m long with a carpet num__75 cm wide at num__45 paise per meter is rs . num__81 . the breadth of the room is : <o> a ) num__7 m <o> b ) num__7.5 metre <o> c ) num__5.5 m <o> d ) num__6.5 m <o> e ) num__8.5 m |
length of the carpet = total cost / rate / m = num__180.0 = num__180 m area of the carpet = num__180 * num__0.75 = num__135 m num__2 breadth of the room = ( area / length ) = num__7.5 = num__7.5 m answer : b <eor> b <eos> |
b |
multiply__0.75__180.0__ divide__135.0__18.0__ round__7.5__ |
multiply__0.75__180.0__ divide__135.0__18.0__ divide__135.0__18.0__ |
| the area of the floor of a room is num__20 m num__2 that of a longer wall num__15 m num__2 and of the shorter wall num__12 m num__2 find the edge of the new cube ? <o> a ) num__40 m num__3 <o> b ) num__50 m num__3 <o> c ) num__60 m num__3 <o> d ) num__90 m num__3 <o> e ) num__40 m num__3 |
explanation : lb = num__20 ; lh = num__15 ; fh = num__12 ( lbh ) num__2 = num__20 * num__15 * num__12 = > lbh = num__60 m num__3 answer : option c <eor> c <eos> |
c |
square_perimeter__15.0__ square_perimeter__15.0__ |
square_perimeter__15.0__ multiply__20.0__3.0__ |
| the price of num__2 sarees and num__4 shirts is rs . num__1600 . with the same money one can buy num__1 saree and num__6 shirts . if one wants to buy num__12 shirts how much shall he have to pay ? <o> a ) num__1200 <o> b ) num__2400 <o> c ) num__4800 <o> d ) can not be determined <o> e ) none |
let the price of a saree and a shirt be rs . x and rs . y respectively . then num__2 x + num__4 y = num__1600 . . . . ( i ) and x + num__6 y = num__1600 . . . . ( ii ) divide equation ( i ) by num__2 we get the below equation . = > x + num__2 y = num__800 . - - - ( iii ) solve the ( ii ) and ( iii ) equations then will get therefore y = num__200 . now apply value of y in ( iii ) = > x + num__2 x num__200 = num__800 = > x + num__400 = num__800 therefore x = num__400 solving ( i ) and ( ii ) we get x = num__400 y = num__200 . cost of num__12 shirts = rs . ( num__12 x num__200 ) = rs . num__2400 . option b <eor> b <eos> |
b |
divide__1600.0__2.0__ divide__800.0__4.0__ multiply__2.0__200.0__ add__1600.0__800.0__ add__1600.0__800.0__ |
divide__1600.0__2.0__ divide__800.0__4.0__ multiply__2.0__200.0__ add__1600.0__800.0__ add__1600.0__800.0__ |
| it costs $ num__2 for the first num__0.166666666667 hour to use the laundry machine at the laundromat . after the first ¼ hour it costs $ num__15 per hour . if a certain customer uses the laundry machine for num__2 hours and num__25 minutes how much will it cost him ? <o> a ) $ num__42.25 . <o> b ) $ num__40.25 . <o> c ) $ num__38.25 . <o> d ) $ num__36.25 . <o> e ) $ num__34.25 |
num__2 hrs num__25 min = num__145 min first num__10 min - - - - - - > $ num__2 time left is num__135 min . . . now num__60 min costs $ num__15 num__1 min costs $ num__0.25 num__145 min costs $ num__0.25 * num__145 = > $ num__36.25 so total cost will be $ num__36.25 + $ num__2 = > $ num__38.25 hence answer will be c <eor> c <eos> |
c |
subtract__25.0__15.0__ subtract__145.0__10.0__ hour_to_min_conversion__ divide__15.0__60.0__ multiply__0.25__145.0__ add__2.0__36.25__ round__38.25__ |
subtract__25.0__15.0__ subtract__145.0__10.0__ hour_to_min_conversion__ divide__15.0__60.0__ multiply__0.25__145.0__ add__2.0__36.25__ add__2.0__36.25__ |
| the events a and b are independent the probability that event a occurs is greater than num__0 and the probability that event a occurs is twice the probability that event b occurs . the probability that at least one of events a and b occurs is num__5 times the probability that both events a and b occur . what is the probability that event a occurs ? <o> a ) num__0.0222222222222 <o> b ) num__0.5 <o> c ) num__0.125 <o> d ) num__0.666666666667 <o> e ) num__0.181818181818 |
let us say probability of a occuring is a . let us say probability of b occuring is b . a = num__2 b probability ( either a or b or both ) = num__5 times probability ( a and b ) a * ( num__1 - b ) + b * ( num__1 - a ) + ab = num__5 * ab substituting a = num__2 b in the second equation : num__2 b * ( num__1 - b ) + b * ( num__1 - num__2 b ) + num__2 b * b = num__5 * num__2 b * b num__3 b - num__2 b ^ num__2 = num__110 b ^ num__2 num__3 b = num__12 b ^ num__2 b = num__0.25 = num__0.25 so a = num__2 b = num__0.5 <eor> b <eos> |
b |
coin_space__ negate_prob__0.0__ choose__3.0__1.0__ negate_prob__0.5__ |
coin_space__ negate_prob__0.0__ choose__3.0__1.0__ negate_prob__0.5__ |
| if a is the set of numbers possible for the eighth term in a sequence of eight consecutive integers that has num__6 as its fourth term which of the following is a ? <o> a ) num__2 and num__11 <o> b ) num__2 and num__10 <o> c ) num__2 and num__9 <o> d ) num__3 and num__9 <o> e ) num__3 and num__10 |
first take num__6 as the num__4 th term from begiining num__8 th term will be num__10 . now take num__6 as the num__4 th last term from the beginning num__8 th last term will be num__2 . hence b . <eor> b <eos> |
b |
add__6.0__4.0__ subtract__6.0__4.0__ subtract__6.0__4.0__ |
add__6.0__4.0__ subtract__6.0__4.0__ subtract__6.0__4.0__ |
| if a b c and d are integers ; w x y and z are prime numbers ; w < x < y < z ; and ( wa ) ( xb ) ( yc ) ( zd ) = num__660 ( wa ) ( xb ) ( yc ) ( zd ) = num__660 what is the value of ( a + b ) – ( c + d ) ? <o> a ) – num__1 <o> b ) num__0 <o> c ) ( a + b ) – ( c + d ) = num__1 <o> d ) num__2 <o> e ) num__3 |
num__660660 = num__2 ^ num__2 x num__3 ^ num__1 x num__5 ^ num__1 x num__11 ^ num__1 w < x < y < z = num__2 < num__3 < num__5 < num__11 so can can say - w = num__2 x = num__3 y = num__5 z = num__11 hence a = num__2 b = c = d = num__1 ( a + b ) – ( c + d ) will be ( num__2 + num__1 ) – ( num__1 + num__1 ) = num__1 answer will be ( c ) <eor> c <eos> |
c |
subtract__3.0__2.0__ add__2.0__3.0__ reverse__1.0__ |
subtract__3.0__2.0__ add__2.0__3.0__ subtract__2.0__1.0__ |
| a goods train runs at the speed of num__72 km / hr and crosses a num__150 m long platform in num__26 sec . what is the length of the goods train ? <o> a ) num__370 <o> b ) num__782 <o> c ) num__278 <o> d ) num__270 <o> e ) num__881 |
speed = num__72 * num__0.277777777778 = num__20 m / sec . time = num__26 sec . let the length of the train be x meters . then ( x + num__150 ) / num__26 = num__20 x = num__370 m . answer : a <eor> a <eos> |
a |
round__370.0__ |
round__370.0__ |
| a man could buy a certain number of notebooks for rs . num__300 . if each notebook cost is rs . num__5 more he could have bought num__10 notebooks less for the same amount . find the price of each notebook ? <o> a ) num__10 <o> b ) num__8 <o> c ) num__6 <o> d ) num__4 <o> e ) num__2 |
let the price of each note book be rs . x . let the number of note books which can be brought for rs . num__300 each at a price of rs . x be y . hence xy = num__300 = > y = num__300 / x ( x + num__5 ) ( y - num__10 ) = num__300 = > xy + num__5 y - num__10 x - num__50 = xy = > num__5 ( num__300 / x ) - num__10 x - num__50 = num__0 = > - num__150 + x num__2 + num__5 x = num__0 multiplying both sides by - num__0.1 x = > x num__2 + num__15 x - num__10 x - num__150 = num__0 = > x ( x + num__15 ) - num__10 ( x + num__15 ) = num__0 = > x = num__10 or - num__15 as x > num__0 x = num__10 . answer : a <eor> a <eos> |
a |
multiply__5.0__10.0__ divide__300.0__150.0__ reverse__10.0__ add__5.0__10.0__ reverse__0.1__ |
multiply__5.0__10.0__ divide__300.0__150.0__ reverse__10.0__ add__5.0__10.0__ reverse__0.1__ |
| if a code word is defined to be a sequence of different letters chosen from the num__10 letters a b c d e f g h i and j what is the ratio of the number of num__5 - letter code words to the number of num__6 - letter code words ? <o> a ) num__1 <o> b ) num__0.25 <o> c ) num__0.2 <o> d ) num__0.333333333333 <o> e ) num__0.5 |
method # num__1 : using the permutation formula # of num__5 letter code words : npr = n ! / ( n – r ) ! = num__10 ! / ( num__10 – num__5 ) ! = num__10 * num__9 * num__8 * num__7 * num__6 # of num__6 letter code words : npr = num__10 ! / ( num__10 – num__6 ) ! = num__10 * num__9 * num__8 * num__7 * num__6 * num__5 # of num__5 words / # of num__4 words = ( num__10 * num__9 * num__8 * num__7 * num__6 ) / ( num__10 * num__9 * num__8 * num__7 * num__6 * num__5 ) = num__0.2 c <eor> c <eos> |
c |
subtract__6.0__5.0__ subtract__10.0__1.0__ subtract__9.0__1.0__ add__6.0__1.0__ subtract__10.0__6.0__ reverse__5.0__ reverse__5.0__ |
subtract__6.0__5.0__ subtract__10.0__1.0__ subtract__9.0__1.0__ add__6.0__1.0__ subtract__10.0__6.0__ reverse__5.0__ reverse__5.0__ |
| a single reservoir supplies the petrol to the whole city while the reservoir is fed by a single pipeline filling the reservoir with the stream of uniform volume . when the reservoir is full and if num__40000 liters of petrol is used daily the suply fails in num__90 days . if num__32000 liters of petrol is used daily it fails in num__60 days . how much petrol can be used daily with out the supply ever failing ? <o> a ) num__64000 liters <o> b ) num__56000 liters of petrol <o> c ) num__78000 liters <o> d ) num__60000 liters <o> e ) none of these |
explanation : let x liter be the per day filling and v litr be the capacity of the reservoir then num__90 x + v = num__40000 × num__90 - - - - - ( num__1 ) num__60 x + v = num__32000 × num__60 - - - - - - ( num__2 ) solving eq . ( num__1 ) and ( num__2 ) we get x = num__56000 hence num__56000 liters per day can be used without the failure of supply . answer : b <eor> b <eos> |
b |
round__56000.0__ |
round__56000.0__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__180 m <o> b ) num__200 m <o> c ) num__240 m <o> d ) num__320 m <o> e ) none |
sol . speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x metres . then x + num__8.33333333333 = num__15 ⇔ x + num__300 = num__540 ⇔ x = num__240 m . answer c <eor> c <eos> |
c |
multiply__20.0__15.0__ divide__300.0__36.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ divide__300.0__36.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
| if p # q denotes the least common multiple of p and q then t = ( ( num__12 # num__16 ) # ( num__18 # num__24 ) ) = ? <o> a ) num__216 <o> b ) num__180 <o> c ) num__144 <o> d ) num__108 <o> e ) num__72 |
there are several ways to find the least common multiple of two numbers . in this case the most efficient method is to use the greatest common factor : ( a * b ) / ( gcf ab ) = lcm ab the greatest common factor of num__12 and num__16 is num__4 . so num__12 # num__16 = num__12 * num__4.0 = num__48 . the greatest common factor of num__18 and num__24 is num__6 . so num__18 # num__24 = num__18 * num__4.0 = num__72 finally the greatest common factor of num__48 and num__72 is num__24 . so t = ( ( num__12 # num__16 ) # ( num__18 # num__24 ) ) = num__48 # num__72 = ( num__48 * num__72 ) / num__24 = num__2 * num__72 = num__144 the correct answer is c . <eor> c <eos> |
c |
subtract__16.0__12.0__ multiply__12.0__4.0__ subtract__18.0__12.0__ multiply__12.0__6.0__ divide__12.0__6.0__ multiply__24.0__6.0__ multiply__24.0__6.0__ |
subtract__16.0__12.0__ multiply__12.0__4.0__ divide__24.0__4.0__ multiply__12.0__6.0__ divide__12.0__6.0__ multiply__24.0__6.0__ multiply__24.0__6.0__ |
| in a certain brick wall each row of bricks above the bottom row contains one less brick than the row just below it . if there are num__4 rows in all and a total of num__134 bricks in the wall how many bricks does the bottom row contain ? <o> a ) num__30 <o> b ) num__31 <o> c ) num__32 <o> d ) num__33 <o> e ) num__34 |
the bottom row has x bricks x + x - num__1 + x - num__2 + x - num__3 = num__134 num__4 x - num__6 = num__134 num__4 x = num__128 x = num__32 answer : c <eor> c <eos> |
c |
subtract__4.0__1.0__ add__4.0__2.0__ subtract__134.0__6.0__ divide__128.0__4.0__ divide__128.0__4.0__ |
subtract__4.0__1.0__ add__4.0__2.0__ subtract__134.0__6.0__ divide__128.0__4.0__ divide__128.0__4.0__ |
| reena took a loan of num__1000 with simple interest for as many years as the rate of interest . if she paid num__40 as interest at the end of the loan period what was the rate of interest ? <o> a ) num__3.8 <o> b ) num__2 <o> c ) num__6 <o> d ) can not be determined <o> e ) none |
explanation : let rate = r % and time = r years . then ( num__1000 x r x r ) / num__100 = num__40 num__10 r  ² = num__40 r  ² = num__4 r = num__2 . answer : option b <eor> b <eos> |
b |
percent__40.0__10.0__ percent__2.0__100.0__ |
percent__40.0__10.0__ percent__2.0__100.0__ |
| if a : b = num__4 : num__1 then find ( a - num__3 b ) / ( num__2 a - b ) ? <o> a ) num__0.857142857143 <o> b ) num__0.142857142857 <o> c ) num__0.714285714286 <o> d ) num__1.5 <o> e ) num__0.571428571429 |
answer : option b a / b = num__4.0 = > a = num__4 b ( a - num__3 b ) / ( num__2 a - b ) = ( num__4 b - num__3 b ) / ( num__8 b - b ) = b / num__7 b = > num__0.142857142857 <eor> b <eos> |
b |
multiply__4.0__2.0__ add__4.0__3.0__ reverse__7.0__ reverse__7.0__ |
multiply__4.0__2.0__ subtract__8.0__1.0__ reverse__7.0__ reverse__7.0__ |
| in a car wheel two spokes cover num__10 degree . then for the entire car how many spokes are there ? <o> a ) num__128 <o> b ) num__168 <o> c ) num__248 <o> d ) num__268 <o> e ) num__288 |
given num__2 spokes cover num__10 degrees so for num__360 degrees . - - - - > num__360 * num__0.2 = num__72 . . so for entire car num__4 * num__72 = num__288 answer : e <eor> e <eos> |
e |
divide__2.0__10.0__ multiply__0.2__360.0__ multiply__4.0__72.0__ multiply__4.0__72.0__ |
divide__2.0__10.0__ multiply__0.2__360.0__ multiply__4.0__72.0__ multiply__4.0__72.0__ |
| if one - third of one - fourth of a number is num__15 then three - tenth of that number is <o> a ) num__35 <o> b ) num__15 <o> c ) num__60 <o> d ) num__54 <o> e ) num__57 |
let the number be x . then num__0.333333333333 of ¼ of x = num__15 x = num__15 x num__12 = num__180 the required number = ( num__0.3 ) * num__180 = num__54 answer d num__54 <eor> d <eos> |
d |
multiply__15.0__12.0__ multiply__0.3__180.0__ multiply__0.3__180.0__ |
multiply__15.0__12.0__ multiply__0.3__180.0__ multiply__0.3__180.0__ |
| a certain company has budgeted $ num__1560 for entertainment expenses for the year divided into num__12 equal monthly allocations . if by the end of the third month the total amount spent on entertainment was $ num__420 how much was the company under budget or over budget ? <o> a ) $ num__60 under budget <o> b ) $ num__30 under budget <o> c ) $ num__30 over budget <o> d ) $ num__60 over budget <o> e ) $ num__180 over budget |
the budget for num__3 months is ( num__0.25 ) * $ num__1560 = $ num__390 the company is $ num__30 over budget . the answer is c . <eor> c <eos> |
c |
divide__3.0__12.0__ multiply__1560.0__0.25__ subtract__420.0__390.0__ subtract__420.0__390.0__ |
divide__3.0__12.0__ multiply__1560.0__0.25__ subtract__420.0__390.0__ subtract__420.0__390.0__ |
| in a consignment of eggs num__552 were cracked which was num__12 percent of the total consignment . how many eggs were in the consignment ? <o> a ) num__4400 <o> b ) num__4800 <o> c ) num__4300 <o> d ) num__4600 <o> e ) num__4900 |
d num__4600 ( num__552 ÷ num__12 ) × num__100 = num__4600 . <eor> d <eos> |
d |
percent__100.0__4600.0__ |
percent__100.0__4600.0__ |
| a certain meter records voltage between num__0 and num__10 volts inclusive . if the average value of num__3 recordings on the meter was num__6 volts what was the smallest possible recording in volts ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
if average of num__3 is num__6 so sum of num__3 should be num__18 num__3 recording can be from num__0 - num__10 inclusive to find one smallest other two should be highest so lets assume three var are a b c say a is smallest and give b and c greatest readings say num__8 and num__8 so a has to be num__2 a <eor> a <eos> |
a |
multiply__3.0__6.0__ subtract__18.0__10.0__ subtract__10.0__8.0__ subtract__10.0__8.0__ |
multiply__3.0__6.0__ subtract__18.0__10.0__ subtract__10.0__8.0__ subtract__10.0__8.0__ |
| a sum of money becomes triple itself in num__5 years at simple interest . how many years will it become nine times at the same rate ? <o> a ) num__12 Â ½ years <o> b ) num__62 Â ½ years <o> c ) num__18 Â ½ years <o> d ) num__72 Â ½ years <o> e ) num__12 Â ½ years |
num__100 - - - - num__200 - - - - num__5 num__200 - - - - num__5 num__100 - - - - num__2 num__0.5 - - - - - - - - - - - - - - - - - - num__900 - - - - num__18 Â ½ years answer : c <eor> c <eos> |
c |
percent__2.0__900.0__ percent__2.0__900.0__ |
percent__2.0__900.0__ percent__2.0__900.0__ |
| a and b start from house at num__10 am . they travel on the mg road at num__20 kmph and num__40 kmph . there is a junction t on their path . a turns left at t junction at num__12 : num__00 noon b reaches t earlier and turns right . both of them continue to travel till num__2 pm . what is the distance between a and b at num__2 pm <o> a ) num__160 kilometre <o> b ) num__170 km <o> c ) num__180 km <o> d ) num__190 km <o> e ) num__150 km |
at num__12 a will travel = num__20 * num__2 = num__40 km b will travel this num__40 km in num__1.0 = num__1 hr i . e . by num__11 am . after t junction for a - distance travelled = num__2 * num__20 = num__40 km for b distance travelled = num__40 * num__3 = num__120 so distance between a & b is = num__120 + num__40 = num__160 km answer : a <eor> a <eos> |
a |
add__10.0__1.0__ add__2.0__1.0__ multiply__10.0__12.0__ add__40.0__120.0__ round__160.0__ |
add__10.0__1.0__ add__2.0__1.0__ multiply__10.0__12.0__ add__40.0__120.0__ add__40.0__120.0__ |
| what is the place value of num__2 in the numeral num__12734 ? <o> a ) num__1500 <o> b ) num__4500 <o> c ) num__2000 <o> d ) num__6000 <o> e ) num__4000 |
option ' c ' num__2 * num__1000 = num__2000 <eor> c <eos> |
c |
multiply__2.0__1000.0__ multiply__2.0__1000.0__ |
multiply__2.0__1000.0__ multiply__2.0__1000.0__ |
| a train num__125 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is : <o> a ) num__40 <o> b ) num__50 <o> c ) num__60 <o> d ) num__70 <o> e ) num__80 |
speed of the train relative to man = ( num__12.5 ) m / sec = ( num__12.5 ) m / sec . [ ( num__12.5 ) * ( num__3.6 ) ] km / hr = num__45 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__45 = = > x = num__50 km / hr . answer : option b <eor> b <eos> |
b |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ round__50.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ multiply__5.0__10.0__ |
| a car gets num__32 miles to the gallon . if it is modified to use a solar panel it will use only num__80 percent as much fuel as it does now . if the fuel tank holds num__12 gallons how many more miles will the car be able to travel per full tank of fuel after it has been modified ? <o> a ) num__84 <o> b ) num__88 <o> c ) num__92 <o> d ) num__96 <o> e ) num__100 |
originally the distance the car could go on a full tank was num__12 * num__32 = num__384 miles . after it has been modified the car can go num__32 / num__0.8 = num__40 miles per gallon . on a full tank the car can go num__12 * num__40 = num__480 miles thus num__96 miles more . the answer is d . <eor> d <eos> |
d |
multiply__32.0__12.0__ divide__32.0__0.8__ multiply__12.0__40.0__ subtract__480.0__384.0__ round__96.0__ |
multiply__32.0__12.0__ divide__32.0__0.8__ multiply__12.0__40.0__ subtract__480.0__384.0__ round__96.0__ |
| when tom works alone he chops num__2 lb . salad in num__3 minutes and when tammy works alone she chops num__3 lb . salad in num__2 minutes . they start working together and after some time finish chopping num__65 lb . of salad . of those num__80 lb . the salad quantity chopped by tammy is what percent greater than the quantifying chopped by tom ? . <o> a ) num__44.0 <o> b ) num__125.0 <o> c ) num__105.0 <o> d ) num__225.0 <o> e ) num__400 % |
tom chops num__4 lbs in num__6 minutes tammy chops num__9 lbs in num__6 minutes so in the same amount of time tammy chops num__125.0 more than tom since num__9 is num__125.0 greater than num__4 . so num__125.0 is the answer . note that the actual time does n ' t matter . if you multiply the time each work by x you ' ll multiply the work each does by x and num__9 x is still num__125.0 greater than num__4 x . ans : b <eor> b <eos> |
b |
multiply__2.0__3.0__ add__3.0__6.0__ round__125.0__ |
multiply__2.0__3.0__ add__3.0__6.0__ round__125.0__ |
| alpha beta and gamma are inner angles in a triangle . if alpha = beta + gamma what ca n ' t be the size of beta ? <o> a ) num__44 degrees <o> b ) num__45 degrees . <o> c ) num__89 degrees <o> d ) num__90 degrees . <o> e ) there is n ' t enough data to determine . |
given alpha = beta + gamma and alpha beta and gamma are inner angles in a triangle . thus alpha + beta + gamma = num__180 beta + gamma + beta + gamma = num__180 = > num__2 ( beta + gamma ) = num__180 = > beta + gamma = num__90 . thus beta will be less than num__90 . hence the answer is d . <eor> d <eos> |
d |
divide__180.0__2.0__ divide__180.0__2.0__ |
divide__180.0__2.0__ divide__180.0__2.0__ |
| mrs . napier has num__45 stickers to give to num__7 students for a reward . how many stickers will each student get ? will there be any stickers left over ? <o> a ) num__2 - num__5 <o> b ) num__6 - num__4 <o> c ) num__2 - num__6 <o> d ) num__2 - num__3 <o> e ) num__2 - num__8 |
num__6.42857142857 = num__6 r num__4 mrs . napier will give each student num__6 stickers and there will be num__4 left over . correct answer b <eor> b <eos> |
b |
divide__45.0__7.0__ round_down__6.4286__ round_down__6.4286__ |
divide__45.0__7.0__ round_down__6.4286__ round_down__6.4286__ |
| robert left from a pvt company . management hold his salary rs . num__15000 / - for one month . earlier robert borrowed rs . num__7280 / - from company . but robert forget that . after one month robert asked his salary and accountant gives rs . num__18500 / - to him . what is the incentive amount given to robert ? <o> a ) num__9500 <o> b ) num__12500 <o> c ) num__10780 <o> d ) num__10500 <o> e ) num__8600 |
total salary = rs . num__15000 / - borrowed money = num__7280 / - balance salary = num__15000 - num__7280 = num__7720 paid amount = num__18500 / - incentive amount = num__18500 - num__7720 = num__10780 / - answer is c <eor> c <eos> |
c |
subtract__15000.0__7280.0__ subtract__18500.0__7720.0__ subtract__18500.0__7720.0__ |
subtract__15000.0__7280.0__ subtract__18500.0__7720.0__ subtract__18500.0__7720.0__ |
| how many seconds will a train num__110 meters long take to cross a railway platform num__165 meters long if the speed of the train is num__132 kmph ? <o> a ) num__6 sec <o> b ) num__7.5 sec <o> c ) num__8 sec <o> d ) num__3 sec <o> e ) num__10 sec |
d = num__110 + num__165 = num__275 s = num__132 * num__0.277777777778 = num__36.6666666667 mps t = num__275 * num__0.0272727272727 = num__7.5 sec b ) num__7.5 sec <eor> b <eos> |
b |
add__110.0__165.0__ divide__275.0__36.6667__ round__7.5__ |
add__110.0__165.0__ divide__275.0__36.6667__ round__7.5__ |
| the ratio between the speeds of two trains is num__3 : num__5 . if the second train runs num__500 kms in num__4 hours then the speed of the first train is <o> a ) num__80 km / h <o> b ) num__75 km / h <o> c ) num__50 km / h <o> d ) num__125 km / h <o> e ) num__60 km / h |
b num__75 km / h <eor> b <eos> |
b |
round__75.0__ |
round__75.0__ |
| two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of num__11 kmph and num__21 kmph respectively . when they meet it is found that one train has traveled num__60 km more than the other one . the distance between the two stations is ? <o> a ) num__477 <o> b ) num__384 <o> c ) num__279 <o> d ) num__276 <o> e ) num__291 |
num__1 h - - - - - num__5 ? - - - - - - num__60 num__12 h rs = num__11 + num__21 = num__32 t = num__12 d = num__32 * num__12 = num__384 . answer : b <eor> b <eos> |
b |
add__11.0__1.0__ add__11.0__21.0__ multiply__32.0__12.0__ round__384.0__ |
add__11.0__1.0__ add__11.0__21.0__ multiply__32.0__12.0__ multiply__32.0__12.0__ |
| the present ages of three persons in proportions num__4 : num__7 : num__9 . eight years ago the sum of their ages was num__16 . find their present ages ( in years ) . <o> a ) num__8 num__14 num__18 <o> b ) num__8 num__28 num__36 <o> c ) num__20 num__35 num__45 <o> d ) num__33 num__4243 <o> e ) num__23 num__42 |
43 |
let their present ages be num__4 x num__7 x and num__9 x years respectively . then ( num__4 x - num__8 ) + ( num__7 x - num__8 ) + ( num__9 x - num__8 ) = num__16 num__20 x = num__40 x = num__2 . their present ages are num__4 x = num__8 years num__7 x = num__14 years and num__9 x = num__18 years respectively . answer : a <eor> a <eos> |
a |
a |
| the parameter of a square is equal to the perimeter of a rectangle of length num__16 cm and breadth num__14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places <o> a ) num__45 <o> b ) num__43 <o> c ) num__42 <o> d ) num__47 <o> e ) num__23.57 |
let the side of the square be a cm . parameter of the rectangle = num__2 ( num__16 + num__14 ) = num__60 cm parameter of the square = num__60 cm i . e . num__4 a = num__60 a = num__15 diameter of the semicircle = num__15 cm circimference of the semicircle = num__0.5 ( ∏ ) ( num__15 ) = num__0.5 ( num__3.14285714286 ) ( num__15 ) = num__23.5714285714 = num__23.57 cm to two decimal places answer : option e <eor> e <eos> |
e |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
| what percent is num__7 gm of num__1 kg ? <o> a ) num__0.7 <o> b ) num__0.5 <o> c ) num__2.5 <o> d ) num__3.5 <o> e ) num__4 % |
num__1 kg = num__1000 gm num__0.007 × num__100 = num__0.7 = num__0.7 = num__0.7 a ) <eor> a <eos> |
a |
percent__100.0__0.7__ |
percent__100.0__0.7__ |
| three consecutive even numbers are such that thrice the first number exceeds double the third number by six . what is the third number ? <o> a ) num__16 <o> b ) num__18 <o> c ) num__20 <o> d ) num__22 <o> e ) num__24 |
let the three numbers be x x + num__2 and x + num__4 . num__3 x = num__2 ( x + num__4 ) + num__6 x = num__14 and then the third number is x + num__4 = num__18 . the answer is b . <eor> b <eos> |
b |
twice__2.0__ triple__2.0__ triple__6.0__ triple__6.0__ |
twice__2.0__ add__2.0__4.0__ add__4.0__14.0__ add__4.0__14.0__ |
| the average weight of a group of boys is num__30 kg . after a boy of weight num__35 kg joins the group the average weight of the group goes up by num__1 kg . find the number of boys in the group originally ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__6 <o> e ) num__7 |
let the number off boys in the group originally be x . total weight of the boys = num__30 x after the boy weighing num__35 kg joins the group total weight of boys = num__30 x + num__35 so num__30 x + num__35 + num__31 ( x + num__1 ) = > x = num__4 . answer : a <eor> a <eos> |
a |
add__30.0__1.0__ subtract__35.0__31.0__ subtract__35.0__31.0__ |
add__30.0__1.0__ subtract__35.0__31.0__ subtract__35.0__31.0__ |
| sum of the numbers from num__1 to num__20 is <o> a ) num__210 <o> b ) num__110 <o> c ) num__220 <o> d ) num__105 <o> e ) num__100 |
explanation : sum of first n natural numbers = num__1 + num__2 + num__3 + . . . . . n = n + num__0.5 substitute n = num__20 . = num__20 * num__10.5 = num__210 answer is a <eor> a <eos> |
a |
add__1.0__2.0__ reverse__2.0__ multiply__20.0__10.5__ multiply__1.0__210.0__ |
add__1.0__2.0__ reverse__2.0__ multiply__20.0__10.5__ multiply__1.0__210.0__ |
| distance between two stations a and b is num__778 km . a train covers the journey from a to b at num__84 km per hour and returns back to a with a uniform speed of num__56 km per hour . find the average speed of the train during the whole journey ? <o> a ) num__69.0 km / hr <o> b ) num__69.2 km / hr <o> c ) num__67.2 km / hr <o> d ) num__67.0 km / hr <o> e ) num__65.0 km / hr |
explanation : if a car covers a certain distance at x kmph and an equal distance at y kmph . then average speed of the whole journey = num__2 xy / x + y kmph . by using the same formula we can find out the average speed quickly . average speed = num__2 × num__84 × num__0.666666666667 + num__56 = num__2 × num__84 × num__0.4 = num__2 × num__21 × num__1.6 = num__2 × num__3 × num__11.2 = num__67.2 = num__67.2 answer : option c <eor> c <eos> |
c |
divide__56.0__84.0__ mile_to_km_conversion__ add__56.0__11.2__ round__67.2__ |
divide__56.0__84.0__ mile_to_km_conversion__ add__56.0__11.2__ add__56.0__11.2__ |
| a ship sails num__4 degrees north then num__13 s . then num__17 n . then num__19 s . and has finally num__11 degrees of south latitude . what was her latitude at starting ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) none of these |
let x = the latitude sought . then marking the northings + and the southings - ; by the statement x + num__4 - num__13 + num__17 - num__19 = - num__11 and x = num__0 . answer a <eor> a <eos> |
a |
sine__0.0__ |
sine__0.0__ |
| find the simple interest on rs . num__73000 at num__16 num__0.666666666667 % per year for num__9 months . <o> a ) num__7500 <o> b ) num__6500 <o> c ) num__8125 <o> d ) num__9125 <o> e ) none of them |
p = rs . num__73000 r = num__16.6666666667 % p . a and t = num__0.75 years = num__0.75 years . simple interest = ( p * r * t ) / num__100 = rs . ( num__73000 * ( num__16.6666666667 ) * ( num__0.75 ) * ( num__0.01 ) ) = rs . num__9125 answer is d . <eor> d <eos> |
d |
percent__100.0__9125.0__ |
percent__100.0__9125.0__ |
| a plant manager must assign num__10 new workers to one of five shifts . she needs a first second and third shift and two alternate shifts . each of the shifts will receive num__2 new workers . how many different ways can she assign the new workers ? <o> a ) num__2430 <o> b ) num__2700 <o> c ) num__3300 <o> d ) num__4860 <o> e ) num__5400 |
my take selecting team of num__2 out of num__10 to assign to the shifts = num__10 c num__2 = num__45 ways . now num__2 out of num__10 means total of num__5 group possible . so putting them in shifts = counting methode : first second third alt alt = num__5 * num__4 * num__3 * num__2 * num__1 = num__120 here alt and alt are the same : so num__60.0 = num__60 ways . total ways of selecting = ( selecting num__2 out of num__10 ) * arranging those teams in shifts = num__45 * num__60 = num__2700 ans : b <eor> b <eos> |
b |
divide__10.0__2.0__ subtract__5.0__2.0__ subtract__3.0__2.0__ hour_to_min_conversion__ multiply__45.0__60.0__ round__2700.0__ |
divide__10.0__2.0__ subtract__5.0__2.0__ subtract__3.0__2.0__ hour_to_min_conversion__ multiply__45.0__60.0__ multiply__45.0__60.0__ |
| sum of num__3 consecutive even no . ' s is num__26 more than the num__1 st no . of the series . find the middle no . ? <o> a ) num__8 <o> b ) twelve ( num__12 ) <o> c ) num__14 <o> d ) num__16 <o> e ) num__18 |
let the numbers be x x + num__2 and x + num__4 then x + x + num__2 + x + num__4 = x + num__26 num__3 x + num__6 = x + num__26 num__2 x = num__20 \ inline \ therefore x = num__10 \ inline \ therefore middle number = x + num__2 = num__10 + num__2 = num__12 b <eor> b <eos> |
b |
subtract__3.0__1.0__ add__3.0__1.0__ multiply__3.0__2.0__ subtract__26.0__6.0__ add__4.0__6.0__ multiply__3.0__4.0__ multiply__3.0__4.0__ |
subtract__3.0__1.0__ add__3.0__1.0__ add__2.0__4.0__ subtract__26.0__6.0__ add__4.0__6.0__ add__2.0__10.0__ add__2.0__10.0__ |
| num__7 ^ num__6 n - num__1 ^ num__6 n when n is an integer > num__0 is divisible by <o> a ) num__112 <o> b ) num__127 <o> c ) num__134 <o> d ) num__154 <o> e ) num__146 |
num__127 b <eor> b <eos> |
b |
multiply__1.0__127.0__ |
multiply__1.0__127.0__ |
| the speed of a boat in still water is num__15 km / hr and the rate of current is num__3 km / hr . the distance travelled downstream in num__12 minutes is <o> a ) num__1.6 km <o> b ) num__2 km <o> c ) num__3.6 km <o> d ) num__4 km <o> e ) none of these |
explanation : speed downstreams = ( num__15 + num__3 ) kmph = num__18 kmph . distance travelled = ( num__18 x num__0.2 ) km = num__3.6 km option c <eor> c <eos> |
c |
add__15.0__3.0__ divide__3.0__15.0__ multiply__18.0__0.2__ round__3.6__ |
add__15.0__3.0__ divide__3.0__15.0__ multiply__18.0__0.2__ round__3.6__ |
| num__90 num__180 num__12 num__50 num__100 num__120 ? num__3 num__30 num__4 num__25 num__2 num__6 num__30 num__3 <o> a ) num__90 <o> b ) num__120 <o> c ) num__180 <o> d ) num__200 <o> e ) num__220 |
num__30 * num__3 = num__90 num__6 * num__30 = num__180 num__6 * num__2 = num__12 num__25 * num__2 = num__50 num__4 * num__25 = num__100 num__30 * num__4 = num__120 num__3 * num__30 = num__90 ans is num__90 answer : a <eor> a <eos> |
a |
subtract__180.0__90.0__ |
multiply__3.0__30.0__ |
| the average student age of a certain class which has num__45 students is num__14 . if one student aged num__15 go to another class and the age of the class teacher is included the average changes to num__14.66 . what is the age of class teacher ? <o> a ) num__30 <o> b ) num__35 <o> c ) num__38 <o> d ) num__40 <o> e ) num__45 |
sum of ages of class before replacement = num__45 x num__14 = num__630 sum of ages of class without teacher = num__630 - num__15 = num__615 sum of ages of class after replacement = num__45 x num__14.66 = num__660 age of teacher = num__660 - num__615 = num__45 answer : e <eor> e <eos> |
e |
multiply__45.0__14.0__ subtract__630.0__15.0__ add__45.0__615.0__ divide__630.0__14.0__ |
multiply__45.0__14.0__ subtract__630.0__15.0__ add__45.0__615.0__ subtract__660.0__615.0__ |
| a satellite is composed of num__24 modular units each of which is equipped with a set of sensors some of which have been upgraded . each unit contains the same number of non - upgraded sensors . if the number of non - upgraded sensors on one unit is num__0.166666666667 the total number of upgraded sensors on the entire satellite what fraction of the sensors on the satellite have been upgraded ? <o> a ) num__0.833333333333 <o> b ) num__0.2 <o> c ) num__0.166666666667 <o> d ) num__0.142857142857 <o> e ) num__0.0416666666667 |
let x be the number of upgraded sensors on the satellite . the number of non - upgraded sensors per unit is x / num__6 . the number of non - upgraded sensors on the whole satellite is num__24 ( x / num__6 ) = num__4 x . the fraction of sensors which have been upgraded is x / ( x + num__4 x ) = x / num__5 x = num__0.2 the answer is b . <eor> b <eos> |
b |
divide__24.0__6.0__ reverse__5.0__ reverse__5.0__ |
divide__24.0__6.0__ reverse__5.0__ reverse__5.0__ |
| tough and tricky questions : word problems . if bill ' s salary increased by num__16 percent he would earn $ num__812 per month . if instead his salary were to increase by only num__10 percent how much money w would he earn per month ? <o> a ) $ num__700 <o> b ) $ num__754 <o> c ) $ num__770 <o> d ) $ num__782 <o> e ) $ num__893 |
official solution : ( c ) if bill ’ s salary increased by num__16.0 he would earn $ num__812 . algebraically this can be written as : $ num__812 = num__1.16 s where s is his current salary . then s = $ num__812 / num__1.16 = $ num__700 . now that we know his current salary is $ num__700 we can calculate what his salary would be if it were increased by num__10.0 . we know that num__10.0 of $ num__700 is $ num__70 so his salary would be : w = $ num__700 + $ num__70 = $ num__770 . the correct answer is choice ( c ) . <eor> c <eos> |
c |
divide__812.0__1.16__ divide__700.0__10.0__ add__70.0__700.0__ add__70.0__700.0__ |
divide__812.0__1.16__ divide__700.0__10.0__ add__70.0__700.0__ add__70.0__700.0__ |
| in year y imported machine tools accounted for num__25 percent of total machine - tool sales in the united states and japanese imports accounted for num__25 percent of the sales of imported machine tools . if the total sales of machine tools imported japan that year was x billion dollars then the total sales of all machine tools in the united states was how many billion dollars ? <o> a ) num__9 x / num__80 <o> b ) num__16 x / num__1 <o> c ) num__8.88888888889 x <o> d ) num__20 x / num__13 <o> e ) num__80 x / num__9 |
imt = num__0.25 t ji = num__0.25 imt num__0.25 imt = $ x b imt = $ x / ( num__0.25 ) b t = num__4 * imt = num__4 x / num__0.25 = num__4 x * num__4.0 = num__16 x ans : ` ` b ' ' <eor> b <eos> |
b |
reverse__0.25__ divide__4.0__0.25__ divide__4.0__0.25__ |
reverse__0.25__ divide__4.0__0.25__ divide__4.0__0.25__ |
| in plutarch enterprises num__70.0 of the employees are marketers num__20.0 are engineers and the rest are managers . marketers make an average salary of $ num__70000 a year and engineers make an average of $ num__80000 . what is the average salary for managers if the average for all employees is also $ num__80000 ? <o> a ) $ num__80000 <o> b ) $ num__150000 <o> c ) $ num__240000 <o> d ) $ num__290000 <o> e ) $ num__320 |
000 |
for sake of ease let ' s say there are num__10 employees : num__7 marketers num__2 engineers and num__1 manager . average company salary * number of employees = total company salary > > > $ num__80000 * num__10 = $ num__800000 subtract the combined salaries for the marketers ( num__7 * $ num__70000 ) and the engineers ( num__2 * $ num__80000 ) > > > $ num__800000 - $ num__490000 - $ num__160000 = $ num__150000 . the correct answer is b . <eor> b <eos> |
b |
b |
| in a consumer survey num__88.0 of those surveyed liked at least one of three products : num__1 num__2 and num__3 . num__55.0 of those asked liked product num__1 num__40.0 liked product num__2 and num__20.0 liked product num__3 . if num__7.0 of the people in the survey liked all three of the products what percentage of the survey participants liked more than one of the three products ? <o> a ) num__15 <o> b ) num__20 <o> c ) num__25 <o> d ) num__30 <o> e ) num__35 |
num__55 + num__40 + num__20 = num__115 this number includes some people counted twice and some people counted three times . num__115 - num__88 = num__27 and this number represents people counted twice or three times . num__7.0 of the people are counted two extra times . the percentage of people counted one extra time is num__27 - num__7 ( num__2 ) = num__13 . the percentage of people who liked more than one product is num__7.0 + num__13.0 = num__20.0 . the answer is b . <eor> b <eos> |
b |
add__20.0__7.0__ subtract__40.0__27.0__ multiply__1.0__20.0__ |
add__20.0__7.0__ subtract__40.0__27.0__ subtract__40.0__20.0__ |
| two pipes can separately fill a tank in num__20 and num__30 hours respectively . both the pipes are opened to fill the tank but when the tank is full a leak develops in the tank through which one - third of water supplied by both the pipes goes out . what is the total time taken to fill the tank ? <o> a ) num__17 hrs <o> b ) num__16 hrs <o> c ) num__21 hrs <o> d ) num__15 hrs <o> e ) num__04 hrs |
num__0.05 + num__0.0333333333333 = num__0.0833333333333 num__1 + num__0.333333333333 = num__1.33333333333 num__1 - - - num__12 num__1.33333333333 - - - ? num__1.33333333333 * num__12 = num__16 hrs answer : b <eor> b <eos> |
b |
add__0.05__0.0333__ multiply__20.0__0.05__ add__1.0__0.3333__ round__16.0__ |
add__0.05__0.0333__ multiply__20.0__0.05__ add__1.0__0.3333__ round__16.0__ |
| how many such num__3 ' s are there in the following number sequence which are immediately preceded by an odd number and immediately followed by an even number ? num__5 num__3 num__8 num__9 num__4 num__3 num__7 num__2 num__3 num__8 num__1 num__3 num__8 num__4 num__2 num__3 num__5 num__7 num__3 num__4 num__2 num__3 num__6 <o> a ) num__1 one <o> b ) num__2 two <o> c ) num__3 three <o> d ) num__4 four <o> e ) more than four |
a is followed by b means : a comes first b comes next a preceded by b means : b comes first a comes next the sets which satisfy the condition is : num__5 num__3 num__8 num__1 num__3 num__8 num__7 num__3 num__4 answer : c <eor> c <eos> |
c |
multiply__3.0__1.0__ |
multiply__3.0__1.0__ |
| if the range q of the six numbers num__4 num__314 num__710 and x is num__12 what is the difference between the greatest possible value of x and least possible value of x ? <o> a ) num__0 <o> b ) num__2 <o> c ) num__12 <o> d ) num__13 <o> e ) num__15 |
the range q of a set is the difference between the largest and smallest elements of a set . without x the difference between the largest and smallest elements of a set is num__14 - num__3 = num__11 < num__12 which means that in order num__12 to be the range of the set x must be either the smallest element so that num__14 - x = num__12 - - - > x = num__2 or x must the largest element so that x - num__3 = num__12 - - > x = num__15 . the the difference between the greatest possible value of x and least possible value of x is num__15 - num__2 = num__13 . answer : d . <eor> d <eos> |
d |
divide__12.0__4.0__ subtract__14.0__3.0__ subtract__14.0__12.0__ add__4.0__11.0__ add__2.0__11.0__ add__2.0__11.0__ |
divide__12.0__4.0__ subtract__14.0__3.0__ subtract__14.0__12.0__ add__4.0__11.0__ subtract__15.0__2.0__ subtract__15.0__2.0__ |
| there are num__10 person among whom num__2 are brother . the total no . of ways in which these persons can be seated around a round table so that exactly num__1 person sit between the brothers is equal to ? <o> a ) num__4 ! * num__2 ! <o> b ) num__7 ! * num__2 ! <o> c ) num__6 ! * num__1 ! <o> d ) num__2 ! * num__1 ! <o> e ) num__5 ! * num__1 ! |
total number of ways = num__7 ! * num__2 ! . b <eor> b <eos> |
b |
multiply__1.0__7.0__ |
multiply__1.0__7.0__ |
| by selling num__50 meters of cloth . i gain the selling price of num__15 meters . find the gain percent ? <o> a ) num__42 num__1.14285714286 % <o> b ) num__42 num__1.5 % <o> c ) num__42 num__0.857142857143 % <o> d ) num__42 num__0.428571428571 % <o> e ) num__42 num__0.666666666667 % |
sp = cp + g num__50 sp = num__50 cp + num__15 sp num__35 sp = num__50 cp num__35 - - - num__15 cp gain num__100 - - - ? = > num__42 num__0.857142857143 % answer : d <eor> d <eos> |
d |
percent__100.0__42.0__ |
percent__100.0__42.0__ |
| the s . i . on a certain sum of money for num__3 years at num__8.0 per annum is half the c . i . on rs . num__4000 for num__2 years at num__10.0 per annum . the sum placed on s . i . is ? <o> a ) rs . num__1550 <o> b ) rs . num__1650 <o> c ) rs . num__1750 <o> d ) rs . num__2000 <o> e ) rs . num__3000 |
c . i . = [ num__4000 * ( num__1 + num__0.1 ) num__2 - num__4000 ] = ( num__4000 * num__1.1 * num__1.1 - num__4000 ) = rs . num__840 . sum = ( num__420 * num__100 ) / ( num__3 * num__8 ) = rs . num__1750 answer : c <eor> c <eos> |
c |
percent__10.0__1.0__ percent__100.0__1750.0__ |
percent__10.0__1.0__ percent__100.0__1750.0__ |
| in a market a dozen eggs cost as much as a pound of rice and a half - liter of kerosene costs as much as num__8 eggs . if the cost of each pound of rice is $ num__0.33 then how many w cents does a liter of kerosene cost ? [ one dollar has num__100 cents . ] <o> a ) num__0.33 <o> b ) num__0.44 <o> c ) num__0.55 <o> d ) num__44 <o> e ) num__55 |
main thing to remember is answer is asked in cents however when we calculate it comes up as num__0.44 $ just multiply by num__100 answer w = num__44 . d <eor> d <eos> |
d |
multiply__100.0__0.44__ multiply__100.0__0.44__ |
multiply__100.0__0.44__ multiply__100.0__0.44__ |
| oak street begins at pine street and runs directly east for num__2 kilometers until it ends when it meets maple street . oak street is intersected every num__400 meters by a perpendicular street and each of those streets other than pine street and maple street is given a number beginning at num__1 st street ( one block east of pine street ) and continuing consecutively ( num__2 nd street num__3 rd street etc . . . ) until the highest - numbered street one block west of maple street . what is the highest - numbered street that intersects oak street ? <o> a ) num__4 th <o> b ) num__5 th <o> c ) num__6 th <o> d ) num__7 th <o> e ) num__8 th |
num__2 km / num__400 m = num__5 . however the street at the num__2 - km mark is not num__5 th street ; it is maple street . therefore the highest numbered street is num__4 th street . the answer is a . <eor> a <eos> |
a |
add__2.0__3.0__ add__1.0__3.0__ round__4.0__ |
add__2.0__3.0__ subtract__5.0__1.0__ divide__4.0__1.0__ |
| the ratio between the present ages of p and q is num__5 : num__7 respectively . if the difference between q ' s present age and p ' s age after num__6 years is num__2 . what is the total of p ' s and q ' s present ages ? <o> a ) num__48 <o> b ) num__277 <o> c ) num__65 <o> d ) num__15 <o> e ) num__12 |
let the present ages of p and q be num__5 x and num__7 x years respectively . then num__7 x - ( num__5 x + num__6 ) = num__2 num__2 x = num__8 = > x = num__4 required sum = num__5 x + num__7 x = num__12 x = num__48 years . answer : a <eor> a <eos> |
a |
add__6.0__2.0__ subtract__6.0__2.0__ add__5.0__7.0__ multiply__6.0__8.0__ multiply__6.0__8.0__ |
add__6.0__2.0__ subtract__6.0__2.0__ add__5.0__7.0__ multiply__6.0__8.0__ multiply__6.0__8.0__ |
| in a certain pet shop the ratio of dogs to cats to bunnies in stock is num__4 : num__7 : num__9 . if the shop carries num__364 dogs and bunnies total in stock how many dogs are there ? <o> a ) num__42 <o> b ) num__66 <o> c ) num__98 <o> d ) num__112 <o> e ) num__154 |
let us assume the number of dogs cats and bunnies to be num__4 x num__7 x and num__9 x total dogs and bunnies = num__13 x . and we are given that num__13 x = num__364 . hence x = num__28 . dogs = num__4 x = num__4 * num__28 = num__112 ( option d ) <eor> d <eos> |
d |
add__4.0__9.0__ multiply__4.0__7.0__ multiply__4.0__28.0__ multiply__4.0__28.0__ |
add__4.0__9.0__ multiply__4.0__7.0__ multiply__4.0__28.0__ multiply__4.0__28.0__ |
| find the area diameter = num__11 m . <o> a ) num__113.00 square meter <o> b ) num__95.07 square meter <o> c ) num__93.08 square meter <o> d ) num__93.24 square meter <o> e ) num__113.43 square meter |
diameter = num__11 meter . radius = diameter / num__2 . = num__5.5 . = num__5.5 meter . area of a circle = Ï € r num__2 . here pi ( Ï € ) = num__3.14 meter radius ( r ) = num__5.5 . area of a circle = num__3.14 Ã — num__5.5 Ã — num__5.5 . . = num__3.14 Ã — num__30.25 . = num__95.07 square meter answer : b <eor> b <eos> |
b |
triangle_area__11.0__5.5__ triangle_area__2.0__95.07__ |
triangle_area__11.0__5.5__ triangle_area__2.0__95.07__ |
| a car going at num__40 miles per hour set out on an num__90 - mile trip at num__9 : num__00 a . m . exactly num__10 minutes later a second car left from the same place and followed the same route . how fast in miles per hour was the second car going if it caught up with the first car at num__10 : num__30 a . m . ? <o> a ) num__45 <o> b ) num__50 <o> c ) num__53 <o> d ) num__55 <o> e ) num__90 |
let car a = car that starts at num__9 am car b = car that starts at num__9 : num__10 am time for which car a travels at speed of num__40 m per hour = num__1.5 hours distance travelled by car a = num__40 * num__1.5 = num__60 miles since car b catches up car a at num__10 : num__30 time = num__90 mins = num__1.5 hour speed of car b = num__60 / ( num__1.5 ) = num__90 miles per hour answer e <eor> e <eos> |
e |
hour_to_min_conversion__ round__90.0__ |
multiply__40.0__1.5__ multiply__9.0__10.0__ |
| if num__10 lions can kill num__10 deers in num__10 minutes how long will it take num__100 lions to kill num__100 deers ? <o> a ) num__1 minutes <o> b ) num__10 minute <o> c ) num__100 minutes <o> d ) num__10000 minutes <o> e ) num__1000 minutes |
we can try the logic of time and work our work is to kill the deers so num__10 ( lions ) * num__10 ( min ) / num__10 ( deers ) = num__100 ( lions ) * x ( min ) / num__100 ( deers ) hence answer is x = num__10 answer : b <eor> b <eos> |
b |
round__10.0__ |
divide__100.0__10.0__ |
| there is a total of num__84 marbles in a box each of which is red green blue or white . if one marble is drawn from the box at random the probability that it will be white is num__0.25 and the probability that it will be green is num__0.285714285714 . what is the probability that the marble will be either red or blue ? <o> a ) num__0.428571428571 <o> b ) num__0.785714285714 <o> c ) num__0.464285714286 <o> d ) num__0.678571428571 <o> e ) num__0.690476190476 |
p ( red or blue ) = num__1 - p ( white ) - p ( green ) = num__1.0 - num__0.25 - num__0.285714285714 = num__0.464285714286 the answer is c . <eor> c <eos> |
c |
multiply__1.0__0.4643__ |
multiply__1.0__0.4643__ |
| in a class of num__5 students average weight of the num__4 lightest students is num__40 kgs average weight of the num__4 heaviest students is num__45 kgs . what is the difference between the the maximum and minimum possible average weight overall ? <o> a ) num__2.8 kgs <o> b ) num__3.2 kgs <o> c ) num__3 kgs <o> d ) num__4 kilograms <o> e ) num__2 kgs |
detailed solution if there are an odd number of terms say num__2 n + num__1 then the median is the middle term . and if average is lesser than the middle term there will at least be n + num__1 terms greater than the average . so there will be more terms above the average than below it . however this need not be the case when there are an even number of terms . when there are num__2 n distinct terms n of them will be greater than the median and n will be lesser than the median . the average of these two terms can be such that there are n terms above the average and n below it . for instance if the numbers are num__0 num__1 num__7 num__7.5 . the median is num__4 average is num__3.875 . average is less than the median . and there are more num__2 numbers above the average and num__2 below the average . correct answer : median is num__4 average is num__3.875 . correct answer : d <eor> d <eos> |
d |
subtract__5.0__4.0__ add__5.0__2.0__ subtract__5.0__1.0__ |
subtract__5.0__4.0__ add__5.0__2.0__ subtract__5.0__1.0__ |
| two trains a and b starting from two points and travelling in opposite directions reach their destinations num__9 hours and num__16 hours respectively after meeting each other . if the train a travels at num__80 kmph find the rate at which the train b runs . <o> a ) num__40 <o> b ) num__160 <o> c ) num__120 <o> d ) num__80 <o> e ) num__100 |
if two objects a and b start simultaneously from opposite points and after meeting reach their destinations in ‘ a ’ and ‘ b ’ hours respectively ( i . e . a takes ‘ a hrs ’ to travel from the meeting point to his destination and b takes ‘ b hrs ’ to travel from the meeting point to his destination ) then the ratio of their speeds is given by : sa / sb = √ ( b / a ) i . e . ratio of speeds is given by the square root of the inverse ratio of time taken . sa / sb = √ ( num__0.25 ) = num__0.5 = num__0.5 this gives us that the ratio of the speed of a : speed of b as num__1 : num__2 . since speed of a is num__80 kmph speed of b must be num__80 * ( num__2.0 ) = num__160 kmph b <eor> b <eos> |
b |
divide__0.5__0.25__ divide__80.0__0.5__ round__160.0__ |
divide__0.5__0.25__ divide__80.0__0.5__ divide__80.0__0.5__ |
| a train of length num__250 m crosses a bridge of length num__150 m in num__20 seconds . what is the speed of train ? <o> a ) num__33 <o> b ) num__27 <o> c ) num__25 <o> d ) num__22 <o> e ) num__72 |
sol : ( length of train + length of bridge ) = speed of train x time ( num__250 + num__150 ) = num__20 x speed speed = num__20.0 = num__20 m / s = num__72 km / h answer = e <eor> e <eos> |
e |
round__72.0__ |
round__72.0__ |
| the perimeter of a rhombus are num__24 cm and num__10 cm the area and the perimeter of the rhombus is <o> a ) num__64 sq . m <o> b ) num__70 sq . m <o> c ) num__78 sq . m <o> d ) num__84 sq . m <o> e ) none |
solution perimeter of the rhombus = num__56 m each side of the rhombus = num__14.0 m = num__14 m . height of the rhombus = num__5 m area = ( num__14 x num__5 ) m ² = num__70 m ² answer b <eor> b <eos> |
b |
subtract__24.0__10.0__ add__56.0__14.0__ round__70.0__ |
subtract__24.0__10.0__ add__56.0__14.0__ round__70.0__ |
| in a series of six consecutive odd numbers the sum of the second and fifth numbers is num__28 . what is the third number ? <o> a ) num__9 <o> b ) num__11 <o> c ) num__17 <o> d ) num__13 <o> e ) num__15 |
let the numbers be x x + num__1 x + num__3 x + num__5 and x + num__7 . given ( x + num__1 ) + ( x + num__7 ) = num__28 = > num__2 x + num__8 = num__28 = > x = num__10 the third number = x + num__3 = num__10 + num__3 = num__13 . answer : d <eor> d <eos> |
d |
subtract__3.0__1.0__ add__1.0__7.0__ multiply__2.0__5.0__ add__3.0__10.0__ multiply__1.0__13.0__ |
subtract__3.0__1.0__ add__1.0__7.0__ add__2.0__8.0__ add__3.0__10.0__ add__3.0__10.0__ |
| ' a ' and ' b ' are positive integers such that their lcm is num__20 and their hcf is num__1 . what is the addition of the maximum and minimum possible values of ' a + b ' ? <o> a ) num__28 <o> b ) num__30 <o> c ) num__22 <o> d ) num__20 <o> e ) num__32 |
possible values of a and b can be num__54 ; num__45 ( which are same for a + b ) and num__120 ; num__201 ( same result for a + b ) so num__21 + num__9 = num__30 . ans b . <eor> b <eos> |
b |
add__20.0__1.0__ subtract__54.0__45.0__ add__21.0__9.0__ multiply__1.0__30.0__ |
add__20.0__1.0__ subtract__54.0__45.0__ add__21.0__9.0__ add__21.0__9.0__ |
| the present population of a town is num__1240 . population increase rate is num__4.0 p . a . find the population of town after num__1 years ? <o> a ) num__990 <o> b ) num__1215 <o> c ) num__1345 <o> d ) num__1142 <o> e ) num__1290 |
p = num__1240 r = num__4.0 required population of town = p * ( num__1 + r / num__100 ) ^ t = num__1240 * ( num__1 + num__0.04 ) = num__1240 * ( num__1.04 ) = num__1290 ( approximately ) answer is e <eor> e <eos> |
e |
percent__4.0__1.0__ percent__100.0__1290.0__ |
percent__4.0__1.0__ percent__100.0__1290.0__ |
| if s t and u are positive real numbers such that s ( t + u ) = num__152 t ( u + s ) = num__162 and u ( s + t ) = num__170 then stu is <o> a ) a ) num__672 <o> b ) b ) num__688 <o> c ) c ) num__704 <o> d ) d ) num__720 <o> e ) e ) num__750 |
st + su = num__152 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - tu + ts = num__162 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - su + ut = num__170 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - num__3 ) re - writing equation num__3 as follows : su + ut = num__162 + num__8 su + ut = tu + ts + num__8 su = ts + num__8 . . . . . . . . . . . . . . . ( num__4 ) adding ( num__1 ) ( num__4 ) num__2 su = num__160 su = num__80 stu has to be multiple of num__80 only num__720 fits in answer = d <eor> d <eos> |
d |
subtract__170.0__162.0__ subtract__4.0__3.0__ subtract__3.0__1.0__ add__152.0__8.0__ divide__160.0__2.0__ multiply__1.0__720.0__ |
subtract__170.0__162.0__ subtract__4.0__3.0__ subtract__3.0__1.0__ add__152.0__8.0__ divide__160.0__2.0__ multiply__1.0__720.0__ |
| the average of num__6 no . ' s is num__3.95 . the average of num__2 of them is num__3.8 while the average of theother num__2 is num__3.85 . what is the average of the remaining num__2 no ' s ? <o> a ) num__4.2 <o> b ) num__4.4 <o> c ) num__4.6 <o> d ) num__5.6 <o> e ) num__5.7 |
sum of the remaining two numbers = ( num__3.95 * num__6 ) - [ ( num__3.8 * num__2 ) + ( num__3.85 * num__2 ) ] = num__8.4 . required average = ( num__8.4 / num__2 ) = num__4.2 . a <eor> a <eos> |
a |
divide__8.4__2.0__ divide__8.4__2.0__ |
divide__8.4__2.0__ divide__8.4__2.0__ |
| find large number from below question the difference of two numbers is num__1365 . on dividing the larger number by the smaller we get num__6 as quotient and the num__15 as remainder <o> a ) num__1234 <o> b ) num__1265 <o> c ) num__1350 <o> d ) num__1467 <o> e ) num__1635 |
let the smaller number be x . then larger number = ( x + num__1365 ) . x + num__1365 = num__6 x + num__15 num__5 x = num__1350 x = num__270 large number = num__270 + num__1365 = num__1635 e <eor> e <eos> |
e |
subtract__1365.0__15.0__ divide__1350.0__5.0__ add__1365.0__270.0__ add__1365.0__270.0__ |
subtract__1365.0__15.0__ divide__1350.0__5.0__ add__1365.0__270.0__ add__1365.0__270.0__ |
| if z = num__343 * num__64 * num__0.142857142857 find the unit digit of z <o> a ) num__5 <o> b ) num__1 <o> c ) num__3 <o> d ) num__2 <o> e ) num__6 |
z = num__343 * num__64 * num__0.142857142857 z = num__7 ^ num__3 * num__4 ^ num__3 * num__3 ^ num__0.666666666667 ^ num__2 * num__7 z = num__7 ^ num__2 * num__4 ^ num__3 z = num__3136 ans : e <eor> e <eos> |
e |
subtract__7.0__3.0__ multiply__2.0__3.0__ |
subtract__7.0__3.0__ multiply__2.0__3.0__ |
| a circular well with a diameter of num__4 metres is dug to a depth of num__24 metres . what is the volume of the earth dug out ? <o> a ) num__32 m num__3 <o> b ) num__36 m num__3 <o> c ) num__40 m num__3 <o> d ) num__44 m num__3 <o> e ) num__301.6 |
solution volume = π r num__2 h ‹ = › ( num__3.14285714286 × num__2 × num__2 × num__24 ) m num__3 ‹ = › num__301.6 m num__3 . answer e <eor> e <eos> |
e |
triangle_area__2.0__301.6__ |
triangle_area__2.0__301.6__ |
| a car traveling at a certain constant speed takes num__10 seconds longer to travel num__1 kilometer than it would take to travel num__1 kilometer at num__40 kilometers per hour . at what speed in kilometers per hour is the car traveling ? <o> a ) num__35 <o> b ) num__35.5 <o> c ) num__36 <o> d ) num__36.5 <o> e ) num__37 |
num__40 * t = num__1 km = > t = num__0.025 km / h v * ( t + num__0.00277777777778 ) = num__1 v ( num__0.025 + num__0.00277777777778 ) = num__1 v ( num__0.0277777777778 ) = num__1 v = num__36 km / h the answer is c . <eor> c <eos> |
c |
divide__1.0__40.0__ add__0.025__0.0028__ round__36.0__ |
divide__1.0__40.0__ add__0.025__0.0028__ multiply__1.0__36.0__ |
| a shopkeeper has num__280 kg of apples . he sells num__40.0 of these at num__20.0 profit and remaining num__60.0 at num__20.0 profit . find his % profit on total . <o> a ) num__24.0 <o> b ) num__20.0 <o> c ) num__26.0 <o> d ) num__28.0 <o> e ) num__35 % |
if the total quantity was num__100 then num__40 x num__20.0 + num__60 x num__20.0 = num__20 this profit will remain same for any total quantity unless the % of products remains the same . hence ' b ' is the answer <eor> b <eos> |
b |
percent__20.0__100.0__ |
percent__20.0__100.0__ |
| a shopkeeper purchased num__70 kg of potatoes for rs . num__420 and sold the whole lot at the rate of rs . num__6.50 per kg . what will be his gain percent ? <o> a ) num__8 num__0.111111111111 % <o> b ) num__8 num__2.33333333333 % <o> c ) num__8 num__0.333333333333 % <o> d ) num__2 num__0.333333333333 % <o> e ) num__8 num__1.0 % |
c . p . of num__1 kg = num__60.0 = rs . num__6 s . p . of num__1 kg = rs . num__6.50 gain % = num__0.50 / num__6 * num__100 = num__8.33333333333 = num__8 num__0.333333333333 % . answer : c <eor> c <eos> |
c |
percent__100.0__8.0__ |
percent__100.0__8.0__ |
| there are ten players in the basketball team . if the average height of the players is num__170 cm what will be the new average height if a num__192 cm player will join the team ? <o> a ) num__168 . <o> b ) num__172.2 <o> c ) num__172 <o> d ) num__181 <o> e ) num__184 |
given ( x num__1 + x num__2 + . . . . . . . . . . x num__10 ) / num__10 = num__170 now ( x num__1 + x num__2 + . . . . . . . . . . x num__10 ) = num__1700 - - - > eq num__1 . now one new person added and the person be a and height is num__192 and the total no of players will be num__11 and let ' s say the new average is b . = > ( x num__1 + x num__2 + . . . . . . . . . . x num__10 + num__192 ) / num__11 = b - - eq num__2 sub x num__1 + x num__2 + . . . . . . . . . . x num__10 from eq num__1 in eq num__2 . = > num__1700 + num__192 = num__11 b b = num__172 . answer : c is the correct option <eor> c <eos> |
c |
multiply__170.0__10.0__ add__1.0__10.0__ add__170.0__2.0__ add__170.0__2.0__ |
multiply__170.0__10.0__ add__1.0__10.0__ add__170.0__2.0__ add__170.0__2.0__ |
| the speed of a boat in upstream is num__60 kmph and the speed of the boat downstream is num__90 kmph . find the speed of the boat in still water and the speed of the stream ? <o> a ) num__15 <o> b ) num__99 <o> c ) num__77 <o> d ) num__55 <o> e ) num__22 |
speed of the boat in still water = ( num__60 + num__90 ) / num__2 = num__75 kmph . speed of the stream = ( num__90 - num__60 ) / num__2 = num__15 kmph . answer : a <eor> a <eos> |
a |
subtract__90.0__75.0__ round__15.0__ |
subtract__90.0__75.0__ subtract__90.0__75.0__ |
| working individually emma can wrap presents for num__6 hours and troy can wrap presents in num__8 hours . if emma and troy work together but independently at the task for num__2 hours at which point troy leaves how many remaining hours will it take emma to complete the task alone ? <o> a ) num__2.5 <o> b ) num__0.416666666667 <o> c ) num__0.75 <o> d ) num__0.166666666667 <o> e ) num__0.6 |
in first num__2 hrs troy will finish num__0.25 = num__0.25 of work and emma will finish num__0.333333333333 work so total num__0.25 + num__0.333333333333 = num__0.583333333333 work is finished and num__1 - num__0.583333333333 = num__0.416666666667 work remaining . now emma will take ( num__0.416666666667 ) * num__6 = num__2.5 hrs to finish it . so answer is a . <eor> a <eos> |
a |
divide__2.0__8.0__ divide__2.0__6.0__ add__0.25__0.3333__ subtract__1.0__0.5833__ round__2.5__ |
divide__2.0__8.0__ divide__2.0__6.0__ add__0.25__0.3333__ subtract__1.0__0.5833__ round__2.5__ |
| find the value of ( num__100 + num__0.222222222222 ) × num__90 <o> a ) num__4520 <o> b ) num__9100 <o> c ) num__9150 <o> d ) num__8120 <o> e ) num__9020 |
( num__9000 + num__20 ) / num__90 * num__90 = num__9020 answer : e <eor> e <eos> |
e |
multiply__100.0__90.0__ add__9000.0__20.0__ add__9000.0__20.0__ |
multiply__100.0__90.0__ add__9000.0__20.0__ add__9000.0__20.0__ |
| a rainstorm increased the amount of water stored in state j reservoirs from num__124 billion gallons to num__140 billion gallons . if the storm increased the amount of water in the reservoirs to num__80 percent of total capacity approximately how many billion gallons of water were the reservoirs short of total capacity prior to the storm ? <o> a ) num__51 <o> b ) num__48 <o> c ) num__55 <o> d ) num__63 <o> e ) num__65 |
let total capacity be x we know num__140 = num__0.80 x x = num__140 / num__0.80 = num__175 prior to storm we had num__124 bn gallons num__175 - num__124 = num__51 answer : a <eor> a <eos> |
a |
divide__140.0__0.8__ subtract__175.0__124.0__ subtract__175.0__124.0__ |
divide__140.0__0.8__ subtract__175.0__124.0__ subtract__175.0__124.0__ |
| if f ( x ) = x - num__4 then ( f o f ) ( num__3 ) = ? <o> a ) - num__5 <o> b ) - num__4 <o> c ) - num__6 <o> d ) - num__3 <o> e ) - num__7 |
( f o f ) ( num__3 ) = f ( f ( num__3 ) ) = f ( num__3 - num__4 ) = f ( - num__1 ) = - num__5 correct answer a <eor> a <eos> |
a |
subtract__4.0__3.0__ add__4.0__1.0__ add__4.0__1.0__ |
subtract__4.0__3.0__ add__4.0__1.0__ add__4.0__1.0__ |
| timmy and tommy pumpernickel are leaving home for college today . peter and patty pumpernickel stand and wave goodbye as their pride and joy head off to schools at opposite ends of the state . if timmy drives a slow num__45 miles / hour and tommy ` ` the lead foot ' ' pumpernickel drives a jaw dropping num__75 miles per hour how long until they are num__500 miles away from one another ? <o> a ) num__4 hours <o> b ) num__4.5 hours <o> c ) num__4.24 hours <o> d ) num__4.17 hours <o> e ) num__5 hours |
the pumpernickel brothers are traveling num__120 miles per hour in opposite directions . num__500 miles / num__120 miles / hour = num__4.17 hours answer is d <eor> d <eos> |
d |
add__45.0__75.0__ round__4.17__ |
add__45.0__75.0__ round__4.17__ |
| there are num__6 boxes numbered num__1 num__2 . . . num__6 . each box is to be filled up either with a red or a green ball in such a way that at least num__1 box contains a green ball and the boxes containing green balls are consecutively numbered . the total number of ways in which this can be done is : <o> a ) num__5 <o> b ) num__6 <o> c ) num__21 <o> d ) num__33 <o> e ) num__60 |
num__6 green - num__1 way num__5 green - num__2 ways num__4 green - num__3 ways num__3 green - num__4 ways num__2 green - num__5 ways num__1 green - num__6 ways total = > num__21 ways answer c <eor> c <eos> |
c |
subtract__6.0__1.0__ subtract__6.0__2.0__ divide__6.0__2.0__ multiply__1.0__21.0__ |
subtract__6.0__1.0__ subtract__6.0__2.0__ subtract__4.0__1.0__ multiply__1.0__21.0__ |
| if num__20 men can build a wall num__112 metres long in num__6 days what length of a similar wall can be built by num__15 men in num__3 days ? <o> a ) num__65 mtr . <o> b ) num__52 mtr <o> c ) num__70 mtr . <o> d ) num__78 mtr . <o> e ) num__42 mtr . |
num__20 men is num__6 days can build num__112 metres num__15 men in num__3 days can build = num__112 * ( num__0.75 ) x ( num__0.5 ) = num__42 meters answer : e . <eor> e <eos> |
e |
divide__15.0__20.0__ divide__3.0__6.0__ round__42.0__ |
divide__15.0__20.0__ divide__3.0__6.0__ round__42.0__ |
| the mark up between ` ` normal plan ' ' price and ` ` premium plan ' ' price is num__30 percent of the ` ` normal plan ' ' price . the markup is what percent of the ` ` premium plan ' ' price ? ( markup = difference between normal and premium price plans ) <o> a ) num__8.0 <o> b ) num__20.0 <o> c ) num__33 num__0.333333333333 % <o> d ) num__45.0 <o> e ) num__56 num__0.666666666667 % |
a = price for normal plan b = price for premium plan mp = num__0.5 a b = a + num__0.5 a = num__1.5 a hence markup = b - a = num__0.5 / num__1.5 b = num__0.333333333333 b . hence mp is num__33.33 of b answer c <eor> c <eos> |
c |
divide__0.5__1.5__ round_down__33.33__ |
divide__0.5__1.5__ round_down__33.33__ |
| car x began traveling at an average speed of num__35 miles per hour . after num__72 minutes car y began traveling at an average speed of num__42 miles per hour . when both cars had traveled the same distance both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ? <o> a ) num__105 <o> b ) num__140 <o> c ) num__175 <o> d ) num__210 <o> e ) num__245 |
in num__72 minutes car x travels num__42 miles . car y gains num__7 miles each hour so it takes num__6 hours to catch car x . in num__6 hours car x travels num__210 miles . the answer is d . <eor> d <eos> |
d |
subtract__42.0__35.0__ divide__42.0__7.0__ multiply__35.0__6.0__ round__210.0__ |
subtract__42.0__35.0__ divide__42.0__7.0__ multiply__35.0__6.0__ round__210.0__ |
| the ratio between the school ages of neelam and shaan is num__5 : num__6 respectively . if the ratio between the one - third age of neelam and half of shaan ' s age of num__5 : num__9 then what is the school age of shaan ? <o> a ) num__17 <o> b ) num__16 <o> c ) num__18 <o> d ) num__15 <o> e ) num__19 |
let the school ages of neelam and shaan be num__5 x and num__6 x years respectively . then ( num__0.333333333333 * num__5 x ) / ( num__0.5 * num__6 x ) = num__0.555555555556 num__15 = num__15 thus shaan ' s age can not be determined . answer : d <eor> d <eos> |
d |
divide__5.0__9.0__ add__6.0__9.0__ add__6.0__9.0__ |
divide__5.0__9.0__ add__6.0__9.0__ add__6.0__9.0__ |
| let p be the product of the positive integers between num__1 and num__7 inclusive . how many distinct prime factors does p have ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
there are num__4 primes between num__1 and num__7 inclusive : num__2 num__3 num__5 and num__7 therefore p = num__1 * num__2 * . . . * num__7 = num__7 ! has num__4 distinct prime factors . answer : d . <eor> d <eos> |
d |
add__1.0__2.0__ add__1.0__4.0__ add__1.0__3.0__ |
add__1.0__2.0__ add__1.0__4.0__ add__1.0__3.0__ |
| if abc ≠ num__1 and the sum of the reciprocals of a b and c equals the reciprocal of the product of a b and c then a = <o> a ) ( num__1 + bc ) / ( b + c ) <o> b ) ( num__1 – bc ) / ( b + c ) <o> c ) ( num__1 + b + c ) / ( bc ) <o> d ) ( num__1 – b – c ) / ( bc ) <o> e ) ( num__1 – b – c ) / ( b + c ) |
answer is c though it took me around num__4 minutes to get that . <eor> c <eos> |
c |
reverse__1.0__ |
reverse__1.0__ |
| if s and t are positive integers such that s / t = num__64.15 which of the following could be the remainder when s is divided by t ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__8 <o> d ) num__20 <o> e ) num__45 |
the remainder is num__0.15 or num__0.15 . you can go one step further and say that x / y = num__0.12 where x and y are whole numbers . plug in all the answer choices for x and see which one makes y a whole number . one thing that jumps out at me is that a b c and d are all even . e on the other hand is odd . why ? maybe i ' ll start plugging in here first . num__45 / y = num__0.15 num__15 y = num__4500 y = num__300 answer : e . <eor> e <eos> |
e |
divide__45.0__0.15__ multiply__0.15__300.0__ |
divide__45.0__0.15__ multiply__0.15__300.0__ |
| on charlie ' s cell phone plan he signed up for num__8 gb of data per num__4 week period . if he went over the num__8 gb per month it would cost him an extra $ num__10.00 per gb . in the first week he used num__2 gb ; in the num__2 nd week he used num__3 gb ; in the num__3 rd week he used num__5 gb ; and in the num__4 th week he used a whopping num__10 gb . how much extra did charlie need to pay on his cell phone bill ? <o> a ) $ num__120 <o> b ) $ num__40 <o> c ) $ num__80 <o> d ) $ num__100 <o> e ) $ num__140 |
to determine how much extra he would need to pay you have to add all num__4 weeks of gb usage ( num__2 + num__3 + num__5 + num__10 = num__20 ) . to determine the extra amount of gb usage you have to subtract the the original amount of gb ( num__8 ) allowed on his plan . num__20 - num__8 = num__12 to determine the extra cost on his cell phone plan you have to multiply $ num__10.00 per extra usage ( num__12 ) for a total of $ num__120.00 . the correct answer is a <eor> a <eos> |
a |
multiply__4.0__5.0__ add__8.0__4.0__ multiply__10.0__12.0__ multiply__10.0__12.0__ |
multiply__4.0__5.0__ add__8.0__4.0__ multiply__10.0__12.0__ multiply__10.0__12.0__ |
| the ratio between the speeds of two trains is num__3 : num__4 . if the second train runs num__400 kms in num__6 hours then the speed of the first train is ? <o> a ) num__50.01 km / hr <o> b ) num__51.01 km / hr <o> c ) num__40.01 km / hr <o> d ) num__52.01 km / hr <o> e ) none of these |
explanation : let the speeds of two trains be num__3 x and num__4 x km / hr . num__4 / x = num__66.6666666667 = > x = num__16.67 km / hr so speed of first train is num__16.67 * num__3 = num__50.01 km / hr option a <eor> a <eos> |
a |
divide__400.0__6.0__ multiply__3.0__16.67__ round__50.01__ |
divide__400.0__6.0__ multiply__3.0__16.67__ multiply__3.0__16.67__ |
| if num__2994 Ã · num__14.5 = num__173 then num__29.94 Ã · num__1.45 = ? <o> a ) num__17.1 <o> b ) num__17.3 <o> c ) num__17.5 <o> d ) num__17.7 <o> e ) num__17.2 |
num__29.94 / num__1.45 = num__299.4 / num__14.5 = ( num__2994 / num__14.5 ) x num__0.1 ) [ here substitute num__173 in the place of num__2994 / num__14.5 ] = num__17.3 = num__17.3 answer is b . <eor> b <eos> |
b |
divide__29.94__299.4__ multiply__173.0__0.1__ multiply__173.0__0.1__ |
divide__29.94__299.4__ multiply__173.0__0.1__ multiply__173.0__0.1__ |
| a volume of num__10936 l water is in a container of sphere . how many hemisphere of volume num__4 l each will be required to transfer all the water into the small hemispheres ? <o> a ) num__2812 <o> b ) num__8231 <o> c ) num__2734 <o> d ) num__4222 <o> e ) num__4254 |
a volume of num__4 l can be kept in num__1 hemisphere therefore a volume of num__10936 l can be kept in ( num__2734.0 ) hemispheres ans . num__2734 answer : c <eor> c <eos> |
c |
multiply__1.0__2734.0__ |
multiply__1.0__2734.0__ |
| the average of four positive integers is num__69 . the highest integer is num__93 and the least integer is num__39 . the difference between the remaining two integers is num__28 . which of the following integers is the higher of the remaining two integers ? <o> a ) num__11 <o> b ) num__86 <o> c ) num__66 <o> d ) num__55 <o> e ) num__44 |
let the four integers be a b c and d where a > b > c > d . ( a + b + c + d ) / num__4 = num__69 = > a + b + c + d = num__276 - - - > ( num__1 ) a = num__93 d = num__39 and b - c = num__28 ( num__1 ) = > b + c = num__276 - ( a + d ) = num__276 - num__132 = num__144 . b + b - num__28 = num__144 b = ( num__144 + num__28 ) / num__2 = num__86 answer : b <eor> b <eos> |
b |
multiply__69.0__4.0__ add__93.0__39.0__ subtract__276.0__132.0__ multiply__1.0__86.0__ |
multiply__69.0__4.0__ add__93.0__39.0__ subtract__276.0__132.0__ divide__86.0__1.0__ |
| num__3 men num__4 women and num__6 children can complete a work in num__7 days . a woman does double the work a man does and a child does half the work a man does . how many women alone can complete this work in num__7 days ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__6 <o> d ) num__5 <o> e ) num__3 |
let num__1 woman ' s num__1 day work = x . then num__1 man ' s num__1 day work = x / num__2 and num__1 child ' s num__1 day work x / num__4 . so ( num__3 x / num__2 + num__4 x + + num__6 x / num__4 ) = num__0.142857142857 num__28 x / num__4 = num__0.142857142857 = > x = num__0.0204081632653 num__1 woman alone can complete the work in num__49 days . so to complete the work in num__7 days number of women required = num__7.0 = num__7 . answer : a <eor> a <eos> |
a |
subtract__4.0__3.0__ subtract__3.0__1.0__ divide__1.0__7.0__ multiply__4.0__7.0__ divide__0.1429__7.0__ round__7.0__ |
subtract__4.0__3.0__ divide__6.0__3.0__ divide__1.0__7.0__ multiply__4.0__7.0__ divide__0.1429__7.0__ add__3.0__4.0__ |
| the perimeter of an isosceles right triangle is num__4 + num__4 sq rt num__2 . what is the length of the hypotenuse of the triangle ? <o> a ) num__2.82 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
side of triangle is a then perimeter = a + a + a . sqrt num__2 ( right angle and pythagorus ) = num__2 a + a . sqrt num__2 = num__4 + num__4 sqrt num__2 or a . ( num__2 + sqrt num__2 ) = num__4 ( num__1 + sqrt num__2 ) a = num__4 . ( num__1 + sqrt num__2 ) / num__2 + sqrt num__2 = num__4 * num__2.414 / num__3.414 = then hypotenuse = num__2.82 a <eor> a <eos> |
a |
triangle_area__2.0__2.82__ |
multiply__1.0__2.82__ |
| if the numerator of a fraction is increased by num__2 and the denominator is increased by num__1 the fraction becomes num__5 ⁄ num__8 and if the numerator of the same fraction is increased by num__3 and the denominator is increased by i the fraction becomes num__3 ⁄ num__4 . what is the original fraction ? <o> a ) data inadequate <o> b ) num__2 ⁄ num__7 <o> c ) num__4 ⁄ num__7 <o> d ) num__3 ⁄ num__7 <o> e ) none of these |
let the original fraction be x ⁄ y . then x + num__2 y + num__1 = num__58 x + num__2 y + num__1 = num__58 or num__8 x – num__5 y = – num__11 . . . . . . . . ( i ) again x + num__3 y + num__1 = num__34 x + num__3 y + num__1 = num__34 or num__4 x – num__3 y = – num__9 . . . . . . . . ( ii ) solving ( i ) and ( ii ) we get x = num__3 and y = num__7 ∴ fraction = num__3 ⁄ num__7 answer d <eor> d <eos> |
d |
add__8.0__3.0__ add__1.0__8.0__ add__2.0__5.0__ add__2.0__1.0__ |
add__8.0__3.0__ add__1.0__8.0__ add__2.0__5.0__ add__2.0__1.0__ |
| a city had num__400 migrants in the year num__2000 . since then the number of migrants in a countrythe city has doubled every num__3 years . if there were num__400 migrants in the country in the year num__2000 what was the increase in the population of migrants during the period from num__2009 to num__2012 ? <o> a ) num__1400 <o> b ) num__2300 <o> c ) num__3000 <o> d ) num__3200 <o> e ) num__3800 |
the population will increase in the following order : num__2000 : num__400 num__2003 : num__800 num__2006 : num__1600 num__2009 : num__3200 num__2012 : num__6400 difference between num__2009 and num__2012 = num__3200 option d <eor> d <eos> |
d |
add__2000.0__3.0__ add__3.0__2003.0__ subtract__2000.0__400.0__ subtract__6400.0__3200.0__ |
add__2000.0__3.0__ add__3.0__2003.0__ subtract__2000.0__400.0__ subtract__6400.0__3200.0__ |
| three numbers are in the ratio num__1 : num__2 : num__3 and their h . c . f is num__62 . the numbers are : <o> a ) num__62 num__124 num__186 <o> b ) num__60 num__124 num__186 <o> c ) num__10 num__20 num__30 <o> d ) num__62 num__24 num__186 <o> e ) num__12 num__24 num__186 |
let the required numbers be x num__2 x and num__3 x . then their h . c . f = x . so x = num__62 . the numbers are num__62 num__124 num__186 . answer : a <eor> a <eos> |
a |
multiply__2.0__62.0__ multiply__3.0__62.0__ multiply__1.0__62.0__ |
multiply__2.0__62.0__ multiply__3.0__62.0__ multiply__1.0__62.0__ |
| working alone at their respective constant rates a can complete a task in ‘ a ’ days and b in ‘ b ’ days . they take turns in doing the task with each working num__2 days at a time . if a starts they finish the task in exactly num__10 days . if b starts they take a day more . how long does it take to complete the task if they both work together ? <o> a ) num__5 <o> b ) num__4.5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
work done by a & b in a day = x & y respectively . when a starts : no . of days when a works = num__6 no . of days when b works = num__4 → num__6 x + num__4 y = num__1 when b starts : no . of days when b works = num__5 no . of days when a works = num__5 → num__5 x + num__5 y = num__1 solving the above two equations for xy x = num__0.1 y = num__0.1 → total work done by ab in a day = num__0.1 + num__0.1 = num__0.2 = num__0.2 → no . of days to complete the work when both work together = num__5 answer : a <eor> a <eos> |
a |
subtract__10.0__6.0__ divide__10.0__2.0__ divide__1.0__10.0__ divide__2.0__10.0__ round__5.0__ |
subtract__10.0__6.0__ add__1.0__4.0__ divide__1.0__10.0__ divide__2.0__10.0__ add__1.0__4.0__ |
| in a mixture num__60 litres the ratio of milk and water num__2 : num__1 . if this ratio is to be num__1 : num__2 then the quanity of water to be further added is : <o> a ) num__20 liters <o> b ) num__30 liters <o> c ) num__40 liters <o> d ) num__60 liters <o> e ) num__70 liters |
quantity of milk = num__60 x num__2 litres = num__40 litres . num__3 quantity of water in it = ( num__60 - num__40 ) litres = num__20 litres . new ratio = num__1 : num__2 let quantity of water to be added further be x litres . then milk : water = num__40 . num__20 + x now num__40 = num__1 num__20 + x num__2 num__20 + x = num__80 x = num__60 . quantity of water to be added = num__60 litres so the correct answer is option d ) <eor> d <eos> |
d |
add__2.0__1.0__ divide__60.0__3.0__ add__60.0__20.0__ multiply__60.0__1.0__ |
add__2.0__1.0__ subtract__60.0__40.0__ add__60.0__20.0__ add__40.0__20.0__ |
| a computer wholesaler sells num__7 different computers and each is priced differently . if the wholesaler chooses three computers for display at a trade show what is the probability ( all things being equal ) that the two most expensive computers will be among the three chosen for display ? <o> a ) num__0.142857142857 <o> b ) num__0.107142857143 <o> c ) num__0.0357142857143 <o> d ) num__0.0178571428571 <o> e ) num__0.00595238095238 |
since two of the choices are prefixed we are free to choose num__1 from the rest of the num__5 avilable . so num__5 c num__1 is the numerator . total no of ways we can choose num__3 from num__7 is num__7 c num__3 which is the denominator . so the probability : num__5 c num__0.142857142857 c num__3 = num__0.142857142857 ans is a . <eor> a <eos> |
a |
reverse__7.0__ reverse__7.0__ |
reverse__7.0__ reverse__7.0__ |
| mike drives his new corvette from san francisco to las vegas a journey of num__640 miles . he drives the first half of the trip at an average rate of num__80 miles per hour but has to slow down for the second half of his journey . if the second half of the trip takes him num__200 percent longer than the first half what is his average rate w in miles per hour for the entire trip ? <o> a ) w = num__26.7 <o> b ) w = num__30.0 <o> c ) w = num__40.0 <o> d ) w = num__53.3 <o> e ) num__60.0 |
veritas prepofficial solution correct answer : c using the formula : time = distance / rate we find that mike takes num__4 hours to cover the first num__320 miles of his trip . since the num__2 nd num__320 miles take num__200.0 longer than the first it takes mike num__8 hours longer or num__12 hours . ( note : num__200.0 longer than the first half is not num__200.0 of the first half . ) the overall time is num__4 hours + num__12 hours or num__16 hours . since the definition of average rate = total distance traveled / total time of travel mike ' s average rate = num__40.0 or num__40 miles per hour . answer choice c is correct . <eor> c <eos> |
c |
multiply__80.0__4.0__ divide__640.0__320.0__ divide__640.0__80.0__ add__4.0__8.0__ multiply__2.0__8.0__ divide__640.0__16.0__ round__40.0__ |
multiply__80.0__4.0__ divide__640.0__320.0__ divide__640.0__80.0__ add__4.0__8.0__ add__4.0__12.0__ divide__640.0__16.0__ divide__640.0__16.0__ |
| a father was as old as his son ' s present at the time of your birth . if the father ' s age is num__36 years now the son ' s age num__5 years back was ? <o> a ) num__10 years <o> b ) num__12 years <o> c ) num__14 years <o> d ) num__13 years <o> e ) num__20 years |
let the son ' s present age be x years . then ( num__36 - x ) = x num__2 x = num__36 . x = num__18 . son ' s age num__5 years back ( num__18 - num__5 ) = num__13 years . d <eor> d <eos> |
d |
divide__36.0__2.0__ subtract__18.0__5.0__ subtract__18.0__5.0__ |
divide__36.0__2.0__ subtract__18.0__5.0__ subtract__18.0__5.0__ |
| a man can row num__6 kmph in still water . when the river is running at num__3 kmph it takes him num__1 hour to row to a place and black . what is the total distance traveled by the man ? <o> a ) num__5.75 <o> b ) num__4.5 <o> c ) num__5.76 <o> d ) num__5.74 <o> e ) num__5.71 |
m = num__6 s = num__3 ds = num__9 us = num__3 x / num__9 + x / num__3 = num__1 x = num__2.25 d = num__2.25 * num__2 = num__4.5 answer : b <eor> b <eos> |
b |
add__6.0__3.0__ divide__6.0__3.0__ multiply__2.25__2.0__ round__4.5__ |
add__6.0__3.0__ divide__6.0__3.0__ multiply__2.25__2.0__ multiply__1.0__4.5__ |
| the difference between the value of a number increased by num__12.5 and the value of the original number decreased by num__25.0 is num__30 . what is the original number t ? <o> a ) num__60 <o> b ) num__80 <o> c ) num__40 <o> d ) num__120 <o> e ) num__160 |
( num__1 + num__0.125 ) x - ( num__1 - num__0.25 ) x = num__30 ( num__1.125 ) x - ( num__0.75 ) x = num__30 x = num__80 = t answer : b <eor> b <eos> |
b |
add__0.125__1.0__ subtract__1.0__0.25__ multiply__1.0__80.0__ |
add__0.125__1.0__ subtract__1.0__0.25__ multiply__1.0__80.0__ |
| a goods train runs at the speed of num__72 kmph and crosses a num__240 m long platform in num__26 seconds . what is the length of the goods train ? <o> a ) num__230 m <o> b ) num__270 m <o> c ) num__643 m <o> d ) num__280 m <o> e ) num__270 m |
speed = ( num__72 x num__0.277777777778 ) m / sec = num__20 m / sec . time = num__26 sec . let the length of the train be x metres . then x + num__9.23076923077 = num__20 x + num__240 = num__520 x = num__280 . answer : d <eor> d <eos> |
d |
divide__240.0__26.0__ multiply__26.0__20.0__ subtract__520.0__240.0__ round__280.0__ |
divide__240.0__26.0__ multiply__26.0__20.0__ subtract__520.0__240.0__ round__280.0__ |
| two trains are moving at num__50 kmph and num__70 kmph in opposite directions . their lengths are num__150 m and num__100 m respectively . the time they will take to pass each other completely is ? <o> a ) num__8 ½ sec <o> b ) num__6 ½ sec <o> c ) num__2 ½ sec <o> d ) num__7 ½ sec <o> e ) num__9 ½ sec |
num__70 + num__50 = num__120 * num__0.277777777778 = num__33.3333333333 mps d = num__150 + num__100 = num__250 m t = num__250 * num__0.03 = num__7.5 = num__7 ½ sec answer : d <eor> d <eos> |
d |
add__50.0__70.0__ add__150.0__100.0__ multiply__0.03__250.0__ round__7.0__ |
add__50.0__70.0__ add__150.0__100.0__ multiply__0.03__250.0__ round__7.0__ |
| a train speeds past a pole in num__15 sec and a platform num__100 m long in num__25 sec its length is ? <o> a ) num__288 <o> b ) num__150 <o> c ) num__188 <o> d ) num__166 <o> e ) num__122 |
let the length of the train be x m and its speed be y m / sec . then x / y = num__15 = > y = x / num__15 ( x + num__100 ) / num__25 = x / num__15 = > x = num__150 m . answer : b <eor> b <eos> |
b |
round__150.0__ |
round__150.0__ |
| at what rate percent on simple interest will rs . num__750 amount to rs . num__900 in num__4 years ? <o> a ) num__6.0 <o> b ) num__2.0 <o> c ) num__4.0 <o> d ) num__5.0 <o> e ) num__3 % |
num__150 = ( num__750 * num__4 * r ) / num__100 r = num__5.0 answer : d <eor> d <eos> |
d |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| two trains are moving at num__90 kmph and num__70 kmph in opposite directions . their lengths are num__150 m and num__100 m respectively . the time they will take to pass each other completely is ? <o> a ) num__8.4 sec <o> b ) num__5.625 sec <o> c ) num__6.66666666667 sec <o> d ) num__6.16666666667 sec <o> e ) num__10.5 sec |
num__70 + num__90 = num__150 * num__0.277777777778 = num__44.4444444444 mps d = num__150 + num__100 = num__250 m t = num__250 * num__0.0225 = num__5.625 sec answer : b <eor> b <eos> |
b |
add__150.0__100.0__ multiply__0.0225__250.0__ multiply__0.0225__250.0__ |
add__150.0__100.0__ multiply__0.0225__250.0__ multiply__0.0225__250.0__ |
| george can fill q cans of paint in num__3 minutes . if there are r cans of paint in one gallon how many gallons can george fill in num__45 minutes ? <o> a ) num__30 r / q . <o> b ) num__15 r / q <o> c ) num__30 q / r <o> d ) num__5 q / r <o> e ) num__15 q / r |
george can fill q cans of paint in num__3 minutes . so in num__45 mins he can fill num__15.0 = num__15 q cans . r cans = num__1 gallons = > num__15 q cans = num__15 q / r gallons . hence the answer is e . <eor> e <eos> |
e |
divide__45.0__3.0__ round__15.0__ |
divide__45.0__3.0__ divide__45.0__3.0__ |
| martha takes a road trip from point a to point b . she drives x percent of the distance at num__70 miles per hour and the remainder at num__50 miles per hour . if martha ' s average speed for the entire trip is represented as a fraction in its reduced form in terms of x which of the following is the numerator ? <o> a ) num__110 <o> b ) num__35000 <o> c ) num__1100 <o> d ) num__3000 <o> e ) num__30 |
000 |
total distance = d total time taken = x / ( num__100 * num__70 ) + ( num__100 - x ) / ( num__100 * num__50 ) speed = distance / time gives numerator = num__35000 answer : b <eor> b <eos> |
b |
b |
| the population of a town is num__8100 . it decreases annually at the rate of num__10.0 p . a . what was its population num__2 years ago ? <o> a ) num__9000 <o> b ) num__8000 <o> c ) num__8500 <o> d ) num__9500 <o> e ) num__10000 |
formula : ( after = num__100 denominator ago = num__100 numerator ) num__8100 × num__1.11111111111 × num__1.11111111111 = num__10000 e ) <eor> e <eos> |
e |
percent__100.0__10000.0__ |
percent__100.0__10000.0__ |
| how many num__2 - inch by num__3 - inch rectangular tiles are required to tile this shaded region ? <o> a ) less than num__10 <o> b ) num__10 — num__100 <o> c ) num__101 — num__1000 <o> d ) num__1001 — num__1500 <o> e ) num__1500 + |
convert all feet into inches as we are looking into tiles which are in inches so total area of garden = num__6 * num__12 * num__12 * num__12 shaded region = num__12 * num__12 ( num__6 * num__12 - num__2 * num__3 ) = num__12 * num__12 * num__2 * num__3 ( num__12 - num__1 ) no of num__2 ' ' * num__3 ' ' tiles required = num__12 ∗ num__12 ∗ num__12 ∗ num__2 ∗ num__3 ∗ num__5.5 ∗ num__3 = num__12 ∗ num__12 ∗ num__11 = num__1584 > num__1500 so ans is num__1500 + answer : e <eor> e <eos> |
e |
multiply__2.0__3.0__ square_perimeter__3.0__ multiply__2.0__5.5__ triangle_area__2.0__1500.0__ |
multiply__2.0__3.0__ multiply__2.0__6.0__ multiply__2.0__5.5__ multiply__1.0__1500.0__ |
| the twonumbers ( num__23 __ ) ^ num__32 and ( num__12 __ ) ^ num__33 has num__6 in the last digit what is the missing digit ? <o> a ) num__4 <o> b ) num__5 <o> c ) six ( num__6 ) <o> d ) num__7 <o> e ) num__8 |
it should be num__6 as last digit in all powers of nos with unit digit num__6 is num__6 . answer : c <eor> c <eos> |
c |
subtract__12.0__6.0__ |
subtract__12.0__6.0__ |
| num__121 x num__5 ^ num__4 = ? <o> a ) num__75625 <o> b ) num__68225 <o> c ) num__72325 <o> d ) num__71225 <o> e ) num__72225 |
explanation : num__121 × num__5 ^ num__4 = num__121 × ( num__5.0 ) ^ num__4 = ( num__121 × num__10000 ) / num__16 = num__7.5625 × num__10000 = num__75625 answer is a <eor> a <eos> |
a |
divide__121.0__16.0__ multiply__7.5625__10000.0__ multiply__7.5625__10000.0__ |
divide__121.0__16.0__ multiply__7.5625__10000.0__ multiply__7.5625__10000.0__ |
| find the area of a parallelogram with base num__48 cm and height num__36 cm ? <o> a ) num__1730 cm num__2 <o> b ) num__1728 cm num__2 <o> c ) num__1870 cm num__2 <o> d ) num__1890 cm num__2 <o> e ) num__668 cm num__2 |
area of a parallelogram = base * height = num__48 * num__36 = num__1728 cm num__2 answer : b <eor> b <eos> |
b |
multiply__48.0__36.0__ multiply__48.0__36.0__ |
multiply__48.0__36.0__ multiply__48.0__36.0__ |
| find the number of different meals of num__4 items that you can get from the given menu of num__6 items and no need to choose different items . <o> a ) num__120 <o> b ) num__126 <o> c ) num__5040 <o> d ) num__15 <o> e ) num__18 |
total item : num__6 we can select : num__4 thus - > num__6 c num__4 which = num__6 c num__2 : : num__6 * num__2.5 * num__1 = num__15 answer : d <eor> d <eos> |
d |
subtract__6.0__4.0__ multiply__6.0__2.5__ multiply__6.0__2.5__ |
subtract__6.0__4.0__ multiply__6.0__2.5__ multiply__6.0__2.5__ |
| calculate num__469111 x num__9999 = ? <o> a ) num__4586970843 <o> b ) num__4686970743 <o> c ) num__4691100843 <o> d ) num__4586870843 <o> e ) num__4690640889 |
answer num__469111 x num__9999 = num__469111 x ( num__10000 - num__1 ) = num__4691110000 - num__469111 = num__4690640889 . option : e <eor> e <eos> |
e |
subtract__10000.0__9999.0__ multiply__469111.0__10000.0__ multiply__469111.0__9999.0__ multiply__469111.0__9999.0__ |
subtract__10000.0__9999.0__ multiply__469111.0__10000.0__ subtract__4691110000.0__469111.0__ subtract__4691110000.0__469111.0__ |
| the first photo shoot takes num__3 minutes long and then the following shoots are taken at a rate of num__24 seconds / shoot as the model is already at the scene . what is the maximum number of photo shoots taken under num__10 minutes ? <o> a ) num__13 <o> b ) num__14 <o> c ) num__15 <o> d ) num__16 <o> e ) num__18 |
a must be an integer as it is the number shoots at a rate of num__24 sec / shoot num__3 * num__60 + num__24 a = num__10 * num__60 num__24 a = num__420 a = num__17 the total number of shoots - - > num__1 + num__17 = num__18 and num__19 th shoot will be taken at num__612 seconds which is above num__10 minutes answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ add__1.0__17.0__ add__1.0__18.0__ round__18.0__ |
hour_to_min_conversion__ add__1.0__17.0__ add__1.0__18.0__ add__1.0__17.0__ |
| a person can walk at a constant rate of num__8 mph and can bike at a rate of num__16 mph . if he wants to travel num__88 miles in num__8 hours using bike and walking at their constant rates how much distance would he require to walk ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__40 <o> d ) num__60 <o> e ) num__72 |
total distance = num__88 distance = speed * time walking speed = s num__1 = num__8 walking time = t num__1 bike speed = s num__2 = num__16 time traveled in bike = t num__2 d num__1 + d num__2 = num__88 s num__1 t num__1 + s num__2 t num__2 = num__88 num__8 * t num__1 + num__16 * t num__2 = num__88 t num__1 + num__2 * t num__2 = num__11 - - - - - ( num__1 ) given : t num__1 + t num__2 = num__8 - - - - - ( num__2 ) ( num__1 ) - ( num__2 ) - - > t num__2 = num__3 and t num__1 = num__8 - num__3 = num__5 walking distance = s num__1 * t num__1 = num__8 * num__5 = num__40 answer : c <eor> c <eos> |
c |
divide__16.0__8.0__ divide__88.0__8.0__ add__1.0__2.0__ subtract__8.0__3.0__ multiply__8.0__5.0__ round__40.0__ |
divide__16.0__8.0__ divide__88.0__8.0__ add__1.0__2.0__ subtract__8.0__3.0__ multiply__8.0__5.0__ multiply__8.0__5.0__ |
| the sum of all the integers k such that − num__26 < k < num__24 is <o> a ) num__0 <o> b ) − num__2 <o> c ) − num__25 <o> d ) − num__49 <o> e ) − num__51 |
since k defines a range between − num__26 < k < num__24 we can set num__0 as the reference point for the negative values and positive values . the negative values will range from - num__25 to num__0 whereas the positive values will range from num__0 - num__23 . we can conclude that for all but - num__25 and - num__24 the number pairs will add to num__0 . so we have left - num__25 - num__24 = - num__49 . answer d . <eor> d <eos> |
d |
add__26.0__23.0__ add__26.0__23.0__ |
add__26.0__23.0__ add__26.0__23.0__ |
| a rectangular - shaped carpet that measures x feet by y feet is priced at $ num__42 . what is the cost of the carpet in dollars per square yard ? ( num__1 square yard = num__9 square feet ) <o> a ) xy / num__360 <o> b ) num__9 xy / num__40 <o> c ) num__40 xy / num__9 <o> d ) num__378 xy <o> e ) num__378 / ( xy ) |
the area of the carpet in feet is xy . the area in square yards is xy / num__9 . the price per square yard is num__42 / ( xy / num__9 ) = num__378 / ( xy ) . the answer is e . <eor> e <eos> |
e |
multiply__42.0__9.0__ multiply__42.0__9.0__ |
multiply__42.0__9.0__ divide__378.0__1.0__ |
| a customer purchased a package of ground beef at a cost of $ num__1.80 per pound . for the same amount of money the customer could have purchased a piece of steak that weighed num__20 percent less than the package of ground beef . what was the cost per pound of the steak ? <o> a ) $ num__2.05 <o> b ) $ num__2.15 <o> c ) $ num__2.25 <o> d ) $ num__2.35 <o> e ) $ num__2.45 |
for simplicity let ' s assume the customer bought num__1 pound of ground beef for $ num__1.80 . let x be the price per pound for the steak . then num__0.8 x = num__180 x = num__180 / num__0.8 = $ num__2.25 the answer is c . <eor> c <eos> |
c |
round_down__1.8__ subtract__1.8__1.0__ divide__1.8__0.8__ divide__1.8__0.8__ |
round_down__1.8__ subtract__1.8__1.0__ divide__1.8__0.8__ divide__1.8__0.8__ |
| the password for a computer account has to consist of exactly eight characters . characters can be chosen from any of the following : letter of the alphabet numerical digits from num__0 to num__9 a hyphen or the exclamation mark . upper - case letters ( e . g . a ) are considered different from lower - case letters ( e . g . a ) and characters can be repeated . given these rules how many different passwords are possible ? <o> a ) num__2 ^ num__9 <o> b ) num__2 ^ num__14 <o> c ) num__2 ^ num__18 <o> d ) num__2 ^ num__40 <o> e ) num__2 ^ num__48 |
num__26 letters in the alphabet num__26 * num__2 = num__52 letters since the password is case - sensitive . accounting for possibilities of the other special characters we have ; num__26 + num__26 + num__10 + num__1 + num__1 = num__64 or num__2 ^ num__6 possibilities per character of the password . total number of possible combinations : ( num__2 ^ num__6 ) ^ num__8 = num__2 ^ num__48 answer : e <eor> e <eos> |
e |
alphabet_space__ coin_space__ card_space__ negate_prob__0.0__ die_space__ coin_space__ |
alphabet_space__ coin_space__ card_space__ negate_prob__0.0__ die_space__ coin_space__ |
| find num__16 th term in the series num__7 num__13 num__19 num__25 . . . <o> a ) num__97 <o> b ) num__98 <o> c ) num__99 <o> d ) num__100 <o> e ) num__101 |
solution : a = num__7 d = num__13 – num__7 = num__6 num__16 th term t num__16 = a + ( n - num__1 ) d = num__7 + ( num__16 – num__1 ) num__6 = num__7 + num__90 = num__97 answer is a <eor> a <eos> |
a |
subtract__13.0__7.0__ subtract__7.0__6.0__ add__7.0__90.0__ add__7.0__90.0__ |
subtract__13.0__7.0__ subtract__7.0__6.0__ add__7.0__90.0__ add__7.0__90.0__ |
| if the wheel is num__14 cm then the number of revolutions to cover a distance of num__1672 cm is ? <o> a ) num__22 <o> b ) num__28 <o> c ) num__19 <o> d ) num__12 <o> e ) num__88 |
num__2 * num__3.14285714286 * num__14 * x = num__1672 = > x = num__19 answer : c <eor> c <eos> |
c |
round__19.0__ |
round__19.0__ |
| during one month at a particular restaurant num__0.166666666667 of the burgers sold were veggie burgers and num__1 of the rest of the burgers sold were double - meat . if x of the burgers sold were double - meat how many were veggie burgers ? <o> a ) x / num__8 <o> b ) x / num__5 <o> c ) num__2 x / num__3 <o> d ) num__3 x / num__4 <o> e ) num__4 x / num__5 |
let y be the number of total burgers . veggie = y / num__6 non veggie = num__5 y / num__6 num__0.25 of the rest of the burgers sold were double - meat = > num__5 y / num__6 * num__1 = double meat = x = > y / num__6 = x / num__5 = veggie hence b <eor> b <eos> |
b |
subtract__6.0__1.0__ multiply__1.0__5.0__ |
subtract__6.0__1.0__ multiply__1.0__5.0__ |
| a can run num__3 km distance in num__2 min while b can run this distance in num__2 min num__30 sec . by how much distance can a beat b ? <o> a ) num__900 m <o> b ) num__600 m <o> c ) num__120 m <o> d ) num__180 m <o> e ) num__190 m |
a takes time num__2 minutes = num__120 sec b takes time num__2.30 minutes = num__150 sec diffrence = num__150 - num__120 = num__30 sec now we are to find distance covered in num__30 sec by b num__150 sec = num__3000 m num__1 sec = num__20 m num__30 sec = num__20 x num__30 = num__600 m answer : b <eor> b <eos> |
b |
add__30.0__120.0__ subtract__3.0__2.0__ divide__3000.0__150.0__ multiply__30.0__20.0__ round__600.0__ |
add__30.0__120.0__ subtract__3.0__2.0__ divide__3000.0__150.0__ multiply__30.0__20.0__ round__600.0__ |
| how many positive integers less than num__9000 are there in which the sum of the digits equals num__5 ? <o> a ) num__56 <o> b ) num__57 <o> c ) num__58 <o> d ) num__59 <o> e ) num__60 |
basically the question asks how many num__4 digit numbers ( including those in the form num__0 xxx num__00 xx and num__000 x ) have digits which add up to num__5 . think about the question this way : we know that there is a total of num__5 to be spread among the num__4 digits we just have to determine the number of ways it can be spread . let x represent a sum of num__1 and | represent a seperator between two digits . as a result we will have num__5 x ' s ( digits add up to the num__5 ) and num__3 | ' s ( num__3 digit seperators ) . so for example : xx | x | x | x = num__2111 | | xxx | xx = num__0032 etc . there are num__8 c num__3 ways to determine where to place the separators . hence the answer is num__8 c num__3 = num__56 . a <eor> a <eos> |
a |
subtract__5.0__4.0__ subtract__4.0__1.0__ add__5.0__3.0__ multiply__1.0__56.0__ |
subtract__5.0__4.0__ subtract__4.0__1.0__ add__5.0__3.0__ multiply__1.0__56.0__ |
| if x ^ num__2 − num__2 x − num__15 = ( x + r ) ( x + s ) for all values of x and if r and s are constants then which of the following is a possible value of r + s ? <o> a ) num__8 <o> b ) num__2 <o> c ) − num__2 <o> d ) − num__3 <o> e ) − num__5 |
we know that given ax ^ num__2 + bx + c = num__0 sum of the roots = - b / a and product of the roots = c / a . the roots here are - r and - s . - r - s = - ( - num__2 ) / num__1 = r + s = - num__2 ( - r ) * ( - s ) = - num__15.0 = rs so one of r and s is - num__5 and the other is num__3 . so r + s could be - num__2 . answer ( c ) <eor> c <eos> |
c |
add__2.0__1.0__ multiply__2.0__1.0__ |
add__2.0__1.0__ multiply__2.0__1.0__ |
| a certain sum of money amounts to rs . num__1008 in num__2 years and to rs . num__1164 in num__3 ½ years . find the sum and rate of interests . <o> a ) num__13.0 <o> b ) num__12.0 <o> c ) num__25.0 <o> d ) num__18.0 <o> e ) num__52 % |
s . i . for num__1 ½ years = rs . ( num__1164 - num__1008 ) = rs . num__156 . s . l . for num__2 years = rs . ( num__156 * ( num__0.666666666667 ) * num__2 ) = rs . num__208 principal = rs . ( num__1008 - num__208 ) = rs . num__800 . now p = num__800 t = num__2 and s . l . = num__208 . rate = ( num__100 * num__208 ) / ( num__800 * num__2 ) % = num__13.0 answer is a . <eor> a <eos> |
a |
percent__100.0__13.0__ |
percent__100.0__13.0__ |
| num__5 num__10 num__13 num__26 num__29 num__58 num__61 ( . . . . ) <o> a ) num__128 <o> b ) num__122 <o> c ) num__64 <o> d ) num__125 <o> e ) num__140 |
num__5 × num__2 = num__10 num__10 + num__3 = num__13 num__13 × num__2 = num__26 num__26 + num__3 = num__29 num__29 × num__2 = num__58 num__58 + num__3 = num__61 num__61 × num__2 = num__122 answer is b <eor> b <eos> |
b |
divide__10.0__5.0__ subtract__5.0__2.0__ multiply__61.0__2.0__ multiply__61.0__2.0__ |
divide__10.0__5.0__ subtract__5.0__2.0__ multiply__61.0__2.0__ multiply__61.0__2.0__ |
| the value of a num__10.5 stock in which an income of rs . num__756 is derived by investing rs . num__9000 brokerage being % is : <o> a ) rs . num__120 <o> b ) rs . num__121 <o> c ) rs . num__124.75 <o> d ) rs . num__130.75 <o> e ) rs . num__167 |
for an income of rs . num__756 investment = rs . num__9000 for an income of rs . investment = = rs . num__125 for a rs . num__100 stock investment = rs . num__125 . market value of rs . num__100 stock = = rs . num__124.75 c <eor> c <eos> |
c |
percent__100.0__124.75__ |
percent__100.0__124.75__ |
| john karen and luke collected cans of vegetables for a food drive . the number of cans that john collected was num__0.25 the number of cans that karen collected and num__0.111111111111 the number of cans that luke collected . the number of cans that karen collected was what fraction of the total number of cans that john karen and luke collected ? <o> a ) num__0.2 <o> b ) num__0.333333333333 <o> c ) num__0.4 <o> d ) num__0.5 <o> e ) num__0.111111111111 |
john = ( num__0.25 ) karen - - > karen = num__4 ( john ) john = num__0.111111111111 ( luke ) - - > luke = num__9 ( john ) total = num__36 ( john ) karen / total = num__0.111111111111 = num__0.111111111111 answer : e <eor> e <eos> |
e |
reverse__0.25__ divide__9.0__0.25__ reverse__9.0__ |
reverse__0.25__ divide__9.0__0.25__ reverse__9.0__ |
| if a father said to his elder son ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is num__25 years now what was the son ' s age four years back ? <o> a ) num__7.5 <o> b ) num__8.5 <o> c ) num__9 <o> d ) num__9.5 <o> e ) num__10 |
let son ' s present age be a years . then ( num__25 − a ) = a ⇒ num__2 a = num__25 ⇒ a = num__12.5 = num__12.5 son ' s age num__4 years back = num__12.5 - num__4 = num__8.5 answer : b <eor> b <eos> |
b |
divide__25.0__2.0__ subtract__12.5__4.0__ subtract__12.5__4.0__ |
divide__25.0__2.0__ subtract__12.5__4.0__ subtract__12.5__4.0__ |
| section - num__1 if num__9 men working num__6 hours a day can do a work in num__88 days . then num__6 men working num__8 hours a day can do it in how many days ? <o> a ) num__22 <o> b ) num__99 <o> c ) num__787 <o> d ) num__66 <o> e ) num__11 |
explanation : if men is fixed work is proportional to time . if work is fixed then time is inversely proportional to men therefore ( m num__1 * t num__1 / w num__1 ) = ( m num__2 * t num__2 / w num__2 ) from the above formula i . e ( m num__1 * t num__1 / w num__1 ) = ( m num__2 * t num__2 / w num__2 ) so ( num__9 * num__6 * num__88.0 ) = ( num__6 * num__8 * d / num__1 ) on solving d = num__99 days . answer : b <eor> b <eos> |
b |
subtract__8.0__6.0__ round__99.0__ |
subtract__8.0__6.0__ multiply__1.0__99.0__ |
| a train num__300 m long passed a pole in num__30 sec . how long will it take to pass a platform num__650 m long ? <o> a ) num__28 sec <o> b ) num__89 sec <o> c ) num__85 sec <o> d ) num__16 sec <o> e ) num__95 sec |
speed = num__10.0 = num__10 m / sec . required time = ( num__300 + num__650 ) / num__10 = num__95 sec . answer : e <eor> e <eos> |
e |
divide__300.0__30.0__ round__95.0__ |
divide__300.0__30.0__ round__95.0__ |
| in a regular week there are num__5 working days and for each day the working hours are num__8 . a man gets rs . num__2.30 per hour for regular work and rs . num__3.20 per hours for overtime . if he earns rs . num__432 in num__4 weeks then how many hours does he work for ? <o> a ) num__160 <o> b ) num__175 <o> c ) num__180 <o> d ) num__200 <o> e ) num__210 |
suppose the man works overtime for x hours . now working hours in num__4 weeks = ( num__5 x num__8 x num__4 ) = num__160 . num__160 x num__2.30 + x x num__3.20 = num__432 x = num__20 . hence total hours of work = ( num__160 + num__20 ) = num__180 . answer : c <eor> c <eos> |
c |
multiply__5.0__4.0__ add__160.0__20.0__ round__180.0__ |
multiply__5.0__4.0__ add__160.0__20.0__ add__160.0__20.0__ |
| if xy + z = x ( y + z ) which of the following must be true ? <o> a ) x = num__2 or z = num__0 <o> b ) x = num__1 or z = num__0 <o> c ) x = num__1 or z = num__1 <o> d ) x = num__0 or z = num__0 <o> e ) x = num__0 or z = num__0 |
xy + z = xy + xz z = xz xz - z = num__0 z ( x - num__1 ) = num__0 means ; either z = num__0 or x - num__1 = num__0 i . e . x = num__1 answer b <eor> b <eos> |
b |
reverse__1.0__ |
reverse__1.0__ |
| how much time will it take for an amount of rs . num__400 to yield rs . num__100 as interest at num__5.0 per annum of simple interest ? <o> a ) num__5 years <o> b ) num__4 years <o> c ) num__7 years <o> d ) num__6 years <o> e ) num__2 years |
explanation : time = ( num__100 x num__100 ) / ( num__400 x num__5 ) years = num__5 years . answer : a <eor> a <eos> |
a |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| length of a rectangular plot is num__50 mtr more than its breadth . if the cost of fencin g the plot at num__26.50 per meter is rs . num__5300 what is the length of the plot in mtr ? <o> a ) num__46 m <o> b ) num__60 m <o> c ) num__58 m <o> d ) num__75 m <o> e ) num__80 m |
let breadth = x metres . then length = ( x + num__50 ) metres . perimeter = num__5300 / num__26.5 m = num__200 m . num__2 [ ( x + num__50 ) + x ] = num__200 num__2 x + num__50 = num__100 num__2 x = num__50 x = num__25 . hence length = x + num__50 = num__75 m d <eor> d <eos> |
d |
divide__5300.0__26.5__ multiply__50.0__2.0__ divide__50.0__2.0__ add__50.0__25.0__ round__75.0__ |
divide__5300.0__26.5__ divide__200.0__2.0__ divide__50.0__2.0__ add__50.0__25.0__ add__50.0__25.0__ |
| the simple interest on a sum in num__2 years is rs . num__96 and the compound interest on the same sum for the sametime is rs . num__99 . find the sum and the rate . <o> a ) num__3.25 <o> b ) num__5.25 <o> c ) num__6.25 <o> d ) num__4.25 <o> e ) num__7.25 % |
p ( r / num__100 ) ^ num__2 = num__3 ie ) pr ^ num__2 = num__30000 s . i = p * num__2 * r / num__100 = num__96 ie ) pr = num__4800 pr ^ num__2 / pr = num__6.25 r = num__6.25 answer : c <eor> c <eos> |
c |
percent__100.0__6.25__ |
percent__100.0__6.25__ |
| the no of revolutions a wheel of diameter num__40 cm makes in traveling a distance of num__176 m is <o> a ) num__122 <o> b ) num__140 <o> c ) num__287 <o> d ) num__128 <o> e ) num__112 |
distance covered in num__1 revolution = = num__2 * ( num__3.14285714286 ) * num__20 = num__125.714285714 cm required no of revolutions = num__17600 * ( num__0.00795454545455 ) = num__140 answer : b <eor> b <eos> |
b |
divide__40.0__2.0__ divide__1.0__125.7143__ divide__17600.0__125.7143__ round__140.0__ |
divide__40.0__2.0__ divide__1.0__125.7143__ divide__17600.0__125.7143__ round__140.0__ |
| a and b can do a work in num__6 days . they both started the work and after three days a left the work . b finished the remaining work in num__15 days . in how many days will a finish the work ? <o> a ) num__7 days <o> b ) num__7.5 days <o> c ) num__8.5 days <o> d ) num__9.5 days <o> e ) num__6.5 days |
a and b ' s num__1 day work i . e a + b = num__0.166666666667 work completed after three days = num__3 * num__0.166666666667 = num__0.5 remaining work num__0.5 i . e num__50 percentage completed by b in num__15 days so b can finish work in num__30 days . . b ' s num__1 day work num__0.0333333333333 a ' s num__1 day work = num__0.166666666667 - num__0.0333333333333 = num__0.133333333333 so a can finish work num__7.5 = num__7.5 days answer : b <eor> b <eos> |
b |
divide__1.0__6.0__ divide__3.0__6.0__ divide__15.0__0.5__ divide__0.5__15.0__ multiply__15.0__0.5__ round__7.5__ |
divide__1.0__6.0__ divide__3.0__6.0__ divide__15.0__0.5__ divide__0.5__15.0__ multiply__15.0__0.5__ multiply__15.0__0.5__ |
| rs . num__1600 is divided into two parts such that if one part is invested at num__6.0 and the other at num__5.0 the whole annual interest from both the sum is rs . num__85 . how much was lent at num__5.0 ? <o> a ) num__1100 <o> b ) num__1200 <o> c ) num__1300 <o> d ) num__1400 <o> e ) num__1700 |
( x * num__5 * num__1 ) / num__100 + [ ( num__1600 - x ) * num__6 * num__1 ] / num__100 = num__85 num__5 x / num__100 + ( num__9600 – num__6 x ) / num__100 = num__85 = > x = num__1100 answer : a <eor> a <eos> |
a |
percent__100.0__1100.0__ |
percent__100.0__1100.0__ |
| a pharmaceutical company received $ num__4 million in royalties on the first $ num__20 million in sales of the generic equivalent of one of its products and then $ num__9 million in royalties on the next $ num__108 million in sales . by approximately what percent did the ratio of royalties to sales decrease from the first $ num__20 million in sales to the next $ num__108 million in sales ? <o> a ) num__8.0 <o> b ) num__15.0 <o> c ) num__45.0 <o> d ) num__52.0 <o> e ) num__58 % |
solution : this is a percent decrease problem . we will use the formula : percent change = ( new – old ) / old x num__100 to calculate the final answer . we first set up the ratios of royalties to sales . the first ratio will be for the first num__20 million in sales and the second ratio will be for the next num__108 million in sales . because all of the sales are in millions we do not have to express all the trailing zeros in our ratios . first num__20 million royalties / sales = num__0.2 = num__0.2 next num__108 million royalties / sales = num__0.0833333333333 = num__0.0833333333333 because each ratio is not an easy number to use we can simplify each one by multiplying each by the lcm of the two denominators which is num__60 . keep in mind that we are able to do this only because our answer choices are expressed in percents . first num__20 million royalties / sales = ( num__0.2 ) x num__60 = num__12 next num__108 million royalties / sales = num__0.0833333333333 = ( num__0.0833333333333 ) x num__60 = num__5 we can plug num__12 and num__5 into our percent change formula : ( new – old ) / old x num__100 [ ( num__5 – num__12 ) / num__12 ] x num__100 - num__58.3333333333 x num__100 at this point we can stop and consider the answer choices . since we know that num__58.3333333333 is just a bit less than ½ we know that - num__58.3333333333 x num__100 is about a num__58.0 decrease . answer e . <eor> e <eos> |
e |
divide__4.0__20.0__ divide__9.0__108.0__ divide__108.0__9.0__ reverse__0.2__ round_down__58.3333__ round_down__58.3333__ |
divide__4.0__20.0__ divide__9.0__108.0__ divide__108.0__9.0__ reverse__0.2__ round_down__58.3333__ round_down__58.3333__ |
| the h . c . f . of two numbers is num__23 and the other two factors of their l . c . m . are num__10 and num__12 . the larger of the two numbers is : <o> a ) num__276 <o> b ) num__299 <o> c ) num__322 <o> d ) num__345 <o> e ) num__395 |
clearly the numbers are ( num__23 x num__10 ) and ( num__23 x num__12 ) . larger number = ( num__23 x num__12 ) = num__276 . answer : option a <eor> a <eos> |
a |
multiply__23.0__12.0__ multiply__23.0__12.0__ |
multiply__23.0__12.0__ multiply__23.0__12.0__ |
| a bat is bought for rs . num__400 and sold at a gain of num__30.0 find its selling price . <o> a ) a ) rs . num__460 / - <o> b ) b ) rs . num__470 / - <o> c ) c ) rs . num__480 / - <o> d ) d ) rs . num__500 / - <o> e ) e ) rs . num__520 / - |
num__100.0 - - - - - - > num__400 ( num__100 * num__4 = num__400 ) num__130.0 - - - - - - > num__520 ( num__130 * num__4 = num__520 ) selling price = rs . num__520 answer : e <eor> e <eos> |
e |
percent__100.0__520.0__ |
percent__100.0__520.0__ |
| think of a number divide it by num__5 and add num__8 to it . the result is num__61 . what is the number thought of ? <o> a ) num__24 <o> b ) num__77 <o> c ) num__297 <o> d ) num__265 <o> e ) num__29 |
explanation : num__61 - num__6 = num__53 num__53 x num__5 = num__265 answer : d <eor> d <eos> |
d |
subtract__61.0__8.0__ multiply__5.0__53.0__ multiply__5.0__53.0__ |
subtract__61.0__8.0__ multiply__5.0__53.0__ multiply__5.0__53.0__ |
| if the salary of an employee is first increased by num__20.0 and the decreased by num__12.0 then what is the change in his salary in percent ? <o> a ) num__5.9 increase <o> b ) num__5.6 increase <o> c ) num__4.6 increase <o> d ) num__1.6 increase <o> e ) num__5.7 increase |
explanation : let the first change be denoted by ‘ x ’ and the second by ‘ y ’ . the net change is given by { x + y + ( x * y ) / num__100 } % therefore the net change in salary is { num__20 + ( - num__12 ) + ( num__20 * ( - num__12 ) ) / num__100 } = num__8 – num__2.4 = num__5.6 ( as the sign is + ve the change is increase ) answer : b <eor> b <eos> |
b |
percent__20.0__12.0__ percent__100.0__5.6__ |
percent__20.0__12.0__ percent__100.0__5.6__ |
| two trains num__121 meters and num__165 meters in length respectively are running in opposite directions one at the rate of num__80 km and the other at the rate of num__65 kmph . in what time will they be completely clear of each other from the moment they meet ? <o> a ) num__7.19 <o> b ) num__7.17 <o> c ) num__7.19 <o> d ) num__7.15 <o> e ) num__7.11 |
t = ( num__121 + num__165 ) / ( num__80 + num__65 ) * num__3.6 t = num__7.15 answer : d <eor> d <eos> |
d |
round__7.15__ |
round__7.15__ |
| reena took a loan of num__1500 with simple interest for as many years as the rate of interest . if she paid num__735 as interest at the end of the loan period what was the rate of interest ? <o> a ) num__3.6 <o> b ) num__7 <o> c ) num__18 <o> d ) can not be determined <o> e ) none |
explanation : let rate = r % and time = r years . then ( num__1500 x r x r ) / num__100 = num__735 num__15 r  ² = num__735 r  ² = num__49 r = num__7 . answer : option b <eor> b <eos> |
b |
percent__100.0__7.0__ |
percent__100.0__7.0__ |
| calculate the percentage gain if a trader bought a bicycle for rs . num__440 and sold it for rs . num__610 ? <o> a ) num__38.64 <o> b ) num__39.64 <o> c ) num__31.64 <o> d ) num__35.64 <o> e ) num__34.64 % |
c . p . = num__440 s . p . = num__610 gain = num__610 - num__440 - - - - num__170.0 gain = > num__0.386363636364 * num__100 = > num__38.64 answer : a <eor> a <eos> |
a |
percent__100.0__38.64__ |
percent__100.0__38.64__ |
| a circular mat with diameter num__14 inches is placed on a square tabletop each of whose sides is num__24 inches long . which of the following is closest to the fraction of the tabletop covered by the mat ? <o> a ) num__0.416666666667 <o> b ) num__0.4 <o> c ) num__0.2 <o> d ) num__0.75 <o> e ) num__0.833333333333 |
so we are looking for the area of the cloth over the area of the table area of the cloth = ( pi ) ( r ) ^ num__2 which is about ( num__3 ) ( num__7 ) ( num__7 ) area of the table = ( num__24 ) ( num__24 ) so the quick way to estimate is looking at the fraction like this : ( num__0.125 ) ( num__1.70833333333 ) i hope this is easy to follow so with some simplification i get ( num__0.125 ) ( num__1.71428571429 ) = num__0.214285714286 = ( num__0.2 ) answer is c <eor> c <eos> |
c |
multiply__0.125__1.7143__ triangle_area__2.0__0.2__ |
multiply__0.125__1.7143__ triangle_area__2.0__0.2__ |
| the price of num__2 sarees and num__4 shirts is rs . num__1600 . with the same money one can buy num__1 saree and num__6 shirts . if one wants to buy num__12 shirts how much shall he have to pay ? <o> a ) rs . num__2400 <o> b ) rs . num__2200 <o> c ) rs . num__2300 <o> d ) rs . num__2500 <o> e ) rs . num__2600 |
let the price of a saree and a shirt be rs . x and rs . y respectively . then num__2 x + num__4 y = num__1600 . . . . ( i ) and x + num__6 y = num__1600 . . . . ( ii ) divide equation ( i ) by num__2 we get the below equation . = > x + num__2 y = num__800 . - - - ( iii ) now subtract ( iii ) from ( ii ) x + num__6 y = num__1600 ( - ) x + num__2 y = num__800 - - - - - - - - - - - - - - - - num__4 y = num__800 - - - - - - - - - - - - - - - - therefore y = num__200 . now apply value of y in ( iii ) = > x + num__2 x num__200 = num__800 = > x + num__400 = num__800 therefore x = num__400 solving ( i ) and ( ii ) we get x = num__400 y = num__200 . cost of num__12 shirts = rs . ( num__12 x num__200 ) = rs . num__2400 . a <eor> a <eos> |
a |
divide__1600.0__2.0__ divide__800.0__4.0__ multiply__2.0__200.0__ add__1600.0__800.0__ add__1600.0__800.0__ |
divide__1600.0__2.0__ divide__800.0__4.0__ multiply__2.0__200.0__ add__1600.0__800.0__ add__1600.0__800.0__ |
| three numbers are in the ratio num__1 : num__2 : num__3 and their h . c . f is num__5 . the numbers are <o> a ) num__12 num__24 num__30 <o> b ) num__12 num__24 num__38 <o> c ) num__12 num__24 num__362 <o> d ) num__5 num__10 num__15 <o> e ) num__12 num__24 num__321 |
explanation : let the required numbers be x num__2 x num__3 x . then their h . c . f = x . so x = num__5 \ inline \ fn _ jvn \ therefore the numbers are num__5 num__10 num__15 answer : d ) num__5 num__10 num__15 <eor> d <eos> |
d |
multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__1.0__5.0__ |
multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__1.0__5.0__ |
| from given equation find the value of x : num__2 x ² + num__9 x − num__5 <o> a ) num__1 <o> b ) num__0.5 <o> c ) num__0.333333333333 <o> d ) num__0.666666666667 <o> e ) num__0.4 |
that quadratic is factored as follows : num__2 x ² + num__9 x − num__5 = ( num__2 x − num__1 ) ( x + num__5 ) . lesson num__17 . now it is easy to see that the second factor will be num__0 when x = − num__5 . as for the value of x that will make num__2 x − num__1 = num__0 we must solve that little equation . ( lesson num__9 . ) we have : num__2 x = num__1 x = num__1 num__2 the solutions are : x = num__0.5 or − num__5 b <eor> b <eos> |
b |
reverse__2.0__ reverse__2.0__ |
reverse__2.0__ reverse__2.0__ |
| a garrison of num__2000 men has provisions for num__40 days . at the end of num__20 days a reinforcement arrives and it is now found that the provisions will last only for num__10 days more . what is the reinforcement ? <o> a ) num__1888 <o> b ) num__2766 <o> c ) num__2999 <o> d ) num__2000 <o> e ) num__1712 |
num__2000 - - - - num__40 num__2000 - - - - num__20 x - - - - - num__10 x * num__10 = num__2000 * num__20 x = num__4000 num__2000 - - - - - - - num__2000 answer : d <eor> d <eos> |
d |
round__2000.0__ |
subtract__4000.0__2000.0__ |
| what is the total number of positive integers that are less than num__200 and that have no positive factor in common with num__200 other than num__1 ? <o> a ) num__60 <o> b ) num__70 <o> c ) num__80 <o> d ) num__90 <o> e ) num__100 |
since num__200 = num__2 ^ num__3 * num__5 ^ num__2 then a number can not have num__2 and / or num__5 as a factor . the odd numbers do not have num__2 as a factor and there are num__100 odd numbers from num__1 to num__200 . we then need to eliminate the num__20 numbers that end with num__5 that is num__5 num__15 num__25 . . . num__195 . there are a total of num__100 - num__20 = num__80 such numbers between num__1 and num__200 . the answer is c . <eor> c <eos> |
c |
add__1.0__2.0__ add__2.0__3.0__ divide__200.0__2.0__ divide__100.0__5.0__ multiply__3.0__5.0__ add__5.0__20.0__ subtract__200.0__5.0__ subtract__100.0__20.0__ multiply__1.0__80.0__ |
add__1.0__2.0__ add__2.0__3.0__ divide__200.0__2.0__ divide__100.0__5.0__ multiply__3.0__5.0__ add__5.0__20.0__ subtract__200.0__5.0__ subtract__100.0__20.0__ multiply__1.0__80.0__ |
| p and q are both positive integers . when p is divided by q the remainder is some positive integer d and when p is divided by ( q + num__4 ) the remainder is also d . if p / q = num__1020.75 and p / ( q + num__4 ) = num__816.6 then which of the following gives the correct set of { d q } ? <o> a ) { num__6 num__12 } <o> b ) { num__6 num__15 } <o> c ) { num__12 num__16 } <o> d ) { num__9 num__15 } <o> e ) { num__15 num__24 } |
p / q = d = p / q + num__4 d = . num__75 q d = . num__6 ( q + num__4 ) we get d = num__12 and q = num__16 . answer = c . <eor> c <eos> |
c |
add__4.0__12.0__ subtract__16.0__4.0__ |
add__4.0__12.0__ subtract__16.0__4.0__ |
| the h . c . f . of two numbers is num__12 and their l . c . m . is num__520 . if one of the numbers is num__480 then the other is : <o> a ) num__13 <o> b ) num__18 <o> c ) num__21 <o> d ) num__24 <o> e ) num__38 |
other number = ( num__12 x num__520 ) / num__480 = num__13 . answer : a <eor> a <eos> |
a |
gcd__520.0__13.0__ |
gcd__520.0__13.0__ |
| the measurement of a rectangular box with lid is num__25 cmx num__18 cmx num__18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of π cm num__3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere ) <o> a ) num__288 <o> b ) num__48 <o> c ) num__72 <o> d ) num__972 <o> e ) num__964 |
d = num__18 r = num__9 ; volume of the largest sphere = num__1.33333333333 π r num__3 = num__1.33333333333 * π * num__9 * num__9 * num__9 = num__972 π cm num__3 answer : d <eor> d <eos> |
d |
round__972.0__ |
round__972.0__ |
| the current of a stream at num__1 kmph . a motor boat goes num__35 km upstream and back to the starting point in num__12 hours . the speed of the motor boat in still water is ? <o> a ) num__6 <o> b ) num__9 <o> c ) num__4 <o> d ) num__5 <o> e ) num__2 |
s = num__1 m = x ds = x + num__1 us = x - num__1 num__35 / ( x + num__1 ) + num__35 / ( x - num__1 ) = num__12 x = num__6 answer : a <eor> a <eos> |
a |
round__6.0__ |
subtract__12.0__6.0__ |
| the present ages of three persons are in the proportion of num__4 : num__7 : num__9 . eight years ago the sum of their ages was num__36 . find their present ages . <o> a ) num__20 num__3545 <o> b ) num__8 num__2028 <o> c ) num__16 num__2836 <o> d ) num__12 num__2127 <o> e ) none of these |
let the present ages of three persons be num__4 k num__7 k and num__9 k respectively . ( num__4 k - num__8 ) + ( num__7 k - num__8 ) + ( num__9 k - num__8 ) = num__36 num__20 k = num__60 k = num__3 therefore then present ages are num__12 num__2127 . answer : d <eor> d <eos> |
d |
subtract__7.0__4.0__ multiply__4.0__3.0__ multiply__4.0__3.0__ |
subtract__7.0__4.0__ add__4.0__8.0__ add__4.0__8.0__ |
| a certain board game is played by rolling a pair of fair six - sided dice and then moving one ' s piece forward the number of spaces indicated by the sum showing on the dice . a player is frozen if her opponent ' s piece comes to rest in the space already occupied by her piece . if player a is about to roll and is currently ten spaces behind player b what is the probability that player b will be frozen after player a rolls ? <o> a ) num__0.0833333333333 <o> b ) num__0.138888888889 <o> c ) num__0.166666666667 <o> d ) num__0.333333333333 <o> e ) num__0.472222222222 |
no . of possible outcomes = num__6 * num__6 = num__36 no . of outcomes that result a total of num__10 ( as a is num__10 spaces behind b ) = num__3 ( ( num__46 ) ( num__55 ) ( num__64 ) ) so the probability = num__0.0833333333333 = num__0.0833333333333 ( option a ) <eor> a <eos> |
a |
add__36.0__10.0__ divide__3.0__36.0__ divide__3.0__36.0__ |
add__36.0__10.0__ divide__3.0__36.0__ divide__3.0__36.0__ |
| the average mark of the students of a class in a particular exam is num__70 . if num__5 students whose average mark in that exam is num__50 are excluded the average mark of the remaining will be num__90 . find the number of students who wrote the exam ? <o> a ) num__20 <o> b ) num__15 <o> c ) num__10 <o> d ) num__12 <o> e ) num__25 |
let the number of students who wrote the exam be x . total marks of students = num__70 x . total marks of ( x - num__5 ) students = num__90 ( x - num__5 ) num__70 x - ( num__5 * num__50 ) = num__90 ( x - num__5 ) num__200 = num__20 x = > x = num__10 answer : c <eor> c <eos> |
c |
subtract__70.0__50.0__ divide__50.0__5.0__ divide__50.0__5.0__ |
subtract__70.0__50.0__ divide__50.0__5.0__ subtract__20.0__10.0__ |
| a railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket . one reserved first class ticket from chennai to trivandrum costs rs . num__216 and one full and one half reserved first class tickets cost rs . num__327 . what is the basic first class full fare and what is the reservation charge ? <o> a ) rs . num__105 and rs . num__6 <o> b ) rs . num__216 and rs . num__12 <o> c ) rs . num__210 and rs . num__12 <o> d ) rs . num__210 and rs . num__6 <o> e ) none |
explanatory answers let half of the full basic fare be rs . x . therefore full basic fare is rs . num__2 x . let the reservation charge be rs . y per ticket . now one full reservation ticket would cost num__2 x ( basic fare ) + y ( reservation charge ) num__2 x + y = num__216 - - - ( num__1 ) the total basic fare for one half and one full ticket = x + num__2 x = num__3 x and the total reservation charge is num__2 y . hence num__3 x + num__2 y = num__327 - - - ( num__2 ) solving ( num__1 ) and ( num__2 ) we get x = num__105 and y = num__6 hence the full basic fare is num__2 x = rs . num__210 and the reservation charge is y = rs . num__6 answer d <eor> d <eos> |
d |
triple__1.0__ triple__2.0__ twice__105.0__ twice__105.0__ |
add__1.0__2.0__ triple__2.0__ subtract__216.0__6.0__ subtract__216.0__6.0__ |
| the sale price of an article including the sales tax is rs . num__616 . the rate of sales tax is num__10.0 . if the shopkeeper has made a profit of num__12.0 then the cost price of the article is : <o> a ) num__280 <o> b ) num__579 <o> c ) num__500 <o> d ) num__400 <o> e ) num__100 |
c num__110.0 of s . p . = num__616 s . p . = ( num__616 * num__100 ) / num__110 = rs . num__560 c . p = ( num__110 * num__560 ) / num__112 = rs . num__500 <eor> c <eos> |
c |
percent__100.0__500.0__ |
percent__100.0__500.0__ |
| given the equation e = f / num__4 + g / num__4 ^ num__2 + h / num__4 ^ num__3 where f g and h are each equal to num__1 or num__0 then e could be any of the values below except : <o> a ) num__0.0625 <o> b ) num__0.3125 <o> c ) num__0.125 <o> d ) num__0.265625 <o> e ) num__0.015625 |
e = f / num__4 + g / num__4 ^ num__2 + h / num__4 ^ num__3 = f / num__4 + g / num__16 + h / num__64 = ( num__16 f + num__4 g + h ) / num__64 depending on whether f g and h take num__0 or num__1 : substitute num__0 or num__1 for the different values of f g and h . answer c <eor> c <eos> |
c |
multiply__4.0__16.0__ divide__2.0__16.0__ |
multiply__4.0__16.0__ divide__2.0__16.0__ |
| a & b throw a dice alternatively till one of them gets a ' num__6 ' and wins the game . what will be a probability ? <o> a ) num__0.363636363636 <o> b ) num__0.454545454545 <o> c ) num__0.545454545455 <o> d ) num__0.636363636364 <o> e ) num__0.727272727273 |
whoever starts first will have a probability of num__0.545454545455 and the other has a probability of num__0.454545454545 . so if a starts as has to be logically inferred from the question pbbl that a wins = num__0.545454545455 and that b wins = num__0.454545454545 . pbbl that the starting person wins = ( num__0.166666666667 ) + ( num__0.833333333333 ) * ( num__0.833333333333 ) * ( num__0.166666666667 ) + ( num__0.833333333333 ) * ( num__0.833333333333 ) * ( num__0.833333333333 ) * ( num__0.833333333333 ) * ( num__0.166666666667 ) + . . . = g . p . with first term ( num__0.166666666667 ) and common ratio ( num__0.833333333333 ) * ( num__0.833333333333 ) = ( num__0.166666666667 ) / [ num__1 - ( num__0.833333333333 ) * ( num__0.833333333333 ) ] = num__0.545454545455 answer : c <eor> c <eos> |
c |
negate_prob__0.5455__ negate_prob__0.1667__ negate_prob__0.4545__ |
negate_prob__0.5455__ negate_prob__0.1667__ negate_prob__0.4545__ |
| if num__20 men can build a water fountain num__56 metres long in num__42 days what length of a similar water fountain can be built by num__35 men in num__3 days ? <o> a ) num__3 m <o> b ) num__4 m <o> c ) num__7 m <o> d ) num__9 m <o> e ) num__10 m |
explanation : let the required length be x metres more men more length built ( direct proportion ) less days less length built ( direct proportion ) men num__20 : num__35 days num__42 : num__3 : : num__56 : x therefore ( num__20 x num__42 x x ) = ( num__35 x num__3 x num__56 ) x = ( num__35 x num__3 x num__56 ) / num__840 = num__7 hence the required length is num__7 m . answer : c <eor> c <eos> |
c |
multiply__20.0__42.0__ subtract__42.0__35.0__ round__7.0__ |
multiply__20.0__42.0__ subtract__42.0__35.0__ round__7.0__ |
| in a certain deck of cards each card has a positive integer written on it in a multiplication game a child draws a card and multiplies the integer on the card with the next large integer . if the each possible product is between num__10 and num__400 then the least and greatest integer on the card would be <o> a ) num__3 and num__15 <o> b ) num__3 and num__20 <o> c ) num__4 and num__13 <o> d ) num__4 and num__14 <o> e ) num__5 and num__14 |
given : num__10 < x ( x + num__1 ) < num__400 . now it ' s better to test the answer choices here rather than to solve : if x = num__3 then x ( x + num__1 ) = num__12 > num__10 - - > so the least value is num__3 . test for the largest value : if x = num__20 then x ( x + num__1 ) = num__20 * num__21 = num__420 > num__400 answer : b . <eor> b <eos> |
b |
add__1.0__20.0__ add__400.0__20.0__ multiply__1.0__3.0__ |
add__1.0__20.0__ add__400.0__20.0__ multiply__1.0__3.0__ |
| the difference between a number and its two - fifth is num__510 . what is num__10.0 of that number ? <o> a ) num__33 <o> b ) num__85 <o> c ) num__37 <o> d ) num__27 <o> e ) num__28 |
explanation : let the number be x . then . num__10.0 num__0 f num__850 = num__85 . answer : b ) num__85 <eor> b <eos> |
b |
divide__850.0__10.0__ divide__850.0__10.0__ |
divide__850.0__10.0__ divide__850.0__10.0__ |
| num__40 percent of the ducks included in a certain side effects study were male . if some of the ducks had side effects during the study and num__25 percent of the ducks who had side effects were male what was the ratio of the side effect rate for the male ducks to the side effect rate for the female ducks ? <o> a ) num__0.25 <o> b ) num__0.642857142857 <o> c ) num__0.5 <o> d ) num__0.875 <o> e ) num__1.14285714286 |
say total male = num__40 total female = num__60 required ratio = ( . num__0.625 ) / ( . num__1.25 ) = num__0.5 c is the answer <eor> c <eos> |
c |
percent__40.0__1.25__ percent__40.0__1.25__ |
percent__40.0__1.25__ percent__40.0__1.25__ |
| this topic is locked . if you want to discuss this question please re - post it in the respective forum . matt and peter can do together a piece of work in num__20 days . after they have worked together for num__12 days matt stops and peter completes the remaining work in num__9 days . in how many days peter complete the work separately . <o> a ) num__26 days <o> b ) num__27.5 days <o> c ) num__22.5 days <o> d ) num__25.35 days <o> e ) num__24 days |
together they complete the job in num__20 days means they complete num__0.6 of the job after num__12 days . peter completes the remaining ( num__0.4 ) of the job in num__9 days which means that the whole job ( num__1 ) can be completed in x days . < = > num__0.4 - > num__9 < = > x = num__9 / ( num__0.4 ) = num__22.5 thus the answer is c . <eor> c <eos> |
c |
km_to_mile_conversion__ add__0.4__0.6__ divide__9.0__0.4__ round__22.5__ |
divide__12.0__20.0__ add__0.4__0.6__ divide__9.0__0.4__ divide__9.0__0.4__ |
| a merchant has num__100 lbs of sugar part of which he sells at num__7.0 profit and the rest at num__12.0 profit . he gains num__10.0 on the whole . find how much is sold at num__7.0 profit ? <o> a ) num__70 lbs <o> b ) num__40 lbs <o> c ) num__30 lbs <o> d ) num__50 lbs <o> e ) num__60 lbs |
these types ofweighted averagequestions can be solved in a variety of ways so you can choose whichever method you find easiest / fastest . here ' s another variation on the weighted average formula : a = # of pounds sold at num__7.0 profit b = # of pounds sold at num__12.0 profit a + b = num__100 pounds ( . num__07 a + . num__12 b ) / ( a + b ) = . num__10 . num__07 a + . num__12 b = . num__1 a + . num__1 b . num__02 b = . num__03 a num__2 b = num__3 a num__0.666666666667 = a / b so for every num__6 pounds of a we have num__4 pounds of b . with num__100 pounds total we have num__40 pounds of a and num__60 pounds of b . option b <eor> b <eos> |
b |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| express num__22 mps in kmph ? <o> a ) num__79.2 kmph <o> b ) num__89.2 kmph <o> c ) num__79.6 kmph <o> d ) num__99.2 kmph <o> e ) num__69.2 kmph |
num__22 * num__3.6 = num__79.2 kmph answer : a <eor> a <eos> |
a |
multiply__22.0__3.6__ round__79.2__ |
multiply__22.0__3.6__ multiply__22.0__3.6__ |
| a number is said to be a “ digifac ” if each of its digits is a factor of the number itself . what is the sum v of the missing digits of the following five - digit digifac : num__9 num__5 num__3 _ _ ? <o> a ) num__5 <o> b ) num__7 <o> c ) num__9 <o> d ) num__10 <o> e ) num__14 |
here the term “ digifac ” should look intimidating . you probably haven ’ t studied digifacs before so how should you approach this problem ? well keep in mind that digifacs aren ’ t being tested ; in fact the author of this question just made that term up and then defined it for you . what makes this question hard is that the non - challenge - seeker ( i think i just made that term up too … ) will see the unfamiliar term “ digifac ” and lose faith immediately . “ i don ’ t know what that is ! ” she who finds the challenge in the gmat fun however will read the definition and think “ got it – i need to find the two digits that ensure that num__9 num__5 and num__3 are both factors of the overall number and that the remaining two digits are also factors ” . and work from there . the number must be divisible by num__5 so the only units digits that work are num__0 or num__5 . and the number must be divisible by num__9 ( and also num__3 ) so we need the sum v of all digits to be a multiple of num__9 . num__9 + num__5 + num__3 = num__17 so our only options are to get the sum to num__18 ( by adding num__1 ) or to num__27 ( by adding num__10 ) . a quick glance at the answer choices shows that num__0 num__1 isn ’ t an option . why not ? that would require num__0 to be one of the digits … and num__0 isn ’ t a factor of anything . so the units digit must be num__5 making the tens digit num__5 and we have num__95355 . that number is a multiple of num__5 num__3 and num__9 so it works : the correct answer is d and more importantly this fun challenge required no “ trivial ” information about digifacs … that term only existed to obscure the link between the given information and the path to the answer . d <eor> d <eos> |
d |
subtract__18.0__17.0__ multiply__9.0__3.0__ add__9.0__1.0__ add__9.0__1.0__ |
subtract__18.0__17.0__ add__9.0__18.0__ add__9.0__1.0__ add__9.0__1.0__ |
| what percent of num__9.6 kg is num__28 gms ? <o> a ) num__25 <o> b ) num__66 <o> c ) num__288 <o> d ) num__29 <o> e ) num__17 |
explanation : required percentage = ( num__0.00291666666667 * num__100 ) % = num__0.29 % = num__0.29 answer : d ) . num__29.0 <eor> d <eos> |
d |
percent__100.0__29.0__ |
percent__100.0__29.0__ |
| the cost of manufacturing a popular model car is made up of three items : cost of raw material labour and overheads - in a year the cost of three items were in the ration of num__4 : num__3 : num__2 . next year the cost of the raw material rose by num__10.0 labour cost increased by num__8.0 but overhead reduced by num__5.0 . then % increase int the price of the car ? <o> a ) num__7.67 <o> b ) num__6.0 <o> c ) num__0.54 <o> d ) num__9.54 <o> e ) num__8.54 % |
before increase total cost = num__4 + num__3 + num__2 = num__9 after increasing the cost = num__9.54 so increase of . num__54 over num__9 equal to num__6.0 answer : b <eor> b <eos> |
b |
add__4.0__5.0__ add__4.0__2.0__ add__4.0__2.0__ |
add__4.0__5.0__ add__4.0__2.0__ add__4.0__2.0__ |
| average expenditure of a person for the first num__3 days of a week is rs . num__350 and for the next num__4 days is rs . num__420 . average expenditure of the man for the whole week is : <o> a ) num__350 <o> b ) num__370 <o> c ) num__390 <o> d ) num__430 <o> e ) none |
explanation : assumed mean = rs . num__350 total excess than assumed mean = num__4 × ( rs . num__420 - rs . num__350 ) = rs . num__280 therefore increase in average expenditure = rs . num__40.0 = rs . num__40 therefore average expenditure for num__7 days = rs . num__350 + rs . num__40 = rs . num__390 correct option : c <eor> c <eos> |
c |
add__3.0__4.0__ add__350.0__40.0__ add__350.0__40.0__ |
add__3.0__4.0__ add__350.0__40.0__ add__350.0__40.0__ |
| a standard veggiematik machine can chop num__35 carrots in num__5 minutes . how many carrots can num__3 standard veggiematik machines chop in num__3 minutes ? <o> a ) num__63 <o> b ) num__87 <o> c ) num__90 <o> d ) num__98 <o> e ) num__112 |
direct relationship : - num__1 standard veggiematik machine - num__35 carrots - num__5 minutes num__1 standard veggiematik machine - num__7 carrots - num__1 minute now num__3 standard veggiematik machine - ? carrots - num__3 minutes hence = num__7 x num__3 x num__3 = num__63 carrots answer a <eor> a <eos> |
a |
divide__35.0__5.0__ round__63.0__ |
divide__35.0__5.0__ round__63.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 sec . what is the length of the train ? <o> a ) num__296 m <o> b ) num__267 m <o> c ) num__297 m <o> d ) num__150 m <o> e ) num__255 m |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__9 = num__150 m answer : d <eor> d <eos> |
d |
round__150.0__ |
round__150.0__ |
| patrick has three daughters named avery bridget & catherine . he decides he wants to make a num__3 - letter password using the a combination of the each of their first initials ( e . g . abc cba etc . ) . how many different permutations can he come up with ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__11 |
permutation calc is : n p r = n ! / ( n - r ) ! num__3 p num__3 = num__3 ! / ( num__3 - num__3 ) ! = num__3 x num__2 x num__1 / num__0 ! = num__6.0 = num__6 correct answer is c ( num__6 ) answer : c <eor> c <eos> |
c |
coin_space__ negate_prob__1.0__ die_space__ die_space__ |
coin_space__ negate_prob__1.0__ die_space__ die_space__ |
| at exactly what time past num__8 : num__00 will the minute and hour hands of an accurate working clock be precisely perpendicular to each other for the first time ? <o> a ) num__20 num__0.619047619048 minutes past num__8 : num__00 <o> b ) num__20 num__0.764705882353 minutes past num__7 : num__00 <o> c ) num__21 num__0.130434782609 minutes past num__7 : num__00 <o> d ) num__21 num__0.818181818182 minutes past num__7 : num__00 <o> e ) num__22 num__0.444444444444 minutes past num__7 : num__00 |
a num__1 = hour angle a num__2 = minute angle a num__1 - a num__2 = num__90 assume the minutes = m a num__1 = ( num__8 * num__60 + m ) / num__12 * num__60 ) * num__360 a num__2 = m * num__6.0 solving all three equation results in the answer a <eor> a <eos> |
a |
hour_to_min_conversion__ subtract__8.0__2.0__ add__8.0__12.0__ |
hour_to_min_conversion__ subtract__8.0__2.0__ add__8.0__12.0__ |
| the weight of every type a widget is the same the weight of every type b widget is the same and the weight of every type c widget is the same . if the weight of num__7 type a widgets is equal to the weight of num__3 type b widgets and the weight of num__5 type b widgets is equal to the weight of num__7 type c widgets . what is the ratio of the total weight of num__1 type a widget and num__1 type b widget to the total weight of num__1 type b widget and num__1 type c widget ? <o> a ) num__1 : num__2 <o> b ) num__2 : num__3 <o> c ) num__3 : num__4 <o> d ) num__4 : num__5 <o> e ) num__5 : num__6 |
num__5 b = num__7 c and so b = num__7 c / num__5 num__7 a = num__3 b and so a = num__3 b / num__7 = num__3 c / num__5 a + b = num__3 c / num__5 + num__7 c / num__5 = num__10 c / num__5 b + c = num__7 c / num__5 + c = num__12 c / num__5 the ratio of a + b : b + c = num__10 : num__12 = num__5 : num__6 the answer is e . <eor> e <eos> |
e |
add__7.0__3.0__ add__7.0__5.0__ subtract__7.0__1.0__ multiply__5.0__1.0__ |
add__7.0__3.0__ add__7.0__5.0__ add__5.0__1.0__ divide__5.0__1.0__ |
| a hollow iron pipe is num__21 cm long and its external diameter is num__8 cm . if the thickness of the pipe is num__1 cm and iron weighs then the weight of the pipe is <o> a ) num__3.496 kg <o> b ) num__3.696 kg <o> c ) num__3.690 kg <o> d ) num__9.696 kg <o> e ) num__3.296 kg |
explanation : external radius = num__4 cm internal radius = num__3 cm . volume of iron = = weight of iron = ( num__462 x num__8 ) gm = num__3696 gm = num__3.696 kg answer : b ) num__3.696 kg <eor> b <eos> |
b |
subtract__4.0__1.0__ multiply__8.0__462.0__ multiply__1.0__3.696__ |
subtract__4.0__1.0__ multiply__8.0__462.0__ multiply__1.0__3.696__ |
| a is half good a work man as b and together they finish a job in num__14 days . in how many days working alone b finish the job ? <o> a ) num__98 days <o> b ) num__21 days <o> c ) num__17 days <o> d ) num__18 days <o> e ) num__19 days |
wc = num__1 : num__2 num__2 x + x = num__0.0714285714286 = > x = num__0.0238095238095 num__2 x = num__0.047619047619 = > num__21 days answer : b <eor> b <eos> |
b |
divide__1.0__14.0__ multiply__2.0__0.0238__ round__21.0__ |
divide__1.0__14.0__ multiply__2.0__0.0238__ round__21.0__ |
| jack invested $ num__2000 in fund a and $ num__2000 in fund b . over the next two years the money in fund a earned a total interest of num__12 percent for the two years combined and the money in fund b earned num__30 percent annual interest compounded annually . two years after jack made these investments . jack ' s investment in fund b was worth how much more than his investment in fund a ? <o> a ) num__1 . $ num__1000 <o> b ) num__2 . $ num__1500 <o> c ) num__3 . $ num__900 <o> d ) num__4 . $ num__1140 <o> e ) num__5 . $ num__700 |
interest on fund a will be num__240 . rate of interest will be num__6.0 per annum simple interest as num__12.0 is for num__2 year . this will make investment a num__2240 at the end of num__2 nd year . interest on fund b will be num__1380 on num__30.0 interest compounded annually . this will make investment b num__3380 at the end of num__2 nd year . difference in investment b and investment a = num__3380 - num__1690 = num__1140 answer is d . <eor> d <eos> |
d |
divide__12.0__6.0__ add__2000.0__240.0__ add__2000.0__1380.0__ divide__3380.0__2.0__ subtract__1380.0__240.0__ subtract__6.0__2.0__ |
divide__12.0__6.0__ add__2000.0__240.0__ add__2000.0__1380.0__ divide__3380.0__2.0__ subtract__1380.0__240.0__ subtract__6.0__2.0__ |
| the captain of a cricket team of num__11 members is num__26 years old and the wicket keeper is num__3 years older . if the ages of these two are excluded the average age of the remaining players is one year less than the average age of the whole team . find out the average age of the team . <o> a ) num__23 years <o> b ) num__20 years <o> c ) num__24 years <o> d ) num__21 years <o> e ) num__25 years |
explanation : number of members in the team = num__11 let the average age of of the team = xx = > sum of the ages of all num__11 members = num__11 x num__11 x age of the captain = num__26 age of the wicket keeper = num__26 + num__3 = num__29 sum of the ages of num__9 members of the team excluding captain and wicket keeper = num__11 x − num__26 − num__29 = num__11 x − num__55 = num__11 x − num__26 − num__29 = num__11 x − num__55 average age of num__9 members of the team excluding captain and wicket keeper = num__11 x − num__559 = num__11 x − num__559 given that num__11 x − num__559 = ( x − num__1 ) num__11 x − num__559 = ( x − num__1 ) ⇒ num__11 x − num__55 = num__9 ( x − num__1 ) ⇒ num__11 x − num__55 = num__9 x − num__9 ⇒ num__2 x = num__46 ⇒ x = num__462 = num__23 years answer is a <eor> a <eos> |
a |
add__26.0__3.0__ add__26.0__29.0__ subtract__11.0__9.0__ subtract__55.0__9.0__ subtract__26.0__3.0__ subtract__26.0__3.0__ |
add__26.0__3.0__ add__26.0__29.0__ subtract__11.0__9.0__ subtract__55.0__9.0__ subtract__26.0__3.0__ subtract__26.0__3.0__ |
| a works twice as fast as b . if b can complete a work in num__12 days independently the number of days in which a and b can together finish the work in : <o> a ) num__8 <o> b ) num__9 <o> c ) num__4 <o> d ) num__18 <o> e ) none |
ratio of rates of working of a and b = num__2 : num__1 so ratio of times taken = num__1 : num__2 b ' s num__1 day ' s work = num__0.0833333333333 therefore a ' s num__1 day ' s work = num__0.166666666667 ( a + b ) ' s num__1 day ' s work = ( num__0.166666666667 + num__0.0833333333333 ) = num__0.25 so a and b together can finish the work in num__4 days . answer : c <eor> c <eos> |
c |
divide__1.0__12.0__ divide__2.0__12.0__ add__0.1667__0.0833__ divide__1.0__0.25__ round__4.0__ |
divide__1.0__12.0__ divide__2.0__12.0__ add__0.1667__0.0833__ divide__1.0__0.25__ round__4.0__ |
| the edge of a cube is num__2 a cm . find its surface ? <o> a ) num__24 a num__8 <o> b ) num__24 a num__4 <o> c ) num__24 a num__1 <o> d ) num__24 a num__2 <o> e ) num__24 a num__7 |
num__6 a num__2 = num__6 * num__2 a * num__2 a = num__24 a num__2 answer : d <eor> d <eos> |
d |
surface_cube__2.0__ surface_cube__2.0__ |
surface_cube__2.0__ surface_cube__2.0__ |
| if num__20.0 of a number is equal to one - third of another number what is the ratio of first number to the second number ? <o> a ) num__2 : num__5 <o> b ) num__1 : num__4 <o> c ) num__5 : num__3 <o> d ) num__6 : num__11 <o> e ) num__2 : num__3 |
let num__20.0 of a = num__0.333333333333 b then num__20 a / num__100 = num__1 b / num__3 a / num__5 = b / num__3 a / b = num__1.66666666667 a : b = num__5 : num__3 answer is c <eor> c <eos> |
c |
divide__100.0__20.0__ divide__5.0__3.0__ multiply__1.0__5.0__ |
divide__100.0__20.0__ divide__5.0__3.0__ divide__100.0__20.0__ |
| find the simple interest on rs . num__580 for num__11 months at num__9 paisa per month ? <o> a ) num__574 <o> b ) num__270 <o> c ) num__566 <o> d ) num__266 <o> e ) num__121 |
i = ( num__580 * num__11 * num__9 ) / num__100 = num__574 answer : a <eor> a <eos> |
a |
percent__100.0__574.0__ |
percent__100.0__574.0__ |
| a car covers a distance of num__624 km in num__6 ½ hours . find its speed ? <o> a ) num__104 kmph <o> b ) num__194 kmph <o> c ) num__109 kmph <o> d ) num__174 kmph <o> e ) num__101 kmph |
num__104.0 = num__104 kmph answer : a <eor> a <eos> |
a |
divide__624.0__6.0__ round__104.0__ |
divide__624.0__6.0__ round__104.0__ |
| if x y and z are positive integers and num__2 x = num__4 y = num__5 z then the least possible value of x + y + z is <o> a ) num__40 <o> b ) num__50 <o> c ) num__55 <o> d ) num__60 <o> e ) num__19 |
given num__2 x = num__4 y = num__5 z x + y + z in terms of x = x + ( num__2 x / num__4 ) + ( num__2 x / num__5 ) = num__38 x / num__20 = num__19 x / num__10 now checking with each of the answers and see which value gives a minimum integer value . a x = num__0.526315789474 * num__40 not an integer b c e can be ruled out similarly . d is minimum value as x = num__19 * num__0.526315789474 = num__10 answer is e <eor> e <eos> |
e |
multiply__4.0__5.0__ divide__38.0__2.0__ multiply__2.0__5.0__ divide__10.0__19.0__ add__2.0__38.0__ divide__38.0__2.0__ |
multiply__4.0__5.0__ divide__38.0__2.0__ multiply__2.0__5.0__ divide__10.0__19.0__ add__2.0__38.0__ divide__38.0__2.0__ |
| num__10 ^ ( num__65 ) Ã · num__10 ^ ( num__64 ) = ? <o> a ) num__1 <o> b ) num__10 <o> c ) num__100 <o> d ) num__1000 <o> e ) num__10000 |
num__10 ^ ( num__65 ) Ã · num__10 ^ ( num__64 ) = num__10 ^ ( num__65 - num__64 ) = num__10 ^ num__1 = num__10 answer : b <eor> b <eos> |
b |
subtract__65.0__64.0__ multiply__10.0__1.0__ |
subtract__65.0__64.0__ multiply__10.0__1.0__ |
| a and b invest num__3000 and num__4000 in a business . a receives num__10 per month out of the profit as a remuneration for running the business and the rest of profit is divided in proportion to the investments . if in a year ‘ a ’ totally receives num__390 what does b receive ? <o> a ) num__375 <o> b ) num__360 <o> c ) num__350 <o> d ) num__260 <o> e ) none of these |
in a year for a total amount as a remuneration = num__10 × num__12 = num__120 ∴ amount of a ’ s profit = num__390 – num__120 = num__270 ratio of investment = num__3 : num__4 let total profit = x then b ’ s profit = ( x – num__270 ) ∴ num__1.0 + num__4 × x = num__270 ⇒ x = num__630 ∴ b ’ s profit = num__630 – num__270 = num__360 answer b <eor> b <eos> |
b |
multiply__10.0__12.0__ subtract__390.0__120.0__ divide__12.0__3.0__ subtract__4.0__3.0__ multiply__3.0__120.0__ multiply__1.0__360.0__ |
multiply__10.0__12.0__ subtract__390.0__120.0__ divide__12.0__3.0__ subtract__4.0__3.0__ multiply__3.0__120.0__ multiply__1.0__360.0__ |
| find the rate at simple interest at which a sum becomes four times of itself in num__15 years . <o> a ) num__10.0 <o> b ) num__20.0 <o> c ) num__30.0 <o> d ) num__40.0 <o> e ) num__50 % |
explanation : let sum be x and rate be r % then ( x * r * num__15 ) / num__100 = num__3 x [ important to note here is that simple interest will be num__3 x not num__4 x beause num__3 x + x = num__4 x ] = > r = num__20.0 option b <eor> b <eos> |
b |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| $ num__3650 is divided among num__4 engineers num__3 mbas and num__5 cas such that num__3 cas get as much as num__2 mbas and num__3 engs as much as num__2 cas . find the share of an mba . <o> a ) num__400 <o> b ) num__150 <o> c ) num__250 <o> d ) num__450 <o> e ) num__350 |
num__4 e + num__3 m + num__5 c = num__3650 num__3 c = num__2 m that is m = num__1.5 c num__3 e = num__2 c that is e = . num__66 c then ( num__4 * num__0.66 c ) + ( num__3 * num__1.5 c ) + num__5 c = num__3650 c = num__3650 / num__12.166 c = num__300 m = num__1.5 c = num__450 . answer : d num__450 <eor> d <eos> |
d |
divide__3.0__2.0__ multiply__1.5__300.0__ multiply__1.5__300.0__ |
divide__3.0__2.0__ multiply__1.5__300.0__ multiply__1.5__300.0__ |
| the present population of a town is num__4032 . population increase rate is num__20.0 p . a . find the population of town before num__2 years ? <o> a ) num__2500 <o> b ) num__2800 <o> c ) num__3500 <o> d ) num__3600 <o> e ) num__2050 |
p = num__4032 r = num__20.0 required population of town = p / ( num__1 + r / num__100 ) ^ t = num__4032 / ( num__1 + num__0.2 ) ^ num__2 = num__4032 / ( num__1.2 ) ^ num__2 = num__2800 ( approximately ) answer is b <eor> b <eos> |
b |
percent__20.0__1.0__ percent__100.0__2800.0__ |
percent__20.0__1.0__ percent__100.0__2800.0__ |
| if x is the sum of first num__50 positive even integers and y is the sum of first num__50 positive odd integers what is the value of x - y ? <o> a ) num__0 <o> b ) num__25 <o> c ) num__50 <o> d ) num__75 <o> e ) num__100 |
first even minus first odd = num__2 - num__1 = num__1 ; the sum of first num__2 even integers minus the sum of first num__2 odd integers = ( num__2 + num__4 ) - ( num__1 + num__3 ) = num__2 ; the sum of first num__3 even integers minus the sum of first num__3 odd integers = ( num__2 + num__4 + num__6 ) - ( num__1 + num__3 + num__5 ) = num__3 ; . . . we can see the patterns here so the sum of first num__50 positive even integers minus the sum of first num__50 positive odd integers will be num__50 . answer : c . <eor> c <eos> |
c |
add__1.0__2.0__ multiply__2.0__3.0__ add__1.0__4.0__ multiply__50.0__1.0__ |
add__1.0__2.0__ add__2.0__4.0__ add__1.0__4.0__ multiply__50.0__1.0__ |
| the difference between the simple interest received from two different sources on rs . num__1600 for num__3 years is rs . num__13.50 . the difference between their rates of interest is <o> a ) num__0.1 <o> b ) num__0.28 <o> c ) num__0.35 <o> d ) num__0.42 <o> e ) num__0.5 % |
( num__1600 xr num__1 x num__3 ) / num__100 - ( num__100 xr num__2 x num__3 ) / num__100 = num__13.50 num__4800 ( r num__1 - r num__2 ) = num__1350 r num__1 - r num__2 = num__0.28125 = num__0.28 answer : b <eor> b <eos> |
b |
percent__100.0__0.28__ |
percent__100.0__0.28__ |
| a train running at num__25 km / hr takes num__20 seconds to pass a platform . next it takes num__12 seconds to pass a man walking at num__5 km / hr in the same direction . find the length of the train and that of the platform . <o> a ) a ) num__66.6666666667 m num__58.3333333333 m <o> b ) b ) num__90 m num__70 m <o> c ) c ) num__50 m num__45 m <o> d ) d ) num__33.3333333333 m num__66.6666666667 <o> e ) e ) none of these |
hi math gurus correct me if i am wrong . i wondered i reached the answer mentioned here i . e num__100 m and num__25 m lets solve it to find out length of train the train which is must cover the static point of the man as well as the distance covered the man during num__12 sec . the man covered the distance during num__12 sec is num__5 * num__0.25 * num__12 = num__15 m so the train covered a distance in num__12 sec = num__25 * num__0.25 * num__12 = num__75 so the length of the train ( lt ) is num__15 + num__75 = num__90 m now crossing the platform means length of train + length of platform ( lt + p ) = num__20 * num__25 * num__0.25 = num__125 m now the length of platform ( lp ) is = num__125 - num__90 = num__35 m so answer is num__90 m and num__35 m must bebeven though its mentioned num__90 m and num__70 m thinking so typo mistake . . . . . . . . . . . <eor> b <eos> |
b |
multiply__20.0__5.0__ divide__25.0__100.0__ subtract__20.0__5.0__ multiply__5.0__15.0__ add__75.0__15.0__ multiply__25.0__5.0__ add__20.0__15.0__ subtract__75.0__5.0__ add__20.0__70.0__ |
multiply__20.0__5.0__ divide__25.0__100.0__ subtract__20.0__5.0__ multiply__5.0__15.0__ add__75.0__15.0__ multiply__25.0__5.0__ add__20.0__15.0__ subtract__75.0__5.0__ add__20.0__70.0__ |
| num__3 x ^ num__2 + num__5 x - num__2 = <o> a ) ( num__3 x - num__1 ) ( x + num__2 ) <o> b ) ( num__3 x - num__4 ) ( x + num__2 ) <o> c ) ( num__3 x - num__4 ) ( x - num__2 ) <o> d ) ( num__3 x - num__2 ) ( x + num__4 ) <o> e ) none of above |
num__3 x ^ num__2 + num__5 x - num__2 = num__3 x ^ num__2 + num__6 x - x - num__2 = num__3 x ( x + num__2 ) - ( x + num__2 ) = ( num__3 x - num__1 ) ( x + num__2 ) option a <eor> a <eos> |
a |
multiply__3.0__2.0__ subtract__3.0__2.0__ multiply__3.0__1.0__ |
multiply__3.0__2.0__ subtract__3.0__2.0__ add__2.0__1.0__ |
| the g . c . d . of num__1.08 num__0.36 and num__0.9 is : <o> a ) num__0.03 <o> b ) num__0.9 <o> c ) g . c . d . = num__0.18 <o> d ) num__0.2 <o> e ) num__0.21 |
given numbers are num__1.08 num__0.36 and num__0.90 . h . c . f . of num__108 num__36 and num__90 is num__18 h . c . f . of given numbers = num__0.18 . answer : option c <eor> c <eos> |
c |
gcd__36.0__90.0__ subtract__1.08__0.9__ subtract__1.08__0.9__ |
gcd__36.0__90.0__ subtract__1.08__0.9__ subtract__1.08__0.9__ |
| in a renowned city the average birth rate is num__7 people every two seconds and the death rate is num__3 people every two seconds . estimate the size of the population net increase that occurs in one day . <o> a ) num__172700 <o> b ) num__172800 <o> c ) num__172900 <o> d ) num__173000 <o> e ) num__173 |
100 |
this question can be modified so that the birth rate is given every m seconds and the death rate is given every n seconds . for this particular question : increase in the population every num__2 seconds = num__7 - num__3 = num__4 people . total num__2 second interval in a day = num__24 * num__60 * num__30.0 = num__43200 population increase = num__43200 * num__4 = num__172800 . hence b . <eor> b <eos> |
b |
b |
| mr yadav spends num__60.0 of his monthly salary on consumable items and num__50.0 of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were num__19008 how much amount per month would he have spent on clothes and transport ? <o> a ) num__4038 <o> b ) num__8076 <o> c ) num__9691.2 <o> d ) num__1584 <o> e ) num__1625 |
∵ amount he have spent in num__1 month on clothes transport = amount spent on saving per month ∵ amount spent on clothes and transport = num__19008 ⁄ num__12 = num__1584 answer d <eor> d <eos> |
d |
divide__19008.0__12.0__ divide__19008.0__12.0__ |
divide__19008.0__12.0__ divide__19008.0__12.0__ |
| a company produces num__72000 bottles of water everyday . if a case can hold num__9 bottles of water . how many cases are required by the company to hold its one day production <o> a ) num__2000 <o> b ) num__4500 <o> c ) num__5000 <o> d ) num__8000 <o> e ) num__9000 |
number of bottles that can be held in a case = num__9 . number of cases required to hold num__72000 bottles = num__8000.0 = num__8000 cases . so the answer is d = num__8000 <eor> d <eos> |
d |
divide__72000.0__9.0__ round__8000.0__ |
divide__72000.0__9.0__ round__8000.0__ |
| the expression num__10 ^ num__14 − num__130 is divisible by all of the following integers except <o> a ) num__9 <o> b ) num__2 <o> c ) num__4 <o> d ) num__8 <o> e ) num__11 |
yeah . i agree this question has two answers which are ( b ) and ( e ) . if num__10 is power to odd number and equal or more than num__5 then the answer choice ( a ) is correct . <eor> a <eos> |
a |
subtract__14.0__5.0__ |
subtract__14.0__5.0__ |
| if num__3 pounds of dried apricots that cost x dollars per pound are mixed with num__2 pounds of prunes that cost y dollars per pound what is the cost in dollars per pound of the mixture ? <o> a ) ( num__3 x + num__2 y ) / num__5 <o> b ) ( num__3 x + num__2 y ) / ( x + y ) <o> c ) ( num__3 x + num__2 y ) / ( xy ) <o> d ) num__5 ( num__3 x + num__2 y ) <o> e ) num__3 x + num__2 y |
total cost = weight ( in pounds ) * price / pound ; to find total cost / pound divide by total pounds . cost of dried apricots = num__3 x ; cost of prunes = num__2 y ; cost per pound = ( num__3 x + num__2 y ) / num__5 ; ans is ( a ) . <eor> a <eos> |
a |
add__3.0__2.0__ subtract__5.0__2.0__ |
add__3.0__2.0__ subtract__5.0__2.0__ |
| a garrison of num__400 men had a provision for num__31 days . after num__20 days num__180 persons re - enforcement leave the garrison . find the number of days for which the remaining ration will be sufficient ? <o> a ) num__65 days <o> b ) num__45 days <o> c ) num__20 days <o> d ) num__16 days <o> e ) num__18 days |
num__400 - - - num__31 num__400 - - - num__11 num__220 - - - ? num__400 * num__11 = num__220 * x = > x = num__20 days . answer : c <eor> c <eos> |
c |
subtract__31.0__20.0__ subtract__400.0__180.0__ divide__400.0__20.0__ |
subtract__31.0__20.0__ subtract__400.0__180.0__ subtract__31.0__11.0__ |
| the average length of the sides of triangle abc is num__12 . what is the perimeter of triangle abc ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__12 <o> d ) num__24 <o> e ) num__36 |
( average ) = ( perimeter ) / num__3 ; num__12 = ( perimeter ) / num__3 ; ( perimeter ) = num__36 . answer : e . <eor> e <eos> |
e |
multiply__12.0__3.0__ multiply__12.0__3.0__ |
multiply__12.0__3.0__ multiply__12.0__3.0__ |
| the speed of a boat in still water is num__60 kmph and the speed of the current is num__20 kmph . find the speed downstream and upstream ? <o> a ) num__15 kmph <o> b ) num__40 kmph <o> c ) num__18 kmph <o> d ) num__19 kmph <o> e ) num__10 kmph |
speed downstream = num__60 + num__20 = num__80 kmph speed upstream = num__60 - num__20 = num__40 kmph answer : b <eor> b <eos> |
b |
add__60.0__20.0__ subtract__60.0__20.0__ round__40.0__ |
add__60.0__20.0__ subtract__60.0__20.0__ subtract__60.0__20.0__ |
| num__5 men are equal to as many women as are equal to num__8 boys . all of them earn rs . num__90 only . men ’ s wages are ? <o> a ) num__6 rs <o> b ) num__7 rs <o> c ) num__8 rs <o> d ) num__4 rs <o> e ) num__1 rs |
num__5 m = xw = num__8 b num__5 m + xw + num__8 b - - - - - num__90 rs . num__5 m + num__5 m + num__5 m - - - - - num__90 rs . num__15 m - - - - - - num__90 rs . = > num__1 m = num__6 rs . answer : a <eor> a <eos> |
a |
add__5.0__1.0__ add__5.0__1.0__ |
add__5.0__1.0__ add__5.0__1.0__ |
| roja and pooja start moving in the opposite directions from a pole . they are moving at the speeds of num__2 km / hr and num__3 km / hr respectively . after num__4 hours what will be the distance between them ? <o> a ) num__21 <o> b ) num__20 <o> c ) num__99 <o> d ) num__277 <o> e ) num__12 |
distance = relative speed * time = ( num__2 + num__3 ) * num__4 = num__20 km [ they are travelling in the opposite direction relative speed = sum of the speeds ] . answer : b <eor> b <eos> |
b |
round__20.0__ |
round__20.0__ |
| a circular mat with diameter num__1.4 inches is placed on a square tabletop each of whose sides is num__4 inches long . which of the following is closest to the fraction of the tabletop covered by the mat ? <o> a ) num__0.375 <o> b ) num__0.875 <o> c ) num__0.2 <o> d ) num__0.428571428571 <o> e ) num__0.777777777778 |
so we are looking for the area of the cloth over the area of the table area of the cloth = ( pi ) ( r ) ^ num__2 which is about ( num__3.14285714286 ) ( num__1.4 ) ( num__1.4 ) area of the table = ( num__4 ) ( num__4 ) so the quick way to estimate is looking at the fraction like this : num__6.16 / num__16 nearest to num__0.375 answer : a <eor> a <eos> |
a |
square_perimeter__4.0__ triangle_area__0.375__2.0__ |
power__4.0__2.0__ triangle_area__0.375__2.0__ |
| the shopkeeper increased the price of a product by num__25.0 so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only num__70.0 of the required amount . what is the net difference in the expenditure on that product ? <o> a ) num__12.5 <o> b ) num__13.0 <o> c ) num__14.5 <o> d ) num__15.0 <o> e ) num__15.7 % |
quantity x rate = price num__1 x num__1 = num__1 num__0.7 x num__1.25 = num__0.875 decrease in price = ( num__0.125 / num__1 ) × num__100 = num__12.5 a ) <eor> a <eos> |
a |
multiply__1.25__0.7__ subtract__1.0__0.875__ divide__70.0__0.7__ multiply__100.0__0.125__ subtract__25.0__12.5__ |
multiply__1.25__0.7__ subtract__1.0__0.875__ divide__70.0__0.7__ multiply__100.0__0.125__ divide__12.5__1.0__ |
| a car covers a distance of num__624 km in num__6 ½ hours . find its speed ? <o> a ) num__104 <o> b ) num__7778 <o> c ) num__266 <o> d ) num__288 <o> e ) num__121 |
num__104.0 = num__104 kmph answer : a <eor> a <eos> |
a |
divide__624.0__6.0__ round__104.0__ |
divide__624.0__6.0__ round__104.0__ |
| on thursday mabel handled num__90 transactions . anthony handled num__10.0 more transactions than mabel cal handled num__0.666666666667 rds of the transactions that anthony handled and jade handled num__16 more transactions than cal . how much transactions did jade handled ? <o> a ) num__33 <o> b ) num__82 <o> c ) num__67 <o> d ) num__28 <o> e ) num__18 |
mabel handled num__90 transactions anthony handled num__10.0 more transactions than mabel anthony = num__90 + num__90 × num__10.0 = num__90 + num__90 × num__0.10 = num__90 + num__9 = num__99 cal handled num__0.666666666667 rds of the transactions than anthony handled cal = num__0.666666666667 × num__99 = num__66 jade handled num__16 more transactions than cal . jade = num__66 + num__16 = num__82 jade handled = num__82 transactions . answer : b <eor> b <eos> |
b |
reverse__10.0__ divide__90.0__10.0__ add__90.0__9.0__ add__16.0__66.0__ add__16.0__66.0__ |
reverse__10.0__ divide__90.0__10.0__ add__90.0__9.0__ add__16.0__66.0__ add__16.0__66.0__ |
| the cost price of num__30 items is equal to the sale price of num__40 items . what is percentage profit ? <o> a ) num__20.0 loss <o> b ) num__25.0 loss <o> c ) num__15.0 loss <o> d ) num__35.0 loss <o> e ) num__5.0 loss |
this results in loss percentage not profit . let cp of num__1 item = num__1 rp . sp of num__40 items = cp of num__30 items = num__30 cp of num__40 items = num__40 sp of num__40 items - num__30 loss = cp - sp = num__10 loss percentage = ( num__0.25 * num__100 ) = num__25.0 answer : b <eor> b <eos> |
b |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| when x is even [ x ] = x / num__2 + num__1 when x is odd [ x ] = num__2 x + num__1 then [ num__5 ] * [ num__4 ] = ? <o> a ) [ num__33 ] <o> b ) [ num__44 ] <o> c ) [ num__45 ] <o> d ) [ num__88 ] <o> e ) [ num__90 ] |
[ num__5 ] * [ num__4 ] = ( num__2 * num__5 + num__1 ) ( num__2.0 + num__1 ) = [ num__33 ] . ans - a <eor> a <eos> |
a |
multiply__1.0__33.0__ |
multiply__1.0__33.0__ |
| what percent of a day is num__8 hours ? <o> a ) num__12 Ã — num__0.5 % <o> b ) num__16 Ã — num__0.5 % <o> c ) num__33 Ã — num__0.333333333333 % <o> d ) num__22 Ã — num__0.5 % <o> e ) none |
solution required percentage = ( num__0.333333333333 Ã — num__100 ) % = ( num__33.3333333333 ) % . = num__33 Ã — num__0.333333333333 % . answer c <eor> c <eos> |
c |
subtract__33.3333__0.3333__ round__33.0__ |
subtract__33.3333__0.3333__ round__33.0__ |
| two trains of length num__100 m and num__200 m are num__100 m apart . they start moving towards each other on parallel tracks at speeds num__54 kmph and num__72 kmph . after how much time will the trains meet ? <o> a ) num__3.85714285714 sec <o> b ) num__2.85714285714 sec <o> c ) num__4.71428571429 sec <o> d ) num__3.0 sec <o> e ) num__4.57142857143 sec |
b num__2.85714285714 sec they are moving in opposite directions relative speed is equal to the sum of their speeds . relative speed = ( num__54 + num__72 ) * num__0.277777777778 = num__7 * num__5 = num__35 mps . the time required = d / s = num__2.85714285714 = num__2.85714285714 sec . <eor> b <eos> |
b |
multiply__5.0__7.0__ divide__100.0__35.0__ |
multiply__5.0__7.0__ divide__100.0__35.0__ |
| num__121 |
112 . . . ? . . num__97 num__9186 <o> a ) num__99 <o> b ) num__104 <o> c ) num__102 <o> d ) num__108 <o> e ) num__100 |
num__121 . num__112 . num__104 . num__97 . num__91 . num__86 the decreasing difference is reducing by num__1 at each step so missing no . will be num__104 answer b <eor> b <eos> |
b |
b |
| convert the num__0.361111111111 m / s into kilometers per hour ? <o> a ) num__2.9 kmph <o> b ) num__9.9 kmph <o> c ) num__1.3 kmph <o> d ) num__1.2 kmph <o> e ) num__5.7 kmph |
num__0.361111111111 m / s = num__0.361111111111 * num__3.6 = num__1.3 = num__1.3 kmph . answer : c <eor> c <eos> |
c |
multiply__0.3611__3.6__ round__1.3__ |
multiply__0.3611__3.6__ multiply__0.3611__3.6__ |
| three candidates contested an election and received num__1136 num__7636 and num__11628 votes respectively . what percentage of the total votes did the winning candidate got <o> a ) num__55.0 <o> b ) num__56.0 <o> c ) num__57.0 <o> d ) num__58.0 <o> e ) num__59 % |
explanation : total number of votes polled = ( num__1136 + num__7636 + num__11628 ) = num__20400 so required percentage = num__0.57 * num__100 = num__57.0 answer is c <eor> c <eos> |
c |
divide__11628.0__20400.0__ multiply__100.0__0.57__ multiply__100.0__0.57__ |
divide__11628.0__20400.0__ multiply__100.0__0.57__ multiply__100.0__0.57__ |
| a num__4 c b num__9 d c num__16 e d num__25 f e num__36 g ? <o> a ) f num__49 g <o> b ) h num__47 j <o> c ) f num__49 h <o> d ) h num__64 i <o> e ) g num__49 i |
a _ c = num__2 ^ num__2 b _ d = num__3 ^ num__2 c _ e = num__4 ^ num__2 d _ f = num__5 ^ num__2 e _ g = num__6 ^ num__2 num__2 ^ num__2 num__3 ^ num__2 num__4 ^ num__2 num__5 ^ num__2 num__6 ^ num__2 - - - > as per this number sequence = > next number = num__7 ^ num__2 = num__49 f _ h = num__49 = > f num__49 h answer : c <eor> c <eos> |
c |
subtract__9.0__4.0__ add__4.0__2.0__ add__4.0__3.0__ power__7.0__2.0__ power__7.0__2.0__ |
subtract__9.0__4.0__ subtract__9.0__3.0__ subtract__9.0__2.0__ power__7.0__2.0__ power__7.0__2.0__ |
| a car crosses a num__600 m long bridge in num__5 minutes . what is the speed of car in km per hr ? <o> a ) num__6 km / hr <o> b ) num__7.2 km / hr <o> c ) num__7.56 km / hr <o> d ) num__7.78 km / hr <o> e ) num__8 km / hr |
speed = num__600 m / sec . num__5 x num__60 = num__2 m / sec . converting m / sec to km / hr ( see important formulas section ) = num__2 x num__18 km / hr num__5 = num__7.2 km / hr b <eor> b <eos> |
b |
hour_to_min_conversion__ round__7.2__ |
hour_to_min_conversion__ round__7.2__ |
| if the price of rice rises from rs . num__6 per kg to rs . num__8 per kg to have no increase in his expenditure on rice a person will have to reduce his consumption of sugar by <o> a ) num__12.0 <o> b ) num__19.0 <o> c ) num__25.0 <o> d ) num__32.0 <o> e ) num__36 % |
explanation : let original consumption = num__100 kg and consumption = y kg . so num__100 x num__6 = y x num__8 y = num__75 kg reduction in consumption = num__25.0 . answer : c <eor> c <eos> |
c |
subtract__100.0__75.0__ subtract__100.0__75.0__ |
subtract__100.0__75.0__ subtract__100.0__75.0__ |
| a person has to cover a distance of num__6 km in num__45 minutes . if he covers one - half of the distance in two - thirds of the total time ; to cover the remaining distance in the remaining time his speed ? . <o> a ) num__6 kilometer / hour <o> b ) num__8 kilometer / hour <o> c ) num__12 kilometer / hour <o> d ) num__15 kilometer / hour <o> e ) none of these |
remaining distance = num__3 km and remaining time = ( num__0.333333333333 * num__45 ) minutes = num__15 minutes = num__0.25 hour . required speed = ( num__3 * num__4 ) kilometer / hour = num__12 km / hour . correct option : c <eor> c <eos> |
c |
divide__45.0__3.0__ divide__3.0__0.25__ round__12.0__ |
divide__45.0__3.0__ divide__3.0__0.25__ divide__3.0__0.25__ |
| at the end of the first quarter the share price of a certain mutual fund was num__30 percent higher than it was at the beginning of the year . at the end of the second quarter the share price was num__50 percent higher than it was at the beginning of the year . what was the percent increase in the share price from the end of the first quarter to the end of the second quarter ? <o> a ) num__15.3 <o> b ) num__25.0 <o> c ) num__30.0 <o> d ) num__33.0 <o> e ) num__40 % |
another method is to use the formula for num__2 successive percentage changes : total = a + b + ab / num__100 num__50 = num__30 + b + num__30 b / num__100 b = num__15.3 answer ( a ) <eor> a <eos> |
a |
percent__15.3__100.0__ |
percent__15.3__100.0__ |
| a sum of money deposited at c . i . amounts to rs . num__2420 in num__2 years and to rs . num__2662 in num__3 years . find the rate percent ? <o> a ) num__11 <o> b ) num__10 <o> c ) num__99 <o> d ) num__88 <o> e ) num__12 |
num__2420 - - - num__242 num__100 - - - ? = > num__10.0 answer : b <eor> b <eos> |
b |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| two goods trains each num__500 m long are running in opposite directions on parallel tracks . their speeds are num__42 km / hr and num__30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? <o> a ) num__22 <o> b ) num__27 <o> c ) num__50 <o> d ) num__99 <o> e ) num__12 |
relative speed = num__42 + num__30 = num__72 km / hr . num__72 * num__0.277777777778 = num__20 m / sec . distance covered = num__500 + num__500 = num__1000 m . required time = num__50.0 = num__50 sec . answer : c <eor> c <eos> |
c |
add__42.0__30.0__ add__30.0__20.0__ round__50.0__ |
add__42.0__30.0__ add__30.0__20.0__ add__30.0__20.0__ |
| if c is a positive integer then num__3 ^ c + num__3 ^ ( c + num__1 ) = <o> a ) num__4 ^ a <o> b ) num__3 ^ a − num__1 <o> c ) num__3 ^ ( num__2 a ) + num__1 <o> d ) num__3 ^ a * ( a − num__1 ) <o> e ) num__4 ( num__3 ^ c ) |
num__3 ^ c + num__3 ^ ( c + num__1 ) = num__3 ^ c + ( num__3 ^ c * num__3 ^ num__1 ) = num__3 ^ c ( num__1 + num__3 ) = num__3 ^ c ( num__4 ) ans . e ) num__4 ( num__3 ^ c ) <eor> e <eos> |
e |
add__3.0__1.0__ add__3.0__1.0__ |
add__3.0__1.0__ add__3.0__1.0__ |
| every student in a certain classroom was given one colored marble . num__0.5 of the students were given a blue marble num__0.333333333333 of the students were given a yellow marble and all of the other students were given a red marble . after lunch num__0.333333333333 of the students who had blue marbles and num__0.25 of the students who had yellow marbles left the classroom ; the other students did not leave . what fraction of all students have either a blue or yellow marble now ? <o> a ) num__0.777777777778 <o> b ) num__0.857142857143 <o> c ) num__0.714285714286 <o> d ) num__0.571428571429 <o> e ) num__0.444444444444 |
lets pick smart numbers . total number of students in the classroom : num__12 blue marble ( num__0.5 ) : num__6 yellow marble ( num__0.333333333333 ) : num__4 red marble ( num__0.166666666667 ) : num__2 after lunch blue marble : num__4 yellow marble : num__3 red marble : num__2 new total number of students in the classroom : num__9 blue and yellow marbles : num__7 answer num__0.777777777778 or a <eor> a <eos> |
a |
multiply__0.5__12.0__ reverse__0.25__ reverse__6.0__ reverse__0.5__ multiply__0.5__6.0__ add__3.0__6.0__ add__3.0__4.0__ divide__7.0__9.0__ divide__7.0__9.0__ |
multiply__0.5__12.0__ reverse__0.25__ reverse__6.0__ reverse__0.5__ multiply__0.5__6.0__ add__3.0__6.0__ add__3.0__4.0__ divide__7.0__9.0__ divide__7.0__9.0__ |
| joe invested a certain sum of money in a simple interest bond whose value grew to $ num__260 at the end of num__3 years and to $ num__360 at the end of another num__5 years . what was the rate of interest in which he invested his sum ? <o> a ) num__6.0 <o> b ) num__8.0 <o> c ) num__10.0 <o> d ) num__12.0 <o> e ) num__15 % |
in num__5 years the value grew $ num__100 so the simple interest was $ num__20 per year . in num__3 years the total interest was num__3 * $ num__20 = $ num__60 the principal is $ num__260 - $ num__60 = num__200 . the interest rate is $ num__20 / $ num__200 = num__10.0 the answer is c . <eor> c <eos> |
c |
percent__5.0__200.0__ percent__5.0__200.0__ |
percent__5.0__200.0__ percent__5.0__200.0__ |
| how many multiples of num__6 are less than num__5000 and also multiples of num__8 ? <o> a ) num__104 <o> b ) num__208 <o> c ) num__625 <o> d ) num__832 <o> e ) num__833 |
lcm of num__6 & num__8 = num__24 tried dividing num__5000 by num__24 got quotient num__208.33 ' so b is answer <eor> b <eos> |
b |
round_down__208.33__ |
round_down__208.33__ |
| a train num__500 meters long completely crosses a num__300 meters long bridge in num__45 seconds . what is the speed of the train is ? <o> a ) num__32 <o> b ) num__28 <o> c ) num__49 <o> d ) num__64 <o> e ) num__21 |
s = ( num__500 + num__300 ) / num__45 = num__17.7777777778 * num__3.6 = num__64 answer : d <eor> d <eos> |
d |
round__64.0__ |
round__64.0__ |
| a gardener grows cabbages in her garden that is in the shape of a square . each cabbage takes num__1 square feet of area in her garden . this year she has increased her output by num__193 cabbages as compared to last year . the shape of the area used for growing the cabbages has remained a square in both these years . how many cabbages did she produce this year ? <o> a ) num__8208 <o> b ) num__9409 <o> c ) num__11424 <o> d ) num__12586 <o> e ) can not be determined |
let the side for growing cabbages this year be x ft . thus the area is x ^ num__2 . let the side for growing cabbages last year be y ft . thus the area was y ^ num__2 . the area would have increased by num__193 sq ft as each cabbage takes num__1 sq ft space . x ^ num__2 - y ^ num__2 = num__193 ( x + y ) ( x - y ) = num__193 num__193 is a prime number and thus it will be ( num__97 + num__96 ) * ( num__97 - num__96 ) . thus x = num__97 and y = num__96 x ^ num__2 = num__97 ^ num__2 = num__9409 the answer is b . <eor> b <eos> |
b |
power__97.0__2.0__ multiply__1.0__9409.0__ |
power__97.0__2.0__ multiply__1.0__9409.0__ |
| there are three rooms in a motel : one single one double and one for four persons . how many ways are there to house seven persons in these rooms ? <o> a ) num__7 ! / num__1 ! num__2 ! num__3 ! <o> b ) num__7 ! <o> c ) num__7 ! / num__3 <o> d ) num__7 ! / num__3 ! <o> e ) num__7 ! / num__4 |
choose num__1 person for the single room & from the remaining choose num__2 for the double room & from the remaining choose num__4 people for the four person room then num__7 c num__1 x num__6 c num__2 x num__4 c num__4 = num__7 ! / num__1 ! num__2 ! num__3 ! answer : a <eor> a <eos> |
a |
coin_space__ die_space__ choose__7.0__6.0__ |
coin_space__ die_space__ choose__7.0__6.0__ |
| the principal that amounts to rs . num__3913 in num__3 years at num__6 num__0.25 % per annum c . i . compounded annually is ? <o> a ) s . num__3096 <o> b ) s . num__4076 <o> c ) s . num__4085 <o> d ) s . num__4096 <o> e ) s . num__5096 |
principal = [ num__4913 / ( num__1 + num__25 / ( num__4 * num__100 ) ) num__3 ] = num__3913 * num__0.941176470588 * num__0.941176470588 * num__0.941176470588 = rs . num__3096 . answer : a <eor> a <eos> |
a |
percent__100.0__3096.0__ |
percent__100.0__3096.0__ |
| a bat is bought for rs . num__440 and sold at a gain of num__20.0 find its selling price <o> a ) s . num__430 / - <o> b ) s . num__480 / - <o> c ) s . num__400 / - <o> d ) s . num__528 / - <o> e ) s . num__600 / - |
num__100.0 - - - - - - > num__440 num__120.0 - - - - - - > num__440 * num__1.2 selling price = rs . num__528 / - d <eor> d <eos> |
d |
percent__100.0__528.0__ |
percent__100.0__528.0__ |
| a man has some hens and cows . if the number of heads be num__48 and the number of feet equals num__140 then the number of hens will be : <o> a ) num__22 <o> b ) num__23 <o> c ) num__24 <o> d ) num__26 <o> e ) num__28 |
let hens be x and cows be y now feet : x * num__2 + y * num__4 = num__140 heads : x * num__1 + y * num__1 = num__48 implies num__2 x + num__4 y = num__140 and x + y = num__48 solving these two equations we get x = num__26 and y = num__22 therefore hens are num__26 . answer : d <eor> d <eos> |
d |
subtract__48.0__26.0__ subtract__48.0__22.0__ |
subtract__48.0__26.0__ multiply__1.0__26.0__ |
| one hour after yolanda started walking from x to y a distance of num__17 miles bob started walking along the same road from y to x . if yolanda ' s walking rate was num__3 miles per hour and bob т ' s was num__4 miles per hour how many miles had bob walked when they met ? <o> a ) num__24 <o> b ) num__23 <o> c ) num__22 <o> d ) num__21 <o> e ) num__8 |
when b started walking y already has covered num__3 miles out of num__17 hence the distance at that time between them was num__17 - num__3 = num__14 miles . combined rate of b and y was num__3 + num__4 = num__7 miles per hour hence they would meet each other in num__2.0 = num__2 hours . in num__6 hours b walked num__2 * num__4 = num__8 miles . answer : e . <eor> e <eos> |
e |
subtract__17.0__3.0__ add__3.0__4.0__ divide__14.0__7.0__ multiply__3.0__2.0__ multiply__4.0__2.0__ round__8.0__ |
subtract__17.0__3.0__ add__3.0__4.0__ divide__14.0__7.0__ multiply__3.0__2.0__ multiply__4.0__2.0__ multiply__4.0__2.0__ |
| find the simple interest on rs . num__70000 at num__16 num__0.666666666667 % per year for num__9 months . <o> a ) num__7500 <o> b ) num__6500 <o> c ) num__8750 <o> d ) num__9500 <o> e ) none of them |
p = rs . num__70000 r = num__16.6666666667 % p . a and t = num__0.75 years = num__0.75 years . simple interest = ( p * r * t ) / num__100 = rs . ( num__70000 * ( num__16.6666666667 ) * ( num__0.75 ) * ( num__0.01 ) ) = rs . num__8750 answer is c . <eor> c <eos> |
c |
percent__100.0__8750.0__ |
percent__100.0__8750.0__ |
| each of the num__26 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs . the three clubs to choose from are the poetry club the history club and the writing club . a total of num__11 students sign up for the poetry club num__17 students for the history club and num__15 students for the writing club . if num__5 students sign up for exactly two clubs how many students sign up for all three clubs ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
the total number in the three clubs is num__11 + num__17 + num__15 = num__43 . all num__26 students signed up for at least one club . num__5 of those students signed up for exactly one more club . num__43 - num__31 = num__12 so num__6 students must have signed up for exactly three clubs . the answer is e . <eor> e <eos> |
e |
add__26.0__17.0__ add__26.0__5.0__ subtract__17.0__5.0__ subtract__11.0__5.0__ subtract__11.0__5.0__ |
add__26.0__17.0__ add__26.0__5.0__ subtract__17.0__5.0__ subtract__11.0__5.0__ subtract__11.0__5.0__ |
| john completes a piece of work in num__10 days rose completes the same work in num__40 days . if both of them work together then the number of days required to complete the work is ? <o> a ) num__6 days <o> b ) num__8 days <o> c ) num__10 days <o> d ) num__12 days <o> e ) num__14 days |
if a can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days . that is the required no . of days = num__10 × num__0.8 = num__8 days b ) <eor> b <eos> |
b |
multiply__10.0__0.8__ round__8.0__ |
multiply__10.0__0.8__ round__8.0__ |
| a pump can fill a tank with water in num__2 hours . because of a leak it took num__2 num__0.333333333333 hours to fill the tank . the leak can drain all the water of the tank in ? <o> a ) num__17 <o> b ) num__19 <o> c ) num__18 <o> d ) num__14 <o> e ) num__11 |
work done by the tank in num__1 hour = ( num__0.5 - num__0.333333333333 ) = num__0.0714285714286 leak will empty the tank in num__14 hrs . answer : d <eor> d <eos> |
d |
divide__1.0__2.0__ round__14.0__ |
divide__1.0__2.0__ round__14.0__ |
| two trains one from howrah to patna and the other from patna to howrah start simultaneously . after they meet the trains reach their destinations after num__25 hours and num__16 hours respectively . the ratio of their speeds is ? <o> a ) num__4 : num__5 <o> b ) num__4 : num__3 <o> c ) num__4 : num__4 <o> d ) num__4 : num__9 <o> e ) num__4 : num__2 |
let us name the trains a and b . then ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ num__16 : √ num__25 = num__4 : num__5 answer : a <eor> a <eos> |
a |
round__4.0__ |
round__4.0__ |
| large medium and small ships are used to bring water . num__4 large ships carry as much water as num__7 small ships ; num__3 medium ships carry as much water as num__2 large ships and num__1 small ship . if num__15 large num__7 medium and num__14 small ships each made num__36 journeys to bring a certain quantity of water then in how many journeys will num__12 large num__14 medium and num__21 small ships bring the same quantity of water ? <o> a ) num__29 <o> b ) num__30 <o> c ) num__20 <o> d ) num__50 <o> e ) num__55 |
explanation : here large medium and small ships are denoted by the symbol l m and s . now according to the question num__4 l = num__7 s . - - - - - - ( i ) num__3 m = num__2 l + s . - - - - - - ( ii ) using above equations the ratios of the capacity of the large medium and small ships are : - num__7 : num__6 : num__4 . let the number of journeys required be x . since quantity of water remains the same so : - = ( ( num__15 × num__7 + num__7 × num__6 + num__14 × num__4 ) num__36 ) = x ( num__12 × num__7 + num__14 × num__6 + num__21 × num__4 ) . = > ( ( num__15 × num__7 + num__7 × num__6 + num__14 × num__4 ) num__36 ) / ( num__12 × num__7 + num__14 × num__6 + num__21 × num__4 ) = x = > x = num__29.0 = > x = num__29 . answer : a <eor> a <eos> |
a |
add__4.0__2.0__ add__15.0__14.0__ multiply__1.0__29.0__ |
add__4.0__2.0__ add__15.0__14.0__ add__15.0__14.0__ |
| bhavan travelled for num__20 hours . he covered the first half of the distance at num__50 kmph and remaining half of the distance at num__40 kmph . find the distance travelled by bhavan ? <o> a ) num__888.89 <o> b ) num__788.89 <o> c ) num__899.89 <o> d ) num__688.89 <o> e ) num__899.99 |
let the distance travelled be x km . total time = ( x / num__2 ) / num__50 + ( x / num__2 ) / num__40 = num__20 = > x / num__100 + x / num__80 = num__20 = > ( num__4 x + num__5 x ) / num__400 = num__20 = > x = num__888.89 km answer : a <eor> a <eos> |
a |
divide__40.0__20.0__ multiply__50.0__2.0__ multiply__40.0__2.0__ divide__80.0__20.0__ divide__20.0__4.0__ multiply__100.0__4.0__ round__888.89__ |
divide__40.0__20.0__ multiply__50.0__2.0__ multiply__40.0__2.0__ divide__80.0__20.0__ divide__20.0__4.0__ multiply__100.0__4.0__ round__888.89__ |
| mary purchased brand x pens for $ num__2 apiece and brand y for $ num__1 apiece . if elena purchased a total of num__12 of these pens for $ num__20.00 how many brand x pens did she purchase ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__5 <o> d ) num__4 <o> e ) num__6 |
num__2 x + y = num__20 - only one positive integers solutions x = num__8 and y = num__4 answer is a <eor> a <eos> |
a |
subtract__20.0__12.0__ subtract__12.0__8.0__ multiply__2.0__4.0__ |
subtract__20.0__12.0__ subtract__12.0__8.0__ subtract__12.0__4.0__ |
| of the families in city x in num__1994 num__35 percent owned a personal computer . the number of families in city x owning a computer in num__1998 was num__20 percent greater than it was in num__1994 and the total number of families in city x was num__4 percent greater in num__1998 than it was in num__1994 . what percent of the families in city x owned a personal computer in num__1998 ? <o> a ) num__50.12 <o> b ) num__40.52 <o> c ) num__40.56 <o> d ) num__40.38 <o> e ) num__40.14 % |
say a num__100 families existed in num__1994 then the number of families owning a computer in num__1994 - num__40 number of families owning computer in num__1998 = num__35 * num__1.2 = num__42 number of families in num__1998 = num__104 the percentage = num__0.403846153846 * num__100 = num__40.38 . option : d <eor> d <eos> |
d |
multiply__35.0__1.2__ add__4.0__100.0__ divide__42.0__104.0__ multiply__100.0__0.4038__ multiply__100.0__0.4038__ |
multiply__35.0__1.2__ add__4.0__100.0__ divide__42.0__104.0__ multiply__100.0__0.4038__ multiply__100.0__0.4038__ |
| a train overtakes two persons who are walking in the same direction to that of the train at num__2 kmph and num__4 kmph and passes them completely in num__9 and num__10 seconds respectively . what is the length of the train ? <o> a ) num__40 m <o> b ) num__10 m <o> c ) num__20 m <o> d ) num__30 m <o> e ) num__50 m |
let speed of the train be v kmph ( v - num__2 ) : ( v - num__4 ) = num__10 : num__9 num__9 v - num__18 = num__10 v - num__40 v = num__22 length of the train = ( num__22 - num__2 ) x num__0.277777777778 x num__9 = num__50 metre answer : e <eor> e <eos> |
e |
multiply__2.0__9.0__ multiply__4.0__10.0__ add__4.0__18.0__ add__10.0__40.0__ round__50.0__ |
multiply__2.0__9.0__ multiply__4.0__10.0__ subtract__40.0__18.0__ add__10.0__40.0__ round__50.0__ |
| six years ago the ratio of the ages of kunal and sagar was num__6 : num__5 . four years hence the ratio of their ages will be num__11 : num__10 . what is sagar ' s age at present ? <o> a ) num__18 years <o> b ) num__16 years <o> c ) num__22 years <o> d ) num__20 years <o> e ) none of these |
the age of kunal and sagar num__6 years ago num__6 x and num__5 x respectively . ( ( num__6 x + num__6 ) + num__4 ) / ( ( num__5 x + num__6 ) + num__4 ) = num__1.1 num__10 ( num__6 x + num__10 ) = num__11 ( num__5 x + num__10 ) num__5 x = num__10 . x = num__2 sagar ' s present age = ( num__5 x + num__6 ) = num__16 years correct answer ( b ) <eor> b <eos> |
b |
subtract__10.0__6.0__ divide__11.0__10.0__ subtract__6.0__4.0__ add__6.0__10.0__ add__6.0__10.0__ |
subtract__10.0__6.0__ divide__11.0__10.0__ divide__10.0__5.0__ add__6.0__10.0__ add__6.0__10.0__ |
| a ’ s speed is num__1.42857142857 times that of b . if a and b run a race what part of the length of the race should a give b as a head start so that the race ends in a dead heat ? <o> a ) num__0.0588235294118 <o> b ) num__0.176470588235 <o> c ) num__0.1 <o> d ) num__0.3 <o> e ) num__0.3 |
we have the ratio of a ’ s speed and b ’ s speed . this means we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers num__20 meters b covers num__14 meters in that time . so if the race is num__20 meters long when a reaches the finish line b would be num__6 meters behind him . if we want the race to end in a dead heat we want b to be at the finish line too at the same time . this means b should get a head start of num__6 meters so that he doesn ’ t need to cover that . in that case the time required by a ( to cover num__20 meters ) would be the same as the time required by b ( to cover num__14 meters ) to reach the finish line . so b should get a head start of num__0.3 th of the race . answer ( d ) <eor> d <eos> |
d |
subtract__20.0__14.0__ divide__6.0__20.0__ divide__6.0__20.0__ |
subtract__20.0__14.0__ divide__6.0__20.0__ divide__6.0__20.0__ |
| a warehouse had a square floor with are num__10000 sq . metres . a rectangular addition was built along one entire side of the warehouse that increased the floor by one - half as much as the original floor . how many metres did the addition extend beyond the original building ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__50 <o> d ) num__200 <o> e ) num__500 |
length = num__100 b = ? area of rectangle = l * b original floor = num__10000 sq . metres increasing the original floor by half of the original floor = num__10000 + num__0.5 ( num__10000 ) = num__15000 so num__15000 - num__10000 = num__5000 ( area of rectangular portion ) l * b = num__5000 l = num__100 so num__100 * b = num__5000 - > b = num__50 so extended part is of num__50 metres answer : c <eor> c <eos> |
c |
multiply__10000.0__0.5__ multiply__0.5__100.0__ multiply__0.5__100.0__ |
multiply__10000.0__0.5__ multiply__0.5__100.0__ multiply__0.5__100.0__ |
| a father said to his son ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is num__38 years now the son ' s age five years back was : <o> a ) num__14 years <o> b ) num__19 years <o> c ) num__33 years <o> d ) num__38 years <o> e ) num__48 years |
let the son ' s present age be x years . then ( num__38 - x ) = x num__2 x = num__38 . x = num__19 . son ' s age num__5 years back ( num__19 - num__5 ) = num__14 years . answer : option a <eor> a <eos> |
a |
divide__38.0__2.0__ subtract__19.0__5.0__ subtract__19.0__5.0__ |
divide__38.0__2.0__ subtract__19.0__5.0__ subtract__19.0__5.0__ |
| there are num__3 numbers a b and c . if a : b = num__0.75 b : c = num__0.8 c : d = num__0.833333333333 then c : d will be ? <o> a ) num__1 : num__2 <o> b ) num__3 : num__5 <o> c ) num__5 : num__7 <o> d ) num__5 : num__6 <o> e ) num__5 : num__3 |
sol . a : b = num__3 : num__4 b : c = num__4 : num__5 c : d = num__5 : num__6 ∴ a ∶ b ∶ c ∶ d = num__3 : num__4 : num__5 : num__6 . thus c : d = num__5 : num__6 d <eor> d <eos> |
d |
divide__3.0__0.75__ divide__4.0__0.8__ divide__4.0__0.8__ |
divide__3.0__0.75__ divide__4.0__0.8__ divide__4.0__0.8__ |
| in the xy plane line m has the equation num__4 x + y = e . line n passes through the origin and is perpendicular to line m . if point p has the coordinates ( r r + num__1 ) and is on both line n and m what is the value of r ? <o> a ) - num__1.33333333333 <o> b ) num__0.25 <o> c ) - num__4 <o> d ) num__0.75 <o> e ) num__1.33333333333 |
the equation of line m is y = - num__4 x + e . now since line n is perpendicular to line m then its slope is negative reciprocal of the slope of line m so the slope of n is num__0.25 . next as line n passes through the origin ( num__0 num__0 ) and has the slope of num__0.25 then its equation is y = num__0.25 * x . point ( r r + num__1 ) lies on line n means that ( r + num__1 ) = num__0.25 * r - - > r = - num__1.33333333333 . answer : a . <eor> a <eos> |
a |
reverse__4.0__ round_down__0.25__ multiply__1.0__1.3333__ |
reverse__4.0__ round_down__0.25__ multiply__1.0__1.3333__ |
| a train crosses a num__120 - metre and a num__70 - metre long plat form in num__17 seconds and num__12 seconds respectively . what is the speed of the train ? <o> a ) num__10 ms - num__1 <o> b ) num__16 ms - num__1 <o> c ) num__20 ms - num__1 <o> d ) num__24 ms - num__1 <o> e ) num__45 ms - num__1 |
explanation : let the length of the train be x ∴ num__120 + x / num__17 = num__70 + x / num__12 num__1440 + num__12 x = num__1190 + num__17 x ∴ x = num__50.0 = num__50 m ∴ speed = num__120 + num__2.94117647059 = num__10.0 = num__10 ms - num__1 answer : option a <eor> a <eos> |
a |
multiply__120.0__12.0__ multiply__70.0__17.0__ subtract__120.0__70.0__ divide__50.0__17.0__ divide__120.0__12.0__ round__10.0__ |
multiply__120.0__12.0__ multiply__70.0__17.0__ subtract__120.0__70.0__ divide__50.0__17.0__ divide__120.0__12.0__ divide__120.0__12.0__ |
| what is the dividend . divisor num__17 the quotient is num__9 and the remainder is num__5 ? <o> a ) num__150 <o> b ) num__54 <o> c ) num__158 <o> d ) num__160 <o> e ) num__180 |
d = d * q + r d = num__17 * num__9 + num__5 d = num__153 + num__5 d = num__158 c ) <eor> c <eos> |
c |
multiply__17.0__9.0__ add__5.0__153.0__ add__5.0__153.0__ |
multiply__17.0__9.0__ add__5.0__153.0__ add__5.0__153.0__ |
| a patient was given a bottle of tablets by the doctor and he was asked to take five tablets in a gap of num__30 minutes . in how much time will he be able to take all the five tablets ? <o> a ) num__1 hour . <o> b ) num__2 hour . <o> c ) num__3 hour . <o> d ) none <o> e ) can not be determined |
suppose he takes the first tablet at num__8 : num__00 pm . then the second will be consumed by him at num__8 : num__30 third at num__9 : num__00 fourth at num__9 : num__30 and fifth at num__10 : num__00 . time = num__2 hour answer b <eor> b <eos> |
b |
subtract__10.0__8.0__ round__2.0__ |
subtract__10.0__8.0__ round__2.0__ |
| if n is a positive integer and n ^ num__2 is divisible by num__150 then what is the largest positive integer that must divide n ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__30 <o> d ) num__40 <o> e ) num__50 |
num__150 = num__2 * num__3 * num__5 ^ num__2 if num__150 divides n ^ num__2 then n must be divisible by num__2 * num__3 * num__5 = num__30 the answer is c . <eor> c <eos> |
c |
add__2.0__3.0__ divide__150.0__5.0__ divide__150.0__5.0__ |
add__2.0__3.0__ divide__150.0__5.0__ divide__150.0__5.0__ |
| a carpenter worked alone for num__1 day on a job that would take him num__6 more days to finish . he and another carpenter completed the job in num__6 more days . how many days would it have taken the second carpenter to do the complete job working alone ? <o> a ) num__4 num__0.666666666667 <o> b ) num__7 <o> c ) num__9 <o> d ) num__24 <o> e ) num__21 |
a carpenter worked only num__1 day on something that takes him num__6 more days . means ; carpenter finishes his work in num__7 days . let his buddy finish the same task in x days . respective rates per day : num__0.142857142857 and num__1 / x to complete num__1 work : first guy worked for num__5 days @ rate = num__0.142857142857 per day . second one worked for num__6 days @ rate = num__1 / x per day expression : days * rate = work num__5 * num__0.142857142857 + num__6 * num__1 / x = num__1 num__5 x + num__42 = num__7 x num__2 x = num__42 x = num__21 days . ans : e <eor> e <eos> |
e |
add__1.0__6.0__ divide__1.0__7.0__ subtract__6.0__1.0__ multiply__6.0__7.0__ subtract__7.0__5.0__ divide__42.0__2.0__ round__21.0__ |
add__1.0__6.0__ divide__1.0__7.0__ subtract__6.0__1.0__ multiply__6.0__7.0__ subtract__7.0__5.0__ divide__42.0__2.0__ multiply__1.0__21.0__ |
| how many factors of num__240 are also multiples of num__3 ? <o> a ) num__5 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__20 |
take factors of num__240 num__240 = num__2 ^ num__4 * num__3 ^ num__1 * num__5 ^ num__1 total factors of num__240 = ( num__4 + num__1 ) * ( num__1 + num__1 ) * ( num__1 + num__1 ) = num__5 * num__2 * num__2 = num__20 out of total factors of num__20 half will have num__0 as power of num__3 and half will have num__1 as power of num__3 . it is because we have considered only num__0 and num__1 as power of num__3 to compute total factors of num__240 . therefore factors that are multiple of num__3 = num__10.0 = num__10 ans - d <eor> d <eos> |
d |
subtract__3.0__2.0__ add__3.0__2.0__ multiply__4.0__5.0__ multiply__2.0__5.0__ gcd__240.0__10.0__ |
subtract__3.0__2.0__ add__3.0__2.0__ multiply__4.0__5.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
| the length of the bridge which a train num__130 metres long and travelling at num__45 km / hr can cross in num__30 seconds is : <o> a ) num__243 m <o> b ) num__240 m <o> c ) num__245 m <o> d ) num__249 m <o> e ) num__242 m |
speed = num__45 x num__0.277777777778 m / sec = num__12.5 m / sec . time = num__30 sec . let the length of bridge be x metres . then num__130 + x / num__30 = num__12.5 num__2 ( num__130 + x ) = num__750 x = num__245 m . answer c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| in a company the manager wants to give some gifts to all of the workers . in each block there are about num__200 workers are there . the total amount for giving the gifts for all the workers is $ num__6000 . the worth of the gift is num__2 $ . how many blocks are there in the company ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__18 <o> d ) num__20 <o> e ) num__15 |
each employee will get a gift worth of = $ num__2 total employees = num__3000.0 = num__3000 total blocks = num__15.0 = num__15 correct option is e <eor> e <eos> |
e |
divide__6000.0__2.0__ divide__3000.0__200.0__ divide__3000.0__200.0__ |
divide__6000.0__2.0__ divide__3000.0__200.0__ divide__3000.0__200.0__ |
| a train covers a distance in num__50 min if it runs at a speed of num__48 kmph on an average . the speed at which the train must run to reduce the time of journey to num__40 min will be <o> a ) num__47 kmph <o> b ) num__60 kmph <o> c ) num__69 kmph <o> d ) num__37 kmph <o> e ) num__80 kmph |
b num__60 kmph time = num__0.833333333333 hr = num__0.833333333333 hr speed = num__48 mph distance = s * t = num__48 * num__0.833333333333 = num__40 km time = num__0.666666666667 hr = num__0.666666666667 hr new speed = num__40 * num__1.5 kmph = num__60 kmph <eor> b <eos> |
b |
hour_to_min_conversion__ divide__50.0__60.0__ divide__40.0__60.0__ add__0.8333__0.6667__ hour_to_min_conversion__ |
hour_to_min_conversion__ divide__50.0__60.0__ divide__40.0__60.0__ add__0.8333__0.6667__ multiply__40.0__1.5__ |
| the average mark of the students of a class in a particular exam is num__80 . if num__5 students whose average mark in that exam is num__30 are excluded the average mark of the remaining will be num__90 . find the number of students who wrote the exam . <o> a ) num__22 <o> b ) num__27 <o> c ) num__25 <o> d ) num__99 <o> e ) num__30 |
let the number of students who wrote the exam be x . total marks of students = num__80 x . total marks of ( x - num__5 ) students = num__90 ( x - num__5 ) num__80 x - ( num__5 * num__30 ) = num__90 ( x - num__5 ) num__300 = num__10 x = > x = num__30 . answer : e <eor> e <eos> |
e |
subtract__90.0__80.0__ divide__300.0__10.0__ |
subtract__90.0__80.0__ divide__300.0__10.0__ |
| sonika deposited rs . num__8000 which amounted to rs . num__9200 after num__3 years at simple interest . had the interest been num__1.5 more . she would get how much ? <o> a ) num__9560 <o> b ) num__96288 <o> c ) num__26667 <o> d ) num__1662 <o> e ) num__2882 |
( num__8000 * num__3 * num__1.5 ) / num__100 = num__360 num__9200 - - - - - - - - num__9560 answer : a <eor> a <eos> |
a |
percent__100.0__9560.0__ |
percent__100.0__9560.0__ |
| a textile worker ’ s wages are increased by num__10.0 and afterwards decreased by num__10.0 . find the change percentage in his wages . <o> a ) num__1.0 decrease <o> b ) num__2.0 decrease <o> c ) num__5.0 decrease <o> d ) num__7.0 decrease <o> e ) num__8.0 decrease |
explanation : let workers wages = x a num__10.0 increase would mean new wages = ( num__1.1 ) x a num__10.0 decrease would mean new wages = ( num__0.9 ) ( num__1.1 x ) = num__0.99 x which means a decrease of num__1.0 answer : a <eor> a <eos> |
a |
multiply__0.9__1.1__ round_down__1.1__ reverse__1.0__ |
multiply__0.9__1.1__ round_down__1.1__ reverse__1.0__ |
| a manufacturer of a certain product can expect that between num__0.4 percent and num__0.5 percent of the units manufactured will be defective . if the retail price is $ num__2500 per unit and the manufacturer offers a full refund for defective units how much money can the manufacturer expect to need to cover the refunds on num__20000 units ? <o> a ) between $ num__15000 and $ num__25000 <o> b ) between $ num__30000 and $ num__50000 <o> c ) between $ num__200000 and $ num__250000 <o> d ) between $ num__150000 and $ num__250000 <o> e ) between $ num__300000 and $ num__500 |
000 |
number of defective units is between = . num__4.0 of num__20000 and . num__5.0 of num__20000 = num__80 and num__100 retail price per unit = num__2500 $ expected price of refund is between = num__2500 x num__80 and num__2500 x num__100 = num__2 num__00000 and num__2 num__50000 dollars answer c <eor> c <eos> |
c |
c |
| out of first num__20 natural numbers one number is selected at random . the probability that it is either an even number or a prime number is <o> a ) num__0.607142857143 <o> b ) num__0.62962962963 <o> c ) num__0.894736842105 <o> d ) probability of num__0.85 <o> e ) num__0.772727272727 |
n ( s ) = num__20 n ( even no ) = num__10 = n ( e ) n ( prime no ) = num__8 = n ( p ) p ( e ᴜ p ) = num__0.5 + num__0.4 - num__0.05 = num__0.85 answer : d <eor> d <eos> |
d |
union_prob__0.5__0.4__0.05__ union_prob__0.5__0.4__0.05__ |
union_prob__0.5__0.4__0.05__ union_prob__0.5__0.4__0.05__ |
| a truck travels num__20 miles due north num__30 miles due east and then num__20 miles due north . how many miles is the truck from the starting point ? <o> a ) num__20.3 <o> b ) num__44.7 <o> c ) num__50 <o> d ) num__70 <o> e ) num__120 |
we have two right angle triangles with sides num__20 and num__15 the distance between starting and end point = sum of the hypotenuse of both the triangles . hypotenuse = [ num__400 + num__225 ] ^ ( num__0.5 ) = num__25 hence the total distance between end point and starting point = num__25 + num__25 = num__50 correct option : c <eor> c <eos> |
c |
divide__15.0__30.0__ add__20.0__30.0__ round__50.0__ |
divide__15.0__30.0__ add__20.0__30.0__ add__20.0__30.0__ |
| suppose you answer the last three questions on this test at random . what is the most likely number of these three questions that you will answer correctly ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) it is impossible to determine |
there are five answers to each question the probabilities you will correctly answer respectively num__0 num__1 num__2 and num__3 are num__0.8 num__0.8 num__0.8 = num__0.512 ; num__3 num__0.2 num__0.8 num__0.8 = num__0.384 ; num__3 num__0.2 num__0.2 num__0.8 = num__0.096 ; num__0.2 num__0.2 num__0.2 = num__0.008 : the first is the most likely possibility . indeed it is more probable than the other three combined correct answer a <eor> a <eos> |
a |
add__1.0__2.0__ power__0.8__3.0__ subtract__1.0__0.8__ power__0.2__3.0__ round_down__0.2__ |
add__1.0__2.0__ power__0.8__3.0__ subtract__1.0__0.8__ power__0.2__3.0__ round_down__0.2__ |
| what profit percent is made by selling an article at a certain price if by selling at num__0.666666666667 rd of that price there would be a loss of num__20.0 ? <o> a ) num__20.0 <o> b ) num__70.0 <o> c ) num__80.0 <o> d ) num__40.0 <o> e ) num__10 % |
explanation : sp num__2 = num__0.666666666667 sp num__1 cp = num__100 sp num__2 = num__80 num__0.666666666667 sp num__1 = num__80 sp num__1 = num__120 num__100 - - - num__20 = > num__20.0 answer : a <eor> a <eos> |
a |
percent__20.0__100.0__ |
percent__20.0__100.0__ |
| a train speeds past a pole in num__15 seconds and a platform num__100 m long in num__25 seconds . its length is : <o> a ) num__100 m <o> b ) num__125 m <o> c ) num__130 m <o> d ) num__150 m <o> e ) num__170 m |
let the length of the train be x meters and its speed be y m / sec . they x / y = num__15 = > y = x / num__15 x + num__4.0 = x / num__15 x = num__150 m . answer : d <eor> d <eos> |
d |
divide__100.0__25.0__ round__150.0__ |
divide__100.0__25.0__ round__150.0__ |
| in covering distance the speed of a & b are in the ratio of num__3 : num__4 . a takes num__30 min more than b to reach the destion . the time taken by a to reach the destinstion is . <o> a ) num__4 hrs <o> b ) num__9 hrs <o> c ) num__3 hrs <o> d ) num__5 hrs <o> e ) num__2 hrs |
ratio of speed = num__3 : num__4 ratio of time = num__4 : num__3 let a takes num__4 x hrs b takes num__3 x hrs then num__4 x - num__3 x = num__0.5 hr x = ½ hr time taken by a to reach the destination is num__4 x = num__4 * ½ = num__2 hr answer is e . <eor> e <eos> |
e |
multiply__4.0__0.5__ round__2.0__ |
multiply__4.0__0.5__ multiply__4.0__0.5__ |
| excluding stoppages the speed of a train is num__45 kmph and including stoppages it is num__35 kmph . of how many minutes does the train stop per hour ? <o> a ) e num__982 <o> b ) num__27 <o> c ) num__12 <o> d ) num__121 <o> e ) num__13.3 |
explanation : t = num__0.222222222222 * num__60 = num__13.3 answer : option e <eor> e <eos> |
e |
hour_to_min_conversion__ round__13.3__ |
hour_to_min_conversion__ round__13.3__ |
| num__5 blue marbles num__3 red marbles and num__4 purple marbles are placed in a bag . if num__4 marbles are drawn without replacement what is the probability that the result will not be num__2 blue and num__2 purpl e marbles ? <o> a ) num__0.121212121212 <o> b ) ( num__0.138888888889 ) ^ num__2 <o> c ) num__0.5 <o> d ) ( num__0.861111111111 ) ^ num__2 <o> e ) num__0.878787878788 |
answer is num__0.878787878788 . the probability of num__2 blue and num__2 purple marbles selected is num__5 c num__2.4 c num__0.166666666667 c num__4 = num__0.121212121212 . subtracting the above from num__1 we get num__0.878787878788 e <eor> e <eos> |
e |
negate_prob__0.8788__ negate_prob__0.1212__ |
negate_prob__0.8788__ negate_prob__0.1212__ |
| find the simple interest on $ num__1200 for num__5 years at num__20.0 per annum ? <o> a ) $ num__1200 <o> b ) $ num__1000 <o> c ) $ num__500 <o> d ) $ num__1100 <o> e ) $ num__1500 |
si = ptr / num__100 = num__1200 * num__5 * num__0.2 = $ num__1200 answer is a <eor> a <eos> |
a |
percent__100.0__1200.0__ |
percent__100.0__1200.0__ |
| three bells commence tolling together and toll at intervals of num__3 num__69 seconds respectively . in num__60 minutes how many times do they toll together ? <o> a ) num__4 <o> b ) num__10 <o> c ) num__15 <o> d ) num__21 <o> e ) num__17 |
l . c . m of num__3 num__69 is num__18 . i . e after each num__3 min they will toll together . so in num__60 min they will toll num__20 times . as they have initially tolled once the answer will be num__20 + num__1 = num__21 . answer : d <eor> d <eos> |
d |
divide__60.0__3.0__ add__3.0__18.0__ round__21.0__ |
divide__60.0__3.0__ add__3.0__18.0__ add__3.0__18.0__ |
| the product of two consecutive negative even integers is num__24 . find the numbers . <o> a ) - num__2 - num__4 <o> b ) - num__10 - num__12 <o> c ) - num__6 - num__8 <o> d ) - num__4 - num__6 <o> e ) - num__12 - num__14 |
( n ) ( n + num__2 ) = num__24 n num__2 + num__2 n = num__24 n num__2 + num__2 n – num__24 = num__0 ( n + num__6 ) ( n – num__4 ) = num__0 then the solutions are n = – num__6 and n = num__4 . since the numbers i am looking for are negative i can ignore the ` ` num__4 ' ' and take n = – num__6 . then the next number is n + num__2 = – num__4 answer : d <eor> d <eos> |
d |
divide__24.0__6.0__ round__4.0__ |
divide__24.0__6.0__ round__4.0__ |
| in how many ways can a three - letter password be chosen using the letters a b c d e and / or f such that at least one letter is repeated within the password ? <o> a ) num__72 <o> b ) num__86 <o> c ) num__90 <o> d ) num__96 <o> e ) num__1 |
296 |
total number of four letter passwords = num__6 * num__6 * num__6 = num__216 - - - - - - ( num__1 ) total number of passwords in which no letter repeats = num__6 c num__3 * num__3 ! = num__20 * num__6 = num__120 - - - - - - ( num__2 ) therefore required value = ( num__1 ) - ( num__2 ) = num__216 - num__120 = num__96 d <eor> d <eos> |
d |
d |
| find value for x from given equation ? x + num__1 = - num__3 <o> a ) num__1 <o> b ) - x = - num__4 <o> c ) - num__3 <o> d ) - num__5 <o> e ) num__5 |
num__1 . subtract num__1 from both sides : x + num__1 - num__1 = - num__3 - num__1 num__2 . simplify both sides : x = - num__4 b <eor> b <eos> |
b |
subtract__3.0__1.0__ add__1.0__3.0__ add__1.0__3.0__ |
subtract__3.0__1.0__ add__1.0__3.0__ add__1.0__3.0__ |
| the population of a town is num__8000 . it decreases annually at the rate of num__20.0 p . a . what will be its population after num__3 years ? <o> a ) num__5100 <o> b ) num__4096 <o> c ) num__5200 <o> d ) num__5400 <o> e ) num__5500 |
formula : ( after = num__100 denominator ago = num__100 numerator ) num__8000 Ã — num__0.8 Ã — num__0.8 x num__0.8 = num__4096 answer : b <eor> b <eos> |
b |
percent__100.0__4096.0__ |
percent__100.0__4096.0__ |
| sum of two number = num__16 and product = num__48 find the difference of the two number ? ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__8 |
( x - y ) num__2 = ( x + y ) num__2 - num__4 xy ( x - y ) num__2 = ( num__16 ) num__2 - num__4 x num__48 ( x - y ) num__2 = num__256 - num__192 ( x - y ) num__2 = num__64 x - y = num__8 answer e <eor> e <eos> |
e |
multiply__48.0__4.0__ add__16.0__48.0__ divide__16.0__2.0__ divide__16.0__2.0__ |
multiply__48.0__4.0__ add__16.0__48.0__ divide__16.0__2.0__ subtract__16.0__8.0__ |
| the average age of a husband and a wife is num__23 years when they were married five years ago but now the average age of the husband wife and child is num__20 years ( the child was born during the interval ) . what is the present age of the child ? <o> a ) num__6 <o> b ) num__5 <o> c ) num__8 <o> d ) num__4 <o> e ) num__2 |
num__28 * num__2 = num__56 num__20 * num__3 = num__60 - - - - - - - - - - - num__4 years . answer : d <eor> d <eos> |
d |
multiply__2.0__28.0__ subtract__23.0__20.0__ multiply__20.0__3.0__ subtract__60.0__56.0__ subtract__60.0__56.0__ |
multiply__2.0__28.0__ subtract__23.0__20.0__ multiply__20.0__3.0__ subtract__60.0__56.0__ subtract__60.0__56.0__ |
| which of the following must be true ? num__1 ) every prime number greater than num__5 can be written as either as num__9 n + num__1 or num__9 n - num__1 num__2 ) every prime number greater than num__5 can be written as num__4 n + num__1 or num__4 n - num__1 num__3 ) every number greater than num__5 of the form num__9 n + num__1 or num__9 n - num__1 is a prime number . <o> a ) num__12 <o> b ) num__23 <o> c ) num__1 num__23 <o> d ) only num__1 <o> e ) only num__2 |
every prime number greater than num__3 can be written in the form of num__9 n + num__1 or num__9 n - num__1 where n is an integer . so num__1 can be possible . num__3 . this is not possible as num__2 and num__3 are being prime we ca n ' t write then in the form of num__9 n + num__1 or num__9 n - num__1 . a <eor> a <eos> |
a |
add__9.0__3.0__ |
add__9.0__3.0__ |
| in a renowned city the average birth rate is num__7 people every two seconds and the death rate is num__1 people every two seconds . estimate the size of the population net increase that occurs in one day . <o> a ) num__32300 <o> b ) num__172800 <o> c ) num__468830 <o> d ) num__338200 <o> e ) num__259 |
200 |
every num__2 seconds num__6 persons are added ( num__7 - num__1 ) . every second num__3 persons are added . in a day num__24 hrs = num__24 * num__60 minutes = num__24 * num__60 * num__60 = num__86400 seconds . num__86400 * num__3 = num__259200 option e <eor> e <eos> |
e |
e |
| if f is the smallest positive integer such that num__3150 multiplied by f is the square of an integer then f must be <o> a ) num__2 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__14 |
solution : this problem is testing us on the rule that when we express a perfect square by its unique prime factors every prime factor ' s exponent is an even number . let ’ s start by prime factorizing num__3150 . num__3150 = num__315 x num__10 = num__5 x num__63 x num__10 = num__5 x num__7 x num__3 x num__3 x num__5 x num__2 num__3150 = num__2 ^ num__1 x num__3 ^ num__2 x num__5 ^ num__2 x num__7 ^ num__1 ( notice that the exponents of both num__2 and num__7 are not even numbers . this tells us that num__3150 itself is not a perfect square . ) we also are given that num__3150 multiplied by f is the square of an integer . we can write this as : num__2 ^ num__1 x num__3 ^ num__2 x num__5 ^ num__2 x num__7 ^ num__1 x f = square of an integer according to our rule we need all unique prime factors ' exponents to be even numbers . thus we need one more num__2 and one more num__7 . therefore y = num__7 x num__2 = num__14 answer is e . <eor> e <eos> |
e |
rectangle_perimeter__2.0__5.0__ multiply__1.0__14.0__ |
rectangle_perimeter__2.0__5.0__ power__14.0__1.0__ |
| num__4 / x + num__5 x = num__5 ( x - num__8 ) <o> a ) - num__9 <o> b ) - num__0.333333333333 <o> c ) - num__0.111111111111 <o> d ) - num__0.1 <o> e ) num__9 |
we can solve - expand the right side multiply by x on both sides and then subtract away the num__5 x ^ num__2 terms : ( num__4 / x ) + num__5 x = num__5 ( x - num__8 ) ( num__4 / x ) + num__5 x = num__5 x - num__40 num__4 + num__5 x ^ num__2 = num__5 x ^ num__2 - num__40 x num__4 = - num__40 x - num__0.1 = x and to confirm you can plug that answer back into the original equation to see that it makes the left and right sides equal . d <eor> d <eos> |
d |
divide__8.0__4.0__ multiply__5.0__8.0__ divide__4.0__40.0__ divide__4.0__40.0__ |
divide__8.0__4.0__ multiply__5.0__8.0__ divide__4.0__40.0__ divide__4.0__40.0__ |
| the average age of three boys is num__15 years and their ages are in proportion num__3 : num__5 : num__7 . what is the age in years of the youngest boy ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__7 <o> d ) num__6 <o> e ) num__5 |
num__3 x + num__5 x + num__7 x = num__45 x = num__3 num__3 x = num__9 answer : b <eor> b <eos> |
b |
multiply__15.0__3.0__ divide__45.0__5.0__ divide__45.0__5.0__ |
multiply__15.0__3.0__ divide__45.0__5.0__ divide__45.0__5.0__ |
| a number is doubled and num__15 is added . if resultant is trebled it becomes num__75 . what is that number <o> a ) num__8 <o> b ) num__10 <o> c ) num__5 <o> d ) num__14 <o> e ) num__7 |
explanation : = > num__3 ( num__2 x + num__15 ) = num__75 = > num__2 x + num__15 = num__25 = > x = num__5 option c <eor> c <eos> |
c |
divide__75.0__3.0__ divide__15.0__3.0__ divide__15.0__3.0__ |
divide__75.0__3.0__ add__2.0__3.0__ add__2.0__3.0__ |
| two trains of length num__100 m and num__160 m are running towards each other on parallel lines at num__42 kmph and num__30 kmph respectively . in what time will they be clear of each other from the moment they meet ? <o> a ) num__18 sec <o> b ) num__70 sec <o> c ) num__13 sec <o> d ) num__20 sec <o> e ) num__19 sec |
relative speed = ( num__42 + num__30 ) * num__0.277777777778 = num__4 * num__5 = num__20 mps . distance covered in passing each other = num__100 + num__160 = num__260 m . the time required = d / s = num__13.0 = num__13 sec . answer : c <eor> c <eos> |
c |
divide__100.0__5.0__ add__100.0__160.0__ divide__260.0__20.0__ round__13.0__ |
divide__100.0__5.0__ add__100.0__160.0__ divide__260.0__20.0__ divide__260.0__20.0__ |
| a farmer has an apple orchard consisting of fuji and gala apple trees . due to high winds this year num__10.0 of his trees cross pollinated . the number of his trees that are pure fuji plus the cross - pollinated ones totals num__204 while num__0.75 of all his trees are pure fuji . how many of his trees are pure gala ? <o> a ) num__22 <o> b ) num__33 <o> c ) num__36 <o> d ) num__77 <o> e ) num__88 |
let f = pure fuji g = pure gala and c - cross pollinated . c = num__10.0 of x where x is total trees . c = . num__1 x also num__3 x / num__4 = f and c + f = num__204 = > . num__1 x + num__0.75 x = num__204 = > x = num__240 num__240 - num__204 = pure gala = num__36 . c <eor> c <eos> |
c |
add__1.0__3.0__ subtract__240.0__204.0__ multiply__1.0__36.0__ |
add__1.0__3.0__ subtract__240.0__204.0__ subtract__240.0__204.0__ |
| two friends a b are running up hill and then to get down ! length if road is num__440 yards a on his return journey met b going up at num__20 yards from top a has finished the race . num__5 minute earlier than b then how much time a had taken to complete the race . <o> a ) num__55 minutes <o> b ) num__95 minutes <o> c ) num__105 minutes <o> d ) num__115 minutes <o> e ) num__125 minutes |
a gains num__20 yards over b in it ' s down the hill journey . so for entire journey a will have a lead of num__40 yards over b . now when a is at destination b is num__40 yards behind which b will cover in num__5 minutes . so by unitary method we get . . for b num__40 yards in num__5 minutes therefore num__1 yard in num__0.125 minutes therefore num__880 yards in ( num__0.125 ) * num__880 = num__110 minutes . now b takes num__110 minutes so a will take num__105 minutes . answer : c <eor> c <eos> |
c |
divide__5.0__40.0__ multiply__0.125__880.0__ subtract__110.0__5.0__ round__105.0__ |
divide__5.0__40.0__ multiply__0.125__880.0__ subtract__110.0__5.0__ round__105.0__ |
| ramu bought an old car for rs . num__42000 . he spent rs . num__13000 on repairs and sold it for rs . num__64900 . what is his profit percent ? <o> a ) num__22 <o> b ) num__77 <o> c ) num__18 <o> d ) num__99 <o> e ) num__88 |
total cp = rs . num__42000 + rs . num__13000 = rs . num__55000 and sp = rs . num__64900 profit ( % ) = ( num__64900 - num__55000 ) / num__55000 * num__100 = num__18.0 answer : c <eor> c <eos> |
c |
percent__18.0__100.0__ |
percent__18.0__100.0__ |
| ann can have her bicycle repaired for $ num__50 or she can trade it in as is and receive $ num__22 credit toward the purchase of a new bicycle that sells for $ num__107 . if ann trades in her current bicycle the cost to her of purchasing the new bicycle is what percent greater than the cost of having her current bicycle repaired ? <o> a ) num__44.0 <o> b ) num__60.0 <o> c ) num__70.0 <o> d ) num__114.0 <o> e ) num__170 % |
the cost for repair is $ num__50 ; the cost of new bicycle is $ num__107 - $ num__22 = $ num__85 ; so the cost to ann of purchasing the new bicycle is num__85 − num__1.0 ∗ num__100 = num__70 greater than the cost of having her current bicycle repaired . answer : c . <eor> c <eos> |
c |
percent__100.0__70.0__ |
percent__100.0__70.0__ |
| the current of a stream at num__1 kmph . a motor boat goes num__35 km upstream and back to the starting point in num__12 hours . the speed of the motor boat in still water is ? <o> a ) num__6 kilometre per hour <o> b ) num__8 kmph <o> c ) num__9 kmph <o> d ) num__10 kmph <o> e ) num__3 kmph |
s = num__1 m = x ds = x + num__1 us = x - num__1 num__35 / ( x + num__1 ) + num__35 / ( x - num__1 ) = num__12 x = num__6 answer : a <eor> a <eos> |
a |
round__6.0__ |
subtract__12.0__6.0__ |
| in a num__1000 m race a beats b by num__50 m and b beats c by num__10 m . in the same race by how many meters does a beat c ? <o> a ) a ) num__60 m <o> b ) b ) num__829 m <o> c ) c ) num__822 m <o> d ) d ) num__929 m <o> e ) e ) num__132 mj |
by the time a covers num__1000 m b covers ( num__1000 - num__50 ) = num__950 m . by the time b covers num__1000 m c covers ( num__1000 - num__10 ) = num__990 m . so the ratio of speeds of a and c = num__1.05263157895 * num__1.0101010101 = num__1.06382978723 so by the time a covers num__1000 m c covers num__940 m . so in num__1000 m race a beats c by num__1000 - num__940 = num__60 m . answer : a <eor> a <eos> |
a |
subtract__1000.0__50.0__ subtract__1000.0__10.0__ divide__1000.0__950.0__ divide__1000.0__990.0__ subtract__950.0__10.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
subtract__1000.0__50.0__ subtract__1000.0__10.0__ divide__1000.0__950.0__ divide__1000.0__990.0__ subtract__950.0__10.0__ subtract__1000.0__940.0__ subtract__1000.0__940.0__ |
| a man goes from a to b at a speed of num__21 kmph and comes back to a at a speed of num__21 kmph . find his average speed for the entire journey ? <o> a ) num__23.3 kmph <o> b ) num__25.3 kmph <o> c ) num__22.5 kmph <o> d ) num__22.3 kmph <o> e ) num__22.9 kmph |
distance from a and b be ' d ' average speed = total distance / total time average speed = ( num__2 d ) / [ ( d / num__21 ) + ( d / num__24 ] = ( num__2 d ) / [ num__15 d / num__168 ) = > num__22.3 kmph . answer : d <eor> d <eos> |
d |
round__22.3__ |
round__22.3__ |
| for four days of work a clerk had daily earnings of $ num__140 $ num__130 $ num__140 and $ num__140 respectively . in which of the following intervals does the standard deviation of the daily earnings lie ? <o> a ) between $ num__10 and $ num__100 <o> b ) between $ num__10 and $ num__50 <o> c ) between $ num__0 and $ num__10 <o> d ) between $ num__10 and $ num__70 <o> e ) between $ num__10 and $ num__90 |
the standard deviation is between num__0 and num__10 . the reason is that none of the numbers in the set is further away from the mean than num__10 ( or rather abs ( num__130 - ( num__140 * num__3 + num__130 ) / num__4 ) ) . i am not fully sure how to calculate the standard deviation ( i simply remember that there were some formula for calculating the average of the squared distances from the mean for all items in the set from statistics classes - variation ) but i believe this will suffice for any gmat question related to standard deviation . answer : c <eor> c <eos> |
c |
subtract__140.0__130.0__ round__0.0__ |
subtract__140.0__130.0__ multiply__140.0__0.0__ |
| the ratio of the radius of two circles is num__3 : num__4 and then the ratio of their areas is ? <o> a ) num__1 : num__7 <o> b ) num__9 : num__16 <o> c ) num__1 : num__9 <o> d ) num__3 : num__7 <o> e ) num__3 : num__4 |
r num__1 : r num__2 = num__3 : num__4 Î r num__1 ^ num__2 : Î r num__2 ^ num__2 r num__1 ^ num__2 : r num__2 ^ num__2 = num__9 : num__16 answer : b <eor> b <eos> |
b |
triangle_area__4.0__1.0__ power__3.0__2.0__ square_perimeter__4.0__ power__3.0__2.0__ |
triangle_area__4.0__1.0__ power__3.0__2.0__ power__4.0__2.0__ power__3.0__2.0__ |
| the sum of ages of num__5 children born at the intervals of num__3 years each is num__50 years . what is the age of the youngest child ? <o> a ) num__4 years <o> b ) num__5 years <o> c ) num__6 years <o> d ) num__7 years <o> e ) none |
let the ages of children be x ( x + num__3 ) ( x + num__6 ) ( x + num__9 ) and ( x + num__12 ) years . then x + ( x + num__3 ) + ( x + num__6 ) + ( x + num__9 ) + ( x + num__12 ) = num__50 num__5 x = num__20 x = num__4 . age of the youngest child = x = num__4 years a ) <eor> a <eos> |
a |
add__3.0__6.0__ add__3.0__9.0__ subtract__9.0__5.0__ subtract__9.0__5.0__ |
add__3.0__6.0__ add__3.0__9.0__ subtract__9.0__5.0__ subtract__9.0__5.0__ |
| a train which has num__410 m long is running num__45 kmph . in what time will it cross a person moving at num__9 kmph in same direction ? <o> a ) num__56 sec <o> b ) num__41 sec <o> c ) num__36 sec <o> d ) num__29 sec . <o> e ) num__19 sec . |
time taken to cross a moving person = length of train / relative speed time taken = num__410 / ( ( num__45 - num__9 ) ( num__0.277777777778 ) = num__11.3888888889 * ( num__0.277777777778 ) = num__41.0 = num__41 sec answer : b <eor> b <eos> |
b |
round__41.0__ |
round__41.0__ |
| two trains running in opposite directions cross a man standing on the platform in num__27 seconds and num__17 seconds respectively . if they cross each other in num__23 seconds what is the ratio of their speeds ? <o> a ) insufficient data <o> b ) num__3 : num__2 <o> c ) num__3 : num__1 <o> d ) num__1 : num__3 <o> e ) num__1 : num__2 |
explanation : let the speed of the trains be x and y respectively length of train num__1 = num__27 x length of train num__2 = num__17 y relative speed = x + y time taken to cross each other = num__23 s = > ( num__27 x + num__17 y ) / ( x + y ) = num__23 = > ( num__27 x + num__17 y ) / = num__23 ( x + y ) = > num__4 x = num__6 y = > x / y = num__1.5 = num__1.5 answer : option b <eor> b <eos> |
b |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ multiply__1.5__2.0__ |
subtract__27.0__23.0__ add__2.0__4.0__ divide__6.0__4.0__ add__1.0__2.0__ |
| excluding stoppages the speed of a bus is num__75 kmph and including stoppages it is num__45 kmph . for how many minutes does the bus stop per hour ? <o> a ) num__15 <o> b ) num__24 <o> c ) num__17 <o> d ) num__19 <o> e ) none of these |
due to stoppages it covers num__30 km less . time taken to cover num__30 km = ( ( num__0.4 ) Ã — num__60 ) = num__24 min . option ( b ) is correct <eor> b <eos> |
b |
subtract__75.0__45.0__ divide__30.0__75.0__ hour_to_min_conversion__ multiply__60.0__0.4__ round__24.0__ |
subtract__75.0__45.0__ divide__30.0__75.0__ hour_to_min_conversion__ multiply__60.0__0.4__ round__24.0__ |
| in traveling from a dormitory to a certain city a student went num__0.25 of the way by foot num__0.5 of the way by bus and the remaining num__10 kilometers by car . what is the distance in kilometers from the dormitory to the city ? <o> a ) num__10 <o> b ) num__45 <o> c ) num__30 <o> d ) num__20 <o> e ) num__40 |
whole trip = distance by foot + distance by bus + distance by car x = num__0.25 x + num__0.5 x + num__10 x - num__0.25 x - num__0.5 x = num__10 x = num__40 km option : e <eor> e <eos> |
e |
divide__10.0__0.25__ round__40.0__ |
divide__10.0__0.25__ round__40.0__ |
| company kw is being sold and both company a and company b were considering the purchase . the price of company kw is num__40.0 more than company a has in assets and this same price is also num__100.0 more than company b has in assets . if companies a and b were to merge and combine their assets the price of company kw would be approximately what percent of these combined assets ? <o> a ) num__66.0 <o> b ) num__75.0 <o> c ) num__82.0 <o> d ) num__116.0 <o> e ) num__150 % |
let the price of company a ' s assets be num__100 price of assets of kw is num__40.0 more than company a ' s assets which is num__140 price of assets of kw is num__100.0 more than company b ' s assets which means price of company b ' s assets is half the price of kw = num__70 a + b = num__170 kw = num__140 kw / ( a + b ) * num__100 = num__0.823529411765 * num__100 = num__82.35 or num__82.0 c <eor> c <eos> |
c |
percent__100.0__82.0__ |
percent__100.0__82.0__ |
| the smallest number when increased by ` ` num__1 ` ` is exactly divisible by num__12 num__18 num__24 num__32 and num__40 is : <o> a ) num__1439 <o> b ) num__1440 <o> c ) num__1459 <o> d ) num__1449 <o> e ) num__1549 |
lcm = num__1440 num__1440 - num__1 = num__1439 answer : a <eor> a <eos> |
a |
subtract__1440.0__1.0__ multiply__1.0__1439.0__ |
subtract__1440.0__1.0__ subtract__1440.0__1.0__ |
| what is the next number of the following sequence num__2 num__510 num__50500 ? <o> a ) num__24000 <o> b ) num__20000 <o> c ) num__30000 <o> d ) num__25000 <o> e ) num__40000 |
num__2 * num__5 = num__10 num__5 * num__10 = num__50 num__10 * num__50 = num__500 num__50 * num__500 = num__25000 therefore next no . is num__25000 answer : d <eor> d <eos> |
d |
multiply__2.0__5.0__ multiply__5.0__10.0__ subtract__510.0__10.0__ multiply__50.0__500.0__ multiply__50.0__500.0__ |
multiply__2.0__5.0__ multiply__5.0__10.0__ multiply__10.0__50.0__ multiply__50.0__500.0__ multiply__50.0__500.0__ |
| a train num__150 m long running at num__72 kmph crosses a platform in num__25 sec . what is the length of the platform ? <o> a ) num__287 <o> b ) num__288 <o> c ) num__350 <o> d ) num__278 <o> e ) num__728 |
d = num__72 * num__0.277777777778 = num__25 = num__500 – num__150 = num__350 answer : c <eor> c <eos> |
c |
subtract__500.0__150.0__ round__350.0__ |
subtract__500.0__150.0__ round__350.0__ |
| what is the value of x if num__1 / ( log num__1.0022675737 x ) + num__1 / ( log num__1.00226244344 x ) + num__1 / ( log num__1.00225733634 x ) + … + num__1 / ( log num__1.00111358575 x ) + num__1 / ( log num__1.00111234705 x ) = num__2 ? <o> a ) num__0.0952380952381 <o> b ) num__1 <o> c ) num__0.07 <o> d ) num__1.42857142857 <o> e ) none of these |
the given equation = num__1 / ( log num__1.0022675737 x ) = ( log num__1.0022675737 ) / logx = logx num__1.0022675737 . = logx num__1.0022675737 + logx num__1.00226244344 + logx num__1.00225733634 + . . . . . . + logx num__1.00111358575 + logx num__1.00111234705 = num__2 . = logx num__2.04081632653 = num__2 = > x num__2 = num__2.04081632653 = > x = num__1.42857142857 = num__1.42857142857 . answer : d <eor> d <eos> |
d |
multiply__1.0__1.4286__ |
divide__1.4286__1.0__ |
| a person is traveling at num__75 km / hr and reached his destiny in num__4 hr then find the distance ? <o> a ) num__300 km <o> b ) num__200 km <o> c ) num__250 km <o> d ) num__400 km <o> e ) num__220 km |
t = num__4 hrs d = t * s = num__75 * num__4 = num__300 km answer is a <eor> a <eos> |
a |
multiply__75.0__4.0__ round__300.0__ |
multiply__75.0__4.0__ multiply__75.0__4.0__ |
| the length of a rectangle is twice its breadth . if its length is decreased by num__5 cm and breadth is increased by num__5 cm the area of the rectangle is increased by num__75 sq . cm . find the length of the rectangle ? <o> a ) num__10 cm <o> b ) num__20 cm <o> c ) num__30 cm <o> d ) num__40 cm <o> e ) num__50 cm |
solution let the breadth = x then length = num__2 x = [ ( num__2 x - num__5 ) ( x + num__5 ) - num__2 x ] x [ x = num__75 ] x = num__20 . length of the rectangle = num__20 cm . answer b <eor> b <eos> |
b |
square_perimeter__5.0__ square_perimeter__5.0__ |
square_perimeter__5.0__ square_perimeter__5.0__ |
| if c is num__20.0 of a and num__10.0 of b what percent of a is b ? <o> a ) num__2.5 <o> b ) num__15.0 <o> c ) num__25.0 <o> d ) num__35.0 <o> e ) num__200 % |
answer = e num__20 a / num__100 = num__10 b / num__100 b = num__20 a / num__10 = num__200 a / num__100 = num__200.0 <eor> e <eos> |
e |
percent__100.0__200.0__ |
percent__100.0__200.0__ |
| in a race of num__4 kms a beats b by num__100 m or num__25 seconds then time taken by a is <o> a ) num__8 min num__15 sec <o> b ) num__10 min num__17 sec <o> c ) num__15 min num__8 sec <o> d ) num__16 min num__15 sec <o> e ) num__17 min num__15 sec |
b covers num__100 m in num__25 seconds b take time = ( num__4000 * num__25 ) / num__100 = num__1000 sec = num__16 min num__40 sec . a takes time = num__1000 sec - num__25 sec = num__975 sec = num__16 min num__25 sec . answer : d <eor> d <eos> |
d |
divide__4000.0__4.0__ divide__4000.0__100.0__ subtract__1000.0__25.0__ round__16.0__ |
divide__4000.0__4.0__ divide__4000.0__100.0__ subtract__1000.0__25.0__ round__16.0__ |
| an article is bought for rs . num__605 and sold for rs . num__900 find the gain percent ? <o> a ) num__30 num__0.333333333333 % <o> b ) num__33 num__0.333333333333 % <o> c ) num__48 num__0.760330578512 % <o> d ) num__35 num__0.333333333333 % <o> e ) num__29 num__0.333333333333 % |
c num__48 num__0.760330578512 num__605.0 - - - - num__295 num__100 - - - - ? = > num__48 num__0.760330578512 % <eor> c <eos> |
c |
percent__48.0__100.0__ |
percent__48.0__100.0__ |
| a box has exactly num__100 balls and each ball is either red blue or white . if the box has num__10 more blue balls than white balls and thrice as many red balls as blue balls how many white balls does the box has ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
x = the number of red balls y = the number of blue balls z = the number of white balls from the first sentence we have equation # num__1 : x + y + z = num__100 . . . the box has num__10 more blue balls than white balls . . . equation # num__2 : y = num__10 + z . . . thrice as many red balls as blue balls . . . equation # num__3 : x = num__3 y solve equation # num__2 for z : z = y - num__10 now we can replace both x and z with y in equation # num__1 num__3 y + y + ( y - num__10 ) = num__100 num__5 y - num__10 = num__100 num__5 y = num__110 y = num__22 there are num__22 blue balls . this is num__10 more than the number of white balls so z = num__12 . that ' s the answer . just as a check x = num__66 and num__66 + num__22 + num__12 = num__100 . answer = num__12 ( c ) <eor> c <eos> |
c |
add__1.0__2.0__ divide__10.0__2.0__ add__100.0__10.0__ divide__110.0__5.0__ add__10.0__2.0__ multiply__3.0__22.0__ add__10.0__2.0__ |
add__1.0__2.0__ add__2.0__3.0__ add__100.0__10.0__ divide__110.0__5.0__ add__10.0__2.0__ multiply__3.0__22.0__ add__10.0__2.0__ |
| the length of a rectangular plot is num__20 metres more than its breadth . if the cost of fencing the plot @ rs . num__26.50 per metre is rs . num__7420 what is the length of the plot in metres ? <o> a ) num__20 <o> b ) num__200 <o> c ) num__300 <o> d ) num__400 <o> e ) num__80 |
let length of plot = l meters then breadth = l - num__20 meters and perimeter = num__2 [ l + l - num__20 ] = [ num__4 l - num__40 ] meters [ num__4 l - num__40 ] * num__26.50 = num__7420 [ num__4 l - num__40 ] = num__7420 / num__26.50 = num__280 num__4 l = num__320 l = num__80.0 = num__80 meters . answer : e <eor> e <eos> |
e |
multiply__20.0__2.0__ divide__7420.0__26.5__ add__40.0__280.0__ multiply__20.0__4.0__ round__80.0__ |
multiply__20.0__2.0__ divide__7420.0__26.5__ add__40.0__280.0__ multiply__20.0__4.0__ multiply__20.0__4.0__ |
| the number of people at ovations bar in the hour from num__12 p . m . to num__1 p . m . was num__20.0 greater than the number of people there from num__11 a . m . to num__12 p . m . the number of the people at the bar from num__11 a . m . to num__12 p . m . was num__10.0 more than the number there from num__10 a . m . to num__11 a . m . if num__242 people were at the bar from num__11 a . m . to num__1 p . m . which of the following is the number of people in the bar from num__10 a . m . to num__11 a . m . ? <o> a ) num__100 <o> b ) num__20 <o> c ) num__30 <o> d ) num__40 <o> e ) num__50 |
no of people from num__10 - num__11 is x no of people from num__11 - num__12 is num__11 x / num__10 no of people from num__12 to num__1 is ( num__1.2 ) ( num__11 x / num__10 ) given that num__11 x / num__10 + ( num__1.2 ) ( num__11 x / num__10 ) = num__121 x / num__50 = num__242 then x = num__100 a <eor> a <eos> |
a |
divide__12.0__10.0__ multiply__1.0__100.0__ |
divide__12.0__10.0__ divide__100.0__1.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__12 sec . what is the length of the train ? <o> a ) num__120 m <o> b ) num__180 m <o> c ) num__200 m <o> d ) can not be determined <o> e ) none of these |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__12 = num__200 m answer : c <eor> c <eos> |
c |
round__200.0__ |
round__200.0__ |
| an inspector rejects num__0.0008 of the meters as defective . how many will be examine to project ? <o> a ) num__2000 <o> b ) num__2300 <o> c ) num__2700 <o> d ) num__250000 <o> e ) num__250 |
let the number of meters to be examined be x . then num__0.0008 of x = num__2 [ ( num__0.0008 ) * ( num__0.01 ) * x ] = num__2 x = [ ( num__2 * num__100 * num__10000 ) / num__8 ] = num__250000 answer is d . <eor> d <eos> |
d |
percent__100.0__250000.0__ |
percent__100.0__250000.0__ |
| if num__4 men can color num__48 m long cloth in num__2 days then num__6 men can color num__36 m long cloth in ? <o> a ) num__1 day <o> b ) num__2 days <o> c ) num__3 days <o> d ) num__4 days <o> e ) num__5 days |
the length of cloth painted by one man in one day = num__12.0 × num__2 = num__6 m no . of days required to paint num__36 m cloth by num__6 men = num__6.0 × num__6 = num__1 day a ) <eor> a <eos> |
a |
divide__48.0__4.0__ round__1.0__ |
divide__48.0__4.0__ round__1.0__ |
| if the ratio of two number is num__2 : num__3 and lcm of the number is num__120 then what is the number . <o> a ) num__15 <o> b ) num__20 <o> c ) num__25 <o> d ) num__30 <o> e ) num__35 |
product of two no = lcm * hcf num__2 x * num__3 x = num__120 * x x = num__20 answer : b <eor> b <eos> |
b |
round__20.0__ |
round__20.0__ |
| a car travels at a speed of num__65 miles per hour . how far will it travel in num__5 hours ? <o> a ) num__255 miles <o> b ) num__325 miles <o> c ) num__645 miles <o> d ) num__121 miles <o> e ) num__333 miles |
during each hour the car travels num__65 miles . for num__5 hours it will travel num__65 + num__65 + num__65 + num__65 + num__65 = num__5 * num__65 = num__325 miles correct answer b <eor> b <eos> |
b |
multiply__65.0__5.0__ round__325.0__ |
multiply__65.0__5.0__ multiply__65.0__5.0__ |
| the average of num__40 results is num__30 and the average of other num__30 results is num__40 . what is the average of all the results ? <o> a ) num__34 <o> b ) num__25 <o> c ) num__48 <o> d ) num__50 <o> e ) none |
answer sum of num__70 result = sum of num__40 result + sum of num__30 result . = num__40 x num__30 + num__30 x num__40 = num__34.2857142857 correct option : a <eor> a <eos> |
a |
add__40.0__30.0__ round_down__34.2857__ |
add__40.0__30.0__ round_down__34.2857__ |
| the price of a book is increased from $ num__300 to $ num__450 . what is the % of increase in its price ? <o> a ) num__10.0 <o> b ) num__20.0 <o> c ) num__40.0 <o> d ) num__50.0 <o> e ) num__60 % |
explanation : change in the price = rs num__450 â € “ rs num__300 = rs num__150 percentage of increase = change in the price initial price * num__100 . percentage increase in price = ( num__150 num__300 ) * num__100 = num__50.0 d <eor> d <eos> |
d |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| from a salary of an officer num__10.0 is deducted as house rent num__15.0 of the rest he spend on children ' s education and num__10.0 of the balance he spends on clothes . then he is left with rs . num__4131 . find his total salary ? <o> a ) num__6000 <o> b ) num__3788 <o> c ) num__9009 <o> d ) num__6788 <o> e ) num__2772 |
suppose his total salary be rs . x then num__90.0 of num__85.0 of num__90.0 of x = num__4131 \ inline \ rightarrow \ frac { num__90 } { num__100 } \ times \ frac { num__85 } { num__100 } \ times \ frac { num__90 } { num__100 } \ times x = num__4131 \ inline \ therefore x = \ frac { num__4131 \ times num__100 \ times num__100 \ times num__100 } { num__90 \ times num__85 \ times num__90 } = num__6000 hence the salary of the officer is rs . num__6000 answer : a <eor> a <eos> |
a |
percent__100.0__6000.0__ |
percent__100.0__6000.0__ |
| what is the largest integral value of ' k ' for which the quadratic equation x num__2 - num__6 x + k = num__0 will have two real and distinct roots ? <o> a ) num__9 <o> b ) num__7 <o> c ) num__3 <o> d ) num__8 <o> e ) num__12 |
explanatory answer any quadratic equation will have real and distinct roots if the discriminant d > num__0 the discriminant ' d ' of a quadratic equation ax num__2 + bx + c = num__0 is given by b num__2 - num__4 ac in this question the value of d = num__62 - num__4 * num__1 * k if d > num__0 then num__36 > num__4 k or k < num__9 . therefore the highest integral value that k can take is num__8 . correct choice is ( d ) <eor> d <eos> |
d |
subtract__6.0__2.0__ divide__36.0__4.0__ add__2.0__6.0__ add__2.0__6.0__ |
subtract__6.0__2.0__ divide__36.0__4.0__ add__2.0__6.0__ add__2.0__6.0__ |
| the factorial expression num__12 ! / num__6 ! is not divisible by which of the following integers ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__13 <o> d ) num__10 <o> e ) num__9 |
num__12 ! / num__6 ! = > num__12 x num__11 x num__10 x num__9 x num__8 x num__7 ( a ) num__3 can divide num__12 or num__9 ( b ) num__5 can divide num__10 ( c ) num__13 none of the multiplicands present ( d ) num__10 can divide num__10 ( e ) num__9 can divide num__9 hence answer will be ( c ) <eor> c <eos> |
c |
subtract__12.0__9.0__ subtract__12.0__7.0__ add__6.0__7.0__ add__6.0__7.0__ |
subtract__12.0__9.0__ subtract__12.0__7.0__ add__6.0__7.0__ add__6.0__7.0__ |
| a train num__800 m long is running at a speed of num__78 km / hr . if it crosses a tunnel in num__1 min then the length of the tunnel is ? <o> a ) num__287 m <o> b ) num__178 m <o> c ) num__500 m <o> d ) num__177 m <o> e ) num__188 m |
speed = num__78 * num__0.277777777778 = num__21.6666666667 m / sec . time = num__1 min = num__60 sec . let the length of the train be x meters . then ( num__800 + x ) / num__60 = num__21.6666666667 x = num__500 m . answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ round__500.0__ |
hour_to_min_conversion__ multiply__1.0__500.0__ |
| the tax on a commodity is diminished by num__40.0 but its consumption is increased by num__25.0 . find the decrease percent in the revenue derived from it ? <o> a ) num__18.0 <o> b ) num__72.0 <o> c ) num__25.0 <o> d ) num__12.0 <o> e ) num__52 % |
num__100 * num__100 = num__10000 num__60 * num__125 = num__7500 num__10000 - - - - - - - num__2500 num__100 - - - - - - - ? = num__25.0 answer : c <eor> c <eos> |
c |
subtract__100.0__40.0__ add__25.0__100.0__ multiply__60.0__125.0__ multiply__25.0__100.0__ divide__2500.0__100.0__ |
subtract__100.0__40.0__ add__25.0__100.0__ multiply__60.0__125.0__ multiply__25.0__100.0__ subtract__125.0__100.0__ |
| the price of commodity p increases by num__40 paise every year while the price of commodity q increases by num__15 paise every year . if in num__2001 the price of commodity p was rs . num__4.20 and that of q was rs . num__6.30 in which year commodity p will cost num__40 paise more than the commodity q ? <o> a ) num__2008 <o> b ) num__2009 <o> c ) num__2010 <o> d ) num__2011 <o> e ) num__2012 |
let the commodity p costs num__40 paise more than the commodity q after n years . price of the commodity p in num__2001 = rs . num__4.20 since the price of the commodity p increases by rs num__0.40 every year price of the commodity p after n years from num__2001 = rs . num__4.20 + ( n × . num__40 ) price of the commodity q in num__2001 = rs . num__6.30 since the price of the commodity q increases by rs num__0.15 every year price of the commodity q after n years from num__2001 = rs . num__6.30 + ( n × . num__15 ) since the commodity p costs rs . num__0.40 more that the commodity q after n years from num__2001 num__4.20 + ( n × . num__40 ) = num__6.30 + ( n × . num__15 ) + num__0.40 = > ( num__40 n - . num__15 n ) = num__6.30 - num__4.20 + num__0.40 = num__2.5 = > . num__25 n = num__2.5 = > n = num__2.5 / . num__25 = num__10 = > commodity p costs rs . num__0.40 more that the commodity q after num__10 years from num__2001 . i . e . in num__2011 answer : d <eor> d <eos> |
d |
reverse__0.4__ subtract__40.0__15.0__ multiply__0.4__25.0__ add__2001.0__10.0__ add__2001.0__10.0__ |
reverse__0.4__ subtract__40.0__15.0__ subtract__25.0__15.0__ add__2001.0__10.0__ add__2001.0__10.0__ |
| if x / num__2 = y / num__5 = z / num__6 then ( x + y + z ) / x is equal to : <o> a ) num__13 <o> b ) num__12 <o> c ) num__7.5 <o> d ) num__6.5 <o> e ) none of these |
explanation : solution : let x / num__2 = y / num__5 = z / num__6 = k = > x = num__2 k y = num__5 k z = num__6 k . . ' . ( x + y + z ) / x = ( num__2 k + num__5 k + num__6 k ) / num__2 k = > num__13 k / num__2 k = num__6.5 answer : d <eor> d <eos> |
d |
divide__13.0__2.0__ divide__13.0__2.0__ |
divide__13.0__2.0__ divide__13.0__2.0__ |
| in an examination the percentage of students qualified to the number of students appeared from school a is num__70.0 . in school b the number of students appeared is num__20.0 more than the students appeared from school a and the number of students qualified from school b is num__50.0 more than the students qualified from school a . what is the percentage of students qualified to the number of students appeared from school b ? <o> a ) num__30.0 <o> b ) num__70.0 <o> c ) num__80.0 <o> d ) num__87.5 <o> e ) none |
solution let number of students appeared from school a = num__100 . then number of students qualified from school a = num__70 . number of students appeared from school b = num__120 . number of students qualified from school b = [ num__1.5 x num__70 ] = num__105 . required percentage = [ num__0.875 x num__100 ] % = num__87.5 answer d <eor> d <eos> |
d |
add__70.0__50.0__ multiply__70.0__1.5__ divide__105.0__120.0__ multiply__0.875__100.0__ multiply__0.875__100.0__ |
add__70.0__50.0__ multiply__70.0__1.5__ divide__105.0__120.0__ multiply__0.875__100.0__ multiply__0.875__100.0__ |
| two goods trains each num__75 m long are running in opposite directions on parallel tracks . their speeds are num__45 km / hr and num__30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? <o> a ) num__2.3 sec <o> b ) num__4.2 sec <o> c ) num__48 sec <o> d ) num__7.2 sec <o> e ) num__9 sec |
relative speed = num__45 + num__30 = num__75 km / hr . num__75 * num__0.277777777778 = num__20.8333333333 m / sec . distance covered = num__75 + num__75 = num__150 m . required time = num__150 * num__0.048 = num__7.2 sec . answer : d <eor> d <eos> |
d |
multiply__0.048__150.0__ round__7.2__ |
multiply__0.048__150.0__ multiply__0.048__150.0__ |
| the average weight of a group of boys is num__30 kg . after a boy of weight num__36 kg joins the group the average weight of the group goes up by num__1 kg . find the number of boys in the group originally ? <o> a ) num__4 <o> b ) num__8 <o> c ) num__5 <o> d ) num__2 <o> e ) num__9 |
let the number off boys in the group originally be x . total weight of the boys = num__30 x after the boy weighing num__36 kg joins the group total weight of boys = num__30 x + num__36 so num__30 x + num__36 = num__31 ( x + num__1 ) = > x = num__5 answer : c <eor> c <eos> |
c |
add__30.0__1.0__ subtract__36.0__31.0__ subtract__36.0__31.0__ |
add__30.0__1.0__ subtract__36.0__31.0__ subtract__36.0__31.0__ |
| a boat running downstream covers a distance of num__16 km in num__2 hours while for covering the same distance upstream it takes num__4 hours . what is the speed of the boat in still water ? <o> a ) num__4 km / hr <o> b ) num__5 km / hr <o> c ) num__6 km / hr <o> d ) num__3 km / hr <o> e ) num__2 km / hr |
rate downstream = num__8.0 kmph = num__8 kmph . rate upstream = num__4.0 kmph = num__4 kmph . speed in still water = num__0.5 ( num__8 + num__4 ) kmph = num__6 kmph . answer : option c <eor> c <eos> |
c |
divide__16.0__2.0__ divide__2.0__4.0__ add__2.0__4.0__ round__6.0__ |
divide__16.0__2.0__ divide__2.0__4.0__ add__2.0__4.0__ add__2.0__4.0__ |
| a motorcyclist goes from bombay to pune a distance of num__192 kms at an average of num__32 kmph speed . another man starts from bombay by car num__2 ½ hours after the first and reaches pune ½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? <o> a ) num__1 : num__2 <o> b ) num__1 : num__5 <o> c ) num__1 : num__3 <o> d ) num__1 : num__1 <o> e ) num__1 : num__7 |
t = num__6.0 = num__6 h t = num__6 - num__3 = num__3 time ratio = num__6 : num__3 = num__2 : num__1 speed ratio = num__1 : num__2 answer : a <eor> a <eos> |
a |
divide__192.0__32.0__ divide__6.0__2.0__ subtract__3.0__2.0__ round__1.0__ |
divide__192.0__32.0__ divide__6.0__2.0__ subtract__3.0__2.0__ subtract__2.0__1.0__ |
| a person is traveling at num__40 km / hr and reached his destiny in num__5 hr find the distance ? <o> a ) a ) num__200 km <o> b ) b ) num__95 km <o> c ) c ) num__135 km <o> d ) d ) num__80 km <o> e ) e ) num__125 km |
speed = num__40 km / hr time = num__5 hr distance = num__40 * num__5 = num__200 km answer is a <eor> a <eos> |
a |
multiply__40.0__5.0__ round__200.0__ |
multiply__40.0__5.0__ multiply__40.0__5.0__ |
| sixty men can stitch num__200 shirts in num__30 days working num__8 hours a day . in how many days can num__45 men stitch num__300 shirts working num__6 hours a day ? <o> a ) num__70 <o> b ) num__90 <o> c ) num__60 <o> d ) num__80 <o> e ) num__10 |
d num__80 we have m num__1 d num__1 h num__1 / w num__1 = m num__2 d num__2 h num__2 / w num__2 ( variation rule ) ( num__60 * num__30 * num__8 ) / num__200 = ( num__45 * d num__2 * num__6 ) / num__300 d num__2 = ( num__60 * num__30 * num__8 * num__300 ) / ( num__200 * num__45 * num__6 ) = > d num__2 = num__80 . <eor> d <eos> |
d |
subtract__8.0__6.0__ hour_to_min_conversion__ round__80.0__ |
subtract__8.0__6.0__ multiply__30.0__2.0__ divide__80.0__1.0__ |
| how many integers from num__40 to num__200 inclusive are divisible by num__3 but not divisible by num__7 ? <o> a ) num__20 <o> b ) num__31 <o> c ) num__35 <o> d ) num__40 <o> e ) num__45 |
we should find # of integers divisible by num__3 but not by num__3 * num__7 = num__21 . # of multiples of num__21 in the range from num__40 to num__200 inclusive is ( num__189 - num__42 ) / num__21 + num__1 = num__8 ; num__53 - num__8 = num__45 . answer : e . <eor> e <eos> |
e |
multiply__3.0__7.0__ add__7.0__1.0__ add__3.0__42.0__ add__3.0__42.0__ |
multiply__3.0__7.0__ add__7.0__1.0__ add__3.0__42.0__ add__3.0__42.0__ |
| the height of two right circular cones are in the ratio num__1 : num__2 and their perimeters of their bases are in the ratio num__3 : num__4 the ratio of their volume is ? <o> a ) num__9 : num__38 <o> b ) num__9 : num__31 <o> c ) num__9 : num__32 <o> d ) num__9 : num__39 <o> e ) num__9 : num__35 |
num__9 : num__32 answer : c <eor> c <eos> |
c |
power__3.0__2.0__ multiply__1.0__9.0__ |
power__3.0__2.0__ multiply__1.0__9.0__ |
| a patient was given a bottle of tablets by the doctor and he was asked to take five tablets in a gap of num__10 minutes . in how much time will he be able to take all the five tablets ? <o> a ) num__1 hour . <o> b ) num__40 min <o> c ) num__50 min <o> d ) none <o> e ) can not be determined |
suppose he takes the first tablet at num__8 : num__00 pm . then the second will be consumed by him at num__8 : num__10 third at num__8 : num__20 fourth at num__8 : num__30 and fifth at num__8 : num__40 . time = num__40 min answer b <eor> b <eos> |
b |
add__10.0__20.0__ add__10.0__30.0__ round__40.0__ |
add__10.0__20.0__ add__10.0__30.0__ round__40.0__ |
| if a choir consists of num__2 boys and num__5 girls in how many ways can the singers be arranged in a row so that all the boys are together ? do not differentiate between arrangements that are obtained by swapping two boys or two girls . <o> a ) num__120 <o> b ) num__30 <o> c ) num__24 <o> d ) num__11 <o> e ) num__6 |
there are num__6 possibilities : bbgggggg gbbggggg ggbbgggg gggbbggg ggggbbgg gggggbbg ggggggbb think of all num__2 boys as a single unit . together with num__5 girls it makes a total of num__6 units . the difference between the arrangements is the position of the boys ( as a single unit ) . so the problem reduces to finding the number of unique patterns generated by changing the position of the boys who can occupy num__1 of num__6 available positions . if the number of available unique positions is num__6 then the number of unique patterns equals num__6 as well . answer : e . <eor> e <eos> |
e |
die_space__ die_space__ |
die_space__ die_space__ |
| if an item that originally sold for z dollars was marked up by x percent and then discounted by y percent which of the following expressions represents the final price of the item ? <o> a ) ( num__10000 z + num__100 z ( x – y ) – xyz ) / num__10000 <o> b ) ( num__10000 z + num__100 z ( y – x ) – xyz ) / num__10000 <o> c ) ( num__100 z ( x – y ) – xyz ) / num__10000 <o> d ) ( num__100 z ( y – x ) – xyz ) / num__10000 <o> e ) num__10000 / ( x – y ) |
z * ( num__1 + ( x / num__100 ) ) * ( num__1 - ( y / num__100 ) ) z * ( num__100 + x / num__100 ) * ( num__100 - y / num__100 ) ( num__100 z + zx ) / num__100 * ( num__100 - y / num__100 ) [ num__10000 z + num__100 zx - num__100 zy - xyz ] / num__10000 [ num__10000 z + num__100 z ( x - y ) - xyz ] / num__10000 choice a <eor> a <eos> |
a |
percent__100.0__10000.0__ |
percent__100.0__10000.0__ |
| what is the rate percent when the simple interest on rs . num__2000 amount to rs . num__320 in num__4 years ? <o> a ) num__4.5 <o> b ) num__4.25 <o> c ) num__4.0 <o> d ) num__4.3 <o> e ) num__4.1 % |
interest for num__1 year = num__80.0 = num__80 interest on rs num__2000 p / a = num__80 interest rate = num__0.04 * num__100 = num__4.0 answer : c <eor> c <eos> |
c |
percent__4.0__2000.0__ percent__4.0__1.0__ percent__4.0__100.0__ |
percent__4.0__2000.0__ percent__4.0__1.0__ percent__4.0__100.0__ |
| a train num__110 m long is running with a speed of num__27 km / hr . in what time will it pass a man who is running at num__6 km / hr in the direction opposite to that in which the train is going ? <o> a ) num__7 sec <o> b ) num__6 sec <o> c ) num__8 sec <o> d ) num__4 sec <o> e ) num__12 sec |
speed of train relative to man = num__27 + num__6 = num__33 km / hr . = num__33 * num__0.277777777778 = num__9.1667 m / sec . time taken to pass the men = num__110 / num__9.1667 = num__12 sec . answer : e <eor> e <eos> |
e |
add__27.0__6.0__ divide__110.0__9.1667__ round__12.0__ |
add__27.0__6.0__ divide__110.0__9.1667__ divide__110.0__9.1667__ |
| the perimeter of a triangle is num__28 cm and the inradius of the triangle is num__2.0 cm . what is the area of the triangle ? <o> a ) num__28 cm num__2 <o> b ) num__27 cm num__2 <o> c ) num__29 cm num__2 <o> d ) num__25 cm num__2 <o> e ) num__35 cm num__2 |
area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = num__2.0 * num__14.0 = num__28 cm num__2 answer : a <eor> a <eos> |
a |
triangle_area__28.0__2.0__ |
multiply__2.0__14.0__ |
| in what time a sum of money double itself at num__3.0 per annum simple interest ? <o> a ) num__36 num__0.333333333333 % <o> b ) num__33 num__0.333333333333 % <o> c ) num__33 num__0.166666666667 % <o> d ) num__63 num__0.333333333333 % <o> e ) num__23 num__0.333333333333 % |
p = ( p * num__3 * r ) / num__100 r = num__33 num__0.333333333333 % answer : b <eor> b <eos> |
b |
percent__33.0__100.0__ |
percent__33.0__100.0__ |
| the distance between two cities a and b is num__1140 km . a train starts from a at num__8 a . m . and travel towards b at num__60 km / hr . another train starts from b at num__9 a . m and travels towards a at num__75 km / hr . at what time do they meet ? <o> a ) num__09 am <o> b ) num__07 am <o> c ) num__11 am <o> d ) num__05 pm <o> e ) num__03 pm |
explanation : suppose they meet x hrs after num__8 a . m then [ distance moved by first in x hrs ] + [ distance moved by second in ( x - num__1 ) hrs ] = num__1140 . therefore num__60 x + num__75 ( x - num__1 ) = num__1140 . = > x = num__9 . so they meet at ( num__8 + num__9 ) i . e num__5 pm answer : d <eor> d <eos> |
d |
subtract__9.0__8.0__ round__5.0__ |
subtract__9.0__8.0__ round__5.0__ |
| car w and car y traveled the same num__80 - mile route . if car w took num__2 hours and car y traveled at an average speed that was num__50 percent faster than the average speed of car w how many hours did it take car y to travel the route ? <o> a ) num__0.666666666667 <o> b ) num__1 <o> c ) num__1.33333333333 <o> d ) num__1.6 <o> e ) num__3 |
the speed of car w is ( distance ) / ( time ) = num__40.0 = num__40 miles per hour . the speed of car y = num__1.5 * num__40 = num__60 miles per hour - - > ( time ) = ( distance ) / ( speed ) = num__1.33333333333 = num__1.33333333333 hours . answer : c . or : to cover the same distance at num__1.5 as fast rate num__0.666666666667 as much time is needed - - > ( time ) * num__0.666666666667 = num__2 * num__0.666666666667 = num__1.33333333333 hours . answer : c . <eor> c <eos> |
c |
divide__80.0__2.0__ hour_to_min_conversion__ divide__80.0__60.0__ subtract__2.0__1.3333__ divide__80.0__60.0__ |
divide__80.0__2.0__ multiply__1.5__40.0__ divide__80.0__60.0__ subtract__2.0__1.3333__ divide__80.0__60.0__ |
| three candidates contested an election and received num__1136 num__7636 and num__11628 votes respectively . what percentage of the total votes did the winning candidate get ? <o> a ) num__56.0 <o> b ) num__57.0 <o> c ) num__54.0 <o> d ) num__53.0 <o> e ) num__52 % |
total number of votes polled = ( num__1136 + num__7636 + num__11628 ) = num__20400 . required percentage = ( num__0.57 * num__100 ) % = num__57.0 answer b ) num__57.0 . <eor> b <eos> |
b |
divide__11628.0__20400.0__ multiply__100.0__0.57__ multiply__100.0__0.57__ |
divide__11628.0__20400.0__ multiply__100.0__0.57__ multiply__100.0__0.57__ |
| a num__11.0 stock yields num__8.0 . the market value of the stock is : <o> a ) rs . num__72 <o> b ) rs . num__137.50 <o> c ) rs . num__112.50 <o> d ) rs . num__116.50 <o> e ) none of these |
solution to obtain rs . num__8 investment = rs . num__100 . to obtain rs . num__11 investment = rs . ( num__12.5 x num__11 ) = rs . num__137.50 ∴ market value of rs . num__100 stock = rs . num__137.50 answer b <eor> b <eos> |
b |
percent__100.0__137.5__ |
percent__100.0__137.5__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__6 seconds . find the length of the train . <o> a ) num__100 <o> b ) num__882 <o> c ) num__772 <o> d ) num__252 <o> e ) num__121 |
speed = num__60 * ( num__0.277777777778 ) m / sec = num__16.6666666667 m / sec length of train ( distance ) = speed * time ( num__16.6666666667 ) * num__6 = num__100 meter . answer : a <eor> a <eos> |
a |
round__100.0__ |
round__100.0__ |
| the lunch menu at a certain restaurant contains num__4 different entrees and num__5 different side dishes . if a meal consists of num__1 entree and num__2 different side dishes how many different meal combinations r could be chosen from this menu ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__40 <o> d ) num__80 <o> e ) num__100 |
r = num__4 c num__1 * num__5 c num__2 = num__4 * ( num__5 * num__4 * num__3 ! ) / ( num__3 ! num__2 ! ) = num__4 * num__10 = num__40 answer - c <eor> c <eos> |
c |
choose__5.0__3.0__ choose__5.0__3.0__ |
choose__5.0__3.0__ choose__5.0__3.0__ |
| num__2 + num__2 + num__2 ² + num__2 ³ . . . + num__2 ^ num__9 <o> a ) num__2 ^ num__9 <o> b ) num__2 ^ num__10 <o> c ) num__2 ^ num__16 <o> d ) num__2 ^ num__35 <o> e ) num__2 ^ num__37 |
num__2 + num__2 = num__2 ^ num__2 num__2 ^ num__2 + num__2 ^ num__2 = ( num__2 ^ num__2 ) * ( num__1 + num__1 ) = num__2 ^ num__3 num__2 ^ num__3 + num__2 ^ num__3 = ( num__2 ^ num__3 ) * ( num__1 + num__1 ) = num__2 ^ num__4 so you can notice the pattern . . . in the end you will have num__2 ^ num__9 + num__2 ^ num__9 which will give you num__2 ^ num__10 answer b <eor> b <eos> |
b |
add__2.0__1.0__ add__1.0__3.0__ add__9.0__1.0__ multiply__2.0__1.0__ |
add__2.0__1.0__ add__1.0__3.0__ add__9.0__1.0__ multiply__2.0__1.0__ |
| if n = num__2.0823 and n * is the decimal obtained by rounding n to the nearest hundredth what is the value of n * – n ? <o> a ) - num__0.0053 <o> b ) - num__0.0023 <o> c ) num__0.0007 <o> d ) num__0.0047 <o> e ) num__0.0153 |
n * = num__2.08 n * - n = num__2.08 - num__2.0823 - num__0.0023 answer : b <eor> b <eos> |
b |
subtract__2.0823__2.08__ subtract__2.0823__2.08__ |
subtract__2.0823__2.08__ subtract__2.0823__2.08__ |
| the ratio of num__2 numbers is num__4 : num__6 and their h . c . f . is num__8 . their l . c . m . is ? <o> a ) num__20 <o> b ) num__24 <o> c ) num__52 <o> d ) num__96 <o> e ) num__60 |
let the numbers be num__4 x and num__6 x their h . c . f . = num__8 so the numbers are num__4 * num__8 num__6 * num__8 = num__3248 l . c . m . = num__96 answer is d <eor> d <eos> |
d |
lcm__2.0__96.0__ |
lcm__2.0__96.0__ |
| a tap can fill a tank in num__6 hours . after half the tank is filled three more similar taps are opened . what is the total time taken to fill the tank completely ? <o> a ) num__3 hr <o> b ) num__3 hr num__45 min <o> c ) num__4 hr num__15 min <o> d ) num__2 hr num__1 min <o> e ) num__1 hr num__45 min |
time taken by one tap to fill the half tank = num__3 hr part filled by the num__4 taps in num__1 hour = num__4 * num__0.166666666667 = num__0.666666666667 remaining part = num__1 - num__0.5 = num__0.5 num__0.666666666667 : num__0.5 : : num__1 : x x = num__0.75 hr = num__45 min total time taken = num__3 hr num__45 min answer is b <eor> b <eos> |
b |
subtract__4.0__3.0__ divide__1.0__6.0__ divide__4.0__6.0__ divide__3.0__6.0__ divide__0.5__0.6667__ round__3.0__ |
subtract__4.0__3.0__ divide__1.0__6.0__ divide__4.0__6.0__ subtract__0.6667__0.1667__ divide__0.5__0.6667__ multiply__6.0__0.5__ |
| in a certain apartment building there are one - bedroom and two - bedroom apartments . the rental prices of the apartment depend on a number of factors but on average two - bedroom apartments have higher rental prices than do one - bedroom apartments . let r be the average rental price for all apartments in the building . if r is $ num__5600 higher than the average rental price for all one - bedroom apartments and if the average rental price for all two - bedroom apartments is $ num__10400 higher that r then what percentage of apartments in the building are two - bedroom apartments ? <o> a ) num__26.0 <o> b ) num__35.0 <o> c ) num__39.0 <o> d ) num__42.0 <o> e ) num__52 % |
ratio of num__2 bedroom apartment : num__1 bedroom apartment = num__5400 : num__104000 - - - - - > num__7 : num__13 let total number of apartments be x no . of num__2 bedroom apartment = ( num__0.35 ) * x percentage of apartments in the building are two - bedroom apartments - - - - > ( num__0.35 ) * num__100 - - - > num__35.0 option b is correct <eor> b <eos> |
b |
multiply__100.0__0.35__ multiply__1.0__35.0__ |
multiply__100.0__0.35__ multiply__1.0__35.0__ |
| at num__1 : num__00 pm a truck left city p and headed toward city q at a constant speed of num__35 km / h . one hour later a car left city q and headed toward city p along the same road at a constant speed of num__40 km / h . if the distance between city p and city q is num__335 km at what time will the truck and the car meet each other ? <o> a ) num__5 : num__00 <o> b ) num__5 : num__30 <o> c ) num__6 : num__00 <o> d ) num__6 : num__30 <o> e ) num__7 : num__00 |
at num__2 : num__00 pm the truck and the car are num__300 km apart . the truck and the car complete a distance of num__75 km each hour . the time it takes to meet is num__4.0 = num__4 hours . they will meet at num__6 : num__00 pm . the answer is c . <eor> c <eos> |
c |
subtract__335.0__35.0__ add__35.0__40.0__ divide__300.0__75.0__ add__2.0__4.0__ round__6.0__ |
subtract__335.0__35.0__ add__35.0__40.0__ divide__300.0__75.0__ add__2.0__4.0__ round__6.0__ |
| a rectangular lawn of dimensions num__80 m * num__60 m has two roads each num__10 m wide running in the middle of the lawn one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . num__5 per sq m ? <o> a ) num__6500 <o> b ) num__2779 <o> c ) num__2779 <o> d ) num__3900 <o> e ) num__2781 |
area = ( l + b â € “ d ) d ( num__80 + num__60 â € “ num__10 ) num__10 = > num__1300 m num__2 num__1300 * num__5 = rs . num__6500 answer : a <eor> a <eos> |
a |
triangle_area__10.0__1300.0__ triangle_area__10.0__1300.0__ |
multiply__5.0__1300.0__ multiply__5.0__1300.0__ |
| what is the area inscribed by the lines y = num__2 x = num__2 y = num__10 - x on an xy - coordinate plane ? <o> a ) a ) num__8 <o> b ) b ) num__10 <o> c ) c ) num__12 <o> d ) d ) num__14 <o> e ) e ) num__18 |
first let ' s graph the lines y = num__2 and x = num__2 at this point we need to find the points where the line y = num__10 - x intersects the other two lines . for the vertical line we know that x = num__2 so we ' ll plug x = num__2 into the equation y = num__10 - x to get y = num__10 - num__2 = num__8 perfect when x = num__2 y = num__8 so one point of intersection is ( num__28 ) for the horizontal line we know that y = num__2 so we ' ll plug y = num__2 into the equation y = num__10 - x to get num__2 = num__10 - x . solve to get : x = num__8 so when y = num__2 x = num__8 so one point of intersection is ( num__82 ) now add these points to our graph and sketch the line y = num__10 - x at this point we can see that we have the following triangle . the base has length num__6 and the height is num__6 area = ( num__0.5 ) ( base ) ( height ) = ( num__0.5 ) ( num__6 ) ( num__6 ) = num__18 answer : e <eor> e <eos> |
e |
subtract__10.0__2.0__ subtract__8.0__2.0__ reverse__2.0__ add__10.0__8.0__ add__10.0__8.0__ |
subtract__10.0__2.0__ subtract__8.0__2.0__ reverse__2.0__ subtract__28.0__10.0__ subtract__28.0__10.0__ |
| what is the ratio whose term differ by num__25 and the measure of which is num__0.285714285714 ? <o> a ) a ) num__32 : num__23 <o> b ) b ) num__16 : num__56 <o> c ) c ) num__71 : num__85 <o> d ) d ) num__32 : num__39 <o> e ) e ) num__10 : num__35 |
let the ratio be x : ( x + num__25 ) then x / ( x + num__25 ) = num__0.285714285714 x = num__10 required ratio = num__10 : num__35 answer is e <eor> e <eos> |
e |
add__25.0__10.0__ subtract__35.0__25.0__ |
add__25.0__10.0__ subtract__35.0__25.0__ |
| a set consists of num__18 numbers all are even or multiple of num__5 . if num__6 numbers are even and num__12 numbers are multiple of num__5 how many numbers is multiple of num__10 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__5 |
{ total } = { even } + { multiple of num__5 } - { both } + { nether } . since { neither } = num__0 ( allare even or multiple of num__5 ) then : num__18 = num__6 + num__12 - { both } + num__0 ; { both } = num__0 ( so num__1 number is both even and multiple of num__5 so it must be a multiple of num__10 ) . answer : a . <eor> a <eos> |
a |
subtract__6.0__5.0__ multiply__18.0__0.0__ |
subtract__6.0__5.0__ multiply__18.0__0.0__ |
| how many seconds will a train num__100 meters long take to cross a bridge num__150 meters long if the speed of the train is num__72 kmph ? <o> a ) num__22 sec <o> b ) num__77 sec <o> c ) num__25 sec <o> d ) num__18 sec <o> e ) num__12.5 sec |
d = num__100 + num__150 = num__250 s = num__72 * num__0.277777777778 = num__20 mps t = num__12.5 = num__12.5 sec answer : e <eor> e <eos> |
e |
add__100.0__150.0__ divide__250.0__20.0__ round__12.5__ |
add__100.0__150.0__ divide__250.0__20.0__ round__12.5__ |
| if the population of a certain country increases at the rate of one person every num__30 seconds by how many persons does the population increase in num__1 hour ? <o> a ) num__100 <o> b ) num__120 <o> c ) num__150 <o> d ) num__180 <o> e ) num__160 |
answer = num__2 * num__60 = num__120 answer is b <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__2.0__60.0__ round__120.0__ |
multiply__30.0__2.0__ multiply__2.0__60.0__ multiply__1.0__120.0__ |
| a tank is filled in num__5 hours by three pipes a b and c . the pipe c is twice as fast as b and b is twice as fast as a . how much time will pipe a alone take to fill the tank ? <o> a ) num__35 hours <o> b ) num__25 hours <o> c ) num__20 hours <o> d ) can not be determined <o> e ) none of these |
let pipe a alone takes to fill the tank = x hours . then pipes b take twice fast as a = x / num__2 and pipes c take twice fast as b = x / num__4 therefore tank filled in num__5 hours by a b c pipes = > num__1 / x + num__2 / x + num__4 / x = num__0.2 = = > num__7 / x = num__0.2 = = > x = num__5 * num__7 x = num__35 hours x = num__35 hrs . answer : a <eor> a <eos> |
a |
subtract__5.0__4.0__ divide__1.0__5.0__ add__5.0__2.0__ multiply__5.0__7.0__ round__35.0__ |
subtract__5.0__4.0__ divide__1.0__5.0__ add__5.0__2.0__ multiply__5.0__7.0__ multiply__5.0__7.0__ |
| a family consists of two grandparents two parents and three grandchildren . the average age of the grandparents is num__67 years that of the parents is num__35 years and that of the grandchildren is num__6 years . what is the average age of family ? <o> a ) num__31 num__0.714285714286 years <o> b ) num__31 <o> c ) num__30 <o> d ) num__33 <o> e ) num__35 |
required average = ( num__67 x num__2 + num__35 x num__2 + num__6 x num__3 ) / ( num__2 + num__2 + num__3 ) = ( num__134 + num__70 + num__18 ) / num__7 = num__31.7142857143 = num__31 num__0.714285714286 years . answer is a <eor> a <eos> |
a |
divide__6.0__2.0__ multiply__67.0__2.0__ add__67.0__3.0__ multiply__6.0__3.0__ round_down__31.7143__ subtract__31.7143__31.0__ round_down__31.7143__ |
divide__6.0__2.0__ multiply__67.0__2.0__ add__67.0__3.0__ multiply__6.0__3.0__ round_down__31.7143__ subtract__31.7143__31.0__ round_down__31.7143__ |
| a car covers a certain distance at aspeed of num__55 kmph in num__6 hours . to cover the same distance in num__2 hrs it must travel at a speed of ? <o> a ) num__567 km / hr <o> b ) num__678 km / hr <o> c ) num__165 km / hr <o> d ) num__789 km / hr <o> e ) num__720 km / hr |
distance = ( num__55 x num__6 ) = num__330 km . speed = distance / time speed = num__165.0 = num__165 kmph answer : c <eor> c <eos> |
c |
multiply__55.0__6.0__ divide__330.0__2.0__ round__165.0__ |
multiply__55.0__6.0__ divide__330.0__2.0__ divide__330.0__2.0__ |
| if m and n are positive integers of j such that m is a factor of n how many positive multiples of m are less than or equal to num__2 n ? <o> a ) num__2 m / n + num__1 <o> b ) num__2 n / m + num__1 <o> c ) num__2 n / ( m + num__1 ) <o> d ) num__2 m / n <o> e ) num__2 n / m |
lets say n = num__10 m = num__5 num__2 n = num__20 . so the answer should be num__4 ( num__4.0 ) lets try to plug in the answers : a - not an integer b - not an integer c - not an integer d - num__1 ( not the answer ) e - num__4 - the answer . ( the only one ) . i would choose e . method num__2 n = m * a ( a is an integer ) so - a = n / m therefore in num__2 n a will be num__2 n / m again - answer is e . <eor> e <eos> |
e |
divide__10.0__2.0__ multiply__2.0__10.0__ divide__20.0__5.0__ subtract__5.0__4.0__ multiply__2.0__1.0__ |
divide__10.0__2.0__ multiply__2.0__10.0__ divide__20.0__5.0__ subtract__5.0__4.0__ multiply__2.0__1.0__ |
| a num__300 m long train crosses a platform in num__45 sec while it crosses a signal pole in num__18 sec . what is the length of the platform ? <o> a ) num__287 <o> b ) num__350 <o> c ) num__828 <o> d ) num__450 <o> e ) num__122 |
speed = num__16.6666666667 = num__16.6666666667 m / sec . let the length of the platform be x meters . then ( x + num__300 ) / num__45 = num__16.6666666667 num__3 x + num__900 = num__2250 = > x = num__450 m . answer : d <eor> d <eos> |
d |
divide__300.0__18.0__ multiply__300.0__3.0__ round__450.0__ |
divide__300.0__18.0__ multiply__300.0__3.0__ round__450.0__ |
| a box contains num__80 balls numbered from num__1 to num__80 . if three balls are selected at random and with replacement from the box what is the probability that the sum of the three numbers on the balls selected from the box will be odd ? <o> a ) num__0.25 <o> b ) num__0.375 <o> c ) num__0.5 <o> d ) num__0.625 <o> e ) num__0.75 |
i do n ' t think order matters in this case because num__2 + num__2 + num__1 = num__2 + num__1 + num__2 my answer is : num__0.25 a <eor> a <eos> |
a |
multiply__1.0__0.25__ |
multiply__1.0__0.25__ |
| two pipes a and b can fill a tank in num__30 and num__60 minutes respectively . if both the pipes are used together then how long will it take to fill the tank ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__30 <o> d ) num__24 <o> e ) num__25 |
part filled by a in num__1 min . = num__0.0333333333333 part filled by b in num__1 min . = num__0.0166666666667 part filled by ( a + b ) in num__1 min . = num__0.0333333333333 + num__0.0166666666667 = num__0.05 . both the pipes can fill the tank in num__20 minutes . answer : b <eor> b <eos> |
b |
divide__1.0__30.0__ divide__1.0__60.0__ add__0.0167__0.0333__ divide__1.0__0.05__ round__20.0__ |
divide__1.0__30.0__ divide__1.0__60.0__ add__0.0167__0.0333__ divide__1.0__0.05__ round__20.0__ |
| if x is a positive odd integer and y is a negative even integer which of the following must be true ? <o> a ) x ^ num__3 + y is a positive odd integer <o> b ) x ^ num__2 + y ^ num__2 is a negative odd integer <o> c ) x ^ num__0 + y ^ num__11 is a negative odd integer <o> d ) x + y is a positive odd integer <o> e ) x + y is a negative odd integer |
anything power num__0 is always num__1 and given that y is even and negative then num__1 + ( - num__2 ) will be negative with odd power values for y . answer : c <eor> c <eos> |
c |
multiply__0.0__1.0__ |
multiply__0.0__1.0__ |
| in what time will a train num__100 metres long cross an electric pole if its speed be num__144 km / hr ? <o> a ) num__2.5 sec <o> b ) num__4.25 sec <o> c ) num__5 sec <o> d ) num__12.5 sec <o> e ) none |
solution speed = ( num__144 x num__0.277777777778 ) m / sec = num__40 m / sec time taken = ( num__2.5 ) sec = num__2.5 sec . answer a <eor> a <eos> |
a |
divide__100.0__40.0__ round__2.5__ |
divide__100.0__40.0__ divide__100.0__40.0__ |
| five people are planning to share equally the cost of a rental car . if one person withdraws from the arrangement and the others share equally the entire cost of the car then the share of each of the remaining persons increased by : <o> a ) num__0.25 <o> b ) num__0.285714285714 <o> c ) num__0.75 <o> d ) num__0.571428571429 <o> e ) none of them |
original share of num__1 person = num__0.2 new share of num__1 person = num__0.25 increase = ( num__0.25 - num__0.2 = num__0.05 ) therefore required fraction = ( num__0.05 ) / ( num__0.2 ) = ( num__0.05 ) x ( num__5.0 ) = num__0.25 answer is a . <eor> a <eos> |
a |
subtract__0.25__0.2__ reverse__0.2__ multiply__0.25__1.0__ |
subtract__0.25__0.2__ reverse__0.2__ divide__0.25__1.0__ |
| a certain roller coaster has num__5 cars and a passenger is equally likely to ride in any num__1 of the num__5 cars each time that passenger rides the roller coaster . if a certain passenger is to ride the roller coaster num__5 times what is the probability that the passenger will ride in each of the num__5 cars ? <o> a ) num__0 <o> b ) num__0.0384 <o> c ) num__0.222222222222 <o> d ) num__0.333333333333 <o> e ) num__1 |
if he is to ride num__5 times and since he can choose any of the num__5 cars each time total number of ways is = num__5 * num__5 * num__5 * num__5 * num__5 = num__3125 now the number of ways if he is to choose a different car each time is = num__5 * num__4 * num__3 * num__2 * num__1 = num__120 so the probability is = num__0.0384 = num__0.0384 answer : b <eor> b <eos> |
b |
subtract__5.0__1.0__ subtract__4.0__1.0__ subtract__5.0__3.0__ divide__120.0__3125.0__ multiply__1.0__0.0384__ |
subtract__5.0__1.0__ subtract__4.0__1.0__ subtract__5.0__3.0__ divide__120.0__3125.0__ multiply__1.0__0.0384__ |
| the average age of num__15 students of a class is num__15 years . out of these the average age of num__5 students is num__14 years and that of the other num__9 students is num__16 years the age of the num__15 th student is <o> a ) num__11 years <o> b ) num__12 years <o> c ) num__13 years <o> d ) num__18 years <o> e ) num__10 years |
age of the num__15 th student = [ num__15 * num__15 - ( num__14 * num__5 + num__16 * num__9 ) ] = ( num__225 - num__214 ) = num__11 years . answer : a <eor> a <eos> |
a |
subtract__16.0__5.0__ subtract__16.0__5.0__ |
subtract__16.0__5.0__ subtract__16.0__5.0__ |
| if $ num__0.30 is the commission for sales of $ num__1000 what percent of the sales amount is the commission ? <o> a ) num__3.0 <o> b ) num__0.3 <o> c ) num__0.03 <o> d ) num__0.003 <o> e ) num__0.0003 % |
% of sales amount of commission = ( commission / total value ) * num__100 = ( num__0.3 / num__1000 ) * num__100 = num__0.03 the answer is c . <eor> c <eos> |
c |
percent__100.0__0.03__ |
percent__100.0__0.03__ |
| a trader bought a car at num__50.0 discount on its original price . he sold it at a num__80.0 increase on the price he bought it . what percent of profit did he make on the original price ? <o> a ) num__17.0 <o> b ) num__62.0 <o> c ) num__12.0 <o> d ) num__19.0 <o> e ) num__71 % |
original price = num__100 cp = num__95 s = num__95 * ( num__1.8 ) = num__112 num__100 - num__171 = num__71.0 answer : e <eor> e <eos> |
e |
percent__100.0__71.0__ |
percent__100.0__71.0__ |
| a train running at the speed of num__90 km / hr crosses a pole in num__9 sec . what is the length of the train ? <o> a ) num__225 m <o> b ) num__150 m <o> c ) num__187 m <o> d ) num__167 m <o> e ) num__197 m |
speed = num__90 * num__0.277777777778 = num__25 m / sec length of the train = speed * time = num__25 * num__9 = num__225 m answer : a <eor> a <eos> |
a |
multiply__9.0__25.0__ round__225.0__ |
multiply__9.0__25.0__ multiply__9.0__25.0__ |
| the difference between the simple interest received from two different sources on rs . num__1500 for num__3 years is rs . num__13.50 . the difference between their rates of interest is ? <o> a ) num__0.8 <o> b ) num__8.3 <o> c ) num__0.3 <o> d ) num__2.3 <o> e ) num__0.4 % |
( num__1500 * r num__1 * num__3 ) / num__100 - ( num__1500 * r num__2 * num__3 ) / num__100 = num__13.50 num__4500 ( r num__1 - r num__2 ) = num__1350 r num__1 - r num__2 = num__0.3 answer : c <eor> c <eos> |
c |
percent__100.0__0.3__ |
percent__100.0__0.3__ |
| a certain telescope increases the visual range at a particular location from num__100 kilometers to num__150 kilometers . by what percent is the visual range increased by using the telescope ? <o> a ) num__30.0 <o> b ) num__33 num__0.5 % <o> c ) num__40.0 <o> d ) num__60.0 <o> e ) num__50 % |
original visual range = num__100 km new visual range = num__150 km percent increase in the visual range by using the telescope = ( num__150 - num__100 ) / num__100 * num__100.0 = num__0.5 * num__100.0 = num__50.0 answer e <eor> e <eos> |
e |
multiply__100.0__0.5__ multiply__100.0__0.5__ |
multiply__100.0__0.5__ multiply__100.0__0.5__ |
| if k is a positive integer which of the following must be divisible by num__21 ? <o> a ) ( k – num__4 ) ( k ) ( k + num__3 ) ( k + num__7 ) <o> b ) ( k – num__4 ) ( k – num__2 ) ( k + num__3 ) ( k + num__5 ) <o> c ) ( k – num__2 ) ( k + num__3 ) ( k + num__5 ) ( k + num__6 ) <o> d ) ( k + num__1 ) ( k + num__3 ) ( k + num__5 ) ( k + num__7 ) <o> e ) ( k – num__3 ) ( k + num__1 ) ( k + num__4 ) ( k + num__6 ) |
num__24 = num__8 * num__3 . note that the product of two consecutive even integers is always divisible by num__8 ( since one of them is divisible by num__4 and another by num__2 ) . only option b offers two consecutive even numbers for any integer value of k : k - num__4 and k - num__2 if k = even or k + num__3 and k + num__5 if k = odd . also from the following num__3 consecutive integers : ( k - num__4 ) ( k - num__3 ) ( k - num__2 ) one must be divisible by num__3 if it ' s not k - num__4 or k - num__2 then it must be k - num__3 ( if it ' s k - num__4 or k - num__2 option b is divisible by num__3 right away ) . but if it ' s k - num__3 then ( k - num__3 ) + num__6 = k + num__3 must also be divisible by num__3 . so option b : ( k – num__4 ) ( k – num__2 ) ( k + num__3 ) ( k + num__5 ) is divisible by num__8 and num__3 in any case . answer : a . <eor> a <eos> |
a |
subtract__24.0__21.0__ divide__8.0__4.0__ add__2.0__3.0__ multiply__2.0__3.0__ subtract__6.0__2.0__ |
subtract__24.0__21.0__ divide__8.0__4.0__ add__2.0__3.0__ multiply__2.0__3.0__ subtract__6.0__2.0__ |
| which expression is the greatest <o> a ) num__0.998464844949 <o> b ) num__0.998555330829 <o> c ) num__0.998609566185 <o> d ) num__0.998507908087 <o> e ) num__0.998669505056 |
options can be re - written as ( x - num__5 ) x = > num__1 - ( num__5 / x ) a ) num__1 - ( num__0.00153515505066 ) b ) num__1 - ( num__0.00144466917076 ) c ) num__1 - ( num__0.00139043381535 ) d ) num__1 - ( num__0.00149209191286 ) e ) num__1 - ( num__0.00133049494412 ) to get the largest among these second half should be the least and so denominator to be largest . hence ' e ' . <eor> e <eos> |
e |
subtract__1.0__0.0013__ |
subtract__1.0__0.0013__ |
| the sum of first three terms of the geometric sequence is num__1.08333333333 and their product is - num__1 . find the common ratio <o> a ) num__0.75 <o> b ) - num__0.75 <o> c ) num__1.33333333333 <o> d ) num__0.5 <o> e ) num__0.125 |
we make the first three terms is of a / r a ar whose sum = a ( num__1 / r + num__1 + r ) = num__1.08333333333 their product = a num__3 = - num__1 hence ( num__3 r + num__4 ) ( num__4 r + num__3 ) = num__0 r = - num__0.75 or - num__1.33333333333 answer b - num__0.75 <eor> b <eos> |
b |
add__1.0__3.0__ divide__3.0__4.0__ reverse__0.75__ reverse__1.3333__ |
add__1.0__3.0__ divide__3.0__4.0__ reverse__0.75__ reverse__1.3333__ |
| thefactor countof an integer n greater than num__1 is the number of distinct prime factors of n . for example the factor count of num__36 is num__2 since num__2 and num__3 are the distinct prime factors of num__36 = num__2 x num__2 x num__3 x num__3 . for which of the following integers is the factor count greatest r ? <o> a ) num__60 <o> b ) num__61 <o> c ) num__62 <o> d ) num__63 <o> e ) num__64 |
num__60 = num__2 * num__2 * num__3 * num__5 factor count = num__3 num__61 = prime number num__62 = num__2 * num__31 factor count = num__2 num__63 = num__3 * num__3 * num__7 factor count = num__2 num__64 = num__2 * num__2 * num__2 * num__2 * num__2 * num__2 factor count = num__1 = r answer is a . <eor> a <eos> |
a |
add__2.0__3.0__ add__1.0__60.0__ add__1.0__61.0__ subtract__36.0__5.0__ add__1.0__62.0__ add__2.0__5.0__ add__1.0__63.0__ multiply__1.0__60.0__ |
add__2.0__3.0__ add__1.0__60.0__ add__1.0__61.0__ subtract__36.0__5.0__ add__1.0__62.0__ add__2.0__5.0__ add__1.0__63.0__ multiply__1.0__60.0__ |
| what is the simplified result of following the steps below in order ? ( num__1 ) add num__5 y to num__2 f ( num__2 ) multiply the sum by num__3 ( num__3 ) subtract f + y from the product <o> a ) num__5 f + num__14 y <o> b ) num__5 x + num__16 y <o> c ) num__5 x + num__5 y <o> d ) num__6 x + num__4 y <o> e ) num__3 x + num__12 y |
num__3 ( num__5 y + num__2 f ) - f - y = num__14 y + num__5 f ' a ' is the answer <eor> a <eos> |
a |
multiply__1.0__5.0__ |
add__2.0__3.0__ |
| last year mrs . long received $ num__160 in dividends on her shares of company x stock all of which she had held for the entire year . if she had had num__12 more shares of the stock last year she would have received $ num__15 more in total annual dividends . how many shares of the stock did she have last year ? <o> a ) num__128 <o> b ) num__140 <o> c ) num__172 <o> d ) num__175 <o> e ) num__200 |
a num__1 . realize that if there were num__12 more shares and num__15 more dollars the shares are worth more than a dollar a share so there would have to be less than num__160 . cancel out c d e . num__2 . now there are at least three ways of doing it all of which depend on who you are and how you think . i ' ll put as many below as i can think of - i think they ' re all fine and expediant . remember - if you ' re spending time looking for the most sophisticated way you could be just wasting time especially if you know how to do it algebraically . method num__1 : algebra num__160 / x = num__175 / ( x + num__12 ) that ' s because each stock is worth the same so if you divide the num__160 by the number of shares or num__175 by the number of shares plus num__12 you get the same answer . solve for x = num__128 . method num__2 : num__12 more shares cost $ num__15 so that means each one cost $ num__1.25 . so num__160 / num__1.25 . to do this quickly convert num__1.25 into num__1.25 ( i love fractions ! ) now you ' ve got num__160 / ( num__1.25 ) = num__160 x num__0.8 = num__32 x num__4 = num__128 . answer is a <eor> a <eos> |
a |
add__160.0__15.0__ divide__160.0__128.0__ reverse__1.25__ subtract__160.0__128.0__ divide__128.0__32.0__ multiply__160.0__0.8__ |
add__160.0__15.0__ divide__160.0__128.0__ reverse__1.25__ subtract__160.0__128.0__ divide__128.0__32.0__ divide__160.0__1.25__ |
| what profit percent is made by selling an article at a certain price if by selling at num__0.666666666667 rd of that price there would be a loss of num__15.0 ? <o> a ) num__20.0 <o> b ) num__27.5 <o> c ) num__10.0 <o> d ) num__80.0 <o> e ) num__90 % |
sp num__2 = num__0.666666666667 sp num__1 cp = num__100 sp num__2 = num__85 num__0.666666666667 sp num__1 = num__85 sp num__1 = num__127.5 num__100 - - - num__27.5 = > num__27.5 answer : b <eor> b <eos> |
b |
percent__100.0__27.5__ |
percent__100.0__27.5__ |
| find the value of y from ( num__12 ) ^ num__3 x num__6 ^ num__4 ÷ num__432 = y ? <o> a ) num__1234 <o> b ) num__2343 <o> c ) num__4546 <o> d ) num__3435 <o> e ) num__5184 |
given exp . = ( num__12 ) num__3 x num__64 = ( num__12 ) num__3 x num__64 = ( num__12 ) num__2 x num__62 = ( num__72 ) num__2 = num__5184 num__432 num__12 x num__62 e <eor> e <eos> |
e |
divide__12.0__6.0__ subtract__64.0__2.0__ multiply__12.0__6.0__ multiply__12.0__432.0__ multiply__12.0__432.0__ |
divide__12.0__6.0__ subtract__64.0__2.0__ multiply__12.0__6.0__ multiply__12.0__432.0__ multiply__12.0__432.0__ |
| by selling an article at rs . num__600 a profit of num__25.0 is made . find its cost price ? <o> a ) rs . num__350 <o> b ) rs . num__400 <o> c ) rs . num__573 <o> d ) rs . num__480 <o> e ) rs . num__580 |
sp = num__600 cp = ( sp ) * [ num__100 / ( num__100 + p ) ] = num__600 * [ num__100 / ( num__100 + num__25 ) ] = num__600 * [ num__0.8 ] = rs . num__480 answer : d <eor> d <eos> |
d |
percent__100.0__480.0__ |
percent__100.0__480.0__ |
| a machine working at a constant rate manufactures num__18 dies in num__25 minutes . how many dies does it make in num__1 hr num__15 min ? <o> a ) num__55 <o> b ) num__53 <o> c ) num__54 <o> d ) num__52 <o> e ) num__50 |
change num__1 hr num__15 min to num__75 min . for this we need to set up a simple proportion of dies per time num__0.72 = s / num__75 the absolutely worst thing you could do at this point in the problem is to cross - multiply . that would be a supremely unstrategic move . we can cancel the common factor of num__25 in the two denominators . num__18.0 = s / num__3 s = num__3 * num__18 s = num__54 the machine would be num__54 dies in num__1 hr num__15 min . answer : c <eor> c <eos> |
c |
divide__18.0__25.0__ subtract__18.0__15.0__ multiply__18.0__3.0__ round__54.0__ |
divide__18.0__25.0__ subtract__18.0__15.0__ multiply__18.0__3.0__ multiply__18.0__3.0__ |
| if it costs d dollars to make the first num__100 copies of a poster and e dollars for each poster after that what is the total cost of num__125 posters ? <o> a ) num__25 d + num__100 e <o> b ) num__100 d + num__25 e <o> c ) num__125 de <o> d ) d + num__25 e <o> e ) num__125 / ( de ) |
dollars needed for first num__100 copies = d dollars needed for each successive copy = e cost of num__125 copies = cost of num__100 copies + cost of num__25 copies = d + num__25 e correct option : d <eor> d <eos> |
d |
subtract__125.0__100.0__ subtract__125.0__100.0__ |
subtract__125.0__100.0__ subtract__125.0__100.0__ |
| if the sides of a cube are in the ratio num__4 : num__3 . what is the ratio of their diagonals ? <o> a ) num__4 : num__3 <o> b ) num__16 : num__9 <o> c ) num__64 : num__27 <o> d ) num__3 : num__4 <o> e ) num__3 : num__8 |
a num__1 : a num__2 = num__4 : num__3 d num__1 : d num__2 = num__4 : num__3 answer : a <eor> a <eos> |
a |
triangle_area__4.0__1.0__ square_perimeter__1.0__ |
triangle_area__4.0__1.0__ square_perimeter__1.0__ |
| rodrick mixes a martini that has a volume of ' n ' ounces having num__38.0 vermouth and num__60.0 gin by volume . he wants to change it so that the martini is num__25.0 vermouth by volume . how many ounces of gin must he add ? <o> a ) n / num__6 <o> b ) n / num__3 <o> c ) num__13 n / num__25 <o> d ) num__5 n / num__6 <o> e ) num__8 n / num__5 |
total v g num__1 ounce num__0.38 num__0.6 n ounce num__0.38 n num__0.6 n - - - - - - - - - - - - - initial expression lets say g ounces of gin is added to this mixture n + g num__0.38 n num__0.6 n + g - - - - - - - - - - - - - - final expression given that after adding g ounces of gin v should become num__25.0 of the total volume . = > volume of v / total volume = num__0.25 = > num__0.38 n / n + g = num__0.25 = > num__1.52 n = n + g = > g = num__3 n / num__5 answer is c . note that after we add pure gin the volume of vermouth will remain the same . based on this set the equation : num__0.38 n = num__0.25 ( n + g ) - - > g = num__13 n / num__25 answer : c . <eor> c <eos> |
c |
divide__38.0__25.0__ divide__3.0__0.6__ subtract__38.0__25.0__ subtract__38.0__25.0__ |
divide__38.0__25.0__ divide__3.0__0.6__ subtract__38.0__25.0__ subtract__38.0__25.0__ |
| the radius of a wheel is num__22.4 cm . what is the distance covered by the wheel in making num__1250 resolutions ? <o> a ) num__794 m <o> b ) num__704 m <o> c ) num__454 m <o> d ) num__1860 m <o> e ) num__1760 m |
in one resolution the distance covered by the wheel is its own circumference . distance covered in num__1250 resolutions . = num__1250 * num__2 * num__3.14285714286 * num__22.4 = num__176000 cm = num__1760 m answer : e <eor> e <eos> |
e |
round__1760.0__ |
round__1760.0__ |
| a man walking at a constant rate of num__9 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of num__15 miles per hour . the woman stops to wait for the man num__3 minutes after passing him while the man continues to walk at his constant rate . how many minutes must the woman wait until the man catches up ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
when the woman passes the man they are aligned ( m and w ) . they are moving in the same direction . after num__5 minutes the woman ( w ) will be ahead the man ( m ) : m - - - - - - m - - - - - - - - - - - - - - - w w in the num__5 minutes after passing the man the woman walks the distance mw = ww which is num__3 * num__0.25 = num__0.75 miles and the man walks the distance mm which is num__3 * num__0.15 = num__0.45 mile . the difference of num__0.75 - num__0.45 = num__0.3 miles ( mw ) will be covered by the man in ( num__0.3 ) / num__9 = num__0.0333333333333 of an hour which is num__2 minutes . answer b . <eor> b <eos> |
b |
divide__15.0__3.0__ multiply__3.0__0.25__ divide__0.75__5.0__ multiply__3.0__0.15__ subtract__0.75__0.45__ divide__0.3__9.0__ subtract__5.0__3.0__ round__2.0__ |
divide__15.0__3.0__ multiply__3.0__0.25__ divide__0.75__5.0__ multiply__3.0__0.15__ subtract__0.75__0.45__ divide__0.3__9.0__ subtract__5.0__3.0__ subtract__5.0__3.0__ |
| let s be the set of all positive integers n such that n ^ num__2 is a multiple of both num__24 and num__108 . which of the following integers are divisors of every integer n in s ? <o> a ) num__12 <o> b ) num__24 <o> c ) num__120 <o> d ) num__72 <o> e ) num__182 |
num__4 = num__2 ^ num__3 * num__3 ; num__108 = num__2 ^ num__2 * num__3 ^ num__3 . the smallest perfect square ( n ^ num__2 ) which is a multiple of both num__24 = num__2 ^ num__3 * num__3 and num__108 = num__2 ^ num__2 * num__3 ^ num__3 is num__2 ^ num__4 * num__3 ^ num__4 thus the smallest n is num__2 ^ num__2 * num__3 ^ num__2 = num__36 . so only num__12 is a divisor of all integers in s . answer : a . <eor> a <eos> |
a |
divide__108.0__3.0__ divide__24.0__2.0__ divide__24.0__2.0__ |
divide__108.0__3.0__ multiply__3.0__4.0__ multiply__3.0__4.0__ |
| an express traveled at an average speed of num__100 km / hr stopping for num__4 min after every num__60 km . how long did it take to reach its destination num__600 km from the starting point ? <o> a ) num__8 hrs num__29 min <o> b ) num__6 hrs num__28 min <o> c ) num__2 hrs num__28 min <o> d ) num__6 hrs num__28 min <o> e ) num__6 hrs num__36 min |
explanation : time taken to cover num__600 km = num__6.0 = num__6 hrs . number of stoppages = num__10.0 - num__1 = num__9 total time of stoppages = num__4 x num__9 = num__36 min hence total time taken = num__6 hrs num__36 min . answer : e <eor> e <eos> |
e |
divide__600.0__100.0__ add__4.0__6.0__ subtract__10.0__1.0__ multiply__4.0__9.0__ divide__60.0__10.0__ |
divide__600.0__100.0__ add__4.0__6.0__ subtract__10.0__1.0__ multiply__4.0__9.0__ subtract__10.0__4.0__ |
| a machine working at a constant rate manufactures num__20 candles in num__40 minutes . how many candles does it make in num__1 hr num__20 min ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__40 <o> d ) num__45 <o> e ) num__50 |
change num__1 hr num__20 min to num__80 min . for this we need to set up a simple proportion of staplers per time num__0.5 = s / num__80 the absolutely worst thing you could do at this point in the problem is to cross - multiply . that would be a supremely unstrategic move . instead cancel before you multiply . for what we can and can ’ t cancel in a proportion seethis post . we can cancel the factor of num__10 in the num__20 and num__40 . num__0.5 = s / num__80 num__0.5 = s / num__80 ; num__0.5 * num__2 = s / num__80 * num__2 num__1.0 = s / num__40 now that the fraction is entirely simplified we can cross - multiply . s = num__1 * num__40 = num__40 the machine would be num__40 candles in num__1 hr num__20 min . answer : c . <eor> c <eos> |
c |
divide__20.0__40.0__ multiply__20.0__0.5__ divide__20.0__10.0__ round__40.0__ |
divide__20.0__40.0__ multiply__20.0__0.5__ divide__20.0__10.0__ divide__20.0__0.5__ |
| when num__52416 is divided by num__312 the quotient is num__168 . what will be the quotient when num__52.416 is divided by num__0.168 ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__4 <o> d ) num__3 <o> e ) num__8 |
for the num__1 st no . there are num__2 digits after decimal for the num__2 nd no . there are num__4 digits after decimal total no . of decimals = num__6 req . no . of digits = ( n - num__1 ) = ( num__6 - num__1 ) = num__5 answer : a <eor> a <eos> |
a |
add__2.0__4.0__ add__1.0__4.0__ add__1.0__4.0__ |
add__2.0__4.0__ subtract__6.0__1.0__ subtract__6.0__1.0__ |
| the average weight of num__5 person ' s increases by num__5.5 kg when a new person comes in place of one of them weighing num__68 kg . what might be the weight of the new person ? <o> a ) num__60 kg <o> b ) num__95.5 kg <o> c ) num__80 kg <o> d ) num__85 kg <o> e ) num__90 kg |
total weight increased = ( num__5 x num__5.5 ) kg = num__27.5 kg . weight of new person = ( num__68 + num__27.5 ) kg = num__95.5 kg option b <eor> b <eos> |
b |
multiply__5.0__5.5__ add__68.0__27.5__ add__68.0__27.5__ |
multiply__5.0__5.5__ add__68.0__27.5__ add__68.0__27.5__ |
| the ratio of the radius of two circles is num__1 : num__3 and then the ratio of their areas is ? <o> a ) num__1 : num__7 <o> b ) num__2 : num__9 <o> c ) num__1 : num__9 <o> d ) num__3 : num__7 <o> e ) num__3 : num__4 |
r num__1 : r num__2 = num__1 : num__3 Π r num__12 : Π r num__22 r num__12 : r num__22 = num__1 : num__9 answer : c <eor> c <eos> |
c |
square_perimeter__3.0__ surface_rectangular_prism__1.0__3.0__2.0__ power__3.0__2.0__ volume_cube__1.0__ |
square_perimeter__3.0__ surface_rectangular_prism__1.0__3.0__2.0__ power__3.0__2.0__ volume_cube__1.0__ |
| how many zeros does num__50 ! end with ? <o> a ) num__20 <o> b ) num__24 <o> c ) num__25 <o> d ) num__12 <o> e ) num__32 |
according to above num__50 ! has num__10.0 + num__2.0 = num__10 + num__2 = num__12 trailing zeros . answer : d . <eor> d <eos> |
d |
add__2.0__10.0__ add__2.0__10.0__ |
add__2.0__10.0__ add__2.0__10.0__ |
| a train running at a speed of num__36 km / h passes an electric pole in num__15 seconds . in how many seconds will the whole train pass a num__370 - meter long platform ? <o> a ) num__46 <o> b ) num__48 <o> c ) num__50 <o> d ) num__52 <o> e ) num__54 |
let the length of the train be x meters . when a train crosses an electric pole the distance covered is its own length x . speed = num__36 km / h = num__36000 m / num__3600 s = num__10 m / s x = num__15 * num__10 = num__150 m . the time taken to pass the platform = ( num__150 + num__370 ) / num__10 = num__52 seconds the answer is d . <eor> d <eos> |
d |
divide__36000.0__3600.0__ multiply__15.0__10.0__ round__52.0__ |
divide__36000.0__3600.0__ multiply__15.0__10.0__ round__52.0__ |
| a bike covers a certain distance at the speed of num__67 km / h in num__8 hrs . if the bike was to cover the same distance in approximately num__6 hrs at what approximate speed should the bike travel ? <o> a ) num__89.33 km / h <o> b ) num__85.34 km / h <o> c ) num__87.67 km / h <o> d ) num__90.25 km / h <o> e ) num__90.64 km / h |
ans . ( a ) sol . total distance = num__67 × num__8 = num__536 km now speed = num__89.3333333333 = num__89.33 km / h <eor> a <eos> |
a |
multiply__67.0__8.0__ divide__536.0__6.0__ round__89.3333__ round__89.3333__ |
multiply__67.0__8.0__ divide__536.0__6.0__ round__89.3333__ round__89.3333__ |
| a school currently maintains a fixed number of students per class . if the ratio of students per class were to be increased by num__1 num__10 fewer classes would be run for a total of num__120 students . what is the current ratio w of students per class ? <o> a ) w = num__3 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__12 |
another way to look at the problem . . . since the total is num__120 ratio * classes = r * c = num__120 . . . . . ( i ) we are looking where ratio increases by num__1 and # of classes decreases by num__10 = ( r + num__1 ) ( c - num__10 ) = rc + c - num__10 r - num__10 = num__120 . . . . ( ii ) ( ii ) - ( i ) . . . . c = num__10 r + num__10 = num__10 ( r + num__1 ) . . . . . . . so # of classes has to be multiple of num__10 as rc = num__120 . . . . num__10 ( r + num__1 ) * r = num__120 . . . . . . . . . . . . . . . . . . . r ( r + num__1 ) = num__12 . . so num__12 is a multiple of consecutive numbers only num__3 * num__4 fits in . . . . . and r = num__3 a <eor> a <eos> |
a |
divide__120.0__10.0__ add__1.0__3.0__ multiply__1.0__3.0__ |
divide__120.0__10.0__ add__1.0__3.0__ multiply__1.0__3.0__ |
| the true discount on rs . num__2562 due num__4 months hence is rs . num__122 . the rate percent is : <o> a ) num__18.0 <o> b ) num__12.0 <o> c ) num__15.0 <o> d ) num__25.0 <o> e ) num__14 % |
explanation : ans . c . num__15.0 the true discount on rs . num__2562 due num__4 months hence is rs . num__122 . therefore present worth is = num__2562 – num__122 = num__2440 . it means that rs . num__122 is interest on rs . num__2440 for num__4 months . therefore rate percent is = ( num__122 × num__100 × num__12 ) / ( num__2440 × num__4 ) = num__15.0 answer : c <eor> c <eos> |
c |
percent__100.0__15.0__ |
percent__100.0__15.0__ |
| a father said to his son ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is num__48 years now the son ' s age five years back was ? <o> a ) num__18 years <o> b ) num__19 years <o> c ) num__20 years <o> d ) num__21 years <o> e ) num__22 years |
let the son ' s present age be x years . then ( num__48 - x ) = x num__2 x = num__48 . x = num__24 . son ' s age num__5 years back ( num__24 - num__5 ) = num__19 years . b ) <eor> b <eos> |
b |
divide__48.0__2.0__ subtract__24.0__5.0__ subtract__24.0__5.0__ |
divide__48.0__2.0__ subtract__24.0__5.0__ subtract__24.0__5.0__ |
| what will be the result of num__6.0 + num__2.0 - num__3 ^ num__5 + num__8 all this multiplied by num__5 ? <o> a ) num__18 <o> b ) num__98 <o> c ) num__5 <o> d ) num__9 <o> e ) num__0 |
the division ( e . g . num__3.0 ) is done first and then the sum ( e . g . num__6 + num__2 ) and subtraction ( e . g . num__2 - num__15 ) at the end you multiply the expression by num__5 and the answer is num__5 option c . <eor> c <eos> |
c |
multiply__3.0__5.0__ add__2.0__3.0__ |
multiply__3.0__5.0__ add__2.0__3.0__ |
| the total of the ages of amar akbar and anthony is num__80 years . what was the total of their ages three years ago ? <o> a ) num__71 <o> b ) num__44 <o> c ) num__66 <o> d ) num__16 <o> e ) num__18 |
explanation : required sum = ( num__80 - num__3 x num__3 ) years = ( num__80 - num__9 ) years = num__71 years . answer : a <eor> a <eos> |
a |
subtract__80.0__9.0__ subtract__80.0__9.0__ |
subtract__80.0__9.0__ subtract__80.0__9.0__ |
| a train num__800 meter long is running with a speed of num__78 km / hr . it crosses a tunnel in num__1 minute . what is the length of the tunnel ( in meters ) ? <o> a ) num__500 <o> b ) num__400 <o> c ) num__300 <o> d ) num__200 <o> e ) num__250 |
distance = num__800 + x meter where x is the length of the tunnel time = num__1 minute = num__60 seconds speed = num__78 km / hr = num__78 × num__0.277777777778 m / s = num__21.6666666667 = num__21.6666666667 m / s distance / time = speed ( num__800 + x ) / num__60 = num__21.6666666667 = > num__800 + x = num__20 × num__65 = num__1300 = > x = num__1300 - num__800 = num__500 meter answer is a . <eor> a <eos> |
a |
hour_to_min_conversion__ multiply__65.0__20.0__ subtract__1300.0__800.0__ round__500.0__ |
hour_to_min_conversion__ multiply__65.0__20.0__ subtract__1300.0__800.0__ divide__500.0__1.0__ |
| for a bake sale simon baked num__2 n more pies than theresa . theresa baked half as many pies as roger who baked num__1313 n pies . no other pies were baked for the sale . what fraction of the total pies for sale did roger bake ? <o> a ) num__0.0625 <o> b ) num__0.125 <o> c ) num__0.1875 <o> d ) num__0.375 <o> e ) num__0.8125 |
s : simon t : theresa r : roger let theresa baked a . as per question we have s = a + num__2 n r = num__2 a = n / num__3 or a = n / num__6 now total = a + ( a + num__2 n ) + num__2 a = num__8 n / num__3 fraction ( n / num__3 ) / ( num__8 n / num__3 ) = num__0.125 answer : b <eor> b <eos> |
b |
multiply__2.0__3.0__ add__2.0__6.0__ reverse__8.0__ reverse__8.0__ |
multiply__2.0__3.0__ add__2.0__6.0__ reverse__8.0__ reverse__8.0__ |
| in assembling a bluetooth device a factory uses one of two kinds of modules . one module costs $ num__15 and the other one that is cheaper costs $ num__10 . the factory holds a $ num__140 worth stock of num__13 modules . how many of the modules in the stock are of the cheaper kind ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__30 <o> d ) num__11 <o> e ) num__35 |
so the number of $ num__10 modules must be num__11 so that the leftover num__2 modules are of $ num__15 which will give a total value $ num__140 . num__11 * num__10 + num__2 * num__15 = num__110 + num__30 = num__140 answer : d <eor> d <eos> |
d |
subtract__15.0__13.0__ multiply__10.0__11.0__ lcm__15.0__10.0__ subtract__13.0__2.0__ |
subtract__15.0__13.0__ multiply__10.0__11.0__ multiply__15.0__2.0__ subtract__13.0__2.0__ |
| a contractor undertakes to built a walls in num__50 days . he employs num__30 peoples for the same . however after num__25 days he finds that only num__40.0 of the work is complete . how many more man need to be employed to complete the work in time ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__45 <o> d ) num__20 <o> e ) none of these |
num__30 men complete num__0.4 work in num__25 days . applying the work rule m num__1 × d num__1 × w num__2 = m num__2 × d num__2 × w num__1 we have num__30 × num__25 × num__0.6 = m num__2 × num__25 × num__0.4 or m num__2 = num__30 × num__25 × num__0.6 / num__25 × num__0.4 = num__45 men answerc <eor> c <eos> |
c |
divide__50.0__25.0__ km_to_mile_conversion__ round__45.0__ |
divide__50.0__25.0__ divide__30.0__50.0__ divide__45.0__1.0__ |
| every day the value of a stock rises by exactly two dollars in the morning and then falls by exactly one dollar in the afternoon . if the stock ' s value at the start of the rst day is $ num__100 on what day will the stock ' s value rst reach $ num__200 ? <o> a ) num__50 <o> b ) num__99 <o> c ) num__100 <o> d ) num__101 <o> e ) num__200 |
note that each day the stock is worth $ num__1 more than on the previous day and that on the rst day it rises to $ num__102 and then falls to $ num__101 . therefore on the nth day the stock ' s value rises to n + num__101 dollars in the morning and falls to n + num__100 dollars in the afternoon . the solution is therefore the smallest n for which n + num__101 = num__200 or n = num__99 . in particular the stock will be $ num__200 in the middle of day num__99 although not at the end of this day . correct answer b <eor> b <eos> |
b |
add__100.0__1.0__ subtract__100.0__1.0__ round__99.0__ |
add__100.0__1.0__ subtract__100.0__1.0__ round__99.0__ |
| a person purchased a tv set for rs . num__16000 and a dvd player for rs . num__6250 . he sold both the items together for rs . num__31150 . what percentage of profit did he make ? <o> a ) num__323 <o> b ) num__88 <o> c ) num__40 <o> d ) num__99 <o> e ) num__81 |
the total cp = rs . num__16000 + rs . num__6250 = rs . num__22250 and sp = rs . num__31150 profit ( % ) = ( num__31150 - num__22250 ) / num__22250 * num__100 = num__40.0 answer : c <eor> c <eos> |
c |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| the ratio of two numbers is num__5 : num__9 . if each number is decreased by num__5 the ratio becomes num__5 : num__11 . find the numbers . <o> a ) num__30 num__19 <o> b ) num__21 num__37 <o> c ) num__15 num__34 <o> d ) num__15 num__27 <o> e ) none of these |
explanation : let the two numbers be num__5 x and num__9 x . ( num__5 x - num__5 ) / ( num__9 x - num__5 ) = num__5 : num__11 ( num__5 x - num__5 ) * num__11 = ( num__9 x - num__5 ) * num__5 num__55 x – num__55 = num__45 x – num__25 num__10 x = num__30 x = num__3 therefore the numbers are num__15 and num__27 . answer : d <eor> d <eos> |
d |
multiply__5.0__11.0__ multiply__5.0__9.0__ subtract__55.0__45.0__ add__5.0__25.0__ divide__30.0__10.0__ multiply__5.0__3.0__ multiply__9.0__3.0__ multiply__5.0__3.0__ |
multiply__5.0__11.0__ multiply__5.0__9.0__ subtract__55.0__45.0__ subtract__55.0__25.0__ divide__30.0__10.0__ multiply__5.0__3.0__ multiply__9.0__3.0__ multiply__5.0__3.0__ |
| a number is increased by num__20.0 and then decreased by num__20.0 . find the net increase or decrease per cent . <o> a ) num__9.0 <o> b ) num__8.0 <o> c ) num__7.0 <o> d ) num__3.0 <o> e ) num__4 % |
let the number be num__100 . increase in the number = num__20.0 = num__20.0 of num__100 = ( num__0.2 Ã — num__100 ) = num__20 therefore increased number = num__100 + num__20 = num__120 this number is decreased by num__20.0 therefore decrease in number = num__20.0 of num__120 = ( num__0.2 Ã — num__120 ) = num__24.0 = num__24 therefore new number = num__120 - num__24 = num__96 thus net decreases = num__100 - num__96 = num__4 hence net percentage decrease = ( num__0.04 Ã — num__100 ) % = ( num__4.0 ) % = num__4.0 answer : e <eor> e <eos> |
e |
divide__20.0__100.0__ add__20.0__100.0__ multiply__0.2__120.0__ subtract__120.0__24.0__ multiply__20.0__0.2__ divide__4.0__100.0__ multiply__20.0__0.2__ |
divide__20.0__100.0__ add__20.0__100.0__ multiply__0.2__120.0__ subtract__120.0__24.0__ subtract__100.0__96.0__ divide__4.0__100.0__ subtract__100.0__96.0__ |
| in an election only two candidates contested . a candidate secured num__70.0 of the valid votes and won by a majority of num__184 votes . find the total number of valid votes ? <o> a ) num__430 <o> b ) num__438 <o> c ) num__436 <o> d ) num__434 <o> e ) num__460 |
let the total number of valid votes be x . num__70.0 of x = num__0.7 * x = num__7 x / num__10 number of votes secured by the other candidate = x - num__7 x / num__100 = num__3 x / num__10 given num__7 x / num__10 - num__3 x / num__10 = num__184 = > num__4 x / num__10 = num__184 = > num__4 x = num__1840 = > x = num__460 . answer : e <eor> e <eos> |
e |
percent__100.0__460.0__ |
percent__100.0__460.0__ |
| a rectangular field is to be fenced on three sides leaving a side of num__20 feet uncovered . if the area of the field is num__210 sq . feet how many feet of fencing will be required ? <o> a ) num__34 <o> b ) num__41 <o> c ) num__68 <o> d ) num__88 <o> e ) none |
explanation we have : l = num__20 ft and lb = num__210 sq . ft . so b = num__10.5 ft . length of fencing = ( l + num__2 b ) = ( num__20 + num__21 ) ft = num__41 ft . answer b <eor> b <eos> |
b |
multiply__10.5__2.0__ triangle_area__2.0__41.0__ |
multiply__10.5__2.0__ triangle_area__2.0__41.0__ |
| a company d has num__15 percent of the employees are secretaries and num__10 percent are salespeople . if there are num__120 other employees of company d how many employees does company d have ? <o> a ) num__200 <o> b ) num__160 <o> c ) num__180 <o> d ) num__152 <o> e ) num__250 |
let the total number of employees in the company be x % of secretaries = num__15.0 % of salespeople = num__10.0 % of of employees other than secretaries and salespeople = num__100 - num__25 = num__75.0 but this number is given as num__120 so num__75.0 of x = num__120 x = num__160 therefore there a total of num__160 employees in the company d correct answer - b <eor> b <eos> |
b |
percent__100.0__160.0__ |
percent__100.0__160.0__ |
| num__1397 x num__1397 = ? <o> a ) num__1961609 <o> b ) num__1851609 <o> c ) num__1951619 <o> d ) num__1951609 <o> e ) num__1851619 |
num__1397 x num__1397 = ( num__1397 ) ^ num__2 = ( num__1400 - num__3 ) ^ num__2 = ( num__1400 ) ^ num__2 + ( num__3 ) ^ num__2 - ( num__2 x num__1400 x num__3 ) = num__1960000 + num__9 - num__8400 = num__1960009 - num__8400 = num__1951609 . answer is d . <eor> d <eos> |
d |
subtract__1400.0__1397.0__ add__1960000.0__9.0__ subtract__1960009.0__8400.0__ subtract__1960009.0__8400.0__ |
subtract__1400.0__1397.0__ add__1960000.0__9.0__ subtract__1960009.0__8400.0__ subtract__1960009.0__8400.0__ |
| mary passed a certain gas station on a highway while traveling west at a constant speed of num__50 miles per hour . then num__15 minutes later paul passed the same gas station while traveling west at a constant speed of num__60 miles per hour . if both drivers maintained their speeds and both remained on the highway for at least num__3 hours how long after he passed the gas station did paul catch up with mary ? <o> a ) num__1.5 <o> b ) num__1.3 <o> c ) num__1.25 <o> d ) num__1.6 <o> e ) num__2 |
d = rt m : r = num__50 mph t = t + num__0.25 hr d = num__50 ( t + num__0.25 ) p : r = num__60 t = t d = num__60 t since they went the same distance : num__50 t + num__12.5 = num__60 t num__10 t = num__12.5 t = num__1.25 or num__1 hr num__15 min c <eor> c <eos> |
c |
divide__15.0__60.0__ multiply__50.0__0.25__ subtract__60.0__50.0__ divide__12.5__10.0__ subtract__1.25__0.25__ round__1.25__ |
divide__15.0__60.0__ multiply__50.0__0.25__ subtract__60.0__50.0__ divide__12.5__10.0__ subtract__1.25__0.25__ add__0.25__1.0__ |
| if the compound interest on a sum of rs . num__5000 at the rate of num__10.0 per annum is rs . num__1050 then time period is ( interest compounded yearly ) <o> a ) num__1 yrs <o> b ) num__10.5 yrs <o> c ) num__3 yrs <o> d ) num__2 yrs <o> e ) num__4 yrs |
amount = num__5000 + num__1050 = num__6050 c . i = p ( num__1 + r / num__100 ) ^ n num__6050 = num__5000 ( num__1 + num__0.1 ) ^ n num__1.21 = ( num__1.1 ) ^ n num__1.21 = ( num__1.1 ) ^ n ( num__1.1 ) ^ num__2 = ( num__1.1 ) ^ n n = num__2 yrs . answer : d <eor> d <eos> |
d |
percent__10.0__1.0__ percent__100.0__2.0__ |
percent__10.0__1.0__ percent__100.0__2.0__ |
| the cube root of . num__000216 is : <o> a ) num__0.6 <o> b ) num__0.06 <o> c ) num__77 <o> d ) num__67 <o> e ) num__87 |
num__0.000216 = num__21.6 ^ num__6 cube root of the above = ( num__21.6 ^ num__6 ) ^ num__0.333333333333 = num__0.06 = . num__06 answer : b <eor> b <eos> |
b |
side_by_diagonal__0.06__0.0002__ |
side_by_diagonal__0.06__0.0002__ |
| the value of ( num__4.7 × num__13.26 + num__4.7 × num__9.43 + num__4.7 × num__77.31 ) is : <o> a ) num__0.47 <o> b ) num__47 <o> c ) num__470 <o> d ) num__4700 <o> e ) none of these |
solution given expression = num__4.7 × ( num__13.26 + num__9.43 + num__77.31 ) = num__4.7 × num__100 = num__470 . answer c <eor> c <eos> |
c |
multiply__4.7__100.0__ multiply__4.7__100.0__ |
multiply__4.7__100.0__ multiply__4.7__100.0__ |
| evaluate : num__1234562 - num__12 * num__3 * num__2 = ? <o> a ) num__32435453 <o> b ) num__1234554 <o> c ) num__76768786 <o> d ) num__97979797 <o> e ) num__75868656 |
according to order of operations num__12 ? num__3 ? num__2 ( division and multiplication ) is done first from left to right num__12 * * num__2 = num__4 * num__2 = num__8 hence num__1234562 - num__12 * num__3 * num__2 = num__1234562 - num__8 = num__1234554 correct answer b <eor> b <eos> |
b |
divide__12.0__3.0__ subtract__12.0__4.0__ subtract__1234562.0__8.0__ subtract__1234562.0__8.0__ |
divide__12.0__3.0__ subtract__12.0__4.0__ subtract__1234562.0__8.0__ subtract__1234562.0__8.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 sec . what is the length of the train ? <o> a ) num__298 m <o> b ) num__150 m <o> c ) num__208 m <o> d ) num__988 m <o> e ) num__299 m |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__9 = num__150 m answer : b <eor> b <eos> |
b |
round__150.0__ |
round__150.0__ |
| the perimeter of one face of a cube is num__40 cm . its volume will be : <o> a ) num__125 cm num__3 <o> b ) num__400 cm num__3 <o> c ) num__250 cm num__3 <o> d ) num__1000 cm num__3 <o> e ) none of these |
explanation : edge of cude = num__10.0 = num__10 cm volume = a * a * a = num__10 * num__10 * num__10 = num__1000 cm cube option d <eor> d <eos> |
d |
volume_cube__10.0__ volume_cube__10.0__ |
volume_cube__10.0__ volume_cube__10.0__ |
| a rectangular grass field is num__75 m * num__55 m it has a path of num__2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . num__7 per sq m ? <o> a ) num__4350 <o> b ) num__4725 <o> c ) num__4328 <o> d ) num__4329 <o> e ) num__4829 |
area = ( l + b + num__2 d ) num__2 d = ( num__75 + num__55 + num__2.5 * num__2 ) num__2 * num__2.5 = > num__675 num__675 * num__7 = rs . num__4725 answer : b <eor> b <eos> |
b |
multiply__7.0__675.0__ multiply__7.0__675.0__ |
multiply__7.0__675.0__ multiply__7.0__675.0__ |
| a certain culture of bacteria quadruples every hour . if a container with these bacteria was half full at num__7 : num__00 a . m . at what time was it one - eighth full ? <o> a ) num__9 : num__00 a . m . <o> b ) num__7 : num__00 a . m . <o> c ) num__6 : num__00 a . m . <o> d ) num__4 : num__00 a . m . <o> e ) num__2 : num__00 a . m . |
to go from one - eighth ( num__0.125 ) full to half ( num__0.5 ) full culture of bacteria should quadruple : num__0.125 * num__4 = num__0.5 as it quadruples every hour then container was one - eighth full at num__7 : num__00 a . m - num__1 hour = num__6 : num__00 a . m . answer : c . <eor> c <eos> |
c |
divide__0.5__0.125__ subtract__7.0__1.0__ round__6.0__ |
divide__0.5__0.125__ subtract__7.0__1.0__ subtract__7.0__1.0__ |
| a basketball team composed of num__12 players scored num__100 points in a particular contest . if none of the individual players scored fewer than num__7 points what is the greatest number of points w that an individual player might have scored ? <o> a ) num__7 <o> b ) num__13 <o> c ) num__16 <o> d ) num__21 <o> e ) num__23 |
general rule for such kind of problems : to maximize one quantity minimize the others ; to minimize one quantity maximize the others . thus to maximize the number of points of one particular player minimize the number of points of all other num__11 players . minimum number of points for a player is num__7 so the minimum number of points of num__11 players is num__7 * num__11 = num__77 . therefore the maximum number of points w for num__12 th player is num__100 - num__77 = num__23 . answer : e . <eor> e <eos> |
e |
multiply__7.0__11.0__ add__12.0__11.0__ add__12.0__11.0__ |
multiply__7.0__11.0__ subtract__100.0__77.0__ subtract__100.0__77.0__ |
| what least value should be replaced by * in num__223431 * so the number become divisible by num__5 <o> a ) num__0 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
explanation : trick : number is divisible by num__5 if last digit is num__0 or num__5 so that least number is num__0 . answer : option a <eor> a <eos> |
a |
reverse__223431.0__ reverse__223431.0__ |
reverse__223431.0__ reverse__223431.0__ |
| if x is the sum of a consecutive positive integers . y is the sum of b consecutive positive integers . for which of the following values of a and b is it impossible that x = y ? <o> a ) a = num__2 ; b = num__6 <o> b ) a = num__3 ; b = num__6 <o> c ) a = num__7 ; b = num__9 <o> d ) a = num__10 ; b = num__4 <o> e ) a = num__10 ; b = num__7 |
i approached this in a different way although may be longer but it ' s certainly more intuitive it is impossible that x = y if one of them is necessarily even while the other is necessarily odd for two consecutive integers the sum is necessarily odd for six consecutive integers the sum is necessarily odd ( sum of num__3 two consecutive integerswhich are all odd ) thus a is incorrect . the sum of num__3 consecutive integers can be even or odd ( an odd integer from the first num__2 + an additional integer which can be even or odd ) . thus b is incorrect . you can do a similar approach for c and e ( it follows that the sum of a specified odd number of consecutive integers can be odd or even ) . leaving us with d to be the only not - wrong answer . to confirm you can easily deduce that the sum of num__4 consecutive integers ( sum of num__2 odd integers ) is necessarily even while the sum of num__10 consecutive integers ( sum of num__5 odd integers ) is necessarily odd . thus d is the right answer . <eor> d <eos> |
d |
add__2.0__3.0__ multiply__2.0__5.0__ |
add__2.0__3.0__ multiply__2.0__5.0__ |
| abcd is a square . f and e are the midpoints of sides ad and cd respectively . the area of triangle fed is num__2 square inches . what is the area of square abcd ( in square inches ) ? <o> a ) num__4 <o> b ) num__8 <o> c ) num__16 <o> d ) num__32 <o> e ) num__64 |
area of fed = num__2 sq inches = num__0.5 * de * fd = num__0.5 * de ^ num__2 because the sides of a square are equal hence half of the sides will also be equal . de ^ num__2 = num__4 de = fd = num__2 hence the side of the square = num__4 area if the square = num__4 * num__4 = num__16 correct option : c <eor> c <eos> |
c |
square_perimeter__4.0__ square_perimeter__4.0__ |
power__2.0__4.0__ power__2.0__4.0__ |
| what is the difference between the largest number and the least number written with the digits num__9 num__2 num__1 num__5 ? <o> a ) num__6084 <o> b ) num__3788 <o> c ) num__2077 <o> d ) num__8262 <o> e ) num__1812 |
explanation : num__1259 num__9521 - - - - - - - - - - - - num__8262 answer : d <eor> d <eos> |
d |
subtract__9521.0__1259.0__ multiply__1.0__8262.0__ |
subtract__9521.0__1259.0__ subtract__9521.0__1259.0__ |
| a tradesman by means of his false balance defrauds to the extent of num__25.0 ? in buying goods as well as by selling the goods . what percent does he gain on his outlay ? <o> a ) num__17.0 <o> b ) num__77.0 <o> c ) num__44.0 <o> d ) num__56.25 <o> e ) num__47 % |
g % = num__25 + num__25 + ( num__25 * num__25 ) / num__100 = num__56.25 answer : d <eor> d <eos> |
d |
percent__56.25__100.0__ |
percent__56.25__100.0__ |
| if the price of a tv is first decreased by num__20.0 and then increased by num__30.0 then the net change in the price will be : <o> a ) num__4.0 increase <o> b ) num__10.0 increase <o> c ) num__10.0 decrease <o> d ) num__6.0 increase <o> e ) none of these |
explanation : solution : let the original price be rs . num__100 . new final price = num__130.0 of ( num__80.0 of num__100 ) = rs . num__1.3 * num__0.8 * num__100 = rs . num__104 . . ' . increase = num__4.0 answer : a <eor> a <eos> |
a |
add__30.0__100.0__ subtract__100.0__20.0__ divide__130.0__100.0__ divide__80.0__100.0__ multiply__130.0__0.8__ subtract__104.0__100.0__ subtract__104.0__100.0__ |
add__30.0__100.0__ subtract__100.0__20.0__ divide__130.0__100.0__ divide__80.0__100.0__ multiply__130.0__0.8__ subtract__104.0__100.0__ subtract__104.0__100.0__ |
| in a family num__15 people eat only vegetarian num__8 people eat only non veg . num__11 people eat both veg and non veg . . how many people eat veg in the family ? <o> a ) num__20 <o> b ) num__11 <o> c ) num__23 <o> d ) num__26 <o> e ) num__21 |
total people eat veg = only veg + both veg and non veg total = num__15 + num__11 = num__26 answer = d <eor> d <eos> |
d |
alphabet_space__ alphabet_space__ |
alphabet_space__ alphabet_space__ |
| the h . c . f . of two numbers is num__23 and the other two factors of their l . c . m . are num__13 and num__14 . the larger of the two numbers is : <o> a ) num__276 <o> b ) num__299 <o> c ) num__322 <o> d ) num__345 <o> e ) num__375 |
explanation : clearly the numbers are ( num__23 x num__13 ) and ( num__23 x num__14 ) . larger number = ( num__23 x num__14 ) = num__322 . answer is c <eor> c <eos> |
c |
multiply__23.0__14.0__ multiply__23.0__14.0__ |
multiply__23.0__14.0__ multiply__23.0__14.0__ |
| if a num__10 percent deposit that has been paid toward the purchase of a certain product is $ num__130 how much more remains to be paid ? <o> a ) $ num__880 <o> b ) $ num__990 <o> c ) $ num__1000 <o> d ) $ num__1170 <o> e ) $ num__1 |
210 |
num__0.1 p = num__130 > > p = num__130 * num__10.0 = num__1300 num__1300 - num__130 = num__1170 answer : d <eor> d <eos> |
d |
d |
| what is the value of x ^ num__2 yz − xyz ^ num__2 if x = − num__2 y = num__1 and z = num__1 ? <o> a ) num__6 <o> b ) num__24 <o> c ) num__30 <o> d ) num__32 <o> e ) num__48 |
num__4 * num__1 * num__1 - ( - num__2 * num__1 * num__1 ) = num__4 + num__2 = num__6 ans : a <eor> a <eos> |
a |
add__2.0__4.0__ add__2.0__4.0__ |
add__2.0__4.0__ add__2.0__4.0__ |
| what is the smallest positive integer which when increased by num__1 is divisible by num__22 num__36 num__26 and num__21 ? <o> a ) num__7271 <o> b ) num__36035 <o> c ) num__122149 <o> d ) num__22362621 <o> e ) num__4109 |
num__22 = num__2 * num__11 num__36 = num__2 * num__2 * num__3 * num__3 num__26 = num__2 * num__13 num__21 = num__3 * num__7 so the answer is ( num__2 * num__2 * num__3 * num__3 * num__7 * num__11 * num__13 ) - num__1 = num__36035 which is option b . <eor> b <eos> |
b |
divide__22.0__2.0__ add__1.0__2.0__ divide__26.0__2.0__ divide__21.0__3.0__ multiply__1.0__36035.0__ |
divide__22.0__2.0__ add__1.0__2.0__ divide__26.0__2.0__ divide__21.0__3.0__ multiply__1.0__36035.0__ |
| a wall photo num__2 inches wide is placed around a rectangular paper with dimensions num__8 inches by num__12 inches . what is the area of the wall photo in square inches ? <o> a ) num__96 <o> b ) num__86 <o> c ) num__108 <o> d ) num__144 <o> e ) num__118 |
this question is an example of a ' punch out ' question - we have to find the area of everything then ' punch out ' the part that we do n ' t want . we ' re told that a wall photo num__2 inches wide is placed around a rectangular paper with dimensions num__8 inches by num__12 inches . we ' re asked for the area of the wall photo in square inches . area of a rectangle = ( length ) ( width ) so the area of the wall photo is . . . ( num__8 ) ( num__12 ) = num__96 the wall photo ' adds ' num__2 inches to the top bottom left and right ' sides ' of the picture so the area of everything is . . . ( num__8 + num__2 + num__2 ) ( num__12 + num__2 + num__2 ) = ( num__12 ) ( num__16 ) = num__196 when we ' punch out ' the area of the paper we ' ll be left with the area of the wall photo : num__192 - num__96 = num__96 final answer : a <eor> a <eos> |
a |
multiply__8.0__12.0__ multiply__2.0__8.0__ rectangle_perimeter__2.0__96.0__ multiply__2.0__96.0__ triangle_area__2.0__96.0__ |
multiply__8.0__12.0__ multiply__2.0__8.0__ rectangle_perimeter__2.0__96.0__ multiply__2.0__96.0__ triangle_area__2.0__96.0__ |
| two trains num__180 m and num__360 m long run at the speed of num__60 kmph and num__30 kmph in opposite directions in parallel tracks . the time which they take to cross each other is ? <o> a ) num__21.6 sec <o> b ) num__8.9 sec <o> c ) num__10.8 sec <o> d ) num__12.6 sec <o> e ) num__15 sec |
relative speed = num__60 + num__30 = num__90 kmph * num__0.277777777778 = num__25 m / s distance covered in crossing each other = num__180 + num__360 = num__540 m required time = num__540 * num__0.04 = num__21.6 sec answer is a <eor> a <eos> |
a |
add__60.0__30.0__ add__180.0__360.0__ multiply__540.0__0.04__ round__21.6__ |
add__60.0__30.0__ add__180.0__360.0__ multiply__540.0__0.04__ multiply__540.0__0.04__ |
| calculate the largest num__4 digit number which is exactly divisible by num__98 ? <o> a ) num__9800 <o> b ) num__9898 <o> c ) num__9702 <o> d ) num__9604 <o> e ) num__9996 |
largest num__4 digit number is num__9999 after doing num__9999 ÷ num__98 we get remainder num__3 hence largest num__4 digit number exactly divisible by num__98 = num__9999 - num__3 = num__9996 e <eor> e <eos> |
e |
subtract__9999.0__3.0__ subtract__9999.0__3.0__ |
subtract__9999.0__3.0__ subtract__9999.0__3.0__ |
| the length of the bridge which a train num__150 metres long and travelling at num__45 km / hr can cross in num__30 seconds is ? <o> a ) num__245 <o> b ) num__777 <o> c ) num__282 <o> d ) num__266 <o> e ) num__225 |
speed = [ num__45 x num__0.277777777778 ] m / sec = [ num__12.5 ] m / sec time = num__30 sec let the length of bridge be x metres . then ( num__150 + x ) / num__30 = num__12.5 = > num__2 ( num__150 + x ) = num__750 = > x = num__225 m . answer : e <eor> e <eos> |
e |
round__225.0__ |
round__225.0__ |
| how many pieces of num__75 cm can be cut from a rope num__57 meters long ? <o> a ) num__30 <o> b ) num__40 <o> c ) num__76 <o> d ) none <o> e ) can not be determined |
explanation : total pieces of num__75 cm that can be cut from a rope of num__57 meters long is = ( num__57 meters ) / ( num__75 cm ) = ( num__57 meters ) / ( num__0.75 meters ) = num__76 answer : c <eor> c <eos> |
c |
divide__57.0__0.75__ round__76.0__ |
divide__57.0__0.75__ divide__57.0__0.75__ |
| each of the products produced yesterday was checked by worker x or worker y . num__0.5 of the products checked by worker x are defective and num__0.8 of the products checked by worker y are defective . if the total defective rate of all the products checked by worker x and worker y is num__0.7 what fraction of the products was checked by worker y ? <o> a ) num__0.666666666667 <o> b ) num__0.833333333333 <o> c ) num__0.875 <o> d ) num__0.8 <o> e ) num__0.625 |
x : num__0.5 is num__0.2 - points from num__0.7 . y : num__0.8 is num__0.1 - points from num__0.7 . therefore the ratio of products checked by y : x is num__2 : num__1 . thus worker y checked num__0.666666666667 of the products . the answer is a . <eor> a <eos> |
a |
subtract__0.7__0.5__ multiply__0.5__0.2__ reverse__0.5__ multiply__0.5__2.0__ multiply__1.0__0.6667__ |
subtract__0.7__0.5__ subtract__0.8__0.7__ reverse__0.5__ multiply__0.5__2.0__ multiply__1.0__0.6667__ |
| a certain college party is attended by both male and female students . the ratio of male to female students is num__4 to num__5 . if num__15 of the male students were to leave the party the ratio would change to num__1 to num__2 . how many total students are at the party ? <o> a ) num__60 <o> b ) num__70 <o> c ) num__80 <o> d ) num__90 <o> e ) num__100 |
the ratio is num__4 : num__5 = num__8 : num__10 so there are num__8 k males and num__10 k females . if num__15 males left the ratio would be num__1 : num__2 = num__5 : num__10 so there would be num__5 k males and num__10 k females . num__8 k - num__5 k = num__15 k = num__5 num__8 k + num__10 k = num__40 + num__50 = num__90 the answer is d . <eor> d <eos> |
d |
multiply__4.0__2.0__ multiply__5.0__2.0__ multiply__4.0__10.0__ multiply__5.0__10.0__ add__40.0__50.0__ multiply__1.0__90.0__ |
multiply__4.0__2.0__ subtract__15.0__5.0__ multiply__4.0__10.0__ add__10.0__40.0__ add__40.0__50.0__ add__40.0__50.0__ |
| the length of a cold storage is double its breadth . its height is num__3 metres . the area of its four walls ( including doors ) is num__108 m num__2 . find its volume . <o> a ) num__206 <o> b ) num__216 <o> c ) num__226 <o> d ) num__246 <o> e ) num__288 |
let xbe breadth then num__2 x is the length then num__2 x * num__3 * num__2 + x * num__3 * num__2 = num__108 num__12 x + num__6 x = num__108 x = num__6.0 = num__6 length num__12 breadth num__6 height num__3 volume = num__12 * num__6 * num__3 = num__216 m ^ num__3 answer : b <eor> b <eos> |
b |
square_perimeter__3.0__ multiply__3.0__2.0__ volume_cube__6.0__ volume_cube__6.0__ |
square_perimeter__3.0__ multiply__3.0__2.0__ multiply__108.0__2.0__ multiply__108.0__2.0__ |
| a number when divided by a divisor leaves a remainder of num__24 . when twice the original number is divided by the same divisor the remainder is num__11 . what is the value of the divisor ? <o> a ) num__37 <o> b ) num__30 <o> c ) num__25 <o> d ) num__28 <o> e ) num__40 |
original no be a let divisor be d & let the quotient of the division of a by d be x = > a / d = x and the remainder is num__24 a = dx + num__24 when twice the original no is divided by d num__2 a is divided by d = > the problem state that ( num__2 dx + num__48 ) / d leaves a remainder of num__11 num__2 dx is perfectly divisible by d and will not leave remainder the remainder of num__11 was obtained by dividing num__48 by d when num__48 is divided by num__37 the remainder that one will obtain is num__11 the divisor is num__37 answer a <eor> a <eos> |
a |
multiply__24.0__2.0__ subtract__48.0__11.0__ subtract__48.0__11.0__ |
multiply__24.0__2.0__ subtract__48.0__11.0__ subtract__48.0__11.0__ |
| two persons ram & lakshman who are at a distance of num__100 km from each other move towards each other from two places p and q at speeds of num__20 kmph and num__25 kmph respectively . lakshman reaches p returns immediately and meets ram at r who started on the return journey to p immediately after reaching q . what is the distance between q and r ? <o> a ) num__33 num__0.333333333333 km <o> b ) num__25 km <o> c ) num__30 km <o> d ) num__27 num__0.333333333333 km <o> e ) num__28 num__0.333333333333 km |
ram takes num__5.0 = num__5 hours to cover the distance from p to q . by that time lakshman covers covers num__5 * num__25 = num__125 km lakshman covers num__25 km more than the distance pq . now the distance between them = num__75 km time taken by them to meet = distance / relative speed = num__75 / ( num__20 + num__25 ) = num__1.66666666667 = num__1.66666666667 hrs . distance between q and r is nothing but the distance covered by ram in num__1.66666666667 hours = num__20 * num__1.66666666667 = num__33.3333333333 km or num__33 num__0.333333333333 km answer : a <eor> a <eos> |
a |
divide__100.0__20.0__ add__100.0__25.0__ subtract__100.0__25.0__ divide__125.0__75.0__ divide__25.0__75.0__ round__33.0__ |
divide__100.0__20.0__ add__100.0__25.0__ subtract__100.0__25.0__ divide__125.0__75.0__ divide__25.0__75.0__ round__33.0__ |
| if f ( x ) = ( x ^ num__2 - num__1 ) / ( x ) what is f ( num__1 / x ) in terms of f ( x ) ? <o> a ) - num__1 * f ( x ) <o> b ) f ( x ) <o> c ) num__1 / f ( x ) <o> d ) - num__1 / f ( x ) <o> e ) num__2 * f ( x ) |
f ( num__1 / x ) = ( ( num__1 / x ) num__2 - num__1 ) / ( ( num__1 / x ) ) = ( ( num__1 / x ^ num__2 ) - num__1 ) / ( num__1 / x ) = ( ( num__1 - x ^ num__2 ) / ( x ^ num__2 ) ) / ( num__1 / x = ( num__1 - x ^ num__2 ) / ( x ^ num__2 ) = - ( ( x ^ num__4 ) - num__1 ) / ( x ) = - num__1 * f ( x ) answer is a . <eor> a <eos> |
a |
reverse__1.0__ |
reverse__1.0__ |
| two watches are set for an alarm at num__10 num__0 ' clock in the night . one watch gains num__2 minutes per hour . when the faster watch shows num__4 : num__12 in the morning when the alarm rings what does the correct watch show ? <o> a ) num__5 : num__00 <o> b ) num__4 : num__00 <o> c ) num__4 : num__10 <o> d ) num__5 : num__20 <o> e ) num__6 : num__00 |
the faster watch gains num__2 minutes per hour the two watches show num__10 : num__00 at num__10 num__0 ' clock at num__11 num__0 ' clock - the correct watch is at num__11 : num__00 the faster watch is at num__11 : num__02 ( num__1 hour + gains num__2 minutes per hour ) at num__12 midnight - the correct watch is at num__12 : num__00 the faster watch is at num__12 : num__04 ( num__2 hours + gains num__4 minutes per num__2 hours ) in num__6 hours the faster watch gains num__6 * num__2 = num__12 minutes and is at num__4 : num__12 and the correct watch is at num__4 : num__00 answer is b <eor> b <eos> |
b |
subtract__12.0__11.0__ subtract__10.0__4.0__ round__4.0__ |
subtract__12.0__11.0__ subtract__10.0__4.0__ subtract__10.0__6.0__ |
| which of the following inequalities indicates the set of all values of d for which the lengths a of the three sides of a triangle can be num__34 and d ? <o> a ) num__0 < d < num__1 <o> b ) num__0 < d < num__5 <o> c ) num__0 < d < num__7 <o> d ) num__1 < d < num__5 <o> e ) num__1 < d < num__7 |
this question is a good way to apply one of the most basic relation between the num__3 sides of a triangle . in a triangle ( any triangle ) any side must be greater than the positive difference of the other two sides and less than than the sum of the other num__2 sides . let the sides of a triangle be a b c . thus | a - b | < c < a + b | b - c | < a < b + c | c - a | < b < a + c thus if the sides of the triangle are num__34 and d num__4 - num__3 < d < num__4 + num__3 - - - > num__1 < d < num__7 . thus e is the correct answer . <eor> e <eos> |
e |
subtract__3.0__2.0__ add__3.0__4.0__ reverse__1.0__ |
subtract__3.0__2.0__ add__3.0__4.0__ subtract__2.0__1.0__ |
| a train num__125 m long passes a man running at num__9 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is : <o> a ) num__35 km / hr <o> b ) num__50 km / hr <o> c ) num__54 km / hr <o> d ) num__65 km / hr <o> e ) num__87 km / hr |
speed of the train relative to man = ( num__12.5 ) m / sec = ( num__12.5 ) m / sec . [ ( num__12.5 ) * ( num__3.6 ) ] km / hr = num__45 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__9 ) km / hr . x - num__9 = num__45 = = > x = num__54 km / hr . answer : c <eor> c <eos> |
c |
divide__125.0__10.0__ multiply__12.5__3.6__ add__9.0__45.0__ round__54.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ add__9.0__45.0__ round__54.0__ |
| simple interest on a certain sum is num__0.64 of the sum . find the rate percent and time if both are numerically equal . <o> a ) num__1 <o> b ) num__8 <o> c ) num__5 <o> d ) num__4 <o> e ) num__2 |
explanation : let sum = x . then s . i = num__16 x / num__25 let rate = r % and time = r years . therefore ( x * r * r ) / num__100 = num__16 x / num__25 or r ^ num__2 = num__64.0 r = num__8.0 = num__8 therefore rate = num__8.0 and time = num__8 years . answer : b ) rate = num__8.0 and time = num__8 years . <eor> b <eos> |
b |
percent__100.0__8.0__ |
percent__100.0__8.0__ |
| if a die has num__1 num__6 and num__3 num__4 and num__2 num__5 opposite each other how many such dies can be made <o> a ) num__10 <o> b ) num__11 <o> c ) num__12 <o> d ) num__13 <o> e ) num__14 |
if one face have num__1 num__32 then the corossponding face will be occupied by respective pair i . e ( num__6 num__45 ) . so there r num__3 place and we have num__3 place soit can be arranged in num__3 ! way i . e num__6 way and every pair will be suffled in num__2 way so ans will be num__6 * num__2 = num__12 . answer : c <eor> c <eos> |
c |
choose__3.0__1.0__ choose__3.0__1.0__ |
choose__3.0__1.0__ choose__3.0__1.0__ |
| two trains are running at num__40 km / hr and num__20 km / hr respectively in the same direction . fast train completely passes a man sitting in the slower train in num__5 seconds . what is the length of the fast train ? <o> a ) num__27 num__0.777777777778 <o> b ) num__28 num__0.777777777778 <o> c ) num__29 num__0.777777777778 <o> d ) num__30 num__0.777777777778 <o> e ) none of these |
explanation : as trains are moving in same direction so relative speed = num__40 - num__20 = num__20 kmph = num__20 * ( num__0.277777777778 ) = num__5.55555555556 m / sec length of train = speed * time length = num__5.55555555556 ∗ num__5 = num__27.7777777778 = num__27 num__0.777777777778 option a <eor> a <eos> |
a |
subtract__27.7778__27.0__ round__27.0__ |
subtract__27.7778__27.0__ subtract__27.7778__0.7778__ |
| if log num__1087.5 = num__6.9421 then the number of digits in ( num__875 ) num__10 is ? <o> a ) num__30 <o> b ) num__28 <o> c ) num__80 <o> d ) num__26 <o> e ) num__25 |
x = ( num__875 ) num__10 = ( num__87.5 x num__10 ) num__10 therefore log num__10 x = num__10 ( log num__1087.5 + num__1 ) = num__10 ( num__6.9421 + num__1 ) = num__10 ( num__7.9421 ) = num__79.421 x = antilog ( num__79.421 ) therefore number of digits in x = num__80 . answer : c <eor> c <eos> |
c |
divide__875.0__10.0__ add__6.9421__1.0__ multiply__10.0__7.9421__ multiply__1.0__80.0__ |
divide__875.0__10.0__ add__6.9421__1.0__ multiply__10.0__7.9421__ multiply__1.0__80.0__ |
| a can do a piece of work in num__12 days . he worked for num__15 days and then b completed the remaining work in num__10 days . both of them together will finish it in . <o> a ) num__11 num__0.5 days <o> b ) num__15 num__0.5 days <o> c ) num__15 num__0.5 days <o> d ) num__16 num__0.5 days <o> e ) num__12 num__0.5 days |
e num__12 num__0.5 days num__0.6 + num__10 / x = num__1 = > x = num__25 num__0.04 + num__0.04 = num__0.08 num__12.5 = num__12 num__0.5 days <eor> e <eos> |
e |
km_to_mile_conversion__ add__15.0__10.0__ divide__1.0__25.0__ divide__0.04__0.5__ add__12.0__0.5__ round__12.0__ |
km_to_mile_conversion__ add__15.0__10.0__ divide__1.0__25.0__ divide__0.04__0.5__ add__12.0__0.5__ divide__12.0__1.0__ |
| the average weight of a group of boys is num__20 kg . after a boy of weight num__33 kg joins the group the average weight of the group goes up by num__1 kg . find the number of boys in the group originally ? <o> a ) num__12 <o> b ) num__14 <o> c ) num__18 <o> d ) num__24 <o> e ) num__10 |
let the number off boys in the group originally be x . total weight of the boys = num__20 x after the boy weighing num__33 kg joins the group total weight of boys = num__20 x + num__33 so num__20 x + num__33 = num__21 ( x + num__1 ) = > x = num__12 . answer : a <eor> a <eos> |
a |
add__20.0__1.0__ subtract__33.0__21.0__ subtract__33.0__21.0__ |
add__20.0__1.0__ subtract__33.0__21.0__ subtract__33.0__21.0__ |
| a num__600 meter long train crosses a signal post in num__20 seconds . how long will it take to cross a num__3 kilometer long bridge at the same speed ? <o> a ) num__4 min <o> b ) num__2 min <o> c ) num__8 min <o> d ) num__9 min <o> e ) num__5 min |
s = num__30.0 = num__1530 mps s = num__120.0 = num__120 sec = num__2 min answer : b <eor> b <eos> |
b |
divide__600.0__20.0__ round__2.0__ |
divide__600.0__20.0__ round__2.0__ |
| if the speed of a man is num__66 km per hour then what is the distance traveled by him in num__30 seconds ? <o> a ) num__550 m <o> b ) num__500 m <o> c ) num__375 m <o> d ) num__420 m <o> e ) num__440 m |
the distance traveled in num__30 sec = num__66 * ( num__0.277777777778 ) * num__30 = num__550 m answer : a <eor> a <eos> |
a |
round__550.0__ |
round__550.0__ |
| in a family gathering there is a basket in which there are oranges bananas and apples . half of the people in the family eat oranges half of the other half eat bananas and the rest eat apples . if the number of people who eat oranges are num__10 less than the total number of people find the number of people in the gathering . <o> a ) num__5 <o> b ) num__10 <o> c ) num__15 <o> d ) num__20 <o> e ) num__25 |
let the number of people who eat oranges be x . then total number of people = x + ( x / num__2 ) + ( x / num__2 ) = num__2 x = > num__2 x - num__10 = x = > x = num__10 = > total number of people were num__2 x = num__20 option d <eor> d <eos> |
d |
multiply__10.0__2.0__ multiply__10.0__2.0__ |
multiply__10.0__2.0__ multiply__10.0__2.0__ |
| if num__4 ( p ' s capital ) = num__6 ( q ' s capital ) = num__10 ( r ' s capital ) then out of the total profit of rs num__3410 r will receive <o> a ) num__660 <o> b ) num__700 <o> c ) num__800 <o> d ) num__900 <o> e ) none of these |
explanation : let p ' s capital = p q ' s capital = q and r ' s capital = r then num__4 p = num__6 q = num__10 r = > num__2 p = num__3 q = num__5 r = > q = num__2 p / num__3 r = num__2 p / num__5 p : q : r = p : num__2 p / num__3 : num__2 p / num__5 = num__15 : num__10 : num__6 r ' s share = num__3410 * ( num__0.193548387097 ) = num__110 * num__6 = num__660 . answer : option a <eor> a <eos> |
a |
subtract__6.0__4.0__ divide__6.0__2.0__ divide__10.0__2.0__ add__10.0__5.0__ multiply__6.0__110.0__ multiply__6.0__110.0__ |
subtract__6.0__4.0__ divide__6.0__2.0__ divide__10.0__2.0__ multiply__3.0__5.0__ multiply__6.0__110.0__ multiply__6.0__110.0__ |
| two horses start trotting towards each other one from a to b and another from b to a . they cross each other after one hour and the first horse reaches b num__0.833333333333 hour before the second horse reaches a . if the distance between a and b is num__50 km . what is the speed of the slower horse ? <o> a ) num__70 km / h <o> b ) num__60 km / h <o> c ) num__40 km / h <o> d ) num__20 km / h <o> e ) num__10 km / h |
explanation : if the speed of the faster horse be \ inline f _ { s } and that of slower horse be \ inline s _ { s } then \ inline f _ { s } + s _ { s } = \ frac { num__50 } { num__1 } = num__50 and \ inline \ frac { num__50 } { s _ { s } } - \ frac { num__50 } { f _ { s } } = \ frac { num__5 } { num__6 } now you can go through options . the speed of slower horse is num__20 km / h since num__20 + num__30 = num__50 and \ inline \ frac { num__50 } { num__20 } - \ frac { num__50 } { num__30 } = \ frac { num__5 } { num__6 } answer : d <eor> d <eos> |
d |
add__1.0__5.0__ subtract__50.0__20.0__ round__20.0__ |
add__1.0__5.0__ subtract__50.0__20.0__ subtract__50.0__30.0__ |
| a gets num__3 times as much money as b gets b gets only rs . num__25 more then what c gets . the three gets rs . num__625 in all . find the share of b ? <o> a ) num__130 <o> b ) num__120 <o> c ) num__218 <o> d ) num__140 <o> e ) num__145 |
a + b + c = num__625 a = num__3 b num__3 b + b + b - num__25 = num__625 num__5 b = num__650 b = num__130 answer : a <eor> a <eos> |
a |
add__25.0__625.0__ divide__650.0__5.0__ divide__650.0__5.0__ |
add__25.0__625.0__ divide__650.0__5.0__ divide__650.0__5.0__ |
| in a certain pet shop the ratio of dogs to cats to bunnies in stock is num__3 : num__7 : num__12 . if the shop carries num__375 dogs and bunnies total in stock how many dogs are there ? <o> a ) num__42 <o> b ) num__66 <o> c ) num__75 <o> d ) num__112 <o> e ) num__154 |
let us assume the number of dogs cats and bunnies to be num__3 x num__7 x and num__12 x total dogs and bunnies = num__15 x . and we are given that num__15 x = num__375 . hence x = num__25 . dogs = num__3 x = num__3 * num__25 = num__75 ( option c ) <eor> c <eos> |
c |
add__3.0__12.0__ divide__375.0__15.0__ multiply__3.0__25.0__ multiply__3.0__25.0__ |
add__3.0__12.0__ divide__375.0__15.0__ multiply__3.0__25.0__ multiply__3.0__25.0__ |
| if f ( a ) = num__5 b what is the value of ( f ( a ) - f ( num__6 b ) ) ? <o> a ) num__32 b <o> b ) num__25 b <o> c ) num__6 a - num__6 b <o> d ) num__3 a + num__2 b <o> e ) num__52 a |
f ( a ) = num__5 b f ( num__6 b ) = num__30 b ( f ( num__6 b ) - f ( a ) ) = num__30 b - num__5 b = num__25 b answer is b <eor> b <eos> |
b |
multiply__5.0__6.0__ subtract__30.0__5.0__ subtract__30.0__5.0__ |
multiply__5.0__6.0__ subtract__30.0__5.0__ subtract__30.0__5.0__ |
| the sum of two numbers is num__16 . the difference is num__2 . what are the two numbers ? <o> a ) num__10 - num__6 <o> b ) num__8 - num__8 <o> c ) num__9 - num__7 <o> d ) num__11 - num__5 <o> e ) num__13 - num__3 |
num__9 + num__7 = num__16 num__9 - num__7 = num__2 the answer is c <eor> c <eos> |
c |
subtract__16.0__9.0__ subtract__16.0__7.0__ |
subtract__16.0__9.0__ subtract__16.0__7.0__ |
| insert the missing number num__4 - num__12 num__36 - num__108 num__324 ( . . . . ) <o> a ) - num__972 <o> b ) - num__972 <o> c ) num__452 <o> d ) - num__452 <o> e ) num__0 |
num__4 * - num__3 = - num__12 - num__12 * - num__3 = num__36 num__36 * - num__3 = - num__108 - num__108 * - num__3 = num__324 num__324 * - num__3 = - num__972 answer : a <eor> a <eos> |
a |
divide__12.0__4.0__ multiply__324.0__3.0__ multiply__324.0__3.0__ |
divide__12.0__4.0__ multiply__324.0__3.0__ multiply__324.0__3.0__ |
| a is twice as good a workman as b and they took num__7 days together to do the work b alone can do it in ? <o> a ) num__25 days <o> b ) num__88 days <o> c ) num__21 days <o> d ) num__11 days <o> e ) num__13 days |
wc = num__2 : num__1 num__2 x + x = num__0.142857142857 x = num__0.047619047619 = > num__21 days answer : c <eor> c <eos> |
c |
divide__1.0__7.0__ round__21.0__ |
divide__1.0__7.0__ round__21.0__ |
| the distance between delhi and mathura is num__110 kms . a starts from delhi with a speed of num__20 kmph at num__7 a . m . for mathura and b starts from mathura with a speed of num__25 kmph at num__8 p . m . from delhi . when will they meet ? <o> a ) num__11 a . m <o> b ) num__10 a . m <o> c ) num__77 a . m <o> d ) num__55 a . m <o> e ) num__65 a . m |
d = num__110 – num__20 = num__90 rs = num__20 + num__25 = num__45 t = num__2.0 = num__2 hours num__8 a . m . + num__2 = num__10 a . m . answer : b <eor> b <eos> |
b |
subtract__110.0__20.0__ add__20.0__25.0__ divide__90.0__45.0__ divide__20.0__2.0__ round__10.0__ |
subtract__110.0__20.0__ add__20.0__25.0__ divide__90.0__45.0__ add__8.0__2.0__ add__8.0__2.0__ |
| the probability that a man will be alive for num__10 more yrs is num__0.25 & the probability that his wife will alive for num__10 more yrs is num__0.333333333333 . the probability that none of them will be alive for num__10 more yrs is <o> a ) num__0.5 <o> b ) num__1 <o> c ) num__0.666666666667 <o> d ) num__0.75 <o> e ) num__0.8 |
sol . required probability = pg . ) x p ( b ) = ( num__1 — d x ( num__1 — i ) = : x num__1 = num__0.5 ans . ( a ) <eor> a <eos> |
a |
divide__0.25__0.5__ |
divide__0.25__0.5__ |
| some articles were bought at num__6 articles for rs . num__5 and sold at num__5 articles for rs . num__6 . gain percent is : <o> a ) num__30.0 <o> b ) num__32.0 <o> c ) num__34.0 <o> d ) num__35.0 <o> e ) num__44 % |
suppose number of articles bought = l . c . m . of num__6 and num__5 = num__30 . c . p . of num__30 articles = rs . num__0.833333333333 x num__30 = rs . num__25 . s . p . of num__30 articles = rs . num__1.2 x num__30 = rs . num__36 . gain % = num__0.44 x num__100.0 = num__44.0 . answer is e . <eor> e <eos> |
e |
percent__100.0__44.0__ |
percent__100.0__44.0__ |
| the speed of a car is num__100 km in the first hour and num__60 km in the second hour . what is the average speed of the car ? <o> a ) num__50 kmph <o> b ) num__65 kmph <o> c ) num__75 kmph <o> d ) num__80 kmph <o> e ) num__90 kmph |
explanation : s = ( num__100 + num__60 ) / num__2 = num__80 kmph d ) <eor> d <eos> |
d |
round__80.0__ |
round__80.0__ |
| when the no . num__7 y num__86038 is exactly divisible by num__11 then the smallest whole no . in place of y ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__8 |
the given number = num__7 y num__86038 sum of the odd places = num__8 + num__0 + num__8 + num__7 = num__23 sum of the even places = num__3 + num__6 + y ( sum of the odd places ) - ( sum of even places ) = number ( exactly divisible by num__11 ) num__23 - ( num__9 + y ) = divisible by num__11 num__14 � y = divisible by num__11 . y must be num__3 to make given number divisible by num__11 . b <eor> b <eos> |
b |
reverse__86038.0__ subtract__11.0__8.0__ add__3.0__6.0__ add__11.0__3.0__ subtract__11.0__8.0__ |
reverse__86038.0__ subtract__11.0__8.0__ add__3.0__6.0__ add__11.0__3.0__ subtract__11.0__8.0__ |
| boy sells a book for rs . num__450 he gets a loss of num__10.0 to gain num__10.0 what should be the sp ? <o> a ) num__430 <o> b ) num__450 <o> c ) num__550 <o> d ) num__590 <o> e ) num__600 |
find selling price to gain num__10.0 . now we are asked to find selling price to gain num__10.0 profit . hint : selling price = ( num__100 + gain % ) × c . p . num__100 selling price = ( num__100 + num__10 ) × num__500 num__100 selling price = ( num__110 ) × num__500 num__100 therefore selling price = rs . num__550 c <eor> c <eos> |
c |
percent__100.0__550.0__ |
percent__100.0__550.0__ |
| john left home and drove at the rate of num__45 mph for num__2 hours . he stopped for lunch then drove for another num__3 hours at the rate of num__50 mph to reach his destination . how many miles did john drive ? <o> a ) num__235 miles . <o> b ) num__245 miles . <o> c ) num__240 miles . <o> d ) num__265 miles . <o> e ) num__275 miles . |
the total distance d traveled by john is given by d = num__45 * num__2 + num__3 * num__50 = num__240 miles . answer c <eor> c <eos> |
c |
round__240.0__ |
round__240.0__ |
| a number whose fifth part increased by num__2 is equal to its fourth part diminished by num__2 is ? <o> a ) num__160 <o> b ) num__80 <o> c ) num__200 <o> d ) num__220 <o> e ) none |
answer let the number be n . then ( n / num__5 ) + num__2 = ( n / num__4 ) - num__2 â ‡ ’ ( n / num__4 ) - ( n / num__5 ) = num__4 â ‡ ’ ( num__5 n - num__4 n ) / num__20 = num__4 â ˆ ´ n = num__80 option : b <eor> b <eos> |
b |
multiply__4.0__5.0__ multiply__20.0__4.0__ multiply__20.0__4.0__ |
multiply__4.0__5.0__ multiply__20.0__4.0__ multiply__20.0__4.0__ |
| a heap of coconuts is divided into groups of num__2 num__3 and num__7 and each time one coconut is left over . the least number of coconuts in the heap is ? a . num__31 b . num__41 c . num__51 d . num__61 <o> a ) a ) num__31 <o> b ) b ) num__43 <o> c ) c ) num__51 <o> d ) d ) num__61 <o> e ) e ) num__71 |
lcm = num__42 = > num__42 + num__1 = num__43 answer : b <eor> b <eos> |
b |
subtract__3.0__2.0__ add__2.0__41.0__ add__2.0__41.0__ |
subtract__3.0__2.0__ add__2.0__41.0__ add__2.0__41.0__ |
| if b is an integer greater than num__6 which of the following must be divisible by num__3 ? <o> a ) b ( b + num__3 ) ( b - num__5 ) <o> b ) b ( b + num__2 ) ( b - num__1 ) <o> c ) b ( b + num__1 ) ( b - num__4 ) <o> d ) b ( b + num__4 ) ( b - num__2 ) <o> e ) b ( b + num__5 ) ( b - num__6 ) |
anything in the form of ( b - num__1 ) ( b ) ( b + num__1 ) is divisible by num__3 . in other word a product of any num__3 consecutive intevers is divisible by num__3 . b ( b + num__1 ) ( b - num__4 ) = b ( b + num__1 ) ( ( b - num__1 ) - num__3 ) is equivalant to ( b - num__1 ) ( b ) ( b + num__1 ) b ( b + num__2 ) ( b - num__1 ) is equivalant to ( b + num__1 ) missing . b ( b + num__3 ) ( b - num__5 ) is equivalant to ( b - num__1 ) missing and b repeating . b ( b + num__4 ) ( b - num__2 ) is equivalant to odd / even consqcutive integers b ( b + num__5 ) ( b - num__6 ) is equivalant to ( b + num__1 ) missing and b repeating . answer : c <eor> c <eos> |
c |
add__3.0__1.0__ divide__6.0__3.0__ subtract__6.0__1.0__ reverse__1.0__ |
add__3.0__1.0__ divide__6.0__3.0__ subtract__6.0__1.0__ reverse__1.0__ |
| if num__144 / num__0.144 = num__14.4 / x then the value of x is : <o> a ) num__2 . <o> b ) num__1 . <o> c ) - num__1 . <o> d ) - num__2 . <o> e ) num__0 |
given expression = [ a ( power num__2 ) - b ( power num__2 ) ] / ( a + b ) ( a - b ) = [ a ( power num__2 - b ( power num__2 ) ] / [ a ( power num__2 ) - b ( power num__2 ) ] = num__1 . answer is b . <eor> b <eos> |
b |
reverse__1.0__ |
reverse__1.0__ |
| a man can row num__6 kmph in still water . when the river is running at num__1.2 kmph it takes him num__1 hour to row to a place and black . how far is the place ? <o> a ) num__2.89 <o> b ) num__2.88 <o> c ) num__2.85 <o> d ) num__2.81 <o> e ) num__2.82 |
answer & m = num__6 s = num__1.2 ds = num__6 + num__1.2 = num__7.2 us = num__6 - num__1.2 = num__4.8 x / num__7.2 + x / num__4.8 = num__1 x = num__2.88 . answer : b <eor> b <eos> |
b |
add__6.0__1.2__ subtract__6.0__1.2__ round__2.88__ |
add__6.0__1.2__ subtract__6.0__1.2__ divide__2.88__1.0__ |
| if f ( x ) = num__3 x − √ x and g ( x ) = x ^ num__2 what is f ( g ( num__2 ) ) ? <o> a ) - num__4 <o> b ) num__10 <o> c ) num__16 <o> d ) num__32 <o> e ) num__44 |
g ( x ) = x ² g ( num__2 ) = num__2 ² = num__4 so f ( g ( num__2 ) ) = f ( num__4 ) f ( x ) = num__3 x − √ x so f ( num__4 ) = num__3 ( num__4 ) − √ num__4 = num__12 - num__2 = num__10 = b <eor> b <eos> |
b |
multiply__3.0__4.0__ subtract__12.0__2.0__ subtract__12.0__2.0__ |
multiply__3.0__4.0__ subtract__12.0__2.0__ subtract__12.0__2.0__ |
| the current birth rate per thousand is num__32 whereas corresponding death rate is num__11 per thousand . the net growth rate in terms of population increase in percent is given by <o> a ) num__0.0021 <o> b ) num__0.021 <o> c ) num__2.1 <o> d ) num__21.0 <o> e ) none |
sol . net growth on num__1000 = ( num__32 - num__11 ) = num__21 . net growth on num__100 = ( num__0.021 × num__100 ) % = num__2.1 . answer c <eor> c <eos> |
c |
percent__100.0__2.1__ |
percent__100.0__2.1__ |
| a b and c can do a work in num__6 days num__8 days and num__12 days respectively . in how many days can all three of them working together complete the work ? <o> a ) num__2 num__0.666666666667 <o> b ) num__2 num__0.25 <o> c ) num__2 num__2 / num__0 <o> d ) num__2 num__1.0 <o> e ) num__2 num__2.0 |
work done by all three of them in one day = num__0.166666666667 + num__0.125 + num__0.0833333333333 = num__0.375 . the number of days required = num__2.66666666667 = num__2 num__0.666666666667 days . answer : a <eor> a <eos> |
a |
subtract__8.0__6.0__ divide__8.0__12.0__ round__2.0__ |
subtract__8.0__6.0__ divide__8.0__12.0__ round__2.0__ |
| a can do a piece of work in num__5 days and b can do it in num__4 days how long will they both work together to complete the work ? <o> a ) num__0.545454545455 <o> b ) num__0.888888888889 <o> c ) num__0.777777777778 <o> d ) num__0.222222222222 <o> e ) num__0.909090909091 |
explanation : a ’ s one day work = num__0.2 b ’ s one day work = num__0.25 ( a + b ) ’ s one day work = num__0.2 + num__0.25 = num__0.45 = > time = num__2.22222222222 = num__2 num__0.222222222222 days answer : option d <eor> d <eos> |
d |
add__0.25__0.2__ subtract__2.2222__2.0__ subtract__2.2222__2.0__ |
add__0.25__0.2__ subtract__2.2222__2.0__ subtract__2.2222__2.0__ |
| the timing of a college is from num__12 p . m to num__4.20 p . m . five lectures are held in the given duration and a break of num__5 minutes after each lecture is given to the students . find the duration of each lecture . <o> a ) num__76 minutes <o> b ) num__66 minutes <o> c ) num__88 minutes <o> d ) num__48 minutes <o> e ) num__218 minutes |
explanation : total time a student spends in college = num__4 hours num__20 minutes = num__260 minutes as there are num__5 lectures the number of breaks between lectures is num__4 . total time of the break = num__20 minutes hence the duration of each lecture is = ( num__260 – num__20 ) / num__5 = num__48 minutes answer : d <eor> d <eos> |
d |
multiply__5.0__4.0__ multiply__12.0__4.0__ round__48.0__ |
multiply__5.0__4.0__ multiply__12.0__4.0__ round__48.0__ |
| roy is now num__6 years older than julia and half of that amount older than kelly . if in num__4 years roy will be twice as old as julia then in num__4 years what would be roy ’ s age multiplied by kelly ’ s age ? <o> a ) num__72 <o> b ) num__84 <o> c ) num__90 <o> d ) num__96 <o> e ) num__108 |
r = j + num__6 = k + num__3 r + num__4 = num__2 ( j + num__4 ) ( j + num__6 ) + num__4 = num__2 j + num__8 j = num__2 r = num__8 k = num__5 in num__4 years ( r + num__4 ) ( k + num__4 ) = num__12 * num__9 = num__108 the answer is e . <eor> e <eos> |
e |
subtract__6.0__4.0__ add__6.0__2.0__ add__2.0__3.0__ multiply__6.0__2.0__ add__6.0__3.0__ multiply__9.0__12.0__ multiply__9.0__12.0__ |
subtract__6.0__4.0__ add__6.0__2.0__ add__2.0__3.0__ multiply__6.0__2.0__ add__6.0__3.0__ multiply__9.0__12.0__ multiply__9.0__12.0__ |
| ahok will arrange num__6 people of num__6 different heights for photograph by placing them in two rows of three so that each person in the first row is standing is standing in front of someone in the second row . the heights of the people within each row must increase from left to right and each person in the second row must be taller than the person standing in front of him or her . how many such arrangements of the num__6 people are possible ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__9 <o> d ) num__24 <o> e ) num__26 |
ahok will arrange num__6 people of num__6 different heights for photograph by placing them in two rows of three so that each person in the first row is standing is standing in front of someone in the second row . person with max height is in the second row person with min height is in the first row . we need to select num__1 person in the middle of each row . . . in the middle of the first row we can put only num__2 num__3 or num__4 . in the middle of the second row we can put only num__3 num__4 num__5 . if we select { num__3 in the sec . row num__2 in the first } { num__42 } { num__52 } { num__43 } { num__53 } { num__54 } . so there are num__0 * num__1 + num__1 * num__1 + num__2 * num__1 + num__1 * num__1 + num__1 * num__1 + num__0 * num__1 = num__5 cases . . . a <eor> a <eos> |
a |
coin_space__ vowel_space__ card_space__ negate_prob__1.0__ vowel_space__ |
coin_space__ vowel_space__ card_space__ negate_prob__1.0__ vowel_space__ |
| tough and tricky questions : number properties . what is the smallest positive integer x such that num__225 x is the cube of a positive integer ? <o> a ) num__5 <o> b ) num__15 <o> c ) num__30 <o> d ) num__60 <o> e ) num__90 |
we want to know the smallest x that will make num__225 x a cube of some number . let ' s call that number y . let ' s first figure out what we ' re working with . the prime factorization of num__225 can be visualized : . . . . . . . . . . . num__225 . . . . . . . . / . . . . . . . \ . . . . . . num__25 . . . . . . . num__9 . . . . . / . . \ . . . . . . / . . . \ . . . num__5 . . . . num__5 . . . num__3 . . . . . num__3 so we have num__5 * num__5 * num__3 * num__3 that can be multiplied together to get num__225 . now we need to figure out what we need to make num__225 * x into a cube of y ( y ^ num__3 = num__225 * x ) . we have two num__5 s and two num__3 s . to arrange these numbers in identical sets ( num__35 ) we need at least one more num__5 and one num__3 . each of these numbers will give us the value of y ( num__3 * num__5 = num__15 ) which multiplied by itself three times gives us num__225 * x . looking at the factors we need to complete the triples we get num__5 * num__3 = num__15 . we know this is the smallest number possible because prime factors by definition can not be broken down any further . therefore we can go with answer choice b . if time permits we can do a sanity check . we calculated that y should be num__3 * num__5 or num__15 . num__15 * num__15 * num__15 = num__3375 . also num__225 * num__15 = num__3375 . answer : b <eor> b <eos> |
b |
multiply__3.0__5.0__ volume_cube__15.0__ multiply__3.0__5.0__ |
multiply__3.0__5.0__ multiply__225.0__15.0__ multiply__3.0__5.0__ |
| by selling num__150 mangoes a fruit - seller gains the selling price of num__30 mangoes . find the gain percent ? <o> a ) num__22.0 <o> b ) num__25.0 <o> c ) num__28.0 <o> d ) num__29.0 <o> e ) num__21 % |
sp = cp + g num__150 sp = num__150 cp + num__30 sp num__120 sp = num__150 cp num__120 - - - num__30 cp num__100 - - - ? = > num__25.0 answer : b <eor> b <eos> |
b |
percent__25.0__100.0__ |
percent__25.0__100.0__ |
| of num__60 children num__30 are happy num__10 are sad and num__20 are neither happy nor sad . there are num__17 boys and num__43 girls . if there are num__6 happy boys and num__4 sad girls how many boys are neither happy nor sad ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
venn diagrams are useful for multiple values of a single variable e . g . state of mind - happy / sad / neither . when you have two or more variables such as here where you have gender - boy / girl too it becomes unwieldy . in this case either use the table or logic . table method is shown above ; here is how you will use logic : there are num__6 happy boys . there are num__4 sad girls but total num__10 sad children . so rest num__6 sad children must be sad boys . we have num__6 happy boys and num__6 sad boys . total we have num__17 boys . so num__17 - num__6 - num__6 = num__5 boys must be neither happy nor sad . answer ( a ) <eor> a <eos> |
a |
vowel_space__ vowel_space__ |
vowel_space__ vowel_space__ |
| in a weight - lifting competition the total weight of joe ' s two lifts was num__900 pounds . if twice the weight of his first lift was num__300 pounds more than the weight of his second lift what was the weight in pounds of his first lift ? <o> a ) num__225 <o> b ) num__275 <o> c ) num__325 <o> d ) num__350 <o> e ) num__400 |
this problem is a general word translation . we first define variables and then set up equations . we can define the following variables : f = the weight of the first lift s = the weight of the second lift we are given that the total weight of joe ' s two lifts was num__900 pounds . we sum the two variables to obtain : f + s = num__900 we are also given that twice the weight of his first lift was num__300 pounds more than the weight of his second lift . we express this as : num__2 f = num__300 + s num__2 f – num__300 = s we can now plug in ( num__2 f – num__300 ) for s into the first equation so we have : f + num__2 f – num__300 = num__900 num__3 f = num__1200 f = num__400 answer is e . <eor> e <eos> |
e |
divide__900.0__300.0__ add__900.0__300.0__ divide__1200.0__3.0__ divide__1200.0__3.0__ |
divide__900.0__300.0__ add__900.0__300.0__ divide__1200.0__3.0__ divide__1200.0__3.0__ |
| find the invalid no . from the following series num__3 num__8 num__15 num__31 num__63 num__127 num__255 <o> a ) num__11 <o> b ) num__8 <o> c ) num__27 <o> d ) num__63 <o> e ) num__127 |
go on multiplying the number by num__2 and adding num__1 to it to get the next number . so num__8 is wrong . b <eor> b <eos> |
b |
subtract__3.0__2.0__ multiply__8.0__1.0__ |
subtract__3.0__2.0__ multiply__8.0__1.0__ |
| num__0.7 of the population of the country of venezia lives in montague province while the rest lives in capulet province . in the upcoming election num__80.0 of montague residents support romeo while num__70.0 of capulet residents support juliet ; each resident of venezia supports exactly one of these two candidates . rounded if necessary to the nearest percent the probability that a juliet supporter chosen at random resides in capulet is <o> a ) num__28.0 <o> b ) num__41.0 <o> c ) num__60.0 <o> d ) num__72.0 <o> e ) num__78 % |
total population = num__100 ( assume ) . num__0.7 * num__100 = num__70 people from montague . num__0.3 * num__100 = num__30 people from capulet . num__0.2 * num__70 = num__14 people from montague support juliet . num__0.7 * num__30 = num__21 people from capulet support juliet . the probability that a juliet supporter chosen at random resides in capulet is num__21 / ( num__14 + num__21 ) = ~ num__60 . answer : c <eor> c <eos> |
c |
percent__70.0__30.0__ percent__100.0__60.0__ |
percent__70.0__30.0__ percent__100.0__60.0__ |
| john has rs num__360 for his expenses . if he exceeds his days by num__4 days he must cut down daily expenses by rs num__3 . the number of days of john ' s tour program is <o> a ) num__10 <o> b ) num__15 <o> c ) num__20 <o> d ) num__24 <o> e ) num__27 |
let john under takes a tour of x days . then expenses for each day = num__360 x num__360 x + num__4 = num__360 x − num__3 x = num__20 and − num__24 hence x = num__20 days . c <eor> c <eos> |
c |
add__4.0__20.0__ round__20.0__ |
add__4.0__20.0__ round__20.0__ |
| tickets numbered num__1 to num__20 are mixed up and then a ticket is drawn at random . what is the probability that the ticket drawn bears a number which is a multiple of num__3 ? <o> a ) num__0.3 <o> b ) num__0.5 <o> c ) num__0.4 <o> d ) num__0.428571428571 <o> e ) num__0.166666666667 |
here s = { num__1 num__23 . . . . . . num__1920 } e = event getting a multiple of num__3 = { num__36 num__912 num__1518 } probability = num__0.3 = num__0.3 correct option is a <eor> a <eos> |
a |
add__20.0__3.0__ multiply__1.0__0.3__ |
add__20.0__3.0__ multiply__1.0__0.3__ |
| a can do a work in num__15 days and b in num__20 days . if they work on it together for num__4 days then the fraction of the work that is left is : <o> a ) num__0.25 <o> b ) num__0.1 <o> c ) num__0.466666666667 <o> d ) num__0.533333333333 <o> e ) num__0.153846153846 |
a ' s num__1 day ' s work = num__0.0666666666667 ; b ' s num__1 day ' s work = num__0.05 ; ( a + b ) ' s num__1 day ' s work = ( num__0.0666666666667 ) + ( num__0.05 ) = num__0.116666666667 ( a + b ) ' s num__4 day ' s work = ( num__0.116666666667 ) * num__4 = num__0.466666666667 therefore remaining work = ( num__1 - ( num__0.466666666667 ) ) = num__0.533333333333 option d <eor> d <eos> |
d |
divide__1.0__15.0__ divide__1.0__20.0__ add__0.05__0.0667__ subtract__1.0__0.4667__ multiply__1.0__0.5333__ |
divide__1.0__15.0__ divide__1.0__20.0__ add__0.05__0.0667__ subtract__1.0__0.4667__ multiply__1.0__0.5333__ |
| if y is the smallest positive integer such that num__4410 multiplied by y is the square of an integer then y must be <o> a ) num__2 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__10 |
num__4410 = num__2 * num__3 ^ num__2 * num__5 * num__7 ^ num__2 to be perfect square we need to multiply by at least num__2 * num__5 = num__10 . the answer is e . <eor> e <eos> |
e |
rectangle_perimeter__2.0__3.0__ rectangle_perimeter__2.0__3.0__ |
multiply__2.0__5.0__ multiply__2.0__5.0__ |
| if the perimeter of a rectangular garden is num__600 m its length when its breadth is num__150 m is ? <o> a ) num__150 m <o> b ) num__778 m <o> c ) num__200 m <o> d ) num__276 m <o> e ) num__971 m |
num__2 ( l + num__150 ) = num__600 = > l = num__150 m answer : a <eor> a <eos> |
a |
round__150.0__ |
round__150.0__ |
| a num__6.0 stock yields num__8.0 . the market value of the stock is : <o> a ) rs . num__48 <o> b ) rs . num__75 <o> c ) rs . num__96 <o> d ) rs . num__133.33 <o> e ) rs . num__122 |
explanation : for an income of rs . num__8 investment = rs . num__100 . for an income of rs . num__6 investment = rs . num__12.5 * num__6 = rs . num__75 . market value of rs . num__100 stock = rs . num__75 . answer is b <eor> b <eos> |
b |
percent__75.0__100.0__ |
percent__75.0__100.0__ |
| for all positive integers m [ m ] = num__3 m when m is odd and [ m ] = ( num__0.5 ) * m when m is even . what is [ num__5 ] * [ num__8 ] equivalent to ? <o> a ) [ num__40 ] <o> b ) [ num__60 ] <o> c ) [ num__80 ] <o> d ) [ num__120 ] <o> e ) [ num__150 ] |
[ num__5 ] * [ num__8 ] = num__15 * num__4 = num__60 = ( num__0.5 ) ( num__120 ) = [ num__120 ] the answer is d . <eor> d <eos> |
d |
multiply__3.0__5.0__ multiply__0.5__8.0__ multiply__4.0__15.0__ multiply__8.0__15.0__ multiply__8.0__15.0__ |
multiply__3.0__5.0__ multiply__0.5__8.0__ multiply__4.0__15.0__ multiply__8.0__15.0__ multiply__8.0__15.0__ |
| if a > num__1 which of the following is equal to ( num__2 a - num__2 ) / ( a ^ num__2 + num__2 a - num__3 ) ? <o> a ) a <o> b ) a + num__3 <o> c ) num__2 / ( a - num__1 ) <o> d ) num__2 / ( a + num__3 ) <o> e ) ( a - num__1 ) / num__2 |
here ' s the algebraic approach : ( num__2 a - num__2 ) / ( a ^ num__2 + num__2 a - num__3 ) can be rewritten as . . . num__2 ( a - num__1 ) / ( a + num__3 ) ( a - num__1 ) we can simplify the fraction which leaves us with . . . num__2 / ( a + num__3 ) ans : d <eor> d <eos> |
d |
multiply__1.0__2.0__ |
divide__2.0__1.0__ |
| four children have small toys . the first child has num__0.1 of the toys the second child has num__12 more toys than the first the third child has one more toy of what the first child has and the fourth child has double the third child . how many toys are there ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__30 <o> d ) num__40 <o> e ) num__50 |
x : the total number of toys x / num__10 : the number of toys for first child x / num__10 + num__12 : the number of toys for second child x / num__10 + num__1 : the number of toys for the third child num__2 ( x / num__10 + num__1 ) : the number of toys for the fourth child x / num__10 + x / num__10 + num__12 + x / num__10 + num__1 + num__2 ( x / num__10 + num__1 ) = x x = num__30 toys : solve for x <eor> c <eos> |
c |
reverse__0.1__ multiply__0.1__10.0__ subtract__12.0__10.0__ multiply__1.0__30.0__ |
reverse__0.1__ multiply__0.1__10.0__ subtract__12.0__10.0__ divide__30.0__1.0__ |
| the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is num__24 kmph find the speed of the stream ? <o> a ) num__18 kmph <o> b ) num__6 kmph <o> c ) num__8 kmph <o> d ) num__10 kmph <o> e ) num__12 kmph |
the ratio of the times taken is num__2 : num__1 . the ratio of the speed of the boat in still water to the speed of the stream = ( num__2 + num__1 ) / ( num__2 - num__1 ) = num__3.0 = num__3 : num__1 speed of the stream = num__8.0 = num__8 kmph . answer : c <eor> c <eos> |
c |
add__1.0__2.0__ divide__24.0__3.0__ round__8.0__ |
add__1.0__2.0__ divide__24.0__3.0__ divide__24.0__3.0__ |
| what range of values of ' x ' will satisfy the inequality num__15 x - num__2 / x > num__1 ? <o> a ) x > num__0.4 <o> b ) x < num__0.333333333333 <o> c ) - num__0.333333333333 < x < num__0.4 x > num__7.5 <o> d ) - num__0.333333333333 < x < num__0 x > num__0.4 <o> e ) x < - - num__0.333333333333 or x > num__0.4 |
this will hold true for the following values of ' x ' : - num__1313 < x < num__0.4 and x < num__0 . combining we get - num__0.333333333333 < x < num__0 . therefore the range of values of x for which the above inequality will hold true is : - num__0.333333333333 < x < num__0 â ˆ ª x > num__0.4 . answert : d <eor> d <eos> |
d |
round_down__0.4__ multiply__1.0__0.3333__ |
round_down__0.4__ multiply__1.0__0.3333__ |
| when num__23 is divided by the positive integer k the remainder is num__3 for how many different values of k is this true ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
this means that num__20 must be a multiple of k . the factors of num__20 are num__1 num__2 num__4 num__5 num__10 and num__20 . out of these k can be num__4 num__5 num__10 and num__20 . the answer is d . <eor> d <eos> |
d |
coin_space__ vowel_space__ choose__4.0__3.0__ |
coin_space__ vowel_space__ choose__4.0__3.0__ |
| tough and tricky questions : ratios num__0.6 of a certain class left on a field trip . num__0.666666666667 of the students who stayed behind did not want to go on the field trip ( all the others did want to go ) . when another vehicle was located num__0.5 of the students who did want to go on the field trip but had been left behind were able to join . what fraction of the class ended up going on the field trip ? <o> a ) num__0.5 <o> b ) num__0.666666666667 <o> c ) num__0.733333333333 <o> d ) num__0.766666666667 <o> e ) num__0.8 |
let total no . of students be num__30 students left on a field trip = num__0.6 ( num__30 ) = num__18 thus num__12 students left behind num__0.666666666667 ( num__12 ) = num__8 students did not want to go . thus num__4 students want to go on the trip num__0.5 ( num__4 ) = num__2 were able to join the other students on the field trip . thus total number of students that went on a field trip = num__18 + num__2 = num__20 thus required fraction = num__0.666666666667 = num__0.666666666667 b <eor> b <eos> |
b |
multiply__0.6__30.0__ subtract__30.0__18.0__ multiply__0.5__8.0__ reverse__0.5__ add__2.0__18.0__ divide__8.0__12.0__ |
multiply__0.6__30.0__ subtract__30.0__18.0__ multiply__0.5__8.0__ reverse__0.5__ add__2.0__18.0__ divide__8.0__12.0__ |
| a car started running at a speed of num__30 km / hr and the speed of the car was increased by num__2 km / hr at the end of every hour . find the total distance covered by the car in the first num__7 hours of the journey . <o> a ) num__342 km <o> b ) num__352 km <o> c ) num__252 km <o> d ) num__742 km <o> e ) num__382 km |
the total distance covered by the car in the first num__7 hours = num__30 + num__32 + num__34 + num__36 + num__38 + num__40 + num__42 = sum of num__7 terms in ap whose first term is num__30 and last term is num__42 = num__3.5 [ num__30 + num__42 ] = num__252 km . answer : c <eor> c <eos> |
c |
add__30.0__2.0__ add__2.0__32.0__ add__2.0__34.0__ add__2.0__36.0__ add__2.0__38.0__ add__2.0__40.0__ divide__7.0__2.0__ multiply__7.0__36.0__ round__252.0__ |
add__30.0__2.0__ add__2.0__32.0__ add__2.0__34.0__ add__2.0__36.0__ add__2.0__38.0__ add__2.0__40.0__ divide__7.0__2.0__ multiply__7.0__36.0__ round__252.0__ |
| john has num__10 pairs of matched socks . if he loses num__11 individual socks what is the greatest number of pairs of matched socks he can have left ? <o> a ) num__7 <o> b ) num__6 <o> c ) num__5 <o> d ) num__4 <o> e ) num__3 |
because we have to maximize the pair of matched socks we will remove num__5 pairs ( num__10 socks ) out of num__10 pairs num__1 sock from the num__6 th pair . thus the no of matching socks pair remaining = num__10 - num__6 = num__4 answer d <eor> d <eos> |
d |
subtract__11.0__10.0__ subtract__11.0__5.0__ subtract__10.0__6.0__ subtract__10.0__6.0__ |
subtract__11.0__10.0__ subtract__11.0__5.0__ subtract__10.0__6.0__ subtract__10.0__6.0__ |
| the price of the jewel passing through three hands rises on the whole by num__62.0 . if the first and the second sellers num__20.0 and num__25.0 profit respectively find the percentage profit earned by the third seller . <o> a ) num__20 <o> b ) num__8 <o> c ) num__5 <o> d ) num__2 <o> e ) num__30 |
let the original price of the jewel be $ p and let the profit earned by the third seller be x % then ( num__100 + x ) % of num__125.0 of num__120.0 of p = num__162.0 of p ( ( num__100 + x ) / num__100 * num__1.25 * num__1.2 * p ) = ( num__1.62 * p ) = = > ( num__100 + x ) = ( num__162 * num__100 * num__100 ) / ( num__125 * num__120 ) = num__108 = > x = num__8.0 answer b ) <eor> b <eos> |
b |
percent__100.0__8.0__ |
percent__100.0__8.0__ |
| what is the value of num__8 num__0.333333333333 % of num__600 + num__37 ½ of num__400 <o> a ) num__100 <o> b ) num__300 <o> c ) num__150 <o> d ) num__200 <o> e ) num__250 |
num__25 * num__2.0 + num__75 * num__2.0 = num__50 + num__150 = num__200 answer : d <eor> d <eos> |
d |
divide__600.0__8.0__ divide__400.0__8.0__ multiply__2.0__75.0__ multiply__8.0__25.0__ multiply__8.0__25.0__ |
divide__600.0__8.0__ multiply__2.0__25.0__ multiply__2.0__75.0__ multiply__8.0__25.0__ multiply__8.0__25.0__ |
| if one - third of one - fourth of a number is num__15 then three - tenth of that number is : <o> a ) num__48 <o> b ) num__40 <o> c ) num__50 <o> d ) num__54 <o> e ) num__64 |
let the number be x . then num__0.333333333333 of num__0.25 of x = num__15 x = num__15 x num__12 = num__180 required number = ( num__0.3 ) x num__180 = num__54 answer : d <eor> d <eos> |
d |
multiply__15.0__12.0__ multiply__0.3__180.0__ multiply__0.3__180.0__ |
multiply__15.0__12.0__ multiply__0.3__180.0__ multiply__0.3__180.0__ |
| sam purchased num__20 dozens of toys at t he rate of rs . num__375 per dozen . he sold each one of them at the rate of rs . num__33 . what was his percentage profit ? <o> a ) num__3.5 <o> b ) num__4.5 <o> c ) num__5.6 percent <o> d ) num__6.5 <o> e ) none |
solution c . p of num__1 toy = rs . ( num__31.25 ) = rs . num__31.25 s . p of num__1 toy = rs . num__33 . therefore profit = ( num__1.75 / num__31.25 × num__100 ) % = ( num__5.6 ) % = num__5.6 . answer c <eor> c <eos> |
c |
percent__100.0__5.6__ |
percent__100.0__5.6__ |
| a pump can fill a tank with water in num__5 hours . because of a leak it took num__7 hours to fill the tank . the leak can drain all the water in ? <o> a ) num__15 hr num__10 min <o> b ) num__16 hr num__20 min <o> c ) num__17 hr num__30 min <o> d ) num__15 hr <o> e ) num__14 hr num__25 min |
work done by the leak in num__1 hour = num__0.2 - num__0.142857142857 = num__0.0571428571429 leak will empty the tank in num__17.5 hrs = num__17 hr num__30 min answer is c <eor> c <eos> |
c |
divide__1.0__5.0__ divide__1.0__7.0__ subtract__0.2__0.1429__ round__17.0__ |
divide__1.0__5.0__ divide__1.0__7.0__ subtract__0.2__0.1429__ round__17.0__ |
| what is the least number . which should be added to num__221 to make it a perfect square ? <o> a ) num__5 <o> b ) num__8 <o> c ) num__4 <o> d ) num__6 <o> e ) num__7 |
num__221 + num__4 = num__225 num__15 ^ num__2 answer : c <eor> c <eos> |
c |
add__221.0__4.0__ subtract__225.0__221.0__ |
add__221.0__4.0__ subtract__225.0__221.0__ |
| how many numbers from num__10 to num__100 are exactly divisible by num__9 ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__16 <o> e ) num__18 |
num__1.11111111111 = num__1 and num__11.1111111111 = num__11 = = > num__11 - num__1 = num__10 . therefore num__10 digits . a ) <eor> a <eos> |
a |
divide__10.0__9.0__ round_down__1.1111__ add__10.0__1.1111__ round_down__11.1111__ multiply__10.0__1.0__ |
divide__10.0__9.0__ subtract__10.0__9.0__ add__10.0__1.1111__ round_down__11.1111__ subtract__11.0__1.0__ |
| a person incurs num__5.0 loss by selling a watch for $ num__1140 . at what price should the watch be sold to earn num__5.0 profit ? <o> a ) num__1160 <o> b ) num__1150 <o> c ) num__1260 <o> d ) num__1260 <o> e ) num__1360 |
let the new s . p be $ x then ( num__100 - loss % ) : ( num__1 st s . p ) = ( num__100 + gain % ) : ( num__2 nd s . p ) ( num__100 - num__5 ) / num__1140 = ( num__100 + num__5 ) / x x = ( num__105 * num__1140 ) / num__95 = $ num__1260 answer c num__1260 <eor> c <eos> |
c |
percent__100.0__1260.0__ |
percent__100.0__1260.0__ |
| a farm has only chickens and pigs . when the manager of the farm counted the heads of the animals on the farm the number totaled up to num__95 . when the number of legs was counted the number totaled up to num__274 . how many more chickens than pigs are there on the farm ? note : each pig has num__4 legs and each chicken had num__2 legs . <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__11 |
let x be the number of chickens . then num__95 - x is the number of pigs . num__2 x + num__4 ( num__95 - x ) = num__274 num__2 x = num__106 x = num__53 ( chickens ) num__95 - x = num__42 ( pigs ) there are num__53 - num__42 = num__11 more chickens on the farm . the answer is e . <eor> e <eos> |
e |
divide__106.0__2.0__ subtract__95.0__53.0__ subtract__106.0__95.0__ subtract__106.0__95.0__ |
divide__106.0__2.0__ subtract__95.0__53.0__ subtract__106.0__95.0__ subtract__106.0__95.0__ |
| in the excel manufacturing company num__46 percent of the employees are men . if num__60 percent of the employees are unionized and num__70 percent of these are men what percent of the non - union employees are women ? <o> a ) num__90.0 <o> b ) num__87.5 <o> c ) num__66.7 <o> d ) num__50.0 <o> e ) num__36 % |
lets say there are total num__100 employees then men = num__46 & women = num__54 unionized emp = num__60 & non unionized emp = num__40 men unionized = . num__7 * num__60 = num__42 women unionized = num__18 women non - unionized = num__54 - num__18 = num__36 percentage of women non unionized = num__0.9 * num__100 = num__90.0 the answer is option a <eor> a <eos> |
a |
subtract__100.0__46.0__ subtract__100.0__60.0__ subtract__60.0__42.0__ subtract__54.0__18.0__ divide__54.0__60.0__ multiply__100.0__0.9__ multiply__100.0__0.9__ |
subtract__100.0__46.0__ subtract__100.0__60.0__ subtract__60.0__42.0__ subtract__54.0__18.0__ divide__54.0__60.0__ multiply__100.0__0.9__ multiply__100.0__0.9__ |
| if an article is sold at num__12.0 profit instead of num__16.0 profit then the profit would be rs . num__105 less . what is the cost price ? <o> a ) rs . num__3000 <o> b ) rs . num__4000 <o> c ) rs . num__3500 <o> d ) rs . num__4500 <o> e ) rs . num__6000 |
explanation : let the cost price of an article be rs . x . ( num__16.0 of x ) - ( num__12.0 of x ) = num__105 num__16 x / num__100 - num__12 x / num__100 = num__105 = > num__4 x = num__160 * num__100 = > x = num__1500 cost price = rs . num__4000 answer : b <eor> b <eos> |
b |
percent__100.0__4000.0__ |
percent__100.0__4000.0__ |
| the average age of an adult class is num__40 years . num__10 new students with an avg age of num__32 years join the class . therefore decreasing the average by num__4 year . find what was the original strength of class ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__16 <o> d ) num__20 <o> e ) num__22 |
let original strength = y then num__40 y + num__10 x num__32 = ( y + num__10 ) x num__36 â ‡ ’ num__40 y + num__320 = num__36 y + num__360 â ‡ ’ num__4 y = num__40 â ˆ ´ y = num__10 a <eor> a <eos> |
a |
subtract__40.0__4.0__ multiply__10.0__32.0__ add__40.0__320.0__ divide__40.0__4.0__ |
add__32.0__4.0__ multiply__10.0__32.0__ add__40.0__320.0__ divide__40.0__4.0__ |
| a certain musical scale has has num__13 notes each having a different frequency measured in cycles per second . in the scale the notes are ordered by increasing frequency and the highest frequency is twice the lowest . for each of the num__12 lower frequencies the ratio of a frequency to the next higher frequency is a fixed constant . if the lowest frequency is num__550 cycles per second then the frequency of the num__7 th note in the scale is how many cycles per second ? <o> a ) num__550 * sqrt num__2 <o> b ) num__550 * sqrt ( num__2 ^ num__7 ) <o> c ) num__550 * sqrt ( num__2 ^ num__12 ) <o> d ) num__550 * the twelfth root of ( num__2 ^ num__7 ) <o> e ) num__550 * the seventh root of ( num__2 ^ num__12 ) |
let the constant be k . f num__1 = num__550 f num__2 = num__550 k f num__3 = num__550 k * k = num__550 * k ^ num__2 f num__13 = num__550 * k ^ num__12 we know f num__13 = num__2 * f num__1 = num__2 * num__550 = num__1100 num__2.0 = k ^ num__12 k = twelfth root of num__2 for f num__7 . . . f num__7 = num__550 * k ^ num__6 ( as we wrote for f num__2 and f num__3 ) f num__7 = num__550 * ( twelfth root of num__2 ) ^ num__6 f num__7 = num__550 * sqrt ( num__2 ) the answer is a . <eor> a <eos> |
a |
subtract__13.0__12.0__ add__1.0__2.0__ multiply__550.0__2.0__ subtract__13.0__7.0__ multiply__550.0__1.0__ |
subtract__13.0__12.0__ add__1.0__2.0__ multiply__550.0__2.0__ multiply__2.0__3.0__ multiply__550.0__1.0__ |
| two cars are driving towards one another . the first car is traveling at a speed of num__120 km / h which is num__28.0 faster than the second car . if the distance between the cars is num__855 km how long will it takes the cars to meet ( in hours ) ? <o> a ) num__2.5 . <o> b ) num__3 . <o> c ) num__3.5 <o> d ) num__4 . <o> e ) num__4.5 . |
let s be the speed of the slower car . then the speed of the faster is num__1.28 s num__1.28 s = num__120 s = num__120 / num__1.28 cumulative speed / rate = s + num__1.28 s = num__2.28 s time taken to meet = num__855 / num__2.28 s = num__855 / num__213.75 = num__4 hrs answer : d <eor> d <eos> |
d |
divide__855.0__213.75__ round__4.0__ |
divide__855.0__213.75__ divide__855.0__213.75__ |
| three friends had dinner at a restaurant . when the bill was received akshitha paid num__0.333333333333 as much as veena paid and veena paid num__0.5 as much as lasya paid . what fraction of the bill did veena pay ? <o> a ) num__0.157894736842 <o> b ) num__0.25 <o> c ) num__0.272727272727 <o> d ) num__0.3 <o> e ) num__0.214285714286 |
let veena paid x so akshitha paid x / num__3 and lasya paid num__2 x so total bill paid is given by x + ( x / num__3 ) + num__2 x = num__1 we get i . e . x = num__0.3 answer : d <eor> d <eos> |
d |
reverse__0.5__ multiply__0.5__2.0__ multiply__1.0__0.3__ |
reverse__0.5__ multiply__0.5__2.0__ divide__0.3__1.0__ |
| a man goes downstream at num__15 kmph and upstream num__8 kmph . the speed of the stream is <o> a ) num__0 kmph <o> b ) num__3.5 kmph <o> c ) num__16 kmph <o> d ) num__2.5 kmph <o> e ) num__26 kmph |
speed of the stream = num__0.5 ( num__15 - num__8 ) kmph = num__3.5 kmph . correct option : b <eor> b <eos> |
b |
round__3.5__ |
round__3.5__ |
| difference between the length & breadth of a rectangle is num__15 m . if its perimeter is num__302 m then its area is ? ? we have : ( l - b ) = num__15 and num__2 ( l + b ) = num__302 or ( l + b ) = num__151 ? <o> a ) num__5446 m ^ num__2 <o> b ) num__5464 m ^ num__2 <o> c ) num__5644 m ^ num__2 <o> d ) num__5434 m ^ num__2 <o> e ) num__5344 m ^ num__2 |
solving the two equations we get : l = num__83 and b = num__68 . area = ( l x b ) = ( num__83 x num__68 ) m num__2 = num__5644 m ^ num__2 c <eor> c <eos> |
c |
subtract__151.0__83.0__ multiply__83.0__68.0__ multiply__83.0__68.0__ |
subtract__151.0__83.0__ multiply__83.0__68.0__ multiply__83.0__68.0__ |
| what is the perimeter of a square with area num__9 p ^ num__0.03125 ? <o> a ) num__3 p / num__2 <o> b ) num__3 p ^ num__0.5 <o> c ) num__3 p <o> d ) num__3 p ^ num__2 <o> e ) num__4 p / num__3 |
area of square ( side ) ^ num__2 = ( num__3 p / num__8 ) ^ num__2 therefore side of the square = num__3 p / num__8 perimeter of square = num__4 * side = num__4 * ( num__3 p / num__8 ) = num__3 p / num__2 answer is a . <eor> a <eos> |
a |
square_perimeter__2.0__ triangle_area__2.0__3.0__ |
power__2.0__3.0__ triangle_area__2.0__3.0__ |
| a train passes a station platform in num__40 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__180 m <o> b ) num__200 m <o> c ) num__240 m <o> d ) num__300 m <o> e ) none |
sol . speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x metres . then x + num__7.5 = num__15 ⇔ x + num__300 = num__600 ⇔ x = num__300 m . answer d <eor> d <eos> |
d |
multiply__20.0__15.0__ divide__300.0__40.0__ multiply__40.0__15.0__ round__300.0__ |
multiply__20.0__15.0__ divide__300.0__40.0__ multiply__40.0__15.0__ multiply__40.0__7.5__ |
| a person purchased a tv set for rs . num__16000 and a dvd player for rs . num__6250 . he sold both the items together for rs . num__37380 . what percentage of profit did he make ? <o> a ) num__80.0 <o> b ) num__68.0 <o> c ) num__40.0 <o> d ) num__70.0 <o> e ) num__90 % |
the total cp = rs . num__16000 + rs . num__6250 = rs . num__22250 and sp = rs . num__31150 profit ( % ) = ( num__37380 - num__22250 ) / num__22250 * num__100 = num__68.0 answer : b <eor> b <eos> |
b |
percent__68.0__100.0__ |
percent__68.0__100.0__ |
| the least number by which num__72 must be multiplied in order to produce a multiple of num__112 is <o> a ) num__6 <o> b ) num__12 <o> c ) num__14 <o> d ) num__18 <o> e ) num__20 |
solution required numbers is divisible by num__72 as well as by num__112 if it is divisible by their lcm which is num__1008 . now num__1008 when divided by num__72 gives quotient = num__14 . required number = num__14 . answer c <eor> c <eos> |
c |
divide__1008.0__72.0__ divide__1008.0__72.0__ |
divide__1008.0__72.0__ divide__1008.0__72.0__ |
| on final examination a student ' s average ( arithmetic mean ) score on num__4 subjects is num__90 . what must be his score on a num__5 th subject for his average score on the num__5 subjects to be num__92 ? <o> a ) num__60 <o> b ) num__70 <o> c ) num__80 <o> d ) num__90 <o> e ) num__100 |
( num__4 * num__90 + x ) / num__5 = num__92 x = ( num__5 * num__92 ) - ( num__4 * num__90 ) x = num__460 - num__360 total score required num__460 - num__360 = num__100 correct answer is e <eor> e <eos> |
e |
multiply__5.0__92.0__ multiply__4.0__90.0__ subtract__460.0__360.0__ subtract__460.0__360.0__ |
multiply__5.0__92.0__ multiply__4.0__90.0__ subtract__460.0__360.0__ subtract__460.0__360.0__ |
| if n is a natural number then ( num__6 n ^ num__2 + num__6 n ) is always divisible by ? <o> a ) num__6 only <o> b ) num__6 and num__12 both <o> c ) num__12 only <o> d ) by num__18 only <o> e ) none of these |
( num__6 n ^ num__2 + num__6 n ) = num__6 n ( n + num__1 ) which is always divisible by num__6 and num__12 both since n ( n + num__1 ) is always even . correct option : b <eor> b <eos> |
b |
multiply__6.0__2.0__ multiply__6.0__1.0__ |
multiply__6.0__2.0__ multiply__6.0__1.0__ |
| running at their respective constant rate machine x takes num__2 days longer to produce w widgets than machines y . at these rates if the two machines together produce num__5 w / num__4 widgets in num__3 days how many days would it take machine x alone to produce num__5 w widgets . <o> a ) num__4 <o> b ) num__6 <o> c ) num__8 <o> d ) num__30 <o> e ) num__12 |
i am getting num__12 . e . hope havent done any calculation errors . . approach . . let y = no . of days taken by y to do w widgets . then x will take y + num__2 days . num__1 / ( y + num__2 ) + num__1 / y = num__0.416666666667 ( num__0.416666666667 is because ( num__1.25 ) w widgets are done in num__3 days . so x widgets will be done in num__2.4 days or num__0.416666666667 th of a widget in a day ) solving we have y = num__4 = > x takes num__6 days to doing x widgets . so he will take num__30 days to doing num__5 w widgets . answer : d <eor> d <eos> |
d |
multiply__4.0__3.0__ subtract__5.0__4.0__ divide__5.0__12.0__ divide__5.0__4.0__ divide__3.0__1.25__ add__2.0__4.0__ multiply__5.0__6.0__ multiply__5.0__6.0__ |
multiply__4.0__3.0__ subtract__5.0__4.0__ divide__5.0__12.0__ divide__5.0__4.0__ divide__3.0__1.25__ add__2.0__4.0__ multiply__5.0__6.0__ divide__30.0__1.0__ |
| in an election contested by two parties party d secured num__12.0 of the total votes more than party r . if party r got num__132000 votes by how many votes did it lose the election ? <o> a ) num__24000 <o> b ) num__36000 votes <o> c ) num__168000 <o> d ) num__240000 <o> e ) num__300 |
000 |
d + r = num__100 - - - - ( num__1 ) d - r = num__12 - - - - ( num__2 ) solving ( num__1 ) and ( num__2 ) d = num__56.0 and r = num__44.0 num__44.0 of total = num__132000 total = num__300000 d = num__168000 difference = num__168000 - num__132000 = num__36000 answer : b <eor> b <eos> |
b |
b |
| factor | percent of respondents | user - friendly | num__56.0 | fast response time | num__48.0 | bargain prices | num__42.0 | the table gives three factors to be considered when choosing an internet service provider and the percent of the num__1600 respondents to a survey who cited that factor as important . if num__30 percent of the respondents cited both “ user - friendly ” and “ fast response time ” what is the maximum possible number of respondents who cited “ bargain prices ” but neither “ user - friendly ” nor “ fast response time ? ” <o> a ) num__312 <o> b ) num__336 <o> c ) num__360 <o> d ) num__384 <o> e ) num__416 |
the way i looked at is is as follows : userfriendly ( uf ) + fastresponse ( fr ) = num__30.0 uf leftover = num__56 - num__30 = num__26.0 fr leftover = num__48 - num__30 = num__18.0 sum these = num__74.0 ( ie . num__74.0 were either uf + fr uf fr ) num__26.0 leftover is the maximum number for bargain price num__0.26 * num__1600 = num__26 * num__16 = num__416 ( e ) . <eor> e <eos> |
e |
percent__26.0__1600.0__ percent__26.0__1600.0__ |
percent__26.0__1600.0__ percent__26.0__1600.0__ |
| a man saves a certain portion of his income during a year and spends the remaining portion on his personal expenses . next year his income increases by num__40.0 but his savings increase by num__100.0 . if his total expenditure in num__2 years is double his expenditure in num__1 st year what % age of his income in the first year did he save ? <o> a ) num__45.0 <o> b ) num__40.0 <o> c ) num__25.0 <o> d ) num__28.0 <o> e ) num__33.33 % |
i year best is to give a number to his income say num__100 . . and let saving be x . . so expenditure = num__100 - x next year - income = num__140 savings = num__2 x expenditure = num__140 - num__2 x . . now num__140 - num__2 x + num__100 - x = num__2 ( num__100 - x ) . . . num__240 - num__3 x = num__200 - num__2 x . . . . . . . . . . . . . . . . x = num__40 . . . saving % = num__0.4 * num__100 = num__40.0 answer : b <eor> b <eos> |
b |
add__40.0__100.0__ add__100.0__140.0__ add__2.0__1.0__ multiply__100.0__2.0__ divide__40.0__100.0__ multiply__40.0__1.0__ |
add__40.0__100.0__ add__100.0__140.0__ add__2.0__1.0__ multiply__100.0__2.0__ divide__40.0__100.0__ multiply__40.0__1.0__ |
| if x is the median of the set { num__4.0 num__3.66666666667 num__3.0 num__4.2 x } x could be <o> a ) num__4 <o> b ) num__3.4 <o> c ) num__3.2 <o> d ) num__4.28571428571 <o> e ) num__4.42857142857 |
the median is the middle number once all the numbers are arranged in increasing / decreasing order . we see that num__3.66666666667 = num__3 . something num__3.11111111111 = num__3 . something num__4.2 = num__4 . something num__4.5 = num__4 . something so x should greater than the smallest two numbers and smaller than the greatest two numbers . we can see that x = num__4 is possible . ( first look at the simplest option or the middle option since options are usually arranged in increasing / decreasing order ) answer ( a ) <eor> a <eos> |
a |
round_down__4.2__ |
round_down__4.2__ |
| if a fair die is rolled three times what is the probability that a num__3 occurs on at least one roll ? <o> a ) num__0.694444444444 <o> b ) num__0.578703703704 <o> c ) num__0.421296296296 <o> d ) num__0.305555555556 <o> e ) num__0.166666666667 |
thus the probability of at least num__1 roll = num__1 - probability of no num__3 s = num__1 - ( num__0.833333333333 ) ( num__0.833333333333 ) ( num__0.833333333333 ) = num__1 - num__0.578703703704 = num__0.421296296296 . num__0.833333333333 is the probability of not getting a num__3 in any num__1 roll with num__5 allowed numbers ( = num__1 num__24 num__56 ) out of a total of num__6 possibilities . answer : c <eor> c <eos> |
c |
negate_prob__0.5787__ vowel_space__ die_space__ negate_prob__0.5787__ |
negate_prob__0.5787__ vowel_space__ die_space__ negate_prob__0.5787__ |
| evaluate permutation num__3 p num__3 <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
explanation : npn = n ! num__3 p num__3 = num__3 ∗ num__2 ∗ num__1 = num__6 option a <eor> a <eos> |
a |
coin_space__ die_space__ die_space__ |
coin_space__ die_space__ die_space__ |
| num__1000 men have provisions for num__15 days . if num__200 more men join them for how many days will the provisions last now ? <o> a ) num__12.9 <o> b ) num__18.9 <o> c ) num__12.5 <o> d ) num__12.2 <o> e ) num__12.1 |
: num__1000 * num__15 = num__1200 * x x = num__12.5 answer : c <eor> c <eos> |
c |
add__1000.0__200.0__ round__12.5__ |
add__1000.0__200.0__ round__12.5__ |
| if r is a positive integer and r ^ num__2 is divisible by num__12 then the largest positive integer that must divide r ^ num__3 is <o> a ) num__2 ^ num__3 <o> b ) num__2 ^ num__6 <o> c ) num__3 ^ num__3 <o> d ) num__6 ^ num__3 <o> e ) num__12 ^ num__2 |
since r is an integer so r can not have a num__2 and sqrt num__3 ( because squaring this will give us a num__2 ^ num__2 and num__3 ( making the product as num__12 and making r ^ num__2 as a multiple of num__12 ) ) r ^ num__2 is divisible by num__12 ( num__12 = num__2 * num__2 * num__3 ) so r should have at least one num__2 and one num__3 so that r ^ num__2 has a num__2 ^ num__2 and two num__3 so r will have a num__2 and a num__3 . or r will be a multiple of num__6 so largest possible integer than should divide r ^ num__3 is num__6 ^ num__3 so answer will be d <eor> d <eos> |
d |
multiply__2.0__3.0__ multiply__2.0__3.0__ |
multiply__2.0__3.0__ multiply__2.0__3.0__ |
| if z is the smallest positive integer that is not prime and not a factor of num__50 ! what is the sum of the factors of z ? <o> a ) num__51 <o> b ) num__162 <o> c ) num__72 <o> d ) num__54 <o> e ) num__50 ! + num__2 |
in fact num__51 = num__3 * num__17 is a factor of num__50 ! . the smallest positive integer that is not prime and not a factor of num__50 ! is num__106 = num__2 * num__53 . the sum of the factors of num__106 is num__162 . answer : b . <eor> b <eos> |
b |
divide__51.0__3.0__ gcd__50.0__106.0__ add__50.0__3.0__ lcm__2.0__162.0__ |
divide__51.0__3.0__ gcd__50.0__106.0__ add__50.0__3.0__ lcm__2.0__162.0__ |
| a certain quantity of num__70.0 solution is replaced with num__25.0 solution such that the new concentration is num__35.0 . what is the fraction of the solution that was replaced ? <o> a ) num__0.25 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__0.777777777778 |
let ' s say that the total original mixture a is num__100 ml the original mixture a thus has num__70 ml of alcohol out of num__100 ml of solution you want to replace some of that original mixture a with another mixture b that contains num__25 ml of alcohol per num__100 ml . thus the difference between num__70 ml and num__25 ml is num__45 ml per num__100 ml of mixture . this means that every time you replace num__100 ml of the original mixture a by num__100 ml of mixture b the original alcohol concentration will decrease by num__45.0 . the question says that the new mixture let ' s call it c must be num__35.0 alcohol a decrease of only num__35.0 . therefore num__35 out of num__45 is num__0.777777777778 and e is the answer . <eor> e <eos> |
e |
percent__100.0__0.7778__ |
percent__100.0__0.7778__ |
| if num__6 x ^ num__2 + x - num__12 = ( ax + w ) ( cx + d ) then | a | + | w | + | c | + | d | = for a complete solution and more practice problems see this blog : http : / / magoosh . com / gmat / num__2012 / algebra - on . . . to - factor / <o> a ) num__10 <o> b ) num__12 <o> c ) num__15 <o> d ) num__18 <o> e ) num__20 |
num__6 x ^ num__2 + x - num__12 = num__6 x ^ num__2 + num__9 x - num__8 x - num__12 = > num__3 x ( num__2 x + num__3 ) - num__4 ( num__2 x + num__3 ) = > ( num__2 x + num__3 ) ( num__3 x - num__4 ) = ( ax + w ) ( cx + d ) hence a = num__2 w = c = num__3 d = - num__4 so num__2 + num__3 + num__3 + | - num__4 | = num__2 + num__3 + num__3 + num__4 = num__12 answer b . <eor> b <eos> |
b |
add__6.0__2.0__ divide__6.0__2.0__ subtract__6.0__2.0__ multiply__6.0__2.0__ |
add__6.0__2.0__ subtract__12.0__9.0__ subtract__6.0__2.0__ add__3.0__9.0__ |
| the average age num__20 members of a committee are the same as it was num__2 years ago because an old number has been replaced by a younger number . find how much younger is the new member than the old number ? <o> a ) num__18 <o> b ) num__99 <o> c ) num__77 <o> d ) num__40 <o> e ) num__12 |
num__20 * num__2 = num__40 years answer : d <eor> d <eos> |
d |
multiply__20.0__2.0__ multiply__20.0__2.0__ |
multiply__20.0__2.0__ multiply__20.0__2.0__ |
| a man has rs . num__480 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__45 <o> b ) num__60 <o> c ) num__75 <o> d ) num__90 <o> e ) num__100 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__480 num__16 x = num__480 x = num__30 . hence total number of notes = num__3 x = num__90 . answer : option d <eor> d <eos> |
d |
divide__480.0__16.0__ divide__30.0__10.0__ multiply__3.0__30.0__ multiply__3.0__30.0__ |
divide__480.0__16.0__ divide__30.0__10.0__ multiply__3.0__30.0__ multiply__3.0__30.0__ |
| x can finish a work in num__18 days . y can finish the same work in num__15 days . yworked for num__10 days and left the job . how many days does x alone need to finish the remaining work ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
work done by x in num__1 day = num__0.0555555555556 work done by y in num__1 day = num__0.0666666666667 work done by y in num__10 days = num__0.666666666667 = num__0.666666666667 remaining work = num__1 – num__0.666666666667 = num__0.333333333333 number of days in which x can finish the remaining work = ( num__0.333333333333 ) / ( num__0.0555555555556 ) = num__6 c <eor> c <eos> |
c |
divide__1.0__18.0__ divide__1.0__15.0__ divide__10.0__15.0__ subtract__1.0__0.6667__ round__6.0__ |
divide__1.0__18.0__ divide__1.0__15.0__ divide__10.0__15.0__ subtract__1.0__0.6667__ divide__6.0__1.0__ |
| in traveling from a dormitory to a certain city a student went num__0.333333333333 of the way by foot num__0.6 of the way by bus and the remaining num__2 kilometers by car . what is the distance in kilometers from the dormitory to the city ? <o> a ) num__30 <o> b ) num__45 <o> c ) num__60 <o> d ) num__90 <o> e ) num__120 |
whole trip = distance by foot + distance by bus + distance by car x = num__0.333333333333 x + num__0.6 x + num__2 x - num__0.333333333333 x - num__0.6 x = num__2 x = num__30 km option : a <eor> a <eos> |
a |
round__30.0__ |
round__30.0__ |
| in measuring the sides of a rectangle one side is taken num__6.0 in excess and other num__5.0 in deficit . find the error percentage in the area calculated from these measurements . <o> a ) num__3.5 <o> b ) num__2.4 <o> c ) num__3.0 <o> d ) num__5.0 <o> e ) num__0.7 % |
say both sides of the rectangle are equal to num__100 ( so consider that we have a square ) . in this case the area is num__100 * num__100 = num__10000 . now the area obtained with wrong measurements would be num__106 * num__95 = num__10070 which is num__0.7 greater than the actual area . answer : e . <eor> e <eos> |
e |
percent__100.0__0.7__ |
percent__100.0__0.7__ |
| the product of two numbers is num__168 and the sum of their squares is num__289 . the sum of the number is ? <o> a ) a ) num__23 <o> b ) b ) num__25 <o> c ) c ) num__27 <o> d ) d ) num__31 <o> e ) e ) num__35 |
let the numbers be x and y . then xy = num__168 and x num__2 + y num__2 = num__289 . ( x + y ) num__2 = x num__2 + y num__2 + num__2 xy = num__289 + ( num__2 x num__168 ) = num__625 x + y = num__25 . option b <eor> b <eos> |
b |
divide__625.0__25.0__ |
divide__625.0__25.0__ |
| at num__1 : num__00 pm a truck left city p and headed toward city q at a constant speed of num__38 km / h . one hour later a car left city q and headed toward city p along the same road at a constant speed of num__42 km / h . if the distance between city p and city q is num__378 km at what time will the truck and the car meet each other ? <o> a ) num__5 : num__30 <o> b ) num__5 : num__45 <o> c ) num__6 : num__00 <o> d ) num__6 : num__15 <o> e ) num__6 : num__30 |
at num__2 : num__00 pm the truck and the car are num__340 km apart . the truck and the car complete a distance of num__80 km each hour . the time it takes to meet is num__4.25 = num__4.25 hours . they will meet at num__6 : num__15 pm . the answer is d . <eor> d <eos> |
d |
subtract__378.0__38.0__ add__38.0__42.0__ divide__340.0__80.0__ round__6.0__ |
subtract__378.0__38.0__ add__38.0__42.0__ divide__340.0__80.0__ round__6.0__ |
| if num__5 a = num__3125 then the value of num__5 ( a - num__3 ) is <o> a ) num__25 <o> b ) num__50 <o> c ) num__250 <o> d ) num__300 <o> e ) none |
sol . num__5 a = num__3125 = num__5 a = num__55 a = num__5 . num__5 ( num__1 - num__3 ) = num__5 ( num__5 - num__3 ) = num__5 ( num__2 ) = num__25 . answer a <eor> a <eos> |
a |
subtract__5.0__3.0__ multiply__1.0__25.0__ |
subtract__5.0__3.0__ multiply__1.0__25.0__ |
| find the ratio in which rice at rs . num__7.00 a kg be mixed with rice at rs . num__5.70 a kg to produce a mixture worth rs . num__6.30 a kg <o> a ) num__6 : num__7 <o> b ) num__2 : num__3 <o> c ) num__2 : num__1 <o> d ) num__2 : num__2 <o> e ) num__2 : num__8 |
by the rule of alligation : cost of num__1 kg rice of num__1 st kind cost of num__1 kg rice of num__2 nd kind required ratio = num__60 : num__70 = num__6 : num__7 answer : a <eor> a <eos> |
a |
round_down__6.3__ round_down__6.3__ |
round_down__6.3__ round_down__6.3__ |
| how many numbers greater than a thousand can be made using the following digits without repetition : num__1 num__0 num__3 num__4 and num__5 : <o> a ) num__96 ^ num__2 <o> b ) num__96 * num__2 is answer <o> c ) num__576 <o> d ) num__24 ^ num__2 <o> e ) num__24 * num__96 |
we have num__5 digits : { num__0 num__1 num__3 num__4 num__5 } we can have num__4 - digit or num__5 - digit numbers greater than num__1000 . # of num__4 - digit numbers greater than num__1000 = num__4 * num__4 * num__3 * num__2 = num__96 . the first digit can take num__4 values : num__1 num__3 num__4 or num__5 . so any value but num__0 since in this case a number will no longer be a num__4 - digit and becomes num__3 - digit . the second digit can take num__4 values ( num__5 values minus num__1 we already used for the first digit ) . . . # of num__5 - digit numbers greater than num__1000 = num__4 * num__4 * num__3 * num__2 * num__1 = num__96 . total : num__96 + num__96 = num__2 * num__96 . answer : b . <eor> b <eos> |
b |
subtract__3.0__1.0__ multiply__1.0__96.0__ |
subtract__3.0__1.0__ multiply__1.0__96.0__ |
| which of the following equations represents a line that is perpendicular to y = num__0.333333333333 * x + num__2 ? <o> a ) y − num__2 x = num__8 <o> b ) num__2 x + num__4 y = num__10 <o> c ) num__3 y + num__6 x = num__12 <o> d ) num__4 y − num__8 x = num__16 <o> e ) num__5 x − num__10 y = num__20 |
for two lines to be perpendicular the product of slopes should be equal to - num__1 . slope of line num__1 = num__0.5 slope of the line perpendicular to line num__1 must be - num__2 . option c can be rewritten as y = - num__2 x + num__4 - - > slope = - num__2 answer : d <eor> d <eos> |
d |
reverse__2.0__ divide__2.0__0.5__ divide__2.0__0.5__ |
reverse__2.0__ divide__2.0__0.5__ divide__2.0__0.5__ |
| walking at num__0.8 of his usual speed a man is num__10 mins too late . find his usual time . <o> a ) num__81 mins <o> b ) num__64 mins <o> c ) num__52 mins <o> d ) num__40 mins <o> e ) none |
speed = s time = t num__0.8 s * ( t + num__10 ) = st t = num__40 answer : d <eor> d <eos> |
d |
round__40.0__ |
round__40.0__ |
| there are num__7 stores in town that had a total of num__21 visitors on a particular day . however only num__11 people went shopping that day ; some people visited more than one store . if num__7 people visited exactly two stores each and everyone visited at least one store what is the largest number of stores anyone could have visited ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__4 <o> d ) num__9 <o> e ) num__2 |
num__7 people visited num__2 stores each for num__14 visits . to maximize the number of stores that one person visited let ' s assume that num__3 people visited num__1 store each . the number of remaining visits is num__21 - num__14 - num__3 = num__4 which is the maximum that one person could have visited . the answer is c . <eor> c <eos> |
c |
multiply__7.0__2.0__ divide__21.0__7.0__ subtract__3.0__2.0__ subtract__7.0__3.0__ subtract__7.0__3.0__ |
subtract__21.0__7.0__ subtract__14.0__11.0__ subtract__3.0__2.0__ subtract__7.0__3.0__ subtract__7.0__3.0__ |
| a certain sum becomes rs . num__20720 in four years and num__24080 in six years at simple interest . find sum and rate of interest ? <o> a ) num__28.0 <o> b ) num__12.0 <o> c ) num__72.0 <o> d ) num__82.0 <o> e ) num__13 % |
let the interest for one year be x . as amount = principal + interest we have p + num__4 x = num__20720 - - - ( num__1 ) ; p + num__6 x = num__24080 - - - ( num__2 ) solving the equations ( num__1 ) and ( num__2 ) we can get p = rs . num__14000 and x = rs . num__1680 interest for one year on rs . num__14000 is rs . num__1680 so r = ( num__100 * num__1680 ) / ( num__14000 * num__1 ) = num__12.0 p . a . answer : b <eor> b <eos> |
b |
percent__12.0__100.0__ |
percent__12.0__100.0__ |
| a train is num__360 meter long is running at a speed of num__45 km / hour . in what time will it pass a bridge of num__140 meter length ? <o> a ) num__22 <o> b ) num__28 <o> c ) num__40 <o> d ) num__27 <o> e ) num__12 |
speed = num__45 km / hr = num__45 * ( num__0.277777777778 ) m / sec = num__12.5 m / sec total distance = num__360 + num__140 = num__500 meter time = distance / speed = num__500 * ( num__0.08 ) = num__40 seconds . answer : c : <eor> c <eos> |
c |
add__360.0__140.0__ divide__500.0__12.5__ round__40.0__ |
add__360.0__140.0__ divide__500.0__12.5__ divide__500.0__12.5__ |
| the circumferences of two circles are num__264 meters and num__352 meters . find the difference between the areas of the larger and the smaller circles ? <o> a ) num__2312 sq m <o> b ) num__2871 sq m <o> c ) num__4312 sq m <o> d ) num__1566 sq m <o> e ) num__7177 sq m |
let the radii of the smaller and the larger circles be s m and l m respectively . num__2 ∏ s = num__264 and num__2 ∏ l = num__352 s = num__132.0 ∏ and l = num__176.0 ∏ difference between the areas = ∏ l num__2 - ∏ s num__2 = ∏ { num__1762 / ∏ num__2 - num__1322 / ∏ num__2 } = num__1762 / ∏ - num__1322 / ∏ = ( num__176 - num__132 ) ( num__176 + num__132 ) / ∏ = ( num__44 ) ( num__308 ) / ( num__3.14285714286 ) = ( num__2 ) ( num__308 ) ( num__7 ) = num__4312 sq m answer : c <eor> c <eos> |
c |
volume_rectangular_prism__2.0__7.0__308.0__ triangle_area__2.0__4312.0__ |
volume_rectangular_prism__2.0__7.0__308.0__ triangle_area__2.0__4312.0__ |
| a company d has num__30 percent of the employees are secretaries and num__45 percent are salespeople . if there are num__50 other employees of company d how many employees does company d have ? <o> a ) num__200 <o> b ) num__162 <o> c ) num__180 <o> d ) num__152 <o> e ) num__250 |
let the total number of employees in the company be x % of secretaries = num__30.0 % of salespeople = num__45.0 % of of employees other than secretaries and salespeople = num__100 - num__75 = num__25.0 but this number is given as num__50 so num__25.0 of x = num__50 x = num__200 therefore there a total of num__200 employees in the company d correct answer - a <eor> a <eos> |
a |
percent__100.0__200.0__ |
percent__100.0__200.0__ |
| the average ( arithmetic mean ) of seven numbers is num__12.2 if the sum of four of these numbers is num__42.8 what is the average of the other num__3 numbers ? <o> a ) ( a ) num__12.4 <o> b ) ( b ) num__14.2 <o> c ) ( c ) num__16.8 <o> d ) ( d ) num__18.6 <o> e ) ( e ) num__19.2 |
we ' re told that the average of num__7 numbers is num__12.2 ( sum of numbers ) / num__7 = num__12.2 sum of numbers = ( num__12.2 ) ( num__7 ) = num__85.4 next we ' re told that the sum of num__4 ( of the num__7 ) numbers is num__42.8 from this information we can calculate the sum of the other num__3 numbers : num__85.4 - num__42.8 = num__42.6 the question asks for the average of the other num__3 numbers . . . . num__42.6 / num__3 = num__14.2 b <eor> b <eos> |
b |
multiply__12.2__7.0__ subtract__7.0__3.0__ subtract__85.4__42.8__ divide__42.6__3.0__ divide__42.6__3.0__ |
multiply__12.2__7.0__ subtract__7.0__3.0__ subtract__85.4__42.8__ divide__42.6__3.0__ divide__42.6__3.0__ |
| the average of ten numbers is num__7 . if each number is multiplied by num__12 then the average of new set of numbers is : <o> a ) num__22 <o> b ) num__29 <o> c ) num__27 <o> d ) num__84 <o> e ) num__19 |
the avg will be = num__12 × num__7 = num__84 answer : d <eor> d <eos> |
d |
multiply__7.0__12.0__ multiply__7.0__12.0__ |
multiply__7.0__12.0__ multiply__7.0__12.0__ |
| how many prime factors in num__25 ^ num__10 * num__36 ^ num__10 * num__20 ^ num__10 ? <o> a ) num__90 <o> b ) num__80 <o> c ) num__70 <o> d ) num__60 <o> e ) num__50 |
num__25 ^ num__10 * num__36 ^ num__10 * num__20 ^ num__10 = ( num__5 * num__5 ) ^ num__10 * ( num__2 * num__2 * num__3 * num__3 ) ^ num__10 * ( num__2 * num__2 * num__5 ) ^ num__10 = ( num__5 ^ num__10 * num__5 ^ num__10 ) * ( num__2 ^ num__10 * num__2 ^ num__10 * num__3 ^ num__10 * num__3 ^ num__10 ) * ( num__2 ^ num__10 * num__2 ^ num__10 * num__5 ^ num__10 ) = num__5 ^ num__30 * num__2 ^ num__40 * num__3 ^ num__20 number of distinct prime factors = num__3 number of prime factors = num__30 + num__40 + num__20 = num__90 answer : a <eor> a <eos> |
a |
subtract__25.0__20.0__ divide__10.0__5.0__ subtract__5.0__2.0__ add__25.0__5.0__ add__10.0__30.0__ multiply__3.0__30.0__ multiply__3.0__30.0__ |
subtract__25.0__20.0__ divide__10.0__5.0__ subtract__5.0__2.0__ add__25.0__5.0__ add__10.0__30.0__ multiply__3.0__30.0__ multiply__3.0__30.0__ |
| two trains each num__180 m in length are running on the same parallel lines in opposite directions with the speed of each num__80 kmph . in what time will they cross each other completely ? <o> a ) num__7.8 sec <o> b ) num__8.1 sec <o> c ) num__8.3 sec <o> d ) num__8.2 sec <o> e ) num__8.4 sec |
d = num__180 m + num__180 m = num__360 m * num__0.001 = num__0.36 kms rs = num__80 + num__80 = num__160 kmph t = ( num__0.36 / num__160 ) * num__3600 = num__8.1 sec answer : b <eor> b <eos> |
b |
multiply__360.0__0.001__ round__8.1__ |
multiply__360.0__0.001__ round__8.1__ |
| the original price of a camera was displayed as a whole dollar amount . after adding sales tax of num__20 percent the final price was also a whole dollar amount . which of the following could be the final price of the camera ? <o> a ) $ num__222 <o> b ) $ num__209 <o> c ) $ num__211 <o> d ) $ num__220 <o> e ) $ num__215 |
final price = ( num__1 + num__0.2 ) * original price = num__1.20 * original price from options given only num__222 is divisible by num__1.20 as it is stated op is whole dollar amount . hence a <eor> a <eos> |
a |
add__1.0__0.2__ multiply__1.0__222.0__ |
add__1.0__0.2__ multiply__1.0__222.0__ |
| the c . p of num__10 pens is equal to the s . p of num__12 pens . find his gain % or loss % ? <o> a ) num__66 num__1.0 % <o> b ) num__16 num__0.666666666667 % <o> c ) num__16 num__1.33333333333 % <o> d ) num__76 num__0.666666666667 % <o> e ) num__66 num__0.666666666667 % |
explanation : num__10 cp = num__12 sp num__12 - - - num__2 cp loss num__100 - - - ? = > num__16 num__0.666666666667 % answer : c <eor> c <eos> |
c |
percent__100.0__16.0__ |
percent__100.0__16.0__ |
| dean winchester has got a long wooden stock of size num__60 feet he need to cut small stock of size num__1 feet long using his axe . he takes num__5 minutes to cut one small stock ( num__1 feet ) how long will he take to make num__60 such small stocks ? <o> a ) num__296 minutes . <o> b ) num__297 minutes . <o> c ) num__295 minutes . <o> d ) num__294 minutes . <o> e ) none |
solution : num__295 minutes . when he cut num__59 stocks the num__60 th stock will remain . num__59 * num__5 = num__295 minutes . answer c <eor> c <eos> |
c |
subtract__60.0__1.0__ round__295.0__ |
subtract__60.0__1.0__ multiply__1.0__295.0__ |
| if each edge of cube increased by num__20.0 the percentage increase in <o> a ) num__42 <o> b ) num__44 <o> c ) num__48 <o> d ) num__45 <o> e ) num__47 |
num__100 × ( num__120 ) / num__100 × ( num__120 ) / num__100 = num__144 = > num__44.0 answer is b . <eor> b <eos> |
b |
percent__100.0__44.0__ |
percent__100.0__44.0__ |
| a total of num__550 players participated in a single tennis knock out tournament . what is the total number of matches played in the tournament ? ( knockout means if a player loses he is out of the tournament ) . no match ends in a tie . <o> a ) num__511 <o> b ) num__512 <o> c ) num__549 <o> d ) num__255 <o> e ) num__1023 |
there are num__550 players only num__1 person wins num__549 players lose . in order to lose you must have lost a game . num__549 games . ans - c <eor> c <eos> |
c |
subtract__550.0__1.0__ subtract__550.0__1.0__ |
subtract__550.0__1.0__ subtract__550.0__1.0__ |
| o ( x ) represents the least odd integer greater than x whereas o ( x ) represents the greatest odd integer less than x . likewise e ( x ) represents the least even integer greater than x whereas e ( x ) represents the greatest even integer less than x . according to these definitions the value of o ( num__11.6 ) + e ( – num__10.4 ) + o ( – num__9.2 ) + e ( num__9.5 ) is : <o> a ) - num__3 <o> b ) - num__1 <o> c ) num__0 <o> d ) num__1 <o> e ) num__3 |
o ( num__11.6 ) + e ( – num__10.4 ) + o ( – num__9.2 ) + e ( num__9.5 ) = num__13 + ( - num__10 ) + ( - num__11 ) + num__8 = num__0 the answer is c . <eor> c <eos> |
c |
round_down__10.4__ round_down__11.6__ multiply__11.6__0.0__ |
round_down__10.4__ round_down__11.6__ multiply__11.6__0.0__ |
| after m students took a test the average score was num__66.0 . if the test has num__50 questions what is the least number of questions that the next student has to get right to bring the average score up to num__68.0 ? <o> a ) num__2 m - num__34 <o> b ) m + num__34 <o> c ) num__0.66 m + num__0.68 <o> d ) num__0.33 m + num__34 <o> e ) num__3 m - num__33 |
the total number of correct answers for m students is ( num__0.66 ) * num__50 * m = num__33 * m for an average of num__68.0 : ( total correct answers ) / ( m + num__1 ) = num__0.68 * num__50 = num__34 let x be the number of correct answers for the next student . ( x + num__33 m ) / m + num__1 = num__34 x + num__33 m = num__34 m + num__34 x = m + num__34 the answer is b . <eor> b <eos> |
b |
multiply__50.0__0.66__ multiply__50.0__0.68__ multiply__50.0__0.68__ |
multiply__50.0__0.66__ multiply__50.0__0.68__ multiply__50.0__0.68__ |
| the number of people at ovations bar in the hour from num__12 p . m . to num__1 p . m . was num__25.0 greater than the number of people there from num__11 a . m . to num__12 p . m . the number of the people at the bar from num__11 a . m . to num__12 p . m . was num__25.0 less than the number there from num__10 a . m . to num__11 a . m . if num__135 people were at the bar from num__11 a . m . to num__1 p . m . which of the following is the number of people in the bar from num__10 a . m . to num__11 a . m . ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__30 <o> d ) num__40 <o> e ) num__80 |
no of people from num__10 - num__11 is x no of people from num__11 - num__12 is num__3 x / num__4 no of people from num__12 to num__1 is ( num__1.25 ) ( num__3 x / num__4 ) given that num__3 x / num__4 + ( num__1.25 ) ( num__3 x / num__4 ) = num__27 x / num__16 = num__135 then x = num__80 e <eor> e <eos> |
e |
divide__12.0__3.0__ add__12.0__4.0__ multiply__1.0__80.0__ |
divide__12.0__3.0__ add__12.0__4.0__ divide__80.0__1.0__ |
| express a speed of num__54 kmph in meters per second ? <o> a ) num__10 mps <o> b ) num__18 mps <o> c ) num__15 mps <o> d ) num__17 mps <o> e ) num__12 mps |
num__54 * num__0.277777777778 = num__15 mps answer : c <eor> c <eos> |
c |
round__15.0__ |
round__15.0__ |
| a completes a work in num__10 days and b complete the same work in num__15 days . if both of them work together then the number of days required to complete the work will be <o> a ) num__6 days <o> b ) num__9 days <o> c ) num__10 days <o> d ) num__12 days <o> e ) num__13 days |
if a can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days therefore here the required number of days = num__10 × num__0.6 = num__6 days . answer : a <eor> a <eos> |
a |
km_to_mile_conversion__ multiply__10.0__0.6__ round__6.0__ |
km_to_mile_conversion__ multiply__10.0__0.6__ round__6.0__ |
| a train num__125 m long passes a man running at num__4 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is ? <o> a ) num__28 <o> b ) num__49 <o> c ) num__88 <o> d ) num__22 <o> e ) num__12 |
speed of the train relative to man = ( num__12.5 ) m / sec = ( num__12.5 ) m / sec . [ ( num__12.5 ) * ( num__3.6 ) ] km / hr = num__45 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__4 ) km / hr . x - num__4 = num__45 = = > x = num__49 km / hr . answer : b <eor> b <eos> |
b |
divide__125.0__10.0__ multiply__12.5__3.6__ add__4.0__45.0__ round__49.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ add__4.0__45.0__ round__49.0__ |
| four people each roll a four die once . find the probability that at least two people will roll the same number ? <o> a ) num__0.722222222222 <o> b ) num__0.764705882353 <o> c ) num__1.18181818182 <o> d ) num__0.684210526316 <o> e ) num__1.08333333333 |
the number of ways of rolling a dice where no two numbers probability that no one rolls the same number = num__6 x num__5 x num__4 x num__3 now total possibilities of rolling a dice = num__6464 the probability that a no one gets the same number = num__6 × num__5 × num__4 × num__364 = num__5186 × num__5 × num__4 × num__364 = num__518 so the probability that at least two people gets same number = num__1 − num__0.277777777778 = num__0.722222222222 answer : a <eor> a <eos> |
a |
die_space__ vowel_space__ negate_prob__0.2778__ negate_prob__0.2778__ |
die_space__ vowel_space__ negate_prob__0.2778__ negate_prob__0.2778__ |
| a farmer has an apple orchard consisting of fuji and gala apple trees . due to high winds this year num__10.0 of his trees cross pollinated . the number of his trees that are pure fuji plus the cross - pollinated ones totals num__153 while num__0.75 of all his trees are pure fuji . how many of his trees are pure gala ? <o> a ) num__27 <o> b ) num__33 <o> c ) num__55 <o> d ) num__77 <o> e ) num__88 |
let f = pure fuji g = pure gala and c - cross pollinated . c = num__10.0 of x where x is total trees . c = . num__1 x also num__3 x / num__4 = f and c + f = num__153 = > . num__1 x + num__0.75 x = num__153 = > x = num__180 num__180 - num__153 = pure gala = num__27 . answer a <eor> a <eos> |
a |
add__1.0__3.0__ subtract__180.0__153.0__ multiply__1.0__27.0__ |
add__1.0__3.0__ subtract__180.0__153.0__ subtract__180.0__153.0__ |
| if the operation Ø is defined for all positive integers x and w by x Ø w = ( num__2 ^ x ) / ( num__2 ^ w ) then ( num__3 Ø num__1 ) Ø num__1 = ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__8 <o> d ) num__16 <o> e ) num__32 |
num__3 Ø num__1 = num__2 ^ num__1.5 ^ num__1 = num__4 num__4 Ø num__1 = num__2 ^ num__2.0 = num__8 the answer is c . <eor> c <eos> |
c |
divide__3.0__2.0__ add__3.0__1.0__ multiply__2.0__4.0__ multiply__2.0__4.0__ |
divide__3.0__2.0__ add__3.0__1.0__ multiply__2.0__4.0__ multiply__2.0__4.0__ |
| a can run num__1 km in num__3 min . num__10 sec . and b can cover the same distance in num__3 min . num__20 sec . by what distance can a beat b ? <o> a ) num__50 m <o> b ) num__18 m <o> c ) num__96 m <o> d ) num__23 m <o> e ) num__13 m |
soln : clearly a beats b by num__10 sec . distance covered by b in num__10 sec . = ( num__1000 x num__10 ) / num__200 m = num__50 m . therefore a beats b by num__50 metres . option a <eor> a <eos> |
a |
multiply__10.0__20.0__ divide__1000.0__20.0__ round__50.0__ |
multiply__10.0__20.0__ divide__1000.0__20.0__ divide__1000.0__20.0__ |
| michael earns $ num__7.00 per hour for the first num__40 hours he works per week and twice this rate for overtime . if michael earned $ num__320 last week how many hours did he work ? <o> a ) num__43 <o> b ) num__44 <o> c ) num__45 <o> d ) num__46 <o> e ) num__47 |
$ num__7 * num__40 + $ num__12 * x = $ num__320 - - > x = num__3 hours . total working hours = num__40 + num__3 = num__43 . answer : a . <eor> a <eos> |
a |
add__40.0__3.0__ round__43.0__ |
add__40.0__3.0__ add__40.0__3.0__ |
| the distance light travels in one year is approximately num__5 num__870000 num__000000 miles . find the distance light travels in num__100 years ? <o> a ) num__587 × num__108 miles <o> b ) num__587 × num__10 - num__12 miles <o> c ) num__587 × num__1010 miles <o> d ) num__587 × num__10 - num__10 miles <o> e ) num__587 × num__1012 miles |
the distance of the light travels in num__100 years is : num__5 num__870000 num__000000 × num__100 miles . = num__587 num__000000 num__000000 miles . = num__587 × num__1012 miles . answer is b . <eor> b <eos> |
b |
round__587.0__ |
round__587.0__ |
| if an object travels at ten feet per second how many feet does it travel in one hour ? <o> a ) num__30 <o> b ) num__300 <o> c ) num__720 <o> d ) num__1800 <o> e ) num__36000 |
speed = num__10 feet per second . num__1 hour = num__60 x num__60 seconds = num__3600 seconds . total no of feet traveled in num__1 hour = num__3600 x num__10 = num__36000 answer e <eor> e <eos> |
e |
hour_to_min_conversion__ multiply__3600.0__10.0__ round__36000.0__ |
hour_to_min_conversion__ multiply__3600.0__10.0__ round__36000.0__ |
| jane makes toy bears . when she works with an assistant she makes num__80 percent more bears per week and works num__10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ? <o> a ) num__20.0 <o> b ) num__80.0 <o> c ) num__100.0 <o> d ) num__180.0 <o> e ) num__200 % |
let ' s assume just jane num__40 bears per num__40 / hrs a week so that is num__1 bear / hr . with an assistant she makes num__72 bears per num__36 hours a week or num__2 bears / hr ( [ num__40 bears * num__1.8 ] / [ num__40 hrs * . num__90 ] ) . [ ( num__2 - num__1 ) / num__1 ] * num__100.0 = num__100.0 answer : c <eor> c <eos> |
c |
divide__80.0__40.0__ divide__72.0__40.0__ add__80.0__10.0__ add__10.0__90.0__ round__100.0__ |
divide__80.0__40.0__ divide__72.0__40.0__ add__80.0__10.0__ add__10.0__90.0__ divide__100.0__1.0__ |
| what is the rate percent when the simple interest on rs . num__900 amount to rs . num__160 in num__4 years ? <o> a ) num__5.0 <o> b ) num__7.0 <o> c ) num__9.0 <o> d ) num__4.4 <o> e ) num__4 % |
num__160 = ( num__900 * num__4 * r ) / num__100 r = num__4.4 answer : d <eor> d <eos> |
d |
percent__4.4__100.0__ |
percent__4.4__100.0__ |
| a school having four classes only have student strength of num__10 num__50 num__30 and num__10 respectively . the pass percentages of these classes are num__20.0 num__30.0 num__60.0 and num__100.0 respectively . what is the pass percentage for the entire school ? <o> a ) num__56.0 <o> b ) num__76.0 <o> c ) num__34.0 <o> d ) num__66.0 <o> e ) num__45 % |
num__20.0 of num__10 + num__30.0 of num__50 + num__60.0 of num__30 + num__100.0 of num__10 = num__2 + num__15 + num__18 + num__10 = now num__45 of num__100 = num__45.0 answer : e <eor> e <eos> |
e |
percent__10.0__20.0__ percent__50.0__30.0__ percent__30.0__60.0__ percent__100.0__45.0__ |
percent__10.0__20.0__ percent__50.0__30.0__ percent__30.0__60.0__ percent__100.0__45.0__ |
| the simple form of the ratio num__0.666666666667 : num__0.4 is <o> a ) num__5 : num__7 <o> b ) num__5 : num__2 <o> c ) num__5 : num__9 <o> d ) num__5 : num__3 <o> e ) num__5 : num__4 |
num__0.666666666667 : num__0.4 = num__10 : num__6 = num__5 : num__3 answer : d <eor> d <eos> |
d |
subtract__10.0__5.0__ |
subtract__10.0__5.0__ |
| in a group of num__68 students each student is registered for at least one of three classes – history math and english . twenty - five students are registered for history twenty - five students are registered for math and thirty - nine students are registered for english . if only three students are registered for all three classes how many students are registered for exactly two classes ? <o> a ) num__13 <o> b ) num__10 <o> c ) num__15 <o> d ) num__8 <o> e ) num__7 |
a u b u c = a + b + c - ab - bc - ac + abc num__68 = num__25 + num__25 + num__39 - ab - bc - ac + num__3 = > ab + bc + ac = num__24 exactly two classes = ab + bc + ac - num__3 abc = num__24 - num__3 * num__3 = num__15 hence c <eor> c <eos> |
c |
subtract__39.0__24.0__ subtract__39.0__24.0__ |
subtract__39.0__24.0__ subtract__39.0__24.0__ |
| what is the place value of num__7 in the numeral num__2734 ? <o> a ) num__450 <o> b ) num__460 <o> c ) num__480 <o> d ) num__500 <o> e ) num__700 |
option ' e ' num__7 * num__100 = num__700 <eor> e <eos> |
e |
multiply__7.0__100.0__ multiply__7.0__100.0__ |
multiply__7.0__100.0__ multiply__7.0__100.0__ |
| si on a certain sum of money for num__3 yrs at num__8 percent / annum is num__0.5 the ci on rs . num__4000 for num__2 yrs at num__10 percent / annum . the sum placed on si is ? <o> a ) rs . num__1650 <o> b ) rs . num__1700 <o> c ) rs . num__1750 <o> d ) rs . num__1950 <o> e ) rs . num__1980 |
c . i . = rs . num__4000 x num__1 + num__10 num__2 - num__4000 num__100 = rs . num__4000 x num__11 x num__11 - num__4000 num__10 num__10 = rs . num__840 . sum = rs . num__420 x num__100 = rs . num__1750 . num__3 x num__8 c <eor> c <eos> |
c |
subtract__3.0__2.0__ add__3.0__8.0__ multiply__0.5__840.0__ multiply__1.0__1750.0__ |
subtract__3.0__2.0__ add__3.0__8.0__ multiply__0.5__840.0__ multiply__1.0__1750.0__ |
| a retailer bought a hat at wholesale and marked it up num__70.0 to its initial price of $ num__34 . by how many more dollars does he need to increase the price to achieve a num__100.0 markup ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
let x be the wholesale price . then num__1.7 x = num__34 and x = num__34 / num__1.7 = num__20 . to achieve a num__100.0 markup the price needs to be $ num__40 . the retailer needs to increase the price by $ num__6 more . the answer is c . <eor> c <eos> |
c |
divide__34.0__1.7__ subtract__40.0__34.0__ subtract__40.0__34.0__ |
divide__34.0__1.7__ subtract__40.0__34.0__ subtract__40.0__34.0__ |
| if you are given $ num__2 and the amount doubles every day how much money will you have after num__12 days ? <o> a ) $ num__256 <o> b ) $ num__512 <o> c ) $ num__1024 <o> d ) $ num__2048 <o> e ) $ num__4096 |
num__2 ^ num__12 = num__4096 the answer is e . <eor> e <eos> |
e |
round__4096.0__ |
round__4096.0__ |
| average of five numbers is - num__10 and the sum of three of the numbers is num__16 wat is the average of the other num__2 numbers ? <o> a ) - num__33 <o> b ) num__33 <o> c ) num__32 <o> d ) - num__31 <o> e ) num__24 |
let the five numbers be a b c d e . then their average is ( a + b + c + d + e num__5 ) = num__10 . ( a + b + c + d + e num__5 ) = num__10 . now three of the numbers have a sum of num__16 say a + b + c = num__16 a + b + c = num__16 . so substitute num__16 for a + b + ca + b + c in the average above : ( num__16 + d + e num__5 ) = num__10 . ( num__16 + d + e num__5 ) = num__10 . solving this equation for d + ed + e gives d + e = − num__66 d + e = − num__66 . finally dividing by num__2 ( to form the average ) gives ( d + e num__2 ) = − num__33 . ( d + e num__2 ) = − num__33 . hence the answer is a : - num__33 <eor> a <eos> |
a |
divide__10.0__2.0__ divide__66.0__2.0__ divide__66.0__2.0__ |
divide__10.0__2.0__ divide__66.0__2.0__ subtract__66.0__33.0__ |
| in covering a distance of num__30 km abhay takes num__2 hours more than sameer . if abhay doubles his speed then he would take num__1 hour less than sameer . abhay ' s speed is : <o> a ) num__5 km / hr <o> b ) num__15 km / hr <o> c ) num__50 km / hr <o> d ) num__20 km / hr <o> e ) num__25 km / hr |
let abhay ' s speed be x km / hr . then ( num__30 / x - num__15.0 x ) = num__3 num__6 x = num__30 x = num__5 km / hr . answer a <eor> a <eos> |
a |
divide__30.0__2.0__ add__2.0__1.0__ multiply__2.0__3.0__ divide__30.0__6.0__ round__5.0__ |
divide__30.0__2.0__ add__2.0__1.0__ multiply__2.0__3.0__ divide__30.0__6.0__ divide__30.0__6.0__ |
| the wages earned by robin is num__30.0 more than that earned by erica . the wages earned by charles is num__60.0 more than that earned by erica . how much % is the wages earned by charles more than that earned by robin ? <o> a ) num__23.0 <o> b ) num__18.75 <o> c ) num__30.0 <o> d ) num__50.0 <o> e ) num__100 % |
explanatory answer let the wages earned by erica be $ num__100 then wages earned by robin and charles will be $ num__130 and $ num__160 respectively . charles earns $ num__30 more than robin who earns $ num__130 . therefore charles ' wage is num__0.230769230769 * num__100 = num__23.07 . correct choice is ( a ) <eor> a <eos> |
a |
add__30.0__100.0__ add__30.0__130.0__ divide__30.0__130.0__ round_down__23.07__ |
add__30.0__100.0__ add__30.0__130.0__ divide__30.0__130.0__ round_down__23.07__ |
| in a certain lottery the probability that a number between num__1 and num__15 inclusive is drawn is num__0.333333333333 . if the probability that a number num__1 or larger is drawn is num__0.666666666667 what is the probability that a number less than or equal to num__15 is drawn ? <o> a ) num__2 by num__3 <o> b ) num__0.333333333333 <o> c ) num__0.25 <o> d ) num__0.166666666667 <o> e ) num__0.4 |
you can simply use sets concept in this question . the formula total = n ( a ) + n ( b ) - n ( a and b ) is applicable here too . set num__1 : number num__1 or larger set num__2 : number num__15 or smaller num__1 = p ( set num__1 ) + p ( set num__2 ) - p ( set num__1 and set num__2 ) ( combined probability is num__1 because every number will be either num__12 or moreor num__20 or less or both ) num__0.666666666667 + p ( set num__2 ) - num__0.333333333333 = num__1 p ( set num__2 ) = num__0.666666666667 answer a <eor> a <eos> |
a |
multiply__1.0__2.0__ |
multiply__1.0__2.0__ |
| when positive integer x is divided by num__8 the quotient is y and the remainder is num__2 . when num__2 x is divided by num__7 the quotient is num__2 y and the remainder is num__4 . what is the value of num__6 y – x ? <o> a ) - num__1 <o> b ) num__1 <o> c ) - num__2 <o> d ) num__2 <o> e ) - num__3 |
( num__1 ) x = num__8 y + num__2 ( num__2 ) num__2 x = num__14 y + num__4 ( num__2 ) - ( num__1 ) : x = num__6 y + num__2 num__6 y - x = - num__2 the answer is c . <eor> c <eos> |
c |
subtract__8.0__7.0__ add__8.0__6.0__ divide__8.0__4.0__ |
subtract__8.0__7.0__ add__8.0__6.0__ subtract__8.0__6.0__ |
| if the range of the set of numbers { num__140 num__90 num__135 num__110 num__170 num__145 x num__100 num__140 } is num__100 which of the following could be x ? <o> a ) num__170 <o> b ) num__190 <o> c ) num__210 <o> d ) num__230 <o> e ) num__250 |
the range of the other num__8 numbers is num__170 - num__90 = num__80 so x must be either the smallest number or the largest number in the set . then x = num__170 - num__100 = num__70 or x = num__90 + num__100 = num__190 the answer is b . <eor> b <eos> |
b |
subtract__170.0__90.0__ subtract__170.0__100.0__ add__90.0__100.0__ add__90.0__100.0__ |
subtract__170.0__90.0__ subtract__170.0__100.0__ add__90.0__100.0__ add__90.0__100.0__ |
| sum of two numbers is num__50 . two times of the difference of first and seceond is num__20 . then the numbers will be ? <o> a ) num__10 num__40 <o> b ) num__20 num__30 <o> c ) num__35 num__15 <o> d ) num__30 num__20 <o> e ) num__15 num__35 |
explanation : x + y = num__50 num__2 x à ¢ â ‚ ¬ â € œ num__2 y = num__20 x = num__30 y = num__20 answer : d <eor> d <eos> |
d |
subtract__50.0__20.0__ subtract__50.0__20.0__ |
subtract__50.0__20.0__ subtract__50.0__20.0__ |
| e is the set of the first n positive odd numbers where n is a positive integer . given that n > k where k is also a positive integer x is the maximum value of the sum of k distinct members of e and y is the minimum value of the sum of k distinct members of e what is x + y ? <o> a ) kn <o> b ) kn + k ^ num__2 <o> c ) kn + num__2 k ^ num__2 <o> d ) num__2 kn – k ^ num__2 <o> e ) num__2 kn |
probably the easiest way to solve this question would be to assume some values for n and k . say n = num__3 so e the set of the first n positive odd numbers would be : e = { num__1 num__3 num__5 } ; say k = num__1 so e the maximum value of the sum of k distinct members of z would simply be num__5 . similarly y the minimum value of the sum of k distinct members of e would simply be num__1 . x + y = num__5 + num__1 = num__6 . now substitute n = num__3 and k = num__1 in the options provided to see which one yields num__6 . only asnwer choice e fits : num__2 kn = num__2 * num__3 * num__1 = num__6 . answer : e . <eor> e <eos> |
e |
add__1.0__5.0__ subtract__3.0__1.0__ multiply__1.0__2.0__ |
add__1.0__5.0__ subtract__3.0__1.0__ multiply__1.0__2.0__ |
| johnny bought six peanut butter cans at an average price ( arithmetic mean ) of num__36.5 ¢ . if johnny returned two cans to the retailer and the average price of the remaining cans was num__30 ¢ then what is the average price in cents of the two returned peanut butter cans ? <o> a ) num__5.5 <o> b ) num__11 <o> c ) num__47.5 <o> d ) num__66 <o> e ) num__49.5 |
total price of six cans = num__6 * num__36.5 = num__219 total price of num__4 cans = num__4 * num__30 = num__120 total rice of two cans = num__219 - num__120 = num__99 average price of two cans = num__49.5 = num__49.5 c another way to do it is this : assume that the four leftover cans were of num__30 c each . the avg was num__36.5 c initially because the two cans were num__36.5 c each and were providing another num__6.5 c of cost to other num__4 cans . so cost of the two cans = num__2 * num__36.5 + num__4 * num__6.5 = num__99 avg cost of the two cans = num__49.5 = num__49.5 c answer ( e ) <eor> e <eos> |
e |
multiply__36.5__6.0__ multiply__30.0__4.0__ subtract__219.0__120.0__ subtract__36.5__30.0__ divide__99.0__49.5__ subtract__99.0__49.5__ |
multiply__36.5__6.0__ multiply__30.0__4.0__ subtract__219.0__120.0__ subtract__36.5__30.0__ subtract__6.0__4.0__ subtract__99.0__49.5__ |
| power saving ( in % ) of a bulb is directly proportional to square root of its efficiency . by how much % is power saving of bulb - num__1 ( efficiency num__0.8 ) greater than power saving of bulb - num__2 ( efficiency num__0.2 ) ? <o> a ) num__60.0 <o> b ) num__70.0 <o> c ) num__80.0 <o> d ) num__90.0 <o> e ) num__100 % |
power p num__1 = k ∗ sqrt ( num__0.8 ) power p num__2 = k ∗ sqrt ( num__0.2 ) difference = p num__1 − p num__2 = k ∗ sqrt ( num__0.8 ) − k ∗ sqrt ( num__0.2 ) % greater = k ∗ sqrt ( num__0.8 ) − k ∗ sqrt ( num__0.2 ) / k ∗ sqrt ( num__0.2 ) ∗ num__100 = num__100.0 . hence e <eor> e <eos> |
e |
multiply__1.0__100.0__ |
divide__100.0__1.0__ |
| the speed at which a man can row a boat in still water is num__18 kmph . if he rows downstream where the speed of current is num__6 kmph what time will he take to cover num__100 metres ? <o> a ) num__6.67 seconds <o> b ) num__18 seconds <o> c ) num__26 seconds <o> d ) num__14.99 seconds <o> e ) num__6 seconds |
speed of the boat downstream = num__18 + num__6 = num__24 kmph = num__24 * num__0.277777777778 = num__6.67 m / s hence time taken to cover num__100 m = num__100 / num__6.67 = num__14.99 seconds . answer : d <eor> d <eos> |
d |
add__18.0__6.0__ round__14.99__ |
add__18.0__6.0__ round__14.99__ |
| a train passes a station platform in num__36 sec and a man standing on the platform in num__20 sec . if the speed of the train is num__72 km / hr . what is the length of the platform ? <o> a ) num__240 <o> b ) num__188 <o> c ) num__177 <o> d ) num__260 <o> e ) num__320 |
speed = num__72 * num__0.277777777778 = num__20 m / sec . length of the train = num__20 * num__20 = num__400 m . let the length of the platform be x m . then ( x + num__400 ) / num__36 = num__20 = > x = num__320 . answer : e <eor> e <eos> |
e |
divide__20.0__72.0__ round__320.0__ |
divide__20.0__72.0__ round__320.0__ |
| divide $ num__500 among y z in the ratio num__2 : num__8 . how many $ that z get ? <o> a ) $ num__100 <o> b ) $ num__200 <o> c ) $ num__300 <o> d ) $ num__500 <o> e ) $ num__400 |
sum of ratio terms = num__2 + num__8 = num__10 z = num__500 * num__0.8 = $ num__400 answer is e <eor> e <eos> |
e |
add__2.0__8.0__ divide__8.0__10.0__ multiply__500.0__0.8__ multiply__500.0__0.8__ |
add__2.0__8.0__ divide__8.0__10.0__ multiply__500.0__0.8__ multiply__500.0__0.8__ |
| a certain bag of gemstones is composed of two - thirds diamonds and one - third rubies . if the probability of randomly selecting two diamonds from the bag without replacement is num__0.416666666667 what is the probability of selecting two rubies from the bag without replacement ? <o> a ) num__0.138888888889 <o> b ) num__0.208333333333 <o> c ) num__0.0833333333333 <o> d ) num__0.166666666667 <o> e ) num__0.25 |
num__0.666666666667 * ( num__2 x - num__1 ) / ( num__3 x - num__1 ) = num__0.416666666667 = > x = num__3 so total gems = num__9 and probability of ruby = num__0.333333333333 * num__0.25 = num__0.0833333333333 answer num__0.0833333333333 c <eor> c <eos> |
c |
add__1.0__2.0__ reverse__3.0__ subtract__0.6667__0.4167__ divide__0.25__3.0__ divide__0.25__3.0__ |
add__1.0__2.0__ reverse__3.0__ subtract__0.6667__0.4167__ divide__0.25__3.0__ divide__0.25__3.0__ |
| at exactly what time past num__9 : num__00 will the minute and hour hands of an accurate working clock be precisely perpendicular to each other for the first time ? <o> a ) num__20 num__0.619047619048 minutes past num__7 : num__00 <o> b ) num__20 num__0.764705882353 minutes past num__7 : num__00 <o> c ) num__21 num__0.130434782609 minutes past num__7 : num__00 <o> d ) num__21 num__0.818181818182 minutes past num__9 : num__00 <o> e ) num__22 num__0.444444444444 minutes past num__7 : num__00 |
num__5.5 is the angle between minute n hour this is what i was taught . . . so should n ' t it be solve by dividing num__90 with num__5.5 ? that would have been the case if your initial difference between the hour and the minute hand was = num__0 degrees or in other words both minute and hour hands were at the same location . but as per the question you are asked for time after num__9 : num__00 . at num__9 : num__00 the angle between the hour and the minute hand is num__210 degrees . you need to take this into account as well . so in order for the difference to decrease to num__90 degrees the minute hand must eat away this difference of num__210 - num__90 = num__120 degree at the rate of num__5.5 degrees per minute - - - > num__120 / num__5.5 = num__21 num__0.818181818182 minutes . thus d is the correct answer . <eor> d <eos> |
d |
subtract__210.0__90.0__ round__21.0__ |
subtract__210.0__90.0__ round__21.0__ |
| three numbers are in the ratio num__3 : num__5 : num__7 . the largest number value is num__63 . find difference between smallest & largest number is ? <o> a ) num__20 <o> b ) num__24 <o> c ) num__30 <o> d ) num__34 <o> e ) num__36 |
= = num__3 : num__5 : num__7 total parts = num__15 = the largest number value is num__63 = the largest number is = num__7 = then num__7 parts - - - - - > num__63 ( num__7 * num__9 = num__63 ) = smallest number = num__3 & largest number = num__7 = difference between smallest number & largest number is = num__7 - num__3 = num__4 = then num__4 parts - - - - - > num__36 ( num__4 * num__9 = num__36 ) e <eor> e <eos> |
e |
multiply__3.0__5.0__ divide__63.0__7.0__ subtract__7.0__3.0__ multiply__4.0__9.0__ multiply__4.0__9.0__ |
multiply__3.0__5.0__ divide__63.0__7.0__ subtract__7.0__3.0__ multiply__4.0__9.0__ multiply__4.0__9.0__ |
| working alone printers x y and z can do a certain printing job consisting of a large number of pages in num__16 num__10 and num__20 hours respectively . what is the ratio of the time it takes printer x to do the job working alone at its rate to the time it takes printers y and z to do the job working together at their individual rates ? <o> a ) num__1.75 <o> b ) num__3.33333333333 <o> c ) num__2.4 <o> d ) num__2.14285714286 <o> e ) num__1.63636363636 |
the time it takes printer x is num__16 hours . the combined rate of y and z is num__0.1 + num__0.05 = num__0.15 the time it takes y and z is num__6.66666666667 the ratio of times is num__16 / ( num__6.66666666667 ) = num__3 * num__0.8 = num__2.4 the answer is c . <eor> c <eos> |
c |
reverse__10.0__ reverse__20.0__ add__0.1__0.05__ reverse__0.15__ multiply__20.0__0.15__ divide__16.0__20.0__ multiply__16.0__0.15__ multiply__16.0__0.15__ |
reverse__10.0__ reverse__20.0__ add__0.1__0.05__ reverse__0.15__ multiply__20.0__0.15__ divide__16.0__20.0__ multiply__16.0__0.15__ multiply__16.0__0.15__ |
| a person is traveling at num__35 km / hr and reached his destiny in num__5 hr find the distance ? <o> a ) a ) num__100 km <o> b ) b ) num__95 km <o> c ) c ) num__175 km <o> d ) d ) num__80 km <o> e ) e ) num__125 km |
speed = num__35 km / hr time = num__5 hr distance = num__35 * num__5 = num__175 km answer is c <eor> c <eos> |
c |
multiply__35.0__5.0__ round__175.0__ |
multiply__35.0__5.0__ multiply__35.0__5.0__ |
| a work which could be finished in num__11 days was finished num__3 days earlier after num__10 more men joined . the number of men employed was ? <o> a ) num__22 <o> b ) num__20 <o> c ) num__88 <o> d ) num__71 <o> e ) num__12 |
x - - - - - - - num__11 ( x + num__10 ) - - - - num__6 x * num__11 = ( x + num__10 ) num__6 x = num__12 \ answer : e <eor> e <eos> |
e |
round__12.0__ |
round__12.0__ |
| in a group of ducks and cows the total number of legs are num__24 more than twice the no . of heads . find the total no . of buffaloes . <o> a ) num__11 <o> b ) num__12 <o> c ) num__14 <o> d ) num__16 <o> e ) num__18 |
let the number of buffaloes be x and the number of ducks be y = > num__4 x + num__2 y = num__2 ( x + y ) + num__24 = > num__2 x = num__24 = > x = num__12 b <eor> b <eos> |
b |
divide__24.0__2.0__ subtract__24.0__12.0__ |
divide__24.0__2.0__ subtract__24.0__12.0__ |
| suraj has a certain average of runs for num__16 innings . in the num__17 th innings he scores num__92 runs thereby increasing his average by num__4 runs . what is his average after the num__17 th innings ? <o> a ) num__48 <o> b ) num__64 <o> c ) num__28 <o> d ) num__72 <o> e ) num__27 |
to improve his average by num__4 runs per innings he has to contribute num__16 x num__4 = num__64 runs for the previous num__16 innings . thus the average after the num__17 th innings = num__92 - num__64 = num__28 . answer : c <eor> c <eos> |
c |
multiply__16.0__4.0__ subtract__92.0__64.0__ subtract__92.0__64.0__ |
multiply__16.0__4.0__ subtract__92.0__64.0__ subtract__92.0__64.0__ |
| a is two years older than b who is twice as old as c . if the total of the ages of a b and c be num__27 the how old is b ? <o> a ) num__10 years <o> b ) num__12 years <o> c ) num__14 years <o> d ) num__16 years <o> e ) num__18 years |
let c ' s age be x years . then b ' s age = num__2 x years . a ' s age = ( num__2 x + num__2 ) years . ( num__2 x + num__2 ) + num__2 x + x = num__27 num__5 x = num__25 x = num__5 . hence b ' s age = num__2 x = num__10 years . a ) <eor> a <eos> |
a |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
| cecilia robbie and briony all bought stamps . the number of stamps cecilia purchased was equal to a single digit . the number of stamps only one of them purchased was divisible by num__6 . the number of stamps one of them bought was an even number . which of the following could represent the numbers of stamps each purchased ? <o> a ) num__3 num__8 num__24 <o> b ) num__7 num__9 num__17 <o> c ) num__6 num__9 num__12 <o> d ) num__5 num__15 num__18 <o> e ) num__9 num__10 num__13 |
i didnt get your point on eliminating a . the number of stamps only one purchased was divisible by num__6 . could that not be cecilia with num__9 stamps . i choose this and marked a . <eor> a <eos> |
a |
subtract__9.0__6.0__ |
subtract__9.0__6.0__ |
| a hiker walked for num__3 days . she walked num__21 miles on the first day walking num__3 miles per hour . on the second day she walked for one less hour but she walked one mile per hour faster than on the first day . on the third day she walked the same number of hours as on the first day but at the same speed as on the second day . how many miles in total did she walk ? <o> a ) num__73 <o> b ) num__44 <o> c ) num__58 <o> d ) num__60 <o> e ) num__62 |
she walked num__21 miles on the first day walking num__3 miles per hour i . e . total time of walk on day - num__1 = num__7.0 = num__7 hours second day time of walk = num__7 - num__1 = num__6 hours and speed = num__3 + num__1 = num__4 miles per hour i . e . distance walked on second day = num__6 * num__4 = num__24 miles third day time of walk = num__7 hours and speed = num__4 miles per hour i . e . distance walked on second day = num__7 * num__4 = num__28 miles total distance travelled on three days = num__21 + num__24 + num__28 = num__73 answer : option a <eor> a <eos> |
a |
divide__21.0__3.0__ subtract__7.0__1.0__ add__3.0__1.0__ add__3.0__21.0__ add__21.0__7.0__ round__73.0__ |
divide__21.0__3.0__ subtract__7.0__1.0__ add__3.0__1.0__ add__3.0__21.0__ add__21.0__7.0__ round__73.0__ |
| a shopkeeper purchased num__85 kg of potatoes for rs . num__1105 and sold the whole lot at the rate of rs . num__15 per kg . what will be his gain percent ? <o> a ) num__18 num__0.111111111111 % <o> b ) num__16 num__3.0 % <o> c ) num__14 num__0.333333333333 % <o> d ) num__12 num__0.333333333333 % <o> e ) num__15 num__0.384615384615 % |
c . p . of num__1 kg = num__13.0 = rs . num__13 s . p . of num__1 kg = rs . num__15 gain % = num__0.153846153846 * num__100 = num__15.3846153846 = num__15 num__0.384615384615 % answer : e <eor> e <eos> |
e |
percent__15.0__100.0__ |
percent__15.0__100.0__ |
| the sum of the present ages of two persons a and b is num__60 . if the age of a is twice that of b find the sum of their ages num__9 years hence ? <o> a ) num__22 <o> b ) num__77 <o> c ) num__78 <o> d ) num__98 <o> e ) num__71 |
a + b = num__60 a = num__2 b num__2 b + b = num__60 = > b = num__20 then a = num__40 . num__9 years their ages will be num__49 and num__29 . sum of their ages = num__49 + num__29 = num__78 . answer : c <eor> c <eos> |
c |
subtract__60.0__20.0__ add__9.0__40.0__ add__9.0__20.0__ add__49.0__29.0__ add__49.0__29.0__ |
subtract__60.0__20.0__ add__9.0__40.0__ add__9.0__20.0__ add__49.0__29.0__ add__49.0__29.0__ |
| x and y are positive integers of h . if num__1 / x + num__1 / y < num__2 which of the following must be true ? <o> a ) x + y > num__4 <o> b ) xy > num__1 <o> c ) x / y + y / x < num__1 <o> d ) ( x - y ) ^ num__2 > num__0 <o> e ) none of the above |
answer is b : num__1 / x + num__1 / y < num__2 the maximum value of num__1 / x is num__1 because if x equals any other number greater than one it will be a fraction . the same is true with num__1 / y . so num__1 / x and num__1 / y will always be less than num__2 as long as both x and y are not both equal to one at the same time . another way of putting it is : x * y > num__1 . b <eor> b <eos> |
b |
reverse__1.0__ |
reverse__1.0__ |
| num__20 . a certain church bell rings the bell twice at half past the hour and four times at the hour plus an additional number of rings equal to what ever time it is . how many rings will the clock make from num__6 : num__20 in the morning to num__08 : num__10 in the morning ? <o> a ) num__27 <o> b ) num__36 . <o> c ) num__42 . <o> d ) num__46 . <o> e ) num__50 . |
@ num__6 : num__30 - num__2 @ num__7 - num__4 + num__7 = num__11 @ num__7 : num__30 - num__2 @ num__8 - num__12 totals to a = num__27 <eor> a <eos> |
a |
add__20.0__10.0__ divide__20.0__10.0__ subtract__6.0__2.0__ add__4.0__7.0__ subtract__20.0__8.0__ add__20.0__7.0__ add__20.0__7.0__ |
add__20.0__10.0__ subtract__8.0__6.0__ subtract__6.0__2.0__ add__4.0__7.0__ subtract__20.0__8.0__ add__20.0__7.0__ add__20.0__7.0__ |
| if the sides of a triangle are num__26 cm num__24 cm and num__10 cm what is its area ? <o> a ) num__108 sq . cm <o> b ) num__112 sq . cm <o> c ) num__116 sq . cm <o> d ) num__120 sq . cm <o> e ) none of these |
explanation : the triangle with sides num__26 cm num__24 cm and num__10 cm is right angled where the hypotenuse is num__26 cm . area of the triangle = num__0.5 x base x height = > num__0.5 x num__24 x num__10 = num__120 sq . cm answer is d <eor> d <eos> |
d |
triangle_area__24.0__10.0__ triangle_area__24.0__10.0__ |
triangle_area__24.0__10.0__ triangle_area__24.0__10.0__ |
| a dozen eggs and num__10 pounds of apples are currently at the same price . if the price of a dozen eggs rises by num__10 percent and the price of apples rises by num__2.0 . how much more will it cost to buy a dozen eggs and num__10 pounds of apples . <o> a ) num__2.0 <o> b ) num__6.0 <o> c ) num__10.0 <o> d ) num__12.0 <o> e ) num__12.2 % |
say currently both a dozen eggs and num__10 pounds of apples cost $ num__50 ( they are at the same price ) . so to buy a dozen eggs and num__10 pounds of apples we need $ num__100 . after the increase the price of a dozen eggs will be $ num__55 and the price of num__10 pounds of apples will be $ num__51 . so after the increase to buy a dozen eggs and num__10 pounds of apples we ' ll need $ num__106 . increase = num__6.0 . answer : b <eor> b <eos> |
b |
percent__100.0__6.0__ |
percent__100.0__6.0__ |
| a perfect square is defined as the square of an integer and a perfect cube is defined as the cube of an integer . how many positive integers n are there such that n is less than num__50000 and at the same time n is a perfect square and a perfect cube ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
if n is a perfect square and a perfect cube then n = a ^ num__6 for some integer a . the numbers are num__1 ^ num__6 = num__1 num__2 ^ num__6 = num__64 num__3 ^ num__6 = num__729 num__4 ^ num__6 = num__4096 num__5 ^ num__6 = num__15625 num__6 ^ = num__46656 . the answer is d . <eor> d <eos> |
d |
power__2.0__6.0__ triangle_area__1.0__6.0__ power__3.0__6.0__ square_perimeter__1.0__ power__64.0__2.0__ power__5.0__6.0__ multiply__64.0__729.0__ surface_cube__1.0__ |
power__2.0__6.0__ triangle_area__1.0__6.0__ power__3.0__6.0__ square_perimeter__1.0__ power__64.0__2.0__ power__5.0__6.0__ multiply__64.0__729.0__ power__6.0__1.0__ |
| ( num__4 + √ num__3 + √ num__5 ) ^ num__2 - ( √ num__3 + √ num__5 ) ^ num__2 = ? <o> a ) num__1 + num__2 √ num__3 + num__2 √ num__5 <o> b ) num__16 + num__8 √ num__3 + num__8 √ num__5 <o> c ) num__1 - √ num__3 + √ num__5 <o> d ) num__1 + √ num__3 - √ num__5 <o> e ) √ num__3 + √ num__5 |
( num__4 + √ num__3 + √ num__5 ) ^ num__2 - ( √ num__3 + √ num__5 ) ^ num__2 ( num__4 + √ num__3 ) ^ num__2 + num__5 + num__2 ( num__4 + √ num__3 ) ( √ num__5 ) - ( num__3 + num__5 + num__2 √ num__3 √ num__5 ) num__16 + num__3 + num__8 √ num__3 + num__5 + num__8 √ num__5 + num__2 √ num__15 - num__3 - num__5 - num__2 √ num__15 num__16 + num__8 √ num__3 + num__8 √ num__5 b is the answer <eor> b <eos> |
b |
multiply__4.0__2.0__ multiply__3.0__5.0__ multiply__2.0__8.0__ |
add__3.0__5.0__ multiply__3.0__5.0__ multiply__2.0__8.0__ |
| a sock drawer contains two socks each one a different solid color . the colors of the socks are indigo and violet . if the socks are laid next to each other how many different arrangements of socks could be made ? <o> a ) num__24 <o> b ) num__240 <o> c ) num__1024 <o> d ) num__2520 <o> e ) num__2 |
i see no condition for arrangement . so num__2 ! ways - imo option e is correct answer . . <eor> e <eos> |
e |
coin_space__ coin_space__ |
coin_space__ coin_space__ |
| a car was initially traveling at an average speed of num__36 miles per hour . during each of ten successive num__5 - minute intervals the car decreased its average speed by num__2 miles per hour . how many miles did the car travel in the fourth num__5 - minute interval ? <o> a ) num__1.32 <o> b ) num__5.32 <o> c ) num__4.56 <o> d ) num__2.33 <o> e ) num__1.44 |
the question does not specifically mentioned whether num__36 miles / h is the speed in first interval or before that . . . so i got confused . if former is the case then average speed of the car in num__4 th interval will be num__28 . this gives num__28 * ( num__0.0833333333333 ) = m = num__2.33 miles correct option is d <eor> d <eos> |
d |
round__2.33__ |
round__2.33__ |
| a combustion reaction forms carbon dioxide . a carbon dioxide molecule contains one carbon and two oxygen atoms . if over a period of num__15 minutes a combustion reaction creates num__15000 molecules of carbon dioxide then approximately how many more atoms of oxygen than carbon are created on average per minute ? <o> a ) num__1000 <o> b ) num__500 <o> c ) num__250 <o> d ) num__50 <o> e ) num__0 |
solution : num__15000 carbon dioxide molecules are created over a period of num__15 minutes . therefore num__1000.0 = num__1000 carbon dioxide molecules are created on average per minute each carbon dioxide molecule contains one carbon atom and two oxygen atoms . so num__1000 carbon dioxide molecules contain num__1 × num__1000 = num__1000 carbon atoms and num__2 × num__1000 = num__2000 oxygen atoms . the difference is num__2000 – num__1000 = num__1000 . the correct answer is a . <eor> a <eos> |
a |
divide__15000.0__15.0__ multiply__2.0__1000.0__ divide__15000.0__15.0__ |
divide__15000.0__15.0__ multiply__2.0__1000.0__ divide__15000.0__15.0__ |
| twenty four men can do a work in num__14 days . how many men are required to complete the work in num__10 days ? <o> a ) num__33.6 <o> b ) num__10.12 <o> c ) num__50 <o> d ) num__14.2 <o> e ) num__60 |
a num__33.6 we have m num__1 d num__1 = m num__2 d num__2 so num__24 * num__14 = m num__2 * num__10 = > m num__2 = num__33.6 . answer is a <eor> a <eos> |
a |
add__14.0__10.0__ round__33.6__ |
add__14.0__10.0__ round__33.6__ |
| boy sells a book for rs . num__540 he gets a loss of num__10.0 to gain num__10.0 what should be the sp ? <o> a ) num__430 <o> b ) num__450 <o> c ) num__550 <o> d ) num__590 <o> e ) num__660 |
cost price = num__6.0 x num__100 = num__600 to gain num__10.0 = num__600 x num__0.1 = num__60 sp = cp + gain = num__600 + num__60 = num__660 answer : e <eor> e <eos> |
e |
percent__10.0__600.0__ percent__100.0__660.0__ |
percent__10.0__600.0__ percent__100.0__660.0__ |
| a is twice as good a workman as b and together they finish a piece in num__18 days . in how many days will a alone finish the work ? <o> a ) num__15 days <o> b ) num__3 days <o> c ) num__12 days <o> d ) num__27 days <o> e ) num__18 days |
sol . ( a ’ s num__1 day ’ s work ) : ) ( b ’ s num__1 days work ) = num__2 : num__1 . ( a + b ) ' s num__1 day ' s work = num__0.0555555555556 divide num__0.0555555555556 in the ratio num__2 : num__1 a ’ s num__1 day ' s work = ( num__0.0555555555556 * num__0.666666666667 ) = num__0.037037037037 hence a alone can finish the work in num__27 days . ans : d <eor> d <eos> |
d |
divide__1.0__18.0__ divide__0.6667__18.0__ round__27.0__ |
divide__1.0__18.0__ divide__0.6667__18.0__ round__27.0__ |
| list r contains five numbers that have an average value of num__50 . if the median of the numbers in the list is equal to the mean and the largest number is equal to num__20 more than two times the smallest number what is the smallest possible value in the list ? <o> a ) num__18 <o> b ) num__22 <o> c ) num__26 <o> d ) num__30 <o> e ) num__34 |
the middle number is num__50 . let the smallest number be x . then the largest number is num__2 x + num__20 . to make x as small as possible let ' s make the remaining numbers as large as possible . so the second largest = the largest = num__2 x + num__20 . the second smallest should be equal to the median . the numbers are x num__50 num__50 num__2 x + num__20 num__2 x + num__20 . x + num__50 + num__50 + num__2 x + num__20 + num__2 x + num__20 = num__5 * num__50 = num__250 num__5 x = num__110 x = num__22 the answer is b . <eor> b <eos> |
b |
multiply__50.0__5.0__ add__20.0__2.0__ add__20.0__2.0__ |
multiply__50.0__5.0__ add__20.0__2.0__ add__20.0__2.0__ |
| a certain family has num__3 sons : richard is num__6 years older than david and david is num__8 years older than scott . if in num__8 years richard will be twice as old as scott then how old was david num__5 years ago ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__9 <o> d ) num__14 <o> e ) num__16 |
let ' s say age of richard is ` ` r ' ' age of david is ` ` d ' ' age of scott is ` ` s ' ' now richard is num__6 years older than david i . e . r = d + num__6 david is num__8 years older than scott i . e . d = s + num__8 if in num__8 years richard will be twice as old as scott i . e . r + num__8 = num__2 x ( s + num__8 ) i . e . r + num__8 = num__2 s + num__16 i . e . r = num__2 s + num__8 but r = d + num__6 = ( s + num__8 ) + num__6 = s + num__14 therefore num__2 s + num__8 = s + num__14 i . e . s = num__6 i . e . r = num__20 i . e . d = num__14 now how old was david num__5 years ago ? i . e . d - num__5 = num__14 - num__5 = num__9 years answer : c <eor> c <eos> |
c |
divide__6.0__3.0__ twice__8.0__ add__6.0__8.0__ add__6.0__14.0__ triple__3.0__ triple__3.0__ |
subtract__8.0__6.0__ twice__8.0__ add__6.0__8.0__ add__6.0__14.0__ add__3.0__6.0__ add__3.0__6.0__ |
| a and b started a business by investing rs . num__4000 / - and rs . num__5000 / - respectively . find the a ’ s share out of a total profit of rs . num__1800 : <o> a ) rs . num__600 / - <o> b ) rs . num__660 / - <o> c ) rs . num__800 / - <o> d ) rs . num__820 / - <o> e ) rs . num__840 / - |
a = rs . num__4000 / - b = rs . num__5000 / - a share num__4 parts & b share num__5 parts total num__9 parts - - - - - > rs . num__1800 / - - - - - > num__1 part - - - - - - - > rs . num__200 / - a share = num__4 parts - - - - - > rs . num__800 / - c <eor> c <eos> |
c |
add__4.0__5.0__ subtract__5.0__4.0__ divide__1800.0__9.0__ divide__4000.0__5.0__ divide__4000.0__5.0__ |
add__4.0__5.0__ subtract__5.0__4.0__ divide__1800.0__9.0__ divide__4000.0__5.0__ divide__4000.0__5.0__ |
| how many num__3 digit integers can be chosen such that none of the digits appear more than twice and none of the digits equal zero ? <o> a ) num__729 <o> b ) num__720 <o> c ) num__648 <o> d ) num__640 <o> e ) num__57 |
number of ways without restriction = num__9 * num__9 * num__9 = num__729 number of ways that break restriction = num__9 * num__1 * num__9 = num__9 so the answer is num__729 - num__9 = num__720 ans : b <eor> b <eos> |
b |
subtract__729.0__9.0__ multiply__720.0__1.0__ |
subtract__729.0__9.0__ multiply__720.0__1.0__ |
| before num__4 years dog a ’ s age was num__4 times of dog b ’ s age and after num__4 years dog a ’ s age will be num__3 times of dog b ’ s age . what is the difference of dog a ’ s age and dog b ’ s now ? <o> a ) num__36 <o> b ) num__42 <o> c ) num__52 <o> d ) num__60 <o> e ) num__64 |
a - num__4 = num__4 ( b - num__4 ) - - > a - num__4 b = - num__12 . . . . . . . . . . . . . num__1 a + num__4 = num__3 ( b + num__4 ) - - > a - num__3 b = num__8 . . . . . . . . . . . . . num__2 ( num__2 ) - ( num__1 ) - - > b = num__20 - - > a = num__3 ( num__24 ) = num__72 a - b = num__72 - num__20 = num__52 answer : c <eor> c <eos> |
c |
multiply__4.0__3.0__ subtract__4.0__3.0__ subtract__12.0__4.0__ subtract__3.0__1.0__ add__8.0__12.0__ add__4.0__20.0__ multiply__3.0__24.0__ subtract__72.0__20.0__ multiply__1.0__52.0__ |
multiply__4.0__3.0__ subtract__4.0__3.0__ subtract__12.0__4.0__ subtract__3.0__1.0__ add__8.0__12.0__ add__4.0__20.0__ multiply__3.0__24.0__ subtract__72.0__20.0__ subtract__72.0__20.0__ |
| if x + y = num__9 and x – y = num__3 then x ^ num__2 - y ^ num__2 = <o> a ) - num__4 <o> b ) num__4 <o> c ) num__10 <o> d ) num__27 <o> e ) num__40 |
the fastest approach has already been shown . here ' s one more option . given : x + y = num__9 x – y = num__3 add the two equations to get : num__2 x = num__12 which means x = num__6 if x = num__6 we can plug that value into either equation to conclude that y = num__3 if x = num__6 and y = num__3 then x ² - y ² = num__6 ² - num__3 ² = num__27 answer : d <eor> d <eos> |
d |
add__9.0__3.0__ subtract__9.0__3.0__ multiply__9.0__3.0__ multiply__9.0__3.0__ |
add__9.0__3.0__ subtract__9.0__3.0__ multiply__9.0__3.0__ multiply__9.0__3.0__ |
| the ages of two persons differ by num__16 years . if num__6 years ago the elder one be num__3 times as old as the younger one find their present ages . <o> a ) num__14 years <o> b ) num__17 years <o> c ) num__24 years <o> d ) num__29 years <o> e ) none |
solution let the age of the younger person be x years . then age of the elder person = ( x + num__16 ) years . therefore ‹ = › num__3 ( x - num__6 ) = ( x + num__16 - num__6 ) ‹ = › num__3 x - num__18 = x + num__10 ‹ = › num__2 x = num__28 x = num__14 years . answer a <eor> a <eos> |
a |
multiply__6.0__3.0__ subtract__16.0__6.0__ divide__6.0__3.0__ add__10.0__18.0__ subtract__16.0__2.0__ subtract__16.0__2.0__ |
multiply__6.0__3.0__ subtract__16.0__6.0__ subtract__18.0__16.0__ add__10.0__18.0__ subtract__16.0__2.0__ subtract__16.0__2.0__ |
| the present ages of three persons in proportions num__2 : num__2 : num__6 . eight years ago the sum of their ages was num__56 . find their present ages ( in years ) . <o> a ) num__8 num__20 num__28 <o> b ) num__16 num__1648 <o> c ) num__20 num__35 num__45 <o> d ) num__18 num__24 num__32 <o> e ) none of these |
explanation : let their present ages be num__2 x num__2 x and num__6 x years respectively . then ( num__2 x - num__8 ) + ( num__2 x - num__8 ) + ( num__6 x - num__8 ) = num__56 num__10 x = num__80 x = num__8 their present ages are num__2 x = num__16 years num__2 x = num__16 years and num__6 x = num__48 years respectively . answer : b <eor> b <eos> |
b |
add__2.0__6.0__ add__2.0__8.0__ multiply__8.0__10.0__ multiply__2.0__8.0__ multiply__6.0__8.0__ multiply__2.0__8.0__ |
add__2.0__6.0__ add__2.0__8.0__ multiply__8.0__10.0__ add__6.0__10.0__ subtract__56.0__8.0__ add__6.0__10.0__ |
| a can do a piece of work in num__5 days and b can do it in num__6 days . how long will they take if both work together ? <o> a ) num__2 num__0.727272727273 <o> b ) num__2 num__0.040404040404 <o> c ) num__2 num__1.14285714286 <o> d ) num__2 num__0.444444444444 <o> e ) num__2 num__0.0714285714286 |
sol . a ’ s num__1 day ’ s work = num__1 th num__5 part of whole work and b ’ s num__1 day ’ s work = num__1 th num__6 part of whole work ? ( a + b ) ’ s one day ’ s work = num__1 num__1 num__11 num__5 num__6 num__30 th part of whole work . so both together will finish the work in num__30 days num__2 num__0.727272727273 days . answer : a <eor> a <eos> |
a |
subtract__6.0__5.0__ add__5.0__6.0__ multiply__5.0__6.0__ round__2.0__ |
subtract__6.0__5.0__ add__5.0__6.0__ multiply__5.0__6.0__ round__2.0__ |
| jack and jill are marathon runners . jack can finish a marathon ( num__42 km ) in num__5.5 hours and jill can run a marathon in num__4.2 hours . what is the ratio of their average running speed ? ( jack : jill ) <o> a ) num__0.933333333333 <o> b ) num__1.07142857143 <o> c ) num__0.8 <o> d ) num__0.763636363636 <o> e ) can not be determined |
average speed of jack = distance / time = num__42 / ( num__5.5 ) = num__7.63636363636 average speed of jill = num__42 / ( num__4.2 ) = num__10 ratio of average speed of jack to jill = ( num__7.63636363636 ) / num__10 = num__0.763636363636 = num__0.763636363636 answer d <eor> d <eos> |
d |
divide__42.0__5.5__ divide__42.0__4.2__ divide__4.2__5.5__ divide__4.2__5.5__ |
divide__42.0__5.5__ divide__42.0__4.2__ divide__4.2__5.5__ divide__4.2__5.5__ |
| by looking at a rectangular box a carpenter estimates that the length of the box is between num__2 to num__2.1 meters inclusive the breadth is between num__1 to num__1.1 meters inclusive and the height is between num__2 to num__2.1 centimeters inclusive . if the actual length breadth and height of the box do indeed fall within the respective ranges estimated by the carpenter which of the following is the closest to the maximum possible magnitude of the percentage error q that the carpenter can make in calculating the volume of the rectangular box ? <o> a ) num__1.0 <o> b ) num__3.0 <o> c ) num__10.0 <o> d ) num__18.0 <o> e ) num__22 % |
the dimensions are between num__2 - num__2.1 num__1 - num__1.1 and num__2 - num__2.1 . . so max error in volume = ? . . important points to note num__1 ) max error would be when actuals and estimate are at the extreme ends . . num__2 ) but what about max % error - - it will be when the base is the lowest . . . so the actuals should be lowest or at lower end and the estimate at the higher end . . solution actuals = num__2 * num__2 * num__1 = num__4 . . estimate = num__2.1 * num__2.1 * num__1.1 = num__4.841 . . max % error q = ( num__4.841 - num__4 ) / num__4 * num__100 = num__84.1 / num__4 = num__21.25 . . num__21.25 is closest to num__22.0 in the choices . . ans e <eor> e <eos> |
e |
multiply__1.0__22.0__ |
multiply__1.0__22.0__ |
| using all the letters of the word ` ` some ' ' how many words can be formed which begin with s and end with e ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
there are four letters in the given word . consider num__4 blanks . . . . the first blank and last blank must be filled with s and e all the remaining two blanks can be filled with the remaining num__2 letters in num__2 ! ways . the number of words = num__2 ! = num__2 . answer : a <eor> a <eos> |
a |
coin_space__ coin_space__ |
coin_space__ coin_space__ |
| in num__140 m race a covers the distance in num__36 seconds and b in num__45 seconds . in this race a beats b by : <o> a ) num__20 m <o> b ) num__28 m <o> c ) num__22.5 m <o> d ) num__9 m <o> e ) num__12 m |
distance covered by b in num__9 sec . = num__3.11111111111 x num__9 m = num__28 m . a beats b by num__28 metres . answer : option b <eor> b <eos> |
b |
subtract__45.0__36.0__ divide__140.0__45.0__ round__28.0__ |
subtract__45.0__36.0__ divide__140.0__45.0__ round__28.0__ |
| if the quantity ( p + num__5 ) is num__5 times the quantity ( q + num__5 ) then what is the relationship between p and q ? <o> a ) p = num__5 q + num__30 <o> b ) p = num__5 q − num__30 <o> c ) num__5 p = num__5 q + num__20 <o> d ) p = num__5 q <o> e ) p = num__5 q − num__20 |
p + num__5 = num__5 ( q + num__5 ) p + num__5 = num__5 q + num__25 p = num__5 q + num__20 option c <eor> c <eos> |
c |
subtract__25.0__5.0__ divide__25.0__5.0__ |
subtract__25.0__5.0__ divide__25.0__5.0__ |
| a man spends num__0.4 of his salary on house rent num__0.3 of his salary on food and num__0.125 of his salary on conveyance . if he has rs . num__1400 left with him find his expenditure on food and conveyance . <o> a ) num__1000 <o> b ) num__2000 <o> c ) num__3000 <o> d ) num__4000 <o> e ) num__5000 |
part of salary left = num__1 - ( num__0.4 + num__0.3 + num__0.125 ) let the monthly salary be rs . x then num__0.175 of x = num__1400 x = ( num__1400 * num__5.71428571429 ) = num__8600 expenditure on food = rs . ( num__0.3 * num__800 ) = rs . num__2400 expenditure on conveyence = rs . ( num__0.125 * num__8000 ) = rs . num__1000 answer is a . <eor> a <eos> |
a |
subtract__0.3__0.125__ reverse__0.175__ divide__1400.0__0.175__ multiply__0.125__8000.0__ multiply__0.125__8000.0__ |
subtract__0.3__0.125__ reverse__0.175__ divide__1400.0__0.175__ multiply__0.125__8000.0__ multiply__0.125__8000.0__ |
| a circular mat with diameter num__7 inches is placed on a square tabletop each of whose sides is num__8 inches long . which of the following is closest to the fraction of the tabletop covered by the mat ? <o> a ) num__0.857142857143 <o> b ) num__0.444444444444 <o> c ) num__2.5 <o> d ) num__1.66666666667 <o> e ) num__0.142857142857 |
so we are looking for the area of the cloth over the area of the table area of the cloth = ( pi ) ( r ) ^ num__2 which is about ( num__3.14285714286 ) ( num__7 ) ( num__7 ) area of the table = ( num__8 ) ( num__8 ) so the quick way to estimate is looking at the fraction like this : num__2.40625 nearest to num__2.5 answer : c <eor> c <eos> |
c |
triangle_area__2.5__2.0__ |
triangle_area__2.5__2.0__ |
| the value of x + x ( xx ) when x = num__5 is : <o> a ) num__130 <o> b ) num__10 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
x + x ( xx ) put the value of x = num__5 in the above expression we get = num__5 + num__5 ( num__5 Ã — num__5 ) = num__5 + num__125 = num__130 answer : a <eor> a <eos> |
a |
add__5.0__125.0__ add__5.0__125.0__ |
add__5.0__125.0__ add__5.0__125.0__ |
| in what time will rs . num__4000 lent at num__3.0 per annum on simple interest earn as much interest as rs . num__5000 will earn in num__5 years at num__4.0 per annum on simple interest ? <o> a ) num__8 num__0.333333333333 <o> b ) num__8 num__0.2 <o> c ) num__8 num__0.166666666667 <o> d ) num__8 num__0.5 <o> e ) num__8 num__0.142857142857 |
( num__4000 * num__3 * r ) / num__100 = ( num__5000 * num__5 * num__4 ) / num__100 r = num__8 num__0.333333333333 . answer : a <eor> a <eos> |
a |
percent__100.0__8.0__ |
percent__100.0__8.0__ |
| two trains are running at num__40 km / hr and num__20 km / hr respectively in the same direction . fast train completely passes a man sitting in the slower train in num__5 sec . what is the length of the fast train ? <o> a ) num__25 <o> b ) num__24 <o> c ) num__26 <o> d ) num__27 num__0.777777777778 <o> e ) num__28 |
relative speed = ( num__40 - num__20 ) = num__20 km / hr . = num__20 * num__0.277777777778 = num__5.55555555556 m / sec . length of faster train = num__5.55555555556 * num__5 = num__27.7777777778 = num__27 num__0.777777777778 m . answer : option d <eor> d <eos> |
d |
subtract__27.7778__27.0__ round__27.0__ |
subtract__27.7778__27.0__ subtract__27.7778__0.7778__ |
| a train num__500 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is : <o> a ) num__45 km / hr <o> b ) num__50 km / hr <o> c ) num__54 km / hr <o> d ) num__55 km / hr <o> e ) num__185 km / hr |
speed of the train relative to man = ( num__50.0 ) m / sec = num__50 m / sec . [ num__50 * ( num__3.6 ) ] km / hr = num__180 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__180 = = > x = num__185 km / hr . answer : e <eor> e <eos> |
e |
divide__500.0__10.0__ multiply__50.0__3.6__ add__5.0__180.0__ round__185.0__ |
divide__500.0__10.0__ multiply__50.0__3.6__ add__5.0__180.0__ round__185.0__ |
| free notebooks were distributed in a class among all the students . each student got notebooks which were num__0.125 th of the number of student . if number of student been half then each child would have received num__16 notebooks in total . find the total number of books distributed among students ? <o> a ) num__430 <o> b ) num__450 <o> c ) num__460 <o> d ) num__480 <o> e ) num__512 |
let suppose total number of students in class are x . then from the question we can conclude it that x ∗ num__18 x = x num__2 ∗ num__16 = > x = num__64 total notebooks = num__18 x num__2 = ( num__18 ∗ num__64 ∗ num__64 ) = num__512 e <eor> e <eos> |
e |
multiply__0.125__16.0__ divide__64.0__0.125__ divide__64.0__0.125__ |
multiply__0.125__16.0__ divide__64.0__0.125__ divide__64.0__0.125__ |
| worker a takes num__5 hours to do a job . worker b takes num__15 hours to do the same job . how long it take both a & b working together but independently to do the same job ? <o> a ) num__3.75 <o> b ) num__4.44444444444 <o> c ) num__5.55555555556 <o> d ) num__6.66666666667 <o> e ) num__8.88888888889 |
one day work of a = num__0.2 one day work of b = num__0.0666666666667 so one day work of a and b together = num__0.2 + num__0.0666666666667 = num__0.266666666667 so total days required = num__3.75 answer : a <eor> a <eos> |
a |
add__0.0667__0.2__ round__3.75__ |
add__0.0667__0.2__ round__3.75__ |
| the slant height of a right circular cone is num__10 m and its height is num__8 m . find the area of its curved surface . <o> a ) num__30 m num__2 <o> b ) num__40 m num__2 <o> c ) num__60 m num__2 <o> d ) num__80 m num__2 <o> e ) num__90 m num__2 |
l = num__10 m h = num__8 m . so r = l num__2 - h num__2 = ( num__10 ) num__2 - num__82 = num__6 m . curved surface area = rl = ( x num__6 x num__10 ) m num__2 = num__60 m num__2 . answer : option c <eor> c <eos> |
c |
side_by_diagonal__10.0__8.0__ multiply__10.0__6.0__ multiply__10.0__6.0__ |
side_by_diagonal__10.0__8.0__ multiply__10.0__6.0__ multiply__10.0__6.0__ |
| a manufacturer of a certain type of screw rejects any screw whose length is less than num__2.5 — num__0.03 centimeters or greater than num__2.53 centimeters . if k represents the length of a screw in centimeters which of the following inequalities specifies all the lengths of screws that are acceptable ? <o> a ) | k + num__0.03 | > num__2.5 <o> b ) | k — num__0.03 | < = num__2.5 <o> c ) | k — num__2.5 | > num__0.03 <o> d ) | num__2 k — num__5 | < = num__0.06 <o> e ) | k — num__2.5 | > = num__0.09 |
so let ' s go through this step by step : rejects any screw whose length is less than num__2.5 — num__0.03 centimeters or greater than num__2.5 + num__0.03 centimeters . in other words any screw that is less than : num__2.50 - num__0.03 = num__2.47 or greater than num__2.53 will be rejected . if k represents the length of a screw in other words kis an acceptable screw that must fall within the acceptable range of num__2.47 to num__2.53 so : num__2.47 ≤ k ≤ num__2.53 you can rule out answers with < or > as opposed to ≤ or ≥ because the length can not be less than num__2.47 or greater than num__2.53 . in other words num__2.47 and num__2.53 are acceptable lengths . let ' s look at ( d ) : | num__2 k — num__5 | < = num__0.06 or dividing by num__2 | k — num__2.5 | < = num__0.03 for the positive case : k - num__2.5 ≤ num__0.03 = = = > k ≤ num__2.53 for the negative case : - ( k - num__2.5 ) ≤ num__0.03 = = = > - k + num__2.5 ≤ num__0.03 = = = > - k ≤ - num__2.47 = = = > k ≥ num__2.47 num__2.47 ≤ k ≤ num__2.53 ( d ) <eor> d <eos> |
d |
subtract__2.5__0.03__ round_down__2.5__ multiply__2.5__2.0__ multiply__0.03__2.0__ round_down__2.5__ |
subtract__2.5__0.03__ round_down__2.5__ add__2.53__2.47__ subtract__2.53__2.47__ round_down__2.5__ |
| a vessel of capacity num__2 litre has num__35.0 of alcohol and another vessel of capacity num__6 litre had num__50.0 alcohol . the total liquid of num__8 litre was poured out in a vessel of capacity num__10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? <o> a ) num__31.0 . <o> b ) num__71.0 . <o> c ) num__49.0 . <o> d ) num__29.0 . <o> e ) num__37.0 . |
num__35.0 of num__2 litres = num__0.7 litres num__50.0 of num__6 litres = num__3 litres therefore total quantity of alcohol is num__3.7 litres . this mixture is in a num__10 litre vessel . hence the concentration of alcohol in this num__10 litre vessel is num__37.0 e <eor> e <eos> |
e |
divide__35.0__50.0__ divide__6.0__2.0__ add__3.0__0.7__ add__2.0__35.0__ add__2.0__35.0__ |
divide__35.0__50.0__ divide__6.0__2.0__ add__3.0__0.7__ add__2.0__35.0__ add__2.0__35.0__ |
| if x and y are positive integers and ( num__5 ^ x ) − ( num__5 ^ y ) = ( num__2 ^ y − num__1 ) ∗ ( num__5 ^ x − num__1 ) what is the value of xy ? <o> a ) num__48 <o> b ) num__36 <o> c ) num__24 <o> d ) num__18 <o> e ) num__12 |
x and y are positive integers which means we will have clean numbers . on the right hand side you have a num__2 as a factor while it is not there on the left hand side . can a num__2 be generated on the left hand side by the subtraction ? here i am thinking that if we take num__5 ^ y common on the left hand side i might be able to get a num__2 . num__56 y ( num__56 ^ x − y − num__1 ) = num__2 ^ y − num__1 ∗ num__5 ^ x − num__1 now i want only num__2 s and num__5 s on the left hand side . if x - y is num__1 then ( num__5 ^ x − y − num__1 ) becomes num__4 which is num__2 ^ num__2 . if instead x - y is num__2 or more i will get factors such as num__3 num__13 too . so let me try putting x - y = num__1 to get num__5 ^ y ( num__2 ^ num__2 ) = num__2 ^ y − num__1 ∗ num__5 ^ x − num__1 this gives me y - num__1 = num__2 y = num__3 x = num__4 check to see that the equations is satisfied with these values . hence xy = num__12 answer ( e ) <eor> e <eos> |
e |
subtract__5.0__1.0__ subtract__5.0__2.0__ multiply__3.0__4.0__ multiply__1.0__12.0__ |
subtract__5.0__1.0__ subtract__5.0__2.0__ subtract__13.0__1.0__ subtract__13.0__1.0__ |
| if the simple interest on a sum of money for num__5 years at num__18.0 per annum is rs . num__900 what is the compound interest on the same sum at the rate and for the same time ? <o> a ) rs . num__1287.76 <o> b ) rs . num__1284.76 <o> c ) rs . num__1587.76 <o> d ) rs . num__1266.76 <o> e ) rs . num__1283.76 |
sum = ( num__900 * num__100 ) / ( num__5 * num__18 ) = rs . num__1 num__000.00 c . i . on rs . rs . num__1 num__000.00 for num__5 years at num__18.0 = rs . num__2 num__287.76 . = rs . num__2 num__287.76 - num__1 num__000.00 = rs . num__1287.76 answer : a <eor> a <eos> |
a |
percent__100.0__1287.76__ |
percent__100.0__1287.76__ |
| if x = - num__2.4 and y = - num__0.857142857143 what is the value of the expression - num__2 x – y ^ num__2 ? <o> a ) num__2.0756302521 <o> b ) num__0.577946768061 <o> c ) num__0.865306122449 <o> d ) num__2.14448669202 <o> e ) - num__0.865306122449 |
x = - num__0.8 and y = - num__0.857142857143 = = > - num__2 ( - num__0.8 ) - ( - num__0.857142857143 ) ^ num__2 = num__1.6 - num__0.734693877551 = num__0.865306122449 ans : c <eor> c <eos> |
c |
subtract__2.4__0.8__ subtract__1.6__0.7347__ subtract__1.6__0.7347__ |
subtract__2.4__0.8__ subtract__1.6__0.7347__ subtract__1.6__0.7347__ |
| what is the difference between the compound interest on rs . num__12000 at num__20.0 p . a . for one year when compounded yearly and half yearly ? <o> a ) num__272 <o> b ) num__120 <o> c ) num__277 <o> d ) num__2898 <o> e ) num__212 |
when compounded annually interest = num__12000 [ num__1 + num__0.2 ] num__1 - num__12000 = rs . num__2400 when compounded semi - annually interest = num__12000 [ num__1 + num__0.1 ] num__2 - num__12000 = rs . num__2520 required difference = num__2520 - num__2400 = rs . num__120 . answer : b <eor> b <eos> |
b |
percent__20.0__1.0__ percent__20.0__12000.0__ percent__1.0__12000.0__ percent__1.0__12000.0__ |
percent__20.0__1.0__ percent__20.0__12000.0__ percent__1.0__12000.0__ percent__1.0__12000.0__ |
| mr . yutaro ’ s class contains num__10 boys and num__6 girls . if two students are chosen one at a time from the class what is the probability that a boy and a girl are chosen ? <o> a ) num__0.1875 <o> b ) num__0.45 <o> c ) num__0.458333333333 <o> d ) num__0.5625 <o> e ) num__0.5 |
probability of a girl being chosen first then a girl : num__0.375 * num__0.666666666667 probability of a boy being chosen first then a boy : num__0.625 * num__0.4 probability of a boy and a girl being chosen : ( num__6 * num__10 + num__6 * num__10 ) / ( num__15 * num__16 ) = num__0.5 answer : e <eor> e <eos> |
e |
negate_prob__0.375__ negate_prob__0.5__ |
negate_prob__0.375__ negate_prob__0.5__ |
| a and b together can do a piece of work in num__8 days . if a alone can do the same work in num__12 days then b alone can do the same work in ? <o> a ) num__76 days <o> b ) num__68 days <o> c ) num__24 days <o> d ) num__97 days <o> e ) num__17 days |
b = num__0.125 – num__0.5 = num__0.0416666666667 = > num__24 days answer : c <eor> c <eos> |
c |
divide__0.5__12.0__ divide__12.0__0.5__ round__24.0__ |
divide__0.5__12.0__ divide__12.0__0.5__ round__24.0__ |
| what is the difference between the local values of num__6 in the number num__56406 ? <o> a ) num__2992 <o> b ) num__2997 <o> c ) num__5994 <o> d ) num__2077 <o> e ) num__5211 |
explanation : num__6000 â € “ num__6 = num__5994 answer : c <eor> c <eos> |
c |
subtract__6000.0__6.0__ subtract__6000.0__6.0__ |
subtract__6000.0__6.0__ subtract__6000.0__6.0__ |
| a sum of money becomes num__1.16666666667 of itself in num__3 years at a certain rate of simple interest . the rate per annum is ? <o> a ) num__5 num__0.555555555556 % <o> b ) num__5 num__0.833333333333 % <o> c ) num__5 num__0.888888888889 % <o> d ) num__1 num__0.555555555556 % <o> e ) num__5 num__0.333333333333 % |
let sum = x . then amount = num__7 x / num__6 s . i . = num__7 x / num__6 - x = x / num__6 ; time = num__3 years . rate = ( num__100 * x ) / ( x * num__6 * num__3 ) = num__5 num__0.555555555556 % . answer : a <eor> a <eos> |
a |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| a bucket contains num__2 mixtures a and b in the ratio num__7 : num__5 . if num__9 liters of the mixture is replaced by num__9 liters of liquid b then the ratio of the two liquids becomes num__7 : num__9 . how much of the liquid a was there in the bucket ? <o> a ) num__15 liters <o> b ) num__21 liters <o> c ) num__12 liters <o> d ) num__18 liters <o> e ) num__25 liters |
num__1 st ratio = num__7 : num__5 num__2 nd ratio = num__7 : num__9 difference of cross products ratio = num__7 * num__9 - num__7 * num__5 = num__28 common factor of the num__1 st ratio = ( quantity replaced / sum of terms in num__1 st ratio ) + ( quantity replaced * term a in num__2 nd ratio / difference ) = ( num__1.28571428571 + num__5 ) + ( num__9 * num__0.25 ) = num__3 quantity of a = num__7 * num__3 = num__21 liters answer is b <eor> b <eos> |
b |
divide__9.0__7.0__ divide__7.0__28.0__ add__2.0__1.0__ multiply__7.0__3.0__ multiply__7.0__3.0__ |
divide__9.0__7.0__ divide__7.0__28.0__ add__2.0__1.0__ multiply__7.0__3.0__ multiply__7.0__3.0__ |
| a can do a piece of work in num__30 days . he works at it for num__5 days and then b finishes it in num__20 days . in what time can a and b together it ? <o> a ) num__15 num__0.333333333333 days <o> b ) num__13 num__0.333333333333 days <o> c ) num__19 num__0.333333333333 days <o> d ) num__11 num__0.333333333333 days <o> e ) num__12 num__0.333333333333 days |
b num__13 num__0.333333333333 days num__0.166666666667 + num__20 / x = num__1 x = num__24 num__0.0333333333333 + num__0.0416666666667 = num__0.075 num__13.3333333333 = num__13 num__0.333333333333 days <eor> b <eos> |
b |
divide__5.0__30.0__ divide__1.0__30.0__ divide__1.0__24.0__ add__0.0417__0.0333__ divide__1.0__0.075__ round__13.0__ |
divide__5.0__30.0__ divide__1.0__30.0__ divide__1.0__24.0__ add__0.0417__0.0333__ divide__1.0__0.075__ divide__13.0__1.0__ |
| the largest number that exactly divides each number of the sequence ( num__15 - num__1 ) ( num__25 - num__2 ) ( num__35 - num__3 ) . . . . ( n num__5 - n ) . . . . . . is <o> a ) num__1 <o> b ) num__15 <o> c ) num__30 <o> d ) num__120 <o> e ) none |
sol . required number = ( num__25 - num__2 ) = ( num__32 - num__2 ) = num__30 . answer c <eor> c <eos> |
c |
power__2.0__5.0__ multiply__15.0__2.0__ multiply__15.0__2.0__ |
subtract__35.0__3.0__ subtract__35.0__5.0__ subtract__35.0__5.0__ |
| f two dice are thrown together the probability of getting an even number on one die and an odd number on the other is ? <o> a ) num__0.111111111111 <o> b ) num__0.5 <o> c ) num__1.0 <o> d ) num__0.2 <o> e ) num__0.125 |
the number of exhaustive outcomes is num__36 . let e be the event of getting an even number on one die and an odd number on the other . let the event of getting either both even or both odd then = num__0.5 = num__0.5 p ( e ) = num__1 - num__0.5 = num__0.5 . answer : b <eor> b <eos> |
b |
negate_prob__0.5__ |
negate_prob__0.5__ |
| amithab ' s average expenditure for the january to june is rs . num__4200 and he spends rs . num__1200 in january and rs . num__1500 in july . the average expenditure for the months of febraury to july is : <o> a ) rs . num__4250 <o> b ) rs . num__4288 <o> c ) rs . num__4227 <o> d ) rs . num__4218 <o> e ) rs . num__4219 |
explanation : amithab ' s total expenditure for jan - june = num__4200 x num__6 = num__25200 expenditure for feb - june = num__25200 - num__1200 = num__24000 expenditure for the months of feb - july = num__24000 + num__1500 = num__25500 the average expenditure = { num__25500 } { num__6 } = num__4250 answer : a <eor> a <eos> |
a |
multiply__4200.0__6.0__ subtract__25200.0__1200.0__ add__1500.0__24000.0__ divide__25500.0__6.0__ divide__25500.0__6.0__ |
multiply__4200.0__6.0__ subtract__25200.0__1200.0__ add__1500.0__24000.0__ divide__25500.0__6.0__ divide__25500.0__6.0__ |
| a boat can move upstream at num__25 kmph and downstream at num__35 kmph then the speed of the current is ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__4 <o> e ) num__32 |
us = num__25 ds = num__35 m = ( num__35 - num__25 ) / num__2 = num__5 answer : a <eor> a <eos> |
a |
round__5.0__ |
divide__25.0__5.0__ |
| to fill a tank num__25 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to num__0.6 of its present ? <o> a ) num__41.66667 <o> b ) num__60.5 <o> c ) num__63.5 <o> d ) num__62.5 <o> e ) num__64.5 |
let capacity of num__1 bucket = x capacity of the tank = num__25 x new capacity of the bucket = num__3 x / num__5 hence number of buckets needed = num__25 x / ( num__3 x / num__5 ) = ( num__25 × num__5 ) / num__3 = num__41.66667 answer is a . <eor> a <eos> |
a |
divide__3.0__0.6__ divide__25.0__0.6__ divide__25.0__0.6__ |
divide__3.0__0.6__ divide__25.0__0.6__ divide__25.0__0.6__ |
| the banker ' s gain on a sum due num__3 years hence at num__12.0 per annum is rs . num__270 . the banker ' s discount is : <o> a ) rs . num__960 <o> b ) rs . num__760 <o> c ) rs . num__840 <o> d ) rs . num__1020 <o> e ) rs . num__1615 |
t . d = ( b . gx num__100 ) / ( r / t ) = ( num__270 x num__100 ) / ( num__12 x num__3 ) = rs . num__750 b . d . = rs . ( num__750 + num__270 ) = rs . num__1020 . answer : d <eor> d <eos> |
d |
percent__100.0__1020.0__ |
percent__100.0__1020.0__ |
| three candidates contested an election and received num__1136 num__7636 and num__11628 votes respectively . what percentage of the total votes did the winning candidate get ? <o> a ) num__57.0 <o> b ) num__58.0 <o> c ) num__60.0 <o> d ) num__65.0 <o> e ) num__68 % |
total number of votes polled = ( num__1136 + num__7636 + num__11628 ) = num__20400 . required % = ( num__0.57 ) * num__100.0 = num__57.0 answer : a <eor> a <eos> |
a |
divide__11628.0__20400.0__ multiply__100.0__0.57__ multiply__100.0__0.57__ |
divide__11628.0__20400.0__ multiply__100.0__0.57__ multiply__100.0__0.57__ |
| what is num__82.0 of num__0.75 ? <o> a ) num__6.9 <o> b ) num__69.0 <o> c ) num__0.6845 <o> d ) num__0.6859 <o> e ) num__0.615 |
num__82.0 * ( num__0.75 ) = num__0.82 * num__0.75 = num__0.615 answer : e <eor> e <eos> |
e |
percent__82.0__0.75__ percent__82.0__0.75__ |
percent__82.0__0.75__ percent__82.0__0.75__ |
| city x has a population num__6 times as great as the population of city y which has a population twice as great as the population of city z . what is the ratio of the population of city x to the population of city z ? <o> a ) num__1 : num__8 <o> b ) num__1 : num__4 <o> c ) num__2 : num__1 <o> d ) num__4 : num__1 <o> e ) num__12 : num__1 |
x = num__6 y y = num__2 * z x : y y : z num__6 : num__1 num__2 : num__1 num__12 : num__2 num__2 : num__1 so x : z = num__12 : num__1 ( e ) <eor> e <eos> |
e |
multiply__6.0__2.0__ multiply__6.0__2.0__ |
multiply__6.0__2.0__ multiply__6.0__2.0__ |
| the average of first num__10 prime numbers is ? <o> a ) num__12.7 <o> b ) num__12.2 <o> c ) num__12.9 <o> d ) num__12.1 <o> e ) num__12.6 |
sum of num__10 prime no . = num__129 average = num__12.9 = num__12.9 answer : c <eor> c <eos> |
c |
divide__129.0__10.0__ divide__129.0__10.0__ |
divide__129.0__10.0__ divide__129.0__10.0__ |
| what is num__15 percent of num__54 ? <o> a ) num__7.3 <o> b ) num__8.1 <o> c ) num__9.5 <o> d ) num__10.2 <o> e ) num__11.6 |
( num__0.15 ) * num__54 = num__8.1 the answer is b . <eor> b <eos> |
b |
percent__15.0__54.0__ percent__15.0__54.0__ |
percent__15.0__54.0__ percent__15.0__54.0__ |
| what is the remainder when num__103 * num__505 * num__607 is divided by num__9 <o> a ) num__6 <o> b ) num__18 <o> c ) num__2 <o> d ) num__4 <o> e ) num__7 |
num__11.4444444444 = = > remainder = num__4 num__56.1111111111 = = > remainder = num__1 num__67.4444444444 = = > remainder = num__4 = num__4 * num__4 * num__1 = num__1.77777777778 = = > remainder = num__7 answer : e <eor> e <eos> |
e |
divide__103.0__9.0__ divide__505.0__9.0__ divide__607.0__9.0__ multiply__1.0__7.0__ |
divide__103.0__9.0__ divide__505.0__9.0__ divide__607.0__9.0__ multiply__1.0__7.0__ |
| in a certain sequence the first term is num__1 and each successive term is num__1 more than the reciprocal of the term that immediately precedes it . what is the fifth term of the sequence ? <o> a ) num__0.6 <o> b ) num__0.625 <o> c ) num__1.6 <o> d ) num__1.66666666667 <o> e ) num__4.5 |
first term = num__1 second term = ( num__1.0 ) + num__1 = num__2 third term = ( num__0.5 ) + num__1 = num__1.5 forth term = [ num__1 / ( num__1.5 ) ] + num__1 = num__1.66666666667 fifth term = [ num__1 / ( num__1.66666666667 ) ] + num__1 = num__1.6 answer : option c <eor> c <eos> |
c |
reverse__2.0__ add__1.0__0.5__ multiply__1.0__1.6__ |
reverse__2.0__ add__1.0__0.5__ divide__1.6__1.0__ |
| nr books bought nr of people num__1 num__5 num__3 num__2 num__4 num__7 num__6 num__3 what is the median of books bought per person ? <o> a ) a ) num__2 <o> b ) b ) num__4 <o> c ) c ) num__6 <o> d ) d ) num__8 <o> e ) e ) num__18 |
num__1 num__11 num__11 num__33 num__44 num__44 num__44 num__46 num__66 so you will observer that the median of the list is num__4 . ans b <eor> b <eos> |
b |
add__5.0__6.0__ multiply__3.0__11.0__ multiply__4.0__11.0__ add__2.0__44.0__ multiply__2.0__33.0__ add__1.0__3.0__ |
add__5.0__6.0__ multiply__3.0__11.0__ multiply__4.0__11.0__ add__2.0__44.0__ multiply__2.0__33.0__ add__1.0__3.0__ |
| the averageincome of m and n is rs . num__5050 . the average monthly income of n and o is rs . num__6250 and the average monthly income of m and ois rs . num__5200 . the income of m is ? <o> a ) rs . num__2000 <o> b ) rs . num__3000 <o> c ) rs . num__4000 <o> d ) rs . num__4500 <o> e ) rs . num__5000 |
let p q and r represent their respective monthly incomes . then we have : p + q = ( num__5050 x num__2 ) = num__10100 . . . . ( i ) q + r = ( num__6250 x num__2 ) = num__12500 . . . . ( ii ) m + o = ( num__5200 x num__2 ) = num__10400 . . . . ( iii ) adding ( i ) ( ii ) and ( iii ) we get : num__2 ( m + n + o ) = num__33000 or m + n + o = num__16500 . . . . ( iv ) subtracting ( ii ) from ( iv ) we get m = num__4000 . m ' s monthly income = rs . num__4000 c <eor> c <eos> |
c |
multiply__5050.0__2.0__ multiply__6250.0__2.0__ multiply__5200.0__2.0__ divide__33000.0__2.0__ subtract__16500.0__12500.0__ subtract__16500.0__12500.0__ |
multiply__5050.0__2.0__ multiply__6250.0__2.0__ multiply__5200.0__2.0__ divide__33000.0__2.0__ subtract__16500.0__12500.0__ subtract__16500.0__12500.0__ |
| in a barrel of juice there is num__40 liters ; in a barrel of beer there are num__80 liters . if the price ratio between barrels of juice to a barrel of beer is num__3 : num__4 what is the price ratio between one liter of juice and one liter of beer ? <o> a ) num__3 : num__2 . <o> b ) num__2 : num__1 . <o> c ) num__3 : num__1 . <o> d ) num__4 : num__3 . <o> e ) num__3 : num__4 . |
juice barrel : beer barrel = num__40 : num__80 i . e . num__3 : num__8 price of juice : price of beer = num__3 : num__4 so for num__0.5 costs num__0.75 ie num__4 j / num__8 b = num__0.75 . solving for j / b we get num__3 : num__2 answer a . <eor> a <eos> |
a |
divide__40.0__80.0__ divide__3.0__4.0__ reverse__0.5__ multiply__4.0__0.75__ |
divide__40.0__80.0__ divide__3.0__4.0__ reverse__0.5__ multiply__4.0__0.75__ |
| the average weight of num__8 person ' s increases by num__6 kg when a new person comes in place of one of them weighing num__40 kg . what might be the weight of the new person ? <o> a ) num__80 kg <o> b ) num__85 kg <o> c ) num__90 kg <o> d ) num__88 kg <o> e ) num__110 kg |
total weight increased = ( num__8 x num__6 ) kg = num__48 kg . weight of new person = ( num__40 + num__48 ) kg = num__88 kg . answer : d <eor> d <eos> |
d |
multiply__8.0__6.0__ add__40.0__48.0__ add__40.0__48.0__ |
add__8.0__40.0__ add__40.0__48.0__ add__40.0__48.0__ |
| in a company the manager wants to give some gifts to all of the workers . in each block there are about num__100 workers are there . the total amount for giving the gifts for all the workers is $ num__4000 . the worth of the gift is $ num__4 . how many blocks are there in the company ? <o> a ) num__12 <o> b ) num__10 <o> c ) num__15 <o> d ) num__17 <o> e ) num__20 |
each employee will get a gift worth of = $ num__4 total employees = num__1000.0 = num__1000 total blocks = num__10.0 = num__10 correct option is b <eor> b <eos> |
b |
divide__4000.0__4.0__ divide__1000.0__100.0__ divide__100.0__10.0__ |
divide__4000.0__4.0__ divide__1000.0__100.0__ divide__100.0__10.0__ |
| mary can do a piece of work in num__28 days . rosy is num__40.0 more efficient than mary . the number of days taken by rosy to do the same piece of work is ? <o> a ) num__22 <o> b ) num__24 <o> c ) num__20 <o> d ) num__25 <o> e ) num__27 |
ratio of times taken by mary and rosy = num__140 : num__100 = num__14 : num__10 suppose rosy takes x days to do the work . num__14 : num__10 : : num__28 : x = > x = num__20 days . hence rosy takes num__20 days to complete the work . answer : c <eor> c <eos> |
c |
subtract__140.0__40.0__ divide__140.0__14.0__ round__20.0__ |
subtract__140.0__40.0__ divide__140.0__14.0__ round__20.0__ |
| a and b started a partnership business investing some amount in the ratio of num__2 : num__5 . c joined then after six months with an amount equal to that of b . in what proportion should the profit at the end of one year be distributed among a b and c ? <o> a ) num__4 : num__10 : num__5 <o> b ) num__5 : num__7 : num__4 <o> c ) num__6 : num__10 : num__5 <o> d ) num__7 : num__9 : num__4 <o> e ) num__5 : num__4 : num__8 |
let the initial investments of a and b be num__2 x and num__5 x . a : b : c = ( num__2 x x num__12 ) : ( num__5 x x num__12 ) : ( num__5 x x num__6 ) = num__24 : num__60 : num__30 = num__4 : num__10 : num__5 . answer : a <eor> a <eos> |
a |
divide__12.0__2.0__ multiply__2.0__12.0__ multiply__5.0__12.0__ multiply__5.0__6.0__ subtract__6.0__2.0__ multiply__2.0__5.0__ subtract__6.0__2.0__ |
divide__12.0__2.0__ multiply__2.0__12.0__ multiply__5.0__12.0__ multiply__5.0__6.0__ subtract__6.0__2.0__ multiply__2.0__5.0__ subtract__6.0__2.0__ |
| myra drove at an average speed of num__30 miles per hour for some time and then at an average speed of num__60 miles / hr for the rest of the journey . if she made no stops during the trip and her average speed for the entire journey was num__50 miles per hour for what fraction of the total time did she drive at num__30 miles / hour ? <o> a ) the average speed is num__0.2 miles per hour . <o> b ) the average speed is num__0.333333333333 miles per hour . <o> c ) the average speed is num__0.4 miles per hour . <o> d ) the average speed is num__0.666666666667 miles per hour . <o> e ) the average speed is num__0.6 miles per hour . |
we know the average speed and must find the fraction of time taken at a particular speed . t num__1 / t num__2 = ( a num__2 – aavg ) / ( aavg – a num__1 ) t num__1 / t num__2 = ( num__60 – num__50 ) / ( num__50 – num__30 ) = num__0.5 so out of a total of num__3 parts of the journey time she drove at num__30 mph for num__1 part and at num__60 mph for num__2 parts of the time . fraction of the total time for which she drove at num__30 mph is num__0.333333333333 . answer ( b ) <eor> b <eos> |
b |
divide__60.0__30.0__ reverse__2.0__ add__1.0__2.0__ reverse__3.0__ reverse__3.0__ |
divide__60.0__30.0__ reverse__2.0__ add__1.0__2.0__ reverse__3.0__ reverse__3.0__ |
| if - num__6 < x < num__7 and - num__6 < y < num__3 which of the following specifies all the possible values of xy ? <o> a ) - num__42 < xy < num__21 <o> b ) - num__42 < xy < num__24 <o> c ) - num__28 < xy < num__18 <o> d ) - num__42 < xy < num__36 <o> e ) - num__24 < xy < num__24 |
the least value of xy is a bit more than num__7 * ( - num__6 ) = - num__42 and the largest value of xy is a bit less than ( - num__6 ) * ( - num__6 ) = num__36 . therefore - num__42 < xy < num__36 . answer : d . <eor> d <eos> |
d |
multiply__6.0__7.0__ subtract__42.0__6.0__ multiply__6.0__7.0__ |
multiply__6.0__7.0__ subtract__42.0__6.0__ multiply__6.0__7.0__ |
| convert num__0.34 in to a vulgar fraction ? <o> a ) num__0.36 <o> b ) num__0.32 <o> c ) num__0.34 <o> d ) num__0.38 <o> e ) none |
answer num__0.34 = num__0.34 = num__0.34 correct option : c <eor> c <eos> |
c |
round__0.34__ |
round__0.34__ |
| the difference between simple and compound interest on rs . num__1200 for one year at num__10.0 per annum reckoned half - yearly is ? <o> a ) num__8 <o> b ) num__3 <o> c ) num__9 <o> d ) num__3 <o> e ) num__2 |
s . i . = ( num__1200 * num__10 * num__1 ) / num__100 = rs . num__120 c . i . = [ num__1200 * ( num__1 + num__0.05 ) num__2 - num__1200 ] = rs . num__123 difference = ( num__123 - num__120 ) = rs . num__3 . answer : b <eor> b <eos> |
b |
percent__10.0__1200.0__ percent__3.0__100.0__ |
percent__10.0__1200.0__ percent__3.0__100.0__ |
| there are num__408 boys and num__312 girls in a school which are to be divided into equal sections of either boys or girls alone . find the total number of sections thus formed . <o> a ) num__31 <o> b ) num__32 <o> c ) num__35 <o> d ) num__30 <o> e ) num__45 |
explanation : hcf ( num__408 num__312 ) = num__24 the number of boys or girls that can be placed in a section = num__24 . thus the total number of sections is given by num__17.0 + num__13.0 = num__17 + num__13 = num__30 answer : d <eor> d <eos> |
d |
divide__408.0__24.0__ divide__312.0__24.0__ add__17.0__13.0__ add__17.0__13.0__ |
divide__408.0__24.0__ divide__312.0__24.0__ add__17.0__13.0__ add__17.0__13.0__ |
| a bucket full of nuts was discovered by the crow living in the basement . the crow eats a fifth of the total number of nuts in num__8 hours . how many hours in total will it take the crow to finish a quarter of the nuts ? <o> a ) num__9 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__14 |
in one hour the crow eats num__0.025 of the nuts . ( num__0.25 ) / ( num__0.025 ) = num__10 hours the answer is b . <eor> b <eos> |
b |
divide__0.25__0.025__ divide__0.25__0.025__ |
divide__0.25__0.025__ divide__0.25__0.025__ |
| a bell curve ( normal distribution ) has a mean of − num__1 and a standard deviation of num__0.125 . how many integer values are within three standard deviations of the mean ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__3 <o> d ) num__6 <o> e ) num__7 |
mean = - num__1 standard deviation = num__0.125 num__1 unit of standard deviation below the mean = - num__1 - num__0.125 = - num__1 num__0.125 num__2 units of standard deviation below the mean = - num__1 - num__0.125 - num__0.125 = - num__1 num__0.25 num__2 units of standard deviation below the mean = - num__1 - num__0.125 - num__0.125 - num__0.125 = - num__1 num__0.375 num__1 unit of standard deviation above the mean = - num__1 + num__0.125 = - num__0.875 num__2 units of standard deviation above the mean = - num__1 + num__0.125 + num__0.125 = - num__0.75 num__3 units of standard deviation above the mean = - num__1 + num__0.125 + num__0.125 + num__0.125 = - num__0.625 so all values from - num__1 num__0.375 to - num__0.625 are within num__3 standard deviations of the mean . within this range there is only num__1 integer value : - num__1 answer : c <eor> c <eos> |
c |
multiply__0.125__2.0__ add__0.125__0.25__ subtract__1.0__0.125__ subtract__1.0__0.25__ add__1.0__2.0__ subtract__1.0__0.375__ add__1.0__2.0__ |
multiply__0.125__2.0__ add__0.125__0.25__ subtract__1.0__0.125__ subtract__1.0__0.25__ add__1.0__2.0__ subtract__1.0__0.375__ add__1.0__2.0__ |
| tough and tricky questions : distance / rate . on a reconnaissance mission a state - of - the - art nuclear powered submarine traveled num__300 miles to reposition itself in the proximity of an aircraft carrier . this journey would have taken num__1 hour less if the submarine had traveled num__10 miles per hour faster . what was the average speed in miles per hour for the actual journey ? <o> a ) num__20 <o> b ) num__40 <o> c ) num__60 <o> d ) num__80 <o> e ) num__50 |
say if speed is num__60 num__6.0 = num__6 hrs and num__5.0 = num__5 hrs ( a reduction of num__1 hr - > correct answer ) answer ( e ) <eor> e <eos> |
e |
divide__60.0__10.0__ divide__300.0__60.0__ divide__300.0__6.0__ |
divide__60.0__10.0__ subtract__6.0__1.0__ subtract__60.0__10.0__ |
| in what time will a railway train num__50 m long moving at the rate of num__36 kmph pass a telegraph post on its way ? <o> a ) num__8 sec <o> b ) num__7 sec <o> c ) num__5 sec <o> d ) num__6 sec <o> e ) num__9 sec |
t = num__1.38888888889 * num__3.6 = num__5 sec answer : c <eor> c <eos> |
c |
divide__50.0__36.0__ multiply__3.6__1.3889__ round__5.0__ |
divide__50.0__36.0__ multiply__3.6__1.3889__ multiply__3.6__1.3889__ |
| the monthly incomes of a and b are in the ratio num__5 : num__2 . b ' s monthly income is num__12.0 more than c ' s monthly income . if c ' s monthly income is rs . num__15000 then find the annual income of a ? <o> a ) rs . num__420000 <o> b ) rs . num__180000 <o> c ) rs . num__201600 <o> d ) rs . num__504000 <o> e ) none of these |
b ' s monthly income = num__15000 * num__1.12 = rs . num__16800 b ' s monthly income = num__2 parts - - - - > rs . num__16800 a ' s monthly income = num__5 parts = num__2.5 * num__16800 = rs . num__42000 a ' s annual income = rs . num__42000 * num__12 = rs . num__504000 answer : d <eor> d <eos> |
d |
multiply__15000.0__1.12__ divide__5.0__2.0__ multiply__16800.0__2.5__ multiply__12.0__42000.0__ multiply__12.0__42000.0__ |
multiply__15000.0__1.12__ divide__5.0__2.0__ multiply__16800.0__2.5__ multiply__12.0__42000.0__ multiply__12.0__42000.0__ |
| abhinav and bhupathi together have rs . num__1210 . if num__0.266666666667 of abhinav ' s amount is equal to num__0.4 of bhupathi ' s amount how much amount does bhupathi have ? <o> a ) rs num__460 <o> b ) rs num__484 <o> c ) rs num__550 <o> d ) rs num__664 <o> e ) none of these |
explanation : let a = abhinav b = bhupathi . num__0.266666666667 a = num__0.4 b = > a = ( num__0.00388349514563 / num__4 ) b = > a = num__1.5 b = > a / b = num__1.5 = > a : b = num__3 : num__2 b ' s share = rs . ( num__2420.4 ) = rs . num__484 . answer : b <eor> b <eos> |
b |
divide__3.0__1.5__ multiply__1210.0__0.4__ multiply__1210.0__0.4__ |
divide__3.0__1.5__ multiply__1210.0__0.4__ multiply__1210.0__0.4__ |
| on a journey across bombay a tourist bus averages num__10 km / h for num__20.0 of the distance num__30 km / h for num__60.0 of it and num__20 km / h for the remainder . the average speed for the whole journey was <o> a ) num__10 km / h <o> b ) num__30 km / h <o> c ) num__5 km / h <o> d ) num__20 kilometre per hour <o> e ) none of these |
let the average speed be x km / h . and total distance = y km . then num__0.2 ⁄ num__10 y + num__0.6 ⁄ num__30 y + num__0.2 ⁄ num__20 y = y ⁄ x ⇒ x = num__1 ⁄ num__0.05 = num__20 km / h answer d <eor> d <eos> |
d |
reverse__20.0__ reverse__0.05__ |
reverse__20.0__ reverse__0.05__ |
| given the two equations num__5 r - num__5 s = num__15 and num__2 r - num__2 s = num__6 by how much does r exceed s ? <o> a ) num__3 <o> b ) num__2 <o> c ) num__11 <o> d ) num__4 <o> e ) num__7 |
solve by simultaneous equations . my answer is a . r = num__6 s = num__3 <eor> a <eos> |
a |
subtract__5.0__2.0__ subtract__5.0__2.0__ |
subtract__5.0__2.0__ subtract__5.0__2.0__ |
| the unit digit in the product num__1258 * num__6754 * num__4512 * num__9783 is ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
unit digit in the given product = unit digit in num__8 * num__4 * num__2 * num__3 = num__2 answer is c <eor> c <eos> |
c |
divide__8.0__4.0__ divide__8.0__4.0__ |
divide__8.0__4.0__ divide__8.0__4.0__ |
| for any number z z * is defined as the greatest positive even integer less than or equal to z . what is the value of num__5.2 – num__5.2 * ? <o> a ) num__0.2 <o> b ) num__1.2 <o> c ) num__1.8 <o> d ) num__2.2 <o> e ) num__4.0 |
since z * is defined as the greatest positive even integer less than or equal to z then num__5.2 * = num__4 ( the greatest positive even integer less than or equal to num__5.2 is num__4 ) . hence num__5.2 – num__5.2 * = num__5.2 - num__4 = num__1.2 answer : b . <eor> b <eos> |
b |
subtract__5.2__4.0__ subtract__5.2__4.0__ |
subtract__5.2__4.0__ subtract__5.2__4.0__ |
| john want to buy a $ num__100 trouser at the store but he think it ’ s too expensive . finally it goes on sale for $ num__40 . what is the percent decrease ? <o> a ) num__20.0 <o> b ) num__30.0 <o> c ) num__40.0 <o> d ) num__70.0 <o> e ) num__60 % |
the is always the difference between our starting and ending points . in this case it ’ s num__100 – num__40 = num__60 . the “ original ” is our starting point ; in this case it ’ s num__100 . ( num__0.6 ) * num__100 = ( num__0.6 ) * num__100 = num__60.0 . e <eor> e <eos> |
e |
subtract__100.0__40.0__ divide__60.0__100.0__ subtract__100.0__40.0__ |
subtract__100.0__40.0__ divide__60.0__100.0__ multiply__100.0__0.6__ |
| increasing the original price of a certain item by num__20 percent and then increasing the new price by num__20 percent is equivalent to increasing the original price by what percent ? <o> a ) num__31.25 <o> b ) num__37.5 <o> c ) num__50.0 <o> d ) num__52.5 <o> e ) num__44.0 |
we ' re told that the original price of an item is increased by num__20.0 and then that price is increased by num__20.0 . . . . if . . . . starting value = $ num__100 + num__20.0 = num__100 + . num__20 ( num__100 ) = num__120 + num__20.0 = num__120 + . num__20 ( num__120 ) = num__120 + num__24 = num__144 the question asks how the final price relates to the original price . this is essentially about percentage change which means we should use the percentage change formula : percentage change = ( new - old ) / old = difference / original doing either calculation will yield the same result : num__0.44 = num__44.0 final answer : e <eor> e <eos> |
e |
percent__20.0__120.0__ percent__100.0__44.0__ |
percent__20.0__120.0__ percent__100.0__44.0__ |
| the average age of a group of persons going for picnic is years . twenty new persons with an average age of num__15 years join the group on the spot due to which their average age becomes num__15.5 years . the number of persons initially going for picnic is <o> a ) num__5 <o> b ) num__10 <o> c ) num__20 <o> d ) num__40 <o> e ) num__50 |
solution let the initial number of persons be x . then num__16 x + num__20 x num__15 = num__15.5 ( x + num__20 ) = num__0.5 x = num__10 x = num__20 . answer c <eor> c <eos> |
c |
subtract__15.5__15.0__ multiply__0.5__20.0__ divide__10.0__0.5__ |
subtract__15.5__15.0__ multiply__0.5__20.0__ divide__10.0__0.5__ |
| x starts a business with rs . num__45000 . y joins in the business after num__4 months with rs . num__30000 . what will be the ratio in which they should share the profit at the end of the year ? <o> a ) num__1 : num__2 <o> b ) num__2 : num__1 <o> c ) num__1 : num__3 <o> d ) num__3 : num__1 <o> e ) num__9 : num__4 |
explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = num__45000 Ã — num__12 : num__30000 Ã — num__8 = num__45 Ã — num__12 : num__30 Ã — num__8 = num__3 Ã — num__12 : num__2 Ã — num__8 = num__9 : num__4 answer : option e <eor> e <eos> |
e |
subtract__12.0__4.0__ divide__12.0__4.0__ divide__8.0__4.0__ subtract__12.0__3.0__ subtract__12.0__3.0__ |
subtract__12.0__4.0__ divide__12.0__4.0__ divide__8.0__4.0__ subtract__12.0__3.0__ subtract__12.0__3.0__ |
| there are num__80 coins among them one coin weighs less compared to other . you are given a physical balance to weigh . in how many weighing the odd coin can be found . <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__4 |
num__80 divide as num__27 num__27 num__26 num__9 num__9 num__9 num__3 num__3 num__3 num__1 num__1 num__1 so weighings are num__4 answer : e <eor> e <eos> |
e |
alphabet_space__ choose__4.0__3.0__ |
alphabet_space__ choose__4.0__3.0__ |
| what is the compound interest paid on a sum of rs . num__3000 for the period of num__2 years at num__10.0 per annum . <o> a ) num__630 <o> b ) num__620 <o> c ) num__610 <o> d ) num__600 <o> e ) none of these |
solution = interest % for num__1 st year = num__10 interest % for num__2 nd year = num__10 + num__10.0 of num__10 = num__10 + num__10 * num__0.1 = num__11 total % of interest = num__10 + num__11 = num__21 total interest = num__21.0 num__3000 = num__3000 * ( num__0.21 ) = num__630 answer a <eor> a <eos> |
a |
percent__10.0__1.0__ percent__1.0__21.0__ percent__21.0__3000.0__ percent__21.0__3000.0__ |
percent__10.0__1.0__ percent__1.0__21.0__ percent__21.0__3000.0__ percent__21.0__3000.0__ |
| a pair of articles was bought for $ num__50 at a discount of num__50.0 . what must be the marked price of each of the article ? <o> a ) $ num__25 <o> b ) $ num__12.50 <o> c ) $ num__29.65 <o> d ) $ num__35.95 <o> e ) $ num__45.62 |
s . p . of each of the article = num__25.0 = $ num__25 let m . p = $ x num__50.0 of x = num__25 x = num__25 * . num__5 = $ num__12.50 answer is b <eor> b <eos> |
b |
percent__50.0__25.0__ percent__50.0__25.0__ |
percent__50.0__25.0__ percent__50.0__25.0__ |
| the g . c . d . of num__1.08 num__0.36 and num__0.9 is ? <o> a ) num__0.03 <o> b ) num__0.9 <o> c ) num__0.18 <o> d ) num__0.108 <o> e ) none of these |
given numbers are num__1.08 num__0.36 and num__0.90 . h . c . f of num__108 num__36 and num__90 is num__18 h . c . f of given numbers = num__0.18 . correct options : c <eor> c <eos> |
c |
gcd__36.0__90.0__ subtract__1.08__0.9__ subtract__1.08__0.9__ |
gcd__36.0__90.0__ subtract__1.08__0.9__ subtract__1.08__0.9__ |
| x ^ num__2 + qx + num__72 = num__0 has two distinct integer roots ; how many values are possible for q ? <o> a ) num__3 <o> b ) num__6 <o> c ) num__8 <o> d ) num__12 <o> e ) num__24 |
for a quadratic equation ax ^ num__2 + qx + c = num__0 we know that - q / a is sum of roots and c / a is product of roots . the quadratic equation here is x ^ num__2 + qx + num__72 = num__0 where product of roots is num__72 . if we find all the factors of num__72 we have the answer . by prime factorization we get num__72 = num__2 ^ num__3 * num__3 ^ num__2 . we know that total factors are ( num__3 + num__1 ) * ( num__2 + num__1 ) = num__12 ( reason : with num__2 ^ n we have n + num__1 possibilities . n ^ num__0 to n ^ n . so n + num__1 ) = d <eor> d <eos> |
d |
subtract__3.0__2.0__ multiply__1.0__12.0__ |
subtract__3.0__2.0__ multiply__1.0__12.0__ |
| a number consists of num__3 digit whose sum is num__10 . the middle digit is equal to the sum of the other two and the number will be increased by num__99 if its digits are reversed . the number is : <o> a ) num__22 <o> b ) num__24 <o> c ) num__77 <o> d ) num__26 <o> e ) num__81 |
explanation : let the ten ' s and unit ' s digit be x and num__8 / x respectively . then ( num__10 x + num__8 / x ) + num__18 = num__10 * num__8 / x + x num__9 x num__2 + num__18 x - num__72 = num__0 x num__2 + num__2 x - num__8 = num__0 ( x + num__4 ) ( x - num__2 ) = num__0 x = num__2 so ten ' s digit = num__2 and unit ' s digit = num__4 . hence required number = num__24 . answer : b <eor> b <eos> |
b |
add__10.0__8.0__ subtract__10.0__8.0__ multiply__8.0__9.0__ divide__8.0__2.0__ multiply__3.0__8.0__ multiply__3.0__8.0__ |
add__10.0__8.0__ subtract__10.0__8.0__ multiply__8.0__9.0__ divide__8.0__2.0__ multiply__3.0__8.0__ multiply__3.0__8.0__ |
| speed of a bus is num__54 kmph and including stoppages it is num__45 kmph . for how many minsdoes the bus stop per hour ? <o> a ) num__10 min <o> b ) num__20 min <o> c ) num__25 min <o> d ) num__30 min <o> e ) num__40 min |
speed of the bus excluding stoppages = num__54 kmph speed of the bus including stoppages = num__45 kmph loss in speed when including stoppages = num__54 - num__45 = num__9 kmph = > in num__1 hour bus covers num__9 km less due to stoppages hence time that the bus stop per hour = time taken to cover num__9 km = distancespeed = num__954 hour = num__16 hour = num__606 min = num__10 min a <eor> a <eos> |
a |
subtract__54.0__45.0__ add__1.0__9.0__ round__10.0__ |
subtract__54.0__45.0__ add__1.0__9.0__ round__10.0__ |
| sebastian bought a meal at a restaurant and left a num__15.0 tip . with the tip he paid exactly $ num__35.19 . how much did the meal cost without the tip ? <o> a ) $ num__28.98 <o> b ) $ num__29.91 <o> c ) $ num__30.15 <o> d ) $ num__30.60 <o> e ) $ num__30.85 |
the tip is a percent increase of num__15.0 which is num__115.0 . let x equal the price before the tip . thus num__115.0 of this price equals $ num__35.19 : num__1.15 x = num__35.19 divide both sides by num__1.15 : = > x = num__35.19 / num__1.15 = num__30.60 correct answer d ) $ num__30.60 <eor> d <eos> |
d |
divide__35.19__1.15__ divide__35.19__1.15__ |
divide__35.19__1.15__ divide__35.19__1.15__ |
| a rocket soars for num__12 seconds at num__150 meters per second . it then plummets num__600 meters in num__3 seconds . what is the average speed of the rocket in meters per second ? <o> a ) num__140 <o> b ) num__150 <o> c ) num__155 <o> d ) num__160 <o> e ) num__175 |
when soaring the rocket travels a total distance of num__1800 m ( num__150 m / s for num__12 s ) while plummeting the rockets travel distance of num__600 m in num__3 s the total distance traveled is num__2400 m ( num__1800 + num__600 ) the time taken to travel this distance is num__15 s ( num__12 + num__3 ) average speed = num__160.0 = num__160 ans : ( option d ) <eor> d <eos> |
d |
multiply__12.0__150.0__ add__600.0__1800.0__ add__12.0__3.0__ divide__2400.0__15.0__ round__160.0__ |
multiply__12.0__150.0__ add__600.0__1800.0__ add__12.0__3.0__ divide__2400.0__15.0__ divide__2400.0__15.0__ |
| five drainage pipes each draining water from a pool at the same constant rate together can drain a certain pool in num__12 days . how many additional pipes k each draining water at the same constant rate will be needed to drain the pool in num__4 days ? <o> a ) num__6 <o> b ) num__9 <o> c ) num__10 <o> d ) num__12 <o> e ) num__15 |
this is an inverse proportional problem . . . . . . num__5 pipes in num__12 days ; so for num__4 days it will be = num__12 x num__1.25 = num__15 so k = num__15 - num__5 = num__10 . c <eor> c <eos> |
c |
divide__5.0__4.0__ multiply__12.0__1.25__ subtract__15.0__5.0__ subtract__15.0__5.0__ |
divide__5.0__4.0__ multiply__12.0__1.25__ subtract__15.0__5.0__ subtract__15.0__5.0__ |
| a num__50 is divided in two parts in such a way its reciprocal is num__0.0833333333333 . find the no ? <o> a ) num__2030 <o> b ) num__2040 <o> c ) num__2010 <o> d ) num__3010 <o> e ) num__30 |
40 |
num__1 / x + num__1 / ( num__50 - x ) = num__0.0833333333333 = > ( x ^ num__2 - num__50 x + num__600 ) = num__0 = > ( x - num__20 ) ( x - num__30 ) = num__0 = > x = num__2030 numbers are num__2030 . answer : a <eor> a <eos> |
a |
a |
| in the x - y plane point ( x y ) is a lattice point if both x and y are integers . an ellipse has a center at ( num__0 num__0 ) and minor / major axes lengths of num__1 num__4 respectively . some points such as the center ( num__0 num__0 ) are inside the ellipse but a point such as ( num__0 num__2 ) is on the ellipse but not in the ellipse . how many lattice points are in the ellipse ? <o> a ) num__26 <o> b ) num__23 <o> c ) num__5 <o> d ) num__4 <o> e ) num__7 |
the lattice points that are in the ellipse that lie on the x axes are ( num__0 num__0 ) ( num__0 num__1 ) ( num__0 num__2 ) ( num__0 num__3 ) ( num__0 - num__1 ) ( num__0 - num__2 ) and ( num__0 - num__3 ) . there are no other points in the ellipse . the only point on y axis in the ellipse is the center which we counted already . there are a total of num__7 lattice points in the circle . answer : e <eor> e <eos> |
e |
add__1.0__2.0__ add__4.0__3.0__ multiply__1.0__7.0__ |
subtract__4.0__1.0__ add__4.0__3.0__ multiply__1.0__7.0__ |
| a man buys an article and sells it at a profit of num__10.0 . if he had bought it at num__10.0 less and sold it for rs . num__55 less he could have gained num__15.0 . what is the cost price ? <o> a ) num__197 <o> b ) num__375 <o> c ) num__846 <o> d ) num__278 <o> e ) num__268 |
cp num__1 = num__100 sp num__1 = num__110 cp num__2 = num__90 sp num__2 = num__90 * ( num__1.15 ) = num__103.5 num__6.5 - - - - - num__100 num__55 - - - - - ? = > num__846 answer : c <eor> c <eos> |
c |
percent__100.0__846.0__ |
percent__100.0__846.0__ |
| a sum of money is to be distributed among a b c d in the proportion of num__5 : num__2 : num__4 : num__3 . if c gets rs . num__3000 more than d what is b ' s share ? <o> a ) a ) num__8239 <o> b ) b ) num__2900 <o> c ) c ) num__6000 <o> d ) d ) num__2393 <o> e ) e ) num__2009 |
let the shares of a b c and d be num__5 x num__2 x num__4 x and num__3 x rs . respectively . then num__4 x - num__3 x = num__3000 = > x = num__3000 . b ' s share = rs . num__2 x = num__2 * num__3000 = rs . num__6000 . answer : c <eor> c <eos> |
c |
multiply__2.0__3000.0__ multiply__2.0__3000.0__ |
multiply__2.0__3000.0__ multiply__2.0__3000.0__ |
| if num__0.125 of a pencil is black ½ of the remaining is white and the remaining num__3 ½ is blue find the total length of the pencil ? <o> a ) num__9 <o> b ) num__5 <o> c ) num__7 <o> d ) num__8 <o> e ) num__4 |
let the total length be xm then black part = x / num__8 cm the remaining part = ( x - x / num__8 ) cm = num__7 x / num__8 cm white part = ( num__0.5 * num__7 x / num__8 ) = num__7 x / num__16 cm remaining part = ( num__7 x / num__8 - num__7 x / num__16 ) = num__7 x / num__16 cm num__7 x / num__16 = num__3.5 x = num__8 cm answer is d . <eor> d <eos> |
d |
reverse__0.125__ divide__8.0__0.5__ add__3.0__0.5__ reverse__0.125__ |
reverse__0.125__ divide__8.0__0.5__ multiply__0.5__7.0__ reverse__0.125__ |
| the volume of a cube is num__3375 cc . find its surface . <o> a ) num__1350 <o> b ) num__1150 <o> c ) num__1456 <o> d ) num__1254 <o> e ) num__1489 |
a num__3 = num__3375 = > a = num__15 num__6 a num__2 = num__6 * num__15 * num__15 = num__1350 answer : a <eor> a <eos> |
a |
surface_cube__15.0__ surface_cube__15.0__ |
surface_cube__15.0__ surface_cube__15.0__ |
| a man swims downstream num__72 km and upstream num__45 km taking num__9 hours each time ; what is the speed of the current ? <o> a ) num__1.5 <o> b ) num__1.8 <o> c ) num__1.2 <o> d ) num__1.1 <o> e ) num__1.9 |
num__72 - - - num__9 ds = num__8 ? - - - - num__1 num__45 - - - - num__9 us = num__5 ? - - - - num__1 s = ? s = ( num__8 - num__5 ) / num__2 = num__1.5 answer : a <eor> a <eos> |
a |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ round__1.5__ |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ divide__1.5__1.0__ |
| ( num__0.333333333333 ) of a number is num__5 more than the ( num__0.166666666667 ) of the same number ? find the no . <o> a ) num__6 <o> b ) num__36 <o> c ) num__30 <o> d ) num__72 <o> e ) num__62 |
x / num__3 = num__5 + x / num__6 x / num__3 - x / num__6 = num__5 x / num__6 = num__5 x = num__30 answer : c <eor> c <eos> |
c |
multiply__5.0__6.0__ multiply__5.0__6.0__ |
multiply__5.0__6.0__ multiply__5.0__6.0__ |
| the product of a positive integer x and num__377910 is divisible by num__3300 then the least value of x is : <o> a ) num__10 <o> b ) num__11 <o> c ) num__55 <o> d ) num__110 <o> e ) num__330 |
num__377 |
910 = num__3 * num__3 * num__10 * num__4199 num__3310 = num__3 * num__10 * num__10 * num__11 so x must have num__10 and num__11 as factors . - - > least value of x is num__110 . answer : d <eor> d <eos> |
d |
d |
| using all the letters of the word ` ` beta ' ' how many words can be formed which begin with b and end with a ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__8 <o> d ) num__9 <o> e ) num__2 |
there are five letters in the given word . consider num__4 blanks . . . . the first blank and last blank must be filled with b and a all the remaining three blanks can be filled with the remaining num__2 letters in num__2 ! ways . the number of words = num__2 ! = num__2 . answer : e <eor> e <eos> |
e |
coin_space__ coin_space__ |
coin_space__ coin_space__ |
| a wheel of a car of radius num__21 cms is rotating at num__400 rpm . what is the speed of the car in km / hr ? <o> a ) num__79.2 km / hr <o> b ) num__47.52 km / hr <o> c ) num__7.92 km / hr <o> d ) num__31.68 km / hr <o> e ) num__3.96 km / hr |
explanatory answer the radius of the wheel measures num__21 cm . in one rotation the wheel will cover a distance which is equal to the circumference of the wheel . ∴ in one rotation this wheel will cover num__2 * π * num__21 = num__132 cm . in a minute the distance covered by the wheel = circumference of the wheel * rpm ∴ this wheel will cover a distance of num__132 * num__400 = num__52800 cm in a minute . in an hour the wheel will cover a distance of num__52600 * num__60 = num__3168000 cm . therefore the speed of the car = num__3168000 cm / hr = num__31.68 km / hr choice d is the correct answer . <eor> d <eos> |
d |
multiply__400.0__132.0__ hour_to_min_conversion__ multiply__52800.0__60.0__ round__31.68__ |
multiply__400.0__132.0__ hour_to_min_conversion__ multiply__52800.0__60.0__ round__31.68__ |
| if num__2.5 of a number is equal to num__0.75 of another number the ratio of the numbers is : <o> a ) num__2 : num__3 <o> b ) num__3 : num__4 <o> c ) num__3 : num__10 <o> d ) num__20 : num__3 <o> e ) num__30 : num__7 |
num__2.5 a = num__0.75 b - > a / b = num__0.75 / num__2.5 = num__0.3 = num__0.3 : . a : b = num__3 : num__10 answer : c <eor> c <eos> |
c |
divide__0.75__2.5__ divide__3.0__0.3__ multiply__10.0__0.3__ |
divide__0.75__2.5__ divide__3.0__0.3__ multiply__10.0__0.3__ |
| one - third of num__600 is what percent of num__120 ? <o> a ) num__313.2 <o> b ) num__30.1 <o> c ) num__12.24 <o> d ) none of these <o> e ) num__166.6 |
answer let one - third of num__600 is n % of num__120 . ∵ num__200.0 = ( n x num__120 ) / num__100 ∴ n = ( num__200 x num__100 ) / num__120 = num__166.6 correct option : e <eor> e <eos> |
e |
percent__100.0__166.6__ |
percent__100.0__166.6__ |
| if num__15 men can reap num__80 hectares in num__24 days then how many hectares can num__36 men reap in num__30 days ? <o> a ) num__127 <o> b ) num__240 <o> c ) num__287 <o> d ) num__450 <o> e ) num__281 |
explanation : let the required no of hectares be x . then men - - - hectares - - - days num__15 - - - - - - - - - num__80 - - - - - - - - - num__24 num__36 - - - - - - - - - x - - - - - - - - - num__30 more men more hectares ( direct proportion ) more days more hectares ( direct proportion ) x = num__2.4 * num__1.25 * num__80 x = num__240 answer : b <eor> b <eos> |
b |
divide__36.0__15.0__ divide__30.0__24.0__ round__240.0__ |
divide__36.0__15.0__ divide__30.0__24.0__ round__240.0__ |
| a train running at the speed of num__108 km / hr crosses a pole in num__7 seconds . find the length of the train . <o> a ) num__150 meter <o> b ) num__145 meter <o> c ) num__140 meter <o> d ) num__135 meter <o> e ) num__210 meter |
explanation : speed = num__108 * ( num__0.277777777778 ) m / sec = num__30 m / sec length of train ( distance ) = speed * time = num__30 * num__7 = num__210 meter option e <eor> e <eos> |
e |
multiply__7.0__30.0__ round__210.0__ |
multiply__7.0__30.0__ multiply__7.0__30.0__ |
| a case contains c cartons . each carton contains b boxes and each box contains num__600 paper clips . how many paper clips are contained in num__2 cases ? <o> a ) num__600 bc <o> b ) num__600 b / c <o> c ) num__1200 bc <o> d ) num__1200 b / c <o> e ) num__1200 / bc |
num__2 cases * c cartons / case * b boxes / carton * num__600 clips / box = num__1200 bc paper clips the answer is c . <eor> c <eos> |
c |
multiply__600.0__2.0__ multiply__600.0__2.0__ |
multiply__600.0__2.0__ multiply__600.0__2.0__ |
| working alone at their respective constant rates a can complete a task in ‘ a ’ days and b in ‘ b ’ days . they take turns in doing the task with each working num__3 days at a time . if a starts they finish the task in exactly num__9 days . if b starts they take a day more . how long does it take to complete the task if they both work together ? <o> a ) num__8 <o> b ) num__6 <o> c ) num__4 <o> d ) num__5 <o> e ) num__4.5 |
work done by ab in a day = xy respectively . when a starts : no . of days when a works = num__6 no . of days when b works = num__3 → num__6 x + num__3 y = num__1 when b starts : no . of days when a works = num__4 no . of days when a works = num__5 → num__5 x + num__4 y = num__1 solving the above two equations for xy x = num__0.111111111111 y = num__0.111111111111 → total work done by ab in a day = num__0.111111111111 + num__0.111111111111 = num__0.222222222222 → no . of days to complete the work when both work together = num__4.5 ; answer : e <eor> e <eos> |
e |
subtract__9.0__3.0__ add__3.0__1.0__ subtract__9.0__4.0__ divide__1.0__9.0__ round__4.5__ |
subtract__9.0__3.0__ add__3.0__1.0__ add__1.0__4.0__ divide__1.0__9.0__ round__4.5__ |
| suraj has a certain average of runs for num__14 innings . in the num__15 th innings he scores num__140 runs thereby increasing his average by num__8 runs . what is his average after the num__15 th innings ? <o> a ) num__48 <o> b ) num__28 <o> c ) num__36 <o> d ) num__72 <o> e ) num__27 |
to improve his average by num__8 runs per innings he has to contribute num__14 x num__8 = num__112 runs for the previous num__14 innings . thus the average after the num__15 th innings = num__140 - num__112 = num__28 . answer : b <eor> b <eos> |
b |
multiply__14.0__8.0__ subtract__140.0__112.0__ subtract__140.0__112.0__ |
multiply__14.0__8.0__ subtract__140.0__112.0__ subtract__140.0__112.0__ |
| total natural no . from num__11 to num__90 which is divisible num__7 . <o> a ) num__9 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__13 |
these no . are num__1421 num__2835 . . . num__84 . these are in a . p series where a = num__14 d = ( num__21 - num__14 ) = num__7 l = num__84 let these no . are n nth no . = a + ( n - num__1 ) d = num__14 + ( n - num__1 ) * num__7 = ( num__7 n + num__7 ) num__7 n + num__7 = num__84 n = num__11 . answer c <eor> c <eos> |
c |
add__7.0__14.0__ multiply__11.0__1.0__ |
add__7.0__14.0__ multiply__11.0__1.0__ |
| when num__17 is divided by the positive integer k the remainder is num__3 for how many different values of k is this true ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
this means that num__14 must be a multiple of k . the factors of num__14 are num__1 num__2 num__7 and num__14 . out of these k can be num__7 and num__14 . the answer is b . <eor> b <eos> |
b |
coin_space__ coin_space__ |
coin_space__ coin_space__ |
| a computer factory produces num__8400 computers per month at a constant rate how many computers are built every num__30 minutes assuming that there are num__28 days in one month ? <o> a ) num__2.25 . <o> b ) num__3.125 . <o> c ) num__4.5 . <o> d ) num__5.225 . <o> e ) num__6.25 . |
number of hours in num__28 days = num__28 * num__24 number of num__30 mins in num__28 days = num__28 * num__24 * num__2 number of computers built every num__30 mins = num__8400 / ( num__28 * num__24 * num__2 ) = num__6.25 answer e <eor> e <eos> |
e |
subtract__30.0__28.0__ round__6.25__ |
subtract__30.0__28.0__ round__6.25__ |
| what is the present worth of rs . num__132 due in num__2 years at num__5.0 simple interest per annum ? <o> a ) rs . num__112 <o> b ) rs . num__118.80 <o> c ) rs . num__120 <o> d ) rs . num__122 <o> e ) none |
solution let the present worth be rs . x . then s . i . = rs . ( num__132 - x ) . ∴ ( x num__5 x num__0.02 ) = num__132 - x = num__10 x = num__13200 - num__100 x = num__110 x = num__13200 = x = num__120 . answer c <eor> c <eos> |
c |
percent__100.0__120.0__ |
percent__100.0__120.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 seconds . find the length of the train . <o> a ) num__150 <o> b ) num__180 <o> c ) num__190 <o> d ) num__170 <o> e ) num__169 |
speed = num__60 * ( num__0.277777777778 ) m / sec = num__16.6666666667 m / sec length of train ( distance ) = speed * time ( num__16.6666666667 ) * num__9 = num__150 meter answer : option a <eor> a <eos> |
a |
round__150.0__ |
round__150.0__ |
| the sum of num__1 hours num__45 minutes and num__2 hours num__55 minutes is approximately what percent of a day ? <o> a ) num__5.0 <o> b ) num__10.0 <o> c ) num__15.0 <o> d ) num__20.0 <o> e ) num__25 % |
since the question is asking for an approximate percentage num__1 : num__45 + num__2 : num__55 ~ num__5 hours % of day = num__5 * num__4.16666666667 ~ num__5 * num__4.0 = num__20.0 answer is d . <eor> d <eos> |
d |
subtract__5.0__1.0__ multiply__4.0__5.0__ round__20.0__ |
subtract__5.0__1.0__ multiply__4.0__5.0__ multiply__1.0__20.0__ |
| num__15 ! is equal to which of the following ? <o> a ) num__1 num__307674 num__368000 <o> b ) num__1 num__307674 num__368242 <o> c ) num__1 num__307674 num__368356 <o> d ) num__1 num__307674 num__368825 <o> e ) num__1 num__307674 num__368 |
624 |
after num__4 ! the units digit of every factorial is num__0 . num__5 ! = num__120 num__6 ! = num__720 etc . . . the answer is a . <eor> a <eos> |
a |
a |
| a b and c are partners . a receives num__0.666666666667 of profits b and c dividing the remainder equally . a ' s income is increased by rs . num__400 when the rate to profit rises from num__5 to num__7 percent . find the capital of b ? <o> a ) num__3999 <o> b ) num__7799 <o> c ) num__2500 <o> d ) num__5000 <o> e ) num__2912 |
a : b : c = num__0.666666666667 : num__0.166666666667 : num__0.166666666667 = num__4 : num__1 : num__1 x * num__0.02 * num__0.666666666667 = num__400 b capital = num__30000 * num__0.166666666667 = num__5000 . answer : d <eor> d <eos> |
d |
subtract__5.0__4.0__ multiply__1.0__5000.0__ |
subtract__5.0__4.0__ multiply__1.0__5000.0__ |
| jackie has two solutions that are num__2 percent sulfuric acid and num__12 percent sulfuric acid by volume respectively . if these solutions are mixed in appropriate quantities to produce num__60 liters of a solution that is num__10 percent sulfuric acid approximately how many liters of the num__2 percent solution will be required ? <o> a ) num__18 <o> b ) num__12 <o> c ) num__24 <o> d ) num__36 <o> e ) num__42 |
let a = amount of num__2.0 acid and b = amount of num__12.0 acid . now the equation translates to num__0.02 a + . num__12 b = . num__1 ( a + b ) but a + b = num__60 therefore . num__02 a + . num__12 b = . num__1 ( num__60 ) = > num__2 a + num__12 b = num__600 but b = num__60 - a therefore num__2 a + num__12 ( num__60 - a ) = num__600 = > num__10 a = num__120 hence a = num__12 . answer : b <eor> b <eos> |
b |
percent__2.0__600.0__ |
percent__2.0__600.0__ |
| out of num__15 students in a class num__8 are wearing blue shirts num__4 are wearing green shirts and num__3 are wearing red shirts . four students are to be selected at random . what is the probability that at least one is wearing a green shirt ? <o> a ) num__0.639344262295 <o> b ) num__0.69014084507 <o> c ) num__0.728395061728 <o> d ) num__0.758241758242 <o> e ) num__0.782178217822 |
total possible ways to choose num__4 students out of num__15 = num__15 c num__4 = num__1365 the number of ways to choose num__4 students with no green shirts = num__11 c num__4 = num__330 p ( no green shirts ) = num__0.241758241758 = num__0.241758241758 p ( at least num__1 green shirt ) = num__1 - num__0.241758241758 = num__0.758241758242 the answer is d . <eor> d <eos> |
d |
negate_prob__0.2418__ negate_prob__0.2418__ |
negate_prob__0.2418__ negate_prob__0.2418__ |
| a computer can perform z calculations in s seconds . how many minutes will it take the computer to perform k calculations ? <o> a ) ks / num__60 z <o> b ) ks / z <o> c ) num__60 ks / z <o> d ) num__60 z / ks <o> e ) k / num__60 zs |
as ' thick ' as this question might look it ' s actually a fairly straight - forward rate question and can be solved by testing values . we ' re told that a computer can perform z calculations in s seconds . let ' s test . . . . z = num__2 s = num__3 num__2 calculations every num__3 seconds = num__40 calculations every num__1 minute we ' re asked how many minutes it will take to perform k calculations . since we already know that the computer can perform num__40 calculations in num__1 minute let ' s test . . . k = num__80 so we ' re looking for an answer that = num__2 when z = num__2 s = num__3 and k = num__80 answer a : ks / num__60 z = ( num__80 ) ( num__3 ) / ( num__60 ) ( num__2 ) = num__2 this is a match answer b : ks / z = ( num__80 ) ( num__3 ) / num__2 = num__120 this is not a match answer c : num__60 ks / z = num__60 ( num__80 ) ( num__3 ) / num__2 = num__7200 this is not a match answer d : num__60 z / ks = num__60 ( num__2 ) / ( num__80 ) ( num__3 ) = num__0.5 this is not a match answer e : k / num__60 zs = num__80 / ( num__60 ) ( num__2 ) ( num__3 ) = a fraction . this is not a match a <eor> a <eos> |
a |
subtract__3.0__2.0__ multiply__2.0__40.0__ hour_to_min_conversion__ multiply__2.0__60.0__ multiply__120.0__60.0__ divide__1.0__2.0__ hour_to_min_conversion__ |
subtract__3.0__2.0__ multiply__2.0__40.0__ hour_to_min_conversion__ multiply__2.0__60.0__ multiply__120.0__60.0__ divide__1.0__2.0__ divide__7200.0__120.0__ |
| in an election between the two candidates the candidates who gets num__70.0 of votes polled is winned by num__360 vote ’ s majority . what is the total number of votes polled ? <o> a ) num__750 <o> b ) num__700 <o> c ) num__900 <o> d ) num__850 <o> e ) none of these |
explanation : note : majority ( num__40.0 ) = difference in votes polled to win ( num__70.0 ) & defeated candidates ( num__30.0 ) num__40.0 = num__70.0 - num__30.0 num__40.0 - - - - - > num__360 ( num__40 * num__9 = num__360 ) num__100.0 - - - - - > num__900 ( num__100 * num__9 = num__900 ) answer : option c <eor> c <eos> |
c |
percent__100.0__900.0__ |
percent__100.0__900.0__ |
| mariah has decided to hire three workers . to determine whom she will hire she has selected a group of num__12 candidates . she plans to have one working interview with num__3 of the num__12 candidates every day to see how well they work together . how many days will it take her to have working interviews with all the different combinations of job candidates ? <o> a ) num__720 <o> b ) num__120 <o> c ) num__300 <o> d ) num__30 <o> e ) num__333 |
num__300 . answer c <eor> c <eos> |
c |
round__300.0__ |
round__300.0__ |
| num__6 people meet for a meeting . each person shakes hands once with each other person present . how many handshakes take place ? <o> a ) num__30 <o> b ) num__21 <o> c ) num__18 <o> d ) num__15 <o> e ) num__10 |
let ' s look at the logic . every person shakes hand with other ' num__5 ' . so num__6 people shake hands with num__6 x num__5 = num__30 people but in this num__30 we have counted two times the total handshakes ( e shakes hand with f and f shakes hand with e ) so just divide num__15.0 = num__15 total handshakes answer is d <eor> d <eos> |
d |
multiply__6.0__5.0__ subtract__30.0__15.0__ |
multiply__6.0__5.0__ subtract__30.0__15.0__ |
| a train passes a station platform in num__34 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__276 m <o> b ) num__279 m <o> c ) num__210 m <o> d ) num__207 m <o> e ) num__202 m |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x meters . then x + num__8.82352941176 = num__15 x + num__300 = num__510 x = num__210 m . answer : c <eor> c <eos> |
c |
multiply__20.0__15.0__ divide__300.0__34.0__ multiply__34.0__15.0__ subtract__510.0__300.0__ round__210.0__ |
multiply__20.0__15.0__ divide__300.0__34.0__ multiply__34.0__15.0__ subtract__510.0__300.0__ round__210.0__ |
| raman mixed num__48 kg of butter at rs . num__150 per kg with num__36 kg butter at the rate of rs . num__125 per kg . at what price per kg should he sell the mixture to make a profit of num__40.0 in the transaction ? <o> a ) num__129 <o> b ) num__287 <o> c ) num__195 <o> d ) num__188 <o> e ) num__112 |
explanation : cp per kg of mixture = [ num__48 ( num__150 ) + num__36 ( num__125 ) ] / ( num__48 + num__36 ) = rs . num__139.28 sp = cp [ ( num__100 + profit % ) / num__100 ] = num__139.28 * [ ( num__100 + num__40 ) / num__100 ] = rs . num__195 answer : c <eor> c <eos> |
c |
percent__100.0__195.0__ |
percent__100.0__195.0__ |
| the profit earned by selling an article for rs num__900 is double the loss incurred when the same article is sold for rs . num__490 . at what price should the article be sold to make num__25.0 profit ? <o> a ) num__715 <o> b ) num__469 <o> c ) num__400 <o> d ) num__750 <o> e ) num__850 |
explanation : let c . p be rs . x num__900 - x = num__2 ( x - num__450 ) = > x = rs . num__600 c . p = num__600 gain required is num__25.0 s . p = [ ( num__100 + num__25 ) × num__600 ] / num__100 = rs . num__750 answer : d <eor> d <eos> |
d |
percent__100.0__750.0__ |
percent__100.0__750.0__ |
| num__0.2 x num__0.7 = ? <o> a ) num__0.0001 <o> b ) num__0.001 <o> c ) num__0.01 <o> d ) num__0.14 <o> e ) none of these |
explanation : num__2 x num__7 = num__14 . sum of decimal places = num__2 num__0.2 x num__0.7 = num__0.14 answer - d <eor> d <eos> |
d |
multiply__2.0__7.0__ multiply__0.2__0.7__ multiply__0.2__0.7__ |
multiply__2.0__7.0__ multiply__0.2__0.7__ multiply__0.2__0.7__ |
| during a certain season a team won num__65 percent of its first num__100 games and num__50 percent of its remaining games . if the team won num__70 percent of its games for the entire season what was the total number of games that the team played ? <o> a ) num__80 <o> b ) num__75 <o> c ) num__56 <o> d ) num__50 <o> e ) num__105 |
we are first given that a team won num__65 percent of its first num__100 games . this means the team won num__0.65 x num__100 = num__65 games out of its first num__100 games . we are next given that the team won num__50 percent of its remaining games . if we use variable t to represent the total number of games in the season then we can say t – num__100 equals the number of remaining games in the season . thus we can say : num__0.5 ( t – num__100 ) = number of wins for remaining games num__0.5 t – num__50 = number of wins for remaining games lastly we are given that team won num__70 percent of all games played in the season . that is they won num__0.7 t games in the entire season . with this we can set up the equation : number of first num__100 games won + number of games won for remaining games = total number of games won in the entire season num__65 + num__0.5 t – num__50 = num__0.7 t num__15 = num__0.2 t num__150 = num__2 t num__75 = t answer is b . <eor> b <eos> |
b |
percent__50.0__150.0__ percent__100.0__75.0__ |
percent__50.0__150.0__ percent__100.0__75.0__ |
| the value of x + x ( x * x * ( in x ) ) when x = num__2 is : <o> a ) num__10 <o> b ) num__16 <o> c ) num__18 <o> d ) num__36 <o> e ) num__64 |
x + x ( xx ) put the value of x = num__2 in the above expression we get num__2 + num__2 ( num__22 ) = num__2 + num__2 ( num__2 × num__2 ) = num__2 + num__2 ( num__4 ) = num__2 + num__8 = num__10 correct answer : a ) num__10 <eor> a <eos> |
a |
multiply__2.0__4.0__ add__2.0__8.0__ add__2.0__8.0__ |
multiply__2.0__4.0__ add__2.0__8.0__ add__2.0__8.0__ |
| in how many different number of ways num__4 boys and num__2 girls can sit on a bench ? <o> a ) num__720 <o> b ) num__740 <o> c ) num__760 <o> d ) num__800 <o> e ) num__810 |
npn = n ! num__6 p num__6 = num__6 × num__5 × num__4 × num__3 × num__2 × num__1 = num__720 a ) <eor> a <eos> |
a |
die_space__ vowel_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
die_space__ vowel_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| if the lcm and hcf of num__4 and another number is num__36 and num__2 respectively . find the other number ? <o> a ) num__18 <o> b ) num__16 <o> c ) num__14 <o> d ) num__13 <o> e ) num__15 |
hcf x lcm = product of numbers num__2 x num__36 = num__4 x the other number other number = ( num__2 x num__36 ) / num__4 other number = num__18 answer : a <eor> a <eos> |
a |
divide__36.0__2.0__ round__18.0__ |
divide__36.0__2.0__ divide__36.0__2.0__ |
| if num__6 log ( num__4 * num__5 ^ num__2 ) = x find x <o> a ) num__10 <o> b ) num__11 <o> c ) num__7 <o> d ) num__12 <o> e ) num__9 |
num__6 ( log num__2 ^ num__2 * num__5 ^ num__2 ) = x num__6 log ( num__5 * num__2 ) ^ num__2 = x num__6 * num__2 log ( num__5 * num__2 ) = x num__12 log num__10 = x log num__10 base num__10 = num__1 so num__12 * num__1 = x x = num__12 answer : d <eor> d <eos> |
d |
multiply__6.0__2.0__ add__6.0__4.0__ subtract__6.0__5.0__ multiply__6.0__2.0__ |
multiply__6.0__2.0__ multiply__5.0__2.0__ subtract__6.0__5.0__ multiply__6.0__2.0__ |
| the edges of a cuboid are num__2 cm num__5 cm and num__3 cm . find the volume of the cuboid ? <o> a ) num__20 <o> b ) num__60 <o> c ) num__80 <o> d ) num__30 <o> e ) num__45 |
num__2 * num__5 * num__3 = num__30 answer : d <eor> d <eos> |
d |
round__30.0__ |
round__30.0__ |
| in an examination a pupil ' s average marks were num__63 per paper . if he had obtained num__20 more marks for his geography paper and num__2 more marks for his history paper his average per paper would have been num__65 . how many papers were there in the examination ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
x ( total marks ) / y ( no . of subjects ) = num__63 ; x + num__20 + num__2 / y = num__65 ; x + num__22 / y = num__65 on solving : y = num__11 answer : d <eor> d <eos> |
d |
add__20.0__2.0__ divide__22.0__2.0__ divide__22.0__2.0__ |
add__20.0__2.0__ divide__22.0__2.0__ divide__22.0__2.0__ |
| in what ratio must water be mixed with milk costing rs . num__12 per litre to obtain a mixture worth of rs . num__8 per litre ? <o> a ) num__1 : num__2 <o> b ) num__2 : num__1 <o> c ) num__4 : num__3 <o> d ) num__5 : num__3 <o> e ) none |
ratio of water to milk = num__4 : num__8 = num__1 : num__2 answer a <eor> a <eos> |
a |
subtract__12.0__8.0__ divide__8.0__4.0__ reverse__1.0__ |
subtract__12.0__8.0__ divide__8.0__4.0__ reverse__1.0__ |
| two numbers are less than third number by num__60.0 and num__74.0 respectively . how much percent is the second number less than by the first ? <o> a ) num__18.0 <o> b ) num__29.0 <o> c ) num__35.0 <o> d ) num__41.0 <o> e ) none of these |
explanation : let the third number is x . then first number = ( num__100 - num__60 ) % of x = num__40.0 of x = num__4 x / num__10 second number is ( num__26 x / num__100 ) difference = num__4 x / num__10 - num__26 x / num__100 = num__14 x / num__10 so required percentage is difference is what percent of first number = > ( num__14 x / num__100 * num__2.5 x * num__100 ) % = num__35.0 answer : c <eor> c <eos> |
c |
percent__35.0__100.0__ |
percent__35.0__100.0__ |
| a sum of rs . num__2665 is lent into two parts so that the interest on the first part for num__8 years at num__3.0 per annum may be equal to the interest on the second part for num__3 years at num__5.0 per annum . find the second sum ? <o> a ) num__2888 <o> b ) num__1640 <o> c ) num__2782 <o> d ) num__833 <o> e ) num__2772 |
( x * num__8 * num__3 ) / num__100 = ( ( num__2665 - x ) * num__3 * num__5 ) / num__100 num__24 x / num__100 = num__399.75 - num__15 x / num__100 num__39 x = num__39975 = > x = num__1025 second sum = num__2665 – num__1025 = num__1640 . answer : b <eor> b <eos> |
b |
percent__100.0__1640.0__ |
percent__100.0__1640.0__ |
| a train num__125 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 sec . the speed of the train is ? <o> a ) num__50 km / hr <o> b ) num__24 km / hr <o> c ) num__8 km / hr <o> d ) num__5 km / hr <o> e ) num__16 km / hr |
explanation : speed of the train relative to man = num__12.5 = num__12.5 m / sec . = num__12.5 * num__3.6 = num__45 km / hr let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__45 = > x = num__50 km / hr . answer : a <eor> a <eos> |
a |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ round__50.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ multiply__5.0__10.0__ |
| $ num__378 is divided among a b and c so that a receives half as much as b and b receives half as much as c . how much money is c ' s share ? <o> a ) $ num__200 <o> b ) $ num__208 <o> c ) $ num__216 <o> d ) $ num__224 <o> e ) $ num__232 |
let the shares for a b and c be x num__2 x and num__4 x respectively . num__7 x = num__378 x = num__54 num__4 x = num__216 the answer is c . <eor> c <eos> |
c |
divide__378.0__7.0__ multiply__4.0__54.0__ multiply__4.0__54.0__ |
divide__378.0__7.0__ multiply__4.0__54.0__ multiply__4.0__54.0__ |
| if on an item a company gives num__25.0 discount and num__25.0 profit it makes . if they now gives num__10.0 discount then the profit % age made by the company will be . . . <o> a ) num__40 <o> b ) num__55 <o> c ) num__35 <o> d ) num__30 <o> e ) num__50 |
let the marked price of the item is x and the cost price be y . for num__25.0 discount the selling price is num__0.75 x . ( num__0.75 x - y ) / y = num__0.25 which gives y = num__0.6 x next if the sp is num__0.9 x then gain % = ( num__0.9 x - y ) / y * num__100 = ( num__0.9 x - num__0.6 x ) / num__0.6 x * num__100 = num__50.0 hence profit % = num__50.0 . answer : e <eor> e <eos> |
e |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| an automobile financier claims to be lending money at simple interest but he includes the interest every six months for calculating the principal . if he is charging an interest of num__10.0 the effective rate of interest becomes : <o> a ) num__10.0 <o> b ) num__10.25 <o> c ) num__10.5 <o> d ) none of these <o> e ) can not be determined |
let the sum be rs num__100 . then s . i . for first num__6 months = ( num__100 × num__10 × num__0.01 × num__2 ) = num__5 s . i . for last num__6 months = ( num__105 × num__10 × num__0.01 × num__2 ) = num__5.25 so amount at the end of num__1 year = ( num__100 + num__5 + num__5.25 ) = num__110.25 . ∴ effective rate = ( num__110.25 – num__100 ) = num__10.25 . answer b <eor> b <eos> |
b |
percent__5.0__105.0__ percent__100.0__10.25__ |
percent__5.0__105.0__ percent__100.0__10.25__ |
| a number is mistakenly divided by num__5 instead of being multiplied by num__5 . find the percentage change in the result due t this mistake . <o> a ) num__96.0 <o> b ) num__95.0 <o> c ) num__2400.0 <o> d ) num__200.0 <o> e ) num__400 % |
lets take a number num__20 num__4.0 = num__4 num__20 * num__5 = num__100 diff = num__100 - num__4 = num__96.0 answer : a <eor> a <eos> |
a |
divide__20.0__5.0__ multiply__5.0__20.0__ subtract__100.0__4.0__ subtract__100.0__4.0__ |
divide__20.0__5.0__ multiply__5.0__20.0__ subtract__100.0__4.0__ subtract__100.0__4.0__ |
| the average height of num__35 boys in a class was calculated as num__181 cm . it has later found that the height of one of the boys in the class was wrongly written as num__166 cm whereas his actual height was num__106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ? <o> a ) num__187.89 cm <o> b ) num__179.29 cm <o> c ) num__123.98 cm <o> d ) num__149.98 cm <o> e ) num__146.89 cm |
calculated average height of num__35 boys = num__181 cm . wrong total height of num__35 boys = num__181 * num__35 cm . this was as a result of an actual height of num__106 cm being wrongly written as num__166 cm . correct total height of num__35 boys = num__181 cm - ( num__166 cm - num__106 cm ) / num__35 = num__181 cm - ( num__166 cm - num__106 cm ) / num__35 = num__181 cm - num__1.71428571429 cm = num__181 cm - num__1.71 cm = num__179.29 cm . answer : b <eor> b <eos> |
b |
subtract__181.0__1.71__ subtract__181.0__1.71__ |
subtract__181.0__1.71__ subtract__181.0__1.71__ |
| ayesha ' s father was num__34 years of age when she was born while her mother was num__28 years old when her brother four years younger to her was born . what is the difference between the ages of her parents ? <o> a ) num__16 <o> b ) num__14 <o> c ) num__15 <o> d ) num__10 <o> e ) num__12 |
mother ' s age when ayesha ' s brother was born = num__28 years . father ' s age when ayesha ' s brother was born = ( num__34 + num__4 ) years = num__38 years . required difference = ( num__38 - num__28 ) years = num__10 years . answer : d <eor> d <eos> |
d |
add__34.0__4.0__ subtract__38.0__28.0__ subtract__38.0__28.0__ |
add__34.0__4.0__ subtract__38.0__28.0__ subtract__38.0__28.0__ |
| shane is playing a board game . for his first turn he moved ahead num__3 spaces for the second num__5 spaces and for the third num__1 space . for his next turn he had to go back num__6 spaces . after that he got a card that said he could move two times the biggest forward move he had done so far . now how many spaces from the beginning is shane ’ s game piece ? <o> a ) num__3 <o> b ) num__13 <o> c ) num__23 <o> d ) num__33 <o> e ) num__43 |
you may want to draw this one . biggest move is num__5 so two times biggest move is num__10 . num__3 + num__5 + num__1 = num__9 num__9 - num__6 = num__3 num__3 + num__10 = num__13 so he is num__13 spaces from the beginning . correct answer b <eor> b <eos> |
b |
add__3.0__6.0__ add__3.0__10.0__ add__3.0__10.0__ |
add__3.0__6.0__ add__3.0__10.0__ add__3.0__10.0__ |
| find the greatest number which leaves the same remainder when it divides num__25 num__62 and num__105 . <o> a ) num__1 <o> b ) num__8 <o> c ) num__12 <o> d ) num__16 <o> e ) none of these |
num__105 - num__62 = num__43 num__62 - num__25 = num__37 num__105 - num__25 = num__80 the h . c . f of num__37 num__43 and num__80 is num__1 . answer : a <eor> a <eos> |
a |
subtract__105.0__62.0__ subtract__62.0__25.0__ subtract__105.0__25.0__ reverse__1.0__ |
subtract__105.0__62.0__ subtract__62.0__25.0__ subtract__105.0__25.0__ reverse__1.0__ |
| water is leaking out from a cylinder container at the rate of num__0.31 m ^ num__3 per minute . after num__10 minutes the water level decreases num__9 meters . what is value of the radius in meters ? <o> a ) num__0.5 <o> b ) num__0.333333333333 <o> c ) num__1.5 <o> d ) num__3 <o> e ) num__6 |
num__10 * num__0.31 = num__3.1 = pi * r ^ num__2 * h r ^ num__2 = num__3.1 / ( pi * num__9 ) which is about num__0.111111111111 r = num__0.333333333333 the answer is b . <eor> b <eos> |
b |
multiply__0.31__10.0__ divide__3.0__9.0__ divide__3.0__9.0__ |
multiply__0.31__10.0__ divide__3.0__9.0__ divide__3.0__9.0__ |
| the city of boston decided to reconstruct its major tunnels . it estimated the job would require num__612 mini projects spread evenly over an num__18 month plan of completion . only num__108 mini projects had been successfully completed after num__5 months . at this time the construction was behind schedule by how many projects ? <o> a ) num__34 <o> b ) num__62 <o> c ) num__198 <o> d ) num__204 <o> e ) num__504 |
project / month : num__34.0 = num__34 project in num__6 month to be completed = num__34 * num__5 = num__170 lag : num__170 - num__108 = num__62 b is the answer <eor> b <eos> |
b |
divide__612.0__18.0__ divide__108.0__18.0__ multiply__5.0__34.0__ subtract__170.0__108.0__ subtract__170.0__108.0__ |
divide__612.0__18.0__ divide__108.0__18.0__ multiply__5.0__34.0__ subtract__170.0__108.0__ subtract__170.0__108.0__ |
| a circle in the coordinate plane passes through points ( - num__3 - num__2 ) and ( num__2 num__4 ) . what is the smallest possible area of that circle ? <o> a ) num__13 π <o> b ) num__15.25 π <o> c ) num__262 √ π <o> d ) num__52 π <o> e ) num__64 π |
the distance between the two points is sqrt ( num__61 ) . radius = sqrt ( num__61 ) / num__2 area = pi * ( sqrt ( num__61 ) / num__2 ) ^ num__2 b . num__15.25 π <eor> b <eos> |
b |
triangle_area__2.0__15.25__ |
triangle_area__2.0__15.25__ |
| the num__8 spokes of a custom circular bicycle wheel radiate from the central axle of the wheel and are arranged such that the sectors formed by adjacent spokes all have different central angles which constitute an arithmetic series of numbers ( that is the difference between any angle and the next largest angle is constant ) . if the largest sector has a central angle of num__73 ° what fraction of the wheel ’ s area is represented by the smallest sector ? <o> a ) num__0.116666666667 <o> b ) num__0.1125 <o> c ) num__0.0916666666667 <o> d ) num__0.0541666666667 <o> e ) num__0.0472222222222 |
the largest angle is num__73 . let d be the difference between any two angles in the progression . the sum of all the angles will be : num__73 + ( num__73 - d ) + ( num__73 - num__2 d ) + . . . + ( num__73 - num__7 d ) = num__584 - num__28 d the sum of all the central angles in a circle = num__360 num__584 - num__28 d = num__360 d = num__8.0 = num__8 the smallest sector is ( num__73 - num__7 d ) = num__73 - num__7 * num__8 = num__17 the fraction of the area covered is num__0.0472222222222 . the answer is e . <eor> e <eos> |
e |
multiply__8.0__73.0__ divide__17.0__360.0__ divide__17.0__360.0__ |
multiply__8.0__73.0__ divide__17.0__360.0__ divide__17.0__360.0__ |
| what no . should be subtracted from x ^ num__3 + num__4 x ^ num__2 − num__7 x + num__12 x ^ num__3 + num__4 x ^ num__2 − num__7 x + num__12 if it is to be perfectly divisible by x + num__3 x + num__3 ? <o> a ) num__39 <o> b ) num__42 <o> c ) num__45 <o> d ) num__47 <o> e ) num__48 |
according to remainder theorem when dfracf ( x ) x + adfracf ( x ) x + a then the remainder is f ( − a ) f ( − a ) . in this case as x + num__3 x + num__3 divides x num__3 + num__4 x num__2 − num__7 x + num__12 – kx num__3 + num__4 x num__2 − num__7 x + num__12 – k perfectly ( kk being the number to be subtracted ) the remainder is num__0 when the value of xx is substituted by - num__3 . i . e . ( − num__3 ) num__3 + num__4 ( − num__3 ) num__2 − num__7 ( − num__3 ) + num__12 − k = num__0 ( − num__3 ) num__3 + num__4 ( − num__3 ) num__2 − num__7 ( − num__3 ) + num__12 − k = num__0 or − num__27 + num__36 + num__21 + num__12 = k − num__27 + num__36 + num__21 + num__12 = k or k = k = num__42 b <eor> b <eos> |
b |
multiply__3.0__12.0__ multiply__3.0__7.0__ multiply__2.0__21.0__ multiply__2.0__21.0__ |
multiply__3.0__12.0__ multiply__3.0__7.0__ multiply__2.0__21.0__ multiply__2.0__21.0__ |
| a person starts walking at a speed of num__5 km / hr through half the distance rest of the distance he covers with aspeed num__4 km / hr . total time of travel is num__9 hours . what is the maximum distance he can cover ? <o> a ) num__20 km <o> b ) num__40 km <o> c ) num__60 km <o> d ) num__80 km <o> e ) num__90 km |
t = d / s so num__9 = x / num__2 * num__0.2 + x / num__2 * num__0.25 ( because half distance with num__5 km / ph and remaining half with num__4 km / hr ) num__9 = x ( num__0.225 ) x = num__40 km answer : b <eor> b <eos> |
b |
divide__9.0__0.225__ round__40.0__ |
divide__9.0__0.225__ divide__9.0__0.225__ |
| if y is not equal to zero and y + num__1 / y = num__2 then what is the value of y ^ num__4 + ( num__1 / y ) ^ num__4 ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__5 <o> d ) num__8 <o> e ) num__0 |
y + num__1 / y = num__2 we square both sides so we have y ^ num__2 + num__1 / y ^ num__2 + num__2 = num__4 or y ^ num__2 + num__1 / y ^ num__2 = num__2 squaring again we have y ^ num__4 + num__1 / y ^ num__4 + num__2 = num__4 or y ^ num__4 + num__1 / y ^ num__4 = num__2 answer = num__2 ( a ) <eor> a <eos> |
a |
multiply__1.0__2.0__ |
divide__2.0__1.0__ |
| what will be the ratio of simple interest earned by certain amount at the same rate of interest for num__2 years and that for num__10 years ? <o> a ) num__3 : num__2 <o> b ) num__1 : num__3 <o> c ) num__2 : num__3 <o> d ) num__1 : num__5 <o> e ) num__2 : num__1 |
explanation : simple interest = prt / num__100 here principal ( p ) and rate of interest ( r ) are constants hence simple interest ∝ t required ratio = simple interest for num__2 years / simple interest for num__10 years = t num__1 / t num__2 = num__0.2 = num__0.2 = num__1 : num__5 answer : option d <eor> d <eos> |
d |
percent__2.0__10.0__ percent__1.0__100.0__ |
percent__2.0__10.0__ percent__1.0__100.0__ |
| a and b have monthly incomes in the ratio num__5 : num__6 and monthly expenditures in the ratio num__3 : num__4 . if they save rs . num__1800 and rs . num__1600 respectively find the monthly income of b <o> a ) s . num__3400 <o> b ) s . num__2700 <o> c ) s . num__1720 <o> d ) rs . num__7200 <o> e ) s . num__8200 |
explanation : incomes of a and b = num__5 x and num__6 x and expenses of a and b = num__3 y and num__4 y then savings of a = num__5 x - num__3 y = num__1800 — ? ( num__1 ) savings of b = num__6 x - num__4 y = num__1600 — ? ( num__2 ) by solving equations ( num__1 ) and ( num__2 ) y = num__1400 monthly income of b = expenses of b + savings of b = num__4 y + num__1600 = num__4 ( num__1400 ) + num__1600 = rs . num__7200 answer : d <eor> d <eos> |
d |
subtract__5.0__4.0__ subtract__5.0__3.0__ multiply__4.0__1800.0__ multiply__4.0__1800.0__ |
subtract__5.0__4.0__ subtract__5.0__3.0__ multiply__4.0__1800.0__ multiply__4.0__1800.0__ |
| a basket of num__1430 apples is divided equally among a group of apple lovers . if num__45 people join the group each apple lover would receive num__9 apples less . how many s apples did each person get before num__45 people joined the feast ? <o> a ) num__20 . <o> b ) num__21 . <o> c ) num__22 . <o> d ) num__23 . <o> e ) num__24 . |
before solving it algebraically let us prime factorize num__1430 = num__2 * num__5 * num__11 * num__13 . since number of apples per person * total persons s = num__1430 the answer should be a factor of num__1430 . only c is . and that ' s your answer . c <eor> c <eos> |
c |
divide__45.0__9.0__ add__9.0__2.0__ add__2.0__11.0__ add__9.0__13.0__ |
divide__45.0__9.0__ add__9.0__2.0__ add__2.0__11.0__ multiply__2.0__11.0__ |
| the angle between the minute hand and the hour hand of a clock when the time is num__4.20 is <o> a ) num__0 ° <o> b ) num__5 ° <o> c ) num__10 ° <o> d ) num__20 ° <o> e ) none |
solution angle traced by hour hand in num__4.33333333333 hrs = ( num__30.0 x num__4.33333333333 ) ° = num__130 ° angle traced by min . hand in num__20 min = ( num__6.0 x num__20 ) ° = num__120 ° required angle = ( num__130 - num__120 ) ° = num__10 ° . answer c <eor> c <eos> |
c |
multiply__6.0__20.0__ subtract__130.0__120.0__ round__10.0__ |
multiply__6.0__20.0__ subtract__130.0__120.0__ subtract__130.0__120.0__ |
| num__36 men can complete a piece of work in num__18 days . in how many days will num__9 men complete the same work ? <o> a ) num__24 <o> b ) num__77 <o> c ) num__88 <o> d ) num__72 <o> e ) num__21 |
explanation : less men means more days { indirect proportion } let the number of days be x then num__9 : num__36 : : num__18 : x x = num__72 answer : d ) num__72 days <eor> d <eos> |
d |
round__72.0__ |
round__72.0__ |
| two airplanes fly the same exact route from atlanta to chicago . plane a flies num__200 mph while plane b flies num__300 mph ( assume that simple model versus actual acceleration / deceleration near the airports ) . if plane a took off exactly num__40 minutes before plane b after how many minutes will plane b overtake plane a ? <o> a ) num__65 <o> b ) num__80 <o> c ) num__90 <o> d ) num__115 <o> e ) num__120 |
we can use some form of the equation d = rt [ distance = rate * time ] rate of plane a : num__200 mph rate of plane b : num__300 mph plane a will be in the sky num__40 minutes longer than plane b ( num__0.666666666667 of an hour ) time of plane a when b overtakes it : t + ( num__0.666666666667 ) use num__0.666666666667 of an hour since the rate is in hours time of plane b when b overtakes a : t at the time that b overtakes a they will have travelled the same distance so rt is equal for each plane : num__200 * ( t + num__0.666666666667 ) = num__300 * t num__200 t + num__133.333333333 = num__300 t num__133.333333333 = num__100 t num__1.33333333333 = t this is num__1 and a third hour so num__60 minutes + num__20 minutes = num__80 minutes b <eor> b <eos> |
b |
divide__200.0__300.0__ subtract__300.0__200.0__ divide__133.3333__100.0__ round_down__1.3333__ subtract__100.0__40.0__ subtract__60.0__40.0__ subtract__100.0__20.0__ multiply__1.0__80.0__ |
divide__200.0__300.0__ subtract__300.0__200.0__ divide__133.3333__100.0__ round_down__1.3333__ subtract__100.0__40.0__ subtract__60.0__40.0__ add__20.0__60.0__ multiply__1.0__80.0__ |
| if the average ( arithmetic mean ) of num__5 positive temperatures is m degrees fahrenheit then the sum of the num__3 greatest of these temperatures in degrees fahrenheit could be : <o> a ) num__6 m <o> b ) num__4 m <o> c ) num__5 m / num__3 <o> d ) num__3 m / num__2 <o> e ) num__3 m / num__5 |
the sum of three greatest should be more than sum of two lowest . the total sum is ; num__5 m a . num__6 m ; num__6 m is more than num__5 m . not possible . b . num__4 m ; num__5 m - num__4 m = m ( possible ) c . num__5 m / num__3 ; num__10 m / num__3 ; num__10 m / num__3 > num__5 m / num__3 . not possible d . num__3 m / num__2 ; num__7 m / num__2 ; num__7 m / num__2 > num__3 m / num__2 . not possible e . num__3 m / num__5 ; num__22 m / num__5 ; num__22 m / num__5 > num__3 m / num__5 . not possible . ans : b <eor> b <eos> |
b |
add__4.0__6.0__ subtract__5.0__3.0__ add__5.0__2.0__ subtract__6.0__2.0__ |
add__4.0__6.0__ subtract__5.0__3.0__ subtract__10.0__3.0__ subtract__6.0__2.0__ |
| a number consists of num__3 digits whose sum is num__10 . the middle digit is equal to the sum of the other two and the number will be increased by num__99 if its digits are reversed . the number is : <o> a ) num__253 <o> b ) num__366 <o> c ) num__298 <o> d ) num__266 <o> e ) num__992 |
explanation : let the middle digit be x . then num__2 x = num__10 or x = num__5 . so the number is either num__253 or num__352 since the number increases on reversing the digits so the hundred ' s digit is smaller than the unit ' s digit . hence required number = num__253 . answer : a ) num__253 <eor> a <eos> |
a |
add__3.0__2.0__ add__99.0__253.0__ subtract__352.0__99.0__ |
add__3.0__2.0__ add__99.0__253.0__ subtract__352.0__99.0__ |
| solve below question num__2 x + num__1 = - num__21 <o> a ) - num__11 <o> b ) - num__9 <o> c ) num__9 <o> d ) num__8 <o> e ) - num__7 |
num__2 x + num__1 = - num__21 x = - num__11 a <eor> a <eos> |
a |
multiply__1.0__11.0__ |
multiply__1.0__11.0__ |
| a man can row downstream at num__18 kmph and upstream at num__10 kmph . find the speed of the man in still water and the speed of stream respectively ? <o> a ) num__14 num__2 <o> b ) num__14 num__9 <o> c ) num__14 num__5 <o> d ) num__14 num__4 <o> e ) num__14 num__1 |
let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = num__18 - - - ( num__1 ) and x - y = num__10 - - - ( num__2 ) from ( num__1 ) & ( num__2 ) num__2 x = num__28 = > x = num__14 y = num__4 . answer : d <eor> d <eos> |
d |
add__18.0__10.0__ divide__28.0__2.0__ subtract__18.0__14.0__ round__14.0__ |
add__18.0__10.0__ divide__28.0__2.0__ subtract__18.0__14.0__ subtract__18.0__4.0__ |
| john makes $ num__60 a week from his job . he earns a raise and now makes $ num__110 a week . what is the % increase ? <o> a ) num__16.0 <o> b ) num__83.33 <o> c ) num__17.0 <o> d ) num__17.61 <o> e ) num__17.56 % |
increase = ( num__0.833333333333 ) * num__100 = ( num__0.833333333333 ) * num__100 = num__83.33 . b <eor> b <eos> |
b |
percent__83.33__100.0__ |
percent__83.33__100.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__3 seconds . what is the length of the train ? <o> a ) num__20 metres <o> b ) num__50 metres <o> c ) num__32 metres <o> d ) num__70 metres <o> e ) num__80 metres |
speed = num__60 x num__0.277777777778 m / sec = num__16.6666666667 m / sec . length of the train = ( speed x time ) . length of the train = num__16.6666666667 x num__3 m = num__50 m . option b <eor> b <eos> |
b |
round__50.0__ |
round__50.0__ |
| a cube of side one meter length is cut into small cubes of side num__10 cm each . how many such small cubes can be obtained ? <o> a ) num__1078 <o> b ) num__1276 <o> c ) num__1000 <o> d ) num__1021 <o> e ) num__1029 |
along one edge the number of small cubes that can be cut = num__10.0 = num__10 along each edge num__10 cubes can be cut . ( along length breadth and height ) . total number of small cubes that can be cut = num__10 * num__10 * num__10 = num__1000 answer : c <eor> c <eos> |
c |
round__1000.0__ |
round__1000.0__ |
| each week harry is paid x dollars per hour for the first num__30 hours and num__1.5 x dollars for each additional hour worked that week . each week james is paid x dollars per hour for the first num__40 hours and num__2 x dollars for each additional hour worked that week . last week james worked a total of num__44 hours . if harry and james were paid the same amount last week how many hours did harry work last week ? <o> a ) num__35 <o> b ) num__36 <o> c ) num__37 <o> d ) num__38 <o> e ) num__42 |
amount earned by james = num__40 * x + num__4 * num__2 x = num__48 x therefore amount earned by james = num__48 x but we know the amount harry earned assuming working y hours ( y > num__30 ) is num__30 * x + ( y - num__30 ) * num__1.5 x [ [ we know y > num__30 because in num__30 h the most harry could earn is num__30 x but he has earned num__48 x ] ] so x * ( num__1.5 y - num__45 + num__30 ) = num__48 x or x * ( num__1.5 y - num__15 ) = num__48 x so num__1.5 y - num__15 = num__48 so num__1.5 y = num__63 so y = num__42 answer is e <eor> e <eos> |
e |
subtract__44.0__40.0__ add__44.0__4.0__ multiply__30.0__1.5__ divide__30.0__2.0__ add__15.0__48.0__ add__40.0__2.0__ add__40.0__2.0__ |
subtract__44.0__40.0__ add__44.0__4.0__ multiply__30.0__1.5__ subtract__45.0__30.0__ add__15.0__48.0__ add__40.0__2.0__ add__40.0__2.0__ |
| a right circular cylinder has a height of num__21 and a radius of num__5 . a rectangular solid with a height of num__15 and a square base is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall . liquid is then poured into the cylinder such that it reaches the rim . what is the volume of the liquid ? <o> a ) num__500 ( π – num__3 ) <o> b ) num__500 ( π – num__2.5 ) <o> c ) num__510 ( π – num__1.5 ) <o> d ) num__500 ( π – num__2 ) <o> e ) num__500 ( π – num__1 ) |
[ quote = bunuel ] a right circular cylinder has a height of num__20 and a radius of num__5 . a rectangular solid with a height of num__15 and a square base is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall . liquid is then poured into the cylinder such that it reaches the rim . what is the volume of the liquid ? the square base has sides of sqrt ( num__50 ) due to the num__45 - num__45 - num__90 triangle num__21 * num__25 * pi - num__15 * sqrt ( num__50 ) ^ num__2 = num__510 ( π – num__1.5 ) c . num__510 ( π – num__1.5 ) <eor> c <eos> |
c |
square_perimeter__5.0__ rectangle_perimeter__5.0__20.0__ triangle_area__2.0__510.0__ |
square_perimeter__5.0__ rectangle_perimeter__5.0__20.0__ triangle_area__2.0__510.0__ |
| a single discount equivalent to the discount series of num__20.0 num__10.0 and num__5.0 is ? <o> a ) num__31.7 <o> b ) num__31.3 <o> c ) num__31.6 <o> d ) num__31.9 <o> e ) num__31.2 |
num__100 * ( num__0.8 ) * ( num__0.9 ) * ( num__0.95 ) = num__68.4 num__100 - num__68.4 = num__31.6 answer : c <eor> c <eos> |
c |
percent__100.0__31.6__ |
percent__100.0__31.6__ |
| the mean of num__50 observations was num__36 . it was found later that an observation num__29 was wrongly taken as num__23 . the corrected new mean is : <o> a ) num__36.1 <o> b ) num__36.5 <o> c ) num__36.22 <o> d ) num__36.12 <o> e ) num__36.18 |
explanation : correct sum = ( num__36 * num__50 + num__29 - num__23 ) = num__1806 . correct mean = = num__36.12 = num__36.12 answer : d ) num__36.12 <eor> d <eos> |
d |
divide__1806.0__50.0__ divide__1806.0__50.0__ |
divide__1806.0__50.0__ divide__1806.0__50.0__ |
| the difference between a number and its two - fifth is num__510 . what is num__5.0 of that number ? <o> a ) num__425 <o> b ) num__300 <o> c ) num__255 <o> d ) num__300 <o> e ) num__400 |
let the number be x . then x - num__0.4 x = num__510 x = ( num__510 * num__5 ) / num__3 = num__850 num__5.0 of num__850 = num__425 . answer : a <eor> a <eos> |
a |
subtract__850.0__425.0__ |
subtract__850.0__425.0__ |
| two ants arthur and amy have discovered a picnic and are bringing crumbs back to the anthill . amy makes twice as many trips and carries one and a half times as many crumbs per trip as arthur . if arthur carries a total of b crumbs to the anthill how many crumbs will amy bring to the anthill in terms of b ? <o> a ) b / num__2 <o> b ) b <o> c ) num__3 b / num__2 <o> d ) num__2 b <o> e ) num__3 b |
lets do it by picking up numbers . let arthur carry num__2 crumbs per trip this means amy carries num__3 crumbs per trip . also let arthur make num__2 trips and so amy makes num__4 trips . thus total crumbs carried by arthur ( b ) = num__2 x num__2 = num__4 total crumbs carried by amy = num__3 x num__4 = num__12 . num__12 is num__3 times num__4 so e <eor> e <eos> |
e |
multiply__3.0__4.0__ divide__12.0__4.0__ |
multiply__3.0__4.0__ divide__12.0__4.0__ |
| a train covers a distance of num__12 km in num__10 min . if it takes num__15 sec to pass a telegraph post then the length of the train is ? <o> a ) num__300 m <o> b ) num__168 m <o> c ) num__120 m <o> d ) num__168 m <o> e ) num__178 m |
speed = ( num__1.2 * num__60 ) km / hr = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . length of the train = num__20 * num__15 = num__120 m . answer : a <eor> a <eos> |
a |
divide__12.0__10.0__ hour_to_min_conversion__ add__12.0__60.0__ multiply__12.0__10.0__ multiply__15.0__20.0__ |
divide__12.0__10.0__ hour_to_min_conversion__ multiply__1.2__60.0__ multiply__12.0__10.0__ multiply__15.0__20.0__ |
| it takes num__10 days for digging a trench of num__100 m long num__50 m broad and num__10 m deep . what length of trench num__25 m broad and num__15 m deep can be dug in num__30 days ? <o> a ) num__400 m <o> b ) num__200 m <o> c ) num__100 m <o> d ) num__89 m <o> e ) num__56 m |
expl : more days more length ( direct ) less breadth more length ( indirect ) more depth less length ( indirect days num__10 : num__30 ; breadth num__25 : num__50 ; : : num__100 : x depth num__15 : num__10 ; : . num__10 * num__25 * num__15 * x = num__30 * num__50 * num__10 * num__100 x = ( num__30 * num__50 * num__10 * num__100 ) / num__10 * num__25 * num__15 = num__400 so the required length = num__400 m answer : a <eor> a <eos> |
a |
round__400.0__ |
round__400.0__ |
| a certain number of men can do a work in num__40 days . if there were num__5 men less it could be finished in num__10 days more . how many men are there ? <o> a ) num__15 men <o> b ) num__25 men <o> c ) num__18 men <o> d ) num__12 men <o> e ) num__14 men |
explanation : x * num__40 = ( x - num__5 ) * num__50 num__5 x - num__4 x = num__25 x = num__25 men answer : option b <eor> b <eos> |
b |
add__40.0__10.0__ divide__40.0__10.0__ round__25.0__ |
multiply__5.0__10.0__ divide__40.0__10.0__ subtract__50.0__25.0__ |
| the sum of two numbers is num__40 and their difference is num__20 . find their product . <o> a ) num__300 <o> b ) num__108 <o> c ) num__114 <o> d ) num__325 <o> e ) none |
sol . let the numbers be x and y . then x + y = num__40 and x - y = num__20 . num__2 x = num__60 = > x = num__30 so y = num__10 xy = num__30 * num__10 = num__300 answer : a <eor> a <eos> |
a |
divide__40.0__20.0__ add__40.0__20.0__ divide__60.0__2.0__ subtract__40.0__30.0__ multiply__10.0__30.0__ multiply__10.0__30.0__ |
divide__40.0__20.0__ add__40.0__20.0__ divide__60.0__2.0__ subtract__40.0__30.0__ multiply__10.0__30.0__ multiply__10.0__30.0__ |
| two stations a and b are num__110 km apart on a straight line . one train starts from a at num__7 a . m . and travels towards b at num__20 kmph . another train starts from b at num__8 a . m . and travels towards a at a speed of num__25 kmph . at what time will they meet ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__13 |
suppose they meet x hours after num__7 a . m . distance covered by a in x hours = num__20 x km . distance covered by b in ( x - num__1 ) hours = num__25 ( x - num__1 ) km . therefore num__20 x + num__25 ( x - num__1 ) = num__110 num__45 x = num__135 x = num__3 . so they meet at num__10 a . m . answer : option b <eor> b <eos> |
b |
subtract__8.0__7.0__ add__20.0__25.0__ add__110.0__25.0__ divide__135.0__45.0__ add__7.0__3.0__ round__10.0__ |
subtract__8.0__7.0__ add__20.0__25.0__ add__110.0__25.0__ divide__135.0__45.0__ add__7.0__3.0__ add__7.0__3.0__ |
| if xy = num__3 and x ^ num__2 + y ^ num__2 = num__12 then x / y + y / x = <o> a ) num__4 <o> b ) num__3 num__0.142857142857 <o> c ) num__5 num__0.333333333333 <o> d ) num__7 <o> e ) num__60 |
we can make simplifying of question and get it in view : ( x ^ num__2 + y ^ num__2 ) / xy and as we know the meaning of this parts : x ^ num__2 + y ^ num__2 = num__12 xy = num__3 we can calculate the answer num__4.0 - > num__4 so answer is a <eor> a <eos> |
a |
divide__12.0__3.0__ divide__12.0__3.0__ |
divide__12.0__3.0__ divide__12.0__3.0__ |
| each of the integers from num__1 to num__18 is written on the a seperate index card and placed in a box . if the cards are drawn from the box at random without replecement how many cards must be drawn to ensure that the product of all the integers drawn is even <o> a ) num__19 <o> b ) num__12 <o> c ) num__11 <o> d ) num__10 <o> e ) num__3 |
out of the num__18 integers : num__9 are odd and num__9 are even . if we need to make sure that the product of all the integers withdrawn is even then we need to make sure that we have at least one even number . in the worst case : num__1 . we will end up picking odd numbers one by one so we will pick all num__9 odd numbers first num__2 . num__10 th number will be the first even number so we need to withdraw at least num__10 numbers to make sure that we get one even number and the product of all the integers picked is even . so answer will be num__10 . ( d ) <eor> d <eos> |
d |
divide__18.0__9.0__ add__1.0__9.0__ add__1.0__9.0__ |
divide__18.0__9.0__ add__1.0__9.0__ add__1.0__9.0__ |
| a polling company surveyed a certain country and it found that num__35.0 of that country ’ s registered voters had an unfavorable impression of both of that state ’ s major political parties and that num__20.0 had a favorable impression only of party t . if one registered voter has a favorable impression of both parties for every two registered voters who have a favorable impression only of party b then what percentage of the country ’ s registered voters have a favorable impression of both parties ( assuming that respondents to the poll were given a choice between favorable and unfavorable impressions only ) ? <o> a ) num__15 <o> b ) num__20 <o> c ) num__30 <o> d ) num__35 <o> e ) num__45 |
s = num__100 not ( t and b ) = num__35 only t = num__20 ( t and b ) / b = num__0.5 let ( t and b ) = x only b = num__2 x so now num__20 + num__35 + x + num__2 x = num__100 x = num__15 a ans <eor> a <eos> |
a |
percent__100.0__15.0__ |
percent__100.0__15.0__ |
| a person purchased a tv set for rs . num__16000 and a dvd player for rs . num__6250 . he sold both the items together for rs . num__31150 . what percentage of profit did he make ? <o> a ) num__40 <o> b ) num__88 <o> c ) num__26 <o> d ) num__18 <o> e ) num__11 |
explanation : the total cp = rs . num__16000 + rs . num__6250 = rs . num__22250 and sp = rs . num__31150 profit ( % ) = ( num__31150 - num__22250 ) / num__22250 * num__100 = num__40.0 answer : a <eor> a <eos> |
a |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| which of the following statement is not correct ? <o> a ) log num__10 num__10 = num__1 <o> b ) log ( num__2 + num__3 ) = log ( num__2 * num__3 ) <o> c ) log num__10 num__1 = num__0 <o> d ) log ( num__1 + num__2 + num__3 ) = log num__1 + log num__2 + log num__3 <o> e ) none |
a ) loga a = num__1 so log num__10 num__10 = num__1 b ) log ( num__2 + num__3 ) = log num__5 and log ( num__2 * num__3 ) = log num__6 = log num__2 + log num__3 therefore log ( num__2 + num__3 ) # log ( num__2 * num__3 ) c ) since loga num__1 = num__0 so log num__10 num__1 = num__0 d ) log ( num__1 + num__2 + num__3 ) = log num__6 = log ( num__1 * num__2 * num__3 ) = log num__1 + log num__2 + log num__3 e ) none correct answer ( b ) <eor> b <eos> |
b |
add__1.0__2.0__ add__2.0__3.0__ add__1.0__5.0__ multiply__1.0__2.0__ |
add__1.0__2.0__ add__2.0__3.0__ add__1.0__5.0__ multiply__1.0__2.0__ |
| the difference between the length and breadth of a rectangle is num__28 m . if its perimeter is num__120 m then its area is : <o> a ) num__704 m num__2 <o> b ) num__720 m num__2 <o> c ) num__620 m num__2 <o> d ) num__820 m num__2 <o> e ) none |
explanation we have : ( l â € “ b ) = num__28 and num__2 ( l + b ) = num__120 or ( l + b ) = num__60 . solving the two equations we get : l = num__44 and b = num__16 . area = ( l x b ) = ( num__44 x num__16 ) m num__2 = num__704 m num__2 . answer a <eor> a <eos> |
a |
rectangle_perimeter__28.0__2.0__ multiply__44.0__16.0__ triangle_area__704.0__2.0__ |
rectangle_perimeter__28.0__2.0__ multiply__44.0__16.0__ triangle_area__704.0__2.0__ |
| the present population of a town is num__300 . population increase rate is num__10.0 p . a . find the population of town after num__1 years ? <o> a ) num__100 <o> b ) num__120 <o> c ) num__200 <o> d ) num__220 <o> e ) num__330 |
p = num__300 r = num__10.0 required population of town = p * ( num__1 + r / num__100 ) ^ t = num__300 * ( num__1 + num__0.1 ) = num__300 * ( num__1.1 ) = num__330 answer is e <eor> e <eos> |
e |
percent__10.0__1.0__ percent__100.0__330.0__ |
percent__10.0__1.0__ percent__100.0__330.0__ |
| in a college the ratio of the numbers of boys to the girls is num__8 : num__5 . if there are num__120 girls the total number of students in the college is ? <o> a ) num__312 <o> b ) num__356 <o> c ) num__452 <o> d ) num__416 <o> e ) num__512 |
let the number of boys and girls be num__8 x and num__5 x then num__5 x = num__120 x = num__24 total number of students = num__13 x = num__13 * num__24 = num__312 answer is a <eor> a <eos> |
a |
divide__120.0__5.0__ add__8.0__5.0__ multiply__13.0__24.0__ multiply__13.0__24.0__ |
divide__120.0__5.0__ add__8.0__5.0__ multiply__13.0__24.0__ multiply__13.0__24.0__ |
| a building is to be completed in num__48 days . to meet the deadline num__54 men were employed and were made to work for num__10 hours a day . after num__30 days num__0.555555555556 th of the work was completed . how many more workers should be employed to meet the deadline if each workers are now made to work num__8 hours a day ? <o> a ) num__36 <o> b ) num__55 <o> c ) num__40 <o> d ) num__45 <o> e ) none of these |
explanation : num__54 men work num__10 hours a day for num__30 days and num__0.555555555556 th of the work is completed . now let x workers be employed to work num__8 hours a day for the rest num__18 days to complete num__0.444444444444 th of the work . m num__1 d num__1 t num__1 w num__2 = m num__2 d num__2 t num__2 w num__1 num__54 * num__30 * num__10 * num__0.444444444444 = x * num__18 * num__8 * num__0.555555555556 x = num__90 hence num__36 more workers should be employed . answer : a <eor> a <eos> |
a |
subtract__48.0__30.0__ divide__8.0__18.0__ add__0.5556__0.4444__ subtract__10.0__8.0__ subtract__54.0__18.0__ round__36.0__ |
subtract__48.0__30.0__ divide__8.0__18.0__ add__0.5556__0.4444__ subtract__10.0__8.0__ multiply__2.0__18.0__ multiply__2.0__18.0__ |
| the ratio of the radius of two circles is num__1 : num__3 and then the ratio of their areas is ? <o> a ) num__1 : num__5 <o> b ) num__1 : num__8 <o> c ) num__1 : num__9 <o> d ) num__1 : num__1 <o> e ) num__1 : num__4 |
r num__1 : r num__2 = num__1 : num__3 Π r num__12 : Π r num__22 r num__12 : r num__22 = num__1 : num__9 answer : c <eor> c <eos> |
c |
square_perimeter__3.0__ surface_rectangular_prism__1.0__3.0__2.0__ power__3.0__2.0__ volume_cube__1.0__ |
square_perimeter__3.0__ surface_rectangular_prism__1.0__3.0__2.0__ power__3.0__2.0__ volume_cube__1.0__ |
| the average age of a husband and his wife was num__29 years at the time of their marriage . after five years they have a one - year old child . the average age of the family now is : <o> a ) num__11 <o> b ) num__23 <o> c ) num__19 <o> d ) num__287 <o> e ) num__27 |
explanation : sum of the present ages of husband wife and child = ( num__29 * num__2 + num__5 * num__2 ) + num__1 = num__69 years . required average = ( num__23.0 ) = num__23 years . answer : b <eor> b <eos> |
b |
multiply__1.0__23.0__ |
multiply__1.0__23.0__ |
| two pipes a and b can separately fill a cistern in num__10 and num__15 minutes respectively . a person opens both the pipes together when the cistern should have been was full he finds the waste pipe open . he then closes the waste pipe and in another num__4 minutes the cistern was full . in what time can the waste pipe empty the cistern when fill ? <o> a ) num__7 min <o> b ) num__8 min <o> c ) num__9 min <o> d ) num__10 min <o> e ) num__12 min |
explanation : num__0.1 + num__0.0666666666667 = num__0.166666666667 * num__4 = num__0.666666666667 num__1 - num__0.666666666667 = num__0.333333333333 num__0.1 + num__0.0666666666667 - num__1 / x = num__0.333333333333 x = num__8 answer is b <eor> b <eos> |
b |
add__0.1__0.0667__ divide__10.0__15.0__ multiply__10.0__0.1__ subtract__1.0__0.6667__ round__8.0__ |
add__0.1__0.0667__ divide__10.0__15.0__ multiply__10.0__0.1__ subtract__1.0__0.6667__ divide__8.0__1.0__ |
| if num__100 < y < num__199 and num__10 < x < num__100 then the product xy can not be equal to : <o> a ) num__19104 <o> b ) num__19303 <o> c ) num__19 num__356.732 <o> d ) num__19910 <o> e ) num__19 |
502 |
correct answer : ( e ) determine the range of xy by multiplying the two extremes of each individual range together . the smallest value of xy must be greater than num__100 * num__10 . the largest value must be less than num__199 * num__100 . this means that num__1000 < xy < num__19900 . ( d ) is outside of this range so it is not a possible product of xy . <eor> d <eos> |
d |
d |
| the length of the bridge which a train num__130 m long and traveling at num__45 km / hr can cross in num__30 sec is ? <o> a ) num__200 m <o> b ) num__225 m <o> c ) num__245 m <o> d ) num__250 m <o> e ) num__270 m |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec . time = num__30 sec let the length of bridge be x meters . then ( num__130 + x ) / num__30 = num__12.5 x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| in a box of num__12 pens a total of num__3 are defective . if a customer buys num__2 pens selected at random from the box what is the probability that neither pen will be defective ? <o> a ) num__0.166666666667 <o> b ) num__0.222222222222 <o> c ) num__0.545454545455 <o> d ) num__0.5625 <o> e ) num__0.75 |
# defective pens = num__3 # good pens = num__9 probability of the num__1 st pen being good = # of favorable outcomes / # of total outcomes = num__0.75 probability of the num__2 nd pen being good = # of remaining favorable outcomes / # of total remaining outcomes = num__0.727272727273 total probability = num__0.75 * num__0.727272727273 = num__0.545454545455 answer will be c <eor> c <eos> |
c |
subtract__12.0__3.0__ subtract__3.0__2.0__ divide__9.0__12.0__ multiply__0.75__0.7273__ multiply__0.75__0.7273__ |
subtract__12.0__3.0__ subtract__3.0__2.0__ divide__9.0__12.0__ multiply__0.75__0.7273__ multiply__0.75__0.7273__ |
| the length of a rectangle is halved while its breadth is tripled . watis the % change in area ? <o> a ) num__20.0 <o> b ) num__30.0 <o> c ) num__50.0 <o> d ) num__70.0 <o> e ) num__80 % |
let original length = x and original breadth = y . original area = xy . new length = x . num__2 new breadth = num__3 y . new area = x x num__3 y = num__3 xy . num__2 num__2 increase % = num__1 xy x num__1 x num__100.0 = num__50.0 . num__2 xy c <eor> c <eos> |
c |
triangle_area__1.0__100.0__ triangle_area__1.0__100.0__ |
triangle_area__1.0__100.0__ triangle_area__1.0__100.0__ |
| a can do a job in num__18 days and b can do it in num__30 days . a and b working together will finish twice the amount of work in - - - - - - - days ? <o> a ) num__22 num__0.125 days <o> b ) num__22 num__0.5 days <o> c ) num__22 num__3.0 days <o> d ) num__72 num__0.5 days <o> e ) num__23 num__0.5 days |
num__0.0555555555556 + num__0.0333333333333 = num__0.0888888888889 = num__0.0888888888889 num__11.25 = num__11 num__0.25 * num__2 = num__22 num__0.5 days answer : b <eor> b <eos> |
b |
add__0.0556__0.0333__ subtract__11.25__11.0__ multiply__11.0__2.0__ multiply__0.25__2.0__ round__22.0__ |
add__0.0556__0.0333__ subtract__11.25__11.0__ multiply__11.0__2.0__ multiply__0.25__2.0__ multiply__11.0__2.0__ |
| a water tank is two - fifth full . pipe a can fill a tank in num__10 minutes and pipe b can empty it in num__6 minutes . if both the pipes are open how long will it take to empty or fill the tank completely ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__6 <o> e ) num__4 |
clearly pipe b is faster than pipe a and so the tank will be emptied . part to be emptied = num__0.4 part emptied by ( a + b ) in num__1 minute = ( num__0.166666666667 - num__0.1 ) = num__0.0666666666667 num__0.0666666666667 : num__0.4 : : num__1 : x x = ( num__0.4 * num__1 * num__15 ) = num__6 min . so the tank will be emptied in num__6 min . answer : a <eor> a <eos> |
a |
divide__1.0__6.0__ divide__1.0__10.0__ divide__0.4__6.0__ divide__6.0__0.4__ round__6.0__ |
divide__1.0__6.0__ divide__1.0__10.0__ multiply__0.4__0.1667__ divide__6.0__0.4__ multiply__0.4__15.0__ |
| jake and kay drive at constant speeds toward each other on a highway . jake drives at a constant speed of num__29 km per hour . at a certain time they pass by each other and then keep driving away from each other maintaining their constant speeds . if jake is num__155 km away from kay at num__6 am and also num__155 km away from kay at num__11 am then how fast is kay driving ( in kilometers per hour ) ? <o> a ) num__29 <o> b ) num__30 <o> c ) num__31 <o> d ) num__32 <o> e ) num__33 |
jake and kay complete num__310 km / num__5 hours = num__62 km / hour kay ' s speed is num__62 - num__29 = num__33 km / hour the answer is e . <eor> e <eos> |
e |
subtract__11.0__6.0__ divide__310.0__5.0__ subtract__62.0__29.0__ round__33.0__ |
subtract__11.0__6.0__ divide__310.0__5.0__ subtract__62.0__29.0__ subtract__62.0__29.0__ |
| in a mixture there is ratio of alcohol and water is num__4 : num__3 . if in this mixture we will add num__3 litre of water . then ratio become num__4 : num__5 . then calculate the amount of alcohol ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
let in mixture alcohol = num__4 x litre and water = num__3 x litre num__4 x / num__3 x + num__3 = num__0.8 x = num__1.5 amount of alcohol = ( num__4 × num__1.5 ) = num__6 litre answer c <eor> c <eos> |
c |
divide__4.0__5.0__ multiply__4.0__1.5__ multiply__4.0__1.5__ |
divide__4.0__5.0__ multiply__4.0__1.5__ multiply__4.0__1.5__ |
| if a farmer wants to plough a farm field on time he must plough num__100 hectares a day . for technical reasons he ploughed only num__85 hectares a day hence he had to plough num__2 more days than he planned and he still has num__40 hectares left . what is the area of the farm field and how many days the farmer planned to work initially ? <o> a ) num__600 <o> b ) num__490 <o> c ) num__720 <o> d ) num__1400 <o> e ) num__1679 |
let x be the number of days in the initial plan . therefore the whole field is num__100 â ‹ … x hectares . the farmer had to work for x + num__2 days and he ploughed num__85 ( x + num__2 ) hectares leaving num__40 hectares unploughed . then we have the equation : num__100 x = num__85 ( x + num__2 ) + num__40 num__15 x = num__210 x = num__14 so the farmer planned to have the work done in num__6 days and the area of the farm field is num__100 ( num__14 ) = num__1400 hectares correct answer d <eor> d <eos> |
d |
subtract__100.0__85.0__ divide__210.0__15.0__ multiply__100.0__14.0__ round__1400.0__ |
subtract__100.0__85.0__ divide__210.0__15.0__ multiply__100.0__14.0__ round__1400.0__ |
| two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of num__26 kmph and num__21 kmph respectively . when they meet it is found that one train has traveled num__60 km more than the other one . the distance between the two stations is ? <o> a ) num__457 km <o> b ) num__444 km <o> c ) num__547 km <o> d ) num__564 km <o> e ) num__453 km |
num__1 h - - - - - num__5 ? - - - - - - num__60 num__12 h rs = num__26 + num__21 = num__47 t = num__12 d = num__47 * num__12 = num__564 answer : d <eor> d <eos> |
d |
subtract__26.0__21.0__ divide__60.0__5.0__ add__26.0__21.0__ multiply__12.0__47.0__ round__564.0__ |
subtract__26.0__21.0__ divide__60.0__5.0__ add__26.0__21.0__ multiply__12.0__47.0__ multiply__12.0__47.0__ |
| the sector of a circle has radius of num__21 cm and central angle num__135 o . find its perimeter ? <o> a ) num__91.5 cm <o> b ) num__91.6 cm <o> c ) num__97.5 cm <o> d ) num__51.5 cm <o> e ) num__91.2 cm |
perimeter of the sector = length of the arc + num__2 ( radius ) = ( num__0.375 * num__2 * num__3.14285714286 * num__21 ) + num__2 ( num__21 ) = num__49.5 + num__42 = num__91.5 cm answer : a <eor> a <eos> |
a |
multiply__21.0__2.0__ add__49.5__42.0__ round__91.5__ |
multiply__21.0__2.0__ add__49.5__42.0__ add__49.5__42.0__ |
| in a circular racetrack of length num__100 m three persons a b and c start together . a and b start in the same direction at speeds of num__10 m / s and num__8 m / s respectively . while c runs in the opposite direction at num__15 m / s . when will all the three meet for the first time on the track after the start ? <o> a ) after num__4 s <o> b ) after num__50 s <o> c ) after num__100 s <o> d ) after num__200 s <o> e ) none of these |
a takes ( num__100 m ) / ( num__10 m / s ) = num__10 s to finish one complete round . b takes ( num__100 m ) / ( num__8 m / s ) = num__12.5 s to finish one complete round . c takes ( num__100 m ) / ( num__15 m / s ) = ( num__6.66666666667 ) s to finish one complete round . for them to meet the time will be lcm of above three times i . e lcm ( num__10 num__12.5 num__6.66666666667 ) = lcm ( num__10 num__2520 ) / gcd ( num__1 num__23 ) = num__100.0 = num__100 sec answer : c <eor> c <eos> |
c |
divide__100.0__8.0__ divide__100.0__15.0__ add__8.0__15.0__ round__100.0__ |
divide__100.0__8.0__ divide__100.0__15.0__ add__8.0__15.0__ divide__100.0__1.0__ |
| two pipes a and b can separately fill a tank in num__20 minutes and num__15 minutes respectively . both the pipes are opened together but num__4 minutes after the start the pipe a is turned off . how much time will it take to fill the tank ? <o> a ) num__12 min <o> b ) num__14 min <o> c ) num__15 min <o> d ) num__13 min <o> e ) num__10 min |
num__0.2 + x / num__15 = num__1 x = num__12 answer : a <eor> a <eos> |
a |
divide__4.0__20.0__ round__12.0__ |
divide__4.0__20.0__ divide__12.0__1.0__ |
| the average weekly salary per head of the entire staff of a factory consisting of supervisors and the laborers is rs num__60 . the average salary per head of the supervisors is rs num__400 and that of the laborers is rs num__56 . given that the number of supervisors is num__12 . find the number of laborers in the factory <o> a ) num__1000 <o> b ) num__1010 <o> c ) num__1020 <o> d ) num__1030 <o> e ) num__1040 |
average salary of laborerrs num__56 average salary of supervisors rs num__400 average salary of entire staff rs num__60 num__340 num__4 number of laborer / number of supervisors = num__85.0 = num__85.0 thus if the number of supervisors is num__1 number of laborers = num__85 . therefore if the number of supervisors is num__12 number of laborers num__85 * num__12 = num__1020 . answer c <eor> c <eos> |
c |
subtract__400.0__60.0__ subtract__60.0__56.0__ divide__340.0__4.0__ multiply__12.0__85.0__ multiply__12.0__85.0__ |
subtract__400.0__60.0__ subtract__60.0__56.0__ divide__340.0__4.0__ multiply__12.0__85.0__ multiply__12.0__85.0__ |
| reduce num__0.613333333333 to the lowest terms <o> a ) num__1.2 <o> b ) num__1.12 <o> c ) num__0.965517241379 <o> d ) num__0.613333333333 <o> e ) none of these |
explanation : we can do it easily by in two steps step num__1 : we get the hcf of num__368 and num__600 which is num__8 step num__2 : divide both by num__8 we will get the answer num__0.613333333333 answer : option d <eor> d <eos> |
d |
multiply__0.6133__1.0__ |
multiply__0.6133__1.0__ |
| an assembly line produces num__36 cogs per hour until an initial order of num__60 cogs is completed . the speed of the assembly line is then immediately increased so that it can produce num__60 cogs per hour until another num__60 cogs are produced . what is the overall average output in cogs per hour for the assembly line during this whole time ? <o> a ) num__38 <o> b ) num__40 <o> c ) num__42 <o> d ) num__45 <o> e ) num__50 |
the time to produce the first num__60 cogs is num__1.66666666667 = num__1.66666666667 hours . the time to produce the next num__60 cogs is num__1.0 = num__1 hour . the average output is num__120 cogs / ( num__2.66666666667 ) hours = num__45 cogs per hour . the answer is d . <eor> d <eos> |
d |
divide__60.0__36.0__ round_down__1.6667__ add__1.0__1.6667__ multiply__1.0__45.0__ |
divide__60.0__36.0__ round_down__1.6667__ add__1.0__1.6667__ divide__45.0__1.0__ |
| two trains each num__100 m long moving in opposite directions cross other in num__12 sec . if one is moving twice as fast the other then the speed of the faster train is ? <o> a ) num__22 <o> b ) num__98 <o> c ) num__60 <o> d ) num__40 <o> e ) num__12 |
let the speed of the slower train be x m / sec . then speed of the train = num__2 x m / sec . relative speed = ( x + num__2 x ) = num__3 x m / sec . ( num__100 + num__100 ) / num__12 = num__3 x = > x = num__5.55555555556 . so speed of the faster train = num__11.1111111111 = num__11.1111111111 * num__3.6 = num__40 km / hr . answer : d <eor> d <eos> |
d |
multiply__11.1111__3.6__ round__40.0__ |
multiply__11.1111__3.6__ multiply__11.1111__3.6__ |
| square a has an area of num__81 square centimeters . square b has a perimeter of num__32 centimeters . if square b is placed within square a and a random point is chosen within square a what is the probability the point is not within square b ? <o> a ) num__0.265625 <o> b ) num__0.20987654321 <o> c ) num__0.64 <o> d ) num__0.6 <o> e ) num__0.24 |
i guess it ' s mean that square b is placed within square aentirely . since the perimeter of b is num__32 then its side is num__8.0 = num__8 and the area is num__4 ^ num__2 = num__64 ; empty space between the squares is num__81 - num__64 = num__17 square centimeters so if a random point is in this area then it wo n ' t be within square b : p = favorable / total = num__0.20987654321 . answer : b <eor> b <eos> |
b |
volume_cube__4.0__ triangle_area__2.0__0.2099__ |
power__8.0__2.0__ triangle_area__2.0__0.2099__ |
| a circular path of num__12 m radius has marginal walk num__2 m wide all round it . find the cost of leveling the walk at num__50 p per m num__2 ? <o> a ) rs . num__49 <o> b ) rs . num__40 <o> c ) rs . num__81.72 <o> d ) rs . num__42 <o> e ) rs . num__43 |
explanation : π ( num__14 ^ num__2 - num__12 ^ num__2 ) = num__3.14285714286 * num__52 = num__163.43 num__163.43 * num__0.5 = rs . num__81.72 answer : option c <eor> c <eos> |
c |
rectangle_perimeter__12.0__14.0__ triangle_area__2.0__81.72__ |
rectangle_perimeter__12.0__14.0__ volume_rectangular_prism__2.0__0.5__81.72__ |
| a man rows his boat num__95 km downstream and num__60 km upstream taking num__2 hours each time . find the speed of the stream ? <o> a ) num__76 kmph <o> b ) num__6 kmph <o> c ) num__9 kmph <o> d ) num__8 kmph <o> e ) num__4 kmph |
speed downstream = d / t = num__95 / ( num__2 ) = num__48 kmph speed upstream = d / t = num__60 / ( num__2 ) = num__30 kmph the speed of the stream = ( num__48 - num__30 ) / num__2 = num__9 kmph answer : c <eor> c <eos> |
c |
divide__60.0__2.0__ round__9.0__ |
divide__60.0__2.0__ round__9.0__ |
| all the students of class are told to sit in circle shape . here the boy at the num__10 th position is exactly opposite to num__40 th boy . total number of boys in the class ? <o> a ) num__50 <o> b ) num__45 <o> c ) num__60 <o> d ) num__55 <o> e ) num__53 |
as half the circle shape consist of num__40 - num__10 = num__30 boys so total number of boys in full circle = num__2 * num__30 = num__60 answer : c <eor> c <eos> |
c |
subtract__40.0__10.0__ multiply__2.0__30.0__ multiply__2.0__30.0__ |
subtract__40.0__10.0__ multiply__2.0__30.0__ multiply__2.0__30.0__ |
| if a tire rotates at num__400 revolutions per minute when the car is traveling num__144 km / h what is the circumference of the tire ? <o> a ) num__2 <o> b ) num__1 <o> c ) num__4 <o> d ) num__6 <o> e ) num__5 |
num__400 rev / minute = num__400 * num__60 rev / num__60 minutes = num__24000 rev / hour num__24000 * c = num__144000 m : c is the circumference c = num__6 meters correct answer d <eor> d <eos> |
d |
hour_to_min_conversion__ multiply__400.0__60.0__ divide__144000.0__24000.0__ round__6.0__ |
hour_to_min_conversion__ multiply__400.0__60.0__ divide__144000.0__24000.0__ divide__144000.0__24000.0__ |
| albert invested rs . num__6500 in a scheme for num__2 years at compound interest rate num__6.5 p . a . how much amount will albert get on maturity of the fixed deposit ? <o> a ) s . num__8829 <o> b ) s . num__7200 <o> c ) s . num__7200.5 <o> d ) s . num__4000 <o> e ) s . num__7372.46 |
amount = [ num__6500 * ( num__1 + num__6.5 / num__100 ) num__2 ] = num__6500 * num__106.5 / num__100 * num__106.5 / num__100 = rs . num__7372.46 answer : e <eor> e <eos> |
e |
percent__100.0__7372.46__ |
percent__100.0__7372.46__ |
| in march kurt ran an average of num__1.2 miles an hour . if by june he had increased his pace by num__10 seconds per mile then which of the following expresses the number of hours it would take kurt to complete one mile in june ? <o> a ) num__59.8333333333 ^ num__2 <o> b ) num__40.1666666667 ^ num__2 <o> c ) num__39.8333333333 ^ num__2 <o> d ) num__59.7666666667 <o> e ) num__49.8333333333 ^ num__2 |
kurt ran at an average of num__1.2 miles / hour in march . so to run num__1 mile he would take num__1 / num__1.2 hours = ( num__60 * num__60 ) / num__1.2 seconds = num__3000 seconds . if he increases his speed by num__10 seconds he will complete a mile in num__2990 seconds . converting in hours = num__2990 / ( num__60 * num__60 ) = num__2990 / ( num__60 ^ num__2 ) answer : e <eor> e <eos> |
e |
round_down__1.2__ subtract__3000.0__10.0__ divide__2990.0__60.0__ |
round_down__1.2__ subtract__3000.0__10.0__ divide__2990.0__60.0__ |
| a father said to his son ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is num__42 years now the son ' s age num__4 years back was : <o> a ) num__12 years . <o> b ) num__14 years . <o> c ) num__17 years . <o> d ) num__16 years . <o> e ) num__19 years . |
let the son ' s present age be x years . then ( num__42 - x ) = x num__2 x = num__42 . x = num__21 son ' s age num__5 years back ( num__21 - num__4 ) = num__17 years . answer : c <eor> c <eos> |
c |
divide__42.0__2.0__ subtract__21.0__4.0__ subtract__21.0__4.0__ |
divide__42.0__2.0__ subtract__21.0__4.0__ subtract__21.0__4.0__ |
| in how many ways can the letters of the word abacus be rearranged such that the vowels always appear together ? <o> a ) num__6 ! / num__2 ! <o> b ) num__3 ! * num__3 ! <o> c ) num__4 ! / num__2 ! <o> d ) ( num__4 ! * num__3 ! ) / num__2 ! <o> e ) ( num__3 ! * num__3 ! ) / num__2 ! |
explanatory answer step num__1 : group the vowels as one unit and rearrange abacus is a num__6 letter word with num__3 of the letters being vowels . because the num__3 vowels have to appear together let us group the aau as one unit . there are num__3 consonants in addition to one unit of vowels . these num__4 elements can be rearranged in num__4 ! ways . step num__2 : rearrange the letters within the unit containing the vowels the num__3 vowels can rearrange among themselves in num__3 ! / num__2 ! ways as ` ` a ' ' appears twice . hence the total number of rearrangements in which the vowels appear together is num__4 ! ∗ num__3 ! / num__2 ! choice d <eor> d <eos> |
d |
die_space__ coin_space__ choose__4.0__3.0__ |
die_space__ coin_space__ choose__4.0__3.0__ |
| a metallic sheet is of rectangular shape with dimensions num__48 m x num__36 m . from each of its corners a square is cut off so as to make an open box . if the length of the square is num__8 m the volume of the box ( in m num__3 ) is : <o> a ) num__4630 <o> b ) num__4920 <o> c ) num__5120 <o> d ) num__7960 <o> e ) num__8960 |
l = ( num__48 - num__16 ) m = num__32 m b = ( num__36 - num__16 ) m = num__20 m h = num__8 m . volume of the box = ( num__32 x num__20 x num__8 ) m num__3 = num__5120 m num__3 . answer : c <eor> c <eos> |
c |
square_perimeter__8.0__ volume_rectangular_prism__8.0__32.0__20.0__ volume_rectangular_prism__8.0__32.0__20.0__ |
square_perimeter__8.0__ volume_rectangular_prism__8.0__32.0__20.0__ volume_rectangular_prism__8.0__32.0__20.0__ |
| which one of the following numbers is exactly divisible by num__12 ? <o> a ) a ) num__12 <o> b ) b ) num__13 <o> c ) c ) num__14 <o> d ) d ) num__15 <o> e ) e ) num__16 |
num__1.0 = num__1 answer is a <eor> a <eos> |
a |
multiply__12.0__1.0__ |
multiply__12.0__1.0__ |
| a certain characteristic in a large population has a distribution that is symmetric about the mean m . num__72 percent of the distribution lies within one standard deviation d of the mean . if the shelf ’ s average life is num__7.6 years and the standard deviation is num__2.6 years what percent of the distribution has more than num__10.2 years as a shelf ’ s average life ? <o> a ) num__17.0 <o> b ) num__16.0 <o> c ) num__15.0 <o> d ) num__14.0 <o> e ) num__13 % |
average = num__7.6 sd = num__2.6 num__7.6 - num__2.6 < num__72.0 of distribution < num__7.6 + num__2.6 num__5.0 < num__72.0 of distribution < num__10.2 num__28.0 is outside this range . given : distribution is symmetric . so num__14.0 of distribution is less than num__5.0 and the other num__14.0 of distribution is greater than num__10.2 . answer : d <eor> d <eos> |
d |
subtract__7.6__2.6__ subtract__28.0__14.0__ |
subtract__7.6__2.6__ subtract__28.0__14.0__ |
| a man swims downstream num__36 km and upstream num__48 km taking num__6 hours each time what is the speed of the man in still water ? <o> a ) num__5 <o> b ) num__24 <o> c ) num__7 <o> d ) num__42 <o> e ) num__6 |
num__36 - - - num__6 ds = num__6 ? - - - - num__1 num__48 - - - - num__6 us = num__8 ? - - - - num__1 m = ? m = ( num__6 + num__8 ) / num__2 = num__7 answer : c <eor> c <eos> |
c |
divide__48.0__6.0__ subtract__8.0__6.0__ add__6.0__1.0__ round__7.0__ |
divide__48.0__6.0__ subtract__8.0__6.0__ add__6.0__1.0__ add__6.0__1.0__ |
| the minute hand of a clock overtakes the hour hand at intervals of num__65 minutes of the correct time . how much a day does the clock gain or lose ? <o> a ) ( num__10 + num__0.0699300699301 ) min <o> b ) ( num__10 + num__0.0291545189504 ) min <o> c ) ( num__10 + num__0.0699300699301 ) min <o> d ) ( num__18 + num__0.0184162062615 ) min <o> e ) ( num__10 + num__0.0134589502019 ) min |
explanation : in a correct clock the minute hand gains num__55 min . spaces over the hour hand in num__60 minutes . to be together again the minute hand must gain num__60 minutes over the hour hand . num__55 minutes are gained in num__60 min . num__60 min . are gained in [ ( num__1.09090909091 ) * num__60 ] min = min . but they are together after num__65 min . therefore gain in num__65 minutes = = min . gain in num__24 hours = = num__10.0699300699 min . therefore the clock gains ( num__10 + num__0.0699300699301 ) minutes in num__24 hours . answer : a ) ( num__10 + num__0.0699300699301 ) min <eor> a <eos> |
a |
hour_to_min_conversion__ divide__60.0__55.0__ subtract__65.0__55.0__ subtract__10.0699__10.0__ round__10.0__ |
hour_to_min_conversion__ divide__60.0__55.0__ subtract__65.0__55.0__ subtract__10.0699__10.0__ round__10.0__ |
| in a circuit board factory all circuit boards that pass a verification process are certified . every board that fails the verification process is indeed faulty but num__0.125 of those that pass are also faulty . approximately how many faulty circuit boards exist in a group of num__800 circuit boards where num__64 fail inspection ? <o> a ) num__72 <o> b ) num__192 <o> c ) num__156 <o> d ) num__256 <o> e ) num__264 |
total of num__800 boards . all that fail verification are indeed faulty . so the num__64 are indeed faulty . num__0.125 those that pass are also faulty . from the num__800 we know num__64 fail . so num__736 must pass . of these num__0.125 are faulty . num__736 divided by num__8 gives you num__92 . what one must do now is to add to the num__92 which were not detected the actually detected faulty ones namely the num__64 . total faulty : num__156 . answer : c <eor> c <eos> |
c |
subtract__800.0__64.0__ reverse__0.125__ multiply__0.125__736.0__ add__64.0__92.0__ add__64.0__92.0__ |
subtract__800.0__64.0__ reverse__0.125__ multiply__0.125__736.0__ add__64.0__92.0__ add__64.0__92.0__ |
| a man covers a distance of num__180 km at num__72 kmph and next num__190 km at num__88 kmph . what is his average speed for his whole journey of num__370 km ? <o> a ) num__79.2 kmph <o> b ) num__79.3 kmph <o> c ) num__77.25 kmph <o> d ) num__78.5 kmph <o> e ) num__65.5 kmph |
formula = num__2 Ã — f . s * s . p / f . s + s . p = num__2 Ã — num__72 Ã — num__1.0 + num__72 = num__79.2 kmph answer is a <eor> a <eos> |
a |
round__79.2__ |
divide__79.2__1.0__ |
| a pump can fill a tank with water in num__6 hours . because of a leak it took num__12 hours to fill the tank . the leak can drain all the water in ? <o> a ) num__12 hr <o> b ) num__11 hr <o> c ) num__13 hr <o> d ) num__14 hr <o> e ) num__16 hr |
work done by the leak in num__1 hour = num__0.166666666667 - num__0.0833333333333 = num__0.0833333333333 leak will empty the tank in num__12 hours answer is a <eor> a <eos> |
a |
divide__1.0__6.0__ divide__1.0__12.0__ round__12.0__ |
divide__1.0__6.0__ divide__1.0__12.0__ round__12.0__ |
| in the mountainside summer camp there are num__50 children . num__90.0 of the children are boys and the rest are girls . the camp administrator decided to make the number of girls only num__5.0 of the total number of children in the camp . how many more boys must she bring to make that happen ? <o> a ) num__50 . <o> b ) num__45 . <o> c ) num__40 . <o> d ) num__30 . <o> e ) num__25 . |
given there are num__50 students in the mountainside summer camp num__90.0 of num__50 = num__45 boys and remaining num__5 girls . now here num__90.0 are boys and num__10.0 are girls . now question is asking about how many boys do we need to add to make the girls percentage to num__5 or num__5.0 . . if we add num__50 to existing num__45 then the count will be num__95 and the girls number will be num__5 as it . now boys are num__95.0 and girls are num__5.0 . ( out of num__100 students = num__95 boys + num__5 girls ) . imo option a is correct . <eor> a <eos> |
a |
subtract__50.0__5.0__ divide__50.0__5.0__ add__50.0__45.0__ add__90.0__10.0__ multiply__5.0__10.0__ |
subtract__50.0__5.0__ divide__50.0__5.0__ add__50.0__45.0__ add__90.0__10.0__ add__5.0__45.0__ |
| city a to city b raja drove for num__1 hour at num__72 mph and for num__3 hours at num__80 mph . what was the average speed for the whole trip ? <o> a ) num__57 <o> b ) num__57.5 <o> c ) num__78 <o> d ) num__59 <o> e ) num__62 |
the total distance is num__1 × num__72 + num__3 × num__80 = num__312 and the total time is num__4 hours . hence average speed = ( total distance / total time ) = num__78.0 = num__78 answer : c <eor> c <eos> |
c |
add__1.0__3.0__ divide__312.0__4.0__ round__78.0__ |
add__1.0__3.0__ divide__312.0__4.0__ divide__78.0__1.0__ |
| in a local school district the high school and middle school each received r dollars toward funding for the student arts program . the high school enrolled num__100 students and the middle school enrolled num__50 students . later the middle school transferred s dollars to the high school so that they would have received the same funding per student . which of the following is equivalent to s ? <o> a ) r / num__3 <o> b ) r / num__5 <o> c ) r / num__7 <o> d ) r / num__9 <o> e ) r / num__2 |
total per head = num__2 r / num__150 after s transfer both schools have total for head . so at high school it will be : r + s = ( num__100 ) * ( num__2 r / num__150 ) = num__4 r / num__3 i . e . s = num__4 r / num__3 - r / num__1 = num__4 r - num__3 r / num__3 = r / num__3 answer : a <eor> a <eos> |
a |
divide__100.0__50.0__ add__100.0__50.0__ divide__150.0__50.0__ subtract__3.0__2.0__ add__1.0__2.0__ |
divide__100.0__50.0__ add__100.0__50.0__ divide__150.0__50.0__ subtract__3.0__2.0__ add__1.0__2.0__ |
| a computer system uses alphanumeric case sensitive characters for its passwords . when the system was created it required users to create passwords having num__4 characters in length . this year it added the option of creating passwords having num__3 characters in length . which of the following gives the expression for the total number of passwords the new computer system can accept ? assume there are num__62 unique alphanumeric case sensitive characters . <o> a ) num__63 ^ num__4 <o> b ) num__62 ^ num__5 <o> c ) num__62 ( num__62 ^ num__4 ) <o> d ) num__63 ( num__62 ^ num__4 ) <o> e ) num__63 ( num__62 ^ num__3 ) |
total number of passwords = number of num__4 character password + number of num__5 character password = num__62 ^ num__4 + num__62 ^ num__3 ( since there is no limitation on repetition each character can be chosen in num__62 ways ) = num__62 ^ num__3 ( num__1 + num__62 ) = num__62 ^ num__3 * num__63 answer e <eor> e <eos> |
e |
subtract__4.0__3.0__ add__62.0__1.0__ add__62.0__1.0__ |
subtract__4.0__3.0__ add__62.0__1.0__ add__62.0__1.0__ |
| if x is to be chosen at random from the set { num__1 num__2 num__3 num__4 } and y is to be chosen at random from the set { num__5 num__6 } what is the probability that xy will be even ? <o> a ) num__0.166666666667 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__0.833333333333 |
probably the best way to solve would be to use num__1 - p ( opposite event ) = num__1 - p ( odd ) = num__1 - p ( odd ) * p ( odd ) = num__1 - num__0.5 * num__0.666666666667 = num__0.666666666667 = num__0.333333333333 . answer : b . <eor> b <eos> |
b |
negate_prob__0.6667__ negate_prob__0.6667__ |
negate_prob__0.6667__ negate_prob__0.6667__ |
| a car covers a distance of num__1260 km in num__10 hours . find its speed ? <o> a ) num__104 <o> b ) num__277 <o> c ) num__298 <o> d ) num__126 <o> e ) num__213 |
num__126.0 = num__126 kmph answer : d <eor> d <eos> |
d |
divide__1260.0__10.0__ round__126.0__ |
divide__1260.0__10.0__ round__126.0__ |
| a sum of money triples itself in twelve years at simple interest . find the rate of interest ? <o> a ) num__16 num__0.25 % <o> b ) num__16 num__2.66666666667 % <o> c ) num__86 num__0.666666666667 % <o> d ) num__16 num__1.66666666667 % <o> e ) num__16 num__0.666666666667 % |
let the pricipal be rs . x then amount = num__3 x ( where r = rate of interest ) = > interest = num__3 x - x = rs . num__2 x r = ( num__100 * num__2 x ) / ( x * num__12 ) = num__16.6666666667 % = num__16 num__0.666666666667 % answer : e <eor> e <eos> |
e |
percent__100.0__16.0__ |
percent__100.0__16.0__ |
| if | num__5 x - num__15 | = num__100 then find the sum of the values of x ? <o> a ) num__1 <o> b ) - num__2 <o> c ) num__6 <o> d ) - num__3 <o> e ) num__4 |
| num__5 x - num__15 | = num__100 num__5 x - num__15 = num__100 or num__5 x - num__15 = - num__100 num__5 x = num__115 or num__5 x = - num__85 x = num__23 or x = - num__17 sum = num__23 - num__17 = num__6 answer is c <eor> c <eos> |
c |
add__15.0__100.0__ subtract__100.0__15.0__ divide__115.0__5.0__ divide__85.0__5.0__ subtract__23.0__17.0__ subtract__23.0__17.0__ |
add__15.0__100.0__ subtract__100.0__15.0__ divide__115.0__5.0__ divide__85.0__5.0__ subtract__23.0__17.0__ subtract__23.0__17.0__ |
| the lunch menu at a certain restaurant contains num__4 different entrees and num__5 different side dishes . if a meal consists of num__1 entree and num__2 different side dishes how many different meal combinations e could be chosen from this menu ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__40 <o> d ) num__80 <o> e ) num__100 |
e = num__4 c num__1 * num__5 c num__2 = num__4 * ( num__5 * num__4 * num__3 ! ) / ( num__3 ! num__2 ! ) = num__4 * num__10 = num__40 answer - c <eor> c <eos> |
c |
choose__5.0__3.0__ choose__5.0__3.0__ |
choose__5.0__3.0__ choose__5.0__3.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__3 sec . what is the length of the train ? <o> a ) num__85 m <o> b ) num__70 m <o> c ) num__115 m <o> d ) num__50 m <o> e ) num__150 m |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__3 = num__50 m answer : d <eor> d <eos> |
d |
round__50.0__ |
round__50.0__ |
| two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of num__25 kmph and num__21 kmph respectively . when they meet it is found that one train has traveled num__60 km more than the other one . the distance between the two stations is ? <o> a ) num__457 km <o> b ) num__444 km <o> c ) num__552 km <o> d ) num__645 km <o> e ) num__453 km |
num__1 h - - - - - num__5 ? - - - - - - num__60 num__12 h rs = num__25 + num__21 = num__46 t = num__12 d = num__46 * num__12 = num__552 answer : c <eor> c <eos> |
c |
divide__60.0__5.0__ add__25.0__21.0__ multiply__12.0__46.0__ round__552.0__ |
divide__60.0__5.0__ add__25.0__21.0__ multiply__12.0__46.0__ multiply__12.0__46.0__ |
| num__25 num__50 num__100 num__175 num__275 ( . . . . ) <o> a ) num__400 <o> b ) num__300 <o> c ) num__425 <o> d ) num__450 <o> e ) num__325 |
explanation : the pattern is num__25 num__50 num__75 num__100 etc . hence num__125 = num__400 answer : a <eor> a <eos> |
a |
add__25.0__50.0__ add__25.0__100.0__ add__275.0__125.0__ add__275.0__125.0__ |
add__25.0__50.0__ add__25.0__100.0__ add__275.0__125.0__ add__275.0__125.0__ |
| a collection of books went on sale and num__0.666666666667 of them were sold for $ num__2.25 each . if none of the num__40 remaining books were sold what was the total amount received for the books that were sold ? <o> a ) $ num__160 <o> b ) $ num__150 <o> c ) $ num__280 <o> d ) $ num__200 <o> e ) $ num__180 |
if num__40 books constitute num__0.333333333333 rd of the total then num__0.666666666667 rd of the total = num__80 books amount received for sold books = num__80 * num__2.25 = $ num__180 answer : e <eor> e <eos> |
e |
multiply__2.25__80.0__ multiply__2.25__80.0__ |
multiply__2.25__80.0__ multiply__2.25__80.0__ |
| by selling num__32 mangoes in a dollar a man loses num__40.0 . how many mangoes must he sell in a dollar to gain num__20.0 profit ? <o> a ) num__16 <o> b ) num__18 <o> c ) num__20 <o> d ) num__25 <o> e ) num__28 |
this is a question that is much easier than it first appears . if you work with the percents directly you can solve it very quickly . if the man sells num__32 mangoes for a dollar he loses num__40.0 . that means he is at num__60.0 of his cost ( num__100.0 - num__40.0 = num__60.0 ) . we are trying to figure out how many mangoes he has to sell in order to make a num__20.0 profit or be at num__120.0 ( num__100.0 + num__20.0 = num__120.0 ) . num__120.0 is double num__60.0 meaning that we simply cut the number of mangoes in half to double our returns ( from num__60.0 to num__120.0 ) yielding num__16.0 = num__16 mangoes or answer choice a . <eor> a <eos> |
a |
percent__100.0__16.0__ |
percent__100.0__16.0__ |
| two trains of equal lengths take num__10 sec and num__15 sec respectively to cross a telegraph post . if the length of each train be num__120 m in what time will they cross other travelling in opposite direction ? <o> a ) num__17 sec <o> b ) num__12 sec <o> c ) num__16 sec <o> d ) num__15 sec <o> e ) num__18 sec |
speed of the first train = num__12.0 = num__12 m / sec . speed of the second train = num__24.0 = num__8 m / sec . relative speed = num__12 + num__8 = num__20 m / sec . required time = ( num__120 + num__120 ) / num__20 = num__12 sec . answer : b <eor> b <eos> |
b |
divide__120.0__10.0__ divide__120.0__15.0__ add__8.0__12.0__ round__12.0__ |
divide__120.0__10.0__ divide__120.0__15.0__ add__8.0__12.0__ divide__120.0__10.0__ |
| how many multiples of num__10 are there between num__81 and num__358 ? <o> a ) num__24 <o> b ) num__25 <o> c ) num__26 <o> d ) num__27 <o> e ) num__28 |
num__10 * num__9 = num__90 num__10 * num__35 = num__350 total no of multiples = ( num__35 - num__9 ) + num__1 = num__27 answer d <eor> d <eos> |
d |
multiply__10.0__9.0__ multiply__10.0__35.0__ subtract__10.0__9.0__ multiply__1.0__27.0__ |
multiply__10.0__9.0__ multiply__10.0__35.0__ subtract__10.0__9.0__ multiply__1.0__27.0__ |
| num__7 ^ num__6 n - num__1 ^ num__6 n when n is an integer > num__0 is divisible by <o> a ) num__120 <o> b ) num__127 <o> c ) num__143 <o> d ) num__164 <o> e ) num__190 |
num__127 b <eor> b <eos> |
b |
multiply__1.0__127.0__ |
multiply__1.0__127.0__ |
| find the middle one when the sum of three consecutive even numbers is num__162 ? <o> a ) num__51 <o> b ) num__50 <o> c ) num__58 <o> d ) num__54 <o> e ) num__56 |
num__3 consecutive numbers can be a - num__1 a a + num__1 so sum of numbers = num__3 a = num__162 . hence a = num__54 . d <eor> d <eos> |
d |
divide__162.0__3.0__ round__54.0__ |
divide__162.0__3.0__ round__54.0__ |
| mangala completes a piece of work in num__30 days raju completes the same work in num__30 days . if both of them work together then the number of days required to complete the work is ? <o> a ) num__8 days <o> b ) num__12 days <o> c ) num__14 days <o> d ) num__15 days <o> e ) num__18 days |
if a can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days . that is the required no . of days = num__30 × num__0.5 = num__15 days d <eor> d <eos> |
d |
multiply__30.0__0.5__ round__15.0__ |
multiply__30.0__0.5__ round__15.0__ |
| if a num__5 percent deposit that has been paid toward the purchase of a certain product is $ num__70 how much more remains to be paid ? <o> a ) $ num__1120 <o> b ) $ num__1190 <o> c ) $ num__1260 <o> d ) $ num__1330 <o> e ) $ num__1400 |
num__95.0 remains to be paid so the remaining amount is num__19 * num__70 = $ num__1330 . the answer is d . <eor> d <eos> |
d |
divide__95.0__5.0__ multiply__70.0__19.0__ multiply__70.0__19.0__ |
divide__95.0__5.0__ multiply__70.0__19.0__ multiply__70.0__19.0__ |
| if a lemon and apple together costs rs . num__12 tomato and a lemon cost rs . num__4 and an apple costs rs . num__8 more than a lemon . what is the cost of lemon ? <o> a ) num__8 <o> b ) num__2 <o> c ) num__3 <o> d ) num__5 <o> e ) num__0 |
l + a = num__12 . . . ( num__1 ) t + l = num__4 . . . . . ( num__2 ) l + num__8 = a taking num__1 and num__3 we get a = num__10 and l = num__2 answer : b <eor> b <eos> |
b |
divide__8.0__4.0__ divide__12.0__4.0__ subtract__12.0__2.0__ subtract__12.0__10.0__ |
divide__8.0__4.0__ add__1.0__2.0__ add__8.0__2.0__ subtract__12.0__10.0__ |
| the average weight of num__19 students is num__15 kg . by the admission of a new student the average weight is reduced to num__14.2 kg . the weight of the new student is ? <o> a ) num__1 kg <o> b ) num__10.8 kg <o> c ) num__11 kg <o> d ) num__14.9 kg <o> e ) none |
answer weight of new student = total weight of all num__20 students - total weight of initial num__19 students = ( num__20 x num__14.2 - num__19 x num__15 ) kg = num__1 kg . correct option : a <eor> a <eos> |
a |
subtract__20.0__19.0__ reverse__1.0__ |
subtract__20.0__19.0__ subtract__20.0__19.0__ |
| the jogging track in a sports complex is num__1000 meters in circumference . deepak and his wife start from the same point and walk in opposite directions at num__20 km / hr and num__15 km / hr respectively . they will meet for the first time in ? <o> a ) num__50 min <o> b ) num__40 min <o> c ) num__35 min <o> d ) num__25 min <o> e ) num__20 min |
clearly the two will meet when they are num__1000 m apart to be num__20 + num__15 = num__35 km apart they take num__1 hour to be num__1000 m apart they take num__35 * num__1.0 = num__35 min . answer is c <eor> c <eos> |
c |
add__20.0__15.0__ add__20.0__15.0__ |
add__20.0__15.0__ add__20.0__15.0__ |
| if the selling price of num__50 articles is equal to the cost price of num__40 articles then the loss or gain percent is : <o> a ) num__10 <o> b ) num__30 <o> c ) num__20 <o> d ) num__60 <o> e ) num__50 |
c num__20 let c . p . of each article be $ num__1 . then c . p . of num__50 articles = $ num__50 ; s . p . of num__50 articles = $ num__40 . loss % = num__0.2 * num__100 = num__20.0 <eor> c <eos> |
c |
percent__50.0__40.0__ percent__1.0__20.0__ percent__50.0__40.0__ |
percent__50.0__40.0__ percent__1.0__20.0__ percent__50.0__40.0__ |
| a trailer carries num__3 num__4 and num__5 crates on a trip . each crate weighs no less than num__150 kg . what is the maximum weight of the crates on a single trip ? <o> a ) num__1250 <o> b ) num__625 <o> c ) num__600 <o> d ) num__750 <o> e ) num__375 |
max no . of crates = num__5 . max weight = num__150 kg max . weight carried = num__5 * num__150 = num__750 kg = d <eor> d <eos> |
d |
multiply__5.0__150.0__ multiply__5.0__150.0__ |
multiply__5.0__150.0__ multiply__5.0__150.0__ |
| a person crosses a num__400 m long street in num__10 minnutes . what is his speed in km per hour ? <o> a ) num__6.4 km / hr <o> b ) num__7.9 km / hr <o> c ) num__1.5 km / hr <o> d ) num__2.4 km / hr <o> e ) num__8 km / hr |
speed = num__40.0 * num__60 = num__0.666666666667 m / sec = num__0.666666666667 * num__3.6 = num__2.4 km / hr answer is d <eor> d <eos> |
d |
divide__400.0__10.0__ hour_to_min_conversion__ divide__40.0__60.0__ round__2.4__ |
divide__400.0__10.0__ hour_to_min_conversion__ divide__40.0__60.0__ round__2.4__ |
| bony has twice as many marbles as tony and tony has thrice as many marbles as vinay . which of the following can not be the sum of marbles with all three of them ? <o> a ) num__72 <o> b ) num__40 <o> c ) num__80 <o> d ) num__92 <o> e ) num__52 |
let each num__1 marble that vinay have tony will have num__3 bony num__6 marbles sum of units = num__1 + num__3 + num__6 = num__10 units therefore the number must be a multiple of num__10 among the choices only num__72 is not divisible by num__10 option a <eor> a <eos> |
a |
triple__1.0__ twice__3.0__ multiply__72.0__1.0__ |
triple__1.0__ twice__3.0__ multiply__72.0__1.0__ |
| how many numbers between num__100 and num__756 are divisible by num__2 num__3 and num__7 together ? <o> a ) num__112 <o> b ) num__77 <o> c ) num__267 <o> d ) num__16 <o> e ) num__99 |
explanation : as the division is by num__2 num__3 num__7 together the numbers are to be divisible by : num__2 * num__3 * num__7 = num__42 the limits are num__100 and num__756 the first number divisible is num__42 * num__3 = num__126 to find out the last number divisible by num__42 within num__756 : num__18.0 = num__18 hence num__42 * num__16 = num__756 is the last number divisible by num__42 within num__756 hence total numbers divisible by num__2 num__3 num__7 together are ( num__18 â € “ num__2 ) = num__16 answer : d <eor> d <eos> |
d |
multiply__3.0__42.0__ divide__756.0__42.0__ subtract__18.0__2.0__ subtract__18.0__2.0__ |
multiply__3.0__42.0__ divide__756.0__42.0__ subtract__18.0__2.0__ subtract__18.0__2.0__ |
| a man can row a boat at num__18 kmph in still water . if the speed of the stream is num__3 kmph what is the time taken to row a distance of num__48 km downstream ? <o> a ) num__1.875 hours <o> b ) num__0.909090909091 hours <o> c ) num__2.28571428571 hours <o> d ) num__1.23076923077 hours <o> e ) num__1.15384615385 hours |
speed downstream = num__18 + num__3 = num__21 kmph . time required to cover num__48 km downstream = d / s = num__2.28571428571 = num__2.28571428571 hours . answer : c <eor> c <eos> |
c |
add__18.0__3.0__ divide__48.0__21.0__ divide__48.0__21.0__ |
add__18.0__3.0__ divide__48.0__21.0__ divide__48.0__21.0__ |
| a student needs num__30.0 of the marks on a test to pass the test . if the student gets num__80 marks and fails the test by num__100 marks find the maximum marks set for the test . <o> a ) num__400 <o> b ) num__600 <o> c ) num__800 <o> d ) num__1000 <o> e ) num__1200 |
num__30.0 = num__180 marks num__1.0 = num__6 marks num__100.0 = num__600 marks the answer is b . <eor> b <eos> |
b |
percent__100.0__600.0__ |
percent__100.0__600.0__ |
| two trains each num__250 m in length are running on the same parallel lines in opposite directions with the speed of num__80 kmph and num__70 kmph respectively . in what time will they cross each other completely ? <o> a ) num__33 <o> b ) num__28 <o> c ) num__12 <o> d ) num__88 <o> e ) num__15 |
d = num__250 m + num__250 m = num__500 m rs = num__80 + num__70 = num__150 * num__0.277777777778 = num__41.6666666667 t = num__500 * num__0.024 = num__12 sec . answer : c <eor> c <eos> |
c |
add__80.0__70.0__ multiply__0.024__500.0__ round__12.0__ |
add__80.0__70.0__ multiply__0.024__500.0__ multiply__0.024__500.0__ |
| a photograph measuring num__2 num__1 ' ' ⁄ num__2 × num__1 num__7 ' ' ⁄ num__8 is to be enlarged so that the length will be num__4 ” . how many inches will the enlarged breadth be ? <o> a ) num__1 num__1 ⁄ num__2 <o> b ) num__2 num__1 ⁄ num__8 <o> c ) num__3 <o> d ) num__3 num__3 ⁄ num__8 <o> e ) none of these |
let enlarged breadth be x inches . then num__5 ⁄ num__2 : num__4 : : num__15 ⁄ num__8 : x ⇒ num__5 ⁄ num__2 x = num__4 × num__15 ⁄ num__8 ⇒ x = num__3 inches . answer c <eor> c <eos> |
c |
add__1.0__4.0__ add__7.0__8.0__ add__2.0__1.0__ add__2.0__1.0__ |
add__1.0__4.0__ add__7.0__8.0__ add__2.0__1.0__ add__2.0__1.0__ |
| on dividing a number by num__56 we get num__29 as remainder . on dividing the same number by num__8 what will be the remainder ? <o> a ) num__2 <o> b ) num__6 <o> c ) num__4 <o> d ) num__5 <o> e ) num__8 |
formula : ( divisor * quotient ) + remainder = dividend . ( num__56 * q ) + num__29 = d - - - - - - - ( num__1 ) d num__8.0 = r - - - - - - - - - - - - - ( num__2 ) from equation ( num__2 ) ( ( num__56 * q ) + num__29 ) num__8.0 = r . = > assume q = num__1 . = > ( num__56 + num__29 ) num__8.0 = r . = > num__85.0 num__8 = r = > num__5 = r . so the required remainder is num__5 . answer d ) num__5 <eor> d <eos> |
d |
add__56.0__29.0__ multiply__1.0__5.0__ |
add__56.0__29.0__ multiply__1.0__5.0__ |
| in a kilometer race if abhishek gives bharti a num__40 m start abhishek wins by num__19 sec . but if abhishek gives bharti a num__30 sec start bharti wins by num__40 m . find the time taken by bharti to run num__5000 m ? <o> a ) num__150 sec <o> b ) num__750 sec <o> c ) num__450 sec <o> d ) num__825 sec <o> e ) num__920 sec |
if abhishek takes x seconds and bharti takes y seconds to run num__1 km then : = > y = num__150 sec and x = num__125 sec = > x + num__19 = num__960 y / num__1000 and = ( num__960 x / num__1000 ) + num__30 = y = > y = num__150 sec and x = num__125 sec answer = ( num__0.15 ) × num__500 = num__750 sec answer : b <eor> b <eos> |
b |
divide__5000.0__40.0__ add__40.0__960.0__ divide__150.0__1000.0__ multiply__5000.0__0.15__ round__750.0__ |
divide__5000.0__40.0__ add__40.0__960.0__ divide__150.0__1000.0__ multiply__5000.0__0.15__ divide__750.0__1.0__ |
| the ratio of the number of females to males at a party was num__1 : num__2 but when num__5 females and num__5 males left the ratio became num__1 : num__3 . how many people were at the party originally ? <o> a ) num__26 <o> b ) num__28 <o> c ) num__30 <o> d ) num__32 <o> e ) num__34 |
the total number of people are x females + num__2 x males . num__3 * ( x - num__5 ) = num__2 x - num__5 x = num__10 there were num__3 x = num__30 people at the party originally . the answer is c . <eor> c <eos> |
c |
multiply__2.0__5.0__ multiply__3.0__10.0__ multiply__1.0__30.0__ |
multiply__2.0__5.0__ multiply__3.0__10.0__ multiply__1.0__30.0__ |
| evaluate : num__30 - | - x + num__6 | for x = num__10 <o> a ) num__23 <o> b ) num__24 <o> c ) num__25 <o> d ) num__26 <o> e ) num__27 |
substitute x by num__10 in the given expression and evaluate num__30 - | - ( num__10 ) + num__6 | = num__30 - | - num__10 + num__6 | = num__30 - | - num__4 | = num__30 - num__4 = num__26 correct answer d ) num__26 <eor> d <eos> |
d |
subtract__10.0__6.0__ subtract__30.0__4.0__ subtract__30.0__4.0__ |
subtract__10.0__6.0__ subtract__30.0__4.0__ subtract__30.0__4.0__ |
| the difference between a two - digit number and the number after interchanging the position of the two digits is num__45 . what is the difference between the two digits of the number ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__5 <o> d ) can not be determined <o> e ) none of these |
let the two - digit no . be l num__0 x + y . then ( num__10 x + y ) â € “ ( num__10 y + x ) = num__45 or num__9 ( x â € “ y ) = num__45 or x â € “ y = num__5 answer c <eor> c <eos> |
c |
divide__45.0__9.0__ divide__45.0__9.0__ |
divide__45.0__9.0__ subtract__10.0__5.0__ |
| a grocer has a sales of euro num__220 euro num__320 euro num__480 euro num__120 and euro num__720 for num__5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of euro num__620 ? <o> a ) uro num__1691 <o> b ) uro num__4591 <o> c ) uro num__1860 <o> d ) uro num__7791 <o> e ) uro num__1991 |
c euro num__1860 total sale for num__5 months = euro ( num__220 + num__320 + num__480 + num__120 + num__720 ) = euro num__1860 . required sale = euro [ ( num__620 x num__6 ) - num__1860 ] = euro ( num__3720 - num__1860 ) = euro num__1860 . answer is c <eor> c <eos> |
c |
divide__720.0__120.0__ multiply__620.0__6.0__ subtract__3720.0__1860.0__ |
divide__720.0__120.0__ multiply__620.0__6.0__ subtract__3720.0__1860.0__ |
| carl is facing very difficult financial times and can only pay the interest on a $ num__30000 loan he has taken . the bank charges him a quarterly compound rate of num__5.0 . what is the approximate interest he pays annually ? <o> a ) $ num__1200 <o> b ) $ num__2000 <o> c ) $ num__2150 <o> d ) $ num__6000 <o> e ) $ num__12000 |
usually you are given the annual rate of interest and it is mentioned that it is annual rate . the bank charges him a quarterly compounded annual rate of num__20.0 . here you find per quarter rate as ( num__5.0 ) % = num__5.0 i have actually never seen a question with quarter rate given but since this question did not mentionannual rate of interestand since the options did not make sense with num__5.0 annual rate of interest it is apparent that the intent was a num__5.0 quarterly rate . so the bank charges num__5.0 every quarter and compounds it in the next quarter . had it been a simple quarterly rate we would have just found num__4 * num__5.0 of num__30000 = $ num__6000 as our answer . but since the interest is compounded it will be a bit more than $ num__6000 . option ( d ) looks correct . <eor> d <eos> |
d |
percent__20.0__30000.0__ percent__20.0__30000.0__ |
percent__20.0__30000.0__ percent__20.0__30000.0__ |
| the value of ( num__4.7 × num__13.26 + num__4.7 × num__9.43 + num__4.7 × num__77.31 ) is <o> a ) num__0.47 <o> b ) num__47 <o> c ) num__470 <o> d ) num__4700 <o> e ) none |
solution given expression = num__4.7 × ( num__13.26 + num__9.43 + num__77.31 ) = num__4.7 × num__100 = num__470 . answer c <eor> c <eos> |
c |
multiply__4.7__100.0__ multiply__4.7__100.0__ |
multiply__4.7__100.0__ multiply__4.7__100.0__ |
| at present the ratio between the ages of arun and deepak is num__4 : num__3 . after num__6 years arun ’ s age will be num__26 years . what is the age of deepak at present ? <o> a ) num__12 years <o> b ) num__15 years <o> c ) num__19 ½ years <o> d ) num__21 years <o> e ) none of these |
solution let the present ages of arun and deepak be num__4 x years and num__3 x years respectively . then num__4 x + num__6 = num__26 ⇔ num__4 x = num__20 ⇔ x = num__5 . ∴ deepak ' s age = num__3 x = num__15 years . answer b <eor> b <eos> |
b |
subtract__26.0__6.0__ divide__20.0__4.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
subtract__26.0__6.0__ divide__20.0__4.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
| tough and tricky questions : work / rate problems . a group of num__4 junior lawyers require num__4 hours to complete a legal research assignment . how many hours would it take a group of three legal assistants to complete the same research assignment assuming that a legal assistant works at two - thirds the rate of a junior lawyer ? source : chili hot gmat <o> a ) num__13 <o> b ) num__10 <o> c ) num__9 <o> d ) num__8 <o> e ) num__5 |
# of people times the # of hours : num__4 * num__4 = num__16 - - > num__4 lawyers do num__16 worksin num__4 hours . num__3 * num__2.66666666667 = num__8.0 = num__8 - - > num__3 assistants do num__8 worksin num__2.66666666667 hours ( num__0.666666666667 * num__5 = num__3.33333333333 ) so since the amount of work the assistants do is half the work the lawyers do the time will be double soans d <eor> d <eos> |
d |
divide__2.6667__4.0__ subtract__8.0__3.0__ subtract__4.0__0.6667__ add__3.0__5.0__ |
divide__2.6667__4.0__ subtract__8.0__3.0__ subtract__4.0__0.6667__ subtract__16.0__8.0__ |
| five people are on an elevator that stops at exactly num__5 floors . what is the probability that exactly one person will push the button for each floor ? <o> a ) num__5 ! / num__5 ^ num__5 <o> b ) num__5 ^ num__1.0 ! <o> c ) num__1.0 ! <o> d ) num__1.0 ^ num__5 <o> e ) num__0.2 ^ num__5 |
each person out of num__5 has num__5 options hence total # of outcomes is num__5 ^ num__5 ; favorable outcomes will be num__5 ! which is # of ways to assign num__5 different buttons to num__5 people so basically # of arrangements of num__5 distinct objects : num__5 ! . p = favorable / total = num__5 ! / num__5 ^ num__5 answer : a . <eor> a <eos> |
a |
vowel_space__ |
vowel_space__ |
| income and expenditure of a person are in the ratio num__5 : num__4 . if the income of the person is rs . num__15000 then find his savings ? <o> a ) num__3600 <o> b ) num__3607 <o> c ) num__3608 <o> d ) num__3602 <o> e ) num__3000 |
let the income and the expenditure of the person be rs . num__5 x and rs . num__4 x respectively . income num__5 x = num__15000 = > x = num__3000 savings = income - expenditure = num__5 x - num__4 x = x so savings = rs . num__3000 . answer : e <eor> e <eos> |
e |
divide__15000.0__5.0__ divide__15000.0__5.0__ |
divide__15000.0__5.0__ divide__15000.0__5.0__ |
| in a jar there are num__3 red gumballs and num__2 blue gumballs . what is the probability of drawing at least one red gumball when drawing two consecutive balls randomly ? <o> a ) num__0.4 <o> b ) num__0.8 <o> c ) num__0.9 <o> d ) num__0.6 <o> e ) num__0.333333333333 |
p ( at least one red ) = num__1 - p ( no red so num__2 blue ) = num__1 - num__0.4 * num__0.25 = num__0.9 . answer : c . <eor> c <eos> |
c |
subtract__3.0__2.0__ round__0.9__ |
subtract__3.0__2.0__ round__0.9__ |
| a man can row num__6 kmph in still water . when the river is running at num__1.2 kmph it takes him num__1 hour to row to a place and black . how far is the place ? <o> a ) num__2.87 <o> b ) num__2.88 <o> c ) num__2.8 <o> d ) num__2.86 <o> e ) num__2.81 |
m = num__6 s = num__1.2 ds = num__6 + num__1.2 = num__7.2 us = num__6 - num__1.2 = num__4.8 x / num__7.2 + x / num__4.8 = num__1 x = num__2.88 answer : b <eor> b <eos> |
b |
add__6.0__1.2__ subtract__6.0__1.2__ round__2.88__ |
add__6.0__1.2__ subtract__6.0__1.2__ divide__2.88__1.0__ |
| the average age of husband wife and their child num__3 years ago was num__27 years and that of wife and the child num__5 years ago was num__20 years . the present age of the husband is ? <o> a ) num__41 years <o> b ) num__40 years <o> c ) num__76 years <o> d ) num__14 years <o> e ) num__16 years |
sum of the present ages of husband wife and child = ( num__27 * num__3 + num__3 * num__3 ) = num__90 years . sum of the present age of wife and child = ( num__20 * num__2 + num__5 * num__2 ) = num__50 years . husband ' s present age = ( num__90 - num__50 ) = num__40 years . answer : b <eor> b <eos> |
b |
subtract__5.0__3.0__ multiply__20.0__2.0__ multiply__20.0__2.0__ |
subtract__5.0__3.0__ multiply__20.0__2.0__ multiply__20.0__2.0__ |
| a and b go around a circular track of length num__200 m on a cycle at speeds of num__36 kmph and num__72 kmph . after how much time will they meet for the first time at the starting point ? <o> a ) num__120 sec <o> b ) num__198 sec <o> c ) num__178 sec <o> d ) num__20 sec <o> e ) num__276 sec |
time taken to meet for the first time at the starting point = lcm { length of the track / speed of a length of the track / speed of b } = lcm { num__200 / ( num__36 * num__0.277777777778 ) num__200 / ( num__72 * num__0.277777777778 ) } = lcm ( num__20 num__10 ) = num__20 sec . answer : d <eor> d <eos> |
d |
divide__200.0__20.0__ round__20.0__ |
divide__200.0__20.0__ divide__200.0__10.0__ |
| the quantities s and t are positive and are related by the equation s = k / t where k is a constant . if the value of s increases by num__40 percent then the value of t decreases by what percent ? <o> a ) num__25.0 <o> b ) num__33 ¹ / ₃ % <o> c ) num__25.0 <o> d ) num__66 ² / ₃ % <o> e ) num__75 % |
we can assign numbers : lets say s = num__2 k = num__6 ( constant ) and t = num__3 ( so that s = k / t ) now increasing s by num__40.0 gives s = num__4 k remains constant so t = num__2 ( num__3.0 = num__3 ) decrease in t = num__1 percent decrease in t = ( num__0.25 ) * num__100 = num__25.0 answer c <eor> c <eos> |
c |
divide__6.0__2.0__ subtract__6.0__2.0__ subtract__3.0__2.0__ reverse__4.0__ multiply__0.25__100.0__ multiply__0.25__100.0__ |
divide__6.0__2.0__ subtract__6.0__2.0__ subtract__3.0__2.0__ reverse__4.0__ multiply__0.25__100.0__ multiply__0.25__100.0__ |
| the radius of a semi circle is num__6.4 cm then its perimeter is ? <o> a ) num__32.9 <o> b ) num__32.4 <o> c ) num__22.4 <o> d ) num__32.8 <o> e ) num__32.1 |
num__5.14285714286 r = num__6.4 = num__32.9 answer : a <eor> a <eos> |
a |
round__32.9__ |
round__32.9__ |
| if num__111 = num__09 num__444 = num__12 num__777 = num__15 then num__888 = ? ? ? <o> a ) num__18 <o> b ) num__11 <o> c ) num__17 <o> d ) num__19 <o> e ) num__15 |
e num__15 one + one + one ( num__3 + num__3 + num__3 ) = num__09 four + four + four ( num__4 + num__4 + num__4 ) = num__12 seven + seven + seven ( num__5 + num__5 + num__5 ) = num__15 therefore eight + eight + eight ( num__5 + num__5 + num__5 ) = num__15 <eor> e <eos> |
e |
subtract__12.0__9.0__ divide__444.0__111.0__ subtract__9.0__4.0__ add__12.0__3.0__ |
subtract__12.0__9.0__ divide__444.0__111.0__ subtract__9.0__4.0__ add__12.0__3.0__ |
| the ratio between the length and the breadth of a rectangular park is num__3 : num__2 . if a man cycling along the boundary of the park at the speed of num__12 km / hr completes one round in num__10 minutes then the area of the park ( in sq . m ) is <o> a ) num__153200 <o> b ) num__240000 <o> c ) num__153600 <o> d ) num__154000 <o> e ) num__154200 |
perimeter = distance covered in num__10 min . = num__12000 x num__10 m = num__2000 m . num__60 let length = num__3 x metres and breadth = num__2 x metres . then num__2 ( num__3 x + num__2 x ) = num__2000 or x = num__200 . length = num__600 m and breadth = num__400 m . area = ( num__600 x num__400 ) m num__2 = num__240000 m num__2 . b <eor> b <eos> |
b |
hour_to_min_conversion__ divide__12000.0__60.0__ multiply__3.0__200.0__ multiply__2.0__200.0__ multiply__400.0__600.0__ round__240000.0__ |
hour_to_min_conversion__ divide__12000.0__60.0__ multiply__3.0__200.0__ multiply__2.0__200.0__ multiply__400.0__600.0__ round__240000.0__ |
| two trains num__140 m and num__160 m long run at the speed of num__60 km / hr and num__40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? <o> a ) num__10.5 <o> b ) num__10.0 <o> c ) num__10.3 <o> d ) num__10.8 <o> e ) num__10.2 |
relative speed = num__60 + num__40 = num__100 km / hr . = num__100 * num__0.277777777778 = num__27.7777777778 m / sec . distance covered in crossing each other = num__140 + num__160 = num__300 m . required time = num__300 * num__0.036 = num__10.8 = num__10.8 sec . answer : d <eor> d <eos> |
d |
subtract__140.0__40.0__ add__140.0__160.0__ multiply__300.0__0.036__ round__10.8__ |
add__60.0__40.0__ add__140.0__160.0__ multiply__300.0__0.036__ multiply__300.0__0.036__ |
| there are three vessels of equal capacity . vessel a contains milk and water in the ratio num__7 : num__3 ; vessel b contains milk and water in the ratio num__2 : num__1 and vessel c contains milk and water in the ratio num__3 : num__2 . if the mixture in all the three vessels is mixed up . what will be the ratio of water to milk ? <o> a ) num__0.525423728814 <o> b ) num__1.90322580645 <o> c ) num__0.290322580645 <o> d ) num__3.44444444444 <o> e ) num__0.349206349206 |
num__7 : num__3 = > num__7 x + num__3 x = num__10 x num__2 : num__1 = > num__2 y + num__1 y = num__3 y num__3 : num__2 = > num__3 z + num__2 z = num__5 z num__10 x = num__3 y = num__5 z take lcm of num__10 num__35 = num__30 or simply ; x = num__3 y = num__10 z = num__6 so ratio of water : milk = ( num__3 x + y + num__2 z ) / ( num__7 x + num__2 y + num__3 z ) = num__0.525423728814 ans : a <eor> a <eos> |
a |
add__7.0__3.0__ subtract__7.0__2.0__ multiply__7.0__5.0__ multiply__3.0__10.0__ subtract__7.0__1.0__ multiply__1.0__0.5254__ |
add__7.0__3.0__ add__3.0__2.0__ multiply__7.0__5.0__ multiply__3.0__10.0__ add__1.0__5.0__ divide__0.5254__1.0__ |
| a scooter costs num__25000 when it is brand new . at the end of each year its value is only num__80.0 of what it was at the beginning of the year . what is the value of the scooter at the end of num__3 years ? <o> a ) num__10000 <o> b ) num__12500 <o> c ) num__12800 <o> d ) num__12000 <o> e ) none of these |
after first year the value of the scooter = num__20000 after second year the value of scooter = num__16000 after third year the value of scooter = num__12800 answer c <eor> c <eos> |
c |
percent__80.0__25000.0__ percent__80.0__20000.0__ percent__80.0__16000.0__ percent__80.0__16000.0__ |
percent__80.0__25000.0__ percent__80.0__20000.0__ percent__80.0__16000.0__ percent__80.0__16000.0__ |
| sakshi can do a piece of work in num__10 days . tanya is num__25.0 more efficient than sakshi . the number of days taken by tanya to do the same piece of work : <o> a ) num__8 <o> b ) num__16 <o> c ) num__18 <o> d ) num__25 <o> e ) num__10 |
solution ratio of times taken by sakshi and tanya = num__125 : num__100 = num__5 : num__4 . suppose tanya taken x days to do the work . num__5 : num__4 : : num__10 : x ⇒ x = ( num__10 x num__0.8 ) ⇒ x = num__8 days . hence tanya takes num__8 days is complete the work . answer a <eor> a <eos> |
a |
subtract__125.0__25.0__ divide__125.0__25.0__ divide__100.0__25.0__ divide__4.0__5.0__ multiply__10.0__0.8__ round__8.0__ |
subtract__125.0__25.0__ divide__125.0__25.0__ divide__100.0__25.0__ divide__4.0__5.0__ multiply__10.0__0.8__ round__8.0__ |
| a man buys an article for $ num__10 . and sells it for $ num__15 . find the gain percent ? <o> a ) num__25.0 <o> b ) num__50.0 <o> c ) num__20.0 <o> d ) num__15.0 <o> e ) num__30 % |
c . p . = $ num__10 s . p . = $ num__15 gain = $ num__5 gain % = num__0.5 * num__100 = num__50.0 answer is b <eor> b <eos> |
b |
percent__10.0__5.0__ percent__100.0__50.0__ |
percent__10.0__5.0__ percent__100.0__50.0__ |
| the number of new words that can be formed by rearranging the letters of the word ' ram ' is ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__10 <o> d ) num__13 <o> e ) num__15 |
number of words which can be formed = num__3 ! - num__1 = num__6 - num__1 = num__5 answer : a <eor> a <eos> |
a |
die_space__ vowel_space__ vowel_space__ |
die_space__ vowel_space__ vowel_space__ |
| find the unit ’ s digit in the product ( num__76 ^ num__41 ) * ( num__41 ^ num__14 ) * ( num__14 ^ num__87 ) * ( num__87 ^ num__76 ) <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
units digit of num__87 ^ num__76 = num__1 because num__87 ^ num__3 n + num__1 has units place as num__1 units digit of num__14 ^ num__87 = num__4 because num__14 ^ ( some odd no . ) has units place as num__4 units digit of num__41 ^ num__14 = num__1 because num__41 ^ ( some even no . ) has units place as num__1 units digit of num__76 ^ num__41 = num__6 because num__76 ^ ( any positive no . ) has units place as num__6 so units digit of the product is = ` ` num__4 ` ` ( units digit of num__1 * num__4 * num__1 * num__6 ) answer : d <eor> d <eos> |
d |
add__1.0__3.0__ add__1.0__3.0__ |
add__1.0__3.0__ add__1.0__3.0__ |
| a straight line is formed using two points a ( num__33 ) and b ( num__66 ) . another point p ( x y ) lies between a and b such that ap / bp = num__7 . what are the coordinates of p ? <o> a ) ( num__0.625 num__0.625 ) <o> b ) ( num__1.875 num__1.875 ) <o> c ) ( num__5.625 num__5.625 ) <o> d ) ( num__2.625 num__2.625 ) <o> e ) ( num__6.625 num__6.625 ) |
the equation of the straight line is y = x so the x - and y - coordinates of p are the same . the x y coordinates are num__0.875 of the distance from num__3 to num__6 . x = num__3 + ( num__6 - num__3 ) * ( num__0.875 ) = num__3 + num__2.625 = num__5.625 = y the answer is c . <eor> c <eos> |
c |
multiply__0.875__3.0__ add__2.625__3.0__ add__2.625__3.0__ |
multiply__0.875__3.0__ add__2.625__3.0__ add__2.625__3.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__3 sec . what is the length of the train ? <o> a ) num__56 <o> b ) num__78 <o> c ) num__27 <o> d ) num__79 <o> e ) num__50 |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__3 = num__50 m answer : e <eor> e <eos> |
e |
round__50.0__ |
round__50.0__ |
| if num__4 num__0.666666666667 is subtracted form num__10 num__0.5 and the difference is multiplied by num__450 then what is the final answer ? <o> a ) num__3389 <o> b ) num__3800 <o> c ) num__2378 <o> d ) num__2203 <o> e ) num__2625 |
final answer = ( num__10 num__0.5 - num__4 num__0.666666666667 ) * num__450 = num__5 num__0.833333333333 * num__450 = num__5 * num__450 + num__0.833333333333 * num__450 = num__2250 + num__375 = num__2625 . answer : e <eor> e <eos> |
e |
multiply__10.0__0.5__ multiply__450.0__5.0__ add__375.0__2250.0__ add__375.0__2250.0__ |
multiply__10.0__0.5__ multiply__450.0__5.0__ add__375.0__2250.0__ add__375.0__2250.0__ |
| the average of ten numbers is num__7 . if each number is multiplied by num__11 then what is the average of the new set of numbers ? <o> a ) num__11 <o> b ) num__18 <o> c ) num__77 <o> d ) num__110 <o> e ) num__121 |
the sum of the ten numbers is num__7 * num__10 = num__70 if each number is multiplied by num__11 the new sum is num__11 * ( num__70 ) . the average is then num__11 * num__7 = num__77 the answer is c . <eor> c <eos> |
c |
multiply__7.0__10.0__ multiply__7.0__11.0__ multiply__7.0__11.0__ |
multiply__7.0__10.0__ multiply__7.0__11.0__ multiply__7.0__11.0__ |
| two cars cover the same distance at the speed of num__80 and num__84 kmps respectively . find the distance traveled by them if the slower car takes num__1 hour more than the faster car ? <o> a ) num__1350 <o> b ) num__1000 <o> c ) num__1200 <o> d ) num__1400 <o> e ) num__1100 |
num__80 ( x + num__1 ) = num__84 x x = num__20 num__60 * num__20 = num__1200 km answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ multiply__20.0__60.0__ round__1200.0__ |
hour_to_min_conversion__ multiply__20.0__60.0__ multiply__1.0__1200.0__ |
| num__8 years ago there were num__5 members in the arthur ' s family and then the average age of the family was num__36 years . mean while arthur got married and gave birth to a child . still the average age of his family is same now . what is the age of his wife at the time of his child ' s birth was . if the difference between the age of her child and herself was num__26 years . <o> a ) num__15 years <o> b ) num__26 years <o> c ) num__16 years <o> d ) num__11 years <o> e ) num__12 years |
explanation : since we know that the difference b / w the age of any two persons remains always constant while the ratio of their ages gets changed as the time changes . so if the age of his child be x ( presently ) then the age of wife be x + num__26 ( presently ) thus the total age = x + ( x + num__26 ) = num__32 [ \ inline \ because num__252 - num__220 = num__32 ] \ inline \ rightarrow x = num__3 \ inline \ therefore the age of her child is num__3 years and her self is num__29 years . hence her age at the time of the birth of her child was num__26 years . answer : b <eor> b <eos> |
b |
subtract__252.0__32.0__ subtract__8.0__5.0__ add__26.0__3.0__ subtract__29.0__3.0__ |
subtract__252.0__32.0__ subtract__8.0__5.0__ add__26.0__3.0__ subtract__29.0__3.0__ |
| how many litres of pure acid are there in num__4 litres of a num__35.0 solution <o> a ) num__1.5 <o> b ) num__1.6 <o> c ) num__1.7 <o> d ) num__1.8 <o> e ) num__1.4 |
explanation : question of this type looks a bit typical but it is too simple as below . . . it will be num__8 * num__0.2 = num__1.4 answer : option e <eor> e <eos> |
e |
percent__4.0__35.0__ percent__4.0__35.0__ |
percent__4.0__35.0__ percent__4.0__35.0__ |
| a car starts from x and moves towards y . at the same time another car starts from y and moves towards x . both travel at a constant speed and meet after num__5 hours . during that time the faster car traveled num__0.666666666667 of the distance xy . how long would it take the slower car to travel the distance xy ? <o> a ) num__6 hours <o> b ) num__8 hours <o> c ) num__10 hours <o> d ) num__12 hours <o> e ) num__15 hours |
if one car travelled num__0.666666666667 then the other car must have travelled only num__0.333333333333 as they are meeting after num__5 hours in a certain point . so : ( num__0.333333333333 ) xy = num__4 * y - - > it took the car num__5 hours to travel num__0.333333333333 of the distance at a constant speed y . so if we solve this last equation : xy = num__3 * num__5 * y = num__15 * y - - > it will take this car num__15 hours in total to reach its final destination . answer : e <eor> e <eos> |
e |
multiply__5.0__3.0__ round__15.0__ |
multiply__5.0__3.0__ multiply__5.0__3.0__ |
| gopi gives rs . num__90 plus one turban as salary to his servant for one year . the servant leaves after num__9 months and receives rs . num__45 and the turban . find the price of the turban . <o> a ) num__27 <o> b ) num__36 <o> c ) num__29 <o> d ) num__90 <o> e ) num__11 |
let the price of turban be x . thus for one year the salary = ( num__90 + x ) for num__9 months he should earn num__3434 ( num__90 + x ) . now he gets one turban and rs . num__45 . thus num__3434 ( num__90 + x ) = num__45 + x or num__270 + num__3 x = num__180 + num__4 x or x = num__90 answer : d <eor> d <eos> |
d |
rectangle_perimeter__90.0__45.0__ square_perimeter__45.0__ triangle_area__45.0__4.0__ |
rectangle_perimeter__90.0__45.0__ square_perimeter__45.0__ triangle_area__45.0__4.0__ |
| what will come in place of the x in the following number series ? num__11 num__14 num__19 num__22 num__27 num__30 x <o> a ) num__23 <o> b ) num__26 <o> c ) num__36 <o> d ) num__35 <o> e ) num__45 |
( d ) the pattern is + num__3 + num__5 + num__3 + num__5 … … … … so the missing term is = num__30 + num__5 = num__35 . <eor> d <eos> |
d |
subtract__14.0__11.0__ subtract__19.0__14.0__ add__30.0__5.0__ add__30.0__5.0__ |
subtract__14.0__11.0__ subtract__19.0__14.0__ add__30.0__5.0__ add__30.0__5.0__ |
| meera purchased two three items from a shop . total price for three items is rs . num__6000 / - she have given rs . num__10000 / - what is the balance amount meera got ? <o> a ) num__6000 <o> b ) num__3500 <o> c ) num__5000 <o> d ) num__4000 <o> e ) num__7500 |
total cost of items : num__6000 / - amount paid : num__10000 / - balance receivable : num__10000 - num__6000 = num__4000 / - answer is d <eor> d <eos> |
d |
subtract__10000.0__6000.0__ subtract__10000.0__6000.0__ |
subtract__10000.0__6000.0__ subtract__10000.0__6000.0__ |
| ifaequals the sum of the even integers from num__2 to num__40 inclusive andbequals the sum of the odd integers from num__1 to num__39 inclusive what is the value of a - b ? <o> a ) num__1 <o> b ) num__10 <o> c ) num__19 <o> d ) num__20 <o> e ) num__21 |
this is a solution from beatthegmat : even numbers : ( num__40 - num__2 ) / num__2 + num__1 = num__20 even integers . ( num__40 + num__2 ) / num__2 = num__21 is the average of the even set . sum = avg * ( # of elements ) = num__21 * num__20 = num__420 = a odd numbers : ( num__39 - num__1 ) / num__2 + num__1 = num__20 odd integers . ( num__39 + num__1 ) / num__2 = num__20 is the average of the odd set . sum = avg * ( # of elements ) = num__20 * num__20 = num__400 = b a - b = num__420 - num__400 = num__20 . answer : d <eor> d <eos> |
d |
divide__40.0__2.0__ add__1.0__20.0__ multiply__20.0__21.0__ subtract__420.0__20.0__ divide__40.0__2.0__ |
divide__40.0__2.0__ add__1.0__20.0__ multiply__20.0__21.0__ subtract__420.0__20.0__ divide__40.0__2.0__ |
| a father said his son ` ` i was as old as you are at present at the time of your birth . ` ` if the father age is num__48 now the son age num__5 years back was <o> a ) num__14 <o> b ) num__17 <o> c ) num__11 <o> d ) num__19 <o> e ) num__99 |
let the son ' s present age be x years . then ( num__48 - x ) = x x = num__24 . son ' s age num__5 years back = ( num__24 - num__5 ) = num__19 years answer : d <eor> d <eos> |
d |
subtract__24.0__5.0__ subtract__24.0__5.0__ |
subtract__24.0__5.0__ subtract__24.0__5.0__ |
| what number has a num__5 : num__1 ratio to the number num__10 ? <o> a ) num__42 <o> b ) num__50 <o> c ) num__55 <o> d ) num__62 <o> e ) num__65 |
explanation : num__5 : num__1 = x : num__10 x = num__50 answer is b <eor> b <eos> |
b |
multiply__5.0__10.0__ multiply__5.0__10.0__ |
multiply__5.0__10.0__ multiply__5.0__10.0__ |
| num__3 - num__3 * num__6 + num__2 = <o> a ) num__2 <o> b ) - num__13 <o> c ) - num__18 <o> d ) - num__17 <o> e ) none of the above |
it is easy and you have to follow bodmas . . num__3 - num__3 * num__6 + num__2 = num__5 - num__18 = - num__13 . . . your oa - num__17 would be if the equation is num__3 - ( num__3 * num__6 + num__2 ) . . . . so the answer is - num__13 . answer : b <eor> b <eos> |
b |
add__3.0__2.0__ multiply__3.0__6.0__ subtract__18.0__5.0__ subtract__18.0__5.0__ |
add__3.0__2.0__ multiply__3.0__6.0__ subtract__18.0__5.0__ subtract__18.0__5.0__ |
| in the parking lot there are num__28 vehicles num__18 of them are buses and the rest are cars . the color of num__14 vehicles is red of which num__11 are buses . how many cars can be found in the parking lot which are not colored red ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__11 |
the number of cars is num__28 - num__18 = num__10 . the number of red cars is num__14 - num__11 = num__3 . the number of cars which are not red is num__10 - num__3 = num__7 . the answer is a . <eor> a <eos> |
a |
subtract__28.0__18.0__ subtract__14.0__11.0__ subtract__18.0__11.0__ subtract__18.0__11.0__ |
subtract__28.0__18.0__ subtract__14.0__11.0__ subtract__18.0__11.0__ subtract__18.0__11.0__ |
| difference between the length & breadth of a rectangle is num__23 m . if its perimeter is num__206 m then its area is ? ? we have : ( l - b ) = num__23 and num__2 ( l + b ) = num__206 or ( l + b ) = num__103 ? <o> a ) num__1245 m ^ num__2 <o> b ) num__1345 m ^ num__2 <o> c ) num__2520 m ^ num__2 <o> d ) num__2560 m ^ num__2 <o> e ) num__2678 m ^ num__2 |
solving the two equations we get : l = num__63 and b = num__40 . area = ( l x b ) = ( num__63 x num__40 ) m num__2 = num__2520 m ^ num__2 c <eor> c <eos> |
c |
subtract__103.0__63.0__ multiply__40.0__63.0__ multiply__40.0__63.0__ |
subtract__103.0__63.0__ multiply__40.0__63.0__ multiply__40.0__63.0__ |
| a train crosses a platform of num__170 m in num__15 sec same train crosses another platform of length num__250 m in num__20 sec . then find the length of the train ? <o> a ) a ) num__150 m <o> b ) b ) num__170 m <o> c ) c ) num__180 m <o> d ) d ) num__70 m <o> e ) e ) num__350 m |
length of the train be ‘ x ’ x + num__11.3333333333 = x + num__12.5 num__4 x + num__680 = num__3 x + num__750 x = num__70 m answer : d <eor> d <eos> |
d |
divide__170.0__15.0__ divide__250.0__20.0__ multiply__170.0__4.0__ multiply__250.0__3.0__ subtract__750.0__680.0__ round__70.0__ |
divide__170.0__15.0__ divide__250.0__20.0__ multiply__170.0__4.0__ multiply__250.0__3.0__ subtract__750.0__680.0__ round__70.0__ |
| in how many different ways can the letters of the word ' eve ' be arranged ? <o> a ) num__3 ways <o> b ) num__20 ways <o> c ) num__30 ways <o> d ) num__60 ways <o> e ) num__40 ways |
totally there are num__3 letters on the word eve . it contains num__2 e num__1 v . the no of ways is num__3 ! / num__2 ! = num__3 ways . answer : a <eor> a <eos> |
a |
coin_space__ choose__3.0__2.0__ |
coin_space__ choose__3.0__2.0__ |
| mohan ' s salary was first increased by num__20.0 and then decreased by num__20.0 . if his present salary is rs . num__7200 then what was his original salary ? <o> a ) num__7508 <o> b ) num__7500 <o> c ) num__7506 <o> d ) num__7566 <o> e ) num__7508 |
let mohan ' s salary be rs . num__100 . when increased by num__20.0 mohan ' s salary = rs . num__120 again when decreased by num__20.0 mohan ' s salary = num__120 - num__24 = rs . num__96 . but present salary is rs . num__7200 for num__96 - - - > num__100 ; num__7200 - - - > ? required salary is num__75.0 * num__100 = rs . num__7500 answer : b <eor> b <eos> |
b |
add__20.0__100.0__ subtract__120.0__24.0__ divide__7200.0__96.0__ multiply__100.0__75.0__ multiply__100.0__75.0__ |
add__20.0__100.0__ subtract__120.0__24.0__ divide__7200.0__96.0__ multiply__100.0__75.0__ multiply__100.0__75.0__ |
| in an election between two candidates num__70.0 of the voters cast their votes out of which num__4.0 of the votes were declared invalid . a candidate got num__65520 votes which were num__75.0 of the total valid votes . find the total number of votes enrolled in that election . <o> a ) num__130000 <o> b ) num__12500 <o> c ) num__14000 <o> d ) num__12000 <o> e ) none of these |
explanation : solution : let the total number of votes enrolled be x . then number of votes cast = num__70.0 of x . valid votes = num__96.0 of ( num__70.0 of x ) . . ' . num__75.0 of ( num__96.0 of ( num__70.0 of of x ) ) = num__65520 . ( num__0.75 * num__0.96 * num__0.7 * x ) = num__65520 . = > x = ( num__65520 * num__100 * num__100 * num__100 ) / ( num__75 * num__96 * num__70 ) = num__130000 answer : a <eor> a <eos> |
a |
percent__100.0__130000.0__ |
percent__100.0__130000.0__ |
| how many values of c in x ^ num__2 - num__5 x + c result in rational roots which are integers ? <o> a ) num__23 <o> b ) num__27 <o> c ) num__25 <o> d ) num__12 <o> e ) num__82 |
explanation : by the quadratic formula the roots of x num__2 − num__5 x + c = num__0 x num__2 − num__5 x + c = num__0 are − ( − num__5 ) ± − num__52 − num__4 ( num__1 ) ( c ) − − − − − − − − − − − √ num__2 ( num__1 ) − ( − num__5 ) ± − num__52 − num__4 ( num__1 ) ( c ) num__2 ( num__1 ) = num__5 ± num__25 − num__4 c − − − − − − √ num__25 ± num__25 − num__4 c num__2 to get rational roots num__25 − num__4 c num__25 − num__4 c should be square of an odd number . why ? because num__5 + odd only divided by num__2 perfectly . now let num__25 - num__4 c = num__1 then c = num__6 if num__25 - num__4 c = num__9 then c = num__4 if num__25 - num__4 c = num__25 then c = num__0 and so on . . . so infinite values are possible . answer : c <eor> c <eos> |
c |
subtract__5.0__4.0__ add__2.0__4.0__ add__5.0__4.0__ multiply__1.0__25.0__ |
subtract__5.0__4.0__ add__2.0__4.0__ add__5.0__4.0__ multiply__1.0__25.0__ |
| ? x num__48 = num__173 x num__240 <o> a ) num__545 <o> b ) num__685 <o> c ) num__865 <o> d ) num__495 <o> e ) num__534 |
let y x num__48 = num__173 x num__240 then y = ( num__173 x num__240 ) / num__48 = num__173 x num__5 = num__865 . answer : c <eor> c <eos> |
c |
divide__240.0__48.0__ multiply__173.0__5.0__ multiply__173.0__5.0__ |
divide__240.0__48.0__ multiply__173.0__5.0__ multiply__173.0__5.0__ |
| what percent of a day is five hours ? <o> a ) num__20.83 <o> b ) num__16 x num__0.5 % <o> c ) num__18 x num__0.666666666667 % <o> d ) num__22 x num__0.5 % <o> e ) none of these |
explanation : required percentage = ( num__0.000207468879668 ) % = ( num__20.8333333333 ) % . = num__20.83 . answer : a <eor> a <eos> |
a |
round__20.8333__ round__20.8333__ |
round__20.8333__ round__20.8333__ |
| a person can swim in still water at num__4 km / h . if the speed of water num__2 km / h how many hours will the man take to swim back against the current for num__6 km ? <o> a ) num__3 <o> b ) num__7 <o> c ) num__5 <o> d ) num__9 <o> e ) num__6 |
m = num__4 s = num__2 us = num__4 - num__2 = num__2 d = num__6 t = num__3.0 = num__3 answer : a <eor> a <eos> |
a |
divide__6.0__2.0__ round__3.0__ |
divide__6.0__2.0__ subtract__6.0__3.0__ |
| train x and train y pass one another traveling in opposite directions . twelve minutes later they are num__100 miles apart . if train x ’ s constant speed is num__30 miles per hour greater than train y ’ s how far does train x travel during that time ? <o> a ) num__28 <o> b ) num__40 <o> c ) num__53 <o> d ) num__72 <o> e ) num__80 |
let the rate of the train y be r then the rate of the train x will be r + num__30 ( given ) as per relative rate concept the rate at which they are increasing the distance between them is r + ( r + num__30 ) [ add the rates ] i . e . num__2 r + num__30 d = num__100 and t = num__12 min i . e num__0.2 hr using rtd table ( num__2 r + num__30 ) * num__0.2 = num__100 = = > r = num__235 miles / hr so the rate of train x is num__265 miles / hr ( since r + num__30 ) the distance traveled by train x in num__12 min is r * t = num__265 * num__0.2 = num__53 miles ( c ) <eor> c <eos> |
c |
add__30.0__235.0__ multiply__0.2__265.0__ round__53.0__ |
add__30.0__235.0__ multiply__0.2__265.0__ multiply__0.2__265.0__ |
| the product of two numbers is num__9375 and the quotient when the larger one is divided by the smaller is num__15 . the sum of the numbers is ? <o> a ) num__200 <o> b ) num__400 <o> c ) num__450 <o> d ) num__500 <o> e ) none |
let the numbers be x and y . then xy = num__9375 and x = num__15 . y xy = num__9375 ( x / y ) num__15 y num__2 = num__625 . y = num__25 . x = num__15 y = ( num__15 x num__25 ) = num__375 . sum of the numbers = x + y = num__375 + num__25 = num__400 . option is b <eor> b <eos> |
b |
divide__9375.0__15.0__ divide__9375.0__25.0__ add__375.0__25.0__ add__375.0__25.0__ |
divide__9375.0__15.0__ divide__9375.0__25.0__ add__375.0__25.0__ add__375.0__25.0__ |
| on a trip a cyclist averaged num__12 miles per hour for the first num__18 miles and num__10 miles per hour for the remaining num__18 miles . if the cyclist returned immediately via the same route and took a total of num__7.3 hours for the round trip what was the average speed ( in miles per hour ) for the return trip ? <o> a ) num__8.4 <o> b ) num__8.6 <o> c ) num__8.8 <o> d ) num__9 <o> e ) num__9.2 |
the time to go num__36 miles was num__1.5 + num__1.8 = num__1.5 + num__1.8 = num__3.3 hours . the average speed for the return trip was num__36 miles / num__4 hours = num__9 mph . the answer is d . <eor> d <eos> |
d |
divide__18.0__12.0__ divide__18.0__10.0__ add__1.5__1.8__ subtract__7.3__3.3__ divide__36.0__4.0__ round__9.0__ |
divide__18.0__12.0__ divide__18.0__10.0__ add__1.5__1.8__ subtract__7.3__3.3__ divide__36.0__4.0__ divide__36.0__4.0__ |
| if the radii of umbra and penumbra cast by an object on a wall are of the ratio num__2 : num__6 what is the area of the penumbra ring around the umbra of the latter ’ s radius is num__40 cms ? <o> a ) num__40288.57 centimeter square <o> b ) num__40388.57 cm ^ num__2 <o> c ) num__40488.57 cm ^ num__2 <o> d ) num__40588.57 cm ^ num__2 <o> e ) num__40688.57 cm ^ num__2 |
et the radius of umbra and penumbra are num__2 k and num__6 k . then as given radius of umbra = num__40 cm so num__2 k = num__40 k = num__20 radius of penumbra = num__20 * num__6 = num__120 area of penumbra ring around the umbra = area of penumbra - area of umbra num__3.14285714286 * [ ( num__120 ) ^ num__2 - ( num__40 ) ^ num__2 ] = num__40288.57 cm ^ num__2 answer : a <eor> a <eos> |
a |
triangle_area__6.0__40.0__ triangle_area__2.0__40288.57__ |
multiply__6.0__20.0__ triangle_area__2.0__40288.57__ |
| a circle in the coordinate plane passes through points ( - num__3 - num__2 ) and ( num__1 - num__4 ) . what is the smallest possible area of that circle ? <o> a ) num__13 π <o> b ) num__26 π <o> c ) num__262 √ π <o> d ) num__5 π <o> e ) num__64 π |
the distance between the two points is sqrt ( num__20 ) . radius = sqrt ( num__20 ) / num__2 area = pi * ( sqrt ( num__20 ) / num__2 ) ^ num__2 d . num__5 π <eor> d <eos> |
d |
triangle_area__2.0__5.0__ |
power__5.0__1.0__ |
| a cube with volume of num__125 cm ^ num__3 is dropped in a fish tank with dimensions num__3 m x num__1 m x num__2 m . if the tank is full of water how much water will remain after the cube is droppped ? <o> a ) num__5.9999999 m ^ num__3 <o> b ) num__5.725 m ^ num__3 <o> c ) num__5.650 m ^ num__3 <o> d ) num__5.999875 m ^ num__3 <o> e ) num__5999.725 m ^ num__3 |
volume of fish tank = num__3 m x num__1 m x num__2 m = num__6 m ^ num__3 num__125 cm ^ num__3 x ( num__1 m ^ num__3.0 num__000000 cm ^ num__3 ) = num__0.000125 m ^ num__3 num__6 - . num__000125 = num__5.999875 m ^ num__3 answer is d <eor> d <eos> |
d |
multiply__3.0__2.0__ subtract__6.0__0.0001__ multiply__1.0__5.9999__ |
multiply__3.0__2.0__ subtract__6.0__0.0001__ subtract__6.0__0.0001__ |
| two pipes can fill a tank in num__20 minutes and num__15 minutes . an outlet pipe can empty the tank in num__10 minutes . if all the pipes are opened when the tank is empty then how many minutes will it take to fill the tank ? <o> a ) num__36 <o> b ) num__42 <o> c ) num__48 <o> d ) num__54 <o> e ) num__60 |
let v be the volume of the tank . the rate per minute at which the tank is filled is : v / num__20 + v / num__15 - v / num__10 = v / num__60 per minute the tank will be filled in num__60 minutes . the answer is e . <eor> e <eos> |
e |
hour_to_min_conversion__ hour_to_min_conversion__ |
hour_to_min_conversion__ hour_to_min_conversion__ |
| the fourteen digits of a credit card are to be written in the boxes shown above . if the sum of every three consecutive digits is num__18 then the value of x is : <o> a ) num__1 <o> b ) num__3 <o> c ) num__8 <o> d ) num__7 <o> e ) num__9 |
sol : a explanation : let us assume right most two squares are a b then sum of all the squares = num__18 x num__4 + a + b . . . . . . . . . . ( num__1 ) also sum of the squares before num__7 = num__18 sum of the squares between num__7 x = num__18 and sum of the squares between x num__8 = num__18 so sum of the num__14 squares = num__18 + num__7 + num__18 + x + num__18 + num__8 + a + b ( num__2 ) equating num__1 and num__2 we get x = num__3 answer : b <eor> b <eos> |
b |
add__1.0__7.0__ subtract__18.0__4.0__ divide__8.0__4.0__ add__1.0__2.0__ add__1.0__2.0__ |
add__1.0__7.0__ subtract__18.0__4.0__ divide__8.0__4.0__ add__1.0__2.0__ add__1.0__2.0__ |
| what is the rate percent when the simple interest on rs . num__800 amount to rs . num__192 in num__4 years ? <o> a ) num__5.0 <o> b ) num__6.0 <o> c ) num__3.0 <o> d ) num__9.0 <o> e ) num__1 % |
num__192 = ( num__800 * num__4 * r ) / num__100 r = num__6.0 answer : b <eor> b <eos> |
b |
percent__6.0__100.0__ |
percent__6.0__100.0__ |
| a pump can fill a tank with a water in num__2 hours . because of a leak it took num__3 and num__0.333333333333 hours to fill the tank . the leak can drain all the water of the full tank in how many hours ? <o> a ) num__5 hrs <o> b ) num__7 hrs <o> c ) num__12 hrs <o> d ) num__14 hrs <o> e ) num__16 hrs |
the rate of the pump + leak = num__0.3 num__0.5 - leak ' s rate = num__0.3 leak ' s rate = num__0.2 the leak will empty the tank in num__5 hours . the answer is a . <eor> a <eos> |
a |
subtract__0.5__0.3__ add__2.0__3.0__ round__5.0__ |
subtract__0.5__0.3__ add__2.0__3.0__ add__2.0__3.0__ |
| a man counted his animals num__80 heads and num__250 legs ( ducks and goats ) . how many goats are there ? <o> a ) num__30 <o> b ) num__40 <o> c ) num__50 <o> d ) num__45 <o> e ) num__70 |
let no of ducks = d and no of goat = g d + g = num__80 ( heads ) - - - - - - - - - - > eq num__1 each duck has num__2 legs and goat has num__4 legs num__2 d + num__4 g = num__250 and divide num__2 d + num__4 g = num__250 by num__2 we get d + num__2 g = num__125 - - - - - - - - - - - > eq num__2 subtract eq num__1 from eq num__2 we get no of goats = num__45 answer : d <eor> d <eos> |
d |
divide__250.0__2.0__ subtract__125.0__80.0__ multiply__1.0__45.0__ |
divide__250.0__2.0__ subtract__125.0__80.0__ subtract__125.0__80.0__ |
| to asphalt num__1 km road num__30 men spent num__12 days working num__8 hours per day . how many days num__20 men will spend to asphalt a road of num__2 km working num__8 hours a day ? <o> a ) num__40 <o> b ) num__36 <o> c ) num__42 <o> d ) num__44 <o> e ) num__46 |
man - hours required to asphalt num__1 km road = num__30 * num__12 * num__8 = num__2880 man - hours required to asphalt num__2 km road = num__2880 * num__2 = num__5760 man - hours available per day = num__20 * num__8 = num__160 therefore number of days = num__24.0 = num__36 days ans = b <eor> b <eos> |
b |
multiply__2.0__2880.0__ multiply__8.0__20.0__ multiply__12.0__2.0__ add__12.0__24.0__ round__36.0__ |
multiply__2.0__2880.0__ multiply__8.0__20.0__ multiply__12.0__2.0__ add__12.0__24.0__ multiply__1.0__36.0__ |
| a watch was sold at a loss of num__10.0 . if it was sold for rs . num__140 more there would have been a gain of num__4.0 . what is the cost price ? <o> a ) num__1000 <o> b ) num__2288 <o> c ) num__2777 <o> d ) num__2999 <o> e ) num__2711 |
num__90.0 num__104.0 - - - - - - - - num__14.0 - - - - num__140 num__100.0 - - - - ? = > rs . num__1000 answer : a <eor> a <eos> |
a |
percent__10.0__140.0__ percent__100.0__1000.0__ |
percent__10.0__140.0__ percent__100.0__1000.0__ |
| one - fourth of certain journey is covered at the rate of num__25 km / h one - third at the rate of num__30 km / h and the rest at num__50 km / h . find the average speed for the whole journey . <o> a ) num__11.320754717 km / h <o> b ) num__22.641509434 km / h <o> c ) num__33.9622641509 km / h <o> d ) num__30.1886792453 km / h <o> e ) none of these |
solution : let distance be num__120 km hence num__30 km is covered by @ num__25 kmph and num__40 km covered by @ num__30 kmph and rest num__50 km has been covered @ num__50 km . now average = ( num__120 / total time taken ) ; = { num__120 / [ ( num__1.2 ) + ( num__1.33333333333 ) + ( num__1.0 ) } ] = num__33.9622641509 = num__33.9622641509 km / h . answer : option c <eor> c <eos> |
c |
divide__30.0__25.0__ divide__40.0__30.0__ round_down__1.3333__ multiply__1.0__33.9623__ |
divide__30.0__25.0__ divide__40.0__30.0__ round_down__1.3333__ divide__33.9623__1.0__ |
| i travel the first part of my journey at num__40 kmph and the second part at num__60 kmph and cover the total distance of num__240 km to my destination in num__5 hours . how long did the first part of my journey last ? <o> a ) num__4 hours <o> b ) num__2 hours <o> c ) num__3 hours <o> d ) num__2 hours num__24 minutes <o> e ) none |
explanatory answer the total time of journey = num__5 hours . let ' x ' hours be the time that i traveled at num__40 kmph therefore num__5 - x hours would be time that i traveled at num__60 kmph . hence i would have covered x * num__40 + ( num__5 - x ) num__60 kms in the num__5 hours = num__240 kms . solving for x in the equation num__40 x + ( num__5 - x ) * num__60 = num__240 we get num__40 x + num__300 - num__60 x = num__240 = > num__20 x = num__60 or x = num__3 hours . answer c <eor> c <eos> |
c |
add__60.0__240.0__ subtract__60.0__40.0__ divide__60.0__20.0__ round__3.0__ |
add__60.0__240.0__ subtract__60.0__40.0__ divide__60.0__20.0__ round__3.0__ |
| a popular website requires users to create a password consisting of the digits { num__1 num__23 num__45 num__67 num__89 } . if no digit may be repeated and each password must be at least num__8 digits long how many passwords are possible ? <o> a ) num__8 ! + num__9 ! <o> b ) num__2 x num__9 ! <o> c ) num__8 ! x num__9 ! <o> d ) num__17 ! <o> e ) num__18 ! |
if we choose the num__9 different digits then they can be arranged in num__9 ! ways . the number of possible num__8 - digit passwords is num__9 * num__8 * . . . * num__2 = num__9 ! the total number of passwords is num__9 ! + num__9 ! = num__2 * num__9 ! the answer is b . <eor> b <eos> |
b |
coin_space__ coin_space__ |
coin_space__ coin_space__ |
| the price of an item is discounted num__5 percent on day num__1 of a sale . on day num__2 the item is discounted another num__5 percent and on day num__3 it is discounted an additional num__10 percent . the price of the item on day num__3 is what percentage of the sale price on day num__1 ? <o> a ) num__80.0 <o> b ) num__82.5 <o> c ) num__83.0 <o> d ) num__84.0 <o> e ) num__85.5 % |
let initial price be num__100 price in day num__1 after num__5.0 discount = num__95 price in day num__2 after num__5.0 discount = num__90.25 price in day num__3 after num__10.0 discount = num__81.23 so price in day num__3 as percentage of the sale price on day num__1 will be = num__81.23 / num__95 * num__100 = > num__85.5 answer will definitely be ( e ) <eor> e <eos> |
e |
percent__100.0__85.5__ |
percent__100.0__85.5__ |
| if the price of a certain computer increased num__20 percent from d dollars to num__351 dollars then num__2 d = <o> a ) num__540 <o> b ) num__585 <o> c ) num__619 <o> d ) num__649 <o> e ) num__700 |
before price increase price = d after num__30.0 price increase price = d + ( num__0.2 ) * d = num__1.2 d = num__351 ( given ) i . e . d = num__351 / num__1.3 = $ num__2925 i . e . num__2 d = num__2 * num__292.5 = num__585 answer : option b <eor> b <eos> |
b |
divide__351.0__1.2__ multiply__2.0__292.5__ multiply__2.0__292.5__ |
divide__351.0__1.2__ multiply__2.0__292.5__ multiply__2.0__292.5__ |
| after successive discounts of num__20.0 num__10.0 and num__5.0 a certain good is sold for rs . num__6500 . find the actual price of the good . <o> a ) s . num__6000 <o> b ) s . num__9000 <o> c ) s . num__10800 <o> d ) s . num__9503 <o> e ) s . num__9980 |
let actual price was num__100 . after three successive discount this will become num__100 = = num__20.0 discount = > num__80 = = num__10.0 discount = > num__72 = = num__5.0 discount = num__68.4 now compare num__68.4 = num__6500 num__1 = num__6500 / num__68.4 num__100 = ( num__6500 * num__100 ) / num__68.4 = rs . num__9503 . answer : option d <eor> d <eos> |
d |
percent__20.0__5.0__ percent__100.0__9503.0__ |
percent__20.0__5.0__ percent__100.0__9503.0__ |
| in the gure below arc sbt is one quarter of a circle with center r and radius r . the length plus the width of rectangle abcr is num__8 and the perimeter of the shaded region is num__10 + num__3 . find the value of r . <o> a ) num__6 <o> b ) num__6 : num__25 <o> c ) num__6 : num__5 <o> d ) num__6 : num__75 <o> e ) num__7 |
set x = rc and y = ar . it must then be x + y = num__8 ( by hypothesis ) and x num__2 + y num__2 = r num__2 ( since the rectangle abcr has one vertex in the circle ) . the perimeter of the shaded area is then as + sbt + tc + ac = ( r y ) + r = num__2 + ( r x ) + p x num__2 + y num__2 = num__3 r + r = num__2 ( x + y ) = num__3 r + r = num__2 num__8 . all we need to so is solve for r in the equation num__18 + num__3 = ( num__3 + num__2 ) r . correct answer a <eor> a <eos> |
a |
side_by_diagonal__10.0__8.0__ |
side_by_diagonal__10.0__8.0__ |
| list of salaries of num__8 employees are listed below . what is the median salary ? $ num__40000 $ num__29000 $ num__35500 $ num__31000 $ num__43000 $ num__30000 $ num__27000 $ num__33000 <o> a ) $ num__32000 <o> b ) $ num__28500 <o> c ) $ num__30200 <o> d ) $ num__30800 <o> e ) $ num__31 |
500 |
ordering the data from least to greatest we get : $ num__27000 $ num__29000 $ num__30000 $ num__31000 $ num__33000 $ num__35500 $ num__40000 $ num__43000 since there is an even number of items in the data set we compute the median by taking the mean of the two middlemost numbers . $ num__31000 + $ num__33000 = $ num__64000 the median salary is $ num__32000 . answer : a <eor> a <eos> |
a |
a |
| tharak covers a part of the journey at num__20 kmph and the balance at num__70 kmph taking total of num__8 hours to cover the distance of num__400 km . how many hours has been driving at num__20 kmph ? <o> a ) num__3 hours num__12 minute <o> b ) num__2 hours <o> c ) num__4 hours num__40 minutes <o> d ) num__3 hours num__20 minutes <o> e ) num__3 hours num__18 minute |
let x be no . of hours and he travelled at num__20 kmph . and balance distance ( num__8 - x ) hr at num__70 kmph . total distance = num__20 x + ( num__8 - x ) hr = num__400 num__20 x + ( num__8 - x ) num__70 = num__400 num__20 x + num__560 - num__70 x = num__400 num__20 x - num__70 x = num__400 - num__560 - num__50 x = - num__160 x = num__3.2 = num__3 hr num__12 minutes answer : a <eor> a <eos> |
a |
multiply__70.0__8.0__ subtract__70.0__20.0__ multiply__20.0__8.0__ divide__160.0__50.0__ subtract__20.0__8.0__ round__3.0__ |
multiply__70.0__8.0__ subtract__70.0__20.0__ subtract__560.0__400.0__ divide__160.0__50.0__ subtract__20.0__8.0__ round__3.0__ |
| find the product of place value and face value of num__5 in num__65231 <o> a ) num__28000 <o> b ) num__25000 <o> c ) num__27000 <o> d ) num__26000 <o> e ) num__28000 |
place value of num__5 in num__65231 = num__5 x num__1000 = num__5000 face value = num__5 required product = num__5000 x num__5 = num__25000 answer : b <eor> b <eos> |
b |
multiply__5.0__1000.0__ multiply__5.0__5000.0__ multiply__5.0__5000.0__ |
multiply__5.0__1000.0__ multiply__5.0__5000.0__ multiply__5.0__5000.0__ |
| the cost of num__4 bags and num__12 purses is rs . num__1520 what is the cost of num__10 bags and num__30 purses ? <o> a ) rs . num__3600 <o> b ) rs . num__3500 <o> c ) rs . num__3800 <o> d ) rs . num__3900 <o> e ) none of these |
explanation : cost of num__4 bags + num__12 purses = rs . num__1520 multiply each term by num__2.5 we get cost of num__10 bags + num__30 purses = rs . num__3800 . answer : option c <eor> c <eos> |
c |
divide__10.0__4.0__ multiply__1520.0__2.5__ multiply__1520.0__2.5__ |
divide__10.0__4.0__ multiply__1520.0__2.5__ multiply__1520.0__2.5__ |
| johnny travels a total of one hour to and from school . on the way there he jogs at num__4 miles per hour and on the return trip he gets picked up by the bus and returns home at num__20 miles per hour . how far is it to the school ? <o> a ) num__2 miles <o> b ) num__6.6 miles <o> c ) num__4.8 miles <o> d ) num__8 miles <o> e ) num__10 miles |
answer : b ) num__6.6 miles . average speed for round trip = num__2 * a * b / ( a + b ) where a b are speeds so average speed was = num__2 * num__4 * num__20 / ( num__4 + num__20 ) = num__6.6 m / hr the distance between schoolhome should be half of that . ie . num__6.6 miles answer b <eor> b <eos> |
b |
round__6.6__ |
round__6.6__ |
| if two numbers are in the ratio num__4 : num__7 . if num__10 is added to both of the numbers then the ratio becomes num__6 : num__10 then find the largest number ? <o> a ) num__80 <o> b ) num__40 <o> c ) num__140 <o> d ) num__100 <o> e ) none of these |
explanation : num__4 : num__7 num__4 x + num__10 : num__7 x + num__10 = num__6 : num__10 num__10 [ num__4 x + num__10 ] = num__6 [ num__7 x + num__10 ] num__40 x + num__100 = num__42 x + num__60 num__100 – num__60 = num__42 x – num__40 x num__2 x = num__40 x = num__20 then the first number is = num__4 num__7 x = num__140 short cut method : a : b = num__4 : num__7 c : d = num__6 : num__10 num__1 . cross multiplication with both ratios a * d ~ b * c = num__4 * num__10 ~ num__7 * num__6 = num__40 ~ num__42 = num__2 num__2 . if num__10 is added both the number means num__10 * num__6 = num__60 and num__10 * num__10 = num__100 then num__60 ~ num__100 = num__40 = = = > num__2 - - - - - > num__40 = = = > num__1 - - - - - > num__20 = = > num__7 - - - - - > num__140 answer : option c <eor> c <eos> |
c |
multiply__4.0__10.0__ multiply__7.0__6.0__ multiply__10.0__6.0__ subtract__6.0__4.0__ multiply__10.0__2.0__ multiply__7.0__20.0__ subtract__7.0__6.0__ multiply__7.0__20.0__ |
multiply__4.0__10.0__ multiply__7.0__6.0__ multiply__10.0__6.0__ subtract__6.0__4.0__ multiply__10.0__2.0__ multiply__7.0__20.0__ subtract__7.0__6.0__ multiply__7.0__20.0__ |
| how many num__3 - digit numbers are completely divisible num__6 ? <o> a ) num__25 <o> b ) num__78 <o> c ) num__93 <o> d ) num__150 <o> e ) num__14 |
explanation : num__3 - digit number divisible by num__6 are : num__102 num__108 num__114 . . . num__996 this is an a . p . in which a = num__102 d = num__6 and l = num__996 let the number of terms be n . then tn = num__996 . a + ( n - num__1 ) d = num__996 num__102 + ( n - num__1 ) x num__6 = num__996 num__6 x ( n - num__1 ) = num__894 ( n - num__1 ) = num__149 n = num__150 number of terms = num__150 . d ) <eor> d <eos> |
d |
add__6.0__102.0__ add__6.0__108.0__ subtract__996.0__102.0__ divide__894.0__6.0__ add__1.0__149.0__ add__1.0__149.0__ |
add__6.0__102.0__ add__6.0__108.0__ subtract__996.0__102.0__ divide__894.0__6.0__ add__1.0__149.0__ add__1.0__149.0__ |
| speed of a boat in standing water is num__10 kmph and speed of the stream is num__1.5 kmph . a man can rows to a place at a distance of num__105 km and comes back to the starting point . the total time taken by him is ? <o> a ) num__20.48 hours <o> b ) num__21.48 hours <o> c ) num__22.48 hours <o> d ) num__23.48 hours <o> e ) num__24.48 hours |
speed upstream = num__8.5 kmph speed downstream = num__11.5 kmph total time taken = num__105 / num__8.5 + num__105 / num__11.5 = num__21.48 hours answer is b <eor> b <eos> |
b |
subtract__10.0__1.5__ add__10.0__1.5__ round__21.48__ |
subtract__10.0__1.5__ add__10.0__1.5__ round__21.48__ |
| how many seconds will a num__750 meter long train moving with a speed of num__63 km / hr take to cross a man walking with a speed of num__3 km / hr in the direction of the train ? <o> a ) num__48 <o> b ) num__36 <o> c ) num__26 <o> d ) num__11 <o> e ) num__45 |
explanation : here distance d = num__750 mts speed s = num__63 - num__3 = num__60 kmph = num__60 x num__0.277777777778 m / s time t = = num__45 sec . answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ round__45.0__ |
subtract__63.0__3.0__ round__45.0__ |
| jane started baby - sitting when she was num__18 years old . whenever she baby - sat for a child that child was no more than half her age at the time . jane is currently num__34 years old and she stopped baby - sitting num__12 years ago . what is the current age of the oldest person for whom jane could have baby - sat ? <o> a ) num__20 <o> b ) num__21 <o> c ) num__22 <o> d ) num__23 <o> e ) num__25 |
check two extreme cases : jane = num__18 child = num__9 years ago = num__34 - num__18 = num__16 - - > child ' s age now = num__9 + num__16 = num__25 ; jane = num__22 child = num__11 years ago = num__34 - num__22 = num__12 - - > child ' s age now = num__11 + num__12 = num__23 . answer : e . <eor> e <eos> |
e |
subtract__34.0__18.0__ subtract__34.0__9.0__ subtract__34.0__12.0__ subtract__34.0__11.0__ subtract__34.0__9.0__ |
subtract__34.0__18.0__ subtract__34.0__9.0__ subtract__34.0__12.0__ subtract__34.0__11.0__ subtract__34.0__9.0__ |
| a candidate got num__35.0 of the votes polled and he lost to his rival by num__2250 votes . how many votes were cast ? <o> a ) num__7500 <o> b ) num__2028 <o> c ) num__2775 <o> d ) num__5496 <o> e ) num__6851 |
num__35.0 - - - - - - - - - - - l num__65.0 - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - num__30.0 - - - - - - - - - - num__2250 num__100.0 - - - - - - - - - ? = > num__7500 answer : a <eor> a <eos> |
a |
percent__100.0__7500.0__ |
percent__100.0__7500.0__ |
| it takes avery num__3 hours to build a brick wall while tom can do it in num__5 hours . if the two start working together and after an hour avery leaves how much time will it take tom to complete the wall on his own ? <o> a ) num__100 <o> b ) num__110 <o> c ) num__120 <o> d ) num__130 <o> e ) num__140 |
avery takes num__3 hours tom takes num__2 hours efficiency of avery is num__0.333333333333 units / hr efficiency of tom is num__0.2 units / hr combined efficiency of tom and avery is num__0.333333333333 + num__0.2 = num__0.533333333333 units / hr since they worked for num__1 hour they completed num__0.533333333333 units of work and num__0.466666666667 units of work is left which is to be completed by tom ( since avery left ) so time taken by tom to complete the remaining work will be num__0.466666666667 / num__0.2 hours = > num__2.33333333333 * num__60 = num__140 minutes . . . answer will be ( e ) <eor> e <eos> |
e |
subtract__5.0__3.0__ add__0.2__0.3333__ subtract__3.0__2.0__ subtract__1.0__0.5333__ add__2.0__0.3333__ hour_to_min_conversion__ round__140.0__ |
subtract__5.0__3.0__ add__0.2__0.3333__ multiply__5.0__0.2__ subtract__1.0__0.5333__ add__2.0__0.3333__ hour_to_min_conversion__ divide__140.0__1.0__ |
| a restaurant meal cost $ num__32.50 and there was no tax . if the tip was more than num__10 percent but less than num__15 percent of the cost of the meal then total amount paid must have been between : <o> a ) $ num__40 and $ num__42 <o> b ) $ num__39 and $ num__41 <o> c ) $ num__38 and num__40 <o> d ) $ num__37 and $ num__39 <o> e ) $ num__36 and $ num__38 |
let tip = t meal cost = num__32.50 range of tip = from num__10.0 of num__32.5 to num__15.0 of num__32.5 = num__3.55 to num__5.325 hence range of amount paid = num__32.5 + t = num__36.05 to num__37.825 answer : e <eor> e <eos> |
e |
add__32.5__3.55__ add__32.5__5.325__ round_down__36.05__ |
add__32.5__3.55__ add__32.5__5.325__ round_down__36.05__ |
| rs num__1000 is invested at num__5.0 pa simple interest . if interest is added to principal after every num__10 yr . then we will get amount num__2000 after how many year ? <o> a ) num__12 years . <o> b ) num__13.66 years . <o> c ) num__14 years . <o> d ) num__16.66 years . <o> e ) num__15 years |
p = num__1000 . r = num__5.0 . t = num__10 years . i = ptr / num__100 = ( num__1000 * num__5 * num__10 ) / num__100 = num__500 . now principal = num__1000 + num__500 = num__1500 . a = num__2000 i = num__500 r = num__5.0 t = ( num__100 * i ) / pr = ( num__100 * num__500 ) / ( num__1500 * num__5 ) = num__6.66 years . total time = num__10 + num__6.66 = num__16.66 years . answer : option d <eor> d <eos> |
d |
percent__5.0__2000.0__ percent__100.0__16.66__ |
percent__5.0__2000.0__ percent__100.0__16.66__ |
| which of the following functions does not intersect with t = num__3 x ^ num__2 + num__2 x + num__1 <o> a ) t = num__3 x ^ num__2 + num__3 x + num__1 <o> b ) t = num__2 x ^ num__2 + num__3 x + num__1 <o> c ) t = num__3 x ^ num__2 + num__2 x + num__3 <o> d ) t = num__4 x ^ num__2 + num__2 x - num__3 <o> e ) t = x ^ num__2 + num__2 x + num__3 |
let ’ s find the actual intersecting point of given function and each choice t . a . num__3 x ^ num__2 + num__2 x + num__1 = num__3 x ^ num__2 + num__3 x + num__1 - - > x = num__0 b . num__3 x ^ num__2 + num__2 x + num__1 = num__2 x ^ num__2 + num__3 x + num__1 - - > x ^ num__2 – x = num__0 - - > x = num__0 or num__1 c . num__3 x ^ num__2 + num__2 x + num__1 = num__3 x ^ num__2 + num__2 x + num__3 - - > num__1 = num__3 ( x ) - - > they don ’ t have any intersect . d . num__3 x ^ num__2 + num__2 x + num__1 = num__4 x ^ num__2 + num__2 x - num__3 - - > x ^ num__2 – num__4 = num__0 - - > x = num__2 or - num__2 e . num__3 x ^ num__2 + num__2 x + num__1 = x ^ num__2 + num__2 x + num__3 - - > num__2 x ^ num__2 – num__2 = num__0 - - > x = num__1 or – num__1 . so only choice ( c ) has no intersecting point . the answer is ( c ) . <eor> c <eos> |
c |
add__3.0__1.0__ multiply__3.0__1.0__ |
add__3.0__1.0__ add__2.0__1.0__ |
| num__40.0 of ram ' s marks is equal to num__20.0 of rahim ' s marks which percent is equal to num__30.0 of robert ' s marks . if robert ' s marks is num__80 then find the average marks of ram and rahim ? <o> a ) num__76 <o> b ) num__56 <o> c ) num__56 <o> d ) num__90 <o> e ) num__15 |
given num__40.0 of ram ' s marks = num__20.0 of rahim ' s marks = num__30.0 of robert ' s marks . given marks of robert = num__80 num__30.0 of num__80 = num__0.3 * num__8 = num__24 given num__40.0 of ram ' s marks = num__24 . = > ram ' s marks = ( num__24 * num__100 ) / num__40 = num__60 also num__20.0 of rahim ' s marks = num__24 = > rahim ' s marks = ( num__24 * num__100 ) / num__20 = num__120 average marks of ram and rahim = ( num__60 + num__120 ) / num__2 = num__90 . answer : d <eor> d <eos> |
d |
multiply__80.0__0.3__ add__20.0__80.0__ add__40.0__20.0__ add__40.0__80.0__ divide__40.0__20.0__ add__30.0__60.0__ add__30.0__60.0__ |
multiply__80.0__0.3__ add__20.0__80.0__ add__40.0__20.0__ add__40.0__80.0__ divide__40.0__20.0__ add__30.0__60.0__ add__30.0__60.0__ |
| the average of numbers num__0.64206 num__0.64207 num__0.64208 and num__0.64209 is ? <o> a ) num__0.64202 <o> b ) num__0.64204 <o> c ) num__0.642022 <o> d ) num__0.642075 <o> e ) none |
answer average = ( num__0.64206 + num__0.64207 + num__0.64208 + num__0.64209 ) / num__4 = num__2.5683 / num__4 = num__0.642075 correct option : d <eor> d <eos> |
d |
divide__2.5683__4.0__ |
divide__2.5683__4.0__ |
| in a class there are num__18 boys who are over num__160 cm tall . if these constitute three - fourths of the boys and the total number of boys is two - thirds of the total number of students in the class what is the number of girls in the class ? <o> a ) num__6 <o> b ) num__12 <o> c ) num__18 <o> d ) num__19 <o> e ) num__22 |
in a class there are num__18 boys who are over num__160 cm tall . if these constitute three - fourths of the boys then total no of boys = num__18 * num__1.33333333333 = num__24 boys . the total number of boys ( num__24 ) is two - thirds of the total number of students in the class then total students = num__24 * num__1.5 = num__36 no . of girls = num__36 - num__24 = num__12 answer : b <eor> b <eos> |
b |
multiply__1.5__24.0__ divide__18.0__1.5__ divide__18.0__1.5__ |
multiply__1.5__24.0__ subtract__36.0__24.0__ subtract__36.0__24.0__ |
| num__0.04 x num__0.0162 is equal to : <o> a ) num__6.48 x num__10 ( power - num__4 ) <o> b ) num__6.48 x num__10 ( power - num__3 ) <o> c ) num__9.48 x num__10 ( power - num__4 ) <o> d ) num__5.48 x num__10 ( power - num__4 ) <o> e ) num__6.08 x num__10 ( power - num__4 ) |
num__4 x num__162 = num__648 . sum of decimal places = num__6 . so num__0.04 x num__0.0162 = num__0.000648 = num__6.48 x num__10 ( power - num__4 ) answer is a . <eor> a <eos> |
a |
multiply__162.0__4.0__ multiply__0.04__0.0162__ multiply__0.04__162.0__ add__4.0__6.0__ multiply__0.04__162.0__ |
multiply__162.0__4.0__ multiply__0.04__0.0162__ multiply__0.04__162.0__ add__4.0__6.0__ multiply__0.04__162.0__ |
| the speed of a boat in still water is num__80 kmph and the speed of the current is num__20 kmph . find the speed downstream and upstream ? <o> a ) num__22 kmph <o> b ) num__40 kmph <o> c ) num__60 kmph <o> d ) num__15 kmph <o> e ) num__23 kmph |
speed downstream = num__80 + num__20 = num__100 kmph speed upstream = num__80 - num__20 = num__60 kmph answer : c <eor> c <eos> |
c |
add__80.0__20.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
add__80.0__20.0__ subtract__80.0__20.0__ subtract__80.0__20.0__ |
| at a certain supplier a machine of type a costs $ num__30000 and a machine of type b costs $ num__70000 . each machine can be purchased by making a num__20 percent down payment and repaying the remainder of the cost and the finance charges over a period of time . if the finance charges are equal to num__40 percent of the remainder of the cost how much less would num__2 machines of type a cost than num__1 machine of type b under this arrangement ? <o> a ) $ num__10000 <o> b ) $ num__11200 <o> c ) $ num__12000 <o> d ) $ num__12800 <o> e ) $ num__13 |
200 |
total cost of num__2 machines of type a = num__20.0 of ( cost of num__2 machine a ) + remainder + num__40.0 remainder = num__20.0 of num__30000 + ( num__30000 - num__20.0 of num__30000 ) + num__40.0 of ( num__40000 - num__20.0 of num__30000 ) = num__79200 total cost of num__1 machine of type b = num__20.0 of ( cost of num__1 machine b ) + remainder + num__40.0 remainder = num__20.0 of num__70000 + ( num__70000 - num__20.0 of num__70000 ) + num__40.0 of ( num__50000 - num__20.0 of num__70000 ) = num__92400 diff = num__92400 - num__79200 = num__13200 hence e . <eor> e <eos> |
e |
e |
| what is the total number of integers between num__20 and num__70 that are divisible by num__2 ? <o> a ) num__19 <o> b ) num__25 <o> c ) num__24 <o> d ) num__26 <o> e ) num__20 |
num__22 num__24 num__26 . . . num__5658 this is an equally spaced list ; you can use the formula : n = ( largest - smallest ) / ( ' space ' ) + num__1 = ( num__70 - num__22 ) / ( num__2 ) + num__1 = num__24.0 + num__1 = num__24 + num__1 = num__25 answer is b <eor> b <eos> |
b |
add__20.0__2.0__ add__2.0__22.0__ add__2.0__24.0__ add__1.0__24.0__ add__1.0__24.0__ |
add__20.0__2.0__ add__2.0__22.0__ add__2.0__24.0__ add__1.0__24.0__ add__1.0__24.0__ |
| the second of two numbers is two less than three times the first . find the numbers if there sum is num__82 . <o> a ) num__7 - num__19 <o> b ) num__8 - num__20 <o> c ) num__10 - num__16 <o> d ) num__15 - num__9 <o> e ) num__21 - num__61 |
we are looking for two numbers . # num__1 - x # num__2 - num__3 x â € “ num__2 the sum is num__82 . # num__1 + # num__2 = num__82 substituting x + num__3 x â € “ num__2 = num__82 num__4 x â € “ num__2 = num__82 num__4 x = num__84 x = num__21 the first number is num__21 the second number is two less than three times num__21 or num__61 . correct answer e <eor> e <eos> |
e |
add__1.0__2.0__ add__1.0__3.0__ add__82.0__2.0__ divide__84.0__4.0__ subtract__82.0__21.0__ subtract__82.0__61.0__ |
add__1.0__2.0__ add__1.0__3.0__ add__82.0__2.0__ divide__84.0__4.0__ subtract__82.0__21.0__ subtract__82.0__61.0__ |
| a room num__4 m num__47 cm long and num__7 m num__77 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor . <o> a ) num__38636 <o> b ) num__38640 <o> c ) num__38647 <o> d ) num__38591 <o> e ) num__38675 |
explanation : area of the room = ( num__447 x num__777 ) cm num__2 . size of largest square tile = h . c . f . of num__447 cm and num__777 cm = num__3 cm . area of num__1 tile = ( num__3 x num__3 ) cm num__2 . number of tiles required = ( num__447 × num__777 ) / ( num__3 × num__3 ) = num__38591 answer : option d <eor> d <eos> |
d |
subtract__7.0__4.0__ subtract__4.0__3.0__ round__38591.0__ |
subtract__7.0__4.0__ subtract__4.0__3.0__ divide__38591.0__1.0__ |
| two numbers are respectively num__20.0 and num__25.0 more than a third number . the percentage that is first of the second is ? <o> a ) num__33 <o> b ) num__71 <o> c ) num__96 <o> d ) num__55 <o> e ) num__77 |
i ii iii num__120 num__125 num__100 num__125 - - - - - - - - - - num__120 num__100 - - - - - - - - - - - ? = > num__96.0 \ answer : c <eor> c <eos> |
c |
percent__96.0__100.0__ |
percent__96.0__100.0__ |
| a certain research group plans to create computer models of x % of a list of num__10000 bacterial species known to inhabit the human body . after a budget cut the group finds it must reduce this selection by ( x − num__7 ) % . in terms of x how many species of bacteria will the group be able to model ? <o> a ) x * x – num__5 x <o> b ) ( x ) * ( num__107 – x ) <o> c ) ( num__100 ) ( num__105 – x ) <o> d ) ( num__100 ) ( num__95 – x ) <o> e ) ( x - num__5 ) / num__100 |
initial : ( x / num__100 ) * num__10000 = num__100 x ( bacterial species ) after reduce by ( x - num__7 ) % the percentage of bacterial species = num__1 - ( x - num__7 ) % = num__1 - ( x - num__7 ) / num__100 = ( num__107 - x ) / num__100 note : difference between reduce to [ means : the remain ] and reduce by [ means : the remain = num__1 - reduce by ] so the number of bacterial species after reducing : num__100 x * ( num__107 - x ) / num__100 = ( x ) * ( num__107 - x ) ans is b . <eor> b <eos> |
b |
percent__100.0__107.0__ |
percent__100.0__107.0__ |
| in each of the following questions a number series is given with one term missing . choose the correct alternative that will continue the same pattern and fill in the blank spaces . num__2 num__7 ? num__23 num__34 num__47 <o> a ) num__31 <o> b ) num__14 <o> c ) num__36 <o> d ) num__31 <o> e ) num__33 |
b num__14 the given sequence is + num__5 + num__7 + num__9 — — ie . num__2 + num__5 = num__7 num__7 + num__7 = num__14 num__14 + num__9 = num__23 <eor> b <eos> |
b |
multiply__2.0__7.0__ subtract__7.0__2.0__ add__2.0__7.0__ multiply__2.0__7.0__ |
multiply__2.0__7.0__ subtract__7.0__2.0__ add__2.0__7.0__ add__5.0__9.0__ |
| the workforce of company samsung is num__60.0 female . the company hired num__20 additional male workers and as a result the percent of female workers dropped to num__50.0 . how many employees did the company have after hiring the additional male workers ? <o> a ) num__180 <o> b ) num__170 <o> c ) num__160 <o> d ) num__150 <o> e ) num__145 |
let x be the total worker then num__0.6 x = female worker and num__0.4 x is male worker then num__20 male worker added num__06 x / ( num__0.4 x + num__20 ) = num__0.5 or num__60 x = num__50 * ( num__0.4 x + num__100 ) = num__20 x + num__5000 or num__40 x = num__5000 x = num__125 total worker = num__125 + num__20 = num__145 e <eor> e <eos> |
e |
percent__100.0__145.0__ |
percent__100.0__145.0__ |
| a computer was sold at a loss of num__20.0 . if it was sold for rs . num__3200 more there would have been a gain of num__12.0 . what is the cost price ? <o> a ) rs . num__12000 <o> b ) rs . num__8000 <o> c ) rs . num__10000 <o> d ) rs . num__20000 <o> e ) rs . num__11000 |
num__80.0 num__112.0 - - - - - - - - num__32.0 - - - - num__3200 num__100.0 - - - - ? = > rs . num__10000 answer : c <eor> c <eos> |
c |
percent__100.0__10000.0__ |
percent__100.0__10000.0__ |
| sum of the squares of three numbers is num__138 and the sum of their products taken two at a time is num__131 . find the sum ? <o> a ) num__20 <o> b ) num__22 <o> c ) num__24 <o> d ) num__26 <o> e ) num__28 |
( a + b + c ) num__2 = a num__2 + b num__2 + c num__2 + num__2 ( ab + bc + ca ) = num__138 + num__2 * num__131 a + b + c = √ num__400 = num__20 a <eor> a <eos> |
a |
divide__400.0__20.0__ |
divide__400.0__20.0__ |
| if a tire rotates at num__400 revolutions per minute when the car is traveling num__96 km / h what is the circumference of the tire ? <o> a ) num__2 <o> b ) num__1 <o> c ) num__4 <o> d ) num__3 <o> e ) num__5 |
num__400 rev / minute = num__400 * num__60 rev / num__60 minutes = num__24000 rev / hour num__24000 * c = num__96000 m : c is the circumference c = num__4 meters correct answer c <eor> c <eos> |
c |
hour_to_min_conversion__ multiply__400.0__60.0__ divide__96000.0__24000.0__ round__4.0__ |
hour_to_min_conversion__ multiply__400.0__60.0__ divide__96000.0__24000.0__ divide__96000.0__24000.0__ |
| how many seconds will a num__700 meter long train moving with a speed of num__63 km / hr take to cross a man walking with a speed of num__3 km / hr in the direction of the train ? <o> a ) num__48 <o> b ) num__36 <o> c ) num__42 <o> d ) num__11 <o> e ) num__18 |
explanation : here distance d = num__700 mts speed s = num__63 - num__3 = num__60 kmph = num__60 x num__0.277777777778 m / s time t = = num__42 sec . answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ round__42.0__ |
subtract__63.0__3.0__ round__42.0__ |
| suppose five circles each num__4 inches in diameter are cut from a rectangular strip of paper num__12 inches long . if the least amount of paper is to be wasted what is the width of the paper strip ? a = sqrt ( num__3 ) <o> a ) num__5 <o> b ) num__4 + num__2 a <o> c ) num__8 <o> d ) num__4 ( num__1 + a ) <o> e ) not enough information |
also if we dont go by num__30 : num__60 : num__90 rule we can arrive at the length of by pythagoras theorem . ( ac ) ^ num__2 + ( bc ) ^ num__2 = ab ^ num__2 ( ac ) ^ num__2 + num__2 ^ num__2 = num__4 ^ num__2 ac = num__2 root b . so the width of paper = num__2 + num__2 + num__2 root num__3 - - > num__4 + num__2 root num__3 . which is b . <eor> b <eos> |
b |
rectangle_perimeter__12.0__3.0__ triangle_area__4.0__30.0__ triangle_area__3.0__60.0__ triangle_area__4.0__2.0__ |
rectangle_perimeter__12.0__3.0__ triangle_area__4.0__30.0__ triangle_area__3.0__60.0__ triangle_area__4.0__2.0__ |
| a women purchased num__3 towels @ rs . num__100 each num__5 towels @ rs . num__150 each and two towels at a certain rate which is now slipped off from his memory . but she remembers that the average price of the towels was rs . num__165 . find the unknown rate of two towels ? <o> a ) a ) num__400 <o> b ) b ) num__450 <o> c ) c ) num__500 <o> d ) d ) num__550 <o> e ) e ) num__600 |
num__10 * num__150 = num__1650 num__3 * num__100 + num__5 * num__150 = num__1050 num__1650 – num__1050 = num__600 e <eor> e <eos> |
e |
multiply__165.0__10.0__ subtract__1650.0__1050.0__ subtract__1650.0__1050.0__ |
multiply__165.0__10.0__ subtract__1650.0__1050.0__ subtract__1650.0__1050.0__ |
| sides of a rectangular park are in the ratio num__3 : num__2 and its area is num__3750 sq m the cost of fencing it at num__80 ps per meter is ? <o> a ) num__287 <o> b ) num__369 <o> c ) num__125 <o> d ) num__200 <o> e ) num__361 |
num__3 x * num__2 x = num__3750 = > x = num__25 num__2 ( num__75 + num__50 ) = num__250 m num__250 * num__0.80 = rs . num__200 answer : d <eor> d <eos> |
d |
multiply__3.0__25.0__ multiply__2.0__25.0__ rectangle_perimeter__75.0__50.0__ square_perimeter__50.0__ square_perimeter__50.0__ |
multiply__3.0__25.0__ multiply__2.0__25.0__ rectangle_perimeter__75.0__50.0__ multiply__0.8__250.0__ multiply__0.8__250.0__ |
| the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square whose area is num__1225 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is num__13 units ? <o> a ) num__140 sq . units <o> b ) num__170 sq . units <o> c ) num__190 sq . units <o> d ) num__940 sq . units <o> e ) num__182 sq . units |
explanation : given that the area of the square = num__1225 sq . units = > side of square = √ num__1225 = num__35 units the radius of the circle = side of the square = num__35 units length of the rectangle = num__0.4 * num__35 = num__14 units given that breadth = num__10 units area of the rectangle = lb = num__14 * num__13 = num__182 sq . units answer : option e <eor> e <eos> |
e |
multiply__35.0__0.4__ multiply__13.0__14.0__ multiply__13.0__14.0__ |
multiply__35.0__0.4__ multiply__13.0__14.0__ multiply__13.0__14.0__ |
| the cost price of num__13 articles is equal to the selling price of num__11 articles . find the profit percent ? <o> a ) num__78 num__0.181818181818 % <o> b ) num__18 num__0.181818181818 % <o> c ) num__88 num__0.181818181818 % <o> d ) num__58 num__0.181818181818 % <o> e ) num__68 num__0.181818181818 % |
explanation : num__13 cp = num__11 sp num__11 - - - num__2 cp num__100 - - - ? = > num__18 num__0.181818181818 % answer : b <eor> b <eos> |
b |
percent__100.0__18.0__ |
percent__100.0__18.0__ |
| the ages of charlie and vaibhav are in the proportion of num__3 : num__5 . after num__9 years the proportion of their ages will be num__3 : num__4 . then the current age of vaibhav is : <o> a ) num__15 years <o> b ) num__16 years <o> c ) num__135 years <o> d ) num__21 years <o> e ) num__18 years |
a num__15 years charlie ’ s age = num__3 a and vaibhav ’ s age = num__5 a { ( num__3 a + num__9 ) / ( num__5 a + num__9 ) } = num__0.75 = > num__4 ( num__3 a + num__9 ) = num__3 ( num__5 a + num__9 ) = > a = num__3 therefore vaibhav ’ s age = num__15 years . <eor> a <eos> |
a |
multiply__3.0__5.0__ divide__3.0__4.0__ multiply__3.0__5.0__ |
multiply__3.0__5.0__ divide__3.0__4.0__ multiply__3.0__5.0__ |
| a tank is num__25 m long num__12 m wide and num__6 m deep . what is the cost of plastering its walls and bottom at the rate of num__75 paise per sq . m ? <o> a ) num__558 <o> b ) num__502 <o> c ) num__516 <o> d ) num__612 <o> e ) num__700 |
total surface s = num__2 lw + num__2 lh + num__2 wh = ( num__25 * num__12 ) + ( num__2 * num__25 * num__6 ) + ( num__2 * num__12 * num__6 ) = num__300 + num__300 + num__144 = num__744 m num__2 therefore cost of plastering = num__744 * num__75 = num__5800 paise = num__558 answer a <eor> a <eos> |
a |
divide__12.0__6.0__ multiply__25.0__12.0__ round__558.0__ |
divide__12.0__6.0__ multiply__25.0__12.0__ round__558.0__ |
| look at this series : num__14 num__28 num__20 num__40 num__32 num__64 . . . what number should come next ? <o> a ) num__56 <o> b ) num__44 <o> c ) num__26 <o> d ) num__48 <o> e ) num__49 |
a num__56 this is an alternating multiplication and subtracting series : first multiply by num__2 and then subtract num__8 . <eor> a <eos> |
a |
divide__28.0__14.0__ subtract__28.0__20.0__ multiply__28.0__2.0__ |
divide__28.0__14.0__ subtract__28.0__20.0__ multiply__28.0__2.0__ |
| determine the equation for the line parallel to r ≡ x + num__2 y − num__2 = num__0 that passes through the point a = ( num__1 num__3 ) . <o> a ) x + num__2 y = num__0 <o> b ) num__2 y − num__7 = num__0 <o> c ) x + num__2 y − num__7 = num__0 <o> d ) x − num__7 = num__0 <o> e ) x - num__2 y + num__7 = num__0 |
the equation for the line parallel to r ≡ x + num__2 y − num__2 = num__0 is x + num__2 y + k = num__0 num__1 + num__2 • num__3 + k = num__0 k = − num__7 x + num__2 y − num__7 = num__0 answer c x + num__2 y − num__7 = num__0 <eor> c <eos> |
c |
multiply__2.0__1.0__ |
multiply__2.0__1.0__ |
| in the sequence s each term after the first is twice the previous term . if the first term of sequence s is num__3 what is the sum of the num__14 th term in sequence s ? <o> a ) num__3 ( num__2 ^ num__13 ) <o> b ) num__9 ( num__2 ^ num__15 ) <o> c ) num__21 ( num__2 ^ num__14 ) <o> d ) num__9 ( num__2 ^ num__14 ) <o> e ) num__21 ( num__2 ^ num__13 ) |
the terms in the sequence can be shown as a ( n ) = num__2 * a ( n - num__1 ) so the sequence will look like : num__3 num__2 * num__3 ( num__2 ^ num__2 ) * num__3 . . . and the nth term will be given as num__2 ^ ( n - num__1 ) * num__3 therefore a ( num__14 ) = ( num__2 ^ num__13 ) * num__3 answer : a <eor> a <eos> |
a |
subtract__3.0__2.0__ subtract__14.0__1.0__ multiply__3.0__1.0__ |
subtract__3.0__2.0__ subtract__14.0__1.0__ multiply__3.0__1.0__ |
| if num__15 men can build a wall of forty two metres long in num__5 days what length of a similar wall can be built by num__25 men in num__4 days <o> a ) num__48 metres <o> b ) num__56 metres <o> c ) num__60 metres <o> d ) num__62 metres <o> e ) none of these |
explanation : solution : let the required length be x metres . more men more length built ( direct proportion ) less days less length built ( direct proportion ) men num__15 : num__25 days num__5 : num__4 . ' . ( num__15 * num__5 * x ) = ( num__25 * num__4 * num__42 ) x = num__56 metres . answer : b <eor> b <eos> |
b |
round__56.0__ |
round__56.0__ |
| if k ^ num__3 is divisible by num__60 what is the least possible value of integer k ? <o> a ) num__12 <o> b ) num__30 <o> c ) num__60 <o> d ) num__90 <o> e ) num__120 |
num__60 = num__2 ^ num__2 * num__3 * num__5 therefore k must include at least num__2 * num__3 * num__5 = num__30 . the answer is b . <eor> b <eos> |
b |
add__3.0__2.0__ divide__60.0__2.0__ divide__60.0__2.0__ |
add__3.0__2.0__ divide__60.0__2.0__ divide__60.0__2.0__ |
| denominator of a number is num__4 less than the numerator . if num__6 is added to the numerator it becomes num__3 times the denominator . find the denominator . <o> a ) num__1 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__9 |
let the numerator be x and denominator be y . then = > x = y + num__4 and num__6 + x = num__3 * y = > num__6 + y + num__4 = num__3 * y = > num__2 y = num__10 = > y = num__5 option c <eor> c <eos> |
c |
subtract__6.0__4.0__ add__4.0__6.0__ add__3.0__2.0__ add__3.0__2.0__ |
subtract__6.0__4.0__ add__4.0__6.0__ add__3.0__2.0__ add__3.0__2.0__ |
| if num__0.2 of a number is equal to num__0.08 of another number the ratio of the numbers is : <o> a ) num__2 : num__3 <o> b ) num__3 : num__4 <o> c ) num__2 : num__5 <o> d ) num__20 : num__3 <o> e ) num__30 : num__7 |
num__0.2 a = num__0.08 b - > a / b = num__0.08 / num__0.20 = num__0.4 = num__0.4 : . a : b = num__2 : num__5 answer : c <eor> c <eos> |
c |
divide__0.08__0.2__ divide__0.4__0.2__ reverse__0.2__ divide__0.4__0.2__ |
divide__0.08__0.2__ divide__0.4__0.2__ reverse__0.2__ divide__0.4__0.2__ |
| a ranch has both horses and ponies . exactly num__0.833333333333 of the ponies have horseshoes and exactly num__0.666666666667 of the ponies with horseshoes are from iceland . if there are num__4 more horses than ponies what is the minimum possible combined number of horses and ponies on the ranch ? <o> a ) num__18 <o> b ) num__21 <o> c ) num__38 <o> d ) num__40 <o> e ) num__57 |
num__0.833333333333 * p have horseshoes so p is a multiple of num__6 . num__0.666666666667 * num__0.833333333333 * p = num__0.555555555556 * p are icelandic ponies with horseshoes so p is a multiple of num__9 . the minimum value of p is num__18 . then h = p + num__4 = num__22 . the minimum number of horses and ponies is num__40 . the answer is d . <eor> d <eos> |
d |
multiply__0.8333__0.6667__ add__4.0__18.0__ add__18.0__22.0__ add__18.0__22.0__ |
multiply__0.8333__0.6667__ add__4.0__18.0__ add__18.0__22.0__ add__18.0__22.0__ |
| if y = num__30 p and p is prime what is the greatest common factor of y and num__10 p in terms of p ? <o> a ) num__10 p <o> b ) num__2 p <o> c ) num__5 p <o> d ) num__7 p <o> e ) p ^ num__2 |
y = num__30 p other number is num__10 p then gcf ( num__30 p num__10 p ) = num__10 p ; a is the correct answer <eor> a <eos> |
a |
gcd__30.0__10.0__ |
gcd__30.0__10.0__ |
| everyone shakes hands with everyone else in a room . total number of handshakes is num__120 . number of persons = ? <o> a ) a . num__14 <o> b ) b . num__12 <o> c ) c . num__11 <o> d ) d . num__15 <o> e ) e . num__16 |
in a room of n people the number of possible handshakes is c ( n num__2 ) or n ( n - num__1 ) / num__2 so n ( n - num__1 ) / num__2 = num__120 or n ( n - num__1 ) = num__240 or n = num__16 answer is ( e ) <eor> e <eos> |
e |
multiply__120.0__2.0__ multiply__1.0__16.0__ |
multiply__120.0__2.0__ divide__16.0__1.0__ |
| the first term in a sequence is num__1 and the second term is num__10 . from the third term on each term is the average ( arithmetic mean ) of all preceding terms . what is the num__29 th term in the sequence ? <o> a ) num__3.5 <o> b ) num__5.5 <o> c ) num__10 <o> d ) num__10.5 <o> e ) num__29 |
the third term is num__5.5 since this is the average of num__1 and num__10 . then the next terms will all be num__5.5 since the average stays at num__5.5 the answer is b . <eor> b <eos> |
b |
multiply__1.0__5.5__ |
multiply__1.0__5.5__ |
| the perimeter of one square is num__48 cm and that of another is num__20 cm . find the perimeter and the diagonal of a square which is equal in area to these two combined ? <o> a ) num__13 √ num__4 <o> b ) num__13 √ num__2 <o> c ) num__23 √ num__2 <o> d ) num__12 √ num__4 <o> e ) num__13 √ num__9 |
num__4 a = num__48 num__4 a = num__20 a = num__12 a = num__5 a num__2 = num__144 a num__2 = num__25 combined area = a num__2 = num__169 = > a = num__13 d = num__13 √ num__2 answer : b <eor> b <eos> |
b |
power__12.0__2.0__ power__5.0__2.0__ triangle_perimeter__20.0__144.0__5.0__ triangle_area__2.0__13.0__ |
power__12.0__2.0__ power__5.0__2.0__ triangle_perimeter__20.0__144.0__5.0__ triangle_area__2.0__13.0__ |
| the average age of a and b is num__50 years . if c were to replace a the average would be num__50 and if c were to replace b the average would be num__40 . what are the age of a b and c ? <o> a ) num__100100 num__80 <o> b ) num__50 num__50 num__40 <o> c ) num__40 num__60 num__40 <o> d ) num__50 num__60 num__40 <o> e ) none of the above |
given a + b = num__100 â € ¦ ( i ) c + b = num__100 â € ¦ ( ii ) a + c = num__80 â € ¦ ( iii ) ( i ) + ( ii ) + ( iii ) â ‡ ’ a + b + c = num__140 â € ¦ ( iv ) from ( i ) and ( iv ) we get c = num__40 years â ˆ ´ b = num__60 years and a = num__40 years answer c <eor> c <eos> |
c |
add__40.0__100.0__ subtract__140.0__80.0__ subtract__80.0__40.0__ |
add__40.0__100.0__ subtract__140.0__80.0__ subtract__80.0__40.0__ |
| if x ^ num__4 > x ^ num__5 > x ^ num__3 which of the following could be the value of x ? <o> a ) - num__3 <o> b ) - num__2 <o> c ) - num__0.666666666667 <o> d ) num__0.666666666667 <o> e ) num__3 |
let ' s break this question down into two parts . i : x num__4 > x num__5 x num__4 > x num__5 and ii : x num__5 > x num__3 x num__5 > x num__3 consider part i : x num__4 > x num__5 x num__4 > x num__5 for this to be true either xx is negative or it is a fraction less than num__1 consider part ii : x num__5 > x num__3 x num__5 > x num__3 if xx is negative then for this to be true xx has to be greater than - num__1 . if xx is a fraction less than num__1 then for this part to be true xx must also be negative . in either case xx can be represented by the inequality num__0 > x > − num__10 > x > − num__1 . the only option that satisfies this condition is c ( - num__0.666666666667 ) answer : c <eor> c <eos> |
c |
subtract__4.0__3.0__ multiply__1.0__0.6667__ |
subtract__4.0__3.0__ multiply__1.0__0.6667__ |
| what is the sum of all the numbers between num__1 and num__16 inclusive ? <o> a ) num__190 <o> b ) num__200 <o> c ) num__185 <o> d ) num__166 <o> e ) num__213 |
all you have to do is add num__1 + num__2 + num__3 + num__4 . . . + num__14 + num__15 + num__16 which is num__190 . final answer : a <eor> a <eos> |
a |
add__1.0__2.0__ add__1.0__3.0__ subtract__16.0__2.0__ add__1.0__14.0__ multiply__1.0__190.0__ |
add__1.0__2.0__ add__1.0__3.0__ subtract__16.0__2.0__ add__1.0__14.0__ multiply__1.0__190.0__ |
| if re . num__250 amounts to rs . num__400 over a period of num__5 years . what is the rate of simple interest ? <o> a ) num__11.0 <o> b ) num__12.0 <o> c ) num__13.0 <o> d ) num__10.0 <o> e ) num__14 % |
num__150 = ( num__250 * num__5 * r ) / num__100 r = num__12.0 answer : b <eor> b <eos> |
b |
percent__12.0__100.0__ |
percent__12.0__100.0__ |
| for the given equation x num__9 + num__5 x num__8 − x num__3 + num__7 x + num__2 = num__0 x num__9 + num__5 x num__8 − x num__3 + num__7 x + num__2 = num__0 how many maximum real roots are possible ? <o> a ) num__3 <o> b ) num__7 <o> c ) num__4 <o> d ) num__0 <o> e ) num__1 |
f ( x ) = x num__9 + num__5 x num__8 − x num__3 + num__7 x + num__2 = num__0 x num__9 + num__5 x num__8 − x num__3 + num__7 x + num__2 = num__0 in f ( x ) there are num__2 changes of sign . so there are two positive roots . f ( - x ) = - x num__9 + num__5 x num__8 − x num__3 + num__7 x + num__2 = num__0 x num__9 + num__5 x num__8 − x num__3 + num__7 x + num__2 = num__0 in f ( - x ) there are num__3 changes of sign . so there are three negative roots . so in all there are num__5 real roots possible ( num__2 positive and num__3 negative ) and as degree of the given equation is num__9 there are total num__9 roots . so remaining num__4 roots will be imaginary . answer : c <eor> c <eos> |
c |
subtract__9.0__5.0__ subtract__9.0__5.0__ |
subtract__9.0__5.0__ subtract__9.0__5.0__ |
| the diagonals of two squares are in the ratio of num__2 : num__5 . find the ratio of their areas . <o> a ) num__5 : num__25 <o> b ) num__4 : num__25 <o> c ) num__3 : num__15 <o> d ) num__7 : num__15 <o> e ) num__6 : num__15 |
let the diagonals of the squares be num__2 x and num__5 x respectively . ratio of their areas = ( num__0.5 ) * ( num__2 x ) num__2 : ( num__0.5 ) * ( num__5 x ) num__2 = num__4 x num__2 : num__25 x num__2 = num__4 : num__25 . answer b num__4 : num__25 <eor> b <eos> |
b |
power__5.0__2.0__ triangle_area__2.0__4.0__ |
power__5.0__2.0__ volume_rectangular_prism__2.0__0.5__4.0__ |
| a wholesaler wishes to sell num__100 pounds of mixed nuts at $ num__1.50 a pound . she mixes peanuts worth $ num__1.50 a pound with cashews worth $ num__4.00 a pound . how many pounds of cashews must she use ? <o> a ) num__40 <o> b ) num__45 <o> c ) num__50 <o> d ) num__55 <o> e ) num__60 |
from the question stem we know that we need a mixture of num__100 pounds of peanuts and cashews . if we represent peanuts as x and cashews as y we get x + y = num__100 . since the wholesaler wants to sell the mixture of num__100 pounds @ $ num__2.50 we can write this as : $ num__1.5 * ( x + y ) = $ num__1.5 x + $ num__4 y from the equation x + y = num__100 we can rewrite y as y = num__100 - x and substitute this into our equation to get : $ num__2.5 * ( x + num__100 - x ) = $ num__1.5 x + $ num__4 ( num__100 - x ) if you solve for x you will get x = num__60 and therefore y = num__40 . so the wholesaler must use num__40 pounds of cashews . you can substitute into the original equation to see that : $ num__250 = $ num__1.5 ( num__60 ) + $ num__4 ( num__40 ) answer is a . <eor> a <eos> |
a |
subtract__4.0__1.5__ divide__100.0__2.5__ multiply__100.0__2.5__ divide__100.0__2.5__ |
subtract__4.0__1.5__ subtract__100.0__60.0__ multiply__100.0__2.5__ subtract__100.0__60.0__ |
| a can finish a piece of work in num__4 days . b can do it in num__8 days . they work together for two days and then a goes away . in how many days will b finish the work ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__2 <o> e ) num__8 |
num__0.5 + ( num__2 + x ) / num__8 = num__1 = > x = num__2 days answer : d <eor> d <eos> |
d |
divide__4.0__8.0__ multiply__4.0__0.5__ multiply__0.5__2.0__ round__2.0__ |
divide__4.0__8.0__ divide__8.0__4.0__ multiply__0.5__2.0__ divide__4.0__2.0__ |
| two trains start from the same station at the same time and travel in opposite directions . one train travels at an average rate of num__40 mph the other at num__65 mph . in how many hours will they be num__315 miles apart ? <o> a ) num__1 hour <o> b ) num__2 hours <o> c ) num__3 hours <o> d ) num__4 hours <o> e ) num__5 hours |
first we ’ ll make the d = rt chart . but we won ’ t fill in the d . d = r x t train num__1 num__40 x train num__2 num__65 x the reason we have an x in the time column is because they left at the same time and will be num__315 at the same time . in other words their times are equal . now the big question . are there distances equal ? since they do not meet the criteria in a type a problem the answer is no . that means the sum of the distances must be equal to a number . d num__1 + d num__2 = # num__40 x + num__65 = num__315 num__105 x = num__315 x = num__3 it will take three hours . correct answer c <eor> c <eos> |
c |
add__40.0__65.0__ divide__315.0__105.0__ round__3.0__ |
add__40.0__65.0__ add__1.0__2.0__ add__1.0__2.0__ |
| in a flight of num__600 km an aircraft was slowed down due to bad weather . its average speed for the trip was reduced by num__200 km / hr and the time of flight increased by num__30 minutes . the duration of the flight is ? <o> a ) num__1 hr <o> b ) num__7 hr <o> c ) num__9 hr <o> d ) num__4 hr <o> e ) num__9 hr |
let the duration of the flight be x hours . then num__600 / x - num__600 / ( x + num__0.5 ) = num__200 x ( num__2 x + num__1 ) = num__3 num__2 x num__2 + x - num__3 = num__0 ( num__2 x + num__3 ) ( x - num__1 ) = num__0 x = num__1 hr . answer : a <eor> a <eos> |
a |
reverse__0.5__ multiply__0.5__2.0__ divide__600.0__200.0__ round_down__0.5__ reverse__1.0__ |
reverse__0.5__ multiply__0.5__2.0__ divide__600.0__200.0__ round_down__0.5__ reverse__1.0__ |
| in a rectangular axis system what is the area of a parallelogram with the coordinates : ( num__44 ) ( num__74 ) ( num__59 ) ( num__89 ) ? <o> a ) num__21 . <o> b ) num__28 . <o> c ) num__5 . <o> d ) num__49 . <o> e ) num__52 . |
delta x will give us the dimension of one side of the parallelogram = num__5 - num__4 = num__1 unit delta y will give us the dimension of the other side of parallelogram = num__9 - num__4 = num__5 unit area of parallelogram = num__1 * num__5 = num__5 answer is c <eor> c <eos> |
c |
multiply__1.0__5.0__ |
multiply__1.0__5.0__ |
| for a certain set the value range of its members is num__84.4 . a new set is created from the members of the old set as follows : num__12 is subtracted from a member of the old set and the result is divided by num__4 . the resulting value is a member of the new set . if this operation is done for each member of the old set what is the range of values of the members of the new set ? <o> a ) num__18.1 <o> b ) num__21.1 <o> c ) num__36.5 <o> d ) num__42.2 <o> e ) num__84.4 |
let x and z be the smallest and largest of the original set respectively . z - x = num__84.4 the smallest and largest members of the new set will be ( x - num__12 ) / num__4 and ( z - num__12 ) / num__4 . then the range is ( z - num__12 ) / num__4 - ( x - num__12 ) / num__4 = ( z - x ) / num__4 = num__84.4 / num__4 = num__21.1 the answer is b . <eor> b <eos> |
b |
divide__84.4__4.0__ divide__84.4__4.0__ |
divide__84.4__4.0__ divide__84.4__4.0__ |
| the interest on a certain deposit at num__4.5 p . a . is rs . num__202.50 in one year . how much will the additional interest in one year be on the same deposit at num__5.0 p . a . ? <o> a ) rs . num__20.25 <o> b ) rs . num__22.50 <o> c ) rs . num__25 <o> d ) rs . num__42.75 <o> e ) none |
solution s . i . = rs . num__202.50 r = num__4.5 t = num__1 year . principal = rs . ( num__100 x num__202.50 / num__4.5 x num__1 ) = rs . num__4500 . now p = rs . num__4500 r = num__5.0 t = num__1 year . s . i . = rs . ( num__4500 x num__5 x num__0.001 = rs . num__225 . ∴ difference in interest = rs . ( num__225 - num__202.50 ) = rs . num__22.50 . answer b <eor> b <eos> |
b |
percent__5.0__4500.0__ percent__100.0__22.5__ |
percent__5.0__4500.0__ percent__100.0__22.5__ |
| if an amount of rs num__42900 is distributed equally amongst num__22 persons how much amount would each person get ? <o> a ) rs num__1950 <o> b ) rs num__2000 <o> c ) rs num__745 <o> d ) rs num__765 <o> e ) none |
required amount = num__1950.0 = rs num__1950 answer a <eor> a <eos> |
a |
divide__42900.0__22.0__ divide__42900.0__22.0__ |
divide__42900.0__22.0__ divide__42900.0__22.0__ |
| when processing flower - nectar into honey bees ' extract a considerable amount of water gets reduced . how much flower - nectar must be processed to yield num__1 kg of honey if nectar contains num__50.0 water and the honey obtained from this nectar contains num__15.0 water ? <o> a ) num__1.2 kg <o> b ) num__1.5 kg <o> c ) num__1.7 kg <o> d ) num__1.9 kg <o> e ) none of these |
explanation : flower - nectar contains num__50.0 of non - water part . in honey this non - water part constitutes num__85.0 ( num__100 - num__15 ) . therefore num__0.5 x amount of flower - nectar = num__0.85 x amount of honey = num__0.85 x num__1 kg therefore amount of flower - nectar needed = ( num__0.85 / num__0.51 ) kg = num__1.7 kgs answer : c <eor> c <eos> |
c |
add__15.0__85.0__ divide__50.0__100.0__ divide__85.0__100.0__ divide__0.85__0.5__ multiply__1.0__1.7__ |
add__15.0__85.0__ divide__50.0__100.0__ divide__85.0__100.0__ divide__0.85__0.5__ divide__1.7__1.0__ |
| what least number should be added to num__1019 so that the sum is completely divisible by num__25 ? <o> a ) num__4 <o> b ) num__3 <o> c ) num__2 <o> d ) num__0 <o> e ) num__6 |
num__1019 Ã · num__25 = num__40 with remainder = num__19 num__19 + num__6 = num__25 . hence num__6 should be added to num__1019 so that the sum will be divisible by num__25 answer : option e <eor> e <eos> |
e |
subtract__25.0__19.0__ subtract__25.0__19.0__ |
subtract__25.0__19.0__ subtract__25.0__19.0__ |
| a num__400 meter long train crosses a platform in num__45 seconds while it crosses a signal pole in num__30 seconds . what is the length of the platform ? <o> a ) num__200 <o> b ) num__400 <o> c ) num__350 <o> d ) num__1800 <o> e ) num__45 |
speed = [ num__13.3333333333 ] m / sec = num__13.3333333333 m / sec . let the length of the platform be x meters . then x + num__8.88888888889 = num__13.3333333333 num__3 ( x + num__400 ) = num__1800 è x = num__200 m . answer : a <eor> a <eos> |
a |
divide__400.0__30.0__ divide__400.0__45.0__ round__200.0__ |
divide__400.0__30.0__ divide__400.0__45.0__ round__200.0__ |
| if num__5 ^ num__5 × num__5 ^ x = ( num__125 ) ^ num__4 then what is the value of x ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
num__5 ^ num__5 × num__5 ^ x = ( num__125 ) ^ num__4 num__5 ^ ( num__5 + x ) = num__5 ^ num__12 since they have the same base we can just set the exponents equal to each other : ( num__5 + x ) = num__12 x = num__7 ans . e ) num__7 <eor> e <eos> |
e |
subtract__12.0__5.0__ subtract__12.0__5.0__ |
subtract__12.0__5.0__ subtract__12.0__5.0__ |
| aravamudhan balakrishnan and chinnan work in a developing software company in starting level positions . however their salaries are different . aravamudhan ' s salary to balakrishnan ' s salary and balakrishnan ' s salary to chinnan ' s salary are in the ratio num__4 : num__3 . if the total salary of all the three employees is rs . num__29230 what is the salary of chinnan ? <o> a ) rs . num__12640 <o> b ) rs . num__9480 <o> c ) rs . num__7110 <o> d ) rs . num__8660 <o> e ) rs . num__9660 |
making them in equal proportion of aravind : bala : cinnana = num__16 : num__12 : num__9 and given total salary num__37 * x = num__29230 x = num__790 chinnana salary is num__9 * num__790 = num__7110 answer : c <eor> c <eos> |
c |
multiply__4.0__3.0__ subtract__12.0__3.0__ divide__29230.0__37.0__ multiply__9.0__790.0__ multiply__9.0__790.0__ |
multiply__4.0__3.0__ subtract__12.0__3.0__ divide__29230.0__37.0__ multiply__9.0__790.0__ multiply__9.0__790.0__ |
| there are two positive integers a and b . what is the probability that a + b is odd / <o> a ) num__1.5 <o> b ) num__0.857142857143 <o> c ) num__0.5 <o> d ) num__3.5 <o> e ) num__0.8 |
s = adding two numbers is ( even + even ) ( even + odd ) ( odd + odd ) ( odd + even ) n ( s ) = num__4 e = ( even + odd ) ( odd + even ) are the points in the event . n ( e ) = num__2 p ( e ) = n ( e ) / n ( s ) = num__0.5 = num__0.5 answer is option c <eor> c <eos> |
c |
reverse__2.0__ reverse__2.0__ |
reverse__2.0__ reverse__2.0__ |
| mary peter and lucy were picking chestnuts . mary picked twice as much chestnuts than peter . lucy picked num__2 kg more than peter . together the three of them picked num__26 kg of chestnuts . how many kilograms did each of them pick ? <o> a ) num__6 num__8 and num__12 <o> b ) num__12 num__8 and num__6 <o> c ) num__8 num__6 and num__12 <o> d ) num__12 num__6 and num__8 <o> e ) num__6 num__12 and num__8 |
let x be the amount peter picked . then mary and lucy picked num__2 x and x + num__2 respectively . so x + num__2 x + x + num__2 = num__26 num__4 x = num__24 x = num__6 therefore peter mary and lucy picked num__6 num__12 and num__8 kg respectively . so answer is e . <eor> e <eos> |
e |
subtract__26.0__2.0__ add__2.0__4.0__ multiply__2.0__6.0__ multiply__2.0__4.0__ add__2.0__4.0__ |
subtract__26.0__2.0__ add__2.0__4.0__ multiply__2.0__6.0__ add__2.0__6.0__ add__2.0__4.0__ |
| a coin is tossed live times . what is the probability that there is at the least one tail ? <o> a ) num__0.96875 <o> b ) num__0.861111111111 <o> c ) num__1.14814814815 <o> d ) num__1.10714285714 <o> e ) num__1.72222222222 |
let p ( t ) be the probability of getting least one tail when the coin is tossed five times . = there is not even a single tail . i . e . all the outcomes are heads . = num__0.03125 ; p ( t ) = num__1 - num__0.03125 = num__0.96875 answer : a <eor> a <eos> |
a |
negate_prob__0.0312__ negate_prob__0.0312__ |
negate_prob__0.0312__ negate_prob__0.0312__ |
| num__3 men and num__8 women complete a task in same time as num__6 men and num__2 women do . how much fraction of work will be finished in same time if num__4 men and num__5 women will do that task . <o> a ) num__0.928571428571 <o> b ) num__1.3 <o> c ) num__0.722222222222 <o> d ) num__0.8125 <o> e ) num__1.18181818182 |
num__3 m + num__8 w = num__6 m + num__2 w num__3 m = num__6 w num__1 m = num__2 w therefore num__3 m + num__8 w = num__14 w num__4 m + num__5 w = num__13 w answer is num__0.928571428571 answer : a <eor> a <eos> |
a |
subtract__3.0__2.0__ add__8.0__6.0__ add__8.0__5.0__ divide__13.0__14.0__ multiply__1.0__0.9286__ |
subtract__3.0__2.0__ add__8.0__6.0__ add__8.0__5.0__ divide__13.0__14.0__ multiply__1.0__0.9286__ |
| find the surface area of a num__12 cm x num__4 cm x num__3 cm brick . <o> a ) num__84 cu cm . <o> b ) num__124 cu cm . <o> c ) num__164 cu cm . <o> d ) num__192 cu cm . <o> e ) none |
sol . surface area = [ num__2 ( num__12 x num__4 + num__4 x num__3 + num__12 x num__3 ) ] = ( num__2 x num__96 ) = num__192 cu cm . answer d <eor> d <eos> |
d |
surface_cube__4.0__ triangle_area__4.0__96.0__ triangle_area__4.0__96.0__ |
surface_cube__4.0__ triangle_area__4.0__96.0__ triangle_area__4.0__96.0__ |
| num__1397 x num__1397 = ? <o> a ) num__1951609 <o> b ) num__1981709 <o> c ) num__18362619 <o> d ) num__2031719 <o> e ) none of them |
= ( num__1397 ) ^ num__2 = ( num__1400 - num__3 ) ^ num__2 = ( num__1400 ) ^ num__2 + num__3 ^ num__2 - num__2 x num__1400 x num__3 = num__1960000 + num__9 - num__8400 = num__1960009 - num__8400 = num__1951609 answer is a <eor> a <eos> |
a |
subtract__1400.0__1397.0__ add__1960000.0__9.0__ subtract__1960009.0__8400.0__ subtract__1960009.0__8400.0__ |
subtract__1400.0__1397.0__ add__1960000.0__9.0__ subtract__1960009.0__8400.0__ subtract__1960009.0__8400.0__ |
| a merchant marks his goods up by num__40.0 and then offers a discount of num__10.0 on the marked price . what % profit does the merchant make after the discount ? <o> a ) num__21.0 <o> b ) num__26.0 <o> c ) num__69.0 <o> d ) num__31.0 <o> e ) num__19 % |
let the price be num__100 . the price becomes num__140 after a num__40.0 markup . now a discount of num__10.0 on num__140 . profit = num__126 - num__100 num__26.0 answer b <eor> b <eos> |
b |
percent__100.0__26.0__ |
percent__100.0__26.0__ |
| two goods trains each num__500 m long are running in opposite directions on parallel tracks . their speeds are num__45 km / hr and num__30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? <o> a ) num__33 <o> b ) num__289 <o> c ) num__48 <o> d ) num__77 <o> e ) num__12 |
relative speed = num__45 + num__30 = num__75 km / hr . num__75 * num__0.277777777778 = num__20.8333333333 m / sec . distance covered = num__500 + num__500 = num__1000 m . required time = num__1000 * num__0.048 = num__48 sec . answer : c <eor> c <eos> |
c |
add__45.0__30.0__ multiply__1000.0__0.048__ round__48.0__ |
add__45.0__30.0__ multiply__1000.0__0.048__ multiply__1000.0__0.048__ |
| the h . c . f . of two numbers is num__20 and the other two factors of their l . c . m . are num__13 and num__14 . the larger of the two numbers is : <o> a ) num__276 <o> b ) num__299 <o> c ) num__322 <o> d ) num__345 <o> e ) num__280 |
clearly the numbers are ( num__20 x num__13 ) and ( num__20 x num__14 ) . larger number = ( num__20 x num__14 ) = num__280 . answer : option e <eor> e <eos> |
e |
multiply__20.0__14.0__ multiply__20.0__14.0__ |
multiply__20.0__14.0__ multiply__20.0__14.0__ |
| a man is num__24 years older than his son . in two years his age will be twice the age of his son . the present age of his son is <o> a ) num__20 years <o> b ) num__21 years <o> c ) num__22 years <o> d ) num__24 years <o> e ) num__26 years |
explanation : let the son ' s present age be x years . then man ' s present age = ( x + num__24 ) years = > ( x + num__24 ) + num__2 = num__2 ( x + num__2 ) = > x + num__26 = num__2 x + num__4 so x = num__22 answer : option c <eor> c <eos> |
c |
add__24.0__2.0__ subtract__24.0__2.0__ subtract__24.0__2.0__ |
add__24.0__2.0__ subtract__24.0__2.0__ subtract__24.0__2.0__ |
| if f ( x ) = num__4 x − num__1 and g ( x ) = num__2 x + num__1 for all integers which of the following is a possible value of g ( f ( x ) ) ? <o> a ) - num__23 <o> b ) - num__17 <o> c ) num__0 <o> d ) num__8 <o> e ) num__20 |
g ( f ( x ) ) = num__2 * f ( x ) + num__1 = num__2 * ( num__4 x - num__1 ) + num__1 = num__8 x - num__1 of these five options only - num__17 has the form num__8 x - num__1 . the answer is b . <eor> b <eos> |
b |
multiply__4.0__2.0__ multiply__1.0__17.0__ |
multiply__4.0__2.0__ multiply__1.0__17.0__ |
| in a tree num__0.2 of the birds are robins while the rest are bluejays . if num__0.5 of the robins are female and num__0.4 of the bluejays are female what fraction of the birds in the tree are male ? <o> a ) num__0.36 <o> b ) num__0.48 <o> c ) num__0.68 <o> d ) num__0.58 <o> e ) num__0.66 |
the fraction of birds that are male robins is ( num__0.5 ) ( num__0.2 ) = num__0.1 . the fraction of birds that are male bluejays is ( num__0.6 ) ( num__0.8 ) = num__0.48 . the total fraction of male birds is num__0.1 + num__0.48 = num__0.58 . the answer is d . <eor> d <eos> |
d |
multiply__0.2__0.5__ add__0.2__0.4__ add__0.2__0.6__ multiply__0.8__0.6__ add__0.1__0.48__ add__0.1__0.48__ |
multiply__0.2__0.5__ add__0.2__0.4__ add__0.2__0.6__ multiply__0.8__0.6__ add__0.1__0.48__ add__0.1__0.48__ |
| the distance between two cities a and b is num__330 km . a train starts from a at num__8 a . m . and travels towards b at num__60 km / hr . another train starts from b at num__9 a . m . and travels towards a at num__75 km / hr . at what time do they meet ? <o> a ) num__12 a . m <o> b ) num__13 a . m <o> c ) num__14 a . m <o> d ) num__11 a . m <o> e ) num__16 a . m |
d num__11 a . m suppose they meet x hrs after num__8 a . m . then ( distance moved by first in x hrs ) + [ distance moved by second in ( x - num__1 ) hrs ] = num__330 num__60 x + num__75 ( x - num__1 ) = num__330 = > x = num__3 so they meet at ( num__8 + num__3 ) i . e . num__11 a . m . <eor> d <eos> |
d |
subtract__9.0__8.0__ subtract__11.0__8.0__ round__11.0__ |
subtract__9.0__8.0__ subtract__11.0__8.0__ add__8.0__3.0__ |
| the avg weight of a b & c is num__50 kg . if d joins the group the avg weight of the group becomes num__53 kg . if another man e who weights is num__3 kg more than d replaces a then the avgof b c d & e becomes num__51 kg . what is the weight of a ? <o> a ) num__56 <o> b ) num__65 <o> c ) num__73 <o> d ) num__89 <o> e ) num__90 |
a + b + c = num__3 * num__50 = num__150 a + b + c + d = num__4 * num__53 = num__212 - - - - ( i ) so d = num__62 & e = num__62 + num__3 = num__65 b + c + d + e = num__51 * num__4 = num__204 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = num__212 – num__204 = num__8 a = e + num__8 = num__65 + num__8 = num__73 answer : c <eor> c <eos> |
c |
multiply__50.0__3.0__ multiply__53.0__4.0__ subtract__212.0__150.0__ add__3.0__62.0__ multiply__51.0__4.0__ subtract__212.0__204.0__ add__65.0__8.0__ add__65.0__8.0__ |
multiply__50.0__3.0__ multiply__53.0__4.0__ subtract__212.0__150.0__ add__3.0__62.0__ multiply__51.0__4.0__ subtract__212.0__204.0__ add__65.0__8.0__ add__65.0__8.0__ |
| ifaequals the sum of the even integers from num__2 to num__120 inclusive andbequals the sum of the odd integers from num__1 to num__119 inclusive what is the value of a - b ? <o> a ) num__60 <o> b ) num__10 <o> c ) num__19 <o> d ) num__20 <o> e ) num__21 |
this is a solution from beatthegmat : even numbers : ( num__120 - num__2 ) / num__2 + num__1 = num__60 even integers . ( num__120 + num__2 ) / num__2 = num__61 is the average of the even set . sum = avg * ( # of elements ) = num__61 * num__60 = num__3660 = a odd numbers : ( num__119 - num__1 ) / num__2 + num__1 = num__60 odd integers . ( num__119 + num__1 ) / num__2 = num__60 is the average of the odd set . sum = avg * ( # of elements ) = num__60 * num__60 = num__3600 = b a - b = num__3660 - num__3600 = num__60 . answer : a <eor> a <eos> |
a |
divide__120.0__2.0__ add__1.0__60.0__ multiply__60.0__61.0__ subtract__3660.0__60.0__ divide__120.0__2.0__ |
divide__120.0__2.0__ add__1.0__60.0__ multiply__60.0__61.0__ subtract__3660.0__60.0__ divide__120.0__2.0__ |
| rebecca ' s annual income is $ num__15000 and jimmy ' s annual income is $ num__18000 . by how much must rebecca ' s annual income increase so that it constitutes num__50.0 of rebecca and jimmy ' s combined income ? <o> a ) num__7000 <o> b ) num__6000 <o> c ) num__5000 <o> d ) num__4000 <o> e ) num__3000 |
total rebecca = x + num__15000 ; total = x + num__15000 + num__18000 x + num__15000 / x + num__33000 = num__0.5 therefore x = num__3000 e <eor> e <eos> |
e |
add__15000.0__18000.0__ subtract__18000.0__15000.0__ subtract__18000.0__15000.0__ |
add__15000.0__18000.0__ subtract__18000.0__15000.0__ subtract__18000.0__15000.0__ |
| num__48 is divided into two parts in such a way that seventh part of first and ninth part of second are equal . find the smallest part ? <o> a ) num__66 <o> b ) num__26 <o> c ) num__42 <o> d ) num__27 <o> e ) num__21 |
x / num__7 = y / num__9 = > x : y = num__7 : num__9 num__0.4375 * num__48 = num__21 answer : e <eor> e <eos> |
e |
multiply__48.0__0.4375__ multiply__48.0__0.4375__ |
multiply__48.0__0.4375__ multiply__48.0__0.4375__ |
| the average amount with a group of seven numbers is rs . num__20 . if the newly joined member has rs . num__50 with him what was the average amount with the group before his joining the group ? <o> a ) rs . num__19 <o> b ) rs . num__29 <o> c ) rs . num__15 <o> d ) rs . num__10 <o> e ) rs . num__13 |
total members in the group = num__7 average amount = rs . num__20 total amount with them = num__7 * num__20 = rs . num__140 one number has rs . num__50 . so the amount with remaining num__6 people = num__140 - num__50 = rs . num__90 the average amount with them = num__15.0 = rs . num__15 . answer : c <eor> c <eos> |
c |
multiply__20.0__7.0__ subtract__140.0__50.0__ divide__90.0__6.0__ divide__90.0__6.0__ |
multiply__20.0__7.0__ subtract__140.0__50.0__ divide__90.0__6.0__ divide__90.0__6.0__ |
| a train num__110 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__7 sec <o> b ) num__6 sec <o> c ) num__4 sec <o> d ) num__5 sec <o> e ) num__9 sec |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__110 * num__0.0545454545455 ] sec = num__6 sec answer : b <eor> b <eos> |
b |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ round__6.0__ |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ divide__110.0__18.3333__ |
| ab + cd = rrr where ab and cd are two - digit numbers and rrr is a three digit number ; a b c and d are distinct positive integers . in the addition problem above what is the value of c ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__7 <o> d ) num__9 <o> e ) can not be determined |
ab and cd are two digit integers their sum can give us only one three digit integer of a kind of rrr it ' s num__111 . so a = num__1 . num__1 b + cd = num__111 now c can not be less than num__9 because no to digit integer with first digit num__1 ( mean that it ' s < num__20 ) can be added to two digit integer less than num__90 to have the sum num__111 ( if cd < num__90 meaning c < num__9 cd + num__1 b < num__111 ) - - > c = num__9 answer : d . <eor> d <eos> |
d |
multiply__1.0__9.0__ |
multiply__1.0__9.0__ |
| working together printer a and printer b would finish the task in num__40 minutes . printer a alone would finish the task in num__60 minutes . how many pages does the task contain if printer b prints num__4 pages a minute more than printer a ? <o> a ) num__250 <o> b ) num__375 <o> c ) num__450 <o> d ) num__480 <o> e ) num__500 |
num__40 * a + num__40 * b = x pages in num__40 mins printer a will print = num__0.666666666667 * x pages = num__0.666666666667 * x pages thus in num__40 mins printer printer b will print x - num__0.666666666667 * x = num__0.333333333333 * x pages also it is given that printer b prints num__4 more pages per min that printer a . in num__40 mins printer b will print num__160 more pages than printer a thus num__0.666666666667 * x - num__0.333333333333 * x = num__160 = > x = num__480 pages answer : d <eor> d <eos> |
d |
divide__40.0__60.0__ multiply__40.0__4.0__ round__480.0__ |
divide__40.0__60.0__ multiply__40.0__4.0__ round__480.0__ |
| a train passes a man standing on the platform . if the train is num__170 meters long and its speed is num__72 kmph how much time it took in doing so ? <o> a ) num__6 ½ sec <o> b ) num__6 ½ sec <o> c ) num__9 ½ sec <o> d ) num__8 ½ sec <o> e ) num__7 ½ sec |
d num__8 ½ sec d = num__170 s = num__72 * num__0.277777777778 = num__20 mps t = num__8.5 = num__8 ½ sec <eor> d <eos> |
d |
divide__170.0__20.0__ round__8.0__ |
divide__170.0__20.0__ round__8.0__ |
| the surface of a cube is num__96 sq cm . find its volume ? <o> a ) num__8 cc <o> b ) num__9 cc <o> c ) num__2 cc <o> d ) num__4 cc <o> e ) num__64 cc |
num__6 a num__2 = num__96 = num__6 * num__16 a = num__4 = > a num__3 = num__64 cc answer : e <eor> e <eos> |
e |
rectangle_perimeter__2.0__6.0__ volume_cube__4.0__ volume_cube__4.0__ |
rectangle_perimeter__2.0__6.0__ multiply__4.0__16.0__ multiply__4.0__16.0__ |
| a meal cost $ num__35.50 and there was no tax . if the tip was more than num__10 pc but less than num__15 pc of the price then the total amount paid should be : <o> a ) num__40 - num__42 <o> b ) num__39 - num__41 <o> c ) num__38 - num__40 <o> d ) num__37 - num__39 <o> e ) num__36 - num__37 |
num__10.0 ( num__35.5 ) = num__3.55 num__15.0 ( num__35.5 ) = num__5.325 total amount could have been num__35.5 + num__3.55 and num__35.5 + num__5.325 = > could have been between num__39.05 and num__40.625 = > approximately between num__39 and num__41 answer is b . <eor> b <eos> |
b |
divide__35.5__10.0__ add__35.5__3.55__ round_down__39.05__ round_down__39.05__ |
divide__35.5__10.0__ add__35.5__3.55__ round_down__39.05__ round_down__39.05__ |
| eight friends met for lunch at an expensive restaurant and everyone decided to contribute equally to the total bill of $ num__290 . if one of the friends had a coupon for num__29.0 off the total bill and if each friend still contributed equally after the coupon was applied to the bill how much did each friend pay ? <o> a ) $ num__90 <o> b ) $ num__70 <o> c ) $ num__45 <o> d ) $ num__37 <o> e ) $ num__25 |
num__290 * num__029 = num__84 num__290 - num__84 = num__206 num__25.75 = num__25 answer e <eor> e <eos> |
e |
subtract__290.0__84.0__ round_down__25.75__ round_down__25.75__ |
subtract__290.0__84.0__ round_down__25.75__ round_down__25.75__ |
| what is the sum of the prime factors of num__3 and the prime factors of num__1656 ? <o> a ) num__35 <o> b ) num__45 <o> c ) num__55 <o> d ) num__25 <o> e ) num__65 |
the prime factors of num__3 are num__3 * num__1 the prime factors of num__1656 are num__2 x num__2 x num__2 x num__3 x num__3 x num__23 num__2 + num__2 + num__2 + num__3 + num__3 + num__23 = num__35 . a is the correct answer <eor> a <eos> |
a |
subtract__3.0__1.0__ multiply__1.0__35.0__ |
subtract__3.0__1.0__ multiply__1.0__35.0__ |
| the perimeter of one square is num__44 cm and that of another is num__20 cm . find the perimeter and the diagonal of a square which is equal in area to these two combined ? <o> a ) num__13 √ num__4 <o> b ) num__12.1 √ num__2 <o> c ) num__23 √ num__2 <o> d ) num__12 √ num__4 <o> e ) num__13 √ num__9 |
num__4 a = num__44 num__4 a = num__20 a = num__11 a = num__5 a num__2 = num__121 a num__2 = num__25 combined area = a num__2 = num__146 = > a = num__12.1 d = num__12.1 √ num__2 answer : b <eor> b <eos> |
b |
power__11.0__2.0__ power__5.0__2.0__ triangle_perimeter__20.0__5.0__121.0__ triangle_area__2.0__12.1__ |
power__11.0__2.0__ power__5.0__2.0__ triangle_perimeter__20.0__5.0__121.0__ triangle_area__2.0__12.1__ |
| three pipes a b and c can fill a tank in num__6 hours . after working at it together for num__2 hours c is closed and a and b can fill the remaining part in num__9 hours . how many hours will take c alone to fill the tank ? <o> a ) num__11.5 <o> b ) num__13.6666666667 <o> c ) num__12.25 <o> d ) num__10.8 <o> e ) num__10.5 |
the rate of a + b + c is num__0.166666666667 of the tank per hour . after num__2 hours the tank is num__0.333333333333 full . the rate of a + b is num__0.666666666667 * num__0.111111111111 = num__0.0740740740741 of a tank per hour . the rate of c is num__0.166666666667 - num__0.0740740740741 = num__0.0925925925926 c can fill the tank in num__10.8 hours . the answer is d . <eor> d <eos> |
d |
divide__2.0__6.0__ divide__6.0__9.0__ multiply__0.1667__0.6667__ multiply__0.1111__0.6667__ subtract__0.1667__0.0741__ round__10.8__ |
divide__2.0__6.0__ divide__6.0__9.0__ multiply__0.1667__0.6667__ multiply__0.1111__0.6667__ subtract__0.1667__0.0741__ round__10.8__ |
| if the reciprocals of two consecutive positive integers are added together what is the sum in terms of the greater integer x ? <o> a ) x / num__3 <o> b ) x ^ num__2 - x <o> c ) num__2 x - num__1 <o> d ) ( num__2 x - num__1 ) / ( x ^ num__2 + x ) <o> e ) ( num__2 x - num__1 ) / ( x ^ num__2 - x ) |
let two consecutive positive integers be x and x - num__1 ( greater integer is x ) so ( num__1 / x ) + [ num__1 / ( x - num__1 ) ] = ( num__2 x - num__1 ) / x ( x - num__1 ) = ( num__2 x - num__1 ) / ( x ^ num__2 - x ) answer e <eor> e <eos> |
e |
multiply__1.0__2.0__ |
divide__2.0__1.0__ |
| a person starting with rs . num__64 & making num__6 bets wins three times and loses num__3 times the wins and loses occurring in random order . the chance for a win is equal to the chance for a loss . if eachwager is for half the money remaining at thetime of the bet then the final result is ? <o> a ) rs . num__37 <o> b ) rs . num__57 <o> c ) rs . num__60 <o> d ) rs . num__64 <o> e ) rs . num__68 |
as the win leads to multiplying the amount by num__1.5 and loss leads to multiplying the amount by num__0.5 we will multiply initial amount by num__1.5 thrice and by num__0.5 thrice ( in any order ) . the overall resultant will remain same . so final amount with the person will be ( in all cases ) : = num__64 ( num__1.5 ) ( num__1.5 ) ( num__1.5 ) ( num__0.5 ) ( num__0.5 ) ( num__0.5 ) = = num__64 ( num__1.5 ) ( num__1.5 ) ( num__1.5 ) ( num__0.5 ) ( num__0.5 ) ( num__0.5 ) = rs num__2727 hence the final result is : num__64 − num__27 = num__3764 − num__27 = num__37 : a loss of rs . num__37 a <eor> a <eos> |
a |
divide__3.0__6.0__ subtract__64.0__27.0__ subtract__64.0__27.0__ |
divide__3.0__6.0__ subtract__64.0__27.0__ subtract__64.0__27.0__ |
| when num__100 is divided by positive integer x the remainder is num__4 . what is the remainder when num__196 is divided by x ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
f num__100 / x leaves a reminder num__4 then ( num__100 - num__4 ) i . e . num__96 is divisible by x so ( num__100 + num__96 ) / x leaves a reminder rem ( num__100 / x ) + rem ( num__96 / x ) = > num__4 + num__0 = num__4 answer : c <eor> c <eos> |
c |
subtract__100.0__4.0__ subtract__100.0__96.0__ |
subtract__100.0__4.0__ subtract__100.0__96.0__ |
| if num__3 persons can do num__3 times of a particular work in num__3 days then num__7 persons can do num__7 times of that work in ? <o> a ) num__2 days <o> b ) num__3 days <o> c ) num__5 days <o> d ) num__7 days <o> e ) num__9 days |
that is num__1 person can do one time of the work in num__3 days . therefore num__7 persons can do num__7 times work in the same num__3 days itself . b ) <eor> b <eos> |
b |
round__3.0__ |
round__3.0__ |
| nicky and cristina are running a race . since cristina is faster than nicky she gives him a num__48 meter head start . if cristina runs at a pace of num__5 meters per second and nicky runs at a pace of only num__3 meters per second how many seconds will nicky have run before cristina catches up to him ? <o> a ) num__15 seconds <o> b ) num__18 seconds <o> c ) num__24 seconds <o> d ) num__30 seconds <o> e ) num__45 seconds |
used pluging in method say t is the time for cristina to catch up with nicky the equation will be as under : for nicky = n = num__3 * t + num__48 for cristina = c = num__5 * t @ t = num__24 n = num__120 c = num__120 right answer ans : c <eor> c <eos> |
c |
multiply__5.0__24.0__ round__24.0__ |
multiply__5.0__24.0__ round__24.0__ |
| paul completes a piece of work in num__80 days rose completes the same work in num__120 days . if both of them work together then the number of days required to complete the work is ? <o> a ) num__42 <o> b ) num__44 <o> c ) num__46 <o> d ) num__48 <o> e ) num__50 |
if a can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days . that is the required no . of days = num__80 × num__0.6 = num__48 days answer is d <eor> d <eos> |
d |
km_to_mile_conversion__ multiply__80.0__0.6__ round__48.0__ |
km_to_mile_conversion__ multiply__80.0__0.6__ round__48.0__ |
| if log num__303 = a log num__305 = b then log num__308 = ? <o> a ) num__3 ( num__1 - a - b ) <o> b ) ( a - b + num__1 ) <o> c ) ( num__1 - a - b ) <o> d ) ( a - b + num__1 ) <o> e ) none of these |
a + b = log num__3015 = log num__30 ( num__15.0 ) = num__1 - log num__302 = > log num__302 = ( num__1 - a - b ) therefore log num__308 = num__3 ( num__1 - a - b ) . answer : a <eor> a <eos> |
a |
subtract__303.0__1.0__ subtract__305.0__302.0__ subtract__305.0__302.0__ |
subtract__303.0__1.0__ subtract__305.0__302.0__ subtract__305.0__302.0__ |
| if a - b = num__6 and a ^ num__2 + b ^ num__2 = num__100 find the value of ab <o> a ) num__10 <o> b ) num__32 <o> c ) num__15 <o> d ) num__18 <o> e ) num__19 |
num__2 ab = ( a ^ num__2 + b ^ num__2 ) - ( a - b ) ^ num__2 = num__100 - num__36 = num__64 = > ab = num__32 answer : b <eor> b <eos> |
b |
subtract__100.0__36.0__ divide__64.0__2.0__ divide__64.0__2.0__ |
subtract__100.0__36.0__ divide__64.0__2.0__ subtract__64.0__32.0__ |
| find the area of a parallelogram with base num__24 cm and height num__16 cm . <o> a ) num__384 cm ^ num__2 <o> b ) num__200 cm ^ num__2 <o> c ) num__250 cm ^ num__2 <o> d ) num__350 cm ^ num__2 <o> e ) num__400 cm ^ num__2 |
area of a parallelogram = base * height = num__24 * num__16 = num__384 cm num__2 answer a <eor> a <eos> |
a |
multiply__24.0__16.0__ multiply__24.0__16.0__ |
multiply__24.0__16.0__ multiply__24.0__16.0__ |
| if q is an odd number and the median of q consecutive integers is num__110 what is the largest of these integers ? <o> a ) ( q - num__1 ) / num__2 + num__110 <o> b ) q / num__2 + num__109 <o> c ) q / num__2 + num__110 <o> d ) ( q + num__109 ) / num__2 <o> e ) ( q + num__110 ) / num__2 |
consider the easiest case say q = num__3 then ; set = { num__109 num__110 num__111 } ; the largest integer = num__111 . now plug q = num__3 into the answers to see which yields num__111 . answer : a . <eor> a <eos> |
a |
subtract__110.0__109.0__ |
subtract__110.0__109.0__ |
| what is the value of num__3 x ^ num__2 − num__2 x + num__0.3 for x = num__0.5 ? <o> a ) − num__0.3 <o> b ) num__0 <o> c ) num__0.05 <o> d ) num__1.08 <o> e ) num__2.46 |
num__3 x ^ num__2 - num__2 x + num__0.3 for x = num__0.6 = num__3 ( num__0.5 * num__0.5 ) - num__4 * num__0.5 * ( num__0.5 ) + num__0.3 = - num__0.5 * num__0.5 + num__0.3 = - num__0.25 + num__0.3 = num__0.05 correct option : c <eor> c <eos> |
c |
multiply__2.0__0.3__ divide__2.0__0.5__ reverse__4.0__ subtract__0.3__0.25__ subtract__0.3__0.25__ |
multiply__2.0__0.3__ divide__2.0__0.5__ reverse__4.0__ subtract__0.3__0.25__ subtract__0.3__0.25__ |
| the avg . age of a group of num__32 students is num__20 years . if num__4 more students join the group the avg age increases by num__1 year . the avg age of the new student is ? <o> a ) num__22 years <o> b ) num__23 years <o> c ) num__24 years <o> d ) num__25 years <o> e ) num__29 years |
total age of num__32 students = num__32 * num__20 = num__640 if total age of num__4 students = x then ( num__640 + x ) / ( num__32 + num__4 ) = ( num__20 + num__1 ) x = num__116 so average age of new students = num__29.0 = num__29 years answer : e <eor> e <eos> |
e |
multiply__32.0__20.0__ divide__116.0__4.0__ multiply__1.0__29.0__ |
multiply__32.0__20.0__ divide__116.0__4.0__ multiply__1.0__29.0__ |
| a train num__120 m long passed a pole in num__20 sec . how long will it take to pass a platform num__780 m long ? <o> a ) num__150 <o> b ) num__170 <o> c ) num__160 <o> d ) num__100 <o> e ) num__120 |
speed = num__6.0 = num__6 m / sec . required time = ( num__120 + num__780 ) / num__6 = num__150 sec . answer : option a <eor> a <eos> |
a |
divide__120.0__20.0__ round__150.0__ |
divide__120.0__20.0__ round__150.0__ |
| raju age after num__9 years will be num__5 times his age num__3 years back what is the present age of raju <o> a ) num__8 years <o> b ) num__7 years <o> c ) num__6 years <o> d ) num__5 years <o> e ) num__4 years |
explanation : clearly x + num__9 = num__5 ( x - num__3 ) < = > num__4 x = num__24 = > x = num__6 option c <eor> c <eos> |
c |
subtract__9.0__5.0__ subtract__9.0__3.0__ subtract__9.0__3.0__ |
subtract__9.0__5.0__ subtract__9.0__3.0__ subtract__9.0__3.0__ |
| the perimeter of a rhombus is num__120 feet and one of its diagonal has a length of num__40 feet . find the area of the rhombus . <o> a ) num__100 sqrt ( num__5 ) ft . sq <o> b ) num__400 sqrt ( num__5 ) ft . sq <o> c ) num__500 sqrt ( num__5 ) ft . sq <o> d ) num__200 sqrt ( num__5 ) ft . sq <o> e ) num__150 sqrt ( num__5 ) ft . sq |
a perimeter of num__120 when divided by num__4 gives the side of the rhombus num__30 feet . the length of the side oc of the right triangle is equal to half the diagonal : num__20 feet . let us now consider the right triangle boc and apply pythagora ' s theorem to find the length of side bo . num__30 ^ num__2 = bo ^ num__2 + num__20 ^ num__2 bo = num__10 sqrt ( num__5 ) feet we now calculate the area of the right triangle boc and multiply it by num__4 to obtain the area of the rhombus . area = num__4 ( num__0.5 ) bo * oc = num__4 ( num__0.5 ) num__10 sqrt ( num__5 ) * num__20 = num__400 sqrt ( num__5 ) ft . sq answer is b <eor> b <eos> |
b |
multiply__40.0__10.0__ multiply__40.0__10.0__ |
multiply__40.0__10.0__ multiply__40.0__10.0__ |
| ( num__0.5 ) ( power num__3 ) - ( num__0.1 ) ( power num__3 ) / ( num__0.5 ) ( power num__2 ) + num__0.05 + ( num__0.1 ) ( power num__2 ) is : <o> a ) num__0.8 <o> b ) num__0.4 <o> c ) num__0.96 <o> d ) num__0.69 <o> e ) num__0.76 |
given expression = ( num__0.5 ) ( power num__3 ) - ( num__0.1 ) ( power num__3 ) / ( num__0.5 ) ( power num__2 ) + ( num__0.5 x num__0.1 ) + ( num__0.1 ) ( power num__2 ) = a ( power num__3 ) - b ( power num__3 ) / a ( power num__2 ) + ab + b ( power num__2 ) = ( a - b ) = ( num__0.5 - num__0.1 ) = num__0.4 answer is b <eor> b <eos> |
b |
subtract__0.5__0.1__ subtract__0.5__0.1__ |
subtract__0.5__0.1__ subtract__0.5__0.1__ |
| set k contains every multiple of num__6 from num__18 to num__306 inclusive . if w is the median of set k and x is the average ( arithmetic mean ) of set k what is the value of w - x ? <o> a ) - num__6 <o> b ) - num__3 <o> c ) num__0 <o> d ) num__3 <o> e ) num__6 |
set k = { num__18 num__2430 . . . . num__306 } let the no . of terms be n . so we know the ap formulae to calculate the the tn term : tn = a + ( n - num__1 ) * d where d = common difference of the terms . num__306 = num__18 + ( n - num__1 ) * num__6 n = num__49 . so the set k consist of n terms . the median of the set having odd nos of elements is ( n + num__1 ) / num__2 . which in this case is num__25 . let ' s find the num__25 th term using the same formulae again : t num__25 = num__18 + ( num__25 - num__1 ) * num__6 t num__25 = num__162 so the median of the set k is num__162 i . e . w = num__162 . now . lets find the average ( arithmetic mean ) of the set . for that we need to find the sum of all the elements first lets call it s . since set k is nothing but a arithmetic progression series having first element ( a ) as num__18 common difference ( d ) as num__6 and no . of terms ( n ) as num__49 . using the formulae to calculate sum of an ap series which is s = n / num__2 [ num__2 a + ( n - num__1 ) * d ] we will calculate the sum . so s = num__24.5 [ num__2 * num__18 + ( num__49 - num__1 ) * num__6 ] this gives us s = num__7938 . now arithmetic mean of set k = num__7938 / no . of terms = num__162.0 = num__162 . so x = num__162 . now ( w - x ) = ( num__162 - num__162 ) = num__0 . therefore correct answer = c <eor> c <eos> |
c |
divide__49.0__2.0__ multiply__162.0__49.0__ multiply__6.0__0.0__ |
divide__49.0__2.0__ multiply__162.0__49.0__ multiply__6.0__0.0__ |
| find out the c . i on rs . num__3000 at num__4.0 p . a . compound half - yearly for num__1 num__0.5 years . <o> a ) a ) rs . num__181.62 <o> b ) b ) rs . num__182.62 <o> c ) c ) rs . num__183.62 <o> d ) d ) rs . num__184.62 <o> e ) e ) rs . num__185.62 |
a = num__3000 ( num__1.02 ) num__3 = num__3183.62 num__3000 - - - - - - - - - - - num__183.62 answer : c <eor> c <eos> |
c |
subtract__4.0__1.0__ subtract__3183.62__3000.0__ multiply__1.0__183.62__ |
subtract__4.0__1.0__ subtract__3183.62__3000.0__ subtract__3183.62__3000.0__ |
| if a man can cover num__20 metres in one second how many kilometres can he cover in num__3 hours num__45 minutes ? <o> a ) num__288 <o> b ) num__162 <o> c ) num__220 <o> d ) num__270 <o> e ) num__122 |
num__20 m / s = num__20 * num__3.6 kmph num__3 hours num__45 minutes = num__3 num__0.75 hours = num__3.75 hours distance = speed * time = num__20 * num__3.6 * num__3.75 km = num__270 km . answer : d <eor> d <eos> |
d |
add__3.0__0.75__ round__270.0__ |
add__3.0__0.75__ round__270.0__ |
| the area of a square field is num__7201 sq m . how long will a lady take to cross the field diagonally at the rate of num__2.4 km / hr ? <o> a ) num__1 min num__15 sec <o> b ) num__1.5 min <o> c ) num__1 min <o> d ) num__2.5 min <o> e ) num__2 min |
area of a square field = num__7201 sq m let the side of square = a a ^ num__2 = num__7201 = > a = num__84.86 diagonal = ( num__2 ) ^ ( num__0.5 ) * a = num__1.414 * num__84.86 = num__120 speed of lady = num__2.4 km / hour = num__2400 m / hour = num__60 m / min time taken by lady to cross the field diagonally = num__2.0 = num__2 min answer e <eor> e <eos> |
e |
multiply__0.5__120.0__ square_perimeter__0.5__ |
multiply__0.5__120.0__ square_perimeter__0.5__ |
| what is the rate percent when the simple interest on rs . num__5000 amount to rs . num__2500 in num__5 years ? <o> a ) num__9.0 <o> b ) num__8.0 <o> c ) num__10.0 <o> d ) num__10.5 <o> e ) num__7.5 % |
interest for num__1 year = num__500.0 = num__500 interest on rs num__5000 p / a = num__500 interest rate = num__0.1 * num__100 = num__10.0 answer : c <eor> c <eos> |
c |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| the speed of light is approximately num__186 * num__10 ^ num__6 miles per second . this approximate speed is how many miles per hour ? <o> a ) num__111 * num__10 ^ num__7 <o> b ) num__670 * num__10 ^ num__7 <o> c ) num__111 * num__10 ^ num__8 <o> d ) num__6 num__70 * num__10 ^ num__9 <o> e ) num__670 * num__10 ^ num__8 |
the easiest way to answer this question is by poe it does not require any calculation ( answered in num__30 seconds ) . in one hour there are num__3600 seconds therefore speed in miles / hour will be . ( num__1.86 * num__10 ^ num__6 ) * num__3600 = ( num__1.86 * num__10 ^ num__6 ) * ( num__3.6 * num__10 ^ num__3 ) = some number * num__10 ^ num__9 . . the key is realizing that thesome numberhas to be bigger thatn num__1.86 ( because num__1.86 * num__3.6 ) and the only answer that fits that is d <eor> d <eos> |
d |
divide__30.0__10.0__ add__6.0__3.0__ round__6.0__ |
divide__30.0__10.0__ add__6.0__3.0__ round__6.0__ |
| if y > num__0 ( num__8 y ) / num__20 + ( num__3 y ) / num__10 is what percent of y ? <o> a ) num__40.0 <o> b ) num__50.0 <o> c ) num__60.0 <o> d ) num__70.0 <o> e ) num__80 % |
can be reduced to num__4 y / num__10 + num__3 y / num__10 = num__7 y / num__10 = num__70.0 answer d <eor> d <eos> |
d |
add__3.0__4.0__ multiply__10.0__7.0__ multiply__10.0__7.0__ |
add__3.0__4.0__ multiply__10.0__7.0__ multiply__10.0__7.0__ |
| if p q and r are positive integers and satisfy x = ( p + q - r ) / r = ( p - q + r ) / q = ( q + r - p ) / p then the value of x is ? <o> a ) - num__1 <o> b ) num__1 <o> c ) num__2 <o> d ) - num__2 <o> e ) - num__3 |
when two or more ratios are equal each of the ratios are equal to sum of the numerators divided by the sum of the denominators provided sum of the denominators is non - zero . hence x = ( p + q - r ) / r = ( p - q + r ) / q = ( q + r - p ) / p = > x = ( p + q - r + p - q + r + q + r - p ) / ( r + q + p ) = > x = ( r + q + p ) / ( r + q + p ) = num__1 p + q + r is non - zero . answer : b <eor> b <eos> |
b |
reverse__1.0__ |
reverse__1.0__ |
| a diagonal is a line segment that connects non - adjacent vertices in a polygon . how many diagonals does an octagon have ? <o> a ) num__20 <o> b ) num__40 <o> c ) num__60 <o> d ) num__90 <o> e ) num__80 |
select one vertex and you can draw num__5 diagonals . if you draw num__5 such diagonals for each of the num__8 vertices you will draw a total of num__5 × num__8 = num__40 diagonals . but you actually draw each diagonal twice once from each of its ends . therefore there are a total of num__40 ÷ num__2 = num__20 different diagonals correct answer a <eor> a <eos> |
a |
multiply__8.0__5.0__ square_perimeter__5.0__ square_perimeter__5.0__ |
multiply__8.0__5.0__ square_perimeter__5.0__ square_perimeter__5.0__ |
| a watch was sold at a loss of num__10.0 . if it was sold for rs . num__140 more there would have been a gain of num__4.0 . what is the cost price ? <o> a ) num__1000 <o> b ) num__2287 <o> c ) num__2677 <o> d ) num__2887 <o> e ) num__2688 |
num__90.0 num__104.0 - - - - - - - - num__14.0 - - - - num__140 num__100.0 - - - - ? = > rs . num__1000 answer : a <eor> a <eos> |
a |
percent__10.0__140.0__ percent__100.0__1000.0__ |
percent__10.0__140.0__ percent__100.0__1000.0__ |
| what least value must be assigned to * so that the number num__197 * num__5462 is divisible by num__9 <o> a ) num__0 <o> b ) num__2 <o> c ) num__4 <o> d ) num__9 <o> e ) num__8 |
let the missing digit be x . sum of digits = ( num__1 + num__9 + num__7 + x + num__5 + num__4 + num__6 + » num__2 ) = ( num__34 + x ) . for ( num__34 + x ) to be divisible by num__9 x must be replaced by num__2 . hence the digit in place of * must be num__2 . answer b num__2 <eor> b <eos> |
b |
subtract__9.0__5.0__ add__1.0__5.0__ subtract__9.0__7.0__ subtract__9.0__7.0__ |
subtract__9.0__5.0__ add__1.0__5.0__ subtract__9.0__7.0__ multiply__1.0__2.0__ |
| a bowl was filled with num__120 ounces of milk and num__0.01 ounce of the milk evaporated each day during a num__60 - day period . what percent of the original amount of milk evaporated during this period ? <o> a ) a ) num__0.005 <o> b ) b ) num__0.05 <o> c ) c ) num__5.0 <o> d ) d ) num__0.5 <o> e ) e ) num__4.2 % |
total amount of milk evaporated each day during a num__60 - day period = . num__01 * num__60 = num__0.6 percent of the original amount of milk evaporated during this period = ( . num__0.05 ) * num__100.0 = num__0.5 answer d <eor> d <eos> |
d |
percent__60.0__1.0__ percent__0.5__100.0__ |
percent__60.0__1.0__ percent__0.5__100.0__ |
| a cistern which could be filled in num__9 hours takes one hour more to be filled owing to a leak in its bottom . if the cistern is full in what time will the leak empty it ? <o> a ) num__22 <o> b ) num__27 <o> c ) num__29 <o> d ) num__90 <o> e ) num__12 |
num__0.111111111111 - num__1 / x = num__0.1 = > num__90 hrs answer : d <eor> d <eos> |
d |
divide__9.0__0.1__ round__90.0__ |
divide__9.0__0.1__ divide__9.0__0.1__ |
| one woman and one man can build a wall together in three hours but the woman would need the help of two girls in order to complete the same job in the same amount of time . if one man and one girl worked together it would take them five hours to build the wall . assuming that rates for men women and girls remain constant how many hours would it take one woman one man and one girl working together to build the wall ? <o> a ) num__2.5 <o> b ) num__1 <o> c ) num__1.42857142857 <o> d ) num__1.71428571429 <o> e ) num__3.14285714286 |
solution : let work done by man women and girl per hour be m w g respectively . then m + w = num__0.333333333333 - - > ( num__1 ) w + num__2 g = num__0.333333333333 - - > ( num__2 ) and m + g = num__0.2 - - > ( num__3 ) . no . of hours it would take forone woman one man and one girl working together to build the wall n = num__1 / m + w + g from ( num__1 ) and ( num__2 ) m = num__2 g and from ( num__3 ) g = num__0.0666666666667 m = num__0.133333333333 and w = num__0.2 . so n = num__1 / ( num__0.4 ) = num__2.5 option a <eor> a <eos> |
a |
add__1.0__2.0__ divide__0.2__3.0__ subtract__0.2__0.0667__ multiply__2.0__0.2__ divide__1.0__0.4__ round__2.5__ |
add__1.0__2.0__ divide__0.2__3.0__ subtract__0.2__0.0667__ add__0.3333__0.0667__ divide__1.0__0.4__ divide__2.5__1.0__ |
| num__5358 x num__61 = ? <o> a ) num__326838 <o> b ) num__323758 <o> c ) num__323298 <o> d ) num__273258 <o> e ) num__327382 |
a num__5358 x num__61 = num__5358 x ( num__60 + num__1 ) = num__5358 x num__60 + num__5358 x num__1 = num__321480 + num__5358 = num__326838 . <eor> a <eos> |
a |
subtract__61.0__60.0__ multiply__5358.0__60.0__ multiply__5358.0__61.0__ multiply__5358.0__61.0__ |
subtract__61.0__60.0__ multiply__5358.0__60.0__ add__5358.0__321480.0__ add__5358.0__321480.0__ |
| if sn is the sum of the first n terms of a certain sequence and if sn = n ( n + num__3 ) for all positive integers n what is the third term of the sequence ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
tn = nth term t num__1 = num__1 st term t num__2 = num__2 nd term and so on . . . sn = sum of first n terms of the sequence sn = num__1 st term + num__2 nd term + num__3 rd term + . . . + nth term sn = t num__1 + t num__2 + t num__3 + . . . . tn you are given here thatsn is the sum of first n terms . . . so you have sn = t num__1 + t num__2 + t num__3 + . . . . tn = n ( n + num__1 ) so s num__1 = t num__1 s num__2 = t num__1 + t num__2 s num__3 = t num__1 + t num__2 + t num__3 and so on s num__1 = t num__1 = num__1 * ( num__1 + num__1 ) = num__2 s num__2 = t num__1 + t num__2 = num__2 + t num__2 = num__2 * ( num__2 + num__1 ) = num__6 so t num__2 = num__4 s num__3 = t num__1 + t num__2 + t num__3 = num__2 + num__4 + t num__3 = num__3 * ( num__3 + num__1 ) = num__12 so t num__3 = num__8 ( this is what we wanted ) the third term is num__8 . d <eor> d <eos> |
d |
subtract__3.0__1.0__ multiply__3.0__2.0__ add__3.0__1.0__ multiply__3.0__4.0__ multiply__2.0__4.0__ multiply__1.0__8.0__ |
subtract__3.0__1.0__ multiply__3.0__2.0__ add__3.0__1.0__ multiply__3.0__4.0__ multiply__2.0__4.0__ multiply__1.0__8.0__ |
| on the independence day bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day num__130 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ? <o> a ) num__260 <o> b ) num__620 <o> c ) num__500 <o> d ) num__520 <o> e ) num__720 |
let the number of children in the school be x . since each child gets num__2 bananas total number of bananas = num__2 x . num__2 x / ( x - num__130 ) = num__2 + num__2 ( extra ) = > num__2 x - num__260 = x = > x = num__260 . answer : a <eor> a <eos> |
a |
multiply__130.0__2.0__ multiply__130.0__2.0__ |
multiply__130.0__2.0__ multiply__130.0__2.0__ |
| a school has received num__60.0 of the amount it needs for a new building by receiving a donation of $ num__300 each from people already solicited . people already solicited represent num__40.0 of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to complete the fund raising exercise ? <o> a ) $ num__200 <o> b ) $ num__177.78 <o> c ) $ num__100 <o> d ) $ num__277.78 <o> e ) $ num__377.78 |
let the amount school needs = x let total people school plans to solicit = t school has received num__60.0 of x = > ( num__0.6 ) x people already solicited = num__40.0 of t = > ( num__0.4 ) t now as per the information given in the question : ( num__0.6 ) x = $ num__400 . ( num__0.4 ) . t - - - - - - - - - - - - - - - - - - - - - - - - - - - num__1 remaning amount is num__40.0 i . e ( num__0.4 ) x - - - - - - because school has already received num__60.0 and the remaining people are num__60.0 i . e ( num__0.6 ) . t - - - - - because num__40.0 of the people are already solicited so average contribution required from the remaining targeted people is ( num__0.4 ) x = ( amount required ) . ( num__0.6 ) . t - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - num__2 divide eqn num__1 by eqn num__2 amount required = $ num__177.78 b <eor> b <eos> |
b |
add__0.4__0.6__ multiply__1.0__177.78__ |
add__0.4__0.6__ multiply__1.0__177.78__ |
| the star running back on our football team got most of his total yardage running . the rest was catching passes . he caught passes for num__60 yards . his total yardage was num__150 yards . the running back for the other team got num__200 yards . how many yards did the star running back on our football team get running ? <o> a ) num__50 yards <o> b ) num__60 yards <o> c ) num__70 yards <o> d ) num__80 yards <o> e ) num__90 yards |
. the other team is extra information . num__150 – num__60 = num__90 he got num__90 yards running . correct answer e <eor> e <eos> |
e |
subtract__150.0__60.0__ subtract__150.0__60.0__ |
subtract__150.0__60.0__ subtract__150.0__60.0__ |
| num__25 people are there they are shaking hands together how many hand shakes possible if they are in pair of cyclic sequence . <o> a ) num__24 <o> b ) num__25 <o> c ) num__44 <o> d ) num__48 <o> e ) num__50 |
n for cyclic hand shake and n - num__1 for linear handshake . so here no of hand shake is num__25 . answer : b <eor> b <eos> |
b |
multiply__25.0__1.0__ |
multiply__25.0__1.0__ |
| a man complete a journey in num__10 hours . he travels first half of the journey at the rate of num__21 km / hr and second half at the rate of num__26 km / hr . find the total journey in km . <o> a ) num__220 km <o> b ) num__224 km <o> c ) num__230 km <o> d ) num__200 km <o> e ) num__234 km |
num__0.5 x / num__10 + num__0.5 x / num__10 = num__20 - - > x / num__10 + x / num__10 = num__40 - - > num__2 x = num__10 x num__40 - - > x = ( num__10 x num__40 ) / num__2 = num__200 km . answer : d . <eor> d <eos> |
d |
divide__10.0__0.5__ divide__20.0__0.5__ divide__40.0__20.0__ multiply__10.0__20.0__ round__200.0__ |
divide__10.0__0.5__ divide__20.0__0.5__ divide__40.0__20.0__ multiply__10.0__20.0__ round__200.0__ |
| arun borrowed a sum for num__4 years on s . i . at num__12.0 . the total interest paid was rs . num__360 . find the principal . <o> a ) rs . num__700 <o> b ) rs . num__650 <o> c ) rs . num__800 <o> d ) rs . num__750 <o> e ) rs . num__850 |
explanation : p = num__100 × si / rt = num__100 × num__30.0 × num__4 = num__100 × num__7.5 = num__25 × num__30 = num__750 answer : option d <eor> d <eos> |
d |
percent__100.0__750.0__ |
percent__100.0__750.0__ |
| jim wishes to buy num__3 gifts that cost num__23 dollars num__9 dollars and num__12 dollars . he has num__0.25 of the money he needs . how much more money must he earn in order to buy the gifts ? <o> a ) $ num__57 <o> b ) $ num__47 <o> c ) $ num__33 <o> d ) $ num__17 <o> e ) $ num__27 |
$ num__23 + $ num__9 + $ num__12 = $ num__44 needed num__0.25 x $ num__44 = $ num__11 on hand $ num__44 - $ num__11 = $ num__33 to earn correct answer c <eor> c <eos> |
c |
subtract__23.0__12.0__ multiply__3.0__11.0__ multiply__3.0__11.0__ |
subtract__23.0__12.0__ subtract__44.0__11.0__ subtract__44.0__11.0__ |
| a tank is filled in num__5 hours by three pipes a b and c . the pipe c is twice as fast as b and b is twice as fast as a . how much time will pipe a alone take to fill the tank ? <o> a ) num__77 hrs <o> b ) num__81 hrs <o> c ) num__35 hrs <o> d ) num__17 hrs <o> e ) num__19 hrs |
suppose pipe a alone takes x hours to fill the tank . then pipes b and c will take x / num__2 and x / num__4 hours respectively to fill the tank . num__1 / x + num__2 / x + num__4 / x = num__0.2 num__7 / x = num__0.2 = > x = num__35 hrs . answer : c <eor> c <eos> |
c |
subtract__5.0__4.0__ divide__1.0__5.0__ add__5.0__2.0__ multiply__5.0__7.0__ round__35.0__ |
subtract__5.0__4.0__ divide__1.0__5.0__ add__5.0__2.0__ divide__7.0__0.2__ divide__35.0__1.0__ |
| what is num__92.0 of num__0.75 ? <o> a ) num__6.9 <o> b ) num__69.0 <o> c ) num__0.6845 <o> d ) num__0.6859 <o> e ) num__0.69 |
soln : - num__120.0 = num__1.2 num__0.625 * num__1.2 = num__0.75 = num__0.75 = num__0.75 = num__0.69 answer : e . <eor> e <eos> |
e |
percent__92.0__0.75__ percent__92.0__0.75__ |
percent__92.0__0.75__ percent__92.0__0.75__ |
| j and m were each paid x dollars in advance to do a certain job together . j worked on the job for num__8 hours and m worked for num__2 hours less than j . if m gave j y dollars of her payment so that they would have received the same hourly wage what was the dollar amount in terms of y that j was paid in advance ? <o> a ) a ) num__4 y <o> b ) b ) num__5 y <o> c ) c ) num__6 y <o> d ) d ) num__7 y <o> e ) e ) num__9 y |
j and m were both paid x dollars if m gave y dollars from her x to j then at the end j will have x + y dollars m will have x - y dollars now since the hourly wages have to be same then hourly wage of j = ( x + y ) / num__8 ( since j worked for num__8 hrs ) hourly wage of m = ( x - y ) / num__6 ( since m worked num__2 hrs less than j ) equating both ( x + y ) / num__8 = ( x - y ) / num__6 on simplifying x = num__7 y answer d <eor> d <eos> |
d |
subtract__8.0__2.0__ round__7.0__ |
subtract__8.0__2.0__ round__7.0__ |
| my grandson is about as many days as my son in weeks and my grandson is as many months as i am in years . my grandson my son and i together are num__120 years . can you tell me my age in years ? <o> a ) num__56 <o> b ) num__72 <o> c ) num__68 <o> d ) num__91 <o> e ) num__85 |
let m be my age in years . if s is my son ' s age in years then my son is num__52 s weeks old . if g is my grandson ' s age in years then my grandson is num__365 g days old . thus num__365 g = num__52 s . since my grandson is num__12 g months old num__12 g = m . since my grandson my son and i together are num__120 years g + s + m = num__120 . the above system of num__3 equations in num__3 unknowns ( g s and m ) can be solved as follows . m / num__12 + num__365 m / ( num__52 x num__12 ) + m = num__120 or num__52 m + num__365 m + num__624 m = num__624 x num__120 or m = num__624 x num__0.115273775216 = num__72 . so i am num__72 years old . correct answer is b ) num__72 <eor> b <eos> |
b |
multiply__12.0__52.0__ round__72.0__ |
multiply__12.0__52.0__ round__72.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 seconds . what is the length of the train ? <o> a ) num__140 m <o> b ) num__150 m <o> c ) num__160 m <o> d ) num__240 m <o> e ) none |
sol . speed = [ num__60 * num__0.277777777778 ] m / sec = [ num__16.6666666667 ] m / sec . length of the train = ( speed * time ) = [ num__16.6666666667 * num__9 ] m = num__150 m . answer b <eor> b <eos> |
b |
round__150.0__ |
round__150.0__ |
| john has both apples and bananas . the ratio of apples to bananas is num__2 to num__5 . if there are num__6 more bananas than apples how many bananas are there ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
let b be the number of bananas and a be the number of apples . since the ratio of a to b is num__2 to num__5 that means that a = num__0.4 b . we also know that a = b - num__6 . therefore we can substitute : b - num__6 = num__0.4 b now we can solve for b : if we isolate b we see that : num__0.6 b = num__6 and so b = num__10 <eor> b <eos> |
b |
divide__2.0__5.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
divide__2.0__5.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
| john ' s bank ' s saving amount is decreased num__20.0 due to loan payment and current balance is rs . num__24000 . find the actual balance before deduction ? <o> a ) num__8000 <o> b ) num__8500 <o> c ) num__9000 <o> d ) num__9500 <o> e ) num__30000 |
num__20.0 decreased num__80.0 balance = num__24000 num__100.0 = num__300.0 * num__100 = num__30000 answer : e <eor> e <eos> |
e |
percent__100.0__30000.0__ |
percent__100.0__30000.0__ |
| a company chauncy co . has an annual travel budget of $ num__60000 . the accounting department estimates that transportation expenses will increase num__5 percent in the coming year and nontransportation travel expenses will increase by num__15 percent in the coming year . last year chauncy co . spent $ num__19500 on transportation - related expenses and $ num__35000 on nontransportation travel expenses . if the annual travel budget does not increase this year and if the accounting department ’ s estimates are correct how far over the annual travel budget will expenses be this year ? <o> a ) expenses will not go over the budget . <o> b ) $ num__725 <o> c ) $ num__4225 <o> d ) $ num__5725 <o> e ) $ num__60 |
725 |
annual travel budget of $ num__60000 let transportation expenses = t = num__19500 and non - transportation expenses = n = num__35000 i . e . increased transportation expenses = num__1.05 t = num__20475 and increased non - transportation expenses = num__1.15 n = num__40250 total expense = num__20475 + num__40250 = num__60725 expense over budget = budget - expense = num__60000 - num__60725 = num__725 answer : option b <eor> b <eos> |
b |
b |
| the perimeter of a triangle is num__32 cm and the inradius of the triangle is num__2.5 cm . what is the area of the triangle ? <o> a ) num__76 <o> b ) num__88 <o> c ) num__66 <o> d ) num__55 <o> e ) num__40 |
area of a triangle = r * s where r is the in radius and s is the semi perimeter of the triangle . area of triangle = num__2.5 * num__16.0 = num__40 cm num__2 answer : e <eor> e <eos> |
e |
triangle_area__32.0__2.5__ triangle_area__32.0__2.5__ |
multiply__2.5__16.0__ multiply__2.5__16.0__ |
| the radius of a semi circle is num__6.83 cm then its perimeter is ? <o> a ) num__32.52 <o> b ) num__32.47 <o> c ) num__34.97 <o> d ) num__32.92 <o> e ) num__32.33 |
num__5.14285714286 r = num__6.3 = num__34.97 answer : c <eor> c <eos> |
c |
round__34.97__ |
round__34.97__ |
| there are num__36 students in a certain geometry class . if one third of the students are boys and three fourths of the boys are under six feet tall how many boys in the class are under six feet tall ? <o> a ) num__9 <o> b ) num__12 <o> c ) num__18 <o> d ) num__24 <o> e ) num__27 |
total students = num__36 one third of the students are boys = num__0.333333333333 * num__36 = num__12 three fourths of the boys are under six feet tall = num__0.75 * num__12 = num__9 . . . therefore number of boys in the class under six feet tall = num__9 . . . answer a . . . . alternatively . . . number of boys in the class under six feet tall = num__0.75 of num__0.333333333333 of total students = num__0.75 * num__0.333333333333 * num__36 = num__9 . . . answer a <eor> a <eos> |
a |
multiply__0.75__12.0__ multiply__0.75__12.0__ |
multiply__0.75__12.0__ multiply__0.75__12.0__ |
| nil and ethan are brothers . they left their home at the same time and drove to the same beach . nil drove at a speed of num__40 miles per hour . ethan drove at a speed of num__20 miles per hour . nil arrived at the beach num__0.5 hour earlier than ethan . what is the distance between their home and the beach ? <o> a ) num__30 miles <o> b ) num__40 miles <o> c ) num__50 miles <o> d ) num__60 miles <o> e ) num__70 miles |
every hour nil gets ahead of ethan num__40 - num__20 = num__20 miles . when nil arrived at the beach ethan is only num__20 × num__0.5 = num__10 miles behind . that tells us they only drove num__1 hour when nil arrived at the beach . the distance between their home and the beach is nil ’ s speed × nil ’ s time = num__40 × num__1 = num__40 miles . correct answer b <eor> b <eos> |
b |
multiply__20.0__0.5__ round__40.0__ |
multiply__20.0__0.5__ round__40.0__ |
| a goods train runs at the speed of num__72 km / hr and crosses a num__250 m long platform in num__26 sec . what is the length of the goods train ? <o> a ) num__287 m <o> b ) num__278 m <o> c ) num__276 m <o> d ) num__270 m <o> e ) num__268 m |
speed = num__72 * num__0.277777777778 = num__20 m / sec . time = num__26 sec . let the length of the train be x meters . then ( x + num__250 ) / num__26 = num__20 x = num__270 m . answer : d <eor> d <eos> |
d |
add__250.0__20.0__ round__270.0__ |
add__250.0__20.0__ add__250.0__20.0__ |
| the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is num__51 kmph find the speed of the stream ? <o> a ) num__77 <o> b ) num__88 <o> c ) num__14 <o> d ) num__12 <o> e ) num__17 |
the ratio of the times taken is num__2 : num__1 . the ratio of the speed of the boat in still water to the speed of the stream = ( num__2 + num__1 ) / ( num__2 - num__1 ) = num__3.0 = num__3 : num__1 speed of the stream = num__17.0 = num__17 kmph . answer : e <eor> e <eos> |
e |
add__1.0__2.0__ divide__51.0__3.0__ round__17.0__ |
add__1.0__2.0__ divide__51.0__3.0__ divide__51.0__3.0__ |
| david was born num__5 years after his father ' s marriage . his mother is num__4 years younger than his father but num__20 years older than david who is num__8 years old . at what age did the father get married ? <o> a ) num__19 years <o> b ) num__20 years <o> c ) num__21 years <o> d ) num__18 years <o> e ) num__17 years |
explanation : david ' s present age = num__8 years . his mother ' s present age = ( num__20 + num__8 ) years = num__28 years . his father ' s present age = ( num__28 + num__4 ) years = num__32 years . his father ' s age at the time of david ' s birth = ( num__32 - num__8 ) years = num__24 years . therefore david ' s father ' s age at the time of marriage = ( num__24 - num__5 ) years = num__19 years . answer : a <eor> a <eos> |
a |
add__20.0__8.0__ multiply__4.0__8.0__ add__4.0__20.0__ subtract__24.0__5.0__ subtract__24.0__5.0__ |
add__20.0__8.0__ add__4.0__28.0__ add__4.0__20.0__ subtract__24.0__5.0__ subtract__24.0__5.0__ |
| the spherical ball of lead num__3 cm in diameter is melted and recast into num__3 spherical balls . the diameters of two of these are num__1 num__0.5 cm and num__2 cm respectively . the diameter of third ball is ? <o> a ) num__2.8 <o> b ) num__2.5 <o> c ) num__2.2 <o> d ) num__2.0 <o> e ) num__2.1 |
num__1.33333333333 π * num__3 * num__3 * num__3 = num__1.33333333333 π [ ( num__1.5 ) num__3 + num__23 + r num__3 ] r = num__1.25 d = num__2.5 answer : b <eor> b <eos> |
b |
multiply__3.0__0.5__ subtract__3.0__0.5__ round__2.5__ |
multiply__3.0__0.5__ add__1.0__1.5__ add__1.0__1.5__ |
| in climbing a round pole of num__80 meters height a monkey climbs num__5 meters in a minute and slips num__2 meters in the alternate minute . to get to the top of the pole the monkey would take : <o> a ) num__51 minutes <o> b ) num__54 minutes <o> c ) num__58 minutes <o> d ) num__61 minutes <o> e ) num__63 minutes |
a monkey climbs num__5 meters in one min & slips num__2 m in other min . therefore the monkey reach num__3 meters in num__2 min num__3 m = num__2 min num__25 * num__3 m = num__25 * num__2 min - > num__75 m = num__50 min hence the remaining num__5 m jumps the monkey in num__1 min to reach original height so num__51 min answer : a <eor> a <eos> |
a |
subtract__5.0__2.0__ subtract__80.0__5.0__ multiply__2.0__25.0__ subtract__3.0__2.0__ add__1.0__50.0__ round__51.0__ |
subtract__5.0__2.0__ subtract__80.0__5.0__ multiply__2.0__25.0__ subtract__3.0__2.0__ add__1.0__50.0__ round__51.0__ |
| kim can do a work in num__3 days while david can do the same work in num__2 days . both of them finish the work together and get rs . num__150 . what is the share of kim ? <o> a ) num__77 <o> b ) num__60 <o> c ) num__99 <o> d ) num__26 <o> e ) num__21 |
kim ' s wages : david ' s wages = kim ' s num__1 day work : david ' s num__1 day work = num__0.333333333333 : num__0.5 = num__2 : num__3 kim ' s share = num__0.4 * num__150 = rs . num__60 answer : b <eor> b <eos> |
b |
subtract__3.0__2.0__ divide__1.0__3.0__ divide__1.0__2.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
subtract__3.0__2.0__ divide__1.0__3.0__ divide__1.0__2.0__ multiply__150.0__0.4__ multiply__150.0__0.4__ |
| on dividing num__22 by a number the quotient is num__7 and the remainder is num__1 . find the divisor . <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__6 <o> e ) num__7 |
d = ( d - r ) / q = ( num__22 - num__1 ) / num__7 = num__3.0 = num__3 c <eor> c <eos> |
c |
multiply__1.0__3.0__ |
divide__3.0__1.0__ |
| an artist wishes to paint a circular region on a square poster that is num__3.4 feet on a side . if the area of the circular region is to be num__0.5 the area of the poster what must be the radius of the circular region in feet ? <o> a ) num__1 / pi <o> b ) sqrt ( num__2 / pi ) <o> c ) num__1 <o> d ) num__2 / sqrt ( pi ) <o> e ) sqrt ( num__5.78 / pi ) |
area of the poster is num__3.4 x num__3.4 = num__11.56 num__0.5 the area = num__5.78 pi * r ^ num__2 = num__5.78 r ^ num__2 = num__5.78 / pi r = sqrt ( num__5.78 / pi ) answer ( e ) <eor> e <eos> |
e |
multiply__0.5__11.56__ square_perimeter__0.5__ multiply__0.5__11.56__ |
multiply__0.5__11.56__ square_perimeter__0.5__ multiply__0.5__11.56__ |
| quentin ' s income is num__60.0 less than dev ' s income and sam ' s income is num__25.0 less than quentin ' s income . if dev gave num__60.0 of his income to sam and num__40.0 of his income to quentin quentin ' s new income would be what fraction of sam ' s new income ? <o> a ) num__0.888888888889 <o> b ) num__0.916666666667 <o> c ) num__0.615384615385 <o> d ) num__0.846153846154 <o> e ) num__0.923076923077 |
we can take some easy numbers and make calculations simpler . let r ( dev ' s income ) = num__100 q ( quentin ' s income ) = num__40.0 r = num__40 s ( sam ' s income ) = num__75.0 q = ( num__0.75 ) * num__40 = num__30 now if dev gives num__40.0 to quentin - - > q = num__40 + num__40 = num__80 num__60.0 given to sam - - > s = num__30 + num__60 = num__90 the ratio is : q / s = num__0.888888888889 = num__0.888888888889 = a <eor> a <eos> |
a |
percent__40.0__75.0__ percent__100.0__0.8889__ |
percent__40.0__75.0__ percent__100.0__0.8889__ |
| if xy = num__5 and x ^ num__2 + y ^ num__2 = num__10 then x / y + y / x = <o> a ) num__2 <o> b ) num__3 num__0.142857142857 <o> c ) num__5 num__0.333333333333 <o> d ) num__7 <o> e ) num__60 |
we can make simplifying of question and get it in view : ( x ^ num__2 + y ^ num__2 ) / xy and as we know the meaning of this parts : x ^ num__2 + y ^ num__2 = num__10 xy = num__5 we can calculate the answer num__2.0 - > num__2 so answer is a <eor> a <eos> |
a |
divide__10.0__5.0__ |
divide__10.0__5.0__ |
| the smallest number which when diminished by num__2 is divisible by num__12 num__16 num__18 num__21 and num__28 is <o> a ) num__1008 <o> b ) num__1010 <o> c ) num__1022 <o> d ) num__1032 <o> e ) num__1043 |
required number = ( l . c . m of num__12 num__16 num__18 num__2128 ) + num__2 = num__1008 + num__2 = num__1010 answer : b <eor> b <eos> |
b |
add__2.0__1008.0__ add__2.0__1008.0__ |
add__2.0__1008.0__ add__2.0__1008.0__ |
| the l ength of a rectangle is halved while its breadth is tripled . watis the % change in area ? <o> a ) num__30.0 <o> b ) num__40.0 <o> c ) num__50 percent <o> d ) num__60.0 <o> e ) num__70 % |
let original length = x and origina l breadth = y . original area = xy . new length = x . num__2 new breadth = num__3 y . new area = x x num__3 y = num__3 xy . num__2 num__2 increase % = num__1 xy x num__1 x num__100.0 = num__50.0 . num__2 xy c <eor> c <eos> |
c |
triangle_area__1.0__100.0__ triangle_area__1.0__100.0__ |
triangle_area__1.0__100.0__ triangle_area__1.0__100.0__ |
| a certain boxer has agreed to pay his opponent a fee of num__2.5 of his total purse for every pound over the specified weight limit he weighs in . if the boxer pays his opponent a fee of $ num__225000 after weighing in seven pounds over the specified limit what was the boxer ' s purse ? <o> a ) $ num__65625 <o> b ) $ num__105000 <o> c ) $ num__150000 <o> d ) $ num__183750 <o> e ) $ num__1 num__285 |
714 |
. num__025 * num__7 = . num__175 num__225000 / . num__175 = $ num__1 num__285714 answer : e <eor> e <eos> |
e |
e |
| calculate how many days it will take for num__10 boys to paint a num__80 m long wall if num__6 boys can paint a num__70 m long wall in num__8 days <o> a ) num__9.48 days <o> b ) num__3.48 days <o> c ) num__7.48 days <o> d ) num__5.48 days <o> e ) num__6.48 days |
the length of wall painted by one boy in one day = num__11.6666666667 * num__0.125 = num__1.46 m no . of days required to paint num__50 m cloth by num__8 boys = num__8.0 * num__1 / num__1.46 = num__5.48 days . d <eor> d <eos> |
d |
divide__70.0__6.0__ divide__10.0__80.0__ multiply__8.0__0.125__ round__5.48__ |
divide__70.0__6.0__ divide__10.0__80.0__ multiply__8.0__0.125__ divide__5.48__1.0__ |
| the dimensions of a room are num__25 feet * num__15 feet * num__12 feet . what is the cost of white washing the four walls of the room at rs . num__6 per square feet if there is one door of dimensions num__6 feet * num__3 feet and three windows of dimensions num__4 feet * num__3 feet each ? <o> a ) num__4000 <o> b ) num__345 <o> c ) num__5436 <o> d ) num__4530 <o> e ) num__4566 |
area of the four walls = num__2 h ( l + b ) since there are doors and windows area of the walls = num__2 * num__12 ( num__15 + num__25 ) - ( num__6 * num__3 ) - num__3 ( num__4 * num__3 ) = num__906 sq . ft . total cost = num__906 * num__6 = rs . num__5436 answer : option c <eor> c <eos> |
c |
triangle_area__12.0__906.0__ triangle_area__12.0__906.0__ |
multiply__6.0__906.0__ multiply__6.0__906.0__ |
| a b and c each working alone can complete a job in num__6 num__8 and num__12 days respectively . if all three of them work together to complete a job and earn $ num__2500 what will be c ' s share of the earnings ? <o> a ) $ num__1200 <o> b ) $ num__1005.8 <o> c ) $ num__555.6 <o> d ) $ num__1009.2 <o> e ) $ num__1490.7 |
the dollars earned will be in the same ratio as amount of work done num__1 day work of c is num__0.0833333333333 ( or num__0.0833333333333 ) num__1 day work of the combined workforce is ( num__0.166666666667 + num__0.125 + num__0.0833333333333 ) = num__0.375 c ' s contribution is num__0.222222222222 of the combined effort translating effort to $ = num__0.222222222222 * num__2500 = $ num__555.6 <eor> c <eos> |
c |
divide__1.0__12.0__ divide__1.0__6.0__ divide__1.0__8.0__ round__555.6__ |
divide__1.0__12.0__ divide__1.0__6.0__ divide__1.0__8.0__ round__555.6__ |
| a boy was asked to multiply a number by num__22 . he instead multiplied the number by num__44 and got the answer num__308 more than the correct answer . what was the number to be multiplied ? <o> a ) num__16 <o> b ) num__10 <o> c ) num__14 <o> d ) num__12 <o> e ) num__8 |
explanation : let the number be x num__22 x + num__308 = num__44 x = > num__44 x - num__22 x = num__308 = > num__22 x = num__308 = > x = num__14.0 = num__14.0 = num__14 answer : c <eor> c <eos> |
c |
divide__308.0__22.0__ divide__308.0__22.0__ |
divide__308.0__22.0__ divide__308.0__22.0__ |
| a rower can row upstream at num__18 kph and downstream at num__34 kph . what is the speed of the rower in still water ? <o> a ) num__24 <o> b ) num__25 <o> c ) num__26 <o> d ) num__27 <o> e ) num__28 |
let v be the rower ' s speed in still water . let s be the speed of the current in the stream v - s = num__18 v + s = num__34 when we add the two equations we get : num__2 v = num__52 then v = num__26 kph . the answer is c . <eor> c <eos> |
c |
add__18.0__34.0__ divide__52.0__2.0__ round__26.0__ |
add__18.0__34.0__ divide__52.0__2.0__ subtract__52.0__26.0__ |
| the average of num__15 numbers is num__15 . if the average of first five numbers is num__14 and that of other num__9 numbers is num__16 then find the middle number . <o> a ) num__12 <o> b ) num__11 <o> c ) num__10 <o> d ) num__9 <o> e ) num__8 |
given : average of num__15 numbers = num__15 average of num__5 numbers = num__14 average of num__9 numbers = num__16 average = total numbers / no . of numbers num__15 = total numbers / num__15 therefore total numbers = num__15 x num__15 = num__225 middle number = ( total numbers ) – [ ( average of num__5 num x no of num ) + ( average of num__9 num x no of num ) ] = ( num__225 ) – [ ( num__14 x num__5 ) + ( num__16 x num__9 ) ] = ( num__225 ) – [ num__214 ] = num__11 therefore the middle number is num__11 answer is b <eor> b <eos> |
b |
subtract__14.0__9.0__ subtract__16.0__5.0__ subtract__16.0__5.0__ |
subtract__14.0__9.0__ subtract__16.0__5.0__ subtract__16.0__5.0__ |
| mary ' s income is num__60.0 more than tim ' s income and tim ' s income is num__60.0 less than juan ' s income . what % of juan ' s income is mary ' s income . <o> a ) num__124.0 <o> b ) b . num__120.0 <o> c ) num__64.0 <o> d ) num__80.0 <o> e ) num__64 % |
even i got num__96.0 j = num__100 t = num__100 * num__0.4 = num__40 m = num__40 * num__1.6 = num__64 if mary ' s income is x percent of j m = j * x / num__100 x = m * num__100 / j = num__64 * num__1.0 = num__64 ans : c <eor> c <eos> |
c |
subtract__100.0__60.0__ divide__96.0__60.0__ multiply__1.6__40.0__ round_down__1.6__ multiply__64.0__1.0__ |
multiply__100.0__0.4__ divide__96.0__60.0__ multiply__1.6__40.0__ round_down__1.6__ multiply__64.0__1.0__ |
| a num__470 m long train is running at a speed of num__55 km / hr . it crossed a platform of length num__520 m in ? <o> a ) num__41.1 sec <o> b ) num__64.8 sec <o> c ) num__31.8 sec <o> d ) num__50.4 sec <o> e ) none of the above |
speed = num__55 km / hr ( to convert km / hr in to m / s ) = num__55 x num__0.277777777778 m / s distance = num__470 m + num__520 m ( if questions is about train crossing a post you need to consider only the length of train ) = num__990 m time = distance / speed = num__990 x num__18 / ( num__5 x num__55 ) = num__64.8 sec ans is : b <eor> b <eos> |
b |
add__470.0__520.0__ divide__990.0__55.0__ round__64.8__ |
add__470.0__520.0__ divide__990.0__55.0__ round__64.8__ |
| a person spent rs . num__5040 from his salary on food and num__5000 on house rent . after that he was left with num__20.0 of his monthly salary . what is his monthly salary ? <o> a ) num__22550 <o> b ) num__32550 <o> c ) num__52550 <o> d ) num__62550 <o> e ) num__12 |
550 |
total money spent on food and house rent = num__5040 + num__5000 = num__10040 which is num__100 - num__20 = num__80.0 of his monthly salary ∴ his salary = num__10040 x num__1.25 = num__12550 answer : e <eor> e <eos> |
e |
e |
| a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . num__59.29 the number of the member is the group is : <o> a ) num__57 <o> b ) num__67 <o> c ) num__77 <o> d ) num__87 <o> e ) num__97 |
money collected = ( num__59.29 x num__100 ) paise = num__5929 paise . number of members = num__5929 = num__77 . answer : option c <eor> c <eos> |
c |
multiply__59.29__100.0__ divide__5929.0__77.0__ |
multiply__59.29__100.0__ divide__5929.0__77.0__ |
| a and b together can complete work in num__8 days . a alone starts working and leaves it after working for num__6 days completing only half of the work . in how many days it can be completed if the remaining job is undertaken by b ? <o> a ) num__4 <o> b ) num__8 <o> c ) num__5 <o> d ) num__12 <o> e ) num__7 |
explanation : ( a + b ) one day work = num__0.125 now a does half of the work in num__6 day so a can complete the whole work in num__12 days a ’ s one day work = num__0.0833333333333 b ’ s one day work = num__0.125 - num__0.0833333333333 = num__0.0416666666667 b alone can complete the work in num__24 days so half of the work in num__12 days answer : option d <eor> d <eos> |
d |
subtract__0.125__0.0833__ round__12.0__ |
subtract__0.125__0.0833__ subtract__24.0__12.0__ |
| find the number which when increased by num__11 is equal to num__60 times the reciprocal of the number <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__9 |
let the number be x . a positive number ( x ) increased by num__11 is equal to num__60 times the reciprocal of the number ( num__1 / x ) x + num__11 = num__60 x x num__2 + num__11 x - num__60 = num__0 x num__2 + num__15 x - num__4 x - num__60 = num__0 x ( x + num__15 ) - num__4 ( x + num__15 ) = num__0 ( x + num__15 ) ( x - num__4 ) = num__0 x = num__4 the positive number ( x ) = num__4 a <eor> a <eos> |
a |
divide__60.0__15.0__ divide__60.0__15.0__ |
divide__60.0__15.0__ divide__60.0__15.0__ |
| if finn was num__20 months old one year ago how old was he in months x months ago ? <o> a ) x − num__30 <o> b ) x − num__12 <o> c ) num__32 − x <o> d ) num__24 − x <o> e ) num__30 − x |
age today = num__20 months + num__12 months ( num__1 year ) = num__32 months x months ago - today ' s age - x = num__32 - x ans - c <eor> c <eos> |
c |
add__20.0__12.0__ add__20.0__12.0__ |
add__20.0__12.0__ add__20.0__12.0__ |
| a man has rs . num__520 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__97.5 <o> b ) num__60 <o> c ) num__75 <o> d ) num__90 <o> e ) num__105 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__520 num__16 x = num__520 x = num__32.5 . hence total number of notes = num__3 x = num__97.5 . answer = a <eor> a <eos> |
a |
divide__520.0__16.0__ multiply__32.5__3.0__ multiply__32.5__3.0__ |
divide__520.0__16.0__ multiply__32.5__3.0__ multiply__32.5__3.0__ |
| at what rate percent per annum will a sum of money double in num__5 years . <o> a ) num__20.0 <o> b ) num__13.5 <o> c ) num__11.5 <o> d ) num__14.5 <o> e ) num__21.5 % |
let principal = p then s . i . = p and time = num__8 years rate = [ ( num__100 x p ) / ( p x num__5 ) ] % = num__20.0 per annum . answer : a <eor> a <eos> |
a |
percent__20.0__100.0__ |
percent__20.0__100.0__ |
| a train num__165 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__165 * num__0.0545454545455 ] sec = num__9 sec answer : e <eor> e <eos> |
e |
add__60.0__6.0__ divide__165.0__18.3333__ round__9.0__ |
add__60.0__6.0__ divide__165.0__18.3333__ divide__165.0__18.3333__ |
| a used - book dealer sells paperback books at num__3 times the dealer ' s cost and hardback books at num__4 times the dealer ' s cost . last week the dealer sold a total of num__130 books each of which had cost the dealer $ num__1 . if the gross profit ( sales revenue minus the dealer ' s cost ) on the sale of all of these books was $ num__300 how many of the books sold were paperbacks ? <o> a ) num__40 <o> b ) num__60 <o> c ) num__75 <o> d ) num__90 <o> e ) num__100 |
the dealer earns a profit of $ num__2 per paperback and $ num__3 per hardback . if all num__130 books were paperbacks the dealer would earn a profit of $ num__260 . since the actual profit was $ num__300 the dealer must have sold num__40 hardbacks . then the number of paperbacks sold was num__130 - num__40 = num__90 . the answer is d . <eor> d <eos> |
d |
subtract__3.0__1.0__ multiply__130.0__2.0__ subtract__300.0__260.0__ subtract__130.0__40.0__ subtract__130.0__40.0__ |
subtract__3.0__1.0__ multiply__130.0__2.0__ subtract__300.0__260.0__ subtract__130.0__40.0__ subtract__130.0__40.0__ |
| a man travelled from the village to the post office at the rate of num__25 kmph and walked backed at the rate of num__4 kmph . if the whole journey took num__5 hours num__48 minutes find the distance of the post office from the village ? <o> a ) num__10 km <o> b ) num__50 km <o> c ) num__30 km <o> d ) num__25 km <o> e ) num__20 km |
average speed = num__2 xy / x + y = num__2 * num__25 * num__0.16 + num__4 = num__6.89655172414 km / hr distance travelled in num__5 hr num__48 mnts = num__22.2222222222 * num__5.8 = num__40 km distance of the post office from the village = num__20.0 = num__20 km answer is e <eor> e <eos> |
e |
divide__4.0__25.0__ subtract__25.0__5.0__ round__20.0__ |
divide__4.0__25.0__ multiply__4.0__5.0__ multiply__4.0__5.0__ |
| yesterday it took robert num__3 hours to drive from city x to city y . today it took robert num__2.5 hours to drive back from city y to city x along the same route . if he had saved num__15 minutes in both trips the speed for the round trip would be num__50 miles per hour . what is the distance between city x and city y ? <o> a ) num__125.5 <o> b ) num__120.5 <o> c ) num__150 <o> d ) num__240.5 <o> e ) num__300 |
num__2 d / num__50 = num__5 ( because time = num__3 + num__2.5 - num__0.5 hrs ) = > d = num__125.5 answer - a <eor> a <eos> |
a |
add__3.0__2.0__ subtract__3.0__2.5__ round__125.5__ |
add__3.0__2.0__ subtract__3.0__2.5__ round__125.5__ |
| two numbers when divided by a divisor leave remainders of num__228 and num__352 respectively . the remainder obtained when the sum of the numbers is divided by the same divisor is num__68 . find the divisor . <o> a ) num__512 <o> b ) num__573 <o> c ) num__634 <o> d ) num__685 <o> e ) num__736 |
let x be the divisor . n num__1 = xa + num__228 n num__2 = xb + num__352 n num__1 + n num__2 = x ( a + b ) + num__580 = xc + num__68 x ( c - a - b ) = num__512 but we know that x > num__352 thus ( c - a - b ) must equal num__1 . the answer is a . <eor> a <eos> |
a |
add__228.0__352.0__ subtract__580.0__68.0__ multiply__1.0__512.0__ |
add__228.0__352.0__ subtract__580.0__68.0__ subtract__580.0__68.0__ |
| a train num__825 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__90 seconds . the speed of the train is : <o> a ) num__33 <o> b ) num__27 <o> c ) num__27 <o> d ) num__50 <o> e ) num__81 |
speed of the train relative to man = ( num__12.5 ) m / sec = ( num__12.5 ) m / sec . [ ( num__12.5 ) * ( num__3.6 ) ] km / hr = num__45 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__45 = = > x = num__50 km / hr . answer : d <eor> d <eos> |
d |
multiply__12.5__3.6__ add__5.0__45.0__ round__50.0__ |
multiply__12.5__3.6__ add__5.0__45.0__ round__50.0__ |
| a man can row num__8 kmph in still water . when the river is running at num__3 kmph it takes him num__1 hour to row to a place and black . what is the total distance traveled by the man ? <o> a ) num__5.75 <o> b ) num__5.7 <o> c ) num__5.76 <o> d ) num__5.74 <o> e ) num__6.8 |
m = num__8 s = num__3 ds = num__11 us = num__5 x / num__11 + x / num__5 = num__1 x = num__3.4 d = num__3.4 * num__2 = num__6.8 answer : e <eor> e <eos> |
e |
add__8.0__3.0__ subtract__8.0__3.0__ subtract__3.0__1.0__ multiply__2.0__3.4__ round__6.8__ |
add__8.0__3.0__ subtract__8.0__3.0__ subtract__3.0__1.0__ multiply__2.0__3.4__ multiply__1.0__6.8__ |
| if x and y are integers what is the least positive number of num__24 x + num__20 y ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
num__24 x + num__20 y = num__4 ( num__6 x + num__5 y ) which will be a minimum positive number when num__6 x + num__5 y = num__1 . num__6 ( num__1 ) + num__5 ( - num__1 ) = num__1 then num__4 ( num__6 x + num__5 y ) can have a minimum positive value of num__4 . the answer is c . <eor> c <eos> |
c |
subtract__24.0__20.0__ divide__24.0__4.0__ divide__20.0__4.0__ subtract__5.0__4.0__ subtract__24.0__20.0__ |
subtract__24.0__20.0__ divide__24.0__4.0__ divide__20.0__4.0__ subtract__5.0__4.0__ subtract__24.0__20.0__ |
| num__100 people are attending a newspaper conference . num__35 of them are writers and more than num__38 are editors . of the people at the conference x are both writers and editors and num__2 x are neither . what is the largest possible number of people who are both writers and editors ? <o> a ) num__21 <o> b ) num__26 <o> c ) num__28 <o> d ) num__30 <o> e ) num__32 |
{ total } = { writers } + { editors } - { both } + { neither } . { total } = num__100 ; { writers } = num__35 ; { editors } > num__38 ; { both } = x ; { neither } = num__2 x ; num__100 = num__35 + { editors } - x + num__2 x - - > x = num__65 - { editors } . we want to maximize x thus we should minimize { editors } minimum possible value of { editors } is num__39 thus x = { both } = num__65 - num__39 = num__26 . answer : b . <eor> b <eos> |
b |
alphabet_space__ alphabet_space__ |
alphabet_space__ alphabet_space__ |
| if the average ( arithmetic mean ) of a and b is num__30 and the average of b and c is num__60 what is the value of c − a ? <o> a ) num__25 <o> b ) num__60 <o> c ) num__90 <o> d ) num__140 <o> e ) it can not be determined from the information given . |
- ( a + b = num__60 ) b + c = num__120 c - a = num__60 b . num__60 <eor> b <eos> |
b |
subtract__120.0__60.0__ |
subtract__120.0__60.0__ |
| a numberais squared and then multiplied by negative num__3 . the result of this operation w is equal to three times the sum of three timesaand two . what is one possible value ofa ? <o> a ) - num__3 <o> b ) - num__2 <o> c ) num__1 <o> d ) num__2 <o> e ) num__3 |
w = - num__3 * a ^ num__2 = num__3 ( num__3 a + num__2 ) a = - num__2 or - num__1 a = - num__2 = b <eor> b <eos> |
b |
subtract__3.0__2.0__ subtract__3.0__1.0__ |
subtract__3.0__2.0__ subtract__3.0__1.0__ |
| the sum of five numbers is num__655 . the average of the first two numbers is num__85 and the third number is num__125 . find the average of the two numbers ? <o> a ) num__180 <o> b ) num__288 <o> c ) num__667 <o> d ) num__299 <o> e ) num__12 |
let the five numbers be p q r s and t . = > p + q + r + s + t = num__655 . ( p + q ) / num__2 = num__85 and r = num__125 p + q = num__170 and r = num__125 p + q + r = num__295 s + t = num__655 - ( p + q + r ) = num__360 average of the last two numbers = ( s + t ) / num__2 = num__180 . answer : a <eor> a <eos> |
a |
multiply__85.0__2.0__ add__125.0__170.0__ subtract__655.0__295.0__ divide__360.0__2.0__ divide__360.0__2.0__ |
multiply__85.0__2.0__ add__125.0__170.0__ subtract__655.0__295.0__ divide__360.0__2.0__ divide__360.0__2.0__ |
| during one month at a particular restaurant num__0.166666666667 of the burgers sold were veggie burgers and num__0.5 of the rest of the burgers sold were double - meat . if x of the burgers sold were double - meat how many were veggie burgers ? <o> a ) x / num__8 <o> b ) x / num__2 <o> c ) num__2 x / num__5 <o> d ) num__3 x / num__4 <o> e ) num__4 x / num__5 |
let y be the number of total burgers . veggie = y / num__6 non veggie = num__5 y / num__6 num__0.25 of the rest of the burgers sold were double - meat = > num__5 y / num__6 * num__0.5 = double meat = x = > y / num__6 = num__2 x / num__5 = veggie hence c <eor> c <eos> |
c |
reverse__0.5__ reverse__0.5__ |
reverse__0.5__ reverse__0.5__ |
| in a tree num__0.428571428571 of the birds are parrots while the rest are toucans . if num__0.666666666667 of the parrots are female and num__0.7 of the toucans are female what fraction of the birds in the tree are male ? <o> a ) num__0.428571428571 <o> b ) num__0.314285714286 <o> c ) num__0.457142857143 <o> d ) num__0.442857142857 <o> e ) num__0.528571428571 |
let x be the number of birds in the tree . the fraction of birds that are male parrots is ( num__0.333333333333 ) ( num__0.428571428571 ) = num__0.142857142857 . the fraction of birds that are male toucans is ( num__0.3 ) ( num__0.571428571429 ) = num__0.171428571429 . the total fraction of male birds is num__0.142857142857 + num__0.171428571429 = num__0.314285714286 . the answer is b . <eor> b <eos> |
b |
multiply__0.4286__0.3333__ multiply__0.4286__0.7__ multiply__0.3__0.5714__ add__0.1714__0.1429__ add__0.1714__0.1429__ |
multiply__0.4286__0.3333__ multiply__0.4286__0.7__ multiply__0.3__0.5714__ add__0.1714__0.1429__ add__0.1714__0.1429__ |
| if f ( x ) = ax ^ num__3 + x ^ num__2 + ax – num__2 x then f ( b ) – f ( - b ) will equal : <o> a ) num__2 ab ^ num__3 - num__2 ab <o> b ) num__2 ab <o> c ) num__2 ab ^ num__3 + num__2 ab - num__4 b <o> d ) - num__4 b <o> e ) num__2 ab ^ num__3 |
f ( x ) = ax ^ num__3 + x ^ num__2 + ax – num__2 x f ( b ) = ab ^ num__3 + b ^ num__2 + ab – num__2 b f ( - b ) = - ab ^ num__3 + b ^ num__2 - ab + num__2 b f ( b ) - f ( - b ) = ab ^ num__3 + b ^ num__2 + ab – num__2 b + ab ^ num__3 - b ^ num__2 + ab – num__2 b = num__2 ab ^ num__3 + num__2 ab - num__4 b answer c <eor> c <eos> |
c |
subtract__4.0__2.0__ |
subtract__4.0__2.0__ |
| the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is num__432 sq m then what is the breadth of the rectangular plot ? <o> a ) num__12 <o> b ) num__17 <o> c ) num__18 <o> d ) num__19 <o> e ) num__14 |
let the breadth of the plot be b m . length of the plot = num__3 b m ( num__3 b ) ( b ) = num__432 num__3 b num__2 = num__432 b num__2 = num__144 b = num__12 m . answer : option a <eor> a <eos> |
a |
square_perimeter__3.0__ square_perimeter__3.0__ |
square_perimeter__3.0__ square_perimeter__3.0__ |
| a number when divided by a divisor leaves a remainder of num__24 . when twice the original number is divided by the same divisor the remainder is num__11 . what is the value of the divisor ? <o> a ) num__13 <o> b ) num__59 <o> c ) num__35 <o> d ) num__37 <o> e ) num__12 |
explanatory answer decoding ` ` a number when divided by a divisor leaves a remainder of num__24 ' ' let the original number be ' a ' . let the divisor be ' d ' . let the quotient of dividing ' a ' by ' d ' be ' x ' . therefore we can write the division as a / d = x and the remainder is num__24 . i . e . a = dx + num__24 decoding ` ` when twice the original number is divided by the same divisor the remainder is num__11 ' ' twice the original number is divided by d means num__2 a is divided by d . we know that a = dx + num__24 . therefore num__2 a = num__2 ( dx + num__48 ) or num__2 a = num__2 dx + num__48 when ( num__2 dx + num__48 ) is divided by ' d ' the remainder is num__11 . num__2 dx is divisible by ' d ' and will therefore not leave a remainder . the remainder of num__11 would be the remainder of dividing num__48 by d . the question is ` ` what number will leave a remainder of num__11 when it divides num__48 ? ' ' when num__37 divides num__48 the remainder is num__11 . hence the divisor is num__37 . choice d <eor> d <eos> |
d |
multiply__24.0__2.0__ subtract__48.0__11.0__ subtract__48.0__11.0__ |
multiply__24.0__2.0__ subtract__48.0__11.0__ subtract__48.0__11.0__ |
| if g and s are positive integers such that ( num__2 ^ g ) ( num__4 ^ s ) = num__16 then num__2 g + s = <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
lets get the eq into simplest orm . . ( num__2 ^ g ) ( num__4 ^ s ) = num__16 . . ( num__2 ^ g ) ( num__2 ^ num__2 s ) = num__2 ^ num__4 . . or g + num__2 s = num__4 . . since g and s are positive integers only g as num__2 and s as num__1 satisfy the equation . . so num__2 g + s = num__2 * num__2 + num__1 = num__5 . . d <eor> d <eos> |
d |
add__4.0__1.0__ add__4.0__1.0__ |
add__4.0__1.0__ add__4.0__1.0__ |
| what will come in place of the x in the following number series ? num__11 num__14 num__19 num__22 num__27 num__30 num__35 num__38 num__43 x <o> a ) num__46 <o> b ) num__26 <o> c ) num__36 <o> d ) num__35 <o> e ) num__45 |
( a ) the pattern is + num__3 + num__5 + num__3 + num__5 … … … … so the missing term is = num__43 + num__3 = num__46 . <eor> a <eos> |
a |
subtract__14.0__11.0__ subtract__19.0__14.0__ add__11.0__35.0__ add__11.0__35.0__ |
subtract__14.0__11.0__ subtract__19.0__14.0__ add__11.0__35.0__ add__11.0__35.0__ |
| if x and y are integers and xy = num__330 x or y must be divisible by which of the following ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__20 <o> e ) num__30 |
xy = num__330 = num__2 * num__3 * num__5 * num__11 one of x or y must be divisible by num__5 . the answer is b . <eor> b <eos> |
b |
add__3.0__2.0__ add__3.0__2.0__ |
add__3.0__2.0__ add__3.0__2.0__ |
| num__60.0 of the employees of a company are men . num__60.0 of the men in the company speak french and num__50.0 of the employees of the company speak french . what is % of the women in the company who do not speak french ? <o> a ) num__4.0 <o> b ) num__10.0 <o> c ) num__96.0 <o> d ) num__90.0 <o> e ) num__65 % |
no of employees = num__100 ( say ) men = num__60 women = num__40 men speaking french = num__0.6 * num__60 = num__36 employees speaking french = num__0.5 * num__100 = num__50 therefore women speaking french = num__50 - num__36 = num__14 and women not speaking french = num__40 - num__14 = num__26.0 of women not speaking french = num__0.65 * num__100 = num__65.0 answer e <eor> e <eos> |
e |
percent__65.0__100.0__ |
percent__65.0__100.0__ |
| how much is num__80.0 of num__40 is greater than num__0.8 of num__25 ? <o> a ) num__15 <o> b ) num__7 <o> c ) num__18 <o> d ) num__12 <o> e ) num__23 |
( num__0.8 ) * num__40 – ( num__0.8 ) * num__25 num__32 - num__20 = num__12 answer : d <eor> d <eos> |
d |
multiply__40.0__0.8__ multiply__0.8__25.0__ subtract__32.0__20.0__ subtract__32.0__20.0__ |
multiply__40.0__0.8__ multiply__0.8__25.0__ subtract__32.0__20.0__ subtract__32.0__20.0__ |
| a thief goes away with a santro car at a speed of num__25 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at num__60 kmph when will the owner over take the thief from the start ? <o> a ) num__0.714285714286 hours <o> b ) num__0.285714285714 hours <o> c ) num__0.666666666667 hours <o> d ) num__0.333333333333 hours <o> e ) num__0.4 hours |
- - - - - - - - - - - num__25 - - - - - - - - - - - - - - - - - - - - | num__60 num__25 d = num__25 rs = num__60 â € “ num__25 = num__35 t = num__0.714285714286 = num__0.714285714286 hours answer : a <eor> a <eos> |
a |
subtract__60.0__25.0__ divide__25.0__35.0__ divide__25.0__35.0__ |
subtract__60.0__25.0__ divide__25.0__35.0__ divide__25.0__35.0__ |
| num__45 x ? = num__25.0 of num__900 <o> a ) num__16.2 <o> b ) num__4 <o> c ) num__5 <o> d ) num__500 <o> e ) none |
answer let num__45 x a = ( num__25 x num__900 ) / num__100 ∴ a = ( num__25 x num__9 ) / num__45 = num__5 correct option : c <eor> c <eos> |
c |
divide__900.0__100.0__ divide__45.0__9.0__ divide__45.0__9.0__ |
divide__900.0__100.0__ divide__45.0__9.0__ divide__45.0__9.0__ |
| there is a sequence a ( n ) such that a ( n + num__1 ) = num__2 a ( n ) - num__1 and a ( num__1 ) = num__3 where n is a positive integer . what is the value of a ( num__66 ) - a ( num__65 ) ? <o> a ) num__2 ^ num__22 + num__1 <o> b ) num__2 ^ num__65 <o> c ) num__2 ^ num__23 + num__1 <o> d ) num__2 ^ num__24 <o> e ) num__2 ^ num__23 + num__1 |
a num__1 = num__3 a num__2 = num__2 * num__3 - num__1 = num__5 a num__3 = num__2 * num__5 - num__1 = num__9 a num__4 = num__2 * num__9 - num__1 = num__17 we can notice that there is a squence a num__2 - a num__1 = num__2 ^ num__1 a num__3 - a num__2 = num__2 ^ num__2 a num__4 - a num__3 = num__2 ^ num__3 hence a num__66 - a num__65 = num__2 ^ num__65 b is the answer <eor> b <eos> |
b |
add__2.0__3.0__ add__1.0__3.0__ multiply__1.0__2.0__ |
add__2.0__3.0__ subtract__5.0__1.0__ multiply__1.0__2.0__ |
| a recycling facility is staffed by num__15 floor workers and one manager . all of the floor workers are paid equal wages but the manager is paid n times as much as a floor worker . if the manager ’ s wages account for num__0.0714285714286 of all wages paid at the facility what is the value of n ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__1.15384615385 <o> e ) num__11 |
say each floor worker is paid $ x then the manager is paid $ xn . total salary would be num__15 x + xn and we are told that it equals to num__14 xn : num__15 x + xn = num__14 xn - - > reduce by x : num__15 + n = num__14 n - - > num__13 n = num__15 n = num__1.15384615385 answer : d <eor> d <eos> |
d |
divide__15.0__13.0__ divide__15.0__13.0__ |
divide__15.0__13.0__ divide__15.0__13.0__ |
| a can do a piece of work in num__20 days ; b can do the same in num__30 days . a started alone but left the work after num__10 days then b worked at it for num__10 days . c finished the remaining work in num__10 days . c alone can do the whole work in ? <o> a ) num__70 days <o> b ) num__65 days <o> c ) num__40 days <o> d ) num__50 days <o> e ) num__60 days |
num__0.5 + num__0.333333333333 + num__10 / x = num__1 x = num__60 days answer : e <eor> e <eos> |
e |
divide__10.0__20.0__ divide__10.0__30.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
divide__10.0__20.0__ divide__10.0__30.0__ divide__30.0__0.5__ divide__30.0__0.5__ |
| in a group of ducks and cows the total number of legs are num__24 more than twice the number of heads . find the total number of cows . <o> a ) a ) num__14 <o> b ) b ) num__12 <o> c ) c ) num__16 <o> d ) d ) num__8 <o> e ) e ) num__6 |
explanation : let the number of ducks be d and number of cows be c then total number of legs = num__2 d + num__4 c = num__2 ( d + num__2 c ) total number of heads = c + d given that total number of legs are num__24 more than twice the number of heads = > num__2 ( d + num__2 c ) = num__24 + num__2 ( c + d ) = > d + num__2 c = num__12 + c + d = > num__2 c = num__12 + c = > c = num__12 answer : b <eor> b <eos> |
b |
divide__24.0__2.0__ subtract__24.0__12.0__ |
divide__24.0__2.0__ subtract__24.0__12.0__ |
| a man purchases an electric heater whose printed price is rs . num__160 . if he received two successive discounts of num__20.0 and num__10.0 ; he paid : <o> a ) rs . num__112 <o> b ) rs . num__129.60 <o> c ) rs . num__119.60 <o> d ) rs . num__115.20 <o> e ) none |
explanation : price after ist discount = num__100.0 of rs . num__160 = rs . num__128 price after num__2 nd discount = num__90.0 of rs . num__128 = rs . num__115.20 correct option : d <eor> d <eos> |
d |
percent__20.0__10.0__ percent__90.0__128.0__ percent__100.0__115.2__ |
percent__20.0__10.0__ percent__90.0__128.0__ percent__100.0__115.2__ |
| what are the last two digits of num__24 * num__62 * num__33 * num__48 * num__39 * num__16 ? <o> a ) num__40 <o> b ) num__08 <o> c ) num__64 <o> d ) num__34 <o> e ) num__12 |
num__24 * num__62 * num__33 * num__48 * num__39 * num__16 ? = we have to focus on the last two digits only so num__24 * num__62 = num__88 * num__33 = num__04 * num__48 = num__92 num__39 * num__92 = num__88 therefore num__88 * num__16 = num__08 hence answer is b <eor> b <eos> |
b |
add__88.0__4.0__ subtract__24.0__16.0__ subtract__24.0__16.0__ |
add__88.0__4.0__ subtract__24.0__16.0__ subtract__24.0__16.0__ |
| the ratio of ages of aman bren and charlie are in the ratio num__5 : num__8 : num__7 respectively . if num__8 years ago the sum of their ages was num__76 what will be the age of bren num__12 years from now ? <o> a ) num__17 <o> b ) num__25 <o> c ) num__27 <o> d ) num__52 <o> e ) num__60 |
let the present ages of aman bren and charlie be num__5 x num__8 x and num__7 x respectively . num__5 x - num__8 + num__8 x - num__8 + num__7 x - num__8 = num__76 x = num__5 present age of bren = num__8 * num__5 = num__40 bren ' s age num__12 years hence = num__40 + num__12 = num__52 answer = e <eor> e <eos> |
e |
multiply__5.0__8.0__ add__12.0__40.0__ multiply__5.0__12.0__ |
multiply__5.0__8.0__ add__12.0__40.0__ multiply__5.0__12.0__ |
| a father told his son ` ` i was as old as you are at present at the time of your birth ' ' . if the father is num__38 years old now then what was the son ' s age five years ago in years ? <o> a ) num__14 <o> b ) num__19 <o> c ) num__38 <o> d ) num__33 <o> e ) num__35 |
et son ' s present age is = x then num__38 - x = x x = num__19 son ' s age num__5 years back is num__19 - num__5 = num__14 years . answer : a <eor> a <eos> |
a |
subtract__19.0__5.0__ subtract__19.0__5.0__ |
subtract__19.0__5.0__ subtract__19.0__5.0__ |
| three factories of conglomerate corporation are capable of manufacturing hubcaps . two of the factories can each produce num__1000 hubcaps in num__15 days . the third factory can produce hubcaps num__30.0 faster . approximately how many days would it take to produce a num__10000 hubcaps with all three factories working simultaneously ? <o> a ) num__38 <o> b ) num__42 <o> c ) num__46 <o> d ) num__50 <o> e ) num__54 |
since third factory is num__30.0 faster thus it will create num__1300 hubcaps in num__15 days in num__15 days total number of hubcaps made by all three will be : num__1000 + num__1000 + num__1300 = num__3300 in num__30 days : num__3300 + num__3300 = num__6600 in num__45 days : num__6600 + num__3300 = num__9900 now only num__100 more hubcaps are required which will take num__1 day thus in all num__46 days . ans : c <eor> c <eos> |
c |
percent__46.0__100.0__ |
percent__46.0__100.0__ |
| there are num__32 balls which are red blue or green . if num__15 balls are green and the sum of red balls and green balls is less than num__23 at most how many red balls are there ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
r + g + b = num__32 g = num__15 r + g < num__23 = > r + num__15 < num__23 = > r < num__8 = > at most num__7 red balls answer : d <eor> d <eos> |
d |
subtract__23.0__15.0__ subtract__15.0__8.0__ subtract__15.0__8.0__ |
subtract__23.0__15.0__ subtract__15.0__8.0__ subtract__15.0__8.0__ |
| a train num__275 m long running with a speed of num__90 km / hr will pass a tree in <o> a ) num__11 sec <o> b ) num__15 sec <o> c ) num__16 sec <o> d ) num__20 sec <o> e ) none |
sol . speed = ( num__90 x num__0.277777777778 ) m / sec . = num__25 m / sec . time taken = ( num__275 x num__0.04 ) sec = num__11 sec answer a <eor> a <eos> |
a |
divide__275.0__25.0__ round__11.0__ |
divide__275.0__25.0__ divide__275.0__25.0__ |
| a coin is tossed four times . what is the probability that there is at the least one tail ? <o> a ) num__0.9375 <o> b ) num__0.837837837838 <o> c ) num__1.72222222222 <o> d ) num__3.1 <o> e ) num__1.63157894737 |
let p ( t ) be the probability of getting least one tail when the coin is tossed four times . = there is not even a single tail . i . e . all the outcomes are heads . = num__0.0625 ; p ( t ) = num__1 - num__0.0625 = num__0.9375 answer : a <eor> a <eos> |
a |
negate_prob__0.0625__ negate_prob__0.0625__ |
negate_prob__0.0625__ negate_prob__0.0625__ |
| convert the num__0.472222222222 m / s into kilometers per hour ? <o> a ) num__1.7 <o> b ) num__1.5 <o> c ) num__1.3 <o> d ) num__1.1 <o> e ) num__1.2 |
num__0.472222222222 m / s = num__0.472222222222 * num__3.6 = num__1.7 = num__1.7 kmph . answer : a <eor> a <eos> |
a |
round__1.7__ |
round__1.7__ |
| a bar over a sequence of digits in a decimal indicates that the sequence repeats indefinitely . what is the value of ( num__10 ^ num__4 - num__10 ^ num__2 ) ( num__0.0017 ) ? <o> a ) num__14.83 <o> b ) num__17.83 <o> c ) num__18.83 <o> d ) num__16.83 <o> e ) num__26.83 |
you get num__10 ^ num__2 ( num__100 - num__1 ) ( . num__0017 ) we know num__0.0017 = . num__0017 num__10 ^ num__2 ( num__99 ) * ( num__0.0017 ) num__10 ^ num__2 = num__100 and num__100 ^ num__2 = num__10000 cancel out the num__100 with the num__10 ^ num__2 left with num__99 ( num__0.17 ) . num__0.99 * num__17 = num__16.83 answer : d <eor> d <eos> |
d |
power__10.0__2.0__ subtract__100.0__1.0__ power__10.0__4.0__ multiply__0.0017__100.0__ divide__99.0__100.0__ multiply__99.0__0.17__ multiply__1.0__16.83__ |
power__10.0__2.0__ subtract__100.0__1.0__ power__10.0__4.0__ multiply__0.0017__100.0__ divide__99.0__100.0__ multiply__99.0__0.17__ multiply__1.0__16.83__ |
| p beats q by num__125 meter in a kilometer race . find q ' s speed if p ' s speed is num__8 meter / sec . <o> a ) num__12.25 meter / sec <o> b ) num__7 meter / sec <o> c ) num__14 meter / sec <o> d ) num__18 meter / sec <o> e ) num__28 meter / sec |
p ' s speed = num__8 m / s p ' s distance = num__1000 m ( as it is a km race ) t = num__125.0 = num__125 sec q ' s distance = num__1000 - num__125 = num__875 m q ' s time to complete this distance is also num__62.5 sec . q ' s speed = dist / time = num__7.0 = num__7 m / s hence b is the answer . <eor> b <eos> |
b |
multiply__125.0__8.0__ subtract__1000.0__125.0__ divide__875.0__125.0__ round__7.0__ |
multiply__125.0__8.0__ subtract__1000.0__125.0__ divide__875.0__125.0__ divide__875.0__125.0__ |
| how long does a train num__110 m long running at the speed of num__72 km / hr takes to cross a bridge num__132 m length ? <o> a ) num__12.7 sec <o> b ) num__12.1 sec <o> c ) num__18.1 sec <o> d ) num__17.1 sec <o> e ) num__12.7 sec |
speed = num__72 * num__0.277777777778 = num__20 m / sec total distance covered = num__110 + num__132 = num__242 m . required time = num__12.1 ' = num__12.1 sec . answer : b <eor> b <eos> |
b |
add__110.0__132.0__ divide__242.0__20.0__ round__12.1__ |
add__110.0__132.0__ divide__242.0__20.0__ divide__242.0__20.0__ |
| the s . i . on a certain sum of money for num__6 years at num__16.0 per annum is half the c . i . on rs . num__8000 for num__2 years at num__20.0 per annum . the sum placed on s . i . is ? <o> a ) num__3000.33 <o> b ) num__2802.33 <o> c ) num__1833.33 <o> d ) num__2990.33 <o> e ) num__2982.33 |
c . i . = [ num__8000 * ( num__1 + num__0.2 ) num__2 - num__8000 ] = ( num__8000 * num__1.2 * num__1.2 - num__8000 ) = rs . num__3520 . sum = ( num__3520 * num__100 ) / ( num__6 * num__16 ) = rs . num__1833.33 answer : c <eor> c <eos> |
c |
percent__20.0__1.0__ percent__6.0__20.0__ percent__100.0__1833.33__ |
percent__20.0__1.0__ percent__6.0__20.0__ percent__100.0__1833.33__ |
| in an examination a student scores num__4 marks for every correct answer and loses num__1 mark for every wrong answer . if he attempts all num__75 questions and secures num__125 marks the number of questions he attempts correctly is : <o> a ) num__38 <o> b ) num__40 <o> c ) num__26 <o> d ) num__25 <o> e ) num__11 |
explanation : let the number of correct answers be x . then numbers of incorrect answers will be num__75 – x we get num__4 x – ( num__75 – x ) × num__1 = num__125 on solving the equation we get x = num__40 answer : b <eor> b <eos> |
b |
round__40.0__ |
round__40.0__ |
| num__3889 + num__12.952 – ? = num__3854.002 <o> a ) num__47.095 <o> b ) num__47.752 <o> c ) num__47.932 <o> d ) num__47.95 <o> e ) none of these |
solution let num__3889 + num__12.952 - x = num__3854.002 . then x = ( num__3889 + num__12.952 ) - num__3854.002 = num__3901.952 - num__3854.002 = num__47.95 . answer d <eor> d <eos> |
d |
add__3889.0__12.952__ subtract__3901.952__3854.002__ subtract__3901.952__3854.002__ |
add__3889.0__12.952__ subtract__3901.952__3854.002__ subtract__3901.952__3854.002__ |
| a satellite is composed of num__24 modular units each of which is equipped with a set of sensors some of which have been upgraded . each unit contains the same number of non - upgraded sensors . if the number of non - upgraded sensors on one unit is num__0.25 the total number of upgraded sensors on the entire satellite what fraction of the sensors on the satellite have been upgraded ? <o> a ) num__0.833333333333 <o> b ) num__0.2 <o> c ) num__0.166666666667 <o> d ) num__0.142857142857 <o> e ) num__0.0416666666667 |
let x be the number of upgraded sensors on the satellite . the number of non - upgraded sensors per unit is x / num__4 . the number of non - upgraded sensors on the whole satellite is num__24 ( x / num__4 ) = num__6 x . the fraction of sensors which have been upgraded is x / ( x + num__6 x ) = x / num__7 x = num__0.142857142857 the answer is d . <eor> d <eos> |
d |
reverse__0.25__ multiply__24.0__0.25__ reverse__7.0__ reverse__7.0__ |
reverse__0.25__ divide__24.0__4.0__ reverse__7.0__ reverse__7.0__ |
| a cargo ships engines failed num__100 miles away from the port . due to the changing wind direction it is moving num__12 miles towards the port and num__6 miles away from the port . if the wind pattern remains same how many miles it will travel before reaching the port ? <o> a ) a - num__179 <o> b ) b - num__240 <o> c ) c - num__280 <o> d ) d - num__100 <o> e ) e - num__155 |
ships overall distance covered per cycle is + num__6 miles take num__100 - num__6 take num__15.6666666667 to lowest divisible number - num__15.0 this means that it will take num__15 ` ` overall ' ' actions to reach the num__90 th mile . - num__15 * num__18 miles added later set cycle start at num__90 . travel num__10 miles and reach the anchor point - add num__10 to total num__15 * num__18 + num__10 = num__280 miles in total answer : c <eor> c <eos> |
c |
multiply__6.0__15.0__ add__12.0__6.0__ subtract__100.0__90.0__ round__280.0__ |
multiply__6.0__15.0__ add__12.0__6.0__ subtract__100.0__90.0__ round__280.0__ |
| two pipes a and b can fill a cistern in num__12 and num__15 minutes respectively . both are opened together but after num__6 minutes a is turned off . after how much more time will the cistern be filled ? <o> a ) num__8 num__0.142857142857 <o> b ) num__8 num__1.0 <o> c ) num__8 num__0.25 <o> d ) num__1 num__0.5 <o> e ) num__8 num__0.428571428571 |
num__0.5 + ( num__6 + x ) / num__15 = num__1 x = num__1 num__0.5 answer : d <eor> d <eos> |
d |
divide__6.0__12.0__ round__1.0__ |
divide__6.0__12.0__ round__1.0__ |
| what is the place value of num__4 in the numeral num__2436 ? <o> a ) num__400 <o> b ) num__500 <o> c ) num__700 <o> d ) num__800 <o> e ) num__840 |
option ' a ' num__4 * num__100 = num__400 <eor> a <eos> |
a |
multiply__4.0__100.0__ multiply__4.0__100.0__ |
multiply__4.0__100.0__ multiply__4.0__100.0__ |
| the sum of how many terms of the series num__6 + num__12 + num__18 + num__24 + . . . is num__1800 ? <o> a ) num__20 <o> b ) num__24 <o> c ) num__28 <o> d ) num__32 <o> e ) num__24 |
this is an a . p . in which a = num__6 d = num__6 and sn = num__1800 then n [ num__2 a + ( n - num__1 ) d ] = num__1800 num__2 n [ num__2 x num__6 + ( n - num__1 ) x num__6 ] = num__1800 num__2 num__3 n ( n + num__1 ) = num__1800 n ( n + num__1 ) = num__600 n num__2 + n - num__600 = num__0 n num__2 + num__25 n - num__24 n - num__600 = num__0 n ( n + num__25 ) - num__24 ( n + num__25 ) = num__0 ( n + num__25 ) ( n - num__24 ) = num__0 n = num__24 number of terms = num__24 . b ) <eor> b <eos> |
b |
divide__12.0__6.0__ divide__6.0__2.0__ divide__1800.0__3.0__ add__24.0__1.0__ add__6.0__18.0__ |
divide__12.0__6.0__ add__1.0__2.0__ divide__1800.0__3.0__ add__24.0__1.0__ add__6.0__18.0__ |
| two pipes can fill a tank in num__20 and num__24 minutes respectively and a waste pipe can empty num__3 gallons per minute . all the three pipes working together can fill the tank in num__15 minutes . the capacity of the tank is : <o> a ) num__228 gallons <o> b ) num__267 gallons <o> c ) num__120 gallons <o> d ) num__767 gallons <o> e ) num__167 gallons |
explanation : work done by the waste pipe in num__1 minute = { \ color { black } \ frac { num__1 } { num__15 } - \ left ( \ frac { num__1 } { num__20 } + \ frac { num__1 } { num__24 } \ right ) = \ left ( \ frac { num__1 } { num__15 } - \ frac { num__11 } { num__120 } \ right ) = - \ frac { num__1 } { num__40 } } [ - ve sign means emptying ] volume of { \ color { black } \ frac { num__1 } { num__40 } } part = num__3 gallons volume of whole = ( num__3 x num__40 ) gallons = num__120 gallons . answer : c <eor> c <eos> |
c |
divide__120.0__3.0__ round__120.0__ |
divide__120.0__3.0__ round__120.0__ |
| by looking at a rectangular box a carpenter estimates that the length of the box is between num__2 to num__2.1 meters inclusive the breadth is between num__1 to num__1.1 meters inclusive and the height is between num__2 to num__2.1 centimeters inclusive . if the actual length breadth and height of the box do indeed fall within the respective ranges estimated by the carpenter which of the following is the closest to the maximum possible magnitude of the percentage error r that the carpenter can make in calculating the volume of the rectangular box ? <o> a ) num__1.0 <o> b ) num__3.0 <o> c ) num__10.0 <o> d ) num__18.0 <o> e ) num__22 % |
the dimensions are between num__2 - num__2.1 num__1 - num__1.1 and num__2 - num__2.1 . . so max error in volume = ? . . important points to note num__1 ) max error would be when actuals and estimate are at the extreme ends . . num__2 ) but what about max % error - - it will be when the base is the lowest . . . so the actuals should be lowest or at lower end and the estimate at the higher end . . solution actuals = num__2 * num__2 * num__1 = num__4 . . estimate = num__2.1 * num__2.1 * num__1.1 = num__4.841 . . max % error r = ( num__4.841 - num__4 ) / num__4 * num__100 = num__84.1 / num__4 = num__21.25 . . num__21.25 is closest to num__22.0 in the choices . . ans e <eor> e <eos> |
e |
multiply__1.0__22.0__ |
multiply__1.0__22.0__ |
| a theater box office sold an average ( arithmetic mean ) of num__64 tickets per staff member to a particular movie . among the daytime staff the average number sold per member was num__76 and among the evening staff the average number sold was num__60 . if there are no other employees what was the ratio of the number of daytime staff members to the number of evening staff members ? <o> a ) num__2 : num__5 <o> b ) num__1 : num__4 <o> c ) num__1 : num__3 <o> d ) num__15 : num__19 <o> e ) num__64 : num__76 |
deviation from the mean for the daytime staff = num__76 - num__64 = num__12 . deviation from the mean for the evening staff = num__64 - num__60 = num__4 . thus the ratio of the number of daytime staff members to the number of evening staff members is num__4 : num__12 = num__1 : num__3 . the answer is c . <eor> c <eos> |
c |
subtract__76.0__64.0__ subtract__64.0__60.0__ subtract__4.0__1.0__ reverse__1.0__ |
subtract__76.0__64.0__ subtract__64.0__60.0__ subtract__4.0__1.0__ subtract__4.0__3.0__ |
| the speed of a car is num__80 km in the first hour and num__40 km in the second hour . what is the average speed of the car ? <o> a ) num__72 kmph <o> b ) num__60 kmph <o> c ) num__30 kmph <o> d ) num__80 kmph <o> e ) num__82 kmph |
s = ( num__80 + num__40 ) / num__2 = num__60 kmph answer : b <eor> b <eos> |
b |
divide__80.0__40.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
divide__80.0__40.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
| initially the men and women in a room were in the ratio of num__4 : num__5 . then num__2 men entered the room and num__3 women left the room . then the number of women doubled . now there are num__14 men in the room . how many q women are currently in the room ? <o> a ) num__12 <o> b ) num__14 <o> c ) num__15 <o> d ) num__24 <o> e ) num__36 |
the number of women doubled means that they have become num__24 from num__12 . . and we have to tell the current strength so num__24 is the answer . . let the number be num__4 x and num__5 x . . given num__4 x + num__2 = num__14 . . so x = num__3 . . women number q = num__5 * num__3 - num__3 = num__12 then doubled = num__24 . . ans d <eor> d <eos> |
d |
multiply__4.0__3.0__ multiply__2.0__12.0__ |
multiply__4.0__3.0__ multiply__2.0__12.0__ |
| nine persons went to a hotel for taking their meals . eight of them spent num__12 each on their meals and the ninth spend num__8 more than the average expenditure of all the nine . what was the total money spent by them ? <o> a ) num__115 <o> b ) num__117 <o> c ) num__119 <o> d ) num__122 <o> e ) none of the above |
let the average expenditure of all the nine be x . then num__12 × num__8 + ( x + num__8 ) = num__9 x or num__8 x = num__104 or x = num__13 . ∴ total money spent = num__9 x = ( num__9 × num__13 ) = num__117 . answer b <eor> b <eos> |
b |
divide__104.0__8.0__ multiply__9.0__13.0__ multiply__9.0__13.0__ |
divide__104.0__8.0__ add__13.0__104.0__ add__13.0__104.0__ |
| if john takes num__11 minutes to eat x raisins how many seconds will it take him to eat z raisins assuming he continues to eat at the same rate ? <o> a ) num__660 x / z <o> b ) num__11 x / z <o> c ) num__660 xz <o> d ) num__660 z / x <o> e ) num__11 z / x |
it will take ( z / x ) ( num__11 ) minutes which is ( z / x ) ( num__11 ) ( num__60 ) seconds = num__660 z / x seconds . the answer is d . <eor> d <eos> |
d |
hour_to_min_conversion__ multiply__11.0__60.0__ round__660.0__ |
hour_to_min_conversion__ multiply__11.0__60.0__ round__660.0__ |
| num__3889 + num__12.998 - ? = num__3854.002 <o> a ) a ) num__47.996 <o> b ) b ) num__47.752 <o> c ) c ) num__47.932 <o> d ) d ) num__47.95 <o> e ) of the above |
let num__3889 + num__12.998 - x = num__3854.002 . then x = ( num__3889 + num__12.998 ) - num__3854.002 = num__3901.998 - num__3854.002 = num__47.996 . answer = a <eor> a <eos> |
a |
add__3889.0__12.998__ subtract__3901.998__3854.002__ subtract__3901.998__3854.002__ |
add__3889.0__12.998__ subtract__3901.998__3854.002__ subtract__3901.998__3854.002__ |
| the difference between the place value and the face value of num__7 in the numeral num__856973 is <o> a ) num__63 <o> b ) num__6973 <o> c ) num__5994 <o> d ) num__6084 <o> e ) none of these |
( place value of num__7 ) - ( face value of num__7 ) = ( num__70 - num__7 ) = num__63 answer : option a <eor> a <eos> |
a |
subtract__70.0__7.0__ subtract__70.0__7.0__ |
subtract__70.0__7.0__ subtract__70.0__7.0__ |
| length of a rectangular plot is num__20 mtr more than its breadth . if the cost of fencin gthe plot at num__26.50 per meter is rs . num__5300 what is the length of the plot in mtr ? <o> a ) num__50 m <o> b ) num__60 m <o> c ) num__80 m <o> d ) num__82 m <o> e ) num__84 m |
let breadth = x metres . then length = ( x + num__20 ) metres . perimeter = num__5300 m = num__200 m . num__26.50 num__2 [ ( x + num__20 ) + x ] = num__200 num__2 x + num__20 = num__100 num__2 x = num__80 x = num__40 . hence length = x + num__20 = num__60 m b <eor> b <eos> |
b |
divide__5300.0__26.5__ divide__200.0__2.0__ subtract__100.0__20.0__ multiply__20.0__2.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
divide__5300.0__26.5__ divide__200.0__2.0__ subtract__100.0__20.0__ multiply__20.0__2.0__ add__20.0__40.0__ add__20.0__40.0__ |
| the measures of the num__2 acute angles of a triangle are in the ratio of num__3 : num__7 . what arethe measures of the num__2 angles ? <o> a ) num__20 ° <o> b ) num__70 ° <o> c ) num__110 ° <o> d ) num__63 ° <o> e ) num__140 ° |
if the ratio of the two angles is num__3 : num__7 then the measures of two angles can be written as num__3 x and num__7 x . also the two acute angles of a triangle is equal to num__90 ° . hence num__3 x + num__7 x = num__90 num__10 x = num__90 x = num__9 measures of the two acute angles are num__3 x = num__3 × num__9 = num__27 ° num__7 x = num__7 × num__9 = num__63 ° d <eor> d <eos> |
d |
rectangle_perimeter__2.0__3.0__ power__3.0__2.0__ volume_cube__3.0__ multiply__7.0__9.0__ triangle_area__2.0__63.0__ |
rectangle_perimeter__2.0__3.0__ power__3.0__2.0__ volume_cube__3.0__ multiply__7.0__9.0__ triangle_area__2.0__63.0__ |
| in how many years will a sum of money doubles itself at num__10.0 per annum on simple interest ? <o> a ) num__70.0 <o> b ) num__20.0 <o> c ) num__10.0 <o> d ) num__28.0 <o> e ) num__20 % |
p = ( p * num__10 * r ) / num__100 r = num__10.0 answer : c <eor> c <eos> |
c |
percent__10.0__100.0__ |
percent__10.0__100.0__ |
| what is the difference between the place value and face value of num__3 in the numeral num__1375 ? <o> a ) num__297 <o> b ) num__300 <o> c ) num__310 <o> d ) num__320 <o> e ) num__322 |
place value of num__3 = num__3 * num__100 = num__300 face value of num__3 = num__3 num__300 - num__3 = num__297 a <eor> a <eos> |
a |
multiply__3.0__100.0__ subtract__300.0__3.0__ subtract__300.0__3.0__ |
multiply__3.0__100.0__ subtract__300.0__3.0__ subtract__300.0__3.0__ |
| a train num__125 m long passes a man running at num__4 km / hr in the same direction in which the train is going in num__10 sec . the speed of the train is ? <o> a ) num__65 km / hr <o> b ) num__17 km / hr <o> c ) num__76 km / hr <o> d ) num__49 km / hr <o> e ) num__15 km / hr |
speed of the train relative to man = num__12.5 = num__12.5 m / sec . = num__12.5 * num__3.6 = num__45 km / hr let the speed of the train be x km / hr . then relative speed = ( x - num__4 ) km / hr . x - num__4 = num__45 = > x = num__49 km / hr . answer : d <eor> d <eos> |
d |
divide__125.0__10.0__ multiply__12.5__3.6__ add__4.0__45.0__ round__49.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ add__4.0__45.0__ round__49.0__ |
| a certain car averages num__25 miles per gallon of gasoline when driven in the city and num__40 miles per gallon when driving on the highway . according to these rates which of the following is closest to the number of miles per gallon that the car averages when it is driven num__10 miles in the city and then num__40 miles on the highway ? <o> a ) num__28 <o> b ) num__30 <o> c ) num__33 <o> d ) num__35 <o> e ) num__38 |
num__10 * num__0.04 gpm + num__40 * num__0.025 gpm = num__1.4 total gallons num__50 total miles / num__1.4 total gallons = num__35.7 average mpg d . num__35 <eor> d <eos> |
d |
reverse__25.0__ reverse__40.0__ add__40.0__10.0__ round_down__35.7__ round_down__35.7__ |
reverse__25.0__ reverse__40.0__ add__40.0__10.0__ add__25.0__10.0__ add__25.0__10.0__ |
| a number increased by num__30.0 gives num__780 . the number is <o> a ) num__250 <o> b ) num__400 <o> c ) num__450 <o> d ) num__600 <o> e ) num__520 |
formula = total = num__100.0 increse = ` ` + ' ' decrease = ` ` - ' ' a number means = num__100.0 that same number increased by num__30.0 = num__130.0 num__130.0 - - - - - - - > num__780 ( num__130 × num__6 = num__780 ) num__100.0 - - - - - - - > num__600 ( num__100 × num__6 = num__600 ) d ) <eor> d <eos> |
d |
percent__100.0__600.0__ |
percent__100.0__600.0__ |
| given that p is a positive even integer with a positive units digit if the units digit of p ^ num__3 minus the units digit of p ^ num__2 is equal to num__0 what is the units digit of p + num__1 ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__11 |
p is a positive even integer with a positive units digit - - > the units digit of p can be num__2 num__4 num__6 or num__8 - - > only in order the units digit of p ^ num__3 - p ^ num__2 to be num__0 the units digit of p ^ num__3 and p ^ num__2 must be the same . i . e num__0 num__15 or num__6 intersection of values is num__6 thus the units digit of p + num__1 is num__6 + num__1 = num__9 . answer : a . <eor> a <eos> |
a |
add__3.0__1.0__ multiply__3.0__2.0__ multiply__2.0__4.0__ add__3.0__6.0__ add__3.0__4.0__ |
add__3.0__1.0__ add__2.0__4.0__ add__2.0__6.0__ add__3.0__6.0__ add__3.0__4.0__ |
| p is able to do a piece of work in num__15 days and q can do the same work in num__20 days . if they can work together for num__4 days what is the fraction of work left ? <o> a ) num__0.533333333333 <o> b ) num__0.6 <o> c ) num__0.733333333333 <o> d ) num__0.764705882353 <o> e ) num__15 |
amount of work p can do in num__1 day = num__0.0666666666667 amount of work q can do in num__1 day = num__0.05 amount of work p and q can do in num__1 day = num__0.0666666666667 + num__0.05 = num__0.116666666667 amount of work p and q can together do in num__4 days = num__4 × ( num__0.116666666667 ) = num__0.466666666667 a ) fraction of work left = num__1 – num__0.466666666667 = num__0.533333333333 <eor> a <eos> |
a |
divide__1.0__15.0__ divide__1.0__20.0__ add__0.05__0.0667__ subtract__1.0__0.4667__ multiply__1.0__0.5333__ |
divide__1.0__15.0__ divide__1.0__20.0__ add__0.05__0.0667__ subtract__1.0__0.4667__ multiply__1.0__0.5333__ |
| in a group of ducks and cows the total number of legs are num__40 more than twice the number of heads . find the total number of cows . <o> a ) num__12 <o> b ) num__14 <o> c ) num__16 <o> d ) num__18 <o> e ) num__20 |
let the number of ducks be d and number of cows be c then total number of legs = num__2 d + num__4 c = num__2 ( d + num__2 c ) total number of heads = c + d given that total number of legs are num__40 more than twice the number of heads = > num__2 ( d + num__2 c ) = num__40 + num__2 ( c + d ) = > d + num__2 c = num__20 + c + d = > num__2 c = num__20 + c = > c = num__20 i . e . total number of cows = num__20 answer is e . <eor> e <eos> |
e |
divide__40.0__2.0__ subtract__40.0__20.0__ |
divide__40.0__2.0__ subtract__40.0__20.0__ |
| a driver covers a certain distance by car driving at num__60 km / hr and returns back to the starting point riding on a scooter at num__20 km / hr . what was the average speed for the whole journey ? <o> a ) num__24 km / h <o> b ) num__26 km / h <o> c ) num__28 km / h <o> d ) num__30 km / h <o> e ) num__32 km / h |
time num__1 = d / num__60 time num__2 = d / num__20 total time = d / num__60 + d / num__20 = num__4 d / num__60 = d / num__15 average speed = total distance / total time = num__2 d / ( d / num__15 ) = num__30 km / h the answer is d . <eor> d <eos> |
d |
divide__60.0__4.0__ divide__60.0__2.0__ divide__60.0__2.0__ |
divide__60.0__4.0__ divide__60.0__2.0__ divide__60.0__2.0__ |
| ram and ramesh can do a work piece of work in num__6 days . ram alone can do it in num__10 days . in how many days can ramesh alone can do it ? <o> a ) num__11 days <o> b ) num__12 days <o> c ) num__15 days <o> d ) num__13 days <o> e ) num__10 days |
explanation : ram and ramesh can do a work in num__6 days . part done by ram and ramesh in num__1 day = num__0.166666666667 ram alone can do it in num__10 days . therefore part done by ram alone in num__1 day = num__0.1 . therefore part can do by rames alone in num__1 day = num__0.166666666667 - num__0.1 = num__0.0666666666667 net work can do by ramesh only = num__15.0 = num__15 answer : c <eor> c <eos> |
c |
divide__1.0__6.0__ divide__1.0__10.0__ subtract__0.1667__0.1__ round__15.0__ |
divide__1.0__6.0__ divide__1.0__10.0__ subtract__0.1667__0.1__ round__15.0__ |
| a deer is standing num__50 meters in from the west end of a tunnel . the deer sees a train approaching from the west at a constant speed ten times the speed the deer can run . the deer reacts by running toward the train and clears the exit when the train is num__80 meters from the tunnel . if the deer ran in the opposite direction it would barely escape out the eastern entrance just as the train came out of the eastern entrance . how long is the tunnel in meters ? <o> a ) num__90 <o> b ) num__100 <o> c ) num__110 <o> d ) num__120 <o> e ) num__130 |
let x be the length of the tunnel . when the deer runs num__50 meters west the train goes num__500 meters to a point num__80 meters from west entrance of the tunnel . when the deer runs east the deer runs x - num__50 meters while the train goes x + num__500 + num__80 meters . x + num__500 + num__80 = num__10 ( x - num__50 ) num__9 x = num__1080 x = num__120 meters the answer is d . <eor> d <eos> |
d |
divide__500.0__50.0__ divide__1080.0__9.0__ round__120.0__ |
divide__500.0__50.0__ divide__1080.0__9.0__ round__120.0__ |
| find the number that fits somewhere into the middle of the series . some of the items involve both numbers and letters look at this series : d num__12 h num__16 __ p num__24 t num__28 . . . what number should fill the blank ? <o> a ) h num__16 <o> b ) i num__17 <o> c ) j num__18 <o> d ) k num__19 <o> e ) l num__20 |
e l num__20 in this series the letters progress by num__4 and the numbers increase by num__4 . <eor> e <eos> |
e |
subtract__16.0__12.0__ add__16.0__4.0__ |
subtract__16.0__12.0__ add__16.0__4.0__ |
| laxmi and prasanna set on a journey . laxmi moves northwards at a speed of num__18 kmph and prasanna moves southward at a speed of num__27 kmph . how far will be prasanna from laxmi after num__60 minutes ? <o> a ) num__11 <o> b ) num__50 <o> c ) num__28 <o> d ) num__45 <o> e ) num__18 |
explanation : we know num__60 min = num__1 hr total northward laxmi ' s distance = num__18 kmph x num__1 hr = num__18 km total southward prasanna ' s distance = num__27 kmph x num__1 hr = num__27 km total distance between prasanna and laxmi is = num__18 + num__27 = num__45 km . answer : d <eor> d <eos> |
d |
add__18.0__27.0__ round__45.0__ |
add__18.0__27.0__ add__18.0__27.0__ |
| a boat can travel with a speed of num__42 km / hr in still water . if the speed of the stream is num__10 km / hr find the time taken by the boat to go num__94 km downstream <o> a ) num__1 hour num__40 min <o> b ) num__2 hour num__40 min <o> c ) num__1 hour num__20 min <o> d ) num__1 hour num__30 min <o> e ) num__1 hour num__50 min |
speed of the boat in still water = num__42 km / hr speed of the stream = num__10 km / hr speed downstream = ( num__42 + num__10 ) = num__52 km / hr distance travelled downstream = num__94 km time taken = distance / speed = num__1.80769230769 = num__1.81 hours = num__1 hour num__50 min . answer : e <eor> e <eos> |
e |
add__42.0__10.0__ divide__94.0__52.0__ round__1.0__ |
add__42.0__10.0__ divide__94.0__52.0__ round__1.0__ |
| the owner of a furniture shop charges his customer num__20.0 more than the cost price . if a customer paid rs . num__8400 for a computer table then what was the cost price of the computer table ? <o> a ) num__7000 <o> b ) num__2699 <o> c ) num__2670 <o> d ) num__6725 <o> e ) num__2601 |
cp = sp * ( num__100 / ( num__100 + profit % ) ) = num__8400 ( num__0.833333333333 ) = rs . num__7000 . answer : a <eor> a <eos> |
a |
percent__100.0__7000.0__ |
percent__100.0__7000.0__ |
| some consecutive natural numbers starting with num__1 are written on the board . now one of the numbers was erased and the average of the remaining numbers is num__20.5128205128 . find the number which was erased . <o> a ) num__24 <o> b ) num__20 <o> c ) num__18 <o> d ) num__16 <o> e ) num__15 |
explanation : we know that average of n consecutive numbes average = n × ( n + num__1 ) / num__2 / n = ( n + num__1 ) / num__2 if the given n is sufficiently large the average does not change much even though we exclude one or two numbers from it . so the approximate number of observations is almost double to the average ( remember : the average of consecutive numbers almost lies in the middle ) the approximate average is num__20.5128205128 = approx num__20 . so the initial numbers may be nearer to num__40 . in this question it is actually num__40 as from the denominator of the new average num__20.5128205128 . the initial numbers are num__40 . sum of num__40 consecutive numbers = num__40 × ( num__40 + num__1 ) / num__2 = num__820 sum of num__39 numbers = average x number of observations = num__20.5128205128 × num__39 = num__800 so the number excluded = num__820 - num__800 = num__20 correct option : b <eor> b <eos> |
b |
round_down__20.5128__ multiply__2.0__20.0__ subtract__40.0__1.0__ multiply__40.0__20.0__ round_down__20.5128__ |
round_down__20.5128__ multiply__2.0__20.0__ subtract__40.0__1.0__ subtract__820.0__20.0__ divide__800.0__40.0__ |
| what is the least common multiple of num__152 and num__190 ? <o> a ) num__304 <o> b ) num__14440 <o> c ) num__760 <o> d ) num__28880 <o> e ) num__1520 |
yes there is a shorter way num__152 = num__2 * num__2 * num__2 * num__19 num__190 = num__2 * num__5 * num__19 i think everyone knows how to do this . then choose num__2 * num__2 * num__2 and choose num__5 and choose num__19 num__2 * num__2 * num__2 * num__5 * num__19 = num__760 answer is c <eor> c <eos> |
c |
multiply__152.0__5.0__ multiply__152.0__5.0__ |
multiply__152.0__5.0__ multiply__152.0__5.0__ |
| the speed of a boat in still water in num__42 km / hr and the rate of current is num__6 km / hr . the distance travelled downstream in num__44 minutes is : <o> a ) num__86.6 km <o> b ) num__46.6 km <o> c ) num__35.2 km <o> d ) num__35.6 km <o> e ) num__26.6 km |
speed downstream = ( num__42 + num__6 ) = num__48 kmph time = num__44 minutes = num__0.733333333333 hour = num__0.733333333333 hour distance travelled = time × speed = num__0.733333333333 × num__48 = num__35.2 km answer : c <eor> c <eos> |
c |
add__42.0__6.0__ round__35.2__ |
add__42.0__6.0__ round__35.2__ |
| in objective test a correct ans score num__4 marks and on a wrong ans num__2 marks are - - - . a student score num__480 marks from num__150 question . how many ans were correct ? <o> a ) num__120 <o> b ) num__130 <o> c ) num__110 <o> d ) num__150 <o> e ) num__180 |
let x be the correct answer and y be the wrong answer so the total number of questions is ( x + y ) = num__150 = > num__4 x - num__2 y = num__480 = > num__6 x = num__780 hence x = num__130 answer : b <eor> b <eos> |
b |
add__4.0__2.0__ divide__780.0__6.0__ divide__780.0__6.0__ |
add__4.0__2.0__ divide__780.0__6.0__ divide__780.0__6.0__ |
| a dishonest dealer professes to sell goods at the cost price but uses a weight of num__600 grams per kg what is his percent ? <o> a ) num__15.0 <o> b ) num__25.0 <o> c ) num__65.0 <o> d ) num__45.0 <o> e ) num__66.7 % |
explanation : num__600 - - - num__400 num__100 - - - ? = > num__66.7 answer : e <eor> e <eos> |
e |
percent__66.7__100.0__ |
percent__66.7__100.0__ |
| two trains start from same place at same time at right angles to each other . their speeds are num__36 km / hr and num__48 km / hr respectively . after num__30 seconds the distance between them will be ? <o> a ) num__270 mts <o> b ) num__190 mts <o> c ) num__100 mts <o> d ) num__500 mts <o> e ) num__110 mts |
explanation : using pythagarous theorem distance travelled by first train = num__36 x num__0.277777777778 x num__30 = num__300 m distance travelled by second train = num__48 x num__0.277777777778 x num__30 = num__400 m so distance between them = â ˆ š ( num__90000 + num__160000 ) = â ˆ š num__250000 = num__500 mts . answer : d <eor> d <eos> |
d |
add__160000.0__90000.0__ round__500.0__ |
add__160000.0__90000.0__ round__500.0__ |
| the length of a rectangle is increased by num__25.0 and its breadth is decreased by num__20.0 . what is the effect on its area ? <o> a ) num__10000 <o> b ) num__18999 <o> c ) num__16990 <o> d ) num__16007 <o> e ) num__16991 |
num__100 * num__100 = num__10000 num__125 * num__80 = num__10000 answer : a <eor> a <eos> |
a |
square_perimeter__25.0__ surface_rectangular_prism__25.0__20.0__100.0__ square_perimeter__20.0__ multiply__80.0__125.0__ |
square_perimeter__25.0__ surface_rectangular_prism__25.0__20.0__100.0__ square_perimeter__20.0__ multiply__80.0__125.0__ |
| how many two - digit numbers yield a remainder of num__2 when divided by both num__7 and num__16 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
easier to start with numbers that are of the form num__16 p + num__2 - - - > num__1834 num__5066 num__8298 . out of these no number is also of the form num__7 q + num__2 . thus num__0 is the answer . a is the correct answer . <eor> a <eos> |
a |
multiply__2.0__0.0__ |
multiply__2.0__0.0__ |
| a shopkeeper sold an article offering a discount of num__5.0 and earned a profit of num__18.75 . what would have been the percentage of profit earned if no discount was offered ? <o> a ) num__60.0 <o> b ) num__23.0 <o> c ) num__25.0 <o> d ) num__56.0 <o> e ) num__73 % |
let c . p . be rs . num__100 . then s . p . = rs . num__118.75 let marked price be rs . x . then num__0.95 x = num__118.75 x = num__125.0 = rs . num__125 now s . p . = rs . num__125 c . p . = rs . num__100 profit % = num__25.0 . answer : c <eor> c <eos> |
c |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| a thief is noticed by a policeman from a distance of num__150 m . the thief starts running and the policeman chases him . the thief and the policeman run at the rate of num__10 km and num__11 km per hour respectively . what is the distance between them after num__6 minutes ? <o> a ) num__50 meter <o> b ) num__100 meter <o> c ) num__110 meter <o> d ) num__120 meter <o> e ) num__130 meter |
explanation : relative speed of the thief and policeman = ( num__11 – num__10 ) km / hr = num__1 km / hr distance covered in num__6 minutes = num__0.0166666666667 ∗ num__6 = num__0.1 = num__100 meters so distance between them after num__6 minutes = num__150 - num__100 = num__50 meters option a <eor> a <eos> |
a |
subtract__11.0__10.0__ divide__1.0__10.0__ divide__10.0__0.1__ subtract__150.0__100.0__ round__50.0__ |
subtract__11.0__10.0__ divide__1.0__10.0__ divide__10.0__0.1__ subtract__150.0__100.0__ subtract__150.0__100.0__ |
| the average of num__11 numbers is num__10.9 . if the average of first six is num__10.5 and that of the last six is num__11.4 the sixth number is ? <o> a ) num__11.9 <o> b ) num__11.0 <o> c ) num__11.3 <o> d ) num__11.5 <o> e ) num__11.1 |
num__1 to num__11 = num__11 * num__10.9 = num__119.9 num__1 to num__6 = num__6 * num__10.5 = num__63 num__6 to num__11 = num__6 * num__11.4 = num__68.4 num__63 + num__68.4 = num__131.4 – num__119.9 = num__11.5 num__6 th number = num__11.5 . answer : d <eor> d <eos> |
d |
multiply__11.0__10.9__ multiply__10.5__6.0__ multiply__11.4__6.0__ add__68.4__63.0__ add__10.5__1.0__ add__10.5__1.0__ |
multiply__11.0__10.9__ multiply__10.5__6.0__ multiply__11.4__6.0__ add__68.4__63.0__ add__10.5__1.0__ add__10.5__1.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__12 sec . what is the length of the train ? <o> a ) num__120 m <o> b ) num__200 m <o> c ) num__115 m <o> d ) num__110 m <o> e ) num__150 m |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__12 = num__200 m answer : b <eor> b <eos> |
b |
round__200.0__ |
round__200.0__ |
| an investment gained an interest payment of $ num__250 each month when the annual interest rate was num__8.0 how much more should we invest annually if we want to gain num__12.0 more per month with a new annual interest rate of num__7.5 ? <o> a ) $ num__9360 <o> b ) $ num__9100 <o> c ) $ num__8250 <o> d ) $ num__7300 <o> e ) $ num__7 |
150 |
answer is d : num__7300 interest per annum with num__8.0 interest rate is = num__250 * num__12 = principal * interest rate ( num__0.08 ) this gives principal as num__37500 $ for an revised interest of num__12.0 more and new interest rate of num__7.5 ie . num__250 * num__12 * num__1.12 = principal * interest rate ( num__7.5 / num__100 ) this gives the new principal as num__44800 . the question is how much more should we invest so num__44800 - num__37500 = num__7300 answer : d <eor> d <eos> |
d |
d |
| if the compound interest on a certain sum of money for num__7 years at num__10.0 per annum be rs . num__993 what would be the simple interest ? <o> a ) rs . num__880 <o> b ) rs . num__890 <o> c ) rs . num__895 <o> d ) rs . num__2100 <o> e ) none |
let p = principal a - amount we have a = p ( num__1 + r / num__100 ) num__3 and ci = a - p atq num__993 = p ( num__1 + r / num__100 ) num__3 - p ? p = num__3000 / - now si @ num__10.0 on num__3000 / - for num__7 yrs = ( num__3000 x num__10 x num__7 ) / num__100 = num__2100 / - answer : d . <eor> d <eos> |
d |
percent__100.0__2100.0__ |
percent__100.0__2100.0__ |
| what is the difference between the largest number and the least number written with the figures num__3 num__4 num__7 num__0 num__3 ? <o> a ) num__98343 <o> b ) num__34389 <o> c ) num__43983 <o> d ) num__43883 <o> e ) num__43700 |
num__74330 largest num__30347 smallest - - - - - - - - - - - - num__43983 answer c <eor> c <eos> |
c |
subtract__74330.0__30347.0__ subtract__74330.0__30347.0__ |
subtract__74330.0__30347.0__ subtract__74330.0__30347.0__ |
| the probability that a man will be alive for num__10 more yrs is num__0.25 & the probability that his wife will alive for num__10 more yrs is num__0.333333333333 . the probability that none of them will be alive for num__10 more yrs is <o> a ) num__0.5 <o> b ) num__1 <o> c ) num__0.666666666667 <o> d ) num__0.75 <o> e ) num__2 |
sol . required probability = pg . ) x p ( b ) = ( num__1 — d x ( num__1 — i ) = : x num__1 = num__0.5 ans . ( a ) <eor> a <eos> |
a |
divide__0.25__0.5__ |
divide__0.25__0.5__ |
| how many numbers from num__19 to num__79 are exactly divisible by num__11 ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__9 <o> d ) num__11 <o> e ) num__12 |
num__1.72727272727 = num__1 and num__7.18181818182 = num__7 = = > num__7 - num__1 = num__6 numbers answer : b <eor> b <eos> |
b |
divide__19.0__11.0__ round_down__1.7273__ divide__79.0__11.0__ round_down__7.1818__ subtract__7.0__1.0__ multiply__1.0__6.0__ |
divide__19.0__11.0__ round_down__1.7273__ divide__79.0__11.0__ round_down__7.1818__ subtract__7.0__1.0__ subtract__7.0__1.0__ |
| miller street begins at baker street and runs directly east for num__4.5 kilometers until it ends when it meets turner street . miller street is intersected every num__250 meters by a perpendicular street and each of those streets other than baker street and turner street is given a number beginning at num__1 st street ( one block east of baker street ) and continuing consecutively ( num__2 nd street num__3 rd street etc . . . ) until the highest - numbered street one block west of turner street . what is the highest - numbered street that intersects miller street ? <o> a ) num__15 th <o> b ) num__16 th <o> c ) num__17 th <o> d ) num__18 th <o> e ) num__19 th |
num__4.5 km / num__250 m = num__18 . however the street at the num__4.5 - km mark is not num__18 th street ; it is turner street . therefore the highest numbered street is num__17 th street . the answer is c . <eor> c <eos> |
c |
subtract__18.0__1.0__ round__17.0__ |
subtract__18.0__1.0__ divide__17.0__1.0__ |
| faiza has num__9 purses she gives num__3 purse as gift . now how many purse did she have ? <o> a ) num__2 <o> b ) num__6 <o> c ) num__11 <o> d ) num__9 <o> e ) num__4 |
num__9 - num__3 = num__6 . answer is b <eor> b <eos> |
b |
subtract__9.0__3.0__ subtract__9.0__3.0__ |
subtract__9.0__3.0__ subtract__9.0__3.0__ |
| a lawyer ' s representation costs $ num__150 for the first hour and $ num__125 for each additional hour . what is the total cost in dollars of her representation for m hours where m is an integer greater than num__1 ? <o> a ) num__150 + num__125 m <o> b ) num__150 + num__125 ( m – num__1 ) <o> c ) num__125 + num__150 m <o> d ) num__125 + num__150 ( m – num__1 ) <o> e ) num__275 ( m – num__1 ) |
for the first hour the cost is num__150 . as first hour is already calculated the remaining is m - num__1 and corresponding cost is num__125 . therefore answer is num__150 + num__125 ( m - num__1 ) . answer b <eor> b <eos> |
b |
round__150.0__ |
round__150.0__ |
| one year ago the ratio between samir and ashok ’ s age was num__4 : num__3 . one year hence the ratio of their ages will be num__5 : num__4 . what is the sum of their present ages in years ? <o> a ) num__12 years <o> b ) num__15 years <o> c ) num__16 years <o> d ) can not be determined <o> e ) none of these |
let their ages one year ago be num__4 x and num__3 x years . ( num__4 x + num__2 ) / ( num__3 x + num__2 ) = num__1.25 = > num__4 ( num__4 x + num__2 ) = num__5 ( num__3 x + num__2 ) = > x = num__2 sum of their present ages = ( num__4 x + num__1 + num__3 x + num__1 ) = ( num__7 x + num__2 ) = num__16 years answer : c <eor> c <eos> |
c |
subtract__5.0__3.0__ divide__5.0__4.0__ round_down__1.25__ add__4.0__3.0__ multiply__1.0__16.0__ |
subtract__5.0__3.0__ divide__5.0__4.0__ round_down__1.25__ add__4.0__3.0__ divide__16.0__1.0__ |
| what is the rate percent when the simple interest on rs . num__4000 amount to rs . num__320 in num__2 years ? <o> a ) num__4.0 <o> b ) num__4.1 <o> c ) num__4.2 <o> d ) num__4.3 <o> e ) num__4.5 % |
interest for num__1 year = num__160.0 = num__160 interest on rs num__4000 p / a = num__160 interest rate = num__0.04 * num__100 = num__4.0 answer : a <eor> a <eos> |
a |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| when n is divided by num__20 the remainder is num__5 . what is the remainder when n + num__16 is divided by num__5 ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
assume n = num__15 remainder ( n / num__20 ) = num__5 n + num__16 = num__31 remainder ( num__6.2 ) = num__1 option a <eor> a <eos> |
a |
subtract__20.0__5.0__ add__16.0__15.0__ divide__31.0__5.0__ subtract__16.0__15.0__ reverse__1.0__ |
subtract__20.0__5.0__ add__16.0__15.0__ divide__31.0__5.0__ subtract__16.0__15.0__ reverse__1.0__ |
| the ratio of a compound by weight consisting only of substances x y and z is num__4 : num__6 : num__10 respectively . due to a dramatic increase in the surrounding temperature the composition of the compound is changed such that the ratio of x to y is halved and the ratio of x to z is tripled . in the changed compound if the total weight is num__174 lbs how much does substance x weigh ? <o> a ) num__48 <o> b ) num__36 <o> c ) num__24 <o> d ) num__12 <o> e ) num__10 |
x : y = num__4 : num__6 if the ratio is halevd then x : y = num__2 : num__6 the old ratio of x to z was num__4 : num__10 . if this ratio is tripled then the new ratio of x to z is num__12 : num__10 . x : y = num__2 : num__6 = num__12 : num__36 ( multiplied the ration with num__1.0 to have a common co eff x in both the rations x : y and x : z ) so x : y : z = num__12 : num__36 : num__10 and we know x + y + z = num__174 lbs from the ratio of x y and z we have x = num__12 k y = num__36 k z = num__10 k put it in the equation we have num__12 k + num__36 k + num__10 k = num__174 k = num__3 hence x = num__12 ( num__3 ) = num__36 ans b <eor> b <eos> |
b |
subtract__6.0__4.0__ multiply__6.0__2.0__ subtract__4.0__1.0__ multiply__1.0__36.0__ |
subtract__6.0__4.0__ add__10.0__2.0__ add__1.0__2.0__ multiply__1.0__36.0__ |
| in how many years rs . num__200 will produce the same interest at num__6.0 as rs . num__800 produce in num__2 years at num__4 ½ % ? <o> a ) num__4 years <o> b ) num__6 years <o> c ) num__8 years <o> d ) num__9 years <o> e ) num__12 years |
explanation : let simple interest for rs . num__200 at num__6.0 for n years = simple interest for rs . num__800 at num__4 ½ % for num__2 years num__200 × num__6 × n / num__100 = num__800 × num__4.5 × num__0.02 num__200 × num__6 × n = num__800 × num__4.5 × num__2 num__200 × num__6 × n = num__800 × num__9 n = num__6 years answer : option b <eor> b <eos> |
b |
percent__4.5__200.0__ percent__6.0__100.0__ |
percent__4.5__200.0__ percent__6.0__100.0__ |
| every disk in a bag is either blue yellow or green . the ratio of blue disks to yellow disks to green disks in this bag is num__3 : num__7 : num__8 . if the total number of disks in the bag is num__144 how many more green disks than blue disks are in the bag ? <o> a ) num__25 <o> b ) num__28 <o> c ) num__30 <o> d ) num__35 <o> e ) num__40 |
let b : y : g = num__3 x : num__7 x : num__8 x . num__3 x + num__7 x + num__8 x = num__18 x = num__144 - - > x = num__8 . g - b = num__8 x - num__3 x = num__5 x = num__40 . the answer is e . <eor> e <eos> |
e |
divide__144.0__8.0__ subtract__8.0__3.0__ multiply__8.0__5.0__ multiply__8.0__5.0__ |
divide__144.0__8.0__ subtract__8.0__3.0__ multiply__8.0__5.0__ multiply__8.0__5.0__ |
| a pupil ’ s marks were wrongly entered as num__83 instead of num__63 . due to that the average marks for the class got increased by half . the number of pupils in the class is : <o> a ) num__10 <o> b ) num__20 <o> c ) num__40 <o> d ) num__73 <o> e ) num__50 |
let there be x pupils in the class . total increase in marks = ( x × num__1 ⁄ num__2 ) = x ⁄ num__2 . ∴ x ⁄ num__2 = ( num__83 - num__63 ) ⇒ x ⁄ num__2 = num__20 ⇒ x = num__40 . answer c <eor> c <eos> |
c |
subtract__83.0__63.0__ multiply__2.0__20.0__ multiply__1.0__40.0__ |
subtract__83.0__63.0__ multiply__2.0__20.0__ multiply__1.0__40.0__ |
| in what ratio must water be added to num__10 litres of milk at rs . num__20 per litre so that cost of mixture is rs . num__16 per litre ? <o> a ) num__3 : num__2 <o> b ) num__1 : num__4 <o> c ) num__2 : num__3 <o> d ) num__4 : num__1 <o> e ) num__4 : num__3 |
c . p of num__1 lt of water = rs . num__0 c . p of num__1 lt of milk = rs . num__20 so o ( cheaper ) num__20 ( dearer ) num__16 ( mixture ) num__4 : num__16 ( dearer - mixture ) ( mixture - cheaper ) so ( note : just cross check ) num__4 : num__16 = num__1 : num__4 answer : b <eor> b <eos> |
b |
subtract__20.0__16.0__ reverse__1.0__ |
subtract__20.0__16.0__ reverse__1.0__ |
| in a function they are distributing noble prize . in how many ways can num__3 prizes be distributed among num__4 boys when no boy gets more than one prize ? <o> a ) num__14 <o> b ) num__19 <o> c ) num__20 <o> d ) num__24 <o> e ) num__26 |
sol . in this case repetitions are not allowed . so the first prize can be given in num__4 ways . the second in num__3 ways and the third in num__2 ways . but fundamental principle ( num__4 x num__3 x num__2 ) ways = num__24 ways num__4 : or num__4 p = — num__4 : - num__4 x num__3 x num__2 x num__1 - num__24 ways d <eor> d <eos> |
d |
coin_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
coin_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
| in a certain district the ratio of the number of registered republicans to the number of registered democrats was num__0.6 . after num__600 additional republicans and num__500 additional democrats registered the ratio was num__0.8 . after these registrations there were how many more voters in the district registered as democrats than as republicans ? <o> a ) num__100 <o> b ) num__300 <o> c ) num__400 <o> d ) num__1000 <o> e ) num__2 |
500 |
num__0.6 republican voters = num__0.6 * x another num__600 register as republican with the total amount being num__80.0 republican ( num__0.8 ) total new voters is num__1100 ( num__500 + num__600 ) so num__0.6 x + num__600 = num__0.8 ( x + num__1100 ) num__0.6 x + num__600 = num__0.8 x + num__880 answer is b <eor> b <eos> |
b |
b |
| the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is num__867 sq m then what is the breadth of the rectangular plot ? <o> a ) num__11 <o> b ) num__17 <o> c ) num__88 <o> d ) num__66 <o> e ) num__12 |
let the breadth of the plot be b m . length of the plot = num__3 b m ( num__3 b ) ( b ) = num__867 num__3 b num__2 = num__867 b num__2 = num__289 = num__172 ( b > num__0 ) b = num__17 m . answer : b <eor> b <eos> |
b |
triangle_area__2.0__17.0__ |
triangle_area__2.0__17.0__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__816 m <o> b ) num__577 m <o> c ) num__240 m <o> d ) num__176 m <o> e ) num__126 m |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x meters . then x + num__8.33333333333 = num__15 x + num__300 = num__540 x = num__240 m . answer : c <eor> c <eos> |
c |
multiply__20.0__15.0__ divide__300.0__36.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ divide__300.0__36.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
| light glows for every num__20 seconds . how many max . times did it glow between num__1 : num__57 : num__58 and num__3 : num__20 : num__47 am . <o> a ) num__380 times <o> b ) num__381 times <o> c ) num__382 times <o> d ) num__248 times <o> e ) num__482 times |
time difference is num__1 hr num__22 min num__49 sec = num__4969 sec . so light glows floor ( num__248.45 ) = num__248 times . answer : d <eor> d <eos> |
d |
divide__4969.0__20.0__ round_down__248.45__ round_down__248.45__ |
divide__4969.0__20.0__ round_down__248.45__ round_down__248.45__ |
| andrew started saving at the beginning of the year and had saved $ num__160 by the end of the year . he continued to save and by the end of num__2 years had saved a total of $ num__540 . which of the following is closest to the percent increase in the amount andrew saved during the second year compared to the amount he saved during the first year ? <o> a ) num__11.0 <o> b ) num__25.0 <o> c ) num__44.0 <o> d ) num__56.0 <o> e ) num__137.5 % |
percent increase in the amount andrew saved during the second year compared to the amount he saved during the first year . percentage change = ( change in value / original value ) * num__100 percentage change = [ ( saving in num__2 nd year - saving in num__1 st year ) / saving in num__1 st year ] * num__100 saving in num__2 nd year + saving in num__1 st year = num__540 saving in num__2 nd year + num__160 = num__540 saving in num__2 nd year = num__540 - num__160 = num__380 change in value = num__380 - num__160 = num__220 original value = num__160 percentage increase = ( num__1.375 ) * num__100 = num__137.5 . . . . . answer e . . . . . <eor> e <eos> |
e |
subtract__540.0__160.0__ subtract__380.0__160.0__ divide__220.0__160.0__ multiply__100.0__1.375__ multiply__1.0__137.5__ |
subtract__540.0__160.0__ subtract__380.0__160.0__ divide__220.0__160.0__ multiply__100.0__1.375__ multiply__1.0__137.5__ |
| all the water in container a which was filled to its brim was poured into two containers b and c . the quantity of water in container b was num__62.5 less than the capacity of container a . if num__148 liters was now transferred from c to b then both the containers would have equal quantities of water . what was the initial quantity of water in container a ? <o> a ) num__1775 liters <o> b ) num__8176 liters <o> c ) num__1467 liters <o> d ) num__1184 liters <o> e ) num__1886 liters |
b has num__62.5 or ( num__0.625 ) of the water in a . therefore let the quantity of water in container a ( initially ) be num__8 k . quantity of water in b = num__8 k - num__5 k = num__3 k . quantity of water in container c = num__8 k - num__3 k = num__5 k container : a b c quantity of water : num__8 k num__3 k num__5 k it is given that if num__148 liters was transferred from container c to container b then both the containers would have equal quantities of water . num__5 k - num__148 = num__3 k + num__148 = > num__2 k = num__296 = > k = num__148 the initial quantity of water in a = num__8 k = num__8 * num__148 = num__1184 liters . answer : d <eor> d <eos> |
d |
multiply__0.625__8.0__ subtract__8.0__5.0__ subtract__5.0__3.0__ multiply__148.0__2.0__ multiply__148.0__8.0__ multiply__148.0__8.0__ |
multiply__0.625__8.0__ subtract__8.0__5.0__ subtract__5.0__3.0__ multiply__148.0__2.0__ multiply__148.0__8.0__ multiply__148.0__8.0__ |
| how many words with or without meaning can be formed by using all the letters of the word â € ˜ ram â € ™ using each letter exactly once ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__5 <o> d ) num__9 <o> e ) num__1 |
the word â € ˜ ram â € ™ contains num__3 different letters . therefore required number of words = number of arrangement of num__3 letters taken all at a time = num__3 p num__3 = num__3 ! = num__3 * num__2 * num__1 = num__6 answer : a <eor> a <eos> |
a |
subtract__3.0__2.0__ multiply__2.0__3.0__ multiply__1.0__6.0__ |
subtract__3.0__2.0__ multiply__2.0__3.0__ multiply__1.0__6.0__ |
| two pipes a and b can separately fill a cistern in num__10 and num__15 minutes respectively . a person opens both the pipes together when the cistern should have been was full he finds the waste pipe open . he then closes the waste pipe and in another num__4 minutes the cistern was full . in what time can the waste pipe empty the cistern when fill ? <o> a ) a ) num__0.125 <o> b ) b ) num__0.333333333333 <o> c ) c ) num__0.111111111111 <o> d ) d ) num__0.5 <o> e ) e ) num__0.25 |
num__0.1 + num__0.0666666666667 = num__0.166666666667 * num__4 = num__0.666666666667 num__1 - num__0.666666666667 = num__0.333333333333 num__0.1 + num__0.0666666666667 - num__1 / x = num__0.333333333333 x = num__8 answer : b <eor> b <eos> |
b |
add__0.1__0.0667__ divide__10.0__15.0__ multiply__10.0__0.1__ subtract__1.0__0.6667__ subtract__1.0__0.6667__ |
add__0.1__0.0667__ divide__10.0__15.0__ multiply__10.0__0.1__ subtract__1.0__0.6667__ subtract__1.0__0.6667__ |
| num__80 pie square inches of material of negligible thickness are required to construct a num__0.0625 scale model of a cylindrical barrel . if the diameter of the base of the barrel is num__160 inches then what is the volume of the barrel to the nearest cubic foot ? <o> a ) num__2 pie <o> b ) num__11 pie <o> c ) num__178 pie <o> d ) num__711 pie <o> e ) num__1280 pie |
radius of barrel r = num__80 radius of model r = num__5.0 = num__5 surface area of the model = num__2 ∗ pi ∗ r ( r + h ) = num__80 ∗ pi which gives h = num__3 and h = num__3 ∗ num__16 = num__48 so volume of the barrel = pi ∗ r num__2 ∗ h = pi ∗ ( num__80 ) num__2 ∗ ( num__48 ) in cubic inches converting in cubic ft results : volume = pi ∗ ( num__80 ) num__2 ∗ ( num__48 ) / ( num__12 ) num__3 = num__178 pi cubic ft approximately answer [ c ] <eor> c <eos> |
c |
multiply__80.0__0.0625__ rectangle_perimeter__3.0__5.0__ multiply__16.0__3.0__ square_perimeter__3.0__ triangle_perimeter__160.0__16.0__2.0__ triangle_area__2.0__178.0__ |
multiply__80.0__0.0625__ rectangle_perimeter__3.0__5.0__ multiply__16.0__3.0__ square_perimeter__3.0__ triangle_perimeter__160.0__16.0__2.0__ triangle_area__2.0__178.0__ |
| the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is num__42 kmph find the speed of the stream ? <o> a ) num__12 kmph <o> b ) num__13 kmph <o> c ) num__14 kmph <o> d ) num__15 kmph <o> e ) num__16 kmph |
the ratio of the times taken is num__2 : num__1 . the ratio of the speed of the boat in still water to the speed of the stream = ( num__2 + num__1 ) / ( num__2 - num__1 ) = num__3.0 = num__3 : num__1 speed of the stream = num__14.0 = num__14 kmph answer : c <eor> c <eos> |
c |
add__1.0__2.0__ divide__42.0__3.0__ round__14.0__ |
add__1.0__2.0__ divide__42.0__3.0__ divide__42.0__3.0__ |
| what is the units digit of ( num__147 ^ num__17 ) ^ num__47 ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__9 |
the units digit of the exponents of num__7 repeat in a cycle of four which is { num__79 num__31 } . the number num__17 has the form num__4 n + num__1 so the units digit is num__7 inside the bracket . the exponent num__47 has the form num__4 n + num__3 so the units digit is num__3 . the answer is b . <eor> b <eos> |
b |
subtract__4.0__1.0__ multiply__1.0__3.0__ |
subtract__4.0__1.0__ multiply__1.0__3.0__ |
| p is three times as old as q . in num__9 years p will be twice as old as q . how old was p three years ago ? <o> a ) num__18 <o> b ) num__21 <o> c ) num__24 <o> d ) num__27 <o> e ) num__30 |
p = num__3 q so q = p / num__3 p + num__9 = num__2 ( q + num__9 ) = num__2 ( p / num__3 + num__9 ) p / num__3 = num__9 p = num__27 three years ago p was num__24 . the answer is c . <eor> c <eos> |
c |
multiply__9.0__3.0__ subtract__27.0__3.0__ subtract__27.0__3.0__ |
multiply__9.0__3.0__ subtract__27.0__3.0__ subtract__27.0__3.0__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__210 <o> b ) num__220 <o> c ) num__240 <o> d ) num__250 <o> e ) num__260 |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x meters . then x + num__8.33333333333 = num__15 x + num__300 = num__540 x = num__240 m . answer : option c <eor> c <eos> |
c |
multiply__20.0__15.0__ divide__300.0__36.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ divide__300.0__36.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
| ramesh can finish a work in num__15 days and sushil in num__25 days . they both work together for num__5 days and then sushil goes away . in how many days will ramesh complete the remaining work ? <o> a ) num__4 days <o> b ) num__7 days <o> c ) num__6 days <o> d ) num__9 days <o> e ) num__5 days |
( num__5 + x ) / num__15 + num__0.2 = num__1 = > x = num__7 days answer : b <eor> b <eos> |
b |
divide__5.0__25.0__ multiply__5.0__0.2__ round__7.0__ |
divide__5.0__25.0__ multiply__5.0__0.2__ divide__7.0__1.0__ |
| a no . when divided by the sum of num__555 and num__445 gives num__2 times their difference as quotient & num__25 as remainder . find the no . is ? <o> a ) num__145646 <o> b ) num__236578 <o> c ) num__645353 <o> d ) num__456546 <o> e ) num__220025 |
( num__555 + num__445 ) * num__2 * num__110 + num__25 = num__220000 + num__25 = num__220025 e <eor> e <eos> |
e |
subtract__555.0__445.0__ add__25.0__220000.0__ add__25.0__220000.0__ |
subtract__555.0__445.0__ add__25.0__220000.0__ add__25.0__220000.0__ |
| a man buys two articles for rs . num__1980 each and he gains num__10.0 on the first and loses num__10.0 on the next . find his total gain or loss percent ? <o> a ) num__9 <o> b ) num__8 <o> c ) num__7 <o> d ) num__6 <o> e ) num__5 |
( num__30 * num__30 ) / num__100 = num__9.0 loss answer : a <eor> a <eos> |
a |
percent__9.0__100.0__ |
percent__9.0__100.0__ |
| a man walking at the rate of num__5 km / hr crosses a bridge in num__15 minutes . the length of the bridge ( in metres ) is <o> a ) num__2347 <o> b ) num__1277 <o> c ) num__2288 <o> d ) num__1250 <o> e ) num__7178 |
explanation : speed = ( num__5 x num__0.277777777778 ) m / sec = num__1.38888888889 m / sec . distance covered in num__15 minutes = ( num__1.38888888889 x num__15 x num__60 ) m = num__1250 m . answer : d ) num__1250 <eor> d <eos> |
d |
hour_to_min_conversion__ round__1250.0__ |
hour_to_min_conversion__ round__1250.0__ |
| when magnified num__1000 times by an electron microscope the image of a certain circular piece of tissue has a diameter of num__0.3 centimeter . the actual diameter of the tissue in centimeters is <o> a ) num__0.005 <o> b ) num__0.002 <o> c ) num__0.001 <o> d ) num__0.0003 <o> e ) num__0.0002 |
it is very easy if x is the diameter then the magnified length is num__1000 x . ince num__1000 x = num__0.3 then x = num__0.3 / num__1000 = num__0.0003 . the answer is d <eor> d <eos> |
d |
volume_cylinder__0.0003__1000.0__ |
volume_cylinder__0.0003__1000.0__ |
| p q and r have rs . num__6000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? <o> a ) rs . num__3000 <o> b ) rs . num__3600 <o> c ) rs . num__2400 <o> d ) rs . num__4000 <o> e ) none of these |
let the amount with r be rs . r r = num__0.666666666667 ( total amount with p and q ) r = num__0.666666666667 ( num__6000 - r ) = > num__3 r = num__12000 - num__2 r = > num__5 r = num__12000 = > r = num__2400 . answer : c <eor> c <eos> |
c |
divide__12000.0__6000.0__ add__2.0__3.0__ divide__12000.0__5.0__ divide__12000.0__5.0__ |
divide__12000.0__6000.0__ add__2.0__3.0__ divide__12000.0__5.0__ divide__12000.0__5.0__ |
| varma can read a book in k minutes . what part of the book can he read in num__4 minutes ? ( k > num__8 ) <o> a ) ( k - num__8 ) / k <o> b ) k / num__8 <o> c ) num__8 + k <o> d ) k - num__8 <o> e ) num__4 / k |
option e explanation : part of the book he can read in num__1 minute = num__1 / k part of the book he can read in num__8 minutes = num__4 / k . <eor> e <eos> |
e |
round__4.0__ |
divide__4.0__1.0__ |
| in an election only two candidates contested . a candidate secured num__70.0 of the valid votes and won by a majority of num__172 votes . find the total number of valid votes ? <o> a ) num__430 <o> b ) num__288 <o> c ) num__761 <o> d ) num__122 <o> e ) num__234 |
let the total number of valid votes be x . num__70.0 of x = num__0.7 * x = num__7 x / num__10 number of votes secured by the other candidate = x - num__7 x / num__100 = num__3 x / num__10 given num__7 x / num__10 - num__3 x / num__10 = num__172 = > num__4 x / num__10 = num__172 = > num__4 x = num__1720 = > x = num__430 . answer : a <eor> a <eos> |
a |
percent__100.0__430.0__ |
percent__100.0__430.0__ |
| a boat can travel num__1.5 times the distance down the stream than up the stream in the same time . if the speed of the current is num__4 kmph find the speed of the boat in still water <o> a ) num__5 <o> b ) num__10 <o> c ) num__15 <o> d ) num__20 <o> e ) num__25 |
simple question messy language . lets say boat ' s speed is b and current ' s speed is c . upstream speed : b - c downstream speed : b + c distance will be the same both times . lets say d . given : time it takes upstream = num__1.5 times time it takes downstream d / ( b + c ) = num__1.5 * [ d / ( b - c ) ] solving b = num__5 c given : c is num__4 kmph so b is num__20 kmph d <eor> d <eos> |
d |
multiply__4.0__5.0__ round__20.0__ |
multiply__4.0__5.0__ multiply__4.0__5.0__ |
| set a consists of all the prime numbers between num__15 and num__36 . what is the range of set a ? <o> a ) num__12 <o> b ) num__13 <o> c ) num__14 <o> d ) num__17 <o> e ) num__23 |
the range of a set of data is the difference between the highest and lowest values in the set in this set highest number = num__31 lowest number = num__17 range = highest - lowest = num__31 - num__17 = num__14 option c <eor> c <eos> |
c |
subtract__31.0__17.0__ subtract__31.0__17.0__ |
subtract__31.0__17.0__ subtract__31.0__17.0__ |
| peter has six times as many dimes as quarters in her piggy bank . she has num__21 coins in her piggy bank totaling $ num__2.55 how many of each type of coin does she have ? <o> a ) num__16 <o> b ) num__17 <o> c ) num__18 <o> d ) num__19 <o> e ) num__20 |
solution let x be the number of quarters . let num__6 x be the number of dimes since one quarter equals num__25 cents x quarters equals x × num__25 cents or num__25 x cents since one dime equals num__10 cents num__6 x dimes equals num__6 x × num__10 cents or num__60 x cents since one num__1 dollar equals num__100 cents num__2.55 dollars equals num__2.55 × num__100 = num__255 cents putting it all together num__25 x cents + num__60 x cents = num__255 cents num__85 x cents = num__255 cents num__85 x cents / num__85 cents = num__255 cents / num__85 cents x = num__3 num__6 x = num__6 × num__3 = num__18 therefore peter has num__3 quarters and num__18 dimes answer c <eor> c <eos> |
c |
multiply__6.0__10.0__ multiply__2.55__100.0__ add__25.0__60.0__ divide__255.0__85.0__ subtract__21.0__3.0__ subtract__21.0__3.0__ |
multiply__6.0__10.0__ multiply__2.55__100.0__ add__25.0__60.0__ divide__255.0__85.0__ subtract__21.0__3.0__ divide__18.0__1.0__ |
| for what value of x is | x – num__5 | + | x + num__5 | + | x | = num__27 ? <o> a ) num__0 <o> b ) num__3 <o> c ) - num__3 <o> d ) num__9 <o> e ) - num__9 |
for what value of x is | x – num__5 | + | x + num__5 | + | x | = num__27 ? it ' s easiest just to plug in answer choices : ( d ) : num__9 | x – num__5 | + | x + num__5 | + | x | = num__27 ? | num__9 - num__5 | + | num__9 + num__5 | + | num__9 | = num__27 ? | num__4 | + | num__14 | + | num__9 | = num__27 ( d ) <eor> d <eos> |
d |
subtract__9.0__5.0__ add__5.0__9.0__ add__5.0__4.0__ |
subtract__9.0__5.0__ add__5.0__9.0__ add__5.0__4.0__ |
| if a - b = num__4 and a num__2 + b num__2 = num__30 find the value of ab . <o> a ) a ) num__7 <o> b ) b ) num__12 <o> c ) c ) num__15 <o> d ) d ) num__18 <o> e ) e ) num__20 |
explanation : num__2 ab = ( a num__2 + b num__2 ) - ( a - b ) num__2 = num__30 - num__16 = num__14 ab = num__7 answer : a <eor> a <eos> |
a |
subtract__30.0__16.0__ divide__14.0__2.0__ divide__14.0__2.0__ |
subtract__30.0__16.0__ divide__14.0__2.0__ subtract__14.0__7.0__ |
| if n is an even integer then which of the following must be an odd integer ? <o> a ) num__2 n <o> b ) num__3 n <o> c ) num__3 n + n ^ num__2 + num__2 <o> d ) n ^ num__2 + num__1 <o> e ) num__2 n + n ^ num__2 |
answer : d a : num__2 n = even b : num__3 n = even c : num__3 n + n ^ num__2 + num__2 = even d : n ^ num__2 + num__1 = odd e : num__2 n + n ^ num__2 = even answer : d <eor> d <eos> |
d |
subtract__3.0__2.0__ multiply__1.0__2.0__ |
subtract__3.0__2.0__ power__2.0__1.0__ |
| three years ago the average age of a family of num__4 members was num__19 years . a boy have been born the average age of the family is the same today . what is the age of the boy ? <o> a ) a ) num__7 <o> b ) b ) num__6 <o> c ) c ) num__5 <o> d ) d ) num__4 <o> e ) e ) num__3 |
num__4 * num__22 = num__88 num__5 * num__19 = num__95 - - - - - - - - - - - - - - num__7 answer : a <eor> a <eos> |
a |
multiply__4.0__22.0__ multiply__19.0__5.0__ subtract__95.0__88.0__ subtract__95.0__88.0__ |
multiply__4.0__22.0__ multiply__19.0__5.0__ subtract__95.0__88.0__ subtract__95.0__88.0__ |
| a car was driving at num__50 km / h for num__30 minutes and then at num__90 km / h for another num__40 minutes . what was its average speed ? <o> a ) num__102 <o> b ) num__80 <o> c ) num__75 <o> d ) num__70 <o> e ) num__65 |
driving at num__50 km / h for num__30 minutes distance covered = num__50 * num__0.5 = num__25 km driving at num__90 km / h for num__40 minutes distance covered = num__90 * num__0.666666666667 = num__60 km average speed = total distance / total time = num__17.0 / num__6 = num__102 km / h answer : a <eor> a <eos> |
a |
multiply__50.0__0.5__ divide__30.0__0.5__ multiply__6.0__17.0__ multiply__6.0__17.0__ |
multiply__50.0__0.5__ divide__30.0__0.5__ multiply__6.0__17.0__ multiply__6.0__17.0__ |
| in what ratio must a grocer mix two varieties of pulses costing rs . num__15 and rs . num__20 per kg respectively so as to get a mixture worth rs . num__16.50 kg ? <o> a ) num__3 : num__7 <o> b ) num__5 : num__7 <o> c ) num__7 : num__3 <o> d ) num__7 : num__5 <o> e ) num__7 : num__4 |
explanation : by the rule of alligation : cost of num__1 kg pulses of num__1 st kindcost of num__1 kg pulses of num__2 nd kind required rate = num__3.50 : num__1.50 = num__7 : num__3 . answer is c <eor> c <eos> |
c |
subtract__20.0__16.5__ subtract__16.5__15.0__ multiply__3.5__2.0__ round_down__3.5__ multiply__3.5__2.0__ |
subtract__20.0__16.5__ subtract__16.5__15.0__ multiply__3.5__2.0__ round_down__3.5__ multiply__3.5__2.0__ |
| solve the equation for x : num__6 x - num__27 + num__3 x = num__4 + num__9 - x <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__1 <o> e ) num__7 |
explanation : num__9 x + x = num__13 + num__27 num__10 x = num__40 = > x = num__4 answer : a <eor> a <eos> |
a |
add__4.0__9.0__ add__6.0__4.0__ add__27.0__13.0__ divide__40.0__10.0__ |
add__4.0__9.0__ add__6.0__4.0__ add__27.0__13.0__ divide__40.0__10.0__ |
| num__365 : num__90 : : num__623 : ? <o> a ) num__36 <o> b ) num__45 <o> c ) num__123 <o> d ) num__63 <o> e ) none of these |
num__365 - - - num__3 * num__6 * num__5 . . . . num__90 num__623 - - - num__6 * num__2 * num__3 . . . . num__36 answer a <eor> a <eos> |
a |
subtract__5.0__3.0__ power__6.0__2.0__ power__6.0__2.0__ |
subtract__5.0__3.0__ power__6.0__2.0__ power__6.0__2.0__ |
| each person in room a is a student and num__0.166666666667 of the students in room a are seniors . each person in room b is a student and num__0.714285714286 of the students in room b are seniors . if num__1 student is chosen at random from room a and num__1 student is chosen at random from room b what is the probability that exactly num__1 of the students chosen is a senior ? <o> a ) num__0.119047619048 <o> b ) num__0.440476190476 <o> c ) num__0.642857142857 <o> d ) num__0.761904761905 <o> e ) num__0.880952380952 |
probability of choosing senior from room a = num__0.166666666667 non senior = num__0.833333333333 probability of choosing senior from room b = num__0.714285714286 non senior = num__0.285714285714 reqd probability = ( num__0.166666666667 * num__0.285714285714 ) + ( num__0.714285714286 * num__0.833333333333 ) = num__0.642857142857 = num__0.642857142857 answer c <eor> c <eos> |
c |
subtract__1.0__0.1667__ subtract__1.0__0.7143__ multiply__1.0__0.6429__ |
subtract__1.0__0.1667__ subtract__1.0__0.7143__ multiply__1.0__0.6429__ |
| if num__0.125 of a pencil is black ½ of the remaining is white and the remaining num__3 ½ is blue find the total length of the pencil ? <o> a ) num__8 <o> b ) num__5 <o> c ) num__7 <o> d ) num__6 <o> e ) num__4 |
let the total length be xm then black part = x / num__8 cm the remaining part = ( x - x / num__8 ) cm = num__7 x / num__8 cm white part = ( num__0.5 * num__7 x / num__8 ) = num__7 x / num__16 cm remaining part = ( num__7 x / num__8 - num__7 x / num__16 ) = num__7 x / num__16 cm num__7 x / num__16 = num__3.5 x = num__8 cm answer is a . <eor> a <eos> |
a |
reverse__0.125__ divide__8.0__0.5__ add__3.0__0.5__ reverse__0.125__ |
reverse__0.125__ divide__8.0__0.5__ multiply__0.5__7.0__ reverse__0.125__ |
| in a class there are num__20 boys whose average age is decreased by num__2 months when one boy aged num__16 years replaced by a new boy . the age of the new boy is ? <o> a ) num__12 years num__8 months <o> b ) num__15 years <o> c ) num__16 years num__4 months <o> d ) num__17 years num__10 months <o> e ) num__17 years |
total decrease = ( num__20 x num__2 ) months = num__3 years num__4 months age of the new boy = num__16 years - num__3 years num__4 months . = num__12 years num__8 months . answer : a <eor> a <eos> |
a |
subtract__20.0__16.0__ subtract__16.0__4.0__ subtract__20.0__12.0__ subtract__20.0__8.0__ |
subtract__20.0__16.0__ subtract__16.0__4.0__ subtract__20.0__12.0__ subtract__20.0__8.0__ |
| sally bought a dozen items at num__120.00 per dozen . what will sally have to sell each item for to generate a num__15.0 profit ? <o> a ) num__10.15 <o> b ) num__11.5 <o> c ) num__15.0 <o> d ) num__21.5 <o> e ) num__150.0 |
c . p . is num__10.0 = num__10 desired profit = num__15.0 the calculation for markup is ( c . p . * desired profit / num__100 ) = num__10 * num__0.15 = num__1.5 the calculation for final sale price = c . p . + markup = num__10 + num__1.5 = num__11.5 answer : b <eor> b <eos> |
b |
percent__15.0__10.0__ percent__11.5__100.0__ |
percent__15.0__10.0__ percent__11.5__100.0__ |
| in an examination a pupil ' s average marks were num__63 per paper . if he had obtained num__22 more marks for his geography paper and num__2 more marks for his history paper his average per paper would have been num__65 . how many papers were there in the examination ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
let the number of papers be x . then num__63 x + num__22 + num__2 = num__65 x = num__2 x = num__24 = x = num__12 . answer : e <eor> e <eos> |
e |
add__22.0__2.0__ divide__24.0__2.0__ divide__24.0__2.0__ |
add__22.0__2.0__ divide__24.0__2.0__ divide__24.0__2.0__ |
| log num__3 n + log num__15 n what is num__3 digit number n that will be whole number <o> a ) num__3375 <o> b ) num__7292 <o> c ) num__8291 <o> d ) num__3929 <o> e ) num__2727 |
no of values n can take is num__1 num__15 ^ num__3 = num__3375 answer : a <eor> a <eos> |
a |
multiply__1.0__3375.0__ |
multiply__1.0__3375.0__ |
| the largest num__4 digit number exactly divisible by num__88 is ? <o> a ) a ) num__9944 <o> b ) b ) num__9954 <o> c ) c ) num__9967 <o> d ) d ) num__9969 <o> e ) e ) num__9970 |
largest num__4 - digit number = num__9999 num__88 ) num__9999 ( num__113 num__88 - - - - num__119 num__88 - - - - num__319 num__264 - - - num__55 - - - required number = ( num__9999 - num__55 ) = num__9944 . a ) <eor> a <eos> |
a |
subtract__319.0__264.0__ multiply__88.0__113.0__ multiply__88.0__113.0__ |
subtract__319.0__264.0__ subtract__9999.0__55.0__ subtract__9999.0__55.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 seconds . what is the length of the train ? <o> a ) num__277 m <o> b ) num__268 m <o> c ) num__276 m <o> d ) num__150 m <o> e ) num__125 m |
speed = ( num__60 * num__0.277777777778 ) m / sec = ( num__16.6666666667 ) m / sec length of the train = ( speed x time ) = ( num__16.6666666667 * num__9 ) m = num__150 m . answer : d <eor> d <eos> |
d |
round__150.0__ |
round__150.0__ |
| find num__10 th term in the series num__1 num__3 num__5 num__7 . . . <o> a ) num__19 <o> b ) num__20 <o> c ) num__21 <o> d ) num__22 <o> e ) num__23 |
solution : a = num__1 d = num__3 – num__1 = num__2 num__10 th term t num__10 = a + ( n - num__1 ) d = num__1 + ( num__10 – num__1 ) num__2 = num__1 + num__18 = num__19 answer is a <eor> a <eos> |
a |
divide__10.0__5.0__ add__1.0__18.0__ add__1.0__18.0__ |
subtract__3.0__1.0__ add__1.0__18.0__ add__1.0__18.0__ |
| the dimensions of a room are num__25 feet * num__15 feet * num__12 feet . what is the cost of white washing the four walls of the room at rs . num__3 per square feet if there is one door of dimensions num__6 feet * num__3 feet and three windows of dimensions num__4 feet * num__3 feet each ? <o> a ) s . num__2718 <o> b ) s . num__4586 <o> c ) s . num__4597 <o> d ) s . num__4530 <o> e ) s . num__4528 |
area of the four walls = num__2 h ( l + b ) since there are doors and windows area of the walls = num__2 * num__12 ( num__15 + num__25 ) - ( num__6 * num__3 ) - num__3 ( num__4 * num__3 ) = num__906 sq . ft . total cost = num__906 * num__3 = rs . num__2718 answer : a <eor> a <eos> |
a |
multiply__3.0__906.0__ multiply__3.0__906.0__ |
multiply__3.0__906.0__ multiply__3.0__906.0__ |
| a sum was put at simple interest at a certain rate for num__3 years had it been put at num__3.0 higher rate it would have fetched num__81 more . find the sum . <o> a ) num__500 <o> b ) num__600 <o> c ) num__700 <o> d ) num__800 <o> e ) num__900 |
difference in s . i . = p × t / num__100 ( r num__1 − r num__2 ) ⇒ num__81 = p × num__3 x num__0.03 ( ∵ r num__1 - r num__2 = num__2 ) ⇒ p = num__81 × num__33.3333333333 x num__3 = num__900 answer e <eor> e <eos> |
e |
percent__3.0__1.0__ percent__100.0__900.0__ |
percent__3.0__1.0__ percent__100.0__900.0__ |
| a car travels the first num__0.333333333333 of certain distance with a speed of num__12 km / hr the next num__0.333333333333 distance with a speed of num__20 km / hr and the last num__0.333333333333 distance with a speed of num__60 km / hr . the average speed of the car for the whole journey is ? <o> a ) num__12 km / hr <o> b ) num__20 km / hr <o> c ) num__89 km / hr <o> d ) num__52 km / hr <o> e ) num__79 km / hr |
let the whole distance travelled be x km and the average speed of the car for he whole journey be y km / hr then ( x / num__3 ) / num__12 + ( x / num__3 ) / num__20 + ( x / num__3 ) / num__60 = x / y x / num__36 + x / num__60 + x / num__180 = x / y num__0.05 y = num__1 y = num__20 km / hr answer ( b ) <eor> b <eos> |
b |
divide__60.0__20.0__ multiply__12.0__3.0__ multiply__60.0__3.0__ reverse__20.0__ multiply__20.0__0.05__ reverse__0.05__ |
divide__60.0__20.0__ multiply__12.0__3.0__ multiply__60.0__3.0__ reverse__20.0__ multiply__20.0__0.05__ reverse__0.05__ |
| if a sum of money doubles itself in num__8 years at simple interest the ratepercent per annum is <o> a ) num__12 <o> b ) num__12.5 <o> c ) num__13 <o> d ) num__13.5 <o> e ) num__14 |
explanation : let sum = x then simple interest = x rate = ( num__100 * x ) / ( x * num__8 ) = num__12.5 option b <eor> b <eos> |
b |
percent__12.5__100.0__ |
percent__12.5__100.0__ |
| jeff has num__252 ounces of peanut butter in num__16 num__28 . and num__40 ounce jars . he has an equal number of each sized jar . how many jars of peanut butter does jeff have ? <o> a ) num__9 <o> b ) num__8 <o> c ) num__7 <o> d ) num__10 <o> e ) num__11 |
let p equal the number of each sized jar then num__16 p + num__28 p + num__40 p = num__252 num__84 p = num__252 p = num__3 therefore the total number of jars of peanut butter jeff has = num__3 p = num__9 answer : a <eor> a <eos> |
a |
divide__252.0__84.0__ divide__252.0__28.0__ divide__252.0__28.0__ |
divide__252.0__84.0__ divide__252.0__28.0__ divide__252.0__28.0__ |
| bricks are sold in packages of eight or thirteen only . if timmy a builder bought num__144 bricks exactly what could be the number of large packs timmy bought ? <o> a ) num__12 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__7 |
let number of packs of eight = e let number of packs of thirteen = t num__8 e + num__13 t = num__144 now we need to test for values of t . since sum num__144 is even and num__8 s will always be even t ca n ' t be odd . now we can test for values e = num__12 num__8 and num__6 num__8 * num__5 + num__13 * num__8 = num__40 + num__104 = num__144 answer d <eor> d <eos> |
d |
subtract__13.0__8.0__ multiply__5.0__8.0__ subtract__144.0__40.0__ subtract__13.0__5.0__ |
subtract__13.0__8.0__ multiply__5.0__8.0__ multiply__8.0__13.0__ subtract__13.0__5.0__ |
| lagaan is levied on the num__60 percent of the cultivated land . the revenue department collected total rs . num__3 num__84000 through the lagaan from the village of mettur . mettur a very rich farmer paid only rs . num__480 as lagaan . the percentage of total land of mettur over the total taxable land of the village is : <o> a ) num__0.20833 <o> b ) num__1.23455 <o> c ) num__2.45466 <o> d ) num__5.54353 <o> e ) num__6.45343 |
total land of sukhiya = \ inline \ frac { num__480 x } { num__0.6 } = num__800 x \ therefore cultivated land of village = num__384000 x \ therefore required percentage = \ inline \ frac { num__800 x } { num__384000 } \ times num__100 = num__0.20833 a <eor> a <eos> |
a |
divide__480.0__0.6__ multiply__480.0__800.0__ divide__60.0__0.6__ divide__100.0__480.0__ divide__100.0__480.0__ |
divide__480.0__0.6__ multiply__480.0__800.0__ divide__60.0__0.6__ divide__100.0__480.0__ divide__100.0__480.0__ |
| two circular frames are kept one above the other . frame x has a diameter of num__16 cm and frame y has a diameter of num__12 cm . what fraction of the surface of frame x is not covered by frame y ? <o> a ) num__0.6 <o> b ) num__0.4375 <o> c ) num__0.75 <o> d ) num__2.0 <o> e ) num__0.666666666667 |
the questions asks us to find the surface which is not covered by the frame y i . e . area of the surface not covered . where as circumference is the length along the edge of the circle num__2 * pi * r implies the length of the curve pi * r ^ num__2 implies area enclosed by that curve . . hence area of the circle is considered for this problem area of the frame y = pi * r ^ num__2 where r = num__6 = > pi * num__36 area of the frame x = pi * num__64 surface not covered by the frame y = pi * num__64 - pi * num__36 = pi * num__28 fraction of frame x ' s surface is not covered by a frame y = pi * num__28 / total area of the frame x = num__0.4375 = num__0.4375 answer is b <eor> b <eos> |
b |
rectangle_perimeter__16.0__2.0__ square_perimeter__16.0__ rectangle_perimeter__12.0__2.0__ triangle_area__0.4375__2.0__ |
power__6.0__2.0__ power__2.0__6.0__ rectangle_perimeter__12.0__2.0__ triangle_area__0.4375__2.0__ |
| the product of two numbers is num__192 and the sum of these two numbers is num__28 . what is the smaller of these two numbers ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
solution let the number be x and ( num__28 - x ) = then x ( num__28 - x ) = num__192 ‹ = › x num__2 - num__28 x + num__192 = num__0 . ‹ = › ( x - num__16 ) ( x - num__12 ) = num__0 ‹ = › x = num__16 or x = num__12 . answer b <eor> b <eos> |
b |
divide__192.0__16.0__ divide__192.0__16.0__ |
subtract__28.0__16.0__ subtract__28.0__16.0__ |
| my grandson is about as many days as my son in weeks and my grandson is as many months as i am in years . my grandson my son and i together are num__120 years . can you tell me my age in years ? <o> a ) num__70 <o> b ) num__69 <o> c ) num__82 <o> d ) num__89 <o> e ) num__72 |
let m be my age in years . if s is my son ' s age in years then my son is num__52 s weeks old . if g is my grandson ' s age in years then my grandson is num__365 g days old . thus num__365 g = num__52 s . since my grandson is num__12 g months old num__12 g = m . since my grandson my son and i together are num__120 years g + s + m = num__120 . the above system of num__3 equations in num__3 unknowns ( g s and m ) can be solved as follows . m / num__12 + num__365 m / ( num__52 x num__12 ) + m = num__120 or num__52 m + num__365 m + num__624 m = num__624 x num__120 or m = num__624 x num__0.115273775216 = num__72 . <eor> e <eos> |
e |
multiply__12.0__52.0__ round__72.0__ |
multiply__12.0__52.0__ round__72.0__ |
| find the unit digit in ( num__544 ) num__102 + ( num__544 ) num__103 <o> a ) num__2 <o> b ) num__4 <o> c ) num__0 <o> d ) num__1 <o> e ) num__3 |
explanation : required digit is = ( num__4 ) num__102 + ( num__4 ) num__103 as ( num__4 ) num__2 gives unit digit num__6 so ( num__4 ) num__102 unit digit is num__6 and ( num__4 ) num__103 unit digit is unit digit of num__6 × num__4 = num__4 so answer will be unit digit of num__6 + num__4 = num__0 option c <eor> c <eos> |
c |
add__2.0__4.0__ multiply__544.0__0.0__ |
add__2.0__4.0__ multiply__544.0__0.0__ |
| the value of x + x ( xx ) when x = num__8 <o> a ) a ) num__520 <o> b ) b ) num__516 <o> c ) c ) num__518 <o> d ) d ) num__536 <o> e ) e ) num__564 |
x + x ( xx ) put the value of x = num__8 in the above expression we get num__8 + num__8 ( num__88 ) = num__8 + num__8 ( num__8 Ã — num__8 ) = num__8 + num__8 ( num__64 ) = num__9 + num__512 = num__520 the answer is ( a ) <eor> a <eos> |
a |
multiply__8.0__64.0__ add__8.0__512.0__ add__8.0__512.0__ |
multiply__8.0__64.0__ add__8.0__512.0__ add__8.0__512.0__ |
| sally has to pick her son up at num__2 p . m . it is now num__12 : num__30 p . m . she has to travel num__53 miles to get her son from camp . what time should she leave to pick her on up with the closest time to num__2 p . m without being late given the distance of num__53 miles with a constant speed of num__50 mph ? <o> a ) she needs to leave at num__12 : num__53 pm <o> b ) she needs to leave at num__12 : num__57 p . m <o> c ) she needs to leave at num__1 : num__03 p . m <o> d ) she can not get there in time . <o> e ) she needs to leave at num__1 : num__15 p . m |
b she is traveling num__50 miles per hour and needs to travel num__53 miles . given that num__0.943396226415 = num__1.03 so approximately num__1 hour and num__3 minutes . num__2 p . m - num__1 hour num__3 minutes = num__12 : num__57 pm . while a gets her there on time it is not as close to num__2 pm . <eor> a <eos> |
a |
divide__50.0__53.0__ add__2.0__1.0__ round__12.0__ |
divide__50.0__53.0__ subtract__53.0__50.0__ round__12.0__ |
| if a and b are digits and num__5 ab is a num__3 - digit number that is divisible by num__10 which of the following is a possible product of a and c ? <o> a ) num__57 <o> b ) num__21 <o> c ) num__52 <o> d ) num__38 <o> e ) num__50 |
key to this question is to remember the fact that a number divisible by num__10 must end with only num__0 ( i . e b ) . if b had to be num__0 product should also be num__0 regardless of a . only one answer choice meets the requirement . ans e . <eor> e <eos> |
e |
multiply__5.0__10.0__ |
multiply__5.0__10.0__ |
| a pipe can fill a cistern in num__20 minutes whereas the cistern when fill can be emptied by a leak in num__28 minutes . when both pipes are opened find when the cistern will be full ? <o> a ) num__65 minutes <o> b ) num__76 minutes <o> c ) num__70 minutes <o> d ) num__62 minutes <o> e ) num__52 minutes |
num__0.05 - num__0.0357142857143 = num__0.0142857142857 num__70 minutes answer : c <eor> c <eos> |
c |
subtract__0.05__0.0357__ round__70.0__ |
subtract__0.05__0.0357__ round__70.0__ |
| if the tens digit x and the units digit y of a positive integer n are reversed the resulting integer is num__27 less than n . what is y in terms of x ? <o> a ) num__10 - x <o> b ) num__9 - x <o> c ) x + num__3 <o> d ) x - num__3 <o> e ) x - num__9 |
original digits = xy i . e . number = num__10 x + y after reversing the digits : digits = yx i . e . number = num__10 y + x num__10 y + x is num__27 less than num__10 x + y num__10 x + y - num__27 = num__10 y + x num__10 x - x - num__27 = num__10 y - y num__9 x - num__27 = num__9 y x - num__3 = y or y = x - num__3 ans : d <eor> d <eos> |
d |
divide__27.0__9.0__ divide__27.0__9.0__ |
divide__27.0__9.0__ divide__27.0__9.0__ |
| a polling company surveyed a certain country and it found that num__35.0 of that country ’ s registered voters had an unfavorable impression of both of that state ’ s major political parties and that num__20.0 had a favorable impression only of party q . if one registered voter has a favorable impression of both parties for every two registered voters who have a favorable impression only of party b then what percentage of the country ’ s registered voters have a favorable impression of both parties ( assuming that respondents to the poll were given a choice between favorable and unfavorable impressions only ) ? <o> a ) num__15 <o> b ) num__20 <o> c ) num__30 <o> d ) num__35 <o> e ) num__45 |
s = num__100 not ( q and b ) = num__35 only q = num__20 ( q and b ) / b = num__0.5 let ( q and b ) = x only b = num__2 x so now num__20 + num__35 + x + num__2 x = num__100 x = num__15 a ans <eor> a <eos> |
a |
percent__100.0__15.0__ |
percent__100.0__15.0__ |
| num__2 + num__2 ^ num__2 + num__2 ^ num__3 + . . . + num__2 ^ num__8 = ? <o> a ) num__510 <o> b ) num__310 <o> c ) num__210 <o> d ) num__410 <o> e ) none of them |
given series is a g . p . with a = num__2 r = num__2 and n = num__8 . sum = a ( r ^ n - num__1 ) / ( r - num__1 ) = num__2 x ( num__2 ^ num__8 – num__1 ) / ( num__2 - num__1 ) = ( num__2 x num__255 ) = num__510 answer is a . <eor> a <eos> |
a |
subtract__3.0__2.0__ multiply__2.0__255.0__ multiply__2.0__255.0__ |
subtract__3.0__2.0__ multiply__2.0__255.0__ divide__510.0__1.0__ |
| a textile manufacturing firm employees num__50 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the num__50 looms is rs num__5 num__00000 and the monthly manufacturing expenses is rs num__1 num__50000 . assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over the number of looms . monthly establishment charges are rs num__75000 . if one loom breaks down and remains idle for one month the decrease in profit is : <o> a ) num__13000 <o> b ) num__7000 <o> c ) num__10000 <o> d ) num__5000 <o> e ) none of these |
explanation : profit = num__5 num__00000 − ( num__1 num__50000 + num__75000 ) = rs . num__2 num__75000 . since such loom contributes equally to sales and manufacturing expenses . but the monthly charges are fixed at rs num__75000 . if one loan breaks down sales and expenses will decrease . new profit : - = > num__500000 × ( num__0.98 ) − num__150000 × ( num__0.98 ) − num__75000 . = > rs num__2 num__68000 . decrease in profit = > num__2 num__75000 − num__2 num__68000 = > rs . num__7000 . answer : b <eor> b <eos> |
b |
multiply__75000.0__2.0__ subtract__75000.0__68000.0__ multiply__1.0__7000.0__ |
multiply__75000.0__2.0__ subtract__75000.0__68000.0__ subtract__75000.0__68000.0__ |
| a no . when divided by the sum of num__555 and num__445 gives num__2 times their difference as quotient & num__70 as remainder . find the no . is ? <o> a ) num__145646 <o> b ) num__236578 <o> c ) num__645353 <o> d ) num__456546 <o> e ) num__220070 |
( num__555 + num__445 ) * num__2 * num__110 + num__70 = num__220000 + num__70 = num__220070 e <eor> e <eos> |
e |
subtract__555.0__445.0__ add__70.0__220000.0__ add__70.0__220000.0__ |
subtract__555.0__445.0__ add__70.0__220000.0__ add__70.0__220000.0__ |
| a can do a job in num__10 days and b can do it in num__30 days . a and b working together will finish twice the amount of work in - - - - - - - days ? <o> a ) num__14 days <o> b ) num__15 days <o> c ) num__22 days <o> d ) num__11 days <o> e ) num__19 days |
b num__0.1 + num__0.0333333333333 = num__0.133333333333 num__7.5 * num__2 = num__15 days <eor> b <eos> |
b |
add__0.1__0.0333__ divide__30.0__2.0__ round__15.0__ |
add__0.1__0.0333__ multiply__7.5__2.0__ multiply__7.5__2.0__ |
| a boat having a length num__3 m and breadth num__2 m is floating on a lake . the boat sinks by num__1 cm when a man gets on it . the mass of the man is : <o> a ) num__12 kg <o> b ) num__60 kg <o> c ) num__72 kg <o> d ) num__96 kg <o> e ) none of these |
explanation : volume of water displaced = ( num__3 x num__2 x num__0.01 ) m num__3 = num__0.06 m num__3 . ∴ mass of man = volume of water displaced x density of water = ( num__0.06 x num__1000 ) kg = num__60 kg . answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ hour_to_min_conversion__ |
hour_to_min_conversion__ hour_to_min_conversion__ |
| a man took loan from a bank at the rate of num__10.0 p . a . simple interest . after num__3 years he had to pay rs . num__3000 interest only for the period . the principal amount borrowed by him was : <o> a ) num__8000 <o> b ) num__10000 <o> c ) num__11000 <o> d ) num__12000 <o> e ) num__13000 |
principal = ( num__100 x num__3000 ) / ( num__10 x num__3 ) = rs . num__10000 answer : b <eor> b <eos> |
b |
percent__100.0__10000.0__ |
percent__100.0__10000.0__ |
| the ratio between the speeds of two trains is num__8 : num__9 . if the second train runs num__270 kms in num__3 hours then the speed of the first train is ? <o> a ) num__85 km / hr <o> b ) num__835 km / hr <o> c ) num__80 km / hr <o> d ) num__82 km / hr <o> e ) none of these |
explanation : let the speeds of two trains be num__8 x and num__9 x km / hr . num__9 / x = num__90.0 = > x = num__10 km / hr so speed of first train is num__10 * num__8 = num__80 km / hr option c <eor> c <eos> |
c |
divide__270.0__3.0__ divide__90.0__9.0__ multiply__8.0__10.0__ round__80.0__ |
divide__270.0__3.0__ divide__90.0__9.0__ multiply__8.0__10.0__ multiply__8.0__10.0__ |
| in an election only two candidates contested . a candidate secured num__70.0 of the valid votes and won by a majority of num__174 votes . find the total number of valid votes ? <o> a ) num__435 <o> b ) num__570 <o> c ) num__480 <o> d ) num__520 <o> e ) num__550 |
let the total number of valid votes be x . num__70.0 of x = num__0.7 * x = num__7 x / num__10 number of votes secured by the other candidate = x - num__7 x / num__100 = num__3 x / num__10 given num__7 x / num__10 - num__3 x / num__10 = num__174 = > num__4 x / num__10 = num__174 = > num__4 x = num__1740 = > x = num__435 . answer : a <eor> a <eos> |
a |
percent__100.0__435.0__ |
percent__100.0__435.0__ |
| a goods train runs at the speed of num__72 kmph and crosses a num__250 m long platform in num__26 seconds . what is the length of the goods train ? <o> a ) num__230 m <o> b ) num__270 m <o> c ) num__643 m <o> d ) num__832 m <o> e ) num__270 m |
speed = ( num__72 x num__0.277777777778 ) m / sec = num__20 m / sec . time = num__26 sec . let the length of the train be x metres . then x + num__9.61538461538 = num__20 x + num__250 = num__520 x = num__270 . answer : b <eor> b <eos> |
b |
divide__250.0__26.0__ multiply__26.0__20.0__ add__250.0__20.0__ round__270.0__ |
divide__250.0__26.0__ multiply__26.0__20.0__ add__250.0__20.0__ add__250.0__20.0__ |
| complete the series num__95 num__115.5 num__138 . . . num__189 <o> a ) num__189 <o> b ) num__198 <o> c ) num__178 <o> d ) num__187 <o> e ) num__193 |
num__95 + num__20.5 = num__115.5 num__115.5 + num__22.5 = num__138 num__138 + num__24.5 = num__162.5 num__162.5 + num__26.5 = num__189 answer is a . <eor> a <eos> |
a |
subtract__115.5__95.0__ subtract__138.0__115.5__ add__138.0__24.5__ subtract__189.0__162.5__ add__162.5__26.5__ |
subtract__115.5__95.0__ subtract__138.0__115.5__ add__138.0__24.5__ subtract__189.0__162.5__ add__162.5__26.5__ |
| evaluate the sum num__1 - num__2 + num__3 - num__4 + num__5 - num__6 + . . . + num__997 - num__998 + num__999 - num__1000 <o> a ) - num__500 <o> b ) - num__1000 <o> c ) - num__999 <o> d ) - num__1001 <o> e ) num__500500 |
one groups the first two terms and each successive two terms to obtain a sum of num__500 expressions each of which is - num__1 . the answer is - num__500 correct answer a <eor> a <eos> |
a |
divide__1000.0__2.0__ multiply__1.0__500.0__ |
divide__1000.0__2.0__ subtract__1000.0__500.0__ |
| a train passes a man standing on the platform . if the train is num__180 meters long and its speed is num__72 kmph how much time it took in doing so ? <o> a ) num__6 Â ½ sec <o> b ) num__6 Â ½ sec <o> c ) num__9 Â ½ sec <o> d ) num__9 sec <o> e ) num__7 Â ½ sec |
d num__9 sec d = num__180 s = num__72 * num__0.277777777778 = num__20 mps t = num__9.0 = num__9 sec <eor> d <eos> |
d |
divide__180.0__9.0__ round__9.0__ |
divide__180.0__9.0__ round__9.0__ |
| insert the missing number . num__8 num__7 num__11 num__12 num__14 num__17 num__17 num__22 num__20 ( . . . . ) <o> a ) num__27 <o> b ) num__20 <o> c ) num__22 <o> d ) num__24 <o> e ) num__26 |
there are two series ( num__8 num__11 num__14 num__17 num__20 ) and ( num__7 num__12 num__17 num__22 num__27 ) increasing by num__3 and num__5 respectively . answer : a <eor> a <eos> |
a |
add__7.0__20.0__ subtract__11.0__8.0__ subtract__8.0__3.0__ add__7.0__20.0__ |
add__7.0__20.0__ subtract__11.0__8.0__ subtract__8.0__3.0__ add__7.0__20.0__ |
| if num__2 / z = num__2 / ( z + num__1 ) + num__2 / ( z + num__25 ) which of these integers could be the value of z ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__5 |
solving for z algebraically in this problem would not be easy . instead we can follow the hint in the question ( “ which of these integers … ” ) and test each answer choice : a . num__2 / num__0 = num__2.0 + num__0.08 incorrect ( division by zero ) b . num__2.0 = num__1.0 + num__0.0769230769231 incorrect c . num__1.0 = num__0.666666666667 + num__0.0740740740741 incorrect d . num__0.666666666667 = num__0.5 + num__0.0714285714286 incorrect e . num__0.5 = num__0.4 + num__0.0666666666667 correct the correct answer is e because it contains the only value that makes the equation work . notice how quickly this strategy worked in this case <eor> e <eos> |
e |
divide__2.0__25.0__ reverse__2.0__ divide__2.0__0.4__ |
divide__2.0__25.0__ reverse__2.0__ divide__2.0__0.4__ |
| a and b together do a work in num__20 days . b and c together in num__15 days and c and a in num__12 days . so a b and c together finish same work in how many days ? <o> a ) num__12 <o> b ) num__15 <o> c ) num__8 <o> d ) num__10 <o> e ) num__11 |
( a + b ) work in num__1 day = num__0.05 ( b + c ) work in num__1 days = num__0.0666666666667 . ( c + a ) work in num__1 days = num__0.0833333333333 ( num__1 ) adding = num__2 [ a + b + c ] in num__1 day work = [ num__0.05 + num__0.0666666666667 + num__0.0833333333333 ] = num__0.2 ( a + b + c ) work in num__1 day = num__0.1 so all three together finish work in num__10 days answer d <eor> d <eos> |
d |
divide__1.0__20.0__ divide__1.0__15.0__ divide__1.0__12.0__ divide__2.0__20.0__ divide__20.0__2.0__ round__10.0__ |
divide__1.0__20.0__ divide__1.0__15.0__ divide__1.0__12.0__ divide__2.0__20.0__ divide__20.0__2.0__ round__10.0__ |
| if x is num__40 percent greater than num__88 then x = <o> a ) num__68 <o> b ) num__70.4 <o> c ) num__123.2 <o> d ) num__105.6 <o> e ) num__108 |
x = num__88 * num__1.4 = num__123.2 so the answer is c . <eor> c <eos> |
c |
multiply__88.0__1.4__ multiply__88.0__1.4__ |
multiply__88.0__1.4__ multiply__88.0__1.4__ |
| elvin ' s monthly telephone bill is the sum of the charge for the calls he made during the month and a fixed monthly charge for internet service . elvin ' s total telephone bill for january was $ num__50 and elvin ' s total telephone bill for february was num__76 $ . if elvin ' s charge for the calls he made in february was twice the charge for the calls he made in january what is elvin ' s fixed monthly charge for internet service ? <o> a ) $ num__5 <o> b ) $ num__10 <o> c ) $ num__14 <o> d ) $ num__24 <o> e ) $ num__28 |
bill = fixed charge + charge of calls made in jan bill = fixed charge ( let y ) + charge of calls made in jan ( let x ) = $ num__50 in feb bill = fixed charge ( let y ) + charge of calls made in feb ( then num__2 x ) = $ num__76 i . e . x + y = num__50 and num__2 x + y = num__76 take the difference if two equations i . e . ( num__2 x + y ) - ( x + y ) = num__76 - num__50 i . e . x = num__26 i . e . fixed monthly charge y = num__26 answer : d <eor> d <eos> |
d |
subtract__76.0__50.0__ subtract__50.0__26.0__ |
subtract__76.0__50.0__ subtract__50.0__26.0__ |
| the average of num__45 students in a class is num__18 years . the average age of num__21 students is num__16 . what is the average age of remaining num__14 students ? <o> a ) num__21 <o> b ) num__16 <o> c ) num__17 <o> d ) num__18 <o> e ) num__19 |
sum of the ages of num__14 students = ( num__18 * num__45 ) - ( num__18 * num__21 ) = num__810 - num__378 = num__432 required average = ( num__11.5652173913 ) = num__18 years . answer : d <eor> d <eos> |
d |
multiply__45.0__18.0__ multiply__18.0__21.0__ subtract__810.0__378.0__ divide__810.0__45.0__ |
multiply__45.0__18.0__ multiply__18.0__21.0__ subtract__810.0__378.0__ divide__810.0__45.0__ |
| a tank is filled by num__3 pipes a b c in num__4 hours . pipe c is twice as fast as b and b is twice as fast as a . how much will pipe a alone take to fill the tank ? <o> a ) num__25 hr <o> b ) num__35 hr <o> c ) num__40 hr <o> d ) num__20 hr <o> e ) num__28 hr |
suppose pipe a alone take x hours to fill the tank then pipe b and c will take x / num__2 and x / num__4 hours respectively to fill the tank . num__1 / x + num__2 / x + num__4 / x = num__0.25 num__7 / x = num__0.25 x = num__28 hours answer is e <eor> e <eos> |
e |
subtract__3.0__2.0__ divide__1.0__4.0__ add__3.0__4.0__ multiply__4.0__7.0__ round__28.0__ |
subtract__3.0__2.0__ divide__1.0__4.0__ add__3.0__4.0__ divide__7.0__0.25__ divide__7.0__0.25__ |
| two cyclist start on a circular track from a given point but in opposite direction with speeds of num__7 m / s and num__8 m / s . if the circumference of the circle is num__300 meters after what time will they meet at the starting point ? <o> a ) num__20 sec <o> b ) num__15 sec <o> c ) num__30 sec <o> d ) num__50 sec <o> e ) num__1 min |
they meet every num__42.8571428571 + num__8 = num__20 sec answer is a <eor> a <eos> |
a |
divide__300.0__7.0__ round__20.0__ |
divide__300.0__7.0__ round__20.0__ |
| a box has exactly num__100 balls and each ball is either red blue or white . if the box has num__15 more blue balls than white balls and thrice as many red balls as blue balls how many white balls does the box has ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
x = the number of red balls y = the number of blue balls z = the number of white balls from the first sentence we have equation # num__1 : x + y + z = num__100 . . . the box has num__15 more blue balls than white balls . . . equation # num__2 : y = num__15 + z . . . thrice as many red balls as blue balls . . . equation # num__3 : x = num__3 y solve equation # num__2 for z : z = y - num__15 now we can replace both x and z with y in equation # num__1 num__3 y + y + ( y - num__15 ) = num__100 num__5 y - num__15 = num__100 num__5 y = num__115 y = num__23 there are num__23 blue balls . this is num__15 more than the number of white balls so z = num__8 . that ' s the answer . just as a check x = num__69 and num__69 + num__23 + num__8 = num__100 . answer = num__8 ( a ) <eor> a <eos> |
a |
add__1.0__2.0__ divide__15.0__3.0__ add__100.0__15.0__ divide__115.0__5.0__ add__3.0__5.0__ multiply__3.0__23.0__ multiply__1.0__8.0__ |
add__1.0__2.0__ add__2.0__3.0__ add__100.0__15.0__ divide__115.0__5.0__ add__3.0__5.0__ multiply__3.0__23.0__ add__3.0__5.0__ |
| there are num__20 people in a room . if each person shakes hands with exactly num__3 other people what is the total number of handshakes ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__30 <o> d ) num__45 <o> e ) num__60 |
note that each handshake includes two people . the total number of handshakes is num__20 * num__1.5 = num__30 the answer is c . <eor> c <eos> |
c |
multiply__20.0__1.5__ multiply__20.0__1.5__ |
multiply__20.0__1.5__ multiply__20.0__1.5__ |
| the food in a camp lasts for num__25 men for num__40 days . if fifteen more men join how many days will the food last ? <o> a ) num__25 days <o> b ) num__30 days <o> c ) num__20 days <o> d ) num__16 days <o> e ) num__27 days |
one man can consume the same food in num__25 * num__40 = num__1000 days . num__15 more men join the total number of men = num__40 the number of days the food will last = num__25.0 = num__30 days . answer : a <eor> a <eos> |
a |
multiply__25.0__40.0__ subtract__40.0__25.0__ round__25.0__ |
multiply__25.0__40.0__ subtract__40.0__25.0__ round__25.0__ |
| in how many different number of ways num__4 boys and num__2 girls can sit on a bench ? <o> a ) num__720 <o> b ) num__730 <o> c ) num__740 <o> d ) num__750 <o> e ) num__800 |
npn = n ! num__6 p num__6 = num__6 × num__5 × num__4 × num__3 × num__2 × num__1 = num__720 a <eor> a <eos> |
a |
die_space__ vowel_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
die_space__ vowel_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| what is the sum of the integers from - num__60 to num__65 inclusive ? <o> a ) num__315 <o> b ) num__215 <o> c ) num__115 <o> d ) num__165 <o> e ) num__765 |
in an arithmetic progression the nth term is given by tn = a + ( n - num__1 ) d here tn = num__65 a = - num__60 d = num__1 hence num__65 = - num__60 + ( n - num__1 ) or n = num__126 sum of n terms can be calculated by sn = n / num__2 ( a + l ) a = first term l = last term n = no . of terms sn = num__126 * ( - num__60 + num__65 ) / num__2 sn = num__126 * num__2.5 = num__315 answer : a <eor> a <eos> |
a |
multiply__2.5__126.0__ multiply__2.5__126.0__ |
multiply__2.5__126.0__ multiply__2.5__126.0__ |
| the new swimming pool needs to be filled with water . if the swimming pool is num__5 m ^ num__3 and the speed to fill the swimming pool is num__2.5 cm ^ num__3 / sec ( num__1 m ^ num__3 = num__10 ^ num__6 cm ^ num__3 ) how long does it take to fill the entire swimming pool ? <o> a ) num__2 * num__10 ^ num__3 <o> b ) num__2 * num__10 ^ num__4 <o> c ) num__2 * num__10 ^ num__5 <o> d ) num__2 * num__10 ^ num__6 <o> e ) num__2 * num__10 ^ num__7 |
from num__1 m ^ num__3 = num__10 ^ num__6 cm ^ num__3 num__5 m ^ num__3 = num__5 * num__10 ^ num__6 cm ^ num__3 time = num__5 * num__10 ^ num__6 cm ^ num__3 / num__2.5 cm ^ num__3 / sec = num__2 * num__10 ^ num__6 seconds . correct option : num__2 * num__10 ^ num__6 answer : d <eor> d <eos> |
d |
subtract__5.0__3.0__ round__2.0__ |
divide__5.0__2.5__ divide__5.0__2.5__ |
| john invested a certain sum of money in a simple interest bond whose value grew to $ num__300 at the end of num__2 years and to $ num__500 at the end of another num__3 years . what was the rate of interest in which he invested his sum ? <o> a ) num__12.0 <o> b ) num__40.0 <o> c ) num__6.67 <o> d ) num__25.0 <o> e ) num__33 % |
lets assume the principal amount ( initial amount invested ) to be p rate of interest to berand time as t . we need to find r now after a time of num__2 years the principal p amounts to $ num__300 and after a time of num__5 years ( question says after another num__3 years so num__2 + num__3 ) p becomes $ num__500 . formulating the above data amount ( a num__1 ) at end of num__2 years a num__1 = p ( num__1 + num__2 r / num__100 ) = num__300 amount ( a num__2 ) at end of num__8 years a num__2 = p ( num__1 + num__5 r / num__100 ) = num__500 dividing a num__2 by a num__1 we get ( num__1 + num__5 r / num__100 ) / ( num__1 + num__2 r / num__100 ) = num__1.66666666667 after cross multiplication we are left with num__5 r = num__100 * num__2 which gives r = num__40.0 option : b <eor> b <eos> |
b |
percent__8.0__500.0__ percent__100.0__40.0__ |
percent__8.0__500.0__ percent__100.0__40.0__ |
| num__100 men started working to complete a work in num__50 days . after working num__8 hours / day they had completed num__0.333333333333 rd of the work in num__25 days . how many men should be hired so that they will finish the work in time by working num__10 hours / day . <o> a ) num__30 men <o> b ) num__60 men <o> c ) num__90 men <o> d ) num__100 men <o> e ) num__120 men |
working num__8 hours / day in num__25 days num__100 men completes num__8 * num__25 * num__100 = num__20000 men hours which is num__0.333333333333 of the work so remaining num__0.666666666667 work to be completed = num__20000 * ( num__0.666666666667 ) / ( num__0.333333333333 ) = num__40000 men hours . if with ' x ' extra men hired remaining work is to be completed in num__50 - num__25 = num__25 days working num__10 hours / day then ( num__100 + x ) * num__25 * num__10 = num__40000 num__25000 + num__250 x = num__40000 x = num__60 men answer : b <eor> b <eos> |
b |
multiply__25.0__10.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
multiply__25.0__10.0__ add__50.0__10.0__ add__50.0__10.0__ |
| what is the total number of integers between num__10 and num__100 ( exclusive ) that are divisible by num__7 ? <o> a ) num__10 <o> b ) num__13 <o> c ) num__15 <o> d ) num__12 <o> e ) num__11 |
num__14 num__21 num__28 . . . num__9198 this is an equally spaced list ; you can use the formula : n = ( largest - smallest ) / ( ' space ' ) + num__1 = ( num__98 - num__14 ) / ( num__7 ) + num__1 = num__12 + num__1 = num__13 answer is b <eor> b <eos> |
b |
add__7.0__14.0__ add__7.0__21.0__ multiply__7.0__14.0__ add__1.0__12.0__ add__1.0__12.0__ |
add__7.0__14.0__ add__7.0__21.0__ multiply__7.0__14.0__ add__1.0__12.0__ add__1.0__12.0__ |
| when a num__192 meters long rod is cut down into small pieces of length num__3.0 meters each . then how many pieces are available ? <o> a ) num__52 <o> b ) num__68 <o> c ) num__62 <o> d ) num__64 <o> e ) num__69 |
answer no . of pieces = total length / length of each piece = num__192 / num__3.0 = num__64 option : d <eor> d <eos> |
d |
divide__192.0__3.0__ round__64.0__ |
divide__192.0__3.0__ divide__192.0__3.0__ |
| num__1000 men have provisions for num__15 days . if num__200 more men join them for how many days will the provisions last now ? <o> a ) num__12.8 <o> b ) num__12.4 <o> c ) num__12.5 <o> d ) num__16.8 <o> e ) num__92.7 |
num__1000 * num__15 = num__1200 * x x = num__12.5 answer : c <eor> c <eos> |
c |
add__1000.0__200.0__ round__12.5__ |
add__1000.0__200.0__ round__12.5__ |
| income and expenditure of a person are in the ratio num__7 : num__6 . if the income of the person is rs . num__14000 then find his savings ? <o> a ) num__500 <o> b ) num__1000 <o> c ) num__1500 <o> d ) num__2000 <o> e ) num__2500 |
let the income and the expenditure of the person be rs . num__7 x and rs . num__6 x respectively . income num__7 x = num__14000 = > x = num__2000 savings = income - expenditure = num__7 x - num__6 x = x so savings = rs . num__2000 . answer : d <eor> d <eos> |
d |
divide__14000.0__7.0__ divide__14000.0__7.0__ |
divide__14000.0__7.0__ divide__14000.0__7.0__ |
| how many times in a day the hands of a clock are straight ? <o> a ) num__22 <o> b ) num__34 <o> c ) num__44 <o> d ) num__54 <o> e ) num__76 |
in num__12 hours the hands coincide or are in opposite direction num__22 times . in num__24 hours the hands coincide or are in opposite direction num__44 times a day . answer : c <eor> c <eos> |
c |
round__44.0__ |
round__44.0__ |
| a train overtakes num__2 girls who are walking inthe opposite direction in which the train is going at the rate of num__3 km / hour & num__6 km / hour and passes them completely in num__36 sec & num__30 sec respectively . find the length of the train is ? <o> a ) num__130 m <o> b ) num__140 m <o> c ) num__150 m <o> d ) num__170 m <o> e ) num__190 m |
let the length of the train e x meter and let the speed of train be y km / h then \ inline x = \ left ( y + num__3 \ right ) \ frac { num__5 } { num__18 } \ times num__36 . . . . . . . . ( num__1 ) and \ inline x = \ left ( y + num__6 \ right ) \ frac { num__5 } { num__18 } \ times num__30 . . . . . . . . ( num__2 ) from eq ( num__1 ) and ( num__2 ) we get \ inline ( y + num__3 ) \ times num__36 = ( y + num__6 ) \ times num__30 y = num__12 km / h \ inline \ therefore \ inline x = ( y + num__3 ) \ times \ frac { num__5 } { num__18 } \ times num__36 x = num__150 m c <eor> c <eos> |
c |
add__2.0__3.0__ multiply__3.0__6.0__ subtract__3.0__2.0__ multiply__2.0__6.0__ multiply__30.0__5.0__ round__150.0__ |
add__2.0__3.0__ divide__36.0__2.0__ subtract__3.0__2.0__ divide__36.0__3.0__ multiply__30.0__5.0__ divide__150.0__1.0__ |
| the ratio of numbers is num__4 : num__5 and their h . c . f is num__4 . their l . c . m is : <o> a ) num__80 <o> b ) num__16 <o> c ) num__24 <o> d ) num__48 <o> e ) num__98 |
let the numbers be num__4 x and num__5 x . then their h . c . f = x . so x = num__4 . so the numbers are num__16 and num__20 . l . c . m of num__16 and num__20 = num__80 . answer : a <eor> a <eos> |
a |
multiply__4.0__5.0__ multiply__4.0__20.0__ multiply__4.0__20.0__ |
multiply__4.0__5.0__ multiply__4.0__20.0__ multiply__4.0__20.0__ |
| if the wheel is num__21 cm then the number of revolutions to cover a distance of num__1056 cm is ? <o> a ) num__8 <o> b ) num__27 <o> c ) num__10 <o> d ) num__7 <o> e ) num__19 |
num__2 * num__3.14285714286 * num__21 * x = num__1056 = > x = num__8 answer : a <eor> a <eos> |
a |
round__8.0__ |
round__8.0__ |
| a fruit seller had some oranges . he sells num__40.0 oranges and still has num__300 oranges . how many oranges he had originally ? <o> a ) num__700 <o> b ) num__710 <o> c ) num__720 <o> d ) num__500 <o> e ) num__740 |
num__60.0 of oranges = num__300 num__100.0 of oranges = ( num__300 × num__100 ) / num__6 = num__500 total oranges = num__500 answer : d <eor> d <eos> |
d |
percent__100.0__500.0__ |
percent__100.0__500.0__ |
| three unbiased coins are tossed . what is the probability of getting at least num__2 tails ? <o> a ) num__0.75 <o> b ) num__0.5 <o> c ) num__0.25 <o> d ) num__0.2 <o> e ) num__0.3 |
explanation : s = { hhh hht hth htt thh tht tth ttt } e = { htt tht tth ttt } n ( s ) = num__8 n ( e ) = num__4 p ( e ) = n ( e ) / n ( s ) = num__0.5 = num__0.5 answer is b <eor> b <eos> |
b |
negate_prob__0.5__ |
negate_prob__0.5__ |
| in a camp there is a meal for num__120 men or num__200 children . if num__150 children have taken the meal how many men will be catered to with remaining meal ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__50 <o> d ) num__10 <o> e ) num__40 |
there is a meal for num__200 children . num__150 children have taken the meal . remaining meal is to be catered to num__50 children . now num__200 children num__120 men . num__50 children ( num__0.6 ) * num__50 = num__30 men answer is b . <eor> b <eos> |
b |
subtract__200.0__150.0__ divide__120.0__200.0__ subtract__150.0__120.0__ subtract__150.0__120.0__ |
subtract__200.0__150.0__ divide__120.0__200.0__ multiply__0.6__50.0__ multiply__0.6__50.0__ |
| find the value of y from ( num__12 ) ^ num__3 x num__6 ^ num__4 ÷ num__432 = y ? <o> a ) num__3484 <o> b ) num__3467 <o> c ) num__3567 <o> d ) num__4567 <o> e ) num__5184 |
given exp . = ( num__12 ) num__3 x num__64 = ( num__12 ) num__3 x num__64 = ( num__12 ) num__2 x num__62 = ( num__72 ) num__2 = num__5184 num__432 num__12 x num__62 e <eor> e <eos> |
e |
divide__12.0__6.0__ subtract__64.0__2.0__ multiply__12.0__6.0__ multiply__12.0__432.0__ multiply__12.0__432.0__ |
divide__12.0__6.0__ subtract__64.0__2.0__ multiply__12.0__6.0__ multiply__12.0__432.0__ multiply__12.0__432.0__ |
| an bus covers a certain distance at aspeed of num__150 kmph in num__5 hours . to cover the same distance in num__1 hr it must travel at a speed of ? <o> a ) num__560 km / h <o> b ) num__450 km / h <o> c ) num__779 km / h <o> d ) num__723 km / h <o> e ) num__720 km / h |
distance = ( num__150 x num__5 ) = num__750 km . speed = distance / time speed = num__750 / ( num__1.66666666667 ) km / hr . [ we can write num__1 hours as num__1.66666666667 hours ] required speed = num__750 x num__0.6 km / hr = num__450 km / hr . b <eor> b <eos> |
b |
multiply__150.0__5.0__ km_to_mile_conversion__ multiply__0.6__750.0__ round__450.0__ |
multiply__150.0__5.0__ divide__1.0__1.6667__ multiply__0.6__750.0__ divide__450.0__1.0__ |
| the average age of num__20 persons in a office is num__15 years . out of these the average age of num__5 of them is num__14 years and that of the other num__9 persons is num__16 years . the age of the num__15 th person is ? <o> a ) num__79 <o> b ) num__86 <o> c ) num__95 <o> d ) num__72 <o> e ) num__80 |
age of the num__15 th student = num__20 * num__15 - ( num__14 * num__5 + num__16 * num__9 ) = num__300 - num__214 = num__86 years answer is b <eor> b <eos> |
b |
multiply__20.0__15.0__ subtract__300.0__214.0__ subtract__300.0__214.0__ |
multiply__20.0__15.0__ subtract__300.0__214.0__ subtract__300.0__214.0__ |
| the rate of spin of a certain gyroscope doubled every num__10 seconds from the moment a particular stopwatch started . if after a minute and a half the gyroscope reached a speed of num__3200 meters per second what was the speed in meters per second when the stopwatch was started ? <o> a ) num__8.33333333333 <o> b ) num__6.25 <o> c ) num__3.125 <o> d ) num__1.5625 <o> e ) num__0.78125 |
let x be the original speed when the stopwatch was started . in num__90 seconds the speed doubled num__9 times . num__2 ^ num__9 * x = num__3200 x = ( num__2 ^ num__7 * num__25 ) / num__2 ^ num__9 = num__6.25 the answer is b . <eor> b <eos> |
b |
divide__90.0__10.0__ subtract__9.0__2.0__ round__6.25__ |
divide__90.0__10.0__ subtract__9.0__2.0__ round__6.25__ |
| in township k num__0.25 of the housing units are equiped with cable tv . if num__0.1 of the housing units including num__0.333333333333 of those that are equiped with cable tv are equipped with videocassette recorders what fraction of the housing units have neither cable tv nor videocassette recorders ? <o> a ) num__0.766666666667 <o> b ) num__0.733333333333 <o> c ) num__0.7 <o> d ) num__0.166666666667 <o> e ) num__0.133333333333 |
num__0.25 - - cable tv ( this includes some data from video cassette recorder ) num__0.1 - - video cassette recorder including num__0.333333333333 ( equiped with cable tv ) i . e . num__0.333333333333 ( num__0.25 ) = num__0.0833333333333 therefore only video cassette recorder = num__0.1 - num__0.0833333333333 = num__0.0166666666667 total = num__0.25 + num__0.0166666666667 + neither cable tv nor videocassette recorders num__1 = num__0.266666666667 + neither cable tv nor videocassette recorders therefore neither cable tv nor videocassette recorders = num__1 - num__0.266666666667 = num__0.733333333333 hence b . <eor> b <eos> |
b |
union_prob__0.25__0.1__0.3333__ union_prob__0.25__0.1__0.0833__ negate_prob__0.2667__ negate_prob__0.2667__ |
union_prob__0.25__0.1__0.3333__ union_prob__0.25__0.1__0.0833__ negate_prob__0.2667__ negate_prob__0.2667__ |
| in num__1982 and num__1983 company b ’ s operating expenses were $ num__11.0 million and $ num__14.0 million respectively and its revenues were $ num__15.6 million and $ num__18.8 million respectively . what was the percent increase in company b ’ s profit ( revenues minus operating expenses ) from num__1982 to num__1983 ? <o> a ) num__3.0 <o> b ) num__16 num__0.666666666667 % <o> c ) num__25.0 <o> d ) num__4 num__0.333333333333 % <o> e ) num__60 % |
profit in num__1982 = num__15.6 - num__11 = num__4.6 million $ profit in num__1983 = num__18.8 - num__14 = num__4.8 million $ percentage increase in profit = ( num__4.8 - num__4.6 ) / num__4.6 * num__100.0 = num__4 num__0.333333333333 % answer d <eor> d <eos> |
d |
subtract__15.6__11.0__ subtract__18.8__14.0__ round_down__4.8__ round_down__4.8__ |
subtract__15.6__11.0__ subtract__18.8__14.0__ round_down__4.8__ round_down__4.8__ |
| how many even multiples of num__55 are there between num__549 and num__1101 ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__9 <o> d ) num__10 <o> e ) num__11 |
num__550 = num__10 * num__55 num__1100 = num__20 * num__55 the even multiples are num__55 multiplied by num__10 num__12 num__14 num__16 num__18 and num__20 for a total of num__6 . the answer is b . <eor> b <eos> |
b |
divide__550.0__55.0__ divide__1100.0__55.0__ subtract__16.0__10.0__ subtract__12.0__6.0__ |
divide__550.0__55.0__ divide__1100.0__55.0__ subtract__16.0__10.0__ subtract__12.0__6.0__ |
| the marks obtained by polly and sandy are in the ratio num__5 : num__6 and those obtained by sandy and willy are in the ratio of num__3 : num__2 . the marks obtained by polly and willy are in the ratio of . . . ? <o> a ) num__3 : num__2 <o> b ) num__5 : num__4 <o> c ) num__7 : num__6 <o> d ) num__9 : num__8 <o> e ) num__11 : num__10 |
polly : sandy = num__5 : num__6 sandy : willy = num__3 : num__2 = num__6 : num__4 polly : sandy : willy = num__5 : num__6 : num__4 polly : willy = num__5 : num__4 the answer is b . <eor> b <eos> |
b |
subtract__6.0__2.0__ add__3.0__2.0__ |
subtract__6.0__2.0__ add__3.0__2.0__ |
| the speed of a railway engine is num__63 km per hour when no compartment is attached and the reduction in speed is directly proportional to the square root of the number of compartments attached . if the speed of the train carried by this engine is num__24 km per hour when num__9 compartments are attached the maximum number of compartments that can be carried by the engine is : <o> a ) num__19 <o> b ) num__18 <o> c ) num__16 <o> d ) num__17 <o> e ) num__23 |
the reduction in speed is directly proportional to the square root of the number of compartments attached does reduction mean amount subtracted ? or percentage decrease ? there are at least two interpretations and the wording does not provide a clear interpretation between them . evidently what the question intends is the subtraction interpretation . what is subtracted from the speed is directly proportional to the square root of the number of compartments attached . in other words if s = speed and n = number of compartments then s = num__63 - k * sqrt ( n ) where k is a constant of the proportionality . in general if a is directly proportional to b we can write a = k * b and solve for k . if n = num__9 then s = num__24 num__24 = num__63 - k * sqrt ( num__9 ) = num__63 - num__3 k k = num__13 now we need to know : what value of n makes s go to zero ? num__0 = num__63 - num__13 * sqrt ( n ) num__13 * sqrt ( n ) = num__63 n = num__23 with num__24 compartments the train does not budge . therefore it would budge if there were one fewer cars . thus num__23 is the maximum number of cars the engine can pull and still move . e <eor> e <eos> |
e |
round__23.0__ |
round__23.0__ |
| a man buys a cycle for rs . num__1200 and sells it at a loss of num__15.0 . what is the selling price of the cycle ? <o> a ) rs . num__1090 <o> b ) rs . num__1020 <o> c ) rs . num__1190 <o> d ) rs . num__1202 <o> e ) none |
solution s . p = num__85.0 of rs . num__1200 = rs . ( num__0.85 × num__1200 ) rs . num__1020 . answer b <eor> b <eos> |
b |
percent__85.0__1200.0__ percent__85.0__1200.0__ |
percent__85.0__1200.0__ percent__85.0__1200.0__ |
| an angry arjun carried some arrows for fighting with bheeshm . with half the arrows he cut down the arrows thrown by bheeshm on him and with six other arrows he killed the chariot driver of bheeshm . with one arrow each he knocked down respectively the chariot the flag and the bow of bheeshm . finally with one more than four times the square root of arrows he laid bheeshm unconscious on an arrow bed . find the total number of arrows arjun had . <o> a ) num__90 <o> b ) num__100 <o> c ) num__110 <o> d ) num__120 <o> e ) num__130 |
x / num__2 + num__6 + num__3 + num__1 + num__4 sqrt ( x ) = x x / num__2 + num__10 + num__4 sqrt ( x ) = x num__4 sqrt ( x ) = x / num__2 - num__10 squaring on both sides num__16 x = x ² / num__4 + num__100 - num__10 x simplifying x ² - num__104 x + num__400 = num__0 x = num__100 num__4 x = num__4 is not possible therefore x = num__100 answer : b <eor> b <eos> |
b |
divide__6.0__2.0__ subtract__3.0__2.0__ add__1.0__3.0__ add__4.0__6.0__ add__6.0__10.0__ add__4.0__100.0__ multiply__4.0__100.0__ multiply__1.0__100.0__ |
divide__6.0__2.0__ subtract__3.0__2.0__ add__1.0__3.0__ add__4.0__6.0__ add__6.0__10.0__ add__4.0__100.0__ multiply__4.0__100.0__ subtract__104.0__4.0__ |
| village p ’ s population is num__800 greater than village q ' s population . if village q ’ s population were reduced by num__500 people then village p ’ s population would be num__3 times as large as village q ' s population . what is village q ' s current population ? <o> a ) num__1100 <o> b ) num__1150 <o> c ) num__1200 <o> d ) num__1250 <o> e ) num__1300 |
p = q + num__800 . p = num__3 ( q - num__500 ) . num__3 ( q - num__500 ) = q + num__800 . num__2 q = num__2300 . q = num__1150 . the answer is b . <eor> b <eos> |
b |
divide__2300.0__2.0__ divide__2300.0__2.0__ |
divide__2300.0__2.0__ subtract__2300.0__1150.0__ |
| how many x ways can you group num__3 people from num__4 sets of twins if no two people from the same set of twins can be chosen ? <o> a ) num__3 <o> b ) num__16 <o> c ) num__28 <o> d ) num__32 <o> e ) num__56 |
ways to select num__3 people from num__8 people ( num__4 twins x num__2 ) = num__8 c num__3 = num__56 ways to select num__1 twin + num__1 people = num__4 c num__1 * num__6 c num__1 = num__24 ways to select a group num__3 people from num__4 sets of twins if no two people from the same set of twins can be chosen x = num__56 - num__24 = num__32 ans : d <eor> d <eos> |
d |
divide__8.0__4.0__ subtract__3.0__2.0__ multiply__3.0__2.0__ multiply__3.0__8.0__ multiply__4.0__8.0__ multiply__4.0__8.0__ |
divide__8.0__4.0__ subtract__3.0__2.0__ multiply__3.0__2.0__ multiply__3.0__8.0__ multiply__4.0__8.0__ multiply__4.0__8.0__ |
| a walks at num__4 kmph and num__4 hours after his start b cycles after him at num__10 kmph . how far from the start does b catch up with a ? <o> a ) num__25 km <o> b ) num__26.7 km <o> c ) num__30 km <o> d ) num__32.5 km <o> e ) num__31.2 km |
suppose after x km from the start b catches up with a . then the difference in the time taken by a to cover x km and that taken by b to cover x km is num__4 hours . x / num__4 - x / num__10 = num__4 x = num__26.7 km answer is b <eor> b <eos> |
b |
round__26.7__ |
round__26.7__ |
| machine a and machine b are each used to manufacture num__440 sprockets . it takes machine a num__10 hours longer to produce num__440 sprockets than machine b . machine b produces num__10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__4 <o> d ) num__3 <o> e ) num__2 |
machine b : takes x hours to produce num__440 sprockets machine a : takes ( x + num__10 ) hours to produce num__440 sprockets machine b : in num__1 hour b makes num__440 / x sprockets machine a : in num__1 hour a makes num__440 / ( x + num__10 ) sprockets equating : num__1.1 ( num__440 / ( x + num__10 ) ) = num__440 / x num__484 / ( x + num__10 ) = num__440 / x num__484 x = num__440 x + num__4400 num__44 x = num__4400 x = num__100 a makes num__440 / ( num__110 ) = num__4 sprockets per hour answer : c <eor> c <eos> |
c |
percent__10.0__440.0__ percent__100.0__4.0__ |
percent__10.0__440.0__ percent__100.0__4.0__ |
| it takes joey the postman num__1 hours to run a num__4 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is num__5 mile / hour what is the speed with which joey returns ? <o> a ) num__5 <o> b ) num__12 <o> c ) num__13 <o> d ) num__14 <o> e ) num__15 |
let his speed for one half of the journey be num__4 miles an hour let the other half be x miles an hour now avg speed = num__5 mile an hour num__2 * num__4 * x / num__4 + x = num__5 num__8 x = num__4 x + num__20 = > x = num__5 a <eor> a <eos> |
a |
multiply__4.0__2.0__ multiply__4.0__5.0__ round__5.0__ |
multiply__4.0__2.0__ multiply__4.0__5.0__ add__1.0__4.0__ |
| what is the greatest positive integer x such that num__5 ^ x is a factor of num__125 ^ num__10 ? <o> a ) num__5 <o> b ) num__9 <o> c ) num__10 <o> d ) num__20 <o> e ) num__30 |
num__125 ^ num__10 = ( num__5 ^ num__3 ) ^ num__10 = num__5 ^ num__30 answer : e <eor> e <eos> |
e |
multiply__10.0__3.0__ multiply__10.0__3.0__ |
multiply__10.0__3.0__ multiply__10.0__3.0__ |
| in how many minimum number of complete years the interest on num__212.50 p at num__3.0 per annum will be in exact number of rupees ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__9 <o> d ) num__7 <o> e ) none of these |
interest for one year = num__212.50 × num__3 ⁄ num__100 × num__1 = num__51 ⁄ num__8 thus in num__8 years the interest is num__51 . answer b <eor> b <eos> |
b |
percent__100.0__8.0__ |
percent__100.0__8.0__ |
| num__300 first - time customers of a fashion store were surveyed for their shopping experience right after leaving the store . num__60.0 of the customers in the survey had purchased clothes for less than $ num__100 . num__40.0 of the customers in the survey reported they were overall satisfied with their purchase . num__65.0 of the customers that had purchased clothes for less than $ num__100 reported they were overall satisfied with their purchase . what percent of the customers surveyed purchased clothes for at least $ num__100 and reported that they were not overall satisfied with their purchase ? <o> a ) num__1 <o> b ) num__25 <o> c ) num__35 <o> d ) num__45 <o> e ) num__75 |
out of num__300 - num__180 purchased for less than num__100 $ num__120 for more out of num__300 - num__120 responded as satisfied and num__180 responded disatisfied out of num__180 ( purchased less than num__100 $ ) - num__65.0 = num__117 responded as satisfied so remaining satisfied are num__120 - num__117 = num__3 so num__3 is what percentage of num__300 - num__1.0 so the answer should be a <eor> a <eos> |
a |
subtract__300.0__180.0__ divide__300.0__100.0__ reverse__1.0__ |
subtract__300.0__180.0__ subtract__120.0__117.0__ reverse__1.0__ |
| when a student joe weighing num__43 kg joins a group of students whose average weight is num__30 kg the average weight goes up by num__1 kg . subsequently if two students excluding joe leave the group the average weight comes back to num__30 kg . what is the difference between the average weight of the two students who left and the weight of joe ? <o> a ) num__5.5 kg <o> b ) num__11 kg <o> c ) num__30 kg <o> d ) num__6.5 kg <o> e ) num__71 kg |
after two persons leave the group the average remains the same . that means the weight of the two persons = num__43 + num__30 = num__73 so the average the two persons = num__36.5 that gives the answer num__43 - num__36.5 = num__6.5 answer d <eor> d <eos> |
d |
add__43.0__30.0__ subtract__43.0__36.5__ subtract__43.0__36.5__ |
add__43.0__30.0__ subtract__43.0__36.5__ subtract__43.0__36.5__ |
| if difference between compound interest and simple interest on a sum at num__10.0 p . a . for num__2 years is rs . num__61 then sum is <o> a ) s . num__6100 <o> b ) s . num__5100 <o> c ) s . num__5800 <o> d ) s . num__6000 <o> e ) s . num__6200 |
p ( r / num__100 ) ^ num__2 = c . i - s . i p ( num__0.1 ) ^ num__2 = num__61 num__6100 answer : a <eor> a <eos> |
a |
percent__100.0__6100.0__ |
percent__100.0__6100.0__ |
| let s be the set of all positive integers that when divided by num__8 have a remainder of num__5 . what is the num__81 th number in this set ? <o> a ) num__605 <o> b ) num__608 <o> c ) num__613 <o> d ) num__616 <o> e ) num__645 |
the set s = { num__5 num__13 num__21 num__29 . . . . . . . . . . . . . . . . . . . . . } num__1 st number = num__8 * num__0 + num__5 = num__5 num__2 nd number = num__8 * num__1 + num__5 = num__13 num__3 rd number = num__8 * num__2 + num__5 = num__21 num__81 th number = num__8 * ( num__81 - num__1 ) + num__5 = num__645 answer = e <eor> e <eos> |
e |
add__8.0__5.0__ add__8.0__13.0__ add__8.0__21.0__ subtract__8.0__5.0__ multiply__1.0__645.0__ |
add__8.0__5.0__ add__8.0__13.0__ add__8.0__21.0__ subtract__8.0__5.0__ multiply__1.0__645.0__ |
| when positive integer n is divided by positive integer j the remainder is num__14 . if n / j = num__134.07 what is value of j ? <o> a ) num__22 <o> b ) num__56 <o> c ) num__78 <o> d ) num__200 <o> e ) num__175 |
num__1 ) we know that decimal part of decimal quotient = { remainder / divisor } so num__0.07 the decimal part of the decimal quotient must equal the remainder num__14 divided by the divisor j . num__0.07 = num__14 / j num__0.07 * j = num__14 j = num__14 / num__0.07 = num__200.0 = num__200 so j = num__200 answer = d . <eor> d <eos> |
d |
divide__14.0__0.07__ divide__14.0__0.07__ |
divide__14.0__0.07__ divide__14.0__0.07__ |
| at what rate percent per annum will the simple interest on a sum of money be num__0.6 of the amount in num__10 years ? <o> a ) num__6.0 <o> b ) num__7.0 <o> c ) num__9.0 <o> d ) num__3.0 <o> e ) num__1 % |
let sum = x . then s . i . = num__2 x / num__5 time = num__10 years . rate = ( num__100 * num__3 x ) / ( x * num__5 * num__10 ) = num__6.0 answer : a <eor> a <eos> |
a |
percent__100.0__6.0__ |
percent__100.0__6.0__ |
| the speed of a car is num__90 km in the first hour and num__60 km in the second hour . what is the average speed of the car ? <o> a ) num__228 <o> b ) num__75 <o> c ) num__299 <o> d ) num__267 <o> e ) num__312 |
s = ( num__90 + num__60 ) / num__2 = num__75 kmph answer : b <eor> b <eos> |
b |
round__75.0__ |
round__75.0__ |
| the function f is defined by subtracting num__25 from the square of a number and the function v is defined as the square root of one - half of a number . if v ( f ( x ) ) = num__10 then which of the following is a possible value of x ? <o> a ) - num__15 <o> b ) - num__5 <o> c ) num__0 <o> d ) num__5 <o> e ) num__25 |
f ( x ) = x ^ num__2 - num__25 v ( x ) = sqrt ( x / num__2 ) not sqrt ( x ) / num__2 because the question clearly says its square root of ( half of the number ) . v ( f ( x ) ) = num__10 v ( x ^ num__2 - num__25 ) = num__10 sqrt ( ( x ^ num__2 - num__25 ) / num__2 ) = num__10 = > ( x ^ num__2 - num__25 ) / num__2 = num__100 = > x ^ num__2 = num__225 = > x = num__15 or - num__15 answer is a . <eor> a <eos> |
a |
subtract__25.0__10.0__ subtract__25.0__10.0__ |
subtract__25.0__10.0__ subtract__25.0__10.0__ |
| john took a bus from home to market that travels at num__50 kmph . while walking back at num__5 kmph halfway through he suddenly realized he was getting late and he cycled back the remaining distance in num__35 kmph . find the average speed . <o> a ) num__8.5 kmph <o> b ) num__16.0 kmph <o> c ) num__22.5 kmph <o> d ) num__18.6 kmph <o> e ) none of these |
let the distance be num__2 x ( one way ) time taken by bus = num__2 x / num__50 by walking = x / num__5 by cycling = x / num__35 hours : . average speed = total distance / total time = num__5 x / x / num__25 + x / num__5 + x / num__35 = num__5 * num__60 / num__2.4 + num__12 + num__1.7 = num__18.6 answer : d <eor> d <eos> |
d |
divide__50.0__2.0__ hour_to_min_conversion__ divide__60.0__25.0__ multiply__5.0__2.4__ round__18.6__ |
divide__50.0__2.0__ add__35.0__25.0__ divide__60.0__25.0__ multiply__5.0__2.4__ round__18.6__ |
| a can do a job in num__10 days and b can do it in num__30 days . a and b working together will finish twice the amount of work in - - - - - - - days ? <o> a ) num__21 ½ days <o> b ) num__22 ½ days <o> c ) num__23 ½ days <o> d ) num__15 days <o> e ) none of these |
explanation : num__0.1 + num__0.0333333333333 = num__0.133333333333 = num__0.133333333333 num__7.5 = num__7.5 * num__2 = num__15 days answer : d <eor> d <eos> |
d |
add__0.1__0.0333__ divide__30.0__2.0__ round__15.0__ |
add__0.1__0.0333__ multiply__7.5__2.0__ multiply__7.5__2.0__ |
| in a sequence of num__12 numbers each term except for the first one is num__12 ^ num__13 less than the previous term . if the greatest term in the sequence is num__12 ^ num__14 what is the smallest term in the sequence ? thus a num__12 = a num__1 + num__12 - num__1 d where d is the difference between each pairs of consecutive terms . since each term is num__1211 less than the previous term d = - num__1211 click individual answer choices for specific explanations . from what i undersltand the formula for an arithmetic sequence represents some - number - in - the - sequence = first - number - in - sequence + the - place - of - that - number - num__1 x the - difference - of - each - # - in - the - sequence <o> a ) − num__12 ^ num__11 <o> b ) num__0 <o> c ) num__12 ^ num__11 <o> d ) num__11 · num__12 ^ num__11 <o> e ) num__12 ^ num__13 |
there are total num__12 terms . andgreatestterm is num__12 ^ num__14 . each term is num__12 ^ num__13 less than previous one . what does this signify ? this shows it is an ap - a decreasing one . ( noticeeach term is num__12 ^ num__11 less than previous one ' ) therefore num__12 ^ num__14 is the first term of the ap and - num__12 ^ num__13 is the difference between successive terms the smallest term should be the last one . last term = first term + ( n - num__1 ) * d therefore last term = num__12 ^ num__14 + num__11 * ( - num__12 ^ num__13 ) = > last term = num__12 ^ num__14 - num__11 * num__12 ^ num__13 = > last term = num__12 ^ num__13 ( num__12 - num__11 ) = num__12 ^ num__13 which is the smallest term . hence ans e it is . <eor> e <eos> |
e |
subtract__12.0__1.0__ multiply__12.0__1.0__ |
subtract__12.0__1.0__ multiply__12.0__1.0__ |
| five years ago the average of the ages of a and b was num__40 years and now the average of the ages of b and c is num__48 years . what will be the age of the b ten years hence ? <o> a ) num__66 <o> b ) num__77 <o> c ) num__66 <o> d ) num__96 <o> e ) num__51 |
let the present ages of a b and c be a b and c respectively . given [ ( a - num__5 ) + ( b - num__5 ) ] / num__2 = num__40 = > a + b = num__90 - - - ( num__1 ) ( b + c ) / num__2 = num__48 = > b + c = num__96 - - - ( num__2 ) from ( num__1 ) and ( num__2 ) we can not find b . answer : d <eor> d <eos> |
d |
multiply__48.0__2.0__ multiply__48.0__2.0__ |
multiply__48.0__2.0__ divide__96.0__1.0__ |
| two numbers are respectively num__20.0 and num__50.0 more than a third number . the ratio of the two numbers is <o> a ) num__4 : num__5 <o> b ) num__5 : num__4 <o> c ) num__3 : num__2 <o> d ) num__4 : num__7 <o> e ) none |
let the third number be x . then first number = num__120.0 of x = num__120 x = num__6 x num__100 num__5 second number = num__150.0 of x = num__150 x = num__3 x num__100 num__2 ratio of first two numbers = num__6 x : num__3 x = num__12 x : num__15 x = num__4 : num__5 . num__5 num__2 a <eor> a <eos> |
a |
divide__120.0__20.0__ subtract__120.0__20.0__ divide__100.0__20.0__ add__50.0__100.0__ divide__150.0__50.0__ divide__100.0__50.0__ multiply__2.0__6.0__ subtract__20.0__5.0__ divide__20.0__5.0__ divide__20.0__5.0__ |
divide__120.0__20.0__ subtract__120.0__20.0__ divide__100.0__20.0__ add__50.0__100.0__ divide__150.0__50.0__ divide__100.0__50.0__ multiply__2.0__6.0__ subtract__20.0__5.0__ divide__20.0__5.0__ divide__20.0__5.0__ |
| find the average of all the numbers between num__11 and num__36 which are divisible by num__5 . <o> a ) num__25 <o> b ) num__77 <o> c ) num__20 <o> d ) num__28 <o> e ) num__10 |
explanation : average = ( num__15 + num__20 + num__25 + num__30 + num__35 ) / num__5 = num__25.0 = num__25 answer : a <eor> a <eos> |
a |
add__5.0__15.0__ subtract__36.0__11.0__ add__5.0__25.0__ add__5.0__30.0__ subtract__36.0__11.0__ |
add__5.0__15.0__ add__5.0__20.0__ add__5.0__25.0__ add__5.0__30.0__ add__5.0__20.0__ |
| in triangle xyz side xy which runs perpendicular to side yz measures num__24 inches in length . if the longest side of the the triangle is num__26 inches what is the area in square inches of triangle xyz ? <o> a ) num__100 <o> b ) num__120 <o> c ) num__140 <o> d ) num__150 <o> e ) num__165 |
okay we know we have a right triangle with hypotenuse num__26 . using pythagoras theory i think : num__24 ^ num__2 + x ^ num__2 = num__26 ^ num__2 so num__576 + x ^ num__2 = num__676 - - - > x ^ num__2 = num__100 - - - > x = num__10 areal of triangle = base * height / num__2 so num__24 * num__5.0 = num__120 so answer choice b <eor> b <eos> |
b |
power__24.0__2.0__ power__26.0__2.0__ rectangle_perimeter__24.0__26.0__ side_by_diagonal__26.0__24.0__ multiply__24.0__5.0__ multiply__24.0__5.0__ |
power__24.0__2.0__ power__26.0__2.0__ rectangle_perimeter__24.0__26.0__ side_by_diagonal__26.0__24.0__ multiply__24.0__5.0__ multiply__24.0__5.0__ |
| the length of minute hand of a clock is num__5.5 cm . what is the area covered by this in num__10 minutes <o> a ) num__15.27 <o> b ) num__16.27 <o> c ) num__17.27 <o> d ) num__19.27 <o> e ) num__15.83 |
area of circle is pi * r ^ num__2 but in num__10 minutes area covered is ( num__0.166666666667 ) * num__360 = num__60 degree so formula is pi * r ^ num__2 * ( angle / num__360 ) = num__3.14 * ( num__5.5 ^ num__2 ) * ( num__0.166666666667 ) = num__15.83 cm ^ num__2 answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ round__15.83__ |
hour_to_min_conversion__ round__15.83__ |
| after spending rs . num__5000 on rent rs . num__1500 on milk rs . num__4500 on groceries rs . num__2500 on childrens education rs . num__2000 on petrol and rs . num__700 on miscellaneous expenses mr . kishore saved num__10.0 of his monthly salary . how much did he save in rs . ? <o> a ) num__2160 <o> b ) num__2350 <o> c ) num__2000 <o> d ) num__2300 <o> e ) num__1800 |
explanation : total exp = num__5000 + num__1500 + num__4500 + num__2500 + num__2000 + num__700 = num__16200 exp in % = num__100 - num__10 = num__90.0 num__16200 = num__90.0 saving = num__10.0 = num__16200 x num__0.111111111111 = rs . num__1800 answer : e <eor> e <eos> |
e |
subtract__100.0__10.0__ divide__10.0__90.0__ subtract__2500.0__700.0__ subtract__2500.0__700.0__ |
subtract__100.0__10.0__ divide__10.0__90.0__ subtract__2500.0__700.0__ subtract__2500.0__700.0__ |
| what is the least number which when divided by num__6 num__9 num__12 and num__18 leaves remainder num__4 in each care ? <o> a ) num__30 <o> b ) num__40 <o> c ) num__36 <o> d ) num__56 <o> e ) num__66 |
explanation : lcm of num__6 num__9 num__12 and num__18 is num__36 required number = num__36 + num__4 = num__40 answer : option b <eor> b <eos> |
b |
multiply__9.0__4.0__ add__4.0__36.0__ add__4.0__36.0__ |
multiply__9.0__4.0__ add__4.0__36.0__ add__4.0__36.0__ |
| in a batch of num__125 students there are num__12 books related with different subjects to be distributed . if the total number of books bought during the time of distribution were num__1511 how many extra books will be lying with the distributor ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__2 <o> d ) num__5 <o> e ) num__3 |
num__125 x num__12 = num__1500 is the total number of books required to be distributed . as the number of books actually bought during the time of distribution num__1511 num__11 extra books were purchased . answer : b <eor> b <eos> |
b |
multiply__125.0__12.0__ subtract__1511.0__1500.0__ subtract__1511.0__1500.0__ |
multiply__125.0__12.0__ subtract__1511.0__1500.0__ subtract__1511.0__1500.0__ |
| in a mixture of num__45 litres the ratio of acid to base is num__4 : num__1 . how much base must be added to make the mixture ratio num__3 : num__2 ? <o> a ) num__72 lts <o> b ) num__24 lts <o> c ) num__15 lts <o> d ) num__1.5 lts <o> e ) num__2.5 lts |
base = num__9 litres acid = num__9 lts to make the ratio of acid : base as num__3 : num__2 num__15 litres of base must be added . answer : c <eor> c <eos> |
c |
divide__45.0__3.0__ divide__45.0__3.0__ |
divide__45.0__3.0__ divide__45.0__3.0__ |
| ( num__7895632 x num__881 ) = ? <o> a ) num__6846381250 <o> b ) num__6584638130 <o> c ) num__6584638135 <o> d ) num__6584638140 <o> e ) num__6956051792 |
num__7895632 x num__881 = num__6956051792 ans e <eor> e <eos> |
e |
multiply__7895632.0__881.0__ multiply__7895632.0__881.0__ |
multiply__7895632.0__881.0__ multiply__7895632.0__881.0__ |
| a plot is sold for rs . num__18700 with a loss of num__15.0 . at what price it should be sold to get profit of num__15.0 . <o> a ) rs num__25300 <o> b ) rs num__22300 <o> c ) rs num__24300 <o> d ) rs num__21300 <o> e ) none of above |
explanation : this type of question can be easily and quickly solved as following : let at rs x it can earn num__15.0 pr num__0 fit num__85 : num__18700 = num__115 : x [ as loss = num__100 - num__15 profit = num__100 + num__15 ] x = ( num__18700 * num__115 ) / num__85 = rs . num__25300 option a <eor> a <eos> |
a |
percent__100.0__25300.0__ |
percent__100.0__25300.0__ |
| two pipes a and b can fill a tank in num__20 and num__30 minutes respectively . if both the pipes are used together then how long will it take to fill the tank ? <o> a ) num__12 min <o> b ) num__15 min <o> c ) num__25 min <o> d ) num__50 min <o> e ) num__20 min |
part filled by a in num__1 min . = num__0.05 part filled by b in num__1 min . = num__0.0333333333333 part filled by ( a + b ) in num__1 min . = num__0.05 + num__0.0333333333333 = num__0.0833333333333 . both the pipes can fill the tank in num__12 minutes . answer : a <eor> a <eos> |
a |
divide__1.0__20.0__ divide__1.0__30.0__ add__0.05__0.0333__ round__12.0__ |
divide__1.0__20.0__ divide__1.0__30.0__ add__0.05__0.0333__ round__12.0__ |
| a retailer bought a hat at wholesale and marked it up num__60.0 to its initial price of $ num__24 . by how many more dollars does he need to increase the price to achieve a num__100.0 markup ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
let x be the wholesale price . then num__1.6 x = num__24 and x = num__24 / num__1.6 = num__15 . to achieve a num__100.0 markup the price needs to be $ num__30 . the retailer needs to increase the price by $ num__6 more . the answer is d . <eor> d <eos> |
d |
divide__24.0__1.6__ subtract__30.0__24.0__ subtract__30.0__24.0__ |
divide__24.0__1.6__ subtract__30.0__24.0__ subtract__30.0__24.0__ |
| if p x y and z are positive integers and ( x ) × ( y ) × ( z ) = p ^ num__2 which of the following could be the values of x y and z ? <o> a ) num__3 num__16 num__25 <o> b ) num__9 num__25 num__24 <o> c ) num__2 num__49 num__32 <o> d ) num__2 num__9 num__16 <o> e ) num__8 num__16 num__36 |
this is one of those questions where using the given options is fastest way to get the solution . we need to find a set where num__3 numbers can be paired into num__2 pairs of some numbers because we need a product which is a square number . c fits the bill . num__2 * num__49 * num__32 = num__2 * num__7 * num__7 * num__8 * num__4 = num__7 * num__7 * num__8 * num__8 ans c <eor> c <eos> |
c |
divide__32.0__8.0__ subtract__4.0__2.0__ |
divide__32.0__8.0__ subtract__4.0__2.0__ |
| if x ^ num__2 - num__9 = num__0 and x > num__0 which of the following must be equal to num__0 ? <o> a ) x ^ num__2 - num__9 x <o> b ) x ^ num__2 - num__9 x + num__20 <o> c ) x ^ num__2 - num__2 x + num__3 <o> d ) x ^ num__2 + num__2 x - num__3 <o> e ) x ^ num__2 - num__5 x + num__6 |
x ^ num__2 - num__9 = num__0 x = + num__3 or x = - num__3 if we substitute x = num__3 in the equation x ^ num__2 - num__5 x + num__6 = num__9 - num__15 + num__6 = num__0 e is also the answer <eor> e <eos> |
e |
add__2.0__3.0__ multiply__2.0__3.0__ add__9.0__6.0__ subtract__5.0__3.0__ |
add__2.0__3.0__ subtract__9.0__3.0__ add__9.0__6.0__ subtract__5.0__3.0__ |
| f n = num__3 ^ num__2 - num__2 ^ num__2 which of the following is not a factor of n ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__4 <o> d ) num__15 <o> e ) num__20 |
num__9 - num__4 = num__5 is prime no . any number not multiply with five is not a factor c <eor> c <eos> |
c |
add__3.0__2.0__ subtract__9.0__5.0__ |
subtract__9.0__4.0__ subtract__9.0__5.0__ |
| which of the following equations has a root in common with x ^ num__2 − num__4 x + num__3 = num__0 ? <o> a ) x ^ num__2 - num__4 x - num__4 = num__0 <o> b ) x ^ num__2 - num__2 x + num__1 = num__0 <o> c ) x ^ num__2 - x - num__1 = num__0 <o> d ) x ^ num__2 - num__10 x - num__1 = num__0 <o> e ) x ^ num__2 + num__1 = num__0 |
if we carefully look at the given equation we can arrange it in the following manner : ( x - num__1 ) ( x - num__3 ) = num__0 so the two roots are num__1 and num__3 . now put num__1 and num__3 in given equations . the equation in which one of them gives value num__0 that will be our answer . ( a ) putting num__1 : we get - num__7 ; putting num__3 we get - num__7 . reject this option . ( b ) putting num__1 : we get num__0 . this is the equation . we are lucky . no need to check other options . b is the answer . <eor> b <eos> |
b |
subtract__4.0__3.0__ add__4.0__3.0__ multiply__2.0__1.0__ |
subtract__4.0__3.0__ add__4.0__3.0__ subtract__4.0__2.0__ |
| num__2 cyclist begin training at same time a complete each lap in num__4 sec b in num__6 minutes how many minutes after start will both cyclist pass at exactly same spot when they begin to cycle <o> a ) num__160 seconds <o> b ) num__260 seconds <o> c ) num__360 seconds <o> d ) num__460 seconds <o> e ) num__560 seconds |
num__6 minutes = num__360 sec lcm of num__4 num__360 is num__360 so they meet after num__360 seconds answer : c <eor> c <eos> |
c |
round__360.0__ |
round__360.0__ |
| the sum of four consecutive odd numbers is equal to the sum of num__3 consecutive even numbers . given that the middle term of the even numbers is greater than num__101 and lesser than num__208 how many such sequences can be formed ? <o> a ) num__26 <o> b ) num__17 <o> c ) num__25 <o> d ) num__33 <o> e ) num__50 |
four consecutive odd numbers : k - num__2 k k + num__2 k + num__4 three consecutive even numbers : n - num__2 n n + num__2 k - num__2 + k + k + num__2 + k + num__4 = n - num__2 + n + n + num__2 num__4 k + num__4 = num__3 n num__4 ( k + num__1 ) = num__3 n k + num__1 = ( num__0.75 ) n k = ( num__0.75 ) n - num__1 all n ' s that ' s divisible by num__4 will have an integral k . so we need to find out how many such n ' s are available within given range : we know num__101 < n < num__208 num__104 < = n < = num__204 count = ( num__204 - num__104 ) / num__4 + num__1 = num__25.0 + num__1 = num__25 + num__1 = num__26 ans : num__26 . <eor> a <eos> |
a |
subtract__3.0__2.0__ divide__3.0__4.0__ add__3.0__101.0__ subtract__208.0__4.0__ add__1.0__25.0__ add__1.0__25.0__ |
subtract__3.0__2.0__ divide__3.0__4.0__ add__3.0__101.0__ subtract__208.0__4.0__ add__1.0__25.0__ add__1.0__25.0__ |
| a person can walk at a constant rate of num__8 mph and can bike at a rate of num__16 mph . if he wants to travel num__80 miles in num__8 hours using bike and walking at their constant rates how much distance would he require to walk ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__48 <o> d ) num__60 <o> e ) num__72 |
total distance = num__80 distance = speed * time walking speed = s num__1 = num__8 walking time = t num__1 bike speed = s num__2 = num__16 time traveled in bike = t num__2 d num__1 + d num__2 = num__80 s num__1 t num__1 + s num__2 t num__2 = num__80 num__8 * t num__1 + num__16 * t num__2 = num__80 t num__1 + num__2 * t num__2 = num__10 - - - - - ( num__1 ) given : t num__1 + t num__2 = num__8 - - - - - ( num__2 ) ( num__1 ) - ( num__2 ) - - > t num__2 = num__2 and t num__1 = num__8 - num__2 = num__6 walking distance = s num__1 * t num__1 = num__8 * num__6 = num__48 answer : c <eor> c <eos> |
c |
divide__16.0__8.0__ add__8.0__2.0__ subtract__8.0__2.0__ multiply__8.0__6.0__ round__48.0__ |
divide__16.0__8.0__ add__8.0__2.0__ subtract__8.0__2.0__ multiply__8.0__6.0__ multiply__8.0__6.0__ |
| calculate num__24 x num__99 <o> a ) num__2976 <o> b ) num__2467 <o> c ) num__2476 <o> d ) num__4276 <o> e ) num__2376 |
= num__24 - num__1 = num__23 = ( decrement each digit of the number obtained from num__9 ) here we got num__23 . now = num__9 - num__2 = num__7 and num__9 - num__3 = num__6 . so we have num__76 just write these numbers together . that is we have num__23 and num__76 . hence answer is num__2376 answer is e . <eor> e <eos> |
e |
subtract__24.0__1.0__ subtract__9.0__2.0__ add__1.0__2.0__ multiply__2.0__3.0__ subtract__99.0__23.0__ multiply__24.0__99.0__ multiply__24.0__99.0__ |
subtract__24.0__1.0__ subtract__9.0__2.0__ add__1.0__2.0__ subtract__7.0__1.0__ subtract__99.0__23.0__ multiply__24.0__99.0__ multiply__24.0__99.0__ |
| nine men went to a hotel . num__8 of them spent num__3 each over their meals and the ninth spent num__4 more than the average expenditure of all the nine . the total money spent by all of them was <o> a ) num__31.5 <o> b ) num__40 <o> c ) num__29.25 <o> d ) num__27 <o> e ) none of the above |
let the average expenditure of all the ninte be x then num__3 × num__8 + x + num__4 = num__9 x ⇒ x = num__3.5 ∴ total money spent = num__9 x = num__9 × num__3.5 = num__31.5 answer a <eor> a <eos> |
a |
multiply__3.5__9.0__ multiply__3.5__9.0__ |
multiply__3.5__9.0__ multiply__3.5__9.0__ |
| if num__1 + num__2 + num__3 + . . . + n = n ( n + num__1 ) then num__3 ( num__1 + num__3 + num__5 + . . . . + num__89 ) = ? <o> a ) num__6150 <o> b ) num__6200 <o> c ) num__6050 <o> d ) num__6075 <o> e ) num__5075 |
explanation : to solve this use the formula of ap sn = ( n / num__2 ) ( a + l ) . . . . . . . . . . . . . . . . ( num__1 ) to find n use = > tn = a + ( n - num__1 ) d = > num__89 = num__1 + ( n - num__1 ) num__2 = > n = num__45 use value of n in ( num__1 ) then sn = ( num__22.5 ) ( num__1 + num__89 ) = num__2025 ans : - num__3 ( sn ) = num__6075 answer : d <eor> d <eos> |
d |
divide__45.0__2.0__ multiply__3.0__2025.0__ multiply__1.0__6075.0__ |
divide__45.0__2.0__ multiply__3.0__2025.0__ divide__6075.0__1.0__ |
| nina has exactly enough money to purchase num__6 widgets . if the cost of each widget were reduced by $ num__1 then nina would have exactly enough money to purchase num__8 widgets . how much money does nina have ? <o> a ) $ num__22 <o> b ) $ num__24 <o> c ) $ num__30 <o> d ) $ num__36 <o> e ) $ num__40 |
b its is . let price = x ( x - num__1 ) num__8 = num__6 x x = num__4 hence total money = num__6 * num__4 = num__24 <eor> b <eos> |
b |
multiply__6.0__4.0__ multiply__6.0__4.0__ |
multiply__6.0__4.0__ multiply__6.0__4.0__ |
| a survey was sent to num__90 customers num__7 of whom responded . then the survey was redesigned and sent to another num__63 customers num__9 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? <o> a ) num__2.0 <o> b ) num__6.0 <o> c ) num__14.0 <o> d ) num__28.0 <o> e ) num__63 % |
case num__1 : ( num__0.0777777777778 ) = x / num__100 x = num__8.0 case num__2 : ( num__0.142857142857 ) = y / num__100 y = num__14.0 so percent increase is = ( y - x ) = ( num__14 - num__8 ) % = num__6.0 answer is b <eor> b <eos> |
b |
percent__100.0__6.0__ |
percent__100.0__6.0__ |
| two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of num__18 kmph and num__21 kmph respectively . when they meet it is found that one train has traveled num__60 km more than the other one . the distance between the two stations is ? <o> a ) num__288 <o> b ) num__468 <o> c ) num__877 <o> d ) num__278 <o> e ) num__178 |
num__1 h - - - - - num__5 ? - - - - - - num__60 num__12 h rs = num__18 + num__21 = num__39 t = num__12 d = num__39 * num__12 = num__468 answer : b <eor> b <eos> |
b |
divide__60.0__5.0__ add__18.0__21.0__ multiply__39.0__12.0__ round__468.0__ |
divide__60.0__5.0__ add__18.0__21.0__ multiply__39.0__12.0__ multiply__39.0__12.0__ |
| the length of a rectangular plot is num__20 metres more than its breadth . if the cost of fencing the plot @ rs . num__26.50 per metre is rs . num__6095 what is the length of the plot in metres ? <o> a ) num__333 <o> b ) num__200 <o> c ) num__230 <o> d ) num__276 <o> e ) num__1999 |
let length of plot = l meters then breadth = l - num__20 meters and perimeter = num__2 [ l + l - num__20 ] = [ num__4 l - num__40 ] meters [ num__4 l - num__40 ] * num__26.50 = num__6095 [ num__4 l - num__40 ] = num__6095 / num__26.50 = num__230 num__4 l = num__270 l = num__67.5 = num__67.5 meters . answer : c <eor> c <eos> |
c |
multiply__20.0__2.0__ divide__6095.0__26.5__ add__230.0__40.0__ divide__270.0__4.0__ round__230.0__ |
multiply__20.0__2.0__ divide__6095.0__26.5__ add__230.0__40.0__ divide__270.0__4.0__ divide__6095.0__26.5__ |
| john ' s grade was num__80 th percentile of the num__100 grades for his class . of the num__60 grades from another class num__20 were higher than john ' s and the rest were lower . if no other grade was the same as john ' s grade then john ' s grade was what percentile of the grades of the two classes of two classes combined ? <o> a ) num__72 nd <o> b ) num__75 th <o> c ) num__83 rd <o> d ) num__85 th <o> e ) num__94 th |
num__72 nd / num__75 th / num__83 rd / num__85 th / num__94 th correct option : b <eor> b <eos> |
b |
percent__100.0__75.0__ |
percent__100.0__75.0__ |
| ajay can walk num__6 km in num__1 hour . in how many hours he can walk num__70 km ? <o> a ) num__5 hrs <o> b ) num__11.6 hrs <o> c ) num__15.6 hrs <o> d ) num__20.1 hrs <o> e ) num__30 hrs |
num__1 hour he walk num__6 km he walk num__70 km in = num__11.6666666667 * num__1 = num__11.6 hours answer is b <eor> b <eos> |
b |
divide__70.0__6.0__ round__11.6__ |
divide__70.0__6.0__ multiply__1.0__11.6__ |
| what is the average ( arithmetic mean ) of all multiples of num__10 from num__10 to num__500 inclusive ? <o> a ) num__190 <o> b ) num__255 <o> c ) num__200 <o> d ) num__205 <o> e ) num__210 |
this question can be solved with the average formula and ' bunching . ' we ' re asked for the average of all of the multiples of num__10 from num__10 to num__500 inclusive . to start we can figure out the total number of terms rather easily : num__1 ( num__10 ) = num__10 num__2 ( num__10 ) = num__20 . . . num__50 ( num__10 ) = num__500 so we know that there are num__50 total numbers . we can now figure out the sum of those numbers with ' bunching ' : num__10 + num__500 = num__510 num__20 + num__490 = num__510 num__30 + num__480 = num__510 etc . since there are num__50 total terms this pattern will create num__25 ' pairs ' of num__510 . thus since the average = ( sum of terms ) / ( number of terms ) we have . . . ( num__25 ) ( num__510 ) / ( num__50 ) = num__255 final answer : b <eor> b <eos> |
b |
multiply__10.0__2.0__ divide__500.0__10.0__ add__10.0__500.0__ subtract__500.0__10.0__ add__10.0__20.0__ subtract__500.0__20.0__ divide__500.0__20.0__ divide__510.0__2.0__ multiply__1.0__255.0__ |
multiply__10.0__2.0__ divide__500.0__10.0__ add__10.0__500.0__ subtract__500.0__10.0__ add__10.0__20.0__ subtract__500.0__20.0__ divide__500.0__20.0__ divide__510.0__2.0__ divide__510.0__2.0__ |
| num__135 liters of a mixture of milk and water contains in the ratio num__3 : num__2 . how much water should now be added so that the ratio of milk and water becomes num__3 : num__4 ? <o> a ) num__12 liters <o> b ) num__32 liters <o> c ) num__41 liters <o> d ) num__54 liters <o> e ) num__34 liters |
milk = num__0.6 * num__135 = num__81 liters water = num__54 liters num__81 : ( num__54 + p ) = num__3 : num__4 num__162 + num__3 p = num__324 = > p = num__54 num__54 liters of water are to be added for the ratio become num__3 : num__4 . answer : d <eor> d <eos> |
d |
multiply__135.0__0.6__ subtract__135.0__81.0__ multiply__3.0__54.0__ multiply__2.0__162.0__ subtract__135.0__81.0__ |
multiply__135.0__0.6__ subtract__135.0__81.0__ multiply__3.0__54.0__ multiply__2.0__162.0__ subtract__135.0__81.0__ |
| find the value of x when the sum of num__20 x ( num__2 x + num__15 ) ( num__3 x - num__12 ) and ( num__12 x + num__5 ) is num__674 ? <o> a ) num__15 <o> b ) num__12 <o> c ) num__18 <o> d ) num__20 <o> e ) num__22 |
sum = num__20 x + ( num__2 x + num__15 ) + ( num__3 x - num__12 ) + ( num__12 x + num__5 ) = num__37 x + num__8 num__37 x + num__8 = num__674 x = num__18.0 = num__18 answer is c <eor> c <eos> |
c |
subtract__20.0__12.0__ subtract__20.0__2.0__ subtract__20.0__2.0__ |
subtract__20.0__12.0__ subtract__20.0__2.0__ subtract__20.0__2.0__ |
| if u > num__1 and v = num__2 ^ ( u − num__1 ) then num__4 ^ u = <o> a ) num__16 v ^ num__2 <o> b ) num__4 v ^ num__2 <o> c ) v ^ num__2 <o> d ) v ^ num__0.5 <o> e ) v ^ num__0.125 |
if u > num__1 and v = num__2 ^ ( u − num__1 ) then num__4 ^ u given u > num__1 so let ' s assume u = num__2 v = num__2 ^ ( u - num__1 ) = num__2 ^ ( num__2 - num__1 ) = num__2 so v = num__2 hence num__4 ^ u = num__4 ^ num__2 = num__16 only num__1 ans . choice can satisfy this : a ) num__16 v ^ num__2 - - > clearly > num__16 b ) num__4 v ^ num__2 - - > num__4 * num__2 ^ num__2 = num__16 ( we can stop after this as there can be only num__1 right answer ) c ) v ^ num__2 - - > clearly < num__16 d ) v ^ num__0.5 - - > clearly < num__16 e ) v ^ num__0.125 - - > clearly < num__16 ans . b ) num__4 v ^ num__2 <eor> b <eos> |
b |
reverse__2.0__ divide__2.0__16.0__ multiply__1.0__4.0__ |
reverse__2.0__ divide__2.0__16.0__ multiply__1.0__4.0__ |
| what is the sum of the squares of the first num__20 natural numbers ( num__1 to num__20 ) ? <o> a ) num__2870 <o> b ) num__2000 <o> c ) num__5650 <o> d ) num__6650 <o> e ) num__7650 |
n ( n + num__1 ) ( num__2 n + num__1 ) / num__6 num__20 ( num__21 ) ( num__21 ) / num__6 = num__2870 answer : a <eor> a <eos> |
a |
surface_cube__1.0__ multiply__1.0__2870.0__ |
surface_cube__1.0__ multiply__1.0__2870.0__ |
| a deer is standing num__60 meters in from the west end of a tunnel . the deer sees a train approaching from the west at a constant speed ten times the speed the deer can run . the deer reacts by running toward the train and clears the exit when the train is num__33 meters from the tunnel . if the deer ran in the opposite direction it would barely escape out the eastern entrance just as the train came out of the eastern entrance . how long is the tunnel in meters ? <o> a ) num__127 <o> b ) num__132 <o> c ) num__137 <o> d ) num__142 <o> e ) num__147 |
let x be the length of the tunnel . when the deer runs num__60 meters west the train goes num__600 meters to a point num__33 meters from the west entrance of the tunnel . when the deer runs east the deer runs x - num__60 meters while the train goes x + num__600 + num__33 meters . x + num__600 + num__33 = num__10 ( x - num__60 ) num__9 x = num__1233 x = num__137 meters the answer is c . <eor> c <eos> |
c |
divide__600.0__60.0__ divide__1233.0__9.0__ round__137.0__ |
divide__600.0__60.0__ divide__1233.0__9.0__ round__137.0__ |
| the value of ( num__10 ^ num__8 - num__10 ^ num__2 ) / ( num__10 ^ num__5 - num__10 ^ num__3 ) is closest to which of the following ? <o> a ) num__1 <o> b ) num__10 <o> c ) num__10 ^ num__2 <o> d ) num__10 ^ num__3 <o> e ) num__10 ^ num__4 |
because the question asks for what value isclosestthe question invites approximation . let ' s look at the numerator : num__10 ^ num__8 - num__10 ^ num__2 num__10 ^ num__8 is huge compared to num__10 ^ num__2 . so num__10 ^ num__8 - num__10 ^ num__2 is very close to num__10 ^ num__8 itself . ( just as num__100 - num__0.0001 is very close to num__100 itself ) . likewise num__10 ^ num__5 is huge compared to num__10 ^ num__3 . so num__10 ^ num__5 - num__10 ^ num__3 is very close to num__10 ^ num__5 . so we have : num__10 ^ num__0.8 ^ num__5 or num__10 ^ ( num__8 - num__5 ) = num__10 ^ num__3 . choose d . <eor> d <eos> |
d |
divide__8.0__10.0__ add__8.0__2.0__ |
divide__8.0__10.0__ add__8.0__2.0__ |
| what is the units digit of ( num__6 ! * num__4 ! + num__6 ! * num__5 ! ) / num__12 ? <o> a ) num__0 <o> b ) num__3 <o> c ) num__2 <o> d ) num__1 <o> e ) num__4 |
( num__6 ! * num__4 ! + num__6 ! * num__5 ! ) / num__12 = num__6 ! ( num__4 ! + num__5 ! ) / num__12 = num__720 ( num__24 + num__120 ) / num__12 = ( num__720 * num__144 ) / num__12 = num__720 * num__12 units digit of the above product will be equal to num__0 answer a <eor> a <eos> |
a |
multiply__6.0__4.0__ multiply__5.0__24.0__ multiply__6.0__24.0__ multiply__6.0__0.0__ |
multiply__6.0__4.0__ multiply__5.0__24.0__ multiply__6.0__24.0__ multiply__6.0__0.0__ |
| a batsman scores num__26 runs and increases his average from num__14 to num__15 . find the runs to be made if he wants top increasing the average to num__19 in the same match ? <o> a ) num__74 <o> b ) num__25 <o> c ) num__27 <o> d ) num__91 <o> e ) num__11 |
number of runs scored more to increse the ratio by num__1 is num__26 - num__14 = num__12 to raise the average by one ( from num__14 to num__15 ) he scored num__12 more than the existing average . therefore to raise the average by five ( from num__14 to num__19 ) he should score num__12 x num__5 = num__60 more than the existing average . thus he should score num__14 + num__60 = num__74 . answer : a <eor> a <eos> |
a |
subtract__15.0__14.0__ subtract__26.0__14.0__ subtract__19.0__14.0__ multiply__5.0__12.0__ add__14.0__60.0__ add__14.0__60.0__ |
subtract__15.0__14.0__ subtract__26.0__14.0__ subtract__19.0__14.0__ multiply__5.0__12.0__ add__14.0__60.0__ add__14.0__60.0__ |
| the difference between the length and breadth of a rectangle is num__24 m . if its perimeter is num__210 m then its area is : <o> a ) num__1520 m num__2 <o> b ) num__2420 m num__2 <o> c ) num__2480 m num__2 <o> d ) num__2520 m num__2 <o> e ) num__2600 m num__2 |
we have : ( l - b ) = num__25 and num__2 ( l + b ) = num__210 or ( l + b ) = num__105 solving the two equations we get : l = num__65 and b = num__40 area = ( l x b ) = ( num__65 x num__40 ) m num__2 = num__2600 m num__2 hence e <eor> e <eos> |
e |
multiply__65.0__40.0__ multiply__65.0__40.0__ |
multiply__65.0__40.0__ multiply__65.0__40.0__ |
| in a race of num__1000 m a can beat by num__100 m in a race of num__800 m b can beat c by num__100 m . by how many meters will a beat c in a race of num__1000 m ? <o> a ) num__122.9 m <o> b ) num__127.5 m . <o> c ) num__122.2 m <o> d ) num__222.9 m <o> e ) num__212.5 m |
when a runs num__1000 m b runs num__900 m and when b runs num__800 m c runs num__700 m . when b runs num__900 m distance that c runs = ( num__900 * num__700 ) / num__800 = num__787.5 = num__787.5 m . in a race of num__1000 m a beats c by ( num__1000 - num__787.5 ) = num__212.5 m to c . in a race of num__1000 m the number of meters by which a beats c = ( num__1000 * num__212.5 ) / num__1000 = num__212.5 m . answer : e <eor> e <eos> |
e |
subtract__1000.0__100.0__ subtract__800.0__100.0__ subtract__1000.0__787.5__ round__212.5__ |
subtract__1000.0__100.0__ subtract__800.0__100.0__ subtract__1000.0__787.5__ subtract__1000.0__787.5__ |
| a grocery shop has a sale of rs . num__6635 rs . num__6927 rs . num__6855 rs . num__7230 and rs . num__6562 for num__5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . num__6500 ? <o> a ) num__4857 <o> b ) num__4184 <o> c ) num__4012 <o> d ) num__4791 <o> e ) num__5291 |
let the sale in the sixth month = x then ( num__6635 + num__6927 + num__6855 + num__7230 + num__6562 + x ) / num__6 = num__6500 = > num__6635 + num__6927 + num__6855 + num__7230 + num__6562 + x = num__6 × num__6500 = > num__34209 + x = num__39000 = > x = num__39000 − num__34209 = num__4791 answer : d <eor> d <eos> |
d |
multiply__6500.0__6.0__ subtract__39000.0__34209.0__ subtract__39000.0__34209.0__ |
multiply__6500.0__6.0__ subtract__39000.0__34209.0__ subtract__39000.0__34209.0__ |
| albert and bob are painting rooms at constant but different rates . albert takes num__1 hour longer than bob to paint n rooms . working side by side they can paint a total of num__3 n / num__4 rooms in num__1.33333333333 hours . how many hours would it take albert to paint num__3 n rooms by himself ? <o> a ) num__7 <o> b ) num__9 <o> c ) num__12 <o> d ) d . num__13 <o> e ) num__15 |
rate of bob to paint n rooms is n / t ( bob ) rate of albert to paint n rooms is n / t ( albert ) . albert paints the room num__1 hour slower than bob so t ( albert ) = t ( bob ) - num__1 together they paint the num__3 n / num__5 rooms in num__1.33333333333 hours . rate is equal to work over time therefore n / x + n / x - num__1 = ( num__3 n / num__5 ) / ( num__1.33333333333 ) = n / x + n / x - num__1 = num__9 n / num__20 . fastest way for me is to think how would make the denominator num__20 . num__4 * num__5 = num__20 and it fits x and x - num__1 or you can solve the quadratic num__4 n / num__20 + num__5 n / num__20 = num__9 n / num__20 . therefore you know it takes albert num__5 hours to paint n room since albert ' s rate is n / num__5 . num__4 * num__3 = num__3 n num__12 = num__3 n . answer is c <eor> c <eos> |
c |
add__1.0__4.0__ add__4.0__5.0__ multiply__4.0__5.0__ multiply__3.0__4.0__ round__12.0__ |
add__1.0__4.0__ add__4.0__5.0__ multiply__4.0__5.0__ multiply__3.0__4.0__ multiply__1.0__12.0__ |
| jar x is num__0.333333333333 full of water . jar y which has half the capacity of jar x is num__0.5 full of water . if the water in jar y is poured into jar x then jar x will be filled to what fraction of its capacity ? <o> a ) num__0.833333333333 <o> b ) num__0.416666666667 <o> c ) num__0.583333333333 <o> d ) num__0.388888888889 <o> e ) num__0.611111111111 |
let p be the capacity of jar x . the amount of water in jar y is num__0.5 * p / num__2 = p / num__4 then the total amount in jar x is p / num__3 + p / num__4 = num__7 p / num__12 the answer is c . <eor> c <eos> |
c |
reverse__0.5__ divide__2.0__0.5__ add__3.0__4.0__ multiply__3.0__4.0__ divide__7.0__12.0__ |
reverse__0.5__ divide__2.0__0.5__ add__3.0__4.0__ multiply__3.0__4.0__ divide__7.0__12.0__ |
| the ratio of the volumes of two cubes is num__729 : num__125 . what is the ratio of their total surface areas ? <o> a ) num__81 : num__21 <o> b ) num__81 : num__22 <o> c ) num__81 : num__24 <o> d ) num__81 : num__25 <o> e ) num__81 : num__29 |
ratio of the sides = num__3 √ num__729 : num__3 √ num__125 = num__9 : num__5 ratio of surface areas = num__9 ^ num__2 : num__5 ^ num__2 = num__81 : num__25 answer : option d <eor> d <eos> |
d |
power__9.0__2.0__ power__5.0__2.0__ triangle_area__2.0__81.0__ |
power__9.0__2.0__ power__5.0__2.0__ power__9.0__2.0__ |
| - num__84 x num__19 + num__100 = ? <o> a ) num__2436 <o> b ) num__2801 <o> c ) - num__1496 <o> d ) - num__2071 <o> e ) none of them |
given exp . = - num__84 x ( num__20 - num__1 ) + num__100 = - ( num__84 x num__20 ) + num__84 + num__100 = - num__1680 + num__184 = - num__1496 answer is c <eor> c <eos> |
c |
subtract__20.0__19.0__ multiply__84.0__20.0__ add__84.0__100.0__ subtract__1680.0__184.0__ multiply__1.0__1496.0__ |
subtract__20.0__19.0__ multiply__84.0__20.0__ add__84.0__100.0__ subtract__1680.0__184.0__ subtract__1680.0__184.0__ |
| a train of num__24 carriages each of num__60 meters length when an engine also of num__60 meters length is running at a speed of num__60 kmph . in what time will the train cross a bridge num__4.5 km long ? <o> a ) num__4 <o> b ) num__3 <o> c ) num__5 <o> d ) num__6 <o> e ) num__9 |
d = num__25 * num__60 + num__4500 = num__6000 m t = num__100.0 * num__3.6 = num__360 sec = num__6 mins answer : d <eor> d <eos> |
d |
divide__6000.0__60.0__ multiply__100.0__3.6__ divide__360.0__60.0__ round__6.0__ |
divide__6000.0__60.0__ multiply__100.0__3.6__ divide__360.0__60.0__ round__6.0__ |
| in an election a candidate who gets num__60.0 of the votes is elected by a majority of num__900 votes . what is the total number of votes polled ? <o> a ) a ) num__4500 <o> b ) b ) num__5200 <o> c ) c ) num__6900 <o> d ) d ) num__7520 <o> e ) e ) num__6000 |
let the total number of votes polled be x then votes polled by other candidate = ( num__100 - num__60 ) % of x = num__40.0 of x num__60.0 of x - num__40.0 of x = num__900 num__20 x / num__100 = num__900 x = num__900 * num__5.0 = num__4500 answer is a <eor> a <eos> |
a |
percent__100.0__4500.0__ |
percent__100.0__4500.0__ |
| the speed of a boat in still water is num__20 km / hr and the rate of current is num__5 km / hr . the distance traveled downstream in num__15 minutes is : <o> a ) num__9.25 <o> b ) num__5.25 <o> c ) num__7.25 <o> d ) num__6.25 <o> e ) num__5.1 |
explanation : speed downstream = ( num__20 + num__5 ) kmph = num__25 kmph distance travelled = ( num__25 * ( num__0.25 ) ) km = num__6.25 km . answer : d <eor> d <eos> |
d |
add__20.0__5.0__ divide__5.0__20.0__ multiply__0.25__25.0__ round__6.25__ |
add__20.0__5.0__ divide__5.0__20.0__ multiply__0.25__25.0__ multiply__0.25__25.0__ |
| the average weight of a b and c is num__45 kg . if the average weight of a and b be num__40 kg and that of b and c be num__43 kg then the weight of b is ? <o> a ) num__65 kg <o> b ) num__26 kg <o> c ) num__16 kg <o> d ) num__31 kg <o> e ) num__18 kg |
let a b c represent their respective weights . then we have : a + b + c = ( num__45 * num__3 ) = num__135 - - - ( i ) a + b = ( num__40 * num__2 ) = num__80 - - - ( ii ) b + c = ( num__43 * num__2 ) = num__86 - - - ( iii ) adding ( ii ) and ( iii ) we get : a + num__2 b + c = num__166 - - - ( iv ) subtracting ( i ) from ( iv ) we get : b = num__31 b ' s weight = num__31 kg . answer : d <eor> d <eos> |
d |
subtract__43.0__40.0__ multiply__45.0__3.0__ subtract__45.0__43.0__ multiply__40.0__2.0__ multiply__43.0__2.0__ add__80.0__86.0__ subtract__166.0__135.0__ subtract__166.0__135.0__ |
subtract__43.0__40.0__ multiply__45.0__3.0__ subtract__45.0__43.0__ multiply__40.0__2.0__ multiply__43.0__2.0__ add__80.0__86.0__ subtract__166.0__135.0__ subtract__166.0__135.0__ |
| one man or two women or three boys can do a work in num__44 days then one man one women and one boy together ? <o> a ) num__24 days <o> b ) num__25 days <o> c ) num__26 days <o> d ) num__27 days <o> e ) num__28 days |
num__1 man or num__2 women or num__3 boys can do in num__44 days . lcm of num__12 & num__3 is = num__6 each one do the unit of work num__63 and num__2 units . add the units ( num__6 + num__3 + num__2 = num__11 ) they can do the work together num__44 * num__0.545454545455 = num__24 days . answer : a <eor> a <eos> |
a |
add__1.0__2.0__ multiply__2.0__3.0__ subtract__12.0__1.0__ divide__6.0__11.0__ multiply__2.0__12.0__ round__24.0__ |
add__1.0__2.0__ multiply__2.0__3.0__ subtract__12.0__1.0__ divide__6.0__11.0__ multiply__2.0__12.0__ multiply__2.0__12.0__ |
| solve this logic number sequence puzzle by the correct digit num__8080 = num__6 num__1357 = num__0 num__2022 = num__1 num__1999 = num__3 num__6666 = ? <o> a ) num__4 <o> b ) num__2 <o> c ) num__5 <o> d ) num__7 <o> e ) num__8 |
a num__4 no of circles in the number . <eor> a <eos> |
a |
add__1.0__3.0__ add__1.0__3.0__ |
add__1.0__3.0__ add__1.0__3.0__ |
| a train num__90 m long is running with a speed of num__60 km / hr . in what time will it pass a man who is running at num__6 km / hr in the direction opposite to that in which the train is going ? <o> a ) num__7 <o> b ) num__6 <o> c ) num__8 <o> d ) num__5 <o> e ) num__4 |
speed of train relative to man = num__60 + num__6 = num__66 km / hr . = num__66 * num__0.277777777778 = num__18.3333333333 m / sec . time taken to pass the men = num__90 * num__0.0545454545455 = num__5 sec . answer d <eor> d <eos> |
d |
add__60.0__6.0__ round__5.0__ |
add__60.0__6.0__ round__5.0__ |
| in the city of san durango num__60 people own cats dogs or rabbits . if num__30 people owned cats num__40 owned dogs num__10 owned rabbits and num__10 owned exactly two of the three types of pet how many people owned all three ? <o> a ) num__2 <o> b ) num__5 <o> c ) num__8 <o> d ) num__12 <o> e ) num__32 |
you are considering a case when cat dogs and rabbit are not exactly equal to num__12 . the solution shall be num__60 = num__30 + answer b <eor> b <eos> |
b |
divide__60.0__12.0__ |
divide__60.0__12.0__ |
| a train passes a station platform in num__36 sec and a man standing on the platform in num__20 sec . if the speed of the train is num__54 km / hr . what is the length of the platform ? <o> a ) num__280 m <o> b ) num__240 m <o> c ) num__152 m <o> d ) num__639 m <o> e ) num__239 m |
speed = num__54 * num__0.277777777778 = num__15 m / sec . length of the train = num__15 * num__20 = num__300 m . let the length of the platform be x m . then ( x + num__300 ) / num__36 = num__15 = > x = num__240 m . answer : b <eor> b <eos> |
b |
multiply__20.0__15.0__ round__240.0__ |
multiply__20.0__15.0__ round__240.0__ |
| how many days are there in x weeks x days ? <o> a ) num__7 x <o> b ) num__8 x <o> c ) num__14 x <o> d ) num__2 x <o> e ) num__1 x |
explanation : x weeks x days = ( num__7 x + x ) days = num__8 x days . answer : b <eor> b <eos> |
b |
round__8.0__ |
round__8.0__ |
| x can finish a work in num__18 days . y can finish the same work in num__15 days . yworked for num__10 days and left the job . how many days does x alone need to finish the remaining work ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
work done by x in num__1 day = num__0.0555555555556 work done by y in num__1 day = num__0.0666666666667 work done by y in num__10 days = num__0.666666666667 = num__0.666666666667 remaining work = num__1 – num__0.666666666667 = num__0.333333333333 number of days in which x can finish the remaining work = ( num__0.333333333333 ) / ( num__0.0555555555556 ) = num__6 c <eor> c <eos> |
c |
divide__1.0__18.0__ divide__1.0__15.0__ divide__10.0__15.0__ subtract__1.0__0.6667__ round__6.0__ |
divide__1.0__18.0__ divide__1.0__15.0__ divide__10.0__15.0__ subtract__1.0__0.6667__ divide__6.0__1.0__ |
| in what time will a train num__100 m long cross an electric pole it its speed be num__144 km / hr ? <o> a ) num__2.5 sec <o> b ) num__3.9 sec <o> c ) num__2.9 sec <o> d ) num__2.7 sec <o> e ) num__2.3 sec |
speed = num__144 * num__0.277777777778 = num__40 m / sec time taken = num__2.5 = num__2.5 sec . answer : a <eor> a <eos> |
a |
divide__100.0__40.0__ round__2.5__ |
divide__100.0__40.0__ divide__100.0__40.0__ |
| brand x coffee costs twice as much as brand y coffee . if a certain blend is num__0.25 brand x and num__0.75 brand y . what fraction of the cost of the blend is brand x ? <o> a ) a . num__0.4 <o> b ) b . num__0.2 <o> c ) num__3 . num__0.5 <o> d ) num__4 . num__0.666666666667 <o> e ) num__5 . num__0.75 |
assume cost of x = cx = num__200 cost of y = cy = num__100 the blend contains num__0.25 x and num__0.75 y - - - > cost of the blend = cbl = num__0.25 * num__200 + num__0.75 * num__100 = num__125 thus fraction of x in the blend = num__0.25 * num__1.6 = num__0.4 . answer : a <eor> a <eos> |
a |
divide__200.0__125.0__ multiply__0.25__1.6__ multiply__0.25__1.6__ |
divide__200.0__125.0__ multiply__0.25__1.6__ multiply__0.25__1.6__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__180 m and they cross each other in num__12 sec then the speed of each train is ? <o> a ) num__54 <o> b ) num__77 <o> c ) num__36 <o> d ) num__88 <o> e ) num__21 |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__180 + num__180 ) / num__12 = > x = num__15 speed of each train = num__10 m / sec . = num__15 * num__3.6 = - num__54 km / hr . answer : a <eor> a <eos> |
a |
divide__180.0__12.0__ subtract__12.0__2.0__ multiply__15.0__3.6__ round__54.0__ |
divide__180.0__12.0__ subtract__12.0__2.0__ multiply__15.0__3.6__ multiply__15.0__3.6__ |
| dave has d books which is num__3 times as many as jercy and num__0.5 as many as pavi . how many books do the three of them have altogether in terms of d ? <o> a ) num__0.833333333333 * d <o> b ) num__2.33333333333 * d <o> c ) num__4.5 * d <o> d ) num__3.5 * d <o> e ) num__3.33333333333 * d |
although we could plug in a real value for d the problem can be just as easily solved by setting up equations . however let ’ s start by defining some variables . since we are given that david has d books we can use variable d to represent how many books david has . number of books dave has = d number of books jercy has = j number of books pavi has = p we are given that dave has num__3 times as many books as jercy . we can now express this in an equation . d = num__3 j d / num__3 = j we are also given that dave has ½ as many books as pavi . we can also express this in an equation . d = ( num__0.5 ) p num__2 d = p notice that we immediately solved forj in terms of d and p in terms of d . getting j and p in terms of d is useful when setting up our final expression . we need to determine in terms of d the sum of the number of books for david jeff and paula . thus we have : d + d / num__3 + num__2 d getting a common denominator of num__3 we have : num__3 d / num__3 + d / num__3 + num__6 d / num__3 = num__10 d / num__3 = num__3.33333333333 * d the answer is e <eor> e <eos> |
e |
reverse__0.5__ divide__3.0__0.5__ divide__10.0__3.0__ divide__10.0__3.0__ |
reverse__0.5__ divide__3.0__0.5__ divide__10.0__3.0__ divide__10.0__3.0__ |
| if ( num__5 ! ) ^ num__2 – ( num__4 ! ) ^ num__2 = num__2 ^ a num__3 ^ b num__13 ^ c and a b and c are integers then what is the value of a + b + c ? <o> a ) num__12 <o> b ) num__10 <o> c ) num__2 <o> d ) num__1 <o> e ) num__0 |
( num__5 ! ) ^ num__2 - ( num__4 ! ) ^ num__2 = num__2 ^ a num__3 ^ b num__13 ^ c = [ num__5 ^ num__2 * ( num__4 ! ) ^ num__2 ] - ( num__4 ! ) ^ num__2 = [ num__25 * ( num__4 ! ) ^ num__2 ] - ( num__4 ! ) ^ num__2 = num__24 * ( num__4 ! ) ^ num__2 = num__24 * ( num__24 ^ num__2 ) = num__24 ^ num__3 = [ num__2 ^ num__3 * num__3 ] ^ num__3 = num__2 ^ num__9 * num__3 ^ num__3 so a = num__9 b = num__3 c = num__0 hence a + b + c = num__9 + num__3 + num__0 = num__12 answer = a <eor> a <eos> |
a |
add__5.0__4.0__ multiply__4.0__3.0__ multiply__4.0__3.0__ |
add__5.0__4.0__ multiply__4.0__3.0__ multiply__4.0__3.0__ |
| a retailer buys num__60 pens at the market price of num__36 pens from a wholesaler if he sells these pens giving a discount of num__1.0 what is the profit % ? <o> a ) num__40 <o> b ) num__65 <o> c ) num__72 <o> d ) num__78 <o> e ) num__20 |
let the market price of each pen be $ num__1 then cost price of num__60 pens = $ num__36 selling price of num__60 pens = num__99.0 of $ num__60 = $ num__59.40 profit % = ( ( num__23.40 * num__100 ) / num__36 ) % = num__65.0 answer b <eor> b <eos> |
b |
percent__60.0__99.0__ percent__100.0__65.0__ |
percent__60.0__99.0__ percent__100.0__65.0__ |
| johnny bought six peanut butter cans at an average price ( arithmetic mean ) of num__36.5 ¢ . if johnny returned two cans to the retailer and the average price of the remaining cans was num__31.5 ¢ then what is the average price in cents of the two returned peanut butter cans ? <o> a ) num__46.5 <o> b ) num__11 <o> c ) num__47.5 <o> d ) num__66 <o> e ) num__67.5 |
total price of six cans = num__6 * num__36.5 = num__219 total price of num__4 cans = num__4 * num__31.5 = num__126 total rice of two cans = num__219 - num__124 = num__93 average price of two cans = num__46.5 = num__46.5 c another way to do it is this : assume that the four leftover cans were of num__31.5 c each . the avg was num__36.5 c initially because the two cans were num__36.5 c each and were providing another num__5 c of cost to other num__4 cans . so cost of the two cans = num__2 * num__36.5 + num__4 * num__5 = num__93 avg cost of the two cans = num__46.5 = num__46.5 c answer ( a ) <eor> a <eos> |
a |
multiply__36.5__6.0__ multiply__31.5__4.0__ subtract__219.0__126.0__ subtract__36.5__31.5__ subtract__6.0__4.0__ divide__93.0__2.0__ |
multiply__36.5__6.0__ multiply__31.5__4.0__ subtract__219.0__126.0__ subtract__36.5__31.5__ subtract__6.0__4.0__ subtract__93.0__46.5__ |
| num__5 n + num__7 > num__12 and num__7 n - num__5 < num__44 ; n must be between which numbers ? <o> a ) num__1 and num__8 <o> b ) num__2 and num__6 <o> c ) num__0 and num__9 <o> d ) num__2 and num__7 <o> e ) num__1 and num__7 |
num__5 n + num__7 > num__12 num__5 n > num__5 n > num__1 num__7 n - num__5 < num__44 num__7 n < num__49 n < num__7 so n must be between num__1 and num__7 num__1 < n < num__7 correct answer e <eor> e <eos> |
e |
add__5.0__44.0__ reverse__1.0__ |
add__5.0__44.0__ reverse__1.0__ |
| tali builds a tower using only red green and blue toy bricks in a ratio of num__4 : num__3 : num__1 . she then removes num__0.5 of the green bricks and adds num__0.333333333333 more blue bricks reducing the size of the tower by num__14 bricks . how many red bricks will she need to add in order to double the total number of bricks used to build the original tower ? <o> a ) num__82 <o> b ) num__96 <o> c ) num__110 <o> d ) num__120 <o> e ) num__192 |
tali want to double the original amount of blocks not just the red blocks the original tower had num__96 bricks the new tower has num__82 bricks and the question is asking how many extra red blocks are needed to build a tower of num__192 bricks ( double the original : num__96 * num__2 ) . num__192 - num__82 = num__110 so num__110 bricks need to be added to the new num__82 brick tower to double the original . they worded the question ashow many red bricks need to be addedbut really it could be any colour just that num__110 more bricks need to be there to equal num__192 = e <eor> e <eos> |
e |
subtract__96.0__14.0__ divide__96.0__0.5__ reverse__0.5__ add__14.0__96.0__ multiply__1.0__192.0__ |
subtract__96.0__14.0__ divide__96.0__0.5__ multiply__4.0__0.5__ subtract__192.0__82.0__ multiply__1.0__192.0__ |
| if ' a ' completes a piece of work in num__3 days which ' b ' completes it in num__5 days and ' c ' takes num__10 days to complete the same work . how long will they take to complete the work if they work together ? <o> a ) num__1.5 days <o> b ) num__4.5 days <o> c ) num__7 days <o> d ) num__9.8 days <o> e ) num__9 days |
explanation : hint : a ' s one day work = num__0.333333333333 b ' s one day work = num__0.2 c ' s one day work = num__0.1 ( a + b + c ) ' s one day work = num__0.333333333333 + num__0.2 + num__0.1 = num__1 / num__1.5 hence a b & c together will take num__1.5 days to complete the work . answer is a <eor> a <eos> |
a |
multiply__5.0__0.2__ round__1.5__ |
multiply__5.0__0.2__ divide__1.5__1.0__ |
| one train is traveling num__45 kmph and other is at num__10 meters a second . ratio of the speed of the two trains is ? <o> a ) num__5 : num__4 <o> b ) num__5 : num__9 <o> c ) num__5 : num__1 <o> d ) num__5 : num__5 <o> e ) num__5 : num__2 |
num__45 * num__0.277777777778 = num__10 num__25 : num__20 = > num__5 : num__4 answer : a <eor> a <eos> |
a |
subtract__45.0__25.0__ subtract__25.0__20.0__ divide__20.0__5.0__ round__5.0__ |
subtract__45.0__25.0__ subtract__25.0__20.0__ divide__20.0__5.0__ round__5.0__ |
| what will come in place of the x in the following number series ? num__612 num__207 num__72 num__27 num__12 x <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
num__612 num__204.0 + num__3 = num__207 num__69.0 + num__3 = num__72 num__24.0 + num__3 = num__27 num__9.0 + num__3 = num__12 num__4.0 + num__3 = num__7 b <eor> b <eos> |
b |
divide__612.0__204.0__ divide__207.0__3.0__ divide__72.0__3.0__ divide__27.0__3.0__ divide__12.0__3.0__ add__3.0__4.0__ add__3.0__4.0__ |
divide__612.0__204.0__ divide__207.0__3.0__ divide__72.0__3.0__ divide__27.0__3.0__ divide__12.0__3.0__ add__3.0__4.0__ add__3.0__4.0__ |
| a man complete a journey in num__30 hours . he travels first half of the journey at the rate of num__20 km / hr and second half at the rate of num__10 km / hr . find the total journey in km . <o> a ) num__220 km <o> b ) num__224 km <o> c ) num__230 km <o> d ) num__400 km <o> e ) num__234 km |
num__0.5 x / num__20 + num__0.5 x / num__10 = num__30 - - > x / num__20 + x / num__10 = num__60 - - > num__3 x = num__20 x num__60 - - > x = ( num__60 x num__20 ) / num__3 = num__400 km . answer : d <eor> d <eos> |
d |
divide__10.0__20.0__ hour_to_min_conversion__ divide__30.0__10.0__ round__400.0__ |
divide__10.0__20.0__ divide__30.0__0.5__ divide__30.0__10.0__ round__400.0__ |
| how many of the positive divisors f of num__120 are also multiples of num__4 not including num__120 ? <o> a ) num__3 . <o> b ) num__4 . <o> c ) num__5 . <o> d ) num__7 . <o> e ) num__8 . |
num__4 num__812 num__2024 num__4060 . ( num__7 ) is the answer other way : factors of num__120 = num__2 ^ num__3 * num__3 * num__5 separate num__2 ^ num__2 ( which means num__4 ) now calculate the number of other factors . f = num__2 * num__3 * num__5 = total positive factors are num__2 * num__2 * num__2 = num__8 this num__8 factors include num__120 so subtract num__1 from num__8 ans is num__7 = d <eor> d <eos> |
d |
subtract__7.0__4.0__ add__2.0__3.0__ multiply__4.0__2.0__ subtract__4.0__3.0__ add__4.0__3.0__ |
subtract__7.0__4.0__ add__2.0__3.0__ multiply__4.0__2.0__ subtract__4.0__3.0__ multiply__1.0__7.0__ |
| find the l . c . m of num__0.133333333333 num__0.3 num__1.2 <o> a ) num__1.2 <o> b ) num__0.666666666667 <o> c ) num__1.4 <o> d ) num__1.6 <o> e ) num__1.8 |
explanation : lcm of fractions = lcm of numerators / hcf of denominators ( lcm of num__2 num__3 num__6 ) / ( hcf of num__15 num__10 num__5 ) = num__1.2 answer : option a <eor> a <eos> |
a |
multiply__2.0__3.0__ divide__3.0__0.3__ add__2.0__3.0__ lcm__0.3__1.2__ |
multiply__2.0__3.0__ divide__3.0__0.3__ divide__6.0__1.2__ divide__6.0__5.0__ |
| if num__0.2 th of a number decreased by num__5 is num__5 then the number is <o> a ) num__25 <o> b ) num__50 <o> c ) num__60 <o> d ) num__75 <o> e ) none |
explanation : let the number be a then a / num__5 − num__5 = num__5 ⇒ a / num__5 = num__10 ⇒ a = num__50 correct option : b <eor> b <eos> |
b |
multiply__5.0__10.0__ multiply__5.0__10.0__ |
divide__10.0__0.2__ divide__10.0__0.2__ |
| in what time will a train num__45 m long cross an electric pole it its speed be num__108 km / hr ? <o> a ) num__0.5 <o> b ) num__1.25 <o> c ) num__1.5 <o> d ) num__2.5 <o> e ) num__3.5 |
speed = num__108 * num__0.277777777778 = num__30 m / sec time taken = num__1.5 = num__1.5 sec . answer : c <eor> c <eos> |
c |
divide__45.0__30.0__ round__1.5__ |
divide__45.0__30.0__ divide__45.0__30.0__ |
| in a function they are distributing noble prize . in how many ways can num__3 prizes be distributed among num__4 boys when no boy gets more than one prize ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__18 <o> d ) num__24 <o> e ) num__29 |
sol . in this case repetitions are not allowed . so the first prize can be given in num__4 ways . the second in num__3 ways and the third in num__2 ways . but fundamental principle ( num__4 x num__3 x num__2 ) ways = num__24 ways num__4 : or num__4 p = — num__4 : - num__4 x num__3 x num__2 x num__1 - num__24 ways d <eor> d <eos> |
d |
coin_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
coin_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
| a jogger running at num__9 km / hr along side a railway track is num__240 m ahead of the engine of a num__120 m long train running at num__45 km / hr in the same direction . in how much time will the train pass the jogger ? <o> a ) num__28 sec <o> b ) num__16 sec <o> c ) num__36 sec <o> d ) num__18 sec <o> e ) num__17 sec |
speed of train relative to jogger = num__45 - num__9 = num__36 km / hr . = num__36 * num__0.277777777778 = num__10 m / sec . distance to be covered = num__240 + num__120 = num__360 m . time taken = num__36.0 = num__36 sec . answer : c <eor> c <eos> |
c |
subtract__45.0__9.0__ add__240.0__120.0__ subtract__45.0__9.0__ |
subtract__45.0__9.0__ add__240.0__120.0__ subtract__45.0__9.0__ |
| mrs . rodger got a weekly raise of $ num__175 . if she gets paid every other week write an integer describing how the raise will affect her paycheck . <o> a ) $ num__204 <o> b ) $ num__231 <o> c ) $ num__156 <o> d ) $ num__175 <o> e ) $ num__200 |
let the num__1 st paycheck be x ( integer ) . mrs . rodger got a weekly raise of $ num__175 . so after completing the num__1 st week she will get $ ( x + num__175 ) . similarly after completing the num__2 nd week she will get $ ( x + num__175 ) + $ num__175 . = $ ( x + num__175 + num__175 ) = $ ( x + num__350 ) so in this way end of every week her salary will increase by $ num__175 . d <eor> d <eos> |
d |
multiply__175.0__2.0__ multiply__175.0__1.0__ |
multiply__175.0__2.0__ multiply__175.0__1.0__ |
| what is the maximum percentage discount that a merchant can offer on her marked price so that she ends up selling at no profit or loss if she had initially marked her goods up by num__50.0 ? <o> a ) num__50.0 <o> b ) num__20.0 <o> c ) num__25.0 <o> d ) num__16.67 <o> e ) num__33.33 % |
explanatory answer quick approach : assume cost price as $ num__100 and solve let the cost price of the goods to be $ num__100 . she had initially marked her goods up by num__50.0 . therefore a num__50.0 mark up would have resulted in her marked price being $ num__100 + num__50.0 of $ num__100 = $ num__100 + $ num__50 = $ num__150 . she finally sells the product at no profit or loss . i . e . she sells the product at cost price which in this case is $ num__100 . therefore she offers a discount of $ num__50 on her marked price of $ num__150 . hence the % discount offered by her = discount / marked price ∗ num__100 = num__0.333333333333 ∗ num__100 = num__33.33 choice e <eor> e <eos> |
e |
percent__100.0__33.33__ |
percent__100.0__33.33__ |
| there are some pigeons and hares in a zoo . if heads are counted there are num__200 . if legs are counted there are num__580 . the number of hares in the zoo is ? <o> a ) num__22 <o> b ) num__73 <o> c ) num__90 <o> d ) num__77 <o> e ) num__12 |
num__200 * num__2 = num__400 num__580 - - - - - num__180 num__1 - - - - num__2 ? - - - - num__180 = num__90 answer : c <eor> c <eos> |
c |
multiply__200.0__2.0__ subtract__580.0__400.0__ divide__180.0__2.0__ multiply__1.0__90.0__ |
multiply__200.0__2.0__ subtract__580.0__400.0__ divide__180.0__2.0__ multiply__1.0__90.0__ |
| varma can read a book in k minutes . what part of the book can he read in num__5 minutes ? ( k > num__8 ) <o> a ) ( k - num__8 ) / k <o> b ) k / num__8 <o> c ) num__5 / k <o> d ) k - num__8 <o> e ) k - num__6 |
option c explanation : part of the book he can read in num__1 minute = num__1 / k part of the book he can read in num__5 minutes = num__5 / k . answer : c <eor> c <eos> |
c |
round__5.0__ |
divide__5.0__1.0__ |
| line k is in the rectangular coordinate system . if the b - intercept of k is - num__2 and the y - intercept is num__3 which of the following is an equation of line k ? <o> a ) - num__3 b + num__2 y = num__6 <o> b ) num__3 b + num__2 y = - num__6 <o> c ) num__3 b - num__2 y = num__6 <o> d ) num__2 b - num__3 y = num__6 <o> e ) - num__2 b - num__3 y = num__6 |
this question can be solved in much simpler way . to find the b intercept put y = num__0 in the equation of the line . to find the y - intercept put b = num__0 in the equation of the line . so sub b = num__0 in the answer choices check whether you are getting y = num__3 . a . - num__3 b + num__2 y = num__6 - - - y = num__3 b . num__3 b + num__2 y = - num__6 - - - y = - num__3 c . num__3 b - num__2 y = num__6 - - - - - y = - num__3 d . num__2 b - num__3 y = num__6 - - - y = - num__2 e . - num__2 b - num__3 y = num__6 - - - y = - num__2 eliminate answer b c d and e . so the answer is a . <eor> a <eos> |
a |
multiply__2.0__3.0__ divide__6.0__2.0__ |
multiply__2.0__3.0__ subtract__6.0__3.0__ |
| a boeing num__757 flies a direct route from dallas tx to phoenix az and then returns to dallas tx . the flight is num__900 miles one way . it took num__2 hrs and num__15 minutes for the flight to phoenix and num__2 hrs for the flight back to dallas . what was the speed of the wind in mph ? <o> a ) num__20 mph <o> b ) num__25 mph <o> c ) num__30 mph <o> d ) num__35 mph <o> e ) num__40 mph |
explanation : change the times in the problem to minutes first . speed of airplane from dallas to phoenix : ( num__6.66666666667 ) x num__60 = num__400 mph speed of airplane from phoenix to dallas : ( num__7.5 ) x num__60 = num__450 mph wind speed : num__0.5 ( num__450 - num__400 ) = num__25.0 = num__25 mph answer : option b <eor> b <eos> |
b |
hour_to_min_conversion__ divide__15.0__2.0__ divide__900.0__2.0__ divide__7.5__15.0__ round__25.0__ |
hour_to_min_conversion__ divide__15.0__2.0__ divide__900.0__2.0__ divide__7.5__15.0__ round__25.0__ |
| num__7.0 of the total quantity of rice is lost in grinding when a country has to import num__6 million tonnes but when only num__7 num__0.75 % is lost it can import num__3 million tonnes . find the quantity of rice grown in the country . <o> a ) num__250 <o> b ) num__300 <o> c ) num__400 <o> d ) num__500 <o> e ) num__600 |
let x be the total grown quantity of wheat . according to the question ( num__7.0 of x ) + num__64 num__31.0 of x ) + num__3 num__7 x num__31 x = > num__100 + num__6 = num__400 + num__3 num__1200 x = — num__400 million tonnes rice grown c <eor> c <eos> |
c |
multiply__3.0__400.0__ divide__1200.0__3.0__ |
multiply__3.0__400.0__ divide__1200.0__3.0__ |
| line k is in the rectangular coordinate system . if the v - intercept of k is - num__2 and the y - intercept is num__3 which of the following is an equation of line k ? <o> a ) - num__3 v + num__2 y = num__6 <o> b ) num__3 v + num__2 y = - num__6 <o> c ) num__3 v - num__2 y = num__6 <o> d ) num__2 v - num__3 y = num__6 <o> e ) - num__2 v - num__3 y = num__6 |
this question can be solved in much simpler way . to find the v intercept put y = num__0 in the equation of the line . to find the y - intercept put v = num__0 in the equation of the line . so sub v = num__0 in the answer choices check whether you are getting y = num__3 . a . - num__3 v + num__2 y = num__6 - - - y = num__3 b . num__3 v + num__2 y = - num__6 - - - y = - num__3 c . num__3 v - num__2 y = num__6 - - - - - y = - num__3 d . num__2 v - num__3 y = num__6 - - - y = - num__2 e . - num__2 v - num__3 y = num__6 - - - y = - num__2 eliminate answer b c d and e . so the answer is a . <eor> a <eos> |
a |
multiply__2.0__3.0__ divide__6.0__2.0__ |
multiply__2.0__3.0__ subtract__6.0__3.0__ |
| meg and bob are among the num__3 participants in a cycling race . if each participant finishes the race and no two participants finish at the same time in how many different possible orders can the participants finish the race so that meg finishes ahead of bob ? <o> a ) num__24 <o> b ) num__3 <o> c ) num__60 <o> d ) num__90 <o> e ) num__120 |
total # of ways the race can be finished is num__3 ! . in half of the cases meg finishes ahead of bob and in other half bob finishes ahead of meg . so ways meg to finish ahead of bob is num__3 ! / num__2 = num__3 . answer : b . <eor> b <eos> |
b |
coin_space__ choose__3.0__2.0__ |
coin_space__ choose__3.0__2.0__ |
| pablo plays num__3 rounds of a game in which his chances of winning each round are num__0.25 num__0.125 and num__1 / n respectively . if n ≠ num__0 what is the probability that pablo wins the first two rounds but loses the third ? <o> a ) num__0.0625 n <o> b ) ( n - num__1 ) / num__32 n <o> c ) num__0.125 n <o> d ) ( n + num__2 ) / num__2 n <o> e ) ( num__3 n - num__2 ) / num__2 n |
num__0.25 * num__0.125 * ( num__1 - num__1 / n ) = num__0.03125 * ( n - num__1 ) / n = ( n - num__1 ) / num__32 n answer - b <eor> b <eos> |
b |
negate_prob__0.0__ |
negate_prob__0.0__ |
| on dividing num__199 by a number the quotient is num__11 and the remainder is num__1 . find the divisor . <o> a ) a ) num__12 <o> b ) b ) num__14 <o> c ) c ) num__16 <o> d ) d ) num__18 <o> e ) e ) num__22 |
d = ( d - r ) / q = ( num__199 - num__1 ) / num__11 = num__18.0 = num__18 d ) <eor> d <eos> |
d |
multiply__1.0__18.0__ |
divide__18.0__1.0__ |
| find the l . c . m . of num__72 num__108 and num__2100 . <o> a ) num__37800 <o> b ) num__37600 <o> c ) num__37200 <o> d ) num__37500 <o> e ) none of them |
num__72 = num__2 ^ num__3 x num__3 ^ num__2 num__108 = num__3 ^ num__3 x num__2 ^ num__2 num__2100 = num__2 ^ num__2 x num__5 ^ num__2 x num__3 x num__7 . l . c . m . = num__2 ^ num__3 x num__3 ^ num__3 x num__5 ^ num__2 x num__7 = num__37800 . answer is a . <eor> a <eos> |
a |
add__2.0__3.0__ add__2.0__5.0__ lcm__72.0__37800.0__ |
add__2.0__3.0__ add__2.0__5.0__ lcm__72.0__37800.0__ |
| a sum of money is to be distributed among a b c d in the proportion of num__5 : num__2 : num__4 : num__3 . if c gets rs . num__1100 more than d what is b ' s share ? <o> a ) rs . num__500 <o> b ) rs . num__1500 <o> c ) rs . num__2200 <o> d ) rs . num__2500 <o> e ) none of the above |
let the shares of a b c and d be rs . num__5 x rs . num__2 x rs . num__4 x and rs . num__3 x respectively . then num__4 x - num__3 x = num__1100 x = num__1100 . b ' s share = rs . num__2 x = rs . ( num__2 x num__1100 ) = rs . num__2200 . answer = c <eor> c <eos> |
c |
multiply__2.0__1100.0__ multiply__2.0__1100.0__ |
multiply__2.0__1100.0__ multiply__2.0__1100.0__ |
| tarun bought a t . v with num__20.0 discount on the labelled price . had he bought it with num__25.0 discount ? he would have saved rs . num__500 . at what price did he buy the t . v ? a . rs . num__5000 b . rs . num__10000 <o> a ) rs . num__5000 <o> b ) rs . num__6000 <o> c ) rs . num__7500 <o> d ) rs . num__80000 <o> e ) rs . num__10000 |
explanation : labelled price be rs . num__100 s . p in num__1 st case = rs . num__80 s . p in num__2 nd case = rs . num__75 saving is rs . num__5 labelled price = rs . num__100 saving is rs . num__500 labelled price = rs . ( num__20.0 × num__500 ) = rs . num__10000 answer : e <eor> e <eos> |
e |
percent__20.0__500.0__ percent__20.0__25.0__ percent__100.0__10000.0__ |
percent__20.0__500.0__ percent__20.0__25.0__ percent__100.0__10000.0__ |
| it is known that no more than num__7 children will be attending a party . what is the smallest number of cookies that must be brought to the party so that each child receives the same number of cookies ? <o> a ) num__220 <o> b ) num__250 <o> c ) num__320 <o> d ) num__420 <o> e ) num__560 |
no more than num__7 children attending the party means the number of children could be num__1 num__23 num__45 num__67 . we need to find a number that is divisible by each of these . out of the choices given only num__420 is divisible by each of these numbers . hence answer is d . <eor> d <eos> |
d |
multiply__1.0__420.0__ |
multiply__1.0__420.0__ |
| if y > num__0 ( num__2 y ) / num__20 + ( num__3 y ) / num__10 is what percent of y ? <o> a ) num__40.0 <o> b ) num__50.0 <o> c ) num__60.0 <o> d ) num__70.0 <o> e ) num__80 % |
soln : - can be reduced to y / num__10 + num__3 y / num__10 = num__2 y / num__5 = num__40.0 answer : a <eor> a <eos> |
a |
add__2.0__3.0__ multiply__2.0__20.0__ multiply__2.0__20.0__ |
add__2.0__3.0__ multiply__2.0__20.0__ multiply__2.0__20.0__ |
| a person covered one - fourth of the total distance at num__26 kmph and remaining distance at num__24 kmph . what is the average speed for the total distance ? <o> a ) num__91 ( num__0.333333333333 ) kmph <o> b ) num__21 ( num__0.333333333333 ) kilometre per hour <o> c ) num__81 ( num__0.333333333333 ) kmph <o> d ) num__21 ( num__0.142857142857 ) kmph <o> e ) num__23 ( num__0.333333333333 ) kmph |
let the total distance be x km total time taken = ( x / num__4 ) / num__16 + ( num__3 x / num__4 ) / num__24 = x / num__64 + x / num__32 = num__3 x / num__64 average speed = x / ( num__3 x / num__64 ) = num__21.3333333333 kmph = num__21 ( num__0.333333333333 ) kmph . answer : b <eor> b <eos> |
b |
multiply__4.0__16.0__ divide__64.0__3.0__ round_down__21.3333__ reverse__3.0__ round_down__21.3333__ |
multiply__4.0__16.0__ divide__64.0__3.0__ round_down__21.3333__ reverse__3.0__ round_down__21.3333__ |
| two pipes a and b can fill a tank in num__15 minutes and num__20 minutes respectively . both the pipes are opened together but after num__4 minutes pipe a is turned off . what is the total time required to fill the tank ? <o> a ) num__10 min . num__20 sec . <o> b ) num__11 min . num__45 sec . <o> c ) num__12 min . num__30 sec . <o> d ) num__14 min . num__40 sec . <o> e ) num__15 min . num__40 sec . |
take lcm of num__15 and num__20 and get the total capacity of tank as num__60 . then a will fill num__4 li in num__1 min and b will fill num__3 li in num__1 min so total will be num__7 . a left after num__4 min so they worked together for num__4 min so num__7 * num__4 = num__28 . now subtract it from num__60 . ( num__60 - num__38 = num__32 ) . this will be filled by b so time taken would be . . num__10 min num__20 sec . answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ subtract__4.0__1.0__ add__4.0__3.0__ multiply__4.0__7.0__ add__4.0__28.0__ add__3.0__7.0__ round__10.0__ |
multiply__15.0__4.0__ subtract__4.0__1.0__ add__4.0__3.0__ multiply__4.0__7.0__ subtract__60.0__28.0__ subtract__38.0__28.0__ subtract__20.0__10.0__ |
| the probability that a man will be alive for num__10 more yrs is num__0.25 & the probability that his wife will alive for num__10 more yrs is num__0.333333333333 . the probability that none of them will be alive for num__10 more yrs is <o> a ) num__0.5 <o> b ) num__2 <o> c ) num__0.666666666667 <o> d ) num__0.8 <o> e ) num__0.714285714286 |
sol . required probability = pg . ) x p ( b ) = ( num__1 — d x ( num__1 — i ) = : x num__1 = num__0.5 ans . ( a ) <eor> a <eos> |
a |
divide__0.25__0.5__ |
divide__0.25__0.5__ |
| the speed of a train is num__90 kmph . what is the distance covered by it in num__10 minutes ? <o> a ) num__15 kmph <o> b ) num__17 kmph <o> c ) num__18 kmph <o> d ) num__19 kmph <o> e ) num__12 kmph |
num__90 * num__0.166666666667 = num__15 kmph answer : a <eor> a <eos> |
a |
round__15.0__ |
round__15.0__ |
| for every order a certain mail - order company charges a shipping fee of $ num__3 plus an additional $ num__2 if the value of the order is over $ num__50 but not over $ num__100 or an additional $ num__3 if the value of the order is over $ num__100 . how much greater are the total shipping fees for num__2 orders of $ num__75 each than the total shipping fee for num__2 order of $ num__150 ? <o> a ) $ num__1 <o> b ) $ num__2 <o> c ) $ num__3 <o> d ) $ num__4 <o> e ) $ num__5 |
total cost involved for each num__75 $ shipment : num__3 + num__2 = num__5 $ . thus for two such orders = num__10 $ total cost involved for a num__150 $ shipment : num__3 + num__3 = num__6 $ . thus for two such orders = num__12 $ the difference = num__2 $ . b . <eor> b <eos> |
b |
add__3.0__2.0__ multiply__2.0__5.0__ multiply__3.0__2.0__ multiply__2.0__6.0__ divide__100.0__50.0__ |
add__3.0__2.0__ multiply__2.0__5.0__ multiply__3.0__2.0__ add__2.0__10.0__ divide__100.0__50.0__ |
| a tank is num__25.0 full of oil . the oil consists of num__25.0 wti crude oil and num__75.0 brent crude oil . num__0.666666666667 rd of the remainder of the tank is filled with num__400 barrels of wti crude oil . how much wti crude oil is in the tank ? <o> a ) num__425 barrels <o> b ) num__437.5 barrels <o> c ) num__450 barrels <o> d ) num__500 barrels <o> e ) num__650 barrels |
the tank was num__25.0 full . then num__0.666666666667 rd of the remainder of the tank so num__0.666666666667 rd of num__75.0 of the tank was filled with num__400 barrels of wti crude oil . num__0.666666666667 rd of num__75.0 is num__50.0 thus we are told that num__50.0 of the tank amounts for num__400 barrels which makes the capacity of the tank num__800 barrels . therefore initially the tank had num__0.25 ∗ num__0.25 ∗ num__800 = num__500 barrels of wti crude oil . after num__400 barrels of wti crude oil were added the amount of wti crude oil became num__50 + num__400 = num__450 barrels . answer : c . <eor> c <eos> |
c |
subtract__75.0__25.0__ add__400.0__50.0__ add__400.0__50.0__ |
subtract__75.0__25.0__ add__400.0__50.0__ add__400.0__50.0__ |
| in the set of positive integers from num__1 to num__70 what is the sum of all the odd multiples of num__5 ? <o> a ) num__180 <o> b ) num__245 <o> c ) num__240 <o> d ) num__200 <o> e ) num__190 |
reduce num__1 - num__70 num__5 - num__15 - num__25 - num__35 - num__45 + num__55 + num__65 are valid multiples num__0 f num__5 . add them - - > num__245 b <eor> b <eos> |
b |
subtract__70.0__25.0__ subtract__70.0__15.0__ subtract__70.0__5.0__ multiply__1.0__245.0__ |
subtract__70.0__25.0__ subtract__70.0__15.0__ subtract__70.0__5.0__ multiply__1.0__245.0__ |
| david has d books which is num__3 times as many as jeff and num__0.666666666667 as many as paula . how many books do the three of them have altogether in terms of d ? <o> a ) num__0.833333333333 * d <o> b ) num__2.33333333333 * d <o> c ) num__8.5 * d <o> d ) num__3.5 * d <o> e ) num__4.5 * d |
although we could plug in a real value for d the problem can be just as easily solved by setting up equations . however let ’ s start by defining some variables . since we are given that david has d books we can use variable d to represent how many books david has . number of books david has = d number of books jeff has = j number of books paula has = p we are given that david has num__3 times as many books as jeff . we can now express this in an equation . d = num__3 j d / num__3 = j we are also given that david has ½ as many books as paula . we can also express this in an equation . d = ( num__0.666666666667 ) p num__3 d / num__2 = p notice that we immediately solved forj in terms of d and p in terms of d . getting j and p in terms of d is useful when setting up our final expression . we need to determine in terms of d the sum of the number of books for david jeff and paula . thus we have : d + d / num__3 + num__3 d / num__2 = num__17 d / num__2 the answer is c . <eor> c <eos> |
c |
divide__17.0__2.0__ |
divide__17.0__2.0__ |
| a can do a piece of work in num__10 days . he works at it for num__4 days and then b finishes it in num__9 days . in how many days can a and b together finish the work ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__12 <o> d ) num__15 <o> e ) num__20 |
num__0.4 + num__9 / x = num__1 / x = num__15 num__0.1 + num__0.0666666666667 = num__0.166666666667 = num__6 days answer a <eor> a <eos> |
a |
divide__4.0__10.0__ subtract__10.0__9.0__ divide__1.0__10.0__ divide__1.0__15.0__ add__0.1__0.0667__ subtract__10.0__4.0__ round__6.0__ |
divide__4.0__10.0__ subtract__10.0__9.0__ divide__1.0__10.0__ divide__1.0__15.0__ add__0.1__0.0667__ subtract__10.0__4.0__ divide__6.0__1.0__ |
| a bat is bought for rs . num__400 and sold at a gain of num__20.0 find its selling price <o> a ) rs . num__480 / - <o> b ) rs . num__500 / - <o> c ) rs . num__520 / - <o> d ) rs . num__540 / - <o> e ) rs . num__560 / - |
num__100.0 - - - - - - > num__400 ( num__100 * num__4 = num__400 ) num__120.0 - - - - - - > num__480 ( num__120 * num__4 = num__480 ) selling price = rs . num__480 / - a <eor> a <eos> |
a |
percent__100.0__480.0__ |
percent__100.0__480.0__ |
| as a treat for her two crying children a mother runs to the freezer in which she has three cherry ice pops three orange ice pops and four lemon - lime ice pops . if she chooses two at random to bring outside to the children but realizes as she runs out the door that she can not bring them different flavors without one invariably being jealous of the other and getting even more upset what is the probability that she has to return to the freezer to make sure that they each receive the same flavor ? <o> a ) num__0.111111111111 <o> b ) num__0.166666666667 <o> c ) num__0.277777777778 <o> d ) num__0.0916666666667 <o> e ) num__0.833333333333 |
probability of not getting the same flavor - > favorable - > cherry - orange [ num__3 c num__1 * num__3 c num__1 [ or simply num__3 * num__3 ] or cherry - lemon [ num__3 * num__4 ] or orange - lemon [ num__3 * num__4 ] prob = ( num__3 * num__3 + num__3 * num__4 + num__3 * num__4 ) / num__9 c num__2 = num__0.0916666666667 = num__0.0916666666667 answer - > d <eor> d <eos> |
d |
add__1.0__3.0__ subtract__3.0__1.0__ multiply__1.0__0.0917__ |
add__1.0__3.0__ subtract__3.0__1.0__ multiply__1.0__0.0917__ |
| if num__4 ( p ' s capital ) = num__6 ( q ' s capital ) = num__10 ( r ' s capital ) then out of the total profit of rs num__4650 how much r will receive ? <o> a ) num__300 <o> b ) num__600 <o> c ) num__900 <o> d ) num__800 <o> e ) num__500 |
let p ' s capital be p q ' s capital be q and r ' s capital be r then num__4 p = num__6 q = num__10 r num__2 p = num__3 q = num__5 r ⋯ ( a ) from ( a ) q = num__2 p / num__3 ⋯ ( num__1 ) r = num__2 p / num__5 ⋯ ( num__2 ) p : q : r = p : num__2 p / num__3 : num__2 p / num__5 = num__15 : num__10 : num__6 r ' s share = num__4650 × num__0.193548387097 = num__150 × num__6 = num__900 answer is c . <eor> c <eos> |
c |
subtract__6.0__4.0__ divide__6.0__2.0__ divide__10.0__2.0__ subtract__4.0__3.0__ add__10.0__5.0__ multiply__10.0__15.0__ multiply__6.0__150.0__ multiply__6.0__150.0__ |
subtract__6.0__4.0__ divide__6.0__2.0__ divide__10.0__2.0__ subtract__4.0__3.0__ add__10.0__5.0__ multiply__10.0__15.0__ multiply__6.0__150.0__ divide__900.0__1.0__ |
| what will be the difference between simple and compound interest at num__10.0 per annum on a sum of rs . num__1000 after num__4 years ? <o> a ) rs . num__64.18 <o> b ) rs . num__64.67 <o> c ) rs . num__64.08 <o> d ) rs . num__64.10 <o> e ) rs . num__64.22 |
s . i . = ( num__1000 * num__10 * num__4 ) / num__100 = rs . num__400 c . i . = [ num__1000 * ( num__1 + num__0.1 ) num__4 - num__1000 ] = rs . num__464.10 difference = ( num__464.10 - num__400 ) = rs . num__64.10 answer : d <eor> d <eos> |
d |
percent__10.0__1000.0__ percent__10.0__1.0__ percent__100.0__64.1__ |
percent__10.0__1000.0__ percent__10.0__1.0__ percent__100.0__64.1__ |
| in a certain animal shelter the ratio of the number of catsto the number of dogs is num__15 to num__7 . if num__16 additional dogs were to be taken in by the shelter the ratio of the number of cats to the number of dogs would be num__15 to num__11 . how many cats are in the shelter ? <o> a ) num__15 <o> b ) num__25 <o> c ) num__30 <o> d ) num__45 <o> e ) num__60 |
this ratio question can be solved in a couple of different ways . here ' s an algebraic approach . . . we ' re told that the ratio of the number of cats to the number of dogs is num__15 : num__7 . we ' re then told that num__16 more dogs are added to this group and the ratio becomes num__15 : num__11 . we ' re asked for the number of cats . algebraically since the number of cats is a multiple of num__15 and the number of dogs is a multiple of num__7 we can write this initial relationship as . . . num__15 x / num__7 x when we add the num__16 cats and factor in the ' ending ratio ' we have an equation . . . . num__15 x / ( num__7 x + num__16 ) = num__1.36363636364 here we have num__1 variable and num__1 equation so we can solve for x . . . . ( num__15 x ) ( num__11 ) = ( num__7 x + num__16 ) ( num__15 ) ( x ) ( num__11 ) = ( num__7 x + num__16 ) ( num__1 ) num__11 x = num__7 x + num__16 num__4 x = num__16 x = num__4 with this x we can figure out the initial number of dogs and cats . . . initial dogs = num__15 x = num__15 ( num__4 ) = num__60 final answer : d <eor> d <eos> |
d |
divide__15.0__11.0__ round_down__1.3636__ subtract__15.0__11.0__ multiply__15.0__4.0__ subtract__60.0__15.0__ |
divide__15.0__11.0__ round_down__1.3636__ subtract__15.0__11.0__ multiply__15.0__4.0__ subtract__60.0__15.0__ |
| in the new budget the price of ghee rose by num__50.0 . by how much percent must a person reduce his consumption so that his expenditure on it does not increase ? <o> a ) num__7.5 <o> b ) num__9.1 <o> c ) num__10.9 <o> d ) num__12.6 <o> e ) num__33.33 % |
reduce in consumption = r / ( num__100 + r ) * num__100.0 = num__0.333333333333 * num__100 = num__33.33 answer is e <eor> e <eos> |
e |
multiply__100.0__0.3333__ multiply__100.0__0.3333__ |
multiply__100.0__0.3333__ multiply__100.0__0.3333__ |
| a train num__400 m long is running at a speed of num__78 km / hr . if it crosses a tunnel in num__1 min then the length of the tunnel is ? <o> a ) num__298 m <o> b ) num__279 m <o> c ) num__500 m <o> d ) num__289 m <o> e ) num__900 m |
speed = num__78 * num__0.277777777778 = num__21.6666666667 m / sec . time = num__1 min = num__60 sec . let the length of the train be x meters . then ( num__400 + x ) / num__60 = num__21.6666666667 x = num__900 m . answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ round__900.0__ |
hour_to_min_conversion__ multiply__1.0__900.0__ |
| the compound interest earned on a sum for the second and the third years are $ num__1400 and $ num__1498 respectively . what is the rate of interest ? <o> a ) num__3.0 <o> b ) num__5.0 <o> c ) num__7.0 <o> d ) num__9.0 <o> e ) num__11 % |
num__1498 - num__1400 = num__98 is the rate of interest on $ num__1400 for one year . the rate of interest = ( num__100 * num__98 ) / ( num__1400 ) = num__7.0 the answer is c . <eor> c <eos> |
c |
percent__100.0__7.0__ |
percent__100.0__7.0__ |
| ashok secured average of num__78 marks in num__6 subjects . if the average of marks in num__5 subjects is num__74 how many marks did he secure in the num__6 th subject ? <o> a ) num__66 <o> b ) num__74 <o> c ) num__78 <o> d ) num__98 <o> e ) none of these |
explanation : number of subjects = num__6 average of marks in num__6 subjects = num__78 therefore total marks in num__6 subjects = num__78 * num__6 = num__468 now no . of subjects = num__5 total marks in num__5 subjects = num__74 * num__5 = num__370 therefore marks in num__6 th subject = num__468 – num__370 = num__98 answer d <eor> d <eos> |
d |
multiply__78.0__6.0__ multiply__5.0__74.0__ subtract__468.0__370.0__ subtract__468.0__370.0__ |
multiply__78.0__6.0__ multiply__5.0__74.0__ subtract__468.0__370.0__ subtract__468.0__370.0__ |
| six bells start ringing together and ring at intervals of num__4 num__8 num__10 num__12 num__15 and num__20 seconds respectively . how many times will they ring together in num__60 minutes ? <o> a ) num__5 <o> b ) num__11 <o> c ) num__21 <o> d ) num__31 <o> e ) num__32 |
lcm of num__4 num__8 num__10 num__12 num__15 and num__20 = num__120 seconds = num__2 minutes that means all the six bells will ring together in every num__2 minutes number of times they will ring together in num__60 minutes = num__1 + num__30.0 = num__31 answer : d <eor> d <eos> |
d |
multiply__8.0__15.0__ divide__8.0__4.0__ add__10.0__20.0__ add__1.0__30.0__ round__31.0__ |
multiply__8.0__15.0__ divide__8.0__4.0__ add__10.0__20.0__ add__1.0__30.0__ add__1.0__30.0__ |
| num__21 ball numbered num__1 to num__21 . a ballis drawn and then another ball is drawn without replacement . <o> a ) num__0.0238095238095 <o> b ) num__0.0731707317073 <o> c ) num__0.214285714286 <o> d ) num__0.113636363636 <o> e ) num__0.106382978723 |
the probability that first toy shows the even number = num__1021 = num__1021 since the toy is not replaced there are now num__9 even numbered toys and total num__20 toys left . hence probability that second toy shows the even number = num__920 = num__920 required probability = ( num__1021 ) × ( num__920 ) = ( num__1021 ) × ( num__920 ) = num__0.214285714286 c <eor> c <eos> |
c |
subtract__21.0__1.0__ multiply__1.0__0.2143__ |
subtract__21.0__1.0__ multiply__1.0__0.2143__ |
| a man whose speed is num__4.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at num__1.5 kmph find his average speed for the total journey ? <o> a ) num__6 <o> b ) num__4 <o> c ) num__8 <o> d ) num__2 <o> e ) num__3 |
m = num__45 s = num__1.5 ds = num__6 us = num__3 as = ( num__2 * num__6 * num__3 ) / num__9 = num__4 answer : b <eor> b <eos> |
b |
add__4.5__1.5__ subtract__4.5__1.5__ divide__3.0__1.5__ multiply__4.5__2.0__ round_down__4.5__ round_down__4.5__ |
add__4.5__1.5__ divide__4.5__1.5__ divide__3.0__1.5__ multiply__4.5__2.0__ divide__6.0__1.5__ divide__6.0__1.5__ |
| a band c are three contestants in a kmrace . if a can give b a start of num__40 m and a can give c a start of num__64 m how many metre ' s start can b give c ? . <o> a ) num__540 m <o> b ) num__178 m <o> c ) num__936 m <o> d ) num__234 m <o> e ) num__713 m |
while a covers num__1000 m b covers ( num__1000 - num__40 ) m = num__960 m and c covers ( num__1000 - num__64 ) m or num__936 m . when b covers num__960 m c covers num__936 m option c <eor> c <eos> |
c |
subtract__1000.0__40.0__ subtract__1000.0__64.0__ round__936.0__ |
subtract__1000.0__40.0__ subtract__1000.0__64.0__ subtract__1000.0__64.0__ |
| a train speeds past a pole in num__15 seconds and a platform num__100 m long in num__25 seconds . its length is <o> a ) num__100 <o> b ) num__120 <o> c ) num__130 <o> d ) num__150 <o> e ) num__140 |
let the length of the train be x meters and its speed be y m / sec . they x / y = num__15 = > y = x / num__15 x + num__4.0 = x / num__15 x = num__150 m . answer : option d <eor> d <eos> |
d |
divide__100.0__25.0__ round__150.0__ |
divide__100.0__25.0__ round__150.0__ |
| a man a woman and a boy can complete a job in num__34 and num__12 days respectively . how many boys must assist num__1 man and num__1 woman to complete the job in ¼ of a day ? <o> a ) num__41 <o> b ) num__50 <o> c ) num__22 <o> d ) num__33 <o> e ) num__14 |
( num__1 man + num__1 woman ) ' s num__1 days work = num__0.333333333333 + num__0.25 = num__0.583333333333 work done by num__1 man and num__1 women n num__0.25 day = ( ( num__0.583333333333 ) * ( num__0.25 ) ) = num__0.145833333333 remaining work = num__1 - num__0.145833333333 = num__0.854166666667 work done by num__1 boy in ¼ day = ( ( num__0.0833333333333 ) * ( num__0.25 ) ) = num__0.0208333333333 therefore number of boys required = ( ( num__0.854166666667 ) * num__48 ) = num__41 days answer a <eor> a <eos> |
a |
add__0.25__0.3333__ multiply__0.25__0.5833__ subtract__1.0__0.1458__ divide__1.0__12.0__ divide__0.25__12.0__ divide__12.0__0.25__ round__41.0__ |
add__0.25__0.3333__ multiply__0.25__0.5833__ subtract__1.0__0.1458__ multiply__0.25__0.3333__ multiply__0.25__0.0833__ divide__12.0__0.25__ multiply__1.0__41.0__ |
| once upon a time in ancient times there was a king who was very found of wines . he had a huge cellar which had num__1000 different varieties of wine all in different caskets ( num__1000 caskets in all ) . in the adjoining kingdom there was a queen who was envious of the king ’ s huge wine collection . after some time when she could not bear it any more show conspired to kill her by poisoning all his wine caskets . so she sent one sentry to poison all the caskets but no sooner had the sentry poisoned only one wine casket he was caught and killed by the royal guards . now the king had a major problem in his hand so as to identify the poisonous casket which he gave to the minister . but the situation had two peculiarities . i : any one who takes even one drop from the poisonous casket will die . ii : he will die only after one month . the king also handed over few prisoners to the minister as “ taster ” of those caskets as their lives was of little value . if the minister is allowed only num__1 month to find out the poisonous casket what is the minimum number of prisoners he should use as “ tasters ” ? <o> a ) num__1023 <o> b ) num__1024 <o> c ) num__1025 <o> d ) num__1026 <o> e ) num__1027 |
sol : study the following table carefully . each of the prisoners drinks wine from whichever casket has a num__1 in his place . if he does not drink there is a num__0 . so if no one dies casket num__1 is poisoned . if p num__1 dies casket num__2 is poisoned . if p num__1 and p num__3 die ⇒ ⇒ casket num__6 is poisoned and so on . so if there are num__3 prisoners we can differentiate upto num__2323 = num__8 caskets . so for num__1000 caskets [ even num__1024 caskets ] we need only num__10 prisoners as num__210210 = num__1024 answer b <eor> b <eos> |
b |
add__1.0__2.0__ multiply__2.0__3.0__ add__2.0__6.0__ add__2.0__8.0__ multiply__1.0__1024.0__ |
add__1.0__2.0__ multiply__2.0__3.0__ add__2.0__6.0__ add__2.0__8.0__ multiply__1.0__1024.0__ |
| three numbers are in the ratio num__3 : num__5 : num__7 . the largest number is num__21 . what is the difference between smallest and largest number ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
the three numbers are num__3 x num__5 x and num__7 x . the largest number is num__21 = num__7 * num__3 so x = num__3 . the smallest number is num__3 * num__3 = num__9 . num__21 - num__9 = num__12 the answer is c . <eor> c <eos> |
c |
add__3.0__9.0__ add__3.0__9.0__ |
subtract__21.0__9.0__ subtract__21.0__9.0__ |
| shipment - - - - - - no . of defective chips / shipment - - - total chips in shipment s num__2 - - - - - - - - - - - - - - num__5 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - num__12000 s num__3 - - - - - - - - - - - - - - num__6 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - num__18000 s num__4 - - - - - - - - - - - - - - num__4 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - num__16000 a computer chip manufacturer expects the ratio of the number of defective chips to be total number of chips in all future shipments equal to the corresponding ratio for shipmemts s num__1 s num__2 s num__3 and s num__4 comined as shown in the table above . what is the expected number t of defective chips in a shipment of num__60000 chips ? <o> a ) num__14 <o> b ) num__20 <o> c ) num__22 <o> d ) num__24 <o> e ) num__25 |
i agree with your solution = num__20 . but the question is : there are different combination to get num__60000 chips . for example : num__1 * s num__3 + num__2 * s num__4 + num__2 * s num__2 . in this way we ship num__60000 chips with only num__6 + num__4 * num__2 + num__2 * num__2 = num__18 defective chips better than the average of num__20 . the question is to find the expected number t of defective chips i guess it assume the minimum # therefore it might not be num__20 . <eor> b <eos> |
b |
multiply__5.0__4.0__ multiply__3.0__6.0__ add__2.0__18.0__ |
multiply__5.0__4.0__ multiply__3.0__6.0__ add__2.0__18.0__ |
| the length of a train and that of a platform are equal . if with a speed of num__90 k / hr the train crosses the platform in one minute then the length of the train ( in metres ) is : <o> a ) num__400 <o> b ) num__525 <o> c ) num__750 <o> d ) num__850 <o> e ) none |
sol . speed = [ num__90 * num__0.277777777778 ] m / sec = num__25 m / sec ; time = num__1 min . = num__60 sec . let the length of the train and that of the platform be x metres . then num__2 x / num__60 = num__25 ⇔ x = num__25 * num__30.0 = num__750 answer c <eor> c <eos> |
c |
hour_to_min_conversion__ subtract__90.0__60.0__ multiply__25.0__30.0__ round__750.0__ |
hour_to_min_conversion__ divide__60.0__2.0__ multiply__25.0__30.0__ divide__750.0__1.0__ |
| num__16 men can do a piece of work in num__16 days . num__4 days after they started the work num__8 more men joined them . how many days will they now take to complete the remaining work <o> a ) num__7 days <o> b ) num__8 days <o> c ) num__9 days <o> d ) num__10 days <o> e ) none of these |
solution : work done by num__1 man in num__1 day = num__1 / ( num__16 * num__16 ) work completed by num__16 men in num__4 days = num__4 [ num__1 / ( num__16 * num__16 ) ] * num__16 = num__0.25 remaining work = num__1 - num__0.25 = num__0.75 total men now = num__16 + num__8 = num__24 num__24 men do num__0.75 work = > num__24 * number of days * work done by num__1 man in num__1 day = num__0.75 = > num__24 * number of days * num__1 / ( num__16 * num__16 ) = num__0.75 = > number of days = num__8 days answer b <eor> b <eos> |
b |
divide__4.0__16.0__ subtract__1.0__0.25__ add__16.0__8.0__ round__8.0__ |
divide__4.0__16.0__ subtract__1.0__0.25__ add__16.0__8.0__ subtract__16.0__8.0__ |
| a rectangular box measures internally num__1.6 m long num__1 m broad and num__60 cm deep . the number of cubical blocks each of edge num__20 cm that can be packed inside the box is <o> a ) num__30 <o> b ) num__60 <o> c ) num__120 <o> d ) num__150 <o> e ) none |
sol . number of blocks = num__160 x num__100 x num__60 num__20 x num__20 x num__20 ] = num__120 . answer c <eor> c <eos> |
c |
divide__160.0__1.6__ add__20.0__100.0__ round__120.0__ |
divide__160.0__1.6__ add__20.0__100.0__ round__120.0__ |
| virginia adrienne and dennis have taught history for a combined total of num__96 years . if virginia has taught for num__9 more years than adrienne and for num__12 fewer years than dennis for how many years has dennis taught ? <o> a ) num__23 <o> b ) num__32 <o> c ) num__35 <o> d ) num__41 <o> e ) num__42 |
let number of years taught by virginia = v number of years taught by adrienne = a number of years taught by dennis = d v + a + d = num__96 v = a + num__9 = > a = v - num__9 v = d - num__12 = > a = ( d - num__12 ) - num__9 = d - num__21 d - num__9 + d - num__21 + d = num__96 = > num__3 d = num__96 + num__30 = num__126 = > d = num__42 answer e <eor> e <eos> |
e |
add__9.0__12.0__ subtract__12.0__9.0__ add__9.0__21.0__ add__96.0__30.0__ add__12.0__30.0__ add__12.0__30.0__ |
add__9.0__12.0__ subtract__12.0__9.0__ add__9.0__21.0__ add__96.0__30.0__ add__12.0__30.0__ add__12.0__30.0__ |
| today jim is twice as old as fred and sam is num__2 years younger than fred . seven years ago jim was num__13 times as old as sam . how old is jim now ? <o> a ) num__8 <o> b ) num__12 <o> c ) num__16 <o> d ) num__20 <o> e ) num__24 |
we ' re asked how old jim is now . we ' re given three facts to work with : num__1 ) today jim is twice as old as fred num__2 ) today sam is num__2 years younger than fred num__3 ) num__7 years ago jim was num__13 times as old as sam . let ' s test answer d : num__20 if . . . . jim is currently num__20 years old . . . . fred is num__10 years old sam is num__8 years old num__7 years ago jim was num__13 and sam was num__1 so jim was num__13 times sam ' s age . this is an exact match for what we were told so this must be the answer . d <eor> d <eos> |
d |
add__2.0__1.0__ add__13.0__7.0__ subtract__13.0__3.0__ add__1.0__7.0__ round__20.0__ |
add__2.0__1.0__ add__13.0__7.0__ subtract__13.0__3.0__ add__1.0__7.0__ round__20.0__ |
| in how many different ways can the letters b e n e f i c i a l be arranged ? <o> a ) num__10 × num__9 × num__8 × num__7 × num__6 × num__5 × num__3 × num__2 <o> b ) num__10 × num__9 × num__8 × num__7 × num__6 × num__5 × num__4 × num__3 <o> c ) num__8 ! <o> d ) num__9 ! <o> e ) num__10 ! |
number of ways of arranging beneficial = num__10 ! / num__2 ! * num__2 ! = num__10 × num__9 × num__8 × num__7 × num__6 × num__5 × num__3 × num__2 . hence a <eor> a <eos> |
a |
coin_space__ die_space__ vowel_space__ choose__10.0__9.0__ |
coin_space__ die_space__ vowel_space__ choose__10.0__9.0__ |
| a computer program assigns consecutive numbers to the days of the week . sunday is num__1 monday is num__2 . . . . . and saturday is num__7 . every day the computer program calculates the value of parameter d according to the following definition : d is the number of the day times the value of d on the previous day . if the computer calculated d to be num__12 on wednesday what would be the value of d calculated on the following saturday ? <o> a ) num__2140 <o> b ) num__2520 <o> c ) num__2960 <o> d ) num__3360 <o> e ) num__3710 |
on saturday the value would be num__7 * num__6 * num__5 * num__12 = num__2520 the answer is b . <eor> b <eos> |
b |
subtract__7.0__1.0__ subtract__7.0__2.0__ round__2520.0__ |
subtract__7.0__1.0__ subtract__7.0__2.0__ multiply__1.0__2520.0__ |
| the monthly rent of a shop of dimension num__20 feet × num__18 feet is rs . num__1440 . what is the annual rent per square foot of the shop ? <o> a ) num__48 <o> b ) num__56 <o> c ) num__68 <o> d ) num__87 <o> e ) num__92 |
sol . monthly rent per square feet = num__1440 / ( num__20 * num__18 ) = num__4 & annual rent per square feet = num__12 * num__4 = num__48 answer : a <eor> a <eos> |
a |
square_perimeter__12.0__ square_perimeter__12.0__ |
multiply__4.0__12.0__ multiply__4.0__12.0__ |
| find the principal which yields a simple interest of rs . num__20 and compound interest of rs . num__21 in two years at the same percent rate per annum ? <o> a ) rs . num__200 <o> b ) rs . num__209 <o> c ) rs . num__202 <o> d ) rs . num__238 <o> e ) rs . num__218 |
explanation : si in num__2 years = rs . num__20 si in num__1 year = rs . num__10 ci in num__2 years = rs . num__21.0 rate per annum = [ ( ci – si ) / ( si in num__1 year ) ] * num__100 = [ ( num__21 – num__20 ) / num__20 ] * num__100 = num__5.0 p . a . let the principal be rs . x time = t = num__2 years % rate = num__5.0 p . a . si = ( prt / num__100 ) num__20 = ( x * num__5 * num__2 ) / num__100 x = rs . num__200 answer : a <eor> a <eos> |
a |
percent__100.0__200.0__ |
percent__100.0__200.0__ |
| if n = num__8 ^ num__9 – num__8 what is the units digit of n ? <o> a ) num__4 <o> b ) num__0 <o> c ) num__1 <o> d ) num__2 <o> e ) num__3 |
num__8 ^ num__9 - num__8 = num__8 ( num__8 ^ num__8 - num__1 ) = = > num__8 ( num__2 ^ num__24 - num__1 ) last digit of num__2 ^ num__24 is num__6 based on what explanation livestronger is saying . num__2 ^ num__24 - num__1 yields num__6 - num__1 = num__5 as the unit digit . now on multiply this with num__8 we get unit digit as num__0 answer : b <eor> b <eos> |
b |
subtract__9.0__8.0__ subtract__8.0__2.0__ subtract__6.0__1.0__ multiply__8.0__0.0__ |
subtract__9.0__8.0__ subtract__8.0__2.0__ subtract__6.0__1.0__ multiply__8.0__0.0__ |
| how often between num__11 o ' clock and num__12 o ' clock are the hands of the clock together at an integral number value ? <o> a ) num__55 <o> b ) num__56 <o> c ) num__4 <o> d ) num__5 <o> e ) num__3 |
solution : at num__11 o ' clock the hour hand is num__5 spaces apart from the minute hand . during the next num__60 minutes i . e . between num__11 ' o clock and num__12 ' o clock the hour hand will move five spaces [ integral values as denoted by the num__56 minute num__57 minute num__58 minute num__59 minute and num__60 minute positions ] . for each of these num__5 positions the minute hand will be at the num__12 th minute num__24 th minute num__36 th minute num__48 th minute and num__60 th minute positions . hence the difference between the positions of the hour hand and the minute hand will have an integral number of minutes between them . i . e . num__5 positions . answer d <eor> d <eos> |
d |
multiply__12.0__5.0__ add__12.0__24.0__ add__12.0__36.0__ divide__60.0__12.0__ |
multiply__12.0__5.0__ add__12.0__24.0__ add__12.0__36.0__ divide__60.0__12.0__ |
| in the manufacture of a certain product num__5 percent of the units produced are defective and num__4 percent of the defective units are shipped for sale . what percent of the units produced are defective units that are shipped for sale ? <o> a ) num__0.125 <o> b ) num__0.2 <o> c ) num__0.8 <o> d ) num__1.25 <o> e ) num__2.0 % |
percent of defective produced = num__5.0 percent of the defective units that are shipped for sale = num__4.0 percent of units produced are defective units that are shipped for sale = ( num__0.04 ) * ( num__0.05 ) * num__100.0 = ( num__0.002 ) * num__100.0 = ( num__0.2 ) % = . num__2.0 answer b <eor> b <eos> |
b |
percent__5.0__0.04__ percent__5.0__4.0__ percent__5.0__4.0__ |
percent__5.0__0.04__ percent__5.0__4.0__ percent__5.0__4.0__ |
| the weight of a glass of jar is num__10.0 of the weight of the jar filled with coffee beans . after some of the beans have been removed the weight of the jar and the remaining beans is num__60.0 of the original total weight . what fraction part of the beans remain in the jar ? <o> a ) num__0.2 <o> b ) num__0.333333333333 <o> c ) num__0.4 <o> d ) num__0.5 <o> e ) num__0.555555555556 |
let weight of jar filled with beans = num__100 g weight of jar = num__10 g weight of coffee beans = num__90 g weight of jar and remaining beans = num__60 g weight of remaining beans = num__50 g fraction remaining = num__0.555555555556 = num__0.555555555556 answer is e . <eor> e <eos> |
e |
percent__100.0__0.5556__ |
percent__100.0__0.5556__ |
| a number is increased by num__50.0 and then decreased by num__50.0 . find the net increase or decrease per cent . <o> a ) num__25.0 <o> b ) num__18.0 <o> c ) num__17.0 <o> d ) num__13.0 <o> e ) num__16 % |
let the number be num__100 . increase in the number = num__50.0 = num__50.0 of num__100 = ( num__0.5 Ã — num__100 ) = num__50 therefore increased number = num__100 + num__50 = num__150 this number is decreased by num__50.0 therefore decrease in number = num__50.0 of num__150 = ( num__0.5 Ã — num__150 ) = num__75.0 = num__75 therefore new number = num__150 - num__75 = num__75 thus net decreases = num__100 - num__75 = num__25 hence net percentage decrease = ( num__0.25 Ã — num__100 ) % = ( num__25.0 ) % = num__25.0 answer : a <eor> a <eos> |
a |
divide__50.0__100.0__ add__50.0__100.0__ multiply__0.5__150.0__ multiply__50.0__0.5__ divide__25.0__100.0__ multiply__50.0__0.5__ |
divide__50.0__100.0__ add__50.0__100.0__ multiply__0.5__150.0__ subtract__100.0__75.0__ divide__25.0__100.0__ subtract__50.0__25.0__ |
| sonika deposited rs . num__6200 which amounted to rs . num__7200 after num__5 years at simple interest . had the interest been num__3.0 more . she would get how much ? <o> a ) num__9680 <o> b ) num__8130 <o> c ) num__8134 <o> d ) num__8556 <o> e ) num__9808 |
( num__6200 * num__5 * num__3 ) / num__100 = num__930 num__7200 - - - - - - - - num__8130 answer : b <eor> b <eos> |
b |
percent__100.0__8130.0__ |
percent__100.0__8130.0__ |
| if x is a positive integer which of the following must be odd ? <o> a ) x + num__1 <o> b ) x ^ num__2 + x <o> c ) x ^ num__2 + x + num__1 <o> d ) x ^ num__2 − num__1 <o> e ) num__3 x ^ num__2 − num__3 |
a . x + num__1 = can be odd or even . since o + o = e or e + o = o b . x ^ num__2 + x = x ( x + num__1 ) . since from the above derivation we already know the term x + num__1 can be odd or even directly substitute here . x ( odd ) = even ( when x is even ) or x ( even ) = even [ when x is odd ] c . here ' s the answer . since we know the term x ^ num__2 + x can always take a even number even + num__1 = odd hence c . <eor> c <eos> |
c |
multiply__1.0__2.0__ |
power__2.0__1.0__ |
| when the positive integer p is divided by num__9 the remainder is num__5 . what is the remainder when num__3 p is divided by num__9 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__3 <o> d ) num__4 <o> e ) num__6 |
lets take the number as p when p is divided by num__9 the remainder is num__5 hence p can be written as p = num__9 k + num__5 multiplying by num__3 will give num__3 p = num__27 k + num__15 we can also write num__3 p = num__27 k + num__9 + num__6 now num__27 k and num__9 are divisible by num__9 leaving the remainder as num__6 hence e is the answer . <eor> e <eos> |
e |
multiply__9.0__3.0__ multiply__5.0__3.0__ subtract__9.0__3.0__ subtract__9.0__3.0__ |
multiply__9.0__3.0__ multiply__5.0__3.0__ subtract__9.0__3.0__ subtract__9.0__3.0__ |
| salaries of ravi and sumit are in the ratio num__2 : num__3 . if the salary of each is increased by rs num__4000 the new ratio becomes num__40 : num__57 . what is sumit present salary . <o> a ) num__32000 <o> b ) num__34000 <o> c ) num__38000 <o> d ) num__40000 <o> e ) none of these |
explanation : let the original salaries of ravi and sumit is num__2 x and num__3 x . so as per question num__2 x + num__1333.33333333 x + num__4000 = num__0.701754385965 = > num__57 ( num__2 x + num__4000 ) = num__40 ( num__3 x + num__4000 ) = > num__6 x = num__68000 = > num__3 x = num__34000 sumit salary = num__3 x + num__4000 num__34000 + num__4000 = num__38000 option c <eor> c <eos> |
c |
divide__4000.0__3.0__ divide__40.0__57.0__ multiply__2.0__3.0__ divide__68000.0__2.0__ add__4000.0__34000.0__ add__4000.0__34000.0__ |
divide__4000.0__3.0__ divide__40.0__57.0__ multiply__2.0__3.0__ divide__68000.0__2.0__ add__4000.0__34000.0__ add__4000.0__34000.0__ |
| a retailer sold an appliance for $ num__100 . if the retailer ' s gross profit on the appliance was num__25 percent of the retailer ' s cost for the appliance how many dollars was the retailer ' s gross profit ? <o> a ) $ num__10 <o> b ) $ num__16 <o> c ) $ num__20 <o> d ) $ num__24 <o> e ) $ num__25 |
let p be the original price paid by the retailer . num__1.25 * p = num__100 p = num__80 the profit is $ num__20 . the answer is c . <eor> c <eos> |
c |
percent__25.0__80.0__ percent__100.0__20.0__ |
percent__25.0__80.0__ percent__100.0__20.0__ |
| a card player holds num__13 cards of four suits of which seven are black and six are red . there are twice as many hearts as clubs and twice as many diamonds as hearts . how many spades does he hold ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
sol . the player holds num__1 club num__2 hearts and num__4 diamonds . as he holds num__13 cards ( or seven black cards ) it follows that there must be num__6 spades . answer : a <eor> a <eos> |
a |
coin_space__ die_space__ die_space__ |
coin_space__ die_space__ die_space__ |
| y and z represents num__2 gas stations which are num__270 km apart . if two cars leave each gas station and drive towards each other at what time will they meet each other given that the first car left the gas station at num__10 a . m . and travels at a speed of num__75 kmph while the second car left the station at num__11 a . m . and travels at a speed of num__55 kmph . <o> a ) num__12.00 p . m <o> b ) num__11.30 p . m <o> c ) num__2.30 p . m <o> d ) num__12.30 p . m <o> e ) num__1.30 p . m |
suppose they meet x hours after num__10 a . m . distance moved by first car in x hours + distance moved by num__2 nd car in x - num__1 hrs = num__270 num__75 x + num__55 ( x - num__1 ) = num__270 num__75 x + num__55 x - num__55 = num__270 num__130 x = num__325 x = num__2.5 so they meet num__2.5 hr after num__10 a . m . which is at num__12.30 p . m . answer is d <eor> d <eos> |
d |
subtract__11.0__10.0__ add__75.0__55.0__ add__270.0__55.0__ divide__325.0__130.0__ round__12.3__ |
subtract__11.0__10.0__ add__75.0__55.0__ add__270.0__55.0__ divide__325.0__130.0__ round__12.3__ |
| what should come in place of question mark ( ? ) in the following equation ? num__5 num__3 ⁄ num__5 ÷ num__3 num__11 ⁄ num__15 + num__5 num__1 ⁄ num__2 = ? <o> a ) num__7 <o> b ) num__8 num__1 ⁄ num__2 <o> c ) num__7 num__1 ⁄ num__2 <o> d ) num__6 num__1 ⁄ num__2 <o> e ) none of these |
? = num__5 num__3 ⁄ num__5 ÷ num__3 num__11 ⁄ num__15 + num__5 num__1 ⁄ num__2 = num__28 ⁄ num__5 ÷ num__56 ⁄ num__15 + num__11 ⁄ num__2 = num__28 ⁄ num__5 × num__15 ⁄ num__56 + num__11 ⁄ num__2 = num__3 ⁄ num__2 + num__11 ⁄ num__2 = num__14 ⁄ num__2 = num__7 answer a <eor> a <eos> |
a |
multiply__2.0__28.0__ add__3.0__11.0__ add__5.0__2.0__ add__5.0__2.0__ |
multiply__2.0__28.0__ add__3.0__11.0__ add__5.0__2.0__ add__5.0__2.0__ |
| the cost of num__2 chairs and num__3 tables is rs . num__1500 . the cost of num__3 chairs and num__2 tables is rs . num__1200 . the cost of each table is more than that of each chair by ? <o> a ) num__228 <o> b ) num__287 <o> c ) num__277 <o> d ) num__188 <o> e ) num__300 |
explanation : num__2 c + num__3 t = num__1500 - - - ( num__1 ) num__3 c + num__2 t = num__1200 - - - ( num__2 ) subtracting num__2 nd from num__1 st we get - c + t = num__300 = > t - c = num__300 answer : e <eor> e <eos> |
e |
subtract__3.0__2.0__ subtract__1500.0__1200.0__ subtract__1500.0__1200.0__ |
subtract__3.0__2.0__ subtract__1500.0__1200.0__ subtract__1500.0__1200.0__ |
| maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is num__30 kilometers maxwell ' s walking speed is num__4 km / h and brad ' s running speed is num__6 km / h what is the distance traveled by brad ? <o> a ) num__16 <o> b ) num__18 <o> c ) num__20 <o> d ) num__24 <o> e ) num__30 |
time taken = total distance / relative speed total distance = num__30 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = num__6 + num__4 = num__10 kms / hr time taken = num__3.0 = num__3 hrs distance traveled by brad = brad ' s speed * time taken = num__6 * num__3 = num__18 kms . . . answer - b <eor> b <eos> |
b |
add__4.0__6.0__ divide__30.0__10.0__ multiply__6.0__3.0__ round__18.0__ |
add__4.0__6.0__ divide__30.0__10.0__ multiply__6.0__3.0__ multiply__6.0__3.0__ |
| if x ^ num__2 + ( num__1 / x ^ num__2 ) = num__3 x ^ num__4 + ( num__1 / x ^ num__4 ) = ? <o> a ) num__7 <o> b ) num__11 <o> c ) num__12 <o> d ) num__14 <o> e ) num__15 |
- > x ^ num__4 + ( num__1 / x ^ num__4 ) = ( x ^ num__2 ) ^ num__2 + ( num__1 / x ^ num__2 ) ^ num__2 = ( x ^ num__2 + num__1 / x ^ num__2 ) ^ num__2 - num__2 x ^ num__2 ( num__1 / x ^ num__2 ) = num__3 ^ num__2 - num__2 = num__7 . thus the answer is a . <eor> a <eos> |
a |
add__3.0__4.0__ multiply__1.0__7.0__ |
add__3.0__4.0__ add__3.0__4.0__ |
| anil grows tomatoes in his backyard which is in the shape of a square . each tomato takes num__1 cm num__2 in his backyard . this year he has been able to grow num__131 more tomatoes than last year . the shape of the backyard remained a square . how many tomatoes did anil produce this year ? <o> a ) num__4225 <o> b ) num__4096 <o> c ) num__4356 <o> d ) insufficient data <o> e ) none of these |
detailed solution let the area of backyard be x num__2 this year and y num__2 last year ∴ x num__2 - y num__2 = num__131 = ) ( x + y ) ( x - y ) = num__131 now num__131 is a prime number ( a unique one too . check out its properties on google ) . also always identify the prime number given in a question . might be helpful in cracking the solution . = ) ( x + y ) ( x - y ) = num__131 x num__1 = ) x + y = num__131 x - y = num__1 = ) num__2 x = num__132 = ) x = num__66 and y = num__65 ∴ number of tomatoes produced this year = num__662 = num__4356 correct choice ( c ) <eor> c <eos> |
c |
add__1.0__131.0__ divide__132.0__2.0__ subtract__131.0__66.0__ round__4356.0__ |
add__1.0__131.0__ divide__132.0__2.0__ subtract__131.0__66.0__ round__4356.0__ |
| pipe a can fill a tank in num__8 hours . due to a leak at the bottom it takes num__12 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? <o> a ) num__5 <o> b ) num__24 <o> c ) num__78 <o> d ) num__90 <o> e ) num__54 |
let the leak can empty the full tank in x hours num__0.125 - num__1 / x = num__0.0833333333333 = > num__1 / x = num__0.125 - num__0.0833333333333 = ( num__3 - num__2 ) / num__24 = num__0.0416666666667 = > x = num__24 . answer : b <eor> b <eos> |
b |
multiply__8.0__0.125__ divide__1.0__12.0__ subtract__3.0__1.0__ multiply__8.0__3.0__ divide__0.125__3.0__ round__24.0__ |
multiply__8.0__0.125__ divide__1.0__12.0__ subtract__3.0__1.0__ divide__3.0__0.125__ divide__0.125__3.0__ divide__3.0__0.125__ |
| a class is num__6 meters num__24 centimeters in length and num__4 meters num__32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the class room ? <o> a ) num__115 <o> b ) num__117 <o> c ) num__116 <o> d ) num__114 <o> e ) none of these |
explanation : length = num__6 m num__24 cm = num__624 cm width = num__4 m num__32 cm = num__432 cm hcf of num__624 and num__432 = num__48 number of square tiles required = ( num__624 x num__432 ) / ( num__48 x num__48 ) = num__13 x num__9 = num__117 . answer is b <eor> b <eos> |
b |
divide__624.0__48.0__ divide__432.0__48.0__ multiply__9.0__13.0__ multiply__9.0__13.0__ |
divide__624.0__48.0__ divide__432.0__48.0__ multiply__9.0__13.0__ multiply__9.0__13.0__ |
| if a - b = num__3 and a num__2 + b num__2 = num__29 find the value of ab . <o> a ) num__10 <o> b ) num__12 <o> c ) num__15 <o> d ) num__18 <o> e ) num__19 |
num__2 ab = ( a num__2 + b num__2 ) - ( a - b ) num__2 = num__29 - num__9 = num__20 ab = num__10 . answer : a <eor> a <eos> |
a |
subtract__29.0__9.0__ divide__20.0__2.0__ divide__20.0__2.0__ |
subtract__29.0__9.0__ divide__20.0__2.0__ subtract__20.0__10.0__ |
| a rectangular field is num__20 yards long and num__18 yards wide . a fence is needed for the perimeter of the field . fencing is also needed to divide the field into three square sections . how many feet of fencing are needed ? ( it is a good idea to make a drawing for this one . ) <o> a ) num__330 ft <o> b ) num__336 ft <o> c ) num__430 ft <o> d ) num__330 ft <o> e ) num__230 ft |
num__20 + num__18 + num__20 + num__18 = num__76 yds . for outside of field . num__18 + num__18 = num__36 yds . for dividing sections num__76 + num__36 = num__112 yds . of fencing num__112 x num__3 = num__336 ft . of fencing correct answer b <eor> b <eos> |
b |
rectangle_perimeter__20.0__18.0__ rectangle_perimeter__20.0__36.0__ multiply__3.0__112.0__ multiply__3.0__112.0__ |
rectangle_perimeter__20.0__18.0__ rectangle_perimeter__20.0__36.0__ multiply__3.0__112.0__ multiply__3.0__112.0__ |
| a does a work in num__20 days and b does the same work in num__5 days . in how many days they together will do the same work ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
a + b can finish = ab / a + b = num__20 * num__5 / ( num__20 + num__5 ) = num__4 days answer is d <eor> d <eos> |
d |
divide__20.0__5.0__ round__4.0__ |
divide__20.0__5.0__ divide__20.0__5.0__ |
| if equation | x / num__5 | + | y / num__5 | = num__2 encloses a certain region on the coordinate plane what is the area of this region ? <o> a ) num__20 <o> b ) num__50 <o> c ) num__200 <o> d ) num__100 <o> e ) num__400 |
the equation can be reduced to intercept form as | x / num__10 | + | y / num__10 | = num__1 so these are lines in four quadrants with x and y intercept as num__10 so it is a rhombus with diagonals of num__20 each and hence area = num__0.5 * d num__1 * d num__2 = num__0.5 * num__20 * num__20 = num__200 . answer c . <eor> c <eos> |
c |
multiply__5.0__2.0__ multiply__2.0__10.0__ reverse__2.0__ multiply__10.0__20.0__ multiply__1.0__200.0__ |
multiply__5.0__2.0__ multiply__2.0__10.0__ reverse__2.0__ multiply__10.0__20.0__ multiply__1.0__200.0__ |
| in traveling from a dormitory to a certain city a student went num__0.2 of the way by foot num__0.666666666667 of the way by bus and the remaining num__15 kilometers by car . what is the distance in kilometers from the dormitory to the city ? <o> a ) num__67 <o> b ) num__45 <o> c ) num__60 <o> d ) num__90 <o> e ) num__120 |
i believe there is a better way to do it . basically one of the options should satisfy the given criteria . num__60 did num__0.2 * num__60 = num__12 num__0.666666666667 * num__60 = num__40 so total distance num__52 + remaining num__15 = num__67 <eor> c <eos> |
c |
hour_to_min_conversion__ multiply__0.2__60.0__ add__40.0__12.0__ add__15.0__52.0__ hour_to_min_conversion__ |
hour_to_min_conversion__ multiply__0.2__60.0__ add__40.0__12.0__ add__15.0__52.0__ hour_to_min_conversion__ |
| rose is two years older than bruce who is twice as old as chris . if the total of the age of rose b and chris be num__27 years then how old is bruce ? <o> a ) num__7 years <o> b ) num__10 years <o> c ) num__12 years <o> d ) num__13 years <o> e ) num__14 years |
let chris ' s age be x years . then bruce ' s age = num__2 x years . rose ' s age = ( num__2 x + num__2 ) years . ( num__2 x + num__2 ) + num__2 x + x = num__27 num__5 x = num__25 x = num__5 . hence bruce ' s age = num__2 x = num__10 years . b <eor> b <eos> |
b |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
| if x is an integer such that num__1 < x < num__9 num__2 < x < num__15 num__7 > x > – num__1 num__4 > x > num__0 and x + num__1 < num__5 then x is <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
num__1 < x < num__9 num__2 < x < num__15 - num__1 < x < num__7 num__0 < x < num__4 x + num__1 < num__5 from above : num__2 < x < num__4 - - > x = num__3 . answer : a . <eor> a <eos> |
a |
add__1.0__2.0__ add__1.0__2.0__ |
add__1.0__2.0__ add__1.0__2.0__ |
| country x imposes a two - tiered tax on imported cars : the first tier imposes a tax of num__30.0 of the car ' s price up to a certain price level . if the car ' s price is higher than the first tier ' s level the tax on the portion of the price that exceeds this value is num__20.0 . if ron imported a $ num__50000 imported car and ended up paying $ num__12000 in taxes what is the first tier ' s price level ? <o> a ) $ num__22000 <o> b ) $ num__20000 <o> c ) $ num__23000 <o> d ) $ num__24000 <o> e ) $ num__25000 |
let t be the tier price p be total price = num__50000 per the given conditions : num__0.30 t + num__0.20 ( p - t ) = num__12000 num__0.30 t + num__0.20 * num__50000 - num__0.20 t = num__12000 num__0.10 t + num__10000 = num__12000 num__0.10 t = num__12000 - num__10000 = num__2000 t = num__2000 / num__0.10 = num__20000 answer b <eor> b <eos> |
b |
subtract__0.3__0.2__ multiply__50000.0__0.2__ subtract__12000.0__10000.0__ divide__2000.0__0.1__ divide__2000.0__0.1__ |
subtract__0.3__0.2__ multiply__50000.0__0.2__ subtract__12000.0__10000.0__ divide__2000.0__0.1__ divide__2000.0__0.1__ |
| the difference between a two digit number and the number obtained by interchanging the positions of its digits is num__36 . what is the difference between the two digits of that number ? <o> a ) num__2 <o> b ) num__3 <o> c ) four <o> d ) num__8 <o> e ) num__9 |
solution let the ten ' s digit be x and units digit be y . then ( num__10 x + y ) - ( num__10 y + x ) = num__36 ‹ = › num__9 ( x - y ) = num__36 ‹ = › x - y = num__4 . answer c <eor> c <eos> |
c |
divide__36.0__9.0__ divide__36.0__9.0__ |
divide__36.0__9.0__ divide__36.0__9.0__ |
| a train num__800 m long is running at a speed of num__78 km / hr . if it crosses a tunnel in num__1 min then the length of the tunnel is ? <o> a ) num__992 <o> b ) num__389 <o> c ) num__500 <o> d ) num__277 <o> e ) num__781 |
speed = num__78 * num__0.277777777778 = num__21.6666666667 m / sec . time = num__1 min = num__60 sec . let the length of the train be x meters . then ( num__800 + x ) / num__60 = num__21.6666666667 x = num__500 m . answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ round__500.0__ |
hour_to_min_conversion__ multiply__1.0__500.0__ |
| the sum of the ages of num__5 children born at the intervals of num__3 year each is num__40 year . what is the age of the youngest child ? <o> a ) num__2 year <o> b ) num__8 year <o> c ) num__10 year <o> d ) none of these <o> e ) num__9 year |
solution let the ages of the children be x ( x + num__3 ) ( x + num__6 ) ( x + num__9 ) and ( x + num__12 ) year . then x + ( x + num__3 ) + ( x + num__6 ) + ( x + num__9 ) + ( x + num__12 ) = num__40 â ‡ ” num__5 x = num__10 â ‡ ” x = num__2 . â ˆ ´ age of the youngest child = x = num__2 years . answer a <eor> a <eos> |
a |
add__3.0__6.0__ add__3.0__9.0__ subtract__5.0__3.0__ subtract__5.0__3.0__ |
add__3.0__6.0__ add__3.0__9.0__ subtract__5.0__3.0__ subtract__5.0__3.0__ |
| num__9 - num__3 ÷ num__0.333333333333 + num__3 = ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__6 <o> d ) num__9 <o> e ) num__12 |
num__9 - num__3 num__0.333333333333 + num__3 = num__9 - num__3 ÷ num__0.333333333333 + num__3 = num__9 - ( num__3 x num__3 ) + num__3 = num__9 - num__9 + num__3 = num__3 correct answer : b <eor> b <eos> |
b |
divide__9.0__3.0__ |
divide__9.0__3.0__ |
| alok ordered num__16 chapatis num__5 plates of rice num__7 plates of mixed vegetable and num__6 ice - cream cups . the cost of each chapati is rs . num__6 that of each plate of rice is rs . num__45 and that of mixed vegetable is rs . num__70 . the amount that alok paid the cashier was rs . num__1051 . find the cost of each ice - cream cup ? <o> a ) num__25 <o> b ) num__40 <o> c ) num__77 <o> d ) num__99 <o> e ) num__91 |
let the cost of each ice - cream cup be rs . x num__16 ( num__6 ) + num__5 ( num__45 ) + num__7 ( num__70 ) + num__6 ( x ) = num__1051 num__96 + num__225 + num__490 + num__6 x = num__1051 num__6 x = num__240 = > x = num__40 . answer : b <eor> b <eos> |
b |
multiply__16.0__6.0__ multiply__5.0__45.0__ multiply__7.0__70.0__ subtract__45.0__5.0__ subtract__45.0__5.0__ |
multiply__16.0__6.0__ multiply__5.0__45.0__ multiply__7.0__70.0__ subtract__45.0__5.0__ subtract__45.0__5.0__ |
| in a certain city num__60 percent of the registered voters are democrats and the rest are republicans . in a mayoral race if num__75 percent of the registered voters who are democrats and num__25 percent of the registered voters who are republicans are expected to vote for candidate a what percent of the registered voters are expected to vote for candidate a ? <o> a ) num__50.0 <o> b ) num__53.0 <o> c ) num__54.0 <o> d ) num__55.0 <o> e ) num__57 % |
say there are total of num__100 registered voters in that city . thus num__60 are democrats and num__40 are republicans . num__60 * num__0.75 = num__45 democrats are expected to vote for candidate a ; num__40 * num__0.25 = num__10 republicans are expected to vote for candidate a . thus total of num__45 + num__10 = num__55 registered voters are expected to vote for candidate a which is num__55.0 of the total number of registered voters . answer : d . <eor> d <eos> |
d |
percent__60.0__75.0__ percent__25.0__40.0__ percent__100.0__55.0__ |
percent__60.0__75.0__ percent__25.0__40.0__ percent__100.0__55.0__ |
| two pipes can fill a tank in num__20 and num__24 minutes respectively and a waste pipe can empty num__6 gallons per minute . all the three pipes working together can fill the tank in num__15 minutes . the capacity of the tank is ? <o> a ) num__60 gallons <o> b ) num__100 gallons <o> c ) num__240 gallons <o> d ) num__180 gallons <o> e ) num__130 gallons |
work done by the waste pipe in num__1 minute = num__0.0666666666667 - ( num__0.05 + num__0.0416666666667 ) = - num__0.025 volume of num__0.025 part = num__6 gallons \ volume of whole = num__6 * num__40 = num__240 gallons . answer : c <eor> c <eos> |
c |
divide__1.0__15.0__ divide__1.0__20.0__ divide__1.0__24.0__ subtract__0.0667__0.0417__ divide__1.0__0.025__ divide__6.0__0.025__ round__240.0__ |
divide__1.0__15.0__ divide__1.0__20.0__ divide__1.0__24.0__ subtract__0.0667__0.0417__ divide__1.0__0.025__ multiply__6.0__40.0__ multiply__6.0__40.0__ |
| in a two - digit number the digit at unit place is num__1 more than twice of the digit at tens place . if the digit at unit and tens place be interchanged then the difference between the new number and original number is less than num__1 to that of original number . what is the original number ? <o> a ) num__52 <o> b ) num__73 <o> c ) num__25 <o> d ) num__49 <o> e ) num__37 |
let the original number be l num__0 x + y y = num__2 x + num__1 . . . . ( i ) and ( l num__0 y + x ) – ( num__10 x + y ) = num__10 x + y – num__1 or num__9 y – num__9 x = l num__0 x + y – num__1 or l num__9 x – num__8 y = num__1 . . . ( ii ) putting the value of ( i ) in equation ( ii ) we get num__19 x – num__8 ( num__2 x + num__1 ) = num__1 or num__19 x – num__16 x – num__8 = num__1 or num__3 x = num__9 or x = num__3 so y = num__2 × num__3 + num__1 = num__7 answer e \ original number = num__10 × num__3 + num__7 = num__37 <eor> e <eos> |
e |
subtract__10.0__1.0__ subtract__9.0__1.0__ add__9.0__10.0__ multiply__2.0__8.0__ add__1.0__2.0__ subtract__8.0__1.0__ multiply__1.0__37.0__ |
subtract__10.0__1.0__ subtract__9.0__1.0__ add__9.0__10.0__ multiply__2.0__8.0__ add__1.0__2.0__ subtract__8.0__1.0__ multiply__1.0__37.0__ |
| a b and c have rs . num__500 between them a and c together have rs . num__200 and b and c rs . num__310 . how much does c have ? <o> a ) num__50 <o> b ) num__78 <o> c ) num__267 <o> d ) num__29 <o> e ) num__10 |
a + b + c = num__500 a + c = num__200 b + c = num__310 - - - - - - - - - - - - - - a + b + num__2 c = num__510 a + b + c = num__500 - - - - - - - - - - - - - - - - c = num__10 answer : e <eor> e <eos> |
e |
add__200.0__310.0__ subtract__510.0__500.0__ subtract__510.0__500.0__ |
add__200.0__310.0__ subtract__510.0__500.0__ subtract__510.0__500.0__ |
| in the formula v = num__1 / ( num__2 r ) ^ num__2 if r is halved then v is multiplied by ? <o> a ) num__64 <o> b ) num__0.125 <o> c ) num__1 <o> d ) num__4 <o> e ) num__0.015625 |
say r = num__2 = > v num__1 = num__0.0625 when r = num__1 ; v num__2 = num__0.25 v num__2 = num__4 * v num__1 . answer : d <eor> d <eos> |
d |
reverse__0.25__ reverse__0.25__ |
reverse__0.25__ multiply__1.0__4.0__ |
| a metallic hemisphere is melted and recat in the shape of a cone with the same base radius ( r ) as that of the hemisphere . if h is the height of the cone then : <o> a ) h = num__2 r <o> b ) h = num__4 r <o> c ) h = √ num__4 r <o> d ) h = num__0.666666666667 r <o> e ) none |
sol . num__0.666666666667 ∏ r ³ = num__0.333333333333 ∏ r ² h ⇒ h = num__2 r . answer a <eor> a <eos> |
a |
rectangle_perimeter__0.6667__0.3333__ rectangle_perimeter__0.6667__0.3333__ |
rectangle_perimeter__0.6667__0.3333__ rectangle_perimeter__0.6667__0.3333__ |
| a train num__385 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__75 sec <o> b ) num__52 sec <o> c ) num__41 sec <o> d ) num__62 sec <o> e ) num__21 sec |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__385 * num__0.0545454545455 ] sec = num__21 sec answer : e <eor> e <eos> |
e |
add__60.0__6.0__ divide__385.0__18.3333__ round__21.0__ |
add__60.0__6.0__ divide__385.0__18.3333__ divide__385.0__18.3333__ |
| the fourth proportional to num__6 num__11 num__36 is : <o> a ) num__64 <o> b ) num__66 <o> c ) num__68 <o> d ) num__70 <o> e ) num__72 |
let the fourth proportional to num__6 num__11 num__36 be x . then num__6 : num__11 : num__36 : x num__6 x = ( num__11 x num__36 ) x = ( num__11 x num__36 ) / num__6 x = num__66.0 = num__66 answer : option b <eor> b <eos> |
b |
multiply__6.0__11.0__ multiply__6.0__11.0__ |
multiply__6.0__11.0__ multiply__6.0__11.0__ |
| ratio of the ages of mahesh and nilesh is num__5 : x . mahesh is num__18 years younger to ramesh . after nine years ramesh will be num__47 years old . if the difference between the ages of mahesh and nilesh is same as the age of ramesh what is the value of x ? <o> a ) num__14.8 <o> b ) num__14.4 <o> c ) num__14.7 <o> d ) num__14.1 <o> e ) num__14.5 |
let the present ages of mahesh nilesh and ramesh be m n and r respectively . m / n = num__5 / x - - - ( num__1 ) m = r - num__18 - - - ( num__2 ) r + num__9 = num__47 - - - ( num__3 ) m - n = r - - - ( num__4 ) ( num__3 ) = > r = num__47 - num__9 = num__38 years ( num__2 ) = > m = num__38 - num__18 = num__20 years ( num__1 ) = > num__20 / n = num__5 / x = > n = num__4 x ( num__4 ) = > num__4 x - num__20 = num__38 = > num__4 x = num__58 = > x = num__14.5 answer : e <eor> e <eos> |
e |
divide__18.0__2.0__ subtract__5.0__2.0__ subtract__5.0__1.0__ subtract__47.0__9.0__ multiply__5.0__4.0__ add__38.0__20.0__ divide__58.0__4.0__ multiply__14.5__1.0__ |
divide__18.0__2.0__ subtract__5.0__2.0__ subtract__5.0__1.0__ subtract__47.0__9.0__ add__18.0__2.0__ add__38.0__20.0__ divide__58.0__4.0__ divide__14.5__1.0__ |
| rahul will arrange num__6 people of num__6 different heights for photograph by placing them in two rows of three so that each person in the first row is standing is standing in front of someone in the second row . the heights of the people within each row must increase from left to right and each person in the second row must be taller than the person standing in front of him or her . how many such arrangements of the num__6 people are possible ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__9 <o> d ) num__24 <o> e ) num__26 |
rahul will arrange num__6 people of num__6 different heights for photograph by placing them in two rows of three so that each person in the first row is standing is standing in front of someone in the second row . person with max height is in the second row person with min height is in the first row . we need to select num__1 person in the middle of each row . . . in the middle of the first row we can put only num__2 num__3 or num__4 . in the middle of the second row we can put only num__3 num__4 num__5 . if we select { num__3 in the sec . row num__2 in the first } { num__42 } { num__52 } { num__43 } { num__53 } { num__54 } . so there are num__0 * num__1 + num__1 * num__1 + num__2 * num__1 + num__1 * num__1 + num__1 * num__1 + num__0 * num__1 = num__5 cases . . . . a <eor> a <eos> |
a |
coin_space__ vowel_space__ card_space__ negate_prob__1.0__ vowel_space__ |
coin_space__ vowel_space__ card_space__ negate_prob__1.0__ vowel_space__ |
| in a certain egg - processing plant every egg must be inspected and is either accepted for processing or rejected . for every num__96 eggs accepted for processing num__4 eggs are rejected . if on a particular day num__12 additional eggs were accepted but the overall number of eggs inspected remained the same the ratio of those accepted to those rejected would be num__99 to num__1 . how many r eggs does the plant process per day ? <o> a ) num__100 <o> b ) num__300 <o> c ) num__400 <o> d ) num__3000 <o> e ) num__4 |
000 |
straight pluggin in for me . as usual i started with c and got the answer . lets ' back calculate and see what we get let us consider eggs processed each day to be num__400 so initial ratio of eggs processed and rejected is num__96 : num__4 or num__24 : num__1 so out of num__400 eggs there will be num__384 eggs processed and num__16 rejected . now if the no . of eggs inspected remain and num__12 more eggs get accepted that means there r = num__384 + num__12 = num__396 eggs accepted and num__4 rejected . . . and the ratio will be num__99 : num__1 bingo . . . this is what the questions says . . . . its always a good idea to start with c . <eor> c <eos> |
c |
c |
| num__15 binders can bind num__1400 books in num__21 days . how many binders will be required to bind num__1600 books in num__20 days ? <o> a ) num__87 <o> b ) num__18 <o> c ) num__17 <o> d ) num__16 <o> e ) num__10 |
binders books days num__15 num__1400 num__21 x num__1600 num__20 x / num__15 = ( num__1.14285714286 ) * ( num__1.05 ) = > x = num__18 answer : b <eor> b <eos> |
b |
divide__1600.0__1400.0__ divide__21.0__20.0__ round__18.0__ |
divide__1600.0__1400.0__ divide__21.0__20.0__ round__18.0__ |
| how many multiples of num__10 are there between num__11 and num__1001 ? <o> a ) num__98 <o> b ) num__99 <o> c ) num__100 <o> d ) num__97 <o> e ) num__95 |
num__10 * num__2 = num__20 num__10 * num__100 = num__1000 total multiples = ( num__100 - num__2 ) + num__1 = num__99 answer b <eor> b <eos> |
b |
multiply__10.0__2.0__ multiply__10.0__100.0__ subtract__11.0__10.0__ subtract__100.0__1.0__ multiply__1.0__99.0__ |
multiply__10.0__2.0__ multiply__10.0__100.0__ subtract__11.0__10.0__ subtract__100.0__1.0__ multiply__1.0__99.0__ |
| a can do a piece of work in num__18 days b in num__27 days c in num__36 days . they start worked together . but only c work till the completion of work . a leaves num__4 days and b leaves num__6 days before the completion of work . in how many days work be completed ? <o> a ) num__12 <o> b ) num__11 <o> c ) num__99 <o> d ) num__27 <o> e ) num__99 |
let the work be completed in x days ( x - num__4 ) days of a + ( x - num__6 ) days of b + x days of c = num__1 \ inline \ rightarrow \ inline \ frac { x - num__4 } { num__18 } + \ frac { x - num__6 } { num__27 } + \ frac { x } { num__36 } = num__1 \ inline \ rightarrow \ frac { num__13 x - num__48 } { num__108 } = num__1 x = num__12 \ inline \ therefore total time = num__12 days answer : a <eor> a <eos> |
a |
multiply__18.0__6.0__ subtract__18.0__6.0__ round__12.0__ |
multiply__18.0__6.0__ subtract__18.0__6.0__ subtract__18.0__6.0__ |
| when positive integer x is divided by positive integer y the remainder is num__1.44 . if x / y = num__96.12 what is the value of y ? <o> a ) num__96 <o> b ) num__75 <o> c ) num__48 <o> d ) num__25 <o> e ) num__12 |
when positive integer x is divided by positive integer y the remainder is num__1.44 - - > x = qy + num__1.44 ; x / y = num__96.12 - - > x = num__96 y + num__0.12 y ( so q above equals to num__96 ) ; num__0.12 y = num__1.44 - - > y = num__12 . answer : e . <eor> e <eos> |
e |
round_down__96.12__ subtract__96.12__96.0__ divide__1.44__0.12__ divide__1.44__0.12__ |
round_down__96.12__ subtract__96.12__96.0__ divide__1.44__0.12__ divide__1.44__0.12__ |
| in num__4 years raj ' s father age twice as raj two years ago raj ' s mother ' s age twice as raj . if raj is num__32 yrs old in eight yrs from now what is the age of raj ' s mother and father ? <o> a ) num__27 <o> b ) num__36 <o> c ) num__28 <o> d ) num__46 <o> e ) num__91 |
raj present age = num__32 - num__8 = num__24 . after num__4 years raj ' s age is num__28 . and raj ' s fathers age is num__28 x num__2 = num__56 and his present age is num__52 . two years ago raj ' s age is num__22 . and his mother ' s age is num__22 x num__2 = num__44 . his mother ' s present age = num__46 answer : c <eor> c <eos> |
c |
divide__32.0__4.0__ subtract__32.0__8.0__ add__4.0__24.0__ divide__8.0__4.0__ add__32.0__24.0__ subtract__56.0__4.0__ subtract__24.0__2.0__ multiply__2.0__22.0__ add__2.0__44.0__ add__4.0__24.0__ |
divide__32.0__4.0__ subtract__32.0__8.0__ subtract__32.0__4.0__ divide__8.0__4.0__ add__32.0__24.0__ subtract__56.0__4.0__ subtract__24.0__2.0__ subtract__52.0__8.0__ add__2.0__44.0__ subtract__32.0__4.0__ |
| in a college the ratio of the number of boys to girls is num__8 : num__5 . if there are num__160 girls the total number of students in the college is <o> a ) num__100 <o> b ) num__250 <o> c ) num__260 <o> d ) num__416 <o> e ) none of these |
explanation : let the number of boys and girls be num__8 x and num__5 x . total number of students = num__13 x = num__13 × num__32 = num__416 . answer : d <eor> d <eos> |
d |
add__8.0__5.0__ divide__160.0__5.0__ multiply__32.0__13.0__ multiply__32.0__13.0__ |
add__8.0__5.0__ divide__160.0__5.0__ multiply__32.0__13.0__ multiply__32.0__13.0__ |
| i chose a number and divide it by num__8 . then i subtracted num__160 from the result and got num__12 . what was the number i chose ? <o> a ) num__1376 <o> b ) num__1800 <o> c ) num__1400 <o> d ) num__2500 <o> e ) num__100 |
let x be the number i chose then x / num__8 − num__160 = num__12 x / num__8 = num__172 x = num__1376 answer is a . <eor> a <eos> |
a |
add__160.0__12.0__ multiply__8.0__172.0__ multiply__8.0__172.0__ |
add__160.0__12.0__ multiply__8.0__172.0__ multiply__8.0__172.0__ |
| raman ' s salary was decreased by num__50.0 and subsequently increased by num__50.0 . how much percent does he loss <o> a ) num__75 <o> b ) num__65 <o> c ) num__45 <o> d ) num__25 <o> e ) num__35 |
explanation : let the origianl salary = rs . num__100 it will be num__150.0 of ( num__50.0 of num__100 ) = ( num__1.5 ) * ( num__0.5 ) * num__100 = num__75 so new salary is num__75 it means his loss is num__25.0 answer : option d <eor> d <eos> |
d |
percent__50.0__150.0__ percent__100.0__25.0__ |
percent__50.0__150.0__ percent__100.0__25.0__ |
| donald plans to invest x dollars in a savings account that pays interest at an annual rate of num__8.0 compounded quarterly . approximately what amount is the minimum that donald will need to invest to earn over $ num__100 in interest within num__6 months ? <o> a ) $ num__1500 <o> b ) $ num__1750 <o> c ) $ num__2000 <o> d ) $ num__2500 <o> e ) $ num__3000 |
solution : r = rate = num__8.0 compounded quarterly . . = num__2.0 per quarter t = num__6 months = num__2 quarters so a = p ( num__1 + num__0.02 ) ^ num__2 = > a = p * num__1.0404 substitute p for num__1500 num__1700 num__2000 . . the a - p should be greater than num__100 ( since interest = amount - principal ) . for num__2500 interest is num__101 . . . hence d <eor> d <eos> |
d |
percent__1.0__2.0__ percent__100.0__2500.0__ |
percent__1.0__2.0__ percent__100.0__2500.0__ |
| an article is bought for rs . num__675 and sold for rs . num__1080 find the gain percent ? <o> a ) num__33 num__0.111111111111 % <o> b ) num__60.0 <o> c ) num__33 num__0.333333333333 % <o> d ) num__33 num__0.666666666667 % <o> e ) num__33 num__0.5 % |
num__675 - - - - num__405 num__100 - - - - ? = > num__60.0 answer : b <eor> b <eos> |
b |
percent__60.0__100.0__ |
percent__60.0__100.0__ |
| two numbers are in the ratio num__3 : num__5 . if num__9 be subtracted from each they are in the ratio of num__2 : num__3 . the first number is : <o> a ) a ) num__27 <o> b ) b ) num__98 <o> c ) c ) num__34 <o> d ) d ) num__35 <o> e ) e ) num__62 |
( num__3 x - num__9 ) : ( num__5 x - num__9 ) = num__2 : num__3 x = num__9 = > num__3 x = num__27 answer : a <eor> a <eos> |
a |
multiply__3.0__9.0__ multiply__3.0__9.0__ |
multiply__3.0__9.0__ multiply__3.0__9.0__ |
| calculate the area of a triangle if the sides of are num__39 cm num__36 cm and num__15 cm what is its area ? <o> a ) num__570 cm num__2 <o> b ) num__370 cm num__2 <o> c ) num__170 cm num__2 <o> d ) num__271 cm num__2 <o> e ) num__270 cm num__2 |
the triangle with sides num__39 cm num__36 cm and num__15 is right angled where the hypotenuse is num__39 cm . area of the triangle = num__0.5 * num__36 * num__15 = num__270 cm num__2 answer : e <eor> e <eos> |
e |
triangle_area__36.0__15.0__ square_perimeter__0.5__ triangle_area__36.0__15.0__ |
volume_rectangular_prism__36.0__15.0__0.5__ square_perimeter__0.5__ volume_rectangular_prism__36.0__15.0__0.5__ |
| a person can swim in still water at num__12 km / h . if the speed of water num__10 km / h how many hours will the man take to swim back against the current for num__12 km ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
m = num__12 s = num__10 us = num__12 - num__10 = num__2 d = num__12 t = num__6.0 = num__6 answer : d <eor> d <eos> |
d |
subtract__12.0__10.0__ divide__12.0__2.0__ round__6.0__ |
subtract__12.0__10.0__ divide__12.0__2.0__ subtract__12.0__6.0__ |
| if num__35.0 of the num__840 students at a certain college are enrolled in biology classes how many students at the college are not enrolled in a biology class ? <o> a ) num__546 <o> b ) num__550 <o> c ) num__555 <o> d ) num__532 <o> e ) num__515 |
we know num__35.0 people study biology therefore the no of people not studying = num__100 - num__35 = num__65.0 > therefore the people not studying biology out of a total num__840 people are = num__65.0 of num__840 > ( num__0.65 ) * num__840 = num__546 people a <eor> a <eos> |
a |
percent__65.0__840.0__ percent__65.0__840.0__ |
percent__65.0__840.0__ percent__65.0__840.0__ |
| the current in a river is num__5 mph . a boat can travel num__20 mph in still water . how far up the river can the boat travel if the round trip is to take num__10 hours ? <o> a ) num__69 miles <o> b ) num__88 miles <o> c ) num__9375 miles <o> d ) num__100 miles <o> e ) num__112 miles |
upstream speed = num__20 - num__5 = num__15 mph downstream speed = num__20 + num__5 = num__25 mph d / num__15 + d / num__25 = num__10 hours solving for d we get d = num__9375 answer : c <eor> c <eos> |
c |
add__5.0__10.0__ add__5.0__20.0__ round__9375.0__ |
add__5.0__10.0__ add__5.0__20.0__ round__9375.0__ |
| how long does a train num__120 m long travelling at num__60 kmph takes to cross a bridge of num__170 m in length ? <o> a ) num__15.8 sec <o> b ) num__14.9 sec <o> c ) num__12.4 sec <o> d ) num__16.8 sec <o> e ) num__17.4 sec |
e num__17.4 sec d = num__120 + num__170 = num__290 m s = num__60 * num__0.277777777778 = num__16.6666666667 t = num__290 * num__0.06 = num__17.4 sec answer is e <eor> e <eos> |
e |
add__120.0__170.0__ divide__290.0__17.4__ divide__17.4__290.0__ round__17.4__ |
add__120.0__170.0__ divide__290.0__17.4__ divide__17.4__290.0__ multiply__290.0__0.06__ |
| a train num__120 m long passed a pole in num__6 sec . how long will it take to pass a platform num__360 m long ? <o> a ) num__20 sec <o> b ) num__21 <o> c ) num__22 <o> d ) num__23 <o> e ) num__24 |
speed = num__20.0 = num__20 m / sec . required time = ( num__120 + num__360 ) / num__20 = num__24 sec . answer : e <eor> e <eos> |
e |
divide__120.0__6.0__ round__24.0__ |
divide__120.0__6.0__ round__24.0__ |
| a family consists of grandparents parents and three grandchildren . the average age of the grandparents is num__67 years that of the parents is num__35 years and that of the grandchildren is num__6 years . what is the average age of the family ? <o> a ) num__31 ( num__0.714285714286 ) <o> b ) num__81 ( num__0.714285714286 ) <o> c ) num__51 ( num__0.714285714286 ) <o> d ) num__41 ( num__0.714285714286 ) <o> e ) num__31 ( num__0.428571428571 ) |
explanation : required average = ( num__67 * num__2 + num__35 * num__2 + num__6 * num__3 ) / ( num__2 + num__2 + num__3 ) = ( num__134 + num__70 + num__18 ) / num__7 = num__31.7142857143 = num__31 ( num__0.714285714286 ) years . answer : a <eor> a <eos> |
a |
divide__6.0__2.0__ multiply__67.0__2.0__ add__67.0__3.0__ multiply__6.0__3.0__ round_down__31.7143__ subtract__31.7143__31.0__ round_down__31.7143__ |
divide__6.0__2.0__ multiply__67.0__2.0__ add__67.0__3.0__ multiply__6.0__3.0__ round_down__31.7143__ subtract__31.7143__31.0__ round_down__31.7143__ |
| the age of father num__8 years ago was twice the age of his son . eight years hence father ' s age will be twice that of his son . the ratio of their present ages is : <o> a ) num__5 : num__3 <o> b ) num__7 : num__3 <o> c ) num__9 : num__2 <o> d ) num__7 : num__2 <o> e ) num__13 : num__4 |
let the ages of father and son num__8 years ago be num__2 x and x years respectively . then ( num__2 x + num__8 ) + num__8 = num__2 [ ( x + num__8 ) + num__8 ] num__2 x + num__16 = num__2 x + num__32 x = num__16 . required ratio = ( num__2 x + num__8 ) : ( x + num__8 ) = num__40 : num__24 = num__5 : num__3 answer : a <eor> a <eos> |
a |
multiply__8.0__2.0__ multiply__2.0__16.0__ add__8.0__32.0__ add__8.0__16.0__ divide__40.0__8.0__ subtract__8.0__5.0__ subtract__8.0__3.0__ |
multiply__8.0__2.0__ multiply__2.0__16.0__ add__8.0__32.0__ add__8.0__16.0__ divide__40.0__8.0__ subtract__8.0__5.0__ add__2.0__3.0__ |
| find out the wrong term in the series num__1 num__10 num__100 num__100 num__1000 num__100000 <o> a ) num__100 num__000000 <o> b ) num__10000 <o> c ) num__100000 <o> d ) num__10 num__000000 <o> e ) num__100000 num__000 |
000 |
explanation : each number with in the series is the product of the two numbers that proceed it . so if ( num__1 * num__10 = num__10 ) then ( num__1000 * num__100000 = num__100 num__000 num__000 ) . num__100000000 should replace num__100000 . answer : a ) num__100 num__000000 <eor> a <eos> |
a |
a |
| if a sum of money trebles itself in num__40 years what is the rate of interest ? <o> a ) num__5.0 <o> b ) num__6.0 <o> c ) num__4.0 <o> d ) num__3.0 <o> e ) none of these |
explanation : let the sum of money be rs . x after num__40 years this becomes num__3 x simple interest = num__3 x - x = num__2 x simple interest = prt / num__100 num__2 x = ( x × r × num__40 ) / num__100 num__2 = ( r × num__40 ) / num__100 num__200 = num__40 r r = num__5.0 answer : option a <eor> a <eos> |
a |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| if the average ( arithmetic mean ) of num__4 x - num__5 num__8 x - num__6 and num__9 x + num__5 is num__279 what is the value of x ? <o> a ) num__18 <o> b ) num__25 <o> c ) num__27 <o> d ) num__29 <o> e ) num__31 |
am of num__4 x - num__5 num__8 x - num__6 and num__9 x + num__5 = num__4 x - num__5 + num__8 x - num__6 + num__9 x + num__1.66666666667 = num__31 x / num__3 given that num__31 x / num__3 = num__279 x = num__27.0 = num__27 x = num__27 answer is c <eor> c <eos> |
c |
divide__279.0__9.0__ subtract__8.0__5.0__ multiply__9.0__3.0__ multiply__9.0__3.0__ |
divide__279.0__9.0__ subtract__8.0__5.0__ subtract__31.0__4.0__ subtract__31.0__4.0__ |
| three medical experts working together at the same constant rate can write an anatomy textbook in num__32 days . how many additional experts working together at this same constant rate are needed to write the textbook in num__6 days ? <o> a ) num__7 <o> b ) num__9 <o> c ) num__11 <o> d ) num__13 <o> e ) num__15 |
each expert can write num__0.0104166666667 of the book per day . to complete the book in num__6 days we need num__16.0 = num__16 experts thus num__13 more experts are needed . the answer is d . <eor> d <eos> |
d |
round__13.0__ |
round__13.0__ |
| what is the remainder when num__54617 ^ ( num__42788234 ) is divided by num__5 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
we need to find the units digit of the number . the units digit of powers of seven repeats num__7 num__9 num__3 and num__1 cyclically . since num__42788234 has the form num__4 a + num__2 the units digit is num__9 . then the remainder when dividing by num__5 is num__4 . the answer is e . <eor> e <eos> |
e |
subtract__5.0__1.0__ subtract__5.0__3.0__ subtract__5.0__1.0__ |
add__1.0__3.0__ subtract__5.0__3.0__ add__1.0__3.0__ |
| a miniature roulette wheel is divided into num__10 equal sectors each bearing a distinct integer from num__1 to num__10 inclusive . each time the wheel is spun a ball randomly determines the winning sector by settling in that sector . if the wheel is spun four times approximately what is the probability that the product of the four winning sectors ’ integers will be even ? <o> a ) num__50.0 <o> b ) num__67.0 <o> c ) num__88.0 <o> d ) num__94.0 <o> e ) num__98 % |
the only way to have an odd product is if all num__4 integers are odd . p ( odd product ) = num__0.5 * num__0.5 * num__0.5 * num__0.5 = num__0.0625 p ( even product ) = num__1 - num__0.0625 = num__0.9375 which is about num__94.0 the answer is d . <eor> d <eos> |
d |
subtract__1.0__0.0625__ multiply__1.0__94.0__ |
subtract__1.0__0.0625__ multiply__1.0__94.0__ |
| in a throw of coin what is the probability of getting head . <o> a ) num__1 <o> b ) num__2 <o> c ) num__0.5 <o> d ) num__0 <o> e ) none of these |
explanation : total cases = [ h t ] - num__2 favourable cases = [ h ] - num__1 so probability of ge ƫ ng head = num__0.5 answer : c <eor> c <eos> |
c |
coin_space__ negate_prob__0.5__ |
coin_space__ negate_prob__0.5__ |
| a manufacturer of a certain type of screw rejects any screw whose length is less than num__2.44 + num__0.03 centimeters or greater than num__2.5 + num__0.03 centimeters . if k represents the length of a screw in centimeters which of the following inequalities specifies all the lengths of screws that are acceptable ? <o> a ) | k + num__0.03 | > num__2.5 <o> b ) | k — num__0.03 | < = num__2.5 <o> c ) | k — num__2.5 | > num__0.03 <o> d ) | k — num__2.5 | > = num__0.06 <o> e ) | k — num__2.5 | < = num__0.03 |
so let ' s go through this step by step : rejects any screw whose length is less than num__2.44 + num__0.03 centimeters or greater than num__2.5 + num__0.03 centimeters . in other words any screw that is less than : num__2.44 + num__0.03 = num__2.47 or greater than num__2.50 + num__0.03 = num__2.53 will be rejected . if k represents the length of a screw in other words kis an acceptable screw that must fall within the acceptable range of num__2.47 to num__2.53 so : num__2.47 ≤ k ≤ num__2.53 you can rule out answers with < or > as opposed to ≤ or ≥ because the length can not be less than num__2.47 or greater than num__2.53 . in other words num__2.47 and num__2.53 are acceptable lengths . let ' s look at ( e ) : | k — num__2.5 | < = num__0.03 for the positive case : k - num__2.5 ≤ num__0.03 = = = > k ≤ num__2.53 for the negative case : - ( k - num__2.5 ) ≤ num__0.03 = = = > - k + num__2.5 ≤ num__0.03 = = = > - k ≤ - num__2.47 = = = > k ≥ num__2.47 num__2.47 ≤ k ≤ num__2.53 ( e ) <eor> e <eos> |
e |
add__2.44__0.03__ add__0.03__2.5__ add__0.03__2.47__ |
add__2.44__0.03__ add__0.03__2.5__ add__0.03__2.47__ |
| a thief goes away with a santro car at a speed of num__40 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at num__60 kmph when will the owner over take the thief from the start ? <o> a ) a ) num__2 <o> b ) b ) num__5 <o> c ) c ) num__7 <o> d ) d ) num__5 <o> e ) e ) num__1 |
explanation : | - - - - - - - - - - - num__20 - - - - - - - - - - - - - - - - - - - - | num__60 num__40 d = num__20 rs = num__60 – num__40 = num__20 t = num__2.0 = num__1 hours answer : option e <eor> e <eos> |
e |
subtract__60.0__40.0__ divide__40.0__20.0__ round__1.0__ |
subtract__60.0__40.0__ divide__40.0__20.0__ subtract__2.0__1.0__ |
| if p > num__1 and q = num__2 ^ ( p − num__1 ) then num__4 ^ p = <o> a ) num__16 q ^ num__2 <o> b ) num__4 q ^ num__2 <o> c ) q ^ num__2 <o> d ) q ^ num__0.5 <o> e ) q ^ num__0.125 |
if p > num__1 and q = num__2 ^ ( p − num__1 ) then num__4 ^ p given p > num__1 so let ' s assume p = num__2 q = num__2 ^ ( p - num__1 ) = num__2 ^ ( num__2 - num__1 ) = num__2 so q = num__2 hence num__4 ^ p = num__4 ^ num__2 = num__16 only num__1 ans . choice can satisfy this : a ) num__16 q ^ num__2 - - > clearly > num__16 b ) num__4 q ^ num__2 - - > num__4 * num__2 ^ num__2 = num__16 ( we can stop after this as there can be only num__1 right answer ) c ) q ^ num__2 - - > clearly < num__16 d ) q ^ num__0.5 - - > clearly < num__16 e ) q ^ num__0.125 - - > clearly < num__16 ans . b ) num__4 q ^ num__2 <eor> b <eos> |
b |
reverse__2.0__ divide__2.0__16.0__ multiply__1.0__4.0__ |
reverse__2.0__ divide__2.0__16.0__ multiply__1.0__4.0__ |
| a scientist used a unique two - color code to identify each of the test subjects involved in a certain study . if the scientist found that choosing from among five colors produced enough color codes to identify all but num__5 of the test subjects how many test subjects were in the study ? ( assume that the order of the colors in the codes does not matter . ) <o> a ) num__7 <o> b ) num__10 <o> c ) num__15 <o> d ) num__17 <o> e ) num__20 |
num__5 c num__2 = num__10 the number of test subjects is num__10 + num__5 = num__15 the answer is c . <eor> c <eos> |
c |
multiply__5.0__2.0__ add__5.0__10.0__ add__5.0__10.0__ |
multiply__5.0__2.0__ add__5.0__10.0__ add__5.0__10.0__ |
| the mass of the sun is approximately num__2 × num__10 ^ num__25 kg and the mass of the moon is approximately num__8 × num__10 ^ num__12 kg . the mass of the sun is approximately how many times the mass of the moon ? <o> a ) num__4.0 × num__10 ^ ( − num__18 ) <o> b ) num__2.5 × num__10 ^ num__12 <o> c ) num__4.0 × num__10 ^ num__18 <o> d ) num__2.5 × num__10 ^ num__19 <o> e ) num__4.0 × num__10 ^ num__42 |
mass of sun = x * mass of moon x = mass of sun / mass of moon = ( num__2 × num__10 ^ num__25 ) / ( num__8 × num__10 ^ num__12 ) = num__2 * num__2 ^ - num__3 * num__10 ^ num__13 = num__10 ^ num__3.25 = num__2.5 * num__10 ^ num__12 ans . b ) num__2.5 × num__10 ^ num__12 <eor> b <eos> |
b |
add__10.0__3.0__ divide__25.0__10.0__ divide__25.0__10.0__ |
subtract__25.0__12.0__ divide__25.0__10.0__ divide__25.0__10.0__ |
| if c + d = num__17 and c and d are positive integers which of the following is a possible value for num__3 c + num__2 d ? <o> a ) num__29 <o> b ) - num__29 <o> c ) num__0 <o> d ) - num__19 <o> e ) num__19 |
c + d = num__11 num__3 c + num__2 d = num__2 c + c + num__2 d = num__2 ( c + d ) + c = num__34 + c = c = - num__34 now this reults in c + d = num__17 = - num__34 + d = num__17 = d = num__51 c = - num__34 ; d = num__51 num__3 c + num__2 d = num__3 ( - num__34 ) + num__2 ( num__51 ) = - num__102 + num__102 = num__0 num__3 c + num__2 d = num__0 hence c is the right answer . <eor> c <eos> |
c |
multiply__17.0__2.0__ multiply__17.0__3.0__ multiply__3.0__34.0__ multiply__17.0__0.0__ |
multiply__17.0__2.0__ add__17.0__34.0__ multiply__3.0__34.0__ multiply__17.0__0.0__ |
| two lorries each num__350 m long are running in opposite directions on parallel tracks . their speeds are num__40 km / hr and num__30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? <o> a ) num__77 sec <o> b ) num__36 sec <o> c ) num__48 sec <o> d ) num__55 sec <o> e ) num__45 sec |
relative speed = num__40 + num__30 = num__70 km / hr . num__70 * num__0.277777777778 = num__19.4444444444 m / sec . distance covered = num__350 + num__350 = num__200 m . required time = num__700 * num__0.0514285714286 = num__36 sec . answer : b <eor> b <eos> |
b |
add__40.0__30.0__ round__36.0__ |
add__40.0__30.0__ round__36.0__ |
| a car crosses a num__600 m long bridge in num__5 minutes . what is the speed of car in km per hour ? <o> a ) num__7 km / hr <o> b ) num__7.2 km / hr <o> c ) num__8 km / hr <o> d ) num__9 km / hr <o> e ) num__11 km / hr |
speed = num__600 m / sec . num__5 x num__60 = num__2 m / sec . converting m / sec to km / hr ( see important formulas section ) = num__2 x num__18 km / hr num__5 = num__7.2 km / hr b <eor> b <eos> |
b |
hour_to_min_conversion__ round__7.2__ |
hour_to_min_conversion__ round__7.2__ |
| if num__5 people undertook a piece of construction work and finished half the job in num__15 days . if two people drop out then the job will be completed in ? <o> a ) num__8 days <o> b ) num__10 days <o> c ) num__18 days <o> d ) num__20 days <o> e ) num__25 days |
that is half the work done = num__5 × num__15 × ½ then num__5 × num__15 × ½ = num__3 × ? × num__0.5 i . e . num__5 × num__15 = num__3 × ? therefore ? ( no . days required ) = num__5 × num__5.0 = num__25 days . e ) <eor> e <eos> |
e |
divide__15.0__5.0__ round__25.0__ |
divide__15.0__5.0__ round__25.0__ |
| a firm is comprised of partners and associates in a ratio of num__2 : num__63 . if num__50 more associates were hired the ratio of partners to associates would be num__1 : num__34 . how many partners are currently in the firm ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__12 <o> d ) num__20 <o> e ) num__25 |
the ratio num__1 : num__34 = num__2 : num__68 so the ratio changed from num__2 : num__63 to num__2 : num__68 . num__68 - num__63 = num__5 which is num__0.1 of the increase in num__50 associates . the ratio changed from num__20 : num__630 to num__20 : num__680 . thus the number of partners is num__20 . the answer is d . <eor> d <eos> |
d |
multiply__2.0__34.0__ subtract__68.0__63.0__ divide__5.0__50.0__ divide__2.0__0.1__ divide__63.0__0.1__ add__50.0__630.0__ divide__2.0__0.1__ |
multiply__2.0__34.0__ subtract__68.0__63.0__ divide__5.0__50.0__ divide__2.0__0.1__ divide__63.0__0.1__ add__50.0__630.0__ divide__2.0__0.1__ |
| simple interest on a certain sum is num__16.2 over num__20 of the sum . find the rate per cent and time if both are equal . <o> a ) num__8.0 and num__8 years <o> b ) num__6.0 and num__6 years <o> c ) num__9.0 and num__9 years <o> d ) num__12.0 and num__12 years <o> e ) none of these |
num__16.2 â „ num__20 p = p à — r à — r / num__100 â ‡ ’ r num__2 = num__1620 â „ num__20 â ‡ ’ num__81 = > r = num__9.0 also time = num__9 years answer c <eor> c <eos> |
c |
percent__100.0__9.0__ |
percent__100.0__9.0__ |
| a certain social security recipient will receive an annual benefit of $ num__12000 provided he has annual earnings of $ num__9360 or less but the benefit will be reduced by $ num__1 for every $ num__3 of annual earnings over $ num__9360 . what amount of total annual earnings would result in a num__55 percent reduction in the recipient ' s annual social security benefit ? ( assume social security benefits are not counted as part of annual earnings . ) <o> a ) $ num__15360 <o> b ) $ num__17360 <o> c ) $ num__25560 <o> d ) $ num__21360 <o> e ) $ num__27 |
360 |
for every $ num__3 earn above $ num__9360 the recipient loses $ num__1 of benefit . or for every $ num__1 loss in the benefit the recipient earns $ num__3 above $ num__9360 if earning is ; num__9360 + num__3 x benefit = num__12000 - x or the vice versa if benefit is num__12000 - x the earning becomes num__9360 + num__3 x he lost num__50.0 of the benefit ; benefit received = num__12000 - num__0.55 * num__12000 = num__12000 - num__6600 x = num__5400 earning becomes num__9360 + num__3 x = num__9360 + num__3 * num__5400 = num__25560 ans : c <eor> c <eos> |
c |
c |
| x and y are positive integers of t . if num__1 / x + num__1 / y < num__2 which of the following must be true ? <o> a ) x + y > num__4 <o> b ) xy > num__1 <o> c ) x / y + y / x < num__1 <o> d ) ( x - y ) ^ num__2 > num__0 <o> e ) none of the above |
answer is b : num__1 / x + num__1 / y < num__2 the maximum value of num__1 / x is num__1 because if x equals any other number greater than one it will be a fraction . the same is true with num__1 / y . so num__1 / x and num__1 / y will always be less than num__2 as long as both x and y are not both equal to one at the same time . another way of putting it is : x * y > num__1 . b <eor> b <eos> |
b |
reverse__1.0__ |
reverse__1.0__ |
| if mango is written as num__41576 and apple is written as num__17735 then grapes would be written as : <o> a ) num__7999340 <o> b ) num__7835610 <o> c ) num__7226710 <o> d ) num__7913350 <o> e ) num__7917510 |
e num__7917510 all letters whose position in the alphabet is below num__10 are being represented by their position number . all letters whose position is num__10 or more are being represented by their position number minus ten . therefore : g is num__7 r is num__18 - num__9 = num__9 a is num__1 p is num__16 - num__9 = num__7 e is num__5 s is num__19 - num__9 = num__10 <eor> e <eos> |
e |
subtract__10.0__9.0__ add__7.0__9.0__ add__1.0__18.0__ multiply__1.0__7917510.0__ |
subtract__10.0__9.0__ add__7.0__9.0__ add__1.0__18.0__ multiply__1.0__7917510.0__ |
| a b and c work on a task . to complete the task alone b takes twice the time that a would take to complete the task alone and num__0.25 rd the time that c would take to complete the task alone . if b actually worked for half the number of days that a worked and num__1.5 times the number of days that c worked what proportion of the total work was completed by b ? <o> a ) num__0.333333333333 <o> b ) num__0.222222222222 <o> c ) num__0.183673469388 <o> d ) num__0.197530864198 <o> e ) num__0.166666666667 |
the first thing to notice is that a is faster than b and b is faster than c . since work is proportional to time in num__1 day lets say if a does num__2 works b does num__1 work and c does num__0.666666666667 rd of a work . if a works for num__2 days b works for num__1 day and c works for only num__0.666666666667 of the day . therefore total work done = ( num__2 * num__2 ) + ( num__1 * num__1 ) + ( num__0.25 * num__0.25 ) = num__5.0625 proportion of work done by b = ( num__1 * num__1 ) / ( num__5.0625 ) = num__0.197530864198 hence answer d . <eor> d <eos> |
d |
round_down__1.5__ reverse__1.5__ reverse__5.0625__ reverse__5.0625__ |
round_down__1.5__ reverse__1.5__ reverse__5.0625__ reverse__5.0625__ |
| two vessels p and q contain num__62.5 and num__87.5 of alcohol respectively . if num__2 litres from vessel p is mixed with num__4 litres from vessel q the ratio of alcohol and water in the resulting mixture is ? <o> a ) num__19 : num__2 <o> b ) num__18 : num__3 <o> c ) num__19 : num__1 <o> d ) num__19 : num__5 <o> e ) num__19 : num__4 |
quantity of alcohol in vessel p = num__62.5 / num__100 * num__2 = num__1.25 litres quantity of alcohol in vessel q = num__87.5 / num__100 * num__4 = num__3.5 litres quantity of alcohol in the mixture formed = num__1.25 + num__3.5 = num__4.75 = num__4.75 litres as num__6 litres of mixture is formed ratio of alcohol and water in the mixture formed = num__4.75 : num__1.25 = num__19 : num__5 . answer : d <eor> d <eos> |
d |
add__1.25__3.5__ add__2.0__4.0__ multiply__4.0__4.75__ multiply__4.0__1.25__ multiply__4.0__4.75__ |
add__1.25__3.5__ add__2.0__4.0__ multiply__4.0__4.75__ multiply__4.0__1.25__ multiply__4.0__4.75__ |
| ram sold two bicycles each for rs . num__990 . if he made num__10.0 profit on the first and num__10.0 loss on the second what is the total cost of both bicycles ? <o> a ) rs . num__2000 <o> b ) rs . num__2029 <o> c ) rs . num__2297 <o> d ) rs . num__2020 <o> e ) rs . num__2293 |
( num__10 * num__10 ) / num__100 = num__1.0 loss num__100 - - - num__99 ? - - - num__1980 = > rs . num__2000 answer : a <eor> a <eos> |
a |
percent__10.0__990.0__ percent__100.0__2000.0__ |
percent__10.0__990.0__ percent__100.0__2000.0__ |
| a person borrowed rs . num__500 at the rate of num__5.0 per annum s . i what amount will he pays to clear the debts after num__4 years ? <o> a ) num__300 <o> b ) num__700 <o> c ) num__500 <o> d ) num__800 <o> e ) num__600 |
explanation : num__500 - - - - - - > num__5 * num__4 = num__20.0 ( num__500 ) = num__100 num__500 + num__100 = num__600 answer : option e <eor> e <eos> |
e |
percent__4.0__500.0__ percent__20.0__500.0__ percent__100.0__600.0__ |
percent__4.0__500.0__ percent__20.0__500.0__ percent__100.0__600.0__ |
| two trains num__140 m and num__210 m long run at the speed of num__60 km / hr and num__40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? <o> a ) num__12.6 sec <o> b ) num__11.8 sec <o> c ) num__27.7 sec <o> d ) num__10.8 sec <o> e ) num__17.8 sec |
relative speed = num__60 + num__40 = num__100 km / hr . = num__100 * num__0.277777777778 = num__27.7777777778 m / sec . distance covered in crossing each other = num__140 + num__210 = num__350 m . required time = num__350 * num__0.036 = num__12.6 sec . answer : a <eor> a <eos> |
a |
subtract__140.0__40.0__ add__140.0__210.0__ multiply__0.036__350.0__ round__12.6__ |
add__60.0__40.0__ add__140.0__210.0__ multiply__0.036__350.0__ multiply__0.036__350.0__ |
| how many zeroes are there at the end of the number n if n = num__60 ! + num__120 ! ? <o> a ) num__12 <o> b ) num__14 <o> c ) num__16 <o> d ) num__18 <o> e ) num__20 |
the number of zeroes at the end of num__60 ! will be less than the number of zeroes at the end of num__120 ! hence it is sufficient to calculate the number of zeroes at the end of num__60 ! the number of zeroes = [ num__12.0 ] + [ num__2.4 ] + [ num__0.48 ] = num__12 + num__2 + num__0 = num__14 the answer is b . <eor> b <eos> |
b |
round_down__2.4__ round_down__0.48__ add__2.0__12.0__ add__2.0__12.0__ |
round_down__2.4__ round_down__0.48__ add__2.0__12.0__ add__2.0__12.0__ |
| a pupil ' s marks were wrongly entered as num__67 instead of num__45 . due to that the average marks for the class got increased by half . the number of pupils in the class is : <o> a ) num__30 <o> b ) num__80 <o> c ) num__44 <o> d ) num__25 <o> e ) num__26 |
let there be x pupils in the class . total increase in marks = ( x * num__0.5 ) = x / num__2 . x / num__2 = ( num__67 - num__45 ) = > x / num__2 = num__22 = > x = num__44 . answer : c <eor> c <eos> |
c |
reverse__0.5__ subtract__67.0__45.0__ multiply__2.0__22.0__ multiply__2.0__22.0__ |
reverse__0.5__ subtract__67.0__45.0__ multiply__2.0__22.0__ multiply__2.0__22.0__ |
| a bag marked at rs . num__80 is sold for rs . num__68 . the rate of discount is : <o> a ) num__12.0 <o> b ) num__13.0 <o> c ) num__15.0 <o> d ) num__18.0 <o> e ) num__19 % |
sol . rate of discount = [ num__0.15 * num__100 ] % = num__15.0 . answer c <eor> c <eos> |
c |
percent__100.0__15.0__ |
percent__100.0__15.0__ |
| if price of t . v set is reduced by num__22.0 then its sale increases by num__86.0 find net effect on sale value <o> a ) num__44 <o> b ) num__45 <o> c ) num__46 <o> d ) num__47 <o> e ) num__48 |
- a + b + ( ( - a ) ( b ) / num__100 ) = - num__22 + num__86 + ( - num__22 * num__86 ) / num__100 = - num__22 + num__86 - num__19 = num__45 answer : b <eor> b <eos> |
b |
percent__100.0__45.0__ |
percent__100.0__45.0__ |
| the average weight of a b and c is num__45 kg . if the average weight of a and b be num__40 kg and that of b and c be num__43 kg then the weight of b is : <o> a ) num__17 <o> b ) num__20 <o> c ) num__26 <o> d ) num__31 <o> e ) none of these |
explanation : let a b c represent their respective weights . then we have : a + b + c = ( num__45 x num__3 ) = num__135 . . . . ( i ) a + b = ( num__40 x num__2 ) = num__80 . . . . ( ii ) b + c = ( num__43 x num__2 ) = num__86 . . . . ( iii ) adding ( ii ) and ( iii ) we get : a + num__2 b + c = num__166 . . . . ( iv ) subtracting ( i ) from ( iv ) we get : b = num__31 . b ' s weight = num__31 kg . answer : d <eor> d <eos> |
d |
subtract__43.0__40.0__ multiply__45.0__3.0__ subtract__45.0__43.0__ multiply__40.0__2.0__ multiply__43.0__2.0__ add__80.0__86.0__ subtract__166.0__135.0__ subtract__166.0__135.0__ |
subtract__43.0__40.0__ multiply__45.0__3.0__ subtract__45.0__43.0__ multiply__40.0__2.0__ multiply__43.0__2.0__ add__80.0__86.0__ subtract__166.0__135.0__ subtract__166.0__135.0__ |
| the average of num__6 no . ' s is num__3.95 . the average of num__2 of them is num__4.2 while the average of theother num__2 is num__3.85 . what is the average of the remaining num__2 no ' s ? <o> a ) num__4.2 <o> b ) num__4.4 <o> c ) num__4.6 <o> d ) num__3.8 <o> e ) num__5.7 |
sum of the remaining two numbers = ( num__3.95 * num__6 ) - [ ( num__4.2 * num__2 ) + ( num__3.85 * num__2 ) ] = num__7.60 . required average = ( num__7.6 / num__2 ) = num__3.8 . d <eor> d <eos> |
d |
divide__7.6__2.0__ divide__7.6__2.0__ |
divide__7.6__2.0__ divide__7.6__2.0__ |
| a charitable association sold an average of num__66 raffle tickets per member . among the female members the average was num__70 raffle tickets . the male to female ratio of the association is num__1 : num__2 . what was the average number w of tickets sold by the male members of the association <o> a ) num__50 <o> b ) num__56 <o> c ) num__58 <o> d ) num__62 <o> e ) num__66 |
given that total average w sold is num__66 male / female = num__0.5 and female average is num__70 . average of male members isx . ( num__70 * f + x * m ) / ( m + f ) = num__66 - > solving this equation after substituting num__2 m = f x = num__58 . ans c . <eor> c <eos> |
c |
reverse__2.0__ multiply__1.0__58.0__ |
reverse__2.0__ multiply__1.0__58.0__ |
| a metallic sphere of radius num__12 cm is melted and drawn into a wire whose radius of cross section is num__16 cm . what is the length of the wire ? <o> a ) num__8 cm <o> b ) num__5 cm <o> c ) num__4 cm <o> d ) num__2 cm <o> e ) num__9 cm |
volume of the wire ( in cylindrical shape ) is equal to the volume of the sphere . π ( num__16 ) num__2 * h = ( num__1.33333333333 ) π ( num__12 ) num__3 = > h = num__9 cm answer : e <eor> e <eos> |
e |
divide__16.0__12.0__ subtract__12.0__3.0__ round__9.0__ |
divide__16.0__12.0__ subtract__12.0__3.0__ round__9.0__ |
| a shopkeeper loses num__15.0 if an article is sold for rs . num__102 . what should be the selling price of the article to gain num__20.0 ? <o> a ) num__228 <o> b ) num__144 <o> c ) num__268 <o> d ) num__277 <o> e ) num__299 |
given that sp = rs . num__102 and loss = num__15.0 cp = [ num__100 ( sp ) ] / ( num__100 - l % ) = ( num__100 * num__102 ) / num__85 = num__20 * num__6 = rs . num__120 . to get num__20.0 profit new sp = [ ( num__100 + p % ) cp ] / num__100 = ( num__120 * num__120 ) / num__100 = rs . num__144 answer : b <eor> b <eos> |
b |
percent__100.0__144.0__ |
percent__100.0__144.0__ |
| a can do a piece of work in num__80 days . he works at it for num__10 days & then b alone finishes the remaining work in num__42 days . in how much time will a and b working together finish the work ? <o> a ) num__20 days <o> b ) num__10 days <o> c ) num__30 days <o> d ) num__60 days <o> e ) num__50 days |
work done by a in num__10 days = num__0.125 = num__0.125 remaining work = ( num__1 - ( num__0.125 ) ) = num__0.875 now work will be done by b in num__42 days . whole work will be done by b in ( num__42 * num__1.14285714286 ) = num__48 days therefore a ' s one day ' s work = num__0.0125 b ’ s one day ' s work = num__0.0208333333333 ( a + b ) ' s one day ' s work = num__0.0125 + num__0.0208333333333 = num__0.0333333333333 = num__0.0333333333333 hence both will finish the work in num__30 days . answer : c . <eor> c <eos> |
c |
divide__10.0__80.0__ subtract__1.0__0.125__ divide__1.0__0.875__ divide__42.0__0.875__ divide__0.125__10.0__ divide__1.0__48.0__ add__0.0125__0.0208__ round__30.0__ |
divide__10.0__80.0__ subtract__1.0__0.125__ divide__1.0__0.875__ divide__42.0__0.875__ divide__0.125__10.0__ divide__1.0__48.0__ add__0.0125__0.0208__ round__30.0__ |
| in a car wheel twenty spokes cover num__120 degrees . then for the entire car how many spokes are there ? <o> a ) num__240 <o> b ) num__340 <o> c ) num__440 <o> d ) num__540 <o> e ) num__640 |
given num__20 spokes cover num__120 degrees so for num__360 degrees . - - - - > num__360 * num__0.166666666667 = num__60 . . so for entire car num__4 * num__60 = num__240 answer : a <eor> a <eos> |
a |
clock_big_arm_angle__4.0__ clock_big_arm_angle__4.0__ |
clock_big_arm_angle__4.0__ clock_big_arm_angle__4.0__ |
| what is the sum of the greatest common factor and the lowest common multiple of num__24 and num__36 ? <o> a ) num__92 <o> b ) num__84 <o> c ) num__24 <o> d ) num__60 <o> e ) num__184 |
prime factorization of num__24 = num__2 x num__2 x num__2 x num__3 prime factorization of num__36 = num__2 x num__2 x num__3 x num__3 gcf = num__12 lcm = num__72 sum = num__84 answer b . <eor> b <eos> |
b |
divide__24.0__2.0__ multiply__24.0__3.0__ add__72.0__12.0__ add__72.0__12.0__ |
divide__24.0__2.0__ multiply__24.0__3.0__ add__72.0__12.0__ add__72.0__12.0__ |
| if a radio is purchased for rs num__490 and sold for rs num__465.50 find the loss % ? <o> a ) num__3.0 <o> b ) num__4.0 <o> c ) num__5 percent <o> d ) num__6.0 <o> e ) none of these |
explanation : loss = num__490 – num__465.5 = num__24.5 loss in % = num__24.5 / num__490 x num__100 = num__5.0 . answer : c <eor> c <eos> |
c |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| analyze the number square carefully and find out what number fits in place of question mark . num__2 num__4 num__6 num__3 num__6 num__9 num__4 num__8 ? <o> a ) num__2 <o> b ) num__8 <o> c ) num__4 <o> d ) num__14 <o> e ) num__20 |
num__20 fits in place of question mark . this is because ( number in first column ) / ( number in second column ) * num__8 = number in third column . here ( left number / middle number ) * num__8 = right number ( num__0.5 ) * num__8 = ( num__0.5 ) * num__8 = num__1 x num__4 = num__4 c <eor> c <eos> |
c |
rectangle_perimeter__2.0__8.0__ multiply__2.0__0.5__ square_perimeter__1.0__ |
rectangle_perimeter__2.0__8.0__ multiply__2.0__0.5__ multiply__4.0__1.0__ |
| the cost to rent a small bus for a trip is x dollars which is to be shared equally among the people taking the trip . if num__10 people take the trip rather than num__18 how many more dollars in terms of x will it cost per person ? <o> a ) x / num__6 <o> b ) x / num__16 <o> c ) x / num__40 <o> d ) num__3 x / num__40 <o> e ) num__3 x / num__80 |
choose x as a multiple of num__16 i chose num__64 : so for num__10 people that ' s num__6.4 each and for num__18 people it ' s num__4 usd each . . . pick one of the options that gives you num__6.4 - num__4 = num__2.4 . . . the answer is b . <eor> b <eos> |
b |
divide__64.0__10.0__ divide__64.0__16.0__ subtract__6.4__4.0__ divide__64.0__4.0__ |
divide__64.0__10.0__ divide__64.0__16.0__ subtract__6.4__4.0__ divide__64.0__4.0__ |
| num__8.008 / num__2.002 <o> a ) num__0.004 <o> b ) num__0.04 <o> c ) num__4 <o> d ) num__40 <o> e ) num__400 |
answer is num__4 move the decimal forward three places for both numerator and denominator or just multiply both by a thousand . the result is num__4.0 = num__4 answer c <eor> c <eos> |
c |
divide__8.008__2.002__ divide__8.008__2.002__ |
divide__8.008__2.002__ divide__8.008__2.002__ |
| a man goes downstream at num__10 kmph and upstream num__8 kmph . the speed of the stream is <o> a ) num__0 kmph <o> b ) num__4 kmph <o> c ) num__1 kmph <o> d ) num__2.5 kmph <o> e ) num__26 kmph |
speed of the stream = num__0.5 ( num__10 - num__8 ) kmph = num__1 kmph . correct option : c <eor> c <eos> |
c |
round__1.0__ |
round__1.0__ |
| a sum of money is to be distributed among a b c d in the proportion of num__5 : num__2 : num__4 : num__3 . if c gets rs . num__2000 more than d what is b ' s share ? <o> a ) rs . num__500 <o> b ) rs . num__1000 <o> c ) rs . num__4000 <o> d ) rs . num__2000 <o> e ) none |
sol . let the shares of a b c and d be rs . num__5 x rs . num__2 x rs . num__4 x and rs . num__3 x respectively . then num__4 x - num__3 x = num__2000 ⇔ x = num__2000 . ∴ b ' s share = rs . num__2 x = rs . ( num__2 x num__2000 ) = rs . num__4000 . answer c <eor> c <eos> |
c |
multiply__2.0__2000.0__ multiply__2.0__2000.0__ |
multiply__2.0__2000.0__ multiply__2.0__2000.0__ |
| num__375 metres long yard num__26 trees are palnted at equal distances one tree being at each end of the yard . what is the distance between num__2 consecutive trees <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__16 <o> e ) num__15 |
num__26 trees have num__25 gaps between them required distance ( num__15.0 ) = num__15 e <eor> e <eos> |
e |
divide__375.0__25.0__ round__15.0__ |
divide__375.0__25.0__ round__15.0__ |
| if a ( a - num__8 ) = num__9 and b ( b - num__8 ) = num__9 where a ≠ b then a + b = <o> a ) − num__48 <o> b ) − num__2 <o> c ) num__8 <o> d ) num__9 <o> e ) num__48 |
i . e . if a = - num__1 then b = num__9 or if a = num__9 then b = - num__1 but in each case a + b = - num__1 + num__9 = num__8 answer : option c <eor> c <eos> |
c |
subtract__9.0__8.0__ multiply__8.0__1.0__ |
subtract__9.0__8.0__ subtract__9.0__1.0__ |
| mary decided to save a certain amount of her monthly salary each month and her salary was unchanged from month to month . if mary ' s savings by the end of the year from these monthly savings were four times the amount she spent per month what should be the fraction of her salary that she spent each month ? <o> a ) num__0.25 <o> b ) num__0.75 <o> c ) num__0.5 <o> d ) num__1 <o> e ) num__1.25 |
saving = s expenditure = e income = i at the end of num__12 months ; num__12 s = num__4 e = num__4 ( i - s ) = > num__16 s = num__4 i = > savings = num__0.25 i hence expenditure = num__1 - ( num__0.25 ) = num__0.75 i answer ( b ) <eor> b <eos> |
b |
add__4.0__12.0__ reverse__4.0__ multiply__0.25__4.0__ subtract__1.0__0.25__ multiply__0.75__1.0__ |
add__4.0__12.0__ reverse__4.0__ multiply__0.25__4.0__ subtract__1.0__0.25__ subtract__1.0__0.25__ |
| if x y and z are three different prime numbers which of the following is the smallest possible value of x + y + z ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__16 <o> e ) num__18 |
the sum of the three small smallest primes numbers is num__2 + num__3 + num__5 = num__10 . the answer is a . <eor> a <eos> |
a |
add__2.0__3.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
add__2.0__3.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
| in the city of san durango num__60 people own cats dogs or rabbits . if num__30 people owned cats num__40 owned dogs num__10 owned rabbits and num__8 owned exactly two of the three types of pet how many people owned all three ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__12 <o> e ) num__32 |
you are considering a case when cat dogs and rabbit are not exactly equal to num__12 . the solution shall be num__60 = num__30 + num__40 + num__10 - num__8 - num__2 x and hence x shall be num__6 answer c <eor> c <eos> |
c |
divide__60.0__30.0__ divide__60.0__10.0__ divide__60.0__10.0__ |
subtract__10.0__8.0__ subtract__8.0__2.0__ subtract__8.0__2.0__ |
| it is being given that ( num__5 ^ num__32 + num__1 ) is completely divisible by a whole number . which of the following numbers is completely divisible by this number ? <o> a ) num__5 ^ num__16 + num__1 <o> b ) num__5 ^ num__16 - num__1 <o> c ) num__7 * num__5 ^ num__33 <o> d ) num__5 ^ num__96 + num__1 <o> e ) none |
solution : let num__5 ^ num__32 = x . then ( num__5 ^ num__32 + num__1 ) = ( x + num__1 ) . let ( x + num__1 ) be completely divisible by the whole number y . then ( num__5 ^ num__96 + num__1 ) = [ ( num__5 ^ num__32 ) ^ num__3 + num__1 ] = > ( x ^ num__3 + num__1 ) = ( x + num__1 ) ( x ^ num__2 - x + num__1 ) which is completely divisible by y . since ( x + num__1 ) is divisible by y . answer d <eor> d <eos> |
d |
divide__96.0__32.0__ subtract__5.0__3.0__ multiply__5.0__1.0__ |
divide__96.0__32.0__ subtract__5.0__3.0__ add__2.0__3.0__ |
| for any positive integer n the sum of the first n positive integers equals n ( n + num__1 ) / num__2 . what is the sum of all the even integers between num__99 and num__301 ? <o> a ) num__10100 <o> b ) num__20200 <o> c ) num__22650 <o> d ) num__40200 <o> e ) num__45 |
150 |
firstly calculate the number of even integers between num__99 and num__301 : no of even integers = ( num__300 - num__100 ) / num__2 + num__1 = num__101 avg of the even integers is ( num__100 + num__300 ) / num__2 = num__200 . the sum would be = no of even integers * avg . = = > num__101 * num__202 = num__20200 answer b . <eor> b <eos> |
b |
b |
| how many numbers from num__32 to num__97 are exactly divisible by num__9 ? <o> a ) num__5 <o> b ) num__7 <o> c ) num__9 <o> d ) num__11 <o> e ) num__12 |
option ' b ' num__3.55555555556 = num__3 and num__10.7777777778 = num__10 = = > num__10 - num__3 = num__7 numbers <eor> b <eos> |
b |
divide__32.0__9.0__ round_down__3.5556__ divide__97.0__9.0__ round_down__10.7778__ subtract__10.0__3.0__ subtract__10.0__3.0__ |
divide__32.0__9.0__ round_down__3.5556__ divide__97.0__9.0__ round_down__10.7778__ subtract__10.0__3.0__ subtract__10.0__3.0__ |
| on dividing a number by num__5 we get num__3 as remainder . what will be the remainder when the square of this number is divided by num__5 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__3 <o> d ) num__4 <o> e ) num__2 |
let the number be x and on dividing x by num__5 we get k as quotient and num__3 as remainder . x = num__5 k + num__3 x ^ num__2 = ( num__5 k + num__3 ) ^ num__2 = ( num__25 k ^ num__2 + num__30 k + num__9 ) = num__5 ( num__5 k ^ num__2 + num__6 k + num__1 ) + num__4 on dividing x ^ num__2 by num__5 we get num__4 as remainder . answer is d <eor> d <eos> |
d |
power__5.0__2.0__ volume_rectangular_prism__5.0__3.0__2.0__ power__3.0__2.0__ multiply__3.0__2.0__ square_perimeter__1.0__ square_perimeter__1.0__ |
power__5.0__2.0__ volume_rectangular_prism__5.0__3.0__2.0__ power__3.0__2.0__ multiply__3.0__2.0__ square_perimeter__1.0__ power__4.0__1.0__ |
| a train with a length of num__100 meters is traveling at a speed of num__72 km / hr . the train enters a tunnel num__3.5 km long . how many minutes does it take the train to pass through the tunnel from the moment the front enters to the moment the rear emerges ? <o> a ) num__3 <o> b ) num__4.2 <o> c ) num__3.4 <o> d ) num__5.5 <o> e ) num__5.7 |
num__72 km / hr = num__1.2 km / min the total distance is num__3.6 km . num__3.6 / num__1.2 = num__3 minutes the answer is a . <eor> a <eos> |
a |
divide__3.6__1.2__ round__3.0__ |
divide__3.6__1.2__ divide__3.6__1.2__ |
| if a card is drawn from a well shuffled pack of cards the probability of drawing a spade or a king is ? <o> a ) num__0.222222222222 <o> b ) num__0.333333333333 <o> c ) num__0.266666666667 <o> d ) num__0.307692307692 <o> e ) num__0.25 |
p ( s ᴜ k ) = p ( s ) + p ( k ) - p ( s ∩ k ) where s denotes spade and k denotes king . p ( s ᴜ k ) = num__0.25 + num__0.0769230769231 - num__0.0192307692308 = num__0.307692307692 answer : d <eor> d <eos> |
d |
union_prob__0.25__0.0769__0.0192__ union_prob__0.25__0.0769__0.0192__ |
union_prob__0.25__0.0769__0.0192__ union_prob__0.25__0.0769__0.0192__ |
| an article is bought for rs . num__475 and sold for rs . num__525 find the gain percent ? <o> a ) num__33 num__0.142857142857 % <o> b ) num__33 num__0.166666666667 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__10.53 <o> e ) num__33 num__0.666666666667 % |
num__475 - - - - num__50 num__100 - - - - ? = > num__10.53 answer : d <eor> d <eos> |
d |
percent__100.0__10.53__ |
percent__100.0__10.53__ |
| in an animal behavior experiment num__40 tagged white pigeons and num__200 tagged gray pigeons were released from a laboratory . within one week num__88 percent of the white pigeons and num__80.5 percent of the gray pigeons had returned to the laboratory . what percent of the total number of pigeons returned to the laboratory ? <o> a ) num__80.5 <o> b ) num__82 <o> c ) num__81.75 <o> d ) num__85 <o> e ) num__86.5 |
weighted mix method is best here . . . . ratio is num__1 : num__5 for num__88 : num__80.5 . . . so the answer required = num__80.5 + ( num__88 - num__80.5 ) * num__0.166666666667 = num__80.5 + num__7.5 * num__0.2 num__80.5 + num__1.25 = num__81.75 c <eor> c <eos> |
c |
divide__200.0__40.0__ subtract__88.0__80.5__ reverse__5.0__ add__80.5__1.25__ add__80.5__1.25__ |
divide__200.0__40.0__ subtract__88.0__80.5__ reverse__5.0__ add__80.5__1.25__ add__80.5__1.25__ |
| a num__300 meter long train crosses a platform in num__39 seconds while it crosses a signal pole in num__18 seconds . what is the length of the platform ? <o> a ) num__769 m <o> b ) num__678 m <o> c ) num__350 m <o> d ) num__168 m <o> e ) num__198 m |
speed = [ num__16.6666666667 ] m / sec = num__16.6666666667 m / sec . let the length of the platform be x meters . then x + num__7.69230769231 = num__16.6666666667 num__3 ( x + num__300 ) = num__1950 è x = num__350 m . answer : c <eor> c <eos> |
c |
divide__300.0__18.0__ divide__300.0__39.0__ round__350.0__ |
divide__300.0__18.0__ divide__300.0__39.0__ round__350.0__ |
| two trains leave a station traveling in the same direction . train a leaves traveling at a constant speed of num__50 mph while train b leaves traveling at a constant speed of num__80 mph . if train b left the station num__30 minutes after train a left in how many minutes will train b overtake train a ? <o> a ) num__30 <o> b ) num__40 <o> c ) num__50 <o> d ) num__60 <o> e ) num__70 |
we can use a form of the equation d = rt [ distance = rate * time ] train a will have traveled for num__30 minutes longer when train b overtakes it so time of train a : t + num__30 minutes = t + num__0.5 hours ( switch to hours since the rates are in hours ) time of train b : t rate of train a : num__50 mph rate of train b : num__80 mph the distance traveled by each will be the same when b overtakes a so set the right side of d = rt equal to each other for the two trains num__50 * ( t + num__0.5 ) = num__80 * t num__50 t + num__25 = num__80 t num__25 = num__30 t num__0.833333333333 = t num__0.833333333333 hours = t which is num__0.833333333333 * num__60 = num__50 minutes c <eor> c <eos> |
c |
multiply__50.0__0.5__ divide__25.0__30.0__ hour_to_min_conversion__ round__50.0__ |
multiply__50.0__0.5__ divide__25.0__30.0__ hour_to_min_conversion__ round__50.0__ |
| num__3 : num__5 = num__1.33333333333 : x . the value of x is ? <o> a ) num__1 <o> b ) num__2.22222222222 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
x * num__3 = num__5 * num__1.33333333333 x * num__3 = num__6.66666666667 x = num__2.22222222222 answer : b <eor> b <eos> |
b |
divide__6.6667__3.0__ divide__6.6667__3.0__ |
divide__6.6667__3.0__ divide__6.6667__3.0__ |
| how many natural numbers between num__23 and num__137 are divisible by num__7 ? <o> a ) num__12 <o> b ) num__17 <o> c ) num__16 <o> d ) num__13 <o> e ) num__15 |
solution : these numbers are num__28 num__35 num__42 … . num__133 . this is in a . p . in which a = num__28 d = ( num__35 - num__28 ) = num__7 and l = num__133 . let the number of there terms be n . then tn = num__133 a + ( n - num__1 ) d = num__133 by solving this we will get n = num__16 . answer c <eor> c <eos> |
c |
add__7.0__28.0__ add__7.0__35.0__ subtract__23.0__7.0__ subtract__23.0__7.0__ |
add__7.0__28.0__ add__7.0__35.0__ subtract__23.0__7.0__ subtract__23.0__7.0__ |
| a train num__300 m long passed a pole in num__30 sec . how long will it take to pass a platform num__1000 m long ? <o> a ) num__90 sec <o> b ) num__100 sec <o> c ) num__110 sec <o> d ) num__120 sec <o> e ) num__130 sec |
speed = num__10.0 = num__10 m / sec . required time = ( num__300 + num__1000 ) / num__10 = num__130 sec answer : e <eor> e <eos> |
e |
divide__300.0__30.0__ round__130.0__ |
divide__300.0__30.0__ round__130.0__ |
| one water pump can fill half of a certain empty tank in num__3 hours . another pump can fill half of the same tank in num__312312 hours . working together how long will it take these two pumps to fill the entire tank ? <o> a ) num__1 num__0.538461538462 <o> b ) num__1 num__0.625 <o> c ) num__3 num__0.25 <o> d ) num__3 num__0.230769230769 <o> e ) num__3 num__0.5 |
one pump can fill a tank in num__3 hours and another in num__3.5 hours so the rate at which both can half fill the tank is ( num__0.333333333333 + num__1 / num__3.5 ) = > num__0.619047619048 thus half of the tank can be filled in num__1.61538461538 so for filling the complete tank = > num__1.61538461538 * num__2 = num__3.23076923077 answer : d <eor> d <eos> |
d |
subtract__3.0__1.0__ multiply__2.0__1.6154__ round__3.0__ |
subtract__3.0__1.0__ multiply__2.0__1.6154__ divide__3.0__1.0__ |
| cost is expressed by the formula tb ^ num__4 . if b is doubled the new cost q is what percent of the original cost ? <o> a ) q = num__200 <o> b ) q = num__600 <o> c ) q = num__800 <o> d ) q = num__1600 <o> e ) q = num__50 |
original cost c num__1 = t num__1 * b num__1 ^ num__4 new cost c num__2 = t num__2 * b num__2 ^ num__4 . . . . only b is doubled so t num__2 = t num__1 and b num__2 = num__2 b num__1 c num__2 = t num__2 * ( num__2 b num__1 ) ^ num__4 = num__16 ( t num__1 * b num__1 ^ num__4 ) = num__16 c num__1 num__16 times c num__1 = > num__1600.0 of c num__1 ans d = num__1600 <eor> d <eos> |
d |
multiply__1.0__1600.0__ |
multiply__1.0__1600.0__ |
| a man walking at the rate of num__5 km / hr crosses a bridge in num__15 minutes . the length of the bridge ( in metres ) is ? <o> a ) num__2267 <o> b ) num__1162 <o> c ) num__2276 <o> d ) num__1250 <o> e ) num__1262 |
explanation : speed = ( num__5 x num__0.277777777778 ) m / sec = num__1.38888888889 m / sec . distance covered in num__15 minutes = ( num__1.38888888889 x num__15 x num__60 ) m = num__1250 m . answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ round__1250.0__ |
hour_to_min_conversion__ round__1250.0__ |
| if a card is drawn from a well shuffled pack of cards the probability of drawing a spade or a king is ? <o> a ) num__0.137931034483 <o> b ) num__0.2 <o> c ) num__0.173913043478 <o> d ) num__0.307692307692 <o> e ) num__0.0434782608696 |
p ( s ᴜ k ) = p ( s ) + p ( k ) - p ( s ∩ k ) where s denotes spade and k denotes king . p ( s ᴜ k ) = num__0.25 + num__0.0769230769231 - num__0.0192307692308 = num__0.307692307692 answer : d <eor> d <eos> |
d |
union_prob__0.25__0.0769__0.0192__ union_prob__0.25__0.0769__0.0192__ |
union_prob__0.25__0.0769__0.0192__ union_prob__0.25__0.0769__0.0192__ |
| in what ratio p : q should the mixture p of milk and water in the ratio of num__2 : num__1 be mixed with another mixture q of milk and water in the ratio num__1 : num__5 so that the resultant mixture contains equal quantities of milk and water ? <o> a ) num__3 : num__2 <o> b ) num__2 : num__1 <o> c ) num__4 : num__3 <o> d ) num__3 : num__5 <o> e ) num__3 : num__1 |
( num__0.666666666667 ) * p + ( num__0.166666666667 ) * q = ( num__0.333333333333 ) * p + ( num__0.833333333333 ) * q num__2 p = num__4 q p / q = num__2.0 the answer is b . <eor> b <eos> |
b |
subtract__1.0__0.6667__ subtract__1.0__0.1667__ subtract__5.0__1.0__ multiply__2.0__1.0__ |
divide__0.6667__2.0__ subtract__1.0__0.1667__ subtract__5.0__1.0__ multiply__2.0__1.0__ |
| what is the simple interest on rs num__7000 at num__10.0 per annum for the period from num__16 th january num__2017 to num__8 th april num__2017 . <o> a ) rs . num__77.8 <o> b ) rs . num__95.8 <o> c ) rs . num__47.8 <o> d ) rs . num__97.8 <o> e ) rs . num__90.8 |
explanation : one thing which is tricky in this question is to calculate the number of days . always remember that the day on which money is deposited is not counted while the day on which money is withdrawn is counted . so lets calculate the number of days now time = ( num__15 + num__28 + num__8 ) days = num__0.139726027397 years p = num__7000 r = num__10.0 s . i . = = num__7000 Ã — num__10 Ã — num__51 / ( num__365 * num__100 ) = num__97.8 answer : d <eor> d <eos> |
d |
percent__100.0__97.8__ |
percent__100.0__97.8__ |
| calculate the percentage gain if a trader bought a bicycle for rs . num__550 and sold it for rs . num__650 ? <o> a ) num__11.18 <o> b ) num__13.18 <o> c ) num__18.18 <o> d ) num__19.18 <o> e ) num__12.18 % |
c . p . = num__500 s . p . = num__650 gain = num__650 - num__550 - - - - num__100.0 gain = > num__0.181818181818 * num__100 = > num__18.18 answer : c <eor> c <eos> |
c |
percent__100.0__18.18__ |
percent__100.0__18.18__ |
| if a certain sample of data has a mean of num__30.0 and a standard deviation of num__3.5 which of the following values is more than num__2.5 standard deviations from the mean ? <o> a ) num__12.0 <o> b ) num__13.5 <o> c ) num__17.0 <o> d ) num__23.5 <o> e ) num__26.5 |
the standard deviation is num__3.5 . num__2.5 standard deviations which is just num__2.5 x num__3.5 is num__8.75 . the only answer more than num__8.75 away from the mean is a . <eor> a <eos> |
a |
multiply__3.5__2.5__ divide__30.0__2.5__ |
multiply__3.5__2.5__ divide__30.0__2.5__ |
| in a graduate physics course num__70 percent of the students are male and num__30 percent of the students are married . if two - sevenths of the male students are married what fraction of the female students is married ? <o> a ) num__0.285714285714 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__0.714285714286 |
let assume there are num__100 students of which num__70 are male and num__30 are females if num__30 are married then num__70 will be single . now its given that two - sevenths of the male students are married that means num__0.285714285714 of num__70 = num__20 males are married if num__30 is the total number of students who are married and out of that num__20 are males then the remaining num__10 will be females who are married . total females = num__30 married females = num__10 then single females = num__30 - num__10 = num__20 we need to find the fraction of female students who are single i . e single female students / total female student = num__0.333333333333 = num__0.333333333333 [ b ] <eor> b <eos> |
b |
percent__100.0__0.3333__ |
percent__100.0__0.3333__ |
| if you cut a num__22 ft piece of wood into two pieces making one piece num__4 ft longer than the other . what size is the smaller piece ? <o> a ) num__4 ft <o> b ) num__6 ft <o> c ) num__8 ft <o> d ) num__10 ft <o> e ) num__9 ft |
total length is num__22 ft one piece is num__4 ft longer ( x + num__4 ) leaving the other piece to figure out ( x ) . ( x ) + ( x + num__4 ) = num__22 x + x + num__4 - num__4 = num__22 - num__4 num__2 x = num__18 num__2 x / num__2 = num__9.0 x = num__9 the piece is e ) num__9 ft . <eor> e <eos> |
e |
subtract__22.0__4.0__ divide__18.0__2.0__ subtract__18.0__9.0__ |
subtract__22.0__4.0__ divide__18.0__2.0__ subtract__18.0__9.0__ |
| a man whose speed is num__4.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at num__1.5 kmph find his average speed for the total journey ? <o> a ) num__7 <o> b ) num__4 <o> c ) num__3 <o> d ) num__2 <o> e ) num__9 |
m = num__45 s = num__1.5 ds = num__6 us = num__3 as = ( num__2 * num__6 * num__3 ) / num__9 = num__4 answer : b <eor> b <eos> |
b |
add__4.5__1.5__ subtract__4.5__1.5__ divide__3.0__1.5__ multiply__4.5__2.0__ round_down__4.5__ round_down__4.5__ |
add__4.5__1.5__ divide__4.5__1.5__ divide__3.0__1.5__ multiply__4.5__2.0__ divide__6.0__1.5__ divide__6.0__1.5__ |
| l num__2 q l num__6 s p num__12 u r num__20 w t num__30 u ? <o> a ) v num__42 y <o> b ) u num__47 y <o> c ) v num__47 z <o> d ) v num__47 y <o> e ) v num__46 x |
l - q = num__2 l - s = num__6 p - u = num__12 r - w = num__20 t - w = num__30 num__1 ^ num__2 + num__1 num__2 ^ num__2 + num__2 num__3 ^ num__2 + num__3 num__4 ^ num__2 + num__4 num__5 ^ num__2 + num__5 - - - > as per number sequence = > next number = num__6 ^ num__2 + num__6 = num__42 v - y = num__42 = > v num__42 y answer : a <eor> a <eos> |
a |
add__2.0__1.0__ subtract__6.0__2.0__ add__2.0__3.0__ add__12.0__30.0__ add__12.0__30.0__ |
add__2.0__1.0__ subtract__6.0__2.0__ add__2.0__3.0__ add__12.0__30.0__ add__12.0__30.0__ |
| if tier is written as num__7163 and brain is written as num__23415 how is rent coded ? <o> a ) num__3653 <o> b ) num__3657 <o> c ) num__2977 <o> d ) num__2790 <o> e ) num__2711 |
explanation : given : letter : t i e r b a n code : num__7 num__1 num__6 num__3 num__2 num__4 num__5 thus the code for rent is num__3657 . answer : b <eor> b <eos> |
b |
subtract__7.0__1.0__ subtract__3.0__1.0__ add__1.0__3.0__ add__1.0__4.0__ multiply__1.0__3657.0__ |
subtract__7.0__1.0__ subtract__3.0__1.0__ add__1.0__3.0__ add__1.0__4.0__ multiply__1.0__3657.0__ |
| a dishonest dealer professes to sell goods at the cost price but uses a weight of num__780 grams per kg what is his percent ? <o> a ) num__28.0 <o> b ) num__25.0 <o> c ) num__55.0 <o> d ) num__28.0 <o> e ) num__28.2 % |
num__780 - - - num__220 num__100 - - - ? = > num__28.2 answer : e <eor> e <eos> |
e |
percent__100.0__28.2__ |
percent__100.0__28.2__ |
| of the final grades received by the students in a certain math course num__0.2 are a ' s num__0.25 are b ' s num__0.5 are c ' s and the remaining num__40 grades are d ' s . what is the number of students in the course ? <o> a ) num__80 <o> b ) num__110 <o> c ) num__160 <o> d ) num__800 <o> e ) num__400 |
we start by creating a variable for the total number of students in the math course . we can say : t = total number of students in the math course next we can use variable t in an equation that we translate from the given information . we are given that of the final grades received by the students in a certain math course num__0.2 are a ' s num__0.25 are b ' s num__0.5 are c ' s and the remaining num__40 grades are d ' s . since this represents all the grades in the class it represents all the students in the class . thus we know : # a ’ s + # b ’ s + # c ’ s + # d ’ s = total number of students in the class num__0.2 ( t ) + ¼ ( t ) + ½ ( t ) + num__40 = t we can multiply the entire equation by num__20 to cancel out the denominators of the fractions and we have : num__4 t + num__5 t + num__10 t + num__800 = num__20 t num__19 t + num__800 = num__20 t num__800 = t there are a total of num__800 students in the math class . answer is d . <eor> d <eos> |
d |
multiply__0.5__40.0__ reverse__0.25__ reverse__0.2__ multiply__0.25__40.0__ multiply__40.0__20.0__ multiply__40.0__20.0__ |
multiply__0.5__40.0__ reverse__0.25__ reverse__0.2__ multiply__0.25__40.0__ multiply__40.0__20.0__ multiply__40.0__20.0__ |
| what percent of num__16 is num__16 percent of num__1 ? <o> a ) num__0.001 <o> b ) num__1 <o> c ) num__0.1 <o> d ) num__100 <o> e ) num__101 . |
num__16.0 of num__1 = ( num__0.16 ) * num__1 = num__0.16 to determine what percentage of num__16 this is : [ num__16 ] [ / num__100 * num__16 ] * num__100 = num__1.0 ans : b <eor> b <eos> |
b |
percent__16.0__1.0__ percent__1.0__100.0__ |
percent__16.0__1.0__ percent__1.0__100.0__ |
| in how many years rs . num__150 will produce the same interest at num__6.0 as rs . num__800 produce in num__2 years at num__4 ½ % ? <o> a ) num__4 years <o> b ) num__6 years <o> c ) num__8 years <o> d ) num__9 years <o> e ) num__10 years |
explanation : let simple interest for rs . num__150 at num__6.0 for n years = simple interest for rs . num__800 at num__4 ½ % for num__2 years ( num__150 × num__6 × n ) / num__100 = ( num__800 × ( num__4.5 ) × num__2 ) / num__100 ( num__150 × num__6 × n ) = ( num__800 × ( num__4.5 ) × num__2 num__150 × num__6 × n = num__800 × num__9 num__3 × num__6 × n = num__16 × num__9 num__6 × n = num__16 × num__3 num__2 × n = num__16 n = num__8 years answer : option c <eor> c <eos> |
c |
percent__6.0__150.0__ percent__2.0__150.0__ percent__2.0__800.0__ percent__100.0__8.0__ |
percent__6.0__150.0__ percent__2.0__150.0__ percent__2.0__800.0__ percent__100.0__8.0__ |
| the cross - section of a cannel is a trapezium in shape . if the cannel is num__14 m wide at the top and num__4 m wide at the bottom and the area of cross - section is num__380 sq m the depth of cannel is ? <o> a ) num__39 <o> b ) num__28 <o> c ) num__27 <o> d ) num__42 <o> e ) num__71 |
num__0.5 * d ( num__14 + num__4 ) = num__380 d = num__42 answer : d <eor> d <eos> |
d |
round__42.0__ |
round__42.0__ |
| a hare and a jackal are running a race . three leaps of the hare are equal to four leaps of the jackal . for every six leaps of the hare the jackal takes num__7 leaps . find the ratio of the speed of the hare to the speed of the jackal . <o> a ) num__64 : num__25 <o> b ) num__8 : num__7 <o> c ) num__5 : num__8 <o> d ) num__25 : num__64 <o> e ) num__6 : num__7 |
the hare takes num__6 leaps and the jackal takes num__7 leaps . num__1 hare leap = num__1.33333333333 jackal leaps thus the hare ' s num__6 leaps = num__6 * ( num__1.33333333333 ) = num__8 jackal leaps . the ratio of their speeds is num__8 : num__7 . the answer is b . <eor> b <eos> |
b |
subtract__7.0__6.0__ add__7.0__1.0__ add__7.0__1.0__ |
subtract__7.0__6.0__ add__7.0__1.0__ multiply__8.0__1.0__ |
| a committee that includes num__6 members is about to be divided into num__2 subcommittees with num__3 members each . on what percent of the possible subcommittees that michael is a member of is david also a member ? <o> a ) num__10.0 <o> b ) num__20.0 <o> c ) num__25.0 <o> d ) num__40.0 <o> e ) num__50 % |
committee # num__1 : num__50.0 chance that david will be a member leaving num__2 possible spots the other num__5 people . michael ' s chances of being one of those num__2 is num__0.4 or num__40.0 . chances of david and michael being on this committee is num__50.0 x num__40.0 = num__20.0 . committee # num__2 : same . num__20.0 + num__20.0 = num__40.0 . answer : d <eor> d <eos> |
d |
vowel_space__ choose__5.0__3.0__ choose__5.0__3.0__ |
vowel_space__ choose__5.0__3.0__ choose__5.0__3.0__ |
| num__50 liters of a mixture contains milk and water in the ratio num__3 : num__2 . if num__5 liters of this mixture be replaced by num__5 liters of milk the ratio of milk to water in the new mixture would be ? <o> a ) num__6 : num__5 <o> b ) num__6 : num__2 <o> c ) num__6 : num__4 <o> d ) num__7 : num__4 <o> e ) num__7 : num__2 |
quantity of milk in num__50 liters if mix = num__50 * num__0.6 = num__30 liters quantity of milk in num__55 liters of new mix = num__30 + num__5 = num__35 liters quantity of water in it = num__55 - num__35 = num__20 liters ratio of milk and water in new mix = num__35 : num__20 = num__7 : num__4 answer is d <eor> d <eos> |
d |
divide__3.0__5.0__ multiply__50.0__0.6__ add__50.0__5.0__ add__5.0__30.0__ subtract__50.0__30.0__ add__2.0__5.0__ subtract__7.0__3.0__ add__3.0__4.0__ |
divide__3.0__5.0__ multiply__50.0__0.6__ add__50.0__5.0__ add__5.0__30.0__ subtract__50.0__30.0__ add__2.0__5.0__ subtract__7.0__3.0__ add__3.0__4.0__ |
| if the cp of num__12 rubbers is equal to the sp of num__8 rubbers the gain % is ? <o> a ) num__30.0 <o> b ) num__40.0 <o> c ) num__50.0 <o> d ) num__60.0 <o> e ) num__70 % |
( explanation : friends we know we will need gain amount to get gain percent right . so lets get gain first . let the cost price of num__1 pen is re num__1 cost of num__8 pens = rs num__8 selling price of num__8 pens = num__12 gain = num__12 - num__8 = num__4 gain % = ( gaincost * num__100 ) % = ( num__48 * num__100 ) % = num__50.0 c <eor> c <eos> |
c |
percent__50.0__100.0__ |
percent__50.0__100.0__ |
| when a number is divided by num__13 the remainder is num__11 . when the same number is divided by num__17 then remainder is num__9 . what is the number ? <o> a ) num__349 <o> b ) num__339 <o> c ) num__400 <o> d ) num__500 <o> e ) num__600 |
x = num__13 p + num__11 and x = num__17 q + num__9 num__13 p + num__11 = num__17 q + num__9 num__17 q - num__13 p = num__2 q = num__2 + num__13 p num__17 the least value of p for which q = num__2 + num__13 p = > p = num__26 num__17 x = ( num__13 x num__26 + num__11 ) = ( num__338 + num__11 ) = num__349 answer a <eor> a <eos> |
a |
subtract__13.0__11.0__ multiply__13.0__2.0__ multiply__13.0__26.0__ add__11.0__338.0__ add__11.0__338.0__ |
subtract__13.0__11.0__ add__17.0__9.0__ multiply__13.0__26.0__ add__11.0__338.0__ add__11.0__338.0__ |
| foodmart customers regularly buy at least one of the following products : milk chicken or apples . num__60.0 of shoppers buy milk num__50.0 buy chicken and num__15.0 buy apples . if num__10.0 of the customers buy all num__3 products what percentage of foodmart customers purchase exactly num__2 of the products listed above ? <o> a ) num__5.0 <o> b ) num__10.0 <o> c ) num__15.0 <o> d ) num__25.0 <o> e ) num__30 % |
the formula is n ( aubuc ) = n ( a ) + n ( b ) + n ( c ) - b + a - n where b is both ( it will sum of ( ab ) ( bc ) ( ca ) ) . and a is all and n is neither so plugging in we get num__100 = num__60 + num__50 + num__15 + num__10 - b - num__0 b = num__35 . exactly two = num__35 - num__3 ( num__10 ) = num__5 subtracting num__10.0 three times as this value is including in all the ( ab ) ( bc ) ( ca ) . answer a <eor> a <eos> |
a |
multiply__50.0__2.0__ subtract__50.0__15.0__ divide__50.0__10.0__ divide__50.0__10.0__ |
multiply__50.0__2.0__ subtract__50.0__15.0__ subtract__15.0__10.0__ subtract__15.0__10.0__ |
| the function f is defined for all positive integers w by the following rule . f ( w ) is the number of positive integers each of which is less than w and has no positive factor in common with w other than num__1 . if p is any prime number then f ( p ) = <o> a ) p - num__1 <o> b ) p - num__2 <o> c ) ( p + num__1 ) / num__2 <o> d ) ( p - num__1 ) / num__2 <o> e ) num__2 |
if not the wording the question would n ' t be as tough as it is now . the gmat often hides some simple concept in complicated way of delivering it . this question for instance basically asks : how many positive integers are less than given prime number p which have no common factor with p except num__1 . well as p is a prime all positive numbers less than p have no common factors with p ( except common factor num__1 ) . so there would be p - num__1 such numbers ( as we are looking number of integers less than p ) . for example : if p = num__4 how many numbers are less than num__4 having no common factors with num__4 : num__1 num__2 num__3 - - > num__4 - num__1 = num__3 . answer : a . <eor> a <eos> |
a |
add__1.0__2.0__ reverse__1.0__ |
subtract__4.0__1.0__ subtract__2.0__1.0__ |
| if p and q are prime numbers how many divisors does the product p ^ num__6 * q ^ num__7 have ? <o> a ) num__40 <o> b ) num__56 <o> c ) num__60 <o> d ) num__62 <o> e ) num__70 |
when a number n = a ^ x * b ^ y where a and b are prime numbers and x y are positive integers the number of divisors of n = ( x + num__1 ) ( y + num__1 ) therefore the answer is b . num__7 * num__8 = num__56 <eor> b <eos> |
b |
subtract__7.0__6.0__ add__7.0__1.0__ multiply__7.0__8.0__ multiply__7.0__8.0__ |
subtract__7.0__6.0__ add__7.0__1.0__ multiply__7.0__8.0__ multiply__7.0__8.0__ |
| if num__9 ^ y = num__3 ^ num__12 what is y ? <o> a ) num__2 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__6 |
num__3 ^ num__2 y = num__3 ^ num__12 num__2 y = num__12 therefore y = num__6 e <eor> e <eos> |
e |
subtract__9.0__3.0__ subtract__9.0__3.0__ |
subtract__9.0__3.0__ subtract__9.0__3.0__ |
| a fruit drink is made of orange watermelon and grape juice where num__15 percent of the drink is orange juice and num__60 percent is watermelon juice . if the drink is made with num__35 ounces of grape juice how many ounces is the drink total ? <o> a ) num__120 <o> b ) num__140 <o> c ) num__160 <o> d ) num__180 <o> e ) num__200 |
let the total number of ounces in the drink be x . % of orange = num__15.0 % of watermelon = num__60.0 % of grape = num__100.0 - num__75.0 = num__25.0 num__0.25 x = num__35 x = num__140 therefore there are a total of num__140 ounces in the drink . the answer is b . <eor> b <eos> |
b |
add__15.0__60.0__ subtract__60.0__35.0__ divide__15.0__60.0__ divide__35.0__0.25__ divide__35.0__0.25__ |
add__15.0__60.0__ subtract__60.0__35.0__ divide__15.0__60.0__ divide__35.0__0.25__ divide__35.0__0.25__ |
| the speeds of three asteroids were compared . asteroids x - num__13 and y - num__14 were observed for identical durations while asteroid z - num__15 was observed for num__2 seconds longer . during its period of observation asteroid y - num__14 traveled three times the distance x - num__13 traveled and therefore y - num__14 was found to be faster than x - num__13 by num__1000 kilometers per second . asteroid z - num__15 had an identical speed as that of x - num__13 but because z - num__15 was observed for a longer period it traveled five times the distance x - num__13 traveled during x - num__13 ' s inspection . asteroid x - num__13 traveled how many kilometers during its observation ? <o> a ) num__250 <o> b ) num__533.333333333 <o> c ) num__1000 <o> d ) num__1500 <o> e ) num__2 |
500 |
x num__13 : ( t d s ) y num__14 : ( t num__3 d s + num__1000 mi / hour ) z num__15 : ( t + num__2 seconds s num__5 d ) d = ? distance = speed * time x num__13 : d = s * t x num__14 : num__3 d = ( s + num__1000 ) * t = = = > num__3 d = ts + num__1000 t z num__15 : num__5 d = s * ( t + num__2 t ) = = = > num__5 d = st + num__2 st = = = > num__5 d - num__2 st = st num__3 d = num__5 d - num__2 st + num__1000 t - num__2 d = - num__2 st + num__1000 t num__2 d = num__2 st - num__1000 t d = st - num__500 t x num__13 : d = s * t st - num__500 t = s * t s - num__500 = s - num__250 = s i got to this point and could n ' t go any further . this seems like a problem where i can set up individual d = r * t formulas and solve but it appears that ' s not the case . for future reference how would i know not to waste my time setting up this problem in the aforementioned way ? thanks ! ! ! the distance of z num__15 is equal to five times the distance of x num__13 ( we established that x num__13 is the baseline and thus it ' s measurements are d s t ) s ( t + num__2 ) = num__5 ( s * t ) what clues would i have to know to set up the equation in this fashion ? is it because i am better off setting two identical distances together ? st + num__2 s = num__5 st t + num__2 = num__5 t num__2 = num__4 t t = num__0.5 we are looking for distance ( d = s * t ) so we need to solve for speed now that we have time . speed y num__14 - speed x num__13 speed = d / t num__3 d / t - d / t = num__1000 ( remember t is the same because both asteroids were observed for the same amount of time ) num__2 d = num__1000 num__2 = num__500 d = s * t d = num__500 * ( num__0.5 ) d = num__250 answer : a <eor> a <eos> |
a |
a |
| three candidates contested an election and received num__1136 num__7636 and num__11628 votes respectively . what percentage of the total votes did the winning candidate got <o> a ) num__55.0 <o> b ) num__56.0 <o> c ) num__57.0 <o> d ) num__58.0 <o> e ) none of these |
explanation : total number of votes polled = ( num__1136 + num__7636 + num__11628 ) = num__20400 so required percentage = num__0.57 * num__100 = num__57.0 answer : option c <eor> c <eos> |
c |
divide__11628.0__20400.0__ multiply__100.0__0.57__ multiply__100.0__0.57__ |
divide__11628.0__20400.0__ multiply__100.0__0.57__ multiply__100.0__0.57__ |
| a train is num__360 meter long is running at a speed of num__45 km / hour . in what time will it pass a bridge of num__140 meter length ? <o> a ) num__77 seconds <o> b ) num__66 seconds <o> c ) num__40 seconds <o> d ) num__55 seconds <o> e ) num__43 seconds |
speed = num__45 km / hr = num__45 * ( num__0.277777777778 ) m / sec = num__12.5 m / sec total distance = num__360 + num__140 = num__500 meter time = distance / speed = num__500 * ( num__0.08 ) = num__40 seconds answer : c <eor> c <eos> |
c |
add__360.0__140.0__ divide__500.0__12.5__ round__40.0__ |
add__360.0__140.0__ divide__500.0__12.5__ divide__500.0__12.5__ |
| in the first round of the elections the only two candidates got exactly the same number of votes . during the second round num__16000 votes switched from the first candidate to the second one . the total number of votes remained the same in both rounds and no other votes switched sides . if in the second round the winning candidate got five times as many votes as the other candidate how many people have voted in each round ? <o> a ) num__15000 <o> b ) num__25000 <o> c ) num__40000 <o> d ) num__48000 <o> e ) num__60 |
000 |
let a be the first candidate and b be the second one . given x + num__16000 = num__5 ( x - num__16000 ) = > x = num__24000 num__2 x = num__48000 imo option d . <eor> d <eos> |
d |
d |
| the owner of a furniture shop charges his customer num__24.0 more than the cost price . if a customer paid rs . num__8091 for a computer table then what was the cost price of the computer table ? <o> a ) num__6727 <o> b ) num__6887 <o> c ) num__6728 <o> d ) num__6725 <o> e ) num__6525 |
: cp = sp * ( num__100 / ( num__100 + profit % ) ) = num__8091 ( num__0.806451612903 ) = rs . num__6525 . answer : e <eor> e <eos> |
e |
percent__100.0__6525.0__ |
percent__100.0__6525.0__ |
| in num__4 years raj ' s father age twice as raj two years ago raj ' s mother ' s age twice as raj . if raj is num__32 yrs old in eight yrs from now what is the age of raj ' s mother and father ? <o> a ) num__27 <o> b ) num__36 <o> c ) num__28 years <o> d ) num__46 years <o> e ) num__91 |
raj present age = num__32 - num__8 = num__24 . after num__4 years raj ' s age is num__28 . and raj ' s fathers age is num__28 x num__2 = num__56 and his present age is num__52 . two years ago raj ' s age is num__22 . and his mother ' s age is num__22 x num__2 = num__44 . his mother ' s present age = num__46 answer : c <eor> c <eos> |
c |
divide__32.0__4.0__ subtract__32.0__8.0__ add__4.0__24.0__ divide__8.0__4.0__ add__32.0__24.0__ subtract__56.0__4.0__ subtract__24.0__2.0__ multiply__2.0__22.0__ add__2.0__44.0__ add__4.0__24.0__ |
divide__32.0__4.0__ subtract__32.0__8.0__ subtract__32.0__4.0__ divide__8.0__4.0__ add__32.0__24.0__ subtract__56.0__4.0__ subtract__24.0__2.0__ subtract__52.0__8.0__ add__2.0__44.0__ subtract__32.0__4.0__ |
| two trains each num__100 m long moving in opposite directions cross other in num__4 sec . if one is moving twice as fast the other then the speed of the faster train is ? <o> a ) num__76 km / hr <o> b ) num__66 km / hr <o> c ) num__60 km / hr <o> d ) num__67 km / hr <o> e ) num__120 km / hr |
let the speed of the slower train be x m / sec . then speed of the train = num__2 x m / sec . relative speed = ( x + num__2 x ) = num__3 x m / sec . ( num__100 + num__100 ) / num__4 = num__3 x = > x = num__16.6666666667 so speed of the faster train = num__33.3333333333 = num__33.3333333333 * num__3.6 = num__120 km / hr . answer : e <eor> e <eos> |
e |
divide__100.0__3.0__ round__120.0__ |
divide__100.0__3.0__ round__120.0__ |
| the ratio of two numbers is num__2 : num__3 and their h . c . f . is num__4 . their l . c . m . is <o> a ) num__48 <o> b ) num__24 <o> c ) num__56 <o> d ) num__27 <o> e ) num__35 |
explanation : let the numbers be num__2 x and num__3 x . then their h . c . f . = x . so x = num__4 . so the numbers num__8 and num__12 . l . c . m . of num__8 and num__12 = num__24 . answer : option b <eor> b <eos> |
b |
multiply__2.0__4.0__ multiply__3.0__4.0__ multiply__2.0__12.0__ multiply__2.0__12.0__ |
multiply__2.0__4.0__ multiply__3.0__4.0__ multiply__2.0__12.0__ multiply__2.0__12.0__ |
| a camera lens filter kit containing num__5 filters sells for $ num__57.50 . if the filters are purchased individually num__2 of them are priced at $ num__10.45 each num__2 at $ num__12.05 each num__1 at $ num__17.50 . the amount saved by purchasing the kit is what percent of the total price of the num__5 filters purchased individually ? <o> a ) num__7.0 <o> b ) num__8.0 <o> c ) num__8.5 <o> d ) num__10.0 <o> e ) num__11 % |
cost of kit = $ num__57.50 if filters are purchased individually - $ num__10.45 * num__2 + $ num__12.05 * num__2 + $ num__17.50 = $ num__62.50 amount saved = $ num__62.50 - $ num__57.50 = $ num__5 required % age = ( $ num__5 / $ num__62.50 ) * num__100 = num__8.0 so the correct answer is b . <eor> b <eos> |
b |
add__5.0__57.5__ multiply__1.0__8.0__ |
add__5.0__57.5__ multiply__1.0__8.0__ |
| the sector of a circle has radius of num__14 cm and its perimeter num__50 cm . find its central angel ? <o> a ) num__180 o <o> b ) num__225 o <o> c ) num__270 o <o> d ) num__150 o <o> e ) num__90 o |
lte central angle = x perimeter of the sector = length of the arc + num__2 ( radius ) num__50 = ( x / num__360 * num__2 * num__3.14285714286 * num__14 ) + num__2 ( num__14 ) num__50 = num__88 x / num__360 + num__28 num__88 x / num__360 = num__22 num__88 x = num__7920 x = num__90 answer : e <eor> e <eos> |
e |
multiply__14.0__2.0__ subtract__50.0__28.0__ multiply__360.0__22.0__ add__2.0__88.0__ round__90.0__ |
multiply__14.0__2.0__ subtract__50.0__28.0__ multiply__360.0__22.0__ add__2.0__88.0__ add__2.0__88.0__ |
| paulson spends num__75.0 of his income . his income is increased by num__20.0 and he increased his expenditure by num__10.0 . find the percentage increase in his savings <o> a ) num__50 <o> b ) num__100 <o> c ) num__20 <o> d ) num__25 <o> e ) num__40 |
let the original income = rs . num__100 . then expenditure = rs . num__75 and savings = rs . num__25 new income = rs . num__120 new expenditure = rs . ( ( num__1.1 ) * num__75 ) = rs . num__82.5 new savings = rs . ( num__120 - ( num__82.5 ) ) = rs . num__37.5 increase in savings = rs . ( ( num__37.5 ) - num__25 ) = rs . num__12.5 increase % = ( ( num__12.5 ) * ( num__0.04 ) * num__100 ) % = num__50.0 . answer a <eor> a <eos> |
a |
subtract__100.0__75.0__ add__20.0__100.0__ multiply__75.0__1.1__ subtract__120.0__82.5__ subtract__37.5__25.0__ reverse__25.0__ subtract__75.0__25.0__ subtract__75.0__25.0__ |
subtract__100.0__75.0__ add__20.0__100.0__ multiply__75.0__1.1__ subtract__120.0__82.5__ subtract__37.5__25.0__ reverse__25.0__ subtract__75.0__25.0__ subtract__75.0__25.0__ |
| tough and tricky questions : probability . medical analysts predict that one over five of all people who are infected by a certain biological agent could be expected to be killed for each day that passes during which they have not received an antidote . what fraction of a group of num__1000 people could be expected to be killed if infected and not treated for three full days ? <o> a ) num__0.197530864198 <o> b ) num__0.296296296296 <o> c ) num__0.666666666667 <o> d ) num__0.703703703704 <o> e ) num__0.488 |
at the end of each day we will have num__0.8 alive . so after num__3 days we will have ( num__0.8 ) ^ num__3 people alive . therefore fraction of dead people will be num__1 - ( num__0.8 ) ^ num__3 = num__0.488 . the correct answer is e . <eor> e <eos> |
e |
multiply__0.488__1.0__ |
multiply__0.488__1.0__ |
| a welder received an order to make a num__1 million liter cube - shaped tank . if he has only num__2 x num__3 meter sheets of metal that can be cut how many metal sheets will be required for this order ? ( num__1 cubic meter = num__1000 liters ) <o> a ) num__92 <o> b ) num__90 <o> c ) num__82 <o> d ) num__78 <o> e ) num__100 |
the question say : a welder received an order to make a num__1 million liter cube - shaped tank . ( num__1 cubic meter = num__1000 liters ) in other words the tank is going to have a volume of num__1000 cubic meters . that would equal num__1 million liters . a cube with a volume of num__1000 cubic meters must be ( num__10 meters ) x ( num__10 meters ) x ( num__10 meters ) . the question does n ' t specify whether this tank should have a closed top or an open top . ( the real gmat is good about specifying things like that . ) here we are going to assume a closed top only because if the top is open we do n ' t use enough metal - - - we get an answer smaller than any of the answer choices . a closed - top cube a full cube has num__6 sides each of which is ( num__10 meters ) x ( num__10 meters ) . that ' s a total surface area of num__600 sq m . the question tells us : he has only num__2 x num__3 meter sheets of metal that can be cut . each sheet has an area of num__6 sq m . how many of these need to fill up num__600 sq m ? num__100.0 = num__100 . he needs num__100 sheets . answer = e <eor> e <eos> |
e |
multiply__2.0__3.0__ divide__1000.0__10.0__ multiply__1.0__100.0__ |
multiply__2.0__3.0__ divide__1000.0__10.0__ multiply__1.0__100.0__ |
| if num__8 < x < num__9 and x ^ num__2 = ( num__10 – y ) ( num__10 + y ) which of the following is a possible value for y ? <o> a ) – num__7 <o> b ) – num__6 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
given : num__8 < x < num__9 = > num__64 < x ^ num__2 < num__81 given x ^ num__2 = num__100 - y ^ num__2 therefore num__64 < num__100 - y ^ num__2 < num__81 - num__36 < - y ^ num__2 < - num__19 = num__19 < y ^ num__2 < num__36 only ( e ) y = num__5 = > y ^ num__2 = num__25 satisfies the above equation . <eor> e <eos> |
e |
subtract__100.0__64.0__ add__9.0__10.0__ divide__10.0__2.0__ divide__10.0__2.0__ |
subtract__100.0__64.0__ subtract__100.0__81.0__ divide__10.0__2.0__ subtract__10.0__5.0__ |
| if num__2 x + num__3 y = num__26 ; num__2 y + z = num__19 and x + num__2 z = num__29 what is the value of x + y + z ? <o> a ) num__18 <o> b ) num__32 <o> c ) num__26 <o> d ) num__22 <o> e ) none of these |
on solving equation we get x = num__7 y = num__4 z = num__11 answer d <eor> d <eos> |
d |
subtract__26.0__19.0__ subtract__7.0__3.0__ add__4.0__7.0__ multiply__2.0__11.0__ |
subtract__26.0__19.0__ subtract__7.0__3.0__ add__4.0__7.0__ multiply__2.0__11.0__ |
| a train num__125 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 sec . the speed of the train is ? <o> a ) num__40 <o> b ) num__50 <o> c ) num__60 <o> d ) num__70 <o> e ) num__80 |
speed of the train relative to man = num__12.5 = num__12.5 m / sec . = num__12.5 * num__3.6 = num__45 km / hr let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__45 = > x = num__50 km / hr . answer : option b <eor> b <eos> |
b |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ round__50.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ multiply__5.0__10.0__ |
| what will be the cost of building a fence around a square plot with area equal to num__49 sq ft if the price per foot of building the fence is rs . num__58 ? <o> a ) num__3944 <o> b ) num__2287 <o> c ) num__1624 <o> d ) num__2668 <o> e ) num__1298 |
let the side of the square plot be a ft . a num__2 = num__49 = > a = num__7 length of the fence = perimeter of the plot = num__4 a = num__28 ft . cost of building the fence = num__28 * num__58 = rs . num__1624 . answer : c <eor> c <eos> |
c |
square_perimeter__7.0__ multiply__58.0__28.0__ multiply__58.0__28.0__ |
multiply__4.0__7.0__ multiply__58.0__28.0__ multiply__58.0__28.0__ |
| a pump can fill a tank with water in num__6 hours . because of a leak it took num__7 hours to fill the tank . the leak can drain all the water in ? <o> a ) num__15 hr num__10 min <o> b ) num__16 hr num__20 min <o> c ) num__17 hr num__30 min <o> d ) num__42 hr <o> e ) num__14 hr num__25 min |
work done by the leak in num__1 hour = num__0.166666666667 - num__0.142857142857 = num__0.0238095238095 leak will empty the tank in num__42 hrs answer is d <eor> d <eos> |
d |
subtract__7.0__6.0__ divide__1.0__6.0__ divide__1.0__7.0__ divide__0.1667__7.0__ multiply__6.0__7.0__ round__42.0__ |
subtract__7.0__6.0__ divide__1.0__6.0__ divide__1.0__7.0__ subtract__0.1667__0.1429__ multiply__6.0__7.0__ round__42.0__ |
| if { x } is the product of all even integers from num__1 to x inclusive what is the greatest prime factor of { num__18 } + { num__16 } ? <o> a ) num__23 <o> b ) num__20 <o> c ) num__11 <o> d ) num__19 <o> e ) num__2 |
soln : { num__18 } + { num__16 } = num__18 * { num__16 } + { num__16 } = num__19 * { num__16 } answer : d <eor> d <eos> |
d |
add__1.0__18.0__ add__1.0__18.0__ |
add__1.0__18.0__ add__1.0__18.0__ |
| a man can row with a speed of num__15 kmph in still water . if the stream flows at num__3 kmph then the speed in downstream is ? <o> a ) num__33 <o> b ) num__77 <o> c ) num__20 <o> d ) num__18 <o> e ) num__71 |
m = num__15 s = num__3 ds = num__15 + num__3 = num__18 answer : d <eor> d <eos> |
d |
add__15.0__3.0__ round__18.0__ |
add__15.0__3.0__ add__15.0__3.0__ |
| a sum is invested at compounded interest payable annually . the interest in the first two successive years was rs . num__400 and rs . num__420 . the sum is <o> a ) rs . num__8000 <o> b ) rs . num__8200 <o> c ) rs . num__7500 <o> d ) rs . num__8500 <o> e ) rs . num__8600 |
explanation : his means that simple interest on rs . num__400 for num__1 year = num__420 - num__400 = num__20 rate = ( num__100 × si ) / pt = ( num__100 × num__20 ) / ( num__400 × num__1 ) = num__5.0 rs . num__400 is the interest on the sum for num__1 st year hence sum = ( num__100 × si ) / rt = ( num__100 × num__400 ) / ( num__5 × num__1 ) = rs . num__8000 answer : option a <eor> a <eos> |
a |
percent__100.0__8000.0__ |
percent__100.0__8000.0__ |
| a train num__330 m long passed a pole in num__11 sec . how long will it take to pass a platform num__760 m long ? <o> a ) num__76 sec <o> b ) num__89 sec <o> c ) num__87 sec <o> d ) num__35 sec <o> e ) num__36 sec |
speed = num__30.0 = num__30 m / sec . required time = ( num__330 + num__760 ) / num__30 = num__36 sec . answer : e <eor> e <eos> |
e |
divide__330.0__11.0__ round__36.0__ |
divide__330.0__11.0__ round__36.0__ |
| the ratio between the length and the breadth of a rectangular park is num__3 : num__2 . if a man cycling alongthe boundary ofthe park at the speed of num__12 km / hr completes one round in num__8 min then the area of the park ( in sq . m ) is ? <o> a ) num__13400 m <o> b ) num__147600 m <o> c ) num__153600 m <o> d ) num__187500 m <o> e ) num__196700 m |
perimeter = distance covered in num__8 min . = num__12000 x num__8 m = num__1600 m . num__60 let length = num__3 x metres and breadth = num__2 x metres . then num__2 ( num__3 x + num__2 x ) = num__1600 or x = num__160 . length = num__480 m and breadth = num__320 m . area = ( num__480 x num__320 ) m num__2 = num__153600 m c <eor> c <eos> |
c |
multiply__3.0__160.0__ multiply__2.0__160.0__ surface_cube__160.0__ surface_cube__160.0__ |
multiply__3.0__160.0__ multiply__2.0__160.0__ surface_cube__160.0__ surface_cube__160.0__ |
| find a positive number which when increased by num__17 is equal to num__60 times the reciprocal of the number . <o> a ) num__3 <o> b ) num__5 <o> c ) num__7 <o> d ) num__12 <o> e ) num__15 |
sol . let the number be x . then x + num__17 = num__60 / x ⇔ x num__2 + num__17 x - num__60 = num__0 ⇔ ( x + num__20 ) ( x - num__3 ) = num__0 ⇔ x = num__3 . answer a <eor> a <eos> |
a |
divide__60.0__20.0__ divide__60.0__20.0__ |
divide__60.0__20.0__ divide__60.0__20.0__ |
| a man can row num__8 kmph in still water . when the river is running at num__1.2 kmph it takes him num__1 hour to row to a place and black . how far is the place ? <o> a ) num__3.58 <o> b ) num__3.68 <o> c ) num__3.78 <o> d ) num__3.88 <o> e ) num__3.98 |
m = num__8 s = num__1.2 ds = num__8 + num__1.2 = num__9.2 us = num__8 - num__1.2 = num__6.8 x / num__9.2 + x / num__6.8 = num__1 x = num__3.68 . answer : b <eor> b <eos> |
b |
add__8.0__1.2__ subtract__8.0__1.2__ round__3.68__ |
add__8.0__1.2__ subtract__8.0__1.2__ divide__3.68__1.0__ |
| a cistern can be filled by a tap in num__4 hours while it can be emptied by another tap in num__9 hours . if both the taps are opened simultaneously then after how much time will the cistern get filled ? <o> a ) num__4.5 hrs <o> b ) num__5 hrs <o> c ) num__6.5 hrs <o> d ) num__7.2 hrs <o> e ) num__7 hrs |
net part filled in num__1 hour = ( num__0.25 - num__0.111111111111 ) = num__0.138888888889 the cistern will be filled in num__7.2 hrs i . e . num__7.2 hrs . answer : d <eor> d <eos> |
d |
divide__1.0__4.0__ divide__1.0__9.0__ subtract__0.25__0.1111__ round__7.2__ |
divide__1.0__4.0__ divide__1.0__9.0__ subtract__0.25__0.1111__ round__7.2__ |
| the units digit of ( num__35 ) ^ ( num__87 ) + ( num__93 ) ^ ( num__53 ) is : <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__0 |
the units digit of powers of num__3 cycles in a group of num__4 : { num__3 num__9 num__7 num__1 } num__53 has the form num__4 k + num__1 so the units digit of num__93 ^ num__53 is num__3 . the units digit of powers of num__5 is always num__5 . num__3 + num__5 = num__8 so the units digit is num__8 . the answer is d . <eor> d <eos> |
d |
add__3.0__4.0__ subtract__4.0__3.0__ divide__35.0__7.0__ add__1.0__7.0__ add__1.0__7.0__ |
add__3.0__4.0__ subtract__4.0__3.0__ add__1.0__4.0__ add__1.0__7.0__ add__1.0__7.0__ |
| a train covers a distance of num__1000 km in num__10 hrs . find the speed of the train ? <o> a ) num__100 km / hr <o> b ) num__120 km / hr <o> c ) num__150 km / hr <o> d ) num__50 km / hr <o> e ) num__75 km / hr |
speed = num__100.0 = num__100 km / hr answer is a <eor> a <eos> |
a |
divide__1000.0__10.0__ round__100.0__ |
divide__1000.0__10.0__ divide__1000.0__10.0__ |
| the distance light travels in one year is approximately num__5 num__870000 num__000000 miles . the distance light in num__100 year is <o> a ) num__587 x num__10 power num__9 miles <o> b ) num__587 x num__10 power num__12 miles <o> c ) num__587 x num__10 power num__8 miles <o> d ) num__587 x num__10 power - num__12 miles <o> e ) num__587 x num__10 power - num__9 miles |
the distance of the light travels in num__100 years is num__5 num__870000 num__000000 x num__100 miles = num__587 num__000000 num__000000 = num__587 x num__10 power num__12 miles correct answer will be ( b ) <eor> b <eos> |
b |
round__587.0__ |
round__587.0__ |
| the breath of a rectangular landscape is num__8 times its length . there is a playground in it whose area is num__3200 square mtr & which is num__0.111111111111 rd of the total landscape . what is the breath of the landscape ? <o> a ) num__420 <o> b ) num__430 <o> c ) num__280 <o> d ) num__360 <o> e ) num__480 |
sol . num__8 x * x = num__9 * num__3200 x = num__60 length = num__8 * num__60 = num__480 e <eor> e <eos> |
e |
multiply__8.0__60.0__ multiply__8.0__60.0__ |
multiply__8.0__60.0__ multiply__8.0__60.0__ |
| if re . num__50 amounts to rs . num__250 over a period of num__30 years . what is the rate of simple interest ? <o> a ) num__13.99 <o> b ) num__13.66 <o> c ) num__13.33 <o> d ) num__13.44 <o> e ) num__13.55 % |
num__200 = ( num__50 * num__30 * r ) / num__100 r = num__13.33 answer : c <eor> c <eos> |
c |
percent__50.0__200.0__ percent__100.0__13.33__ |
percent__50.0__200.0__ percent__100.0__13.33__ |
| a train num__440 meters long is running with a speed of num__45 kmph . the time taken by it to cross a tunnel num__180 meters long is ? <o> a ) num__40.6 sec <o> b ) num__41.6 sec <o> c ) num__42.6 sec <o> d ) num__49.6 sec <o> e ) num__44.6 sec |
d = num__440 + num__180 = num__620 s = num__45 * num__0.277777777778 = num__12.5 mps t = num__620 / num__12.5 = num__49.6 sec answer : d <eor> d <eos> |
d |
add__440.0__180.0__ divide__620.0__12.5__ round__49.6__ |
add__440.0__180.0__ divide__620.0__12.5__ divide__620.0__12.5__ |
| the price of commodity x increases by num__45 cents every year while the price of commodity y increases by num__20 cents every year . in num__2001 the price of commodity x was $ num__5.20 and the price of commodity y was $ num__7.30 . in which year will the price of commodity x be num__10 cents less than the price of commodity y ? <o> a ) num__2008 <o> b ) num__2009 <o> c ) num__2010 <o> d ) num__2011 <o> e ) num__2012 |
the price of commodity x increases num__25 cents each year relative to commodity y . the price difference is $ num__2.10 and commodity x needs to be num__10 cents less than commodity y . $ num__2.00 / num__25 cents = num__8 years the answer is num__2001 + num__8 years = num__2009 . the answer is b . <eor> b <eos> |
b |
subtract__45.0__20.0__ subtract__7.3__5.2__ round_down__2.1__ subtract__10.0__2.0__ add__2001.0__8.0__ add__2001.0__8.0__ |
subtract__45.0__20.0__ subtract__7.3__5.2__ divide__20.0__10.0__ subtract__10.0__2.0__ add__2001.0__8.0__ add__2001.0__8.0__ |
| if num__2994 ÷ num__14.5 = num__172 then num__29.94 ÷ num__1.45 = ? <o> a ) num__172 <o> b ) num__0.172 <o> c ) num__1.72 <o> d ) num__17.2 <o> e ) none |
solution ( num__29.94 / num__1.45 ) = num__299.4 / num__14.5 = ( num__2994 / num__14.5 × num__0.1 ) = num__17.2 = num__17.2 . answer d <eor> d <eos> |
d |
divide__29.94__299.4__ multiply__172.0__0.1__ multiply__172.0__0.1__ |
divide__29.94__299.4__ multiply__172.0__0.1__ multiply__172.0__0.1__ |
| john makes $ num__45 a week from his job . he earns a raise andnow makes $ num__75 a week . what is the % increase ? <o> a ) num__16.0 <o> b ) num__46.66 <o> c ) num__44.44 <o> d ) num__36.98 <o> e ) num__17 % |
increase = ( num__0.444444444444 ) * num__100 = ( num__0.444444444444 ) * num__100 = num__44.44 . c <eor> c <eos> |
c |
percent__100.0__44.44__ |
percent__100.0__44.44__ |
| a man walks at the speed of num__5 km / hr and runs at the speed of num__10 km / hr . how much time will the man require to cover the distance of num__28 km if he covers half ( first num__14 km ) of his journey walking and half of his journey running ? <o> a ) num__3.2 hrs <o> b ) num__5.2 hrs <o> c ) num__4.2 hrs <o> d ) num__8.2 hrs <o> e ) num__9.2 hrs |
c num__4.2 hrs total time required = num__2.8 + num__1.4 num__28 + num__1.4 = num__4.2 hrs <eor> c <eos> |
c |
divide__28.0__10.0__ divide__14.0__10.0__ round__4.2__ |
divide__28.0__10.0__ divide__14.0__10.0__ add__2.8__1.4__ |
| during a certain two - week period num__72 percent of the movies rented from a video store were comedies and of the remaining movies rented there were num__6 times as many dramas as action movies . if no other movies were rented during that two - week period and there were a action movies rented then how many comedies in terms of a were rented during that two - week period ? <o> a ) num__0.05 a <o> b ) num__10 a <o> c ) num__15 a <o> d ) num__18 a <o> e ) num__23 a |
movies : num__72.0 comedies . num__28.0 remaining genre . now in this num__28.0 there are only num__2 categories . action movies and drama movies . if action = x ; drama movies = num__6 x . total num__7 x . num__7 x = num__28 ; x = num__4 action movies : num__4.0 drama movies : num__24.0 we can say that out of num__100 z : comedies : num__72 z action : num__4 z drama : num__24 z now action movies were a this means : a = num__4 z . z = ( a / num__4 ) comedies : num__72 z = num__72 * ( a / num__4 ) num__18 a d is the answer . <eor> d <eos> |
d |
subtract__6.0__2.0__ multiply__6.0__4.0__ add__72.0__28.0__ divide__72.0__4.0__ divide__72.0__4.0__ |
divide__28.0__7.0__ multiply__6.0__4.0__ add__72.0__28.0__ divide__72.0__4.0__ divide__72.0__4.0__ |
| the profit earned by selling an article for num__832 is equal to the loss incurred when the same article is sold for num__448 . what should be the sale price of the article for making num__50 per cent profit ? <o> a ) num__960 <o> b ) num__1060 <o> c ) num__1200 <o> d ) num__920 <o> e ) none of these |
let the profit or loss be x and num__832 – x = num__448 + x or x = num__384 ⁄ num__2 = num__192 \ cost price of the article = num__832 – x = num__448 + x = num__640 \ sp of the article = num__640 × num__150 ⁄ num__100 = num__960 answer a <eor> a <eos> |
a |
percent__50.0__384.0__ percent__100.0__960.0__ |
percent__50.0__384.0__ percent__100.0__960.0__ |
| twenty percent of country y ' s yearly exports come from fruit exports . one - sixth of all fruit exports from country y are orange exports . if country y generates $ num__2.5 million from its orange exports how much money does it generate from its yearly total of exports ? <o> a ) $ num__21.25 m <o> b ) $ num__75 m <o> c ) $ num__106.25 m <o> d ) $ num__127.5 m <o> e ) $ num__153 m |
num__0.2 * num__0.166666666667 * ( total ) = num__2.5 num__0.0333333333333 * ( total ) = num__2.5 ( total ) = num__2.5 * num__30 = num__75 answer : b . <eor> b <eos> |
b |
subtract__0.2__0.1667__ multiply__2.5__30.0__ multiply__2.5__30.0__ |
multiply__0.2__0.1667__ multiply__2.5__30.0__ multiply__2.5__30.0__ |
| set a = { num__1 num__2 num__3 num__4 num__5 num__6 e } which of the following possible values for e would cause set a to have the smallest standard deviation ? <o> a ) num__1 <o> b ) num__2.5 <o> c ) num__3 <o> d ) num__3.5 <o> e ) num__7 |
i agree . the mean of the set e num__12 num__34 num__56 is num__3.5 . now if we add one extra number to the set in order for the standard deviation to be minimum that number must be as close as possible to the original set ( num__12 num__34 num__56 ) . therefore we have to choose the number closest to num__35 from the options we have leaving d as the best choice . answer d . <eor> d <eos> |
d |
multiply__2.0__6.0__ add__1.0__34.0__ multiply__1.0__3.5__ |
multiply__2.0__6.0__ add__1.0__34.0__ multiply__1.0__3.5__ |
| johnny makes $ num__6.75 per hour at his work . if he works num__10 hours how much money will he earn ? <o> a ) $ num__67.50 <o> b ) $ num__54 <o> c ) $ num__28.50 <o> d ) $ num__12.50 <o> e ) $ num__9.60 |
num__6.75 * num__10 = num__67.50 . answer is a . <eor> a <eos> |
a |
multiply__6.75__10.0__ round__67.5__ |
multiply__6.75__10.0__ multiply__6.75__10.0__ |
| in a bag there are a certain number of black balls and blue balls . the probability of picking up exactly num__1 blue ball when num__2 balls are randomly drawn is num__0.5 . which of the following is the ratio of the number of black balls to blue balls in the bag <o> a ) num__1 : num__2 <o> b ) num__1 : num__4 <o> c ) num__1 : num__1 <o> d ) num__1 : num__5 <o> e ) num__1 : num__6 |
since the probability of drawing a blue ball out of two picks is num__0.5 . the ratio of the blue ball to black balls should be num__1 : num__1 answer : c <eor> c <eos> |
c |
reverse__1.0__ |
reverse__1.0__ |
| if x ^ num__2 + ( num__1 / x ^ num__2 ) = num__6 x ^ num__4 + ( num__1 / x ^ num__4 ) = ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__12 <o> d ) num__34 <o> e ) num__15 |
- > x ^ num__4 + ( num__1 / x ^ num__4 ) = ( x ^ num__2 ) ^ num__2 + ( num__1 / x ^ num__2 ) ^ num__2 = ( x ^ num__2 + num__1 / x ^ num__2 ) ^ num__2 - num__2 x ^ num__2 ( num__1 / x ^ num__2 ) = num__6 ^ num__2 - num__2 = num__34 . thus the answer is d . <eor> d <eos> |
d |
multiply__1.0__34.0__ |
divide__34.0__1.0__ |
| if the length of a rectangle is doubled and its breadth is tripled what is the percentage change in its area ? <o> a ) num__200.0 increase <o> b ) num__345.0 increase <o> c ) num__400.0 increase <o> d ) num__125.0 increase <o> e ) num__25.0 decrease |
length is halved . i . e . length is increased by num__200.0 breadth is tripled i . e . breadth is increased by num__200.0 change in area = ( − num__200 + num__200 + ( num__200 × num__200 ) / num__100 ) % = num__400.0 i . e . area is increased by num__400.0 answer : c <eor> c <eos> |
c |
square_perimeter__100.0__ square_perimeter__100.0__ |
square_perimeter__100.0__ square_perimeter__100.0__ |
| a batsman makes a score of num__87 runs in the num__17 th inning and thus increases his average by num__3 . find his average after num__17 th inning ? <o> a ) num__39 <o> b ) num__88 <o> c ) num__3 <o> d ) num__7 <o> e ) num__62 |
let the average after num__7 th inning = x then average after num__16 th inning = x - num__3 \ inline \ fn _ jvn \ therefore num__16 ( x - num__3 ) + num__87 = num__17 x \ inline \ fn _ jvn \ therefore x = num__87 - num__48 = num__39 answer : a <eor> a <eos> |
a |
multiply__3.0__16.0__ subtract__87.0__48.0__ subtract__87.0__48.0__ |
multiply__3.0__16.0__ subtract__87.0__48.0__ subtract__87.0__48.0__ |
| a jar contains num__3 red marbles num__7 green marbles and num__10 white marbles . if a marble is drawn from the jar at random what is the probability that this marble is white ? <o> a ) num__0.2 <o> b ) num__0.3 <o> c ) num__0.5 <o> d ) num__0.7 <o> e ) num__0.4 |
color frequency red num__3 green num__7 white num__10 let p ( e ) be the event of getting white marble p ( e ) = frequency for white color / total frequencies in the above table = num__10 / ( num__3 + num__7 + num__10 ) = num__0.5 = num__0.5 my answer is ( c ) <eor> c <eos> |
c |
negate_prob__0.5__ |
negate_prob__0.5__ |
| points a b c and d in that order lie on a line . if ab = num__3 cm ac = num__4 cm and bd = num__8 cm what is cd in centimeters ? <o> a ) num__7 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
putting a value to each point lets use the following : a - num__0 b - num__3 ( ab = num__3 ) c - num__4 ( ac = num__4 ) d - num__11 ( bd = num__8 ) cd is num__11 - num__4 = num__7 . ans a <eor> a <eos> |
a |
add__3.0__8.0__ add__3.0__4.0__ round__7.0__ |
add__3.0__8.0__ subtract__11.0__4.0__ subtract__11.0__4.0__ |
| find the odd man out . num__1 num__9 num__25 num__4 num__36 num__65 num__49 <o> a ) num__1 <o> b ) num__25 <o> c ) num__9 <o> d ) num__65 <o> e ) num__49 |
explanation : each of the numbers except num__65 is a perfect squares . answer : d ) num__65 <eor> d <eos> |
d |
multiply__1.0__65.0__ |
multiply__1.0__65.0__ |
| here are num__6 periods in each working day of a school . in how many ways can one organize num__5 subjects such that each subject is allowed at least one period ? <o> a ) num__2100 <o> b ) num__1800 <o> c ) num__1245 <o> d ) num__3210 <o> e ) num__4521 |
num__5 subjects can be arranged in num__6 periods in num__6 p num__5 ways . any of the num__5 subjects can be organized in the remaining period ( num__5 c num__1 ways ) . two subjects are alike in each of the arrangement . so we need to divide by num__2 ! to avoid overcounting . total number of arrangements = num__6 p num__5 × num__5 c num__0.5 ! = num__1800 ans : b <eor> b <eos> |
b |
subtract__6.0__5.0__ divide__1.0__2.0__ round__1800.0__ |
subtract__6.0__5.0__ divide__1.0__2.0__ round__1800.0__ |
| a train travels from new york to chicago a distance of approximately num__540 miles at an average rate of num__60 miles per hour and arrives in chicago at num__6 : num__00 in evening chicago time . at what hour in the morning new york time did the train depart for chicago ? ( note : chicago time is one hour earlier than new york time ) <o> a ) num__3 : num__00 <o> b ) num__4 : num__00 <o> c ) num__5 : num__00 <o> d ) num__6 : num__00 <o> e ) num__10 : num__00 |
num__6 : num__00 in evening in chicago = num__7 : num__00 in evening in new york . so the train was in chicago num__7 : num__00 in the evening new york time . the trip took t = d / r = num__9.0 = num__9 hours . therefore the train depart from new york at num__7 : num__00 - num__9 hours = num__10 : num__00 in the morning new york time . answer : e . <eor> e <eos> |
e |
divide__540.0__60.0__ divide__60.0__6.0__ round__10.0__ |
divide__540.0__60.0__ divide__60.0__6.0__ divide__60.0__6.0__ |
| a train num__540 meters long is running with a speed of num__54 kmph . the time taken by it to cross a tunnel num__180 meters long is ? <o> a ) num__287 sec <o> b ) num__288 sec <o> c ) num__48 sec <o> d ) num__16 sec <o> e ) num__17 sec |
d = num__540 + num__180 = num__720 s = num__54 * num__0.277777777778 = num__15 mps t = num__48.0 = num__48 sec answer : c <eor> c <eos> |
c |
add__540.0__180.0__ divide__720.0__15.0__ round__48.0__ |
add__540.0__180.0__ divide__720.0__15.0__ round__48.0__ |
| given f ( x ) = num__3 x – num__5 for what value of x does num__2 * [ f ( x ) ] + num__2 = f ( num__3 x – num__6 ) <o> a ) num__0 <o> b ) num__4 <o> c ) num__5 <o> d ) num__7 <o> e ) num__13 |
answer = c = num__5 f ( x ) = num__3 x – num__5 num__2 * [ f ( x ) ] + num__2 = f ( num__3 x – num__6 ) num__2 ( num__3 x - num__5 ) + num__2 = num__3 ( num__3 x - num__6 ) - num__5 num__6 x - num__8 = num__9 x - num__23 x = num__5 <eor> c <eos> |
c |
add__3.0__5.0__ add__3.0__6.0__ add__3.0__2.0__ |
add__3.0__5.0__ add__3.0__6.0__ add__3.0__2.0__ |
| last year for every num__100 million vehicles that travelled on a certain highway num__94 vehicles were involved in accidents . if num__3 billion vehicles travelled on the highway last year how many of those vehicles were involved in accidents ? ( num__1 billion = num__1000 num__000000 ) <o> a ) num__288 <o> b ) num__320 <o> c ) num__2820 <o> d ) num__3200 <o> e ) num__28 |
800 |
to solve we will set up a proportion . we know that “ num__100 million vehicles is to num__94 accidents as num__3 billion vehicles is to x accidents ” . to express everything in terms of “ millions ” we can use num__3000 million rather than num__3 billion . creating a proportion we have : num__1.06382978723 = num__3000 / x cross multiplying gives us : num__100 x = num__3000 * num__94 x = num__30 * num__94 = num__2820 correct answer is c . <eor> c <eos> |
c |
c |
| x is able to do a piece of work in num__14 days and y can do the same work in num__20 days . if they can work together for num__5 days what is the fraction of work completed ? <o> a ) num__0.535714285714 <o> b ) num__0.607142857143 <o> c ) num__0.464285714286 <o> d ) num__0.678571428571 <o> e ) num__0.392857142857 |
explanation : amount of work x can do in num__1 day = num__0.0714285714286 amount of work y can do in num__1 day = num__0.05 amount of work x and y can do in num__1 day = num__0.0714285714286 + num__0.05 = num__0.121428571429 amount of work x and y can together do in num__5 days = num__5 × ( num__0.121428571429 ) = num__0.607142857143 answer : option b <eor> b <eos> |
b |
divide__1.0__14.0__ divide__1.0__20.0__ add__0.0714__0.05__ multiply__1.0__0.6071__ |
divide__1.0__14.0__ divide__1.0__20.0__ add__0.0714__0.05__ multiply__1.0__0.6071__ |
| # p is defined as num__2 p + num__20 for any number p . what is p if # ( # ( # p ) ) = num__12 ? <o> a ) – num__16 <o> b ) – num__44 <o> c ) num__10 <o> d ) num__16 <o> e ) num__18 |
# p = num__2 p + num__20 - - - > # ( # p ) = num__2 ( num__2 p + num__20 ) + num__20 = num__4 p + num__60 and thus # ( num__4 p + num__60 ) = num__2 ( num__4 p + num__60 ) + num__20 = num__8 p + num__140 = num__12 - - - > num__8 p = - num__128 - - - > p = - num__16 a is the correct answer . <eor> a <eos> |
a |
multiply__2.0__4.0__ subtract__140.0__12.0__ multiply__2.0__8.0__ multiply__2.0__8.0__ |
subtract__20.0__12.0__ subtract__140.0__12.0__ subtract__20.0__4.0__ subtract__20.0__4.0__ |
| two motor bikes cover the same distance at the speed of num__60 and num__64 kmps respectively . find the distance traveled by them if the slower bike takes num__1 hour more than the faster bike ? <o> a ) num__860 km <o> b ) num__870 km <o> c ) num__960 km <o> d ) num__260 km <o> e ) num__840 km |
explanation : num__60 ( x + num__1 ) = num__64 x x = num__15 num__60 x num__16 = num__960 km answer : c <eor> c <eos> |
c |
add__1.0__15.0__ multiply__60.0__16.0__ round__960.0__ |
add__1.0__15.0__ multiply__60.0__16.0__ round__960.0__ |
| two pipes can fill a tank in num__20 and num__24 minutes respectively and a waste pipe can empty num__1 gallons per minute . all the three pipes working together can fill the tank in num__15 minutes . the capacity of the tank is ? <o> a ) num__40 gallons <o> b ) num__100 gallons <o> c ) num__120 gallons <o> d ) num__180 gallons <o> e ) num__130 gallons |
work done by the waste pipe in num__1 minute = num__0.0666666666667 - ( num__0.05 + num__0.0416666666667 ) = - num__0.025 volume of num__0.025 part = num__1 gallons \ volume of whole = num__1 * num__40 = num__40 gallons . answer : a <eor> a <eos> |
a |
divide__1.0__15.0__ divide__1.0__20.0__ divide__1.0__24.0__ subtract__0.0667__0.0417__ divide__1.0__0.025__ round__40.0__ |
divide__1.0__15.0__ divide__1.0__20.0__ divide__1.0__24.0__ subtract__0.0667__0.0417__ divide__1.0__0.025__ round__40.0__ |
| a contractor is engaged for num__30 days on the condition thathe receives rs . num__25 for each day he works & is fined rs . num__7.50 for each day is absent . he gets rs . num__620 in all . for how many days was he absent ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__15 <o> d ) num__7 <o> e ) num__4 |
num__30 * num__25 = num__750 num__620 - - - - - - - - - - - num__130 num__25 + num__7.50 = num__32.5 num__130 / num__32.5 = num__4 e <eor> e <eos> |
e |
multiply__30.0__25.0__ subtract__750.0__620.0__ add__25.0__7.5__ divide__30.0__7.5__ round__4.0__ |
multiply__30.0__25.0__ subtract__750.0__620.0__ add__25.0__7.5__ divide__30.0__7.5__ divide__30.0__7.5__ |
| a sum of money lent out at s . i . amounts to rs . num__720 after num__2 years and to rs . num__1020 after a further period of num__5 years . the sum is ? <o> a ) num__299 <o> b ) num__600 <o> c ) num__77 <o> d ) num__266 <o> e ) num__222 |
s . i for num__5 years = ( num__1020 - num__720 ) = rs . num__300 . s . i . for num__2 years = num__60.0 * num__2 = rs . num__120 . principal = ( num__720 - num__120 ) = rs . num__600 . answer : b <eor> b <eos> |
b |
subtract__1020.0__720.0__ divide__300.0__5.0__ multiply__2.0__60.0__ subtract__720.0__120.0__ subtract__720.0__120.0__ |
subtract__1020.0__720.0__ divide__300.0__5.0__ multiply__2.0__60.0__ subtract__720.0__120.0__ subtract__720.0__120.0__ |
| there is num__7 friends ( a num__1 a num__2 a num__3 . . . . a num__7 ) . if a num__1 have to have shake with all without repeat . how many handshakes possible ? <o> a ) num__6 <o> b ) num__21 <o> c ) num__28 <o> d ) num__7 <o> e ) num__5 |
a num__1 with ( a num__2 a num__3 a num__4 a num__5 a num__6 a num__7 ) num__6 handshakes answer : a <eor> a <eos> |
a |
subtract__7.0__3.0__ subtract__7.0__2.0__ subtract__7.0__1.0__ subtract__7.0__1.0__ |
subtract__7.0__3.0__ subtract__7.0__2.0__ subtract__7.0__1.0__ subtract__7.0__1.0__ |
| a box measuring num__27 inches long by num__15 inches wide by num__6 inches deep is to be filled entirely with identical cubes . no space is to be left unfilled . what is the smallest number of cubes that can accomplish this objective ? <o> a ) num__90 <o> b ) num__180 <o> c ) num__210 <o> d ) num__240 <o> e ) num__864 |
least number of cubes will be required when the cubes that could fit in are biggest . num__3 is the biggest number that could divide all three num__27 num__15 and num__6 . thus side of cube must be num__3 and total number of cubes = num__9.0 * num__5.0 * num__2.0 = num__90 ans a it is . <eor> a <eos> |
a |
triangle_area__6.0__3.0__ multiply__15.0__6.0__ multiply__15.0__6.0__ |
triangle_area__6.0__3.0__ multiply__15.0__6.0__ multiply__15.0__6.0__ |
| the difference between a two - digit number and the number obtained by interchanging the digit is num__36 . what is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is num__1 : num__2 ? <o> a ) num__4 <o> b ) num__8 <o> c ) num__16 <o> d ) num__24 <o> e ) none of these |
since the number is greater than the number obtained on reversing the digits so the ten ' s digit is greater than the unit ' s digit . let the ten ' s and unit ' s digits be num__2 x and x respectively . then ( num__10 * num__2 x + x ) - ( num__10 x + num__2 x ) = num__36 num__9 x = num__36 x = num__4 required difference = ( num__2 x + x ) - ( num__2 x - x ) = num__2 x = num__8 . answer : b <eor> b <eos> |
b |
subtract__10.0__1.0__ divide__36.0__9.0__ multiply__2.0__4.0__ multiply__1.0__8.0__ |
subtract__10.0__1.0__ divide__36.0__9.0__ multiply__2.0__4.0__ multiply__1.0__8.0__ |
| there are num__870 male and female participants in a meeting . half the female participants and one - quarterof the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ? <o> a ) num__75 <o> b ) num__100 <o> c ) num__125 <o> d ) num__175 <o> e ) num__145 |
let m be the number of male participants and f be the number of female articipants in the meeting . thetotal number of participants is given as num__870 . hence we have m + f = num__870 now we have that half the female participants and one - quarter of the male participants are democrats . let d equal the number of the democrats . then we have the equation f / num__2 + m / num__4 = d now we have that one - third of the total participants are democrats . hence we have the equation d = num__290.0 = num__290 solving the three equations yields the solution f = num__290 m = num__580 and d = num__290 . the number of female democratic participants equals half the female participants equals num__145.0 = num__145 . answer : e <eor> e <eos> |
e |
twice__2.0__ twice__290.0__ divide__580.0__4.0__ divide__580.0__4.0__ |
twice__2.0__ subtract__870.0__290.0__ divide__580.0__4.0__ divide__580.0__4.0__ |
| a goods train runs at the speed of num__72 km / hr and crosses a num__250 m long platform in num__26 sec . what is the length of the goods train ? <o> a ) num__49 sec <o> b ) num__50 sec <o> c ) num__48 second <o> d ) num__43 sec <o> e ) num__32 sec |
d = num__540 + num__180 = num__720 s = num__54 * num__0.277777777778 = num__15 mps t = num__48.0 = num__48 sec answer : c <eor> c <eos> |
c |
add__180.0__540.0__ divide__720.0__15.0__ round__48.0__ |
add__180.0__540.0__ divide__720.0__15.0__ round__48.0__ |
| what is the probability of getting a number less than num__4 when a die is rolled ? <o> a ) num__0.5 <o> b ) num__0.166666666667 <o> c ) num__0.333333333333 <o> d ) num__0.25 <o> e ) num__0.2 |
total number of outcomes possible when a die is rolled = num__6 ( â ˆ µ any one face out of the num__6 faces ) i . e . n ( s ) = num__6 e = getting a number less than num__4 = { num__1 num__2 num__3 } hence n ( e ) = num__3 p ( e ) = n ( e ) / n ( s ) = num__0.5 = num__0.5 answer : a <eor> a <eos> |
a |
die_space__ coin_space__ negate_prob__0.5__ |
die_space__ coin_space__ negate_prob__0.5__ |
| the numbers { num__1 num__3 num__6 num__7 num__7 num__7 } are used to form three num__2 - digit numbers . if the sum of these three numbers is a prime number p what is the largest possible value of p ? <o> a ) num__97 <o> b ) num__151 <o> c ) num__209 <o> d ) num__211 <o> e ) num__219 |
what is the largest possible sum of these three numbers that we can form ? maximize the first digit : num__76 + num__73 + num__71 = num__220 = even so not a prime . let ' s try next largest sum switch digits in num__76 and we ' ll get : num__67 + num__73 + num__71 = num__211 = prime . answer : d . <eor> d <eos> |
d |
subtract__76.0__3.0__ subtract__73.0__2.0__ subtract__73.0__6.0__ multiply__1.0__211.0__ |
subtract__76.0__3.0__ subtract__73.0__2.0__ subtract__73.0__6.0__ multiply__1.0__211.0__ |
| a rower whose speed is num__8 km / hr in still water rows to a certain point upstream and back to the starting point in a river which flows at num__4 km / hr . what is the rower ' s average speed ( in km / hr ) for the total journey ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
time upstream = d / num__4 time downstream = d / num__12 total time = d / num__4 + d / num__12 = d / num__3 average speed = num__2 d / ( d / num__3 ) = num__6 km / hr the answer is b . <eor> b <eos> |
b |
add__8.0__4.0__ divide__12.0__4.0__ divide__8.0__4.0__ subtract__8.0__2.0__ subtract__8.0__2.0__ |
add__8.0__4.0__ divide__12.0__4.0__ divide__8.0__4.0__ add__4.0__2.0__ add__4.0__2.0__ |
| mike drives his new corvette from san francisco to las vegas a journey of num__640 miles . he drives the first half of the trip at an average rate of num__80 miles per hour but has to slow down for the second half of his journey . if the second half of the trip takes him num__200 percent longer than the first half what is his average rate j in miles per hour for the entire trip ? <o> a ) num__26.7 <o> b ) j = num__30.0 <o> c ) j = num__40.0 <o> d ) j = num__53.3 <o> e ) num__60.0 |
veritas prepofficial solution correct answer : c using the formula : time = distance / rate we find that mike takes num__4 hours to cover the first num__320 miles of his trip . since the num__2 nd num__320 miles take num__200.0 longer than the first it takes mike num__8 hours longer or num__12 hours . ( note : num__200.0 longer than the first half is not num__200.0 of the first half . ) the overall time is num__4 hours + num__12 hours or num__16 hours . since the definition of average rate = total distance traveled / total time of travel mike ' s average rate = num__40.0 or num__40 miles per hour . answer choice c is correct . <eor> c <eos> |
c |
multiply__80.0__4.0__ divide__640.0__320.0__ divide__640.0__80.0__ add__4.0__8.0__ multiply__2.0__8.0__ divide__640.0__16.0__ round__40.0__ |
multiply__80.0__4.0__ divide__640.0__320.0__ divide__640.0__80.0__ add__4.0__8.0__ add__4.0__12.0__ divide__640.0__16.0__ divide__640.0__16.0__ |
| a man buys oranges at rs num__5 a dozen and an equal number at rs num__4 a dozen . he sells them at rs num__5.50 a dozen and makes a profit of rs num__50 . how many oranges does he buy ? <o> a ) num__30 dozens <o> b ) num__40 dozens <o> c ) num__50 dozens <o> d ) num__60 dozens <o> e ) num__70 dozens |
explanation : cost price of num__2 dozen oranges rs . ( num__5 + num__4 ) = rs . num__9 . sell price of num__2 dozen oranges = rs . num__11 . if profit is rs num__2 oranges bought = num__2 dozen . if profit is rs . num__50 oranges bought = ( num__1.0 ) * num__50 dozens = num__50 dozens . answer : c <eor> c <eos> |
c |
add__5.0__4.0__ multiply__5.5__2.0__ subtract__5.0__4.0__ multiply__50.0__1.0__ |
add__5.0__4.0__ multiply__5.5__2.0__ subtract__5.0__4.0__ multiply__50.0__1.0__ |
| five years ago the average age of a b c and d was num__45 years . with e joining them now the average of all the five is num__48 years . the age of e is ? <o> a ) num__45 <o> b ) num__47 <o> c ) num__48 <o> d ) num__49 <o> e ) num__40 |
solution num__5 years ago average age of a b c d = num__45 years = > num__5 years ago total age of a b c d = num__45 x num__4 = num__180 years = > total present age of a b c d = num__180 + num__5 x num__4 = num__200 years if e ' s present age is x years = num__200 + x / num__5 = num__48 x = num__40 years . answer e <eor> e <eos> |
e |
multiply__45.0__4.0__ subtract__45.0__5.0__ subtract__45.0__5.0__ |
multiply__45.0__4.0__ divide__200.0__5.0__ divide__200.0__5.0__ |
| a side of beef lost num__35 percent of its weight in processing . if the side of beef weighed num__570 pounds after processing how many pounds did it weigh before processing ? <o> a ) num__191 <o> b ) num__355 <o> c ) num__737 <o> d ) num__876 <o> e ) num__1 |
560 |
let weight of side of beef before processing = x ( num__0.65 ) * x = num__570 = > x = ( num__570 * num__100 ) / num__65 = num__876 answer d <eor> d <eos> |
d |
d |
| the ratio of three numbers is num__5 : num__1 : num__4 and their sum is num__1000 . the last number of the three numbers is ? <o> a ) num__24 <o> b ) num__26 <o> c ) num__27 <o> d ) num__400 <o> e ) num__30 |
num__5 : num__1 : num__4 total parts = num__10 num__10 parts - - > num__1000 num__1 part - - - - > num__100 the last number of the three numbers is = num__4 * num__100 = num__400 answer : d <eor> d <eos> |
d |
divide__1000.0__10.0__ multiply__4.0__100.0__ multiply__1.0__400.0__ |
divide__1000.0__10.0__ multiply__4.0__100.0__ multiply__1.0__400.0__ |
| there were num__45 students in a hostel if the numbers of students increased by num__7 the expenses of the mess were increased by rs . num__39 per day while the average expenditure per head diminished by re . num__1 . what is the original expenditure of the mess ? <o> a ) num__575 <o> b ) num__585 <o> c ) num__595 <o> d ) num__625 <o> e ) num__555 |
explanation : let the original expenditure be rs . x original average expenditure = x / num__45 new average expenditure = ( x + num__39 ) / num__52 so ( x / num__45 ) – ( ( x + num__39 ) / num__52 ) = num__1 so x = num__585 so original expenditure is rs num__585 answer : b <eor> b <eos> |
b |
add__45.0__7.0__ multiply__1.0__585.0__ |
add__45.0__7.0__ divide__585.0__1.0__ |
| a train covers a distance of num__12 km in num__10 minutes . if it takes num__4 seconds to pass a telegraph post then the length of the train is <o> a ) num__80 m <o> b ) num__120 m <o> c ) num__140 m <o> d ) num__160 m <o> e ) num__170 cm |
explanation : speed = num__1.2 x num__60 km / hr = num__72 x num__0.277777777778 m / sec = num__20 m / sec . length of the train = ( speed x time ) = ( num__20 x num__4 ) m = num__80 m answer : option a <eor> a <eos> |
a |
divide__12.0__10.0__ hour_to_min_conversion__ add__12.0__60.0__ multiply__4.0__20.0__ round__80.0__ |
divide__12.0__10.0__ hour_to_min_conversion__ add__12.0__60.0__ multiply__4.0__20.0__ round__80.0__ |
| what least number should be subtracted from num__997 so that the remainder when divided by num__5 num__9 and num__11 will leave in each case the same remainder num__3 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
the lcm of num__59 and num__11 is num__495 . the next multiple is num__2 * num__495 = num__990 . num__990 + { remainder } = num__990 + num__3 = num__993 which is num__4 less than num__997 . answer : d . <eor> d <eos> |
d |
subtract__5.0__3.0__ multiply__2.0__495.0__ add__3.0__990.0__ subtract__997.0__993.0__ subtract__997.0__993.0__ |
subtract__5.0__3.0__ multiply__2.0__495.0__ add__3.0__990.0__ subtract__997.0__993.0__ subtract__997.0__993.0__ |
| bert left the house with n dollars . he spent num__0.25 of this at the hardware store then $ num__8 at the dry cleaners and then half of what was left at the grocery store . when he got home he had $ num__10 left in his pocket . what was the value of n ? <o> a ) $ num__32 <o> b ) $ num__36 <o> c ) $ num__52 <o> d ) $ num__60 <o> e ) $ num__68 |
started to test answer b if he had num__36 then he spent num__9 at hardware store now he was left with num__27 $ he spent num__8 dollars on cleaning thus he remained with num__19 $ he then spent num__0.5 of num__19 or num__9.50 hence the only option that can be right is a . ans a <eor> a <eos> |
a |
multiply__0.25__36.0__ subtract__36.0__9.0__ add__10.0__9.0__ subtract__10.0__0.5__ divide__8.0__0.25__ |
multiply__0.25__36.0__ subtract__36.0__9.0__ add__10.0__9.0__ subtract__10.0__0.5__ divide__8.0__0.25__ |
| two trains one from howrah to patna and the other from patna to howrah start simultaneously . after they meet the trains reach their destinations after num__25 hours and num__16 hours respectively . the ratio of their speeds is ? <o> a ) num__4 : num__6 <o> b ) num__4 : num__3 <o> c ) num__4 : num__9 <o> d ) num__4 : num__5 <o> e ) num__4 : num__2 |
let us name the trains a and b . then ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š num__16 : â ˆ š num__25 = num__4 : num__5 . answer : d <eor> d <eos> |
d |
round__4.0__ |
round__4.0__ |
| set # num__1 = { a b g f e } set # num__2 = { k l m n o q } there are these two sets of letters and you are going to pick exactly one letter from each set . what is the probability of picking at least one vowel ? <o> a ) num__0.5 <o> b ) num__0.333333333333 <o> c ) num__0.25 <o> d ) num__0.166666666667 <o> e ) num__0.142857142857 |
at least questions are best solved by taking the opposite scenario and subtracting it from num__1 . probability of choosing no vowel from set num__1 is num__0.6 and set num__2 is num__0.833333333333 . multiply these to get num__0.5 . therefore probability of picking at least one vowel = num__1 - num__0.5 = num__0.5 . <eor> a <eos> |
a |
reverse__2.0__ reverse__2.0__ |
reverse__2.0__ subtract__1.0__0.5__ |
| the length of a train and that of a platform are equal . if with a speed of num__90 k / hr the train crosses the platform in one minute then the length of the train ( in meters ) is : <o> a ) num__850 <o> b ) num__525 <o> c ) num__550 <o> d ) num__750 <o> e ) num__950 |
speed = [ num__90 * num__0.277777777778 ] m / sec = num__25 m / sec ; time = num__1 min . = num__60 sec . let the length of the train and that of the platform be x meters . then num__2 x / num__60 = num__25 è x = num__25 * num__30.0 = num__750 answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ subtract__90.0__60.0__ multiply__25.0__30.0__ round__750.0__ |
hour_to_min_conversion__ divide__60.0__2.0__ multiply__25.0__30.0__ divide__750.0__1.0__ |
| ramu crosses a num__1000 m long street in num__10 minutes . what is his speed in km per hour ? <o> a ) num__5.6 km / hr <o> b ) num__7.2 km / hr <o> c ) num__3.6 km / hr <o> d ) num__5.9 km / hr <o> e ) num__10 km / hr |
speed = num__100.0 * num__60 = num__1.66 m / sec converting m / sec to km / hr . = num__1.66 * num__3.6 = num__5.9 km / hr answer is d <eor> d <eos> |
d |
divide__1000.0__10.0__ hour_to_min_conversion__ round__5.9__ |
divide__1000.0__10.0__ hour_to_min_conversion__ round__5.9__ |
| a goods train runs at the speed of num__72 km / hr and crosses a num__250 m long platform in num__26 sec . what is the length of the goods train ? <o> a ) num__228 <o> b ) num__277 <o> c ) num__288 <o> d ) num__270 <o> e ) num__128 |
speed = num__72 * num__0.277777777778 = num__20 m / sec . time = num__26 sec . let the length of the train be x meters . then ( x + num__250 ) / num__26 = num__20 x = num__270 m . answer : d <eor> d <eos> |
d |
add__250.0__20.0__ round__270.0__ |
add__250.0__20.0__ add__250.0__20.0__ |
| a train num__150 m long is running at a speed of num__68 kmph . how long does it take to pass a man who is running at num__8 kmph in the same direction as the train ? <o> a ) num__5 sec <o> b ) num__9 sec <o> c ) num__12 sec <o> d ) num__15 sec <o> e ) num__18 sec |
speed of the train relative to man = ( num__68 - num__8 ) kmph = ( num__60 * num__0.277777777778 ) m / sec = ( num__16.6666666667 ) m / sec time taken by the train to cross the man = time taken by it to cover num__150 m at num__16.6666666667 m / sec = num__150 * num__0.06 sec = num__9 sec answer : b . <eor> b <eos> |
b |
hour_to_min_conversion__ divide__150.0__16.6667__ round__9.0__ |
subtract__68.0__8.0__ divide__150.0__16.6667__ divide__150.0__16.6667__ |
| working alone a can complete a certain kind of job in num__12 hours . a and d working together at their respective rates can complete one of these jobs in num__4 hours . in how many hours can d working alone complete one of these jobs ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__4 <o> d ) num__6 <o> e ) num__9 |
let total time taken by s to complete the job = s total time taken by a to complete the job = num__12 work done by a in an hour num__1 / a = num__0.0833333333333 working together a and d can complete the job in num__4 hours num__1 / a + num__1 / d = num__0.25 = > num__1 / d = num__0.25 - num__0.0833333333333 = num__0.25 - num__0.0833333333333 = num__0.166666666667 = > d = num__6 hours answer d <eor> d <eos> |
d |
divide__1.0__12.0__ divide__1.0__4.0__ subtract__0.25__0.0833__ round__6.0__ |
divide__1.0__12.0__ divide__1.0__4.0__ subtract__0.25__0.0833__ subtract__12.0__6.0__ |
| a train passes a man standing on the platform . if the train is num__170 meters long and its speed is num__72 kmph how much time it took in doing so ? <o> a ) num__9 ½ sec <o> b ) num__5 ½ sec <o> c ) num__6 ½ sec <o> d ) num__8 ½ sec <o> e ) num__2 ½ sec |
d = num__170 s = num__72 * num__0.277777777778 = num__20 mps t = num__8.5 = num__8 ½ sec answer : d <eor> d <eos> |
d |
divide__170.0__20.0__ round__8.0__ |
divide__170.0__20.0__ round__8.0__ |
| difference between two numbers is num__5 six times of the smaller lacks by num__6 from the four times of the greater . find the numbers ? <o> a ) num__9 <o> b ) num__8 <o> c ) num__7 <o> d ) num__6 <o> e ) num__4 |
explanation : x – y = num__5 num__4 x – num__6 y = num__6 x = num__12 y = num__7 answer : option c <eor> c <eos> |
c |
subtract__12.0__5.0__ subtract__12.0__5.0__ |
subtract__12.0__5.0__ subtract__12.0__5.0__ |
| how many two letter words are formed using the letters of the word machine ? <o> a ) num__52 <o> b ) num__56 <o> c ) num__38 <o> d ) num__42 <o> e ) num__48 |
the number of letters in the given word is seven . the number of two letter words that can be formed using these seven letters is num__7 p num__2 = num__7 * num__6 = num__42 . answer : d <eor> d <eos> |
d |
multiply__6.0__7.0__ multiply__6.0__7.0__ |
multiply__6.0__7.0__ multiply__6.0__7.0__ |
| if num__9 persons can do num__9 times of a particular work in num__9 days then num__7 persons can do num__7 times of that work in ? <o> a ) num__2 days <o> b ) num__3 days <o> c ) num__5 days <o> d ) num__7 days <o> e ) num__9 days |
that is num__1 person can do one time of the work in num__9 days . therefore num__7 persons can do num__7 times work in the same num__9 days itself . e ) <eor> e <eos> |
e |
round__9.0__ |
round__9.0__ |
| a boat running up stram takes num__6 hours to cover a certain distance while it takes num__9 hours to cover the same distance running down stream . what is the ratio between the speed of the boat and the speed of water current respectively ? <o> a ) num__2 : num__3 <o> b ) num__5 : num__1 <o> c ) num__4 : num__5 <o> d ) num__7 : num__1 <o> e ) num__8 : num__1 |
explanation : let speed of boat is x km / h and speed stream is y km / hr num__6 ( x + y ) = num__9 ( x - y ) num__6 x + num__6 y = num__9 x - num__9 y num__15 y = num__3 x num__5 y = x x / y = num__5.0 num__5 : num__1 answer : option b <eor> b <eos> |
b |
add__6.0__9.0__ subtract__9.0__6.0__ divide__15.0__3.0__ subtract__6.0__5.0__ round__5.0__ |
add__6.0__9.0__ subtract__9.0__6.0__ divide__15.0__3.0__ subtract__6.0__5.0__ subtract__6.0__1.0__ |
| at the end of the first quarter the share price of a certain mutual fund was num__20 percent higher than it was at the beginning of the year . at the end of the second quarter the share price was num__50 percent higher than it was at the beginning of the year . what was the percent increase in the share price from the end of the first quarter to the end of the second quarter ? <o> a ) num__20.0 <o> b ) num__25.0 <o> c ) num__30.0 <o> d ) num__33.0 <o> e ) num__40 % |
say price at the beginning of year = num__100 end of num__1 st quarter = num__100 + num__20 = num__120 end of num__2 nd quarter = num__100 + num__50 = num__150 percentage increase between num__1 st & num__2 nd quarter = num__150 − num__1.0 ∗ num__100 = num__25 answer = b <eor> b <eos> |
b |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| a cistern can be filled by pipes a x in num__3 minutes num__45 seconds . if the larger pipe takes num__4 minutes less than the smaller pipe to fill the cistern how much time will the smaller pipe take to fill the tank ? <o> a ) num__12 min <o> b ) num__6 min <o> c ) num__10 min <o> d ) num__8 min <o> e ) num__9 min |
let a is larger pipe and x is smaller then let time taken to x = t then time taken to a = t - num__4 so efficiency of x = t - num__4 and efficiency of a = t using formula time = total work / total efficiency total work = t ( t - num__4 ) total time = num__3 + num__0.75 = num__3.75 using formula num__3.75 = t ( t - num__4 ) / ( t + t - num__4 ) by solving it t = num__10 min so smaller pipe take num__10 mins . answer : c <eor> c <eos> |
c |
divide__3.0__4.0__ add__3.0__0.75__ round__10.0__ |
divide__3.0__4.0__ add__3.0__0.75__ round__10.0__ |
| a man misses a bus by num__40 minutes if he travels at num__30 kmph . if he travels at num__40 kmph then also he misses the bus by num__10 minutes . what is the minimum speed required to catch the bus on time ? <o> a ) num__17 kmph <o> b ) num__45 kmph <o> c ) num__86 kmph <o> d ) num__197 kmph <o> e ) num__15 kmph |
let the distance to be travelled to catch the bus be x km x / num__30 - x / num__40 = num__0.5 = > ( num__4 x - num__3 x ) / num__120 = num__0.5 = > x = num__60 km by traavelling num__30 kmph time taken = num__2.0 = num__2 hours by taking num__2 hours he is late by num__40 min . so he has to cover num__60 km in at most speed = num__60 / ( num__1.33333333333 ) = num__45 kmph . answer : b <eor> b <eos> |
b |
divide__40.0__10.0__ divide__30.0__10.0__ multiply__40.0__3.0__ hour_to_min_conversion__ multiply__0.5__4.0__ divide__40.0__30.0__ round__45.0__ |
divide__40.0__10.0__ divide__30.0__10.0__ multiply__40.0__3.0__ divide__30.0__0.5__ divide__120.0__60.0__ divide__40.0__30.0__ round__45.0__ |
| a set of consecutive positive integers beginning with num__1 is written on the blackboard . a student came along and erased one number . the average of the remaining numbers is num__35 * num__0.411764705882 . what was the number erased ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__6 <o> d ) num__5 <o> e ) num__4 |
explanation : let the higher number be n and x be the number erased . then ( ( n ( n + num__1 ) / num__2 ) + x ) / ( n + num__1 ) = num__35 * num__0.411764705882 = num__35.4117647059 hence n = num__69 and x = num__7 satisfy the above conditions . answer : a <eor> a <eos> |
a |
add__35.0__0.4118__ multiply__1.0__7.0__ |
add__35.0__0.4118__ multiply__1.0__7.0__ |
| a man has rs . num__640 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__110 <o> b ) num__115 <o> c ) num__120 <o> d ) num__125 <o> e ) num__130 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__640 num__16 x = num__640 x = num__40 . hence total number of notes = num__3 x = num__120 answer : c <eor> c <eos> |
c |
divide__640.0__16.0__ multiply__3.0__40.0__ multiply__3.0__40.0__ |
divide__640.0__16.0__ multiply__3.0__40.0__ multiply__3.0__40.0__ |
| a trader sells num__92 meters of cloth for rs . num__9890 at the profit of rs . num__24 per metre of cloth . what is the cost price of one metre of cloth ? <o> a ) num__22 <o> b ) num__89.5 <o> c ) num__76.9 <o> d ) num__83.5 <o> e ) num__11 |
sp of num__1 m of cloth = num__107.5 = rs . num__107.5 cp of num__1 m of cloth = sp of num__1 m of cloth - profit on num__1 m of cloth = rs . num__107.5 - rs . num__24 = rs . num__83.5 answer : d <eor> d <eos> |
d |
divide__9890.0__92.0__ subtract__107.5__24.0__ round__83.5__ |
divide__9890.0__92.0__ subtract__107.5__24.0__ subtract__107.5__24.0__ |
| in a num__500 m race dishu beats abhishek by num__100 m or num__5 seconds . in another race on the same track at the same speeds . abhishek and prashant start at one end while dishu starts at the opposite end . how many metres would abhishek have covered by the time dishu meets prashant given that dishu ' s speed is num__10 m / sec more than that of prashant . <o> a ) num__250 m <o> b ) num__560 m <o> c ) num__100 m <o> d ) num__320 m <o> e ) num__240 m |
explanation : abhishek ' s speed = num__20.0 = num__20 m / s . time taken by abhishek to cover num__500 m = num__25.0 = num__25 seconds . dishu ' s speed = num__25.0 or num__25 m / s prashant ' s speed = num__15 m / s time taken by dishu to meet prashant in num__500 m race in opposite direction : distance covered by abhishek : = > num__500 / ( num__15 + num__25 ) . = > num__12.5 sec . = > num__12.5 ( num__20 ) = num__250 m . answer : a <eor> a <eos> |
a |
divide__100.0__5.0__ divide__500.0__20.0__ add__5.0__10.0__ multiply__10.0__25.0__ round__250.0__ |
divide__100.0__5.0__ divide__500.0__20.0__ add__5.0__10.0__ multiply__10.0__25.0__ round__250.0__ |
| rs . num__4500 amounts to rs . num__5544 in two years at compound interest compounded annually . if the rate of the interest for the first year is num__12.0 find the rate of interest for the second year ? <o> a ) num__10.0 <o> b ) num__18.0 <o> c ) num__19.0 <o> d ) num__12.0 <o> e ) num__13 % |
let the rate of interest during the second year be r % . given num__4500 * { ( num__100 + num__12 ) / num__100 } * { ( num__100 + r ) / num__100 } = num__5544 r = num__10.0 answer : a <eor> a <eos> |
a |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| the speed of a boat in upstream is num__20 kmph and the speed of the boat downstream is num__80 kmph . find the speed of the boat in still water and the speed of the stream ? <o> a ) num__10 kmph . <o> b ) num__17 kmph . <o> c ) num__18 kmph . <o> d ) num__41 kmph . <o> e ) num__30 kmph . |
speed of the boat in still water = ( num__20 + num__80 ) / num__2 = num__50 kmph . speed of the stream = ( num__80 - num__20 ) / num__2 = num__30 kmph . answer : e <eor> e <eos> |
e |
subtract__80.0__50.0__ round__30.0__ |
subtract__80.0__50.0__ subtract__80.0__50.0__ |
| in what proportion water to be mixed with spirit to gain num__12.5 by selling it at cost price ? <o> a ) num__1 : num__2 <o> b ) num__1 : num__3 <o> c ) num__1 : num__4 <o> d ) num__1 : num__8 <o> e ) num__1 : num__16 |
if the total quantity = num__100 actual quantity of spirit to be sold to get num__12.5 gain = num__100 * num__100 / num__112.5 = num__88.8888888889 so the water = num__100 - ( num__88.8888888889 ) = num__11.1111111111 proportion of water : spirit = num__11.1111111111 : num__88.8888888889 = num__1 : num__8 answer : d <eor> d <eos> |
d |
percent__12.5__88.8889__ percent__12.5__8.0__ |
percent__12.5__88.8889__ percent__12.5__8.0__ |
| the length of a rectangle is doubled while its width is doubled . what is the % change in area ? <o> a ) num__250.0 <o> b ) num__300.0 <o> c ) num__500.0 <o> d ) num__650.0 <o> e ) num__700 % |
the original area is l * w the new area is num__2 l * num__2 w = num__4 * l * w = l * w + num__3 * l * w the area increased by num__300.0 . the answer is b . <eor> b <eos> |
b |
triangle_area__300.0__2.0__ |
triangle_area__300.0__2.0__ |
| bhanu spends num__30.0 of his income on petrol on scooter num__30.0 of the remaining on house rent and the balance on food . if he spends rs . num__300 on petrol then what is the expenditure on house rent ? <o> a ) num__2287 <o> b ) num__140 <o> c ) num__128 <o> d ) num__210 <o> e ) num__123 |
given num__30.0 ( income ) = num__300 ⇒ ⇒ income = num__1000 after having spent rs . num__300 on petrol he left with rs . num__700 . his spending on house rent = num__30.0 ( num__700 ) = rs . num__210 answer : d <eor> d <eos> |
d |
percent__30.0__700.0__ percent__30.0__700.0__ |
percent__30.0__700.0__ percent__30.0__700.0__ |
| a boat crossed a lake from north to south at the speed of num__4 km / h entered a river and covered twice as much distance going upstream at num__3 km / h . it then turned around and stopped at the south shore of the lake . if it averaged num__4 km / h that day what was its approximate downstream speed ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
another way will be : total time for the trip = distance covered in lake / speed in still water + distance covered upstream / upstream speed + distance covered downstream / downstream speed if num__5 d is the total distance then distance covered in lake = d distance covered upstream = num__2 d and distance covered downstream = num__2 d = > num__5 d / num__4 = d / num__4 + num__2 d / num__3 + num__2 d / x ( where x is the downstream speed ) = > num__1.25 = num__0.25 + num__0.666666666667 + num__2 / x ( div . both sides by d ) = > num__1.25 = . num__25 + . num__66 + . num__33 / x = > x = num__5.99 ~ num__6 option c <eor> c <eos> |
c |
subtract__5.0__3.0__ divide__5.0__4.0__ reverse__4.0__ divide__2.0__3.0__ divide__66.0__2.0__ add__4.0__2.0__ add__4.0__2.0__ |
subtract__5.0__3.0__ divide__5.0__4.0__ reverse__4.0__ divide__2.0__3.0__ divide__66.0__2.0__ add__4.0__2.0__ add__4.0__2.0__ |
| a batsman makes a score of num__80 runs in the num__11 th inning and thus increases his average by num__5 . find his average after num__11 th inning . <o> a ) num__40 <o> b ) num__50 <o> c ) num__62 <o> d ) num__30 <o> e ) num__56 |
let the average after num__11 th inning = x then average after num__10 th inning = x - num__5 num__10 ( x - num__5 ) + num__80 = num__11 x x = num__80 - num__50 = num__30 answer is d <eor> d <eos> |
d |
multiply__5.0__10.0__ subtract__80.0__50.0__ subtract__80.0__50.0__ |
multiply__5.0__10.0__ subtract__80.0__50.0__ subtract__80.0__50.0__ |
| for what values of k is the system of equations independent ? the equations are as follows : kx + num__5 y = num__3 ; num__3 x + num__4 y = num__9 <o> a ) num__1.66666666667 <o> b ) num__7.5 <o> c ) num__15.0 <o> d ) num__3.75 <o> e ) num__2.14285714286 |
for the equations to be independent conditions is a num__1 a num__2 ≠ b num__1 b num__2 a num__1 a num__2 ≠ b num__1 b num__2 k num__3 ≠ num__54 k / num__3 ≠ num__1.25 or k ≠ num__3.75 answer : d <eor> d <eos> |
d |
subtract__5.0__4.0__ subtract__5.0__3.0__ divide__5.0__4.0__ subtract__5.0__1.25__ subtract__5.0__1.25__ |
subtract__5.0__4.0__ subtract__5.0__3.0__ divide__5.0__4.0__ subtract__5.0__1.25__ divide__3.75__1.0__ |
| if it is num__4 : num__37 in the evening on a certain day what time in the morning was it exactly num__2 num__880717 minutes earlier ? ( assume standard time in one location . ) <o> a ) num__4 : num__40 <o> b ) num__4 : num__44 <o> c ) num__4 : num__47 <o> d ) num__4 : num__30 <o> e ) num__4 : num__32 |
num__4 : num__37 minus num__2 num__880717 in any way must end with num__0 the only answer choice which ends with num__0 is a . answer : a . <eor> a <eos> |
a |
divide__4.0__880717.0__ round__4.0__ |
divide__4.0__880717.0__ round__4.0__ |
| a train num__700 m long can cross an electric pole in num__20 sec and then find the speed of the train ? <o> a ) num__72 <o> b ) num__67 <o> c ) num__28 <o> d ) num__126 <o> e ) num__27 |
length = speed * time speed = l / t s = num__35.0 s = num__35 m / sec speed = num__35 * num__3.6 ( to convert m / sec in to kmph multiply by num__3.6 ) speed = num__126 kmph answer : d <eor> d <eos> |
d |
divide__700.0__20.0__ multiply__3.6__35.0__ round__126.0__ |
divide__700.0__20.0__ multiply__3.6__35.0__ multiply__3.6__35.0__ |
| if e : f = num__3 : num__1 and f : g = num__5 : num__2 then e : f : g is <o> a ) num__15 : num__5 : num__3 <o> b ) num__15 : num__5 : num__2 <o> c ) num__16 : num__5 : num__2 <o> d ) num__15 : num__4 : num__2 <o> e ) num__5 : num__5 : num__2 |
solution : e / f = num__3.0 ; f / g = num__2.5 ; e : f : g = num__3 * num__5 : num__1 * num__5 : num__1 * num__2 = num__15 : num__5 : num__2 . answer : option b <eor> b <eos> |
b |
divide__5.0__2.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
divide__5.0__2.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
| a motorboat whose speed in num__15 km / hr in still water goes num__30 km downstream and comes back in a total of num__4 hours num__30 minutes . the speed of the stream ( in km / hr ) is <o> a ) num__2 km / hr <o> b ) num__3 km / hr <o> c ) num__4 km / hr <o> d ) num__5 km / hr <o> e ) none of these |
explanation : let the speed of the stream be x km / hr . then speed downstream = ( num__15 + x ) km / hr speed upstream = ( num__15 - x ) km / hr so we know from question that it took num__4 ( num__0.5 ) hrs to travel back to same point . so num__2.0 + x − num__2.0 − x = num__4 num__0.5 = > num__4.0 − x num__2 = num__4.5 = > num__9 x num__2 = num__225 = > x = num__5 km / hr option d <eor> d <eos> |
d |
divide__15.0__30.0__ divide__30.0__15.0__ add__4.0__0.5__ divide__4.5__0.5__ add__0.5__4.5__ round__5.0__ |
divide__15.0__30.0__ divide__30.0__15.0__ add__4.0__0.5__ divide__4.5__0.5__ add__0.5__4.5__ add__0.5__4.5__ |
| a fair coin is tossed num__5 times . what is the probability that it lands heads up at least twice ? <o> a ) num__0.0625 <o> b ) num__0.3125 <o> c ) num__0.4 <o> d ) num__0.8125 <o> e ) num__0.84375 |
fair coin is tossed num__5 times . hence total number of outcomes = num__2 ^ num__5 = num__32 . problem asks for probability of getting atleast heads twice . hence if we calculate probability of getting heads exactly once and probability of not getting heads at all and subract it from the total probability of the event which is num__1 ( as total probability of certain event will be always num__1 ) we can get the probability of getting atleast heads twice . probability of getting exactly one head and no heads = ( number of possible outcomes [ htttt thttt tthtt tttht tttth ttttt ] = num__6 ) / ( total possible outcomes = num__32 ) = > num__0.1875 = num__0.1875 hence probability of getting atleast heads twice = num__1 - ( num__0.1875 ) = num__0.8125 = > choice [ d ] <eor> d <eos> |
d |
coin_space__ die_space__ negate_prob__0.1875__ negate_prob__0.1875__ |
coin_space__ die_space__ negate_prob__0.1875__ negate_prob__0.1875__ |
| if num__8 a = num__9 b and ab ≠ num__0 what is the ratio of a / num__9 to b / num__8 ? <o> a ) num__0.79012345679 <o> b ) num__0.888888888889 <o> c ) num__1 <o> d ) num__1.125 <o> e ) num__1.265625 |
if ab ≠ num__0 then a and b has two integer sets of pair if a = num__9 then b = num__8 and if a = - num__9 then b = - num__8 also in fraction if a = num__0.125 then b = num__0.111111111111 any of the pair we check the ratio num__8 a / num__9 b = num__1 answer : c <eor> c <eos> |
c |
reverse__8.0__ reverse__9.0__ multiply__8.0__0.125__ reverse__1.0__ |
reverse__8.0__ reverse__9.0__ subtract__9.0__8.0__ reverse__1.0__ |
| the avg weight of a b & c is num__70 kg . if d joins the group the avg weight of the group becomes num__70 kg . if another man e who weights is num__3 kg more than d replaces a then the avgof b c d & e becomes num__68 kg . what is the weight of a ? <o> a ) num__81 <o> b ) num__65 <o> c ) num__75 <o> d ) num__89 <o> e ) num__90 |
a + b + c = num__3 * num__70 = num__210 a + b + c + d = num__4 * num__70 = num__280 - - - - ( i ) so d = num__70 & e = num__70 + num__3 = num__73 b + c + d + e = num__68 * num__4 = num__272 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = num__280 – num__272 = num__8 a = e + num__8 = num__73 + num__8 = num__81 answer : a <eor> a <eos> |
a |
multiply__70.0__3.0__ multiply__70.0__4.0__ add__70.0__3.0__ multiply__68.0__4.0__ subtract__280.0__272.0__ add__8.0__73.0__ add__8.0__73.0__ |
multiply__70.0__3.0__ multiply__70.0__4.0__ add__70.0__3.0__ multiply__68.0__4.0__ subtract__280.0__272.0__ add__8.0__73.0__ add__8.0__73.0__ |
| how many legs are there if each sheep has num__4 legs and there are total num__250 of that kind ? <o> a ) a ) num__700 <o> b ) b ) num__800 <o> c ) c ) num__1000 <o> d ) d ) num__10000 <o> e ) of the above |
sol . total of num__250 each has num__4 legs = num__250 * num__4 = num__1000 answer : c <eor> c <eos> |
c |
multiply__4.0__250.0__ multiply__4.0__250.0__ |
multiply__4.0__250.0__ multiply__4.0__250.0__ |
| a garrison of num__400 men had a provision for num__31 days . after num__28 days num__280 persons re - enforcement leave the garrison . find the number of days for which the remaining ration will be sufficient ? <o> a ) num__10 days <o> b ) num__20 days <o> c ) num__5 days <o> d ) num__25 days <o> e ) num__15 days |
num__400 - - - num__31 num__400 - - - num__3 num__120 - - - ? num__400 * num__3 = num__120 * x = > x = num__10 days answer a <eor> a <eos> |
a |
subtract__31.0__28.0__ subtract__400.0__280.0__ divide__280.0__28.0__ divide__280.0__28.0__ |
subtract__31.0__28.0__ subtract__400.0__280.0__ divide__280.0__28.0__ divide__280.0__28.0__ |
| if num__14 men working num__6 hours a day can do a work in num__88 days . then num__6 men working num__8 hours a day can do it in how many days ? <o> a ) num__78 days . <o> b ) num__154 days . <o> c ) num__149 days . <o> d ) num__148 days . <o> e ) num__167 days . |
b num__154 days . from the above formula i . e ( m num__1 * t num__1 / w num__1 ) = ( m num__2 * t num__2 / w num__2 ) so ( num__14 * num__6 * num__88.0 ) = ( num__6 * num__8 * d / num__1 ) on solving d = num__154 days . <eor> b <eos> |
b |
subtract__8.0__6.0__ round__154.0__ |
subtract__8.0__6.0__ divide__154.0__1.0__ |
| anil can do a work in num__15 days while sunil can do it in num__25 days . how long will they take if both work together ? a . num__3 num__0.444444444444 days <o> a ) num__9 num__0.428571428571 days <o> b ) num__9 num__0.875 days <o> c ) num__9 num__0.375 days <o> d ) num__9 num__0.375 days <o> e ) num__2 num__0.375 days |
num__0.0666666666667 + num__0.04 = num__0.106666666667 num__9.375 = num__9 num__0.375 days answer : c <eor> c <eos> |
c |
add__0.0667__0.04__ divide__9.375__25.0__ round__9.0__ |
add__0.0667__0.04__ divide__9.375__25.0__ round__9.0__ |
| a certain essay consists of num__12 paragraphs . each paragraph contains at least num__110 words but not more than num__120 words . which of the following could be the total number of words in the essay ? <o> a ) num__1430 <o> b ) num__1540 <o> c ) num__1640 <o> d ) num__1740 <o> e ) num__1840 |
lets consider minimum no of words . . . num__12 * num__110 = num__1320 lets consider maximum no of words . . . . num__12 * num__120 = num__1440 answer should be one of these two values or should lie between these two values . . . only num__1 option a - num__1430 <eor> a <eos> |
a |
multiply__12.0__110.0__ multiply__12.0__120.0__ add__110.0__1320.0__ add__110.0__1320.0__ |
multiply__12.0__110.0__ multiply__12.0__120.0__ add__110.0__1320.0__ multiply__1.0__1430.0__ |
| a square marked num__1 num__2 num__3 and num__4 on its four sides . four colors pink purple brown and yellow are used to paint the four sides of the square . if the adjacent sides are painted with the different colors in how many ways can the sides be painted ? <o> a ) num__4 <o> b ) num__12 <o> c ) num__9 <o> d ) num__27 <o> e ) num__6 |
if the base of the square is pink then in order the adjacent faces to be painted with the different colors the top must also be pink . num__4 side faces can be painted in brown - purple - yellow brown - purple - yellow or purple - yellow - brown - purple - yellow - brown ( num__2 options ) . but we can have the base painted in either of the four colors thus the total number of ways to paint the square is num__2 * num__2 = num__4 . answer : a . <eor> a <eos> |
a |
choose__4.0__3.0__ |
choose__4.0__3.0__ |
| the ages of raju and biju are in the ratio num__3 : num__1 . fifteen years hence the ratio will be num__2 : num__1 . their present ages are : <o> a ) num__30 yrs num__10 yrs <o> b ) num__45 yrs num__15 yrs <o> c ) num__21 yrs num__7 yrs <o> d ) num__60 yrs num__20 yrs <o> e ) none of these |
expl : let the ages of raju and biju is num__3 x and x years respectively . then ( num__3 x + num__15 ) / ( x + num__15 ) = num__2.0 ; - > num__2 x + num__30 = num__3 x + num__15 - > x = num__15 so raju ’ s age = num__3 * num__15 = num__45 and biju ’ s age = num__15 years answer : b <eor> b <eos> |
b |
multiply__2.0__15.0__ multiply__3.0__15.0__ multiply__3.0__15.0__ |
multiply__2.0__15.0__ multiply__3.0__15.0__ multiply__3.0__15.0__ |
| if a number e between num__0 and num__1 is selected at random which of the following will the number most likely be between ? <o> a ) num__0 and num__0.15 <o> b ) num__0.15 and num__0.2 <o> c ) num__0.2 and num__0.25 <o> d ) num__0.25 and num__0.3 <o> e ) num__0.3 and num__0.5 |
the number e will most likely be between the largest range . a . num__0 and num__0.15 - - > range num__0.15 ; b . num__0.15 and num__0.2 - - > range num__0.05 ; c . num__0.2 and num__0.25 - - > range num__0.05 ; d . num__0.25 and num__0.3 - - > range num__0.05 ; e . num__0.3 and num__0.5 - - > range num__0.2 . answer : e . <eor> e <eos> |
e |
subtract__0.2__0.15__ add__0.2__0.05__ add__0.25__0.05__ add__0.2__0.3__ multiply__1.0__0.3__ |
subtract__0.2__0.15__ add__0.2__0.05__ add__0.25__0.05__ add__0.2__0.3__ subtract__0.5__0.2__ |
| a standard veggiematik machine can chop num__36 carrots in num__4 minutes . how many carrots can num__6 standard veggiematik machines chop in num__9 minutes ? <o> a ) num__486 <o> b ) num__54 <o> c ) num__108 <o> d ) num__216 <o> e ) num__324 |
one veggiematik machine can chop num__36 carrots in num__4 minutes so one veggiematik machine can chop ( num__1.5 ) ( num__36 ) carrots in ( num__1.5 ) ( num__4 ) minutes in other words one veggiematik machine can chop num__54 carrots in num__4 minutes so sixveggiematik machines can chop ( num__9 ) ( num__54 ) carrots in num__9 minutes in other words six veggiematik machines can chop num__486 carrots in num__9 minutes answer a <eor> a <eos> |
a |
divide__6.0__4.0__ multiply__36.0__1.5__ multiply__9.0__54.0__ round__486.0__ |
divide__6.0__4.0__ multiply__36.0__1.5__ multiply__9.0__54.0__ round__486.0__ |
| a wheel has a diameter of x inches and a second wheel has a diameter of y inches . the first wheel covers a distance of d feet in num__60 revolutions . how many revolutions does the second wheel make in covering d feet ? <o> a ) num__60 x - y <o> b ) num__30 x + y <o> c ) num__30 x - y <o> d ) num__60 x + y <o> e ) num__60 x / y |
a wheel covers num__2 π r distance in one revolution . where r = diameter / num__2 first wheel covers d feet in num__60 rev . = > d = num__60 * num__2 π * ( x / num__2 ) second wheel covers d feet in let ' s say p revolutions ; = > d = p * num__2 π * ( y / num__2 ) comparing both equations : - = > p = ( num__60 ∗ num__2 π ∗ x / num__2 ) / ( num__2 π ∗ y / num__2 ) = > num__60 x / y answer : - e <eor> e <eos> |
e |
hour_to_min_conversion__ |
hour_to_min_conversion__ |
| num__12 buckets of water fill a tank when the capacity of each tank is num__13.5 litres . how many buckets will be needed to fill the same tank if the capacity of each bucket is num__9 litres ? <o> a ) num__18 <o> b ) num__15 <o> c ) num__16 <o> d ) num__12 <o> e ) none |
sol . capacity of each bucket = num__9 litres . number of buckets needed = ( num__18.0 ) = num__18 . answer a <eor> a <eos> |
a |
round__18.0__ |
round__18.0__ |
| if log num__1087.5 = num__4.9421 then the number of digits in ( num__875 ) num__10 is ? <o> a ) num__60 <o> b ) num__28 <o> c ) num__27 <o> d ) num__26 <o> e ) num__25 |
x = ( num__875 ) num__10 = ( num__87.5 x num__10 ) num__10 therefore log num__10 x = num__10 ( log num__4087.5 + num__1 ) = num__10 ( num__4.9421 + num__1 ) = num__10 ( num__5.9421 ) = num__59.421 x = antilog ( num__59.421 ) therefore number of digits in x = num__60 . answer : a <eor> a <eos> |
a |
divide__875.0__10.0__ add__4.9421__1.0__ multiply__10.0__5.9421__ multiply__1.0__60.0__ |
divide__875.0__10.0__ add__4.9421__1.0__ multiply__10.0__5.9421__ multiply__1.0__60.0__ |
| last year company x paid out a total of $ num__1 num__050000 in salaries to its num__21 employees . if no employee earned a salary that is more than num__25.0 greater than any other employee what is the lowest possible salary that any one employee earned ? <o> a ) $ num__40 num__384.61 <o> b ) $ num__41667 <o> c ) $ num__42000 <o> d ) $ num__50000 <o> e ) $ num__60 |
000 |
employee num__1 earned $ x ( say ) employee num__2 will not earn more than $ num__1.25 x therfore to minimize the salary of any one employee we need to maximize the salaries of the other num__20 employees ( num__1.25 x * num__20 ) + x = num__1 num__050000 solving for x = $ num__40 num__384.61 answer a <eor> a <eos> |
a |
a |
| a man buys num__56 pens at marked price of num__46 pens from a whole seller . if he sells these pens giving a discount of num__1.0 what is the profit percent ? <o> a ) num__7.6 <o> b ) num__7.7 <o> c ) num__20.52 <o> d ) num__23.62 <o> e ) num__27.82 % |
explanation : let marked price be re . num__1 each c . p . of num__56 pens = rs . num__46 s . p . of num__56 pens = num__99.0 of rs . num__56 = rs . num__55.44 profit % = ( profit / c . p . ) x num__100 profit % = ( num__9.44 ) x num__100 = num__20.52 answer : c <eor> c <eos> |
c |
percent__56.0__99.0__ percent__100.0__20.52__ |
percent__56.0__99.0__ percent__100.0__20.52__ |
| what is the rate percent when the simple interest on rs . num__800 amount to rs . num__400 in num__2 years ? <o> a ) num__5.0 <o> b ) num__6.0 <o> c ) num__2.0 <o> d ) num__25.0 <o> e ) num__1 % |
interest for num__2 yrs = num__400 interest for num__1 yr = num__200 interest rate = num__0.25 x num__100 = num__25.0 answer : d <eor> d <eos> |
d |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| excluding stoppages the speed of a bus is num__54 kmph and including stoppages it is num__45 kmph . for how many minutes does the bus stop per hour ? <o> a ) num__9 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__13 |
due to stoppages it covers num__9 km less . time taken to cover num__9 km = num__0.166666666667 x num__60 min = num__10 min . answer : b <eor> b <eos> |
b |
subtract__54.0__45.0__ divide__9.0__54.0__ hour_to_min_conversion__ round__10.0__ |
subtract__54.0__45.0__ divide__9.0__54.0__ hour_to_min_conversion__ round__10.0__ |
| robet traveled a distance of num__30 km covering the first num__10 km in x minutes the next num__10 km in y minutes and the last num__10 km in z minutes . if he totally took num__3 y minutes to cover the whole distance then which of the following can not be true ? assume x y and z are different . <o> a ) z = num__3 x <o> b ) x = num__3 z <o> c ) y = num__2 x <o> d ) x = num__2 y <o> e ) y = num__3 x |
robet travelled for x y and for z min . total time : x + y + z which is equal to : num__3 y equating both sides we get x + y + z = num__3 y = > x + z = num__2 y . . . . . . . . eqn num__1 looking out at options d says x = num__2 y using it in eqn num__1 num__2 y + z = num__2 y = > z = num__0 mins which i guess is not possible . the ans isd <eor> d <eos> |
d |
subtract__3.0__2.0__ subtract__3.0__1.0__ |
subtract__3.0__2.0__ subtract__3.0__1.0__ |
| spandy travels from x to y at a speed od num__40 kmph and returns by increasing his speed by num__50.0 . what is his average speed for the whole journey ? <o> a ) num__36 kmph <o> b ) num__45 kmph <o> c ) num__48 kmps <o> d ) num__45 kmps <o> e ) num__42 kmps |
explanation : let the distance between x and y be x km . time taken from x to y @ num__40 km / h = ( x / num__40 ) hours time taken from y to x @ num__60 km / h = ( x / num__60 ) hours average speed = num__2 x / ( ( x / num__40 ) + ( x / num__60 ) ) = num__48 km / h . answer : c <eor> c <eos> |
c |
subtract__50.0__2.0__ subtract__50.0__2.0__ |
subtract__50.0__2.0__ subtract__50.0__2.0__ |
| num__20 beavers working together in a constant pace can build a dam in num__3 hours . how many v hours will it take num__12 beavers that work at the same pace to build the same dam ? <o> a ) num__2 . <o> b ) v = num__4 . <o> c ) v = num__5 . <o> d ) v = num__6 . <o> e ) num__8 . |
c . num__5 hrs if there were num__10 beavers it qould have taken double v = num__6 hrs . . so closest to that option is num__5 . <eor> c <eos> |
c |
round__5.0__ |
round__5.0__ |
| if num__20.0 of a = b then b % of num__20 is the same as : <o> a ) none of these <o> b ) num__10.0 of a <o> c ) num__4.0 of a <o> d ) num__20.0 of a <o> e ) none of these |
explanation : num__20.0 of a = b ⇒ b = num__0.2 a b % of num__20 = ( b / num__100 ) num__20 = ( num__0.2 a ) / num__100 x num__20 = num__20 × num__20 × a / num__100 × num__100 = ( num__4 a / num__100 ) = num__4.0 of a answer : option c <eor> c <eos> |
c |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| if num__1 / ( x + num__4 ) + num__1 / ( x - num__4 ) = num__1 / ( x - num__4 ) what is the value of x ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__1 <o> d ) num__2 <o> e ) num__3 |
if we solve the question we get x = num__4 . option : b <eor> b <eos> |
b |
multiply__1.0__4.0__ |
multiply__1.0__4.0__ |
| three years ago the average age of m and n was num__18 years . with p joining them the average age becomes num__22 years how old is p now ? <o> a ) num__24 years <o> b ) num__27 years <o> c ) num__28 years <o> d ) num__30 years <o> e ) num__40 years |
explanation : present age of ( m + n ) = ( num__18 x num__2 + num__3 x num__2 ) years = num__42 years . present age of ( m + n + p ) = ( num__22 x num__3 ) years = num__66 years . p ' s age = ( num__66 - num__42 ) years = num__24 years . answer : a <eor> a <eos> |
a |
multiply__22.0__3.0__ add__22.0__2.0__ add__22.0__2.0__ |
multiply__22.0__3.0__ add__22.0__2.0__ add__22.0__2.0__ |
| num__21 ball numbered num__1 to num__21 . a ballis drawn and then another ball is drawn without replacement . <o> a ) num__0.19512195122 <o> b ) num__0.186046511628 <o> c ) num__0.214285714286 <o> d ) num__0.191489361702 <o> e ) num__0.243243243243 |
the probability that first toy shows the even number = num__1021 = num__1021 since the toy is not replaced there are now num__9 even numbered toys and total num__20 toys left . hence probability that second toy shows the even number = num__920 = num__920 required probability = ( num__1021 ) × ( num__920 ) = ( num__1021 ) × ( num__920 ) = num__0.214285714286 c <eor> c <eos> |
c |
subtract__21.0__1.0__ multiply__1.0__0.2143__ |
subtract__21.0__1.0__ multiply__1.0__0.2143__ |
| a can do a piece of work in num__10 days . he works at it for num__4 days and then b finishes it in num__9 days . in how many days can a and b together finish the work ? <o> a ) num__6 days <o> b ) num__8 days <o> c ) num__5 days <o> d ) num__3 days <o> e ) num__4 days |
num__0.4 + num__9 / x = num__1 = > x = num__15 num__0.1 + num__0.0666666666667 = num__0.166666666667 = > num__6 days answer : a <eor> a <eos> |
a |
divide__4.0__10.0__ subtract__10.0__9.0__ divide__1.0__10.0__ divide__1.0__15.0__ add__0.1__0.0667__ subtract__10.0__4.0__ round__6.0__ |
divide__4.0__10.0__ subtract__10.0__9.0__ divide__1.0__10.0__ divide__1.0__15.0__ add__0.1__0.0667__ subtract__10.0__4.0__ divide__6.0__1.0__ |
| the num__8 spokes of a custom circular bicycle wheel radiate from the central axle of the wheel and are arranged such that the sectors formed by adjacent spokes all have different central angles which constitute an arithmetic series of numbers ( that is the difference between any angle and the next largest angle is constant ) . if the largest sector has a central angle of num__66 ° what fraction of the wheel ’ s area is represented by the smallest sector ? <o> a ) num__0.0416666666667 <o> b ) num__0.05 <o> c ) num__0.0666666666667 <o> d ) num__0.0833333333333 <o> e ) num__0.1 |
the largest angle is num__66 . let d be the difference between any two angles in the progression . the sum of all the angles will be : num__66 + ( num__66 - d ) + ( num__66 - num__2 d ) + . . . + ( num__66 - num__7 d ) = num__528 - num__28 d the sum of all the central angles in a circle = num__360 num__528 - num__28 d = num__360 d = num__6.0 = num__6 the smallest sector is ( num__66 - num__7 d ) = num__66 - num__7 * num__6 = num__24 the fraction of the area covered is num__0.0666666666667 = num__0.0666666666667 . the answer is c . <eor> c <eos> |
c |
multiply__8.0__66.0__ subtract__8.0__2.0__ divide__24.0__360.0__ divide__24.0__360.0__ |
multiply__8.0__66.0__ subtract__8.0__2.0__ divide__24.0__360.0__ divide__24.0__360.0__ |
| if dale works at full efficiency he needs a break of num__1 day after every num__2 days of work and he completes a job in a total of num__8 days . if he works at reduced efficiency he can work without break and finish that same job in num__8 days . dale ' s output at reduced efficiency is what fraction of his output at full efficiency ? <o> a ) num__0.333333333333 <o> b ) num__0.75 <o> c ) num__1.2 <o> d ) num__0.916666666667 <o> e ) num__0.833333333333 |
we ' re told that there are num__2 ways for dale to complete a job : num__1 ) full efficiency : num__2 days of work followed by num__1 dayofffor a total of num__8 days . num__2 ) reduced efficiency : num__8 straight days with no days off . working at full efficiency creates the following pattern : num__2 days on num__1 day off num__2 days on num__1 day off num__2 days on num__1 day off num__2 days on = num__2 + num__1 + num__2 + num__1 + num__2 = num__8 days totals : num__6 days on num__2 days off reduced efficiency means that dale will do num__6 days of work in num__8 days thus those reduceddaysare num__0.75 = num__0.75 of full efficiency . answer : b <eor> b <eos> |
b |
subtract__8.0__2.0__ divide__6.0__8.0__ round__0.75__ |
subtract__8.0__2.0__ divide__6.0__8.0__ round__0.75__ |
| the ratio of oranges to apples in a fruit basket is num__3 to num__4 . however after mary and alan eat four apples the ratio changes to num__3 to num__2 . how many oranges are there in the fruit basket ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__6 <o> d ) num__9 <o> e ) num__12 |
o / a = num__0.75 = > o = num__0.75 ( a ) . now o / a - num__4 = num__1.5 . now sub o value in this . . . we get num__3 a / num__4 a - num__16 = num__1.5 = > a = num__8 and o = num__6 . ans option c <eor> c <eos> |
c |
divide__3.0__4.0__ divide__3.0__2.0__ multiply__4.0__2.0__ multiply__3.0__2.0__ multiply__3.0__2.0__ |
divide__3.0__4.0__ divide__3.0__2.0__ divide__16.0__2.0__ subtract__8.0__2.0__ subtract__8.0__2.0__ |
| a total of num__40 percent of the geese included in a certain migration study were male . if some of the geese migrated during the study and num__20 percent of the migrating geese were male what was the ratio of the migration rate for the male geese to the migration rate for the female geese ? [ migration rate for geese of a certain sex = ( number of geese of that sex migrating ) / ( total number of geese of that sex ) ] <o> a ) num__0.375 <o> b ) num__0.583333333333 <o> c ) num__0.666666666667 <o> d ) num__0.875 <o> e ) num__1.14285714286 |
let ' take the number of geese to be num__100 . male = num__40 . female = num__60 . now the second part of the q let ' s take the number migrated to be num__20 . so we have num__20 geese that migrated and out of that num__20.0 are male i . e num__0.2 * num__20 = num__4 geese ( males ) and now we know out of the total num__20 geese num__4 are male then num__16 have to be female . now the ratio part male geese ratios = num__0.1 = num__0.1 . - a female geese ratios = num__0.266666666667 = num__0.266666666667 - b cross multiply equations a and b and you get = num__0.375 . ans a <eor> a <eos> |
a |
add__40.0__20.0__ divide__20.0__100.0__ multiply__20.0__0.2__ subtract__20.0__4.0__ divide__4.0__40.0__ divide__16.0__60.0__ divide__0.1__0.2667__ divide__0.1__0.2667__ |
subtract__100.0__40.0__ divide__20.0__100.0__ multiply__20.0__0.2__ subtract__20.0__4.0__ divide__4.0__40.0__ divide__16.0__60.0__ divide__0.1__0.2667__ divide__0.1__0.2667__ |
| by selling a bicycle for rs . num__2850 a shopkeeper gains num__14.0 . if the profit is reduced to num__8.0 then the selling price will be : <o> a ) rs . num__2600 <o> b ) rs . num__2700 <o> c ) rs . num__2800 <o> d ) rs . num__3000 <o> e ) none |
solution : let cost price was x . x + num__14.0 of x = num__2850 x + num__14 x / num__100 = num__2850 x + num__0.14 x = num__2850 num__1.14 x = num__2850 x = num__2500 . so cost price = rs . num__2500 . now selling price when profit remains at num__8.0 = num__2500 + num__8.0 of num__2500 = rs . num__2700 . short - cut cp of bicycle = num__0.877192982456 * num__2850 = rs . num__2500 ; sp for a profit of num__8.0 = num__1.08 * num__2500 = rs . num__2700 . answer : option b <eor> b <eos> |
b |
percent__100.0__2700.0__ |
percent__100.0__2700.0__ |
| a train num__280 m long running with a speed of num__63 km / hr will pass a tree in ? <o> a ) num__15 <o> b ) num__16 <o> c ) num__17 <o> d ) num__18 <o> e ) num__19 |
speed = num__63 * num__0.277777777778 = num__17.5 m / sec time taken = num__280 * num__0.0571428571429 = num__16 sec answer : option b <eor> b <eos> |
b |
divide__280.0__17.5__ round__16.0__ |
divide__280.0__17.5__ divide__280.0__17.5__ |
| if “ * ” is called “ + ” “ / ” is called “ * ” “ - ” is called “ / ” “ + ” is called “ - ” . num__2.0 – num__5 * num__10 + num__5 = ? <o> a ) num__165 <o> b ) num__276 <o> c ) num__269 <o> d ) num__657 <o> e ) num__251 |
explanation : given : num__2.0 – num__5 * num__10 + num__5 = ? substituting the coded symbols for mathematical operations we get num__40 * num__4.0 + num__10 – num__5 = ? num__40 * num__4 + num__10 – num__5 = ? num__160 + num__10 – num__5 = ? num__170 – num__5 = num__165 answer : a <eor> a <eos> |
a |
divide__40.0__10.0__ multiply__4.0__40.0__ add__10.0__160.0__ add__5.0__160.0__ add__5.0__160.0__ |
divide__40.0__10.0__ multiply__4.0__40.0__ add__10.0__160.0__ add__5.0__160.0__ add__5.0__160.0__ |
| two trains running in opposite directions cross a man standing on the platform in num__35 seconds and num__28 seconds respectively and they cross each other in num__32 seconds . the ratio of their speeds is : <o> a ) num__3 : num__1 <o> b ) num__4 : num__3 <o> c ) num__3 : num__8 <o> d ) num__3 : num__25 <o> e ) num__3 : num__4 |
let the speeds of the two trains be x m / sec and y m / sec respectively . then length of the first train = num__35 x meters and length of the second train = num__28 y meters . ( num__35 x + num__28 y ) / ( x + y ) = num__32 = = > num__35 x + num__28 y = num__32 x + num__32 y = = > num__3 x = num__4 y = = > x / y = num__1.33333333333 answer : option b <eor> b <eos> |
b |
subtract__35.0__32.0__ subtract__32.0__28.0__ divide__4.0__3.0__ round__4.0__ |
subtract__35.0__32.0__ subtract__32.0__28.0__ divide__4.0__3.0__ round__4.0__ |
| if $ num__500 amounts to $ num__583.20 in two years compounded annually find the rate of interest per annum . <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__5 <o> e ) num__4 |
principal = rs . num__500 ; amount = rs . num__583.20 ; time = num__2 years . let the rate be r % per annum . then [ num__500 ( num__1 + ( r / num__100 ) ^ num__2 ] = num__583.20 or [ num__1 + ( r / num__100 ) ] ^ num__2 = num__1.1664 = num__1.1664 [ num__1 + ( r / num__100 ) ] ^ num__2 = ( num__1.08 ) ^ num__2 or num__1 + ( r / num__100 ) = num__1.08 or r = num__8 answer c . so rate = num__8.0 p . a . <eor> c <eos> |
c |
percent__100.0__8.0__ |
percent__100.0__8.0__ |
| if the true discount on s sum due num__2 years hence at num__14.0 per annum be rs . num__154 the sum due is : <o> a ) s . num__704 <o> b ) s . num__968 <o> c ) s . num__1960 <o> d ) s . num__2400 <o> e ) s . num__2800 |
td = pw * r * t / num__100 so num__154 = pw * num__14 * num__0.02 so pw = num__550 sum = pw + td . . sum = num__550 + num__154 = num__704 answer : a <eor> a <eos> |
a |
percent__100.0__704.0__ |
percent__100.0__704.0__ |
| a man buys a cycle for rs . num__1400 and sells it at a loss of num__10.0 . what is the selling price of the cycle ? <o> a ) s . num__1090 <o> b ) s . num__1160 <o> c ) s . num__1190 <o> d ) s . num__1260 <o> e ) s . num__1204 |
since c . p = num__1400 loss % = ( c . p - s . p ) / c . p * num__100 num__10 = ( num__1400 - s . p ) / num__1400 * num__100 so after solving answer = num__1260 . answer : d <eor> d <eos> |
d |
percent__100.0__1260.0__ |
percent__100.0__1260.0__ |
| a train running at the speed of num__30 km / hr crosses a pole in num__9 seconds . find the length of the train ? <o> a ) num__75 <o> b ) num__65 <o> c ) num__25 <o> d ) num__288 <o> e ) num__212 |
speed = num__30 * ( num__0.277777777778 ) m / sec = num__8.33333333333 m / sec length of train ( distance ) = speed * time ( num__8.33333333333 ) * num__9 = num__75 meter answer : a <eor> a <eos> |
a |
round__75.0__ |
round__75.0__ |
| train a leaves new york at num__7 : num__00 am traveling to boston at num__80 mph . train b leaves boston at num__7 : num__45 am traveling to new york at num__70 mph on a parallel track . if the distance between new york and boston is num__210 miles at what time will the two trains pass each other ? <o> a ) num__8 : num__15 am <o> b ) num__8 : num__45 am <o> c ) num__9 : num__00 am <o> d ) num__9 : num__30 am <o> e ) can not be determined from the information given |
train a will cover num__60 kms in num__45 min at num__7 : num__45 the time when train b starts . . . distance left to cover is num__210 - num__60 = num__150 . . combined speed = num__80 + num__70 = num__150 . . so the trains meet in num__1 hour i . e . num__7 : num__45 + num__1 hr = num__8 : num__45 answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ add__80.0__70.0__ add__7.0__1.0__ round__8.0__ |
hour_to_min_conversion__ add__80.0__70.0__ add__7.0__1.0__ add__7.0__1.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 seconds . what is the length of the train ? <o> a ) num__816 m <o> b ) num__167 m <o> c ) num__156 m <o> d ) num__150 m <o> e ) num__178 m |
speed = ( num__60 * num__0.277777777778 ) m / sec = ( num__16.6666666667 ) m / sec length of the train = ( speed x time ) = ( num__16.6666666667 * num__9 ) m = num__150 m . answer : d <eor> d <eos> |
d |
round__150.0__ |
round__150.0__ |
| ajay sells goods to a customer at a profit of k % over the marked price besides it he cheats his customer by giving num__880 g only instead of num__1 kg . thus his overall profit percentage is num__25.0 . find the value of k . <o> a ) num__8.33 <o> b ) num__8.25 <o> c ) num__10.0 <o> d ) num__12.5 <o> e ) none of these |
let c . p . of num__1000 g = rs . num__1000 profit % = k % profit = k / num__100 x num__1000 = rs . num__10 k he then cheats his customer by giving num__880 g only instead of num__1 kg so c . p of num__880 g = rs . num__880 s . p . of num__880 g = num__1000 + num__10 k profit % = [ ( num__1000 + num__10 k - num__880 ) / num__880 ] x num__100 = num__25 k = num__10.0 answer : c <eor> c <eos> |
c |
percent__1.0__1000.0__ percent__1.0__1000.0__ |
percent__1.0__1000.0__ percent__1.0__1000.0__ |
| johnny travels a total of one hour to and from school . on the way there he jogs at num__5 miles per hour and on the return trip he gets picked up by the bus and returns home at num__25 miles per hour . how far is it to the school ? <o> a ) num__2 miles <o> b ) num__4 miles <o> c ) num__6.6 miles <o> d ) num__8 miles <o> e ) num__10 miles |
answer : c ) num__6.6 miles . average speed for round trip = num__2 * a * b / ( a + b ) where a b are speeds so average speed was = num__2 * num__5 * num__25 / ( num__5 + num__25 ) = num__6.6 m / hr the distance between schoolhome should be half of that . ie . num__6.6 miles answer c <eor> c <eos> |
c |
round__6.6__ |
round__6.6__ |
| two successive discounts of num__20.0 and num__20.0 are equivalent to a single discount of <o> a ) num__42.0 <o> b ) num__40.0 <o> c ) num__36.0 <o> d ) num__34.0 <o> e ) none |
answer given r num__1 = num__20.0 and r num__2 = num__20.0 ∴ single discount equal to r num__1 and r num__2 = ( r num__1 + r num__2 - r num__1 x r num__2 ) / num__100.0 = ( num__20 + num__20 - num__20 x num__20 ) / num__100 = num__40 - num__4 = num__36.0 correct option : c <eor> c <eos> |
c |
percent__100.0__36.0__ |
percent__100.0__36.0__ |
| a series of five integers have the following criteria : i . the median of the integers is num__6 ii . the only mode is num__2 iii . the mean is num__5 which of the following is true ? <o> a ) the sum of the five integers is num__30 . <o> b ) the sum of the five integers is num__10 <o> c ) the median of the three largest integers is num__6 . <o> d ) the largest integer is num__9 . <o> e ) the largest integer is num__8 . |
as per the given statements we can infer ( num__2 + num__2 + num__6 + x + y ) / num__5 = num__5 ; x + y = num__15 since num__6 is the median x and y will have to be greater than num__6 . only num__7 + num__8 can satisfy this condition . hence e is the correct answer . <eor> e <eos> |
e |
add__2.0__5.0__ add__6.0__2.0__ add__6.0__2.0__ |
add__2.0__5.0__ add__6.0__2.0__ add__6.0__2.0__ |
| a train sets off at num__9 : num__00 am at the speed of num__70 km / h . another train starts at num__10 : num__30 am in the same direction at the rate of num__80 km / h . at what time will the second train catch the first train ? <o> a ) num__7 : num__00 <o> b ) num__8 : num__00 <o> c ) num__9 : num__00 <o> d ) num__10 : num__00 <o> e ) num__11 : num__00 |
in one hour and thirty minutes the first train travels num__105 km . the second train catches the first train at a rate of num__80 km / h - num__70 km / h = num__10 km / h . the second train will catch the first train in num__10.5 = num__10.5 hours so at num__9 : num__00 pm . the answer is c . <eor> c <eos> |
c |
divide__105.0__10.0__ round__9.0__ |
divide__105.0__10.0__ round__9.0__ |
| let f ( x y ) be defined as the remainder when ( x – y ) ! is divided by x . if x = num__45 what is the maximum value of y for which f ( x y ) = num__0 ? <o> a ) num__32 <o> b ) num__35 <o> c ) num__39 <o> d ) num__40 <o> e ) num__42 |
the question is finding y such that ( num__45 - y ) ! is a multiple of num__45 . that means we need to have num__5 * num__3 ^ num__2 in ( num__45 - y ) ! num__6 ! is the smallest factorial number with num__5 * num__3 ^ num__2 as a factor . num__45 - y = num__6 y = num__39 the answer is c . <eor> c <eos> |
c |
subtract__5.0__3.0__ multiply__2.0__3.0__ subtract__45.0__6.0__ subtract__45.0__6.0__ |
subtract__5.0__3.0__ multiply__2.0__3.0__ subtract__45.0__6.0__ subtract__45.0__6.0__ |
| if rs . num__10 be allowed as true discount on a bill of rs . num__110 at the end of a certain time then the discount allowed on the same sum due at the end of double the time is ? <o> a ) num__68.33 <o> b ) num__18.33 <o> c ) num__28.33 <o> d ) num__48.33 <o> e ) num__98.33 |
explanation : present worth = amount - truediscount = num__110 - num__10 = rs . num__100 si on rs . num__100 for a certain time = rs . num__10 si on rs . num__100 for doube the time = rs . num__20 truediscount on rs . num__120 = num__120 - num__100 = rs . num__20 truediscount on rs . num__110 = = rs . num__18.33 answer : b <eor> b <eos> |
b |
percent__100.0__18.33__ |
percent__100.0__18.33__ |
| a num__64 gallon solution of salt and water is num__10.0 salt . how many gallons of water must be added to the solution in order to decrease the salt to num__8.0 of the volume ? <o> a ) num__8 <o> b ) num__12 <o> c ) num__13 <o> d ) num__14 <o> e ) num__16 |
amount of salt = num__6.4 assume x gallons of water are added . num__6.4 / num__64 + x = num__0.08 num__640 = num__8 x + num__512 num__8 x = num__128 x = num__16 correct option : e <eor> e <eos> |
e |
divide__64.0__10.0__ multiply__64.0__10.0__ multiply__64.0__8.0__ subtract__640.0__512.0__ divide__128.0__8.0__ divide__128.0__8.0__ |
divide__64.0__10.0__ multiply__64.0__10.0__ multiply__64.0__8.0__ subtract__640.0__512.0__ divide__128.0__8.0__ divide__128.0__8.0__ |
| two pipes a and b can separately fill a tank num__12 minutes and num__15 minutes respectively . both the pipes are opened together but num__4 minutes after the start the pipe a is turned off . how much time will it take to fill the tank ? <o> a ) num__5 min <o> b ) num__6 min <o> c ) num__10 min <o> d ) num__8 min <o> e ) num__9 min |
num__0.333333333333 + x / num__15 = num__1 x = num__10 answer : c <eor> c <eos> |
c |
divide__4.0__12.0__ round__10.0__ |
divide__4.0__12.0__ divide__10.0__1.0__ |
| if num__2 x - num__1 + num__2 x + num__1 = num__1280 then find the value of x . <o> a ) num__0 <o> b ) num__2 <o> c ) num__4 <o> d ) num__9 <o> e ) num__6 |
sol . num__2 x - num__1 + num__2 x + num__1 = num__1280 ⇔ num__2 x - num__1 ( num__1 + num__22 ) = num__1280 ⇔ num__2 x - num__1 = num__256.0 = num__256 = num__28 ⇔ x - num__1 = num__8 ⇔ x = num__9 . hence x = num__9 . answer d <eor> d <eos> |
d |
add__1.0__8.0__ add__1.0__8.0__ |
add__1.0__8.0__ add__1.0__8.0__ |
| what should come in the place of the question mark ( ? ) in the following equation ? num__21 ⁄ num__25 ÷ num__9 ⁄ num__20 × num__5 ⁄ num__12 ÷ num__10 ⁄ num__17 <o> a ) num__7 num__77 ⁄ num__125 <o> b ) num__11 num__9 ⁄ num__10 <o> c ) num__119 ⁄ num__450 <o> d ) num__1 num__29 ⁄ num__90 <o> e ) none of these |
? = num__21 ⁄ num__25 × num__20 ⁄ num__9 × num__5 ⁄ num__12 ÷ num__17 ⁄ num__10 = num__119 ⁄ num__90 = num__129 ⁄ num__90 answer d <eor> d <eos> |
d |
multiply__9.0__10.0__ add__10.0__119.0__ subtract__21.0__20.0__ |
multiply__9.0__10.0__ add__10.0__119.0__ subtract__21.0__20.0__ |
| nicole cycles at a constant rate of num__15 kilometers per hour and is passed by jessica who cycles at a constant rate of num__30 kilometers per hour . if jessica cycles at her constant rate for x minutes after passing nicole then stops to wait for her how many minutes will jessica have to wait for nicole to catch up to her ? <o> a ) a ) x minutes <o> b ) b ) x / num__2 minutes <o> c ) c ) num__2 x / num__3 minutes <o> d ) d ) num__3 x / num__2 minutes <o> e ) e ) num__2 x minutes |
speed of nicole = num__20 km / h or num__0.333333333333 km / min = num__0.333333333333 km / min . once jessica passed the nicole the distance between the nicole and jessica will increase at the rate of ( num__30 - num__20 ) = num__10 km / h or num__0.166666666667 km / min now jessica is cycling for x minutes after passing the nicole so in those x minutes distance between jessica and nicole would be ( num__0.166666666667 ) * x = x / num__6 km . so the time taken by nicole to travel x / num__6 km = ( x / num__6 ) / ( num__0.333333333333 ) = num__2 x . hence answer should be e . <eor> e <eos> |
e |
subtract__30.0__20.0__ divide__30.0__15.0__ round__2.0__ |
subtract__30.0__20.0__ divide__30.0__15.0__ divide__30.0__15.0__ |
| the sequence of numbers a num__1 a num__2 a num__3 . . . an is defined by an = num__1 / n - num__1 / ( n + num__2 ) for each integer n > = num__1 . what is the sum of the first num__60 terms of this sequence ? <o> a ) ( num__1 + num__0.5 ) / ( num__0.0196078431373 + num__0.0192307692308 ) <o> b ) ( num__1 + num__0.5 ) – ( num__0.016393442623 + num__0.0161290322581 ) <o> c ) ( num__1 + num__0.5 ) * ( num__0.0196078431373 + num__0.0192307692308 ) <o> d ) ( num__1 + num__0.5 ) + ( num__0.0196078431373 + num__0.0192307692308 ) <o> e ) ( num__0.0196078431373 + num__0.0192307692308 ) |
the answer would most certainly be [ b ] . but the question needs a slight modification . n > = num__1 since the answer does consider a num__1 under the sum . the sequence is : a num__1 = num__1 - num__0.333333333333 a num__2 = num__0.5 - num__0.25 a num__3 = num__0.333333333333 - num__0.2 . . . . we can observe that the third term in the sequence cancels the negative term in the first . a similar approach can be seen on all the terms and we would be left with num__1 + num__0.5 from a num__1 and a num__2 along with - num__0.0161290322581 and - num__0.016393442623 from a num__60 and a num__59 term which could not be cancelled . hence the sum = ( num__1 + num__0.5 ) – ( num__0.016393442623 + num__0.0161290322581 ) answer : b <eor> b <eos> |
b |
reverse__3.0__ reverse__2.0__ divide__0.5__2.0__ subtract__60.0__1.0__ reverse__1.0__ |
reverse__3.0__ reverse__2.0__ divide__0.5__2.0__ subtract__60.0__1.0__ subtract__2.0__1.0__ |
| num__5.511 / num__10.02 = <o> a ) num__0.35 <o> b ) num__0.55 <o> c ) num__0.3509 <o> d ) num__0.351 <o> e ) num__0.3527 |
num__5.511 / num__10.02 num__5.5 = num__5.5 move the comma two places to the left as we have num__2 decimal places from the divisor : num__0.55 . answer : b <eor> b <eos> |
b |
divide__5.511__10.02__ divide__5.511__10.02__ |
divide__5.511__10.02__ divide__5.511__10.02__ |
| the radius of a semi circle is num__6.3 cm then its perimeter is ? <o> a ) num__32.7 <o> b ) num__32.4 <o> c ) num__22.4 <o> d ) num__32.8 <o> e ) num__32.1 |
num__5.14285714286 r = num__6.3 = num__32.4 answer : b <eor> b <eos> |
b |
round__32.4__ |
round__32.4__ |
| a train num__360 m long runs with a speed of num__45 km / hr . what time will it take to pass a platform of num__240 m long ? <o> a ) num__38 sec <o> b ) num__35 sec <o> c ) num__44 sec <o> d ) num__48 sec <o> e ) num__50 sec |
explanation : speed = num__45 km / hr = num__45 × ( num__0.277777777778 ) m / s = num__12.5 = num__12.5 = num__12.5 m / s total distance = length of the train + length of the platform = num__360 + num__240 = num__600 meter time taken to cross the platform = num__600 / ( num__12.5 ) = num__600 × num__0.08 = num__48 seconds answer : option d <eor> d <eos> |
d |
add__360.0__240.0__ divide__600.0__12.5__ round__48.0__ |
add__360.0__240.0__ divide__600.0__12.5__ divide__600.0__12.5__ |
| a man sells two articles for rs . num__3600 each and he gains num__30.0 on the first and loses num__30.0 on the next . find his total gain or loss ? <o> a ) num__9.0 loss <o> b ) num__10.0 gain <o> c ) num__450 <o> d ) num__12.0 loss <o> e ) num__12.0 gain |
( num__30 * num__30 ) / num__100 = num__9.0 loss answer a <eor> a <eos> |
a |
percent__9.0__100.0__ |
percent__9.0__100.0__ |
| akshay buys a motorcycle for rs . num__25000 . if he decides to sell the motorcycle for a profit of num__7.0 find the selling price . <o> a ) num__27000 <o> b ) num__26750 <o> c ) num__26000 <o> d ) num__25750 <o> e ) none of these |
explanation : profit = % profit / num__100 * cost price p = num__0.07 * num__25000 = num__1750 . selling price = cost price + profit = num__25000 + num__1750 = rs . num__26750 answer b <eor> b <eos> |
b |
percent__7.0__25000.0__ percent__100.0__26750.0__ |
percent__7.0__25000.0__ percent__100.0__26750.0__ |
| which number replace question mark in the series ? num__16 num__8 num__40 num__20 num__100 num__50 num__250 ? num__625 <o> a ) num__254 <o> b ) num__125 <o> c ) num__354 <o> d ) num__199 <o> e ) num__540 |
explanation : ( alternate series of / num__2 and & num__5 ) num__8.0 = num__8 num__8 * num__5 = num__40 num__20.0 = num__20 num__20 * num__5 = num__100 num__50.0 = num__50 num__50 * num__5 = num__250 num__125.0 = num__125 num__125 * num__5 = num__625 correct answer is b ) num__125 <eor> b <eos> |
b |
divide__16.0__8.0__ divide__40.0__8.0__ divide__250.0__2.0__ divide__250.0__2.0__ |
divide__16.0__8.0__ divide__40.0__8.0__ divide__250.0__2.0__ divide__250.0__2.0__ |
| num__2 is what percent of num__50 <o> a ) num__2.0 <o> b ) num__4.0 <o> c ) num__6.0 <o> d ) num__8.0 <o> e ) num__10 % |
explanation : num__0.04 * num__100 = num__0.04 * num__100 = num__4.0 answer : option b <eor> b <eos> |
b |
percent__4.0__100.0__ |
percent__4.0__100.0__ |
| a work which could be finished in num__9 days was finished num__3 days earlier after num__10 more men joined . the number of men employed was ? <o> a ) num__86 <o> b ) num__20 <o> c ) num__78 <o> d ) num__27 <o> e ) num__28 |
x - - - - - - - num__9 ( x + num__10 ) - - - - num__6 x * num__9 = ( x + num__10 ) num__6 x = num__20 answer : b <eor> b <eos> |
b |
subtract__9.0__3.0__ round__20.0__ |
subtract__9.0__3.0__ round__20.0__ |
| in an election only two candidates contested . a candidate secured num__70.0 of the valid votes and won by a majority of num__188 votes . find the total number of valid votes ? <o> a ) num__430 <o> b ) num__438 <o> c ) num__436 <o> d ) num__470 <o> e ) num__422 |
let the total number of valid votes be x . num__70.0 of x = num__0.7 * x = num__7 x / num__10 number of votes secured by the other candidate = x - num__7 x / num__100 = num__3 x / num__10 given num__7 x / num__10 - num__3 x / num__10 = num__188 = > num__4 x / num__10 = num__188 = > num__4 x = num__1880 = > x = num__470 . answer : d <eor> d <eos> |
d |
percent__100.0__470.0__ |
percent__100.0__470.0__ |
| if num__4 x + y + z = num__80 num__2 x - y - z = num__4003 x + y - z = num__20 for integers of x y and z find x = ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__15 <o> d ) num__26 <o> e ) num__18 |
num__4 x + y + z = num__80 - - - - - - - - - - num__1 ) num__2 x - y - z = num__40 - - - - - - - - - - - - - num__2 ) num__3 x + y - z = num__20 - - - - - - - - - - - - num__3 ) from num__1 ) and num__2 ) num__6 x = num__120 x = num__20 answer is b <eor> b <eos> |
b |
divide__80.0__2.0__ subtract__4.0__1.0__ add__4.0__2.0__ add__80.0__40.0__ divide__80.0__4.0__ |
divide__80.0__2.0__ subtract__4.0__1.0__ add__4.0__2.0__ add__80.0__40.0__ subtract__40.0__20.0__ |
| a man has rs . num__480 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__90 <o> b ) num__70 <o> c ) num__50 <o> d ) num__80 <o> e ) num__60 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__480 num__16 x = num__480 x = num__30 . hence total number of notes = num__3 x = num__90 . answer is a . <eor> a <eos> |
a |
divide__480.0__16.0__ divide__30.0__10.0__ multiply__3.0__30.0__ multiply__3.0__30.0__ |
divide__480.0__16.0__ divide__30.0__10.0__ multiply__3.0__30.0__ multiply__3.0__30.0__ |
| a num__300 m long train crosses a platform in num__39 sec while it crosses a signal pole in num__18 sec . what is the length of the platform ? <o> a ) num__288 <o> b ) num__350 <o> c ) num__277 <o> d ) num__2651 <o> e ) num__212 |
speed = num__16.6666666667 = num__16.6666666667 m / sec . let the length of the platform be x meters . then ( x + num__300 ) / num__39 = num__16.6666666667 num__3 x + num__900 = num__1950 = > x = num__350 m . answer : b <eor> b <eos> |
b |
divide__300.0__18.0__ multiply__300.0__3.0__ round__350.0__ |
divide__300.0__18.0__ multiply__300.0__3.0__ round__350.0__ |
| find the si on rs . num__2400 for num__8 yrs num__4 months at num__6.0 per annum rate of interest ? <o> a ) rs num__1000 <o> b ) rs num__1200 <o> c ) rs num__1300 <o> d ) rs num__1400 <o> e ) rs num__1500 |
time period is num__8 years num__4 months = num__8 num__4 num__12 years = num__25 num__3 years simple interest = ptr num__100 = num__2400 x num__25 num__3 x num__6 num__100 = rs num__1200 b <eor> b <eos> |
b |
percent__12.0__25.0__ percent__100.0__1200.0__ |
percent__12.0__25.0__ percent__100.0__1200.0__ |
| num__65 bags of nuts are to be divided among num__13 students . each bag contains num__15 nuts . how many nuts will each student receive ? <o> a ) num__75 <o> b ) num__65 <o> c ) num__43 <o> d ) num__52 <o> e ) num__78 |
num__65 ÷ num__13 = num__5 bags per student num__5 x num__15 = num__75 nuts per student correct answer a <eor> a <eos> |
a |
divide__65.0__13.0__ multiply__15.0__5.0__ multiply__15.0__5.0__ |
divide__65.0__13.0__ multiply__15.0__5.0__ multiply__15.0__5.0__ |
| a candidate got num__35.0 of the votes polled and he lost to his rival by num__2250 votes . how many votes were cast ? <o> a ) num__7500 <o> b ) num__2888 <o> c ) num__2666 <o> d ) num__2999 <o> e ) num__2661 |
num__35.0 - - - - - - - - - - - l num__65.0 - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - num__30.0 - - - - - - - - - - num__2250 num__100.0 - - - - - - - - - ? = > num__7500 answer : a <eor> a <eos> |
a |
percent__100.0__7500.0__ |
percent__100.0__7500.0__ |
| rary spent num__2 ⁄ num__5 of her money on new clothes and then deposited num__1 ⁄ num__2 of what remained into her savings account . if she then had $ num__21 left how much did she have at the start ? <o> a ) $ num__30 <o> b ) $ num__35 <o> c ) $ num__70 <o> d ) $ num__105 <o> e ) $ num__210 |
rary spent num__2 ⁄ num__5 of her money on new clothes but if you see num__21 is half of num__0.6 money so num__210 is too huge to be correct . . solution . . let the money be x . . num__0.4 is used . . . num__0.5 of remaining num__0.6 is put in savings account and remaining half is num__21 . . so num__0.5 * num__0.6 * x = num__21 . . x = num__21 * num__2 * num__1.66666666667 = num__70 . . c <eor> c <eos> |
c |
divide__2.0__5.0__ reverse__2.0__ reverse__0.6__ multiply__1.0__70.0__ |
divide__2.0__5.0__ reverse__2.0__ reverse__0.6__ multiply__1.0__70.0__ |
| one woman and one man can build a wall together in four hours but the woman would need the help of two girls in order to complete the same job in the same amount of time . if one man and one girl worked together it would take them eight hours to build the wall . assuming that rates for men women and girls remain constant how many hours would it take one woman one man and one girl working together to build the wall ? <o> a ) num__0.714285714286 <o> b ) num__1 <o> c ) num__1.42857142857 <o> d ) num__1.71428571429 <o> e ) num__3.42857142857 |
solution : let work done by man women and girl per hour be m w g respectively . then m + w = num__0.25 - - > ( num__1 ) w + num__2 g = num__0.25 - - > ( num__2 ) and m + g = num__0.125 - - > ( num__3 ) . no . of hours it would take forone woman one man and one girl working together to build the wall n = num__1 / m + w + g from ( num__1 ) and ( num__2 ) m = num__2 g and from ( num__3 ) g = num__0.0416666666667 m = num__0.0833333333333 and w = num__0.166666666667 . so n = num__1 / ( num__0.291666666667 ) = num__3.42857142857 option e <eor> e <eos> |
e |
divide__0.25__2.0__ add__1.0__2.0__ divide__0.125__3.0__ divide__0.25__3.0__ subtract__0.25__0.0833__ add__0.25__0.0417__ multiply__1.0__3.4286__ |
divide__0.25__2.0__ add__1.0__2.0__ divide__0.125__3.0__ divide__0.25__3.0__ subtract__0.25__0.0833__ add__0.25__0.0417__ divide__3.4286__1.0__ |
| the ratio between the radii of two spheres is num__1 : num__3 . find the ratio between their volumes ? <o> a ) num__1 : num__28 <o> b ) num__1 : num__27 <o> c ) num__1 : num__18 <o> d ) num__1 : num__21 <o> e ) num__1 : num__12 |
r num__1 : r num__2 = num__1 : num__3 r num__13 : r num__23 = num__1 : num__27 answer : b <eor> b <eos> |
b |
volume_cube__3.0__ volume_cube__1.0__ |
volume_cube__3.0__ volume_cube__1.0__ |
| three pipes of same capacity can fill a tank in num__8 hours . if there are only two pipes of same capacity the tank can be filled in . <o> a ) num__22 <o> b ) num__12 <o> c ) num__88 <o> d ) num__99 <o> e ) num__77 |
the part of the tank filled by three pipes in one hour = num__0.125 = > the part of the tank filled by two pipes in num__1 hour = num__0.666666666667 * num__0.125 = num__0.0833333333333 . the tank can be filled in num__12 hours . answer : b <eor> b <eos> |
b |
multiply__8.0__0.125__ multiply__0.125__0.6667__ round__12.0__ |
multiply__8.0__0.125__ multiply__0.125__0.6667__ round__12.0__ |
| look at this series : num__201 num__202 num__204 num__207 . . . what number should come next ? <o> a ) num__211 <o> b ) num__205 <o> c ) num__224 <o> d ) num__267 <o> e ) num__269 |
in this addition series num__1 is added to the first number ; num__2 is added to the second number ; num__3 is added to the third number ; num__4 is added to the fourth number ; and go on . answer a <eor> a <eos> |
a |
subtract__202.0__201.0__ subtract__204.0__202.0__ subtract__204.0__201.0__ add__1.0__3.0__ add__207.0__4.0__ |
subtract__202.0__201.0__ subtract__204.0__202.0__ subtract__204.0__201.0__ add__1.0__3.0__ add__207.0__4.0__ |
| find the number difference between number and its num__0.6 is num__58 . <o> a ) num__140 <o> b ) num__143 <o> c ) num__144 <o> d ) num__145 <o> e ) num__146 |
explanation : let the number = x then x - ( num__0.6 ) x = num__58 = > ( num__0.4 ) x = num__58 = > num__2 x = num__58 * num__5 = > x = num__145 answer : option d <eor> d <eos> |
d |
divide__2.0__0.4__ divide__58.0__0.4__ divide__58.0__0.4__ |
divide__2.0__0.4__ divide__58.0__0.4__ divide__58.0__0.4__ |
| two trains started at the same time one from a to b and the other from b to a . if they arrived at b and a respectively num__9 hours and num__16 hours after they passed each other the ratio of the speeds of the two trains was <o> a ) num__2 : num__1 <o> b ) num__3 : num__2 <o> c ) num__4 : num__3 <o> d ) num__5 : num__4 <o> e ) num__1 : num__2 |
explanation : note : if two trains ( or bodies ) start at the same time from points a and b towards each other and after crossing they take a and b sec in reaching b and a respectively then : ( a ' s speed ) : ( b ' s speed ) = ( b : a ) therefore ratio of the speeds of two trains = = num__4 : num__3 . answer : c <eor> c <eos> |
c |
round__4.0__ |
round__4.0__ |
| excluding stoppages the speed of a bus is num__60 kmph and including stoppages it is num__50 kmph . for how many minutes does the bus stop per hour ? <o> a ) num__10 min <o> b ) num__15 min <o> c ) num__12 min <o> d ) num__8 min <o> e ) num__16 min |
due to stoppages it covers num__10 km less . time taken to cover num__10 km = ( num__0.166666666667 x num__60 ) min = num__10 min answer : a <eor> a <eos> |
a |
subtract__60.0__50.0__ divide__10.0__60.0__ round__10.0__ |
subtract__60.0__50.0__ divide__10.0__60.0__ round__10.0__ |
| the sum of the first num__70 positive even integers is num__4970 . what is the sum of the first num__70 odd integers ? <o> a ) num__4500 <o> b ) num__4600 <o> c ) num__4300 <o> d ) num__4900 <o> e ) num__5000 |
sum of first n even numbers = n ( n + num__1 ) = num__4970 sum of first n odd numbers = n ^ num__2 = num__70 * num__70 = num__4900 ( here n = num__70 ) answer : d <eor> d <eos> |
d |
subtract__4970.0__70.0__ subtract__4970.0__70.0__ |
subtract__4970.0__70.0__ multiply__1.0__4900.0__ |
| when a certain tree was first planted it was num__4 feet tall and the height of the tree increased by a constant amount each year for the next num__6 years . at the end of the num__6 th year the tree was num__0.166666666667 taller than it was at the end of the num__4 th year . by how many feet did the height of the tree increase each year ? <o> a ) num__0.3 <o> b ) num__0.4 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__1.2 |
say the tree grows by x feet every year . then num__4 + num__6 x = ( num__1 + num__0.166666666667 ) ( num__4 + num__4 x ) or x = num__0.5 answer c <eor> c <eos> |
c |
subtract__1.0__0.5__ |
subtract__1.0__0.5__ |
| suppose you work for a manufacturing plant that pays you $ num__12.50 an hour plus $ num__0.16 for each widget you make . how many widgets must you produce in a num__40 hour week to earn $ num__620 ( before payroll deductions ) ? <o> a ) num__670 <o> b ) num__710 <o> c ) num__750 <o> d ) num__790 <o> e ) num__830 |
total pay = num__40 * $ num__12.50 + $ num__0.16 * x = num__620 x = num__120 / num__0.16 = num__750 the answer is c . <eor> c <eos> |
c |
divide__120.0__0.16__ round__750.0__ |
divide__120.0__0.16__ divide__120.0__0.16__ |
| an analyst will recommend a combination of num__2 industrial stocks num__3 transportation stocks and num__3 utility stocks . if the analyst can choose from num__6 industrial stocks num__3 transportation stocks and num__4 utility stocks how many different combinations of num__9 stocks are possible ? <o> a ) num__12 <o> b ) num__19 <o> c ) num__60 <o> d ) num__180 <o> e ) num__720 |
num__6 c num__2 * num__3 c num__3 * num__4 c num__3 = num__15 * num__1 * num__4 = num__60 . answer : c <eor> c <eos> |
c |
choose__6.0__4.0__ choose__6.0__4.0__ |
choose__6.0__4.0__ choose__6.0__4.0__ |
| if num__5 x * x + num__14 * x + k = num__0 having two reciprocal roots find the value of k . <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
product of reciprocal roots = num__1 product of roots of quadratic eqn ax ^ num__2 + bx + c = num__0 is c / a . hence k / num__5 = num__1 k = num__5 answer : e <eor> e <eos> |
e |
multiply__5.0__1.0__ |
divide__5.0__1.0__ |
| if num__1 * num__3 * num__5 = num__16 num__3 * num__5 * num__7 = num__38 then find num__5 * num__7 * num__10 = ? <o> a ) num__65 <o> b ) num__68 <o> c ) num__72 <o> d ) num__80 <o> e ) num__82 |
( num__10 * num__7 ) + num__5 = num__82 e <eor> e <eos> |
e |
multiply__1.0__82.0__ |
multiply__1.0__82.0__ |
| a palindrome is a word or a number that reads the same forward and backward . for example num__2442 and num__111 are palindromes . if num__5 - digit palindromes are formed using one or more of the digits num__1 num__2 num__3 and num__4 how many palindromes are possible ? <o> a ) num__32 <o> b ) num__64 <o> c ) num__96 <o> d ) num__128 <o> e ) num__256 |
there are num__4 choices for each of the first three digits . the number of possible palindromes is num__4 ^ num__3 = num__64 . the answer is b . <eor> b <eos> |
b |
multiply__1.0__64.0__ |
multiply__1.0__64.0__ |
| three pipes of same capacity can fill a tank in num__8 hours . if there are only two pipes of same capacity the tank can be filled in . <o> a ) num__17 <o> b ) num__13 <o> c ) num__15 <o> d ) num__16 <o> e ) num__12 |
he part of the tank filled by three pipes in one hour = num__0.125 = > the part of the tank filled by two pipes in num__1 hour = num__0.666666666667 * num__0.125 = num__0.0833333333333 . the tank can be filled in num__12 hours . answer e <eor> e <eos> |
e |
multiply__8.0__0.125__ multiply__0.125__0.6667__ round__12.0__ |
multiply__8.0__0.125__ multiply__0.125__0.6667__ round__12.0__ |
| a rectangular grass field is num__75 m * num__55 m it has a path of num__2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . num__2 per sq m ? <o> a ) num__1350 <o> b ) num__1357 <o> c ) num__1328 <o> d ) num__1329 <o> e ) num__1829 |
area = ( l + b + num__2 d ) num__2 d = ( num__75 + num__55 + num__2.5 * num__2 ) num__2 * num__2.5 = > num__675 num__675 * num__2 = rs . num__1350 answer : a <eor> a <eos> |
a |
multiply__2.0__675.0__ multiply__2.0__675.0__ |
multiply__2.0__675.0__ multiply__2.0__675.0__ |
| a cistern can be filled by a tap in num__4 hours while it can be emptied by another tap in num__8 hours . if both the taps are opened simultaneously then after how much time will the cistern get filled ? <o> a ) num__2.9 hrs <o> b ) num__8.9 hrs <o> c ) num__2.9 hrs <o> d ) num__7.2 hrs <o> e ) num__8 hrs |
net part filled in num__1 hour = ( num__0.25 - num__0.125 ) = num__0.125 the cistern will be filled in num__8 hrs i . e . num__8 hrs . answer : e <eor> e <eos> |
e |
divide__1.0__4.0__ divide__1.0__8.0__ round__8.0__ |
divide__1.0__4.0__ divide__1.0__8.0__ round__8.0__ |
| in a certain game a large container is filled with red yellow green and blue beads worth respectively num__7 num__5 num__3 and num__2 points each . a number of beads are then removed from the container . if the product of the point values of the removed beads is num__10 num__290000 how many red beads were removed ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
num__10 num__290000 = num__2 ^ num__4 * num__5 ^ num__4 * num__1029 = num__2 ^ num__4 * num__3 * num__5 ^ num__4 * num__343 = num__2 ^ num__4 * num__3 * num__5 ^ num__4 * num__7 ^ num__3 the answer is b . <eor> b <eos> |
b |
subtract__7.0__3.0__ divide__1029.0__3.0__ subtract__7.0__4.0__ |
subtract__7.0__3.0__ divide__1029.0__3.0__ subtract__7.0__4.0__ |
| by selling an article at rs . num__700 a shopkeeper makes a profit of num__25.0 . at what price should he sell the article so as to make a loss of num__25.0 ? <o> a ) s . num__600 <o> b ) s . num__480 <o> c ) s . num__500 <o> d ) s . num__450 <o> e ) s . num__420 |
sp = num__700 profit = num__25.0 cp = ( sp ) * [ num__100 / ( num__100 + p ) ] = num__700 * [ num__0.8 ] = num__560 loss = num__25.0 = num__25.0 of num__560 = rs . num__140 sp = cp - loss = num__560 - num__140 = rs . num__420 answer : e <eor> e <eos> |
e |
percent__25.0__560.0__ percent__100.0__420.0__ |
percent__25.0__560.0__ percent__100.0__420.0__ |
| if the area of circle is num__706 sq cm then its circumference ? <o> a ) num__21 <o> b ) num__88 <o> c ) num__66 <o> d ) num__94 <o> e ) num__90 |
num__3.14285714286 r num__2 = num__706 = > r = num__15 num__2 * num__3.14285714286 * num__15 = num__94 answer : d <eor> d <eos> |
d |
triangle_area__2.0__94.0__ |
triangle_area__2.0__94.0__ |
| two trains of equal length are running on parallel lines in the same direction at num__46 km / hr and num__36 km / hr . the faster train catches and completely passes the slower train in num__54 seconds . what is the length of each train ( in meters ) ? <o> a ) num__45 <o> b ) num__55 <o> c ) num__65 <o> d ) num__75 <o> e ) num__85 |
the relative speed = num__46 - num__36 = num__10 km / hr = num__10 * num__0.277777777778 = num__2.77777777778 m / s in num__54 seconds the relative difference in distance traveled is num__54 * num__2.77777777778 = num__150 meters this distance is twice the length of each train . the length of each train is num__75.0 = num__75 meters the answer is d . <eor> d <eos> |
d |
subtract__46.0__36.0__ divide__10.0__36.0__ round__75.0__ |
subtract__46.0__36.0__ divide__10.0__36.0__ subtract__150.0__75.0__ |
| a car travels uphill at num__30 km / hr and downhill at num__50 km / hr . it goes num__100 km uphill and num__50 km downhill . find the average speed of the car ? <o> a ) num__32 kmph <o> b ) num__33 kmph <o> c ) num__34 kmph <o> d ) num__35 kmph <o> e ) num__36 kmph |
avg speed = total distance / total time . total distance traveled = num__100 + num__50 = num__150 km ; time taken for uphill journey = num__3.33333333333 = num__3.33333333333 ; time taken for down hill journey = num__1.0 = num__1.0 ; avg speed = num__150 / ( num__3.33333333333 + num__1.0 ) = num__35 kmph answer : d <eor> d <eos> |
d |
add__50.0__100.0__ divide__100.0__30.0__ multiply__1.0__35.0__ |
add__50.0__100.0__ divide__100.0__30.0__ divide__35.0__1.0__ |
| a circle passes through two adjacent vertices of a square and is tangent to one side of the square . if the side length of the square is num__2 what is the radius of the circle ? <o> a ) num__1.5 <o> b ) num__1.33333333333 <o> c ) num__1.25 <o> d ) num__1.2 <o> e ) none of these |
let a and b be the vertices through which the circle passes . letm be the midpoint of the line segment ab . the center c of the circle is on the line joining m to the midpoint of the opposite side which is also on the circle ( see diagram below ) . let r be the radius of the circle . the length of the line segment joiningm to c is r - ( num__2 r - num__2 ) = num__2 - r as shown below . then num__12 + ( num__2 - r ) num__2 = r num__2 by the pythagorean theorem . this gives num__1 + num__4 - num__4 r + r num__2 = r num__2 which simplifies to num__5 - r = num__0 so r = num__1.25 correct answer c <eor> c <eos> |
c |
square_perimeter__1.0__ triangle_area__2.0__1.25__ |
square_perimeter__1.0__ triangle_area__2.0__1.25__ |
| n is a positive integer and k is the product of all integers from num__1 to n inclusive . if k is a multiple of num__1800 then what is the smallest possible value of n ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__15 <o> e ) num__18 |
num__1800 = num__2 ^ num__3 * num__3 ^ num__2 * num__5 ^ num__2 num__5 ^ num__2 means that n must be at least num__10 . that is we need num__5 and num__10 . the answer is b . <eor> b <eos> |
b |
add__1.0__2.0__ add__2.0__3.0__ multiply__2.0__5.0__ multiply__1.0__10.0__ |
add__1.0__2.0__ add__2.0__3.0__ multiply__2.0__5.0__ multiply__1.0__10.0__ |
| one hour after yolanda started walking from x to y a distance of num__80 miles bob started walking along the same road from y to x . if yolanda â s walking rate was num__8 miles per hour and bob â s was num__9 miles per hour how many miles had bob walked when they met ? <o> a ) num__38.07 <o> b ) num__40 <o> c ) num__42 <o> d ) num__39 <o> e ) num__40.07 |
let t be the number of hours that bob had walked when he met yolanda . then when they met bob had walked num__4 t miles and yolanda had walked num__8 ( t + num__1 ) miles . these distances must sum to num__80 miles so num__9 t + num__8 ( t + num__1 ) = num__80 which may be solved for t as follows num__9 t + num__8 ( t + num__1 ) = num__80 num__9 t + num__8 t + num__8 = num__80 num__17 t = num__72 t = num__4.23 ( hours ) therefore bob had walked num__9 t = num__9 ( num__4.23 ) = num__38.07 miles when they met . the best answer is a . <eor> a <eos> |
a |
subtract__9.0__8.0__ add__8.0__9.0__ subtract__80.0__8.0__ multiply__9.0__4.23__ round__38.07__ |
subtract__9.0__8.0__ add__8.0__9.0__ subtract__80.0__8.0__ multiply__9.0__4.23__ round__38.07__ |
| there are deer and peacocks in a zoo . by counting heads they are num__80 . the number of their legs is num__200 . how many peacocks are there ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__50 <o> d ) num__60 peacocks <o> e ) num__80 |
explanation : let x and y be the number of deer and peacocks in the zoo respectively . then x + y = num__80 . . . ( i ) and num__4 x + num__2 y = num__200 or num__2 x + y = num__100 . . . ( ii ) solving ( i ) and ( ii ) we get ) x = num__20 y = num__60 . answer : d <eor> d <eos> |
d |
divide__200.0__2.0__ divide__80.0__4.0__ subtract__80.0__20.0__ subtract__80.0__20.0__ |
divide__200.0__2.0__ divide__80.0__4.0__ subtract__80.0__20.0__ subtract__80.0__20.0__ |
| if the price of a book is first decreased by num__25.0 and then increased by num__20.0 then the decrease in the price will be ? <o> a ) num__20.0 <o> b ) num__10.0 <o> c ) num__25.0 <o> d ) num__18.0 <o> e ) num__30 % |
let the original price be $ num__100 new final price = num__120.0 of ( num__75.0 of $ num__100 ) = num__1.2 * num__0.75 * num__100 = $ num__90 decrease is num__10.0 answer is b <eor> b <eos> |
b |
add__20.0__100.0__ subtract__100.0__25.0__ divide__120.0__100.0__ divide__75.0__100.0__ multiply__0.75__120.0__ subtract__100.0__90.0__ subtract__20.0__10.0__ |
add__20.0__100.0__ subtract__100.0__25.0__ divide__120.0__100.0__ divide__75.0__100.0__ multiply__0.75__120.0__ subtract__100.0__90.0__ subtract__20.0__10.0__ |
| a person covers a distance in num__21 minutes . if runs at a speed of num__12 km per hour on an average . find the speed at which he must run to reduce the time of journey to num__5 minutes . <o> a ) num__18 m / s <o> b ) num__4 m / s <o> c ) num__14 m / s <o> d ) num__10 m / s <o> e ) num__5 m / s |
explanation : t = num__21 m spees = num__12 kmph = num__12 x num__0.277777777778 = num__3.33333333333 m / s let new speed be ` ` x ' ' num__21 x ( num__3.33333333333 ) = num__5 x x = num__14 m / s answer : option c <eor> c <eos> |
c |
round__14.0__ |
round__14.0__ |
| a dishonest shopkeeper professes to sell pulses at the cost price but he uses a false weight of num__960 gm . for a kg . his gain is … % . <o> a ) num__4.16 <o> b ) num__5.36 <o> c ) num__4.26 <o> d ) num__6.26 <o> e ) num__7.26 % |
his percentage gain is num__100 * num__0.0416666666667 as he is gaining num__40 units for his purchase of num__960 units . so num__4.16 . answer : a <eor> a <eos> |
a |
percent__100.0__4.16__ |
percent__100.0__4.16__ |
| the average age of a group of persons going for picnic is num__16 years . twelve new persons with an average age of num__15 years join the group on the spot due to which their average age becomes num__15.5 years . the number of persons initially going for picnic is <o> a ) num__5 <o> b ) num__10 <o> c ) num__12 <o> d ) num__40 <o> e ) num__50 |
solution let the initial number of persons be x . then num__16 x + num__12 x num__15 = num__15.5 ( x + num__20 ) = num__0.5 x = num__6 x = num__12 . answer c <eor> c <eos> |
c |
subtract__16.0__15.5__ multiply__0.5__12.0__ divide__6.0__0.5__ |
subtract__16.0__15.5__ multiply__0.5__12.0__ divide__6.0__0.5__ |
| what percent is num__36 paisa ' s of num__12 rupees ? <o> a ) num__2.0 <o> b ) num__3.0 <o> c ) num__4.0 <o> d ) num__5.0 <o> e ) num__6 % |
num__12 rupees = num__1200 paisa ' s num__0.03 × num__100 = num__0.25 num__4.0 = num__3.0 b <eor> b <eos> |
b |
percent__0.25__1200.0__ percent__0.25__1200.0__ |
percent__0.25__1200.0__ percent__0.25__1200.0__ |
| in a function they are distributing noble prize . in how many ways can num__3 prizes be distributed among num__4 boys when no boy gets more than one prize ? <o> a ) num__10 <o> b ) num__15 <o> c ) num__17 <o> d ) num__24 <o> e ) num__26 |
sol . in this case repetitions are not allowed . so the first prize can be given in num__4 ways . the second in num__3 ways and the third in num__2 ways . but fundamental principle ( num__4 x num__3 x num__2 ) ways = num__24 ways num__4 : or num__4 p = — num__4 : - num__4 x num__3 x num__2 x num__1 - num__24 ways d <eor> d <eos> |
d |
coin_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
coin_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
| to fill a tank num__25 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two - fifth of its present ? <o> a ) num__60.5 <o> b ) num__35.5 <o> c ) num__25.5 <o> d ) num__62 <o> e ) num__62.5 |
let the capacity of num__1 bucket = x . then the capacity of tank = num__25 x . new capacity of bucket = ( num__0.4 ) x required number of buckets = num__25 x / ( num__2 x / num__5 ) = num__25 x * num__2.5 x = num__62.5 = num__62.5 answer is e . <eor> e <eos> |
e |
divide__2.0__0.4__ divide__1.0__0.4__ multiply__25.0__2.5__ round__62.5__ |
divide__2.0__0.4__ divide__1.0__0.4__ multiply__25.0__2.5__ multiply__25.0__2.5__ |
| in num__12 pumps can raise num__1218 tons of water in num__11 days of num__9 hrs each how many pumps will raise num__2030 tons of water in num__12 days of num__11 hrs each ? <o> a ) num__12 <o> b ) num__15 <o> c ) num__18 <o> d ) num__21 <o> e ) num__22 |
explanation : pumps work time num__12 num__1218 num__99 x num__2030 num__132 = > num__1218 / ( num__912 * num__99 ) = num__2020 / ( x × num__132 ) = > x = num__15 pumps answer : option b <eor> b <eos> |
b |
multiply__11.0__9.0__ multiply__12.0__11.0__ round__15.0__ |
multiply__11.0__9.0__ multiply__12.0__11.0__ round__15.0__ |
| a metallic sheet is of rectangular shape with dimensions num__48 m x num__36 m . from each of its corners a square is cut off so as to make an open box . if the length of the square is num__8 m the volume of the box ( in m ^ num__3 ) is : <o> a ) num__2130 <o> b ) num__3620 <o> c ) num__4512 <o> d ) num__5620 <o> e ) num__5120 |
clearly l = ( num__48 - num__16 ) m = num__32 m b = ( num__36 - num__16 ) m = num__20 m h = num__8 m . volume of the box = ( num__32 x num__20 x num__8 ) m num__3 = num__5120 m ^ num__3 . answer e <eor> e <eos> |
e |
square_perimeter__8.0__ volume_rectangular_prism__8.0__32.0__20.0__ volume_rectangular_prism__8.0__32.0__20.0__ |
square_perimeter__8.0__ volume_rectangular_prism__8.0__32.0__20.0__ volume_rectangular_prism__8.0__32.0__20.0__ |
| weights of two friends ram and shyam are in the ratio num__3 : num__5 . if ram ' s weight is increased by num__10.0 and total weight of ram and shyam become num__82.8 kg with an increases of num__15.0 . by what percent did the weight of shyam has to be increased ? <o> a ) num__18.0 <o> b ) num__10.0 <o> c ) num__21.0 <o> d ) num__16.0 <o> e ) none |
solution : given ratio of ram and shayam ' s weight = num__3 : num__5 hence ( x - num__15 ) / ( num__15 - num__10 ) = num__0.6 or x = num__18.0 . answer : option a <eor> a <eos> |
a |
divide__3.0__5.0__ add__3.0__15.0__ add__3.0__15.0__ |
divide__3.0__5.0__ add__3.0__15.0__ add__3.0__15.0__ |
| the average age of num__20 students in a class is num__5 years . if teacher ' s age is also included then average increases num__1 year then find the teacher ' s age ? <o> a ) num__28 <o> b ) num__24 <o> c ) num__20 <o> d ) num__22 <o> e ) num__18 |
total age of num__50 students = num__20 * num__5 = num__100 total age of num__51 persons = num__20 * num__6 = num__120 age of teacher = num__120 - num__100 = num__20 years answer is c <eor> c <eos> |
c |
multiply__20.0__5.0__ add__1.0__50.0__ add__5.0__1.0__ add__20.0__100.0__ multiply__20.0__1.0__ |
multiply__20.0__5.0__ add__1.0__50.0__ add__5.0__1.0__ multiply__20.0__6.0__ multiply__20.0__1.0__ |
| if n is a positive integer such that n ! / ( n - num__2 ) ! = num__342 find n . <o> a ) num__17 <o> b ) num__18 <o> c ) num__19 <o> d ) num__20 <o> e ) num__21 |
we write n ! = n * ( n - num__1 ) ( n - num__2 ! ) therefore n ! / ( n - num__2 ) ! = n ( n - num__1 ) * ( n - num__2 ) ! / ( n - num__2 ) ! = n ( n - num__1 ) . - - > n ( n - num__1 ) = num__342 - - > n ^ num__2 - n - num__342 = num__0 - - > n ^ num__2 - num__19 n + num__18 n - num__342 = num__0 - - > n ( n - num__19 ) + num__18 ( n - num__19 ) = num__0 - - > ( n - num__19 ) ( n + num__18 ) = num__0 therefore n - num__19 = num__0 ; n + num__18 = num__0 ; ( i . e ) n = num__19 ; n = - num__18 we want positive integer . so n = num__19 . answer : c <eor> c <eos> |
c |
divide__342.0__19.0__ divide__342.0__18.0__ |
divide__342.0__19.0__ divide__342.0__18.0__ |
| the sum of the first n positive perfect squares where n is a positive integer is given by the formula n ^ num__1.0 + c * n ^ num__2 + n / num__6 where c is a constant . what is the sum of the first num__19 positive perfect squares ? <o> a ) num__1010 <o> b ) num__1164 <o> c ) num__1240 <o> d ) num__1316 <o> e ) num__2 |
470 |
first we need to find the constant ' c ' . the easiest way to find this is for the sum of the first two perfect squares for num__1 and num__2 = num__1 and num__4 respectively . hence lhs = num__1 + num__4 and plug n = num__2 for rhs and simplify to get c = num__0.5 . plug values of n = num__19 and c = num__0.5 into the equation and simplify to get the answer num__2470 . option e . <eor> e <eos> |
e |
e |
| a merchant marks his goods up by num__40.0 and then offers a discount of num__20.0 on the marked price . what % profit does the merchant make after the discount ? <o> a ) num__12.0 <o> b ) num__10.0 <o> c ) num__21.0 <o> d ) num__15.0 <o> e ) num__17 % |
let the price be num__100 . the price becomes num__140 after a num__40.0 markup . now a discount of num__20.0 on num__140 . profit = num__112 - num__100 num__12.0 answer a <eor> a <eos> |
a |
percent__100.0__12.0__ |
percent__100.0__12.0__ |
| the radius of a cone is num__49 m slant height is num__35 m . find the curved surface area ? <o> a ) num__5160 <o> b ) num__5390 <o> c ) num__6430 <o> d ) num__6720 <o> e ) num__7280 |
cone curved surface area = Ï € rl num__3.14285714286 Ã — num__49 Ã — num__35 = num__154 Ã — num__35 = num__5390 m ( power num__2 ) answer is b . <eor> b <eos> |
b |
multiply__35.0__154.0__ multiply__35.0__154.0__ |
multiply__35.0__154.0__ multiply__35.0__154.0__ |
| at a certain amateur bands rock show the ratio of freaks to nerds to geeks is num__1 : num__2 : num__3 . if these are the only types of people in the crowd and each person in the crowd has the same chance of yelling ` ` rock on ! ` ` what are the chances that the next person who yells ` ` rock on ! ' ' will be a geek ? <o> a ) num__0.166666666667 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__0.833333333333 |
the ratio of freaks to nerds to geeks is num__1 : num__2 : num__3 . so assuming the number as x num__2 x and num__3 x for freaks nerds and geeks respc . probability of next person being geek = num__3 x / total or num__3 x / num__6 x = num__0.5 . hence c . <eor> c <eos> |
c |
multiply__2.0__3.0__ reverse__2.0__ reverse__2.0__ |
multiply__2.0__3.0__ reverse__2.0__ reverse__2.0__ |
| the product of n ^ num__2 consecutive positive integers is always divisible by <o> a ) n ^ num__2 - num__1 <o> b ) ( n + num__1 ) ! <o> c ) num__2 n + num__1 <o> d ) n ^ num__2 + num__1 <o> e ) n ! |
plug in some vlues and check - product of first num__2 number is num__2 * num__1 product of first num__3 number is num__3 * num__2 * num__1 product of first num__4 number is num__4 * num__3 * num__2 * num__1 product of first num__5 number is num__5 * num__4 * num__3 * num__2 * num__1 so the product of first n natural numbers is always divisible by n ! answer will be ( a ) <eor> a <eos> |
a |
add__2.0__1.0__ add__1.0__3.0__ add__2.0__3.0__ multiply__2.0__1.0__ |
add__2.0__1.0__ add__1.0__3.0__ add__2.0__3.0__ multiply__2.0__1.0__ |
| if a boat goes num__5 km upstream in num__25 minutes and the speed of the stream is num__7 kmph then the speed of the boat in still water is ? <o> a ) num__18 <o> b ) num__19 <o> c ) num__20 <o> d ) num__21 <o> e ) num__22 |
rate upsteram = ( num__0.2 * num__60 ) kmph = num__12 kmph . speed of the stream = num__7 kmph let speed in still water be xkm / hr . then speed upstream = ( x - num__7 ) km / hr . x - num__7 = num__12 = = > x = num__19 km / hr answer ( b ) <eor> b <eos> |
b |
divide__5.0__25.0__ hour_to_min_conversion__ add__5.0__7.0__ add__7.0__12.0__ round__19.0__ |
divide__5.0__25.0__ hour_to_min_conversion__ divide__60.0__5.0__ add__7.0__12.0__ round__19.0__ |
| write num__450000 num__000000 in scientific notation . <o> a ) num__400.50 * num__10 num__11 <o> b ) num__400.5 * num__10 num__11 <o> c ) num__40.5 * num__10 num__11 <o> d ) num__4.5 * num__10 num__11 <o> e ) num__450.00 * num__10 num__11 |
a * num__10 n where a is a real number such that num__1 * | a | < num__10 and n is an integer . num__450000 num__000000 = num__4.5 * num__100000 num__000000 = num__4.5 * num__10 num__11 * * correct answer d <eor> d <eos> |
d |
divide__450000.0__4.5__ add__1.0__10.0__ divide__450000.0__100000.0__ |
divide__450000.0__4.5__ add__1.0__10.0__ multiply__1.0__4.5__ |
| the racing magic takes num__60 seconds to circle the racing track once . the charging bull makes num__40 rounds of the track in an hour . if they left the starting point together how many minutes will it take for them to meet at the starting point for the second time ? <o> a ) num__3 <o> b ) num__6 <o> c ) num__9 <o> d ) num__12 <o> e ) num__15 |
time taken by racing magic to make one circle = num__60 seconds time taken bycharging bullto make one circle = num__60 mins / num__40 = num__1.5 mins = num__90 seconds lcm of num__90 and num__60 seconds = num__180 seconds time taken for them to meet at the starting point for the second time = num__180 * num__2 = num__360 seconds = num__6 mins answer b <eor> b <eos> |
b |
divide__60.0__40.0__ multiply__60.0__1.5__ divide__180.0__90.0__ multiply__2.0__180.0__ divide__360.0__60.0__ round__6.0__ |
divide__60.0__40.0__ multiply__60.0__1.5__ divide__180.0__90.0__ multiply__2.0__180.0__ divide__360.0__60.0__ divide__360.0__60.0__ |
| the owner of a furniture shop charges his customer num__24.0 more than the cost price . if a customer paid rs . num__8339 for a computer table then what was the cost price of the computer table ? <o> a ) rs . num__6727 <o> b ) rs . num__6735 <o> c ) rs . num__6721 <o> d ) rs . num__6725 <o> e ) rs . num__6729 |
cp = sp * ( num__100 / ( num__100 + profit % ) ) = num__8339 ( num__0.806451612903 ) = rs . num__6725 . answer : d <eor> d <eos> |
d |
percent__100.0__6725.0__ |
percent__100.0__6725.0__ |
| the radius of a wheel is num__22.4 cm . what is the distance covered by the wheel in making num__2500 resolutions ? <o> a ) num__1187 m <o> b ) num__1704 m <o> c ) num__2179 m <o> d ) num__3520 m <o> e ) num__4297 m |
in one resolution the distance covered by the wheel is its own circumference . distance covered in num__2500 resolutions . = num__2500 * num__2 * num__3.14285714286 * num__22.4 = num__352000 cm = num__3520 m answer : d <eor> d <eos> |
d |
round__3520.0__ |
round__3520.0__ |
| the present population of a town is num__500 . population increase rate is num__10.0 p . a . find the population of town before num__2 years ? <o> a ) num__512 <o> b ) num__615 <o> c ) num__810 <o> d ) num__413 <o> e ) num__123 |
p = num__500 r = num__10.0 required population of town = p / ( num__1 + r / num__100 ) ^ t = num__500 / ( num__1 + num__0.1 ) ^ num__2 = num__500 / ( num__1.1 ) ^ num__2 = num__413 ( approximately ) answer is d <eor> d <eos> |
d |
percent__10.0__1.0__ percent__100.0__413.0__ |
percent__10.0__1.0__ percent__100.0__413.0__ |
| the sum of the present ages of a father and his son is num__60 years . six years ago father ' s age was five times the age of the son . after num__6 years son ' s age will be : <o> a ) num__12 years <o> b ) num__14 years <o> c ) num__18 years <o> d ) num__20 years <o> e ) num__22 years |
let the present ages of son and father be x and ( num__60 - x ) years respectively . then ( num__60 - x ) - num__6 = num__5 ( x - num__6 ) num__54 - x = num__5 x - num__30 num__6 x = num__84 x = num__14 . son ' s age after num__6 years = ( x + num__6 ) = num__20 years . . answer : option d <eor> d <eos> |
d |
subtract__60.0__6.0__ multiply__6.0__5.0__ add__54.0__30.0__ divide__84.0__6.0__ add__6.0__14.0__ add__6.0__14.0__ |
subtract__60.0__6.0__ multiply__6.0__5.0__ add__54.0__30.0__ divide__84.0__6.0__ add__6.0__14.0__ add__6.0__14.0__ |
| one urn contains two pieces of candy { one green and one red . a second urn contains four pieces of candy { one green and three red . for each urn each piece of candy is equally likely of being picked . you pick a piece of candy from each urn and eat the two chosen candies . if you eat exactly one piece of green candy you draw a second piece of candy from the urn still containing a green piece of candy . you now eat the candy you just chose . what is the probability that you ate two pieces of green candy ? <o> a ) num__0.125 <o> b ) num__0.25 <o> c ) num__0.375 <o> d ) num__0.5 <o> e ) num__0.625 |
there are four possibilities for the rst round : you pick a green piece of candy from each urn . ( probability num__0.5 * num__0.25 = num__0.125 ) you pick the green piece of candy from the rst urn and the red piece of candy from the second . ( probability num__0.5 * num__0.75 = num__0.375 : ) you have a num__0.333333333333 chance of eating the green piece of candy on the second round ( since three pieces of candy remain in the second urn and one of them is green . you pick the red piece of candy from the rst urn and the green piece of candy from the second . ( probability num__0.5 * num__0.25 = num__0.125 : ) there is only one more piece of candy in the rst urn so you eat it on the second round . you pick a red piece of candy from each urn . ( probability num__0.5 * num__0.75 = num__0.375 . ) your chance of eating two pieces of green candy is num__0.125 + num__0.375 num__0.333333333333 + num__0.125 = num__0.375 correct answer c <eor> c <eos> |
c |
negate_prob__0.25__ union_prob__0.5__0.125__0.25__ union_prob__0.5__0.25__0.375__ |
negate_prob__0.25__ union_prob__0.5__0.125__0.25__ union_prob__0.5__0.25__0.375__ |
| an athlete runs num__200 meters in num__24 seconds . his speed is ? <o> a ) num__10 km / hr <o> b ) num__17 k / hr <o> c ) num__27 km / hr <o> d ) num__30 km / hr <o> e ) none of these |
explana Ɵ on : speed = distancetime = num__20024 m / sec = num__253 m / sec num__253 ∗ num__185 km / hr = num__30 km / hr answer : d <eor> d <eos> |
d |
round__30.0__ |
round__30.0__ |
| in an election between the two candidates the candidates who gets num__60.0 of votes polled is wined by num__280 votes majority . what is the total number of votes polled ? <o> a ) num__1400 <o> b ) num__1500 <o> c ) num__1550 <o> d ) num__1600 <o> e ) num__1650 |
note : majority ( num__20.0 ) = difference in votes polled to win ( num__60.0 ) & defeated candidates ( num__40.0 ) num__20.0 = num__60.0 - num__40.0 num__20.0 - - - - - > num__280 ( num__20 × num__14 = num__280 ) num__100.0 - - - - - > num__1400 ( num__100 × num__14 = num__1400 ) a <eor> a <eos> |
a |
percent__100.0__1400.0__ |
percent__100.0__1400.0__ |
| two numbers are in the ratio num__3 : num__4 . if their l . c . m . is num__48 . what is sum of the numbers ? <o> a ) num__28 <o> b ) num__56 <o> c ) num__76 <o> d ) num__84 <o> e ) num__85 |
explanation : let the numbers be num__3 x and num__4 x lcm of num__3 x and num__4 x = num__12 x ( since lcm of num__3 and num__4 is num__12 . hence lcm of num__3 x and num__4 x is num__12 x ) given that lcm of num__3 x and num__4 x is num__48 = > num__12 x = num__48 = > x = num__4.0 = num__4 sum of the numbers = num__3 x + num__4 x = num__7 x = num__7 x num__4 = num__28 answer : option a <eor> a <eos> |
a |
multiply__3.0__4.0__ add__3.0__4.0__ multiply__4.0__7.0__ multiply__4.0__7.0__ |
multiply__3.0__4.0__ add__3.0__4.0__ multiply__4.0__7.0__ multiply__4.0__7.0__ |
| evaluate the expression [ x : ( y - num__3 ) ] ⋅ ( - num__4 ) - [ xy + ( - num__3 ) ] : - num__1 = ? when x = - num__5 and y = - num__2 <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) none |
solution : evaluate the expression [ x : ( y - num__3 ) ] ⋅ ( - num__4 ) - [ xy + ( - num__3 ) ] : - num__1 = [ ( - num__5 ) : ( - num__2 - num__3 ) ] ⋅ ( - num__4 ) - [ ( - num__5 ) ⋅ ( - num__2 ) + ( - num__3 ) ] : - num__1 = [ ( - num__5 ) : ( - num__5 ) ] ⋅ ( - num__4 ) - [ num__10 + ( - num__3 ) ] : - num__1 = ( + num__1 ) ⋅ ( - num__4 ) - ( + num__7 ) : - num__1 = - num__4 - ( - num__7 ) = - num__4 + num__7 = num__3 answer a <eor> a <eos> |
a |
multiply__5.0__2.0__ add__3.0__4.0__ multiply__3.0__1.0__ |
multiply__5.0__2.0__ add__3.0__4.0__ subtract__4.0__1.0__ |
| if the given two numbers are respectively num__7.0 and num__28.0 of a third number then what percentage is the first of the second ? <o> a ) num__20.0 <o> b ) num__25.0 <o> c ) num__18.0 <o> d ) num__30.0 <o> e ) none of these |
here l = num__7 and m = num__28 therefore first number = l / m x num__100.0 of second number = num__0.25 x num__100.0 of second number = num__25.0 of second number answer : b <eor> b <eos> |
b |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| two trains each num__250 m in length are running on the same parallel lines in opposite directions with the speed of num__80 kmph and num__70 kmph respectively . in what time will they cross each other completely ? <o> a ) num__11 sec <o> b ) num__15 sec <o> c ) num__19 sec <o> d ) num__12 seconds <o> e ) num__20 sec |
d d = num__250 m + num__250 m = num__500 m rs = num__80 + num__70 = num__150 * num__0.277777777778 = num__41.6666666667 t = num__500 * num__0.024 = num__12 sec <eor> d <eos> |
d |
add__80.0__70.0__ multiply__0.024__500.0__ round__12.0__ |
add__80.0__70.0__ multiply__0.024__500.0__ multiply__0.024__500.0__ |
| if y = num__35 p and p is prime what is the greatest common factor of y and num__14 p in terms of p ? <o> a ) p <o> b ) num__2 p <o> c ) num__5 p <o> d ) num__7 p <o> e ) p ^ num__2 |
y = num__35 p = num__5 * num__7 * p num__14 p = num__2 * num__7 * p the greatest common factor of num__35 p and num__14 p is the product of all the common prime factors using the lower power of repeated factors . the greatest common factor is num__7 * p = num__7 p the answer is d . <eor> d <eos> |
d |
gcd__35.0__14.0__ divide__14.0__7.0__ gcd__35.0__14.0__ |
gcd__35.0__14.0__ divide__14.0__7.0__ gcd__35.0__14.0__ |
| a man can do a piece of work in num__6 days but with the help of his son he can finish it in num__3 days . in what time can the son do it alone ? <o> a ) num__5 <o> b ) num__5 num__0.5 <o> c ) num__7 num__0.5 <o> d ) num__6 <o> e ) num__9 num__0.5 |
son ' s num__1 day work = num__0.333333333333 - num__0.166666666667 = num__0.166666666667 son alone can do the work in num__6 days = num__6 days answer is d <eor> d <eos> |
d |
divide__1.0__3.0__ divide__1.0__6.0__ round__6.0__ |
divide__1.0__3.0__ divide__1.0__6.0__ round__6.0__ |
| a circular mat with diameter num__20 inches is placed on a square tabletop each of whose sides is num__30 inches long . which of the following is closest to the fraction of the tabletop covered by the mat ? <o> a ) num__0.416666666667 <o> b ) num__0.3489 <o> c ) num__0.5 <o> d ) num__0.75 <o> e ) num__0.833333333333 |
c . it is a circle inscribed in a square . square side = num__30 - - - > square ( table ) area = num__30 ^ num__2 circle diameter = num__20 - - - > circle area = pir ^ num__2 = num__100 pi ( where pi = ~ num__3.14 ) covered fraction = num__100 * num__3.14 / num__30 * num__30 = ~ num__10.4666666667 * num__30 = num__3.14 / num__9 = num__0.3489 b <eor> b <eos> |
b |
rectangle_perimeter__20.0__30.0__ triangle_area__2.0__0.3489__ |
rectangle_perimeter__20.0__30.0__ triangle_area__2.0__0.3489__ |
| if apples are bought at the rate of num__30 for rs . num__100 . how many apples must be sold for rs . num__100 so as to gain num__20.0 ? <o> a ) num__28 <o> b ) num__25 <o> c ) num__20 <o> d ) num__22 <o> e ) num__23 |
sp for num__30 apples to gain num__20.0 = num__100 * num__1.2 = num__120 so num__30 apples must be sold for rs num__120 rs num__120 = num__30 apples rs num__100 = ( num__0.25 ) * num__100 = num__25 apple answer : b <eor> b <eos> |
b |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| how long does a train num__110 m long running at the speed of num__72 km / hr takes to cross a bridge num__132 m length ? <o> a ) num__13.1 sec <o> b ) num__12.1 sec <o> c ) num__22.1 sec <o> d ) num__32.1 sec <o> e ) num__13.2 sec |
speed = num__72 * num__0.277777777778 = num__20 m / sec total distance covered = num__110 + num__132 = num__242 m . required time = num__12.1 = num__12.1 sec . answer : b <eor> b <eos> |
b |
add__110.0__132.0__ divide__242.0__20.0__ round__12.1__ |
add__110.0__132.0__ divide__242.0__20.0__ divide__242.0__20.0__ |
| nicky and cristina are running a num__300 meter race . since cristina is faster than nicky she gives him a num__12 second head start . if cristina runs at a pace of num__5 meters per second and nicky runs at a pace of only num__3 meters per second how many seconds will nicky have run before cristina catches up to him ? <o> a ) num__40 <o> b ) num__50 <o> c ) num__30 <o> d ) num__60 <o> e ) num__45 |
the distance traveled by both of them is the same at the time of overtaking . num__3 ( t + num__12 ) = num__5 t t = num__18 . cristina will catch up nicky in num__18 seconds . so in num__18 seconds cristina would cover = num__18 * num__5 = num__90 meter . now time taken my nicky to cover num__90 meter = num__30.0 = num__30 seconds . c <eor> c <eos> |
c |
multiply__5.0__18.0__ add__12.0__18.0__ round__30.0__ |
multiply__5.0__18.0__ add__12.0__18.0__ add__12.0__18.0__ |
| ifaequals the sum of the even integers from num__2 to num__20 inclusive andbequals the sum of the odd integers from num__1 to num__19 inclusive what is the value of ( a - b ) / num__2 ? <o> a ) num__1 <o> b ) num__10 <o> c ) num__19 <o> d ) num__5 <o> e ) num__2 |
answer is num__10 yes ! there is really a faster way to solve it . sum of consecutive odd or even integers = ( no . of odd or even ints ) * ( first int + last int ) / num__2 here a = sum of even ints from num__2 to num__20 inclusive number of even ints = num__10 first int + last int = num__2 + num__20 = num__22 a = num__10 * num__11.0 = num__110 b = sum of odd ints from num__1 to num__19 inclusive number of odd ints = num__10 first int + last int = num__1 + num__19 = num__20 a = num__10 * num__10.0 = num__100 ( a - b ) / num__2 = num__110 - num__100 = num__5.0 = num__5 ans : d <eor> d <eos> |
d |
divide__20.0__2.0__ add__2.0__20.0__ add__1.0__10.0__ multiply__10.0__11.0__ subtract__110.0__10.0__ divide__100.0__20.0__ multiply__1.0__5.0__ |
divide__20.0__2.0__ add__2.0__20.0__ add__1.0__10.0__ multiply__10.0__11.0__ subtract__110.0__10.0__ divide__100.0__20.0__ multiply__1.0__5.0__ |
| a train num__500 m long can cross an electric pole in num__10 sec and then find the speed of the train ? <o> a ) num__140 <o> b ) num__150 <o> c ) num__160 <o> d ) num__170 <o> e ) num__180 |
length = speed * time speed = l / t s = num__50.0 s = num__50 m / sec speed = num__50 * num__3.6 ( to convert m / sec in to kmph multiply by num__3.6 ) speed = num__180 kmph answer : e <eor> e <eos> |
e |
divide__500.0__10.0__ multiply__50.0__3.6__ round__180.0__ |
divide__500.0__10.0__ multiply__50.0__3.6__ multiply__50.0__3.6__ |
| a sum at simple interests at num__13 ½ % per annum amounts to rs . num__2502.50 after num__4 years find the sum . <o> a ) num__1525 <o> b ) num__1625 <o> c ) num__1725 <o> d ) num__1825 <o> e ) num__1925 |
let the sum be x = p then sum ( p ) + ( p * r * t ) / num__100 = num__2502.50 therefore x + ( x * num__4 * num__13.5 ) / num__100 = num__2502.50 x = num__1625 answer : b <eor> b <eos> |
b |
percent__100.0__1625.0__ |
percent__100.0__1625.0__ |
| chang li has pet cockroaches that multiply at the same rate every week . the first week he had num__5 the next week he had num__10 the second week he had num__20 and the fourth week he had num__80 . how many did he have on week num__3 ? <o> a ) num__30 <o> b ) num__25 <o> c ) num__40 <o> d ) num__60 <o> e ) num__50 |
the answer is c ) num__40 because the cockroaches are doubling in size every week - half of num__80 is num__40 and num__20 doubled is num__40 . <eor> c <eos> |
c |
subtract__80.0__40.0__ |
subtract__80.0__40.0__ |
| two trains start from p and q respectively and travel towards each other at a speed of num__50 km / hr and num__40 km / hr respectively . by the time they meet the first train has traveled num__100 km more than the second . the distance between p and q is ? <o> a ) num__767 km <o> b ) num__975 km <o> c ) num__678 km <o> d ) num__900 km <o> e ) num__546 km |
at the time of meeting let the distance traveled by the second train be x km . then distance covered by the first train is ( x + num__100 ) km . x / num__40 = ( x + num__100 ) / num__50 num__50 x = num__40 x + num__4000 = > x = num__400 so distance between p and q = ( x + x + num__100 ) km = num__900 km . answer : d <eor> d <eos> |
d |
multiply__40.0__100.0__ round__900.0__ |
multiply__40.0__100.0__ round__900.0__ |
| danny bought num__2 q steaks for w dollars . jerome buys r steaks for a num__50.0 discount how much will the steaks cost him in cents ? <o> a ) num__50 rw / q . <o> b ) num__50 qr / w . <o> c ) num__25 rq / w . <o> d ) num__25 rw / q . <o> e ) rw / ( num__4 q ) . |
danny bought num__2 q steaks for w dollars so num__1 steak = w / num__2 q jerome buys r steaks for a num__50.0 discount : r * ( w / num__4 q ) in cents the answer will be : r * ( num__100 w / num__4 q ) = num__25 rw / q <eor> d <eos> |
d |
percent__2.0__50.0__ percent__25.0__100.0__ |
percent__2.0__50.0__ percent__25.0__100.0__ |
| the average salary of all the workers in a workshop is rs . num__8000 . the average salary of num__7 technicians is rs . num__12000 and the average salary of the rest is rs . num__6000 . the total number of workers in the workshop is ? <o> a ) num__23 <o> b ) num__21 <o> c ) num__52 <o> d ) num__56 <o> e ) num__12 |
let the total number of workers be x . then num__8000 x = ( num__12000 * num__7 ) + num__6000 ( x - num__7 ) = num__2000 x = num__42000 = x = num__21 . answer : b <eor> b <eos> |
b |
subtract__8000.0__6000.0__ multiply__7.0__6000.0__ divide__42000.0__2000.0__ divide__42000.0__2000.0__ |
subtract__8000.0__6000.0__ multiply__7.0__6000.0__ divide__42000.0__2000.0__ divide__42000.0__2000.0__ |
| when a certain perfect square is increased by num__148 the result is another perfect square . what is the value of the original perfect square ? <o> a ) num__1296 <o> b ) num__1369 <o> c ) num__1681 <o> d ) num__1764 <o> e ) num__2500 |
let ’ s call the two perfect squares x ^ num__2 and y ^ num__2 respectively . then the given information translates as x ^ num__2 + num__148 = y ^ num__2 . subtracting x ^ num__2 gives num__148 = y ^ num__2 − x ^ num__2 a difference of squares . this in turn factors as ( y + x ) ( y − x ) = num__148 . the next step is tricky . it begins with factoring num__148 which breaks down as num__2 ∗ num__2 ∗ num__37 . since we ’ re dealing with perfect squares x and y are positive integers and ( y + x ) and ( y − x ) must be paired integer factors of num__148 . the options are num__148 ∗ num__174 ∗ num__2 and num__37 ∗ num__4 . but our number properties establish that ( y + x ) and ( y − x ) must be either both odd or both even so only num__74 ∗ num__2 is an actual possibility . and because for any positive integers ( y + x ) > ( y − x ) we can conclude that y + x = num__74 and y – x = num__2 . solving by elimination num__2 y = num__76 y = num__38 and x = num__36 . finally we just need to square num__36 . but rather than multiplying it out note that num__36 ^ num__2 ends in num__6 – as does only one answer choice a . this answer must be the one we want . <eor> a <eos> |
a |
multiply__2.0__37.0__ power__36.0__2.0__ |
multiply__2.0__37.0__ power__36.0__2.0__ |
| how many bricks each measuring num__25 cm x num__11.25 cm x num__6 cm will be needed to build a wall of num__8 m x num__6 m x num__22.5 cm ? <o> a ) num__1245 <o> b ) num__4175 <o> c ) num__5412 <o> d ) num__4712 <o> e ) num__6400 |
number of bricks = volume of the wall / volume of num__1 brick = ( num__800 x num__600 x num__22.5 ) / ( num__25 x num__11.25 x num__6 ) = num__6400 . ans : e <eor> e <eos> |
e |
multiply__8.0__800.0__ round__6400.0__ |
multiply__8.0__800.0__ divide__6400.0__1.0__ |
| an uneducated retailer marks all his goods at num__40.0 above the cost price and thinking that he will still make num__25.0 profit offers a discount of num__25.0 on the marked price . what is his actual profit on the sales ? <o> a ) num__12.5 <o> b ) num__13.5 <o> c ) num__5.0 <o> d ) num__14.5 <o> e ) none |
sol . let c . p . = rs . num__100 . then marked price = rs . num__140 . s . p . = num__75.0 of rs . num__140 = rs . num__105 . ∴ gain % = num__5.0 . answer c <eor> c <eos> |
c |
percent__75.0__140.0__ percent__100.0__5.0__ |
percent__75.0__140.0__ percent__100.0__5.0__ |
| num__21 ball numbered num__1 to num__21 . a ballis drawn and then another ball is drawn without replacement . <o> a ) num__0.0666666666667 <o> b ) num__0.0566037735849 <o> c ) num__0.214285714286 <o> d ) num__0.4 <o> e ) num__0.583333333333 |
the probability that first toy shows the even number = num__1021 = num__1021 since the toy is not replaced there are now num__9 even numbered toys and total num__20 toys left . hence probability that second toy shows the even number = num__920 = num__920 required probability = ( num__1021 ) × ( num__920 ) = ( num__1021 ) × ( num__920 ) = num__0.214285714286 c <eor> c <eos> |
c |
subtract__21.0__1.0__ multiply__1.0__0.2143__ |
subtract__21.0__1.0__ multiply__1.0__0.2143__ |
| the maximum numbers of students among them num__451 pens and num__410 toys can be distributed in such a way that each student gets the same number of pens and same number of toys is <o> a ) num__41 <o> b ) num__910 <o> c ) num__1001 <o> d ) num__1911 <o> e ) none |
olution required number of students . = h . c . f of num__451 and num__410 . â € ¹ = â € º num__41 . answer a <eor> a <eos> |
a |
subtract__451.0__410.0__ subtract__451.0__410.0__ |
subtract__451.0__410.0__ subtract__451.0__410.0__ |
| if you cut a num__18 ft piece of wood into two pieces making one piece num__4 ft longer than the other . what size is the smaller piece ? <o> a ) num__4 ft <o> b ) num__6 ft <o> c ) num__8 ft <o> d ) num__7 ft <o> e ) num__12 ft |
total length is num__18 ft one piece is num__4 ft longer ( x + num__4 ) leaving the other piece to figure out ( x ) . ( x ) + ( x + num__4 ) = num__18 x + x + num__4 - num__4 = num__18 - num__4 num__2 x = num__14 num__2 x / num__2 = num__7.0 x = num__7 the piece is d ) num__7 ft . <eor> d <eos> |
d |
subtract__18.0__4.0__ divide__14.0__2.0__ divide__14.0__2.0__ |
subtract__18.0__4.0__ divide__14.0__2.0__ divide__14.0__2.0__ |
| the average age of a husband wife and their child num__5 years ago was num__27 years and that of wife and the child num__5 years ago was num__20 years . the present age of the husband is ? <o> a ) num__10.1 years <o> b ) num__54 years <o> c ) num__11.4 years <o> d ) num__12.6 years <o> e ) num__14 years |
sum of the present ages of husband wife and child = ( num__23 * num__2 + num__5 * num__2 ) = num__57 years . required average = num__11.4 = num__11.4 years . answer : c <eor> c <eos> |
c |
divide__57.0__5.0__ divide__57.0__5.0__ |
divide__57.0__5.0__ divide__57.0__5.0__ |
| if the selling price of num__50 articles is equal to the cost price of num__30 articles then the loss or gain percent is : <o> a ) num__40.0 <o> b ) num__23.0 <o> c ) num__20.0 <o> d ) num__60.0 <o> e ) num__56 % |
c . p . of each article be re . num__1 . then c . p . of num__50 articles = rs . num__50 ; s . p . of num__50 articles = rs . num__30 . loss % = num__0.4 * num__100 = num__40.0 answer a <eor> a <eos> |
a |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| if num__4 spiders make num__3 webs in num__9 days then how many days are needed for num__1 spider to make num__1 web ? <o> a ) num__9 <o> b ) num__12 <o> c ) num__27 <o> d ) num__24 <o> e ) num__16 |
explanation : let num__1 spider make num__1 web in x days . more spiders less days ( indirect proportion ) more webs more days ( direct proportion ) hence we can write as ( spiders ) num__4 : num__1 ( webs ) num__1 : num__9 } : : x : num__3 â ‡ ’ num__4 Ã — num__1 Ã — num__9 = num__1 Ã — num__3 Ã — x â ‡ ’ x = num__12 answer : option b <eor> b <eos> |
b |
multiply__4.0__3.0__ round__12.0__ |
multiply__4.0__3.0__ round__12.0__ |
| find out the c . i on rs . num__9000 at num__4.0 p . a . compound half - yearly for num__1 num__0.5 years <o> a ) num__550.87 <o> b ) num__506.07 <o> c ) num__506.04 <o> d ) num__506.03 <o> e ) num__306.01 |
a = num__9000 ( num__1.02 ) num__3 = num__9550.87 num__9000 - - - - - - - - - - - num__550.87 answer : a <eor> a <eos> |
a |
subtract__4.0__1.0__ subtract__9550.87__9000.0__ multiply__1.0__550.87__ |
subtract__4.0__1.0__ subtract__9550.87__9000.0__ subtract__9550.87__9000.0__ |
| the population of a town increased from num__75000 to num__2 num__25000 in a decade . the average percent increase of population per year is : <o> a ) num__20.0 <o> b ) num__5.0 <o> c ) num__6.0 <o> d ) num__8.75 <o> e ) none of these |
solution increase in num__10 year = ( num__225000 - num__75000 ) = num__150000 . increase % = ( num__2.0 x num__100 ) % = num__200.0 â ˆ ´ required average = ( num__20.0 ) % = num__20.0 answer a <eor> a <eos> |
a |
multiply__75000.0__2.0__ multiply__2.0__100.0__ multiply__2.0__10.0__ multiply__2.0__10.0__ |
subtract__225000.0__75000.0__ multiply__2.0__100.0__ multiply__2.0__10.0__ multiply__2.0__10.0__ |
| the side of a square is increased by num__10.0 then how much % does its area increases ? <o> a ) num__40.0 <o> b ) num__10.0 <o> c ) num__25.0 <o> d ) num__21.0 <o> e ) num__15.00 % |
a = num__100 a num__2 = num__10000 a = num__110 a num__2 = num__12100 - - - - - - - - - - - - - - - - num__10000 - - - - - - - - - num__2100 num__100 - - - - - - - ? = > num__21.0 answer : d <eor> d <eos> |
d |
power__100.0__2.0__ power__110.0__2.0__ triangle_area__2.0__21.0__ |
power__100.0__2.0__ power__110.0__2.0__ triangle_area__2.0__21.0__ |
| what is the sum of the different positive prime factors of num__550 ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__15 <o> d ) num__16 <o> e ) num__18 |
i think answer is e : num__27 num__550 = num__55 * num__10 = ( num__5 * num__11 ) * ( num__2 * num__5 ) sum of thedifferentpositive prime factors = num__2 + num__5 + num__11 = num__18 <eor> e <eos> |
e |
divide__550.0__55.0__ gcd__10.0__55.0__ divide__55.0__5.0__ divide__10.0__5.0__ lcm__2.0__18.0__ |
divide__550.0__55.0__ gcd__10.0__55.0__ divide__55.0__5.0__ divide__10.0__5.0__ lcm__2.0__18.0__ |
| the average ( arithmetic mean ) of y numbers is x . if num__10 is added to the set of numbers then the average will be x - num__5 . what is the value of y in terms of x ? <o> a ) x / num__6 - num__6 <o> b ) x / num__6 - num__5 <o> c ) x / num__7 - num__5 <o> d ) x / num__5 - num__3 <o> e ) x / num__5 - num__6 |
( a num__1 + a num__2 + . . + ay ) / y = x ( a num__1 + a num__2 + . . + ay + num__10 ) / ( y + num__1 ) = x - num__5 = > ( xy + num__10 ) / ( y + num__1 ) = x - num__5 = > xy + num__10 = yx - num__5 y + x - num__5 = > num__15 = x - num__5 y = > num__5 y = x - num__15 = > y = x / num__5 - num__3 answer - d <eor> d <eos> |
d |
divide__10.0__5.0__ add__10.0__5.0__ subtract__5.0__2.0__ subtract__10.0__5.0__ |
divide__10.0__5.0__ add__10.0__5.0__ subtract__5.0__2.0__ subtract__10.0__5.0__ |
| the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square whose area is num__3025 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is num__10 units ? <o> a ) num__140 <o> b ) num__150 <o> c ) num__160 <o> d ) num__170 <o> e ) num__220 |
given that the area of the square = num__3025 sq . units = > side of square = √ num__3025 = num__55 units the radius of the circle = side of the square = num__55 units length of the rectangle = num__0.4 * num__55 = num__22 units given that breadth = num__10 units area of the rectangle = lb = num__22 * num__10 = num__220 sq . units answer : option e <eor> e <eos> |
e |
multiply__0.4__55.0__ square_perimeter__55.0__ square_perimeter__55.0__ |
multiply__0.4__55.0__ multiply__10.0__22.0__ multiply__10.0__22.0__ |
| the mean of num__50 observations was num__36 . it was found later that an observation num__48 was wrongly taken as num__23 . the corrected new mean is ? <o> a ) num__36.3 <o> b ) num__36.7 <o> c ) num__36.5 <o> d ) num__36.2 <o> e ) num__36.1 |
correct sum = ( num__36 * num__50 + num__48 - num__23 ) = num__1825 . correct mean = num__36.5 = num__36.5 answer : c <eor> c <eos> |
c |
divide__1825.0__50.0__ divide__1825.0__50.0__ |
divide__1825.0__50.0__ divide__1825.0__50.0__ |
| how many kgs of tea worth rs . num__20 / kg must be blended with num__30 kgs of tea worth rs . num__30 / kg so that by selling the blended variety at rs . num__30 / kg there should be a gain of num__10.0 <o> a ) num__3 <o> b ) num__4 <o> c ) num__6 <o> d ) num__5 <o> e ) num__2 |
( x + num__30 ) * num__30 = ( num__1.1 ) ( num__20 x + num__30 * num__30 ) solving we get num__3 . answer is a . <eor> a <eos> |
a |
percent__30.0__10.0__ percent__30.0__10.0__ |
percent__30.0__10.0__ percent__30.0__10.0__ |
| if the reciprocals of two consecutive positive integers are added together what is the sum in terms of the greater integer d ? <o> a ) ( num__3 d - num__1 ) / ( d ^ num__2 - d ) <o> b ) ( num__2 d - num__1 ) / ( d ^ num__4 - d ) <o> c ) d ^ num__2 - d <o> d ) num__2 d - num__1 <o> e ) ( num__2 d - num__1 ) / ( d ^ num__2 - d ) |
let two consecutive positive integers be d and d - num__1 ( greater integer is d ) so ( num__1 / d ) + [ num__1 / ( d - num__1 ) ] = ( num__2 d - num__1 ) / d ( d - num__1 ) = ( num__2 d - num__1 ) / ( d ^ num__2 - d ) answer : e <eor> e <eos> |
e |
multiply__1.0__2.0__ |
divide__2.0__1.0__ |
| find the natural number nearest to num__9217 which completely divides num__88 without giving any remainder . <o> a ) num__9240 <o> b ) num__9064 <o> c ) num__9184 <o> d ) num__9152 <o> e ) num__9169 |
explanation : these types of questions are to be solved by trial and error . make sure minimum amount of time is wasted and your calculations are quick . devote maximum num__1 minute to such questions . here in this case clearly num__9240 is the closest number which divides num__88 . answer : a <eor> a <eos> |
a |
multiply__9240.0__1.0__ |
multiply__9240.0__1.0__ |
| the difference between a number and its two - fifth is num__510 . what is num__40.0 of that number ? <o> a ) num__342 <o> b ) num__340 <o> c ) num__344 <o> d ) num__346 <o> e ) num__348 |
let the number be x . then x - num__0.4 x = num__510 x = ( num__510 * num__5 ) / num__3 = num__850 num__40.0 of num__850 = num__340 . answer : b <eor> b <eos> |
b |
multiply__0.4__850.0__ multiply__0.4__850.0__ |
multiply__0.4__850.0__ multiply__0.4__850.0__ |
| if a boat goes num__7 km upstream in num__42 minutes and the speed of the stream is num__4 kmph then the speed of the boat in still water is <o> a ) num__12 kmph <o> b ) num__13 kmph <o> c ) num__14 kmph <o> d ) num__15 kmph <o> e ) none of these |
explanation : rate upstream = ( num__0.166666666667 ) * num__60 kmh = num__10 kmph . speed of stream = num__4 kmph . let speed in still water is x km / hr then speed upstream = ( x ï ¿ ½ num__4 ) km / hr . x - num__4 = num__10 or x = num__14 kmph answer : c <eor> c <eos> |
c |
divide__7.0__42.0__ hour_to_min_conversion__ add__4.0__10.0__ round__14.0__ |
divide__7.0__42.0__ hour_to_min_conversion__ add__4.0__10.0__ round__14.0__ |
| if a : b = num__0.333333333333 : num__0.5 b : c = num__0.333333333333 : num__0.5 then a : b : c ? <o> a ) num__4 : num__6 : num__9 <o> b ) num__6 : num__6 : num__9 <o> c ) num__7 : num__6 : num__9 <o> d ) num__6 : num__8 : num__9 <o> e ) num__8 : num__4 : num__9 |
a : b = num__0.333333333333 : num__0.5 = num__2 : num__3 b : c = num__0.333333333333 : num__0.5 = num__2 : num__3 - - - - - - - - - - - - - - - - - - - - a : b : c = num__4 : num__6 : num__9 answer : a <eor> a <eos> |
a |
reverse__0.5__ divide__2.0__0.5__ multiply__2.0__3.0__ add__3.0__6.0__ divide__2.0__0.5__ |
reverse__0.5__ divide__2.0__0.5__ multiply__2.0__3.0__ add__3.0__6.0__ subtract__6.0__2.0__ |
| there num__3 kinds of books in the library physics chemistry and biology . ratio of physics to chemistry is num__3 to num__2 ; ratio of chemistry to biology is num__4 to num__3 and the total of the books is more than num__3000 . which one of following can be the total r of the book ? <o> a ) num__3003 <o> b ) num__3027 <o> c ) num__3024 <o> d ) num__3021 <o> e ) num__3018 |
first you have to find the common ratio for all num__3 books . you have : p : c : b num__3 : num__2 - - > multiply by num__2 ( gives you row num__3 ) num__4 : num__6 num__6 : num__4 : num__3 hence : p : c : b : t ( total ) r num__6 : num__4 : num__3 : num__13 - - - - > this means the total number must be a multiple of num__13 . answer a is correct since num__299 is divisible by num__13 hence is num__2990 and so is num__3003 ( num__2990 + num__13 ) . <eor> a <eos> |
a |
multiply__3.0__2.0__ add__3.0__3000.0__ add__3.0__3000.0__ |
add__2.0__4.0__ add__3.0__3000.0__ add__3.0__3000.0__ |
| riya and priya set on a journey . riya moves eastward at a speed of num__25 kmph and priya moves westward at a speed of num__40 kmph . how far will be priya from riya after num__15 minutes <o> a ) num__18 kms <o> b ) num__16 kms <o> c ) num__50 kms <o> d ) num__30 kms <o> e ) num__40 kms |
total eastward distance = num__25 kmph * num__0.25 hr = num__6.25 km total westward distance = num__40 kmph * num__0.25 hr = num__10 km total distn betn them = num__6.25 + num__10 = num__16.25 km ans num__16 km answer : b <eor> b <eos> |
b |
multiply__25.0__0.25__ subtract__25.0__15.0__ add__6.25__10.0__ subtract__16.25__0.25__ round__16.0__ |
multiply__25.0__0.25__ multiply__40.0__0.25__ add__6.25__10.0__ subtract__16.25__0.25__ round__16.0__ |
| a number of num__53 marbles is to be divided and contain with boxes . if each box is to contain num__3 num__4 or num__5 marbles what is the largest possible number of boxes ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__15 <o> d ) num__16 <o> e ) num__17 |
to maximize # of boxes we should minimize marbles per box : num__16 * num__3 + num__1 * num__5 = num__53 - - > num__16 + num__1 = num__17 . answer : e . <eor> e <eos> |
e |
subtract__4.0__3.0__ add__1.0__16.0__ add__1.0__16.0__ |
subtract__4.0__3.0__ add__1.0__16.0__ add__1.0__16.0__ |
| the hen lays num__3 eggs then she lays num__6 eggs more . totally how many eggs she lays ? <o> a ) num__3 <o> b ) num__9 <o> c ) num__19 <o> d ) num__11 <o> e ) num__23 |
num__3 + num__6 = num__9 . answer is b . <eor> b <eos> |
b |
add__3.0__6.0__ add__3.0__6.0__ |
add__3.0__6.0__ add__3.0__6.0__ |
| a num__12.0 stock yielding num__10.0 is quoted at : <o> a ) num__83.33 <o> b ) num__110 <o> c ) num__112 <o> d ) num__120 <o> e ) num__160 |
solution to earn rs . num__10 money invested = rs . num__100 . to earn rs . num__12 money invested = rs . ( num__10.0 x num__12 ) = rs . num__120 . ∴ market value of rs . num__100 stock = rs . num__120 answer d <eor> d <eos> |
d |
percent__100.0__120.0__ |
percent__100.0__120.0__ |
| sheik abdullah decides to buy num__2 new cars for his collection . if he has to choose between nine different cars what is the number of purchasing possibilities he has ? <o> a ) a ) num__1312 <o> b ) b ) num__1412 <o> c ) c ) num__1512 <o> d ) d ) num__1612 <o> e ) e ) num__1712 |
sheik abdullah decides to buy num__2 new cars for his collection . he has to choose between nine different cars num__8 c num__3 = num__9 * num__8 * num__7 * num__6 / ( num__2 * num__1 ) = num__1512 ans : c <eor> c <eos> |
c |
power__3.0__2.0__ subtract__9.0__2.0__ multiply__2.0__3.0__ subtract__3.0__2.0__ multiply__1.0__1512.0__ |
power__3.0__2.0__ subtract__9.0__2.0__ multiply__2.0__3.0__ subtract__3.0__2.0__ multiply__1.0__1512.0__ |
| if p = | | q – num__3 | – num__2 | for how many values of q is m = num__6 ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__3 <o> d ) num__2 <o> e ) num__1 |
p = | | q – num__3 | – num__2 | can be num__4 only and only when q - num__3 = + / - num__8 . so there are num__2 values of q answer : d <eor> d <eos> |
d |
subtract__6.0__2.0__ add__2.0__6.0__ round__2.0__ |
subtract__6.0__2.0__ add__2.0__6.0__ divide__6.0__3.0__ |
| how many odd three - digit integers greater than num__400 are there such that all their digits are different ? <o> a ) num__140 <o> b ) num__156 <o> c ) num__172 <o> d ) num__181 <o> e ) num__200 |
num__5 ( hundred ' s digit ) * num__9 ( ten ' s digit ) * num__8 ( unit ' s digit ) = num__360 now take numbers in the range num__400 - num__900 . total numbers where all digits are different = num__360 ( as before ) [ highlight ] number of odd numbers = num__5 * num__8 * num__5 = num__200 [ / highlight ] ( now there are num__5 possibilities for the unit ' s digit ) e <eor> e <eos> |
e |
subtract__400.0__200.0__ |
subtract__400.0__200.0__ |
| in a can there is a mixture of milk and water in the ratio num__4 : num__5 . if it is filled with an additional num__8 litres of milk the can would be full and ratio of milk and water would become num__6 : num__5 . find the capacity of the can ? a . num__40 <o> a ) num__87 <o> b ) num__8 <o> c ) num__46 <o> d ) num__44 <o> e ) num__64 |
explanation : let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = num__0.444444444444 ( t - num__8 ) after adding milk quantity of milk in the mixture = num__0.545454545455 t . num__6 t / num__11 - num__8 = num__0.444444444444 ( t - num__8 ) num__10 t = num__792 - num__352 = > t = num__44 . answer : option d <eor> d <eos> |
d |
add__5.0__6.0__ add__4.0__6.0__ add__4.0__40.0__ add__4.0__40.0__ |
add__5.0__6.0__ divide__40.0__4.0__ divide__352.0__8.0__ divide__352.0__8.0__ |
| if ( n + num__2 ) ! / n ! = num__110 n = ? <o> a ) num__0.0152671755725 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
( n + num__2 ) ! / n ! = num__110 rewrite as : [ ( n + num__2 ) ( n + num__1 ) ( n ) ( n - num__1 ) ( n - num__2 ) . . . . ( num__3 ) ( num__2 ) ( num__1 ) ] / [ ( n ) ( n - num__1 ) ( n - num__2 ) . . . . ( num__3 ) ( num__2 ) ( num__1 ) ] = num__132 cancel out terms : ( n + num__2 ) ( n + num__1 ) = num__110 from here we might just test the answer choices . since ( num__11 ) ( num__10 ) = num__110 we can see that n = num__9 b <eor> b <eos> |
b |
add__2.0__1.0__ divide__110.0__11.0__ subtract__10.0__1.0__ multiply__1.0__9.0__ |
add__2.0__1.0__ divide__110.0__11.0__ subtract__10.0__1.0__ divide__9.0__1.0__ |
| in a right angled triangle two sides are consecutive whole number in which one side is hypotenuse . what could be the possible length of third side ? <o> a ) num__277 <o> b ) num__361 <o> c ) num__269 <o> d ) num__171 <o> e ) num__112 |
explanation : pythagorean triplets are generated with each ` ` odd number ' ' greater than num__1 by using a formula . if n is an odd number then pythagorean triplet = n n num__2 − num__12 n num__2 + num__12 n n num__2 − num__12 n num__2 + num__12 . here num__361 is an odd number . so the triplet is num__361 num__65160 num__65161 . answer : b <eor> b <eos> |
b |
multiply__1.0__361.0__ |
multiply__1.0__361.0__ |
| the length of a rectangle is double its width . if the length is diminished by num__5 cm and the width is increased by num__5 cm then its area is increased by num__75 cm square . what is the length of the rectangle ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__40 <o> d ) num__56 <o> e ) num__65 |
sol . according to question ( l - num__5 ) ( b + num__5 ) - lb = num__75 and l = num__2 b so b = num__20 l = num__40 ans . ( c ) <eor> c <eos> |
c |
square_perimeter__5.0__ multiply__2.0__20.0__ triangle_area__40.0__2.0__ |
square_perimeter__5.0__ multiply__2.0__20.0__ triangle_area__40.0__2.0__ |
| the two roots of the quadratic equation x num__2 - num__85 x + c = num__0 are prime numbers . what is the value of c ? <o> a ) num__84 <o> b ) num__166 <o> c ) num__332 <o> d ) num__664 <o> e ) num__1328 |
assuming that two prime numbers x num__1 and x num__2 are the solutions then x num__1 + x num__2 = num__85 . since num__85 is an odd number either x num__1 or x num__2 must be an even number . the only even prime number is num__2 . therefore one number must be num__2 and the other num__83 . hence c = num__2 * num__83 = num__166 . correct answer b <eor> b <eos> |
b |
subtract__85.0__2.0__ multiply__2.0__83.0__ multiply__2.0__83.0__ |
subtract__85.0__2.0__ multiply__2.0__83.0__ multiply__2.0__83.0__ |
| num__5 years ago kate was num__5 times as old as her son . num__5 years hence her age will be num__8 less than three times the corresponding age of her son . find the age of son ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__11 |
let x be the mother ' s age and y be the son ' s age x - num__5 = num__5 ( y - num__5 ) x + num__5 = num__3 ( y + num__5 ) - num__8 by solving this we get x = num__35 and y = num__11 answer : e <eor> e <eos> |
e |
subtract__8.0__5.0__ add__8.0__3.0__ add__8.0__3.0__ |
subtract__8.0__5.0__ add__8.0__3.0__ add__8.0__3.0__ |
| if the selling price of num__50 articles is equal to the cost price of num__30 articles then the loss or gain percent is : <o> a ) num__10.0 <o> b ) num__40.0 <o> c ) num__30.0 <o> d ) num__25.0 <o> e ) num__35 % |
let c . p . of each article be re . num__1 . then c . p . of num__50 articles = rs . num__50 ; s . p . of num__50 articles = rs . num__30 . loss % = num__0.4 * num__100 = num__40.0 answer : b <eor> b <eos> |
b |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| p alone can complete a job in num__12 days . the work done by q alone in one day is equal to one - half of the work done by p alone in one day . in how many days can the work be completed if p and q work together ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
p ' s rate is num__0.0833333333333 q ' s rate is num__0.0416666666667 the combined rate is num__0.0833333333333 + num__0.0416666666667 = num__0.125 if they work together the job will take num__8 days . the answer is d . <eor> d <eos> |
d |
add__0.0417__0.0833__ round__8.0__ |
add__0.0417__0.0833__ round__8.0__ |
| jane makes toy bears . when she works with an assistant she makes num__70 percent more bears per week and works num__10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ? <o> a ) num__20.0 <o> b ) num__80.0 <o> c ) num__100.0 <o> d ) num__180.0 <o> e ) num__200 % |
we can use fractional equivalents here to solve the problem num__80.0 = num__0.8 ; this means that in num__1 st case if she prepares num__5 bears in num__2 nd case she prepares num__9 bears num__10.0 = num__0.1 ; this means that in num__1 st case if she needs num__10 hours in num__2 nd case she needs num__9 hours now we come to productivity based on above fractional values the productivity in num__1 st case is num__0.5 bears / hour and in the num__2 nd case it is num__1 bear / hour hence the productivity is double with the assistant i . e . the increase in productivity is num__80.0 b <eor> b <eos> |
b |
add__70.0__10.0__ divide__10.0__5.0__ subtract__10.0__1.0__ divide__1.0__10.0__ divide__1.0__2.0__ round__80.0__ |
add__70.0__10.0__ divide__10.0__5.0__ subtract__10.0__1.0__ divide__1.0__10.0__ divide__1.0__2.0__ divide__80.0__1.0__ |
| maya purchase num__15 jars in a store then she returned num__5 jars to store . now how many jars she had ? <o> a ) a ) num__5 <o> b ) b ) num__11 <o> c ) c ) num__19 <o> d ) d ) num__10 <o> e ) e ) num__4 |
num__15 - num__5 = num__10 . answer is d <eor> d <eos> |
d |
subtract__15.0__5.0__ subtract__15.0__5.0__ |
subtract__15.0__5.0__ subtract__15.0__5.0__ |
| the principal that amounts to rs . num__4893 in num__3 years at num__6 num__0.25 % per annum c . i . compounded annually is ? <o> a ) s . num__3096 <o> b ) s . num__4076 <o> c ) s . num__4085 <o> d ) s . num__4096 <o> e ) s . num__5096 |
principal = [ num__4913 / ( num__1 + num__25 / ( num__4 * num__100 ) ) num__3 ] = num__4893 * num__0.941176470588 * num__0.941176470588 * num__0.941176470588 = rs . num__4085 . answer : c <eor> c <eos> |
c |
percent__100.0__4085.0__ |
percent__100.0__4085.0__ |
| look at this series : num__15 num__15 num__27 num__27 num__39 num__39 num__51 num__51 . . . what number should fill the blank ? <o> a ) num__51 <o> b ) num__39 <o> c ) num__63 <o> d ) num__15 <o> e ) none |
explanation : in this simple addition with repetition series each number in the series repeats itself and then increases by num__12 to arrive at the next number . answer : option c <eor> c <eos> |
c |
subtract__27.0__15.0__ add__51.0__12.0__ |
subtract__27.0__15.0__ add__51.0__12.0__ |
| a and b together can do a work in num__7 days . if a alone can do it in num__56 days . in how many days can b alone do it ? <o> a ) num__11 <o> b ) num__8 <o> c ) num__21 <o> d ) num__20 <o> e ) num__25 |
b num__8 num__0.142857142857 â € “ num__0.0178571428571 = num__0.125 = > num__8 <eor> b <eos> |
b |
divide__56.0__7.0__ divide__8.0__56.0__ divide__0.1429__8.0__ divide__7.0__56.0__ round__8.0__ |
divide__56.0__7.0__ divide__8.0__56.0__ divide__0.1429__8.0__ divide__7.0__56.0__ round__8.0__ |
| pipes a and b can fill a cistern in num__8 and num__56 minutes respectively . they are opened an alternate minutes . find how many minutes the cistern shall be full ? <o> a ) num__13 <o> b ) num__14 <o> c ) num__16 <o> d ) num__18 <o> e ) num__19 |
: num__0.125 + num__0.0178571428571 = num__0.142857142857 num__7 * num__2 = num__14 . answer : b <eor> b <eos> |
b |
divide__8.0__56.0__ divide__56.0__8.0__ multiply__2.0__7.0__ round__14.0__ |
add__0.125__0.0179__ multiply__56.0__0.125__ multiply__2.0__7.0__ multiply__2.0__7.0__ |
| the salary of all officers is increased twice successively by num__20.0 . what is the net percentage increase in their salaries ? <o> a ) num__20.0 <o> b ) num__40.0 <o> c ) num__21.0 <o> d ) num__44.0 <o> e ) num__48 % |
m . f = num__1.2 * num__1.2 = num__1.44 net % increase = ( m . f - num__1 ) * num__100 = ( num__1.44 - num__1 ) * num__100 num__0.44 * num__100 = num__44.0 answer : d <eor> d <eos> |
d |
percent__100.0__44.0__ |
percent__100.0__44.0__ |
| a sum of money invested at c . i . amounts to rs . num__800 in num__3 years to rs . num__860 in num__4 years . the rate of interest per annum is ? <o> a ) num__2 num__0.5 % <o> b ) num__4.0 <o> c ) num__5.0 <o> d ) num__6 num__0.666666666667 % <o> e ) num__7 num__0.5 % |
s . i . on rs . num__800 for num__1 year = ( num__860 - num__800 ) = rs . num__60 rate = ( num__100 * num__60 ) / ( num__800 * num__1 ) = num__7 num__0.5 % answer : e <eor> e <eos> |
e |
percent__100.0__7.0__ |
percent__100.0__7.0__ |
| the area of a sector of a circle of radius num__5 cm formed by an arc of length num__3.5 cm is ? <o> a ) num__0.35 cm num__2 <o> b ) num__17.5 cm num__2 <o> c ) num__8.75 cm num__2 <o> d ) num__55 cm num__2 <o> e ) num__65 cm num__2 |
explanation : ( num__5 * num__3.5 ) / num__2 = num__8.75 answer is c <eor> c <eos> |
c |
triangle_area__5.0__3.5__ triangle_area__5.0__3.5__ |
triangle_area__5.0__3.5__ triangle_area__5.0__3.5__ |
| the angle between the minute hand and the hour hand of a clock when the time is num__8.30 is : <o> a ) num__75 <o> b ) num__60 <o> c ) num__35 <o> d ) num__45 <o> e ) num__50 |
hr hand in num__8.5 hrs = ( num__30.0 * num__8.5 ) = num__255 hr hand in num__30 min = ( num__6.0 * num__30 ) = num__180 req angle = ( num__255 - num__180 ) = num__75 d answer a <eor> a <eos> |
a |
multiply__8.5__30.0__ multiply__6.0__30.0__ subtract__255.0__180.0__ round__75.0__ |
multiply__8.5__30.0__ multiply__6.0__30.0__ subtract__255.0__180.0__ subtract__255.0__180.0__ |
| the radius of a semi circle is num__6.8 cm then its perimeter is ? <o> a ) num__32.7 <o> b ) num__32.4 <o> c ) num__22.4 <o> d ) num__34.9 <o> e ) num__35.1 |
num__5.14285714286 r = num__6.8 = num__34.9 answer : d <eor> d <eos> |
d |
round__34.9__ |
round__34.9__ |
| in a num__1000 m race a beats b by num__90 m and b beats c by num__100 m . in the same race by how many meters does a beat c ? <o> a ) num__145 m <o> b ) num__176 m <o> c ) num__181 m <o> d ) num__159 m <o> e ) num__218 m |
by the time a covers num__1000 m b covers ( num__1000 - num__90 ) = num__910 m . by the time b covers num__1000 m c covers ( num__1000 - num__100 ) = num__900 m . so the ratio of speeds of a and c = num__1.0989010989 * num__1.11111111111 = num__1.221001221 so by the time a covers num__1000 m c covers num__819 m . so in num__1000 m race a beats c by num__1000 - num__819 = num__181 m . answer : c <eor> c <eos> |
c |
subtract__1000.0__90.0__ subtract__1000.0__100.0__ divide__1000.0__910.0__ divide__1000.0__900.0__ multiply__1.0989__1.1111__ subtract__1000.0__819.0__ round__181.0__ |
subtract__1000.0__90.0__ subtract__1000.0__100.0__ divide__1000.0__910.0__ divide__1000.0__900.0__ multiply__1.0989__1.1111__ subtract__1000.0__819.0__ subtract__1000.0__819.0__ |
| a train num__210 m long passed a pole in num__21 sec . how long will it take to pass a platform num__310 m long ? <o> a ) num__19 <o> b ) num__32 <o> c ) num__52 <o> d ) num__74 <o> e ) num__83 |
speed = num__10.0 = num__10 m / sec . required time = ( num__210 + num__310 ) / num__10 = num__52 sec . answer : c <eor> c <eos> |
c |
divide__210.0__21.0__ round__52.0__ |
divide__210.0__21.0__ round__52.0__ |
| if x num__2 + kx - num__3 is divisible by ( x - num__1 ) what is the value of k <o> a ) - num__1 <o> b ) num__1 <o> c ) num__2 <o> d ) - num__1 <o> e ) num__0 |
explanation : if x num__2 + kx - num__3 is divisible by ( x - num__1 ) then x = num__1 ( num__1 ) num__2 + num__1 ( k ) - num__3 = num__0 num__1 + k - num__3 = num__0 k = num__2 answer : option c <eor> c <eos> |
c |
multiply__2.0__1.0__ |
subtract__3.0__1.0__ |
| two trains one from howrah to patna and the other from patna to howrah start simultaneously . after they meet the trains reach their destinations after num__9 hours and num__49 hours respectively . the ratio of their speeds is ? <o> a ) num__4 : num__5 <o> b ) num__7 : num__3 <o> c ) num__4 : num__4 <o> d ) num__7 : num__8 <o> e ) num__7 : num__1 |
let us name the trains a and b . then ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ num__49 : √ num__9 = num__7 : num__3 answer : b <eor> b <eos> |
b |
round__7.0__ |
round__7.0__ |
| in how much time will a train of length num__100 m moving at num__36 kmph cross an electric pole ? <o> a ) num__15 sec <o> b ) num__12 sec <o> c ) num__10 sec <o> d ) num__14 sec <o> e ) num__11 sec |
c num__10 sec convert kmph to mps . num__36 kmph = num__36 * num__0.277777777778 = num__10 mps . the distance to be covered is equal to the length of the train . required time t = d / s = num__10.0 = num__10 sec . <eor> c <eos> |
c |
divide__10.0__36.0__ round__10.0__ |
divide__10.0__36.0__ divide__100.0__10.0__ |
| what distance ( in meters ) will be covered by a bus moving at num__54 km / hr in num__15 seconds ? <o> a ) num__175 <o> b ) num__200 <o> c ) num__225 <o> d ) num__250 <o> e ) num__275 |
num__54 km / hr = num__54 * num__0.277777777778 = num__15 m / s distance = num__15 * num__15 = num__225 meters the answer is c . <eor> c <eos> |
c |
divide__15.0__54.0__ round__225.0__ |
divide__15.0__54.0__ round__225.0__ |
| out of three consecutive odd numbers nine times the first number is equal to addition of twice the third number and adding num__9 to twice the second . what is the first number ? <o> a ) num__0.952380952381 <o> b ) num__0.92 <o> c ) num__1.04347826087 <o> d ) num__4.4 <o> e ) num__4.2 |
description : = > num__9 x = num__2 ( x + num__2 ) + num__9 + num__2 ( x + num__4 ) = > num__9 x = num__4 x + num__21 = > num__5 x = num__21 x = num__4.2 answer e <eor> e <eos> |
e |
subtract__9.0__4.0__ divide__21.0__5.0__ divide__21.0__5.0__ |
subtract__9.0__4.0__ divide__21.0__5.0__ divide__21.0__5.0__ |
| how many integers between num__50 and num__200 are there such that their unit digit is even ? <o> a ) num__50 <o> b ) num__75 <o> c ) num__100 <o> d ) num__150 <o> e ) num__200 |
num__150 numbers between num__50 and num__200 out of which half would be even half would be odd . number of even unit digit number = num__75 . the correct option is b <eor> b <eos> |
b |
subtract__200.0__50.0__ subtract__150.0__75.0__ |
subtract__200.0__50.0__ subtract__150.0__75.0__ |
| an electric pump can fill a tank in num__5 hours . because of a leak in the tank it took num__10 hours to fill the tank . if the tank is full how much time will the leak take to empty it ? <o> a ) num__10 hours <o> b ) num__12 hours <o> c ) num__8 hours <o> d ) num__5 hours <o> e ) num__15 hours |
work done by the leak in num__1 hour = num__0.2 - num__0.1 = num__0.1 the leak will empty the tank in num__10 hours answer is a <eor> a <eos> |
a |
divide__1.0__5.0__ divide__1.0__10.0__ round__10.0__ |
divide__1.0__5.0__ divide__1.0__10.0__ round__10.0__ |
| let the second number be num__3 x so that the first number is num__6 x and the third number is num__2 x . <o> a ) num__22 <o> b ) num__77 <o> c ) num__98 <o> d ) num__27 <o> e ) num__36 |
∴ num__6 x + num__3 x + num__2 x = num__132 ⇒ x = num__12 . second number = num__3 x = num__3 × num__12 = num__36 . answer : e <eor> e <eos> |
e |
multiply__6.0__2.0__ multiply__3.0__12.0__ round__36.0__ |
multiply__6.0__2.0__ multiply__3.0__12.0__ round__36.0__ |
| there are num__3 pairs of socks and num__2 socks are worn from that such that the pair of socks worn are not of the same pair . what is the number of pair that can be formed . <o> a ) num__5 <o> b ) num__1 <o> c ) num__3 <o> d ) num__6 <o> e ) num__7 |
first of all you should remember that there is a difference in left and right sock . now no . of way to select any of the sock = num__3 and for second = num__2 so total methods = num__3 * num__2 = num__6 answer : d <eor> d <eos> |
d |
die_space__ die_space__ |
die_space__ die_space__ |
| three partners a b c in a business invested money such that num__4 ( a â € ™ s capital ) = num__2 ( b â € ™ s capital ) = num__8 ( c â € ™ s capital ) then the ratio of their capitals is <o> a ) num__63 : num__45 : num__34 <o> b ) num__63 : num__54 : num__34 <o> c ) num__36 : num__54 : num__28 <o> d ) num__2 : num__4 : num__1 <o> e ) none of these |
explanation : let num__4 ( a â € ™ s capital ) = num__2 ( b â € ™ s capital ) = num__8 ( c â € ™ s capital ) = rs . x then a â € ™ s capital = rs x / num__4 b â € ™ s capital = rs . x / num__2 and c â € ™ s capital = rs . x / num__8 . a : b : c = x / num__4 : x / num__2 : x / num__8 num__2 : num__4 : num__1 answer : option d <eor> d <eos> |
d |
subtract__4.0__2.0__ |
divide__4.0__2.0__ |
| a gardener grows cabbages in her garden that is in the shape of a square . each cabbage takes num__1 square feet of area in her garden . this year she has increased her output by num__199 cabbages as compared to last year . the shape of the area used for growing the cabbages has remained a square in both these years . how many cabbages did she produce this year ? <o> a ) num__10000 <o> b ) num__11025 <o> c ) num__14400 <o> d ) num__12696 <o> e ) can not be determined |
let the side for growing cabbages this year be x ft . thus the area is x ^ num__2 . let the side for growing cabbages last year be y ft . thus the area was y ^ num__2 . the area would have increased by num__199 sq ft as each cabbage takes num__1 sq ft space . x ^ num__2 - y ^ num__2 = num__199 ( x + y ) ( x - y ) = num__199 num__199 is a prime number and thus it will be ( num__100 + num__99 ) * ( num__100 - num__99 ) . thus x = num__100 and y = num__99 x ^ num__2 = num__100 ^ num__2 = num__10000 the answer is a . <eor> a <eos> |
a |
power__100.0__2.0__ multiply__1.0__10000.0__ |
power__100.0__2.0__ multiply__1.0__10000.0__ |
| in a certain school num__55 percent of student are females and the rest are males . in school elections if num__60 percent of males and num__80 percent of females are expected to vote for a candidate ( a ) what percent of students are expected to vote for candidate ( a ) ? <o> a ) num__58.0 <o> b ) num__65.0 <o> c ) num__69.0 <o> d ) num__71.0 <o> e ) num__73 % |
say there are a total of num__100 students in this school num__55 are females and num__45 are males num__55 * num__0.8 = num__44 female are expected to vote num__45 * num__0.6 = num__27 males are expected to vote thus total of num__44 + num__27 = num__71 are expected to vote which is num__71 percent of students . answer : d <eor> d <eos> |
d |
percent__55.0__80.0__ percent__60.0__45.0__ percent__100.0__71.0__ |
percent__55.0__80.0__ percent__60.0__45.0__ percent__100.0__71.0__ |
| a car gets num__40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel num__120 kilometers ? <o> a ) num__4.5 <o> b ) num__5.5 <o> c ) num__6.5 <o> d ) num__3 <o> e ) num__8.5 |
each num__40 kilometers num__1 gallon is needed . we need to know how many num__40 kilometers are there in num__120 kilometers ? num__120 ÷ num__40 = num__3 × num__1 gallon = num__3 gallons correct answer d <eor> d <eos> |
d |
divide__120.0__40.0__ round__3.0__ |
divide__120.0__40.0__ round__3.0__ |
| nil and ethan are brothers . they left their home at the same time and drove to the same beach . nil drove at a speed of num__70 miles per hour . ethan drove at a speed of num__20 miles per hour . nil arrived at the beach num__0.5 hour earlier than ethan . what is the distance between their home and the beach ? <o> a ) num__80 miles <o> b ) num__70 miles <o> c ) num__60 miles <o> d ) num__50 miles <o> e ) num__40 miles |
every hour nil gets ahead of ethan num__70 - num__20 = num__50 miles . when nil arrived at the beach ethan is only num__20 × num__0.5 = num__10 miles behind . that tells us they only drove num__1 hour when nil arrived at the beach . the distance between their home and the beach is nil ’ s speed × nil ’ s time = num__70 × num__1 = num__70 miles . correct answer b <eor> b <eos> |
b |
subtract__70.0__20.0__ multiply__20.0__0.5__ round__70.0__ |
subtract__70.0__20.0__ multiply__20.0__0.5__ round__70.0__ |
| a person traveled num__3 equal distances num__4 km / hr num__5 km / hr num__6 km / hr then find its avg speed ? <o> a ) num__5.2 km / hr <o> b ) num__3.18 km / hr <o> c ) num__4.86 km / hr <o> d ) num__2.16 km / hr <o> e ) num__6.32 km / hr |
avg speed = num__3 xyz / ( xy + yz + zx ) required answer is = num__3 * num__4 * num__5 * num__6 / ( num__20 + num__30 + num__24 ) = num__4.86 km / hr answer is c <eor> c <eos> |
c |
multiply__4.0__5.0__ multiply__5.0__6.0__ multiply__4.0__6.0__ round__4.86__ |
multiply__4.0__5.0__ multiply__5.0__6.0__ multiply__4.0__6.0__ round__4.86__ |
| the ratio of num__3 seconds to num__30 hours <o> a ) num__1 : num__80 <o> b ) num__1 : num__36000 <o> c ) num__1 : num__600 <o> d ) num__1 : num__400 <o> e ) num__1 : num__500 |
num__1 hour = num__3600 sec then num__30 hours = num__30 * num__3600 = num__108000 so num__3 : num__108000 = num__1 : num__36000 answer : b <eor> b <eos> |
b |
multiply__30.0__3600.0__ divide__108000.0__3.0__ round__1.0__ |
multiply__30.0__3600.0__ divide__108000.0__3.0__ round__1.0__ |
| a and b go around a circular track of length num__600 m on a cycle at speeds of num__36 kmph and num__54 kmph . after how much time will they meet for the first time at the starting point ? <o> a ) num__110 sec <o> b ) num__130 sec <o> c ) num__120 sec <o> d ) num__140 sec <o> e ) num__150 sec |
c num__120 sec time taken to meet for the first time at the starting point = lcm { length of the track / speed of a length of the track / speed of b } = lcm { num__600 / ( num__36 * num__0.277777777778 ) num__600 / ( num__54 * num__0.277777777778 ) } = lcm ( num__60 num__40 ) = num__120 sec . <eor> c <eos> |
c |
hour_to_min_conversion__ round__120.0__ |
hour_to_min_conversion__ round__120.0__ |
| when num__60 per cent of a number is added to another number the second number increases to its num__140 per cent . what is the ratio between the first and the second number ? <o> a ) num__3 : num__4 <o> b ) num__4 : num__3 <o> c ) num__2 : num__3 <o> d ) data inadequate <o> e ) none of these |
let the first and the second numbers be x and y respect then y + num__60.0 of x = num__140.0 of y or y + num__0.6 x = num__1.4 y or num__0.6 x = num__0.4 y ∴ x : y = num__0.4 : num__0.6 = num__2 : num__3 answer c <eor> c <eos> |
c |
add__0.6__1.4__ add__0.6__1.4__ |
add__0.6__1.4__ add__0.6__1.4__ |
| find out the wrong number in the given sequence of numbers . num__1 num__2 num__6 num__15 num__31 num__56 num__91 <o> a ) num__2 <o> b ) num__6 <o> c ) num__15 <o> d ) num__31 <o> e ) num__91 |
num__1 ( result ) + ( num__1 * num__1 ) = num__2 . num__2 ( result ) + ( num__2 * num__2 ) = num__6 . num__6 ( result ) + ( num__3 * num__3 ) = num__15 . num__15 ( result ) + ( num__4 * num__4 ) = num__31 . num__31 ( result ) + ( num__5 * num__5 ) = num__56 . num__56 ( result ) + ( num__6 * num__6 ) = num__92 . now we are getting num__92 not num__91 . . so num__91 is the wrong number of the given . answer : e <eor> e <eos> |
e |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ add__1.0__91.0__ multiply__1.0__91.0__ |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ add__1.0__91.0__ multiply__1.0__91.0__ |
| a boy was asked to multiply a number by num__25 . he instead multiplied the next number by num__52 and got the answers num__324 more than the correct answer . the number to be multiplied was <o> a ) num__10 <o> b ) num__12 <o> c ) num__15 <o> d ) num__25 <o> e ) num__35 |
let the number be a num__52 xa = num__25 xa + num__324 num__52 a = num__25 a + num__324 num__52 a - num__25 a = num__324 num__27 a = num__324 a = num__12 . answer : b <eor> b <eos> |
b |
subtract__52.0__25.0__ divide__324.0__27.0__ divide__324.0__27.0__ |
subtract__52.0__25.0__ divide__324.0__27.0__ divide__324.0__27.0__ |
| in an examination it is required to get num__30.0 of minimum marks to pass . a student got num__170 marks and was declared failed by num__10 marks difference . then what is total marks : ? <o> a ) num__550 <o> b ) num__500 <o> c ) num__600 <o> d ) num__700 <o> e ) none of these |
explanation : num__30.0 - - - - - - - > num__180 ( num__30 * num__6 = num__180 ) num__100.0 - - - - - - - > num__600 ( num__100 * num__6 = num__600 ) answer : option c <eor> c <eos> |
c |
add__170.0__10.0__ divide__180.0__30.0__ multiply__100.0__6.0__ multiply__100.0__6.0__ |
add__170.0__10.0__ divide__180.0__30.0__ multiply__100.0__6.0__ multiply__100.0__6.0__ |
| the ratio of num__2 numbers is num__3 : num__7 and their h . c . f . is num__10 . their l . c . m . is ? <o> a ) num__220 <o> b ) num__210 <o> c ) num__252 <o> d ) num__236 <o> e ) num__260 |
let the numbers be num__3 x and num__7 x their h . c . f . = num__10 so the numbers are num__3 * num__10 num__7 * num__10 = num__3070 l . c . m . = num__210 answer is b <eor> b <eos> |
b |
lcm__2.0__210.0__ |
lcm__2.0__210.0__ |
| a man spends num__1810 for buying bed sheets at num__200 each and pillows at num__70 each . what will be the ratio of bed sheets to pillows when maximum number of bed sheets are bought ? <o> a ) num__3 : num__8 <o> b ) num__8 : num__3 <o> c ) num__9 : num__1 <o> d ) num__1 : num__9 <o> e ) none of these |
the man can ’ t purchase more than num__8 bedsheets ∴ cost of num__8 bedsheets = num__8 × num__200 = num__1600 remaining amount = num__1810 – num__1600 = num__210 and in num__210 the man can purchase num__210 ⁄ num__70 = num__3 pillows reqd ratio = num__8 : num__3 answer b <eor> b <eos> |
b |
multiply__200.0__8.0__ subtract__1810.0__1600.0__ divide__210.0__70.0__ divide__1600.0__200.0__ |
multiply__200.0__8.0__ subtract__1810.0__1600.0__ divide__210.0__70.0__ divide__1600.0__200.0__ |
| look at this series : num__8 num__22 num__8 num__28 num__8 . . . what number should come next ? <o> a ) num__9 <o> b ) num__29 <o> c ) num__32 <o> d ) num__34 <o> e ) num__36 |
explanation : this is a simple addition series with a random number num__8 interpolated as every other number . in the series num__6 is added to each number except num__8 to arrive at the next number . answer : option d <eor> d <eos> |
d |
subtract__28.0__22.0__ add__28.0__6.0__ |
subtract__28.0__22.0__ add__28.0__6.0__ |
| a student chose a number multiplied it by num__2 then subtracted num__138 from the result and got num__102 . what was the number he chose ? <o> a ) num__120 <o> b ) num__240 <o> c ) num__360 <o> d ) num__480 <o> e ) num__520 |
let x be the number he chose then num__2 ⋅ x − num__138 = num__102 num__2 x = num__240 x = num__120 correct answer a <eor> a <eos> |
a |
add__138.0__102.0__ divide__240.0__2.0__ divide__240.0__2.0__ |
add__138.0__102.0__ divide__240.0__2.0__ divide__240.0__2.0__ |
| num__2 trains starting at the same time from num__2 stations num__200 km apart and going in opposite direction cross each other ata distance of num__110 km from one of the stations . what is the ratio of their speeds . <o> a ) num__10 : num__9 <o> b ) num__11 : num__9 <o> c ) num__9 : num__9 <o> d ) num__9 : num__10 <o> e ) num__5 : num__6 |
in same time they cover num__110 km & num__90 km respectively so ratio of their speed = num__110 : num__90 = num__11 : num__9 answer is b . <eor> b <eos> |
b |
subtract__200.0__110.0__ subtract__11.0__2.0__ round__11.0__ |
subtract__200.0__110.0__ subtract__11.0__2.0__ round__11.0__ |
| there are two examinations rooms a and b . if num__10 students are sent from a to b then the number of students in each room is the same . if num__20 candidates are sent from b to a then the number of students in a is double the number of students in b . the number of students in room a is ? <o> a ) num__128 <o> b ) num__276 <o> c ) num__100 <o> d ) num__277 <o> e ) num__110 |
let the number of students in rooms a and b be x and y respectively . then x - num__10 = y + num__10 ⇒ x - y = num__20 . . . . ( i ) and x + num__20 = num__2 ( y - num__20 ) ⇒ x - num__2 y = - num__60 . . . . ( ii ) solving ( i ) and ( ii ) we get : x = num__100 y = num__80 . ∴ the required answer a = num__100 . answer : c <eor> c <eos> |
c |
divide__20.0__10.0__ add__20.0__60.0__ add__20.0__80.0__ |
divide__20.0__10.0__ add__20.0__60.0__ add__20.0__80.0__ |
| what profit percent is made by selling an article at a certain price if by selling at num__0.666666666667 rd of that price there would be a loss of num__20.0 ? <o> a ) num__85.0 <o> b ) num__26.0 <o> c ) num__10.0 <o> d ) num__20.0 <o> e ) num__15 % |
explanation : sp num__2 = num__0.666666666667 sp num__1 cp = num__100 sp num__2 = num__80 num__0.666666666667 sp num__1 = num__80 sp num__1 = num__120 num__100 - - - num__20 = > num__20.0 answer : d <eor> d <eos> |
d |
percent__20.0__100.0__ |
percent__20.0__100.0__ |
| of the families in city x in num__1998 num__30 percent owned a personal computer . the number of families in city x owning a computer in num__2002 was num__20 percent greater than it was in num__1998 and the total number of families in city x was num__8 percent greater in num__2002 than it was in num__1998 . what percent of the families in city x owned a personal computer in num__2002 ? <o> a ) num__50.12 <o> b ) num__52.66 <o> c ) num__56.33 <o> d ) num__33.33 <o> e ) num__74.12 % |
say a num__100 families existed in num__1998 then the number of families owning a computer in num__1998 - num__30 number of families owning computer in num__2002 = num__30 * num__1.2 = num__36 number of families in num__2002 = num__108 the percentage = num__0.333333333333 * num__100 = num__33.33 . option : d <eor> d <eos> |
d |
multiply__30.0__1.2__ add__8.0__100.0__ divide__36.0__108.0__ multiply__100.0__0.3333__ multiply__100.0__0.3333__ |
multiply__30.0__1.2__ add__8.0__100.0__ divide__36.0__108.0__ multiply__100.0__0.3333__ multiply__100.0__0.3333__ |
| a car after covering ½ of a journey of num__100 km develops engine trouble and later travels at ½ of its original speed . as a result it arrives num__2 hours late than its normal time . what is the normal speed of the car is ? <o> a ) num__111 <o> b ) num__105 <o> c ) num__888 <o> d ) num__266 <o> e ) num__882 |
x : y : z = num__100 : num__45 : num__30 num__20 : num__9 : num__6 num__9 - - - num__27 num__35 - - - ? = > num__105 answer : b <eor> b <eos> |
b |
round__105.0__ |
round__105.0__ |
| a dishonest dealer professes to sell his goods at cost price but still gets num__20.0 profit by using a false weight . what weight does he substitute for a kilogram ? <o> a ) num__833 num__0.142857142857 <o> b ) num__833 num__0.333333333333 <o> c ) num__833 num__0.222222222222 <o> d ) num__833 num__1.0 <o> e ) num__833 num__0.25 |
if the cost price is rs . num__100 then to get a profit of num__20.0 the selling price should be rs . num__120 . if num__120 kg are to be sold and the dealer gives only num__100 kg to get a profit of num__20.0 . how many grams he has to give instead of one kilogram ( num__1000 gm ) . num__120 gm - - - - - - num__100 gm num__1000 gm - - - - - - ? ( num__1000 * num__100 ) / num__120 = num__833.333333333 = num__833 num__0.333333333333 grams . answer : b <eor> b <eos> |
b |
percent__100.0__833.0__ |
percent__100.0__833.0__ |
| ronald and elan are working on an assignment . ronald takes num__6 hrs to type num__32 pages on a computer while elan takes num__5 hrs to type num__40 pages . how much time will they take working together on two different computers to type an assignment of num__110 pages ? <o> a ) num__8 hrs num__19 min <o> b ) num__8 hrs num__18 min <o> c ) num__8 hrs num__15 min <o> d ) num__8 hrs num__65 min <o> e ) num__8 hrs num__18 min |
number of pages typed by ronald in num__1 hour = num__5.33333333333 = num__5.33333333333 number of pages typed by elan in num__1 hour = num__8.0 = num__8 number of pages typed by both in num__1 hour = ( num__5.33333333333 + num__8 ) = num__13.3333333333 time taken by both to type num__110 pages = ( num__110 * num__0.075 ) = num__8 num__0.25 = num__8 hrs num__15 min answer : c <eor> c <eos> |
c |
subtract__6.0__5.0__ divide__32.0__6.0__ subtract__40.0__32.0__ add__5.3333__8.0__ reverse__13.3333__ divide__8.0__32.0__ multiply__32.0__0.25__ |
subtract__6.0__5.0__ divide__32.0__6.0__ subtract__40.0__32.0__ add__5.3333__8.0__ reverse__13.3333__ divide__8.0__32.0__ multiply__32.0__0.25__ |
| bruno and sacha are running in the same direction around a stadium . sacha runs at a constant speed of num__6 meters per second and bruno runs at a constant speed of num__5 meters per second . at a certain point sacha overtakes bruno . if two minute afterward sacha stops and waits for bruno to reach him then how many seconds does he have to wait ? <o> a ) num__12 <o> b ) num__24 <o> c ) num__36 <o> d ) num__60 <o> e ) num__72 |
the difference of the speed is num__1 m per second so in two minute sacha will be num__120 m ahead of bruno . . bruno will cover this in num__24.0 = num__24 secs . . b <eor> b <eos> |
b |
subtract__6.0__5.0__ divide__120.0__5.0__ round__24.0__ |
subtract__6.0__5.0__ divide__120.0__5.0__ round__24.0__ |
| a team of six entered for a shooting competition . the best marks man scored num__85 points . if he had scored num__92 points the average scores for . the team would have been num__84 . how many points altogether did the team score ? <o> a ) num__288 <o> b ) num__497 <o> c ) num__168 <o> d ) num__127 <o> e ) num__664 |
explanation : num__6 * num__84 = num__504 - num__7 = num__497 answer : b <eor> b <eos> |
b |
multiply__84.0__6.0__ subtract__92.0__85.0__ subtract__504.0__7.0__ subtract__504.0__7.0__ |
multiply__84.0__6.0__ subtract__92.0__85.0__ subtract__504.0__7.0__ subtract__504.0__7.0__ |
| a group of people dogs and birds has seventy legs thirty heads and twenty tails . how many cats are among this group ? ( assume all birds have two legs and a tail . ) <o> a ) num__0 <o> b ) num__5 <o> c ) num__13 <o> d ) num__15 <o> e ) num__20 |
write p c and b for the number of people dogs and birds respectively so that we have num__2 p + num__4 c + num__2 b = num__70 p + c + b = num__30 c + b = num__20 : comparing the last two equations we obtain p = num__10 and the first equation simplifies to num__4 c + num__2 b = num__50 or num__2 c + b = num__25 . therefore c = ( num__2 c + b ) - ( c + b ) = num__25 - num__5 = num__5 correct answer b <eor> b <eos> |
b |
divide__20.0__2.0__ subtract__70.0__20.0__ divide__50.0__2.0__ divide__10.0__2.0__ divide__10.0__2.0__ |
subtract__30.0__20.0__ subtract__70.0__20.0__ divide__50.0__2.0__ subtract__25.0__20.0__ subtract__10.0__5.0__ |
| a man buys an article and sells it at a profit of num__20.0 . if he had bought it at num__20.0 less and sold it for rs . num__15 less he could have gained num__25.0 . what is the cost price ? <o> a ) num__197 <o> b ) num__375 <o> c ) num__279 <o> d ) num__278 <o> e ) num__75 |
cp num__1 = num__100 sp num__1 = num__120 cp num__2 = num__80 sp num__2 = num__80 * ( num__1.25 ) = num__100 num__20 - - - - - num__100 num__15 - - - - - ? = > num__75 answer : e <eor> e <eos> |
e |
percent__100.0__75.0__ |
percent__100.0__75.0__ |
| a tank is filled in num__5 hours by three pipes p q and r . the pipe r is twice as fast as q and q is twice as fast as p . how much time will pipe p alone take to fill the tank ? <o> a ) num__35 hrs <o> b ) num__20 hrs <o> c ) num__25 hrs <o> d ) num__50 hrs <o> e ) num__28 hrs |
pipe p alone - x hrs to fill then q and r - > x / num__2 and x / num__4 hrs then num__1 / x + num__2 / x + num__4 / x = num__0.2 num__7 / x = num__0.2 x = num__33 hrs answer a <eor> a <eos> |
a |
subtract__5.0__4.0__ divide__1.0__5.0__ add__5.0__2.0__ multiply__5.0__7.0__ |
subtract__5.0__4.0__ divide__1.0__5.0__ add__5.0__2.0__ add__2.0__33.0__ |
| a man buys an article for rs . num__27.50 and sells it for rs . num__28.50 . find his gain % . <o> a ) num__2.0 <o> b ) num__3.0 <o> c ) num__4.0 <o> d ) num__5.0 <o> e ) num__6 % |
sol . cp = rs num__27.50 sp = rs num__28.50 gain = rs ( num__28.50 â € “ num__27.50 ) = rs num__1.10 so gain % = { ( num__1.10 / num__27.50 ) * num__100 } = num__4.0 answer c <eor> c <eos> |
c |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| a sporting goods store sold num__64 frisbees in one week some for $ num__3 and the rest for $ num__4 each . if receipts from frisbee sales for the week totaled $ num__196 what is the fewest number of $ num__4 frisbees that could have been sold ? <o> a ) num__24 <o> b ) num__12 <o> c ) num__8 <o> d ) num__2 <o> e ) num__4 |
in this question however because we are told that exactly num__64 frisbees have been sold and revenue was exactly $ num__204 there is only one possible solution for the number of $ num__3 and $ num__4 frisbees sold . to solve we have num__2 equations and num__2 unknowns let x = number of $ num__3 frisbees sold let y = number of $ num__4 frisbees sold x + y = num__64 num__3 x + num__4 y = num__196 x = num__64 - y num__3 ( num__64 - y ) + num__4 y = num__196 num__192 - num__3 y + num__4 y = num__196 y = num__4 answer : e <eor> e <eos> |
e |
multiply__64.0__3.0__ subtract__196.0__192.0__ |
subtract__196.0__4.0__ subtract__196.0__192.0__ |
| a rectangular lawn of dimensions num__70 m * num__60 m has two roads each num__10 m wide running in the middle of the lawn one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . num__3 per sq m ? <o> a ) num__2288 <o> b ) num__2779 <o> c ) num__2779 <o> d ) num__3900 <o> e ) num__3600 |
area = ( l + b â € “ d ) d ( num__70 + num__60 â € “ num__10 ) num__10 = > num__1200 m num__2 num__1200 * num__3 = rs . num__3600 answer : e <eor> e <eos> |
e |
power__60.0__2.0__ power__60.0__2.0__ |
multiply__3.0__1200.0__ multiply__3.0__1200.0__ |
| john bought a total of num__20 mangoes and oranges . each mango costs num__80 cents and each orange costs num__60 cents . if the average price of the num__20 mangoes and oranges that john originally purchased was num__65 cents then how many oranges needs to return to raise the average price of his purchase to num__72 cents ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__9 <o> e ) num__12 |
let number of mangoes be x number of oranges be num__12 - x num__0.80 x + ( num__20 - x ) num__0.60 / num__20 = num__0.65 solving for x we get x = num__5 - - > mangoes num__5 oranges num__15 now number of oranges to be returned be y num__0.80 * num__5 + ( num__15 - y ) * num__0.60 / num__20 - y = num__0.72 solving for y y = num__12 ans : e <eor> e <eos> |
e |
subtract__72.0__60.0__ divide__12.0__20.0__ divide__60.0__12.0__ subtract__20.0__5.0__ multiply__20.0__0.6__ |
subtract__72.0__60.0__ divide__12.0__20.0__ divide__60.0__12.0__ subtract__20.0__5.0__ multiply__20.0__0.6__ |
| if w = { num__1 num__7 num__18 num__20 num__29 num__33 } how much less is the mean of the numbers in w than the median of the numbers in w ? <o> a ) num__1.0 <o> b ) num__1.5 <o> c ) num__2.0 <o> d ) num__2.5 <o> e ) num__3.0 |
this is a good question to understand the difference between mean and median . mean : average of all the numbers . ( sum of all the elements divided by the number of elements ) median : arrange the elements of the set in increasing order . if the number of terms is odd the middle term is the median . if the number of terms is even the average of middle two terms is the median coming to this question mean = ( num__1 + num__7 + num__18 + num__20 + num__29 + num__33 ) / num__6 = num__18 median = ( num__18 + num__20 ) / num__2 = num__19 difference = num__1 option a <eor> a <eos> |
a |
subtract__7.0__1.0__ subtract__20.0__18.0__ add__1.0__18.0__ reverse__1.0__ |
subtract__7.0__1.0__ subtract__20.0__18.0__ add__1.0__18.0__ reverse__1.0__ |
| if a and b are two prime numbers bigger than num__2 which of the following can be true ? ( num__1 ) a + b is an even number . ( num__2 ) a x b is also a prime number . ( num__3 ) a - b is also a even number . <o> a ) num__1 only . <o> b ) num__1 and num__3 only . <o> c ) num__3 only . <o> d ) num__2 and num__3 only . <o> e ) num__1 num__2 and num__3 . |
answer is d : num__2 and num__3 only ( num__1 ) a + b is an even number . : all primes beyond num__2 are odd so odd + odd = even so true ( num__2 ) a x b is also a prime number . : beats the property of primes so false ( num__3 ) a ^ b is also a prime number . : true ans : b <eor> b <eos> |
b |
reverse__1.0__ |
reverse__1.0__ |
| foodmart customers regularly buy at least one of the following products : milk chicken or apples . num__60.0 of shoppers buy milk num__50.0 buy chicken and num__35.0 buy apples . if num__16.0 of the customers buy all num__3 products what percentage of foodmart customers purchase exactly num__2 of the products listed above ? <o> a ) num__5.0 <o> b ) num__17.0 <o> c ) num__16.0 <o> d ) num__13.0 <o> e ) num__30 % |
num__60 - ( x + num__16 + z ) + num__50 - ( x + num__16 + y ) + num__35 - ( z + num__16 + y ) + x + y + z + num__16 = num__100 where x = people who bought milkchicken y = people who bought chickenapples z = people who bought milk and apples x + y + z = the number of people who bought just exactly two products . hence solving the above equation we get num__113 - ( x + y + z ) = num__100 thus x + y + z = num__13 answer : d <eor> d <eos> |
d |
multiply__50.0__2.0__ subtract__16.0__3.0__ subtract__16.0__3.0__ |
multiply__50.0__2.0__ subtract__16.0__3.0__ subtract__16.0__3.0__ |
| { num__1 / ( num__4 + √ num__15 ) } = <o> a ) a . num__16 + num__16 √ num__15 <o> b ) b . num__31 - num__8 √ num__15 <o> c ) c . num__31 + num__8 √ num__15 <o> d ) d . num__4 - √ num__15 <o> e ) e . num__32 + num__4 √ num__15 |
= num__1 / ( num__4 + root num__15 ) multiply numeratordenominator by ( num__4 - root num__15 ) = ( num__4 - root num__15 ) / ( num__16 - num__15 ) = ( num__4 - root num__15 ) answer = d <eor> d <eos> |
d |
add__1.0__15.0__ multiply__1.0__4.0__ |
add__1.0__15.0__ divide__4.0__1.0__ |
| nicky and cristina are running a race . since cristina is faster than nicky she gives him a num__36 meter head start . if cristina runs at a pace of num__4 meters per second and nicky runs at a pace of only num__3 meters per second how many seconds will nicky have run before cristina catches up to him ? <o> a ) num__15 seconds <o> b ) num__18 seconds <o> c ) num__25 seconds <o> d ) num__36 seconds <o> e ) num__45 seconds |
used pluging in method say t is the time for cristina to catch up with nicky the equation will be as under : for nicky = n = num__3 * t + num__36 for cristina = c = num__5 * t @ t = num__36 n = num__144 c = num__144 right answer ans : d <eor> d <eos> |
d |
multiply__36.0__4.0__ round__36.0__ |
multiply__36.0__4.0__ round__36.0__ |
| two trains of length num__100 m and num__200 m are num__200 m apart . they start moving towards each other on parallel tracks at speeds num__54 kmph and num__72 kmph . after how much time will the trains meet ? <o> a ) num__2.33333333333 sec <o> b ) num__10.6666666667 sec <o> c ) num__2.85714285714 sec <o> d ) num__10.6666666667 sec <o> e ) num__5.71428571429 sec |
they are moving in opposite directions relative speed is equal to the sum of their speeds . relative speed = ( num__54 + num__72 ) * num__0.277777777778 = num__7 * num__5 = num__35 mps . the time required = d / s = num__5.71428571429 = num__5.71428571429 sec . answer : e <eor> e <eos> |
e |
multiply__5.0__7.0__ divide__200.0__35.0__ divide__200.0__35.0__ |
multiply__5.0__7.0__ divide__200.0__35.0__ divide__200.0__35.0__ |
| calculate the speed of a boat in still water ( in km / hr ) if in one hour the boat goes num__15 km / hr downstream and num__7 km / hr upstream . <o> a ) num__12 kmph <o> b ) num__13 kmph <o> c ) num__14 kmph <o> d ) num__15 kmph <o> e ) num__11 kmph |
speed in still water = ( num__15 + num__7 ) num__0.5 kmph = num__11 kmph . answer : e <eor> e <eos> |
e |
round__11.0__ |
round__11.0__ |
| the length of the bridge which a train num__130 m long and traveling at num__45 km / hr can cross in num__30 sec is ? <o> a ) num__878 <o> b ) num__2878 <o> c ) num__221 <o> d ) num__276 <o> e ) num__245 |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec . time = num__30 sec let the length of bridge be x meters . then ( num__130 + x ) / num__30 = num__12.5 x = num__245 m . answer : e <eor> e <eos> |
e |
round__245.0__ |
round__245.0__ |
| if a and b are two events such that p ( a ) = num__0.75 p ( b ) = num__0.5 and p ( a n b ) = num__0.375 find p ( not a and not b ) . <o> a ) num__0.111111111111 <o> b ) num__0.125 <o> c ) num__0.0909090909091 <o> d ) num__0.0769230769231 <o> e ) num__0.4 |
p ( not a and not b ) = num__1 - ( p ( a ) + p ( b ) - p ( ab ) ) which you might find somewhere in your text . substituting in our probabilities we get : p ( not a and not b ) = num__1 - ( num__0.75 + num__0.5 - num__0.375 ) p ( not a and not b ) = num__1 - ( num__0.875 ) p ( not a and not b ) = num__0.125 . b <eor> b <eos> |
b |
add__0.5__0.375__ subtract__0.5__0.375__ subtract__0.5__0.375__ |
add__0.5__0.375__ subtract__0.5__0.375__ subtract__0.5__0.375__ |
| given num__2 x + y + z = num__3 and num__5 x + num__3 y + z = num__9 what is the value of x + y - z ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
( num__1 ) num__5 x + num__3 y + z = num__9 if num__2 x + y + z = num__3 then ( num__2 ) num__4 x + num__2 y + num__2 z = num__6 let ' s subtract equation ( num__2 ) for equation ( num__1 ) . x + y - z = num__3 the answer is a . <eor> a <eos> |
a |
subtract__3.0__2.0__ add__3.0__1.0__ multiply__2.0__3.0__ add__2.0__1.0__ |
subtract__3.0__2.0__ add__3.0__1.0__ add__2.0__4.0__ add__2.0__1.0__ |
| a mixture contains alcohol and water in the ratio num__4 : num__3 . if num__5 litres of water is added to the mixture the ratio becomes num__4 : num__5 . find the quantity of alcohol in the given mixture <o> a ) num__20 litres <o> b ) num__30 litres <o> c ) num__15 litres <o> d ) num__10 litres <o> e ) num__25 litres |
let the quantity of alcohol and water be num__4 x litres and num__3 x litres respectively num__4 x / ( num__3 x + num__5 ) = num__0.8 num__20 x = num__4 ( num__3 x + num__5 ) num__8 x = num__20 x = num__2.5 quantity of alcohol = ( num__4 x num__2.5 ) litres = num__10 litres . option d <eor> d <eos> |
d |
divide__4.0__5.0__ multiply__4.0__5.0__ add__3.0__5.0__ divide__20.0__8.0__ multiply__4.0__2.5__ multiply__4.0__2.5__ |
divide__4.0__5.0__ multiply__4.0__5.0__ add__3.0__5.0__ divide__20.0__8.0__ divide__8.0__0.8__ divide__8.0__0.8__ |
| rani has only pennies dimes and nickels in a jar . the jar has at least num__1 but no more than num__4 pennies . if the jar has at least num__1 nickel and num__1 dime which of the following could not be the total amount of money in the jar ? <o> a ) num__62 <o> b ) num__63 <o> c ) num__64 <o> d ) num__65 <o> e ) num__61 |
let there be a pennies b nickels and c dimes so total amount can be num__1 + num__5 b + num__10 c cents to num__4 + num__5 b + num__10 c as you can see the equation of total whenever divided by num__5 would leave a remainder from num__1 to num__4 ( as pennies can only be from num__1 to num__4 and hence a is limited to values from num__1 to num__4 ) so the total can never be divisible by num__5 and hence only num__65 that is c is the option which is divisible by num__5 . so answer is d <eor> d <eos> |
d |
add__1.0__4.0__ multiply__1.0__65.0__ |
add__1.0__4.0__ multiply__1.0__65.0__ |
| murali travelled from city a to city b at a speed of num__40 kmph and from city b to city c at num__60 kmph . what is the average speed of murali from a to c given that the ratio of distances between a to b and b to c is num__2 : num__5 ? a . num__48 kmph <o> a ) num__33 <o> b ) num__52.5 <o> c ) num__28 <o> d ) num__27 <o> e ) num__16.6 |
let the distances between city a to b and b to c be num__2 x km and num__5 x km respectively . total time taken to cover from a to c = ( num__2 x ) / num__40 + ( num__5 x ) / num__60 = ( num__6 x + num__10 x ) / num__120 = num__16 x / num__120 average speed = ( num__2 x + num__5 x ) / ( num__16 x / num__120 ) = num__52.5 kmph . answer : b <eor> b <eos> |
b |
divide__60.0__6.0__ multiply__60.0__2.0__ add__6.0__10.0__ round__52.5__ |
divide__60.0__6.0__ multiply__60.0__2.0__ add__6.0__10.0__ round__52.5__ |
| mathew is planning a vacation trip to london next year from today for num__5 days he has calculated he would need about $ num__3000 for expenses including a round - trip plane ticket from l . a to london . he nets around $ num__1500 monthly in gross income after all bills are paid he is left with about $ num__350 each month free for whatever he desires . how much money would mathew need to evenly save from his $ num__350 to have $ num__3000 in his bank within num__12 months ? <o> a ) $ num__240 <o> b ) $ num__350 <o> c ) $ num__217 <o> d ) $ num__250 <o> e ) $ num__340 |
answer is ( d ) . if mathew is left with about $ num__350 after all expenses each month he would need to divide the total expense budget to london ( $ num__3000 ) by num__12 months to determine how much he would need to put away every single month to hit his target . $ num__250.0 = $ num__250 . <eor> d <eos> |
d |
divide__3000.0__12.0__ divide__3000.0__12.0__ |
divide__3000.0__12.0__ divide__3000.0__12.0__ |
| an unbiased die is tossed . find the probability of getting a multiple of num__2 . <o> a ) num__0.5 <o> b ) num__0.333333333333 <o> c ) num__0.666666666667 <o> d ) num__0.25 <o> e ) num__0.166666666667 |
here s = { num__12 num__34 num__56 } let e be the getting of a multiple of num__2 . then e = { num__2 num__46 } probability = num__0.5 = num__0.5 correct option is a <eor> a <eos> |
a |
negate_prob__0.5__ |
negate_prob__0.5__ |
| a b and c are partners . a receives num__0.666666666667 of profits b and c dividing the remainder equally . a ' s income is increased by rs . num__500 when the rate to profit rises from num__5 to num__7 percent . find the capital of b ? <o> a ) num__3999 <o> b ) num__7799 <o> c ) num__2500 <o> d ) num__2772 <o> e ) num__6250 |
a : b : c = num__0.666666666667 : num__0.166666666667 : num__0.166666666667 = num__4 : num__1 : num__1 x * num__0.02 * num__0.666666666667 = num__500 b capital = num__37500 * num__0.166666666667 = num__6250 . answer : e <eor> e <eos> |
e |
subtract__5.0__4.0__ multiply__1.0__6250.0__ |
subtract__5.0__4.0__ multiply__1.0__6250.0__ |
| a train num__130 meters long travels at a speed of num__45 km / hr crosses a bridge in num__30 seconds . the length of the bridge is <o> a ) num__235 m <o> b ) num__245 m <o> c ) num__247 m <o> d ) num__249 m <o> e ) num__252 m |
length of the bridge = x meter total distance covered = num__130 + x meter total time taken = num__30 s speed = total distance covered / total time taken = ( num__130 + x ) / num__30 m / s = num__45 × ( num__0.277777777778 ) = ( num__130 + x ) / num__30 = num__45 × num__10 × num__0.833333333333 = num__130 + x = num__45 × num__10 × num__0.833333333333 = num__130 + x = num__15 × num__10 × num__2.5 = num__130 + x = num__15 × num__25 = num__130 + x = num__375 = x = num__375 - num__130 = num__245 <eor> b <eos> |
b |
subtract__45.0__30.0__ multiply__2.5__10.0__ multiply__15.0__25.0__ subtract__375.0__130.0__ round__245.0__ |
subtract__45.0__30.0__ add__10.0__15.0__ multiply__15.0__25.0__ subtract__375.0__130.0__ subtract__375.0__130.0__ |
| the area of a square is equal to five times the area of a rectangle of dimensions num__125 cm * num__64 cm . what is the perimeter of the square ? <o> a ) num__187 cm <o> b ) num__800 cm <o> c ) num__216 cm <o> d ) num__278 cm <o> e ) num__272 cm |
area of the square = s * s = num__5 ( num__125 * num__64 ) = > s = num__25 * num__8 = num__200 cm perimeter of the square = num__4 * num__200 = num__800 cm . answer : b <eor> b <eos> |
b |
multiply__8.0__25.0__ square_perimeter__200.0__ square_perimeter__200.0__ |
multiply__8.0__25.0__ multiply__4.0__200.0__ multiply__4.0__200.0__ |
| a train num__350 m long passed a pole in num__35 sec . how long will it take to pass a platform num__650 m long ? <o> a ) num__100 sec <o> b ) num__89 sec <o> c ) num__85 sec <o> d ) num__16 sec <o> e ) num__15 sec |
speed = num__10.0 = num__10 m / sec . required time = ( num__350 + num__650 ) / num__10 = num__100 sec . answer : a <eor> a <eos> |
a |
divide__350.0__35.0__ round__100.0__ |
divide__350.0__35.0__ round__100.0__ |
| a and b are working on an assignment . a takes num__6 hours to type num__32 pages on a computer while b takes num__5 hours to type num__40 pages . how much time will they take working together on two different computers to type an assignment of num__110 pages ? <o> a ) num__5 hours <o> b ) num__6 hours <o> c ) num__7 hours <o> d ) num__8 hours <o> e ) num__9 hours |
number of pages typed by a in one hour = num__5.33333333333 = num__5.33333333333 number of pages typed by b in one hour = num__8.0 = num__8 number of pages typed by both in one hour = ( ( num__5.33333333333 ) + num__8 ) = num__13.3333333333 time taken by both to type num__110 pages = num__110 * num__0.075 = num__8 hours . answer : d . <eor> d <eos> |
d |
divide__32.0__6.0__ subtract__40.0__32.0__ add__8.0__5.3333__ round__8.0__ |
divide__32.0__6.0__ subtract__40.0__32.0__ add__8.0__5.3333__ round__8.0__ |
| if num__2 x + y = num__7 and x + num__2 y = num__5 then num__10 xy / num__3 = ? <o> a ) a ) num__10 <o> b ) b ) num__2 <o> c ) c ) num__3.4 <o> d ) d ) num__3.6 <o> e ) e ) num__4 |
num__2 * ( x + num__2 y = num__5 ) equals num__2 x + num__4 y = num__10 num__2 x + num__4 y = num__10 - num__2 x + y = num__7 = num__3 y = num__3 therefore y = num__1 plug and solve . . . num__2 x + num__1 = num__7 num__2 x = num__6 x = num__3 ( num__10 * num__3 * num__1 ) / num__3 = num__10.0 = num__10 a <eor> a <eos> |
a |
subtract__7.0__3.0__ subtract__5.0__4.0__ multiply__2.0__3.0__ multiply__2.0__5.0__ |
subtract__7.0__3.0__ subtract__5.0__4.0__ multiply__2.0__3.0__ multiply__2.0__5.0__ |
| the perimeter of a rhombus is num__68 cm and one of its diagonals is num__16 cm . find its area ? <o> a ) num__247 <o> b ) num__240 <o> c ) num__249 <o> d ) num__244 <o> e ) num__241 |
num__4 a = num__68 = > a = num__17 num__172 – num__82 = num__152 num__0.5 * num__16 * num__30 = num__240 answer : b <eor> b <eos> |
b |
triangle_area__16.0__30.0__ triangle_area__16.0__30.0__ |
volume_rectangular_prism__16.0__0.5__30.0__ volume_rectangular_prism__16.0__0.5__30.0__ |
| a boat running upstream takes num__8 hours num__48 minutes to cover a certain distance while it takes num__4 hours to cover the same distance running downstream . what is the ratio between the speed of the boat and speed of the water current respectively ? <o> a ) num__8 : num__1 <o> b ) num__8 : num__2 <o> c ) num__8 : num__7 <o> d ) num__8 : num__3 <o> e ) num__8 : num__9 |
let the man ' s rate upstream be x kmph and that downstream be y kmph . then distance covered upstream in num__8 hrs num__48 min = distance covered downstream in num__4 hrs . num__8 : num__7 answer : c <eor> c <eos> |
c |
round__8.0__ |
round__8.0__ |
| if the arithmetic mean of seventy - five numbers is calculated it is num__35 . if each number is increased by num__5 then mean of new numbers is : <o> a ) num__22 <o> b ) num__40 <o> c ) num__88 <o> d ) num__27 <o> e ) num__11 |
explanation : let the numbers are by hypothesis . . . . . . ( i ) mean of new numbers = by ( i ) answer : b ) num__40 <eor> b <eos> |
b |
add__35.0__5.0__ add__35.0__5.0__ |
add__35.0__5.0__ add__35.0__5.0__ |
| tough and tricky questions : percents . over the course of a year a certain microbrewery increased its beer output by num__20 percent . at the same time it decreased its total working hours by num__30 percent . by what percent did this factory increase its output per hour ? <o> a ) num__171.4 <o> b ) num__171.5 <o> c ) num__171.6 <o> d ) num__171.7 <o> e ) none of the above |
lets assume the initial production was num__100 litres of beer for num__100 hr . with the num__20.0 increase the total amount of beer production will be num__120 litres and with num__30.0 decrease in total hours will be reduced to num__70 hr . num__100 hr - - - - > num__100 lts num__1 hr - - - - - > num__1 lts num__70 hr - - - - - > num__120 lts num__1 hr - - - - - > num__1.714 lts total increase in production for num__1 hr = num__171.4 answer a <eor> a <eos> |
a |
add__20.0__100.0__ subtract__100.0__30.0__ multiply__100.0__1.714__ multiply__1.0__171.4__ |
add__20.0__100.0__ subtract__100.0__30.0__ multiply__100.0__1.714__ multiply__1.0__171.4__ |
| a number is doubled and num__9 is added . if the resultant is trebled it becomes num__51 . what is that number ? <o> a ) num__3.5 <o> b ) num__6 <o> c ) num__8 <o> d ) num__7 <o> e ) num__4 |
let the number be x . then num__3 ( num__2 x + num__9 ) = num__51 num__2 x = num__8 = > x = num__4 answer : e <eor> e <eos> |
e |
divide__8.0__2.0__ divide__8.0__2.0__ |
divide__8.0__2.0__ divide__8.0__2.0__ |
| all the students of class are told to sit in circle shape . here the boy at the num__10 th position is exactly opposite to num__45 th boy . total number of boys in the class ? <o> a ) num__65 <o> b ) num__68 <o> c ) num__72 <o> d ) num__70 <o> e ) num__58 |
as half the circle shape consist of num__45 - num__10 = num__35 boys so total number of boys in full circle = num__2 * num__35 = num__70 answer : d <eor> d <eos> |
d |
subtract__45.0__10.0__ multiply__2.0__35.0__ multiply__2.0__35.0__ |
subtract__45.0__10.0__ multiply__2.0__35.0__ multiply__2.0__35.0__ |
| ron cycles his way up on a hill side and back . he takes the exact same route both ways . on the trip out he cycles an average speed of num__5 miles per hour . on the trip back downhill cycling at an average speed of num__100 miles per hour he is quick to return back . what is his approximate average speed for the round trip in miles per hour ? <o> a ) num__8 <o> b ) num__9.52 <o> c ) num__10 <o> d ) num__11.7 <o> e ) num__20 |
average speed = total distance / total time here distance to hill top and back is same as he takes the same route . d = num__5 t num__1 and d = num__100 t num__2 therefore t num__1 = d / num__5 and t num__2 = d / num__100 t num__1 + t num__2 = num__21 d / num__100 therefore average speed = num__2 d / num__21 d / num__100 = num__2 d * num__4.7619047619 d answer is num__9.52 since the options are far away we need not solve till the decimal points so correct answer is option b <eor> b <eos> |
b |
divide__100.0__21.0__ round__9.52__ |
divide__100.0__21.0__ divide__9.52__1.0__ |
| a certain maroon paint contains num__70 percent blue pigment and num__30 percent red pigment by weight . a certain green paint contains num__10 percent blue pigment and num__90 percent yellow pigment . when these paints are mixed to produce a brown paint the brown paint contains num__40 percent blue pigment . if the brown paint weighs num__10 grams then the red pigment contributes how many grams of that weight ? <o> a ) num__1.5 <o> b ) num__2 <o> c ) num__2.5 <o> d ) num__3 <o> e ) num__3.5 |
num__10 grams of combined mixture and num__40.0 blue pigment means that the mixtures were mixed num__50.0 each . thus num__5 grams a piece . out of the num__5 grams of the dark blue paint num__60.0 is red . therefore num__5 * . num__3 = num__1.5 grams of red pigment <eor> a <eos> |
a |
percent__10.0__50.0__ percent__30.0__10.0__ percent__30.0__5.0__ percent__30.0__5.0__ |
percent__10.0__50.0__ percent__30.0__10.0__ percent__30.0__5.0__ percent__30.0__5.0__ |
| what is the perimeter of an isosceles right angled triangle . the area of the isosceles right angled triangle is num__50 . <o> a ) num__25 <o> b ) num__36 <o> c ) num__25 <o> d ) num__30 <o> e ) num__34.14 |
an isosceles right angled triangle has both the base and height same . the area of an isosceles triangle = num__0.5 * base * height since height = base num__0.5 * height * height = num__0.5 * height ^ num__2 = num__50 height ^ num__2 = num__100 height = num__10 = base since we know the base and height . we will calculate the hypotenuse using pythagoras base ^ num__2 + height ^ num__2 = hypotenuse ^ num__2 num__10 ^ num__2 + num__10 ^ num__2 = hypotenuse ^ num__2 hypotenuse ^ num__2 = num__200 hypotenuse = num__14.14 perimeter of the triangle = num__10 + num__10 + num__14.14 = num__34.14 ans : e <eor> e <eos> |
e |
square_perimeter__0.5__ multiply__50.0__2.0__ power__100.0__0.5__ square_perimeter__50.0__ triangle_area__2.0__34.14__ |
square_perimeter__0.5__ multiply__50.0__2.0__ power__100.0__0.5__ multiply__2.0__100.0__ volume_rectangular_prism__0.5__2.0__34.14__ |
| the total number of plums that grow during each year on a certain plum tree is equal to the number of plums that grew during the previous year less the age of the tree in years ( rounded down to the nearest integer ) . during its num__5 th year the plum tree grew num__50 plums . if this trend continues how many plums will it grow during its num__7 th year ? <o> a ) num__36 <o> b ) num__37 <o> c ) num__38 <o> d ) num__39 <o> e ) num__40 |
num__1 st year : num__0 - num__1 ( age ) we take age = num__0 ( as the question says that we have to ( rounded down to the nearest integer ) ) num__2 ndyear : num__1 - num__2 ( age ) we take age = num__1 num__3 rd year : num__2 - num__3 ( age ) we take age = num__2 num__4 th year : num__3 - num__4 ( age ) we take age = num__3 num__5 th year : num__4 - num__5 ( age ) we take age = num__4 num__6 th year : num__5 - num__6 ( age ) we take age = num__5 num__7 th year : num__6 - num__7 ( age ) we take age = num__6 thus for the num__5 th year = num__50 num__6 th year = num__50 - num__5 = num__45 num__7 th year = num__45 - num__6 = num__39 the correct answer is d . <eor> d <eos> |
d |
subtract__7.0__5.0__ subtract__5.0__2.0__ subtract__5.0__1.0__ add__5.0__1.0__ subtract__50.0__5.0__ subtract__45.0__6.0__ multiply__1.0__39.0__ |
subtract__7.0__5.0__ subtract__5.0__2.0__ subtract__5.0__1.0__ subtract__7.0__1.0__ subtract__50.0__5.0__ subtract__45.0__6.0__ subtract__45.0__6.0__ |
| the sum of the numbers is num__98 . if the ratio between the first and the second be num__2 : num__3 and that between the second and third be num__5 : num__8 then find the second number ? <o> a ) num__24 <o> b ) num__30 <o> c ) num__36 <o> d ) num__42 <o> e ) num__50 |
given ratios num__2 : num__3 num__5 : num__8 num__10 : num__15 : num__24 the second number = num__98 / ( num__10 + num__15 + num__24 ) * num__15 = num__30 answer is b <eor> b <eos> |
b |
multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__3.0__8.0__ multiply__2.0__15.0__ multiply__2.0__15.0__ |
multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__3.0__8.0__ multiply__2.0__15.0__ multiply__2.0__15.0__ |
| a man can row num__30 km downstream and num__20 km upstream in num__4 hours . he can row num__45 km downstream and num__40 km upstream in num__7 hours . find the speed of man in still water ? <o> a ) num__12.9 kmph <o> b ) num__62.5 kmph <o> c ) num__10.5 kmph <o> d ) num__12.5 kmph <o> e ) num__16.5 kmph |
let the speed of the man in still water be a kmph and let the speed of the stream be b kmph . now num__30 / ( a + b ) + num__20 / ( a - b ) = num__4 and num__45 / ( a + b ) + num__40 / ( a - b ) = num__7 solving the equation the speed of man in still water is num__12.5 kmph . answer : d <eor> d <eos> |
d |
round__12.5__ |
round__12.5__ |
| out of num__9 people working on a project num__4 are graduates . if num__3 people are selected at random what is the probability that there is at least one graduate among them ? <o> a ) num__0.652173913043 <o> b ) num__0.75 <o> c ) num__0.815789473684 <o> d ) num__0.880952380952 <o> e ) num__0.767857142857 |
p ( no graduates ) = num__5 c num__0.333333333333 c num__3 = num__0.119047619048 p ( at least one graduate ) = num__1 - num__0.119047619048 = num__0.880952380952 the answer is d . <eor> d <eos> |
d |
vowel_space__ negate_prob__0.119__ negate_prob__0.119__ |
vowel_space__ negate_prob__0.119__ negate_prob__0.119__ |
| on a certain road num__10.0 of the motorists exceed the posted speed limit and receive speeding tickets but num__20.0 of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on that road exceed the posted speed limit ? <o> a ) num__10.5 <o> b ) num__12.5 <o> c ) num__15.0 <o> d ) num__22.0 <o> e ) num__30 % |
suppose there are x motorists . num__10.0 of them exceeded the speed limit and received the ticket i . e . x / num__10 . again suppose total no . of motorists who exceeded the speed limit are y . num__20.0 of y exceeded the speed limit but did n ' t received the ticket i . e . y / num__5 . it means num__4 y / num__5 received the ticket . hence num__4 y / num__5 = x / num__10 or y / x = num__0.125 or y / x * num__100 = num__0.125 * num__100 = num__12.5 answer : b <eor> b <eos> |
b |
percent__12.5__100.0__ |
percent__12.5__100.0__ |
| the l . c . m of two numbers is num__2310 and their h . c . f is num__30 . if one number is num__210 the other is <o> a ) num__330 <o> b ) num__300 <o> c ) num__270 <o> d ) num__250 <o> e ) num__350 |
the other number = l . c . m * h . c . f / given number = num__2310 * num__0.142857142857 = num__330 answer is a . <eor> a <eos> |
a |
divide__30.0__210.0__ gcd__2310.0__330.0__ |
divide__30.0__210.0__ gcd__2310.0__330.0__ |
| if a can complete a work in num__36 days and b can do the same work in num__42 days . if a after doing num__6 days leaves the work find in how many days b will do the remaining work ? <o> a ) num__35 days <o> b ) num__15 days <o> c ) num__30 days <o> d ) num__25 days <o> e ) num__27 days |
explanation : a ’ s one day work = num__0.0277777777778 num__6 day work = num__6 * num__0.0277777777778 = num__0.166666666667 remaining work = num__1 - num__0.166666666667 = num__0.833333333333 b completes the work in num__0.833333333333 * num__42 = num__35 days answer : option a <eor> a <eos> |
a |
divide__6.0__36.0__ subtract__1.0__0.1667__ subtract__36.0__1.0__ round__35.0__ |
divide__6.0__36.0__ subtract__1.0__0.1667__ subtract__36.0__1.0__ subtract__36.0__1.0__ |
| a fruit vendor purchased num__20 dozens of bananas at rs . num__15 per dozen . but one - fourth of the bananas were rotten and had to be thrown away . he sold two - third of the remaining bananas at rs . num__22.50 per dozen . at what price per dozen should he sell the remaining bananas to make neither a profit nor a loss ? <o> a ) num__18 <o> b ) num__15 <o> c ) num__16 <o> d ) num__19 <o> e ) num__11 |
cp of num__20 dozen of bananas = num__15 * num__20 = rs . num__300 number of bananas which are rotten = num__0.25 * num__20 = num__5 dozen . sp of two - third of remaining bananas = ( num__0.666666666667 * num__15 ) * num__22.5 = rs . num__225 sp of remaining num__5 dozens of bananas to make no profit and no loss = ( num__300 - num__225 ) = rs . num__75 . sp of num__1 dozen bananas = num__15.0 = rs . num__15 . answer : b <eor> b <eos> |
b |
percent__20.0__5.0__ percent__20.0__75.0__ |
percent__20.0__5.0__ percent__20.0__75.0__ |
| express a speed of num__42 kmph in meters per second ? <o> a ) num__10 mps <o> b ) num__18 mps <o> c ) num__19 mps <o> d ) num__17 mps <o> e ) num__12 mps |
num__42 * num__0.277777777778 = num__12 mps answer : e <eor> e <eos> |
e |
round__12.0__ |
round__12.0__ |
| in a dairy farm num__40 cows eat num__40 bags of husk in num__40 days . in how many days one cow will eat one bag of husk ? <o> a ) num__33 <o> b ) num__389 <o> c ) num__40 <o> d ) num__38 <o> e ) num__22 |
explanation : let the required number of days be x . less cows more days ( indirect proportion ) less bags less days ( direct proportion ) { \ color { blue } \ left . \ begin { matrix } cows & num__1 : num__40 \ \ bags & num__40 : num__1 \ end { matrix } \ right \ } : : num__40 : x } { \ color { blue } \ therefore } num__1 x num__40 x x = num__40 x num__1 x num__40 { \ color { blue } \ rightarrow } x = num__40 . answer : c ) num__40 <eor> c <eos> |
c |
round__40.0__ |
round__40.0__ |
| on the coordinate plane points m and q are defined by the coordinates ( - num__10 ) and ( num__33 ) respectively and are connected to form a chord of a circle which also lies on the plane . if the area of the circle is ( num__6.25 ) π what are the coordinates of the center of the circle ? <o> a ) ( num__1.5 num__1 ) <o> b ) ( num__2 - num__5 ) <o> c ) ( num__00 ) <o> d ) ( num__1 num__1.5 ) <o> e ) ( num__22 ) |
although it took me num__3 mins to solve this question using all those equations later i thought this question can be solved easily using options . one property to keep in mind - a line passing through the centre of the circle bisects the chord ( or passes from the mid point of the chord ) . now mid point of chord here is ( - num__1 + num__3 ) / num__2 ( num__3 + num__0 ) / num__2 i . e . ( num__1 num__1.5 ) now luckily we have this in our ans . choice . so definitely this is the ans . it also indictaes that mq is the diameter of the circle . there can be a case when mq is not a diameter but in that case also the y - coordinate will remain same as it is the midpoint of the chord and we are moving up in the st . line to locate the centre of the circle . if ans choices are all distinct ( y cordinates ) only check for y cordinate and mark the ans = d <eor> d <eos> |
d |
triangle_area__1.0__3.0__ volume_cube__1.0__ |
triangle_area__1.0__3.0__ volume_cube__1.0__ |
| how many multiples of num__4 are there between num__1 and num__100 inclusive ? <o> a ) num__25 <o> b ) num__22 <o> c ) num__23 <o> d ) num__30 <o> e ) num__36 |
num__25 multiples of num__4 between num__1 and num__100 inclusive . from num__4 * num__1 upto num__4 * num__25 ( num__12 num__34 . . . num__25 ) . hence num__25 multiples ! correct option is a <eor> a <eos> |
a |
divide__100.0__4.0__ multiply__1.0__25.0__ |
divide__100.0__4.0__ multiply__1.0__25.0__ |
| num__21 ball numbered num__1 to num__21 . a ballis drawn and then another ball is drawn without replacement . <o> a ) num__0.0731707317073 <o> b ) num__0.0444444444444 <o> c ) num__0.214285714286 <o> d ) num__0.108695652174 <o> e ) num__0.106382978723 |
the probability that first toy shows the even number = num__1021 = num__1021 since the toy is not replaced there are now num__9 even numbered toys and total num__20 toys left . hence probability that second toy shows the even number = num__920 = num__920 required probability = ( num__1021 ) × ( num__920 ) = ( num__1021 ) × ( num__920 ) = num__0.214285714286 c <eor> c <eos> |
c |
subtract__21.0__1.0__ multiply__1.0__0.2143__ |
subtract__21.0__1.0__ multiply__1.0__0.2143__ |
| if taxi fares were $ num__1.00 for the first num__0.2 mile and $ num__0.45 for each num__0.2 mile there after then the taxi fare for a num__3 - mile ride was <o> a ) $ num__1.56 <o> b ) $ num__2.40 <o> c ) $ num__3.80 <o> d ) $ num__4.20 <o> e ) $ num__7.30 |
in num__3 miles initial num__0.2 mile charge is $ num__1 rest of the distance = num__3 - ( num__0.2 ) = num__2.8 rest of the distance charge = num__14 ( num__0.45 ) = $ num__6.3 ( as the charge is num__0.45 for every num__0.2 mile ) = > total charge for num__3 miles = num__1 + num__6.3 = num__7.3 answer is e . <eor> e <eos> |
e |
subtract__3.0__0.2__ divide__2.8__0.2__ multiply__0.45__14.0__ add__1.0__6.3__ multiply__1.0__7.3__ |
subtract__3.0__0.2__ divide__2.8__0.2__ multiply__0.45__14.0__ add__1.0__6.3__ add__1.0__6.3__ |
| if a : b = num__0.5 : num__0.25 b : c = num__0.5 : num__0.25 then a : b : c ? <o> a ) num__12 : num__24 : num__36 <o> b ) num__1 : num__2 : num__3 <o> c ) num__4 : num__14 : num__24 <o> d ) num__12 : num__8 : num__16 <o> e ) num__16 : num__8 : num__4 |
a : b = num__0.5 : num__0.25 = num__4 : num__2 b : c = num__0.5 : num__0.25 = num__4 : num__2 - - - - - - - - - - - - - - - - - - - - a : b : c = num__16 : num__8 : num__4 answer : e <eor> e <eos> |
e |
reverse__0.25__ reverse__0.5__ divide__4.0__0.25__ multiply__0.5__16.0__ multiply__2.0__8.0__ |
reverse__0.25__ reverse__0.5__ divide__4.0__0.25__ multiply__0.5__16.0__ multiply__2.0__8.0__ |
| a family consists of grandparents parents and three grand children . the average age of the grandparents is num__67 years that of the parents is num__39 years and that of the grandchildren is num__6 years . what is the average age of the family ? <o> a ) num__28 num__4 â „ num__7 years <o> b ) num__31 num__5 â „ num__7 years <o> c ) num__32 num__6 â „ num__7 years <o> d ) num__27 num__1 â „ num__2 years <o> e ) none of these |
required average = ( num__67 Ã — num__2 + num__39 Ã — num__2 + num__6 Ã — num__1.5 + num__2 + num__3 ) = num__32.8571428571 = num__32 num__6 â „ num__7 years answer c <eor> c <eos> |
c |
divide__6.0__2.0__ round_down__32.8571__ subtract__39.0__32.0__ round_down__32.8571__ |
divide__6.0__2.0__ round_down__32.8571__ subtract__39.0__32.0__ round_down__32.8571__ |
| num__66 cubic centimeters of silver is drawn into a wire num__1 mm in diameter . the length of the wire in metres will be : <o> a ) num__44 m <o> b ) num__84 m . <o> c ) num__89 m <o> d ) num__83 m <o> e ) num__42 m |
let the length of the wire be h . radius = num__0.5 mm = num__0.05 cm . then = num__3.14285714286 x num__0.05 x num__0.05 x h = num__66 . h = num__66 x num__20 x num__20 x num__0.318181818182 = num__8400 cm = num__84 m . answer : b <eor> b <eos> |
b |
divide__1.0__0.05__ divide__1.0__3.1429__ round__84.0__ |
divide__1.0__0.05__ divide__1.0__3.1429__ round__84.0__ |
| a manufacturing plant produces num__2 products ( bolts and screws ) . num__3.0 of all products are defective . if the ratio of products made is num__3 : num__2 respectively and the percentage of bolts defective num__5.0 what percentage of screws are defective ? <o> a ) num__0.333333333333 % <o> b ) num__3.0 <o> c ) num__5.0 <o> d ) num__2.5 % <o> e ) num__1.66666666667 % |
let b and s be number bolts and screws respectively and d represent defective . b + s = total products . b : s = num__2 : num__3 bd / b = num__5.0 sd / s = ? [ ( num__2 ) ( bd / b ) + ( num__3 ) ( sd / s ) ] / ( num__2 + num__3 ) = num__3.0 [ ( num__2 ) ( num__5.0 ) + ( num__3 ) ( sd / s ) ] / num__5 = num__3.0 sd / s = num__1.66666666667 % answer : e <eor> e <eos> |
e |
divide__5.0__3.0__ divide__5.0__3.0__ |
divide__5.0__3.0__ divide__5.0__3.0__ |
| an employer pays rs . num__20 for each day a worker works and for feits rs . num__3 for each day is ideal at the end of sixty days a worker gets rs . num__280 . for how many days did the worker remain ideal ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__40 <o> d ) num__50 <o> e ) num__60 |
if he work for num__20 days then he get = num__20 * num__20 = rs . num__400 and left num__40 days he will not work so for this employer feits rs . num__120 now total earning of a worker = num__400 - num__120 = num__280 num__40 days he will be ideal answer : c <eor> c <eos> |
c |
multiply__3.0__40.0__ round__40.0__ |
multiply__3.0__40.0__ round__40.0__ |
| if there are num__200 questions in a num__3 hr examination . among these questions are num__50 type a problems which requires twice as much as time be spent than the rest of the type b problems . how many minutes should be spent on type a problems ? <o> a ) num__72 min <o> b ) num__62 min <o> c ) num__70 min <o> d ) num__74 min <o> e ) num__76 min |
x = time for type b prolems num__2 x = time for type a problem total time = num__3 hrs = num__180 min num__150 x + num__50 * num__2 x = num__180 x = num__0.72 x = num__0.72 time taken for type a problem = num__50 * num__2 * num__0.72 = num__72 min answer : a <eor> a <eos> |
a |
subtract__200.0__50.0__ round__72.0__ |
multiply__3.0__50.0__ round__72.0__ |
| the weight of a glass of jar is num__15.0 of the weight of the jar filled with coffee beans . after some of the beans have been removed the weight of the jar and the remaining beans is num__60.0 of the original total weight . what fraction part of the beans remain in the jar ? <o> a ) num__0.2 <o> b ) num__0.333333333333 <o> c ) num__0.4 <o> d ) num__0.529411764706 <o> e ) num__0.391304347826 |
let weight of jar filled with beans = num__100 g weight of jar = num__15 g weight of coffee beans = num__85 g weight of jar and remaining beans = num__60 g weight of remaining beans = num__45 g fraction remaining = num__0.529411764706 = num__0.529411764706 answer is d . <eor> d <eos> |
d |
percent__100.0__0.5294__ |
percent__100.0__0.5294__ |
| a batsman scored num__130 runs which included num__3 boundaries and num__8 sixes . what percent of his total score did he make by running between the wickets ? <o> a ) num__53.85 <o> b ) num__40.0 <o> c ) num__60.0 <o> d ) num__70.0 <o> e ) num__80 % |
number of runs made by running = num__130 - ( num__3 x num__4 + num__8 x num__6 ) = num__130 - ( num__60 ) = num__70 now we need to calculate num__70 is what percent of num__130 . = > num__0.538461538462 x num__100 = num__53.85 answer : a <eor> a <eos> |
a |
subtract__130.0__60.0__ divide__70.0__130.0__ multiply__0.5385__100.0__ multiply__0.5385__100.0__ |
subtract__130.0__60.0__ divide__70.0__130.0__ multiply__0.5385__100.0__ multiply__0.5385__100.0__ |
| in common wealth games an athlete runs num__200 meters in num__36 seconds . hisspeed is ? <o> a ) num__25 km / hr <o> b ) num__27 km / hr <o> c ) num__30 km / hr <o> d ) num__20 km / hr <o> e ) none of these |
explanation : ( num__0.2 ) / ( num__0.01 ) = num__20 km / hr answer : d <eor> d <eos> |
d |
divide__0.2__0.01__ round__20.0__ |
divide__0.2__0.01__ divide__0.2__0.01__ |
| a train passes a platform in num__25 seconds . the same train passes a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr the length of the platform is <o> a ) num__75 <o> b ) num__25 <o> c ) num__26 <o> d ) num__23 <o> e ) num__22 |
speed of the train = num__54 km / hr = ( num__54 × num__10 ) / num__36 m / s = num__15 m / s length of the train = speed × time taken to cross the man = num__15 × num__20 = num__300 m let the length of the platform = l time taken to cross the platform = ( num__300 + l ) / num__15 = > ( num__300 + l ) / num__15 = num__20 = > num__300 + l = num__15 × num__25 = num__375 = > l = num__375 - num__300 = num__75 meter answer is a . <eor> a <eos> |
a |
subtract__25.0__10.0__ multiply__20.0__15.0__ multiply__25.0__15.0__ subtract__375.0__300.0__ round__75.0__ |
subtract__25.0__10.0__ multiply__20.0__15.0__ multiply__25.0__15.0__ subtract__375.0__300.0__ subtract__375.0__300.0__ |
| in store a there are num__30 pairs of pants for every num__60 store b has . the price ratio between the pants in store b and the pants in store a is num__3 : num__4 . if all the pants were sold in both places until the stock ran out what is the ratio between the total amount stores a earned to the total amount store b earned ? <o> a ) num__3 : num__16 . <o> b ) num__2 : num__3 . <o> c ) num__1 : num__3 . <o> d ) num__3 : num__4 . <o> e ) num__2 : num__5 . |
num__1 st statement : ratio of pants store a : store b num__30 x : num__60 x x : num__2 x price : num__4 y : num__3 y total revenue num__4 xy : num__6 xy num__2 : num__3 answer is b <eor> b <eos> |
b |
subtract__4.0__3.0__ divide__60.0__30.0__ multiply__3.0__2.0__ divide__60.0__30.0__ |
subtract__4.0__3.0__ divide__60.0__30.0__ multiply__3.0__2.0__ divide__60.0__30.0__ |
| a certain company ’ s profit in num__1996 was num__10 percent greater than its profit in num__1995 and its profit in num__1997 was num__20 percent greater than its profit in num__1996 . the company ’ s profit in num__1997 was what percent greater than its profit in num__1995 ? <o> a ) num__5.0 <o> b ) num__18.0 <o> c ) num__32.0 <o> d ) num__35.0 <o> e ) num__38 % |
profit in num__1995 - num__100 profit in num__1996 - num__110.0 increae profit in num__1997 in comparison to num__1995 = num__10 + num__110 * num__20.0 = num__32 correct option : c <eor> c <eos> |
c |
percent__32.0__100.0__ |
percent__32.0__100.0__ |
| best friends sprite and icey have are playing a game of marbles . they invite num__1 of their friends to play with them . there are num__36 marbles in the bag . if all num__3 people are playing how many marbles does each person get ? <o> a ) num__3 <o> b ) num__12 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
answer = b num__12.0 = num__12 answer = b <eor> b <eos> |
b |
divide__36.0__3.0__ multiply__1.0__12.0__ |
divide__36.0__3.0__ multiply__1.0__12.0__ |
| a b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = num__2 : num__3 and b : c = num__2 : num__5 . if the total runs scored by all of them are num__50 the runs scored by b are ? a . num__15 b . num__18 <o> a ) num__12 <o> b ) num__18 <o> c ) num__99 <o> d ) num__77 <o> e ) num__24 |
a : b = num__2 : num__3 b : c = num__2 : num__5 a : b : c = num__4 : num__6 : num__15 num__0.24 * num__50 = num__12 answer : a <eor> a <eos> |
a |
multiply__2.0__3.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
multiply__2.0__3.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
| find the largest num__4 digit number which isexactly divisible by num__88 ? <o> a ) num__3454 <o> b ) num__4586 <o> c ) num__9944 <o> d ) num__10564 <o> e ) num__18664 |
largest num__4 digit number is num__9999 after doing num__9999 ÷ num__88 we get remainder num__55 hence largest num__4 digit number exactly divisible by num__88 = num__9999 - num__55 = num__9944 c <eor> c <eos> |
c |
subtract__9999.0__55.0__ subtract__9999.0__55.0__ |
subtract__9999.0__55.0__ subtract__9999.0__55.0__ |
| after num__6 games team b had an average of num__75 points per game . if it got only num__47 points in game num__7 how many more points does it need to score to get its total above num__500 ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__7 <o> d ) num__3 <o> e ) num__8 |
( num__6 * num__75 ) + num__47 + x > num__500 num__450 + num__47 + x > num__500 num__497 + x > num__500 = > x > num__3 option d <eor> d <eos> |
d |
multiply__6.0__75.0__ add__47.0__450.0__ subtract__500.0__497.0__ subtract__6.0__3.0__ |
multiply__6.0__75.0__ add__47.0__450.0__ subtract__500.0__497.0__ subtract__6.0__3.0__ |
| if the area of a circle is num__81 pi square feet find its circumference . <o> a ) num__8 <o> b ) num__28 <o> c ) num__18 <o> d ) num__48 <o> e ) num__38 |
the area is given by pi × r × r . hence pi × r × r = num__81 pi r × r = num__81 ; hence r = num__81 feet the circumference is given by num__2 × pi × r = num__2 × pi × num__9 = num__18 pi feet correct answer c <eor> c <eos> |
c |
multiply__2.0__9.0__ triangle_area__2.0__18.0__ |
multiply__2.0__9.0__ triangle_area__2.0__18.0__ |
| the average age num__7 members of a committee are the same as it was num__2 years ago because an old number has been replaced by a younger number . find how much younger is the new member than the old number ? <o> a ) num__14 years <o> b ) num__17 years <o> c ) num__18 years <o> d ) num__12 years <o> e ) num__11 years |
num__7 * num__2 = num__14 answer : a <eor> a <eos> |
a |
multiply__7.0__2.0__ multiply__7.0__2.0__ |
multiply__7.0__2.0__ multiply__7.0__2.0__ |
| mike needs num__30.0 to pass . if he scored num__212 marks and falls short by num__13 marks what was the maximum marks he could have got ? <o> a ) num__343 <o> b ) num__377 <o> c ) num__750 <o> d ) num__367 <o> e ) num__232 |
if mike had scored num__13 marks more he could have scored num__30.0 therefore mike required num__212 + num__13 = num__225 marks let the maximum marks be m . then num__30.0 of m = num__225 ( num__0.3 ) × m = num__225 m = ( num__225 × num__100 ) / num__30 m = num__750.0 m = num__750 answer : c <eor> c <eos> |
c |
percent__100.0__750.0__ |
percent__100.0__750.0__ |
| pankajam bought an electric drill at num__80.0 of the regular price . she paid $ num__32.80 for the drill . what was the regular price ? <o> a ) $ num__28.69 <o> b ) $ num__18.69 <o> c ) $ num__38.69 <o> d ) $ num__58.69 <o> e ) $ num__41.00 |
regular price = num__32.80 / num__80 * num__100 = num__41 answer : e <eor> e <eos> |
e |
percent__100.0__41.0__ |
percent__100.0__41.0__ |
| a cubical block of metal weighs num__5 pounds . how much will another cube of the same metal weigh if its sides are twice as long ? <o> a ) num__48 <o> b ) num__40 <o> c ) num__24 <o> d ) num__18 <o> e ) num__12 |
for example our cube have a side num__1 meter so we have num__1 cubical meter in this cube and this cubical meter weigth num__5 pounds if we take cube with side num__2 meters we will have num__8 cubical meters in this cube num__8 meters * num__5 pounds = num__40 pounds so answer is b and similar but more theoretical approach : if we have sides a and b than they have equal ration with their areas : a / b = a ^ num__2 / b ^ num__2 and they have equal ration with their volumes : a / b = a ^ num__3 / b ^ num__3 we have two sides num__0.5 so their volume will be in ratio num__0.125 weight of one cube * volume of another cube num__5 * num__8 = num__40 so answer is b <eor> b <eos> |
b |
square_perimeter__2.0__ multiply__5.0__8.0__ volume_cube__0.5__ multiply__5.0__8.0__ |
square_perimeter__2.0__ multiply__5.0__8.0__ power__0.5__3.0__ multiply__5.0__8.0__ |
| marla starts running around a circular track at the same time nick starts walking around the same circular track . marla completes num__20 laps around the track per hour and nick completes num__10 laps around the track per hour . how many minutes after marla and nick begin moving will marla have completed num__4 more laps around the track than nick ? <o> a ) num__5 <o> b ) num__8 <o> c ) num__24 <o> d ) num__15 <o> e ) num__20 |
maria ' s rate - num__20 laps per hour - - > num__0.333333333333 laps / min nick ' s rate - num__10 laps per hour - - > num__0.166666666667 laps / min lets set equations : num__0.333333333333 * t = num__4 ( since maria had to run num__4 laps before nick would start ) num__0.166666666667 * t = num__0 ( hick has just started and has n ' t run any lap yet ) ( num__0.333333333333 - num__0.166666666667 ) * t = num__4 - num__0 ( since nick was chasing maria ) t = num__24 min needed maria to run num__4 laps answer : c <eor> c <eos> |
c |
add__20.0__4.0__ round__24.0__ |
add__20.0__4.0__ round__24.0__ |
| the length of the bridge which a train num__130 metres long and travelling at num__45 km / hr can cross in num__30 seconds is : <o> a ) num__200 m <o> b ) num__225 m <o> c ) num__245 m <o> d ) num__250 m <o> e ) num__240 m |
speed = [ num__45 x num__0.277777777778 ] m / sec = [ num__12.5 ] m / sec time = num__30 sec let the length of bridge be x metres . then ( num__130 + x ) / num__30 = num__12.5 = > num__2 ( num__130 + x ) = num__750 = > x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| the inverse ratio of num__3 : num__7 : num__1 is ? <o> a ) num__2 : num__3 : num__7 <o> b ) num__2 : num__3 : num__9 <o> c ) num__2 : num__3 : num__2 <o> d ) num__2 : num__3 : num__6 <o> e ) num__2 : num__3 : num__1 |
num__0.333333333333 : num__0.142857142857 : num__1.0 = num__7 : num__3 : num__21 answer : e <eor> e <eos> |
e |
reverse__3.0__ reverse__7.0__ multiply__3.0__7.0__ subtract__3.0__1.0__ |
reverse__3.0__ reverse__7.0__ multiply__3.0__7.0__ subtract__3.0__1.0__ |
| maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is num__40 kilometers maxwell ' s walking speed is num__3 km / h and brad ' s running speed is num__5 km / h . what is the distance traveled by maxwell when they meet in the middle ? <o> a ) num__16 <o> b ) num__17 <o> c ) num__18 <o> d ) num__15 <o> e ) num__14 |
consider max starts from point a and brad starts from point b and move towards each other . assume they shall meet at point o after time ' t ' . the question asks us to find oa . from the question stem we can make out : - distance oa = num__50 km - distance ob = > num__3 xt = num__40 - num__5 xt ( i . e distance = speed x time ) = > num__8 t = num__40 hence t = num__5 oa = num__3 x num__5 = num__15 km answer : d <eor> d <eos> |
d |
divide__40.0__5.0__ multiply__3.0__5.0__ round__15.0__ |
divide__40.0__5.0__ multiply__3.0__5.0__ round__15.0__ |
| the average of four consecutive even numbers is num__27 . find the largest of thesenumbers ? <o> a ) num__24 <o> b ) num__26 <o> c ) num__28 <o> d ) num__30 <o> e ) num__32 |
let the numbers be x x + num__2 x + num__4 andx + num__6 . then ( x + ( x + num__2 ) + ( x + num__4 ) + ( x + num__6 ) ) / num__4 ) = num__27 ( num__4 x + num__12 ) / num__4 = num__27 x + num__3 = num__27 x = num__24 . therefore the largest number = ( x + num__6 ) = num__24 + num__6 = num__30 . answer d num__30 <eor> d <eos> |
d |
add__2.0__4.0__ multiply__2.0__6.0__ divide__6.0__2.0__ subtract__27.0__3.0__ add__27.0__3.0__ add__27.0__3.0__ |
add__2.0__4.0__ multiply__2.0__6.0__ divide__6.0__2.0__ subtract__27.0__3.0__ add__27.0__3.0__ add__27.0__3.0__ |
| there are num__200 female managers in a certain company . find the total number of female employees q in the company if num__0.4 of all the employees are managers and num__0.4 of all male employees are managers . <o> a ) q = num__300 <o> b ) q = num__400 <o> c ) q = num__500 <o> d ) q = num__600 <o> e ) none of these |
{ managers } = { female managers } + { male managers } ; we are told that the total number of managers in the company is num__0.4 of all the employees thus { managers } = num__0.4 ( m + f ) where m and f are number of female and male employees respectively . also we know that num__0.4 of all male employees are managers : { male managers } = num__0.4 * mas well as there are total of num__200 female managers : { female managers } = num__200 ; thus : num__0.4 ( m + f ) = num__200 + num__0.4 * m - - > f = num__500 . answer : c . <eor> c <eos> |
c |
divide__200.0__0.4__ divide__200.0__0.4__ |
divide__200.0__0.4__ divide__200.0__0.4__ |
| a train num__440 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__72 sec <o> b ) num__24 sec <o> c ) num__44 sec <o> d ) num__62 sec <o> e ) num__21 sec |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__440 * num__0.0545454545455 ] sec = num__8 sec answer : b <eor> b <eos> |
b |
add__60.0__6.0__ divide__440.0__18.3333__ |
add__60.0__6.0__ divide__440.0__18.3333__ |
| jack rode his bicycle at an average speed of num__5 mph for some time and then at an average speed of num__15 mph for the rest of the journey . if he made no stops during the trip and his average speed for the entire journey was num__10 miles per hour for what fraction of the total time did he ride at num__15 mph ? <o> a ) num__0.2 <o> b ) num__0.333333333333 <o> c ) num__0.666666666667 <o> d ) num__0.4 <o> e ) num__0.5 |
we do n ' t need to get into calculations for solving this question . we can use the concept of weighted averages . the average speed for the entire journey is num__10 mph ; so he rode at num__5 mph and num__15 mph for an equal duration of time . difference is num__5 and num__5 respectively . num__5 - - - - num__10 - - - - num__15 this shows that you can divide the entire journey into num__2 equal parts . thus num__0.5 of the journey he rode at num__10 mph and num__0.5 of the journey he rode at num__15 mph . answer : e <eor> e <eos> |
e |
divide__10.0__5.0__ divide__5.0__10.0__ round__0.5__ |
divide__10.0__5.0__ divide__5.0__10.0__ round__0.5__ |
| in a game of billiards a can give b num__10 points in num__30 and he can give c num__15 points in num__30 . how many points can b give c in a game of num__50 ? <o> a ) num__12.5 <o> b ) num__25 <o> c ) num__20 <o> d ) num__22 <o> e ) num__23 |
a scores num__30 while b score num__20 and c scores num__15 the number of points that c scores when b scores num__50 = ( num__50 * num__15 ) / num__20 = num__37.5 in a game of num__50 points b gives ( num__50 - num__37.5 ) = num__12.5 points to c answer : a <eor> a <eos> |
a |
subtract__30.0__10.0__ subtract__50.0__37.5__ subtract__50.0__37.5__ |
subtract__30.0__10.0__ subtract__50.0__37.5__ subtract__50.0__37.5__ |
| which of the following numbers completes the sequence num__8 num__14 num__21 num__29 num__38 ___ ? <o> a ) num__35 <o> b ) num__36 <o> c ) num__37 <o> d ) num__48 <o> e ) num__69 |
the numbers increase at a somewhat steady rate so you have to figure out how much you have to add to each number to produce the next in the sequence : num__8 + num__6 = num__14 ; num__14 + num__7 = num__21 ; num__21 + num__8 = num__29 num__29 + num__9 = num__38 and so on . the rule for the sequence is to add successively larger numbers to each number ; therefore the next number is num__38 + num__10 = num__48 correct answer d ) num__48 <eor> d <eos> |
d |
subtract__14.0__8.0__ subtract__21.0__14.0__ subtract__38.0__29.0__ multiply__8.0__6.0__ multiply__8.0__6.0__ |
subtract__14.0__8.0__ subtract__21.0__14.0__ subtract__38.0__29.0__ add__38.0__10.0__ add__38.0__10.0__ |
| the perimeter of a semi circle is num__144 cm then the radius is ? <o> a ) num__87 <o> b ) num__28 cm <o> c ) num__26 <o> d ) num__27 <o> e ) num__25 |
num__5.14285714286 r = num__144 = > r = num__28 answer : b <eor> b <eos> |
b |
round__28.0__ |
round__28.0__ |
| how many workers are required for completing the construction work in num__10 days ? i . num__20.0 of the work can be completed by num__8 workers in num__8 days . ii . num__20 workers can complete the work in num__16 days . iii . one - eighth of the work can be completed by num__8 workers in num__5 days . <o> a ) i only <o> b ) ii and iii only <o> c ) iii only <o> d ) i and iii only <o> e ) any one of the three |
explanation i . num__0.2 work can be completed by ( num__8 x num__8 ) workers in num__1 day . ⇒ whole work can be completed by ( num__8 x num__8 x num__5 ) workers in num__1 day . ⇒ num__8 x num__8 x num__0.5 workers in num__10 days = num__32 workers in num__10 days . ii . ( num__20 x num__16 ) workers can finish it in num__1 day . ⇒ ( num__20 x num__16 ) / num__10 workers can finish it in num__10 days . ⇒ num__32 workers can finish it in num__10 days . iii . num__0.125 work can be completed by ( num__8 x num__5 ) workers in num__1 day . ⇒ whole work can be completed by ( num__8 x num__5 x num__8 ) workers in num__1 day . ⇒ num__8 x num__5 x num__0.8 workers in num__10 days = num__32 workers in num__10 days . any one of the three gives the answer . answer e <eor> e <eos> |
e |
multiply__5.0__0.2__ divide__10.0__20.0__ divide__16.0__0.5__ divide__1.0__8.0__ divide__8.0__10.0__ round__1.0__ |
multiply__5.0__0.2__ divide__10.0__20.0__ divide__16.0__0.5__ divide__1.0__8.0__ divide__8.0__10.0__ round__1.0__ |
| in a certain supermarket a triangular display of cans is arranged in num__10 rows numbered num__1 through num__10 from top to bottom . each successively numbered row contains num__3 more cans than the row immediately above it . if there are fewer than num__150 cans in the entire display how many cans are in the sixth row ? <o> a ) num__14 <o> b ) num__16 <o> c ) num__18 <o> d ) num__20 <o> e ) num__22 |
let x be the number of cans in row num__1 . the total number of cans is x + ( x + num__3 ) + . . . + ( x + num__27 ) = num__10 x + num__3 ( num__1 + num__2 + . . . + num__9 ) = num__10 x + num__3 ( num__9 ) ( num__10 ) / num__2 = num__10 x + num__135 since the total is less than num__150 x must equal num__1 . the number of cans in the num__6 th row is num__1 + num__3 ( num__5 ) = num__16 the answer is b . <eor> b <eos> |
b |
subtract__3.0__1.0__ subtract__10.0__1.0__ multiply__3.0__2.0__ divide__10.0__2.0__ add__10.0__6.0__ add__10.0__6.0__ |
subtract__3.0__1.0__ divide__27.0__3.0__ multiply__3.0__2.0__ divide__10.0__2.0__ add__10.0__6.0__ add__10.0__6.0__ |
| num__40 men took a dip in a water tank num__40 m long and num__20 m broad on a religious day . if the average displacement of water by a man is num__4 m num__3 then the rise in the water level in the tank will be : <o> a ) num__25 cm <o> b ) num__20 cm <o> c ) num__35 cm <o> d ) num__50 cm <o> e ) none of these |
explanation : total volume of water displaced = ( num__4 x num__40 ) m num__3 = num__160 m num__3 rise in water level = num__4.0 Ã — num__20 = num__0.2 m = num__20 cm answer : b <eor> b <eos> |
b |
multiply__40.0__4.0__ divide__4.0__20.0__ round__20.0__ |
multiply__40.0__4.0__ divide__4.0__20.0__ round__20.0__ |
| the parameter of a square is double the perimeter of a rectangle . the area of the rectangle is num__480 sq cm . find the area of the square . <o> a ) num__378 <o> b ) num__277 <o> c ) num__389 <o> d ) num__480 <o> e ) num__881 |
let the side of the square be a cm . let the length and the breadth of the rectangle be l cm and b cm respectively . num__4 a = num__2 ( l + b ) num__2 a = l + b l . b = num__480 we can not find ( l + b ) only with the help of l . b . therefore a can not be found . area of the square can not be found . answer : d <eor> d <eos> |
d |
triangle_area__480.0__2.0__ |
triangle_area__480.0__2.0__ |
| if num__7 persons can do num__7 times of a particular work in num__7 days then num__9 persons can do num__9 times of that work in ? <o> a ) num__3 days <o> b ) num__8 days <o> c ) num__9 days <o> d ) num__11 days <o> e ) num__7 days |
that is num__1 person can do one time of the work in num__7 days . therefore num__9 persons can do num__9 times work in the same num__7 days itself . option ' e ' <eor> e <eos> |
e |
round__7.0__ |
round__7.0__ |
| find the value for x from below equation : x / num__3 = - num__5 ? <o> a ) - num__66 <o> b ) num__13 <o> c ) - num__15 <o> d ) - num__31 <o> e ) num__40 |
num__1 . multiply both sides by num__3 : x * num__1.0 = - num__1.66666666667 simplify both sides : x = - num__15 c <eor> c <eos> |
c |
divide__5.0__3.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
divide__5.0__3.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
| find the principal which yields a simple interest of rs . num__20 and compound interest of rs . num__28 in two years at the same percent rate per annum ? <o> a ) s . num__25 <o> b ) s . num__48 <o> c ) s . num__42 <o> d ) s . num__20 <o> e ) s . num__60 |
explanation : si in num__2 years = rs . num__20 si in num__1 year = rs . num__10 ci in num__2 years = rs . num__28.0 rate per annum = [ ( ci – si ) / ( si in num__1 year ) ] * num__100 = [ ( num__28 – num__20 ) / num__20 ] * num__100 = num__40.0 p . a . let the principal be rs . x time = t = num__2 years % rate = num__40.0 p . a . si = ( prt / num__100 ) num__20 = ( x * num__40 * num__2 ) / num__100 x = rs . num__25 answer : a <eor> a <eos> |
a |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| in how many years does a sum of rs . num__6000 yield a simple interest of rs . num__15000 at num__15.0 p . a . ? <o> a ) num__16.66 years <o> b ) num__17.66 years <o> c ) num__18.66 years <o> d ) num__19.66 years <o> e ) num__20.66 years |
t = ( num__100 * num__15000 ) / ( num__15 * num__6000 ) = num__16.66 years answer : a <eor> a <eos> |
a |
percent__16.66__100.0__ |
percent__16.66__100.0__ |
| a ball is bought for rs . num__400 and sold at a gain of num__30.0 find its selling price ? <o> a ) a ) rs . num__480 / - <o> b ) b ) rs . num__490 / - <o> c ) c ) rs . num__500 / - <o> d ) d ) rs . num__520 / - <o> e ) e ) rs . num__540 / - |
num__100.0 - - - - - - > num__400 ( num__100 * num__4 = num__400 ) num__130.0 - - - - - - > num__520 ( num__130 * num__4 = num__520 ) selling price = rs . num__520 / - d <eor> d <eos> |
d |
percent__100.0__520.0__ |
percent__100.0__520.0__ |
| num__6 num__10 num__18 num__30 num__46 ? <o> a ) num__66 <o> b ) num__56 <o> c ) num__74 <o> d ) num__57 <o> e ) num__32 |
the pattern is + num__4 + num__8 + num__12 + num__16 + num__20 answer : a . <eor> a <eos> |
a |
subtract__10.0__6.0__ subtract__18.0__10.0__ subtract__18.0__6.0__ add__6.0__10.0__ subtract__30.0__10.0__ add__46.0__20.0__ |
subtract__10.0__6.0__ subtract__18.0__10.0__ add__8.0__4.0__ add__6.0__10.0__ add__8.0__12.0__ add__46.0__20.0__ |
| if â € œ * â € is called â € œ + â € â € œ / â € is called â € œ * â € â € œ - â € is called â € œ / â € â € œ + â € is called â € œ - â € . num__3.0 â € “ num__60 * num__40 - num__10 = ? <o> a ) num__355 <o> b ) num__341 <o> c ) num__323.2 <o> d ) num__425.6 <o> e ) num__225.7 |
explanation : given : num__3.0 â € “ num__60 * num__40 - num__10 = ? substituting the coded symbols for mathematical operations we get num__240 * num__1.33333333333 + num__4.0 = ? num__240 * num__1.33 + num__4 = ? num__319.2 + num__4 = num__323.2 answer : c <eor> c <eos> |
c |
divide__40.0__10.0__ multiply__240.0__1.33__ add__4.0__319.2__ add__4.0__319.2__ |
divide__40.0__10.0__ multiply__240.0__1.33__ add__4.0__319.2__ add__4.0__319.2__ |
| the two trains of lengths num__400 m num__600 m respectively running at same directions . the faster train can cross the slower train in num__180 sec the speed of the slower train is num__48 km . then find the speed of the faster train ? <o> a ) num__68 <o> b ) num__75 <o> c ) num__58 <o> d ) num__55 <o> e ) num__40 |
length of the two trains = num__600 m + num__400 m speed of the first train = x speed of the second train = num__48 kmph num__1000 / x - num__48 = num__180 num__1000 / x - num__48 * num__0.277777777778 = num__180 num__50 = num__9 x - num__120 x = num__68 kmph answer a <eor> a <eos> |
a |
add__400.0__600.0__ round__68.0__ |
add__400.0__600.0__ round__68.0__ |
| it takes num__40 identical printing presses num__9 hours to print num__500000 papers . how many hours would it take num__30 of these printing presses to print num__500000 papers ? <o> a ) num__9 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__13 |
num__40 printing presses can do num__0.111111111111 of the job each hour . num__30 printing presses can do num__0.75 * num__0.111111111111 = num__0.0833333333333 of the job each hour . the answer is d . <eor> d <eos> |
d |
divide__30.0__40.0__ divide__0.75__9.0__ divide__9.0__0.75__ |
divide__30.0__40.0__ multiply__0.75__0.1111__ divide__9.0__0.75__ |
| which of the following is equal to the average ( arithmetic mean ) of ( y + num__4 ) ^ num__2 and ( y - num__4 ) ^ num__2 ? <o> a ) x ^ num__2 <o> b ) x ^ num__2 + num__2 <o> c ) x ^ num__2 + num__12 <o> d ) x ^ num__2 + num__21 <o> e ) y ^ num__2 + num__16 |
avg = [ ( y + num__4 ) ^ num__2 + ( y - num__4 ) ^ num__2 ] / num__2 expanding and simplifying ( y ^ num__2 + num__8 y + num__16 + y ^ num__2 - num__8 y + num__16 ) / num__2 = y ^ num__2 + num__16 answer e . <eor> e <eos> |
e |
multiply__4.0__2.0__ multiply__2.0__8.0__ subtract__4.0__2.0__ |
multiply__4.0__2.0__ multiply__2.0__8.0__ subtract__4.0__2.0__ |
| if the annual rate of simple interest increases from num__10.0 to num__12 num__0.5 % a man ' s yearly income increases by rs . num__1250 . his principal in rs . is ? <o> a ) num__50009 <o> b ) num__50000 <o> c ) num__50011 <o> d ) num__50077 <o> e ) num__50027 |
let the sum be rs . x . then ( x * num__12.5 * num__0.01 ) - ( x * num__10 * num__1 ) / num__100 = num__1250 num__25 x - num__20 x = num__250000 x = num__50000 answer : b <eor> b <eos> |
b |
percent__20.0__250000.0__ percent__100.0__50000.0__ |
percent__20.0__250000.0__ percent__100.0__50000.0__ |
| find the smallest number which should be multiplied with num__560 to make it a perfect square . <o> a ) num__60 <o> b ) num__55 <o> c ) num__45 <o> d ) num__25 <o> e ) num__35 |
num__560 = num__2 * num__2 * num__2 * num__2 * num__7 * num__5 required smallest number = num__7 * num__5 = num__35 num__35 is the smallest number which should be multiplied with num__560 to make it a perfect square . answer : e <eor> e <eos> |
e |
multiply__5.0__7.0__ triangle_area__2.0__35.0__ |
multiply__5.0__7.0__ multiply__5.0__7.0__ |
| one side of a rectangular field is num__15 m and one of its diagonal is num__17 m . find the area of field ? <o> a ) num__128 <o> b ) num__120 <o> c ) num__298 <o> d ) num__112 <o> e ) num__129 |
other side = [ ( num__17 x num__17 ) - ( num__15 x num__15 ) ] = ( num__289 - num__225 ) = num__8 m area = num__15 x num__8 = num__120 sq . m answer : b <eor> b <eos> |
b |
side_by_diagonal__17.0__15.0__ multiply__15.0__8.0__ multiply__15.0__8.0__ |
side_by_diagonal__17.0__15.0__ multiply__15.0__8.0__ multiply__15.0__8.0__ |
| [ ( num__3.242 x num__16 ) / num__100 ] = ? <o> a ) num__0.045388 <o> b ) num__4.5388 <o> c ) num__0.051872 <o> d ) num__473.88 <o> e ) none of these |
answer multiplying num__3.242 x num__16 = num__5.1872 now divide num__5.1872 by num__100 so num__5.1872 ÷ num__100 = num__0.051872 ∴ shift the decimal two places to the left as num__100 correct option : c <eor> c <eos> |
c |
divide__5.1872__100.0__ divide__5.1872__100.0__ |
divide__5.1872__100.0__ divide__5.1872__100.0__ |
| how many prime numbers between num__1 and num__100 are factors of num__10010 ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__3 <o> d ) num__2 <o> e ) num__1 |
factor of num__10010 = num__2 * num__5 * num__7 * num__11 * num__13 - - - num__5 prime numbers a <eor> a <eos> |
a |
add__2.0__5.0__ add__2.0__11.0__ multiply__1.0__5.0__ |
add__2.0__5.0__ add__2.0__11.0__ multiply__1.0__5.0__ |
| the cost price of a radio is rs . num__1500 and it was sold for rs . num__1245 find the loss % ? <o> a ) num__18 <o> b ) num__16 <o> c ) num__26 <o> d ) num__17 <o> e ) num__11 |
explanation : num__1500 - - - - num__255 num__100 - - - - ? = > num__17.0 answer : d <eor> d <eos> |
d |
percent__100.0__17.0__ |
percent__100.0__17.0__ |
| in a garden there are num__10 rows and num__12 columns of mango trees . the distance between the two trees is num__2 metres and a distance of three metres is left from all sides of the boundary of the garden . what is the length of the garden ? <o> a ) num__22 <o> b ) num__24 <o> c ) num__26 <o> d ) num__28 <o> e ) num__30 |
between the num__12 mango trees there are num__11 gaps and each gap has num__2 meter length also num__3 meter is left from all sides of the boundary of the garden . hence length of the garden = ( num__11 Ã — num__2 ) + num__3 + num__3 = num__28 meter answer is d . <eor> d <eos> |
d |
round__28.0__ |
round__28.0__ |
| a man buys an article and sells it at a profit of num__20.0 . if he had bought it at num__20.0 less and sold it for rs . num__75 less he could have gained num__25.0 . what is the cost price ? <o> a ) num__197 <o> b ) num__375 <o> c ) num__279 <o> d ) num__278 <o> e ) num__268 |
cp num__1 = num__100 sp num__1 = num__120 cp num__2 = num__80 sp num__2 = num__80 * ( num__1.25 ) = num__100 num__20 - - - - - num__100 num__75 - - - - - ? = > num__375 answer : b <eor> b <eos> |
b |
percent__100.0__375.0__ |
percent__100.0__375.0__ |
| here ' s an easy question of averages but let ' s try to see innovative ways of solving this . a class has num__12 boys and x girls . average score of boys and girls is num__81 and num__92 respectively . the average of the whole class is num__86 what is the value of x ? <o> a ) a ) num__6 <o> b ) b ) num__7 <o> c ) c ) num__8 <o> d ) d ) num__10 <o> e ) e ) num__12 |
num__12 ( num__81 ) + num__92 x / num__12 + x = num__86 num__972 + num__92 x / num__12 + x = num__86 num__972 + num__92 x = num__86 ( num__12 + x ) num__972 + num__92 x = num__1032 + num__86 x x ' s one side numbers one side we get num__92 x - num__86 x = num__1032 - num__972 num__6 x = num__60 hence x = num__10 answer d <eor> d <eos> |
d |
multiply__12.0__81.0__ multiply__12.0__86.0__ subtract__92.0__86.0__ subtract__1032.0__972.0__ divide__60.0__6.0__ divide__60.0__6.0__ |
multiply__12.0__81.0__ multiply__12.0__86.0__ subtract__92.0__86.0__ subtract__1032.0__972.0__ divide__60.0__6.0__ divide__60.0__6.0__ |
| the perimeter of a polygon with sides of integer length is num__45 . if the smallest side of the polygon is num__5 and the longest side of the polygon is num__10 then the number of sides could be any number from <o> a ) num__5 to num__7 <o> b ) num__5 to num__8 <o> c ) num__5 to num__9 <o> d ) num__6 to num__8 <o> e ) num__6 to num__9 |
the sum of the length of the remaining sides must be num__45 - ( num__5 + num__10 ) = num__30 . we can have num__5 - sided polygon { num__510 num__10 num__1010 } ( we can not increase the length of any side to reduce the number of sides ) ; we can have num__6 - sided polygon { num__55 num__5 num__10 num__1010 } ; we can have num__7 - sided polygon { num__55 num__5 num__5 num__5 num__1010 } ; we can have num__8 - sided polygon { num__55 num__5 num__5 num__5 num__5 num__510 } ( we can not decrease the length of any side to increase the number of sides ) . answer : b . <eor> b <eos> |
b |
divide__30.0__5.0__ add__45.0__10.0__ round__5.0__ |
divide__30.0__5.0__ add__45.0__10.0__ subtract__10.0__5.0__ |
| a train num__75 m long takes num__6 sec to cross a man walking at num__5 kmph in a direction opposite to that of the train . find the speed of the train ? <o> a ) num__35 kmph <o> b ) num__40 kmph <o> c ) num__45 kmph <o> d ) num__50 kmph <o> e ) num__55 kmph |
let the speed of the train be x kmph speed of the train relative to man = x + num__5 = ( x + num__5 ) * num__0.277777777778 m / sec num__75 / [ ( x + num__5 ) * num__0.277777777778 ] = num__6 num__30 ( x + num__5 ) = num__1350 x = num__40 kmph answer is b <eor> b <eos> |
b |
multiply__6.0__5.0__ round__40.0__ |
multiply__6.0__5.0__ round__40.0__ |
| a b c together can do a work in num__16 days . a alone can do the work in num__64 days and b alone can do the same work in num__80 days . find in what time c alone can do that work ? <o> a ) num__24 days <o> b ) num__29.50 days <o> c ) num__33.59 days <o> d ) num__29.09 days <o> e ) num__38.80 days |
the required answer is = num__16 * num__64 * num__1.25 * num__80 - num__16 ( num__64 + num__80 ) = num__29.0909090909 = num__29.09 days answer is d <eor> d <eos> |
d |
divide__80.0__64.0__ round__29.0909__ round__29.0909__ |
divide__80.0__64.0__ round__29.0909__ round__29.0909__ |
| a baseball card decreased in value num__20.0 in its first year and num__20.0 in its second year . what was the total percent decrease of the card ' s value over the two years ? <o> a ) num__28.0 <o> b ) num__30.0 <o> c ) num__32.0 <o> d ) num__36.0 <o> e ) num__72 % |
consider the initial value of the baseball card as $ num__100 after first year price = num__100 * num__0.8 = num__80 after second year price = num__80 * num__0.8 = num__64 final decrease = [ ( num__100 - num__64 ) / num__100 ] * num__100 = num__36.0 correct answer - d <eor> d <eos> |
d |
percent__100.0__36.0__ |
percent__100.0__36.0__ |
| look at this series : num__80 num__10 num__70 num__15 num__60 . . . what number should come next ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__30 <o> d ) num__50 <o> e ) none |
explanation : this is an alternating addition and subtraction series . in the first pattern num__10 is subtracted from each number to arrive at the next . in the second num__5 is added to each number to arrive at the next . answer : option a <eor> a <eos> |
a |
subtract__15.0__10.0__ subtract__80.0__60.0__ |
subtract__15.0__10.0__ subtract__80.0__60.0__ |
| ravi invested certain amount for two rates of simple interests at num__6.0 p . a . and num__7.0 p . a . what is the ratio of ravi ' s investments if the interests from those investments are equal ? <o> a ) num__7 : num__2 <o> b ) num__7 : num__8 <o> c ) num__7 : num__4 <o> d ) num__7 : num__6 <o> e ) num__7 : num__1 |
let x be the investment of ravi in num__6.0 and y be in num__7.0 x ( num__6 ) ( n ) / num__100 = y ( num__7 ) ( n ) / num__100 = > x / y = num__1.16666666667 x : y = num__7 : num__6 answer : d <eor> d <eos> |
d |
percent__7.0__100.0__ |
percent__7.0__100.0__ |
| how many halves are there in six - fourth ? <o> a ) num__8 <o> b ) num__12 <o> c ) num__32 <o> d ) num__7 <o> e ) num__3 |
divide num__1.5 by num__0.5 = num__1.5 ÷ num__0.5 = num__1.5 * num__2.0 = num__3.0 = num__3 . answer is e . <eor> e <eos> |
e |
reverse__0.5__ divide__1.5__0.5__ divide__1.5__0.5__ |
reverse__0.5__ multiply__1.5__2.0__ multiply__1.5__2.0__ |
| mudit ' s age num__26 years hence will be thrice his age four years ago . find mudit ' s present age ? <o> a ) num__12 <o> b ) num__22 <o> c ) num__19 <o> d ) num__18 <o> e ) num__15 |
explanation : let mudit ' s present age be ' m ' years . m + num__26 = num__3 ( m - num__4 ) = > num__2 m = num__38 = > m = num__19 years . answer : c <eor> c <eos> |
c |
divide__38.0__2.0__ divide__38.0__2.0__ |
divide__38.0__2.0__ subtract__38.0__19.0__ |
| what is the unit digit of ( num__6 ! * num__6 ! / num__6 ! * num__3 ! ) ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
( num__6 ! * num__6 ! / num__6 ! * num__3 ! ) = ( num__6 ! / num__3 ! ) = num__120.0 = num__120 units digit of the above product will be equal to num__0 answer a <eor> a <eos> |
a |
multiply__6.0__0.0__ |
multiply__6.0__0.0__ |
| an article is bought for rs . num__675 and sold for rs . num__900 find the gain percent ? <o> a ) num__33 num__0.125 % <o> b ) num__33 num__1.0 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__33 num__0.666666666667 % <o> e ) num__32 num__0.333333333333 % |
num__675 - - - - num__225 num__100 - - - - ? = > num__33 num__0.333333333333 % answer : c <eor> c <eos> |
c |
percent__100.0__33.0__ |
percent__100.0__33.0__ |
| the perimeter of a triangle is num__28 cm and the inradius of the triangle is num__2.5 cm . what is the area of the triangle ? <o> a ) num__76 <o> b ) num__88 <o> c ) num__66 <o> d ) num__55 <o> e ) num__35 |
area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = num__2.5 * num__14.0 = num__35 cm num__2 answer : e <eor> e <eos> |
e |
triangle_area__28.0__2.5__ triangle_area__28.0__2.5__ |
multiply__2.5__14.0__ multiply__2.5__14.0__ |
| the average length of num__6 bars is num__80 cm . if the average length of one third of the barsis num__70 cm what is the average of the other bars ? <o> a ) num__75 . <o> b ) num__85 . <o> c ) num__90 . <o> d ) num__94 . <o> e ) num__100 . |
edit : given ( x num__1 + x num__2 . . . + x num__6 ) / num__6 = num__80 ( x num__1 + x num__2 . . . + x num__6 ) = num__480 - - > eq num__1 . now given avg length of one third barsis num__70 . that means out num__2.0 = num__2 bars . let the avg length of two barsbe ( x num__1 + x num__2 ) / num__2 = num__70 . ( x num__1 + x num__2 ) = num__140 . - - > eq num__2 . now we are asked to find the average of the remaining i . e . ( x num__3 + x num__4 + x num__5 + x num__6 ) substitute eq num__2 in eq num__1 then we get num__140 + x num__3 + x num__4 + x num__5 + x num__6 = num__480 = > x num__3 + x num__4 + x num__5 + x num__6 = num__340 now divide num__340 by num__4 we get num__85 . = > ( x num__3 + x num__4 + x num__5 + x num__6 ) / num__4 = num__85 = avg length of remaining bars . imo correct option is b . <eor> b <eos> |
b |
multiply__6.0__80.0__ multiply__70.0__2.0__ divide__6.0__2.0__ subtract__6.0__2.0__ subtract__6.0__1.0__ subtract__480.0__140.0__ add__80.0__5.0__ add__80.0__5.0__ |
multiply__6.0__80.0__ multiply__70.0__2.0__ divide__6.0__2.0__ subtract__6.0__2.0__ subtract__6.0__1.0__ subtract__480.0__140.0__ add__80.0__5.0__ add__80.0__5.0__ |
| two boys started running simultaneously around a circular track of length num__4800 m from the same point at speeds of num__60 km / hr and num__100 km / hr . when will they meet for the first time any where on the track if they are moving in opposite directions ? <o> a ) num__105 <o> b ) num__106 <o> c ) num__107 <o> d ) num__108 <o> e ) num__109 |
time taken to meet for the first time anywhere on the track = length of the track / relative speed = num__4800 / ( num__60 + num__100 ) num__0.277777777778 = num__4800 * num__0.1125 * num__5 = num__108 seconds . answer : d <eor> d <eos> |
d |
round__108.0__ |
round__108.0__ |
| a rectangular farm has to be fenced one long side one short side and the diagonal . if the cost of fencing is rs . num__14 per meter . the area of farm is num__1200 m num__2 and the short side is num__30 m long . how much would the job cost ? <o> a ) num__1276 <o> b ) num__1680 <o> c ) num__2832 <o> d ) num__1299 <o> e ) num__1236 |
explanation : l * num__30 = num__1200 è l = num__40 num__40 + num__30 + num__50 = num__120 num__120 * num__14 = num__1680 answer : option b <eor> b <eos> |
b |
square_perimeter__30.0__ multiply__14.0__120.0__ multiply__14.0__120.0__ |
square_perimeter__30.0__ multiply__14.0__120.0__ multiply__14.0__120.0__ |
| two pipes a and b can fill a cistern in num__20 and num__30 minutes respectively and a third pipe c can empty it in num__40 minutes . how long will it take to fill the cistern if all the three are opened at the same time ? <o> a ) num__17 num__0.166666666667 <o> b ) num__17 num__0.111111111111 <o> c ) num__17 num__0.142857142857 <o> d ) num__17 num__0.2 <o> e ) num__17 num__1.0 |
num__0.05 + num__0.0333333333333 - num__0.025 = num__0.0583333333333 num__17.1428571429 = num__17 num__0.142857142857 answer : c <eor> c <eos> |
c |
add__0.025__0.0333__ subtract__17.1429__17.0__ round__17.0__ |
add__0.025__0.0333__ subtract__17.1429__17.0__ subtract__17.1429__0.1429__ |
| in a forest num__250 deer were caught tagged with electronic markers then released . a week later num__50 deer were captured in the same forest . of these num__50 deer it was found that num__5 had been tagged with the electronic markers . if the percentage of tagged deer in the second sample approximates the percentage of tagged deer in the forest and if no deer had either left or entered the forest over the preceding week what is the approximate number of deer in the forest ? <o> a ) num__150 <o> b ) num__750 <o> c ) num__1250 <o> d ) num__1500 <o> e ) num__2 |
500 |
the percentage of tagged deer in the second sample = num__0.1 * num__100 = num__10.0 . so num__250 tagged deers comprise num__10.0 of total # of deers - - > total # of deers = num__250 * num__10 = num__2500 . answer : e <eor> e <eos> |
e |
e |
| if y = num__30 p and p is prime what is the greatest common factor of y and num__5 p in terms of p ? <o> a ) p <o> b ) num__2 p <o> c ) num__5 p <o> d ) num__7 p <o> e ) p ^ num__2 |
y = num__30 p other number is num__5 p then gcf ( num__30 p num__5 p ) = num__5 p ; c is the correct answer <eor> c <eos> |
c |
gcd__30.0__5.0__ |
gcd__30.0__5.0__ |
| sebastian bought a meal at a restaurant and left a num__15.0 tip . with the tip he paid exactly $ num__36.57 . how much did the meal cost without the tip ? <o> a ) $ num__31.80 <o> b ) $ num__29.91 <o> c ) $ num__30.15 <o> d ) $ num__30.60 <o> e ) $ num__30.85 |
the tip is a percent increase of num__15.0 which is num__115.0 . let x equal the price before the tip . thus num__115.0 of this price equals $ num__36.57 : num__1.15 x = num__36.57 divide both sides by num__1.15 : = > x = num__36.57 / num__1.15 = num__31.80 correct answer a ) $ num__31.80 <eor> a <eos> |
a |
divide__36.57__1.15__ divide__36.57__1.15__ |
divide__36.57__1.15__ divide__36.57__1.15__ |
| the sum of five numbers is num__655 . the average of the first two numbers is num__85 and the third number is num__125 . find the average of the two numbers ? <o> a ) num__115 <o> b ) num__135 <o> c ) num__155 <o> d ) num__175 <o> e ) num__195 |
let the five numbers be p q r s and t . = > p + q + r + s + t = num__645 . ( p + q ) / num__2 = num__85 and r = num__125 p + q = num__170 and r = num__125 p + q + r = num__295 s + t = num__645 - ( p + q + r ) = num__350 average of the last two numbers = ( s + t ) / num__2 = num__175 . answer : d <eor> d <eos> |
d |
multiply__85.0__2.0__ add__125.0__170.0__ subtract__645.0__295.0__ divide__350.0__2.0__ divide__350.0__2.0__ |
multiply__85.0__2.0__ add__125.0__170.0__ subtract__645.0__295.0__ divide__350.0__2.0__ divide__350.0__2.0__ |
| a mixture contains alcohol and water in the ratio num__4 : num__3 . if num__10 litres of water is added to the mixture the ratio becomes num__4 : num__5 . find the quantity of alcohol in the given mixture <o> a ) num__15 litres <o> b ) num__10 litres <o> c ) num__30 litres <o> d ) num__20 litres <o> e ) num__8 litres |
let the quantity of alcohol and water be num__4 x litres and num__3 x litres respectively num__4 x / ( num__3 x + num__10 ) = num__0.8 num__20 x = num__4 ( num__3 x + num__10 ) num__8 x = num__40 x = num__5 quantity of alcohol = ( num__4 x num__5 ) litres = num__20 litres . answer is d . <eor> d <eos> |
d |
divide__4.0__5.0__ multiply__4.0__5.0__ add__3.0__5.0__ multiply__4.0__10.0__ multiply__4.0__5.0__ |
divide__4.0__5.0__ multiply__4.0__5.0__ add__3.0__5.0__ multiply__4.0__10.0__ multiply__4.0__5.0__ |
| ice - coldice - cream factory produces only tricolor ice - cream products where each ice - cream has three stripes of different colors . the factory uses the colors pink purple orange silver blue and white . how many different ice - cream products have at least one stripe out of the following colors : pink purple or orange ( assume that the order of the stripes in a single ice - cream does not matter ) ? <o> a ) num__12 <o> b ) num__14 <o> c ) num__18 <o> d ) num__19 <o> e ) num__20 |
imo : d - num__19 there are num__6 c num__3 = num__20 ways to create different colored ice cream products . out of these num__20 only num__1 ( silver blue white ) will not contain at least one of the colors pink purple or orange . num__20 - num__1 = num__19 . the other way would be to calculate the number of ice cream products that contain at least one of the colors pink purple or orange ( ppo ) . # num__1 : pick one out of ppo and two out of sbp : num__3 c num__1 * num__3 c num__2 = num__3 * num__3 = num__9 # num__2 : pick two out of ppo and one out of sbp : num__3 c num__2 * num__3 c num__1 = num__3 * num__3 = num__9 # num__3 : pick three out of ppo : num__3 c num__3 = num__1 num__9 + num__9 + num__1 = num__19 answer is d <eor> d <eos> |
d |
subtract__20.0__19.0__ subtract__3.0__1.0__ add__3.0__6.0__ multiply__1.0__19.0__ |
subtract__20.0__19.0__ subtract__3.0__1.0__ add__3.0__6.0__ multiply__1.0__19.0__ |
| when a certain shoe store reduces the price of its best - selling style of shoe by num__10 percent the weekly sales of this style increase by num__30 percent . which of the following best describes the resulting change in the store ' s weekly revenue from sales of this style of shoe ? <o> a ) revenue decreases by num__10.0 <o> b ) revenue decreases by num__1.0 <o> c ) there is no change in revenue <o> d ) revenue increases by num__17.0 <o> e ) revenue increases by num__10 % |
old price = num__10 old sales = num__10 - - > old revenue = num__100 ; new price = num__9 new sales = num__13 - - > new revenue = num__117 . revenue increased by num__17.0 . answer : d . <eor> d <eos> |
d |
multiply__9.0__13.0__ subtract__30.0__13.0__ subtract__30.0__13.0__ |
multiply__9.0__13.0__ subtract__30.0__13.0__ subtract__30.0__13.0__ |
| num__75.0 of the guestrooms at the stagecoach inn have a queen - sized bed and each of the remaining rooms has a king - sized bed . of the non - smoking rooms num__60.0 have a queen - sized bed . if num__28.0 of the rooms at the stagecoach inn are non - smoking rooms with king - sized beds what percentage of the rooms permit smoking ? <o> a ) num__25.0 <o> b ) num__30.0 <o> c ) num__50.0 <o> d ) num__55.0 <o> e ) num__75 % |
let x be the number of non - smoking rooms . let n be the total number of rooms at the inn . num__40.0 of non - smoking rooms have a king - sized bed . num__0.4 x = num__0.28 n x = num__0.7 n the percentage of rooms which permit smoking is num__1 - num__0.7 = num__30.0 the answer is b . <eor> b <eos> |
b |
percent__75.0__40.0__ percent__75.0__40.0__ |
percent__75.0__40.0__ percent__75.0__40.0__ |
| a certain factory produces buttons and buckles at a uniform weight . if the total weight of num__2 buttons and num__2 buckles is one third of num__11 buckles and num__3 buttons then the weight of num__3 buttons and num__6 buckles is how many times that of num__5 buckles and num__6 buttons ? <o> a ) num__0.466666666667 <o> b ) num__0.4 <o> c ) num__0.533333333333 <o> d ) num__0.6 <o> e ) num__0.733333333333 |
let x be the weight of a button and let y be the weight of a buckle . num__2 x + num__2 y = ( num__0.333333333333 ) ( num__3 x + num__11 y ) num__3 x = num__5 y x = num__5 y / num__3 num__3 x + num__6 y = a ( num__6 x + num__5 y ) num__11 y = a ( num__15 y ) a = num__0.733333333333 the answer is e . <eor> e <eos> |
e |
reverse__3.0__ multiply__3.0__5.0__ divide__11.0__15.0__ divide__11.0__15.0__ |
reverse__3.0__ multiply__3.0__5.0__ divide__11.0__15.0__ divide__11.0__15.0__ |
| calculate the ratio between x and y if num__80.0 of x equal to num__20.0 of y ? <o> a ) num__1 : num__1 <o> b ) num__3 : num__1 <o> c ) num__4 : num__5 <o> d ) num__4 : num__3 <o> e ) num__4 : num__1 |
explanation : num__80 x = num__20 y x : y = num__80 : num__20 = num__4 : num__1 answer : e <eor> e <eos> |
e |
divide__80.0__20.0__ divide__80.0__20.0__ |
divide__80.0__20.0__ divide__80.0__20.0__ |
| a clock is set at num__5 a . m . the clock loses num__16 minutes in num__24 hours . what will be the true time when the clock indicates num__10 p . m . on num__4 th day ? <o> a ) num__9 p . m <o> b ) num__10 p . m <o> c ) num__11 p . m <o> d ) num__12 p . m <o> e ) none |
solution time from num__5 a . m on a day to num__10 p . m . on num__4 th day = num__89 hours . now num__23 hrs num__44 min . of this clock = num__24 hours of correct clock . therefore num__23.7333333333 hrs of this clock = num__24 hours of correct clock . num__89 hrs of this clock = ( num__24 x num__0.0421348314607 x num__89 ) hrs = num__90 hrs so the correct time is num__11 p . m . answer c <eor> c <eos> |
c |
subtract__16.0__5.0__ round__11.0__ |
subtract__16.0__5.0__ round__11.0__ |
| there are num__8 teams in a certain league and each team plays each of the other teams exactly twice . if each game is played by num__2 teams what is the total number of games played ? <o> a ) num__15 <o> b ) num__16 <o> c ) num__28 <o> d ) num__56 <o> e ) num__64 |
every team plays with num__7 teams . . . so total no of matches = num__8 x num__7 = num__56 . now each match is played twice = > num__56 x num__2 but num__2 teams play a match = > num__56 x num__1.0 = num__56 . ans : d <eor> d <eos> |
d |
multiply__8.0__7.0__ subtract__8.0__7.0__ multiply__8.0__7.0__ |
multiply__8.0__7.0__ subtract__8.0__7.0__ multiply__8.0__7.0__ |
| the product of two numbers is num__9375 and the quotient when the larger one is divided by the smaller is num__15 . the sum of the numbers is : <o> a ) num__380 <o> b ) num__395 <o> c ) num__400 <o> d ) num__425 <o> e ) num__482 |
let the numbers be x and y . then xy = num__9375 and x = num__15 . y xy = num__9375 ( x / y ) num__15 y num__2 = num__625 . y = num__25 . x = num__15 y = ( num__15 x num__25 ) = num__375 . sum of the numbers = x + y = num__375 + num__25 = num__400 . answer : c <eor> c <eos> |
c |
divide__9375.0__15.0__ divide__9375.0__25.0__ add__375.0__25.0__ add__375.0__25.0__ |
divide__9375.0__15.0__ divide__9375.0__25.0__ add__375.0__25.0__ add__375.0__25.0__ |
| a person purchased a tv set for rs . num__16000 and a dvd player for rs . num__6250 . he sold both the items together for rs . num__31150 . what percentage of profit did he make ? <o> a ) num__80.0 <o> b ) num__49.0 <o> c ) num__40.0 <o> d ) num__70.0 <o> e ) num__90 % |
the total cp = rs . num__16000 + rs . num__6250 = rs . num__22250 and sp = rs . num__31150 profit ( % ) = ( num__31150 - num__22250 ) / num__22250 * num__100 = num__40.0 answer : c <eor> c <eos> |
c |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| in what time will a train num__175 m long cross an electric pole it its speed be num__144 km / hr ? <o> a ) num__2.58 sec <o> b ) num__2.91 sec <o> c ) num__4.37 sec <o> d ) num__2.9 sec <o> e ) num__1.8 sec |
speed = num__144 * num__0.277777777778 = num__40 m / sec time taken = num__4.375 = num__4.37 sec . answer : c <eor> c <eos> |
c |
divide__175.0__40.0__ round__4.37__ |
divide__175.0__40.0__ round__4.37__ |
| out of first num__20 natural numbers one number is selected at random . the probability that it is either an even number or a prime number is <o> a ) num__0.5 <o> b ) num__0.842105263158 <o> c ) num__0.8 <o> d ) num__0.85 <o> e ) num__0.6 |
explanation : n ( s ) = num__20 n ( even no ) = num__10 = n ( e ) n ( prime no ) = num__8 = n ( p ) p ( e á ´ œ p ) = num__0.5 + num__0.4 - num__0.05 = num__0.85 answer is d <eor> d <eos> |
d |
union_prob__0.5__0.4__0.05__ union_prob__0.5__0.4__0.05__ |
union_prob__0.5__0.4__0.05__ union_prob__0.5__0.4__0.05__ |
| what is the equation of the line that goes through ( – num__2 num__3 ) and ( num__5 – num__4 ) ? <o> a ) y = – x + num__1 <o> b ) y = x + num__5 <o> c ) y = – num__3 x / num__7 + num__2.14285714286 <o> d ) y = – num__4 x / num__3 + num__0.333333333333 <o> e ) y = num__9 x / num__5 + num__6.6 |
here we will follow the procedure we demonstrated in the last section . call ( – num__2 num__3 ) the “ first ” point and ( num__5 – num__4 ) the “ second . ” rise = – num__4 – num__3 = – num__7 . run = num__5 – ( – num__2 ) = num__7 . slope = rise / run = – num__1.0 = – num__1 . yes it makes sense that the slope is negative . we have the slope so plug m = – num__1 and ( x y ) = ( – num__2 num__3 ) into y = mx + b : num__3 = ( – num__1 ) * ( – num__2 ) + b num__3 = num__2 + b num__1 = b so plugging in m = – num__1 and b = num__1 we get an equation y = – x + num__1 . answer = a <eor> a <eos> |
a |
add__2.0__5.0__ subtract__3.0__2.0__ reverse__1.0__ |
add__2.0__5.0__ subtract__3.0__2.0__ reverse__1.0__ |
| in a certain office num__0.333333333333 of the workers are women ½ of the women are married and num__0.333333333333 of the married women have children if ¾ of the men are married and num__0.666666666667 of the married men have children what part of workers are without children ? <o> a ) num__0.833333333333 <o> b ) num__0.388888888889 <o> c ) num__0.944444444444 <o> d ) num__0.333333333333 <o> e ) num__0.611111111111 |
let the total no of workers be x no of women = x / num__3 no of men = x - ( x / num__3 ) = num__2 x / num__3 no of women having children = num__0.333333333333 of ½ ofx / num__3 = x / num__18 no of men having children = num__0.666666666667 of ¾ of num__2 x / num__3 = x / num__3 no of workers having children = x / num__8 + x / num__3 = num__7 x / num__18 workers having no children = x - num__7 x / num__18 = num__11 x / num__18 = num__0.611111111111 of all workers answer is e . <eor> e <eos> |
e |
add__3.0__8.0__ divide__11.0__18.0__ divide__11.0__18.0__ |
add__3.0__8.0__ divide__11.0__18.0__ divide__11.0__18.0__ |
| if g > x > y > z on the number line y is halfway between x and z and x is halfway between w and z then ( y - x ) / ( y - g ) = <o> a ) num__0.25 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.75 <o> e ) num__1 |
let y - z = t - - - > since y is halfway between x and z and x > y we have x - y = t . moreover x - z = ( x - y ) + ( y - z ) = num__2 t . similarly since x is halfway between w and z we have g - x = num__2 t . so y - x = - t y - g = - num__3 t . - - - > ( y - x ) / ( y - g ) = num__0.333333333333 . the answer is ( b ) . <eor> b <eos> |
b |
reverse__3.0__ reverse__3.0__ |
reverse__3.0__ reverse__3.0__ |
| four staff members at a certain company worked on a project . the amounts of time that the four staff members worked on the project were in the ratio num__2 to num__3 to num__5 to num__6 . if one of the four staff members worked on the project for num__45 hours which of the following can not be the total number of hours that the four staff members worked on the project ? <o> a ) num__80 <o> b ) num__360 <o> c ) num__120 <o> d ) num__144 <o> e ) num__240 |
a : b : c : d = num__2 x : num__3 x : num__5 x : num__6 x for some positive number x . total time num__2 x + num__3 x + num__5 x + num__6 x = num__16 x . if num__2 x = num__45 then num__16 x = num__360 ; if num__3 x = num__45 then num__16 x = num__240 ; if num__5 x = num__45 then num__16 x = num__144 ; if num__6 x = num__45 then num__16 x = num__120 ; only answer choices which is not obtained is num__80 . answer : a . <eor> a <eos> |
a |
divide__240.0__2.0__ multiply__5.0__16.0__ round__80.0__ |
divide__240.0__2.0__ multiply__5.0__16.0__ round__80.0__ |
| one hour after yolanda started walking from x to y a distance of num__24 miles bob started walking along the same road from y to x . if yolanda ' s walking rate was num__3 miles per hour and bob т ' s was num__4 miles per hour how many miles had bob walked when they met ? <o> a ) num__24 <o> b ) num__23 <o> c ) num__22 <o> d ) num__21 <o> e ) num__12 |
when b started walking y already has covered num__3 miles out of num__24 hence the distance at that time between them was num__24 - num__3 = num__21 miles . combined rate of b and y was num__3 + num__4 = num__7 miles per hour hence they would meet each other in num__3.0 = num__3 hours . in num__6 hours b walked num__3 * num__4 = num__12 miles . answer : e . <eor> e <eos> |
e |
subtract__24.0__3.0__ add__3.0__4.0__ divide__24.0__4.0__ multiply__3.0__4.0__ round__12.0__ |
subtract__24.0__3.0__ add__3.0__4.0__ divide__24.0__4.0__ multiply__3.0__4.0__ subtract__24.0__12.0__ |
| if w x and y are consecutive odd positive integers and w < x < y which of the following could be equal to y - x - w ? <o> a ) - num__4 <o> b ) - num__2 <o> c ) - num__1 <o> d ) num__0 <o> e ) num__3 |
given : w x and y are consecutive odd positive integers and w < x < y let take value of x as a ; value of y = x + num__2 ( consecutive odd integer ) ; and w = x - num__2 so ; y - x - w = x + num__2 - x - ( x + num__2 ) therefore y - x - w = num__4 - x now x is odd and x > = num__3 so x can assume values = num__3 num__57 . . . therefore when x = num__5 num__4 - num__5 = - num__1 answer : c <eor> c <eos> |
c |
add__2.0__3.0__ subtract__3.0__2.0__ reverse__1.0__ |
add__2.0__3.0__ subtract__3.0__2.0__ subtract__2.0__1.0__ |
| it rained as much as on wednesday as on all the other days of the week combined . if the average rainfall for the whole week was num__2.5 cms how much did it rain on wednesday ? <o> a ) num__10.9 <o> b ) num__10.5 <o> c ) num__8.75 <o> d ) num__10.1 <o> e ) num__10.6 |
explanation : let the rainfall on wednesday = num__6 x . â ˆ ´ rainfall on the remaining days = num__6 x given ( num__6 x + num__6 x ) / num__7 = num__2.5 â ‡ ’ num__12 x = num__17.5 â ‡ ’ num__6 x = num__8.75 answer : c <eor> c <eos> |
c |
multiply__2.5__7.0__ subtract__17.5__8.75__ |
multiply__2.5__7.0__ subtract__17.5__8.75__ |
| rs . num__1100 is divided into three parts a b and c . how much a is more than c if their ratio is num__0.5 : num__0.333333333333 : num__0.25 ? <o> a ) num__300 <o> b ) num__992 <o> c ) num__400 <o> d ) num__552 <o> e ) num__312 |
num__0.5 : num__0.25 : num__0.166666666667 = num__6 : num__3 : num__2 num__0.181818181818 * num__1100 = num__200 num__600 - num__200 = num__400 answer : c <eor> c <eos> |
c |
subtract__0.5__0.3333__ multiply__0.5__6.0__ reverse__0.5__ multiply__3.0__200.0__ multiply__2.0__200.0__ multiply__2.0__200.0__ |
subtract__0.5__0.3333__ multiply__0.5__6.0__ reverse__0.5__ multiply__3.0__200.0__ multiply__2.0__200.0__ multiply__2.0__200.0__ |
| if n = num__9 ^ num__11 – num__9 what is the units digit of n ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
always divide the power ( incase num__11 ) by num__4 and use the remainder as the new power . the question now becomes num__9 ^ num__3 - num__9 . now num__9 ^ num__3 has last digit num__9 . we subtract num__9 from num__9 thus num__9 - num__9 = num__0 is the answer . option a <eor> a <eos> |
a |
multiply__9.0__0.0__ |
multiply__9.0__0.0__ |
| solve below question num__6 x - num__1 = - num__13 <o> a ) - num__8 <o> b ) - num__2 <o> c ) - num__5 <o> d ) - num__4 <o> e ) num__1 |
num__1 . subtract num__1 from both sides : num__6 x - num__1 + num__1 = - num__13 + num__1 num__2 . simplify both sides : num__6 x = - num__12 num__3 . divide both sides by num__6 : num__4 . simplify both sides : x = - num__2 b <eor> b <eos> |
b |
multiply__6.0__2.0__ divide__6.0__2.0__ subtract__6.0__2.0__ divide__6.0__3.0__ |
subtract__13.0__1.0__ add__1.0__2.0__ subtract__6.0__2.0__ subtract__6.0__4.0__ |
| in a party num__65.0 of women wearing black tshirts num__50.0 of the total people in party wearing black t shirts then what is the ratio of male is to female . num__35.0 of the man wearing black t shirs . there are num__120 people in the party . <o> a ) num__1 : num__1 <o> b ) num__1 : num__2 <o> c ) num__1 : num__3 <o> d ) num__1 : num__4 <o> e ) num__1 : num__5 |
suppose male = m female = f so num__65.0 of f + num__35.0 of m = num__50.0 ( f + m ) solving num__65.0 of f + num__35.0 of m = num__50.0 of f + num__50.0 of m num__15.0 of f = num__15.0 of m f / m = num__15.0 / num__15.0 = num__1 : num__1 answer : a <eor> a <eos> |
a |
subtract__65.0__50.0__ reverse__1.0__ |
subtract__65.0__50.0__ reverse__1.0__ |
| this is how edward ’ s lotteries work . first num__8 different numbers are selected . tickets with exactly num__5 of the num__8 numbers randomly selected are printed such that no two tickets have the same set of numbers . finally the winning ticket is the one containing the num__5 numbers drawn from the num__8 randomly . there is exactly one winning ticket in the lottery system . how many tickets can the lottery system print ? <o> a ) num__9 p num__6 <o> b ) num__8 p num__3 <o> c ) num__9 c num__9 <o> d ) num__8 c num__6 <o> e ) num__8 c num__5 |
since we have to select random num__5 numbers from num__8 and they all are distinct . i think it should be num__8 c num__5 or e <eor> e <eos> |
e |
round__8.0__ |
round__8.0__ |
| one - third of num__1206 is what percent of num__134 ? <o> a ) num__3 <o> b ) num__30 <o> c ) num__300 <o> d ) none of these <o> e ) can not be determined |
answer let one - third of num__1206 is n % of num__134 . ∵ num__402.0 = ( n x num__134 ) / num__100 ∴ n = ( num__402 x num__100 ) / num__134 = num__300 correct option : c <eor> c <eos> |
c |
percent__100.0__300.0__ |
percent__100.0__300.0__ |
| winson traveled the entire num__60 miles trip . if he did the first num__10 miles of at a constant rate num__20 miles per hour and the remaining trip of at a constant rate num__48 miles per hour what is the his average speed in miles per hour ? <o> a ) num__20 mph <o> b ) num__24 mph <o> c ) num__30 mph <o> d ) num__32 mph <o> e ) num__40 mph |
average speed = sum of distance / sum of time . if he travelled the first num__10 miles at num__20 miles / hr it would take num__0.5 hr . for the remaining trip if he went at num__48 miles / num__1 hr it would take num__1 hour . then the average speed is num__60 miles / ( num__0.5 + num__1 ) hrs = num__40 miles / num__1 hr . therefore the answer is e . <eor> e <eos> |
e |
divide__10.0__20.0__ subtract__60.0__20.0__ round__40.0__ |
divide__10.0__20.0__ divide__20.0__0.5__ divide__20.0__0.5__ |
| raj invested an amount of rs . num__17400 for two years . find the rate of compound interest that will fetch him an amount of rs . num__1783.50 at the end of two years ? <o> a ) num__4 <o> b ) num__3 <o> c ) num__2 <o> d ) num__5 <o> e ) num__6 |
let the rate of interest be r % p . a . num__17400 [ num__1 + r / num__100 ] num__2 = num__17400 + num__1783.50 [ num__1 + r / num__100 ] num__2 = ( num__17400 + num__1783.50 ) / num__17400 = num__1 + num__0.1025 = num__1 + num__0.1025 = num__1.1025 = [ num__1.05 ] num__2 [ num__1 + r / num__100 ] = num__1.05 r / num__100 = num__0.05 therefore r = num__5 answer : d <eor> d <eos> |
d |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| the profit earned by selling an article for rs num__900 is double the loss incurred when the same article is sold for rs . num__490 . at what price should the article be sold to make num__25.0 profit ? <o> a ) num__715 <o> b ) num__469 <o> c ) num__400 <o> d ) num__750 <o> e ) num__560 |
let c . p be rs . x num__900 - x = num__2 ( x - num__450 ) = > x = rs . num__600 c . p = num__600 gain required is num__25.0 s . p = [ ( num__100 + num__25 ) * num__600 ] / num__100 = rs . num__750 answer : d <eor> d <eos> |
d |
percent__100.0__750.0__ |
percent__100.0__750.0__ |
| r campers fished for num__7 hours . if m of the campers caught num__2 fish apiece and the rest caught a combined total of n fish how many fish did the r campers catch per hour ? <o> a ) num__2 m + n ( r - m ) <o> b ) ( num__2 m + n ( r - m ) ) / num__3 <o> c ) ( num__2 m + n ( m - r ) ) / num__3 <o> d ) ( num__2 m + n ) / num__7 <o> e ) ( num__2 m + n ) / r |
m of the campers caught num__2 fish apiece in num__7 hours - - > total fish caught by m campers = num__2 m ; the rest caught a combined total of n fish . total fish caught in num__7 hours = num__2 m + n - - > per hour ( num__2 m + n ) / num__7 . answer : d . <eor> d <eos> |
d |
round__2.0__ |
round__2.0__ |
| a can give b num__100 meters start and c num__200 meters start in a kilometer race . how much start can b give c in a kilometer race ? <o> a ) num__111.12 <o> b ) num__111.18 <o> c ) num__111.15 <o> d ) num__111.13 <o> e ) num__191.18 |
a runs num__1000 m while b runs num__900 m and c runs num__800 m . the number of meters that c runs when b runs num__1000 m = ( num__1000 * num__800 ) / num__900 = num__888.888888889 = num__888.88 m . b can give c = num__1000 - num__888.88 = num__111.12 m . answer : a <eor> a <eos> |
a |
subtract__1000.0__100.0__ subtract__900.0__100.0__ subtract__1000.0__888.88__ round__111.12__ |
subtract__1000.0__100.0__ subtract__900.0__100.0__ subtract__1000.0__888.88__ subtract__1000.0__888.88__ |
| a contractor isengaged for num__30 days on the condition that he receives rs . num__25 for eachday he works & fined rs . num__7.50 for each day is absent . he gets rs . num__425 in all . for how many days was heabsent ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__15 <o> d ) num__17 <o> e ) num__21 |
num__30 * num__25 = num__750 num__425 - - - - - - - - - - - num__325 num__25 + num__7.50 = num__32.5 num__325 / num__32.5 = b <eor> b <eos> |
b |
multiply__30.0__25.0__ subtract__750.0__425.0__ add__25.0__7.5__ divide__325.0__32.5__ |
multiply__30.0__25.0__ subtract__750.0__425.0__ add__25.0__7.5__ divide__325.0__32.5__ |
| what is the least number that should be added to num__1076 so the sum of the number is divisible by num__23 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
( num__46.7826086957 ) gives a remainder num__18 so we need to add num__5 . the answer is e . <eor> e <eos> |
e |
divide__1076.0__23.0__ subtract__23.0__18.0__ subtract__23.0__18.0__ |
divide__1076.0__23.0__ subtract__23.0__18.0__ subtract__23.0__18.0__ |
| dick and jane each saved $ num__2500 in num__1989 . in num__1990 dick saved num__9 percent more than in num__1989 and together he and jane saved a total of $ num__5200 . approximately what percent less did jane save in num__1990 than in num__1989 ? <o> a ) num__1.0 <o> b ) num__4.0 <o> c ) num__6.0 <o> d ) num__8.0 <o> e ) num__9 % |
num__1990 dick saved = $ num__2725 jane saved = $ num__2475 ( jane saved $ num__25 less than she did the prior year ) jane saved approximately $ num__25 / $ num__2500 ( num__1.0 ) less in num__1990 answer : a <eor> a <eos> |
a |
subtract__5200.0__2725.0__ subtract__2500.0__2475.0__ subtract__1990.0__1989.0__ reverse__1.0__ |
subtract__5200.0__2725.0__ subtract__2500.0__2475.0__ subtract__1990.0__1989.0__ reverse__1.0__ |
| a number y is chosen at random from the numbers num__1 - num__3 - num__1 num__0 num__1 num__2 num__3 . what is the probability that | x | < num__3 ? <o> a ) num__0.333333333333 <o> b ) num__0.666666666667 <o> c ) num__0.428571428571 <o> d ) num__0.272727272727 <o> e ) none |
| x | | x | can take num__7 values . to get | x | < num__2 | x | < num__2 ( i . e . − num__2 < x < + num__2 − num__2 < x ⇒ p ( | x | < num__2 ) = favourable casestotal casesp ( | x | < num__2 ) = favourable casestotal cases = num__0.428571428571 c <eor> c <eos> |
c |
divide__3.0__7.0__ multiply__1.0__0.4286__ |
divide__3.0__7.0__ multiply__1.0__0.4286__ |
| a certain sum amounts to rs . num__2250 in num__5 years and rs . num__2550 in num__7 years . find the rate % per annum ? <o> a ) num__10.0 <o> b ) num__11.0 <o> c ) num__9.0 <o> d ) num__8.0 <o> e ) num__7 % |
num__5 - - - num__2250 num__7 - - - num__2550 - - - - - - - - - - - - - - num__2 - - - num__300 n = num__1 i = num__150 r = ? p = num__2250 - num__750 = num__1500 num__150 = ( num__1500 * num__1 * r ) / num__100 r = num__10.0 answer : a <eor> a <eos> |
a |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| in a sequence of num__40 numbers each term except for the first one is num__7 less than the previous term . if the greatest term in the sequence is num__282 what is the smallest term in the sequence ? <o> a ) num__2 <o> b ) - num__2 <o> c ) num__0 <o> d ) num__8 <o> e ) num__9 |
which term is the greatest ? the first or the last ? it is given to you that every term is num__7 less than the previous term . hence as you go on your terms keep becoming smaller and smaller . the first term is the greatest term . an = num__282 + ( num__40 - num__1 ) * ( - num__7 ) an = num__282 - num__273 = num__9 e <eor> e <eos> |
e |
subtract__282.0__273.0__ subtract__282.0__273.0__ |
subtract__282.0__273.0__ subtract__282.0__273.0__ |
| what is the least number of digits ( including repetitions ) needed to express num__10 ^ num__200 in decimal notation ? <o> a ) a ) num__201 <o> b ) b ) num__100 <o> c ) c ) num__101 <o> d ) d ) num__1000 <o> e ) e ) num__1001 |
num__10 ^ n is a decimal number with a num__1 followed by n zeros . so num__10 ^ num__200 will include num__200 num__0 ' s + num__1 digit for num__1 = num__201 so the answer is a <eor> a <eos> |
a |
add__200.0__1.0__ add__200.0__1.0__ |
add__200.0__1.0__ add__200.0__1.0__ |
| num__00 a num__3 + num__4 b ___ num__96 if a and b represent positive single digits in the correctly worked computation above what is the value of the two digit integer ab ? <o> a ) num__43 <o> b ) num__35 <o> c ) num__45 <o> d ) num__43 <o> e ) num__46 |
num__0 + num__3 + b = num__6 b = num__3 num__0 + a + num__4 = num__8 a = num__4 ab = num__43 ans = a <eor> a <eos> |
a |
round__43.0__ |
round__43.0__ |
| a is twice as good a workman as b and they took num__7 days together to do the work b alone can do it in . <o> a ) num__22 <o> b ) num__77 <o> c ) num__21 <o> d ) num__77 <o> e ) num__88 |
wc = num__2 : num__1 num__2 x + x = num__0.142857142857 x = num__0.047619047619 = > num__21 days answer : c <eor> c <eos> |
c |
divide__1.0__7.0__ round__21.0__ |
divide__1.0__7.0__ round__21.0__ |
| if a : b = num__0.166666666667 : num__0.333333333333 b : c = num__0.5 : num__0.333333333333 then a : b : c ? <o> a ) num__9 : num__6 : num__6 <o> b ) num__9 : num__6 : num__8 <o> c ) num__9 : num__18 : num__12 <o> d ) num__9 : num__6 : num__4 <o> e ) num__9 : num__6 : num__1 |
a : b = num__0.166666666667 : num__0.333333333333 = num__3 : num__6 b : c = num__0.5 : num__0.333333333333 = num__3 : num__2 - - - - - - - - - - - - - - - - - - - - a : b : c = num__9 : num__18 : num__12 answer : c <eor> c <eos> |
c |
divide__3.0__0.5__ reverse__0.5__ add__3.0__6.0__ multiply__2.0__9.0__ multiply__2.0__6.0__ multiply__0.5__18.0__ |
divide__3.0__0.5__ reverse__0.5__ add__3.0__6.0__ multiply__2.0__9.0__ subtract__18.0__6.0__ subtract__12.0__3.0__ |
| yesterday it took robert num__6 hours to drive from city a to city b . today it took robert num__4.5 hours to drive back from city В to city a along the same route . if he had saved num__30 minutes in both trips the speed for the round trip would be num__90 miles per hour . what is the distance between city a and city b ? <o> a ) num__90 <o> b ) num__120 <o> c ) num__150 <o> d ) num__440 <o> e ) num__430 |
num__2 d / num__90 = num__9.5 ( because time = num__6 + num__4.5 - num__1 hrs ) = > d = num__430 answer - e <eor> e <eos> |
e |
round__430.0__ |
divide__430.0__1.0__ |
| what annual installment will discharge a debt of rs . num__1092 due in num__3 years at num__12.0 simple interest ? <o> a ) num__315 <o> b ) num__345 <o> c ) num__325 <o> d ) num__335 <o> e ) none of them |
let each installment be rs . x then ( x + ( ( x * num__12 * num__1 ) / num__100 ) ) + ( x + ( ( x * num__12 * num__2 ) / num__100 ) ) + x = num__1092 = ( ( num__28 x / num__25 ) + ( num__31 x / num__25 ) + x ) = num__1092 ( num__28 x + num__31 x + num__25 x ) = ( num__1092 * num__25 ) x = ( num__1092 * num__25 ) / num__84 = rs . num__325 . therefore each installment = rs . num__325 . answer is c . <eor> c <eos> |
c |
percent__100.0__325.0__ |
percent__100.0__325.0__ |
| what is the difference between the place value and face value of num__3 in the numeral num__1375 ? <o> a ) num__280 <o> b ) num__290 <o> c ) num__297 <o> d ) num__333 <o> e ) num__340 |
place value of num__3 = num__3 * num__100 = num__300 face value of num__3 = num__3 num__300 - num__3 = num__297 c ) <eor> c <eos> |
c |
multiply__3.0__100.0__ subtract__300.0__3.0__ subtract__300.0__3.0__ |
multiply__3.0__100.0__ subtract__300.0__3.0__ subtract__300.0__3.0__ |
| the sum of two numbers is num__12 and the product of the numbers is num__35 . find sum of the squares of that numbers . <o> a ) num__84 <o> b ) num__87 <o> c ) num__96 <o> d ) num__79 <o> e ) num__74 |
let a and b be the two numbers ( a + b ) ^ num__2 = a ^ num__2 + num__2 ab + b ^ num__2 given ( a + b ) = num__12 ab = num__35 so num__12 ^ num__2 = a ^ num__2 + b ^ num__2 + num__2 * num__35 num__144 = a ^ num__2 + b ^ num__2 + num__70 a ^ num__2 + b ^ num__2 = num__74 ans e <eor> e <eos> |
e |
multiply__35.0__2.0__ subtract__144.0__70.0__ subtract__144.0__70.0__ |
multiply__35.0__2.0__ subtract__144.0__70.0__ subtract__144.0__70.0__ |
| a trader mixes num__26 kg of rice at rs . num__20 per kg with num__30 kg of rice of other variety at rs . num__36 per kg and sells the mixture at rs . num__30 per kg . his profit percent is : <o> a ) no profit no loss <o> b ) num__5.0 <o> c ) num__8.0 <o> d ) num__10.0 <o> e ) num__14 % |
c . p . of num__56 kg rice = rs . ( num__26 x num__20 + num__30 x num__36 ) = rs . ( num__520 + num__1080 ) = rs . num__1600 . s . p . of num__56 kg rice = rs . ( num__56 x num__30 ) = rs . num__1680 . gain = ( num__0.05 ) x num__100.0 = num__5.0 . answer : b <eor> b <eos> |
b |
percent__5.0__100.0__ |
percent__5.0__100.0__ |
| a pipe can fill a cistern in num__20 minutes whereas the cistern when fill can be emptied by a leak in num__28 minutes . when both pipes are opened find when the cistern will be full ? <o> a ) num__76 minutes <o> b ) num__29 minutes <o> c ) num__70 minutes <o> d ) num__17 minutes <o> e ) num__98 minutes |
num__0.05 - num__0.0357142857143 = num__0.0142857142857 num__70 minutes answer : c <eor> c <eos> |
c |
subtract__0.05__0.0357__ round__70.0__ |
subtract__0.05__0.0357__ round__70.0__ |
| a shopkeeper sells his goods at cost price but uses a faulty meter that weighs num__800 grams . find the profit percent . <o> a ) num__5.88235294118 % <o> b ) num__4.34782608696 % <o> c ) num__5.26315789474 % <o> d ) num__25.0 % <o> e ) none of these |
explanation : ( num__100 + g ) / ( num__100 + x ) = true measure / faulty measure x = num__0 true measure = num__1000 faulty measure = num__800 num__100 + g / num__100 + num__0 = num__1.25 num__100 + g = num__1.25 * num__100 g = num__25.0 % answer : d <eor> d <eos> |
d |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| for num__6 - digit integer num__46 x y num__12 when x and y are drawn at random from { num__3 num__58 } what is the probability w that the integer drawn is divisible by num__8 ? * a solution will be posted in two days . <o> a ) num__0.166666666667 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__0.833333333333 |
in general the remainder divided by integer n is the same as the remainder that up to tens of n is divided by num__8 . thus in order for num__46 x y num__12 to be divided by num__8 y num__12 should be divided by num__8 . amongst num__3 num__58 num__3 and num__5 can be divided by num__8 . therefore w = two out of three can be divided and the answer is d . <eor> d <eos> |
d |
subtract__8.0__3.0__ divide__8.0__12.0__ |
subtract__8.0__3.0__ divide__8.0__12.0__ |
| num__31 of the scientists that attended a certain workshop were wolf prize laureates and num__16 of these num__31 were also nobel prize laureates . of the scientists that attended that workshop and had not received the wolf prize the number of scientists that had received the nobel prize was num__3 greater than the number of scientists that had not received the nobel prize . if num__50 of the scientists attended that workshop how many of them were nobel prize laureates ? <o> a ) a ) num__11 <o> b ) b ) num__18 <o> c ) c ) num__24 <o> d ) d ) num__27 <o> e ) d ) num__36 |
lets solve by creating equation . . w = num__31 . . total = num__50 . . not w = num__50 - num__31 = num__19 . . now let people who were neither be x so out of num__19 who won nobel = x + num__3 . . so x + x + num__3 = num__19 or x = num__8 . . so who won nobel but not wolf = x + num__3 = num__11 . . but people who won both w and n = num__13 . . so total who won n = num__11 + num__16 = num__27 . . d <eor> d <eos> |
d |
add__16.0__3.0__ add__3.0__8.0__ subtract__16.0__3.0__ add__16.0__11.0__ round__27.0__ |
add__16.0__3.0__ add__3.0__8.0__ subtract__16.0__3.0__ add__16.0__11.0__ add__16.0__11.0__ |
| the sum of even numbers between num__1 and num__31 is <o> a ) num__16 <o> b ) num__128 <o> c ) num__240 <o> d ) num__512 <o> e ) none |
solution required numbers are num__2 num__4 num__6 . . num__30 . this is an a . p containing num__15 terms . required sum = n / num__2 ( first term + last term ) ‹ = › num__7.5 ( num__2 + num__30 ) ‹ = › num__240 . answer c <eor> c <eos> |
c |
add__2.0__4.0__ subtract__31.0__1.0__ divide__30.0__2.0__ divide__15.0__2.0__ multiply__1.0__240.0__ |
add__2.0__4.0__ subtract__31.0__1.0__ divide__30.0__2.0__ divide__15.0__2.0__ divide__240.0__1.0__ |
| five machines at a certain factory operate at the same constant rate . if four of these machines operating simultaneously take num__25 hours to fill a certain production order how many fewer hours does it take all five machines operating simultaneously to fill the same production order ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__16 <o> e ) num__24 |
total work = num__4 * num__25 machine hrs time taken = num__4 * num__5.0 = > num__20 hours . . . thus all five machines operating simultaneously will take ( num__25 - num__20 ) num__5 hours . . answer will be ( b ) num__5 <eor> b <eos> |
b |
subtract__25.0__5.0__ gcd__25.0__5.0__ |
subtract__25.0__5.0__ subtract__25.0__20.0__ |
| teas worth rs . num__126 per kg and rs . num__135 per kg are mixed with a third variety in the ratio num__1 : num__1 : num__2 . if the mixture is worth rs num__157 per kg the price of the third variety per kg will be ? <o> a ) rs . num__147.50 <o> b ) rs . num__785.50 <o> c ) rs . num__179.50 <o> d ) rs . num__258.50 <o> e ) none of these |
explanation : since first and second varieties are mixed in equal proportions . so their average price = rs . ( num__126 + num__135 ) / num__2 . = > rs . num__130.50 . so the mixture is formed by mixing two varieties one at rs . num__130.50 per kg and the other at say rs . x per kg in the ratio num__2 : num__2 i . e . num__1 : num__1 . we have to find x . by the rule of alligation we have : cost of num__1 kg cost of num__1 kg of num__1 st kind of num__2 nd kind ( rs . num__130.50 ) ( rs . x ) \ / mean price ( rs . num__157 ) / \ x â ˆ ’ num__157 num__22.50 = > x â ˆ ’ ( num__157 / num__22.50 ) = num__1 . = > x â ˆ ’ num__157 = num__22.50 . = > x = num__179.50 rs . answer : c <eor> c <eos> |
c |
add__157.0__22.5__ multiply__1.0__179.5__ |
add__157.0__22.5__ add__157.0__22.5__ |
| a shopkeeper has num__280 kg of apples . he sells num__35.0 of these at num__20.0 profit and remaining num__60.0 at num__30.0 profit . find his % profit on total . <o> a ) num__24.0 <o> b ) num__25.0 <o> c ) num__26.0 <o> d ) num__28.0 <o> e ) num__35 % |
if the total quantity was num__100 then num__35 x num__20.0 + num__60 x num__30.0 = num__25 this profit will remain same for any total quantity unless the % of products remains the same . hence ' b ' is the answer <eor> b <eos> |
b |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| company c produces toy trucks at a cost of $ num__5.00 each for the first num__100 trucks and $ num__3.50 for each additional truck . if num__500 toy trucks were produced by company c and sold for $ num__7.00 each what was company c ’ s gross profit ? <o> a ) $ num__2250 <o> b ) $ num__2500 <o> c ) $ num__1600 <o> d ) $ num__1250 <o> e ) $ num__1 |
500 |
cost of num__500 trucks : ( num__100 * num__5 ) + ( num__400 * num__3.5 ) = num__500 + num__1400 = $ num__1900 revenue : num__500 * num__7 = $ num__3500 profit : num__3500 - num__1900 = $ num__1600 option c is correct <eor> c <eos> |
c |
c |
| the speed of the boat in still water in num__12 kmph . it can travel downstream through num__42 kms in num__3 hrs . in what time would it cover the same distance upstream ? <o> a ) num__8 hours <o> b ) num__6 hours <o> c ) num__4.2 hours <o> d ) num__5 hours <o> e ) num__6 hours |
still water = num__12 km / hr downstream = num__14.0 = num__14 km / hr upstream = > > still water = ( u + v / num__2 ) = > > num__12 = u + num__7.0 = num__10 km / hr so time taken in upstream = num__4.2 = num__4.2 hrs answer : c <eor> c <eos> |
c |
divide__42.0__3.0__ subtract__14.0__12.0__ divide__14.0__2.0__ subtract__12.0__2.0__ divide__42.0__10.0__ round__4.2__ |
divide__42.0__3.0__ subtract__14.0__12.0__ divide__14.0__2.0__ add__3.0__7.0__ divide__42.0__10.0__ divide__42.0__10.0__ |
| the tax on a commodity is diminished by num__20.0 and its consumption increased by num__15.0 . the effect on revenue is ? <o> a ) num__9.0 decrease <o> b ) num__8.0 decrease in revenue <o> c ) num__6.0 decrease <o> d ) num__1.0 decrease <o> e ) num__2.0 decrease |
num__100 * num__100 = num__10000 num__80 * num__115 = num__9200 - - - - - - - - - - - num__10000 - - - - - - - - - - - num__800 num__100 - - - - - - - - - - - ? = > num__8.0 decrease answer : b <eor> b <eos> |
b |
subtract__100.0__20.0__ add__15.0__100.0__ multiply__80.0__115.0__ subtract__10000.0__9200.0__ divide__800.0__100.0__ divide__800.0__100.0__ |
subtract__100.0__20.0__ add__15.0__100.0__ multiply__80.0__115.0__ subtract__10000.0__9200.0__ divide__800.0__100.0__ divide__800.0__100.0__ |
| a man sells an article at a profit of num__25.0 . if he had bought it at num__20.0 less and sold it for rs . num__16.80 less he would have gained num__30.0 . find the cost of the article . <o> a ) num__30 <o> b ) num__40 <o> c ) num__50 <o> d ) num__60 <o> e ) num__80 |
let c . p = num__100 gain = num__25.0 s . p = num__125 supposed c . p = num__80 gain = num__30.0 s . p = ( num__130 * num__80 ) / num__100 = num__104 diff = ( num__125 - num__104 ) = num__21 diff num__21 when c . p = num__100 then diff num__16.80 when c . p = ( num__100 * num__16.80 ) / num__21 = num__50 answer : e <eor> e <eos> |
e |
percent__80.0__130.0__ percent__16.8__125.0__ percent__100.0__80.0__ |
percent__80.0__130.0__ percent__16.8__125.0__ percent__100.0__80.0__ |
| square rstu shown above is rotated in a plane about its center in a clockwise direction the minimum number of degrees necessary for s to be in the position where u is now shown . the number of degrees through which rstu is rotated is <o> a ) num__180 degree <o> b ) num__160 degree <o> c ) num__225 degree <o> d ) num__270 degree <o> e ) num__315 degree |
from the options i am assuming the positioning of u and s relative to each other to be as shown . to replace u by s focus on os . say you rotate os clockwise ( and with it the entire square ) and bring it in place of ou . how many degrees did you go ? you covered num__2 right angles i . e . num__180 degrees . answer : a <eor> a <eos> |
a |
straight_angle__ straight_angle__ |
straight_angle__ straight_angle__ |
| shri prakash walked num__40 metres facing towards north . from there he walked num__50 metres after turning to his left . after this he walked num__40 metres after turning to his left . how far and in what direction is he now from his starting point ? <o> a ) num__40 m north <o> b ) num__50 m west <o> c ) num__10 m east <o> d ) num__10 m west <o> e ) none of these |
num__50 m west answer : b <eor> b <eos> |
b |
round__50.0__ |
round__50.0__ |
| james was hiking on a num__10 - mile loop trail at a rate of num__2 miles per hour . two hours into james ’ hike john started hiking from the same starting point on the loop trail at num__4 miles per hour . what is the shortest time that john could hike on the trail in order to meet up with james ? <o> a ) num__0.5 hours <o> b ) num__1 hour <o> c ) num__2 hours <o> d ) num__3 hours <o> e ) num__4 hours |
we do n ' t need to assume here the direction . we need to find which will take less time so there are num__2 cases num__1 ) both moving in same direction . . . then time would be num__2 hours num__2 ) both moving in opposite direction then time would be num__1 hour as we need the shortest time it would be the second case answer : b <eor> b <eos> |
b |
round__1.0__ |
round__1.0__ |
| let f ( x y ) be defined as the remainder when ( x – y ) ! is divided by x . if x = num__24 what is the maximum value of y for which f ( x y ) = num__0 ? <o> a ) num__12 <o> b ) num__15 <o> c ) num__18 <o> d ) num__20 <o> e ) num__21 |
the question is finding y such that ( num__24 - y ) ! is a multiple of num__24 . that means we need to have num__2 ^ num__3 * num__3 in ( num__24 - y ) ! num__4 ! is the smallest factorial number with num__2 ^ num__3 * num__3 as a factor . num__24 - y = num__4 y = num__20 the answer is d . <eor> d <eos> |
d |
subtract__24.0__4.0__ subtract__24.0__4.0__ |
subtract__24.0__4.0__ subtract__24.0__4.0__ |
| the average age of a class of num__39 students is num__15 years . if the age of the teacher be included then the average increases by num__3 months . find the age of the teacher . <o> a ) num__22 <o> b ) num__25 <o> c ) num__26 <o> d ) num__32 <o> e ) num__34 |
sol . total age of num__39 persons = ( num__39 x num__15 ) years = num__585 years . average age of num__40 persons = num__15 yrs num__3 months = num__15.25 years . total age of num__40 persons = ( _ ( num__15.25 ) x num__40 ) years = num__610 years . : . age of the teacher = ( num__610 - num__585 ) years = num__25 years . answer b <eor> b <eos> |
b |
multiply__39.0__15.0__ multiply__15.25__40.0__ subtract__610.0__585.0__ subtract__610.0__585.0__ |
multiply__39.0__15.0__ multiply__15.25__40.0__ subtract__610.0__585.0__ subtract__610.0__585.0__ |
| the captain of a cricket team of num__11 members is num__26 years old and the wicket keeper is num__7 years older . if the ages of these two are excluded the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? <o> a ) num__23 years <o> b ) num__24 years <o> c ) num__25 years <o> d ) num__26 years <o> e ) num__27 years |
explanation let the average age of the whole team by x years . num__11 x â € “ ( num__26 + num__33 ) = num__9 ( x - num__1 ) num__11 x â € “ num__9 x = num__50 num__2 x = num__50 x = num__25 . so average age of the team is num__25 years . answer c <eor> c <eos> |
c |
add__26.0__7.0__ subtract__11.0__9.0__ subtract__26.0__1.0__ subtract__26.0__1.0__ |
add__26.0__7.0__ subtract__11.0__9.0__ subtract__26.0__1.0__ subtract__26.0__1.0__ |
| a man rows his boat num__95 km downstream and num__45 km upstream taking num__2 num__0.5 hours each time . find the speed of the stream ? <o> a ) num__10 kmph <o> b ) num__6 kmph <o> c ) num__5 kmph <o> d ) num__8 kmph <o> e ) num__7 kmph |
speed downstream = d / t = num__95 / ( num__2 num__0.5 ) = num__38 kmph speed upstream = d / t = num__45 / ( num__2 num__0.5 ) = num__18 kmph the speed of the stream = ( num__38 - num__18 ) / num__2 = num__10 kmph answer : a <eor> a <eos> |
a |
round__10.0__ |
round__10.0__ |
| apple inc recently conducted a survey and found that num__50000 of its customers live in rural areas . if the number of it customers who live in urban areas is num__200 percents grater than the numbers of customers who live in rural areas how many of apple inc customers live in urban areas ? <o> a ) num__150000 <o> b ) num__216000 <o> c ) num__360000 <o> d ) num__480000 <o> e ) num__540.0 |
let ' s say that we have a starting value of x . . . num__100.0 of x = x num__100.0 more than x = x + x = num__2 x num__200.0 more than x = x + num__2 x = num__3 x num__300.0 more than x = x + num__3 x = num__4 x etc . here we ' re told that num__50000 customers live in rural areas and the number who live in urban areas is num__200.0 greater than the number who live in rural areas . that means that the number who are in urban areas is num__3 times the number in rural areas . . . num__3 ( num__50000 ) = num__150000 final answer : a <eor> a <eos> |
a |
percent__2.0__200.0__ percent__100.0__150000.0__ |
percent__2.0__200.0__ percent__100.0__150000.0__ |
| the average salary of a person for the months of january february march and april is rs . num__8000 and that for the months february march april and may is rs . num__8700 . if his salary for the month of may is rs . num__6500 find his salary for the month of january ? <o> a ) s . num__3700 <o> b ) s . num__4570 <o> c ) s . num__4500 <o> d ) s . num__4550 <o> e ) s . num__2500 |
sum of the salaries of the person for the months of january february march and april = num__4 * num__8000 = num__32000 - - - - ( num__1 ) sum of the salaries of the person for the months of february march april and may = num__4 * num__8700 = num__34800 - - - - ( num__2 ) ( num__2 ) - ( num__1 ) i . e . may - jan = num__2800 salary of may is rs . num__6500 salary of january = rs . num__3700 answer : a <eor> a <eos> |
a |
multiply__8000.0__4.0__ multiply__8700.0__4.0__ subtract__34800.0__32000.0__ subtract__6500.0__2800.0__ subtract__6500.0__2800.0__ |
multiply__8000.0__4.0__ multiply__8700.0__4.0__ subtract__34800.0__32000.0__ subtract__6500.0__2800.0__ subtract__6500.0__2800.0__ |
| from a container having pure milk num__20.0 is replaced by water and the process is repeated thrice . at the end of the third operation the milk is <o> a ) num__40.0 pure <o> b ) num__50.0 pure <o> c ) num__51.2 pure <o> d ) num__58.8 pure <o> e ) none |
solution let total quantity of original milk = num__100 gm . milk after first operation = num__80.0 of num__1000 ‹ = › num__800 gm . milk after second operation = num__80.0 of num__800 ‹ = › num__640 gm . milk after third operation = num__80.0 of num__640 ‹ = › num__512 gm . strength of final mixtures = num__51.2 . answer c <eor> c <eos> |
c |
percent__80.0__1000.0__ percent__80.0__800.0__ percent__80.0__640.0__ percent__100.0__51.2__ |
percent__80.0__1000.0__ percent__80.0__800.0__ percent__80.0__640.0__ percent__100.0__51.2__ |
| if y is the smallest positive integer that is not prime and not a factor of num__50 ! what is the sum of the factors of y ? <o> a ) num__162 <o> b ) num__54 <o> c ) num__72 <o> d ) num__51 <o> e ) num__50 ! + num__2 |
in fact num__51 = num__3 * num__17 is a factor of num__50 ! . the smallest positive integer that is not prime and not a factor of num__50 ! is num__106 = num__2 * num__53 . the sum of the factors of num__106 is num__162 . answer : a . <eor> a <eos> |
a |
divide__51.0__3.0__ gcd__50.0__106.0__ add__50.0__3.0__ lcm__2.0__162.0__ |
divide__51.0__3.0__ gcd__50.0__106.0__ add__50.0__3.0__ lcm__2.0__162.0__ |
| how long does a train num__110 m long traveling at num__60 kmph takes to cross a bridge of num__170 m in length ? <o> a ) num__12.8 sec <o> b ) num__12.9 sec <o> c ) num__13.9 sec <o> d ) num__16.8 sec <o> e ) num__12.3 sec |
d = num__110 + num__170 = num__280 m s = num__60 * num__0.277777777778 = num__16.6666666667 t = num__280 * num__0.06 = num__16.8 sec answer : d <eor> d <eos> |
d |
add__110.0__170.0__ multiply__0.06__280.0__ round__16.8__ |
add__110.0__170.0__ multiply__0.06__280.0__ multiply__0.06__280.0__ |
| what is the least number which when divided by num__7 num__9 num__12 and num__18 leaves remainder num__4 in each care ? <o> a ) num__230 <o> b ) num__240 <o> c ) num__236 <o> d ) num__256 <o> e ) num__266 |
explanation : lcm of num__7 num__9 num__12 and num__18 is num__252 required number = num__252 + num__4 = num__256 answer : option d <eor> d <eos> |
d |
add__4.0__252.0__ add__4.0__252.0__ |
add__4.0__252.0__ add__4.0__252.0__ |
| how many multiples of num__10 are there between num__10 and num__290 ? <o> a ) num__8 <o> b ) num__18 <o> c ) num__27 <o> d ) num__51 <o> e ) num__60 |
it should be mentioned whether num__10 and num__290 are inclusive . if num__10 and num__290 are inclusive then the answer is ( num__290 - num__10 ) / num__10 + num__1 = num__29 . if num__10 and num__290 are not inclusive then the answer is ( num__280 - num__20 ) / num__10 + num__1 = num__27 . since oa is c then we have not inclusive case . <eor> c <eos> |
c |
divide__290.0__10.0__ subtract__290.0__10.0__ multiply__1.0__27.0__ |
divide__290.0__10.0__ subtract__290.0__10.0__ divide__27.0__1.0__ |
| the length e of a rectangle is decreased by num__15.0 and its width is increased by num__40.0 . does the area of the rectangle decrease or increase and by what percent ? <o> a ) decreases by num__19.0 <o> b ) decreases by num__25.0 <o> c ) increases by num__6.0 <o> d ) increases by num__19.0 <o> e ) increases by num__25 % |
let the length e of the rectangle be num__100 x and width be num__100 y . area = num__100 x * num__100 y = num__10000 xy now after the change length = num__85 x and width = num__140 y . area = num__11900 xy % change = ( num__11900 xy - num__10000 xy ) / ( num__10000 xy ) = num__19.0 increase . hence d . <eor> d <eos> |
d |
percent__100.0__19.0__ |
percent__100.0__19.0__ |
| p can do a work in num__24 days . q can do the same work in num__9 days and r can do the same in num__12 days . q and r start the work and leave after num__3 days . p finishes the remaining work in - - - days . <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__11 |
explanation : work done by p in num__1 day = num__0.0416666666667 work done by q in num__1 day = num__0.111111111111 work done by r in num__1 day = num__0.0833333333333 work done by q and r in num__1 day = num__0.111111111111 + num__0.0833333333333 = num__0.194444444444 work done by q and r in num__3 days = num__3 × num__0.194444444444 = num__0.583333333333 remaining work = num__1 – num__0.583333333333 = num__0.416666666667 number of days in which p can finish the remaining work = ( num__0.416666666667 ) / ( num__0.0416666666667 ) = num__10 answer : option d <eor> d <eos> |
d |
divide__1.0__24.0__ divide__1.0__9.0__ divide__1.0__12.0__ add__0.1111__0.0833__ subtract__1.0__0.5833__ add__9.0__1.0__ round__10.0__ |
divide__1.0__24.0__ divide__1.0__9.0__ divide__1.0__12.0__ add__0.1111__0.0833__ subtract__1.0__0.5833__ add__9.0__1.0__ add__9.0__1.0__ |
| find the mean proportional between num__64 & num__81 ? <o> a ) a ) num__59 <o> b ) b ) num__61 <o> c ) c ) num__63 <o> d ) d ) num__72 <o> e ) e ) num__67 |
formula = √ a × b a = num__64 and b = num__81 √ num__64 × num__81 = num__8 × num__9 = num__72 d <eor> d <eos> |
d |
add__64.0__8.0__ add__64.0__8.0__ |
add__64.0__8.0__ add__64.0__8.0__ |
| set a { num__3 num__33 num__45 num__55 } has a standard deviation of num__1 . what will the standard deviation be if every number in the set is multiplied by num__2 ? <o> a ) a ) num__1 <o> b ) b ) num__2 <o> c ) c ) num__4 <o> d ) d ) num__8 <o> e ) e ) num__16 |
points to remember - num__1 . if oneadd / subtractthe same amont from every term in a set sd does n ' t change . num__2 . if onemultiply / divideevery term by the same number in a set sd changes by same number . hence the answer to the above question is b <eor> b <eos> |
b |
subtract__3.0__1.0__ |
subtract__3.0__1.0__ |
| in the xy - plane the points ( c d ) ( c - d ) and ( - c - d ) are three vertices of a certain square . if c < num__0 and d > num__0 which of the following points n is in the same quadrant as the fourth vertex of the square ? <o> a ) ( - num__5 - num__3 ) <o> b ) ( - num__5 num__3 ) <o> c ) ( num__5 - num__3 ) <o> d ) ( num__3 - num__5 ) <o> e ) ( num__3 num__5 ) |
the question : in the xy - plane the points n = ( c d ) ( c - d ) and ( - c - d ) are three vertices of a certain square . if c < num__0 and d > num__0 which of the following points is in the same quadrant as the fourth vertex of the square ? i marked the tricky part in red . it seems c is anegativenumber and d is a positive number . this means vertex # num__1 = ( c d ) is in qii ( that is negative x and positive y ) vertex # num__2 = ( c - d ) is in qiii ( that is both xy negative ) vertex # num__3 = ( - c - d ) is in qiv ( that is y is negative but x is positive ) that means the last vertex should be in the first quadrant - - - the only first quadrant point is ( num__5 num__3 ) answer = e . <eor> e <eos> |
e |
add__1.0__2.0__ add__2.0__3.0__ add__1.0__2.0__ |
add__1.0__2.0__ add__2.0__3.0__ subtract__5.0__2.0__ |
| num__0.05 x num__0.0173 is equal to : <o> a ) num__8.65 x num__10 ( power - num__4 ) <o> b ) num__6.84 x num__10 ( power - num__4 ) <o> c ) num__4.68 x num__10 ( power - num__4 ) <o> d ) num__5.48 x num__10 ( power - num__4 ) <o> e ) none of them |
num__5 x num__173 = num__865 . sum of decimal places = num__6 . so num__0.05 x num__0.0173 = num__0.000865 = num__8.65 x num__10 ( power - num__4 ) answer is a . <eor> a <eos> |
a |
multiply__5.0__173.0__ multiply__0.05__0.0173__ multiply__0.05__173.0__ subtract__10.0__6.0__ multiply__0.05__173.0__ |
multiply__5.0__173.0__ multiply__0.05__0.0173__ multiply__0.05__173.0__ subtract__10.0__6.0__ multiply__0.05__173.0__ |
| a loan of rs . num__700 is made at num__9.5 simple interest for num__4 months . how much interest is owed when the loan is due ? <o> a ) num__22.167 <o> b ) num__23.167 <o> c ) num__24.167 <o> d ) num__25.167 <o> e ) num__26.167 |
si = p * r * t / num__100 = num__700 * num__9.5 * num__0.04 * num__12 = num__22.167 answer : a <eor> a <eos> |
a |
percent__100.0__22.167__ |
percent__100.0__22.167__ |
| the current of a stream at num__1 kmph . a motor boat goes num__35 km upstream and back to the starting point in num__12 hours . the speed of the motor boat in still water is ? <o> a ) num__12 <o> b ) num__77 <o> c ) num__66 <o> d ) num__88 <o> e ) num__94 |
s = num__1 m = x ds = x + num__1 us = x - num__1 num__35 / ( x + num__1 ) + num__35 / ( x - num__1 ) = num__12 x = num__6 answer : a <eor> a <eos> |
a |
round__12.0__ |
divide__12.0__1.0__ |
| if x ^ num__4 + y ^ num__4 = num__20 then the greatest possible value of x is between : <o> a ) num__0 to num__2 <o> b ) num__2 to num__4 <o> c ) num__4 to num__6 <o> d ) num__6 to num__8 <o> e ) num__8 to num__10 |
the answers to this question provide a great ' hint ' as to how to go about solving it ; since they ' re all essentially ' ranges ' you can use them to figure out which solution contains the maximum value of x . we ' re told that x ^ num__4 + y ^ num__4 = num__20 . to maximize the value of x we need to minimize the value of y ^ num__4 . the smallest that y ^ num__4 could be is num__0 ( when y = num__0 ) so we ' ll have . . . . x ^ num__4 = num__20 looking at the answers it makes sense to see what num__2 ^ num__4 equals . . . . num__2 ^ num__4 = num__16 since that is below num__20 and num__3 ^ num__4 will clearly be more than num__20 we have the correct answer . final answer : a <eor> a <eos> |
a |
subtract__20.0__4.0__ multiply__4.0__0.0__ |
subtract__20.0__4.0__ multiply__4.0__0.0__ |
| a number is doubled and num__5 is added . if the resultant is trebled it becomes num__105 . what is that number ? <o> a ) num__9 <o> b ) num__15 <o> c ) num__18 <o> d ) none of these <o> e ) can not be determined |
explanation : let the number be x . therefore num__3 ( num__2 x + num__5 ) = num__105 num__6 x + num__15 = num__105 num__6 x = num__90 x = num__15 answer : b <eor> b <eos> |
b |
subtract__5.0__3.0__ multiply__2.0__3.0__ multiply__5.0__3.0__ subtract__105.0__15.0__ multiply__5.0__3.0__ |
subtract__5.0__3.0__ multiply__2.0__3.0__ multiply__5.0__3.0__ subtract__105.0__15.0__ multiply__5.0__3.0__ |
| a num__270 metres long train running at the speed of num__120 kmph crosses another train running in opposite direction at the speed of num__80 kmph in num__9 seconds . what is the length of the other train ? <o> a ) num__253 m <o> b ) num__239 m <o> c ) num__236 m <o> d ) num__240 m <o> e ) num__230 m |
relative speed = ( num__120 + num__80 ) km / hr num__200 x num__0.277777777778 = num__55.5555555556 m / sec let the length of the other train be x metres . then x + num__30.0 = num__55.5555555556 x + num__270 = num__500 x = num__230 m answer e <eor> e <eos> |
e |
add__120.0__80.0__ divide__270.0__9.0__ add__200.0__30.0__ round__230.0__ |
add__120.0__80.0__ divide__270.0__9.0__ add__200.0__30.0__ add__200.0__30.0__ |
| a jar full of whisky contains num__40.0 alcohol . a part of this whisky is replaced by another containg num__19.0 alcohol and now the percentage of alcohol was found to be num__25.0 . what quantity of whisky is replaced ? <o> a ) num__0.333333333333 <o> b ) num__0.666666666667 <o> c ) num__0.4 <o> d ) num__0.6 <o> e ) num__0.8 |
let us assume the total original amount of whiskey = num__10 ml - - - > num__4 ml alcohol and num__6 ml non - alcohol . let x ml be the amount removed - - - > total alcohol left = num__4 - num__0.4 x new quantity of whiskey added = x ml out of which num__0.19 is the alcohol . thus the final quantity of alcohol = num__4 - num__0.4 x + num__0.19 x - - - - > ( num__4 - num__0.21 x ) / num__10 = num__0.26 - - - > x = num__6.66666666667 ml . per the question you need to find the x ml removed as a ratio of the initial volume - - - > ( num__6.66666666667 ) / num__10 = num__0.4 . hence c is the correct answer . <eor> c <eos> |
c |
percent__40.0__25.0__ percent__40.0__10.0__ percent__4.0__10.0__ percent__4.0__10.0__ |
percent__40.0__25.0__ percent__40.0__10.0__ percent__4.0__10.0__ percent__4.0__10.0__ |
| two trains start from two opposite directions towards each other . the stations from which they start are num__50 miles apart . both the trains start at the same time on a single track . a falcon which is sitting on one train . starts at the same time towards the other train . as soon as it reaches the second one it flies back to the first train and so on and so forth . it continues to do so flying backwards and forwards from one train to the other until the trains meet . both the trains travel at a speed of num__25 miles per hour and the bird flies at num__100 miles per hour . how many miles will the falcon have flown before the trains meet ? <o> a ) num__100 miles <o> b ) num__140 miles <o> c ) num__107 miles <o> d ) num__230 miles <o> e ) num__321 miles |
a num__100 miles the train travel at num__25 miles per hour . therefore they will meet after travelling for one hour and the falcon also must have been flying for one hour . since it travel at num__100 miles per hour the bird must have flown num__100 miles . <eor> a <eos> |
a |
round__100.0__ |
round__100.0__ |
| the edges of a cuboid are num__2 cm num__5 cm and num__8 cm . find the volume of the cuboid ? <o> a ) num__90 <o> b ) num__80 <o> c ) num__40 <o> d ) num__120 <o> e ) num__70 |
num__2 * num__5 * num__8 = num__80 answer : b <eor> b <eos> |
b |
round__80.0__ |
round__80.0__ |
| a computer system uses alphanumeric case sensitive characters for its passwords . when the system was created it required users to create passwords having num__6 characters in length . this year it added the option of creating passwords having num__5 characters in length . which of the following gives the expression for the total number of passwords the new computer system can accept ? assume there are num__62 unique alphanumeric case sensitive characters . <o> a ) num__63 ^ num__4 <o> b ) num__62 ^ num__5 <o> c ) num__62 ( num__62 ^ num__4 ) <o> d ) num__63 ( num__62 ^ num__5 ) <o> e ) num__63 ( num__62 ^ num__6 ) |
total number of passwords = number of num__6 character password + number of num__5 character password = num__62 ^ num__6 + num__62 ^ num__5 ( since there is no limitation on repetition each character can be chosen in num__62 ways ) = num__62 ^ num__5 ( num__1 + num__62 ) = num__62 ^ num__5 * num__63 answer d <eor> d <eos> |
d |
subtract__6.0__5.0__ add__62.0__1.0__ add__62.0__1.0__ |
subtract__6.0__5.0__ add__62.0__1.0__ add__62.0__1.0__ |
| num__6 in covering a distance of num__30 km abhay takes num__2 hours more than sameer . if abhay doubles his speed then he would take num__1 hour less than sameer . abhay ' s speed is : <o> a ) num__5 kmph <o> b ) num__6 kmph <o> c ) num__6.25 kmph <o> d ) num__7.5 kmph <o> e ) num__7.8 kmph |
let abhay ' s speed be x km / hr . then num__30 / x - num__15.0 x = num__3 num__6 x = num__30 x = num__5 km / hr . answer : option a <eor> a <eos> |
a |
divide__30.0__2.0__ divide__6.0__2.0__ subtract__6.0__1.0__ round__5.0__ |
divide__30.0__2.0__ divide__6.0__2.0__ subtract__6.0__1.0__ subtract__6.0__1.0__ |
| the milk level in a rectangular box measuring num__60 feet by num__25 feet is to be lowered by num__6 inches . how many gallons of milk must be removed ? ( num__1 cu ft = num__7.5 gallons ) <o> a ) num__100 <o> b ) num__250 <o> c ) num__750 <o> d ) num__5625 <o> e ) num__5635 |
num__6 inches = num__0.5 feet ( there are num__12 inches in a foot . ) so num__60 * num__25 * num__0.5 = num__750 feet ^ num__3 of milk must be removed which equals to num__750 * num__7.5 = num__5625 gallons . answer : d . <eor> d <eos> |
d |
divide__6.0__0.5__ multiply__6.0__0.5__ multiply__7.5__750.0__ multiply__1.0__5625.0__ |
divide__6.0__0.5__ multiply__6.0__0.5__ multiply__7.5__750.0__ multiply__1.0__5625.0__ |
| if { x } is the product of all even integers from num__1 to x inclusive what is the greatest prime factor of { num__12 } + { num__10 } ? <o> a ) num__23 <o> b ) num__20 <o> c ) num__13 <o> d ) num__5 <o> e ) num__2 |
soln : { num__12 } + { num__10 } = num__12 * { num__10 } + { num__10 } = num__13 * { num__10 } answer : c <eor> c <eos> |
c |
add__1.0__12.0__ add__1.0__12.0__ |
add__1.0__12.0__ add__1.0__12.0__ |
| a batsman scored num__120 runs which included num__5 boundaries and num__5 sixes . what percent of his total score did he make by running between the wickets ? <o> a ) num__45 <o> b ) num__58.33 <o> c ) num__54.5454545455 <o> d ) num__55 <o> e ) num__35 |
explanation : number of runs made by running = > num__120 − ( num__5 × num__4 + num__5 × num__6 ) . = > num__120 − ( num__50 ) . = > num__70 . hence the required percentage is : - = > num__0.583333333333 * num__100 = > num__58.33 answer : b <eor> b <eos> |
b |
subtract__120.0__50.0__ divide__70.0__120.0__ multiply__0.5833__100.0__ multiply__0.5833__100.0__ |
subtract__120.0__50.0__ divide__70.0__120.0__ multiply__0.5833__100.0__ multiply__0.5833__100.0__ |
| if | x ^ num__2 − num__2 | = x which of the following could be the value of x ? <o> a ) - num__2 <o> b ) - num__1 <o> c ) num__0 <o> d ) num__1 <o> e ) num__4 |
the lhs is not negative so the rhs is also not negative . thus x > = num__0 . first let ' s assume that x ^ num__2 - num__2 is negative . - ( x ^ num__2 - num__2 ) = x x ^ num__2 + x - num__2 = num__0 ( x + num__2 ) ( x - num__1 ) = num__0 x = num__1 or x = - num__2 ( however x can not be negative . ) then x = num__1 is a possible value for x . the answer is d . <eor> d <eos> |
d |
reverse__1.0__ |
subtract__2.0__1.0__ |
| log num__3 n + log num__12 n what is num__3 digit number n that will be whole number <o> a ) num__6291 <o> b ) num__7292 <o> c ) num__1728 <o> d ) num__1929 <o> e ) num__1727 |
no of values n can take is num__1 num__12 ^ num__3 = num__1728 answer : c <eor> c <eos> |
c |
multiply__1728.0__1.0__ |
multiply__1728.0__1.0__ |
| a cycle is bought for rs . num__900 and sold for rs . num__1080 find the gain percent ? <o> a ) num__27.0 <o> b ) num__20.0 <o> c ) num__80.0 <o> d ) num__30.0 <o> e ) num__24 % |
num__900 - - - - num__180 num__100 - - - - ? = > num__20.0 answer : b <eor> b <eos> |
b |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| lisa and robert have taken the same number of photos on their school trip . lisa has taken num__3 times as many photos as claire and robert has taken num__24 more photos than claire . how many photos has claire taken ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__12 <o> e ) num__14 |
l = r l = num__3 c r = c + num__24 num__3 c = c + num__24 c = num__12 the answer is d . <eor> d <eos> |
d |
subtract__24.0__12.0__ |
subtract__24.0__12.0__ |
| if r = num__2 ^ num__4 * num__5 ^ num__2 * num__7 and s = num__2 ^ num__2 * num__3 ^ num__2 * num__5 which of the following is equal to the greatest common divisor of r and s ? <o> a ) num__2 * num__5 <o> b ) num__2 ^ num__2 * num__5 <o> c ) num__2 ^ num__3 * num__5 ^ num__2 <o> d ) num__2 * num__3 * num__5 * num__7 <o> e ) num__2 ^ num__3 * num__3 ^ num__2 * num__5 ^ num__2 * num__7 |
gcd = product of prime factors raised to the least power = num__2 ^ num__2 * num__5 the answer is b . <eor> b <eos> |
b |
subtract__4.0__2.0__ |
subtract__4.0__2.0__ |
| a train speeds past a pole in num__25 seconds and a platform num__200 m long in num__30 seconds . its length is : <o> a ) num__1200 m <o> b ) num__2000 m <o> c ) num__1500 m <o> d ) num__1000 m <o> e ) num__4000 m |
let the length of the train be x meters and its speed be y m / sec . they x / y = num__25 = > y = x / num__25 x + num__6.66666666667 = x / num__25 x = num__1000 m . answer : option d <eor> d <eos> |
d |
divide__200.0__30.0__ round__1000.0__ |
divide__200.0__30.0__ round__1000.0__ |
| eight cubes each with a volume of num__512 cm ^ num__3 are joined to form one large cube . what is the surface area of the large cube ? <o> a ) num__4096 sq cm <o> b ) num__1536 sq cm <o> c ) num__1024 sq cm <o> d ) num__2048 sq cm <o> e ) num__512 sq cm |
volume = a x a x a = a ^ num__3 volume of cube = num__8 x num__512 = num__4096 cm ^ num__3 num__4096 = a ^ num__3 a = num__16 surface area of cube = num__6 a ^ num__2 = num__6 x num__16 ^ num__2 = num__1536 answer is b <eor> b <eos> |
b |
multiply__512.0__8.0__ surface_cube__16.0__ surface_cube__16.0__ |
multiply__512.0__8.0__ surface_cube__16.0__ surface_cube__16.0__ |
| what amount does kiran get if he invests rs . num__9800 at num__12.0 p . a . compound interest for three years compounding done annually ? <o> a ) num__13456 <o> b ) num__13487 <o> c ) num__18978 <o> d ) num__13456 <o> e ) num__12308 |
a = p { num__1 + r / num__100 } n = > num__9600 { num__1 + num__0.12 } num__3 = rs . num__13487 answer : b <eor> b <eos> |
b |
percent__12.0__1.0__ percent__100.0__13487.0__ |
percent__12.0__1.0__ percent__100.0__13487.0__ |
| a and b together can do a work in num__3 days . if a alone can do it in num__6 days . in how many days can b alone do it ? <o> a ) num__10 <o> b ) num__99 <o> c ) num__77 <o> d ) num__55 <o> e ) num__6 |
num__0.333333333333 – num__0.166666666667 = num__0.166666666667 = > num__6 answer : e <eor> e <eos> |
e |
round__6.0__ |
round__6.0__ |
| w q and r are positive integers . if w q and r are assembled into the six - digit number wqrwqr which one of the following must be a factor of wqrwqr ? <o> a ) num__23 <o> b ) num__19 <o> c ) num__17 <o> d ) num__7 <o> e ) none of the above |
one short way - wqrwqr = num__1000 wqr + wqr = ( num__1000 + num__1 ) wqr = num__1001 wqr therefore any factor of num__1001 is a factor of wqrwqr num__7 is a factor of num__1001 so d <eor> d <eos> |
d |
add__1000.0__1.0__ multiply__1.0__7.0__ |
add__1000.0__1.0__ multiply__1.0__7.0__ |
| express a speed of num__84 kmph in meters per second ? <o> a ) num__13.33 mps <o> b ) num__23.33 mps <o> c ) num__33.33 mps <o> d ) num__25.33 mps <o> e ) num__43.33 mps |
num__84 * num__0.277777777778 = num__23.33 mps answer : b <eor> b <eos> |
b |
round__23.33__ |
round__23.33__ |
| a car averages num__30 miles per hour for the first num__5 hours of a trip and averages num__42 miles per hour for each additional hour of travel time . if the average speed for the entire trip is num__38 miles per hour how many hours long is the trip ? <o> a ) num__13 <o> b ) num__14 <o> c ) num__15 <o> d ) num__16 <o> e ) num__17 |
let t be the total time of the trip . num__30 * num__5 + num__42 ( t - num__5 ) = num__38 t num__4 t = num__210 - num__150 t = num__15 the answer is c . <eor> c <eos> |
c |
subtract__42.0__38.0__ multiply__5.0__42.0__ multiply__30.0__5.0__ round__15.0__ |
subtract__42.0__38.0__ multiply__5.0__42.0__ multiply__30.0__5.0__ subtract__30.0__15.0__ |
| what approximate value should come in place of question mark ( ? ) in the following equation ? num__33 num__1 ⁄ num__3.0 of num__768.9 + num__25.0 of num__161.2 – num__65.12 = ? <o> a ) num__230 <o> b ) num__225 <o> c ) num__235 <o> d ) num__233 <o> e ) num__240 |
? = num__33 num__1 ⁄ num__3.0 of num__768.9 + num__25.0 of num__161.2 – num__65.12 = num__1 ⁄ num__3 of num__768.9 + num__1 ⁄ num__4 of num__161.2 – num__65.12 = num__256.3 + num__40.3 – num__65.12 ≈ num__233 answer d <eor> d <eos> |
d |
add__1.0__3.0__ divide__768.9__3.0__ divide__161.2__4.0__ multiply__1.0__233.0__ |
add__1.0__3.0__ divide__768.9__3.0__ divide__161.2__4.0__ multiply__1.0__233.0__ |
| a cashier mentally reversed the digits of one customer ' s correct amount of change and thus gave the customer an incorrect amount of change . if the cash register contained num__63 cents more than it should have as a result of this error which of the following could have been the correct amount of change in cents ? <o> a ) num__14 <o> b ) num__45 <o> c ) num__54 <o> d ) num__65 <o> e ) num__81 |
just check the answers and reverse the numbers until you get num__63 . num__81 - num__18 = num__63 answer e <eor> e <eos> |
e |
subtract__81.0__63.0__ add__63.0__18.0__ |
subtract__81.0__63.0__ add__63.0__18.0__ |
| the distance light travels in one year is approximately num__5 num__870000 num__000000 miles . the distance light travels in num__100 years is : <o> a ) num__587 × num__10 ^ num__8 miles <o> b ) num__587 × num__10 ^ num__10 miles <o> c ) num__587 × num__10 ^ - num__10 miles <o> d ) num__587 × num__10 ^ num__12 ml <o> e ) num__587 × num__10 ^ - num__12 miles |
solution : the distance of the light travels in num__100 years is : num__5 num__870000 num__000000 × num__100 miles . = num__587 num__000000 num__000000 miles . = num__587 × num__10 ^ num__12 miles . answer : ( d ) <eor> d <eos> |
d |
round__587.0__ |
round__587.0__ |
| num__16 machines can do a work in num__10 days . how many machines are needed to complete the work in num__40 days ? <o> a ) num__10 <o> b ) num__6 <o> c ) num__4 <o> d ) num__7 <o> e ) num__5 |
required number of machines = num__16 * num__0.25 = num__4 answer is c <eor> c <eos> |
c |
divide__10.0__40.0__ multiply__16.0__0.25__ round__4.0__ |
divide__10.0__40.0__ multiply__16.0__0.25__ multiply__16.0__0.25__ |
| rs . num__385 were divided among x y z in such a way that x had rs . num__20 more than y and z had rs num__15 more than x . how much was y ’ s share ? <o> a ) rs . num__110 <o> b ) rs . num__145 <o> c ) rs . num__150 <o> d ) rs . num__176 <o> e ) rs . num__181 |
let y gets rs x . then we can say x gets rs ( x + num__20 ) and y gets rs ( x + num__35 ) . x + num__20 + x + x + num__35 = num__385 num__3 x = num__330 x = num__110 . r ’ s share = rs ( num__110 + num__35 ) = rs . num__145 b <eor> b <eos> |
b |
add__20.0__15.0__ divide__330.0__3.0__ add__110.0__35.0__ add__110.0__35.0__ |
add__20.0__15.0__ divide__330.0__3.0__ add__110.0__35.0__ add__110.0__35.0__ |
| on a certain day orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ num__0.60 per glass on the first day what was the price e per glass on the second day ? <o> a ) $ num__015 <o> b ) $ num__0.20 <o> c ) $ num__0.30 <o> d ) $ num__0.40 <o> e ) $ num__0.45 |
on the first day num__1 unit of orange juice and num__1 unit of water was used to make num__2 units of orangeade ; on the second day num__1 unit of orange juice and num__2 units of water was used to make num__3 units of orangeade ; so the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is num__2 to num__3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is num__2 to num__3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from num__2 glasses on the first day equals to the revenue from num__3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then num__2 * num__0.6 = num__3 * x - - > x = $ num__0.4 . answer : d . <eor> d <eos> |
d |
add__1.0__2.0__ subtract__1.0__0.6__ subtract__1.0__0.6__ |
add__1.0__2.0__ subtract__1.0__0.6__ subtract__1.0__0.6__ |
| the cross - section of a cannel is a trapezium in shape . if the cannel is num__10 m wide at the top and num__6 m wide at the bottom and the area of cross - section is num__640 sq m the depth of cannel is ? <o> a ) num__29 <o> b ) num__19 <o> c ) num__17 <o> d ) num__80 <o> e ) num__14 |
num__0.5 * d ( num__10 + num__6 ) = num__640 d = num__80 answer : d <eor> d <eos> |
d |
round__80.0__ |
round__80.0__ |
| which of the following values of a satisfies the inequality ? a ( a – num__2 ) = num__5 a – num__10 <o> a ) num__2 < a = num__5 <o> b ) num__2 < a = num__8 <o> c ) num__2 < a = num__6 <o> d ) num__2 < a = num__1 <o> e ) num__2 < a = num__2 |
explanation : a ( a – num__2 ) = num__5 a – num__10 a ( a – num__2 ) = num__5 ( a – num__2 ) a = num__5 if a – num__2 > num__0 a = num__5 a > num__2 i . e . num__2 < a = num__5 answer : a <eor> a <eos> |
a |
divide__10.0__5.0__ |
divide__10.0__5.0__ |
| out of the three numbers a b and c a exceeds b by num__20 and c exceeds a by num__55 . if the sum of all the numbers is num__230 what is the difference between the largest and the smallest number ? <o> a ) num__52 <o> b ) num__58 <o> c ) num__63 <o> d ) num__75 <o> e ) num__76 |
explanation : we have a = b + num__20 & c = a + num__55 = b + num__75 thus ( b + num__20 ) + ( b ) + ( b + num__75 ) = num__230 b = num__45 thus a = num__65 & c = num__120 . hence ( c ) – ( b ) = num__120 – num__45 = num__75 short - cut method : since c is larger than a by num__55 and a is larger than b by num__20 we get the required difference = num__55 + num__20 = num__75 ( without calculating the actual values of a b and c ) . answer d <eor> d <eos> |
d |
add__20.0__55.0__ add__20.0__45.0__ add__55.0__65.0__ add__20.0__55.0__ |
add__20.0__55.0__ add__20.0__45.0__ add__55.0__65.0__ add__20.0__55.0__ |
| the ratio of number of boys and girls in a school is num__2 : num__5 . if there are num__350 students in the school find the number of girls in the school ? <o> a ) num__150 <o> b ) num__250 <o> c ) num__300 <o> d ) num__370 <o> e ) num__280 |
let the number of boys and girls be num__2 x and num__5 x total students = num__350 number of girls in the school = num__5 * num__50.0 = num__250 answer is b <eor> b <eos> |
b |
multiply__5.0__50.0__ multiply__5.0__50.0__ |
multiply__5.0__50.0__ multiply__5.0__50.0__ |
| two numbers are in the ratio num__3 : num__5 . if num__9 be subtracted from each they are in the ratio of num__9 : num__17 . the first number is ? <o> a ) num__36 <o> b ) num__77 <o> c ) num__89 <o> d ) num__16 <o> e ) num__22 |
( num__3 x - num__9 ) : ( num__5 x - num__9 ) = num__9 : num__17 x = num__12 = > num__3 x = num__36 answer : a <eor> a <eos> |
a |
add__3.0__9.0__ multiply__3.0__12.0__ multiply__3.0__12.0__ |
subtract__17.0__5.0__ multiply__3.0__12.0__ multiply__3.0__12.0__ |
| a can give b num__100 meters start and c num__200 meters start in a kilometer race . how much start can b give c in a kilometer race ? <o> a ) num__111.12 <o> b ) num__111.18 <o> c ) num__111.11 <o> d ) num__111.98 <o> e ) num__111.15 |
a runs num__1000 m while b runs num__900 m and c runs num__800 m . the number of meters that c runs when b runs num__1000 m = ( num__1000 * num__800 ) / num__900 = num__888.888888889 = num__888.88 m . b can give c = num__1000 - num__888.88 = num__111.12 m . answer : a <eor> a <eos> |
a |
subtract__1000.0__100.0__ subtract__900.0__100.0__ subtract__1000.0__888.88__ round__111.12__ |
subtract__1000.0__100.0__ subtract__900.0__100.0__ subtract__1000.0__888.88__ subtract__1000.0__888.88__ |
| the sum of the digits of a two - digit number is num__12 . the difference of the digits is num__6 . find the number ? <o> a ) num__12 <o> b ) num__76 <o> c ) num__21 <o> d ) num__26 <o> e ) num__39 |
explanation : let the two - digit number be num__10 a + b a + b = num__12 - - - ( num__1 ) if a > b a - b = num__6 if b > a b - a = num__6 if a - b = num__6 adding it to equation ( num__1 ) we get num__2 a = num__18 = > a = num__9 so b = num__12 - a = num__3 number would be num__93 . if b - a = num__6 adding it to the equation ( num__1 ) we get num__2 b = num__18 = > b = num__9 a = num__12 - b = num__3 . number would be num__39 . there fore number would be num__39 or num__93 . answer : e <eor> e <eos> |
e |
divide__12.0__6.0__ add__12.0__6.0__ subtract__10.0__1.0__ subtract__12.0__9.0__ multiply__1.0__39.0__ |
subtract__12.0__10.0__ add__12.0__6.0__ subtract__10.0__1.0__ subtract__12.0__9.0__ multiply__1.0__39.0__ |
| a person ' s salary is getting reduced by num__20.0 . what percentage should be added to get back his original salary ? <o> a ) num__25.0 <o> b ) num__50.0 <o> c ) num__75.0 <o> d ) num__85.0 <o> e ) num__95 % |
salary he get after reduced num__20.0 = num__100 - num__20.0 = num__80.0 so percentage should be added to get back his salary = ( num__0.25 ) * num__100 = num__25.0 answer : a <eor> a <eos> |
a |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| calculate the circumference of a circular field whose radius is num__10 centimeters . <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__20 |
circumference c is given by c = num__2 Ï € r = num__2 Ï € * num__10 = num__20 Ï € cm correct answer e <eor> e <eos> |
e |
multiply__10.0__2.0__ multiply__10.0__2.0__ |
multiply__10.0__2.0__ multiply__10.0__2.0__ |
| if num__10 men do a work in num__80 days in how many days will num__20 men do it ? <o> a ) num__18 days <o> b ) num__38 days <o> c ) num__40 days <o> d ) num__48 days <o> e ) num__50 days |
num__10 * num__80 = num__20 * x x = num__40 days answer : c <eor> c <eos> |
c |
round__40.0__ |
round__40.0__ |
| a train is num__360 meter long is running at a speed of num__45 km / hour . in what time will it pass a bridge of num__140 meter length ? <o> a ) num__65 seconds <o> b ) num__86 seconds <o> c ) num__40 seconds <o> d ) num__43 seconds <o> e ) num__15 seconds |
speed = num__45 km / hr = num__45 * ( num__0.277777777778 ) m / sec = num__12.5 m / sec total distance = num__360 + num__140 = num__500 meter time = distance / speed = num__500 * ( num__0.08 ) = num__40 seconds answer : c <eor> c <eos> |
c |
add__360.0__140.0__ divide__500.0__12.5__ round__40.0__ |
add__360.0__140.0__ divide__500.0__12.5__ divide__500.0__12.5__ |
| pooja is twice as efficient as aarti and takes num__90 days less than aarti to complete the job . find the time in which they can finish the job together . <o> a ) num__30 days <o> b ) num__45 days <o> c ) num__60 days <o> d ) num__90 days <o> e ) none of these |
explanation : hint : assume that pooja completes the job in ' x ' days . so aarti will take ' num__2 x ' days to complete the same job . as pooja takes num__90 days less than aarti we get x = num__2 x – num__90 by solving this equation we get x = num__90 . thus num__2 x = num__2 x num__90 = num__180 part of job done by pooja in num__1 day = num__0.0111111111111 part of job done by aarti in num__1 day = num__0.00555555555556 ( part of job done together in num__1 day ) = ( part of job done by pooja in num__1 day ) + ( part of job done by aarti in num__1 day ) = ( num__0.0111111111111 ) + ( num__0.00555555555556 ) = num__0.0166666666667 = num__0.0166666666667 ( num__0.0166666666667 ) th part of whole job will be completed by pooja and aarti together in one day . therefore the whole job will be completed in num__60 days together . answer is c <eor> c <eos> |
c |
multiply__90.0__2.0__ divide__1.0__90.0__ divide__1.0__180.0__ add__0.0056__0.0111__ hour_to_min_conversion__ hour_to_min_conversion__ |
multiply__90.0__2.0__ divide__1.0__90.0__ divide__1.0__180.0__ add__0.0056__0.0111__ hour_to_min_conversion__ hour_to_min_conversion__ |
| haris can wash all the windows of his house in num__10 hours . his wife maggie can wash all the windows in num__5 hours . how many hours will it take for both of them working together to wash all the windows ? <o> a ) num__2 <o> b ) num__3 num__0.333333333333 <o> c ) num__4 num__0.25 <o> d ) num__5 <o> e ) num__6 |
work hrs = ab / ( a + b ) = num__3.33333333333 = num__3 num__0.333333333333 answer is b <eor> b <eos> |
b |
divide__10.0__3.3333__ divide__3.3333__10.0__ round__3.0__ |
divide__10.0__3.3333__ divide__3.3333__10.0__ divide__10.0__3.3333__ |
| what is the smallest positive integer x such that √ num__392 x is an integer ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__7 <o> d ) num__8 <o> e ) num__14 |
for a perfect square the powers of prime factors should be even . num__392 can be broken up as num__2 ^ num__3 * num__7 ^ num__2 . you just need num__2 to be multiplied to this number so that the number finally becomes num__2 ^ num__4 * num__7 ^ num__2 which is a perfect square . therefore x = num__2 . answer : a <eor> a <eos> |
a |
subtract__7.0__3.0__ subtract__4.0__2.0__ |
subtract__7.0__3.0__ subtract__4.0__2.0__ |
| if the cost price of num__110 chairs is equal to the selling price of num__88 chairs the gain percent is <o> a ) num__25 <o> b ) num__30 <o> c ) num__35 <o> d ) num__40 <o> e ) num__45 |
let c . p . of each pen be re . num__1 . then c . p . of num__88 chairs = rs . num__88 ; s . p . of num__88 chairs = rs . num__110 . gain % = num__0.25 * num__100 = num__25.0 answer : a <eor> a <eos> |
a |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| if two dice are thrown together the probability of getting an even number on one die and an odd number on the other is ? <o> a ) num__0.142857142857 <o> b ) num__0.5 <o> c ) num__0.111111111111 <o> d ) num__1.0 <o> e ) num__0.333333333333 |
the number of exhaustive outcomes is num__36 . let e be the event of getting an even number on one die and an odd number on the other . let the event of getting either both even or both odd then = num__0.5 = num__0.5 p ( e ) = num__1 - num__0.5 = num__0.5 . answer : b <eor> b <eos> |
b |
negate_prob__0.5__ |
negate_prob__0.5__ |
| sum of the squares of three numbers is num__225 and the sum of their products taken two at a time is num__200 . find the sum ? <o> a ) num__20 <o> b ) num__22 <o> c ) num__25 <o> d ) num__26 <o> e ) num__29 |
( a + b + c ) num__2 = a num__2 + b num__2 + c num__2 + num__2 ( ab + bc + ca ) = num__225 + num__2 * num__200 a + b + c = √ num__625 = num__25 c <eor> c <eos> |
c |
subtract__225.0__200.0__ subtract__225.0__200.0__ |
subtract__225.0__200.0__ subtract__225.0__200.0__ |
| if num__2 / z = num__2 / ( z + num__1 ) + num__2 / ( z + num__16 ) which of these integers could be the value of z ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
solving for z algebraically in this problem would not be easy . instead we can follow the hint in the question ( “ which of these integers … ” ) and test each answer choice : a . num__2 / num__0 = num__2.0 + num__0.125 incorrect ( division by zero ) a . num__2.0 = num__1.0 + num__0.117647058824 incorrect a . num__1.0 = num__0.666666666667 + num__0.111111111111 incorrect a . num__0.666666666667 = num__0.5 + num__0.105263157895 incorrect a . num__0.5 = num__0.4 + num__0.1 correct the correct answer is e because it contains the only value that makes the equation work . notice how quickly this strategy worked in this case <eor> e <eos> |
e |
divide__2.0__16.0__ reverse__2.0__ subtract__0.5__0.4__ divide__2.0__0.5__ |
divide__2.0__16.0__ reverse__2.0__ subtract__0.5__0.4__ divide__2.0__0.5__ |
| if ram and gohul can do a job in num__10 days and num__15 days independently . how many days would they take to complete the same job working simultaneously ? <o> a ) num__5 <o> b ) num__8 <o> c ) num__4 <o> d ) num__7 <o> e ) num__6 |
if total work is x . ram rate of working = x / num__10 per day . gohul rate of working = x / num__15 per day . rate of work = ( x / num__10 ) + ( x / num__15 ) = num__30 x / num__5 x = num__6 days answer is option e <eor> e <eos> |
e |
subtract__15.0__10.0__ divide__30.0__5.0__ round__6.0__ |
subtract__15.0__10.0__ divide__30.0__5.0__ divide__30.0__5.0__ |
| the average temperature of the town in the first four days of a month was num__58 degrees . the average for the second third fourth and fifth days was num__60 degrees . if the temperatures of the first and fifth days were in the ratio num__7 : num__8 then what is the temperature on the fifth day ? <o> a ) num__33 <o> b ) num__64 <o> c ) num__38 <o> d ) num__37 <o> e ) num__29 |
sum of temperatures on num__1 st num__2 nd num__3 rd and num__4 th days = ( num__58 * num__4 ) = num__232 degrees . . . ( num__1 ) sum of temperatures on num__2 nd num__3 rd num__4 th and num__5 th days - ( num__60 * num__4 ) = num__240 degrees . . . . ( num__2 ) subtracting ( num__1 ) from ( num__2 ) we get : temp on num__5 th day - temp on num__1 st day = num__8 degrees . let the temperatures on num__1 st and num__5 th days be num__7 x and num__8 x degrees respectively . then num__8 x - num__7 x = num__8 or x = num__8 . temperature on the num__5 th day = num__8 x = num__64 degrees . answer : b <eor> b <eos> |
b |
subtract__8.0__7.0__ subtract__60.0__58.0__ add__1.0__2.0__ subtract__7.0__3.0__ multiply__58.0__4.0__ subtract__7.0__2.0__ multiply__60.0__4.0__ add__60.0__4.0__ add__60.0__4.0__ |
subtract__8.0__7.0__ subtract__60.0__58.0__ add__1.0__2.0__ subtract__7.0__3.0__ multiply__58.0__4.0__ subtract__7.0__2.0__ multiply__60.0__4.0__ add__60.0__4.0__ multiply__64.0__1.0__ |
| a rectangular grassy plot num__100 m . by num__70 m has a gravel path num__2.5 m wide all round it on the inside . find the cost of gravelling the path at num__90 paise per sq . metre . <o> a ) rs . num__700 <o> b ) rs . num__708.50 <o> c ) rs . num__732.50 <o> d ) rs . num__742.50 <o> e ) none of these |
explanation : area of the plot = ( num__100 x num__70 ) m num__2 = num__7000 m num__2 area of the plot excluding the path = [ ( num__100 - num__5 ) * ( num__70 - num__5 ) ] m num__2 = num__6175 m num__2 . area of the path = ( num__7000 - num__6175 ) m num__2 = num__825 m num__2 . cost of gravelling the path = rs . num__825 * ( num__0.9 ) = rs . num__742.50 answer : option d <eor> d <eos> |
d |
multiply__100.0__70.0__ multiply__2.5__2.0__ subtract__7000.0__6175.0__ divide__90.0__100.0__ multiply__0.9__825.0__ round__742.5__ |
multiply__100.0__70.0__ multiply__2.5__2.0__ subtract__7000.0__6175.0__ divide__90.0__100.0__ multiply__0.9__825.0__ multiply__0.9__825.0__ |
| a man buys an article for num__13.0 less than its value and sells it for num__13.0 more than its value . his gain or loss percent is : <o> a ) no profit no loss <o> b ) num__20.0 profit <o> c ) less than num__25.0 profit <o> d ) more than num__25.0 profit <o> e ) none |
let the article be worth rs . x . c . p . num__87.0 of rs . x = rs . num__87 x / num__100 s . p . = num__113.0 of rs . x = rs . num__113 x / num__100 gain = ( num__113 x / num__100 - num__87 x / num__100 ) = rs . num__13 x / num__50 gain % = num__13 x / num__50 * num__1.14942528736 x * num__100 = num__29 num__0.885057471264 % > num__25.0 answer : d <eor> d <eos> |
d |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| the length of the bridge which a train num__130 m long and traveling at num__45 km / hr can cross in num__30 sec is ? <o> a ) num__5.0 <o> b ) num__8.33333333333 <o> c ) num__12.5 <o> d ) num__25.0 <o> e ) num__2.77777777778 |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec . time = num__30 sec let the length of bridge be x meters . then ( num__130 + x ) / num__30 = num__12.5 x = num__245 m . answer : c <eor> c <eos> |
c |
round__12.5__ |
round__12.5__ |
| a coin is tossed two times . what is the probability that there is at the least one tail ? <o> a ) num__0.75 <o> b ) num__0.837837837838 <o> c ) num__1.72222222222 <o> d ) num__3.1 <o> e ) num__1.63157894737 |
let p ( t ) be the probability of getting least one tail when the coin is tossed two times . = there is not even a single tail . i . e . all the outcomes are heads . = num__0.25 ; p ( t ) = num__1 - num__0.25 = num__0.75 answer : a <eor> a <eos> |
a |
negate_prob__0.25__ negate_prob__0.25__ |
negate_prob__0.25__ negate_prob__0.25__ |
| integer x is equal to the product of all even numbers from num__2 to num__60 inclusive . if q is the smallest prime number that is also a factor of x - num__1 then which of the following expressions must be true ? <o> a ) num__0 < y < num__4 <o> b ) num__4 < y < num__10 <o> c ) num__10 < y < num__20 <o> d ) num__20 < y < num__30 <o> e ) y > num__30 |
q smallest prime factor is greater than num__29 . so answer is e <eor> e <eos> |
e |
divide__60.0__2.0__ |
divide__60.0__2.0__ |
| p and q can complete a work in num__20 days and num__12 days respectively . p alone started the work and q joined him after num__4 days till the completion of the work . how long did the work last ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__12 <o> d ) num__15 <o> e ) num__17 |
work done by p in num__1 day = num__0.05 work done by q in num__1 day = num__0.0833333333333 work done by p in num__4 days = num__4 × ( num__0.05 ) = num__0.2 remaining work = num__1 – num__0.2 = num__0.8 work done by p and q in num__1 day = num__0.05 + num__0.0833333333333 = num__0.133333333333 = num__0.133333333333 number of days p and q take to complete the remaining work = ( num__0.8 ) / ( num__0.133333333333 ) = num__6 total days = num__4 + num__6 = num__10 answer is b . <eor> b <eos> |
b |
divide__1.0__20.0__ divide__1.0__12.0__ divide__4.0__20.0__ multiply__4.0__0.2__ add__0.05__0.0833__ add__4.0__6.0__ round__10.0__ |
divide__1.0__20.0__ divide__1.0__12.0__ divide__4.0__20.0__ multiply__4.0__0.2__ add__0.05__0.0833__ add__4.0__6.0__ add__4.0__6.0__ |
| [ ( num__4.5 ÷ num__0.5 × num__4.5 ) ] / [ ( num__4.5 × num__0.5 ÷ num__4.5 ) ] = ? <o> a ) num__71 <o> b ) num__61 <o> c ) num__51 <o> d ) num__81 <o> e ) num__41 |
explanation : [ ( num__4.5 ÷ num__0.5 × num__4.5 ) ] / [ ( num__4.5 × num__0.5 ÷ num__4.5 ) ] = [ num__4.5 × num__2.0 × num__4.5 ] / [ num__4.5 × num__0.5 × num__0.222222222222 ] = num__40.5 / num__0.5 = num__40.5 x num__2.0 = num__81 answer : option d <eor> d <eos> |
d |
reverse__0.5__ reverse__4.5__ multiply__2.0__40.5__ multiply__2.0__40.5__ |
reverse__0.5__ reverse__4.5__ divide__40.5__0.5__ divide__40.5__0.5__ |
| a fruit - salad mixture consists of apples peaches and grapes in the ratio num__6 : num__5 : num__2 respectively by weight . if num__78 pounds of the mixture is prepared the mixture includes how many more pounds of apples than grapes ? <o> a ) num__24 <o> b ) num__12 <o> c ) num__9 <o> d ) num__6 <o> e ) num__4 |
we can first set up our ratio using variable multipliers . we are given that a fruit - salad mixture consists of apples peaches and grapes in the ratio of num__6 : num__5 : num__2 respectively by weight . thus we can say : apples : peaches : grapes = num__6 x : num__5 x : num__2 x we are given that num__39 pounds of the mixture is prepared so we can set up the following question and determine a value for x : num__6 x + num__5 x + num__2 x = num__78 num__13 x = num__78 x = num__6 now we can determine the number of pounds of apples and of grapes . pounds of grapes = ( num__2 ) ( num__6 ) = num__12 pounds of apples = ( num__6 ) ( num__6 ) = num__36 thus we know that there are num__36 - num__12 = num__24 more pounds of apples than grapes . answer is a . <eor> a <eos> |
a |
divide__78.0__2.0__ divide__78.0__6.0__ multiply__6.0__2.0__ multiply__2.0__12.0__ multiply__2.0__12.0__ |
divide__78.0__2.0__ divide__78.0__6.0__ multiply__6.0__2.0__ subtract__36.0__12.0__ subtract__36.0__12.0__ |
| a number when divided by a divisor leaves a remainder of num__24 . when twice the original number is divided by the same divisor the remainder is num__11 . what is the value of the divisor ? <o> a ) num__35 <o> b ) num__37 <o> c ) num__41 <o> d ) num__43 <o> e ) num__44 |
let the original number be ' a ' let the divisor be ' d ' let the quotient of the division of aa by dd be ' x ' therefore we can write the relation as a / d = x and the remainder is num__24 . i . e . a = dx + num__24 when twice the original number is divided by d num__2 a is divided by d . we know that a = dx + num__24 . therefore num__2 a = num__2 dx + num__48 the problem states that ( num__2 dx + num__48 ) / d leaves a remainder of num__11 . num__2 dx num__2 dx is perfectly divisible by d and will therefore not leave a remainder . the remainder of num__11 was obtained by dividing num__48 by d . when num__48 is divided by num__37 the remainder that one will obtain is num__11 . hence the divisor is num__37 . b <eor> b <eos> |
b |
multiply__24.0__2.0__ subtract__48.0__11.0__ subtract__48.0__11.0__ |
multiply__24.0__2.0__ subtract__48.0__11.0__ subtract__48.0__11.0__ |
| in three years janice will be three times as old as her daughter . six years ago her age was her daughter ’ s age squared . how old is janice ' s daughter ? <o> a ) num__12 <o> b ) num__36 <o> c ) num__40 <o> d ) num__42 <o> e ) num__45 |
let jane ' s age be j and daughters ' age be d . . . given j + num__3 = num__3 ( d + num__3 ) = > j - num__6 = num__3 d - > eq num__1 . given j - num__6 = ( d - num__6 ) ^ num__2 - - > eq num__2 . . sub j - num__6 value in eq num__2 . . . num__3 d = d ^ num__2 - num__12 d + num__36 . num__0 = d ^ num__2 - num__15 d + num__36 = > d = num__12 or d = num__3 . when d = num__12 we get from eq num__1 . . . j + num__3 = num__45 = > j = num__42 . . when d = num__3 . . we get from eq num__1 . . j + num__3 = num__18 = > j = num__15 . . imo option a is correct answer . . <eor> a <eos> |
a |
subtract__3.0__1.0__ multiply__2.0__6.0__ multiply__3.0__12.0__ add__3.0__12.0__ multiply__3.0__15.0__ add__36.0__6.0__ multiply__3.0__6.0__ multiply__1.0__12.0__ |
subtract__3.0__1.0__ multiply__2.0__6.0__ multiply__3.0__12.0__ add__3.0__12.0__ multiply__3.0__15.0__ add__36.0__6.0__ add__3.0__15.0__ subtract__15.0__3.0__ |
| the cost of num__3 kg of sugar is $ num__12 . what will the cost of num__8 kg of sugar be ? <o> a ) num__32 <o> b ) num__48 <o> c ) num__64 <o> d ) num__120 <o> e ) num__180 |
num__3 kg costs $ num__12 num__1 kg costs $ num__4 num__8 kg costs $ num__32 the answer is a . <eor> a <eos> |
a |
add__3.0__1.0__ multiply__8.0__4.0__ multiply__8.0__4.0__ |
add__3.0__1.0__ multiply__8.0__4.0__ multiply__8.0__4.0__ |
| a boat goes num__8 km upstream in num__24 minutes . the speed of stream is num__4 km / hr . the speed of boat in still water is : <o> a ) num__25 km / hr <o> b ) num__26 km / hr <o> c ) num__22 km / hr <o> d ) num__24 km / hr <o> e ) num__28 km / hr |
rate upstream = num__0.333333333333 * num__60 = num__20 kmph speed of stream = num__4 kmph let speed of boat in still water = xkmph speed upstream = x - num__4 kmph x - num__4 = num__20 x = num__24 kmph answer : d <eor> d <eos> |
d |
divide__8.0__24.0__ hour_to_min_conversion__ subtract__24.0__4.0__ round__24.0__ |
divide__8.0__24.0__ hour_to_min_conversion__ subtract__24.0__4.0__ round__24.0__ |
| a person borrows rs . num__5000 for num__2 years at num__4.0 p . a . simple interest . he immediately lends it to another person at num__5.0 p . a for num__2 years . find his gain in the transaction per year . <o> a ) num__50 <o> b ) num__150 <o> c ) num__225 <o> d ) num__112.5 <o> e ) num__212.5 |
explanation : the person borrows rs . num__5000 for num__2 years at num__4.0 p . a . simple interest simple interest that he needs to pay = prt / num__100 = num__5000 × num__4 × num__0.02 = num__400 he also lends it at num__5.0 p . a for num__2 years simple interest that he gets = prt / num__100 = num__5000 × num__5 × num__0.02 = num__500 his overall gain in num__2 years = rs . num__500 - rs . num__400 = rs . num__100 his overall gain in num__1 year = num__50.0 = rs . num__50 answer : option a <eor> a <eos> |
a |
percent__2.0__5000.0__ percent__0.02__5000.0__ percent__1.0__5000.0__ percent__1.0__5000.0__ |
percent__2.0__5000.0__ percent__0.02__5000.0__ percent__1.0__5000.0__ percent__1.0__5000.0__ |
| if x is equal to the sum of the integers from num__30 to num__50 inclusive and y is the number of even integers from num__30 to num__50 inclusive what is the value of x + y ? <o> a ) num__810 <o> b ) num__811 <o> c ) num__830 <o> d ) num__850 <o> e ) x + y = num__851 |
num__21 integers from num__30 to num__50 inclusive so num__11 th integer is mean it is num__40 num__21 * num__40 = num__840 from num__21 integers num__11 ones is even num__840 + num__11 = num__851 answer : e <eor> e <eos> |
e |
multiply__40.0__21.0__ add__11.0__840.0__ add__11.0__840.0__ |
multiply__40.0__21.0__ add__11.0__840.0__ add__11.0__840.0__ |
| a chair is bought for rs . num__500 / - and sold at a loss of num__20.0 find its selling price <o> a ) rs . num__500 / - <o> b ) rs . num__600 / - <o> c ) rs . num__400 / - <o> d ) rs . num__700 / - <o> e ) rs . num__800 / - |
loss = num__500 * num__0.2 = num__100 s . p = c . p - loss = num__500 - num__100 = rs . num__400 / - answer : c <eor> c <eos> |
c |
percent__20.0__500.0__ percent__100.0__400.0__ |
percent__20.0__500.0__ percent__100.0__400.0__ |
| a man can row a boat at num__26 kmph in still water . if the speed of the stream is num__6 kmph what is the time taken to row a distance of num__60 km downstream ? <o> a ) num__0.361445783133 hours <o> b ) num__6.15384615385 hours <o> c ) num__1.15384615385 hours <o> d ) num__0.842105263158 hours <o> e ) num__1.875 hours |
speed downstream = num__26 + num__6 = num__32 kmph . time required to cover num__60 km downstream = d / s = num__1.875 = num__1.875 hours . answer : e <eor> e <eos> |
e |
add__26.0__6.0__ divide__60.0__32.0__ divide__60.0__32.0__ |
add__26.0__6.0__ divide__60.0__32.0__ divide__60.0__32.0__ |
| let n ~ be defined for all positive integers n as the remainder when ( n - num__1 ) ! is divided by n . what is the value of num__30 ~ ? <o> a ) num__1 <o> b ) num__0 <o> c ) num__2 <o> d ) num__8 <o> e ) num__31 |
n ~ = ( n - num__1 ) ! so num__30 ~ = ( num__30 - num__1 ) ! = num__29 ! when num__29 ! / num__30 we have num__16 * num__2 inside num__29 ! hence num__30 gets cancelled and we get remainder as num__0 b <eor> b <eos> |
b |
subtract__30.0__1.0__ multiply__1.0__0.0__ |
subtract__30.0__1.0__ multiply__1.0__0.0__ |
| if x ^ num__2 − num__2 x − num__15 = ( x + r ) ( x + s ) for all values of x and if r and s are constants then which of the following is a possible value of ( r − s ) ^ num__2 ? <o> a ) num__8 <o> b ) num__2 <o> c ) num__64 <o> d ) − num__3 <o> e ) − num__5 |
we know that given ax ^ num__2 + bx + c = num__0 sum of the roots = - b / a and product of the roots = c / a . the roots here are - r and - s . - r - s = - ( - num__2 ) / num__1 = r + s = - num__2 ( - r ) * ( - s ) = - num__15.0 = rs so one of r and s is - num__5 and the other is num__3 . so ( r - s ) ^ num__2 could be num__64 . answer ( c ) <eor> c <eos> |
c |
add__2.0__1.0__ multiply__1.0__64.0__ |
add__2.0__1.0__ multiply__1.0__64.0__ |
| notebooks are sold in packages of four or seven only . if wilson bought num__69 notebooks exactly what could be the number of large packs wilson bought ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
let number of packs of four = f let number of packs of seven = s num__4 f + num__7 s = num__69 now we need to test for values of s . since sum num__69 is odd and num__4 f will always be even e ca n ' t be even . now we can test for values e = num__3 num__7 and num__9 num__4 * num__5 + num__7 * num__7 = num__20 + num__49 = num__69 answer c <eor> c <eos> |
c |
subtract__7.0__4.0__ subtract__9.0__4.0__ multiply__4.0__5.0__ subtract__69.0__20.0__ add__3.0__4.0__ |
subtract__7.0__4.0__ subtract__9.0__4.0__ multiply__4.0__5.0__ subtract__69.0__20.0__ add__3.0__4.0__ |
| a and b start walking towards each other at num__6 pm at speed of num__4 kmph and num__6 kmph . they were initially num__18 km apart . at what time do they meet ? <o> a ) num__8 pm <o> b ) num__6 pm <o> c ) num__7 pm <o> d ) num__10 pm <o> e ) num__5 pm |
time of meeting = distance / relative speed = num__3.0 + num__3 = num__2.0 = num__2 hrs after num__6 pm = num__8 pm answer is a <eor> a <eos> |
a |
divide__18.0__6.0__ subtract__6.0__4.0__ add__6.0__2.0__ round__8.0__ |
divide__18.0__6.0__ divide__6.0__3.0__ add__6.0__2.0__ add__6.0__2.0__ |
| tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of num__7 meters and a circumference of num__8 meters and the interior of tank b has a height of num__8 meters and a circumference of num__10 meters . the capacity of tank a is what percent of the capacity of tank b ? <o> a ) num__56.0 <o> b ) num__80.0 <o> c ) num__100.0 <o> d ) num__120.0 <o> e ) num__125 % |
for a r = num__4.0 pi . its capacity = ( num__4 pi ) ^ num__2 * num__7 = num__112 pi for b r = num__10 / pi . its capacity = ( num__5 pi ) ^ num__2 * num__8 = num__200 pi a / b = num__112 pi / num__200 pi = num__0.56 a <eor> a <eos> |
a |
divide__8.0__4.0__ subtract__7.0__2.0__ divide__112.0__200.0__ multiply__7.0__8.0__ |
divide__8.0__4.0__ divide__10.0__2.0__ divide__112.0__200.0__ multiply__7.0__8.0__ |
| the sum of first nine prime numbers is ? <o> a ) num__12 <o> b ) num__100 <o> c ) num__24 <o> d ) num__28 <o> e ) num__30 |
required sum = ( num__2 + num__3 + num__5 + num__7 + num__11 + num__13 + num__17 + num__19 + num__23 ) = num__100 . note : num__1 is not a prime number . definition : a prime number ( or a prime ) is a natural number that has exactly two distinct natural number divisors : num__1 and itself . b <eor> b <eos> |
b |
add__2.0__3.0__ add__2.0__5.0__ add__2.0__11.0__ add__2.0__17.0__ subtract__3.0__2.0__ multiply__1.0__100.0__ |
add__2.0__3.0__ add__2.0__5.0__ add__2.0__11.0__ add__2.0__17.0__ subtract__3.0__2.0__ multiply__1.0__100.0__ |
| in a certain store the profit is num__320.0 of the cost . if the cost increases by num__25.0 but the selling price remains constant approximately what percentage of the selling price is the profit ? <o> a ) num__30.0 <o> b ) num__70.0 <o> c ) num__100.0 <o> d ) num__250.0 <o> e ) num__150 % |
let c . p . = rs . num__100 . then profit = rs . num__320 s . p . = rs . num__420 . new c . p . = num__125.0 of rs . num__100 = rs . num__125 new s . p . = rs . num__420 . profit = rs . ( num__420 - num__125 ) = rs . num__295 . required percentage = num__0.702380952381 x num__100.0 = num__70.2380952381 % = num__70.0 answer is b . <eor> b <eos> |
b |
percent__70.0__100.0__ |
percent__70.0__100.0__ |
| if num__15 percent of the students at a certain school went to a camping trip and took more than $ num__100 and num__75 percent of the students who went to the camping trip did not take more than $ num__100 what percentage of the students at the school went to the camping trip ? <o> a ) num__95 <o> b ) num__90 <o> c ) num__85 <o> d ) num__80 <o> e ) num__60 |
let x be the number of students in the school . num__0.15 x students went to the trip and took more than num__100 $ . they compose ( num__100 - num__75 ) = num__25.0 of all students who went to the trip . therefore the toal of num__0.15 x / num__0.25 = num__0.6 x students went to the camping which is num__60.0 . the answer is e <eor> e <eos> |
e |
divide__15.0__100.0__ subtract__100.0__75.0__ divide__25.0__100.0__ divide__15.0__25.0__ divide__15.0__0.25__ divide__15.0__0.25__ |
divide__15.0__100.0__ subtract__100.0__75.0__ divide__25.0__100.0__ divide__15.0__25.0__ divide__15.0__0.25__ divide__15.0__0.25__ |
| the product of two numbers is num__2028 and their h . c . f . is num__13 . the number of such pairs is : <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
explanation let the numbers num__13 a and num__13 b . then num__13 a x num__13 b = num__2028 ab = num__12 . now the co - primes with product num__12 are ( num__1 num__12 ) and ( num__3 num__4 ) . [ note : two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than num__1 or equivalently if their greatest common divisor is num__1 ] so the required numbers are ( num__13 x num__1 num__13 x num__12 ) and ( num__13 x num__3 num__13 x num__4 ) . clearly there are num__2 such pairs . answer b <eor> b <eos> |
b |
subtract__13.0__12.0__ add__1.0__3.0__ subtract__3.0__1.0__ gcd__2028.0__2.0__ |
subtract__13.0__12.0__ add__1.0__3.0__ subtract__3.0__1.0__ subtract__3.0__1.0__ |
| the ratio of the earnings of p and q is num__9 : num__10 . if the earnings of p increases by one - fourth and the earnings of q decreases by one - fourth then find the new ratio of their earnings ? <o> a ) num__0.571428571429 <o> b ) num__3 by num__2 <o> c ) num__0.6 <o> d ) num__2.25 <o> e ) num__3.0 |
let the earnings of p and q be num__9 x and num__10 x respectively . new ratio = [ num__9 x + num__0.25 ( num__9 x ) ] / [ num__10 x - num__0.25 ( num__10 x ) ] = > num__9 * ( num__1 + num__0.25 ) / num__10 * ( num__1 - num__0.25 ) = > num__0.9 * ( num__1.25 ) / ( num__0.75 ) = num__1.5 answer : b <eor> b <eos> |
b |
subtract__10.0__9.0__ divide__9.0__10.0__ add__0.25__1.0__ subtract__1.0__0.25__ add__0.25__1.25__ divide__0.75__0.25__ |
subtract__10.0__9.0__ divide__9.0__10.0__ add__0.25__1.0__ subtract__1.0__0.25__ add__0.25__1.25__ divide__0.75__0.25__ |
| a man decided to cover a distance of num__80 km in num__8 hrs . he can walk at a speed of num__6 km / hr and cycle at num__8 km / hr successively . how long will he walk and how long will he cycle ? <o> a ) num__26.6 km num__53.3 km <o> b ) num__27.6 km num__53.3 km <o> c ) num__26.6 km num__54.3 km <o> d ) num__26.6 km num__55.3 km <o> e ) num__28.6 km num__53.3 km |
as speed in num__80 km / num__8 hrs = num__10 km / hr as ques solve through mixture and alligation method num__6 km / hr and num__8 km / hr num__10 km / hr num__10 - num__8 num__10 - num__6 num__2 km / hr num__4 km / hr so the ratio is num__1 : num__2 ahd we have to find the distance through walk and through cycle num__0.333333333333 * num__80 = num__26.6 km num__0.666666666667 * num__80 = num__53.3 km answer : a <eor> a <eos> |
a |
divide__80.0__8.0__ subtract__8.0__6.0__ divide__8.0__2.0__ divide__2.0__6.0__ subtract__1.0__0.3333__ round__26.6__ |
divide__80.0__8.0__ subtract__8.0__6.0__ divide__8.0__2.0__ divide__2.0__6.0__ subtract__1.0__0.3333__ divide__26.6__1.0__ |
| a corporation that had $ num__115.19 billion in profits for the year paid out $ num__230.10 million in employee benefits . approximately what percent of the profits were the employee benefits ? ( note : num__1 billion = num__10 ^ num__9 ) <o> a ) num__50.0 <o> b ) num__20.0 <o> c ) num__5.0 <o> d ) num__2.0 <o> e ) num__0.2 % |
required answer = [ employee benefit / profit ] * num__100 = [ ( num__230.10 million ) / ( num__115.19 billion ) ] * num__100 = [ ( num__230.10 * num__10 ^ num__6 ) / ( num__115.19 * num__10 ^ num__9 ) ] * num__100 = ( num__0.002 ) * num__100 = num__0.2 so answer is ( e ) <eor> e <eos> |
e |
multiply__100.0__0.002__ multiply__1.0__0.2__ |
multiply__100.0__0.002__ multiply__1.0__0.2__ |
| a mixture contains alcohol and water in the ratio num__4 : num__3 . if num__5 litres of water is added to the mixture the ratio becomes num__4 : num__5 . find the quantity of alcohol in the given mixture <o> a ) num__15 litres <o> b ) num__10 litres <o> c ) num__30 litres <o> d ) num__22 litres <o> e ) num__8 litres |
let the quantity of alcohol and water be num__4 x litres and num__3 x litres respectively num__4 x / ( num__3 x + num__5 ) = num__0.8 num__20 x = num__4 ( num__3 x + num__5 ) num__8 x = num__20 x = num__2.5 quantity of alcohol = ( num__4 x num__2.5 ) litres = num__10 litres . answer is b . <eor> b <eos> |
b |
divide__4.0__5.0__ multiply__4.0__5.0__ add__3.0__5.0__ divide__20.0__8.0__ multiply__4.0__2.5__ multiply__4.0__2.5__ |
divide__4.0__5.0__ multiply__4.0__5.0__ add__3.0__5.0__ divide__20.0__8.0__ divide__8.0__0.8__ divide__8.0__0.8__ |
| a man spend num__0.2 of his salary on food num__0.1 of his salary on house rent and num__0.6 salary on clothes . he still has $ num__18000 left with him . find salary . . <o> a ) $ num__10800 <o> b ) $ num__18000 <o> c ) $ num__180000 <o> d ) $ num__1800 <o> e ) none |
[ num__1 / ( x num__1 / y num__1 + x num__2 / y num__2 + x num__3 / y num__3 ) ] * total amount = balance amount [ num__1 - ( num__0.2 + num__0.1 + num__0.6 ) } * total salary = $ num__18000 = [ num__1 - num__0.9 ] * total salary = $ num__18000 total salary = $ num__18000 * num__10 = $ num__180000 correct answer ( c ) <eor> c <eos> |
c |
divide__0.2__0.1__ divide__0.6__0.2__ subtract__1.0__0.1__ reverse__0.1__ divide__18000.0__0.1__ divide__18000.0__0.1__ |
divide__0.2__0.1__ divide__0.6__0.2__ subtract__1.0__0.1__ reverse__0.1__ divide__18000.0__0.1__ divide__18000.0__0.1__ |
| the cost c in dollars to remove p percent of a certain pollutant from a pond is estimated by using the formula c = num__100000 p / ( num__100 - p ) . according to this estimate how much more would it cost to remove num__90 percent of the pollutant from the pond than it would cost to remove num__50 percent of the pollutant ? <o> a ) $ num__400000 <o> b ) $ num__100000 <o> c ) $ num__50000 <o> d ) $ num__10000 <o> e ) $ num__5 |
000 |
to remove num__90.0 of pollutant c = num__100000 ( num__90 ) / ( num__100 - num__90 ) = num__900000 to remove num__50.0 of pollutant c ' = num__100000 ( num__80 ) / ( num__100 - num__80 ) = num__500000 difference = c - c ' = num__400000 answer - a <eor> a <eos> |
a |
a |
| the average age of num__8 men is increased by years when two of them whose ages are num__21 years and num__23 years are replaced by two new men . the average age of the two new men is <o> a ) num__38 <o> b ) num__30 <o> c ) num__39 <o> d ) num__12 <o> e ) num__11 |
explanation : total age increased = ( num__8 * num__2 ) years = num__16 years . sum of ages of two new men = ( num__21 + num__23 + num__16 ) years = num__60 years average age of two new men = ( num__30.0 ) years = num__30 years . answer : b <eor> b <eos> |
b |
subtract__23.0__21.0__ multiply__8.0__2.0__ divide__60.0__2.0__ divide__60.0__2.0__ |
subtract__23.0__21.0__ multiply__8.0__2.0__ divide__60.0__2.0__ divide__60.0__2.0__ |
| working independently x takes num__10 hours to finish a certain work . he works for num__9 hours . the rest of the work is finished by y whose rate is num__0.1 th of x . in how much time does y finish his work ? <o> a ) num__2 hrs <o> b ) num__4 hrs <o> c ) num__6 hrs <o> d ) num__8 hrs <o> e ) num__10 hrs |
machine x will do num__0.9 in num__9 hrs so x does num__0.9 of the work . . therefore y will do the remaining num__0.1 th work alone . . as the speed of y is num__0.1 rate of x y will do the num__0.1 th work in same time that x takes to complete full job . . . ans num__10 e <eor> e <eos> |
e |
divide__9.0__10.0__ reverse__0.1__ |
divide__9.0__10.0__ reverse__0.1__ |
| the probability that a speaks truth is num__0.6 and that of b speaking truth is num__0.571428571429 . what is the probability that they agree in stating the same fact ? <o> a ) num__0.514285714286 <o> b ) num__0.818181818182 <o> c ) num__0.620689655172 <o> d ) num__1.5 <o> e ) num__0.545454545455 |
if both agree stating the same fact either both of them speak truth of both speak false . probability = num__0.6 * num__0.571428571429 + num__0.4 * num__0.428571428571 = num__0.342857142857 + num__0.171428571429 = num__0.514285714286 answer : a <eor> a <eos> |
a |
negate_prob__0.6__ negate_prob__0.5714__ union_prob__0.6__0.3429__0.4286__ union_prob__0.6__0.4286__0.5143__ |
negate_prob__0.6__ negate_prob__0.5714__ union_prob__0.6__0.3429__0.4286__ union_prob__0.6__0.4286__0.5143__ |
| the sum of two numbers is num__20 and their product is num__375 . what will be the sum of their reciprocals ? <o> a ) num__0.025 <o> b ) num__0.0533333333333 <o> c ) num__18.75 <o> d ) num__9.375 <o> e ) num__12.5 |
( num__1 / a ) + ( num__1 / b ) = ( a + b ) / ab = num__0.0533333333333 = num__0.0533333333333 answer : b <eor> b <eos> |
b |
divide__20.0__375.0__ divide__20.0__375.0__ |
divide__20.0__375.0__ divide__20.0__375.0__ |
| num__31 num__29 num__24 num__22 num__17 ? <o> a ) num__12 <o> b ) num__14 <o> c ) num__15 <o> d ) num__17 <o> e ) num__20 |
c num__15 this is a simple alternating subtraction series with a pattern - num__2 - num__5 - num__2 - num__5 . . . . <eor> c <eos> |
c |
subtract__31.0__29.0__ subtract__29.0__24.0__ subtract__17.0__2.0__ |
subtract__31.0__29.0__ subtract__29.0__24.0__ subtract__17.0__2.0__ |
| in the coordinate plane points ( x num__1 ) and ( num__18 y ) are on line k . if line k passes through the origin and has slope num__0.5 then x + y = <o> a ) num__4.5 <o> b ) num__7 <o> c ) num__8 <o> d ) num__11 <o> e ) num__12 |
line k passes through the origin and has slope num__0.5 means that its equation is y = num__0.5 * x . thus : ( x num__1 ) = ( num__2 num__1 ) and ( num__18 y ) = ( num__189 ) - - > x + y = num__2 + num__9 = num__11 . answer : d . <eor> d <eos> |
d |
reverse__0.5__ multiply__18.0__0.5__ add__2.0__9.0__ multiply__1.0__11.0__ |
reverse__0.5__ multiply__18.0__0.5__ add__2.0__9.0__ multiply__1.0__11.0__ |
| in how many ways can the letters of the word abacus be rearranged such that the vowels always appear together ? <o> a ) num__6 ! / num__2 ! <o> b ) num__3 ! * num__3 ! <o> c ) num__4 ! / num__2 ! <o> d ) num__4 ! * num__3 ! / num__2 ! <o> e ) num__3 ! * num__3 ! / num__2 |
in the word abacus there are num__3 vowels - num__2 a ' s and u number of ways the letters of word abacus be rearranged such that the vowels always appear together = ( num__4 ! * num__3 ! ) / num__2 ! we can consider the the num__3 vowels as a single unit and there are num__3 ways to arrange them . but since num__2 elements of vowel group are identical we divide by num__2 ! . the entire vowel group is considered as a single group . answer d <eor> d <eos> |
d |
coin_space__ choose__4.0__3.0__ |
coin_space__ choose__4.0__3.0__ |
| num__4 ! / ( num__4 - num__3 ) ! = ? <o> a ) num__24 <o> b ) num__34 <o> c ) num__35 <o> d ) num__36 <o> e ) num__39 |
num__4 ! / ( num__4 - num__3 ) ! = num__4 ! / num__1 ! = num__4 * num__3 * num__2 = num__24 . hence the correct answer is a . <eor> a <eos> |
a |
subtract__4.0__3.0__ subtract__3.0__1.0__ multiply__24.0__1.0__ |
subtract__4.0__3.0__ subtract__3.0__1.0__ multiply__24.0__1.0__ |
| a chain smoker had spent all the money he had . he had no money to buy his cigarettes . hence he resorted to join the stubs and to smoke them . he needed num__4 stubs to make a single cigarette . if he got a pack of num__10 cigarettes as a gift then how many cigarettes could he smoke in all ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__12 <o> d ) num__13 <o> e ) num__14 |
explanation : num__10 cigarettes will produce num__10 stubs . since out of num__10 cigarettes = > num__8 cigarettes ' stubs gives num__2 cigarette and num__2 stubs are still left . = > these num__2 gigarette will give num__2 stubs . now he will make num__1 more cigarette by using num__2 stubs + num__2 stubs . hence total number of cigarettes will be num__10 + num__2 + num__1 i . e num__13 . answer : d <eor> d <eos> |
d |
subtract__10.0__8.0__ multiply__1.0__13.0__ |
subtract__10.0__8.0__ multiply__1.0__13.0__ |
| the sector of a circle has radius of num__21 cm and central angle num__135 o . find its perimeter ? <o> a ) num__91.5 <o> b ) num__92 <o> c ) num__93 <o> d ) num__94 <o> e ) num__95 |
perimeter of the sector = length of the arc + num__2 ( radius ) = ( num__0.375 * num__2 * num__3.14285714286 * num__21 ) + num__2 ( num__21 ) = num__49.5 + num__42 = num__91.5 cm answer : option a <eor> a <eos> |
a |
multiply__21.0__2.0__ add__49.5__42.0__ round__91.5__ |
multiply__21.0__2.0__ add__49.5__42.0__ add__49.5__42.0__ |
| if a truck is traveling at a constant rate of num__36 kilometers per hour how many seconds will it take the truck to travel a distance of num__600 meters ? ( num__1 kilometer = num__1000 meters ) <o> a ) num__18 <o> b ) num__24 <o> c ) num__60 <o> d ) num__36 <o> e ) num__48 |
speed = num__36 km / hr = > num__36000 m / hr in one minute = > num__600.0 = num__600 meters in one sec = > num__10.0 = num__10 meters time = total distance need to be covered / avg . speed = > num__60.0 = num__60 and hence the answer : c <eor> c <eos> |
c |
multiply__36.0__1000.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
multiply__36.0__1000.0__ divide__600.0__10.0__ divide__600.0__10.0__ |
| vijay bought num__160 shirts at the rate of rs . num__225 per shirt . the transport expenditure was rs . num__1400 . he paid an octroi at the rate of rs . num__1.75 per shirt and labour charges were rs . num__320 . what should be the selling price of one shirt if he wants a profit of num__20.0 ? <o> a ) num__338 <o> b ) num__2677 <o> c ) num__288 <o> d ) num__285 <o> e ) num__261 |
total cp per shirt = num__225 + num__8.75 + num__1.75 + num__2.0 = rs . num__237.5 sp = cp [ ( num__100 + profit % ) / num__100 ] = num__237.5 * [ ( num__100 + num__20 ) / num__100 ] = rs . num__285 . answer : d <eor> d <eos> |
d |
percent__100.0__285.0__ |
percent__100.0__285.0__ |
| if a man reduces the selling price of a fan from num__400 to num__380 his loss increases by num__20.0 . what is the cost price of fan <o> a ) num__300 <o> b ) num__400 <o> c ) num__500 <o> d ) num__600 <o> e ) num__700 |
increase in loss = num__20.0 consider loss is x initially so final loss will increase by . num__2 x which is equal to num__20 . . num__2 x = num__20 x = num__100 so total cost = selling price + loss = num__400 + num__100 = num__500 . so ans is num__500 . answer : c <eor> c <eos> |
c |
percent__100.0__500.0__ |
percent__100.0__500.0__ |
| john and steve are speed walkers in a race . john is num__10 meters behind steve when he begins his final push . john blazes to the finish at a pace of num__4.2 m / s while steve maintains a blistering num__3.7 m / s speed . if john finishes the race num__2 meters ahead of steve how long was john ’ s final push ? <o> a ) num__13 seconds <o> b ) num__17 seconds <o> c ) num__24 seconds <o> d ) num__34 seconds <o> e ) num__51 seconds |
let t be the time that john spent for his final push . thus per the question num__4.2 t = num__3.7 t + num__10 + num__2 - - - > num__0.5 t = num__12 - - - > t = num__24 seconds . c is the correct answer . <eor> c <eos> |
c |
subtract__4.2__3.7__ add__10.0__2.0__ multiply__2.0__12.0__ round__24.0__ |
subtract__4.2__3.7__ add__10.0__2.0__ multiply__2.0__12.0__ round__24.0__ |
| the length of rectangle is thrice its breadth and its perimeter is num__56 m find the area of the rectangle ? <o> a ) num__432 <o> b ) num__252 <o> c ) num__299 <o> d ) num__276 <o> e ) num__111 |
num__2 ( num__3 x + x ) = num__96 l = num__36 b = num__7 lb = num__36 * num__7 = num__252 b <eor> b <eos> |
b |
multiply__36.0__7.0__ triangle_area__2.0__252.0__ |
multiply__36.0__7.0__ multiply__36.0__7.0__ |
| the speed of a boat in still water is num__10 km / hr . if it can travel num__78 km downstream and num__42 km upstream in the same time the speed of the stream is <o> a ) num__3 km / hr <o> b ) num__12 km / hr <o> c ) num__1.5 km / hr <o> d ) num__4.4 km / hr <o> e ) none of these |
explanation : let the speed of the stream be x km / hr . then speed upstream = ( num__10 - x ) km / hr speed downstream = ( num__10 + x ) km / hr time taken to travel num__78 km downstream = time taken to travel num__42 km upstream = > num__78 / ( num__10 + x ) = num__42 / ( num__10 − x ) = > num__26 / ( num__10 + x ) = num__14 / ( num__10 − x ) = > num__13 / ( num__10 + x ) = num__7 / ( num__10 − x ) = > num__130 − num__13 x = num__70 + num__7 x = > num__20 x = num__60 = > x = num__3 km / hr answer : option a <eor> a <eos> |
a |
multiply__10.0__13.0__ multiply__10.0__7.0__ add__7.0__13.0__ hour_to_min_conversion__ subtract__10.0__7.0__ round__3.0__ |
multiply__10.0__13.0__ multiply__10.0__7.0__ add__7.0__13.0__ subtract__130.0__70.0__ subtract__10.0__7.0__ subtract__10.0__7.0__ |
| two cylindrical candles of the same height but different diameters are lit at the same time . the first is consumed in num__4 hours and the second in num__3 hours . assuming that they burn at a constant rate how long after being lit was the first candle twice the height of the second candle ? <o> a ) num__1 hr <o> b ) num__1 hr num__12 min <o> c ) num__2 hr <o> d ) num__2 hr num__12 min <o> e ) num__2 hr num__24 min |
assume each candle is num__1 unit high . since the burn rates are num__0.333333333333 and num__0.25 units per hour after t hours they have been consumed by an amount of t / num__3 and t / num__4 . we want num__1 - t = num__4 = num__2 ( num__1 - t = num__3 ) . this occurs when t = num__2.4 = num__2 * num__0.4 hrs . correct answer e <eor> e <eos> |
e |
subtract__4.0__3.0__ divide__1.0__3.0__ divide__1.0__4.0__ subtract__3.0__1.0__ subtract__2.4__2.0__ round__2.0__ |
subtract__4.0__3.0__ divide__1.0__3.0__ divide__1.0__4.0__ subtract__3.0__1.0__ subtract__2.4__2.0__ divide__4.0__2.0__ |
| if m and n are positive integers and m ^ num__2 + n ^ num__2 = num__20 what is the value of m ^ num__3 + n ^ num__3 ? <o> a ) num__72 <o> b ) num__224 <o> c ) num__320 <o> d ) num__512 <o> e ) num__1 |
600 |
you need to integers which squared are equal num__20 . which could it be ? let ' s start with the first integer : num__1 ^ num__2 = num__1 num__2 ^ num__2 = num__4 num__3 ^ num__2 = num__9 num__4 ^ num__2 = num__16 stop . the integers ca n ' t be greater than num__4 or we will score above num__20 . the second integer need to be picked up the same way . num__1 ^ num__2 = num__1 num__2 ^ num__2 = num__4 num__3 ^ num__2 = num__9 num__4 ^ num__2 = num__16 the only pair that matches is num__4 ^ num__2 + num__2 ^ num__2 = num__20 . so num__4 ^ num__3 + num__2 ^ num__3 = num__72 . answer a . ) <eor> a <eos> |
a |
a |
| q . how many num__4 - digit even numbers less than num__3 num__000000 can be formed using the following digits : num__1 num__2 num__2 num__3 ? <o> a ) num__180 <o> b ) num__240 <o> c ) num__210 <o> d ) num__270 <o> e ) num__300 |
the number is less than num__3000000 so we will have the digit on the left most side to be either num__1 or num__2 i . e your number can be num__1 _ _ _ _ _ _ or num__2 _ _ _ _ _ _ case num__1 > num__1 _ _ _ _ _ _ here the units digit can be either num__2 or num__6 when the units digit is num__2 i . e num__1 _ _ _ _ _ num__2 number of ways this can be done would be num__5 ! / num__2 ! ( as num__5 is repeated twice ) = num__60 when the units digit is num__6 i . e . num__1 _ _ _ _ _ num__6 number of ways would be num__5 ! / ( num__2 ! * num__2 ! ) { both num__2 and num__5 repeat twice } = num__30 case num__2 > num__2 _ _ _ _ _ _ ( here units digit may be num__2 or num__6 ) number of ways this can be done would be num__5 ! / ( num__2 ! ) for num__2 = num__60 and num__5 ! / num__2 ! for num__6 . . . = num__60 adding all of these . . . gives the answer num__300 . . . the explanation looks difficult but this technique helps if you are familiar with the concept of permutations when the numbers / alphabets repeat e <eor> e <eos> |
e |
add__4.0__2.0__ add__4.0__1.0__ multiply__5.0__6.0__ multiply__5.0__60.0__ multiply__1.0__300.0__ |
multiply__3.0__2.0__ add__4.0__1.0__ multiply__5.0__6.0__ multiply__5.0__60.0__ multiply__1.0__300.0__ |
| this is a famous probability puzzle in which you have to choose the correct answer at random from the four options below . can you tell us whats the probability of choosing correct answer in this random manner . <o> a ) num__0.0 <o> b ) num__2.0 <o> c ) num__3.0 <o> d ) num__6.0 <o> e ) num__1 % |
a num__0.0 explanation : num__1 ) why cant be num__0.25 : if the answer is num__0.25 then as we know two out of four answer choices is ' num__0.25 ' the answer has be num__0.5 . this is a contradiction so the answer can not be num__0.25 . num__2 ) why cant be num__0.5 : if the answer is num__0.5 then because answer : ' ' num__0.5 ' ' is num__1 out of num__4 answer choices the answer must be num__0.25 . this is also a contradiction . so the answer can not be num__0.5 . num__3 ) why cant be num__1 : if the answer is num__1 then because answer : ' ' num__1 ' ' is num__1 out of num__4 answer choices the answer must be num__0.25 . again the same contradiction and therefore answer can not be num__1 . <eor> a <eos> |
a |
negate_prob__0.0__ coin_space__ choose__2.0__0.0__ negate_prob__1.0__ |
negate_prob__0.0__ coin_space__ choose__2.0__0.0__ negate_prob__1.0__ |
| by selling an article at rs . num__800 a shopkeeper makes a profit of num__25.0 . at what price should he sell the article so as to make a loss of num__25.0 ? <o> a ) rs . num__429 <o> b ) rs . num__480 <o> c ) rs . num__429 <o> d ) rs . num__128 <o> e ) rs . num__419 |
sp = num__800 profit = num__25.0 cp = ( sp ) * [ num__100 / ( num__100 + p ) ] = num__800 * [ num__0.8 ] = num__640 loss = num__25.0 = num__25.0 of num__640 = rs . num__160 sp = cp - loss = num__640 - num__160 = rs . num__480 answer : b <eor> b <eos> |
b |
percent__25.0__640.0__ percent__100.0__480.0__ |
percent__25.0__640.0__ percent__100.0__480.0__ |
| the average price of an antique car increases over the years . if from num__1990 to num__1996 the price of the car increased by num__13.0 and from num__1996 to num__2001 it increased by num__20.0 what is the price of the car in num__2001 if the price in num__1990 was $ num__11500 ? <o> a ) $ num__15594 <o> b ) $ num__15322 . <o> c ) $ num__14786 <o> d ) $ num__14543 <o> e ) $ num__12 |
988 |
price of the car in num__1990 = $ num__11500 from num__1990 to num__1996 increase = num__13.0 which is ( num__13 * num__115 ) price of the car in num__1996 = $ num__12995 from num__1996 to num__2001 increase = num__20.0 which is ( num__12995 ) / num__5 = num__2599 price of the car in num__2001 = num__15594 answer : a <eor> a <eos> |
a |
a |
| if a dealer wants to earn num__20.0 profit on an article after offering num__30.0 discount to the customer by what percentage should he increase his marked price to arrive at the label price ? <o> a ) num__71 num__0.428571428571 % <o> b ) num__50.0 <o> c ) num__75.0 <o> d ) num__60 num__0.666666666667 % <o> e ) num__90 % |
profit = num__20.0 discount = num__30.0 required answer = ( num__20 + num__0.3 - num__30 ) * num__100 = num__71 num__0.428571428571 % answer is a <eor> a <eos> |
a |
percent__100.0__71.0__ |
percent__100.0__71.0__ |
| the area of a square is equal to five times the area of a rectangle of dimensions num__36 cm * num__20 cm . what is the perimeter of the square ? <o> a ) num__289 cm <o> b ) num__800 cm <o> c ) num__829 cm <o> d ) num__240 cm <o> e ) num__289 cm |
area of the square = s * s = num__5 ( num__36 * num__20 ) = > s = num__60 = num__60 cm perimeter of the square = num__4 * num__60 = num__240 cm . answer : d <eor> d <eos> |
d |
square_perimeter__60.0__ square_perimeter__60.0__ |
multiply__4.0__60.0__ multiply__4.0__60.0__ |
| find the value of x from the below equation ? : num__3 x ^ num__2 + num__6 x + num__3 = num__0 <o> a ) - num__0.833333333333 <o> b ) num__0.833333333333 <o> c ) num__5 <o> d ) num__6 <o> e ) num__25 |
a = num__3 b = num__6 c = num__3 x num__12 = ( - num__5 Â ± â ˆ š ( num__6 ^ num__2 - num__4 Ã — num__3 Ã — num__3 ) ) / ( num__2 Ã — num__3 ) = ( - num__5 Â ± â ˆ š ( num__36 - num__36 ) ) / num__6 = ( - num__5 Â ± num__0 ) / num__6 x num__1 = x num__2 = - num__0.833333333333 a <eor> a <eos> |
a |
multiply__2.0__6.0__ add__3.0__2.0__ subtract__6.0__2.0__ multiply__3.0__12.0__ subtract__3.0__2.0__ divide__5.0__6.0__ multiply__1.0__0.8333__ |
multiply__2.0__6.0__ add__3.0__2.0__ subtract__6.0__2.0__ multiply__3.0__12.0__ subtract__3.0__2.0__ divide__5.0__6.0__ divide__5.0__6.0__ |
| you collect coins . suppose you start out with num__11 . since you ' re nice you give sharon num__3 coins . since you ' re nice you give steven num__4 coins . how many coins do you have at the end ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
solution start with num__11 coins . sharon takes num__3 coins from you : num__11 - num__3 = num__8 coins . steven takes num__4 coins from you : num__8 - num__4 = num__4 coins . so you have num__4 at the end . correct answer : c <eor> c <eos> |
c |
choose__4.0__3.0__ |
choose__4.0__3.0__ |
| in what time will a train num__150 m long cross an electric pole it its speed be num__144 km / hr ? <o> a ) num__2.55 sec <o> b ) num__2.9 sec <o> c ) num__1.3 sec <o> d ) num__3.75 sec <o> e ) num__1.8 sec |
speed = num__144 * num__0.277777777778 = num__40 m / sec time taken = num__2.5 = num__3.75 sec . answer : d <eor> d <eos> |
d |
divide__150.0__40.0__ round__3.75__ |
divide__150.0__40.0__ divide__150.0__40.0__ |
| find the value of sin num__75 ° + sin num__15 ° = ? <o> a ) num__0 <o> b ) num__0.5 <o> c ) num__1 <o> d ) num__1.5 <o> e ) num__2 |
sin a + sin b = num__2 sin [ ( a + b ) / num__2 ] cos [ ( a - b ) / num__2 ] = num__2 sin [ ( num__75 + num__15 ) / num__2 ] cos [ ( num__75 - num__15 ) / num__2 ] = num__2 sin num__90 cos num__60 = num__2 * num__1 * ( num__0.5 ) = num__1 answer : c <eor> c <eos> |
c |
add__75.0__15.0__ subtract__75.0__15.0__ reverse__2.0__ reverse__1.0__ |
add__75.0__15.0__ subtract__75.0__15.0__ reverse__2.0__ reverse__1.0__ |
| which of the following can not be the sum of num__4 different prime numbers ? <o> a ) num__21 <o> b ) num__17 <o> c ) num__29 <o> d ) num__32 <o> e ) num__30 |
option a : num__21 = num__2 + num__3 + num__5 + num__11 . sum of num__4 different prime numbers option b : num__17 = num__2 + num__3 + num__5 + num__7 . sum of num__4 different prime numbers option c : num__29 = num__2 + num__3 + num__7 + num__17 . sum of num__4 different prime numbers option d : num__32 = num__5 + num__3 + num__11 + num__13 . sum of num__4 different prime numbers option e : num__30 = this is not a sum of num__4 different prime numbers correct option : e <eor> e <eos> |
e |
add__2.0__3.0__ subtract__21.0__4.0__ add__4.0__3.0__ power__2.0__5.0__ add__2.0__11.0__ subtract__32.0__2.0__ subtract__32.0__2.0__ |
add__2.0__3.0__ subtract__21.0__4.0__ add__4.0__3.0__ add__3.0__29.0__ add__2.0__11.0__ add__13.0__17.0__ add__13.0__17.0__ |
| what is the distance from ( – num__7 num__2 ) to ( num__5 – num__3 ) ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__12 <o> d ) num__13 <o> e ) num__14 |
solutions suppose we made a right triangle as in the example above . the difference in the x - coordinates is ( – num__7 ) – num__5 = – num__12 so the horizontal leg would have a length of num__12 ( length has to be positive ) . the difference in the y - coordinates is num__2 – ( – num__3 ) = num__5 so this is the length of the vertical leg . we recognize that we have a common pythagorean triplet so we don ’ t even need to perform a calculation : we know the length of the hypotenuse and hence the distance between the two points is num__13 . answer = d <eor> d <eos> |
d |
add__7.0__5.0__ round__13.0__ |
add__7.0__5.0__ round__13.0__ |
| a large box contains num__20 small boxes and each small box contains num__32 chocolate bars . how many chocolate bars are in the large box ? <o> a ) a ) num__220 <o> b ) b ) num__490 <o> c ) c ) num__380 <o> d ) d ) num__450 <o> e ) e ) num__640 |
the only thing you have to do is a simple multiply we have : num__20 * num__32 = num__640 correct answer is : e ) num__640 <eor> e <eos> |
e |
multiply__20.0__32.0__ multiply__20.0__32.0__ |
multiply__20.0__32.0__ multiply__20.0__32.0__ |
| in an athletic event with n participants if num__0.2 could not complete num__100 meter race within cut off time of num__10 seconds . and of those who didnt clear num__100 meter race within cut off time num__0.333333333333 also did not qualify in long jump . how many athletes cleared both events ? <o> a ) n / num__15 <o> b ) num__4 n / num__15 <o> c ) num__14 n / num__15 <o> d ) n / num__5 <o> e ) num__4 n / num__5 |
num__0.2 did not clear running event thus n / num__5 did not clear . of those who didnt clear running event num__0.333333333333 also did not clear long jump event thus num__0.333333333333 * n / num__5 = n / num__15 had both not cleared both events . therefore n - n / num__15 = num__14 n / num__15 cleared both events . answer : c . <eor> c <eos> |
c |
add__10.0__5.0__ round__14.0__ |
add__10.0__5.0__ round__14.0__ |
| simple question but big one on average age . sth like a b c weighted separately num__1 st a b c then a & b then b & c then c & a at last abc the last weight was num__167 then what will be the average weight of the num__7 reading ? <o> a ) num__32 <o> b ) num__92 <o> c ) num__94 <o> d ) num__95 <o> e ) num__100 |
avg = a + b + c + ( a + b ) + ( b + c ) + ( c + a ) + ( a + b + c ) / num__7 now num__4 a + num__4 b + num__4 c / num__7 = num__4 ( num__167 ) / num__7 ( given abc = num__167 ) = num__95.43 answer : d <eor> d <eos> |
d |
round_down__95.43__ |
round_down__95.43__ |
| in an election between the two candidates the candidates who gets num__60.0 of votes polled is winned by num__280 votes majority . what is the total number of votes polled ? <o> a ) num__1400 <o> b ) num__1600 <o> c ) num__1800 <o> d ) num__2000 <o> e ) num__2100 |
note : majority ( num__20.0 ) = difference in votes polled to win ( num__60.0 ) & defeated candidates ( num__40.0 ) num__20.0 = num__60.0 - num__40.0 num__20.0 - - - - - > num__280 ( num__20 × num__14 = num__280 ) num__100.0 - - - - - > num__1400 ( num__100 × num__14 = num__1400 ) a ) <eor> a <eos> |
a |
percent__100.0__1400.0__ |
percent__100.0__1400.0__ |
| a can complete a work in num__15 days and b can do the same work in num__7 days . if a after doing num__3 days leaves the work find in how many days b will do the remaining work ? <o> a ) num__2 days <o> b ) num__5 num__0.5 days <o> c ) num__6 num__0.5 days <o> d ) num__7 num__0.5 days <o> e ) num__10 days |
the required answer = ( num__15 - num__3 ) * num__0.466666666667 = num__5.6 = num__5 num__0.5 days answer is b <eor> b <eos> |
b |
divide__7.0__15.0__ divide__15.0__3.0__ round__5.0__ |
divide__7.0__15.0__ divide__15.0__3.0__ round__5.0__ |
| three times the first of three consecutive odd integers is num__3 more than twice the third . the third integer is <o> a ) num__8 <o> b ) num__9 <o> c ) num__13 <o> d ) num__15 <o> e ) num__17 |
solution let the three numbers be x x + num__2 x + num__4 then num__3 x = num__2 ( x + num__4 ) + num__3 ‹ = › x = num__11 third integer = x + num__4 = num__15 . answer d <eor> d <eos> |
d |
add__11.0__4.0__ add__11.0__4.0__ |
add__11.0__4.0__ add__11.0__4.0__ |
| if | x ^ num__2 − num__6 | = x which of the following could be the value of x ? <o> a ) – num__3 <o> b ) – num__2 <o> c ) num__0 <o> d ) num__1 <o> e ) num__2 |
the lhs is not negative so the rhs is also not negative . thus x > = num__0 . first let ' s assume that x ^ num__2 - num__6 is negative . - ( x ^ num__2 - num__6 ) = x x ^ num__2 + x - num__6 = num__0 ( x + num__3 ) ( x - num__2 ) = num__0 x = num__2 or x = - num__3 ( however x can not be negative . ) then x = num__2 is a possible value for x . the answer is e . <eor> e <eos> |
e |
divide__6.0__2.0__ divide__6.0__3.0__ |
divide__6.0__2.0__ divide__6.0__3.0__ |
| when the smallest of num__3 consecutive odd integers is added to four times the largest it produces a result num__727 more than num__4 times the middle integer . find the numbers ? <o> a ) num__650 <o> b ) num__678 <o> c ) num__727 <o> d ) num__710 <o> e ) num__729 |
x + num__4 ( x + num__4 ) = num__727 + num__4 ( x + num__2 ) solve for x and find all three numbers x + num__4 x + num__16 = num__727 + num__4 x + num__8 x = num__719 x + num__2 = num__721 x + num__4 = num__723 check : the smallest is added to four times the largest num__719 + num__4 * num__723 = num__3611 four times the middle num__4 * num__721 = num__2884 num__3611 is more than num__2884 by num__3611 - num__2884 = num__727 c <eor> c <eos> |
c |
multiply__4.0__2.0__ subtract__727.0__8.0__ add__2.0__719.0__ subtract__727.0__4.0__ multiply__4.0__721.0__ add__4.0__723.0__ |
multiply__4.0__2.0__ subtract__727.0__8.0__ add__2.0__719.0__ subtract__727.0__4.0__ multiply__4.0__721.0__ add__4.0__723.0__ |
| two trains of equal lengths are running at speeds of num__30 kmph and num__60 kmph . the two trains crossed each other in num__30 seconds when travelling in opposite direction . in what time will they cross each other when travelling in the same direction ? <o> a ) num__90 sec <o> b ) num__30 sec <o> c ) num__10 sec <o> d ) num__70 sec <o> e ) num__30 sec |
let the length of each train be x m . ( x + x ) / ( num__60 + num__30 ) num__0.277777777778 = ( num__750 * num__18 ) / ( num__30 * num__5 ) = num__90 sec . answer : a <eor> a <eos> |
a |
add__30.0__60.0__ round__90.0__ |
add__30.0__60.0__ add__30.0__60.0__ |
| for a certain article if discount is num__25.0 the profit is num__25.0 . if the discount is num__10.0 then the profit is <o> a ) num__10.0 <o> b ) num__20.0 <o> c ) num__35.0 <o> d ) num__50.0 <o> e ) none of these |
for same article num__100 − d num__0.01 − d num__2 = num__100 + g num__0.01 + g num__2 ⇒ num__100 − num__0.25 − num__10 = num__100 + num__0.25 + g num__2 ⇒ num__0.833333333333 = num__1.25 + g num__2 ⇒ num__100 + g num__2 = num__90 × num__1.66666666667 = num__150 ⇒ g num__2 = num__50.0 answer d <eor> d <eos> |
d |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| the sum of first five prime numbers is ? <o> a ) num__11 <o> b ) num__18 <o> c ) num__26 <o> d ) num__28 <o> e ) none |
required sum = ( num__2 + num__3 + num__5 + num__7 + num__11 ) = num__28 . note : num__1 is not a prime number . definition : a prime number ( or a prime ) is a natural number that has exactly two distinct natural number divisors : num__1 and itself . a ) <eor> a <eos> |
a |
add__2.0__3.0__ add__2.0__5.0__ subtract__3.0__2.0__ multiply__1.0__11.0__ |
add__2.0__3.0__ add__2.0__5.0__ subtract__3.0__2.0__ multiply__1.0__11.0__ |
| a jar is filled with red white and blue tokens that are equivalent except for their color . the chance of randomly selecting a red token replacing it then randomly selecting a white token is the same as the chance of randomly selecting a blue token . if the number of tokens of every color is a multiple of num__6 what is the smallest possible total number of tokens in the jar ? <o> a ) num__9 <o> b ) num__12 <o> c ) num__15 <o> d ) num__18 <o> e ) num__21 |
( red / total coins ) * ( white / total coins ) = ( blue / total coins ) i . e . red * white = blue * total coins let red = num__3 a white = num__3 b blue = num__3 c total coins = num__3 ( a + b + c ) i . e . num__3 a * num__3 b = num__3 c * num__3 ( a + b + c ) i . e . a * b = c * ( a + b + c ) for smallest values of a b and c num__2 * num__3 = num__1 * ( num__1 + num__2 + num__3 ) i . e . minimum total coins = num__3 * ( num__1 + num__2 + num__3 ) = num__21 answer : option e <eor> e <eos> |
e |
divide__6.0__3.0__ subtract__3.0__2.0__ multiply__1.0__21.0__ |
divide__6.0__3.0__ subtract__3.0__2.0__ multiply__1.0__21.0__ |
| a coin is tossed seven times . what is the probability that there is at the least one tail ? <o> a ) num__0.9921875 <o> b ) num__0.224637681159 <o> c ) num__0.185628742515 <o> d ) num__0.264957264957 <o> e ) num__0.2421875 |
let p ( t ) be the probability of getting least one tail when the coin is tossed seven times . = there is not even a single tail . i . e . all the outcomes are heads . = num__0.0078125 ; p ( t ) = num__1 - num__0.0078125 = num__0.9921875 answer : a <eor> a <eos> |
a |
negate_prob__0.0078__ negate_prob__0.0078__ |
negate_prob__0.0078__ negate_prob__0.0078__ |
| a man is standing on a railway bridge which is num__180 m long . he finds that a train crosses the bridge in num__20 seconds but himself in num__10 seconds . find the length of the train and its speed . <o> a ) num__10 m / sec <o> b ) num__15 m / sec <o> c ) num__18 m / sec <o> d ) num__19 m / sec <o> e ) num__25 m / sec |
if length of train is l and speed of train is s then l / s = num__10 ( l + num__180 ) / s = num__20 ( l + num__180 ) / l = num__2.0 = num__2 l + num__180 = num__2 l l = num__180 l = num__180 mtrs s = l / num__10 = num__22.5 = num__10 m / sec answer : a <eor> a <eos> |
a |
divide__20.0__10.0__ round__10.0__ |
divide__20.0__10.0__ divide__20.0__2.0__ |
| if m and n are whole numbers such that mn = num__121 the value of ( m − num__1 ) n + num__1 is <o> a ) num__1 <o> b ) num__10 <o> c ) num__100 <o> d ) num__1000 <o> e ) none of these |
explanation : we know that ( num__11 ) num__2 = num__121 so putting values in said equation we get ( num__11 − num__1 ) num__2 + num__1 = ( num__10 ) num__3 = num__1000 option d <eor> d <eos> |
d |
subtract__11.0__1.0__ add__1.0__2.0__ multiply__1.0__1000.0__ |
subtract__11.0__1.0__ add__1.0__2.0__ multiply__1.0__1000.0__ |
| in a survey of political preferences num__78.0 of those asked were in favour of at least one of the proposals : i ii and iii . num__50.0 of those asked favoured proposal i num__30.0 favoured proposal ii and num__20.0 favoured proposal iii . if num__5.0 of those asked favoured all three of the proposals what a percentage of those asked favoured more than one of the num__3 proposals . <o> a ) num__10 <o> b ) num__12 <o> c ) num__17 <o> d ) num__22 <o> e ) num__30 |
bunuel my answer for exactly num__2 people was num__17 and this was my approach : num__100.0 = ( a + b + c ) - ( anb + anc + bnc ) - num__5.0 + num__22.0 which leads me to a = num__100.0 = ( num__50 + num__30 + num__20 ) - ( at least num__2 people ) - num__5.0 + num__22.0 . c <eor> c <eos> |
c |
percent__100.0__17.0__ |
percent__100.0__17.0__ |
| a computer is programmed to multiply consecutive even integers num__2 * num__4 * num__6 * num__8 * … * n until the product is divisible by num__1881 what is the value of n ? <o> a ) num__22 <o> b ) num__38 <o> c ) num__114 <o> d ) num__122 <o> e ) num__672 |
factorise num__1881 . . num__3 * num__11 * num__57 . . so n has to be a multiple of largest prime number num__61 . . so n = num__2 * num__57 = num__114 . . ans : c <eor> c <eos> |
c |
divide__6.0__2.0__ add__8.0__3.0__ add__4.0__57.0__ multiply__2.0__57.0__ multiply__2.0__57.0__ |
divide__6.0__2.0__ add__8.0__3.0__ add__4.0__57.0__ multiply__2.0__57.0__ multiply__2.0__57.0__ |
| find the value of ( x ) in the given equation ? num__35.0 of num__1500 + x = num__45.0 of num__4200 – num__320 <o> a ) answer : num__1049 <o> b ) answer : num__1048 <o> c ) answer : num__1041 <o> d ) answer : num__1045 <o> e ) answer : num__10412 |
explanation : num__35.0 of num__1500 + x = num__45.0 of num__4200 – num__320 ( num__0.35 * num__1500 ) + x = ( num__0.45 * num__4200 ) – num__320 num__525 + x = num__1890 – num__320 x = num__1890 – num__320 – num__525 x = num__1890 – num__845 x = num__1045 answer : d <eor> d <eos> |
d |
multiply__1500.0__0.35__ multiply__4200.0__0.45__ add__320.0__525.0__ subtract__1890.0__845.0__ subtract__1890.0__845.0__ |
multiply__1500.0__0.35__ multiply__4200.0__0.45__ add__320.0__525.0__ subtract__1890.0__845.0__ subtract__1890.0__845.0__ |
| ram goes to market with a speed of num__3 km / h and comes back home with a speed of num__2 km / h . if he takes num__5 hours in all find the distance between home and market . <o> a ) num__8 km <o> b ) num__4 km <o> c ) num__6 km <o> d ) num__5 km <o> e ) num__7 km |
let the required distance be x km . then time to go to the market = x / num__3 hrs . time taken to go come back = x / num__2 hrs . total time taken = ( x / num__3 + x / num__2 ) hrs . = num__5 x / num__6 here num__5 x / num__6 = num__5 ie ) x = num__6 hence the distance between home and market = num__6 km answer : c <eor> c <eos> |
c |
multiply__3.0__2.0__ round__6.0__ |
multiply__3.0__2.0__ round__6.0__ |
| in what time will a railway train num__60 m long moving at the rate of num__36 kmph pass a telegraph post on its way ? <o> a ) num__9 sec <o> b ) num__6 sec <o> c ) num__8 sec <o> d ) num__10 sec <o> e ) num__11 sec |
t = num__1.66666666667 * num__3.6 = num__6 sec answer b <eor> b <eos> |
b |
divide__60.0__36.0__ round__6.0__ |
divide__60.0__36.0__ round__6.0__ |
| an athlete runs num__200 meters race in num__24 sec . his speed is ? <o> a ) num__20 km / hr <o> b ) num__15 km / hr <o> c ) num__30 km / hr <o> d ) num__25 km / hr <o> e ) num__40 km / hr |
speed = num__8.33333333333 = num__8.33333333333 m / sec = num__8.33333333333 * num__6.0 km / hr = num__30 km / hr answer is c <eor> c <eos> |
c |
divide__200.0__24.0__ add__24.0__6.0__ round__30.0__ |
divide__200.0__24.0__ add__24.0__6.0__ round__30.0__ |
| if num__16.0 of num__40.0 of a number is num__6 then the number is <o> a ) num__200 <o> b ) num__225 <o> c ) num__93.75 <o> d ) num__320 <o> e ) none of these |
explanation : let num__0.16 × num__0.4 × a = num__6 a = num__6 × num__100 × num__6.25 × num__40 = num__93.75 correct option : c <eor> c <eos> |
c |
percent__93.75__100.0__ |
percent__93.75__100.0__ |
| how much num__60.0 of num__50 is greater than num__40.0 of num__30 ? <o> a ) num__18 <o> b ) num__55 <o> c ) num__33 <o> d ) num__22 <o> e ) num__11 |
( num__0.6 ) * num__50 – ( num__0.4 ) * num__30 num__30 - num__12 = num__18 answer : a <eor> a <eos> |
a |
percent__40.0__30.0__ percent__60.0__30.0__ percent__60.0__30.0__ |
percent__40.0__30.0__ percent__60.0__30.0__ percent__60.0__30.0__ |
| for every positive odd integer n the function h ( n ) is defined to be the product of all the odd integers from num__1 to n inclusive . if p is the smallest prime factor of h ( num__100 ) + num__2 ( num__100 ) + num__1 then p is ? <o> a ) between num__1 and num__20 <o> b ) between num__10 and num__20 <o> c ) between num__20 and num__30 <o> d ) between num__30 and num__40 <o> e ) num__1 |
h ( num__100 ) will have num__0 in its unit digit because while doing the calculation of the function we are multiplying by num__100 . so h ( num__100 ) + num__2 ( num__100 ) + num__1 will have num__1 in the units digit so smallest prime factor is num__1 . answer : e <eor> e <eos> |
e |
reverse__1.0__ |
reverse__1.0__ |
| if n is a positive integer and n ^ num__2 is divisible by num__72 then the largest positive integer that must divide n is ? . <o> a ) num__6 <o> b ) num__12 <o> c ) num__24 <o> d ) num__36 <o> e ) num__48 |
num__72 x num__1 = num__72 num__72 x num__2 = num__144 num__14 ^ num__4 = num__12 ^ num__2 largest possible which can divide num__12 is num__12 answer = b <eor> b <eos> |
b |
multiply__2.0__72.0__ subtract__14.0__2.0__ multiply__1.0__12.0__ |
multiply__2.0__72.0__ subtract__14.0__2.0__ multiply__1.0__12.0__ |
| in a mixture of num__13 litres the ratio of milk and water is num__3 : num__2 . if num__3 liters of this mixture is replaced by num__3 liters of milk then what will be the ratio of milk and water in the newly formed mixture ? <o> a ) num__9 : num__1 <o> b ) num__9 : num__2 <o> c ) num__9 : num__7 <o> d ) num__9 : num__4 <o> e ) num__9 : num__3 |
given : total quantity of mixture = num__13 liters num__3 litres of mixture is removed from the container – so let ' s forget this altogether ! now you are left with only num__10 litres of mixture in num__3 : num__2 ratio . milk in num__10 litres mix = num__10 x num__3 = num__6 litres ( num__2 + num__3 ) water in num__10 litres mix = num__10 x num__2 = num__4 litres ( num__2 + num__3 ) we add num__3 litres milk to this . so milk in new mix is = num__6 liters + num__3 litres = num__9 litres water = num__4 litres ratio of milk : water = num__9 : num__4 correct option : d <eor> d <eos> |
d |
subtract__13.0__3.0__ multiply__3.0__2.0__ subtract__6.0__2.0__ subtract__13.0__4.0__ subtract__13.0__4.0__ |
subtract__13.0__3.0__ multiply__3.0__2.0__ subtract__6.0__2.0__ add__3.0__6.0__ add__3.0__6.0__ |
| a circular mat with diameter num__7 inches is placed on a square tabletop each of whose sides is num__10 inches long . which of the following is closest to the fraction of the tabletop covered by the mat ? <o> a ) num__0.333333333333 <o> b ) num__0.333333333333 <o> c ) num__0.166666666667 <o> d ) num__1.5 <o> e ) num__0.142857142857 |
so we are looking for the area of the cloth over the area of the table area of the cloth = ( pi ) ( r ) ^ num__2 which is about ( num__3.14285714286 ) ( num__7 ) ( num__7 ) area of the table = ( num__10 ) ( num__10 ) so the quick way to estimate is looking at the fraction like this : num__1.54 nearest to num__1.5 answer : d <eor> d <eos> |
d |
triangle_area__1.5__2.0__ |
triangle_area__1.5__2.0__ |
| peter invested a certain sum of money in a simple interest bond whose value grew to $ num__300 at the end of num__3 years and further to $ num__400 at the end of another num__5 years . what was the rate of interest in which he invested his sum ? <o> a ) num__12.0 <o> b ) num__12.5 <o> c ) num__6.67 <o> d ) num__6.25 <o> e ) num__8.33 % |
answer initial amount invested = $ x amount at the end of year num__3 = $ num__300 amount at the end of year num__8 ( another num__5 years ) = $ num__400 therefore the interest earned for the num__5 year period between the num__3 rd year and num__8 th year = $ num__400 - $ num__300 = $ num__100 as the simple interest earned for a period of num__5 years is $ num__100 interest earned per year = $ num__20 . therefore interest earned for num__3 years = num__3 * num__20 = $ num__60 . hence initial amount invested x = amount after num__3 years - interest for num__3 years = num__300 - num__60 = $ num__240 . rate of interest = ( interest per year / principal ) * num__100 = num__0.0833333333333 * num__100 = num__8.33 choice is ( e ) <eor> e <eos> |
e |
percent__5.0__400.0__ percent__20.0__300.0__ percent__60.0__400.0__ percent__100.0__8.33__ |
percent__5.0__400.0__ percent__20.0__300.0__ percent__60.0__400.0__ percent__100.0__8.33__ |
| a gardener wants to plant trees in his garden in such a way that the number of trees in each row should be the same . if there are num__8 rows or num__6 rows or num__4 rows then no tree will be left . find the least number of trees required <o> a ) num__22 <o> b ) num__60 <o> c ) num__24 <o> d ) num__76 <o> e ) num__21 |
explanation : the least number of trees that are required = lcm ( num__8 num__64 ) = num__24 . answer : c <eor> c <eos> |
c |
multiply__6.0__4.0__ multiply__6.0__4.0__ |
multiply__6.0__4.0__ multiply__6.0__4.0__ |
| in an increasing sequence of num__10 consecutive integers the sum of the first num__5 integers is num__565 . what is the sum of the last num__5 integers in the sequence ? <o> a ) num__585 <o> b ) num__580 <o> c ) num__575 <o> d ) num__590 <o> e ) num__565 |
all num__5 integers are num__5 numbers larger than in the first sum ( eg . num__1 becomes num__6 num__2 num__7 . . . ) . num__5 * num__5 = num__25 + num__565 = num__590 d <eor> d <eos> |
d |
add__5.0__1.0__ divide__10.0__5.0__ add__5.0__2.0__ power__5.0__2.0__ add__565.0__25.0__ add__565.0__25.0__ |
add__5.0__1.0__ divide__10.0__5.0__ add__5.0__2.0__ power__5.0__2.0__ add__565.0__25.0__ add__565.0__25.0__ |
| ( ab ) x − num__2 = ( ba ) x − num__7 . what is the value of x ? <o> a ) num__1.5 <o> b ) num__4.5 <o> c ) num__7.5 <o> d ) num__9.5 <o> e ) num__8.7 |
explanation : ( a / b ) x − num__2 = ( b / a ) x − num__7 ⇒ ( a / b ) x − num__2 = ( a / b ) − ( x − num__7 ) ⇒ x − num__2 = − ( x − num__7 ) ⇒ x − num__2 = − x + num__7 ⇒ x − num__2 = − x + num__7 ⇒ num__2 x = num__9 ⇒ x = num__92 = num__4.5 option b <eor> b <eos> |
b |
add__2.0__7.0__ divide__9.0__2.0__ divide__9.0__2.0__ |
add__2.0__7.0__ divide__9.0__2.0__ divide__9.0__2.0__ |
| if the price of num__100 toys is num__2500 then what will the price of num__50 toys ? <o> a ) num__144 <o> b ) num__1250 <o> c ) num__117 <o> d ) num__287 <o> e ) num__112 |
one toy price = num__25.0 = num__25 num__50 toy price = num__50 * num__25 = num__1250 answer : b <eor> b <eos> |
b |
divide__2500.0__100.0__ multiply__50.0__25.0__ subtract__2500.0__1250.0__ |
divide__2500.0__100.0__ multiply__50.0__25.0__ multiply__50.0__25.0__ |
| on a sum of money the s . i . for num__2 years is $ num__600 while the c . i . is $ num__615 the rate of interest being the same in both the cases . the rate of interest is ? <o> a ) num__4.0 <o> b ) num__5.0 <o> c ) num__6.0 <o> d ) num__7.0 <o> e ) num__8 % |
difference in c . i . and s . i for num__2 years = $ num__615 - $ num__600 = $ num__15 s . i for one year = $ num__300 s . i . on $ num__300 for num__1 year = $ num__15 rate = ( num__100 * num__15 ) / ( num__300 ) = num__5.0 the answer is b . <eor> b <eos> |
b |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| the least number of complete years in which a sum of money put out at num__20.0 c . i . will be more than doubled is ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
p ( num__1 + num__0.2 ) n > num__2 p or ( num__1.2 ) n > num__2 now ( num__1.2 * num__1.2 * num__1.2 * num__1.2 ) > num__2 . so n = num__4 years . answer : b <eor> b <eos> |
b |
add__1.0__0.2__ multiply__20.0__0.2__ multiply__20.0__0.2__ |
add__1.0__0.2__ multiply__20.0__0.2__ multiply__20.0__0.2__ |
| two trains are moving in opposite directions at num__60 km / hr and num__90 km / hr . their lengths are num__1.10 km and num__0.9 km respectively . the time taken by the slower train to cross the faster train in seconds is ? <o> a ) num__22 <o> b ) num__37 <o> c ) num__48 <o> d ) num__266 <o> e ) num__121 |
relative speed = num__60 + num__90 = num__150 km / hr . = num__150 * num__0.277777777778 = num__41.6666666667 m / sec . distance covered = num__1.10 + num__0.9 = num__2 km = num__2000 m . required time = num__2000 * num__0.024 = num__48 sec . answer : c <eor> c <eos> |
c |
add__60.0__90.0__ add__1.1__0.9__ multiply__2000.0__0.024__ round__48.0__ |
add__60.0__90.0__ add__1.1__0.9__ multiply__2000.0__0.024__ multiply__2000.0__0.024__ |
| the sum of the first num__80 positive even integers is num__6480 . what is the sum of the first num__80 odd integers ? <o> a ) num__6200 <o> b ) num__6300 <o> c ) num__6400 <o> d ) num__6500 <o> e ) num__6600 |
sum of first n even numbers = n ( n + num__1 ) = num__6480 sum of first n odd numbers = n ^ num__2 = num__80 * num__80 = num__6400 ( here n = num__80 ) answer : c <eor> c <eos> |
c |
subtract__6480.0__80.0__ subtract__6480.0__80.0__ |
subtract__6480.0__80.0__ multiply__1.0__6400.0__ |
| the food in a camp lasts for num__20 men for num__40 days . if ten more men join how many days will the food last ? <o> a ) num__40 days <o> b ) num__20 days <o> c ) num__30 days <o> d ) num__50 days <o> e ) num__45 days |
one man can consume the same food in num__20 * num__40 = num__800 days . num__10 more men join the total number of men = num__20 the number of days the food will last = num__40.0 = num__40 days . answer : a <eor> a <eos> |
a |
multiply__20.0__40.0__ round__40.0__ |
multiply__20.0__40.0__ round__40.0__ |
| a trouser in bazaar was marked down num__40.0 the first week and another num__10.0 the next week . what percent of the regular price was the final cost of the trouser after the two markdowns ? <o> a ) num__30.0 <o> b ) num__36.0 <o> c ) num__40.0 <o> d ) num__50.0 <o> e ) num__54 % |
if the price of the trouser is num__100 and it is marked down by num__40.0 and then by num__10.0 or charged num__60.0 of the price first and num__90.0 of the price of reduced price the final price is : - num__100 * num__60 * num__0.009 = num__54 num__54.0 is the final price of old price . e is the answer <eor> e <eos> |
e |
percent__90.0__60.0__ percent__100.0__54.0__ |
percent__90.0__60.0__ percent__100.0__54.0__ |
| evaluate num__4.3 * num__4.3 * num__4.3 + num__1 / num__4.3 * num__4.3 - num__4.3 + num__1 <o> a ) num__14.3 <o> b ) num__52.3 <o> c ) num__5.3 <o> d ) num__42.3 <o> e ) num__32.3 |
given expression = a num__3 + b num__3 / a num__2 - ab + b num__2 = ( a + b ) = ( num__4.3 + num__1 ) = num__5.3 answer : c <eor> c <eos> |
c |
subtract__3.0__1.0__ add__4.3__1.0__ add__4.3__1.0__ |
subtract__3.0__1.0__ add__4.3__1.0__ add__4.3__1.0__ |
| the compounded ratio of ( num__2 : num__3 ) ( num__6 : num__11 ) and ( num__11 : num__2 ) is <o> a ) num__2 : num__0 <o> b ) num__2 : num__1 <o> c ) num__2 : num__9 <o> d ) num__2 : num__2 <o> e ) num__2 : num__5 |
answer : b ) num__2 : num__1 <eor> b <eos> |
b |
subtract__3.0__2.0__ multiply__2.0__1.0__ |
subtract__3.0__2.0__ multiply__2.0__1.0__ |
| a and b together can do a piece of work in num__10 days . b alone can finish it in num__20 days . in how many days can a alone finish the work ? <o> a ) num__20 days <o> b ) num__10 days <o> c ) num__30 days <o> d ) num__40 days <o> e ) num__50 days |
a num__20 days time taken by a to finish the work = xy / ( y - x ) num__10 x num__20 / ( num__20 - num__10 ) num__20.0 num__20 days <eor> a <eos> |
a |
round__20.0__ |
round__20.0__ |
| if k is a positive integer which of the following must be divisible by num__26 ? <o> a ) ( k – num__4 ) ( k ) ( k + num__3 ) ( k + num__7 ) <o> b ) ( k – num__4 ) ( k – num__2 ) ( k + num__3 ) ( k + num__5 ) <o> c ) ( k – num__2 ) ( k + num__3 ) ( k + num__5 ) ( k + num__6 ) <o> d ) ( k + num__1 ) ( k + num__3 ) ( k + num__5 ) ( k + num__7 ) <o> e ) ( k – num__3 ) ( k + num__1 ) ( k + num__4 ) ( k + num__6 ) |
num__24 = num__8 * num__3 . note that the product of two consecutive even integers is always divisible by num__8 ( since one of them is divisible by num__4 and another by num__2 ) . only option b offers two consecutive even numbers for any integer value of k : k - num__4 and k - num__2 if k = even or k + num__3 and k + num__5 if k = odd . also from the following num__3 consecutive integers : ( k - num__4 ) ( k - num__3 ) ( k - num__2 ) one must be divisible by num__3 if it ' s not k - num__4 or k - num__2 then it must be k - num__3 ( if it ' s k - num__4 or k - num__2 option b is divisible by num__3 right away ) . but if it ' s k - num__3 then ( k - num__3 ) + num__6 = k + num__3 must also be divisible by num__3 . so option b : ( k – num__4 ) ( k – num__2 ) ( k + num__3 ) ( k + num__5 ) is divisible by num__8 and num__3 in any case . answer : d . <eor> d <eos> |
d |
divide__24.0__8.0__ subtract__26.0__24.0__ add__2.0__3.0__ multiply__2.0__3.0__ subtract__3.0__2.0__ |
divide__24.0__8.0__ subtract__26.0__24.0__ add__2.0__3.0__ multiply__2.0__3.0__ subtract__3.0__2.0__ |
| num__15 men take num__21 days of num__8 hrs . each to do a piece of work . how many days of num__9 hrs . each would it take for num__21 women if num__3 women do as much work as num__2 men ? <o> a ) num__30 <o> b ) num__20 <o> c ) num__19 <o> d ) num__29 <o> e ) num__39 |
let num__1 man does num__1 unit / hr of work num__15 m in num__21 days of num__8 hrs will do ( num__15 * num__21 * num__8 ) units num__3 w = num__2 m num__1 w = ( num__0.666666666667 ) units / hr num__21 w with num__9 hrs a day will take ( num__15 * num__21 * num__8 ) / ( num__21 * num__9 * ( num__0.666666666667 ) ) days = > num__20 days answer : b <eor> b <eos> |
b |
subtract__9.0__8.0__ divide__2.0__3.0__ subtract__21.0__1.0__ round__20.0__ |
subtract__9.0__8.0__ divide__2.0__3.0__ subtract__21.0__1.0__ divide__20.0__1.0__ |
| average of five numbers is num__8 and the sum of three of the numbers is num__21 what is the average of the other num__2 numbers ? <o> a ) num__9.1 <o> b ) num__5.2 <o> c ) num__9.5 <o> d ) num__7.8 <o> e ) num__5.6 |
let the five numbers be a b c d e . then their average is ( a + b + c + d + e ) / num__5 = num__8 a + b + c = num__21 ( num__21 + d + e ) / num__5 = num__8 num__21 + d + e = num__40 d + e = num__19 average = num__9.5 = num__9.5 ans c <eor> c <eos> |
c |
multiply__8.0__5.0__ subtract__21.0__2.0__ divide__19.0__2.0__ divide__19.0__2.0__ |
multiply__8.0__5.0__ subtract__21.0__2.0__ divide__19.0__2.0__ divide__19.0__2.0__ |
| num__5 men are equal to as many women as are equal to num__8 boys . all of them earn rs . num__90 only . men â € ™ s wages are ? <o> a ) rs . num__6 <o> b ) rs . num__6.5 <o> c ) rs . num__8 <o> d ) rs . num__5 <o> e ) rs . num__2 |
answer : option a num__5 m = xw = num__8 b num__5 m + xw + num__8 b - - - - - num__90 rs . num__5 m + num__5 m + num__5 m - - - - - num__90 rs . num__15 m - - - - - - num__90 rs . = > num__1 m = num__6 rs . <eor> a <eos> |
a |
add__5.0__1.0__ add__5.0__1.0__ |
add__5.0__1.0__ add__5.0__1.0__ |
| if a : b = num__4 : num__9 and b : c = num__5 : num__9 then a : b : c is : <o> a ) num__20 : num__35 : num__63 <o> b ) num__35 : num__36 : num__63 <o> c ) num__30 : num__35 : num__65 <o> d ) num__20 : num__45 : num__81 <o> e ) none of these |
expl : a : b = num__4 : num__9 b : c = num__5 : num__9 = num__5 * num__1.8 : num__9 * num__1.8 = num__9 : num__16.2 a : b : c = num__4 : num__9 : num__16.2 = num__20 : num__45 : num__81 answer : d <eor> d <eos> |
d |
divide__9.0__5.0__ multiply__9.0__1.8__ multiply__4.0__5.0__ multiply__9.0__5.0__ multiply__5.0__16.2__ multiply__4.0__5.0__ |
divide__9.0__5.0__ multiply__9.0__1.8__ multiply__4.0__5.0__ multiply__9.0__5.0__ multiply__5.0__16.2__ multiply__4.0__5.0__ |
| calculate the % profit or loss if the cost price of num__9 balls is equal to the selling price of num__14 balls . <o> a ) num__35.7 <o> b ) num__25.7 <o> c ) num__15.7 <o> d ) num__55.7 <o> e ) num__45.7 % |
let the cost price of num__1 balls be rs num__1 cost of num__14 balls = rs num__14 selling price of num__14 balls = num__9 gain / loss = num__14 - num__9 = num__5.0 = num__0.357142857143 â ˆ — num__100 = num__35.7 answer : a <eor> a <eos> |
a |
percent__100.0__35.7__ |
percent__100.0__35.7__ |
| if num__2 m + n = num__7 and m + num__2 n = num__5 then ( num__2 m + n ) / num__2 = <o> a ) num__1 <o> b ) num__3.5 <o> c ) num__3.4 <o> d ) num__3.6 <o> e ) num__4 |
num__2 * ( m + num__2 m = num__5 ) equals num__2 m + num__4 n = num__10 num__2 m + num__4 n = num__10 - num__2 m + n = num__7 = num__3 n = num__3 therefore n = num__1 plug and solve . . . num__2 m + num__1 = num__7 num__2 m = num__6 m = num__3 ( num__6 + num__1 ) / num__2 = num__3.5 b <eor> b <eos> |
b |
multiply__2.0__5.0__ subtract__7.0__4.0__ subtract__5.0__4.0__ multiply__2.0__3.0__ divide__7.0__2.0__ divide__7.0__2.0__ |
multiply__2.0__5.0__ subtract__7.0__4.0__ subtract__5.0__4.0__ multiply__2.0__3.0__ divide__7.0__2.0__ divide__7.0__2.0__ |
| if an object travels num__80 feet in num__4 seconds what is the object ’ s approximate speed in miles per hour ? ( note : num__1 mile = num__5280 feet ) <o> a ) num__7.25 <o> b ) num__9.47 <o> c ) num__11.58 <o> d ) num__13.64 <o> e ) num__15.92 |
num__80 feet / num__4 seconds = num__20 feet / second ( num__20 feet / second ) * ( num__3600 seconds / hour ) * ( num__1 mile / num__5280 feet ) = num__13.64 miles / hour ( approximately ) the answer is d . <eor> d <eos> |
d |
divide__80.0__4.0__ round__13.64__ |
divide__80.0__4.0__ multiply__1.0__13.64__ |
| a man takes num__6 hours num__15 minutes in walking a distance and riding back to the starting place . he could walk both ways in num__7 hours num__45 minutes . the time taken by him to ride both ways is <o> a ) num__4 hours <o> b ) num__4 hours num__30 minutes <o> c ) num__4 hours num__45 minutes <o> d ) num__5 hours <o> e ) none of these |
explanation : time taken in walking both ways = num__7 hours num__45 minutes - - - - - - - - ( i ) time taken in walking one way and riding back = num__6 hours num__15 minutes - - - - - - - ( ii ) by equation ( ii ) * num__2 - ( i ) we have time taken to man ride both ways = num__12 hours num__30 minutes - num__7 hours num__45 minutes = num__4 hours num__45 answer : c <eor> c <eos> |
c |
multiply__6.0__2.0__ multiply__15.0__2.0__ subtract__6.0__2.0__ round__4.0__ |
multiply__6.0__2.0__ multiply__15.0__2.0__ subtract__6.0__2.0__ subtract__6.0__2.0__ |
| if x ^ num__2 + ( num__1 / x ^ num__2 ) = num__9 x ^ num__4 + ( num__1 / x ^ num__4 ) = ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__12 <o> d ) num__14 <o> e ) num__79 |
- > x ^ num__4 + ( num__1 / x ^ num__4 ) = ( x ^ num__2 ) ^ num__2 + ( num__1 / x ^ num__2 ) ^ num__2 = ( x ^ num__2 + num__1 / x ^ num__2 ) ^ num__2 - num__2 x ^ num__2 ( num__1 / x ^ num__2 ) = num__9 ^ num__2 - num__2 = num__79 . thus the answer is e . <eor> e <eos> |
e |
multiply__1.0__79.0__ |
divide__79.0__1.0__ |
| there are num__100 men in town . out of which num__85.0 were married num__70.0 have a phone num__75.0 own a car num__80.0 own a house . what is the maximum number of people who are married own a phone own a car and own a house ? <o> a ) num__50 <o> b ) num__60 <o> c ) num__70 <o> d ) num__80 <o> e ) num__90 |
total persons = num__100 no . peoplehaving phone = num__70 no . people having car = num__75 no . people having house = num__80 no . people got marriage = num__85 people got married having car house phone = num__70 answer : c <eor> c <eos> |
c |
percent__100.0__70.0__ |
percent__100.0__70.0__ |
| a person purchased a tv set for rs . num__15000 and a dvd player for rs . num__6000 . he sold both the items together for rs . num__31500 . what percentage of profit did he make ? <o> a ) num__48.0 <o> b ) num__50.0 <o> c ) num__40.0 <o> d ) num__45.0 <o> e ) num__20 % |
the total cp = rs . num__15000 + rs . num__6000 = rs . num__21000 and sp = rs . num__31500 profit ( % ) = ( num__31500 - num__21000 ) / num__21000 * num__100 = num__50.0 answer : b <eor> b <eos> |
b |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| a train num__125 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is : <o> a ) num__56 <o> b ) num__50 <o> c ) num__72 <o> d ) num__27 <o> e ) num__25 |
speed of the train relative to man = ( num__12.5 ) m / sec = ( num__12.5 ) m / sec . [ ( num__12.5 ) * ( num__3.6 ) ] km / hr = num__45 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__45 = = > x = num__50 km / hr . answer : b <eor> b <eos> |
b |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ round__50.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ multiply__5.0__10.0__ multiply__5.0__10.0__ |
| train a and train b began traveling towards each other from opposite ends of a num__480 - mile long track at num__1 : num__00 pm . if train a traveled at num__35 miles per hour and train b traveled at num__25 miles per hour at what time did the trains meet ? <o> a ) num__5 : num__40 pm <o> b ) num__6 : num__00 pm <o> c ) num__7 : num__20 pm <o> d ) num__9 : num__00 pm <o> e ) num__9 : num__20 pm |
this question tests your concepts of relative speed : if two trains are moving in opposite direction then distance will reduce at a rate of ( speed of num__1 + speed of num__2 ) if two trains are moving in same direction then distance will reduce at a rate of ( speed of num__1 - speed of num__2 ) here the trains are moving towards each other hence in opposite direction . trains will cover the num__500 mile track in a time of num__500 / ( num__25 + num__35 ) = num__8.0 = num__8 hours trains started at num__1 : num__00 pm there they will meet at num__9 : num__00 pm option d <eor> d <eos> |
d |
add__1.0__8.0__ round__9.0__ |
add__1.0__8.0__ add__1.0__8.0__ |
| what is the product of all positive factors of num__12 ? <o> a ) num__39 <o> b ) num__324 <o> c ) num__1728 <o> d ) num__3042 <o> e ) num__5832 |
positive factors of num__12 are num__1 num__2 num__3 num__4 num__6 num__12 . so product is : num__12 * num__6 * num__4 * num__3 * num__2 * num__1 = num__1728 answer : c <eor> c <eos> |
c |
add__1.0__2.0__ divide__12.0__3.0__ divide__12.0__2.0__ lcm__12.0__1728.0__ |
add__1.0__2.0__ divide__12.0__3.0__ multiply__2.0__3.0__ multiply__1728.0__1.0__ |
| a wheel rotates num__20 times every minute and moves num__15 cm during each rotation . how many metres does the wheel move in one hour ? <o> a ) num__6 metre <o> b ) num__18 metre <o> c ) num__180 metre <o> d ) num__1200 metre <o> e ) num__130 metre |
number of times wheel moves in num__1 hour = num__12 * num__60 = num__1200 : . distance moves = ( num__1200 * num__15 ) cms = num__18000 cms in metres = num__180 metre answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ multiply__20.0__60.0__ multiply__15.0__1200.0__ multiply__15.0__12.0__ round__180.0__ |
hour_to_min_conversion__ multiply__20.0__60.0__ multiply__15.0__1200.0__ multiply__15.0__12.0__ multiply__15.0__12.0__ |
| num__16 men can complete a piece of work in num__30 days . in how many days can num__20 men complete that piece of work ? <o> a ) num__55 days <o> b ) num__77 days <o> c ) num__20 days <o> d ) num__24 days <o> e ) num__44 days |
num__16 * num__30 = num__20 * x = > x = num__24 days answer : d <eor> d <eos> |
d |
round__24.0__ |
round__24.0__ |
| the rate of increase of the price of sugar is observed to be two percent more than the inflation rate expressed in percentage . the price of sugar on january num__1 num__1994 is rs . num__20 per kg . the inflation rate for the years num__1994 and num__1995 are expected to be num__8.0 each . the expected price of sugar on january num__1 num__1996 would be <o> a ) num__23.6 <o> b ) num__24.0 <o> c ) num__24.2 <o> d ) num__24.6 <o> e ) none of these |
explanation : increase in the price of sugar = ( num__8 + num__2 ) = num__10.0 hence price of the sugar on jan num__1 num__1996 = > ( num__20 * num__110 * num__110 ) / ( num__100 * num__100 ) = rs num__24.20 . answer : c <eor> c <eos> |
c |
subtract__1996.0__1994.0__ divide__20.0__2.0__ subtract__110.0__10.0__ multiply__1.0__24.2__ |
subtract__1996.0__1994.0__ divide__20.0__2.0__ subtract__110.0__10.0__ multiply__1.0__24.2__ |
| if x and y are integers and num__3 ^ ( x - num__4 ) = num__3 ^ ( y + num__2 ) what is x in terms of y ? <o> a ) y - num__6 <o> b ) y - num__5 <o> c ) y - num__2 <o> d ) y + num__2 <o> e ) y + num__6 |
answer : a num__3 ^ ( x - num__4 ) = num__3 ^ ( y + num__2 ) since the base is the same i isolated the equations and solved for y . x - num__4 = y + num__2 x = y + num__6 ans : e <eor> e <eos> |
e |
multiply__3.0__2.0__ multiply__3.0__2.0__ |
add__4.0__2.0__ add__4.0__2.0__ |
| num__1 num__35 num__79 . . num__50 find term of sequnce for this . <o> a ) num__1345 <o> b ) num__1567 <o> c ) num__1243 <o> d ) num__2767 <o> e ) num__2500 |
this is an arithmetic progression and we can write down a = num__1 a = num__1 d = num__2 d = num__2 n = num__50 n = num__50 . we now use the formula so that sn = num__12 n ( num__2 a + ( n − num__1 ) l ) sn = num__12 n ( num__2 a + ( n − num__1 ) l ) s num__50 = num__12 × num__50 × ( num__2 × num__1 + ( num__50 − num__1 ) × num__2 ) s num__50 = num__12 × num__50 × ( num__2 × num__1 + ( num__50 − num__1 ) × num__2 ) = num__25 × ( num__2 + num__49 × num__2 ) = num__25 × ( num__2 + num__49 × num__2 ) = num__25 × ( num__2 + num__98 ) = num__25 × ( num__2 + num__98 ) = num__2500 = num__2500 . e <eor> e <eos> |
e |
divide__50.0__2.0__ subtract__50.0__1.0__ multiply__2.0__49.0__ power__50.0__2.0__ multiply__1.0__2500.0__ |
divide__50.0__2.0__ subtract__50.0__1.0__ multiply__2.0__49.0__ power__50.0__2.0__ multiply__1.0__2500.0__ |
| if ( num__6.75 ) x - num__18 = num__3 x + num__27 what is the value of x ? <o> a ) num__12 <o> b ) - num__12 <o> c ) num__0.0833333333333 <o> d ) - num__0.0833333333333 <o> e ) num__3 |
( num__6.75 ) x - num__18 = num__3 x + num__27 = > num__27 x - num__72 = num__12 x + num__108 = > num__15 x = num__180 = > x = num__12 answer : a <eor> a <eos> |
a |
subtract__18.0__3.0__ add__72.0__108.0__ subtract__27.0__15.0__ |
subtract__18.0__3.0__ add__72.0__108.0__ subtract__27.0__15.0__ |
| two trains are moving in opposite directions at num__60 km / hr and num__90 km / hr . their lengths are num__1.10 km and num__0.9 km respectively . the time taken by the slower train to cross the faster train in seconds is ? <o> a ) num__48 sec <o> b ) num__77 <o> c ) num__55 <o> d ) num__27 <o> e ) num__21 |
relative speed = num__60 + num__90 = num__150 km / hr . = num__150 * num__0.277777777778 = num__41.6666666667 m / sec . distance covered = num__1.10 + num__0.9 = num__2 km = num__2000 m . required time = num__2000 * num__0.024 = num__48 sec . answer : a <eor> a <eos> |
a |
add__60.0__90.0__ add__1.1__0.9__ multiply__2000.0__0.024__ round__48.0__ |
add__60.0__90.0__ add__1.1__0.9__ multiply__2000.0__0.024__ multiply__2000.0__0.024__ |
| when positive integer x is divided by positive integer y the remainder is num__11.52 . if x / y = num__96.12 what is the value of y ? <o> a ) num__96 <o> b ) num__75 <o> c ) num__48 <o> d ) num__25 <o> e ) num__12 |
when positive integer x is divided by positive integer y the remainder is num__11.52 - - > x = qy + num__11.52 ; x / y = num__96.12 - - > x = num__96 y + num__0.12 y ( so q above equals to num__96 ) ; num__0.12 y = num__11.52 - - > y = num__96 . answer : a . <eor> a <eos> |
a |
round_down__96.12__ divide__11.52__96.0__ round_down__96.12__ |
round_down__96.12__ divide__11.52__96.0__ divide__11.52__0.12__ |
| three independent strategies a b and c have been initiated for cost cutting in a company producing respectively num__30.0 num__40.0 and num__10.0 savings . assuming that they operate independently what is the net saving achieved ? <o> a ) num__56.0 <o> b ) num__64.0 <o> c ) num__62.2 <o> d ) num__68.0 <o> e ) num__61 % |
if initial cost is rs num__100 then final cost will be num__100 * num__0.7 * num__0.6 * num__0.9 = rs . num__37.8 savings = num__100 - num__37.8 = num__62.2 so num__62.2 answer : c <eor> c <eos> |
c |
percent__100.0__62.2__ |
percent__100.0__62.2__ |
| the area of a square garden is q square feet and the perimeter is p feet . if q = num__2 p + num__33 what is the perimeter of the garden in feet ? <o> a ) num__28 <o> b ) num__32 <o> c ) num__36 <o> d ) num__40 <o> e ) num__44 |
let x be the length of one side of the square garden . x ^ num__2 = num__8 x + num__33 x ^ num__2 - num__8 x - num__33 = num__0 ( x - num__11 ) ( x + num__3 ) = num__0 x = num__11 - num__3 p = num__4 ( num__11 ) = num__44 the answer is e . <eor> e <eos> |
e |
square_perimeter__2.0__ rectangle_perimeter__2.0__0.0__ square_perimeter__11.0__ square_perimeter__11.0__ |
square_perimeter__2.0__ rectangle_perimeter__2.0__0.0__ square_perimeter__11.0__ square_perimeter__11.0__ |
| in a certain experiment the data collected is the number of organisms per sample and this data follows a normal distribution . if the sample of data has a mean of num__50 and a standard deviation of num__10 which of the following is exactly num__1.75 standard deviations more than the mean ? <o> a ) a ) num__48 <o> b ) b ) num__67.5 <o> c ) c ) num__72 <o> d ) d ) num__77.5 <o> e ) e ) num__81 |
standard deviation is a relatively rare category in the quant section although you ' re like to be tested on it num__1 time on test day . you ' ll never be asked to calculate sd though so you really just need to learn the basic ' concepts ' behind it . here we ' re told two things about a group of numbers : num__1 ) the average of the group is num__50 num__2 ) the standard deviation of the group is num__10 if you go num__1 sdupfrom the average you hit . . . . . num__50 + num__10 = num__60 if you go num__1 sddownfrom the average you hit . . . . . num__50 - num__10 = num__40 if you go num__2 sdsupfrom the average you hit . . . . . num__50 + num__2 ( num__10 ) = num__70 if you go num__2 sdsdownfrom the average you hit . . . . . num__50 - num__2 ( num__10 ) = num__30 etc . here we ' re asked for the number that is exactly num__1.75 sds above the mean . . . . num__1.75 sdsupwould be . . . . . num__50 + num__1.75 ( num__10 ) = num__67.5 b <eor> b <eos> |
b |
round_down__1.75__ add__50.0__10.0__ subtract__50.0__10.0__ add__10.0__60.0__ subtract__70.0__40.0__ multiply__1.0__67.5__ |
round_down__1.75__ add__50.0__10.0__ subtract__50.0__10.0__ add__10.0__60.0__ subtract__70.0__40.0__ multiply__1.0__67.5__ |
| what is the perimeter of a square with area num__9 p ^ num__0.5 ? <o> a ) num__3 p / num__4 <o> b ) num__3 p ^ num__0.5 <o> c ) num__6 p <o> d ) num__3 p ^ num__2 <o> e ) num__4 p / num__3 |
area of square ( side ) ^ num__2 = ( num__3 p / num__2 ) ^ num__2 therefore side of the square = num__3 p / num__2 perimeter of square = num__4 * side = num__4 * ( num__3 p / num__2 ) = num__6 p answer is c . <eor> c <eos> |
c |
square_perimeter__0.5__ power__9.0__0.5__ multiply__2.0__3.0__ multiply__2.0__3.0__ |
square_perimeter__0.5__ power__9.0__0.5__ multiply__2.0__3.0__ multiply__2.0__3.0__ |
| if num__18888 – n is divisible by num__11 and num__0 < n < num__11 what is n ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__9 |
num__18 |
888 / num__11 = num__1717 with a remainder of num__1 . we need to subtract the remainder to get a multiple of num__11 . the answer is a . <eor> a <eos> |
a |
a |
| the speed at which a man can row a boat in still water is num__15 kmph . if he rows downstream where the speed of current is num__3 kmph what time will he take to cover num__60 metres ? <o> a ) num__16 seconds <o> b ) num__34 seconds <o> c ) num__14 seconds <o> d ) num__12 seconds <o> e ) num__15 seconds |
speed of the boat downstream = num__15 + num__3 = num__18 kmph = num__18 * num__0.277777777778 = num__5 m / s hence time taken to cover num__60 m = num__12.0 = num__12 seconds . answer : d <eor> d <eos> |
d |
add__15.0__3.0__ divide__15.0__3.0__ subtract__15.0__3.0__ round__12.0__ |
add__15.0__3.0__ divide__15.0__3.0__ divide__60.0__5.0__ divide__60.0__5.0__ |
| what is the present worth of rs . num__143 due in num__2 years at num__5.0 simple interest per annum <o> a ) num__110 <o> b ) num__120 <o> c ) num__130 <o> d ) num__140 <o> e ) none of these |
explanation : let the present worth be rs . x then s . i . = rs . ( num__143 - x ) = ( x * num__5 * num__0.02 ) = num__143 - x = num__10 x = num__14300 - num__100 x = num__110 x = num__14300 x = num__130 answer : c <eor> c <eos> |
c |
percent__100.0__130.0__ |
percent__100.0__130.0__ |
| a man gets a simple interest of rs . num__1000 on a certain principal at the rate of num__5.0 p . a in two years . find the compound interest the man will get on twice the principal in two years at the same rate . <o> a ) rs . num__500 <o> b ) rs . num__250 <o> c ) rs . num__5012.50 <o> d ) rs . num__512.5 <o> e ) none of these . |
let the principal be rs . p s . i at num__5.0 p . a in num__8 years on rs . p = rs . num__1000 ( p ) ( num__8 ) ( num__5 ) / num__100 = num__1000 p = num__2500 c . i on rs . num__2 p i . e . rs . num__5000 at num__5.0 p . a in two years = num__5000 { [ num__1 + num__0.05 ] num__2 - num__1 } = num__5000 { num__212 - num__1.0 } = rs . num__512.5 answer : d <eor> d <eos> |
d |
percent__5.0__1.0__ percent__100.0__512.5__ |
percent__5.0__1.0__ percent__100.0__512.5__ |
| a train num__180 m long running at num__75 kmph crosses a platform in num__40 sec . what is the length of the platform ? <o> a ) num__687 <o> b ) num__638 <o> c ) num__683 <o> d ) num__726 <o> e ) num__267 |
d = num__75 * num__0.277777777778 = num__40 = num__833 â € “ num__150 = num__683 answer : c <eor> c <eos> |
c |
subtract__833.0__150.0__ round__683.0__ |
subtract__833.0__150.0__ round__683.0__ |
| a train num__575 m long crosses a tunnel of length num__325 in num__90 sec . what is the speed of the train in kmph . <o> a ) num__28 <o> b ) num__32 <o> c ) num__36 <o> d ) num__24 <o> e ) num__42 |
explanation : total distance traveled = length of train + length of tunnel = num__575 + num__325 = num__900 time taken to cross the tunnel = num__90 sec . speed in kmph = distance / time * num__3.6 = num__10.0 * num__3.6 = num__36.0 = num__36 kmph answer : c <eor> c <eos> |
c |
add__575.0__325.0__ divide__900.0__90.0__ multiply__10.0__3.6__ round__36.0__ |
add__575.0__325.0__ divide__900.0__90.0__ multiply__10.0__3.6__ multiply__10.0__3.6__ |
| a train with num__120 wagons crosses john who is going in the same direction in num__36 seconds . it travels for half an hour from the time it starts overtaking the john ( he is riding on the horse ) before it starts overtaking the mike ( who is also riding on his horse ) coming from the opposite direction in num__24 seconds . in how much time ( in secs ) after the train has crossed the mike do the john meets to mike ? <o> a ) num__3476 s <o> b ) num__3500 s <o> c ) num__3650 s <o> d ) num__3670 s <o> e ) num__3576 s |
let the length of the train be l metres and speeds of the train arun and sriram be r a and s respectively then - - - - - - - - - - ( i ) and - - - - - - - - - ( ii ) from eq . ( i ) and ( ii ) num__3 ( r - a ) = num__2 ( r + k ) r = num__3 a + num__2 k in num__30 minutes ( i . e num__1800 seconds ) the train covers num__1800 r ( distance ) but the arun also covers num__1800 a ( distance ) in the same time . therefore distance between arun and sriram when the train has just crossed sriram = num__1800 ( r - a ) - num__24 ( a + k ) time required = = ( num__3600 - num__24 ) = num__3576 s e <eor> e <eos> |
e |
multiply__120.0__30.0__ subtract__3600.0__24.0__ round__3576.0__ |
multiply__120.0__30.0__ subtract__3600.0__24.0__ subtract__3600.0__24.0__ |
| if two sides of a triangle are num__6 and num__8 which of the following could be the area of triangle ? num__1 . num__35 num__2 . num__48 num__3 . num__56 <o> a ) a ) num__1 only <o> b ) b ) num__1 and num__2 only <o> c ) c ) num__1 and num__3 only <o> d ) d ) num__2 and num__3 only <o> e ) e ) num__1 num__2 and num__3 only |
when two sides of a triangle are known the maximum area occurs when the angle between those two sides is num__90 . lets say base = num__12 height = num__8 ( angle = num__90 ) maximum area = ( num__0.5 ) * num__12 * num__8 = num__48 ( this is the same maximum area even when base is num__8 and height is num__12 ) . if we fix the base and keep lowering the angle it will result in a lower height . hence the resulting area will always be < num__48 . c d and e are ruled out . num__1 are possible areas as their areas are less than num__48 . hence the answer is a . <eor> a <eos> |
a |
square_perimeter__3.0__ volume_cube__1.0__ |
multiply__6.0__2.0__ multiply__2.0__0.5__ |
| two trains num__170 m and num__160 m long run at the speed of num__60 km / hr and num__40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? <o> a ) num__10.7 <o> b ) num__10.9 <o> c ) num__10.2 <o> d ) num__10.88 <o> e ) num__11.88 |
relative speed = num__60 + num__40 = num__100 km / hr . = num__100 * num__0.277777777778 = num__27.7777777778 m / sec . distance covered in crossing each other = num__170 + num__160 = num__330 m . required time = num__330 * num__0.036 = num__11.88 sec . answer : e <eor> e <eos> |
e |
subtract__160.0__60.0__ add__170.0__160.0__ multiply__330.0__0.036__ round__11.88__ |
add__60.0__40.0__ add__170.0__160.0__ multiply__330.0__0.036__ multiply__330.0__0.036__ |
| find the curved surface area if the radius of a cone is num__28 m and slant height is num__30 m ? <o> a ) num__2550 <o> b ) num__2640 <o> c ) num__3960 <o> d ) num__4280 <o> e ) num__5490 |
cone curved surface area = Ï € rl num__3.14285714286 Ã — num__28 Ã — num__30 = num__88 Ã — num__30 = num__2640 m ( power num__2 ) answer is b . <eor> b <eos> |
b |
multiply__30.0__88.0__ multiply__30.0__88.0__ |
multiply__30.0__88.0__ multiply__30.0__88.0__ |
| it takes ten minutes to load a certain video on a cellphone and fifteen seconds to load that same video on a laptop . if the two devices were connected so that they operated in concert at their respective rates how many seconds would it take them to load the video rounded to the nearest hundredth ? <o> a ) num__13.42 <o> b ) num__13.86 <o> c ) num__14.25 <o> d ) num__14.63 <o> e ) num__14.88 |
the laptop can load the video at a rate of num__0.0666666666667 of the video per second . the phone can load the video at a rate of num__1 / ( num__60 * num__10 ) = num__0.00166666666667 of the video per second . the combined rate is num__0.0666666666667 + num__0.00166666666667 = num__0.0683333333333 of the video per second . the time required to load the video is num__14.6341463415 = num__14.63 seconds . the answer is d . <eor> d <eos> |
d |
hour_to_min_conversion__ round__14.63__ |
hour_to_min_conversion__ divide__14.63__1.0__ |
| two horses start trotting towards each other one from a to b and another from b to a . they cross each other after one hour and the first horse reaches b num__0.833333333333 hour before the second horse reaches a . if the distance between a and b is num__50 km . what is the speed of the slower horse ? <o> a ) num__22 km / h <o> b ) num__88 km / h <o> c ) num__11 km / h <o> d ) num__20 km / h <o> e ) num__87 km / h |
explanation : if the speed of the faster horse be \ inline f _ { s } and that of slower horse be \ inline s _ { s } then \ inline f _ { s } + s _ { s } = \ frac { num__50 } { num__1 } = num__50 and \ inline \ frac { num__50 } { s _ { s } } - \ frac { num__50 } { f _ { s } } = \ frac { num__5 } { num__6 } now you can go through options . the speed of slower horse is num__20 km / h since num__20 + num__30 = num__50 and \ inline \ frac { num__50 } { num__20 } - \ frac { num__50 } { num__30 } = \ frac { num__5 } { num__6 } answer : d ) num__20 km / h <eor> d <eos> |
d |
add__1.0__5.0__ subtract__50.0__20.0__ round__20.0__ |
add__1.0__5.0__ subtract__50.0__20.0__ subtract__50.0__30.0__ |
| a can finish a piece of work in num__5 days . b can do it in num__16 days . they work together for two days and then a goes away . in how many days will b finish the work ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
num__0.5 + ( num__2 + x ) / num__16 = num__1 = > x = num__6 days answer : c <eor> c <eos> |
c |
multiply__0.5__2.0__ add__5.0__1.0__ round__6.0__ |
multiply__0.5__2.0__ add__5.0__1.0__ add__5.0__1.0__ |
| a plane was originally flying at an altitude of x feet when it ascended num__2000 feet and then descended num__5000 feet . if the plane ' s altitude after these two changes was num__0.166666666667 its original altitude then the solution of which of the following equations gives the plane ' s original altitude in feet ? <o> a ) x + num__2000 = num__0.333333333333 * ( x - num__3000 ) <o> b ) num__0.333333333333 * ( x - num__3000 ) = x <o> c ) x + num__3000 = num__0.333333333333 * x <o> d ) x - num__3000 = num__0.166666666667 * x <o> e ) x - num__3000 = num__0.333333333333 * x |
plane ' s original altitude = x plane ' s new altitude after ascending num__2000 ft = x + num__2000 plane ' s new altitude after descending num__5000 ft from previous altitude = x + num__2000 - num__5000 = x - num__3000 so after two changes plane is at num__0.166666666667 its original altitude = > x - num__3000 = x / num__6 answer ( d ) <eor> d <eos> |
d |
subtract__5000.0__2000.0__ subtract__5000.0__2000.0__ |
subtract__5000.0__2000.0__ subtract__5000.0__2000.0__ |
| bat is bought for rs . num__400 and sold at a gain of num__20.0 find its selling price ? <o> a ) rs . num__320 <o> b ) rs . num__350 <o> c ) rs . num__370 <o> d ) rs . num__480 <o> e ) none of these |
explanation : num__100.0 = num__400 ( num__100 * num__4 = num__400 ) num__120.0 = num__480 ( num__120 * num__4 = num__480 ) selling price = rs . num__480 / - answer : d <eor> d <eos> |
d |
percent__100.0__480.0__ |
percent__100.0__480.0__ |
| ray writes a two digit number . he sees that the number exceeds num__4 times the sum of its digits by num__3 . if the number is increased by num__18 the result is the same as the number formed by reversing the digits . find the number . <o> a ) num__2 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
solution : let the two digit number be xy . num__4 ( x + y ) + num__3 = num__10 x + y . . . . . . . ( num__1 ) num__10 x + y + num__18 = num__10 y + x . . . . ( num__2 ) solving num__1 st equation we get num__2 x - y = num__1 . . . . . ( num__3 ) solving num__2 nd equation we get y - x = num__2 . . . . . ( num__4 ) solving num__3 and num__4 we get x = num__3 and y = num__5 answer : b <eor> b <eos> |
b |
subtract__4.0__3.0__ subtract__3.0__1.0__ add__4.0__1.0__ add__4.0__1.0__ |
subtract__4.0__3.0__ subtract__3.0__1.0__ add__4.0__1.0__ add__4.0__1.0__ |
| a train takes num__18 seconds to pass completely through a station num__162 m long and num__15 seconds through another station num__120 m long . the length of the train is <o> a ) num__70 m <o> b ) num__80 m <o> c ) num__90 m <o> d ) num__100 m <o> e ) num__110 m |
let l be the length of the train . the train covers a distance of ( num__162 + l ) in num__18 seconds . the rate is then ( num__162 + l ) / num__18 . the train also covers a distance of ( num__120 + l ) in num__15 seconds . the rate is then ( num__120 + l ) / num__15 . presumably the train is traveling at a constant rate . therefore set the two rates equal to each other and solve for l . ( num__162 + l ) / num__18 = ( num__120 + l ) / num__15 i . e . num__90 answer : c <eor> c <eos> |
c |
round__90.0__ |
round__90.0__ |
| what is the rate percent when the simple interest on rs . num__800 amount to rs . num__160 in num__4 years ? <o> a ) num__5.0 <o> b ) num__7.0 <o> c ) num__2.0 <o> d ) num__1.0 <o> e ) num__9 % |
num__160 = ( num__180 * num__4 * r ) / num__100 r = num__5.0 answer : a <eor> a <eos> |
a |
percent__5.0__100.0__ |
percent__5.0__100.0__ |
| difference between the length & breadth of a rectangle is num__23 m . if its perimeter is num__206 m then its area is ? ? we have : ( l - b ) = num__23 and num__2 ( l + b ) = num__206 or ( l + b ) = num__103 ? <o> a ) num__2500 m ^ num__2 <o> b ) num__2510 m ^ num__2 <o> c ) num__2520 m ^ num__2 <o> d ) num__2600 m ^ num__2 <o> e ) num__2640 m ^ num__2 |
solving the two equations we get : l = num__63 and b = num__40 . area = ( l x b ) = ( num__63 x num__40 ) m num__2 = num__2520 m ^ num__2 c <eor> c <eos> |
c |
subtract__103.0__63.0__ multiply__40.0__63.0__ multiply__40.0__63.0__ |
subtract__103.0__63.0__ multiply__40.0__63.0__ multiply__40.0__63.0__ |
| a man rows his boat num__85 km downstream and num__45 km upstream taking num__2 num__0.5 hours each time . find the speed of the stream ? <o> a ) num__7 kmph <o> b ) num__4 kmph <o> c ) num__4 kmph <o> d ) num__8 kmph <o> e ) num__5 kmph |
speed downstream = d / t = num__85 / ( num__2 num__0.5 ) = num__34 kmph speed upstream = d / t = num__45 / ( num__2 num__0.5 ) = num__18 kmph the speed of the stream = ( num__34 - num__18 ) / num__2 = num__8 kmph answer : d <eor> d <eos> |
d |
round__8.0__ |
round__8.0__ |
| jacob ' s age num__18 years hence will be thrice his age four years ago . find jacob ' s present age ? <o> a ) num__16 <o> b ) num__20 <o> c ) num__15 <o> d ) num__11 <o> e ) num__16 |
c num__15 let jacob ' s present age be ' m ' years . m + num__18 = num__3 ( m - num__4 ) = > num__2 m = num__30 = > m = num__15 years . <eor> c <eos> |
c |
subtract__18.0__15.0__ multiply__2.0__15.0__ subtract__18.0__3.0__ |
subtract__18.0__15.0__ multiply__2.0__15.0__ subtract__18.0__3.0__ |
| tough and tricky questions : exponents . if num__3 ^ m num__3 ^ m num__3 ^ m = num__6 ^ n then m / n = <o> a ) num__0.333333333333 <o> b ) num__0.666666666667 <o> c ) num__1 <o> d ) num__1.5 <o> e ) num__3 |
i also soved it but a bit differently . thank you for reminding me of this much easier and logical way ! what i did was to test powers of num__3 that would lead to num__9 . so we have num__3 raised to the same power num__3 times and num__9 raised to another power one time : num__3 ^ m num__3 ^ m num__3 ^ m = num__9 ^ n what i did was test num__3 raised to the power of num__2 like this : num__3 ^ num__2 * num__3 ^ num__2 * num__3 ^ num__2 = num__9 * num__9 * num__9 = num__9 ^ num__3 . this means that m = num__2 and n = num__3 . so m / n = num__3 . e <eor> e <eos> |
e |
add__3.0__6.0__ divide__6.0__3.0__ subtract__6.0__3.0__ |
add__3.0__6.0__ divide__6.0__3.0__ divide__6.0__2.0__ |
| the ratio of two numbers is num__5 : num__6 . if a number is added to both the numbers the ratio becomes num__7 : num__8 . if the larger number exceeds the smaller number by num__10 . find the number added . <o> a ) num__30 <o> b ) num__10 <o> c ) num__20 <o> d ) num__40 <o> e ) can not be determined |
let the numbers are num__5 x and num__6 x . let the number added is y so num__5 x + y / num__6 x + y = num__0.875 and num__6 x - num__5 x = num__10 implies x = num__10 taking x = num__10 num__50 + y / num__60 + y = num__0.875 implies num__400 + num__8 y = num__420 + num__7 y or y = num__20 answer : c <eor> c <eos> |
c |
divide__7.0__8.0__ multiply__5.0__10.0__ multiply__6.0__10.0__ multiply__8.0__50.0__ multiply__7.0__60.0__ subtract__420.0__400.0__ subtract__420.0__400.0__ |
divide__7.0__8.0__ multiply__5.0__10.0__ add__10.0__50.0__ multiply__8.0__50.0__ multiply__7.0__60.0__ subtract__420.0__400.0__ subtract__420.0__400.0__ |
| if the wheel is num__12 cm then the number of revolutions to cover a distance of num__1240 cm is ? <o> a ) num__11 <o> b ) num__14 <o> c ) num__16 <o> d ) num__12 <o> e ) num__19 |
num__2 * num__3.14285714286 * num__12 * x = num__1240 = > x = num__16 answer : c <eor> c <eos> |
c |
round__16.0__ |
round__16.0__ |
| jane covered a distance of num__350 miles between city a and city b taking a total of num__5 hours . if part of the distance was covered at num__60 miles per hour speed and the balance at num__80 miles per hour speed how many hours did she travel at num__60 miles per hour ? <o> a ) num__3 hours num__30 minutes <o> b ) num__3 hours <o> c ) num__2 hours <o> d ) num__1 hour num__45 minutes <o> e ) none of these |
answer let jane travel ' x ' at num__60 miles per hour . as the total time taken to cover num__340 miles is num__5 hours jane would have traveled ( num__5 - x ) hours at num__80 miles per hour . distance covered at num__60 miles per hour = speed * time = num__60 * x = num__60 x miles . distance covered at num__80 miles per hour = speed * time = num__80 ( num__5 - x ) = num__400 - num__80 x miles . total distance covered = distance covered at num__60 miles per hour + distance covered at num__80 miles per hour . therefore total distance = num__60 x + num__400 - num__80 x . total distance travelled = num__340 miles . therefore num__350 = num__60 x + num__400 - num__80 x num__20 x = num__70 or x = num__3 hours num__30 minutes . choice a <eor> a <eos> |
a |
multiply__5.0__80.0__ subtract__80.0__60.0__ divide__350.0__5.0__ divide__60.0__20.0__ round__3.0__ |
multiply__5.0__80.0__ subtract__80.0__60.0__ divide__350.0__5.0__ divide__60.0__20.0__ round__3.0__ |
| sachin is younger than robert by num__4 years . if their ages are in the respective ratio of num__7 : num__9 how old is sachin ? <o> a ) num__24.5 <o> b ) num__32 <o> c ) num__25 <o> d ) num__28 <o> e ) num__30 |
let robert ' s age be x years then sachin ' s age = x - num__7 years ( x - num__7 ) / x = num__0.777777777778 x = num__31.5 sachin ' s age = x - num__7 = num__24.5 years answer is a <eor> a <eos> |
a |
divide__7.0__9.0__ subtract__31.5__7.0__ subtract__31.5__7.0__ |
divide__7.0__9.0__ subtract__31.5__7.0__ subtract__31.5__7.0__ |
| icici bank offers an interest of num__5.0 per annum compounded annually on all its deposits . if $ num__10000 is deposited what will be the ratio of the interest earned in the num__4 th year to the interest earned in the num__5 th year ? <o> a ) num__1 : num__5 <o> b ) num__625 : num__3125 <o> c ) num__100 : num__105 <o> d ) num__100 ^ num__4 : num__105 ^ num__4 <o> e ) num__725 : num__3225 |
hi bunuel here is my approach : is this correct ? icici bank offers an interest of num__5.0 per annum compounded annually on all its deposits . interest earned in num__4 year = num__10000 ( num__1 + num__0.05 ) ^ num__4 interest earned in num__5 year = num__10000 ( num__1 + num__0.05 ) ^ num__5 ratio = { num__10000 ( num__1.05 ) ^ num__4 } / { num__10000 ( num__1.05 ^ num__5 ) } = > num__1.05 ^ num__4 / num__1.05 ^ num__5 = > num__1 / num__1.05 multiplied by num__100 in both numerator and denominator gives num__100 : num__105 hence ans : c <eor> c <eos> |
c |
percent__5.0__1.0__ percent__1.0__10000.0__ percent__1.05__10000.0__ percent__1.0__10000.0__ |
percent__5.0__1.0__ percent__1.0__10000.0__ percent__1.05__10000.0__ percent__1.0__10000.0__ |
| the profit earned by selling an article for num__852 is equal to the loss incurred when the same article is sold for num__448 . what should be the sale price of the article for making num__50 per cent profit ? <o> a ) num__960 <o> b ) num__975 <o> c ) num__1200 <o> d ) num__920 <o> e ) none of these |
let the profit or loss be x and num__852 – x = num__448 + x or x = num__404 ⁄ num__2 = num__202 \ cost price of the article = num__852 – x = num__448 + x = num__650 \ sp of the article = num__650 × num__150 ⁄ num__100 = num__975 answer b <eor> b <eos> |
b |
percent__50.0__404.0__ percent__100.0__975.0__ |
percent__50.0__404.0__ percent__100.0__975.0__ |
| a train is num__100 meter long and is running at the speed of num__30 km per hour . find the time it will take to pass a man standing at a crossing . <o> a ) num__10 seconds <o> b ) num__12 seconds <o> c ) num__14 seconds <o> d ) num__16 seconds <o> e ) num__18 seconds |
explanation : as we need to get answer in seconds so never forget to convert speed into meter per second . speed = num__30 km / hr = num__30 * num__0.277777777778 m / sec = num__8.33333333333 m / sec distance = length of train = num__100 meter required time = distance / speed = num__4.0 / num__3 = num__100 ∗ num__0.12 = num__12 sec option b <eor> b <eos> |
b |
multiply__100.0__0.12__ round__12.0__ |
multiply__100.0__0.12__ multiply__100.0__0.12__ |
| anil can do a work in num__15 days while sunil can do it in num__25 days . how long will they take if both work together ? <o> a ) num__9 num__0.428571428571 days <o> b ) num__9 num__0.875 days <o> c ) num__9 num__0.375 days <o> d ) num__4 num__0.375 days <o> e ) num__9 num__0.75 days |
num__0.0666666666667 + num__0.04 = num__0.106666666667 num__9.375 = num__9 num__0.375 days answer : c <eor> c <eos> |
c |
add__0.0667__0.04__ divide__9.375__25.0__ round__9.0__ |
add__0.0667__0.04__ divide__9.375__25.0__ round__9.0__ |
| charan can do a piece o work in num__24 days which hari alone can finish in num__19 days . both together work for num__7 days and then hari leaves off . how many days will caran take to finish remaining work ? <o> a ) num__22 days <o> b ) num__22 ½ days <o> c ) num__21 ½ days <o> d ) num__20 days <o> e ) num__15 days |
explanation : caran and hari num__7 days work = num__7 ( num__0.0416666666667 + num__0.0526315789474 ) = num__0.094298245614 remaining work = ( num__1 - num__0.094298245614 ) = num__0.905701754386 charan num__1 days work = num__0.0416666666667 num__0.905701754386 work will be done by caran in num__24 * num__0.905701754386 = num__22 days ( approx ) answer : option a <eor> a <eos> |
a |
add__0.0417__0.0526__ subtract__1.0__0.0943__ round__22.0__ |
add__0.0417__0.0526__ subtract__1.0__0.0943__ round__22.0__ |
| if num__10 men and num__2 boys working together can do nine times as much work per hour as a man and a boy together . find the ratio of the work done by a man and that of a boy for a given time ? <o> a ) num__7 : num__5 <o> b ) num__7 : num__3 <o> c ) num__7 : num__8 <o> d ) num__7 : num__1 <o> e ) num__7 : num__2 |
num__10 m + num__2 b = num__9 ( num__1 m + num__1 b ) num__10 m + num__2 b = num__9 m + num__9 b num__1 m = num__7 b the required ratio of work done by a man and a boy = num__7 : num__1 answer : d <eor> d <eos> |
d |
subtract__10.0__9.0__ subtract__9.0__2.0__ round__7.0__ |
subtract__10.0__9.0__ subtract__9.0__2.0__ round__7.0__ |
| mira ’ s expenditure and sayings are in the ratio num__3 : num__2 . her income increases by num__10 percent . her expenditure also increases by num__12 percent . by how much per cent does her saving increase ? <o> a ) num__7.0 <o> b ) num__10.0 <o> c ) num__9.0 <o> d ) num__13.0 <o> e ) num__15 % |
we get two values of x num__7 and num__13 . but to get a viable answer we must keep in mind that the central value ( num__10 ) must lie between x and num__12 . thus the value of x should be num__7 and not num__13 thus required % increase = num__7.0 answer : a <eor> a <eos> |
a |
subtract__10.0__3.0__ add__3.0__10.0__ subtract__10.0__3.0__ |
subtract__10.0__3.0__ add__3.0__10.0__ subtract__10.0__3.0__ |
| three friends had dinner at a restaurant . when the bill was received akshitha paid num__0.2 as much as veena paid and veena paid num__0.5 as much as lasya paid . what fraction of the bill did veena pay ? <o> a ) num__0.157894736842 <o> b ) num__0.3125 <o> c ) num__0.272727272727 <o> d ) num__0.3 <o> e ) num__0.214285714286 |
let veena paid x so akshitha paid x / num__5 and lasya paid num__2 x so total bill paid is given by x + ( x / num__5 ) + num__2 x = num__1 we get i . e . x = num__0.3125 answer : b <eor> b <eos> |
b |
reverse__0.2__ reverse__0.5__ multiply__0.2__5.0__ multiply__0.3125__1.0__ |
reverse__0.2__ reverse__0.5__ multiply__0.2__5.0__ divide__0.3125__1.0__ |
| a box contains three red marbles and num__1 green marble . if the marbles are removed from the box one at a time in random order what is the probability that all three red marbles are removed before the green marble ? <o> a ) num__0.015625 <o> b ) num__0.0416666666667 <o> c ) num__0.0833333333333 <o> d ) num__0.25 <o> e ) num__0.5 |
there are num__3 red marbles and num__4 marbles total so the probability of drawing a red marble on the first draw is num__0.75 . on the second draw there are num__2 red marbles and three marbles total so the probability of drawing a red marble is num__0.666666666667 . on the third draw there is only one red marble and one green marble so the probability of drawing the red marble is num__0.5 . the probability of all three events happening is the product of the probabilities num__0.75 ∗ num__0.666666666667 ∗ num__0.5 or num__0.25 answer d <eor> d <eos> |
d |
coin_space__ negate_prob__0.75__ negate_prob__0.75__ |
coin_space__ negate_prob__0.75__ negate_prob__0.75__ |
| a small experimental plane has three engines one of which is redundant . that is as long as two of the engines are working the plane will stay in the air . over the course of a typical flight there is a num__0.333333333333 chance that engine one will fail . there is a num__70.0 probability that engine two will work . the third engine works only half the time . what is the probability that the plane will crash in any given flight ? <o> a ) num__0.583333333333 <o> b ) num__0.25 <o> c ) num__0.5 <o> d ) num__0.291666666667 <o> e ) num__0.708333333333 |
in probability questions the trap answer is just the multiple of the numbers in the question . i . e . if you multiply num__0.333333333333 * num__0.25 * num__0.5 = num__0.0416666666667 is trap answer the other trap answer could be num__0.666666666667 * num__0.75 * num__0.5 = num__0.25 is trap answer so lets say you have num__30 secsand you want to guess the answer then b c are ruled out because they can be traps . you best guess is a d e . so you have num__33.0 chances of being correct . e <eor> e <eos> |
e |
negate_prob__0.3333__ negate_prob__0.25__ union_prob__0.25__0.5__0.0417__ |
negate_prob__0.3333__ negate_prob__0.25__ union_prob__0.25__0.5__0.0417__ |
| ayush was born two years after his father ' s marriage . his mother is five years younger than his father but num__30 years older than ayush who is num__10 years old . at what age did the father get married ? <o> a ) num__30 years <o> b ) num__31 years <o> c ) num__32 years <o> d ) num__33 years <o> e ) num__34 years |
explanation : ayush ' s present age = num__10 years . his mother ' s present age = ( num__10 + num__30 ) years = num__40 years . ayush ' s father ' s present age = ( num__40 + num__5 ) years = num__45 years . ayush ' s father ' s age at the time of ayush ' s birth = ( num__45 - num__10 ) years = num__35 years . therefore ayush ' s father ' s age at the time of marriage = ( num__35 - num__2 ) years = num__33 years . answer : d ) num__33 year <eor> d <eos> |
d |
add__30.0__10.0__ add__5.0__40.0__ add__30.0__5.0__ divide__10.0__5.0__ subtract__35.0__2.0__ subtract__35.0__2.0__ |
add__30.0__10.0__ add__5.0__40.0__ add__30.0__5.0__ divide__10.0__5.0__ subtract__35.0__2.0__ subtract__35.0__2.0__ |
| find the cost of fencing around a circular field of diameter num__22 m at the rate of rs . num__2.50 a meter ? <o> a ) num__188 <o> b ) num__172 <o> c ) num__278 <o> d ) num__279 <o> e ) num__222 |
num__2 * num__3.14285714286 * num__11 = num__69 num__69 * num__2 num__0.5 = rs . num__172 answer : b <eor> b <eos> |
b |
divide__22.0__2.0__ subtract__2.5__2.0__ round__172.0__ |
divide__22.0__2.0__ subtract__2.5__2.0__ round__172.0__ |
| rs . num__1000 is divided into three parts a b and c . how much a is more than c if their ratio is num__0.5 : num__0.5 : num__0.25 ? <o> a ) num__200 <o> b ) num__992 <o> c ) num__772 <o> d ) num__552 <o> e ) num__312 |
num__0.5 : num__0.5 : num__0.25 = num__2 : num__2 : num__1 num__0.2 * num__1000 = num__200 num__400 - num__200 = num__200 answer : a <eor> a <eos> |
a |
reverse__0.5__ multiply__0.5__2.0__ multiply__1000.0__0.2__ multiply__2.0__200.0__ multiply__1000.0__0.2__ |
reverse__0.5__ multiply__0.5__2.0__ multiply__1000.0__0.2__ multiply__2.0__200.0__ multiply__1000.0__0.2__ |
| an air cooler is available for $ num__39000 cash or $ num__17000 as down payment followed by five equal monthly instalments of $ num__4800 each . the simple rate of interest per annum under the instalment plan would be <o> a ) num__18.0 <o> b ) num__19.0 <o> c ) num__21.2 <o> d ) num__21.81 <o> e ) num__22.07 % |
simple interest si = ( p * r * t ) / num__100 where p is the principal amount r is the rate of interest and t is time in years the way i see it : air cooler on down payment of num__17000 customer is not paying any interest . it is the remaining sum which will be paid for num__5 months that will bear an interest . therefore the principal amount for which interest is being charged is num__39000 - num__17000 = num__22000 for this num__22000 a total sum of num__5 * num__4800 = num__24000 was pain . ( time is five months so t = num__0.416666666667 as t is in years . ) thus si = num__2000 or num__2000 = ( p * r * t ) / num__100 num__2000 = ( num__22000 * r * num__5 ) / ( num__100 * num__12 ) r = ( num__2000 * num__12 * num__100 ) / num__22000 * num__5 r = num__21.81 therefore answer is d <eor> d <eos> |
d |
percent__21.81__100.0__ |
percent__21.81__100.0__ |
| the average mark obtained by num__22 candidates in an examination is num__48 . the average of the first ten is num__55 while the last eleven is num__40 . the marks obtained by the num__11 th candidate is ? <o> a ) num__22 <o> b ) num__0 <o> c ) num__49 <o> d ) num__58 <o> e ) num__66 |
it is clear that num__22 x num__48 = num__10 x num__55 + k + num__11 x num__40 ⇒ ⇒ k = num__66 answer : e <eor> e <eos> |
e |
add__55.0__11.0__ add__55.0__11.0__ |
add__55.0__11.0__ add__55.0__11.0__ |
| what is the probability of flipping a fair coin eight times and the coin landing on heads on at least two flips ? <o> a ) num__0.953125 <o> b ) num__0.9140625 <o> c ) num__0.9609375 <o> d ) num__0.96484375 <o> e ) num__0.98828125 |
the number of possible outcomes is num__2 ^ num__8 = num__256 . num__0 heads : there is num__1 way to have all tails . num__1 head : there are num__8 ways to have one head . p ( num__0 or num__1 head ) = num__0.03515625 p ( at least num__2 heads ) = num__1 - num__0.03515625 = num__0.96484375 the answer is d . <eor> d <eos> |
d |
coin_space__ negate_prob__0.0__ negate_prob__0.0352__ negate_prob__0.0352__ |
coin_space__ negate_prob__0.0__ negate_prob__0.0352__ negate_prob__0.0352__ |
| num__14 men can complete a piece of work in num__22 days . in how many days can num__18 men complete that piece of work ? <o> a ) num__23 days <o> b ) num__26 days <o> c ) num__17 days <o> d ) num__29 days <o> e ) num__20 days |
c num__17 days num__14 * num__22 = num__18 * x = > x = num__17 days <eor> c <eos> |
c |
round__17.0__ |
round__17.0__ |
| the measurement of a rectangular box with lid is num__25 cmx num__6 cmx num__18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of π cm num__3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere ) <o> a ) num__288 <o> b ) num__48 <o> c ) num__36 <o> d ) num__864 <o> e ) num__964 |
d = num__6 r = num__2 ; volume of the largest sphere = num__1.33333333333 π r num__3 = num__1.33333333333 * π * num__3 * num__3 * num__3 = num__36 π cm num__3 answer : c <eor> c <eos> |
c |
divide__6.0__3.0__ multiply__18.0__2.0__ round__36.0__ |
divide__6.0__3.0__ multiply__18.0__2.0__ multiply__18.0__2.0__ |
| a river num__2 m deep and num__45 m wide is flowing at the rate of num__5 kmph the amount of water that runs into the sea per minute is ? <o> a ) num__4500 m num__3 <o> b ) num__4580 m num__3 <o> c ) num__7500 m num__3 <o> d ) num__4900 m num__3 <o> e ) num__4700 m num__3 |
explanation : ( num__5000 * num__2 * num__45 ) / num__60 = num__7500 m num__3 answer : option c <eor> c <eos> |
c |
hour_to_min_conversion__ subtract__5.0__2.0__ round__7500.0__ |
hour_to_min_conversion__ subtract__5.0__2.0__ round__7500.0__ |
| a train passes a train coming from opposite direction speed num__32 kmph in num__25 sec . if the speed of the train is num__40 km / hr . what is the length of train ? <o> a ) num__300 <o> b ) num__400 <o> c ) num__500 <o> d ) num__350 <o> e ) num__200 |
speed = ( num__40 + num__32 ) * num__0.277777777778 = num__72 * num__0.277777777778 = num__20 m / sec . length of the train = num__25 * num__20 = num__500 m . answer : c <eor> c <eos> |
c |
add__32.0__40.0__ multiply__25.0__20.0__ round__500.0__ |
add__32.0__40.0__ multiply__25.0__20.0__ multiply__25.0__20.0__ |
| a certain animal in the zoo has consumed num__39 pounds of food in six days . if it continues to eat at the same rate in how many more days will its total consumption be num__104 pounds ? <o> a ) num__8 <o> b ) num__7 <o> c ) num__16 <o> d ) num__10 <o> e ) none of the above |
ans is d : num__39 pounds - - > num__6 days num__104 pounds - - > x days x = num__104 * num__0.153846153846 = num__16 the animal has already consumed food in num__6 days so the the number of days for it ' s total consumption be num__104 pounds is num__16 - num__6 = num__10 <eor> d <eos> |
d |
divide__6.0__39.0__ subtract__16.0__6.0__ subtract__16.0__6.0__ |
divide__6.0__39.0__ subtract__16.0__6.0__ subtract__16.0__6.0__ |
| a motorcyclist goes from nagpur to pune a distance of num__500 kms at an average of num__100 kmph speed . another man starts from nagpur by car num__3  ½ hours after the first and reaches pune  ½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? <o> a ) num__1 : num__2 <o> b ) num__1 : num__7 <o> c ) num__3 : num__5 <o> d ) num__1 : num__5 <o> e ) num__1 : num__1 |
t = num__5.0 = num__10 h t = num__10 - num__4 = num__6 time ratio = num__10 : num__6 = num__5 : num__3 speed ratio = num__3 : num__5 answer : c <eor> c <eos> |
c |
divide__500.0__100.0__ subtract__10.0__4.0__ round__3.0__ |
divide__500.0__100.0__ subtract__10.0__4.0__ subtract__6.0__3.0__ |
| rs . num__1500 is divided into two parts such that if one part is invested at num__6.0 and the other at num__5.0 the whole annual interest from both the sum is rs . num__86 . how much was lent at num__5.0 ? <o> a ) num__228 <o> b ) num__299 <o> c ) num__400 <o> d ) num__500 <o> e ) num__188 |
( x * num__5 * num__1 ) / num__100 + [ ( num__1500 - x ) * num__6 * num__1 ] / num__100 = num__85 num__5 x / num__100 + num__90 â € “ num__6 x / num__100 = num__86 x / num__100 = num__4 = > x = num__400 answer : c <eor> c <eos> |
c |
percent__6.0__1500.0__ percent__100.0__400.0__ |
percent__6.0__1500.0__ percent__100.0__400.0__ |
| a cistern num__4 meters long and num__4 meters wide contains water up to a depth of num__1 meter num__25 cm . what is the total area of the wet surface ? <o> a ) num__28 <o> b ) num__32 <o> c ) num__36 <o> d ) num__40 <o> e ) num__44 |
area of the wet surface = [ num__2 ( lb + bh + lh ) - lb ] = num__2 ( bh + lh ) + lb = [ num__2 ( num__4 x num__1.25 + num__4 x num__1.25 ) + num__4 x num__4 ] = num__36 the answer is c . <eor> c <eos> |
c |
round__36.0__ |
round__36.0__ |
| a cubical tank is filled with water to a level of num__1 foot . if the water in the tank occupies num__16 cubic feet to what fraction of its capacity is the tank filled with water ? <o> a ) num__0.5 <o> b ) num__0.333333333333 <o> c ) num__0.25 <o> d ) num__0.2 <o> e ) num__0.166666666667 |
the volume of water in the tank is h * l * b = num__16 cubic feet . since h = num__1 then l * b = num__16 and l = b = num__4 . since the tank is cubical the capacity of the tank is num__4 * num__4 * num__4 = num__64 . the ratio of the water in the tank to the capacity is num__0.25 = num__0.25 the answer is c . <eor> c <eos> |
c |
multiply__16.0__4.0__ divide__1.0__4.0__ round__0.25__ |
multiply__16.0__4.0__ divide__1.0__4.0__ round__0.25__ |
| a man cheats while buying as well as while selling . while buying he takes num__10.0 more than what he pays for and while selling he gives num__20.0 less than what he claims to . find the profit percent if he sells at num__10.0 below the cost price of the claimed weight . <o> a ) num__19.81 <o> b ) num__20.0 <o> c ) num__37.5 <o> d ) num__23.75 <o> e ) num__37.5 % |
there is a one step calculation method too . it requires more thought but is faster . the man takes num__10.0 more than what he pays for . so if he claims to take num__100 pounds he pays $ num__100 but he actually takes num__110 pounds for which he will take from the customer $ num__110 . hence in effect there is a num__10.0 mark up . while selling he sells num__20.0 less . this means he claims to sell num__100 pounds and gets $ num__100 but actually sells only num__80 pounds and should have got only $ num__80 for it . so this is again a mark up of $ num__20 on $ num__80 which is num__25.0 . but he also sells at num__10.0 less ( num__1 + m num__1.0 ) ( num__1 + m num__2.0 ) ( num__1 - d % ) = ( num__1 + p % ) num__1.1 * num__1.25 * num__0.9 = ( num__1 + p % ) profit % = num__23.75 d <eor> d <eos> |
d |
percent__10.0__20.0__ percent__1.0__110.0__ percent__100.0__23.75__ |
percent__10.0__20.0__ percent__1.0__110.0__ percent__100.0__23.75__ |
| find the area of a parallelogram with base num__24 cm and height num__16 cm ? <o> a ) num__297 cm num__2 <o> b ) num__384 cm num__2 <o> c ) num__672 cm num__2 <o> d ) num__267 cm num__2 <o> e ) num__286 cm num__2 |
area of a parallelogram = base * height = num__24 * num__16 = num__384 cm num__2 answer : b <eor> b <eos> |
b |
multiply__24.0__16.0__ multiply__24.0__16.0__ |
multiply__24.0__16.0__ multiply__24.0__16.0__ |
| the average age of students of a class is num__15.8 years . the average age of boys in the class is num__16.2 years and that of the girls is num__15.4 years . the ration of the number of boys to the number of girls in the class is : <o> a ) num__2 : num__5 <o> b ) num__2 : num__3 <o> c ) num__2 : num__4 <o> d ) num__1 : num__1 <o> e ) num__2 : num__9 |
let the ratio be k : num__1 . then k * num__16.2 + num__1 * num__15.4 = ( k + num__1 ) * num__15.8 = ( num__16.2 - num__15.8 ) k = ( num__15.8 - num__15.4 ) = k = num__0.4 / num__0.4 = num__1.0 required ratio = num__1.0 : num__1 = num__1 : num__1 . answer : d <eor> d <eos> |
d |
subtract__15.8__15.4__ reverse__1.0__ |
subtract__15.8__15.4__ reverse__1.0__ |
| an article is bought for rs . num__600 and sold for rs . num__500 find the loss percent ? <o> a ) num__16 num__0.25 % <o> b ) num__11 num__0.666666666667 % <o> c ) num__16 num__3.0 % <o> d ) num__16 num__0.666666666667 % <o> e ) num__16 num__0.4 % |
num__600 - - - - num__100 num__100 - - - - ? = > num__16 num__0.666666666667 % answer : d <eor> d <eos> |
d |
percent__16.0__100.0__ |
percent__16.0__100.0__ |
| a triangle and a parallelogram are constructed on the same base such that their areas are equal . if the altitude of the parallelogram is num__100 m then the altitude of the triangle is . <o> a ) num__200 m <o> b ) num__150 m <o> c ) num__148 m <o> d ) num__140 m <o> e ) none of these |
explanation : let the triangle and parallelogram have common base b let the altitude of triangle is h num__1 and of parallelogram is h num__2 ( which is equal to num__100 m ) then area of triangle = num__0.5 ∗ b ∗ h num__1 area of rectangle = b ∗ h num__2 as per question num__0.5 ∗ b ∗ h num__1 = b ∗ h num__2 num__0.5 ∗ b ∗ h num__1 = b ∗ num__100 h num__1 = num__100 ∗ num__2 = num__200 m option a <eor> a <eos> |
a |
multiply__100.0__2.0__ multiply__100.0__2.0__ |
multiply__100.0__2.0__ multiply__100.0__2.0__ |
| a car finishes a journey in ten hours at the speed of num__50 km / hr . if the same distance is to be covered in five hours how much more speed does the car have to gain ? <o> a ) num__8 km / hr <o> b ) num__10 km / hr <o> c ) num__12 km / hr <o> d ) num__16 km / hr <o> e ) num__50 km / hr |
distance covered by the car = num__50 × num__10 = num__500 km \ speed = num__500 ⁄ num__5 = num__100 km / hr \ speed gain = num__100 – num__50 = num__50 km / hr answer e <eor> e <eos> |
e |
multiply__50.0__10.0__ divide__50.0__10.0__ divide__500.0__5.0__ round__50.0__ |
multiply__50.0__10.0__ divide__50.0__10.0__ divide__500.0__5.0__ divide__500.0__10.0__ |
| at + cd = aaa where at and cd are two - digit numbers and aaa is a three digit number ; a t c and d are distinct positive integers . in the addition problem above what is the value of c ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__7 <o> d ) num__9 <o> e ) can not be determined |
since at and cd are two - digit integers their sum can give us only one three digit integer of a kind of aaa : num__111 . so a = num__1 and we have num__1 t + cd = num__111 now c can not be less than num__9 because no two - digit integer with first digit num__1 ( num__1 t < num__20 ) can be added to two - digit integer less than num__90 so that to have the sum num__111 ( if cd < num__90 so if c < num__9 cd + num__1 t < num__111 ) . hence c = num__9 . answer : d . <eor> d <eos> |
d |
multiply__1.0__9.0__ |
multiply__1.0__9.0__ |
| an article was sold after a discount of num__20.0 and there was a gain of num__20.0 . if the profit made on it was rs . num__6 less than the discount offered on it find its selling price ? <o> a ) rs . num__72 <o> b ) rs . num__90 <o> c ) rs . num__66 <o> d ) rs . num__96 <o> e ) none of these |
let cp = rs . num__100 x sp = rs . num__120 x mp = num__120 x / num__80 * num__100 = rs . num__150 x d = rs . num__150 x - rs . num__120 x = rs . num__30 x d - p = num__30 x - num__20 x = rs . num__6 num__10 x = rs . num__6 num__120 x = num__12.0 * num__6 = rs . num__72 answer : a <eor> a <eos> |
a |
percent__20.0__150.0__ percent__10.0__120.0__ percent__100.0__72.0__ |
percent__20.0__150.0__ percent__10.0__120.0__ percent__100.0__72.0__ |
| a train crosses a platform of num__120 m in num__15 sec same train crosses another platform of length num__180 m in num__18 sec . then find the length of the train ? <o> a ) num__170 <o> b ) num__180 <o> c ) num__190 <o> d ) num__120 <o> e ) num__130 |
length of the train be ‘ x ’ x + num__8.0 = x + num__10.0 num__6 x + num__720 = num__5 x + num__900 x = num__180 m answer : option b <eor> b <eos> |
b |
divide__120.0__15.0__ divide__180.0__18.0__ multiply__120.0__6.0__ subtract__15.0__10.0__ multiply__180.0__5.0__ round__180.0__ |
divide__120.0__15.0__ divide__180.0__18.0__ multiply__120.0__6.0__ subtract__15.0__10.0__ add__180.0__720.0__ round__180.0__ |
| the original price of a camera was displayed as a whole dollar amount . after adding sales tax of num__12 percent the final price was also a whole dollar amount . which of the following could be the final price of the camera ? <o> a ) $ num__223 <o> b ) $ num__225 <o> c ) $ num__224 <o> d ) $ num__213 <o> e ) $ num__215 |
final price = ( num__1 + num__0.12 ) * original price = num__1.12 * original price from options given only num__224 is divisible by num__1.12 as it is stated op is whole dollar amount . hence c <eor> c <eos> |
c |
add__1.0__0.12__ multiply__224.0__1.0__ |
add__1.0__0.12__ multiply__224.0__1.0__ |
| if the sides of a square are multiplied by num__10 the area of the original square is how many times as large as the area of the resultant square ? <o> a ) num__1.0 <o> b ) num__10.0 <o> c ) num__100.0 <o> d ) num__500.0 <o> e ) num__1000 % |
let x be the original length of one side . then the original area is x ^ num__2 . the new square has sides of length num__10 x so the area is num__100 x ^ num__2 . the area of the original square is num__0.01 = num__1.0 times the area of the new square . the answer is a . <eor> a <eos> |
a |
power__10.0__2.0__ multiply__100.0__0.01__ volume_cube__1.0__ |
power__10.0__2.0__ multiply__100.0__0.01__ power__1.0__10.0__ |
| at a game of billiards a can give b num__15 points in num__60 & a can give c num__20 points in num__60 . how many points can b give c in a game of num__90 ? <o> a ) num__30 points <o> b ) num__20 points <o> c ) num__10 points <o> d ) num__5 points <o> e ) num__0 points |
num__10 points answer : c <eor> c <eos> |
c |
subtract__20.0__10.0__ |
subtract__20.0__10.0__ |
| a person walks at a speed of num__4 km / hr and runs at a speed of num__8 km / hr . how many hours will the person require to cover a distance of num__20 km if the person completes half of the distance by walking and the other half by running ? <o> a ) num__2.75 <o> b ) num__3.25 <o> c ) num__3.75 <o> d ) num__4.25 <o> e ) num__4.75 |
time = num__2.5 + num__1.25 = num__3.75 = num__3.75 hours the answer is c . <eor> c <eos> |
c |
divide__20.0__8.0__ add__2.5__1.25__ round__3.75__ |
divide__20.0__8.0__ add__2.5__1.25__ add__2.5__1.25__ |
| a certain city with population of num__132000 is to be divided into num__11 voting districts and no district is to have a population that is more than num__10 percent greater than the population of any other district . what is the minimum possible population that the least populated district could have ? <o> a ) num__10700 <o> b ) num__10800 <o> c ) num__10900 <o> d ) num__11000 <o> e ) num__11 |
100 |
population that is num__10.0 greater than the population of any other district . . num__11000 . . num__10.0 of num__11000 = num__1100 . . so the next population will be num__12100 . . num__10 * num__12100 + num__11000 = num__132000 . d <eor> d <eos> |
d |
d |
| a man purchased num__3 blankets @ rs . num__100 each num__5 blankets @ rs . num__150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . num__162 . find the unknown rate of two blankets ? <o> a ) num__420 <o> b ) num__550 <o> c ) num__490 <o> d ) num__570 <o> e ) num__457 |
num__10 * num__162 = num__1620 num__3 * num__100 + num__5 * num__150 = num__1050 num__1620 – num__1050 = num__570 answer : d <eor> d <eos> |
d |
multiply__162.0__10.0__ subtract__1620.0__1050.0__ subtract__1620.0__1050.0__ |
multiply__162.0__10.0__ subtract__1620.0__1050.0__ subtract__1620.0__1050.0__ |
| on thrusday mabel handled num__90 transactions . anthony handled num__10.0 more transaction than mabel cal handled num__0.666666666667 rds of the transactions that anthony handled and jade handled num__16 more transaction than cal . how much transaction did jade handled ? <o> a ) num__92 <o> b ) num__80 <o> c ) num__72 <o> d ) num__82 <o> e ) num__28 |
mabel handled num__90 transaction anthony num__10.0 more transaction the mabel . anthony = num__90 + num__90 * num__10.0 = num__90 + num__90 * num__0.10 = num__90 + num__9 = num__99 cal handled num__0.666666666667 rds than anthony handled cal = num__0.666666666667 * num__99 = num__66 jade handled num__16 more than cal . jade = num__66 + num__16 = num__82 jade handled = num__82 tranasactions . correct answer : ( d ) <eor> d <eos> |
d |
reverse__10.0__ divide__90.0__10.0__ add__90.0__9.0__ add__16.0__66.0__ add__16.0__66.0__ |
reverse__10.0__ multiply__90.0__0.1__ add__90.0__9.0__ add__16.0__66.0__ add__16.0__66.0__ |
| which of the following statements is equivalent to num__8 + num__2 x < num__18 − num__6 x < num__23 + num__2 x num__8 + num__2 x < num__18 − num__6 x < num__23 + num__2 x ? <o> a ) num__0.8 < x < num__2.5 . <o> b ) num__0.8 < x < num__1.6 . <o> c ) - num__0.625 < x < num__1.25 . <o> d ) num__2.5 < x < num__3.5 . <o> e ) num__5 < num__8 x < num__12 . |
num__8 + num__2 x < num__18 − num__6 x < num__23 + num__2 x num__8 < num__18 - num__8 x < num__23 ( add - num__2 x on all sides ) - num__10 < - num__8 x < num__5 num__10 > num__8 x > - num__5 ( multiply by - num__1 on all sides ) num__1.25 > x > - num__0.625 c is the answer <eor> c <eos> |
c |
add__8.0__2.0__ subtract__23.0__18.0__ subtract__6.0__5.0__ divide__10.0__8.0__ divide__1.25__2.0__ divide__1.25__2.0__ |
add__8.0__2.0__ subtract__23.0__18.0__ subtract__6.0__5.0__ divide__10.0__8.0__ divide__1.25__2.0__ subtract__1.25__0.625__ |
| two numbers are in ratio num__4 : num__5 and their lcm is num__180 . the smaller number is <o> a ) num__9 <o> b ) num__15 <o> c ) num__36 <o> d ) num__45 <o> e ) none |
solution : let two numbers be num__4 x and num__5 x ; their lcm = num__180 and hcf = x ; now num__1 st number * num__2 nd number = lcm * hcf or num__4 x * num__5 x = num__180 * x ; or num__20 x = num__180 ; or x = num__9 ; then the smaller number = num__4 * num__9 = num__36 . answer : option c <eor> c <eos> |
c |
subtract__5.0__4.0__ multiply__4.0__5.0__ add__4.0__5.0__ multiply__4.0__9.0__ round__36.0__ |
subtract__5.0__4.0__ multiply__4.0__5.0__ add__4.0__5.0__ multiply__4.0__9.0__ multiply__4.0__9.0__ |
| x + ( num__1 / x ) = num__2 find x ^ num__2 + ( num__1 / x ^ num__2 ) <o> a ) num__2 <o> b ) num__3.25 <o> c ) num__4.25 <o> d ) num__5.25 <o> e ) num__6.25 |
squaring on both sides ( x + num__1 / x ) ^ num__2 = num__2 ^ num__2 x ^ num__2 + num__1 / x ^ num__2 = num__4 - num__2 x ^ num__2 + num__1 / x ^ num__2 = num__2 answer : a <eor> a <eos> |
a |
multiply__1.0__2.0__ |
divide__2.0__1.0__ |
| a man ' s age is num__125.0 of what it was num__10 years ago but num__83 num__0.333333333333 % of what it will be after num__10 years . what is his present age ? <o> a ) num__50 <o> b ) num__70 <o> c ) num__80 <o> d ) num__55 <o> e ) num__40 |
num__125 x / num__100 = x + num__10 x = num__40 present age = x + num__10 = num__40 + num__10 = num__50 answer a <eor> a <eos> |
a |
add__10.0__40.0__ add__10.0__40.0__ |
add__10.0__40.0__ add__10.0__40.0__ |
| a small table has a length of num__12 inches and a breadth of b inches . cubes are placed on the surface of the table so as to cover the entire surface . the maximum side of such cubes is found to be num__4 inches . also a few such tables are arranged to form a square . the minimum length of side possible for such a square is num__80 inches . find b . <o> a ) num__8 <o> b ) num__16 <o> c ) num__24 <o> d ) num__32 <o> e ) num__48 |
from the info that the maximum sides of the cubes is num__4 we know that the gcf of num__12 ( = num__2 ^ num__2 * num__3 ) andbis num__4 ( = num__2 ^ num__2 ) sob = num__2 ^ x where x > = num__2 . from the second premise we know that the lcm of num__12 ( num__2 ^ num__2 * num__3 ) andbis num__80 ( num__2 ^ num__4 * num__5 ) sob = num__2 ^ num__4 or num__2 ^ num__4 * num__5 ( num__16 or num__80 ) . combining num__2 premises shows the answer is b ( num__16 ) . <eor> b <eos> |
b |
square_perimeter__4.0__ square_perimeter__4.0__ |
power__4.0__2.0__ power__4.0__2.0__ |
| kim bought a total of $ num__4.65 worth of postage stamps in four denominations . if she bought an equal number of num__10 - cent and num__25 - cent stamps and twice as many num__20 - cent stamps as num__5 - cent stamps what is the least number of num__1 - cent stamps she could have bought ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__75 <o> d ) num__90 <o> e ) num__95 |
let : # of num__10 and num__25 cents stamps = n # of num__20 cent stamps = num__2 n # of num__1 cent stamps = m therefore : num__10 n + num__25 n + num__20 ( num__2 n ) + m = num__465 cents num__75 n + m = num__465 forleastnumber of num__1 cents stamps maximize num__50 n . hence num__75 * num__5 + m = num__465 m = num__90 i . e . num__90 stamps num__1 cents each answer : d <eor> d <eos> |
d |
divide__10.0__5.0__ multiply__10.0__5.0__ multiply__1.0__90.0__ |
divide__10.0__5.0__ multiply__10.0__5.0__ multiply__1.0__90.0__ |
| for any number z z * is defined as the greatest positive even integer less than or equal to y . what is the value of num__6.25 – num__6.25 * ? <o> a ) num__1.5 <o> b ) num__0.5 <o> c ) num__6.25 <o> d ) num__0.25 <o> e ) num__6.0 |
since z * is defined as the greatest positive even integer less than or equal to z then num__6.25 * = num__6 ( the greatest positive even integer less than or equal to num__6.25 is num__6 ) . hence num__6.25 – num__6.25 * = num__6.25 - num__6 = num__0.25 answer : d . <eor> d <eos> |
d |
round_down__6.25__ subtract__6.25__6.0__ subtract__6.25__6.0__ |
round_down__6.25__ subtract__6.25__6.0__ subtract__6.25__6.0__ |
| how many num__9 ' s are there between num__1 and num__100 ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__22 <o> d ) num__18 <o> e ) num__27 |
b num__20 <eor> b <eos> |
b |
multiply__1.0__20.0__ |
multiply__1.0__20.0__ |
| speed of a boat in standing water is num__9 kmph and the speed of the stream is num__1.5 kmph . a man rows to place at a distance of num__105 km and comes back to the starting point . the total time taken by him is : <o> a ) num__12 hours <o> b ) num__24 hours <o> c ) num__36 hours <o> d ) num__48 hours <o> e ) none |
sol . speed upstream = num__7.5 kmph ; speed downstream = num__10.5 kmph . ∴ total time taken = [ num__105 / num__7.5 + num__105 / num__10.5 ] hours = num__24 hours . answer b <eor> b <eos> |
b |
subtract__9.0__1.5__ add__9.0__1.5__ round__24.0__ |
subtract__9.0__1.5__ add__9.0__1.5__ round__24.0__ |
| the parameter of a square is equal to the perimeter of a rectangle of length num__14 cm and breadth num__10 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) ? <o> a ) num__23.56 <o> b ) num__23.59 <o> c ) num__23.55 <o> d ) num__18.86 <o> e ) num__23.57 |
let the side of the square be a cm . parameter of the rectangle = num__2 ( num__14 + num__10 ) = num__48 cm parameter of the square = num__48 cm i . e . num__4 a = num__48 a = num__12 diameter of the semicircle = num__12 cm circimference of the semicircle = num__0.5 ( â ˆ ) ( num__12 ) = num__0.5 ( num__3.14285714286 ) ( num__12 ) = num__18.8571428571 = num__18.86 cm to two decimal places answer : d <eor> d <eos> |
d |
rectangle_perimeter__14.0__10.0__ rectangle_perimeter__2.0__4.0__ triangle_area__2.0__18.86__ |
rectangle_perimeter__14.0__10.0__ rectangle_perimeter__2.0__4.0__ triangle_area__2.0__18.86__ |
| find the average of all the numbers between num__6 and num__34 which are divisible by num__5 ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__25 <o> d ) num__30 <o> e ) num__15 |
average = ( num__10 + num__15 + num__20 + num__25 + num__30 ) num__5 = num__20.0 = num__20 answer is b <eor> b <eos> |
b |
add__5.0__10.0__ add__5.0__15.0__ add__5.0__20.0__ multiply__6.0__5.0__ add__5.0__15.0__ |
add__5.0__10.0__ add__5.0__15.0__ add__5.0__20.0__ add__5.0__25.0__ add__5.0__15.0__ |
| working at their respective constant rates paul abdul and adam alone can finish a certain work in num__3 num__4 and num__5 hours respectively . if all three work together to finish the work what fraction k of the work will be done by adam ? <o> a ) num__0.25 <o> b ) num__0.255319148936 <o> c ) num__0.333333333333 <o> d ) num__0.416666666667 <o> e ) num__0.425531914894 |
let the total work be num__60 units . pual does num__20.0 = num__20 units of work per hr . abdul does num__15 units per hr and adam does num__12 units per hr . if all work together they do ( num__20 + num__15 + num__12 ) units per hr = num__47 units per hr . so the time taken to finish the work = num__1.27659574468 hrs . adam will do num__1.27659574468 * num__12 units of work in num__1.27659574468 hr . fraction of work adam does = work done by adam / total work k > ( num__1.27659574468 * num__12 ) / num__60 = num__0.255319148936 answer b <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__4.0__5.0__ multiply__3.0__5.0__ multiply__3.0__4.0__ divide__60.0__47.0__ divide__1.2766__5.0__ divide__1.2766__5.0__ |
hour_to_min_conversion__ multiply__4.0__5.0__ multiply__3.0__5.0__ multiply__3.0__4.0__ divide__60.0__47.0__ divide__1.2766__5.0__ divide__1.2766__5.0__ |
| a b and c can independently complete a piece of work in num__812 and num__6 days respectively . a and b work together for num__4 days and leave . how long will c take to finish the remaining work ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
a and b work together for num__4 days and completed num__4 * ( num__0.125 + num__0.0833333333333 ) = num__0.833333333333 work . balance num__0.166666666667 work will be completed by c in one day . answer : a <eor> a <eos> |
a |
add__0.1667__0.8333__ |
add__0.1667__0.8333__ |
| a train running at a speed of num__36 kmph crosses an electric pole in num__12 seconds . in how much time will it cross a num__370 m long platform ? <o> a ) num__37 min <o> b ) num__55 min <o> c ) num__47 min <o> d ) num__67 min <o> e ) num__49 min |
e num__49 min let the length of the train be x m . when a train crosses an electric pole the distance covered is its own length . so x = num__12 * num__36 * num__0.277777777778 m = num__120 m . time taken to cross the platform = ( num__120 + num__370 ) / num__36 * num__0.277777777778 = num__49 min . <eor> e <eos> |
e |
round__49.0__ |
round__49.0__ |
| in num__2008 the profits of company n were num__10 percent of revenues . in num__2009 the revenues of company n fell by num__20 percent but profits were num__12 percent of revenues . the profits in num__2009 were what percent of the profits in num__2008 ? <o> a ) num__96.0 <o> b ) num__105.0 <o> c ) num__120.0 <o> d ) num__124.2 <o> e ) num__138 % |
x = profits r = revenue x / r = num__01 x = num__10 r = num__100 num__2009 : r = num__80 x / num__80 = num__012 = num__0.12 x = num__80 * num__0.12 x = num__9.6 num__9.6 / num__10 = num__0.96 = num__96.0 answer a <eor> a <eos> |
a |
percent__12.0__1.0__ percent__12.0__80.0__ percent__10.0__9.6__ percent__96.0__100.0__ |
percent__12.0__1.0__ percent__12.0__80.0__ percent__10.0__9.6__ percent__96.0__100.0__ |
| the value of ( ( x – y ) ³ + ( y - z ) ³ + ( z – x ) ³ ) / ( num__15 ( x – y ) ( y – z ) ( z – x ) ) is equal to : <o> a ) num__0 <o> b ) num__0.0833333333333 <o> c ) num__1 <o> d ) num__0.2 <o> e ) num__0.333333333333 |
since ( x – y ) + ( y – z ) + ( z – x ) = num__0 so ( x – y ) ³ + ( y – z ) ³ + ( z – x ) ³ = num__3 ( x – y ) ( y – z ) ( z – x ) . ( num__3 ( x – y ) ( y – z ) ( z – x ) ) / ( num__15 ( x – y ) ( y – z ) ( z – x ) ) = num__0.2 . answer : d <eor> d <eos> |
d |
divide__3.0__15.0__ divide__3.0__15.0__ |
divide__3.0__15.0__ divide__3.0__15.0__ |
| if x = y + num__3 + num__4.5 x and y = x + num__33 what is the value of x / y ? <o> a ) - num__1.2 . <o> b ) - num__0.32 <o> c ) num__0.25 . <o> d ) num__0.833333333333 . <o> e ) num__1.5 . |
( num__1 ) x = y + num__3 + num__4.5 x substitute y for y = x + num__33 ( eq . we are given ) x = x + num__33 + num__3 + num__4.5 x combine like terms x = num__5.5 x + num__36 - - > - num__4.5 x = num__36 - - > x = - num__8 ( num__2 ) y = x + num__18 substitute x for x = - num__8 y = ( - num__8 ) + num__33 = num__25 ( num__3 ) x / y = ? substitute x and y we found above x / y = ( - num__8 ) / ( num__25 ) = - num__0.32 my answer : b <eor> b <eos> |
b |
add__4.5__1.0__ add__3.0__33.0__ divide__36.0__4.5__ subtract__3.0__1.0__ divide__36.0__2.0__ subtract__33.0__8.0__ divide__8.0__25.0__ multiply__1.0__0.32__ |
add__4.5__1.0__ add__3.0__33.0__ divide__36.0__4.5__ subtract__3.0__1.0__ divide__36.0__2.0__ subtract__33.0__8.0__ divide__8.0__25.0__ divide__8.0__25.0__ |
| working alone at its constant rate machine k took num__3 hours to produce num__0.166666666667 of the units produced last friday . then machine m started working and the two machines working simultaneously at their respective constant rates took num__6 hours to produce the rest of the units produced last friday . how many hours would it have taken machine m working alone at its constant rate to produce all of the units produced last friday ? <o> a ) num__8 <o> b ) num__12 <o> c ) num__16 <o> d ) num__24 <o> e ) num__30 |
machine k works at a rate of num__0.0555555555556 of the units per hour . the rate of k + m together is num__0.833333333333 * num__0.166666666667 = num__0.138888888889 of the units per hour . the rate of machine m is num__0.138888888889 - num__0.0555555555556 = num__0.0833333333333 . it would have taken machine m a total time of num__12 hours . the answer is b . <eor> b <eos> |
b |
divide__0.1667__3.0__ multiply__0.1667__0.8333__ subtract__0.1389__0.0556__ round__12.0__ |
divide__0.1667__3.0__ multiply__0.1667__0.8333__ subtract__0.1389__0.0556__ round__12.0__ |
| let s be the set of all positive integers that when divided by num__8 have a remainder of num__5 . what is the num__72 th number in this set ? <o> a ) num__573 <o> b ) num__608 <o> c ) num__613 <o> d ) num__616 <o> e ) num__621 |
the set s = { num__5 num__13 num__21 num__29 . . . . . . . . . . . . . . . . . . . . . } num__1 st number = num__8 * num__0 + num__5 = num__5 num__2 nd number = num__8 * num__1 + num__5 = num__13 num__3 rd number = num__8 * num__2 + num__5 = num__21 num__72 th number = num__8 * ( num__72 - num__1 ) + num__5 = num__573 answer = a <eor> a <eos> |
a |
add__8.0__5.0__ add__8.0__13.0__ add__8.0__21.0__ subtract__8.0__5.0__ multiply__1.0__573.0__ |
add__8.0__5.0__ add__8.0__13.0__ add__8.0__21.0__ subtract__8.0__5.0__ multiply__1.0__573.0__ |
| what is the least number that should be added to num__1077 so the sum of the number is divisible by num__23 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
( num__46.8260869565 ) gives a remainder num__19 so we need to add num__4 . the answer is d . <eor> d <eos> |
d |
divide__1077.0__23.0__ subtract__23.0__19.0__ subtract__23.0__19.0__ |
divide__1077.0__23.0__ subtract__23.0__19.0__ subtract__23.0__19.0__ |
| look at this series : num__3 num__1 ( num__0.333333333333 ) ( num__0.111111111111 ) . . . what number should come next ? <o> a ) num__0.037037037037 <o> b ) num__0.0555555555556 <o> c ) num__0.016393442623 <o> d ) num__0.0740740740741 <o> e ) none |
explanation : this is a simple division series ; each number is divisible by num__3 of the previous number . in other terms to say the number is divisible by num__3 successively to get the next result . num__3.0 = num__3 num__1.0 = num__1 num__0.333333333333 = num__0.333333333333 ( num__0.333333333333 ) / num__3 = num__0.111111111111 ( num__0.111111111111 ) / num__3 = num__0.037037037037 and so on . answer : option a <eor> a <eos> |
a |
power__0.3333__3.0__ multiply__1.0__0.037__ |
divide__0.1111__3.0__ divide__0.1111__3.0__ |
| three numbers are in the ratio num__1 : num__2 : num__3 and their h . c . f is num__12 . the numbers are <o> a ) num__12 num__24 num__30 <o> b ) num__12 num__24 num__38 <o> c ) num__12 num__24 num__362 <o> d ) num__12 num__24 num__36 <o> e ) num__12 num__24 num__321 |
explanation : let the required numbers be x num__2 x num__3 x . then their h . c . f = x . so x = num__12 \ inline \ fn _ jvn \ therefore the numbers are num__12 num__24 num__36 answer : d ) num__12 num__24 num__36 <eor> d <eos> |
d |
multiply__2.0__12.0__ multiply__3.0__12.0__ multiply__1.0__12.0__ |
multiply__2.0__12.0__ multiply__3.0__12.0__ multiply__1.0__12.0__ |
| num__78 num__64 num__48 num__30 num__10 ( . . . ) <o> a ) - num__12 <o> b ) - num__14 <o> c ) num__2 <o> d ) num__8 <o> e ) num__6 |
explanation : num__78 - num__14 = num__64 num__64 - num__16 = num__48 num__48 - num__18 = num__30 num__30 - num__20 = num__10 num__10 - num__22 = - num__12 answer : option a <eor> a <eos> |
a |
subtract__78.0__64.0__ subtract__64.0__48.0__ subtract__48.0__30.0__ subtract__30.0__10.0__ subtract__30.0__18.0__ subtract__30.0__18.0__ |
subtract__78.0__64.0__ subtract__64.0__48.0__ subtract__48.0__30.0__ subtract__30.0__10.0__ subtract__30.0__18.0__ subtract__30.0__18.0__ |
| according to the direction on a can of frozen orange juice concentrate is to be mixed with num__3 cans of water to make orange juice . how many num__10 - ounce cans of the concentrate are required to prepare num__200 num__6 - ounce servings of orange juice ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__50 <o> d ) num__67 <o> e ) num__100 |
orange juice concentrate : water : : num__1 : num__3 total quantity of orange juice = num__200 * num__6 = num__1200 oz so orange juice concentrate : water : : num__300 oz : num__900 oz no . of num__10 oz can = num__300 oz / num__10 oz = num__30 answer b num__30 cans <eor> b <eos> |
b |
multiply__200.0__6.0__ multiply__3.0__300.0__ multiply__3.0__10.0__ multiply__3.0__10.0__ |
multiply__200.0__6.0__ multiply__3.0__300.0__ multiply__3.0__10.0__ multiply__3.0__10.0__ |
| the ratio of sum of squares of first n natural numbers to square of sum of first n natural numbers is num__17 : num__325 . the value of n is <o> a ) num__15 <o> b ) n = num__25 <o> c ) num__35 <o> d ) num__30 <o> e ) none of these |
sum of num__1 st n natural no . s is n ( n + num__1 ) / num__2 sum of sqaures of num__1 st n natural no . s is n ( n + num__1 ) ( num__2 n + num__1 ) / num__6 the ratio of sum of squares of first n natural numbers to square of sum of first n natural numbers = n ( n + num__1 ) ( num__2 n + num__1 ) * num__2 * num__2 / ( n * n * ( n + num__1 ) * ( n + num__1 ) ) = num__2 ( num__2 n + num__1 ) / ( num__3 n ( n + num__1 ) ) now by hit and trial for the given options if we put num__25 then this will give num__0.0523076923077 therefore the answer is num__25 answer : b <eor> b <eos> |
b |
add__1.0__2.0__ divide__17.0__325.0__ multiply__1.0__25.0__ |
add__1.0__2.0__ divide__17.0__325.0__ multiply__1.0__25.0__ |
| how many pieces of num__0.85 meteres can be cut from a rod num__42.5 meteres long <o> a ) num__30 <o> b ) num__40 <o> c ) num__50 <o> d ) num__60 <o> e ) num__70 |
explanation : we need so simple divide num__42.5 / num__0.85 = ( num__50.0 ) = num__50 option c <eor> c <eos> |
c |
divide__42.5__0.85__ round__50.0__ |
divide__42.5__0.85__ divide__42.5__0.85__ |
| two trains each num__150 m in length each are running on two parallel lines in opposite directions . if one goes at the speed of num__95 km / h while the other travels at num__85 km / h . how long will it take for them to pass each other completely . <o> a ) num__25 sec <o> b ) num__23 sec <o> c ) num__34 sec <o> d ) num__22 sec <o> e ) num__6 sec |
explanation : d = num__150 m + num__150 m = num__300 m rs = num__95 + num__85 = num__180 * num__0.277777777778 = num__50 t = num__300 * num__0.02 = num__6 sec answer : option e <eor> e <eos> |
e |
add__95.0__85.0__ divide__300.0__50.0__ round__6.0__ |
add__95.0__85.0__ multiply__300.0__0.02__ multiply__300.0__0.02__ |
| the duplicate ratio of num__6 : num__4 is ? <o> a ) num__1 : num__2 <o> b ) num__9 : num__4 <o> c ) num__1 : num__8 <o> d ) num__1 : num__18 <o> e ) num__1 : num__13 |
num__36 ^ num__2 : num__4 ^ num__2 = num__36 : num__16 = num__9 : num__4 answer : b <eor> b <eos> |
b |
subtract__6.0__4.0__ divide__36.0__4.0__ divide__36.0__4.0__ |
subtract__6.0__4.0__ divide__36.0__4.0__ divide__36.0__4.0__ |
| a leak in the bottom of a tank can empty the full tank in num__6 hours . an inlet pipe fills water at the rate of num__4 liters per minute . when the tank is full in inlet is opened and due to the leak the tank is empties in num__8 hours . the capacity of the tank is ? <o> a ) num__2877 <o> b ) num__5760 <o> c ) num__1797 <o> d ) num__1797 <o> e ) num__2972 |
explanation : num__1 / x - num__0.166666666667 = - num__0.125 x = num__24 hrs num__24 * num__60 * num__4 = num__5760 answer : b <eor> b <eos> |
b |
divide__1.0__6.0__ divide__1.0__8.0__ multiply__6.0__4.0__ hour_to_min_conversion__ round__5760.0__ |
divide__1.0__6.0__ divide__1.0__8.0__ multiply__6.0__4.0__ hour_to_min_conversion__ divide__5760.0__1.0__ |
| three unbiased coins are tossed what is the probability of getting at least num__2 tails ? <o> a ) num__0.2 <o> b ) num__0.142857142857 <o> c ) num__0.125 <o> d ) num__0.5 <o> e ) num__0.1 |
total cases are = num__2 ^ num__3 = num__8 favoured cases = [ tth tht htt ttt ] = num__4 probability = num__0.5 = num__0.5 answer d <eor> d <eos> |
d |
negate_prob__0.5__ |
negate_prob__0.5__ |
| the diameter of the driving wheel of a bus in num__140 cm . how many revolutions per minute must the wheel make in order to keep a speed of num__14 kmph ? <o> a ) num__53.03 <o> b ) num__50.01 <o> c ) num__23.23 <o> d ) num__24.23 <o> e ) num__41.23 |
distance covered in num__1 min = ( num__14 * num__1000 ) / num__60 = num__233.33 m circumference of the wheel = ( num__2 * ( num__3.14285714286 ) * . num__70 ) = num__4.4 m no of revolution per min = num__233.33 / num__4.4 = num__53.03 answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ divide__140.0__2.0__ round__53.03__ |
hour_to_min_conversion__ divide__140.0__2.0__ divide__53.03__1.0__ |
| find number which is num__70.0 less than num__80 . <o> a ) num__18 <o> b ) num__22 <o> c ) num__24 <o> d ) num__26 <o> e ) num__98 |
explanation : num__70.0 less is num__30.0 of the given number therefore num__30.0 of num__80 is num__24 . answer : c <eor> c <eos> |
c |
percent__80.0__30.0__ percent__80.0__30.0__ |
percent__80.0__30.0__ percent__80.0__30.0__ |
| the sum of money at compound interest amounts to thrice itself in num__2 years . in how many years will it be num__6 times itself ? <o> a ) num__8 years <o> b ) num__9 years <o> c ) num__2 years <o> d ) num__6 years <o> e ) num__4 years |
num__100 - - - - num__300 - - - num__2 num__600 - - - num__2 - - - - num__4 years answer : e <eor> e <eos> |
e |
multiply__2.0__300.0__ subtract__6.0__2.0__ subtract__6.0__2.0__ |
multiply__2.0__300.0__ subtract__6.0__2.0__ subtract__6.0__2.0__ |
| three pounds of num__05 grass seed contain num__1 percent herbicide . a different type of grass seed num__20 which contains num__20 percent herbicide will be mixed with three pounds of num__05 grass seed . how much grass seed of type num__20 should be added to the three pounds of num__05 grass seed so that the mixture contains num__15 percent herbicide ? <o> a ) num__3 <o> b ) num__3.75 <o> c ) num__4.5 <o> d ) num__8.4 <o> e ) num__9 |
num__05 grass seed contains num__5.0 herbicide and its amount is num__3 pound num__20 grass seed contains num__20.0 herbicide and its amount is x when these two types of grass seeds are mixed their average becomes num__15.0 thus we have num__3 ( num__1 ) + x ( num__20 ) / ( x + num__3 ) = num__15 num__3 + num__20 x = num__15 x + num__45 num__5 x = num__42 or x = num__8.4 d <eor> d <eos> |
d |
divide__15.0__5.0__ multiply__15.0__3.0__ subtract__45.0__3.0__ divide__42.0__5.0__ multiply__1.0__8.4__ |
divide__15.0__5.0__ multiply__15.0__3.0__ subtract__45.0__3.0__ divide__42.0__5.0__ divide__42.0__5.0__ |
| a traveler changes num__150 pounds into rupees at the rate of rs . num__7000 for num__100 pounds . he spends rs . num__9060 and changes the remaining amount back to pounds at the rate of num__100 pounds to rs . num__7200 . how many pounds will he get ? <o> a ) num__18 pounds <o> b ) num__20 pounds <o> c ) num__25 pounds <o> d ) num__30 pounds <o> e ) none of these |
explanation : amount of rupees for exchange of num__150 pounds = ( num__1.5 ) * ( num__7000.0 ) = rs . num__10500 out of these rs . num__10500 traveler spent rs . num__9060 remaining balance in rupees = num__10500 – num__9060 = rs . num__1440 now these rs . num__1440 are exchanged back to pounds with the rate of num__100 pounds for rs . num__7200 amount in pounds = ( num__1440 * num__100 ) / num__7200 = num__20 pounds answer : b <eor> b <eos> |
b |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| bill downloads the movierevenge of the avengersto his computer in num__2.5 hours using a download manager that downloads from num__3 sources marked a b and c . each source provides download at a constant rate but the rates of different sources are not necessarily identical . if the movie was downloaded from sources a and c alone it would take num__4 hours to complete the download . the next day source b is available but the other sources are inactive . how long will it take to download the trailer of the movie a file that is num__80 times smaller from source b alone ? <o> a ) num__6 hours and num__40 minutes <o> b ) num__15 minutes <o> c ) num__12 minutes <o> d ) num__10 minutes <o> e ) num__5 minutes |
let the movie size be num__400 u . given a + c = num__4 hrs . a + c = num__100 u / hr and a + b + c = num__2.5 hrs or num__400 / num__2.5 = num__160 u / hr b alone = num__160 - num__100 = num__60 u / hr trailer = num__80 times smaller or num__5.0 = num__5 u b will take num__0.0833333333333 hrs or num__5 minutes . ans e <eor> e <eos> |
e |
divide__400.0__4.0__ divide__400.0__2.5__ subtract__160.0__100.0__ divide__400.0__80.0__ divide__5.0__60.0__ divide__400.0__80.0__ |
divide__400.0__4.0__ divide__400.0__2.5__ subtract__160.0__100.0__ divide__400.0__80.0__ divide__5.0__60.0__ divide__400.0__80.0__ |
| a is twice as good workman as b and together they complete a work in num__15 days . in how many days can the work be complete by b alone ? <o> a ) num__35 days <o> b ) num__65 days <o> c ) num__45 days <o> d ) num__75 days <o> e ) num__95 days |
c num__45 days <eor> c <eos> |
c |
round__45.0__ |
round__45.0__ |
| in one alloy there is num__12.0 chromium while in another alloy it is num__8.0 . num__15 kg of the first alloy was melted together with num__35 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . <o> a ) num__8.8 <o> b ) num__9.0 <o> c ) num__9.2 <o> d ) num__8.6 <o> e ) num__8.4 % |
the amount of chromium in the new num__15 + num__35 = num__50 kg alloy is num__0.12 * num__15 + num__0.08 * num__35 = num__4.6 kg so the percentage is num__4.6 / num__50 * num__100 = num__9.2 . answer : c . <eor> c <eos> |
c |
percent__100.0__9.2__ |
percent__100.0__9.2__ |
| a shopkeeper purchased num__70 kg of potatoes for rs . num__420 and sold the whole lot at the rate of rs . num__6.70 per kg . what will be his gain percent ? <o> a ) num__8 num__0.142857142857 % <o> b ) num__2 num__0.333333333333 % <o> c ) num__11 num__0.666666666667 % <o> d ) num__8 num__0.125 % <o> e ) num__8 num__0.111111111111 % |
c . p . of num__1 kg = num__6.0 = rs . num__6 s . p . of num__1 kg = rs . num__6.70 gain % = num__0.70 / num__6 * num__100 = num__11.6666666667 = num__11 num__0.666666666667 % answer : c <eor> c <eos> |
c |
percent__70.0__1.0__ percent__100.0__11.0__ |
percent__70.0__1.0__ percent__100.0__11.0__ |
| the guests at a football banquet consumed a total of num__327 pounds of food . if no individual guest consumed more than num__2 pounds of food what is the minimum number of guests that could have attended the banquet ? <o> a ) num__160 <o> b ) num__161 <o> c ) num__162 <o> d ) num__163 <o> e ) num__164 |
to minimize one quantity maximize other . num__163 * num__2 ( max possible amount of food a guest could consume ) = num__326 pounds so there must be more than num__163 guests next integer is num__164 . answer : e . <eor> e <eos> |
e |
multiply__2.0__163.0__ subtract__327.0__163.0__ subtract__327.0__163.0__ |
multiply__2.0__163.0__ subtract__327.0__163.0__ subtract__327.0__163.0__ |
| a goods train runs at the speed of num__72 km / hr and crosses a num__270 m long platform in num__26 sec . what is the length of the goods train ? <o> a ) num__278 <o> b ) num__166 <o> c ) num__151 <o> d ) num__250 <o> e ) num__109 |
speed = num__72 * num__0.277777777778 = num__20 m / sec . time = num__26 sec . let the length of the train be x meters . then ( x + num__270 ) / num__26 = num__20 x = num__250 m . answer : d <eor> d <eos> |
d |
subtract__270.0__20.0__ round__250.0__ |
subtract__270.0__20.0__ round__250.0__ |
| in how many different ways can the letters of the word ' leading ' be arranged in such a way that the vowels always come together ? <o> a ) num__720 ways <o> b ) num__735 ways <o> c ) num__767 ways <o> d ) num__723 ways <o> e ) num__730 ways |
the word ' leading ' has num__7 different letters . when the vowels eai are always together they can be supposed to form one letter . then we have to arrange the letters lndg ( eai ) . now num__5 ( num__4 + num__1 = num__5 ) letters can be arranged in num__5 ! = num__120 ways . the vowels ( eai ) can be arranged among themselves in num__3 ! = num__6 ways . required number of ways = ( num__120 x num__6 ) = num__720 . answer : a <eor> a <eos> |
a |
vowel_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
vowel_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| a rectangular farm has to be fenced one long side one short side and the diagonal . if the cost of fencing is rs . num__13 per meter . the area of farm is num__1200 m num__2 and the short side is num__30 m long . how much would the job cost ? <o> a ) num__1276 <o> b ) num__1560 <o> c ) num__2832 <o> d ) num__1299 <o> e ) num__1236 |
explanation : l * num__30 = num__1200 è l = num__40 num__40 + num__30 + num__50 = num__120 num__120 * num__13 = num__1560 answer : option b <eor> b <eos> |
b |
square_perimeter__30.0__ multiply__13.0__120.0__ multiply__13.0__120.0__ |
square_perimeter__30.0__ multiply__13.0__120.0__ multiply__13.0__120.0__ |
| difference between the length & breadth of a rectangle is num__10 m . if its perimeter is num__60 m then its area is ? ? <o> a ) num__2000 m ^ num__2 <o> b ) num__200 m ^ num__2 <o> c ) num__2520 m ^ num__2 <o> d ) num__2556 m ^ num__2 <o> e ) num__2534 m ^ num__2 |
l - b = num__10 num__2 ( l + b ) = num__60 = > l + b = num__30 num__2 l = num__40 = > l = num__20 so b = num__10 area of the rectangle = l x b = num__20 x num__10 = num__200 m ^ num__2 answer : b <eor> b <eos> |
b |
square_perimeter__10.0__ multiply__10.0__2.0__ triangle_area__10.0__40.0__ triangle_area__10.0__40.0__ |
square_perimeter__10.0__ multiply__10.0__2.0__ triangle_area__10.0__40.0__ triangle_area__10.0__40.0__ |
| the banker â € ™ s gain on a sum due num__3 years hence at num__10.0 per annum is rs . num__60 . the banker â € ™ s discount is <o> a ) rs . num__960 <o> b ) rs . num__840 <o> c ) rs . num__1020 <o> d ) rs . num__760 <o> e ) rs . num__260 |
solution t . d = ( b . g x num__100 / r x t ) = rs . ( num__60 x num__10.0 x num__3 ) = rs . num__200 . b . d = rs ( num__200 + num__60 ) = rs . num__260 . answer e <eor> e <eos> |
e |
percent__100.0__260.0__ |
percent__100.0__260.0__ |
| shawn invested one half of his savings in a bond that paid simple interest for num__2 years and received $ num__550 as interest . he invested the remaining in a bond that paid compound interest interest being compounded annually for the same num__2 years at the same rate of interest and received $ num__605 as interest . what was the value of his total savings before investing in these two bonds ? <o> a ) $ num__5500 <o> b ) $ num__11000 <o> c ) $ num__22000 <o> d ) $ num__2750 <o> e ) $ num__44000 |
explanatory answer shawn received an extra amount of ( $ num__605 - $ num__550 ) $ num__55 on his compound interest paying bond as the interest that he received in the first year also earned interest in the second year . the extra interest earned on the compound interest bond = $ num__55 the interest for the first year = $ num__275.0 = $ num__275 therefore the rate of interest = ( num__0.2 ) * num__100 = num__20.0 p . a . num__20.0 interest means that shawn received num__20.0 of the amount he invested in the bonds as interest if num__20.0 of his investment in one of the bonds = $ num__275 then his total investment in each of the bonds = ( num__13.75 ) * num__100 = $ num__1375 as he invested equal sums in both the bonds his total savings before investing = num__2 * num__1375 = $ num__2750 . correct choice is ( d ) <eor> d <eos> |
d |
percent__100.0__2750.0__ |
percent__100.0__2750.0__ |
| adam sat with his friends in the chinnaswamy stadium at madurai to watch the num__100 metres running race organized by the asian athletics association . five rounds were run . after every round half the teams were eliminated . finally one team wins the game . how many teams participated in the race ? <o> a ) num__32 <o> b ) num__77 <o> c ) num__279 <o> d ) num__27 <o> e ) num__91 |
total five rounds were run . so in the final round num__2 teams must have participated . in the penultimate round num__4 teams and num__3 rd round num__8 num__2 nd round num__16 and in the first round num__32 teams must have participated as in each round half of the teams got eliminated . answer : a <eor> a <eos> |
a |
multiply__2.0__4.0__ multiply__2.0__8.0__ multiply__2.0__16.0__ round__32.0__ |
multiply__2.0__4.0__ multiply__2.0__8.0__ multiply__2.0__16.0__ round__32.0__ |
| a sporting goods store sold num__64 frisbees in one week some for $ num__3 and the rest for $ num__4 each . if receipts from frisbee sales for the week totaled $ num__204 what is the fewest number of $ num__3 frisbees that could have been sold ? <o> a ) num__24 <o> b ) num__12 <o> c ) num__8 <o> d ) num__52 <o> e ) num__2 |
in this question however because we are told that exactly num__64 frisbees have been sold and revenue was exactly $ num__204 there is only one possible solution for the number of $ num__3 and $ num__4 frisbees sold . to solve we have num__2 equations and num__2 unknowns let x = number of $ num__3 frisbees sold let y = number of $ num__4 frisbees sold x + y = num__64 num__3 x + num__4 y = num__204 x = num__64 - y num__3 ( num__64 - y ) + num__4 y = num__204 num__192 - num__3 y + num__4 y = num__204 y = num__12 x = num__64 - num__12 = num__52 d <eor> d <eos> |
d |
multiply__64.0__3.0__ multiply__3.0__4.0__ subtract__64.0__12.0__ subtract__64.0__12.0__ |
multiply__64.0__3.0__ subtract__204.0__192.0__ subtract__64.0__12.0__ subtract__64.0__12.0__ |
| if the tens digit of positive integers x y are num__6 how many values of the tens digit of num__2 ( x + y ) can be there ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
if x = y = num__60 num__2 ( x + y ) = num__240 is derived . if x = y = num__69 num__2 ( x + y ) = num__276 is derived which makes num__45 num__67 possible for the tens digit . therefore the answer is c . <eor> c <eos> |
c |
subtract__69.0__2.0__ subtract__6.0__2.0__ |
subtract__69.0__2.0__ subtract__6.0__2.0__ |
| three - fourth of a positive number and num__0.115740740741 of its reciprocal are equal . the number is : <o> a ) num__5 / num__12.72 <o> b ) num__2.4 <o> c ) num__0.173611111111 <o> d ) num__5.76 <o> e ) num__5.84 |
let the number be x . then num__0.75 x = num__0.115740740741 * num__1 / x x num__2 = num__0.115740740741 * num__1.33333333333 = num__0.154320987654 x = num__5 / num__12.72 answer : a <eor> a <eos> |
a |
reverse__0.75__ divide__0.1157__0.75__ multiply__1.0__5.0__ |
reverse__0.75__ divide__0.1157__0.75__ multiply__1.0__5.0__ |
| the dimensions of a room are num__25 feet * num__15 feet * num__12 feet . what is the cost of white washing the four walls of the room at rs . num__5 per square feet if there is one door of dimensions num__6 feet * num__3 feet and three windows of dimensions num__4 feet * num__3 feet each ? <o> a ) rs . num__4589 <o> b ) rs . num__4528 <o> c ) rs . num__4528 <o> d ) rs . num__4530 <o> e ) rs . num__4537 |
area of the four walls = num__2 h ( l + b ) since there are doors and windows area of the walls = num__2 * num__12 ( num__15 + num__25 ) - ( num__6 * num__3 ) - num__3 ( num__4 * num__3 ) = num__906 sq . ft . total cost = num__906 * num__5 = rs . num__4530 answer : d <eor> d <eos> |
d |
multiply__5.0__906.0__ multiply__5.0__906.0__ |
multiply__5.0__906.0__ multiply__5.0__906.0__ |
| adding num__20.0 of x to x is equivalent to multiplying x by which of the following ? <o> a ) num__12.5 <o> b ) num__1.05 <o> c ) num__1.15 <o> d ) num__1.2 <o> e ) num__1.25 |
num__120 x / num__100 = num__1.2 * x answer : d <eor> d <eos> |
d |
subtract__120.0__20.0__ divide__120.0__100.0__ divide__120.0__100.0__ |
subtract__120.0__20.0__ divide__120.0__100.0__ divide__120.0__100.0__ |
| find the simple interest on rs . num__500 for num__9 months at num__6 paisa per month ? <o> a ) num__278 <o> b ) num__270 <o> c ) num__876 <o> d ) num__279 <o> e ) num__367 |
i = ( num__500 * num__9 * num__6 ) / num__100 = num__270 answer : b <eor> b <eos> |
b |
percent__100.0__270.0__ |
percent__100.0__270.0__ |
| . ram sold two bicycles each for rs . num__990 . if he made num__10.0 profit on the first and num__10.0 loss on the second what is the total cost of both bicycles ? <o> a ) num__2000 <o> b ) num__2009 <o> c ) num__2007 <o> d ) num__20054 <o> e ) num__2002 |
( num__10 * num__10 ) / num__100 = num__1.0 loss num__100 - - - num__99 ? - - - num__1980 = > rs . num__2000 answer : a <eor> a <eos> |
a |
percent__10.0__990.0__ percent__100.0__2000.0__ |
percent__10.0__990.0__ percent__100.0__2000.0__ |
| at num__6 ′ o a clock ticks num__6 times . the time between first and last ticks is num__20 seconds . how long does it tick at num__12 ′ o clock <o> a ) num__47 <o> b ) num__76 <o> c ) num__44 <o> d ) num__66 <o> e ) num__11 |
explanation : for ticking num__6 times there are num__5 intervals . each interval has time duration of num__4.0 = num__4 secs at num__12 o ' clock there are num__11 intervals so total time for num__11 intervals = num__11 × num__4 = num__44 secs . answer : c <eor> c <eos> |
c |
divide__20.0__5.0__ add__6.0__5.0__ multiply__4.0__11.0__ round__44.0__ |
divide__20.0__5.0__ add__6.0__5.0__ multiply__4.0__11.0__ round__44.0__ |
| a and b together can do a work in num__6 days . if a alone can do it in num__15 days . in how many days can b alone do it ? <o> a ) num__10 <o> b ) num__15 <o> c ) num__12 <o> d ) num__16 <o> e ) num__18 |
explanation : num__0.166666666667 â € “ num__0.0666666666667 = num__0.1 = > num__10 answer is a <eor> a <eos> |
a |
subtract__0.1667__0.0667__ round__10.0__ |
subtract__0.1667__0.0667__ round__10.0__ |
| the number of natural numbers n such that ( n + num__1 ) ^ num__2 / ( n + num__7 ) is an integer is ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
we can use ( n + num__1 ) ^ num__2 greater than ( n + num__7 ) which gives n > num__2 - num__3 . . . . . . . . now there are four integers - num__2 . - num__1 num__01 in between - num__3 and num__2 . so answer is num__4 . answer : a <eor> a <eos> |
a |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__3.0__ |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__3.0__ |
| mary peter and lucy were picking chestnuts . mary picked twice as much chestnuts than peter . lucy picked num__2 kg more than peter . together the three of them picked num__26 kg of chestnuts . how many kilograms did each of them pick ? <o> a ) num__5 num__12 and num__8 kg <o> b ) num__6 num__12 and num__8 kg <o> c ) num__7 num__12 and num__8 kg <o> d ) num__8 num__12 and num__8 kg <o> e ) none |
solution : let x be the amount peter picked . then mary and lucy picked num__2 x and x + num__2 x + num__2 respectively . so x + num__2 x + x + num__2 = num__26 num__4 x = num__24 x = num__6 therefore peter mary and lucy picked num__6 num__12 and num__8 kg respectively . answer b <eor> b <eos> |
b |
subtract__26.0__2.0__ add__2.0__4.0__ multiply__2.0__6.0__ multiply__2.0__4.0__ add__2.0__4.0__ |
subtract__26.0__2.0__ add__2.0__4.0__ multiply__2.0__6.0__ add__2.0__6.0__ add__2.0__4.0__ |
| if a = num__105 and a ^ num__3 = num__21 * num__25 * num__315 * b what is the value of b ? <o> a ) num__35 <o> b ) num__7 <o> c ) num__45 <o> d ) num__49 <o> e ) num__54 |
first step will be to break down all the numbers into their prime factors . num__105 = num__3 * num__5 * num__7 num__21 = num__7 * num__3 num__25 = num__5 * num__5 num__315 = num__3 * num__3 * num__5 * num__7 so ( num__105 ) ^ num__3 = num__3 * num__7 * num__5 * num__5 * num__3 * num__3 * num__5 * num__7 * b therefore ( num__3 * num__5 * num__7 ) ^ num__3 = num__3 ^ num__3 * num__5 ^ num__3 * num__7 ^ num__2 * b therefore b = num__3 ^ num__3 * num__5 ^ num__3 * num__7 ^ num__1.0 ^ num__3 * num__5 ^ num__3 * num__7 ^ num__2 b = num__7 correct answer b . <eor> b <eos> |
b |
divide__105.0__21.0__ divide__21.0__3.0__ subtract__5.0__3.0__ subtract__3.0__2.0__ divide__21.0__3.0__ |
divide__105.0__21.0__ divide__21.0__3.0__ subtract__5.0__3.0__ subtract__3.0__2.0__ multiply__1.0__7.0__ |
| a chair is bought for rs . num__600 / - and sold at a loss of num__10.0 find its selling price ? <o> a ) rs . num__540 / - <o> b ) rs . num__560 / - <o> c ) rs . num__580 / - <o> d ) rs . num__590 / - <o> e ) rs . num__600 / - |
num__100.0 - - - - - - > num__600 ( num__100 * num__6 = num__600 ) num__90.0 - - - - - - > num__540 ( num__90 * num__6 = num__540 ) selling price = rs . num__540 / - option ' a ' <eor> a <eos> |
a |
percent__90.0__600.0__ percent__90.0__600.0__ |
percent__90.0__600.0__ percent__90.0__600.0__ |
| a single discount equivalent to the discount series of num__15.0 num__10.0 and num__5.0 is ? <o> a ) num__27.325 <o> b ) num__31.0 <o> c ) num__31.6 <o> d ) num__31.1 <o> e ) num__31.5 |
num__100 * ( num__0.85 ) * ( num__0.9 ) * ( num__0.95 ) = num__68.4 num__100 - num__72.675 = num__27.325 answer : a <eor> a <eos> |
a |
percent__27.325__100.0__ |
percent__27.325__100.0__ |
| if n is a prime number which of the following could be true <o> a ) n ^ n = n <o> b ) n ^ num__0.5 = is even <o> c ) ( n ) ( n ^ n ) = is negative <o> d ) n ^ n / num__4 = num__1 ^ ( n - num__1 ) <o> e ) n ^ num__2 + n ^ num__3 = n ^ num__5 |
d should be the answer . num__2 ^ num__0.5 = num__1 ^ ( num__2 - num__1 ) <eor> d <eos> |
d |
reverse__2.0__ multiply__0.5__2.0__ divide__2.0__0.5__ |
reverse__2.0__ multiply__0.5__2.0__ divide__2.0__0.5__ |
| what is the product of the greatest num__2 digit multiple of num__13 and the greatest num__2 digit prime number ? <o> a ) num__9312 <o> b ) num__9408 <o> c ) num__9506 <o> d ) num__8827 <o> e ) num__9 |
702 |
the greatest num__2 digit multiple of num__13 : num__91 the greatest num__2 digit prime numebr : num__97 num__97 * num__91 . num__8827 d <eor> d <eos> |
d |
d |
| pipe a can fill a tank in num__12 hours . due to a leak at the bottom it takes num__18 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? <o> a ) num__36 <o> b ) num__87 <o> c ) num__40 <o> d ) num__37 <o> e ) num__86 |
let the leak can empty the full tank in x hours num__0.0833333333333 - num__1 / x = num__0.0555555555556 = > num__1 / x = num__0.0833333333333 - num__0.0555555555556 = ( num__3 - num__2 ) / num__36 = num__0.0277777777778 = > x = num__36 . answer : a <eor> a <eos> |
a |
divide__1.0__18.0__ subtract__3.0__1.0__ multiply__12.0__3.0__ divide__1.0__36.0__ round__36.0__ |
divide__1.0__18.0__ subtract__3.0__1.0__ multiply__12.0__3.0__ divide__1.0__36.0__ divide__36.0__1.0__ |
| a man has rs . num__160 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__45 <o> b ) num__60 <o> c ) num__30 <o> d ) num__90 <o> e ) num__95 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__160 num__16 x = num__160 x = num__10 . hence total number of notes = num__3 x = num__30 . answer : option c <eor> c <eos> |
c |
divide__160.0__10.0__ multiply__3.0__10.0__ multiply__3.0__10.0__ |
divide__160.0__10.0__ multiply__3.0__10.0__ multiply__3.0__10.0__ |
| two cars cover the same distance at the speed of num__50 and num__63 kmps respectively . find the distance traveled by them if the slower car takes num__1 hour more than the faster car . <o> a ) num__212 km <o> b ) num__214 km <o> c ) num__224 km <o> d ) num__216 km <o> e ) num__231 km |
num__50 ( x + num__1 ) = num__63 x x = num__3.85 num__60 * num__3.85 = num__231 km answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ multiply__3.85__60.0__ round__231.0__ |
hour_to_min_conversion__ multiply__3.85__60.0__ multiply__1.0__231.0__ |
| a pump can fill a tank with water in num__2 hours . because of a leak it took num__2 num__0.333333333333 hours to fill the tank . the leak can drain all the water in ? <o> a ) num__10 hr <o> b ) num__14 hr <o> c ) num__12 hr <o> d ) num__9 hr <o> e ) num__15 hr |
work done by the leak in num__1 hour = num__0.5 - num__0.428571428571 = num__0.0714285714286 leak will empty the tank in num__14 hrs answer is b <eor> b <eos> |
b |
divide__1.0__2.0__ subtract__0.5__0.4286__ round__14.0__ |
divide__1.0__2.0__ subtract__0.5__0.4286__ round__14.0__ |
| two trains are running in opposite directions in the same speed . the length of each train is num__120 meter . if they cross each other in num__12 seconds the speed of each train ( in km / hr ) is <o> a ) num__42 <o> b ) num__36 <o> c ) num__28 <o> d ) num__20 <o> e ) num__24 |
explanation : distance covered = num__120 + num__120 = num__240 m time = num__12 s let the speed of each train = v . then relative speed = v + v = num__2 v num__2 v = distance / time = num__20.0 = num__20 m / s speed of each train = v = num__10.0 = num__10 m / s = num__10 × num__3.6 km / hr = num__36 km / hr answer : option b <eor> b <eos> |
b |
divide__240.0__120.0__ divide__240.0__12.0__ divide__120.0__12.0__ multiply__10.0__3.6__ round__36.0__ |
divide__240.0__120.0__ divide__240.0__12.0__ divide__120.0__12.0__ multiply__10.0__3.6__ round__36.0__ |
| if num__5 a + num__7 b = l where a and b are positive integers what is the largest possible value of l for which exactly one pair of integers ( a b ) makes the equation true ? <o> a ) num__35 <o> b ) num__48 <o> c ) num__69 <o> d ) num__70 <o> e ) num__74 |
num__5 * a num__1 + num__7 * b num__1 = l num__5 * a num__2 + num__7 * b num__2 = l num__5 * ( a num__1 - a num__2 ) = num__7 * ( b num__2 - b num__1 ) since we are dealing with integers we can assume that a num__1 - a num__2 = num__7 * q and b num__2 - b num__1 = num__5 * q where q is integer so whenever we get a pair for ( a ; b ) we can find another one by simply adding num__7 to a and subtracting num__5 from b or vice versa subtracting num__7 from a and adding num__5 to b . lets check how it works for our numbers starting from the largest : e ) num__74 = num__5 * num__12 + num__7 * num__2 ( a num__1 = num__12 b num__1 = num__2 ) subtract num__7 fromaand add num__5 tobrespectively so a num__2 = num__5 and b num__2 = num__7 second pair - bad d ) num__70 = num__5 * num__7 + num__7 * num__5 ( a num__1 = num__7 b num__1 = num__5 ) if we add num__7 toawe will have to subtract num__5 from b but b ca n ' t be num__0 so - no pair if we subtract num__7 froma we ' ll get a = num__0 which also is n ' t allowed - no pair thus this is the only pair for ( a ; b ) that works good ! thus d is the answer <eor> d <eos> |
d |
subtract__7.0__5.0__ add__5.0__7.0__ multiply__1.0__70.0__ |
subtract__7.0__5.0__ add__5.0__7.0__ multiply__1.0__70.0__ |
| how long will a boy take to run round a square field of side num__30 meters if he runs at the rate of num__12 km / hr ? <o> a ) num__52 sec <o> b ) num__45 sec <o> c ) num__60 sec <o> d ) num__25 sec <o> e ) num__36 sec |
speed = num__12 km / hr = num__12 * num__0.277777777778 = num__3.33333333333 m / sec distance = num__30 * num__4 = num__120 m time taken = num__120 * num__0.3 = num__36 sec answer is e <eor> e <eos> |
e |
percent__30.0__120.0__ percent__30.0__120.0__ |
percent__30.0__120.0__ percent__30.0__120.0__ |
| num__46 - num__3 * num__4 ^ num__2 + num__15 = ? <o> a ) num__287 <o> b ) - num__13 <o> c ) - num__73 <o> d ) num__13 <o> e ) none |
num__46 - num__3 * num__4 ^ num__2 + num__15 = num__46 - num__3 * num__16 + num__15 = num__20 - num__48 + num__15 = num__13 . . d <eor> d <eos> |
d |
add__4.0__16.0__ add__46.0__2.0__ subtract__15.0__2.0__ subtract__15.0__2.0__ |
add__4.0__16.0__ add__46.0__2.0__ subtract__15.0__2.0__ subtract__15.0__2.0__ |
| a num__300 m long train crosses a platform in num__39 sec while it crosses a signal pole in num__18 sec . what is the length of the platform ? <o> a ) num__350 <o> b ) num__300 <o> c ) num__400 <o> d ) num__450 <o> e ) num__500 |
speed = num__16.6666666667 = num__16.6666666667 m / sec . let the length of the platform be x meters . then ( x + num__300 ) / num__39 = num__16.6666666667 num__3 x + num__900 = num__1950 = > x = num__350 m . answer a <eor> a <eos> |
a |
divide__300.0__18.0__ multiply__300.0__3.0__ round__350.0__ |
divide__300.0__18.0__ multiply__300.0__3.0__ round__350.0__ |
| victor gets num__80.0 marks in examinations . if these are num__240 marks find the maximum marks . <o> a ) num__334 <o> b ) num__300 <o> c ) num__376 <o> d ) num__288 <o> e ) num__271 |
let the maximum marks be m then num__80.0 of m = num__240 ⇒ num__0.8 × m = num__240 ⇒ m = ( num__240 × num__100 ) / num__80 ⇒ m = num__300.0 ⇒ m = num__300 therefore maximum marks in the examinations are num__300 answer : b <eor> b <eos> |
b |
percent__100.0__300.0__ |
percent__100.0__300.0__ |
| mr . smith calculated the average of num__10 three digit numbers . but due to a mistake he reversed the digits of a number and thus his average increased by num__59.4 . the difference between the unit digit and hundreds digit of that number is : <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
since the average increased by num__59.4 and there were a total of num__10 numbers it means the incorrect number was num__297 greater than the correct number . say the correct number was abc ( where a b and c are the digits of the num__3 digit number ) then the incorrect number was cba . num__100 c + num__10 b + a - ( num__100 a + num__10 b + c ) = num__594 num__99 c - num__99 a = num__99 ( c - a ) = num__594 num__594 = num__99 * num__6 = num__99 ( c - a ) so c - a = num__6 answer ( d ) <eor> d <eos> |
d |
multiply__10.0__59.4__ divide__297.0__3.0__ divide__594.0__99.0__ divide__594.0__99.0__ |
multiply__10.0__59.4__ divide__297.0__3.0__ divide__594.0__99.0__ divide__594.0__99.0__ |
| if a train travelling at a speed of num__90 kmph crosses a pole in num__5 sec then the length of train is ? <o> a ) num__128 m <o> b ) num__125 m <o> c ) num__198 m <o> d ) num__276 m <o> e ) num__279 m |
d = num__90 * num__0.277777777778 * num__5 = num__125 m answer : b <eor> b <eos> |
b |
round__125.0__ |
round__125.0__ |
| john has taken four ( num__8 ) tests that have an average of num__82 . in order to bring his course grade up to a ‘ b ’ he will need to have a final average of num__87 . what will he need to average on his final two tests to achieve this grade ? <o> a ) num__87 <o> b ) num__90 <o> c ) num__92 <o> d ) num__107 <o> e ) num__97 |
num__87 - num__82 = num__5 * num__8 = num__40 points num__2 * num__87 = num__174 num__174 + num__40 = num__214 points to be scored in the num__2 tests . avg of num__2 tests = num__107.0 = num__107 answer : d traditional method : total scored till now num__82 * num__8 = num__656 total score to avg num__87 in num__10 tests = num__87 * num__10 = num__870 total to be scored on num__2 tests = num__870 - num__656 = num__214 avg on num__2 tests = num__107.0 = num__107 answer d <eor> d <eos> |
d |
subtract__87.0__82.0__ multiply__8.0__5.0__ multiply__87.0__2.0__ add__174.0__40.0__ divide__214.0__2.0__ multiply__8.0__82.0__ add__8.0__2.0__ multiply__87.0__10.0__ divide__214.0__2.0__ |
subtract__87.0__82.0__ multiply__8.0__5.0__ multiply__87.0__2.0__ add__174.0__40.0__ divide__214.0__2.0__ multiply__8.0__82.0__ add__8.0__2.0__ multiply__87.0__10.0__ subtract__214.0__107.0__ |
| if shreehari walks in the speed of num__4.5 km / hr from his house in what time will he reach his school which is num__750 m long from his house ? <o> a ) num__5 <o> b ) num__30 <o> c ) num__10 <o> d ) num__12 <o> e ) num__15 |
speed = num__4.5 * num__0.277777777778 = num__1.25 m / sec time taken = num__750 / num__1.25 = num__600 sec ie . num__10 mins . answer : c <eor> c <eos> |
c |
divide__750.0__1.25__ round__10.0__ |
divide__750.0__1.25__ round__10.0__ |
| at company k num__15 percent of the employees are secretaries and num__60 percent are salespeople . if there are num__60 other employees of company k how many employees does company k have ? <o> a ) num__240 <o> b ) num__180 <o> c ) num__190 <o> d ) num__200 <o> e ) num__400 |
let the total number of employees in the company be x % of secretaries = num__15.0 % of salespeople = num__60.0 % of of employees other than secretaries and salespeople = num__100 - num__75 = num__25.0 but this number is given as num__60 so num__25.0 of x = num__60 x = num__240 therefore there a total of num__240 employees in the company k correct answer - a <eor> a <eos> |
a |
percent__100.0__240.0__ |
percent__100.0__240.0__ |
| two goods trains each num__500 m long are running in opposite directions on parallel tracks . their speeds are num__45 km / hr and num__30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? <o> a ) num__40 <o> b ) num__45 <o> c ) num__48 <o> d ) num__51 <o> e ) num__44 |
relative speed = num__45 + num__30 = num__75 km / hr . num__75 * num__0.277777777778 = num__20.8333333333 m / sec . distance covered = num__500 + num__500 = num__1000 m . required time = num__1000 * num__0.048 = num__48 sec . answer : option c <eor> c <eos> |
c |
add__45.0__30.0__ multiply__1000.0__0.048__ round__48.0__ |
add__45.0__30.0__ multiply__1000.0__0.048__ multiply__1000.0__0.048__ |
| num__96.0 of the population of a village is num__23040 . the total population of the village is ? <o> a ) num__24000 <o> b ) num__25000 <o> c ) num__15000 <o> d ) num__1750 <o> e ) num__2589 |
explanation : x * ( num__0.96 ) = num__23040 x = num__240 * num__100 x = num__24000 a <eor> a <eos> |
a |
divide__23040.0__96.0__ divide__96.0__0.96__ divide__23040.0__0.96__ divide__23040.0__0.96__ |
divide__23040.0__96.0__ divide__96.0__0.96__ multiply__240.0__100.0__ multiply__240.0__100.0__ |
| i remember during the school days the teacher asked the class ` ` can you tell me the sum of the first num__50 odd numbers ? ` ` . i ran quickly to the teacher and told her ` ` the answer is num__2500 ' ' . the teacher replied ` ` lucky guess ' ' . she then asked me ` ` can you tell me the sum of first num__75 odd numbers ? ` ` . i wait for approx num__10 seconds and replied with the correct answer . how can i answer so quickly and whats the correct answer ? <o> a ) num__8715 <o> b ) num__0152 <o> c ) num__3581 <o> d ) num__5625 <o> e ) num__9126 |
d num__5625 n ^ num__1 num__75 * num__75 = num__5625 ( sum of first num__75 odd numbers ) . num__50 * num__50 = num__2500 ( sum of first num__50 odd numbers ) . <eor> d <eos> |
d |
round__5625.0__ |
round__5625.0__ |
| how many two - digit numbers are there such that all two digits are different and the first digit is not zero ? <o> a ) num__50 <o> b ) num__64 <o> c ) num__72 <o> d ) num__79 <o> e ) num__81 |
all two digits are different and first digit is not zero . so first digit can be filled in num__9 ways . and second digit can be filled in num__8 ways . total ways = num__9 * num__8 = num__72 hence option ( c ) . <eor> c <eos> |
c |
multiply__8.0__9.0__ multiply__8.0__9.0__ |
multiply__8.0__9.0__ multiply__8.0__9.0__ |
| the length of the bridge which a train num__130 metres long and travelling at num__45 km / hr can cross in num__30 seconds is ? <o> a ) num__288 <o> b ) num__277 <o> c ) num__245 <o> d ) num__266 <o> e ) num__261 |
speed = [ num__45 x num__0.277777777778 ] m / sec = [ num__12.5 ] m / sec time = num__30 sec let the length of bridge be x metres . then ( num__130 + x ) / num__30 = num__12.5 = > num__2 ( num__130 + x ) = num__750 = > x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| if g ( x ) = ax ^ num__5 + bx ^ num__3 + num__2 and g ( num__5 ) = num__10 then g ( – num__5 ) = ? <o> a ) – num__10 <o> b ) – num__8 <o> c ) – num__6 <o> d ) num__0 <o> e ) num__4 |
g ( num__5 ) = num__10 or num__5 ^ num__5 a + num__5 ^ num__3 b + num__2 = num__10 or num__5 ^ num__5 a + num__5 ^ num__3 b = num__8 g ( - num__5 ) = - num__5 ^ num__5 a - num__5 ^ num__3 b + num__2 = - ( num__5 ^ num__5 a + num__5 ^ num__3 b ) + num__2 = - num__8 + num__2 = - num__6 = ( c ) <eor> c <eos> |
c |
add__5.0__3.0__ multiply__3.0__2.0__ multiply__3.0__2.0__ |
add__5.0__3.0__ subtract__8.0__2.0__ subtract__8.0__2.0__ |
| a num__70 cm long wire is to be cut into two pieces so that one piece will be num__0.4 th of the other how many centimeters will the shorter piece be ? <o> a ) num__21 <o> b ) num__20 <o> c ) num__17 <o> d ) num__12 <o> e ) num__18 |
num__1 : num__0.4 = num__5 : num__2 num__0.285714285714 * num__70 = num__20 answer : b <eor> b <eos> |
b |
multiply__0.4__5.0__ round__20.0__ |
multiply__0.4__5.0__ round__20.0__ |
| if num__20 typists can type num__50 letters in num__30 minutes then how many letters will num__40 typists working at the same rate complete in num__1 hour ? <o> a ) num__400 <o> b ) num__500 <o> c ) num__350 <o> d ) num__250 <o> e ) num__200 |
no . of letters typing by num__20 typists in num__30 minutes = num__50 no . of letters typing by num__20 typists in num__60 minutes = num__50 * num__2 = num__100 no . of letters typing by num__40 typists in num__60 minutes = num__10.0 * num__40 = num__400 answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ divide__40.0__20.0__ multiply__50.0__2.0__ divide__20.0__2.0__ multiply__40.0__10.0__ round__400.0__ |
hour_to_min_conversion__ divide__40.0__20.0__ multiply__50.0__2.0__ divide__20.0__2.0__ multiply__40.0__10.0__ multiply__40.0__10.0__ |
| analyze the number square carefully and find out what number fits in place of question mark . num__35 num__20 num__14 num__27 num__12 num__18 num__6 num__4 ? <o> a ) num__6 <o> b ) num__12 <o> c ) num__24 <o> d ) num__36 <o> e ) num__8 |
num__20 fits in place of question mark . this is because ( number in first column ) / ( number in second column ) * num__8 = number in third column . here ( left number / middle number ) * num__8 = right number ( num__1.5 ) * num__8 = num__6 * num__2 = num__12 b <eor> b <eos> |
b |
triangle_area__12.0__2.0__ |
multiply__6.0__2.0__ |
| if a b and c together can finish a piece of work in num__4 days . a alone in num__12 days and b in num__18 days then c alone can do it in ? <o> a ) num__20 <o> b ) num__22 <o> c ) num__30 <o> d ) num__10 <o> e ) num__9 |
c = num__0.25 - num__0.0833333333333 â € “ num__0.0555555555556 = num__0.111111111111 = > num__9 days answer e <eor> e <eos> |
e |
round__9.0__ |
subtract__18.0__9.0__ |
| to reach her destination jeanette must drive num__90 miles . if she drives num__5 miles every num__6 minutes how much time will it take her to reach her destination ? <o> a ) num__2 hours and num__2 minutes <o> b ) num__1 hours and num__48 minutes <o> c ) num__2 hours and num__10 minutes <o> d ) num__2 hours and num__12 minutes <o> e ) num__2 hours and num__15 minutes |
num__6 minute for num__5 miles how much for num__90 ? minutes for num__1 ( num__1.2 ) mile multiplied by num__90 miles . num__6 * num__18.0 = num__108 > num__1 hours num__48 min answer is b <eor> b <eos> |
b |
subtract__6.0__5.0__ divide__6.0__5.0__ divide__90.0__5.0__ multiply__90.0__1.2__ round__1.0__ |
subtract__6.0__5.0__ divide__6.0__5.0__ divide__90.0__5.0__ multiply__90.0__1.2__ round__1.0__ |
| if the number of num__194 n num__70 where n represents the hundreds digit is a multiple of num__9 then the value of n could be which of the following ? <o> a ) num__6 <o> b ) num__5 <o> c ) num__4 <o> d ) num__2 <o> e ) num__1 |
divisibility rule of num__9 : the sum of the digits must be divisible by num__9 i . e num__1 + num__9 + num__4 + n + num__7 + num__0 = multiple of num__9 or num__21 + n = multiple of num__9 . if n = num__6 num__21 + n = num__27 which is a multiple of num__9 . hence answer is a . <eor> a <eos> |
a |
subtract__7.0__1.0__ add__6.0__21.0__ multiply__1.0__6.0__ |
subtract__7.0__1.0__ add__6.0__21.0__ multiply__1.0__6.0__ |
| how many two letter words are formed using the letters of the word book ? <o> a ) num__12 <o> b ) num__10 <o> c ) num__6 <o> d ) num__3 <o> e ) num__9 |
the number of letters in the given word is four . the number of two letter words that can be formed using these four letters is num__4 p num__2 = num__4 * num__3 = num__12 . answer : a <eor> a <eos> |
a |
multiply__3.0__4.0__ multiply__3.0__4.0__ |
multiply__3.0__4.0__ multiply__3.0__4.0__ |
| a gambler bought $ num__4000 worth of chips at a casino in denominations of $ num__20 and $ num__100 . that evening the gambler lost num__16 chips and then cashed in the remainder . if the number of $ num__20 chips lost was num__2 more or num__2 less than the number of $ num__100 chips lost what is the largest amount of money that the gambler could have received back ? <o> a ) $ num__2040 <o> b ) $ num__2120 <o> c ) $ num__1960 <o> d ) $ num__3120 <o> e ) $ num__1 |
400 |
in order to maximize the amount of money that the gambler kept we should maximize # of $ num__20 chips lost and minimize # of $ num__100 chips lost which means that # of $ num__20 chips lost must be num__2 more than # of $ num__100 chips lost . so if # of $ num__20 chips lost is x then # of $ num__100 chips lost should be x - num__2 . now given that total # of chips lost is num__16 : x + x - num__2 = num__16 - - > x = num__9 : num__9 $ num__20 chips were lost and num__9 - num__2 = num__7 $ num__100 chips were lost . total worth of chips lost is num__9 * num__20 + num__7 * num__100 = $ num__880 so the gambler kept $ num__4000 - $ num__880 = $ num__3120 . answer : d <eor> d <eos> |
d |
d |
| a garrison of num__400 men had a provision for num__31 days . after num__28 days num__280 persons re - enforcement leave the garrison . find the number of days for which the remaining ration will be sufficient ? <o> a ) num__16 days <o> b ) num__34 days <o> c ) num__10 days <o> d ) num__16 days <o> e ) num__15 days |
num__400 - - - num__31 num__400 - - - num__3 num__120 - - - ? num__400 * num__3 = num__120 * x = > x = num__10 days answer : c <eor> c <eos> |
c |
subtract__31.0__28.0__ subtract__400.0__280.0__ divide__280.0__28.0__ divide__280.0__28.0__ |
subtract__31.0__28.0__ subtract__400.0__280.0__ divide__280.0__28.0__ divide__280.0__28.0__ |
| the temperature of a certain cup of coffee num__2 minutes after it was poured was num__120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = num__120 * num__2 ^ ( - at ) + num__60 where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee num__30 minutes after it was poured was how many degrees fahrenheit ? <o> a ) num__65 <o> b ) num__60.003662 <o> c ) num__80.2 <o> d ) num__85 <o> e ) num__90 |
first we have to find a . we know that after t = num__2 minutes the temperature f = num__120 degrees . hence : num__120 = num__120 * ( num__2 ^ - num__2 a ) + num__60 num__60 = num__120 * ( num__2 ^ - num__2 a ) num__0.5 = num__2 ^ - num__2 a num__0.5 = num__2 ^ - num__2 a num__2 ^ - num__1 = num__2 ^ - num__2 a - num__1 = - num__2 a num__0.5 = a now we need to find f after t = num__30 minutes : f = num__120 * ( num__2 ^ - num__0.5 * num__30 ) + num__60 f = num__120 * ( num__2 ^ - num__15 ) + num__60 f = num__120 * ( num__0.5 ^ num__15 ) + num__60 f = num__120 * 3.0517578125e-05 + num__60 f = num__0.003662 + num__60 = num__60.003662 answer b ! <eor> b <eos> |
b |
reverse__2.0__ multiply__2.0__0.5__ divide__30.0__2.0__ add__60.0__0.0037__ add__60.0__0.0037__ |
reverse__2.0__ multiply__2.0__0.5__ multiply__30.0__0.5__ add__60.0__0.0037__ add__60.0__0.0037__ |
| a number when divided by num__5 gives a number which is num__8 more than the remainder obtained on dividing the same number by num__34 . such a least possible number m is <o> a ) num__74 <o> b ) m = num__75 <o> c ) m = num__175 <o> d ) m = num__680 <o> e ) num__690 |
i solved this question by plugging in numbers from the answer choices . a . ) num__74 starting with answer choice a i immediately eliminated it because num__74 is not even divisible by num__5 . b . ) num__75 i divide num__15.0 and get num__15 as an answer . i divide num__2.20588235294 and get a remainder of num__7 . num__15 - num__7 = num__8 so i know the correct answer isb <eor> b <eos> |
b |
divide__75.0__5.0__ divide__75.0__34.0__ subtract__15.0__8.0__ multiply__5.0__15.0__ |
divide__75.0__5.0__ divide__75.0__34.0__ subtract__15.0__8.0__ multiply__5.0__15.0__ |
| if the population of a certain country increases at the rate of one person every num__30 seconds by how many persons does the population increase in num__10 minutes ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__15 <o> d ) num__10 <o> e ) num__80 |
answer = num__2 * num__10 = num__20 answer = a <eor> a <eos> |
a |
subtract__30.0__10.0__ round__20.0__ |
multiply__10.0__2.0__ multiply__10.0__2.0__ |
| length of a rectangular plot is num__20 mtr more than its breadth . if the cost of fencin gthe plot at num__26.50 per meter is rs . num__5300 what is the length of the plot in mtr ? <o> a ) num__45 m <o> b ) num__60 m <o> c ) num__57 m <o> d ) num__69 m <o> e ) num__78 m |
let breadth = x metres . then length = ( x + num__20 ) metres . perimeter = num__5300 m = num__200 m . num__26.50 num__2 [ ( x + num__20 ) + x ] = num__200 num__2 x + num__20 = num__100 num__2 x = num__80 x = num__40 . hence length = x + num__20 = num__60 m b <eor> b <eos> |
b |
divide__5300.0__26.5__ divide__200.0__2.0__ subtract__100.0__20.0__ multiply__20.0__2.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
divide__5300.0__26.5__ divide__200.0__2.0__ subtract__100.0__20.0__ multiply__20.0__2.0__ add__20.0__40.0__ add__20.0__40.0__ |
| a train traveling at num__100 kmph overtakes a motorbike traveling at num__64 kmph in num__85 seconds . what is the length of the train in meters ? <o> a ) num__400 meters <o> b ) num__1111 meters <o> c ) num__160 meters <o> d ) num__850 meters <o> e ) none of these |
train overtakes a bike means that we are talking about total length of the train . ( train ' s head is close to bike when it started and its tail crosses the bike when it overtakes the bike ) relative speed = num__100 - num__64 = num__36 km / h = num__36000 m / h time = num__85 seconds distance = speed * time num__36000 * num__0.0236111111111 = num__850 meters . d is the answer . <eor> d <eos> |
d |
subtract__100.0__64.0__ round__850.0__ |
subtract__100.0__64.0__ round__850.0__ |
| the ratio of the incomes of lakshmi and latha is num__5 : num__4 and the ratio of their expenditure is num__4 : num__3 . if at the end of the year each saves $ num__6000 then the income of lakshmi is ? <o> a ) $ num__26000 <o> b ) $ num__20000 <o> c ) $ num__23000 <o> d ) $ num__25000 <o> e ) $ num__22000 |
let the income of lakshmi and latha be $ num__5 x and $ num__4 x let their expenditures be $ num__4 y and $ num__3 y num__5 x - num__4 y = num__6000 - - - - - - - num__1 ) num__4 x - num__3 y = num__6000 - - - - - - - num__2 ) from num__1 ) and num__2 ) x = num__6000 lakshmi ' s income = num__5 x = num__5 * num__5000 = $ num__25000 answer is d <eor> d <eos> |
d |
subtract__5.0__4.0__ subtract__5.0__3.0__ multiply__5.0__5000.0__ multiply__5.0__5000.0__ |
subtract__5.0__4.0__ subtract__5.0__3.0__ multiply__5.0__5000.0__ multiply__5.0__5000.0__ |
| one half of a two digit number exceeds its one third by num__7 . what is the sum of the digits of the number ? <o> a ) a ) num__3 <o> b ) b ) num__5 <o> c ) c ) num__7 <o> d ) d ) num__9 <o> e ) e ) num__10 |
x / num__2 – x / num__3 = num__7 = > x = num__46 num__4 + num__6 = num__10 answer : e <eor> e <eos> |
e |
subtract__7.0__3.0__ multiply__2.0__3.0__ add__7.0__3.0__ add__7.0__3.0__ |
subtract__7.0__3.0__ add__2.0__4.0__ add__7.0__3.0__ add__7.0__3.0__ |
| doug who runs track for his high school was challenged to a race by his younger brother matt . matt started running first and doug didn ’ t start running until matt had finished a quarter - mile lap on the school track . doug passed matt as they both finished their sixth lap . if both boys ran at a constant speed with doug running num__2 miles an hour faster than matt what was matt ’ s speed ? <o> a ) num__10.5 miles per hour <o> b ) num__10 miles per hour <o> c ) num__9 miles per hour <o> d ) num__8 miles per hour <o> e ) num__7.5 miles per hour |
doug runs num__2 miles an hour faster than matt so let matt ’ s speed equal x miles per hour . then doug ’ s speed equals x + num__2 miles per hour . each lap is one - quarter of a mile so doug runs num__1.5 miles in the time it takes matt to run num__1.25 miles . place this information in a chart : dough rate : x - num__2 time : num__1.5 / num__1 distance : num__1.5 matt rate : x time : num__1.25 / x distance : num__1.25 the two boys took the same amount of time from the time doug started so make an equation by setting the two times in the chart equal to each other and then solve for x : num__1.5 / x + num__2 = num__1.25 / x num__1.5 x = num__1.25 ( x + num__2 ) num__1.5 x = num__1.25 x + num__2.5 num__0.25 x = num__2.5 x = num__10 so matt ran at num__10 miles per hour . correct answer b ) num__10 miles per hour <eor> b <eos> |
b |
multiply__2.0__1.25__ subtract__1.5__1.25__ divide__2.5__0.25__ round__10.0__ |
add__1.5__1.0__ subtract__1.5__1.25__ divide__2.5__0.25__ divide__10.0__1.0__ |
| a person plans to start a business with x % of a list of num__10000 products in his inventory . next year after a business loss the person wants to reduce the list by ( x − num__4 ) % . in terms of x how many products will be in the inventory ? <o> a ) x * x – num__4 x <o> b ) ( x ) ( num__104 – x ) <o> c ) ( num__100 ) ( num__104 – x ) <o> d ) ( num__100 ) ( num__96 – x ) <o> e ) ( x - num__4 ) / num__100 |
based on the answer choices and the question this question begs the use of x = num__4 as a sample number . initial = num__4.0 * num__10000 = num__400 reduction = num__4 - num__4 = num__0.0 so no math required here to calculate the reduction ; just make sure that you can calculate num__400 in your answer . a . x * x – num__4 x = num__0 ; no b . ( x ) ( num__104 – x ) = num__400 ; winner ! c . ( num__100 ) ( num__104 – x ) > num__400 ; no d . ( num__100 ) ( num__96 – x ) > num__400 ; no e . ( x - num__4 ) / num__100 = num__0 ; no b <eor> b <eos> |
b |
percent__4.0__10000.0__ percent__100.0__104.0__ |
percent__4.0__10000.0__ percent__100.0__104.0__ |
| a grocer has num__400 pounds of coffee in stock num__20 percent of which is decaffeinated . if the grocer buys another num__100 pounds of coffee of which num__60 percent is decaffeinated what percent by weight of the grocer ’ s stock of coffee is decaffeinated ? <o> a ) num__28.0 <o> b ) num__30.0 <o> c ) num__32.0 <o> d ) num__34.0 <o> e ) num__40 % |
num__1 . num__20.0 of num__400 = num__80 pounds of decaffeinated coffee num__2 . num__60.0 of num__100 = num__60 pounds of decaffeinated coffee num__3 . wt have num__140 pounds of decaffeinated out of num__500 pounds that means num__0.28 * num__100.0 = num__28.0 . the correct answer is a . <eor> a <eos> |
a |
percent__20.0__400.0__ percent__20.0__140.0__ percent__20.0__140.0__ |
percent__20.0__400.0__ percent__20.0__140.0__ percent__20.0__140.0__ |
| a father was as old as his son ' s present at the time of your birth . if the father ' s age is num__38 years now the son ' s age num__5 years back was ? <o> a ) num__10 years <o> b ) num__12 years <o> c ) num__14 years <o> d ) num__15 years <o> e ) num__16 years |
let the son ' s present age be x years . then ( num__38 - x ) = x num__2 x = num__38 . x = num__19 . son ' s age num__5 years back ( num__19 - num__5 ) = num__14 years . c <eor> c <eos> |
c |
divide__38.0__2.0__ subtract__19.0__5.0__ subtract__19.0__5.0__ |
divide__38.0__2.0__ subtract__19.0__5.0__ subtract__19.0__5.0__ |
| n is a whole number which when divided by num__4 gives num__3 as remainder . what will be the remainder when num__2 n is divided by num__4 ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__7 <o> d ) num__9 <o> e ) num__11 |
let n = num__4 q + num__3 . then num__2 n = num__8 q + num__6 = num__4 ( num__2 q + num__1 ) + num__2 . thus when num__2 n is divided by num__4 the remainder is num__2 . option a <eor> a <eos> |
a |
multiply__4.0__2.0__ add__4.0__2.0__ subtract__4.0__3.0__ subtract__4.0__2.0__ |
multiply__4.0__2.0__ add__4.0__2.0__ subtract__4.0__3.0__ subtract__4.0__2.0__ |
| find the value for m ? num__19 ( m + n ) + num__17 = num__19 ( - m + n ) - num__59 <o> a ) num__0 <o> b ) - num__1 <o> c ) num__1 <o> d ) num__2 <o> e ) - num__2 |
num__19 m + num__19 n + num__17 = - num__19 m + num__19 n - num__59 num__38 m = - num__76 = > m = - num__2 e <eor> e <eos> |
e |
add__17.0__59.0__ subtract__19.0__17.0__ subtract__19.0__17.0__ |
add__17.0__59.0__ subtract__19.0__17.0__ subtract__19.0__17.0__ |
| in what time will a train num__300 m long cross an electric pole it its speed be num__144 km / hr ? <o> a ) num__2.5 sec <o> b ) num__9.7 sec <o> c ) num__3.5 sec <o> d ) num__2.9 sec <o> e ) num__7.5 sec |
speed = num__144 * num__0.277777777778 = num__40 m / sec time taken = num__7.5 = num__7.5 sec . answer : e <eor> e <eos> |
e |
divide__300.0__40.0__ round__7.5__ |
divide__300.0__40.0__ divide__300.0__40.0__ |
| num__24 = num__3 a + num__5 b num__3 > | – a | if ‘ a ’ and ‘ b ’ are both integers then what is the highest possible values of ‘ b ’ that can be used to solve the above equation . <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
let us understand the meaning of num__5 > | - a | num__1 ) when we write | - x | this means the same as | x | . num__2 ) mod is very easy concept if you solve mod question by considering as a distance . when a mod is written as | x - ( a ) | = b this means the distance from point ' a ' ( both side left and right of ' a ' on number line ) is b . | x - ( a ) | < b means the distance is between the two extreme distance ( left and right side of ' a ' on number line considering the max distance is ' b ' from ' a ' - as per this scenario . . . . . hence the value of ' a ' must be between these two extremes . | x - ( a ) | > b means the distance is greater than the distance of ' b ' . . i . e the value of a could be anywhere more than ' b ' . now come to the question . first its given | - a | < num__3 = = > | a | < num__3 = = = > | a - num__0 | < num__3 = = > the distance from zero is less than num__3 . so the point ‘ a ’ will be - num__2 - num__10 num__12 as distance from num__0 to these values is less than num__3 . now lets move to equation num__3 a + num__5 b = num__24 = = > b = ( num__24 – ( num__3 * a ) ) / num__5 . according to question b is an integer hence to make ‘ b ’ integer ( num__24 – ( num__3 * a ) ) must be divisible by num__5 . now remove the value which can ’ t make ( num__24 - ( num__3 * a ) ) divisible by num__5 from the possible values of a . it will remain is - num__2 . with only one value of ‘ a ’ we can have only one value of ‘ b ’ and that is b = ( num__24 – num__3 * ( - num__2 ) ) / num__5 = = > b = ( num__24 + num__6 ) / num__5 = = > b = num__6 . hence the highest value that b can take is num__6 and the answer is e ( value num__6 ) . <eor> e <eos> |
e |
subtract__3.0__1.0__ multiply__5.0__2.0__ divide__24.0__2.0__ multiply__3.0__2.0__ multiply__3.0__2.0__ |
subtract__3.0__1.0__ multiply__5.0__2.0__ divide__24.0__2.0__ multiply__3.0__2.0__ multiply__3.0__2.0__ |
| two taps can separately fill a cistern num__10 minutes and num__15 minutes respectively and when the waste pipe is open they can together fill it in num__15 minutes . the waste pipe can empty the full cistern in ? <o> a ) num__10 min <o> b ) num__7 min <o> c ) num__5 min <o> d ) num__9 min <o> e ) num__4 min |
num__0.1 + num__0.0666666666667 - num__1 / x = num__0.1 x = num__10 answer : a <eor> a <eos> |
a |
multiply__10.0__0.1__ round__10.0__ |
multiply__10.0__0.1__ divide__10.0__1.0__ |
| in august a cricket team that played num__120 matches won num__28.0 of the games it played . after a continuous winning streak this team raised its average to num__52.0 . how many matches did the team win to attain this average ? <o> a ) num__40 <o> b ) num__52 <o> c ) num__60 <o> d ) num__80 <o> e ) num__98 |
let the no of matches played more = x so ( num__120 + x ) * num__0.52 = num__33.6 + x by solving we get x = num__60 answer : c <eor> c <eos> |
c |
subtract__120.0__60.0__ |
subtract__120.0__60.0__ |
| if num__13 lions can kill num__13 deers in num__13 minutes how long will it take num__100 lions to kill num__100 deers ? <o> a ) num__1 minutes <o> b ) num__13 minute <o> c ) num__100 minutes <o> d ) num__10000 minutes <o> e ) num__1000 minutes |
we can try the logic of time and work our work is to kill the deers so num__13 ( lions ) * num__13 ( min ) / num__13 ( deers ) = num__100 ( lions ) * x ( min ) / num__100 ( deers ) hence answer is x = num__13 answer : b <eor> b <eos> |
b |
round__13.0__ |
round__13.0__ |
| in a class the average age of num__30 boys is num__13 years and the average of num__20 girls is num__12 years . what is the average age of the whole class ? <o> a ) num__12 years <o> b ) num__12.3 years <o> c ) num__12.6 years <o> d ) num__13 years <o> e ) num__15 years |
total age of num__50 students ( num__30 x num__13 + num__20 x num__12 ) = num__630 average = num__12.6 = num__12.6 years c <eor> c <eos> |
c |
add__30.0__20.0__ divide__630.0__50.0__ divide__630.0__50.0__ |
add__30.0__20.0__ divide__630.0__50.0__ divide__630.0__50.0__ |
| how many positive three - digit integers are divisible by both num__3 and num__5 ? <o> a ) num__50 <o> b ) num__60 <o> c ) num__65 <o> d ) num__70 <o> e ) num__80 |
a number to be divisible by both num__3 and num__5 should be divisible by the least common multiple of num__3 and num__5 so by num__15 . multiples of num__15 between num__100 and num__999 inclusive is ( last - first ) / multiple + num__1 = ( num__990 - num__105 ) / num__15 + num__1 = num__59 + num__1 = num__60 answer : b . <eor> b <eos> |
b |
multiply__3.0__5.0__ add__5.0__100.0__ add__1.0__59.0__ add__1.0__59.0__ |
multiply__3.0__5.0__ add__5.0__100.0__ add__1.0__59.0__ add__1.0__59.0__ |
| townville has num__100 residents and num__20 of them are females . if half of townville ’ s male residents are smokers and num__0.5 of the female residents are smokers which of the following represents the total number of townville ’ s residents who are not smokers ? <o> a ) num__40 <o> b ) num__45 <o> c ) num__50 <o> d ) num__55 <o> e ) num__60 |
number of people = num__100 females = num__20 men = num__100 - num__20 = num__80 half of the men are smoker num__40.0 = num__40 that means the remaining men num__80 - num__40 = num__40 are non smokers . num__0.5 females are smoker . i . e num__0.5 * num__20 = num__10 num__20 - num__10 = num__10 females are non smokers so total number of non smokers in the town are num__40 + num__10 = num__50 answer : c <eor> c <eos> |
c |
subtract__100.0__20.0__ divide__20.0__0.5__ multiply__20.0__0.5__ multiply__100.0__0.5__ multiply__100.0__0.5__ |
subtract__100.0__20.0__ multiply__0.5__80.0__ multiply__20.0__0.5__ multiply__100.0__0.5__ multiply__100.0__0.5__ |
| two goods trains each num__500 m long are running in opposite directions on parallel tracks . their speeds are num__45 km / hr and num__30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? <o> a ) num__12 sec <o> b ) num__24 sec <o> c ) num__48 sec <o> d ) num__60 sec <o> e ) num__62 sec |
relative speed = num__45 + num__30 = num__75 km / hr . num__75 * num__0.277777777778 = num__20.8333333333 m / sec . distance covered = num__500 + num__500 = num__1000 m . required time = num__1000 * num__0.048 = num__48 sec . answer : c <eor> c <eos> |
c |
add__45.0__30.0__ multiply__1000.0__0.048__ round__48.0__ |
add__45.0__30.0__ multiply__1000.0__0.048__ multiply__1000.0__0.048__ |
| a jeep takes num__3 hours to cover a distance of num__440 km . how much should the speed in kmph be maintained to cover the same direction in num__0.5 th of the previous time ? <o> a ) num__48 kmph <o> b ) num__252 kmph <o> c ) num__293 kmph <o> d ) num__263 kmph <o> e ) num__265 kmph |
time = num__3 distance = num__340 num__0.5 of num__3 hours = num__6 * num__0.5 = num__1.5 hours required speed = num__440 / num__1.5 = num__293 kmph c ) <eor> c <eos> |
c |
divide__3.0__0.5__ multiply__3.0__0.5__ round__293.0__ |
divide__3.0__0.5__ multiply__3.0__0.5__ round__293.0__ |
| there are num__3 people a b & c . rs . num__700 is divided among a b & c so that a receives num__0.5 as much as b and b num__0.5 as much as c . what is the c ' s share is ? <o> a ) rs . num__100 <o> b ) rs . num__200 <o> c ) rs . num__400 <o> d ) rs . num__500 <o> e ) rs . num__550 |
let c ' s share = rs . x . then b ' s share = rs . x num__2 and a ' s share = rs . x num__4 a : b : c = x num__4 : x num__2 : x = num__1 : num__2 : num__4 hence c ' s share = rs . ( num__700 × num__47 ) = rs . num__400 c <eor> c <eos> |
c |
reverse__0.5__ divide__2.0__0.5__ subtract__3.0__2.0__ multiply__1.0__400.0__ |
reverse__0.5__ divide__2.0__0.5__ subtract__3.0__2.0__ multiply__1.0__400.0__ |
| foodmart customers regularly buy at least one of the following products : milk chicken or apples . num__60.0 of shoppers buy milk num__50.0 buy chicken and num__35.0 buy apples . if num__17.0 of the customers buy all num__3 products what percentage of foodmart customers purchase exactly num__2 of the products listed above ? <o> a ) num__5.0 <o> b ) num__13.0 <o> c ) num__21.0 <o> d ) num__11.0 <o> e ) num__30 % |
num__60 - ( x + num__17 + z ) + num__50 - ( x + num__17 + y ) + num__35 - ( z + num__17 + y ) + x + y + z + num__17 = num__100 where x = people who bought milkchicken y = people who bought chickenapples z = people who bought milk and apples x + y + z = the number of people who bought just exactly two products . hence solving the above equation we get num__111 - ( x + y + z ) = num__100 thus x + y + z = num__11 answer : d <eor> d <eos> |
d |
multiply__50.0__2.0__ subtract__111.0__100.0__ subtract__111.0__100.0__ |
multiply__50.0__2.0__ subtract__111.0__100.0__ subtract__111.0__100.0__ |
| for any real number x the operatoris defined as : ( x ) = x ( num__3 − x ) if p + num__1 = ( p + num__1 ) then p = <o> a ) − num__3 <o> b ) num__0 <o> c ) num__1 <o> d ) num__2 <o> e ) num__3 |
( x ) = x ( num__1 − x ) ( p + num__1 ) = ( p + num__1 ) ( num__3 - p - num__1 ) = ( num__2 - p ) ( p + num__1 ) we are given that p + num__1 = ( p + num__1 ) therefore ( num__2 - p ) ( p + num__1 ) = ( p + num__1 ) or ( p + num__1 ) + ( p - num__2 ) ( p + num__1 ) = num__0 ( p + num__1 ) ( p - num__2 ) = num__0 p = - num__1 p = num__2 option d <eor> d <eos> |
d |
subtract__3.0__1.0__ subtract__3.0__1.0__ |
subtract__3.0__1.0__ subtract__3.0__1.0__ |
| a train traveled the first d miles of its journey it an average speed of num__60 miles per hour the next d miles of its journey at an average speed of y miles per hour and the final d miles of its journey at an average speed of num__160 miles per hour . if the train ’ s average speed over the total distance was num__98 miles per hour what is the value of y ? <o> a ) num__68 <o> b ) num__84 <o> c ) num__90 <o> d ) num__120 <o> e ) num__125 |
average speed = total distance traveled / total time taken num__3 d / d / num__60 + d / y + d / num__160 = num__98 solving for d and y num__15 y = num__11 y + num__480 num__4 y = num__500 y = num__125 answer e <eor> e <eos> |
e |
multiply__160.0__3.0__ divide__60.0__15.0__ divide__500.0__4.0__ round__125.0__ |
multiply__160.0__3.0__ divide__60.0__15.0__ divide__500.0__4.0__ divide__500.0__4.0__ |
| the ratio between the length and the breadth of a rectangular park is num__3 : num__2 . if a man cycling along the boundary of the park at the speed of num__12 km / hr completes one round in num__9 minutes then the area of the park ( in sq . m ) is : <o> a ) num__153601 <o> b ) num__153600 <o> c ) num__153602 <o> d ) num__153603 <o> e ) num__194400 |
perimeter = distance covered in num__9 min . = ( num__200.0 ) x num__9 m = num__1800 m . let length = num__3 x metres and breadth = num__2 x metres . then num__2 ( num__3 x + num__2 x ) = num__1800 or x = num__180 . length = num__540 m and breadth = num__360 m . area = ( num__540 x num__360 ) m num__2 = num__194400 m num__2 . answer : e <eor> e <eos> |
e |
multiply__9.0__200.0__ multiply__3.0__180.0__ multiply__2.0__180.0__ multiply__360.0__540.0__ round__194400.0__ |
multiply__9.0__200.0__ multiply__3.0__180.0__ multiply__2.0__180.0__ multiply__360.0__540.0__ round__194400.0__ |
| a tank is filled by three pipes with uniform flow . the first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone . the second pipe fills the tank num__5 hours faster than the first pipe and num__4 hours slower than the third pipe . the time required by the first pipe is : <o> a ) num__15 hrs <o> b ) num__20 hrs <o> c ) num__25 hrs <o> d ) num__30 hrs <o> e ) num__10 hrs |
first pipe = x ssecond and third ( x - num__5 ) and ( x - num__9 ) num__1 / x + num__1 / ( x - num__5 ) = num__1 / ( x - num__9 ) = > x = num__15 answer a <eor> a <eos> |
a |
add__5.0__4.0__ subtract__5.0__4.0__ round__15.0__ |
add__5.0__4.0__ subtract__5.0__4.0__ divide__15.0__1.0__ |
| num__6 women can do num__75 units of work in num__8 days by working num__5 hrs / day . in how many days can num__4 women do num__30 units of work by working num__8 hrs / day ? <o> a ) num__3 days <o> b ) num__4 days <o> c ) num__5 days <o> d ) num__6 days <o> e ) num__7 days |
num__6 x num__8 x num__0.0666666666667 = num__4 x num__8 xa / num__30 solve this for a then we get a = num__3 days answer : a <eor> a <eos> |
a |
divide__5.0__75.0__ subtract__8.0__5.0__ round__3.0__ |
divide__5.0__75.0__ subtract__8.0__5.0__ round__3.0__ |
| if a - b = num__3 and a ^ num__2 + b ^ num__2 = num__29 find the value of ab . <o> a ) num__12 <o> b ) num__15 <o> c ) num__10 <o> d ) num__18 <o> e ) num__13 |
num__2 ab = ( a ^ num__2 + b ^ num__2 ) - ( a - b ) ^ num__2 = num__29 - num__9 = num__20 ab = num__10 . answer is c . <eor> c <eos> |
c |
subtract__29.0__9.0__ divide__20.0__2.0__ divide__20.0__2.0__ |
subtract__29.0__9.0__ divide__20.0__2.0__ subtract__20.0__10.0__ |
| in an exam c scored num__35 percent d scored num__65 percent and e num__50 percent . the maximum score awarded in the exam is num__2000 . find the average mark scored by all the three candidates ? <o> a ) num__100 <o> b ) num__500 <o> c ) num__1000 <o> d ) num__1500 <o> e ) num__2000 |
average mark scored by all the three boys = [ num__0.35 ( num__2000 ) + num__0.65 ( num__2000 ) + num__0.5 ( num__2000 ) ] / num__3 = num__1000 answer : c <eor> c <eos> |
c |
multiply__2000.0__0.5__ multiply__2000.0__0.5__ |
multiply__2000.0__0.5__ multiply__2000.0__0.5__ |
| on multiplying a number by num__7 the product is a number each of whose digits is num__3 . the smallest such number is : <o> a ) num__47619 <o> b ) num__49619 <o> c ) num__47719 <o> d ) num__57619 <o> e ) num__47611 |
last digit should be num__9 . when multiplying num__7 with num__9 we get num__63 . place num__3 at unit place we get num__6 as carry . next digit of number should be such that we get product with num__7 as a number whose last digit is num__13 - num__6 = num__7 . such number is num__1 . when num__7 is multiplied with num__1 we get num__7 . add carry num__6 to it and we get num__13 . so num__3 will get at tens place and num__1 carry for next place . so next digit should give last digit as num__2 ( num__3 - num__1 = num__2 ) . such digit is num__6 . . . . . . so continuing like this we get desired number as num__47619 answer : a <eor> a <eos> |
a |
multiply__7.0__9.0__ subtract__9.0__3.0__ add__7.0__6.0__ subtract__7.0__6.0__ subtract__3.0__1.0__ multiply__1.0__47619.0__ |
multiply__7.0__9.0__ subtract__9.0__3.0__ add__7.0__6.0__ subtract__7.0__6.0__ subtract__3.0__1.0__ multiply__1.0__47619.0__ |
| what is the sum of the odd integers from num__55 to num__65 inclusive ? <o> a ) num__495 <o> b ) num__360 <o> c ) num__555 <o> d ) num__600 <o> e ) num__605 |
the mean is num__60 . sum = mean ( # of elements ) there are num__6 odd numbers between num__55 - num__65 inclusive . num__6 * num__60 = num__360 b <eor> b <eos> |
b |
multiply__60.0__6.0__ multiply__60.0__6.0__ |
multiply__60.0__6.0__ multiply__60.0__6.0__ |
| if x = - num__5 and y = num__8 what is the value of num__2 ( x - y ) ^ num__2 - xy ? <o> a ) num__358 <o> b ) num__348 <o> c ) num__368 <o> d ) num__388 <o> e ) num__378 |
x = - num__5 and y = num__8 x - y = - num__5 - num__8 = - num__13 x * y = - num__5 * num__8 = - num__40 now we apply it in the equation num__2 ( x - y ) ^ num__2 - xy = num__2 ( - num__13 ) ^ num__2 - ( - num__40 ) = = > num__2 * num__169 + num__40 = num__338 + num__40 = num__378 answer : e <eor> e <eos> |
e |
add__5.0__8.0__ multiply__5.0__8.0__ multiply__2.0__169.0__ add__40.0__338.0__ add__40.0__338.0__ |
add__5.0__8.0__ multiply__5.0__8.0__ multiply__2.0__169.0__ add__40.0__338.0__ add__40.0__338.0__ |
| a cistern can be filled by a tap in num__4 hours while it can be emptied by another tap in num__11 hours . if both the taps are opened simultaneously then after how much time will the cistern get filled ? <o> a ) num__6.24 <o> b ) num__7.4 <o> c ) num__7.92 <o> d ) num__6.28 <o> e ) num__7.24 |
net part filled in num__1 hour = ( num__0.25 - num__0.0909090909091 ) = num__0.159090909091 the cistern will be filled in num__6.28571428571 hrs i . e . num__6.28 hrs . answer : d <eor> d <eos> |
d |
divide__1.0__4.0__ divide__1.0__11.0__ subtract__0.25__0.0909__ round__6.28__ |
divide__1.0__4.0__ divide__1.0__11.0__ subtract__0.25__0.0909__ round__6.28__ |
| a room is num__6 meters num__24 centimeters in length and num__4 meters num__32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the room ? <o> a ) num__110 <o> b ) num__124 <o> c ) num__96 <o> d ) num__117 <o> e ) num__127 |
length = num__6 m num__24 cm = num__624 cm width = num__4 m num__32 cm = num__432 cm hcf of num__624 and num__432 = num__48 number of square tiles required = ( num__624 * num__432 ) / ( num__48 * num__48 ) = num__13 * num__9 = num__117 answer : d <eor> d <eos> |
d |
divide__624.0__48.0__ divide__432.0__48.0__ multiply__9.0__13.0__ multiply__9.0__13.0__ |
divide__624.0__48.0__ divide__432.0__48.0__ multiply__9.0__13.0__ multiply__9.0__13.0__ |
| if the cp of num__12 rubbers is equal to the sp of num__8 rubbers the gain % is ? <o> a ) num__25.0 <o> b ) num__30.0 <o> c ) num__50.0 <o> d ) num__70.0 <o> e ) num__80 % |
( explanation : friends we know we will need gain amount to get gain percent right . so lets get gain first . let the cost price of num__1 pen is re num__1 cost of num__8 pens = rs num__8 selling price of num__8 pens = num__12 gain = num__12 - num__8 = num__4 gain % = ( gaincost * num__100 ) % = ( num__48 * num__100 ) % = num__50.0 c <eor> c <eos> |
c |
percent__50.0__100.0__ |
percent__50.0__100.0__ |
| if a : b : : num__3 : num__7 then what is ( num__5 a + num__6 b ) : ( a - num__3 b ) ? <o> a ) - num__57 : num__2 <o> b ) - num__57 : num__11 <o> c ) num__11 : num__10 <o> d ) - num__11 : num__10 <o> e ) - num__1 : num__10 |
a / b = num__0.428571428571 dividing numerator & denominator of ' ( num__5 a + num__6 b ) / ( a - num__2 b ) ' by b [ num__5 ( a / b ) + num__6 ] / [ ( a / b ) - num__3 ] = [ num__5 * ( num__0.428571428571 ) + num__6 ] / [ ( num__0.428571428571 ) - num__3 ] = - num__28.5 answer : a <eor> a <eos> |
a |
divide__3.0__7.0__ subtract__7.0__5.0__ multiply__28.5__2.0__ |
divide__3.0__7.0__ subtract__7.0__5.0__ multiply__28.5__2.0__ |
| the speed of a train is num__90 kmph . what is the distance covered by it in num__10 minutes ? <o> a ) num__15 kmph <o> b ) num__8 kmph <o> c ) num__9 kmph <o> d ) num__1 kmph <o> e ) num__2 kmph |
explanation : distance = speed * time num__90 * num__0.166666666667 = num__15 kmph answer : a <eor> a <eos> |
a |
round__15.0__ |
round__15.0__ |
| there are num__690 male and female participants in a meeting . half the female participants and one - quarter of the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ? <o> a ) num__75 <o> b ) num__100 <o> c ) num__115 <o> d ) num__175 <o> e ) num__225 |
female = x male = num__690 - x x / num__2 + num__690 - x / num__4 = num__0.333333333333 * ( num__690 ) = num__230 x = num__230 x / num__2 = num__115 is supposed to be the answer m is missing something correct option c <eor> c <eos> |
c |
twice__2.0__ divide__230.0__2.0__ divide__230.0__2.0__ |
twice__2.0__ divide__230.0__2.0__ divide__230.0__2.0__ |
| the ratio of the present ages of p and q is num__3 : num__4 . num__5 years ago the ratio of their ages was num__5 : num__7 . find the their present ages ? <o> a ) num__30 num__40 <o> b ) num__30 num__46 <o> c ) num__30 num__49 <o> d ) num__30 num__43 <o> e ) num__30 num__43 |
their present ages be num__3 x and num__4 x . num__5 years age the ratio of their ages was num__5 : num__7 then ( num__3 x - num__5 ) : ( num__4 x - num__5 ) = num__5 : num__7 x = num__35 - num__25 = > x = num__10 . their present ages are : num__30 num__40 . . answer : a <eor> a <eos> |
a |
multiply__5.0__7.0__ add__3.0__7.0__ multiply__3.0__10.0__ multiply__4.0__10.0__ multiply__3.0__10.0__ |
multiply__5.0__7.0__ subtract__35.0__25.0__ subtract__35.0__5.0__ multiply__4.0__10.0__ subtract__35.0__5.0__ |
| two cars cover the same distance at the speed of num__62 and num__64 kmps respectively . find the distance traveled by them if the slower car takes num__1 hour more than the faster car . <o> a ) num__1873 <o> b ) num__1848 <o> c ) num__1984 <o> d ) num__1838 <o> e ) num__1939 |
explanation : num__62 ( x + num__1 ) = num__64 x x = num__31 num__62 * num__32 = num__1984 km answer : option c <eor> c <eos> |
c |
add__1.0__31.0__ multiply__62.0__32.0__ round__1984.0__ |
add__1.0__31.0__ multiply__62.0__32.0__ multiply__62.0__32.0__ |
| in a num__100 m race a can beat b by num__25 m and b can beat c by num__4 m . in the same race a can beat c by : <o> a ) num__21 <o> b ) num__22 <o> c ) num__28 <o> d ) num__25 <o> e ) num__30 |
a : b = num__100 : num__75 b : c = num__100 : num__96 a : c = num__1.38888888889 = num__100 : num__72 a beats ( num__100 - num__72 ) = num__28 m answer c <eor> c <eos> |
c |
subtract__100.0__25.0__ subtract__100.0__4.0__ subtract__100.0__72.0__ round__28.0__ |
subtract__100.0__25.0__ subtract__100.0__4.0__ subtract__100.0__72.0__ subtract__100.0__72.0__ |
| one half of a two digit number exceeds its one third by num__6 . what is the sum of the digits of the number ? <o> a ) num__9 <o> b ) num__3 <o> c ) num__2 <o> d ) num__4 <o> e ) num__5 |
x / num__2 â € “ x / num__3 = num__6 = > x = num__6 num__3 + num__6 = num__9 answer a <eor> a <eos> |
a |
divide__6.0__2.0__ add__6.0__3.0__ add__6.0__3.0__ |
divide__6.0__2.0__ add__6.0__3.0__ add__6.0__3.0__ |
| in what ratio must rice at $ num__9.30 / kg be mixed with rice at $ num__10.80 / kg so that the mixture is worth $ num__10 / kg ? <o> a ) num__1 : num__3 <o> b ) num__9 : num__4 <o> c ) num__8 : num__7 <o> d ) num__11 : num__9 <o> e ) num__11 : num__7 |
num__9.30 x + num__10.80 y = num__10 ( x + y ) . num__80 y = . num__70 x x / y = num__1.14285714286 c is the answer <eor> c <eos> |
c |
subtract__80.0__10.0__ divide__80.0__70.0__ divide__80.0__10.0__ |
subtract__80.0__10.0__ divide__80.0__70.0__ divide__80.0__10.0__ |
| what is the difference between local value & face value of num__9 in the numeral num__65693 ? <o> a ) num__84 <o> b ) num__89 <o> c ) num__81 <o> d ) num__90 <o> e ) num__10000 |
( local value of num__9 ) - ( face value of num__9 ) = ( num__90 - num__9 ) = num__81 c <eor> c <eos> |
c |
subtract__90.0__9.0__ subtract__90.0__9.0__ |
subtract__90.0__9.0__ subtract__90.0__9.0__ |
| anand finishes a work in num__7 days bittu finishes the same job in num__8 days and chandu in num__6 days . they take turns to finish the work . anand on the first day bittu on the second and chandu on the third day and then anand again and so on . on which day will the work get over ? <o> a ) num__3 rd <o> b ) num__6 th <o> c ) num__9 th <o> d ) num__7 th <o> e ) num__8 th |
they completed num__0.869047619048 work in num__6 days working one by one on eacg day . on num__7 th day num__0.130952380952 work is left which will be completed by anand on num__7 th day . answer : d <eor> d <eos> |
d |
round__7.0__ |
round__7.0__ |
| if num__3 x - y = num__6 and x + num__2 y = num__8 then what is the value of num__4 x + y ? <o> a ) num__0.142857142857 <o> b ) num__14 <o> c ) num__15 <o> d ) num__7.42857142857 <o> e ) num__8.57142857143 |
num__3 x - y = num__6 . . . equation num__1 x + num__2 y = num__8 . . . equation num__2 adding both the equations num__4 x + y = num__14 correct answer option b <eor> b <eos> |
b |
subtract__3.0__2.0__ add__6.0__8.0__ add__6.0__8.0__ |
subtract__3.0__2.0__ add__6.0__8.0__ add__6.0__8.0__ |
| a group of num__55 adults and num__70 children go for trekking . if there is meal for either num__70 adults or num__90 children and if num__21 adults have their meal find the total number of children that can be catered with the remaining food . <o> a ) num__33 <o> b ) num__54 <o> c ) num__63 <o> d ) num__17 <o> e ) num__01 |
explanation : as there is meal for num__70 adults and num__21 have their meal the meal left can be catered to num__49 adults . now num__70 adults = num__90 children num__7 adults = num__9 children therefore num__49 adults = num__63 children hence the meal can be catered to num__63 children . answer : c <eor> c <eos> |
c |
subtract__70.0__21.0__ subtract__70.0__7.0__ subtract__70.0__7.0__ |
subtract__70.0__21.0__ subtract__70.0__7.0__ subtract__70.0__7.0__ |
| if the number of num__3729 n where n represents the ones digit is a multiple of num__3 then the value of n could be which of the following ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
then for any number to be a multiple of num__3 the sum of its digits should be the multiple of num__3 . i . e num__3 + num__7 + num__2 + num__9 + n = multiple of three or num__21 + n = multiple of num__3 . if n = num__3 num__21 + n = num__24 which is a multiple of num__3 . hence answer is c . <eor> c <eos> |
c |
power__3.0__2.0__ multiply__3.0__7.0__ add__3.0__21.0__ divide__9.0__3.0__ |
add__2.0__7.0__ multiply__3.0__7.0__ add__3.0__21.0__ divide__9.0__3.0__ |
| each book on a certain shelf is labeled by a single category . for every num__2 history books there are num__7 fantasy books and for every num__3 fantasy books there are num__8 reference books . if the proportion of history to reference books is doubled while the proportion of fantasy to reference books is maintained which of the following could be the number of history books if there are fewer than num__60 fantasy books on the shelf after the changes ? <o> a ) num__12 <o> b ) num__21 <o> c ) num__27 <o> d ) num__35 <o> e ) num__36 |
for every num__2 history books there are num__7 fantasy books : originally h : f = num__2 : num__7 = num__6 : num__21 . for every num__3 fantasy books there are num__8 reference books : f : r = num__3 : num__8 = num__21 : num__56 h : f : r = num__6 : num__21 : num__56 after the changes h : f : r = num__12 : num__21 : num__56 there are fewer than num__60 fantasy books so there are num__21 or num__42 fantasy books . the number of history books could be num__12 or num__24 . the answer is a . <eor> a <eos> |
a |
multiply__2.0__3.0__ multiply__7.0__3.0__ multiply__7.0__8.0__ multiply__2.0__6.0__ multiply__2.0__21.0__ multiply__2.0__12.0__ multiply__2.0__6.0__ |
multiply__2.0__3.0__ multiply__7.0__3.0__ multiply__7.0__8.0__ multiply__2.0__6.0__ multiply__2.0__21.0__ multiply__2.0__12.0__ multiply__2.0__6.0__ |
| a wheel has a diameter of x inches and a second wheel has a diameter of y inches . the first wheel covers a distance of d feet in num__300 revolutions . how many revolutions does the second wheel make in covering d feet ? <o> a ) num__300 xy <o> b ) num__300 x / y <o> c ) num__300 x - y <o> d ) num__200 y / x <o> e ) num__600 x / y |
first wheel with diameter x inches - - - - > so the circumference = x * pi - - - - - - - - - ( num__1 ) second wheel with diameter y inches - - - - > so the circumference = y * pi - - - - - - - - - ( num__2 ) revolutions = distance / circumference so from equation ( num__1 ) num__300 = d / ( x * pi ) d = num__300 * x * pi - - - - - - - - - - ( num__3 ) from equation ( num__2 ) revolutions = d / ( y * pi ) = ( num__300 * x * pi ) / ( y * pi ) = num__300 x / y answer ( b ) <eor> b <eos> |
b |
add__1.0__2.0__ round__300.0__ |
add__1.0__2.0__ divide__300.0__1.0__ |
| dawson has to secure num__30.0 marks to clear his exam of class num__8 th . he got num__30 marks and failed by num__36 marks . what is the maximum marks ? <o> a ) num__180 <o> b ) num__190 <o> c ) num__200 <o> d ) num__210 <o> e ) num__220 |
e num__220 to pass the exam ravish needs num__30 + num__36 = num__66 marks . = > ( num__2.2 ) * num__100 = num__220 <eor> e <eos> |
e |
percent__30.0__220.0__ percent__100.0__220.0__ |
percent__30.0__220.0__ percent__100.0__220.0__ |
| rs . num__800 amounts to rs . num__920 in num__3 years at simple interest . if the interest is increased by num__3.0 it would amount to how much ? <o> a ) num__226 <o> b ) num__992 rs <o> c ) num__665 <o> d ) num__771 <o> e ) num__211 |
( num__800 * num__3 * num__3 ) / num__100 = num__72 num__920 + num__72 = num__992 answer : b <eor> b <eos> |
b |
percent__100.0__992.0__ |
percent__100.0__992.0__ |
| a person travels equal distances with speeds of num__3 km / hr num__4 km / hr and num__5 km / hr and takes a total time of num__50 minutes . the total distance is ? <o> a ) num__1 km <o> b ) num__2 km <o> c ) num__3.19 km <o> d ) num__4 km <o> e ) num__5 km |
c num__3 km let the total distance be num__3 x km . then x / num__3 + x / num__4 + x / num__5 = num__0.833333333333 num__47 x / num__60 = num__0.833333333333 = > x = num__1.06 . total distance = num__3 * num__1.06 = num__3.19 km . <eor> c <eos> |
c |
subtract__50.0__3.0__ hour_to_min_conversion__ round__3.19__ |
subtract__50.0__3.0__ hour_to_min_conversion__ round__3.19__ |
| working alone at its constant rate machine a produces x boxes in num__10 minutes and working alone at its constant rate machine b produces num__2 x boxes in num__5 minutes . how many minutes does it take machines a and b working simultaneously at their respective constant rates to produce num__7 x boxes ? <o> a ) num__3 minutes <o> b ) num__4 minutes <o> c ) num__5 minutes <o> d ) num__14 minutes <o> e ) num__12 minutes |
rate = work / time given rate of machine a = x / num__10 min machine b produces num__2 x boxes in num__5 min hence machine b produces num__4 x boxes in num__10 min . rate of machine b = num__4 x / num__10 we need tofind the combined time that machines a and b working simultaneouslytakeat their respective constant rates let ' s first find the combined rate of machine a and b rate of machine a = x / num__10 min + rate of machine b = num__4 x / num__10 = num__5 x / num__10 now combine time = combine work needs to be done / combine rate = num__7 x / num__5 x * num__10 = num__14 min ans : d <eor> d <eos> |
d |
add__10.0__4.0__ round__14.0__ |
add__10.0__4.0__ add__10.0__4.0__ |
| a sales staff is composed of a sales manager and two sales people all of whom earn commission as a percentage of sales . each sales person earns num__5.0 commission on sales . in a given week the sales staff earned a total of $ num__2750 in commissions on $ num__5000 worth of sales . what commission rate did the sales manager earn during that week ? <o> a ) num__25.0 <o> b ) num__30.0 <o> c ) num__35.0 <o> d ) num__40.0 <o> e ) num__45 % |
e for me . let managers comminsion rate be m . m * num__5000 + num__2 * num__0.05 * num__5000 = num__2750 num__5000 * m = num__2250 m = num__0.45 = num__45.0 <eor> e <eos> |
e |
percent__2.0__2250.0__ percent__2.0__2250.0__ |
percent__2.0__2250.0__ percent__2.0__2250.0__ |
| if the sides of a square are doubled in length the area of the original square is now how many times as large as the area of the resultant square ? <o> a ) num__25.0 <o> b ) num__50.0 <o> c ) num__100.0 <o> d ) num__200.0 <o> e ) num__400 % |
let the original square have sides of length l and the new square have sides of length num__2 l . the resulting areas are l ^ num__2 for the original square and num__4 l ^ num__2 for the new square ( as the length of the side is square to get the area ) . the original square ' s area is l ^ num__0.5 l ^ num__2 or num__25.0 of the new square ' s area . choice a <eor> a <eos> |
a |
triangle_area__25.0__2.0__ |
triangle_area__25.0__2.0__ |
| the price of cooking oil has increased by num__25.0 . the percentage of reduction that a family should effect in the use of cooking oil so as not to increase the expenditure on this account is : <o> a ) num__15.0 <o> b ) num__20.0 <o> c ) num__25.0 <o> d ) num__30.0 <o> e ) none |
explanation : reduction in consumption : = ( num__0.2 × num__100 ) % = num__20.0 correct option : b <eor> b <eos> |
b |
multiply__100.0__0.2__ multiply__100.0__0.2__ |
multiply__100.0__0.2__ multiply__100.0__0.2__ |
| the perimeter of a semi circle is num__144 cm then the radius is ? <o> a ) num__29 <o> b ) num__28 <o> c ) num__277 <o> d ) num__24 <o> e ) num__21 |
num__5.14285714286 r = num__144 = > r = num__28 answer : b <eor> b <eos> |
b |
round__28.0__ |
round__28.0__ |
| sound is said to travel in air at about num__1100 feet per second . a man hears the axe striking the tree num__2.2 seconds after he sees it strike the tree . how far is the man from the wood chopper ? <o> a ) num__2197 feet <o> b ) num__2420 ft <o> c ) num__2500 feet <o> d ) num__2629 feet <o> e ) none of these |
distance = ( num__1100 * num__2.2 ) feet = num__2420 feet . correct option : b <eor> b <eos> |
b |
multiply__1100.0__2.2__ multiply__1100.0__2.2__ |
multiply__1100.0__2.2__ multiply__1100.0__2.2__ |
| a train num__125 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__15 seconds . the speed of the train is ? <o> a ) num__35 <o> b ) num__50 <o> c ) num__88 <o> d ) num__66 <o> e ) num__22 |
speed of the train relative to man = ( num__8.33333333333 ) m / sec = ( num__8.33333333333 ) m / sec . [ ( num__8.33333333333 ) * ( num__3.6 ) ] km / hr = num__30 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__30 = = > x = num__35 km / hr . answer : a <eor> a <eos> |
a |
divide__125.0__15.0__ add__5.0__30.0__ round__35.0__ |
divide__125.0__15.0__ add__5.0__30.0__ round__35.0__ |
| on a store counter are exactly two boxes containing only purple marbles and yellow marbles . box a has num__30 purple marbles and num__20 yellow marbles ; box b has num__15 purple marbles and num__35 yellow marbles . if melanie randomly selects one of the boxes and then randomly selects one marble from the box what is the probability that the marble selected will be purple ? <o> a ) num__0.15 <o> b ) num__0.3 <o> c ) num__0.45 <o> d ) num__0.6 <o> e ) num__0.9 |
we are given that box a has num__30 purple marbles and num__20 yellow marbles and that box b has num__15 purple marbles and num__35 yellow marbles . we need to determine the probability when selecting one marble from the box of selecting a purple marble . since we have two boxes a and b there are multiple scenarios to account for when selecting the marbles . we must account for the probability of first selecting either box and then secondly for selecting a purple marble . let ’ s start with box a . p ( selecting box a ) = ½ p ( selecting a purple marble in box a ) = num__0.6 = ⅗ thus the probability of selecting a purple marble from box a is ½ x ⅗ = num__0.3 next we can determine the probability of selecting a purple marble from box b . p ( selecting box b ) = ½ p ( selecting a purple marble in box b ) = num__0.3 = num__0.3 thus the probability of selecting a purple marble from box b is ½ x num__0.3 = num__0.15 now we can determine the probability of selecting a purple marble from box a or box b : num__0.3 + num__0.15 = num__0.3 + num__0.15 = num__0.45 . answer : c <eor> c <eos> |
c |
union_prob__0.15__0.6__0.3__ union_prob__0.15__0.6__0.3__ |
union_prob__0.15__0.6__0.3__ union_prob__0.15__0.6__0.3__ |
| $ represents addition and * represents subtraction what is the value of the following expression ? num__15 * num__5 $ num__15 $ num__10 $ num__30 * num__25 * num__35 * num__25 $ num__20 * num__25 * num__20 <o> a ) - num__15 <o> b ) - num__35 <o> c ) - num__55 <o> d ) - num__45 <o> e ) - num__65 |
num__15 - num__5 + num__15 + num__10 + num__30 - num__25 - num__35 - num__25 + num__20 - num__25 - num__20 = num__10 + num__15 + num__10 + num__30 - num__25 - num__35 - num__25 + num__20 - num__25 - num__20 = num__25 + num__10 + num__30 - num__25 - num__35 - num__25 + num__20 - num__25 - num__20 = num__35 + num__30 - num__25 - num__35 - num__25 + num__20 - num__25 - num__20 = num__65 - num__25 - num__35 - num__25 + num__20 - num__25 - num__20 = num__40 - num__35 - num__25 + num__20 - num__25 - num__20 = num__5 - num__25 + num__20 - num__25 - num__20 = - num__20 + num__20 - num__25 - num__20 = num__0 - num__25 - num__20 = - num__45 answer : d <eor> d <eos> |
d |
add__30.0__35.0__ add__15.0__25.0__ add__15.0__30.0__ add__15.0__30.0__ |
add__30.0__35.0__ add__15.0__25.0__ add__15.0__30.0__ add__15.0__30.0__ |
| in a tournament prize of num__1 st num__2 nd and num__3 rd is given in same interval . if total amount of prize rs num__4800 and prize of num__1 st is rs num__2000 find the interval ? ? <o> a ) num__600 <o> b ) num__450 <o> c ) num__350 <o> d ) num__550 <o> e ) num__400 |
let the interval is x . prize of num__2 nd and num__3 rd is num__2000 - x and num__2000 - num__2 x num__2000 + num__2000 - x + num__2000 - num__2 x = num__4800 x = num__400 answer : e <eor> e <eos> |
e |
multiply__1.0__400.0__ |
multiply__1.0__400.0__ |
| a player holds num__13 cards of four suits of which seven are black and six are red . there are twice as many diamonds as spades and twice as many hearts as diamonds . how many clubs does he hold <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
explanation : clearly the black cards are either clubs or spades while the red cards are either diamonds or hearts . let the number of spades be x . then number of clubs = ( num__7 - x ) . number of diamonds = num__2 x number of spades = num__2 x ; number of hearts = num__2 x number of diamonds = num__4 x . total number of cards = x + num__2 x + num__4 x + num__7 - x = num__6 x + num__7 . therefore num__6 x + num__7 = num__13 num__6 x = num__6 x - num__1 . hence number of clubs = ( num__7 - x ) = num__6 . answer : c <eor> c <eos> |
c |
coin_space__ die_space__ die_space__ |
coin_space__ die_space__ die_space__ |
| the length of the bridge which a train num__160 meters long and travelling at num__45 km / hr can cross in num__30 seconds is : <o> a ) num__239 <o> b ) num__277 <o> c ) num__215 <o> d ) num__88 <o> e ) num__232 |
speed = ( num__45 * num__0.277777777778 ) m / sec = ( num__12.5 ) m / sec . time = num__30 sec . let the length of bridge be x meters . then ( num__160 + x ) / num__30 = num__12.5 = = > num__2 ( num__160 + x ) = num__750 = = > x = num__215 m . answer : c <eor> c <eos> |
c |
round__215.0__ |
round__215.0__ |
| a rower can row upstream at num__22 kph and downstream at num__34 kph . what is the speed of the rower in still water ? <o> a ) num__28 <o> b ) num__29 <o> c ) num__30 <o> d ) num__31 <o> e ) num__32 |
let v be the rower ' s speed in still water . let s be the speed of the current in the stream v - s = num__22 v + s = num__34 when we add the two equations we get : num__2 v = num__56 then v = num__28 kph . the answer is a . <eor> a <eos> |
a |
add__22.0__34.0__ divide__56.0__2.0__ round__28.0__ |
add__22.0__34.0__ divide__56.0__2.0__ subtract__56.0__28.0__ |
| a truck driver drove for num__2 days . on the second day he drove num__3 hours longer and at an average speed of num__15 miles per hour faster than he drove on the first day . if he drove a total of num__1000 miles and spent num__21 hours driving during the num__2 days what was his average speed on the first day in miles per hour ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__35 <o> d ) num__39 <o> e ) num__45 |
day num__1 num__2 t + num__3 = num__21 t = num__9 day num__2 t + num__3 num__9 + num__3 = num__12 num__9 r + num__12 ( r + num__15 ) = num__1000 r = num__39 answer : d <eor> d <eos> |
d |
subtract__3.0__2.0__ add__3.0__9.0__ round__39.0__ |
subtract__3.0__2.0__ add__3.0__9.0__ round__39.0__ |
| if two of the four expressions x + y x + num__5 y x - y and num__5 x - y are chosen at random what is the probability r that their product will be of the form of x ^ num__2 - ( by ) ^ num__2 where b is an integer ? <o> a ) num__0.5 <o> b ) num__0.333333333333 <o> c ) num__0.25 <o> d ) num__0.2 <o> e ) r = num__0.166666666667 |
only ( x + y ) ( x - y ) pair will give the form x ^ num__2 - ( by ) ^ num__2 the probability of selecting these two pairs are num__0.25 * num__0.333333333333 = num__0.0833333333333 assuming x + y is picked first then x - y but x - y can be picked first followed by x + y . so the probability r = num__0.0833333333333 * num__2 = num__0.166666666667 ans e <eor> e <eos> |
e |
multiply__0.25__0.3333__ subtract__0.25__0.0833__ subtract__0.25__0.0833__ |
multiply__0.25__0.3333__ subtract__0.25__0.0833__ subtract__0.25__0.0833__ |
| a student travels from his house to school at num__10 km / hr and reaches school num__1 hour late . the next day he travels num__12 km / hr and reaches school num__1 hour early . what is the distance between his house and the school ? <o> a ) num__100 <o> b ) num__105 <o> c ) num__110 <o> d ) num__115 <o> e ) num__120 |
let x be the distance from his house to the school . x / num__10 = x / num__12 + num__2 num__6 x = num__5 x + num__120 x = num__120 km the answer is e . <eor> e <eos> |
e |
subtract__12.0__10.0__ divide__12.0__2.0__ divide__10.0__2.0__ multiply__10.0__12.0__ round__120.0__ |
subtract__12.0__10.0__ divide__12.0__2.0__ divide__10.0__2.0__ multiply__10.0__12.0__ divide__120.0__1.0__ |
| on a two - dimensional coordinate plane the line z = x ^ num__2 - x ^ num__3 touches the x - axis in how many places ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
apparently it ' s z = x ^ num__2 - x ^ num__3 instead of z = x ^ num__2 - z ^ num__3 . in this case : the x - intercept is the value ( s ) of x for z = num__0 . num__0 = x ^ num__2 - x ^ num__3 ; num__0 = x ^ num__2 ( num__1 - x ) ; x = num__0 or x = num__1 . answer : c . <eor> c <eos> |
c |
subtract__3.0__2.0__ multiply__2.0__1.0__ |
subtract__3.0__2.0__ subtract__3.0__1.0__ |
| . in a certain city num__60 percent of the registered voters are democrats and the rest are republicans . in a mayoral race if num__80 percent of the registered voters who are democrats and num__20 percent of the registered voters who are republicans are expected to vote for candidate a what percent of the registered voters are expected to vote for candidate a ? <o> a ) num__50.0 <o> b ) num__53.0 <o> c ) num__54.0 <o> d ) num__55.0 <o> e ) num__56 % |
registered voters = num__100 d = num__60 r = num__40 num__80.0 of d ( num__60 ) = num__48 num__20.0 of r ( num__40 ) = num__8 total voter % of registered voters = num__0.56 num__56.0 ans e <eor> e <eos> |
e |
percent__60.0__80.0__ percent__20.0__40.0__ percent__100.0__56.0__ |
percent__60.0__80.0__ percent__20.0__40.0__ percent__100.0__56.0__ |
| a man in train notice that he can count num__51 telephone posts in num__1 minute . if they are known to be num__30 meters apart then at what speed is the train traveling ? <o> a ) num__100 km / hr <o> b ) num__110 km / hr <o> c ) num__50 km / hr <o> d ) num__150 km / hr <o> e ) num__90 km / hr |
number of gaps between num__51 telephone posts = num__50 distance traveled in num__1 minute = num__50 * num__30 = num__1500 m = num__1.5 km speed = num__60 * num__1.5 = num__900 km / hr answer is e <eor> e <eos> |
e |
subtract__51.0__1.0__ multiply__30.0__50.0__ hour_to_min_conversion__ add__30.0__60.0__ |
subtract__51.0__1.0__ multiply__30.0__50.0__ hour_to_min_conversion__ multiply__1.5__60.0__ |
| out of three consecutive odd numbers eleven times the first number is equal to addition of thrice the third number and adding num__8 to five times the second . what is the first number ? <o> a ) num__10 <o> b ) num__6 <o> c ) num__11 <o> d ) num__8 <o> e ) num__9 |
description : = > num__11 x = num__5 ( x + num__2 ) + num__8 + num__3 ( x + num__4 ) = > num__11 x = num__8 x + num__30 = > num__3 x = num__30 x = num__10 answer a <eor> a <eos> |
a |
subtract__8.0__5.0__ divide__8.0__2.0__ add__8.0__2.0__ add__8.0__2.0__ |
subtract__8.0__5.0__ divide__8.0__2.0__ add__8.0__2.0__ add__8.0__2.0__ |
| two employees x and y are paid a total of rs . num__506 per week by their employer . if x is paid num__120 percent of the sum paid to y how much is y paid per week ? <o> a ) s . num__250 <o> b ) s . num__280 <o> c ) s . num__290 <o> d ) s . num__299 <o> e ) s . num__230 |
let the amount paid to x per week = x and the amount paid to y per week = y then x + y = num__506 but x = num__120.0 of y = num__120 y / num__100 = num__12 y / num__10 ∴ num__12 y / num__10 + y = num__506 ⇒ y [ num__1.2 + num__1 ] = num__506 ⇒ num__22 y / num__10 = num__506 ⇒ num__22 y = num__5060 ⇒ y = num__230.0 = num__230.0 = rs . num__230 e <eor> e <eos> |
e |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__10.0__12.0__ multiply__506.0__10.0__ divide__5060.0__22.0__ multiply__1.0__230.0__ |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__10.0__12.0__ multiply__506.0__10.0__ divide__5060.0__22.0__ divide__230.0__1.0__ |
| a research team is to consist of num__3 scientists from company a and num__3 scientists from company b . if the pool of available scientists from company a is num__5 and the pool of available scientists from company b is num__4 scientists how many different research teams are possible ? <o> a ) num__84 <o> b ) num__40 <o> c ) num__30 <o> d ) num__20 <o> e ) num__12 |
c num__35 ∗ c num__34 = num__40 answer : b . <eor> b <eos> |
b |
choose__5.0__3.0__ choose__5.0__3.0__ |
choose__5.0__3.0__ choose__5.0__3.0__ |
| what is the remainder if num__7 ^ num__12 is divided by num__100 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
num__7 ^ num__16 can be written as ( num__7 ^ num__4 ) ^ num__3 if we divide num__7 ^ num__4 by num__100 the reminder is num__1 so ( num__7 ^ num__4 ) ^ num__3 by num__100 the reminder is num__1 ^ num__3 = num__1 answer : a <eor> a <eos> |
a |
subtract__16.0__12.0__ subtract__7.0__4.0__ subtract__4.0__3.0__ reverse__1.0__ |
subtract__16.0__12.0__ subtract__7.0__4.0__ subtract__4.0__3.0__ reverse__1.0__ |
| a retail item is offered at a discount of d percent ( where d > num__10 ) with a num__5.0 state sales tax assessed on the discounted purchase price . if the state sales tax were not assessed what percent discount from the item ’ s original retail price in terms of d would result in the same final price ? <o> a ) d + num__5 / num__1.05 <o> b ) d / num__1.05 + num__5 <o> c ) num__1.05 d - num__5 <o> d ) d - num__5 / num__1.05 <o> e ) num__1.05 ( p – num__5 ) |
let x be the price of the item . final price after discount and sales tax = x * ( num__1 - d / num__100 ) * num__1.05 let e be the percent discount which would result in the same final price . then x * ( num__1 - d / num__100 ) * num__1.05 = x * ( num__1 - e / num__100 ) = > num__1.05 - num__1.05 d / num__100 = num__1 - e / num__100 = > e / num__100 = num__1.05 d / num__100 - . num__05 = > e = num__1.05 d - num__5 hence option c is correct . <eor> c <eos> |
c |
percent__1.05__100.0__ |
percent__1.05__100.0__ |
| if a fair die is rolled once what is the probability that a num__3 occurs on at least one roll ? <o> a ) num__0.694444444444 <o> b ) num__0.578703703704 <o> c ) num__0.166666666667 <o> d ) num__0.0277777777778 <o> e ) num__0.833333333333 |
questions such as these that talk aboutat leastormaximumorminimumin probability questions should make sure realize that probability of any event ( n ) to occur = num__1 - p ( not n ) thus the probability of at least num__1 roll = num__1 - probability of no num__3 s = num__1 - ( num__0.833333333333 ) = num__1 - num__0.833333333333 = num__0.166666666667 num__0.833333333333 is the probability of not getting a num__3 in any num__1 roll with num__5 allowed numbers ( = num__1 num__24 num__56 ) out of a total of num__6 possibilities . c is thus the correct answer . <eor> c <eos> |
c |
negate_prob__0.8333__ vowel_space__ die_space__ negate_prob__0.8333__ |
negate_prob__0.8333__ vowel_space__ die_space__ negate_prob__0.8333__ |
| when usha was thrice as old as nisha her sister asha was num__25 when nisha was half as old as asha then sister usha was num__34 . their ages add to num__100 . how old is usha ? <o> a ) num__40 <o> b ) num__41 <o> c ) num__42 <o> d ) num__43 <o> e ) num__45 |
let the age of usha is num__3 x then nisha is x and asha is num__25 also usha num__34 nisha y and asha num__2 y . we know that num__3 x - num__34 = x - num__2 y = num__25 - num__2 y solving above three equations we get x = num__9 y = num__16 their ages are num__34 num__16 num__32 . whose sum = num__82 . so after num__18 years their ages will be equal to num__100 . so usha age is num__34 + num__6 = num__40 answer : a <eor> a <eos> |
a |
subtract__34.0__25.0__ subtract__25.0__9.0__ subtract__34.0__2.0__ subtract__34.0__16.0__ multiply__3.0__2.0__ add__34.0__6.0__ add__34.0__6.0__ |
subtract__34.0__25.0__ subtract__25.0__9.0__ subtract__34.0__2.0__ subtract__34.0__16.0__ subtract__9.0__3.0__ add__34.0__6.0__ add__34.0__6.0__ |
| for every positive integer n the nth term of a sequence is the total sum of three consecutive integers starting at n . what is the total sum of terms num__1 through num__80 of this series ? <o> a ) num__8990 <o> b ) num__9380 <o> c ) num__9630 <o> d ) num__9960 <o> e ) num__10 |
240 |
each term of the series has the form ( n + n + num__1 + n + num__2 ) = num__3 n + num__3 since the series goes from num__1 to num__80 the sum of the series is : num__3 ( num__1 + num__2 + . . . + num__80 ) + num__80 ( num__3 ) = num__3 ( num__80 ) ( num__81 ) / num__2 + num__80 ( num__3 ) = num__120 * num__81 + num__240 = num__9960 the answer is d . <eor> d <eos> |
d |
d |
| a certain board game is played by rolling a pair of fair six - sided dice and then moving one ' s piece forward the number of spaces indicated by the sum showing on the dice . a player is ` ` frozen ' ' if her opponent ' s piece comes to rest in the space already occupied by her piece . if player a is about to roll and is currently six spaces behind player b what is the probability that player b will be frozen after player a rolls ? <o> a ) num__0.0833333333333 <o> b ) num__0.138888888889 <o> c ) num__0.166666666667 <o> d ) num__0.333333333333 <o> e ) num__0.472222222222 |
no . of possible outcomes = num__6 * num__6 = num__36 no . of outcomes that result a total of num__6 ( as a is num__6 spaces behind b ) = num__5 ( ( num__15 ) ( num__24 ) ( num__33 ) ( num__42 ) ( num__51 ) ) so the probability = num__0.138888888889 ( option b ) <eor> b <eos> |
b |
add__36.0__6.0__ add__36.0__15.0__ divide__5.0__36.0__ divide__5.0__36.0__ |
add__36.0__6.0__ add__36.0__15.0__ divide__5.0__36.0__ divide__5.0__36.0__ |
| two spherical balls lie on the ground touching . if one of the balls has a radius of num__6 cm and the point of contact is num__8 cm above the ground what is the radius of the other ball ? <o> a ) num__12 cm <o> b ) num__16 cm <o> c ) num__18 cm <o> d ) num__20 cm <o> e ) none of the these |
similar triangle properties . . num__2 / r + num__6 = num__6 / r - num__6 giving r = num__12 . answer : a <eor> a <eos> |
a |
subtract__8.0__6.0__ multiply__6.0__2.0__ round__12.0__ |
subtract__8.0__6.0__ multiply__6.0__2.0__ round__12.0__ |
| what is the difference between the place value and the face value of num__9 in the numeral num__296827 ? <o> a ) none of these <o> b ) num__89993 <o> c ) num__89991 <o> d ) num__89994 <o> e ) num__89995 |
explanation : place value of num__9 = num__90000 face value of num__9 = num__9 difference = num__9000 - num__9 = num__89991 answer : c <eor> c <eos> |
c |
subtract__90000.0__9.0__ subtract__90000.0__9.0__ |
subtract__90000.0__9.0__ subtract__90000.0__9.0__ |
| a whale goes on a feeding frenzy that lasts for num__5 hours . for the first hour he catches and eats x kilos of plankton . in every hour after the first it consumes num__3 kilos of plankton more than it consumed in the previous hour . if by the end of the frenzy the whale will have consumed a whopping accumulated total num__450 kilos of plankton how many kilos did he consume on the third hour ? <o> a ) num__38 <o> b ) num__47 <o> c ) num__50 <o> d ) num__53 <o> e ) num__93 |
if you list the amount eaten each hour you ' ll get an equally spaced list increasing by num__3 each hour . in any equally spaced list the median equals the mean . here the mean is num__90.0 = num__90 so the median is also num__90 and that is the amount eaten in the num__2 nd hour . we need to add num__3 to find the total eaten in the next hour so the answer is num__93 option e <eor> e <eos> |
e |
divide__450.0__5.0__ subtract__5.0__3.0__ add__3.0__90.0__ add__3.0__90.0__ |
divide__450.0__5.0__ subtract__5.0__3.0__ add__3.0__90.0__ add__3.0__90.0__ |
| if the average ( arithmetic mean ) of seven consecutive integers is k - num__2 then the product of the greatest and least integer is <o> a ) k num__2 - num__9 <o> b ) k num__2 - num__2 k + num__1 <o> c ) k num__2 - num__4 k - num__5 <o> d ) k num__2 + num__6 k + num__9 <o> e ) k num__2 + num__4 k - num__12 |
n = smallest number n + num__6 = largest number ( n + n + num__6 ) / num__2 = k - num__2 = > ( num__2 n + num__6 ) / num__2 = k - num__2 = > n + num__3 = k - num__2 = > n = k - num__5 so product of n and n + num__6 = ( k - num__5 ) ( k + num__1 ) = k ^ num__2 - num__5 k + k - num__5 = k ^ num__2 - num__4 k - num__5 answer - c <eor> c <eos> |
c |
divide__6.0__2.0__ add__2.0__3.0__ subtract__3.0__2.0__ add__1.0__3.0__ multiply__2.0__1.0__ |
divide__6.0__2.0__ add__2.0__3.0__ subtract__3.0__2.0__ add__1.0__3.0__ divide__2.0__1.0__ |
| find the odd man out . num__3 num__8 num__18 num__38 num__78 num__158 num__316 <o> a ) num__38 <o> b ) num__158 <o> c ) num__316 <o> d ) num__8 <o> e ) none of these |
explanation : num__3 num__3 × num__2 + num__2 = num__8 num__8 × num__2 + num__2 = num__18 num__18 × num__2 + num__2 = num__38 num__38 × num__2 + num__2 = num__78 num__78 × num__2 + num__2 = num__158 num__158 × num__2 + num__2 = num__318 hence num__316 is wrong and num__318 should have come in place of that . answer : option c <eor> c <eos> |
c |
divide__316.0__158.0__ add__316.0__2.0__ multiply__158.0__2.0__ |
divide__316.0__158.0__ add__316.0__2.0__ multiply__158.0__2.0__ |
| johnson has a corporate proposal . the probability that vice - president adams will approve the proposal is num__0.7 . the probability that vice - president baker will approve the proposal is num__0.5 . the probability that vice - president corfu will approve the proposal is num__0.4 . the approvals of the three ( vice presidents ) / vps are entirely independent of one another . suppose the johnson must get vp adam ’ s approval as well as the approval of at least one of the other vps baker or corfu to win funding . what is the probability that johnson ’ s proposal is funded ? <o> a ) num__0.5 <o> b ) num__0.6 <o> c ) num__0.7 <o> d ) num__0.49 <o> e ) num__0.8 |
probability of adams approval = num__0.7 baker approval = num__0.5 corfu approval = num__0.4 proabability of approval is must adam approval and atleast one approval = adam approval * baker approval * corfu approval ( approval of both remaining ) + adam approval * baker approval * corfu not approval ( atleast one remaining approval ) + adam approval * baker not approval * coffu approval ( atleast one remaining approval = num__0.7 * num__0.5 * num__0.4 + num__0.7 * num__0.5 * num__0.6 + num__0.7 * num__0.5 * num__0.4 = num__0.14 + num__0.21 + num__0.14 = num__0.49 d <eor> d <eos> |
d |
subtract__0.7__0.21__ subtract__0.7__0.21__ |
subtract__0.7__0.21__ subtract__0.7__0.21__ |
| a train num__130 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__4 <o> b ) num__7 <o> c ) num__5 <o> d ) num__9 <o> e ) num__8 |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__130 * num__0.0545454545455 ] sec = num__7 sec answer : b <eor> b <eos> |
b |
add__60.0__6.0__ round__7.0__ |
add__60.0__6.0__ round__7.0__ |
| a hall num__36 m long and num__15 m broad is to be paved with stones each measuring num__2 dm by num__5 dm . the number of stones required is : <o> a ) num__180 <o> b ) num__1800 <o> c ) num__5400 <o> d ) num__18000 <o> e ) num__1.8 |
area of the hall = num__3600 * num__1500 area of each stone = ( num__20 * num__50 ) therefore number of stones = ( num__3600 * num__75.0 * num__50 ) = num__5400 answer : c <eor> c <eos> |
c |
square_perimeter__5.0__ rectangle_perimeter__5.0__20.0__ multiply__15.0__5.0__ volume_rectangular_prism__36.0__2.0__75.0__ triangle_area__2.0__5400.0__ |
square_perimeter__5.0__ rectangle_perimeter__5.0__20.0__ multiply__15.0__5.0__ volume_rectangular_prism__36.0__2.0__75.0__ volume_rectangular_prism__36.0__2.0__75.0__ |
| a motorcyclist goes from bombay to pune a distance of num__192 kms at an average of num__32 kmph speed . another man starts from bombay by car num__2 ½ hours after the first and reaches pune ½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? <o> a ) num__1 : num__2 <o> b ) num__1 : num__5 <o> c ) num__1 : num__9 <o> d ) num__1 : num__3 <o> e ) num__1 : num__1 |
t = num__6.0 = num__6 h t = num__6 - num__3 = num__3 time ratio = num__6 : num__3 = num__2 : num__1 speed ratio = num__1 : num__2 answer : a <eor> a <eos> |
a |
divide__192.0__32.0__ divide__6.0__2.0__ subtract__3.0__2.0__ round__1.0__ |
divide__192.0__32.0__ divide__6.0__2.0__ subtract__3.0__2.0__ subtract__2.0__1.0__ |
| car q and car y traveled the same num__80 - mile route . if car q took num__2 hours and car y traveled at an average speed that was num__50 percent faster than the average speed of car q how many hours did it take car y to travel the route ? <o> a ) num__0.666666666667 <o> b ) num__1 <o> c ) num__1.33333333333 <o> d ) num__1.6 <o> e ) num__3 |
the speed of car q is ( distance ) / ( time ) = num__40.0 = num__40 miles per hour . the speed of car y = num__1.5 * num__40 = num__60 miles per hour - - > ( time ) = ( distance ) / ( speed ) = num__1.33333333333 = num__1.33333333333 hours . answer : c . or : to cover the same distance at num__1.5 as fast rate num__0.666666666667 as much time is needed - - > ( time ) * num__0.666666666667 = num__2 * num__0.666666666667 = num__1.33333333333 hours . answer : c . <eor> c <eos> |
c |
divide__80.0__2.0__ hour_to_min_conversion__ divide__80.0__60.0__ subtract__2.0__1.3333__ divide__80.0__60.0__ |
divide__80.0__2.0__ multiply__1.5__40.0__ divide__80.0__60.0__ subtract__2.0__1.3333__ divide__80.0__60.0__ |
| the price of stock increased by num__8.0 last year and decreased by num__6.0 this year . what is the net percentage change in the price of the stock ? <o> a ) num__0.2 <o> b ) num__0.8 <o> c ) num__1.2 <o> d ) num__1.52 <o> e ) num__2 % |
( num__100.0 + num__8.0 ) * ( num__100.0 - num__6.0 ) = num__1.08 * num__0.94 = num__1.0152 = num__101.52 . the net percentage change in the price of the stock is ( + ) num__1.52 the answer is d <eor> d <eos> |
d |
multiply__0.94__1.08__ multiply__100.0__1.0152__ subtract__101.52__100.0__ subtract__101.52__100.0__ |
multiply__0.94__1.08__ multiply__100.0__1.0152__ subtract__101.52__100.0__ subtract__101.52__100.0__ |
| a trader sells num__85 meters of cloth for rs . num__8925 at the profit of rs . num__15 per metre of cloth . what is the cost price of one metre of cloth ? <o> a ) num__22 <o> b ) num__28 <o> c ) num__90 <o> d ) num__27 <o> e ) num__23 |
sp of num__1 m of cloth = num__105.0 = rs . num__105 cp of num__1 m of cloth = sp of num__1 m of cloth - profit on num__1 m of cloth = rs . num__105 - rs . num__15 = rs . num__90 . answer : c <eor> c <eos> |
c |
divide__8925.0__85.0__ subtract__105.0__15.0__ round__90.0__ |
divide__8925.0__85.0__ subtract__105.0__15.0__ subtract__105.0__15.0__ |
| the distance between delhi and mathura is num__110 kms . a starts from delhi with a speed of num__20 kmph at num__7 a . m . for mathura and b starts from mathura with a speed of num__25 kmph at num__8 p . m . from delhi . when will they meet ? <o> a ) num__11 <o> b ) num__77 <o> c ) num__16 <o> d ) num__10 <o> e ) num__98 |
d = num__110 – num__20 = num__90 rs = num__20 + num__25 = num__45 t = num__2.0 = num__2 hours num__8 a . m . + num__2 = num__10 a . m . . answer : d <eor> d <eos> |
d |
subtract__110.0__20.0__ add__20.0__25.0__ divide__90.0__45.0__ divide__20.0__2.0__ round__10.0__ |
subtract__110.0__20.0__ add__20.0__25.0__ divide__90.0__45.0__ add__8.0__2.0__ add__8.0__2.0__ |
| three is the largest number that can be divided evenly into num__27 and the positive integer x while num__10 is the largest number that can be divided evenly into both num__100 and x . which of the following is the largest possible number that could be divided into x and num__1500 <o> a ) num__30 <o> b ) num__70 <o> c ) num__210 <o> d ) num__300 <o> e ) num__150 |
num__27 = num__3 * num__3 * num__3 and its gcf with x is num__3 which implies that x has one factor of num__3 but not two factors of num__3 . num__100 and x have a gcf of num__10 which implies that x has one factor of num__10 but not two factors of num__10 . then we want to know what is the largest possible gcf of x and num__2100 . well num__1500 = num__3 * num__5 * num__10 * num__10 we want x to include as many factors in common with num__2100 as possible to make the gcf with num__2100 as big as possible . we know x has one factor of num__3 but not two factors - - - that takes the num__3 . we know x has one factor of num__10 but not two factors - - - we can take one of those num__10 ' s but we have to leave the other no other restrictions so we can also grab that factor of num__7 - - - nothing saying that we ca n ' t and it ' s there for the taking . num__3 * num__5 * num__10 = num__150 if we allow x to include as many factors as possible within the constraints given that is the most x could have in common with num__2100 . e <eor> e <eos> |
e |
subtract__10.0__3.0__ divide__1500.0__10.0__ divide__1500.0__10.0__ |
subtract__10.0__3.0__ divide__1500.0__10.0__ divide__1500.0__10.0__ |
| two trains running in opposite directions cross a man standing on the platform in num__27 seconds and num__17 seconds respectively and they cross each other in num__23 seconds . the ratio of their speeds is ? <o> a ) num__0.5 <o> b ) num__1.5 <o> c ) num__0.75 <o> d ) num__3.0 <o> e ) num__0.5 |
let the speeds of the two trains be x m / sec and y m / sec respectively . then length of the first train = num__27 x meters and length of the second train = num__17 y meters . ( num__27 x + num__17 y ) / ( x + y ) = num__23 = = > num__27 x + num__17 y = num__23 x + num__23 y = = > num__4 x = num__6 y = = > x / y = num__1.5 . answer : b <eor> b <eos> |
b |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ round__1.5__ |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ divide__6.0__4.0__ |
| the h . c . f of two numbers is num__23 and the other two factors of their l . c . m are num__14 and num__15 . the larger of the two numbers is : <o> a ) num__338 <o> b ) num__278 <o> c ) num__345 <o> d ) num__231 <o> e ) num__121 |
clearly the numbers are ( num__23 * num__14 ) and ( num__23 * num__15 ) . larger number = ( num__23 * num__15 ) = num__345 . answer : c <eor> c <eos> |
c |
multiply__23.0__15.0__ multiply__23.0__15.0__ |
multiply__23.0__15.0__ multiply__23.0__15.0__ |
| if the ratio of the sum of the first num__6 terms of a g . p . to the sum of the first num__3 terms of the g . p . is num__126 what is the common ratio of the g . p ? <o> a ) num__3 <o> b ) num__0.333333333333 <o> c ) num__2 <o> d ) num__5 <o> e ) num__0.2 |
num__126 = ( a num__1 + a num__2 + a num__3 + a num__4 + a num__5 + a num__6 ) / ( a num__1 + a num__2 + a num__3 ) factorize the same terms num__126 = num__1 + ( a num__4 + a num__5 + a num__6 ) / ( a num__1 + a num__2 + a num__3 ) write every term with respect to r a num__1 = a num__1 a num__2 = a num__1 * r ^ num__1 a num__3 = a num__1 * r ^ num__2 . . . . . . . . . num__126 = num__1 + ( a num__1 ( r ^ num__3 + r ^ num__4 + r ^ num__5 ) ) / ( a num__1 ( num__1 + r ^ num__1 + r ^ num__2 ) ) num__125 = ( r ^ num__3 ( num__1 + r ^ num__1 + r ^ num__2 ) ) / ( ( num__1 + r ^ num__1 + r ^ num__2 ) ) num__125 = r ^ num__3 r = num__5 d <eor> d <eos> |
d |
divide__6.0__3.0__ subtract__6.0__2.0__ subtract__6.0__1.0__ subtract__126.0__1.0__ subtract__6.0__1.0__ |
divide__6.0__3.0__ add__3.0__1.0__ add__3.0__2.0__ subtract__126.0__1.0__ add__3.0__2.0__ |
| a train covers a certain distance at a speed of num__320 kmph in num__6 hours . to cover the same distance in num__1 num__0.666666666667 hours it must travel at a speed of <o> a ) num__1152 km / hr <o> b ) num__1120 km / hr <o> c ) num__1650 km / hr <o> d ) num__1750 km / hr <o> e ) num__1850 km / hr |
explanation : distance = num__320 Ã — num__6 = num__1920 km required speed = ( num__1920 Ã — num__0.6 ) = num__1152 km / hr answer : option a <eor> a <eos> |
a |
multiply__320.0__6.0__ km_to_mile_conversion__ multiply__1920.0__0.6__ round__1152.0__ |
multiply__320.0__6.0__ km_to_mile_conversion__ multiply__1920.0__0.6__ divide__1152.0__1.0__ |
| the diameters of two spheres are in the ratio num__1 : num__2 what is the ratio of their volumes ? <o> a ) num__1 : num__7 <o> b ) num__1 : num__3 <o> c ) num__1 : num__8 <o> d ) num__1 : num__1 <o> e ) num__1 : num__2 |
num__1 : num__8 answer : c <eor> c <eos> |
c |
square_perimeter__2.0__ volume_cube__1.0__ |
square_perimeter__2.0__ volume_cube__1.0__ |
| how many integers are there between num__42 and num__97 inclusive ? <o> a ) num__51 <o> b ) num__55 <o> c ) num__56 <o> d ) num__57 <o> e ) num__58 |
i guess the easiest way to answer this is - say you have two positive integers x and y where y > x then the number of integers between x and y is given by - ( y - x ) + num__1 in this case it ' s ( num__97 - num__42 ) + num__1 = num__56 option c <eor> c <eos> |
c |
multiply__56.0__1.0__ |
multiply__56.0__1.0__ |
| during a trip francine traveled x percent of the total distance at an average speed of num__20 miles per hour and the rest of the distance at an average speed of num__80 miles per hour . in terms of x what was francine â € ™ s average speed for the entire trip ? <o> a ) ( num__180 - x ) / num__2 <o> b ) ( x + num__60 ) / num__4 <o> c ) ( num__300 - x ) / num__5 <o> d ) num__600 / ( num__115 - x ) <o> e ) num__8000 / ( num__400 - num__4 x ) |
e . . . . x % traveeled at num__80 mph num__100 - x % traveeled at num__20 mph total timt taken = x / num__80 + ( num__100 - x ) / num__20 = ( num__400 - num__4 x ) / num__80 total distance = x + ( num__100 - x ) = num__100 av speed = tot dis / time taken which is e <eor> e <eos> |
e |
add__20.0__80.0__ divide__80.0__20.0__ multiply__20.0__400.0__ |
add__20.0__80.0__ divide__80.0__20.0__ multiply__20.0__400.0__ |
| which of the following describes all values of x for which num__2 – x ^ num__2 > = num__0 ? <o> a ) x > = num__1 <o> b ) x < = – num__1 <o> c ) num__0 < = x < = num__2 <o> d ) x < = – num__1 or x > = num__1 <o> e ) – num__2 < = x < = num__2 |
which of the following describes all values of x for which num__2 – x ^ num__2 > = num__0 ? ( a ) x > = num__1 plugged in num__2 . num__2 – ( num__3 ) ^ num__2 > = num__0 - num__7 > = num__0 ? no . wrong ( b ) x < = – num__1 plugged in - num__3 . num__2 – ( - num__3 ) ^ num__2 > = num__0 - num__7 > = num__0 ? no . wrong ( c ) num__0 < = x < = num__2 plugged in num__0 num__1 and num__2 . all of them work . but e is better because it describes all the values of x ( d ) x < = – num__1 or x > = num__1 a and b answer this . wrong . ( e ) – num__2 < = x < = num__2 x is a positive or negative fraction or a zero . x = - num__2 x = num__2 x = num__0 plug all of them . they work . answer is e . <eor> e <eos> |
e |
add__2.0__1.0__ multiply__2.0__1.0__ |
add__2.0__1.0__ subtract__3.0__1.0__ |
| a man buys an item at rs . num__800 and sells it at the loss of num__20 percent . then what is the selling price of that item <o> a ) rs . num__640 <o> b ) rs . num__760 <o> c ) rs . num__860 <o> d ) rs . num__960 <o> e ) none of these |
explanation : here always remember when ever x % loss it means s . p . = ( num__100 - x ) % of c . p when ever x % profit it means s . p . = ( num__100 + x ) % of c . p so here will be ( num__100 - x ) % of c . p . = num__80.0 of num__800 = num__0.8 * num__800 = num__640 option a <eor> a <eos> |
a |
percent__80.0__800.0__ percent__80.0__800.0__ |
percent__80.0__800.0__ percent__80.0__800.0__ |
| a freight elevator can carry a maximum load of num__1200 pounds . sean who weighs num__200 pounds is in the elevator with two packages weighing num__150 pounds and num__280 pounds . if he needs to fit three more packages in the elevator that weigh as much as possible without exceeding the elevator limit what is the difference between their average and the average of the two packages already in the elevator ? <o> a ) num__25 <o> b ) num__85 <o> c ) num__190 <o> d ) num__215 <o> e ) num__210 |
the average of existing num__2 package is ( num__150 + num__280 ) / num__2 = num__215.0 = num__215 remaining allowed weight = num__1200 - num__200 - num__430 = num__570 . allowed per package = num__190.0 = num__190 so difference in average of existing and allowable = num__215 - num__190 = num__25 hence a <eor> a <eos> |
a |
add__150.0__280.0__ subtract__215.0__190.0__ subtract__215.0__190.0__ |
add__150.0__280.0__ subtract__215.0__190.0__ subtract__215.0__190.0__ |
| among num__250 viewers interviewed who watch at least one of the three tv channels namely a bc . num__116 watch a num__127 watch c while num__107 watch b . if num__60 watch exactly two channels . how many watch exactly one channel ? <o> a ) num__185 <o> b ) num__180 <o> c ) num__170 <o> d ) num__190 <o> e ) num__195 |
num__250 = n ( exactly num__1 channel ) + n ( exactly num__2 channels ) + n ( exactly num__3 channels ) num__250 = n ( exactly num__1 channel ) + num__60 + n ( exactly num__3 channels ) let ' s find the value of n ( exactly num__3 channels ) = x num__250 = n ( a ) + n ( b ) + n ( c ) - n ( a and b ) - n ( b and c ) - n ( c and a ) + n ( a and b and c ) note that each of n ( a and b ) is the sum of ' number of people watching exactly two channels a and b ' and ' number of people watching all three channels ' . num__250 = num__116 + num__127 + num__107 - n ( exactly num__2 channels ) - num__3 x + x num__250 = num__116 + num__127 + num__107 - num__60 - num__2 x x = num__25 num__250 = n ( exactly num__1 channel ) + num__60 + num__25 n ( exactly num__1 channel ) = num__170 answer ( c ) <eor> c <eos> |
c |
add__1.0__2.0__ multiply__1.0__170.0__ |
add__1.0__2.0__ multiply__1.0__170.0__ |
| in a dairy farm num__40 cows eat num__40 bags of husk in num__40 days . in how many days one cow will eat one bag of husk ? <o> a ) num__32 <o> b ) num__36 <o> c ) num__38 <o> d ) num__40 <o> e ) num__50 |
assume that in x days one cow will eat one bag of husk . more cows less days ( indirect proportion ) more bags more days ( direct proportion ) hence we can write as ( cows ) num__40 : num__1 ( bags ) num__1 : num__40 x : num__40 = num__40 × num__1 × num__40 = num__1 × num__40 × x = x = num__40 answer : d <eor> d <eos> |
d |
round__40.0__ |
round__40.0__ |
| a grocer has a sale of rs . num__6435 rs . num__6927 rs . num__6855 rs . num__7230 and rs . num__6562 for num__5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . num__6500 ? <o> a ) rs . num__4991 <o> b ) rs . num__5991 <o> c ) rs . num__6001 <o> d ) rs . num__6991 <o> e ) rs . num__7991 |
total sale for num__5 months = rs . ( num__6435 + num__6927 + num__6855 + num__7230 + num__6562 ) = rs . num__34009 . required sale = rs . [ ( num__6500 x num__6 ) - num__34009 ] = rs . ( num__39000 - num__34009 ) = rs . num__4991 . answer : a <eor> a <eos> |
a |
multiply__6500.0__6.0__ subtract__39000.0__34009.0__ subtract__39000.0__34009.0__ |
multiply__6500.0__6.0__ subtract__39000.0__34009.0__ subtract__39000.0__34009.0__ |
| if the wages of num__6 men for num__15 days be rs . num__2100 then find the wages of for num__12 days . <o> a ) rs . num__2530 <o> b ) rs . num__2520 <o> c ) rs . num__2500 <o> d ) rs . num__1520 <o> e ) rs . num__3520 |
let the required wages be rs . x . more men more wages ( direct proportion ) less days less wages ( direct proportion ) men num__6 : num__9 : : num__2100 : x days num__15 : num__12 therefore ( num__6 x num__15 x x ) = ( num__9 x num__12 x num__2100 ) < = > x = ( num__9 x num__12 x num__2100 ) / ( num__6 x num__15 ) = num__2520 hence the required wages are rs . num__2520 . answer is b . <eor> b <eos> |
b |
subtract__15.0__6.0__ round__2520.0__ |
subtract__15.0__6.0__ round__2520.0__ |
| a and b together can complete one - ninth of a job in a day . what is the difference between the maximum and minimum number of days a could take to complete the job alone if a and b take whole number of days to complete the job alone ? <o> a ) num__45 <o> b ) b . num__60 <o> c ) c . num__80 <o> d ) d . num__90 <o> e ) be determined |
a and b together can complete one - ninth of a job in a day means that ab together does one whole work in num__9 days . num__1 / a + num__1 / b = num__0.111111111111 the minimum value a can take is num__10 days ( anything less than num__10 days is not possible because in other case no of days taken by b will become negative which is not possible practically ) when a takes least time ( num__10 days ) then b would take the max time ( num__90 days ) to complete the work alone . in order to find the max time taken by a find out the min time taken by b as we did earlier . the minimum value b can take is num__10 days ( anything less than num__10 days is not possible because in other case no of days taken by a will become negative which is not possible practically ) when b takes least time ( num__10 days ) then a would take the max time ( num__90 days ) to complete the work alone . so the required difference = max time taken - min time taken = num__90 - num__10 = num__80 days option c <eor> c <eos> |
c |
reverse__9.0__ add__1.0__9.0__ multiply__9.0__10.0__ subtract__90.0__10.0__ multiply__1.0__80.0__ |
reverse__9.0__ add__1.0__9.0__ multiply__9.0__10.0__ subtract__90.0__10.0__ divide__80.0__1.0__ |
| what is the rate percent when the simple interest on rs . num__910 amount to rs . num__260 in num__4 years ? <o> a ) num__7.14 <o> b ) num__6.14 <o> c ) num__5.14 <o> d ) num__3.14 <o> e ) num__2.14 |
num__260 = ( num__910 * num__4 * r ) / num__100 r = num__7.14 answer : a <eor> a <eos> |
a |
percent__7.14__100.0__ |
percent__7.14__100.0__ |
| evaluate permutation num__8 p num__8 <o> a ) num__40290 <o> b ) num__40300 <o> c ) num__40310 <o> d ) num__40320 <o> e ) num__40330 |
explanation : npn = n ! num__8 p num__8 = num__8 * num__7 * num__6 * num__5 ∗ num__4 ∗ num__3 ∗ num__2 ∗ num__1 = num__40320 option d <eor> d <eos> |
d |
die_space__ vowel_space__ coin_space__ choose__8.0__4.0__ choose__8.0__4.0__ |
die_space__ vowel_space__ coin_space__ choose__8.0__4.0__ choose__8.0__4.0__ |
| num__1 . if the product of num__6 integers is negative at most how many of the integers can be negative ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
we ' re told that the product of num__6 numbers is negative ( the fact that they ' re integers is actually irrelevant ) . we ' re asked for the maximum number of those num__6 that could be negative . we ' ll start with the largest and work our way down . . num__6 are negative : ( - num__1 ) ( - num__1 ) ( - num__1 ) ( - num__1 ) ( - num__1 ) ( - num__1 ) = + num__1 . . . which is not a negative product num__5 are negative : ( - num__1 ) ( - num__1 ) ( - num__1 ) ( - num__1 ) ( - num__1 ) ( + num__1 ) = - num__1 . . . which is a negative product d <eor> d <eos> |
d |
subtract__6.0__1.0__ multiply__1.0__5.0__ |
subtract__6.0__1.0__ subtract__6.0__1.0__ |
| the ratio of the incomes of rajan and balan is num__7 : num__6 and the ratio of their expenditure is num__6 : num__5 . if at the end of the year each saves $ num__1000 then the income of rajan is ? <o> a ) $ num__7600 <o> b ) $ num__7000 <o> c ) $ num__8000 <o> d ) $ num__9000 <o> e ) $ num__5000 |
let the income of rajan and balan be $ num__7 x and $ num__6 x let their expenditures be $ num__6 y and $ num__5 y num__7 x - num__6 y = num__1000 - - - - - - - num__1 ) num__6 x - num__5 y = num__1000 - - - - - - - num__2 ) from num__1 ) and num__2 ) x = num__1000 rajan ' s income = num__7 x = num__7 * num__1000 = $ num__7000 answer is b <eor> b <eos> |
b |
subtract__7.0__6.0__ subtract__7.0__5.0__ multiply__7.0__1000.0__ multiply__7.0__1000.0__ |
subtract__7.0__6.0__ subtract__7.0__5.0__ multiply__7.0__1000.0__ multiply__7.0__1000.0__ |
| in a apartment num__30.0 of the people speak english num__20.0 speak hindi and num__10.0 speak both . if a people is selected at random what is the probability that he has speak english or hindi ? <o> a ) num__0.333333333333 <o> b ) num__0.666666666667 <o> c ) num__0.4 <o> d ) num__0.222222222222 <o> e ) num__0.428571428571 |
p ( e ) = num__0.3 = num__0.3 p ( h ) = num__0.2 = num__0.2 and p ( e ∩ h ) = num__0.1 = num__0.1 p ( e or h ) = p ( e u h ) = p ( e ) + p ( h ) - p ( e ∩ h ) = ( num__0.3 ) + ( num__0.2 ) - ( num__0.1 ) = num__0.4 = num__0.4 c <eor> c <eos> |
c |
union_prob__0.3__0.2__0.1__ union_prob__0.3__0.2__0.1__ |
union_prob__0.3__0.2__0.1__ union_prob__0.3__0.2__0.1__ |
| if a and b are integers and ( a * b ) ^ num__5 = num__128 y y could be : <o> a ) num__1 <o> b ) num__2 <o> c ) num__4 <o> d ) num__8 <o> e ) num__16 |
distribute the exponent . a ^ num__5 * b ^ num__5 = num__128 y find the prime factorization of num__96 . this is num__2 ^ num__5 * num__2 ^ num__2 . we need num__2 ^ num__3 ( or some other power of num__2 that will give us a multiple of num__2 ^ num__5 as our second term ) . num__2 ^ num__3 = num__8 the answer is d . <eor> d <eos> |
d |
subtract__5.0__2.0__ add__5.0__3.0__ add__5.0__3.0__ |
subtract__5.0__2.0__ add__5.0__3.0__ add__5.0__3.0__ |
| if num__34 men completed num__0.4 th of a work in num__8 days working num__9 hours a day . how many more man should be engaged to finish the rest of the work in num__6 days working num__9 hours a day ? <o> a ) num__100 <o> b ) num__102 <o> c ) num__104 <o> d ) num__106 <o> e ) num__108 |
using the chain rule m num__1 x t num__1 = m num__2 x t num__2 thus ( num__34 x num__8 x num__9 ) / num__0.4 = ( m num__2 x num__6 x num__9 ) / num__0.6 m num__2 = num__68 . therefore required number of men = ( num__68 + num__34 ) = num__102 answer : b <eor> b <eos> |
b |
subtract__9.0__8.0__ subtract__8.0__6.0__ km_to_mile_conversion__ multiply__34.0__2.0__ add__34.0__68.0__ round__102.0__ |
subtract__9.0__8.0__ subtract__8.0__6.0__ km_to_mile_conversion__ multiply__34.0__2.0__ add__34.0__68.0__ add__34.0__68.0__ |
| what is the average of first num__21 multiples of num__7 ? <o> a ) num__49 <o> b ) num__77 <o> c ) num__66 <o> d ) num__57 <o> e ) num__50 |
required average = num__7 ( num__1 + num__2 + . . . . + num__21 ) / num__21 ( num__0.333333333333 ) x ( ( num__21 x num__22 ) / num__2 ) ( because sum of first num__21 natural numbers ) = num__77 answer is b <eor> b <eos> |
b |
divide__7.0__21.0__ add__21.0__1.0__ multiply__1.0__77.0__ |
divide__7.0__21.0__ add__21.0__1.0__ divide__77.0__1.0__ |
| a recipe for a large batch of fruit juice at a factory calls for num__10.0 oranges num__20.0 lemons and num__70.0 limes . if num__200 lemons are used how many limes are needed ? <o> a ) num__100 <o> b ) num__200 <o> c ) num__300 <o> d ) num__700 <o> e ) num__1400 |
num__20.0 = num__200 fruit num__100.0 = num__1000 fruit num__70.0 of num__1000 fruit = num__700 fruit num__700 limes answer : d <eor> d <eos> |
d |
multiply__10.0__100.0__ multiply__10.0__70.0__ multiply__10.0__70.0__ |
multiply__10.0__100.0__ multiply__10.0__70.0__ multiply__10.0__70.0__ |
| what is the compound interest paid on a sum of rs . num__4500 for the period of num__2 years at num__10.0 per annum . <o> a ) num__945 <o> b ) num__620 <o> c ) num__610 <o> d ) num__600 <o> e ) none of these |
solution = interest % for num__1 st year = num__10 interest % for num__2 nd year = num__10 + num__10.0 of num__10 = num__10 + num__10 * num__0.1 = num__11 total % of interest = num__10 + num__11 = num__21 total interest = num__21.0 num__4500 = num__4500 * ( num__0.21 ) = num__945 answer a <eor> a <eos> |
a |
percent__10.0__1.0__ percent__1.0__21.0__ percent__21.0__4500.0__ percent__21.0__4500.0__ |
percent__10.0__1.0__ percent__1.0__21.0__ percent__21.0__4500.0__ percent__21.0__4500.0__ |
| combine terms : num__8 a + num__26 b - num__4 b – num__16 a . <o> a ) num__4 a + num__22 b <o> b ) - num__28 a + num__30 b <o> c ) - num__8 a + num__22 b <o> d ) num__28 a + num__30 b <o> e ) num__5 a + num__30 b |
solution : num__8 a + num__26 b - num__4 b – num__16 a . = num__8 a – num__16 a + num__26 b – num__4 b . = - num__8 a + num__22 b . answer : ( c ) <eor> c <eos> |
c |
subtract__26.0__4.0__ subtract__16.0__8.0__ |
subtract__26.0__4.0__ subtract__16.0__8.0__ |
| num__8 cups of milk are to be poured into a num__4 - cup bottle and a num__8 - cup bottle . if each bottle is to be filled to the same fraction of its capacity how many cups of milk should be poured into the num__8 - cup bottle ? <o> a ) num__5.33333333333 <o> b ) num__2.33333333333 <o> c ) num__2.5 <o> d ) num__1.33333333333 <o> e ) num__3 |
let x be the # of cups going into the num__8 cup bottle . so . . . . x / num__8 = ( ( num__8 - x ) / num__4 ) num__64 - num__8 x = num__4 x num__64 = num__12 x x = num__5.33333333333 . answer : a <eor> a <eos> |
a |
add__8.0__4.0__ divide__64.0__12.0__ divide__64.0__12.0__ |
add__8.0__4.0__ divide__64.0__12.0__ divide__64.0__12.0__ |
| which number should replace both the asterisks in ( * / num__20 ) x ( * / num__80 ) = num__1 ? <o> a ) num__20 <o> b ) num__40 <o> c ) num__60 <o> d ) num__80 <o> e ) num__120 |
let ( y / num__20 ) x ( y / num__80 ) = num__1 y ^ num__2 = num__20 x num__80 = num__20 x num__20 x num__4 y = ( num__20 x num__2 ) = num__40 the answer is b . <eor> b <eos> |
b |
divide__80.0__20.0__ multiply__20.0__2.0__ multiply__20.0__2.0__ |
divide__80.0__20.0__ divide__80.0__2.0__ divide__80.0__2.0__ |
| mrs . rodger got a weekly raise of $ num__148 . if she gets paid every other week write an integer describing how the raise will affect her paycheck . <o> a ) $ num__140 <o> b ) $ num__141 <o> c ) $ num__142 <o> d ) $ num__148 <o> e ) $ num__145 |
let the num__1 st paycheck be x ( integer ) . mrs . rodger got a weekly raise of $ num__148 . so after completing the num__1 st week she will get $ ( x + num__148 ) . similarly after completing the num__2 nd week she will get $ ( x + num__148 ) + $ num__148 . = $ ( x + num__148 + num__148 ) = $ ( x + num__296 ) so in this way end of every week her salary will increase by $ num__148 . correct answer d ) $ num__148 <eor> d <eos> |
d |
multiply__148.0__2.0__ multiply__148.0__1.0__ |
multiply__148.0__2.0__ multiply__148.0__1.0__ |
| the average temperature in lincoln in july is num__85 degrees . last wednesday it was num__90 degrees . today it was num__15 degrees cooler than last wednesday . what was the temperature today ? <o> a ) num__15 degrees <o> b ) num__25 degrees <o> c ) num__75 degrees <o> d ) num__45 degrees <o> e ) num__55 degrees |
the july temperature is extra information . num__90 – num__15 = num__75 it was num__75 degrees today . correct answer c <eor> c <eos> |
c |
subtract__90.0__15.0__ subtract__90.0__15.0__ |
subtract__90.0__15.0__ subtract__90.0__15.0__ |
| boys and girls in a class are writing letters . there are twice as many girls as boys in the class and each girl writes num__3 more letters than each boy . if boys write num__24 of the num__90 total letters written by the class how many letters does each boy write ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__12 |
there are twice as many girls as boys in the class - - > g = num__2 b . each girl writes num__3 more letters than each boy - - > boys write x letters girls write x + num__3 letters . boys write num__24 letters - - > bx = num__24 . girls write num__90 - num__24 = num__66 letters - - > ( num__2 b ) ( x + num__3 ) = num__66 - - > num__2 bx + num__6 b = num__66 - - > num__2 * num__24 + num__6 b = num__66 - - > b = num__3 . bx = num__24 - - > num__3 x = num__24 - - > x = num__8 . answer : d . <eor> d <eos> |
d |
subtract__90.0__24.0__ multiply__3.0__2.0__ divide__24.0__3.0__ divide__24.0__3.0__ |
subtract__90.0__24.0__ multiply__3.0__2.0__ add__2.0__6.0__ add__2.0__6.0__ |
| in a num__100 m race a runs at num__8 km per hour . if a gives b a start of num__4 m and still him by num__15 seconds what is the speed of b ? <o> a ) num__5.56 km / hr . <o> b ) num__5.06 km / hr . <o> c ) num__5.76 km / hr . <o> d ) num__6.76 km / hr . <o> e ) num__5.72 km / hr . |
time taken by a to cover num__100 m = ( num__60 x num__0.0075 ) x num__100 sec = num__45 sec . b covers ( num__100 - num__4 ) m = num__96 m in ( num__45 + num__15 ) sec = num__60 sec . b ' s speed = ( num__96 x num__60 x num__60 ) / ( num__60 x num__1000 ) km / hr = num__5.76 km / hr . answer is c <eor> c <eos> |
c |
hour_to_min_conversion__ subtract__60.0__15.0__ subtract__100.0__4.0__ round__5.76__ |
hour_to_min_conversion__ subtract__60.0__15.0__ subtract__100.0__4.0__ round__5.76__ |
| a boat running up stram takes num__4 hours to cover a certain distance while it takes num__8 hours to cover the same distance running down stream . what is the ratio between the speed of the boat and the speed of water current respectively ? <o> a ) num__2 : num__3 <o> b ) num__5 : num__6 <o> c ) num__4 : num__5 <o> d ) num__3 : num__1 <o> e ) num__8 : num__1 |
explanation : let speed of boat is x km / h and speed stream is y km / hr num__4 ( x + y ) = num__8 ( x - y ) num__4 x + num__4 y = num__8 x - num__8 y num__12 y = num__4 x num__3 y = x x / y = num__3.0 num__3 : num__1 answer : option d <eor> d <eos> |
d |
add__4.0__8.0__ divide__12.0__4.0__ subtract__4.0__3.0__ round__3.0__ |
add__4.0__8.0__ divide__12.0__4.0__ subtract__4.0__3.0__ subtract__4.0__1.0__ |
| a man travelled from the village to the post - office at the rate of num__25 kmph and walked back at the rate of num__4 kmph . if the whole journey took num__5 hours num__48 minutes find the distance of the post - office from the village <o> a ) num__20 <o> b ) num__87 <o> c ) num__276 <o> d ) num__198 <o> e ) num__171 |
explanation : average speed = ( num__2 × a × b ) / ( a + b ) here a = num__25 b = num__4 average speed = num__2 × num__25 × num__4 / ( num__25 + num__4 ) = num__6.89655172414 kmph . distance covered in num__5 hours num__48 minutes = speed × time distance = ( num__6.89655172414 ) × ( num__5.8 ) = num__40 kms distance covered in num__5 hours num__48 minutes = num__40 kms distance of the post office from the village = ( num__20.0 ) = num__20 km . answer : a <eor> a <eos> |
a |
subtract__25.0__5.0__ round__20.0__ |
divide__40.0__2.0__ divide__40.0__2.0__ |
| the mean of num__50 observations is num__200 . but later he found that there is decrements of num__15 from each observations . what is the the updated mean is ? <o> a ) num__165 <o> b ) num__185 <o> c ) num__198 <o> d ) num__221 <o> e ) num__240 |
num__185 answer is b <eor> b <eos> |
b |
subtract__200.0__15.0__ subtract__200.0__15.0__ |
subtract__200.0__15.0__ subtract__200.0__15.0__ |
| set a contains all the even numbers between num__12 and num__60 inclusive . set b contains all the even numbers between num__62 and num__110 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? <o> a ) num__850 <o> b ) num__1250 <o> c ) num__1650 <o> d ) num__2050 <o> e ) num__2450 |
each term in set b is num__50 more than the corresponding term in set a . the difference of the sums = num__25 * num__50 = num__1250 . the answer is b . <eor> b <eos> |
b |
subtract__62.0__12.0__ multiply__25.0__50.0__ multiply__25.0__50.0__ |
subtract__62.0__12.0__ multiply__25.0__50.0__ multiply__25.0__50.0__ |
| the average ( arithmetic mean ) of four different positive integers is num__10 . if the first of these integers in num__3 times the second integer and the second integer is num__2 less than the third integer what is the least possible value of the fourth integer ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__3 <o> d ) num__2 <o> e ) num__0 |
let the second integer be x and the fourth be a . then [ num__3 x + x + ( x + num__2 ) + a ] / num__4 = num__10 = > num__5 x + num__2 + a = num__40 = > num__5 x + a = num__40 = > a = num__40 - num__5 x from the above equation we can see that a is minimum when x is maximum provided both are positive the maximum value that x can take in the above equation while still keeping a positive is x = num__8 this gives us a = num__40 - num__40 = num__0 therefore the minimum value that the fourth integer can have is num__0 . option e . <eor> e <eos> |
e |
divide__10.0__2.0__ multiply__10.0__4.0__ subtract__10.0__2.0__ multiply__10.0__0.0__ |
divide__10.0__2.0__ multiply__10.0__4.0__ subtract__10.0__2.0__ divide__0.0__10.0__ |
| machine a and machine r are each used to manufacture num__660 sprockets . it takes machine a num__10 hours longer to produce num__660 sprockets than machine r . machine r produces num__10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ? <o> a ) num__6 <o> b ) num__6.6 <o> c ) num__60 <o> d ) num__100 <o> e ) num__110 |
[ reveal ] spoiler : timer : num__660 / x timea : [ num__660 / x + num__10 ] num__660 / x = [ num__660 / x + num__10 ] * num__1.1 num__660 / x = num__66 * num__11 / x + num__10 num__660 x + num__10 = num__66 * num__11 * x num__660 x + num__6600 = num__66 * num__11 * x x = num__100 plug in back to timea num__6.6 + num__10 = > num__6.0 = num__6 <eor> a <eos> |
a |
percent__10.0__660.0__ percent__10.0__66.0__ percent__100.0__6.0__ |
percent__10.0__660.0__ percent__10.0__66.0__ percent__100.0__6.0__ |
| how many positive even integers less than num__100 contain digits num__3 or num__9 ? <o> a ) num__16 <o> b ) num__17 <o> c ) num__18 <o> d ) num__19 <o> e ) num__10 |
two digit numbers : num__3 at tens place : num__30 num__3234 num__3638 num__9 at tens place : num__90 num__9294 num__9698 if num__3 and num__7 is at units place the number cant be even total : num__5 + num__5 = num__10 answer e <eor> e <eos> |
e |
multiply__3.0__30.0__ subtract__100.0__90.0__ divide__100.0__10.0__ |
multiply__3.0__30.0__ add__3.0__7.0__ add__3.0__7.0__ |
| for any a and b that satisfy | a – b | = b – a and a > num__0 then | a + num__2 | + | - b | + | b – a | + | ab | = <o> a ) - ab + num__3 <o> b ) ab + num__2 b + num__2 <o> c ) ab + num__2 b – num__2 a – num__3 <o> d ) - ab + num__2 b + num__3 <o> e ) ab + num__3 |
observation - num__1 : | a – b | = b – a which is possible only when signs of a and b are same since given a > num__0 so we figure out that a and b are both positive observation - num__2 : | a – b | must be non - negative and so should be the value of b - a which is possible only when absolute value of b is greater than or equal to absolute value of a now you may choose the values of a and b based on above observations e . g . b = num__2 and a = num__1 and check the value of given functions and options | a + num__2 | + | - b | + | b – a | + | ab | = | num__1 + num__2 | + | - num__2 | + | num__2 – num__1 | + | num__1 * num__2 | = num__8 - ab + num__3 = - num__1 * num__2 + num__3 = num__1 ab + num__2 b + num__2 = num__1 * num__2 + num__4 + num__2 = num__8 ab + num__2 b – num__2 a – num__3 = num__1 * num__2 + num__2 * num__2 - num__2 * num__1 - num__3 = num__1 - ab + num__2 b + num__3 = num__5 ab + num__3 = num__5 answer : b <eor> b <eos> |
b |
add__2.0__1.0__ add__1.0__3.0__ add__2.0__3.0__ multiply__2.0__1.0__ |
add__2.0__1.0__ add__1.0__3.0__ add__2.0__3.0__ multiply__2.0__1.0__ |
| rathi took a loan for num__6 years at the rate of num__5.0 per annum on simple interest if the total interest paid was rs . num__1230 find the principal amount ? <o> a ) num__4000 <o> b ) num__3500 <o> c ) num__3100 <o> d ) num__4100 <o> e ) num__5000 |
p = si * num__100 / r * t p = num__1230 * num__16.6666666667 * num__5 = = > num__4100 answer d <eor> d <eos> |
d |
percent__100.0__4100.0__ |
percent__100.0__4100.0__ |
| winson will arrange num__6 people of num__6 different heights for photograph by placing them in two rows of three so that each person in the first row is standing is standing in front of someone in the second row . the heights of the people within each row must increase from left to right and each person in the second row must be taller than the person standing in front of him or her . how many such arrangements of the num__6 people are possible ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__9 <o> d ) num__24 <o> e ) num__26 |
winson will arrange num__6 people of num__6 different heights for photograph by placing them in two rows of three so that each person in the first row is standing is standing in front of someone in the second row . person with max height is in the second row person with min height is in the first row . we need to select num__1 person in the middle of each row . . . in the middle of the first row we can put only num__2 num__3 or num__4 . in the middle of the second row we can put only num__3 num__4 num__5 . if we select { num__3 in the sec . row num__2 in the first } { num__42 } { num__52 } { num__43 } { num__53 } { num__54 } . so there are num__0 * num__1 + num__1 * num__1 + num__2 * num__1 + num__1 * num__1 + num__1 * num__1 + num__0 * num__1 = num__5 cases . . . a <eor> a <eos> |
a |
coin_space__ vowel_space__ card_space__ negate_prob__1.0__ vowel_space__ |
coin_space__ vowel_space__ card_space__ negate_prob__1.0__ vowel_space__ |
| two employees a and b are paid a total of rs . num__550 per week by their employer . if a is paid num__150 percent of the sum paid to b how much is b paid per week ? <o> a ) num__130 <o> b ) num__140 <o> c ) num__150 <o> d ) num__220 <o> e ) num__280 |
let the amount paid to a per week = x and the amount paid to b per week = y then x + y = num__550 but x = num__150.0 of y = num__150 y / num__100 = num__15 y / num__10 ∴ num__15 y / num__10 + y = num__550 ⇒ y [ num__1.5 + num__1 ] = num__550 ⇒ num__25 y / num__10 = num__550 ⇒ num__25 y = num__5500 ⇒ y = num__220.0 = rs . num__220 d ) <eor> d <eos> |
d |
divide__150.0__15.0__ divide__150.0__100.0__ round_down__1.5__ add__10.0__15.0__ multiply__550.0__10.0__ divide__5500.0__25.0__ multiply__1.0__220.0__ |
divide__150.0__15.0__ divide__150.0__100.0__ round_down__1.5__ add__10.0__15.0__ multiply__550.0__10.0__ divide__5500.0__25.0__ divide__220.0__1.0__ |
| if x > num__6 and y < - num__6 then which of the following must be true ? <o> a ) x / y > num__1 <o> b ) x / y < - num__1 <o> c ) x / y < num__0 <o> d ) x + y > num__0 <o> e ) xy > num__0 |
pick x = num__3 y = - num__3 a ) x / y > num__1 - incorrect as x / y = - num__1 b ) x / y < - num__1 - incorrect as x / y = - num__1 c ) x / y < num__0 - correct . it will hold for all values x > num__2 and y < - num__2 as x / y = - num__1 < num__0 d ) x + y > num__0 - incorrect . x + y = num__0 e ) xy > num__0 - incorrect . xy = - num__9 which is less than zero . d should be the answer . <eor> d <eos> |
d |
divide__6.0__3.0__ add__6.0__3.0__ multiply__6.0__0.0__ |
divide__6.0__3.0__ add__6.0__3.0__ divide__0.0__6.0__ |
| if the probability of rain on any given day in city x is num__50.0 what is probability that it rains on exactly num__3 days in a num__4 day period ? <o> a ) num__0.5 <o> b ) num__0.25 <o> c ) num__0.125 <o> d ) num__0.0625 <o> e ) num__0.1875 |
one possible way is rain - rain - rain - no rain . the probability of this is num__0.5 * num__0.5 * num__0.5 * num__0.5 = num__0.0625 the number of possible ways is num__4 c num__3 = num__4 so we multiply this by num__4 . p ( rain on exactly num__3 days ) = num__4 * num__0.0625 = num__0.25 = num__0.25 the answer is b . <eor> b <eos> |
b |
percent__50.0__0.5__ percent__50.0__0.5__ |
percent__50.0__0.5__ percent__50.0__0.5__ |
| a bullet train passes a station platform in num__36 seconds and a man standing on the platform in num__24 seconds . if the speed of the bullet train is num__54 km / hr what is the length of the platform ? <o> a ) num__240 m <o> b ) num__180 m <o> c ) num__260 m <o> d ) num__224 m <o> e ) num__239 m |
a num__180 m speed = num__54 x num__0.277777777778 = num__15 m / s length of the bullet train = ( num__15 x num__24 ) m = num__360 m . let the length of the platform be x metres . then ( x + num__360 ) / num__36 = num__15 - - > x + num__360 = num__540 x = num__180 m . b <eor> b <eos> |
b |
multiply__24.0__15.0__ multiply__36.0__15.0__ round__180.0__ |
multiply__24.0__15.0__ add__360.0__180.0__ subtract__360.0__180.0__ |
| a and b tanks are there . num__0.125 th of the tank b is filled in num__22 hrs . what is time to fill the full tank <o> a ) num__175 <o> b ) num__176 <o> c ) num__177 <o> d ) num__178 <o> e ) num__179 |
num__0.125 th = num__22 hrs therefore entire tank is num__8 x num__22 . num__176 hrs full tank ll be filled in num__176 hrs answer : b <eor> b <eos> |
b |
divide__22.0__0.125__ round__176.0__ |
divide__22.0__0.125__ round__176.0__ |
| jerry is mixing up a salad dressing . regardless of the number of servings the recipe requires that num__0.625 of the finished dressing mix be olive oil num__0.25 mint and the remainder an even mixture of salt pepper and sugar . if jerry accidentally doubles the mint and forgets the sugar altogether what proportion of the botched dressing will be olive oil ? <o> a ) num__0.51724137931 <o> b ) num__0.625 <o> c ) num__0.3125 <o> d ) num__0.5 <o> e ) num__0.481481481481 |
olive oil = num__0.625 = num__0.625 - - > num__15 parts out of num__24 ; mint = num__0.25 = num__0.25 - - > num__6 parts out of num__24 ; salt + pepper + sugar = num__1 - ( num__0.625 + num__0.25 ) = num__0.125 so each = num__0.0416666666667 - - > num__1 part out of num__24 each ; if mint = num__12 ( instead of num__6 ) and sugar = num__0 ( instead of num__1 ) then total = num__15 + num__12 + num__1 + num__1 + num__0 = num__29 parts out of which num__15 parts are olive oil - - > proportion = num__0.51724137931 . answer : a . <eor> a <eos> |
a |
divide__15.0__0.625__ multiply__0.25__24.0__ reverse__24.0__ round_down__0.625__ divide__15.0__29.0__ multiply__1.0__0.5172__ |
divide__15.0__0.625__ multiply__0.25__24.0__ reverse__24.0__ round_down__0.625__ divide__15.0__29.0__ multiply__1.0__0.5172__ |
| a = { num__2 num__3 num__4 num__5 } b = { num__4 num__5 num__6 num__7 num__8 } two integers will be randomly selected from the sets above one integer from set a and one integer from set b . what is the probability that the sum of the two integers will equal num__9 ? <o> a ) num__0.15 <o> b ) num__0.2 <o> c ) num__0.25 <o> d ) num__0.3 <o> e ) num__0.33 |
the total number of pairs possible is num__4 * num__5 = num__20 . out of these num__20 pairs only num__4 sum up to num__9 : ( num__2 num__7 ) ; ( num__3 num__6 ) ( num__4 num__5 ) and ( num__5 num__4 ) . the probability thus is num__0.2 = num__0.2 . answer : b . <eor> b <eos> |
b |
multiply__4.0__5.0__ reverse__5.0__ reverse__5.0__ |
multiply__4.0__5.0__ reverse__5.0__ reverse__5.0__ |
| the perimeter of a triangle is num__28 cm and the inradius of the triangle is num__2.5 cm . what is the area of the triangle ? <o> a ) num__77 cm num__2 <o> b ) num__66 cm num__2 <o> c ) num__54 cm num__2 <o> d ) num__44 cm num__2 <o> e ) num__35 cm num__2 |
area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = num__2.5 * num__14.0 = num__35 cm num__2 answer : e <eor> e <eos> |
e |
triangle_area__28.0__2.5__ triangle_area__28.0__2.5__ |
multiply__2.5__14.0__ multiply__2.5__14.0__ |
| in a game of billiards a can give b num__20 points in num__60 and he can give c num__30 points in num__60 . how many points can b give c in a game of num__100 ? <o> a ) num__22 <o> b ) num__17 <o> c ) num__25 <o> d ) num__27 <o> e ) num__12 |
a scores num__60 while b score num__40 and c scores num__30 . the number of points that c scores when b scores num__100 = ( num__100 * num__30 ) / num__40 = num__25 * num__3 = num__75 . in a game of num__100 points b gives ( num__100 - num__75 ) = num__25 points to c . answer : c <eor> c <eos> |
c |
subtract__60.0__20.0__ divide__60.0__20.0__ subtract__100.0__25.0__ subtract__100.0__75.0__ |
subtract__60.0__20.0__ divide__60.0__20.0__ subtract__100.0__25.0__ subtract__100.0__75.0__ |
| a mixture of num__150 liters of wine and water contains num__20.0 water . how much more water should be added so that water becomes num__25.0 of the new mixture ? <o> a ) num__65 liters <o> b ) num__88 liters <o> c ) num__10 liters <o> d ) num__45 liters <o> e ) num__8 liters |
number of liters of water in num__150 liters of the mixture = num__20.0 of num__150 = num__0.2 * num__150 = num__30 liters . p liters of water added to the mixture to make water num__25.0 of the new mixture . total amount of water becomes ( num__30 + p ) and total volume of mixture is ( num__150 + p ) . ( num__30 + p ) = num__0.25 * ( num__150 + p ) num__120 + num__4 p = num__150 + p = > p = num__10 liters . answer : c <eor> c <eos> |
c |
multiply__150.0__0.2__ subtract__150.0__30.0__ reverse__0.25__ subtract__30.0__20.0__ subtract__20.0__10.0__ |
multiply__150.0__0.2__ subtract__150.0__30.0__ multiply__20.0__0.2__ subtract__30.0__20.0__ subtract__20.0__10.0__ |
| list d consists of num__12 consecutive integers . if - num__4 is the least integer in list d what is the range of positive integers in list d ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__11 <o> e ) num__12 |
since - num__4 is the least integer in list d then num__7 is the largest integer in that list . thus the range of the positive integers in the list is num__7 - num__1 = num__6 . answer : b . <eor> b <eos> |
b |
subtract__7.0__1.0__ subtract__12.0__6.0__ |
subtract__7.0__1.0__ subtract__12.0__6.0__ |
| mrs . rodger got a weekly raise of $ num__145 . if she gets paid every other week write an integer describing how the raise will affect her paycheck . <o> a ) num__140 <o> b ) num__130 <o> c ) num__120 <o> d ) num__150 <o> e ) num__145 |
let the num__1 st paycheck be x ( integer ) . mrs . rodger got a weekly raise of $ num__145 . so after completing the num__1 st week she will get $ ( x + num__145 ) . similarly after completing the num__2 nd week she will get $ ( x + num__145 ) + $ num__145 . = $ ( x + num__145 + num__145 ) = $ ( x + num__290 ) so in this way end of every week her salary will increase by $ num__145 . answer is e . <eor> e <eos> |
e |
multiply__145.0__2.0__ multiply__145.0__1.0__ |
multiply__145.0__2.0__ multiply__145.0__1.0__ |
| for each num__6 - month period during a light bulb ' s life span the odds of it not burning out from over - use are half what they were in the previous num__6 - month period . if the odds of a light bulb burning out during the first num__6 - month period following its purchase are num__0.285714285714 what are the odds of it burning out during the period from num__6 months to num__1 year following its purchase ? <o> a ) num__0.185185185185 <o> b ) num__0.222222222222 <o> c ) num__0.5 <o> d ) num__0.444444444444 <o> e ) num__0.666666666667 |
p ( of not burning out in a six mnth period ) = num__0.5 of p ( of not burning out in prev num__6 mnth period ) p ( of burning out in num__1 st num__6 mnth ) = num__0.285714285714 - - - > p ( of not burning out in num__1 st num__6 mnth ) = num__1 - num__0.285714285714 = num__0.714285714286 - - - - > p ( of not burning out in a six mnth period ) = num__0.5 * num__0.714285714286 = num__0.333333333333 - - - > p ( of burning out in a six mnth period ) = num__1 - num__0.333333333333 = num__0.666666666667 now p ( of burning out in num__2 nd six mnth period ) = p ( of not burning out in num__1 st six mnth ) * p ( of burning out in a six mnth ) = num__0.714285714286 * num__0.666666666667 = num__0.5 ans c <eor> c <eos> |
c |
subtract__1.0__0.2857__ subtract__1.0__0.3333__ reverse__0.5__ reverse__2.0__ |
subtract__1.0__0.2857__ subtract__1.0__0.3333__ reverse__0.5__ subtract__1.0__0.5__ |
| if $ num__2000 was invested at an annual interest rate of num__5.6 compounded annually which of the following represents the amount the investment was worth after num__3 years ? <o> a ) num__2000 ( num__1.056 ) ( num__3 ) <o> b ) num__2000 ( num__1.056 ) ^ num__3 <o> c ) num__2000 ( num__1 + num__3 ( num__0.056 ) ) <o> d ) num__2000 ( num__1 + ( num__0.056 ) ^ num__3 ) <o> e ) num__2000 ( num__3 + num__1.056 ) |
the formula is ci = p ( num__1 + r / num__100 ) ^ t in this case i think b is the ans <eor> b <eos> |
b |
percent__100.0__2000.0__ |
percent__100.0__2000.0__ |
| the heights of three individuals are in the ratio num__4 : num__5 : num__6 . if the sum of the heights of the heaviest and the lightest boy is num__150 cm more than the height of the third boy what is the weight of the lightest boy ? <o> a ) num__120 cm <o> b ) num__150 cm <o> c ) num__160 cm <o> d ) num__190 cm <o> e ) of these |
let the heights of the three boys be num__4 k num__5 k and num__6 k respectively . num__4 k + num__6 k = num__5 k + num__150 = > num__5 k = num__150 = > k = num__30 therefore the height of the lightest boy = num__4 k = num__4 ( num__30 ) = num__120 cm . answer : a <eor> a <eos> |
a |
multiply__5.0__6.0__ multiply__4.0__30.0__ multiply__4.0__30.0__ |
multiply__5.0__6.0__ multiply__4.0__30.0__ multiply__4.0__30.0__ |
| the sum of number of boys and girls in a school is num__400 . if the number of boys is x then the number of girls becomes x % of the total number of students . the number of boys is ? <o> a ) num__50 <o> b ) num__40 <o> c ) num__60 <o> d ) num__80 <o> e ) num__70 |
we have x + x % of num__400 = num__400 x + x / num__100 * num__400 = num__400 num__5 * x = num__400 x = num__80 answer is d <eor> d <eos> |
d |
divide__400.0__5.0__ divide__400.0__5.0__ |
divide__400.0__5.0__ divide__400.0__5.0__ |
| given the two equations num__5 r + s = num__26 and r + num__5 s = num__10 by how much does r exceed s ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
solve by simultaneous equations . my answer is b . r = num__5 s = num__1 <eor> b <eos> |
b |
subtract__5.0__1.0__ |
subtract__5.0__1.0__ |
| the sum of first six prime numbers is ? <o> a ) num__12 <o> b ) num__23 <o> c ) num__24 <o> d ) num__28 <o> e ) num__41 |
required sum = ( num__2 + num__3 + num__5 + num__7 + num__11 + num__13 ) = num__41 . note : num__1 is not a prime number . definition : a prime number ( or a prime ) is a natural number that has exactly two distinct natural number divisors : num__1 and itself . e <eor> e <eos> |
e |
add__2.0__3.0__ add__2.0__5.0__ add__2.0__11.0__ subtract__3.0__2.0__ multiply__1.0__41.0__ |
add__2.0__3.0__ add__2.0__5.0__ add__2.0__11.0__ subtract__3.0__2.0__ multiply__1.0__41.0__ |
| the average speed of a car is num__1 num__0.8 times the avg speed of a bike . a tractor covers num__575 km in num__23 hrs . how much distance will the car cover in num__4 hrs if the speed of the bike is twice speed of the tractor ? <o> a ) num__300 km <o> b ) num__320 km <o> c ) num__360 km <o> d ) num__430 km <o> e ) num__670 km |
sol . average speed of a tractor = num__25 km / h the speed of a bus in an hour = num__25 × num__2 = num__50 km the speed of a car in an hour = num__1.8 * num__50 = num__90 km so the distance covered by car in num__4 h is num__90 × num__4 = num__360 km ans . ( c ) <eor> c <eos> |
c |
divide__575.0__23.0__ subtract__25.0__23.0__ multiply__2.0__25.0__ add__1.0__0.8__ multiply__50.0__1.8__ multiply__4.0__90.0__ multiply__1.0__360.0__ |
divide__575.0__23.0__ subtract__25.0__23.0__ multiply__2.0__25.0__ add__1.0__0.8__ multiply__50.0__1.8__ multiply__4.0__90.0__ multiply__1.0__360.0__ |
| set s consists of integers { num__12 num__34 } . if two numbers are chosen from set s at random what is the probability that the sum of num__2 numbers is more than num__6 ? <o> a ) num__0.333333333333 <o> b ) p ( e ) = num__0.166666666667 <o> c ) num__0.666666666667 <o> d ) num__0.4 <o> e ) num__0.428571428571 |
number of ways you can choose num__2 from num__4 = num__4 c num__2 = num__6 e = event of getting the num__2 numbers is more than num__6 = ( num__34 ) = num__1 probability = num__0.166666666667 answer is b <eor> b <eos> |
b |
subtract__6.0__2.0__ reverse__6.0__ reverse__6.0__ |
subtract__6.0__2.0__ reverse__6.0__ reverse__6.0__ |
| three hoses work to fill a tub at at different rate . hose p and b working together can fill the tub in num__1.2 of an hour . hoses p and c can fill it in num__1.5 an hour . houses b and c can fill it in num__2 hours . how long does it take all num__3 hoses working together to fill the tub ? edited for accurate solution <o> a ) num__0.3 <o> b ) num__0.4 <o> c ) num__0.5 <o> d ) num__1 <o> e ) num__1.2 |
convert the given time to rate and you will be able to add it up . total rate of p and b = rate of p + rate of b = num__1 / ( num__1.2 ) = num__0.833333333333 total rate of p and c = rate of p + rate of c = num__1 / ( num__1.5 ) = num__0.666666666667 total rate of b and c = rate of b + rate of c = num__0.5 adding all three num__2 ( rate of p + rate of b + rate of c ) = num__0.833333333333 + num__0.666666666667 + num__0.5 = num__2 rate of p + rate of b + rate of c = num__1 tub / hour time taken by all three together to fill up the tub is num__1 hour = d <eor> d <eos> |
d |
subtract__3.0__2.0__ divide__1.0__1.2__ subtract__1.5__0.8333__ divide__1.5__3.0__ round__1.0__ |
subtract__3.0__2.0__ divide__1.0__1.2__ divide__2.0__3.0__ divide__1.5__3.0__ round__1.0__ |
| in a rectangular coordinate system points o ( num__20 ) p ( num__28 ) and q ( num__80 ) represent the sites of three proposed housing developments . if a fire station can be built at any point in the coordinate system at which point would it be equidistant from all three developments ? <o> a ) ( num__35 ) <o> b ) ( num__43 ) <o> c ) ( num__54 ) <o> d ) ( num__44 ) <o> e ) ( num__53 ) |
all points equidistant from o and q lie on the line x = num__5 so the fire station should lie on this line . all points equidistant from o and p lie on the line y = num__4 so the fire station should lie on this line . these two intersect at ( num__54 ) and that will be the point equidistant from all num__3 points . the answer is c . <eor> c <eos> |
c |
side_by_diagonal__5.0__4.0__ surface_cube__3.0__ |
side_by_diagonal__5.0__4.0__ surface_cube__3.0__ |
| the population of a town is num__10000 . it increases annually at the rate of num__10.0 p . a . what will be its population after num__2 years ? <o> a ) num__12300 <o> b ) num__12350 <o> c ) num__12100 <o> d ) num__12500 <o> e ) num__12600 |
formula : ( after = num__100 denominator ago = num__100 numerator ) num__10000 Ã — num__1.1 Ã — num__1.1 = num__12100 c <eor> c <eos> |
c |
percent__100.0__12100.0__ |
percent__100.0__12100.0__ |
| the hcf and lcm of two numbers m and n are respectively num__6 and num__210 . if m + n = num__72 then num__1 / m + num__1 / n is equal to <o> a ) num__0.0285714285714 <o> b ) num__0.0857142857143 <o> c ) num__0.135135135135 <o> d ) num__0.0571428571429 <o> e ) none |
answer we have m x n = num__6 x num__210 = num__1260 ∴ num__1 / m + num__1 / n = ( m + n ) / mn = num__0.0571428571429 = num__0.0571428571429 = num__0.0571428571429 correct option : d <eor> d <eos> |
d |
multiply__6.0__210.0__ divide__72.0__1260.0__ divide__72.0__1260.0__ |
multiply__6.0__210.0__ divide__72.0__1260.0__ divide__72.0__1260.0__ |
| the area of a triangle will be when a = num__1 m b = num__2 m c = num__3 m a b c being lengths of respective sides ? <o> a ) num__3 <o> b ) num__6 <o> c ) num__4 <o> d ) num__9 <o> e ) num__1 |
s = ( num__1 + num__2 + num__3 ) / num__2 = num__3 answer : a <eor> a <eos> |
a |
multiply__1.0__3.0__ |
multiply__1.0__3.0__ |
| if a + b = x and a - b = y then num__3 ab = can someone explain ! <o> a ) ( num__3 x ^ num__2 - num__3 y ^ num__2 ) / num__4 <o> b ) ( num__3 y ^ num__2 - num__3 x ^ num__2 ) / num__4 <o> c ) num__3 x + num__3 y / num__2 <o> d ) num__3 x - num__3 y / num__2 <o> e ) num__3 x ^ num__2 - num__3 y ^ num__1.0 |
plugging numbers is quite fast in this case : let a = num__1 b = num__2 then ; x = num__1 + num__2 = > num__3 and y = num__1 - num__2 = > - num__1 question asks num__3 ab = . . . . . . . . . as our number num__3 ab = num__3 * num__1 * num__2 = num__6 now plug the value of x and y in the answer choices . option a gives num__6 and that is the number we need . hence : ans is a . <eor> a <eos> |
a |
subtract__3.0__1.0__ multiply__3.0__2.0__ multiply__3.0__1.0__ |
subtract__3.0__1.0__ multiply__3.0__2.0__ multiply__3.0__1.0__ |
| set a contains all the even numbers between num__32 and num__80 inclusive . set b contains all the even numbers between num__62 and num__110 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? <o> a ) num__450 <o> b ) num__550 <o> c ) num__650 <o> d ) num__750 <o> e ) num__850 |
each term in set b is num__30 more than the corresponding term in set a . the difference of the sums = num__25 * num__30 = num__750 . the answer is d . <eor> d <eos> |
d |
subtract__62.0__32.0__ multiply__25.0__30.0__ multiply__25.0__30.0__ |
subtract__62.0__32.0__ multiply__25.0__30.0__ multiply__25.0__30.0__ |
| sally has $ num__100 in pennies . she wants to stack her pennies into three large piles and five small piles . the large piles are all the same size and the small piles are all the same size . additionally sally wants all of the small piles put together to equal one of the large piles . how many pennies does sally put in one of the small piles ? <o> a ) num__2500 pennies <o> b ) num__500 pennies <o> c ) num__250 pennies <o> d ) num__100 pennies <o> e ) num__50 pennies |
b . each large pile gets num__0.25 of the total pennies . each small pile gets num__0.2 of the remaining num__0.25 of the total pennies . so each small pile gets num__0.05 of the total pennies . $ num__100 is num__10000 pennies . num__0.05 of num__10000 is num__500 so the correct answer is b : num__500 pennies . <eor> b <eos> |
b |
subtract__0.25__0.2__ divide__100.0__0.2__ divide__100.0__0.2__ |
subtract__0.25__0.2__ divide__100.0__0.2__ divide__100.0__0.2__ |
| if num__32.5 of the num__880 students at a certain college are enrolled in biology classes how many students at the college are not enrolled in a biology class ? <o> a ) num__110 <o> b ) num__330 <o> c ) num__550 <o> d ) num__594 <o> e ) num__880 |
students enrolled in biology are num__32.5 and therefore not enrolled are num__67.5 . so of num__880 is num__880 * . num__675 = num__594 answer is d num__594 <eor> d <eos> |
d |
percent__67.5__880.0__ percent__67.5__880.0__ |
percent__67.5__880.0__ percent__67.5__880.0__ |
| the average of temperatures at noontime from monday to friday is num__50 ; the lowest one is num__45 what is the possible maximum range of the temperatures ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__40 <o> d ) num__45 <o> e ) num__75 |
average = num__50 sum of temperatures = num__50 * num__5 = num__250 as the min temperature is num__45 max would be num__250 - num__4 * num__45 = num__70 - - > the range = num__70 ( max ) - num__45 ( min ) = num__25 answer : b . <eor> b <eos> |
b |
subtract__50.0__45.0__ multiply__50.0__5.0__ subtract__70.0__45.0__ subtract__50.0__25.0__ |
subtract__50.0__45.0__ multiply__50.0__5.0__ subtract__70.0__45.0__ subtract__50.0__25.0__ |
| a certain sum is invested at simple interest at num__15.0 p . a . for two years instead of investing at num__12.0 p . a . for the same time period . therefore the interest received is more by rs . num__900 . find the sum ? <o> a ) s . num__7000 <o> b ) s . num__9000 <o> c ) s . num__15000 <o> d ) s . num__17000 <o> e ) s . num__27000 |
let the sum be rs . x . ( x * num__15 * num__2 ) / num__100 - ( x * num__12 * num__2 ) / num__100 = num__900 = > num__30 x / num__100 - num__24 x / num__100 = num__900 = > num__6 x / num__100 = num__900 = > x = num__15000 . answer : c <eor> c <eos> |
c |
percent__100.0__15000.0__ |
percent__100.0__15000.0__ |
| if an integer n is divisible by num__3 num__5 num__12 . what is the next larger integer divisible by all these numbers <o> a ) n + num__3 <o> b ) n + num__5 <o> c ) n + num__12 <o> d ) n + num__60 <o> e ) n + num__15 |
if n is divisible by num__3 num__5 and num__12 it must a multiple of the lcm of num__3 num__5 and num__12 which is num__60 . n = num__60 k n + num__60 is also divisible by num__60 since n + num__60 = num__60 k + num__60 = num__60 ( k + num__1 ) the answer is d . <eor> d <eos> |
d |
multiply__5.0__12.0__ multiply__5.0__12.0__ |
multiply__5.0__12.0__ multiply__5.0__12.0__ |
| if it is num__6 : num__12 in the evening on a certain day what time in the morning was it exactly num__1 num__440711 minutes earlier ? ( assume standard time in one location . ) <o> a ) num__6 : num__12 <o> b ) num__6 : num__15 <o> c ) num__6 : num__18 <o> d ) num__6 : num__21 <o> e ) num__6 : num__24 |
num__6 : num__12 minus num__1 num__440711 must end with num__1 the only answer choice which ends with num__1 is d . answer : d . <eor> d <eos> |
d |
round__6.0__ |
round__6.0__ |
| x ^ num__2 - y ^ num__2 = num__16 and num__2 . xy = num__4 so find out x + y = ? tell that both statement are required to find out the value of x + y <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__5 |
x ^ num__2 - y ^ num__2 = num__16 = num__25 - num__9 = num__5 ^ num__2 - num__3 ^ num__2 = > ( x + y ) * ( x - y ) = ( num__5 + num__3 ) * ( num__5 - num__3 ) = > ( x + y ) = ( num__5 + num__3 ) = num__8 answer : c <eor> c <eos> |
c |
subtract__25.0__16.0__ subtract__9.0__4.0__ subtract__5.0__2.0__ multiply__2.0__4.0__ multiply__2.0__4.0__ |
subtract__25.0__16.0__ subtract__9.0__4.0__ subtract__5.0__2.0__ multiply__2.0__4.0__ multiply__2.0__4.0__ |
| a number increased by num__20.0 gives num__480 . the number is <o> a ) num__250 <o> b ) num__400 <o> c ) num__450 <o> d ) num__500 <o> e ) num__520 |
formula = total = num__100.0 increse = ` ` + ' ' decrease = ` ` - ' ' a number means = num__100.0 that same number increased by num__20.0 = num__120.0 num__120.0 - - - - - - - > num__480 ( num__120 × num__4 = num__480 ) num__100.0 - - - - - - - > num__400 ( num__100 × num__4 = num__400 ) b ) <eor> b <eos> |
b |
percent__100.0__400.0__ |
percent__100.0__400.0__ |
| n is a whole number which when divided by num__4 gives num__3 as remainder . what will be the remainder when num__2 n is divided by num__4 ? <o> a ) num__6 <o> b ) num__3 <o> c ) num__2 <o> d ) num__1 <o> e ) num__8 |
let n = num__4 q + num__3 . then num__2 n = num__8 q + num__6 = num__4 ( num__2 q + num__1 ) + num__2 . thus when num__2 n is divided by num__4 the remainder is num__2 . answer is c <eor> c <eos> |
c |
multiply__4.0__2.0__ add__4.0__2.0__ subtract__4.0__3.0__ subtract__4.0__2.0__ |
multiply__4.0__2.0__ add__4.0__2.0__ subtract__4.0__3.0__ subtract__4.0__2.0__ |
| the area of a triangle will be when a = num__1 m b = num__2 m c = num__3 m a b c being lengths of respective sides ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
s = ( num__1 + num__2 + num__3 ) / num__2 = num__3 = > no triangle exists answer : a <eor> a <eos> |
a |
multiply__1.0__3.0__ |
multiply__1.0__3.0__ |
| mr . depak has $ num__459774 in his bank account . what is the least amount of money ( in whole number of dollars ) that he must add to his account if he wants to split this money evenly among her six best friends ? options : <o> a ) $ num__0 <o> b ) $ num__2 <o> c ) $ num__3 <o> d ) $ num__4 <o> e ) $ num__6 |
to find the least amount deepak should add to his saving account to split the money evenly among his num__6 best friends he needs to make the total divisible by num__6 simply add the individual digits of the total = num__4 + num__5 + num__9 + num__7 + num__7 + num__4 = num__36 if you add num__0 the number is divisible by num__6 ( num__36 + num__0 ) correct option : a <eor> a <eos> |
a |
add__5.0__4.0__ multiply__9.0__4.0__ reverse__459774.0__ reverse__459774.0__ |
add__5.0__4.0__ multiply__9.0__4.0__ reverse__459774.0__ reverse__459774.0__ |
| a person standing on a railway platform noticed that a train took num__21 seconds to completely pass through the platform which was num__84 m long and it took num__9 seconds in passing him . find the speed of the train ? <o> a ) num__6 m / sec <o> b ) num__5 m / sec <o> c ) num__12 m / sec <o> d ) num__7 m / sec <o> e ) none |
let the length of the train be x m . then then train covers x meters in num__9 seconds and ( x + num__84 ) m in num__21 seconds . therefore x / num__9 = ( x + num__84 ) / num__21 = num__63 m so the length of the train = num__63 m speed of the train = num__7.0 = num__7 m / sec answer : option d <eor> d <eos> |
d |
subtract__84.0__21.0__ divide__63.0__9.0__ round__7.0__ |
subtract__84.0__21.0__ divide__63.0__9.0__ divide__63.0__9.0__ |
| if a is a positive integer which of the following must be odd ? <o> a ) a + num__1 <o> b ) a ^ num__2 + a <o> c ) a ^ num__2 + a + num__1 <o> d ) x ^ num__2 − num__1 <o> e ) num__3 x ^ num__2 − num__3 |
a . a + num__1 = can be odd or even . since o + o = e or e + o = o b . a ^ num__2 + a = a ( a + num__1 ) . since from the above derivation we already know the term a + num__1 can be odd or even directly substitute here . a ( odd ) = even ( when a is even ) or a ( even ) = even [ when a is odd ] c . here ' s the answer . since we know the term a ^ num__2 + a can always take a even number even + num__1 = odd hence c . <eor> c <eos> |
c |
multiply__1.0__2.0__ |
power__2.0__1.0__ |
| a train crosses a platform of num__150 m in num__15 sec same train crosses another platform of length num__250 m in num__20 sec . then find the length of the train ? <o> a ) num__150 <o> b ) num__788 <o> c ) num__267 <o> d ) num__266 <o> e ) num__123 |
length of the train be ‘ x ’ x + num__10.0 = x + num__12.5 num__4 x + num__600 = num__3 x + num__750 x = num__150 m answer : a <eor> a <eos> |
a |
divide__150.0__15.0__ divide__250.0__20.0__ multiply__150.0__4.0__ add__150.0__600.0__ round__150.0__ |
divide__150.0__15.0__ divide__250.0__20.0__ multiply__150.0__4.0__ add__150.0__600.0__ round__150.0__ |
| a cow was standing on a bridge num__5 m away from the middle of the bridge . a train was coming towards the bridge from the end nearest to the cow . seeing this cow ran towards the train and managed to escape when the train was num__2 m away from the bridge . if it had run in opposite direction ( i . e away from the train ) it would have been hit by the train num__2 m before the end of the bridge . what is the length of bridge in meters assuming speed of the train is num__4 times that of the cow . <o> a ) num__32 <o> b ) num__36 <o> c ) num__40 <o> d ) num__44 <o> e ) num__50 |
explanation : explation : let length of bridge is x then middle of bridge is x / num__2 . and cow is standing at x / num__2 - num__5 from the end so after seeing the train it would run opposite direction and cover distance of x / num__2 - num__5 when the train is num__2 m away from other side . so the bridge from one side to cow is { ( x / num__2 - num__5 ) + ( x / num__2 - num__5 ) } = x - num__10 so remaining distance is num__10 m the cow is hit before num__2 m i . e cow covers num__8 m and train will cover num__32 m . answer : a <eor> a <eos> |
a |
multiply__5.0__2.0__ multiply__2.0__4.0__ multiply__4.0__8.0__ multiply__4.0__8.0__ |
multiply__5.0__2.0__ subtract__10.0__2.0__ multiply__4.0__8.0__ multiply__4.0__8.0__ |
| a fort had provision of food for num__150 men for num__30 days . after num__10 days num__25 men left the fort . the number of days for which the remaining food will last is : <o> a ) num__29 num__0.2 <o> b ) num__24 <o> c ) num__42 <o> d ) num__54 <o> e ) num__48 |
after num__10 days : num__150 men had food for num__20 days . suppose num__125 men had food for x days . now less men more days ( indirect proportion ) therefore num__125 : num__150 : : num__20 : x < = > num__125 x x = num__150 x num__20 = > x = num__150 x num__0.16 = > x = num__24 . correct answer is b <eor> b <eos> |
b |
subtract__30.0__10.0__ subtract__150.0__25.0__ divide__20.0__125.0__ multiply__150.0__0.16__ round__24.0__ |
subtract__30.0__10.0__ subtract__150.0__25.0__ divide__20.0__125.0__ multiply__150.0__0.16__ round__24.0__ |
| the number of diagonals of a polygon of n sides is given by the formula q = n ( n - num__3 ) / num__2 . if a polygon has twice as many diagonals as sides how many sides does it have ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
q = n ( n - num__3 ) q = num__2 * n num__2 n = n ( n - num__3 ) = > num__2 = n - num__3 = > n = num__5 answer b <eor> b <eos> |
b |
add__3.0__2.0__ add__3.0__2.0__ |
add__3.0__2.0__ add__3.0__2.0__ |
| a family i know has several children . each boy in this family has as many sisters as brothers but each girl has twice as many brothers as sisters . how many brothers and sisters are there ? <o> a ) num__4 brothers and num__2 sisters <o> b ) num__4 brothers and num__3 sisters <o> c ) num__3 brothers and num__3 sisters <o> d ) num__4 brothers and num__5 sisters <o> e ) num__4 brothers and num__6 sisters |
let no . of girls = g n boys = b b = g + num__1 . . . . . . . . . . ( i ) as each boy have equal no . of sisters and brothers and he is not included in brothers . so num__1 is for him in g + num__1 . now b = num__2 ( g - num__1 ) . . . . . . . . ( ii ) as each girl have g - num__1 sisters . solving ( i ) and ( ii ) we get g = num__3 and b = num__4 answer : b <eor> b <eos> |
b |
twice__1.0__ triple__1.0__ twice__2.0__ twice__2.0__ |
twice__1.0__ add__1.0__2.0__ add__1.0__3.0__ add__1.0__3.0__ |
| the ages of old and young total num__48 . old is twice as old as young was when old was half as old as young will be when young is three times as old was when old was three times as old as young . how old is old ? <o> a ) num__23 <o> b ) num__27 <o> c ) num__12 <o> d ) num__43 <o> e ) num__37 |
explanation : from the options itself we can see that option c old = num__30 young = num__18 num__30 + num__18 = num__48 by reducing this years only by one before num__6 years old = num__24 ( half of young ) young = num__12 ( twice of old ) answer : c <eor> c <eos> |
c |
subtract__48.0__30.0__ add__6.0__18.0__ subtract__18.0__6.0__ subtract__18.0__6.0__ |
subtract__48.0__30.0__ add__6.0__18.0__ subtract__18.0__6.0__ subtract__18.0__6.0__ |
| he population of a city is num__110000 . if it increases at the rate of num__6.0 per annum then what will be its population num__2 years hence ? <o> a ) num__123596 <o> b ) num__1237723 <o> c ) num__123776 <o> d ) num__123728 <o> e ) num__123718 |
population after n years = p [ num__1 + ( r / num__100 ) ] num__2 population after num__2 years = num__110000 * [ num__1 + num__0.06 ] num__2 = ( num__110000 x num__1.06 x num__1.06 ) = num__123596 answer : a <eor> a <eos> |
a |
percent__6.0__1.0__ percent__100.0__123596.0__ |
percent__6.0__1.0__ percent__100.0__123596.0__ |
| combine terms : num__6 a + num__26 b - num__4 b – num__16 a . <o> a ) num__4 a + num__22 b <o> b ) - num__28 a + num__30 b <o> c ) - num__10 a + num__22 b <o> d ) num__28 a + num__30 b <o> e ) num__5 a + num__30 b |
solution : num__6 a + num__26 b - num__4 b – num__16 a . = num__6 a – num__16 a + num__26 b – num__4 b . = - num__10 a + num__22 b . answer : ( c ) <eor> c <eos> |
c |
add__6.0__4.0__ add__6.0__16.0__ add__6.0__4.0__ |
add__6.0__4.0__ add__6.0__16.0__ add__6.0__4.0__ |
| saturn travels around the sun at an approximate speed of num__5.82 miles per second . this speed is how many kilometers per hour ? ( num__1 km = num__0.6 miles ) <o> a ) num__30560 <o> b ) num__31650 <o> c ) num__32740 <o> d ) num__33830 <o> e ) num__34 |
920 |
the speed is num__5.82 miles / s . then num__5.82 / num__0.6 = num__9.7 km / s num__9.7 * num__3600 = num__34920 kph the answer is e . <eor> e <eos> |
e |
e |
| a sum of money is sufficient to pay a ' s wages for num__21 days and b ' s wages for num__28 days . the same money is sufficient to pay the wages of both for ? <o> a ) num__10 days <o> b ) num__12 days <o> c ) num__15 days <o> d ) num__20 days <o> e ) num__18 days |
let the total money be $ x a ' s num__1 day work = $ x / num__21 b ' s num__1 day work = $ x / num__28 a + b num__1 day work = $ x / num__12 the money is sufficient to pay the wages of both for num__12 days answer is b <eor> b <eos> |
b |
multiply__1.0__12.0__ |
divide__12.0__1.0__ |
| the bear alarm at grizzly ’ s peak ski resort sounds an average of once every thirty days but the alarm is so sensitively calibrated that it sounds an average of ten false alarms for every undetected bear . despite this the alarm only sounds for three out of four bears that actually appear at the resort . if the alarm sounds what is the probability that a bear has actually been sighted ? <o> a ) num__0.25 <o> b ) num__0.230769230769 <o> c ) num__0.519230769231 <o> d ) num__0.75 <o> e ) num__0.769230769231 |
how many times the alarm goes off - - > num__10 false + num__3 true how many of these alarms are true - - > num__3 true alarms therefore p = true alarms / total alarms = num__0.230769230769 or we can calculate the p ( of alarm sounding but no bear detected i . e . false alarm ) num__10 times when the alarm goes off - no bear is seen num__13 times in all the alarm goes off so num__0.769230769231 is the p that alarm is a false alarm subtracting num__0.769230769231 from num__1 is the p that it ' s a true alarm - - > ( num__1 - num__0.769230769231 ) = num__0.230769230769 . . . . answer b <eor> b <eos> |
b |
negate_prob__0.2308__ negate_prob__0.7692__ |
negate_prob__0.2308__ negate_prob__0.7692__ |
| find the number when num__15 is subtracted from num__7 times the number the result is num__10 more than twice of the number <o> a ) num__5 <o> b ) num__15 <o> c ) num__7.5 <o> d ) num__4 <o> e ) num__3 |
explanation : let the number be x . num__7 x - num__15 = num__2 x + num__10 = > num__5 x = num__25 = > x = num__5 option a <eor> a <eos> |
a |
subtract__15.0__10.0__ add__15.0__10.0__ subtract__15.0__10.0__ |
subtract__15.0__10.0__ add__15.0__10.0__ subtract__15.0__10.0__ |
| if a person walks at num__14 km / hr instead of num__10 km / hr he would have walked num__20 km more . what is the actual distance travelled by him ? <o> a ) num__50 <o> b ) num__60 <o> c ) num__40 <o> d ) num__30 <o> e ) num__25 |
speed = x / num__10 - - - num__1 x + num__20 km if travelled at num__14 km / hr speed = x + num__1.42857142857 - - - num__2 from num__1 & num__2 x = num__50 answer a <eor> a <eos> |
a |
divide__20.0__14.0__ divide__20.0__10.0__ multiply__1.0__50.0__ |
divide__20.0__14.0__ divide__20.0__10.0__ divide__50.0__1.0__ |
| the average marks of a class of num__30 students is num__40 and that of another class of num__50 students is num__60 . find the average marks of all the students ? <o> a ) num__52.3 <o> b ) num__52.2 <o> c ) num__52.1 <o> d ) num__52.5 <o> e ) num__52.4 |
sum of the marks for the class of num__30 students = num__30 * num__40 = num__1200 sum of the marks for the class of num__50 students = num__50 * num__60 = num__3000 sum of the marks for the class of num__80 students = num__1200 + num__3000 = num__4200 average marks of all the students = num__52.5 = num__52.5 answer : d <eor> d <eos> |
d |
multiply__30.0__40.0__ multiply__50.0__60.0__ add__30.0__50.0__ add__1200.0__3000.0__ divide__4200.0__80.0__ divide__4200.0__80.0__ |
multiply__30.0__40.0__ multiply__50.0__60.0__ add__30.0__50.0__ add__1200.0__3000.0__ divide__4200.0__80.0__ divide__4200.0__80.0__ |
| what is the greatest prime factor of ( num__10 ! num__9 ! ) + ( num__10 ! num__11 ! ) ? <o> a ) num__23 <o> b ) num__29 <o> c ) num__31 <o> d ) num__37 <o> e ) num__41 |
num__10 ! num__9 ! + num__10 ! num__11 ! = num__10 ! num__9 ! ( num__1 + num__10 * num__11 ) = num__12 ! num__11 ! ( num__111 ) = num__12 ! num__11 ! ( num__3 * num__37 ) the greatest prime factor is num__37 . the answer is d . <eor> d <eos> |
d |
subtract__10.0__9.0__ add__11.0__1.0__ gcd__9.0__12.0__ divide__111.0__3.0__ multiply__1.0__37.0__ |
subtract__10.0__9.0__ add__11.0__1.0__ gcd__9.0__12.0__ divide__111.0__3.0__ multiply__1.0__37.0__ |
| in a box of num__10 pens a total of num__2 are defective . if a customer buys num__2 pens selected at random from the box what is the probability that neither pen will be defective ? <o> a ) num__0.583333333333 <o> b ) num__0.55 <o> c ) num__0.566666666667 <o> d ) num__0.622222222222 <o> e ) num__0.566666666667 |
# defective pens = num__2 # good pens = num__8 probability of the num__1 st pen being good = num__0.8 probability of the num__2 nd pen being good = num__0.777777777778 total probability = num__0.8 * num__0.777777777778 = num__0.622222222222 the answer is d . <eor> d <eos> |
d |
subtract__10.0__2.0__ divide__8.0__10.0__ multiply__0.8__0.7778__ multiply__1.0__0.6222__ |
subtract__10.0__2.0__ divide__8.0__10.0__ multiply__0.8__0.7778__ multiply__1.0__0.6222__ |
| an article is bought for rs . num__1280 and sold for rs . num__900 find the loss percent ? <o> a ) num__28.0 <o> b ) num__29.0 <o> c ) num__16.0 <o> d ) num__26.0 <o> e ) num__32 % |
num__1280 - - - - num__380 num__100 - - - - ? = > num__29.0 answer : b <eor> b <eos> |
b |
percent__100.0__29.0__ |
percent__100.0__29.0__ |
| if num__0.465909090909 = num__0.46590 what is the num__77 th digit to the right of the decimal point of the fraction ? <o> a ) num__6 <o> b ) num__2 <o> c ) num__5 <o> d ) num__0 <o> e ) num__9 |
we are not concerned what num__0.465909090909 means . . we have to look at the decimal . . num__0.6590 means num__0.465909090 . . . . so leaving girst second and third digit to the right of decimal all odd numbered are num__0 and all even numbered are num__9 . . here num__77 is odd so ans is num__0 answer is d <eor> d <eos> |
d |
round_down__0.4659__ round_down__0.4659__ |
round_down__0.4659__ round_down__0.4659__ |
| milk contains num__10.0 water . what content of pure milk should be added to num__20 liters of milk to reduce this to num__5.0 ? <o> a ) num__20 liters <o> b ) num__25 liters <o> c ) num__30 liters <o> d ) num__15 liters <o> e ) num__18 liters |
quantity of water in num__20 liters = num__10.0 of num__20 liters = num__2 liters let x liters of pure milk be added . then num__2 / ( num__20 + x ) = num__0.05 num__5 x = num__100 x = num__20 liters answer is a <eor> a <eos> |
a |
percent__10.0__20.0__ percent__20.0__100.0__ |
percent__10.0__20.0__ percent__20.0__100.0__ |
| in a race of num__200 m a can beat b by num__31 m and c by num__18 m . in a race of num__350 m c will beat b by ? <o> a ) num__25 m <o> b ) num__75 m <o> c ) num__85 m <o> d ) num__05 m <o> e ) num__15 m |
explanation : a : b = num__200 : num__169 a : c = num__200 : num__182 = = > num__182 : num__169 when c covers num__182 m b covers num__169 m when c covers num__350 m b covers = > num__325 m so c beats b by ( num__350 - num__325 ) = num__25 m . answer : a <eor> a <eos> |
a |
subtract__200.0__31.0__ subtract__200.0__18.0__ subtract__350.0__325.0__ round__25.0__ |
subtract__200.0__31.0__ subtract__200.0__18.0__ subtract__350.0__325.0__ subtract__350.0__325.0__ |
| a train passes a station platform in num__39 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__328 <o> b ) num__279 <o> c ) num__240 <o> d ) num__285 <o> e ) num__231 |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x meters . then x + num__7.69230769231 = num__15 x + num__300 = num__585 x = num__285 m . answer : d <eor> d <eos> |
d |
subtract__54.0__39.0__ multiply__20.0__15.0__ divide__300.0__39.0__ multiply__39.0__15.0__ subtract__585.0__300.0__ round__285.0__ |
subtract__54.0__39.0__ multiply__20.0__15.0__ divide__300.0__39.0__ multiply__39.0__15.0__ subtract__585.0__300.0__ round__285.0__ |
| the length of a rectangular plot is num__10 mtr more than its width . the cost of fencing the plot along its perimeter at the rate of rs . num__6.5 mtr is rs . num__1950 . the perimeter of the plot is ? <o> a ) num__126 <o> b ) num__156 <o> c ) num__190 <o> d ) num__300 <o> e ) num__260 |
sol . let width = x length = ( num__10 + x ) perimeter = num__2 ( x + ( num__10 + x ) ) = num__2 ( num__2 x = num__10 ) & num__2 ( num__2 x + num__10 ) * num__6.5 = num__1950 x = num__70 required perimeter = num__2 ( num__70 + num__80 ) = num__300 d <eor> d <eos> |
d |
rectangle_perimeter__70.0__80.0__ triangle_area__2.0__300.0__ |
rectangle_perimeter__70.0__80.0__ triangle_area__2.0__300.0__ |
| it takes num__2 team of farm workers num__12 days to completely prepare a piece of land for planting . if both team of workers were to work separately one of them can complete the work num__10 days earlier than the other how many days will it take each of them to separately complete the work ? <o> a ) num__12 and num__22 <o> b ) num__11 and num__21 <o> c ) num__15 and num__25 <o> d ) num__9 and num__19 <o> e ) num__20 and num__30 |
work = ( a ) ( b ) / ( a + b ) where a and b are the individual times of each entity . here we ' re told that ( working together ) the two team of workers would complete a job in num__12 days . this means that ( individually ) each of team would take more than num__12 days to do the job . answers d a and b are illogical since the individual times must both be greater than num__12 days . so we can test the values for answers c and e . using the values for answers b and d . . . answer b : ( num__15 ) ( num__25 ) / ( num__15 + num__25 ) = num__9.375 = num__9.4 this is a match answer e : ( num__20 ) ( num__30 ) / ( num__20 + num__30 ) = num__12 final answer : e <eor> e <eos> |
e |
add__10.0__15.0__ multiply__2.0__10.0__ multiply__2.0__15.0__ round__20.0__ |
add__10.0__15.0__ multiply__2.0__10.0__ add__10.0__20.0__ round__20.0__ |
| if f ( x ) = num__5 x ^ num__3 - num__4 x + num__8 and g ( y ) = num__6 y - num__4 then g ( f ( x ) ) = <o> a ) num__11 x ^ num__2 + num__4 x + num__4 <o> b ) num__11 x ^ num__2 - num__12 x + num__44 <o> c ) num__30 x ^ num__3 - num__24 x + num__44 <o> d ) num__30 x ^ num__3 + num__4 x + num__4 <o> e ) num__30 x ^ num__3 - num__12 x + num__44 |
g ( f ( x ) ) = num__6 ( f ( x ) ) - num__4 = num__6 ( num__5 x ^ num__3 - num__4 x + num__8 ) - num__4 = num__30 x ^ num__3 - num__24 x + num__44 = > c <eor> c <eos> |
c |
multiply__5.0__6.0__ multiply__3.0__8.0__ multiply__5.0__6.0__ |
multiply__5.0__6.0__ subtract__30.0__6.0__ add__6.0__24.0__ |
| by selling an article for $ num__195 a person gains $ num__45 . what is the gain % ? <o> a ) num__25.0 <o> b ) num__30.0 <o> c ) num__50.0 <o> d ) num__20.0 <o> e ) num__10 % |
s . p . = $ num__195 gain = $ num__45 c . p . = num__195 - num__45 = num__150 gain % = num__0.3 * num__100.0 = num__30.0 answer is b <eor> b <eos> |
b |
percent__100.0__30.0__ |
percent__100.0__30.0__ |
| harold and millicent are getting married and need to combine their already - full libraries . if harold who has num__0.5 as many books as millicent brings num__0.333333333333 of his books to their new home then millicent will have enough room to bring num__0.2 of her books to their new home . what fraction of millicent ' s old library capacity is the new home ' s library capacity ? <o> a ) num__0.5 <o> b ) num__0.666666666667 <o> c ) num__0.75 <o> d ) num__0.366666666667 <o> e ) num__0.833333333333 |
because we see h willbring num__0.333333333333 of his booksto the new home - - > try to pick a number that isdivisible by num__3 . before : assume h = num__30 books h = num__0.5 m - - > m = num__60 books after : h ' = num__0.333333333333 h = num__10 books m ' = num__0.2 m = num__12 books total = num__22 books m ' = num__22 = num__0.366666666667 * num__60 ratio : num__0.366666666667 ans : d <eor> d <eos> |
d |
divide__30.0__0.5__ divide__30.0__3.0__ multiply__0.2__60.0__ add__10.0__12.0__ divide__22.0__60.0__ divide__22.0__60.0__ |
divide__30.0__0.5__ divide__30.0__3.0__ multiply__0.2__60.0__ add__10.0__12.0__ divide__22.0__60.0__ divide__22.0__60.0__ |
| a herd of goats consists of a males and b females . in that herd half of the goats that have horns are females . if c goats of the herd do not have horns then the total number of male goats in the herd that do not have horns in terms of a b and c is - <o> a ) ( a − b + c ) / num__2 <o> b ) ( a − b − c ) / num__2 <o> c ) ( a + b + c ) / num__2 <o> d ) ( a + b − c ) / num__2 <o> e ) ( num__2 a − b + c ) / num__2 |
male goats = a female goats = b num__0.5 of horn goats = females = x ( say ) therefore other half will be male = x now male goats without horn = a - x - - ( num__1 ) also total goats without horn = c = ( a - x ) + ( b - x ) or x = ( a + b - c ) / num__2 substituting the value of x in equation ( num__1 ) we will have a - x = ( a − b + c ) / num__2 . answer a . <eor> a <eos> |
a |
reverse__0.5__ reverse__0.5__ |
reverse__0.5__ reverse__0.5__ |
| find the value of num__3 + num__2 • ( num__8 – num__3 ) <o> a ) num__25 <o> b ) num__13 <o> c ) num__17 <o> d ) num__24 <o> e ) num__15 |
num__3 + num__2 • ( num__8 – num__3 ) = num__3 + num__2 ( num__5 ) = num__3 + num__2 * num__5 = num__3 + num__10 = num__13 correct answer b <eor> b <eos> |
b |
add__3.0__2.0__ add__2.0__8.0__ add__3.0__10.0__ add__3.0__10.0__ |
add__3.0__2.0__ add__2.0__8.0__ add__3.0__10.0__ add__3.0__10.0__ |
| if num__200 lb of a mixture contain num__80.0 husk and num__20.0 sand . then how much husk needs to be extracted in order to have num__75.0 concentration of husk ? <o> a ) num__0.25 <o> b ) num__6.66666666667 <o> c ) num__0.5 <o> d ) num__40 <o> e ) num__60 |
we need to extract num__40 lb of husk . num__80.0 of num__200 = num__160 num__20.0 of num__200 = num__40 now we have t remove husk to have num__75.0 of husk ( key point sand is not supposed to be removed ) let x be the new quantity after removing husk so num__40 = num__25.0 of x x = num__160 husk will num__120 lbs hence we need to remove num__0.25 of the total husk or num__40 lbs answer : d <eor> d <eos> |
d |
subtract__200.0__40.0__ subtract__200.0__80.0__ divide__20.0__80.0__ subtract__200.0__160.0__ |
subtract__200.0__40.0__ subtract__200.0__80.0__ divide__20.0__80.0__ subtract__200.0__160.0__ |
| while flying over the pacific an airplane makes a num__25 ° turn to the right to avoid a storm . if as a result the airplane is traveling in a direction num__3 ° east of north in what direction was it originally flying ? <o> a ) ( a ) num__30 ° west of north <o> b ) ( b ) num__30 ° east of north <o> c ) ( c ) num__22 ° west of north <o> d ) ( d ) num__22 ° east of north <o> e ) ( e ) num__5 ° west of north |
if after a turn of num__25 ° you are num__3 ° neast with a num__22 ° turn you would be perfectly pointing at north . so you were before the turn num__22 ° to the other side ( west ) . answer c ) <eor> c <eos> |
c |
subtract__25.0__3.0__ round__22.0__ |
subtract__25.0__3.0__ round__22.0__ |
| triangle atriangle b are similar triangles with areas num__1536 units square and num__2166 units square respectively . the ratio of there corresponding height would be <o> a ) num__9 : num__10 <o> b ) num__17 : num__19 <o> c ) num__23 : num__27 <o> d ) num__13 : num__17 <o> e ) num__16 : num__19 |
let x be the height of triangle a and y be the height of triangle of b . since triangles are similar ratio of area of a and b is in the ratio of x ^ num__2 / y ^ num__2 therefore ( x ^ num__2 / y ^ num__2 ) = num__0.709141274238 ( x ^ num__2 / y ^ num__2 ) = ( num__16 * num__16 * num__6 ) / ( num__19 * num__19 * num__6 ) ( x ^ num__2 / y ^ num__2 ) = num__17 ^ num__0.105263157895 ^ num__2 x / y = num__0.842105263158 ans = e <eor> e <eos> |
e |
rectangle_perimeter__2.0__6.0__ |
rectangle_perimeter__2.0__6.0__ |
| you collect balls . suppose you start out with num__7 . john takes half of one more than the number of balls you have . since his mother makes balls peter decides to triple your balls . how many balls do you have at the end ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
solution start with num__7 balls . john takes half of one more than the number of balls you have . so john takes half of num__7 + num__1 which is num__4 so you ' re left with num__7 - num__4 = num__3 . peter triples the number of balls you have : num__3 × num__3 = num__9 balls . so you have num__9 at the end . correct answer : d <eor> d <eos> |
d |
triple__1.0__ triple__3.0__ triple__3.0__ |
subtract__7.0__4.0__ triple__3.0__ triple__3.0__ |
| a circular rim num__21 inches in diameter rotates the same number of inches per second as a circular rim num__35 inches in diameter . if the smaller rim makes x revolutions per second how many revolutions per minute does the larger rim makes in terms of x ? <o> a ) num__48 pi / x <o> b ) num__75 x <o> c ) num__48 x <o> d ) num__45 x <o> e ) x / num__75 |
let ' s try the explanation . we have two wheels . one with num__21 pi and the other one with num__35 pi . they have the same speed . in the smaller wheel it ' s num__21 pi * x which must be equal to the speed of the bigger one ( num__35 pi * a number of revolutions ) . they are asking that number of revolutions ( but in minutes which makes the question even harder ) . anyway we have num__21 pi * x = num__35 pi * a . ( num__21 pi * x ) / ( num__35 pi ) . as i said that ' s in seconds . so to convert it to minutes we multiply by num__60 and we get the result num__45 x . ans : d <eor> d <eos> |
d |
hour_to_min_conversion__ round__45.0__ |
hour_to_min_conversion__ round__45.0__ |
| a group of men decided to do a work in num__20 days but num__10 of them became absent . if the rest of the group did the work in num__40 days find the original number of men ? <o> a ) num__20 <o> b ) num__50 <o> c ) num__40 <o> d ) num__100 <o> e ) num__25 |
original number of men = num__10 * num__40 / ( num__40 - num__20 ) = num__20 answer is a <eor> a <eos> |
a |
round__20.0__ |
subtract__40.0__20.0__ |
| two cubes have their volumes in the ratio num__1 : num__27 . find the ratio of their surface areas . <o> a ) num__1 : num__9 <o> b ) num__2 : num__9 <o> c ) num__3 : num__9 <o> d ) num__5 : num__9 <o> e ) none of them |
let their edges be a and b . then = a ^ num__3 / b ^ num__3 = num__0.037037037037 ( or ) ( a / b ) ^ num__3 = ( num__0.333333333333 ) num__3 ( or ) ( a / b ) = ( num__0.333333333333 ) . therefore ratio of their surface area = num__6 a ^ num__0.333333333333 b ^ num__2 = a ^ num__2 / b ^ num__2 = ( a / b ) ^ num__2 = num__0.111111111111 i . e . num__1 : num__9 . answer is a . <eor> a <eos> |
a |
surface_cube__1.0__ power__0.3333__2.0__ power__3.0__2.0__ volume_cube__1.0__ |
surface_cube__1.0__ power__0.3333__2.0__ power__3.0__2.0__ power__1.0__2.0__ |
| the average age of num__15 students of a class is num__15 years . out of these the average age of num__5 students is num__14 years and that of the other num__9 students is num__16 years the age of the num__15 th student is <o> a ) num__11 <o> b ) num__12 <o> c ) num__13 <o> d ) num__14 <o> e ) num__15 |
explanation : age of the num__15 th student = [ num__15 * num__15 - ( num__14 * num__5 + num__16 * num__9 ) ] = ( num__225 - num__214 ) = num__11 years . answer : a <eor> a <eos> |
a |
subtract__16.0__5.0__ subtract__16.0__5.0__ |
subtract__16.0__5.0__ subtract__16.0__5.0__ |
| the length of an edge of a hollow cube open at one face is √ num__3 metres . what is the length of the largest pole that it can accommodate ? <o> a ) num__2 metres <o> b ) num__3 metres <o> c ) num__4 metres <o> d ) num__3 √ num__3 metres <o> e ) none |
solution required length = diagonal ‹ = › √ num__3 a ‹ = › ( √ num__3 × √ num__3 ) m ‹ = › num__3 m . answer b <eor> b <eos> |
b |
round__3.0__ |
round__3.0__ |
| the volume of cube is equal to the surface area of that cube . then find the distance of side of the cube ? <o> a ) num__7 <o> b ) num__10 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
cube volume = a ( power ) num__3 cubic units surface area = num__6 a ( power ) num__2 sq . units a ( power ) num__3 = num__6 a ( power ) num__2 a = num__6 m answer is d . <eor> d <eos> |
d |
multiply__2.0__3.0__ |
multiply__2.0__3.0__ |
| how long does a train num__120 m long running at the speed of num__70 km / hr takes to cross a bridge num__145 m length ? <o> a ) num__13.9 sec <o> b ) num__12.1 sec <o> c ) num__17.9 sec <o> d ) num__13.9 sec <o> e ) num__47.98 sec |
speed = num__70 * num__0.277777777778 = num__19 m / sec total distance covered = num__120 + num__145 = num__265 m . required time = num__13.9473684211 = num__13.9 sec . answer : d <eor> d <eos> |
d |
add__120.0__145.0__ divide__265.0__19.0__ round__13.9__ |
add__120.0__145.0__ divide__265.0__19.0__ round__13.9__ |
| a man misses a bus by num__40 minutes if he travels at num__30 kmph . if he travels at num__40 kmph then also he misses the bus by num__10 minutes . what is the minimum speed required to catch the bus on time ? <o> a ) num__11 <o> b ) num__45 <o> c ) num__66 <o> d ) num__72 <o> e ) num__32 |
let the distance to be travelled to catch the bus be x km x / num__30 - x / num__40 = num__0.5 = > ( num__4 x - num__3 x ) / num__120 = num__0.5 = > x = num__60 km by traavelling num__30 kmph time taken = num__2.0 = num__2 hours by taking num__2 hours he is late by num__40 min . so he has to cover num__60 km in at most speed = num__60 / ( num__1.33333333333 ) = num__45 kmph . answer : b <eor> b <eos> |
b |
divide__40.0__10.0__ divide__30.0__10.0__ multiply__40.0__3.0__ hour_to_min_conversion__ multiply__0.5__4.0__ divide__40.0__30.0__ round__45.0__ |
divide__40.0__10.0__ divide__30.0__10.0__ multiply__40.0__3.0__ divide__30.0__0.5__ divide__120.0__60.0__ divide__40.0__30.0__ round__45.0__ |
| a circular mat with diameter num__22 inches is placed on a square tabletop each of whose sides is num__24 inches long . which of the following is closest to the fraction of the tabletop covered by the mat ? <o> a ) num__0.416666666667 <o> b ) num__0.4 <o> c ) num__1.5 <o> d ) num__0.75 <o> e ) num__0.625 |
so we are looking for the area of the cloth over the area of the table area of the cloth = ( pi ) ( r ) ^ num__2 which is about ( num__3 ) ( num__11 ) ( num__11 ) area of the table = ( num__24 ) ( num__24 ) so the quick way to estimate is looking at the fraction like this : ( num__0.125 ) ( num__5.04166666667 ) i hope this is easy to follow so with some simplification i get ( num__0.125 ) ( num__5 ) = ( num__0.625 ) answer is e <eor> e <eos> |
e |
multiply__0.125__5.0__ multiply__0.125__5.0__ |
multiply__0.125__5.0__ multiply__0.125__5.0__ |
| there are num__28 stations between ernakulam and chennai . how many second class tickets have to be printed so that a passenger can travel from one station to any other station ? <o> a ) num__800 <o> b ) num__820 <o> c ) num__850 <o> d ) num__870 <o> e ) num__900 |
the total number of stations = num__30 from num__30 stations we have to choose any two stations and the direction of travel ( ernakulam to chennai is different from chennai to ernakulam ) in num__30 p num__2 ways . num__30 p num__2 = num__30 * num__29 = num__870 answer : d <eor> d <eos> |
d |
subtract__30.0__28.0__ multiply__29.0__30.0__ round__870.0__ |
subtract__30.0__28.0__ multiply__29.0__30.0__ multiply__29.0__30.0__ |
| the total marks obtained by a student in physics chemistry and mathematics is num__150 more than the marks obtained by him in physics . what is the average mark obtained by him in chemistry and mathematics ? <o> a ) num__75 <o> b ) num__267 <o> c ) num__299 <o> d ) num__266 <o> e ) num__21 |
let the marks obtained by the student in physics chemistry and mathematics be p c and m respectively . p + c + m = num__150 + p c + m = num__150 average mark obtained by the student in chemistry and mathematics = ( c + m ) / num__2 = num__75.0 = num__75 . answer : a <eor> a <eos> |
a |
divide__150.0__2.0__ divide__150.0__2.0__ |
divide__150.0__2.0__ divide__150.0__2.0__ |
| how many words can be formed by using all letters of the word ' pig ' ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__9 <o> d ) num__5 <o> e ) num__7 |
the word pig contains num__3 different letters required number of words = num__3 p num__3 = num__3 ! = num__3 * num__2 * num__1 = num__6 answer is a <eor> a <eos> |
a |
coin_space__ die_space__ die_space__ |
coin_space__ die_space__ die_space__ |
| if a man can cover num__12 metres in one second how many kilometres can he cover in num__3 hours num__45 minutes ? <o> a ) num__132 km <o> b ) num__167 km <o> c ) num__143 km <o> d ) num__162 kilometres <o> e ) num__245 km |
d num__12 m / s = num__12 * num__3.6 kmph num__3 hours num__45 minutes = num__3 num__0.75 hours = num__3.75 hours distance = speed * time = num__12 * num__3.6 * num__3.75 km = num__162 km . <eor> d <eos> |
d |
add__3.0__0.75__ multiply__45.0__3.6__ round__162.0__ |
divide__45.0__12.0__ multiply__45.0__3.6__ multiply__45.0__3.6__ |
| if x and y are integers such that | y + num__3 | ≤ num__3 and num__2 y – num__3 x + num__6 = num__0 what is the least possible value r of the product xy ? <o> a ) - num__12 <o> b ) - num__3 <o> c ) num__0 <o> d ) num__2 <o> e ) none of the above |
how to deal with inequalities involving absolute values ? first example shows us the so callednumber case in this case we have | y + num__3 | ≤ num__3 which is generalized | something | ≤ some number . first we solve as if there were no absolute value brackets : y + num__3 ≤ num__3 y ≤ num__0 so y is num__0 or negative second scenario - remove the absolute value brackets . put a negative sign around the other side of the inequality andflip the sign : y + num__3 > = - num__3 y > = - num__6 therefore we have a possible range for y : - num__6 = < y < = num__0 ok so far so good we ' re half way through . what about x ? here ' s the formula : num__2 y – num__3 x + num__6 = num__0 rewrite it as num__2 y + num__6 = num__3 x . you can say that num__2 y + num__6 is a multiple of num__3 ( = num__3 x ) . so all values which must be integer must also satisfy this constraint . i ' m just saying that so it ' s easier to evaluate all the possible numbers ( - num__6 - num__3 num__0 ) . if you plug in y = num__0 x will be num__2 and xy = num__0 as the lowest possible value r . hence answer choice c is the one to go . <eor> c <eos> |
c |
multiply__3.0__0.0__ |
multiply__3.0__0.0__ |
| a cube is divided into num__512 identical cubelets . each cut is made parallel to some surface of the cube . but before doing that the cube is painted with green on one set of opposite faces red on another set of opposite faces and blue on the third set of opposite faces . how many cubelets are painted with exactly one colour ? <o> a ) num__216 <o> b ) num__264 <o> c ) num__296 <o> d ) num__312 <o> e ) num__324 |
each face of the cube has num__8 x num__8 = num__64 cubelets . on each face only the interior cubelets are painted one colour . on each side num__6 x num__6 = num__36 cubelets are painted one colour . since the cube has six sides the number of cubes with one colour is num__6 * num__36 = num__216 the answer is a . <eor> a <eos> |
a |
volume_cube__6.0__ volume_cube__6.0__ |
multiply__36.0__6.0__ multiply__36.0__6.0__ |
| a sum of rs num__12500 amounts to rs . num__15500 in the num__4 years at the rate of simple interest . find the rate percent <o> a ) num__6.0 <o> b ) num__7.0 <o> c ) num__8.0 <o> d ) num__9.0 <o> e ) num__10 % |
explanation : s . i . = p ∗ r ∗ t / num__100 = > r = s . i . ∗ num__100 / p ∗ t so s . i = num__15500 - num__12500 = num__3000 . = > r = num__3000 ∗ num__0.008 ∗ num__4 = num__6.0 option a <eor> a <eos> |
a |
percent__100.0__6.0__ |
percent__100.0__6.0__ |
| a luxury liner queen marry ii is transporting several cats as well as the crew ( sailors a cook and one - legged captain ) to a nearby port . altogether these passengers have num__14 heads and num__41 legs . how many cats does the ship host ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
sa ' s + co + ca + cats = num__14 . sa ' s + num__1 + num__1 + cats = num__14 or sa ' s + cats = num__12 . sa ' s ( num__2 ) + num__2 + num__1 + cats * num__4 = num__41 sa ' s * num__2 + cats * num__4 = num__38 or sa ' s + cats * num__2 = num__19 or num__12 - cats + cat * num__2 = num__19 then cats = num__7 c <eor> c <eos> |
c |
subtract__14.0__12.0__ divide__38.0__2.0__ divide__14.0__2.0__ divide__14.0__2.0__ |
subtract__14.0__12.0__ divide__38.0__2.0__ subtract__19.0__12.0__ subtract__14.0__7.0__ |
| if i earn a profit num__50 $ in a day and i have num__9 employees . after keeping num__10 percent profit for me and distribute remianing money equally among my employees then how much each employee would get ? <o> a ) $ num__3 <o> b ) $ num__7 <o> c ) $ num__1 <o> d ) $ num__5 <o> e ) $ num__9 |
since i earned $ num__50 sp num__10 percent would be num__5 $ and num__45 $ would remain . dividing num__45 by num__9 each employee would get num__5 $ each and hince option d is correct . <eor> d <eos> |
d |
percent__50.0__10.0__ percent__50.0__10.0__ |
percent__50.0__10.0__ percent__50.0__10.0__ |
| a box contains six bulbs out of which num__4 are defective . if four bulbs are chosen at random find the probability that atleast one bulb is good . <o> a ) num__0.748502994012 <o> b ) num__0.933333333333 <o> c ) num__0.992063492063 <o> d ) num__0.672043010753 <o> e ) num__0.9765625 |
required probability = num__1 - num__0.0666666666667 = num__0.933333333333 answer : b <eor> b <eos> |
b |
negate_prob__0.0667__ negate_prob__0.0667__ |
negate_prob__0.0667__ negate_prob__0.0667__ |
| one day a car rental agency rented num__0.75 of its cars including num__0.6 of its cars with cd players . if num__0.6 of its cars have cd players what percent of the cars that were not rented had cd players ? <o> a ) num__0.966666666667 <o> b ) num__0.96 <o> c ) num__0.95 <o> d ) num__0.933333333333 <o> e ) num__0.9 |
the cars with cd players which were not rented is ( num__0.4 ) ( num__0.6 ) = num__0.24 of all the cars . the cars which were not rented is num__0.25 of all the cars . the percent of non - rented cars which had cd players is ( num__0.24 ) / ( num__0.25 ) = num__0.96 the answer is b . <eor> b <eos> |
b |
multiply__0.6__0.4__ divide__0.24__0.25__ divide__0.24__0.25__ |
multiply__0.6__0.4__ divide__0.24__0.25__ divide__0.24__0.25__ |
| in a certain sequence the first term is num__3 and each successive term is num__1 more than the reciprocal of the term that immediately precedes it . what is the fifth term in this sequence ? <o> a ) num__1.625 <o> b ) num__1.61538461538 <o> c ) num__1.6 <o> d ) num__0.625 <o> e ) num__1.63636363636 |
let five terms in the sequence be a b c d e a = num__3 b = num__1 + num__0.333333333333 = num__1.33333333333 c = num__1 + num__0.75 = num__1.75 d = num__1 + num__0.571428571429 = num__1.57142857143 e = num__1 + num__0.636363636364 = num__1.63636363636 hence answer should be e . <eor> e <eos> |
e |
reverse__3.0__ add__1.0__0.3333__ reverse__1.3333__ add__1.0__0.75__ reverse__1.75__ add__1.0__0.5714__ reverse__1.5714__ add__1.0__0.6364__ add__1.0__0.6364__ |
reverse__3.0__ add__1.0__0.3333__ reverse__1.3333__ add__1.0__0.75__ reverse__1.75__ add__1.0__0.5714__ reverse__1.5714__ add__1.0__0.6364__ add__1.0__0.6364__ |
| mrs . napier has num__23 stickers to give to num__9 students for a reward . how many stickers will each student get ? will there be any stickers left over ? <o> a ) num__2 - num__5 <o> b ) num__2 - num__4 <o> c ) num__2 - num__6 <o> d ) num__2 - num__3 <o> e ) num__2 - num__8 |
num__2.55555555556 = num__2 r num__5 mrs . napier will give each student num__2 stickers and there will be num__5 left over . correct answer a <eor> a <eos> |
a |
divide__23.0__9.0__ round_down__2.5556__ round_down__2.5556__ |
divide__23.0__9.0__ round_down__2.5556__ round_down__2.5556__ |
| a rectangular garden is num__12 m by num__5 m what is its area ? <o> a ) num__12 square meters <o> b ) num__5 square meters <o> c ) num__44 square meters <o> d ) num__60 square meters <o> e ) num__22 square meters |
area of a rectangle : a = w × h d ) num__60 square meters <eor> d <eos> |
d |
multiply__12.0__5.0__ multiply__12.0__5.0__ |
multiply__12.0__5.0__ multiply__12.0__5.0__ |
| if the true discount on s sum due num__2 years hence at num__14.0 per annum be rs . num__140 the sum due is : <o> a ) s . num__768 <o> b ) s . num__640 <o> c ) s . num__1960 <o> d ) s . num__2400 <o> e ) s . num__2800 |
td = pw * r * t / num__100 so num__140 = pw * num__14 * num__0.02 so pw = num__500 sum = pw + td . . sum = num__500 + num__140 = num__640 answer : b <eor> b <eos> |
b |
percent__100.0__640.0__ |
percent__100.0__640.0__ |
| souju ' s age is num__131.0 of what it was num__10 years ago but num__80 num__0.666666666667 % of what it will be after num__10 years . what is her present age ? <o> a ) num__46 years <o> b ) num__27 years <o> c ) num__36 years <o> d ) num__42 years <o> e ) num__54 years |
let the age before num__10 years = x . then num__131 x / num__100 = x + num__10 â ‡ ’ num__131 x = num__100 x + num__1000 â ‡ ’ x = num__32.2580645161 = num__32 present age = x + num__10 = num__32 + num__10 = num__42 answer : d <eor> d <eos> |
d |
multiply__10.0__100.0__ round_down__32.2581__ add__10.0__32.0__ add__10.0__32.0__ |
multiply__10.0__100.0__ round_down__32.2581__ add__10.0__32.0__ add__10.0__32.0__ |
| by selling num__175 apples a shopkeeper gains the price of num__25 apples . his gain per cent is <o> a ) num__18.66 <o> b ) num__17.66 <o> c ) num__16.66 <o> d ) num__15.66 <o> e ) num__14.66 % |
suppose s . p of num__175 apples = rs num__175 therefore gain = rs num__25 = > c . p = rs num__150 gain % = num__0.166666666667 x num__100 = num__16.66 gain = num__16.66 answer : c <eor> c <eos> |
c |
percent__100.0__16.66__ |
percent__100.0__16.66__ |
| a grocer has a sale of rs . num__4435 rs . num__4927 rs . num__4855 rs . num__5230 and rs . num__4562 for num__5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . num__4500 ? <o> a ) s . num__1991 <o> b ) s . num__2991 <o> c ) s . num__3991 <o> d ) s . num__4991 <o> e ) s . num__5991 |
explanation : total sale for num__5 months = rs . ( num__4435 + num__4927 + num__4855 + num__5230 + num__4562 ) = rs . num__24009 . required sale = rs . [ ( num__4500 x num__6 ) â € “ num__24009 ] = rs . ( num__27000 â € “ num__24009 ) = rs . num__2991 . answer b <eor> b <eos> |
b |
multiply__4500.0__6.0__ subtract__27000.0__24009.0__ subtract__27000.0__24009.0__ |
multiply__4500.0__6.0__ subtract__27000.0__24009.0__ subtract__27000.0__24009.0__ |
| a straight line in the xy - plane has a slope of num__2 and a y - intercept of num__2 . on this line what is the x - coordinate of the point whose y - coordinate is num__480 ? <o> a ) num__249 <o> b ) num__498 <o> c ) num__239 <o> d ) num__823 <o> e ) num__1 |
002 |
slope of num__2 and a y - intercept of num__2 y - coordinate is num__480 y = num__2 x + num__2 num__478 = num__2 x x = num__239 answer : c . num__239 <eor> c <eos> |
c |
c |
| a man can row a boat at num__20 kmph in still water . if the speed of the stream is num__8 kmph what is the time taken to row a distance of num__60 km downstream ? <o> a ) num__1.875 hours <o> b ) num__6.15384615385 hours <o> c ) num__0.41095890411 hours <o> d ) num__2.14285714286 hours <o> e ) num__3.84615384615 hours |
speed downstream = num__20 + num__8 = num__28 kmph . time required to cover num__60 km downstream = d / s = num__2.14285714286 = num__2.14285714286 hours . answer : d <eor> d <eos> |
d |
add__20.0__8.0__ divide__60.0__28.0__ divide__60.0__28.0__ |
add__20.0__8.0__ divide__60.0__28.0__ divide__60.0__28.0__ |
| a train num__108 m long moving at a speed of num__50 km / hr crosses a train num__112 m long coming from opposite direction in num__6 seconds . the speed of the second train is : <o> a ) num__48 km / hr <o> b ) num__54 km / hr <o> c ) num__66 km / hr <o> d ) num__82 km / hr <o> e ) num__85 km / hr |
explanation : let the speed of the second train be x km / hr . relative speed = ( x + num__50 ) km / hr = ( x + num__50 ) x num__0.277777777778 m / sec = num__250 + num__5 x / num__18 m / sec distance covered = ( num__108 + num__112 ) = num__220 m . num__220 / [ ( num__250 + num__5 x ) / num__18 ] = num__6 num__250 + num__5 x = num__660 x = num__82 km / hr . answer is d <eor> d <eos> |
d |
divide__250.0__50.0__ divide__108.0__6.0__ add__108.0__112.0__ round__82.0__ |
divide__250.0__50.0__ divide__108.0__6.0__ add__108.0__112.0__ round__82.0__ |
| if each side of a square is increased by num__25.0 find the percentage change in its area . <o> a ) num__26.5 <o> b ) num__36.25 <o> c ) num__46.25 <o> d ) num__56.25 <o> e ) num__66.25 % |
let each side of the square be a . then area = a num__2 . new side = ( num__125 a / num__100 ) = ( num__5 a / num__4 ) . new area = ( num__5 a / num__4 ) num__2 = ( num__25 a num__2 ) / num__16 . increase in area = ( ( num__25 a num__2 ) / num__16 ) - a num__2 = ( num__9 a num__2 ) / num__16 . increase % = [ ( ( num__9 a num__2 ) / num__16 ) * ( num__1 / a num__2 ) * num__100 ] % = num__56.25 . answer d num__56.25 <eor> d <eos> |
d |
percent__25.0__4.0__ percent__56.25__100.0__ |
percent__25.0__4.0__ percent__56.25__100.0__ |
| in a camp there is a meal for num__120 men or num__200 children . if num__150 children have taken the meal how many men will be catered to with remaining meal ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__40 <o> d ) num__50 <o> e ) num__60 |
here given num__120 men = num__200 children therefore num__1 child = num__0.6 of man - - - - - - step i now given that the food was consumed by num__150 children from step i we get num__150 children = num__90 men ( num__1 child = num__0.6 man ie num__150 x num__0.6 = num__90 ) so the food is for ( num__120 - num__90 ) men = num__30 men answer : b <eor> b <eos> |
b |
divide__120.0__200.0__ multiply__150.0__0.6__ subtract__120.0__90.0__ subtract__120.0__90.0__ |
divide__120.0__200.0__ multiply__150.0__0.6__ subtract__120.0__90.0__ subtract__120.0__90.0__ |
| a rectangular circuit board is designed to have a width of w inches a length of l inches a perimeter of p inches and an area of d square inches . which of the following equations must be true ? <o> a ) num__2 w ^ num__2 + pw + num__2 d = num__0 <o> b ) num__2 w ^ num__2 − pw − num__2 d = num__0 <o> c ) num__2 w ^ num__2 − pw + num__2 d = num__0 <o> d ) w ^ num__2 + pw + d = num__0 <o> e ) w ^ num__2 − pw + num__2 d = num__0 |
p = num__2 ( l + w ) - - - - - - - - - - - - - - - - - num__1 ) a = lw - - - - - - - - - - - - - - - - - - - - - - - - num__2 ) option a is not possible why ? ? because all the terms are positive . lets try option b put value of p and a from num__1 and num__2 we have num__2 w ^ num__2 - num__2 ( l + w ) w + num__2 ( lw ) num__2 w ^ num__2 - num__2 lw - num__2 w ^ num__2 + num__2 lw = num__0 . hence answer is c . <eor> c <eos> |
c |
rectangle_perimeter__0.0__1.0__ |
power__2.0__1.0__ |
| rs . num__2500 is divided into two parts such that if one part be put out at num__5.0 simple interest and the other at num__6.0 the yearly annual income may be rs . num__140 . how much was lent at num__5.0 ? <o> a ) num__2888 <o> b ) num__2999 <o> c ) num__2799 <o> d ) num__1000 <o> e ) num__2881 |
( x * num__5 * num__1 ) / num__100 + [ ( num__2500 - x ) * num__6 * num__1 ] / num__100 = num__140 x = num__1000 answer : d <eor> d <eos> |
d |
percent__100.0__1000.0__ |
percent__100.0__1000.0__ |
| when a cylindrical tank is filled with water at a rate of num__22 cubic meters per hour the level of water in the tank rises at a rate of num__0.7 meters per hour . which of the following best approximates the radius of the tank in meters ? <o> a ) √ num__5.0 <o> b ) √ num__10 <o> c ) num__4 <o> d ) num__5 <o> e ) num__10 |
v = pi ( r ^ num__2 ) ( h ) we have the volume ( num__22 cubic meters ) and the height ( num__0.7 meters ) . . . num__22 = pi ( r ^ num__2 ) ( num__0.7 ) pi = about num__3.14 . num__7 ( pi ) = about num__2.2 num__22 = num__2.2 ( r ^ num__2 ) num__10 = r ^ num__2 \ sqrt { num__10 } = r final answer : b <eor> b <eos> |
b |
divide__22.0__2.2__ round__10.0__ |
divide__22.0__2.2__ round__10.0__ |
| what is greatest positive integer n such that num__2 ^ n is a factor of num__12 ^ num__14 ? <o> a ) a ) num__10 <o> b ) b ) num__12 <o> c ) c ) num__16 <o> d ) d ) num__20 <o> e ) e ) num__28 |
the given number is num__12 ^ num__14 = ( num__2 * num__2 * num__3 ) ^ num__14 = ( num__2 ^ num__28 ) * ( num__3 ^ num__14 ) so the greatest possible value for n such that num__2 ^ n can be factor of given number is num__28 . answer e <eor> e <eos> |
e |
multiply__2.0__14.0__ multiply__2.0__14.0__ |
multiply__2.0__14.0__ multiply__2.0__14.0__ |
| can n and can В are both right circular cylinders . the radius of can n is twice the radius of can b while the height of can n is half the height of can b . if it costs $ num__4.00 to fill half of can b with a certain brand of gasoline how much would it cost to completely fill can n with the same brand of gasoline ? <o> a ) $ num__1 <o> b ) $ num__2 <o> c ) $ num__4 <o> d ) $ num__8 <o> e ) $ num__16 |
let x be the radius of b and num__2 h be the height of b . therefore radius of n = num__2 x and height = h vol of b = num__3.14 * x ^ num__2 * num__2 h vol of a = num__3.14 * num__4 x ^ num__2 * h cost to fill half of b = $ num__4 - - > cost to fill full b = $ num__8 - - > num__3.14 * x ^ num__2 * num__2 h = num__8 - - > num__3.14 * x ^ num__2 * h = num__4 - - > num__4 * ( num__3.14 * x ^ num__2 * h ) = $ num__16 ans e <eor> e <eos> |
e |
square_perimeter__2.0__ square_perimeter__4.0__ square_perimeter__4.0__ |
multiply__4.0__2.0__ power__4.0__2.0__ power__4.0__2.0__ |
| if the wheel is num__14 cm then the number of revolutions to cover a distance of num__880 cm is ? <o> a ) a ) num__15 <o> b ) b ) num__10 <o> c ) c ) num__14 <o> d ) d ) num__12 <o> e ) e ) num__11 |
num__2 * num__3.14285714286 * num__14 * x = num__880 = > x = num__10 answer : b <eor> b <eos> |
b |
round__10.0__ |
round__10.0__ |
| a desalination process is able to remove num__0.666666666667 of salt in each cycle . if we have num__1 liter of sea water that has num__5.0 salt ( i . e . num__5 g in a liter ) how much salt would have been removed from the liter of water after num__3 cycles of the desalination process ? <o> a ) num__4.15 <o> b ) num__4.21 <o> c ) num__4.52 <o> d ) num__4.81 <o> e ) num__4.93 |
x is the initial quantity of salt in the water . after a cycle of desalination the quantity of salt left would be x num__1 = x - num__0.666666666667 x i . e . x num__1 = num__0.333333333333 x same thing for x num__2 x num__3 respectively representing the quantity of salt left after num__2 num__3 cycles of desalination . so we have : x num__1 = num__0.333333333333 x x num__2 = num__0.333333333333 x num__1 x num__3 = num__0.333333333333 x num__2 by multiplying both ends of the equations we get : x num__1 * x num__2 * x num__3 = ( num__0.333333333333 ) ^ num__3 x * x num__1 * x num__2 simplification leaves us with : x num__3 = ( num__0.333333333333 ) ^ num__3 x x num__3 is the quantity of salt left after three cycles . so the cumulative quantity of salt ( q num__3 ) which has been removed after num__3 cycles is : q num__3 = x - x num__3 = x - ( num__0.333333333333 ) ^ num__3 x i . e . q num__3 = x ( num__1 - ( num__0.333333333333 ) ^ num__3 ) in this case given x was : x = num__5 g so we have q num__3 = num__5 * ( num__1 - num__0.037037037037 ) = num__5 * num__0.962962962963 = num__4.814 g rounded q num__3 = num__4.81 answer : d <eor> d <eos> |
d |
reverse__3.0__ subtract__5.0__3.0__ subtract__1.0__0.037__ multiply__1.0__4.81__ |
subtract__1.0__0.6667__ subtract__5.0__3.0__ subtract__1.0__0.037__ multiply__1.0__4.81__ |
| two trains are moving in opposite directions with speed of num__110 km / hr and num__90 km / hr respectively . their lengths are num__1.10 km and num__0.9 km respectively . the slower train cross the faster train in - - - seconds <o> a ) num__26 <o> b ) num__48 <o> c ) num__47 <o> d ) num__36 <o> e ) num__25 |
explanation : relative speed = num__110 + num__90 = num__200 km / hr ( since both trains are moving in opposite directions ) total distance = num__1.1 + . num__9 = num__2 km time = num__0.01 hr = num__0.01 hr = num__36.0 = num__36 seconds answer : option d <eor> d <eos> |
d |
add__110.0__90.0__ add__1.1__0.9__ divide__1.1__110.0__ round__36.0__ |
add__110.0__90.0__ add__1.1__0.9__ divide__1.1__110.0__ round__36.0__ |
| a father said to his son ` ` i was as old as you are at present at the time of your birth . ' ' if the father ' s age is num__42 years now the son ' s age num__5 years back was ? <o> a ) num__12 yr <o> b ) num__15 yr <o> c ) num__14 yr <o> d ) num__16 yr <o> e ) num__20 yr |
let the son ' s present age be x years then num__42 - x = x x = num__21 son ' s age num__5 years back = num__21 - num__5 = num__16 years answer is d <eor> d <eos> |
d |
subtract__21.0__5.0__ subtract__21.0__5.0__ |
subtract__21.0__5.0__ subtract__21.0__5.0__ |
| the sum of the present ages of two persons a and b is num__60 . if the age of a is twice that of b find the sum of their ages num__5 years hence ? <o> a ) num__100 <o> b ) num__80 <o> c ) num__70 <o> d ) num__20 <o> e ) num__40 |
a + b = num__60 a = num__2 b num__2 b + b = num__60 = > b = num__20 then a = num__40 . num__5 years their ages will be num__45 and num__25 . sum of their ages = num__45 + num__25 = num__70 . answer is c . <eor> c <eos> |
c |
subtract__60.0__20.0__ add__5.0__40.0__ add__5.0__20.0__ add__45.0__25.0__ add__45.0__25.0__ |
subtract__60.0__20.0__ add__5.0__40.0__ add__5.0__20.0__ add__45.0__25.0__ add__45.0__25.0__ |
| a person crosses a num__600 m long street in num__4 minnutes . what is his speed in km per hour ? <o> a ) num__5.8 km / hr <o> b ) num__7.2 km / hr <o> c ) num__9 km / hr <o> d ) num__2.5 km / hr <o> e ) num__3 km / hr |
speed = num__150.0 * num__60 = num__2.5 m / sec = num__2.5 * num__3.6 = num__9 km / hr answer is c <eor> c <eos> |
c |
divide__600.0__4.0__ hour_to_min_conversion__ divide__150.0__60.0__ multiply__2.5__3.6__ round__9.0__ |
divide__600.0__4.0__ hour_to_min_conversion__ divide__150.0__60.0__ multiply__2.5__3.6__ multiply__2.5__3.6__ |
| a man rows his boat num__90 km downstream and num__5 ` km upstream taking num__3 hours each time . find the speed of the stream ? <o> a ) num__7 kmph <o> b ) num__5 kmph <o> c ) num__2 kmph <o> d ) num__8 kmph <o> e ) num__1 kmph |
speed downstream = d / t = num__90 / ( num__3 ) = num__30 kmph speed upstream = d / t = num__51 / ( num__3 ) = num__17 kmph the speed of the stream = ( num__30 - num__17 ) / num__2 = num__7 kmph answer : a <eor> a <eos> |
a |
divide__90.0__3.0__ divide__51.0__3.0__ subtract__5.0__3.0__ add__5.0__2.0__ round__7.0__ |
divide__90.0__3.0__ divide__51.0__3.0__ subtract__5.0__3.0__ add__5.0__2.0__ round__7.0__ |
| two persons start running simultaneously around a circular track of length num__200 m from the same point at speeds of num__15 km / hr and num__25 km / hr . when will they meet for the first time any where on the track if they are moving in opposite directions ? <o> a ) num__11 <o> b ) num__10 <o> c ) num__18 <o> d ) num__27 <o> e ) num__12 |
time taken to meet for the first time anywhere on the track = length of the track / relative speed = num__200 / ( num__15 + num__25 ) num__0.277777777778 = num__200 * num__0.45 * num__5 = num__18 seconds . answer : c <eor> c <eos> |
c |
round__18.0__ |
round__18.0__ |
| if you drew a dot on the edge of a wheel and traced the path of the dot as the wheel rolled one complete revolution along a line then the path formed would be called a cycloid combining both forward and circular motion . what is the length of the path formed by one complete revolution ? assume the wheel has a radius of r . <o> a ) num__5 r <o> b ) num__7 r <o> c ) num__4 r <o> d ) num__8 r <o> e ) num__9 r |
imagine the circle resting on coordinate ( num__00 ) and moving east . also magine the point on the circle to start at ( num__00 ) . let t be the angle between the point on the circle and the center of the circle . the position of the point on the circle relative to t is : x = rt - r × sin ( t ) y = r - r × cos ( t ) taking the derivatives : dx / dt = r - r × cos ( t ) dy / dt = r × sin ( t ) the change in arc length can be defined as ( ( dx / dt ) num__2 + ( dy / dt ) num__2 ) num__0.5 . so the total arc length is the integral from num__0 to num__2 pi of ( ( dx / dt ) num__2 + ( dy / dt ) num__2 ) num__0.5 . after a few steps this integral becomes : r × num__10.5 × ( num__1 - cos ( t ) ) num__0.5 . using the hint : r × num__10.5 × num__10.5 × integral of sin ( t / num__2 ) dt from num__0 to num__2 × pi = num__2 × r × ( - num__2 × cos ( t / w ) from num__2 × pi to num__0 ) = num__8 r hint : ( num__1 - cos ( t ) ) num__0.5 = num__20.5 × sin ( t / num__2 ) <eor> d <eos> |
d |
rectangle_perimeter__0.0__0.5__ square_perimeter__2.0__ triangle_perimeter__2.0__10.5__8.0__ square_perimeter__2.0__ |
rectangle_perimeter__0.0__0.5__ square_perimeter__2.0__ triangle_perimeter__2.0__10.5__8.0__ square_perimeter__2.0__ |
| diane find num__6 and a half cans of paint are just enough to paint one third of her room . how many more cans of paint will she need to finish her room and paint a second room of the same size ? <o> a ) num__5 <o> b ) num__7 and a half <o> c ) num__10 <o> d ) num__32 and a half <o> e ) num__35 |
she will need num__13 cans to paint the rest of this room and num__19 num__0.5 for the next room for a total of num__32 num__0.5 cans . d <eor> d <eos> |
d |
add__6.0__13.0__ add__19.0__13.0__ round__32.0__ |
add__6.0__13.0__ add__19.0__13.0__ round__32.0__ |
| a restaurant meal cost $ num__31.50 and there was no tax . if the tip was more than num__10 percent but less than num__15 percent of the cost of the meal then total amount paid must have been between : <o> a ) $ num__40 and $ num__42 <o> b ) $ num__39 and $ num__41 <o> c ) $ num__38 and num__40 <o> d ) $ num__37 and $ num__39 <o> e ) $ num__35 and $ num__37 |
let tip = t meal cost = num__31.50 range of tip = from num__10.0 of num__31.5 to num__15.0 of num__35.5 = num__3.55 to num__5.325 hence range of amount paid = num__31.5 + t = num__35.05 to num__36.825 answer : e <eor> e <eos> |
e |
divide__35.5__10.0__ add__31.5__3.55__ add__31.5__5.325__ round_down__35.5__ |
divide__35.5__10.0__ add__31.5__3.55__ add__31.5__5.325__ round_down__35.5__ |
| a bullock cart has to cover a distance of num__80 km in num__10 hrs . if it covers half of the journey in num__0.6 th time . what should be its speed to cover the remaining distance in the time left . <o> a ) num__8 km / h <o> b ) num__10 km / h <o> c ) num__12 km / h <o> d ) num__14 km / h <o> e ) num__16 km / h |
time left = num__10 - num__0.6 * num__10 = num__4 hr num__10 km / h speed = num__40 km / num__4 hr = num__10 kmph answer : b . <eor> b <eos> |
b |
multiply__10.0__4.0__ round__10.0__ |
multiply__10.0__4.0__ divide__40.0__4.0__ |
| ada and paul received their scores on three tests . on the first test ada ' s score was num__10 points higher than paul ' s score . on the second test ada ' s score was num__4 points higher than paul ' s score . if paul ' s average ( arithmetic mean ) score on the three tests was num__5 points higher than ada ' s average score on the three tests then paul ' s score on the third test was how many points higher than ada ' s score ? <o> a ) num__9 <o> b ) num__14 <o> c ) num__17 <o> d ) num__29 <o> e ) num__25 |
my take is option d ( num__23 ) i followed a simple approach ( explained below ) : test num__1 : ada ' s score = paul ' s score + num__10 test num__2 : ada ' s score = paul ' s score + num__4 avg . of paul ' s score = num__5 points higher than avg . of ada ' s score this implies that : sum of paul ' s score [ num__3 tests ] = num__15 points higher than sum of ada ' s score [ num__3 tests ] ( num__9 points higher since num__3 points were given in terms of average of num__3 scores ) so paul needs to score num__29 points higher than ada in test num__3 since paul needs to compensate for the lower score in test num__1 and test num__2 ( num__29 = num__10 + num__4 + num__15 ) d <eor> d <eos> |
d |
subtract__5.0__4.0__ divide__10.0__5.0__ subtract__4.0__1.0__ add__10.0__5.0__ subtract__10.0__1.0__ multiply__1.0__29.0__ |
subtract__5.0__4.0__ divide__10.0__5.0__ add__1.0__2.0__ add__10.0__5.0__ add__4.0__5.0__ multiply__1.0__29.0__ |
| if n is a positive integer and the product of all integers from num__1 to n inclusive is a multiple of num__420 what is the least possible value of n ? <o> a ) num__5 <o> b ) num__7 <o> c ) num__9 <o> d ) num__11 <o> e ) num__12 |
num__420 = num__2 * num__2 * num__3 * num__5 * num__7 so n must be at least num__7 . the answer is b . <eor> b <eos> |
b |
add__1.0__2.0__ add__2.0__3.0__ add__2.0__5.0__ multiply__1.0__7.0__ |
add__1.0__2.0__ add__2.0__3.0__ add__2.0__5.0__ multiply__1.0__7.0__ |
| how many num__3 digit positive integers exist that when divided by num__7 leave a remainder of num__5 ? <o> a ) num__128 <o> b ) num__142 <o> c ) num__143 <o> d ) num__141 <o> e ) num__129 |
explanatory answer step num__1 : find the first and last term of the series the smallest num__3 - digit positive integer that leaves a remainder of num__5 when divided by num__7 is num__103 . the largest num__3 - digit positive integer that leaves a remainder of num__5 when divided by num__7 is num__999 . the series of numbers that satisfy the condition that the number should leave a remainder of num__5 when divided by num__7 is an a . p ( arithmetic progression ) with the first term being num__103 and the last term being num__999 . the common difference of the sequence is num__7 . step num__2 : compute the number of terms in an a . p the last term l = a + ( n - num__1 ) * d where ' a ' is the first term ' n ' is the number of terms of the series and ' d ' is the common difference . therefore num__999 = num__103 + ( n - num__1 ) * num__7 or num__999 - num__103 = ( n - num__1 ) * num__7 or num__896 = ( n - num__1 ) * num__7 so n - num__1 = num__128 or n = num__129 choice e is the correct answer . <eor> e <eos> |
e |
subtract__3.0__1.0__ subtract__999.0__103.0__ divide__896.0__7.0__ add__1.0__128.0__ multiply__1.0__129.0__ |
subtract__3.0__1.0__ subtract__999.0__103.0__ divide__896.0__7.0__ add__1.0__128.0__ multiply__1.0__129.0__ |
| a certain company that sells only cars and trucks reported that revenues from car sales in num__1997 were down num__11 percent from num__1996 and revenues from truck sales were up num__7 percent from num__1996 . if total revenues from car sales and truck sales in num__1997 were up num__1 percent from num__1996 what is the ratio q of revenue from car sales in num__1996 to revenue from truck sales in num__1996 ? <o> a ) num__1 : num__2 <o> b ) num__4 : num__5 <o> c ) num__1 : num__1 <o> d ) num__3 : num__2 <o> e ) num__5 : num__3 |
a . . i have probably solved this question num__3 - num__4 times by now . . remember the answer . . num__1 : num__2 <eor> a <eos> |
a |
subtract__11.0__7.0__ subtract__3.0__1.0__ reverse__1.0__ |
subtract__11.0__7.0__ subtract__3.0__1.0__ subtract__1997.0__1996.0__ |
| if p and q are prime numbers how many divisors does the product p ^ num__3 * q ^ num__7 have ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__32 <o> d ) num__35 <o> e ) num__40 |
when a number n = a ^ x * b ^ y where a and b are prime numbers and x y are positive integers the number of divisors of n = ( x + num__1 ) ( y + num__1 ) therefore the answer is c . num__4 * num__8 = num__32 <eor> c <eos> |
c |
add__3.0__1.0__ add__7.0__1.0__ multiply__8.0__4.0__ multiply__8.0__4.0__ |
add__3.0__1.0__ add__7.0__1.0__ multiply__8.0__4.0__ multiply__8.0__4.0__ |
| in the land of oz only one or two - letter words are used . the local language has num__64 different letters . the parliament decided to forbid the use of the seventh letter . how many words have the people of oz lost because of the prohibition ? <o> a ) num__65 <o> b ) num__66 <o> c ) num__67 <o> d ) num__131 <o> e ) num__128 |
the answer to the question is indeed e . the problem with above solutions is that they do not consider words like aa bb . . . the number of num__1 letter words ( x ) that can be made from num__64 letters is num__64 ; the number of num__2 letter words ( xx ) that can be made from num__64 letters is num__64 * num__64 since each x can take num__64 values . total : num__64 + num__64 * num__64 . similarly : the number of num__1 letter words ( x ) that can be made from num__63 letters is num__63 ; the number of num__2 letter words ( xx ) that can be made from num__63 letters is num__63 * num__63 since each x can take num__63 values . total : num__63 + num__63 * num__63 . the difference is ( num__64 + num__64 * num__64 ) - ( num__63 + num__63 * num__63 ) = num__128 . answer : e . <eor> e <eos> |
e |
subtract__64.0__1.0__ multiply__64.0__2.0__ multiply__64.0__2.0__ |
subtract__64.0__1.0__ multiply__64.0__2.0__ multiply__64.0__2.0__ |
| two trains a and b starting from two points and travelling in opposite directions reach their destinations num__9 hours and num__4 hours respectively after meeting each other . if the train a travels at num__70 kmph find the rate at which the train b runs . <o> a ) num__40 <o> b ) num__60 <o> c ) num__120 <o> d ) num__105 <o> e ) num__100 |
if two objects a and b start simultaneously from opposite points and after meeting reach their destinations in ‘ a ’ and ‘ b ’ hours respectively ( i . e . a takes ‘ a hrs ’ to travel from the meeting point to his destination and b takes ‘ b hrs ’ to travel from the meeting point to his destination ) then the ratio of their speeds is given by : sa / sb = √ ( b / a ) i . e . ratio of speeds is given by the square root of the inverse ratio of time taken . sa / sb = √ ( num__0.444444444444 ) = num__0.666666666667 this gives us that the ratio of the speed of a : speed of b as num__2 : num__3 . since speed of a is num__70 kmph speed of b must be num__70 * ( num__1.5 ) = num__105 kmph answer d <eor> d <eos> |
d |
divide__4.0__9.0__ divide__3.0__2.0__ multiply__70.0__1.5__ round__105.0__ |
divide__4.0__9.0__ divide__3.0__2.0__ multiply__70.0__1.5__ multiply__70.0__1.5__ |
| a monkey ascends a greased pole num__6 meters high . he ascends num__2 meters in the first minute and then slips down num__1 meter in the alternate minute . if this pattern continues until he climbs the pole in how many minutes would he reach at the top of the pole ? <o> a ) num__9 th minute <o> b ) num__21 st minute <o> c ) num__11 th minute <o> d ) num__22 nd minute <o> e ) num__13 th minute |
the money is climbing num__1 meter in num__2 min . this pattern will go on till he reaches num__4 meters . i mean this will continue for first num__4 * num__2 = num__8 mins . he would have reached num__4 meters . after that he will climb num__2 meters and he will reach the pole . so total time taken = num__8 + num__1 = num__9 mins . so asnwer will be a <eor> a <eos> |
a |
subtract__6.0__2.0__ add__6.0__2.0__ add__1.0__8.0__ round__9.0__ |
subtract__6.0__2.0__ add__6.0__2.0__ add__1.0__8.0__ add__1.0__8.0__ |
| raja ' s bank ' s saving amount is decreased num__40.0 due to loan payment and current balance is rs . num__30000 . find the actual balance before deduction ? <o> a ) num__8000 <o> b ) num__8500 <o> c ) num__50000 <o> d ) num__9500 <o> e ) num__10000 |
num__40.0 decreased num__60.0 balance = num__30000 num__100.0 = num__500.0 * num__100 = num__50000 answer : c <eor> c <eos> |
c |
percent__100.0__50000.0__ |
percent__100.0__50000.0__ |
| if each side of a square is increased by num__325.0 in length find the percentage change in its area ? <o> a ) num__54.0 <o> b ) num__957.0 <o> c ) num__956.25 <o> d ) num__958.25 <o> e ) num__959.50 % |
let each side of the square be a then area = a x a new side = num__325 a / num__100 = num__13 a / num__4 new area = ( num__13 a x num__13 a ) / ( num__4 x num__4 ) = ( num__25 a ² / num__16 ) increased area = = ( num__169 a ² / num__16 ) - a ² increase % = [ ( num__153 a ² / num__16 ) x ( num__1 / a ² ) x num__100 ] % = num__956.25 answer : c <eor> c <eos> |
c |
percent__4.0__25.0__ percent__100.0__956.25__ |
percent__4.0__25.0__ percent__100.0__956.25__ |
| the present population of a town is num__220 . population increase rate is num__10.0 p . a . find the population of town after num__1 years ? <o> a ) num__100 <o> b ) num__120 <o> c ) num__200 <o> d ) num__242 <o> e ) num__250 |
p = num__220 r = num__10.0 required population of town = p * ( num__1 + r / num__100 ) ^ t = num__220 * ( num__1 + num__0.1 ) = num__220 * ( num__1.1 ) = num__242 answer is d <eor> d <eos> |
d |
percent__10.0__1.0__ percent__100.0__242.0__ |
percent__10.0__1.0__ percent__100.0__242.0__ |
| the average age of a group of num__10 friends is num__15 yr . if one friend leaves the group the average becomes num__16 yr . find the age of the friend who left the group . <o> a ) num__5 yr <o> b ) num__4 yr <o> c ) num__6 yr <o> d ) none of these <o> e ) can not be determined |
answer age of friend left = previous average of the age - number of persons present x increase in average age = num__15 - ( num__10 - num__1 ) x ( num__16 - num__15 ) = num__15 - num__9 x num__1 = num__15 - num__9 = num__6 yr correct option : c <eor> c <eos> |
c |
subtract__16.0__15.0__ subtract__10.0__1.0__ subtract__15.0__9.0__ subtract__15.0__9.0__ |
subtract__16.0__15.0__ subtract__10.0__1.0__ subtract__15.0__9.0__ subtract__15.0__9.0__ |
| how many four - digit numbers that do not contain the digits num__2 num__3 or num__6 are there ? <o> a ) num__2058 <o> b ) num__3584 <o> c ) num__4096 <o> d ) num__5040 <o> e ) num__7200 |
the num__1 st digit can be filled up by the numbers : { num__14 num__57 num__89 } = num__6 ways the num__2 nd digit can be filled up by the numbers : { num__0 num__14 num__57 num__89 } = num__7 ways the num__3 rd digit can be filled up by the numbers : { num__0 num__14 num__57 num__89 } = num__7 ways the num__4 th digit can be filled up by the numbers : { num__0 num__14 num__57 num__89 } = num__7 ways the total number of such four - digit numbers is num__6 * num__7 * num__7 * num__7 = num__2058 the answer is a . <eor> a <eos> |
a |
subtract__3.0__2.0__ add__6.0__1.0__ add__3.0__1.0__ multiply__1.0__2058.0__ |
subtract__3.0__2.0__ add__6.0__1.0__ subtract__6.0__2.0__ multiply__1.0__2058.0__ |
| all of the citizens of a country have a seven - character or eight - character national identification code that is created using the num__26 letters of the alphabet and the num__10 digits from num__0 to num__9 . what is the maximum number of citizens who can be designated with these codes ? <o> a ) num__36 ^ num__15 <o> b ) num__35 ( num__36 ^ num__7 ) <o> c ) num__35 ( num__36 ^ num__8 ) <o> d ) num__37 ( num__36 ^ num__7 ) <o> e ) num__37 ( num__36 ^ num__8 ) |
the number of possible num__7 - digit codes is num__36 ^ num__7 . the number of possible num__8 - digit codes is num__36 ^ num__8 . the total sum is num__36 ^ num__7 + num__36 ^ num__8 = num__36 ^ num__7 ( num__1 + num__36 ) = num__37 ( num__36 ^ num__7 ) the answer is d . <eor> d <eos> |
d |
add__26.0__10.0__ subtract__10.0__9.0__ add__1.0__36.0__ add__1.0__36.0__ |
add__26.0__10.0__ subtract__10.0__9.0__ add__1.0__36.0__ add__1.0__36.0__ |
| a girl swims downstream num__112 km and upstream num__32 km taking num__8 hours each time what is the speed of the girl in still water ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
num__112 - - - num__8 ds = num__14 ? - - - - num__1 num__32 - - - - num__8 us = num__4 ? - - - - num__1 m = ? m = ( num__14 + num__4 ) / num__2 = num__9 answer : b <eor> b <eos> |
b |
divide__112.0__8.0__ divide__32.0__8.0__ divide__8.0__4.0__ add__8.0__1.0__ round__9.0__ |
divide__112.0__8.0__ divide__32.0__8.0__ divide__8.0__4.0__ add__8.0__1.0__ add__8.0__1.0__ |
| a train num__125 m long crosses a platform num__125 m long in num__14 sec ; find the speed of the train ? <o> a ) num__87 kmph <o> b ) num__64 kmph <o> c ) num__54 kmph <o> d ) num__16 kmph <o> e ) num__18 kmph |
d = num__125 + num__125 = num__250 t = num__14 s = num__17.8571428571 * num__3.6 = num__64 kmph answer : b <eor> b <eos> |
b |
divide__250.0__14.0__ round__64.0__ |
divide__250.0__14.0__ round__64.0__ |
| last year the price of a vacation package was p . at the beginning of this year the price went up num__30.0 . lucas used a travel voucher to purchase the vacation package at num__20.0 off this year ’ s price . in terms of p how much did lucas pay ? <o> a ) p + num__10 <o> b ) num__1.1 * p <o> c ) num__1.12 * p <o> d ) num__0.04 * p <o> e ) num__1.04 * p |
last year price = p ; this year price = num__1.3 p ; lucas used a travel voucher to purchase the vacation package at num__20.0 off this year ’ s price thus he paid ( num__1 - num__0.2 ) * num__1.3 p = num__1.04 p . answer : e . <eor> e <eos> |
e |
round_down__1.3__ multiply__1.04__1.0__ |
round_down__1.3__ multiply__1.04__1.0__ |
| the largest four - digit number which when divided by num__4 num__7 or num__13 leaves a remainder of num__3 in each case is : <o> a ) num__8739 <o> b ) num__9831 <o> c ) num__9834 <o> d ) num__9893 <o> e ) num__8976 |
solution greatest number of num__4 digits is num__9999 . l . c . m . of num__4 num__7 and num__13 = num__364 . on dividing num__9999 by num__364 remainder obtained is num__171 . so greatest number of num__4 digits divisible by num__47 and num__13 = ( num__9999 - num__171 ) = num__9828 . hence required number = ( num__9828 + num__3 ) = num__9831 . answer b <eor> b <eos> |
b |
subtract__9999.0__171.0__ add__3.0__9828.0__ add__3.0__9828.0__ |
subtract__9999.0__171.0__ add__3.0__9828.0__ add__3.0__9828.0__ |
| if the average ( arithmetic mean ) of a and b is num__120 and the average of b and c is num__150 what is the value of a − c ? <o> a ) − num__60 <o> b ) − num__100 <o> c ) num__100 <o> d ) num__135 <o> e ) it can not be determined from the information given |
a + b = num__240 b + c = num__300 a - c = - num__60 . imo option a . <eor> a <eos> |
a |
subtract__300.0__240.0__ subtract__120.0__60.0__ |
subtract__300.0__240.0__ subtract__120.0__60.0__ |
| a person can swim in still water at num__10 km / h . if the speed of water num__8 km / h how many hours will the man take to swim back against the current for num__16 km ? <o> a ) num__3 <o> b ) num__7 <o> c ) num__5 <o> d ) num__8 <o> e ) num__6 |
m = num__10 s = num__8 us = num__10 - num__8 = num__2 d = num__16 t = num__8.0 = num__8 answer : d <eor> d <eos> |
d |
subtract__10.0__8.0__ round__8.0__ |
subtract__10.0__8.0__ subtract__10.0__2.0__ |
| a polling company reports that there is a num__40.0 chance that a certain candidate will win the next election . if the candidate wins there is a num__60.0 chance that she will sign bill x and no other bills . if she decides not to sign bill x she will sign either bill y or bill z chosen randomly . what is the chance that the candidate will sign bill z ? <o> a ) num__10 <o> b ) num__8 <o> c ) num__6 <o> d ) num__4 <o> e ) num__5 |
fairly straight forward . . . num__40.0 - candidate elected num__100.0 - num__60.0 = num__40.0 - candidate does not sigh bill x num__50.0 - candidate randomly chooses between two bills . these are multiplicative : num__40.0 x num__40.0 x num__50.0 num__0.4 x num__0.4 x num__0.5 = num__0.08 = num__8.0 answer ( b ) <eor> b <eos> |
b |
percent__100.0__8.0__ |
percent__100.0__8.0__ |
| given that num__1 + num__3 x > num__4 and num__2 x - num__3 < num__5 all values of x must be between which of the following pairs of integers ? <o> a ) num__3 and num__8 <o> b ) num__1 and num__4 <o> c ) num__3 and num__12 <o> d ) num__1.33333333333 and num__2.5 <o> e ) - num__5 and num__1 |
given num__1 + num__3 x > num__4 and num__2 x - num__3 < num__5 i . e . num__3 x > num__3 and num__2 x < num__8 i . e . x > num__1 and x < num__4 answer : option b <eor> b <eos> |
b |
add__3.0__5.0__ reverse__1.0__ |
add__3.0__5.0__ subtract__3.0__2.0__ |
| find the odd man out num__1 num__2 num__6 num__24 num__120 num__620 <o> a ) num__2 <o> b ) num__620 <o> c ) num__120 <o> d ) num__24 <o> e ) num__6 |
num__1.1 = num__1 num__1 * num__2 = num__2 num__2 * num__3 = num__6 num__6 * num__4 = num__24 num__24 * num__5 = num__120 num__120 * num__6 = num__720 answer : b <eor> b <eos> |
b |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ multiply__6.0__120.0__ multiply__1.0__620.0__ |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ multiply__6.0__120.0__ multiply__1.0__620.0__ |
| num__5216 x num__51 = ? <o> a ) num__266016 <o> b ) num__212016 <o> c ) num__266436 <o> d ) num__216314 <o> e ) none of these |
explanation : normal way of multiplication may take time . here are one alternative . num__5216 x num__51 = ( num__5216 x num__50 ) + num__5216 = ( num__5216 x num__50.0 ) + num__5216 = num__260800.0 + num__5216 = num__260800 + num__5216 = num__266016 . answer : option a <eor> a <eos> |
a |
multiply__5216.0__50.0__ multiply__5216.0__51.0__ multiply__5216.0__51.0__ |
multiply__5216.0__50.0__ add__5216.0__260800.0__ add__5216.0__260800.0__ |
| a cycle is bought for rs . num__900 and sold for rs . num__1120 find the gain percent ? <o> a ) num__11 <o> b ) num__24 <o> c ) num__99 <o> d ) num__77 <o> e ) num__18 |
num__900 - - - - num__220 num__100 - - - - ? = > num__24.0 answer : b <eor> b <eos> |
b |
percent__24.0__100.0__ |
percent__24.0__100.0__ |
| a & b are two towns . a person covers the distance from a to b on cycle at num__17 kmph and returns to a by a boat running at a uniform speed of num__8 kmph . his average speed for the whole journey is <o> a ) num__09.88 kmph <o> b ) num__12.88 kmph <o> c ) num__10.88 kmph <o> d ) num__16.45 kmph <o> e ) num__13.71 kmph |
c num__10.88 kmph when same distance is covered with different speeds then the average speed = num__2 xy / x + y = num__10.88 kmph <eor> c <eos> |
c |
round__10.88__ |
round__10.88__ |
| the sector of a circle has radius of num__21 cm and central angle num__135 o . find its perimeter ? <o> a ) num__23.9 cm <o> b ) num__62.9 cm <o> c ) num__91.5 cm <o> d ) num__26.8 cm <o> e ) num__16.8 cm |
perimeter of the sector = length of the arc + num__2 ( radius ) = ( num__0.375 * num__2 * num__3.14285714286 * num__21 ) + num__2 ( num__21 ) = num__49.5 + num__42 = num__91.5 cm answer : c <eor> c <eos> |
c |
multiply__21.0__2.0__ add__49.5__42.0__ round__91.5__ |
multiply__21.0__2.0__ add__49.5__42.0__ add__49.5__42.0__ |
| a right triangle abc has to be constructed in the xy - plane so that the right angle is at a and ab is parallel to x axis . the coordinates of a b and c are to satisfy the inequalities - num__3 ≤ x ≤ num__5 and num__2 ≤ y ≤ num__11 and x and y are integers . the number of different triangles that can be constructed with these properties are ? <o> a ) num__100 <o> b ) num__6480 <o> c ) num__2320 <o> d ) num__1500 <o> e ) num__9000 |
we have the rectangle with dimensions num__9 * num__10 ( num__9 horizontal dots and num__10 vertical ) . ab is parallel to x - axis and ac is parallel to y - axis . choose the ( x y ) coordinates for vertex a : num__9 c num__1 * num__10 c num__1 = num__90 ; choose the x coordinate for vertex b ( as y coordinate is fixed by a ) : num__8 c num__1 ( num__9 - num__1 = num__8 as num__1 horizontal dot is already occupied by a ) ; choose the y coordinate for vertex c ( as x coordinate is fixed by a ) : num__9 c num__1 ( num__10 - num__1 = num__9 as num__1 vertical dot is already occupied by a ) . num__9 c num__1 * num__10 c num__1 * num__8 c num__1 * num__9 c num__1 = num__6480 . answer : b . <eor> b <eos> |
b |
subtract__11.0__2.0__ multiply__5.0__2.0__ subtract__3.0__2.0__ multiply__9.0__10.0__ add__3.0__5.0__ multiply__1.0__6480.0__ |
subtract__11.0__2.0__ multiply__5.0__2.0__ subtract__3.0__2.0__ multiply__9.0__10.0__ subtract__11.0__3.0__ multiply__1.0__6480.0__ |
| solution y is num__30 percent liquid x and num__70 percent water . if num__4 kilograms of water evaporate from num__8 kilograms of solution y and num__4 kilograms of solution y are added to the remaining num__6 kilograms of liquid what percent of this new solution is liquid x ? <o> a ) num__40.0 <o> b ) num__41 num__0.5 % <o> c ) num__44.0 <o> d ) num__45.0 <o> e ) num__46 % |
in num__8 kilograms of solution y there are num__0.3 * num__8 = num__2.4 kilograms of solution x ; after num__4 kilograms of water are replaced by num__4 kilograms of solution y to the existing num__2.4 kilograms of solution x num__0.3 * num__4 = num__1.2 kilograms of solution x are added so in the new solution of num__8 kilograms there are num__2.4 + num__1.2 = num__3.6 kilograms of solution x which is num__3.6 / num__8 * num__100 = num__45.0 of this new solution . answer : d . <eor> d <eos> |
d |
percent__30.0__8.0__ percent__30.0__4.0__ percent__100.0__45.0__ |
percent__30.0__8.0__ percent__30.0__4.0__ percent__100.0__45.0__ |
| a shopkeeper sells two articles at rs . num__1000 each making a profit of num__20.0 on the first article and a loss of num__20.0 on the second article . find the net profit or loss that he makes ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__65 |
sp of first article = num__1000 profit = num__20.0 cp = ( sp ) * [ num__100 / ( num__100 + p ) ] = num__833.333333333 = num__833.333333333 sp of second article = num__1000 loss = num__20.0 cp = ( sp ) * [ num__100 / ( num__100 - l ) ] = num__1250.0 = num__1250 total sp = num__2000 total cp = num__833.333333333 + num__1250 = num__2083.33333333 cp is more than sp he makes a loss . loss = cp - sp = ( num__2083.33333333 ) - num__2000 = num__83.3333333333 loss percent = [ ( num__83.3333333333 ) / ( num__2083.33333333 ) ] * num__100 = num__0.04 * num__100 = num__4.0 answer : d <eor> d <eos> |
d |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| which of the following must yield an integer when divided by num__9 ? <o> a ) the sum of nine consecutive positive integers . <o> b ) the square of a prime number . <o> c ) the sum of two odd integers . <o> d ) the product of three consecutive odd numbers . <o> e ) the difference between a multiple of num__8 and a multiple of num__3 . |
for option a : let the numbers be ( x - num__4 ) ( x - num__3 ) ( x - num__2 ) ( x - num__1 ) ( x ) ( x + num__1 ) ( x + num__2 ) ( x + num__3 ) ( x + num__4 ) . now ( x - num__4 ) + ( x - num__3 ) + ( x - num__2 ) + ( x - num__1 ) + ( x ) + ( x + num__1 ) + ( x + num__2 ) + ( x + num__3 ) + ( x + num__4 ) = num__9 ( x ) . hence a is true . no need to check other answers since there can only be one correct answer . so ans ( a ) <eor> a <eos> |
a |
subtract__3.0__2.0__ multiply__9.0__1.0__ |
subtract__3.0__2.0__ multiply__9.0__1.0__ |
| the area of a field in the shape of a trapezium measures num__1440 sq m . the perpendicular distance between its parallel sides is num__24 cm . if the ratio of the sides is num__5 : num__3 the length of the longer parallel side is : <o> a ) num__75 <o> b ) num__25 <o> c ) num__26 <o> d ) num__76 <o> e ) num__19 |
explanation : area of field = = num__96 x sq . m num__96 x = num__1440 x = = num__15 hence the length of longer parallel side = num__5 x = num__75 m answer : a ) num__75 <eor> a <eos> |
a |
divide__1440.0__96.0__ multiply__5.0__15.0__ round__75.0__ |
divide__1440.0__96.0__ multiply__5.0__15.0__ round__75.0__ |
| two trains each num__100 m long moving in opposite directions cross other in num__12 sec . if one is moving twice as fast the other then the speed of the faster train is ? <o> a ) num__76 km / hr <o> b ) num__66 km / hr <o> c ) num__40 km / hr <o> d ) num__67 km / hr <o> e ) num__22 km / hr |
let the speed of the slower train be x m / sec . then speed of the train = num__2 x m / sec . relative speed = ( x + num__2 x ) = num__3 x m / sec . ( num__100 + num__100 ) / num__12 = num__3 x = > x = num__5.55555555556 so speed of the faster train = num__11.1111111111 = num__11.1111111111 * num__3.6 = num__40 km / hr . answer : c <eor> c <eos> |
c |
multiply__11.1111__3.6__ round__40.0__ |
multiply__11.1111__3.6__ multiply__11.1111__3.6__ |
| ayayai electronics is a company manufacturing mp num__3 players . ayayai manufactures only two players : ime and imine . either player is assembled using a lcd screen or an oled screen . excluding the expensive oled - imine model ayayai manufactures num__21000 players per month . every month ayayai manufactures num__11000 players with an oled screen and num__12000 imine models . if ayayai manufactures num__8000 lcd ime models monthly how many players ( of all models ) does ayayai manufacture per month ? <o> a ) num__24000 <o> b ) num__25000 <o> c ) num__26000 <o> d ) num__27000 <o> e ) num__28 |
000 |
let x be the imine lcd players let y be the ime oled players num__8000 + x + y = num__21000 x + y = num__13000 assume num__60007000 for x and y . num__6000 + imine oled = num__11000 . imine oled = num__5000 . hence total number of players = num__8000 + x + y + num__5000 = num__26000 ans : ( option c ) <eor> c <eos> |
c |
c |
| a grocer has a sale of rs . num__5420 rs . num__5660 rs . num__6200 rs . num__6350 and rs . num__6500 for num__5 consecutive months . find the sale he should have in the sixth month so that he gets an average sale of rs . num__6300 ? <o> a ) rs . num__5870 <o> b ) rs . num__5991 <o> c ) rs . num__6020 <o> d ) rs . num__7670 <o> e ) none of these |
explanation : total sale for num__5 months = rs . ( num__5420 + num__5660 + num__6200 + num__6350 + num__6500 ) = rs . num__30130 therefore required sale = rs . [ ( num__6300 * num__6 ) – num__30130 ] = rs . ( num__37800 – num__30130 ) = rs . num__7670 answer d <eor> d <eos> |
d |
multiply__6300.0__6.0__ subtract__37800.0__30130.0__ subtract__37800.0__30130.0__ |
multiply__6300.0__6.0__ subtract__37800.0__30130.0__ subtract__37800.0__30130.0__ |
| how many litres of pure acid are there in num__5 litres of a num__30.0 solution <o> a ) num__1.5 <o> b ) num__1.6 <o> c ) num__1.7 <o> d ) num__1.8 <o> e ) num__1.9 |
explanation : question of this type looks a bit typical but it is too simple as below . . . it will be num__5 * num__0.3 = num__1.5 answer : option a <eor> a <eos> |
a |
percent__5.0__30.0__ percent__5.0__30.0__ |
percent__5.0__30.0__ percent__5.0__30.0__ |
| certain stocks in january were num__20.0 less than they were in february and num__30.0 greater than they were in march . what was the percentage decrease in the stocks from february to march ? <o> a ) num__5.7 <o> b ) num__11.0 <o> c ) num__20.0 <o> d ) num__25.3 <o> e ) num__38.5 % |
let stocks value in feb = num__100 = > in jan = num__80 and march = num__80 * num__0.769230769231 = num__61.5 thus percent decrease in stocks from feb to march = num__100 - num__61.5 = num__38.5 hence answer is e <eor> e <eos> |
e |
subtract__100.0__20.0__ subtract__100.0__61.5__ subtract__100.0__61.5__ |
subtract__100.0__20.0__ subtract__100.0__61.5__ subtract__100.0__61.5__ |
| pipes a and b can fill a cistern in num__8 and num__24 minutes respectively . they are opened an alternate minutes . find how many minutes the cistern shall be full ? <o> a ) num__16 <o> b ) num__12 <o> c ) num__18 <o> d ) num__17 <o> e ) num__13 |
num__0.125 + num__0.0416666666667 = num__0.166666666667 num__6 * num__2 ' = num__12 answer : b <eor> b <eos> |
b |
add__0.125__0.0417__ subtract__8.0__6.0__ divide__24.0__2.0__ round__12.0__ |
add__0.125__0.0417__ subtract__8.0__6.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
| a man can row with a speed of num__14 kmph in still water . if the stream flows at num__10 kmph then the speed in downstream is ? <o> a ) num__28 <o> b ) num__26 <o> c ) num__20 <o> d ) num__87 <o> e ) num__24 |
m = num__14 s = num__10 ds = num__14 + num__10 = num__24 answer : e <eor> e <eos> |
e |
add__14.0__10.0__ round__24.0__ |
add__14.0__10.0__ add__14.0__10.0__ |
| if a certain sample of data has a mean of num__27.0 and a standard deviation of num__2.0 which of the following pairs contain two values that are each at least num__2.5 standard deviations from the mean ? <o> a ) ( num__24.0 ; num__28.5 ) <o> b ) ( num__24.5 ; num__31.0 ) <o> c ) ( num__25.0 ; num__29.5 ) <o> d ) ( num__19.5 ; num__33.0 ) <o> e ) ( num__24.0 ; num__29.5 ) |
the standard deviation is num__2 so num__2.5 standard deviations would be ( num__2.5 ) ( num__2 ) = num__5 . the values between num__22 and num__32 are within num__2.5 standard deviations from the mean . the answer is d . <eor> d <eos> |
d |
multiply__2.0__2.5__ subtract__27.0__5.0__ add__27.0__5.0__ subtract__22.0__2.5__ |
multiply__2.0__2.5__ subtract__27.0__5.0__ add__27.0__5.0__ subtract__22.0__2.5__ |
| ajit has a certain average for num__9 innings . in the tenth innings he scores num__100 runs thereby increasing his average by num__8 runs . his new average is : <o> a ) num__20 <o> b ) num__21 <o> c ) num__28 <o> d ) num__32 <o> e ) none |
solution : let ajit ' s average be x for num__9 innings . so ajit scored num__9 x run in num__9 innings . in the num__10 th inning he scored num__100 runs then average became ( x + num__8 ) . and he scored ( x + num__8 ) * num__10 runs in num__10 innings . now = > num__9 x + num__100 = num__10 * ( x + num__8 ) or num__9 x + num__100 = num__10 x + num__80 or x = num__100 - num__80 or x = num__20 new average = ( x + num__8 ) = num__28 runs answer : option c <eor> c <eos> |
c |
multiply__8.0__10.0__ subtract__100.0__80.0__ add__8.0__20.0__ add__8.0__20.0__ |
multiply__8.0__10.0__ subtract__100.0__80.0__ add__8.0__20.0__ add__8.0__20.0__ |
| a manufacturer is using glass as the surface for the multi - touch screen of its smartphone . the glass on the manufactured phone has a num__3.0 probability of not passing quality control tests . the quality control manager bundles the smartphone in groups of num__10 . if that bundle has any smartphone that does not pass the quality control test the entire bundle of num__10 is rejected . what is the probability that a smartphone bundle that will be rejected by quality control ? <o> a ) num__0.25 <o> b ) . num__05 ^ num__10 <o> c ) num__1 - num__0.95 ^ num__10 <o> d ) num__1 - num__0.97 ^ num__10 <o> e ) num__0.95 ^ num__10 |
find the probability of the opposite event and subtract from num__1 . the opposite event is that bundle will not be rejected by quality control which will happen if all num__10 phones pass the test so p ( all num__10 phones pass test ) = num__0.97 ^ num__10 . p ( at least one phone do not pass the test ) = num__1 - p ( all num__10 phones pass test ) = num__1 - num__0.97 ^ num__10 . answer : d . <eor> d <eos> |
d |
reverse__1.0__ |
reverse__1.0__ |
| a barman ' s train rails across an open track at num__250 kilometers per hour . a regular passenger train travels at num__68.0 of the barman ' s train speed . if the two trains start moving from the same station at the same time how much time longer will it take the passenger train than the barman ' s to travel num__850 kilometers ? <o> a ) num__1 hour and num__24 minutes <o> b ) num__1 hour and num__36 minutes <o> c ) num__2 hours and num__24 minutes . <o> d ) num__2 hours and num__36 minutes <o> e ) num__5 hours |
b train speed num__250 p train speed num__250 ( num__0.68 ) = num__170 difference of speed b - p = num__80 km / hrs total distance = num__850 additional time taken num__10.625 num__1 h num__36 min answer : b <eor> b <eos> |
b |
multiply__250.0__0.68__ subtract__250.0__170.0__ divide__850.0__80.0__ round__1.0__ |
multiply__250.0__0.68__ subtract__250.0__170.0__ divide__850.0__80.0__ round__1.0__ |
| in how many ways can a person post num__5 letters in num__4 letter boxes ? <o> a ) num__480 <o> b ) num__1024 <o> c ) num__54 <o> d ) num__4 ^ num__5 <o> e ) num__5 ^ num__5 |
there are num__4 places ( abcd ) that need to be filled . letter num__1 can be posted in either a b c or d i . e . num__4 ways letter num__2 can be posted in either a b c or d i . e . num__4 ways letter num__3 can be posted in either a b c or d i . e . num__4 ways letter num__4 can be posted in either a b c or d i . e . num__4 ways letter num__5 can be posted in either a b c or d i . e . num__4 ways so the total no of ways in which num__5 letters can be posted in num__4 boxes are = num__4 x num__4 x num__4 x num__4 x num__4 = num__4 ^ num__5 thus the answer is d <eor> d <eos> |
d |
coin_space__ choose__4.0__3.0__ |
coin_space__ choose__4.0__3.0__ |
| the population of a bacteria culture doubles every num__5 minutes . approximately how many minutes will it take for the population to grow from num__1000 to num__500000 bacteria <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__16 <o> e ) num__45 |
this one ' s easy . num__1000 * num__2 ^ t = num__500000 num__2 ^ t = num__500 now gauging since num__2 ^ num__8 = num__256 then num__2 ^ num__9 = num__512 so t = num__9 but be careful ' t ' is in time intervals of num__5 minutes so answer is num__9 * num__5 = num__45 minutes answer ( e ) <eor> e <eos> |
e |
divide__1000.0__2.0__ multiply__256.0__2.0__ multiply__5.0__9.0__ round__45.0__ |
divide__1000.0__2.0__ multiply__256.0__2.0__ multiply__5.0__9.0__ multiply__5.0__9.0__ |
| if a * b * c = ( √ ( a + num__2 ) ( b + num__3 ) ) / ( c + num__1 ) find the value of num__6 * num__15 * num__7 . <o> a ) num__8 <o> b ) num__5 <o> c ) num__11 <o> d ) num__3 <o> e ) num__1.5 |
num__6 * num__15 * num__3 = ( √ ( num__6 + num__2 ) ( num__15 + num__3 ) ) / ( num__7 + num__1 ) = ( √ num__8 * num__18 ) / num__8 = ( √ num__144 ) / num__8 = num__1.5 = num__1.5 answer is e <eor> e <eos> |
e |
add__2.0__6.0__ multiply__3.0__6.0__ multiply__8.0__18.0__ divide__3.0__2.0__ divide__3.0__2.0__ |
add__2.0__6.0__ multiply__3.0__6.0__ multiply__8.0__18.0__ divide__3.0__2.0__ divide__3.0__2.0__ |
| the value of x is to be randomly selected from the integers from num__1 to num__6 inclusive and then substituted into the equation y = x ^ num__2 - num__4 x + num__3 . what is the probability that the value of y will be negative ? <o> a ) num__0.333333333333 <o> b ) num__0.25 <o> c ) num__0.2 <o> d ) num__0.166666666667 <o> e ) num__0.142857142857 |
y will only be negative for x = num__2 . ( we can check the values from num__1 to num__6 to be certain . ) p ( y is negative ) = num__0.166666666667 the answer is d . <eor> d <eos> |
d |
reverse__6.0__ reverse__6.0__ |
reverse__6.0__ reverse__6.0__ |
| some students are standing in a circle in which num__6 th and the num__16 th student are standing opposite to each other . find how many students were present there . <o> a ) num__10 <o> b ) num__15 <o> c ) num__20 <o> d ) num__25 <o> e ) num__30 |
( n / num__2 ) - x = num__6 n - x = num__16 solving these two = > n = number of students = num__20 answer : c <eor> c <eos> |
c |
triangle_area__2.0__20.0__ |
triangle_area__2.0__20.0__ |
| a trader mixes num__25 kg of rice at rs . num__30 per kg with num__25 kg of rice of other variety at rs . num__40 per kg and sells the mixture at rs . num__40 per kg . his profit percent is : <o> a ) no profit no loss <o> b ) num__5.0 <o> c ) num__14.29 <o> d ) num__10.0 <o> e ) none of these |
c . p . of num__50 kg rice = rs . ( num__25 x num__30 + num__25 x num__40 ) = rs . ( num__750 + num__1000 ) = rs . num__1750 s . p . of num__50 kg rice = rs . ( num__50 x num__40 ) = rs . num__2000 gain = num__0.142857142857 x num__100.0 = num__14.29 . answer : option c <eor> c <eos> |
c |
percent__100.0__14.29__ |
percent__100.0__14.29__ |
| the ratio of present ages of nisha and shilpa is num__7 : num__8 respectively . four years hence this ratio becomes num__9 : num__10 respectively . what is nisha ’ s present age in years ? <o> a ) num__18 <o> b ) num__14 <o> c ) num__17 <o> d ) data inadequate <o> e ) none of these |
present age of nisha = num__4 × ( num__10 − num__9 ) / num__9 × num__8 − num__10 × num__7 × num__7 = num__14 years answer b <eor> b <eos> |
b |
add__10.0__4.0__ add__10.0__4.0__ |
add__10.0__4.0__ add__10.0__4.0__ |
| the speed of a car is num__90 km in the first hour and num__40 km in the second hour . what is the average speed of the car ? <o> a ) num__12 <o> b ) num__75 <o> c ) num__88 <o> d ) num__65 <o> e ) num__15 |
s = ( num__90 + num__40 ) / num__2 = num__65 kmph answer : d <eor> d <eos> |
d |
round__65.0__ |
round__65.0__ |
| a cake is cut into num__16 pieces . whats amount of cake must be eaten to make cake cut into half . <o> a ) num__4 <o> b ) num__2 <o> c ) num__9 <o> d ) num__8 <o> e ) num__3 |
the cake is cut into num__16 equal pieces . half of num__16 is num__8 . so num__8 pieces must be removed from num__16 pieces to make it half . answer : d <eor> d <eos> |
d |
subtract__16.0__8.0__ |
subtract__16.0__8.0__ |
| on a scale of a map num__0.6 cm represents num__6.6 km . if the distance between two points on the map is num__80.5 cm what is the the actual distance between these points ? <o> a ) num__885.5 km <o> b ) num__860 km <o> c ) num__892.5 km <o> d ) num__825 km <o> e ) num__845 km |
explanation : num__0.6 cm in map ≡ actual distance of num__6.6 km num__1 cm in map ≡ num__6.6 / . num__6 km num__80.5 cm in map ≡ num__80.5 × num__6.6 / . num__6 km = num__885.5 km answer : option a <eor> a <eos> |
a |
subtract__6.6__0.6__ round__885.5__ |
subtract__6.6__0.6__ divide__885.5__1.0__ |
| shyam purchased a cow for rs num__20000 and num__2 goats for num__10000 . he sold the cow at a loss of num__5.0 and goats at a profit of num__15.0 . overall he make a . ? <o> a ) num__500 <o> b ) num__400 <o> c ) num__300 <o> d ) num__350 <o> e ) num__550 |
let the sp of the cow and the goats be rs . c and rs . g respectively . c = num__20000 ( num__1 - num__0.05 ) = num__20000 - num__1000 g = num__10000 ( num__1 + num__0.15 ) = num__10000 + num__1500 total sp - total cp = c + g - ( num__20000 + num__10000 ) = - num__1000 + num__1500 = rs . num__500 as this is positive an overall profit of rs . num__500 was made . answer : a <eor> a <eos> |
a |
percent__5.0__1.0__ percent__5.0__20000.0__ percent__15.0__1.0__ percent__15.0__10000.0__ percent__5.0__10000.0__ percent__5.0__10000.0__ |
percent__5.0__1.0__ percent__5.0__20000.0__ percent__15.0__1.0__ percent__15.0__10000.0__ percent__5.0__10000.0__ percent__5.0__10000.0__ |
| an article is bought for rs . num__500 and sold for rs . num__700 find the gain percent ? <o> a ) num__33 num__0.142857142857 % <o> b ) num__33 num__0.166666666667 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__38 num__0.333333333333 % <o> e ) num__40 % |
num__500 - - - - num__200 num__100 - - - - ? = > num__40.0 answer : e <eor> e <eos> |
e |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| if a number between num__0 and num__0.5 is selected at random which of the following will the number most likely be between <o> a ) num__0 and num__0.15 <o> b ) num__0.15 and num__0.2 <o> c ) num__0.2 and num__0.25 <o> d ) num__0.25 and num__0.3 <o> e ) num__0.25 and num__0.5 |
e . num__0.25 and num__0.5 definitely not . i am always happy to explain a given question or issue if i can . however i sometime donot put my explanation if i am answering first because i want to put others first for their explanation . now you have excellent explanantion byscthakur . in fact the question is asking the likelyness of or high chances / probability of having a given number . hope you already got it . e <eor> e <eos> |
e |
subtract__0.5__0.25__ |
subtract__0.5__0.25__ |
| y = num__248 - num__398 x which of the following values of x gives the greatest value of y in the equation above ? <o> a ) num__200 <o> b ) num__100 <o> c ) num__0.5 <o> d ) num__0 <o> e ) - num__1 |
explanation : this is quite simple . just by looking at the values they are all positive except two . if you enter any of the values instead of x you will get a negative y . by entering num__0 you already get to positive = num__248 . but the correct answer is - num__1 since this equals y = num__248 - ( num__398 * ( - num__1 ) ) = num__248 + num__398 . this is the greatest value . answer : option e <eor> e <eos> |
e |
reverse__1.0__ |
reverse__1.0__ |
| num__17 . num__2 x num__40 = ? <o> a ) num__188.0 <o> b ) num__688.0 <o> c ) num__288.0 <o> d ) num__388.0 <o> e ) num__488.0 |
multiply num__17.2 x num__40 answer : num__688.0 correct answer b <eor> b <eos> |
b |
multiply__40.0__17.2__ multiply__40.0__17.2__ |
multiply__40.0__17.2__ multiply__40.0__17.2__ |
| a group of hikers is planning a trip that will take them up a mountain using one route and back down using another route . they plan to travel down the mountain at a rate of one and a half times the rate they will use on the way up but the time each route will take is the same . if they will go up the mountain at a rate of num__8 miles per day and it will take them two days how many miles long is the route down the mountain ? <o> a ) num__18 <o> b ) num__20 <o> c ) num__22 <o> d ) num__24 <o> e ) num__25 |
on the way down the rate is num__1.5 * num__8 = num__12 miles per day . the distance of the route down the mountain is num__2 * num__12 = num__24 miles . the answer is d . <eor> d <eos> |
d |
multiply__8.0__1.5__ multiply__2.0__12.0__ round__24.0__ |
multiply__8.0__1.5__ multiply__2.0__12.0__ multiply__2.0__12.0__ |
| if x : y = num__9 : num__4 then ( num__7 x - num__3 y ) : ( num__7 x + num__3 y ) = ? <o> a ) num__51 : num__76 <o> b ) num__6 : num__13 <o> c ) num__51 : num__75 <o> d ) num__51 : num__76 <o> e ) none of these |
explanation : x / y = num__2.25 ( given ) ( num__7 x − num__3 y ) / ( num__7 x + num__3 y ) num__7 ( x / y ) − num__0.428571428571 ( x / y ) + num__3 ( on dividing nr and dr by y ) ( num__7 ( num__2.25 ) − num__3 ) / ( num__7 ( num__2.25 ) + num__3 ) ( num__15.75 − num__3.0 ) / ( num__15.75 + num__3.0 ) [ ( num__63 − num__12 ) / num__4 ] / [ ( num__63 + num__12 ) / num__4 ] = num__0.68 ( num__7 x - num__3 y ) : ( num__7 x + num__3 y ) = num__51 : num__75 answer : option c <eor> c <eos> |
c |
divide__9.0__4.0__ divide__3.0__7.0__ multiply__7.0__2.25__ multiply__9.0__7.0__ add__9.0__3.0__ subtract__63.0__12.0__ add__12.0__63.0__ multiply__0.68__75.0__ |
divide__9.0__4.0__ divide__3.0__7.0__ multiply__7.0__2.25__ multiply__9.0__7.0__ add__9.0__3.0__ subtract__63.0__12.0__ add__12.0__63.0__ subtract__63.0__12.0__ |
| the sum of three numbers is num__294 . if the ratio of the first to the second is num__2 : num__3 and that of the second to the third is num__5 : num__8 then the second number is : <o> a ) num__20 <o> b ) num__30 <o> c ) num__38 <o> d ) num__90 <o> e ) none of these |
a : b = num__2 : num__3 = num__2 × num__5 : num__3 × num__5 = num__10 : num__15 and b : c = num__5 : num__8 = num__5 × num__3 : num__8 × num__3 = num__15 : num__24 therefore a : b : c = num__10 : num__15 : num__24 ∴ a : b : c = num__10 : num__15 : num__24 let the number be num__10 x num__15 x and num__24 x . then num__10 x + num__15 x + num__24 x = num__294 or num__49 x = num__294 or x = num__6 ⇒ second number = num__15 x = num__15 × num__6 = num__90 answer d <eor> d <eos> |
d |
multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__3.0__8.0__ divide__294.0__49.0__ multiply__15.0__6.0__ multiply__15.0__6.0__ |
add__2.0__8.0__ add__5.0__10.0__ multiply__3.0__8.0__ divide__294.0__49.0__ multiply__15.0__6.0__ multiply__15.0__6.0__ |
| a sporting goods store sold num__60 frisbees in one week some for $ num__3 and the rest for $ num__4 each . if receipts from frisbee sales for the week totaled $ num__204 what is the fewest number of $ num__4 frisbees that could have been sold ? <o> a ) num__24 <o> b ) num__12 <o> c ) num__8 <o> d ) num__4 <o> e ) num__2 |
in this question however because we are told that exactly num__64 frisbees have been sold and revenue was exactly $ num__204 there is only one possible solution for the number of $ num__3 and $ num__4 frisbees sold . to solve we have num__2 equations and num__2 unknowns let x = number of $ num__3 frisbees sold let y = number of $ num__4 frisbees sold x + y = num__60 num__3 x + num__4 y = num__204 x = num__60 - y num__3 ( num__60 - y ) + num__4 y = num__204 num__180 - num__3 y + num__4 y = num__204 y = num__24 answer : a <eor> a <eos> |
a |
add__60.0__4.0__ multiply__60.0__3.0__ subtract__204.0__180.0__ subtract__204.0__180.0__ |
add__60.0__4.0__ multiply__60.0__3.0__ subtract__204.0__180.0__ subtract__204.0__180.0__ |
| in a num__1100 m race usha beats shiny by num__50 m . in the same race by what time margin shiny beat mercy who runs at num__4 m / s ? <o> a ) num__100 sec . <o> b ) num__55 sec <o> c ) num__25 sec <o> d ) data not sufficient <o> e ) none of these |
speed of shiny = num__5.0 = num__5 m / s time taken by shiny to complete the race is b = num__220.0 = num__220 sec . time taken by baley to complete the race is d = num__250.0 = num__275 sec . hence d - b = num__55 sec answer : b <eor> b <eos> |
b |
divide__1100.0__5.0__ multiply__50.0__5.0__ divide__1100.0__4.0__ add__50.0__5.0__ round__55.0__ |
divide__1100.0__5.0__ multiply__50.0__5.0__ divide__1100.0__4.0__ subtract__275.0__220.0__ subtract__275.0__220.0__ |
| workers at a campaign office have num__2000 fliers to send out . if they send out num__0.25 of them in the morning and num__0.2 of the remaining ones out during the afternoon how many are left for the next day ? <o> a ) num__300 <o> b ) num__800 <o> c ) num__1100 <o> d ) num__1200 <o> e ) num__1180 |
( num__0.25 ) * num__2000 = num__500 remaining = num__2000 - num__500 = num__1500 ( num__0.2 ) of remaining = ( num__0.2 ) * num__1600 = num__320 remaining now = num__1500 - num__320 = num__1180 answer : option e <eor> e <eos> |
e |
multiply__2000.0__0.25__ subtract__2000.0__500.0__ multiply__0.2__1600.0__ subtract__1500.0__320.0__ round__1180.0__ |
multiply__2000.0__0.25__ subtract__2000.0__500.0__ multiply__0.2__1600.0__ subtract__1500.0__320.0__ subtract__1500.0__320.0__ |
| a present value of a machine is $ num__40000 . its value depletiation rate is num__5.0 per annum then find the machine value after num__2 years ? <o> a ) $ num__36540 <o> b ) $ num__36100 <o> c ) $ num__29580 <o> d ) $ num__31256 <o> e ) $ num__41250 |
p = $ num__40000 r = num__5.0 t = num__2 years machine value after num__3 years = p / ( num__1 - r / num__100 ) ^ t = num__40000 * num__0.95 * num__0.95 * num__0.95 = $ num__36100 answer is b <eor> b <eos> |
b |
percent__100.0__36100.0__ |
percent__100.0__36100.0__ |
| the sum of two numbers is num__29 and the difference of their squares is num__145 . the difference between the number is <o> a ) num__13 <o> b ) num__5 <o> c ) num__8 <o> d ) num__11 <o> e ) num__12 |
explanation : let the numbers be x and y . we know that ( x − y ) = x num__2 − y num__2 / ( x + y ) = num__5.0 = num__5 correct option : b <eor> b <eos> |
b |
divide__145.0__29.0__ divide__145.0__29.0__ |
divide__145.0__29.0__ divide__145.0__29.0__ |
| the sum of num__3 rd and num__9 th term of a . p is num__8 . find the sum of num__1 st num__11 terms of a . p <o> a ) num__33 <o> b ) num__22 <o> c ) num__44 <o> d ) num__55 <o> e ) num__66 |
a . p . series is a a + d a + num__2 d a + num__3 d a + num__4 d . . . . . . . . . . . . . . nth terms where a = first term d = common difference we have to given sum of num__3 rd and num__9 th terms = num__8 ( a + num__2 d ) + ( a + num__8 d ) = num__8 num__2 a + num__10 d = num__8 a + num__5 d = num__4 . . . . . . . . . . . . . ( num__1 ) formula sum of num__1 st num__11 terms of ap s = n [ num__2 a + ( n - num__1 ) d ] / num__2 = num__11 [ num__2 a + ( num__11 - num__1 ) d ] / num__2 = num__11 [ a + num__5 d ] = num__11 [ num__4 ] ( from eq . . . . ( num__1 ) ) s = num__44 answer : c <eor> c <eos> |
c |
subtract__3.0__1.0__ add__3.0__1.0__ add__9.0__1.0__ add__3.0__2.0__ multiply__11.0__4.0__ multiply__1.0__44.0__ |
subtract__3.0__1.0__ add__3.0__1.0__ add__9.0__1.0__ add__3.0__2.0__ multiply__11.0__4.0__ divide__44.0__1.0__ |
| tough and tricky questions : decimals . the value of x is derived by summing a b and c and then rounding the result to the tenths place . the value of y is derived by first rounding a b and c to the tenths place and then summing the resulting values . if a = num__5.58 b = num__2.95 and c = num__3.65 what is y – x ? <o> a ) - num__0.1 <o> b ) num__0.2 <o> c ) num__0.05 <o> d ) num__0.1 <o> e ) num__0.2 |
to find x we first sum a b and c then round to the tenths place . num__5.58 + num__2.95 + num__3.65 = num__12.18 which rounds to num__12.2 . to find y we first round a b and c to the tenths place and them sum them . so num__5.5 + num__2.9 + num__3.6 = num__12.0 . we are looking for y - x which gives us num__12.5 - num__12.4 = num__0.2 or answer choice b . b <eor> b <eos> |
b |
round_down__12.18__ subtract__12.4__12.2__ subtract__12.4__12.2__ |
round_down__12.18__ subtract__12.4__12.2__ subtract__12.4__12.2__ |
| an agency wants to buy furnitures of x % of a list of num__10000 furnitures in a catalogue . after a market price increase the agency must reduce this selection by ( x − num__2 ) % . in terms of x how many furnitures will the agency be able to buy ? <o> a ) x * x – num__2 x <o> b ) ( x ) ( num__102 – x ) <o> c ) ( num__100 ) ( num__102 – x ) <o> d ) ( num__100 ) ( num__98 – x ) <o> e ) ( x - num__2 ) / num__100 |
based on the answer choices and the question this question begs the use of x = num__2 as a sample number . initial = num__2.0 * num__10000 = num__200 reduction = num__2 - num__2 = num__0.0 so no math required here to calculate the reduction ; just make sure that you can calculate num__200 in your answer . a . x * x – num__2 x = num__0 ; no b . ( x ) ( num__102 – x ) = num__200 ; winner ! c . ( num__100 ) ( num__102 – x ) > num__200 ; no d . ( num__100 ) ( num__98 – x ) > num__200 ; no e . ( x - num__2 ) / num__100 = num__0 ; no b <eor> b <eos> |
b |
subtract__102.0__2.0__ subtract__100.0__2.0__ add__2.0__100.0__ |
subtract__102.0__2.0__ subtract__100.0__2.0__ subtract__200.0__98.0__ |
| a fires num__5 shots to b ' s num__3 but a kills only once in num__3 shots while b kills once in num__2 shots . when b has missed num__27 times a has killed : <o> a ) num__30 birds <o> b ) num__22 birds <o> c ) num__18 birds <o> d ) num__38 birds <o> e ) num__28 birds |
let total number of shots = x shots fired by a = num__5 x / num__8 shots fired by b = num__3 x / num__8 killing shots by a = ( num__5 x / num__8 ) × ( num__0.333333333333 ) = num__5 x / num__24 missing shots by b = ( num__3 x / num__8 ) × ( num__0.5 ) = num__3 x / num__16 b has missed num__27 times ⇒ num__3 x / num__16 = num__27 ⇒ x = ( num__27 × num__16 ) / num__3 = num__144 hence killing shots by a = num__5 x / num__24 = ( num__5 × num__144 ) / num__24 = num__30 i . e . a has killed num__30 birds answer : a <eor> a <eos> |
a |
add__5.0__3.0__ reverse__3.0__ multiply__3.0__8.0__ reverse__2.0__ multiply__2.0__8.0__ add__3.0__27.0__ add__3.0__27.0__ |
add__5.0__3.0__ reverse__3.0__ multiply__3.0__8.0__ reverse__2.0__ divide__8.0__0.5__ add__3.0__27.0__ add__3.0__27.0__ |
| num__125 liters of a mixture of milk and water contains in the ratio num__3 : num__2 . how much water should now be added so that the ratio of milk and water becomes num__3 : num__4 ? <o> a ) num__3 : num__7 <o> b ) num__3 : num__9 <o> c ) num__3 : num__1 <o> d ) num__3 : num__4 <o> e ) num__3 : num__2 |
milk = num__0.6 * num__125 = num__75 liters water = num__50 liters num__75 : ( num__50 + p ) = num__3 : num__4 num__150 + num__3 p = num__400 = > p = num__50 num__50 liters of water are to be added for the ratio become num__3 : num__4 . answer : d <eor> d <eos> |
d |
multiply__125.0__0.6__ subtract__125.0__75.0__ multiply__3.0__50.0__ divide__150.0__50.0__ |
multiply__125.0__0.6__ subtract__125.0__75.0__ multiply__3.0__50.0__ divide__150.0__50.0__ |
| a and b can do a work in num__4 hours and num__12 hours respectively . a starts the work at num__6 am and they work alternately for one hour each . when will the work be completed ? <o> a ) num__8 days <o> b ) num__5 days <o> c ) num__4 days <o> d ) num__6 days <o> e ) num__2 days |
work done by a and b in the first two hours working alternately = first hour a + second hour b = num__0.25 + num__0.0833333333333 = num__0.333333333333 . total time required to complete the work = num__2 * num__3 = num__6 days . answer : d <eor> d <eos> |
d |
divide__4.0__12.0__ divide__12.0__6.0__ divide__12.0__4.0__ round__6.0__ |
add__0.25__0.0833__ divide__12.0__6.0__ multiply__12.0__0.25__ add__4.0__2.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 seconds . find the length of the train ? <o> a ) num__150 meter <o> b ) num__178 meter <o> c ) num__209 meter <o> d ) num__279 meter <o> e ) num__269 meter |
speed = num__60 * ( num__0.277777777778 ) m / sec = num__16.6666666667 m / sec length of train ( distance ) = speed * time ( num__16.6666666667 ) * num__9 = num__150 meter answer : a <eor> a <eos> |
a |
round__150.0__ |
round__150.0__ |
| a train num__90 meters long completely crosses a num__200 meters long bridge in num__36 seconds . what is the speed of the train is ? <o> a ) num__29 <o> b ) num__40 <o> c ) num__25 <o> d ) num__66 <o> e ) num__09 |
s = ( num__90 + num__200 ) / num__36 = num__8.05555555556 * num__3.6 = num__29 answer : a <eor> a <eos> |
a |
round__29.0__ |
round__29.0__ |
| in a game of num__80 points a can give b num__5 points and c num__15 points . then how many points b can give c in a game of num__60 ? <o> a ) num__7 points <o> b ) num__8 points <o> c ) num__9 points <o> d ) num__10 points <o> e ) num__11 points |
in a game of num__80 points a give b num__5 points and c num__15 points means b = num__75 c = num__65 in num__75 points b give num__10 since num__75 - num__65 = num__10 in num__60 points b gives = num__60 * num__0.133333333333 = num__8 so ans num__8 points answer : b <eor> b <eos> |
b |
subtract__80.0__5.0__ subtract__80.0__15.0__ subtract__15.0__5.0__ divide__10.0__75.0__ divide__80.0__10.0__ divide__80.0__10.0__ |
subtract__80.0__5.0__ subtract__80.0__15.0__ subtract__15.0__5.0__ divide__10.0__75.0__ divide__80.0__10.0__ divide__80.0__10.0__ |
| in num__20 minutes the minute hand gains over the hour hand by <o> a ) num__16 ° <o> b ) num__80 ° <o> c ) num__88 ° <o> d ) num__96 ° <o> e ) num__110 ° |
in one hour the hour hand moves num__30 ° which is num__0.5 ° each minute . in one hour the minute hand moves num__360 ° which is num__6 ° each minutes . the minute hand gains num__5.5 ° each minute . in num__20 minutes the minute hand gains num__20 * num__5.5 ° = num__110 ° . the answer is e . <eor> e <eos> |
e |
subtract__6.0__0.5__ multiply__20.0__5.5__ round__110.0__ |
subtract__6.0__0.5__ multiply__20.0__5.5__ multiply__20.0__5.5__ |
| the length of a rectangle is halved while its breadth is tripled . watis the % change in area ? <o> a ) num__30.0 <o> b ) num__40.0 <o> c ) num__50.0 <o> d ) num__60.0 <o> e ) num__70 % |
let original length = x and original breadth = y . original area = xy . new length = x . num__2 new breadth = num__3 y . new area = x x num__3 y = num__3 xy . num__2 num__2 increase % = num__1 xy x num__1 x num__100.0 = num__50.0 . num__2 xy c <eor> c <eos> |
c |
triangle_area__1.0__100.0__ triangle_area__1.0__100.0__ |
triangle_area__1.0__100.0__ triangle_area__1.0__100.0__ |
| on her annual road trip to visit her family in seal beach california traci stopped to rest after she traveled num__1 ⁄ num__2 of the total distance and again after she traveled num__1 ⁄ num__4 of the distance remaining between her first stop and her destination . she then drove the remaining num__400 miles and arrived safely at her destination . what was the total distance in miles from traci ’ s starting point to seal beach ? <o> a ) num__250 <o> b ) num__1066.66666667 <o> c ) num__1050.0 <o> d ) num__1334.0 <o> e ) num__550 |
let d = total distance traci traveled num__0.5 = d / num__2 i . e . remaining distance = d / num__2 she traveled num__0.25 th of d / num__2 = d / num__8 thus : d = ( d / num__2 ) + ( d / num__8 ) + num__400 d = num__1066.66666667 answer : b <eor> b <eos> |
b |
divide__1.0__2.0__ divide__1.0__4.0__ multiply__2.0__4.0__ multiply__1.0__1066.6667__ |
divide__1.0__2.0__ divide__1.0__4.0__ divide__2.0__0.25__ divide__1066.6667__1.0__ |
| pipes a and b can fill a tank in num__3 and num__6 hours respectively . pipe c can empty it in num__12 hours . if all the three pipes are opened together then the tank will be filled in ? <o> a ) num__3 num__0.0927835051546 hrs <o> b ) num__3 num__0.116883116883 hrs <o> c ) num__3 num__0.529411764706 hrs <o> d ) num__3 num__0.333333333333 hrs <o> e ) num__2 num__0.4 hrs |
net part filled in num__1 hour = num__0.333333333333 + num__0.166666666667 - num__0.0833333333333 = num__0.416666666667 the tank will be full in num__2.4 hrs i . e . num__2 num__0.4 hrs . answer : e <eor> e <eos> |
e |
divide__1.0__3.0__ divide__1.0__6.0__ divide__1.0__12.0__ subtract__3.0__1.0__ divide__2.4__6.0__ round__2.0__ |
divide__1.0__3.0__ divide__1.0__6.0__ divide__1.0__12.0__ subtract__3.0__1.0__ subtract__2.4__2.0__ subtract__3.0__1.0__ |
| the cost price of a radio is rs . num__1500 and it was sold for rs . num__1245 find the loss % ? <o> a ) num__17.0 <o> b ) num__16.0 <o> c ) num__19.0 <o> d ) num__78.0 <o> e ) num__28 % |
num__1500 - - - - num__255 num__100 - - - - ? = > num__17.0 answer : a <eor> a <eos> |
a |
percent__100.0__17.0__ |
percent__100.0__17.0__ |
| if the annual rate of simple interest increases from num__10.0 to num__12 num__0.5 % a man ' s yearly income increases by rs . num__1250 . his principal in rs . is ? <o> a ) num__29879 <o> b ) num__50000 <o> c ) num__72900 <o> d ) num__27992 <o> e ) num__89282 |
let the sum be rs . x . then ( x * num__12.5 * num__0.01 ) - ( x * num__10 * num__1 ) / num__100 = num__1250 num__25 x - num__20 x = num__250000 x = num__50000 answer : b <eor> b <eos> |
b |
percent__20.0__250000.0__ percent__100.0__50000.0__ |
percent__20.0__250000.0__ percent__100.0__50000.0__ |
| two pipes a and b can fill a cistern in num__37 num__0.5 minutes and num__45 minutes respectively . both pipes are opened . the cistern will be filled in just half an hour if pipe b is turned off after <o> a ) num__5 min <o> b ) num__9 min <o> c ) num__10 min <o> d ) num__15 min <o> e ) num__17 min |
explanation : part filled by pipe a in num__1 minute = num__2 num__1.4 part filled by pipe b in num__1 minute = num__1 num__0.8 suppose pipe b is closed after x minutes . then num__2 num__1.4 × num__30 + num__1 num__0.8 × x = num__1 num__0.8 + x / num__45 = num__1 x = num__9 answer : option b <eor> b <eos> |
b |
divide__1.0__0.5__ round__9.0__ |
divide__1.0__0.5__ divide__9.0__1.0__ |
| the sum of three consecutive odd natural numbers each divisible by num__3 is num__90 . what is the largest among them ? <o> a ) num__21 <o> b ) num__24 <o> c ) num__27 <o> d ) num__33 <o> e ) num__45 |
let the num__3 consecutive odd numbers each divisible by num__3 be x x + num__3 x + num__6 given x + x + num__3 + x + num__6 = num__90 num__3 x = num__81 x = num__27 the numbers are num__27 num__30 num__33 therefore the largest number is num__33 answer d . <eor> d <eos> |
d |
divide__81.0__3.0__ add__3.0__27.0__ add__3.0__30.0__ add__3.0__30.0__ |
divide__81.0__3.0__ add__3.0__27.0__ add__3.0__30.0__ add__3.0__30.0__ |
| a watch was sold at a loss of num__16.0 . if it was sold for rs . num__140 more there would have been a gain of num__4.0 . what is the cost price ? <o> a ) num__700 <o> b ) num__882 <o> c ) num__799 <o> d ) num__778 <o> e ) num__901 |
num__84.0 num__104.0 - - - - - - - - num__20.0 - - - - num__140 num__100.0 - - - - ? = > rs . num__700 answer : a <eor> a <eos> |
a |
percent__100.0__700.0__ |
percent__100.0__700.0__ |
| a b and c can do a work in num__18 num__30 and num__45 days respectively . if they work together in how many days will they complete the work ? <o> a ) num__9 <o> b ) num__16 <o> c ) num__17 <o> d ) num__18 <o> e ) num__11 |
one days ' s work of a b and c = num__0.0555555555556 + num__0.0333333333333 + num__0.0222222222222 = ( num__5 + num__3 + num__2 ) / num__90 = num__0.111111111111 a b and c together can do the work in num__9 days . answer : a <eor> a <eos> |
a |
subtract__5.0__3.0__ multiply__18.0__5.0__ divide__2.0__18.0__ divide__18.0__2.0__ round__9.0__ |
subtract__5.0__3.0__ multiply__18.0__5.0__ divide__2.0__18.0__ divide__18.0__2.0__ divide__18.0__2.0__ |
| in a full deck of num__52 cards there are num__13 spades . a game is played whereby a card from is chosen at random from the deck . if the card is a spade a winner is declared . if the card is not a spade it is returned to the deck the deck is reshuffled and another card is drawn . this process is repeated until a spade is drawn . what is the probability that one will pick the first spade on the second draw or later ? <o> a ) num__0.125 <o> b ) num__0.25 <o> c ) num__0.5 <o> d ) num__0.75 <o> e ) num__0.875 |
favorable case = the spade is picked in the third draw or later unfavorable case = the spade is picked in either first draw or in second draw probability = favorable outcomes / total out comes also probability = num__1 - ( unfavorable outcomes / total out comes ) unfavorable case : probability of spade picked in first draw = num__0.25 = num__0.25 i . e . favorable probability = num__1 - ( num__0.25 ) = num__0.75 answer option : d <eor> d <eos> |
d |
negate_prob__0.25__ negate_prob__0.25__ |
negate_prob__0.25__ negate_prob__0.25__ |
| in a game of billiards a can give b num__10 points in num__50 and he can give c num__30 points in num__50 . how many points can b give c in a game of num__100 ? <o> a ) num__65 <o> b ) num__60 <o> c ) num__70 <o> d ) num__50 <o> e ) num__75 |
a scores num__50 while b score num__40 and c scores num__20 . the number of points that c scores when b scores num__100 = ( num__100 * num__20 ) / num__40 = num__50 . in a game of num__100 points b gives ( num__100 - num__50 ) = num__50 points to c answer : d <eor> d <eos> |
d |
add__10.0__30.0__ subtract__50.0__30.0__ add__10.0__40.0__ |
subtract__50.0__10.0__ subtract__50.0__30.0__ subtract__100.0__50.0__ |
| a person got rs . num__55 more when he invested a certain sum at compound interest instead of simple interest for two years at num__9.0 p . a . find the sum ? <o> a ) num__6790 <o> b ) num__6890 <o> c ) num__6700 <o> d ) num__2667 <o> e ) num__2671 |
p = ( d * num__1002 ) / r num__2 = > ( num__55 * num__100 * num__100 ) / num__9 * num__9 = rs . num__6790 answer : a <eor> a <eos> |
a |
percent__100.0__6790.0__ |
percent__100.0__6790.0__ |
| rs num__70000 is divided into two parts one part is given to a person with num__10.0 interest and another part is given to a person with num__20.0 interest . at the end of first year he gets profit num__8000 find money given by num__10.0 ? <o> a ) num__30000 <o> b ) num__40000 <o> c ) num__50000 <o> d ) num__60000 <o> e ) num__70000 |
let first parrt is x and second part is y then x + y = num__70000 - - - - - - - - - - eq num__1 total profit = profit on x + profit on y num__8000 = ( x * num__10 * num__1 ) / num__100 + ( y * num__20 * num__1 ) / num__100 num__80000 = x + num__2 y - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - eq num__2 num__80000 = num__70000 + y so y = num__10000 then x = num__70000 - num__10000 = num__60000 first part = num__60000 answer : d <eor> d <eos> |
d |
percent__10.0__20.0__ percent__100.0__60000.0__ |
percent__10.0__20.0__ percent__100.0__60000.0__ |
| two carpenters working in the same pace can build num__2 desks in two hours and a half . how many desks can num__4 carpenters build in num__1 hours ? <o> a ) num__1.6 . <o> b ) num__3.6 . <o> c ) num__4.2 . <o> d ) num__5.5 . <o> e ) num__6.4 |
w = num__2 desks t = num__2.5 hrs rate of num__2 carpenters = num__2 × r rate = work done / time num__2 xr = num__2 / num__2.5 r = num__1 / num__2.5 = num__0.4 ( this is the rate of each carpenter ) work done by num__4 carpenters in num__1 hrs = num__4 × rate of each carpenter x time = num__4 × num__0.4 × num__1 = num__1.6 desks a is the correct answer . <eor> a <eos> |
a |
divide__1.0__2.5__ mile_to_km_conversion__ mile_to_km_conversion__ |
divide__1.0__2.5__ divide__4.0__2.5__ divide__4.0__2.5__ |
| ten years ago the age of anand was one - third the age of bala at that time . the present age of bala is num__12 years more than the present age of anand . find the present age of anand ? <o> a ) num__14 <o> b ) num__16 <o> c ) num__18 <o> d ) num__20 <o> e ) num__22 |
let the present ages of anand and bala be ' a ' and ' b ' respectively . a - num__10 = num__0.333333333333 ( b - num__10 ) - - - ( num__1 ) b = a + num__12 substituting b = a + num__12 in first equation a - num__10 = num__0.333333333333 ( a + num__2 ) = > num__3 a - num__30 = a + num__2 = > num__2 a = num__32 = > a = num__16 . answer : b <eor> b <eos> |
b |
subtract__12.0__10.0__ add__1.0__2.0__ multiply__3.0__10.0__ add__2.0__30.0__ divide__32.0__2.0__ divide__32.0__2.0__ |
subtract__12.0__10.0__ add__1.0__2.0__ multiply__3.0__10.0__ add__2.0__30.0__ divide__32.0__2.0__ subtract__32.0__16.0__ |
| a trader sells num__85 meters of cloth for rs . num__8925 at the profit of rs . num__15 per metre of cloth . what is the cost price of one metre of cloth ? <o> a ) num__22 <o> b ) num__77 <o> c ) num__90 <o> d ) num__77 <o> e ) num__21 |
sp of num__1 m of cloth = num__105.0 = rs . num__105 cp of num__1 m of cloth = sp of num__1 m of cloth - profit on num__1 m of cloth = rs . num__105 - rs . num__15 = rs . num__90 . answer : c <eor> c <eos> |
c |
divide__8925.0__85.0__ subtract__105.0__15.0__ round__90.0__ |
divide__8925.0__85.0__ subtract__105.0__15.0__ subtract__105.0__15.0__ |
| an express traveled at an average speed of num__100 km / hr stopping for num__4 min after every num__75 km . how long did it take to reach its destination num__300 km from the starting point ? <o> a ) num__8 hrs num__29 min <o> b ) num__6 hrs num__28 min <o> c ) num__3 hrs num__12 min <o> d ) num__6 hrs num__28 min <o> e ) num__1 hrs num__28 min |
explanation : time taken to cover num__300 km = num__3.0 = num__3 hrs . number of stoppages = num__4.0 - num__1 = num__3 total time of stoppages = num__4 x num__3 = num__12 min hence total time taken = num__3 hrs num__12 min . answer : c <eor> c <eos> |
c |
divide__300.0__100.0__ subtract__4.0__3.0__ multiply__4.0__3.0__ subtract__4.0__1.0__ |
divide__300.0__100.0__ subtract__4.0__3.0__ multiply__4.0__3.0__ subtract__4.0__1.0__ |
| the angle between the minute hand and the hour hand of a clock when the time is num__11.30 is <o> a ) num__35 ° <o> b ) num__65 ° <o> c ) num__45 ° <o> d ) num__165 ° <o> e ) num__95 ° |
angle between hands of a clock when the minute hand is behind the hour hand the angle between the two hands at m minutes past h ' o clock = num__30 ( h − m / num__5 ) + m / num__2 degree when the minute hand is ahead of the hour hand the angle between the two hands at m minutes past h ' o clock = num__30 ( m / num__5 − h ) − m / num__2 degree here h = num__11 m = num__30 and minute hand is behind the hour hand . hence the angle = num__30 ( h − m / num__5 ) + m / num__2 = num__30 ( num__11 − num__6.0 ) + num__15.0 = num__30 ( num__11 − num__6 ) + num__15 = num__30 × num__5 + num__15 = num__165 ° answer is d . <eor> d <eos> |
d |
subtract__11.0__5.0__ divide__30.0__2.0__ multiply__11.0__15.0__ round__165.0__ |
divide__30.0__5.0__ divide__30.0__2.0__ multiply__11.0__15.0__ round__165.0__ |
| alice ’ s take - home pay last year was the same each month and she saved the same fraction of her take - home pay each month . the total amount of money that she had saved at the end of the year was num__4 times the amount of that portion of her monthly take - home pay that she did not save . if all the money that she saved last year was from her take - home pay what fraction of her take - home pay did she save each month ? <o> a ) num__0.142857142857 <o> b ) num__0.2 <o> c ) num__0.25 <o> d ) num__0.333333333333 <o> e ) num__0.5 |
suppose the portion of her salary that she saved is x and the remaining portion is y x + y = total take home salary given num__12 x = num__4 y or num__3 x = y total take - home salary = x + num__3 x = num__4 x hence she saved num__0.25 - th portion of her take home salary c <eor> c <eos> |
c |
divide__12.0__4.0__ reverse__4.0__ reverse__4.0__ |
divide__12.0__4.0__ reverse__4.0__ reverse__4.0__ |
| the average of runs of a batsman of seven innings was num__38 . how many runs must be made in his next innings so as to increase his average of runs by num__3 ? <o> a ) num__72 <o> b ) num__54 <o> c ) num__70 <o> d ) num__62 <o> e ) none of these |
explanation : average after num__8 innings = num__41 required number of runs = ( num__41 num__8 ) - ( num__38 num__7 ) = num__328 - num__266 = num__62 answer : d <eor> d <eos> |
d |
add__38.0__3.0__ multiply__8.0__41.0__ multiply__38.0__7.0__ subtract__328.0__266.0__ subtract__328.0__266.0__ |
add__38.0__3.0__ multiply__8.0__41.0__ multiply__38.0__7.0__ subtract__328.0__266.0__ subtract__328.0__266.0__ |
| a num__25 cm wide path is to be made around a circular garden having a diameter of num__4 meters . approximate area of the path is square meters is <o> a ) num__3.34 <o> b ) num__45 <o> c ) num__44 <o> d ) num__33 <o> e ) num__77 |
area of the path = area of the outer circle - area of the inner circle = ∏ { num__2.0 + num__0.25 } num__2 - ∏ [ num__2.0 ] num__2 = ∏ [ num__2.252 - num__22 ] = ∏ ( num__0.25 ) ( num__4.25 ) { ( a num__2 - b num__2 = ( a - b ) ( a + b ) } = ( num__3.14 ) ( num__0.25 ) ( num__4.25 ) = num__53.38 / num__16 = num__3.34 sq m answer : option a <eor> a <eos> |
a |
volume_rectangular_prism__4.0__4.25__3.14__ square_perimeter__4.0__ triangle_area__2.0__3.34__ |
volume_rectangular_prism__4.0__4.25__3.14__ square_perimeter__4.0__ triangle_area__2.0__3.34__ |
| a train running at the speed of num__300 km / hr crosses a pole in num__33 seconds . what is the length of the train ? <o> a ) num__2750 <o> b ) num__2850 <o> c ) num__2950 <o> d ) num__3050 <o> e ) num__3150 |
speed = ( num__300 x ( num__0.277777777778 ) m / sec = ( num__83.3333333333 ) m / sec . length of the train = ( speed x time ) . length of the train = ( ( num__83.3333333333 ) x num__33 ) m = num__2750 m a <eor> a <eos> |
a |
round__2750.0__ |
round__2750.0__ |
| two pipes a and b can fill a tank in num__6 and num__12 minutes respectively . if both the pipes are used together then how long will it take to fill the tank ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__8 <o> d ) num__10 <o> e ) num__12 |
part filled by a in num__1 min . = num__0.166666666667 part filled by b in num__1 min . = num__0.0833333333333 part filled by ( a + b ) in num__1 min . = num__0.166666666667 + num__0.0833333333333 = num__0.25 . both the pipes can fill the tank in num__4 minutes . answer : a <eor> a <eos> |
a |
divide__1.0__6.0__ divide__1.0__12.0__ add__0.1667__0.0833__ divide__1.0__0.25__ round__4.0__ |
divide__1.0__6.0__ divide__1.0__12.0__ add__0.1667__0.0833__ divide__1.0__0.25__ round__4.0__ |
| the sum of two numbers is num__40 and their difference is num__28 . find their product . <o> a ) num__104 <o> b ) num__108 <o> c ) num__114 <o> d ) num__204 <o> e ) none |
sol . let the numbers be x and y . then x + y = num__40 and x - y = num__28 num__2 x = num__68 = > x = num__34 y = num__6 xy = num__34 * num__6 = num__204 answer : d <eor> d <eos> |
d |
add__40.0__28.0__ divide__68.0__2.0__ subtract__40.0__34.0__ multiply__6.0__34.0__ multiply__6.0__34.0__ |
add__40.0__28.0__ divide__68.0__2.0__ subtract__40.0__34.0__ multiply__6.0__34.0__ multiply__6.0__34.0__ |
| an integer n that is greater than num__1 is said to be ` ` prime - saturated ' ' if it has no prime factor greater than or equal to n √ n . which of the following integers is prime saturated ? <o> a ) num__6 <o> b ) num__35 <o> c ) num__46 <o> d ) num__66 <o> e ) num__75 |
according to the definition integer is said to be ` ` prime - saturated ' ' if the square of its largest prime is smaller than the integer itself . only e satisfies this : num__75 = num__3 * num__5 ^ num__2 - - > the largest prime is num__5 - - > num__5 ^ num__2 = num__25 < num__75 . answer : e . <eor> e <eos> |
e |
subtract__3.0__1.0__ divide__75.0__3.0__ multiply__1.0__75.0__ |
subtract__3.0__1.0__ divide__75.0__3.0__ multiply__1.0__75.0__ |
| in the x - y plane there are num__4 points ( num__00 ) ( num__04 ) ( num__74 ) and ( num__70 ) . if these num__4 points makes a rectangle what is the probability that x + y < num__4 ? <o> a ) num__0.666666666667 <o> b ) num__0.6 <o> c ) num__0.285714285714 <o> d ) num__0.444444444444 <o> e ) num__0.7 |
the line y = - x + num__4 intersects the rectangle and these three points of intersection ( num__00 ) ( num__04 ) and ( num__40 ) form a triangle . the points below the line y = - x + num__4 satisfy x + y < num__4 . the area of this triangle is ( num__0.5 ) ( num__4 ) ( num__4 ) = num__8 the area of the rectangle is num__28 . p ( x + y < num__4 ) = num__0.285714285714 = num__0.285714285714 the answer is c . <eor> c <eos> |
c |
divide__4.0__0.5__ divide__8.0__28.0__ divide__8.0__28.0__ |
divide__4.0__0.5__ divide__8.0__28.0__ divide__8.0__28.0__ |
| a father said his son ` ` i was as old as you are at present at the time of your birth . ` ` if the father age is num__38 now the son age num__5 years back was : <o> a ) num__14 <o> b ) num__36 <o> c ) num__28 <o> d ) num__29 <o> e ) num__11 |
explanation : let the son ' s present age be x years . then ( num__38 - x ) = x x = num__19 . son ' s age num__5 years back = ( num__19 - num__5 ) = num__14 years answer : a ) num__14 <eor> a <eos> |
a |
subtract__19.0__5.0__ subtract__19.0__5.0__ |
subtract__19.0__5.0__ subtract__19.0__5.0__ |
| at what rate percent on simple interest will a sum of money double itself in num__30 years ? <o> a ) num__3 num__0.333333333333 % <o> b ) num__3 num__0.166666666667 % <o> c ) num__3 num__0.2 % <o> d ) num__3 num__1.66666666667 % <o> e ) num__7 num__0.333333333333 % |
p = ( p * num__30 * r ) / num__100 r = num__3 num__0.333333333333 % answer : a <eor> a <eos> |
a |
percent__3.0__100.0__ |
percent__3.0__100.0__ |
| a certain sum is invested at simple interest at num__18.0 p . a . for two years instead of investing at num__12.0 p . a . for the same time period . therefore the interest received is more by rs . num__840 . find the sum ? <o> a ) num__7000 <o> b ) num__7029 <o> c ) num__7290 <o> d ) num__7010 <o> e ) num__7102 |
let the sum be rs . x . ( x * num__18 * num__2 ) / num__100 - ( x * num__12 * num__2 ) / num__100 = num__840 = > num__36 x / num__100 - num__24 x / num__100 = num__840 = > num__12 x / num__100 = num__840 = > x = num__7000 . answer : a <eor> a <eos> |
a |
percent__100.0__7000.0__ |
percent__100.0__7000.0__ |
| out of three consecutive odd numbers ten times the first number is equal to addition of thrice the third number and adding num__8 to thrice the second . what is the first number ? <o> a ) num__0.833333333333 <o> b ) num__1.40625 <o> c ) num__4.66666666667 <o> d ) num__6.5 <o> e ) num__1.875 |
description : = > num__10 x = num__3 ( x + num__2 ) + num__8 + num__3 ( x + num__4 ) = > num__10 x = num__6 x + num__26 = > num__4 x = num__26 x = num__6.5 = num__6.5 answer d <eor> d <eos> |
d |
subtract__10.0__8.0__ divide__8.0__2.0__ subtract__8.0__2.0__ divide__26.0__4.0__ divide__26.0__4.0__ |
subtract__10.0__8.0__ divide__8.0__2.0__ add__2.0__4.0__ divide__26.0__4.0__ divide__26.0__4.0__ |
| a number x is multiplied with itself and then added to the product of num__4 and x . if the result of these two operations is num__5 what is the value of x ? <o> a ) - num__4 <o> b ) - num__2 <o> c ) num__2 <o> d ) num__1 and - num__5 <o> e ) can not be determined . |
a number x is multiplied with itself - - > x ^ num__2 added to the product of num__4 and x - - > x ^ num__2 + num__4 x if the result of these two operations is - num__4 - - > x ^ num__2 + num__4 x = num__5 i . e x ^ num__2 + num__4 x - num__5 = num__0 is the quadratic equation which needs to be solved . ( x - num__1 ) ( x + num__5 ) = num__0 hence x = num__1 . c = - num__5 imo d <eor> d <eos> |
d |
subtract__5.0__4.0__ reverse__1.0__ |
subtract__5.0__4.0__ subtract__5.0__4.0__ |
| in a school with num__604 students the average age of the boys is num__12 years and that of the girls is num__11 years . if the average age of the school is num__11 years num__9 months then the number of girls in the school is <o> a ) num__150 <o> b ) num__151 <o> c ) num__250 <o> d ) num__350 <o> e ) none |
sol . let the number of grils be x . then number of boys = ( num__600 - x ) . then ( num__11 num__0.75 × num__604 ) ⇔ num__11 x + num__12 ( num__604 - x ) ⇔ x = num__7248 - num__7097 ⇔ num__151 . answer b <eor> b <eos> |
b |
divide__9.0__12.0__ multiply__604.0__12.0__ subtract__7248.0__7097.0__ subtract__7248.0__7097.0__ |
divide__9.0__12.0__ multiply__604.0__12.0__ subtract__7248.0__7097.0__ subtract__7248.0__7097.0__ |
| num__60 + num__5 * num__12 / ( num__60.0 ) = ? <o> a ) num__22 <o> b ) num__77 <o> c ) num__29 <o> d ) num__61 <o> e ) num__21 |
num__60 + num__5 * num__12 / ( num__60.0 ) = num__60 + num__5 * num__12 / ( num__60 ) = num__60 + ( num__5 * num__12 ) / num__60 = num__60 + num__1 = num__61 . answer : d <eor> d <eos> |
d |
add__60.0__1.0__ add__60.0__1.0__ |
add__60.0__1.0__ add__60.0__1.0__ |
| find large no . from below question the difference of two numbers is num__1365 . on dividing the larger number by the smaller we get num__6 as quotient and the num__10 as remainder <o> a ) num__1235 <o> b ) num__1456 <o> c ) num__1567 <o> d ) num__1636 <o> e ) num__1635 |
let the smaller number be x . then larger number = ( x + num__1365 ) . x + num__1365 = num__6 x + num__10 num__5 x = num__1355 x = num__271 large number = num__271 + num__1365 = num__1636 d <eor> d <eos> |
d |
subtract__1365.0__10.0__ divide__1355.0__5.0__ add__1365.0__271.0__ add__1365.0__271.0__ |
subtract__1365.0__10.0__ divide__1355.0__5.0__ add__1365.0__271.0__ add__1365.0__271.0__ |
| a train num__110 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__7 sec <o> b ) num__6 sec <o> c ) num__4 sec <o> d ) num__2 sec <o> e ) num__1 sec |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__110 * num__0.0545454545455 ] sec = num__6 sec answer : b <eor> b <eos> |
b |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ round__6.0__ |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ divide__110.0__18.3333__ |
| a rectangular park num__60 m long and num__40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn . if the area of the lawn is num__2109 sq . m then what is the width of the road ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
area of the park = ( num__60 x num__40 ) m num__2 = num__2400 m num__2 . area of the lawn = num__2109 m num__2 . area of the crossroads = ( num__2400 - num__2109 ) m num__2 = num__291 m num__2 . let the width of the road be x metres . then num__60 x + num__40 x - x num__2 = num__291 x num__2 - num__100 x + num__291 = num__0 ( x - num__97 ) ( x - num__3 ) = num__0 x = num__3 . a <eor> a <eos> |
a |
multiply__60.0__40.0__ triangle_area__2.0__3.0__ |
multiply__60.0__40.0__ triangle_area__2.0__3.0__ |
| in what time will two trains cross each other completely which are running on the same parallel lines in opposite directions each train running with a speed of num__90 kmph being num__160 m and num__140 m in length respectively ? <o> a ) num__5.6 sec <o> b ) num__6.2 sec <o> c ) num__6.0 sec <o> d ) num__7.2 sec <o> e ) num__6.8 sec |
d = num__160 m + num__140 m = num__300 m * num__0.001 = num__0.3 kms rs = num__90 + num__90 = num__180 kmph t = ( num__0.3 / num__180 ) * num__3600 = num__6.0 sec answer : c <eor> c <eos> |
c |
add__160.0__140.0__ divide__90.0__300.0__ round__6.0__ |
add__160.0__140.0__ divide__90.0__300.0__ round__6.0__ |
| by selling an article at rs . num__800 a shopkeeper makes a profit of num__25.0 . at what price should he sell the article so as to make a loss of num__25.0 ? <o> a ) rs . num__429 <o> b ) rs . num__480 <o> c ) rs . num__429 <o> d ) rs . num__427 <o> e ) rs . num__482 |
sp = num__800 profit = num__25.0 cp = ( sp ) * [ num__100 / ( num__100 + p ) ] = num__800 * [ num__0.8 ] = num__640 loss = num__25.0 = num__25.0 of num__640 = rs . num__160 sp = cp - loss = num__640 - num__160 = rs . num__480 answer : b <eor> b <eos> |
b |
percent__25.0__640.0__ percent__100.0__480.0__ |
percent__25.0__640.0__ percent__100.0__480.0__ |
| the jogging track in a sports complex is num__1000 meters in circumference . deepak and his wife start from the same point and walk in opposite directions at num__20 km / hr and num__12 km / hr respectively . they will meet for the first time in ? <o> a ) num__32 min <o> b ) num__40 min <o> c ) num__35 min <o> d ) num__25 min <o> e ) num__20 min |
clearly the two will meet when they are num__1000 m apart to be num__20 + num__12 = num__32 km apart they take num__1 hour to be num__1000 m apart they take num__32 * num__1.0 = num__32 min . answer is a <eor> a <eos> |
a |
add__20.0__12.0__ add__20.0__12.0__ |
add__20.0__12.0__ add__20.0__12.0__ |
| what is num__0.625 as a percent <o> a ) num__62.5 <o> b ) num__63.0 <o> c ) num__63.5 <o> d ) num__64.5 <o> e ) num__65 % |
num__0.625 as a percent = ( num__0.625 ) x num__100 = num__0.625 x num__100 = num__62.5 answer : a i . e . num__62.5 <eor> a <eos> |
a |
percent__62.5__100.0__ |
percent__62.5__100.0__ |
| if num__4 x + y = num__8 and y − num__3 x = num__7 then what is the value of x + num__2 y ? <o> a ) num__0.142857142857 <o> b ) num__3 <o> c ) num__15 <o> d ) num__7.42857142857 <o> e ) num__8.57142857143 |
num__4 x + y = num__8 − num__3 x + y = num__7 adding both the equations - - > x + num__2 y = num__15 answer : c <eor> c <eos> |
c |
add__8.0__7.0__ add__8.0__7.0__ |
add__8.0__7.0__ add__8.0__7.0__ |
| num__7 ^ num__9 - num__7 ^ num__2 is approximation of which of the following ? <o> a ) num__7 ^ num__9 <o> b ) num__9 ^ num__8 <o> c ) num__9 ^ num__9 <o> d ) num__9 ^ num__11 <o> e ) num__9 ^ num__10 |
= > since num__9 ^ num__2 is a small number compared to num__7 ^ num__9 it can be disregarded . thus num__7 ^ num__9 - num__7 ^ num__2 = num__7 ^ num__9 is derived and the answer is a . <eor> a <eos> |
a |
subtract__9.0__2.0__ |
subtract__9.0__2.0__ |
| mike jim and bob are all professional fisherman . mike can catch num__30 fish in one hour jim can catch twice as much and bob can catch num__50.0 more than jim . if the three started to fish together and after num__40 minutes mike and bob left how many fish did the three fishermen catch in one hour ? <o> a ) num__110 <o> b ) num__120 <o> c ) num__140 <o> d ) num__130 <o> e ) num__112 |
all of them catch fishes in relation to number num__30 . . . . num__0.666666666667 * num__30 + num__2 * num__30 + num__2 * num__1.5 * num__30 * num__0.666666666667 = num__140 answer is c <eor> c <eos> |
c |
round__140.0__ |
round__140.0__ |
| a boat can travel with a speed of num__22 km / hr in still water . if the speed of the stream is num__5 km / hr find the time taken by the boat to go num__135 km downstream <o> a ) num__5 hours <o> b ) num__4 hours <o> c ) num__3 hours <o> d ) num__2 hours <o> e ) none of these |
explanation : speed of the boat in still water = num__22 km / hr speed of the stream = num__5 km / hr speed downstream = ( num__22 + num__5 ) = num__27 km / hr distance travelled downstream = num__135 km time taken = distance / speed = num__5.0 = num__5 hours . answer : option a <eor> a <eos> |
a |
add__22.0__5.0__ round__5.0__ |
add__22.0__5.0__ divide__135.0__27.0__ |
| a tradesman by means of his false balance defrauds to the extent of num__26.0 ? in buying goods as well as by selling the goods . what percent does he gain on his outlay ? <o> a ) num__58.76 <o> b ) num__49.89 <o> c ) num__84.67 <o> d ) num__24.65 <o> e ) num__45 % |
g % = num__26 + num__26 + ( num__26 * num__26 ) / num__100 = num__58.76 answer : a <eor> a <eos> |
a |
percent__58.76__100.0__ |
percent__58.76__100.0__ |
| how many positive four - digit integers do not contain any digits other than num__1 or num__0 and are such that the sum of the first two digits equals the sum of the last two digits ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__8 <o> d ) num__10 <o> e ) num__12 |
the sum of the first two ( last two ) digits can not be num__0 because num__0000 is not a four - digit integer . if the sum of the first two digits is num__1 two variants are possible : num__1001 and num__1010 ( we have to remember that num__0 can not be the first digit ) . if the sum of the first two digits is num__2 one variant is possible : num__1111 . in all there are three numbers that satisfy the given conditions : num__1001 num__1010 and num__1111 . answer : a <eor> a <eos> |
a |
add__1.0__2.0__ |
add__1.0__2.0__ |
| two pipes a and b can separately fill a cistern in num__10 and num__15 minutes respectively . a person opens both the pipes together when the cistern should have been was full he finds the waste pipe open . he then closes the waste pipe and in another num__4 minutes the cistern was full . in what time can the waste pipe empty the cistern when fill ? <o> a ) num__0.333333333333 <o> b ) num__0.111111111111 <o> c ) num__0.25 <o> d ) num__0.5 <o> e ) num__0.166666666667 |
num__0.1 + num__0.0666666666667 = num__0.166666666667 * num__4 = num__0.666666666667 num__1 - num__0.666666666667 = num__0.333333333333 num__0.1 + num__0.0666666666667 - num__1 / x = num__0.333333333333 x = num__8 . answer : a <eor> a <eos> |
a |
add__0.1__0.0667__ divide__10.0__15.0__ multiply__10.0__0.1__ subtract__1.0__0.6667__ subtract__1.0__0.6667__ |
add__0.1__0.0667__ divide__10.0__15.0__ multiply__10.0__0.1__ subtract__1.0__0.6667__ subtract__1.0__0.6667__ |
| all the faces of cubes are painted with red colour . it is then cut into num__64 equal small cubes . find how many small cubes have no faces coloured ? <o> a ) num__24 <o> b ) num__8 <o> c ) num__16 <o> d ) num__0 <o> e ) num__4 |
there are num__64 small cubes hence one side side of the big cube is num__3 √ num__64 = num__4 cm number of small cubes having only one faces coloured = ( x - num__2 ) ^ num__3 here x = side of big cube / side of small cube x = num__4.0 x = num__4 required number = ( num__4 - num__2 ) ^ num__3 = num__8 answer : b <eor> b <eos> |
b |
square_perimeter__2.0__ square_perimeter__2.0__ |
power__2.0__3.0__ power__2.0__3.0__ |
| for what values of ' k ' will the pair of equations num__3 x + num__4 y = num__12 and kx + num__12 y = num__30 not have a unique solution ? <o> a ) num__12 <o> b ) num__9 <o> c ) num__3 <o> d ) num__7.5 <o> e ) num__2.5 |
explanatory answer condition for unique solution to linear equations a system of linear equations ax + by + c = num__0 and dx + ey + g = num__0 will have a unique solution if the two lines represented by the equations ax + by + c = num__0 and dx + ey + g = num__0 intersect at a point . i . e . if the two lines are neither parallel nor coincident . essentially the slopes of the two lines should be different . what does that translate into ? ax + by + c = num__0 and dx + ey + g = num__0 will intersect at one point if their slopes are different . express both the equations in the standardized y = mx + c format where ' m ' is the slope of the line and ' c ' is the y - intercept . ax + by + c = num__0 can be written as y = − abx − ca − abx − ca and dx + ey + g = num__0 can be written as y = − dex − ge − dex − ge slope of the first line is − ab − ab and that of the second line is − de − de for a unique solution the slopes of the lines should be different . ∴ − ab ≠ − de − ab ≠ − de or ad ≠ bead ≠ be condition for the equations to not have a unique solution the slopes should be equal or ad = bead = be apply the condition in the given equations to find k in the question given above a = num__3 b = num__4 d = k and e = num__12 . therefore num__3 k = num__4123 k = num__412 or ' k ' should be equal to num__9 for the pair of equations to not have a unique solution . when k = num__9 the system of equations will represent a pair of parallel lines ( their y - intercepts are different ) . so there will be no solution . answer b <eor> b <eos> |
b |
subtract__12.0__3.0__ subtract__12.0__3.0__ |
subtract__12.0__3.0__ subtract__12.0__3.0__ |
| if two positive numbers are in the ratio num__0.166666666667 : num__0.2 then by what percent is the second number more than the first ? <o> a ) num__70.0 <o> b ) num__90.0 <o> c ) num__60.0 <o> d ) num__50.0 <o> e ) num__20 % |
given ratio = num__0.166666666667 : num__0.2 = num__5 : num__6 let first number be num__5 x and the second number be num__9 x . the second number is more than first number by num__1 x . required percentage = x / num__5 x * num__100 = num__20.0 . answer : e <eor> e <eos> |
e |
reverse__0.2__ multiply__0.2__5.0__ multiply__0.2__100.0__ multiply__0.2__100.0__ |
reverse__0.2__ multiply__0.2__5.0__ multiply__0.2__100.0__ multiply__0.2__100.0__ |
| two trains of length num__100 m and num__200 m are num__90 m apart . they start moving towards each other on parallel tracks at speeds num__54 kmph and num__72 kmph . after how much time will the trains meet ? <o> a ) num__2.33333333333 sec <o> b ) num__10.6666666667 sec <o> c ) num__2.57142857143 sec <o> d ) num__10.6666666667 sec <o> e ) num__26.5 sec |
they are moving in opposite directions relative speed is equal to the sum of their speeds . relative speed = ( num__54 + num__72 ) * num__0.277777777778 = num__7 * num__5 = num__35 mps . the time required = d / s = num__2.57142857143 = num__2.57142857143 sec . answer : c <eor> c <eos> |
c |
multiply__5.0__7.0__ divide__90.0__35.0__ divide__90.0__35.0__ |
multiply__5.0__7.0__ divide__90.0__35.0__ divide__90.0__35.0__ |
| if a / num__2 = b / num__3 = c / num__5 then the value of ( a + b + c ) / c is : <o> a ) num__1 / √ num__5 <o> b ) √ num__2 <o> c ) num__2 <o> d ) num__5 <o> e ) num__6 |
a / num__2 = b / num__3 = c / num__5 = k ( say ) . then a = num__2 k b = num__3 k c = num__5 k . ( a + b + c ) / c = ( num__2 k + num__3 k + num__5 k ) / num__5 k = num__10 k / num__5 k = num__2 . answer : c <eor> c <eos> |
c |
multiply__2.0__5.0__ subtract__5.0__3.0__ |
multiply__2.0__5.0__ divide__10.0__5.0__ |
| the ratio between the radii of two spheres is num__1 : num__3 . find the ratio between their volumes ? <o> a ) num__1 : num__28 <o> b ) num__1 : num__27 <o> c ) num__1 : num__22 <o> d ) num__1 : num__21 <o> e ) num__1 : num__29 |
r num__1 : r num__2 = num__1 : num__3 r num__13 : r num__23 = num__1 : num__27 answer : b <eor> b <eos> |
b |
volume_cube__3.0__ volume_cube__1.0__ |
volume_cube__3.0__ volume_cube__1.0__ |
| the speed of a boat in still water is num__20 km / hr and the rate of current is num__5 km / hr . the distance travelled downstream in num__21 minutes is : <o> a ) num__9.75 <o> b ) num__5.75 <o> c ) num__8.75 <o> d ) num__6.75 <o> e ) num__5.15 |
explanation : speed downstream = ( num__20 + num__5 ) kmph = num__25 kmph distance travelled = ( num__25 * ( num__0.35 ) ) km = num__8.75 km . answer : c <eor> c <eos> |
c |
add__20.0__5.0__ multiply__25.0__0.35__ round__8.75__ |
add__20.0__5.0__ multiply__25.0__0.35__ multiply__25.0__0.35__ |
| the simple form of the ratio num__1.33333333333 : num__0.4 is ? <o> a ) num__10 : num__6 <o> b ) num__10 : num__3 <o> c ) num__15 : num__3 <o> d ) num__25 : num__3 <o> e ) num__30 : num__3 |
num__1.33333333333 : num__0.4 = num__20 : num__6 = num__10 : num__3 answer : b <eor> b <eos> |
b |
subtract__20.0__10.0__ |
subtract__20.0__10.0__ |
| during a sale the price of a pair of shoes is marked down num__15.0 from the regular price . after the sale ends the price goes back to the original price . what is the percent of increase to the nearest percent from the sale price back to the regular price for the shoes ? <o> a ) num__9.0 <o> b ) num__10.0 <o> c ) num__11.0 <o> d ) num__18.0 <o> e ) num__90 % |
assume the price = num__100 price during sale = num__85 price after sale = num__100 percent increase = num__0.176470588235 * num__100 = num__18.0 approx . correct option : d <eor> d <eos> |
d |
percent__18.0__100.0__ |
percent__18.0__100.0__ |
| the train num__160 m long is running with a speed of num__80 km / hr . in what time will it pass a man running in same direction with the speed of num__8 km / hr . <o> a ) num__16 seconds <o> b ) num__8 seconds <o> c ) num__15 seconds <o> d ) num__6 seconds <o> e ) num__28 seconds |
man and train are running in same direction relative speed = num__80 - num__8 = num__72 kmph . = ( num__72 * num__5 ) / num__18 = num__20 m / sec length of the train = num__160 m . time taken to pass the man = num__8.0 = num__8 seconds . answer : option b <eor> b <eos> |
b |
subtract__80.0__8.0__ divide__160.0__8.0__ round__8.0__ |
subtract__80.0__8.0__ divide__160.0__8.0__ divide__160.0__20.0__ |
| if the perimeter of a rectangular garden is num__950 m its length when its breadth is num__100 m is ? <o> a ) num__338 m <o> b ) num__778 m <o> c ) num__375 m <o> d ) num__276 m <o> e ) num__971 m |
num__2 ( l + num__100 ) = num__950 = > l = num__375 m answer : c <eor> c <eos> |
c |
round__375.0__ |
round__375.0__ |
| a man is walking at a speed of num__10 km per hour . after every kilometre he takes rest for num__5 minutes . how much time will be take to cover a distance of num__5 kilometres ? <o> a ) num__48 min . <o> b ) num__50 min . <o> c ) num__45 min . <o> d ) num__55 min <o> e ) none of these |
explanation rest time = number of rest × time for each rest = num__4 × num__5 = num__20 minutes total time to cover num__5 km = ( num__5 ⁄ num__10 × num__60 ) minutes + num__20 minutes = num__50 minutes answer b <eor> b <eos> |
b |
multiply__5.0__4.0__ hour_to_min_conversion__ multiply__10.0__5.0__ round__50.0__ |
multiply__5.0__4.0__ hour_to_min_conversion__ multiply__10.0__5.0__ round__50.0__ |
| tea worth rs . num__126 per kg and rs . num__135 per kg are mixed with a third variety of tea in the ratio num__1 : num__1 : num__2 . if the mixture is worth rs . num__153 per kg what is the price of the third variety per kg ? <o> a ) rs . num__182.50 <o> b ) rs . num__170.5 <o> c ) rs . num__175.50 <o> d ) rs . num__180 <o> e ) rs . num__190 |
explanation : tea worth rs . num__126 per kg and rs . num__135 per kg are mixed in the ratio num__1 : num__1 so their average price = ( num__126 + num__135 ) / num__2 = num__130.5 hence let ' s consider that the mixture is formed by mixing two varieties of tea one at rs . num__130.50 per kg and the other at rs . x per kg in the ratio num__2 : num__2 i . e . num__1 : num__1 . now let ' s find out x by rule of alligation ( x â ˆ ’ num__153 ) : num__22.5 = num__1 : num__1 â ‡ ’ x â ˆ ’ num__153 = num__22.50 â ‡ ’ x = num__153 + num__22.5 = num__175.5 answer is c <eor> c <eos> |
c |
subtract__153.0__130.5__ add__153.0__22.5__ multiply__1.0__175.5__ |
subtract__153.0__130.5__ add__153.0__22.5__ add__153.0__22.5__ |
| at a certain paint store forest green is made by mixing num__4 parts blue paint with num__3 parts yellow paint . verdant green is made by mixing num__4 parts yellow paint with num__3 parts blue paint . how many liters of yellow paint must be added to num__35 liters of forest green to change it to verdant green ? <o> a ) num__0.333333333333 <o> b ) num__0.666666666667 <o> c ) num__1.33333333333 <o> d ) num__11.6666666667 <o> e ) num__3.33333333333 |
num__35 liter of forset green have num__20 liter of blue and num__15 liter of yellow suppose we add x liter of yellow to make it a verdant green so the ratio of blue to yellow in verdant green is ¾ so the equation is blue / yellow = num__20 / ( num__15 + x ) = ¾ num__45 + num__3 x = num__80 = > x = num__11.6666666667 answer : d <eor> d <eos> |
d |
subtract__35.0__20.0__ multiply__3.0__15.0__ multiply__4.0__20.0__ divide__35.0__3.0__ divide__35.0__3.0__ |
subtract__35.0__20.0__ multiply__3.0__15.0__ add__35.0__45.0__ divide__35.0__3.0__ divide__35.0__3.0__ |
| which of these numbers is not divisible by num__3 ? <o> a ) num__339 <o> b ) num__342 <o> c ) num__552 <o> d ) num__1111 <o> e ) num__672 |
one may answer this question using a calculator and test for divisibility by num__3 . however we can also test for divisibilty by adding the digits and if the result is divisible by num__3 then the number is divisible by num__3 . num__3 + num__3 + num__9 = num__15 divisible by num__3 . num__3 + num__4 + num__2 = num__9 divisible by num__3 . num__5 + num__5 + num__2 = num__12 divisible by num__3 . num__1 + num__1 + num__1 + num__1 = num__4 not divisible by num__3 . the number num__1111 is not divisible by num__3 the answer is d . <eor> d <eos> |
d |
add__3.0__2.0__ multiply__3.0__4.0__ subtract__3.0__2.0__ multiply__1.0__1111.0__ |
add__3.0__2.0__ add__3.0__9.0__ subtract__3.0__2.0__ multiply__1.0__1111.0__ |
| if ( t - num__8 ) is a factor of t ^ num__2 - kt - num__40 then k = <o> a ) num__16 <o> b ) num__1 <o> c ) num__2 <o> d ) num__6 <o> e ) num__14 |
t ^ num__2 - kt - num__48 = ( t - num__8 ) ( t + m ) where m is any positive integer . if num__6.0 = num__6 then we know as a matter of fact that : m = + num__6 and thus k = num__8 - num__6 = num__1 t ^ num__2 - kt - m = ( t - a ) ( t + m ) where a > m t ^ num__2 + kt - m = ( t - a ) ( t + m ) where a < m t ^ num__2 - kt + m = ( t - a ) ( t - m ) t ^ num__2 + kt + m = ( t + a ) ( t + m ) b <eor> b <eos> |
b |
add__8.0__40.0__ subtract__8.0__2.0__ reverse__1.0__ |
add__8.0__40.0__ subtract__8.0__2.0__ subtract__2.0__1.0__ |
| a man buys a cycle for rs . num__1750 and sells it at a loss of num__8.0 . what is the selling price of the cycle ? <o> a ) s . num__1090 <o> b ) s . num__1610 <o> c ) s . num__1190 <o> d ) s . num__1202 <o> e ) s . num__1092 |
s . p . = num__92.0 of rs . num__1750 = rs . num__92 x num__17.5 = rs . num__1610 answer : option b <eor> b <eos> |
b |
percent__92.0__1750.0__ percent__92.0__1750.0__ |
percent__92.0__1750.0__ percent__92.0__1750.0__ |
| a train passes a station platform in num__42 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__123 <o> b ) num__166 <o> c ) num__240 <o> d ) num__157 <o> e ) num__330 |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x meters . then ( x + num__300 ) / num__42 = num__15 x + num__300 = num__630 x = num__330 m . answer : e <eor> e <eos> |
e |
multiply__20.0__15.0__ multiply__42.0__15.0__ subtract__630.0__300.0__ round__330.0__ |
multiply__20.0__15.0__ multiply__42.0__15.0__ subtract__630.0__300.0__ round__330.0__ |
| the value of underrot num__72 pi / num__9 + underrot num__4 is most nearly equal to which of the following integers ? ( please refer to the picture below for the exact version of the question ) <o> a ) num__27 <o> b ) num__14 <o> c ) num__18 <o> d ) num__19 <o> e ) num__21 |
under root ( num__72 pi / num__9 ) ~ num__25 under root ( num__4 ) = num__2 hence num__9 + num__1 = num__10 is the approx answer . a is the correct answer . <eor> a <eos> |
a |
add__9.0__1.0__ add__2.0__25.0__ |
add__9.0__1.0__ add__2.0__25.0__ |
| a boy swims downstream num__91 km and upstream num__21 km taking num__7 hours each time what is the speed of the boy in still water ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
num__91 - - - num__7 ds = num__13 ? - - - - num__1 num__21 - - - - num__7 us = num__3 ? - - - - num__1 m = ? m = ( num__13 + num__3 ) / num__2 = num__8 answer : c <eor> c <eos> |
c |
divide__91.0__7.0__ divide__21.0__7.0__ subtract__3.0__1.0__ subtract__21.0__13.0__ round__8.0__ |
divide__91.0__7.0__ divide__21.0__7.0__ subtract__3.0__1.0__ subtract__21.0__13.0__ subtract__21.0__13.0__ |
| pipes a and b can fill a tank in num__5 hours and num__6 hours respectively . pipe c can empty it in num__12 hours . if all the three pipes are opened together then the tank will be filled in . <o> a ) num__3 num__1.8 <o> b ) num__3 num__0.529411764706 <o> c ) num__3 num__1.4 <o> d ) num__3 num__0.411764705882 <o> e ) none of these |
explanation : net part filled in num__1 hour = ( num__0.2 + num__0.166666666667 - num__0.0833333333333 ) = num__0.283333333333 hrs = num__3 num__0.529411764706 answer is b <eor> b <eos> |
b |
subtract__6.0__5.0__ divide__1.0__5.0__ divide__1.0__6.0__ divide__1.0__12.0__ add__0.2__0.0833__ round__3.0__ |
subtract__6.0__5.0__ divide__1.0__5.0__ divide__1.0__6.0__ divide__1.0__12.0__ add__0.2__0.0833__ subtract__6.0__3.0__ |
| if num__1 / ( num__6 + num__3 / x ) = num__1 then x = <o> a ) - num__0.6 <o> b ) num__1 <o> c ) num__0.333333333333 <o> d ) - num__0.333333333333 <o> e ) - num__3 |
the expression num__1 / ( num__6 + num__3 / x ) = num__1 / x ) should have been equal to something . if num__1 / ( num__6 + num__3 / x ) = num__1 = > x / ( num__6 x + num__3 ) = num__1 = > x = num__6 x + num__3 = > - num__5 x = num__3 = > x = - num__0.6 correct option : a <eor> a <eos> |
a |
subtract__6.0__1.0__ divide__3.0__5.0__ multiply__1.0__0.6__ |
subtract__6.0__1.0__ divide__3.0__5.0__ divide__3.0__5.0__ |
| a is the hundreds digit of the three digit integer x b is the tens digit of x and c is the units digit of x . num__4 a = num__2 b = c and a > num__0 . what is the difference between the two greatest possible values of x ? tip : dont stop till you have exhausted all answer choices to arrive at the correct one . <o> a ) num__124 <o> b ) num__297 <o> c ) num__394 <o> d ) num__421 <o> e ) num__842 |
ratio of a : b : c = num__1 : num__2 : num__4 two possible greatest single digit values for c are num__8 and num__4 if c is num__8 then x = num__248 if c is num__4 then x = num__124 difference = num__248 - num__124 = num__124 a is the answer <eor> a <eos> |
a |
multiply__4.0__2.0__ divide__248.0__2.0__ multiply__1.0__124.0__ |
multiply__4.0__2.0__ divide__248.0__2.0__ subtract__248.0__124.0__ |
| if ( c - a ) / ( c - b ) = num__1 then ( num__5 b - num__2 a ) / ( c - a ) = <o> a ) num__0.5 <o> b ) num__1 <o> c ) num__1.5 <o> d ) num__2 <o> e ) num__6 |
let ' s say c = num__3 b = num__2 a = num__2 so that our num__1 st expression holds true . now ibsert those numbers in the second expression and we ' ll get num__6 answer e ( hopefully ) ) ) <eor> e <eos> |
e |
add__1.0__2.0__ add__1.0__5.0__ add__1.0__5.0__ |
add__1.0__2.0__ add__1.0__5.0__ add__1.0__5.0__ |
| michael cashed a check for $ num__1300 and received only $ num__10 and $ num__50 bills in return . during the course of a day he used num__15 bills and then lost the rest of the money . if the number of $ num__10 bills used was either one more or one less than the number of $ num__50 bills used what is the minimum possible amount of money that was lost ? <o> a ) $ num__830 <o> b ) $ num__800 <o> c ) $ num__770 <o> d ) $ num__730 <o> e ) $ num__700 |
num__1 . num__15 bills spent num__2 . number of $ num__10 bills is one more or one less than number of $ num__50 bills . so num__2 choices num__1 . num__7 * num__10 + num__8 * num__50 = $ num__470 num__2 . num__8 * num__10 + num__7 * num__50 = $ num__430 least money lost implies most money spent . i . e $ num__470 therefore lost money = num__1300 - num__470 = num__830 . answer a <eor> a <eos> |
a |
subtract__10.0__2.0__ subtract__1300.0__470.0__ subtract__1300.0__470.0__ |
subtract__10.0__2.0__ subtract__1300.0__470.0__ subtract__1300.0__470.0__ |
| in the coordinate plane line a has a slope of - num__2 and an x - intercept of num__2 . line b has a slope of num__5 and a y - intercept of - num__10 . if the two lines intersect at the point ( a b ) what is the sum a + b ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
the equation of line a is y = - num__2 x + num__4 the equation of line b is y = num__5 x - num__10 num__5 x - num__10 = - num__2 x + num__4 x = num__2 y = num__0 the point of intersection is ( num__20 ) and then a + b = num__2 . the answer is c . <eor> c <eos> |
c |
multiply__2.0__10.0__ divide__10.0__5.0__ |
multiply__2.0__10.0__ subtract__4.0__2.0__ |
| a train num__100 m long crosses a platform num__125 m long in num__15 sec ; find the speed of the train ? <o> a ) num__29 <o> b ) num__28 <o> c ) num__54 <o> d ) num__27 <o> e ) num__21 |
d = num__100 + num__125 = num__225 t = num__15 s = num__15.0 * num__3.6 = num__54 kmph . answer : c <eor> c <eos> |
c |
add__100.0__125.0__ multiply__15.0__3.6__ round__54.0__ |
add__100.0__125.0__ multiply__15.0__3.6__ multiply__15.0__3.6__ |
| a and b started a partnership business investing some amount in the ratio of num__3 : num__5 . c joined then after six months with an amount equal to that of b . in what proportion should the profit at the end of one year be distributed among a b and c ? <o> a ) num__3 : num__5 : num__2 <o> b ) num__3 : num__5 : num__5 <o> c ) num__6 : num__10 : num__5 <o> d ) data inadequate <o> e ) none |
explanation let the initial investments of a and b be num__3 x and num__5 x . a : b : c = ( num__3 x x num__12 ) : ( num__5 x x num__12 ) : ( num__5 x x num__6 ) = num__36 : num__60 : num__30 = num__6 : num__10 : num__5 . answer c <eor> c <eos> |
c |
multiply__3.0__12.0__ multiply__5.0__12.0__ multiply__5.0__6.0__ divide__60.0__6.0__ divide__36.0__6.0__ |
multiply__3.0__12.0__ multiply__5.0__12.0__ multiply__5.0__6.0__ divide__60.0__6.0__ divide__36.0__6.0__ |
| during a trip on an expressway don drove a total of x miles . his average speed on a certain num__5 - mile section of the expressway was num__20 miles per hour and his average speed for the remainder of the trip was num__60 miles per hour . his travel time for the x - mile trip was what percent greater than it would have been if he had traveled at a constant rate of num__60 miles per hour for the entire trip ? <o> a ) num__8.5 <o> b ) num__50.0 <o> c ) x / num__12.0 <o> d ) num__300 / x % <o> e ) num__500 / x % |
say x = num__5 miles ( so no remainder of the trip ) . time to cover x = num__5 miles at num__20 miles per hour = ( time ) = ( distance ) / ( rate ) = num__0.166666666667 = num__0.166666666667 hours = num__10 minutes . time to cover x = num__5 miles at num__60 miles per hour = ( time ) = ( distance ) / ( rate ) = num__0.0833333333333 = num__0.0833333333333 hours = num__5 minutes . ( or simply half rate will result in doubling the time . ) so we can see that the time to cover x = num__3 miles at num__30 miles per hour ( num__10 minutes ) is num__100.0 greater than the time to cover x = num__3 miles at num__60 miles per hour ( num__5 minutes ) . now plug x = num__3 miles into the answer choices to see which one yields num__100.0 . only answer e works . answer : d . <eor> d <eos> |
d |
divide__5.0__60.0__ divide__60.0__20.0__ add__20.0__10.0__ multiply__5.0__20.0__ multiply__5.0__60.0__ |
divide__5.0__60.0__ divide__60.0__20.0__ add__20.0__10.0__ multiply__5.0__20.0__ multiply__5.0__60.0__ |
| the equation x = num__2 y ^ num__2 + num__5 y - num__16 describes a parabola in the xy coordinate plane . if line l with slope of num__3 intersects the parabola in the upper - left quadrant at x = - num__5 the equation for l is <o> a ) num__3 x + y + num__15 = num__0 <o> b ) y + num__3 x - num__1 = num__0 <o> c ) - num__3 x + y - num__16.5 = num__0 <o> d ) - num__2 x - y - num__7 = num__0 <o> e ) - num__3 x + y + num__13.5 = num__0 |
the line has a point ( - num__5 y ) on it where y is positive ( since the point lies in upper left quadrant ) . in options b if you put x = - num__5 you get - ve value for y co - ordinate . answer must be ( b ) <eor> b <eos> |
b |
subtract__5.0__2.0__ |
subtract__5.0__2.0__ |
| in the xy - coordinate plane the graph of y = - x ^ num__2 + num__9 intersects line l at ( p num__5 ) and ( t num__7 ) . what is the least possible value of the slope of line l ? <o> a ) num__6 <o> b ) num__2 <o> c ) - num__2 <o> d ) num__1 <o> e ) - num__10 |
we need to find out the value of p and l to get to the slope . line l and graph y intersect at point ( p num__5 ) . hence x = p and y = num__5 should sactisfy the graph . soliving num__5 = - p num__2 + num__9 p num__2 = num__4 p = + or - num__2 simillarly point ( t num__7 ) should satisfy the equation . hence x = t and y = num__7 . - num__7 = - t num__2 + num__9 t = + or - num__4 considering p = - num__2 and t = num__4 the least slope is ( num__7 - num__5 ) / ( num__4 - num__2 ) = num__1 imo option d is correct answer . <eor> d <eos> |
d |
subtract__9.0__5.0__ subtract__5.0__4.0__ reverse__1.0__ |
subtract__9.0__5.0__ subtract__5.0__4.0__ reverse__1.0__ |
| a wheel of a car of radius num__21 cms is rotating at num__800 rpm . what is the speed of the car in km / hr ? <o> a ) num__79.2 km / hr <o> b ) num__47.52 km / hr <o> c ) num__7.92 km / hr <o> d ) num__39.6 km / hr <o> e ) num__63.36 km / hr |
explanatory answer the radius of the wheel measures num__21 cm . in one rotation the wheel will cover a distance which is equal to the circumference of the wheel . ∴ in one rotation this wheel will cover num__2 * π * num__21 = num__132 cm . in a minute the distance covered by the wheel = circumference of the wheel * rpm ∴ this wheel will cover a distance of num__132 * num__800 = num__105600 cm in a minute . in an hour the wheel will cover a distance of num__105600 * num__60 = num__6336000 cm . therefore the speed of the car = num__6336000 cm / hr = num__63.36 km / hr choice e is the correct answer . <eor> e <eos> |
e |
multiply__800.0__132.0__ hour_to_min_conversion__ multiply__105600.0__60.0__ round__63.36__ |
multiply__800.0__132.0__ hour_to_min_conversion__ multiply__105600.0__60.0__ round__63.36__ |
| the c . p of num__10 pens is equal to the s . p of num__13 pens . find his gain % or loss % ? <o> a ) loss num__16 num__0.25 % <o> b ) loss num__16 num__2.0 % <o> c ) loss num__23.08 <o> d ) loss num__18 num__0.666666666667 % <o> e ) loss num__16 num__25 % |
num__10 cp = num__13 sp num__13 - - - num__3 cp loss num__100 - - - ? = > num__23.08 loss answer : c <eor> c <eos> |
c |
percent__100.0__23.08__ |
percent__100.0__23.08__ |
| a boat moves upstream at the rate of num__1 km in num__40 minutes and down stream num__1 km in num__12 minutes . then the speed of the current is : <o> a ) num__1 kmph <o> b ) num__2 kmph <o> c ) num__1.75 kmph <o> d ) num__2.5 kmph <o> e ) num__3.5 kmph |
rate upstream = ( num__0.025 * num__60 ) = num__1.5 kmph rate dowm stream = num__0.0833333333333 * num__60 = num__5 kmph rate of the current = ½ ( num__5 - num__1.5 ) = num__1.75 kmph answer : c <eor> c <eos> |
c |
divide__1.0__40.0__ hour_to_min_conversion__ multiply__0.025__60.0__ divide__1.0__12.0__ divide__60.0__12.0__ round__1.75__ |
divide__1.0__40.0__ hour_to_min_conversion__ multiply__0.025__60.0__ divide__1.0__12.0__ divide__60.0__12.0__ multiply__1.0__1.75__ |
| for all real numbers v an operation is defined by the equation v * = v - v / num__3 . if ( v * ) * = num__12 then v = <o> a ) num__15 <o> b ) num__18 <o> c ) num__21 <o> d ) num__24 <o> e ) num__27 |
( v * ) * = ( v - v / num__3 ) - ( v - v / num__3 ) / num__3 num__12 = num__2 v / num__3 - num__2 v / num__9 = num__4 v / num__9 v = num__27 the answer is e . <eor> e <eos> |
e |
subtract__12.0__3.0__ divide__12.0__3.0__ multiply__3.0__9.0__ multiply__3.0__9.0__ |
subtract__12.0__3.0__ divide__12.0__3.0__ multiply__3.0__9.0__ multiply__3.0__9.0__ |
| if x = - num__0.125 and y = - num__0.5 what is the value of the expression - num__2 x – y ^ num__2 ? <o> a ) - num__1.5 <o> b ) - num__1 <o> c ) num__0 <o> d ) num__1.5 <o> e ) num__1.75 |
- num__2 x - y ^ num__2 - num__2 x - num__0.125 - ( - num__0.5 ) ^ num__2 num__0.25 - num__0.25 num__0 c correct <eor> c <eos> |
c |
divide__0.125__0.5__ round_down__0.125__ round_down__0.125__ |
divide__0.125__0.5__ round_down__0.125__ round_down__0.125__ |
| num__18 times a positive integer is more than its square by num__80 then the positive integer is <o> a ) num__13 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__14 |
explanation : let the number be x . then num__18 x = x num__2 + num__80 = > x num__2 - num__18 x + num__80 = num__0 = > ( x - num__10 ) ( x - num__8 ) = num__0 = > x = num__10 or num__8 answer : b <eor> b <eos> |
b |
subtract__18.0__10.0__ subtract__18.0__8.0__ |
subtract__18.0__10.0__ subtract__18.0__8.0__ |
| at num__1 : num__00 pm there were num__10.0 grams of bacteria . the bacteria increased to x grams at num__4 : num__00 pm and num__25.6 grams at num__7 : num__00 pm . if the amount of bacteria present increased by the same fraction during each of the num__3 - hour periods how many grams of bacteria were present at num__4 : num__00 pm ? <o> a ) num__15.1 <o> b ) num__15.4 <o> c ) num__15.7 <o> d ) num__16.0 <o> e ) num__16.3 |
let x be the factor by which the bacteria increases every three hours . at num__4 : num__00 pm the amount of bacteria was num__10 x and at num__7 : num__00 pm it was num__10 x ^ num__2 . num__10 x ^ num__2 = num__25.6 x ^ num__2 = num__2.56 x = num__1.6 at num__4 : num__00 pm the amount of bacteria was num__10 ( num__1.6 ) = num__16 grams . the answer is d . <eor> d <eos> |
d |
subtract__3.0__1.0__ divide__25.6__10.0__ multiply__10.0__1.6__ multiply__1.0__16.0__ |
subtract__3.0__1.0__ divide__25.6__10.0__ multiply__10.0__1.6__ multiply__1.0__16.0__ |
| if the number num__653 xy is divisible by num__90 then ( x + y ) = ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__6 <o> e ) num__5 |
num__90 = num__10 x num__9 clearly num__653 xy is divisible by num__10 so y = num__0 now num__653 x num__0 is divisible by num__9 . so ( num__6 + num__5 + num__3 + x + num__0 ) = ( num__14 + x ) is divisible by num__9 . so x = num__4 . hence ( x + y ) = ( num__4 + num__0 ) = num__4 . answer : c <eor> c <eos> |
c |
divide__90.0__10.0__ subtract__9.0__6.0__ add__5.0__9.0__ subtract__9.0__5.0__ subtract__9.0__5.0__ |
divide__90.0__10.0__ subtract__9.0__6.0__ add__5.0__9.0__ subtract__9.0__5.0__ subtract__9.0__5.0__ |
| a and b enter into partnership with capitals in the ratio num__3 : num__4 . at the end of num__10 months a withdraws and the profits now are divided in the ratio of num__3 : num__2 . find how long b remained in the business ? <o> a ) num__5 months <o> b ) num__8 months <o> c ) num__6 months <o> d ) num__7 months <o> e ) none of these |
initially a ’ s investment = num__3 x and b ’ s investment = num__4 x let b remain in the business for ‘ n ’ months . ⇒ num__3 x × num__10 : num__4 x × n = num__3 : num__2 ∴ num__3 x × num__10 × num__2 = num__4 x × n × num__3 ⇒ n = num__5 answer a <eor> a <eos> |
a |
add__3.0__2.0__ add__3.0__2.0__ |
add__3.0__2.0__ add__3.0__2.0__ |
| there was two trains from calcutta to kanyakumari one train is fast service travels with a speed num__75 km per hour another travels with a speed of num__44 km per hour the time taken to reach from calcutta to kanyakumari is num__4 hours less than the first train . . . find the distance b / w calcatta to kanyakumari <o> a ) num__415.8 km <o> b ) num__425.8 km <o> c ) num__435.8 km <o> d ) num__445.8 km <o> e ) num__455.8 km |
let distance b / w calcutta to kanyakumari is = x km then eqn is x / num__44 - x / num__75 = num__4 = > x * ( num__75 - num__44 ) = num__4 * num__75 * num__44 = > x = num__425.8 km answer : b <eor> b <eos> |
b |
round__425.8__ |
round__425.8__ |
| the arithmetic mean and standard deviation of a certain normal distribution are num__16.2 and num__2.3 respectively . what value is exactly num__2 standard deviations less than the mean ? <o> a ) num__10.5 <o> b ) num__11 <o> c ) num__11.6 <o> d ) num__12 <o> e ) num__12.5 |
mean = num__16.2 two standard deviations is num__2.3 + num__2.3 = num__4.6 there could be two calues for this . mean + two standard deviations = num__20.8 mean - two standard deviations = num__11.6 answer choice has num__11.6 and so c is the answer . <eor> c <eos> |
c |
multiply__2.3__2.0__ add__16.2__4.6__ subtract__16.2__4.6__ subtract__16.2__4.6__ |
multiply__2.3__2.0__ add__16.2__4.6__ subtract__16.2__4.6__ subtract__16.2__4.6__ |
| if num__4 a = num__16 b and num__9 b = num__11 c find a : b : c ? <o> a ) num__52 : num__13 : num__7 <o> b ) num__44 : num__11 : num__9 <o> c ) num__7 : num__13 : num__52 <o> d ) num__7 : num__13 : num__54 <o> e ) none of these |
explanation : ( num__4 a = num__16 b = = > a / b = num__4.0 ) and ( num__9 b = num__11 c = = > b / c = num__1.22222222222 ) = = > a : b = num__16 : num__4 and b : c = num__11 : num__9 a : b : c = num__44 : num__11 : num__9 answer : option b <eor> b <eos> |
b |
divide__11.0__9.0__ multiply__4.0__11.0__ multiply__4.0__11.0__ |
divide__11.0__9.0__ multiply__4.0__11.0__ multiply__4.0__11.0__ |
| the speed of the boat in still water in num__12 kmph . it can travel downstream through num__39 kms in num__3 hrs . in what time would it cover the same distance upstream ? <o> a ) num__3.5 hours <o> b ) num__6 hours <o> c ) num__4 hours <o> d ) num__5 hours <o> e ) num__6 hours |
still water = num__12 km / hr downstream = num__13.0 = num__13 km / hr upstream = > > still water = ( u + v / num__2 ) = > > num__12 = u + num__6.5 = num__11 km / hr so time taken in upstream = num__3.54545454545 = num__3.5 hrs answer : a <eor> a <eos> |
a |
divide__39.0__3.0__ divide__13.0__2.0__ subtract__13.0__2.0__ divide__39.0__11.0__ subtract__6.5__3.0__ round__3.5__ |
divide__39.0__3.0__ divide__13.0__2.0__ subtract__13.0__2.0__ divide__39.0__11.0__ subtract__6.5__3.0__ round__3.5__ |
| two pipes p and q can fill a cistern in num__12 and num__15 minutes respectively . both are opened together but at the end of num__3 minutes the first is turned off . how much longer will the cistern take to fill ? <o> a ) num__11 num__0.2 <o> b ) num__11 num__0.25 <o> c ) num__11 num__0.2 <o> d ) num__11 num__0.333333333333 <o> e ) num__11 num__0.666666666667 |
num__0.25 + x / num__15 = num__1 x = num__11 num__0.25 answer : b <eor> b <eos> |
b |
divide__3.0__12.0__ subtract__12.0__1.0__ round__11.0__ |
divide__3.0__12.0__ subtract__12.0__1.0__ divide__11.0__1.0__ |
| an article is bought for rs . num__575 and sold for rs . num__900 find the gain percent ? <o> a ) num__73 num__0.333333333333 % <o> b ) num__63 num__0.333333333333 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__93 num__0.333333333333 % <o> e ) num__56.5 % |
explanation : num__575 - - - - num__325 num__100 - - - - ? = > num__56.5 answer : e <eor> e <eos> |
e |
percent__56.5__100.0__ |
percent__56.5__100.0__ |
| a batsman scored num__120 runs which included num__3 boundaries and num__8 sixes . what % of his total score did he make by running between the wickets <o> a ) num__30.0 <o> b ) num__50.0 <o> c ) num__60.0 <o> d ) num__80.0 <o> e ) num__90 % |
number of runs made by running = num__110 - ( num__3 x num__4 + num__8 x num__6 ) = num__120 - ( num__60 ) = num__60 now we need to calculate num__60 is what percent of num__120 . = > num__0.5 * num__100 = num__50.0 b <eor> b <eos> |
b |
divide__3.0__6.0__ multiply__0.5__100.0__ multiply__0.5__100.0__ |
divide__3.0__6.0__ multiply__0.5__100.0__ multiply__0.5__100.0__ |
| the average age of a group of num__20 students was num__20 . the average age increased by num__2 years when two new students joined the group . what is the average age of the two new students who joined the group ? <o> a ) num__22 years <o> b ) num__30 years <o> c ) num__42 years <o> d ) num__32 years <o> e ) none of these |
answer the average age of a group of num__20 students is num__20 . therefore the sum of the ages of all num__20 of them = num__20 * num__20 = num__400 when two new students join the group the average age increases by num__2 . new average = num__22 . now there are num__22 students . therefore the sum of the ages of all num__22 of them = num__22 * num__22 = num__484 therefore the sum of the ages of the two new students who joined = num__484 - num__400 = num__84 and the average age of each of the two new students = num__42.0 = num__42 years . answer c <eor> c <eos> |
c |
add__20.0__2.0__ subtract__484.0__400.0__ add__20.0__22.0__ add__20.0__22.0__ |
add__20.0__2.0__ subtract__484.0__400.0__ add__20.0__22.0__ subtract__84.0__42.0__ |
| a train traveled the first d miles of its journey it an average speed of num__40 miles per hour the next d miles of its journey at an average speed of y miles per hour and the final d miles of its journey at an average speed of num__160 miles per hour . if the train ’ s average speed over the total distance was num__96 miles per hour what is the value of y ? <o> a ) num__68 <o> b ) num__84 <o> c ) num__100 <o> d ) num__120 <o> e ) num__135 |
average speed = total distance traveled / total time taken num__3 d / d / num__40 + d / y + d / num__160 = num__96 solving for d and y num__15 y = num__11 y + num__480 num__4 y = num__400 y = num__100 answer c <eor> c <eos> |
c |
multiply__160.0__3.0__ divide__160.0__40.0__ add__96.0__4.0__ round__100.0__ |
multiply__160.0__3.0__ divide__160.0__40.0__ add__96.0__4.0__ add__96.0__4.0__ |
| when ticket sales began pat was the mth customer in line for a ticket and customers purchased their tickets at the rate of x customers per minute . of the following which best approximates the time in minutes that pat had to wait in line from the moment ticket sales began ? <o> a ) ( m - num__1 ) x <o> b ) m + x – num__1 <o> c ) ( m - num__1 ) / x <o> d ) x / ( m - num__1 ) <o> e ) m / ( x - num__1 ) |
c . ( m - num__1 ) / x <eor> c <eos> |
c |
reverse__1.0__ |
reverse__1.0__ |
| of the num__800 employees of company x num__70 percent have been with the company for at least ten years . if y of theselong - termmembers were to retire and no other employee changes were to occur what value of y would reduce the percent oflong - termemployees in the company to num__40 percent ? <o> a ) num__200 <o> b ) num__160 <o> c ) num__400 <o> d ) num__80 <o> e ) num__56 |
the # oflong - termemployees is num__70.0 * num__800 = num__560 . after y of them retire new # oflong - termemployees would become num__560 - y . total # of employees would become num__800 - y . we want num__560 - y to be num__40.0 of num__800 - y - - > num__560 - y = ( num__800 - y ) * num__40.0 - - > y = num__400 . answer : c . <eor> c <eos> |
c |
subtract__800.0__400.0__ |
subtract__800.0__400.0__ |
| the product of two numbers is num__4107 . if the h . c . f . of these numbers is num__37 then the greater number is : explanation : <o> a ) num__101 <o> b ) num__107 <o> c ) num__111 <o> d ) num__117 <o> e ) num__121 |
let the numbers be num__37 a and num__37 b . then num__37 a x num__37 b = num__4107 ab = num__3 . now co - primes with product num__3 are ( num__1 num__3 ) . so the required numbers are ( num__37 x num__1 num__37 x num__3 ) i . e . ( num__37 num__111 ) . greater number = num__111 . answer : option c <eor> c <eos> |
c |
divide__4107.0__37.0__ divide__4107.0__37.0__ |
divide__4107.0__37.0__ divide__4107.0__37.0__ |
| john and lewis leave city a for city b simultaneously at num__6 a . m in the morning driving in two cars at speeds of num__40 mph and num__60 mph respectively . as soon as lewis reaches city b he returns back to city a along the same route and meets john on the way back . if the distance between the two cities is num__240 miles how far from city a did john and lewis meet ? <o> a ) num__150 miles <o> b ) num__160 miles <o> c ) num__140 miles <o> d ) num__170 miles <o> e ) num__200 miles |
time taken by lewis to reach city b = num__4.0 = num__4 hours in num__4 hours john travels num__40 * num__4 = num__160 miles so distance at which they meet should be greater than num__160 miles . only b satisfies . answer is b . <eor> b <eos> |
b |
divide__240.0__60.0__ multiply__40.0__4.0__ round__160.0__ |
divide__240.0__60.0__ multiply__40.0__4.0__ multiply__40.0__4.0__ |
| rahul ' s mathematics test had num__85 problems num__10 arithmetic num__30 algebra num__35 geometry problems . although he answered num__70.0 of arithmetic num__40.0 of arithmetic and num__60.0 of geometry problems correctly still he got less than num__60.0 problems right . how many more questions he would have to answer more to get passed <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__11 |
explanation : number of questions attempted correctly = ( num__70.0 of num__10 + num__40.0 of num__30 + num__60.0 of num__35 ) = num__7 + num__12 + num__21 = num__40 . questions to be answered correctly for num__60.0 = num__60.0 of total quations = num__60.0 of num__85 = num__51 . he would have to answer num__51 - num__40 = num__11 answer : option e <eor> e <eos> |
e |
divide__70.0__10.0__ add__30.0__21.0__ subtract__51.0__40.0__ subtract__51.0__40.0__ |
divide__70.0__10.0__ add__30.0__21.0__ subtract__51.0__40.0__ subtract__51.0__40.0__ |
| find the number difference between number and its num__0.6 is num__50 . <o> a ) num__110 <o> b ) num__125 <o> c ) num__120 <o> d ) num__140 <o> e ) num__145 |
no = x then x - num__0.6 x = num__50 ( num__0.4 ) x = num__50 = = > num__2 x = num__50 * num__5 x = num__125 answer b <eor> b <eos> |
b |
divide__2.0__0.4__ divide__50.0__0.4__ divide__50.0__0.4__ |
divide__2.0__0.4__ divide__50.0__0.4__ divide__50.0__0.4__ |
| every monday marina eats one croissant and every tuesday she eats two croissants . on each subsequent day of the week she eats a number of croissants equal to the sum of the croissants eaten on the two previous days with the exception that if she eats more than four croissants on any particular day the next day she will eat only one croissant . at the end of the week ( which runs from monday through sunday ) the cycle resets and marina goes back to eating one croissant on monday two on tuesday and so forth . if a particular month begins on a tuesday how many croissants will marina eat on the num__24 th of that month ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__5 <o> e ) num__6 |
she eat as follow mon - num__1 tue - num__2 wed - num__3 thr - num__5 fri - num__1 ( since she had more than num__4 previous day ) sat - num__6 sunday - num__1 ( since she had more than num__4 previous day ) so num__24 th day of month she will have num__5 . answer is d <eor> d <eos> |
d |
add__1.0__2.0__ add__2.0__3.0__ add__1.0__3.0__ divide__24.0__4.0__ add__1.0__4.0__ |
add__1.0__2.0__ add__2.0__3.0__ subtract__5.0__1.0__ divide__24.0__4.0__ subtract__6.0__1.0__ |
| how long will it take a sum of money invested at num__20.0 p . a . s . i . to increase double the value ? <o> a ) num__12 years . <o> b ) num__8 years . <o> c ) num__15 years . <o> d ) num__13 years . <o> e ) num__10 years . |
sol . let the sum be x . then s . i . = num__200.0 of x = num__2 x ; rate = num__20.0 . â ˆ ´ time = [ num__100 * num__2 x / num__5 * num__1 / x * num__5 ] = num__10 years . answer e <eor> e <eos> |
e |
percent__20.0__5.0__ percent__5.0__200.0__ percent__100.0__10.0__ |
percent__20.0__5.0__ percent__5.0__200.0__ percent__100.0__10.0__ |
| one tap can fill a water tank four times as fast as another tap . if together the two taps can fill the water tank in num__30 minutes then the slower tap alone will be able to fill the water tank in _______ . <o> a ) num__120 min <o> b ) num__150 min <o> c ) num__170 min <o> d ) num__190 min <o> e ) num__250 min |
let the slower pipe alone fill the water tank in t minutes . then faster pipe will fill it in t / num__4 minutes . therefore num__1 / t + num__4 / t = num__0.0333333333333 = > num__5 / t = num__0.0333333333333 now solve for t . = > t = num__150 min answer : b <eor> b <eos> |
b |
divide__1.0__30.0__ add__1.0__4.0__ multiply__30.0__5.0__ round__150.0__ |
divide__1.0__30.0__ add__1.0__4.0__ multiply__30.0__5.0__ divide__150.0__1.0__ |
| the speed of a train is num__102 kmph . what is the distance covered by it in num__30 minutes ? <o> a ) num__50 kmph <o> b ) num__51 kmph <o> c ) num__41 kmph <o> d ) num__61 kmph <o> e ) num__21 kmph |
num__102 * num__0.5 = num__51 kmph answer : b <eor> b <eos> |
b |
multiply__102.0__0.5__ round__51.0__ |
multiply__102.0__0.5__ multiply__102.0__0.5__ |
| how much simple interest per annum can i negotiate with my bank so my family can pay back in num__40 years at most double of the money we will borrow . <o> a ) num__4 <o> b ) num__5 <o> c ) num__2.5 <o> d ) num__4.5 <o> e ) num__2 |
p = ( p * si * num__40 ) / num__100 num__100 p = p * si * num__40 num__100 = num__40 si therefore si = num__2.5 c = num__2.5 <eor> c <eos> |
c |
percent__2.5__100.0__ |
percent__2.5__100.0__ |
| a certain pilot flew num__400 miles to city k at an average speed of num__550 miles per hour with the wind and made the trip back at an average speed of num__250 miles per hour against the wind . which of the following is closest to the pilot ’ s average speed in miles per hour for the round - trip ? <o> a ) num__280 <o> b ) num__290 <o> c ) num__300 <o> d ) num__310 <o> e ) num__340 |
avg speed = total distance / total time total distance = num__800 total time = num__0.727272727273 + num__1.6 = num__2.32727272727 = > avg speed = ( num__800 * num__55 ) / num__128 = num__340 ( approx ) ans is e <eor> e <eos> |
e |
add__550.0__250.0__ divide__400.0__550.0__ mile_to_km_conversion__ add__1.6__0.7273__ round__340.0__ |
add__550.0__250.0__ divide__400.0__550.0__ divide__400.0__250.0__ add__1.6__0.7273__ round__340.0__ |
| it is required to fix a pipe such that water flowing through it at a speed of num__7 metres per minute fills a tank of capacity num__1760 cubic metres in num__10 minutes . the inner radius of the pipe should be : <o> a ) √ num__2 m <o> b ) num__2 m <o> c ) num__1 ⁄ num__2 m <o> d ) num__2 √ num__2 m <o> e ) none of these |
let inner radius of the pipe be r . then num__1760 = num__22 ⁄ num__7 × r num__2 × num__7 × num__10 or r num__2 = num__80.0 × num__10 = num__8 or r = num__2 √ num__2 m answer d <eor> d <eos> |
d |
divide__1760.0__22.0__ subtract__10.0__2.0__ round__2.0__ |
divide__1760.0__22.0__ subtract__10.0__2.0__ round__2.0__ |
| if x and y are positive integers and num__18 x = num__12 y what is the least possible value of xy ? <o> a ) num__14 <o> b ) num__28 <o> c ) num__36 <o> d ) num__84 <o> e ) num__252 |
num__18 x = num__12 y = > x / y = num__0.666666666667 = > num__3 x = num__2 y num__3 ( num__3 ) = num__2 ( num__3 ) = > x * y = num__9 but it is not given num__3 ( num__6 ) = num__2 ( num__6 ) = > x * y = num__36 c <eor> c <eos> |
c |
divide__12.0__18.0__ divide__18.0__2.0__ subtract__18.0__12.0__ multiply__18.0__2.0__ multiply__18.0__2.0__ |
divide__12.0__18.0__ divide__18.0__2.0__ divide__18.0__3.0__ multiply__18.0__2.0__ multiply__18.0__2.0__ |
| this topic is locked . if you want to discuss this question please re - post it in the respective forum . the sum of the first num__54 positive even integers is num__2550 . what is the sum of the even integers from num__94 to num__200 inclusive ? <o> a ) num__5100 <o> b ) num__7938 <o> c ) num__10100 <o> d ) num__15500 <o> e ) num__20 |
100 |
even numbers from num__94 to num__200 = num__53.0 + num__1 = num__54 sum = ( num__200 + num__94 ) x num__27.0 = num__7938 so b is the ans . <eor> b <eos> |
b |
b |
| three numbers are in the ratio num__1 : num__2 : num__3 and their h . c . f is num__6 . the numbers are <o> a ) num__12 num__24 num__30 <o> b ) num__12 num__24 num__38 <o> c ) num__12 num__24 num__362 <o> d ) num__6 num__12 num__18 <o> e ) num__12 num__24 num__321 |
explanation : let the required numbers be x num__2 x num__3 x . then their h . c . f = x . so x = num__6 \ inline \ fn _ jvn \ therefore the numbers are num__6 num__12 num__18 answer : d ) num__6 num__12 num__18 <eor> d <eos> |
d |
multiply__2.0__6.0__ multiply__3.0__6.0__ multiply__1.0__6.0__ |
multiply__2.0__6.0__ multiply__3.0__6.0__ multiply__1.0__6.0__ |
| the c . p of num__40 pens is equal to the s . p of num__60 pens . find his gain % or loss % <o> a ) num__25 num__0.666666666667 <o> b ) num__32 num__0.333333333333 <o> c ) num__33 num__0.333333333333 <o> d ) num__25 num__0.333333333333 <o> e ) num__33 num__0.666666666667 |
num__40 cp = num__60 sp num__60 - - - num__20 cp loss num__100 - - - ? = > num__33 num__0.333333333333 % answer : c <eor> c <eos> |
c |
percent__33.0__100.0__ |
percent__33.0__100.0__ |
| find the value for m ? num__19 ( m + n ) + num__17 = num__19 ( - m + n ) - num__21 <o> a ) num__0 <o> b ) - num__1 <o> c ) num__1 <o> d ) num__2 <o> e ) num__3 |
num__19 m + num__19 n + num__17 = - num__19 m + num__19 n - num__21 num__38 m = - num__38 = > m = - num__1 b <eor> b <eos> |
b |
add__17.0__21.0__ reverse__1.0__ |
add__17.0__21.0__ reverse__1.0__ |
| in the interior of a forest a certain number of apes equal to the square of one - eighth of the total number are playing and having great fun . the remaining sixteen apes are on a hill and the echo of their shrieks by the adjoining hills frightens them . they came and join the apes in the forest and play with enthusiasm . what is the total number of apes ? <o> a ) num__32 <o> b ) num__16 <o> c ) num__64 <o> d ) num__80 <o> e ) num__16 or num__48 |
let total number be x no in the interior = ( x / num__8 ) ^ num__2 no outside = num__16 so : x - ( x / num__8 ) ^ num__2 = num__16 x ^ num__2 - num__64 x + num__16 * num__64 = num__0 ( x - num__32 ) ^ num__2 = num__0 x = num__32 a <eor> a <eos> |
a |
multiply__2.0__8.0__ divide__64.0__2.0__ subtract__64.0__32.0__ |
multiply__2.0__8.0__ divide__64.0__2.0__ subtract__64.0__32.0__ |
| evaluate : num__1024 x num__237 + num__976 x num__237 <o> a ) num__486000 <o> b ) num__568000 <o> c ) num__378000 <o> d ) num__474000 <o> e ) none of them |
num__1024 x num__237 + num__976 x num__237 = num__237 x ( num__1024 + num__976 ) = num__237 x num__2000 = num__474000 . answer is d . <eor> d <eos> |
d |
add__1024.0__976.0__ multiply__237.0__2000.0__ multiply__237.0__2000.0__ |
add__1024.0__976.0__ multiply__237.0__2000.0__ multiply__237.0__2000.0__ |
| a tradesman by means of his false balance defrauds to the extent of num__30.0 ? in buying goods as well as by selling the goods . what percent does he gain on his outlay ? <o> a ) num__79.0 <o> b ) num__58.0 <o> c ) num__44.0 <o> d ) num__69.0 <o> e ) num__43 % |
g % = num__30 + num__30 + ( num__30 * num__30 ) / num__100 = num__69.0 answer : d <eor> d <eos> |
d |
percent__100.0__69.0__ |
percent__100.0__69.0__ |
| if the cost price of num__12 articles is same as the selling price of num__25 articles . find the gain or loss percentage ? <o> a ) num__30.0 gain <o> b ) num__30.0 loss <o> c ) num__40.0 gain <o> d ) num__40.0 loss <o> e ) num__52.0 loss |
explanation : num__12 cp = num__25 sp cost price cp = num__25 selling price sp = num__12 formula = ( sp - cp ) / cp * num__100 = ( num__12 - num__25 ) / num__25 * num__100 = num__52.0 loss answer : option e <eor> e <eos> |
e |
percent__100.0__52.0__ |
percent__100.0__52.0__ |
| in a sample of high school students num__20 percent are sophomores and num__65 percent are not freshmen . what percentage of the high school students are juniors or seniors ? <o> a ) num__15 <o> b ) num__25 <o> c ) num__35 <o> d ) num__45 <o> e ) num__55 |
num__65 percent are not freshman - - > this means num__65 percent are either sophomores juniors or seniors already we know % that are sophomores = num__20.0 so % of juniors and seniors = num__65.0 - num__20.0 = num__45.0 . . = > ( d ) <eor> d <eos> |
d |
subtract__65.0__20.0__ subtract__65.0__20.0__ |
subtract__65.0__20.0__ subtract__65.0__20.0__ |
| how long does a train num__110 m long running at the speed of num__72 km / hr takes to cross a bridge num__132 m length ? <o> a ) num__12.5 <o> b ) num__12.1 <o> c ) num__12.9 <o> d ) num__12.3 <o> e ) num__12.2 |
speed = num__72 * num__0.277777777778 = num__20 m / sec total distance covered = num__110 + num__132 = num__242 m . required time = num__12.1 = num__12.1 sec . answer : b <eor> b <eos> |
b |
add__110.0__132.0__ divide__242.0__20.0__ round__12.1__ |
add__110.0__132.0__ divide__242.0__20.0__ divide__242.0__20.0__ |
| the average marks scored by ganesh in english science mathematics and history is less than num__14 from that scored by him in english history geography and mathematics . what is the difference of marks in science and geography scored by him ? <o> a ) num__40 <o> b ) num__56 <o> c ) num__60 <o> d ) data inadequate <o> e ) none of these |
e + h + g + m / num__4 â ˆ ’ e + s + m + h / num__4 = num__14 â ‡ ’ g - s = num__56 answer b <eor> b <eos> |
b |
multiply__14.0__4.0__ multiply__14.0__4.0__ |
multiply__14.0__4.0__ multiply__14.0__4.0__ |
| a train num__150 metres long is moving at a speed of num__25 kmph . it will cross a man coming from the opposite direction at a speed of num__2 km per hour in : <o> a ) num__20 sec <o> b ) num__32 sec <o> c ) num__36 sec <o> d ) num__38 sec <o> e ) num__40 sec |
relative speed = ( num__25 + num__2 ) km / hr = num__27 km / hr = ( num__27 × num__0.277777777778 ) m / sec = num__7.5 m / sec . time taken by the train to pass the man = ( num__150 × num__0.133333333333 ) sec = num__20 sec answer : a <eor> a <eos> |
a |
add__25.0__2.0__ divide__150.0__7.5__ round__20.0__ |
add__25.0__2.0__ divide__150.0__7.5__ divide__150.0__7.5__ |
| if num__12 men do a work in num__80 days in how many days will num__16 men do it ? <o> a ) num__60 <o> b ) num__30 <o> c ) num__50 <o> d ) num__40 <o> e ) num__20 |
num__12 * num__80 = num__16 * x x = num__60 days answer a <eor> a <eos> |
a |
hour_to_min_conversion__ hour_to_min_conversion__ |
hour_to_min_conversion__ hour_to_min_conversion__ |
| the average age of num__36 students in a group is num__14 years . when teacher ' s age is included to it the average increases by one . find out the teacher ' s age in years ? <o> a ) num__51 years <o> b ) num__49 years <o> c ) num__53 years <o> d ) num__50 years <o> e ) num__52 years |
explanation : average age of num__36 students in a group is num__14 sum of the ages of num__36 students = num__36 × num__14 when teacher ' s age is included to it the average increases by one = > average = num__15 sum of the ages of num__36 students and the teacher = num__37 × num__15 hence teachers age = num__37 × num__15 - num__36 × num__14 = num__37 × num__15 - num__14 ( num__37 - num__1 ) = num__37 × num__15 - num__37 × num__14 + num__14 = num__37 ( num__15 - num__14 ) + num__14 = num__37 + num__14 = num__51 answer : option a <eor> a <eos> |
a |
subtract__37.0__36.0__ add__36.0__15.0__ add__36.0__15.0__ |
subtract__37.0__36.0__ add__36.0__15.0__ add__36.0__15.0__ |
| if num__36 men can do a piece of work in num__25 hours in how mwny hours will num__15 men do it ? <o> a ) num__33 <o> b ) num__38 <o> c ) num__60 <o> d ) num__88 <o> e ) num__22 |
explanation : let the required no of hours be x . then less men more hours ( indirct proportion ) \ inline \ fn _ jvn \ therefore num__15 : num__36 : : num__25 : x \ inline \ fn _ jvn \ leftrightarrow ( num__15 x x ) = ( num__36 x num__25 ) \ inline \ fn _ jvn \ leftrightarrow \ inline \ fn _ jvn x = \ frac { num__36 \ times num__25 } { num__15 } = num__60 hence num__15 men can do it in num__60 hours . answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ hour_to_min_conversion__ |
hour_to_min_conversion__ hour_to_min_conversion__ |
| find the area of trapezium whose parallel sides are num__20 cm and num__18 cm long and the distance between them is num__15 cm . <o> a ) num__456 <o> b ) num__234 <o> c ) num__285 <o> d ) num__342 <o> e ) num__213 |
area of a trapezium = num__0.5 ( sum of parallel sides ) * ( perpendicular distance between them ) = num__0.5 ( num__20 + num__18 ) * ( num__15 ) = num__285 cm num__2 answer : option c <eor> c <eos> |
c |
subtract__20.0__18.0__ round__285.0__ |
subtract__20.0__18.0__ round__285.0__ |
| a trader sells two brands of petrol ; one is extra premium and other one is speed . he mixes num__12 litres extra premium with num__3 litres of speed and by selling this mixture at the price of extra premium he gets the profit of num__9.09 . if the price of extra premium rs . num__48 per litre then the price of speed is : <o> a ) rs . num__38 per litre <o> b ) rs . num__42 per litre <o> c ) rs . num__28 per litre <o> d ) rs . num__28 per litre <o> e ) none of these |
solution : selling price mix ep = ( cp + num__9.09 of cp ) num__48 = num__1.09090909091 of cp cp = num__44 no using alligation method mcq aptitude profit and loss num__11 num__4 : num__1 thus x = num__44 thus the price of speed brand is rs . num__28 / litre . answer : option c <eor> c <eos> |
c |
divide__12.0__3.0__ subtract__12.0__11.0__ round__28.0__ |
divide__12.0__3.0__ subtract__12.0__11.0__ divide__28.0__1.0__ |
| a man swims downstream num__72 km and upstream num__45 km taking num__9 hours each time ; what is the speed of the current ? <o> a ) num__3 kmph <o> b ) num__1.5 kmph <o> c ) num__13 kmph <o> d ) num__6.5 kmph <o> e ) num__7.5 kmph |
num__72 - - - num__9 ds = num__8 ? - - - - num__1 num__45 - - - - num__9 us = num__5 ? - - - - num__1 s = ? s = ( num__8 - num__5 ) / num__2 = num__1.5 answer : b <eor> b <eos> |
b |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ round__1.5__ |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ divide__1.5__1.0__ |
| a man have num__8 t - shirt and some pant . he can dress in num__72 ways . then tell me the no . of pant that man have . <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
let he have x pant . by fundamental principle total no . of ways in which he can dress = num__8 x num__8 x = num__72 x = num__9 pants answer b <eor> b <eos> |
b |
divide__72.0__8.0__ divide__72.0__8.0__ |
divide__72.0__8.0__ divide__72.0__8.0__ |
| the line r ≡ num__3 x + ny − num__7 = num__0 passes through the point a = ( num__3 num__2 ) and is parallel to the line s ≡ mx + num__2 y − num__13 = num__0 . calculate the values of m and n . <o> a ) - num__6 - num__1 <o> b ) - num__3 - num__2 <o> c ) num__32 <o> d ) num__61 <o> e ) num__7 |
2 |
r ≡ num__3 x + ny − num__7 = num__0 passes through the point a = ( num__3 num__2 ) num__3 . num__3 + nn . num__2 - num__7 = num__0 n = - num__1 num__3 / m = - num__0.5 m = num__6 answer : a - num__6 - num__1 <eor> a <eos> |
a |
a |
| present ages of abi and suji are in the ratio of num__5 : num__4 respectively . three years hence the ratio of their ages will become num__11 : num__9 respectively . what is suji ' s present age in years ? <o> a ) num__20 <o> b ) num__21 <o> c ) num__22 <o> d ) num__23 <o> e ) num__24 |
present ages = num__5 x : num__4 x three hence = num__5 x + num__3 : num__4 x + num__3 num__5 x + num__3 : num__4 x + num__3 = num__11 : num__9 weget x = num__6 suji age is = num__4 x = num__4 * num__6 = num__24 answer : e <eor> e <eos> |
e |
subtract__11.0__5.0__ multiply__4.0__6.0__ multiply__4.0__6.0__ |
subtract__11.0__5.0__ multiply__4.0__6.0__ multiply__4.0__6.0__ |
| a man whose speed is num__4.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at num__1.5 kmph find his average speed for the total journey ? <o> a ) num__7 kmph <o> b ) num__4 kmph <o> c ) num__8 kmph <o> d ) num__3 kmph <o> e ) num__9 kmph |
m = num__45 s = num__1.5 ds = num__6 us = num__3 as = ( num__2 * num__6 * num__3 ) / num__9 = num__4 answer : b <eor> b <eos> |
b |
add__4.5__1.5__ subtract__4.5__1.5__ divide__3.0__1.5__ multiply__4.5__2.0__ round_down__4.5__ round_down__4.5__ |
add__4.5__1.5__ divide__4.5__1.5__ divide__3.0__1.5__ multiply__4.5__2.0__ divide__6.0__1.5__ divide__6.0__1.5__ |
| num__1 quarter kg of potatoes cost num__60 paise . calculate the cost of num__200 grams <o> a ) num__48 paise <o> b ) num__77 paise <o> c ) num__96 paise <o> d ) num__66 paise <o> e ) num__55 paise |
explanation : let num__200 gm potato cost x paise according to question cost of num__0.25 kg potato = num__60 paise = > cost of num__250 gm potato = num__60 paise = > num__1 kg = num__1000 gm = > num__0.25 kg = num__250.0 gm = num__250 gm by direct proportion more quantity more cost num__200 grams : num__250 grams = x : num__60 = > num__200 * num__60 = num__250 x = > num__12000 = num__250 x = > x = num__48 = > thus the cost of num__200 grams is num__48 paise . answer : a <eor> a <eos> |
a |
quarter__ divide__250.0__0.25__ multiply__60.0__200.0__ divide__12000.0__250.0__ multiply__1.0__48.0__ |
quarter__ divide__250.0__0.25__ multiply__60.0__200.0__ divide__12000.0__250.0__ multiply__1.0__48.0__ |
| a and b are two stations num__390 km apart . a train starts from a at num__10 a . m . and travels towards b at num__65 kmph . another train starts from b at num__11 a . m . and travels towards a at num__35 kmph . at what time do they meet ? <o> a ) num__2.15 p . m <o> b ) num__1.15 p . m <o> c ) num__4.15 p . m <o> d ) num__3.15 p . m <o> e ) num__12.15 p . m |
suppose they meet x hours after num__10 a . m . then ( distance moved by first in x hrs ) + [ distance moved by second in ( x - num__1 ) hrs ] = num__390 . num__65 x + num__35 ( x - num__1 ) = num__390 = > num__100 x = num__425 = > x = num__4.25 so they meet num__4 hrs . num__15 min . after num__10 a . m i . e . at num__2.15 p . m . ans : a <eor> a <eos> |
a |
subtract__11.0__10.0__ add__65.0__35.0__ add__390.0__35.0__ divide__425.0__100.0__ add__11.0__4.0__ round__2.15__ |
subtract__11.0__10.0__ add__65.0__35.0__ add__390.0__35.0__ divide__425.0__100.0__ add__11.0__4.0__ round__2.15__ |
| the number of ways in which six boys and six girls can be seated in a row for a photograph so that no two girls sit together is ? <o> a ) num__9 ! * ⁷ p ₆ <o> b ) num__6 ! * ⁷ p ₆ <o> c ) num__9 ! * ⁷ p ₆ <o> d ) num__6 ! * ⁷ num__9 ₆ <o> e ) num__7 ! * ⁷ p ₆ |
we can initially arrange the six boys in num__6 ! ways . having done this now three are seven places and six girls to be arranged . this can be done in ⁷ p ₆ ways . hence required number of ways = num__6 ! * ⁷ p ₆ answer : b <eor> b <eos> |
b |
die_space__ die_space__ |
die_space__ die_space__ |
| solve below question num__2 x + num__1 = - num__15 <o> a ) - num__8 <o> b ) - num__9 <o> c ) num__9 <o> d ) num__8 <o> e ) - num__7 |
num__2 x + num__1 = - num__15 x = - num__8 a <eor> a <eos> |
a |
multiply__1.0__8.0__ |
multiply__1.0__8.0__ |
| a sum of rs . num__2665 is lent into two parts so that the interest on the first part for num__8 years at num__3.0 per annum may be equal to the interest on the second part for num__3 years at num__5.0 per annum . find the second sum ? <o> a ) num__2387 <o> b ) num__1640 <o> c ) num__2781 <o> d ) num__2778 <o> e ) num__1781 |
( x * num__8 * num__3 ) / num__100 = ( ( num__2665 - x ) * num__3 * num__5 ) / num__100 num__24 x / num__100 = num__399.75 - num__15 x / num__100 num__39 x = num__39975 = > x = num__1025 second sum = num__2665 – num__1025 = num__1640 answer : b <eor> b <eos> |
b |
percent__100.0__1640.0__ |
percent__100.0__1640.0__ |
| the product of two numbers is num__120 and the sum of their squares is num__289 . the sum of the number is ? <o> a ) num__23 <o> b ) num__25 <o> c ) num__27 <o> d ) num__31 <o> e ) num__35 |
let the numbers be x and y . then xy = num__120 and x num__2 + y num__2 = num__289 . ( x + y ) num__2 = x num__2 + y num__2 + num__2 xy = num__289 + ( num__2 x num__120 ) = num__529 x + y = num__529 = num__23 . option a <eor> a <eos> |
a |
divide__529.0__23.0__ |
divide__529.0__23.0__ |
| two trains running in opposite directions cross a man standing on the platform in num__27 seconds and num__17 seconds respectively . if they cross each other in num__23 seconds what is the ratio of their speeds ? <o> a ) insufficient data <o> b ) num__3 : num__1 <o> c ) num__1 : num__3 <o> d ) num__3 : num__2 <o> e ) num__2 : num__3 |
let the speed of the trains be x and y respectively length of train num__1 = num__27 x length of train num__2 = num__17 y relative speed = x + y time taken to cross each other = num__23 s = > ( num__27 x + num__17 y ) / ( x + y ) = num__23 = > ( num__27 x + num__17 y ) = num__23 ( x + y ) = > num__4 x = num__6 y = > x / y = num__1.5 = num__1.5 answer : d <eor> d <eos> |
d |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ multiply__1.5__2.0__ |
subtract__27.0__23.0__ add__2.0__4.0__ divide__6.0__4.0__ add__1.0__2.0__ |
| num__5100 − ( num__102 ÷ num__20.4 ) = ? <o> a ) num__5200 <o> b ) num__5150 <o> c ) num__5250 <o> d ) num__6150 <o> e ) num__5095 |
explanation : = num__5100 − ( num__0.5 × num__10 ) = num__5100 − num__5 = num__5095 option e <eor> e <eos> |
e |
divide__102.0__20.4__ subtract__5100.0__5.0__ subtract__5100.0__5.0__ |
divide__102.0__20.4__ subtract__5100.0__5.0__ subtract__5100.0__5.0__ |
| two trains num__130 m and num__160 m long run at the speed of num__60 km / hr and num__40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? <o> a ) num__10.9 <o> b ) num__10.7 <o> c ) num__10.3 <o> d ) num__10.44 <o> e ) num__10.8 |
relative speed = num__60 + num__40 = num__100 km / hr . = num__100 * num__0.277777777778 = num__27.7777777778 m / sec . distance covered in crossing each other = num__130 + num__160 = num__290 m . required time = num__290 * num__0.036 = num__10.44 = num__10.44 sec . ' answer : d <eor> d <eos> |
d |
subtract__160.0__60.0__ add__130.0__160.0__ multiply__290.0__0.036__ round__10.44__ |
add__60.0__40.0__ add__130.0__160.0__ multiply__290.0__0.036__ multiply__290.0__0.036__ |
| a car traveling at a certain constant speed takes num__2 seconds longer to travel num__1 kilometer than it would take to travel num__1 kilometer at num__900 kilometers per hour . at what speed in kilometers per hour is the car traveling ? <o> a ) num__671.5 <o> b ) num__600 <o> c ) num__672.5 <o> d ) num__673 <o> e ) num__773.5 |
many approaches are possible one of them : let the distance be num__1 kilometer . time to cover this distance at num__900 kilometers per hour is num__0.00111111111111 hours = num__4.0 seconds = num__4 seconds ; time to cover this distance at regular speed is num__900 + num__2 = num__902 seconds = num__902 / num__3600 hours = num__0.00166666666667 hours ; so we get that to cover num__1 kilometer num__0.00166666666667 hours is needed - - > regular speed num__600 kilometers per hour ( rate is a reciprocal of time or rate = distance / time ) . answer : b . <eor> b <eos> |
b |
divide__1.0__900.0__ add__2.0__900.0__ multiply__900.0__4.0__ round__600.0__ |
divide__1.0__900.0__ add__2.0__900.0__ multiply__900.0__4.0__ divide__600.0__1.0__ |
| in a sequence of num__11 numbers each term except for the first one is num__11 ^ num__11 less than the previous term . if the greatest term in the sequence is num__11 ^ num__12 what is the smallest term in the sequence ? thus a num__12 = a num__1 + num__12 - num__1 d where d is the difference between each pairs of consecutive terms . since each term is num__1211 less than the previous term d = - num__1211 click individual answer choices for specific explanations . from what i undersltand the formula for an arithmetic sequence represents some - number - in - the - sequence = first - number - in - sequence + the - place - of - that - number - num__1 x the - difference - of - each - # - in - the - sequence <o> a ) − num__12 ^ num__11 <o> b ) num__0 <o> c ) num__11 ^ num__11 <o> d ) num__11 · num__12 ^ num__11 <o> e ) num__12 ^ num__12 |
there are total num__11 terms . andgreatestterm is num__11 ^ num__12 . each term is num__11 ^ num__11 less than previous one . what does this signify ? this shows it is an ap - a decreasing one . ( noticeeach term is num__11 ^ num__11 less than previous one ' ) therefore num__11 ^ num__12 is the first term of the ap and - num__11 ^ num__11 is the difference between successive terms the smallest term should be the last one . last term = first term + ( n - num__1 ) * d therefore last term = num__11 ^ num__12 + num__10 * ( - num__11 ^ num__11 ) = > last term = num__11 ^ num__12 - num__10 * num__11 ^ num__11 = > last term = num__11 ^ num__11 ( num__11 - num__10 ) = num__11 ^ num__11 which is the smallest term . hence ans c it is . <eor> c <eos> |
c |
subtract__11.0__1.0__ multiply__11.0__1.0__ |
subtract__11.0__1.0__ multiply__11.0__1.0__ |
| at a certain restaurant the ratio of the number of cooks to the number of waiters is num__3 to num__10 . when num__12 more waiters are hired the ratio of the number of cooks to the number of waiters changes to num__3 to num__13 . how many cooks does the restaurant have ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__9 <o> d ) num__12 <o> e ) num__15 |
originally there were num__3 k cooks and num__10 k waiters . num__13 k = num__10 k + num__12 k = num__4 there are num__12 cooks . the answer is d . <eor> d <eos> |
d |
divide__12.0__3.0__ multiply__3.0__4.0__ |
divide__12.0__3.0__ multiply__3.0__4.0__ |
| how many num__5 letter words ( with or without meaning ) can be formed using all the following num__5 letters p u r s and t so that letter p is to the left of letter r ? <o> a ) num__60 <o> b ) num__30 <o> c ) num__90 <o> d ) num__120 <o> e ) num__75 |
two positions can be selected out of num__5 positions in num__5 c num__2 ways = num__10 ways . and the remaining num__3 letters can be placed in num__3 posions in num__3 ! ways = num__6 ways . so the final answer = number of ways letters p and r can be arranged x number of ways the other num__3 letters can be arranged final answer = num__10 x num__6 = num__60 ways a <eor> a <eos> |
a |
multiply__5.0__2.0__ subtract__5.0__2.0__ multiply__2.0__3.0__ multiply__6.0__10.0__ multiply__6.0__10.0__ |
multiply__5.0__2.0__ subtract__5.0__2.0__ multiply__2.0__3.0__ multiply__6.0__10.0__ multiply__6.0__10.0__ |
| if the product num__4864 * num__9 p num__2 is divisible by num__12 the value of p ? <o> a ) num__1 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__9 |
explanation : clearly num__4864 is divisible by num__4 so num__9 p num__2 must be divisible by num__3 . so ( num__9 + p + num__2 ) must be divisible by num__3 . so p = num__1 . answer : a ) num__1 <eor> a <eos> |
a |
subtract__12.0__9.0__ subtract__3.0__2.0__ reverse__1.0__ |
subtract__12.0__9.0__ subtract__3.0__2.0__ reverse__1.0__ |
| the average ( arithmetic mean ) of eight numbers is num__44.1 . if the sum of half of these numbers is num__156.4 what is the average of the other half ? <o> a ) num__12.8 <o> b ) num__24.2 <o> c ) num__49.1 <o> d ) num__72.1 <o> e ) num__96.8 |
the average of this half is num__156.4 / num__4 = num__39.1 this is num__5 below the overall average thus the average of the other half of the numbers must be num__5 above the overall age that is num__44.1 + num__5 = num__49.1 the answer is c . <eor> c <eos> |
c |
divide__156.4__4.0__ subtract__44.1__39.1__ add__44.1__5.0__ add__44.1__5.0__ |
divide__156.4__4.0__ subtract__44.1__39.1__ add__44.1__5.0__ add__44.1__5.0__ |
| a train num__110 m long is running with a speed of num__60 km / hr . in what time will it pass a man who is running at num__6 km / hr in the direction opposite to that in which the train is going ? <o> a ) num__7 sec . <o> b ) num__6 sec <o> c ) num__3 sec . <o> d ) num__1 sec . <o> e ) num__8 sec . |
speed of train relative to man = num__60 + num__6 = num__66 km / hr . = num__66 * num__0.277777777778 = num__18.3333333333 m / sec . time taken to pass the men = num__110 * num__0.0545454545455 = num__6 sec . answer : b <eor> b <eos> |
b |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ round__6.0__ |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ divide__110.0__18.3333__ |
| a car traveled the first quarter of a certain distance at five times the speed it traveled the remaining distance . what proportion of the total time traveled was the time taken to travel the first quarter of the distance ? <o> a ) num__0.0714285714286 <o> b ) num__0.0666666666667 <o> c ) num__0.0625 <o> d ) num__0.0588235294118 <o> e ) num__0.0555555555556 |
these problems can be solved through algebra or sly number picking . being a big fan of solving problems with numbers let ' s pick a total distance divisible by num__4 ( say num__40 ) so we can break it up into quarters and a speed that can easily be five times say num__10 . each quarter is thus num__10 kilometers ( or miles or feet or angstroms for all it matters ) and the runner ' s speed is num__10 km / h for the first quarter and num__2 km / h for the remaining quarters . he ' ll take num__1 hour to do the first quarter and then num__15 hours for the second third and fourth quarter . on total he will take num__16 hours to complete this race of which num__1 hour was spent on the first quarter . so num__0.0625 . answer c . <eor> c <eos> |
c |
divide__40.0__4.0__ add__1.0__15.0__ divide__1.0__16.0__ divide__0.0625__1.0__ |
divide__40.0__4.0__ add__1.0__15.0__ divide__1.0__16.0__ divide__0.0625__1.0__ |
| a train num__200 meters long completely crosses a num__300 meters long bridge in num__45 seconds . what is the speed of the train is ? <o> a ) num__32 <o> b ) num__28 <o> c ) num__40 <o> d ) num__27 <o> e ) num__21 |
s = ( num__200 + num__300 ) / num__45 = num__11.1111111111 * num__3.6 = num__40 answer : c <eor> c <eos> |
c |
multiply__11.1111__3.6__ round__40.0__ |
multiply__11.1111__3.6__ multiply__11.1111__3.6__ |
| num__30 men can do a work in num__40 days . when should num__20 men leave the work so that the entire work is completed in num__40 days after they leave the work ? <o> a ) num__11 <o> b ) num__10 <o> c ) num__19 <o> d ) num__187 <o> e ) num__12 |
total work to be done = num__30 * num__40 = num__1200 let num__20 men leave the work after ' p ' days so that the remaining work is completed in num__40 days after they leave the work . num__40 p + ( num__20 * num__40 ) = num__1200 num__40 p = num__400 = > p = num__10 days . answer : b <eor> b <eos> |
b |
multiply__30.0__40.0__ subtract__30.0__20.0__ round__10.0__ |
multiply__30.0__40.0__ subtract__30.0__20.0__ round__10.0__ |
| a baseball player just signed a new contract in which he will earn a num__12.0 raise in base pay over last year . last year he earned no performance bonus but this year he will have the opportunity to earn up to $ num__50000 in extra pay through performance . if he earns the full performance bonus his salary will effectively have been raised by num__12.5 from the last year . assuming he earns the full performance bonus what will his new salary be ? <o> a ) num__5 num__150000 <o> b ) num__10 num__550000 <o> c ) num__11 num__250000 <o> d ) num__11 num__450000 <o> e ) num__12 num__250 |
000 |
let x = the old salary the new base pay + num__50000 = the new pay ( num__1 + . num__12 ) * x + num__50000 = ( num__1 + . num__125 ) * x num__1.12 x + num__50000 = num__1.125 x num__50000 = . num__005 x num__10 num__000000 = x thus the new pay = ( num__1 + . num__125 ) * num__10 num__000000 = num__11 num__250000 c <eor> c <eos> |
c |
c |
| the average of num__6 observations is num__12 . a new observation is included and the new average is decreased by num__1 . the seventh observation is ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
let seventh observation = x . then according to the question we have = > ( num__72 + x ) / num__7 = num__11 = > x = num__5 . hence the seventh observation is num__5 . answer : c <eor> c <eos> |
c |
multiply__6.0__12.0__ add__6.0__1.0__ subtract__12.0__1.0__ subtract__6.0__1.0__ subtract__6.0__1.0__ |
multiply__6.0__12.0__ add__6.0__1.0__ subtract__12.0__1.0__ subtract__6.0__1.0__ divide__5.0__1.0__ |
| the sum of the squares of three numbers is num__138 while the sum of their products taken two at a time is num__131 . their sum is : <o> a ) num__20 <o> b ) num__30 <o> c ) num__40 <o> d ) num__50 <o> e ) none of these |
let the numbers be a b and c . then a num__2 + b num__2 + c num__2 = num__138 and ( ab + bc + ca ) = num__131 . ( a + b + c ) num__2 = a num__2 + b num__2 + c num__2 + num__2 ( ab + bc + ca ) = num__138 + num__2 x num__131 = num__400 . ( a + b + c ) = num__400 = num__20 answer : a <eor> a <eos> |
a |
divide__400.0__20.0__ |
divide__400.0__20.0__ |
| a cylindrical water tower with radius num__28 m and height num__10 m is num__0.75 full at noon . every minute . num__08 π m num__3 is drawn from tank while . num__03 π m num__3 is added . additionally starting at num__1 pm and continuing each hour on the hour there is a periodic drain of num__4 π m num__3 . from noon how many hours will it take to drain the entire tank ? <o> a ) num__840 num__0.285714285714 <o> b ) num__820 num__0.857142857143 <o> c ) num__821 <o> d ) num__821 num__0.428571428571 <o> e ) num__840 |
initial volume = ( num__0.75 ) × ∏ × num__28 ² × num__10 = num__5880 ∏ relative drain / min = . num__08 ∏ - . num__03 ∏ = . num__05 ∏ m ³ / min drain relative drain / hour = . num__05 ∏ × num__60 = num__3 ∏ m ³ / hr every one hour starting from num__1 pm num__4 ∏ m ³ of water is drained . it means that only at the hour the water is drained and not “ in that num__1 hour “ so after num__1 hr the relative drain would be num__3 ∏ + num__4 ∏ = num__7 ∏ m ³ of water drain what i did initially was formed an equation num__5880 ∏ = num__7 ∏ × n ( n is the number of hrs ) so ended up with num__840 . e <eor> e <eos> |
e |
subtract__8.0__3.0__ hour_to_min_conversion__ divide__28.0__4.0__ divide__5880.0__7.0__ round__840.0__ |
subtract__8.0__3.0__ hour_to_min_conversion__ divide__28.0__4.0__ divide__5880.0__7.0__ divide__840.0__1.0__ |
| a new apartment complex purchased num__60 toilets and num__20 shower heads . if the price of a toilet is three times the price of a shower head what percent of the total cost was the cost of all the shower heads ? <o> a ) num__9.0 <o> b ) num__10.0 <o> c ) num__11.0 <o> d ) num__13.0 <o> e ) num__15 % |
let price of shower heads = x then tiolets = num__3 x totalcost = num__20 * x + num__60 * num__3 x = num__200 x % of total cost was the cost of all the shower heads = ( num__20 x / num__200 x ) * num__100 = num__10.0 answer : b <eor> b <eos> |
b |
divide__60.0__20.0__ divide__200.0__20.0__ subtract__20.0__10.0__ |
divide__60.0__20.0__ divide__200.0__20.0__ divide__100.0__10.0__ |
| for the symbol m ” n = n ^ num__2 − m for all values of m and n . what is the value of num__4 ” num__2 ? <o> a ) num__5 <o> b ) num__3 <o> c ) num__2 <o> d ) num__1 <o> e ) num__0 |
num__4 ” num__2 = num__4 - num__4 = num__0 answer : e <eor> e <eos> |
e |
multiply__2.0__0.0__ |
multiply__2.0__0.0__ |
| if num__7 spiders make num__7 webs in num__7 days then how many days are needed for num__1 spider to make num__1 web ? <o> a ) num__1 <o> b ) num__7 <o> c ) num__3 <o> d ) num__14 <o> e ) num__15 |
explanation : let num__1 spider make num__1 web in x days . more spiders less days ( indirect proportion ) more webs more days ( direct proportion ) ⇒ num__7 × num__1 × num__7 = num__1 × num__7 × x ⇒ x = num__7 . answer : option b <eor> b <eos> |
b |
round__7.0__ |
round__7.0__ |
| in a party every person shakes hands with every other person . if there were a total of num__190 handshakes in the party then what is the number of persons present in the party ? <o> a ) num__15 <o> b ) num__16 <o> c ) num__17 <o> d ) num__18 <o> e ) num__20 |
explanation : let the number of persons be n â ˆ ´ total handshakes = nc num__2 = num__190 n ( n - num__1 ) / num__2 = num__190 â ˆ ´ n = num__20 answer : option e <eor> e <eos> |
e |
multiply__1.0__20.0__ |
divide__20.0__1.0__ |
| a man swims downstream num__30 km and upstream num__18 km taking num__3 hours each time what is the speed of the man in still water ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__5 <o> d ) num__6 <o> e ) num__3 |
num__30 - - - num__3 ds = num__10 ? - - - - num__1 num__18 - - - - num__3 us = num__6 ? - - - - num__1 m = ? m = ( num__10 + num__6 ) / num__2 = num__8 answer : b <eor> b <eos> |
b |
divide__30.0__3.0__ divide__18.0__3.0__ subtract__3.0__1.0__ subtract__18.0__10.0__ round__8.0__ |
divide__30.0__3.0__ divide__18.0__3.0__ subtract__3.0__1.0__ subtract__18.0__10.0__ subtract__18.0__10.0__ |
| a trader marks his articles num__20.0 more than the cost price . if he allows num__20.0 discount then find his gain or loss percent ? <o> a ) num__6.0 <o> b ) num__8.0 <o> c ) num__5.0 <o> d ) num__2.0 <o> e ) num__4 % |
let cp of an article = rs . num__100 mp = rs . num__120 discount = num__20.0 sp = m [ ( num__100 - d % ) / num__100 ] = num__120 ( num__0.8 ) = rs . num__96 clearly the trader gets num__4.0 loss . answer : e <eor> e <eos> |
e |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| the difference between the place values of two threes in the numerical num__40378273 is <o> a ) num__29997 <o> b ) num__299997 <o> c ) num__297 <o> d ) num__0 <o> e ) none of them |
required difference = ( num__300000 - num__3 ) = num__299997 . answer is b <eor> b <eos> |
b |
subtract__300000.0__3.0__ subtract__300000.0__3.0__ |
subtract__300000.0__3.0__ subtract__300000.0__3.0__ |
| what is the units digit of ( num__5 ! * num__5 ! + num__6 ! * num__5 ! ) / num__3 ? <o> a ) num__4 <o> b ) num__3 <o> c ) num__2 <o> d ) num__1 <o> e ) num__0 |
( num__5 ! * num__5 ! + num__6 ! * num__5 ! ) / num__3 = num__5 ! ( num__5 ! + num__6 ! ) / num__3 = num__120 ( num__120 + num__720 ) / num__3 = ( num__120 * num__840 ) / num__3 = num__120 * num__280 units digit of the above product will be equal to num__0 answer e <eor> e <eos> |
e |
multiply__6.0__120.0__ add__720.0__120.0__ divide__840.0__3.0__ multiply__5.0__0.0__ |
multiply__6.0__120.0__ add__720.0__120.0__ divide__840.0__3.0__ multiply__5.0__0.0__ |
| a can finish a work in num__18 days and b can do the same work in num__15 days . b worked for num__10 days and left the job . in how many days a alone can finish the remaining work ? <o> a ) num__3 days . <o> b ) num__2 days . <o> c ) num__4 days . <o> d ) num__1 days . <o> e ) num__6 days . |
b ' s num__10 day ' s work = ( num__0.0666666666667 x num__10 ) = num__0.666666666667 . remaining work = ( num__1 - num__0.666666666667 ) = num__0.333333333333 . now num__0.0555555555556 work is done by a in num__1 day . num__0.333333333333 work is done by a in ( num__18 x num__0.333333333333 ) = num__6 days . answer : option e <eor> e <eos> |
e |
divide__10.0__15.0__ subtract__1.0__0.6667__ divide__1.0__18.0__ round__6.0__ |
divide__10.0__15.0__ subtract__1.0__0.6667__ divide__1.0__18.0__ round__6.0__ |
| if x < num__0 and num__0 < y < num__1 which of the following has the greatest value w ? <o> a ) x ^ num__2 <o> b ) ( xy ) ^ num__2 <o> c ) ( x / y ) ^ num__2 <o> d ) x ^ num__2 / y <o> e ) x ^ num__2 * y |
given x < num__0 and num__0 < y < num__1 let x = - num__2 and y = num__0.5 a . x ^ num__2 = ( - num__2 ) ^ num__2 = num__4 b . ( xy ) ^ num__2 = ( - num__2 * num__0.5 ) ^ num__2 = num__1 c . ( x / y ) ^ num__2 = { - num__2 / ( num__0.5 ) } ^ num__2 = ( - num__4 ) ^ num__2 = num__16 d . x ^ num__2 / y = ( - num__2 ) ^ num__2 / ( num__0.5 ) = num__4 * num__2 = num__8 e . x ^ num__2 * y = ( - num__2 ) ^ num__2 * ( num__0.5 ) = num__2 answer : option c <eor> c <eos> |
c |
reverse__2.0__ divide__2.0__0.5__ power__2.0__4.0__ multiply__2.0__4.0__ reverse__0.5__ |
reverse__2.0__ divide__2.0__0.5__ power__2.0__4.0__ multiply__2.0__4.0__ reverse__0.5__ |
| a class average mark in an exam is num__70 . the average of students who scored below num__60 is num__50 . the average of students who scored num__60 or more is num__75 . if the total number of students in this class is num__20 how many students scored below num__60 ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__15 <o> d ) num__17 <o> e ) num__20 |
let n the number of students who scored below num__60 and n the number of students who scored num__60 or more . xi the grades below num__60 and yi the grades num__60 or above . [ sum ( xi ) + sum ( yi ) ] / num__20 = num__70 : class average sum ( xi ) / n = num__50 : average for less that num__60 sum ( yi ) / n = num__75 : average for num__60 or more num__50 n + num__75 n = num__1400 : combine the above equations n + n = num__20 : total number of students n = num__4 and n = num__16 : solve the above system <eor> e <eos> |
e |
multiply__70.0__20.0__ subtract__20.0__4.0__ subtract__70.0__50.0__ |
multiply__70.0__20.0__ subtract__20.0__4.0__ add__4.0__16.0__ |
| the true discount on rs . num__1760 due after a certain time at num__12.0 per annum is rs . num__160 . the time after which it is due is : <o> a ) num__6 months <o> b ) num__8 months <o> c ) num__9 months <o> d ) num__10 months <o> e ) none |
solution p . w = rs . ( num__1760 - num__160 ) = rs . num__1600 . ∴ s . i on rs . num__1600 ar num__12.0 is rs . num__160 . ∴ time = num__100 x num__0.1 x num__12 ) = num__0.833333333333 years = ( num__0.833333333333 x num__12 ) months = num__10 months . answer d <eor> d <eos> |
d |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| a train num__800 m long is running at a speed of num__78 km / hr . if it crosses a tunnel in num__1 min then the length of the tunnel is ? <o> a ) num__532 m <o> b ) num__537 m <o> c ) num__500 m <o> d ) num__846 m <o> e ) num__536 m |
speed = num__78 * num__0.277777777778 = num__21.6666666667 m / sec . time = num__1 min = num__60 sec . let the length of the train be x meters . then ( num__800 + x ) / num__60 = num__21.6666666667 x = num__500 m . answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ round__500.0__ |
hour_to_min_conversion__ multiply__1.0__500.0__ |
| amithab ' s average expenditure for the january to june is rs . num__4200 and he spends rs . num__1200 in january and rs . num__1500 in july . the average expenditure for the months of febraury to july is <o> a ) rs . num__4250 <o> b ) rs . num__4228 <o> c ) rs . num__4128 <o> d ) rs . num__4988 <o> e ) rs . num__4192 |
amithab ' s total expenditure for jan - june = num__4200 x num__6 = num__25200 expenditure for feb - june = num__25200 - num__1200 = num__24000 expenditure for the months of feb - july = num__24000 + num__1500 = num__25500 the average expenditure = { num__25500 } / { num__6 } = num__4250 answer : a <eor> a <eos> |
a |
multiply__4200.0__6.0__ subtract__25200.0__1200.0__ add__1500.0__24000.0__ divide__25500.0__6.0__ divide__25500.0__6.0__ |
multiply__4200.0__6.0__ subtract__25200.0__1200.0__ add__1500.0__24000.0__ divide__25500.0__6.0__ divide__25500.0__6.0__ |
| a taxi driver states that his cab number is divisible by the no num__2 num__34 num__56 with a remainder of num__1 and when the number is divided by num__11 it does not give any remainder <o> a ) num__101 <o> b ) num__11 <o> c ) num__121 <o> d ) num__111 <o> e ) num__131 |
on finding the lcm of num__23 num__45 & num__6 we get the value as num__60 on adding num__1 to num__60 satisfy the first condition now num__61 not divisible by num__11 then multiply num__60 by num__2 we get num__120 : num__120 + num__1 num__121 is exactly divisible by num__11 answer : c <eor> c <eos> |
c |
subtract__34.0__11.0__ add__34.0__11.0__ add__1.0__60.0__ multiply__2.0__60.0__ add__1.0__120.0__ add__1.0__120.0__ |
subtract__34.0__11.0__ add__34.0__11.0__ add__1.0__60.0__ multiply__2.0__60.0__ add__1.0__120.0__ add__1.0__120.0__ |
| company s produces two kinds of radios : basic and deluxe . of the radios produced by company s last month num__0.666666666667 were basic and the rest were deluxe . if it takes num__1.4 as many hours to produce a deluxe radio as it does to produce a basic radio then the number of hours it took to produce the deluxe radio last month was what fraction of the total number of hours it took to produce all the radios ? <o> a ) num__0.411764705882 <o> b ) num__0.666666666667 <o> c ) num__0.470588235294 <o> d ) num__0.352941176471 <o> e ) num__0.75 |
no of basic radios was num__0.666666666667 of total and # of deluxe radios was num__0.333333333333 of total let ' s assume total = num__12 then basic = num__8 and deluxe = num__4 . now if time needed to produce one deluxe rdio is num__1 unit than time needed to produce one basic radio would be num__1.4 units . total time for basic would be num__8 * num__1 = num__8 and total time for deluxe would be num__4 * num__1.4 = num__5.6 - - > total time for both of them would be num__8 + num__5.6 - - > num__13.6 deluxe / total = num__5.6 / num__13.6 = num__0.411764705882 = num__0.411764705882 answer : a . <eor> a <eos> |
a |
subtract__12.0__8.0__ round_down__1.4__ multiply__1.4__4.0__ add__5.6__8.0__ divide__5.6__13.6__ divide__5.6__13.6__ |
subtract__12.0__8.0__ add__0.6667__0.3333__ multiply__1.4__4.0__ add__5.6__8.0__ divide__5.6__13.6__ divide__5.6__13.6__ |
| a train num__100 meters long completely crosses a num__300 meters long bridge in num__12 seconds . what is the speed of the train is ? <o> a ) num__32 kmph <o> b ) num__76 kmph <o> c ) num__120 kmph <o> d ) num__43 kmph <o> e ) num__40 kmph |
s = ( num__100 + num__300 ) / num__12 = num__33.3333333333 * num__3.6 = num__120 answer : c <eor> c <eos> |
c |
round__120.0__ |
round__120.0__ |
| in how many ways can be num__5 boys and num__5 girls sit around circular table so that no two boys sit next to each other ? <o> a ) ( num__5 ! ) ^ num__2 <o> b ) ( num__6 ! ) ^ num__2 <o> c ) num__5 ! num__4 ! <o> d ) num__11 ! <o> e ) ( num__5 ! ) ^ num__2 * num__6 ! |
first fix one boy and place other num__4 in alt seats so total ways is num__4 ! now place each girl between a pair of boys . . . total ways of seating arrangement of girls num__5 ! total is num__5 ! * num__4 ! ans c <eor> c <eos> |
c |
vowel_space__ |
vowel_space__ |
| a waiter ' s salary consists of his salary and tips . during one week his tips were num__1.25 of his salary . what fraction of his income came from tips ? <o> a ) num__0.444444444444 <o> b ) num__1.25 <o> c ) num__0.625 <o> d ) num__0.555555555556 <o> e ) num__0.666666666667 |
income = salary ( s ) + tips = s + s * num__1.25 = s * num__2.25 tips = s * num__1.25 fraction of his income came from tips = ( s * num__1.25 ) / ( s * num__2.25 ) = num__0.555555555556 answer : d <eor> d <eos> |
d |
divide__1.25__2.25__ divide__1.25__2.25__ |
divide__1.25__2.25__ divide__1.25__2.25__ |
| mangala completes a piece of work in num__30 days raju completes the same work in num__45 days . if both of them work together then the number of days required to complete the work is <o> a ) num__18 days <o> b ) num__6 days <o> c ) num__8 days <o> d ) num__10 days <o> e ) num__11 days |
if a can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days . that is the required no . of days = num__30 × num__0.6 = num__18 days . answer : a <eor> a <eos> |
a |
km_to_mile_conversion__ multiply__30.0__0.6__ round__18.0__ |
km_to_mile_conversion__ multiply__30.0__0.6__ round__18.0__ |
| in what time will a train num__120 m long cross an electric pole it its speed be num__144 km / hr ? <o> a ) num__2.5 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
speed = num__144 * num__0.277777777778 = num__40 m / sec time taken = num__3.0 = num__3 sec . answer : option b <eor> b <eos> |
b |
divide__120.0__40.0__ round__3.0__ |
divide__120.0__40.0__ divide__120.0__40.0__ |
| two vessels a and b contain milk and water mixed in the ratio num__8 : num__5 and num__5 : num__2 respectively . the ratio in which these two mixtures be mixed to get a new mixture containing milk and water in the ratio num__9 : num__4 <o> a ) num__2 : num__7 <o> b ) num__5 : num__2 <o> c ) num__3 : num__5 <o> d ) num__5 : num__7 <o> e ) num__6 : num__7 |
soln : - required ratio = ( num__0.714285714286 − num__0.692307692308 ) / ( num__0.692307692308 − num__0.615384615385 ) = num__0.285714285714 answer : a <eor> a <eos> |
a |
divide__8.0__4.0__ |
divide__8.0__4.0__ |
| there are num__6 magazines lying on a table ; num__3 are fashion magazines and the other num__3 are sports magzines . if num__2 magazines are to be selected at random from the num__8 magazines what is the probability that at least one of the fashion magazine will be selected ? <o> a ) num__0.8 <o> b ) num__0.888888888889 <o> c ) num__0.666666666667 <o> d ) num__0.5 <o> e ) num__0.5 |
num__1 - ( num__3 c num__0.333333333333 c num__2 ) = num__1 - num__0.2 = num__0.8 answer : a <eor> a <eos> |
a |
negate_prob__0.2__ negate_prob__0.2__ |
negate_prob__0.2__ negate_prob__0.2__ |
| two employees x and y are paid a total of rs . num__660 per week by their employer . if x is paid num__120 percent of the sum paid to y how much is y paid per week ? <o> a ) s . num__150 <o> b ) s . num__200 <o> c ) s . num__250 <o> d ) s . num__350 <o> e ) s . num__300 |
let the amount paid to x per week = x and the amount paid to y per week = y then x + y = num__660 but x = num__120.0 of y = num__120 y / num__100 = num__12 y / num__10 â ˆ ´ num__12 y / num__10 + y = num__660 â ‡ ’ y [ num__1.2 + num__1 ] = num__660 â ‡ ’ num__22 y / num__10 = num__660 â ‡ ’ num__22 y = num__6600 â ‡ ’ y = num__300.0 = num__300.0 = rs . num__300 e <eor> e <eos> |
e |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__12.0__10.0__ multiply__660.0__10.0__ divide__6600.0__22.0__ multiply__1.0__300.0__ |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__12.0__10.0__ multiply__660.0__10.0__ divide__6600.0__22.0__ divide__6600.0__22.0__ |
| a certain sum of money at simple interest amounted rs . num__850 in num__10 years at num__3.0 per annum find the sum ? <o> a ) num__651.85 <o> b ) num__652.85 <o> c ) num__653.85 <o> d ) num__654.85 <o> e ) num__655.85 |
num__850 = p [ num__1 + ( num__10 * num__3 ) / num__100 ] p = num__653.85 answer : c <eor> c <eos> |
c |
percent__100.0__653.85__ |
percent__100.0__653.85__ |
| a person can swim in still water at num__4 km / h . if the speed of water num__2 km / h how many hours will the man take to swim back against the current for num__14 km ? <o> a ) num__3 <o> b ) num__6 <o> c ) num__8 <o> d ) num__9 <o> e ) num__7 |
m = num__4 s = num__2 us = num__4 - num__2 = num__2 d = num__14 t = num__7.0 = num__7 answer : e <eor> e <eos> |
e |
divide__14.0__2.0__ round__7.0__ |
divide__14.0__2.0__ subtract__14.0__7.0__ |
| if the selling price of num__50 articles is equal to the cost price of num__25 articles then the loss or gain percent is : <o> a ) num__45.0 <o> b ) num__23.0 <o> c ) num__20.0 <o> d ) num__60.0 <o> e ) num__50 % |
c . p . of each article be re . num__1 . then c . p . of num__50 articles = rs . num__50 ; s . p . of num__50 articles = rs . num__25 . loss % = num__0.5 * num__100 = num__50.0 answer e <eor> e <eos> |
e |
percent__50.0__1.0__ percent__50.0__100.0__ |
percent__50.0__1.0__ percent__50.0__100.0__ |
| num__3 women and a few men participated in a chess tournament . each player played two matches with each of the other players . if the number of matches that men played among themselves is num__78 more than those they played with the women how many more men than women participated in the tournament ? <o> a ) num__20 <o> b ) num__15 <o> c ) num__11 <o> d ) num__10 <o> e ) num__19 |
let x be the number of men . number of matches men play among themselves is num__2 * ( xc num__2 ) number of matches men play with women is num__2 * ( num__3 x ) num__2 * ( xc num__2 ) - num__2 * ( num__3 x ) = num__78 x = num__13 difference between men and women is num__13 - num__3 = num__10 . answer d <eor> d <eos> |
d |
subtract__13.0__3.0__ subtract__13.0__3.0__ |
subtract__13.0__3.0__ subtract__13.0__3.0__ |
| a train num__110 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__110 * num__0.0545454545455 ] sec = num__6 sec answer : option b <eor> b <eos> |
b |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ round__6.0__ |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ divide__110.0__18.3333__ |
| in the infinite sequence num__50 num__60 num__70 num__80 num__90 . . . where each term is num__10 greater than the previous term the num__100 th term is <o> a ) num__1330 <o> b ) num__1807 <o> c ) num__1910 <o> d ) num__1040 <o> e ) num__2060 |
t ( n ) = a + ( n - num__1 ) d here d = num__10 a = num__50 n = num__100 solving we get t ( n ) = num__1040 . answer : d <eor> d <eos> |
d |
multiply__1040.0__1.0__ |
multiply__1040.0__1.0__ |
| x starts a business with rs . num__45000 . y joins in the business after num__3 months with rs . num__30000 . what will be the ratio in which they should share the profit at the end of the year ? <o> a ) num__1 : num__2 <o> b ) num__2 : num__1 <o> c ) num__1 : num__3 <o> d ) num__3 : num__1 <o> e ) num__4 : num__2 |
explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = num__45000 Ã — num__12 : num__30000 Ã — num__9 = num__45 Ã — num__12 : num__30 Ã — num__9 = num__3 Ã — num__12 : num__2 Ã — num__9 = num__2 : num__1 answer is b <eor> b <eos> |
b |
subtract__12.0__3.0__ subtract__3.0__2.0__ subtract__3.0__1.0__ |
subtract__12.0__3.0__ subtract__3.0__2.0__ subtract__3.0__1.0__ |
| the radius of a wheel is num__22.4 cm . what is the distance covered by the wheel in making num__600 resolutions ? <o> a ) num__844.8 m <o> b ) num__704 m <o> c ) num__774 m <o> d ) num__714 m <o> e ) num__744 m |
in one resolution the distance covered by the wheel is its own circumference . distance covered in num__600 resolutions . = num__600 * num__2 * num__3.14285714286 * num__22.4 = num__84480 cm = num__844.8 m answer : a <eor> a <eos> |
a |
round__844.8__ |
round__844.8__ |
| how many num__5 letter words ( with or without meaning ) can be formed using all the following num__5 letters p q r v and t so that letter p is to the left of letter r ? <o> a ) num__90 <o> b ) num__60 <o> c ) num__120 <o> d ) num__144 <o> e ) num__160 |
two positions can be selected out of num__5 positions in num__5 c num__2 ways = num__10 ways . and the remaining num__3 letters can be placed in num__3 posions in num__3 ! ways = num__6 ways . so the final answer = number of ways letters p and r can be arranged x number of ways the other num__3 letters can be arranged final answer = num__10 x num__6 = num__60 ways b <eor> b <eos> |
b |
multiply__5.0__2.0__ subtract__5.0__2.0__ multiply__2.0__3.0__ multiply__6.0__10.0__ multiply__6.0__10.0__ |
multiply__5.0__2.0__ subtract__5.0__2.0__ multiply__2.0__3.0__ multiply__6.0__10.0__ multiply__6.0__10.0__ |
| a num__300 m long train crosses a platform in num__39 sec while it crosses a signal pole in num__24 sec . what is the length of the platform ? <o> a ) num__389 m <o> b ) num__350 m <o> c ) num__187.5 m <o> d ) num__299 m <o> e ) num__219.5 m |
speed = num__12.5 = num__12.5 m / sec . let the length of the platform be x meters . then ( x + num__300 ) / num__39 = num__12.5 = > x = num__187.5 m . answer : c <eor> c <eos> |
c |
divide__300.0__24.0__ round__187.5__ |
divide__300.0__24.0__ round__187.5__ |
| line k is in the rectangular coordinate system . if the c - intercept of k is - num__2 and the y - intercept is num__3 which of the following is an equation of line k ? <o> a ) - num__3 c + num__2 y = num__6 <o> b ) num__3 c + num__2 y = - num__6 <o> c ) num__3 c - num__2 y = num__6 <o> d ) num__2 c - num__3 y = num__6 <o> e ) - num__2 c - num__3 y = num__6 |
this question can be solved in much simpler way . to find the c intercept put y = num__0 in the equation of the line . to find the y - intercept put c = num__0 in the equation of the line . so sub c = num__0 in the answer choices check whether you are getting y = num__3 . a . - num__3 c + num__2 y = num__6 - - - y = num__3 b . num__3 c + num__2 y = - num__6 - - - y = - num__3 c . num__3 c - num__2 y = num__6 - - - - - y = - num__3 d . num__2 c - num__3 y = num__6 - - - y = - num__2 e . - num__2 c - num__3 y = num__6 - - - y = - num__2 eliminate answer b c d and e . so the answer is a . <eor> a <eos> |
a |
multiply__2.0__3.0__ divide__6.0__2.0__ |
multiply__2.0__3.0__ subtract__6.0__3.0__ |
| the sale price of an article including the sales tax is rs . num__616 . the rate of sales tax is num__10.0 . if the shopkeeper has made a profit of num__12.0 then the cost price of the article is : <o> a ) rs . num__500 <o> b ) rs . num__515 <o> c ) rs . num__550 <o> d ) rs . num__600 <o> e ) rs . num__700 |
num__110.0 of s . p . = num__616 s . p . = ( num__616 * num__100 ) / num__110 = rs . num__560 c . p = ( num__110 * num__560 ) / num__112 = rs . num__500 answer : a <eor> a <eos> |
a |
percent__100.0__500.0__ |
percent__100.0__500.0__ |
| in an election between two candidates one got num__60.0 of the total valid votes num__30.0 of the votes were invalid . if the total number of votes was num__9000 the number of valid votes that the other candidate got was : <o> a ) num__2500 <o> b ) num__2520 <o> c ) num__3000 <o> d ) num__3100 <o> e ) none of these |
explanation : number of valid votes = num__70.0 of num__9000 = num__6300 . valid votes polled by other candidates = num__40.0 of num__6300 ( 3.97495776607e-05 ) = num__2520 . harsh mishra a year ago num__0 upvotes answer : b <eor> b <eos> |
b |
percent__70.0__9000.0__ percent__40.0__6300.0__ percent__40.0__6300.0__ |
percent__70.0__9000.0__ percent__40.0__6300.0__ percent__40.0__6300.0__ |
| a rectangular grassy plot num__110 m . by num__65 m has a gravel path num__2.5 m wide all round it on the inside . find the cost of gravelling the path at num__50 paise per sq . metre <o> a ) s num__425 <o> b ) s num__780 <o> c ) s num__880 <o> d ) s num__480 <o> e ) s num__980 |
area of the plot = num__110 m * num__65 m = num__7150 sq . m area of plot excluding gravel = num__105 m * num__60 m = num__6300 sq . m area of gravel = num__7150 sq . m - num__6300 sq . m = num__850 sq . m cost of building it = num__850 sq . m * num__50 = num__42500 p in rs = num__425.0 = rs num__425 answer : a <eor> a <eos> |
a |
multiply__110.0__65.0__ hour_to_min_conversion__ multiply__60.0__105.0__ subtract__7150.0__6300.0__ multiply__50.0__850.0__ round__425.0__ |
multiply__110.0__65.0__ subtract__110.0__50.0__ multiply__60.0__105.0__ subtract__7150.0__6300.0__ multiply__50.0__850.0__ subtract__850.0__425.0__ |
| an article is bought for rs . num__685 and sold for rs . num__900 find the gain percent ? <o> a ) num__33 num__0.125 % <o> b ) num__33 num__2.66666666667 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__31 num__0.387096774194 % <o> e ) num__32 num__0.333333333333 % |
num__685 - - - - num__215 num__100 - - - - ? = > num__31 num__0.387096774194 % answer : d <eor> d <eos> |
d |
percent__31.0__100.0__ |
percent__31.0__100.0__ |
| num__1 / num__0.04 is equal to <o> a ) num__25.5 <o> b ) num__2.5 <o> c ) num__25 <o> d ) . num__25 <o> e ) none of these |
explanation : num__1 / num__0.04 = ( num__1 * num__100 ) / num__4 = num__25.0 = num__25 option c <eor> c <eos> |
c |
multiply__0.04__100.0__ reverse__0.04__ reverse__0.04__ |
multiply__0.04__100.0__ reverse__0.04__ reverse__0.04__ |
| the difference of two numbers is num__1365 . on dividing the larger number by the smaller we get num__6 as quotient and the num__15 as remainder . what is the smaller number ? <o> a ) num__226 <o> b ) num__523 <o> c ) num__775 <o> d ) num__890 <o> e ) num__270 |
e num__270 let the smaller number be x . then larger number = ( x + num__1365 ) . x + num__1365 = num__6 x + num__15 num__5 x = num__1350 x = num__270 smaller number = num__270 . <eor> e <eos> |
e |
subtract__1365.0__15.0__ divide__1350.0__5.0__ |
subtract__1365.0__15.0__ divide__1350.0__5.0__ |
| ayesha ' s father was num__38 years of age when she was born while her mother was num__36 years old when her brother seven years younger to her was born . what is the difference between the ages of her parents ? <o> a ) num__2 years <o> b ) num__4 years <o> c ) num__6 years <o> d ) num__7 years <o> e ) num__9 years |
mother ' s age when ayesha ' s brother was born = num__36 years . father ' s age when ayesha ' s brother was born = ( num__38 + num__7 ) years = num__45 years . required difference = ( num__45 - num__36 ) years = num__9 years . answer : option e <eor> e <eos> |
e |
add__38.0__7.0__ subtract__45.0__36.0__ subtract__45.0__36.0__ |
add__38.0__7.0__ subtract__45.0__36.0__ subtract__45.0__36.0__ |
| a sum of money is to be distributed among a b c d in the proportion of num__5 : num__2 : num__4 : num__3 . if c gets rs . num__800 more than d what is b ' s share ? <o> a ) rs . num__1600 <o> b ) rs . num__1500 <o> c ) rs . num__2000 <o> d ) rs . num__2500 <o> e ) none of the above |
let the shares of a b c and d be rs . num__5 x rs . num__2 x rs . num__4 x and rs . num__3 x respectively . then num__4 x - num__3 x = num__800 x = num__800 . b ' s share = rs . num__2 x = rs . ( num__2 x num__800 ) = rs . num__1600 . answer = a <eor> a <eos> |
a |
multiply__2.0__800.0__ multiply__2.0__800.0__ |
multiply__2.0__800.0__ multiply__2.0__800.0__ |
| how much time does a train num__150 metres long running at num__90 km / hr take to pass a pole ? <o> a ) num__7.9 s <o> b ) num__6 s <o> c ) num__7.5 s <o> d ) num__7.6 s <o> e ) num__7.4 s |
explanation : num__90 km / hr = num__90 * num__0.277777777778 = num__25 m / s speed = distance / time ; v = d / t num__25 = num__150 / t t = num__6 s answer : b <eor> b <eos> |
b |
divide__150.0__25.0__ divide__150.0__25.0__ |
divide__150.0__25.0__ divide__150.0__25.0__ |
| karthikeyan arranged a party to celebrate his birth day . num__20 of his close friends were invited . one friend left before the cake was cut due to an urgent phone call . the big cake was drawing the attention of the party visitors . this cake is to be divided among friends and of course karthikeyan will get his share . a man eats num__3 pieces a woman eats two pieces and a child eats half a piece of cake . including karthikeyan count the number of men women and children in the party . there are num__20 pieces of cake in all . <o> a ) num__6 women num__2 men and num__12 children <o> b ) num__7 women num__1 men and num__12 children <o> c ) num__5 women num__1 men and num__14 children <o> d ) num__4 women num__2 men and num__14 children <o> e ) none of these |
let the number of men be m number of women be w and number of children be c total number of pieces = pieces ate by men + pieces ate by women + pieces ate by children there are num__20 total pieces . also a man eats num__3 pieces a woman eats num__2 and children half . therefore above equation becomes . num__20 = num__3 m + num__2 w + num__0.5 c in the above equation you have to substitute the values given in options to find which one satisfies the equation . option c with m = num__1 c = num__14 and w = num__5 perfectly satisfies the equation as follows num__20 = num__3 ( num__1 ) + num__2 ( num__5 ) + ( num__0.5 ) ( num__14 ) num__20 = num__20 i e our equation is satisfied perfectly by option c answer : c <eor> c <eos> |
c |
reverse__2.0__ subtract__3.0__2.0__ add__3.0__2.0__ add__3.0__2.0__ |
reverse__2.0__ subtract__3.0__2.0__ add__3.0__2.0__ add__3.0__2.0__ |
| toby is four years younger than debby . thrice the sum of the ages of toby and debby equals their mother ’ s age . if the age of the mother is num__60 find the ages of toby and debby ? <o> a ) num__8 and num__12 <o> b ) num__5 and num__9 <o> c ) num__6 and num__10 <o> d ) num__5 and num__10 <o> e ) num__12 and num__16 |
let the age of debby be x and toby be x - num__4 num__3 ( x + x - num__4 ) = num__60 x = num__12 the ages of toby and debby are num__8 and num__12 . answer : a <eor> a <eos> |
a |
multiply__3.0__4.0__ subtract__12.0__4.0__ subtract__12.0__4.0__ |
multiply__3.0__4.0__ subtract__12.0__4.0__ subtract__12.0__4.0__ |
| in the youth summer village there are num__100 people num__50 of them are not working num__25 of them have families and num__75 of them like to sing in the shower . what is the largest possible number of people in the village which are working that do n ' t have families and that are singing in the shower ? <o> a ) num__100 <o> b ) num__75 <o> c ) num__25 <o> d ) num__50 <o> e ) num__125 |
total = num__100 not working = num__50 having family = num__25 like to sing in shower = num__75 working = num__100 - num__50 = num__50 not having family = num__100 - num__25 = num__75 like to sing in shower = num__75 largest possible number is the lowest possible among the above thus num__50 d <eor> d <eos> |
d |
subtract__100.0__50.0__ |
subtract__100.0__50.0__ |
| ajay and vijay have some marbles with them . ajay told vijay ` ` if you give me ' x ' marbles both of us will have equal number of marbles ' ' . vijay then told ajay ` ` if you give me twice as many marbles i will have num__30 more marbles than you would ' ' . find ' x ' ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__7 |
if vijay gives ' x ' marbles to ajay then vijay and ajay would have v - x and a + x marbles . v - x = a + x - - - ( num__1 ) if ajay gives num__2 x marbles to vijay then ajay and vijay would have a - num__2 x and v + num__2 x marbles . v + num__2 x - ( a - num__2 x ) = num__30 = > v - a + num__4 x = num__30 - - - ( num__2 ) from ( num__1 ) we have v - a = num__2 x substituting v - a = num__2 x in ( num__2 ) num__6 x = num__30 = > x = num__5 . answer : b <eor> b <eos> |
b |
add__2.0__4.0__ divide__30.0__6.0__ divide__30.0__6.0__ |
add__2.0__4.0__ add__1.0__4.0__ add__1.0__4.0__ |
| if $ num__3000 was invested at an annual interest rate of num__5.6 compounded annually which of the following represents the amount the investment was worth after num__3 years ? <o> a ) num__3000 ( num__1.056 ) ^ num__3 <o> b ) num__3000 ( num__3 + num__1.056 ) <o> c ) num__3000 ( num__1 + num__3 ( num__0.056 ) ) <o> d ) num__3000 ( num__1 + ( num__0.056 ) ^ num__3 ) <o> e ) num__3000 ( num__1.056 ) ( num__3 ) |
the formula is ci = p ( num__1 + r / num__100 ) ^ t in this case i think a is the ans <eor> a <eos> |
a |
percent__100.0__3000.0__ |
percent__100.0__3000.0__ |
| working alone machine x can manufacture num__1000 nails in num__12 hours . working together machines x and y can manufacture num__1000 nails in num__5 hours . how many hours does it take machine y to manufacture num__800 nails working alone ? <o> a ) num__6 num__0.857142857143 <o> b ) num__7 num__0.333333333333 <o> c ) num__7 <o> d ) num__8 num__0.2 <o> e ) num__9 num__0.571428571429 |
rate of machine x = num__83.3333333333 ( num__83.3333333333 + y ) num__5 = num__1000 y = num__116.666666667 num__116.666666667 * t = num__800 t = num__6 num__0.857142857143 a . <eor> a <eos> |
a |
divide__1000.0__12.0__ round__6.0__ |
divide__1000.0__12.0__ round__6.0__ |
| four packages have an average weight of num__12.5 pounds . what is the minimum possible weight of the heaviest package in pounds if the median is num__8 pounds ? <o> a ) num__18 <o> b ) num__20 <o> c ) num__22 <o> d ) num__24 <o> e ) num__26 |
let us denote the weights of the packages in pounds by a b c d naming from the lightest one to the heaviest one . the median is num__8 pounds . therefore ( b + c ) / num__2 = num__8 . b + c = num__16 the average is num__12.5 pounds . therefore ( a + b + c + d ) / num__4 = num__12.5 . a + ( b + c ) + d = num__50 a + num__16 + d = num__50 a + d = num__34 the weight a must be no greater than num__8 since num__8 is the median . therefore the minimum possible weight of the heaviest package is num__34 – num__8 = num__26 pounds ( all the other packages would weigh num__8 pounds in this case ) . answer : e <eor> e <eos> |
e |
multiply__8.0__2.0__ divide__8.0__2.0__ multiply__12.5__4.0__ subtract__50.0__16.0__ subtract__34.0__8.0__ subtract__34.0__8.0__ |
multiply__8.0__2.0__ divide__8.0__2.0__ multiply__12.5__4.0__ subtract__50.0__16.0__ subtract__34.0__8.0__ subtract__34.0__8.0__ |
| in what ratio must rice of rs . num__16 per kg be mixed with rice of rs . num__10 per kg so that cost of mixture is rs . num__12 per kg ? <o> a ) num__2 : num__3 <o> b ) num__5 : num__8 <o> c ) num__5 : num__6 <o> d ) num__3 : num__4 <o> e ) num__1 : num__2 |
( num__12 - num__10 ) / ( num__16 - num__12 ) = num__0.5 = num__1 : num__2 answer : e <eor> e <eos> |
e |
reverse__0.5__ reverse__1.0__ |
reverse__0.5__ reverse__1.0__ |
| a retailer buys num__120 pens at the market price of num__36 pens from a wholesaler if he sells these pens giving a discount of num__1.0 what is the profit % ? <o> a ) num__210 <o> b ) num__215 <o> c ) num__212 <o> d ) num__230 <o> e ) num__320 |
let the market price of each pen be $ num__1 then cost price of num__120 pens = $ num__36 selling price of num__120 pens = num__99.0 of $ num__120 = $ num__118.80 profit % = ( ( num__82.80 * num__100 ) / num__36 ) % = num__230.0 answer d <eor> d <eos> |
d |
percent__99.0__120.0__ percent__100.0__230.0__ |
percent__99.0__120.0__ percent__100.0__230.0__ |
| ( num__169 ) ^ num__2 - ( num__168 ) ^ num__2 = <o> a ) num__1 <o> b ) num__100 <o> c ) num__229 <o> d ) num__329 <o> e ) num__337 |
using the formula : ( a + num__1 ) ^ num__2 - a ^ num__2 = num__2 a + num__1 so answer = num__168 * num__2 + num__1 = num__336 + num__1 = num__337 = answer = e <eor> e <eos> |
e |
subtract__169.0__168.0__ multiply__2.0__168.0__ add__169.0__168.0__ add__169.0__168.0__ |
subtract__169.0__168.0__ multiply__2.0__168.0__ add__169.0__168.0__ add__169.0__168.0__ |
| if num__3 : num__12 : : x : num__16 then find the value of x <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
explanation : treat num__3 : num__12 as num__0.25 and x : num__16 as x / num__16 treat : : as = so we get num__0.25 = x / num__16 = > num__12 x = num__48 = > x = num__4 option b <eor> b <eos> |
b |
divide__3.0__12.0__ multiply__3.0__16.0__ reverse__0.25__ multiply__12.0__0.25__ |
divide__3.0__12.0__ divide__12.0__0.25__ reverse__0.25__ divide__12.0__4.0__ |
| the average weight of a group of num__30 friends increases by num__1 kg when the weight of their football coach was added . if average weight of the group after including the weight of the football coach is num__31 kg what is the weight of their football coach ? <o> a ) num__31 kg <o> b ) num__61 kg <o> c ) num__60 kg <o> d ) num__62 kg <o> e ) num__91 kg |
explanatory answer the new average weight of the group after including the football coach = num__31 kg . as the new average is num__1 kg more than the old average old average without including the football coach = num__30 kg . the total weight of the num__30 friends without including the football coach = num__30 * num__30 = num__900 . after including the football coach the number people in the group increases to num__31 and the average weight of the group increases by num__1 kg . therefore the total weight of the group after including the weight of the football coach = num__31 * num__31 = num__961 kg . therefore the weight of the football coach = num__961 - num__900 = num__61 kg . answer b <eor> b <eos> |
b |
add__30.0__31.0__ add__30.0__31.0__ |
subtract__961.0__900.0__ multiply__1.0__61.0__ |
| at the wholesale store you can buy an num__8 - pack of hot dogs for $ num__1.55 a num__20 - pack for $ num__3.05 and a num__250 - pack for $ num__22.95 . what is the greatest number of hot dogs you can buy at this store with $ num__300 ? <o> a ) num__3108 <o> b ) num__3100 <o> c ) num__3108 <o> d ) num__3124 <o> e ) num__3 |
250 |
to maximize number of hot dogs with num__300 $ total number of hot dogs bought in num__250 - pack = num__22.95 * num__13 = num__298.35 $ amount remaining = num__300 - num__298.35 = num__1.65 $ this amount is too less to buy any num__8 - pack . greatest number of hot dogs one can buy with num__300 $ = num__250 * num__13 = num__3250 answer e <eor> e <eos> |
e |
e |
| if one - third of one - fourth of a number is num__15 then three - tenth of that number is <o> a ) num__35 <o> b ) num__36 <o> c ) num__45 <o> d ) num__54 <o> e ) num__57 |
let the number be x . then num__0.333333333333 of ¼ of x = num__15 x = num__15 x num__12 = num__180 the required number = ( num__0.3 ) * num__180 = num__54 answer d num__54 <eor> d <eos> |
d |
multiply__15.0__12.0__ multiply__0.3__180.0__ multiply__0.3__180.0__ |
multiply__15.0__12.0__ multiply__0.3__180.0__ multiply__0.3__180.0__ |
| a man has rs . num__528 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__90 <o> b ) num__94 <o> c ) num__96 <o> d ) num__97 <o> e ) num__99 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__528 num__16 x = num__528 x = num__33 . hence total number of notes = num__3 x = num__99 . e <eor> e <eos> |
e |
divide__528.0__16.0__ multiply__3.0__33.0__ multiply__3.0__33.0__ |
divide__528.0__16.0__ multiply__3.0__33.0__ multiply__3.0__33.0__ |
| the average monthly income of p and q is rs . num__5050 . the average monthly income of q and r is rs . num__6250 and the average monthly income of p and r is rs . num__5200 . the monthly income of p is : <o> a ) num__3500 <o> b ) num__4000 <o> c ) num__4050 <o> d ) num__5000 <o> e ) num__5500 |
explanation : let p q and r represent their respective monthly incomes . then we have : p + q = ( num__5050 x num__2 ) = num__10100 . . . . ( i ) q + r = ( num__6250 x num__2 ) = num__12500 . . . . ( ii ) p + r = ( num__5200 x num__2 ) = num__10400 . . . . ( iii ) adding ( i ) ( ii ) and ( iii ) we get : num__2 ( p + q + r ) = num__33000 or p + q + r = num__16500 . . . . ( iv ) subtracting ( ii ) from ( iv ) we get p = num__4000 . p ' s monthly income = rs . num__4000 . answer is b <eor> b <eos> |
b |
multiply__5050.0__2.0__ multiply__6250.0__2.0__ multiply__5200.0__2.0__ divide__33000.0__2.0__ subtract__16500.0__12500.0__ subtract__16500.0__12500.0__ |
multiply__5050.0__2.0__ multiply__6250.0__2.0__ multiply__5200.0__2.0__ divide__33000.0__2.0__ subtract__16500.0__12500.0__ subtract__16500.0__12500.0__ |
| a man swims downstream num__96 km and upstream num__64 km taking num__4 hours each time ; what is the speed of the current ? <o> a ) num__6 <o> b ) num__4 <o> c ) num__2 <o> d ) num__8 <o> e ) num__5 |
num__96 - - - num__4 ds = num__24 ? - - - - num__1 num__64 - - - - num__4 us = num__16 ? - - - - num__1 s = ? s = ( num__24 - num__16 ) / num__2 = num__4 answer : b <eor> b <eos> |
b |
divide__96.0__4.0__ divide__64.0__4.0__ round__4.0__ |
divide__96.0__4.0__ divide__64.0__4.0__ divide__96.0__24.0__ |
| five gallons of a solution of vinegar and water with num__8.0 vinegar is to be diluted with water to make a num__4.0 vinegar mixture . how many gallons of water should be added ? <o> a ) num__10 <o> b ) num__5 <o> c ) num__4 <o> d ) num__2 <o> e ) num__1 |
in these types of mixture questions if the answers are numbers then it ' s likely that you can test the answers to quickly get to the solution . here ' s how : first we need to know about the original mixture of vinegar and water . we ' re told that a num__5 gallon mixture is num__8.0 vinegar . . . . ( num__5 gallons ) ( . num__08 ) = num__0.4 gallons vinegar the rest = num__4.6 gallons water we ' re told to add water until the mixture goes from num__8.0 vinegar to num__4.0 vinegar . since the answer choices are numbers we know that one of them will lead us to a num__4.0 mix when added to the current mix . changing num__8.0 to num__4.0 means that we have to add a significant amount of water ( relative to what we started with ) so i would test one of the bigger numbers first . let ' s test b if num__5 gallons of water were added then we ' d have . . . . num__0.4 gallons vinegar num__9.6 gallons of water num__0.4 / num__10 = num__4.0 vinegar . this is exactly what we ' re looking for so b must be the answer . final answer : b <eor> b <eos> |
b |
subtract__5.0__0.4__ add__5.0__4.6__ divide__4.0__0.4__ add__0.4__4.6__ |
subtract__5.0__0.4__ add__5.0__4.6__ divide__4.0__0.4__ add__0.4__4.6__ |
| a company has a job to prepare certain number of cans and there are three machines a b and c for this job . a can complete the job in num__2 days b can complete the job in num__8 days and c can complete the job in num__10 days . how many days the company will it take to complete the job if all the machines are used simultaneously ? <o> a ) num__0.148148148148 days <o> b ) num__1.37931034483 days <o> c ) num__0.75 days <o> d ) num__0.5 days <o> e ) num__0.25 days |
let the total number of cans to be prepared be num__40 . the number of cans prepared by a in num__1 day = num__20 . the number of cans prepared by b in num__1 day = num__5 . the number of cans prepared by c in num__1 day = num__4 . thus the total number of cans that can be prepared by all the machines working simultaneously in a single day = num__29 . therefore the number of days taken to complete the whole work = num__1.37931034483 days . answer : b <eor> b <eos> |
b |
multiply__2.0__10.0__ divide__10.0__2.0__ divide__8.0__2.0__ divide__40.0__29.0__ multiply__1.0__1.3793__ |
multiply__2.0__10.0__ divide__10.0__2.0__ divide__8.0__2.0__ divide__40.0__29.0__ multiply__1.0__1.3793__ |
| three hoses work to fill a tub at at different rate . hose e and b working together can fill the tub in num__1.2 of an hour . hoses e and c can fill it in num__1.5 an hour . houses b and c can fill it in num__2 hours . how long does it take all num__3 hoses working together to fill the tub ? edited for accurate solution <o> a ) num__0.3 <o> b ) num__0.4 <o> c ) num__0.5 <o> d ) num__1 <o> e ) num__1.2 |
convert the given time to rate and you will be able to add it up . total rate of e and b = rate of e + rate of b = num__1 / ( num__1.2 ) = num__0.833333333333 total rate of e and c = rate of e + rate of c = num__1 / ( num__1.5 ) = num__0.666666666667 total rate of b and c = rate of b + rate of c = num__0.5 adding all three num__2 ( rate of e + rate of b + rate of c ) = num__0.833333333333 + num__0.666666666667 + num__0.5 = num__2 rate of e + rate of b + rate of c = num__1 tub / hour time taken by all three together to fill up the tub is num__1 hour = d <eor> d <eos> |
d |
subtract__3.0__2.0__ divide__1.0__1.2__ subtract__1.5__0.8333__ divide__1.5__3.0__ round__1.0__ |
subtract__3.0__2.0__ divide__1.0__1.2__ divide__2.0__3.0__ divide__1.5__3.0__ round__1.0__ |
| if the price of an item is decreased by num__30.0 and then increased by num__30.0 the net effect on the price of the item is <o> a ) a decrease of num__99.0 <o> b ) no change <o> c ) a decrease of num__1.0 <o> d ) a decrease of num__9.0 <o> e ) none |
initially assume num__100 rupees num__30.0 discount in num__100 gives price of num__70 rupees then num__30.0 raise in num__70 is only num__21 rupees . therefore total price = num__91 rupees . hence num__9.0 is the loss answer : d <eor> d <eos> |
d |
subtract__100.0__30.0__ add__70.0__21.0__ subtract__30.0__21.0__ subtract__30.0__21.0__ |
subtract__100.0__30.0__ add__70.0__21.0__ subtract__30.0__21.0__ subtract__30.0__21.0__ |
| in how much time will a train of length num__100 m moving at num__36 kmph cross an electric pole ? <o> a ) num__11 <o> b ) num__66 <o> c ) num__19 <o> d ) num__10 <o> e ) num__77 |
convert kmph to mps . num__36 kmph = num__36 * num__0.277777777778 = num__10 mps . the distance to be covered is equal to the length of the train . required time t = d / s = num__10.0 = num__10 sec . answer : d <eor> d <eos> |
d |
round__10.0__ |
divide__100.0__10.0__ |
| let q represent a set of two distinct prime numbers . if the sum of the numbers in q is even and x is a member of q then what is the least possible value that x can be ? <o> a ) num__1 <o> b ) num__5 <o> c ) num__3 <o> d ) num__2 <o> e ) num__7 |
q = p num__1 + p num__2 = even ( and all primes are distinct ) if the least prime is num__2 then we have sum of q = even . ans . d . num__2 <eor> d <eos> |
d |
multiply__1.0__2.0__ |
multiply__1.0__2.0__ |
| a can run num__128 metre in num__28 seconds and b in num__32 seconds . by what distance a beat b ? <o> a ) num__38 metre <o> b ) num__28 metre <o> c ) num__23 metre <o> d ) num__16 metre <o> e ) num__28 metre |
clearly a beats b by num__4 seconds now find out how much b will run in these num__4 seconds speed of b = distance / time taken by b = num__4.0 = num__4 m / s distance covered by b in num__4 seconds = speed à — time = num__4 à — num__4 = num__16 metre i . e . a beat b by num__16 metre answer is d <eor> d <eos> |
d |
divide__128.0__32.0__ round__16.0__ |
divide__128.0__32.0__ round__16.0__ |
| a sum of rs . num__2769 is lent into two parts so that the interest on the first part for num__8 years at num__3.0 per annum may be equal to the interest on the second part for num__3 years at num__5.0 per annum . find the second sum ? <o> a ) num__1642 <o> b ) num__1640 <o> c ) num__1632 <o> d ) num__2789 <o> e ) num__1704 |
( x * num__8 * num__3 ) / num__100 = ( ( num__2769 - x ) * num__3 * num__5 ) / num__100 num__24 x / num__100 = num__415.35 - num__15 x / num__100 num__39 x = num__41535 = > x = num__1065 second sum = num__2769 – num__1065 = num__1704 answer : e <eor> e <eos> |
e |
percent__100.0__1704.0__ |
percent__100.0__1704.0__ |
| the length of a rectangle is twice its breadth if its length is decreased by num__5 cm and breadth is increased by num__5 cm the area of the rectangle is increased by num__75 sq cm . find the length of the rectangle . <o> a ) num__22 <o> b ) num__40 <o> c ) num__38 <o> d ) num__278 <o> e ) num__29 |
explanation : let length = num__2 x and breadth = x then ( num__2 x - num__5 ) ( x + num__5 ) = ( num__2 x * x ) + num__75 num__5 x - num__25 = num__75 = > x = num__20 length of the rectangle = num__40 cm answer : b ) num__40 <eor> b <eos> |
b |
power__5.0__2.0__ square_perimeter__5.0__ multiply__2.0__20.0__ triangle_area__2.0__40.0__ |
power__5.0__2.0__ square_perimeter__5.0__ multiply__2.0__20.0__ multiply__2.0__20.0__ |
| in company b the total monthly payroll for the num__20 factory workers is $ num__36000 and the total monthly payroll for the num__50 office workers is $ num__110000 . by how much does the average ( arithmetic mean ) monthly salary of an office worker exceed that of a factory worker in this company ? <o> a ) $ num__400 <o> b ) $ num__450 <o> c ) $ num__500 <o> d ) $ num__550 <o> e ) $ num__600 |
the average monthly salary of a factory worker is : $ num__1800.0 = $ num__1800 . the average monthly salary of an office worker is : $ num__2200.0 = $ num__2200 . the difference in average salary is : $ num__2200 - $ num__1800 = $ num__400 . the answer is a . <eor> a <eos> |
a |
divide__36000.0__20.0__ divide__110000.0__50.0__ subtract__2200.0__1800.0__ subtract__2200.0__1800.0__ |
divide__36000.0__20.0__ divide__110000.0__50.0__ subtract__2200.0__1800.0__ subtract__2200.0__1800.0__ |
| a boatman goes num__3 km against the current of the stream in num__3 hour and goes num__1 km along the current in num__30 minutes . how long will it take to go num__3 km in stationary water ? <o> a ) num__5 hours <o> b ) num__4 hours <o> c ) num__3 hours <o> d ) num__2 hours <o> e ) num__1 hours |
explanation : speed upstream = num__1.0 = num__1 km / hr speed downstream = num__1 / ( num__0.5 ) = num__2 km / hr speed in still water = num__0.5 ( num__2 + num__1 ) = num__1.5 km / hr time taken to travel num__3 km in still water = num__3 / ( num__1.5 ) = num__2.0 = num__2 hours answer : option d <eor> d <eos> |
d |
subtract__3.0__1.0__ multiply__3.0__0.5__ round__2.0__ |
divide__1.0__0.5__ divide__3.0__2.0__ divide__3.0__1.5__ |
| what is the ratio between perimeters of two squares one having num__5 times the diagonal then the other ? <o> a ) num__3 : num__8 <o> b ) num__3 : num__6 <o> c ) num__3 : num__7 <o> d ) num__5 : num__1 <o> e ) num__3 : num__3 |
d = num__5 d d = d a √ num__2 = num__5 d a √ num__2 = d a = num__5 d / √ num__2 a = d / √ num__2 = > num__5 : num__1 answer : d <eor> d <eos> |
d |
multiply__5.0__1.0__ |
multiply__5.0__1.0__ |
| what is the rate percent when the simple interest on rs . num__600 amount to rs . num__160 in num__4 years ? <o> a ) num__6.66 <o> b ) num__7.66 <o> c ) num__6.06 <o> d ) num__6.16 <o> e ) num__8.66 % |
num__160 = ( num__600 * num__4 * r ) / num__100 r = num__6.66 answer : a <eor> a <eos> |
a |
percent__100.0__6.66__ |
percent__100.0__6.66__ |
| consider two postmen a and b respectively . a is young and can deliver num__20 parcels in num__3 hours while b is older than a and can deliver only num__15 parcels in num__4 hours . if the total number of parcels to deliver is num__60 how long they will take working together . <o> a ) num__6.54545454545 <o> b ) num__5.76 <o> c ) num__6.85714285714 <o> d ) num__0.651583710407 <o> e ) num__0.657534246575 |
work done by num__1 st in num__1 hour = num__6.66666666667 parcels / hour work done by num__2 nd om num__1 hour = num__3.75 parcels / hour total work done by both together per hour = num__6.66666666667 + num__3.75 = num__10.4166666667 parcels / hour time to do num__60 unit work ( ie parcels ) = num__60 ÷ num__10.4166666667 = num__60 × num__0.096 = num__5.76 hours . answer : b <eor> b <eos> |
b |
subtract__4.0__3.0__ divide__20.0__3.0__ subtract__3.0__1.0__ divide__15.0__4.0__ add__3.75__6.6667__ divide__1.0__10.4167__ divide__60.0__10.4167__ round__5.76__ |
subtract__4.0__3.0__ divide__20.0__3.0__ subtract__3.0__1.0__ divide__15.0__4.0__ add__3.75__6.6667__ divide__1.0__10.4167__ divide__60.0__10.4167__ divide__60.0__10.4167__ |
| in what time will a train num__175 m long cross an electric pole it its speed be num__180 km / hr ? <o> a ) num__7.5 <o> b ) num__6.5 <o> c ) num__5.5 <o> d ) num__4.5 <o> e ) num__3.5 |
speed = num__180 * num__0.277777777778 = num__50 m / sec time taken = num__3.5 = num__3.5 sec . answer : e <eor> e <eos> |
e |
divide__175.0__50.0__ round__3.5__ |
divide__175.0__50.0__ divide__175.0__50.0__ |
| twenty five percent of country y ' s yearly exports come from fruit exports . one - fifth of all fruit exports from country y are orange exports . if country y generates $ num__10 million from its orange exports how much money does it generate from its yearly total of exports ? <o> a ) $ num__2125 m <o> b ) $ num__25 m <o> c ) $ num__200 m <o> d ) $ num__300 m <o> e ) $ num__153 m |
num__0.25 * num__0.2 * ( total ) = num__10 num__0.05 * ( total ) = num__10 ( total ) = num__10 * num__20 = num__200 answer : c <eor> c <eos> |
c |
subtract__0.25__0.2__ reverse__0.05__ divide__10.0__0.05__ divide__10.0__0.05__ |
multiply__0.25__0.2__ reverse__0.05__ multiply__10.0__20.0__ multiply__10.0__20.0__ |
| point ( m n ) is on the circle represented by m ^ num__2 + n ^ num__2 = num__10 and m n are integers . how many such points are possible ? <o> a ) num__0 <o> b ) num__2 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
m ^ num__2 + n ^ num__2 = num__10 and m n are integers means that num__10 is the sum of two perfect squares . num__10 is the sum of only one pair of perfect squares num__1 and num__9 . so there can be num__8 such points num__4 in each quadrant : ( num__1 num__3 ) ; ( num__1 - num__3 ) ; ( - num__1 num__3 ) ; ( - num__1 - num__3 ) ; ( num__3 num__1 ) ; ( num__3 - num__1 ) ; ( - num__3 num__1 ) ; ( - num__3 - num__1 ) . answer : e . <eor> e <eos> |
e |
subtract__10.0__1.0__ subtract__10.0__2.0__ divide__8.0__2.0__ add__2.0__1.0__ multiply__2.0__4.0__ |
subtract__10.0__1.0__ subtract__10.0__2.0__ divide__8.0__2.0__ add__2.0__1.0__ subtract__10.0__2.0__ |
| in a mixture num__60 litres the ratio of milk and water num__2 : num__1 . if this ratio is to be num__1 : num__2 then the quanity of water to be further added is <o> a ) num__50 liters <o> b ) num__55 liters <o> c ) num__60 liters <o> d ) num__70 liters <o> e ) num__75 liters |
quantity of milk = num__60 x num__2 litres = num__40 litres . num__3 quantity of water in it = ( num__60 - num__40 ) litres = num__20 litres . new ratio = num__1 : num__2 let quantity of water to be added further be x litres . then milk : water = num__40 . num__20 + x now num__40 = num__1 num__20 + x num__2 num__20 + x = num__80 x = num__60 . quantity of water to be added = num__60 liters . c <eor> c <eos> |
c |
add__2.0__1.0__ divide__60.0__3.0__ add__60.0__20.0__ multiply__60.0__1.0__ |
add__2.0__1.0__ subtract__60.0__40.0__ add__60.0__20.0__ add__40.0__20.0__ |
| in a school num__10.0 of the boys are same in number as num__0.25 th of the girls . what is the ratio of boys to the girls in the school ? <o> a ) num__5 : num__2 <o> b ) num__2 : num__3 <o> c ) num__1 : num__4 <o> d ) num__3 : num__7 <o> e ) num__2 : num__5 |
num__10.0 of b = num__0.25 g num__10 b / num__100 = g / num__4 b = num__5 g / num__2 b / g = num__2.5 b : g = num__5 : num__2 answer is a <eor> a <eos> |
a |
reverse__0.25__ divide__10.0__5.0__ multiply__10.0__0.25__ divide__10.0__2.0__ |
reverse__0.25__ divide__10.0__5.0__ divide__10.0__4.0__ divide__10.0__2.0__ |
| the length of the rectangular field is double its width . inside the field there is square shaped pond num__8 m long . if the area of the pond is num__0.125 of the area of the field . what is the length of the field ? <o> a ) num__31 m <o> b ) num__32 m <o> c ) num__16 m <o> d ) num__20 m <o> e ) num__22 m |
a / num__8 = num__8 * num__8 = > a = num__8 * num__8 * num__8 x * num__2 x = num__8 * num__8 * num__8 x = num__16 = > num__2 x = num__32 answer : b <eor> b <eos> |
b |
multiply__8.0__2.0__ square_perimeter__8.0__ square_perimeter__8.0__ |
multiply__8.0__2.0__ multiply__16.0__2.0__ multiply__16.0__2.0__ |
| at a meeting of the num__3 joint chiefs of staff the chief of naval operations does not want to sit next to the chief of the national guard bureau . how many ways can the num__3 chiefs of staff be seated around a circular table ? <o> a ) num__120 <o> b ) num__480 <o> c ) num__960 <o> d ) num__2520 <o> e ) num__5040 |
bunuel i ' m also a little confused with the number of arrangements of n distinct objects in a circle . why is it given by ( n - num__1 ) ! . in theveritasanswer they say : answer c ( num__960 ) should be the number of ways to arrange all num__7 without the seating restriction given . is this incorrect ? c <eor> c <eos> |
c |
multiply__960.0__1.0__ |
multiply__960.0__1.0__ |
| two trains are running in opposite directions in the same speed . the length of each train is num__120 meter . if they cross each other in num__12 seconds the speed of each train ( in km / hr ) is <o> a ) num__42 <o> b ) num__36 <o> c ) num__28 <o> d ) num__20 <o> e ) num__15 |
explanation : distance covered = num__120 + num__120 = num__240 m time = num__12 s let the speed of each train = v . then relative speed = v + v = num__2 v num__2 v = distance / time = num__20.0 = num__20 m / s speed of each train = v = num__10.0 = num__10 m / s = num__10 × num__3.6 km / hr = num__36 km / hr . answer : b <eor> b <eos> |
b |
divide__240.0__120.0__ divide__240.0__12.0__ divide__120.0__12.0__ multiply__10.0__3.6__ round__36.0__ |
divide__240.0__120.0__ divide__240.0__12.0__ divide__120.0__12.0__ multiply__10.0__3.6__ round__36.0__ |
| in a hostel of num__12 boys and some girls an average consumption of rice per month is num__12 kg while the average consumption for boys is num__16 kg per head and for girls num__8 kg per head . the number of girls in the hostel ? <o> a ) num__12 <o> b ) num__13 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
explanation : number of girls — g ( num__12 + g ) × num__12 = ( num__12 × num__16 ) + ( num__8 × g ) num__144 + num__12 g = num__192 + num__8 g num__12 g − num__8 g = num__192 − num__144 num__4 g = num__48 g = num__12 . answer a <eor> a <eos> |
a |
multiply__12.0__16.0__ subtract__12.0__8.0__ multiply__12.0__4.0__ subtract__16.0__4.0__ |
multiply__12.0__16.0__ subtract__12.0__8.0__ multiply__12.0__4.0__ add__8.0__4.0__ |
| the speed of a boat in still water is num__12 km / hr and the rate of current is num__4 km / hr . the distance travelled downstream in num__18 minutes is <o> a ) num__1.6 km <o> b ) num__2 km <o> c ) num__3.6 km <o> d ) num__4.8 km <o> e ) none of these |
explanation : speed downstreams = ( num__12 + num__4 ) kmph = num__16 kmph . distance travelled = ( num__16 x num__0.3 ) km = num__4.8 km option d <eor> d <eos> |
d |
add__12.0__4.0__ multiply__16.0__0.3__ round__4.8__ |
add__12.0__4.0__ multiply__16.0__0.3__ round__4.8__ |
| if the cost price of num__11 pens is equal to the selling price of num__10 pens the gain percent is : <o> a ) num__10.0 <o> b ) num__26.0 <o> c ) num__50.0 <o> d ) num__80.0 <o> e ) num__34 % |
solution : let cost price of the price be rs . num__1 num__10 pens c . p . = num__10 num__11 pens s . p . = num__11 gain = num__0.1 * num__100 = num__10.0 answer : a <eor> a <eos> |
a |
percent__10.0__1.0__ percent__10.0__100.0__ |
percent__10.0__1.0__ percent__10.0__100.0__ |
| by walking at num__0.75 th of his usual speed a man reaches office num__20 minutes later than usual . what is his usual time ? <o> a ) num__40 min <o> b ) num__50 min <o> c ) num__65 min <o> d ) num__60 min <o> e ) num__70 min |
let t be the usual time . time spent = num__4 t / num__3 therefore num__4 t / num__3 = t + num__20 num__4 t = num__3 t + num__60 . . . therefor t = num__60 min answer : d <eor> d <eos> |
d |
multiply__0.75__4.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
multiply__0.75__4.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
| the average of num__6 no . ' s is num__3.95 . the average of num__2 of them is num__3.4 while the average of theother num__2 is num__3.85 . what is the average of the remaining num__2 no ' s ? <o> a ) num__3.5 <o> b ) num__3.9 <o> c ) num__4.2 <o> d ) num__4.3 <o> e ) num__4.6 |
sum of the remaining two numbers = ( num__3.95 * num__6 ) - [ ( num__3.4 * num__2 ) + ( num__3.85 * num__2 ) ] = num__23.70 - ( num__6.8 + num__7.7 ) = num__23.70 - num__14.5 = num__9.20 . required average = ( num__9.2 / num__2 ) = num__4.6 . e <eor> e <eos> |
e |
multiply__6.0__3.95__ multiply__2.0__3.4__ multiply__2.0__3.85__ add__7.7__6.8__ subtract__23.7__14.5__ divide__9.2__2.0__ divide__9.2__2.0__ |
multiply__6.0__3.95__ multiply__2.0__3.4__ multiply__2.0__3.85__ add__7.7__6.8__ subtract__23.7__14.5__ divide__9.2__2.0__ divide__9.2__2.0__ |
| a flagstaff num__17.5 m high casts a shadow of length num__40.25 m . the height of the building which casts a shadow of length num__28.75 m under similar conditions will be : <o> a ) num__10 m <o> b ) num__12.5 m <o> c ) num__17.5 m <o> d ) num__21.25 m <o> e ) none of these |
solution num__14.8 let the height of the building be x meters . less lengthy shadow less is the height ( direct proportion ) ∴ num__40.25 : num__28.75 : : num__17.5 : x ⇔ num__40.25 × x = num__28.75 × num__17.5 ⇔ x = ( num__28.75 x num__17.50 ) / num__40.25 ⇔ x = num__12.5 answer b <eor> b <eos> |
b |
round__12.5__ |
round__12.5__ |
| the contents of a bag of candy consist of num__11 green pieces of candy and num__20 blue pieces of candy . how many blue pieces of candy must be eaten so that num__55.0 of the candies in the bag will be green ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__11 <o> d ) num__13 <o> e ) num__15 |
this question is rooted in basic arithmetic . since the question asks how many marbles need to be removed and the answers are numbers we can test the answers until we find the one thatfits . to start we have a bag of candy . . . . num__11 green num__20 blue num__31 total we ' re asked to eat enough blue candies until the green candies make up num__55.0 of the bag . . . . right now green marbles make up num__0.354838709677 which equals about num__35.0 of the bag we need to eat blue candy to achieve green candies making up for num__55.0 of the bag . we can test our answers let ' s test b . . . . if num__10 blue candies were eaten then we ' d have . . . num__11 green candies num__10 blue candies num__21 total num__0.52380952381 is about num__52.0 . let ' s test e . . . if num__11 blue candies were eaten then we ' d have . . . num__11 green candies num__9 blue candies num__20 total num__0.55 is num__55.0 so this must be the answer ! final answer : c <eor> c <eos> |
c |
percent__20.0__55.0__ |
percent__20.0__55.0__ |
| john professes to sell his goods at the cost price but he made use of num__800 grams instead of a kg what is the gain percent ? <o> a ) num__12 num__0.333333333333 % <o> b ) num__12 num__0.25 % <o> c ) num__12 num__0.666666666667 % <o> d ) num__12 num__2.0 % <o> e ) num__12 num__0.5 % |
num__800 - - - num__100 num__100 - - - ? = > num__12 num__0.5 % answer : e <eor> e <eos> |
e |
percent__12.0__100.0__ |
percent__12.0__100.0__ |
| if a card is drawn from a well shuffled pack of cards the probability of drawing a spade or a king is ? <o> a ) num__0.25 <o> b ) num__0.210526315789 <o> c ) num__0.333333333333 <o> d ) num__0.307692307692 <o> e ) num__0.363636363636 |
p ( s ᴜ k ) = p ( s ) + p ( k ) - p ( s ∩ k ) where s denotes spade and k denotes king . p ( s ᴜ k ) = num__0.25 + num__0.0769230769231 - num__0.0192307692308 = num__0.307692307692 answer : d <eor> d <eos> |
d |
union_prob__0.25__0.0769__0.0192__ union_prob__0.25__0.0769__0.0192__ |
union_prob__0.25__0.0769__0.0192__ union_prob__0.25__0.0769__0.0192__ |
| a cistern is filled by a tap in num__3 num__0.5 hours . due to leak in the bottom of the cistern it takes half an hour longer to fill the cistern . if the cistern is full how long will it take the leak to empty it ? <o> a ) num__7 hours <o> b ) num__8 hours <o> c ) num__14 hours <o> d ) num__28 hours <o> e ) num__24 hours |
num__0.285714285714 - num__1 / x = num__0.25 x = num__28 answer : d <eor> d <eos> |
d |
round__28.0__ |
divide__28.0__1.0__ |
| eighteen years ago a father was three times as old as his son . now the father is only twice as old his son . then the sum of the present ages of the son and the father is : <o> a ) num__54 <o> b ) num__72 <o> c ) num__105 <o> d ) num__108 <o> e ) num__51 |
answer : option d let the present ages of the father and son be num__2 x and x years respectively . then ( num__2 x - num__18 ) = num__3 ( x - num__18 ) = > x = num__36 required sum = ( num__2 x + x ) = num__108 years . <eor> d <eos> |
d |
multiply__2.0__18.0__ multiply__3.0__36.0__ multiply__3.0__36.0__ |
multiply__2.0__18.0__ multiply__3.0__36.0__ multiply__3.0__36.0__ |
| the speed of a boat in still water in num__42 km / hr and the rate of current is num__4 km / hr . the distance travelled downstream in num__44 minutes is : <o> a ) num__86.6 km <o> b ) num__46.6 km <o> c ) num__33.7 km <o> d ) num__35.6 km <o> e ) num__26.6 km |
speed downstream = ( num__42 + num__4 ) = num__46 kmph time = num__44 minutes = num__0.733333333333 hour = num__0.733333333333 hour distance travelled = time × speed = num__0.733333333333 × num__46 = num__33.7 km answer : c <eor> c <eos> |
c |
add__42.0__4.0__ round__33.7__ |
add__42.0__4.0__ round__33.7__ |
| it takes num__60 identical printing machines num__4 hours to print num__200000 cards . how long would it take num__50 of these machines to print the same total ? <o> a ) num__10 hr <o> b ) num__5 hr <o> c ) num__6 hr <o> d ) num__4 hr num__30 min <o> e ) num__5 hr num__12 min |
num__50.0 of num__60 = num__40.0 or num__0.30 num__4 hr x num__60 min = num__240 min num__240 min x num__0.30 = num__72 min or num__1 hr and num__12 min num__1 hr and num__12 min + num__4 hr = num__5 hr and num__12 min answer is e <eor> e <eos> |
e |
vowel_space__ vowel_space__ |
vowel_space__ vowel_space__ |
| a train running at the speed of num__70 km / hr crosses a pole in num__9 sec . what is the length of the train ? <o> a ) num__298 m <o> b ) num__175 m <o> c ) num__208 m <o> d ) num__988 m <o> e ) num__299 m |
speed = num__70 * num__0.277777777778 = num__19.4444444444 m / sec length of the train = speed * time = num__19.4444444444 * num__9 = num__175 m answer : b <eor> b <eos> |
b |
round__175.0__ |
round__175.0__ |
| a garrison of num__1000 men has provisions for num__60 days . at the end of num__15 days a reinforcement arrives and it is now found that the provisions will last only for num__20 days more . what is the reinforcement ? <o> a ) num__1000 <o> b ) num__2250 <o> c ) num__60 <o> d ) num__45 <o> e ) num__1250 |
num__1000 - - - - num__60 num__1000 - - - - num__45 x - - - - - num__20 x * num__20 = num__1000 * num__45 x = num__2250 num__1000 - - - - - - - num__1250 answer : e <eor> e <eos> |
e |
subtract__60.0__15.0__ subtract__2250.0__1000.0__ round__1250.0__ |
subtract__60.0__15.0__ subtract__2250.0__1000.0__ subtract__2250.0__1000.0__ |
| two trains of equal lengths take num__10 sec and num__8 sec respectively to cross a telegraph post . if the length of each train be num__120 m in what time will they cross other travelling in opposite direction ? <o> a ) num__22 <o> b ) num__12 <o> c ) num__9 <o> d ) num__99 <o> e ) num__21 |
speed of the first train = num__12.0 = num__12 m / sec . speed of the second train = num__15.0 = num__15 m / sec . relative speed = num__12 + num__15 = num__27 m / sec . required time = ( num__120 + num__120 ) / num__27 = num__9 sec . answer : c <eor> c <eos> |
c |
divide__120.0__10.0__ divide__120.0__8.0__ add__12.0__15.0__ round__9.0__ |
divide__120.0__10.0__ divide__120.0__8.0__ add__12.0__15.0__ round__9.0__ |
| the average weight of num__7 person ' s increases by num__6.2 kg when a new person comes in place of one of them weighing num__76 kg . what might be the weight of the new person ? <o> a ) num__160 kg <o> b ) num__175 kg <o> c ) num__180.7 kg <o> d ) num__119.4 kg <o> e ) num__190.8 kg |
total weight increased = ( num__7 x num__6.2 ) kg = num__43.4 kg . weight of new person = ( num__76 + num__43.4 ) kg = num__119.4 kg option d <eor> d <eos> |
d |
multiply__7.0__6.2__ add__76.0__43.4__ add__76.0__43.4__ |
multiply__7.0__6.2__ add__76.0__43.4__ add__76.0__43.4__ |
| a boatman can row num__96 km downstream in num__8 hr . if the speed of the current is num__4 km / hr then find in what time he will be able to cover num__8 km upstream ? <o> a ) num__1.5 hrs <o> b ) num__1 hrs <o> c ) num__2.5 hrs <o> d ) num__2 hrs <o> e ) num__3 hrs |
explanation : speed in downstream = num__12.0 = num__12 kmph speed of current = num__4 km / hr speed of the boatman in still water = num__12 â € “ num__4 = num__8 kmph speed in upstream = num__8 â € “ num__4 = num__4 kmph time taken to cover num__8 km upstream = num__2.0 = num__2 hours . answer is d <eor> d <eos> |
d |
divide__96.0__8.0__ divide__8.0__4.0__ round__2.0__ |
divide__96.0__8.0__ divide__8.0__4.0__ divide__8.0__4.0__ |
| how many integer values t are there for x such that num__1 < num__3 x + num__5 < num__17 ? <o> a ) two <o> b ) three <o> c ) four <o> d ) five <o> e ) six |
num__1 < num__3 x + num__5 < num__17 = > - num__4 < num__3 x < num__12 = > - num__1.33333333333 < x < num__4 x can take integer values t - num__10 num__1 num__2 num__3 answer d <eor> d <eos> |
d |
add__1.0__3.0__ multiply__3.0__4.0__ divide__4.0__3.0__ subtract__3.0__1.0__ multiply__1.0__5.0__ |
add__1.0__3.0__ subtract__17.0__5.0__ divide__4.0__3.0__ subtract__3.0__1.0__ add__1.0__4.0__ |
| in a kilometer race a beats b by num__80 meters or num__10 seconds . what time does a take to complete the race ? <o> a ) num__199 sec <o> b ) num__190 sec <o> c ) num__726 sec <o> d ) num__127 sec <o> e ) num__115 sec |
time taken by b run num__1000 meters = ( num__1000 * num__10 ) / num__80 = num__125 sec . time taken by a = num__125 - num__10 = num__115 sec . answer : e <eor> e <eos> |
e |
subtract__125.0__10.0__ round__115.0__ |
subtract__125.0__10.0__ subtract__125.0__10.0__ |
| what is the maximum number of pieces of birthday cake of size num__5 ” by num__5 ” that can be cut from a cake num__15 ” by num__15 ” ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__16 <o> d ) num__20 <o> e ) num__9 |
the prompt is essentially asking for the maximum number of num__5 x num__5 squares that can be cut from a larger num__15 by num__15 square . since each ' row ' and each ' column ' of the larger square can be sub - divided into num__3 ' pieces ' each we have ( num__3 ) ( num__3 ) = num__9 total smaller squares ( at maximum ) . e <eor> e <eos> |
e |
divide__15.0__5.0__ round__9.0__ |
divide__15.0__5.0__ round__9.0__ |
| the average of marks obtained by num__120 boys was num__36 . if the average of marks of passed boys was num__39 and that of failed boys was num__15 the number of boys who passed the examination is ? <o> a ) num__105 <o> b ) num__110 <o> c ) num__120 <o> d ) num__130 <o> e ) num__140 |
let the number of boys who passed = x . then num__39 x x + num__15 x ( num__120 - x ) = num__120 x num__36 num__24 x = num__4320 - num__1800 = > x = num__105.0 x = num__105 . hence the number of boys passed = num__105 . answer : a <eor> a <eos> |
a |
subtract__39.0__15.0__ multiply__120.0__36.0__ multiply__120.0__15.0__ subtract__120.0__15.0__ subtract__120.0__15.0__ |
subtract__39.0__15.0__ multiply__120.0__36.0__ multiply__120.0__15.0__ subtract__120.0__15.0__ subtract__120.0__15.0__ |
| a bike covers a certain distance at the speed of num__66 km / h in num__8 hrs . if the bike was to cover the same distance in approximately num__6 hrs at what approximate speed should the bike travel ? <o> a ) num__85 km / h <o> b ) num__85.34 km / h <o> c ) num__87.67 km / h <o> d ) num__90.25 km / h <o> e ) num__88 km / h |
ans . ( e ) sol . total distance = num__66 × num__8 = num__528 km now speed = num__88.0 = num__88 km / h <eor> e <eos> |
e |
multiply__66.0__8.0__ divide__528.0__6.0__ round__88.0__ |
multiply__66.0__8.0__ divide__528.0__6.0__ divide__528.0__6.0__ |
| the distance light travels in one year is approximately num__5 num__860000 num__000000 miles . the distance light travels in num__100 years is <o> a ) a ) num__587 × num__108 miles <o> b ) b ) num__587 × num__1010 miles <o> c ) c ) num__587 × num__10 - num__10 miles <o> d ) d ) num__586 × num__1012 miles <o> e ) e ) num__587 × num__10 - num__12 miles |
the distance of the light travels in num__100 years is : num__5 num__860000 num__000000 × num__100 miles . = num__586 num__000000 num__000000 miles . = num__586 × num__1012 miles . the answer is ( d ) <eor> d <eos> |
d |
round__586.0__ |
round__586.0__ |
| the difference between the simple interest received from two different sources on rs . num__1900 for num__3 years is rs . num__13.50 . the difference between their rates of interest is <o> a ) num__0.13 <o> b ) num__0.23 <o> c ) num__0.33 <o> d ) num__0.43 <o> e ) num__0.53 % |
( num__1900 xr num__1 x num__3 ) / num__100 - ( num__100 xr num__2 x num__3 ) / num__100 = num__13.50 num__5700 ( r num__1 - r num__2 ) = num__1350 r num__1 - r num__2 = num__0.236842105263 = num__0.23 answer : b <eor> b <eos> |
b |
percent__100.0__0.23__ |
percent__100.0__0.23__ |
| how many numbers from num__29 to num__79 are exactly divisible by num__11 ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__8 <o> d ) num__5 <o> e ) num__7 |
numbers = ( num__7.18181818182 ) - ( num__2.63636363636 ) = num__7 - num__2 = num__5 numbers option d is answer <eor> d <eos> |
d |
divide__79.0__11.0__ divide__29.0__11.0__ round_down__7.1818__ round_down__2.6364__ subtract__7.0__2.0__ subtract__7.0__2.0__ |
divide__79.0__11.0__ divide__29.0__11.0__ round_down__7.1818__ round_down__2.6364__ subtract__7.0__2.0__ subtract__7.0__2.0__ |
| what is the smallest positive integer x such that num__80 - x is the cube of a positive integer <o> a ) num__16 <o> b ) num__12 <o> c ) num__47 <o> d ) num__9 <o> e ) num__12 |
given num__80 - x is a perfect cube so we will take num__64 = num__4 * num__4 * num__4 num__80 - x = num__64 x = num__80 - num__64 = num__16 correct option is a <eor> a <eos> |
a |
subtract__80.0__64.0__ subtract__80.0__64.0__ |
subtract__80.0__64.0__ subtract__80.0__64.0__ |
| a river num__2 m deep and num__45 m wide is flowing at the rate of num__3 kmph the amount of water that runs into the sea per minute is ? <o> a ) num__4500 <o> b ) num__3889 <o> c ) num__2777 <o> d ) num__1299 <o> e ) num__2791 |
( num__3000 * num__2 * num__5 ) / num__60 = num__4500 m num__3 answer : a <eor> a <eos> |
a |
add__2.0__3.0__ hour_to_min_conversion__ round__4500.0__ |
add__2.0__3.0__ hour_to_min_conversion__ round__4500.0__ |
| a trader sells num__85 meters of cloth for rs . num__8925 at the profit of rs . num__15 per metre of cloth . what is the cost price of one metre of cloth ? <o> a ) num__21 <o> b ) num__28 <o> c ) num__90 <o> d ) num__26 <o> e ) num__11 |
explanation : sp of num__1 m of cloth = num__105.0 = rs . num__105 cp of num__1 m of cloth = sp of num__1 m of cloth - profit on num__1 m of cloth = rs . num__105 - rs . num__15 = rs . num__90 . answer : c <eor> c <eos> |
c |
divide__8925.0__85.0__ subtract__105.0__15.0__ round__90.0__ |
divide__8925.0__85.0__ subtract__105.0__15.0__ subtract__105.0__15.0__ |
| if ( num__0.2 ) ^ m * ( num__0.25 ) ^ num__2 = num__1 / ( ( num__10 ) ^ num__4 ) then m = <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__3 <o> e ) num__2 |
num__0.2 ^ m * num__0.25 ^ num__2 = num__0.5 ^ num__4 * num__5 ^ num__4 num__2 ^ num__4 * num__5 ^ num__2.0 ^ num__4 = num__5 ^ m or num__5 ^ num__4 = num__5 ^ m therefore m = num__4 a <eor> a <eos> |
a |
reverse__2.0__ reverse__0.2__ reverse__0.25__ |
multiply__0.25__2.0__ multiply__10.0__0.5__ reverse__0.25__ |
| how many num__1 ' s are there preceded by num__2 but not followed by num__7 ? num__5 num__9 num__3 num__2 num__1 num__7 num__4 num__2 num__6 num__9 num__7 num__4 num__2 num__1 num__3 num__2 num__8 num__7 num__4 num__1 num__3 num__8 num__3 num__2 num__5 num__6 num__7 num__4 num__3 num__9 num__5 num__8 num__2 num__0 num__1 num__8 num__7 num__4 num__6 num__3 <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__1 <o> e ) num__9 |
num__2 num__1 num__3 only at these places num__1 is preceded by num__2 but not followed by num__7 answer : d <eor> d <eos> |
d |
reverse__1.0__ |
reverse__1.0__ |
| what least number must be added to num__1056 so that the sum is completely divisible by num__23 ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__18 <o> d ) num__21 <o> e ) none of these |
num__23 ) num__1056 ( num__45 num__92 - - - num__136 num__115 - - - num__21 - - - required number = ( num__23 - num__21 ) = num__2 . answer : a <eor> a <eos> |
a |
add__23.0__92.0__ subtract__136.0__115.0__ subtract__23.0__21.0__ subtract__23.0__21.0__ |
add__23.0__92.0__ subtract__136.0__115.0__ subtract__23.0__21.0__ subtract__23.0__21.0__ |
| in a group of num__25 factory workers num__17 have brown eyes . six of the women do not have brown eyes . how many of the num__11 men have brown eyes . <o> a ) num__10 <o> b ) num__9 <o> c ) num__8 <o> d ) num__7 <o> e ) num__4 |
total number of worker ( m + w ) : num__25 no . of men ( m ) : num__11 ( inferred fromhow many of the num__11 men ) no . of women ( w ) : num__14 total no . of workers who have brown eyes ( b ) : num__17 no . of women who do not have brown eyes : num__6 therefore no . of women who have brown eyes : w - num__6 = num__14 - num__6 = num__8 remaining num__9 are men . ( b - num__8 = num__17 - num__8 = num__9 ) so num__9 out of num__11 men have brown eyes . b <eor> b <eos> |
b |
die_space__ choose__9.0__8.0__ |
die_space__ choose__9.0__8.0__ |
| the ratio of two numbers is num__3 : num__4 and their sum is num__28 . the greater of the two numbers is ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__16 <o> d ) num__18 <o> e ) num__22 |
num__3 : num__4 total parts = num__7 = num__7 parts - - > num__28 ( num__7 × num__4 = num__28 ) = num__1 part - - - - > num__4 ( num__1 × num__4 = num__4 ) = the greater of the two number is = num__4 = num__4 parts - - - - > num__16 ( num__4 × num__4 = num__16 ) c <eor> c <eos> |
c |
add__3.0__4.0__ subtract__4.0__3.0__ multiply__1.0__16.0__ |
add__3.0__4.0__ subtract__4.0__3.0__ multiply__1.0__16.0__ |
| if q is a positive integer which of the following must be odd ? <o> a ) q + num__1 <o> b ) q ^ num__2 + q <o> c ) q ^ num__2 + q + num__1 <o> d ) x ^ num__2 − num__1 <o> e ) num__3 x ^ num__2 − num__3 |
a . q + num__1 = can be odd or even . since o + o = e or e + o = o b . q ^ num__2 + q = q ( q + num__1 ) . since from the above derivation we already know the term q + num__1 can be odd or even directly substitute here . q ( odd ) = even ( when q is even ) or q ( even ) = even [ when q is odd ] c . here ' s the answer . since we know the term q ^ num__2 + q can always take a even number even + num__1 = odd hence c . <eor> c <eos> |
c |
multiply__1.0__2.0__ |
power__2.0__1.0__ |
| set s consists of all prime numbers less than num__10 . if two numbers are chosen from et s at random what is the probability that the product of these numbers will be greater than the product of the numbers which were not chosen ? <o> a ) num__0.333333333333 <o> b ) num__0.666666666667 <o> c ) num__0.5 <o> d ) num__0.7 <o> e ) num__0.8 |
prime integers below num__10 : num__2 num__3 num__5 num__7 total outcomes : number of ways you can choose num__2 from num__4 = num__4 c num__2 = num__6 pairs num__4 c num__2 = num__6 pairs no . of ways for event : pairs that will have products greater than product of pairs not selected i . e . ( num__35 ) ( num__37 ) ( num__57 ) = num__3 pairs = num__3 pairs probability = number of ways / total outcomes = num__0.5 = num__0.5 = num__0.5 = num__0.5 hence choice ( c ) is correct . <eor> c <eos> |
c |
divide__10.0__2.0__ subtract__10.0__3.0__ subtract__7.0__3.0__ subtract__10.0__4.0__ multiply__5.0__7.0__ add__2.0__35.0__ reverse__2.0__ reverse__2.0__ |
divide__10.0__2.0__ subtract__10.0__3.0__ subtract__7.0__3.0__ subtract__10.0__4.0__ multiply__5.0__7.0__ add__2.0__35.0__ reverse__2.0__ reverse__2.0__ |
| a paint store mixes num__0.75 pint of red paint and num__0.666666666667 pint of white paint to make a new paint color called perfect pink . how many pints of red paint would be needed to make num__36 pints of perfect pink paint ? <o> a ) num__9 <o> b ) num__16 <o> c ) num__19 <o> d ) num__25 num__0.333333333333 <o> e ) num__28 num__0.5 |
num__0.75 pint is required to make num__0.75 + num__0.666666666667 = num__1.41666666667 pint of perfect pink so num__1.41666666667 pint requires num__0.75 pint of red . . num__1 pint will require num__0.75 * num__0.705882352941 = num__0.529411764706 . . num__36 pints will require num__0.529411764706 * num__36 = num__19 pints . . c <eor> c <eos> |
c |
add__0.75__0.6667__ round_down__1.4167__ reverse__1.4167__ multiply__0.75__0.7059__ multiply__1.0__19.0__ |
add__0.75__0.6667__ round_down__1.4167__ reverse__1.4167__ multiply__0.75__0.7059__ multiply__1.0__19.0__ |
| there are num__6 more women than there are men on a local co - ed softball team . if there are a total of num__16 players on the team what is the ratio of men to women ? <o> a ) num__0.625 <o> b ) num__0.375 <o> c ) num__0.25 <o> d ) num__0.6 <o> e ) num__0.454545454545 |
w = m + num__6 w + m = num__16 m + num__6 + m = num__16 num__2 m = num__10 m = num__5 w = num__11 ratio : num__5 : num__11 ans : e <eor> e <eos> |
e |
subtract__16.0__6.0__ divide__10.0__2.0__ add__6.0__5.0__ divide__5.0__11.0__ |
subtract__16.0__6.0__ divide__10.0__2.0__ add__6.0__5.0__ divide__5.0__11.0__ |
| a quantity of tea is sold at rs . num__5.75 per kilogram . the total gain by selling the tea at this rate is rs . num__60 . find the quantity of tea being sold if a profit of num__15.0 is made on the deal . <o> a ) num__80 kg . <o> b ) num__90 kg . <o> c ) num__70 kg . <o> d ) num__60 kg . <o> e ) num__50 kg . |
say total cost price of tea is x . then total profit at a rate of num__15.0 is = ( num__15 x / num__100 ) according to question num__15 x / num__100 = num__60 so x = num__400 c . p of the tea is num__400 . so total selling price will be = ( num__400 + num__60 ) = num__460 so the quantity of the tea will be = ( num__460 / num__5.75 ) = num__80 kg . answer : a <eor> a <eos> |
a |
percent__80.0__100.0__ |
percent__80.0__100.0__ |
| a train num__280 m long running with a speed of num__63 km / hr will pass a tree in ? <o> a ) num__17 sec <o> b ) num__16 sec <o> c ) num__18 sec <o> d ) num__14 sec <o> e ) num__12 sec |
speed = num__63 * num__0.277777777778 = num__17.5 m / sec time taken = num__280 * num__0.0571428571429 = num__16 sec answer : b <eor> b <eos> |
b |
divide__280.0__17.5__ round__16.0__ |
divide__280.0__17.5__ divide__280.0__17.5__ |
| what approximate value should come in the place of question mark ( ? ) in the following equation ? num__98.98 ÷ num__11.03 + num__7.014 × num__15.99 = ( ? ) num__2 <o> a ) num__131 <o> b ) num__144 <o> c ) num__12 <o> d ) num__121 <o> e ) num__11 |
num__98.98 ÷ num__11.03 + num__7.014 × num__15.99 = ( ? ) num__2 suppose ? = x then num__99 ÷ num__11 + num__7 × num__16 + ≈ num__121 ( taking approximate value ) ∴ x = num__11 answer e <eor> e <eos> |
e |
round_down__11.03__ round_down__7.014__ round_down__11.03__ |
round_down__11.03__ round_down__7.014__ round_down__11.03__ |
| sum of the numbers from num__1 to num__45 is <o> a ) num__1789 <o> b ) num__1035 <o> c ) num__1025 <o> d ) num__1009 <o> e ) none of these |
explanation : sum of first n natural numbers = num__1 + num__2 + num__3 + . . . . . n = n ( n + num__1 ) / num__2 substitute n = num__45 . so s num__20 = num__45 Ã — num__23.0 = num__1035 correct option : b <eor> b <eos> |
b |
add__1.0__2.0__ add__3.0__20.0__ multiply__45.0__23.0__ multiply__1.0__1035.0__ |
add__1.0__2.0__ add__3.0__20.0__ multiply__45.0__23.0__ divide__1035.0__1.0__ |
| maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is num__60 kilometers maxwell ' s walking speed is num__4 km / h and brad ' s running speed is num__6 km / h what is the distance traveled by brad ? <o> a ) num__16 <o> b ) num__36 <o> c ) num__20 <o> d ) num__24 <o> e ) num__30 |
time taken = total distance / relative speed total distance = num__60 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = num__6 + num__4 = num__10 kms / hr time taken = num__6.0 = num__6 hrs distance traveled by brad = brad ' s speed * time taken = num__6 * num__6 = num__36 kms . . . answer - b <eor> b <eos> |
b |
divide__60.0__6.0__ round__36.0__ |
divide__60.0__6.0__ round__36.0__ |
| an woman swims downstream num__81 km and upstream num__36 km taking num__9 hours each time ; what is the speed of the current ? <o> a ) num__0.5 <o> b ) num__1.5 <o> c ) num__2.5 <o> d ) num__3.5 <o> e ) num__4.5 |
num__81 - - - num__9 ds = num__9 ? - - - - num__1 num__36 - - - - num__9 us = num__4 ? - - - - num__1 s = ? s = ( num__9 - num__4 ) / num__2 = num__2.5 answer : c <eor> c <eos> |
c |
divide__36.0__9.0__ round__2.5__ |
divide__36.0__9.0__ divide__2.5__1.0__ |
| a train speeds past a pole in num__15 sec and a platform num__140 m long in num__25 sec its length is ? <o> a ) num__50 m <o> b ) num__150 m <o> c ) num__210 m <o> d ) num__300 m <o> e ) none of these |
let the length of the train be x m and its speed be y m / sec . then x / y = num__15 = > y = x / num__15 ( x + num__140 ) / num__25 = x / num__15 = > x = num__210 m . answer : c <eor> c <eos> |
c |
round__210.0__ |
round__210.0__ |
| the average of six number is num__3.99 . the average of two of them is num__3.4 while the average of the other two is num__3.85 . what is the average of the remaining two number ? <o> a ) num__4.5 <o> b ) num__4.6 <o> c ) num__4.72 <o> d ) num__4.8 <o> e ) none of these |
solution : sum of the remaining two numbers = ( num__3.99 × num__6 ) - [ ( num__3.4 × num__2 ) + ( num__3.85 × num__2 ) ] = num__23.94 - ( num__6.8 + num__7.7 ) = num__23.94 - num__14.5 = num__9.44 . ∴ required average = num__9.44 / num__2 = num__4.72 answer c <eor> c <eos> |
c |
multiply__3.99__6.0__ multiply__3.4__2.0__ multiply__3.85__2.0__ add__7.7__6.8__ subtract__23.94__14.5__ divide__9.44__2.0__ divide__9.44__2.0__ |
multiply__3.99__6.0__ multiply__3.4__2.0__ multiply__3.85__2.0__ add__7.7__6.8__ subtract__23.94__14.5__ divide__9.44__2.0__ divide__9.44__2.0__ |
| num__50 men can complete a work in num__65 days . five days after started the work num__20 men left the group . in how many days can the remaining work be completed ? <o> a ) num__277 <o> b ) num__268 <o> c ) num__260 <o> d ) num__276 <o> e ) num__100 |
after num__5 days the following situation prevails . num__50 men can complete the work in num__60 days . num__30 men can complete the work in ? days . m num__1 d num__1 = m num__2 d num__2 = > num__50 * num__60 = num__30 * d num__2 = > d num__2 = ( num__50 * num__60 ) / num__30 = num__100 days . answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ subtract__50.0__20.0__ divide__60.0__30.0__ multiply__50.0__2.0__ round__100.0__ |
hour_to_min_conversion__ subtract__50.0__20.0__ divide__60.0__30.0__ multiply__50.0__2.0__ multiply__50.0__2.0__ |
| in how many ways can an answer key for a quiz be written if the quiz contains num__3 true - false questions followed by num__3 multiple - choice questions with num__4 answer choices each if the correct answers to all true - false questions can not be the same ? <o> a ) num__164 <o> b ) num__224 <o> c ) num__280 <o> d ) num__384 <o> e ) num__476 |
there are num__2 ^ num__3 = num__8 possibilities for the true - false answers . however we need to remove two cases for ttt and fff . there are num__4 * num__4 * num__4 = num__64 possibilities for the multiple choice questions . the total number of possibilities is num__6 * num__64 = num__384 . the answer is d . <eor> d <eos> |
d |
multiply__4.0__2.0__ multiply__3.0__2.0__ multiply__64.0__6.0__ multiply__64.0__6.0__ |
multiply__4.0__2.0__ multiply__3.0__2.0__ multiply__64.0__6.0__ multiply__64.0__6.0__ |
| merry and mich play a card game . in the beginning of the game they have an equal number of cards . each player at her turn gives the other a third of her cards . mich plays first giving merry a third of her cards . merry plays next and mich follows . then the game ends . merry ended up with num__14 more cards than mich . how many cards did each player have originally ? <o> a ) num__60 <o> b ) num__58 <o> c ) num__56 <o> d ) num__55 <o> e ) num__54 |
game mich merry initially num__54 num__54 assume after game num__1 num__36 num__72 after game num__2 num__60 num__48 after game num__3 num__40 num__68 now merry has num__28 cards more than mich . this option gives us exactly what number of cards they had initially . so the answer is e <eor> e <eos> |
e |
divide__72.0__36.0__ add__1.0__2.0__ subtract__54.0__14.0__ add__14.0__54.0__ multiply__14.0__2.0__ add__14.0__40.0__ |
divide__72.0__36.0__ add__1.0__2.0__ subtract__54.0__14.0__ add__14.0__54.0__ multiply__14.0__2.0__ add__14.0__40.0__ |
| num__40 people attend a party . num__11 men are single and the rest are there with their wives . there are no children in the party . in all num__22 women are present . then the number of married men at the party is ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__11 <o> e ) num__9 |
total people = number of men + number of women num__40 = num__7 + number of married men + num__22 number of married men = num__40 - num__22 - num__7 = num__11 men answer : d <eor> d <eos> |
d |
subtract__22.0__11.0__ |
subtract__22.0__11.0__ |
| the length of a rectangle is num__2 cm more than the width of the rectangle . the perimeter of the rectangle is num__20 cm . find the length and the width of the rectangle . <o> a ) l = num__2 w = num__9 <o> b ) l = num__5 w = num__8 <o> c ) l = num__6 w = num__4 <o> d ) l = num__1 w = num__7 <o> e ) l = num__2 w = num__3 |
let length l = x width w = x − num__2 and perimeter = p ∴ p = num__2 l + num__2 w = num__2 x + num__2 ( x − num__2 ) num__20 = num__2 x + num__2 x − num__4 num__4 x = num__24 x = num__6 l = num__6 cm and w = l − num__2 = num__4 cm answer is c . <eor> c <eos> |
c |
add__20.0__4.0__ add__2.0__4.0__ round__6.0__ |
add__20.0__4.0__ add__2.0__4.0__ add__2.0__4.0__ |
| if f ( x ) = num__5 x ^ num__2 - num__2 x + num__6 and g ( y ) = num__3 y - num__7 then g ( f ( x ) ) <o> a ) num__82 x ^ num__2 - num__9 x + num__38 <o> b ) num__15 x ^ num__2 - num__6 x + num__11 <o> c ) num__58 x ^ num__2 - num__4 x + num__58 <o> d ) num__87 x ^ num__2 - num__5 x + num__96 <o> e ) num__98 x ^ num__2 - num__7 x + num__94 |
g ( f ( x ) ) = num__3 ( num__5 x ^ num__2 - num__2 x + num__6 ) - num__7 = num__15 x ^ num__2 - num__6 x + num__18 - num__7 = num__15 x ^ num__2 - num__6 x + num__11 the answer is b <eor> b <eos> |
b |
multiply__5.0__3.0__ multiply__6.0__3.0__ add__5.0__6.0__ multiply__5.0__3.0__ |
multiply__5.0__3.0__ add__3.0__15.0__ add__5.0__6.0__ subtract__18.0__3.0__ |
| the average age of a group of person going for picnic is num__16 years . twenty new persons with an average age of num__15 years join the group on the spot due to which their average becomes num__15.5 years . find the number of persons initially going for picnic . <o> a ) num__20 <o> b ) num__18 <o> c ) num__22 <o> d ) num__19 <o> e ) none of the above |
let the number of persons initially going for picnic = x ∴ sum of their ages = num__16 x also num__16 x + num__15 × num__20 / x + num__20 = num__15.5 ⇒ num__0.5 x = num__10 ⇒ x = num__20 years . answer a <eor> a <eos> |
a |
subtract__16.0__15.5__ multiply__0.5__20.0__ divide__10.0__0.5__ |
subtract__16.0__15.5__ multiply__0.5__20.0__ divide__10.0__0.5__ |
| i spend num__40 hours a week ( num__5 days ) at work and like to organize my time so that i spend an equal number of hours on the two tasks i currently have . currently i am spending num__5 hours a day on task num__1 and num__3 on task num__2 . how many hours a week less do i need to spend on task num__1 in order to evenly distribute my time ? <o> a ) num__5 <o> b ) num__3 <o> c ) num__7 <o> d ) num__1 <o> e ) num__6 |
num__5 x num__5 = num__25 num__25 - num__5 = num__20 the answer is a . <eor> a <eos> |
a |
divide__40.0__2.0__ round__5.0__ |
subtract__25.0__5.0__ subtract__25.0__20.0__ |
| if the radius of a cylinder is doubled and height num__5 times what is the new volume of the cylinder divided by the old one ? <o> a ) num__8 . <o> b ) num__2 . <o> c ) num__6 . <o> d ) num__20 . <o> e ) num__10 . |
let v and v ' be the original and the changed volume now v = pir ^ num__2 h v ' = pi ( num__2 r ) ^ num__2 ( num__5 h ) v ' = num__20 v d ) num__20 <eor> d <eos> |
d |
square_perimeter__5.0__ square_perimeter__5.0__ |
square_perimeter__5.0__ square_perimeter__5.0__ |
| a train speeds past a pole in num__15 seconds and a platform num__120 meters long in num__25 seconds . what is the length of the train ( in meters ) ? <o> a ) num__100 m <o> b ) num__140 m <o> c ) num__130 m <o> d ) num__180 m <o> e ) num__170 m |
let the length of the train be x meters . the speed of the train is x / num__15 . then x + num__120 = num__25 * ( x / num__15 ) num__10 x = num__1800 x = num__180 meters the answer is d . <eor> d <eos> |
d |
subtract__25.0__15.0__ multiply__15.0__120.0__ divide__1800.0__10.0__ round__180.0__ |
subtract__25.0__15.0__ multiply__15.0__120.0__ divide__1800.0__10.0__ divide__1800.0__10.0__ |
| a can be divided by num__11 with no remainder . which of the following expressions could be divided by num__11 leaving a remainder of num__1 ? <o> a ) a - num__20 <o> b ) a - num__12 <o> c ) a - num__9 <o> d ) a - num__10 <o> e ) a - num__13 |
since a is a multiple of num__11 we could very well say a = num__11 k ( where k is num__12 . . . n ) . the condition here should now be num__11 k + num__1 = < operation > such that it gives us an integer num__11 k + num__1 = a - num__20 = > a = num__11 k + num__21 not divisible by num__11 num__11 k + num__1 = a - num__12 = > a = num__11 k + num__31 not divisible by num__11 num__11 k + num__1 = a - num__9 = > a = num__11 k + num__10 not divisible by num__11 num__11 k + num__1 = a - num__10 = > a = num__11 k + num__11 divisible by num__11 num__11 k + num__1 = a - num__13 = > a = num__11 k + num__14 not divisible by num__11 answer therefore is a - num__10 answer : d <eor> d <eos> |
d |
add__11.0__1.0__ add__1.0__20.0__ add__11.0__20.0__ subtract__20.0__11.0__ subtract__11.0__1.0__ add__1.0__12.0__ add__1.0__13.0__ subtract__11.0__1.0__ |
add__11.0__1.0__ add__1.0__20.0__ add__11.0__20.0__ subtract__20.0__11.0__ subtract__11.0__1.0__ add__1.0__12.0__ add__1.0__13.0__ subtract__11.0__1.0__ |
| on a certain day orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ num__0.40 per glass on the first day what was the price per glass on the second day ? <o> a ) $ num__0.15 <o> b ) $ num__0.26 <o> c ) $ num__0.30 <o> d ) $ num__0.40 <o> e ) $ num__0.45 |
on the first day num__1 unit of orange juice and num__1 unit of water was used to make num__2 units of orangeade ; on the second day num__1 unit of orange juice and num__2 units of water was used to make num__3 units of orangeade ; so the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is num__2 to num__3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is num__2 to num__3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from num__2 glasses on the first day equals to the revenue from num__3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then num__2 * num__0.4 = num__3 * x - - > x = $ num__0.26 . answer : b . <eor> b <eos> |
b |
add__1.0__2.0__ multiply__0.26__1.0__ |
add__1.0__2.0__ multiply__0.26__1.0__ |
| the banker ' s gain on a sum due num__6 years hence at num__12.0 per annum is rs . num__612 . what is the banker ' s discount ? <o> a ) num__1240 <o> b ) num__1120 <o> c ) num__1190 <o> d ) num__1462 <o> e ) none of these |
explanation : td = ( bg × num__100 ) / tr = ( num__612 × num__100 ) / ( num__6 × num__12 ) = rs . num__850 bg = bd – td = > num__612 = bd - num__850 = > bd = num__612 + num__850 = num__1462 answer : option d <eor> d <eos> |
d |
percent__100.0__1462.0__ |
percent__100.0__1462.0__ |
| two employees x and y are paid a total of rs . num__500 per week by their employer . if x is paid num__120 percent of the sum paid to y how much is y paid per week ? <o> a ) s . num__227 <o> b ) s . num__287 <o> c ) s . num__297 <o> d ) s . num__300 <o> e ) s . num__380 |
let the amount paid to x per week = x and the amount paid to y per week = y then x + y = num__500 but x = num__120.0 of y = num__120 y / num__100 = num__12 y / num__10 ∴ num__12 y / num__10 + y = num__500 ⇒ y [ num__1.2 + num__1 ] = num__500 ⇒ num__22 y / num__10 = num__500 ⇒ num__22 y = num__5000 ⇒ y = num__227.272727273 = rs . num__227.77 a ) <eor> a <eos> |
a |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__12.0__10.0__ multiply__500.0__10.0__ divide__5000.0__22.0__ round_down__227.2727__ |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__12.0__10.0__ multiply__500.0__10.0__ divide__5000.0__22.0__ round_down__227.2727__ |
| what is the sixth digit to the right of the decimal point in the decimal expansion of ( num__0.3 ) ^ num__6 ? <o> a ) num__7 <o> b ) num__9 <o> c ) num__3 <o> d ) num__6 <o> e ) num__5 |
step num__1 convert fraction to decimal with two decimal places num__0.3 = num__0.30 then convert to scientific number = num__3 * num__10 ^ - num__1 step num__2 multiply num__3 ^ num__6 by num__10 ^ - num__6 step num__3 num__3 has the sequence num__3 num__9 num__27 answer is b <eor> b <eos> |
b |
divide__3.0__0.3__ divide__6.0__3.0__ add__6.0__3.0__ multiply__3.0__9.0__ add__6.0__3.0__ |
divide__3.0__0.3__ subtract__3.0__1.0__ subtract__10.0__1.0__ multiply__3.0__9.0__ multiply__1.0__9.0__ |
| a can do a work in num__15 days and b in num__20 days . if they work on it together then the number of days the whole work can be finished is : <o> a ) num__0.175 <o> b ) num__0.116666666667 <o> c ) num__5.71428571429 <o> d ) num__8.57142857143 <o> e ) num__7.14285714286 |
a ' s num__1 day ' s work = num__0.0666666666667 b ' s num__1 day ' s work = num__0.05 ( a + b ) ' s num__1 day ' s work = ( num__0.0666666666667 + num__0.05 ) = num__0.116666666667 the whole work can be finished in num__8.57142857143 days . answer : d <eor> d <eos> |
d |
divide__1.0__15.0__ divide__1.0__20.0__ add__0.0667__0.05__ multiply__1.0__8.5714__ |
divide__1.0__15.0__ divide__1.0__20.0__ add__0.0667__0.05__ multiply__1.0__8.5714__ |
| p q and r have $ num__9000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? <o> a ) num__2400 <o> b ) num__3600 <o> c ) num__3998 <o> d ) num__2539 <o> e ) num__1930 |
b num__3600 let the amount with r be $ r r = num__0.666666666667 ( total amount with p and q ) r = num__0.666666666667 ( num__9000 - r ) = > num__3 r = num__18000 - num__2 r = > num__5 r = num__18000 = > r = num__3600 . <eor> b <eos> |
b |
divide__18000.0__9000.0__ divide__18000.0__3600.0__ divide__18000.0__5.0__ |
divide__18000.0__9000.0__ divide__18000.0__3600.0__ divide__18000.0__5.0__ |
| amy ' s sells kale at x dollar per pound for the first num__20 pounds and . num__8 x for every subsequent pound . lucia ' s price is x per pound for the first num__14 pounds and . num__9 x for subsequent pounds . what is the minimum number of pounds over num__20 for which amy ' s becomes an equal or better deal ? <o> a ) num__26 <o> b ) num__25 <o> c ) num__24 <o> d ) num__27 <o> e ) num__28 |
for amy ' s deal to be better the cost has to be less or equal to lucia ' s assuming ' n ' is the number of pounds of kale the equation is num__20 x + ( n - num__20 ) ( num__0.8 x ) < = num__14 x + ( n - num__14 ) ( num__0.9 x ) resolve it : = = > num__20 x + num__0.8 nx - num__16 x < = num__14 x + num__0.9 nx - num__12.6 x = = > num__2.6 x < = num__0.1 nx = = > num__26 x < = nx = = > x ( n - num__26 ) > = num__0 as x can not be num__0 = = > n - num__26 > = num__0 = = > n > = num__26 so the minimum value is num__26 ' a ' would be the correct answer <eor> a <eos> |
a |
multiply__20.0__0.8__ multiply__14.0__0.9__ divide__0.8__8.0__ divide__2.6__0.1__ round_down__0.1__ divide__2.6__0.1__ |
multiply__20.0__0.8__ multiply__14.0__0.9__ subtract__0.9__0.8__ divide__2.6__0.1__ round_down__0.1__ divide__2.6__0.1__ |
| a circle circumscribes three unit circles that touch each other . what is the area of the larger circle ? note that p is the ratio of the circumference to the diameter of a circle ( num__3.14159265 ) . <o> a ) num__3.14 × ( num__7 + num__43 – √ num__0.333333333333 ) <o> b ) num__3.14 × ( num__7 + num__43 – √ num__0.428571428571 ) <o> c ) num__3.14 × ( num__7 + num__43 – √ num__0.5 ) <o> d ) num__3.14 × ( num__7 + num__43 – √ num__1.0 ) <o> e ) num__3.14 × ( num__7 + num__43 – √ num__3.0 ) |
by joining centers of num__3 unit circles we will get an equilateral triangle of length num__2 unit . we have to find the length of the orange line . and center of the equilateral triangle will be the center of the big circle . so radius of the big circle will be = ( num__1 + circum radius of the equilateral triagle ) formula for circul radius of the equilateral triangle = num__23 × ( num__3 – √ num__2 a ) num__23 × ( num__32 a ) here num__3 – √ num__2 a num__32 a is the height of the triangle . a is the side of the triangle circum radius of equilateral triangle = num__23 × num__3 – √ num__2 × num__2 = num__23 – √ num__23 × num__32 × num__2 = num__23 area of big circle will be = π r num__2 = num__3.14 × ( num__1 + num__23 – √ ) num__2 π r num__2 = num__3.14 × ( num__1 + num__23 ) num__2 = num__3.14 × ( num__1 + num__43 – √ + num__43 ) num__3.14 × ( num__1 + num__43 + num__43 ) = num__3.14 × ( num__1 + num__43 – √ + num__43 ) = num__3.14 × ( num__1 + num__43 + num__43 ) = num__3.14 × ( num__73 + num__43 – √ ) num__3.14 × ( num__73 + num__43 ) = num__3.14 × ( num__7 + num__43 – √ num__1.0 ) answer : d <eor> d <eos> |
d |
multiply__1.0__3.14__ |
multiply__1.0__3.14__ |
| assume all cds are equal in time . if num__21 cds have x hours of music how many minutes will y cds have ? <o> a ) num__21 xy / num__60 <o> b ) num__0.116666666667 xy <o> c ) num__0.35 xy <o> d ) num__20 xy / num__7 <o> e ) num__0.35 xy |
num__21 cds have x hours or num__60 * x minutes num__1 cd has time = num__60 * x / num__21 = num__20 x / num__7 y cd time in minutes = y * num__20 * x / num__7 = num__20 xy / num__7 hence the answer is d . <eor> d <eos> |
d |
subtract__21.0__1.0__ subtract__21.0__1.0__ |
subtract__21.0__1.0__ multiply__1.0__20.0__ |
| a company pays num__12.5 dividend to its investors . if an investor buys rs . num__60 shares and gets num__25.0 on investment at what price did the investor buy the shares ? <o> a ) num__25 <o> b ) num__66 <o> c ) num__30 <o> d ) num__19 <o> e ) num__01 |
explanation : dividend on num__1 share = ( num__12.5 * num__60 ) / num__100 = rs . num__7.5 rs . num__25 is income on an investment of rs . num__100 rs . num__7.5 is income on an investment of rs . ( num__7.5 * num__100 ) / num__25 = rs . num__30 answer : c <eor> c <eos> |
c |
percent__12.5__60.0__ percent__100.0__30.0__ |
percent__12.5__60.0__ percent__100.0__30.0__ |
| a train passes a platform in num__36 seconds . the same train passes a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr the length of the platform is <o> a ) num__280 meter <o> b ) num__240 meter <o> c ) num__200 meter <o> d ) num__260 meter <o> e ) none of these |
explanation : speed of the train = num__54 km / hr = ( num__54 × num__10 ) / num__36 m / s = num__15 m / s length of the train = speed × time taken to cross the man = num__15 × num__20 = num__300 m let the length of the platform = l time taken to cross the platform = ( num__300 + l ) / num__15 = > ( num__300 + l ) / num__15 = num__36 = > num__300 + l = num__15 × num__36 = num__540 = > l = num__540 - num__300 = num__240 meter answer : option b <eor> b <eos> |
b |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ subtract__540.0__300.0__ |
| a pipe x is num__30 meters and num__45.0 longer than another pipe y . find the length of the pipe y . <o> a ) num__20.12 <o> b ) num__20.68 <o> c ) num__20 <o> d ) num__20.5 <o> e ) none of these |
explanation : length of pipe x = num__30 meter given that pipe x is num__45.0 longer than y let the length of pipe y = y then length of pipe x = y × ( num__100 + num__45 ) / num__100 ⇒ num__30 = y × ( num__1.45 ) = > y = num__30 × num__0.689655172414 = num__6 × num__3.44827586207 = num__20.6896551724 = num__20.68 answer : option b <eor> b <eos> |
b |
percent__100.0__20.68__ |
percent__100.0__20.68__ |
| the sum of the non - prime numbers between num__30 and num__40 non - inclusive is <o> a ) num__202 <o> b ) num__217 <o> c ) num__232 <o> d ) num__247 <o> e ) num__262 |
sum of consecutive integers from num__31 to num__39 inclusive = = = = > ( a num__1 + an ) / num__2 * # of terms = ( num__31 + num__39 ) / num__2 * num__9 = num__35 * num__9 = num__315 sum of non - prime numbers b / w num__30 and num__40 non inclusive = = = > num__315 - num__68 ( i . e . num__31 + num__37 being the prime # s in the range ) = num__247 answer : d <eor> d <eos> |
d |
subtract__40.0__39.0__ subtract__40.0__31.0__ multiply__35.0__9.0__ add__2.0__35.0__ subtract__315.0__68.0__ multiply__1.0__247.0__ |
subtract__40.0__39.0__ subtract__40.0__31.0__ multiply__35.0__9.0__ add__2.0__35.0__ subtract__315.0__68.0__ multiply__1.0__247.0__ |
| a dishonest grocer professes to sell pure butter at cost price but he mixes it with adulterated fat and thereby gains num__45.0 . find the percentage of adulterated fat in the mixture assuming that adulterated fat is freely available ? <o> a ) num__20.0 <o> b ) num__25.0 <o> c ) num__33.33 <o> d ) num__40.0 <o> e ) num__35 % |
say num__100 g pure butter costs $ num__100 but ; $ num__100 is his num__25.0 profit because he just used num__80 gram pure butter . num__80 g would have costed him $ num__80 and he sold it for $ num__100 by adding num__20 g of freely available fat . $ num__100 is num__1.45 ( num__65 ) so ; total weight = num__100 g fat = num__35 g num__35.0 e <eor> e <eos> |
e |
percent__80.0__25.0__ percent__35.0__100.0__ |
percent__80.0__25.0__ percent__35.0__100.0__ |
| john bought num__9.25 m of cloth for $ num__434.75 . find the cost price per metre . <o> a ) num__46 <o> b ) num__47 <o> c ) num__58 <o> d ) num__56 <o> e ) num__54 |
cloth bought by john = num__9.25 m cost of num__9.25 m = $ num__434.75 cost per metre = num__434.75 ÷ num__9.25 the cost of the cloth per metre = $ num__47 answers : b <eor> b <eos> |
b |
divide__434.75__9.25__ round__47.0__ |
divide__434.75__9.25__ round__47.0__ |
| all the water in container a which was filled to its brim was poured into two containers b and c . the quantity of water in container b was num__62.5 less than the capacity of container a . if num__152 liters was now transferred from c to b then both the containers would have equal quantities of water . what was the initial quantity of water in container a ? <o> a ) num__1289 <o> b ) num__1198 <o> c ) num__281 <o> d ) num__1216 <o> e ) num__282 |
explanation : b has num__62.5 or ( num__0.625 ) of the water in a . therefore let the quantity of water in container a ( initially ) be num__8 k . quantity of water in b = num__8 k - num__5 k = num__3 k . quantity of water in container c = num__8 k - num__3 k = num__5 k container : a b c quantity of water : num__8 k num__3 k num__5 k it is given that if num__152 liters was transferred from container c to container b then both the containers would have equal quantities of water . num__5 k - num__152 = num__3 k + num__152 = > num__2 k = num__304 = > k = num__152 the initial quantity of water in a = num__8 k = num__8 * num__152 = num__1216 liters . answer : option d <eor> d <eos> |
d |
multiply__0.625__8.0__ subtract__8.0__5.0__ subtract__5.0__3.0__ multiply__152.0__2.0__ multiply__152.0__8.0__ multiply__152.0__8.0__ |
multiply__0.625__8.0__ subtract__8.0__5.0__ subtract__5.0__3.0__ multiply__152.0__2.0__ multiply__152.0__8.0__ multiply__152.0__8.0__ |
| evaluate : num__17 + sqrt ( - num__7 + num__10 Ã — num__6 Ã · num__5 ) = ? <o> a ) num__40 <o> b ) num__42 <o> c ) num__44 <o> d ) num__46 <o> e ) num__48 |
according to order of operations inner brackets first where num__10 x num__6 Ã · num__5 is first calculated since it has a multiplication and a division . num__10 x num__6 Ã · num__5 = num__60 Ã · num__5 = num__12 hence num__17 + sqrt ( - num__7 + num__10 Ã — num__6 Ã · num__5 ) = num__17 + sqrt ( - num__7 + num__12 ) = num__17 + sqrt ( num__5 ) = num__17 + num__25 = num__42 correct answer b ) num__42 <eor> b <eos> |
b |
multiply__10.0__6.0__ subtract__17.0__5.0__ add__17.0__25.0__ add__17.0__25.0__ |
multiply__10.0__6.0__ subtract__17.0__5.0__ add__17.0__25.0__ add__17.0__25.0__ |
| a num__25.0 stock yielding num__20.0 is quoted at : <o> a ) s . num__83.33 <o> b ) s . num__110 <o> c ) s . num__112 <o> d ) s . num__125 <o> e ) s . num__140 |
income of rs num__20 on investment of rs num__100 income of rs num__25 on investment of ? = ( num__25 * num__100 ) / num__20 = num__125 answer : d <eor> d <eos> |
d |
percent__100.0__125.0__ |
percent__100.0__125.0__ |
| a table is bought for rs . num__200 / - and sold at rs . num__150 / - find gain or loss percentage <o> a ) num__15.0 loss <o> b ) num__20.0 loss <o> c ) num__25.0 loss <o> d ) num__30.0 loss <o> e ) none |
formula = ( selling price ~ cost price ) / cost price * num__100 = ( num__150 - num__200 ) / num__200 = num__25.0 loss c <eor> c <eos> |
c |
percent__25.0__100.0__ |
percent__25.0__100.0__ |
| in plutarch enterprises num__60.0 of the employees are marketers num__30.0 are engineers and the rest are managers . marketers make an average salary of $ num__50000 a year and engineers make an average of $ num__80000 . what is the average salary for managers if the average for all employees is also $ num__80000 ? <o> a ) $ num__80000 <o> b ) $ num__130000 <o> c ) $ num__260000 <o> d ) $ num__290000 <o> e ) $ num__320 |
000 |
for sake of ease let ' s say there are num__10 employees : num__6 marketers num__3 engineers and num__1 manager . average company salary * number of employees = total company salary > > > $ num__80000 * num__10 = $ num__800000 subtract the combined salaries for the marketers ( num__6 * $ num__50000 ) and the engineers ( num__3 * $ num__80000 ) > > > $ num__800000 - $ num__300000 - $ num__240000 = $ num__260000 . the correct answer is c . <eor> c <eos> |
c |
c |
| what will come in place of the x in the following number series ? num__18 x num__30 num__46 num__78 num__142 <o> a ) num__35 <o> b ) num__56 <o> c ) num__22 <o> d ) num__87 <o> e ) num__98 |
( c ) the pattern is + num__4 + num__8 + num__16 + num__32 + num__64 so the missing term is = num__22 . <eor> c <eos> |
c |
subtract__46.0__30.0__ subtract__78.0__46.0__ add__18.0__46.0__ add__18.0__4.0__ add__18.0__4.0__ |
subtract__46.0__30.0__ subtract__78.0__46.0__ add__18.0__46.0__ add__18.0__4.0__ add__18.0__4.0__ |
| three pipes a b & c are attached to a tank . a & b can fill it in num__20 & num__30 minutes respectively while c can empty it in num__15 minutes . if a b & c are kept open successively for num__1 minute each how soon will the tank be filled ? <o> a ) num__2 hours <o> b ) num__4 hours <o> c ) num__3 hours <o> d ) num__5 hours <o> e ) num__6 hours |
in three minute num__0.05 + num__0.0333333333333 - num__0.0666666666667 = num__0.0166666666667 part is filled num__3 min - - - - - - - - num__0.0166666666667 parts x min - - - - - - - - - num__1 part ( full ) x = num__180 min = num__3 hours answer : c <eor> c <eos> |
c |
divide__1.0__20.0__ divide__1.0__30.0__ divide__1.0__15.0__ subtract__0.05__0.0333__ round__3.0__ |
divide__1.0__20.0__ divide__1.0__30.0__ divide__1.0__15.0__ subtract__0.05__0.0333__ round__3.0__ |
| a taxi company charges $ num__2.5 for the first quarter of a mile and fifteen cents for each additional quarter of a mile . what is the maximum distance someone could travel with $ num__4.90 ? <o> a ) num__4 miles <o> b ) num__4 num__0.25 miles <o> c ) num__4 num__0.75 miles <o> d ) num__5 num__0.5 miles <o> e ) num__6 num__0.25 miles |
if we start out with $ num__4.90 and have to spend $ num__2.5 for the first quarter - mile we will have $ num__2.40 left to spend on quarter - mile intervals . since $ num__2.40 / $ num__0.15 = num__16 we can buy num__16 more quarter - miles and will travel num__17 quarter miles in all : num__17 × num__0.25 = num__4 num__0.25 miles . the correct answer is choice ( b ) . <eor> b <eos> |
b |
subtract__4.9__2.5__ divide__2.4__0.15__ multiply__0.25__16.0__ round__4.0__ |
subtract__4.9__2.5__ divide__2.4__0.15__ multiply__0.25__16.0__ divide__16.0__4.0__ |
| the value of x + x ( xx ) when x = num__3 is : <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__14 <o> e ) num__30 |
x + x ( xx ) put the value of x = num__3 in the above expression we get num__3 + num__3 ( num__33 ) = num__3 + num__3 ( num__3 Ã — num__3 ) = num__3 + num__3 ( num__9 ) = num__3 + num__27 = num__30 answer : e <eor> e <eos> |
e |
multiply__3.0__9.0__ add__3.0__27.0__ add__3.0__27.0__ |
multiply__3.0__9.0__ add__3.0__27.0__ add__3.0__27.0__ |
| how much time will it take for an amount of num__900 to yield num__81 as interest at num__4.5 per annum of simple interest ? <o> a ) num__3 years <o> b ) num__4 years <o> c ) num__6 years <o> d ) num__5 years <o> e ) num__2 years |
time = ( num__100 x num__81 ) / ( num__900 x num__4.5 ) years = num__2 years . answer e <eor> e <eos> |
e |
percent__100.0__2.0__ |
percent__100.0__2.0__ |
| sam earns $ num__6.00 per hour for the first num__60 hours he works per week and twice this rate for overtime . if michael earned $ num__358 last week how many hours did he work ? <o> a ) num__43 <o> b ) num__44 <o> c ) num__45 <o> d ) num__46 <o> e ) num__69 |
$ num__6 * num__40 + $ num__12 * x = $ num__358 - - > x = num__9 hours . total working hours = num__60 + num__9 = num__69 . answer : e . <eor> e <eos> |
e |
add__60.0__9.0__ round__69.0__ |
add__60.0__9.0__ add__60.0__9.0__ |
| find the cost of fencing around a circular field of diameter num__28 m at the rate of rs . num__1.50 a meter ? <o> a ) num__178 <o> b ) num__132 <o> c ) num__279 <o> d ) num__265 <o> e ) num__801 |
num__2 * num__3.14285714286 * num__14 = num__88 num__88 * num__1 num__0.5 = rs . num__132 answer : b <eor> b <eos> |
b |
divide__28.0__2.0__ subtract__1.5__1.0__ multiply__1.5__88.0__ round__132.0__ |
divide__28.0__2.0__ subtract__1.5__1.0__ multiply__1.5__88.0__ multiply__1.5__88.0__ |
| a watch was sold at a loss of num__10.0 . if it was sold for rs . num__280 more there would have been a gain of num__4.0 . what is the cost price ? <o> a ) s : num__2000 <o> b ) s : num__1067 <o> c ) s : num__1278 <o> d ) s : num__1028 <o> e ) s : num__1027 |
num__90.0 num__104.0 - - - - - - - - num__14.0 - - - - num__280 num__100.0 - - - - ? = > rs : num__2000 answer : a <eor> a <eos> |
a |
percent__100.0__2000.0__ |
percent__100.0__2000.0__ |
| if rs . num__10 be allowed as true discount on a bill of rs . num__110 due at the end of a certain time then the discount allowed on the same sum due at the end of double the time is : <o> a ) rs . num__20 <o> b ) rs . num__21.81 <o> c ) rs . num__22 <o> d ) rs . num__18.33 <o> e ) none of these |
solution s . i . on rs . ( num__110 - num__10 ) for a certain time = rs . num__10 . s . i . on rs . num__100 for double the time = rs . num__20 . t . d . on rs . num__120 = rs . ( num__120 - num__100 ) = rs . num__20 . t . d . on rs . num__110 = rs . left ( num__0.166666666667 x num__120 ) = rs . num__18.33 . answer d <eor> d <eos> |
d |
percent__100.0__18.33__ |
percent__100.0__18.33__ |
| find highest power of num__2 in num__1 ! + num__2 ! + num__3 ! + num__4 ! . . . . . . . . . . . + num__600 ! <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
unit digit of num__1 ! + num__2 ! + num__3 ! + num__4 ! . . . . . . . . . . . + num__600 ! is odd i . e the given sum gives an odd no . so there is no factor of num__2 . highest power of num__2 = num__2 ^ num__0 hence answer is num__0 . answer : a <eor> a <eos> |
a |
multiply__2.0__0.0__ |
multiply__2.0__0.0__ |
| m is the sum of the reciprocals of the consecutive integers from num__401 to num__500 inclusive . which of the following is true ? <o> a ) num__0.333333333333 < m < num__0.5 <o> b ) num__0.2 < m < num__0.25 <o> c ) num__0.142857142857 < m < num__0.2 <o> d ) num__0.111111111111 < m < num__0.142857142857 <o> e ) num__0.0833333333333 < m < num__0.111111111111 |
m = num__0.00249376558603 + num__0.25 + num__0.00248138957816 + . . . . . . + num__0.002 if we replace the first num__99 terms by num__0.002 then we get a sum = num__0.2 = num__0.2 . since the actual terms are larger than num__0.002 the sum is larger than num__0.2 . if we replace the all the num__100 terms by num__0.0025 we get a sum = num__0.25 = num__0.25 . since the actual terms are smaller than num__0.0025 the sum is less than num__0.25 . therefore num__0.2 < m < num__0.25 choice b <eor> b <eos> |
b |
reverse__401.0__ reverse__500.0__ subtract__500.0__401.0__ multiply__500.0__0.2__ divide__100.0__500.0__ |
reverse__401.0__ reverse__500.0__ subtract__500.0__401.0__ multiply__500.0__0.2__ divide__100.0__500.0__ |
| num__8 identical machines working alone and at their constant rates take num__6 hours to complete a job lot . how long would it take for num__2 such machines to perform the same job ? <o> a ) num__2.25 hours <o> b ) num__8.75 hours <o> c ) num__12 hours <o> d ) num__24 hours <o> e ) num__16 hours |
let each machine do num__1 unit of work for num__1 hour num__8 machines - - > num__8 units of work in num__1 hour for num__6 hours = num__8 * num__6 = num__48 units of total work is done . now this num__48 units of total work must be done by num__2 machines num__2 units of work ( num__2 machines ) - - - > num__1 hour for num__48 units of work num__2 * num__24 - - - > num__1 * num__24 hours d num__24 hours <eor> d <eos> |
d |
multiply__8.0__6.0__ divide__48.0__2.0__ round__24.0__ |
multiply__8.0__6.0__ divide__48.0__2.0__ subtract__48.0__24.0__ |
| working at constant rate pump x pumped out half of the water in a flooded basement in num__6 hours . the pump y was started and the two pumps working independently at their respective constant rates pumped out rest of the water in num__3 hours . how many hours would it have taken pump y operating alone at its own constant rate to pump out all of the water that was pumped out of the basement ? <o> a ) a . num__10 <o> b ) b . num__12 <o> c ) c . num__14 <o> d ) d . num__28 <o> e ) e . num__24 |
rate of x = num__0.125 rate of x + y = num__0.166666666667 rate of y = num__0.166666666667 - num__0.125 = num__0.0416666666667 num__28 hours d <eor> d <eos> |
d |
divide__0.125__3.0__ round__28.0__ |
subtract__0.1667__0.125__ round__28.0__ |
| look at this series : num__7 num__10 num__8 num__11 num__9 num__12 . . . what number should come next ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
this is a simple alternating addition and subtraction series . in the first pattern num__3 is added ; in the second num__2 is subtracted . answer a <eor> a <eos> |
a |
subtract__10.0__7.0__ subtract__10.0__8.0__ add__7.0__3.0__ |
subtract__10.0__7.0__ subtract__10.0__8.0__ add__7.0__3.0__ |
| a can do a work in num__8 days . b can do the same work in num__24 days . if both a & b are working together in how many days they will finish the work ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__4 <o> d ) num__2 <o> e ) num__6 |
a rate = num__0.125 b rate = num__0.0416666666667 ( a + b ) rate = ( num__0.125 ) + ( num__0.0416666666667 ) = num__0.166666666667 a & b finish the work in num__6 days correct option is e <eor> e <eos> |
e |
add__0.125__0.0417__ round__6.0__ |
add__0.125__0.0417__ round__6.0__ |
| a certain furniture can be repaired for $ num__15.50 and will last for num__1 year . a same kind of furniture can be purchased new for $ num__40.00 and will last for num__2 years . the average cost per year of the new shoes is what percent greater than the cost of repairing the used shoes ? <o> a ) num__28.9 <o> b ) num__31.4 <o> c ) num__27.8 <o> d ) num__29.03 <o> e ) num__30 |
num__1 ) cost of repairing = num__15.5 ( for one year ) therefore for num__2 years it would be $ num__31 . num__2 ) cost of new furniture which will last for num__2 years is $ num__40 . percentage change formula = ( final value - initial value ) / ( initial value ) * num__100 . in this case the final value would be the price of new furniture value would be the cost of repairing the old shoe . i . e ( num__40 - num__31 ) / ( num__31 ) * num__100 = num__29.03 . ans is d . <eor> d <eos> |
d |
multiply__15.5__2.0__ multiply__1.0__29.03__ |
multiply__15.5__2.0__ multiply__1.0__29.03__ |
| the sum of the mean the median and the range of the set { num__5 num__67 } equals which one of the following values ? <o> a ) num__18 <o> b ) num__14 <o> c ) num__16 <o> d ) num__12 <o> e ) num__10 |
here mean = > num__5 + num__6 + num__2.33333333333 = > num__6 median = > num__6 and range = > num__7 - num__1 = > num__6 hence sum = > num__6 + num__6 + num__6 = > num__18 answer : a <eor> a <eos> |
a |
subtract__6.0__5.0__ multiply__1.0__18.0__ |
subtract__6.0__5.0__ multiply__1.0__18.0__ |
| the food in a camp lasts for num__20 men for num__40 days . if ten more men join the food will last for how many days ? <o> a ) num__40 days <o> b ) num__27 days <o> c ) num__37 days <o> d ) num__50 days <o> e ) num__45 days |
one man can consume the same food in num__20 * num__40 = num__800 days . num__10 more men join the total number of men = num__20 the number of days the food will last = num__26.6666666667 = num__27 days . answer : b <eor> b <eos> |
b |
multiply__20.0__40.0__ round__27.0__ |
multiply__20.0__40.0__ round__27.0__ |
| in a sports club with num__42 members num__20 play badminton and num__23 play tennis and num__6 do not play either . how many members play both badminton and tennis ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
num__20 + num__23 = num__43 but where as total number is num__42 - num__6 = num__36 therefore answer is num__43 - num__36 = num__7 hence answer is d <eor> d <eos> |
d |
choose__7.0__6.0__ |
choose__7.0__6.0__ |
| the probability that b can shoot a target num__3 out of num__3 times is num__0.25 . what is the probability that the target will be missed by b immediately after such two shots ? <o> a ) num__0.25 <o> b ) num__0.5 <o> c ) num__0.4 <o> d ) num__0.75 <o> e ) num__0.8 |
suppose there are total ' x ' shots . b shoots num__3 out of num__3 times . means out of x shots ( x > num__3 ) therefore num__3 / x = num__0.25 ( given ) the target will be missed by b immediately after such two shots : this means he can shot just twice . . . the third shot can not happen which means he missed ( x - num__3 ) shots . thus the probabilty of missing just after num__2 shots is ( x - num__3 ) / x . ( x - num__3 ) / x = num__1 - num__3 / x = num__1 - num__0.25 = num__0.75 answer : d <eor> d <eos> |
d |
coin_space__ negate_prob__0.25__ negate_prob__0.25__ |
coin_space__ negate_prob__0.25__ negate_prob__0.25__ |
| a train passes a station platform in num__18 seconds and a man standing on the platform in num__12 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__100 <o> b ) num__150 <o> c ) num__96 <o> d ) num__88 <o> e ) num__90 |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__12 ) m = num__180 m . let the length of the platform be x meters . then x + num__10.0 = num__15 x + num__180 = num__270 x = num__90 m . answer : e <eor> e <eos> |
e |
multiply__12.0__15.0__ divide__180.0__18.0__ multiply__18.0__15.0__ subtract__270.0__180.0__ round__90.0__ |
multiply__12.0__15.0__ divide__180.0__18.0__ multiply__18.0__15.0__ subtract__270.0__180.0__ round__90.0__ |
| of two numbers num__4 times the smaller one is less then num__3 times the num__1 arger one by num__5 . if the sum of the numbers is larger than num__6 times their difference by num__6 find the two numbers . <o> a ) num__59 and num__43 <o> b ) num__55 and num__44 <o> c ) num__58 and num__43 <o> d ) num__59 and num__42 <o> e ) num__57 and num__40 |
let the numbers be x and y such that x > y then num__3 x - num__4 y = num__5 . . . ( i ) and ( x + y ) - num__6 ( x - y ) = num__6 = > - num__5 x + num__7 y = num__6 … ( ii ) solving ( i ) and ( ii ) we get : x = num__59 and y = num__43 . hence the required numbers are num__59 and num__43 . answer is a . <eor> a <eos> |
a |
add__4.0__3.0__ multiply__1.0__59.0__ |
add__4.0__3.0__ multiply__1.0__59.0__ |
| the distance between delhi and mathura is num__125 kms . a starts from delhi with a speed of num__25 kmph at num__9 a . m . for mathura and b starts from mathura with a speed of num__35 kmph at num__10 a . m . from delhi . when will they meet ? <o> a ) num__11 num__0.5 <o> b ) num__77 num__0.666666666667 <o> c ) num__16 num__0.75 <o> d ) num__10 <o> e ) num__98 |
d = num__125 – num__25 = num__100 rs = num__35 + num__25 = num__60 t = num__1.66666666667 = num__1.6 hours num__10 a . m . + num__1.6 = num__11 num__0.5 a . m . . answer : a <eor> a <eos> |
a |
subtract__125.0__25.0__ hour_to_min_conversion__ divide__100.0__60.0__ mile_to_km_conversion__ round__11.0__ |
subtract__125.0__25.0__ add__25.0__35.0__ divide__100.0__60.0__ mile_to_km_conversion__ round__11.0__ |
| find the num__10 th term of an arithmetic progression whose first term is num__8 and the common difference is num__2 . <o> a ) num__45 <o> b ) num__26 <o> c ) num__44 <o> d ) num__40 <o> e ) num__46 |
n th term of a . p = a + ( n - num__1 ) * d = num__8 + ( num__10 - num__1 ) * num__2 = num__8 + num__18 = num__26 . answer : b <eor> b <eos> |
b |
add__10.0__8.0__ add__8.0__18.0__ add__8.0__18.0__ |
add__10.0__8.0__ add__8.0__18.0__ add__8.0__18.0__ |
| two vessels p and q contain num__62.5 and num__87.5 of alcohol respectively . if num__2 litres from vessel p is mixed with num__4 litres from vessel q the ratio of alcohol and water in the resulting mixture is ? <o> a ) num__13 : num__9 <o> b ) num__12 : num__2 <o> c ) num__19 : num__3 <o> d ) num__19 : num__5 <o> e ) num__19 : num__7 |
quantity of alcohol in vessel p = num__62.5 / num__100 * num__2 = num__1.25 litres quantity of alcohol in vessel q = num__87.5 / num__100 * num__4 = num__3.5 litres quantity of alcohol in the mixture formed = num__1.25 + num__3.5 = num__4.75 = num__4.75 litres as num__6 litres of mixture is formed ratio of alcohol and water in the mixture formed = num__4.75 : num__1.25 = num__19 : num__5 . answer : d <eor> d <eos> |
d |
add__1.25__3.5__ add__2.0__4.0__ multiply__4.0__4.75__ multiply__4.0__1.25__ multiply__4.0__4.75__ |
add__1.25__3.5__ add__2.0__4.0__ multiply__4.0__4.75__ multiply__4.0__1.25__ multiply__4.0__4.75__ |
| in a class of num__40 students num__2 students did not borrow any books from the library num__12 students each borrowed num__1 book num__14 students each borrowed num__2 books and the rest borrowed at least num__3 books . if the average number of books per student was num__2 what is the maximum number of books any single student could have borrowed ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
the class borrowed a total of num__40 * num__2 = num__80 books . the num__28 students who borrowed num__0 num__1 or num__2 books borrowed a total of num__12 + num__14 * num__2 = num__40 . to maximize the number of books borrowed by num__1 student let ' s assume that num__11 students borrowed num__3 books and num__1 student borrowed the rest . num__80 - num__40 - num__3 * num__11 = num__7 the maximum number of books borrowed by any student is num__7 . the answer is e . <eor> e <eos> |
e |
multiply__40.0__2.0__ subtract__40.0__12.0__ subtract__12.0__1.0__ divide__14.0__2.0__ multiply__1.0__7.0__ |
multiply__40.0__2.0__ subtract__40.0__12.0__ subtract__12.0__1.0__ divide__14.0__2.0__ multiply__1.0__7.0__ |
| a man is num__24 years older than his son . in two years his age will be twice the age of his son . the present age of his son is : <o> a ) num__20 <o> b ) num__21 <o> c ) num__22 <o> d ) num__23 <o> e ) num__24 |
let the son ' s present age be x years . then man ' s present age = ( x + num__24 ) years . ( x + num__24 ) + num__2 = num__2 ( x + num__2 ) x + num__26 = num__2 x + num__4 x = num__22 . answer : a <eor> a <eos> |
a |
add__24.0__2.0__ subtract__24.0__2.0__ subtract__24.0__4.0__ |
add__24.0__2.0__ subtract__24.0__2.0__ subtract__24.0__4.0__ |
| if dale works at full efficiency he needs a break of num__1 day after every num__2 days of work and he completes a job in a total of num__20 days . if he works at reduced efficiency he can work without break and finish that same job in num__20 days . dale ' s output at reduced efficiency is what fraction of his output at full efficiency ? <o> a ) num__2.33333333333 <o> b ) num__2.5 <o> c ) num__1.2 <o> d ) num__0.2 <o> e ) num__1.8 |
we ' re told that there are num__2 ways for dale to complete a job : num__1 ) full efficiency : num__2 days of work followed by num__1 dayofffor a total of num__20 days . num__2 ) reduced efficiency : num__20 straight days with no days off . working at full efficiency creates the following pattern : num__2 days on num__1 day off num__2 days on num__1 day off num__2 days on num__1 day off num__2 days on = num__2 + num__1 + num__2 + num__1 + num__2 + num__1 + num__2 + num__1 + num__2 + num__1 + num__2 + num__1 + num__2 = num__20 days totals : num__14 days on num__6 days off reduced efficiency means that dale will do num__14 days of work in num__20 days thus those reduceddaysare num__2.33333333333 = num__2.33333333333 of full efficiency . answer : a <eor> a <eos> |
a |
subtract__20.0__14.0__ divide__14.0__6.0__ multiply__1.0__2.3333__ |
subtract__20.0__14.0__ divide__14.0__6.0__ multiply__1.0__2.3333__ |
| the length of a train and that of a platform are equal . if with a speed of num__126 k / hr the train crosses the platform in one minute then the length of the train ( in meters ) is ? <o> a ) num__757 <o> b ) num__758 <o> c ) num__1050 <o> d ) num__750 <o> e ) num__738 |
speed = [ num__126 * num__0.277777777778 ] m / sec = num__35 m / sec ; time = num__1 min . = num__60 sec . let the length of the train and that of the platform be x meters . then num__2 x / num__60 = num__35 è x = num__35 * num__30.0 = num__1050 answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ divide__60.0__2.0__ multiply__35.0__30.0__ round__1050.0__ |
hour_to_min_conversion__ divide__60.0__2.0__ multiply__35.0__30.0__ multiply__35.0__30.0__ |
| at a certain conference num__32.0 of the attendees registered at least two weeks in advance and paid their conference fee in full . if num__50.0 of the attendees who paid their conference fee in full did not register at least two weeks in advance what percent of conference attendees registered at least two weeks in advance ? <o> a ) num__18.0 <o> b ) num__62.0 <o> c ) num__79.2 <o> d ) num__64.0 <o> e ) num__82.0 % |
refer to the table in the attachment : let x = no . of members who have paid in full num__50.0 members paid in full and did not register in advance = num__0.5 x num__32.0 registerd in advance and paid in full . so if total no . of members = num__100 then num__32 members paid full and registered in advance . hence total members who paid full amount = num__0.5 x + num__32 = x num__0.5 x = num__32 hence x = num__64 i . e . num__64 out of num__100 or num__64.0 ans . d <eor> d <eos> |
d |
percent__100.0__64.0__ |
percent__100.0__64.0__ |
| in the xy - plane a line has slope num__4 and x - intercept num__3 . what is the y - intercept of the line ? <o> a ) - num__12 <o> b ) - num__3 <o> c ) num__0 <o> d ) num__3 <o> e ) num__9 |
let the line be represented by a general equation y = mx + b where m = slope ( num__4 ) and b = y intercept . we are also given the value of x - intercept num__3 . theory : y intercept represents the point on the line where the x = num__0 and x intercept represents the point on the line where the y = num__0 . putting these values in the equation : num__0 = num__4 * num__3 + b = > b = - num__12 . hence a . <eor> a <eos> |
a |
multiply__4.0__3.0__ multiply__4.0__3.0__ |
multiply__4.0__3.0__ multiply__4.0__3.0__ |
| the average ( arithmetic mean ) of six numbers is num__8 . if num__3 is subtracted from each of four of the numbers what is the new average ? <o> a ) num__6 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__4.5 |
sum of num__6 numbers = num__6 * num__8 = num__48 if num__3 is subtracted from each of four of the numbers we subtract num__3 * num__4 = num__12 from the total sum sum of num__6 number after subtracting num__3 from each of four of the numbers = num__48 - num__12 = num__36 new average = num__6.0 = num__6 answer a <eor> a <eos> |
a |
multiply__8.0__6.0__ add__8.0__4.0__ multiply__3.0__12.0__ subtract__12.0__6.0__ |
multiply__8.0__6.0__ multiply__3.0__4.0__ multiply__3.0__12.0__ subtract__12.0__6.0__ |
| the average of first five multiples of num__5 is ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__9 <o> d ) num__15 <o> e ) num__7 |
average = num__5 ( num__1 + num__2 + num__3 + num__4 + num__5 ) / num__5 = num__15.0 = num__15 . answer : d <eor> d <eos> |
d |
subtract__5.0__2.0__ subtract__5.0__1.0__ multiply__5.0__3.0__ multiply__5.0__3.0__ |
add__1.0__2.0__ add__1.0__3.0__ multiply__5.0__3.0__ divide__15.0__1.0__ |
| for positive integers q and r which of the following can be written as r ^ num__2 ? <o> a ) q ^ num__2 - num__12 <o> b ) q ^ num__2 - num__11 <o> c ) q ^ num__2 - num__10 <o> d ) q ^ num__2 - num__9 <o> e ) q ^ num__2 - num__14 |
plug values : if q = num__5 then num__5 ^ num__2 - num__9 = num__4 the question asks which of the following can be written as r ^ num__2 . if q = num__16 then q ^ num__2 - num__9 can be written as num__2 ^ num__2 . answer : d <eor> d <eos> |
d |
subtract__9.0__5.0__ power__2.0__4.0__ subtract__4.0__2.0__ |
subtract__9.0__5.0__ power__2.0__4.0__ subtract__4.0__2.0__ |
| what is the area of a square field whose diagonal of length num__20 m ? <o> a ) num__238 <o> b ) num__379 <o> c ) num__200 <o> d ) num__328 <o> e ) num__397 |
d num__1.0 = ( num__20 * num__20 ) / num__2 = num__200 answer : c <eor> c <eos> |
c |
multiply__200.0__1.0__ |
multiply__200.0__1.0__ |
| calculate how many days it will take for num__7 boys to wash a num__35 m long wall if num__5 boys can wash a num__25 m long wall in num__4 days <o> a ) num__6 days <o> b ) num__3 days <o> c ) num__4 days <o> d ) num__2 days <o> e ) num__5 days |
the length of wall painted by one boy in one day = num__5.0 * num__0.25 = num__1.25 m no . of days required to paint num__50 m cloth by num__8 boys = num__5.0 * num__1 / num__1.25 = num__4 days . c <eor> c <eos> |
c |
divide__5.0__4.0__ subtract__5.0__4.0__ round__4.0__ |
divide__5.0__4.0__ multiply__4.0__0.25__ divide__5.0__1.25__ |
| a reduction of num__20.0 in the price of salt enables a lady to obtain num__10 kgs more for rs . num__600 find the original price per kg ? <o> a ) s . num__11 <o> b ) s . num__14 <o> c ) s . num__15 <o> d ) s . num__21 <o> e ) s . num__29 |
num__100 * ( num__0.2 ) = num__20 - - - num__10 ? - - - num__1 = > rs . num__2 num__600 - - - num__80 ? - - - num__2 = > rs . num__15 answer : c <eor> c <eos> |
c |
percent__20.0__10.0__ percent__100.0__15.0__ |
percent__20.0__10.0__ percent__100.0__15.0__ |
| in how many years rs num__200 will produce the same interest at num__10.0 as rs . num__400 produce in num__5 years at num__12.0 <o> a ) num__16 <o> b ) num__13 <o> c ) num__12 <o> d ) num__14 <o> e ) num__15 |
explanation : clue : firstly we need to calculate the si with prinical num__400 time num__5 years and rate num__12.0 it will be rs . num__240 then we can get the time as time = ( num__100 * num__240 ) / ( num__200 * num__10 ) = num__12 option c <eor> c <eos> |
c |
percent__5.0__240.0__ |
percent__5.0__240.0__ |
| a merchant marks his goods up by num__30.0 and then offers a discount of num__10.0 on the marked price . what % profit does the merchant make after the discount ? <o> a ) num__8.0 <o> b ) num__10.0 <o> c ) num__21.0 <o> d ) num__15.0 <o> e ) num__17 % |
let the price be num__100 . the price becomes num__130 after a num__30.0 markup . now a discount of num__10.0 on num__130 . profit = num__117 - num__100 num__17.0 answer e <eor> e <eos> |
e |
percent__100.0__17.0__ |
percent__100.0__17.0__ |
| ram sold two bicycles each for rs . num__990 . if he made num__10.0 profit on the first and num__10.0 loss on the second what is the total cost of both bicycles ? <o> a ) num__2000 <o> b ) num__2388 <o> c ) num__2798 <o> d ) num__2666 <o> e ) num__1081 |
( num__10 * num__10 ) / num__100 = num__1.0 loss num__100 - - - num__99 ? - - - num__1980 = > rs . num__2000 answer : a <eor> a <eos> |
a |
percent__10.0__990.0__ percent__100.0__2000.0__ |
percent__10.0__990.0__ percent__100.0__2000.0__ |
| which of the following is closest to ( - num__0.833333333333 ) ^ num__199 ? <o> a ) - num__1 <o> b ) - num__0.5 <o> c ) num__0 <o> d ) num__1 <o> e ) num__2 |
( - num__0.833333333333 ) ^ num__4 = num__0.48225308642 which is already less than num__0.5 . for larger exponents the expression will get closer and closer to zero . the answer is c . <eor> c <eos> |
c |
round_down__0.8333__ |
round_down__0.8333__ |
| a can finish a piece of work in num__5 days . b can do it in num__10 days . they work together for two days and then a goes away . in how many days will b finish the work ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__76 <o> e ) num__7 |
num__0.4 + ( num__2 + x ) / num__10 = num__1 = > x = num__4 days \ answer : a <eor> a <eos> |
a |
multiply__5.0__0.4__ subtract__5.0__1.0__ round__4.0__ |
divide__10.0__5.0__ subtract__5.0__1.0__ divide__4.0__1.0__ |
| a person travels equal distances with speeds of num__3 km / hr num__4 km / hr and num__5 km / hr and takes a total time of num__45 minutes . the total distance is ? <o> a ) num__1 km <o> b ) num__2.87 km <o> c ) num__3 km <o> d ) num__4 km <o> e ) num__5 km |
b num__3 km let the total distance be num__3 x km . then x / num__3 + x / num__4 + x / num__5 = num__0.75 num__47 x / num__60 = num__0.75 = > x = num__0.95 . total distance = num__3 * num__0.95 = num__2.87 km . <eor> b <eos> |
b |
divide__3.0__4.0__ hour_to_min_conversion__ round__2.87__ |
divide__3.0__4.0__ divide__45.0__0.75__ round__2.87__ |
| a certain social security recipient will receive an annual benefit of $ num__12000 provided he has annual earnings of $ num__9360 or less but the benefit will be reduced by $ num__1 for every $ num__3 of annual earnings over $ num__9360 . what amount of total annual earnings would result in a num__45 percent reduction in the recipient ' s annual social security benefit ? ( assume social security benefits are not counted as part of annual earnings . ) <o> a ) $ num__15360 <o> b ) $ num__17360 <o> c ) $ num__18000 <o> d ) $ num__21360 <o> e ) $ num__29 |
160 |
for every $ num__3 earn above $ num__9360 the recipient loses $ num__1 of benefit . or for every $ num__1 loss in the benefit the recipient earns $ num__3 above $ num__9360 if earning is ; num__9360 + num__3 x benefit = num__12000 - x or the vice versa if benefit is num__12000 - x the earning becomes num__9360 + num__3 x he lost num__50.0 of the benefit ; benefit received = num__12000 - num__0.45 * num__12000 = num__12000 - num__5400 x = num__6600 earning becomes num__9360 + num__3 x = num__9360 + num__3 * num__6600 = num__29160 ans : e <eor> e <eos> |
e |
e |
| if the range of the set of numbers { num__150 num__90 num__125 num__110 num__170 num__155 x num__100 num__140 } is num__90 which of the following could be x ? <o> a ) num__80 <o> b ) num__95 <o> c ) num__110 <o> d ) num__125 <o> e ) num__140 |
the range of the other num__8 numbers is num__170 - num__90 = num__80 so x must be either the smallest number or the largest number in the set . then x = num__170 - num__90 = num__80 or x = num__90 + num__90 = num__180 the answer is a . <eor> a <eos> |
a |
subtract__170.0__90.0__ add__100.0__80.0__ subtract__170.0__90.0__ |
subtract__170.0__90.0__ add__100.0__80.0__ subtract__170.0__90.0__ |
| age of a father and his son is in the ratio of num__5 : num__2 . after two years the ratio becomes num__19 : num__8 . what are the present ages of father and son respectively ? <o> a ) num__45 and num__12 <o> b ) num__35 and num__12 <o> c ) num__40 and num__16 <o> d ) num__45 and num__18 <o> e ) num__55 and num__22 |
let present ages of father and son be num__5 x : num__2 x . after two years : ( num__5 x + num__2 ) / ( num__2 x + num__2 ) = num__2.375 num__40 x + num__16 = num__38 x + num__38 = > num__2 x = num__22 = > x = num__11 . father = num__5 x = num__5 x num__11 = num__55 ; son = num__2 x = num__2 x num__11 = num__22 option e <eor> e <eos> |
e |
divide__19.0__8.0__ multiply__5.0__8.0__ multiply__2.0__8.0__ multiply__2.0__19.0__ subtract__38.0__16.0__ subtract__19.0__8.0__ multiply__5.0__11.0__ multiply__5.0__11.0__ |
divide__19.0__8.0__ multiply__5.0__8.0__ multiply__2.0__8.0__ multiply__2.0__19.0__ subtract__38.0__16.0__ divide__22.0__2.0__ multiply__5.0__11.0__ multiply__5.0__11.0__ |
| a goods train runs at a speed of num__72 kmph and crosses a num__250 m long platform in num__26 seconds . what is the length of the goods train ? <o> a ) num__230 m <o> b ) num__240 m <o> c ) num__260 m <o> d ) num__270 m <o> e ) num__250 m |
s = num__250 + x / t num__72 * num__0.277777777778 = num__250 + x / num__26 x = num__270 answer : d <eor> d <eos> |
d |
round__270.0__ |
round__270.0__ |
| find the odd man out num__1 num__514 num__3050 num__5591 . <o> a ) num__5 <o> b ) num__50 <o> c ) num__55 <o> d ) num__91 <o> e ) num__1 |
num__50 is the odd man . . diff between sucessive no . s is a perfect square answer : b <eor> b <eos> |
b |
multiply__1.0__50.0__ |
multiply__1.0__50.0__ |
| the edge of three cubes of metal is num__3 dm num__4 dm and num__5 dm . they are melted and formed into a single cube . find the edge of the new cube ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__12 <o> e ) num__14 |
explanation : num__33 + num__43 + num__53 = a num__3 = > a = num__6 a ) <eor> a <eos> |
a |
triangle_area__3.0__4.0__ triangle_area__3.0__4.0__ |
triangle_area__3.0__4.0__ triangle_area__3.0__4.0__ |
| a sum of money invested at c . i . amounts to rs . num__800 in num__3 years to rs . num__870 in num__4 years . the rate of interest per annum is ? <o> a ) num__2 num__0.5 % <o> b ) num__8 num__0.75 % <o> c ) num__5.0 <o> d ) num__6 num__0.666666666667 % <o> e ) num__6 % |
s . i . on rs . num__800 for num__1 year = ( num__870 - num__800 ) = rs . num__70 rate = ( num__100 * num__70 ) / ( num__800 * num__1 ) = num__8 num__0.75 % answer : b <eor> b <eos> |
b |
percent__1.0__800.0__ percent__1.0__800.0__ |
percent__1.0__800.0__ percent__1.0__800.0__ |
| a waitress ' s income consists of her salary and tips . during one week her tips were num__0.5 of her salary . what fraction of her income for the week came from tips ? <o> a ) num__0.111111111111 <o> b ) num__0.166666666667 <o> c ) num__0.333333333333 <o> d ) num__0.444444444444 <o> e ) num__0.555555555556 |
her tips were num__0.5 of her salary . let ' s say her salary = $ num__4 this mean her tips = ( num__0.5 ) ( $ num__4 ) = $ num__2 so her total income = $ num__4 + $ num__2 = $ num__6 what fraction of her income for the week came from tips $ num__2 / $ num__6 = num__0.333333333333 = c <eor> c <eos> |
c |
reverse__0.5__ add__2.0__4.0__ divide__2.0__6.0__ divide__2.0__6.0__ |
reverse__0.5__ add__2.0__4.0__ divide__2.0__6.0__ divide__2.0__6.0__ |
| a father tells his son ` ` i was of your present age when you were born ' ' . if the father is num__36 now how old was the boy five years back ? <o> a ) num__13 <o> b ) num__15 <o> c ) num__17 <o> d ) num__20 <o> e ) num__22 |
f is fathers age and b is boys age f - b = b f = num__2 b num__2 b = num__36 b = num__18 boys age num__5 years back would be num__13 answer : a <eor> a <eos> |
a |
divide__36.0__2.0__ subtract__18.0__5.0__ subtract__18.0__5.0__ |
divide__36.0__2.0__ subtract__18.0__5.0__ subtract__18.0__5.0__ |
| the c . p of num__10 pens is equal to the s . p of num__12 pens . find his gain % or loss % ? <o> a ) num__16 num__0.222222222222 % <o> b ) num__16 num__1.33333333333 % <o> c ) num__16 num__0.666666666667 % <o> d ) num__16 num__1.0 % <o> e ) num__12 num__0.666666666667 % |
num__10 cp = num__12 sp num__12 - - - num__2 cp loss num__100 - - - ? = > num__16 num__0.666666666667 % answer : c <eor> c <eos> |
c |
percent__100.0__16.0__ |
percent__100.0__16.0__ |
| a car travels from b at a speed of num__20 km / hr . the bus travel starts from a at a time of num__6 a . m . there is a bus for every half an hour interval . the car starts at num__12 noon . each bus travels at a speed of num__25 km / hr . distance between a and b is num__100 km . during its journey the number of buses that the car encounter is ? <o> a ) num__15 <o> b ) num__16 <o> c ) num__17 <o> d ) num__18 <o> e ) num__19 |
total journey = num__100 km . the bus travel starts from a at a time of num__6 a . m . and each bus travels at a speed of num__25 km / hr so time required for the bus to complete the journey = num__4 hrs . so the bus starts at num__8 : num__30 will meet with the car moving in opposite direction definitely . car travels from b at a speed of num__20 km / hr so time of arrival of car at a is num__5 p . m . it means the last bus which will meet with the car is that starts at num__4 : num__30 p . m . so total bus ranges from num__8 : num__30 p . m . to num__4 : num__30 p . m . = num__17 answer : c <eor> c <eos> |
c |
divide__100.0__25.0__ subtract__20.0__12.0__ divide__20.0__4.0__ add__12.0__5.0__ round__17.0__ |
divide__100.0__25.0__ subtract__20.0__12.0__ divide__20.0__4.0__ add__12.0__5.0__ round__17.0__ |
| john makes $ num__40 a week from his job . he earns a raise and now makes $ num__60 a week . what is the % increase ? <o> a ) num__16.0 <o> b ) num__16.66 <o> c ) num__18.0 <o> d ) num__21.0 <o> e ) num__50 % |
increase = ( num__0.5 ) * num__100 = num__50.0 . e <eor> e <eos> |
e |
percent__50.0__100.0__ |
percent__50.0__100.0__ |
| if num__14 lions can kill num__14 deers in num__14 minutes how long will it take num__100 lions to kill num__100 deers ? <o> a ) num__1 minutes <o> b ) num__14 minute <o> c ) num__100 minutes <o> d ) num__10000 minutes <o> e ) num__1000 minutes |
we can try the logic of time and work our work is to kill the deers so num__14 ( lions ) * num__14 ( min ) / num__14 ( deers ) = num__100 ( lions ) * x ( min ) / num__100 ( deers ) hence answer is x = num__14 answer : b <eor> b <eos> |
b |
round__14.0__ |
round__14.0__ |
| the remainder obtained by the division of two numbers by a common divisor are num__33 and num__47 respectively . however when the sum of these two numbers is divided by the same divisor the remainder obtained is only num__30 . what is the divisor ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__40 <o> d ) num__50 <o> e ) num__65 |
remainders are num__33 & num__47 . so directly rule out the possibility of the divisor being num__20 num__30 & num__40 option a b c ruled out consider option d - num__50 num__50 + num__33 = num__83 & num__50 + num__47 = num__97 num__83 + num__97 = num__180 num__3.6 gives remainder = num__30 answer = d <eor> d <eos> |
d |
add__30.0__20.0__ add__33.0__50.0__ add__47.0__50.0__ add__97.0__83.0__ divide__180.0__50.0__ add__30.0__20.0__ |
add__30.0__20.0__ add__33.0__50.0__ add__47.0__50.0__ add__97.0__83.0__ divide__180.0__50.0__ add__30.0__20.0__ |
| in a camp there is a meal for num__120 men or num__200 children . if num__150 children have taken the meal how many men will be catered to with remaining meal ? <o> a ) num__37 <o> b ) num__28 <o> c ) num__30 <o> d ) num__19 <o> e ) num__12 |
explanation : there is a meal for num__200 children . num__150 children have taken the meal . remaining meal is to be catered to num__50 children . now num__200 children = num__120 men . num__50 children = { \ color { blue } \ left ( \ frac { num__120 } { num__200 } \ right ) \ times num__50 } = num__30 men answer : c <eor> c <eos> |
c |
subtract__200.0__150.0__ subtract__150.0__120.0__ subtract__150.0__120.0__ |
subtract__200.0__150.0__ subtract__150.0__120.0__ subtract__150.0__120.0__ |
| a dragon has num__100 heads . a soldier can cut off num__15 num__17 num__20 or num__5 heads respectively with one blow of his sword . in each of these cases num__24 num__2 num__14 num__17 new heads grow on its shoulders . if all heads are cut off the dragon dies . can the dragon ever die ? <o> a ) num__61 <o> b ) num__3 <o> c ) num__5 <o> d ) num__6 <o> e ) num__1 |
here the difference in the cutoff heads and newly grown heads is num__3 . num__100 is not divisible by num__3 . the dragon never die . answer : b <eor> b <eos> |
b |
choose__3.0__2.0__ |
choose__3.0__2.0__ |
| find the next number in the series num__1 num__6 num__13 num__22 num__33 . . . . . . <o> a ) num__44 <o> b ) num__45 <o> c ) num__46 <o> d ) num__47 <o> e ) num__48 |
num__1 + num__5 = num__6 num__6 + num__7 = num__13 num__13 + num__9 = num__22 num__22 + num__11 = num__33 num__33 + num__13 = num__46 answer : c <eor> c <eos> |
c |
subtract__6.0__1.0__ add__1.0__6.0__ subtract__22.0__13.0__ add__6.0__5.0__ add__13.0__33.0__ multiply__1.0__46.0__ |
subtract__6.0__1.0__ add__1.0__6.0__ subtract__22.0__13.0__ add__6.0__5.0__ add__13.0__33.0__ add__13.0__33.0__ |
| which of the following is between num__0.111111111111 and num__0.238095238095 ? <o> a ) num__0.153846153846 <o> b ) num__0.375 <o> c ) num__0.5 <o> d ) num__0.625 <o> e ) num__0.777777777778 |
i see this as a poe ( process of elimination ) and ballparking ( estimation ) question . not sure if this is the most efficient but it worked : num__1 ) i estimate num__0.111111111111 to be ~ num__0.1 + ( approximately slightly greater than num__0.75 ) num__2 ) i estimate num__0.238095238095 to be ~ num__0.2 + ( approximately slightly greater than num__0.2 ) num__3 ) so now i ' m looking for an answer choice that is . num__1 < x < . num__2 answer a - num__0.153846153846 ~ . . num__15 which is greater than . num__1 and less than . num__2 <eor> a <eos> |
a |
multiply__0.1__2.0__ add__1.0__2.0__ divide__3.0__0.2__ multiply__1.0__0.1538__ |
multiply__0.1__2.0__ add__1.0__2.0__ divide__3.0__0.2__ multiply__1.0__0.1538__ |
| what will come in place of the question mark ( ? ) in the following equation ? num__25 ( num__7.5 ) × num__5 ( num__2.5 ) ÷ num__125 ( num__1.5 ) = num__5 ? <o> a ) num__16 <o> b ) num__17.5 <o> c ) num__8.5 <o> d ) num__13 <o> e ) none of these |
num__25 ( num__7.5 ) × num__5 ( num__2.5 ) ÷ num__125 ( num__1.5 ) = num__5 ? or num__5 ( num__2 × num__7.5 ) × num__5 ( num__2.5 ) ÷ num__5 ( num__3 × num__1.5 ) = num__5 ? or num__5 ( num__15 ) × num__5 ( num__2.5 ) × num__1 ⁄ num__54.5 = num__5 ? or num__5 ( num__13 ) = num__5 ? or ? = num__13 answer d <eor> d <eos> |
d |
round_down__2.5__ divide__7.5__2.5__ multiply__7.5__2.0__ round_down__1.5__ subtract__15.0__2.0__ multiply__1.0__13.0__ |
round_down__2.5__ divide__7.5__2.5__ multiply__7.5__2.0__ round_down__1.5__ subtract__15.0__2.0__ multiply__1.0__13.0__ |
| two numbers are in the ratio num__3 : num__5 . if num__9 be subtracted from each they are in the ratio of num__9 : num__17 . the first number is ? <o> a ) num__36 <o> b ) num__67 <o> c ) num__87 <o> d ) num__56 <o> e ) num__51 |
( num__3 x - num__9 ) : ( num__5 x - num__9 ) = num__9 : num__17 x = num__12 = > num__3 x = num__36 answer : a <eor> a <eos> |
a |
add__3.0__9.0__ multiply__3.0__12.0__ multiply__3.0__12.0__ |
subtract__17.0__5.0__ multiply__3.0__12.0__ multiply__3.0__12.0__ |
| for num__1 rs num__6 p interest wat will be for num__1000 rs ? <o> a ) num__45 rs <o> b ) num__50 rs <o> c ) num__40 rs <o> d ) num__30 rs <o> e ) num__60 rs |
for num__1 rs num__4 p interest for num__1000 rs x x = num__1000.0 * num__6 p = = > num__6000 paise to express in rs num__60.0 = num__60 rs answer : e <eor> e <eos> |
e |
percent__1.0__6000.0__ percent__1.0__6000.0__ |
percent__1.0__6000.0__ percent__1.0__6000.0__ |
| three numbers are in the ratio num__1 : num__2 : num__3 and their h . c . f is num__12 . the numbers are <o> a ) num__4 num__8 num__12 <o> b ) num__5 num__10 num__15 <o> c ) num__10 num__20 num__30 <o> d ) num__12 num__24 num__36 <o> e ) none of these |
explanation : let the required numbers be x num__2 x num__3 x . then their h . c . f = x . so x = num__12 the numbers are num__12 num__24 num__36 . answer : d <eor> d <eos> |
d |
multiply__2.0__12.0__ multiply__3.0__12.0__ multiply__1.0__12.0__ |
multiply__2.0__12.0__ multiply__3.0__12.0__ multiply__1.0__12.0__ |
| in a barrel of juice there is num__30 liters ; in a barrel of beer there are num__80 liters . if the price ratio between barrels of juice to a barrel of beer is num__3 : num__4 what is the price ratio between one liter of juice and one liter of beer ? <o> a ) num__3 : num__2 . <o> b ) num__2 : num__1 . <o> c ) num__3 : num__1 . <o> d ) num__4 : num__3 . <o> e ) num__3 : num__4 |
price of num__30 l juice = num__3 x num__1 l = num__3 x / num__30 price of num__80 l beer = num__4 x num__1 l = num__4 x / num__80 ratio of num__1 l price = num__3 x / num__7.5 x / num__80 = num__2 : num__1 b is the answer <eor> b <eos> |
b |
subtract__4.0__3.0__ divide__30.0__4.0__ subtract__3.0__1.0__ subtract__3.0__1.0__ |
subtract__4.0__3.0__ divide__30.0__4.0__ subtract__3.0__1.0__ divide__4.0__2.0__ |
| ayesha ' s father was num__38 years of age when she was born while her mother was num__36 years old when her brother four years younger to her was born . what is the difference between the ages of her parents ? <o> a ) num__2 years <o> b ) num__4 years <o> c ) num__6 years <o> d ) num__7 years <o> e ) num__6 years |
explanation : mother ' s age when ayesha ' s brother was born = num__36 years . father ' s age when ayesha ' s brother was born = ( num__38 + num__4 ) = num__42 years . required difference = ( num__42 - num__36 ) = num__6 years . answer : option c <eor> c <eos> |
c |
add__38.0__4.0__ subtract__42.0__36.0__ divide__36.0__6.0__ |
add__38.0__4.0__ subtract__42.0__36.0__ subtract__42.0__36.0__ |
| how many positive integers will divide evenly into num__410 ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__8 <o> d ) num__12 <o> e ) num__16 |
the question is asking how many factors num__410 has . num__410 = num__2 * num__5 * num__41 the number of factors is num__2 ^ num__3 = num__8 the answer is c . <eor> c <eos> |
c |
subtract__5.0__2.0__ add__3.0__5.0__ add__3.0__5.0__ |
subtract__5.0__2.0__ add__3.0__5.0__ add__3.0__5.0__ |
| if the average ( mean ) of num__9 positive temperatures is x degrees fahrenheit then the sum of the num__3 greatest of these temperatures in degrees fahrenheit could be <o> a ) num__8 x <o> b ) num__4 x <o> c ) num__5 x / num__3 <o> d ) num__3 x / num__2 <o> e ) num__3 x / num__5 |
let the num__5 numbers be num__1 num__23 num__45 . . num__9 ( since no restrictions are given ) . there mean is num__4 ( x ) . now the sum of greatest three would be num__7 + num__8 + num__9 = num__24 so the answer has to be num__8 x . . . . that is option a <eor> a <eos> |
a |
multiply__9.0__5.0__ subtract__9.0__5.0__ add__3.0__4.0__ subtract__9.0__1.0__ multiply__3.0__8.0__ subtract__9.0__1.0__ |
multiply__9.0__5.0__ add__3.0__1.0__ add__3.0__4.0__ add__3.0__5.0__ add__1.0__23.0__ add__3.0__5.0__ |
| factor : num__2 x num__4 y num__3 â € “ num__32 y num__3 <o> a ) a ) num__3 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__2 ) <o> b ) b ) num__2 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__2 ) <o> c ) c ) num__3 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__3 ) <o> d ) d ) num__3 y num__3 ( x num__2 + num__4 ) ( x + num__3 ) ( x - num__2 ) <o> e ) e ) num__3 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__4 ) |
num__2 x num__4 y num__3 â € “ num__32 y num__3 . = num__2 y num__3 ( x num__4 â € “ num__16 ) . = num__2 y num__3 [ ( x num__2 ) num__2 - num__42 ] . = num__2 y num__3 ( x num__2 + num__4 ) ( x num__2 - num__4 ) . = num__2 y num__3 ( x num__2 + num__4 ) ( x num__2 - num__22 ) . = num__2 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__2 ) . answer : ( b ) num__2 y num__3 ( x num__2 + num__4 ) ( x + num__2 ) ( x - num__2 ) <eor> b <eos> |
b |
divide__32.0__2.0__ gcd__2.0__4.0__ |
divide__32.0__2.0__ subtract__4.0__2.0__ |
| a runner runs the num__40 miles from marathon to athens at a constant speed . halfway through the run she injures her foot and continues to run at half her previous speed . if the second half takes her num__11 hours longer than the first half how many hours did it take the runner to run the second half ? <o> a ) num__14 <o> b ) num__15 <o> c ) num__18 <o> d ) num__22 <o> e ) num__33 |
the runner runs the first num__20 miles at speed v and the second num__20 miles at speed v / num__2 . the time t num__2 to run the second half must be twice the time t num__1 to run the first half . t num__2 = num__2 * t num__1 = t num__1 + num__11 t num__1 = num__11 and so t num__2 = num__22 . the answer is d . <eor> d <eos> |
d |
divide__40.0__20.0__ multiply__11.0__2.0__ round__22.0__ |
divide__40.0__20.0__ multiply__11.0__2.0__ multiply__11.0__2.0__ |
| on a scale that measures the intensity of a certain phenomenon a reading of f + num__1 corresponds to an intensity that is num__10 times the intensity corresponding to a reading of f . on that scale the intensity corresponding to a reading of num__8 is how many times as great as the intensity corresponding to a reading of num__3 ? <o> a ) num__5 <o> b ) num__50 <o> c ) num__10 ^ num__5 <o> d ) num__5 ^ num__10 <o> e ) num__8 ^ num__10 - num__3 ^ num__10 |
to solve this problem we need to examine the information in the first sentence . we are told that “ a reading of f + num__1 corresponds to an intensity that is num__10 times the intensity corresponding to a reading of f . ” let ’ s practice this idea with some real numbers . let ’ s say f is num__2 . this means that f + num__1 = num__3 . with the information we were given we can say that a reading of num__3 is ten times as great as the intensity of a reading of num__2 . furthermore we can say that a reading of num__4 is actually num__10 x num__10 = num__10 ^ num__2 times as great as the intensity of a reading of num__2 . increasing one more unit we can say that a reading of num__5 is num__10 x num__10 x num__10 = num__10 ^ num__3 times as great as the intensity of a reading of num__2 . we have found a pattern which can be applied to the problem presented in the stem : num__3 is “ one ” unit away from num__2 and thus a reading of num__3 is num__10 ^ num__1 times as great as the intensity of a reading of num__2 . num__4 is “ two ” units away from num__2 and thus a reading of num__4 is num__10 ^ num__2 times as great as the intensity of a reading of num__2 . num__5 is “ three ” units away from num__2 and thus a reading of num__5 is num__10 ^ num__3 times as great as the intensity of a measure of num__2 . we can use this pattern to easily answer the question . here we are being asked for the number of times the intensity corresponding to a reading of num__8 is as great as the intensity corresponding to a reading of num__3 . because num__8 is num__5 units greater than num__3 a reading of num__8 is num__10 ^ num__5 times as great as the intensity corresponding to a reading of num__3 . answer c . <eor> c <eos> |
c |
subtract__10.0__8.0__ add__1.0__3.0__ add__1.0__4.0__ multiply__1.0__10.0__ |
subtract__10.0__8.0__ add__1.0__3.0__ add__1.0__4.0__ add__8.0__2.0__ |
| find the expenditure on digging a well num__14 m deep and of num__3 m diameter at rs . num__17 per cubic meter ? <o> a ) num__2998 <o> b ) num__2799 <o> c ) num__2890 <o> d ) num__1683 <o> e ) num__2780 |
num__3.14285714286 * num__14 * num__1.5 * num__1.5 = num__99 m num__2 num__99 * num__17 = num__1683 answer : d <eor> d <eos> |
d |
divide__3.0__1.5__ multiply__17.0__99.0__ round__1683.0__ |
divide__3.0__1.5__ multiply__17.0__99.0__ multiply__17.0__99.0__ |
| in a certain sequence of numbers a num__1 a num__2 a num__3 . . . an the average ( arithmetic mean ) of the first m consecutive terms starting with a num__1 is m for any positive integer m . if a num__1 = num__1 what is a num__14 ? <o> a ) num__100 <o> b ) num__55 <o> c ) num__27 <o> d ) num__19 <o> e ) num__1 |
as stated above ( a num__1 + a num__2 + a num__3 . . . . . . + a num__10 ) / num__10 = num__10 therefore a num__1 + a num__2 + a num__3 . . . . . . . a num__14 = num__196 ( num__1 ) using the same logic we got a num__1 + a num__2 + a num__3 . . . . . . . . . . + a num__13 = num__169 ( num__2 ) ( num__2 ) - ( num__1 ) we got a num__14 = num__27 c <eor> c <eos> |
c |
add__3.0__10.0__ add__14.0__13.0__ multiply__1.0__27.0__ |
add__3.0__10.0__ add__14.0__13.0__ add__14.0__13.0__ |
| if s and t are positive integers and num__3 is not a factor of t then t may be which of the following ? <o> a ) ( s − num__1 ) · s · ( s + num__1 ) <o> b ) ( s − num__3 ) · ( s − num__1 ) · ( s + num__1 ) <o> c ) ( s − num__2 ) · s · ( s + num__2 ) <o> d ) ( s − num__1 ) · s · ( s + num__2 ) <o> e ) ( s − num__3 ) · ( s + num__1 ) · ( s + num__2 ) |
plugged in values : let s = num__5 a . ( s − num__1 ) · s · ( s + num__1 ) - - > divisible by num__3 b . ( s − num__3 ) · ( s − num__1 ) · ( s + num__1 ) - - > divisible by num__3 c . ( s − num__2 ) · s · ( s + num__2 ) - - > divisible by num__3 d . ( s − num__1 ) · s · ( s + num__2 ) - - > not divisible . hence the answer . <eor> d <eos> |
d |
subtract__3.0__1.0__ reverse__1.0__ |
subtract__3.0__1.0__ subtract__3.0__2.0__ |
| a b and c can do a piece of work in num__7 days num__14 days and num__28 days respectively . how long will they taken if all the three work together ? <o> a ) num__3 days <o> b ) num__4 days <o> c ) num__5 days <o> d ) num__6 days <o> e ) num__7 days |
explanation : num__0.142857142857 + num__0.0714285714286 + num__0.0357142857143 = num__0.25 = num__0.25 = > num__4 days answer : b <eor> b <eos> |
b |
divide__7.0__28.0__ divide__28.0__7.0__ round__4.0__ |
divide__7.0__28.0__ divide__28.0__7.0__ round__4.0__ |
| in triangle pqr the angle q = num__90 degree pq = num__7 cm qr = num__8 cm . x is a variable point on pq . the line through x parallel to qr intersects pr at y and the line through y parallel to pq intersects qr at z . find the least possible length of xz <o> a ) num__5.6 cm <o> b ) num__2.4 cm <o> c ) num__4.8 cm <o> d ) num__2.16 cm <o> e ) num__3.2 cm |
look at the diagram below : now in case when qy is perpendicular to pr two right triangles pqr and pqy are similar : qy : qp = qr : pr - - > qy : num__7 = num__8 : num__10 - - > qy = num__5.6 . answer : a . <eor> a <eos> |
a |
round__5.6__ |
round__5.6__ |
| what is y if y + num__2 x = num__10 and x = num__3 ? <o> a ) a ) num__10 <o> b ) b ) num__8 <o> c ) c ) num__6 <o> d ) d ) num__4 <o> e ) e ) num__2 |
y = num__10 - num__2 x y = num__10 - num__8 . y = num__2 answer : d <eor> d <eos> |
d |
subtract__10.0__2.0__ divide__8.0__2.0__ |
subtract__10.0__2.0__ divide__8.0__2.0__ |
| the h . c . f . of two numbers is num__23 and the other two factors of their l . c . m . are num__13 and num__14 . the larger of the two numbers is : <o> a ) num__276 <o> b ) num__299 <o> c ) num__322 <o> d ) num__345 <o> e ) num__367 |
the numbers are ( num__23 x num__13 ) and ( num__23 x num__14 ) . larger number = ( num__23 x num__14 ) = num__322 . answer : option c <eor> c <eos> |
c |
multiply__23.0__14.0__ multiply__23.0__14.0__ |
multiply__23.0__14.0__ multiply__23.0__14.0__ |
| find the average of all numbers between num__1 and num__13 which are divisible by num__2 <o> a ) num__15 <o> b ) num__21 <o> c ) num__25 <o> d ) num__30 <o> e ) num__35 |
explanation : average = ( num__2 + num__4 + num__6 + num__8 + num__10 + num__12 ) / num__2 = num__21.0 = num__21 option b <eor> b <eos> |
b |
add__2.0__4.0__ multiply__2.0__4.0__ add__2.0__8.0__ subtract__13.0__1.0__ add__13.0__8.0__ multiply__1.0__21.0__ |
add__2.0__4.0__ add__2.0__6.0__ add__2.0__8.0__ add__2.0__10.0__ add__13.0__8.0__ add__13.0__8.0__ |
| if x is equal to the sum of the integers from num__30 to num__40 inclusive and y is the number of even integers from num__30 to num__40 inclusive what is the value of x + y ? <o> a ) num__171 <o> b ) num__281 <o> c ) num__391 <o> d ) num__591 <o> e ) num__601 |
sum s = n / num__2 { num__2 a + ( n - num__1 ) d } = num__5.5 { num__2 * num__30 + ( num__11 - num__1 ) * num__1 } = num__11 * num__35 = num__385 = x number of even number = ( num__40 - num__30 ) / num__2 + num__1 = num__6 = y x + y = num__385 + num__6 = num__391 c <eor> c <eos> |
c |
multiply__5.5__2.0__ multiply__35.0__11.0__ add__6.0__385.0__ multiply__1.0__391.0__ |
multiply__5.5__2.0__ multiply__35.0__11.0__ add__6.0__385.0__ multiply__1.0__391.0__ |
| there are num__24 students in a seventh grade class . they decided to plant birches and roses at the school ' s backyard . while each girl planted num__3 roses every three boys planted num__1 birch . by the end of the day they planted num__2424 plants . how many birches were planted ? <o> a ) num__2 <o> b ) num__5 <o> c ) num__8 <o> d ) num__6 <o> e ) num__4 |
let x be the number of roses . then the number of birches is num__24 − x and the number of boys is num__3 × ( num__24 − x ) . if each girl planted num__3 roses there are x num__3 girls in the class . we know that there are num__24 students in the class . therefore x num__3 + num__3 ( num__24 − x ) = num__24 x + num__9 ( num__24 − x ) = num__3 ⋅ num__24 x + num__216 − num__9 x = num__72 num__216 − num__72 = num__8 x num__1448 = x num__1 x = num__18 so students planted num__18 roses and num__24 - x = num__24 - num__18 = num__6 birches . correct answer is d ) num__6 <eor> d <eos> |
d |
multiply__24.0__9.0__ multiply__24.0__3.0__ divide__24.0__3.0__ subtract__24.0__18.0__ round__6.0__ |
multiply__24.0__9.0__ multiply__24.0__3.0__ subtract__9.0__1.0__ subtract__24.0__18.0__ subtract__24.0__18.0__ |
| a and b started a partnership business investing capital in the ratio of num__3 : num__5 . c joined in the partnership after six months with an amount equal to that of b . at the end of one year the profit should be distributed among a b and c in - - - proportion . <o> a ) num__10 : num__5 : num__4 <o> b ) num__5 : num__3 : num__4 <o> c ) num__3 : num__4 : num__5 <o> d ) num__6 : num__10 : num__5 <o> e ) none of these |
explanation : initial investment capital ratio of a and b = num__3 : num__5 hence we can assume that initial capital of a and b are num__3 x and num__5 x respectively . amount that c invest after num__6 months = num__5 x ( since it is equal to b ' s investment ) ratio in which profit should be distributed after num__1 year = num__3 x * num__12 : num__5 x * num__12 : num__5 x * num__6 = > num__3 * num__12 : num__5 * num__12 : num__5 * num__6 = num__6 : num__10 : num__5 . answer : option d <eor> d <eos> |
d |
subtract__6.0__5.0__ add__5.0__1.0__ |
subtract__6.0__5.0__ multiply__1.0__6.0__ |
| a dealer purchases num__15 articles for rs . num__25 and sells num__12 articles for rs . num__36 . find the profit percentage ? <o> a ) num__80.0 <o> b ) num__50.0 <o> c ) num__59.0 <o> d ) num__40.0 <o> e ) num__53 % |
l . c . m of num__15 and num__12 = num__60 cp of num__60 articles = rs . num__100 ( num__25 * num__4 ) sp of num__60 articles = rs . num__180 ( num__36 * num__5 ) profit percentage = ( num__180 - num__100 ) / num__100 * num__100 = num__80.0 answer : a <eor> a <eos> |
a |
percent__100.0__80.0__ |
percent__100.0__80.0__ |
| in a fuel station the service costs $ num__1.75 per car every liter of fuel costs num__0.65 $ . assuming that a company owns num__12 cars and that every fuel tank contains num__60 liters and they are all empty how much money total will it cost to fuel all cars ? <o> a ) num__320 $ <o> b ) num__380 $ <o> c ) num__420 $ <o> d ) num__450 $ <o> e ) num__489 $ |
total cost = ( num__1.75 * num__12 ) + ( num__0.65 * num__12 * num__60 ) = num__21 + num__468 = > num__489 hence answer will be ( e ) num__489 <eor> e <eos> |
e |
multiply__1.75__12.0__ add__468.0__21.0__ add__468.0__21.0__ |
multiply__1.75__12.0__ add__468.0__21.0__ add__468.0__21.0__ |
| if x and y are positive integers which of the following can not be the greatest common divisor of num__35 x and num__20 y ? <o> a ) num__5 <o> b ) num__5 ( x – y ) <o> c ) num__20 x <o> d ) num__20 y <o> e ) num__35 x |
looking at the list of answer choices there are some that we can quickly eliminate . . . if x = num__1 y = num__1 then we have num__35 and num__20 so the gcd = num__5 eliminate a . x = num__20 y = num__1 then we have num__700 and num__20 so the gcd = num__20 eliminate d . x = num__1 y = num__35 then we have num__35 and num__700 so the gcd = num__35 eliminate e . with the remaining num__2 answers we have to think a little more . to get a simple example of num__5 ( x - y ) to be the gcd we probably need num__35 x to be odd if . . . x = num__3 y = num__2 then we have num__105 and num__40 so the gcd = num__5 num__5 ( x - y ) = num__5 ( num__3 - num__2 ) = num__5 this can also be the gcd eliminate b . answer : c <eor> c <eos> |
c |
multiply__35.0__20.0__ add__1.0__2.0__ multiply__35.0__3.0__ add__35.0__5.0__ multiply__20.0__1.0__ |
multiply__35.0__20.0__ subtract__5.0__2.0__ multiply__35.0__3.0__ add__35.0__5.0__ subtract__40.0__20.0__ |
| the sum of three numbers is num__98 . if the ratio of the first to the second is num__2 : num__3 and that of the second to the third is num__5 : num__8 then the second number is : <o> a ) num__20 <o> b ) num__30 <o> c ) num__38 <o> d ) num__48 <o> e ) none of these |
a : b = num__2 : num__3 = num__2 × num__5 : num__3 × num__5 = num__10 : num__15 and b : c = num__5 : num__8 = num__5 × num__3 : num__8 × num__3 = num__15 : num__24 therefore a : b : c = num__10 : num__15 : num__24 ∴ a : b : c = num__10 : num__15 : num__24 let the number be num__10 x num__15 x and num__24 x . then num__10 x + num__15 x + num__24 x = num__98 or num__49 x = num__98 or x = num__2 ⇒ second number = num__15 x = num__15 × num__2 = num__30 answer b <eor> b <eos> |
b |
multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__3.0__8.0__ divide__98.0__2.0__ multiply__2.0__15.0__ multiply__2.0__15.0__ |
add__2.0__8.0__ add__5.0__10.0__ multiply__3.0__8.0__ divide__98.0__2.0__ multiply__2.0__15.0__ multiply__2.0__15.0__ |
| in an election between two candidates one got num__55.0 of the total valid votes num__20.0 of the votes were invalid . if the total number of votes was num__6500 the number of valid votes that the other candidate got was : <o> a ) num__2800 <o> b ) num__2700 <o> c ) num__2900 <o> d ) num__2340 <o> e ) num__2300 |
d number of valid votes = num__80.0 of num__6500 = num__5200 . valid votes polled by other candidate = num__45.0 of num__5200 = ( num__0.45 x num__5200 ) = num__2340 . <eor> d <eos> |
d |
percent__80.0__6500.0__ percent__45.0__5200.0__ percent__45.0__5200.0__ |
percent__80.0__6500.0__ percent__45.0__5200.0__ percent__45.0__5200.0__ |
| how many terminating zeroes r does num__200 ! have ? <o> a ) num__40 <o> b ) num__48 <o> c ) num__49 <o> d ) num__55 <o> e ) num__64 |
you have num__40 multiples of num__5 num__8 of num__25 and num__1 of num__125 . this will give num__49 zeros . c <eor> c <eos> |
c |
divide__200.0__40.0__ divide__40.0__5.0__ divide__200.0__8.0__ multiply__5.0__25.0__ multiply__1.0__49.0__ |
divide__200.0__40.0__ divide__40.0__5.0__ divide__200.0__8.0__ multiply__5.0__25.0__ multiply__1.0__49.0__ |
| num__75 men working num__8 hours per day dig num__50 m deep . how many extra men should be put to dig to a depth of num__70 m working num__6 hours per day ? <o> a ) num__75 <o> b ) num__45 <o> c ) num__55 <o> d ) num__65 <o> e ) num__35 |
( num__75 * num__8 ) / num__50 = ( x * num__6 ) / num__70 = > x = num__140 num__140 â € “ num__65 = num__65 answer : d <eor> d <eos> |
d |
subtract__140.0__75.0__ round__65.0__ |
subtract__140.0__75.0__ round__65.0__ |
| if the sum of two numbers is num__60 and the h . c . f and l . c . m of these numbers are num__3 and num__120 respectively then the sum of the reciprocal of the numbers is equal to : <o> a ) num__0.222222222222 <o> b ) num__0.8 <o> c ) num__0.166666666667 <o> d ) num__1.0 <o> e ) num__0.333333333333 |
let the numbers be a and b . then a + b = num__60 and ab = num__3 * num__120 = num__360 . required sum = num__1 / a + num__1 / b = ( a + b ) / ab = num__0.166666666667 = num__0.166666666667 . answer : c <eor> c <eos> |
c |
multiply__3.0__120.0__ divide__60.0__360.0__ divide__60.0__360.0__ |
multiply__3.0__120.0__ divide__60.0__360.0__ divide__60.0__360.0__ |
| a certain bus driver is paid a regular rate of $ num__15 per hour for any number of hours that does not exceed num__40 hours per week . for any overtime hours worked in excess of num__40 hours per week the bus driver is paid a rate that is num__75.0 higher than his regular rate . if last week the bus driver earned $ num__976 in total compensation how many total hours did he work that week ? <o> a ) num__56 <o> b ) num__51 <o> c ) num__59 <o> d ) num__54 <o> e ) num__52 |
for num__40 hrs = num__40 * num__15 = num__600 excess = num__976 - num__600 = num__376 for extra hours = . num__75 ( num__15 ) = num__11.25 + num__15 = num__26.25 number of extra hrs = num__376 / num__26.25 = num__14.3 = num__14 approx . total hrs = num__40 + num__14 = num__54 answer d <eor> d <eos> |
d |
multiply__15.0__40.0__ subtract__976.0__600.0__ add__15.0__11.25__ round_down__14.3__ add__40.0__14.0__ add__40.0__14.0__ |
multiply__15.0__40.0__ subtract__976.0__600.0__ add__15.0__11.25__ round_down__14.3__ add__40.0__14.0__ add__40.0__14.0__ |
| if the true discount on s sum due num__2 years hence at num__14.0 per annum be rs . num__168 the sum due is ? <o> a ) rs . num__948 <o> b ) rs . num__876 <o> c ) rs . num__768 <o> d ) rs . num__658 <o> e ) none of these |
explanation : p . w . = ( num__100 * t . d ) / r * t = ( num__100 * num__168 ) / num__14 * num__2 = num__600 . sum = ( p . w . + t . d . ) = rs . ( num__600 + num__168 ) = rs . num__768 . answer is c <eor> c <eos> |
c |
percent__100.0__768.0__ |
percent__100.0__768.0__ |
| to fill a tank num__10 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two - fifth of its present ? <o> a ) num__25 <o> b ) num__50 <o> c ) num__75 <o> d ) num__100 <o> e ) num__125 |
let the capacity of num__1 bucket = x . then the capacity of tank = num__10 x . new capacity of bucket = num__0.4 x therefore required number of buckets = ( num__10 x ) / ( num__2 x / num__5 ) = ( num__10 x ) x num__2.5 x = num__25.0 = num__25 answer is a . <eor> a <eos> |
a |
divide__10.0__2.0__ divide__1.0__0.4__ multiply__10.0__2.5__ round__25.0__ |
divide__10.0__2.0__ divide__1.0__0.4__ divide__10.0__0.4__ divide__10.0__0.4__ |
| the youngest of num__4 children has siblings who are num__3 num__5 and num__8 years older than she is . if the average ( arithmetic mean ) age of the num__4 siblings is num__21 what is the age of the youngest sibling ? <o> a ) num__17 <o> b ) num__18 <o> c ) num__19 <o> d ) num__21 <o> e ) num__22 |
total age of the num__4 sibling is num__21 x num__4 = num__84 years . . we already have the total age of all the children is num__4 y + num__16 so num__4 y + num__16 = num__84 or num__4 y = num__68 or y = num__17 so age of the youngest child is num__17 years . answer : a <eor> a <eos> |
a |
multiply__4.0__21.0__ subtract__21.0__5.0__ subtract__84.0__16.0__ subtract__21.0__4.0__ subtract__21.0__4.0__ |
multiply__4.0__21.0__ subtract__21.0__5.0__ subtract__84.0__16.0__ subtract__21.0__4.0__ subtract__21.0__4.0__ |
| the average temperature on wednesday in chicago st . louis and pittsburgh was num__27 Â ° c . the average temperature that same day of st . louis pittsburgh and jersey city was num__30 Â ° c . if temperature in jersey city was num__32 Â ° c what was the temperature on wednesday in chicago ? <o> a ) num__22 Â ° c <o> b ) num__23 Â ° c <o> c ) num__24 Â ° c <o> d ) num__25 Â ° c <o> e ) num__26 Â ° c |
solution : average temperature for chicago st . louis and pittsburgh = num__27 Â ° c total temperature = num__3 * num__27 = num__81 average temperature for st . louis pittsburgh and jersey city = num__30 Â ° c total temperature = num__3 * num__30 = num__90 temperature in jersey city = num__32 Â ° c now ( chicago + st . louis + pittsburgh ) - ( st . louis + pittsburgh + jersey city ) = num__81 - num__90 chicago - jersey city = - num__9 chicago = jersey city - num__9 = num__23 Â ° c answer : option b <eor> b <eos> |
b |
subtract__30.0__27.0__ multiply__27.0__3.0__ multiply__30.0__3.0__ divide__27.0__3.0__ subtract__32.0__9.0__ round__23.0__ |
subtract__30.0__27.0__ multiply__27.0__3.0__ multiply__30.0__3.0__ subtract__90.0__81.0__ subtract__32.0__9.0__ subtract__32.0__9.0__ |
| at what rate of compound interest per annum will a sum of rs . num__1200 become rs . num__1348.32 in num__2 years <o> a ) num__3.0 <o> b ) num__4.0 <o> c ) num__5.0 <o> d ) num__6.0 <o> e ) none of these |
explanation : let rate will be r num__1200.0 ( num__1 + r num__100 ) num__2 = num__134832100 ( num__1 + r num__100 ) num__2 = num__134832120000 ( num__1 + r num__100 ) num__2 = num__1123610000 ( num__1 + r num__100 ) = num__106100 = > r = num__6.0 answer : d <eor> d <eos> |
d |
percent__100.0__6.0__ |
percent__100.0__6.0__ |
| a train speeds past a pole in num__15 seconds and a platform num__110 meters long in num__25 seconds . what is the length of the train ? <o> a ) num__100 m <o> b ) num__125 m <o> c ) num__130 m <o> d ) num__150 m <o> e ) num__165 m |
let the length of the train be x meters . the speed of the train is x / num__15 . then x + num__110 = num__25 * ( x / num__15 ) num__10 x = num__1650 x = num__165 meters the answer is e . <eor> e <eos> |
e |
subtract__25.0__15.0__ multiply__15.0__110.0__ divide__1650.0__10.0__ round__165.0__ |
subtract__25.0__15.0__ multiply__15.0__110.0__ divide__1650.0__10.0__ divide__1650.0__10.0__ |
| there are num__3 numbers a b and c . if a : b = num__0.75 b : c = num__0.8 c : d = num__0.833333333333 then a : d will be ? <o> a ) num__1 : num__2 <o> b ) num__2 : num__3 <o> c ) num__3 : num__5 <o> d ) num__3 : num__7 <o> e ) num__3 : num__8 |
sol . a : b = num__3 : num__4 b : c = num__4 : num__5 c : d = num__5 : num__6 ∴ a ∶ b ∶ c ∶ d = num__3 : num__4 : num__5 : num__6 . thus a : d = num__3 : num__6 or num__1 : num__2 a <eor> a <eos> |
a |
divide__3.0__0.75__ divide__4.0__0.8__ subtract__4.0__3.0__ subtract__3.0__1.0__ reverse__1.0__ |
divide__3.0__0.75__ divide__4.0__0.8__ subtract__4.0__3.0__ subtract__3.0__1.0__ reverse__1.0__ |
| a train num__500 m long can cross an electric pole in num__20 sec and then find the speed of the train ? <o> a ) num__17 kmph <o> b ) num__78 kmph <o> c ) num__90 kmph <o> d ) num__18 kmph <o> e ) num__19 kmph |
length = speed * time speed = l / t s = num__25.0 s = num__25 m / sec speed = num__25 * num__3.6 ( to convert m / sec in to kmph multiply by num__3.6 ) speed = num__90 kmph answer : c <eor> c <eos> |
c |
divide__500.0__20.0__ multiply__25.0__3.6__ round__90.0__ |
divide__500.0__20.0__ multiply__25.0__3.6__ multiply__25.0__3.6__ |
| two spherical balls lie on the ground touching . if one of the balls has a radius of num__3 cm and the point of contact is num__5 cm above the ground what is the radius of the other ball ? <o> a ) num__3 cm <o> b ) num__9 cm <o> c ) num__12 cm <o> d ) num__15 cm <o> e ) none of the these |
similar triangle properties . . num__2 / r + num__3 = num__3 / r - num__3 giving r = num__15 . answer : d <eor> d <eos> |
d |
subtract__5.0__3.0__ multiply__3.0__5.0__ round__15.0__ |
subtract__5.0__3.0__ multiply__3.0__5.0__ round__15.0__ |
| three workers have a productivity ratio of num__1 to num__3 to num__5 . all three workers are working on a job for num__4 hours . at the beginning of the num__5 th hour the slowest worker takes a break . the slowest worker comes back to work at the beginning of the num__9 th hour and begins working again . the job is done in ten hours . what was the ratio of the work performed by the fastest worker as compared to the slowest ? <o> a ) num__25 to num__3 <o> b ) num__6 to num__1 <o> c ) num__5 to num__1 <o> d ) num__1 to num__6 <o> e ) num__1 to num__5 |
the fastest worker who does num__5 units of job worked for all num__10 hours so he did num__5 * num__10 = num__50 units of job ; the slowest worker who does num__1 unit of job worked for only num__4 + num__2 = num__6 hours ( first num__4 hours and last num__2 hours ) so he did num__1 * num__6 = num__6 units of job ; the ratio thus is num__50 to num__6 or num__25 to num__3 . answer : a <eor> a <eos> |
a |
add__1.0__9.0__ multiply__5.0__10.0__ subtract__3.0__1.0__ add__1.0__5.0__ divide__50.0__2.0__ round__25.0__ |
add__1.0__9.0__ multiply__5.0__10.0__ subtract__3.0__1.0__ add__1.0__5.0__ divide__50.0__2.0__ multiply__1.0__25.0__ |
| the sum and the product of two numbers are num__24 and num__23 respectively the difference of the number is ? <o> a ) num__1 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__22 |
explanation : x + y = num__24 xy = num__23 ( x - y ) num__2 = ( x + y ) num__2 - num__4 xy ( x - y ) num__2 = num__576 - num__92 = > ( x - y ) = num__22 answer : e <eor> e <eos> |
e |
multiply__23.0__4.0__ subtract__24.0__2.0__ subtract__24.0__2.0__ |
multiply__23.0__4.0__ subtract__24.0__2.0__ subtract__24.0__2.0__ |
| in a certain store the profit is num__1150.0 of the cost . if the cost increases by num__25.0 but the selling price remains constant approximately what percentage of the selling price is the profit ? <o> a ) num__30.0 <o> b ) num__70.0 <o> c ) num__900.0 <o> d ) num__250.0 <o> e ) num__120 % |
let c . p . = rs . num__100 . then profit = rs . num__1150 s . p . = rs . num__1250 new c . p . = num__125.0 of rs . num__100 = rs . num__125 . new s . p . = rs . num__1250 profit = num__1250 - num__125 = rs . num__1125 required percentage = num__0.9 * num__100 = num__900.0 answer : c <eor> c <eos> |
c |
percent__100.0__900.0__ |
percent__100.0__900.0__ |
| at a certain food stand the price of each apple is num__40 ¢ and the price of each orange is num__60 ¢ . mary selects a total of num__10 apples and oranges from the food stand and the average ( arithmetic mean ) price of the num__10 pieces of fruit is num__54 ¢ . how many oranges must mary put back so that the average price of the pieces of fruit that she keeps is num__45 ¢ ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
if the average price is num__54 then mary has num__7 oranges and num__3 apples ( a ratio of num__7 : num__3 ) . for the average price to be num__45 the ratio should be num__1 : num__3 . mary must put back num__6 oranges . the answer is e . <eor> e <eos> |
e |
subtract__10.0__7.0__ divide__60.0__10.0__ divide__60.0__10.0__ |
subtract__10.0__7.0__ divide__60.0__10.0__ divide__60.0__10.0__ |
| two pipes can fill a tank in num__15 minutes and num__12 minutes . the outlet pipe can empty the tank in num__20 minutes . if all the pipes are opened when the tank is empty then in how many minutes will it take to fill the tank ? <o> a ) num__12 <o> b ) num__13 <o> c ) num__11 <o> d ) num__10 <o> e ) num__9 |
part of tank filled by all three pipes in one minute = num__0.0666666666667 + num__0.0833333333333 – num__0.05 = ( num__8 + num__10 - num__6 ) / num__120 = num__18 - num__0.05 = num__0.1 so the tank becomes full in num__10 minutes . answer : d <eor> d <eos> |
d |
subtract__20.0__12.0__ multiply__15.0__8.0__ add__12.0__6.0__ divide__12.0__120.0__ round__10.0__ |
subtract__20.0__12.0__ divide__6.0__0.05__ add__12.0__6.0__ divide__12.0__120.0__ subtract__20.0__10.0__ |
| jake can dig a well in num__16 days . paul can dig the same well in num__24 days . jake paul and hari together dig the well in num__8 days . hari alone can dig the well in <o> a ) num__48 <o> b ) num__99 <o> c ) num__22 <o> d ) num__27 <o> e ) num__92 |
explanation : simple one . let the total work to be done is num__48 meters . now jake can dig num__3 mts paul can dig num__2 mts a day . now all of them combined dug in num__8 days so per day they dug num__6.0 = num__6 mts . so of these num__8 mts hari capacity is num__1 mt . so he takes num__48.0 = num__48 days to complete the digging job . answer : a <eor> a <eos> |
a |
divide__24.0__8.0__ divide__16.0__8.0__ subtract__8.0__2.0__ subtract__3.0__2.0__ round__48.0__ |
divide__24.0__8.0__ divide__16.0__8.0__ subtract__8.0__2.0__ subtract__3.0__2.0__ round__48.0__ |
| a train num__150 m long pass a telegraph pole in num__5 seconds . find the speed of the train <o> a ) num__66 km / hr <o> b ) num__108 km / hr <o> c ) num__72 km / hr <o> d ) num__79.2 km / hr <o> e ) none |
sol . speed = [ num__30.0 ] m / sec = [ num__30 * num__3.6 ] km / hr = num__108 km / hr . answer b <eor> b <eos> |
b |
divide__150.0__5.0__ multiply__3.6__30.0__ round__108.0__ |
divide__150.0__5.0__ multiply__3.6__30.0__ multiply__3.6__30.0__ |
| a certain number of men can do a work in num__65 days working num__6 hours a day . if the number of men are decreased by one - fourth then for how many hours per day should they work in order to complete the work in num__40 days ? <o> a ) num__33 <o> b ) num__13 <o> c ) num__88 <o> d ) num__77 <o> e ) num__12 |
let the number of men initially be x . we have m num__1 d num__1 h num__1 = m num__2 d num__2 h num__2 so x * num__65 * num__6 = ( num__3 x ) / num__4 * num__40 * h num__2 = > h num__2 = ( num__65 * num__6 * num__4 ) / ( num__3 * num__40 ) = num__13 . answer : b <eor> b <eos> |
b |
divide__6.0__2.0__ subtract__6.0__2.0__ round__13.0__ |
divide__6.0__2.0__ subtract__6.0__2.0__ divide__13.0__1.0__ |
| the average age of a class of num__36 students is num__16 yrs . if the teacher ' s age is also included the average increases by one year . find the age of the teacher <o> a ) num__45 years <o> b ) num__53 years <o> c ) num__49 years <o> d ) num__52 years <o> e ) num__54 years |
total age of students is num__36 x num__16 = num__576 years total age inclusive of teacher = num__37 x ( num__16 + num__1 ) = num__629 so teacher ' s age i num__629 - num__576 = num__53 yrs there is a shortcut for these type of problems teacher ' s age is num__16 + ( num__36 x num__1 ) = num__53 years answer : b <eor> b <eos> |
b |
multiply__36.0__16.0__ subtract__37.0__36.0__ add__16.0__37.0__ add__16.0__37.0__ |
multiply__36.0__16.0__ subtract__37.0__36.0__ add__16.0__37.0__ add__16.0__37.0__ |
| in a police selections the weight of arun is marked as num__64 kg instead of num__48 kg . find the percentage of correction required . <o> a ) num__30.0 <o> b ) num__25.0 <o> c ) num__24.0 <o> d ) num__29.0 <o> e ) num__34 % |
explanation : % correction = ( num__64 - num__48 ) / num__64 x num__100 = num__25.0 answer : b <eor> b <eos> |
b |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| express a speed of num__36 kmph in meters per second ? <o> a ) num__10 <o> b ) num__88 <o> c ) num__66 <o> d ) num__81 <o> e ) num__121 |
num__36 * num__0.277777777778 = num__10 mps answer : a <eor> a <eos> |
a |
round__10.0__ |
round__10.0__ |
| if num__375 * s is a perfect square then s can be which of the following numbers ? <o> a ) num__21 <o> b ) num__35 <o> c ) num__54 <o> d ) num__15 <o> e ) num__150 |
my solution : for a number to be a perfect square all the prime factors must be in pairs . so prime factors of num__375 are num__5 * num__5 * num__5 * num__3 num__1 ) num__21 = num__3 * num__7 num__2 ) num__35 = num__5 * num__7 num__3 ) num__54 = num__2 * num__3 * num__3 * num__3 num__4 ) num__15 = num__3 * num__5 num__5 ) num__150 = num__2 * num__3 * num__5 * num__5 so only answer choice d completes the pairing of num__3 num__5 . it adds a num__3 a num__5 and a num__7 to the prime factors of num__375 ( num__3 * num__5 * num__5 * num__5 ) and makes it a perfect square . d <eor> d <eos> |
d |
divide__21.0__3.0__ subtract__3.0__1.0__ multiply__5.0__7.0__ add__1.0__3.0__ multiply__3.0__5.0__ multiply__1.0__15.0__ |
divide__21.0__3.0__ subtract__3.0__1.0__ multiply__5.0__7.0__ add__1.0__3.0__ multiply__3.0__5.0__ multiply__1.0__15.0__ |
| the average age of num__20 students of a class is num__20 years . out of these the average age of num__5 students is num__14 years and that of the other num__9 students is num__16 years the age of the num__20 th student is <o> a ) num__186 <o> b ) num__120 <o> c ) num__152 <o> d ) num__220 <o> e ) num__220 |
explanation : age of the num__20 th student = [ num__20 * num__20 - ( num__14 * num__5 + num__16 * num__9 ) ] = ( num__400 - num__214 ) = num__186 years . answer : a <eor> a <eos> |
a |
subtract__400.0__214.0__ subtract__400.0__214.0__ |
subtract__400.0__214.0__ subtract__400.0__214.0__ |
| the sum of four consecutive even numbers is num__84 . what would be the smallest number ? <o> a ) num__33 <o> b ) num__88 <o> c ) num__70 <o> d ) num__123 <o> e ) num__18 |
let the four consecutive even numbers be num__2 ( x - num__2 ) num__2 ( x - num__1 ) num__2 x num__2 ( x + num__1 ) their sum = num__8 x - num__4 = num__84 = > x = num__11 smallest number is : num__2 ( x - num__2 ) = num__18 . answer : e <eor> e <eos> |
e |
divide__8.0__2.0__ round__18.0__ |
divide__8.0__2.0__ round__18.0__ |
| a man spend num__0.2 of his salary on food num__0.1 of his salary on house rent and num__0.6 salary on clothes . he still has $ num__15000 left with him . find salary . . <o> a ) $ num__150000 <o> b ) $ num__18000 <o> c ) $ num__180000 <o> d ) $ num__1800 <o> e ) none |
[ num__1 / ( x num__1 / y num__1 + x num__2 / y num__2 + x num__3 / y num__3 ) ] * total amount = balance amount [ num__1 - ( num__0.2 + num__0.1 + num__0.6 ) } * total salary = $ num__15000 = [ num__1 - num__0.9 ] * total salary = $ num__15000 total salary = $ num__15000 * num__10 = $ num__150000 correct answer ( a ) <eor> a <eos> |
a |
divide__0.2__0.1__ divide__0.6__0.2__ subtract__1.0__0.1__ reverse__0.1__ divide__15000.0__0.1__ divide__15000.0__0.1__ |
divide__0.2__0.1__ divide__0.6__0.2__ subtract__1.0__0.1__ reverse__0.1__ divide__15000.0__0.1__ divide__15000.0__0.1__ |
| by selling num__99 pens a trader gains the cost of num__33 pens . find his gain percentage ? <o> a ) num__33 num__0.333333333333 % <o> b ) num__68 num__0.333333333333 % <o> c ) num__28 num__0.333333333333 % <o> d ) num__18 num__0.333333333333 % <o> e ) num__11 num__0.333333333333 % |
explanation : let the cp of each pen be rs . num__1 . cp of num__99 pens = rs . num__99 profit = cost of num__33 pens = rs . num__33 profit % = num__0.333333333333 * num__100 = num__33 num__0.333333333333 % answer : a <eor> a <eos> |
a |
percent__33.0__100.0__ |
percent__33.0__100.0__ |
| a thief steals at a car at num__3.30 p . m . and drives it at num__60 km / hr . the theft is discovered at num__3 p . m . and the owner sets off in another car at num__75 km / hr . when will he overtake the thief ? <o> a ) num__1 p . m <o> b ) num__3 p . m <o> c ) num__4 p . m <o> d ) num__5 p . m <o> e ) num__6 p . m |
e num__6 p . m suppose the thief is overtaken x hrs after num__3.30 p . m . then distance covered by the owner in ( x - num__0.5 ) hrs . num__60 x = num__75 ( x - num__0.5 ) = > x = num__2.5 hrs . so the thief is overtaken at num__6 p . m . <eor> e <eos> |
e |
divide__3.0__6.0__ subtract__3.0__0.5__ divide__3.0__0.5__ |
divide__3.0__6.0__ subtract__3.0__0.5__ divide__3.0__0.5__ |
| teas worth rs . num__126 per kg and rs . num__135 per kg are mixed with a third variety in the ratio num__1 : num__1 : num__2 . if the mixture is worth rs num__154 per kg the price of the third variety per kg will be ? <o> a ) rs . num__147.50 <o> b ) rs . num__785.50 <o> c ) rs . num__176.50 <o> d ) rs . num__258.50 <o> e ) none of these |
explanation : since first and second varieties are mixed in equal proportions . so their average price = rs . ( num__126 + num__135 ) / num__2 . = > rs . num__130.50 . so the mixture is formed by mixing two varieties one at rs . num__130.50 per kg and the other at say rs . x per kg in the ratio num__2 : num__2 i . e . num__1 : num__1 . we have to find x . by the rule of alligation we have : cost of num__1 kg cost of num__1 kg of num__1 st kind of num__2 nd kind ( rs . num__130.50 ) ( rs . x ) \ / mean price ( rs . num__154 ) / \ x â ˆ ’ num__154 num__22.50 = > x â ˆ ’ ( num__154 / num__22.50 ) = num__1 . = > x â ˆ ’ num__154 = num__22.50 . = > x = num__176.50 rs . answer : c <eor> c <eos> |
c |
add__154.0__22.5__ multiply__1.0__176.5__ |
add__154.0__22.5__ add__154.0__22.5__ |
| which is the following is divisible by num__11 <o> a ) a ) num__4305 <o> b ) b ) num__4825 <o> c ) c ) num__1635 <o> d ) d ) num__4805 <o> e ) e ) num__4906 |
explanation : sum of first ' n ' natural numbers = n ( n + num__1 ) / num__2 sum of first num__9 natural numbers = num__5 ( num__5 + num__1 ) / / num__2 = num__5 x num__3 = num__15 sum of first num__99 natural numbers = num__55 ( num__55 + num__1 ) / / num__2 = num__55 x num__30 = num__1650 num__1650 - num__15 = num__1635 answer : option c <eor> c <eos> |
c |
subtract__11.0__2.0__ add__1.0__2.0__ multiply__3.0__5.0__ multiply__11.0__9.0__ multiply__11.0__5.0__ multiply__2.0__15.0__ multiply__55.0__30.0__ subtract__1650.0__15.0__ multiply__1.0__1635.0__ |
subtract__11.0__2.0__ add__1.0__2.0__ multiply__3.0__5.0__ multiply__11.0__9.0__ multiply__11.0__5.0__ multiply__2.0__15.0__ multiply__55.0__30.0__ subtract__1650.0__15.0__ divide__1635.0__1.0__ |
| if one third of one fourth of number is num__15 then three tenth of number is <o> a ) num__34 <o> b ) num__44 <o> c ) num__54 <o> d ) num__64 <o> e ) num__74 |
explanation : let the number is x num__0.333333333333 of num__0.25 ∗ x = num__15 = > x = num__15 × num__12 = num__180 = > so num__0.3 × x = num__0.3 × num__180 = num__54 option c <eor> c <eos> |
c |
multiply__15.0__12.0__ multiply__0.3__180.0__ multiply__0.3__180.0__ |
multiply__15.0__12.0__ multiply__0.3__180.0__ multiply__0.3__180.0__ |
| what is the percentage increase in the area of a rectangle if each of its sides is increased by num__20.0 ? <o> a ) num__45.0 <o> b ) num__44.0 <o> c ) num__40.0 <o> d ) num__42.0 <o> e ) num__50 % |
( num__20 + num__20 + num__20 * num__0.2 ) % = num__44.0 area increased by num__44.0 then original area num__10 * num__10 = num__100 length is incresed by num__20.0 new length = num__10 + num__2 = num__12 breadth is increased num__20.0 new breadth = num__10 + num__2 - num__12 new area = num__12 * num__12 = num__144 increase in area num__144 - num__100 = num__44 percentage increase in area = increase in area / original area = num__0.44 * num__100 = num__44 answer b <eor> b <eos> |
b |
percent__20.0__10.0__ percent__100.0__44.0__ |
percent__20.0__10.0__ percent__100.0__44.0__ |
| a and b together can do a piece of work in num__8 days . if a alone can do the same work in num__12 days then b alone can do the same work in ? <o> a ) num__24 <o> b ) num__30 <o> c ) num__35 <o> d ) num__12 <o> e ) num__10 |
b = num__188 - num__0.5 = num__0.0416666666667 = > num__24 days answer a <eor> a <eos> |
a |
divide__0.5__12.0__ divide__12.0__0.5__ round__24.0__ |
divide__0.5__12.0__ divide__12.0__0.5__ round__24.0__ |
| the radius of a wheel is num__22.4 cm . what is the distance covered by the wheel in making num__650 resolutions ? <o> a ) num__794 m <o> b ) num__704 m <o> c ) num__454 m <o> d ) num__186 m <o> e ) num__915.2 m |
in one resolution the distance covered by the wheel is its own circumference . distance covered in num__650 resolutions . = num__650 * num__2 * num__3.14285714286 * num__22.4 = num__91520 cm = num__915.2 m answer : e <eor> e <eos> |
e |
round__915.2__ |
round__915.2__ |
| a train covers a distance of num__20 km in num__20 min . if it takes num__9 sec to pass a telegraph post then the length of the train is ? <o> a ) num__150 m <o> b ) num__200 m <o> c ) num__120 m <o> d ) num__225 m <o> e ) num__160 m |
speed = ( num__1.0 * num__60 ) km / hr = ( num__60 * num__0.277777777778 ) m / sec = num__16.6666666667 m / sec . length of the train = num__16.6666666667 * num__9 = num__150 m . answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ round__150.0__ |
hour_to_min_conversion__ divide__150.0__1.0__ |
| a and b can do a piece of work in num__8 days . b and c can do it in num__12 days and a and c in num__16 days . working together they will complete the work in how many days ? <o> a ) num__7.39 days <o> b ) num__7.78 days <o> c ) num__7.34 days <o> d ) num__7.38 days <o> e ) num__3.38 days |
a + b = num__0.125 b + c = num__0.0833333333333 c + a = num__0.0625 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - num__2 ( a + b + c ) = num__0.125 + num__0.0833333333333 + num__0.0625 = num__0.270833333333 a + b + c = num__0.135416666667 num__7.38461538462 = num__7.38 days answer : d <eor> d <eos> |
d |
divide__16.0__8.0__ divide__0.2708__2.0__ round__7.38__ |
divide__16.0__8.0__ divide__0.2708__2.0__ round__7.38__ |
| an accurate clock shows num__8 o ' clock in the morning . through how may degrees will the hour hand rotate when the clock shows num__2 o ' clock in the afternoon ? <o> a ) num__112 <o> b ) num__180 <o> c ) num__288 <o> d ) num__278 <o> e ) num__292 |
explanation : angle traced by the hour hand in num__6 hours = ( num__30.0 ) * num__6 answer : b ) num__180 <eor> b <eos> |
b |
straight_angle__ straight_angle__ |
straight_angle__ straight_angle__ |
| a wood maker constructed a rectangular sandbox with a capacity of num__10 cubic feet . if the carpenter were to make a similar sandbox twice as long twice as wide and twice as high as the first sandbox what would be the capacity in cubic feet of the second sandbox ? <o> a ) num__20 <o> b ) num__40 <o> c ) num__60 <o> d ) num__80 <o> e ) num__100 |
a quick note on doubling . when you double a length you have num__2 * l num__1 . when you double all lengths of a rectangle you have ( num__2 * l num__1 ) ( num__2 * l num__2 ) = a . an increase of num__2 ^ num__2 or num__4 . when you double all lengths of a rectangular prism you have ( num__2 * l num__1 ) ( num__2 * l num__2 ) ( num__2 * l num__3 ) = v . an increase of num__2 ^ num__3 or num__8 . this leads to the basic relationship : line : num__2 * original size rectangle : num__4 * original size rectangular prism : num__8 * original size answer is d <eor> d <eos> |
d |
square_perimeter__1.0__ square_perimeter__2.0__ multiply__10.0__8.0__ |
square_perimeter__1.0__ power__2.0__3.0__ multiply__10.0__8.0__ |
| the length of a train and that of a platform are equal . if with a speed of num__144 k / hr the train crosses the platform in one minute then the length of the train ( in meters ) is ? <o> a ) num__1200 <o> b ) num__299 <o> c ) num__276 <o> d ) num__750 <o> e ) num__211 |
speed = [ num__144 * num__0.277777777778 ] m / sec = num__40 m / sec ; time = num__1 min . = num__60 sec . let the length of the train and that of the platform be x meters . then num__2 x / num__60 = num__40 è x = num__40 * num__30.0 = num__1200 answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ divide__60.0__2.0__ multiply__40.0__30.0__ round__1200.0__ |
hour_to_min_conversion__ divide__60.0__2.0__ multiply__40.0__30.0__ multiply__40.0__30.0__ |
| ab + cd = xxx where ab and cd are two - digit numbers and xxx is a three digit number ; a b c and d are distinct positive integers . in the addition problem above what is the value of c ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__7 <o> d ) num__9 <o> e ) can not be determined |
ab and cd are two digit integers their sum can give us only one three digit integer of a kind of xxx it ' s num__111 . so a = num__1 . num__1 b + cd = num__111 now c can not be less than num__9 because no to digit integer with first digit num__1 ( mean that it ' s < num__20 ) can be added to two digit integer less than num__90 to have the sum num__111 ( if cd < num__90 meaning c < num__9 cd + num__1 b < num__111 ) - - > c = num__9 answer : d . <eor> d <eos> |
d |
multiply__1.0__9.0__ |
multiply__1.0__9.0__ |
| - num__88 * num__49 + num__100 = ? <o> a ) - num__4212 <o> b ) num__4601 <o> c ) - num__4801 <o> d ) - num__3471 <o> e ) none of these |
= > - num__88 * ( num__50 - num__1 ) + num__100 ; = > - ( num__88 * num__50 ) + num__88 + num__100 ; = > - num__4400 + num__188 = - num__4212 . correct option : a <eor> a <eos> |
a |
subtract__50.0__49.0__ multiply__88.0__50.0__ add__88.0__100.0__ subtract__4400.0__188.0__ multiply__1.0__4212.0__ |
subtract__50.0__49.0__ multiply__88.0__50.0__ add__88.0__100.0__ subtract__4400.0__188.0__ multiply__1.0__4212.0__ |
| a sum of money invested at compound interest to rs . num__800 in num__3 years and to rs num__860 in num__4 years . the rate on interest per annum is . <o> a ) num__4.5 <o> b ) num__5.0 <o> c ) num__6.0 <o> d ) num__7.5 <o> e ) num__8 % |
explanation : s . i . on rs num__800 for num__1 year = num__60 rate = ( num__100 * num__60 ) / ( num__800 * num__1 ) = num__7.5 answer : d <eor> d <eos> |
d |
percent__100.0__7.5__ |
percent__100.0__7.5__ |
| in an intercollegiate competition that lasted for num__3 days num__157 students took part on day num__1 num__111 on day num__2 and num__98 on day num__3 . if num__89 took part on day num__1 and day num__2 and num__56 took part on day num__2 and day num__3 and num__32 took part on all three days how many students took part only on day num__1 ? <o> a ) num__40 <o> b ) num__44 <o> c ) num__35 <o> d ) num__49 <o> e ) num__38 |
day num__1 & num__2 = num__89 ; only day num__1 & num__2 ( num__89 - num__32 ) = num__57 day num__2 & num__3 = num__56 ; only day num__2 & num__3 ( num__56 - num__32 ) = num__24 only day num__1 = num__157 - ( num__57 + num__24 + num__32 ) = num__44 answer : b <eor> b <eos> |
b |
add__1.0__56.0__ subtract__56.0__32.0__ round__44.0__ |
add__1.0__56.0__ subtract__56.0__32.0__ round__44.0__ |
| the ratio of the present ages of p and q is num__3 : num__4 . num__5 years ago the ratio of their ages was num__5 : num__7 . find the their present ages ? <o> a ) num__30 num__40 <o> b ) num__25 num__30 <o> c ) num__50 num__60 <o> d ) num__20 num__40 <o> e ) num__30 num__20 |
explanation : their present ages be num__3 x and num__4 x . num__5 years age the ratio of their ages was num__5 : num__7 then ( num__3 x - num__5 ) : ( num__4 x - num__5 ) = num__5 : num__7 x = num__35 - num__25 = > x = num__10 . their present ages are : num__30 num__40 . answer is a <eor> a <eos> |
a |
multiply__5.0__7.0__ add__3.0__7.0__ multiply__3.0__10.0__ multiply__4.0__10.0__ multiply__3.0__10.0__ |
multiply__5.0__7.0__ subtract__35.0__25.0__ subtract__35.0__5.0__ multiply__4.0__10.0__ subtract__35.0__5.0__ |
| dacid obtained num__72 num__45 num__72 num__77 and num__75 marks ( out of num__100 ) in english mathematics physics chemistry and biology . what are his average marks ? <o> a ) num__67 <o> b ) num__68 <o> c ) num__87 <o> d ) num__26 <o> e ) num__75 |
average = ( num__72 + num__45 + num__72 + num__77 + num__75 ) / num__5 = num__68.2 = num__68 . answer : b <eor> b <eos> |
b |
subtract__77.0__72.0__ round_down__68.2__ round_down__68.2__ |
subtract__77.0__72.0__ round_down__68.2__ round_down__68.2__ |
| a certain industrial loom weaves num__0.128 meters of cloth every second . approximately how many seconds will it take for the loom to weave num__15 meters of cloth ? <o> a ) num__114 <o> b ) num__115 <o> c ) num__116 <o> d ) num__117 <o> e ) num__118 |
let the required number of seconds be x more cloth more time ( direct proportion ) hence we can write as ( cloth ) num__0.128 : num__15 : : num__1 : x = > num__0.128 * x = num__15 = > x = num__15 / num__0.128 = > x = num__117 answer : d <eor> d <eos> |
d |
round__117.0__ |
divide__117.0__1.0__ |
| the ratio of incomes of two person p num__1 and p num__2 is num__5 : num__4 and the ratio of their expenditures is num__3 : num__2 . if at the end of the year each saves rs . num__2000 then what is the income of p num__1 ? <o> a ) s . num__5000 <o> b ) s . num__2400 <o> c ) s . num__4000 <o> d ) s . num__3200 <o> e ) s . num__4200 |
let the income of p num__1 and p num__2 be rs . num__5 x and rs . num__4 x respectively and let their expenditures be rs . num__3 y and num__2 y respectively . then num__5 x – num__3 y = num__2000 … ( i ) and num__4 x – num__2 y = num__2000 … … . . ( ii ) on multiplying ( i ) by num__2 ( ii ) by num__3 and subtracting we get : num__2 x = num__2000 - > x = num__1000 p num__1 ’ s income = rs num__5 * num__1000 = rs . num__5000 answer : a <eor> a <eos> |
a |
divide__2000.0__2.0__ multiply__5.0__1000.0__ multiply__1.0__5000.0__ |
divide__2000.0__2.0__ multiply__5.0__1000.0__ multiply__1.0__5000.0__ |
| at num__12 : num__30 the hour hand and the minute hand of a clock form an angle of <o> a ) num__120 ° <o> b ) num__135 ° <o> c ) num__125 ° <o> d ) num__150 ° <o> e ) num__180 ° |
answer : num__180 degree answer : e <eor> e <eos> |
e |
round__180.0__ |
round__180.0__ |
| how many times in a day the hands of a clock are straight ? <o> a ) num__12 <o> b ) num__24 <o> c ) num__44 <o> d ) num__56 <o> e ) none |
sol . in num__12 hours the hands coincide or are in opposite direction num__22 times . ∴ in num__24 hours the hands coincide or are in opposite direction num__44 times a day . answer c <eor> c <eos> |
c |
round__44.0__ |
round__44.0__ |
| num__0 num__0 num__1 num__2 num__2 num__4 num__3 num__6 num__4 _ can you find out which number will complete the above series ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
solution : actually there are two series in the question . if you separate the alternate places you will find that the first series is : num__0 num__1 num__2 num__3 num__4 where the digits are multiple of num__1 . the second series is : num__0 num__2 num__4 num__6 __ here the digits are progressing with the multiples of num__2 . therefore the next digit will be num__8 . answer a <eor> a <eos> |
a |
multiply__2.0__4.0__ multiply__1.0__8.0__ |
multiply__2.0__4.0__ multiply__1.0__8.0__ |
| arabica coffee costs $ num__0.4 per ounce while robusta coffee costs $ num__0.2 per ounce . if the blend of arabica and robusta costs $ num__0.33 per ounce what is the share of arabica in this blend ? <o> a ) num__20.0 <o> b ) num__65.0 <o> c ) num__30.0 <o> d ) num__33.0 <o> e ) num__40 % |
a = amount of arabica coffee num__1 - a = amount of robusta coffee . because if you subtract a from the num__1 ounce the remaining amount is robusta therefore : . num__4 a + . num__2 ( num__1 - a ) = . num__33 . num__4 a + . num__2 - . num__2 a = . num__33 a = . num__65 therefore : . num__65.0 ounce = num__65.0 . therefore the answer should be b <eor> b <eos> |
b |
divide__0.4__0.2__ multiply__1.0__65.0__ |
divide__0.4__0.2__ multiply__1.0__65.0__ |
| a train passes a station platform in num__38 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__177 m <o> b ) num__176 m <o> c ) num__270 m <o> d ) num__187 m <o> e ) num__186 m |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x meters . then x + num__7.89473684211 = num__15 x + num__300 = num__570 x = num__270 m . answer : c <eor> c <eos> |
c |
multiply__20.0__15.0__ divide__300.0__38.0__ multiply__38.0__15.0__ subtract__570.0__300.0__ round__270.0__ |
multiply__20.0__15.0__ divide__300.0__38.0__ multiply__38.0__15.0__ subtract__570.0__300.0__ round__270.0__ |
| what is the greatest positive integer w such that num__3 ^ w is a factor of num__9 ^ num__10 ? <o> a ) num__5 <o> b ) w = num__9 <o> c ) w = num__10 <o> d ) w = num__20 <o> e ) w = num__30 |
what is the greatest positive integer w such that num__3 ^ w is a factor of num__9 ^ num__10 ? num__9 ^ num__10 = ( num__3 ^ num__2 ) ^ num__10 = num__3 ^ num__20 d . num__20 <eor> d <eos> |
d |
multiply__10.0__2.0__ lcm__10.0__20.0__ |
multiply__10.0__2.0__ lcm__10.0__20.0__ |
| a man took loan from a bank at the rate of num__12.0 p . a . simple interest . after num__3 years he had to pay rs . num__5400 interest only for the period . the principal amount borrowed by him was : <o> a ) rs . num__10000 <o> b ) rs . num__12000 <o> c ) rs . num__14000 <o> d ) rs . num__15000 <o> e ) rs . num__16 |
000 |
principal = rs . ( num__100 x num__5400 ) / ( num__12 x num__3 ) = rs . num__15000 answer : d <eor> d <eos> |
d |
d |
| ramu bought an old car for rs . num__42000 . he spent rs . num__10000 on repairs and sold it for rs . num__64900 . what is his profit percent ? <o> a ) num__12.0 <o> b ) num__16.0 <o> c ) num__18.0 <o> d ) num__24.8 <o> e ) num__23 % |
total cp = rs . num__42000 + rs . num__10000 = rs . num__52000 and sp = rs . num__64900 profit ( % ) = ( num__64900 - num__52000 ) / num__52000 * num__100 = num__24.8 answer : d <eor> d <eos> |
d |
percent__24.8__100.0__ |
percent__24.8__100.0__ |
| out of first num__25 natural numbers one number is selected at random . the probability that it is either an even number or a prime number is ? <o> a ) num__0.235294117647 <o> b ) num__0.571428571429 <o> c ) num__0.444444444444 <o> d ) num__0.8 <o> e ) num__1.33333333333 |
n ( s ) = num__25 n ( even no ) = num__12 = n ( e ) n ( prime no ) = num__9 = n ( p ) p ( e ᴜ p ) = num__0.48 + num__0.36 - num__0.04 = num__0.8 = num__0.8 answer : d <eor> d <eos> |
d |
union_prob__0.36__0.48__0.04__ union_prob__0.36__0.48__0.04__ |
union_prob__0.36__0.48__0.04__ union_prob__0.36__0.48__0.04__ |
| two trains of length num__200 m and num__280 m are running towards each other on parallel lines at num__42 kmph and num__30 kmph respectively . in what time will they be clear of each other from the moment they meet ? <o> a ) num__22 sec <o> b ) num__77 sec <o> c ) num__76 sec <o> d ) num__20 sec <o> e ) num__24 sec |
relative speed = ( num__42 + num__30 ) * num__0.277777777778 = num__4 * num__5 = num__20 mps . distance covered in passing each other = num__200 + num__280 = num__480 m . the time required = d / s = num__24.0 = num__24 sec . answer : e <eor> e <eos> |
e |
multiply__4.0__5.0__ add__200.0__280.0__ divide__480.0__20.0__ round__24.0__ |
multiply__4.0__5.0__ add__200.0__280.0__ divide__480.0__20.0__ divide__480.0__20.0__ |
| in a tailoring num__8 seamstresses sew a large order of pants doing a third of the work in num__6 days . if the owner hires num__4 seamstresses more how many days will take to finish the order since the first day ? <o> a ) num__12 days <o> b ) num__10 days <o> c ) num__9 days <o> d ) num__8 days <o> e ) num__7 days |
we have that : num__8 seamstresses - - - - - num__6 days ( num__8 + num__4 ) s - - - - - x d that is : ( x d / num__8 s ) = ( num__6 d / num__12 s ) then : x d = ( num__6 d / num__12 s ) num__8 s = num__4 d . is to say the num__12 seamstresses do the same quantity in num__4 days ; as it is only num__0.333333333333 of the work all the order would be sew in : p = num__6 d + num__4 d = num__10 days . answer b . <eor> b <eos> |
b |
add__8.0__4.0__ divide__4.0__12.0__ add__6.0__4.0__ round__10.0__ |
add__8.0__4.0__ divide__4.0__12.0__ add__6.0__4.0__ add__6.0__4.0__ |
| if x is the interest on y and y is the interest on z the rate and time is the same on both the cases . what is the relation between x y and z ? <o> a ) y num__2 = xy <o> b ) y num__2 = yz <o> c ) y num__2 = xz <o> d ) x num__2 = xz <o> e ) z num__2 = xz |
x = ( y * nr ) / num__100 y = ( z * nr ) / num__100 x / y = nr / num__100 y / z = nr / num__100 x / y = y / z y num__2 = xz answer : c <eor> c <eos> |
c |
percent__2.0__100.0__ |
percent__2.0__100.0__ |
| two employees x and y are paid a total of rs . num__550 per week by their employer . if x is paid num__120 percent of the sum paid to y how much is y paid per week ? <o> a ) rs . num__250 <o> b ) rs . num__300 <o> c ) rs . num__350 <o> d ) rs . num__450 <o> e ) rs . num__550 |
let the amount paid to x per week = x and the amount paid to y per week = y then x + y = num__550 but x = num__120.0 of y = num__120 y / num__100 = num__12 y / num__10 ∴ num__12 y / num__10 + y = num__550 ⇒ y [ num__1.2 + num__1 ] = num__550 ⇒ num__22 y / num__10 = num__550 ⇒ num__22 y = num__5500 ⇒ y = num__250.0 = num__250.0 = rs . num__250 a ) <eor> a <eos> |
a |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__10.0__12.0__ multiply__550.0__10.0__ divide__5500.0__22.0__ multiply__1.0__250.0__ |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__10.0__12.0__ multiply__550.0__10.0__ divide__5500.0__22.0__ divide__250.0__1.0__ |
| a sum of money is to be distributed among a b c d in the proportion of num__5 : num__4 : num__5 : num__3 . if c gets rs . num__10000 more than d what is b ' s share ? <o> a ) rs . num__500 <o> b ) rs . num__20000 <o> c ) rs . num__1500 <o> d ) rs . num__2000 <o> e ) none |
sol . let the shares of a b c and d be rs . num__5 x rs . num__4 x rs . num__5 x and rs . num__3 x respectively . then num__5 x - num__3 x = num__10000 ⇔ num__2 x = num__10000 ⇔ x = num__5000 . ∴ b ' s share = rs . num__4 x = rs . ( num__4 x num__5000 ) = rs . num__20000 . answer b <eor> b <eos> |
b |
subtract__5.0__3.0__ divide__10000.0__2.0__ multiply__4.0__5000.0__ multiply__4.0__5000.0__ |
subtract__5.0__3.0__ divide__10000.0__2.0__ multiply__4.0__5000.0__ multiply__4.0__5000.0__ |
| the wages earned by robin is num__30.0 more than that earned by erica . the wages earned by charles is num__60.0 more than that earned by erica . how much percent is the wages earned by charles more than that earned by robin ? <o> a ) num__18.75 <o> b ) num__23.0 <o> c ) num__30.0 <o> d ) num__50.0 <o> e ) num__100 % |
let wage of erica = num__10 wage of robin = num__1.3 * num__10 = num__13 wage of charles = num__1.6 * num__10 = num__16 percentage by which wage earned by charles is more than that earned by robin = ( num__16 - num__13 ) / num__13 * num__100.0 = num__0.230769230769 * num__100.0 = num__23.0 answer b <eor> b <eos> |
b |
multiply__10.0__1.3__ multiply__1.6__10.0__ add__10.0__13.0__ add__10.0__13.0__ |
multiply__10.0__1.3__ multiply__1.6__10.0__ add__10.0__13.0__ add__10.0__13.0__ |
| a and b together can do a work in num__5 days . if a alone can do it in num__15 days . in how many days can b alone do it ? <o> a ) num__12.5 <o> b ) num__7.5 <o> c ) num__8.5 <o> d ) num__9.5 <o> e ) num__5.5 |
num__0.2 – num__0.0666666666667 = num__0.133333333333 = > num__7.5 answer : b <eor> b <eos> |
b |
subtract__0.2__0.0667__ round__7.5__ |
subtract__0.2__0.0667__ round__7.5__ |
| a person ' s present age is two - fifth of the age of his mother . after num__4 years he will be one - half of the age of his mother . how old is the mother at present ? <o> a ) a ) num__25 <o> b ) b ) num__40 <o> c ) c ) num__20 <o> d ) d ) num__45 <o> e ) e ) num__28 |
let the mother ' s present age be x years then the person ' s present age = num__2 x / num__5 ( num__2 x / num__5 ) + num__4 = num__0.5 ( x + num__4 ) num__2 ( num__2 x + num__20 ) = num__5 ( x + num__4 ) x = num__20 answer is c <eor> c <eos> |
c |
reverse__2.0__ multiply__4.0__5.0__ multiply__4.0__5.0__ |
reverse__2.0__ multiply__4.0__5.0__ multiply__4.0__5.0__ |
| angela has num__12 pairs of matched socks . if she loses num__7 individual socks which of the following is not a possible number of matched pairs she has left ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
i think if angela loses num__7 individual socks then she is bound to have lost minimum of num__3 pairs and num__1 individual sock in this way she is left with only num__8 pairs of socks ( num__12 - ( num__3 + num__1 ) ) . hence num__9 can not be the answer as maximum is num__8 . hence option ( b ) <eor> b <eos> |
b |
add__7.0__1.0__ subtract__12.0__3.0__ subtract__12.0__3.0__ |
add__7.0__1.0__ subtract__12.0__3.0__ subtract__12.0__3.0__ |
| an aeroplane covers a certain distance at a speed of num__240 kmph in num__5 hours . to cover the same distance in num__1 num__0.666666666667 hours it must travel at a speed of : <o> a ) num__300 kmph <o> b ) num__360 kmph <o> c ) num__600 kmph <o> d ) num__720 kmph <o> e ) num__650 kmph |
distance = ( num__240 x num__5 ) = num__1200 km . speed = distance / time speed = num__1200 / ( num__1.66666666667 ) km / hr . [ we can write num__1 num__0.666666666667 hours as num__1.66666666667 hours ] required speed = ( num__1200 x num__0.6 ) km / hr = num__720 km / hr . answer : d <eor> d <eos> |
d |
multiply__240.0__5.0__ add__1.0__0.6667__ km_to_mile_conversion__ multiply__1200.0__0.6__ round__720.0__ |
multiply__240.0__5.0__ add__1.0__0.6667__ divide__1.0__1.6667__ multiply__1200.0__0.6__ divide__720.0__1.0__ |
| if num__3 a = num__4 b and ab ≠ num__0 what is the ratio of a / num__4 to b / num__3 ? <o> a ) num__2.37037037037 <o> b ) num__0.5625 <o> c ) num__1 <o> d ) num__0.75 <o> e ) num__1.33333333333 |
a nice fast approach is the first find a pair of numbers that satisfy the given equation : num__3 a = num__4 b here ' s one pair : a = num__4 and b = num__3 what is the ratio of a / num__9 to b / num__8 ? in other words what is the value of ( a / num__4 ) / ( b / num__3 ) ? plug in values to get : ( a / num__4 ) / ( b / num__3 ) = ( num__1.0 ) / ( num__1.0 ) = num__1.0 = num__1 c <eor> c <eos> |
c |
subtract__4.0__3.0__ reverse__1.0__ |
subtract__4.0__3.0__ reverse__1.0__ |
| look at this series : num__404 num__369 num__334 num__299 num__264 num__229 num__194 . . . what number should come next ? <o> a ) num__194 <o> b ) num__108 <o> c ) num__106 <o> d ) num__107 <o> e ) num__104 |
num__194 this is a simple subtraction series . each number is num__35 less than the previous number . a <eor> a <eos> |
a |
subtract__404.0__369.0__ subtract__229.0__35.0__ |
subtract__404.0__369.0__ subtract__229.0__35.0__ |
| jane and ashley take num__5 num__0.714285714286 days and num__40 days respectively to complete a project when they work on it alone . they thought if they worked on the project together they would take fewer days to complete it . during the period that they were working together jane took an eight day leave from work . this led to jane ' s working for four extra days on her own to complete the project . how long did it take to finish the project ? <o> a ) num__12.5 days <o> b ) num__13.5 days <o> c ) num__16 days <o> d ) num__18 days <o> e ) num__20 days |
let us assume that the work is laying num__40 bricks . jane = num__7 bricks per day ashley = num__1 brick per day together = num__8 bricks per day let ' s say first num__8 days ashley works alone no of bricks = num__8 last num__4 days jane works alone no . of bricks = num__28 remaining bricks = num__40 - num__36 = num__4 so together they would take num__0.5 = num__0.5 total no . of days = num__8 + num__4 + num__0.5 = num__12.5 answer is a <eor> a <eos> |
a |
divide__40.0__5.0__ subtract__5.0__1.0__ multiply__4.0__7.0__ subtract__40.0__4.0__ divide__4.0__8.0__ round__12.5__ |
add__1.0__7.0__ subtract__5.0__1.0__ multiply__4.0__7.0__ subtract__40.0__4.0__ divide__4.0__8.0__ round__12.5__ |
| a watch was sold at a loss of num__10.0 . if it was sold for rs . num__180 more there would have been a gain of num__5.0 . what is the cost price ? <o> a ) rs . num__1200 <o> b ) rs . num__1100 <o> c ) rs . num__1300 <o> d ) rs . num__1400 <o> e ) rs . num__1000 |
num__90.0 num__105.0 - - - - - - - - num__15.0 - - - - num__180 num__100.0 - - - - ? = > rs . num__1200 answer : a <eor> a <eos> |
a |
percent__100.0__1200.0__ |
percent__100.0__1200.0__ |
| on a map num__2.5 inches represent num__40 miles . how many miles approximately is the distance if you measured num__155 centimeters assuming that num__1 - inch is num__2.54 centimeters ? <o> a ) num__990.4 <o> b ) num__970 <o> c ) num__972 <o> d ) num__976.4 <o> e ) num__975 |
num__1 inch = num__2.54 cm num__2.5 inch = num__2.54 * num__2.5 cm num__6.35 cm = num__40 miles num__155 cms = num__40 / num__6.35 * num__155 = num__976.4 miles answer : d <eor> d <eos> |
d |
multiply__2.5__2.54__ round__976.4__ |
multiply__2.5__2.54__ multiply__1.0__976.4__ |
| last year ’ s receipts from the sale of greeting cards during the week before mother ’ s day totaled $ num__189 million which represented num__10 percent of total greeting card sales for the year . total greeting card sales for the year totaled how many million dollars ? <o> a ) num__17010 <o> b ) num__2100 <o> c ) num__1890 <o> d ) num__1701 <o> e ) num__210 |
num__10.0 - - - - num__189 millions for num__100.0 = > ( num__189 * num__100.0 ) / num__10.0 = num__1890 . option c . <eor> c <eos> |
c |
percent__100.0__1890.0__ |
percent__100.0__1890.0__ |
| how many seconds will a num__700 meter long train take to cross a man walking with a speed of num__3 km / hr in the direction of the moving train if the speed of the train is num__63 km / hr ? <o> a ) num__287 <o> b ) num__288 <o> c ) num__600 <o> d ) num__277 <o> e ) num__121 |
let length of tunnel is x meter distance = num__700 + x meter time = num__1 minute = num__60 seconds speed = num__78 km / hr = num__78 * num__0.277777777778 m / s = num__21.6666666667 m / s distance = speed * time num__700 + x = ( num__21.6666666667 ) * num__60 num__700 + x = num__20 * num__65 = num__1300 x = num__1300 - num__700 = num__600 meters answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ divide__60.0__3.0__ multiply__65.0__20.0__ subtract__1300.0__700.0__ round__600.0__ |
subtract__63.0__3.0__ divide__60.0__3.0__ multiply__65.0__20.0__ subtract__1300.0__700.0__ divide__600.0__1.0__ |
| p and q can complete a work in num__40 days and num__24 days respectively . p alone started the work and q joined him after num__16 days till the completion of the work . how long did the work last ? <o> a ) num__5 days <o> b ) num__10 days <o> c ) num__14 days <o> d ) num__22 days <o> e ) num__25 days |
explanation : work done by p in num__1 day = num__0.025 work done by q in num__1 day = num__0.0416666666667 work done by p in num__16 days = num__16 Ã — ( num__0.025 ) = num__0.4 remaining work = num__1 â € “ num__0.4 = num__0.6 work done by p and q in num__1 day = num__0.025 + num__0.0416666666667 = num__0.0666666666667 number of days p and q take to complete the remaining work = ( num__0.6 ) / ( num__0.0666666666667 ) = num__9 total days = num__16 + num__9 = num__25 answer : option e <eor> e <eos> |
e |
divide__1.0__40.0__ divide__1.0__24.0__ divide__16.0__40.0__ km_to_mile_conversion__ add__0.025__0.0417__ add__24.0__1.0__ round__25.0__ |
divide__1.0__40.0__ divide__1.0__24.0__ divide__16.0__40.0__ divide__24.0__40.0__ add__0.025__0.0417__ add__24.0__1.0__ add__24.0__1.0__ |
| a num__300 m long train crosses a platform in num__39 sec while it crosses a signal pole in num__18 sec . what is the length of the platform ? <o> a ) num__400 m <o> b ) num__700 m <o> c ) num__120 m <o> d ) num__240 m <o> e ) num__350 m |
explanation : speed = num__16.6666666667 = num__16.6666666667 m / sec . let the length of the platform be x meters . then ( x + num__300 ) / num__39 = num__16.6666666667 num__3 x + num__900 = num__1950 = > x = num__350 m . answer : e <eor> e <eos> |
e |
divide__300.0__18.0__ multiply__300.0__3.0__ round__350.0__ |
divide__300.0__18.0__ multiply__300.0__3.0__ round__350.0__ |
| if the average ( mean ) of num__13 positive temperatures is x degrees fahrenheit then the sum of the num__3 greatest of these temperatures in degrees fahrenheit could be <o> a ) num__6 x <o> b ) num__12 x <o> c ) num__5 x / num__3 <o> d ) num__3 x / num__2 <o> e ) num__3 x / num__5 |
let the num__5 numbers be num__1 num__23 num__45 . . . num__13 ( since no restrictions are given ) . there mean is num__7 ( x ) . now the sum of greatest three would be num__11 + num__12 + num__13 = num__36 so the answer has to be num__12 x . . . . that is option b <eor> b <eos> |
b |
subtract__13.0__1.0__ add__13.0__23.0__ subtract__13.0__1.0__ |
add__1.0__11.0__ add__13.0__23.0__ add__1.0__11.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 seconds . find the length of the train . <o> a ) num__150 <o> b ) num__882 <o> c ) num__772 <o> d ) num__252 <o> e ) num__121 |
speed = num__60 * ( num__0.277777777778 ) m / sec = num__16.6666666667 m / sec length of train ( distance ) = speed * time ( num__16.6666666667 ) * num__9 = num__150 meter . answer : a <eor> a <eos> |
a |
round__150.0__ |
round__150.0__ |
| the average of num__45 results is num__25 and the average of other num__25 results is num__45 . what is the average of all the results ? <o> a ) num__24 <o> b ) num__32 <o> c ) num__48 <o> d ) num__50 <o> e ) none |
answer sum of num__70 result = sum of num__45 result + sum of num__25 result . = num__45 x num__25 + num__25 x num__45 = num__32.1428571429 correct option : b <eor> b <eos> |
b |
add__45.0__25.0__ round_down__32.1429__ |
add__45.0__25.0__ round_down__32.1429__ |
| the age of a man is twice the sum of the ages of his two sons . five years hence his age will be num__16 years more than the sum of the ages of his sons . what is the father ' s present age ? <o> a ) num__42 years <o> b ) num__40 years <o> c ) num__45 years <o> d ) num__48 years <o> e ) num__55 years |
let man ' s age be m and sons ages will be x & y given that m = num__2 ( x + y ) = x + y = m / num__2 m + num__5 = x + num__5 + y + num__5 + num__16 m = x + y + num__21 m = m / num__2 + num__21 m = num__42 answer : a <eor> a <eos> |
a |
add__16.0__5.0__ multiply__2.0__21.0__ multiply__2.0__21.0__ |
add__16.0__5.0__ multiply__2.0__21.0__ multiply__2.0__21.0__ |
| two identical machines have the ability to produce both nuts and bolts . however it takes num__1 second to produce a bolt but num__2 seconds to produce a nut . what is the fastest the two machines working together can produce num__2000 nuts and num__2000 bolts ? <o> a ) num__1250 seconds <o> b ) num__1500 seconds <o> c ) num__1750 seconds <o> d ) num__2000 seconds <o> e ) num__3000 seconds |
i used the rate formula ( r = \ frac { num__1 } { t } ) rate of num__1 st machine = rate of making num__1 bolt + rate of making num__1 nut = num__1 + \ frac { num__1 } { num__2 } rate of num__1 st machine = rate of num__2 nd machine after this i got lost . please can you help how to approach using algebra ? i would n ' t complicate : num__1 machine needs num__2000 * num__1 seconds to produce num__2000 bolts so num__2 need half of that so num__1000.0 seconds . num__1 machine needs num__2000 * num__2 seconds to produce num__2000 nuts so num__2 need half of that so num__2000 * num__1.0 seconds . e <eor> e <eos> |
e |
divide__2000.0__2.0__ add__2000.0__1000.0__ |
divide__2000.0__2.0__ add__2000.0__1000.0__ |
| a b c and d are four consecutive numbers . if the sum of a and dis num__107 what is the product of b and c ? <o> a ) num__2862 <o> b ) num__2578 <o> c ) num__2534 <o> d ) num__3354 <o> e ) num__4234 |
a num__2862 here d = a + num__3 a + a + num__3 = num__107 num__2 a = num__104 a = num__52 so numbers are num__52 num__53 num__54 and num__55 ∴ b × c = num__53 × num__54 = num__2862 <eor> a <eos> |
a |
subtract__107.0__3.0__ divide__104.0__2.0__ subtract__107.0__53.0__ subtract__107.0__52.0__ multiply__53.0__54.0__ |
subtract__107.0__3.0__ divide__104.0__2.0__ add__2.0__52.0__ add__2.0__53.0__ multiply__53.0__54.0__ |
| a person can swim in still water at num__4 km / h . if the speed of water num__2 km / h how many hours will the man take to swim back against the current for num__6 km ? <o> a ) num__3 <o> b ) num__8 <o> c ) num__7 <o> d ) num__6 <o> e ) num__2 |
m = num__4 s = num__2 us = num__4 - num__2 = num__2 d = num__6 t = num__3.0 = num__3 answer : a <eor> a <eos> |
a |
divide__6.0__2.0__ round__3.0__ |
divide__6.0__2.0__ subtract__6.0__3.0__ |
| num__150 liters of a mixture of milk and water contains in the ratio num__3 : num__2 . how much water should now be added so that the ratio of milk and water becomes num__3 : num__4 ? <o> a ) num__90 liters <o> b ) num__20 liters <o> c ) num__50 liters <o> d ) num__20 liters <o> e ) num__70 liters |
milk = num__0.6 * num__150 = num__90 liters water = num__50 liters num__90 : ( num__50 + p ) = num__3 : num__4 num__150 + num__3 p = num__360 = > p = num__70 num__50 liters of water are to be added for the ratio become num__3 : num__4 . answer : c <eor> c <eos> |
c |
multiply__150.0__0.6__ divide__150.0__3.0__ multiply__4.0__90.0__ divide__150.0__3.0__ |
multiply__150.0__0.6__ divide__150.0__3.0__ multiply__4.0__90.0__ divide__150.0__3.0__ |
| find large number from below question the difference of two numbers is num__1365 . on dividing the larger number by the smaller we get num__6 asquotient and the num__15 as remainder ? <o> a ) num__1245 <o> b ) num__1567 <o> c ) num__1678 <o> d ) num__1335 <o> e ) num__1635 |
let the smaller number be x . then larger number = ( x + num__1365 ) . x + num__1365 = num__6 x + num__15 num__5 x = num__1350 x = num__270 large number = num__270 + num__1365 = num__1635 e <eor> e <eos> |
e |
subtract__1365.0__15.0__ divide__1350.0__5.0__ add__1365.0__270.0__ add__1365.0__270.0__ |
subtract__1365.0__15.0__ divide__1350.0__5.0__ add__1365.0__270.0__ add__1365.0__270.0__ |
| in a jar there are balls in different colors : blue red green and yellow . the probability of drawing a blue ball is num__0.125 . the probability of drawing a red ball is num__0.6 . the probability of drawing a green ball is num__0.1 . if a jar can not contain more than num__50 balls how many yellow balls are in the jar ? <o> a ) num__33 <o> b ) num__7 <o> c ) num__14 <o> d ) num__21 <o> e ) num__9 |
answer of num__1 st just add the given probabilities p ( blue ) + p ( red ) + p ( green ) i . e num__0.125 + num__0.6 + num__0.1 = num__5 + num__24 + num__0.1 = num__0.825 now we know p ( blue ) + p ( red ) + p ( green ) + p ( yellow ) = num__1 num__0.825 + p ( yellow ) = num__1 p ( yellow ) = num__0.175 i . e why yellow balls are num__7 . b <eor> b <eos> |
b |
multiply__0.1__50.0__ subtract__1.0__0.825__ multiply__1.0__7.0__ |
multiply__0.1__50.0__ subtract__1.0__0.825__ multiply__1.0__7.0__ |
| hari spends num__50.0 of his income . his income is increased by num__10.0 and he increased his expenditure by num__5.0 . find the percentage increase in his savings ? <o> a ) num__10.0 <o> b ) num__15.0 <o> c ) num__20.0 <o> d ) num__25.0 <o> e ) num__30 % |
let original income = $ num__100 expenditure = $ num__50 savings = $ num__50 new income = $ num__110 new expenditure = $ num__1.05 * num__50 = $ num__52.5 new savings = num__110 - num__52.5 = $ num__57.5 increase in savings = num__57.5 - num__50 = $ num__7.5 increase % = num__7.5 * num__0.02 * num__100 = num__15.0 answer is b <eor> b <eos> |
b |
add__10.0__100.0__ multiply__50.0__1.05__ add__5.0__52.5__ subtract__57.5__50.0__ reverse__50.0__ add__10.0__5.0__ add__10.0__5.0__ |
add__10.0__100.0__ multiply__50.0__1.05__ subtract__110.0__52.5__ subtract__57.5__50.0__ reverse__50.0__ add__10.0__5.0__ add__10.0__5.0__ |
| mary peter and lucy were picking chestnuts . mary picked twice as much chestnuts than peter . lucy picked num__4 kg more than peter . together the three of them picked num__28 kg of chestnuts . how many kilograms of chestnuts did mary peter and lucy pick respectively ? <o> a ) num__6 num__10 and num__12 <o> b ) num__12 num__10 and num__6 <o> c ) num__10 num__6 and num__12 <o> d ) num__12 num__6 and num__10 <o> e ) num__6 num__12 and num__10 |
m = num__2 p l = p + num__4 m + p + l = num__28 num__2 p + p + ( p + num__4 ) = num__28 p = num__6 m = num__12 l = num__10 therefore mary peter and lucy picked num__12 num__6 and num__10 kg respectively . the answer is d . <eor> d <eos> |
d |
add__4.0__2.0__ multiply__2.0__6.0__ add__4.0__6.0__ multiply__2.0__6.0__ |
add__4.0__2.0__ multiply__2.0__6.0__ add__4.0__6.0__ add__2.0__10.0__ |
| if num__16.0 of num__40.0 of a number is num__4 then the number is <o> a ) num__200 <o> b ) num__225 <o> c ) num__62.5 <o> d ) num__320 <o> e ) none of these |
explanation : let num__0.16 × num__0.4 × a = num__4 a = num__4 × num__100 × num__6.25 × num__40 = num__62.5 correct option : c <eor> c <eos> |
c |
percent__62.5__100.0__ |
percent__62.5__100.0__ |
| a restaurant meal cost $ num__30.00 and there was no tax . if the tip was more than num__10 percent but less than num__15 percent of the cost of the meal then the total amount paid must have been between <o> a ) $ num__40 and $ num__42 <o> b ) $ num__39 and $ num__41 <o> c ) $ num__38 and $ num__40 <o> d ) $ num__37 and $ num__39 <o> e ) $ num__33 and $ num__35 |
the total amount for the meal was between num__30.0 * num__1.1 = num__33.00 and num__30.0 * num__1.15 = num__34.5 . only option which covers all possible values of the meal is e . answer : e . <eor> e <eos> |
e |
multiply__30.0__1.1__ multiply__30.0__1.15__ multiply__30.0__1.1__ |
multiply__30.0__1.1__ multiply__30.0__1.15__ multiply__30.0__1.1__ |
| two heavily loaded sixteen - wheeler transport trucks are num__690 kilometers apart sitting at two rest stops on opposite sides of the same highway . driver a begins heading down the highway driving at an average speed of num__90 kilometers per hour . exactly one hour later driver b starts down the highway toward driver a maintaining an average speed of num__80 kilometers per hour . how many kilometers farther than driver b will driver a have driven when they meet and pass each other on the highway ? <o> a ) num__120 <o> b ) num__130 <o> c ) num__150 <o> d ) num__320 <o> e ) num__450 |
i ' ve been reading the website for a while and i ' m always keen to see different approaches so i would like to share one that works for me : short version : truck a travels for an hour . distance remaining = num__690 - num__90 = num__600 k ratio of speeds num__9 : num__8 - > num__30.0 = num__30 truck a = num__90 + num__30 * num__9 = num__360 truck b = num__30 * num__8 = num__240 delta = num__120 km answer a <eor> a <eos> |
a |
subtract__690.0__90.0__ multiply__8.0__30.0__ add__90.0__30.0__ add__90.0__30.0__ |
subtract__690.0__90.0__ multiply__8.0__30.0__ add__90.0__30.0__ add__90.0__30.0__ |
| one person tharak speed with the current is num__10 km / hr and the speed of the current is num__5 km / hr . tharak ' s speed against the current is : <o> a ) num__0 km / hr <o> b ) num__1 km / hr <o> c ) num__2 km / hr <o> d ) num__3 km / hr <o> e ) num__4 km / hr |
man ' s speed with the current = num__10 km / hr = > speed of the man + speed of the current = num__10 km / hr speed of the current is num__5 km / hr hence speed of the man = num__10 - num__5 = num__5 km / hr man ' s speed against the current = speed of the man - speed of the current = num__5 - num__5 = num__0 km / hr answer : a <eor> a <eos> |
a |
percent__10.0__0.0__ |
percent__10.0__0.0__ |
| how many positive integer solutions does the equation num__2 x + num__5 y = num__100 have ? <o> a ) num__50 <o> b ) num__33 <o> c ) num__16 <o> d ) num__35 <o> e ) num__10 |
formula : ( constant ) / ( lcm of two nos ) = num__100 / ( num__2 * num__5 ) = num__10 answer : e <eor> e <eos> |
e |
multiply__2.0__5.0__ multiply__2.0__5.0__ |
multiply__2.0__5.0__ multiply__2.0__5.0__ |
| the compound ratio of num__5 : num__6 num__3 : num__2 and num__6 : num__5 ? <o> a ) num__1 : num__1 <o> b ) num__1 : num__87 <o> c ) num__1 : num__6 <o> d ) num__3 : num__2 <o> e ) num__1 : num__2 |
num__0.833333333333 * num__1.5 * num__1.2 = num__1.5 num__3 : num__2 answer : d <eor> d <eos> |
d |
divide__5.0__6.0__ divide__3.0__2.0__ reverse__0.8333__ subtract__5.0__2.0__ |
divide__5.0__6.0__ divide__3.0__2.0__ reverse__0.8333__ multiply__2.0__1.5__ |
| real - estate salesman z is selling a house at a num__20 percent discount from its retail price . real - estate salesman x vows to match this price and then offers an additional num__10 percent discount . real - estate salesman y decides to average the prices of salesmen z and x then offer an additional num__25 percent discount . salesman y ' s final price is what fraction of salesman x ' s final price ? <o> a ) num__0.9 <o> b ) num__0.72 <o> c ) num__0.7125 <o> d ) num__0.791666666667 <o> e ) num__0.76 |
let retail price = $ num__100 . z sells with num__20.0 discount . so z sells at $ num__80 . x matches the price and gives additional num__10.0 . so x sells at num__80 - ( num__10.0 of num__80 ) = num__80 - num__8 = $ num__72 y avgs x and z [ ( num__72 + num__80 ) / num__2 = num__76 ] . y gives additional num__25.0 discount . so y sells at num__76 - ( num__25.0 of num__76 ) = num__76 - num__19 = $ num__57 now ratio of y to x = num__0.791666666667 = num__0.791666666667 answer will be d . <eor> d <eos> |
d |
subtract__100.0__20.0__ divide__80.0__10.0__ subtract__80.0__8.0__ divide__20.0__10.0__ subtract__76.0__19.0__ divide__57.0__72.0__ divide__57.0__72.0__ |
subtract__100.0__20.0__ divide__80.0__10.0__ subtract__80.0__8.0__ divide__20.0__10.0__ subtract__76.0__19.0__ divide__57.0__72.0__ divide__57.0__72.0__ |
| how many four - digit numbers that do not contain the digits num__2 num__3 num__4 num__5 or num__6 are there ? <o> a ) num__200 <o> b ) num__400 <o> c ) num__500 <o> d ) num__600 <o> e ) num__800 |
the num__1 st digit can be filled up by the numbers : { num__17 num__89 } = num__4 ways the num__2 nd digit can be filled up by the numbers : { num__0 num__17 num__89 } = num__5 ways the num__3 rd digit can be filled up by the numbers : { num__0 num__17 num__89 } = num__5 ways the num__4 th digit can be filled up by the numbers : { num__0 num__17 num__89 } = num__5 ways the total number of such four - digit numbers is num__4 * num__5 * num__5 * num__5 = num__500 the answer is c . <eor> c <eos> |
c |
subtract__3.0__2.0__ multiply__1.0__500.0__ |
subtract__3.0__2.0__ multiply__1.0__500.0__ |
| an employer pays rs . num__30 for each day a worker works and forfeits rs . num__5 for each day he is idle . at the end of num__60 days a worker gets rs . num__500 . for how many days did the worker remain idle ? <o> a ) num__52 <o> b ) num__27 <o> c ) num__99 <o> d ) num__61 <o> e ) num__11 |
explanation : suppose the worker remained idle for m days . then he worked for ( num__60 - m ) days . num__30 ( num__60 - m ) – num__5 m = num__500 num__1800 – num__25 m = num__500 num__25 m = num__1300 m = num__52 so the worker remained idle for num__52 days . answer : a <eor> a <eos> |
a |
multiply__30.0__60.0__ subtract__30.0__5.0__ subtract__1800.0__500.0__ divide__1300.0__25.0__ round__52.0__ |
multiply__30.0__60.0__ subtract__30.0__5.0__ subtract__1800.0__500.0__ divide__1300.0__25.0__ round__52.0__ |
| in what time will a train num__100 m long cross an electric pole it its speed be num__144 km / hr ? <o> a ) num__2.5 sec <o> b ) num__4.25 sec <o> c ) num__5 sec <o> d ) num__12.5 sec <o> e ) num__6 sec |
answer : option a speed = num__144 * num__0.277777777778 = num__40 m / sec time taken = num__2.5 = num__2.5 sec . <eor> a <eos> |
a |
divide__100.0__40.0__ round__2.5__ |
divide__100.0__40.0__ divide__100.0__40.0__ |
| gold is num__10 times as heavy as water and copper is num__5 times as heavy as water . in what ratio should these be mixed to get an alloy num__6 times as heavy as water ? <o> a ) num__3 : num__2 <o> b ) num__1 : num__4 <o> c ) num__3 : num__1 <o> d ) num__5 : num__2 <o> e ) num__4 : num__3 |
g = num__10 w c = num__5 w let num__1 gm of gold mixed with x gm of copper to get num__1 + x gm of the alloy num__1 gm gold + x gm copper = x + num__1 gm of alloy num__10 w + num__5 wx = x + num__1 * num__6 w num__10 + num__5 x = num__6 ( x + num__1 ) x = num__4 ratio of gold with copper = num__1 : num__4 = num__1 : num__4 answer is b <eor> b <eos> |
b |
subtract__6.0__5.0__ subtract__10.0__6.0__ reverse__1.0__ |
subtract__6.0__5.0__ subtract__10.0__6.0__ reverse__1.0__ |
| a certain nyc taxi driver has decided to start charging a rate of r cents per person per mile . how much in dollars would it cost num__10 people to travel x miles if he decides to give them a num__50.0 discount ? <o> a ) num__3 xr / num__2 <o> b ) num__3 x / num__200 r <o> c ) num__3 r / num__200 x <o> d ) num__3 xr / num__200 <o> e ) xr / num__20 |
num__10 xr / num__2 is in cents - num__10 xr / num__200 dollars = xr / num__20 answer : e <eor> e <eos> |
e |
percent__10.0__200.0__ percent__10.0__200.0__ |
percent__10.0__200.0__ percent__10.0__200.0__ |
| num__6 persons standing in queue with different age group after two years their average age will be num__43 and seventh person joined with them . hence the current average age has become num__45 . find the age of seventh person ? <o> a ) num__65 <o> b ) num__67 <o> c ) num__68 <o> d ) num__69 <o> e ) num__50 |
explanation : let the sum of current ages of num__6 persons = x given average age of num__6 person after num__2 years = num__43 = > x + num__6 ( num__2 ) / num__6 = num__43 = > x + num__12 = num__258 = x = num__246 let the seventh ' s person age will be y given current average age of num__7 persons = num__45 [ sum of current num__6 person ' s age ( x ) + seventh person ' s age ( y ) ] / num__7 = num__45 = > num__246 + y = num__45 ( num__7 ) = > y = num__315 - num__246 = > y = num__69 hence seventh person ' s age = num__69 hence ( d ) is the correct answer . answer : d <eor> d <eos> |
d |
subtract__45.0__43.0__ multiply__6.0__2.0__ multiply__6.0__43.0__ subtract__258.0__12.0__ multiply__45.0__7.0__ subtract__315.0__246.0__ subtract__315.0__246.0__ |
subtract__45.0__43.0__ multiply__6.0__2.0__ multiply__6.0__43.0__ subtract__258.0__12.0__ multiply__45.0__7.0__ subtract__315.0__246.0__ subtract__315.0__246.0__ |
| if a b and c together can finish a piece of work in num__4 days . a alone in num__12 days and b in num__18 days then c alone can do it in ? <o> a ) num__1 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__6 |
c = num__0.25 - num__0.0833333333333 – num__0.0555555555556 = num__0.111111111111 = > num__9 days ' answer : d <eor> d <eos> |
d |
round__9.0__ |
subtract__18.0__9.0__ |
| a rectangular floor that measures num__8 meters by num__10 meters is to be covered with carpet squares that each measure num__2 meters by num__2 meters . if the carpet squares cost $ num__10 apiece what is the total cost for the number of carpet squares needed to cover the floor ? <o> a ) $ num__200 <o> b ) $ num__240 <o> c ) $ num__480 <o> d ) $ num__960 <o> e ) $ num__1 |
920 |
the width of the rectangular floor ( num__8 m ) is a multiple of one side of the square ( num__2 m ) and the length of the floor ( num__10 m ) is also a multiple of the side of the square . so the number of carpets to cover the floor is ( num__4.0 ) * ( num__5.0 ) = num__20 . the total cost is num__20 * num__10 = $ num__200 . the answer is therefore a . <eor> a <eos> |
a |
a |
| if num__6 women can color num__360 m long cloth in num__3 days then num__5 women can color num__100 m long cloth in ? <o> a ) num__2 days <o> b ) num__3 days <o> c ) num__2.5 days <o> d ) num__4 days <o> e ) num__1 day |
the length of cloth painted by one woman in one day = num__60.0 × num__3 = num__20 m no . of days required to paint num__100 m cloth by num__5 women = num__20.0 × num__20 = num__1 day answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ divide__100.0__5.0__ subtract__6.0__5.0__ round__1.0__ |
hour_to_min_conversion__ divide__100.0__5.0__ subtract__6.0__5.0__ round__1.0__ |
| if num__25.0 of ( x - y ) = num__15.0 of ( x + y ) then what percent of x is y ? <o> a ) num__20.0 <o> b ) num__15.0 <o> c ) num__30.0 <o> d ) num__25.0 <o> e ) none of these |
explanation : solution : num__25.0 of ( x - y ) = num__15.0 of ( x + y ) num__25 ( x - y ) / num__100 = num__15 ( x + y ) / num__100 num__5 ( x - y ) = num__3 ( x + y ) x = num__4 y . . ' . required percentage = ( y * num__100 / x ) % = y * num__25.0 y = num__25.0 answer : d <eor> d <eos> |
d |
divide__15.0__5.0__ divide__100.0__25.0__ divide__100.0__4.0__ |
divide__15.0__5.0__ divide__100.0__25.0__ divide__100.0__4.0__ |
| when a certain number x is divided by num__54 the remainder is num__22 . what is the remainder when x is divided by num__18 ? <o> a ) num__3 <o> b ) num__1 <o> c ) num__2 <o> d ) num__4 <o> e ) num__5 |
let possible value of x is num__76 least possible value of x / num__18 is num__4.22222222222 = > num__4 quotient with remainder num__4 thus answer is ( d ) num__4 <eor> d <eos> |
d |
add__54.0__22.0__ divide__76.0__18.0__ round_down__4.2222__ round_down__4.2222__ |
add__54.0__22.0__ divide__76.0__18.0__ round_down__4.2222__ round_down__4.2222__ |
| if y is num__60.0 greater than x than x is what % less than y ? <o> a ) num__29.3 <o> b ) num__33.4 <o> c ) num__37.5 <o> d ) num__41.6 <o> e ) num__45.7 % |
y = num__1.6 x x = y / num__1.6 = num__10 y / num__16 = num__5 y / num__8 x is num__0.375 less which is num__37.5 less than y . the answer is c . <eor> c <eos> |
c |
multiply__1.6__10.0__ multiply__5.0__1.6__ divide__60.0__1.6__ divide__60.0__1.6__ |
multiply__1.6__10.0__ multiply__5.0__1.6__ divide__60.0__1.6__ divide__60.0__1.6__ |
| ayesha ’ s father was num__38 years of age when she was born while her mother was num__31 years old when her brother four years younger to her was born . what is the difference between the ages of her parents ? <o> a ) num__2 years <o> b ) num__4 years <o> c ) num__11 years <o> d ) num__8 years <o> e ) none |
explanation mother ’ s age when ayesha ’ s brother was born = num__31 years . father ’ s age when ayesha ’ s brother was born = ( num__38 + num__4 ) years = num__42 years . required difference = ( num__42 – num__31 ) years = num__11 years . answer c <eor> c <eos> |
c |
add__38.0__4.0__ subtract__42.0__31.0__ subtract__42.0__31.0__ |
add__38.0__4.0__ subtract__42.0__31.0__ subtract__42.0__31.0__ |
| the difference between the ages of two persons is num__11 years . ten years ago the elder one was twice as old as the younger one . the present age of the elder person is <o> a ) num__18 <o> b ) num__19 <o> c ) num__20 <o> d ) num__17 <o> e ) num__16 |
let their ages of x years and ( x + num__11 ) years respectively . then ( x + num__11 ) - num__10 = num__2 ( x - num__10 ) ⇔ x + num__1 = num__2 x - num__20 ⇔ num__3 x = num__21 . ∴ present age of the elder person = ( num__7 + num__11 ) = num__18 years . answer : num__18 years <eor> a <eos> |
a |
subtract__11.0__10.0__ multiply__2.0__10.0__ add__1.0__2.0__ add__11.0__10.0__ subtract__10.0__3.0__ add__11.0__7.0__ add__11.0__7.0__ |
subtract__11.0__10.0__ multiply__2.0__10.0__ add__1.0__2.0__ add__11.0__10.0__ subtract__10.0__3.0__ add__11.0__7.0__ add__11.0__7.0__ |
| there is num__66 x num__33 m rectangular area . ram is num__1.375 times faster than krishna . both of them started walking at opposite ends and they met at some point then ram said ` ` see you in the other end ' ' then they continued walking . after some time ram thought he will have tea so he turned back walked back num__15 meters then he changed his mind again and continued walking . how much krishna has traveled by the time they meet ? <o> a ) num__93 <o> b ) num__34 <o> c ) num__95 <o> d ) num__96 <o> e ) num__85 |
let x be the distance between the point they first met total time taken by the two willbe equal so ( x + num__30 ) * num__0.727272727273 = ( num__198 - x ) x = num__102 m distance travelled by krishna = num__198 - num__102 = num__96 m answer : d <eor> d <eos> |
d |
reverse__1.375__ add__66.0__30.0__ add__66.0__30.0__ |
reverse__1.375__ add__66.0__30.0__ add__66.0__30.0__ |
| a boy runs num__200 metres in num__30 seconds . what is his speed ? <o> a ) num__20 km / hr <o> b ) num__24 km / hr <o> c ) num__30 km / hr <o> d ) num__32 km / hr <o> e ) num__34 km / hr |
num__6.66666666667 * num__3.6 = num__24 km / hr answer : b <eor> b <eos> |
b |
divide__200.0__30.0__ round__24.0__ |
divide__200.0__30.0__ round__24.0__ |
| if y = num__2 + num__2 k and y ≠ num__0 y ≠ num__0 then num__1 / y + num__1 / y + num__1 / y + num__1 / y = ? <o> a ) num__1 / ( num__8 + num__8 k ) <o> b ) num__2 / ( num__1 + k ) <o> c ) num__1 / ( num__8 + k ) <o> d ) num__4 / ( num__8 + k ) <o> e ) num__4 / ( num__1 + k ) |
num__1 / y + num__1 / y + num__1 / y + num__1 / y = num__4 / y = num__4 / ( num__2 + num__2 k ) = num__2 / ( num__1 + k ) answer : b <eor> b <eos> |
b |
multiply__2.0__1.0__ |
divide__2.0__1.0__ |
| one fourth of one third of two fifth of a number is num__16 . what will be num__40.0 of that number <o> a ) num__140 <o> b ) num__150 <o> c ) num__180 <o> d ) num__192 <o> e ) num__250 |
explanation : ( num__0.25 ) * ( num__0.333333333333 ) * ( num__0.4 ) * x = num__16 then x = num__16 * num__30 = num__480 num__40.0 of num__480 = num__192 answer : option d <eor> d <eos> |
d |
percent__40.0__480.0__ percent__40.0__480.0__ |
percent__40.0__480.0__ percent__40.0__480.0__ |
| the temperature of a certain cup of coffee num__30 minutes after it was poured was num__120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = num__120 * num__2 ^ ( - at ) + num__60 where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee num__30 minutes after it was poured was how many degrees fahrenheit ? <o> a ) num__65 <o> b ) num__120 <o> c ) num__80 <o> d ) num__85 <o> e ) num__90 |
first we have to find a . we know that after t = num__30 minutes the temperature f = num__120 degrees . hence : num__120 = num__120 * ( num__2 ^ - num__30 a ) + num__60 num__60 = num__120 * ( num__2 ^ - num__30 a ) num__0.5 = num__2 ^ - num__30 a num__0.5 = num__2 ^ - num__30 a num__2 ^ - num__1 = num__2 ^ - num__30 a - num__1 = - num__30 a num__0.0333333333333 = a now we need to find f after t = num__30 minutes : f = num__120 * ( num__2 ^ - num__0.0333333333333 * num__30 ) + num__60 f = num__120 * ( num__2 ^ - num__1 ) + num__60 f = num__120 * ( num__0.5 ^ num__1 ) + num__60 f = num__120 * num__0.5 + num__60 f = num__60 + num__60 = num__120 answer b ! <eor> b <eos> |
b |
reverse__2.0__ multiply__2.0__0.5__ reverse__30.0__ multiply__120.0__1.0__ |
reverse__2.0__ multiply__2.0__0.5__ reverse__30.0__ multiply__120.0__1.0__ |
| a circular path of num__13 m radius has marginal walk num__2 m wide all round it . find the cost of leveling the walk at num__25 p per m num__2 ? <o> a ) rs . num__45 <o> b ) rs . num__78 <o> c ) rs . num__44 <o> d ) rs . num__40 <o> e ) rs . num__46 |
π ( num__152 - num__132 ) = num__176 num__176 * num__0.25 = rs . num__44 answer : c <eor> c <eos> |
c |
multiply__176.0__0.25__ triangle_area__2.0__44.0__ |
multiply__176.0__0.25__ multiply__176.0__0.25__ |
| if each participant of a chess tournament plays exactly one game with each of the remaining participants then num__171 games will be played during the tournament . find the number of participants . <o> a ) num__15 <o> b ) num__16 <o> c ) num__17 <o> d ) num__18 <o> e ) num__19 |
let p be the number of participants . pc num__2 = num__171 ( p ) ( p - num__1 ) = num__342 = num__19 * num__18 p = num__19 the answer is e . <eor> e <eos> |
e |
multiply__171.0__2.0__ subtract__19.0__1.0__ add__1.0__18.0__ |
multiply__171.0__2.0__ subtract__19.0__1.0__ add__1.0__18.0__ |
| a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by num__15.0 while buying and by num__20.0 while selling . what is his percentage profit ? <o> a ) num__10.22 <o> b ) num__20.22 <o> c ) num__21.22 <o> d ) num__43.75 <o> e ) ca n ' t be calculated |
the owner buys num__100 kg but actually gets num__115 kg ; the owner sells num__100 kg but actually gives num__80 kg ; profit : ( num__115 - num__80 ) / num__80 * num__100 = ~ num__43.75 answer : d . <eor> d <eos> |
d |
percent__43.75__100.0__ |
percent__43.75__100.0__ |
| two trains of equal lengths take num__10 sec and num__18 sec respectively to cross a telegraph post . if the length of each train be num__120 m in what time will they cross other travelling in opposite direction ? <o> a ) num__11 <o> b ) num__9 <o> c ) num__13 <o> d ) num__14 <o> e ) num__12.8 |
speed of the first train = num__12.0 = num__12 m / sec . speed of the second train = num__6.66666666667 = num__6.7 m / sec . relative speed = num__12 + num__6.7 = num__18.7 m / sec . required time = ( num__120 + num__120 ) / num__18.7 = num__12.8 sec . answer : option e <eor> e <eos> |
e |
divide__120.0__10.0__ divide__120.0__18.0__ add__12.0__6.7__ round__12.8__ |
divide__120.0__10.0__ divide__120.0__18.0__ add__12.0__6.7__ round__12.8__ |
| a furniture manufacturer has two machines but only one can be used at a time . machine w is utilized during the first shift and machine b during the second shift while both work half of the third shift . if machine w can do the job in num__12 days working two shifts and machine b can do the job in num__15 days working two shifts how many days will it take to do the job with the current work schedule ? <o> a ) num__14 <o> b ) num__13 <o> c ) num__11 <o> d ) num__9 <o> e ) num__7 |
' approximately ' could actually make such a question ambiguous . not this one though but a similar question with the answer as num__9.2 days . you round off num__8.89 days as num__9 days and everything is fine in this question . what do you do when you get num__9.2 days ? do you need num__9 days or num__10 days ? can you round off num__9.2 as num__9 even though that is what you do with numbers ? no because in num__9 days your work is not over . you do need num__10 days . to finish a work machine w say you need to work full num__9 days and a part of the num__10 th day . if i ask you how many days do you need to complete the work will you say num__9 or num__10 ? you will say num__10 even if you do n ' t use the num__10 th day fully = d <eor> d <eos> |
d |
round__9.0__ |
round__9.0__ |
| num__6 men num__8 women and num__6 children complete a job for a sum of rs . num__950 . if their individual wages are in the ratio num__4 : num__3 : num__2 the total money earned by the children is <o> a ) rs . num__150 <o> b ) rs . num__170 <o> c ) rs . num__180 <o> d ) rs . num__190 <o> e ) none of these |
explanation : ratio of wages of num__6 men num__8 women and num__6 children = num__6 x num__4 : num__8 x num__3 : num__6 x num__2 = num__24 : num__24 : num__12 = num__2 : num__2 : num__1 money earned by children = num__950 x ( num__0.2 ) = rs . num__190 answer : option d <eor> d <eos> |
d |
multiply__6.0__4.0__ multiply__6.0__2.0__ subtract__4.0__3.0__ multiply__950.0__0.2__ multiply__950.0__0.2__ |
multiply__6.0__4.0__ multiply__6.0__2.0__ subtract__4.0__3.0__ multiply__950.0__0.2__ multiply__950.0__0.2__ |
| second saturday and every sunday is a holiday . how many working days will be there in a month of num__30 days beginning on a saturday ? <o> a ) num__24 <o> b ) num__23 <o> c ) num__18 <o> d ) num__21 <o> e ) num__22 |
explanation : mentioned month begins on a saturday and has num__30 days sundays = num__2 nd num__9 th num__16 th num__23 rd num__30 th = > total sundays = num__5 every second saturday is holiday . num__1 second saturday in every month total days in the month = num__30 total working days = num__30 - ( num__5 + num__1 ) = num__24 . answer : option a <eor> a <eos> |
a |
add__1.0__23.0__ round__24.0__ |
add__1.0__23.0__ add__1.0__23.0__ |
| if x = the product of eight distinct prime numbers how many factors does x have besides num__1 and itself ? <o> a ) num__166 <o> b ) num__188 <o> c ) num__210 <o> d ) num__232 <o> e ) num__254 |
since x has num__8 distinct prime factors x has a total of num__2 ^ num__8 = num__256 factors . besides num__1 and itself x has num__254 factors . the answer is e . <eor> e <eos> |
e |
subtract__256.0__2.0__ multiply__1.0__254.0__ |
subtract__256.0__2.0__ multiply__1.0__254.0__ |
| a combustion reaction forms carbon dioxide . a carbon dioxide molecule contains one carbon and two oxygen atoms . if over a period of num__10 minutes a combustion reaction creates num__10000 molecules of carbon dioxide then approximately how many more atoms of oxygen than carbon are created on average per minute ? <o> a ) num__1000 <o> b ) num__500 <o> c ) num__250 <o> d ) num__50 <o> e ) num__0 |
solution : num__10000 carbon dioxide molecules are created over a period of num__10 minutes . therefore num__1000.0 = num__1000 carbon dioxide molecules are created on average per minute each carbon dioxide molecule contains one carbon atom and two oxygen atoms . so num__1000 carbon dioxide molecules contain num__1 × num__1000 = num__1000 carbon atoms and num__2 × num__1000 = num__2000 oxygen atoms . the difference is num__2000 – num__1000 = num__1000 . the correct answer is a . <eor> a <eos> |
a |
divide__10000.0__10.0__ multiply__2.0__1000.0__ divide__10000.0__10.0__ |
divide__10000.0__10.0__ multiply__2.0__1000.0__ divide__10000.0__10.0__ |
| using all the letters of the word ` ` nokia ' ' how many words can be formed which begin with n and end with a ? <o> a ) num__9 <o> b ) num__8 <o> c ) num__6 <o> d ) num__45 <o> e ) num__2 |
there are five letters in the given word . consider num__5 blanks . . . . the first blank and last blank must be filled with n and a all the remaining three blanks can be filled with the remaining num__3 letters in num__3 ! ways . the number of words = num__3 ! = num__6 . answer : c <eor> c <eos> |
c |
vowel_space__ die_space__ die_space__ |
vowel_space__ die_space__ die_space__ |
| the area of a square is num__4096 sq cm . find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is num__24 cm less than the side of the square ? <o> a ) num__5 : num__17 <o> b ) num__5 : num__14 <o> c ) num__5 : num__10 <o> d ) num__5 : num__11 <o> e ) num__5 : num__16 |
let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a num__2 = num__4096 = num__212 a = ( num__212 ) num__0.5 = num__26 = num__64 l = num__2 a and b = a - num__24 b : l = a - num__24 : num__2 a = num__40 : num__128 = num__5 : num__16 answer : e <eor> e <eos> |
e |
power__4096.0__0.5__ rectangle_perimeter__24.0__40.0__ rectangle_perimeter__0.5__2.0__ triangle_area__0.5__64.0__ rectangle_perimeter__0.5__2.0__ |
power__4096.0__0.5__ rectangle_perimeter__24.0__40.0__ rectangle_perimeter__0.5__2.0__ triangle_area__0.5__64.0__ rectangle_perimeter__0.5__2.0__ |
| if two dice are thrown together the probability of getting an even number on one die and an odd number on the other is ? <o> a ) num__0.2 <o> b ) num__0.5 <o> c ) num__0.333333333333 <o> d ) num__0.111111111111 <o> e ) num__0.5 |
the number of exhaustive outcomes is num__36 . let e be the event of getting an even number on one die and an odd number on the other . let the event of getting either both even or both odd then = num__0.5 = num__0.5 p ( e ) = num__1 - num__0.5 = num__0.5 . answer : b <eor> b <eos> |
b |
negate_prob__0.5__ |
negate_prob__0.5__ |
| the perimeters of two squares are num__40 cm and num__32 cm . find the perimeter of a third square whose area is equal to the difference of the areas of the two squares . <o> a ) num__22 <o> b ) num__24 <o> c ) num__77 <o> d ) num__99 <o> e ) num__11 |
explanation : side of first square = num__10.0 = num__10 cm side of second square = num__8.0 = num__8 cm area of third squre = num__10 ² - num__8 ² = num__36 sq cm side of third square = \ inline { \ color { black } \ sqrt { num__36 } } = num__6 cm required perimeter = num__6 * num__4 = num__24 cm answer : b ) num__24 <eor> b <eos> |
b |
rectangle_perimeter__8.0__10.0__ side_by_diagonal__10.0__8.0__ square_perimeter__6.0__ square_perimeter__6.0__ |
rectangle_perimeter__8.0__10.0__ side_by_diagonal__10.0__8.0__ multiply__4.0__6.0__ multiply__4.0__6.0__ |
| the cricket team of num__11 members is num__26 yrs old & the wicket keeper is num__3 yrs older . if the ages ofthese num__2 are excluded the average age of theremaining players is num__1 year less than the average age of the whole team . what is the average age of the team ? <o> a ) num__19 <o> b ) num__21 <o> c ) num__23 <o> d ) num__27 <o> e ) num__30 |
let the average age of the whole team be x years . num__11 x - ( num__26 + num__29 ) = num__9 ( x - num__1 ) = > num__11 x - num__9 x = num__46 = > num__2 x = num__46 = > x = num__23 . so average age of the team is num__23 years . c <eor> c <eos> |
c |
add__26.0__3.0__ subtract__11.0__2.0__ subtract__26.0__3.0__ subtract__26.0__3.0__ |
add__26.0__3.0__ subtract__11.0__2.0__ subtract__26.0__3.0__ subtract__26.0__3.0__ |
| excluding stoppages the speed of a bus is num__50 kmph and including stoppages it is num__35 kmph . for how many minutes does the bus stop per hour ? <o> a ) num__17 minutes <o> b ) num__18 minutes <o> c ) num__19 minutes <o> d ) num__16 minutes <o> e ) none of these |
explanation : in one hour due to stoppages it covers num__15 km less . time taken to cover num__15 km = [ num__0.3 * num__60 ] min = num__18 min answer : b <eor> b <eos> |
b |
subtract__50.0__35.0__ divide__15.0__50.0__ hour_to_min_conversion__ multiply__60.0__0.3__ round__18.0__ |
subtract__50.0__35.0__ divide__15.0__50.0__ hour_to_min_conversion__ multiply__60.0__0.3__ multiply__60.0__0.3__ |
| all possible two factors products are formed from the numbers num__1 num__2 num__3 num__4 . . . . . num__200 . the number of factors out of total obtained which are multiples of num__5 is <o> a ) num__5040 <o> b ) num__7180 <o> c ) num__8150 <o> d ) num__7280 <o> e ) none of these |
the total number of two factor products = num__200 c num__2 . the number of numbers from num__1 to num__200 which are not multiples of num__5 is num__160 . therefore the total number of two factor products which are not multiple of num__5 is num__160 c num__2 . hence the required number of factors which are multiples of num__5 = num__200 c num__2 – num__160 c num__2 = num__7180 . answer b <eor> b <eos> |
b |
multiply__1.0__7180.0__ |
multiply__1.0__7180.0__ |
| henrikh lives x blocks from his office . it takes him num__1 minute per block to walk to work and num__20 seconds per block to ride his bicycle to work . if it takes him exactly num__8 minutes more to walk to work than to ride his bicycle then x equals <o> a ) num__4 <o> b ) num__7 <o> c ) num__12 <o> d ) num__15 <o> e ) num__20 |
the best way would be . . . per km he takes num__40 secs extra if he walks so he will take num__8 mins or num__8 * num__60 secs in num__8 * num__1.5 = num__12 km answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ divide__60.0__40.0__ subtract__20.0__8.0__ round__12.0__ |
hour_to_min_conversion__ divide__60.0__40.0__ multiply__8.0__1.5__ multiply__1.0__12.0__ |
| if num__1 * num__3 * num__5 = num__16 num__3 * num__5 * num__7 = num__38 then find num__2 * num__7 * num__9 = ? <o> a ) num__65 <o> b ) num__68 <o> c ) num__72 <o> d ) num__80 <o> e ) num__77 |
( num__9 * num__7 ) + num__2 = num__65 a <eor> a <eos> |
a |
multiply__1.0__65.0__ |
multiply__1.0__65.0__ |
| using all the letters of the word ` ` nokiased ' ' how many words can be formed which begin with n and end with a ? <o> a ) num__800 <o> b ) num__720 <o> c ) num__900 <o> d ) num__300 <o> e ) num__100 |
there are five letters in the given word . consider num__8 blanks . . . . the first blank and last blank must be filled with n and a all the remaining three blanks can be filled with the remaining num__3 letters in num__6 ! ways . the number of words = num__6 ! = num__720 . answer : b <eor> b <eos> |
b |
die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| a man has rs . num__320 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__37 <o> b ) num__38 <o> c ) num__60 <o> d ) num__90 <o> e ) num__28 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__320 num__16 x = num__320 x = num__20 . hence total number of notes = num__3 x = num__60 . answer : c <eor> c <eos> |
c |
divide__320.0__16.0__ multiply__3.0__20.0__ multiply__3.0__20.0__ |
divide__320.0__16.0__ multiply__3.0__20.0__ multiply__3.0__20.0__ |
| the surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are num__6 cm each . the radius of the sphere is <o> a ) num__3 cm <o> b ) num__4 cm <o> c ) num__6 cm <o> d ) num__8 cm <o> e ) none |
solution num__4 Î r num__2 = num__2 Î num__3 x num__6 â ‡ ’ r num__2 = ( num__3 x num__3.0 ) â ‡ ’ num__9 â ‡ ’ r = num__3 cm . answer a <eor> a <eos> |
a |
triangle_area__6.0__3.0__ triangle_area__2.0__3.0__ |
triangle_area__6.0__3.0__ triangle_area__2.0__3.0__ |
| a man can row upstream at num__15 kmph and downstream at num__25 kmph and then find the speed of the man in still water ? <o> a ) num__27 <o> b ) num__29 <o> c ) num__30 <o> d ) num__20 <o> e ) num__24 |
us = num__15 ds = num__25 m = ( num__15 + num__25 ) / num__2 = num__20 answer : d <eor> d <eos> |
d |
round__20.0__ |
round__20.0__ |
| perimeter of the back wheel = num__9 feet front wheel = num__7 feet on a certain distance the front wheel gets num__10 revolutions more than the back wheel . what is the distance ? <o> a ) num__315 <o> b ) num__325 <o> c ) num__335 <o> d ) num__345 <o> e ) num__305 |
for num__1 revolution : front wheel goes num__7 ft . and back wheel goes num__9 feet . let distance is x feet . x / num__7 = ( x / num__9 ) + num__10 or num__2 x / num__63 = num__10 or x = ( num__10 * num__63 ) / num__2 therefore x = num__315 answer : a <eor> a <eos> |
a |
subtract__10.0__9.0__ subtract__9.0__7.0__ multiply__9.0__7.0__ round__315.0__ |
subtract__10.0__9.0__ subtract__9.0__7.0__ multiply__9.0__7.0__ divide__315.0__1.0__ |
| let n ~ be defined for all positive integers n as the remainder when ( n - num__1 ) ! is divided by n . what is the value of num__31 ~ ? <o> a ) num__2 <o> b ) num__1 <o> c ) num__0 <o> d ) num__8 <o> e ) num__31 |
n ~ = ( n - num__1 ) ! so num__31 ~ = ( num__31 - num__1 ) ! = num__30 ! when num__30 ! / num__31 we have num__16 * num__2 inside num__30 ! hence num__31 gets cancelled and we get remainder as num__0 c <eor> c <eos> |
c |
subtract__31.0__1.0__ multiply__1.0__0.0__ |
subtract__31.0__1.0__ multiply__1.0__0.0__ |
| a and b started a partnership business investing some amount in the ratio of num__3 : num__5 . c joined then after six months with an amount equal to that of b . in what proportion should the profit at the end of one year be distributed among a b and c ? <o> a ) num__3 : num__5 : num__2 <o> b ) num__3 : num__5 : num__5 <o> c ) num__6 : num__10 : num__5 <o> d ) num__5 : num__8 : num__10 <o> e ) data inadequate |
let the initial investments of a and b be num__3 x and num__5 x . a : b : c = ( num__3 x x num__12 ) : ( num__5 x x num__12 ) : ( num__5 x x num__6 ) = num__36 : num__60 : num__30 = num__6 : num__10 : num__5 . <eor> c <eos> |
c |
multiply__3.0__12.0__ multiply__5.0__12.0__ multiply__5.0__6.0__ divide__60.0__6.0__ divide__36.0__6.0__ |
multiply__3.0__12.0__ multiply__5.0__12.0__ multiply__5.0__6.0__ divide__60.0__6.0__ divide__36.0__6.0__ |
| john and andy start a two - length swimming race at the same moment but from opposite ends of the pool . they swim in lanes at uniform speeds but john is faster than andy . they num__1 st pass at a point num__18.5 m from the deep end and having completed one length each num__1 is allowed to rest on the edge for exactly num__45 sec . after setting off on the return length the swimmers pass for the num__2 nd time just num__10.5 m from the shallow end . how long is the pool ? <o> a ) num__45 <o> b ) num__50 <o> c ) num__55 <o> d ) num__60 <o> e ) num__65 |
let x = length of pool at first meeting combined distance = x at second meeting combined distance = num__3 x if andy swims num__18.5 m of x then he will swim num__3 * num__18.5 = num__55.5 m of num__3 x andy ' s total distance to second meeting = x + num__10.5 m x + num__10.5 = num__55.5 m x = num__45 m a <eor> a <eos> |
a |
add__1.0__2.0__ multiply__18.5__3.0__ round__45.0__ |
add__1.0__2.0__ multiply__18.5__3.0__ multiply__1.0__45.0__ |
| two pipes a and b can separately fill a tank in num__2 minutes and num__15 minutes respectively . both the pipes are opened together but num__4 minutes after the start the pipe a is turned off . how much time will it take to fill the tank ? <o> a ) num__9 min <o> b ) num__10 min <o> c ) num__11 min <o> d ) num__12 min <o> e ) num__13 min |
num__0.333333333333 + x / num__15 = num__1 x = num__10 answer : b <eor> b <eos> |
b |
round__10.0__ |
divide__10.0__1.0__ |
| the product of two numbers is num__120 and the sum of their squares is num__289 . the sum of the number is : <o> a ) num__23 <o> b ) num__20 <o> c ) num__12 <o> d ) num__128 <o> e ) num__171 |
let the numbers be x and y . then xy = num__120 and x num__2 + y num__2 = num__289 . ( x + y ) num__2 = x num__2 + y num__2 + num__2 xy = num__289 + ( num__2 x num__120 ) = num__529 x + y = num__529 = num__23 . answer a <eor> a <eos> |
a |
divide__529.0__23.0__ |
divide__529.0__23.0__ |
| if the digits num__29 in the decimal num__0.00029 repeat indefinitely what is the value of ( num__10 ^ num__5 - num__10 ^ num__3 ) ( num__0.00029 ) ? <o> a ) num__2.9 e - num__06 <o> b ) num__2.9 e - num__05 <o> c ) num__0.0029 <o> d ) num__29 <o> e ) num__0.29 |
num__99 * num__0.29 = num__28.71 approx . num__29 answer : d <eor> d <eos> |
d |
subtract__29.0__0.29__ add__0.29__28.71__ |
multiply__99.0__0.29__ add__0.29__28.71__ |
| by selling an article at rs . num__600 a shopkeeper makes a profit of num__25.0 . at what price should he sell the article so as to make a loss of num__25.0 ? <o> a ) s . num__360 <o> b ) s . num__480 <o> c ) s . num__500 <o> d ) s . num__450 <o> e ) s . num__550 |
sp = num__600 profit = num__25.0 cp = ( sp ) * [ num__100 / ( num__100 + p ) ] = num__600 * [ num__0.8 ] = num__480 loss = num__25.0 = num__25.0 of num__480 = rs . num__120 sp = cp - loss = num__480 - num__120 = rs . num__360 answer : a <eor> a <eos> |
a |
percent__25.0__480.0__ percent__100.0__360.0__ |
percent__25.0__480.0__ percent__100.0__360.0__ |
| by selling num__15 pencils for a rupee a man loses num__25.0 . how many for a rupee should he sell in order to gain num__25.0 ? <o> a ) num__8 <o> b ) num__7 <o> c ) num__9 <o> d ) num__4 <o> e ) num__11 |
num__75.0 - - - num__12 num__125.0 - - - ? num__0.6 * num__15 = num__9 answer : c <eor> c <eos> |
c |
percent__75.0__12.0__ percent__75.0__12.0__ |
percent__75.0__12.0__ percent__75.0__12.0__ |
| a person purchased a tv set for rs . num__16000 and a dvd player for rs . num__6250 . he sold both the items together for rs . num__31150 . what percentage of profit did he make ? <o> a ) num__28 <o> b ) num__27 <o> c ) num__40 <o> d ) num__99 <o> e ) num__82 |
the total cp = rs . num__16000 + rs . num__6250 = rs . num__22250 and sp = rs . num__31150 profit ( % ) = ( num__31150 - num__22250 ) / num__22250 * num__100 = num__40.0 answer : c <eor> c <eos> |
c |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| what is the area of a square field whose diagonal of length num__20 m ? <o> a ) num__300 sq m <o> b ) num__250 sq m <o> c ) num__200 sq m <o> d ) num__400 sq m <o> e ) num__450 sq m |
explanation : d num__1.0 = ( num__20 * num__20 ) / num__2 = num__200 answer is c <eor> c <eos> |
c |
multiply__200.0__1.0__ |
multiply__200.0__1.0__ |
| mr . johnson was to earn £ num__300 and a free holiday for seven weeks ' work . he worked for only num__4 weeks and earned £ num__30 and a free holiday . what was the value of the holiday ? <o> a ) £ num__300 <o> b ) £ num__330 <o> c ) £ num__360 <o> d ) £ num__420 <o> e ) £ num__460 |
num__7 week = num__300 + holiday num__1 week = ( num__300 + holiday ) / num__7 num__4 week = ( num__300 + hliday ) * num__0.571428571429 - - eqn num__1 bt num__4 week = num__30 + holiday ( given ) - - eqn num__2 equate eqn num__1 and eqn num__2 to get th ecost of holiday = num__330 answer : b <eor> b <eos> |
b |
divide__4.0__7.0__ add__300.0__30.0__ round__330.0__ |
divide__4.0__7.0__ add__300.0__30.0__ add__300.0__30.0__ |
| four years ago p was half of q ' s age . if the ratio of their present ages is num__3 : num__4 what will be the total of their present ages ? <o> a ) num__14 <o> b ) num__17 <o> c ) num__29 <o> d ) num__31 <o> e ) num__35 |
let present age of p and q be num__3 x num__3 x and num__4 x num__4 x respectively . ten years ago p was half of q ' s age â ‡ ’ ( num__3 x â ˆ ’ num__4 ) = num__0.5 ( num__4 x â ˆ ’ num__4 ) â ‡ ’ num__6 x â ˆ ’ num__8 = num__4 x â ˆ ’ num__4 â ‡ ’ num__2 x = num__4 â ‡ ’ x = num__2 total of their present ages = num__3 x + num__4 x = num__7 x = num__7 Ã — num__2 = num__14 a <eor> a <eos> |
a |
divide__3.0__0.5__ divide__4.0__0.5__ reverse__0.5__ add__3.0__4.0__ multiply__2.0__7.0__ multiply__2.0__7.0__ |
divide__3.0__0.5__ divide__4.0__0.5__ reverse__0.5__ add__3.0__4.0__ add__6.0__8.0__ add__6.0__8.0__ |
| num__122 * num__252 - num__12234 = ? <o> a ) num__15310 <o> b ) num__18870 <o> c ) num__24510 <o> d ) num__18510 <o> e ) num__42510 |
d ? = num__122 * num__252 - num__12234 = num__30744 - num__12234 = num__18510 <eor> d <eos> |
d |
multiply__122.0__252.0__ subtract__30744.0__12234.0__ subtract__30744.0__12234.0__ |
multiply__122.0__252.0__ subtract__30744.0__12234.0__ subtract__30744.0__12234.0__ |
| anil can do a work in num__12 days while sunil can do it in num__25 days . how long will they take if both work together ? <o> a ) num__10 num__0.1 days <o> b ) num__18 num__0.1 days <o> c ) num__7 num__0.1 days <o> d ) num__8 num__0.1 days <o> e ) num__9 num__0.1 days |
num__0.0833333333333 + num__0.04 = num__0.123333333333 num__8.10810810811 = num__8 num__0.1 days answer : d <eor> d <eos> |
d |
add__0.04__0.0833__ round__8.0__ |
add__0.04__0.0833__ round__8.0__ |
| the original price of a television is num__350 $ . what percent of the original price has been taken off if the retailer offers a discount of num__12250 $ on the television ? <o> a ) num__35.0 <o> b ) num__30.0 <o> c ) num__40.0 <o> d ) num__45.0 <o> e ) num__25 % |
explanation : total cost of the television : num__350 required percentage = ( num__3.5 ) Ã — num__100 = num__35.0 answer : option a <eor> a <eos> |
a |
percent__35.0__100.0__ |
percent__35.0__100.0__ |
| num__4 baseball players each stand at different corners of a baseball diamond . the sides of the diamond are all of equal length . two arrangements of baseball players are considered different only when the relative positions of the players differ . how many different ways can the baseball players arrange themselves around the diamond ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__16 <o> d ) num__24 <o> e ) num__256 |
it is like a circular arrangement . total ways for n people to arrange in a circle is = factorial ( n - num__1 ) in this case n = num__4 hence ans = num__3 factorial = num__6 hence b <eor> b <eos> |
b |
die_space__ die_space__ |
die_space__ die_space__ |
| a boat travels upstream from b to a and downstream from a to b in num__3 hours . if the speed of the boat in still water is num__9 km / hour and the speed of the current is num__3 km / hour the distance between a and b is <o> a ) num__4 km <o> b ) num__8 km <o> c ) num__6 km <o> d ) num__12 km <o> e ) none of these |
let d is the distance travelled . then d / ( num__9 + num__3 ) + d / ( num__9 - num__3 ) = num__3 = > d = num__12 answer : d <eor> d <eos> |
d |
add__3.0__9.0__ round__12.0__ |
add__3.0__9.0__ add__3.0__9.0__ |
| a num__300 meter long train crosses a platform in num__51 seconds while it crosses a signal pole in num__18 seconds . what is the length of the platform ? <o> a ) num__550 <o> b ) num__289 <o> c ) num__350 <o> d ) num__882 <o> e ) num__281 |
speed = [ num__16.6666666667 ] m / sec = num__16.6666666667 m / sec . let the length of the platform be x meters . then x + num__5.88235294118 = num__16.6666666667 num__3 ( x + num__300 ) = num__2550 è x = num__550 m . answer : a <eor> a <eos> |
a |
divide__300.0__18.0__ divide__300.0__51.0__ round__550.0__ |
divide__300.0__18.0__ divide__300.0__51.0__ round__550.0__ |
| num__12.5 of num__192 = num__50.0 of ? <o> a ) num__48 <o> b ) num__96 <o> c ) num__24 <o> d ) none of these <o> e ) can not be determined |
answer let num__12.5 of num__192 = num__50.0 of a ⇒ ( num__12.5 x num__192 ) / num__100 = ( num__50 x a ) / num__100 ∴ a = ( num__12.5 x num__192 ) / num__50 = num__48 correct option : a <eor> a <eos> |
a |
percent__100.0__48.0__ |
percent__100.0__48.0__ |
| the speeds of three motor bikes are in the ratio num__24 : num__12 : num__6 . the ratio between the time taken by them to travel the same distance is : <o> a ) num__1 : num__3 : num__4 <o> b ) num__1 : num__2 : num__6 <o> c ) num__1 : num__4 : num__4 <o> d ) num__1 : num__2 : num__3 <o> e ) num__1 : num__2 : num__4 |
ratio of time taken : num__0.0416666666667 : num__0.0833333333333 : num__0.166666666667 = num__1 : num__2 : num__4 answer : e <eor> e <eos> |
e |
divide__24.0__12.0__ divide__24.0__6.0__ round__1.0__ |
divide__24.0__12.0__ divide__24.0__6.0__ round__1.0__ |
| a team of eight entered for a shooting competition . the best marks man scored num__82 points . if he had scored num__92 points the average scores for . the team would have been num__87 . how many points altogether did the team score ? <o> a ) num__686 <o> b ) num__672 <o> c ) num__652 <o> d ) num__642 <o> e ) num__721 |
num__8 * num__87 = num__696 – num__10 = num__686 answer : a <eor> a <eos> |
a |
multiply__87.0__8.0__ subtract__92.0__82.0__ subtract__696.0__10.0__ subtract__696.0__10.0__ |
multiply__87.0__8.0__ subtract__92.0__82.0__ subtract__696.0__10.0__ subtract__696.0__10.0__ |
| mary ' s income is num__50 percent more than tim ' s income and tim ' s income is num__40 percent less than juan ' s income . what percent of juan ' s income is mary ' s income ? <o> a ) num__124.0 <o> b ) num__120.0 <o> c ) num__96.0 <o> d ) num__90.0 <o> e ) num__64 % |
juan ' s income = num__100 ( assume ) ; tim ' s income = num__60 ( num__40 percent less than juan ' s income ) ; mary ' s income = num__90 ( num__50 percent more than tim ' s income ) . thus mary ' s income ( num__90 ) is num__90.0 of juan ' s income ( num__100 ) . answer : d . <eor> d <eos> |
d |
subtract__100.0__40.0__ add__50.0__40.0__ add__50.0__40.0__ |
subtract__100.0__40.0__ add__50.0__40.0__ add__50.0__40.0__ |
| a person is traveling at num__25 km / hr and reached his destiny in num__5 hr find the distance ? <o> a ) num__100 km <o> b ) num__95 km <o> c ) num__135 km <o> d ) num__80 km <o> e ) num__125 km |
speed = num__25 km / hr time = num__5 hr distance = num__25 * num__5 = num__125 km answer is e <eor> e <eos> |
e |
multiply__25.0__5.0__ round__125.0__ |
multiply__25.0__5.0__ multiply__25.0__5.0__ |
| in country z the unemployment rate among construction workers dropped from num__16 percent on september num__1 num__1992 to num__9 percent on september num__1 num__1996 . if the number of construction workers was num__20 percent greater on september num__1 num__1996 than on september num__1 num__1992 what was the approximate percent change in the number of unemployed construction workers over this period ? <o> a ) num__50.0 decrease <o> b ) num__30.0 decrease <o> c ) num__15.0 decrease <o> d ) num__30.0 increase <o> e ) num__55.0 increase |
country z num__1992 num__1996 no of construction workers num__100 num__120 unemployment rate num__16.0 num__9.0 unemployed workers num__16 num__11.0 change in unemployed workers = ( num__16 - num__11 ) = num__0.3125 = ~ num__33.0 decrease closest ans = num__30.0 decrease ans = b <eor> b <eos> |
b |
percent__100.0__30.0__ |
percent__100.0__30.0__ |
| num__12 buckets of water fill a tank when the capacity of each bucket is num__55 litres . how many buckets will be needed to fill the same tank if the capacity of each bucket is num__5 litres ? <o> a ) num__142 <o> b ) num__130 <o> c ) num__132 <o> d ) num__123 <o> e ) num__120 |
capacity of the tank = ( num__12 Ã — num__55 ) litre number of buckets required of capacity of each bucket is num__17 litre = num__12 Ã — num__11.0 = num__12 Ã — num__11 = num__132 answer is c <eor> c <eos> |
c |
add__12.0__5.0__ divide__55.0__5.0__ multiply__12.0__11.0__ round__132.0__ |
add__12.0__5.0__ divide__55.0__5.0__ multiply__12.0__11.0__ round__132.0__ |
| a man took a loan at rate of num__12.0 per annum simple interest . after num__3 years he had to pay num__3600 interest . the principal amount borrowed by him was . <o> a ) rs num__10000 <o> b ) rs num__15000 <o> c ) rs num__16000 <o> d ) rs num__17000 <o> e ) none of these |
explanation : s . i . = p â ˆ — r â ˆ — t / num__100 = > p = s . i . â ˆ — num__100 / r â ˆ — t = > p = num__3600 â ˆ — num__8.33333333333 â ˆ — num__3 = rs num__10000 option a <eor> a <eos> |
a |
percent__100.0__10000.0__ |
percent__100.0__10000.0__ |
| a train covers a certain distance at a speed of num__230 kmph in num__4 hours . to cover the same distance in num__2 num__0.666666666667 hours it must travel at a speed of <o> a ) num__220 km / hr <o> b ) num__320 km / hr <o> c ) num__345 km / hr <o> d ) num__450 km / hr <o> e ) num__650 km / hr |
explanation : distance = num__230 Ã — num__4 = num__920 km required speed = ( num__920 Ã — num__0.375 ) = num__345 km / hr answer : option c <eor> c <eos> |
c |
multiply__230.0__4.0__ multiply__920.0__0.375__ round__345.0__ |
multiply__230.0__4.0__ multiply__920.0__0.375__ round__345.0__ |
| what is the smallest five digit number that is divisible by num__16 num__36 num__40 and num__54 ? <o> a ) num__10200 <o> b ) num__10500 <o> c ) num__10800 <o> d ) num__11000 <o> e ) num__11400 |
num__16 = num__2 ^ num__4 num__36 = num__2 ^ num__2 * num__3 ^ num__2 num__40 = num__2 ^ num__3 * num__5 num__54 = num__2 * num__3 ^ num__3 lcm = num__2 ^ num__4 * num__3 ^ num__3 * num__5 = num__2160 the smallest five - digit number that is a multiple of num__2160 is num__5 * num__2160 = num__10800 the answer is c . <eor> c <eos> |
c |
subtract__40.0__36.0__ add__2.0__3.0__ multiply__40.0__54.0__ multiply__2160.0__5.0__ multiply__2160.0__5.0__ |
subtract__40.0__36.0__ add__2.0__3.0__ multiply__40.0__54.0__ multiply__2160.0__5.0__ multiply__2160.0__5.0__ |
| how many positive integers less than num__100 are multiples of a prime number ? . <o> a ) num__91 <o> b ) num__92 <o> c ) num__94 <o> d ) num__96 <o> e ) num__98 |
there are a total of num__99 positive integers less than num__100 and except the no . num__1 every other no . is a multiple of some prime number so answer should be num__98 answer : option e <eor> e <eos> |
e |
subtract__100.0__99.0__ subtract__99.0__1.0__ multiply__1.0__98.0__ |
subtract__100.0__99.0__ subtract__99.0__1.0__ multiply__1.0__98.0__ |
| an article is bought for rs . num__675 and sold for rs . num__945 find the gain percent ? <o> a ) num__40.0 <o> b ) num__33 num__2.66666666667 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__33 num__0.666666666667 % <o> e ) num__33 num__0.5 % |
num__675 - - - - num__270 num__100 - - - - ? = > num__40.0 answer : a <eor> a <eos> |
a |
percent__40.0__100.0__ |
percent__40.0__100.0__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__288 <o> b ) num__240 <o> c ) num__277 <o> d ) num__127 <o> e ) num__922 |
speed = ( num__54 * num__0.277777777778 ) m / sec = num__15 m / sec . length of the train = ( num__15 x num__20 ) m = num__300 m . let the length of the platform be x meters . then ( x + num__300 ) / num__36 = num__15 = = > x + num__300 = num__540 = = > x = num__240 m . answer : b <eor> b <eos> |
b |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
| a cistern which could be filled in num__7 hours takes one hour more to be filled owing to a leak in its bottom . if the cistern is full in what time will the leak empty it ? <o> a ) num__76 hrs <o> b ) num__23.3333333333 hrs <o> c ) num__55 hrs <o> d ) num__90 hrs <o> e ) num__11 hrs |
num__0.142857142857 - num__1 / x = num__0.1 = > num__23.3333333333 hrs answer : b <eor> b <eos> |
b |
divide__23.3333__1.0__ |
divide__23.3333__1.0__ |
| the speed of the man in still water is num__40 kmph . if he can row upstream at num__20 kmph find the downstream . <o> a ) num__40 <o> b ) num__50 <o> c ) num__55 <o> d ) num__60 <o> e ) num__65 |
m = ( ds + us ) / num__2 ds = num__2 m - us = num__2 * num__40 - num__20 = num__60 answer : d <eor> d <eos> |
d |
divide__40.0__20.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
divide__40.0__20.0__ add__40.0__20.0__ add__40.0__20.0__ |
| how many bricks each measuring num__25 cm x num__11.25 cm x num__6 cm will be needed to build a wall of num__8 m x num__6 m x num__22.5 cm ? <o> a ) num__6400 <o> b ) num__6410 <o> c ) num__6440 <o> d ) num__6500 <o> e ) num__6800 |
number of bricks = volume of wall / volume of bricks = num__800 x num__600 x num__22.5 / num__21 x num__11.25 x num__6 = = num__6400 answer : a <eor> a <eos> |
a |
multiply__8.0__800.0__ round__6400.0__ |
multiply__8.0__800.0__ round__6400.0__ |
| what number should replace the question mark ? num__16 num__23 num__19 num__19 num__22 num__15 num__25 ? <o> a ) num__11 <o> b ) num__12 <o> c ) num__13 <o> d ) num__14 <o> e ) num__15 |
a num__11 there are two alternate sequences + num__3 and - num__4 <eor> a <eos> |
a |
subtract__19.0__16.0__ subtract__23.0__19.0__ subtract__22.0__11.0__ |
subtract__19.0__16.0__ subtract__23.0__19.0__ subtract__22.0__11.0__ |
| a man walks at a speed of num__8 km / hr and runs at a speed of num__7 km / hr . how much time will the man require to cover a distance of num__10 num__0.5 km if he completes half of the distance i . e . ( num__5 num__0.25 ) km on foot and the other half by running ? <o> a ) num__1 num__2.0 hours <o> b ) num__1 num__0.4 hours <o> c ) num__2 num__0.166666666667 hours <o> d ) num__2 num__1.0 hours <o> e ) num__2 num__0.5 hours |
required time = ( num__5 num__0.25 ) / num__8 + ( num__5 num__0.25 ) / num__7 = num__1 num__0.4 hours . answer : b <eor> b <eos> |
b |
subtract__8.0__7.0__ round__1.0__ |
subtract__8.0__7.0__ round__1.0__ |
| consider a sequence of numbers given by the expression num__7 + ( h - num__1 ) * num__5 where n runs from num__1 to num__80 . what is the sum of this series ? <o> a ) num__409 <o> b ) num__1636 <o> c ) num__16360 <o> d ) num__16000 <o> e ) num__15360 |
terms in this sequence are num__7 num__1217 - - - - num__402 now since this is an a . p . with a common difference of num__5 . therefore its sum can be given as h ( a + l ) / num__2 - - - - - - - - - - - - - - - - num__1 ) h = total no . of terms = num__80 a = first term = num__7 l = last term = num__402 subsituting values in the expression num__1 we have num__80 ( num__7 + num__402 ) / num__2 = num__40 ( num__409 ) = num__16360 = c <eor> c <eos> |
c |
subtract__7.0__5.0__ divide__80.0__2.0__ add__7.0__402.0__ multiply__40.0__409.0__ multiply__1.0__16360.0__ |
subtract__7.0__5.0__ divide__80.0__2.0__ add__7.0__402.0__ multiply__40.0__409.0__ divide__16360.0__1.0__ |
| a sum of rs . num__2704 is lent into two parts so that the interest on the first part for num__8 years at num__3.0 per annum may be equal to the interest on the second part for num__3 years at num__5.0 per annum . find the second sum ? <o> a ) num__1672 <o> b ) num__1664 <o> c ) num__1677 <o> d ) num__1698 <o> e ) num__1679 |
( x * num__8 * num__3 ) / num__100 = ( ( num__2704 - x ) * num__3 * num__5 ) / num__100 num__24 x / num__100 = num__405.6 - num__15 x / num__100 num__39 x = num__40560 = > x = num__1040 second sum = num__2704 â € “ num__1040 = num__1664 answer : b <eor> b <eos> |
b |
percent__100.0__1664.0__ |
percent__100.0__1664.0__ |
| a sum of money is distributed among a b c d in the proportion of num__1 : num__3 : num__4 : num__2 . if c gets $ num__500 more than d what is the b ' s share ? <o> a ) $ num__450 <o> b ) $ num__500 <o> c ) $ num__750 <o> d ) $ num__800 <o> e ) $ num__840 |
let the shares of a b c d are x num__3 x num__4 x num__2 x num__4 x - num__2 x = num__500 x = num__250 b ' s share = num__3 x = $ num__750 answer is c <eor> c <eos> |
c |
divide__500.0__2.0__ multiply__3.0__250.0__ multiply__1.0__750.0__ |
divide__500.0__2.0__ multiply__3.0__250.0__ multiply__1.0__750.0__ |
| a number consists of two digits . if num__0.6 of num__0.2 of the number is num__9 . find the sum of its two digits ? <o> a ) num__33 <o> b ) num__28 <o> c ) num__13 <o> d ) num__12 <o> e ) num__11 |
explanation : x * num__0.6 * num__0.2 = num__9 x = num__75 = > num__7 + num__5 = num__12 answer : d <eor> d <eos> |
d |
reverse__0.2__ add__5.0__7.0__ add__5.0__7.0__ |
reverse__0.2__ add__5.0__7.0__ add__5.0__7.0__ |
| at what rate percent per annum will sum of money double in num__20 years ? <o> a ) num__1.2 <o> b ) num__2.0 <o> c ) num__4.0 <o> d ) num__5.0 <o> e ) none of these |
explanation : hint : if sum of money becomes ( z times ) in ( t ) years at simple interest then rate of interest ( r ) can be calculated using the formula : rate of interest ( r ) % = num__100 ( z – num__1 ) / t here principal amount is not given . hence we can directly use the trick to calculate the rate of interest . rate of interest ( r ) % = num__100 ( num__2 – num__1 ) / num__20 rate of interest ( r ) % = num__5.0 p . a . answer is d <eor> d <eos> |
d |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| if dy + z = d ( y + z ) which of the following must be true ? <o> a ) d = num__0 and z = num__0 <o> b ) d = num__1 and y = num__1 <o> c ) y = num__1 and z = num__0 <o> d ) d = num__1 or y = num__0 <o> e ) d = num__1 or z = num__0 |
dy + z = dy + dz z = dz case num__1 : z not = num__0 d = z / z = num__1 case num__2 : z = num__0 num__0 = d num__0 = num__0 combining num__2 cases : d = num__1 or z = num__0 e . is the answer . <eor> e <eos> |
e |
reverse__1.0__ |
reverse__1.0__ |
| in the xy plane the point ( - num__2 - num__3 ) is the center of a circle . the point ( - num__21 ) lies inside the circle and the point ( num__4 - num__3 ) lies outside the circle . if the radius r of the circle is an integer then r = <o> a ) num__6 <o> b ) num__5 <o> c ) num__4 <o> d ) num__3 <o> e ) num__2 |
this question might look intimidating because of its language but once you start solving it you will realise that the options are given to you in such a way that you reach the correct answer easily . the radius will lie somewhere between the distance of centre from the inner point and the distance from the outer point . distance between centre and inner point = distance between ( - num__2 - num__3 ) and ( - num__2 num__1 ) we can solve for the distance by using the formula for distance between two points . but that is not required here . if one of the co - ordinates is same then the distance between two points is simply the difference between the other coordinate . in this case distance = num__1 - ( - num__3 ) = num__4 distance between centre and utter point = distance between ( - num__2 - num__3 ) and ( num__4 - num__3 ) = num__4 - ( - num__2 ) = num__6 the radius has to be between num__4 and num__6 on looking at the options only num__5 satisfies correct option : b <eor> b <eos> |
b |
subtract__3.0__2.0__ multiply__2.0__3.0__ add__2.0__3.0__ add__2.0__3.0__ |
subtract__3.0__2.0__ multiply__2.0__3.0__ subtract__6.0__1.0__ subtract__6.0__1.0__ |
| one copy machine can make num__15 copies a minute and a second copy machine makes num__10 copies a minute . if the two copiers work together how long would it take them to make num__1000 copies ? <o> a ) num__20 minutes <o> b ) num__30 minutes <o> c ) num__40 minutes <o> d ) num__50 minutes <o> e ) num__60 minutes |
total work done by both machines in a minute = num__15 + num__10 = num__25 copies total number of copies required = num__1000 time = num__40.0 = num__40 mins answer c <eor> c <eos> |
c |
add__15.0__10.0__ add__15.0__25.0__ round__40.0__ |
add__15.0__10.0__ add__15.0__25.0__ add__15.0__25.0__ |
| if num__5 people contributed a total of $ num__20.00 toward a gift and each of them contributed at least $ num__3.00 then the maximum possible amount any one person could have contributed is <o> a ) $ num__1.00 <o> b ) $ num__1.25 <o> c ) $ num__5.00 <o> d ) $ num__8.00 <o> e ) $ num__20.00 |
d for me num__4 people with num__3 $ each - > maximum = num__8 <eor> d <eos> |
d |
divide__20.0__5.0__ add__5.0__3.0__ add__5.0__3.0__ |
divide__20.0__5.0__ add__5.0__3.0__ add__5.0__3.0__ |
| a retailer bought a shirt at wholesale and marked it up num__80.0 to its initial price of $ num__27 . by how many more dollars does he need to increase the price to achieve a num__100.0 markup ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
let x be the wholesale price . then num__1.8 x = num__27 and x = num__27 / num__1.8 = num__15 . to achieve a num__100.0 markup the price needs to be $ num__30 . the retailer needs to increase the price by $ num__3 more . the answer is c . <eor> c <eos> |
c |
divide__27.0__1.8__ subtract__30.0__27.0__ subtract__30.0__27.0__ |
divide__27.0__1.8__ subtract__30.0__27.0__ subtract__30.0__27.0__ |
| at what % above c . p must an article be marked so as to gain num__33.0 after allowing a customer a discount of num__5.0 ? <o> a ) num__49.0 <o> b ) num__40.0 <o> c ) num__48.0 <o> d ) num__45.0 <o> e ) num__38 % |
let c . p be rs num__100 . then s . p be rs num__133 let the market price be rs x then num__90.0 of x = num__133 = > num__95 x / num__100 = num__133 = > x = ( num__133 * num__1.05263157895 ) = num__140 market price = num__40.0 above c . p answer is b . <eor> b <eos> |
b |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| an article is bought for rs . num__675 and sold for rs . num__775 find the gain percent ? <o> a ) num__73.0 <o> b ) num__14.8 <o> c ) num__33 num__0.333333333333 % <o> d ) num__93 num__0.333333333333 % <o> e ) num__33 num__0.333333333333 % |
explanation : num__675 - - - - num__100 num__100 - - - - ? = > num__14.8 answer : b <eor> b <eos> |
b |
percent__100.0__14.8__ |
percent__100.0__14.8__ |
| what percent is num__20.0 of num__25.0 ? <o> a ) num__20.0 <o> b ) num__30.0 <o> c ) num__50.0 <o> d ) num__60.0 <o> e ) num__80 % |
required percentage = num__20.0 / num__25.0 * num__100 = num__0.8 * num__100 = num__80.0 answer is e <eor> e <eos> |
e |
percent__80.0__100.0__ |
percent__80.0__100.0__ |
| a boy pays rs . num__369 for an article marked at rs . num__600 by enjoying two successive discounts . if the first discount is of num__25.0 how much should be the second discount ? <o> a ) num__20.0 <o> b ) num__10.0 <o> c ) num__25.0 <o> d ) num__18.0 <o> e ) none of these |
explanation : first discount = num__25.0 of num__600 = rs . num__150 . thus the reduced price = num__600 – num__150 = rs . num__450 . since the person actually paid rs . num__369 the value of the second discount must be equal to rs . num__81 ( num__450 – num__369 ) . let the second discount be x thus we get num__81 = x of num__450 ( num__0.18 ) * num__100 = num__18.0 answer d <eor> d <eos> |
d |
percent__25.0__600.0__ percent__100.0__18.0__ |
percent__25.0__600.0__ percent__100.0__18.0__ |
| when rahul was born his father was num__26 years older than his brother and his mother was num__23 years older than his sister . if rahul ' s brother is num__4 years older than him and his mother is num__3 years younger than his father how old was rahul ' s sister when he was born ? <o> a ) num__7 years <o> b ) num__5 years <o> c ) num__4 years <o> d ) num__6 years <o> e ) num__8 years |
when rahul was born his brother ' s age = num__4 years ; his father ' s age = ( num__4 + num__26 ) years = num__30 years his mother ' s age = ( num__30 - num__3 ) years = num__27 years ; his sister ' s age = ( num__27 - num__23 ) years = num__4 years . answer : option c <eor> c <eos> |
c |
add__26.0__4.0__ add__23.0__4.0__ subtract__27.0__23.0__ |
add__26.0__4.0__ add__23.0__4.0__ subtract__27.0__23.0__ |
| the length of a train and that of a platform are equal . if with a speed of num__72 k / hr the train crosses the platform in one minute then the length of the train ( in meters ) is ? <o> a ) num__227 <o> b ) num__299 <o> c ) num__276 <o> d ) num__600 <o> e ) num__211 |
speed = [ num__72 * num__0.277777777778 ] m / sec = num__20 m / sec ; time = num__1 min . = num__60 sec . let the length of the train and that of the platform be x meters . then num__2 x / num__60 = num__20 è x = num__20 * num__30.0 = num__600 answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ divide__60.0__2.0__ multiply__20.0__30.0__ round__600.0__ |
hour_to_min_conversion__ divide__60.0__2.0__ multiply__20.0__30.0__ multiply__20.0__30.0__ |
| when positive integer n is divided by num__3 the remainder is num__1 . when n is divided by num__11 the remainder is num__9 . what is the smallest positive integer k such that k + n is a multiple of num__33 ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
n = num__3 p + num__1 = num__11 q + num__9 n + num__2 = num__3 p + num__3 = num__11 q + num__11 n + num__2 is a multiple of num__3 and num__11 so it is a multiple of num__33 . the answer is a . <eor> a <eos> |
a |
subtract__3.0__1.0__ subtract__3.0__1.0__ |
subtract__3.0__1.0__ subtract__3.0__1.0__ |
| a man can row upstream at num__37 kmph and downstream at num__53 kmph and then find the speed of the man in still water ? <o> a ) num__29 <o> b ) num__92 <o> c ) num__30 <o> d ) num__32 <o> e ) num__45 |
us = num__37 ds = num__53 m = ( num__53 + num__37 ) / num__2 = num__45 answer : e <eor> e <eos> |
e |
round__45.0__ |
round__45.0__ |
| a girl scout sold num__5 macadamia cookies num__10 peanut butter cookies and num__15 butter cookies . what is the ratio of the peanut butter cookies to the total sold ? <o> a ) num__0.5 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__0.333333333333 |
total # of cookies is num__5 + num__10 + num__15 = num__30 ; ratio of peanut butter cookies sold is num__0.333333333333 = num__0.333333333333 answer : b <eor> b <eos> |
b |
divide__5.0__15.0__ divide__5.0__15.0__ |
divide__5.0__15.0__ divide__5.0__15.0__ |
| if the volume of two cubes are in the ratio num__125 : num__1 the ratio of their edges is : <o> a ) num__3 : num__1 <o> b ) num__5 : num__1 <o> c ) num__3 : num__5 <o> d ) num__3 : num__7 <o> e ) none of these |
explanation : let the edges be a and b of two cubes then a num__3 / b num__3 = num__125.0 = > ( a / b ) num__3 = ( num__5.0 ) num__3 a / b = num__5.0 = > a : b = num__5 : num__1 option b <eor> b <eos> |
b |
multiply__1.0__5.0__ |
multiply__1.0__5.0__ |
| at what percentage above the c . p must an article be marked so as to gain num__33.0 after allowing a customer a discount of num__5.0 ? <o> a ) num__38.0 <o> b ) num__40.0 <o> c ) num__43.0 <o> d ) num__48.0 <o> e ) num__50 % |
let c . p = rs . num__100 then s . p = rs . num__133 . let the marked price be x then num__95.0 of x = num__133 num__95 x / num__100 = num__133 x = num__133 * num__1.05263157895 = num__140 marked price = num__40.0 above c . p answer : b <eor> b <eos> |
b |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| if each term in the sum a num__1 + a num__2 + a num__3 + . . . . . + an is either num__7 or num__77 and the sum equals num__336 which of the following could be equal to n ? <o> a ) num__38 <o> b ) num__39 <o> c ) num__40 <o> d ) num__41 <o> e ) num__42 |
since there is no num__48 in the answer choices ( num__48.0 = num__48 ) we know there is at least one num__77 . num__336 - num__77 = num__259 num__37.0 = num__37 num__37 + num__1 = num__38 if num__38 was n ' t there i would have subtracted num__77 from num__259 and continued in a similar way . ans . a <eor> a <eos> |
a |
divide__336.0__7.0__ subtract__336.0__77.0__ divide__259.0__7.0__ add__1.0__37.0__ add__1.0__37.0__ |
divide__336.0__7.0__ subtract__336.0__77.0__ divide__259.0__7.0__ add__1.0__37.0__ add__1.0__37.0__ |
| a company has two models of computers model m and model n . operating at a constant rate a model m computer can complete a certain task in num__36 minutes and a model n computer can complete the same task in num__18 minutes . if the company used the same number of each model of computer to complete the task in num__1 minute how many model m computers were used ? <o> a ) num__12 <o> b ) num__11 <o> c ) num__10 <o> d ) num__9 <o> e ) num__8 |
let ' s say num__1 work is processing num__36 gb of data . model m : num__1 gb per min model n : num__2 gb per min working together num__1 m and num__1 n = num__3 gb per min so num__12 times as many computers would work at num__36 gb per min . so no . of m = num__12 answer is a <eor> a <eos> |
a |
divide__36.0__18.0__ add__1.0__2.0__ divide__36.0__3.0__ round__12.0__ |
divide__36.0__18.0__ add__1.0__2.0__ divide__36.0__3.0__ round__12.0__ |
| how many positive integer solutions does the equation num__5 x + num__10 y = num__100 have ? <o> a ) num__2 <o> b ) num__33 <o> c ) num__38 <o> d ) num__35 <o> e ) num__14 |
formula : ( constant ) / ( lcm of two nos ) = num__100 / ( num__5 * num__10 ) = num__2 answer : a <eor> a <eos> |
a |
divide__10.0__5.0__ divide__10.0__5.0__ |
divide__10.0__5.0__ divide__10.0__5.0__ |
| two trains num__200 m and num__300 m long run at the speed of num__70 kmph and num__50 kmph in opposite directions in parallel tracks . the time which they take to cross each other is ? <o> a ) num__10 sec <o> b ) num__15 sec <o> c ) num__18 sec <o> d ) num__20 sec <o> e ) num__22 sec |
relative speed = num__70 + num__50 = num__120 kmph * num__0.277777777778 = num__33.3333333333 m / s distance covered in crossing each other = num__200 + num__300 = num__500 m required time = num__500 * num__0.03 = num__15 sec answer is b <eor> b <eos> |
b |
add__70.0__50.0__ add__200.0__300.0__ multiply__0.03__500.0__ round__15.0__ |
add__70.0__50.0__ add__200.0__300.0__ multiply__0.03__500.0__ multiply__0.03__500.0__ |
| a person travels through num__5 cities - a b c d e . cities e is num__2 km west of d . d is num__3 km north - east of a . c is num__5 km north of b and num__4 km west of a . if this person visits these citiesin the sequence b - c - a - e - d what is the effective distance between cities b and d ? <o> a ) num__12 <o> b ) num__13 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
make diagram first then traverse according to b c a e d u can find num__5 + num__4 = num__9 ( b to a ) a to e is not given but u can see it as num__1.5 or num__2 km so b to e is num__10 km now e to d is num__2 km total = num__5 + num__4 + num__1.5 + num__2 = num__12.5 or num__5 + num__4 + num__2 + num__2 = num__13 answer : b <eor> b <eos> |
b |
add__5.0__4.0__ divide__3.0__2.0__ multiply__5.0__2.0__ add__3.0__10.0__ add__3.0__10.0__ |
add__5.0__4.0__ divide__3.0__2.0__ multiply__5.0__2.0__ add__3.0__10.0__ add__3.0__10.0__ |
| in a group of ducks and cows the total number of legs are num__16 more than twice the no . of heads . find the total no . of buffaloes . <o> a ) num__10 <o> b ) num__12 <o> c ) num__8 <o> d ) num__12 <o> e ) num__14 |
let the number of buffaloes be x and the number of ducks be y = > num__4 x + num__2 y = num__2 ( x + y ) + num__16 = > num__2 x = num__16 = > x = num__8 c <eor> c <eos> |
c |
divide__16.0__2.0__ subtract__16.0__8.0__ |
divide__16.0__2.0__ subtract__16.0__8.0__ |
| what profit percent is made by selling an article at a certain price if by selling at num__0.666666666667 rd of that price there would be a loss of num__12.0 ? <o> a ) num__20.0 <o> b ) num__29.0 <o> c ) num__32.0 <o> d ) num__80.0 <o> e ) num__90 % |
sp num__2 = num__0.666666666667 sp num__1 cp = num__100 sp num__2 = num__88 num__0.666666666667 sp num__1 = num__88 sp num__1 = num__132 num__100 - - - num__32 = > num__32.0 answer : c <eor> c <eos> |
c |
percent__32.0__100.0__ |
percent__32.0__100.0__ |
| if the cost price of num__12 pencils is equal to the selling price of num__8 pencils the gain percent is : <o> a ) num__30 <o> b ) num__40 <o> c ) num__50 <o> d ) num__60 <o> e ) num__70 |
c num__50.0 let c . p . of each pencil be $ num__1 . then c . p . of num__8 pencils = $ num__8 ; s . p . of num__8 pencils = $ num__12 . gain % = num__0.5 * num__100 = num__50.0 <eor> c <eos> |
c |
percent__1.0__50.0__ percent__100.0__50.0__ |
percent__1.0__50.0__ percent__100.0__50.0__ |
| a box contains num__100 balls numbered from num__1 to num__100 . if three balls are selected at random and with replacement from the box what is the probability that the sum of the three numbers on the balls selected from the box will be odd ? <o> a ) num__0.25 <o> b ) num__0.375 <o> c ) num__0.5 <o> d ) num__0.625 <o> e ) num__0.75 |
the sum of the three numbers on the balls selected from the box to be odd one should select either three odd numbered balls ( odd + odd + odd = odd ) or two even numbered balls and one odd numbered ball ( even + even + odd = odd ) ; p ( ooo ) = ( num__0.5 ) ^ num__3 ; p ( eeo ) = num__3 * ( num__0.5 ) ^ num__2 * num__0.5 = num__0.375 ( you should multiply by num__3 as the scenario of two even numbered balls and one odd numbered ball can occur in num__3 different ways : eeo eoe or oee ) ; so finally p = num__0.125 + num__0.375 = num__0.5 . answer : c . <eor> c <eos> |
c |
reverse__0.5__ subtract__0.5__0.375__ reverse__2.0__ |
reverse__0.5__ subtract__0.5__0.375__ add__0.375__0.125__ |
| the units digit of ( num__10 ) ^ ( num__87 ) + ( num__5 ) ^ ( num__46 ) is : <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__0 |
any power of anything ending in num__5 always has a units digit of num__5 . so the first term has a units digit of num__5 . done . the second term anything power to num__10 unit digit will be zero then num__5 + num__0 = num__5 e <eor> e <eos> |
e |
multiply__10.0__0.0__ |
multiply__10.0__0.0__ |
| if a num__6 cm cube is cut into num__1 cm cubes then what is the percentage increase in the surface area of the resulting cubes ? <o> a ) num__300.0 <o> b ) num__400.0 <o> c ) num__500.0 <o> d ) num__600.0 <o> e ) num__750 % |
the area a of the large cube is num__6 * num__6 * num__6 = num__216 square cm . the area of the num__216 small cubes is num__216 * num__6 = num__6 a an increase of num__500.0 . the answer is c . <eor> c <eos> |
c |
volume_cube__6.0__ multiply__1.0__500.0__ |
volume_cube__6.0__ multiply__1.0__500.0__ |
| two goods trains each num__500 m long are running in opposite directions on parallel tracks . their speeds are num__45 km / hr and num__45 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? <o> a ) num__40 <o> b ) num__93 <o> c ) num__26 <o> d ) num__23 <o> e ) num__12 |
relative speed = num__45 + num__45 = num__90 km / hr . num__90 * num__0.277777777778 = num__25 m / sec . distance covered = num__500 + num__500 = num__1000 m . required time = num__40.0 = num__40 sec . answer : a <eor> a <eos> |
a |
divide__1000.0__25.0__ round__40.0__ |
divide__1000.0__25.0__ divide__1000.0__25.0__ |
| stacy and heather are num__5 miles apart and walk towards each other along the same route . stacy walks at constant rate that is num__1 mile per hour fast than heather ' s constant rate of num__5 miles / hour . if heather starts her journey num__24 minutes after stacy how far from the original destination has heather walked when the two meet ? . <o> a ) num__2.5 mile <o> b ) num__1.9 mile <o> c ) num__1.7 mile <o> d ) num__1.5 mile <o> e ) num__1.18 mile |
ss - stacy ' s speed = num__6 m / hr sh - heather ' s speed = num__5 m / hr in num__24 minutes stacy will cover = ( num__0.4 ) * num__6 = num__2.4 miles now since both are walking in opposite directions add their speeds - num__6 + num__5 = num__11 m / hr and distance to cover is num__5 - num__2.4 = num__2.6 time taken = distance / speed = num__2.6 / num__11 = num__0.23 hrs heather will cover = num__5 * num__0.23 = num__1.18 miles . answer e <eor> e <eos> |
e |
add__5.0__1.0__ multiply__0.4__6.0__ add__5.0__6.0__ subtract__5.0__2.4__ round__1.18__ |
add__5.0__1.0__ multiply__0.4__6.0__ add__5.0__6.0__ subtract__5.0__2.4__ multiply__1.0__1.18__ |
| express a speed of num__36 kmph in meters per second ? <o> a ) num__10 mps <o> b ) num__17 mps <o> c ) num__97 mps <o> d ) num__17 mps <o> e ) num__18 mps |
num__36 * num__0.277777777778 = num__10 mps answer : a <eor> a <eos> |
a |
round__10.0__ |
round__10.0__ |
| num__16 ltr of water is added with num__24 ltr of a solution containing num__90.0 of alcohol in the water . the % of alcohol in the new mixture is ? <o> a ) num__48.0 <o> b ) num__52.0 <o> c ) num__54.0 <o> d ) num__60.0 <o> e ) num__70 % |
we have a num__24 litre solution containing num__90.0 of alcohol in the water . = > quantity of alcohol in the solution = num__24 Ã — num__0.9 now num__16 litre of water is added to the solution . = > total quantity of the new solution = num__24 + num__16 = num__40 percentage of alcohol in the new solution = num__24 Ã — num__0.9 num__40 Ã — num__100 = num__24 Ã — num__9010040 Ã — num__100 = num__24 Ã — num__2.25 / num__100 = num__54.0 c <eor> c <eos> |
c |
add__16.0__24.0__ divide__90.0__0.9__ divide__90.0__40.0__ multiply__24.0__2.25__ multiply__24.0__2.25__ |
add__16.0__24.0__ divide__90.0__0.9__ divide__90.0__40.0__ multiply__24.0__2.25__ multiply__24.0__2.25__ |
| robert ' s salary was decreased by num__50.0 and subsequently increased by num__50.0 . how much percentage does he lose ? <o> a ) num__10.0 <o> b ) num__20.0 <o> c ) num__25.0 <o> d ) num__30.0 <o> e ) num__50 % |
let original salary be $ num__100 new final salary = num__150.0 of ( num__50.0 of $ num__100 ) = ( num__1.5 ) * ( num__0.5 ) * num__100 = $ num__75 decrease = num__25.0 correct option is c <eor> c <eos> |
c |
add__50.0__100.0__ divide__150.0__100.0__ divide__50.0__100.0__ multiply__50.0__1.5__ multiply__50.0__0.5__ multiply__50.0__0.5__ |
add__50.0__100.0__ divide__150.0__100.0__ divide__50.0__100.0__ multiply__50.0__1.5__ multiply__50.0__0.5__ multiply__50.0__0.5__ |
| the positive difference between sam and lucy ’ s ages is b and the sum of their ages is x . if lucy is older than sam then which of the following represents lucy ’ s age ? <o> a ) ( x - b ) / num__2 <o> b ) b - x / num__2 <o> c ) num__2 b + x <o> d ) ( x + b ) / num__2 <o> e ) ( b - x ) / num__2 |
let lucy ' s age be l and sam ' s age be s as given l - s = b - - num__1 l + s = x - - num__2 adding both the equations num__2 l = a + z l = ( b + x ) / num__2 answer is d <eor> d <eos> |
d |
multiply__1.0__2.0__ |
divide__2.0__1.0__ |
| in an election candidate a got num__55.0 of the total valid votes . if num__15.0 of the total votes were declared invalid and the total numbers of votes is num__560000 find the number of valid vote polled in favor of candidate ? <o> a ) num__261800 <o> b ) num__355800 <o> c ) num__356500 <o> d ) num__356800 <o> e ) num__357000 |
total number of invalid votes = num__15.0 of num__560000 = num__0.15 × num__560000 = num__84000.0 = num__84000 total number of valid votes num__560000 – num__84000 = num__476000 percentage of votes polled in favour of candidate a = num__55.0 therefore the number of valid votes polled in favour of candidate a = num__55.0 of num__476000 = num__0.55 × num__476000 = num__261800.0 = num__261800 a ) <eor> a <eos> |
a |
percent__15.0__560000.0__ percent__55.0__476000.0__ percent__55.0__476000.0__ |
percent__15.0__560000.0__ percent__55.0__476000.0__ percent__55.0__476000.0__ |
| mike earns $ num__17.5 per hour and phil earns $ num__10.5 per hour . approximately how much less as a percentage does phil earn than mike per hour ? <o> a ) num__25.0 <o> b ) num__32.5 <o> c ) num__37.0 <o> d ) num__37.5 <o> e ) num__40 % |
what % less of num__17.5 is num__10.5 let it be x % less then = num__17.5 ( num__1 - x / num__100 ) = num__10.5 num__1 - x / num__100 = num__10.5 / num__17.5 x = num__40.0 x = num__40.0 ans e <eor> e <eos> |
e |
multiply__40.0__1.0__ |
divide__40.0__1.0__ |
| two trains are moving at num__60 kmph and num__70 kmph in opposite directions . their lengths are num__150 m and num__100 m respectively . the time they will take to pass each other completely is ? <o> a ) num__6 num__0.923076923077 sec <o> b ) num__7 num__0.142857142857 sec <o> c ) num__7 num__1.0 sec <o> d ) num__8 num__0.5 sec <o> e ) num__7 num__0.111111111111 sec |
num__70 + num__60 = num__130 * num__0.277777777778 = num__36.1111111111 mps d = num__150 + num__100 = num__250 m t = num__250 * num__0.0276923076923 = num__6.92307692308 = num__6 num__0.923076923077 sec answer : a <eor> a <eos> |
a |
add__60.0__70.0__ add__150.0__100.0__ divide__250.0__36.1111__ subtract__6.9231__6.0__ round__6.0__ |
add__60.0__70.0__ add__150.0__100.0__ divide__250.0__36.1111__ subtract__6.9231__6.0__ round__6.0__ |
| ratio of ravi ' s and ashok ' s present ages is num__4 : num__5 after num__6 years it will be num__5 : num__6 . what is ravi ' s present age ? <o> a ) num__16 years <o> b ) num__15 years <o> c ) num__20 years <o> d ) num__24 years <o> e ) num__25 years |
let present ages are num__4 x and num__5 x after num__6 years hence num__4 x + num__1.2 x + num__6 = num__0.833333333333 after solving above eqn x = num__6 so ravis present age become num__4 * num__6 = num__24 answer : d <eor> d <eos> |
d |
divide__6.0__5.0__ reverse__1.2__ multiply__4.0__6.0__ multiply__4.0__6.0__ |
divide__6.0__5.0__ reverse__1.2__ multiply__4.0__6.0__ multiply__4.0__6.0__ |
| what is the remainder when num__46 * num__49 is divided by num__8 ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__4 <o> d ) num__6 <o> e ) num__7 |
we can make use of the rule : remainder of { ( a * b ) / n } } = remainder of ( a / n ) * remainder of ( b / n ) here remainder of { num__46 * num__49 ) / num__8 } } = remainder of ( num__5.75 ) * remainder of ( num__6.125 ) = num__6 * num__1 = num__6 answer : d <eor> d <eos> |
d |
divide__46.0__8.0__ divide__49.0__8.0__ round_down__6.125__ round_down__6.125__ |
divide__46.0__8.0__ divide__49.0__8.0__ round_down__6.125__ divide__6.0__1.0__ |
| nick ' s average ( arithmetic mean ) test score on num__4 tests is num__78 . what must be the student ' s score on a num__5 th test for the nick ' s average score on the num__5 tests to be num__80 ? <o> a ) num__80 <o> b ) num__82 <o> c ) num__84 <o> d ) num__86 <o> e ) num__88 |
e . num__88 nick must score at least an num__80 for sure . if he scores an num__8 then he will need to score num__2 pots for each of the num__4 other tests tomake upthe difference . they each were at num__78 ( at least the average is but this is a small point and does n ' t matter to the answer ) . so num__4 tests that were each num__2 points short of the num__80 average that is desired means the next test must be num__8 points higher than the desired average so num__80 + num__8 = num__88 . <eor> e <eos> |
e |
subtract__88.0__80.0__ subtract__80.0__78.0__ add__80.0__8.0__ |
subtract__88.0__80.0__ subtract__80.0__78.0__ add__80.0__8.0__ |
| guadalupe owns num__2 rectangular tracts of land . one is num__300 m by num__500 m and the other is num__250 m by num__630 m . the combined area of these num__2 tracts is how many square meters ? <o> a ) num__3360 <o> b ) num__307500 <o> c ) num__621500 <o> d ) num__704000 <o> e ) num__2 num__816 |
000 |
one is num__300 m by num__500 m and the other is num__250 m by num__630 m : ( num__300 * num__500 ) + ( num__250 * num__630 ) = num__150000 + num__157500 num__307500 answer : b <eor> b <eos> |
b |
b |
| what is the ratio between perimeters of two squares one having num__3 times the diagonal then the other ? <o> a ) num__3 : num__5 <o> b ) num__3 : num__3 <o> c ) num__3 : num__9 <o> d ) num__3 : num__1 <o> e ) num__3 : num__2 |
d = num__3 d d = d a √ num__2 = num__3 d a √ num__2 = d a = num__3 d / √ num__2 a = d / √ num__2 = > num__3 : num__1 answer : d <eor> d <eos> |
d |
multiply__3.0__1.0__ |
multiply__3.0__1.0__ |
| sound is said to travel in air at about num__1100 feet per second . a man hears the axe striking the tree num__2.2 seconds after he sees it strike the tree . how far is the man from the wood chopper ? <o> a ) num__3449 <o> b ) num__3777 <o> c ) num__2998 <o> d ) num__2420 <o> e ) num__2778 |
speed of the sound = num__1100 ft / s time = num__2.2 second distance = speed × time = num__1100 × num__115 = num__220 × num__11 = num__2420 ft answer : d <eor> d <eos> |
d |
multiply__1100.0__2.2__ multiply__1100.0__2.2__ |
multiply__1100.0__2.2__ multiply__1100.0__2.2__ |
| which of the following is closest to num__10 ^ num__150 – num__10 ^ num__30 ? <o> a ) num__10 ^ num__210 <o> b ) num__10 ^ num__180 <o> c ) num__10 ^ num__150 <o> d ) num__10 ^ num__90 <o> e ) num__10 ^ num__6 |
num__10 ^ num__150 – num__10 ^ num__30 num__10 ^ num__30 * ( num__10 ^ num__120 – num__1 ) as we know num__10 ^ num__2 - num__1 means num__100 - num__1 and we get num__99 which is approximately num__100 . hence ( num__10 ^ num__120 – num__1 ) would remain as num__10 ^ num__120 . and num__10 ^ num__30 * num__10 ^ num__120 = num__10 ^ num__150 . answer is c . <eor> c <eos> |
c |
subtract__150.0__30.0__ power__10.0__2.0__ subtract__100.0__1.0__ multiply__10.0__1.0__ |
subtract__150.0__30.0__ power__10.0__2.0__ subtract__100.0__1.0__ multiply__10.0__1.0__ |
| in some quantity of ghee num__60.0 of pure ghee and num__40.0 of is vanaspati . if num__10 kg of pure ghee is added then the strength of vanaspati ghee becomes num__20.0 . the original quantity was ? <o> a ) num__5 kg <o> b ) num__10 kg <o> c ) num__15 kg <o> d ) num__20 kg <o> e ) num__25 kg |
let the original quantity be x kg vanaspati ghee in x kg = num__0.4 * x = num__2 x / num__5 kg ( num__2 x / num__5 ) / ( x + num__10 ) = num__0.2 num__2 x / ( num__5 x + num__50 ) = num__0.2 num__5 x = num__50 x = num__10 kg answer is b <eor> b <eos> |
b |
percent__10.0__20.0__ percent__10.0__2.0__ percent__20.0__50.0__ |
percent__10.0__20.0__ percent__10.0__2.0__ percent__20.0__50.0__ |
| the present population of a town is num__3600 . population increase rate is num__20.0 p . a . find the population of town before num__2 years ? <o> a ) num__2500 <o> b ) num__2100 <o> c ) num__3500 <o> d ) num__3600 <o> e ) num__2050 |
p = num__3600 r = num__20.0 required population of town = p / ( num__1 + r / num__100 ) ^ t = num__3600 / ( num__1 + num__0.2 ) ^ num__2 = num__3600 / ( num__1.2 ) ^ num__2 = num__2500 ( approximately ) answer is a <eor> a <eos> |
a |
percent__20.0__1.0__ percent__100.0__2500.0__ |
percent__20.0__1.0__ percent__100.0__2500.0__ |
| anita indu and geeta can do a piece of work in num__18 days num__27 days and num__36 days respectively . they start working together . after working for num__4 days . anita goes away and indu leaves num__7 days before the work is finished . only geeta remains at work from beginning to end . in how many days was the whole work done ? <o> a ) num__16 days <o> b ) num__66 days <o> c ) num__77 days <o> d ) num__44 days <o> e ) num__33 days |
: num__0.222222222222 + ( x - num__7 ) / num__27 + x / num__36 = num__1 x = num__16 days answer : a <eor> a <eos> |
a |
divide__4.0__18.0__ round__16.0__ |
divide__4.0__18.0__ divide__16.0__1.0__ |
| the sides of a cube measures num__6.5 cm . if the surface area of the cube is the same as a ball what is the radius of the ball ? round to the nearest whole number . <o> a ) num__9 <o> b ) num__7 <o> c ) num__4 <o> d ) num__8 <o> e ) num__6 |
first calculate the surface area of the cube . the cube is made of num__6 squares with the sides measuring num__6.5 cm each . the surface area of num__1 side of the cube is num__42.25 cm . multiply the surface area of the num__1 side by num__6 will give you the cube ' s total surface area . the cubes surface area is num__253.5 cm . now we know that the surface area of the ball is num__253.5 cm . to find out the surface area of a sphere you need to square the radius then multiply by pi and multiply by the number num__4 . going backwards to find the radius - you will need to dived the surface area by num__4 . then divide by pi . then determine the square root of the answer . the radius is num__4.460 . rounding to the nearest whole number the radius is num__4 . the correct answer is ( c ) . <eor> c <eos> |
c |
surface_cube__6.5__ square_perimeter__1.0__ square_perimeter__1.0__ |
surface_cube__6.5__ square_perimeter__1.0__ square_perimeter__1.0__ |
| if the side length of square b is four times that of square a the area of square b is how many times the area of square a ? <o> a ) num__20 <o> b ) num__16 <o> c ) num__12 <o> d ) num__8 <o> e ) num__4 |
let x be the side length of square a . then the area of square a is x ^ num__2 . the area of square b is ( num__4 x ) ^ num__2 = num__16 x ^ num__2 . the answer is b . <eor> b <eos> |
b |
square_perimeter__4.0__ square_perimeter__4.0__ |
power__2.0__4.0__ power__2.0__4.0__ |
| in a simultaneous throw of a pair of dice find the probability of getting a total more than num__10 <o> a ) num__0.384615384615 <o> b ) num__0.133333333333 <o> c ) num__1 by num__12 <o> d ) num__0.176470588235 <o> e ) num__0.315789473684 |
total number of cases = num__6 * num__6 = num__36 favourable cases = [ ( num__56 ) ( num__65 ) ( num__66 ) ] = num__3 so probability = num__0.0833333333333 = num__0.0833333333333 answer is c <eor> c <eos> |
c |
add__10.0__56.0__ divide__3.0__36.0__ subtract__66.0__65.0__ |
add__10.0__56.0__ divide__3.0__36.0__ subtract__66.0__65.0__ |
| what is the sum of the first num__24 natural numbers ( starting from num__1 ) ? <o> a ) num__250 <o> b ) num__300 <o> c ) num__350 <o> d ) num__400 <o> e ) num__450 |
the sum of n natural numbers = n * ( n + num__1 ) / num__2 = num__24 * num__12.5 = num__300 the answer is b . <eor> b <eos> |
b |
multiply__24.0__12.5__ multiply__24.0__12.5__ |
multiply__24.0__12.5__ multiply__24.0__12.5__ |
| evaluate : num__20 - num__12 ÷ num__4 × num__2 = <o> a ) a ) num__12 <o> b ) b ) num__24 <o> c ) c ) num__36 <o> d ) d ) num__48 <o> e ) e ) num__60 |
according to order of operations num__12 ÷ num__4 × num__2 ( division and multiplication ) is done first from left to right num__12 ÷ num__4 × num__2 = num__3 × num__2 = num__6 hence num__20 - num__12 ÷ num__4 × num__2 = num__20 - num__6 = num__14 correct answer is b ) num__22 <eor> b <eos> |
b |
divide__12.0__4.0__ divide__12.0__2.0__ subtract__20.0__6.0__ add__20.0__2.0__ add__20.0__4.0__ |
divide__12.0__4.0__ divide__12.0__2.0__ subtract__20.0__6.0__ add__20.0__2.0__ add__20.0__4.0__ |
| the average of num__11 results is num__53 if the average of first six results is num__49 and that of the last six is num__52 . find the sixth result ? <o> a ) a ) num__46 <o> b ) b ) num__23 <o> c ) c ) num__66 <o> d ) d ) num__76 <o> e ) e ) num__74 |
num__1 to num__11 = num__11 * num__53 = num__583 num__1 to num__6 = num__6 * num__49 = num__294 num__6 to num__11 = num__6 * num__52 = num__312 num__6 th = num__294 + num__312 – num__583 = num__23 answer : b <eor> b <eos> |
b |
subtract__53.0__52.0__ multiply__11.0__53.0__ multiply__49.0__6.0__ multiply__52.0__6.0__ multiply__1.0__23.0__ |
subtract__53.0__52.0__ multiply__11.0__53.0__ multiply__49.0__6.0__ multiply__52.0__6.0__ multiply__1.0__23.0__ |
| some ladies can do a piece of work in num__12 days . two times the number of such ladies will do half of that work in : <o> a ) num__6 days <o> b ) num__4 days <o> c ) num__12 days <o> d ) num__3 days <o> e ) num__2 days |
expl : let x ladies can do the work in num__12 days . more ladies less days ( indirect ) less work less days ( direct ) ladies num__2 x : x work num__1 : ½ num__2 x : x num__1 : ½ : : num__12 : y : . num__2 x * num__1 * y = x * ½ * num__12 or y = num__3 hence the required number of days = num__3 answer : d <eor> d <eos> |
d |
add__1.0__2.0__ round__3.0__ |
add__1.0__2.0__ round__3.0__ |
| a sum of money becomes triple itself in num__5 years at simple interest . how many years will it become eight times at the same rate ? <o> a ) num__12 Â ½ years <o> b ) num__16 Â ½ years <o> c ) num__18 Â ½ years <o> d ) num__72 Â ½ years <o> e ) num__12 Â ½ years |
num__100 - - - - num__200 - - - - num__5 num__200 - - - - num__5 num__100 - - - - num__2 num__0.5 - - - - - - - - - - - - - - - - - - num__800 - - - - num__16 Â ½ years answer : b <eor> b <eos> |
b |
percent__2.0__800.0__ percent__2.0__800.0__ |
percent__2.0__800.0__ percent__2.0__800.0__ |
| num__1240 men have provisions for num__12 days . if num__300 more men join them for how many days will the provisions last now ? <o> a ) num__12.9 <o> b ) num__12.0 <o> c ) num__9.7 <o> d ) num__8.6 <o> e ) num__12.1 |
num__1240 * num__12 = num__1540 * x x = num__9.7 answer : c <eor> c <eos> |
c |
add__1240.0__300.0__ round__9.7__ |
add__1240.0__300.0__ round__9.7__ |
| all of the stocks on the over - the - counter market are designated by either a num__3 - letter or a num__4 - letter code that is created by using the num__26 letters of the alphabet . which of the following gives the maximum number of different stocks that can be designated with these codes ? <o> a ) a . num__2 ( num__26 ^ num__5 ) <o> b ) b . num__26 ( num__26 ^ num__4 ) <o> c ) c . num__27 ( num__26 ^ num__3 ) <o> d ) d . num__26 ( num__26 ^ num__5 ) <o> e ) e . num__27 ( num__26 ^ num__5 ) |
no of different stocks for num__3 - letter code = num__26 ^ num__3 no of different stocks for num__4 - letter code = num__26 ^ num__4 total = num__26 ^ num__3 + num__26 ^ num__4 = num__26 ^ num__3 ( num__26 + num__1 ) = num__27 ( num__26 ^ num__3 ) [ reveal ] spoiler : option c <eor> c <eos> |
c |
subtract__4.0__3.0__ add__26.0__1.0__ add__26.0__1.0__ |
subtract__4.0__3.0__ add__26.0__1.0__ add__26.0__1.0__ |
| if m > num__0 and z is m percent of y then in terms of m y is what percent of z ? <o> a ) a ) num__100 m <o> b ) b ) num__0.01 m <o> c ) c ) num__1 / m <o> d ) d ) num__10000 / m <o> e ) e ) num__10 / m |
z = m / num__100 * y so y = num__100 / m * z so y = ( num__10000 / m ) / num__100 * z d <eor> d <eos> |
d |
percent__100.0__10000.0__ |
percent__100.0__10000.0__ |
| the number of new words that can be formed by rearranging the letters of the word ' can ' is ? <o> a ) num__3 <o> b ) num__6 <o> c ) num__10 <o> d ) num__8 <o> e ) num__5 |
number of words which can be formed = num__3 ! - num__1 = num__6 - num__1 = num__5 . answer : e <eor> e <eos> |
e |
die_space__ vowel_space__ vowel_space__ |
die_space__ vowel_space__ vowel_space__ |
| the largest num__4 digit number exactly divisible by num__88 is ? <o> a ) num__9935 <o> b ) num__9939 <o> c ) num__9944 <o> d ) num__9954 <o> e ) num__9960 |
largest num__4 - digit number = num__9999 num__88 ) num__9999 ( num__113 num__88 - - - - num__119 num__88 - - - - num__319 num__264 - - - num__55 - - - required number = ( num__9999 - num__55 ) = num__9944 . c ) <eor> c <eos> |
c |
subtract__319.0__264.0__ multiply__88.0__113.0__ multiply__88.0__113.0__ |
subtract__319.0__264.0__ subtract__9999.0__55.0__ subtract__9999.0__55.0__ |
| find the l . c . m of num__2 ^ num__3 * num__3 ^ num__2 * num__5 * num__7 ^ num__3 * num__11 ^ num__3 num__2 ^ num__4 * num__3 ^ num__2 * num__5 ^ num__3 * num__7 ^ num__3 * num__11 num__2 ^ num__2 * num__3 * num__5 ^ num__3 * num__7 ^ num__3 * num__11 <o> a ) num__2 ^ num__4 * num__3 ^ num__2 * num__5 ^ num__3 * num__7 ^ num__3 * num__11 ^ num__3 <o> b ) num__2 ^ num__2 * num__3 ^ num__3 * num__5 ^ num__2 * num__7 ^ num__4 * num__11 <o> c ) num__2 ^ num__2 * num__3 ^ num__4 * num__5 ^ num__3 * num__7 ^ num__4 * num__11 <o> d ) num__2 ^ num__2 * num__3 ^ num__3 * num__5 ^ num__3 * num__7 ^ num__4 * num__11 ^ num__3 <o> e ) num__2 ^ num__3 * num__3 ^ num__3 * num__5 ^ num__3 * num__7 ^ num__4 * num__11 ^ num__2 |
l . c . m = product of the highest powers of num__23 num__57 and num__11 = num__2 ^ num__4 * num__3 ^ num__2 * num__5 ^ num__3 * num__7 ^ num__3 * num__11 ^ num__3 answer ( a ) <eor> a <eos> |
a |
subtract__5.0__3.0__ |
subtract__5.0__3.0__ |
| when processing flower - nectar into honey bees ' extract a considerable amount of water gets reduced . how much flower - nectar must be processed to yield num__1 kg of honey if nectar contains num__50.0 water and the honey obtained from this nectar contains num__20.0 water ? <o> a ) num__1.2 kg <o> b ) num__1.5 kg <o> c ) num__1.7 kg <o> d ) num__1.6 kg <o> e ) none of these |
explanation : flower - nectar contains num__50.0 of non - water part . in honey this non - water part constitutes num__80.0 ( num__100 - num__20 ) . therefore num__0.5 x amount of flower - nectar = num__0.80 x amount of honey = num__0.80 x num__1 kg therefore amount of flower - nectar needed = ( num__0.80 / num__0.51 ) kg = num__1.6 kgs answer : d <eor> d <eos> |
d |
add__20.0__80.0__ divide__50.0__100.0__ divide__80.0__100.0__ divide__0.8__0.5__ multiply__1.0__1.6__ |
add__20.0__80.0__ divide__50.0__100.0__ divide__80.0__100.0__ divide__0.8__0.5__ divide__1.6__1.0__ |
| find the area of trapezium whose parallel sides are num__20 cm and num__18 cm long and the distance between them is num__12 cm <o> a ) num__178 cm num__2 <o> b ) num__179 cm num__2 <o> c ) num__285 cm num__2 <o> d ) num__167 cm num__2 <o> e ) num__228 cm num__2 |
area of a trapezium = num__0.5 ( sum of parallel sides ) * ( perpendicular distance between them ) = num__0.5 ( num__20 + num__18 ) * ( num__12 ) = num__228 cm num__2 answer : e <eor> e <eos> |
e |
subtract__20.0__18.0__ round__228.0__ |
subtract__20.0__18.0__ round__228.0__ |
| p alone can complete a piece of work in num__6 days . work done by q alone in one day is equal to one - third of the work done by p alone in one day . in how many days can the work be completed if p and q work together ? <o> a ) num__7 <o> b ) num__4 ( num__0.5 ) <o> c ) num__65 <o> d ) num__9 <o> e ) num__1 |
work done by p alone in one day = num__0.166666666667 th of the total work done by q alone in one day = num__0.333333333333 ( of that done by p in one day ) = num__0.333333333333 ( num__0.166666666667 of the total ) = num__0.0555555555556 of the total . work done by p and q working together in one day = num__0.166666666667 + num__0.0555555555556 = num__0.222222222222 = num__0.222222222222 of the total they would take num__4.5 days = num__4 ( num__0.5 ) days to complete the work working together . answer : b <eor> b <eos> |
b |
multiply__0.1667__0.3333__ subtract__4.5__4.0__ round__4.0__ |
multiply__0.1667__0.3333__ add__0.1667__0.3333__ round__4.0__ |
| num__15 java programmers working in a constant pace finish a web page in num__3 days . if after one day num__9 programmers quit how many more days are needed to finish the remainder of the job ? <o> a ) num__5 . <o> b ) num__2 . <o> c ) num__8 . <o> d ) num__4 . <o> e ) num__6 . |
given : num__3 day ' s work : x num__1 day ' s work : x / num__3 given : num__15 people complete the work in num__3 days . after num__1 day ' s work ( x / num__3 ) num__9 people left . remaining work = x - x / num__3 = num__2 x / num__3 now num__1 day ' s work of num__15 persons = x / num__3 num__1 day ' s work of num__1 person = x / num__45 . num__1 day ' s work of num__6 persons ( num__15 - num__9 ) = num__2 x / num__15 . num__2 x / num__15 work completed in num__1 day . num__2 x / num__3 work completed in = num__5 days question asks how many more days needed to complete the remaining work therefore num__5 more days needed to complete the remaining work . answer is a . <eor> a <eos> |
a |
subtract__3.0__1.0__ multiply__15.0__3.0__ subtract__15.0__9.0__ divide__15.0__3.0__ round__5.0__ |
subtract__3.0__1.0__ multiply__15.0__3.0__ subtract__15.0__9.0__ divide__15.0__3.0__ divide__15.0__3.0__ |
| how many two - digit whole numbers yield a remainder of num__2 when divided by num__12 and also yield a remainder of num__2 when divided by num__6 ? <o> a ) none <o> b ) five <o> c ) six <o> d ) seven <o> e ) eight |
the possible number n can be written as follow : n = multiple of lcm ( num__126 ) + num__1 st such number n = num__12 x + num__2 possible values = num__2 num__2638 num__5062 num__8698 answer : num__6 such num__2 digit number . c . <eor> c <eos> |
c |
divide__12.0__2.0__ |
divide__12.0__2.0__ |
| of the num__150 employees at company x num__50 are full - time and num__100 have worked at company x for at least a year . there are num__20 employees at company x who aren ’ t full - time and haven ’ t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__50 <o> d ) num__80 <o> e ) num__100 |
num__150 employees num__50 are full - time num__100 have worked at company x for at least a year num__20 employees at company x who aren ’ t full - time and haven ’ t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ? num__150 - num__50 = num__100 employees not full time num__100 - num__20 = num__80 employees not full time who worked over a year num__100 employees have worked at company x for at least a year - num__80 employees not full time who worked over a year = num__20 full - time employees of company x have worked at the company for at least a year ans a <eor> a <eos> |
a |
subtract__100.0__20.0__ subtract__100.0__80.0__ |
subtract__100.0__20.0__ subtract__100.0__80.0__ |
| set s contains exactly num__10 numbers and has an average ( arithmetic mean ) of num__6.2 . if one of the numbers in set s is increased by num__3 while all other numbers remain the same what is the new average of set s ? <o> a ) num__6.5 <o> b ) num__6.7 <o> c ) num__6.8 <o> d ) num__6.85 <o> e ) num__6.9 |
old set s - total is avg * no of elements = num__6.2 * num__10 = num__62 if one number is increased by num__3 then total increased to num__62 + num__3 = num__65 new avg - num__6.5 = num__6.5 . hence answer is a . <eor> a <eos> |
a |
multiply__10.0__6.2__ add__3.0__62.0__ divide__65.0__10.0__ divide__65.0__10.0__ |
multiply__10.0__6.2__ add__3.0__62.0__ divide__65.0__10.0__ divide__65.0__10.0__ |
| each shelf of a bookcase contained num__12 books . if the librarian took out num__22 books and rearranged the remaining books so that all shelves but one contained num__8 books and the last shelf contained num__10 books how many shelves does the bookcase have ? <o> a ) num__8 <o> b ) num__6 <o> c ) num__7 <o> d ) num__5 <o> e ) num__4 |
let x be the number of shelves . no of books in the beginning = num__12 * x after taking out num__21 books no of remaining books = num__12 x - num__21 when the remaining books are rearranged each shelf except the last shelf has num__8 books and the last shelf has num__11 books . so no of remaining books is also equal to num__8 * ( x - num__1 ) + num__10 num__12 x - num__22 = num__8 ( x - num__1 ) + num__10 num__4 x = num__24 x = num__6 . answer b <eor> b <eos> |
b |
die_space__ die_space__ |
die_space__ die_space__ |
| of all the homes on gotham street num__0.333333333333 are termite - ridden and num__0.625 of these are collapsing . what fraction of the homes are termite - ridden but not collapsing ? <o> a ) num__0.333333333333 <o> b ) num__0.25 <o> c ) num__0.2 <o> d ) num__0.166666666667 <o> e ) num__0.125 |
the fraction of homes which are termite - infested but not collapsing is num__0.375 * num__0.333333333333 = num__0.125 the answer is e . <eor> e <eos> |
e |
multiply__0.3333__0.375__ multiply__0.3333__0.375__ |
multiply__0.3333__0.375__ multiply__0.3333__0.375__ |
| alex takes a loan of $ num__8000 to buy a used truck at the rate of num__9.0 simple interest . calculate the annual interest to be paid for the loan amount . <o> a ) num__680 <o> b ) num__700 <o> c ) num__720 <o> d ) num__730 <o> e ) num__750 |
from the details given in the problem principle = p = $ num__8000 and r = num__9.0 or num__0.09 expressed as a decimal . as the annual interest is to be calculated the time period t = num__1 . plugging these values in the simple interest formula i = p x t x r = num__8000 x num__1 x num__0.09 = num__720.00 annual interest to be paid = $ num__720 answer : c <eor> c <eos> |
c |
percent__9.0__8000.0__ percent__9.0__8000.0__ |
percent__9.0__8000.0__ percent__9.0__8000.0__ |
| a shopkeeper sold an article offering a discount of num__5.0 and earned a profit of num__23.5 . what would have been the percentage of profit earned if no discount was offered ? <o> a ) num__17.0 <o> b ) num__18.0 <o> c ) num__30.0 <o> d ) num__17.0 <o> e ) num__12 % |
let c . p . be rs . num__100 . then s . p . = rs . num__123.50 let the marked price be rs . x . then num__0.95 x = num__123.50 x = num__130.0 = rs . num__130 now s . p . = rs . num__130 c . p . = rs . num__100 profit % = num__30.0 . answer : c <eor> c <eos> |
c |
percent__100.0__30.0__ |
percent__100.0__30.0__ |
| pipe a fills a tank in num__20 minutes . pipe b can fill the same tank num__4 times as fast as pipe a . if both the pipes are kept open when the tank is empty how many minutes will it take to fill the tank ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
a ' s rate is num__0.05 and b ' s rate is num__0.2 . the combined rate is num__0.05 + num__0.2 = num__0.25 the pipes will fill the tank in num__4 minutes . the answer is c . <eor> c <eos> |
c |
divide__4.0__20.0__ add__0.2__0.05__ round__4.0__ |
divide__4.0__20.0__ add__0.2__0.05__ round__4.0__ |
| the function g ( j ) is defined for integers j such that if j is even g ( j ) = j / num__2 and if j is odd g ( j ) = j + num__5 . given that g ( g ( g ( g ( g ( j ) ) ) ) ) = num__19 how many possible values for j would satisfy this equation ? <o> a ) num__1 <o> b ) num__5 <o> c ) num__7 <o> d ) num__8 <o> e ) num__11 |
let me define terms : in g ( j ) = r j is argument r is result g ( ) is function in g ( g ( g ( g ( g ( j ) ) ) ) ) g num__1 is inner most g num__5 is outermost for identification . from definition of function g we can deduce that : if result is even then two possibilities for argument = num__1 even num__1 odd if result is odd then one possibility for argument = num__1 even since final result = num__19 = odd possibilities : g num__1 : num__1 even g num__2 : num__1 * ( even odd ) = num__1 even num__1 odd g num__3 : num__1 * ( even odd ) + num__1 even = num__2 even num__1 odd g num__4 : num__2 * ( even odd ) + num__1 even = num__3 even num__2 odd g num__5 : num__3 * ( even odd ) + num__2 even = num__5 even num__3 odd = total num__8 ans d it is ! <eor> d <eos> |
d |
add__2.0__1.0__ subtract__5.0__1.0__ multiply__2.0__4.0__ multiply__2.0__4.0__ |
add__2.0__1.0__ add__1.0__3.0__ multiply__2.0__4.0__ multiply__2.0__4.0__ |
| num__13 buckets of water fill a tank when the capacity of each bucket is num__42 litres . how many buckets will be needed to fill the same tank if the capacity of each bucket is num__6 litres ? <o> a ) num__91 <o> b ) num__80 <o> c ) num__96 <o> d ) num__98 <o> e ) num__90 |
capacity of the tank = ( num__13 Ã — num__42 ) litre number of buckets required of capacity of each bucket is num__17 litre = num__13 Ã — num__7.0 = num__13 Ã — num__7 = num__91 answer is a <eor> a <eos> |
a |
subtract__13.0__6.0__ multiply__13.0__7.0__ round__91.0__ |
subtract__13.0__6.0__ multiply__13.0__7.0__ round__91.0__ |
| a certain lab experiments with white and brown mice only . in one experiment num__0.666666666667 of the mice are white . if there are num__14 white mice in the experiment how many brown mice are in the experiment ? <o> a ) num__12 <o> b ) num__8 <o> c ) num__28 <o> d ) num__7 <o> e ) num__27 |
let total number of mice = m number of white mice = num__0.666666666667 m = num__14 m = num__21 number of brown mice = num__0.333333333333 m = num__0.333333333333 * num__21 = > brown mice = num__7 answer d <eor> d <eos> |
d |
subtract__21.0__14.0__ subtract__14.0__7.0__ |
subtract__21.0__14.0__ subtract__14.0__7.0__ |
| the area of one square is x ^ num__2 + num__10 x + num__25 and the area of another square is num__4 x ^ num__2 − num__12 x + num__9 . if the sum of the perimeters of both squares is num__32 what is the value of x ? <o> a ) num__0 <o> b ) num__2 <o> c ) num__2.5 <o> d ) num__4.67 <o> e ) num__10 |
spotting the pattern of equations both are in form of ( x + c ) ^ num__2 so a num__1 = ( x + num__5 ) ^ num__2 & a num__2 = ( num__2 x - num__3 ) ^ num__2 l num__1 = x + num__5 & l num__2 = num__2 x - num__3 p num__1 = num__4 ( x + num__5 ) & p num__2 = num__4 ( num__2 x - num__3 ) p num__1 + p num__2 = num__32 num__4 ( x + num__5 ) + num__4 ( num__2 x - num__3 ) = num__32 . . . . . . . . . . . . . . > x = num__2 answer : b <eor> b <eos> |
b |
subtract__10.0__9.0__ divide__10.0__2.0__ add__2.0__1.0__ multiply__2.0__1.0__ |
subtract__10.0__9.0__ add__4.0__1.0__ add__2.0__1.0__ subtract__4.0__2.0__ |
| in a num__500 m race the ratio of the speeds of two contestants a and b is num__3 : num__4 . a has a start of num__140 m . then a wins by : <o> a ) num__20 m <o> b ) num__40 m <o> c ) num__60 m <o> d ) num__45 m <o> e ) num__55 m |
to reach the winning post a will have to cover a distance of ( num__500 - num__140 ) m i . e . num__360 m . while a covers num__3 m b covers num__4 m . while a covers num__360 m b covers ( num__1.33333333333 * num__360 ) = num__480 m when a reaches the winning post b covers num__480 m and num__20 m behind answer a <eor> a <eos> |
a |
subtract__500.0__140.0__ divide__4.0__3.0__ subtract__500.0__480.0__ round__20.0__ |
subtract__500.0__140.0__ divide__4.0__3.0__ subtract__500.0__480.0__ subtract__500.0__480.0__ |
| a manufacturer of a certain product can expect that between num__0.2 percent and num__0.5 percent of the units manufactured will be defective . if the retail price is $ num__2500 per unit and the manufacturer offers a full refund for defective units how much money can the manufacturer expect to need to cover the refunds on num__20000 units ? <o> a ) between $ num__15000 and $ num__25000 <o> b ) between $ num__30000 and $ num__50000 <o> c ) between $ num__60000 and $ num__100000 <o> d ) between $ num__100000 and $ num__250000 <o> e ) between $ num__300000 and $ num__500 |
000 |
number of defective units is between = . num__2.0 of num__20000 and . num__5.0 of num__20000 = num__40 and num__100 retail price per unit = num__2500 $ expected price of refund is between = num__2500 x num__40 and num__2500 x num__100 = num__1 num__00000 and num__2 num__50000 dollars answer d <eor> d <eos> |
d |
d |
| the difference of the areas of two squares drawn on two line segments in num__32 sq . cm . find the length of the greater line segment if one is longer than the other by num__2 cm <o> a ) num__9 cm <o> b ) num__8 cm <o> c ) num__7 cm <o> d ) num__6 cm <o> e ) num__5 cm |
explanation : let the lengths of the line segments be x and x + num__2 cm then ( x + num__2 ) num__2 − x num__2 = num__32 x num__2 + num__4 x + num__4 − x num__2 = num__324 x = num__28 x = num__7 cm option c <eor> c <eos> |
c |
triangle_area__2.0__7.0__ |
triangle_area__2.0__7.0__ |
| a man ' s speed with the current is num__20 kmph and speed of the current is num__4 kmph . the man ' s speed against the current will be <o> a ) num__11 kmph <o> b ) num__12 kmph <o> c ) num__14 kmph <o> d ) num__17 kmph <o> e ) none of these |
explanation : speed with current is num__20 speed of the man + it is speed of the current speed in s Ɵ ll water = num__20 - num__4 = num__16 now speed against the current will be speed of the man - speed of the current = num__16 - num__4 = num__12 kmph answer : b <eor> b <eos> |
b |
subtract__20.0__4.0__ subtract__16.0__4.0__ round__12.0__ |
subtract__20.0__4.0__ subtract__16.0__4.0__ subtract__16.0__4.0__ |
| the parameter of a square is double the perimeter of a rectangle . the area of the rectangle is num__480 sq cm . find the area of the square ? <o> a ) num__482 <o> b ) num__268 <o> c ) num__260 <o> d ) num__480 <o> e ) num__632 |
let the side of the square be a cm . let the length and the breadth of the rectangle be l cm and b cm respectively . num__4 a = num__2 ( l + b ) num__2 a = l + b l . b = num__480 we can not find ( l + b ) only with the help of l . b . therefore a can not be found . area of the square can not be found . answer : d <eor> d <eos> |
d |
triangle_area__480.0__2.0__ |
triangle_area__480.0__2.0__ |
| zachary is helping his younger brother sterling learn his multiplication tables . for every question that sterling answers correctly zachary gives him num__3 pieces of candy . for every question that sterling answers incorrectly zachary takes away two pieces of candy . after num__8 questions if sterling had answered num__2 more questions correctly he would have earned num__31 pieces of candy . how many of the num__8 questions did zachary answer correctly ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
i got two equations : num__3 x - num__2 y = num__25 x + y = num__8 num__3 x - num__2 ( num__8 - x ) = num__25 num__3 x - num__16 + num__2 x = num__25 num__5 x = num__41 x = num__8.2 or between num__8 and num__9 . num__9 ans c ) <eor> c <eos> |
c |
multiply__8.0__2.0__ add__3.0__2.0__ add__16.0__25.0__ divide__41.0__5.0__ subtract__25.0__16.0__ round_down__8.2__ |
multiply__8.0__2.0__ add__3.0__2.0__ add__16.0__25.0__ divide__41.0__5.0__ subtract__25.0__16.0__ add__3.0__5.0__ |
| the difference between simple and compound interests compounded annually on a certain sum of money for num__2 years at num__4.0 per annum is re . num__1 . the sum ( in rs . ) is ? <o> a ) num__625 <o> b ) num__298 <o> c ) num__269 <o> d ) num__200 <o> e ) num__267 |
let the sum be rs . x . then [ x ( num__1 + num__0.04 ) num__2 - x ] = ( num__1.0816 x - x ) = num__0.0816 x s . i . = ( x * num__4 * num__2 ) / num__100 = num__2 x / num__25 num__51 x / num__625 - num__2 x / num__25 = num__1 or x = num__625 . answer : a <eor> a <eos> |
a |
percent__4.0__1.0__ percent__100.0__625.0__ |
percent__4.0__1.0__ percent__100.0__625.0__ |
| if k m and n are positive real numbers such that k ( m + n ) = num__152 m ( n + k ) = num__162 and n ( k + m ) = num__170 then kmn is <o> a ) a ) num__672 <o> b ) b ) num__688 <o> c ) c ) num__704 <o> d ) d ) num__720 <o> e ) e ) num__750 |
km + kn = num__152 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - num__1 ) mn + mk = num__162 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - num__2 ) kn + nm = num__170 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - num__3 ) re - writing equation num__3 as follows : kn + nm = num__162 + num__8 kn + nm = mn + mk + num__8 kn = mk + num__8 . . . . . . . . . . . . . . . ( num__4 ) adding ( num__1 ) ( num__4 ) num__2 kn = num__160 kn = num__80 kmn has to be multiple of num__80 only num__720 fits in answer = d <eor> d <eos> |
d |
add__1.0__2.0__ subtract__170.0__162.0__ add__1.0__3.0__ add__152.0__8.0__ divide__160.0__2.0__ multiply__1.0__720.0__ |
add__1.0__2.0__ subtract__170.0__162.0__ add__1.0__3.0__ add__152.0__8.0__ divide__160.0__2.0__ multiply__1.0__720.0__ |
| arjun started a business with rs . num__5000 and is joined afterwards by anoop with rs . num__10 num__000 . after how many months did anoop join if the profits at the end of the year are divided equally ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
suppose anoop joined after num__3 months . then num__5000 * num__12 = num__10000 * ( num__12 – x ) = > x = num__6 . answer : d <eor> d <eos> |
d |
subtract__12.0__6.0__ |
subtract__12.0__6.0__ |
| the average of num__1 st num__3 of num__4 numbers is num__16 and of the last num__3 are num__15 . if the sum of the first and the last number is num__13 . what is the last numbers ? <o> a ) num__9 <o> b ) num__8 <o> c ) num__5 <o> d ) num__4 <o> e ) num__3 |
a + b + c = num__48 b + c + d = num__45 a + d = num__13 a – d = num__3 a + d = num__13 num__2 d = num__10 d = num__5 answer : c <eor> c <eos> |
c |
multiply__3.0__16.0__ multiply__3.0__15.0__ subtract__3.0__1.0__ subtract__13.0__3.0__ add__1.0__4.0__ add__1.0__4.0__ |
multiply__3.0__16.0__ multiply__3.0__15.0__ subtract__3.0__1.0__ subtract__13.0__3.0__ add__1.0__4.0__ add__1.0__4.0__ |
| present ages of sameer and anand are in the ratio of num__5 : num__4 respectively . five years hence the ratio of their ages will become num__11 : num__9 respectively . what is anand ' s present age in years ? <o> a ) ca n ' t be determined <o> b ) num__40 <o> c ) num__27 <o> d ) num__24 <o> e ) none of these |
explanation : let the present ages of sameer and anand be num__5 x years and num__4 x years respectively . then ( num__5 x + num__5 ) / ( num__4 x + num__5 ) = num__1.22222222222 ⇒ num__45 x + num__45 = num__44 x + num__55 ⇒ num__9 ( num__5 x + num__5 ) = num__11 ( num__4 x + num__5 ) ⇒ num__45 x - num__44 x = num__55 - num__45 ⇒ x = num__10 . anand ' s present age = num__4 x = num__40 years . answer : b <eor> b <eos> |
b |
divide__11.0__9.0__ multiply__5.0__9.0__ multiply__4.0__11.0__ multiply__5.0__11.0__ subtract__55.0__45.0__ multiply__4.0__10.0__ multiply__4.0__10.0__ |
divide__11.0__9.0__ multiply__5.0__9.0__ multiply__4.0__11.0__ add__11.0__44.0__ subtract__55.0__45.0__ subtract__44.0__4.0__ subtract__44.0__4.0__ |
| if n is an integer f ( n ) = f ( n - num__1 ) - n and f ( num__4 ) = num__14 . what is the value of f ( num__6 ) ? <o> a ) num__3 <o> b ) num__0 <o> c ) num__1 <o> d ) num__2 <o> e ) num__4 |
since f ( n ) = f ( n - num__1 ) - n then : f ( num__6 ) = f ( num__5 ) - num__6 and f ( num__5 ) = f ( num__4 ) - num__5 . as given that f ( num__4 ) = num__14 then f ( num__5 ) = num__14 - num__5 = num__9 - - > substitute the value of f ( num__5 ) back into the first equation : f ( num__6 ) = f ( num__5 ) - num__6 = num__9 - num__6 = num__3 . answer : a . questions on funtions to practice : <eor> a <eos> |
a |
add__1.0__4.0__ add__4.0__5.0__ subtract__4.0__1.0__ multiply__1.0__3.0__ |
subtract__6.0__1.0__ subtract__14.0__5.0__ subtract__4.0__1.0__ subtract__4.0__1.0__ |
| if the area of a square with sides of length num__9 centimeters is equal to the area of a rectangle with a width of num__3 centimeters what is the length of the rectangle in centimeters ? <o> a ) num__4 <o> b ) num__27 <o> c ) num__12 <o> d ) num__16 <o> e ) num__18 |
let length of rectangle = l num__9 ^ num__2 = l * num__3 = > l = num__27.0 = num__27 answer b <eor> b <eos> |
b |
volume_cube__3.0__ volume_cube__3.0__ |
multiply__9.0__3.0__ multiply__9.0__3.0__ |
| last year department store x had a sales total for december that was num__6 times the average ( arithmetic mean ) of the monthly sales totals for january through november . the sales total for december was what fraction of the sales total for the year ? <o> a ) num__0.352941176471 <o> b ) num__0.266666666667 <o> c ) num__0.333333333333 <o> d ) num__0.363636363636 <o> e ) num__0.8 |
let avg for num__11 mos . = num__10 therefore dec = num__60 year total = num__11 * num__10 + num__60 = num__170 answer = num__0.352941176471 = num__0.352941176471 = a <eor> a <eos> |
a |
multiply__6.0__10.0__ divide__60.0__170.0__ divide__60.0__170.0__ |
multiply__6.0__10.0__ divide__60.0__170.0__ divide__60.0__170.0__ |
| if r is even and s is odd which of the following is odd ? <o> a ) rs <o> b ) num__5 rs <o> c ) num__6 ( r ^ num__2 ) s <o> d ) num__5 r + num__6 s <o> e ) num__6 r + num__5 s |
just checking options ( in case you have difficulty then choose values of r = num__2 and t = num__1 ) r = even s = odd a . rs = even * odd = evenincorrect b . num__5 rs = odd * even * odd = evenincorrect c . num__6 ( r ^ num__2 ) s = num__6 * ( even ^ num__2 ) * odd = evenincorrect d . num__5 r + num__6 s = num__5 * even + num__6 * odd = even + even = evenincorrect e . num__6 r + num__5 s = num__6 * even + num__5 * odd = even + odd = oddcorrect answer : option e <eor> e <eos> |
e |
add__1.0__5.0__ add__1.0__5.0__ |
add__1.0__5.0__ add__1.0__5.0__ |
| during num__2001 a stock lost num__80 percent of its value . during the following year the stock ' s value increased by num__100 percent . which of the following is the percent change in the stock ' s value during those two years ? <o> a ) num__60 percent decrease <o> b ) num__40 percent decrease <o> c ) num__20 percent increase <o> d ) num__60 percent increase <o> e ) num__140 percent increase |
let the value of the stock in num__2000 be num__100 then in num__2001 its value would be num__100 * ( num__1 - num__0.8 ) = num__100 * num__0.2 = num__20 and in num__2002 its value would be num__20 * ( num__1 + num__1.0 ) = num__40 . hence during those two years the value of the stock decreased from num__100 to num__40 so by num__60.0 . answer : a . <eor> a <eos> |
a |
subtract__2001.0__2000.0__ divide__80.0__100.0__ subtract__1.0__0.8__ subtract__100.0__80.0__ add__2001.0__1.0__ subtract__80.0__20.0__ subtract__80.0__20.0__ |
subtract__2001.0__2000.0__ divide__80.0__100.0__ subtract__1.0__0.8__ subtract__100.0__80.0__ add__2001.0__1.0__ subtract__80.0__20.0__ subtract__80.0__20.0__ |
| one night a certain hotel rented num__0.75 of its rooms . including num__0.666666666667 of their air conditioned rooms . if num__0.6 of its rooms were air conditioned what percent of the rooms that were not rented were air conditioned ? <o> a ) num__20 <o> b ) num__33 num__0.333333333333 <o> c ) num__35 <o> d ) num__40 <o> e ) num__80 |
consider total # of rooms to be num__100 ; as num__0.6 of the rooms are air conditioned then # of rooms that are air conditioned is num__0.6 * num__100 = num__60 ; num__0.75 rooms were rented - - > num__0.25 * num__100 = num__25 were not rented ; num__0.666666666667 of air conditioned rooms were rented - - > num__0.333333333333 * num__60 = num__20 air conditioned room were not rented ; num__0.8 = num__0.8 = num__80.0 . answer : e . <eor> e <eos> |
e |
percent__100.0__80.0__ |
percent__100.0__80.0__ |
| in town x num__64 percent of the population are employed and num__46 percent of the population are employed males . what percent of the employed people in town x are females ? <o> a ) num__16.0 <o> b ) num__27.0 <o> c ) num__32.0 <o> d ) num__40.0 <o> e ) num__52 % |
answer b male - employed - num__64.0 female - employed - num__18.0 total employed num__64.0 means total unemployed = num__36.0 therefore d and e are not correct because female unemployed has to be less than num__36.0 female - unemployed = num__32.0 male unemployed = num__4.0 num__48.0 + num__4.0 = num__52.0 num__16.0 + num__32.0 = num__48.0 plug in a and b in place of num__32.0 and the sum is not num__100.0 num__0.28125 = num__27 b <eor> b <eos> |
b |
percent__27.0__100.0__ |
percent__27.0__100.0__ |
| the age of father num__6 years ago was four times the age of his son . five years hence father ' s age will be twice that of his son . the ratio of their present ages is : <o> a ) num__55 : num__23 <o> b ) num__43 : num__23 <o> c ) num__51 : num__23 <o> d ) num__53 : num__23 <o> e ) num__56 : num__23 |
let the ages of father and son num__6 years ago be num__3 x and x years respectively . then ( num__4 x + num__6 ) + num__5 = num__2 [ ( x + num__6 ) + num__5 ] num__4 x + num__11 = num__2 x + num__22 x = num__5.5 . required ratio = ( num__4 x + num__6 ) : ( x + num__6 ) = num__28 : num__11.5 = num__56 : num__23 . answer : option e <eor> e <eos> |
e |
divide__6.0__3.0__ add__6.0__5.0__ multiply__2.0__11.0__ divide__11.0__2.0__ add__6.0__22.0__ add__6.0__5.5__ multiply__2.0__28.0__ multiply__11.5__2.0__ multiply__2.0__28.0__ |
divide__6.0__3.0__ add__6.0__5.0__ multiply__2.0__11.0__ divide__11.0__2.0__ add__6.0__22.0__ add__6.0__5.5__ multiply__2.0__28.0__ multiply__11.5__2.0__ multiply__2.0__28.0__ |
| the difference between a two - digit number and the number obtained by interchanging the positions of its digits is num__36 . what is the difference between the two digits of that number ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) none |
let the ten ' s digit be x and unit ' s digit be y . then ( num__10 x + y ) - ( num__10 y + x ) = num__36 num__9 ( x - y ) = num__36 x - y = num__4 . option d <eor> d <eos> |
d |
divide__36.0__9.0__ divide__36.0__9.0__ |
divide__36.0__9.0__ divide__36.0__9.0__ |
| little texas drilling company has three wells each producing oil at a constant rate . well a produces one barrel every two minutes . well b produces one barrel every three minutes . well c produces one barrel every four minutes . how many hours does it take little texas drilling company to produce num__205 barrels of oil ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__18 <o> e ) num__180 |
well a produces num__0.5 a barrel per minute . wells b and c produce num__0.333333333333 + num__0.25 = num__0.583333333333 ( a little more than half ) a barrel per minute . so all three wells combined produce a little more than num__1 barrel per minute . so for num__205 barrels they will take a bit less than num__205 mins which will be more than num__3 hrs but less than num__4 hrs . answer ( c ) <eor> c <eos> |
c |
add__0.25__0.3333__ add__3.0__1.0__ round__4.0__ |
add__0.25__0.3333__ add__3.0__1.0__ add__3.0__1.0__ |
| what amount does kiran get if he invests rs . num__4500 at num__10.0 p . a . compound interest for three years compounding done annually ? <o> a ) num__2999 <o> b ) num__2787 <o> c ) num__2097 <o> d ) num__5989 <o> e ) num__2886 |
a = p { num__1 + r / num__100 } n = > num__4500 { num__1 + num__0.1 } num__3 = rs . num__5989 answer : d <eor> d <eos> |
d |
percent__10.0__1.0__ percent__100.0__5989.0__ |
percent__10.0__1.0__ percent__100.0__5989.0__ |
| a number when divided by a divisor leaves a remainder of num__24 . when twice the original number is divided by the same divisor the remainder is num__11 . what is the value of the divisor ? <o> a ) num__37 <o> b ) num__39 <o> c ) num__42 <o> d ) num__44 <o> e ) num__45 |
let the original number be ' a ' let the divisor be ' d ' let the quotient of the division of aa by dd be ' x ' therefore we can write the relation as a / d = x and the remainder is num__24 . i . e . a = dx + num__24 when twice the original number is divided by d num__2 a is divided by d . we know that a = dx + num__24 . therefore num__2 a = num__2 dx + num__48 the problem states that ( num__2 dx + num__48 ) / d leaves a remainder of num__11 . num__2 dx num__2 dx is perfectly divisible by d and will therefore not leave a remainder . the remainder of num__11 was obtained by dividing num__48 by d . when num__48 is divided by num__37 the remainder that one will obtain is num__11 . hence the divisor is num__37 . a <eor> a <eos> |
a |
multiply__24.0__2.0__ subtract__48.0__11.0__ subtract__48.0__11.0__ |
multiply__24.0__2.0__ subtract__48.0__11.0__ subtract__48.0__11.0__ |
| a goods train runs at the speed of num__72 kmph and crosses a num__250 m long platform in num__24 seconds . what is the length of the goods train ? <o> a ) num__230 m <o> b ) num__240 m <o> c ) num__260 m <o> d ) num__270 m <o> e ) none of these |
explanation : speed = [ num__72 x ( num__0.277777777778 ) ] m / sec = num__20 m / sec . time = num__24 sec . let the length of the train be x metres . then [ ( x + num__250 ) / num__24 ] = num__20 = > x + num__250 = num__480 = > x = num__230 . answer : a <eor> a <eos> |
a |
multiply__24.0__20.0__ subtract__250.0__20.0__ round__230.0__ |
multiply__24.0__20.0__ subtract__250.0__20.0__ round__230.0__ |
| the ratio of two numbers is num__3 : num__4 and their sum is num__49 . the greater of the two numbers is ? <o> a ) num__12 <o> b ) num__14 <o> c ) num__16 <o> d ) num__17 <o> e ) num__28 |
num__3 : num__4 total parts = num__7 = num__7 parts - - > num__49 ( num__7 Ã — num__7 = num__49 ) = num__1 part - - - - > num__7 ( num__1 Ã — num__7 = num__7 ) = the greater of the two number is = num__4 = num__4 parts - - - - > num__28 ( num__7 Ã — num__4 = num__28 ) e <eor> e <eos> |
e |
add__3.0__4.0__ subtract__4.0__3.0__ multiply__4.0__7.0__ multiply__4.0__7.0__ |
add__3.0__4.0__ subtract__4.0__3.0__ multiply__4.0__7.0__ multiply__4.0__7.0__ |
| ( num__51 + num__52 + num__53 + . . . + num__100 ) = ? <o> a ) num__3775 <o> b ) num__3795 <o> c ) num__3800 <o> d ) num__3900 <o> e ) num__4000 |
this is an a . p . in which a = num__51 l = num__100 and n = num__50 . sum = n ( a + l ) = num__50 x ( num__51 + num__100 ) = ( num__25 x num__151 ) = num__3775 . num__2 num__2 a ) <eor> a <eos> |
a |
add__51.0__100.0__ multiply__151.0__25.0__ subtract__52.0__50.0__ multiply__151.0__25.0__ |
add__51.0__100.0__ multiply__151.0__25.0__ subtract__52.0__50.0__ multiply__151.0__25.0__ |
| find compound interest on rs . num__8000 at num__15.0 per annum for num__2 years num__4 months compounded annually <o> a ) num__3358 <o> b ) num__8952 <o> c ) num__3309 <o> d ) num__3109 <o> e ) num__3528 |
time = num__2 years num__4 months = num__2 ( num__0.333333333333 ) years = num__2 ( num__0.333333333333 ) years . amount = rs ' . [ num__8000 x ( num__1 + ( num__0.15 ) ) ^ num__2 x ( num__1 + ( ( num__0.333333333333 ) * num__15 ) / num__100 ) ] = rs . [ num__8000 * ( num__1.15 ) * ( num__1.15 ) * ( num__1.05 ) ] = rs . num__11109 . . : . c . i . = rs . ( num__11109 - num__8000 ) = rs . num__3109 . answer : d <eor> d <eos> |
d |
percent__15.0__1.0__ percent__100.0__3109.0__ |
percent__15.0__1.0__ percent__100.0__3109.0__ |
| on dividing a number by num__56 we get num__29 as remainder . on dividing the same number by num__8 what will be the remainder ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
explanation : formula : ( divisor * quotient ) + remainder = dividend . soln : ( num__56 * q ) + num__29 = d - - - - - - - ( num__1 ) d num__8.0 = r - - - - - - - - - - - - - ( num__2 ) from equation ( num__2 ) ( ( num__56 * q ) + num__29 ) num__8.0 = r . = > assume q = num__1 . = > ( num__56 + num__29 ) num__8.0 = r . = > num__85.0 num__8 = r = > num__5 = r . answer b <eor> b <eos> |
b |
add__56.0__29.0__ multiply__1.0__5.0__ |
add__56.0__29.0__ multiply__1.0__5.0__ |
| total dinning bill for num__5 people was $ num__211.00 . if they add num__15.0 tip and divided the bill evenly approximate . what was each persons find share <o> a ) $ num__30.14 <o> b ) num__48.53 <o> c ) num__34.66 <o> d ) num__32.29 <o> e ) num__33.16 |
num__211 * num__15 = num__31.65 = num__31.65 num__211 + num__31.65 = num__242.65 num__242.65 / num__5 = num__48.53 answer : b <eor> b <eos> |
b |
add__211.0__31.65__ divide__242.65__5.0__ divide__242.65__5.0__ |
add__211.0__31.65__ divide__242.65__5.0__ divide__242.65__5.0__ |
| a grocer stacked oranges in a pile . the bottom layer was rectangular with num__4 rows of num__8 oranges each . in the second layer from the bottom each orange rested on num__4 oranges from the bottom layer and in the third layer each orange rested on num__4 oranges from the second layer . which of the following is the maximum number of oranges that could have been in the third layer ? <o> a ) num__14 <o> b ) num__12 <o> c ) num__10 <o> d ) num__7 <o> e ) num__4 |
bottom layer = num__8 x num__4 = num__32 iind layer = ( num__8 - num__1 ) x ( num__4 - num__1 ) = num__21 iiird layer = ( num__7 - num__1 ) x ( num__3 - num__1 ) = num__12 answer = num__12 = b <eor> b <eos> |
b |
multiply__4.0__8.0__ subtract__8.0__1.0__ subtract__4.0__1.0__ add__4.0__8.0__ add__4.0__8.0__ |
multiply__4.0__8.0__ subtract__8.0__1.0__ subtract__4.0__1.0__ add__4.0__8.0__ add__4.0__8.0__ |
| a trumpet section includes four musicians . if they hold an audition to determine chair order within the section and there are to be no ties how many distinct chair orders can result ? <o> a ) num__240 <o> b ) num__125 <o> c ) num__120 <o> d ) num__100 <o> e ) num__24 |
we have a word problem using permutations . we must identify the number of total items in the group . there are four possible musicians who could be placed in the first chair three possible musicians for the second two for the third one for the fourth . to determine the number of possible orderings we multiply these numbers which we can express as num__4 ! . now num__4 ! = ( num__4 ) ( num__3 ) ( num__2 ) ( num__1 ) = num__24 = e <eor> e <eos> |
e |
coin_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
coin_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
| a room is num__6 meters num__24 centimeters in length and num__4 meters num__32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the room . <o> a ) num__017 <o> b ) num__117 <o> c ) num__127 <o> d ) num__137 <o> e ) num__147 |
let us calculate both the length and width of the room in centimeters . length = num__6 meters and num__24 centimeters = num__624 cm width = num__4 meters and num__32 centimeters = num__432 cm as we want the least number of square tiles required it means the length of each square tile should be as large as possible . further the length of each square tile should be a factor of both the length and width of the room . hence the length of each square tile will be equal to the hcf of the length and width of the room = hcf of num__624 and num__432 = num__48 thus the number of square tiles required = ( num__624 x num__432 ) / ( num__48 x num__48 ) = num__13 x num__9 = num__117 answer : b <eor> b <eos> |
b |
divide__624.0__48.0__ divide__432.0__48.0__ multiply__9.0__13.0__ multiply__9.0__13.0__ |
divide__624.0__48.0__ divide__432.0__48.0__ multiply__9.0__13.0__ multiply__9.0__13.0__ |
| in country t the unemployment rate among construction workers dropped from num__16 percent on september num__1 num__1992 to num__9 percent on september num__1 num__1996 . if the number of construction workers was num__20 percent greater on september num__1 num__1996 than on september num__1 num__1992 what was the approximate percent change in the number of unemployed construction workers over this period ? <o> a ) num__50.0 decrease <o> b ) num__30.0 decrease <o> c ) num__15.0 decrease <o> d ) num__30.0 increase <o> e ) num__55.0 increase |
country t num__1992 num__1996 no of construction workers num__100 num__120 unemployment rate num__16.0 num__9.0 unemployed workers num__16 num__11.0 change in unemployed workers = ( num__16 - num__11 ) = num__0.3125 = ~ num__33.0 decrease closest ans = num__30.0 decrease ans = b <eor> b <eos> |
b |
percent__100.0__30.0__ |
percent__100.0__30.0__ |
| it is being given that ( num__2 ^ num__32 + num__1 ) is completely divisible by a whole number . which of the following numbers is completely divisible by this number ? <o> a ) ( num__2 ^ num__16 + num__1 ) <o> b ) ( num__2 ^ num__16 - num__1 ) <o> c ) ( num__7 x num__2 ^ num__23 ) <o> d ) ( num__2 ^ num__96 + num__1 ) <o> e ) ( num__2 ^ num__98 + num__1 ) |
let num__2 ^ num__32 = x . then ( num__2 ^ num__32 + num__1 ) = ( x + num__1 ) . let ( x + num__1 ) be completely divisible by the natural number n . then ( num__2 ^ num__96 + num__1 ) = [ ( num__2 ^ num__32 ) ^ num__3 + num__1 ] = ( x ^ num__3 + num__1 ) = ( x + num__1 ) ( x ^ num__2 - x + num__1 ) which is completely divisible by n since ( x + num__1 ) is divisible by n . answer : d <eor> d <eos> |
d |
add__2.0__1.0__ multiply__2.0__1.0__ |
add__2.0__1.0__ subtract__3.0__1.0__ |
| a certain set of numbers has an average ( arithmetic mean ) of num__50 and a standard deviation of num__50.5 . if c and n two numbers in the set are both within num__2 standard deviations from the average then which of the following could be the sum of c and n ? <o> a ) - num__200 <o> b ) - num__130 <o> c ) - num__104 <o> d ) num__51 <o> e ) num__305 |
num__2 standard deviations from the average is frommean - num__2 * sdtomean + num__2 * sd thus from num__50 - num__2 * num__50.5 = num__51 to num__50 + num__2 * num__50.5 = num__151 : - num__51 < c < num__151 - num__51 < n < num__151 - num__102 < c + n < num__302 . only option d is in this range . answer : d . <eor> d <eos> |
d |
multiply__2.0__51.0__ multiply__2.0__151.0__ divide__102.0__2.0__ |
multiply__2.0__51.0__ multiply__2.0__151.0__ subtract__102.0__51.0__ |
| seven machines at a certain factory operate at the same constant rate . if six of these machines operating simultaneously take num__42 hours to fill a certain production order how many fewer hours does it take all seven machines operating simultaneously to fill the same production order ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
the total work is num__6 * num__42 = num__252 machine hours the time required for seven machines is num__36.0 = num__36 hours thus num__6 fewer hours . the answer is d . <eor> d <eos> |
d |
multiply__42.0__6.0__ subtract__42.0__6.0__ subtract__42.0__36.0__ |
multiply__42.0__6.0__ subtract__42.0__6.0__ subtract__42.0__36.0__ |
| how many cubes of num__5 cm edge can be put in a cubical box of num__1 m edge . <o> a ) num__1770 cm <o> b ) num__8000 cm <o> c ) num__7860 cm <o> d ) num__6170 cm <o> e ) num__1870 cm |
number of cubes = num__100 â ˆ — num__100 â ˆ — num__20.0 * num__5 * num__5 = num__8000 note : num__1 m = num__100 cm answer : b <eor> b <eos> |
b |
divide__100.0__5.0__ round__8000.0__ |
divide__100.0__5.0__ multiply__1.0__8000.0__ |
| x starts a business with rs . num__45000 . y joins in the business after num__7 months with rs . num__30000 . what will be the ratio in which they should share the profit at the end of the year ? <o> a ) num__1 : num__2 <o> b ) num__18 : num__5 <o> c ) num__1 : num__5 <o> d ) num__3 : num__1 <o> e ) num__1 : num__1 |
explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = num__45000 Ã — num__12 : num__30000 Ã — num__5 = num__45 Ã — num__12 : num__30 Ã — num__5 = num__3 Ã — num__12 : num__2 Ã — num__5 = num__18 : num__5 answer : option b <eor> b <eos> |
b |
subtract__12.0__7.0__ subtract__7.0__5.0__ subtract__30.0__12.0__ subtract__30.0__12.0__ |
subtract__12.0__7.0__ subtract__7.0__5.0__ subtract__30.0__12.0__ subtract__30.0__12.0__ |
| brenda caught three less than twice as many fish as olga . together they caught num__33 fish . how many fish did olga catch ? <o> a ) num__21 <o> b ) num__20 <o> c ) num__13 <o> d ) num__12 <o> e ) num__15 |
if olga = f brenda = num__2 ( f ) - num__3 brenda + olga = num__33 ( num__2 f - num__3 ) + ( f ) = num__33 num__2 f + f - num__3 + num__3 = num__33 + num__3 num__3 f = num__36 f = num__12 olga has num__12 fish the answer is d ) <eor> d <eos> |
d |
add__33.0__3.0__ divide__36.0__3.0__ divide__36.0__3.0__ |
add__33.0__3.0__ divide__36.0__3.0__ divide__36.0__3.0__ |
| a train num__150 m long running at num__72 kmph crosses a platform in num__25 sec . what is the length of the platform ? <o> a ) num__287 <o> b ) num__298 <o> c ) num__350 <o> d ) num__726 <o> e ) num__267 |
d = num__72 * num__0.277777777778 = num__25 = num__500 – num__150 = num__350 answer : c <eor> c <eos> |
c |
subtract__500.0__150.0__ round__350.0__ |
subtract__500.0__150.0__ round__350.0__ |
| a train speeds past a pole in num__11 seconds and a platform num__120 m long in num__22 seconds . its length is ? <o> a ) num__128 <o> b ) num__177 <o> c ) num__199 <o> d ) num__120 <o> e ) num__150 |
let the length of the train be x meters and its speed be y m / sec . they x / y = num__11 = > y = x / num__11 x + num__5.45454545455 = x / num__11 x = num__120 m . answer : d <eor> d <eos> |
d |
divide__120.0__22.0__ round__120.0__ |
divide__120.0__22.0__ round__120.0__ |
| pavan travelled for num__15 hours . he covered the first half of the distance at num__30 kmph and remaining half of the distance at num__25 kmph . find the distance travelled by pavan . <o> a ) num__408 km <o> b ) num__409 km <o> c ) num__410 km <o> d ) num__412 km <o> e ) num__419 km |
let the distance travelled be x km . total time = ( x / num__2 ) / num__30 + ( x / num__2 ) / num__25 = num__15 = > x / num__60 + x / num__50 = num__15 = > ( num__5 x + num__6 x ) / num__300 = num__15 = > x = num__409 km answer : b <eor> b <eos> |
b |
divide__30.0__15.0__ hour_to_min_conversion__ multiply__25.0__2.0__ subtract__30.0__25.0__ divide__30.0__5.0__ multiply__5.0__60.0__ round__409.0__ |
divide__30.0__15.0__ hour_to_min_conversion__ multiply__25.0__2.0__ subtract__30.0__25.0__ divide__30.0__5.0__ multiply__5.0__60.0__ round__409.0__ |
| sonika deposited rs . num__8000 which amounted to rs . num__9200 after num__3 years at simple interest . had the interest been num__2.0 more . she would get how much ? <o> a ) num__9680 <o> b ) num__2277 <o> c ) num__2999 <o> d ) num__2774 <o> e ) num__1212 |
( num__8000 * num__3 * num__2 ) / num__100 = num__480 num__9200 - - - - - - - - num__9680 answer : a <eor> a <eos> |
a |
percent__100.0__9680.0__ |
percent__100.0__9680.0__ |
| a contractor hires num__100 men to finish a job in num__50 days . after num__24 days n men leave . after some more days the contractor hires num__2 n more men to complete the job on time . for how many days did these num__2 n men work ? <o> a ) num__20 <o> b ) num__18 <o> c ) num__16 <o> d ) num__13 <o> e ) num__8 |
step num__1 : find each day what % of work is finished ; it requires num__50 days to complete which means each day num__0.02 = num__2.0 work will be completed . step num__2 : after num__24 days = num__24 * num__2 = num__48.0 of work is finished . remaining is num__52.0 ( which means if same num__100 men were about to complete the job they would have required num__26 more days ) step num__3 : let us assume all people have left job . therefore contractor hires num__2 n people . initially it was taking num__26 days to compete for n people . at present num__2 n people are present to finish job . hence num__13 more days . d <eor> d <eos> |
d |
divide__2.0__100.0__ subtract__50.0__2.0__ subtract__100.0__48.0__ subtract__50.0__24.0__ add__2.0__1.0__ divide__26.0__2.0__ round__13.0__ |
divide__2.0__100.0__ multiply__24.0__2.0__ subtract__100.0__48.0__ subtract__50.0__24.0__ add__2.0__1.0__ divide__26.0__2.0__ round__13.0__ |
| three people with different speeds of num__4.5 kmph num__0.5 kmph and num__8 kmph move around a circular track of num__11 km . at what time they meet each other ? <o> a ) num__21 hours <o> b ) num__22 hours <o> c ) num__20 hours <o> d ) num__23 hours <o> e ) num__24 hours |
time taken by num__1 st person to cover the circular track = num__11 / num__4.5 = num__2.44444444444 time taken by num__2 nd person to cover the circular track = num__11 / num__0.5 = num__22 time taken by num__3 rd person to cover the circular track = num__1.375 they meet for the num__1 st time after time t = lcm ( num__2.44444444444 num__276.375 ) = num__22 hours answer : b <eor> b <eos> |
b |
divide__11.0__4.5__ divide__1.0__0.5__ divide__11.0__0.5__ subtract__11.0__8.0__ divide__11.0__8.0__ round__22.0__ |
divide__11.0__4.5__ divide__1.0__0.5__ divide__11.0__0.5__ subtract__11.0__8.0__ divide__11.0__8.0__ divide__11.0__0.5__ |
| what is the smallest number which when diminished by num__16 is divisible num__4 num__6 num__8 and num__10 ? <o> a ) num__136 <o> b ) num__192 <o> c ) num__198 <o> d ) num__122 <o> e ) num__142 |
required number = ( lcm of num__4 num__68 and num__10 ) + num__16 = num__120 + num__16 = num__136 option a <eor> a <eos> |
a |
add__16.0__120.0__ add__16.0__120.0__ |
add__16.0__120.0__ add__16.0__120.0__ |
| to save money arkadelphia cream cheese will reduce each dimension of its rectangular box container ( which is entirely full of cream cheese ) by num__60.0 and reduce the price it charges its consumers by num__60.0 as well . by what percentage does this increase the price - per - cubic - inch that each consumer will pay for cream cheese ? <o> a ) num__1 . no change <o> b ) num__2 . num__50.0 <o> c ) num__3 . num__525.0 <o> d ) num__4 . num__300.0 <o> e ) num__5 . num__400 % |
take smart numbers let l = num__20 : b = num__10 : h = num__10 of initial box and price = num__100 $ therefore price / cubic inch = num__100 / ( num__20 * num__10 * num__10 ) = num__0.05 now when dimensions are reduced by num__60.0 and price also reduced by num__60.0 l = num__8 ; b = num__4 ; h = num__4 and price = num__40 $ therefore price / cubic inch = num__40 / ( num__8 * num__4 * num__4 ) = num__0.3125 percentage change = ( num__0.3125 - num__0.05 ) * num__100 / num__0.05 = num__525.0 answer is c <eor> c <eos> |
c |
reverse__20.0__ subtract__60.0__20.0__ multiply__60.0__0.05__ |
reverse__20.0__ subtract__60.0__20.0__ multiply__60.0__0.05__ |
| a computer factory produces num__5376 computers per month at a constant rate how many computers are built every num__30 minutes assuming that there are num__28 days in one month ? <o> a ) num__2.25 . <o> b ) num__3.125 . <o> c ) num__4.5 . <o> d ) num__4.00 . <o> e ) num__6.25 . |
number of hours in num__28 days = num__28 * num__24 number of num__30 mins in num__28 days = num__28 * num__24 * num__2 number of computers built every num__30 mins = num__5376 / ( num__28 * num__24 * num__2 ) = num__4.00 answer d <eor> d <eos> |
d |
subtract__30.0__28.0__ subtract__28.0__24.0__ round__4.0__ |
subtract__30.0__28.0__ subtract__28.0__24.0__ round__4.0__ |
| a mixture of num__150 liters of wine and water contains num__20.0 water . how much more water should be added so that water becomes num__25.0 of the new mixture ? <o> a ) num__65 liters <o> b ) num__44 <o> c ) num__10 liters water <o> d ) num__45 liters <o> e ) num__8 liters |
number of liters of water in num__150 liters of the mixture = num__20.0 of num__150 = num__0.2 * num__150 = num__30 liters . p liters of water added to the mixture to make water num__25.0 of the new mixture . total amount of water becomes ( num__30 + p ) and total volume of mixture is ( num__150 + p ) . ( num__30 + p ) = num__0.25 * ( num__150 + p ) num__120 + num__4 p = num__150 + p = > p = num__10 liters . answer : c <eor> c <eos> |
c |
multiply__150.0__0.2__ subtract__150.0__30.0__ reverse__0.25__ subtract__30.0__20.0__ subtract__20.0__10.0__ |
multiply__150.0__0.2__ subtract__150.0__30.0__ multiply__20.0__0.2__ subtract__30.0__20.0__ subtract__20.0__10.0__ |
| if x > num__7 which of the following is equal to ( x ^ num__2 + num__10 x + num__25 ) / ( x ^ num__2 - num__25 ) ? <o> a ) ( x + num__5 ) / ( x - num__5 ) <o> b ) ( x - num__4 ) / ( x + num__4 ) <o> c ) ( x - num__2 ) / ( x + num__4 ) <o> d ) ( x + num__4 ) / ( x - num__4 ) <o> e ) ( x - num__8 ) / ( x - num__4 ) |
( x ^ num__2 + num__10 x + num__25 ) / ( x ^ num__2 - num__25 ) = ( x + num__5 ) ( x + num__5 ) / ( x + num__5 ) ( x - num__5 ) = ( x + num__5 ) / ( x - num__5 ) a . ( x + num__5 ) / ( x - num__5 ) <eor> a <eos> |
a |
subtract__7.0__2.0__ subtract__7.0__2.0__ |
subtract__7.0__2.0__ subtract__7.0__2.0__ |
| for any positive integer n n > num__1 thelengthof n is the number of positive primes ( not necessary distinct ) whose product is n . for ex the length of num__40 is num__3 since num__50 = num__2 x num__5 x num__5 . what is the greatest possible length of a positive integer less than num__1000 . <o> a ) num__10 <o> b ) num__9 <o> c ) num__8 <o> d ) num__7 <o> e ) num__6 |
you are missing something in your post : for any positive integer n n > num__1 thelengthof n is the number of positive primes ( not distinct ) whose product is n . for example the length of num__50 is num__3 since num__50 = ( num__2 ) ( num__5 ) ( num__5 ) the lenght of num__1000 = ( num__2 ) ( num__5 ) ( num__2 ) ( num__5 ) ( num__2 ) ( num__5 ) = num__6 but we need n < num__1000 using num__2 as the base = ( num__2 ) ( num__2 ) ( num__2 ) ( num__2 ) ( num__2 ) ( num__2 ) ( num__2 ) ( num__2 ) ( num__2 ) = num__6 the length of num__512 . the answer is ( e ) <eor> e <eos> |
e |
add__1.0__5.0__ add__1.0__5.0__ |
add__1.0__5.0__ add__1.0__5.0__ |
| a certain sum of money at simple interest amounted rs . num__900 in num__9 years at num__5.0 per annum find the sum ? <o> a ) num__337 <o> b ) num__268 <o> c ) num__198 <o> d ) num__620 <o> e ) num__168 |
num__900 = p [ num__1 + ( num__9 * num__5 ) / num__100 ] p = num__620 ' answer : d <eor> d <eos> |
d |
percent__100.0__620.0__ |
percent__100.0__620.0__ |
| in an election candidate douglas won num__66 percent of the total vote in counties x and y . he won num__74 percent of the vote in county x . if the ratio of people who voted in county x to county y is num__2 : num__1 what percent of the vote did candidate douglas win in county y ? <o> a ) num__25.0 <o> b ) num__30.0 <o> c ) num__50.0 <o> d ) num__75.0 <o> e ) num__80 % |
given voters in ratio num__2 : num__1 let x has num__200 votersy has num__100 voters for x num__66.0 voted means num__74 * num__200 = num__148 votes combined for xy has num__300 voters and voted num__66.0 so total votes = num__198 balance votes = num__198 - num__148 = num__50 as y has num__100 voters so num__50 votes means num__50.0 of votes required ans c <eor> c <eos> |
c |
divide__200.0__2.0__ multiply__74.0__2.0__ add__100.0__200.0__ subtract__200.0__2.0__ divide__100.0__2.0__ multiply__1.0__50.0__ |
divide__200.0__2.0__ multiply__74.0__2.0__ add__100.0__200.0__ subtract__200.0__2.0__ subtract__198.0__148.0__ multiply__1.0__50.0__ |
| how many seconds will a num__620 metre long train take to cross a man running with a speed of num__8 km / hr in the direction of the moving train if the speed of the train is num__80 km / hr ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__31 <o> d ) num__45 <o> e ) num__60 |
explanation : speed of train relatively to man = ( num__80 - num__8 ) km / hr = num__72 km / hr = ( num__72 x num__0.277777777778 ) m / sec = num__20 m / sec time taken to pass the man = ( num__31.0 ) sec = num__31 sec . answer : c <eor> c <eos> |
c |
subtract__80.0__8.0__ divide__620.0__20.0__ round__31.0__ |
subtract__80.0__8.0__ divide__620.0__20.0__ divide__620.0__20.0__ |
| if a bicyclist in motion increases his speed by num__30 percent and then increases this speed by num__10 percent what percent of the original speed is the total increase in speed ? <o> a ) num__10.0 <o> b ) num__40.0 <o> c ) num__43.0 <o> d ) num__64.0 <o> e ) num__140 % |
let the sped be num__100 an increase of num__30.0 the speed now is num__130 a further increase of num__10.0 on num__130 = num__13 total increase = num__43 on num__100 = num__43.0 c <eor> c <eos> |
c |
add__30.0__100.0__ divide__130.0__10.0__ add__30.0__13.0__ add__30.0__13.0__ |
add__30.0__100.0__ divide__130.0__10.0__ add__30.0__13.0__ add__30.0__13.0__ |
| a train num__280 m long passed a pole in num__28 sec . how long will it take to pass a platform num__650 m long ? <o> a ) num__93 sec <o> b ) num__89 sec <o> c ) num__54 sec <o> d ) num__27 sec <o> e ) num__22 sec |
speed = num__10.0 = num__10 m / sec . required time = ( num__280 + num__650 ) / num__10 = num__93 sec . answer : a <eor> a <eos> |
a |
divide__280.0__28.0__ round__93.0__ |
divide__280.0__28.0__ round__93.0__ |
| what is the remainder when you divide num__2 ^ num__200 - num__3 by num__7 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
answer is a . this is very similar to the num__32 ^ num__32 ^ num__32 - num__3 divided by num__7 question . <eor> a <eos> |
a |
subtract__3.0__2.0__ |
subtract__3.0__2.0__ |
| a particular library has num__75 books in a special collection all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if by the end of the month num__60 percent of books that were loaned out are returned and there are num__65 books in the special collection at that time how many books of the special collection were loaned out during that month ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__35 <o> d ) num__40 <o> e ) num__55 |
total = num__75 books . num__60.0 of books that were loaned out are returned - - > num__100.0 - num__60.0 = num__40.0 of books that were loaned out are not returned . now there are num__68 books thus num__75 - num__65 = num__10 books are not returned . { loaned out } * num__0.4 = num__10 - - > { loaned out } = num__25 . answer : b . <eor> b <eos> |
b |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| if the sum of the first n positive integers is s what is the sum of the first n positive integers divisible by num__4 in terms of s ? <o> a ) s / num__2 <o> b ) s <o> c ) num__2 s <o> d ) num__2 s + num__2 <o> e ) num__4 s |
answer is e given sum of num__1 num__2 num__3 . . . . . n = s sum of num__4 num__8 num__12 . . . . num__2 n = num__4 [ sum ( num__1 num__2 num__3 . . . . n ) ] = num__4 s <eor> e <eos> |
e |
subtract__4.0__1.0__ multiply__4.0__2.0__ multiply__4.0__3.0__ multiply__4.0__1.0__ |
subtract__4.0__1.0__ multiply__4.0__2.0__ multiply__4.0__3.0__ multiply__4.0__1.0__ |
| in what proportion water must be added to spirit to gain num__20.0 by selling it at the cost price ? <o> a ) num__1 : num__9 <o> b ) num__1 : num__1 <o> c ) num__1 : num__5 <o> d ) num__1 : num__3 <o> e ) num__1 : num__4 |
let the c . p of spirit be = rs . num__10 per litre s . p of the mixture = rs . num__10 per litre profit = num__20 c . p of the mixture = rs . = rs . per litre ratio of water and spirit = num__1 : num__5 answer : c <eor> c <eos> |
c |
percent__20.0__5.0__ |
percent__20.0__5.0__ |
| num__12 men working num__8 hours per day complete a piece of work in num__10 days . to complete the same work in num__8 days working num__15 hours a day the number of men required is <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__7 |
solution let the required number of men be x . less days more men ( indirect propertion ) more working hrs per days less men ( indirect propertion ) ∴ num__8 × num__15 × x = num__10 × num__8 × num__12 ⇔ x = num__10 x num__8 x num__1.5 x num__15 = x = num__8 . answer d <eor> d <eos> |
d |
divide__12.0__8.0__ round__8.0__ |
divide__12.0__8.0__ round__8.0__ |
| num__40.0 of ram ' s marks is equal to num__20.0 of rahim ' s marks which percent is equal to num__30.0 of robert ' s marks . if robert ' s marks is num__80 then find the average marks of ram and rahim ? <o> a ) num__87 <o> b ) num__87 <o> c ) num__87 <o> d ) num__90 <o> e ) num__77 |
given num__40.0 of ram ' s marks = num__20.0 of rahim ' s marks = num__30.0 of robert ' s marks . given marks of robert = num__80 num__30.0 of num__80 = num__0.3 * num__8 = num__24 given num__40.0 of ram ' s marks = num__24 . = > ram ' s marks = ( num__24 * num__100 ) / num__40 = num__60 also num__20.0 of rahim ' s marks = num__24 = > rahim ' s marks = ( num__24 * num__100 ) / num__20 = num__120 average marks of ram and rahim = ( num__60 + num__120 ) / num__2 = num__90 . answer : d <eor> d <eos> |
d |
multiply__80.0__0.3__ add__20.0__80.0__ add__40.0__20.0__ add__40.0__80.0__ divide__40.0__20.0__ add__30.0__60.0__ add__30.0__60.0__ |
multiply__80.0__0.3__ add__20.0__80.0__ add__40.0__20.0__ add__40.0__80.0__ divide__40.0__20.0__ add__30.0__60.0__ add__30.0__60.0__ |
| a space shuttle orbits the earth at about num__2 kilometers per second . this speed is equal to how many kilometers per hour ? <o> a ) num__7200 <o> b ) num__8880 <o> c ) num__10800 <o> d ) num__28800 <o> e ) num__48 |
000 |
seconds in num__1 hours : num__60 s in num__1 min num__60 min in num__1 hr num__60 * num__60 = num__3600 sec in num__1 hr num__2 * num__3600 = num__28800 answer : a <eor> a <eos> |
a |
a |
| which of the following is closer to ( num__23 ! + num__22 ! ) / ( num__23 ! - num__22 ! ) ? <o> a ) num__49 <o> b ) num__36 <o> c ) num__25 <o> d ) num__16 <o> e ) num__1 |
notice that num__23 ! = ( num__23 ) ( num__22 ! ) so we can factor out num__22 ! from top and bottom . ( num__23 ! + num__22 ! ) / ( num__23 ! - num__22 ! ) = [ num__22 ! ( num__23 + num__1 ) ] / [ num__22 ! ( num__23 - num__1 ) ] = ( num__23 + num__1 ) / ( num__22 - num__1 ) = num__1.09090909091 ≈ num__1 answer : e <eor> e <eos> |
e |
subtract__23.0__22.0__ reverse__1.0__ |
subtract__23.0__22.0__ reverse__1.0__ |
| num__36 * num__36 * num__36 = num__6 ^ ? <o> a ) num__3 <o> b ) num__2 <o> c ) num__6 <o> d ) num__1 <o> e ) num__5 |
num__6 ^ num__2 * num__6 ^ num__2 * num__6 ^ num__2 = num__6 ^ ( num__2 + num__2 + num__2 ) = num__6 ^ num__6 answer : num__6 answer : c <eor> c <eos> |
c |
divide__36.0__6.0__ |
divide__36.0__6.0__ |
| a train num__480 m in length crosses a telegraph post in num__16 seconds . the speed of the train is ? <o> a ) num__22 kmph <o> b ) num__108 kmph <o> c ) num__54 kmph <o> d ) num__71 kmph <o> e ) num__88 kmph |
s = num__30.0 * num__3.6 = num__108 kmph answer : b <eor> b <eos> |
b |
divide__480.0__16.0__ multiply__3.6__30.0__ round__108.0__ |
divide__480.0__16.0__ multiply__3.6__30.0__ multiply__3.6__30.0__ |
| a certain hall contains two cuckoo clocks . if the first clock chirps num__20 times per hour and the second clock chirps num__8 times per hour and both clocks chirp for the first time at num__2 : num__03 pm at what time will the first clock have chirped three times as many times as the second clock can chirp per hour ? <o> a ) num__2 : num__40 pm <o> b ) num__2 : num__55 pm <o> c ) num__3 : num__12 pm <o> d ) num__3 : num__24 pm <o> e ) num__3 : num__45 pm |
the question is asking when the first clock will have chirped num__24 times ( num__3 * num__8 chirps ) . the first clock chirps once every num__3 minutes . if the first chirp is at num__2 : num__03 then the num__24 th chirp is at num__72 minutes after num__2 : num__00 . the answer is c . <eor> c <eos> |
c |
multiply__8.0__3.0__ multiply__3.0__24.0__ round__3.0__ |
multiply__8.0__3.0__ multiply__3.0__24.0__ round__3.0__ |
| the g . c . d . of num__1.08 num__0.36 and num__0.3 is ? <o> a ) num__0.06 <o> b ) num__0.9 <o> c ) num__0.18 <o> d ) num__0.108 <o> e ) none of these |
given numbers are num__1.08 num__0.36 and num__0.30 . h . c . f of num__108 num__36 and num__30 is num__6 h . c . f of given numbers = num__0.06 . correct options : a <eor> a <eos> |
a |
gcd__36.0__30.0__ subtract__0.36__0.3__ subtract__0.36__0.3__ |
gcd__36.0__30.0__ subtract__0.36__0.3__ subtract__0.36__0.3__ |
| if n is an integer and n ^ num__4 is divisible by num__8 which of the following could be the remainder when n is divided by num__8 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__5 <o> e ) num__6 |
option num__1 remainder num__1 means num__8 + num__1 = num__9 n ^ num__4 = num__9 ^ num__4 = num__9 * num__9 * num__9 * num__1.125 is not an integer option num__2 remainder num__2 means num__8 + num__2 = num__10 n ^ num__4 = num__10 * num__10 * num__10 * num__1.25 is an integer means n ^ num__4 is divisible by num__8 if we take n = num__2 so b is correct <eor> b <eos> |
b |
add__8.0__1.0__ divide__9.0__8.0__ divide__8.0__4.0__ add__8.0__2.0__ divide__10.0__8.0__ subtract__4.0__2.0__ |
add__8.0__1.0__ divide__9.0__8.0__ divide__8.0__4.0__ add__8.0__2.0__ divide__10.0__8.0__ multiply__1.0__2.0__ |
| in a num__200 m race if a gives b a start of num__25 metres then a wins the race by num__10 seconds . alternatively if a gives b a start of num__45 metres the race ends in a dead heat . how long does a take to run num__200 m ? <o> a ) num__100 seconds <o> b ) num__112.5 seconds <o> c ) num__77.5 seconds <o> d ) num__87.5 seconds <o> e ) none |
explanatory answer a gives b a start of num__25 metres and still wins the race by num__10 seconds . alternatively if a gives b a start of num__45 metres then the race ends in a dead heat . therefore the additional num__20 metres start given to b compensates for the num__10 seconds . i . e . b runs num__20 metres in num__10 seconds . hence b will take num__100 seconds to run num__200 metres . we know that a gives b a start of num__45 metres . b will take num__22.5 seconds to run this num__45 metres as b runs num__20 metres in num__10 seconds or at the speed of num__2 m / s . hence a will take num__22.5 seconds lesser than b or num__100 - num__22.5 = num__77.5 seconds to complete the race . answer c <eor> c <eos> |
c |
divide__200.0__10.0__ divide__200.0__100.0__ subtract__100.0__22.5__ round__77.5__ |
divide__200.0__10.0__ divide__200.0__100.0__ subtract__100.0__22.5__ subtract__100.0__22.5__ |
| a bus can hold num__40 passengers . if there are num__10 rows of seats on the bus how many seats are in each row ? <o> a ) num__0 <o> b ) num__2 <o> c ) num__4 <o> d ) num__6 <o> e ) num__5 |
no . of seats = num__4.0 = num__4 there are num__4 seats in each row answer : c <eor> c <eos> |
c |
divide__40.0__10.0__ divide__40.0__10.0__ |
divide__40.0__10.0__ divide__40.0__10.0__ |
| the first three terms of a proportion are num__2 num__8 and num__16 . the fourth term is ? <o> a ) num__22 <o> b ) num__64 <o> c ) num__67 <o> d ) num__60 <o> e ) num__88 |
( num__8 * num__16 ) / num__2 = num__64 answer : b <eor> b <eos> |
b |
power__8.0__2.0__ power__8.0__2.0__ |
power__8.0__2.0__ power__8.0__2.0__ |
| if ( t - num__8 ) is a factor of t ^ num__2 - kt - num__47 then k = <o> a ) num__16 <o> b ) num__12 <o> c ) num__2 <o> d ) num__6 <o> e ) num__14 |
t ^ num__2 - kt - num__48 = ( t - num__8 ) ( t + m ) where m is any positive integer . if num__6.0 = num__6 then we know as a matter of fact that : m = + num__6 and thus k = num__8 - num__6 = num__6 t ^ num__2 - kt - m = ( t - a ) ( t + m ) where a > m t ^ num__2 + kt - m = ( t - a ) ( t + m ) where a < m t ^ num__2 - kt + m = ( t - a ) ( t - m ) t ^ num__2 + kt + m = ( t + a ) ( t + m ) d <eor> d <eos> |
d |
subtract__8.0__2.0__ subtract__8.0__2.0__ |
subtract__8.0__2.0__ subtract__8.0__2.0__ |
| there are num__3 red chips and num__2 blue ones . when arranged in a row they form a certain color pattern for example rbrrb . how many color patterns ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__24 <o> d ) num__60 <o> e ) num__100 |
there are num__3 red chips and num__2 blue ones . when arranged in a row they form a certain color pattern for example rbrrb . how many color patterns ? num__10 num__12 num__24 num__60 num__100 soln : total number of patterns is num__5 ! since num__3 red chips are identical and num__2 blue ones are identical thus we have = num__5 ! / ( num__2 ! * num__3 ! ) = num__10 such different patterns a <eor> a <eos> |
a |
add__2.0__10.0__ multiply__2.0__12.0__ add__3.0__2.0__ multiply__2.0__5.0__ |
add__2.0__10.0__ multiply__2.0__12.0__ divide__10.0__2.0__ multiply__2.0__5.0__ |
| what annual payment dischargea debit of rs . num__12900 due in num__4 yrs . at num__5.0 rate ? <o> a ) num__2000 <o> b ) num__2200 <o> c ) num__3000 <o> d ) num__3400 <o> e ) num__3500 |
a . p . = ( num__200 x num__12900 ) / [ num__4 ( num__200 + num__5 x num__3 ) ] = num__3000 ans alternative num__100 + num__105 + num__110 + num__115 = num__12900 num__430 = num__12900 num__100 = num__12900 / ( num__430 ) x num__100 = num__3000 c <eor> c <eos> |
c |
percent__100.0__3000.0__ |
percent__100.0__3000.0__ |
| a convoy of num__7 trucks each num__22 meters in length pass through a construction zone num__0.5 km long . through the construction zone they travel num__3 m / s maintaining a distance of five meters between them . how long does it take for the entire convoy to make it through the zone ? <o> a ) num__3 min num__22 sec <o> b ) num__3 min num__48 sec <o> c ) num__4 min num__12 sec <o> d ) num__4 min num__20 sec <o> e ) num__4 min num__36 sec |
b num__3 min num__48 sec there are num__7 trucks of num__22 m with six num__5 - m gaps between them passing a distance of num__500 m . d = num__7 * num__22 m + num__6 * num__5 m + num__500 m = num__684 m t = num__684 m / num__3 m / s = num__228 s = num__3 min num__48 sec <eor> b <eos> |
b |
divide__3.0__0.5__ divide__684.0__3.0__ round__3.0__ |
divide__3.0__0.5__ divide__684.0__3.0__ multiply__0.5__6.0__ |
| a pharmaceutical company received $ num__4 million in royalties on the first $ num__22 million in sales of generic equivalent of one of its products and then $ num__8 million in royalties on the next $ num__88 million in sales . by approximately what percent did the ratio of royalties to sales decrease from the first $ num__22 million in sales to the next $ num__92 million in sales ? <o> a ) num__20.0 <o> b ) num__25.0 <o> c ) num__30.0 <o> d ) num__45.0 <o> e ) num__50 % |
change in ratio of royalties to sales = num__0.181818181818 - num__0.0909090909091 = num__0.0909090909091 % decrease = ( num__0.0909090909091 ) / ( num__0.181818181818 ) * num__100 = num__50.0 ( approx ) answer : e ) <eor> e <eos> |
e |
divide__4.0__22.0__ divide__8.0__88.0__ add__8.0__92.0__ subtract__100.0__50.0__ |
divide__4.0__22.0__ divide__8.0__88.0__ add__8.0__92.0__ subtract__100.0__50.0__ |
| square rstu shown above is rotated in a plane about its center in a clockwise direction the minimum number of degrees necessary for u to be in the position where r is now shown . the number of degrees through which rstu is rotated is <o> a ) num__135 degree <o> b ) num__180 degree <o> c ) num__90 degree <o> d ) num__270 degree <o> e ) num__315 degree |
from the options i am assuming the positioning of r and u relative to each other to be as shown . to replace r by u focus on ou . say you rotate ou clockwise ( and with it the entire square ) and bring it in place of or . how many degrees did you go ? you covered num__1 right angles i . e . num__90 degrees . answer : c <eor> c <eos> |
c |
right_angle__ right_angle__ |
right_angle__ right_angle__ |
| if num__2 tables and num__3 chairs cost rs num__3500 and num__3 tables and num__2 chairs cost rs . num__4000 then how much does a table cost ? <o> a ) num__500 <o> b ) num__1000 <o> c ) num__1500 <o> d ) num__2000 <o> e ) none of these |
explanation : let the cost of a table and that of a chair be rs . x and rs y respectively . then num__2 x + num__3 y = num__3500 . . . ( i ) and num__3 x + num__2 y = num__4000 . . . . . ( ii ) solving ( i ) and ( ii ) we get x = num__1000 y = num__500 answer : b <eor> b <eos> |
b |
subtract__4000.0__3500.0__ multiply__2.0__500.0__ |
subtract__4000.0__3500.0__ multiply__2.0__500.0__ |
| a present value of a machine is $ num__400 . its value depletiation rate is num__25.0 per annum then find the machine value after num__2 years ? <o> a ) $ num__125 <o> b ) $ num__100 <o> c ) $ num__200 <o> d ) $ num__215 <o> e ) $ num__225 |
p = $ num__400 r = num__25.0 t = num__2 years machine value after num__2 years = p [ ( num__1 - r / num__100 ) ^ t ] = num__400 * num__0.75 * num__0.75 = $ num__225 approximately answer is e <eor> e <eos> |
e |
percent__25.0__400.0__ percent__100.0__225.0__ |
percent__25.0__400.0__ percent__100.0__225.0__ |
| in a competitive examination in state a num__6.0 candidates got selected from the total appeared candidates . state b had an equal number of candidates appeared and num__7.0 candidates got selected with num__84 more candidates got selected than a . what was the number of candidates appeared from each state ? <o> a ) num__7000 <o> b ) num__8400 <o> c ) num__6000 <o> d ) num__5000 <o> e ) num__4000 |
state a and state b had an equal number of candidates appeared . in state a num__6.0 candidates got selected from the total appeared candidates in state b num__7.0 candidates got selected from the total appeared candidates but in state b num__84 more candidates got selected than state a from these it is clear that num__1.0 of the total appeared candidates in state b = num__84 = > total appeared candidates in state b = num__84 x num__100 = num__8400 = > total appeared candidates in state a = total appeared candidates in state b = num__8400 <eor> b <eos> |
b |
subtract__7.0__6.0__ multiply__84.0__100.0__ multiply__84.0__100.0__ |
subtract__7.0__6.0__ multiply__84.0__100.0__ multiply__84.0__100.0__ |
| a clock is set right at num__5 a . m . the clock loses num__16 minutes in num__24 hours . what will be the true time when the clock indicates num__10 p . m . on num__4 th day ? <o> a ) num__11 <o> b ) num__27 <o> c ) num__62 <o> d ) num__62 <o> e ) num__82 |
time from num__5 am . on a day to num__10 pm . on num__4 th day = num__89 hours . now num__23 hrs num__44 min . of this clock = num__24 hours of correct clock . num__23.7333333333 hrs of this clock = num__24 hours of correct clock num__89 hrs of this clock = ( num__24 x num__31556 x num__89 ) hrs of correct clock . = num__90 hrs of correct clock . so the correct time is num__11 p . m . answer : a <eor> a <eos> |
a |
subtract__16.0__5.0__ round__11.0__ |
subtract__16.0__5.0__ round__11.0__ |
| the roof of an apartment building is rectangular and its length is num__4 times longer than its width . if the area of the roof is num__900 feet squared what is the difference between the length and the width of the roof ? <o> a ) num__38 . <o> b ) num__40 . <o> c ) num__42 . <o> d ) num__45 . <o> e ) num__46 . |
answer is d : num__45 let w be the width so length is num__4 w . therefore : w * num__4 w = num__900 solving for w = num__15 so num__4 w - w = num__3 w = num__3 * num__15 = num__45 <eor> d <eos> |
d |
multiply__3.0__15.0__ |
multiply__3.0__15.0__ |
| ( num__786 × num__74 ) ÷ ? = num__1938.8 <o> a ) a ) num__4.8 <o> b ) b ) num__48 <o> c ) c ) num__30 <o> d ) d ) num__68 <o> e ) e ) num__48 |
explanation : num__58164 / x = num__1938.8 = > x = num__58164 / num__1938.8 = num__30 answer : option c <eor> c <eos> |
c |
multiply__786.0__74.0__ divide__58164.0__1938.8__ divide__58164.0__1938.8__ |
multiply__786.0__74.0__ divide__58164.0__1938.8__ divide__58164.0__1938.8__ |
| find the value of y from ( num__12 ) ^ num__3 x num__6 ^ num__4 ÷ num__432 = y ? <o> a ) num__4363 <o> b ) num__4765 <o> c ) num__4879 <o> d ) num__6789 <o> e ) num__5184 |
given exp . = ( num__12 ) num__3 x num__64 = ( num__12 ) num__3 x num__64 = ( num__12 ) num__2 x num__62 = ( num__72 ) num__2 = num__5184 num__432 num__12 x num__62 e <eor> e <eos> |
e |
divide__12.0__6.0__ subtract__64.0__2.0__ multiply__12.0__6.0__ multiply__12.0__432.0__ multiply__12.0__432.0__ |
divide__12.0__6.0__ subtract__64.0__2.0__ multiply__12.0__6.0__ multiply__12.0__432.0__ multiply__12.0__432.0__ |
| a num__600 meter long train crosses a signal post in num__60 seconds . how long will it take to cross a num__3 kilometer long bridge at the same speed ? <o> a ) num__4 min <o> b ) num__2 min <o> c ) num__6 min <o> d ) num__9 min <o> e ) num__5 min |
s = num__10.0 = num__10 mps s = num__360.0 = num__360 sec = num__6 min answer : c <eor> c <eos> |
c |
divide__600.0__60.0__ divide__60.0__10.0__ round__6.0__ |
divide__600.0__60.0__ divide__60.0__10.0__ round__6.0__ |
| it takes num__4 tons of copper ore and num__6 tons of iron ore to make one ton of alloy a . how many tons of alloy a can be made from num__60 tons of copper ore and num__90 tons of iron ore ? <o> a ) num__12 <o> b ) num__15 <o> c ) num__20 <o> d ) num__30 <o> e ) num__50 |
yes you need copper ore : iron ore in the ratio num__4 : num__6 . total num__10 tons of the mix in this ratio will give num__1 ton of alloy a . if you have num__60 tons of copper ore it is enough for num__15.0 = num__15 tons of alloy a . if you have num__90 tons of iron ore it is enough for num__15.0 = num__15 tons of alloy a . since iron ore is available for only num__15 tons of alloy a you can make only num__15 tons of alloy a . the leftover copper ore alone can not make any alloy a and hence will be leftover only . answer must be num__15 . ( b ) <eor> b <eos> |
b |
add__4.0__6.0__ divide__60.0__4.0__ divide__60.0__4.0__ |
add__4.0__6.0__ divide__60.0__4.0__ divide__60.0__4.0__ |
| the area of one square is x ^ num__2 + num__6 x + num__9 and the area of another square is num__4 x ^ num__2 − num__28 x + num__49 . if the sum of the perimeters of both squares is num__56 what is the value of x ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
the areas are ( x + num__3 ) ^ num__2 and ( num__2 x - num__7 ) ^ num__2 . the lengths of the sides are x + num__3 and num__2 x - num__7 . if we add the two perimeters : num__4 ( x + num__3 ) + num__4 ( num__2 x - num__7 ) = num__56 num__12 x = num__72 x = num__6 the answer is d . <eor> d <eos> |
d |
divide__6.0__2.0__ subtract__9.0__2.0__ multiply__2.0__6.0__ multiply__6.0__12.0__ add__2.0__4.0__ |
subtract__9.0__6.0__ subtract__9.0__2.0__ add__9.0__3.0__ multiply__6.0__12.0__ add__2.0__4.0__ |
| antonio works in a bakery . he made cookies that cost $ num__3 and made $ num__300 . how many customer did he have ? <o> a ) num__100 customers <o> b ) num__185 customers <o> c ) num__250 customers <o> d ) num__200 customers <o> e ) num__270 customers |
a cookie costs $ num__3 adding another one is $ num__6 . num__300 divided by num__6 is num__50 x num__2 is num__100 . he had num__100 customers . the correct answer is a . <eor> a <eos> |
a |
divide__300.0__6.0__ divide__6.0__3.0__ divide__300.0__3.0__ round__100.0__ |
divide__300.0__6.0__ divide__6.0__3.0__ divide__300.0__3.0__ round__100.0__ |
| a goods train runs at the speed of num__72 kmph and crosses a num__250 m long platform in num__26 seconds . what is the length of the goods train ? <o> a ) num__230 m <o> b ) num__240 m <o> c ) num__260 m <o> d ) num__270 m <o> e ) num__250 m |
speed = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . then ( x + num__250 ) / num__26 = num__20 x + num__250 = num__520 x = num__520 - num__250 then x = num__270 answer : d <eor> d <eos> |
d |
multiply__26.0__20.0__ add__250.0__20.0__ round__270.0__ |
multiply__26.0__20.0__ add__250.0__20.0__ add__250.0__20.0__ |
| if an average hard drive had a capacity of num__0.1 tb in num__2000 and average hard drive capacities double every num__5 years what will be the average hard drive capacity in num__2050 ? <o> a ) num__256 <o> b ) num__512 <o> c ) num__768 <o> d ) num__1024 <o> e ) num__1536 |
num__0.1 * num__2 ^ num__10 = num__0.1 * num__1024 = num__102.4 the answer is d . <eor> d <eos> |
d |
reverse__0.1__ multiply__0.1__1024.0__ divide__102.4__0.1__ |
multiply__5.0__2.0__ multiply__0.1__1024.0__ multiply__102.4__10.0__ |
| a box contains num__24 electric bulbs out of which num__4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is <o> a ) num__0.210526315789 <o> b ) num__0.368421052632 <o> c ) num__0.631578947368 <o> d ) num__0.311688311688 <o> e ) none |
solution p ( none is defective ) = num__20 c num__0.0833333333333 c num__2 = num__0.688311688312 . p ( at least one is defective ) = ( num__1 - num__0.688311688312 ) = num__0.311688311688 . answer d <eor> d <eos> |
d |
coin_space__ negate_prob__0.6883__ negate_prob__0.6883__ |
coin_space__ negate_prob__0.6883__ negate_prob__0.6883__ |
| a and b can do a piece of work in num__45 days and num__40 days respectively . they began to do the work together but a leaves after some days and then b completed the remaining work in num__23 days . the number of days after which a left the work was ? <o> a ) num__12 <o> b ) num__11 <o> c ) num__10 <o> d ) num__9 <o> e ) num__8 |
let the total units of work to be done be num__360 . the units of work done by a in a single day = num__8 similarly the units of work done by b in a single day = num__9 . a and b ’ s one day work = num__17 units a and b worked together for some days = num__17 x ( assume ) b ’ s work alone for num__23 days = num__23 x num__9 = num__207 so the work done by a and b together = ( num__360 - num__207 ) = num__153 units therefore num__17 x = num__153 = > num__9 units therefore the number of days after which a left the work was num__9 days . answer : d <eor> d <eos> |
d |
divide__360.0__45.0__ divide__360.0__40.0__ subtract__40.0__23.0__ multiply__23.0__9.0__ multiply__17.0__9.0__ round__9.0__ |
divide__360.0__45.0__ divide__360.0__40.0__ subtract__40.0__23.0__ multiply__23.0__9.0__ subtract__360.0__207.0__ subtract__17.0__8.0__ |
| three cricket players - - ramesh suresh and ganesh play for three different cricket teams haryana delhi and rajasthan respectively . ramesh was born in chennai but brought up in karnal haryana . suresh was born in delhi and brought up in delhi . ganesh ' s fore - fathers were from madurai but settled in rajasthan . all the three participated in a cricket tournament . ramesh ' s runs to suresh ' s runs and suresh ' s runs to ganesh ' s runs are in the ratio num__5 : num__7 . if the total runs scored by all the three players in the tournament is num__327 find the total runs scored by suresh in the tournament ? <o> a ) num__95 <o> b ) num__85 <o> c ) num__65 <o> d ) num__105 <o> e ) num__115 |
ramesh : suresh = num__5 : num__7 and suresh : ganesh = num__5 : num__7 by making suresh to be equal in two ratios we get r : s = num__25 : num__35 and s : g = num__35 : num__49 so total runs scored by them = num__109 x = num__327 x = num__3.0 = num__3 total runs scored by suresh = num__35 * x = num__35 * num__3 = num__105 answer : d <eor> d <eos> |
d |
multiply__5.0__7.0__ divide__327.0__109.0__ multiply__35.0__3.0__ multiply__35.0__3.0__ |
multiply__5.0__7.0__ divide__327.0__109.0__ multiply__35.0__3.0__ multiply__35.0__3.0__ |
| two trains each num__100 m long moving in opposite directions cross other in num__8 sec . if one is moving twice as fast the other then the speed of the faster train is ? <o> a ) num__40 <o> b ) num__50 <o> c ) num__60 <o> d ) num__70 <o> e ) num__80 |
let the speed of the slower train be x m / sec . then speed of the train = num__2 x m / sec . relative speed = ( x + num__2 x ) = num__3 x m / sec . ( num__100 + num__100 ) / num__8 = num__3 x = > x = num__8.33333333333 . so speed of the faster train = num__16.6666666667 = num__16.6666666667 * num__3.6 = num__60 km / hr . answer : option c <eor> c <eos> |
c |
hour_to_min_conversion__ hour_to_min_conversion__ |
hour_to_min_conversion__ hour_to_min_conversion__ |
| carol and jordan draw rectangles of equal area . if carol ' s rectangle measures num__8 inches by num__15 inches and jordan ' s rectangle is num__4 inches long how wide is jordan ' s rectangle in inches ? <o> a ) num__24 <o> b ) num__26 <o> c ) num__28 <o> d ) num__30 <o> e ) num__32 |
area of first rectangle is num__8 * num__15 = num__120 hence area of second would be num__4 x = num__120 x x = num__30 answer is d <eor> d <eos> |
d |
multiply__8.0__15.0__ triangle_area__15.0__4.0__ triangle_area__15.0__4.0__ |
multiply__8.0__15.0__ triangle_area__15.0__4.0__ triangle_area__15.0__4.0__ |
| peter rolls two dice at the same time . what is the probability that the total sum of numbers on the dices is num__3 ? <o> a ) a ) num__0.0833333333333 <o> b ) b ) num__0.0277777777778 <o> c ) c ) num__0.166666666667 <o> d ) d ) num__0.0555555555556 <o> e ) e ) num__0.333333333333 |
we can have num__6 different outcomes on a dice . total outcomes on two dices = num__6 * num__6 we need only the cases where sum of the outcomes is num__3 they can be { num__1 num__2 } { num__2 num__1 } a total of num__2 outcomes probability = favorable outcomes / total outcomes = num__0.0555555555556 = num__0.0555555555556 . option d <eor> d <eos> |
d |
subtract__3.0__1.0__ multiply__1.0__0.0556__ |
divide__6.0__3.0__ multiply__1.0__0.0556__ |
| let a = { x : x is a natural number and a factor of num__18 } b = { x : x is a natural number and less than num__6 } find a ∪ b and a ∩ b <o> a ) a ∩ b = { num__1 num__23 } <o> b ) a ∩ b = { num__2 num__54 } <o> c ) a ∪ b = { num__5 num__79 } <o> d ) a ∪ b = { num__1 num__23 } <o> e ) a ∪ b = { num__2 num__54 } |
a = { num__12 num__36 num__918 } b = { num__1 num__23 num__45 } therefore correct answer : a ∩ b = { num__1 num__23 } ( a ) <eor> a <eos> |
a |
subtract__18.0__6.0__ reverse__1.0__ |
subtract__18.0__6.0__ reverse__1.0__ |
| look at this series : num__1000 num__200 num__40 . . . what number should come next ? <o> a ) num__17 <o> b ) num__10 <o> c ) num__15 <o> d ) num__18 <o> e ) num__8 |
e num__8 this is a simple division series . each number is divided by num__5 . <eor> e <eos> |
e |
divide__1000.0__200.0__ divide__40.0__5.0__ |
divide__1000.0__200.0__ divide__40.0__5.0__ |
| due to construction the speed limit along an num__5 - mile section of highway is reduced from num__60 miles per hour to num__40 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? <o> a ) a ) num__3.12 <o> b ) b ) num__2.5 <o> c ) c ) num__10 <o> d ) d ) num__15 <o> e ) e ) num__24 |
old time in minutes to cross num__5 miles stretch = num__5 * num__1.0 = num__5 * num__1.0 = num__5 new time in minutes to cross num__5 miles stretch = num__5 * num__1.5 = num__5 * num__1.5 = num__7.5 time difference = num__2.5 ans : b <eor> b <eos> |
b |
divide__60.0__40.0__ multiply__5.0__1.5__ add__1.5__1.0__ round__2.5__ |
divide__60.0__40.0__ multiply__5.0__1.5__ add__1.5__1.0__ round__2.5__ |
| two trains num__140 m and num__170 m long run at the speed of num__60 km / hr and num__40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? <o> a ) num__10.9 sec <o> b ) num__11.16 sec <o> c ) num__53.8 sec <o> d ) num__10.8 sec <o> e ) num__10.4 sec |
relative speed = num__60 + num__40 = num__100 km / hr . = num__100 * num__0.277777777778 = num__27.7777777778 m / sec . distance covered in crossing each other = num__140 + num__170 = num__310 m . required time = num__310 * num__0.036 = num__11.16 sec . answer : b : <eor> b <eos> |
b |
subtract__140.0__40.0__ add__140.0__170.0__ multiply__0.036__310.0__ round__11.16__ |
add__60.0__40.0__ add__140.0__170.0__ multiply__0.036__310.0__ multiply__0.036__310.0__ |
| if the perimeter of a rectangular garden is num__600 m its length when its breadth is num__150 m is ? <o> a ) num__150 m <o> b ) num__899 m <o> c ) num__200 m <o> d ) num__166 m <o> e ) num__187 m |
num__2 ( l + num__150 ) = num__600 = > l = num__150 m answer : a <eor> a <eos> |
a |
round__150.0__ |
round__150.0__ |
| what is the least number that should be added to num__1053 so the sum of the number is divisible by num__23 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
( num__45.7826086957 ) gives a remainder num__18 so we need to add num__5 . the answer is e . <eor> e <eos> |
e |
divide__1053.0__23.0__ subtract__23.0__18.0__ subtract__23.0__18.0__ |
divide__1053.0__23.0__ subtract__23.0__18.0__ subtract__23.0__18.0__ |
| find a sum for num__1 st num__5 prime number ' s ? <o> a ) num__25 <o> b ) num__28 <o> c ) num__30 <o> d ) num__34 <o> e ) num__36 |
required sum = ( num__2 + num__3 + num__5 + num__7 + num__11 ) = num__28 note : num__1 is not a prime number option b <eor> b <eos> |
b |
add__1.0__2.0__ add__5.0__2.0__ multiply__1.0__28.0__ |
add__1.0__2.0__ add__5.0__2.0__ multiply__1.0__28.0__ |
| in a shop the cost of num__4 shirts num__4 pairs of trousers and num__2 hats is $ num__560 . the cost of num__9 shirts num__9 pairs of trousers and num__6 hats is $ num__1290 . what is the total cost of num__1 shirt num__1 pair of trousers and num__1 hat ? <o> a ) num__110 $ <o> b ) num__120 $ <o> c ) num__130 $ <o> d ) num__140 $ <o> e ) num__150 $ |
et x be the price of one shirt y be the price of one pair of trousers and z be the price of one hat . num__4 x + num__4 y + num__2 z = num__560 : num__9 x + num__9 y + num__6 z = num__1290 num__3 x + num__3 y + num__2 z = num__430 : divide all terms of equation c by num__3 x + y = num__130 : subtract equation d from equation b num__3 ( x + y ) + num__2 z = num__430 : equation d with factored terms . num__3 * num__130 + num__2 z = num__430 z = num__20 : solve for z x + y + z = num__130 + num__20 = $ num__150 <eor> e <eos> |
e |
subtract__4.0__1.0__ divide__1290.0__3.0__ subtract__560.0__430.0__ add__130.0__20.0__ multiply__1.0__150.0__ |
add__2.0__1.0__ divide__1290.0__3.0__ subtract__560.0__430.0__ add__130.0__20.0__ multiply__1.0__150.0__ |
| a cube of side num__8 meter length is cut into small cubes of side num__16 cm each . how many such small cubes can be obtained ? <o> a ) num__10780 <o> b ) num__127600 <o> c ) num__125000 <o> d ) num__152000 <o> e ) num__10000 |
along one edge the number of small cubes that can be cut = num__50.0 = num__50 along each edge num__50 cubes can be cut . ( along length breadth and height ) . total number of small cubes that can be cut = num__50 * num__50 * num__50 = num__125000 answer : c <eor> c <eos> |
c |
round__125000.0__ |
round__125000.0__ |
| set x consists of the integers from num__0 to num__12 inclusive while set y consists of the integers from num__6 to num__10 inclusive . how many distinct integers do belong to the both sets at the same time ? <o> a ) num__4 <o> b ) num__10 <o> c ) num__5 <o> d ) num__9 <o> e ) num__8 |
x = { num__0 num__12 num__34 num__56 num__7 num__8 num__9 num__10 num__11 num__12 } y = { num__6 num__7 num__8 num__9 num__10 } common elements = { num__6 num__7 num__8 num__9 num__10 } = num__5 elements answer : option c . <eor> c <eos> |
c |
vowel_space__ vowel_space__ |
vowel_space__ vowel_space__ |
| maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is num__110 kilometers maxwell ' s walking speed is num__4 km / h and brad ' s running speed is num__7 km / h what is the distance traveled by brad ? <o> a ) num__16 <o> b ) num__18 <o> c ) num__20 <o> d ) num__70 <o> e ) num__30 |
time taken = total distance / relative speed total distance = num__110 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = num__7 + num__4 = num__11 kms / hr time taken = num__10.0 = num__10 hrs distance traveled by brad = brad ' s speed * time taken = num__7 * num__10 = num__70 kms . . . answer - d <eor> d <eos> |
d |
add__4.0__7.0__ divide__110.0__11.0__ multiply__7.0__10.0__ round__70.0__ |
add__4.0__7.0__ divide__110.0__11.0__ multiply__7.0__10.0__ multiply__7.0__10.0__ |
| mudit ' s age num__30 years hence will be thrice his age four years ago . find mudit ' s present age ? <o> a ) num__12 <o> b ) num__19 <o> c ) num__27 <o> d ) num__21 <o> e ) num__15 |
explanation : let mudit ' s present age be ' m ' years . m + num__30 = num__3 ( m - num__4 ) = > num__2 m = num__42 = > m = num__21 years . answer : d <eor> d <eos> |
d |
divide__42.0__2.0__ divide__42.0__2.0__ |
divide__42.0__2.0__ subtract__42.0__21.0__ |
| the probability that a man will be alive for num__10 more yrs is num__0.25 & the probability that his wife will alive for num__10 more yrs is num__0.333333333333 . the probability that none of them will be alive for num__10 more yrs is <o> a ) num__0.5 <o> b ) num__1 <o> c ) num__1.5 <o> d ) num__0.6 <o> e ) num__0.333333333333 |
sol . required probability = pg . ) x p ( b ) = ( num__1 — d x ( num__1 — i ) = : x num__1 = num__0.5 ans . ( a ) <eor> a <eos> |
a |
divide__0.25__0.5__ |
divide__0.25__0.5__ |
| if the average ( arithmetic mean ) of the four numbers k + num__1 num__2 k + num__3 num__3 k – num__5 and num__5 k + num__1 is num__63 what is the value of k ? <o> a ) num__11 <o> b ) num__15 num__0.75 <o> c ) num__22 <o> d ) num__23 <o> e ) num__22 num__0.909090909091 |
k + num__1 + num__2 k + num__3 + num__3 k - num__5 + num__5 k + num__1 = num__11 k ( num__11 k ) / num__4 = num__63 num__11 k = num__63 * num__4 = num__252 k = num__22.9090909091 = num__22 num__0.909090909091 answer e . <eor> e <eos> |
e |
add__1.0__3.0__ multiply__63.0__4.0__ divide__252.0__11.0__ round_down__22.9091__ subtract__22.9091__22.0__ round_down__22.9091__ |
add__1.0__3.0__ multiply__63.0__4.0__ divide__252.0__11.0__ multiply__2.0__11.0__ subtract__22.9091__22.0__ multiply__1.0__22.0__ |
| a wheel rotates num__10 times every minute and moves num__30 cm during each rotation . how many metres does the wheel move in one hour ? <o> a ) num__6 metre <o> b ) num__18 metre <o> c ) num__180 metre <o> d ) num__1800 metre <o> e ) num__2200 metre |
expl : number of times wheel moves in num__1 hour = num__10 * num__60 = num__600 : . distance moves = ( num__600 * num__30 ) cm = num__18000 cm in metres = num__180 metre answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ multiply__10.0__60.0__ multiply__30.0__600.0__ round__180.0__ |
hour_to_min_conversion__ multiply__10.0__60.0__ multiply__30.0__600.0__ round__180.0__ |
| a committee is composed of w women and m men . if num__3 women and num__2 men are added to the committee and if one person is selected at random from the enlarged committee then the probability that a woman is selected can be represented by <o> a ) w / m <o> b ) w / ( w + m ) <o> c ) ( w + num__3 ) / ( m + num__2 ) <o> d ) ( w + num__3 ) / ( w + m + num__3 ) <o> e ) ( w + num__3 ) / ( w + m + num__5 ) |
the total number of members on the enlarged committee is ( w + num__3 ) + ( m + num__2 ) = w + m + num__5 ; the total number of women on the enlarged committee is w + num__3 . thus the probability that a woman is selected is p = favorable / total = ( w + num__3 ) / ( w + m + num__5 ) . answer : e . <eor> e <eos> |
e |
vowel_space__ choose__3.0__2.0__ |
vowel_space__ choose__3.0__2.0__ |
| the output of a factory was increased by num__10.0 to keep up with rising demand . to handle the holiday rush this new output was increased by num__30.0 . by approximately what percent would the output now have to be decreased in order to restore the original output ? <o> a ) num__20.0 <o> b ) num__24.0 <o> c ) num__30.0 <o> d ) num__32.0 <o> e ) num__79 % |
the original output increases by num__10.0 and then num__30.0 . total % change = a + b + ab / num__100 total % change = num__10 + num__30 + num__10 * num__0.3 = num__43.0 now you want to change it to num__0 so num__0 = num__43 + x + num__43 x / num__100 x = - num__43 ( num__100 ) / num__143 = num__30.0 approximately answer is c <eor> c <eos> |
c |
divide__30.0__100.0__ round_down__0.3__ add__100.0__43.0__ multiply__100.0__0.3__ |
divide__30.0__100.0__ round_down__0.3__ add__100.0__43.0__ multiply__100.0__0.3__ |
| what is the product of all the possible values of x if x ^ num__2 ( x + num__2 ) + num__7 x ( x + num__2 ) + num__6 ( x + num__2 ) = num__0 ? <o> a ) – num__29 <o> b ) – num__12 <o> c ) num__12 <o> d ) num__29 <o> e ) num__168 |
x ^ num__2 ( x + num__2 ) + num__7 x ( x + num__2 ) + num__6 ( x + num__2 ) = num__0 = > ( x + num__2 ) ( x ^ num__2 + num__7 x + num__6 ) = num__0 = > ( x + num__2 ) ( x + num__1 ) ( x + num__6 ) = num__0 = > x = - num__1 or - num__2 or - num__6 product = - num__12 answer option - num__12 answer : b <eor> b <eos> |
b |
subtract__7.0__6.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
subtract__7.0__6.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
| a jogger running at num__9 kmph alongside a railway track in num__240 meters ahead of the engine of a num__120 metres long train running at num__45 kmph in the same direction . in how much time will the train pass the jogger ? <o> a ) num__18 seconds <o> b ) num__26 seconds <o> c ) num__36 seconds <o> d ) num__16 seconds <o> e ) num__46 seconds |
total distance ( d ) = num__240 + num__120 = num__360 meter relative speed ( s ) = num__45 - num__9 = num__36 kmph = num__36 * num__0.277777777778 = num__10 mps d = s * t num__360 = num__10 * t t = num__36 seconds answer : c <eor> c <eos> |
c |
add__240.0__120.0__ subtract__45.0__9.0__ divide__360.0__36.0__ round__36.0__ |
add__240.0__120.0__ subtract__45.0__9.0__ divide__360.0__36.0__ subtract__45.0__9.0__ |
| the side of a square is increased by num__15.0 then how much % does its area increases ? <o> a ) num__52.65 <o> b ) num__56.25 <o> c ) num__50.75 <o> d ) num__32.25 <o> e ) num__52.75 |
a = num__100 a num__2 = num__10000 a = num__115 a num__2 = num__13225 - - - - - - - - - - - - - - - - num__10000 - - - - - - - - - num__3225 num__100 - - - - - - - ? = > num__32.25 answer : d <eor> d <eos> |
d |
power__100.0__2.0__ power__115.0__2.0__ triangle_area__32.25__2.0__ |
power__100.0__2.0__ power__115.0__2.0__ triangle_area__32.25__2.0__ |
| tanks p and b are each in the shape of a right circular cylinder . the interior of tank p has a height of num__10 meters and a circumference of num__8 meters and the interior of tank b has a height of num__8 meters and a circumference of num__10 meters . the capacity of tank p is what percent of the capacity of tank b ? <o> a ) num__75.0 <o> b ) num__80.0 <o> c ) num__100.0 <o> d ) num__120.0 <o> e ) num__125 % |
b . for p r = num__4.0 pi . its capacity = ( num__4 pi ) ^ num__2 * num__10 = num__160 pi for b r = num__10 / pi . its capacity = ( num__5 pi ) ^ num__2 * num__8 = num__200 pi p / b = num__160 pi / num__200 pi = num__0.8 <eor> b <eos> |
b |
subtract__10.0__8.0__ divide__10.0__2.0__ divide__8.0__10.0__ multiply__10.0__8.0__ |
divide__8.0__4.0__ divide__10.0__2.0__ divide__8.0__10.0__ multiply__10.0__8.0__ |
| the sector of a circle has radius of num__21 cm and central angle num__135 o . find its perimeter ? <o> a ) num__91.5 cm <o> b ) num__91.8 cm <o> c ) num__99.5 cm <o> d ) num__91.6 cm <o> e ) num__91.3 cm |
perimeter of the sector = length of the arc + num__2 ( radius ) = ( num__0.375 * num__2 * num__3.14285714286 * num__21 ) + num__2 ( num__21 ) = num__49.5 + num__42 = num__91.5 cm answer : a <eor> a <eos> |
a |
multiply__21.0__2.0__ add__49.5__42.0__ round__91.5__ |
multiply__21.0__2.0__ add__49.5__42.0__ add__49.5__42.0__ |
| chris wants to pay back his debt to the bank . he has already paid back num__60.0 of the initial debt . his family offered to help by paying num__33.0 of the initial debt so now he only needs an additional $ num__105 in order to pay off the debt . what was the initial amount of the debt chris owed to the bank ? <o> a ) $ num__1500 <o> b ) $ num__2000 <o> c ) $ num__2500 <o> d ) $ num__3000 <o> e ) none |
answer chris already has both num__60.0 and num__33.0 of the debt : = num__60.0 + num__33.0 = num__93.0 the remaining amount that he will need to pay is equal to : = num__100.0 - num__93.0 = num__7.0 we know that this additional amount is $ num__105 ; this allows us to find the initial total amount of his debt : num__7.0 = $ num__105 $ num__15.0 = $ num__105 / num__0.07 = $ num__1500 answer is a . <eor> a <eos> |
a |
percent__100.0__1500.0__ |
percent__100.0__1500.0__ |
| the length of a rectangular garden is num__2 feet longer than num__3 times its width . if the perimeter of the garden is num__100 feet find the length of the garden . <o> a ) num__42 <o> b ) num__38 <o> c ) num__28 <o> d ) num__57 <o> e ) num__49 |
let l and w be the length and width of the garden . the statement ` ` the length of a rectangular garden is num__2 feet longer than num__3 times its width ' ' may be formulated by l = num__2 + num__3 w the formula for the perimeter is given by p = num__2 l + num__2 w substitute p and l in the above equation by num__100 and num__2 + num__3 w respectively to obtain num__100 = num__2 ( num__2 + num__3 w ) + num__2 w solve for w and l w = num__12 and l = num__2 + num__3 w = num__38 . check that the perimeter of the rectangular garden is num__100 p = num__2 l + num__2 w = num__76 + num__24 = num__100 correct answer b <eor> b <eos> |
b |
square_perimeter__3.0__ multiply__2.0__38.0__ surface_cube__2.0__ triangle_area__2.0__38.0__ |
square_perimeter__3.0__ multiply__2.0__38.0__ surface_cube__2.0__ triangle_area__2.0__38.0__ |
| worker a takes num__8 hours to do a job . worker b takes num__10 hours to do a job . how long should it take both a and b working together to do same job . <o> a ) num__0.444444444444 <o> b ) num__2 num__0.444444444444 <o> c ) num__3 num__0.444444444444 <o> d ) num__4 num__0.444444444444 <o> e ) none of these |
explanation : in this type of questions first we need to calculate num__1 hours work then their collective work as a ' s num__1 hour work is num__0.125 b ' s num__1 hour work is num__0.1 ( a + b ) ' s num__1 hour work = num__0.125 + num__0.1 = num__0.225 so both will finish the work in num__4.44444444444 hours = num__4 num__0.444444444444 option d <eor> d <eos> |
d |
divide__1.0__8.0__ divide__1.0__10.0__ add__0.125__0.1__ divide__1.0__0.225__ multiply__0.1__4.4444__ round__4.0__ |
divide__1.0__8.0__ divide__1.0__10.0__ add__0.125__0.1__ divide__1.0__0.225__ multiply__0.1__4.4444__ round__4.0__ |
| a is two years older than b who is twice as old as c . if the total of the ages of a b and c be num__47 the how old is b ? <o> a ) num__7 <o> b ) num__9 <o> c ) num__18 <o> d ) num__11 <o> e ) num__10 |
explanation : let c ' s age be x years . then b ' s age = num__2 x years . a ' s age = ( num__2 x + num__2 ) years . ( num__2 x + num__2 ) + num__2 x + x = num__47 ⇒ num__5 x = num__45 ⇒ x = num__9 . hence b ' s age = num__2 x = num__18 years . answer : c <eor> c <eos> |
c |
subtract__47.0__2.0__ divide__45.0__5.0__ multiply__2.0__9.0__ multiply__2.0__9.0__ |
subtract__47.0__2.0__ divide__45.0__5.0__ multiply__2.0__9.0__ multiply__2.0__9.0__ |
| train t leaves new york at num__7 : num__00 am traveling to boston at num__80 mph . train b leaves boston at num__7 : num__45 am traveling to new york at num__70 mph on a parallel track . if the distance between new york and boston is num__210 miles at what time will the two trains pass each other ? <o> a ) num__8 : num__15 am <o> b ) num__8 : num__45 am <o> c ) num__9 : num__00 am <o> d ) num__9 : num__30 am <o> e ) can not be determined from the information given |
train t will cover num__60 kms in num__45 min at num__7 : num__45 the time when train b starts . . . distance left to cover is num__210 - num__60 = num__150 . . combined speed = num__80 + num__70 = num__150 . . so the trains meet in num__1 hour i . e . num__7 : num__45 + num__1 hr = num__8 : num__45 b <eor> b <eos> |
b |
hour_to_min_conversion__ add__80.0__70.0__ add__7.0__1.0__ round__8.0__ |
hour_to_min_conversion__ add__80.0__70.0__ add__7.0__1.0__ add__7.0__1.0__ |
| how many seconds will a train num__150 meters long take to cross a bridge num__200 meters long if the speed of the train is num__54 kmph ? <o> a ) num__17 sec <o> b ) num__21 sec <o> c ) num__23.3 sec <o> d ) num__27.5 sec <o> e ) num__29 sec |
d = num__150 + num__200 = num__350 s = num__54 * num__0.277777777778 = num__15 mps t = num__23.3333333333 = num__23.3 sec c ) num__23.3 sec <eor> c <eos> |
c |
add__150.0__200.0__ divide__350.0__15.0__ round__23.3__ |
add__150.0__200.0__ divide__350.0__15.0__ round__23.3__ |
| mike took a taxi to the airport and paid $ num__2.50 to start plus $ num__0.25 per mile . annie took a different route to the airport and paid $ num__2.50 plus $ num__5.00 in bridge toll fees plus $ num__0.25 per mile . if each was charged exactly the same amount and annie ' s ride was num__26 miles how many miles was mike ' s ride ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__36 <o> d ) num__46 <o> e ) num__54 |
the cost of annie ' s ride was num__2.5 + num__5 + ( num__0.25 * num__26 ) = $ num__14 let x be the distance of mike ' s ride . the cost of mike ' s ride is num__2.5 + ( num__0.25 * x ) = num__14 num__0.25 * x = num__11.5 x = num__46 miles the answer is d . <eor> d <eos> |
d |
subtract__14.0__2.5__ divide__11.5__0.25__ divide__11.5__0.25__ |
subtract__14.0__2.5__ divide__11.5__0.25__ divide__11.5__0.25__ |
| evaluate : num__10111 - num__10 * num__2 * num__5 = ? <o> a ) num__10101 <o> b ) num__10110 <o> c ) num__11100 <o> d ) num__11001 <o> e ) num__11011 |
according to order of operations num__10 ? num__2 ? num__5 ( division and multiplication ) is done first from left to right num__5.0 = num__5 * num__2 = num__10 hence num__10111 - num__10 * num__2 * num__5 = num__10111 - num__10 = num__10101 correct answer a <eor> a <eos> |
a |
subtract__10111.0__10.0__ subtract__10111.0__10.0__ |
subtract__10111.0__10.0__ subtract__10111.0__10.0__ |
| find the average of the first num__20 natural numbers ? <o> a ) num__18.5 <o> b ) num__10.0 <o> c ) num__10.4 <o> d ) num__10.5 <o> e ) num__10.2 |
average of the first ' n ' natural numbers = ( n + num__1 ) / num__2 we get ( num__20 + num__1 ) / num__2 = num__10.5 . answer : d <eor> d <eos> |
d |
multiply__10.5__1.0__ |
divide__10.5__1.0__ |
| what will come in place of the question mark ( ? ) in the following equation ? num__25 ( num__7.5 ) × num__5 ( num__2.5 ) ÷ num__125 ( num__1.5 ) = num__5 ? <o> a ) num__16 <o> b ) num__17.5 <o> c ) num__8.5 <o> d ) ? = num__13 <o> e ) none of these |
num__25 ( num__7.5 ) × num__5 ( num__2.5 ) ÷ num__125 ( num__1.5 ) = num__5 ? or num__5 ( num__2 × num__7.5 ) × num__5 ( num__2.5 ) ÷ num__5 ( num__3 × num__1.5 ) = num__5 ? or num__5 ( num__15 ) × num__5 ( num__2.5 ) × num__1 ⁄ num__54.5 = num__5 ? or num__5 ( num__13 ) = num__5 ? or ? = num__13 answer d <eor> d <eos> |
d |
round_down__2.5__ divide__7.5__2.5__ multiply__7.5__2.0__ round_down__1.5__ subtract__15.0__2.0__ multiply__1.0__13.0__ |
round_down__2.5__ divide__7.5__2.5__ multiply__7.5__2.0__ round_down__1.5__ subtract__15.0__2.0__ multiply__1.0__13.0__ |
| two men can fill the tank in num__5 hr and num__4 hr respectively while the third empty it in num__20 hr . if all pipes are opened simultaneously then the tank will be filled in <o> a ) num__2 <o> b ) num__2.5 <o> c ) num__3 <o> d ) num__3.5 <o> e ) num__3.7 |
work done by all the men working together in num__1 hour . num__0.2 + num__0.25 − num__0.05 = num__0.4 hence tank will be filled in num__2.5 = num__2.5 hour b <eor> b <eos> |
b |
subtract__5.0__4.0__ divide__4.0__20.0__ divide__5.0__20.0__ divide__0.25__5.0__ divide__1.0__0.4__ round__2.5__ |
subtract__5.0__4.0__ divide__4.0__20.0__ divide__5.0__20.0__ divide__0.25__5.0__ divide__1.0__0.4__ round__2.5__ |
| num__0.6 of all married couples have more than one child . num__0.5 of all married couples have more than num__3 children . what fraction of all married couples have num__2 or num__3 children ? <o> a ) num__0.2 <o> b ) num__0.1 <o> c ) num__0.35 <o> d ) num__0.6 <o> e ) it can not be determined from the given information . |
plug in simple numbers . take num__100 couples for example . num__0.6 of num__100 couples have more than one child = num__60 couples . num__0.5 of num__100 couples have more than num__3 kids = num__50 couples . this implies that num__50 couples are a subset of num__60 couples and the complement of num__60 couples within those num__100 couples which equals num__40 couples have either one or no kids at all . we need to find couples that have num__2 or num__3 kids so essentially it is num__60 - num__50 = num__10 . fraction will be num__0.1 = num__0.1 . option b <eor> b <eos> |
b |
multiply__0.6__100.0__ multiply__0.5__100.0__ subtract__100.0__60.0__ subtract__50.0__40.0__ reverse__10.0__ reverse__10.0__ |
multiply__0.6__100.0__ multiply__0.5__100.0__ subtract__100.0__60.0__ subtract__50.0__40.0__ subtract__0.6__0.5__ subtract__0.6__0.5__ |
| the number of degrees that the hour hand of a clock moves through between noon and num__2.30 in the afternoon of the same day is <o> a ) num__720 <o> b ) num__180 <o> c ) num__75 <o> d ) num__65 <o> e ) num__60 |
explanation : the hour hand moves from pointing to num__12 to pointing to half way between num__2 and num__3 . the angle covered between each hour marking on the clock is num__30.0 = num__30 . since the hand has covered num__2.5 of these divisions the angle moved through is num__75 . answer is c <eor> c <eos> |
c |
divide__30.0__12.0__ multiply__2.5__30.0__ round__75.0__ |
divide__30.0__12.0__ multiply__2.5__30.0__ round__75.0__ |
| on dividing a number by num__5 we get num__3 as remainder . what will the remainder when the square of the this number is divided by num__5 ? <o> a ) num__0 remainder <o> b ) num__1 remainder <o> c ) num__2 remainder <o> d ) num__4 remainder <o> e ) num__5 remainder |
explanation let the number be x and on dividing x by num__5 we get k as quotient and num__3 as remainder . x = num__5 k + num__3 x num__2 = ( num__5 k + num__3 ) num__2 = ( num__25 k num__2 + num__30 k + num__9 ) = num__5 ( num__5 k num__2 + num__6 k + num__1 ) + num__4 on dividing x num__2 by num__5 we get num__4 as remainder . remainder = num__4 answer d <eor> d <eos> |
d |
power__5.0__2.0__ volume_rectangular_prism__5.0__3.0__2.0__ power__3.0__2.0__ multiply__3.0__2.0__ square_perimeter__1.0__ square_perimeter__1.0__ |
power__5.0__2.0__ volume_rectangular_prism__5.0__3.0__2.0__ power__3.0__2.0__ multiply__3.0__2.0__ square_perimeter__1.0__ square_perimeter__1.0__ |
| five a - list actresses are vying for the three leading roles in the new film ` ` catfight in denmark . ' ' the actresses are julia robards merly strep sally fieldstone lauren bake - all and hallie strawberry . assuming that no actresses has any advantage in getting any role what is the probability that julia and hallie will star in the film together ? <o> a ) num__0.3 <o> b ) num__0.024 <o> c ) num__0.6 <o> d ) num__0.6 <o> e ) num__0.008 |
pick hallie and julia . now you can pick the third actress in num__3 ways so total number of ways of picking num__3 actresses ( including hallie and julia ) = num__3 total number of ways of picking any num__3 actresses out of num__5 = num__5 * num__4 * num__1.0 ! = num__10 probability that both hallie and julia will be picked = num__0.3 ans : a <eor> a <eos> |
a |
subtract__4.0__3.0__ divide__3.0__10.0__ multiply__1.0__0.3__ |
subtract__4.0__3.0__ divide__3.0__10.0__ multiply__1.0__0.3__ |
| given the two equations num__7 r + num__2 s = num__43 and num__4 r - num__3 s = num__8 by how much does r exceed s ? <o> a ) num__3 <o> b ) num__1 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
solve by simultaneous equations . my answer is b . r = num__5 s = num__4 <eor> b <eos> |
b |
subtract__7.0__2.0__ subtract__4.0__3.0__ |
subtract__7.0__2.0__ subtract__4.0__3.0__ |
| if num__3 workers collect num__48 kg of cotton in num__4 days how many kg of cotton will num__9 workers collect in num__2 days ? <o> a ) num__87 kg <o> b ) num__67 kg <o> c ) num__72 kg <o> d ) num__87 kg <o> e ) num__46 kg |
( num__3 * num__4 ) / num__48 = ( num__9 * num__2 ) / x x = num__72 kg answer : c <eor> c <eos> |
c |
round__72.0__ |
round__72.0__ |
| two trains each num__100 m long moving in opposite directions cross other in num__10 sec . if one is moving twice as fast the other then the speed of the faster train is ? <o> a ) num__22 <o> b ) num__98 <o> c ) num__48 <o> d ) num__88 <o> e ) num__12 |
let the speed of the slower train be x m / sec . then speed of the train = num__2 x m / sec . relative speed = ( x + num__2 x ) = num__3 x m / sec . ( num__100 + num__100 ) / num__10 = num__3 x = > x = num__6.66666666667 . so speed of the faster train = num__13.3333333333 = num__13.3333333333 * num__3.6 = num__48 km / hr . answer : c <eor> c <eos> |
c |
round__48.0__ |
round__48.0__ |
| what is the simplified result of following the steps below in order ? ( num__1 ) add num__5 y to num__2 v ( num__2 ) multiply the sum by num__3 ( num__3 ) subtract v + y from the product <o> a ) num__5 v + num__14 y <o> b ) num__5 x + num__16 y <o> c ) num__5 x + num__5 y <o> d ) num__6 x + num__4 y <o> e ) num__3 x + num__12 y |
num__3 ( num__5 y + num__2 v ) - v - y = num__14 y + num__5 v ' a ' is the answer <eor> a <eos> |
a |
multiply__1.0__5.0__ |
add__2.0__3.0__ |
| if two resistors a ( r num__1 ) and b ( r num__2 ) stand in parallel with each other in electrical wire the total resistor appears as r num__1 r num__2 / ( r num__1 + r num__2 ) . if three resistors a ( r num__1 ) b ( r num__2 ) and c ( num__2 r num__2 ) stand in parallel in electrical wire what is the ratio v of the resistors ’ sum of a and c to the resistors ’ sum of a and b ? <o> a ) num__2 ( r num__1 + r num__2 ) : ( r num__1 + num__2 r num__2 ) <o> b ) ( r num__1 + r num__2 ) : ( r num__1 + num__2 r num__2 ) <o> c ) ( num__2 r num__1 + r num__2 ) : ( r num__1 + num__2 r num__2 ) <o> d ) num__2 ( r num__1 + r num__2 ) : ( num__2 r num__1 + r num__2 ) <o> e ) num__2 ( r num__1 - r num__2 ) : ( r num__1 + num__2 r num__2 ) |
two resistors a ( r num__1 ) and b ( r num__2 ) . total or sum of two resistors appear as r num__1 r num__2 / r num__1 + r num__2 . it is looks like inversion of sum of rates . num__1 / r num__1 + num__1 / r num__2 = r num__1 + r num__2 / r num__1 r num__2 . same way sum of a ( r num__1 ) and c ( num__2 r num__2 ) = num__1 / r num__1 + num__0.5 r num__2 = num__2 r num__2 + r num__1 / r num__12 r num__2 . inversion rate = r num__12 r num__1.0 r num__2 + r num__1 . ratio v of sum of a and c / sum of a and b = num__2 r num__2 r num__0.5 r num__2 + r num__1 * r num__1 + r num__2 / r num__1 r num__2 = num__2 ( r num__1 + r num__2 ) / num__2 r num__2 + r num__1 . a <eor> a <eos> |
a |
reverse__2.0__ reverse__0.5__ |
reverse__2.0__ reverse__0.5__ |
| how many seconds will a num__220 metre long train take to cross a man running with a speed of num__8 km / hr in the direction of the moving train if the speed of the train is num__80 km / hr ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__40 <o> d ) num__11 <o> e ) num__60 |
explanation : speed of train relatively to man = ( num__80 - num__8 ) km / hr = num__72 km / hr = ( num__72 x num__0.277777777778 ) m / sec = num__20 m / sec time taken to pass the man = ( num__11.0 ) sec = num__11 sec . answer : d <eor> d <eos> |
d |
subtract__80.0__8.0__ divide__220.0__20.0__ round__11.0__ |
subtract__80.0__8.0__ divide__220.0__20.0__ divide__220.0__20.0__ |
| num__4 bells first begin to toll together with an intervals of num__5 num__10 num__15 & num__20 sec . how many times does they toll together in an hr ? <o> a ) num__40 <o> b ) num__60 <o> c ) num__80 <o> d ) num__90 <o> e ) num__110 |
num__5 | num__5 num__10 num__15 num__20 num__2 | num__1 num__2 num__3 num__4 | num__1 num__1 num__3 num__2 l . c . m of num__5 num__10 num__15 num__20 is num__60 num__60 times b <eor> b <eos> |
b |
divide__10.0__5.0__ subtract__5.0__4.0__ subtract__4.0__1.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
divide__10.0__5.0__ subtract__5.0__4.0__ subtract__4.0__1.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
| a man rides at the rate of num__30 km / hr . but stops num__30 minutes to change horses at the end of every num__25 th kilometer . how long will he take to go a distance of num__150 kilometers ? <o> a ) num__5 hr <o> b ) num__7 hr num__30 min <o> c ) num__8 hr num__10 min <o> d ) num__7 hr num__20 min <o> e ) num__9 hr |
speed of man = num__30 km / hr number of rests = ( num__6.0 ) - num__1 = num__5 time taken for the man = ( num__5.0 ) + num__5 * ( num__0.5 ) = num__7 hr num__30 min answer is b <eor> b <eos> |
b |
divide__150.0__25.0__ subtract__30.0__25.0__ add__1.0__6.0__ round__7.0__ |
divide__150.0__25.0__ subtract__30.0__25.0__ add__1.0__6.0__ add__1.0__6.0__ |
| num__39 persons can repair a road in num__12 days working num__5 hours a day . in how many days will num__15 persons working num__6 hours a day complete the work ? <o> a ) num__10 <o> b ) num__13 <o> c ) num__14 <o> d ) num__26 <o> e ) num__16 |
let the required number of days be x . less persons more days ( indirect proportion ) more working hours per day less days ( indirect proportion ) persons num__15 : num__39 : : num__12 : x working hours / day num__6 : num__5 num__15 x num__6 x x = num__39 x num__5 x num__12 x = ( num__39 x num__5 x num__12 ) / ( num__15 x num__6 ) x = num__26 . answer : d <eor> d <eos> |
d |
round__26.0__ |
round__26.0__ |
| a man can row upstream at num__5 kmph and downstream at num__25 kmph and then find the speed of the man in still water ? <o> a ) num__15 <o> b ) num__77 <o> c ) num__30 <o> d ) num__88 <o> e ) num__34 |
us = num__5 ds = num__25 m = ( num__5 + num__25 ) / num__2 = num__15 answer : a <eor> a <eos> |
a |
round__15.0__ |
round__15.0__ |
| right triangle pqr is the base of the prism in the figure above . if pq = pr = â ˆ š num__5 and the height of the prism is num__10 what is the volume of the prism ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__15 <o> d ) num__20 <o> e ) num__25 |
volume of prism = area of base * height = num__0.5 * ( square root of num__5 ) * ( square root of num__5 ) * num__10 = num__25 answer : e <eor> e <eos> |
e |
triangle_area__5.0__10.0__ triangle_area__5.0__10.0__ |
volume_rectangular_prism__5.0__10.0__0.5__ volume_rectangular_prism__5.0__10.0__0.5__ |
| the average weight of a group of boys is num__20 kg . after a boy of weight num__32 kg joins the group the average weight of the group goes up by num__1 kg . find the number of boys in the group originally ? <o> a ) num__12 <o> b ) num__11 <o> c ) num__18 <o> d ) num__24 <o> e ) num__10 |
let the number off boys in the group originally be x . total weight of the boys = num__20 x after the boy weighing num__32 kg joins the group total weight of boys = num__20 x + num__32 so num__20 x + num__32 = num__21 ( x + num__1 ) = > x = num__11 . answer : b <eor> b <eos> |
b |
add__20.0__1.0__ subtract__32.0__21.0__ subtract__32.0__21.0__ |
add__20.0__1.0__ subtract__32.0__21.0__ subtract__32.0__21.0__ |
| a shopkeeper buys mangoes at the rate of num__5 a rupee and sells them at num__3 a rupee . find his net profit or loss percent ? <o> a ) num__33 num__1.33333333333 % <o> b ) num__33 num__0.142857142857 % <o> c ) num__40.0 <o> d ) num__32 num__0.333333333333 % <o> e ) num__50 % |
the total number of mangoes bought by the shopkeeper be num__15 . if he buys num__5 a rupee his cp = num__3 he selling at num__3 a rupee his sp = num__5 profit = sp - cp = num__5 - num__3 = num__2 profit percent = num__0.4 * num__100 = num__40.0 answer : c <eor> c <eos> |
c |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| a does a work in num__10 days and b does the same work in num__15 days . in how many days they together will do the same work ? <o> a ) num__5 days <o> b ) num__6 days <o> c ) num__7 days <o> d ) num__8 days <o> e ) num__9 days |
explanation : firstly we will find num__1 day work of both a and b then by adding we can get collective days for them so a ' s num__1 day work = num__0.1 b ' s num__1 day work = num__0.0666666666667 ( a + b ) ' s num__1 day work = ( num__0.1 + num__0.0666666666667 ) = ( num__3 + num__0.0666666666667 ) = num__0.166666666667 so together they can complete work in num__6 days . option b <eor> b <eos> |
b |
divide__1.0__10.0__ divide__1.0__15.0__ add__0.1__0.0667__ round__6.0__ |
divide__1.0__10.0__ divide__1.0__15.0__ add__0.1__0.0667__ round__6.0__ |
| num__12 times a positive integer is more than its square by num__36 then the positive integer is <o> a ) num__6 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__14 |
explanation : let the number be x . then num__12 x = x num__2 + num__36 = > x num__2 - num__12 x + num__36 = num__0 = > ( x - num__6 ) ( x - num__6 ) = num__0 = > x = num__6 answer : a <eor> a <eos> |
a |
divide__12.0__2.0__ divide__12.0__2.0__ |
divide__12.0__2.0__ subtract__12.0__6.0__ |
| pipes p and q would fill a cistern num__18 and num__24 minutes respectively . both pipes being opened find when the first pipe must be turned off so that the cistern may be just filled in num__12 minutes ? <o> a ) num__2 <o> b ) num__9 <o> c ) num__8 <o> d ) num__7 <o> e ) num__5 |
x / num__18 + num__0.5 = num__1 x = num__9 answer : b <eor> b <eos> |
b |
divide__12.0__24.0__ multiply__18.0__0.5__ round__9.0__ |
divide__12.0__24.0__ multiply__18.0__0.5__ divide__9.0__1.0__ |
| machines p and q are two different machines that cover jars in machine shop . when machine p works alone it covers num__5 jars in m hours . when machines p and q work simultaneously at their respective rates they cover num__5 jars in n hours . in terms of m and n how many hours does it take machine q working alone at its constant rate to cover num__5 jars ? <o> a ) m <o> b ) n <o> c ) m + n <o> d ) num__5 n / num__25 - mn <o> e ) m - n |
working rate p = num__5 / m ; working rate pq = ( num__1 / p + num__1 / q ) * n = num__5 or mn / num__5 + n / q = num__5 num__0 r n / q = num__5 - mn / num__5 = num__25 - mn / num__5 q = num__5 n / num__25 - mn d <eor> d <eos> |
d |
round__5.0__ |
divide__5.0__1.0__ |
| free notebooks were distributed equally among children of a class . the number of notebooks each child got was one - eighth of the number of children . had the number of children been half each child would have got num__16 notebooks . total how many notebooks were distributed ? <o> a ) num__412 <o> b ) num__512 <o> c ) num__312 <o> d ) num__112 <o> e ) num__122 |
let total number of children be x . then ( x ) x num__0.125 x = ( x / num__2 ) x num__16 = x = num__64 . number of notebooks = num__0.125 x num__2 = ( num__0.125 x num__64 x num__64 ) = num__512 . answer is b . <eor> b <eos> |
b |
multiply__16.0__0.125__ divide__64.0__0.125__ divide__64.0__0.125__ |
multiply__16.0__0.125__ divide__64.0__0.125__ divide__64.0__0.125__ |
| a man sells two articles for rs . num__2850 each and he gains num__32.0 on the first and loses num__18.0 on the next . find his total gain or loss ? <o> a ) num__9.0 loss <o> b ) num__2.0 loss <o> c ) num__7.0 loss <o> d ) num__6.0 loss <o> e ) num__1.0 loss |
explanation : ( num__32 * num__18 ) / num__100 = num__6.0 loss answer : d <eor> d <eos> |
d |
percent__100.0__6.0__ |
percent__100.0__6.0__ |
| five bells first begin to toll together and then at intervals of num__5 num__10 num__15 num__20 and num__25 seconds respectively . after what interval of time will they toll again together ? <o> a ) num__5 min <o> b ) num__5.5 min <o> c ) num__5.2 min <o> d ) num__6.2 min <o> e ) none |
lcm = num__5.0 = num__5 min answer : a <eor> a <eos> |
a |
percent__20.0__25.0__ |
percent__20.0__25.0__ |
| num__16 men can complete a piece of work in num__25 days . in how many days can num__20 men complete that piece of work ? <o> a ) num__27 <o> b ) num__29 <o> c ) num__20 <o> d ) num__22 <o> e ) num__12 |
num__16 * num__25 = num__20 * x = > x = num__20 days answer : c <eor> c <eos> |
c |
round__20.0__ |
round__20.0__ |
| two cars start at the same time from opposite ends of a highway that is num__333 miles long . one car is riding at num__54 mph and the second car is riding at num__57 mph . how long after they begin will they meet ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
as cars are moving in opposite directions their speeds will be added . so their relative speeds : num__57 + num__54 = num__111 mph total distance to be covered = num__333 miles . time taken would be : num__333 miles / num__111 mph = num__3 hours c is the answer . <eor> c <eos> |
c |
add__54.0__57.0__ divide__333.0__111.0__ round__3.0__ |
add__54.0__57.0__ divide__333.0__111.0__ divide__333.0__111.0__ |
| you have a rock solid dark chocolate bar of size num__2 x num__8 . you need to break it and get num__1 x num__1 pieces of that chocolate . if you can break that bar only along its length or breadth how many times will you have to break to get the num__1 x num__1 pieces ? <o> a ) num__12 times . <o> b ) num__13 times . <o> c ) num__14 times . <o> d ) num__15 times . <o> e ) num__16 times . |
solution : there is a simple logic to solve this question . the size of the chocolate is num__2 x num__8 . thus you need to have num__2 x num__8 = num__16 pieces . every time you break the chocolate you will get one extra piece . thus to get num__16 pieces you must break it ( num__16 - num__1 ) = num__15 times . answer d <eor> d <eos> |
d |
multiply__2.0__8.0__ subtract__16.0__1.0__ multiply__1.0__15.0__ |
multiply__2.0__8.0__ subtract__16.0__1.0__ subtract__16.0__1.0__ |
| the price of an item is discounted num__5 percent on day num__1 of a sale . on day num__2 the item is discounted another num__10 percent and on day num__3 it is discounted an additional num__15 percent . the price of the item on day num__3 is what percentage of the sale price on day num__1 ? <o> a ) num__28.0 <o> b ) num__40.0 <o> c ) num__64.8 <o> d ) num__76.5 <o> e ) num__78 % |
let initial price be num__1000 price in day num__1 after num__5.0 discount = num__950 price in day num__2 after num__10.0 discount = num__855 price in day num__3 after num__15.0 discount = num__726.75 so price in day num__3 as percentage of the sale price on day num__1 will be = num__726.75 / num__950 * num__100 = > num__76.5 answer will definitely be ( d ) <eor> d <eos> |
d |
percent__10.0__1000.0__ percent__100.0__76.5__ |
percent__10.0__1000.0__ percent__100.0__76.5__ |
| a and b started a partnership business investing some amount in the ratio of num__7 : num__5 . c joined then after six months with an amount equal to that of b . in what proportion should the profit at the end of one year be distributed among a b and c ? <o> a ) num__4 : num__9 : num__8 <o> b ) num__5 : num__7 : num__4 <o> c ) num__6 : num__10 : num__5 <o> d ) num__14 : num__10 : num__5 <o> e ) num__5 : num__4 : num__8 |
let the initial investments of a and b be num__7 x and num__5 x . a : b : c = ( num__7 x x num__12 ) : ( num__5 x x num__12 ) : ( num__5 x x num__6 ) = num__84 : num__60 : num__30 = num__14 : num__10 : num__5 . answer : d <eor> d <eos> |
d |
add__7.0__5.0__ multiply__7.0__12.0__ multiply__5.0__12.0__ multiply__5.0__6.0__ divide__84.0__6.0__ divide__60.0__6.0__ divide__84.0__6.0__ |
add__7.0__5.0__ multiply__7.0__12.0__ multiply__5.0__12.0__ multiply__5.0__6.0__ divide__84.0__6.0__ divide__60.0__6.0__ divide__84.0__6.0__ |
| a person can row at num__10 kmph in still water . if the velocity of the current is num__2 kmph and it takes him num__30 hour to row to a place and come back how far is the place ? <o> a ) num__144 km <o> b ) num__30 km <o> c ) num__48 km <o> d ) num__12 km <o> e ) num__15 km |
speed of down stream = num__10 + num__2 = num__12 kmph speed of upstream = num__10 - num__2 = num__8 kmph let the required distance be xkm x / num__12 + x / num__8 = num__30 num__2 x + num__3 x = num__720 x = num__144 km answer is a <eor> a <eos> |
a |
add__10.0__2.0__ subtract__10.0__2.0__ divide__30.0__10.0__ round__144.0__ |
add__10.0__2.0__ subtract__10.0__2.0__ divide__30.0__10.0__ round__144.0__ |
| what will be the ratio of simple interest earned by certain amount at the same rate of interest for num__5 years and that for num__15 years ? <o> a ) num__3 : num__2 <o> b ) num__1 : num__3 <o> c ) num__2 : num__3 <o> d ) num__3 : num__1 <o> e ) num__2 : num__1 |
explanation : simple interest = prt / num__100 here principal ( p ) and rate of interest ( r ) are constants hence simple interest ∝ t required ratio = simple interest for num__5 years / simple interest for num__15 years = t num__1 / t num__2 = num__0.333333333333 = num__0.333333333333 = num__1 : num__3 answer : option b <eor> b <eos> |
b |
percent__1.0__100.0__ |
percent__1.0__100.0__ |
| which of the following can not be the range of a set consisting of num__5 odd multiples of num__11 ? <o> a ) num__44 <o> b ) num__88 <o> c ) num__45 <o> d ) num__176 <o> e ) num__352 |
solution : sequence of num__11 lets say sequence is num__11 n num__11 ( n + num__1 ) num__11 ( n + num__2 ) num__11 ( n + num__3 ) num__11 ( n + num__4 ) so range is num__11 n + num__44 - num__11 n = num__44 if we put the different values of n we will get different but the range will be multiple of num__44 and only num__45 is not multiple of num__44 answer : c <eor> c <eos> |
c |
subtract__5.0__2.0__ subtract__5.0__1.0__ multiply__11.0__4.0__ add__1.0__44.0__ add__1.0__44.0__ |
subtract__5.0__2.0__ subtract__5.0__1.0__ multiply__11.0__4.0__ add__1.0__44.0__ add__1.0__44.0__ |
| a new tower has just been built at the verbico military hospital ; the number of beds available for patients at the hospital is now num__2 times the number available before the new tower was built . currently num__0.333333333333 of the hospital ' s original beds as well as num__0.2 of the beds in the new tower are occupied . for the purposes of renovating the hospital ' s original wing all of the patients in the hospital ' s original beds must be transferred to beds in the new tower . if patients are neither admitted nor discharged during the transfer what fraction of the beds in the new tower will be unoccupied once the transfer is complete ? <o> a ) num__0.366666666667 <o> b ) num__0.483333333333 <o> c ) num__0.566666666667 <o> d ) num__0.633333333333 <o> e ) num__0.466666666667 |
i think e - num__0.466666666667 is the correct answer . here goes : lets assume originally the number of beds = x after the new tower the total combined no of beds = num__2 x so old = x new = x now num__0.333333333333 of x are occupied and num__0.2 of x are occupied which simplifies to ( num__0.2 ) x we are shifting num__0.333333333333 of x to the new ward so there will now be : num__0.333333333333 of x plus num__0.2 of x occupied in the new ward . add them up to get num__0.533333333333 of x there are x beds in new tower so ratio is : ( num__0.533333333333 ) x / x = num__0.533333333333 of x subtract that from num__1.0 of x and you get the number of un - occupied beds to total capacity of new towerb = num__0.466666666667 . e <eor> e <eos> |
e |
add__0.3333__0.2__ add__0.5333__0.4667__ subtract__1.0__0.5333__ |
add__0.3333__0.2__ add__0.5333__0.4667__ subtract__1.0__0.5333__ |
| if p is a positive integer and num__10 p / num__96 is an integer then the minimum number of unique prime factors p could have is b : <o> a ) b = num__1 <o> b ) b = num__2 <o> c ) b = num__3 <o> d ) b = num__4 <o> e ) b = num__5 |
num__10 p / num__96 should resolve to a integer = > num__5 p / num__48 should resolve to an integer . hence p / num__48 should resolve to a integer . num__48 - - > num__2 ^ num__4 * num__3 . hence p should have atleast num__2 ^ num__4 * num__3 for p / num__48 to resolve to a int . the unique prime factors in num__2 ^ num__4 * num__3 is num__2 num__3 and hence the answer is b ( num__2 ) . <eor> b <eos> |
b |
gcd__10.0__96.0__ subtract__5.0__2.0__ gcd__10.0__96.0__ |
divide__10.0__5.0__ subtract__5.0__2.0__ divide__10.0__5.0__ |
| george baked a total of num__150 pizzas for num__7 straight days beginning on saturday . he baked num__0.6 of the pizzas the first day and num__0.6 of the remaining pizzas the second day . if each successive day he baked fewer pizzas than the previous day what is the maximum number of pizzas he could have baked on wednesday ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__4 <o> d ) num__3 <o> e ) num__2 |
num__0.6 of the num__150 pizzas cooked on saturday = num__90 pizzas num__0.6 of the remaining pizzas on sunday = num__36 pizzas we ' re left with ( num__150 - num__90 - num__36 ) = num__24 pizzas for the remaining num__5 days . the prompt tells us that each day has fewer pizzas than the day before it so we ca n ' t have duplicate numbers . m t w th f num__8 num__7 num__6 num__2 num__1 = num__24 w = num__6 b <eor> b <eos> |
b |
multiply__150.0__0.6__ subtract__7.0__5.0__ subtract__7.0__6.0__ subtract__7.0__1.0__ |
multiply__150.0__0.6__ subtract__7.0__5.0__ subtract__7.0__6.0__ subtract__7.0__1.0__ |
| of the people who responded to a market survey num__200 preferred brand x and the rest preferred brand y . if the respondents indicated a preference for brand x over brand y by ratio of num__4 to num__1 how many people responded to the survey ? <o> a ) num__80 <o> b ) num__160 <o> c ) num__250 <o> d ) num__350 <o> e ) num__480 |
ratio = num__4 : num__1 = > num__4 x respondents preferred brand x and x preferred brand y since no . of respondents who preferred brand x = num__200 = > num__4 x = num__200 = > x = num__50 hence total no . of respondents = num__200 + num__50 = num__250 hence c is the answer . <eor> c <eos> |
c |
divide__200.0__4.0__ add__200.0__50.0__ add__200.0__50.0__ |
divide__200.0__4.0__ add__200.0__50.0__ add__200.0__50.0__ |
| a wheel makes num__1000 revolutions in covering a distance of num__88 km . find the radius of the wheel . <o> a ) num__14 <o> b ) num__11 <o> c ) num__28 <o> d ) num__26 <o> e ) num__18 |
explanation : distance covered in one revolution = = num__88 m . answer : a ) num__14 <eor> a <eos> |
a |
round__14.0__ |
round__14.0__ |
| in a group of num__800 people num__0.2 play at least one instrument num__128 play two or more . what is the probability that one student play exactly one instrument ? <o> a ) num__0.016 <o> b ) num__0.024 <o> c ) c ) num__0.08 <o> d ) num__0.12 <o> e ) num__0.04 |
p ( playing num__2 or more instruments ) = num__0.16 = num__0.16 . then the probability of playing exactly one instrument is given by : p ( playing num__1 or more instruments ) - p ( playing num__2 or more instruments ) = num__0.2 - num__0.16 = num__0.04 . answer e . <eor> e <eos> |
e |
divide__128.0__800.0__ subtract__0.2__0.16__ subtract__0.2__0.16__ |
divide__128.0__800.0__ subtract__0.2__0.16__ subtract__0.2__0.16__ |
| the average of num__20 numbers is zero . of them at the most how many may be greater than zero ? <o> a ) num__0 <o> b ) num__9 <o> c ) num__8 <o> d ) num__7 <o> e ) num__6 |
average of num__20 numbers = num__0 . sum of num__20 numbers = ( num__0 * num__20 ) = num__0 . it is quite possible that num__19 of these numbers may be positive and if their sum is a then num__20 th number is ( - a ) . answer : a <eor> a <eos> |
a |
multiply__20.0__0.0__ |
multiply__20.0__0.0__ |
| the time taken by mr . dhoni to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is num__72 kmph find the speed of the stream ? <o> a ) num__24 kmph <o> b ) num__25 kmph <o> c ) num__26 kmph <o> d ) num__30 kmph <o> e ) num__28 kmph |
the ratio of the times taken is num__2 : num__1 . the ratio of the speed of the boat in still water to the speed of the stream = ( num__2 + num__1 ) / ( num__2 - num__1 ) = num__3.0 = num__3 : num__1 speed of the stream = num__24.0 = num__24 kmph . answer : a <eor> a <eos> |
a |
add__1.0__2.0__ divide__72.0__3.0__ round__24.0__ |
add__1.0__2.0__ divide__72.0__3.0__ divide__72.0__3.0__ |
| p software has coding line num__5.0 more than n n software has coding line num__0.5 more than m . m software has num__100 lines of coding . find p lines . <o> a ) num__106 <o> b ) num__107 <o> c ) num__108 <o> d ) num__109 <o> e ) num__158 |
m s / w has num__100 line of code n s / w has = num__100 + num__100 * num__0.5 = num__150 line of code p s / w num__5.0 more n ' code num__150 + num__7.5 = num__157.5 or num__158 line of code answer : e <eor> e <eos> |
e |
add__7.5__150.0__ add__0.5__157.5__ add__0.5__157.5__ |
add__7.5__150.0__ add__0.5__157.5__ add__0.5__157.5__ |
| a dishonest dealer professes to sell goods at the cost price but uses a weight of num__850 grams per kg what is his percent ? <o> a ) num__28.0 <o> b ) num__25.0 <o> c ) num__55.0 <o> d ) num__17.6 <o> e ) num__55 % |
num__850 - - - num__150 num__100 - - - ? = > num__17.6 answer : d <eor> d <eos> |
d |
percent__100.0__17.6__ |
percent__100.0__17.6__ |
| num__475.124 x num__15.98 ÷ num__8.001 + num__33.33 = ? <o> a ) num__983.578 <o> b ) num__659.121 <o> c ) num__781.189 <o> d ) num__656.112 <o> e ) num__456.512 |
explanation : ? = num__475.124 x ( num__15.98 ÷ num__8.001 ) + num__33.33 ≈ ( num__475.124 x num__2.0 ) + num__33.33 ≈ num__950.248 + num__33.33 ≈ num__983.578 answer : option a <eor> a <eos> |
a |
multiply__475.124__2.0__ add__33.33__950.248__ add__33.33__950.248__ |
multiply__475.124__2.0__ add__33.33__950.248__ add__33.33__950.248__ |
| can you deduce the pattern and find the next number in the series ? num__3 num__2481 num__192 __ ? <o> a ) num__276 <o> b ) num__277 <o> c ) num__278 <o> d ) num__279 <o> e ) num__375 |
solution : num__375 explanation : num__3 = num__1 ^ num__1 ^ num__1 + num__1 ^ num__1 ^ num__1 + num__1 ^ num__1 ^ num__1 num__24 = num__2 ^ num__2 ^ num__2 + num__2 ^ num__2 ^ num__2 + num__2 ^ num__2 ^ num__2 num__81 = num__3 ^ num__3 ^ num__3 + num__3 ^ num__3 ^ num__3 + num__3 ^ num__3 ^ num__3 num__192 = num__4 ^ num__4 ^ num__4 + num__4 ^ num__4 ^ num__4 + num__4 ^ num__4 ^ num__4 thus the next number will be num__375 = num__5 ^ num__5 ^ num__5 + num__5 ^ num__5 ^ num__5 + num__5 ^ num__5 ^ num__5 answer e <eor> e <eos> |
e |
subtract__3.0__1.0__ add__3.0__1.0__ add__3.0__2.0__ multiply__1.0__375.0__ |
subtract__3.0__1.0__ add__3.0__1.0__ add__3.0__2.0__ power__375.0__1.0__ |
| if a : b = num__17 : num__5 b : c = num__13 : num__11 find a : b : c ? <o> a ) num__21 : num__65 : num__55 <o> b ) num__121 : num__65 : num__55 <o> c ) num__221 : num__65 : num__55 <o> d ) num__221 : num__75 : num__55 <o> e ) num__221 : num__65 : num__45 |
a : b = num__17 : num__5 b : c = num__13 : num__11 a : b : c = num__221 : num__65 : num__55 answer : c <eor> c <eos> |
c |
multiply__17.0__13.0__ multiply__5.0__13.0__ multiply__5.0__11.0__ multiply__17.0__13.0__ |
multiply__17.0__13.0__ multiply__5.0__13.0__ multiply__5.0__11.0__ multiply__17.0__13.0__ |
| two goods trains each num__820 m long are running in opposite directions on parallel tracks . their speeds are num__45 km / hr and num__30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? <o> a ) num__228 <o> b ) num__27.32 <o> c ) num__76.6 <o> d ) num__78.72 <o> e ) num__21 |
relative speed = num__45 + num__30 = num__75 km / hr . num__75 * num__0.277777777778 = num__20.8333333333 m / sec . distance covered = num__820 + num__820 = num__1640 m . required time = num__1640 * num__0.048 = num__78.72 sec . answer : d <eor> d <eos> |
d |
add__45.0__30.0__ multiply__1640.0__0.048__ round__78.72__ |
add__45.0__30.0__ multiply__1640.0__0.048__ multiply__1640.0__0.048__ |
| a car is traveling num__54 kilometers per hour . how many meters does the car travel in one minute ? <o> a ) num__1250 meters / minute <o> b ) num__1350 meters / minute <o> c ) num__900 meters / minute <o> d ) num__950 meters / minute <o> e ) num__1650 meters / minute |
convert hour into minutes ( num__1 hour = num__60 minutes ) and kilometers into meters ( num__1 km = num__1000 m ) and simplify num__54 kilometers per hour = num__54 km / hr = ( num__54 * num__1000 meters ) / ( num__60 minutes ) = num__900 meters / minute correct answer c <eor> c <eos> |
c |
hour_to_min_conversion__ round__900.0__ |
hour_to_min_conversion__ divide__900.0__1.0__ |
| the mean of num__50 observations is num__100 . but later he found that there is decrements of num__13 from each observations . what is the the updated mean is ? <o> a ) num__87 <o> b ) num__97 <o> c ) num__67 <o> d ) num__57 <o> e ) num__46 |
num__87 answer is a <eor> a <eos> |
a |
subtract__100.0__13.0__ subtract__100.0__13.0__ |
subtract__100.0__13.0__ subtract__100.0__13.0__ |
| the tax on a commodity is diminished by num__20.0 and its consumption increased by num__15.0 . the effect on revenue is ? <o> a ) num__7.0 decrease <o> b ) num__8.0 decrease <o> c ) num__9.0 decrease <o> d ) num__6.0 decrease <o> e ) num__4.0 decrease |
num__100 * num__100 = num__10000 num__80 * num__115 = num__9200 - - - - - - - - - - - num__10000 - - - - - - - - - - - num__800 num__100 - - - - - - - - - - - ? = > num__8.0 decrease answer : b <eor> b <eos> |
b |
subtract__100.0__20.0__ add__15.0__100.0__ multiply__80.0__115.0__ subtract__10000.0__9200.0__ divide__800.0__100.0__ divide__800.0__100.0__ |
subtract__100.0__20.0__ add__15.0__100.0__ multiply__80.0__115.0__ subtract__10000.0__9200.0__ divide__800.0__100.0__ divide__800.0__100.0__ |
| two - third of a positive number and num__0.296296296296 of its reciprocal are equal . the number is : <o> a ) num__0.416666666667 <o> b ) num__2.4 <o> c ) num__0.173611111111 <o> d ) num__0.666666666667 <o> e ) num__5.84 |
let the number be x . then num__0.666666666667 x = num__0.296296296296 * num__1 / x x num__2 = num__0.296296296296 * num__1.5 = num__0.444444444444 = num__0.444444444444 x = num__0.666666666667 answer : d <eor> d <eos> |
d |
divide__0.2963__0.6667__ reverse__1.5__ |
divide__0.2963__0.6667__ reverse__1.5__ |
| in num__1986 the book value of a certain car was num__0.666666666667 of the original price and in num__1988 its book value was num__0.5 of the original purchase price . by what percent did the book value for this car decre ase from num__1986 to num__1988 ? <o> a ) num__16 num__0.666666666667 % <o> b ) num__25 percent <o> c ) num__33 num__0.333333333333 % <o> d ) num__50.0 <o> e ) num__75 % |
num__1986 ; num__0.666666666667 num__1988 ; num__0.5 % decrease = change / original * num__100 num__0.666666666667 − num__0.5 / num__0.666666666667 ∗ num__100 num__0.166666666667 ∗ num__1.5 ∗ num__100 = num__14 ∗ num__100 = num__25 answer : ` ` b ' ' <eor> b <eos> |
b |
percent__25.0__100.0__ |
percent__25.0__100.0__ |
| in how many ways can the letters of the word ‘ technology ’ be arranged ? <o> a ) num__1804400 <o> b ) num__1814400 <o> c ) num__1714400 <o> d ) num__1704400 <o> e ) num__1724400 |
the word ‘ technology ’ contains num__10 letters namely num__2 o num__1 t num__1 e num__1 c num__1 h num__1 n num__1 l num__1 g num__1 y . therefore required number of ways = num__10 ! / ( num__2 ! ) ( num__1 ! ) ( num__1 ! ) ( num__1 ! ) ( num__1 ! ) ( num__1 ! ) ( num__1 ! ) ( num__1 ! ) ( num__1 ! ) = num__10 ! / num__2 ! = num__10 * num__9 * num__8 * num__7 * num__6 * num__5 * num__4 * num__3 * num__2 * num__0.5 * num__1 = num__1814400 answer : b <eor> b <eos> |
b |
coin_space__ die_space__ vowel_space__ choose__9.0__4.0__ choose__9.0__4.0__ |
coin_space__ die_space__ vowel_space__ choose__9.0__4.0__ choose__9.0__4.0__ |
| a man has rs . num__480 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__45 <o> b ) num__60 <o> c ) num__75 <o> d ) num__90 <o> e ) num__105 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__480 num__16 x = num__480 x = num__30 . hence total number of notes = num__3 x = num__90 . answer = d <eor> d <eos> |
d |
divide__480.0__16.0__ divide__30.0__10.0__ multiply__3.0__30.0__ multiply__3.0__30.0__ |
divide__480.0__16.0__ divide__30.0__10.0__ multiply__3.0__30.0__ multiply__3.0__30.0__ |
| if f ( x ) = num__3 x ^ num__4 - num__4 x ^ num__3 - num__2 x ^ num__2 + num__5 x then f ( - num__1 ) = <o> a ) - num__4 <o> b ) - num__2 <o> c ) num__0 <o> d ) num__2 <o> e ) num__4 |
f ( - num__1 ) = num__3 ( - num__1 ) ^ num__4 - num__4 ( - num__1 ) ^ num__3 - num__2 ( - num__1 ) ^ num__2 + num__5 ( - num__1 ) = num__3 + num__4 - num__2 - num__5 = num__0 the answer is c . <eor> c <eos> |
c |
multiply__3.0__0.0__ |
multiply__3.0__0.0__ |
| the speed of the boat in still water in num__12 kmph . it can travel downstream through num__45 kms in num__3 hrs . in what time would it cover the same distance upstream ? <o> a ) num__6 hours <o> b ) num__5 hours <o> c ) num__9 hours <o> d ) num__15 hours <o> e ) num__23 hours |
speed of the boat in still water = num__12 km / hr speed downstream = num__45 ⁄ num__3 = num__15 km / hr speed of the stream = num__15 - num__12 = num__3 km / hr speed upstream = num__12 - num__3 = num__9 km / hr answer is b time taken to cover num__45 km upstream = num__45 ⁄ num__9 = num__5 hours <eor> b <eos> |
b |
add__12.0__3.0__ subtract__12.0__3.0__ divide__45.0__9.0__ round__5.0__ |
divide__45.0__3.0__ subtract__12.0__3.0__ divide__45.0__9.0__ divide__45.0__9.0__ |
| of num__440 surveyed students num__20.0 of those who read book a also read book b and num__25.0 of those who read book b also read book a . if each student read at least one of the books what is the difference between the number of students who read only book a and the number of students who read only book b ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__30 <o> d ) num__35 <o> e ) num__55 |
say the number of students who read book a is a and the number of students who read book b is b . given that num__20.0 of those who read book a also read book b and num__25.0 of those who read book b also read book a so the number of students who read both books is num__0.2 a = num__0.25 b - - > a = num__1.25 b . since each student read at least one of the books then { total } = { a } + { b } - { both } - - > num__440 = num__1.25 b + b - num__0.25 b - - > b = num__220 a = num__1.25 b = num__275 and { both } = num__0.25 b = num__55 . the number of students who read only book a is { a } - { both } = num__125 - num__25 = num__100 ; the number of students who read only book b is { b } - { both } = num__275 - num__55 - num__220 ; the difference is num__220 - num__55 = num__165 . answer : e . <eor> e <eos> |
e |
percent__20.0__275.0__ percent__20.0__275.0__ |
percent__20.0__275.0__ percent__20.0__275.0__ |
| the ratio of boarders to day students at a school was originally num__5 to num__12 . however after a number of new boarders join the initial num__220 boarders the ratio changed to num__1 to num__2 . if no boarders became day students and vice versa and no students left the school how many new boarders joined the school ? <o> a ) num__44 <o> b ) num__64 <o> c ) num__70 <o> d ) num__80 <o> e ) num__84 |
let x be the number of new boarders . the ratio changed from num__5 : num__12 up to num__1 : num__2 = num__6 : num__12 . num__220 / ( num__220 + x ) = num__0.833333333333 x = num__44 the answer is a . <eor> a <eos> |
a |
add__5.0__1.0__ divide__5.0__6.0__ divide__220.0__5.0__ round__44.0__ |
add__5.0__1.0__ divide__5.0__6.0__ divide__220.0__5.0__ divide__220.0__5.0__ |
| if each side of a square is increased by num__20.0 find the percentage change in its area ? <o> a ) num__40.0 <o> b ) num__42.0 <o> c ) num__46.0 <o> d ) num__44.0 <o> e ) num__38 % |
let each side of the square be a then area = a x a new side = num__120 a / num__100 = num__6 a / num__5 new area = ( num__6 a x num__6 a ) / ( num__5 x num__5 ) = ( num__36 a ² / num__25 ) increased area = = ( num__36 a ² / num__25 ) - a ² increase % = [ ( num__11 a ² / num__25 ) x ( num__1 / a ² ) x num__100 ] % = num__44.0 answer : d <eor> d <eos> |
d |
percent__20.0__5.0__ percent__44.0__100.0__ |
percent__20.0__5.0__ percent__44.0__100.0__ |
| a father and his son are waiting at a bus stop in the evening . there is a lamp post behind them . the lamp post the father and his son stand on the same straight line . the father observes that the shadows of his head and his son ’ s head are incident at the same point on the ground . if the heights of the lamp post the father and his son are num__6 metres num__1.8 metres and num__0.9 metres respectively and the father is standing num__2.1 metres away from the post then how far ( in metres ) is the son standing from his father ? <o> a ) num__0.9 <o> b ) num__0.75 <o> c ) num__0.6 <o> d ) num__0.45 <o> e ) none of these |
explanation : l is the lamp post position f is father and s is son ’ s position . x is the point where the shadow falls . ld = num__0.9 = son ’ s height lb = num__1.8 = father ’ s height . so ab = num__6 – num__1.8 = num__4.2 also bc = lf = num__2.1 we observe that ∆ abc ~ ∆ ade ( two triangles are similar ) . hence the corresponding sides are proportional . so ab / ad = dc / de . num__4.2 / num__5.1 = num__2.1 / de . de = num__5.1 * num__2.1 / num__4.2 = num__5.1 / num__2 = num__2.55 . ls = de = num__2.55 fs = ls – lf = num__2.55 – num__2.1 = num__0.45 . answer : d <eor> d <eos> |
d |
subtract__6.0__1.8__ subtract__6.0__0.9__ divide__1.8__0.9__ divide__5.1__2.0__ divide__0.9__2.0__ round__0.45__ |
subtract__6.0__1.8__ subtract__6.0__0.9__ divide__1.8__0.9__ divide__5.1__2.0__ divide__0.9__2.0__ divide__0.9__2.0__ |
| two goods trains each num__500 m long are running in opposite directions on parallel tracks . their speeds are num__45 km / hr and num__30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? <o> a ) num__77 sec <o> b ) num__66 sec <o> c ) num__48 sec <o> d ) num__55 sec <o> e ) num__45 sec |
relative speed = num__45 + num__30 = num__75 km / hr . num__75 * num__0.277777777778 = num__20.8333333333 m / sec . distance covered = num__500 + num__500 = num__1000 m . required time = num__1000 * num__0.048 = num__48 sec . answer : c <eor> c <eos> |
c |
add__45.0__30.0__ multiply__1000.0__0.048__ round__48.0__ |
add__45.0__30.0__ multiply__1000.0__0.048__ multiply__1000.0__0.048__ |
| linda spent num__0.75 of her savings on furniture and the rest on a tv . if the tv cost her $ num__300 what were her original savings ? <o> a ) $ num__1200 <o> b ) $ num__1300 <o> c ) $ num__1400 <o> d ) $ num__1800 <o> e ) $ num__1900 |
if linda spent num__0.75 of her savings on furnitute the rest num__1.0 - num__0.75 = num__0.25 on a tv but the tv cost her $ num__300 . so num__0.25 of her savings is $ num__300 . so her original savings are num__4 times $ num__300 = $ num__1200 correct answer a <eor> a <eos> |
a |
subtract__1.0__0.75__ reverse__0.25__ multiply__300.0__4.0__ multiply__300.0__4.0__ |
subtract__1.0__0.75__ reverse__0.25__ multiply__300.0__4.0__ multiply__300.0__4.0__ |
| if a : b = num__12 : num__6 b : c = num__9 : num__7 find a : b : c ? <o> a ) num__118 : num__53 : num__66 <o> b ) num__138 : num__73 : num__46 <o> c ) num__128 : num__43 : num__36 <o> d ) num__108 : num__63 : num__56 <o> e ) num__148 : num__83 : num__76 |
a : b = num__12 : num__7 b : c = num__9 : num__8 a : b : c = num__108 : num__63 : num__56 answer : d <eor> d <eos> |
d |
multiply__12.0__9.0__ multiply__9.0__7.0__ multiply__7.0__8.0__ multiply__12.0__9.0__ |
multiply__12.0__9.0__ multiply__9.0__7.0__ multiply__7.0__8.0__ multiply__12.0__9.0__ |
| the average salary per head of the entire staff of an office including the officers and clerks is rs . num__90 . the average salary of officers is rs . num__500 and that of the clerks is rs . num__84 . if the number of officers is num__2 find the number of officers in the office ? <o> a ) num__620 <o> b ) num__720 <o> c ) num__820 <o> d ) num__920 <o> e ) num__1020 |
num__500 num__84 \ / num__90 / \ num__6 num__410 num__3 : num__205 num__3 - > num__12 num__205 - > ? num__820 answer : c <eor> c <eos> |
c |
subtract__90.0__84.0__ subtract__500.0__90.0__ divide__6.0__2.0__ divide__410.0__2.0__ multiply__2.0__6.0__ multiply__2.0__410.0__ multiply__2.0__410.0__ |
subtract__90.0__84.0__ subtract__500.0__90.0__ divide__6.0__2.0__ divide__410.0__2.0__ multiply__2.0__6.0__ multiply__2.0__410.0__ multiply__2.0__410.0__ |
| how many different subsets of the set { num__0 num__1 num__2 num__3 } do not contain num__0 ? <o> a ) a . num__3 <o> b ) b . num__7 <o> c ) c . num__8 <o> d ) d . num__2 <o> e ) e . num__4 |
number of subset since we have num__3 digits other than num__0 we can take any numbers from the set of num__3 to make a subset . also it is a matter of selection and not arrangement . so we will consider combinations . num__3 c num__1 + num__3 c num__2 + num__3 c num__3 = num__7 and one set is the null set having no elements in it so num__7 + num__1 = num__8 . answer c . <eor> c <eos> |
c |
choose__8.0__7.0__ |
choose__8.0__7.0__ |
| if two numbers are in the ratio num__3 : num__4 . if num__10 is added to both of the numbers then the ratio becomes num__4 : num__5 then find the smallest number ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__30 <o> d ) num__40 <o> e ) num__50 |
explanation : num__3 : num__4 num__3 x + num__10 : num__4 x + num__10 = num__4 : num__5 num__5 [ num__3 x + num__10 ] = num__4 [ num__4 x + num__10 ] num__15 x + num__50 = num__16 x + num__40 num__16 x - num__15 x = num__50 - num__40 x = num__10 then smallest number is = num__3 num__3 x = num__30 answer : option c <eor> c <eos> |
c |
multiply__3.0__5.0__ multiply__10.0__5.0__ multiply__4.0__10.0__ multiply__3.0__10.0__ multiply__3.0__10.0__ |
add__10.0__5.0__ multiply__10.0__5.0__ subtract__50.0__10.0__ subtract__40.0__10.0__ subtract__40.0__10.0__ |
| the average weight of num__8 person ' s increases by num__2.5 kg when a new person comes in place of one of them weighing num__40 kg . what is the weight of the new person ? <o> a ) num__75 <o> b ) num__65 <o> c ) num__85 <o> d ) num__95 <o> e ) num__60 |
total increase in weight = num__8 × num__2.5 = num__20 if x is the weight of the new person total increase in weight = x − num__40 = > num__20 = x - num__40 = > x = num__20 + num__40 = num__60 answer is e . <eor> e <eos> |
e |
multiply__8.0__2.5__ add__40.0__20.0__ add__40.0__20.0__ |
multiply__8.0__2.5__ add__40.0__20.0__ add__40.0__20.0__ |
| a rectangular box has two sides whose lengths are num__3 centimeters and num__9 centimeters and a volume of num__135 cm num__3 . what is the area of its largest side ? <o> a ) num__27 cm num__2 * square <o> b ) num__36 cm num__2 * square <o> c ) num__39 cm num__2 * square <o> d ) num__45 centimeter square <o> e ) num__48 cm num__2 * square |
the box has dimensions of num__3 and num__9 and a volume of num__135 so plug these values into the formula for the volume of a box : v = lwh = > num__135 = ( num__3 ) ( num__9 ) h = > num__135 = num__27 h = > num__5 h so the remaining dimension of the box is num__5 . the two longest dimensions are num__5 and num__9 so the area of the largest side is num__5 × num__9 = num__45 correct answer d ) num__45 cm num__2 * square <eor> d <eos> |
d |
volume_cube__3.0__ multiply__9.0__5.0__ multiply__9.0__5.0__ |
multiply__3.0__9.0__ multiply__9.0__5.0__ multiply__9.0__5.0__ |
| one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill tank in num__39 min then the slower pipe alone will be able to fill the tank in ? <o> a ) num__229 <o> b ) num__156 <o> c ) num__144 <o> d ) num__128 <o> e ) num__121 |
let the slower pipe alone fill the tank in x min . then faster pipe will fill it in x / num__3 min . num__1 / x + num__3 / x = num__0.025641025641 num__4 / x = num__0.025641025641 = > x = num__156 min . answer : b <eor> b <eos> |
b |
divide__1.0__39.0__ add__1.0__3.0__ multiply__39.0__4.0__ round__156.0__ |
divide__1.0__39.0__ add__1.0__3.0__ multiply__39.0__4.0__ divide__156.0__1.0__ |
| a boy is travelling from his home to school at num__5 km / hr and reached num__7 min late . next day he traveled at num__10 km / hr and reached num__8 min early . distance between home and school ? <o> a ) num__2.5 km <o> b ) num__3.5 km <o> c ) num__4.5 km <o> d ) num__5.5 km <o> e ) num__6.5 km |
let the distance be x t num__1 = x / num__5 hr t num__2 = x / num__10 hr difference in time = num__7 + num__8 = num__15 = num__0.25 hr x / num__5 - x / num__10 = num__0.25 x / num__10 = num__0.25 x = num__2.5 km answer is a <eor> a <eos> |
a |
subtract__8.0__7.0__ subtract__7.0__5.0__ add__5.0__10.0__ divide__2.0__8.0__ divide__5.0__2.0__ round__2.5__ |
subtract__8.0__7.0__ subtract__7.0__5.0__ add__5.0__10.0__ divide__2.0__8.0__ divide__5.0__2.0__ divide__5.0__2.0__ |
| two spherical balls lie on the ground touching . if one of the balls has a radius of num__8 cm and the point of contact is num__12 cm above the ground what is the radius of the other ball ( in centimeters ) ? <o> a ) num__20 <o> b ) num__24 <o> c ) num__28 <o> d ) num__32 <o> e ) num__36 |
a straight line will join the two centers and the point of contact thus making similar triangles . num__0.5 = ( r - num__12 ) / r num__4 r = num__8 r - num__96 r = num__24 the answer is b . <eor> b <eos> |
b |
multiply__8.0__0.5__ multiply__8.0__12.0__ divide__12.0__0.5__ round__24.0__ |
subtract__12.0__8.0__ multiply__8.0__12.0__ divide__12.0__0.5__ divide__12.0__0.5__ |
| simplify num__70 â ˆ ’ [ num__5 â ˆ ’ ( num__6 + num__2 ( num__7 â ˆ ’ num__8 â ˆ ’ num__5 Â ¯ Â ¯ Â ¯ Â ¯ Â ¯ Â ¯ Â ¯ Â ¯ Â ¯ Â ¯ Â ¯ ) ) ] <o> a ) num__23 <o> b ) num__25 <o> c ) num__28 <o> d ) num__79 <o> e ) num__32 |
explanation : = num__70 â ˆ ’ [ num__5 â ˆ ’ ( num__6 + num__2 ( num__7 â ˆ ’ num__8 + num__5 ) ) ] ( please check due to overline sign has been changed ) = num__70 â ˆ ’ [ num__5 â ˆ ’ ( num__6 + num__2 Ã — num__4 ) ) ] = num__70 â ˆ ’ [ â ˆ ’ num__9 ] = num__70 + num__9 = num__79 option d <eor> d <eos> |
d |
subtract__6.0__2.0__ add__5.0__4.0__ add__70.0__9.0__ add__70.0__9.0__ |
subtract__6.0__2.0__ add__5.0__4.0__ add__70.0__9.0__ add__70.0__9.0__ |
| find the ratio in which rice at rs . num__7.20 a kg be mixed with rice at rs . num__5.70 a kg to produce a mixture worth rs . num__6.30 a kg . <o> a ) num__1 : num__3 <o> b ) num__2 : num__3 <o> c ) num__3 : num__4 <o> d ) num__4 : num__5 <o> e ) num__5 : num__6 |
cost of num__1 kg of num__1 st kind num__720 p cost of num__1 kg of num__2 nd kind num__570 p mean price num__630 p required ratio = num__60 : num__90 = num__2 : num__3 . answer : b <eor> b <eos> |
b |
subtract__630.0__570.0__ subtract__720.0__630.0__ add__1.0__2.0__ multiply__1.0__2.0__ |
subtract__630.0__570.0__ subtract__720.0__630.0__ add__1.0__2.0__ multiply__1.0__2.0__ |
| find out the wrong term . num__2 num__5 num__10 num__50 num__500 num__5000 <o> a ) num__5 <o> b ) num__10 <o> c ) num__50 <o> d ) num__5000 <o> e ) num__500 |
num__2 * num__5 = num__10 num__5 * num__10 = num__50 num__10 * num__50 = num__500 num__50 * ? = num__5000 answer : d <eor> d <eos> |
d |
multiply__10.0__500.0__ |
multiply__10.0__500.0__ |
| what is the average of num__0.5 num__0.4 and num__0.1 ? <o> a ) num__0.2 <o> b ) num__0.3 <o> c ) num__0.4 <o> d ) num__0.5 <o> e ) num__0.266666666667 |
sum of num__1 / num__22 / num__5.1 = num__0.8 average = num__0.8 * num__0.333333333333 = num__0.266666666667 answer e num__0.266666666667 <eor> e <eos> |
e |
divide__0.4__0.5__ multiply__0.2667__1.0__ |
divide__0.4__0.5__ multiply__0.2667__1.0__ |
| num__10 ^ num__23 - num__7 is divided by num__6 remainder is ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
anything power by num__10 we got num__10 and some zeros after that . ( i . e num__1000000000000000 . . . . ) on that we minus num__7 means we must get last two digits is num__93 . and num__93 is divided by num__6 we get reminder num__3 . answer : a <eor> a <eos> |
a |
subtract__10.0__7.0__ subtract__10.0__7.0__ |
subtract__10.0__7.0__ subtract__10.0__7.0__ |
| arjun started a business with rs . num__20000 and is joined afterwards by anoop with rs . num__30 num__000 . after how many months did anoop join if the profits at the end of the year are divided equally ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
suppose anoop joined after num__3 months . then num__20000 * num__12 = num__30000 * ( num__12 – x ) = > num__120000 = num__30000 x = > x = num__4 . answer : b <eor> b <eos> |
b |
divide__120000.0__30000.0__ divide__120000.0__30000.0__ |
divide__120000.0__30000.0__ divide__120000.0__30000.0__ |
| john ' s bank ' s saving amount is decreased num__12.0 due to loan payment and current balance is rs . num__44000 . find the actual balance before deduction ? <o> a ) num__8000 <o> b ) num__50000 <o> c ) num__9000 <o> d ) num__9500 <o> e ) num__10000 |
num__12.0 decreased num__88.0 balance = num__44000 num__100.0 = num__500.0 * num__100 = num__50000 answer : b <eor> b <eos> |
b |
percent__100.0__50000.0__ |
percent__100.0__50000.0__ |
| in a two - digit if it is known that its unit ' s digit exceeds its ten ' s digit by num__2 and that the product of the given number and the sum of its digits is equal to num__144 then the number is ? <o> a ) num__12 <o> b ) num__24 <o> c ) num__36 <o> d ) num__38 <o> e ) num__40 |
let the ten ' s digit be x . then unit ' s digit = x + num__2 . number = num__10 x + ( x + num__2 ) = num__11 x + num__2 . sum of digits = x + ( x + num__2 ) = num__2 x + num__2 . ( num__11 x + num__2 ) ( num__2 x + num__2 ) = num__144 num__22 x num__2 + num__26 x - num__140 = num__0 num__11 x num__2 + num__13 x - num__70 = num__0 ( x - num__2 ) ( num__11 x + num__35 ) = num__0 x = num__2 . hence required number = num__11 x + num__2 = num__24 . option b <eor> b <eos> |
b |
multiply__2.0__11.0__ add__2.0__11.0__ divide__140.0__2.0__ divide__70.0__2.0__ add__2.0__22.0__ add__2.0__22.0__ |
multiply__2.0__11.0__ add__2.0__11.0__ divide__140.0__2.0__ add__13.0__22.0__ add__2.0__22.0__ add__2.0__22.0__ |
| num__3 a = num__4 b = num__5 c a : b : c = ? <o> a ) num__12 : num__15 : num__22 <o> b ) num__20 : num__15 : num__12 <o> c ) num__15 : num__20 : num__12 <o> d ) num__12 : num__20 : num__15 <o> e ) none of these |
explanation : solution : num__3 a = num__4 b = num__5 c a : b : c = num__0.333333333333 : num__0.25 : num__0.2 = > num__0.333333333333 : num__0.25 : num__0.2 = num__20 : num__15 : num__12 answer : b <eor> b <eos> |
b |
reverse__3.0__ reverse__4.0__ reverse__5.0__ multiply__4.0__5.0__ multiply__3.0__5.0__ multiply__3.0__4.0__ multiply__4.0__5.0__ |
reverse__3.0__ reverse__4.0__ reverse__5.0__ multiply__4.0__5.0__ multiply__3.0__5.0__ multiply__3.0__4.0__ multiply__4.0__5.0__ |
| find the value of y from ( num__12 ) ^ num__3 x num__6 ^ num__4 ÷ num__432 = y ? <o> a ) num__2435 <o> b ) num__1243 <o> c ) num__1509 <o> d ) num__3456 <o> e ) num__5184 |
given exp . = ( num__12 ) num__3 x num__64 = ( num__12 ) num__3 x num__64 = ( num__12 ) num__2 x num__62 = ( num__72 ) num__2 = num__5184 num__432 num__12 x num__62 e <eor> e <eos> |
e |
divide__12.0__6.0__ subtract__64.0__2.0__ multiply__12.0__6.0__ multiply__12.0__432.0__ multiply__12.0__432.0__ |
divide__12.0__6.0__ subtract__64.0__2.0__ multiply__12.0__6.0__ multiply__12.0__432.0__ multiply__12.0__432.0__ |
| reena took a loan of $ . num__1200 with simple interest for as many years as the rate of interest . if she paid $ num__300 as interest at the end of the loan period what was the rate of interest ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__18 <o> d ) can not be determined <o> e ) none of these |
let rate = r % and time = r years . then num__1200 x r x r / num__100 = num__300 num__12 r num__2 = num__300 r num__2 = num__25 r = num__5 . answer : a <eor> a <eos> |
a |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| the sum of the ages of num__5 children born at the intervals of num__3 years each is num__50 years . what is the age of the youngest child ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__1 <o> d ) num__7 <o> e ) num__4 |
let the ages of the children be x ( x + num__3 ) ( x + num__6 ) ( x + num__9 ) and ( x + num__12 ) years . then x + ( x + num__3 ) + ( x + num__6 ) + ( x + num__9 ) + ( x + num__12 ) = num__50 num__5 x = num__20 = > x = num__4 . age of youngest child = x = num__4 years . answer is e . <eor> e <eos> |
e |
add__3.0__6.0__ add__3.0__9.0__ subtract__9.0__5.0__ subtract__9.0__5.0__ |
add__3.0__6.0__ add__3.0__9.0__ subtract__9.0__5.0__ subtract__9.0__5.0__ |
| the angle between the minute hand and the hour hand of a clock when the time is num__8.30 is <o> a ) num__35 ° <o> b ) num__65 ° <o> c ) num__45 ° <o> d ) num__75 ° <o> e ) num__95 ° |
angle between hands of a clock when the minute hand is behind the hour hand the angle between the two hands at m minutes past h ' o clock = num__30 ( h − m / num__5 ) + m / num__2 degree when the minute hand is ahead of the hour hand the angle between the two hands at m minutes past h ' o clock = num__30 ( m / num__5 − h ) − m / num__2 degree here h = num__8 m = num__30 and minute hand is behind the hour hand . hence the angle = num__30 ( h − m / num__5 ) + m / num__2 = num__30 ( num__8 − num__6.0 ) + num__15.0 = num__30 ( num__8 − num__6 ) + num__15 = num__30 × num__2 + num__15 = num__75 ° answer is d . <eor> d <eos> |
d |
subtract__8.0__2.0__ divide__30.0__2.0__ multiply__5.0__15.0__ round__75.0__ |
divide__30.0__5.0__ divide__30.0__2.0__ multiply__5.0__15.0__ round__75.0__ |
| the sum of the two numbers is num__12 and their product is num__35 . what is the sum of the reciprocals of these numbers ? <o> a ) num__0.342857142857 <o> b ) num__0.0285714285714 <o> c ) num__4.375 <o> d ) num__0.21875 <o> e ) none of these |
let the numbers be a and b . then a + b = num__12 and ab = num__35 . a + b / ab = num__0.342857142857 ; ( num__1 / b + num__1 / a ) = num__0.342857142857 sum of reciprocals of given numbers = num__0.342857142857 . correct option : a <eor> a <eos> |
a |
divide__12.0__35.0__ divide__12.0__35.0__ |
divide__12.0__35.0__ divide__12.0__35.0__ |
| john has num__4 friends who want to ride in his new car that can accommodate only num__3 people at a time ( john plus num__2 passengers ) . how many different combinations of num__2 passengers can be formed from john ' s num__4 friends ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__15 <o> e ) num__18 |
num__4 c num__2 = num__6 the answer is a . <eor> a <eos> |
a |
die_space__ die_space__ |
die_space__ die_space__ |
| three partners a b c in a business invested money such that num__5 ( a ’ s capital ) = num__7 ( b ’ s capital ) = num__9 ( c ’ s capital ) then the ratio of their capitals is <o> a ) num__63 : num__45 : num__34 <o> b ) num__63 : num__54 : num__34 <o> c ) num__36 : num__54 : num__28 <o> d ) num__63 : num__45 : num__35 <o> e ) none of these |
explanation : let num__5 ( a ’ s capital ) = num__7 ( b ’ s capital ) = num__9 ( c ’ s capital ) = rs . x then a ’ s capital = rs x / num__5 b ’ s capital = rs . x / num__7 and c ’ s capital = rs . x / num__9 . a : b : c = x / num__5 : x / num__7 : x / num__9 num__63 : num__45 : num__35 answer : option d <eor> d <eos> |
d |
multiply__7.0__9.0__ multiply__5.0__9.0__ multiply__5.0__7.0__ multiply__7.0__9.0__ |
multiply__7.0__9.0__ multiply__5.0__9.0__ multiply__5.0__7.0__ multiply__7.0__9.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 seconds . what is the length of the train ? <o> a ) num__165 m <o> b ) num__167 m <o> c ) num__987 m <o> d ) num__150 m <o> e ) num__168 m |
speed = ( num__60 * num__0.277777777778 ) m / sec = ( num__16.6666666667 ) m / sec length of the train = ( speed x time ) = ( num__16.6666666667 * num__9 ) m = num__150 m . answer : d <eor> d <eos> |
d |
round__150.0__ |
round__150.0__ |
| a man owns num__0.666666666667 of market reserch beauro buzness and sells num__0.75 of his shares for num__75000 rs what is the value of buzness ? <o> a ) num__150000 <o> b ) num__160000 <o> c ) num__170000 <o> d ) num__190000 <o> e ) num__250000 |
if value of business = x total sell ( num__2 x / num__3 ) ( num__0.75 ) = num__75000 - > x = num__150000 answer : a <eor> a <eos> |
a |
multiply__75000.0__2.0__ multiply__75000.0__2.0__ |
multiply__75000.0__2.0__ multiply__75000.0__2.0__ |
| if num__15.0 of a is the same as num__30.0 of b then a : b is : <o> a ) num__1 : num__4 <o> b ) num__4 : num__3 <o> c ) num__6 : num__7 <o> d ) num__3 : num__5 <o> e ) num__2 : num__1 |
expl : num__15.0 of a i = num__30.0 of b = num__15 a / num__100 = num__30 b / num__100 = num__2.0 = num__2 : num__1 answer : e <eor> e <eos> |
e |
percent__2.0__100.0__ |
percent__2.0__100.0__ |
| if num__76 is divided into four parts proportional to num__7 num__5 num__3 num__4 then the smallest part is <o> a ) num__12 <o> b ) num__15 <o> c ) num__16 <o> d ) num__19 <o> e ) num__20 |
sol . given ratio = num__7 : num__5 : num__3 : num__4 sum of ratio terms = num__19 . ∴ smallest part = [ num__76 x num__0.157894736842 ] = num__12 . answer a <eor> a <eos> |
a |
divide__76.0__4.0__ divide__3.0__19.0__ add__7.0__5.0__ add__7.0__5.0__ |
divide__76.0__4.0__ divide__3.0__19.0__ add__7.0__5.0__ add__7.0__5.0__ |
| the h . c . f of two numbers is num__36 and their l . c . m is num__1600 . if one of the numbers is num__160 find the other ? <o> a ) num__320 <o> b ) num__350 <o> c ) num__360 <o> d ) num__330 <o> e ) num__320 |
explanation : the other number is = num__36 * num__10.0 = num__360 answer : option c <eor> c <eos> |
c |
divide__1600.0__160.0__ multiply__36.0__10.0__ lcm__36.0__360.0__ |
divide__1600.0__160.0__ multiply__36.0__10.0__ multiply__36.0__10.0__ |
| find the area of a parallelogram with base num__34 cm and height num__18 cm ? <o> a ) num__198 cm num__2 <o> b ) num__384 cm num__2 <o> c ) num__510 cm num__2 <o> d ) num__612 cm num__2 <o> e ) num__680 cm num__2 |
area of a parallelogram = base * height = num__34 * num__18 = num__612 cm num__2 answer : d <eor> d <eos> |
d |
multiply__34.0__18.0__ multiply__34.0__18.0__ |
multiply__34.0__18.0__ multiply__34.0__18.0__ |
| an order was placed for the supply of a carper whose length and breadth were in the ratio of num__3 : num__2 . subsequently the dimensions of the carpet were altered such that its length and breadth were in the ratio num__7 : num__3 but were was no change in its parameter . find the ratio of the areas of the carpets in both the cases ? <o> a ) num__8 : num__8 <o> b ) num__8 : num__7 <o> c ) num__8 : num__1 <o> d ) num__8 : num__2 <o> e ) num__8 : num__6 |
let the length and breadth of the carpet in the first case be num__3 x units and num__2 x units respectively . let the dimensions of the carpet in the second case be num__7 y num__3 y units respectively . from the data . num__2 ( num__3 x + num__2 x ) = num__2 ( num__7 y + num__3 y ) = > num__5 x = num__10 y = > x = num__2 y required ratio of the areas of the carpet in both the cases = num__3 x * num__2 x : num__7 y : num__3 y = num__6 x num__2 : num__21 y num__2 = num__6 * ( num__2 y ) num__2 : num__21 y num__2 = num__6 * num__4 y num__2 : num__21 y num__2 = num__8 : num__7 answer : b <eor> b <eos> |
b |
vowel_space__ die_space__ choose__8.0__7.0__ |
vowel_space__ die_space__ choose__8.0__7.0__ |
| after taking n tests each containing num__100 questions john had an average of num__70.0 of correct answers . how much does john need to score on the next test to make his average equal num__74.0 ? m num__13 - num__03 . <o> a ) n − num__35 <o> b ) n + num__72 <o> c ) num__2 n + num__76 <o> d ) num__2 n + num__70 <o> e ) num__2 n − num__35 |
say n = num__1 . so after num__1 test john has num__70 correct answers . in num__2 tests so in num__200 questions he needs to have num__0.74 * num__200 = num__148 correct answers so in the second test he must get num__148 - num__70 = num__78 questions correctly . now plug n = num__1 into the answer choices to see which one yields num__78 . only option d fits . answer : c . <eor> c <eos> |
c |
subtract__3.0__1.0__ multiply__100.0__2.0__ divide__74.0__100.0__ multiply__74.0__2.0__ subtract__148.0__70.0__ subtract__3.0__1.0__ |
subtract__3.0__1.0__ multiply__100.0__2.0__ divide__74.0__100.0__ multiply__74.0__2.0__ subtract__148.0__70.0__ subtract__3.0__1.0__ |
| what is the units ' digit of the following expression ( num__13 ) ^ num__5 * ( num__15 ) ^ num__4 * ( num__17 ) ^ num__5 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__3 <o> d ) num__5 <o> e ) num__9 |
the unit digit of num__3 ^ num__5 is num__3 . the unit digit of num__5 ^ num__4 is num__5 the unit digit of num__7 ^ num__5 is num__7 . the unit digit of num__3 ∗ num__5 ∗ num__7 is num__5 . answer d . <eor> d <eos> |
d |
divide__15.0__5.0__ add__4.0__3.0__ divide__15.0__3.0__ |
divide__15.0__5.0__ add__4.0__3.0__ divide__15.0__3.0__ |
| by selling an article for $ num__180 a person gains $ num__30 . what is the gain % ? <o> a ) num__25.0 <o> b ) num__30.0 <o> c ) num__50.0 <o> d ) num__20.0 <o> e ) num__10 % |
s . p . = $ num__180 gain = $ num__30 c . p . = num__180 - num__30 = num__150 gain % = num__0.2 * num__100 = num__20.0 answer is d <eor> d <eos> |
d |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| what will come in place of the x in the following number series ? num__6 num__12 num__21 x num__48 <o> a ) num__33 <o> b ) num__45 <o> c ) num__49 <o> d ) num__53 <o> e ) num__61 |
( a ) the pattern is + num__6 + num__9 + num__12 + num__15 … … … . . so the missing term is = num__21 + num__12 = num__33 <eor> a <eos> |
a |
subtract__21.0__12.0__ add__6.0__9.0__ add__12.0__21.0__ add__12.0__21.0__ |
subtract__21.0__12.0__ add__6.0__9.0__ add__12.0__21.0__ add__12.0__21.0__ |
| the total price of a basic computer and printer are $ num__2500 . if the same printer had been purchased with an enhanced computer whose price was $ num__500 more than the price of the basic computer then the price of the printer would have been num__0.4 of that total . what was the price of the basic computer ? <o> a ) num__1300 <o> b ) num__1600 <o> c ) num__1750 <o> d ) num__1900 <o> e ) num__2000 |
let the price of basic computer be c and the price of the printer be p : c + p = $ num__2500 . the price of the enhanced computer will be c + num__500 and total price for that computer and the printer will be num__2500 + num__500 = $ num__3000 . now we are told that the price of the printer is num__0.4 of that new total price : p = num__0.4 * $ num__3000 = $ num__1200 . plug this value in the first equation : c + num__1200 = $ num__2500 - - > c = $ num__1300 . answer : a . <eor> a <eos> |
a |
add__2500.0__500.0__ multiply__0.4__3000.0__ subtract__2500.0__1200.0__ subtract__2500.0__1200.0__ |
add__2500.0__500.0__ multiply__0.4__3000.0__ subtract__2500.0__1200.0__ subtract__2500.0__1200.0__ |
| a train num__400 m long can cross an electric pole in num__20 sec and then find the speed of the train ? <o> a ) num__17 kmph <o> b ) num__78 kmph <o> c ) num__72 kmph <o> d ) num__18 kmph <o> e ) num__19 kmph |
length = speed * time speed = l / t s = num__20.0 s = num__20 m / sec speed = num__20 * num__3.6 ( to convert m / sec in to kmph multiply by num__3.6 ) speed = num__72 kmph answer : c <eor> c <eos> |
c |
multiply__20.0__3.6__ round__72.0__ |
multiply__20.0__3.6__ multiply__20.0__3.6__ |
| which one of the following numbers is the greatest positive integer x such that num__3 ^ x is a factor of num__729 ^ num__5 ? <o> a ) num__5 <o> b ) num__8 <o> c ) num__10 <o> d ) num__15 <o> e ) num__30 |
num__729 = num__3 * num__3 * num__3 * num__3 * num__3 * num__3 = = num__3 ^ num__36 num__729 ^ num__5 = ( num__3 ^ num__6 ) ^ num__5 = num__3 ^ num__30 we have to find the greatest positive integer value for x when x = num__30 it divides exactly once and that is the least positive integer value of the quotient therefore the correct option is e <eor> e <eos> |
e |
multiply__5.0__6.0__ multiply__5.0__6.0__ |
multiply__5.0__6.0__ multiply__5.0__6.0__ |
| solve below question num__2 x + num__1 = - num__11 <o> a ) - num__8 <o> b ) - num__6 <o> c ) num__9 <o> d ) num__8 <o> e ) - num__7 |
num__2 x + num__1 = - num__11 x = - num__6 b <eor> b <eos> |
b |
multiply__1.0__6.0__ |
multiply__1.0__6.0__ |
| a man can hit a target once in num__4 shots . if he fires num__4 shots in succession what is the probability that he will hit his target ? <o> a ) num__1 <o> b ) num__0.00390625 <o> c ) num__0.31640625 <o> d ) num__0.68359375 <o> e ) num__2 |
solution : the man will hit the target even if he hits it once or twice or thrice or all four times in the four shots that he takes . so the only case where the man will not hit the target is when he fails to hit the target even in one of the four shots that he takes . the probability that he will not hit the target in one shot = num__1 - probability that he will hit target in exact one shot = num__1 - num__0.25 = num__0.75 therefore the probability that he will not hit the target in all the four shots = ( num__0.75 ) Ã — ( num__0.75 ) Ã — ( num__0.75 ) Ã — ( num__0.75 ) = num__0.31640625 hence the probability that he will hit the target at least in one of the four shots : = num__1 - num__0.31640625 = num__0.68359375 answer is d <eor> d <eos> |
d |
negate_prob__0.25__ negate_prob__0.3164__ negate_prob__0.3164__ |
negate_prob__0.25__ negate_prob__0.3164__ negate_prob__0.3164__ |
| john ' s bank ' s saving amount is decreased num__50.0 due to loan payment and current balance is rs . num__50000 . find the actual balance before deduction ? <o> a ) num__100000 <o> b ) num__8500 <o> c ) num__9000 <o> d ) num__9500 <o> e ) num__10000 |
num__50.0 decreased num__50.0 balance = num__50000 num__100.0 = num__1000.0 * num__100 = num__100000 answer : a <eor> a <eos> |
a |
percent__100.0__100000.0__ |
percent__100.0__100000.0__ |
| a can do a work in num__20 days b in num__30 days . a alone start the work after num__10 days he left work . in how many days b can finish the work ? <o> a ) num__30 days <o> b ) num__25 days <o> c ) num__20 days <o> d ) num__10 days <o> e ) num__15 days |
num__10 days work of a = num__0.5 = num__0.5 remaining work = num__1 - num__0.5 = num__0.5 b can finish num__0.5 work = num__30 * num__0.5 = num__15 days answer is e <eor> e <eos> |
e |
divide__10.0__20.0__ multiply__30.0__0.5__ round__15.0__ |
divide__10.0__20.0__ multiply__30.0__0.5__ multiply__30.0__0.5__ |
| if a is a positive integer and if the units digit of a ^ num__2 is num__9 and the units digit of ( a + num__1 ) ^ num__2 is num__4 what is the units u digit of ( a + num__2 ) ^ num__2 ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__5 <o> d ) num__6 <o> e ) c . num__14 |
i also got a . by punching in numers : u . . . num__7 ^ num__2 = . . . num__9 . . . num__8 ^ num__2 = . . . num__4 . . . num__9 ^ num__2 = . . . num__1 . a <eor> a <eos> |
a |
subtract__9.0__2.0__ multiply__2.0__4.0__ reverse__1.0__ |
subtract__9.0__2.0__ multiply__2.0__4.0__ reverse__1.0__ |
| how many of the positive divisors g of num__120 are also multiples of num__4 not including num__120 ? <o> a ) num__3 . <o> b ) num__4 . <o> c ) num__5 . <o> d ) num__7 . <o> e ) num__8 . |
num__4 num__812 num__2024 num__4060 . ( num__7 ) is the answer other way : factors of num__120 = num__2 ^ num__3 * num__3 * num__5 separate num__2 ^ num__2 ( which means num__4 ) now calculate the number of other factors . g = num__2 * num__3 * num__5 = total positive factors are num__2 * num__2 * num__2 = num__8 this num__8 factors include num__120 so subtract num__1 from num__8 ans is num__7 = d <eor> d <eos> |
d |
subtract__7.0__4.0__ add__2.0__3.0__ multiply__4.0__2.0__ subtract__4.0__3.0__ add__4.0__3.0__ |
subtract__7.0__4.0__ add__2.0__3.0__ multiply__4.0__2.0__ subtract__4.0__3.0__ multiply__1.0__7.0__ |
| look at this series : num__439404 num__369 num__334 num__299 num__264 . . . what number should come next ? <o> a ) num__206 <o> b ) num__208 <o> c ) num__206 <o> d ) num__229 <o> e ) num__204 |
num__229 this is a simple subtraction series . each number is num__35 less than the previous number . d <eor> d <eos> |
d |
subtract__369.0__334.0__ subtract__264.0__35.0__ |
subtract__369.0__334.0__ subtract__264.0__35.0__ |
| the side of a square is increased by num__25.0 then how much % does its area increases ? <o> a ) num__56.75 <o> b ) num__56.25 <o> c ) num__53.85 <o> d ) num__57.25 <o> e ) num__56.21 % |
a = num__100 a num__2 = num__10000 a = num__125 a num__2 = num__15625 - - - - - - - - - - - - - - - - num__10000 - - - - - - - - - num__5625 num__100 - - - - - - - ? = > num__56.25 answer : b <eor> b <eos> |
b |
square_perimeter__25.0__ power__100.0__2.0__ volume_cube__25.0__ triangle_area__56.25__2.0__ |
square_perimeter__25.0__ power__100.0__2.0__ volume_cube__25.0__ triangle_area__56.25__2.0__ |
| a number x is multiplied with itself and then added to the product of num__4 and x . if the result of these two operations is - num__3 what is the value of x ? <o> a ) - num__4 <o> b ) - num__1 and - num__3 <o> c ) num__2 <o> d ) num__4 <o> e ) can not be determined . |
a number x is multiplied with itself - - > x ^ num__2 added to the product of num__4 and x - - > x ^ num__2 + num__4 x if the result of these two operations is - num__3 - - > x ^ num__2 + num__4 x = - num__3 i . e x ^ num__2 + num__4 x + num__3 = num__0 is the quadratic equation which needs to be solved . ( x + num__1 ) ( x + num__3 ) = num__0 hence x = - num__1 . x = - num__3 imo b <eor> b <eos> |
b |
subtract__4.0__3.0__ reverse__1.0__ |
subtract__4.0__3.0__ subtract__4.0__3.0__ |
| when y is divided by num__276 the remainder is num__44 . what is the remainder when the same y is divided by num__23 ? <o> a ) num__19 <o> b ) num__20 <o> c ) num__21 <o> d ) num__22 <o> e ) num__23 |
y = num__276 * a + num__44 = ( num__23 * num__12 ) * a + num__23 + num__21 the answer is c . <eor> c <eos> |
c |
divide__276.0__23.0__ subtract__44.0__23.0__ subtract__44.0__23.0__ |
divide__276.0__23.0__ subtract__44.0__23.0__ subtract__44.0__23.0__ |
| if the average of num__5 positive integers is num__45 and the difference between the largest and the smallest of these num__5 numbers is num__10 what is the maximum value possible for the largest of these num__5 integers ? <o> a ) num__50 <o> b ) num__53 <o> c ) num__49 <o> d ) num__48 <o> e ) num__44 |
sum of num__5 integer ( a b c d e ) = num__5 * num__45 = num__225 e - a = num__10 i . e . e = a + num__10 for e to be maximum remaining num__4 must be as small as possible since smallest of num__5 numbers is a so to minimize other numbers we can take them equal to the smallest of num__5 numbers i . e . a + a + a + a + ( a + num__10 ) = num__225 i . e . num__5 a = num__215 i . e . a = num__43 i . e . largest e = num__43 + num__10 = num__53 answer : option b <eor> b <eos> |
b |
multiply__5.0__45.0__ subtract__225.0__10.0__ divide__215.0__5.0__ add__10.0__43.0__ add__10.0__43.0__ |
multiply__5.0__45.0__ subtract__225.0__10.0__ divide__215.0__5.0__ add__10.0__43.0__ add__10.0__43.0__ |
| an order was placed for the supply of a carper whose length and breadth were in the ratio of num__3 : num__2 . subsequently the dimensions of the carpet were altered such that its length and breadth were in the ratio num__7 : num__3 but were was no change in its parameter . find the ratio of the areas of the carpets in both the ? <o> a ) num__8 : num__6 <o> b ) num__8 : num__7 <o> c ) num__8 : num__1 <o> d ) num__8 : num__2 <o> e ) num__8 : num__4 |
let the length and breadth of the carpet in the first case be num__3 x units and num__2 x units respectively . let the dimensions of the carpet in the second case be num__7 y num__3 y units respectively . from the data . num__2 ( num__3 x + num__2 x ) = num__2 ( num__7 y + num__3 y ) = > num__5 x = num__10 y = > x = num__2 y required ratio of the areas of the carpet in both the cases = num__3 x * num__2 x : num__7 y : num__3 y = num__6 x num__2 : num__21 y num__2 = num__6 * ( num__2 y ) num__2 : num__21 y num__2 = num__6 * num__4 y num__2 : num__21 y num__2 = num__8 : num__7 answer : b <eor> b <eos> |
b |
vowel_space__ die_space__ choose__8.0__7.0__ |
vowel_space__ die_space__ choose__8.0__7.0__ |
| num__3889 + num__12.543 - ? = num__3854.002 <o> a ) a ) num__47.095 <o> b ) b ) num__47.752 <o> c ) c ) num__47.932 <o> d ) d ) num__47.541 <o> e ) of the above |
let num__3889 + num__12.543 - x = num__3854.002 . then x = ( num__3889 + num__12.543 ) - num__3854.002 = num__3901.543 - num__3854.002 = num__47.541 . answer = d <eor> d <eos> |
d |
add__3889.0__12.543__ subtract__3901.543__3854.002__ subtract__3901.543__3854.002__ |
add__3889.0__12.543__ subtract__3901.543__3854.002__ subtract__3901.543__3854.002__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 sec . what is the length of the train ? <o> a ) num__176 m <o> b ) num__167 m <o> c ) num__178 m <o> d ) num__198 m <o> e ) num__150 m |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__9 = num__150 m answer : e <eor> e <eos> |
e |
round__150.0__ |
round__150.0__ |
| each machine of type a has num__3 steel parts and num__2 chrome parts . each machine of type b has num__2 steel parts and num__4 chrome parts . if a certain group of type a and type b machines has a total of num__50 steel parts and num__66 chrome parts how many machines are in the group <o> a ) num__11 <o> b ) num__21 <o> c ) num__31 <o> d ) num__61 <o> e ) num__51 |
look at the below representation of the problem : steel chrome total a num__3 num__2 num__50 > > no . of type a machines = num__10.0 = num__10 b num__2 num__4 num__66 > > no . of type b machines = num__11.0 = num__11 so the answer is num__21 i . e b . hope its clear . <eor> b <eos> |
b |
add__10.0__11.0__ add__10.0__11.0__ |
add__10.0__11.0__ add__10.0__11.0__ |
| divide rs . num__1500 among a b and c so that a receives num__0.333333333333 as much as b and c together and b receives num__0.666666666667 as a and c together . a ' s share is ? <o> a ) s . num__360 <o> b ) s . num__400 <o> c ) s . num__375 <o> d ) s . num__500 <o> e ) s . num__900 |
a + b + c = num__1500 a = num__0.333333333333 ( b + c ) ; b = num__0.666666666667 ( a + c ) a / ( b + c ) = num__0.333333333333 a = num__0.25 * num__1500 = > num__375 answer : c <eor> c <eos> |
c |
multiply__1500.0__0.25__ multiply__1500.0__0.25__ |
multiply__1500.0__0.25__ multiply__1500.0__0.25__ |
| two pipes can separately fill a tank in num__20 and num__30 hours respectively . both the pipes are opened to fill the tank but when the tank is full a leak develops in the tank through which one - third of water supplied by both the pipes goes out . what is the total time taken to fill the tank ? <o> a ) num__22 hrs <o> b ) num__16 hrs <o> c ) num__77 hrs <o> d ) num__99 hrs <o> e ) num__65 hrs |
num__0.05 + num__0.0333333333333 = num__0.0833333333333 num__1 + num__0.333333333333 = num__1.33333333333 num__1 - - - num__12 num__1.33333333333 - - - ? num__1.33333333333 * num__12 = num__16 hrs answer : b <eor> b <eos> |
b |
add__0.05__0.0333__ multiply__20.0__0.05__ add__1.0__0.3333__ round__16.0__ |
add__0.05__0.0333__ multiply__20.0__0.05__ add__1.0__0.3333__ round__16.0__ |
| in a recent election ms . robinson received num__8000 voters cast by independent voters that is voters not registered with a specific political party . she also received num__25 percent of the votes cast by those voters registered with a political party . if n is the total number of votes cast in the election and num__40 percent of the votes cast were cast by independent voters which of the following represents the number of votes that mrs . robbins received ? <o> a ) num__0.06 n + num__3200 <o> b ) num__0.1 n + num__7200 <o> c ) num__0.4 n + num__7200 <o> d ) num__0.15 n + num__8000 <o> e ) num__0.06 n + num__8000 |
num__40.0 of n are independent voters hence num__60.0 of n are not independent voters . from this group she received num__25.0 votes so num__0.25 * num__0.6 * n plus num__8000 votes from independents : total = num__0.25 * num__0.6 * n + num__8000 = num__0.15 * n + num__8000 . answer : d . <eor> d <eos> |
d |
multiply__0.25__0.6__ multiply__0.25__0.6__ |
multiply__0.25__0.6__ multiply__0.25__0.6__ |
| the parameter of a square is equal to the perimeter of a rectangle of length num__16 cm and breadth num__14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places <o> a ) num__34 <o> b ) num__35 <o> c ) num__56 <o> d ) num__67 <o> e ) num__23.57 |
let the side of the square be a cm . parameter of the rectangle = num__2 ( num__16 + num__14 ) = num__60 cm parameter of the square = num__60 cm i . e . num__4 a = num__60 a = num__15 diameter of the semicircle = num__15 cm circimference of the semicircle = num__0.5 ( ∏ ) ( num__15 ) = num__0.5 ( num__3.14285714286 ) ( num__15 ) = num__23.5714285714 = num__23.57 cm to two decimal places answer : option e <eor> e <eos> |
e |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
| two trains of length num__250 m and num__120 m are num__50 m apart . they start moving towards each other on parallel tracks at speeds num__64 kmph and num__42 kmph . after how much time will the trains meet ? <o> a ) num__1.5 <o> b ) num__1.7 <o> c ) num__1.6 <o> d ) num__1.8 <o> e ) num__1.6 |
they are moving in opposite directions relative speed is equal to the sum of their speeds . relative speed = ( num__64 + num__42 ) * num__0.277777777778 = num__7 * num__5 = num__29.4 mps . the time required = d / s = num__50 / num__29.4 = num__1.7 sec . answer : b <eor> b <eos> |
b |
divide__250.0__50.0__ round__1.7__ |
divide__250.0__50.0__ round__1.7__ |
| the vertices of a rectangle in the standard ( x y ) coordinate place are ( num__00 ) ( num__03 ) ( num__60 ) and ( num__63 ) . if a line through ( num__22 ) partitions the interior of this rectangle into num__2 regions that have equal areas what is the slope of this line ? <o> a ) num__0.5 <o> b ) num__0.4 <o> c ) num__0.571428571429 <o> d ) num__1 <o> e ) num__1.75 |
concept : the readers need to know that a rectangle can be divided into two equal area by a straight line only when the straight line passes through the centre of the rectangle ( intersection of its two diagonals ) draw a figure and know it for yourself . the point of intersections of the diagonals will be the midpoint of any diagonal i . e . midpoint of ( num__00 ) and ( num__63 ) or midpoint of ( num__03 ) and ( num__60 ) i . e . either [ ( num__0 + num__6 ) / num__2 ( num__0 + num__3 ) / num__2 ] or [ ( num__0 + num__6 ) / num__2 ( num__3 + num__0 ) / num__2 ] = [ num__3 num__1.5 ] slope of line passing through points ( num__22 ) and ( num__3 num__1.5 ) = ( num__2 - num__1.5 ) / ( num__3 - num__2 ) = num__0.5 / num__1 = num__0.5 a <eor> a <eos> |
a |
multiply__3.0__2.0__ divide__3.0__2.0__ reverse__2.0__ round_down__1.5__ reverse__2.0__ |
multiply__3.0__2.0__ divide__3.0__2.0__ reverse__2.0__ subtract__3.0__2.0__ reverse__2.0__ |
| the sum of ages of num__5 children born at the intervals of num__8 years each is num__90 years . what is the age of the youngest child ? <o> a ) num__2 <o> b ) num__9 <o> c ) num__6 <o> d ) num__4 <o> e ) num__0 |
let the ages of children be x ( x + num__8 ) ( x + num__16 ) ( x + num__24 ) and ( x + num__32 ) years . then x + ( x + num__8 ) + ( x + num__16 ) + ( x + num__24 ) + ( x + num__32 ) = num__90 num__5 x = num__10 x = num__2 . age of the youngest child = x = num__2 years . answer : a <eor> a <eos> |
a |
add__8.0__16.0__ add__8.0__24.0__ divide__32.0__16.0__ divide__32.0__16.0__ |
add__8.0__16.0__ add__8.0__24.0__ divide__32.0__16.0__ divide__32.0__16.0__ |
| sally has a gold credit card with a certain spending limit and a platinum card with twice the spending limit of the gold card . currently she has a balance on her gold card that is num__0.333333333333 of the spending limit on that card and she has a balance on her platinum card that is num__0.166666666667 of the spending limit on that card . if sally transfers the entire balance on her gold card to her platinum card what portion of her limit on the platinum card will remain unspent ? <o> a ) num__0.666666666667 <o> b ) num__0.483333333333 <o> c ) num__0.566666666667 <o> d ) num__0.633333333333 <o> e ) num__0.733333333333 |
let s assume the platinum card spending limit = x gold card spending limit will be = x / num__2 balance on gold card is = x / num__2 * num__0.333333333333 = x / num__6 platinum card unspent limit is = x - num__0.166666666667 x = num__0.833333333333 x so if gold card balance is transferred then the rest unspent will be num__0.833333333333 x - x / num__6 = num__0.666666666667 x so the ans is a <eor> a <eos> |
a |
coin_space__ die_space__ negate_prob__0.1667__ negate_prob__0.3333__ negate_prob__0.3333__ |
coin_space__ die_space__ negate_prob__0.1667__ negate_prob__0.3333__ negate_prob__0.3333__ |
| the average age of a family of num__6 members is num__28 years . if the age of the youngest member is num__7 years what was the average age of the family at the birth of the youngest member ? <o> a ) num__15 <o> b ) num__18 <o> c ) num__21 <o> d ) num__12 <o> e ) num__19 |
present age of total members = num__6 x num__28 = num__168 num__7 yrs back their ages were = num__6 x num__7 = num__42 ages at the birth of youngest member = num__168 - num__42 = num__126 therefore avg age at the birth of youngest member = num__21.0 = num__21 . answer : c <eor> c <eos> |
c |
multiply__6.0__28.0__ multiply__6.0__7.0__ subtract__168.0__42.0__ subtract__28.0__7.0__ subtract__28.0__7.0__ |
multiply__6.0__28.0__ multiply__6.0__7.0__ subtract__168.0__42.0__ subtract__28.0__7.0__ subtract__28.0__7.0__ |
| there are some passengers riding on a bus . at the first stop half the passengers get off and nobody gets on the bus . this pattern continues as the bus goes to the next stops . if only one passenger gets off at stop number num__7 how many passengers were on the bus originally ? <o> a ) num__128 <o> b ) num__64 <o> c ) num__32 <o> d ) num__16 <o> e ) num__8 |
before stop num__7 there were num__2 passengers on the bus . before stop num__6 there were num__4 passengers on the bus . etc . . . before stop x there were num__2 ^ ( num__8 - x ) passengers on the bus . before stop num__1 there were num__2 ^ num__7 = num__128 passengers on the bus . the answer is a . <eor> a <eos> |
a |
subtract__6.0__2.0__ multiply__2.0__4.0__ subtract__7.0__6.0__ multiply__128.0__1.0__ |
subtract__6.0__2.0__ multiply__2.0__4.0__ subtract__7.0__6.0__ multiply__128.0__1.0__ |
| if √ num__2 n = num__64 then the value of n is : <o> a ) num__6 <o> b ) num__12 <o> c ) num__24 <o> d ) num__48 <o> e ) none |
sol . √ num__2 n = num__64 ⇔ ( num__2 n ) num__0.5 ⇔ num__2 n / num__2 = num__26 ⇔ n / num__2 = num__6 ⇔ n = num__12 . answer b <eor> b <eos> |
b |
reverse__2.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
reverse__2.0__ divide__6.0__0.5__ divide__6.0__0.5__ |
| the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is num__867 sq m then what is the breadth of the rectangular plot ? <o> a ) num__18 m <o> b ) num__17 m <o> c ) num__26 m <o> d ) num__27 m <o> e ) num__15 m |
let the breadth of the plot be b m . length of the plot = num__3 b m ( num__3 b ) ( b ) = num__867 num__3 b num__2 = num__867 b num__2 = num__289 = num__172 ( b > num__0 ) b = num__17 m . answer : b <eor> b <eos> |
b |
triangle_area__2.0__17.0__ |
triangle_area__2.0__17.0__ |
| a rectangular box measures internally num__1.6 m long i m broad and num__60 cm deep . the number of cubical blocks each of edge num__20 cm that can be packed inside the box is <o> a ) num__30 <o> b ) num__53 <o> c ) num__60 <o> d ) num__120 <o> e ) none |
solution number of blocks = ( num__160 × num__100 × num__3.0 × num__20 × num__20 ) = num__120 . answer d <eor> d <eos> |
d |
divide__160.0__1.6__ divide__60.0__20.0__ add__20.0__100.0__ round__120.0__ |
divide__160.0__1.6__ divide__60.0__20.0__ add__20.0__100.0__ round__120.0__ |
| if a committee of num__4 people is to be selected from among num__5 married couples so that the committee does not include two people who are married to each other how many such committees are possible ? <o> a ) num__20 <o> b ) num__40 <o> c ) num__50 <o> d ) num__80 <o> e ) num__120 |
each couple can send only onerepresentativeto the committee . let ' s see in how many ways we can choose num__4 couples ( as there should be num__4 members ) each to send onerepresentativeto the committee : num__5 c num__4 = num__5 . but these num__4 chosen couples can send two persons ( either husband or wife ) : num__2 * num__2 * num__2 = num__2 ^ num__3 = num__8 . total # of ways : num__5 c num__3 * num__2 ^ num__3 = num__40 . answer : b . <eor> b <eos> |
b |
coin_space__ choose__5.0__3.0__ choose__5.0__3.0__ |
coin_space__ choose__5.0__3.0__ choose__5.0__3.0__ |
| a is the hundreds digit of the three digit integer x b is the tens digit of x and c is the units digit of x . num__2 a = b = num__4 c and a > num__0 . what is the difference between the two greatest possible values of x ? tip : dont stop till you have exhausted all answer choices to arrive at the correct one . <o> a ) num__124 <o> b ) num__241 <o> c ) num__394 <o> d ) num__421 <o> e ) num__842 |
ratio of a : b : c = num__2 : num__4 : num__1 two possible greatest single digit values for b are num__8 and num__4 if b is num__8 then x = num__482 if b is num__4 then x = num__241 difference = num__482 - num__241 = num__241 b is the answer <eor> b <eos> |
b |
multiply__2.0__4.0__ divide__482.0__2.0__ multiply__1.0__241.0__ |
multiply__2.0__4.0__ divide__482.0__2.0__ subtract__482.0__241.0__ |
| there are num__8 teams in a certain league and each team plays each of the other teams exactly once . if each game is played by num__2 teams what is the total number of games played ? <o> a ) num__15 <o> b ) num__16 <o> c ) num__28 <o> d ) num__56 <o> e ) num__64 |
number of ways num__2 teams can be selected out of num__8 teams : c num__4.0 = num__8 ∗ num__3.5 = num__28 ans : ' ' c ' ' <eor> c <eos> |
c |
divide__8.0__2.0__ multiply__8.0__3.5__ multiply__8.0__3.5__ |
divide__8.0__2.0__ multiply__8.0__3.5__ multiply__8.0__3.5__ |
| the list price of an article is rs . num__65 . a customer pays rs . num__56.16 for it . he was given two successive discounts one of them being num__10.0 . the other discount is ? <o> a ) num__8.0 <o> b ) num__4.0 <o> c ) num__5.0 <o> d ) num__3.0 <o> e ) num__2 % |
num__65 * ( num__0.9 ) * ( ( num__100 - x ) / num__100 ) = num__56.16 x = num__4.0 answer : b <eor> b <eos> |
b |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| a line that passes through ( – num__1 – num__4 ) and ( num__5 k ) has a slope = k . what is the value of k ? <o> a ) num__0.75 <o> b ) num__1 <o> c ) num__0.8 <o> d ) num__2 <o> e ) num__3.5 |
slope = ( y num__2 - y num__1 ) / ( x num__2 - x num__1 ) = > k = ( k + num__4 ) / ( num__5 + num__1 ) = > num__6 k = k + num__4 = > k = num__0.8 ans c it is ! <eor> c <eos> |
c |
add__1.0__5.0__ divide__4.0__5.0__ multiply__1.0__0.8__ |
add__1.0__5.0__ divide__4.0__5.0__ divide__4.0__5.0__ |
| tom wants to buy items costing $ num__25.35 $ num__70.69 and $ num__85.96 . he earns $ num__6.50 an hour doing odd jobs . if ten percent of his income is put aside for other purposes how many hours must he work to earn the money he needs for his purchases ? round your answer to the nearest whole hour . <o> a ) num__8 hours <o> b ) num__48 hours <o> c ) num__31 hours <o> d ) num__18 hours <o> e ) num__28 hours |
$ num__6.50 x . num__10 = $ . num__65 is num__10.0 of his hourly income $ num__6.50 - . num__65 = $ num__5.85 hourly amount available to spend $ num__25.35 + $ num__70.69 + $ num__85.96 = $ num__182 total needed $ num__182 ÷ $ num__5.85 = num__31.11 . . . rounds to num__31 hours correct answer c <eor> c <eos> |
c |
multiply__6.5__10.0__ round__31.0__ |
multiply__6.5__10.0__ round__31.0__ |
| two trains running in opposite directions cross a man standing on the platform in num__27 seconds and num__17 seconds respectively and they cross each other in num__23 seconds . the ratio of their speeds is ? <o> a ) num__0.6 <o> b ) num__1.5 <o> c ) num__0.333333333333 <o> d ) num__0.5 <o> e ) num__3.0 |
let the speeds of the two trains be x m / sec and y m / sec respectively . then length of the first train = num__27 x meters and length of the second train = num__17 y meters . ( num__27 x + num__17 y ) / ( x + y ) = num__23 = = > num__27 x + num__17 y = num__23 x + num__23 y = = > num__4 x = num__6 y = = > x / y = num__1.5 . answer : b <eor> b <eos> |
b |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ round__1.5__ |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ divide__6.0__4.0__ |
| find the ratio in which rice at rs . num__7.30 a kg be mixed with rice at rs . num__5.70 a kg to produce a mixture worth rs . num__6.30 a kg <o> a ) num__2 : num__0 <o> b ) num__2 : num__3 <o> c ) num__3 : num__5 <o> d ) num__2 : num__2 <o> e ) num__2 : num__8 |
by the rule of alligation : cost of num__1 kg rice of num__1 st kind cost of num__1 kg rice of num__2 nd kind required ratio = num__60 : num__100 = num__3 : num__5 answer : c <eor> c <eos> |
c |
subtract__7.3__6.3__ add__1.0__2.0__ round_down__5.7__ add__1.0__2.0__ |
subtract__7.3__6.3__ add__1.0__2.0__ round_down__5.7__ add__1.0__2.0__ |
| if s + a < s - a < a - s which of the following is correct ? <o> a ) a < s < num__0 . <o> b ) s < a < num__0 . <o> c ) s < num__0 < a . <o> d ) num__0 < b < a . <o> e ) b > a > num__0 . |
check first inequality s + a < s - a a < - a a is - ve check another equality s - a < a - s - num__2 a < - num__2 s a > s combining both num__0 > a > s option b fits only ans b <eor> b <eos> |
b |
multiply__0.0__2.0__ |
multiply__0.0__2.0__ |
| a man took loan from a bank at the rate of num__12.0 p . a . simple interest . after num__3 years he had to pay dollar num__5400 interest only for the period . the principal amount borrowed by him was : <o> a ) dollar num__2000 <o> b ) dollar num__10000 <o> c ) dollar num__15000 <o> d ) dollar num__20000 <o> e ) none of these |
explanation : principal = dollar ( num__100 x num__5400 ) / ( num__12 x num__3 ) = dollar num__15000 . answer : option c <eor> c <eos> |
c |
percent__100.0__15000.0__ |
percent__100.0__15000.0__ |
| there are num__6 orators a b c d e f . in how many ways can the arrangements be made so that a always comes before b and b always comes before c . <o> a ) num__6 ! / num__3 ! <o> b ) num__8 ! / num__6 ! <o> c ) num__5 ! x num__3 ! <o> d ) num__8 ! / ( num__5 ! x num__3 ! ) <o> e ) num__8 ! / ( num__5 ! x num__4 ! ) |
select any three places for a b and c . they need no arrangement amongst themselves as a would always come before b and b would always come before c . the remaining num__3 people have to be arranged in num__3 places . thus num__6 c num__3 x num__3 ! = num__6 ! / num__3 ! answer : a <eor> a <eos> |
a |
die_space__ |
die_space__ |
| in a group of num__3 boys and num__2 girls can be formed out of a total of num__4 boys and num__4 girls ? <o> a ) num__20 <o> b ) num__22 <o> c ) num__24 <o> d ) num__26 <o> e ) num__29 |
solution : total number of ways = ( select num__3 boys from group of num__4 boys ) * ( num__2 girls from group of num__4 girls ) = > ( num__4 c num__3 ) * ( num__4 c num__2 ) = num__24 c <eor> c <eos> |
c |
choose__4.0__2.0__ choose__4.0__2.0__ |
choose__4.0__2.0__ choose__4.0__2.0__ |
| a sun is divided among x y and z in such a way that for each num__2 rupees x gets y gets num__1 rupee num__50 paisa and z gets num__1 rupee num__20 paisa . if the share of x is rs . num__125 what is the total amount ? <o> a ) num__283.75 <o> b ) num__193.75 <o> c ) num__293.75 <o> d ) num__290.75 <o> e ) num__313.75 |
x : y : z = num__200 : num__150 : num__120 num__20 : num__15 : num__12 num__20 - - - num__125 num__47 - - - ? = > num__293.75 answer : c <eor> c <eos> |
c |
subtract__200.0__50.0__ multiply__1.0__293.75__ |
subtract__200.0__50.0__ multiply__1.0__293.75__ |
| the average temperature for monday tuesday wednesday and thursday was num__48 degrees and for tuesday wednesday thursday and friday was num__46 degrees . if the temperature on monday was num__42 degrees . find the temperature on friday ? <o> a ) num__227 <o> b ) num__268 <o> c ) num__127 <o> d ) num__150 <o> e ) num__187 |
m + tu + w + th = num__4 * num__48 = num__192 tu + w + th + f = num__4 * num__46 = num__184 m = num__42 tu + w + th = num__192 - num__42 = num__150 f = num__184 – num__150 = num__34 answer : d <eor> d <eos> |
d |
subtract__46.0__42.0__ multiply__48.0__4.0__ multiply__46.0__4.0__ subtract__192.0__42.0__ subtract__184.0__150.0__ subtract__192.0__42.0__ |
subtract__46.0__42.0__ multiply__48.0__4.0__ multiply__46.0__4.0__ subtract__192.0__42.0__ subtract__184.0__150.0__ subtract__192.0__42.0__ |
| a group of n students can be divided into equal groups of num__4 with num__1 student left over or equal groups of num__7 with num__2 students left over . what is the sum of the two smallest possible values of n ? <o> a ) num__42 . <o> b ) num__46 . <o> c ) num__50 . <o> d ) num__54 . <o> e ) num__58 . |
n = num__4 k + num__1 = num__7 j + num__2 let ' s start at num__1 = num__4 ( num__0 ) + num__1 and keep adding num__4 until we find a number in the form num__7 j + num__2 . num__1 num__5 num__9 = num__7 ( num__1 ) + num__2 the next such number is num__9 + num__4 * num__7 = num__37 . num__9 + num__37 = num__46 the answer is b . <eor> b <eos> |
b |
add__4.0__1.0__ add__4.0__5.0__ add__9.0__37.0__ multiply__1.0__46.0__ |
add__4.0__1.0__ add__4.0__5.0__ add__9.0__37.0__ multiply__1.0__46.0__ |
| the average of six numbers is num__3.95 . the average of two of them is num__3.4 while the average of the other two is num__3.85 . what is the average of the remaining two numbers ? <o> a ) num__4.7 <o> b ) num__4.6 <o> c ) num__4.2 <o> d ) num__4.1 <o> e ) num__4.9 |
explanation : sum of the remaining two numbers = ( num__3.95 * num__6 ) - [ ( num__3.4 * num__2 ) + ( num__3.85 * num__2 ) ] = num__23.70 - ( num__6.8 + num__7.7 ) = num__23.70 - num__14.5 = num__9.20 . required average = ( num__9.2 / num__2 ) = num__4.6 . answer : b ) num__4.6 <eor> b <eos> |
b |
multiply__3.95__6.0__ multiply__3.4__2.0__ multiply__3.85__2.0__ add__7.7__6.8__ subtract__23.7__14.5__ divide__9.2__2.0__ divide__9.2__2.0__ |
multiply__3.95__6.0__ multiply__3.4__2.0__ multiply__3.85__2.0__ add__7.7__6.8__ subtract__23.7__14.5__ divide__9.2__2.0__ divide__9.2__2.0__ |
| what is the least number . which should be added to num__0.0355 to make it a perfect square ? <o> a ) num__0.0005 <o> b ) num__0.0016 <o> c ) num__0.0056 <o> d ) num__0.0066 <o> e ) num__0.0006 |
num__0.0355 + num__0.0006 = num__0.0361 ( num__0.19 ) ^ num__2 answer : e <eor> e <eos> |
e |
add__0.0355__0.0006__ subtract__0.0361__0.0355__ |
add__0.0355__0.0006__ subtract__0.0361__0.0355__ |
| a and b can do a piece of work in num__3 days b and c in num__4 days c and a in num__6 days . how long will c take to do it ? <o> a ) num__23 days <o> b ) num__16 days <o> c ) num__11 days <o> d ) num__24 days <o> e ) num__35 days |
d num__24 days num__2 c = ¼ + num__0.166666666667 – num__0.333333333333 = num__0.0833333333333 c = num__0.0416666666667 = > num__24 days <eor> d <eos> |
d |
multiply__4.0__6.0__ divide__6.0__3.0__ divide__4.0__24.0__ divide__2.0__6.0__ divide__2.0__24.0__ divide__0.1667__4.0__ round__24.0__ |
multiply__4.0__6.0__ divide__6.0__3.0__ divide__4.0__24.0__ divide__2.0__6.0__ divide__2.0__24.0__ divide__0.1667__4.0__ round__24.0__ |
| if n is an integer f ( n ) = f ( n - num__1 ) - n and f ( num__4 ) = num__15 . what is the value of f ( num__6 ) ? <o> a ) num__4 <o> b ) num__0 <o> c ) num__1 <o> d ) num__2 <o> e ) num__24 |
since f ( n ) = f ( n - num__1 ) - n then : f ( num__6 ) = f ( num__5 ) - num__6 and f ( num__5 ) = f ( num__4 ) - num__5 . as given that f ( num__4 ) = num__15 then f ( num__5 ) = num__15 - num__5 = num__10 - - > substitute the value of f ( num__5 ) back into the first equation : f ( num__6 ) = f ( num__5 ) - num__6 = num__10 - num__6 = num__4 . answer : a . questions on funtions to practice : <eor> a <eos> |
a |
add__1.0__4.0__ add__4.0__6.0__ multiply__1.0__4.0__ |
subtract__6.0__1.0__ subtract__15.0__5.0__ subtract__5.0__1.0__ |
| tough and tricky questions : exponents . if num__7 ^ ( num__3 x - num__1 ) * num__3 ^ ( num__4 y - num__3 ) = num__49 ^ x * num__27 ^ y then x + y = <o> a ) num__3 <o> b ) num__4 <o> c ) num__2 <o> d ) num__1 <o> e ) num__5 |
here is my solution . num__7 ^ ( num__3 x - num__1 ) * num__3 ^ ( num__4 y - num__3 ) = num__49 ^ x * num__27 ^ y here rhs num__49 ^ x * num__27 ^ y = num__7 ^ ( num__2 x ) * num__3 ^ ( num__3 y ) equating powers on both sides - - > num__3 x - num__1 = num__2 x thus x = num__1 and num__4 y - num__3 = num__3 y giving y = num__3 so x + y = num__4 option : b <eor> b <eos> |
b |
subtract__3.0__1.0__ subtract__7.0__3.0__ |
subtract__3.0__1.0__ subtract__7.0__3.0__ |
| an error num__3.0 in excess is made while measuring the side of a square . what is the percentage of error in the calculated area of the square ? <o> a ) num__9.09 <o> b ) num__4.02 <o> c ) num__4.0 <o> d ) num__3.0 <o> e ) num__2 % |
percentage error in calculated area = ( num__3 + num__3 + ( num__3 Ã — num__3 ) / num__100 ) % = num__9.09 answer : a <eor> a <eos> |
a |
percent__100.0__9.09__ |
percent__100.0__9.09__ |
| the temperature of a certain cup of coffee num__6 minutes after it was poured was num__120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = num__120 * num__2 ^ ( - at ) + num__60 where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee num__30 minutes after it was poured was how many degrees fahrenheit ? <o> a ) num__65 <o> b ) num__63.75 <o> c ) num__80 <o> d ) num__85 <o> e ) num__90 |
first we have to find a . we know that after t = num__6 minutes the temperature f = num__120 degrees . hence : num__120 = num__120 * ( num__2 ^ - num__6 a ) + num__60 num__60 = num__120 * ( num__2 ^ - num__6 a ) num__0.5 = num__2 ^ - num__6 a num__0.5 = num__2 ^ - num__6 a num__2 ^ - num__1 = num__2 ^ - num__6 a - num__1 = - num__6 a num__0.166666666667 = a now we need to find f after t = num__30 minutes : f = num__120 * ( num__2 ^ - num__0.166666666667 * num__30 ) + num__60 f = num__120 * ( num__2 ^ - num__5 ) + num__60 f = num__120 * ( num__0.5 ^ num__5 ) + num__60 f = num__120 * num__0.03125 + num__60 f = num__3.75 + num__60 = num__63.75 answer b ! <eor> b <eos> |
b |
reverse__2.0__ multiply__2.0__0.5__ reverse__6.0__ subtract__6.0__1.0__ add__60.0__3.75__ add__60.0__3.75__ |
reverse__2.0__ multiply__2.0__0.5__ reverse__6.0__ subtract__6.0__1.0__ add__60.0__3.75__ add__60.0__3.75__ |
| what is the greatest positive integer n such that num__3 ^ n is a factor of num__36 ^ num__450 ? <o> a ) num__100 <o> b ) num__200 <o> c ) num__300 <o> d ) num__600 <o> e ) num__900 |
num__36 = num__3 ^ num__2 * num__2 ^ num__2 . num__36 ^ num__450 = num__3 ^ num__900 * num__2 ^ num__900 the answer is e . <eor> e <eos> |
e |
lcm__36.0__450.0__ lcm__3.0__900.0__ |
multiply__450.0__2.0__ multiply__450.0__2.0__ |
| if x < y < z and y - x > num__5 where x is an even integer and y and z are odd integers what is the least possible value s of z - x ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
x < y < z to find the least possible value for z - x ; we need to find the values for z and x that can be closest to each other . if x is some even number then what could be minimum possible odd z . if x is some even number y - x > num__5 ; y > x + num__5 ; minimum value for y = x + num__5 + num__2 = x + num__7 [ note : x + num__5 is as even + odd = odd and nearest odd greater than x + num__5 is x + num__5 + num__2 ] minimum value for z = y + num__2 = x + num__7 + num__2 = x + num__9 [ note : z = y + num__2 because both z and y are odd . difference between two odd numbers is num__2 ] s = z - x = x + num__9 - x = num__9 ans : d <eor> d <eos> |
d |
add__5.0__2.0__ add__2.0__7.0__ add__2.0__7.0__ |
add__5.0__2.0__ add__2.0__7.0__ add__2.0__7.0__ |
| what is the place value of num__7 in the numeral num__2734 ? <o> a ) num__450 <o> b ) num__500 <o> c ) num__700 <o> d ) num__800 <o> e ) num__840 |
option ' c ' num__7 * num__100 = num__700 <eor> c <eos> |
c |
multiply__7.0__100.0__ multiply__7.0__100.0__ |
multiply__7.0__100.0__ multiply__7.0__100.0__ |
| a bowl was half full of water . num__4 cups of water were then added to the bowl filling the bowl to num__70.0 of its capacity . how many cups of water are now in the bowl ? <o> a ) num__14 cups <o> b ) num__15 <o> c ) num__16 <o> d ) num__17 <o> e ) num__18 |
lets say total volume of the container = v initial water content is half of total volume = v / num__2 then num__4 cups of water were added . current water content = ( v / num__2 ) + num__4 cups = ( num__0.7 ) v = > v = num__20 cups = > current water content is equivalent to = v / num__2 + num__4 cups = num__10.0 + num__4 = num__14 cups ; answer : a <eor> a <eos> |
a |
percent__70.0__20.0__ percent__70.0__20.0__ |
percent__70.0__20.0__ percent__70.0__20.0__ |
| what is the greatest value of n such that num__17 ^ n is a factor of num__17 ! ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
another way to solve this question : we know that num__18 = num__2 * num__3 ^ num__2 . hence we need to find total number of num__3 s in num__18 ! ( as out of num__3 and num__2 we will have count of num__3 least ) no . of num__3 s = num__6.0 = num__2.0 = num__2 . add the numbers in bold you will get total number of num__3 s = num__8 . hence total number of num__2 ^ num__2 = num__2.0 = num__2 . hence answer is num__2 . b <eor> b <eos> |
b |
divide__18.0__3.0__ add__2.0__6.0__ gcd__8.0__18.0__ |
multiply__3.0__2.0__ add__2.0__6.0__ gcd__8.0__18.0__ |
| two trains running in opposite directions cross a man standing on the platform in num__27 seconds and num__17 seconds respectively and they cross each other in num__23 seconds . the ratio of their speeds is ? <o> a ) num__0.375 <o> b ) num__1.5 <o> c ) num__0.6 <o> d ) num__0.333333333333 <o> e ) num__3.0 |
let the speeds of the two trains be x m / sec and y m / sec respectively . then length of the first train = num__27 x meters and length of the second train = num__17 y meters . ( num__27 x + num__17 y ) / ( x + y ) = num__23 = = > num__27 x + num__17 y = num__23 x + num__23 y = = > num__4 x = num__6 y = = > x / y = num__1.5 answer : b : <eor> b <eos> |
b |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ round__1.5__ |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ divide__6.0__4.0__ |
| a man can row num__6 kmph in still water . when the river is running at num__1.2 kmph it takes him num__1 hour to row to a place and black . what is the total distance traveled by the man ? <o> a ) num__6.24 km <o> b ) num__6 km <o> c ) num__5.76 km <o> d ) num__5.66 km <o> e ) num__10 km |
m = num__6 s = num__1.2 ds = num__7.2 us = num__4.8 x / num__7.2 + x / num__4.8 = num__1 x = num__2.88 d = num__2.88 * num__2 = num__5.76 answer : c <eor> c <eos> |
c |
add__6.0__1.2__ subtract__6.0__1.2__ multiply__1.2__4.8__ round__5.76__ |
add__6.0__1.2__ subtract__6.0__1.2__ multiply__1.2__4.8__ multiply__1.2__4.8__ |
| a profit of rs . num__900 is divided between x and y in the ratio of num__0.5 : num__0.333333333333 . what is the difference between their profit shares ? <o> a ) s . num__280 <o> b ) s . num__180 <o> c ) s . num__380 <o> d ) s . num__50 <o> e ) s . num__90 |
a profit of rs . num__900 is divided between x and y in the ratio of num__0.5 : num__0.333333333333 or num__3 : num__2 . so profits are num__540 and num__360 . difference in profit share = num__540 - num__360 = num__180 answer : b <eor> b <eos> |
b |
reverse__0.5__ subtract__900.0__540.0__ multiply__0.5__360.0__ multiply__0.5__360.0__ |
reverse__0.5__ subtract__900.0__540.0__ subtract__540.0__360.0__ subtract__360.0__180.0__ |
| a train speeds past a pole in num__15 sec and a platform num__100 m long in num__25 sec its length is ? <o> a ) num__155 m <o> b ) num__150 m <o> c ) num__187 m <o> d ) num__177 m <o> e ) num__186 m |
let the length of the train be x m and its speed be y m / sec . then x / y = num__15 = > y = x / num__15 ( x + num__100 ) / num__25 = x / num__15 = > x = num__150 m . answer : b <eor> b <eos> |
b |
round__150.0__ |
round__150.0__ |
| what will come in place of the x in the following number series ? num__6 num__12 num__21 x num__48 <o> a ) num__33 <o> b ) num__36 <o> c ) num__45 <o> d ) num__50 <o> e ) num__52 |
( a ) the pattern is + num__6 + num__9 + num__12 + num__15 … … … . . so the missing term is = num__21 + num__12 = num__33 <eor> a <eos> |
a |
subtract__21.0__12.0__ add__6.0__9.0__ add__12.0__21.0__ add__12.0__21.0__ |
subtract__21.0__12.0__ add__6.0__9.0__ add__12.0__21.0__ add__12.0__21.0__ |
| what is the remainder when ( num__63 ) ( num__65 ) is divided by num__8 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__4 <o> d ) num__5 <o> e ) num__7 |
( num__63 ) ( num__65 ) = ( num__64 - num__1 ) ( num__64 + num__1 ) = num__64 ^ num__2 - num__1 which is num__1 less than a multiple of num__8 . then the remainder will be num__7 . the answer is e . <eor> e <eos> |
e |
subtract__65.0__64.0__ subtract__65.0__63.0__ subtract__8.0__1.0__ subtract__8.0__1.0__ |
subtract__65.0__64.0__ subtract__65.0__63.0__ subtract__8.0__1.0__ subtract__8.0__1.0__ |
| tara bought num__19 cartons of ice cream and num__4 cartons of yoghurt . each carton of ice cream cost $ num__7 and each carton of yoghurt cost $ num__1 . how much more did tara spend on ice cream than on yoghurt ? <o> a ) $ num__144 <o> b ) $ num__130 <o> c ) $ num__80 <o> d ) $ num__110 <o> e ) $ num__129 |
step num__1 : find the cost of the ice cream . num__19 × $ num__7 = $ num__133 step num__2 : find the cost of the yoghurt . num__4 × $ num__1 = $ num__4 step num__3 : find how much more the ice cream cost than the yoghurt . $ num__133 – $ num__4 = $ num__129 tara spent $ num__129 more on ice cream . answer is e . <eor> e <eos> |
e |
multiply__19.0__7.0__ subtract__4.0__1.0__ subtract__133.0__4.0__ multiply__1.0__129.0__ |
multiply__19.0__7.0__ subtract__4.0__1.0__ subtract__133.0__4.0__ multiply__1.0__129.0__ |
| if k and w are the dimensions of a rectangle that has area num__21 and if k and w are integers such that k > w what is the total number of possible values of k ? <o> a ) two <o> b ) three <o> c ) four <o> d ) five <o> e ) six |
kw = num__21 = num__21 * num__1 = num__7 * num__3 - - > k can take num__2 values answer : a <eor> a <eos> |
a |
multiply__1.0__2.0__ |
multiply__1.0__2.0__ |
| when num__0.222222222222 of the votes on a certain resolution have been counted num__0.75 of those counted are in favor of the resolution . what fraction e of the remaining votes must be against the resolution so that the total count will result in a vote of num__2 to num__1 against the resolution ? <o> a ) num__0.785714285714 <o> b ) num__0.722222222222 <o> c ) num__0.571428571429 <o> d ) num__0.428571428571 <o> e ) num__0.214285714286 |
if we use variable for total votes there will be too many fractions to manipulate with so pick some smart # : let set total # of votes is num__18 . num__0.222222222222 of the votes on a certain resolution have been counted - - > num__4 counted and num__18 - num__4 = num__14 votes left to be counted ; num__0.75 of those counted are in favor of the resolution - - > num__3 in favor and num__1 against ; ratio of those who voted against to those who voted for to be num__2 to num__1 there should be total of num__18 * num__0.666666666667 = num__12 people who voted against so in the remaining num__14 votes there should be num__12 - num__1 = num__11 people who voted against . thus e = num__0.785714285714 of the remaining votes must be against . answer : a . <eor> a <eos> |
a |
subtract__18.0__4.0__ multiply__0.75__4.0__ divide__2.0__3.0__ multiply__3.0__4.0__ subtract__12.0__1.0__ divide__11.0__14.0__ multiply__1.0__0.7857__ |
subtract__18.0__4.0__ multiply__0.75__4.0__ divide__2.0__3.0__ multiply__3.0__4.0__ subtract__12.0__1.0__ divide__11.0__14.0__ multiply__1.0__0.7857__ |
| of num__60 children num__30 are happy num__10 are sad and num__20 are neither happy nor sad . there are num__18 boys and num__42 girls . if there are num__6 happy boys and num__4 sad girls how many boys are neither happy nor sad ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
venn diagrams are useful for multiple values of a single variable e . g . state of mind - happy / sad / neither . when you have two or more variables such as here where you have gender - boy / girl too it becomes unwieldy . in this case either use the table or logic . table method is shown above ; here is how you will use logic : there are num__6 happy boys . there are num__4 sad girls but total num__10 sad children . so rest num__6 sad children must be sad boys . we have num__6 happy boys and num__6 sad boys . total we have num__18 boys . so num__18 - num__6 - num__6 = num__6 boys must be neither happy nor sad . answer ( c ) <eor> c <eos> |
c |
die_space__ |
die_space__ |
| which of the following is closest to ( num__7 ! - num__6 ! ) / ( num__7 ! + num__6 ! ) ? <o> a ) num__0.75 <o> b ) num__1.25 <o> c ) num__0.92 <o> d ) num__0.86 <o> e ) num__1 |
( num__7 ! - num__6 ! ) / ( num__7 ! + num__6 ! ) num__6 ! ( num__7 - num__1 ) / num__6 ! ( num__7 + num__1 ) num__0.75 = num__0.75 = num__0.75 a is the answer <eor> a <eos> |
a |
subtract__7.0__6.0__ multiply__0.75__1.0__ |
subtract__7.0__6.0__ divide__0.75__1.0__ |
| sum of num__3 consecutive even no . ' s is num__26 more than the num__1 st no . of the series . find the middle no . ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__17 <o> e ) num__18 |
let the numbers be x x + num__2 and x + num__4 then x + x + num__2 + x + num__4 = x + num__26 num__3 x + num__6 = x + num__26 num__2 x = num__20 \ inline \ therefore x = num__10 \ inline \ therefore middle number = x + num__2 = num__10 + num__2 = num__12 b <eor> b <eos> |
b |
subtract__3.0__1.0__ add__3.0__1.0__ multiply__3.0__2.0__ subtract__26.0__6.0__ add__4.0__6.0__ multiply__3.0__4.0__ multiply__3.0__4.0__ |
subtract__3.0__1.0__ add__3.0__1.0__ add__2.0__4.0__ subtract__26.0__6.0__ add__4.0__6.0__ add__2.0__10.0__ add__2.0__10.0__ |
| the distance between num__2 cities a and b is num__1000 km . a train starts from a at num__9 a . m . and travels towards b at num__100 km / hr . another starts from b at num__10 a . m . and travels towards a at num__150 km / hr . at what time do they meet ? <o> a ) num__11 am . <o> b ) num__12 p . m . <o> c ) num__3 pm . <o> d ) num__2 p . m . <o> e ) num__1 p . m . |
suppose they meet x hrs after num__9 a . m . distance moved by first in x hrs + distance moved by second in ( x - num__1 ) hrs = num__1000 num__100 x + num__150 ( x - num__1 ) = num__1000 x = num__4.60 = num__5 hrs they meet at num__9 + num__5 = num__2 p . m . answer is d <eor> d <eos> |
d |
subtract__10.0__9.0__ divide__10.0__2.0__ round__2.0__ |
subtract__10.0__9.0__ divide__10.0__2.0__ round__2.0__ |
| from the given equation find the value of z : z  ² + num__5 z + num__6 <o> a ) num__2 <o> b ) num__1 <o> c ) - num__1 <o> d ) - num__3 <o> e ) num__3 |
( z + num__2 ) ( z + num__3 ) z = - num__2 or - num__3 answer is d <eor> d <eos> |
d |
subtract__5.0__2.0__ subtract__5.0__2.0__ |
subtract__5.0__2.0__ subtract__5.0__2.0__ |
| a bell curve ( normal distribution ) has a mean of − num__1 and a standard deviation of num__0.125 . how many integer values z are within three standard deviations of the mean ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__3 <o> d ) num__6 <o> e ) num__7 |
got the question correctly - - the second item - - [ highlight ] bthe list of elements in the set is required . [ / highlight ] is not required . with the new information there is only one integer value ( - num__1 ) that is between ( - num__1.375 - num__0.625 ) i . e . falls within the three num__3 sd range . b <eor> b <eos> |
b |
reverse__1.0__ |
reverse__1.0__ |
| in smithtown the ratio of right - handed people to left - handed people is num__3 to num__1 and the ratio of men to women is num__3 to num__2 . if the number of right - handed men is maximized then what percent z of all the people in smithtown are left - handed women ? <o> a ) num__50.0 <o> b ) num__40.0 <o> c ) num__25.0 <o> d ) num__20.0 <o> e ) num__10 % |
looking at the ratio we can take total number of people = num__20 . . ans num__0.25 or num__25.0 c <eor> c <eos> |
c |
multiply__1.0__25.0__ |
multiply__1.0__25.0__ |
| the average age of num__11 boys is num__50 if the average age of first six boys is num__49 and that of the last six is num__52 . find the boy age ? <o> a ) num__48 <o> b ) num__50 <o> c ) num__56 <o> d ) num__62 <o> e ) num__64 |
num__1 to num__11 = num__11 * num__50 = num__550 num__1 to num__6 = num__6 * num__49 = num__294 num__6 to num__11 = num__6 * num__52 = num__312 num__6 th = num__294 + num__312 – num__550 = num__56 c <eor> c <eos> |
c |
subtract__50.0__49.0__ multiply__11.0__50.0__ multiply__49.0__6.0__ multiply__52.0__6.0__ add__50.0__6.0__ add__50.0__6.0__ |
subtract__50.0__49.0__ multiply__11.0__50.0__ multiply__49.0__6.0__ multiply__52.0__6.0__ add__50.0__6.0__ add__50.0__6.0__ |
| a bank offers num__5.0 c . i . calculated on half - yearly basis . a customer deposits rs . num__1600 each on num__1 st january and num__1 st july of a year . at the end of the year the amount he would have gained by way of interest is ? <o> a ) num__288 <o> b ) num__121 <o> c ) num__772 <o> d ) num__992 <o> e ) num__212 |
amount = [ num__1600 * ( num__1 + num__5 / ( num__2 * num__100 ) num__2 + num__1600 * ( num__1 + num__5 / ( num__2 * num__100 ) ] = [ num__1600 * num__1.025 ( num__1.025 + num__1 ) = [ ( num__1600 * num__41 * num__81 ) / ( num__40 * num__40 ) ] = rs . num__3321 . c . i . = num__3321 - num__3200 = rs . num__121 . answer : b <eor> b <eos> |
b |
percent__100.0__121.0__ |
percent__100.0__121.0__ |
| if x + ( num__1 / x ) = num__5 what is the value of t = x ^ num__2 + ( num__1 / x ) ^ num__2 ? <o> a ) t = num__21 <o> b ) t = num__22 <o> c ) t = num__23 <o> d ) num__24 <o> e ) num__27 |
squaring on both sides x ^ num__2 + ( num__1 / x ) ^ num__2 + num__2 ( x ) ( num__1 / x ) = num__5 ^ num__2 x ^ num__2 + ( num__1 / x ) ^ num__2 = num__23 answer : c <eor> c <eos> |
c |
multiply__1.0__23.0__ |
divide__23.0__1.0__ |
| in how many ways can an answer key for a quiz be written if the quiz contains num__4 true - false questions followed by num__2 multiples - choice questions with num__4 answer choices each if the correct answers to all true - false questions can not be the same ? <o> a ) num__120 <o> b ) num__190 <o> c ) num__224 <o> d ) num__298 <o> e ) num__256 |
there are num__2 ^ num__4 = num__16 possibilities for the true - false answers . however we need to remove two cases for tttt and ffff . there are num__4 * num__4 = num__16 possibilities for the multiple choice questions . the total number of possibilities is num__14 * num__16 = num__224 . the answer is c . <eor> c <eos> |
c |
subtract__16.0__2.0__ multiply__16.0__14.0__ multiply__16.0__14.0__ |
subtract__16.0__2.0__ multiply__16.0__14.0__ multiply__16.0__14.0__ |
| a trained covered x km at num__40 kmph and another num__2 x km at num__20 kmph . find the average speed of the train in covering the entire num__3 x km . <o> a ) num__28 kmph <o> b ) num__26 kmph <o> c ) num__24 kmph <o> d ) num__84 kmph <o> e ) num__74 kmph |
total time taken = x / num__40 + num__2 x / num__20 hours = num__5 x / num__40 = x / num__8 hours average speed = num__3 x / ( x / num__8 ) = num__24 kmph answer : c <eor> c <eos> |
c |
add__2.0__3.0__ divide__40.0__5.0__ multiply__3.0__8.0__ multiply__3.0__8.0__ |
add__2.0__3.0__ divide__40.0__5.0__ multiply__3.0__8.0__ multiply__3.0__8.0__ |
| the amount of time that three people worked on a special project was in the ratio of num__2 to num__3 to num__4 . if the project took num__90 hours how many more hours did the hardest working person work than the person who worked the least ? <o> a ) num__18 hours <o> b ) num__22 hours <o> c ) num__25 hours <o> d ) num__26 hours <o> e ) num__20 hours |
let the persons be a b c . hours worked : a = num__2 * num__10.0 = num__20 hours b = num__3 * num__10.0 = num__30 hours c = num__4 * num__10.0 = num__40 hours c is the hardest worker and a worked for the least number of hours . so the difference is num__40 - num__20 = num__20 hours . answer : e <eor> e <eos> |
e |
multiply__2.0__10.0__ multiply__3.0__10.0__ multiply__2.0__20.0__ round__20.0__ |
multiply__2.0__10.0__ multiply__3.0__10.0__ multiply__2.0__20.0__ multiply__2.0__10.0__ |
| in a dairy farm num__20 cows eat num__10 bags of husk in num__40 days . in how many days one cow will eat one bag of husk ? <o> a ) num__20 <o> b ) num__40 <o> c ) num__80 <o> d ) num__30 <o> e ) num__60 |
one cow eats num__0.5 = num__0.5 bags of husk in num__40 days . then it eats num__1 bag in num__40 / num__0.5 = num__80 days answer : c <eor> c <eos> |
c |
divide__20.0__40.0__ divide__40.0__0.5__ round__80.0__ |
divide__20.0__40.0__ divide__40.0__0.5__ divide__40.0__0.5__ |
| hcf of two numbers is num__15 and their lcm is num__180 . if their sum is num__105 then the numbers are <o> a ) num__30 and num__75 <o> b ) num__35 and num__70 <o> c ) num__40 and num__65 <o> d ) num__45 and num__60 <o> e ) num__55 and num__70 |
explanation : let the numbers be num__15 a and num__15 b . then num__15 a + num__15 b = num__105 or a + b = num__7 . . ( i ) lcm = num__15 ab = num__180 ab = num__12 . . ( ii ) solving equations ( i ) and ( ii ) we get a = num__4 b = num__3 so the numbers are num__15 × num__4 and num__15 × num__3 i . e . num__60 and num__45 answer : d <eor> d <eos> |
d |
divide__105.0__15.0__ divide__180.0__15.0__ subtract__15.0__12.0__ hour_to_min_conversion__ multiply__15.0__3.0__ round__45.0__ |
divide__105.0__15.0__ divide__180.0__15.0__ subtract__15.0__12.0__ hour_to_min_conversion__ multiply__15.0__3.0__ round__45.0__ |
| anna and carol buy cds and tapes at a music store that sells each of its cds for a certain price and each of its tapes for a certain price . anna spends twice as much as carol spends buying three times as many cds and the same number of tapes . if carol spends $ num__60.00 on four cds and five tapes how much does one tape cost ? <o> a ) $ num__5.00 <o> b ) $ num__6.00 <o> c ) $ num__12.00 <o> d ) $ num__25.00 <o> e ) $ num__100.00 |
cd = > c type = > t carol : num__4 c + num__5 t = num__60 anna : num__12 c + num__5 t = num__120 num__12 c - num__4 c = num__60 num__8 c = num__60 c = num__7.5 num__4 c + num__5 t = num__60 num__30 + num__5 t = num__60 num__5 t = num__30 t = num__6 answer is b <eor> b <eos> |
b |
divide__60.0__5.0__ subtract__12.0__4.0__ divide__60.0__8.0__ multiply__7.5__4.0__ divide__30.0__5.0__ subtract__12.0__6.0__ |
divide__60.0__5.0__ subtract__12.0__4.0__ divide__60.0__8.0__ multiply__7.5__4.0__ divide__30.0__5.0__ subtract__12.0__6.0__ |
| one out of five employees are capable of doing a certain task . num__50 percent of the five employees including the one who are capable are assigned to a project involving this task . what percentage of employees assigned to the project are not capable ? <o> a ) num__43.33 <o> b ) num__33.33 <o> c ) num__50.0 <o> d ) num__38.33 <o> e ) num__23.33 % |
given num__40.0 of num__5 employees including num__1 who are capable of doing task . num__40.0 of num__5 employeees = num__0.4 * num__5 = num__2 employees = = = > num__1 employees who are capable of doing the task and one employee who is not capable . percentage of employees assigned who are not capable = num__0.5 * num__100 = num__50.0 answer : c <eor> c <eos> |
c |
percent__1.0__40.0__ percent__5.0__40.0__ percent__50.0__1.0__ percent__50.0__100.0__ |
percent__1.0__40.0__ percent__5.0__40.0__ percent__50.0__1.0__ percent__50.0__100.0__ |
| a man spends rs . num__1600 per month on an average for the first three months rs num__1550 for next four months and rs . num__1800 per month for the last five months and saves rs . num__4000 a year . what is his average monthly income ? <o> a ) num__2000 <o> b ) num__2010 <o> c ) num__2100 <o> d ) num__2200 <o> e ) num__2300 |
total expenditure for the first num__3 months = num__3 × num__1600 = num__4800 total expenditure for num__4 months = num__4 × num__1550 = num__6200 total expenditure for num__5 months = num__5 × num__1800 = num__9000 total expenditure and saving ( which is income for one year ) = num__4800 + num__6200 + num__9000 + num__4000 = rs . num__24000 so average monthly income = num__2000.0 = rs . num__2000 a <eor> a <eos> |
a |
multiply__1600.0__3.0__ multiply__1550.0__4.0__ multiply__1800.0__5.0__ multiply__4800.0__5.0__ subtract__4000.0__2000.0__ |
multiply__1600.0__3.0__ multiply__1550.0__4.0__ multiply__1800.0__5.0__ multiply__4800.0__5.0__ subtract__4000.0__2000.0__ |
| at num__1 : num__00 pm train x departed from station a on the road to station b . at num__1 : num__10 pm train y departed station b on the same road for station a . if station a and station b are p miles apart train x ’ s speed is r miles per hour and train y ’ s speed is s miles per hour how many hours after num__1 : num__00 pm in terms of p r and s do the two trains pass each other ? <o> a ) num__0.5 + ( p - num__0.5 s ) / ( r + s ) <o> b ) ( p - num__0.5 s ) / ( r + s ) <o> c ) num__0.5 + ( p - num__0.5 r ) / r <o> d ) ( p - num__0.5 r ) / ( r + s ) <o> e ) num__0.5 + ( p - num__0.5 r ) / ( r + s ) |
the distance a is going to cover between num__1 : num__00 and num__1 : num__30 = . num__5 r now the distance between the two trains = ( p - . num__5 r ) the relative velocity = ( r - ( - s ) ) = r + s from num__1 : num__30 time is going to take when they meet = ( p - . num__5 r ) / ( r + s ) so the ans is . num__5 + ( ( p - . num__5 r ) / ( r + s ) ) [ . num__5 is added for the time from num__1 : num__00 to num__1 : num__30 ] ans is a <eor> a <eos> |
a |
divide__5.0__10.0__ |
divide__5.0__10.0__ |
| the average age of num__17 students of a class is num__17 years . out of these the average age of num__5 students is num__14 years and that of the other num__9 students is num__16 years the age of the num__17 th student is <o> a ) num__62 <o> b ) num__66 <o> c ) num__18 <o> d ) num__75 <o> e ) num__12 |
explanation : age of the num__17 th student = [ num__17 * num__17 - ( num__14 * num__5 + num__16 * num__9 ) ] = ( num__289 - num__214 ) = num__75 years . answer : d <eor> d <eos> |
d |
subtract__289.0__214.0__ subtract__289.0__214.0__ |
subtract__289.0__214.0__ subtract__289.0__214.0__ |
| if num__1000 microns = num__1 decimeter and num__1 num__000000 angstroms = num__1 decimeter how many angstroms equal num__1 micron ? <o> a ) num__100 <o> b ) num__10 <o> c ) num__1000 <o> d ) num__10000 <o> e ) num__100 |
000 |
given that num__1000 microns = num__1 decimeter = num__1 num__000000 angstroms so num__1 micron = num__1 num__000000 / num__1000 = num__1000 answer : c <eor> c <eos> |
c |
c |
| kate and danny each have $ num__10 . together they flip a fair coin num__5 times . every time the coin lands on heads kate gives danny $ num__1 . every time the coin lands on tails danny gives kate $ num__1 . after the five coin flips what is the probability m that kate has more than $ num__10 but less than $ num__15 ? <o> a ) num__0.3125 <o> b ) num__0.5 <o> c ) m = num__0.4 <o> d ) m = num__0.46875 <o> e ) m = num__0.375 |
for num__21 heads kate will end up with < $ num__10 and we want her to win . therefore only possibilities are num__3 or num__4 heads . i made an educated guess and it worked fine . ans - ' d ' mgmat ' s anagram helped here as well . hhhht = num__5 ! / num__4 ! * num__1 ! = num__5 hhhtt = num__5 ! / num__3 ! * num__2 ! = num__10 total acceptable cases = num__15 total cases = num__32 p = num__0.46875 . d <eor> d <eos> |
d |
divide__15.0__5.0__ subtract__5.0__1.0__ divide__10.0__5.0__ divide__15.0__32.0__ multiply__1.0__0.4688__ |
divide__15.0__5.0__ subtract__5.0__1.0__ divide__10.0__5.0__ divide__15.0__32.0__ multiply__1.0__0.4688__ |
| a cyclist traveled for two days . on the second day the cyclist traveled num__4 hours longer and at an average speed num__10 mile per hour slower than she traveled on the first day . if during the two days she traveled a total of num__280 miles and spent a total of num__10 hours traveling what was her average speed on the second day ? <o> a ) num__5 mph <o> b ) num__10 mph <o> c ) num__20 mph <o> d ) num__30 mph <o> e ) num__40 mph |
solution : d = num__280 mi t = num__12 hrs day num__1 time = t num__1 day num__2 time = t num__2 t num__2 - t num__1 = num__4 hrs - - - - - ( i ) t num__1 + t num__2 = num__12 hrs - - - - - ( ii ) adding i and ii t num__2 = num__8 hrs and t num__1 = num__4 hrs day num__1 rate = r num__1 day num__2 rate = r num__2 r num__1 - r num__2 = num__10 mph i . e . r num__1 = num__10 + r num__2 num__280 = num__8 r num__2 + num__4 r num__1 i . e . num__280 = num__8 r num__2 + num__4 ( num__10 + r num__2 ) i . e . r num__2 = num__20 mph answer : e <eor> e <eos> |
e |
subtract__12.0__10.0__ multiply__4.0__2.0__ multiply__10.0__2.0__ multiply__4.0__10.0__ |
subtract__12.0__10.0__ subtract__10.0__2.0__ add__8.0__12.0__ multiply__4.0__10.0__ |
| the price of rice falls by num__20.0 . how much rice can be bought now with the money that was sufficient to buy num__20 kg of rice previously ? <o> a ) num__5 kg <o> b ) num__15 kg <o> c ) num__25 kg <o> d ) num__30 kg <o> e ) none |
solution : let rs . num__100 be spend on rice initially for num__20 kg . as the price falls by num__20.0 new price for num__20 kg rice = ( num__100 - num__20.0 of num__100 ) = num__80 new price of rice = num__4.0 = rs . num__4 per kg . rice can bought now at = num__25.0 = num__25 kg . answer : option c <eor> c <eos> |
c |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| if n = num__1000 x + num__100 y + num__10 z where x y and z are different positive integers less than num__4 the remainder when n is divided by num__9 is <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
since we can not have more than one correct answers in ps questions then pick some numbers for x y and z and find the reminder when num__1000 x + num__100 y + num__10 z is divided by num__9 . say x = num__1 y = num__2 and z = num__3 then num__1000 x + num__100 y + num__10 z = num__1230 - - > num__1230 divide by num__9 yields the remainder of num__6 ( num__1224 is divisible by num__9 since the sum of its digit is a multiple of num__9 thus num__1230 which is num__6 more than a multiple of num__9 yields the remainder of num__6 when divided by num__9 ) . answer ; c . <eor> c <eos> |
c |
subtract__10.0__9.0__ subtract__4.0__1.0__ subtract__10.0__4.0__ subtract__1230.0__6.0__ subtract__10.0__4.0__ |
subtract__10.0__9.0__ subtract__4.0__1.0__ subtract__10.0__4.0__ subtract__1230.0__6.0__ subtract__10.0__4.0__ |
| three times a number is num__20.0 more than twice another number when increased by num__105 . if twice the first number increased by num__36 is num__20.0 less than three times of the second number then what is the first number ? <o> a ) num__150 <o> b ) num__162 <o> c ) num__180 <o> d ) num__170 <o> e ) num__160 |
let the two numbers be x and y . num__3 x = num__1.2 ( num__2 y + num__105 ) num__3 x = num__2.4 y + num__126 … … … … … … . . ( num__1 ) num__2 x + num__36 = num__0.8 ( num__3 y ) num__2 x + num__36 = num__2.4 y … … … … … … . . ( num__2 ) x = num__162 answer : b <eor> b <eos> |
b |
multiply__2.0__1.2__ multiply__105.0__1.2__ round_down__1.2__ subtract__2.0__1.2__ add__36.0__126.0__ add__36.0__126.0__ |
multiply__2.0__1.2__ multiply__105.0__1.2__ round_down__1.2__ subtract__2.0__1.2__ add__36.0__126.0__ add__36.0__126.0__ |
| ram sold two bicycles each for rs . num__990 . if he made num__10.0 profit on the first and num__10.0 loss on the second what is the total cost of both bicycles ? <o> a ) num__2000 <o> b ) num__2879 <o> c ) num__2700 <o> d ) num__2797 <o> e ) num__2761 |
( num__10 * num__10 ) / num__100 = num__1.0 loss num__100 - - - num__99 ? - - - num__1980 = > rs . num__2000 answer : a <eor> a <eos> |
a |
percent__10.0__990.0__ percent__100.0__2000.0__ |
percent__10.0__990.0__ percent__100.0__2000.0__ |
| a license plate in the country kerrania consists of four digits followed by two letters . the letters a b and c are used only by government vehicles while the letters d through z are used by non - government vehicles . kerrania ' s intelligence agency has recently captured a message from the country gonzalia indicating that an electronic transmitter has been installed in a kerrania government vehicle with a license plate starting with num__79 . if it takes the police num__14 minutes to inspect each vehicle what is the probability that the police will find the transmitter within three hours ? <o> a ) num__0.227848101266 <o> b ) num__0.166666666667 <o> c ) num__0.04 <o> d ) num__0.02 <o> e ) num__0.0155555555556 |
everything is correct except that you assumed the letters ca n ' t be repeated . it can be repeated . aa bb cc ab ac bc ba ca cb . thus ; total government vehicles = ( num__10 ) ^ num__2 * ( num__3 ) ^ num__2 = num__100 * num__9 = num__900 vehicles inspected within num__3 hours = num__60 * num__0.214285714286 = num__14 p = num__0.0155555555556 = num__0.0155555555556 ans : e <eor> e <eos> |
e |
multiply__9.0__100.0__ hour_to_min_conversion__ divide__3.0__14.0__ divide__14.0__900.0__ divide__14.0__900.0__ |
multiply__9.0__100.0__ hour_to_min_conversion__ divide__3.0__14.0__ divide__14.0__900.0__ divide__14.0__900.0__ |
| a can empty the tank in num__28 hrs b can fill the tank in num__14 hrs and c takes num__7 hrs for filling . if all three pipes are opened simultaneously how long it will take to fill the tank ? <o> a ) num__4 hr num__36 minutes <o> b ) num__5 hr num__36 minutes <o> c ) num__6 hr num__36 minutes <o> d ) num__7 hr num__36 minutes <o> e ) none of these |
solution : time taken = ( num__28 * num__14 * num__7 * ) / ( num__28 * num__14 - num__14 * num__7 + num__28 * num__7 ) to solve this quicker find out lcm ( num__28 num__147 ) we get num__28 time taken = num__28 / ( - num__1.0 + num__2.0 + num__4.0 ) = num__28 / ( - num__1 + num__2 + num__4 ) = num__5.6 = num__5.6 hrs = num__5 hr num__36 minutes answer b <eor> b <eos> |
b |
divide__28.0__14.0__ divide__28.0__7.0__ divide__28.0__5.6__ round__5.0__ |
divide__28.0__14.0__ divide__28.0__7.0__ divide__28.0__5.6__ divide__28.0__5.6__ |
| vijay bought num__160 shirts at the rate of rs . num__225 per shirt . the transport expenditure was rs . num__1400 . he paid an octroi at the rate of rs . num__1.75 per shirt and labour charges were rs . num__320 . what should be the selling price of one shirt if he wants a profit of num__60.0 ? <o> a ) num__282 <o> b ) num__229 <o> c ) num__277 <o> d ) num__380 <o> e ) num__2812 |
total cp per shirt = num__225 + num__8.75 + num__1.75 + num__2.0 = rs . num__237.5 sp = cp [ ( num__100 + profit % ) / num__100 ] = num__237.5 * [ ( num__100 + num__60 ) / num__100 ] = rs . num__380 . answer : d <eor> d <eos> |
d |
percent__100.0__380.0__ |
percent__100.0__380.0__ |
| the two trains of lengths num__400 m num__600 m respectively running at same directions . the faster train can cross the slower train in num__180 sec the speed of the slower train is num__48 km . then find the speed of the faster train ? <o> a ) num__28 <o> b ) num__68 <o> c ) num__66 <o> d ) num__51 <o> e ) num__11 |
length of the two trains = num__600 m + num__400 m speed of the first train = x speed of the second train = num__48 kmph num__1000 / x - num__48 = num__180 num__1000 / x - num__48 * num__0.277777777778 = num__180 num__50 = num__9 x - num__120 x = num__68 kmph answer : b <eor> b <eos> |
b |
add__400.0__600.0__ round__68.0__ |
add__400.0__600.0__ round__68.0__ |
| ramu bought an old car for rs . num__42000 . he spent rs . num__8000 on repairs and sold it for rs . num__64900 . what is his profit percent ? <o> a ) num__29.8 <o> b ) num__16.0 <o> c ) num__18.0 <o> d ) num__82.0 <o> e ) num__23 % |
total cp = rs . num__42000 + rs . num__8000 = rs . num__50000 and sp = rs . num__64900 profit ( % ) = ( num__64900 - num__50000 ) / num__50000 * num__100 = num__29.8 answer : a <eor> a <eos> |
a |
percent__29.8__100.0__ |
percent__29.8__100.0__ |
| there are num__11 person among whom num__2 are brother . the total no . of ways in which these persons can be seated around a round table so that exactly num__1 person sit between the brothers is equal to ? <o> a ) num__4 ! * num__2 ! <o> b ) num__7 ! * num__2 ! <o> c ) num__6 ! * num__1 ! <o> d ) num__2 ! * num__1 ! <o> e ) num__8 ! * num__2 ! |
total number of ways = num__8 ! * num__2 ! . e <eor> e <eos> |
e |
multiply__1.0__8.0__ |
multiply__1.0__8.0__ |
| in how many years will a sum of money doubles itself at num__5.0 per annum on simple interest ? <o> a ) num__80.0 <o> b ) num__20.0 <o> c ) num__29.0 <o> d ) num__30.0 <o> e ) num__60 % |
p = ( p * num__5 * r ) / num__100 r = num__20.0 answer : b <eor> b <eos> |
b |
percent__20.0__100.0__ |
percent__20.0__100.0__ |
| two trains num__121 meters and num__165 meters in length respectively are running in opposite directions one at the rate of num__80 km and the other at the rate of num__65 kmph . in what time will they be completely clear of each other from the moment they meet ? <o> a ) num__7.16 <o> b ) num__7.12 <o> c ) num__7.11 <o> d ) num__7.15 <o> e ) num__7.13 |
t = ( num__121 + num__165 ) / ( num__80 + num__65 ) * num__3.6 t = num__7.15 answer : d <eor> d <eos> |
d |
round__7.15__ |
round__7.15__ |
| pipe a usually fills a tank in num__2 hours . on account of a leak at the bottom of the tank it takes pipe a num__30 more minutes to fill the tank . how long will the leak take to empty a full tank if pipe a is shut ? <o> a ) num__7 hours <o> b ) num__8 hours <o> c ) num__9 hours <o> d ) num__10 hours <o> e ) num__11 hours |
pipe a fills the tank normally in num__2 hours . therefore it will fill ½ of the tank in an hour . let the leak take x hours to empty a full tank when pipe a is shut . therefore the leak will empty of the tank in an hour . the net amount of water that gets filled in the tank in an hour when pipe a is open and when there is a leak = of the tank . — ( num__1 ) when there is a leak the problem states that pipe a takes two and a half hours to fill the tank . i . e . hours . therefore in an hour of the tank gets filled . – ( num__2 ) equating ( num__1 ) and ( num__2 ) we get = > = > x = num__10 hours . the problem can also be mentally done as follows . pipe a takes num__2 hours to fill the tank . therefore it fills half the tank in an hour or num__50.0 of the tank in an hour . when there is a leak it takes num__2 hours num__30 minutes for the tank to fill . i . e hours to fill the tank or or num__40.0 of the tank gets filled . on account of the leak ( num__50 - num__40 ) % = num__10.0 of the water gets wasted every hour . therefore the leak will take num__10 hours to drain a full tank . answer : d <eor> d <eos> |
d |
add__30.0__10.0__ round__10.0__ |
subtract__50.0__10.0__ subtract__40.0__30.0__ |
| a telephone number contains num__10 digit including a num__3 - digit area code . bob remembers the area code and the next num__5 digits of the number . he also remembers that the remaining digits are not num__1 num__6 or num__7 . if bob tries to find the number by guessing the remaining digits at random the probability that he will be able to find the correct number in at most num__2 attempts is closest to which of the following ? <o> a ) num__0.142857142857 <o> b ) num__0.0714285714286 <o> c ) num__0.0357142857143 <o> d ) num__0.0408163265306 <o> e ) num__0.0246913580247 |
there are num__7 * num__7 = num__49 possibilities . bob has num__2 chances so the probability is num__0.0408163265306 . the answer is d . <eor> d <eos> |
d |
divide__2.0__49.0__ multiply__1.0__0.0408__ |
divide__2.0__49.0__ multiply__1.0__0.0408__ |
| num__5 n + num__7 > num__12 and num__7 n - num__5 < num__44 ; n must be between which numbers ? <o> a ) num__1 and num__8 <o> b ) num__2 and num__6 <o> c ) num__0 and num__9 <o> d ) num__1 and num__7 <o> e ) num__2 and num__9 |
num__5 n > num__5 - - > n > num__1 num__7 n < num__49 - - > n < num__7 num__1 < n < num__7 answer : d <eor> d <eos> |
d |
add__5.0__44.0__ reverse__1.0__ |
add__5.0__44.0__ reverse__1.0__ |
| a tank is filled in num__5 hours by three pipes a b and c . the pipe c is twice as fast as b and b is twice as fast as a . how much time will pipe a alone take to fill the tank ? <o> a ) num__35 hrs <o> b ) num__20 hrs <o> c ) num__10 hrs <o> d ) num__15 hrs <o> e ) num__50 hrs |
pipe a alone take x hrs then pipe b and c will take x / num__2 and x / num__4 hrs = > num__1 / x + num__2 / x + num__4 / x = num__1 / x = > num__7 / x = num__0.2 x = num__35 hrs answer a <eor> a <eos> |
a |
subtract__5.0__4.0__ add__5.0__2.0__ divide__1.0__5.0__ multiply__5.0__7.0__ round__35.0__ |
subtract__5.0__4.0__ add__5.0__2.0__ divide__1.0__5.0__ divide__7.0__0.2__ divide__35.0__1.0__ |
| the weight of four dogs is determined to be num__25 pounds num__31 pounds num__35 pounds and num__33 pounds respectively . the weight of a fifth dog is determined to be y pounds . if the average ( arithmetic mean ) weight of the first four dogs is the same as that of all five dogs what is the value of y ? <o> a ) num__31 <o> b ) num__33 <o> c ) num__35 <o> d ) num__37 <o> e ) num__39 |
total weight of the num__4 dogs = ( num__25 + num__31 + num__35 + num__33 ) = num__124 avg = num__31.0 = num__31 total weight of num__5 dogs = num__124 + y or num__4 ( num__31 ) + y average of num__5 dogs as per question = num__31 equation : num__4 ( num__31 ) + y = num__5 ( num__31 ) or y = num__31 . choose a <eor> a <eos> |
a |
subtract__35.0__31.0__ multiply__31.0__4.0__ subtract__35.0__4.0__ |
subtract__35.0__31.0__ multiply__31.0__4.0__ subtract__35.0__4.0__ |
| the sum of three consecutive even numbers is num__42 . find the middle number of the three ? <o> a ) num__5 <o> b ) num__8 <o> c ) num__96 <o> d ) num__32 <o> e ) num__14 |
explanation : three consecutive even numbers ( num__2 p - num__2 ) num__2 p ( num__2 p + num__2 ) . ( num__2 p - num__2 ) + num__2 p + ( num__2 p + num__2 ) = num__42 num__6 p = num__42 = > p = num__7 . the middle number is : num__2 p = num__14 . answer e <eor> e <eos> |
e |
divide__42.0__6.0__ multiply__2.0__7.0__ round__14.0__ |
divide__42.0__6.0__ multiply__2.0__7.0__ round__14.0__ |
| num__200 liters of a mixture contains milk and water in the ratio num__4 : num__1 . if num__20 liters of this mixture be replaced by num__20 liters of milk the ratio of milk to water in the new mixture would be ? <o> a ) num__6 : num__1 <o> b ) num__1 : num__6 <o> c ) num__2 : num__3 <o> d ) num__3 : num__2 <o> e ) num__4 : num__1 |
quantity of milk in num__200 liters if mix = num__200 * num__0.8 = num__160 liters quantity of milk in num__210 liters of new mix = num__160 + num__20 = num__180 liters quantity of water in it = num__210 - num__180 = num__30 liters ratio of milk and water in new mix = num__180 : num__30 = num__6 : num__1 answer is a <eor> a <eos> |
a |
multiply__200.0__0.8__ subtract__200.0__20.0__ subtract__210.0__180.0__ divide__180.0__30.0__ multiply__1.0__6.0__ |
multiply__200.0__0.8__ subtract__200.0__20.0__ subtract__210.0__180.0__ divide__180.0__30.0__ multiply__1.0__6.0__ |
| operation # is defined as adding a randomly selected two digit multiple of num__14 to a randomly selected two digit prime number and reducing the result by half . if operation # is repeated num__10 times what is the probability that it will yield at least two integers ? <o> a ) num__0.0 <o> b ) num__10.0 <o> c ) num__20.0 <o> d ) num__30.0 <o> e ) num__40 % |
any multiple of num__14 is even . any two - digit prime number is odd . ( even + odd ) / num__2 is not an integer . thus # does not yield an integer at all . therefore p = num__0 . answer : a . <eor> a <eos> |
a |
multiply__14.0__0.0__ |
divide__0.0__14.0__ |
| an bus covers a certain distance at aspeed of num__180 kmph in num__5 hours . to cover the same distance in num__1 hr it must travel at a speed of ? <o> a ) num__560 km / h <o> b ) num__567 km / h <o> c ) num__779 km / h <o> d ) num__723 km / h <o> e ) num__540 km / h |
distance = ( num__180 x num__5 ) = num__900 km . speed = distance / time speed = num__900 / ( num__1.66666666667 ) km / hr . [ we can write num__1 hours as num__1.66666666667 hours ] required speed = num__900 x num__0.6 km / hr = num__540 km / hr . e <eor> e <eos> |
e |
multiply__180.0__5.0__ km_to_mile_conversion__ multiply__900.0__0.6__ round__540.0__ |
multiply__180.0__5.0__ divide__1.0__1.6667__ multiply__900.0__0.6__ divide__540.0__1.0__ |
| kim finds a num__4 - meter tree branch and marks it off in thirds and fifths . she then breaks the branch along all the markings and removes one piece of every distinct length . what fraction of the original branch remains ? <o> a ) num__0.4 <o> b ) num__1.4 <o> c ) num__0.6 <o> d ) num__0.533333333333 <o> e ) num__0.5 |
num__3 pieces of num__0.2 length and two piece each of num__0.0666666666667 and num__0.133333333333 lengths . removing one piece each from pieces of each kind of lengths the all that will remain will be num__2 pieces of num__0.2 i . e num__0.4 num__1 piece of num__0.0666666666667 and num__1 piece of num__0.133333333333 which gives us num__0.4 + num__0.0666666666667 + num__0.133333333333 - - - - - > num__0.6 answer is c <eor> c <eos> |
c |
divide__0.2__3.0__ subtract__0.2__0.0667__ multiply__2.0__0.2__ subtract__4.0__3.0__ km_to_mile_conversion__ km_to_mile_conversion__ |
divide__0.2__3.0__ subtract__0.2__0.0667__ multiply__2.0__0.2__ subtract__4.0__3.0__ subtract__1.0__0.4__ subtract__1.0__0.4__ |
| a num__750 m long train crosses a platform in num__39 sec while it crosses a signal pole in num__18 sec . what is the length of the platform ? <o> a ) num__300 <o> b ) num__875 <o> c ) num__360 <o> d ) num__770 <o> e ) num__380 |
speed = num__41.6666666667 = num__41.6666666667 m / sec . let the length of the platform be x meters . then ( x + num__750 ) / num__39 = num__41.6666666667 = > x = num__1625 m . l = num__1625 - num__750 = num__875 answer : option b <eor> b <eos> |
b |
divide__750.0__18.0__ subtract__1625.0__750.0__ round__875.0__ |
divide__750.0__18.0__ subtract__1625.0__750.0__ subtract__1625.0__750.0__ |
| the average age of a group of num__12 students was num__20 . the average age increased by num__2 years when two new students joined the group . what is the average age of the two new students who joined the group ? <o> a ) num__22 years <o> b ) num__30 years <o> c ) num__34 years <o> d ) num__32 years <o> e ) none of these |
answer the average age of a group of num__12 students is num__20 . therefore the sum of the ages of all num__12 of them = num__12 * num__20 = num__240 when two new students join the group the average age increases by num__2 . new average = num__22 . now there are num__14 students . therefore the sum of the ages of all num__14 of them = num__14 * num__22 = num__308 therefore the sum of the ages of the two new students who joined = num__308 - num__240 = num__68 and the average age of each of the two new students = num__34.0 = num__34 years . answer c <eor> c <eos> |
c |
multiply__12.0__20.0__ add__20.0__2.0__ add__12.0__2.0__ multiply__14.0__22.0__ subtract__308.0__240.0__ add__12.0__22.0__ add__12.0__22.0__ |
multiply__12.0__20.0__ add__20.0__2.0__ add__12.0__2.0__ multiply__14.0__22.0__ subtract__308.0__240.0__ add__12.0__22.0__ subtract__68.0__34.0__ |
| a man has some hens and cows . if the number of heads be num__48 and the number of feet equals num__140 then the number of hens will be : <o> a ) num__22 <o> b ) num__23 <o> c ) num__24 <o> d ) num__26 <o> e ) num__27 |
explanation let the number of hens be x and the number of cows be y . then x + y = num__48 … . ( i ) and num__2 x + num__4 y = num__140 x + num__2 y = num__70 … . ( ii ) solving ( i ) and ( ii ) we get : x = num__26 y = num__22 . the required answer = num__26 . answer d <eor> d <eos> |
d |
divide__140.0__2.0__ subtract__48.0__26.0__ subtract__48.0__22.0__ |
divide__140.0__2.0__ subtract__48.0__26.0__ add__4.0__22.0__ |
| lionel left his house and walked towards walt ' s house num__48 miles away . two hours later walt left his house and ran towards lionel ' s house . if lionel ' s speed was num__3 miles per hour and walt ' s num__3 miles per hour how many miles had lionel walked when he met walt ? <o> a ) num__12 <o> b ) num__16 <o> c ) num__20 <o> d ) num__27 <o> e ) num__28 |
in the first num__2 hours lionel at the rate of num__3 miles per hour covered distance = rate * time = num__3 * num__2 = num__6 miles . so the distance between him and walt was num__48 - num__6 = num__42 miles when walt left his house . now their combined rate to cover this distance was num__3 + num__3 = num__6 miles per hour hence they will meet ( they will cover that distance ) in time = distance / rate = num__7.0 = num__7 hours . total time that lionel was walking is num__2 + num__7 = num__9 hours which means that he covered in that time interval distance = rate * time = num__3 * num__9 = num__27 miles . answer : d . <eor> d <eos> |
d |
multiply__3.0__2.0__ subtract__48.0__6.0__ divide__42.0__6.0__ add__3.0__6.0__ multiply__3.0__9.0__ round__27.0__ |
multiply__3.0__2.0__ subtract__48.0__6.0__ divide__42.0__6.0__ add__3.0__6.0__ multiply__3.0__9.0__ multiply__3.0__9.0__ |
| a and b can do a piece of work in num__18 days ; band c can do it in num__24 days a and c can do it in num__36 days . in how many days will a band c finish it together ? <o> a ) num__16 days <o> b ) num__12 days <o> c ) num__3 days <o> d ) num__5 days <o> e ) num__9 days |
sol . ( a + b ) ' s num__1 day ' s work = ( num__0.0555555555556 ) ( b + c ) ' s num__1 day ' s work = ( num__0.0416666666667 ) and ( a + c ) ' s num__1 day ' s work = ( num__0.0277777777778 ) adding we get : num__2 ( a + b + c ) ' s num__1 day ' s work = ¬ ( num__0.0555555555556 + num__0.0416666666667 + num__0.0277777777778 ) = num__0.125 = num__0.125 ( a + b + c ) ' s num__1 day ' s work = num__0.0625 thus a band c together can finish the work in num__16 days . ans : a <eor> a <eos> |
a |
divide__1.0__18.0__ divide__1.0__24.0__ divide__1.0__36.0__ divide__36.0__18.0__ divide__0.125__2.0__ subtract__18.0__2.0__ round__16.0__ |
divide__1.0__18.0__ divide__1.0__24.0__ divide__1.0__36.0__ divide__36.0__18.0__ divide__0.125__2.0__ subtract__18.0__2.0__ round__16.0__ |
| after the typist writes num__12 letters and addresses num__12 envelopes she inserts the letters randomly into the envelopes ( num__1 letter per envelope ) . what is the probability that exactly num__1 letter is inserted in an improper envelope ? <o> a ) num__0.0833333333333 <o> b ) num__0 <o> c ) num__0.0566037735849 <o> d ) num__0.916666666667 <o> e ) num__0.846153846154 |
num__0 if one letter is in wrong envelope the letter corresponding to that letter will also be in wrong envelope . answer : b <eor> b <eos> |
b |
multiply__12.0__0.0__ |
multiply__12.0__0.0__ |
| in an election between two candidates one got num__55.0 of the total valid votes num__20.0 of the votes were invalid . if the total number of votes was num__7500 the number of valid votes that the other candidate got was <o> a ) num__2500 <o> b ) num__2700 <o> c ) num__3000 <o> d ) num__3100 <o> e ) nobe |
solution number of valid votes = num__80.0 of num__7500 = num__6000 . valid votes polled by other candidates = num__45.0 of num__6000 ( num__0.45 × num__6000 ) = num__2700 . answer b <eor> b <eos> |
b |
percent__80.0__7500.0__ percent__45.0__6000.0__ percent__45.0__6000.0__ |
percent__80.0__7500.0__ percent__45.0__6000.0__ percent__45.0__6000.0__ |
| the radius of a wheel is num__22.4 cm . what is the distance covered by the wheel in making num__500 resolutions ? <o> a ) num__287 m <o> b ) num__704 m <o> c ) num__278 m <o> d ) num__298 m <o> e ) num__286 m |
in one resolution the distance covered by the wheel is its own circumference . distance covered in num__500 resolutions . = num__500 * num__2 * num__3.14285714286 * num__22.4 = num__70400 cm = num__704 m answer : b <eor> b <eos> |
b |
round__704.0__ |
round__704.0__ |
| victor ' s job requires him to complete a series of identical jobs . if victor is supervised at work he finishes each job three days faster than if he is unsupervised . if victor works for num__144 days and is supervised for half the time he will finish a total of num__36 jobs . how long t would it take victor to complete num__10 jobs without any supervision ? <o> a ) num__34 <o> b ) num__52 <o> c ) num__60 <o> d ) num__70 <o> e ) num__92 |
rate when supervised = ( job ) / ( time ) = num__1 / t . rate when unsupervised = ( job ) / ( time ) = num__1 / ( t + num__3 ) . for num__72.0 = num__72 days he is supervised and for num__72.0 = num__72 days he is unsupervised and does num__36 jobs : num__72 / t + num__72 / ( t + num__3 ) = num__36 - - > t = num__3 days - - > t + num__3 = num__6 days . victor to complete num__10 jobs without any supervision will need t num__10 ( t + num__3 ) = num__60 days . answer : c . <eor> c <eos> |
c |
hour_to_min_conversion__ hour_to_min_conversion__ |
hour_to_min_conversion__ divide__60.0__1.0__ |
| how many three digit number are thre in which if one digit is num__3 then it must be follwed by num__7 . repetetion is not allowed <o> a ) num__466 <o> b ) num__463 <o> c ) num__14 <o> d ) num__17 <o> e ) num__18 |
num__1 st case : three digits no without considering num__3 first place can be filled using num__8 numbers ( excluding num__0 and num__3 ) = num__8 second place can be filled using num__8 numbers ( excluding num__3 and digit filled in first place ) = num__8 last place can be filled using num__7 numbers ( excluding num__3 and digit filled in first place & second place ) = num__7 therefore three digits formed are = num__8 * num__8 * num__7 = num__448 num__2 nd case : now consider number using num__3 if num__3 and num__7 are first two digit then last digit can be filled using num__8 digits if num__3 and num__7 are last two digits then first place can be filled using num__7 digits therefore total num__3 digit no including num__3 are = num__8 + num__7 = num__15 hence total num__3 digits numbers are = num__448 + num__15 = num__463 answer : b <eor> b <eos> |
b |
add__7.0__1.0__ subtract__3.0__1.0__ add__7.0__8.0__ add__448.0__15.0__ add__448.0__15.0__ |
add__7.0__1.0__ subtract__3.0__1.0__ add__7.0__8.0__ add__448.0__15.0__ add__448.0__15.0__ |
| there are two inlets and one outlet to a cistern . one of the inlets takes num__2.5 hours to fill up the cistern and the other inlet takes twice as much time to fill up the same cistern . both of the inlets are turned on at num__9 : num__00 am with the cistern completely empty and at num__10 : num__00 am the outlet is turned on and it takes num__1 more hour to fill the cistern completely . how much time does the outlet working alone takes to empty the cistern when the cistern is full ? <o> a ) num__2 hours <o> b ) num__2.5 hours <o> c ) num__3 hours <o> d ) num__3.5 hours <o> e ) num__5 hours |
the combined inflow rate of the two inlets is num__0.4 + num__0.2 = num__0.6 cistern / hour . thus working together it takes num__1.66666666667 hours ( time is reciprocal of rate ) to fill the cistern . from num__9 : num__00 am to num__10 : num__00 am so in num__1 hour the inlet pipes will fill ( time ) * ( rate ) = num__1 * num__0.6 = num__0.6 th of the cistern . then the outlet is turned on and the remaining num__0.4 th of the cistern is filled in num__1 hour . letting x to be the rate of the outlet we would have : num__0.6 - x = num__0.4 - - > x = num__0.2 cistern / hour which means that it takes num__5 hours the outlet working alone to empty the cistern . answer : e . <eor> e <eos> |
e |
divide__1.0__2.5__ km_to_mile_conversion__ divide__1.0__0.6__ divide__1.0__0.2__ round__5.0__ |
divide__1.0__2.5__ subtract__1.0__0.4__ divide__1.0__0.6__ divide__1.0__0.2__ subtract__10.0__5.0__ |
| in a num__4 person race medals are awarded to the fastest num__3 runners . the first - place runner receives a gold medal the second - place runner receives a silver medal and the third - place runner receives a bronze medal . in the event of a tie the tied runners receive the same color medal . ( for example if there is a two - way tie for first - place the top two runners receive gold medals the next - fastest runner receives a silver medal and no bronze medal is awarded ) . assuming that exactly three medals are awarded and that the three medal winners stand together with their medals to form a victory circle how many d different victory circles are possible ? <o> a ) num__24 <o> b ) num__52 <o> c ) num__96 <o> d ) num__144 <o> e ) num__648 |
possible scenarios are : num__1 . gold / silver / bronze / no medal ( no ties ) - num__4 ! = num__24 ; num__2 . gold / gold / silver / no medal - num__4 ! / num__2 ! = num__12 ; num__3 . gold / silver / silver / no medal - num__4 ! / num__2 ! = num__12 ; num__4 . gold / gold / gold / no medal - num__4 ! / num__3 ! = num__4 . total : num__24 + num__12 + num__12 + num__4 = num__52 answer : b . <eor> b <eos> |
b |
subtract__4.0__3.0__ subtract__3.0__1.0__ multiply__4.0__3.0__ multiply__1.0__52.0__ |
subtract__4.0__3.0__ subtract__3.0__1.0__ divide__24.0__2.0__ divide__52.0__1.0__ |
| diana is painting statues . she has num__0.5 of a gallon of paint remaining . each statue requires num__0.0625 gallon of paint . how many statues can she paint ? <o> a ) num__8 <o> b ) num__20 <o> c ) num__28 <o> d ) num__14 <o> e ) num__19 |
number of statues = all the paint ÷ amount used per statue = num__0.5 ÷ num__0.0625 = num__0.5 * num__16.0 = num__8.0 = num__8 answer is a . <eor> a <eos> |
a |
reverse__0.0625__ divide__0.5__0.0625__ divide__0.5__0.0625__ |
reverse__0.0625__ multiply__0.5__16.0__ multiply__0.5__16.0__ |
| if num__50.0 of x is num__25 less than num__25.0 of num__2500 then x is ? <o> a ) num__1880 <o> b ) num__2160 <o> c ) num__1200 <o> d ) num__8400 <o> e ) num__1210 |
num__50.0 of x = x / num__2 ; num__25.0 of num__2500 = num__0.25 * num__2500 = num__625 given that x / num__2 = num__625 - num__25 = > x / num__2 = num__600 = > x = num__1200 . answer : c <eor> c <eos> |
c |
divide__50.0__25.0__ multiply__2500.0__0.25__ subtract__625.0__25.0__ multiply__2.0__600.0__ multiply__2.0__600.0__ |
divide__50.0__25.0__ multiply__2500.0__0.25__ subtract__625.0__25.0__ multiply__2.0__600.0__ multiply__2.0__600.0__ |
| the owner of a local jewellery store hired three watchmen to guard his diamonds but a thief still got in and stole some diamonds . on the way out the thief met each watchman one at a time . to each he gave num__0.5 of the diamonds he had then and num__2 more besides . he escaped with one diamond . how many did he steal originally ? <o> a ) num__40 <o> b ) num__36 <o> c ) num__25 <o> d ) none of these <o> e ) can not be determined |
explanation : since thief escaped with num__1 diamond before num__3 rd watchman he had ( num__1 + num__2 ) x num__2 = num__6 diamonds . before num__2 nd watchman he had ( num__6 + num__2 ) x num__2 = num__16 diamonds . before num__1 st watchman he had ( num__16 + num__2 ) x num__2 = num__36 diamonds . answer : b <eor> b <eos> |
b |
multiply__0.5__2.0__ add__2.0__1.0__ multiply__2.0__3.0__ multiply__1.0__36.0__ |
multiply__0.5__2.0__ add__2.0__1.0__ multiply__2.0__3.0__ multiply__1.0__36.0__ |
| train x crosses a stationary train y in num__60 seconds and a pole in num__25 seconds with the same speed . the length of the train x is num__300 m . what is the length of the stationary train y ? <o> a ) num__360 m <o> b ) num__420 m <o> c ) num__460 m <o> d ) num__320 m <o> e ) num__380 m |
explanation : let the length of the stationary train y be ly given that length of train x lx = num__300 m let the speed of train x be v . since the train x crosses train y and a pole in num__60 seconds and num__25 seconds respectively . = > num__300 / v = num__25 - - - > ( num__1 ) ( num__300 + ly ) / v = num__60 - - - > ( num__2 ) from ( num__1 ) v = num__12.0 = num__12 m / sec . from ( num__2 ) ( num__300 + ly ) / num__12 = num__60 = > num__300 + ly = num__60 ( num__12 ) = num__720 = > ly = num__720 - num__300 = num__420 m length of the stationary train = num__420 m answer is b <eor> b <eos> |
b |
divide__300.0__25.0__ multiply__60.0__12.0__ subtract__720.0__300.0__ round__420.0__ |
divide__300.0__25.0__ multiply__60.0__12.0__ subtract__720.0__300.0__ divide__420.0__1.0__ |
| find the highest common factor of num__36 and num__84 . <o> a ) num__4 <o> b ) num__6 <o> c ) num__12 <o> d ) num__18 <o> e ) num__19 |
num__36 = num__22 x num__32 num__84 = num__22 x num__3 x num__7 h . c . f . = num__22 x num__3 = num__12 . answer : option c <eor> c <eos> |
c |
gcd__36.0__84.0__ gcd__36.0__84.0__ |
gcd__36.0__84.0__ gcd__36.0__84.0__ |
| percentage of profit earned by selling a book for $ num__2200 is equal to the percentage loss incurred by selling the same book for $ num__1800 . what price should the book be sold to make num__25.0 profit ? <o> a ) $ num__2250 <o> b ) $ num__2100 <o> c ) $ num__2500 <o> d ) $ num__2700 <o> e ) $ num__3000 |
let c . p . be rs . x . then ( num__2200 - x ) x num__100 = ( x - num__1800 ) x num__100 num__2200 - x = x - num__1800 num__2 x = num__4000 x = num__2000 required s . p . = num__125.0 of rs . num__2000 = num__1.25 x num__2000 = $ num__2500 answer : c <eor> c <eos> |
c |
percent__100.0__2500.0__ |
percent__100.0__2500.0__ |
| two trains of equal length are running on parallel lines in the same direction at num__46 km / hr and num__36 km / hr . the faster train catches and completely passes the slower train in num__18 seconds . what is the length of each train ( in meters ) ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__35 <o> d ) num__50 <o> e ) num__65 |
the relative speed = num__46 - num__36 = num__10 km / hr = num__10 * num__0.277777777778 = num__2.77777777778 m / s in num__18 seconds the relative difference in distance traveled is num__18 * num__2.77777777778 = num__50 meters this distance is twice the length of each train . the length of each train is num__25.0 = num__25 meters the answer is b . <eor> b <eos> |
b |
subtract__46.0__36.0__ divide__10.0__36.0__ round__25.0__ |
subtract__46.0__36.0__ divide__10.0__36.0__ subtract__50.0__25.0__ |
| if taxi fares were $ num__3.00 for the first num__0.2 mile and $ num__0.20 for each num__0.2 mile there after then the taxi fare for a num__4 - mile ride was <o> a ) $ num__6.80 <o> b ) $ num__6.50 <o> c ) $ num__16.80 <o> d ) $ num__6.85 <o> e ) $ num__61.80 |
in num__4 miles initial num__0.2 mile charge is $ num__3 rest of the distance = num__4 - ( num__0.2 ) = num__3.8 rest of the distance charge = num__19 ( num__0.2 ) = $ num__3.8 ( as the charge is num__0.2 for every num__0.2 mile ) = > total charge for num__4 miles = num__3 + num__3.8 = num__6.8 answer is a <eor> a <eos> |
a |
subtract__4.0__0.2__ divide__3.8__0.2__ add__3.0__3.8__ add__3.0__3.8__ |
subtract__4.0__0.2__ divide__3.8__0.2__ add__3.0__3.8__ add__3.0__3.8__ |
| num__0.333333333333 - num__0.4 + num__0.5 - num__0.833333333333 + num__0.2 + num__0.25 - num__0.45 = <o> a ) num__0 <o> b ) num__0.133333333333 <o> c ) num__0.4 <o> d ) num__0.45 <o> e ) num__0.833333333333 |
we need to determine the result of num__0.333333333333 + num__0.5 - num__0.833333333333 + num__0.2 + num__0.25 - num__0.45 let ’ s add the given fractions in two groups . in the group of the first three fractions notice that num__0.333333333333 and num__0.5 share a common denominator of num__6 with num__0.833333333333 . num__0.5 + num__0.333333333333 = num__0.5 + num__0.333333333333 = num__0.833333333333 thus num__0.833333333333 – num__0.833333333333 = num__0 looking at the num__2 nd group of the fractions ( num__0.2 num__0.25 and num__0.45 ) notice that num__0.2 and num__0.25 share a common denominator of num__20 with num__0.45 . num__0.2 + num__0.25 = num__0.2 + num__0.25 = num__0.45 thus num__0.45 – num__0.45 = num__0 . thus the result of num__0.333333333333 + num__0.5 – num__0.833333333333 + num__0.2 + num__0.25 – num__0.45 is num__0.4 . answer : c <eor> c <eos> |
c |
round_down__0.3333__ reverse__0.5__ divide__0.3333__0.8333__ |
round_down__0.3333__ reverse__0.5__ divide__0.3333__0.8333__ |
| how many ways are there to award a gold silver bronze and platinum medal to num__10 contending teams ? <o> a ) num__10 × num__9 × num__8 <o> b ) num__10 ! / ( num__3 ! num__7 ! ) <o> c ) num__10 ! / num__3 ! <o> d ) num__10 ! num__7 <o> e ) num__10 * num__9 * num__8 * num__7 |
we clearly know that there can be only one winning team which deserves the gold medal . we can do the selection in num__10 ways if gold medal is given to num__1 of the teams then only possible num__9 teams can be considered for silver medal . we can do selection in num__9 ways similarly if gold and silver medals are awarded then only remaining num__8 teams can be considered for a bronze medal . we can do the selection in num__8 ways similarly if gold and silver medals are awarded then only remaining num__7 teams can be considered for a bronze medal . we can do the selection in num__7 ways total number of ways to select the num__4 possible medal winners = num__10 * num__9 * num__8 * num__7 correct answer - e <eor> e <eos> |
e |
choose__10.0__9.0__ |
choose__10.0__9.0__ |
| how many numbers amongst the numbers num__28 to num__42 are there which are exactly divisible by num__7 but not by num__5 ? <o> a ) nil <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) more than num__3 |
num__28 num__3542 are divisible by num__7 . num__35 is divisible by num__5 also . so there are num__2 numbers exactly divisible by num__7 but not by num__5 answer : c <eor> c <eos> |
c |
add__28.0__7.0__ subtract__7.0__5.0__ subtract__7.0__5.0__ |
add__28.0__7.0__ subtract__7.0__5.0__ subtract__7.0__5.0__ |
| the price of an item is discounted num__4 percent on day num__1 of a sale . on day num__2 the item is discounted another num__4 percent and on day num__3 it is discounted an additional num__10 percent . the price of the item on day num__3 is what percentage of the sale price on day num__1 ? <o> a ) num__82.5 <o> b ) num__89.9 <o> c ) num__87.7 <o> d ) num__86.4 <o> e ) num__83.3 % |
let initial price be num__100 price in day num__1 after num__4.0 discount = num__96 price in day num__2 after num__4.0 discount = num__92.16 price in day num__3 after num__10.0 discount = num__82.94 so price in day num__3 as percentage of the sale price on day num__1 will be = num__82.94 / num__96 * num__100 = > num__86.4 answer will definitely be ( d ) <eor> d <eos> |
d |
percent__100.0__86.4__ |
percent__100.0__86.4__ |
| a train crosses a platform of num__150 m in num__15 sec same train crosses another platform of length num__250 m in num__20 sec . then find the length of the train ? <o> a ) num__150 <o> b ) num__160 <o> c ) num__170 <o> d ) num__180 <o> e ) num__190 |
length of the train be ‘ x ’ x + num__10.0 = x + num__12.5 num__4 x + num__600 = num__3 x + num__750 x = num__150 m answer : option a <eor> a <eos> |
a |
divide__150.0__15.0__ divide__250.0__20.0__ multiply__150.0__4.0__ add__150.0__600.0__ round__150.0__ |
divide__150.0__15.0__ divide__250.0__20.0__ multiply__150.0__4.0__ add__150.0__600.0__ round__150.0__ |
| the average marks of a class of num__30 students is num__40 and that of another class of num__50 students is num__60 . find the average marks of all the students ? <o> a ) num__52.5 <o> b ) num__52.9 <o> c ) num__52.1 <o> d ) num__52.3 <o> e ) num__42.5 |
sum of the marks for the class of num__30 students = num__30 * num__40 = num__1200 sum of the marks for the class of num__50 students = num__50 * num__60 = num__3000 sum of the marks for the class of num__80 students = num__1200 + num__3000 = num__4200 average marks of all the students = num__52.5 = num__52.5 answer : a <eor> a <eos> |
a |
multiply__30.0__40.0__ multiply__50.0__60.0__ add__30.0__50.0__ add__1200.0__3000.0__ divide__4200.0__80.0__ divide__4200.0__80.0__ |
multiply__30.0__40.0__ multiply__50.0__60.0__ add__30.0__50.0__ add__1200.0__3000.0__ divide__4200.0__80.0__ divide__4200.0__80.0__ |
| a table is bought for rs . num__400 / - and sold at rs . num__450 / - find gain or loss percentage <o> a ) num__10.0 loss <o> b ) num__12.5 gain <o> c ) num__10.0 gain <o> d ) num__15.0 loss <o> e ) num__20.0 loss |
formula = ( selling price ~ cost price ) / cost price * num__100 = ( num__450 - num__400 ) / num__400 = num__12.5 gain b <eor> b <eos> |
b |
percent__12.5__100.0__ |
percent__12.5__100.0__ |
| ram works in a bakery . he made cookies that cost $ num__1.75 and made $ num__325 . how many customer did he have ? <o> a ) num__200 customers <o> b ) num__186 customers <o> c ) num__250 customers <o> d ) num__100 customers <o> e ) num__170 customers |
a cookie costs $ num__1.75 adding another one is $ num__3.5 . num__325 divided by num__3.5 is num__92.86 x num__2 is num__186 . he had num__186 customers . the correct answer is b . <eor> b <eos> |
b |
divide__3.5__1.75__ round__186.0__ |
divide__3.5__1.75__ round__186.0__ |
| george went to a fruit market with certain amount of money . with this money he can buy either num__40 oranges or num__40 mangoes . he retains num__10.0 of the money for taxi fare and buys num__20 mangoes . how many oranges can he buy ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__20 <o> d ) num__16 <o> e ) num__12 |
let the amount of money be num__200 let cost of num__1 orange be num__4 let cost of num__1 mango be num__4 he decides to retain num__10.0 of num__200 = num__20 for taxi fare so he is left with num__180 he buys num__20 mangoes ( @ num__4 ) so he spends num__80 money left is num__100 ( num__180 - num__80 ) no of oranges he can buy = num__25.0 = > num__25 so george can buy num__25 oranges . a <eor> a <eos> |
a |
multiply__10.0__20.0__ divide__40.0__10.0__ subtract__200.0__20.0__ multiply__20.0__4.0__ add__20.0__80.0__ divide__100.0__4.0__ multiply__1.0__25.0__ |
multiply__10.0__20.0__ divide__40.0__10.0__ subtract__200.0__20.0__ multiply__20.0__4.0__ subtract__180.0__80.0__ divide__100.0__4.0__ multiply__1.0__25.0__ |
| there is a num__1 km long wire placed on some number of poles which are in equal distance . if the number of poles is reduced by num__1 then the distance of wire between each poles increases num__1 ( num__0.666666666667 ) . how many poles are there initially . <o> a ) num__24 <o> b ) num__25 <o> c ) num__26 <o> d ) num__27 <o> e ) num__28 |
num__26 answer : c <eor> c <eos> |
c |
multiply__1.0__26.0__ |
multiply__1.0__26.0__ |
| working alone printers x y and z can do a certain printing job consisting of a large number of pages in num__12 num__10 and num__20 hours respectively . what is the ratio of the time it takes printer x to do the job working alone at its rate to the time it takes printers y and z to do the job working together at their individual rates ? <o> a ) num__1.66666666667 <o> b ) num__1.75 <o> c ) num__1.8 <o> d ) num__1.83333333333 <o> e ) num__1.85714285714 |
the time it takes printer x is num__12 hours . the combined rate of y and z is num__0.1 + num__0.05 = num__0.15 the time it takes y and z is num__6.66666666667 the ratio of times is num__12 / ( num__6.66666666667 ) = num__3 * num__0.6 = num__1.8 the answer is c . <eor> c <eos> |
c |
reverse__10.0__ reverse__20.0__ add__0.1__0.05__ reverse__0.15__ multiply__20.0__0.15__ divide__12.0__20.0__ multiply__12.0__0.15__ multiply__12.0__0.15__ |
reverse__10.0__ reverse__20.0__ add__0.1__0.05__ reverse__0.15__ multiply__20.0__0.15__ divide__12.0__20.0__ multiply__12.0__0.15__ multiply__12.0__0.15__ |
| julian owes his classmate jenny num__20 dollars . if he borrows num__8 dollars how much will he owe her ? <o> a ) num__28 <o> b ) num__12 <o> c ) - num__28 <o> d ) num__12 <o> e ) num__0 |
add how he has already borrowed and how much he is going to borrow - num__20 + - num__8 correct answer c ) - num__28 <eor> c <eos> |
c |
add__20.0__8.0__ add__20.0__8.0__ |
add__20.0__8.0__ add__20.0__8.0__ |
| a girl was asked to multiply a certain number by num__43 . she multiplied it by num__34 and got his answer less than the correct one by num__1224 . find the number to be multiplied . <o> a ) num__130 <o> b ) num__132 <o> c ) num__134 <o> d ) num__136 <o> e ) num__138 |
let the required number be x . then num__43 x – num__34 x = num__1224 or num__9 x = num__1224 or x = num__136 . required number = num__136 . answer : d <eor> d <eos> |
d |
subtract__43.0__34.0__ divide__1224.0__9.0__ divide__1224.0__9.0__ |
subtract__43.0__34.0__ divide__1224.0__9.0__ divide__1224.0__9.0__ |
| a train num__240 m long passed a pole in num__24 sec . how long will it take to pass a platform num__650 m long ? <o> a ) num__29 <o> b ) num__89 <o> c ) num__77 <o> d ) num__55 <o> e ) num__12 |
speed = num__10.0 = num__10 m / sec . required time = ( num__240 + num__650 ) / num__10 = num__89 sec . answer : b <eor> b <eos> |
b |
divide__240.0__24.0__ round__89.0__ |
divide__240.0__24.0__ round__89.0__ |
| a num__300 m long train crosses a platform in num__39 sec while it crosses a signal pole in num__18 sec . what is the length of the platform ? <o> a ) num__266 <o> b ) num__350 <o> c ) num__327 <o> d ) num__267 <o> e ) num__256 |
speed = num__16.6666666667 = num__16.6666666667 m / sec . let the length of the platform be x meters . then ( x + num__300 ) / num__39 = num__16.6666666667 num__3 x + num__900 = num__1950 = > x = num__350 m . answer : b <eor> b <eos> |
b |
divide__300.0__18.0__ multiply__300.0__3.0__ round__350.0__ |
divide__300.0__18.0__ multiply__300.0__3.0__ round__350.0__ |
| a is twice as good a work man as b and together they finish the work in num__10 days . in how many days a alone can finish the work ? <o> a ) num__23 <o> b ) num__15 <o> c ) num__77 <o> d ) num__92 <o> e ) num__61 |
wc = num__2 : num__1 num__2 x + x = num__0.1 = > x = num__0.0333333333333 num__2 x = num__0.0666666666667 a can do the work in num__15 days . answer : b <eor> b <eos> |
b |
divide__1.0__10.0__ subtract__0.1__0.0333__ round__15.0__ |
divide__1.0__10.0__ subtract__0.1__0.0333__ round__15.0__ |
| the present average age of a couple and their daughter is num__35 years . fifteen years from now the age of the mother will be equal to the sum of present ages of the father and the daughter . find the present age of mother ? <o> a ) num__22 years <o> b ) num__77 years <o> c ) num__76 years <o> d ) num__45 years <o> e ) num__56 years |
( f + m + d ) / num__3 = num__35 = > f + m + d = num__105 - - - ( num__1 ) m + num__15 = f + d substituting f + d as m + num__15 in ( num__1 ) we get num__2 m + num__15 = num__105 num__2 m = num__90 = > m = num__45 years . answer : d <eor> d <eos> |
d |
multiply__35.0__3.0__ subtract__3.0__1.0__ subtract__105.0__15.0__ multiply__15.0__3.0__ multiply__1.0__45.0__ |
multiply__35.0__3.0__ subtract__3.0__1.0__ subtract__105.0__15.0__ divide__90.0__2.0__ divide__45.0__1.0__ |
| find four consecutive even integers so that the sum of the first two added to twice the sum of the last two is equal to num__766 . <o> a ) num__124 num__126 num__128 num__130 <o> b ) num__120 num__122 num__124 num__128 <o> c ) num__120 num__121 num__122 num__123 <o> d ) num__123 num__125 num__127 num__129 <o> e ) none of these |
let x x + num__2 x + num__4 and x + num__6 be the four integers . the sum of the first two x + ( x + num__2 ) twice the sum of the last two is written as num__2 ( ( x + num__4 ) + ( x + num__6 ) ) = num__4 x + num__20 sum of the first two added to twice the sum of the last two is equal to num__766 is written as x + ( x + num__2 ) + num__4 x + num__20 = num__766 solve for x and find all four numbers x = num__124 x + num__2 = num__126 x + num__4 = num__128 x + num__6 = num__130 answer a <eor> a <eos> |
a |
add__2.0__4.0__ add__2.0__124.0__ add__2.0__126.0__ add__128.0__2.0__ subtract__128.0__4.0__ |
add__2.0__4.0__ add__2.0__124.0__ add__2.0__126.0__ add__128.0__2.0__ subtract__128.0__4.0__ |
| if dev works alone he will take num__20 more hours to complete a task than if he worked with tina to complete the task . if tina works alone she will take num__5 more hours to complete the complete the task then if she worked with dev to complete the task ? what is the ratio of the time taken by dev to that taken by tina if each of them worked alone to complete the task ? <o> a ) num__4 : num__1 <o> b ) num__2 : num__1 <o> c ) num__10 : num__1 <o> d ) num__3 : num__1 <o> e ) num__1 : num__2 |
let time taken by dev to complete the work alone be x days and that by tina be y days work done by dev in num__1 day = num__1 / x work done by tina in num__1 day = num__1 / y work done by devtina in num__1 day = num__1 / x + num__1 / y = ( x + y ) / xy thus working together they will complete the work in xy / ( x + y ) days acc . to the ques : num__1 ) if dev works alone he will take num__20 more hours to complete a task than if he worked with tina ' x - xy / ( x + y ) = num__20 = > ( x ^ num__2 ) / ( x + y ) = num__20 . . . . . . . . . . . . . ( i ) num__2 ) if tina works alone she will take num__5 more hours to complete the complete the task then if she worked with dev y - xy / ( x + y ) = num__5 = > ( y ^ num__2 ) / ( x + y ) = num__5 . . . . . . . . . . . . . . ( ii ) dividing ( i ) by ( ii ) x ^ num__2 / y ^ num__2 = num__4 = > x / y = num__2 ans : b hope it helps <eor> b <eos> |
b |
divide__20.0__5.0__ round__2.0__ |
divide__20.0__5.0__ divide__2.0__1.0__ |
| num__6 x – num__5 y + num__3 z = num__27 num__4 x + num__8 y – num__11 z = num__7 num__5 x – num__6 y + num__2 z = num__12 given the equations above x + y + z = ? <o> a ) num__11 <o> b ) num__12 <o> c ) num__13 <o> d ) num__14 <o> e ) num__15 |
( num__6 x – num__5 y + num__3 z ) - ( num__5 x – num__6 y + num__2 z ) = num__27 - num__12 or x + y + z = num__15 option e is the ans <eor> e <eos> |
e |
multiply__5.0__3.0__ multiply__5.0__3.0__ |
add__3.0__12.0__ add__3.0__12.0__ |
| a num__300 meter long train crosses a platform in num__39 seconds while it crosses a signal pole in num__18 seconds . what is the length of the platform ? <o> a ) num__350 m <o> b ) num__380 m <o> c ) num__350 m <o> d ) num__320 m <o> e ) num__150 m |
speed = [ num__16.6666666667 ] m / sec = num__16.6666666667 m / sec . let the length of the platform be x meters . then x + num__7.69230769231 = num__16.6666666667 num__3 ( x + num__300 ) = num__1950 è x = num__350 m . answer : c <eor> c <eos> |
c |
divide__300.0__18.0__ divide__300.0__39.0__ round__350.0__ |
divide__300.0__18.0__ divide__300.0__39.0__ round__350.0__ |
| the average of num__10 numbers was calculated as num__17 . it is discovered later on that while calculating the average one number namely num__56 was incorrectly read as num__26 . what is the correct average ? <o> a ) num__19 <o> b ) num__20 <o> c ) num__21 <o> d ) num__22 <o> e ) num__23 |
num__10 * num__17 - num__26 + num__56 = num__200 num__20.0 = num__20 the answer is b . <eor> b <eos> |
b |
divide__200.0__10.0__ divide__200.0__10.0__ |
divide__200.0__10.0__ divide__200.0__10.0__ |
| two trains running in opposite directions cross a man standing on the platform in num__27 seconds and num__17 seconds respectively and they cross each other in num__23 seconds . the ratio of their speeds is ? <o> a ) num__1 : num__3 <o> b ) num__3 : num__2 <o> c ) num__3 : num__5 <o> d ) num__3 : num__7 <o> e ) none of these |
explanation : let the speeds of the two trains be x m / sec and y m / sec respectively . then length of the first train = num__27 x metres length of the second train = num__17 y metres . [ because distance = speed * time ] num__27 x + num__17 y / x + y = num__23 = > num__27 x + num__17 y = num__23 x + num__23 y = > num__4 x = num__6 y = > x / y = num__1.5 so ratio of the speeds of train is num__3 : num__2 option b <eor> b <eos> |
b |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ divide__3.0__1.5__ round__3.0__ |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ divide__3.0__1.5__ multiply__1.5__2.0__ |
| if two numbers are in the ratio num__5 : num__2 . if num__10 is added to both of the numbers then the ratio becomes num__5 : num__4 then find the largest number ? <o> a ) num__3 <o> b ) num__7 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
num__5 : num__2 num__5 x + num__10 : num__2 x + num__10 = num__5 : num__4 num__5 [ num__2 x + num__10 ] = num__4 [ num__5 x + num__10 ] num__10 x + num__50 = num__20 x + num__40 num__20 x - num__10 x = num__50 - num__40 x = num__1 then the first number is = num__5 x num__5 x = num__5 option ' d ' <eor> d <eos> |
d |
multiply__5.0__10.0__ multiply__5.0__4.0__ multiply__2.0__20.0__ subtract__5.0__4.0__ multiply__5.0__1.0__ |
multiply__5.0__10.0__ multiply__5.0__4.0__ subtract__50.0__10.0__ subtract__5.0__4.0__ subtract__10.0__5.0__ |
| find the length of the wire required to go num__15 times round a square field containing num__69696 m ^ num__2 . <o> a ) num__15840 <o> b ) num__26480 <o> c ) num__15642 <o> d ) num__15640 <o> e ) num__15849 |
a ^ num__2 = num__69696 = > a = num__264 num__4 a = num__1056 num__1056 * num__15 = num__15840 answer : a <eor> a <eos> |
a |
multiply__4.0__264.0__ multiply__15.0__1056.0__ multiply__15.0__1056.0__ |
multiply__4.0__264.0__ multiply__15.0__1056.0__ multiply__15.0__1056.0__ |
| excluding stoppages the speed of a train is num__48 kmph and including stoppages it is num__40 kmph . of how many minutes does the train stop per hour ? <o> a ) num__16 <o> b ) num__17 <o> c ) num__15 <o> d ) num__10 <o> e ) num__12 |
t = num__0.166666666667 * num__60 = num__10 answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ round__10.0__ |
hour_to_min_conversion__ round__10.0__ |
| a cyclist traveled for two days . on the second day the cyclist traveled num__4 hours longer and at an average speed num__10 mile per hour slower than she traveled on the first day . if during the two days she traveled a total of num__240 miles and spent a total of num__12 hours traveling what was her average speed on the second day ? <o> a ) num__5 mph <o> b ) num__10 mph <o> c ) num__20 mph <o> d ) num__30 mph <o> e ) num__40 mph |
solution : d = num__280 mi t = num__12 hrs Đ â y num__1 time = t num__1 d â y num__2 time = t num__2 t num__2 - t num__1 = num__4 hrs - - - - - ( i ) t num__1 + t num__2 = num__12 hrs - - - - - ( ii ) adding i and ii t num__2 = num__8 hrs and t num__1 = num__4 hrs d à y num__1 rate = r num__1 d â y num__2 rate = r num__2 r num__1 - r num__2 = num__10 mph í . ẹ . r num__1 = num__10 + r num__2 num__280 = num__8 r num__2 + num__4 r num__1 í . ẹ . num__280 = num__8 r num__2 + num__4 ( num__10 + r num__2 ) í . ẹ . r num__2 = num__20 mph answer : b <eor> b <eos> |
b |
subtract__12.0__10.0__ multiply__4.0__2.0__ multiply__10.0__2.0__ round__10.0__ |
subtract__12.0__10.0__ subtract__10.0__2.0__ add__12.0__8.0__ subtract__12.0__2.0__ |
| if there are only num__2 wheelers and num__4 wheelers parked in a school located at the heart of the city find the number of num__4 wheelers parked there if the total number of wheels is num__54 ? <o> a ) num__11 <o> b ) num__12 <o> c ) num__13 <o> d ) num__14 <o> e ) num__15 |
four wheeler = num__13 * num__4 = num__52 ( max ) num__2 wheel = num__1 so no of num__4 wheeler = num__13 answer : c <eor> c <eos> |
c |
multiply__4.0__13.0__ divide__52.0__4.0__ |
multiply__4.0__13.0__ multiply__13.0__1.0__ |
| if a b and c together can finish a piece of work in num__4 days . a alone in num__12 days and b in num__18 days then c alone can do it in ? <o> a ) num__20 <o> b ) num__21 <o> c ) num__30 <o> d ) num__9 <o> e ) num__7 |
c = num__0.25 - num__0.0833333333333 â € “ num__0.0555555555556 = num__0.111111111111 = > num__9 days answer d <eor> d <eos> |
d |
round__9.0__ |
subtract__18.0__9.0__ |
| a train of length num__150 metres takes num__40.5 seconds to cross a tunnel of length num__300 metres . what is the speed of the train in km / hr ? <o> a ) num__13 num__33 <o> b ) num__26.67 <o> c ) num__40 <o> d ) num__66.67 <o> e ) none |
solution speed = ( num__150 + num__300 / num__40.5 ) m / sec = ( num__450 / num__40.5 x num__3.6 ) km / hr = num__40 km / hr answer c <eor> c <eos> |
c |
add__150.0__300.0__ round__40.0__ |
add__150.0__300.0__ round__40.0__ |
| how many multiples of num__7 are there between num__14 and num__252 ? <o> a ) num__33 <o> b ) num__46 <o> c ) num__59 <o> d ) num__66 <o> e ) num__80 |
it should be mentioned whether num__14 and num__252 are inclusive . if num__14 and num__252 are inclusive then the answer is ( num__252 - num__14 ) / num__7 + num__1 = num__35 . if num__14 and num__252 are not inclusive then the answer is ( num__245 - num__21 ) / num__7 + num__1 = num__33 . since oa is a then we have not inclusive case . <eor> a <eos> |
a |
multiply__7.0__35.0__ add__7.0__14.0__ multiply__1.0__33.0__ |
subtract__252.0__7.0__ add__7.0__14.0__ divide__33.0__1.0__ |
| if a train travelling at a speed of num__90 kmph crosses a pole in num__5 sec then the length of train is ? <o> a ) num__167 m <o> b ) num__125 m <o> c ) num__761 m <o> d ) num__143 m <o> e ) num__132 m |
d = num__90 * num__0.277777777778 * num__5 = num__125 m answer : b <eor> b <eos> |
b |
round__125.0__ |
round__125.0__ |
| if n = num__4 p where p is a prime number greater than num__2 how many different positive even divisors does n have without n value ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
we ' re told that n = num__4 p and that p is a prime number greater than num__2 . let ' s test p = num__3 ; so n = num__12 the question now asks how many different positive even divisors does num__12 have including num__12 ? num__12 : num__112 num__26 num__34 how many of these divisors are even ? num__2 num__4 num__6 … . . num__3 even divisors . answer : c <eor> c <eos> |
c |
alphabet_space__ die_space__ choose__3.0__2.0__ |
alphabet_space__ die_space__ choose__3.0__2.0__ |
| ritesh and co . generated revenue of rs . num__1600 in num__2006 . this was num__12.5 of its gross revenue . in num__2007 the gross revenue grew by rs . num__2500 . what is the percentage increase in the revenue in num__2007 ? <o> a ) num__12.5 <o> b ) num__19.53 <o> c ) num__25.0 <o> d ) num__50.0 <o> e ) none of these |
explanation : given ritesh and co . generated revenue of rs . num__1600 in num__2006 and that this was num__12.5 of the gross revenue . hence if num__1600 is num__12.5 of the revenue then num__100.0 ( gross revenue ) is : = > ( num__100 / num__12.5 ) × num__1600 . = > num__12800 . hence the total revenue by end of num__2007 is rs . num__12800 . in num__2006 revenue grew by rs . num__2500 . this is a growth of : = > ( num__0.1953125 ) × num__100 . = > num__19.53 . answer : b <eor> b <eos> |
b |
percent__100.0__19.53__ |
percent__100.0__19.53__ |
| ( num__112 x num__54 ) = ? <o> a ) num__70000 <o> b ) num__71000 <o> c ) num__72000 <o> d ) num__73000 <o> e ) num__74000 |
( num__112 x num__54 ) = num__112 x num__10 num__4 = num__112 x num__104 = num__1120000 = num__70000 num__2 num__24 num__16 a ) <eor> a <eos> |
a |
divide__1120000.0__70000.0__ divide__1120000.0__16.0__ |
divide__1120000.0__70000.0__ divide__1120000.0__16.0__ |
| a man can swim in still water at num__4.5 km / h but takes twice as long to swim upstream than downstream . the speed of the stream is ? <o> a ) num__1.8 <o> b ) num__1.6 <o> c ) num__1.3 <o> d ) num__1.5 <o> e ) num__1.1 |
m = num__4.5 s = x ds = num__4.5 + x us = num__4.5 + x num__4.5 + x = ( num__4.5 - x ) num__2 num__4.5 + x = num__9 - num__2 x num__3 x = num__4.5 x = num__1.5 answer : d <eor> d <eos> |
d |
multiply__4.5__2.0__ divide__4.5__3.0__ round__1.5__ |
multiply__4.5__2.0__ subtract__4.5__3.0__ subtract__4.5__3.0__ |
| p : q = num__7 : num__6 q : r = num__4 : num__7 p : q : r = ? <o> a ) num__28 : num__24 : num__42 <o> b ) num__28 : num__24 : num__40 <o> c ) num__28 : num__24 : num__52 <o> d ) num__28 : num__24 : num__41 <o> e ) num__28 : num__24 : num__48 |
p : q = num__7 : num__6 q : r = num__4 : num__7 p : q : r = num__28 : num__24 : num__42 answer : a <eor> a <eos> |
a |
multiply__7.0__4.0__ multiply__6.0__4.0__ multiply__7.0__6.0__ multiply__7.0__4.0__ |
multiply__7.0__4.0__ multiply__6.0__4.0__ multiply__7.0__6.0__ multiply__7.0__4.0__ |
| a person buys an article at $ num__500 . at what price should he sell the article so as to make a profit of num__20.0 ? <o> a ) num__560 <o> b ) num__449 <o> c ) num__600 <o> d ) num__740 <o> e ) num__460 |
c num__600 cost price = $ num__500 profit = num__20.0 of num__500 = $ num__100 selling price = cost price + profit = num__500 + num__100 = num__600 <eor> c <eos> |
c |
percent__20.0__500.0__ percent__100.0__600.0__ |
percent__20.0__500.0__ percent__100.0__600.0__ |
| the average age of husband wife and their child num__3 years ago was num__27 years and that of wife and the child num__5 years ago was num__20 years . what is the present age of the husband ? <o> a ) num__30 <o> b ) num__40 years <o> c ) num__20 <o> d ) num__50 <o> e ) num__25 |
let the present age of the husband = h present age of the wife = w present age of the child = c num__3 years ago average age of husband wife and their child = num__27 = > sum of age of husband wife and their child before num__3 years = num__3 × num__27 = num__81 = > ( h - num__3 ) + ( w - num__3 ) + ( c - num__3 ) = num__81 = > h + w + c = num__81 + num__9 = num__90 - - - equation ( num__1 ) num__5 years ago average age of wife and child = num__20 = > sum of age of wife and child before num__5 years = num__2 × num__20 = num__40 = > ( w - num__5 ) + ( c - num__5 ) = num__40 = > w + c = num__40 + num__10 = num__50 - - - equation ( num__2 ) substituting equation ( num__2 ) in equation ( num__1 ) = > h + num__50 = num__90 = > h = num__90 - num__50 = num__40 i . e . present age of the husband = num__40 answer is b . <eor> b <eos> |
b |
multiply__3.0__27.0__ divide__27.0__3.0__ add__9.0__81.0__ subtract__3.0__1.0__ multiply__20.0__2.0__ multiply__5.0__2.0__ multiply__5.0__10.0__ multiply__20.0__2.0__ |
multiply__3.0__27.0__ divide__27.0__3.0__ add__9.0__81.0__ subtract__3.0__1.0__ multiply__20.0__2.0__ add__1.0__9.0__ add__40.0__10.0__ subtract__50.0__10.0__ |
| a man divides $ num__6800 among num__4 sons num__4 daughters and num__2 nephews . if each daughter receives four times as much as each nephews and each son receives five times as much as each nephews how much does each daughter receive ? <o> a ) a ) $ num__200 <o> b ) b ) $ num__1000 <o> c ) c ) $ num__800 <o> d ) d ) $ num__1200 <o> e ) e ) $ num__400 |
let the share of each nephews be $ x . then share of each daughter = $ num__4 x share of each son = $ num__5 x . so num__4 * num__4 x + num__4 * num__4 x + num__2 * x = num__6800 num__16 x + num__16 x + num__2 x = num__8600 num__34 x = num__6800 x = num__200 . daughter receives four times of nephew so num__4 * num__200 = num__800 . so each daughter receives $ num__800 . answer is option c ) $ num__800 . <eor> c <eos> |
c |
divide__6800.0__34.0__ multiply__4.0__200.0__ multiply__4.0__200.0__ |
divide__6800.0__34.0__ multiply__4.0__200.0__ multiply__4.0__200.0__ |
| a and b finish the job in num__15 days . while a b and c can finish it in num__11 days . c alone will finish the job in <o> a ) num__41.25 days <o> b ) num__30.25 days <o> c ) num__60.25 days <o> d ) num__71.25 days <o> e ) num__51.25 days |
explanation : num__11 = ( num__15 * x ) / ( num__15 + x ) num__165 + num__11 x = num__15 x num__4 x = num__165 x = num__41.25 answer : option a <eor> a <eos> |
a |
multiply__15.0__11.0__ subtract__15.0__11.0__ divide__165.0__4.0__ round__41.25__ |
multiply__15.0__11.0__ subtract__15.0__11.0__ divide__165.0__4.0__ divide__165.0__4.0__ |
| kim purchased n items from a catalog for $ num__8 each . postage and handling charges consisted of $ num__2 for the first item and $ num__1 for each additional item . which of the following gives the total dollar amount for kim ’ s purchase including postage and handling in terms of n ? <o> a ) num__8 n + num__2 <o> b ) num__9 n + num__1 <o> c ) num__9 n + num__2 <o> d ) num__9 n + num__3 <o> e ) num__9 n + num__4 |
its c n items for $ num__8 each total price $ num__8 n postage and handling of $ num__3 for num__1 st item and $ num__1 for the rest total postage and handling = $ [ num__2 + ( n - num__1 ) ] = $ n + num__1 total cost num__8 n + n + num__1 = num__9 n + num__1 b <eor> b <eos> |
b |
add__2.0__1.0__ add__8.0__1.0__ add__8.0__1.0__ |
add__2.0__1.0__ add__8.0__1.0__ add__8.0__1.0__ |
| which of the following is satisfied with | x - num__4 | + | x - num__3 | < num__7 ? <o> a ) num__0 < x < num__7 <o> b ) num__2 < x < num__5 <o> c ) num__2.5 < x < num__4.5 <o> d ) num__2.5 < x < num__4 <o> e ) num__3 < x < num__4 |
- - > if there is addition when there are num__2 absolute values you can just ignore the middle . that is | x - num__4 | + | x - num__3 | < num__7 - > | x - num__4 + x - num__3 | < num__7 - > | num__2 x - num__7 | < num__7 - num__7 < num__2 x - num__7 < num__7 num__0 < num__2 x < num__14 num__0 < x < num__7 therefore the answer is a . <eor> a <eos> |
a |
multiply__7.0__2.0__ multiply__4.0__0.0__ |
multiply__7.0__2.0__ multiply__4.0__0.0__ |
| a rectangular photograph is surrounded by a border that is num__1 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been num__6 inches wide on each side the total area would have been ( m + num__200 ) square inches . what is the perimeter of the photograph in inches ? <o> a ) num__8 <o> b ) num__12 <o> c ) num__16 <o> d ) num__20 <o> e ) num__24 |
let x and y be the width and length of the photograph . ( x + num__2 ) ( y + num__2 ) = m and so ( num__1 ) xy + num__2 x + num__2 y + num__4 = m ( x + num__12 ) ( y + num__12 ) = m and so ( num__2 ) xy + num__12 x + num__12 y + num__144 = m + num__200 let ' s subtract equation ( num__1 ) from equation ( num__2 ) . num__10 x + num__10 y + num__140 = num__200 num__2 x + num__2 y = num__12 which is the perimeter of the photograph . the answer is b . <eor> b <eos> |
b |
square_perimeter__1.0__ multiply__6.0__2.0__ power__12.0__2.0__ rectangle_perimeter__1.0__4.0__ multiply__1.0__12.0__ |
square_perimeter__1.0__ multiply__6.0__2.0__ power__12.0__2.0__ rectangle_perimeter__1.0__4.0__ multiply__1.0__12.0__ |
| find the one which does not belong to that group ? <o> a ) num__16 <o> b ) num__28 <o> c ) num__36 <o> d ) num__64 <o> e ) num__4 |
explanation : num__16 num__36 num__64 and num__4 are perfect squares but not num__28 . answer is c <eor> c <eos> |
c |
divide__64.0__16.0__ subtract__64.0__36.0__ subtract__64.0__28.0__ |
divide__64.0__16.0__ subtract__64.0__36.0__ subtract__64.0__28.0__ |
| the average of num__25 results is num__18 . the average of first num__12 of those is num__14 and the average of last num__12 is num__17 . what is the num__13 th result ? <o> a ) num__74 <o> b ) num__75 <o> c ) num__69 <o> d ) num__78 <o> e ) num__45 |
solution : sum of num__1 st num__12 results = num__12 * num__14 sum of last num__12 results = num__12 * num__17 num__13 th result = x ( let ) now num__12 * num__14 + num__12 * num__17 + x = num__25 * num__18 or x = num__78 . answer : option d <eor> d <eos> |
d |
subtract__18.0__17.0__ multiply__1.0__78.0__ |
subtract__18.0__17.0__ multiply__1.0__78.0__ |
| wendy begins sanding a kitchen floor by herself and works for num__8 hours . she is then joined by bruce and together the two of them finish sanding the floor in num__2 hours . if bruce can sand the floor by himself in num__20 hours how long would it take wendy to sand the floor by herself ? <o> a ) num__11.1020408163 hours <o> b ) num__0.15 hours <o> c ) num__6.66666666667 hours <o> d ) num__8.88888888889 hours <o> e ) num__10 hours |
let wendy finishes sanding the floor alone in w hours while b be the hours for bruce . thus in num__1 hour wendy finishes num__1 / w of the work while bruce finishes num__1 / b of the work . if wendy works for num__8 hours and is then joined by bruce to finish the work in num__2 more hours num__8 / w + num__2 / w + num__2 / b = num__1 ( num__1 denotes the total amount of work ) num__10 / w + num__2 / b = num__1 and given b = num__20 hours . thus w = num__11.1020408163 hours a is the correct answer . <eor> a <eos> |
a |
add__8.0__2.0__ divide__11.102__1.0__ |
add__8.0__2.0__ divide__11.102__1.0__ |
| in what time a sum of money double itself at num__6.0 per annum simple interest ? <o> a ) num__33 num__0.125 % <o> b ) num__16 num__0.666666666667 % <o> c ) num__33 num__2.33333333333 % <o> d ) num__32 num__0.333333333333 % <o> e ) num__23 num__0.333333333333 % |
p = ( p * num__6 * r ) / num__100 r = num__16 num__0.666666666667 % answer : b <eor> b <eos> |
b |
percent__16.0__100.0__ |
percent__16.0__100.0__ |
| in an election only two candidates contested . a candidate secured num__70.0 of the valid votes and won by a majority of num__172 votes . find the total number of valid votes ? <o> a ) num__430 <o> b ) num__437 <o> c ) num__435 <o> d ) num__431 <o> e ) num__433 |
let the total number of valid votes be x . num__70.0 of x = num__0.7 * x = num__7 x / num__10 number of votes secured by the other candidate = x - num__7 x / num__100 = num__3 x / num__10 given num__7 x / num__10 - num__3 x / num__10 = num__172 = > num__4 x / num__10 = num__172 = > num__4 x = num__1720 = > x = num__430 . answer : a <eor> a <eos> |
a |
percent__100.0__430.0__ |
percent__100.0__430.0__ |
| num__6 men and num__2 boys working together can do four times as much work as a man and a boy . working capacity of man and boy is in the ratio <o> a ) num__1 : num__1 <o> b ) num__1 : num__3 <o> c ) num__2 : num__1 <o> d ) num__2 : num__3 <o> e ) none of these |
explanation : let num__1 man num__1 day work = x num__1 boy num__1 day work = y then num__6 x + num__2 y = num__4 ( x + y ) = > num__2 x = num__2 y = > x / y = num__1.0 = > x : y = num__1 : num__1 option a <eor> a <eos> |
a |
subtract__6.0__2.0__ reverse__1.0__ |
subtract__6.0__2.0__ reverse__1.0__ |
| what is the least integer greater than – num__3 + num__0.5 ? <o> a ) – num__2 <o> b ) – num__1 <o> c ) num__0 <o> d ) num__1 <o> e ) num__2 |
this question is just about doing careful arithmetic and remembering what makes a numberbiggerorsmallercompared to another number . first let ' s take care of the arithmetic : ( - num__3 ) + ( num__0.5 ) = - num__2.5 on a number line since we ' re adding + . num__5 to a number the total moves to the right ( so we ' re moving from - num__3 to - num__2.5 ) . next the question asks for the least integer that is greater than - num__2.5 again we can use a number line . numbers become greater as you move to the right . the first integer to the right of - num__2.5 is - num__2 . final answer : a <eor> a <eos> |
a |
subtract__3.0__0.5__ divide__2.5__0.5__ reverse__0.5__ reverse__0.5__ |
subtract__3.0__0.5__ divide__2.5__0.5__ subtract__2.5__0.5__ subtract__2.5__0.5__ |
| if a speaks the truth num__65.0 of the times b speaks the truth num__60.0 of the times . what is the probability that they tell the truth at the same time <o> a ) num__0.39 <o> b ) num__0.48 <o> c ) num__0.41 <o> d ) num__0.482 <o> e ) num__0.411 |
explanation : probability that a speaks truth is num__0.65 = num__0.65 probability that b speaks truth is num__0.6 = num__0.6 since both a and b are independent of each other so probability of a intersection b is p ( a ) × p ( b ) = num__0.65 × num__0.6 = num__0.39 answer : a <eor> a <eos> |
a |
percent__65.0__0.6__ percent__65.0__0.6__ |
percent__65.0__0.6__ percent__65.0__0.6__ |
| a trader mixes num__80 kg of tea at num__15 per kg with num__20 kg of tea at cost price of num__20 per kg . in order to earn a profit of num__25.0 what should be the sale price of the mixed tea ? <o> a ) num__23.75 <o> b ) num__22 <o> c ) num__20 <o> d ) num__19.2 <o> e ) none of these |
c . p . of mixture = num__80 × num__15 + num__20 × num__0.25 + num__20 = num__16 ∴ s . p . = ( num__100 + num__25 ) / num__100 × num__16 = num__20 answer c <eor> c <eos> |
c |
percent__80.0__20.0__ percent__80.0__25.0__ |
percent__80.0__20.0__ percent__80.0__25.0__ |
| if a num__3 - member subcommittee is to be formed from a certain num__8 - member committee how many different such subcommittee are possible ? <o> a ) num__6 <o> b ) num__18 <o> c ) num__40 <o> d ) num__108 <o> e ) num__216 |
another way : num__1 st member can be selected in num__8 ways num__2 nd can be selected in num__6 ways num__3 rd can be selected in num__5 ways so total ways : num__240 but to avoid the similar scenarios num__80.0 ! = num__40 c <eor> c <eos> |
c |
coin_space__ die_space__ vowel_space__ choose__5.0__3.0__ choose__5.0__3.0__ |
coin_space__ die_space__ vowel_space__ choose__5.0__3.0__ choose__5.0__3.0__ |
| there are num__5 burglars and once went to a bakery to rob it obviously . the first guy ate num__0.5 of the total bread and num__0.5 of the bread . the second guy ate num__0.5 of the remaining and num__0.5 of the bread . the third guy fourth guy and fifth guy did the same . after fifth guy there is no bread left out . how many bread are there ? <o> a ) num__30 <o> b ) num__31 <o> c ) num__32 <o> d ) num__33 <o> e ) num__34 |
( num__0.5 + num__0.5 ) = num__1 num__2 ( num__1 + num__0.5 ) = num__3 num__2 ( num__3 + num__0.5 ) = num__7 num__2 ( num__7 + num__0.5 ) = num__15 num__2 ( num__15 + num__0.5 ) = num__31 answer : b <eor> b <eos> |
b |
reverse__0.5__ subtract__5.0__2.0__ add__5.0__2.0__ multiply__5.0__3.0__ multiply__1.0__31.0__ |
reverse__0.5__ add__1.0__2.0__ add__5.0__2.0__ multiply__5.0__3.0__ multiply__1.0__31.0__ |
| if i walk at num__3 kmph i miss the train by num__2 min if however i walk at num__4 kmph . i reach the station num__2 min before the arrival of the train . how far do i walk to reach the station ? <o> a ) num__0.8 km <o> b ) num__0.5 km <o> c ) num__0.4 km <o> d ) num__0.666666666667 km <o> e ) num__0.6 km |
x / num__3 – x / num__4 = num__0.0666666666667 x = num__0.8 km answer : a <eor> a <eos> |
a |
round__0.8__ |
round__0.8__ |
| in a group of dogs and people the number of legs was num__28 more than twice the number of heads . how many dogs were there ? [ assume none of the people or dogs is missing a leg . ] <o> a ) num__4 <o> b ) num__7 <o> c ) num__12 <o> d ) num__14 <o> e ) num__28 |
if there were only people there would be exactly twice the number of legs as heads . each dog contributes two extra legs over and above this number . so num__28 extra legs means num__14 dogs . correct answer d <eor> d <eos> |
d |
subtract__28.0__14.0__ |
subtract__28.0__14.0__ |
| a and b start from opladen and cologne respectively at the same time and travel towards each other at constant speeds along the same route . after meeting at a point between opladen and cologne a and b proceed to their destinations of cologne and opladen respectively . a reaches cologne num__40 minutes after the two meet and b reaches opladen num__60 minutes after their meeting . how long did a take to cover the distance between opladen and cologne ? <o> a ) num__1 hour <o> b ) num__1 hour num__20 minutes <o> c ) num__2 hours num__30 minutes <o> d ) num__1 hour num__40 minutes <o> e ) num__2 hours num__10 minutes |
v num__1 and v num__2 are speeds . v num__1 . t / num__60 = v num__2 v num__2 . t / num__40 = v num__1 v num__1 / v num__2 = num__1.5 which train a would num__60 . num__0.666666666667 mins to cover the same distance num__40 + num__40 = num__80 mins ( ans b ) <eor> b <eos> |
b |
divide__60.0__40.0__ divide__40.0__60.0__ multiply__40.0__2.0__ round__1.0__ |
divide__60.0__40.0__ divide__40.0__60.0__ multiply__40.0__2.0__ round__1.0__ |
| if a mixture is num__1 ⁄ num__7 alcohol by volume and num__2 ⁄ num__7 water by volume what is the ratio of the volume of alcohol to the volume of water in this mixture ? <o> a ) num__2.0 <o> b ) num__1.5 <o> c ) num__0.5 <o> d ) num__2.5 <o> e ) num__3.5 |
should be a sub - num__600 level q . . volume = { num__0.142857142857 } / { num__0.285714285714 } = num__0.5 c <eor> c <eos> |
c |
reverse__7.0__ divide__2.0__7.0__ reverse__2.0__ reverse__2.0__ |
reverse__7.0__ divide__2.0__7.0__ reverse__2.0__ reverse__2.0__ |
| if num__1 + x ^ num__4 + x ^ num__3 + x ^ num__2 + x = num__80 then the average ( arithmetic mean ) r of x x ^ num__2 x ^ num__3 x ^ num__4 and x ^ num__5 is equal to which of the following ? <o> a ) num__12 x <o> b ) num__13 x <o> c ) num__14 x <o> d ) num__16 x <o> e ) num__20 x |
num__1 + x ^ num__4 + x ^ num__3 + x ^ num__2 + x = num__81 i . e . num__1 + x + x ^ num__2 + x ^ num__3 + x ^ num__4 = num__81 x + x ^ num__2 + x ^ num__3 + x ^ num__4 = num__80 x ( num__1 + x + x ^ num__2 + x ^ num__3 ) = num__80 x ( num__81 - x ^ num__4 ) = num__80 num__81 x - x ^ num__5 = num__80 x ^ num__5 = num__81 x - num__80 now x + x ^ num__2 + x ^ num__3 + x ^ num__4 + x ^ num__5 = num__80 + num__81 x - num__80 = num__81 x average of r { x x ^ num__2 x ^ num__3 x ^ num__4 x ^ num__5 } = num__81 x / num__5 ~ num__16 x answerd <eor> d <eos> |
d |
add__1.0__80.0__ divide__80.0__5.0__ multiply__1.0__16.0__ |
add__1.0__80.0__ divide__80.0__5.0__ divide__80.0__5.0__ |
| if num__9 engines consume num__24 metric tonnes of coal when each is working num__8 hoursa day how much coal will be required for num__8 engines each running num__13 hours a day it is being given that num__3 engines of former type consume as much as num__4 engines of latter type ? <o> a ) num__23 <o> b ) num__25 <o> c ) num__26 <o> d ) num__28 <o> e ) num__30 |
explanation : let num__3 engines of former type consume num__1 unit in num__1 hour . then num__4 engines of latter type consume num__1 unit in num__1 hour . therefore num__1 engine of former type consumes ( num__0.333333333333 ) unit in num__1 hour . num__1 engine of latter type consumes ( num__0.25 ) unit in num__1 hour . let the required consumption of coal be x units . less engines less coal consumed ( direct proportion ) more working hours more coal consumed ( direct proportion ) less rate of consumption less coal consumed ( direct proportion ) number of engines num__9 : num__8 working hours num__8 : num__13 } : : num__24 : x rate of consumption ( num__0.333333333333 ) : ( num__0.25 ) ( num__9 * num__8 * ( num__0.333333333333 ) * x ) = ( num__8 * num__13 * ( num__0.25 ) * num__24 ) < = > num__24 x = num__624 < = > x = num__26 . answer : c <eor> c <eos> |
c |
subtract__9.0__8.0__ divide__8.0__24.0__ divide__1.0__4.0__ divide__624.0__24.0__ round__26.0__ |
subtract__9.0__8.0__ divide__8.0__24.0__ divide__1.0__4.0__ divide__624.0__24.0__ round__26.0__ |
| what is the greater of the two numbers whose product is num__2688 given that the sum of the two numbers exceeds their difference by num__64 ? <o> a ) a ) num__84 <o> b ) b ) num__104 <o> c ) c ) num__110 <o> d ) d ) num__72 <o> e ) of these |
let the greater and the smaller number be g and s respectively . gs = num__2560 g + s exceeds g - s by num__64 i . e . g + s - ( g - s ) = num__64 i . e . num__2 s = num__64 = > s = num__32 . g = num__2688 / s = num__84 . answer : a <eor> a <eos> |
a |
divide__64.0__2.0__ divide__2688.0__32.0__ divide__2688.0__32.0__ |
divide__64.0__2.0__ divide__2688.0__32.0__ divide__2688.0__32.0__ |
| for any numbers a and b a # b = ab ( num__6 − b ) . if a and a # b both represent positive numbers which of the following could be a value of b ? <o> a ) num__5 <o> b ) num__54 <o> c ) num__5 num__43 <o> d ) num__5 num__43 num__21 <o> e ) num__54 num__3 |
2 |
a # b positive then b may be positive or negative . if positive then b < num__6 may be num__54 num__32 or num__1 and if negative then it is more than or equal to - num__1 negative . ab will be - ve which is not possible as a and a # b should be positive . ans num__54 num__32 and num__1 d <eor> d <eos> |
d |
d |
| if each side of a square is increased by num__25.0 find the percentage change in its area ? <o> a ) num__54.0 <o> b ) num__56.0 <o> c ) num__56.25 <o> d ) num__58.25 <o> e ) num__59 % |
let each side of the square be a then area = a x a new side = num__125 a / num__100 = num__5 a / num__4 new area = ( num__5 a x num__5 a ) / ( num__4 x num__4 ) = ( num__25 a ² / num__16 ) increased area = = ( num__25 a ² / num__16 ) - a ² increase % = [ ( num__9 a ² / num__16 ) x ( num__1 / a ² ) x num__100 ] % = num__56.25 answer : c <eor> c <eos> |
c |
percent__25.0__4.0__ percent__56.25__100.0__ |
percent__25.0__4.0__ percent__56.25__100.0__ |
| rs . num__600 amounts to rs . num__900 in num__3 years at simple interest . if the interest is increased by num__4.0 it would amount to how much ? <o> a ) num__672 <o> b ) num__246 <o> c ) num__258 <o> d ) num__856 <o> e ) num__653 |
( num__600 * num__3 * num__4 ) / num__100 = num__72 num__600 + num__72 = num__672 answer : a <eor> a <eos> |
a |
percent__100.0__672.0__ |
percent__100.0__672.0__ |
| the average ( arithmetic mean ) monthly income of four workers is $ num__2000 . after one worker ’ s income increases by num__25 percent the new average income is $ num__2250 . what was the original income of the worker whose monthly income increased ? <o> a ) $ num__3000 <o> b ) $ num__2000 <o> c ) $ num__4000 <o> d ) $ num__5000 <o> e ) $ num__6 |
000 |
increase in total income was num__250 * num__4 = $ num__1000 we know that this increase was num__25.0 ( num__0.25 th ) of the workers original income thus his / her original income was num__1000 * num__4 = $ num__4000 answer : c <eor> c <eos> |
c |
c |
| four packages have an average weight of num__12.5 pounds . what is the minimum possible weight of the heaviest package in pounds if the median is num__12 pounds ? <o> a ) num__12 <o> b ) num__13 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
let us denote the weights of the packages in pounds by a b c d naming from the lightest one to the heaviest one . the median is num__12 pounds . therefore ( b + c ) / num__2 = num__12 . b + c = num__24 the average is num__12.5 pounds . therefore ( a + b + c + d ) / num__4 = num__12.5 . a + ( b + c ) + d = num__50 a + num__24 + d = num__50 a + d = num__26 the weight a must be no greater than num__12 since num__12 is the median . therefore the minimum possible weight of the heaviest package is num__26 – num__12 = num__14 pounds ( all the other packages would weigh num__12 pounds in this case ) . answer : c <eor> c <eos> |
c |
multiply__12.0__2.0__ multiply__12.5__4.0__ add__2.0__24.0__ add__12.0__2.0__ add__12.0__2.0__ |
multiply__12.0__2.0__ multiply__12.5__4.0__ add__2.0__24.0__ add__12.0__2.0__ add__12.0__2.0__ |
| a man can row with a speed of num__15 kmph in still water . if the stream flows at num__5 kmph then the speed in downstream is ? <o> a ) num__27 <o> b ) num__23 <o> c ) num__20 <o> d ) num__28 <o> e ) num__12 |
m = num__15 s = num__5 ds = num__15 + num__5 = num__20 answer : c <eor> c <eos> |
c |
add__15.0__5.0__ round__20.0__ |
add__15.0__5.0__ add__15.0__5.0__ |
| f a card is drawn from a well shuffled pack of cards the probability of drawing a spade or a king is - . <o> a ) num__0.266666666667 <o> b ) num__0.4 <o> c ) num__0.222222222222 <o> d ) num__0.307692307692 <o> e ) num__0.363636363636 |
p ( s ᴜ k ) = p ( s ) + p ( k ) - p ( s ∩ k ) where s denotes spade and k denotes king . p ( s ᴜ k ) = num__0.25 + num__0.0769230769231 - num__0.0192307692308 = num__0.307692307692 answer : d <eor> d <eos> |
d |
union_prob__0.25__0.0769__0.0192__ union_prob__0.25__0.0769__0.0192__ |
union_prob__0.25__0.0769__0.0192__ union_prob__0.25__0.0769__0.0192__ |
| p can do the work in num__10 days and q can do the same work in num__12 days . if they work together for num__4 days what is the fraction of work that is left ? <o> a ) num__0.333333333333 <o> b ) num__0.2 <o> c ) num__0.4 <o> d ) num__0.266666666667 <o> e ) num__0.466666666667 |
num__0.4 + num__0.333333333333 = num__0.733333333333 = num__0.733333333333 ( completed work ) the work that is left is num__1 - num__0.733333333333 = num__0.266666666667 the answer is d . <eor> d <eos> |
d |
divide__4.0__10.0__ divide__4.0__12.0__ add__0.4__0.3333__ subtract__1.0__0.7333__ divide__0.2667__1.0__ |
divide__4.0__10.0__ divide__4.0__12.0__ add__0.4__0.3333__ subtract__1.0__0.7333__ subtract__1.0__0.7333__ |
| a fort had provision of food for num__150 men for num__60 days . after num__10 days num__25 men left the fort . the number of days for which the remaining food will last is : <o> a ) num__29 num__0.2 <o> b ) num__37 num__0.25 <o> c ) num__42 <o> d ) num__54 <o> e ) num__60 |
after num__10 days : num__150 men had food for num__50 days . suppose num__125 men had food for x days . now less men more days ( indirect proportion ) therefore num__125 : num__150 : : num__50 : x < = > num__125 x x = num__150 x num__50 = > x = num__150 x num__0.4 = > x = num__60 . correct answer is e <eor> e <eos> |
e |
subtract__60.0__10.0__ subtract__150.0__25.0__ divide__60.0__150.0__ hour_to_min_conversion__ |
subtract__60.0__10.0__ subtract__150.0__25.0__ divide__60.0__150.0__ hour_to_min_conversion__ |
| m and n are the x and y coordinates respectively of a point in the coordinate plane . if the points ( m n ) and ( m + p n + num__9 ) both lie on the line defined by the equation x = ( y / num__3 ) - ( num__0.4 ) what is the value of p ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
x = ( y / num__3 ) - ( num__0.4 ) and so y = num__3 x + num__1.2 . the slope is num__3 . ( n + num__9 - n ) / ( m + p - m ) = num__3 p = num__3 the answer is c . <eor> c <eos> |
c |
multiply__3.0__0.4__ divide__9.0__3.0__ |
multiply__3.0__0.4__ divide__9.0__3.0__ |
| length of a rectangular plot is num__20 mtr more than its breadth . if the cost of fencing the plot at num__26.50 per meter is rs . num__5300 what is the length of the plot in mtr ? <o> a ) num__59 m <o> b ) num__60 m <o> c ) num__80 m <o> d ) num__82 m <o> e ) num__84 m |
let breadth = x metres . then length = ( x + num__18 ) metres . perimeter = num__5300 m = num__200 m . num__26.50 num__2 [ ( x + num__18 ) + x ] = num__200 num__2 x + num__18 = num__100 num__2 x = num__82 x = num__41 . hence length = x + num__18 = num__59 m a <eor> a <eos> |
a |
divide__5300.0__26.5__ subtract__20.0__18.0__ divide__200.0__2.0__ subtract__100.0__18.0__ divide__82.0__2.0__ subtract__100.0__41.0__ round__59.0__ |
divide__5300.0__26.5__ subtract__20.0__18.0__ divide__200.0__2.0__ subtract__100.0__18.0__ divide__82.0__2.0__ add__41.0__18.0__ add__41.0__18.0__ |
| for the positive numbers n n + num__1 n + num__2 n + num__7 and n + num__10 the mean is how much greater than the median ? <o> a ) num__0 <o> b ) num__1 <o> c ) n + l <o> d ) num__2 <o> e ) n + num__3 |
let ’ s first calculate the mean ( arithmetic average ) . mean = sum / quantity mean = ( n + n + num__1 + n + num__2 + n + num__7 + n + num__10 ) / num__5 mean = ( num__5 n + num__20 ) / num__5 mean = n + num__4 next we determine the median . the median is the middle value when the terms are ordered from least to greatest . the terms ordered from least to greatest are as follows : n n + num__1 n + num__2 n + num__7 n + num__10 the median is n + num__2 . finally we are asked how much greater the mean is than the median . to determine the difference we can subtract the smaller value ( the median ) from the larger value ( the mean ) and we get : n + num__4 – ( n + num__2 ) = n + num__4 – n – num__2 = num__2 the answer is d <eor> d <eos> |
d |
subtract__7.0__2.0__ multiply__2.0__10.0__ subtract__5.0__1.0__ multiply__1.0__2.0__ |
divide__10.0__2.0__ multiply__2.0__10.0__ divide__20.0__5.0__ divide__2.0__1.0__ |
| the average age of num__8 men is increased by years when two of them whose ages are num__21 years and num__23 years are replaced by two new men . the average age of the two new men is <o> a ) num__22 <o> b ) num__30 <o> c ) num__99 <o> d ) num__28 <o> e ) num__17 |
explanation : total age increased = ( num__8 * num__2 ) years = num__16 years . sum of ages of two new men = ( num__21 + num__23 + num__16 ) years = num__60 years average age of two new men = ( num__30.0 ) years = num__30 years answer : b <eor> b <eos> |
b |
subtract__23.0__21.0__ multiply__8.0__2.0__ divide__60.0__2.0__ divide__60.0__2.0__ |
subtract__23.0__21.0__ multiply__8.0__2.0__ divide__60.0__2.0__ divide__60.0__2.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 sec . what is the length of the train ? <o> a ) num__158 <o> b ) num__159 <o> c ) num__155 <o> d ) num__112 <o> e ) num__150 |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__9 = num__150 m answer : e <eor> e <eos> |
e |
round__150.0__ |
round__150.0__ |
| ab + ba ___ num__161 in the addition problem above a and b represent digits in two different two - digit numbers . what is the sum of a and b ? <o> a ) num__6 <o> b ) num__13 <o> c ) num__9 <o> d ) num__11 <o> e ) num__14 |
two ways to do it . . num__1 ) straight logic . . we can see that ones digit is num__1 when units digit b and a are added . . only num__13 fits in . . b <eor> b <eos> |
b |
multiply__1.0__13.0__ |
multiply__1.0__13.0__ |
| in an office totally there are num__1800 employees and num__60.0 of the total employees are males . num__30.0 of the males in the office are at - least num__50 years old . find the number of males aged below num__50 years ? <o> a ) num__756 <o> b ) num__750 <o> c ) num__600 <o> d ) num__760 <o> e ) none of these |
number of male employees = num__1800 * num__0.6 = num__1080 required number of male employees who are less than num__50 years old = num__1080 * ( num__100 - num__30 ) % = num__1080 * num__0.7 = num__756 . answer : a <eor> a <eos> |
a |
divide__30.0__50.0__ multiply__1800.0__0.6__ divide__60.0__0.6__ multiply__1080.0__0.7__ multiply__1080.0__0.7__ |
divide__30.0__50.0__ multiply__1800.0__0.6__ divide__60.0__0.6__ multiply__1080.0__0.7__ multiply__1080.0__0.7__ |
| a can complete a project in num__20 days while b can complete same project in num__30 days . if a and b start working together and a leaves the work num__5 days before completion of project then in how many days the project will be completed ? <o> a ) num__15 <o> b ) num__19 <o> c ) num__20 <o> d ) num__21 <o> e ) num__22 |
a ' s num__1 day work = num__0.05 ; b ' s num__1 day work = num__0.0333333333333 ; ( a + b ) num__1 day work = ( num__0.05 + num__0.0333333333333 ) = num__0.0833333333333 ; it is given that a leaves the work num__5 days before completion of the project . . thus b alone does the remaining job in num__5 days . so in num__5 days b can do num__0.166666666667 w . . thus ( a + b ) have worked ( num__1 - num__0.166666666667 ) = num__0.833333333333 w . . ( a + b ) can do num__0.0833333333333 work in num__1 day . . . they did num__0.833333333333 w in num__10 days . total days = ( num__10 + num__5 ) = num__15 answer : a <eor> a <eos> |
a |
divide__1.0__20.0__ divide__1.0__30.0__ add__0.05__0.0333__ divide__5.0__30.0__ subtract__1.0__0.1667__ subtract__30.0__20.0__ subtract__20.0__5.0__ round__15.0__ |
divide__1.0__20.0__ divide__1.0__30.0__ add__0.05__0.0333__ divide__5.0__30.0__ subtract__1.0__0.1667__ subtract__30.0__20.0__ subtract__20.0__5.0__ subtract__20.0__5.0__ |
| charles walks over a railway - bridge . at the moment that he is just ten meters away from the middle of the bridge he hears a train coming from behind . at that moment the train which travels at a speed of num__90 km / h is exactly as far away from the bridge as the bridge measures in length . without hesitation charles rushes straight towards the train to get off the bridge . in this way he misses the train by just four meters ! if charles had rushed exactly as fast in the other direction the train would have hit him eight meters before the end of the bridge . what is the length of the railway - bridge ? <o> a ) num__37 <o> b ) num__44 <o> c ) num__73 <o> d ) num__52 <o> e ) num__48 |
b num__44 m let the length of the bridge be x meters . running towards the train charles covers num__0.5 x - num__10 meters in the time that the train travels x - num__4 meters . running away from the train charles covers num__0.5 x + num__2 meters in the time that the train travels num__2 x - num__8 meters . because their speeds are constant the following holds : ( num__0.5 x - num__10 ) / ( x - num__4 ) = ( num__0.5 x + num__2 ) / ( num__2 x - num__8 ) which can be rewritten to num__0.5 x num__2 - num__24 x + num__88 = num__0 using the quadratic formula we find that x = num__44 so the railway - bridge has a length of num__44 meters . <eor> b <eos> |
b |
reverse__0.5__ multiply__2.0__4.0__ subtract__90.0__2.0__ round_down__0.5__ multiply__0.5__88.0__ |
reverse__0.5__ divide__4.0__0.5__ subtract__90.0__2.0__ round_down__0.5__ divide__88.0__2.0__ |
| the averge score of a cricketer for ten matches is num__38.9 runs . if the average for the first six matches is num__42 . then find the average for the last four matches ? <o> a ) num__33.25 <o> b ) num__33.5 <o> c ) num__34.25 <o> d ) num__35 <o> e ) num__36 |
sum of last num__4 matches = ( num__10 × num__38.9 ) – ( num__6 × num__42 ) = num__389 – num__252 = num__137 average = num__34.25 = num__34.25 answer : c <eor> c <eos> |
c |
subtract__10.0__4.0__ multiply__38.9__10.0__ multiply__42.0__6.0__ subtract__389.0__252.0__ divide__137.0__4.0__ divide__137.0__4.0__ |
subtract__10.0__4.0__ multiply__38.9__10.0__ multiply__42.0__6.0__ subtract__389.0__252.0__ divide__137.0__4.0__ divide__137.0__4.0__ |
| a man swims downstream num__70 km and upstream num__30 km taking num__10 hours each time ; what is the speed of the current ? <o> a ) num__1 <o> b ) num__4 <o> c ) num__2 <o> d ) num__6 <o> e ) num__8 |
num__70 - - - num__10 ds = num__7 ? - - - - num__1 num__30 - - - - num__10 us = num__3 ? - - - - num__1 s = ? s = ( num__7 - num__3 ) / num__2 = num__2 answer : c <eor> c <eos> |
c |
divide__70.0__10.0__ divide__30.0__10.0__ subtract__3.0__1.0__ round__2.0__ |
divide__70.0__10.0__ divide__30.0__10.0__ subtract__3.0__1.0__ divide__2.0__1.0__ |
| a mixture contains milk and water in the ratio num__3 : num__2 . on adding num__10 liters of water the ratio of milk to water becomes num__2 : num__3 . total quantity of milk & water before adding water to it ? <o> a ) num__30 <o> b ) num__40 <o> c ) num__20 <o> d ) num__10 <o> e ) num__25 |
milk : water = num__3 : num__2 after adding num__10 liters of water milk : water = num__2 : num__3 olny water patrs increase when mixture of water milk : wate = num__3 : num__2 = num__2 * ( num__3 : num__2 ) = num__6 : num__4 after adding num__10 liters of water milk : water = num__2 : num__3 = num__3 * ( num__2 : num__3 ) = num__6 : num__9 option ' c ' <eor> c <eos> |
c |
multiply__3.0__2.0__ subtract__10.0__6.0__ add__3.0__6.0__ multiply__2.0__10.0__ |
multiply__3.0__2.0__ subtract__10.0__6.0__ add__3.0__6.0__ multiply__2.0__10.0__ |
| the average of first num__50 non - zero positive integers is <o> a ) num__11.5 <o> b ) num__25.5 <o> c ) num__22 <o> d ) num__25 <o> e ) num__27 |
explanation : sum of first n non - zero positive integers n ( n + num__1 ) / num__2 so average of first n non - zero positive integers n ( n + num__1 ) / num__2 n = ( n + num__1 ) / num__2 = > ( num__50 + num__1 ) / num__2 = num__25.5 answer : b <eor> b <eos> |
b |
multiply__25.5__1.0__ |
divide__25.5__1.0__ |
| a tank is filled by three pipes with uniform flow . the first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone . the second pipe fills the tank num__5 hours faster than the first pipe and num__4 hours slower than the third pipe . find the time required by the first pipe to fill the tank ? <o> a ) num__10 hours <o> b ) num__15 hours <o> c ) num__17 hours <o> d ) num__18 hours <o> e ) num__19 hours |
explanation : suppose first pipe alone takes x hours to fill the tank . then second and third pipes will take ( x - num__5 ) and ( x - num__9 ) hours respectively to fill the tank . as per question we get num__1 / x + num__1 / x - num__5 = num__1 / x - num__9 = > x â ˆ ’ num__5 + x / x ( x â ˆ ’ num__5 ) = num__1 / x - num__9 = > ( num__2 x â ˆ ’ num__5 ) ( x â ˆ ’ num__9 ) = x ( x â ˆ ’ num__5 ) = > x num__2 â ˆ ’ num__18 x + num__45 = num__0 after solving this euation we get ( x - num__15 ) ( x + num__3 ) = num__0 as value can not be negative so x = num__15 answer is b <eor> b <eos> |
b |
add__5.0__4.0__ subtract__5.0__4.0__ multiply__2.0__9.0__ multiply__5.0__9.0__ subtract__5.0__2.0__ round__15.0__ |
add__5.0__4.0__ subtract__5.0__4.0__ multiply__2.0__9.0__ multiply__5.0__9.0__ subtract__5.0__2.0__ divide__45.0__3.0__ |
| a foreign language club at washington middle school consists of n students num__0.4 of whom are boys . all of the students in the club study exactly one foreign language . num__0.333333333333 of the girls in the club study spanish and num__0.75 of the remaining girls study french . if the rest of the girls in the club study german how many girls in the club in terms of n study german ? <o> a ) num__2 n / num__5 <o> b ) n / num__3 <o> c ) n / num__5 <o> d ) num__2 n / num__15 <o> e ) n / num__10 |
num__0.4 of the students are boys thus num__0.6 of the students are girls . num__0.333333333333 of the girls in the club study spanish and num__0.75 of the remaining girls study french . thus num__0.25 of num__0.666666666667 = num__0.166666666667 of the girls study german . since girls comprise num__0.6 of the students then num__0.6 * num__0.166666666667 = num__0.1 of students are girls who study german . the answer is e . <eor> e <eos> |
e |
multiply__0.3333__0.75__ divide__0.4__0.6__ multiply__0.25__0.6667__ multiply__0.4__0.25__ reverse__0.1__ |
multiply__0.3333__0.75__ divide__0.4__0.6__ multiply__0.25__0.6667__ multiply__0.4__0.25__ reverse__0.1__ |
| a semicircular cubicle has a radius of num__14 . what is the approximate perimeter of the cubicle ? <o> a ) num__55 <o> b ) num__86 <o> c ) num__25 <o> d ) num__72 <o> e ) num__35 |
perimeter of a circle = num__2 pi * r perimeter of a semicircle = pi * r + num__2 r aprox perimiter = num__3.14 * num__14 + num__2 * num__14 = num__71.96 approximately num__72 answer d <eor> d <eos> |
d |
triangle_area__72.0__2.0__ |
triangle_area__72.0__2.0__ |
| jack went on a diet num__6 months ago when he weighed num__222 pounds . if he now weighs num__198 pounds and continues to lose at the same average monthly rate in approximately how many months will he weigh num__180 pounds ? <o> a ) num__3 <o> b ) num__35 <o> c ) num__4 <o> d ) num__45 <o> e ) num__5 |
joe lost his weigh num__24 pounds during num__6 months . that means that he lost num__4 pounds per month . he should lose more num__18 pounds to be num__180 pounds . so it would take num__4.5 = num__4.5 months . correct answer is d <eor> d <eos> |
d |
subtract__222.0__198.0__ divide__24.0__6.0__ subtract__198.0__180.0__ divide__18.0__4.0__ divide__180.0__4.0__ |
subtract__222.0__198.0__ divide__24.0__6.0__ subtract__198.0__180.0__ divide__18.0__4.0__ divide__180.0__4.0__ |
| seven different books ( a b c d e f g and h ) are to be arranged on a shelf . books a and b are to be arranged first and second starting from the right of the shelf . the number of different orders in which books c d e f and g may be arranged is <o> a ) num__9 ! <o> b ) num__12 ! <o> c ) num__5 ! <o> d ) num__25 ! <o> e ) num__18 ! |
solution since books a and b are arranged first and second only books c d e f and g will change order . therefore it an arrangement problem involving num__5 items and the number of different order is given by num__5 ! answer c <eor> c <eos> |
c |
vowel_space__ vowel_space__ |
vowel_space__ vowel_space__ |
| if two resistors a ( r num__1 ) and b ( r num__2 ) stand in parallel with each other in electrical wire the total resistor appears as r num__1 r num__2 / ( r num__1 + r num__2 ) . if three resistors a ( r num__1 ) b ( r num__2 ) and c ( num__2 r num__2 ) stand in parallel in electrical wire what is the ratio q of the resistors ’ sum of a and c to the resistors ’ sum of a and b ? <o> a ) num__2 ( r num__1 + r num__2 ) : ( r num__1 + num__2 r num__2 ) <o> b ) ( r num__1 + r num__2 ) : ( r num__1 + num__2 r num__2 ) <o> c ) ( num__2 r num__1 + r num__2 ) : ( r num__1 + num__2 r num__2 ) <o> d ) num__2 ( r num__1 + r num__2 ) : ( num__2 r num__1 + r num__2 ) <o> e ) num__2 ( r num__1 - r num__2 ) : ( r num__1 + num__2 r num__2 ) |
two resistors a ( r num__1 ) and b ( r num__2 ) . total or sum of two resistors appear as r num__1 r num__2 / r num__1 + r num__2 . it is looks like inversion of sum of rates . num__1 / r num__1 + num__1 / r num__2 = r num__1 + r num__2 / r num__1 r num__2 . same way sum of a ( r num__1 ) and c ( num__2 r num__2 ) = num__1 / r num__1 + num__0.5 r num__2 = num__2 r num__2 + r num__1 / r num__12 r num__2 . inversion rate = r num__12 r num__1.0 r num__2 + r num__1 . ratio of sum of a and c / sum of a and b = num__2 r num__2 r num__0.5 r num__2 + r num__1 * r num__1 + r num__2 / r num__1 r num__2 q = num__2 ( r num__1 + r num__2 ) / num__2 r num__2 + r num__1 . a <eor> a <eos> |
a |
reverse__2.0__ reverse__0.5__ |
reverse__2.0__ reverse__0.5__ |
| what is the median from the below series num__5 num__8 num__11 num__6 num__10 num__4 num__18 num__16 num__13 num__12 and num__14 <o> a ) num__11 <o> b ) num__5 <o> c ) num__12 <o> d ) num__13 <o> e ) num__10 |
ordering the data from least to greatest we get : num__4 num__5 num__6 num__8 num__10 num__11 num__12 num__13 num__14 num__16 num__18 the median number was num__11 . ( five numbers were higher than num__11 and five were lower . ) a <eor> a <eos> |
a |
add__5.0__6.0__ |
add__5.0__6.0__ |
| boring a well num__60 m depth the cost is rs . num__4 / – for the first meter . the cost of each successive meter increases by rs . num__3 / – . what is the cost of boring the entire well ? <o> a ) num__5550 <o> b ) num__5450 <o> c ) num__5650 <o> d ) num__5780 <o> e ) num__5980 |
as it is increased by num__3 rupees so num__4 + num__7 + num__10 + num__13 . . . . . . . . first term a = num__4 difference = d = num__3 then num__60 th term = a + num__59 d = num__181 total cost = n / num__2 ( a + l ) = num__30.0 ( num__4 + num__181 ) = num__5550 answer : a <eor> a <eos> |
a |
add__4.0__3.0__ add__3.0__7.0__ add__3.0__10.0__ divide__60.0__2.0__ round__5550.0__ |
add__4.0__3.0__ add__3.0__7.0__ add__3.0__10.0__ divide__60.0__2.0__ round__5550.0__ |
| num__7 people ( a b c d e f and g ) go to a movie and sit next to each other in num__7 adjacent seats in the front row of the theater . how many different arrangements are possible ? if a will not sit to the left of f and f will not sit to the left of e . how many different arrangements are possible . <o> a ) num__7 ! / num__2 <o> b ) num__7 ! / num__3 <o> c ) num__7 ! / num__4 <o> d ) num__7 ! / num__5 <o> e ) num__7 ! / num__6 |
num__7 people can be arranged in a row in num__7 ! ways . now three people among those num__7 can be arranged in num__3 ! = num__6 ways : afe aef eaf efa fae fea from the num__6 arrangements above only efa is possible ( a is not to the left of f and f is not to the left of e ) so out of total num__7 ! ways only in num__0.166666666667 th of the arrangements they are sitting as they want . answer : e ( num__7 ! / num__6 ) . <eor> e <eos> |
e |
die_space__ choose__7.0__6.0__ |
die_space__ choose__7.0__6.0__ |
| five years ago the average age of a and b was num__15 years . average age of a b and c today is num__20 years . how old will c be after num__14 years ? <o> a ) num__30 <o> b ) num__34 <o> c ) num__40 <o> d ) num__50 <o> e ) num__60 |
explanation : ( a + b ) five years ago = ( num__15 * num__2 ) = num__30 years . ( a + b ) now = ( num__30 + num__5 * num__2 ) years = num__40 years . ( a + b + c ) now = ( num__20 x num__3 ) years = num__60 years . c now = ( num__60 - num__40 ) years = num__20 years . c after num__14 years = ( num__20 + num__14 ) years = num__34 years . answer : b <eor> b <eos> |
b |
multiply__15.0__2.0__ subtract__20.0__15.0__ multiply__20.0__2.0__ divide__15.0__5.0__ multiply__20.0__3.0__ add__20.0__14.0__ add__20.0__14.0__ |
multiply__15.0__2.0__ subtract__20.0__15.0__ multiply__20.0__2.0__ subtract__5.0__2.0__ multiply__20.0__3.0__ add__20.0__14.0__ add__20.0__14.0__ |
| a soccer store typically sells replica jerseys at a discount of num__30 percent to num__50 percent off list price . during the annual summer sale everything in the store is an additional num__20 percent off the original list price . if a replica jersey ' s list price is $ num__80 approximately what w percent of the list price is the lowest possible sale price ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__30 <o> d ) num__40 <o> e ) num__50 |
let the list price be num__2 x for min sale price the first discount given should be num__50.0 num__2 x becomes x here now during summer sale additional num__20.0 off is given ie sale price becomes num__0.8 x it is given lise price is $ num__80 = > num__2 x = num__80 = > x = num__40 and num__0.8 x = num__32 so lowest sale price is num__32 which w is num__40.0 of num__80 hence d is the answer <eor> d <eos> |
d |
multiply__50.0__0.8__ add__30.0__2.0__ multiply__50.0__0.8__ |
multiply__50.0__0.8__ add__30.0__2.0__ multiply__50.0__0.8__ |
| num__1200 boys and num__800 girls are examined in a test ; num__4.0 of the boys and num__30.0 of the girls pass . the percentage of the total who failed is ? <o> a ) a ) num__63.3 <o> b ) b ) num__52.4 <o> c ) c ) num__62.8 <o> d ) d ) num__75.4 <o> e ) e ) num__69.1 % |
total number of students = num__1200 + num__800 = num__2000 number of students passed = ( num__42.0 of num__1200 + num__30.0 of num__800 ) = num__504 + num__240 = num__744 number of failures = num__1256 * num__0.05 = num__62.8 answer is c <eor> c <eos> |
c |
add__1200.0__800.0__ add__240.0__504.0__ subtract__2000.0__744.0__ multiply__0.05__1256.0__ multiply__0.05__1256.0__ |
add__1200.0__800.0__ add__240.0__504.0__ subtract__2000.0__744.0__ multiply__0.05__1256.0__ multiply__0.05__1256.0__ |
| find the missing number in the given sequence : num__16 num__1116 ? num__2631 ? <o> a ) num__20 & num__35 <o> b ) num__19 & num__34 <o> c ) num__18 & num__33 <o> d ) num__21 & num__36 <o> e ) num__25 & num__35 |
num__1 + num__5 = num__6 num__6 + num__5 = num__11 num__11 + num__5 = num__16 num__16 + num__5 = num__21 num__21 + num__5 = num__26 num__26 + num__5 = num__31 num__31 + num__5 = num__36 answer : d <eor> d <eos> |
d |
add__1.0__5.0__ subtract__16.0__5.0__ add__16.0__5.0__ add__5.0__21.0__ add__5.0__26.0__ divide__1116.0__31.0__ add__16.0__5.0__ |
add__1.0__5.0__ add__5.0__6.0__ add__16.0__5.0__ add__5.0__21.0__ add__5.0__26.0__ add__5.0__31.0__ add__16.0__5.0__ |
| if a square mirror has a num__20 - inch diagonal what is the approximate perimeter r of the mirror in inches ? <o> a ) num__40 <o> b ) num__60 <o> c ) num__80 <o> d ) num__100 <o> e ) num__120 |
if you draw the square and diagonal inside the square . u can see square becomes part of two triangles opposite to each other . and we know the property of the triangle addition of two sides of triangle must be greater than its diagonal in order to complete the triangle . and each side must be less than num__20 and perimeter r must be less than num__80 so we can eliminate answer choice c d and e . so side num__1 + side num__2 > num__20 that means side num__1 or side num__2 must be > num__10 . so we can eliminate the answer choice a . now we are left with is b <eor> b <eos> |
b |
square_perimeter__20.0__ triangle_area__20.0__1.0__ rectangle_perimeter__20.0__10.0__ |
square_perimeter__20.0__ triangle_area__20.0__1.0__ rectangle_perimeter__20.0__10.0__ |
| the number num__50 can be written as the sum of squares of num__3 integers . which of the following could be the difference between the largest and smallest integers of the num__3 ? <o> a ) num__2 <o> b ) num__5 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
notice that the question asks which of the followingcouldbe the difference between the largest and smallest integers not must be . the num__3 integers could be : + / - num__3 + / - num__4 and + / - num__5 so the difference could be num__2 - num__2 or num__1 . since only one of them is among the choices then it must be the correct answer . answer : a <eor> a <eos> |
a |
subtract__5.0__3.0__ subtract__3.0__2.0__ subtract__3.0__1.0__ |
subtract__5.0__3.0__ subtract__3.0__2.0__ subtract__3.0__1.0__ |
| at what rate percent per annum will a sum of money double in num__7 years . <o> a ) num__12.5 <o> b ) num__13.5 <o> c ) num__14.2 <o> d ) num__14.5 <o> e ) num__21.5 % |
let principal = p then s . i . = p and time = num__8 years rate = [ ( num__100 x p ) / ( p x num__8 ) ] % = num__14.2 per annum . answer : c <eor> c <eos> |
c |
percent__14.2__100.0__ |
percent__14.2__100.0__ |
| a certain basketball team that has played num__0.666666666667 of its games has a record of num__13 wins and num__7 losses . what is the greatest number of the remaining games that the team can lose and still win at least num__0.75 of all of its games ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__4 <o> e ) num__3 |
num__13 wins num__7 losses - total num__20 games played . the team has played num__0.666666666667 rd of all games so total number of games = num__30 num__0.75 th of num__30 is num__22.5 so the team must win num__23 games and can afford to lose at most num__7 total games . it has already lost num__7 games so it can lose another num__0 at most . answer ( a ) <eor> a <eos> |
a |
add__13.0__7.0__ multiply__0.75__30.0__ subtract__30.0__7.0__ round_down__0.6667__ round_down__0.6667__ |
add__13.0__7.0__ multiply__0.75__30.0__ subtract__30.0__7.0__ round_down__0.6667__ round_down__0.6667__ |
| the price of commodity x increases by num__40 paise every year while the price of commodity y increases by num__15 paise every year . if in num__2001 the price of commodity x was rs . num__4.20 and that of y was rs . num__6.30 in which year commodity x will cost num__40 paise more than the commodity y ? <o> a ) num__2010 <o> b ) num__2011 <o> c ) num__2012 <o> d ) num__2013 <o> e ) num__2014 |
suppose commodity x will cost num__40 paise more than y after z years . then ( num__4.20 + num__0.40 z ) - ( num__6.30 + num__0.15 z ) = num__0.40 num__0.25 z = num__0.40 + num__2.10 z = num__2.50 / num__0.25 z = num__10.0 z = num__10 x will cost num__40 paise more than y num__10 years after num__2001 i . e . num__2011 . answer = b <eor> b <eos> |
b |
subtract__0.4__0.15__ subtract__6.3__4.2__ reverse__0.4__ multiply__40.0__0.25__ add__2001.0__10.0__ add__2001.0__10.0__ |
subtract__0.4__0.15__ subtract__6.3__4.2__ reverse__0.4__ divide__2.5__0.25__ add__2001.0__10.0__ add__2001.0__10.0__ |
| the h . c . f of two numbers is num__13 and their l . c . m is num__2200 . if one of the numbers is num__286 then the other is ? <o> a ) num__85 <o> b ) num__90 <o> c ) num__95 <o> d ) num__100 <o> e ) num__105 |
other number = ( num__13 * num__2200 ) / num__286 = num__100 . answer : d <eor> d <eos> |
d |
gcd__2200.0__100.0__ |
gcd__2200.0__100.0__ |
| find the area of a rhombus one side of which measures num__20 cm and one diagonal is num__26 cm . <o> a ) num__600 cm num__2 <o> b ) num__500 cm num__2 <o> c ) num__516 cm num__2 <o> d ) num__416 cm num__2 <o> e ) num__100 cm num__2 |
explanation : let other diagonal = num__2 x cm . since diagonals of a rhombus bisect each other at right angles we have : ( num__20 ) num__2 = ( num__12 ) num__2 + ( x ) num__2 = > x = √ ( num__20 ) num__2 – ( num__12 ) num__2 = √ num__256 = num__16 cm . _ i so other diagonal = num__32 cm . area of rhombus = ( num__0.5 ) x ( product of diagonals ) = ( num__0.5 × num__26 x num__32 ) cm num__2 = num__416 cm num__2 answer : option d <eor> d <eos> |
d |
side_by_diagonal__20.0__12.0__ multiply__2.0__16.0__ triangle_area__26.0__32.0__ triangle_area__26.0__32.0__ |
side_by_diagonal__20.0__12.0__ multiply__2.0__16.0__ triangle_area__26.0__32.0__ triangle_area__26.0__32.0__ |
| a man is num__20 years older than his son . in two years his age will be twice the age of his son . the present age of his son is <o> a ) num__18 years <o> b ) num__21 years <o> c ) num__22 years <o> d ) num__24 years <o> e ) num__26 years |
explanation : let the son ' s present age be x years . then man ' s present age = ( x + num__20 ) years = > ( x + num__20 ) + num__2 = num__2 ( x + num__2 ) = > x + num__22 = num__2 x + num__4 so x = num__18 answer : option a <eor> a <eos> |
a |
add__20.0__2.0__ subtract__20.0__2.0__ subtract__20.0__2.0__ |
add__20.0__2.0__ subtract__20.0__2.0__ subtract__20.0__2.0__ |
| rice weighing num__12.5 pounds was divided equally and placed in num__4 containers . how many ounces of rice were in each container ? ( note that num__1 pound = num__16 ounces ) <o> a ) num__40 <o> b ) num__50 <o> c ) num__60 <o> d ) num__70 <o> e ) num__80 |
num__12.5 ÷ num__4 = num__3.125 pounds in each container num__3.125 pounds * num__16 ounces / pound = num__50 ounces in each container the answer is b . <eor> b <eos> |
b |
divide__12.5__4.0__ multiply__12.5__4.0__ multiply__12.5__4.0__ |
divide__12.5__4.0__ multiply__12.5__4.0__ multiply__12.5__4.0__ |
| a certain sum is invested at simple interest at num__18.0 p . a . for two years instead of investing at num__12.0 p . a . for the same time period . therefore the interest received is more by rs . num__480 . find the sum ? <o> a ) num__7000 <o> b ) num__7029 <o> c ) num__4000 <o> d ) num__2800 <o> e ) num__2791 |
let the sum be rs . x . ( x * num__18 * num__2 ) / num__100 - ( x * num__12 * num__2 ) / num__100 = num__480 = > num__36 x / num__100 - num__24 x / num__100 = num__480 = > num__12 x / num__100 = num__480 = > x = num__4000 . answer : c <eor> c <eos> |
c |
percent__100.0__4000.0__ |
percent__100.0__4000.0__ |
| if a new town has num__150 residents and the population doubles every num__10 years what will be its population after num__75 years ? <o> a ) num__6034 residents <o> b ) num__9051 residents <o> c ) num__12068 residents <o> d ) num__15075 residents <o> e ) num__27153 residents |
num__150 * num__2 ^ ( num__7.5 ) = num__150 * num__2 ^ num__7.5 = num__150 * num__181.02 = num__27153 the answer is e . <eor> e <eos> |
e |
divide__150.0__75.0__ divide__75.0__10.0__ multiply__150.0__181.02__ multiply__150.0__181.02__ |
divide__150.0__75.0__ divide__75.0__10.0__ multiply__150.0__181.02__ multiply__150.0__181.02__ |
| the tax on a commodity is diminished by num__20.0 but its consumption is increased by num__5.0 . find the decrease percent in the revenue derived from it ? <o> a ) num__12.0 <o> b ) num__14.0 <o> c ) num__16.0 <o> d ) num__20.0 <o> e ) num__22 % |
explanation : num__100 * num__100 = num__10000 num__80 * num__105 = num__8400 num__10000 - - - - - - - num__1600 num__100 - - - - - - - ? = num__16.0 c ) <eor> c <eos> |
c |
multiply__20.0__5.0__ subtract__100.0__20.0__ add__5.0__100.0__ multiply__80.0__105.0__ multiply__20.0__80.0__ divide__1600.0__100.0__ divide__1600.0__100.0__ |
multiply__20.0__5.0__ subtract__100.0__20.0__ add__5.0__100.0__ multiply__80.0__105.0__ multiply__20.0__80.0__ divide__1600.0__100.0__ divide__1600.0__100.0__ |
| the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square whose area is num__2025 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is num__10 units ? <o> a ) num__140 <o> b ) num__150 <o> c ) num__160 <o> d ) num__170 <o> e ) num__180 |
given that the area of the square = num__2025 sq . units = > side of square = √ num__2025 = num__45 units the radius of the circle = side of the square = num__45 units length of the rectangle = num__0.4 * num__45 = num__18 units given that breadth = num__10 units area of the rectangle = lb = num__18 * num__10 = num__180 sq . units answer : option e <eor> e <eos> |
e |
multiply__0.4__45.0__ square_perimeter__45.0__ square_perimeter__45.0__ |
multiply__0.4__45.0__ multiply__10.0__18.0__ multiply__10.0__18.0__ |
| a bus started its journey from mumbai and reached pune in num__65 min with its average speed of num__60 km / hr . if the average speed of the bus is increased by num__5 km / hr how much time will it take to cover the same distance ? <o> a ) num__2 hrs <o> b ) num__3 hrs <o> c ) num__1 hrs <o> d ) num__1 num__0.5 hrs <o> e ) num__2 num__0.5 hrs |
sol . distance between ramgarh and devgarh = ( num__60 * num__65 ) / num__60 = num__60 average speed of the bus is increased by num__5 km / hr then the speed of the bus = num__60 km / hr required time = num__1.0 = num__1 hr c <eor> c <eos> |
c |
reverse__1.0__ |
reverse__1.0__ |
| shawn and raquel want to have gardens that are the same size . shawn plants a garden num__25 feet by num__24 feet . raquel wants one side of his garden to be num__30 feet . how long should the other side be ? <o> a ) num__22 <o> b ) num__29 <o> c ) num__20 <o> d ) num__15 <o> e ) num__18 |
s num__1 s num__2 = r num__1 r num__2 so num__25 * num__24 = num__30 * r num__2 = > r num__2 = num__20 answer : c <eor> c <eos> |
c |
subtract__25.0__24.0__ multiply__1.0__20.0__ |
subtract__25.0__24.0__ multiply__1.0__20.0__ |
| the average of num__11 numbers is num__10.9 . if the average of first six is num__10.5 and that of the last six is num__11.4 the sixth number is ? <o> a ) num__11.4 <o> b ) num__11.3 <o> c ) num__11.8 <o> d ) num__11.5 <o> e ) num__11.1 |
num__1 to num__11 = num__11 * num__10.9 = num__119.9 num__1 to num__6 = num__6 * num__10.5 = num__63 num__6 to num__11 = num__6 * num__11.4 = num__68.4 num__63 + num__68.4 = num__131.4 – num__119.9 = num__11.5 num__6 th number = num__11.5 answer : d <eor> d <eos> |
d |
multiply__11.0__10.9__ multiply__10.5__6.0__ multiply__11.4__6.0__ add__68.4__63.0__ add__10.5__1.0__ add__10.5__1.0__ |
multiply__11.0__10.9__ multiply__10.5__6.0__ multiply__11.4__6.0__ add__68.4__63.0__ add__10.5__1.0__ add__10.5__1.0__ |
| a watch was sold at a loss of num__10.0 . if it was sold for rs . num__210 more there would have been a gain of num__4.0 . what is the cost price ? <o> a ) s . num__1000 <o> b ) s . num__1009 <o> c ) s . num__1007 <o> d ) s . num__1006 <o> e ) s . num__1500 |
explanation : num__90.0 num__104.0 - - - - - - - - num__14.0 - - - - num__210 num__100.0 - - - - ? = > rs . num__1500 answer : e <eor> e <eos> |
e |
percent__100.0__1500.0__ |
percent__100.0__1500.0__ |
| what is the num__4 digit number in which the num__1 st digit is num__0.333333333333 of the second the num__3 rd is the sum of the num__1 st and num__2 nd & the last is three times the second ? <o> a ) num__1234 <o> b ) num__1567 <o> c ) no = num__1349 <o> d ) num__1456 <o> e ) num__1356 |
first digit is num__0.333333333333 second digit = > the numbers can be num__1 & num__3 num__2 & num__6 num__3 & num__9 . first + second = third = > we can eliminate num__3 & num__9 since num__3 + num__9 = num__12 . last is num__3 times the second = > we can eliminate option num__2 & num__6 since num__3 * num__6 = num__18 . hence the number is num__1349 c <eor> c <eos> |
c |
add__4.0__2.0__ add__3.0__6.0__ multiply__4.0__3.0__ multiply__3.0__6.0__ multiply__1.0__1349.0__ |
add__4.0__2.0__ add__3.0__6.0__ multiply__4.0__3.0__ multiply__3.0__6.0__ multiply__1.0__1349.0__ |
| a woman is num__4 times as old as her daughter . in num__3 years she will be num__3 times as old as her daughter . how old is the woman now ? <o> a ) num__3 years <o> b ) num__6 years <o> c ) num__12 years <o> d ) num__24 years <o> e ) num__36 years |
let woman ' s present age = w and daughter ' s present age = d w = num__4 d - - - equation num__1 w + num__3 = num__3 ( d + num__3 ) = > w = num__3 d + num__6 - - - equation num__2 from num__1 and num__2 we get d = num__6 w = num__24 answer d <eor> d <eos> |
d |
subtract__4.0__3.0__ subtract__3.0__1.0__ multiply__4.0__6.0__ multiply__4.0__6.0__ |
subtract__4.0__3.0__ subtract__3.0__1.0__ multiply__4.0__6.0__ multiply__4.0__6.0__ |
| if p is a prime number greater than num__3 find the remainder when p ^ num__2 + num__15 is divided by num__12 . <o> a ) num__6 <o> b ) num__4 <o> c ) num__0 <o> d ) num__8 <o> e ) num__7 |
every prime number greater than num__3 can be written num__6 n + num__1 or num__6 n - num__1 . if p = num__6 n + num__1 then p ^ num__2 + num__15 = num__36 n ^ num__2 + num__12 n + num__1 + num__15 = num__36 n ^ num__2 + num__12 n + num__12 + num__4 if p = num__6 n - num__1 then p ^ num__2 + num__15 = num__36 n ^ num__2 - num__12 n + num__1 + num__15 = num__36 n ^ num__2 - num__12 n + num__12 + num__4 when divided by num__12 it must leave a remainder of num__4 . the answer is b . <eor> b <eos> |
b |
multiply__3.0__2.0__ subtract__3.0__2.0__ multiply__3.0__12.0__ add__3.0__1.0__ add__3.0__1.0__ |
multiply__3.0__2.0__ subtract__3.0__2.0__ multiply__3.0__12.0__ add__3.0__1.0__ add__3.0__1.0__ |
| a boat can travel with a speed of num__16 km / hr in still water . if the rate of stream is num__5 km / hr then find the time taken by the boat to cover distance of num__84 km downstream . <o> a ) num__4 hours <o> b ) num__5 hours <o> c ) num__6 hours <o> d ) num__7 hours <o> e ) num__8 hours |
explanation : it is very important to check if the boat speed given is in still water or with water or against water . because if we neglect it we will not reach on right answer . i just mentioned here because mostly mistakes in this chapter are of this kind only . lets see the question now . speed downstream = ( num__16 + num__5 ) = num__21 kmph time = distance / speed = num__4.0 = num__4 hours answer is a <eor> a <eos> |
a |
add__16.0__5.0__ divide__84.0__21.0__ round__4.0__ |
add__16.0__5.0__ divide__84.0__21.0__ divide__16.0__4.0__ |
| the u . s . defense department has decided that the pentagon is an obsolete building and that it must be replaced with an upgraded version : the hexagon . the secretary of defense wants a building that is exactly num__70 feet high and num__200 feet on a side and that has a hexagonal bull ' s - eye cutout in the center ( somewhat like the current one ) that is num__60 feet on a side . what will be the volume of the new building in cubic feet ? <o> a ) num__3 num__937500 cubic feet <o> b ) num__6 num__619898 cubic feet <o> c ) num__11550 √ num__3 cubic feet <o> d ) num__15750 √ num__3 cubic feet <o> e ) num__3 num__937500 √ num__3 cubic feet |
volume of the hexagon with side num__200 and height num__70 = area * height = num__6 * ( sqrt ( num__3 ) / num__4 ) ( num__200 ^ num__2 ) ( num__70 ) volume of the center bull ' s eye that is similar in shape of a hexagon but side num__60 = num__6 * ( sqrt ( num__3 ) / num__4 ) ( num__60 ^ num__2 ) * num__70 volume of the building = num__6 * ( sqrt ( num__3 ) / num__4 ) ( num__200 ^ num__2 ) ( num__70 ) - num__6 * ( sqrt ( num__3 ) / num__4 ) ( num__60 ^ num__2 ) * num__70 = num__6 num__619898 answer is b <eor> b <eos> |
b |
multiply__2.0__3.0__ |
multiply__2.0__3.0__ |
| abhishek starts to paint a fence on one day . on the second day two more friends of abhishek join him . on the third day num__3 more friends of him join him and so on . if the fence is completely painted this way in exactly num__20 days then find the number of days in which num__10 girls painting together can paint the fence completely given that every girl can paint twice as fast as abhishek and his friends ( boys ) ? ( assume that the friends of abhishek are all boys ) . <o> a ) num__20 <o> b ) num__40 <o> c ) num__45 <o> d ) num__77 <o> e ) num__87 |
number of men working on first day = num__1 number of men working on second day = num__3 number of men working on second day = num__6 and so on . . total number of boys till the end of the work = [ n ( n + num__1 ) ( n + num__2 ) ] / num__6 = [ num__20 x num__21 x num__22 ] / num__6 = num__1540 given that every girl paints twice as fast as abhishek ’ s friends . hence num__20 girls work is being done . thus the number of days taken to paint the fence = num__77.0 = num__77 . answer : d <eor> d <eos> |
d |
subtract__3.0__1.0__ add__20.0__1.0__ add__20.0__2.0__ divide__1540.0__20.0__ multiply__1.0__77.0__ |
divide__20.0__10.0__ add__20.0__1.0__ add__20.0__2.0__ divide__1540.0__20.0__ divide__1540.0__20.0__ |
| henry eats x scones in x percent of the time it takes rachel to eat y scones . if rachel eats four scones in ten minutes then the number of minutes it takes henry to eat num__10 scones must be equal to which of the following ? <o> a ) y / num__4 <o> b ) num__200 / y <o> c ) num__100 y / ( num__6 x ) <o> d ) xy / num__200 <o> e ) y / ( num__5 x ) |
rachel eats num__4 scones / num__10 minutes = num__0.4 scones per minute the time for rachel to eat num__1 scone is num__2.5 minutes . the time for rachel to eat y scones is num__5 y / num__2 minutes . the time for henry to eat x scones is num__5 yx / num__200 minutes . the time for henry to eat num__1 scone is num__5 y / num__200 minutes . the time for henry to eat num__10 scones is num__50 y / num__200 = y / num__4 minutes . the answer is a . <eor> a <eos> |
a |
divide__4.0__10.0__ reverse__0.4__ add__1.0__4.0__ round_down__2.5__ multiply__10.0__5.0__ divide__10.0__2.5__ |
divide__4.0__10.0__ reverse__0.4__ add__1.0__4.0__ divide__10.0__5.0__ divide__200.0__4.0__ divide__10.0__2.5__ |
| a man buys an article and sells it at a profit of num__20.0 . if he had bought it at num__20.0 less and sold it for rs . num__75 less he could have gained num__25.0 . what is the cost price ? <o> a ) num__337 <o> b ) num__375 <o> c ) num__297 <o> d ) num__266 <o> e ) num__291 |
cp num__1 = num__100 sp num__1 = num__120 cp num__2 = num__80 sp num__2 = num__80 * ( num__1.25 ) = num__100 num__20 - - - - - num__100 num__75 - - - - - ? = > num__375 answer : b <eor> b <eos> |
b |
percent__100.0__375.0__ |
percent__100.0__375.0__ |
| a train num__110 m long is running with a speed of num__80 km / h . in how many seconds will the train pass a man who is running at num__8 km / h in the direction opposite to that in which the train is going ? <o> a ) num__3.5 <o> b ) num__4.5 <o> c ) num__5.5 <o> d ) num__6.5 <o> e ) num__7.5 |
the speed of the train relative to the man = num__80 + num__8 = num__88 km / h . num__88000 m / h * num__1 h / num__3600 s = ( num__24.4444444444 ) m / s ( num__110 m ) / ( num__24.4444444444 m / s ) = ( num__110 * num__36 ) / num__880 = num__4.5 = num__4.5 seconds the answer is b . <eor> b <eos> |
b |
add__80.0__8.0__ divide__88000.0__3600.0__ multiply__110.0__8.0__ divide__110.0__24.4444__ round__4.5__ |
add__80.0__8.0__ divide__88000.0__3600.0__ multiply__110.0__8.0__ divide__110.0__24.4444__ divide__110.0__24.4444__ |
| the cost of carpeting a room num__18 m long with a carpet num__75 cm wide at num__45 paise per meter is rs . num__81 . the breadth of the room is : <o> a ) num__7 m <o> b ) num__7.5 m <o> c ) num__5.5 m <o> d ) num__6.5 m <o> e ) num__8.5 m |
length of the carpet = total cost / rate / m = num__180.0 = num__180 m area of the carpet = num__180 * num__0.75 = num__135 m num__2 breadth of the room = ( area / length ) = num__7.5 = num__7.5 m answer : b <eor> b <eos> |
b |
multiply__0.75__180.0__ divide__135.0__18.0__ round__7.5__ |
multiply__0.75__180.0__ divide__135.0__18.0__ divide__135.0__18.0__ |
| a bus has num__40 seats and the passengers agree to share the total bus fare among themselves equally . if the total fair is num__80.67 find the total no of the seats unoccupied <o> a ) num__34 <o> b ) num__35 <o> c ) num__36 <o> d ) num__37 <o> e ) num__38 |
passengers will occupy num__3 seats per head amount payable will be num__26.89 . rupees num__80.67 can not be divisible other than num__3 so the number of occupied seats will be num__3 unoccupied seats will be num__40 - num__3 = num__37 answer : d <eor> d <eos> |
d |
divide__80.67__3.0__ subtract__40.0__3.0__ subtract__40.0__3.0__ |
divide__80.67__3.0__ subtract__40.0__3.0__ subtract__40.0__3.0__ |
| eighty percent of the lights at hotel california are on at num__8 p . m . a certain evening . however forty percent of the lights that are supposed to be off are actually on and ten percent of the lights that are supposed to be on are actually off . what percent z of the lights that are on are supposed to be off ? <o> a ) num__22 ( num__0.222222222222 ) % <o> b ) num__16 ( num__0.666666666667 ) % <o> c ) num__11 ( num__0.111111111111 ) % <o> d ) num__10.0 <o> e ) num__5 % |
is the answer d . let me try . . let the light which are supposed to be off = so let the light which are supposed to be on = sn let the light which are actually off = ao let the light which are actually on = an let the total no . of lights be num__100 so actually on lights = num__80 and actually off lights = num__20 also given > > forty percent of the lights that are supposed to off are actually on > > > ( num__0.4 ) * so are actually on it means > > > ( num__0.6 ) * so are actually off also given > > ten percent of the lights that are supposed to be on are actually off > > > ( num__0.1 ) * sn are actually off it means > > > ( num__0.9 ) * sn are actually on so total actually on lights = ( num__0.4 ) * so + ( num__0.9 ) * sn = num__80 and total actually off lights = ( num__0.6 ) * so + ( num__0.1 ) * sn = num__80 from here we get so = num__20 we need to find : what percent of the lights that are on are supposed to be off > > > so light actually on are num__80 and light which are actually on which are supposed to be off = ( num__0.4 ) * so = num__8 . so ( num__0.1 ) * num__100 z = num__10.0 . d <eor> d <eos> |
d |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| ferrari s . p . a is an italian sports car manufacturer based in maranello italy . founded by enzo ferrari in num__1928 as scuderia ferrari the company sponsored drivers and manufactured race cars before moving into production of street - legal vehicles in num__1947 as ferrari s . p . a . throughout its history the company has been noted for its continued participation in racing especially in formula one where it has employed great success . rohit once bought a ferrari . it could go num__4 times as fast as mohan ’ s old mercedes . if the speed of mohan ’ s mercedes is num__35 km / hr and the distance traveled by the ferrari is num__490 km find the total time taken for rohit to drive that distance . <o> a ) num__3.5 <o> b ) num__3.9 <o> c ) num__3.2 <o> d ) num__3.1 <o> e ) num__3.7 |
as ferrari ' s speed is four times that of the mercedes its speed is num__35 x num__4 = num__140 so time taken by the ferrari = num__3.5 = num__3.5 hours answer : a <eor> a <eos> |
a |
multiply__4.0__35.0__ divide__490.0__140.0__ divide__490.0__140.0__ |
multiply__4.0__35.0__ divide__490.0__140.0__ divide__490.0__140.0__ |
| the table below shows how many coaches work with each of the major sports teams at kristensen school . although no single coach works with all three teams num__3 coaches work with both the track and tennis teams num__2 coaches work with both the track and baseball teams and num__1 coach works with both the tennis and baseball teams . how many different coaches work with these three teams ? sports no of coaches track num__9 tennis num__5 baseball num__4 <o> a ) num__6 <o> b ) num__9 <o> c ) num__11 <o> d ) num__12 <o> e ) num__17 |
x = num__9 + num__5 + num__4 - ( num__3 + num__2 + num__1 ) = num__12 answer is d <eor> d <eos> |
d |
choose__3.0__1.0__ choose__3.0__1.0__ |
choose__3.0__1.0__ choose__3.0__1.0__ |
| how many quarters are equal to num__3 dollars ? <o> a ) num__1 <o> b ) num__8 <o> c ) num__12 <o> d ) num__9 <o> e ) num__7 |
num__3 * num__4 = num__12 quarters answer : c <eor> c <eos> |
c |
multiply__3.0__4.0__ multiply__3.0__4.0__ |
multiply__3.0__4.0__ multiply__3.0__4.0__ |
| in how many years does a sum of rs . num__5000 yield a simple interest of rs . num__16500 at num__15.0 p . a . ? <o> a ) num__22 years <o> b ) num__77 years <o> c ) num__66 years <o> d ) num__55 years <o> e ) num__44 years |
t = ( num__100 * num__16500 ) / ( num__15 * num__5000 ) = num__22 years . answer : a <eor> a <eos> |
a |
percent__100.0__22.0__ |
percent__100.0__22.0__ |
| the sequence x num__1 x num__2 x num__3 . . . is such that xn = num__1 / n - ( num__1 / ( n + num__1 ) ) . what is the sum of the first num__10 terms of the sequence ? <o> a ) num__2.01 <o> b ) num__0.99 <o> c ) num__0.990099009901 <o> d ) num__0.0001 <o> e ) num__0.909090909091 |
easy task and accomplish x num__1 = num__1 - num__0.5 x num__2 = num__0.5 - num__0.333333333333 x num__3 = num__0.333333333333 - num__0.25 . . . . . x num__10 = num__0.1 - num__0.0909090909091 sum = x num__1 + x num__2 + x num__3 + . . . . x num__10 = num__1 - num__0.5 + num__0.5 - num__0.333333333333 + . . . . . . . num__0.010101010101 - num__0.01 + num__0.1 - num__0.0909090909091 = num__1 - num__0.0909090909091 = num__0.909090909091 e is the answer <eor> e <eos> |
e |
reverse__2.0__ reverse__3.0__ divide__0.5__2.0__ reverse__10.0__ divide__0.1__10.0__ subtract__1.0__0.0909__ multiply__1.0__0.9091__ |
reverse__2.0__ reverse__3.0__ divide__0.5__2.0__ reverse__10.0__ divide__0.1__10.0__ subtract__1.0__0.0909__ subtract__1.0__0.0909__ |
| a person can swim in still water at num__4 km / h . if the speed of water num__2 km / h how many hours will the man take to swim back against the current for num__6 km ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__1 <o> d ) num__5 <o> e ) num__2 |
m = num__4 s = num__2 us = num__4 - num__2 = num__2 d = num__6 t = num__3.0 = num__3 answer : a <eor> a <eos> |
a |
divide__6.0__2.0__ round__3.0__ |
divide__6.0__2.0__ subtract__6.0__3.0__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__180 m and they cross each other in num__12 sec then the speed of each train is ? <o> a ) num__78 <o> b ) num__54 <o> c ) num__36 <o> d ) num__34 <o> e ) num__23 |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__180 + num__180 ) / num__12 = > x = num__15 speed of each train = num__15 m / sec . = num__15 * num__3.6 = num__54 km / hr . answer : option b <eor> b <eos> |
b |
divide__180.0__12.0__ multiply__15.0__3.6__ round__54.0__ |
divide__180.0__12.0__ multiply__15.0__3.6__ multiply__15.0__3.6__ |
| a began a business with rs . num__85000 . he was joined afterwards by b with ks . num__42500 . for how much period does b join if the profits at the end of the year are divided in the ratio of num__3 : num__1 ? <o> a ) num__6 months <o> b ) num__3 months <o> c ) num__2 months <o> d ) num__8 months <o> e ) num__9 months |
suppose b joined for x months . then ( num__85000 * num__12 ) / ( num__42500 * x ) = num__3 . or x = ( num__85000 * num__12 ) / ( num__42500 * num__3 ) = num__8 . so b joined for num__8 months . answer : d <eor> d <eos> |
d |
multiply__1.0__8.0__ |
multiply__1.0__8.0__ |
| if ( num__1 - x ) y = y which of the following must be true ? <o> a ) x = - num__0 or y = num__0 <o> b ) x = - num__1 or y = num__0 <o> c ) x = num__1 or y = num__1 <o> d ) x = - num__1 or y = - num__1 <o> e ) x = num__0 or y = num__1 |
we have ( num__1 - x ) y = y . if x = num__0 y = y . true if y = num__0 ( num__1 - x ) * num__0 = num__0 true so a will be the answer . <eor> a <eos> |
a |
multiply__1.0__0.0__ |
multiply__1.0__0.0__ |
| the average ( arithmetic mean ) of eight numbers is num__41.5 . if the sum of half of these numbers is num__154.4 what is the average of the other half ? <o> a ) num__39.9 <o> b ) num__41.1 <o> c ) num__44.4 <o> d ) num__49.9 <o> e ) num__56.6 |
the average of this half is num__154.4 / num__4 = num__38.6 this is num__2.9 below the overall average thus the average of the other half of the numbers must be num__2.9 above the overall age that is num__41.5 + num__2.9 = num__44.4 the answer is c . <eor> c <eos> |
c |
divide__154.4__4.0__ subtract__41.5__38.6__ add__41.5__2.9__ add__41.5__2.9__ |
divide__154.4__4.0__ subtract__41.5__38.6__ add__41.5__2.9__ add__41.5__2.9__ |
| find a sum for num__1 st num__9 prime number ' s ? <o> a ) num__25 <o> b ) num__28 <o> c ) num__30 <o> d ) num__34 <o> e ) num__100 |
required sum = ( num__2 + num__3 + num__5 + num__7 + num__11 + num__13 + num__17 + num__19 + num__23 ) = num__100 note : num__1 is not a prime number option e <eor> e <eos> |
e |
add__1.0__2.0__ add__2.0__3.0__ subtract__9.0__2.0__ add__9.0__2.0__ add__2.0__11.0__ add__2.0__17.0__ multiply__1.0__100.0__ |
add__1.0__2.0__ add__2.0__3.0__ add__2.0__5.0__ add__9.0__2.0__ add__2.0__11.0__ add__2.0__17.0__ multiply__1.0__100.0__ |
| john and amanda stand at opposite ends of a straight road and start running towards each other at the same moment . their rates are randomly selected in advance so that john runs at a constant rate of num__3 num__4 or num__5 miles per hour and amanda runs at a constant rate of num__3 num__4 num__5 num__6 or num__7 miles per hour . what is the probability that john has traveled farther than amanda by the time they meet ? <o> a ) num__0.5 <o> b ) num__0.333333333333 <o> c ) num__0.2 <o> d ) num__0.133333333333 <o> e ) num__0.266666666667 |
john will run farther if he runs at num__5 mph and amanda runs at num__4 mph or num__3 mph . in this case p ( john runs farther ) = num__0.333333333333 * num__0.4 = num__0.133333333333 john will run farther if he runs at num__4 mph and amanda runs at num__3 mph . in this case p ( john runs farther ) = num__0.333333333333 * num__0.2 = num__0.0666666666667 p ( john runs farther ) = num__0.133333333333 + num__0.0666666666667 = num__0.2 = num__0.2 the answer is c . <eor> c <eos> |
c |
divide__0.4__3.0__ subtract__0.3333__0.1333__ divide__0.4__6.0__ round__0.2__ |
multiply__0.4__0.3333__ subtract__0.3333__0.1333__ multiply__0.2__0.3333__ add__0.1333__0.0667__ |
| the product z of two prime numbers is between num__16 and num__40 . if one of the prime numbers is greater than num__2 but less than num__6 and the other is greater than num__12 but less than num__30 then what is z ? <o> a ) num__21 <o> b ) num__22 <o> c ) num__28 <o> d ) num__35 <o> e ) num__39 |
the smallest possible product is num__39 which is num__3 * num__13 . all other products are too big . the answer is e . <eor> e <eos> |
e |
divide__6.0__2.0__ subtract__16.0__3.0__ multiply__3.0__13.0__ |
divide__6.0__2.0__ subtract__16.0__3.0__ multiply__3.0__13.0__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__120 m and they cross each other in num__12 sec then the speed of each train is ? <o> a ) num__17 km / hr <o> b ) num__89 km / hr <o> c ) num__36 km / hr <o> d ) num__89 km / hr <o> e ) num__82 km / hr |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__120 + num__120 ) / num__12 = > x = num__10 speed of each train = num__10 m / sec . = num__10 * num__3.6 = - num__36 km / hr . answer : c <eor> c <eos> |
c |
divide__120.0__12.0__ multiply__10.0__3.6__ round__36.0__ |
divide__120.0__12.0__ multiply__10.0__3.6__ multiply__10.0__3.6__ |
| in each series look for the degree and direction of change between the numbers . in other words do the numbers increase or decrease and by how much look at this series : num__1.5 num__2.3 num__3.1 num__3.9 . . . what number should come next ? <o> a ) num__4.7 <o> b ) num__3.6 <o> c ) num__4.9 <o> d ) num__5.0 <o> e ) num__4.4 |
a num__4.7 in this simple addition series each number increases by num__0.8 . <eor> a <eos> |
a |
subtract__2.3__1.5__ add__3.9__0.8__ |
subtract__2.3__1.5__ add__3.9__0.8__ |
| ben is driving on the highway at z miles per hour . ( one mile equals num__5280 feet . ) ben ' s tires have a circumference of y feet . which of the following expressions gives the number of revolutions each wheel turns in one hour ? <o> a ) num__5280 ( z / y ) <o> b ) num__5280 ( y / x ) <o> c ) num__5280 ( xy ) <o> d ) num__5280 / ( xy ) <o> e ) ( xy ) / num__5 |
280 |
in one hour at the rate of z miles per hour ben covers z miles so num__5280 z feet . the number of revolutions = distance / circumference = num__5280 z / y . answer : a . <eor> a <eos> |
a |
a |
| the grade point average of one third of the classroom is num__15 ; the grade point average of the rest is num__18 . what is the grade point average of the whole class ? <o> a ) num__17 <o> b ) num__19 <o> c ) num__21 <o> d ) num__23 <o> e ) num__25 |
let n = total students in class total points for num__0.333333333333 class = num__15 n / num__3 = num__5 n total points for num__0.666666666667 class = num__18 * num__2 n / num__3 = num__12 n total points for whole class = num__5 n + num__12 n = num__17 n num__17 n total class points / n total students = num__17 grade point average for total class answer : a <eor> a <eos> |
a |
subtract__18.0__15.0__ divide__15.0__3.0__ subtract__5.0__3.0__ subtract__15.0__3.0__ add__15.0__2.0__ add__15.0__2.0__ |
subtract__18.0__15.0__ divide__15.0__3.0__ subtract__5.0__3.0__ subtract__15.0__3.0__ add__15.0__2.0__ add__15.0__2.0__ |
| the percentage profit earned by selling an article for rs . num__1920 is equal to the percentage loss incurred by selling the same article for rs . num__1280 . at what price should the article be sold to make num__15.0 profit ? <o> a ) num__2000 <o> b ) num__2778 <o> c ) num__1840 <o> d ) num__2778 <o> e ) num__2771 |
let c . p . be rs . x . then ( num__1920 - x ) / x * num__100 = ( x - num__1280 ) / x * num__100 num__1920 - x = x - num__1280 num__2 x = num__3200 = > x = num__1600 required s . p . = num__115.0 of rs . num__1600 = num__1.15 * num__1600 = rs . num__1840 . answer : c <eor> c <eos> |
c |
percent__100.0__1840.0__ |
percent__100.0__1840.0__ |
| company c produces toy trucks at a cost of $ num__5.00 each for the first num__100 trucks and $ num__3.50 for each additional truck . if num__500 toy trucks were produced by company c and sold for $ num__12.00 each what was company c ’ s gross profit ? <o> a ) $ num__2250 <o> b ) $ num__2500 <o> c ) $ num__4100 <o> d ) $ num__3250 <o> e ) $ num__4 |
500 |
cost of num__500 trucks : ( num__100 * num__5 ) + ( num__400 * num__3.5 ) = num__500 + num__1400 = $ num__1900 revenue : num__500 * num__12 = $ num__6000 profit : num__6000 - num__1900 = $ num__4100 option c is correct <eor> c <eos> |
c |
c |
| the total of company c ' s assets in num__1994 was num__300.0 greater than the total in num__1993 which in turn was num__500.0 greater than the total in num__1992 . if the total of company c ' s assets in in num__1992 was n dollars which one of the following represents company c ' s assets in num__1994 : <o> a ) num__7 n <o> b ) num__8 n <o> c ) num__9 n <o> d ) num__24 <o> e ) num__20 |
let ' s use actual numbers . starting in num__1992 let ' s say company c had $ num__100 in assets . in num__1993 the total assets were num__500.0 greater which means num__1992 plus num__500.0 of num__1992 : $ num__100 + num__5 x $ num__100 = $ num__100 + $ num__500 = $ num__600 in num__1994 the total assets were num__300.0 greater than they were in num__1993 which means num__1993 plus num__300.0 of num__1993 : $ num__600 + num__3 x $ num__600 = $ num__600 + $ num__1800 = $ num__2400 this is num__24 times the num__1992 number so the correct answer is num__24 n . d <eor> d <eos> |
d |
divide__500.0__100.0__ add__500.0__100.0__ divide__300.0__100.0__ multiply__3.0__600.0__ add__1800.0__600.0__ divide__2400.0__100.0__ divide__2400.0__100.0__ |
divide__500.0__100.0__ add__500.0__100.0__ divide__300.0__100.0__ multiply__3.0__600.0__ add__1800.0__600.0__ divide__2400.0__100.0__ divide__2400.0__100.0__ |
| a sum of money amounts to rs . num__9800 after num__5 years and rs . num__12005 after num__8 years at the same rate of simple interest . the rate of interest per annum is : <o> a ) num__5.0 <o> b ) num__8.0 <o> c ) num__12.0 <o> d ) num__15.0 <o> e ) num__18 % |
s . i . for num__3 years = rs . ( num__12005 - num__9800 ) = rs . num__2205 . s . i . for num__5 years = rs . ( num__2205 x num__5 ) / num__3 = rs . num__3675 principal = rs . ( num__9800 - num__3675 ) = rs . num__6125 . hence rate = ( num__100 x num__3675 ) / ( num__6125 x num__5 ) % = num__12.0 answer : c <eor> c <eos> |
c |
percent__100.0__12.0__ |
percent__100.0__12.0__ |
| a num__25 cm wide path is to be made around a circular garden having a diameter of num__4 meters . approximate area of the path is square meters is <o> a ) num__3.34 <o> b ) num__3.36 <o> c ) num__3.76 <o> d ) num__3.88 <o> e ) num__3.22 |
explanation : area of the path = area of the outer circle - area of the inner circle = ∏ { num__2.0 + num__0.25 } num__2 - ∏ [ num__2.0 ] num__2 = ∏ [ num__2.252 - num__22 ] = ∏ ( num__0.25 ) ( num__4.25 ) { ( a num__2 - b num__2 = ( a - b ) ( a + b ) } = ( num__3.14 ) ( num__0.25 ) ( num__4.25 ) = num__53.38 / num__16 = num__3.34 sq m answer : option a <eor> a <eos> |
a |
volume_rectangular_prism__4.0__4.25__3.14__ square_perimeter__4.0__ triangle_area__2.0__3.34__ |
volume_rectangular_prism__4.0__4.25__3.14__ square_perimeter__4.0__ triangle_area__2.0__3.34__ |
| on a certain day orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ num__0.60 per glass on the first day what was the price w per glass on the second day ? <o> a ) $ num__015 <o> b ) $ num__0.20 <o> c ) $ num__0.30 <o> d ) $ num__0.40 <o> e ) $ num__0.45 |
on the first day num__1 unit of orange juice and num__1 unit of water was used to make num__2 units of orangeade ; on the second day num__1 unit of orange juice and num__2 units of water was used to make num__3 units of orangeade ; so the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is num__2 to num__3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is num__2 to num__3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from num__2 glasses on the first day equals to the revenue from num__3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then num__2 * num__0.6 = num__3 * x - - > x = $ num__0.4 . answer : d . <eor> d <eos> |
d |
add__1.0__2.0__ subtract__1.0__0.6__ subtract__1.0__0.6__ |
add__1.0__2.0__ subtract__1.0__0.6__ subtract__1.0__0.6__ |
| if the length and breadth of a rectangular room are each increased by num__1 m then the area of floor is increased by num__21 sq . m . if the length is creased by num__1 m and breadth is decreased by num__1 m then the area is decreased by num__5 sq . m . the perimeter of the floor is : <o> a ) num__27 <o> b ) num__35 <o> c ) num__40 <o> d ) num__49 <o> e ) num__38 |
let length = x meters and breadth = y meters . then ( x + num__1 ) ( y + num__1 ) - xy = num__21 x + y = num__20 Ã ¢ â ‚ ¬ Â ¦ Ã ¢ â ‚ ¬ Â ¦ Ã ¢ â ‚ ¬ Â ¦ Ã ¢ â ‚ ¬ Â ¦ Ã ¢ â ‚ ¬ Â ¦ Ã ¢ â ‚ ¬ Â ¦ num__1 and xy - [ ( x + num__1 ) ( y - num__1 ) ] = num__5 x - y = num__6 Ã ¢ â ‚ ¬ Â ¦ Ã ¢ â ‚ ¬ Â ¦ Ã ¢ â ‚ ¬ Â ¦ Ã ¢ â ‚ ¬ Â ¦ Ã ¢ â ‚ ¬ Â ¦ Ã ¢ â ‚ ¬ Â ¦ . . num__2 solving ( i ) and ( ii ) we get : x = num__13 and y = num__7 so length = num__13 m and breadth = num__7 m . perimeter = [ num__2 ( num__13 + num__7 ) ] m = num__40 m . answer : c <eor> c <eos> |
c |
square_perimeter__5.0__ surface_cube__1.0__ triangle_perimeter__5.0__2.0__6.0__ multiply__2.0__20.0__ multiply__1.0__40.0__ |
square_perimeter__5.0__ surface_cube__1.0__ triangle_perimeter__5.0__2.0__6.0__ multiply__2.0__20.0__ multiply__1.0__40.0__ |
| the sale price of an article including the sales tax is rs . num__616 . the rate of sales tax is num__10.0 . if the shopkeeper has made a profit of num__12.0 then the cost price of the article is : <o> a ) num__500 <o> b ) num__334 <o> c ) num__555 <o> d ) num__664 <o> e ) num__5598 |
num__110.0 of s . p . = num__616 s . p . = ( num__616 * num__100 ) / num__110 = rs . num__560 c . p = ( num__110 * num__560 ) / num__112 = rs . num__500 answer : option a <eor> a <eos> |
a |
percent__100.0__500.0__ |
percent__100.0__500.0__ |
| james took a num__3 - hour bike ride . in the second hour he travlled num__24 miles which was num__20 percent farther than he traveled the first hour . if he traveled num__25 percent farther in the third hour than he did in the second hour how many miles did jose travel during the entire ride ? <o> a ) num__74.0 <o> b ) num__54.9 <o> c ) num__55.5 <o> d ) num__57.0 <o> e ) num__63.0 |
let the distance travelled in the first hour be x . thus num__1.2 x = num__24 x = num__20 . now the distance travelled in the num__3 rd hour = num__24 + num__0.25 â ˆ — num__24 = num__30 . = num__24 + num__20 + num__30 = num__74 answer : a <eor> a <eos> |
a |
divide__24.0__20.0__ multiply__25.0__1.2__ round__74.0__ |
divide__24.0__20.0__ multiply__25.0__1.2__ round__74.0__ |
| a can do a piece of work in num__10 days and b can do it in num__15 days and c can do it num__20 days . they started the work together and a leaves after num__2 days and b leaves after num__4 days from the beginning . how long will work lost ? <o> a ) num__10 num__0.4 <o> b ) num__10 num__2 / num__0 <o> c ) num__10 num__0.666666666667 <o> d ) num__10 num__2.0 <o> e ) num__10 num__0.666666666667 |
num__0.2 + num__0.266666666667 + x / num__20 = num__1 x = num__10.6666666667 = num__10 num__0.666666666667 answer : c <eor> c <eos> |
c |
divide__2.0__10.0__ divide__4.0__15.0__ divide__10.0__15.0__ round__10.0__ |
divide__2.0__10.0__ divide__4.0__15.0__ divide__10.0__15.0__ divide__10.0__1.0__ |
| if q is the smallest positive integer that is not prime and not a factor of num__50 ! what is the sum of the factors of q ? <o> a ) num__51 <o> b ) num__54 <o> c ) num__72 <o> d ) num__50 ! + num__2 <o> e ) num__162 |
in fact num__51 = num__3 * num__17 is a factor of num__50 ! . the smallest positive integer that is not prime and not a factor of num__50 ! is num__106 = num__2 * num__53 . the sum of the factors of num__106 is num__162 . answer : e . <eor> e <eos> |
e |
divide__51.0__3.0__ gcd__50.0__106.0__ add__50.0__3.0__ lcm__2.0__162.0__ |
divide__51.0__3.0__ gcd__50.0__106.0__ add__50.0__3.0__ lcm__2.0__162.0__ |
| a rectangular - shaped carpet that measures x feet by y feet is priced at $ num__35 . what is the cost of the carpet in dollars per square yard ? ( num__1 square yard = num__9 square feet ) <o> a ) num__180 xy <o> b ) num__180 / ( xy ) <o> c ) num__20 xy <o> d ) num__315 / ( xy ) <o> e ) xy / num__180 |
the area of the carpet in feet is xy . the area in square yards is xy / num__9 . the price per square yard is num__35 / ( xy / num__9 ) = num__315 / ( xy ) . the answer is d . <eor> d <eos> |
d |
multiply__35.0__9.0__ multiply__35.0__9.0__ |
multiply__35.0__9.0__ divide__315.0__1.0__ |
| if x and b are integers and ( num__15 ^ x + num__15 ^ ( x + num__1 ) ) / num__4 ^ b = num__15 ^ b what is the value of x ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) can not be determined |
\ frac { num__15 ^ x ( num__1 + num__15 ) } { num__4 ^ b } = num__15 ^ b ( num__15 ^ x . num__4 ^ num__2 ) / num__4 ^ b = num__15 ^ b . num__4 ^ num__0 num__15 ^ x . num__4 ^ ( num__2 - b ) = num__15 ^ b . num__4 ^ num__0 equating the powers x = b ; num__2 - b = num__0 ; so x = b = num__2 = a <eor> a <eos> |
a |
multiply__1.0__2.0__ |
subtract__4.0__2.0__ |
| a worker is paid a regular rate of rs . num__30 for completing a survey . the worker completes num__100 surveys per week . for any survey involving the use of her cellphone she is paid at a rate of that is num__20.0 higher than her regular rate . if she completed num__50 surveys involving the use of her cellphone how much did she get that week ? <o> a ) num__2200 <o> b ) num__1100 <o> c ) num__2500 <o> d ) num__2800 <o> e ) num__3300 |
amount earned using her cell phone = num__50 * num__36 = num__1800 earned for remaining surveys = num__50 * num__30 = num__1500 total earning = num__3300 answer : e <eor> e <eos> |
e |
percent__100.0__3300.0__ |
percent__100.0__3300.0__ |
| there are num__24 students in a class one of them who is num__18 years old left the class and a new comer filled his place . if the avg age of the class was there by lowered by one month . the age of new comer is <o> a ) num__14 years <o> b ) num__15 years <o> c ) num__16 years <o> d ) num__17 years <o> e ) num__18 years |
let new comer age was x . average age decreased by = num__1 month = num__0.0833333333333 years . you can use this formula for these type of problems . no . of students in class * aged decreased = difference of replacement num__24 * num__0.0833333333333 = num__18 - x num__2 = num__18 - x x = num__16 years . age of new comer age was num__16 years . answer : option c <eor> c <eos> |
c |
subtract__18.0__2.0__ subtract__18.0__2.0__ |
subtract__18.0__2.0__ subtract__18.0__2.0__ |
| a and b can do a piece of work in num__6 num__0.666666666667 days and num__5 days respectively . they work together for num__2 days and then a leaves . in how many days after that b will complete the work alone . <o> a ) num__1 num__0.125 days <o> b ) num__1 num__0.5 days <o> c ) num__3 num__0.5 days <o> d ) num__1 num__0.2 days <o> e ) num__1 num__2.5 days |
num__0.15 * num__2 + ( num__2 + x ) / num__5 = num__1 x = num__1 num__0.5 days answer : b <eor> b <eos> |
b |
subtract__6.0__5.0__ divide__1.0__2.0__ round__1.0__ |
subtract__6.0__5.0__ divide__1.0__2.0__ multiply__2.0__0.5__ |
| if num__40.0 of a number is equal to two - third of another number what is the ratio of first number to the second number ? <o> a ) num__5 : num__3 <o> b ) num__2 : num__6 <o> c ) num__5 : num__6 <o> d ) num__6 : num__4 <o> e ) num__2 : num__5 |
let num__40.0 of a = ( num__0.666666666667 ) b then num__40 a / num__100 = num__2 b / num__3 num__2 a / num__5 = num__2 b / num__3 a / b = ( num__0.666666666667 x num__2.5 ) = num__1.66666666667 a : b = num__5 : num__3 . answer a <eor> a <eos> |
a |
add__2.0__3.0__ divide__100.0__40.0__ multiply__2.5__0.6667__ multiply__2.5__2.0__ |
add__2.0__3.0__ divide__100.0__40.0__ divide__5.0__3.0__ multiply__2.5__2.0__ |
| find the principle on a certain sum of money at num__5.0 per annum for num__3 num__0.2 years if the amount being rs . num__2030 ? <o> a ) rs . num__2000 <o> b ) rs . num__1750 <o> c ) rs . num__2010 <o> d ) rs . num__2005 <o> e ) none of these |
explanation : num__2030 = p [ num__1 + ( num__5 * num__3.2 ) / num__100 ] p = num__1750 answer : option b <eor> b <eos> |
b |
multiply__5.0__0.2__ add__3.0__0.2__ multiply__1.0__1750.0__ |
multiply__5.0__0.2__ add__3.0__0.2__ multiply__1.0__1750.0__ |
| num__5555 × num__9999 = ? <o> a ) num__55500005 <o> b ) num__55511115 <o> c ) num__55522225 <o> d ) num__55533335 <o> e ) num__55544445 |
e num__55544445 num__5555 × num__9999 = num__5555 ( num__10000 - num__1 ) = num__5555 × num__10000 - num__5555 × num__1 = num__55550000 - num__5555 = num__55544445 <eor> e <eos> |
e |
multiply__5555.0__9999.0__ subtract__10000.0__9999.0__ multiply__5555.0__10000.0__ multiply__5555.0__9999.0__ |
multiply__5555.0__9999.0__ subtract__10000.0__9999.0__ multiply__5555.0__10000.0__ subtract__55550000.0__5555.0__ |
| a train num__350 m long running with a speed of num__63 km / hr will pass a tree in ? <o> a ) num__15 sec <o> b ) num__16 sec <o> c ) num__18 sec <o> d ) num__20 sec <o> e ) num__25 sec |
speed = num__63 * num__0.277777777778 = num__17.5 m / sec time taken = num__350 * num__0.0571428571429 = num__20 sec answer : d <eor> d <eos> |
d |
divide__350.0__17.5__ round__20.0__ |
divide__350.0__17.5__ divide__350.0__17.5__ |
| the average age of students of a class is num__15.8 years . the average age of boys in the class is num__16.4 years and that of the girls is num__15.4 years the ratio of the number of boys to the number of girls in the class is <o> a ) num__2 : num__9 <o> b ) num__2 : num__3 <o> c ) num__2 : num__2 <o> d ) num__2 : num__1 <o> e ) num__2 : num__7 |
let the ratio be k : num__1 . then k * num__16.4 + num__1 * num__15.4 = ( k + num__1 ) * num__15.8 < = > ( num__16.4 - num__15.8 ) k = ( num__15.8 - num__15.4 ) < = > k = num__0.4 / num__0.6 = num__0.666666666667 . required ratio = num__0.666666666667 : num__1 = num__2 : num__3 . answer : b <eor> b <eos> |
b |
subtract__16.4__15.4__ subtract__15.8__15.4__ subtract__16.4__15.8__ divide__0.4__0.6__ add__1.0__2.0__ multiply__1.0__2.0__ |
subtract__16.4__15.4__ subtract__15.8__15.4__ subtract__16.4__15.8__ divide__0.4__0.6__ add__1.0__2.0__ multiply__1.0__2.0__ |
| two trains each num__100 m long moving in opposite directions cross other in num__8 sec . if one is moving twice as fast the other then the speed of the faster train is ? <o> a ) num__30 km / hr <o> b ) num__45 km / hr <o> c ) num__60 km / hr <o> d ) num__75 km / hr <o> e ) num__85 km / hr |
explanation : let the speed of the slower train be x m / sec . then speed of the train = num__2 x m / sec . relative speed = ( x + num__2 x ) = num__3 x m / sec . ( num__100 + num__100 ) / num__8 = num__3 x = > x = num__8.33333333333 . so speed of the faster train = num__16.6666666667 = num__16.6666666667 * num__3.6 = num__60 km / hr . answer is c <eor> c <eos> |
c |
hour_to_min_conversion__ hour_to_min_conversion__ |
hour_to_min_conversion__ hour_to_min_conversion__ |
| if a - b = num__6 and a num__2 + b num__2 = num__48 find the value of ab . <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__6 <o> e ) num__19 |
num__2 ab = ( a num__2 + b num__2 ) - ( a - b ) num__2 = num__48 - num__36 = num__12 ab = num__6 . answer : d <eor> d <eos> |
d |
multiply__6.0__2.0__ subtract__12.0__6.0__ |
subtract__48.0__36.0__ subtract__12.0__6.0__ |
| the average monthly income of a and b is rs . num__5050 . the average monthly income of b and c is rs . num__6250 and the average monthly income of a and c is rs . num__5200 . what is the monthly income of a ? <o> a ) num__2000 <o> b ) num__3000 <o> c ) num__4000 <o> d ) num__5000 <o> e ) num__6000 |
explanation : let monthly income of a = a monthly income of b = b monthly income of c = c a + b = num__2 × num__5050 . . . . ( equation num__1 ) b + c = num__2 × num__6250 . . . . ( equation num__2 ) a + c = num__2 × num__5200 . . . . ( equation num__3 ) ( equation num__1 ) + ( equation num__3 ) - ( equation num__2 ) = > a + b + a + c - ( b + c ) = ( num__2 × num__5050 ) + ( num__2 × num__5200 ) - ( num__2 × num__6250 ) = > num__2 a = num__2 ( num__5050 + num__5200 - num__6250 ) = > a = num__4000 i . e . monthly income of a = num__4000 answer : option c <eor> c <eos> |
c |
add__1.0__2.0__ multiply__4000.0__1.0__ |
add__1.0__2.0__ multiply__4000.0__1.0__ |
| express the ratio num__3 num__0.666666666667 : num__7 num__0.333333333333 in its simplest form . <o> a ) num__1 : num__21 <o> b ) num__1 : num__1 <o> c ) num__2 : num__1 <o> d ) num__1 : num__2 <o> e ) num__2 : num__11 |
solution we first convert the mixed numbers num__3 num__0.666666666667 and num__7 num__0.333333333333 into fractions num__3 num__0.666666666667 = num__3 * num__1.0 + num__0.666666666667 = num__3.66666666667 num__7 num__0.333333333333 = num__7 * num__1.0 + num__0.333333333333 = num__7.33333333333 the ratio num__3 num__0.666666666667 : num__7 num__0.333333333333 can be expressed as num__3.66666666667 ÷ num__7.33333333333 = num__3.66666666667 × num__0.136363636364 simplify = num__0.5 = num__0.5 the ratio is num__0.5 or num__1 : num__2 answer is d <eor> d <eos> |
d |
add__0.6667__0.3333__ add__3.0__0.6667__ add__7.0__0.3333__ reverse__7.3333__ divide__3.6667__7.3333__ reverse__0.5__ reverse__1.0__ |
add__0.6667__0.3333__ add__3.0__0.6667__ add__7.0__0.3333__ reverse__7.3333__ divide__3.6667__7.3333__ reverse__0.5__ add__0.6667__0.3333__ |
| if k is a positive integer which of the following must be divisible by num__27 ? <o> a ) ( k – num__4 ) ( k ) ( k + num__3 ) ( k + num__7 ) <o> b ) ( k – num__4 ) ( k – num__2 ) ( k + num__3 ) ( k + num__5 ) <o> c ) ( k – num__2 ) ( k + num__3 ) ( k + num__5 ) ( k + num__6 ) <o> d ) ( k + num__1 ) ( k + num__3 ) ( k + num__5 ) ( k + num__7 ) <o> e ) ( k – num__3 ) ( k + num__1 ) ( k + num__4 ) ( k + num__6 ) |
num__24 = num__8 * num__3 . note that the product of two consecutive even integers is always divisible by num__8 ( since one of them is divisible by num__4 and another by num__2 ) . only option b offers two consecutive even numbers for any integer value of k : k - num__4 and k - num__2 if k = even or k + num__3 and k + num__5 if k = odd . also from the following num__3 consecutive integers : ( k - num__4 ) ( k - num__3 ) ( k - num__2 ) one must be divisible by num__3 if it ' s not k - num__4 or k - num__2 then it must be k - num__3 ( if it ' s k - num__4 or k - num__2 option b is divisible by num__3 right away ) . but if it ' s k - num__3 then ( k - num__3 ) + num__6 = k + num__3 must also be divisible by num__3 . so option b : ( k – num__4 ) ( k – num__2 ) ( k + num__3 ) ( k + num__5 ) is divisible by num__8 and num__3 in any case . answer : e . <eor> e <eos> |
e |
subtract__27.0__24.0__ divide__8.0__4.0__ add__2.0__3.0__ multiply__2.0__3.0__ subtract__27.0__24.0__ |
subtract__27.0__24.0__ divide__8.0__4.0__ add__2.0__3.0__ multiply__2.0__3.0__ subtract__27.0__24.0__ |
| two trains are moving in opposite directions at num__60 km / hr and num__90 km / hr . their lengths are num__2.10 km and num__1.4 km respectively . the time taken by the slower train to cross the faster train in seconds is ? <o> a ) num__99 <o> b ) num__277 <o> c ) num__84 <o> d ) num__96 <o> e ) num__22 |
relative speed = num__60 + num__90 = num__150 km / hr . = num__150 * num__0.277777777778 = num__41.6666666667 m / sec . distance covered = num__2.10 + num__1.4 = num__3.5 km = num__3500 m . required time = num__3500 * num__0.024 = num__84 sec . answer : c <eor> c <eos> |
c |
add__60.0__90.0__ add__2.1__1.4__ multiply__60.0__1.4__ round__84.0__ |
add__60.0__90.0__ add__2.1__1.4__ multiply__60.0__1.4__ multiply__60.0__1.4__ |
| father is num__24 years older than his son . in two years his age will be twice the age of his son . the present age of his son is : <o> a ) num__20 years <o> b ) num__21 years <o> c ) num__22 years <o> d ) num__28 years <o> e ) num__25 years |
age consider as x ( x + num__24 ) + num__2 = num__2 ( x + num__2 ) x + num__26 = num__2 x + num__4 x = num__22 . answer c <eor> c <eos> |
c |
add__24.0__2.0__ subtract__24.0__2.0__ subtract__24.0__2.0__ |
add__24.0__2.0__ subtract__24.0__2.0__ subtract__24.0__2.0__ |
| the ratio between the length and the breadth of a rectangular park is num__3 : num__2 . if a man cycling along theboundary of the park at the speed of num__12 km / hr completes one round in num__8 min then the area of the park ( in sq . m ) is ? <o> a ) num__143530 m <o> b ) num__145600 m <o> c ) num__153600 m <o> d ) num__134500 m <o> e ) num__155600 m |
perimeter = distance covered in num__8 min . = num__12000 x num__8 m = num__1600 m . num__60 let length = num__3 x metres and breadth = num__2 x metres . then num__2 ( num__3 x + num__2 x ) = num__1600 or x = num__160 . length = num__480 m and breadth = num__320 m . area = ( num__480 x num__320 ) m num__2 = num__153600 m c <eor> c <eos> |
c |
multiply__3.0__160.0__ multiply__2.0__160.0__ surface_cube__160.0__ surface_cube__160.0__ |
multiply__3.0__160.0__ multiply__2.0__160.0__ surface_cube__160.0__ surface_cube__160.0__ |
| a can give b num__150 meters start and c num__300 meters start in a kilometer race . how much start can b give c in a kilometer race ? <o> a ) num__111.12 <o> b ) num__111.67 <o> c ) num__111.64 <o> d ) num__111.11 <o> e ) num__176.47 |
a runs num__1000 m while b runs num__850 m and c runs num__700 m . the number of meters that c runs when b runs num__1000 m = ( num__1000 * num__700 ) / num__850 = num__823.53 m . b can give c = num__1000 - num__823.53 = num__176.47 m . answer : e <eor> e <eos> |
e |
subtract__1000.0__150.0__ subtract__1000.0__300.0__ subtract__1000.0__823.53__ round__176.47__ |
subtract__1000.0__150.0__ subtract__1000.0__300.0__ subtract__1000.0__823.53__ subtract__1000.0__823.53__ |
| it is being given that ( num__232 + num__1 ) is completely divisible by a whole number . which of the following numbers is completely divisible by this number ? <o> a ) ( num__2 ^ num__96 + num__1 ) <o> b ) ( num__2 ^ num__16 + num__1 ) <o> c ) ( num__2 ^ num__95 + num__1 ) <o> d ) ( num__2 ^ num__95 - num__1 ) <o> e ) ( num__2 ^ num__75 + num__1 ) |
let num__232 = x . then ( num__232 + num__1 ) = ( x + num__1 ) . let ( x + num__1 ) be completely divisible by the natural number n . then ( num__296 + num__1 ) = [ ( num__232 ) num__3 + num__1 ] = ( x num__3 + num__1 ) = ( x + num__1 ) ( x num__2 - x + num__1 ) which is completely divisible by n since ( x + num__1 ) is divisible by n . a ) <eor> a <eos> |
a |
subtract__3.0__1.0__ multiply__1.0__2.0__ |
subtract__3.0__1.0__ subtract__3.0__1.0__ |
| the population of a town is num__10000 . it increases annually at the rate of num__20.0 p . a . what will be its population after num__2 years ? <o> a ) num__14000 <o> b ) num__14400 <o> c ) num__14500 <o> d ) num__14600 <o> e ) num__14700 |
formula : ( after = num__100 denominator ago = num__100 numerator ) num__10000 × num__1.2 × num__1.2 = num__14400 b <eor> b <eos> |
b |
percent__100.0__14400.0__ |
percent__100.0__14400.0__ |
| a = √ [ num__2 √ num__63 + num__2 / ( num__8 + num__3 √ num__7 ) ] = <o> a ) num__8 + num__3 √ num__7 <o> b ) num__4 + num__3 √ num__7 <o> c ) num__8 <o> d ) num__4 <o> e ) √ num__7 |
in these type of question u multiply the nominator and denominator with conjugate . . . conjugate of num__8 + num__3 sqrt ( num__7 ) is num__8 - num__3 sqrt ( num__7 ) sqrt [ num__2 sqrt ( num__63 ) + num__2 { num__8 - num__3 sqrt ( num__7 ) } / { num__64 - num__63 } ] a = sqrt [ num__2 sqrt ( num__63 ) + num__16 - num__2 sqrt ( num__63 ) ] = num__4 answer is num__4 . d <eor> d <eos> |
d |
multiply__2.0__8.0__ divide__8.0__2.0__ divide__8.0__2.0__ |
multiply__2.0__8.0__ divide__8.0__2.0__ divide__8.0__2.0__ |
| a train num__100 m long crosses a platform num__180 m long in num__14 sec ; find the speed of the train ? <o> a ) num__72 kmph <o> b ) num__58 kmph <o> c ) num__54 kmph <o> d ) num__94 kmph <o> e ) num__59 kmph |
d = num__100 + num__180 = num__280 t = num__14 s = num__20.0 * num__3.6 = num__72 kmph answer : a <eor> a <eos> |
a |
add__100.0__180.0__ divide__280.0__14.0__ multiply__3.6__20.0__ round__72.0__ |
add__100.0__180.0__ divide__280.0__14.0__ multiply__3.6__20.0__ multiply__3.6__20.0__ |
| what is the number of num__7 - element subsets of the set { num__1 num__2 num__3 num__4 num__5 num__6 num__7 num__8 num__9 } for which the sum of those num__7 elements is a multiple of num__3 ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__12 number <o> d ) num__13 <o> e ) num__14 |
we describe num__2 solutions . let s = { num__1 num__2 . . . num__9 } . observe that the sum of the elements of s is divisible by num__3 . so the problem is the same as asking for the number of ways that num__2 elements of s can be chosen so that their sum is divisible by num__3 ( the other num__7 elements of s correspond to a num__7 - element set as in the problem ) . choosing num__2 elements of s with sum divisible by num__3 corresponds to either choosing num__2 numbers divisible by num__3 ( which can be done in num__3 ways ) or choosing num__1 number that is one more than a multiple of num__3 and num__1 number that is one less than a multiple of num__3 ( which can be done in num__3 · num__3 = num__9 ways ) . hence the answer is num__3 + num__9 = num__12 . correct answer c <eor> c <eos> |
c |
add__7.0__5.0__ add__7.0__5.0__ |
add__7.0__5.0__ add__7.0__5.0__ |
| if n is an integer greater than num__6 which of the following must be divisible by num__2 ? <o> a ) num__1 . n ( n + num__1 ) ( n - num__4 ) <o> b ) num__2 . n ( n + num__2 ) ( n - num__1 ) <o> c ) num__3 . n ( n + num__3 ) ( n - num__5 ) <o> d ) num__4 . n ( n + num__4 ) ( n - num__2 ) <o> e ) num__5 . n ( n + num__5 ) ( n - num__6 ) |
we need to find out the number which is divisible by three in every num__3 consecutive integers there must contain num__1 multiple of num__3 . so n + num__4 and n + num__1 are same if we need to find out the num__3 ' s multiple . replace all the numbers which are more than or equal to three num__1 . n ( n + num__1 ) ( n - num__4 ) = > n ( n + num__1 ) ( n - num__1 ) = > ( n - num__1 ) n ( n + num__1 ) num__2 . n ( n + num__2 ) ( n - num__1 ) = > n ( n + num__2 ) ( n - num__1 ) = > ( n - num__1 ) n ( n + num__1 ) num__3 . n ( n + num__3 ) ( n - num__5 ) = > n ( n + num__0 ) ( n - num__2 ) = > ( n - num__2 ) n ( n ) num__4 . n ( n + num__4 ) ( n - num__2 ) = > n ( n + num__1 ) ( n - num__2 ) = > ( n - num__2 ) n ( n + num__1 ) num__5 . n ( n + num__5 ) ( n - num__6 ) = > n ( n + num__2 ) ( n - num__0 ) = > ( n ) n ( n + num__2 ) from the above onlyoption ais product of consecutive three numbers . b <eor> b <eos> |
b |
divide__6.0__2.0__ subtract__3.0__2.0__ subtract__6.0__2.0__ subtract__6.0__1.0__ divide__6.0__3.0__ |
divide__6.0__2.0__ subtract__3.0__2.0__ subtract__6.0__2.0__ subtract__6.0__1.0__ subtract__6.0__4.0__ |
| a certain manufacturer produces items for which the production costs consist of annual fixed costs totaling $ num__130000 and variables costs averaging $ num__8 per item . if the manufacturer ’ s selling price per item is $ num__12 how many items the manufacturer produce and sell to earn an annual profit of $ num__140000 ? <o> a ) num__2858 <o> b ) num__8667 <o> c ) num__21429 <o> d ) num__35000 <o> e ) num__67 |
500 |
let the items manufactured or sold bex num__130000 + num__8 x = num__12 x - num__140000 num__4 x = num__270000 x = num__67500 ans : e <eor> e <eos> |
e |
e |
| kanul spent $ num__3000 in buying raw materials $ num__1000 in buying machinery and num__30.0 of the total amount he had as cash with him . what was the total amount ? <o> a ) $ num__5825.16 <o> b ) $ num__5725.26 <o> c ) $ num__5714.28 <o> d ) $ num__5912.52 <o> e ) $ num__5614.46 |
let the total amount be x then ( num__100 - num__30 ) % of x = num__3000 + num__1000 num__70.0 of x = num__4000 num__70 x / num__100 = num__4000 x = $ num__5714.28571429 x = $ num__5714.28 answer is c <eor> c <eos> |
c |
percent__100.0__5714.28__ |
percent__100.0__5714.28__ |
| the parameter of a square is equal to the perimeter of a rectangle of length num__16 cm and breadth num__14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) ? <o> a ) num__23.77 cm <o> b ) num__23.47 cm <o> c ) num__83.57 cm <o> d ) num__23.57 cm <o> e ) num__23.56 cm |
let the side of the square be a cm . parameter of the rectangle = num__2 ( num__16 + num__14 ) = num__60 cm parameter of the square = num__60 cm i . e . num__4 a = num__60 a = num__15 diameter of the semicircle = num__15 cm circimference of the semicircle = num__0.5 ( ∏ ) ( num__15 ) = num__0.5 ( num__3.14285714286 ) ( num__15 ) = num__23.5714285714 = num__23.57 cm to two decimal places answer : d <eor> d <eos> |
d |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
rectangle_perimeter__16.0__14.0__ triangle_area__2.0__23.57__ |
| if d is an even integer and d / num__18 is an odd integer which of the following is not an even integer ? <o> a ) ( d ^ num__2 ) / num__54 <o> b ) ( d ^ num__2 ) / num__20 <o> c ) ( d ^ num__2 ) / num__9 <o> d ) ( d ^ num__2 ) / num__6 <o> e ) ( d ^ num__2 ) / num__2 |
since d is an even integer and d / num__18 is an odd integer . thus maximum power of num__2 in a is one . why ? ? because if the power of num__2 in the expression has been num__2 or greater then the d / num__18 must have been an even integer . ( num__18 = num__2 * num__3 ^ num__2 . as num__18 contain only one num__2 in the expression ) now any expression in which we have num__4 at the denominator will be odd . out of the given options only num__12 is a multiple of num__4 . hence answer should be b <eor> b <eos> |
b |
multiply__3.0__4.0__ subtract__4.0__2.0__ |
multiply__3.0__4.0__ divide__4.0__2.0__ |
| the largest num__4 digit number exactly divisible by num__88 is : <o> a ) num__3377 <o> b ) num__2255 <o> c ) num__7732 <o> d ) num__9944 <o> e ) num__9932 |
d num__9944 divide largest four digit number num__9999 by num__88 . you get num__113.625 . obviously num__113 would be exactly divisible so we want to know what that number is . we get this by multiplying num__113 with num__88 = num__9944 <eor> d <eos> |
d |
divide__9999.0__88.0__ round_down__113.625__ multiply__88.0__113.0__ |
divide__9999.0__88.0__ round_down__113.625__ multiply__88.0__113.0__ |
| which of the following data sets has the smallest standard deviation ? <o> a ) { num__3 num__3 num__3 num__3 num__3 } <o> b ) { num__2 num__3 num__3 num__3 num__4 } <o> c ) { num__2 num__2 num__2 num__4 num__5 } <o> d ) { num__0 num__2 num__3 num__4 num__6 } <o> e ) { - num__1 num__1 num__3 num__5 num__7 } |
since we are asked to compare ' standard deviation ' we should look out for similarity between all the sets . . the similarity is - the sum of all sets is equal = num__15 . . and hence average = num__3.0 = num__3 . . . now we have num__5 different sets which have avg as num__3 so we will look for the spread of the other elements of set around the average . . clearly e has a range of num__7 - ( - num__1 ) or num__8 and has the highest standard deviation . . next is d which has a range of num__6 - num__0 or num__6 . . and ( a ) { num__3 num__3 num__3 num__3 num__3 } has smallest standard of deviation . ans a <eor> a <eos> |
a |
divide__15.0__3.0__ add__1.0__7.0__ add__1.0__5.0__ multiply__1.0__3.0__ |
divide__15.0__3.0__ subtract__15.0__7.0__ subtract__7.0__1.0__ subtract__6.0__3.0__ |
| a cow is tethered in the middle of a field with a num__14 feet long rope . if the cow grazes num__100 sq . ft . per day then approximately what time will be taken by the cow to graze the whole field ? <o> a ) num__5 days <o> b ) num__6 days <o> c ) num__7 days <o> d ) num__8 days <o> e ) num__9 days |
area of the field grazed = [ num__3.14285714286 * num__14 * num__14 ] sq . ft . = num__616 sq . ft . number of days taken to graze the field = num__6.16 days = > num__6 days answer : b <eor> b <eos> |
b |
divide__616.0__100.0__ round__6.0__ |
divide__616.0__100.0__ round__6.0__ |
| the digit in the unit ’ s place of a number is equal to the digit in the ten ’ s place of half of that number and the digit in the ten ’ s place of that number is less than the digit in unit ’ s place of half of the number by num__1 . if the sum of the digits of number is num__7 then what is the number <o> a ) num__34 <o> b ) num__49 <o> c ) num__162 <o> d ) num__52 <o> e ) num__79 |
let the ten ’ s digit be x and unit ’ s digit be y then ( num__10 x + y ) / num__2 = num__10 y + ( x + num__1 ) num__8 x - num__19 y = num__2 and x + y = num__7 x = num__5 and y = num__2 then reqired number is num__52 option d <eor> d <eos> |
d |
add__1.0__7.0__ subtract__7.0__2.0__ multiply__1.0__52.0__ |
add__1.0__7.0__ subtract__7.0__2.0__ divide__52.0__1.0__ |
| a motorcyclist goes from bombay to pune a distance of num__256 kms at an average of num__32 kmph speed . another man starts from bombay by car num__2 ½ hours after the first and reaches pune ½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? <o> a ) num__1 : num__2 <o> b ) num__1 : num__5 <o> c ) num__1 : num__4 <o> d ) num__1 : num__1 <o> e ) num__5 : num__8 |
t = num__8.0 = num__8 h t = num__8 - num__3 = num__5 time ratio = num__8 : num__5 speed ratio = num__5 : num__8 answer : e <eor> e <eos> |
e |
divide__256.0__32.0__ add__2.0__3.0__ round__5.0__ |
divide__256.0__32.0__ subtract__8.0__3.0__ subtract__8.0__3.0__ |
| if r is an integer greater than num__6 which of the following must be divisible by num__3 ? <o> a ) r ( r + num__1 ) ( r - num__4 ) <o> b ) n ( n + num__2 ) ( n - num__1 ) <o> c ) n ( n + num__3 ) ( n - num__5 ) <o> d ) n ( n + num__4 ) ( n - num__2 ) <o> e ) n ( n + num__5 ) ( n - num__6 ) |
now take r = num__3 k r = num__3 k + num__1 r = num__3 k + num__2 . . put in all the choices . if by putting all the values of r we get it is divisible by num__3 then it is correct answer choice . a is correct . it will hardy take num__10 sec per choice as we have to consider only num__3 k + num__1 and num__3 k + num__2 . <eor> a <eos> |
a |
divide__6.0__3.0__ reverse__1.0__ |
divide__6.0__3.0__ reverse__1.0__ |
| the c . p of num__10 pens is equal to the s . p of num__12 pens . find his gain % or loss % ? <o> a ) num__16 num__0.333333333333 % <o> b ) num__16 num__3.0 % <o> c ) num__16 num__0.666666666667 % <o> d ) num__16 num__1.0 % <o> e ) num__12 num__0.666666666667 % |
num__10 cp = num__12 sp num__12 - - - num__2 cp loss num__100 - - - ? = > num__16 num__0.666666666667 % answer : c <eor> c <eos> |
c |
percent__100.0__16.0__ |
percent__100.0__16.0__ |
| if two numbers x a perfect square and y a perfect cube are added results a two digit number whose digits if reversed difference is num__72 find x and y ? <o> a ) x = num__4 y = num__8 <o> b ) x = num__4 y = num__9 <o> c ) x = num__16 y = num__64 <o> d ) x = num__4 y = num__9 <o> e ) x = num__5 y = num__8 |
num__16 + num__64 = num__80 when reversed num__08 num__80 - num__8 = num__72 x = num__16 y = num__64 answer : c <eor> c <eos> |
c |
add__16.0__64.0__ subtract__72.0__64.0__ subtract__80.0__64.0__ |
add__16.0__64.0__ subtract__72.0__64.0__ subtract__80.0__64.0__ |
| a box contains nine bulbs out of which num__4 are defective . if four bulbs are chosen at random find the probability that atleast one bulb is good ? <o> a ) num__0.968992248062 <o> b ) num__1.02459016393 <o> c ) num__0.992063492063 <o> d ) num__1.03305785124 <o> e ) num__1.0 |
required probability = num__1 - num__0.00793650793651 = num__0.992063492063 answer : c <eor> c <eos> |
c |
negate_prob__0.0079__ negate_prob__0.0079__ |
negate_prob__0.0079__ negate_prob__0.0079__ |
| concentrated grapes juice comes inside a cylinder tube with a radius of num__2.5 inches and a height of num__15 inches . the tubes are packed into wooden boxes each with dimensions of num__11 inches by num__10 inches by num__31 inches . how many tubes of concentrated grapes juice at the most can fit into num__3 wooden boxes ? <o> a ) num__24 . <o> b ) num__28 . <o> c ) num__36 . <o> d ) num__42 . <o> e ) num__48 . |
concentrated grapes juice comes inside a cylinder tube since height of a tube is num__15 inches the tubes can fit only in one way . now diameter of each tube = num__5 inches therefore num__4 * num__2 can be put in each wooden box in num__3 boxes num__3 * num__4 * num__2 can be accommodated = num__24 = a <eor> a <eos> |
a |
side_by_diagonal__5.0__3.0__ surface_cube__2.0__ surface_cube__2.0__ |
side_by_diagonal__5.0__3.0__ volume_rectangular_prism__3.0__2.0__4.0__ volume_rectangular_prism__3.0__2.0__4.0__ |
| if x dollars is invested at num__10 percent for one year and y dollars is invested at num__8 percent for one year the annual income from the num__10 percent investment will exceed the annual income from the num__8 percent investment by $ num__74 . if $ num__2000 is the total amount invested how much is invested at num__8 percent ? <o> a ) $ num__600 <o> b ) $ num__700 <o> c ) $ num__800 <o> d ) $ num__900 <o> e ) $ num__1000 |
num__0.1 x = num__0.08 ( num__2000 - x ) + num__74 num__0.18 x = num__234 x = num__1300 then the amount invested at num__8.0 is $ num__2000 - $ num__1300 = $ num__700 the answer is b . <eor> b <eos> |
b |
reverse__10.0__ add__0.1__0.08__ divide__234.0__0.18__ subtract__2000.0__1300.0__ subtract__2000.0__1300.0__ |
reverse__10.0__ add__0.1__0.08__ divide__234.0__0.18__ subtract__2000.0__1300.0__ subtract__2000.0__1300.0__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__120 m <o> b ) num__240 m <o> c ) num__300 m <o> d ) num__200 m <o> e ) none of these |
explanation : speed = ( num__54 * num__0.277777777778 ) m / sec = num__15 m / sec . length of the train = ( num__15 x num__20 ) m = num__300 m . let the length of the platform be x meters . then ( x + num__300 ) / num__36 = num__15 = = > x + num__300 = num__540 = = > x = num__240 m . answer is b <eor> b <eos> |
b |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
| it takes beth num__9 minutes to do one oil change and it takes logan num__12 minutes . at num__4 : num__56 pm they both finish an oil change simultaneously . if beth and logan both began doing oil changes at exactly the same time and worked without pausing when did they both start doing oil changes ? <o> a ) num__4 : num__11 pm <o> b ) num__4 : num__20 pm <o> c ) num__4 : num__29 pm <o> d ) num__4 : num__38 pm <o> e ) num__4 : num__47 pm |
since beth finishes s oil changes after s × num__9 minutes and logan finishes c oil changes after c × num__12 minutes they both finished doing an oil change at the same time when s × num__9 = c × num__12 . since s and c must be integers ( they represent the number of oil changes finished ) this question is asking you to find a common multiple of num__9 and num__12 . the question asks for the first time they began a car simultaneously so you must find the least common multiple and subtract . the least common multiple of num__9 and num__12 is num__36 so in the context of the question this would be num__36 minutes . therefore they began changing the oil of a car at the same time at num__4 : num__56 - num__36 minutes or num__4 : num__20 p . m . the answer is ( b ) . <eor> b <eos> |
b |
multiply__9.0__4.0__ subtract__56.0__36.0__ round__4.0__ |
multiply__9.0__4.0__ subtract__56.0__36.0__ round__4.0__ |
| a man can row with a speed of num__30 kmph in still water . if the stream flows at num__5 kmph then the speed in downstream is ? <o> a ) a ) num__63 kmph <o> b ) b ) num__63 kmph <o> c ) c ) num__35 kmph <o> d ) d ) num__62 kmph <o> e ) e ) num__74 kmph |
explanation : m = num__30 s = num__5 ds = num__30 + num__5 = num__35 answer : option c <eor> c <eos> |
c |
add__30.0__5.0__ round__35.0__ |
add__30.0__5.0__ add__30.0__5.0__ |
| rs . num__1500 is divided into two parts such that if one part is invested at num__6.0 and the other at num__5.0 the whole annual interest from both the sum is rs . num__83 . how much was lent at num__5.0 ? <o> a ) num__700 <o> b ) num__299 <o> c ) num__266 <o> d ) num__500 <o> e ) num__188 |
( x * num__5 * num__1 ) / num__100 + [ ( num__1500 - x ) * num__6 * num__1 ] / num__100 = num__83 num__5 x / num__100 + num__90 â € “ num__6 x / num__100 = num__83 x / num__100 = num__7 = > x = num__700 answer : a <eor> a <eos> |
a |
percent__6.0__1500.0__ percent__100.0__700.0__ |
percent__6.0__1500.0__ percent__100.0__700.0__ |
| what do you get if you add num__7 to num__290 three times ? <o> a ) num__305 <o> b ) num__296 <o> c ) num__289 <o> d ) num__297 <o> e ) num__309 |
d num__297 num__297 num__297 <eor> d <eos> |
d |
add__7.0__290.0__ add__7.0__290.0__ |
add__7.0__290.0__ add__7.0__290.0__ |
| a train num__275 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__15 <o> e ) num__9 |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__275 * num__0.0545454545455 ] sec = num__15 sec answer : d <eor> d <eos> |
d |
add__60.0__6.0__ divide__275.0__18.3333__ round__15.0__ |
add__60.0__6.0__ divide__275.0__18.3333__ divide__275.0__18.3333__ |
| a rectangular circuit board is designed to have width w inches perimeter p inches and area q square inches . which of the following equations must be true ? <o> a ) w ^ num__2 + pw + q = num__0 <o> b ) w ^ num__2 - pw + num__2 q = num__0 <o> c ) num__2 w ^ num__2 + pw + num__2 q = num__0 <o> d ) num__2 w ^ num__2 - pw - num__2 q = num__0 <o> e ) num__2 w ^ num__2 - pw + num__2 q = num__0 |
notice that we can discard options a and c right away . the sum of num__3 positive values can not be num__0 . now assume : width = w = num__1 inchand length = num__1 inch ; perimeter = p = num__4 inches ; area = q = num__1 square inches . plug the values of w p and q into the answer choices : only for e num__2 w ^ num__2 - pw + num__2 q = num__2 - num__4 + num__2 = num__0 . answer : e . <eor> e <eos> |
e |
square_perimeter__1.0__ rectangle_perimeter__0.0__1.0__ rectangle_perimeter__0.0__1.0__ |
square_perimeter__1.0__ rectangle_perimeter__0.0__1.0__ power__2.0__1.0__ |
| num__15 people can write num__90 book in num__20 days working num__7 hour a day . then in how many day num__240 can be written by num__72 people ? <o> a ) num__26.6666666667 <o> b ) num__11.0357142857 <o> c ) num__11.1111111111 <o> d ) num__22.2222222222 <o> e ) num__13.0384615385 |
work per day epr hour per person = num__90 / ( num__20 * num__7 * num__15 ) / / eq - num__1 people = num__72 ; let suppose day = p ; per day work for num__8 hours acc . to condition work per day epr hour per person = num__240 / ( p * num__7 * num__72 ) / / eq - num__2 eq - num__1 = = eq - num__2 ; p = num__11.1111111111 answer : c <eor> c <eos> |
c |
subtract__15.0__7.0__ multiply__1.0__11.1111__ |
subtract__15.0__7.0__ multiply__1.0__11.1111__ |
| a rectangular - shaped carpet remnant that measures x feet by y feet is priced at $ num__80 . what is the cost of the carpet in dollars per square yard ? ( num__9 square feet = num__1 square yard ) <o> a ) num__80 xy <o> b ) num__720 xy <o> c ) xy / num__9 <o> d ) xy / num__80 <o> e ) num__720 / ( xy ) |
xy sq ft = $ num__80 num__1 sq ft = $ num__80 / xy multiplying by num__9 on both side num__9 sq ft = $ num__720 / xy or num__1 sq yard = $ num__720 / xy hence e . <eor> e <eos> |
e |
multiply__80.0__9.0__ multiply__80.0__9.0__ |
multiply__80.0__9.0__ divide__720.0__1.0__ |
| a num__12 month project had a total budget of $ num__24000 . after seven months the project had spent $ num__12500 . at this point how much was the project under budget ? <o> a ) $ num__1500 <o> b ) $ num__1600 <o> c ) $ num__1700 <o> d ) $ num__1800 <o> e ) $ num__1900 |
each month the project should spend $ num__2000.0 = $ num__2000 . in num__7 months the project should spend num__7 * $ num__2000 = $ num__14000 . the project is under budget by $ num__14000 - $ num__12500 = $ num__1500 . the answer is a . <eor> a <eos> |
a |
divide__24000.0__12.0__ multiply__2000.0__7.0__ subtract__14000.0__12500.0__ subtract__14000.0__12500.0__ |
divide__24000.0__12.0__ multiply__2000.0__7.0__ subtract__14000.0__12500.0__ subtract__14000.0__12500.0__ |
| a cube has two of its faces painted half red and half white . the other faces are completely painted white . what is the ratio between the red painted areas and the white painted areas of the cube ? <o> a ) num__1 : num__5 <o> b ) num__3 : num__6 <o> c ) num__1 : num__2 <o> d ) num__2 : num__9 <o> e ) num__1 : num__3 |
let x be the area of each face of the cube . the area painted red is num__2 ( x / num__2 ) = x the area painted white is num__2 ( x / num__2 ) + num__4 x = num__5 x the ratio of red to white is x : num__5 x which is num__1 : num__5 . the answer is a . <eor> a <eos> |
a |
volume_cube__1.0__ |
volume_cube__1.0__ |
| in a bank investment compounds annually at an interest rate of num__34.1 what is the smallest investment period by which time the investment will more than triple in value ? <o> a ) num__1.2 <o> b ) num__1 <o> c ) num__4 <o> d ) num__2 <o> e ) num__3 |
assume initial amount is x annual interest is num__34.1 so after num__1 year the amount will become x * ( num__100 + num__34.1 ) / num__100 = > x * num__1.33333333333 now we need to find n for x * ( num__1.33333333333 ) ^ n = num__3 x or in other words n = num__4 c <eor> c <eos> |
c |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| by selling num__50 meters of cloth . i gain the selling price of num__15 meters . find the gain percent ? <o> a ) num__42 num__1.2 <o> b ) num__42 num__6 / num__0 <o> c ) num__42 num__2.0 <o> d ) num__42 num__0.857142857143 <o> e ) num__42 num__3.0 |
sp = cp + g num__50 sp = num__50 cp + num__15 sp num__35 sp = num__50 cp num__35 - - - num__15 cp gain num__100 - - - ? = > num__42 num__0.857142857143 % answer : d <eor> d <eos> |
d |
percent__100.0__42.0__ |
percent__100.0__42.0__ |
| the average age of a committee of num__8 members is num__36 years . a member aged num__55 years retired and his place was taken by another member aged num__39 years . the average age of present committee is ; <o> a ) num__39 years <o> b ) num__34 years <o> c ) num__36 years <o> d ) num__35 years <o> e ) num__31 years |
exp . total age of the committee = num__36 * num__8 = num__288 total age when a member is retired and a new one was joined = num__288 - num__55 + num__39 = num__272 average age of present committee = num__34.0 = num__34 . answer : b <eor> b <eos> |
b |
multiply__8.0__36.0__ divide__272.0__8.0__ divide__272.0__8.0__ |
multiply__8.0__36.0__ divide__272.0__8.0__ divide__272.0__8.0__ |
| set s contains exactly num__10 numbers and has an average ( arithmetic mean ) of num__6.2 . if one of the numbers in set s is increased by num__6 while all other numbers remain the same what is the new average of set s ? <o> a ) num__6.6 <o> b ) num__6.7 <o> c ) num__6.8 <o> d ) num__6.85 <o> e ) num__6.9 |
old set s - total is avg * no of elements = num__6.2 * num__10 = num__62 if one number is increased by num__6 then total increased to num__62 + num__6 = num__68 new avg - num__6.8 = num__6.8 . hence answer is c . <eor> c <eos> |
c |
multiply__10.0__6.2__ add__6.0__62.0__ divide__68.0__10.0__ divide__68.0__10.0__ |
multiply__10.0__6.2__ add__6.0__62.0__ divide__68.0__10.0__ divide__68.0__10.0__ |
| car a is num__10 miles behind car b which is traveling in the same direction along the same route as car a . car a is traveling at a constant speed of num__58 miles per hour and car bis traveling at a constant speed of num__50 miles per hour . how many hours will it take for car a to overtake and drive num__8 miles ahead of car b ? <o> a ) num__1.5 <o> b ) num__2.25 <o> c ) num__2.75 <o> d ) num__2.5 <o> e ) num__3.5 |
relative speed of car a is num__58 - num__50 = num__8 miles per hour to catch up num__10 miles and drive num__8 miles ahead so to drive num__18 miles it ' ll need num__2.25 = num__2.25 hours . answer : b <eor> b <eos> |
b |
add__10.0__8.0__ divide__18.0__8.0__ round__2.25__ |
add__10.0__8.0__ divide__18.0__8.0__ round__2.25__ |
| the positive two - digit integers p and q have the same digits but in reverse order . which of the following must be a factor of p + q ? <o> a ) num__66 <o> b ) num__9 <o> c ) num__10 <o> d ) num__12 <o> e ) num__14 |
remember : when you take the difference between the two it will always be num__9 . e . g num__23 - num__32 = num__9 num__89 - num__98 = num__9 and when you add both integers the sum will always be a multiple of num__11 e . g num__23 + num__32 = num__55 num__89 + num__98 = num__187 num__66 is multiple of num__11 answer : a <eor> a <eos> |
a |
add__9.0__23.0__ add__9.0__89.0__ add__32.0__23.0__ add__98.0__89.0__ subtract__98.0__32.0__ subtract__98.0__32.0__ |
add__9.0__23.0__ add__9.0__89.0__ add__32.0__23.0__ add__98.0__89.0__ subtract__98.0__32.0__ subtract__98.0__32.0__ |
| reena took a loan of rs . num__1200 with simple interest for as many years as the rate of interest . if she paid rs . num__432 as interest at the end of the loan period what was the rate of interest ? <o> a ) num__7 <o> b ) num__6 <o> c ) num__5 <o> d ) num__4 <o> e ) num__3 |
let rate = r % and time = r years . then ( num__1200 * r * r ) / num__100 = num__432 num__12 r num__2 = num__432 r num__2 = num__36 = > r = num__6 . answer : b <eor> b <eos> |
b |
percent__100.0__6.0__ |
percent__100.0__6.0__ |
| if x and y are three - digit integers such that x > num__400 and y < num__700 which of the following is closest to the maximum possible value of xy ? <o> a ) num__698301 <o> b ) num__2800 <o> c ) num__45000 <o> d ) num__76000 <o> e ) num__28 |
000 |
to get the maximum possible value of xy we need maximize the value of both x and y max x can be = num__999 ( given x is a two - digit integer > num__400 ) max y can be = num__699 ( given y is a two - digit integer < num__700 ) xy = num__999 * num__699 = num__698301 answer ( a ) <eor> a <eos> |
a |
a |
| a person covers a distance in num__12 minutes . if runs at a speed of num__12 km per hour on an average . find the speed at which he must run to reduce the time of journey to num__5 minutes . <o> a ) num__8 m / s <o> b ) num__4 m / s <o> c ) num__15 m / s <o> d ) num__10 m / s <o> e ) num__5 m / s |
explanation : t = num__12 m spees = num__12 kmph = num__12 x num__0.277777777778 = num__3.33333333333 m / s let new speed be ` ` x ' ' num__12 x ( num__3.33333333333 ) = num__5 x x = num__8 m / s answer : option a <eor> a <eos> |
a |
round__8.0__ |
round__8.0__ |
| the distance between num__2 cities a and b is num__1000 km . a train starts from a at num__11 a . m . and travels towards b at num__100 km / hr . another starts from b at num__12 p . m . and travels towards a at num__150 km / hr . at what time do they meet ? <o> a ) num__11 am . <o> b ) num__12 p . m . <o> c ) num__3 pm . <o> d ) num__4 p . m . <o> e ) num__1 p . m . |
suppose they meet x hrs after num__11 a . m . distance moved by first in x hrs + distance moved by second in ( x - num__1 ) hrs = num__1000 num__100 x + num__150 ( x - num__1 ) = num__1000 x = num__4.60 = num__5 hrs they meet at num__11 + num__5 = num__4 p . m . answer is d <eor> d <eos> |
d |
subtract__12.0__11.0__ subtract__5.0__1.0__ round__4.0__ |
subtract__12.0__11.0__ subtract__5.0__1.0__ subtract__5.0__1.0__ |
| one third of a two digit number exceeds its one fourth by num__3 . what is the sum of the digits of the number ? <o> a ) seven <o> b ) eight <o> c ) nine <o> d ) ten <o> e ) eleven |
explanation : x / num__3 – x / num__4 = num__3 = > x = num__36 num__3 + num__6 = num__9 c <eor> c <eos> |
c |
add__3.0__6.0__ add__3.0__6.0__ |
add__3.0__6.0__ add__3.0__6.0__ |
| find the one which does not belong to that group ? <o> a ) num__36 <o> b ) num__77 <o> c ) num__28 <o> d ) num__99 <o> e ) num__22 |
explanation : num__16 num__36 num__64 and num__4 are perfect squares but not num__28 . answer : c <eor> c <eos> |
c |
divide__64.0__16.0__ subtract__64.0__36.0__ subtract__64.0__36.0__ |
divide__64.0__16.0__ subtract__64.0__36.0__ subtract__64.0__36.0__ |
| on a certain road num__20.0 of the motorists exceed the posted speed limit and receive speeding tickets but num__20.0 of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on that road exceed the posted speed limit ? <o> a ) num__10.5 <o> b ) num__12.5 <o> c ) num__15.0 <o> d ) num__25.0 <o> e ) num__30 % |
suppose there are x motorists . num__20.0 of them exceeded the speed limit and received the ticket i . e . x / num__5 . again suppose total no . of motorists who exceeded the speed limit are y . num__20.0 of y exceeded the speed limit but did n ' t received the ticket i . e . y / num__5 . it means num__4 y / num__5 received the ticket . hence num__4 y / num__5 = x / num__5 or y / x = num__0.25 or y / x * num__100 = num__0.25 * num__100 = num__25.0 d <eor> d <eos> |
d |
percent__25.0__100.0__ |
percent__25.0__100.0__ |
| two trains of equal lengths take num__10 sec and num__15 sec respectively to cross a telegraph post . if the length of each train be num__120 m in what time will they cross other travelling in opposite direction ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__15 <o> d ) num__20 <o> e ) num__5 |
answer : option b speed of the first train = num__12.0 = num__12 m / sec . speed of the second train = num__24.0 = num__8 m / sec . relative speed = num__12 + num__8 = num__20 m / sec . required time = ( num__120 + num__120 ) / num__20 = num__12 sec . <eor> b <eos> |
b |
divide__120.0__10.0__ divide__120.0__15.0__ add__8.0__12.0__ round__12.0__ |
divide__120.0__10.0__ divide__120.0__15.0__ add__8.0__12.0__ divide__120.0__10.0__ |
| the present ages of three persons in proportions num__10 : num__1 : num__5 . eight years ago the sum of their ages was num__56 . find their present ages ( in years ) . <o> a ) num__50 num__5 num__25 <o> b ) num__16 num__28 num__36 <o> c ) num__16 num__28 num__35 <o> d ) num__16 num__28 num__34 <o> e ) num__16 num__28 num__33 |
let their present ages be num__10 x num__1 x and num__5 x years respectively . then ( num__10 x - num__8 ) + ( num__1 x - num__8 ) + ( num__5 x - num__8 ) = num__56 num__16 x = num__80 x = num__5 . their present ages are num__10 x = num__50 years num__1 x = num__5 years and num__5 x = num__25 years respectively . answer : a <eor> a <eos> |
a |
multiply__10.0__8.0__ multiply__10.0__5.0__ multiply__10.0__5.0__ |
multiply__10.0__8.0__ multiply__10.0__5.0__ multiply__10.0__5.0__ |
| for what value of x is | x – num__3 | + | x + num__1 | + | x | = num__8 ? <o> a ) num__0 <o> b ) num__3 <o> c ) - num__3 <o> d ) num__4 <o> e ) - num__2 |
for what value of x is | x – num__3 | + | x + num__1 | + | x | = num__8 ? it ' s easiest just to plug in answer choices : ( e ) : - num__2 | x – num__3 | + | x + num__1 | + | x | = num__8 ? | - num__2 - num__3 | + | - num__2 + num__1 | + | - num__2 | = num__8 ? | num__1 | + | num__5 | + | num__4 | = num__8 ( e ) <eor> e <eos> |
e |
subtract__3.0__1.0__ add__3.0__2.0__ add__3.0__1.0__ subtract__3.0__1.0__ |
subtract__3.0__1.0__ add__3.0__2.0__ add__3.0__1.0__ subtract__3.0__1.0__ |
| a boat crossed a lake from north to west at the speed of num__5 km / h entered a river and covered twice as much distance going upstream at num__4 km / h . it then turned around and stopped at the south shore of the lake . if it averaged num__3.6 km / h that day what was its approximate downstream speed ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
one way of solving this is : speed of boat on still water ( lake ) = num__5 kmph speed upstream = num__4 kmph = speed in still water - speed of river = > speed of river = num__1 kmph = > speed downstream = speed in still water + speed of river = num__5 + num__1 = num__6 kmph ans is c <eor> c <eos> |
c |
subtract__5.0__4.0__ add__5.0__1.0__ add__5.0__1.0__ |
subtract__5.0__4.0__ add__5.0__1.0__ add__5.0__1.0__ |
| in racing over a given distance d at uniform speed a can beat b by num__20 meters b can beat c by num__5 meters and a can beat c by num__24 meters . what is the distance d in meters ? <o> a ) num__70 <o> b ) num__80 <o> c ) num__90 <o> d ) num__100 <o> e ) num__120 |
when a is at the finish line b is num__20 meters back and c is num__24 meters back . when b runs another num__20 meters to the finish line c is num__5 meters back . so b gains num__1 meter on c every num__20 meters covered . to gain a final difference of num__5 meters b runs a total of num__100 meters . the answer is d . <eor> d <eos> |
d |
multiply__20.0__5.0__ round__100.0__ |
multiply__20.0__5.0__ round__100.0__ |
| the average monthly income of p and q is rs . num__5050 . the average monthly income of q and r is num__6250 and the average monthly income of p and r is rs . num__5500 . the monthly income of p is ? <o> a ) a ) rs . num__4078 <o> b ) b ) rs . num__4000 <o> c ) c ) rs . num__4029 <o> d ) d ) rs . num__4300 <o> e ) e ) rs . num__4020 |
let p q and r represent their respective monthly incomes . then we have : p + q = ( num__5050 * num__2 ) = num__10100 - - - ( i ) q + r = ( num__6250 * num__2 ) = num__12500 - - - ( ii ) p + r = ( num__5500 * num__2 ) = num__11000 - - - ( iii ) adding ( i ) ( ii ) and ( iii ) we get : num__2 ( p + q + r ) = num__33600 = p + q + r = num__16800 - - - ( iv ) subtracting ( ii ) from ( iv ) we get p = num__4300 . p ' s monthly income = rs . num__4300 . answer : d <eor> d <eos> |
d |
multiply__5050.0__2.0__ multiply__6250.0__2.0__ multiply__5500.0__2.0__ divide__33600.0__2.0__ subtract__16800.0__12500.0__ subtract__16800.0__12500.0__ |
multiply__5050.0__2.0__ multiply__6250.0__2.0__ multiply__5500.0__2.0__ divide__33600.0__2.0__ subtract__16800.0__12500.0__ subtract__16800.0__12500.0__ |
| equal weights of two alloys containing tin copper and lead in the ratio num__3 : num__2 : num__7 and num__4 : num__11 : num__3 are melted and mixed together . what is the ratio of tin copper and lead in the resultant alloy ? <o> a ) num__41 : num__81 : num__37 <o> b ) num__33 : num__91 : num__81 <o> c ) num__17 : num__28 : num__27 <o> d ) num__51 : num__86 : num__89 <o> e ) num__41 : num__86 : num__89 |
let the weight of the two alloys be w each required ratio = ( num__3 w / num__12 + num__4 w / num__18 ) : ( num__2 w / num__12 + num__11 w / num__18 ) : ( num__7 w / num__12 + num__3 w / num__18 ) = num__17 w / num__36 : num__28 w / num__36 : num__27 w / num__36 = num__17 : num__28 : num__27 answer : c <eor> c <eos> |
c |
multiply__3.0__4.0__ add__7.0__11.0__ multiply__3.0__12.0__ multiply__7.0__4.0__ subtract__28.0__11.0__ |
multiply__3.0__4.0__ add__7.0__11.0__ multiply__3.0__12.0__ add__11.0__17.0__ subtract__28.0__11.0__ |
| the diameter of a garden roller is num__1.2 m and it is num__2 m long . how much area will it cover in num__5 revolutions ? ( use Ï € = num__22 â „ num__7 ) <o> a ) num__40 m num__2 <o> b ) num__44 m num__2 <o> c ) num__48 m num__2 <o> d ) num__37.7 m num__2 <o> e ) none of these |
required area covered in num__5 revolutions = num__5 Ã — num__2 Ï € rh = num__5 Ã — num__2 Ã — num__22 â „ num__7 Ã — num__0.6 Ã — num__2 = num__37.7 m num__2 answer d <eor> d <eos> |
d |
km_to_mile_conversion__ round__37.7__ |
km_to_mile_conversion__ round__37.7__ |
| two trains are moving in opposite directions with speed of num__90 km / hr and num__90 km / hr respectively . their lengths are num__1.10 km and num__0.9 km respectively . the slower train cross the faster train in - - - seconds <o> a ) num__56 <o> b ) num__40 <o> c ) num__47 <o> d ) num__26 <o> e ) num__25 |
explanation : relative speed = num__90 + num__90 = num__180 km / hr ( since both trains are moving in opposite directions ) total distance = num__1.1 + . num__9 = num__2 km time = num__0.0111111111111 hr = num__0.0111111111111 hr = num__40.0 seconds = num__40 seconds answer : option b <eor> b <eos> |
b |
add__1.1__0.9__ divide__2.0__180.0__ round__40.0__ |
add__1.1__0.9__ divide__2.0__180.0__ round__40.0__ |
| in a caravan in addition to num__60 hens there are num__35 goats and num__6 camels with some keepers . if the total number of feet be num__193 more than the number of heads in the caravan the number of keepers is <o> a ) num__5 <o> b ) num__10 <o> c ) num__15 <o> d ) num__20 <o> e ) num__25 |
explanation : let number of keepers be x . then total number of feet = num__2 x num__60 + num__4 x num__35 + num__4 x num__6 + num__2 x = num__2 x + num__284 . total number of heads = num__60 + num__35 + num__6 + x = num__101 + x . therefore ( num__2 x + num__284 ) = ( num__101 + x ) + num__193 or x = num__10 . answer : b <eor> b <eos> |
b |
subtract__6.0__2.0__ divide__60.0__6.0__ divide__60.0__6.0__ |
subtract__6.0__2.0__ add__6.0__4.0__ add__6.0__4.0__ |
| if a b c d and x are nonzero numbers which of the following conditions must be true for ( ax ) ^ num__61 ( bx ) ^ num__21 ( cx ) ^ num__14 ( dx ) ^ num__6 > num__0 ? <o> a ) x > num__0 <o> b ) a × b > num__0 <o> c ) c × d > num__0 <o> d ) a × d > num__0 <o> e ) a × b < num__0 |
the question becomes a ^ ( num__61 ) * b ^ ( num__21 ) * c ^ ( num__14 ) * d ^ num__6 * x ^ ( num__102 ) > num__0 c ^ num__14 d ^ num__6 x ^ num__102 are positive no matter what the sign of c d and x respectively = > a ^ num__61 * b ^ num__21 must me positive = > a * b should be positive ans b <eor> b <eos> |
b |
multiply__61.0__0.0__ |
multiply__61.0__0.0__ |
| a cricket player whose bowling average was num__23.50 runs per wicket takes num__5 wicket for num__52 runs in a match . due to this his average decreases by num__0.50 . what will be the number of wickets taken by him till the last match ? <o> a ) num__64 <o> b ) num__72 <o> c ) num__131 <o> d ) num__456 <o> e ) num__108 |
average = total runs / total wickets total runs after last match = num__23.5 w + num__52 total wickets after last match = w + num__5 ( num__23.5 w + num__52 ) / ( w + num__5 ) = num__23.5 - num__0.5 = num__23 w = num__126 so total wickets aftr last match = w + num__5 = num__131 answer : c <eor> c <eos> |
c |
round_down__23.5__ add__5.0__126.0__ add__5.0__126.0__ |
subtract__23.5__0.5__ add__5.0__126.0__ add__5.0__126.0__ |
| the list price of an article is rs . num__70 . a customer pays rs . num__56.16 for it . he was given two successive discounts one of them being num__10.0 . the other discount is ? <o> a ) num__10.85 <o> b ) num__14.85 <o> c ) num__15.85 <o> d ) num__16.85 <o> e ) num__17.85 % |
num__70 * ( num__0.9 ) * ( ( num__100 - x ) / num__100 ) = num__56.16 x = num__10.85 answer : a <eor> a <eos> |
a |
percent__10.85__100.0__ |
percent__10.85__100.0__ |
| a certain galaxy is known to comprise approximately num__4 x num__10 ^ num__10 stars . of every num__50 million of these stars one is larger in mass than our sun . approximately how many stars in this galaxy are larger than the sun ? <o> a ) num__800 <o> b ) num__1250 <o> c ) num__8000 <o> d ) num__12000 <o> e ) num__80 |
000 |
total no . of stars on galaxy = num__4 * num__10 ^ num__10 of every num__50 million stars num__1 is larger than sun . num__1 million = num__10 ^ num__6 therofore num__50 million = num__50 * num__10 ^ num__6 total no . of stars larger than sun = num__4 * num__10 ^ num__0.2 * num__10 ^ num__6 = num__40 * num__10 ^ num__0.4 = num__800 therefore answer is a <eor> a <eos> |
a |
a |
| a train num__125 m long passes a man running at num__20 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is ? <o> a ) num__36 <o> b ) num__50 <o> c ) num__28 <o> d ) num__26 <o> e ) num__65 |
speed of the train relative to man = ( num__12.5 ) m / sec = ( num__12.5 ) m / sec . [ ( num__12.5 ) * ( num__3.6 ) ] km / hr = num__45 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__20 ) km / hr . x - num__20 = num__45 = = > x = num__65 km / hr answer : e <eor> e <eos> |
e |
divide__125.0__10.0__ multiply__12.5__3.6__ add__20.0__45.0__ round__65.0__ |
divide__125.0__10.0__ multiply__12.5__3.6__ add__20.0__45.0__ round__65.0__ |
| a sum of money deposited at c . i . amounts to rs . num__2420 in num__2 years and to rs . num__2662 in num__3 years . find the rate percent ? <o> a ) num__10 <o> b ) num__99 <o> c ) num__17 <o> d ) num__11 <o> e ) num__12 |
explanation : num__2420 - - - num__242 num__100 - - - ? = > num__10.0 answer : a <eor> a <eos> |
a |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| thin rectangular slab of potato was cut into two pieces for an osmosis lab . one piece is num__50 mm greater than the other . if the original uncut slab is num__600 mm what is the length of the other piece of the potato after it is cut . <o> a ) num__550 <o> b ) num__275 <o> c ) num__50 <o> d ) num__300 <o> e ) num__250 |
piece one as p num__1 and piece two as p num__2 . p num__2 + num__50 = p num__1 as eq . a p num__1 + p num__2 = num__600 as eq . b substitute eq . a into eq . b resulting in ( p num__2 + num__50 ) + p num__2 = num__600 mm . group like terms p num__2 + p num__2 = num__600 - num__50 ( num__2 * p num__1.0 ) = num__275.0 solves to p num__2 = num__275 . ansewr is b . <eor> b <eos> |
b |
multiply__1.0__275.0__ |
multiply__1.0__275.0__ |
| length of a rectangular plot is num__20 mtr more than its breadth . if the cost of fencing gthe plot at num__26.50 per meter is rs . num__5300 what is the length of the plot in mtr ? <o> a ) num__40 m <o> b ) num__60 m <o> c ) num__70 m <o> d ) num__90 m <o> e ) num__100 m |
let breadth = x metres . then length = ( x + num__20 ) metres . perimeter = num__5300 m = num__200 m . num__26.50 num__2 [ ( x + num__20 ) + x ] = num__200 num__2 x + num__20 = num__100 num__2 x = num__80 x = num__40 . hence length = x + num__20 = num__60 m b <eor> b <eos> |
b |
divide__5300.0__26.5__ divide__200.0__2.0__ subtract__100.0__20.0__ multiply__20.0__2.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
divide__5300.0__26.5__ divide__200.0__2.0__ subtract__100.0__20.0__ multiply__20.0__2.0__ add__20.0__40.0__ add__20.0__40.0__ |
| two trains num__210 m and num__260 m long run at the speed of num__60 km / hr and num__40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? <o> a ) num__16.92 sec <o> b ) num__14.8 sec <o> c ) num__10.68 sec <o> d ) num__10.8 sec <o> e ) num__20.8 sec |
relative speed = num__60 + num__40 = num__100 km / hr . = num__100 * num__0.277777777778 = num__27.7777777778 m / sec . distance covered in crossing each other = num__210 + num__260 = num__470 m . required time = num__470 * num__0.036 = num__16.92 = num__16.92 sec . answer : a <eor> a <eos> |
a |
add__60.0__40.0__ add__210.0__260.0__ multiply__0.036__470.0__ round__16.92__ |
add__60.0__40.0__ add__210.0__260.0__ multiply__0.036__470.0__ multiply__0.036__470.0__ |
| marcella has num__23 pairs of shoes . if she loses num__9 individual shoes what is the greatest number of matching pairs she could have left ? <o> a ) num__21 <o> b ) num__20 <o> c ) num__19 <o> d ) num__18 <o> e ) num__15 |
marcella has num__23 pairs of shoes and loses num__9 shoes . to minimize the loss of identical pairs of shoes we want marcella to lose as many identical pairs as possible . this would yield num__4 identical pairs and num__1 additional shoe ( destroying num__5 pairs of shoes ) . the num__23 pairs of shoes minus the num__5 ' destroyed ' pairs yields num__20 pairs that still fulfill the requirements . answer : d <eor> d <eos> |
d |
subtract__9.0__4.0__ multiply__4.0__5.0__ subtract__23.0__5.0__ |
subtract__9.0__4.0__ multiply__4.0__5.0__ subtract__23.0__5.0__ |
| one hour after matthew started waking from q to y a distance of num__45 km johnny started walking along the same road from y to q . matthew ' s walking rate was num__3 km per hour and johnny ' s was num__4 km per hour how many km had johnny walked when they met ? <o> a ) num__24 <o> b ) num__23 <o> c ) num__22 <o> d ) num__21 <o> e ) num__19.5 |
just an alternative method . . . after the first hour the distance q is num__42 km ( num__45 - num__3 ) . now the problem can be treated as if bothof them started at the same time . since the speeds are in the ratio num__3 : num__4 the distances will also be in the same ratio . splitting num__42 in that ratio we get num__18 : num__24 . so answer is num__24 . hence a . <eor> a <eos> |
a |
subtract__45.0__3.0__ subtract__42.0__18.0__ round__24.0__ |
subtract__45.0__3.0__ subtract__42.0__18.0__ subtract__42.0__18.0__ |
| num__63 persons can repair a road in num__12 days working num__5 hours a day . in how many days will num__30 persons working num__6 hours a day complete the work ? <o> a ) num__21 <o> b ) num__23 <o> c ) num__24 <o> d ) num__25 <o> e ) num__26 |
according to the chain rule m num__1 x t num__1 = m num__2 x t num__2 therefore num__63 x num__12 x num__5 = num__30 x num__6 x x x = num__21 hence the number of days = num__21 . answer : a <eor> a <eos> |
a |
subtract__6.0__5.0__ divide__12.0__6.0__ round__21.0__ |
subtract__6.0__5.0__ divide__12.0__6.0__ round__21.0__ |
| if the length of the side of a square is doubled what is the ratio of the areas of the original square to the area of the new square ? <o> a ) num__0.25 <o> b ) num__0.5 <o> c ) num__0.75 <o> d ) num__0.2 <o> e ) num__0.4 |
if x be the side of the original square then its area is equal to x num__2 * square if x is doubled to num__2 x then the new area is equal to ( num__2 x ) num__2 * = num__4 x num__2 * the ratio of the areas of the original square to the area of the new square x num__2 * / ( num__4 x num__2 * ) = num__0.25 or num__1 : num__4 correct answer is a ) num__0.25 <eor> a <eos> |
a |
square_perimeter__0.25__ multiply__0.25__1.0__ |
multiply__0.25__4.0__ multiply__0.25__1.0__ |
| the average salary of all the workers in a workshop is rs . num__9000 . the average salary of num__6 technicians is rs . num__12000 and the average salary of the rest is rs . num__6000 . the total number of workers in the workshop is ? <o> a ) a ) num__23 <o> b ) b ) num__21 <o> c ) c ) num__52 <o> d ) d ) num__56 <o> e ) e ) num__12 |
let the total number of workers be x . then num__9000 x = ( num__12000 * num__6 ) + num__6000 ( x - num__6 ) = > num__3000 x = num__36000 = x = num__12 . answer : e <eor> e <eos> |
e |
subtract__9000.0__6000.0__ multiply__6.0__6000.0__ divide__36000.0__3000.0__ divide__36000.0__3000.0__ |
subtract__9000.0__6000.0__ multiply__6.0__6000.0__ divide__36000.0__3000.0__ divide__36000.0__3000.0__ |
| susan drove an average speed of num__30 miles per hour for the first num__40 miles of a trip then at a average speed of num__15 miles / hr for the remaining num__40 miles of the trip if she made no stops during the trip what was susan ' s avg speed in miles / hr for the entire trip <o> a ) num__35 <o> b ) num__20 <o> c ) num__45 <o> d ) num__50 <o> e ) num__55 |
avg . speed = total distance / total time total distance = num__80 miles total time = num__1.33333333333 + num__2.66666666667 = num__4 avg . speed = num__20 . answer - b <eor> b <eos> |
b |
divide__40.0__30.0__ divide__40.0__15.0__ add__1.3333__2.6667__ divide__80.0__4.0__ round__20.0__ |
divide__40.0__30.0__ divide__40.0__15.0__ add__1.3333__2.6667__ divide__80.0__4.0__ subtract__40.0__20.0__ |
| in num__10 years p will be as old as q is now . forty years ago q was twice as old as p was then . how old is p now ? <o> a ) num__30 <o> b ) num__35 <o> c ) num__40 <o> d ) num__45 <o> e ) num__50 |
q = p + num__10 q - num__40 = num__2 ( p - num__40 ) p - num__30 = num__2 p - num__80 p = num__50 the answer is e . <eor> e <eos> |
e |
triple__10.0__ twice__40.0__ add__10.0__40.0__ add__10.0__40.0__ |
subtract__40.0__10.0__ twice__40.0__ add__10.0__40.0__ add__10.0__40.0__ |
| two numbers are in the ratio num__3 : num__5 . if num__9 be subtracted from each they are in the ratio of num__9 : num__17 . the first number is : <o> a ) num__24.4 <o> b ) num__18.4 <o> c ) num__12.4 <o> d ) num__14.4 <o> e ) num__15.4 |
( num__3 x - num__9 ) : ( num__4 x - num__9 ) = num__9 : num__17 x = num__4.8 = > num__3 x = num__14.4 answer : d <eor> d <eos> |
d |
subtract__9.0__5.0__ multiply__3.0__4.8__ multiply__3.0__4.8__ |
subtract__9.0__5.0__ multiply__3.0__4.8__ multiply__3.0__4.8__ |
| the bankers gain of a certain sum due num__2 years hence at num__10.0 per annum is rs num__24 . the percent worth is <o> a ) num__600 <o> b ) num__398 <o> c ) num__279 <o> d ) num__211 <o> e ) num__331 |
t . d = ( b . g * num__100 ) / ( rate * time ) ( num__24 * num__100 ) / ( num__10 * num__2 ) = num__120 . p . w = ( num__100 * t . d ) / ( rate * time ) = ( num__100 * num__120 ) / ( num__10 * num__2 ) = num__600 answer : a <eor> a <eos> |
a |
percent__100.0__600.0__ |
percent__100.0__600.0__ |
| if num__2994 Ã · num__14.5 = num__175 then num__29.94 Ã · num__1.45 = ? <o> a ) num__17.1 <o> b ) num__17.3 <o> c ) num__17.5 <o> d ) num__17.7 <o> e ) num__17.2 |
num__29.94 / num__1.45 = num__299.4 / num__14.5 = ( num__2994 / num__14.5 ) x num__0.1 ) [ here substitute num__175 in the place of num__2994 / num__14.5 ] = num__17.5 = num__17.5 answer is c . <eor> c <eos> |
c |
divide__29.94__299.4__ multiply__175.0__0.1__ multiply__175.0__0.1__ |
divide__29.94__299.4__ multiply__175.0__0.1__ multiply__175.0__0.1__ |
| a cupcake recipe calls for num__2 cups of sugar and makes num__2 dozen cupcakes . a caterer will supply cupcakes for num__150 guests at a birthday party in which there should be at least num__2 cupcakes per guest . what is the minimum amount of sugar needed ? <o> a ) num__24 cups <o> b ) num__75 cups <o> c ) num__2 cups <o> d ) num__15 cups <o> e ) num__250 cups |
the proportion equation can be used . let c = sugar needed . note : num__12 = num__1 dozen proportion : a / b = c / d num__2 cups of sugar ( a ) makes num__24 cupcakes ( b ) . ( num__2 ) ( num__150 guests ) = num__300 ( d ) is the least number cupcakes needed . proportion : num__0.0833333333333 = c / num__300 c = ( num__2 ) ( num__300 ) / num__24 = num__25.0 = num__24 . at least num__24 cups are needed . answer is a <eor> a <eos> |
a |
multiply__2.0__12.0__ multiply__2.0__150.0__ divide__2.0__24.0__ add__1.0__24.0__ round__24.0__ |
multiply__2.0__12.0__ multiply__2.0__150.0__ divide__2.0__24.0__ divide__300.0__12.0__ divide__24.0__1.0__ |
| the least number which must be subtracted from num__820 to make it exactly divisible by num__9 is : <o> a ) a ) num__2 <o> b ) b ) num__3 <o> c ) c ) num__1 <o> d ) d ) num__5 <o> e ) e ) num__6 |
on dividing num__820 by num__9 we get remainder = num__1 therefore required number to be subtracted = num__1 answer : c <eor> c <eos> |
c |
reverse__1.0__ |
reverse__1.0__ |
| a man buys s cycle for $ num__100 and sells it at a loss of num__50.0 . what is the selling price of the cycle ? <o> a ) $ num__50 <o> b ) $ num__60 <o> c ) $ num__70 <o> d ) $ num__80 <o> e ) $ num__90 |
s . p . = num__50.0 of the $ num__100 = num__0.5 * num__100 = $ num__50 answer is a <eor> a <eos> |
a |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| what is the value of the num__201 st term of asequence if the first term of the sequence is num__2 and each successive term is num__4 more thanthe term immediately preceding it ? <o> a ) num__801 <o> b ) num__802 <o> c ) num__803 <o> d ) num__804 <o> e ) num__805 |
here the first term ` ` a ' ' = num__2 and the common difference ` ` d ' ' = num__4 and the sequence is an a . p . finding the num__201 st term use the following formula tn = a + ( n - num__1 ) d = num__2 + ( num__201 - num__1 ) * num__4 = num__802 answer : b <eor> b <eos> |
b |
multiply__802.0__1.0__ |
multiply__802.0__1.0__ |
| in x game of billiards x can give y num__20 points in num__60 and he can give z num__30 points in num__60 . how many points can y give z in x game of num__100 ? <o> a ) num__30 <o> b ) num__20 <o> c ) num__25 <o> d ) num__40 <o> e ) num__50 |
c num__25 x scores num__60 while y score num__40 and z scores num__30 . the number of points that z scores when y scores num__100 = ( num__100 * num__30 ) / num__40 = num__25 * num__3 = num__75 . in x game of num__100 points y gives ( num__100 - num__75 ) = num__25 points to c . <eor> c <eos> |
c |
subtract__60.0__20.0__ divide__60.0__20.0__ subtract__100.0__25.0__ subtract__100.0__75.0__ |
subtract__60.0__20.0__ divide__60.0__20.0__ subtract__100.0__25.0__ subtract__100.0__75.0__ |
| at a loading dock each worker on the night crew loaded num__0.75 as many boxes as each worker on the day crew . if the night crew has num__0.75 as many workers as the day crew what fraction of all the boxes loaded by the two crews did the day crew load ? <o> a ) num__0.6 <o> b ) num__0.8 <o> c ) num__0.733333333333 <o> d ) num__0.64 <o> e ) num__0.76 |
let x be the number of workers on the day crew . let y be the number of boxes loaded by each member of the day crew . then the number of boxes loaded by the day crew is xy . the number of boxes loaded by the night crew is ( num__3 x / num__4 ) ( num__3 y / num__4 ) = num__9 xy / num__16 the total number of boxes is xy + num__9 xy / num__16 = num__25 xy / num__16 the fraction loaded by the day crew is xy / ( num__25 xy / num__16 ) = num__0.64 the answer is d . <eor> d <eos> |
d |
divide__3.0__0.75__ add__9.0__16.0__ divide__16.0__25.0__ round__0.64__ |
divide__3.0__0.75__ add__9.0__16.0__ divide__16.0__25.0__ divide__16.0__25.0__ |
| two pipes a and b can fill a tank in num__6 hours and num__4 hours respectively . if they are opened on alternate hours and if pipe a is opened first in how many hours the tank shall be full ? <o> a ) num__3 hours <o> b ) num__5 hours <o> c ) num__7 hours <o> d ) num__10 hours <o> e ) num__15 hours |
explanation : ( a + b ) ' s num__2 hour ' s work when opened = num__0.166666666667 + num__0.25 = num__0.416666666667 ( a + b ) ′ s num__4 hour ' s work = num__0.416666666667 ∗ num__2 = num__0.833333333333 remaining work = num__1 − num__0.833333333333 = num__0.166666666667 now its a turn in num__5 th hour num__0.166666666667 work will be done by a in num__1 hourtotal time = num__4 + num__1 = num__5 hours option b <eor> b <eos> |
b |
subtract__6.0__4.0__ add__0.25__0.1667__ multiply__4.0__0.25__ subtract__6.0__1.0__ round__5.0__ |
subtract__6.0__4.0__ add__0.25__0.1667__ add__0.8333__0.1667__ add__4.0__1.0__ add__4.0__1.0__ |
| the number of years of service of the eight employees in a production department are num__15 num__10 num__9 num__17 num__6 num__3 num__14 and num__16 . what is the range in the number of years of service of the eight employees ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__12 <o> d ) num__13 <o> e ) num__14 |
= num__17 - num__3 = num__14 answer e <eor> e <eos> |
e |
subtract__17.0__3.0__ |
subtract__17.0__3.0__ |
| num__1397 x num__1397 = ? <o> a ) num__1951609 <o> b ) num__1951601 <o> c ) num__1951602 <o> d ) num__1951603 <o> e ) num__1951604 |
num__1397 x num__1397 = ( num__1397 ) num__2 = ( num__1400 - num__3 ) num__2 = ( num__1400 ) num__2 + ( num__3 ) num__2 - ( num__2 x num__1400 x num__3 ) = num__1960000 + num__9 - num__8400 = num__1960009 - num__8400 = num__1951609 . answer : a <eor> a <eos> |
a |
subtract__1400.0__1397.0__ add__1960000.0__9.0__ subtract__1960009.0__8400.0__ subtract__1960009.0__8400.0__ |
subtract__1400.0__1397.0__ add__1960000.0__9.0__ subtract__1960009.0__8400.0__ subtract__1960009.0__8400.0__ |
| if it would take one machine num__8 minutes to fill a large production order and another machine num__12 minutes to fill the same order how many minutes would it take both machines working together at their respective rates to fill the order ? <o> a ) num__4 num__0.8 <o> b ) num__5 <o> c ) num__5 num__0.454545454545 <o> d ) num__5 num__0.5 <o> e ) num__11 |
since machine num__1 - m num__1 take num__8 mins to fill the order the work done by m num__1 in a min = num__0.125 machine num__2 - m num__2 take num__12 mins to fill the same order and work done by m num__2 in a min = num__0.0833333333333 total work done by m num__1 and m num__2 in a min = num__0.125 + num__0.0833333333333 = num__0.208333333333 time needed for m num__1 and m num__2 to complete the order = num__4.8 = num__4 num__0.8 answer a <eor> a <eos> |
a |
divide__1.0__8.0__ divide__1.0__12.0__ add__0.125__0.0833__ divide__8.0__2.0__ subtract__4.8__4.0__ round__4.0__ |
divide__1.0__8.0__ divide__1.0__12.0__ add__0.125__0.0833__ subtract__12.0__8.0__ subtract__4.8__4.0__ subtract__8.0__4.0__ |
| if $ num__5000 is invested in an account at a simple annual rate of r percent the interest is $ num__250 . when $ num__15000 is invested at the same interest rate what is the interest from the investment ? <o> a ) $ num__700 <o> b ) $ num__850 <o> c ) $ num__800 <o> d ) $ num__750 <o> e ) $ num__900 |
- > num__250 / num__5000 = num__5.0 and num__15000 * num__5.0 = num__750 . thus d is the answer . <eor> d <eos> |
d |
percent__5.0__15000.0__ percent__5.0__15000.0__ |
percent__5.0__15000.0__ percent__5.0__15000.0__ |
| the area of sector of a circle whose radius is num__12 metro and whose angle at the center is num__42 ° is ? <o> a ) num__52.7 m num__2 <o> b ) num__57.8 m num__2 <o> c ) num__52.8 m num__2 <o> d ) num__72.8 m num__2 <o> e ) num__52.8 m num__2 |
num__0.116666666667 * num__3.14285714286 * num__12 * num__12 = num__52.8 m num__2 answer : c <eor> c <eos> |
c |
triangle_area__52.8__2.0__ |
triangle_area__52.8__2.0__ |
| a num__300 m long train crosses a platform in num__39 sec while it crosses a signal pole in num__18 sec . what is the length of the platform ? <o> a ) num__298 m <o> b ) num__350 m <o> c ) num__267 m <o> d ) num__276 m <o> e ) num__268 m |
speed = num__16.6666666667 = num__16.6666666667 m / sec . let the length of the platform be x meters . then ( x + num__300 ) / num__39 = num__16.6666666667 num__3 x + num__900 = num__1950 = > x = num__350 m . answer : b <eor> b <eos> |
b |
divide__300.0__18.0__ multiply__300.0__3.0__ round__350.0__ |
divide__300.0__18.0__ multiply__300.0__3.0__ round__350.0__ |
| amount of bacteria present time amount num__1 : num__00 p . m . num__14.0 grams num__4 : num__00 p . m . x grams num__7 : num__00 p . m . num__18.4 grams data for a certain biology experiment are given in the table above . if the amount of bacteria present increased by the same fraction during each of the two num__3 - hour periods shown how many grams of bacteria were present at num__4 : num__00 p . m . ? <o> a ) num__14.0 <o> b ) num__16.1 <o> c ) num__16.0 <o> d ) num__12.3 <o> e ) num__12.4 |
the question says that bacteria increased by same fraction not by same amount in num__2 intervals of num__3 hours . let x represent the amount of bacteria present at num__4 : num__00 pm . since the fractional increase must remain constant from num__1 to num__4 pm as it is from num__4 pm to num__7 pm : fractional increase from num__1 pm to num__4 pm = x / num__14.0 fractional increase from num__4 pm to num__7 pm = num__18.4 / x x \ num__14 = num__18.4 \ x x ^ num__2 = num__18.4 * num__14 x = num__16 answer : c <eor> c <eos> |
c |
divide__14.0__7.0__ add__14.0__2.0__ multiply__1.0__16.0__ |
divide__14.0__7.0__ add__14.0__2.0__ multiply__1.0__16.0__ |
| car x began traveling at an average speed of num__35 miles per hour . after num__72 minutes car y began traveling at an average speed of num__65 miles per hour . when both cars had traveled the same distance both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ? <o> a ) num__15 <o> b ) num__20 <o> c ) num__40 <o> d ) num__47 <o> e ) num__49 |
car y began travelling after num__72 minutes or num__1.2 hours . let t be the time for which car y travelled before it stopped . both cars stop when they have travelled the same distance . so num__35 ( t + num__1.2 ) = num__65 t t = num__1.4 distance traveled by car x from the time car y began traveling until both cars stopped is num__35 x num__1.4 = num__49 miles answer : - e <eor> e <eos> |
e |
multiply__35.0__1.4__ round__49.0__ |
multiply__35.0__1.4__ round__49.0__ |
| a tank is filled by three pipes with uniform flow . the first two pipes operating simultaneously fill the tank in the same during which the tank is filled by the third pipe alone . the second pipe fills the tank num__5 hours faster than the first pipe and num__4 hours slower than the third pipe . the time required by the first pipe is ? <o> a ) num__1 <o> b ) num__16 <o> c ) num__15 <o> d ) num__88 <o> e ) num__19 |
suppose first pipe alone takes x hours to fill the tank . then second and third pipes will take ( x - num__5 ) and ( x - num__9 ) hours respectively to fill the tank . num__1 / x + num__1 / ( x - num__5 ) = num__1 / ( x - num__9 ) ( num__2 x - num__5 ) ( x - num__9 ) = x ( x - num__5 ) x num__2 - num__18 x + num__45 = num__0 ( x - num__15 ) ( x - num__3 ) = num__0 = > x = num__15 . answer : c <eor> c <eos> |
c |
add__5.0__4.0__ subtract__5.0__4.0__ multiply__2.0__9.0__ multiply__5.0__9.0__ subtract__5.0__2.0__ round__15.0__ |
add__5.0__4.0__ subtract__5.0__4.0__ multiply__2.0__9.0__ multiply__5.0__9.0__ subtract__5.0__2.0__ divide__45.0__3.0__ |
| bert left the house with n dollars . he spent num__0.25 of this at the hardware store then $ num__9 at the dry cleaners and then half of what was left at the grocery store . when he got home he had $ num__12 left in his pocket . what was the value of n ? <o> a ) $ num__36 <o> b ) $ num__44 <o> c ) $ num__52 <o> d ) $ num__60 <o> e ) $ num__68 |
started to test answer b if he had num__44 then he spent num__11 at hardware store now he was left with num__33 $ he spent num__9 dollars on cleaning thus he remained with num__24 $ he then spent num__0.5 of num__24 or num__12 and was left with num__12 . hence the only option that can be right is b <eor> b <eos> |
b |
multiply__0.25__44.0__ subtract__44.0__11.0__ subtract__33.0__9.0__ divide__12.0__24.0__ add__33.0__11.0__ |
multiply__0.25__44.0__ subtract__44.0__11.0__ subtract__33.0__9.0__ divide__12.0__24.0__ add__33.0__11.0__ |
| a person travels equal distances with speeds of num__4 km / hr num__5 km / hr and num__6 km / hr and takes a total time of num__47 minutes . the total distance is ? <o> a ) num__2 km <o> b ) num__4 km <o> c ) num__7 km <o> d ) num__9 km <o> e ) num__5 km |
let the total distance be num__3 x km . then x / num__4 + x / num__5 + x / num__6 = num__0.783333333333 num__37 x / num__60 = num__0.783333333333 = > x = num__1.27 total distance = num__3 * num__1.27 = num__3.81 km . answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__3.0__1.27__ round__4.0__ |
hour_to_min_conversion__ multiply__3.0__1.27__ round__4.0__ |
| jackson consulting reduced headcount by laying off num__20.0 of its employees . if the earnings of each of the remaining employees were the same before and after the layoffs what is the maximum percentage by which the average earnings of the employees at the company could have increased from its level before the layoffs to its level after the layoffs ? <o> a ) num__15 <o> b ) num__16 <o> c ) num__20 <o> d ) num__25 <o> e ) num__40 |
since we are looking to maximize the average with no change in earnings . . let all the laid off employees were ding social service : wink : that is they were working without any pay . . so if all num__100.0 say num__100 earlier were earning num__100 with average = num__1 . . now num__80.0 are earning num__100 so average = num__1.25 = num__1.25 . . increase = num__1.25 - num__1 = . num__25.0 = . num__25.0 * num__100 = num__25.0 answer : d <eor> d <eos> |
d |
subtract__100.0__20.0__ divide__100.0__80.0__ multiply__20.0__1.25__ multiply__20.0__1.25__ |
subtract__100.0__20.0__ divide__100.0__80.0__ multiply__20.0__1.25__ multiply__20.0__1.25__ |
| length of train is num__130 meters and speed of train is num__45 km / hour . this train can pass a bridge in num__30 seconds then find the length of the bridge <o> a ) num__230 meters <o> b ) num__235 meters <o> c ) num__240 meters <o> d ) num__245 meters <o> e ) none of these |
explanation : let the length of bridge is x [ as always we do : ) ] speed of train is = num__45 * ( num__0.277777777778 ) m / sec = num__12.5 m / sec time = num__30 seconds total distance = num__130 + x we know speed = distance / time so num__130 + x / num__30 = num__12.5 = > num__2 ( num__130 + x ) = num__750 x = num__245 meters so length of the bridge is num__245 meters option d <eor> d <eos> |
d |
round__245.0__ |
round__245.0__ |
| kevin professes to sell his goods at the cost price but he made use of num__300 grams instead of a kg what is the gain percent ? <o> a ) num__33 num__0.333333333333 % <o> b ) num__12 num__0.333333333333 % <o> c ) num__66 num__0.666666666667 % <o> d ) num__32 num__0.333333333333 % <o> e ) num__34 num__0.333333333333 % |
num__300 - - - num__100 num__100 - - - ? = > num__33 num__0.333333333333 % answer : a <eor> a <eos> |
a |
percent__100.0__33.0__ |
percent__100.0__33.0__ |
| a man buys an article for $ num__100 . and sells it for $ num__125 . find the gain percent ? <o> a ) num__10.0 <o> b ) num__15.0 <o> c ) num__25.0 <o> d ) num__20.0 <o> e ) num__30 % |
c . p . = $ num__100 s . p . = $ num__125 gain = $ num__25 gain % = num__0.25 * num__100 = num__25.0 answer is c <eor> c <eos> |
c |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| a woman traveled num__20 kilometer in num__14 meter per second and another num__20 kilometer in num__18 miles per second . what is the average speed of the entire trip ? <o> a ) num__12.25 <o> b ) num__13.75 <o> c ) num__15.75 <o> d ) num__18.25 <o> e ) num__19.75 |
since the distance traveled is same we can apply direct formula = num__2 xs num__1 xs num__2 / ( s num__1 + s num__2 ) num__2 x num__14 x num__0.5625 = num__15.75 . ' c ' is the answer . <eor> c <eos> |
c |
subtract__20.0__18.0__ round__15.75__ |
subtract__20.0__18.0__ divide__15.75__1.0__ |
| the average age of husband wife and their child num__3 years ago was num__27 years and that of wife and the child num__5 years ago was num__21 years . the present age of the husband is <o> a ) num__40 <o> b ) num__38 <o> c ) num__45 <o> d ) num__55 <o> e ) num__65 |
explanation : sum of the present ages of husband wife and child = ( num__27 * num__3 + num__3 * num__3 ) years = num__90 years . sum of the present ages of wife and child = ( num__21 * num__2 + num__5 * num__2 ) years = num__52 years . husband ' s present age = ( num__90 - num__52 ) years = num__38 years option b <eor> b <eos> |
b |
subtract__5.0__3.0__ subtract__90.0__52.0__ subtract__90.0__52.0__ |
subtract__5.0__3.0__ subtract__90.0__52.0__ subtract__90.0__52.0__ |
| some of num__45.0 - intensity red paint is replaced with num__25.0 solution of red paint such that the new paint intensity is num__40.0 . what fraction of the original paint was replaced ? <o> a ) num__0.4 <o> b ) num__0.25 <o> c ) num__0.2 <o> d ) num__0.142857142857 <o> e ) num__0.666666666667 |
let total paint = num__1 let amount replaced = x num__45 ( num__1 - x ) + num__25 x = num__40 x = num__0.25 answer : b <eor> b <eos> |
b |
percent__25.0__1.0__ percent__25.0__1.0__ |
percent__25.0__1.0__ percent__25.0__1.0__ |
| a train running at a speed of num__60 kmph crosses a pole in num__15 seconds . what is the length of the train ? <o> a ) num__120 m <o> b ) num__180 m <o> c ) num__190 m <o> d ) num__250 m <o> e ) num__160 m |
num__60 kmph = num__16.6666666667 m / sec num__16.6666666667 * num__15 = num__250 m answer : d <eor> d <eos> |
d |
round__250.0__ |
round__250.0__ |
| on a test consisting of num__30 questions . no . of wrong answers is num__50.0 less than number of right answers . each answer was either right or wrong . what is the ratio of right answers to wrong answers ? <o> a ) num__1 : num__3 <o> b ) num__2 : num__3 <o> c ) num__2 : num__1 <o> d ) num__3 : num__5 <o> e ) num__3 : num__7 |
sol . let the number of wrong answer - x and number of right answer = num__30 - x x = num__0.5 of ( num__30 - x ) x = num__10 required ratio = num__20 : num__10 = num__2 : num__1 c <eor> c <eos> |
c |
subtract__30.0__10.0__ reverse__0.5__ multiply__0.5__2.0__ reverse__0.5__ |
subtract__30.0__10.0__ reverse__0.5__ multiply__0.5__2.0__ reverse__0.5__ |
| a car traveled num__560 miles per tankful of gasoline on the highway and num__336 miles per tankful of gasoline in the city . if the car traveled num__6 fewer miles per gallon in the city than on the highway how many miles per gallon did the car travel in the city ? <o> a ) num__9 <o> b ) num__16 <o> c ) num__21 <o> d ) num__22 <o> e ) num__27 |
let the speed in highway be h mpg and in city be c mpg . h = c + num__6 h miles are covered in num__1 gallon num__462 miles will be covered in num__462 / h . similarly c miles are covered in num__1 gallon num__336 miles will be covered in num__336 / c . both should be same ( as car ' s fuel capacity does not change with speed ) = > num__336 / c = num__560 / h = > num__336 / c = num__560 / ( c + num__6 ) = > num__336 c + num__336 * num__6 = num__560 c = > c = num__336 * num__0.0267857142857 = num__9 answer a . <eor> a <eos> |
a |
round__9.0__ |
divide__9.0__1.0__ |
| the sum of four consecutive even numbers is num__292 . what would be the smallest number ? <o> a ) num__49 <o> b ) num__68 <o> c ) num__60 <o> d ) num__57 <o> e ) number num__70 |
e num__70 let the four consecutive even numbers be num__2 ( x - num__2 ) num__2 ( x - num__1 ) num__2 x num__2 ( x + num__1 ) their sum = num__8 x - num__4 = num__292 = > x = num__37 smallest number is : num__2 ( x - num__2 ) = num__70 . <eor> e <eos> |
e |
divide__8.0__2.0__ round__70.0__ |
divide__8.0__2.0__ round__70.0__ |
| the average of num__6 no . ' s is num__2.5 . the average of num__2 of them is num__1.1 while the average of the other num__2 is num__1.4 . what is the average of the remaining num__2 no ' s ? <o> a ) num__2.3 <o> b ) num__2.6 <o> c ) num__3.6 <o> d ) num__4.5 <o> e ) num__5 |
sum of the remaining two numbers = ( num__2.5 * num__6 ) - [ ( num__1.1 * num__2 ) + ( num__1.4 * num__2 ) ] = num__15 - ( num__2.2 + num__2.8 ) = num__15 - num__5 = num__10 required average = ( num__5.0 ) = num__5 answer : e <eor> e <eos> |
e |
multiply__6.0__2.5__ multiply__2.0__1.1__ multiply__2.0__1.4__ multiply__2.5__2.0__ multiply__2.0__5.0__ multiply__2.5__2.0__ |
multiply__6.0__2.5__ multiply__2.0__1.1__ multiply__2.0__1.4__ multiply__2.5__2.0__ multiply__2.0__5.0__ multiply__2.5__2.0__ |
| in a recent election ms . robbins received num__5000 votes cast by independent voters that is voters not registered with a specific political party . she also received num__10 percent of the votes cast by those voters registered with a political party . if n is the total number of votes cast in the election and num__40 percent of the votes cast were cast by independent voters which of the following represents the number of votes that ms . robbins received ? <o> a ) num__0.06 n + num__3200 <o> b ) num__0.06 n + num__5000 <o> c ) num__0.4 n + num__7200 <o> d ) num__0.1 n + num__8000 <o> e ) num__0.06 n + num__8 |
000 |
total vote = n vote cast by independent voters = num__0.4 n vote cast by registered voters = num__0.6 n vote received by ms . robbins = num__5000 + num__10.0 of vote be registered voters = num__5000 + num__0.06 n answer choice b <eor> b <eos> |
b |
b |
| if y = num__30 p and p is prime what is the greatest common factor of y and num__6 p in terms of p ? <o> a ) p <o> b ) num__2 p <o> c ) num__5 p <o> d ) num__6 p <o> e ) p ^ num__2 |
y = num__30 p other number is num__6 p then gcf ( num__30 p num__6 p ) = num__6 p ; d is the correct answer <eor> d <eos> |
d |
gcd__30.0__6.0__ |
gcd__30.0__6.0__ |
| m = { - num__6 - num__5 - num__4 - num__3 - num__2 } t = { - num__2 - num__1 num__1 num__2 num__3 num__4 } if an integer is to be randomly selected from set m above and an integer is to be randomly selected from set t above what is the probability that the product of the two integers will be negative ? <o> a ) num__0 <o> b ) num__0.333333333333 <o> c ) num__0.4 <o> d ) num__0.5 <o> e ) num__0.8 |
answer e . total # of outcomes : num__5 * num__6 = num__30 # of outcomes where product is - ve : ( - num__61 ) ( - num__62 ) ( - num__63 ) . . . hence total : num__24 probability : num__0.8 = num__0.8 <eor> e <eos> |
e |
multiply__6.0__5.0__ add__1.0__61.0__ add__2.0__61.0__ multiply__6.0__4.0__ divide__4.0__5.0__ divide__4.0__5.0__ |
multiply__6.0__5.0__ add__1.0__61.0__ add__2.0__61.0__ multiply__6.0__4.0__ divide__4.0__5.0__ multiply__1.0__0.8__ |
| in an election contested by two parties party x secured num__20.0 of the total votes more than party y . if party y got num__130000 votes by how many votes did it lose the election ? <o> a ) num__26000 <o> b ) num__40000 <o> c ) num__65000 <o> d ) num__220000 <o> e ) num__100 |
000 |
x + y = num__100 - - - - ( num__1 ) x - y = num__20 - - - - ( num__2 ) solving ( num__1 ) and ( num__2 ) x = num__60.0 and y = num__40.0 num__45.0 of total = num__130000 total = num__325000 x = num__195000 difference = num__195000 - num__130000 = num__65000 answer : c <eor> c <eos> |
c |
c |
| the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is num__45 kmph find the speed of the stream ? <o> a ) num__15 <o> b ) num__19 <o> c ) num__14 <o> d ) num__18 <o> e ) num__16 |
the ratio of the times taken is num__2 : num__1 . the ratio of the speed of the boat in still water to the speed of the stream = ( num__2 + num__1 ) / ( num__2 - num__1 ) = num__3.0 = num__3 : num__1 speed of the stream = num__15.0 = num__15 kmph . answer : a <eor> a <eos> |
a |
add__1.0__2.0__ divide__45.0__3.0__ round__15.0__ |
add__1.0__2.0__ divide__45.0__3.0__ divide__45.0__3.0__ |
| carina has num__130 ounces of coffee divided into num__5 - and num__10 - ounce packages . if she has num__2 more num__5 - ounce packages than num__10 - ounce packages how many num__10 - ounce packages does she have ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
lets say num__5 and num__10 ounce packages be x and y respectively . given that num__5 x + num__10 y = num__130 and x = y + num__2 . what is the value of y . substituting the x in first equation num__5 y + num__10 + num__10 y = num__130 - > y = num__8.0 . = num__8 d <eor> d <eos> |
d |
subtract__10.0__2.0__ subtract__10.0__2.0__ |
subtract__10.0__2.0__ subtract__10.0__2.0__ |
| the speed of the boat in still water in num__12 kmph . it can travel downstream through num__45 kms in num__3 hrs . in what time would it cover the same distance upstream ? <o> a ) num__8 hours <o> b ) num__6 hours <o> c ) num__4 hours <o> d ) num__5 hours <o> e ) num__6 hours |
still water = num__12 km / hr downstream = num__15.0 = num__15 km / hr upstream = > > still water = ( u + v / num__2 ) = > > num__12 = u + num__7.5 = num__9 km / hr so time taken in upstream = num__5.0 = num__5 hrs answer : d <eor> d <eos> |
d |
add__12.0__3.0__ divide__15.0__2.0__ subtract__12.0__3.0__ divide__45.0__9.0__ round__5.0__ |
add__12.0__3.0__ divide__15.0__2.0__ subtract__12.0__3.0__ divide__45.0__9.0__ divide__45.0__9.0__ |
| a man is num__24 years older than his son . in two years his age will be twice the age of his son . the present age of his son is <o> a ) num__28 <o> b ) num__77 <o> c ) num__22 <o> d ) num__44 <o> e ) num__12 |
let the son ' s present age be x years . then man ' s present age = ( x + num__24 ) years = > ( x + num__24 ) + num__2 = num__2 ( x + num__2 ) = > x + num__26 = num__2 x + num__4 so x = num__22 answer : c <eor> c <eos> |
c |
add__24.0__2.0__ subtract__24.0__2.0__ subtract__24.0__2.0__ |
add__24.0__2.0__ subtract__24.0__2.0__ subtract__24.0__2.0__ |
| the speeds of num__3 trains in the ratio num__10 : num__15 : num__20 . the ratio between time taken by them to travel the same distance is ? <o> a ) num__6 : num__4 : num__3 <o> b ) num__1 : num__2 : num__3 <o> c ) num__4 : num__8 : num__9 <o> d ) num__6 : num__5 : num__2 <o> e ) num__4 : num__8 : num__7 |
ratio of time taken is = num__0.1 : num__0.0666666666667 : num__0.05 = num__6 : num__4 : num__3 answer is a <eor> a <eos> |
a |
subtract__10.0__6.0__ round__6.0__ |
subtract__10.0__6.0__ round__6.0__ |
| a fruit seller had some oranges . he sells num__40.0 oranges and still has num__420 oranges . how many oranges he had originally ? <o> a ) num__700 <o> b ) num__710 <o> c ) num__720 <o> d ) num__730 <o> e ) num__740 |
num__60.0 of oranges = num__420 num__100.0 of oranges = ( num__420 × num__100 ) / num__6 = num__700 total oranges = num__700 answer : a <eor> a <eos> |
a |
percent__100.0__700.0__ |
percent__100.0__700.0__ |
| a b and c can do a work in num__15 num__2045 days respectively . in how many days they can complete the work together . <o> a ) num__5.2 days <o> b ) num__9.2 days <o> c ) num__8.2 days <o> d ) num__6.2 days <o> e ) num__7.2 days |
lcm = num__180 no of days = [ num__180 / ( num__12.0 + num__9.0 + num__4.0 ) = [ num__180 / ( num__12 + num__9 + num__4 ) ] = [ num__7.2 ] = num__7.2 days answer : e <eor> e <eos> |
e |
divide__180.0__15.0__ round__7.2__ |
divide__180.0__15.0__ round__7.2__ |
| in a garden there are num__10 rows and num__12 columns of mango trees . the distance between the two trees is num__2 metres and a distance of six metres is left from all sides of the boundary of the garden . what is the length of the garden ? <o> a ) num__32 <o> b ) num__34 <o> c ) num__26 <o> d ) num__28 <o> e ) num__30 |
between the num__12 mango trees there are num__11 gaps and each gap has num__2 meter length also num__6 meter is left from all sides of the boundary of the garden . hence length of the garden = ( num__11 Ã — num__2 ) + num__6 + num__6 = num__34 meter answer is b . <eor> b <eos> |
b |
divide__12.0__2.0__ round__34.0__ |
divide__12.0__2.0__ round__34.0__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__2387 <o> b ) num__209 <o> c ) num__240 <o> d ) num__278 <o> e ) num__121 |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x meters . then x + num__8.33333333333 = num__15 x + num__300 = num__540 x = num__240 m . answer : c <eor> c <eos> |
c |
multiply__20.0__15.0__ divide__300.0__36.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ divide__300.0__36.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
| an empty bottle weighs num__0.166666666667 th of the full bottle . when a certain percent of water was removed and the bottle was weighed the weight of the bottle turned out to be num__0.333333333333 rd of the bottle when it was full . what is the percent of water removed ? <o> a ) num__70.0 <o> b ) num__85.0 <o> c ) num__80.0 <o> d ) num__75.0 <o> e ) num__65 % |
explanation : let the weight of full bottle be num__6 kg . therefore weight of empty bottle is num__1 kg and that of water is num__5 kg . if x % of water is removed the weight of the bottle becomes num__2 kg . therefore the amount of water removed is num__4 kg . % of water removed = num__0.8 * num__100 = num__80.0 answer c <eor> c <eos> |
c |
percent__80.0__100.0__ |
percent__80.0__100.0__ |
| by selling num__100 pens a trader gains the cost of num__40 pens . find his gain percentage ? <o> a ) num__40.0 <o> b ) num__50.0 <o> c ) num__60.0 <o> d ) num__70.0 <o> e ) num__30 % |
let the cp of each pen be rs . num__1 . cp of num__100 pens = rs . num__100 profit = cost of num__40 pens = rs . num__40 profit % = num__0.4 * num__100 = num__40.0 answer : a <eor> a <eos> |
a |
percent__40.0__1.0__ percent__100.0__40.0__ |
percent__40.0__1.0__ percent__100.0__40.0__ |
| a train passes a man standing on the platform . if the train is num__150 meters long and its speed is num__54 kmph how much time it took in doing so ? <o> a ) num__10 sec <o> b ) num__13 sec <o> c ) num__14 sec <o> d ) num__15 sec <o> e ) num__16 sec |
d = num__150 s = num__54 * num__0.277777777778 = num__15 mps t = num__10.0 = num__10 sec answer : a <eor> a <eos> |
a |
divide__150.0__15.0__ round__10.0__ |
divide__150.0__15.0__ round__10.0__ |
| in the construction of a house some persons can do a piece of work in num__8 days . four times the number of such persons will do half of that work in : <o> a ) num__1 day <o> b ) num__2 days <o> c ) num__3 days <o> d ) num__4 days <o> e ) none of these |
explanation : solution : let x men can do the work in num__8 days and the required number of days be z . more men less days ( ip ) less work less days ( dp ) men num__4 x : x } : : num__8 : z work num__1 : num__0.5 . ' . ( num__4 x * num__1 * z ) = ( x * num__0.5 * num__8 ) = > z = num__1 answer : a <eor> a <eos> |
a |
divide__4.0__8.0__ round__1.0__ |
divide__4.0__8.0__ round__1.0__ |
| albert invested an amount of rs . num__8000 in a fd scheme for num__2 yrs at ci rate num__5 p . c . p . a . how much amount will albert get on maturity of the fd ? <o> a ) rs . num__7000 <o> b ) rs . num__7020 <o> c ) rs . num__7900 <o> d ) rs . num__8000 <o> e ) rs . num__8820 |
amount = rs . num__8000 x num__1 + num__5 num__2 num__100 = rs . num__8000 x num__21 x num__21 num__20 num__20 = rs . num__8820 . e <eor> e <eos> |
e |
percent__100.0__8820.0__ |
percent__100.0__8820.0__ |
| how many multiples of num__13 are less than num__6000 and also multiples of num__16 ? <o> a ) num__30 <o> b ) num__29 <o> c ) num__34 <o> d ) num__32 <o> e ) num__33 |
lcm of num__13 & num__16 = num__208 tried dividing num__6000 by num__208 got quotient num__28.84 ' so b is answer <eor> b <eos> |
b |
multiply__13.0__16.0__ add__13.0__16.0__ |
multiply__13.0__16.0__ add__13.0__16.0__ |
| a man whose speed is num__4.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at num__1.5 kmph find his average speed for the total journey ? <o> a ) num__7 <o> b ) num__4 <o> c ) num__6 <o> d ) num__9 <o> e ) num__3 |
m = num__45 s = num__1.5 ds = num__6 us = num__3 as = ( num__2 * num__6 * num__3 ) / num__9 = num__4 answer : b <eor> b <eos> |
b |
add__4.5__1.5__ subtract__4.5__1.5__ divide__3.0__1.5__ multiply__4.5__2.0__ round_down__4.5__ round_down__4.5__ |
add__4.5__1.5__ divide__4.5__1.5__ divide__3.0__1.5__ multiply__4.5__2.0__ divide__6.0__1.5__ divide__6.0__1.5__ |
| x starts a business with rs . num__45000 . y joins in the business after num__5 months with rs . num__30000 . what will be the ratio in which they should share the profit at the end of the year ? <o> a ) num__1 : num__2 <o> b ) num__2 : num__1 <o> c ) num__1 : num__3 <o> d ) num__3 : num__1 <o> e ) num__18 : num__7 |
explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = num__45000 Ã — num__12 : num__30000 Ã — num__7 = num__45 Ã — num__12 : num__30 Ã — num__7 = num__3 Ã — num__12 : num__2 Ã — num__7 = num__18 : num__7 answer : option e <eor> e <eos> |
e |
subtract__12.0__5.0__ subtract__5.0__3.0__ subtract__30.0__12.0__ subtract__30.0__12.0__ |
subtract__12.0__5.0__ subtract__5.0__3.0__ subtract__30.0__12.0__ subtract__30.0__12.0__ |
| each of the three people individually can complete a certain job in num__4 num__6 and num__8 hours respectively . what is the lowest fraction of the job that can be done in num__1 hour by num__2 of the people working together at their respective rates ? <o> a ) num__0.375 <o> b ) num__0.416666666667 <o> c ) num__0.291666666667 <o> d ) num__0.305555555556 <o> e ) num__0.270833333333 |
the two slowest people work at rates of num__0.166666666667 and num__0.125 of the job per hour . the sum of these rates is num__0.166666666667 + num__0.125 = num__0.291666666667 of the job per hour . the answer is c . <eor> c <eos> |
c |
divide__1.0__6.0__ divide__1.0__8.0__ add__0.1667__0.125__ multiply__1.0__0.2917__ |
divide__1.0__6.0__ divide__1.0__8.0__ add__0.1667__0.125__ add__0.1667__0.125__ |
| a volunteer organization is recruiting new members . in the fall they manage to increase their number by num__6.0 . by the spring however membership falls by num__19.0 . what is the total change in percentage from fall to spring ? <o> a ) - num__16.16 <o> b ) - num__15.15 <o> c ) - num__14.14 <o> d ) - num__13.13 <o> e ) - num__12.12 % |
( num__100.0 + num__6.0 ) * ( num__100.0 - num__19.0 ) = num__1.06 * . num__81 = num__0.8586 . num__1 - num__0.8586 = num__14.14 lost = - num__14.14 the answer is c the organization has lost num__14.14 of its total volunteers from fall to spring . <eor> c <eos> |
c |
subtract__100.0__19.0__ round_down__1.06__ multiply__1.0__14.14__ |
subtract__100.0__19.0__ round_down__1.06__ multiply__1.0__14.14__ |
| a circle graph shows how the megatech corporation allocates its research and development budget : num__14.0 microphotonics ; num__25.0 home electronics ; num__15.0 food additives ; num__29.0 genetically modified microorganisms ; num__8.0 industrial lubricants ; and the remainder for basic astrophysics . if the arc of each sector of the graph is proportional to the percentage of the budget it represents how many degrees of the circle are used to represent basic astrophysics research ? <o> a ) num__8 ° <o> b ) num__10 ° <o> c ) num__18 ° <o> d ) num__33 ° <o> e ) num__52 ° |
here all percentage when summed we need to get num__100.0 . as per data num__14 + num__25 + num__15 + num__29 + num__8 = num__91.0 . so remaining num__9.0 is the balance for the astrophysics . since this is a circle all percentage must be equal to num__360 degrees . num__100.0 - - - - num__360 degrees then num__9.0 will be num__33 degrees . . imo option d . <eor> d <eos> |
d |
subtract__100.0__91.0__ add__25.0__8.0__ add__25.0__8.0__ |
subtract__100.0__91.0__ add__25.0__8.0__ add__25.0__8.0__ |
| the kiljaro highway is marked with milestones denoting the distance to the town of kiljaro . tommy left kiljaro and drove the highway passing the num__60 km milestone at num__8 : num__30 . some time afterwards tommy got a phone call asking him to return home and he made a u - turn at the num__160 km milestone . at num__09 : a num__0 tommy passed the milestone marking num__70 km to kiljaro . the variable a represents the tens digit of the minutes in num__09 : a num__0 . assuming tommy maintained the same constant speed during the entire drive how many kilometers did tommy travel in one minute ? <o> a ) num__190 / ( num__30 + num__10 a ) <o> b ) num__190 / ( num__30 + num__60 a ) <o> c ) num__6.12903225806 a <o> d ) num__220 / ( num__30 + num__10 a ) <o> e ) num__5.5 a |
since we are dealing with the variables in the answer choices the best possible method according to me would be substitution . substitute a with num__3 . meaning tommy would have travelled a distance of ( ( num__160 - num__60 ) + ( num__160 - num__70 ) ) in num__60 minutes . num__190 kms in num__60 minutes = = > num__3.1 km / hr . substitute a with num__3 in the answer options . option a <eor> a <eos> |
a |
add__30.0__160.0__ round__190.0__ |
add__30.0__160.0__ add__30.0__160.0__ |
| mr . suresh is on tour and he has num__360 for his expenses . if he exceeds his tour by num__4 days he must cut down daily expenses by num__3 . the number of days of mr . suresh ’ s tour programme is : <o> a ) num__20 days <o> b ) num__24 days <o> c ) num__40 days <o> d ) num__42 days <o> e ) none of these |
let suresh undertakes a tour of x days . then expenses for each day = num__360 ⁄ x now num__360 / x + num__4 = num__360 / x − num__3 or num__360 ( num__1 / x − num__1 / x + num__4 ) = num__33 or x num__2 + num__4 x - num__480 = num__0 or x = – num__24 or x = num__20 since x ≠ - num__24 we have x = num__20 answer a <eor> a <eos> |
a |
subtract__4.0__3.0__ subtract__3.0__1.0__ divide__480.0__24.0__ round__20.0__ |
subtract__4.0__3.0__ subtract__3.0__1.0__ divide__480.0__24.0__ divide__480.0__24.0__ |
| a bat is bought for rs . num__400 and sold at a gain of num__20.0 find its selling price <o> a ) rs . num__400 / - <o> b ) rs . num__420 / - <o> c ) rs . num__440 / - <o> d ) rs . num__460 / - <o> e ) rs . num__480 / - |
num__100.0 - - - - - - > num__400 ( num__100 * num__4 = num__400 ) num__120.0 - - - - - - > num__480 ( num__120 * num__4 = num__480 ) selling price = rs . num__480 / - e <eor> e <eos> |
e |
percent__100.0__480.0__ |
percent__100.0__480.0__ |
| lindsay can paint num__1 / x of a certain room in num__40 minutes . what fraction of the same room can joseph paint in num__40 minutes if the two of them can paint the room in an hour working together at their respective rates ? <o> a ) num__0.666666666667 x <o> b ) num__3 x / ( x – num__3 ) <o> c ) ( num__2 x – num__3 ) / num__3 x <o> d ) num__1 / ( x – num__3 ) <o> e ) ( x – num__3 ) / x |
lindsey and joseph together can paint the room in num__1 hour in num__40 mins they can paint num__0.666666666667 part of the room . lindsay alone paints num__1 / x part of the room in num__40 mins take joseph paints num__1 / y parts in num__40 mins num__1 / x + num__1 / y = num__0.666666666667 num__1 / y = num__0.666666666667 - num__1 / x = ( num__2 x - num__3 ) / num__3 x ans c <eor> c <eos> |
c |
add__1.0__2.0__ round__2.0__ |
add__1.0__2.0__ divide__2.0__1.0__ |
| if a stationery store owner buys num__50.0 more identically - priced calendars than she usually purchases she will be given a num__20.0 discount off the standard price . her total cost would then be num__120 times the dollar value of the standard price of one calendar . how many calendars does she usually purchase ? <o> a ) num__40 <o> b ) num__80 <o> c ) num__100 <o> d ) num__120 <o> e ) num__140 |
let the price is x per item and the final number of items she bought is y items so to buy y items the cost will be . num__8 x ( num__80.0 of original price since there is a num__20.0 discount on it ) num__120 * x = y * num__0.8 * x solving we get y = num__150 now to get the discount she bought num__50.0 more than what she usually buys so the number of items she usually buys is num__50.0 less than num__150 i . e . original number of items she usually buys is = num__150 * ( num__0.666666666667 ) = num__100 hence answer is ( c ) <eor> c <eos> |
c |
divide__120.0__0.8__ divide__80.0__120.0__ add__20.0__80.0__ add__20.0__80.0__ |
divide__120.0__0.8__ divide__80.0__120.0__ add__20.0__80.0__ add__20.0__80.0__ |
| a combustion reaction forms carbon dioxide . a carbon dioxide molecule contains one carbon and two oxygen atoms . if over a period of num__12 minutes a combustion reaction creates num__12000 molecules of carbon dioxide then approximately how many more atoms of oxygen than carbon are created on average per minute ? <o> a ) num__20 <o> b ) num__60 <o> c ) num__600 <o> d ) num__1000 <o> e ) num__1 |
200 |
solution : num__12000 carbon dioxide molecules are created over a period of num__12 minutes . therefore num__1000.0 = num__1000 carbon dioxide molecules are created on average per minute each carbon dioxide molecule contains one carbon atom and two oxygen atoms . so num__1000 carbon dioxide molecules contain num__1 × num__1000 = num__1000 carbon atoms and num__2 × num__1000 = num__2000 oxygen atoms . the difference is num__2000 – num__1000 = num__1000 . the correct answer is d . <eor> d <eos> |
d |
d |
| in a dice game a player wins if he rolls the same number on his second roll as he rolls on his first . if jason rolls a total of num__4 on the first roll what is the probability that he wins on his next roll ? in this game each roll is made with num__2 fair six - sided dice . <o> a ) num__0.166666666667 <o> b ) num__0.111111111111 <o> c ) num__0.0833333333333 <o> d ) num__0.0666666666667 <o> e ) num__0.0555555555556 |
there are num__3 ways to roll a num__4 : num__1 and num__3 num__3 and num__1 and num__2 and num__2 . there are num__6 * num__6 = num__36 ways to roll two dice . thus the probability of rolling a num__4 is num__0.0833333333333 = num__0.0833333333333 c <eor> c <eos> |
c |
subtract__4.0__3.0__ add__4.0__2.0__ divide__3.0__36.0__ multiply__1.0__0.0833__ |
subtract__4.0__3.0__ multiply__2.0__3.0__ divide__3.0__36.0__ multiply__1.0__0.0833__ |
| a positive number when decreased by num__4 is equal to num__21 times the reciprocal of the number . the number is : <o> a ) num__2 <o> b ) num__5 <o> c ) num__9 <o> d ) num__7 <o> e ) num__6 |
let the number be x x - num__4 = num__21 / x x ^ num__2 - num__4 x - num__21 = num__0 ( x - num__7 ) ( x + num__3 ) = num__0 x = num__7 correct answer d <eor> d <eos> |
d |
divide__21.0__7.0__ add__4.0__3.0__ |
divide__21.0__7.0__ add__4.0__3.0__ |
| if p ( a ) = num__0.2 and p ( b ) = num__0.4 find p ( a n b ) if a and b are independent events . <o> a ) num__0.28 <o> b ) num__0.12 <o> c ) num__0.32 <o> d ) num__0.08 <o> e ) num__0.176470588235 |
p ( a n b ) = p ( a ) . p ( b ) p ( a n b ) = num__0.2 . num__0.4 p ( a n b ) = num__0.08 . d <eor> d <eos> |
d |
multiply__0.2__0.4__ multiply__0.2__0.4__ |
multiply__0.2__0.4__ multiply__0.2__0.4__ |
| a man invested rs . num__4455 in rs . num__10 shares quoted at rs . num__8.25 . if the rate of dividend be num__12.0 his annual income is : <o> a ) num__233 <o> b ) num__648 <o> c ) num__366 <o> d ) num__266 <o> e ) num__991 |
explanation : number of shares = = num__540 . face value = rs . ( num__540 x num__10 ) = rs . num__5400 . annual income = = rs . num__648 . answer : b ) num__648 <eor> b <eos> |
b |
percent__12.0__5400.0__ percent__12.0__5400.0__ |
percent__12.0__5400.0__ percent__12.0__5400.0__ |
| the average of four consecutive even numbers is num__27 . find the largest of these numbers . <o> a ) num__28 <o> b ) num__30 <o> c ) num__32 <o> d ) num__34 <o> e ) num__36 |
given : consider the consecutive even numbers as : x ( x + num__2 ) ( x + num__4 ) and ( x + num__6 ) average = sum of quantities / number of quantities = [ x + ( x + num__2 ) + ( x + num__4 ) + ( x + num__6 ) ] / num__4 = ( num__4 x + num__12 ) / num__4 = num__27 simplifying we get x = num__24 therefore largest number = ( x + num__6 ) = ( num__24 + num__6 ) = num__30 smallest number = num__24 answer is b <eor> b <eos> |
b |
add__2.0__4.0__ multiply__2.0__6.0__ multiply__2.0__12.0__ add__6.0__24.0__ add__6.0__24.0__ |
add__2.0__4.0__ multiply__2.0__6.0__ multiply__2.0__12.0__ add__6.0__24.0__ add__6.0__24.0__ |
| a man can row upstream at num__25 kmph and downstream at num__35 kmph and then find the speed of the man in still water ? <o> a ) num__86 <o> b ) num__67 <o> c ) num__30 <o> d ) num__15 <o> e ) num__17 |
us = num__25 ds = num__35 m = ( num__35 + num__25 ) / num__2 = num__30 answer : c <eor> c <eos> |
c |
round__30.0__ |
round__30.0__ |
| the length and breadth of a rectangle is increased by num__10.0 and num__25.0 respectively . what is the increase in the area ? <o> a ) num__37.2 <o> b ) num__37.5 <o> c ) num__57.5 <o> d ) num__37.4 <o> e ) num__27.2 % |
num__100 * num__100 = num__10000 num__110 * num__125 = num__13750 - - - - - - - - - - - num__3750 num__10000 - - - - - - num__3750 num__100 - - - - - - - ? = > num__37.5 num__27 . the side of a rhombus is num__2 answer : b <eor> b <eos> |
b |
square_perimeter__25.0__ triangle_area__10.0__25.0__ multiply__110.0__125.0__ surface_cube__25.0__ triangle_area__37.5__2.0__ |
square_perimeter__25.0__ triangle_area__10.0__25.0__ multiply__110.0__125.0__ surface_cube__25.0__ triangle_area__37.5__2.0__ |
| if sy + z = s ( y + z ) which of the following must be true ? <o> a ) x = num__0 and z = num__0 <o> b ) s = num__1 and y = num__1 <o> c ) s = num__1 and z = num__0 <o> d ) s = num__1 or y = num__0 <o> e ) s = num__1 or z = num__0 |
sy + z = sy + sz z = sz case num__1 : z not = num__0 s = z / z = num__1 case num__2 : z = num__0 num__0 = s num__0 = num__0 combining num__2 cases : s = num__1 or z = num__0 e . is the answer . <eor> e <eos> |
e |
reverse__1.0__ |
reverse__1.0__ |
| find the least number which when divided by num__26 num__36 and num__46 leaves the remainders num__12 num__22 and num__32 respectively . <o> a ) num__10570 <o> b ) num__10750 <o> c ) num__17050 <o> d ) num__10075 <o> e ) num__10085 |
explanation : the difference between any divisor and the corresponding remainder is num__14 l . c . m of num__26 num__3646 - num__14 = num__10764 - num__14 = num__10750 answer : option b <eor> b <eos> |
b |
subtract__26.0__12.0__ subtract__10764.0__14.0__ subtract__10764.0__14.0__ |
subtract__26.0__12.0__ subtract__10764.0__14.0__ subtract__10764.0__14.0__ |
| six years ago ram was p times as old as shyam was . if ram is now num__17 years old how old is shyam now in terms of p ? <o> a ) num__11 / p + num__6 <o> b ) p / num__11 + num__6 <o> c ) num__17 - p / num__6 <o> d ) num__17 / p <o> e ) num__11.5 p |
let ' s call shyam ' s age six years ago is x so six years ago ram was : num__17 - num__6 = num__11 years old and six years ago ram was p times as old as shyam was = > num__11 / x = p = > x = num__11 / p = > now shyam ' s age will be : x + num__6 = num__11 / p + num__6 = > the answer is a <eor> a <eos> |
a |
subtract__17.0__6.0__ subtract__17.0__6.0__ |
subtract__17.0__6.0__ subtract__17.0__6.0__ |
| if x is the interest on y and y is the interest on z the rate and time is the same on both the cases . what is the relation between x y and z ? <o> a ) num__9810 <o> b ) num__1002 <o> c ) num__2799 <o> d ) num__1000 <o> e ) num__2729 |
( x * num__5 * num__1 ) / num__100 + [ ( num__2500 - x ) * num__6 * num__1 ] / num__100 = num__140 x = num__1000 answer : d <eor> d <eos> |
d |
percent__100.0__1000.0__ |
percent__100.0__1000.0__ |
| the average age of a family of num__6 members is num__26 years . if the age of the youngest member is num__10 years what was the average age of the family at the birth of the youngest member ? <o> a ) num__15 <o> b ) num__18 <o> c ) num__16 <o> d ) num__12 <o> e ) num__19 |
present age of total members = num__6 x num__26 = num__156 num__10 yrs back their ages were = num__6 x num__10 = num__60 ages at the birth of youngest member = num__156 - num__60 = num__96 therefore avg age at the birth of youngest member = num__16.0 = num__16 . answer : c <eor> c <eos> |
c |
multiply__6.0__26.0__ multiply__6.0__10.0__ subtract__156.0__60.0__ add__6.0__10.0__ add__6.0__10.0__ |
multiply__6.0__26.0__ multiply__6.0__10.0__ subtract__156.0__60.0__ subtract__26.0__10.0__ subtract__26.0__10.0__ |
| what amount does kiran get if he invests rs . num__16000 at num__15.0 p . a . simple interest for four years ? <o> a ) num__29893 <o> b ) num__27773 <o> c ) num__25600 <o> d ) num__28800 <o> e ) num__29883 |
simple interest = ( num__16000 * num__4 * num__15 ) / num__100 = rs . num__9600 amount = p + i = num__16000 + num__9600 = rs . num__25600 answer : c <eor> c <eos> |
c |
percent__100.0__25600.0__ |
percent__100.0__25600.0__ |
| the average of five numbers id num__27 . if one number is excluded the average becomes num__25 . what is the excluded number ? <o> a ) num__30 <o> b ) num__40 <o> c ) num__32.5 <o> d ) num__35 <o> e ) num__37 |
explanation : sum of num__5 numbers = num__5 × num__27 sum of num__4 numbers after excluding one number = num__4 × num__25 excluded number = num__5 × num__27 - num__4 × num__25 = num__135 - num__100 = num__35 answer : option d <eor> d <eos> |
d |
multiply__27.0__5.0__ multiply__25.0__4.0__ subtract__135.0__100.0__ subtract__135.0__100.0__ |
multiply__27.0__5.0__ multiply__25.0__4.0__ subtract__135.0__100.0__ subtract__135.0__100.0__ |
| if ( num__2 to the x ) - ( num__2 to the ( x - num__2 ) ) = num__3 ( num__2 to the num__10 ) what is the value of x ? <o> a ) num__9 <o> b ) num__11 <o> c ) num__13 <o> d ) num__12 <o> e ) num__17 |
( num__2 to the power x ) - ( num__2 to the power ( x - num__2 ) ) = num__3 ( num__2 to the power num__10 ) num__2 ^ x - num__2 ^ ( x - num__2 ) = num__3 . num__2 ^ num__10 hence x = num__12 . answer is d <eor> d <eos> |
d |
add__2.0__10.0__ add__2.0__10.0__ |
add__2.0__10.0__ add__2.0__10.0__ |
| the price of an item is discounted num__10 percent on day num__1 of a sale . on day num__2 the item is discounted another num__10 percent and on day num__3 it is discounted an additional num__25 percent . the price of the item on day num__3 is what percentage of the sale price on day num__1 ? <o> a ) num__28.0 <o> b ) num__67.5 <o> c ) num__64.8 <o> d ) num__70.0 <o> e ) num__72 % |
original price = num__100 day num__1 discount = num__10.0 price = num__100 - num__10 = num__90 day num__2 discount = num__10.0 price = num__90 - num__9 = num__81 day num__3 discount = num__25.0 price = num__81 - num__20.25 = num__60.75 which is num__60.75 / num__90 * num__100 of the sale price on day num__1 = ~ num__67.5 answer b <eor> b <eos> |
b |
percent__10.0__90.0__ percent__25.0__81.0__ percent__100.0__67.5__ |
percent__10.0__90.0__ percent__25.0__81.0__ percent__100.0__67.5__ |
| john read the quarter of the time that tom read . tom read only two - fifth of the time that sasha read . sasha read twice as long as mike . if mike read num__5 hours how long did john read ? <o> a ) num__1 hour <o> b ) num__2 hour <o> c ) num__3 hour <o> d ) num__4 hour <o> e ) num__5 hour |
mike read num__5 hours . sasha read twice as long as mike . hence sasha read : num__2 ? num__5 = num__10 hours tom read two - fifths of the time that sasha read . hence tom read : ( num__0.4 ) ? num__10 = num__4 hours john read the quarter of the time that tom read . hence john read : ( num__0.25 ) ? num__4 = num__1 hour correct answer a <eor> a <eos> |
a |
twice__5.0__ divide__2.0__5.0__ twice__2.0__ quarter__ subtract__5.0__4.0__ subtract__5.0__4.0__ |
twice__5.0__ divide__2.0__5.0__ twice__2.0__ quarter__ subtract__5.0__4.0__ subtract__5.0__4.0__ |
| a circular ground whose diameter is num__34 metres has a num__2 metre - broad garden around it . what is the area of the garden in square metres ? <o> a ) num__226.3 <o> b ) num__226.2 <o> c ) num__228.2 <o> d ) num__227 <o> e ) num__226 |
req . area = Ï € [ ( num__19 ) num__2 â € “ ( num__17 ) num__2 ] = num__22 â „ num__7 Ã — ( num__36 Ã — num__2 ) [ since a num__2 - b num__2 = ( a + b ) ( a - b ) ] = ( num__22 Ã — num__36 Ã — num__2 ) / num__7 = num__226.3 sq m answer a <eor> a <eos> |
a |
triangle_area__2.0__226.3__ |
triangle_area__2.0__226.3__ |
| what is the least number of square tiles required to pave the floor of a room num__15 m num__17 cm long and num__9 m num__2 cm broad ? <o> a ) num__724 <o> b ) num__804 <o> c ) num__814 <o> d ) num__844 <o> e ) none |
solution length of largest tile = h . c . f . of num__1517 cm & num__902 cm = num__41 cm . area of each tile = ( num__41 x num__41 ) cm num__2 ∴ required number of tiles = [ num__1517 x num__22.0 x num__41 ] = num__814 . answer c <eor> c <eos> |
c |
divide__902.0__41.0__ round__814.0__ |
divide__902.0__41.0__ round__814.0__ |
| in may mrs lee ' s earnings were num__30 percent of the lee family ' s total income . in june mrs lee earned num__20 percent more than in may . if the rest of the family ' s income was the same both months then in june mrs lee ' s earnings were approximately what percent of the lee family ' s total income ? <o> a ) num__34 <o> b ) num__67 <o> c ) num__85 <o> d ) num__25 <o> e ) num__38 |
lets say the family income is num__100 in may lee earned num__30 family income is num__70 in june lee earned num__20.0 more than may so it is ( num__30 + num__20 * num__0.3 = num__36 ) family income is same num__70 in june lee ' s income percent is num__36 * num__0.943396226415 ~ num__34 ans is a <eor> a <eos> |
a |
subtract__100.0__30.0__ divide__30.0__100.0__ subtract__70.0__36.0__ subtract__70.0__36.0__ |
subtract__100.0__30.0__ divide__30.0__100.0__ subtract__70.0__36.0__ subtract__70.0__36.0__ |
| the average speed of a car is num__1 num__0.8 times the avg speed of a bike . a tractor covers num__575 km in num__25 hrs . how much distance will the car cover in num__4 hrs if the speed of the bike is twice speed of the tractor ? <o> a ) num__400 km <o> b ) num__500 km <o> c ) num__331.2 km <o> d ) num__550 km <o> e ) num__600 km |
sol . average speed of a tractor = num__23 km / h the speed of a bus in an hour = num__23 × num__2 = num__46 km the speed of a car in an hour = num__1.8 * num__46 = num__82.8 km so the distance covered by car in num__4 h is num__82.8 × num__4 = num__331.2 km ans . ( c ) <eor> c <eos> |
c |
divide__575.0__25.0__ subtract__25.0__23.0__ multiply__2.0__23.0__ add__1.0__0.8__ multiply__46.0__1.8__ multiply__4.0__82.8__ multiply__1.0__331.2__ |
divide__575.0__25.0__ subtract__25.0__23.0__ multiply__2.0__23.0__ add__1.0__0.8__ multiply__46.0__1.8__ multiply__4.0__82.8__ multiply__1.0__331.2__ |
| mary ’ s annual income is $ num__15000 and john ’ s annual income is $ num__18000 . by how much must mary ’ s annual income increase so that it constitutes num__80.0 of mary and john ’ s combined income ? <o> a ) $ num__3000 <o> b ) $ num__4000 <o> c ) $ num__7000 <o> d ) $ num__11000 <o> e ) $ num__57 |
000 |
let mary ' s income increase by x then the equation will be num__15000 + x = ( num__0.8 ) * ( num__15000 + x + num__18000 ) num__15000 + x = ( num__0.8 ) * ( num__33000 + x ) num__75000 + num__5 x = num__4 x + num__132000 x = num__57000 x = num__57000 so answer will be e <eor> e <eos> |
e |
e |
| riya and priya set on a journey . riya moves eastward at a speed of num__21 kmph and priya moves westward at a speed of num__22 kmph . how far will be priya from riya after num__60 minutes <o> a ) num__43 kms <o> b ) num__45 kms <o> c ) num__50 kms <o> d ) num__30 kms <o> e ) num__40 kms |
total eastward distance = num__21 kmph * num__1 hr = num__21 km total westward distance = num__22 kmph * num__1 hr = num__22 km total distn betn them = num__21 + num__22 = num__43 km ans num__43 km answer : a <eor> a <eos> |
a |
subtract__22.0__21.0__ add__21.0__22.0__ round__43.0__ |
subtract__22.0__21.0__ add__21.0__22.0__ add__21.0__22.0__ |
| a rectangular lawn of dimensions num__70 m * num__60 m has two roads each num__10 m wide running in the middle of the lawn one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . num__3 per sq m ? <o> a ) a ) num__3600 <o> b ) b ) num__930 <o> c ) c ) num__9309 <o> d ) d ) num__3900 <o> e ) e ) num__8302 |
explanation : area = ( l + b – d ) d ( num__70 + num__60 – num__10 ) num__10 = > num__1200 m num__2 num__1200 * num__3 = rs . num__3600 answer : option a <eor> a <eos> |
a |
power__60.0__2.0__ power__60.0__2.0__ |
multiply__3.0__1200.0__ multiply__3.0__1200.0__ |
| the length of rectangle is thrice its breadth and its perimeter is num__96 m find the area of the rectangle ? <o> a ) num__432 sq m <o> b ) num__356 sq m <o> c ) num__452 sq m <o> d ) num__428 sq m <o> e ) num__525 sq m |
num__2 ( num__3 x + x ) = num__96 l = num__36 b = num__12 lb = num__36 * num__12 = num__432 answer : a <eor> a <eos> |
a |
square_perimeter__3.0__ multiply__36.0__12.0__ triangle_area__2.0__432.0__ |
square_perimeter__3.0__ multiply__36.0__12.0__ multiply__36.0__12.0__ |
| if num__5 x = num__7 y and xy ≠ num__0 what is the ratio of num__0.2 * x to num__0.166666666667 * y ? <o> a ) num__4.16666666667 <o> b ) num__1.68 <o> c ) num__1.2 <o> d ) num__0.833333333333 <o> e ) num__0.694444444444 |
num__5 x = num__7 y = > x / y = num__1.4 num__0.2 * x to num__0.166666666667 * y = x / y * num__1.2 = ( num__1.4 ) * ( num__1.2 ) = num__1.68 ans : b <eor> b <eos> |
b |
divide__7.0__5.0__ subtract__1.4__0.2__ multiply__1.2__1.4__ multiply__1.2__1.4__ |
divide__7.0__5.0__ subtract__1.4__0.2__ multiply__1.2__1.4__ multiply__1.2__1.4__ |
| num__3 / [ ( num__1 / num__0.03 ) + ( num__1 / num__0.37 ) ] = ? <o> a ) num__0.004 <o> b ) num__0.08333 <o> c ) num__2.775 <o> d ) num__3.6036 <o> e ) num__36.036 |
approximate . num__1 / . num__03 = num__33.3333333333 = num__33 num__1 / . num__37 = num__2.7027027027 = num__3 denominator becomes num__33 + num__3 = num__36 num__0.0833333333333 = . num__08 something answer ( b ) <eor> b <eos> |
b |
reverse__0.03__ round_down__33.3333__ reverse__0.37__ add__3.0__33.0__ divide__3.0__36.0__ divide__3.0__36.0__ |
reverse__0.03__ round_down__33.3333__ reverse__0.37__ add__3.0__33.0__ divide__3.0__36.0__ divide__3.0__36.0__ |
| what amount does kiran get if he invests rs . num__18000 at num__15.0 p . a . simple interest for four years ? <o> a ) num__228777 <o> b ) num__26788 <o> c ) num__26088 <o> d ) num__28800 <o> e ) num__67881 |
simple interest = ( num__18000 * num__4 * num__15 ) / num__100 = rs . num__10800 amount = p + i = num__18000 + num__10800 = rs . num__28800 answer : d <eor> d <eos> |
d |
percent__100.0__28800.0__ |
percent__100.0__28800.0__ |
| how many halves are there in two - fourth ? <o> a ) num__8 <o> b ) num__1 <o> c ) num__32 <o> d ) num__7 <o> e ) num__3 |
divide num__0.5 by num__0.5 = num__0.5 ÷ num__0.5 = num__0.5 * num__2.0 = num__1.0 = num__1 . answer is b . <eor> b <eos> |
b |
reverse__0.5__ multiply__0.5__2.0__ reverse__1.0__ |
reverse__0.5__ multiply__0.5__2.0__ multiply__0.5__2.0__ |
| if num__213 × num__16 = num__3408 then num__1.6 × num__213.0 is equal to : <o> a ) num__0.3408 <o> b ) num__3.408 <o> c ) num__34.08 <o> d ) num__340.8 <o> e ) none of these |
solution num__1.6 × num__21.3 = ( num__1.6 x num__213 ) = ( num__16 x num__21.3 ) = num__340.8 = num__340.8 . answer d <eor> d <eos> |
d |
multiply__213.0__1.6__ multiply__213.0__1.6__ |
multiply__213.0__1.6__ multiply__213.0__1.6__ |
| a father said to his son ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is num__42 years now the son ' s age five years back was : <o> a ) num__14 <o> b ) num__16 <o> c ) num__19 <o> d ) num__22 <o> e ) num__24 |
let the son ' s present age be x years . then ( num__42 - x ) = x num__2 x = num__42 . x = num__21 . son ' s age num__5 years back ( num__21 - num__5 ) = num__16 years . answer : b <eor> b <eos> |
b |
divide__42.0__2.0__ subtract__21.0__5.0__ subtract__21.0__5.0__ |
divide__42.0__2.0__ subtract__21.0__5.0__ subtract__21.0__5.0__ |
| how many cuboids of length num__6 m width num__4 m and height num__3 m can be farmed from a cuboid of num__18 m length num__15 m width and num__2 m height . <o> a ) num__2.6 <o> b ) num__5.8 <o> c ) num__6.2 <o> d ) num__7.5 <o> e ) num__1.2 |
( num__18 Ã — num__15 Ã — num__2 ) / ( num__6 Ã — num__4 Ã — num__3 ) = num__7.5 answer is d . <eor> d <eos> |
d |
divide__15.0__2.0__ round__7.5__ |
divide__15.0__2.0__ divide__15.0__2.0__ |
| if c : d = num__2 : num__5 and d : f = num__4 : num__7 then c : d : f is <o> a ) num__8 : num__20 : num__35 <o> b ) num__6 : num__20 : num__35 <o> c ) num__8 : num__20 : num__33 <o> d ) num__8 : num__22 : num__35 <o> e ) num__5 : num__20 : num__35 |
solution : c / d = num__0.4 ; d / f = num__0.571428571429 ; c : d : f = num__2 * num__4 : num__5 * num__4 : num__5 * num__7 = num__8 : num__20 : num__35 . answer : option a <eor> a <eos> |
a |
divide__2.0__5.0__ divide__4.0__7.0__ multiply__2.0__4.0__ multiply__5.0__4.0__ multiply__5.0__7.0__ multiply__2.0__4.0__ |
divide__2.0__5.0__ divide__4.0__7.0__ multiply__2.0__4.0__ multiply__5.0__4.0__ multiply__5.0__7.0__ multiply__2.0__4.0__ |
| when positive integer x is divided by positive integer y the remainder is num__9 . if x / y = num__96.45 what is the value of y ? <o> a ) num__96 <o> b ) num__75 <o> c ) num__48 <o> d ) num__20 <o> e ) num__12 |
guys one more simple funda . num__2.5 = num__2.5 now . num__5 x num__2 = num__1 is the remainder num__6.25 = num__6.25 now . num__25 x num__4 = num__1 is the remainder num__6.4 = num__6.4 now . num__4 x num__5 = num__2 is the remainder given x / y = num__96.45 and remainder is num__9 so . num__45 x y = num__9 hence y = num__20 ans d <eor> d <eos> |
d |
round_down__2.5__ subtract__9.0__5.0__ multiply__9.0__5.0__ multiply__4.0__5.0__ multiply__1.0__20.0__ |
divide__5.0__2.5__ divide__25.0__6.25__ multiply__9.0__5.0__ multiply__4.0__5.0__ divide__20.0__1.0__ |
| a train requires num__5 seconds to pass a pole while it requires num__25 seconds to cross a stationary train which is num__360 mtrs long . find the speed of the train . <o> a ) num__66.8 kmph <o> b ) num__65.8 kmph <o> c ) num__54.8 kmph <o> d ) num__64.8 kmph <o> e ) num__44.8 kmph |
in num__5 s the train crosses the pole and in num__25 sec the train crosses one more stationary train in num__20 sec the train travels a distance of num__360 mtrs speed = num__18.0 = num__18 m / s = num__18 ( num__3.6 ) = num__18 * num__3.6 = num__64.8 kmph answer : d <eor> d <eos> |
d |
subtract__25.0__5.0__ divide__360.0__20.0__ divide__18.0__5.0__ multiply__18.0__3.6__ round__64.8__ |
subtract__25.0__5.0__ divide__360.0__20.0__ divide__18.0__5.0__ multiply__18.0__3.6__ multiply__18.0__3.6__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__277 <o> b ) num__267 <o> c ) num__240 <o> d ) num__882 <o> e ) num__212 |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x meters . then x + num__8.33333333333 = num__15 x + num__300 = num__540 x = num__240 m . answer : c <eor> c <eos> |
c |
multiply__20.0__15.0__ divide__300.0__36.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ divide__300.0__36.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
| there are two examinations rooms a and b . if num__10 students are sent from a to b then the number of students in each room is the same . if num__20 candidates are sent from b to a then the number of students in a is double the number of students in b . the number of students in room a is : <o> a ) num__50 <o> b ) num__100 <o> c ) num__150 <o> d ) num__200 <o> e ) num__250 |
let the number of students in rooms a and b be x and y respectively . then x - num__10 = y + num__10 x - y = num__20 . . . . ( i ) and x + num__20 = num__2 ( y - num__20 ) x - num__2 y = - num__60 . . . . ( ii ) solving ( i ) and ( ii ) we get : x = num__100 y = num__80 . the required answer a = num__100 . answer is b . <eor> b <eos> |
b |
divide__20.0__10.0__ add__20.0__60.0__ add__20.0__80.0__ |
divide__20.0__10.0__ add__20.0__60.0__ add__20.0__80.0__ |
| when leo imported a certain item he paid a num__7 percent import tax on the portion of the total value of the item in excess of $ num__1000 . if the amount of the import tax that leo paid was $ num__87.50 what was the total value of the item ? <o> a ) $ num__1600 <o> b ) $ num__1850 <o> c ) $ num__2250 <o> d ) $ num__2400 <o> e ) $ num__2 |
750 |
let the value of the item be $ x then num__0.07 ( x − num__1000 ) = num__87.5 - - > x − num__1000 = num__1250 x − num__1000 = num__1250 - - > x = num__2250 . answer : c . <eor> c <eos> |
c |
c |
| mike took a taxi to the airport and paid $ num__2.50 to start plus $ num__0.25 per mile . annie took a different route to the airport and paid $ num__2.50 plus $ num__5.00 in bridge toll fees plus $ num__0.25 per mile . if each was charged exactly the same amount and annie ' s ride was num__16 miles how many miles was mike ' s ride ? <o> a ) num__32 <o> b ) num__36 <o> c ) num__40 <o> d ) num__44 <o> e ) num__48 |
the cost of annie ' s ride was num__2.5 + num__5 + ( num__0.25 * num__16 ) = $ num__11.50 let x be the distance of mike ' s ride . the cost of mike ' s ride is num__2.5 + ( num__0.25 * x ) = num__11.5 num__0.25 * x = num__9 x = num__36 miles the answer is b . <eor> b <eos> |
b |
subtract__11.5__2.5__ divide__9.0__0.25__ divide__9.0__0.25__ |
subtract__11.5__2.5__ divide__9.0__0.25__ divide__9.0__0.25__ |
| in the xy - coordinate system if ( m n ) and ( m + num__2 n + k ) are two points on the line perpendicular to x = num__2 y + num__5 then k = ? <o> a ) - num__5 <o> b ) - num__8 <o> c ) num__2 <o> d ) - num__4 <o> e ) num__1 |
the slope = k / num__2 the line perpendicular to x = num__2 y + num__5 or y = num__0.5 x - num__2.5 has a slope = - num__1 / ( num__0.5 ) = - num__2 - - - > k / num__2 = - num__2 - - > k = - num__4 answer is d <eor> d <eos> |
d |
reverse__2.0__ add__2.0__0.5__ multiply__2.0__0.5__ divide__2.0__0.5__ divide__2.0__0.5__ |
reverse__2.0__ add__2.0__0.5__ multiply__2.0__0.5__ divide__2.0__0.5__ divide__2.0__0.5__ |
| a chair is bought for rs . num__600 / - and sold at a loss of num__10.0 find its selling price <o> a ) rs . num__500 / - <o> b ) rs . num__540 / - <o> c ) rs . num__600 / - <o> d ) rs . num__640 / - <o> e ) rs . num__660 / - |
num__100.0 - - - - - - > num__600 ( num__100 * num__6 = num__600 ) num__90.0 - - - - - - > num__540 ( num__90 * num__6 = num__540 ) selling price = rs . num__540 / - b <eor> b <eos> |
b |
percent__90.0__600.0__ percent__90.0__600.0__ |
percent__90.0__600.0__ percent__90.0__600.0__ |
| out of seven given numbers the average of the first four numbers is num__4 and that of the last four numbers is also num__4 . if the average of all the seven numbers is num__3 fourth number is <o> a ) num__3 <o> b ) num__4 <o> c ) num__7 <o> d ) fourth no = num__11 <o> e ) num__12 |
( a + b + c + d ) / num__4 = num__4 = > ( a + b + c + d ) = num__16 . . . . . . . . . . . ( num__1 ) ( d + e + f + g ) / num__4 = num__4 = > ( d + e + f + g ) = num__16 = > ( e + f + g ) = num__16 - d . . . . . . . . . . . ( num__2 ) also ( a + b + c + d + e + f + g ) / num__7 = num__3 ( given ) using ( num__1 ) and ( num__2 ) equations num__16 + num__16 - d = num__21 = > d = num__11 answer : d <eor> d <eos> |
d |
subtract__4.0__3.0__ subtract__3.0__1.0__ add__4.0__3.0__ multiply__3.0__7.0__ add__4.0__7.0__ add__4.0__7.0__ |
subtract__4.0__3.0__ subtract__3.0__1.0__ add__4.0__3.0__ multiply__3.0__7.0__ add__4.0__7.0__ add__4.0__7.0__ |
| find large number from below question the difference of two numbers is num__1365 . on dividing the larger number by the smaller we get num__6 as quotient and the num__15 as remainder <o> a ) num__1235 <o> b ) num__1345 <o> c ) num__1678 <o> d ) num__1767 <o> e ) num__1635 |
let the smaller number be x . then larger number = ( x + num__1365 ) . x + num__1365 = num__6 x + num__15 num__5 x = num__1350 x = num__270 large number = num__270 + num__1365 = num__1635 e <eor> e <eos> |
e |
subtract__1365.0__15.0__ divide__1350.0__5.0__ add__1365.0__270.0__ add__1365.0__270.0__ |
subtract__1365.0__15.0__ divide__1350.0__5.0__ add__1365.0__270.0__ add__1365.0__270.0__ |
| three pounds of num__05 grass seed contain num__10 percent herbicide . a different type of grass seed num__20 which contains num__20 percent herbicide will be mixed with three pounds of num__05 grass seed . how much grass seed of type num__20 should be added to the three pounds of num__05 grass seed so that the mixture contains num__15 percent herbicide ? <o> a ) num__3 <o> b ) num__3.75 <o> c ) num__4.5 <o> d ) num__6 <o> e ) num__9 |
num__05 grass seed contains num__5.0 herbicide and its amount is num__3 pound num__20 grass seed contains num__20.0 herbicide and its amount is x when these two types of grass seeds are mixed their average becomes num__15.0 thus we have num__3 ( num__10 ) + x ( num__20 ) / ( x + num__3 ) = num__15 num__30 + num__20 x = num__15 x + num__45 num__5 x = num__15 or x = num__3 a <eor> a <eos> |
a |
divide__15.0__5.0__ add__10.0__20.0__ multiply__15.0__3.0__ divide__15.0__5.0__ |
divide__15.0__5.0__ add__10.0__20.0__ add__15.0__30.0__ divide__15.0__5.0__ |
| the average temperature for wednesday thursday and friday was num__40 oc . the average for thursday friday and saturday was num__41 oc . if temperature on saturday was num__42 oc what was the temperature on wednesday ? <o> a ) num__39 oc <o> b ) num__44 oc <o> c ) num__38 oc <o> d ) num__41 oc <o> e ) none |
solution : average temperature for wednesday thursday and friday = num__40 oc total temperature = num__3 * num__40 = num__120 oc average temperature for thursday friday and saturday = num__41 oc total temperature = num__41 * num__3 = num__123 oc temperature on saturday = num__42 oc now ( thursday + friday + saturday ) - ( wednesday + thursday + friday ) = num__123 - num__120 ; saturday - wednesday = num__3 wednesday = num__42 - num__3 = num__39 oc . answer : option a <eor> a <eos> |
a |
multiply__40.0__3.0__ multiply__41.0__3.0__ subtract__42.0__3.0__ subtract__42.0__3.0__ |
multiply__40.0__3.0__ multiply__41.0__3.0__ subtract__42.0__3.0__ subtract__42.0__3.0__ |
| a can finish a work in num__24 days b n num__9 days and c in num__12 days . b & c start the work but are forced to leave after num__3 days . the remaining work done by a in : <o> a ) num__12 days <o> b ) num__13 days <o> c ) num__11 days <o> d ) num__10 days <o> e ) num__9 days |
( b + c ) ' s one day ' s work = num__0.111111111111 + num__0.0833333333333 = num__0.194444444444 work done by b & c in num__3 days = num__3 * num__0.194444444444 = num__0.583333333333 remaining work = num__1 - ( num__0.583333333333 ) = num__0.416666666667 now num__0.0416666666667 work is done by a in num__1 day . so num__0.416666666667 work is done by a in num__24 * num__0.416666666667 = num__10 days answer d num__10 days <eor> d <eos> |
d |
add__0.1111__0.0833__ subtract__1.0__0.5833__ divide__1.0__24.0__ add__9.0__1.0__ round__10.0__ |
add__0.1111__0.0833__ subtract__1.0__0.5833__ divide__1.0__24.0__ add__9.0__1.0__ add__9.0__1.0__ |
| the h . c . f . of two numbers is num__23 and the other two factors of their l . c . m . are num__13 and num__14 . the larger of the two numbers is : <o> a ) num__276 <o> b ) num__299 <o> c ) num__322 <o> d ) num__336 <o> e ) num__347 |
the numbers are ( num__23 x num__13 ) and ( num__23 x num__14 ) . larger number = ( num__23 x num__14 ) = num__322 . answer : option c <eor> c <eos> |
c |
multiply__23.0__14.0__ multiply__23.0__14.0__ |
multiply__23.0__14.0__ multiply__23.0__14.0__ |
| two pipes a and b can separately fill a cistern in num__24 minutes and num__48 minutes respectively . there is a third pipe in the bottom of the cistern to empty it . if all the three pipes are simultaneously opened then the cistern is full in num__12 minutes . in how much time the third pipe alone can empty the cistern ? <o> a ) num__90 min <o> b ) num__48 min <o> c ) num__54 min <o> d ) num__120 min <o> e ) num__130 min |
num__0.0833333333333 - ( num__0.0416666666667 + num__0.0208333333333 ) = - num__0.0208333333333 third pipe can empty in num__48 minutes answer : b <eor> b <eos> |
b |
round__48.0__ |
round__48.0__ |
| there are num__6 people in the elevator . their average weight is num__154 lbs . another person enters the elevator and increases the average weight to num__151 lbs . what is the weight of the num__7 th person . <o> a ) num__133 <o> b ) num__168 <o> c ) num__189 <o> d ) num__190 <o> e ) num__200 |
solution average of num__7 people after the last one enters = num__151 . â ˆ ´ required weight = ( num__7 x num__151 ) - ( num__6 x num__154 ) = num__1057 - num__924 = num__133 . answer a <eor> a <eos> |
a |
multiply__151.0__7.0__ multiply__6.0__154.0__ subtract__1057.0__924.0__ subtract__1057.0__924.0__ |
multiply__151.0__7.0__ multiply__6.0__154.0__ subtract__1057.0__924.0__ subtract__1057.0__924.0__ |
| what will come in place of the x in the following number series ? num__279936 num__46656 num__7776 num__1296 num__216 x <o> a ) num__23 <o> b ) num__34 <o> c ) num__36 <o> d ) num__42 <o> e ) num__45 |
go on dividing by num__6 to the next number c ) <eor> c <eos> |
c |
divide__279936.0__46656.0__ divide__279936.0__7776.0__ |
divide__279936.0__46656.0__ divide__279936.0__7776.0__ |
| a baseball card decreased in value num__40.0 in its first year and num__10.0 in its second year . what was the total percent decrease of the card ' s value over the two years ? <o> a ) num__46.0 <o> b ) num__30.0 <o> c ) num__32.0 <o> d ) num__36.0 <o> e ) num__72 % |
let the initial value of baseball card = num__100 after first year value of baseball card = ( num__1 - num__0.4 ) * num__100 = num__60 after second year value of baseball card = ( num__1 - num__0.1 ) * num__60 = num__54 total percent decrease of the card ' s value over the two years = ( num__100 - num__54 ) / num__100 * num__100.0 = num__46.0 answer a <eor> a <eos> |
a |
percent__40.0__1.0__ percent__10.0__1.0__ percent__100.0__46.0__ |
percent__40.0__1.0__ percent__10.0__1.0__ percent__100.0__46.0__ |
| a student scored an average of num__75 marks in num__3 subjects : physics chemistry and mathematics . if the average marks in physics and mathematics is num__90 and that in physics and chemistry is num__70 what are the marks in physics ? <o> a ) num__86 <o> b ) num__16 <o> c ) num__76 <o> d ) num__95 <o> e ) num__26 |
given m + p + c = num__75 * num__3 = num__225 - - - ( num__1 ) m + p = num__90 * num__2 = num__180 - - - ( num__2 ) p + c = num__70 * num__2 = num__140 - - - ( num__3 ) where m p and c are marks obtained by the student in mathematics physics and chemistry . p = ( num__2 ) + ( num__3 ) - ( num__1 ) = num__180 + num__140 - num__225 = num__95 answer : d <eor> d <eos> |
d |
multiply__75.0__3.0__ subtract__3.0__1.0__ multiply__90.0__2.0__ multiply__70.0__2.0__ multiply__1.0__95.0__ |
multiply__75.0__3.0__ subtract__3.0__1.0__ multiply__90.0__2.0__ multiply__70.0__2.0__ multiply__1.0__95.0__ |
| the total price of a basic computer and printer are $ num__2500 . if the same printer had been purchased with an enhanced computer whose price was $ num__500 more than the price of the basic computer then the price of the printer would have been num__0.166666666667 of that total . what was the price of the basic computer ? <o> a ) num__1500 <o> b ) num__1600 <o> c ) num__1750 <o> d ) num__1900 <o> e ) num__2000 |
let the price of basic computer be c and the price of the printer be p : c + p = $ num__2500 . the price of the enhanced computer will be c + num__500 and total price for that computer and the printer will be num__2500 + num__500 = $ num__3000 . now we are told that the price of the printer is num__0.166666666667 of that new total price : p = num__0.166666666667 * $ num__3000 = $ num__500 . plug this value in the first equation : c + num__500 = $ num__2500 - - > c = $ num__2000 answer : e . <eor> e <eos> |
e |
add__2500.0__500.0__ subtract__2500.0__500.0__ subtract__2500.0__500.0__ |
add__2500.0__500.0__ subtract__2500.0__500.0__ subtract__2500.0__500.0__ |
| a number consists of num__3 digits whose sum is num__10 . the middle digit is equal to the sum of the other two and the number will be increased by num__99 if its digits are reversed . the number is : <o> a ) num__145 <o> b ) num__253 <o> c ) num__370 <o> d ) num__352 <o> e ) num__382 |
we can easily solve the ans using the result all options are fullfill num__1 st condition ie the sum of digit is num__10 now only b and d are full fill the num__2 nd condition ie middle digit = sum of the side digit now finally b is the result bcze if we add two number the number must be increase but here if we reverse num__352 then it become num__253 but if we add num__99 the itnever be a small number so thats why result is num__253 answer : b <eor> b <eos> |
b |
subtract__3.0__1.0__ subtract__352.0__99.0__ subtract__352.0__99.0__ |
subtract__3.0__1.0__ subtract__352.0__99.0__ subtract__352.0__99.0__ |
| there are two numbers . if num__55.0 of the first number is added to the second number then the second number increases to its five - fourth . find the ratio of the first number to the second number ? <o> a ) a ) num__0.833333333333 <o> b ) b ) num__0.625 <o> c ) c ) num__0.454545454545 <o> d ) d ) num__0.857142857143 <o> e ) e ) num__7 |
let the two numbers be x and y . num__0.55 * x + y = num__1.25 y = > num__0.55 x = num__0.25 y = > x / y = num__0.454545454545 c ) <eor> c <eos> |
c |
divide__0.25__0.55__ divide__0.25__0.55__ |
divide__0.25__0.55__ divide__0.25__0.55__ |
| in how many years rs num__100 will produce the same interest at num__5.0 as rs . num__600 produce in num__4 years at num__10.0 <o> a ) num__44 <o> b ) num__52 <o> c ) num__50 <o> d ) num__46 <o> e ) num__48 |
explanation : clue : firstly we need to calculate the si with prinical num__600 time num__4 years and rate num__10.0 it will be rs . num__240 then we can get the time as time = ( num__100 * num__240 ) / ( num__100 * num__5 ) = num__48 option e <eor> e <eos> |
e |
percent__100.0__48.0__ |
percent__100.0__48.0__ |
| a worker ' s take - home pay last year was the same each month and she saved the same fraction of her take - home pay each month . the total amount of money that she had saved at the end of the year was num__4 times the amount of that portion of her monthly take - home pay that she did not save . if all the money that she saved last year was from her take - home pay what fraction of her take - home pay did she save each month ? <o> a ) num__0.333333333333 <o> b ) num__0.25 <o> c ) num__0.2 <o> d ) num__0.166666666667 <o> e ) num__0.142857142857 |
let x be the fraction of her take - home pay that the worker saved . let p be the monthly pay . num__12 xp = num__4 ( num__1 - x ) p num__12 xp = num__4 p - num__4 xp num__16 xp = num__4 p x = num__0.25 the answer is b . <eor> b <eos> |
b |
add__4.0__12.0__ reverse__4.0__ reverse__4.0__ |
add__4.0__12.0__ reverse__4.0__ reverse__4.0__ |
| how many digits are required to number a book containing num__250 pages ? <o> a ) num__756 <o> b ) num__642 <o> c ) num__492 <o> d ) num__372 <o> e ) num__250 |
num__9 pages from num__1 to num__9 will require num__9 digits . num__90 pages from num__10 to num__99 will require num__90 * num__2 = num__180 digits . num__250 - ( num__90 + num__9 ) = num__151 pages will require num__151 * num__3 = num__453 digits . the total number of digits is num__9 + num__180 + num__453 = num__642 . the answer is b . <eor> b <eos> |
b |
add__1.0__9.0__ add__9.0__90.0__ multiply__2.0__90.0__ subtract__250.0__99.0__ add__1.0__2.0__ multiply__3.0__151.0__ multiply__1.0__642.0__ |
add__1.0__9.0__ add__9.0__90.0__ multiply__2.0__90.0__ subtract__250.0__99.0__ add__1.0__2.0__ multiply__3.0__151.0__ multiply__1.0__642.0__ |
| in how many ways can the letters of the word actual be rearranged such that the vowels always appear together ? <o> a ) num__6 ! / num__2 ! <o> b ) num__3 ! * num__3 ! <o> c ) num__4 ! / num__2 ! <o> d ) ( num__4 ! * num__3 ! ) / num__2 ! <o> e ) num__3 ! * num__3 ! / num__2 |
in the word abacus there are num__3 vowels - num__2 a ' s and u number of ways the letters of word actual be rearranged such that the vowels always appear together = ( num__4 ! * num__3 ! ) / num__2 ! we can consider the the num__3 vowels as a single unit and there are num__3 ways to arrange them . but since num__2 elements of vowel group are identical we divide by num__2 ! . the entire vowel group is considered as a single group . answer d <eor> d <eos> |
d |
coin_space__ choose__4.0__3.0__ |
coin_space__ choose__4.0__3.0__ |
| there is a total of num__84 marbles in a box each of which is red green blue or white . if one marble is drawn from the box at random the probability that it will be white is num__0.25 and the probability that it will be green is num__0.142857142857 . what is the probability that the marble will be either red or blue ? <o> a ) num__0.571428571429 <o> b ) num__0.642857142857 <o> c ) num__0.785714285714 <o> d ) num__0.607142857143 <o> e ) num__0.821428571429 |
p ( red or blue ) = num__1 - p ( white ) - p ( green ) = num__1.0 - num__0.25 - num__0.142857142857 = num__0.607142857143 the answer is d . <eor> d <eos> |
d |
multiply__1.0__0.6071__ |
multiply__1.0__0.6071__ |
| a worker makes a toy in every num__1 h . if he works for num__100 h then how many toys will he make ? <o> a ) num__40 <o> b ) num__54 <o> c ) num__45 <o> d ) num__39 <o> e ) num__100 |
no . of toys = num__100.0 = num__100 answer : e <eor> e <eos> |
e |
round__100.0__ |
round__100.0__ |
| a chessboard is an num__8 × num__8 array of identically sized squares . each square has a particular designation depending on its row and column . an l - shaped card exactly the size of four squares on the chessboard is laid on the chessboard as shown covering exactly four squares . this l - shaped card can be moved around rotated and even picked up and turned over to give the mirror - image of an l . in how many different ways can this l - shaped card cover exactly four squares on the chessboard ? <o> a ) num__256 <o> b ) num__336 <o> c ) num__424 <o> d ) num__512 <o> e ) num__672 |
select num__3 squares which are in a line together first . we will call them a block of num__3 squares . consider just the vertical arrangement for now ( for horizontal we will multiply everything by num__2 at the end ) . consider a block of num__3 squares lying vertically on the left edge ( the first column of the chess board ) . you can place another square on the right to make an l at either extreme of the block . so for each such block you can make an l in num__2 ways . there will be num__12 such blocks ( num__6 on either edge ) . you get num__12 * num__2 ls . now consider the blocks of num__3 squares lying vertically in columns num__2 to num__7 . you can make an l by placing a square on left or right at either end . so for each block of num__3 squares you can make num__4 ls . there are num__6 such blocks in each of the num__6 columns you get num__6 * num__6 * num__4 ls . total you get num__12 * num__2 + num__6 * num__6 * num__4 = num__168 ls . now you just multiply it by num__2 to account for the ls lying horizontally too . since it is a square the number of ls found vertically will be the same as the number of ls found horizontally . total = num__168 * num__2 = num__336 answer ( b ) <eor> b <eos> |
b |
subtract__8.0__2.0__ divide__8.0__2.0__ multiply__2.0__168.0__ multiply__2.0__168.0__ |
multiply__2.0__3.0__ divide__8.0__2.0__ multiply__2.0__168.0__ multiply__2.0__168.0__ |
| a train speeds past a pole in num__15 sec and a platform num__150 m long in num__25 sec its length is ? <o> a ) num__238 <o> b ) num__150 <o> c ) num__988 <o> d ) num__177 <o> e ) num__225 |
let the length of the train be x m and its speed be y m / sec . then x / y = num__15 = > y = x / num__15 ( x + num__150 ) / num__25 = x / num__15 = > x = num__225 m . answer : e <eor> e <eos> |
e |
round__225.0__ |
round__225.0__ |
| the number num__94 can be written as the sum of squares of num__3 integers . which of the following could be the difference between the largest and smallest integers of the num__3 ? <o> a ) num__2 <o> b ) num__5 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
notice that the question asks which of the followingcouldbe the difference between the largest and smallest integers not must be . the num__3 integers could be : + / - num__2 + / - num__3 and + / - num__9 so the difference could be num__7 num__11 or num__12 . since only one of them is among the choices then it must be the correct answer . answer : c <eor> c <eos> |
c |
subtract__9.0__2.0__ add__2.0__9.0__ add__3.0__9.0__ subtract__9.0__2.0__ |
subtract__9.0__2.0__ add__2.0__9.0__ add__3.0__9.0__ subtract__9.0__2.0__ |
| two trains of equal length are running on parallel lines in the same direction at num__46 km / hr and num__36 km / hr . the faster train catches and completely passes the slower train in num__72 seconds . what is the length of each train ( in meters ) ? <o> a ) num__70 <o> b ) num__80 <o> c ) num__90 <o> d ) num__100 <o> e ) num__110 |
the relative speed = num__46 - num__36 = num__10 km / hr = num__10 * num__0.277777777778 = num__2.77777777778 m / s in num__72 seconds the relative difference in distance traveled is num__72 * num__2.77777777778 = num__200 meters this distance is twice the length of each train . the length of each train is num__100.0 = num__100 meters the answer is d . <eor> d <eos> |
d |
subtract__46.0__36.0__ divide__10.0__36.0__ round__100.0__ |
subtract__46.0__36.0__ divide__10.0__36.0__ subtract__200.0__100.0__ |
| a certain article of clothing was discounted during a special sale to num__0.666666666667 of its original retail price . when the clothing did n ' t sell it was discounted even further to num__0.5 of its original retail price during a second sale . by what percent did the price of this article of clothing decrease from the first sale to the second sale ? <o> a ) num__50.0 <o> b ) num__22.22 <o> c ) num__25.0 <o> d ) num__16.66 <o> e ) num__12.5 % |
say the original retail price of the item was $ num__200 . the price after the first sale = num__0.666666666667 * $ num__200 = $ num__133.333333333 . the price after the second sale = num__0.5 * $ num__200 = $ num__100 . the percent change from the first sale to the second = ( num__133.333333333 - num__100 ) / num__150 = num__0.333333333333 = num__22.22 . answer : b . <eor> b <eos> |
b |
percent__100.0__22.22__ |
percent__100.0__22.22__ |
| in a num__500 m race ravi beats ramesh by num__5 seconds or num__100 m . they decide to run another race and this time ravi gives ramesh a head tart of num__200 m . if ravi ’ s speed is twice his previous speed and ramesh ’ s speed is one and half times his previous speed how far from the starting point should the winning post be so that they finish at the same time ? <o> a ) num__500 m <o> b ) num__1000 m <o> c ) num__1500 m <o> d ) num__2000 m <o> e ) num__2500 m |
ravi beats ramesh by num__5 sec or num__100 mtr . . . . means ramesh cover num__100 mtr in num__5 sec . . . . by this we can get ramesh speed in num__20 m / sec num__20 . . . . . . . . . . . . . in num__1 sec num__500 . . . . . . . . in num__25.0 = num__25 sec ( ramesh ' s time ) so ravi take num__20 sec to cover num__500 mtr . . . . . . by this we can get that ravis speed is num__25 m / sec now assume x be the distance from staring line where we need to put the winning post . according to question ravi ’ s speed will be num__25 * num__2 = num__50 m / sec & ramesh ’ s speed will be num__20 * num__1.5 = num__30 m / sec now assume x be the distance from staring line where we need to put the winning post . they will take same time to reach the wining post . as ravi gives ramesh a num__200 m ahead start so he has to accede x meter and ramesh has to accede ( x - num__200 ) meter hence we may write x / num__50 = ( x - num__200 ) / num__30 x = num__500 meter answer : a <eor> a <eos> |
a |
divide__100.0__5.0__ divide__500.0__20.0__ divide__200.0__100.0__ divide__100.0__2.0__ add__5.0__25.0__ round__500.0__ |
divide__100.0__5.0__ divide__500.0__20.0__ divide__200.0__100.0__ divide__100.0__2.0__ multiply__1.5__20.0__ divide__500.0__1.0__ |
| out of first num__20 natural numbers one number is selected at random . the probability that it is either an even number or a prime number is ? <o> a ) num__0.62962962963 <o> b ) num__0.68 <o> c ) num__0.223684210526 <o> d ) num__0.85 <o> e ) num__0.739130434783 |
n ( s ) = num__20 n ( even no ) = num__10 = n ( e ) n ( prime no ) = num__8 = n ( p ) p ( e ᴜ p ) = num__0.5 + num__0.4 - num__0.05 = num__0.85 answer : d <eor> d <eos> |
d |
union_prob__0.5__0.4__0.05__ union_prob__0.5__0.4__0.05__ |
union_prob__0.5__0.4__0.05__ union_prob__0.5__0.4__0.05__ |
| johnny has num__4 classes . he averaged a mark of num__83 in those classes . if johnny takes night school and scores an num__88 in the course what will his new average be ? <o> a ) num__83 <o> b ) num__84 <o> c ) num__85 <o> d ) num__85.5 <o> e ) num__86 |
( sum of the num__4 marks ) / num__4 = num__83 sum of the num__4 marks = num__332 new sum = num__332 + num__88 = num__420 new average = num__84.0 = num__84 ans : b <eor> b <eos> |
b |
multiply__4.0__83.0__ add__88.0__332.0__ subtract__88.0__4.0__ subtract__88.0__4.0__ |
multiply__4.0__83.0__ add__88.0__332.0__ subtract__88.0__4.0__ subtract__88.0__4.0__ |
| a student chose a number multiplied it by num__7 then subtracted num__150 from the result and got num__130 . what was the number he chose ? <o> a ) num__40 <o> b ) num__42 <o> c ) num__44 <o> d ) num__46 <o> e ) num__48 |
let x be the number he chose then num__7 ⋅ x − num__150 = num__130 num__7 x = num__280 x = num__40 correct answer a <eor> a <eos> |
a |
add__150.0__130.0__ divide__280.0__7.0__ divide__280.0__7.0__ |
add__150.0__130.0__ divide__280.0__7.0__ divide__280.0__7.0__ |
| the surface area of a cube is num__1734 sq . cm . find its volume <o> a ) num__2334 cubic . cm <o> b ) num__3356 cubic . cm <o> c ) num__4913 cubic . cm <o> d ) num__3478 cubic . cm <o> e ) none of these |
explanation : let the edge of the cube bea . then num__6 a num__2 = num__1734 = > a = num__17 cm . volume = a num__3 = num__173 = num__4193 cm num__3 answer : c <eor> c <eos> |
c |
volume_cube__17.0__ |
volume_cube__17.0__ |
| a certain pilot flew num__400 miles to city k at an average speed of num__350 miles per hour with the wind and made the trip back at an average speed of num__150 miles per hour against the wind . which of the following is closest to the pilot ’ s average speed in miles per hour for the round - trip ? <o> a ) num__280 <o> b ) num__210 <o> c ) num__300 <o> d ) num__310 <o> e ) num__320 |
avg speed = total distance / total time total distance = num__800 total time = num__1.14285714286 + num__2.66666666667 = num__3.80952380952 = > avg speed = ( num__800 * num__21 ) / num__80 = num__210 ( approx ) ans is b <eor> b <eos> |
b |
divide__400.0__350.0__ divide__400.0__150.0__ round__210.0__ |
divide__400.0__350.0__ divide__400.0__150.0__ round__210.0__ |
| a toy store regularly sells all stock at a discount of num__10 percent . if an additional num__10 percent were deducted from the discount price during a special sale what would be the lowest possible price of a toy costing $ num__10 before any discount ? <o> a ) $ num__7.20 <o> b ) $ num__8.00 <o> c ) $ num__8.10 <o> d ) $ num__8.27 <o> e ) $ num__9.00 |
sale price = ( original price - num__10.0 decrease ) - num__10.0 decrease price = [ num__10 * ( num__0.9 ) ] * ( num__0.9 ) = num__10 * ( . num__90 ) * ( . num__90 ) = num__9 * . num__90 = num__8.10 the answer is c . <eor> c <eos> |
c |
percent__10.0__90.0__ percent__9.0__90.0__ percent__9.0__90.0__ |
percent__10.0__90.0__ percent__9.0__90.0__ percent__9.0__90.0__ |
| in a certain store the profit is num__320.0 of the cost . if the cost increases by num__25.0 but the selling price remains constant approximately what percentage of the selling price is the profit ? <o> a ) num__30.0 <o> b ) num__70 percent <o> c ) num__100.0 <o> d ) num__120.0 <o> e ) none |
solution let c . p . = rs . num__100 . then profit = rs . num__320 s . p . = num__420 . let c . p . = rs . num__125.0 of rs . num__100 = rs . num__125 . new s . p . = rs . num__420 . profit = rs . ( num__420 - num__125 ) = rs . num__295 . ∴ required percentage = ( num__0.702380952381 x num__100 ) % = num__70.2380952381 % = num__70.0 . answer b <eor> b <eos> |
b |
percent__70.0__100.0__ |
percent__70.0__100.0__ |
| the ratio num__25 : num__50 expressed as percent equals to <o> a ) num__50.0 <o> b ) num__85.0 <o> c ) num__25.0 <o> d ) num__75.0 <o> e ) none of above |
explanation : actually it means num__25 is what percent of num__50 which can be calculated as ( num__0.5 ) * num__100 = num__25 * num__2 = num__50 answer : option a <eor> a <eos> |
a |
divide__25.0__50.0__ divide__50.0__0.5__ reverse__0.5__ divide__25.0__0.5__ |
divide__25.0__50.0__ divide__50.0__0.5__ reverse__0.5__ multiply__25.0__2.0__ |
| consider a quarter of a circle of radius num__36 . let r be the radius of the circle inscribed in this quarter of a circle . find r . <o> a ) num__16 * ( sqr num__2 - num__1 ) <o> b ) num__8 * ( sqr num__3 - num__1 ) <o> c ) num__4 * ( sqr num__7 - num__1 ) <o> d ) num__36 * ( sqr num__2 - num__1 ) <o> e ) none of these |
i got num__36 / ( sqr num__2 + num__1 ) and just forgot to multiply by ( sqr num__2 - num__1 ) . answer is d <eor> d <eos> |
d |
multiply__36.0__1.0__ |
multiply__36.0__1.0__ |
| dan ' s age after num__16 years will be num__4 times his age num__8 years ago . what is the present age of dan ? <o> a ) num__12 <o> b ) num__16 <o> c ) num__20 <o> d ) num__24 <o> e ) num__28 |
let dan ' s present age be x . x + num__16 = num__4 ( x - num__8 ) num__3 x = num__48 x = num__16 the answer is b . <eor> b <eos> |
b |
multiply__16.0__3.0__ divide__48.0__3.0__ |
multiply__16.0__3.0__ divide__48.0__3.0__ |
| a train covers a distance of num__18 km in num__10 min . if it takes num__6 sec to pass a telegraph post then the length of the train is ? <o> a ) num__120 m <o> b ) num__180 m <o> c ) num__240 m <o> d ) num__220 m <o> e ) num__280 m |
speed = ( num__1.8 * num__60 ) km / hr = ( num__108 * num__0.277777777778 ) m / sec = num__30 m / sec . length of the train = num__30 * num__6 = num__180 m . answer : b <eor> b <eos> |
b |
divide__18.0__10.0__ hour_to_min_conversion__ multiply__18.0__6.0__ multiply__18.0__10.0__ round__180.0__ |
divide__18.0__10.0__ multiply__10.0__6.0__ multiply__18.0__6.0__ multiply__18.0__10.0__ multiply__18.0__10.0__ |
| use distributive property to solve the problem below : maria bought num__10 notebooks and num__5 pens costing num__3 dollars each . how much did maria pay ? <o> a ) num__30 dollars <o> b ) num__40 dollars <o> c ) num__45 dollars <o> d ) num__60 dollars <o> e ) num__70 dollars |
solution num__3 × ( num__10 + num__5 ) = num__3 × num__10 + num__3 × num__5 = num__30 + num__15 = num__45 dollars answer c <eor> c <eos> |
c |
multiply__10.0__3.0__ add__10.0__5.0__ multiply__3.0__15.0__ multiply__3.0__15.0__ |
multiply__10.0__3.0__ add__10.0__5.0__ add__15.0__30.0__ add__15.0__30.0__ |
| the present ages of three persons in proportions num__4 : num__7 : num__5 . eight years ago the sum of their ages was num__56 . find their present ages ( in years ) . <o> a ) num__8 num__20 num__28 <o> b ) num__16 num__28 num__36 <o> c ) num__20 num__35 num__25 <o> d ) num__16 num__28 num__34 <o> e ) num__16 num__28 num__33 |
let their present ages be num__4 x num__7 x and num__5 x years respectively . then ( num__4 x - num__8 ) + ( num__7 x - num__8 ) + ( num__5 x - num__8 ) = num__56 num__16 x = num__5 . their present ages are num__4 x = num__20 years num__7 x = num__35 years and num__5 x = num__25 years respectively . answer : c <eor> c <eos> |
c |
divide__56.0__7.0__ multiply__4.0__5.0__ multiply__7.0__5.0__ add__5.0__20.0__ multiply__4.0__5.0__ |
divide__56.0__7.0__ add__4.0__16.0__ multiply__7.0__5.0__ add__5.0__20.0__ add__4.0__16.0__ |
| a man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream . the ratio of the speed of the boat ( in still water ) and the stream is <o> a ) num__1 : num__2 <o> b ) num__1 : num__3 <o> c ) num__2 : num__1 <o> d ) num__3 : num__1 <o> e ) none |
sol . let man ’ s rate upstream be x kmph . then his rate downstream = num__2 x kmph . ∴ ( speed in still water ) : ( speed of stream ) = [ num__2 x + x / num__2 ] : [ num__2 x - x / num__2 ] num__3 x / num__2 : x / num__2 = num__3 : num__1 . answer d <eor> d <eos> |
d |
subtract__3.0__2.0__ round__3.0__ |
subtract__3.0__2.0__ add__1.0__2.0__ |
| a shopkeeper sells two articles at rs . num__1000 each making a profit of num__20.0 on the first article and a loss of num__20.0 on the second article . find the net profit or loss that he makes ? <o> a ) num__8.0 <o> b ) num__2.0 <o> c ) num__3.0 <o> d ) num__4.0 <o> e ) num__6 % |
sp of first article = num__1000 profit = num__20.0 cp = ( sp ) * [ num__100 / ( num__100 + p ) ] = num__833.333333333 = num__833.333333333 sp of second article = num__1000 loss = num__20.0 cp = ( sp ) * [ num__100 / ( num__100 - l ) ] = num__1250.0 = num__1250 total sp = num__2000 total cp = num__833.333333333 + num__1250 = num__2083.33333333 cp is more than sp he makes a loss . loss = cp - sp = ( num__2083.33333333 ) - num__2000 = num__83.3333333333 loss percent = [ ( num__83.3333333333 ) / ( num__2083.33333333 ) ] * num__100 = num__0.04 * num__100 = num__4.0 answer : d <eor> d <eos> |
d |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| the smallest number when increased by ` ` num__1 ` ` is exactly divisible by num__618 num__3648 num__60 is : <o> a ) num__724 <o> b ) num__721 <o> c ) num__720 <o> d ) num__719 <o> e ) num__700 |
lcm = num__720 num__720 - num__1 = num__719 answer : d <eor> d <eos> |
d |
subtract__720.0__1.0__ multiply__1.0__719.0__ |
subtract__720.0__1.0__ subtract__720.0__1.0__ |
| a cyclist traveled for two days . on the second day the cyclist traveled num__3 hours longer and at an average speed num__10 mile per hour slower than she traveled on the first day . if during the two days she traveled a total of num__280 miles and spent a total of num__12 hours traveling what was her average speed on the second day ? <o> a ) num__5 mph <o> b ) num__10 mph <o> c ) num__20 mph <o> d ) num__30 mph <o> e ) num__40 mph |
solution : d = num__280 mi t = num__12 hrs day num__1 time = t num__1 day num__2 time = t num__2 t num__2 - t num__1 = num__4 hrs - - - - - ( i ) t num__1 + t num__2 = num__12 hrs - - - - - ( ii ) adding i and ii t num__2 = num__8 hrs and t num__1 = num__4 hrs day num__1 rate = r num__1 day num__2 rate = r num__2 r num__1 - r num__2 = num__10 mph i . e . r num__1 = num__10 + r num__2 num__280 = num__8 r num__2 + num__4 r num__1 i . e . num__280 = num__8 r num__2 + num__4 ( num__10 + r num__2 ) i . e . r num__2 = num__20 mph answer : b <eor> b <eos> |
b |
subtract__3.0__1.0__ add__3.0__1.0__ subtract__10.0__2.0__ multiply__10.0__2.0__ round__10.0__ |
subtract__3.0__1.0__ add__3.0__1.0__ subtract__10.0__2.0__ add__12.0__8.0__ subtract__12.0__2.0__ |
| in a group of hats consisting of only blue hats green hats and purple hats the ratio of blue hats to green hats to purple hats is num__7 : num__4 : num__12 . if there are a total of num__161 hats in this group how many of these hats are not blue ? <o> a ) num__28 <o> b ) num__42 <o> c ) num__48 <o> d ) num__112 <o> e ) num__76 |
since the hats blue green and purple are in ratio of num__7 : num__4 : num__12 . . . the total no . of balls will be num__7 x + num__4 x + num__12 x = num__161 or num__23 x = num__161 . . . here num__7 x num__4 x and num__12 x represent hats of each type num__23 x = num__161 so x = num__7 . . . blue hats = num__7 * num__7 = num__49 . . . not blue will be num__161 - num__49 = num__112 . ans d <eor> d <eos> |
d |
divide__161.0__7.0__ subtract__161.0__49.0__ subtract__161.0__49.0__ |
divide__161.0__7.0__ subtract__161.0__49.0__ subtract__161.0__49.0__ |
| by selling an article at rs . num__600 a profit of num__25.0 is made . find its cost price ? <o> a ) num__288 <o> b ) num__992 <o> c ) num__678 <o> d ) num__480 <o> e ) num__131 |
sp = num__600 cp = ( sp ) * [ num__100 / ( num__100 + p ) ] = num__600 * [ num__100 / ( num__100 + num__25 ) ] = num__600 * [ num__0.8 ] = rs . num__480 answer : d <eor> d <eos> |
d |
percent__100.0__480.0__ |
percent__100.0__480.0__ |
| a person can swim in still water at num__4 km / h . if the speed of water num__2 km / h how many hours will the man take to swim back against the current for num__6 km ? <o> a ) num__3 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
m = num__4 s = num__2 us = num__4 - num__2 = num__2 d = num__6 t = num__3.0 = num__3 answer : a <eor> a <eos> |
a |
divide__6.0__2.0__ round__3.0__ |
divide__6.0__2.0__ subtract__6.0__3.0__ |
| x starts a business with rs . num__45000 . y joins in the business after num__3 months with rs . num__30000 . what will be the ratio in which they should share the profit at the end of the year ? <o> a ) num__1 : num__2 <o> b ) num__1 : num__3 <o> c ) num__2 : num__1 <o> d ) num__1 : num__4 <o> e ) num__1 : num__5 |
ratio in which they should share the profit = ratio of the investments multiplied by the time period = num__45000 × num__12 : num__30000 × num__9 = num__45 × num__12 : num__30 × num__9 = num__3 × num__12 : num__2 × num__9 = num__2 : num__1 answer is c . <eor> c <eos> |
c |
subtract__12.0__3.0__ subtract__3.0__2.0__ subtract__3.0__1.0__ |
subtract__12.0__3.0__ subtract__3.0__2.0__ subtract__3.0__1.0__ |
| find num__356 x num__936 - num__356 x num__836 = ? <o> a ) num__35600 <o> b ) num__34500 <o> c ) num__49630 <o> d ) num__93600 <o> e ) none |
answer num__356 x num__936 - num__356 x num__836 = num__356 x ( num__936 - num__836 ) = num__356 x num__100 = num__35600 . option : a <eor> a <eos> |
a |
subtract__936.0__836.0__ multiply__356.0__100.0__ multiply__356.0__100.0__ |
subtract__936.0__836.0__ multiply__356.0__100.0__ multiply__356.0__100.0__ |
| the milk level in a rectangular box measuring num__64 feet by num__25 feet is to be lowered by num__6 inches . how many gallons of milk must be removed ? ( num__1 cu ft = num__7.5 gallons ) <o> a ) num__100 <o> b ) num__250 <o> c ) num__750 <o> d ) num__5625 <o> e ) num__6000 |
num__6 inches = num__0.5 feet ( there are num__12 inches in a foot . ) so num__64 * num__25 * num__0.5 = num__800 feet ^ num__3 of milk must be removed which equals to num__800 * num__7.5 = num__6000 gallons . answer : e . <eor> e <eos> |
e |
divide__6.0__0.5__ multiply__6.0__0.5__ multiply__7.5__800.0__ multiply__1.0__6000.0__ |
divide__6.0__0.5__ multiply__6.0__0.5__ multiply__7.5__800.0__ multiply__1.0__6000.0__ |
| p can do a certain work in num__16 days . q is num__60.0 more efficient than p . how many days does q alone take to do the same job ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
each day p can do num__0.0625 of the work . each day q can do num__1.6 / num__16 of the work = num__0.1 of the work the time taken by b alone to complete the total work is num__10.0 = num__10 days . the answer is c . <eor> c <eos> |
c |
mile_to_km_conversion__ multiply__0.0625__1.6__ divide__16.0__1.6__ round__10.0__ |
mile_to_km_conversion__ divide__1.6__16.0__ divide__16.0__1.6__ divide__16.0__1.6__ |
| a can do a piece of work in num__15 days and b in num__20 days . they began the work together but num__5 days before the completion of the work a leaves . the work was completed in ? <o> a ) num__11 num__0.333333333333 <o> b ) num__19 num__0.428571428571 <o> c ) num__11 num__0.6 <o> d ) num__11 num__0.428571428571 <o> e ) num__11 num__3.0 |
( x – num__5 ) / num__15 + x / num__20 = num__1 x = num__11 num__0.428571428571 days answer : d <eor> d <eos> |
d |
round__11.0__ |
divide__11.0__1.0__ |
| in march kurt ran an average of num__1.2 miles an hour . if by june he had increased his pace by num__20 seconds per mile then which of the following expresses the number of hours it would take kurt to complete one mile in june ? <o> a ) num__59.8333333333 ^ num__2 <o> b ) num__49.6666666667 ^ num__2 <o> c ) num__39.8333333333 ^ num__2 <o> d ) num__59.7666666667 <o> e ) num__60 ^ num__0.000557103064067 |
kurt ran at an average of num__1.2 miles / hour in march . so to run num__1 mile he would take num__1 / num__1.2 hours = ( num__60 * num__60 ) / num__1.2 seconds = num__3000 seconds . if he increases his speed by num__20 seconds he will complete a mile in num__2980 seconds . converting in hours = num__2980 / ( num__60 * num__60 ) = num__2980 / ( num__60 ^ num__2 ) answer : b <eor> b <eos> |
b |
round_down__1.2__ subtract__3000.0__20.0__ divide__2980.0__60.0__ |
round_down__1.2__ subtract__3000.0__20.0__ divide__2980.0__60.0__ |
| find the area of a parallelogram with base num__12 cm and height num__8 cm ? <o> a ) num__96 cm num__2 <o> b ) num__104 cm num__2 <o> c ) num__78 cm num__2 <o> d ) num__86 cm num__2 <o> e ) num__76 cm num__2 |
area of a parallelogram = base * height = num__12 * num__8 = num__96 cm num__2 answer : a <eor> a <eos> |
a |
multiply__12.0__8.0__ multiply__12.0__8.0__ |
multiply__12.0__8.0__ multiply__12.0__8.0__ |
| a train num__200 m long is running at the speed of num__30 km / hr . find the time taken by it to pass a man standing near the railway line in seconds <o> a ) num__12 sec <o> b ) num__6 sec <o> c ) num__7 sec <o> d ) num__8 sec <o> e ) num__9 sec |
explanation : speed of the train = ( num__60 x num__0.277777777778 m / sec = num__16.6666666667 m / sec . distance moved in passing the standing man = num__100 m . required time taken = num__200 / ( num__16.6666666667 ) = ( num__200 Ã — ( num__0.06 ) ) sec = num__12 sec answer : option a <eor> a <eos> |
a |
hour_to_min_conversion__ divide__200.0__16.6667__ round__12.0__ |
hour_to_min_conversion__ divide__200.0__16.6667__ divide__200.0__16.6667__ |
| in a store the profit is num__320.0 of the cost . if the cost increases by num__25.0 but the sp remains constant approximately what % of the sp is the profit ? <o> a ) num__50.0 <o> b ) num__70.0 <o> c ) num__90.0 <o> d ) num__100.0 <o> e ) num__140 % |
let c . p . = rs . num__100 . then profit = rs . num__320 s . p . = rs . num__420 new c . p . = num__125.0 of rs . num__100 = rs . num__125 . new s . p . = rs . num__420 profit = num__420 - num__125 = rs . num__295 required percentage = num__0.702380952381 * num__100 = num__70.2380952381 = num__70.0 b <eor> b <eos> |
b |
percent__70.0__100.0__ |
percent__70.0__100.0__ |
| if a - b = num__5 and a num__2 + b num__2 = num__29 find the value of ab . <o> a ) a ) num__2 <o> b ) b ) num__12 <o> c ) c ) num__15 <o> d ) d ) num__18 <o> e ) e ) num__20 |
explanation : num__2 ab = ( a num__2 + b num__2 ) - ( a - b ) num__2 = num__29 - num__25 = num__4 ab = num__2 . answer : a <eor> a <eos> |
a |
subtract__29.0__25.0__ subtract__4.0__2.0__ |
subtract__29.0__25.0__ subtract__4.0__2.0__ |
| the average age of a husband and his wife was num__23 years at the time of their marriage . after five years they have a one year old child . what is the average age of the family ? <o> a ) num__21 years <o> b ) num__20 years <o> c ) num__18 years <o> d ) num__19 years <o> e ) num__17 years |
after num__5 years the average age of the two will be num__28 . now the child is one year old total age = num__28 * num__2 + num__1 = num__57 average age = total age / family members ( num__3 ) = > num__19.0 = num__19 answer : d <eor> d <eos> |
d |
add__23.0__5.0__ add__1.0__2.0__ divide__57.0__3.0__ multiply__1.0__19.0__ |
add__23.0__5.0__ add__1.0__2.0__ divide__57.0__3.0__ multiply__1.0__19.0__ |
| if n = num__8 ^ num__8 – num__7 what is the units digit of n ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__7 <o> e ) num__4 |
num__8 ^ num__8 - num__8 = num__8 ( num__8 ^ num__7 - num__1 ) = = > num__8 ( num__2 ^ num__21 - num__1 ) last digit of num__2 ^ num__21 is num__2 based on what explanation livestronger is saying . num__2 ^ num__24 - num__1 yields num__2 - num__1 = num__1 as the unit digit . now on multiply this with num__7 we get unit digit as num__7 answer : d <eor> d <eos> |
d |
subtract__8.0__7.0__ subtract__8.0__1.0__ |
subtract__8.0__7.0__ subtract__8.0__1.0__ |
| num__9.009 / num__5.005 = <o> a ) num__1.08 <o> b ) num__1.4 <o> c ) num__1.8018 <o> d ) num__1.4014 <o> e ) num__1.8 |
num__9.009 / num__5.005 = num__1.8 = num__9 ( num__1001 ) / num__5 ( num__1001 ) = num__1.8 = num__1.8 the answer is e . <eor> e <eos> |
e |
divide__9.009__5.005__ round_down__9.009__ round_down__5.005__ divide__9.009__5.005__ |
divide__9.009__5.005__ round_down__9.009__ divide__9.0__1.8__ divide__9.009__5.005__ |
| which of the following describes all values of x for which num__36 – x ^ num__2 > = num__0 ? <o> a ) x > = num__1 <o> b ) x < = – num__1 <o> c ) - num__6 < = x < = num__6 <o> d ) x < = – num__1 or x > = num__1 <o> e ) – num__1 < = x < = num__1 |
num__36 - x ^ num__2 > = num__0 means x ^ num__2 - num__36 < = num__0 = > ( x - num__6 ) ( x + num__6 ) < = num__0 = > - num__6 < = x < = num__6 answer - c <eor> c <eos> |
c |
divide__36.0__6.0__ |
divide__36.0__6.0__ |
| if num__15 men working num__9 hours a day can reap a field in num__16 days in how many days will num__18 men reap the field working num__8 hours a day ? <o> a ) num__10 <o> b ) num__15 <o> c ) num__20 <o> d ) num__25 <o> e ) none of them |
let the required number of days be x . more men less days ( indirect proportion ) less hours per day more days ( indirect proportion ) men num__18 : num__15 hours per day num__8 : num__9 } : : num__16 : x ( num__18 x num__8 x x ) = ( num__15 x num__9 x num__16 ) = x = ( num__44 x num__15 ) num__144 = num__15 hence required number of days = num__15 . answer is b . <eor> b <eos> |
b |
multiply__9.0__16.0__ round__15.0__ |
multiply__9.0__16.0__ round__15.0__ |
| the average of num__50 numbers id num__44 . if two numbers namely num__45 and num__55 are discarded the average of the remaining numbers is : <o> a ) num__22.35 <o> b ) num__33.25 <o> c ) num__22.25 <o> d ) num__11.35 <o> e ) num__43.75 |
explanation : total of num__50 numbers = ( num__50 × num__44 ) = num__2200 total of num__48 numbers = ( num__2200 - ( num__45 + num__55 ) ] = num__2100 required average = num__43.75 = num__43.75 answer : e <eor> e <eos> |
e |
multiply__50.0__44.0__ divide__2100.0__48.0__ divide__2100.0__48.0__ |
multiply__50.0__44.0__ divide__2100.0__48.0__ divide__2100.0__48.0__ |
| num__10.0 of inhabitants of a village having died of cholera a panic set in during which num__25.0 of the remaining inhabitants let the village . the population is then reduced to num__4050 . find the original inhabitants <o> a ) num__5500 <o> b ) num__6000 <o> c ) num__6500 <o> d ) num__7000 <o> e ) num__8000 |
explanation : let the total number is x then ( num__100 - num__25 ) % of ( num__100 - num__10 ) % x = num__4050 = > num__75.0 of num__90.0 of x = num__4050 = > num__0.75 * num__0.9 * x = num__4050 = > x = ( num__4050 * num__50 ) / num__27 = num__6000 answer : option b <eor> b <eos> |
b |
percent__100.0__6000.0__ |
percent__100.0__6000.0__ |
| the average age of num__15 students of a class is num__15 years . out of these the average age of num__5 students is num__14 years and that of the other num__9 students is num__16 years . the age of the num__15 th student is ? <o> a ) num__9 <o> b ) num__11 <o> c ) num__15 <o> d ) num__12 <o> e ) num__20 |
age of the num__15 th student = num__15 * num__15 - ( num__14 * num__5 + num__16 * num__9 ) = num__225 - num__214 = num__11 years answer is b <eor> b <eos> |
b |
subtract__16.0__5.0__ subtract__16.0__5.0__ |
subtract__16.0__5.0__ subtract__16.0__5.0__ |
| what is the units digit of num__6 ^ m ( num__2 ^ num__7 + num__1 ) for a positive integer m ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__4 <o> d ) num__7 <o> e ) num__0 |
equation can be re - written as num__6 ^ m ( num__2 ^ num__7 + num__1 ) = num__6 ^ m ( num__128 + num__1 ) = num__6 ^ m ( num__129 ) now putting m = num__0 ; eqn = num__129 putting m = num__1 ; eqn = num__129 . num__6 putting m = num__2 ; eqn = num__129 . num__36 putting m = num__3 ; eqn = num__129 . num__216 we can see that there lies a pattern while multiplying num__129 with multiples of num__6 units digit is always going to be num__4 answer : c <eor> c <eos> |
c |
add__1.0__128.0__ divide__6.0__2.0__ multiply__6.0__36.0__ subtract__6.0__2.0__ subtract__6.0__2.0__ |
add__1.0__128.0__ add__2.0__1.0__ multiply__6.0__36.0__ subtract__6.0__2.0__ subtract__6.0__2.0__ |
| arun and tarun can do a work in num__10 days . after num__4 days tarun went to his village . how many days are required to complete the remaining work by arun alone . arun can do the work alone in num__20 days . <o> a ) num__12 days . <o> b ) num__17 days . <o> c ) num__18 days . <o> d ) num__19 days . <o> e ) num__20 days . |
they together completed num__0.4 work in num__4 days . balance num__0.6 work will be completed by arun alone in num__20 * num__0.6 = num__12 days . answer : a <eor> a <eos> |
a |
divide__4.0__10.0__ km_to_mile_conversion__ multiply__20.0__0.6__ round__12.0__ |
divide__4.0__10.0__ km_to_mile_conversion__ multiply__20.0__0.6__ multiply__20.0__0.6__ |
| a person covered one - fourth of the total distance at num__26 kmph and remaining distance at num__24 kmph . what is the average speed for the total distance ? <o> a ) num__21 ( num__0.125 ) kmph <o> b ) num__21 ( num__0.333333333333 ) kmph <o> c ) num__21 ( num__2.33333333333 ) kmph <o> d ) num__27 ( num__0.333333333333 ) kmph <o> e ) num__21 ( num__0.125 ) kmph |
let the total distance be x km total time taken = ( x / num__4 ) / num__16 + ( num__3 x / num__4 ) / num__24 = x / num__64 + x / num__32 = num__3 x / num__64 average speed = x / ( num__3 x / num__64 ) = num__21.3333333333 kmph = num__21 ( num__0.333333333333 ) kmph . answer : b <eor> b <eos> |
b |
multiply__4.0__16.0__ divide__64.0__3.0__ round_down__21.3333__ reverse__3.0__ round_down__21.3333__ |
multiply__4.0__16.0__ divide__64.0__3.0__ round_down__21.3333__ reverse__3.0__ round_down__21.3333__ |
| a team of four entered for a shooting competition . the best marks man scored num__85 points . if he had scored num__92 points the average scores for . the team would have been num__84 . how many points altogether did the team score ? <o> a ) num__288 <o> b ) num__329 <o> c ) num__168 <o> d ) num__127 <o> e ) num__664 |
explanation : num__4 * num__84 = num__336 - num__7 = num__329 answer : b <eor> b <eos> |
b |
multiply__84.0__4.0__ subtract__92.0__85.0__ subtract__336.0__7.0__ subtract__336.0__7.0__ |
multiply__84.0__4.0__ subtract__92.0__85.0__ subtract__336.0__7.0__ subtract__336.0__7.0__ |
| product of two natural numbers is num__5 . then the sum of reciprocals of their squares is <o> a ) num__1.00346020761 <o> b ) num__1.04 <o> c ) num__3.22222222222 <o> d ) num__15.2631578947 <o> e ) none of these |
explanation : if the numbers are a b then ab = num__5 as num__17 is a prime number so a = num__1 b = num__5 . num__1 / a num__2 + num__1 / b num__2 = num__1.0 ( num__2 ) + num__0.2 ( num__2 ) = num__1.04 option b <eor> b <eos> |
b |
reverse__5.0__ multiply__1.0__1.04__ |
reverse__5.0__ divide__1.04__1.0__ |
| raj and roshan has some money with them in the ratio num__5 : num__4 respectively . if raj has rs . num__45 . how much money raj has more than roshan ? <o> a ) num__5 <o> b ) num__9 <o> c ) num__16 <o> d ) num__12 <o> e ) num__18 |
let the money roshan has be x then ( num__45 / x ) = num__1.25 upon cross multiplication and solving for x we get x = num__36 so raj has rs . num__9 more than roshan answer : b <eor> b <eos> |
b |
divide__5.0__4.0__ divide__45.0__1.25__ add__5.0__4.0__ add__5.0__4.0__ |
divide__5.0__4.0__ divide__45.0__1.25__ divide__45.0__5.0__ divide__45.0__5.0__ |
| the average of seven numbers is num__20 . the average of first three numbers is num__14 and the average of last three numbers is num__19 . what is the middle number ? <o> a ) num__27 <o> b ) num__29 <o> c ) num__31 <o> d ) num__41 <o> e ) num__35 |
the total of seven numbers = num__7 x num__20 = num__140 the total of first num__3 and last num__3 numbers is = num__3 x num__14 + num__3 x num__19 = num__99 so the middle number is ( num__140 - num__99 ) = num__41 d <eor> d <eos> |
d |
multiply__20.0__7.0__ subtract__140.0__99.0__ subtract__140.0__99.0__ |
multiply__20.0__7.0__ subtract__140.0__99.0__ subtract__140.0__99.0__ |
| a boat takes num__19 hours for travelling downstream from point a to point b and coming back to a point c midway between a and b . if the velocity of the stream is num__4 km / h and the speed of the boat in still water is num__14 km / h what is the distance between a and b ? <o> a ) num__200 km <o> b ) num__180 km <o> c ) num__160 km <o> d ) num__220 km <o> e ) none of these |
speed of boat for downstream = num__14 + num__4 = num__18 km / hr speed of boat for upstream = num__14 – num__4 = num__10 km / hr distance = x x / num__18 + ( x / num__2 ) / num__10 = num__19 x = num__180 km answer : b <eor> b <eos> |
b |
add__4.0__14.0__ subtract__14.0__4.0__ multiply__10.0__18.0__ round__180.0__ |
add__4.0__14.0__ subtract__14.0__4.0__ multiply__10.0__18.0__ round__180.0__ |
| with the money i have i can buy num__50 pens or num__150 pencils . i kept num__10.0 aside for taxi fare . with the remaining i purchased num__54 pencils and p pens . what is the value of p ? <o> a ) num__32 <o> b ) num__30 <o> c ) num__27 <o> d ) num__25 <o> e ) num__23 |
since cost of num__50 pens = num__150 pencils with the cost of num__3 pencils i can buy num__1 pen . after putting aside num__10.0 for taxi i was left with num__90.0 of the money with which i can buy num__135 pencils ( num__90.0 of num__150 ) or num__45 ( num__90.0 of num__50 ) pens i bought num__54 pencils and p pens or i could have bought ( num__54 + num__3 p ) pencils num__54 + num__3 p = num__135 num__3 p = num__135 - num__54 = num__81 p = num__27 answer : c <eor> c <eos> |
c |
percent__90.0__150.0__ percent__50.0__90.0__ percent__54.0__150.0__ percent__50.0__54.0__ percent__50.0__54.0__ |
percent__90.0__150.0__ percent__50.0__90.0__ percent__54.0__150.0__ percent__50.0__54.0__ percent__50.0__54.0__ |
| after decreasing num__24.0 in the price of an article costs rs . num__912 . find the actual cost of an article ? <o> a ) num__2992 <o> b ) num__2882 <o> c ) num__1200 <o> d ) num__2999 <o> e ) num__2123 |
cp * ( num__0.76 ) = num__912 cp = num__12 * num__100 = > cp = num__1200 answer : c <eor> c <eos> |
c |
percent__100.0__1200.0__ |
percent__100.0__1200.0__ |
| pipes a and b can fill a tank in num__5 and num__6 hours respectively . pipe c can empty it in num__12 hours . if all the three pipes are opened together then the tank will be filled in ? <o> a ) num__3 num__0.5625 hrs <o> b ) num__3 num__0.473684210526 hrs <o> c ) num__3 num__0.529411764706 hrs <o> d ) num__3 num__0.176470588235 hrs <o> e ) num__4 num__0.529411764706 hrs |
net part filled in num__1 hour = num__0.2 + num__0.166666666667 - num__0.0833333333333 = num__0.283333333333 the tank will be full in num__3.52941176471 hrs i . e . num__3 num__0.529411764706 hrs . answer : c <eor> c <eos> |
c |
subtract__6.0__5.0__ divide__1.0__5.0__ divide__1.0__6.0__ divide__1.0__12.0__ add__0.2__0.0833__ subtract__3.5294__3.0__ round__3.0__ |
subtract__6.0__5.0__ divide__1.0__5.0__ divide__1.0__6.0__ divide__1.0__12.0__ add__0.2__0.0833__ subtract__3.5294__3.0__ subtract__6.0__3.0__ |
| what is the product of all the prime factors of num__21 ? <o> a ) num__20 <o> b ) num__21 <o> c ) num__18 <o> d ) num__17 <o> e ) num__16 |
factors : num__13 num__721 num__3 * num__7 = num__21 answer : b <eor> b <eos> |
b |
gcd__21.0__721.0__ lcm__21.0__3.0__ |
gcd__21.0__721.0__ multiply__3.0__7.0__ |
| in the land of oz only one or two - letter words are used . the local language has num__67 different letters . the parliament decided to forbid the use of the seventh letter . how many words have the people of oz lost because of the prohibition ? <o> a ) num__65 <o> b ) num__66 <o> c ) num__67 <o> d ) num__131 <o> e ) num__134 |
the answer to the question is indeed e . the problem with above solutions is that they do not consider words like aa bb . . . the number of num__1 letter words ( x ) that can be made from num__67 letters is num__67 ; the number of num__2 letter words ( xx ) that can be made from num__67 letters is num__67 * num__67 since each x can take num__67 values . total : num__67 + num__67 * num__67 . similarly : the number of num__1 letter words ( x ) that can be made from num__66 letters is num__66 ; the number of num__2 letter words ( xx ) that can be made from num__66 letters is num__66 * num__66 since each x can take num__66 values . total : num__66 + num__66 * num__66 . the difference is ( num__67 + num__67 * num__67 ) - ( num__66 + num__66 * num__66 ) = num__134 . answer : e . <eor> e <eos> |
e |
subtract__67.0__1.0__ multiply__67.0__2.0__ multiply__67.0__2.0__ |
subtract__67.0__1.0__ multiply__67.0__2.0__ multiply__67.0__2.0__ |
| two whales are moving in the same direction at num__18 mps and num__15 mps . the faster whale crosses the slow whale in num__15 seconds . what is the length of the slower whale in meters ? <o> a ) num__45 m <o> b ) num__56 m <o> c ) num__39 m <o> d ) num__33 m <o> e ) num__62 m |
relative speed = ( num__18 - num__15 ) = num__3 mps . distance covered in num__15 sec = num__15 * num__3 = num__45 m . the length of the faster train = num__45 m . answer : a <eor> a <eos> |
a |
subtract__18.0__15.0__ multiply__15.0__3.0__ round__45.0__ |
subtract__18.0__15.0__ multiply__15.0__3.0__ multiply__15.0__3.0__ |
| little krish had $ num__200.50 . he spent $ num__35.25 on sweets and gave to his two friends $ num__25.20 each . how much money was left ? <o> a ) $ num__220.85 <o> b ) $ num__214.85 <o> c ) $ num__114.85 <o> d ) $ num__314.85 <o> e ) $ num__104.85 |
krish spent and gave to his two friends a total of num__35.25 + num__25.20 + num__25.20 = $ num__85.65 money left num__200.50 - num__85.65 = $ num__114.85 correct answer is c ) $ num__114.85 <eor> c <eos> |
c |
subtract__200.5__85.65__ subtract__200.5__85.65__ |
subtract__200.5__85.65__ subtract__200.5__85.65__ |
| find the area of trapezium whose parallel sides are num__20 cm and num__18 cm long and the distance between them is num__14 cm <o> a ) num__178 cm num__2 <o> b ) num__179 cm num__2 <o> c ) num__266 cm num__2 <o> d ) num__167 cm num__2 <o> e ) num__197 cm num__2 |
area of a trapezium = num__0.5 ( sum of parallel sides ) * ( perpendicular distance between them ) = num__0.5 ( num__20 + num__18 ) * ( num__14 ) = num__266 cm num__2 answer : c <eor> c <eos> |
c |
subtract__20.0__18.0__ round__266.0__ |
subtract__20.0__18.0__ round__266.0__ |
| in one hour a boat goes num__14 km / hr along the stream and num__8 km / hr against the stream . the speed of the boat in still water ( in km / hr ) is : <o> a ) num__14 kmph <o> b ) num__11 kmph . <o> c ) num__51 kmph . <o> d ) num__61 kmph . <o> e ) num__71 kmph . |
speed in still water = num__0.5 ( num__14 + num__8 ) kmph = num__11 kmph . answer : b <eor> b <eos> |
b |
round__11.0__ |
round__11.0__ |
| a tank is filled by three pipes with uniform flow . the first two pipes operating simultaneously fill the tank in the same during which the tank is filled by the third pipe alone . the second pipe fills the tank num__5 hours faster than the first pipe and num__4 hours slower than the third pipe . the time required by the first pipe is ? <o> a ) num__12 <o> b ) num__14 <o> c ) num__15 <o> d ) num__19 <o> e ) num__10 |
suppose first pipe alone takes x hours to fill the tank . then second and third pipes will take ( x - num__5 ) and ( x - num__9 ) hours respectively to fill the tank . num__1 / x + num__1 / ( x - num__5 ) = num__1 / ( x - num__9 ) ( num__2 x - num__5 ) ( x - num__9 ) = x ( x - num__5 ) x num__2 - num__18 x + num__45 = num__0 ( x - num__15 ) ( x - num__3 ) = num__0 = > x = num__15 . answer : c <eor> c <eos> |
c |
add__5.0__4.0__ subtract__5.0__4.0__ multiply__2.0__9.0__ multiply__5.0__9.0__ subtract__5.0__2.0__ round__15.0__ |
add__5.0__4.0__ subtract__5.0__4.0__ multiply__2.0__9.0__ multiply__5.0__9.0__ subtract__5.0__2.0__ divide__45.0__3.0__ |
| num__9 . the least number which should be added to num__28523 so that the sum is exactly divisible by num__3 num__5 num__7 and num__8 is <o> a ) num__41 <o> b ) num__42 <o> c ) num__32 <o> d ) num__37 <o> e ) num__39 |
lcm of num__3 num__5 num__7 and num__8 = num__840 num__28523 ÷ num__840 = num__33 remainder = num__803 hence the least number which should be added = num__840 - num__803 = num__37 answer : option d <eor> d <eos> |
d |
subtract__840.0__803.0__ subtract__840.0__803.0__ |
subtract__840.0__803.0__ subtract__840.0__803.0__ |
| suraj has a certain average of runs for num__12 innings . in the num__13 th innings he scores num__96 runs thereby increasing his average by num__5 runs . what is his average after the num__13 th innings ? <o> a ) num__48 <o> b ) num__64 <o> c ) num__36 <o> d ) num__72 <o> e ) num__27 |
to improve his average by num__5 runs per innings he has to contribute num__12 x num__5 = num__60 runs for the previous num__12 innings . thus the average after the num__13 th innings = num__96 - num__60 = num__36 . answer : c <eor> c <eos> |
c |
multiply__12.0__5.0__ subtract__96.0__60.0__ subtract__96.0__60.0__ |
multiply__12.0__5.0__ subtract__96.0__60.0__ subtract__96.0__60.0__ |
| if a + b = − num__9 and a = num__30 / b what is the value of a ^ num__2 + b ^ num__2 ? <o> a ) num__31 <o> b ) num__21 <o> c ) num__41 <o> d ) num__61 <o> e ) num__51 |
a ^ num__2 + b ^ num__2 should make you think of these formulas : ( a + b ) ( a + b ) = a ^ num__2 + b ^ num__2 + num__2 ab we already know ( a + b ) = - num__9 and a * b = num__30 ( a + b ) ( a + b ) = ( - num__9 ) ( - num__9 ) = a ^ num__2 + b ^ num__2 + num__2 * ( num__30 ) a ^ num__2 + b ^ num__2 = num__81 - num__60 = num__21 answer : b <eor> b <eos> |
b |
multiply__30.0__2.0__ subtract__30.0__9.0__ subtract__30.0__9.0__ |
multiply__30.0__2.0__ subtract__30.0__9.0__ subtract__30.0__9.0__ |
| the speed of a boat in upstream is num__70 kmph and the speed of the boat downstream is num__80 kmph . find the speed of the boat in still water and the speed of the stream ? <o> a ) num__10 kmph . <o> b ) num__05 kmph . <o> c ) num__18 kmph . <o> d ) num__11 kmph . <o> e ) num__12 kmph . |
speed of the boat in still water = ( num__70 + num__80 ) / num__2 = num__75 kmph . speed of the stream = ( num__80 - num__70 ) / num__2 = num__5 kmph . answer : b <eor> b <eos> |
b |
subtract__80.0__75.0__ round__5.0__ |
subtract__80.0__75.0__ subtract__80.0__75.0__ |
| if f ( f ( n ) ) + f ( n ) = num__2 n + num__3 f ( num__0 ) = num__1 then f ( num__2017 ) = ? <o> a ) num__2018 <o> b ) num__2088 <o> c ) num__270 <o> d ) num__1881 <o> e ) num__1781 |
f ( f ( num__0 ) ) + f ( num__0 ) = num__2 ( num__0 ) + num__3 ⇒ ⇒ f ( num__1 ) = num__3 - num__1 = num__2 f ( num__1 ) = num__2 f ( f ( num__1 ) ) + f ( num__1 ) = num__2 ( num__1 ) + num__3 ⇒ ⇒ f ( num__2 ) = num__5 - num__2 = num__3 f ( num__2 ) = num__3 f ( f ( num__2 ) ) + f ( num__2 ) = num__2 ( num__2 ) + num__3 ⇒ ⇒ f ( num__3 ) = num__7 - num__3 = num__4 f ( num__3 ) = num__4 . . . . . . . . . . . . . . f ( num__2017 ) = num__2018 ans : a <eor> a <eos> |
a |
add__2.0__3.0__ add__2.0__5.0__ add__3.0__1.0__ add__1.0__2017.0__ add__1.0__2017.0__ |
add__2.0__3.0__ add__2.0__5.0__ add__3.0__1.0__ add__1.0__2017.0__ add__1.0__2017.0__ |
| what sum of money will produce rs . num__70 as simple interest in num__3 years at num__3 num__0.5 percent ? <o> a ) num__263 <o> b ) num__500 <o> c ) num__367 <o> d ) num__368 <o> e ) num__666.7 |
num__70 = ( p * num__3 * num__3.5 ) / num__100 p = num__666.7 answer : e <eor> e <eos> |
e |
percent__100.0__666.7__ |
percent__100.0__666.7__ |
| the greatest number of four digits which is divisible by num__15 num__25 num__40 and num__75 is : <o> a ) num__9000 <o> b ) num__9400 <o> c ) num__9600 is greatest number <o> d ) num__9670 <o> e ) num__9800 |
greatest number of num__4 - digits is num__9999 . l . c . m . of num__15 num__25 num__40 and num__75 is num__600 . on dividing num__9999 by num__600 the remainder is num__399 . required number ( num__9999 - num__399 ) = num__9600 . answer : option c <eor> c <eos> |
c |
multiply__15.0__40.0__ subtract__9999.0__399.0__ subtract__9999.0__399.0__ |
multiply__15.0__40.0__ subtract__9999.0__399.0__ subtract__9999.0__399.0__ |
| a rectangular grassy plot num__110 m . by num__65 m has a gravel path num__2.5 m wide all round it on the inside . find the cost of gravelling the path at num__30 paise per sq . metre <o> a ) s num__255 <o> b ) s num__780 <o> c ) s num__880 <o> d ) s num__480 <o> e ) s num__980 |
area of the plot = num__110 m * num__65 m = num__7150 sq . m area of plot excluding gravel = num__105 m * num__60 m = num__6300 sq . m area of gravel = num__7150 sq . m - num__6300 sq . m = num__850 sq . m cost of building it = num__850 sq . m * num__30 = num__25500 p in rs = num__255.0 = rs num__255 answer : a <eor> a <eos> |
a |
multiply__110.0__65.0__ hour_to_min_conversion__ multiply__60.0__105.0__ subtract__7150.0__6300.0__ multiply__30.0__850.0__ round__255.0__ |
multiply__110.0__65.0__ hour_to_min_conversion__ multiply__60.0__105.0__ subtract__7150.0__6300.0__ multiply__30.0__850.0__ round__255.0__ |
| vijay bought num__160 shirts at the rate of rs . num__235 per shirt . the transport expenditure was rs . num__1400 . he paid an octroi at the rate of rs . num__1.75 per shirt and labour charges were rs . num__320 . what should be the selling price of one shirt if he wants a profit of num__20.0 ? <o> a ) rs . num__297 <o> b ) rs . num__270 <o> c ) rs . num__277.5 <o> d ) rs . num__285 <o> e ) none of these |
total cp per shirt = num__235 + num__8.75 + num__1.75 + num__2.0 = rs . num__247.5 sp = cp [ ( num__100 + profit % ) / num__100 ] = num__247.5 * [ ( num__100 + num__20 ) / num__100 ] = rs . num__297 . answer : a <eor> a <eos> |
a |
percent__100.0__297.0__ |
percent__100.0__297.0__ |
| when is | x - num__3 | = num__3 - x ? <o> a ) x < = num__3 <o> b ) x = num__3 <o> c ) x > num__3 <o> d ) x = num__0 <o> e ) x < num__0 |
when is | x - num__3 | = num__3 - x ? choice a : x = num__3 it is true but x can not be always num__3 choice b : x = num__0 it is also true but x can not be always num__0 choice c : x > num__3 it is false for e . g . x = num__6 then one side of equation is num__2 and the other side is - num__2 choice d : x < = num__3 this choice encapsulate choice a choice b and for all other conditions and is true for above said equation . hence the answer choice is a . <eor> a <eos> |
a |
divide__6.0__3.0__ subtract__6.0__3.0__ |
divide__6.0__3.0__ subtract__6.0__3.0__ |
| a person travels equal distances with speeds of num__3 km / hr num__4 km / hr and num__5 km / hr and takes a total time of num__47 minutes . the total distance ( in km ) is : <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
sol . let the total distance be num__3 x km . then x / num__3 + x / num__4 + x / num__5 = num__0.783333333333 ⇔ num__47 x / num__60 = num__0.783333333333 ⇔ x = num__1 . ∴ total distance = ( num__3 * num__1 ) km = num__3 km . answer c <eor> c <eos> |
c |
hour_to_min_conversion__ subtract__4.0__3.0__ round__3.0__ |
hour_to_min_conversion__ subtract__4.0__3.0__ divide__3.0__1.0__ |
| what least value should be replaced by * in num__2631 * num__4 so the number become divisible by num__3 <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__2 <o> e ) num__7 |
explanation : trick : number is divisible by num__3 if sum of all digits is divisible by num__3 so ( num__2 + num__6 + num__3 + num__1 + * + num__4 ) = num__16 + * should be divisible by num__9 num__16 + num__2 will be divisible by num__2 so that least number is num__2 . answer : option d <eor> d <eos> |
d |
add__4.0__2.0__ subtract__4.0__3.0__ add__3.0__6.0__ subtract__4.0__2.0__ |
add__4.0__2.0__ subtract__4.0__3.0__ add__3.0__6.0__ multiply__1.0__2.0__ |
| in a certain game a large bag is filled with blue green purple and red chips worth num__1 num__5 x and num__11 points each respectively . the purple chips are worth more than the green chips but less than the red chips . a certain number of chips are then selected from the bag . if the product of the point values of the selected chips is num__11000 how many purple chips were selected ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
num__11 |
000 = num__1 * num__5 ^ num__3 * num__8 * num__11 the factor of num__8 must come from the purple point value so there is num__1 purple chip . the answer is a . <eor> a <eos> |
a |
a |
| num__39 ! is divided by num__41 then what is the remainder ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__4 <o> d ) num__6 <o> e ) num__7 |
substituting p = num__41 in the wilson ' s theorem we get num__40 ! + num__141 = num__0 num__40 × num__39 ! + num__141 = num__0 − num__1 × num__39 ! num__41 = − num__1 cancelling - num__1 on both sides num__39 ! num__41 = num__1 a <eor> a <eos> |
a |
subtract__41.0__40.0__ reverse__1.0__ |
subtract__41.0__40.0__ subtract__41.0__40.0__ |
| a trader sells num__80 meters of cloth for rs . num__10000 at the profit of rs . num__7 per metre of cloth . what is the cost price of one metre of cloth ? <o> a ) num__118 <o> b ) num__88 <o> c ) num__90 <o> d ) num__42 <o> e ) num__22 |
sp of num__1 m of cloth = num__105.0 = rs . num__125 cp of num__1 m of cloth = sp of num__1 m of cloth - profit on num__1 m of cloth = rs . num__125 - rs . num__7 = rs . num__118 answer : a <eor> a <eos> |
a |
divide__10000.0__80.0__ subtract__125.0__7.0__ round__118.0__ |
divide__10000.0__80.0__ subtract__125.0__7.0__ subtract__125.0__7.0__ |
| f b and c are positive integers . if f b and c are assembled into the six - digit number fbcfbc which one of the following must be a factor of fbcfbc ? <o> a ) num__16 <o> b ) num__13 <o> c ) num__5 <o> d ) num__3 <o> e ) none of the above |
plug in some values and check - fbcfbc = num__123123 not divisible by num__16 and num__5 let fbcfbc = num__125125 not divisible by num__3 only option ( b ) and ( e ) is left in both the cases . . . check once more to marke ( b ) as correct answer let fbcfbc = num__135135 again divisible by num__13 so mark answer as ( b ) num__13 <eor> b <eos> |
b |
subtract__16.0__3.0__ subtract__16.0__3.0__ |
subtract__16.0__3.0__ subtract__16.0__3.0__ |
| on a certain day orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ num__0.70 per glass on the first day what was the price per glass on the second day ? <o> a ) $ num__0.15 <o> b ) $ num__0.20 <o> c ) $ num__0.30 <o> d ) $ num__0.40 <o> e ) $ num__0.46 |
on the first day num__1 unit of orange juice and num__1 unit of water was used to make num__2 units of orangeade ; on the second day num__1 unit of orange juice and num__2 units of water was used to make num__3 units of orangeade ; so the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is num__2 to num__3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is num__2 to num__3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from num__2 glasses on the first day equals to the revenue from num__3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then num__2 * num__0.7 = num__3 * x - - > x = $ num__0.46 . answer : e . <eor> e <eos> |
e |
add__1.0__2.0__ multiply__1.0__0.46__ |
add__1.0__2.0__ multiply__1.0__0.46__ |
| in a certain school the ratio of boys to girls is num__5 to num__17 . if there are num__72 more girls than boys how many boys are there ? <o> a ) num__27 <o> b ) num__36 <o> c ) num__45 <o> d ) num__72 <o> e ) num__108 |
the ratio of b to g is num__5 : num__13 and the other data point is g are more than boys by num__72 . . . looking at the ratio we can say that the num__12 ( num__17 - num__5 ) extra parts caused this diff of num__72 . so num__1 part corresponds to num__9.0 = num__9 and so num__5 parts correspond to num__12 * num__9 = num__108 . e <eor> e <eos> |
e |
subtract__17.0__5.0__ subtract__13.0__12.0__ multiply__9.0__12.0__ multiply__1.0__108.0__ |
subtract__17.0__5.0__ subtract__13.0__12.0__ multiply__9.0__12.0__ multiply__1.0__108.0__ |
| a certain collage has total of num__450 seniors each majoring in exactly one of six subjects . a minimum of num__20 seniors major in each six subjects . if three quarter of the seniors major in one of four subjects what is the greatest possible number of seniors majoring in one of the other two subjects ? <o> a ) num__130 <o> b ) num__80 <o> c ) num__75 <o> d ) num__60 <o> e ) num__50 |
answer is a . six majors are as follows : a + b + c + d + e + f each has at least num__20 seniors : num__20 + num__20 + num__20 + num__20 + num__20 + num__20 num__4 of the majors have num__300 seniors : a + b + c + d = num__300 there are a total of num__450 seniors . so e + f = num__150 . one of the subjects has to be num__20 for the other to be maximized . if e = num__20 then f = num__130 . a <eor> a <eos> |
a |
subtract__450.0__300.0__ subtract__150.0__20.0__ subtract__150.0__20.0__ |
subtract__450.0__300.0__ subtract__150.0__20.0__ subtract__150.0__20.0__ |
| find the surface area of a num__10 cm x num__4 cm x num__3 cm brick <o> a ) num__84 <o> b ) num__124 <o> c ) num__164 <o> d ) num__180 <o> e ) num__120 |
explanation : surface area = [ num__2 ( num__10 x num__4 + num__4 x num__3 + num__10 x num__3 ) ] cm num__2 = num__164 cm num__2 answer : c <eor> c <eos> |
c |
surface_rectangular_prism__10.0__4.0__3.0__ triangle_area__164.0__2.0__ |
surface_rectangular_prism__10.0__4.0__3.0__ triangle_area__164.0__2.0__ |
| a number is doubled and num__5 is added . if the resultant is trebled it becomes num__123 . what is that number ? <o> a ) num__12 <o> b ) num__29 <o> c ) num__27 <o> d ) num__18 <o> e ) num__99 |
explanation : let the number be x . therefore num__3 ( num__2 x + num__5 ) = num__123 num__6 x + num__15 = num__123 num__6 x = num__108 x = num__18 answer : d <eor> d <eos> |
d |
subtract__5.0__3.0__ multiply__2.0__3.0__ multiply__5.0__3.0__ subtract__123.0__15.0__ multiply__3.0__6.0__ multiply__3.0__6.0__ |
subtract__5.0__3.0__ multiply__2.0__3.0__ multiply__5.0__3.0__ subtract__123.0__15.0__ add__3.0__15.0__ add__3.0__15.0__ |
| the timing of a college is from num__12 p . m to num__4.10 p . m . five lectures are held in the given duration and a break of num__5 minutes after each lecture is given to the students . find the duration of each lecture . <o> a ) num__52 minutes <o> b ) num__45 minutes <o> c ) num__46 minutes <o> d ) num__48 minutes <o> e ) num__44 minutes |
explanation : total time a student spends in college = num__4 hours num__10 minutes = num__250 minutes as there are num__5 lectures the number of breaks between lectures is num__4 . total time of the break = num__20 minutes hence the duration of each lecture is = ( num__250 â € “ num__20 ) / num__5 = num__46 minutes answer c <eor> c <eos> |
c |
multiply__5.0__4.0__ round__46.0__ |
multiply__5.0__4.0__ round__46.0__ |
| if num__15 toys cost rs . num__545 what do num__37 toys cost ? <o> a ) rs . num__1344 <o> b ) rs . num__1349 <o> c ) rs . num__1346 <o> d ) rs . num__1341 <o> e ) rs . num__1343 |
explanation : let the required cost be rs . x . more toys more cost ( direct proportion ) therefore num__15 : num__37 : : num__545 : x ( num__15 * x ) = ( num__37 * num__545 ) x = ( num__37 * num__545 ) / num__15 = num__1344.33 hence the cost of num__37 toys is rs . num__1344.33 answer : a <eor> a <eos> |
a |
round_down__1344.33__ |
round_down__1344.33__ |
| if it takes num__10 kids num__6 hours to wear out their teacher how long would it take num__15 kids ? <o> a ) num__5 hours <o> b ) num__4 hours <o> c ) num__3 hours <o> d ) num__2 hours <o> e ) num__1 hour |
num__10 * num__6 = num__15 * x x = num__4 answer : b <eor> b <eos> |
b |
subtract__10.0__6.0__ round__4.0__ |
subtract__10.0__6.0__ round__4.0__ |
| a motorcyclist goes from bombay to pune a distance of num__128 kms at an average of num__32 kmph speed . another man starts from bombay by car num__2 ½ hours after the first and reaches pune ½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? <o> a ) num__1 : num__3 <o> b ) num__1 : num__5 <o> c ) num__1 : num__4 <o> d ) num__1 : num__1 <o> e ) num__1 : num__8 |
t = num__4.0 = num__4 h t = num__4 - num__3 = num__1 time ratio = num__4 : num__1 = num__4 : num__1 speed ratio = num__1 : num__4 answer : c <eor> c <eos> |
c |
divide__128.0__32.0__ subtract__3.0__2.0__ round__1.0__ |
divide__128.0__32.0__ subtract__3.0__2.0__ subtract__2.0__1.0__ |
| three pipes a b and c can fill a tank in num__6 hours . after working at it together for num__2 hours . c is closed a and b can fill the remaining par in num__7 hours . the number of hours taken by c alone to fill the tank is ? <o> a ) num__15 hours <o> b ) num__13 hours <o> c ) num__14 hours <o> d ) num__12 hours <o> e ) num__19 hours |
part filled in num__2 hours = num__0.333333333333 = num__0.333333333333 . remaining part = num__1 - num__0.333333333333 = num__0.666666666667 ( a + b ) ' s num__1 hour work = num__0.0952380952381 c ' s num__1 hour work = [ ( a + b + c ) ' s num__1 hour work - ( a + b ) ' s num__1 hour work ] = ( num__0.166666666667 - num__0.0952380952381 ) = num__0.0714285714286 c alone can fill the tank in num__14 hours . answer : c <eor> c <eos> |
c |
divide__2.0__6.0__ subtract__7.0__6.0__ subtract__1.0__0.3333__ divide__0.6667__7.0__ divide__1.0__6.0__ multiply__2.0__7.0__ round__14.0__ |
divide__2.0__6.0__ subtract__7.0__6.0__ subtract__1.0__0.3333__ divide__0.6667__7.0__ divide__1.0__6.0__ multiply__2.0__7.0__ round__14.0__ |
| a can run num__4 times as fast as b and gives b a start of num__72 m . how long should the race course be so that a and b might reach in the same time ? <o> a ) num__70 m <o> b ) num__60 m <o> c ) num__80 m <o> d ) num__65 m <o> e ) num__96 m |
speed of a : speed of b = num__4 : num__1 means in a race of num__4 m a gains num__3 m . then in a race of num__72 m he gains num__72 * ( num__1.33333333333 ) i . e num__96 m answer : e <eor> e <eos> |
e |
subtract__4.0__1.0__ divide__4.0__3.0__ multiply__96.0__1.0__ |
subtract__4.0__1.0__ divide__4.0__3.0__ multiply__96.0__1.0__ |
| how many integers n are there such that num__1 < num__5 n + num__4 < num__24 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
num__1 < num__5 n + num__4 < num__24 we first subtract num__4 from all three parts of the inequality and we obtain : - num__3 < num__5 n < num__20 next we divide both sides of the inequality by num__5 and we get : - num__0.6 < n < num__4 the answer is d <eor> d <eos> |
d |
subtract__4.0__1.0__ multiply__5.0__4.0__ divide__3.0__5.0__ multiply__1.0__4.0__ |
subtract__4.0__1.0__ subtract__24.0__4.0__ divide__3.0__5.0__ add__1.0__3.0__ |
| the radius of a semi circle is num__6.3 cm then its perimeter is ? <o> a ) num__32.8 <o> b ) num__32.4 <o> c ) num__32.1 <o> d ) num__32.2 <o> e ) num__32.9 |
num__5.14285714286 r = num__6.3 = num__32.4 answer : b <eor> b <eos> |
b |
round__32.4__ |
round__32.4__ |
| what is the sum of the multiples of num__5 from num__80 to num__140 inclusive ? <o> a ) num__560 <o> b ) num__780 <o> c ) num__990 <o> d ) num__1008 <o> e ) num__1200 |
the formula we want to use in this type of problem is this : average * total numbers = sum first find the average by taking the sum of the f + l number and divide it by num__2 : a = ( f + l ) / num__2 second find the total numbers in our range by dividing our f and l numbers by num__5 and add num__1 . ( num__28.0 ) - ( num__16.0 ) + num__1 multiply these together so what we show average * total numbers = sum ( num__80 + num__140 ) / num__2 * ( num__28.0 ) - ( num__16.0 ) + num__1 = sum num__110 * num__9 = num__990 c <eor> c <eos> |
c |
divide__140.0__5.0__ divide__80.0__5.0__ multiply__9.0__110.0__ multiply__1.0__990.0__ |
divide__140.0__5.0__ divide__80.0__5.0__ multiply__9.0__110.0__ multiply__1.0__990.0__ |
| if x and y are prime numbers such that x > y > num__4 then x ^ num__2 − y ^ num__2 must be divisible by which one of the following numbers ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__5 <o> d ) num__9 <o> e ) num__13 |
if x = num__7 and y = num__5 then x ^ num__2 - y ^ num__2 = num__24 and num__24 is divisible only by num__8 from the options thus it must be correct . answer : b . <eor> b <eos> |
b |
subtract__7.0__2.0__ multiply__4.0__2.0__ multiply__4.0__2.0__ |
subtract__7.0__2.0__ multiply__4.0__2.0__ multiply__4.0__2.0__ |
| a gets num__3 times as much money as b gets b gets only rs . num__25 more then what c gets . the three gets rs . num__655 in all . find the share of b ? <o> a ) num__130 <o> b ) num__120 <o> c ) num__218 <o> d ) num__140 <o> e ) num__136 |
a + b + c = num__655 a = num__3 b num__3 b + b + b - num__25 = num__655 num__5 b = num__680 b = num__136 answer : e <eor> e <eos> |
e |
add__25.0__655.0__ divide__680.0__5.0__ divide__680.0__5.0__ |
add__25.0__655.0__ divide__680.0__5.0__ divide__680.0__5.0__ |
| if the average ( arithmetic mean ) of five consecutive negative integers is num__2 k – num__1 what is the difference e between the greatest and least of the five integers ? <o> a ) num__4 k <o> b ) num__4 k ^ num__2 - num__4 k <o> c ) num__4 - num__4 k <o> d ) num__4 <o> e ) num__4 k + num__4 |
( n - num__2 + n - num__1 + n + n + num__1 + n + num__2 ) / num__5 = num__2 k - num__1 = > n = num__10 k - num__5 greatest = n + num__2 = num__10 k - num__5 + num__2 = num__10 k - num__3 least = n - num__2 = num__10 k - num__5 - num__2 = > difference e = num__10 k - num__3 - ( num__10 k - num__7 ) = num__4 answer - d <eor> d <eos> |
d |
multiply__2.0__5.0__ add__2.0__1.0__ add__2.0__5.0__ add__1.0__3.0__ add__1.0__3.0__ |
multiply__2.0__5.0__ add__2.0__1.0__ add__2.0__5.0__ add__1.0__3.0__ add__1.0__3.0__ |
| a birthday child reaches into a bag filled with toys ( num__1 yo - yo num__1 ball num__1 whistle and num__1 box crayons ) . if the child randomly picks three toys what is the probability the child will pick the yo - yo and a ball as two of the three toys ? <o> a ) num__0.333333333333 <o> b ) num__0.5 <o> c ) num__0.25 <o> d ) num__0.2 <o> e ) num__0.4 |
combination formula : ncr = n ! / ( r ! ( n - r ) ! ) where n is the population / set and r is the sample / subset . total number possible = num__4 c num__3 - num__4 ! / ( num__3 ! ( num__4 - num__3 ) ! ) = num__4 number yo - yo possible = num__1 c num__1 = num__1 number ball possible = num__1 c num__1 = num__1 number other possible = num__2 c num__1 = num__2 probability formula : p ( a ) = ( number favorable outcomes ) / ( total number possible outcomes ) p ( yb ) = ( num__1 c num__1 ) ( num__1 c num__1 ) ( num__2 c num__1 ) / num__4 c num__3 = num__0.5 = num__0.5 answer : b <eor> b <eos> |
b |
subtract__4.0__1.0__ subtract__3.0__1.0__ divide__1.0__2.0__ round__0.5__ |
subtract__4.0__1.0__ subtract__3.0__1.0__ divide__1.0__2.0__ subtract__1.0__0.5__ |
| two airplanes take off from one airfield at noon . one flies due east at num__206 miles per hour while the other flies directly northeast at num__283 miles per hour . approximately how many miles apart are the airplanes at num__2 p . m . ? <o> a ) num__412 <o> b ) num__332 <o> c ) num__400 <o> d ) num__483 <o> e ) num__566 |
a in two hours : the plane flying east will be num__412 miles away from airport . the other plane will be num__566 miles away from airport . num__1.37378640777 = ~ num__1.4 = ~ sqrt ( num__2 ) this means that planes formed a right isocheles triangle = > sides of such triangles relate as num__1 : num__1 : sqrt ( num__2 ) = > the planes are num__412 miles apart . a <eor> a <eos> |
a |
multiply__206.0__2.0__ multiply__283.0__2.0__ divide__283.0__206.0__ round__412.0__ |
multiply__206.0__2.0__ multiply__283.0__2.0__ divide__283.0__206.0__ round__412.0__ |
| if x > num__7 which of the following is equal to ( x ^ num__2 - num__8 x + num__16 ) / ( x ^ num__2 - num__16 ) ? <o> a ) ( x + num__4 ) / ( num__4 ( x - num__4 ) ) <o> b ) ( x - num__4 ) / ( x + num__4 ) <o> c ) ( x - num__2 ) / ( x + num__4 ) <o> d ) ( x + num__4 ) / ( x - num__4 ) <o> e ) ( x - num__8 ) / ( x - num__4 ) |
( x ^ num__2 - num__8 x + num__16 ) / ( x ^ num__2 - num__16 ) = ( x - num__4 ) ( x - num__4 ) / ( x + num__4 ) ( x - num__4 ) = ( x - num__4 ) / ( x + num__4 ) b . ( x - num__4 ) / ( x + num__4 ) <eor> b <eos> |
b |
divide__8.0__2.0__ divide__8.0__2.0__ |
divide__8.0__2.0__ divide__8.0__2.0__ |
| num__12 years ago the age of anand was one - third the age of bala at that time . the present age of bala is num__12 years more than the present age of anand . find the present age of anand ? <o> a ) num__13 <o> b ) num__14 <o> c ) num__15 <o> d ) num__16 <o> e ) num__18 |
let the present ages of anand and bala be ' a ' and ' b ' respectively . a - num__12 = num__0.333333333333 ( b - num__12 ) - - - ( num__1 ) b = a + num__12 substituting b = a + num__12 in first equation a - num__12 = num__0.333333333333 ( a + num__0 ) = > num__3 a - num__36 = a = > num__2 a = num__36 = > a = num__18 . answer : e <eor> e <eos> |
e |
round_down__0.3333__ multiply__12.0__3.0__ subtract__3.0__1.0__ divide__36.0__2.0__ multiply__1.0__18.0__ |
round_down__0.3333__ multiply__12.0__3.0__ subtract__3.0__1.0__ divide__36.0__2.0__ subtract__36.0__18.0__ |
| a can do a piece of work in num__30 days . he works at it for num__5 days and then b finishes it in num__20 days . in what time can a and b together it ? <o> a ) num__13 num__0.333333333333 days <o> b ) num__8 days <o> c ) num__9 days <o> d ) num__63 days <o> e ) num__15 days |
explanation : num__0.166666666667 + num__20 / x = num__1 x = num__24 num__0.0333333333333 + num__0.0416666666667 = num__0.075 num__13.3333333333 = num__13 num__0.333333333333 days answer a <eor> a <eos> |
a |
divide__5.0__30.0__ divide__1.0__30.0__ divide__1.0__24.0__ add__0.0417__0.0333__ divide__1.0__0.075__ subtract__13.3333__13.0__ round__13.0__ |
divide__5.0__30.0__ divide__1.0__30.0__ divide__1.0__24.0__ add__0.0417__0.0333__ divide__1.0__0.075__ subtract__13.3333__13.0__ divide__13.0__1.0__ |
| the average salary of all the workers in a workshop is rs . num__8000 . the average salary of num__7 technicians is rs . num__12000 and the average salary of the rest is rs . num__6000 . the total number of workers in the workshop is ? <o> a ) num__65 <o> b ) num__21 <o> c ) num__78 <o> d ) num__67 <o> e ) num__51 |
let the total number of workers be x . then num__8000 x = ( num__12000 * num__7 ) + num__6000 ( x - num__7 ) = num__2000 x = num__42000 = x = num__21 . answer : b <eor> b <eos> |
b |
subtract__8000.0__6000.0__ multiply__7.0__6000.0__ divide__42000.0__2000.0__ divide__42000.0__2000.0__ |
subtract__8000.0__6000.0__ multiply__7.0__6000.0__ divide__42000.0__2000.0__ divide__42000.0__2000.0__ |
| two cyclists are moving towards each other at num__10 miles / hour . they are now num__50 miles apart . at this instance a fly starts from one cyclist and move towards other and moves to and fro till the two cyclist meet each other . if the fly is moving at num__15 miles / hour what is the total distance covered by the fly ? <o> a ) num__37.5 miles <o> b ) num__38.5 miles <o> c ) num__39.5 miles <o> d ) num__36.5 miles <o> e ) num__35.5 miles |
time taken by cyclists to meet = num__2.5 num__2.5 hrs distance covered by fly during num__2.5 hrs = num__15 * num__2.5 = num__37.5 miles answer : a <eor> a <eos> |
a |
multiply__15.0__2.5__ round__37.5__ |
multiply__15.0__2.5__ multiply__15.0__2.5__ |
| p can lay railway track between two stations in num__16 days . q can do the same job in num__12 days . with the help of r they completes the job in num__4 days . how much days does it take for r alone to complete the work ? <o> a ) num__8 ( num__0.6 ) days <o> b ) num__9 ( num__0.6 ) days <o> c ) num__7 ( num__0.6 ) days <o> d ) num__6 ( num__0.6 ) days <o> e ) num__5 ( num__0.6 ) days |
amount of work p can do in num__1 day = num__0.0625 amount of work q can do in num__1 day = num__0.0833333333333 amount of work p q and r can together do in num__1 day = num__0.25 amount of work r can do in num__1 day = num__0.25 - ( num__0.0625 + num__0.0833333333333 ) = num__0.1875 – num__0.0833333333333 = num__0.104166666667 = > hence r can do the job on num__9.6 days = num__9 ( num__0.6 ) days answer is b . <eor> b <eos> |
b |
divide__1.0__16.0__ divide__1.0__12.0__ divide__4.0__16.0__ subtract__0.25__0.0625__ subtract__0.1875__0.0833__ km_to_mile_conversion__ round__9.0__ |
divide__1.0__16.0__ divide__1.0__12.0__ divide__4.0__16.0__ subtract__0.25__0.0625__ subtract__0.1875__0.0833__ subtract__9.6__9.0__ subtract__9.6__0.6__ |
| during a certain two - week period num__60 percent of the movies rented from a video store were comedies and of the remaining movies rented there were num__4 times as many dramas as action movies . if no other movies were rented during that two - week period and there were a action movies rented then how many comedies in terms of a were rented during that two - week period ? <o> a ) num__14 <o> b ) num__12 <o> c ) num__10 <o> d ) num__8 <o> e ) num__7.5 |
total movies = num__100 . comedies = num__60 . action + drama = num__40 . since there were num__5 times as many dramas as action movies then action + num__4 * action = num__40 - - > action = a = num__8 . comedies = num__60 = num__7.5 a . answer : e <eor> e <eos> |
e |
subtract__100.0__60.0__ divide__40.0__5.0__ divide__60.0__8.0__ divide__60.0__8.0__ |
subtract__100.0__60.0__ divide__40.0__5.0__ divide__60.0__8.0__ divide__60.0__8.0__ |
| bus x is num__12 miles ahead of bus y which is traveling the same direction along the same route as bus x . if bus x is traveling at an average speed of num__41 miles per hour and bus y is traveling at an average speed of num__57 miles per hour how long will it take bus y to overtake and drive num__4 miles ahead of bus x ? <o> a ) num__2 hours <o> b ) num__1 hour <o> c ) num__2 hours num__20 minutes <o> d ) num__3 hours num__45 minutes <o> e ) num__5 hours |
relative speed = num__57 - num__41 = num__16 miles per hour dist required = num__12 + num__4 = num__16 miles time taken to overtake = num__1.0 = num__1 hour . b is the answer . <eor> b <eos> |
b |
add__12.0__4.0__ round__1.0__ |
add__12.0__4.0__ round__1.0__ |
| how many digits are in ( num__8 × num__10 ^ num__12 ) ( num__10 × num__10 ^ num__9 ) ? <o> a ) num__20 <o> b ) num__23 <o> c ) num__26 <o> d ) num__27 <o> e ) num__28 |
the question simplfies to ( num__8 × num__10 ^ num__12 ) ( num__10 ^ num__10 ) = > num__8 * num__10 ^ num__22 = > will contain num__22 zeros + num__1 digit num__8 = > num__23 ans b <eor> b <eos> |
b |
add__10.0__12.0__ subtract__10.0__9.0__ add__1.0__22.0__ add__1.0__22.0__ |
add__10.0__12.0__ subtract__10.0__9.0__ add__1.0__22.0__ add__1.0__22.0__ |
| the divisor is num__12 the quotient is num__4 and the remainder is num__1 . what is the dividend ? <o> a ) num__48 <o> b ) num__49 <o> c ) num__96 <o> d ) num__97 <o> e ) num__72 |
d = d * q + r d = num__12 * num__4 + num__1 d = num__48 + num__1 d = num__49 answer : b <eor> b <eos> |
b |
multiply__12.0__4.0__ add__1.0__48.0__ add__1.0__48.0__ |
multiply__12.0__4.0__ add__1.0__48.0__ add__1.0__48.0__ |
| if a person walks at num__5 km / hr instead of num__4 km / hr he would have walked num__6 km more . the actual distance traveled by him is ? <o> a ) num__14 <o> b ) num__20 <o> c ) num__18 <o> d ) num__24 <o> e ) num__21 |
let the actual distance traveled be x km . then x / num__4 = ( x + num__6 ) / num__5 x - num__24 = > x = num__24 km . answer : d <eor> d <eos> |
d |
multiply__4.0__6.0__ multiply__4.0__6.0__ |
multiply__4.0__6.0__ multiply__4.0__6.0__ |
| during a certain two - week period num__50 percent of the movies rented from a video store were comedies and of the remaining movies rented there were num__9 times as many dramas as action movies . if no other movies were rented during that two - week period and there were a action movies rented then how many comedies in terms of a were rented during that two - week period ? <o> a ) num__5 a <o> b ) num__10 a <o> c ) num__20 a <o> d ) num__25 a <o> e ) num__28 a |
movies : num__50.0 comedies . num__50.0 remaining genre . now in this num__50.0 there are only num__2 categories . action movies and drama movies . if action = x ; drama movies = num__9 x . total num__10 x . num__10 x = num__50 ; x = num__5 action movies : num__5.0 drama movies : num__45.0 we can say that out of num__100 z : comedies : num__50 z action : num__5 z drama : num__45 z now action movies were a this means : a = num__5 z . z = ( a / num__5 ) comedies : num__50 z = num__50 * ( a / num__5 ) num__10 a b is the answer . <eor> b <eos> |
b |
divide__50.0__10.0__ subtract__50.0__5.0__ multiply__50.0__2.0__ divide__50.0__5.0__ |
divide__50.0__10.0__ multiply__9.0__5.0__ multiply__50.0__2.0__ divide__50.0__5.0__ |
| two trains travelling in the same direction at num__80 and num__26 kmph completely pass off another in num__1 minute . if the length of the first train is num__400 m what is the length of the second train ? <o> a ) num__500 <o> b ) num__266 <o> c ) num__175 <o> d ) num__99 <o> e ) num__171 |
rs = num__80 â € “ num__26 = num__54 * num__0.277777777778 = num__15 mps t = num__60 sec d = num__15 * num__60 = num__900 m num__400 - - - - - - - - num__500 m answer : a <eor> a <eos> |
a |
subtract__80.0__26.0__ hour_to_min_conversion__ multiply__15.0__60.0__ subtract__900.0__400.0__ round__500.0__ |
subtract__80.0__26.0__ hour_to_min_conversion__ multiply__15.0__60.0__ subtract__900.0__400.0__ multiply__1.0__500.0__ |
| a certain car averages num__25 miles per gallon when driving in the city and num__40 miles per gallon when driving on the highway . according to these rates which of the following is closest to the number of miles per gallon that the car averages when it is driven num__10 miles in the city and then num__45 miles on the highway ? <o> a ) num__28 <o> b ) num__30 <o> c ) num__36 <o> d ) num__35 <o> e ) num__38 |
answer c no of miles per gallon = total no of miles / total no of gallons = num__10 + num__45 / ( num__0.4 + num__1.125 ) ~ num__36.06 <eor> c <eos> |
c |
divide__10.0__25.0__ divide__45.0__40.0__ round_down__36.06__ |
divide__10.0__25.0__ divide__45.0__40.0__ round_down__36.06__ |
| the length of a room is num__5.5 m and width is num__3.75 m . find the cost of paying the floor by slabs at the rate of rs . num__800 per sq . metre . <o> a ) rs . num__15000 <o> b ) rs . num__15500 <o> c ) rs . num__15600 <o> d ) rs . num__16500 <o> e ) none |
solution area of the floor = ( num__5.5 x num__3.75 ) m ² = num__20.635 m ² cost of paying = rs . ( num__800 x num__20.625 ) = rs . num__16500 . answer d <eor> d <eos> |
d |
multiply__5.5__3.75__ multiply__800.0__20.625__ round__16500.0__ |
multiply__5.5__3.75__ multiply__800.0__20.625__ round__16500.0__ |
| num__20 binders can bind num__1400 books in num__21 days . how many binders will be required to bind num__1600 books in num__20 days ? <o> a ) num__24 <o> b ) num__18 <o> c ) num__17 <o> d ) num__16 <o> e ) num__10 |
binders books days num__20 num__1400 num__21 x num__1600 num__20 x / num__20 = ( num__1.14285714286 ) * ( num__1.05 ) = > x = num__24 answer : a <eor> a <eos> |
a |
divide__1600.0__1400.0__ divide__21.0__20.0__ round__24.0__ |
divide__1600.0__1400.0__ divide__21.0__20.0__ round__24.0__ |
| the average age of a b and c is num__28 years . if the average age of a and c is num__29 years what is the age of b in years ? <o> a ) num__19 <o> b ) num__26 <o> c ) num__20 <o> d ) num__32 <o> e ) num__21 |
age of b = age of ( a + b + c ) â € “ age of ( a + c ) = num__28 Ã — num__3 â € “ num__29 Ã — num__2 = num__84 â € “ num__58 = num__26 years b <eor> b <eos> |
b |
multiply__28.0__3.0__ multiply__29.0__2.0__ subtract__28.0__2.0__ subtract__28.0__2.0__ |
multiply__28.0__3.0__ multiply__29.0__2.0__ subtract__28.0__2.0__ subtract__28.0__2.0__ |
| in num__70 liters of a mixture of milk and water the quantity of water is num__10.0 . how much water should be added so that new mixture may num__25.0 water . <o> a ) num__14 <o> b ) num__15 <o> c ) num__16 <o> d ) num__17 <o> e ) num__18 |
there are num__10.0 of water in num__70 l - - > there are num__7 l of waters ; after x liters of water is added amount of water becomes ( num__7 + x ) liters which on the other hand is num__0.25 th ( num__25.0 ) of new ( num__70 + x ) liters of mixture : num__7 + x = num__0.25 * ( num__70 + x ) - - > num__28 + num__4 x = num__70 + x - - > x = num__14 answer : a <eor> a <eos> |
a |
divide__70.0__10.0__ divide__7.0__0.25__ reverse__0.25__ add__10.0__4.0__ add__10.0__4.0__ |
divide__70.0__10.0__ divide__7.0__0.25__ reverse__0.25__ add__10.0__4.0__ add__10.0__4.0__ |
| water consists of hydrogen and oxygen and the approximate ratio by mass of hydrogen to oxygen is num__2 : num__16 . approximately how many grams of oxygen are there in num__162 grams of water ? <o> a ) num__16 <o> b ) num__72 <o> c ) num__112 <o> d ) num__128 <o> e ) num__144 |
solution : we are given that the ratio of hydrogen to oxygen in water by mass is num__2 : num__16 . using our ratio multiplier we can re - write this as num__2 x : num__16 x . we can now use these expressions to determine how much oxygen is in num__162 grams of water . num__2 x + num__16 x = num__162 num__18 x = num__162 x = num__9 since x is num__9 we know that there are num__16 x num__9 = num__144 grams of oxygen in num__162 grams of water . answer e . <eor> e <eos> |
e |
add__2.0__16.0__ divide__162.0__18.0__ multiply__16.0__9.0__ multiply__16.0__9.0__ |
add__2.0__16.0__ divide__162.0__18.0__ subtract__162.0__18.0__ subtract__162.0__18.0__ |
| a hall is num__15 m long and num__12 m broad . if the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls the volume of the hall is : <o> a ) num__720 <o> b ) num__900 <o> c ) num__1200 <o> d ) num__1800 <o> e ) num__1900 |
num__2 ( num__15 + num__12 ) x h = num__2 ( num__15 x num__12 ) h = num__6.66666666667 m = num__6.66666666667 m . volume = num__15 x num__12 x num__20 m num__3 = num__1200 m num__3 . answer : c <eor> c <eos> |
c |
triangle_area__2.0__1200.0__ |
triangle_area__2.0__1200.0__ |
| the cross - section of a cannel is a trapezium in shape . if the cannel is num__10 m wide at the top and num__6 m wide at the bottom and the area of cross - section is num__640 sq m the depth of cannel is ? <o> a ) num__28 <o> b ) num__77 <o> c ) num__99 <o> d ) num__80 <o> e ) num__21 |
num__0.5 * d ( num__10 + num__6 ) = num__640 d = num__80 answer : d <eor> d <eos> |
d |
round__80.0__ |
round__80.0__ |
| if a man can cover num__12 metres in one second how many kilometres can he cover in num__3 hours num__45 minutes ? <o> a ) num__12 km <o> b ) num__162 km <o> c ) num__6 km <o> d ) num__87 km <o> e ) num__15 km |
explanation : num__12 m / s = num__12 * num__3.6 kmph num__3 hours num__45 minutes = num__3 num__0.75 hours = num__3.75 hours distance = speed * time = num__12 * num__3.6 * num__3.75 km = num__162 km . b ) <eor> b <eos> |
b |
add__3.0__0.75__ multiply__45.0__3.6__ round__162.0__ |
divide__45.0__12.0__ multiply__45.0__3.6__ multiply__45.0__3.6__ |
| two pipes a and b can separately fill a tank in num__2 minutes and num__15 minutes respectively . both the pipes are opened together but num__4 minutes after the start the pipe a is turned off . how much time will it take to fill the tank ? <o> a ) num__33 <o> b ) num__10 <o> c ) num__99 <o> d ) num__73 <o> e ) num__23 |
num__0.333333333333 + x / num__15 = num__1 x = num__10 answer : b <eor> b <eos> |
b |
round__10.0__ |
divide__10.0__1.0__ |
| water is poured into a tank so that the tank is being filled at the rate of num__4 cubic feet per hour . if the empty rectangular tank is num__6 feet long num__4 feet wide and num__3 feet deep approximately how many hours does it take to fill the tank ? <o> a ) num__16 <o> b ) num__18 <o> c ) num__20 <o> d ) num__22 <o> e ) num__24 |
the volume the tank is : length * width * depth = num__6 * num__4 * num__3 = num__72 cubic feet . num__72 cubic feet / num__4 cubic feet per hour = num__18 hours . it will take num__18 hours to fill the tank . the answer is b . <eor> b <eos> |
b |
multiply__6.0__3.0__ round__18.0__ |
multiply__6.0__3.0__ multiply__6.0__3.0__ |
| a man can row with a speed of num__15 kmph in still water . if the stream flows at num__5 kmph then the speed in downstream is ? <o> a ) num__29 <o> b ) num__378 <o> c ) num__20 <o> d ) num__27 <o> e ) num__121 |
m = num__15 s = num__5 ds = num__15 + num__5 = num__20 answer : c <eor> c <eos> |
c |
add__15.0__5.0__ round__20.0__ |
add__15.0__5.0__ add__15.0__5.0__ |
| at a special sale num__9 tickets can be purchased for the price of num__6 tickets . if num__9 tickets are purchased at the sale the amount saved will be what percent of the original price of the num__9 tickets ? <o> a ) num__20.0 <o> b ) num__33.3 <o> c ) num__40.0 <o> d ) num__60.0 <o> e ) num__66.6 % |
let the price of a ticket be rs . num__100 so num__6 tickets cost num__600 & num__9 tickets cost num__900 num__9 tickets purchased at price of num__6 tickets ie . for num__600 so amount saved s rs . num__300 % of num__9 tickets = ( num__0.333333333333 ) * num__100 = num__33.3 answer : b <eor> b <eos> |
b |
percent__33.3__100.0__ |
percent__33.3__100.0__ |
| alex takes a loan of $ num__5000 to buy a used truck at the rate of num__2.0 simple interest . calculate the annual interest to be paid for the loan amount . <o> a ) num__100 <o> b ) num__150 <o> c ) num__720 <o> d ) num__200 <o> e ) num__750 |
from the details given in the problem principle = p = $ num__5000 and r = num__2.0 or num__0.02 expressed as a decimal . as the annual interest is to be calculated the time period t = num__1 . plugging these values in the simple interest formula i = p x t x r = num__5000 x num__1 x num__0.02 = num__100.00 annual interest to be paid = $ num__100 answer : a <eor> a <eos> |
a |
percent__0.02__5000.0__ percent__2.0__5000.0__ percent__2.0__5000.0__ |
percent__0.02__5000.0__ percent__2.0__5000.0__ percent__2.0__5000.0__ |
| a can run a kilometer race in num__4 num__0.5 min while b can run same race in num__5 min . how many meters start can a give b in a kilometer race so that the race mat end in a dead heat ? <o> a ) num__887 m <o> b ) num__998 m <o> c ) num__55 m <o> d ) num__100 m <o> e ) num__44 m |
a can give b ( num__5 min - num__4 num__0.5 min ) = num__30 sec start . the distance covered by b in num__5 min = num__1000 m . distance covered in num__30 sec = ( num__1000 * num__30 ) / num__300 = num__100 m . a can give b num__100 m start . answer : d <eor> d <eos> |
d |
round__100.0__ |
round__100.0__ |
| the cubic f ( x ) = x num__3 + bx num__2 + cx + d satisfies f ( num__1 ) = num__3 f ( num__2 ) = num__6 and f ( num__4 ) = num__12 . compute f ( num__3 ) . <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
note that f ( x ) - num__3 x = ( x - num__1 ) ( x - num__2 ) ( x - num__4 ) so f ( x ) = ( x - num__1 ) ( x - num__2 ) ( x - num__4 ) + num__3 x and therefore f ( num__3 ) = num__2 * num__1 * ( - num__1 ) + num__9 = num__7 . correct answer b <eor> b <eos> |
b |
add__3.0__6.0__ add__3.0__4.0__ add__3.0__4.0__ |
add__3.0__6.0__ add__3.0__4.0__ add__3.0__4.0__ |
| the edge of a cube is num__6 a cm . find its surface ? <o> a ) num__216 a num__2 <o> b ) num__24 a num__4 <o> c ) num__24 a num__1 <o> d ) num__24 a num__2 <o> e ) num__24 a num__7 |
num__6 a num__2 = num__6 * num__6 a * num__6 a = num__216 a num__2 answer : a <eor> a <eos> |
a |
volume_cube__6.0__ volume_cube__6.0__ |
volume_cube__6.0__ volume_cube__6.0__ |
| the difference of the squares of two consecutive even integers is divisible by which of the following integers ? <o> a ) num__5 <o> b ) num__18 <o> c ) num__7 <o> d ) num__45 <o> e ) num__4 |
explanation : let the two consecutive even integers be num__2 n and ( num__2 n + num__2 ) . then ( num__2 n + num__2 ) ^ num__2 = ( num__2 n + num__2 num__2 n ) ( num__2 n num__2 - num__2 n ) = num__2 ( num__4 n + num__2 ) = num__4 ( num__2 n + num__1 ) which is divisible by num__4 . answer : e <eor> e <eos> |
e |
multiply__1.0__4.0__ |
multiply__1.0__4.0__ |
| num__2 num__7 num__14 num__23 ? num__47 <o> a ) num__31 <o> b ) num__28 <o> c ) num__34 <o> d ) num__33 <o> e ) num__32 |
the given sequence is + num__5 + num__7 + num__9 — — ie . num__2 + num__5 = num__7 num__7 + num__7 = num__14 num__14 + num__9 = num__23 missing number = num__23 + num__11 = num__34 . answer : c <eor> c <eos> |
c |
subtract__7.0__2.0__ add__2.0__7.0__ add__2.0__9.0__ add__23.0__11.0__ add__23.0__11.0__ |
subtract__7.0__2.0__ add__2.0__7.0__ add__2.0__9.0__ add__23.0__11.0__ add__23.0__11.0__ |
| if a certain sample of data has a mean of num__20.0 and a standard deviation of num__2.0 which of the following values is more than num__2.5 standard deviations from the mean ? <o> a ) num__13.5 <o> b ) num__12.0 <o> c ) num__17.0 <o> d ) num__23.5 <o> e ) num__26.5 |
the standard deviation is num__2.0 . num__2.5 standard deviations which is just num__2.5 x num__2.0 is num__5.0 . the only answer more than num__6.0 away from the mean is b . <eor> b <eos> |
b |
multiply__2.0__2.5__ multiply__2.0__6.0__ |
multiply__2.0__2.5__ multiply__2.0__6.0__ |
| henry answered num__0.6 of the questions he was given for homework in the first hour num__0.263157894737 of the remaining questions in the second hour and the remaining x questions in the third hour where x is an integer . how many questions was henry given for homework ? <o> a ) x <o> b ) ( num__13.5714285714 ) * x <o> c ) ( num__6.78571428571 ) * x <o> d ) ( num__4.52380952381 ) * x <o> e ) ( num__3.39285714286 ) * x |
let q be the number of homework questions . q = num__3 q / num__5 + ( num__0.263157894737 ) ( num__2 q / num__5 ) + x q = num__57 q / num__95 + num__10 q / num__95 + x num__28 q / num__95 = x q = ( num__3.39285714286 ) * x the answer is e . <eor> e <eos> |
e |
divide__3.0__0.6__ subtract__5.0__3.0__ divide__57.0__0.6__ multiply__2.0__5.0__ divide__95.0__28.0__ divide__95.0__28.0__ |
divide__3.0__0.6__ subtract__5.0__3.0__ divide__57.0__0.6__ multiply__2.0__5.0__ divide__95.0__28.0__ divide__95.0__28.0__ |
| a grazing land have length num__200 m and breadth num__50 m . there is grass num__0.5 kg / m num__2 . a cow graze num__20 kg grass daily . if cow graze for num__20 days . so after num__20 days calculate the percentage of grass filled area . <o> a ) num__90 <o> b ) num__92 <o> c ) num__95 <o> d ) num__85 <o> e ) num__80 |
area of land = num__200 * num__50 = num__10000 m num__2 . total grass grazed = num__400 kg . area grazed = num__400 / num__0.5 = num__800 m num__2 . % of grass filled area = num__0.92 * num__100 = num__92.0 answer b <eor> b <eos> |
b |
multiply__200.0__50.0__ divide__200.0__0.5__ multiply__2.0__400.0__ multiply__200.0__0.5__ multiply__100.0__0.92__ multiply__100.0__0.92__ |
multiply__200.0__50.0__ divide__200.0__0.5__ multiply__2.0__400.0__ multiply__200.0__0.5__ multiply__100.0__0.92__ multiply__100.0__0.92__ |
| the product of three consecutive numbers is num__120 . then the sum of the smallest two numbers is ? <o> a ) num__9 <o> b ) num__15 <o> c ) num__20 <o> d ) num__38 <o> e ) num__56 |
product of three numbers = num__120 num__120 = num__2 * num__2 * num__5 * num__6 = num__4 * num__5 * num__6 so the three numbers are num__45 and num__6 . and sum of smallest of these two = num__4 + num__5 = num__9 . answer : option a <eor> a <eos> |
a |
subtract__6.0__2.0__ add__4.0__5.0__ add__4.0__5.0__ |
subtract__6.0__2.0__ add__4.0__5.0__ add__4.0__5.0__ |
| if a * b denotes the greatest common divisor of a and b then ( ( num__16 * num__20 ) * ( num__18 * num__24 ) ) = ? <o> a ) num__24 <o> b ) num__12 <o> c ) num__6 <o> d ) num__4 <o> e ) num__2 |
the greatest common divisor of num__16 and num__20 is num__4 . hence num__16 * num__20 = num__4 ( note that * here denotes the function not multiplication ) . the greatest common divisor of num__18 and num__24 is num__6 . hence num__18 * num__24 = num__6 . hence ( ( num__16 * num__20 ) * ( num__18 * num__24 ) ) = num__4 * num__6 . the greatest common divisor of num__4 and num__6 is num__2 . answer ; e . <eor> e <eos> |
e |
subtract__20.0__16.0__ subtract__24.0__18.0__ subtract__20.0__18.0__ subtract__20.0__18.0__ |
subtract__20.0__16.0__ subtract__24.0__18.0__ subtract__20.0__18.0__ subtract__20.0__18.0__ |
| if the sum of the num__4 th term and the num__12 th term of an arithmetic progression is num__8 what is the sum of the first num__15 terms of the progression ? <o> a ) num__60 <o> b ) num__120 <o> c ) num__160 <o> d ) num__240 <o> e ) num__840 |
let a = first term d = common difference . num__4 th term = a + num__3 d num__12 th term = a + num__11 d a + num__3 d + a + num__11 d = num__8 num__2 a + num__14 d = num__8 a + num__7 d = num__4 . sum of the first num__15 terms = num__7.5 ( num__2 a + num__14 d ) = num__15 ( a + num__7 d ) = num__15 ( num__4 ) = num__60 . ans ( a ) <eor> a <eos> |
a |
divide__12.0__4.0__ add__8.0__3.0__ divide__8.0__4.0__ add__12.0__2.0__ add__4.0__3.0__ divide__15.0__2.0__ multiply__4.0__15.0__ multiply__4.0__15.0__ |
divide__12.0__4.0__ add__8.0__3.0__ divide__8.0__4.0__ add__12.0__2.0__ add__4.0__3.0__ divide__15.0__2.0__ multiply__4.0__15.0__ multiply__4.0__15.0__ |
| a certain roller coaster has num__2 cars and a passenger is equally likely to ride in any num__1 of the num__2 cars each time that passenger rides the roller coaster . if a certain passenger is to ride the roller coaster num__2 times what is the probability that the passenger will ride in each of the num__2 cars ? <o> a ) num__0 <o> b ) num__0.5 <o> c ) num__0.222222222222 <o> d ) num__0.333333333333 <o> e ) num__1 |
if he is to ride num__2 times and since he can choose any of the num__2 cars each time total number of ways is = num__2 * num__2 = num__4 now the number of ways if he is to choose a different car each time is = num__2 * num__1 = num__2 so the probability is = num__0.5 = num__0.5 answer : b <eor> b <eos> |
b |
reverse__2.0__ reverse__2.0__ |
reverse__2.0__ reverse__2.0__ |
| a train passes a bridge in num__40 sec and a man standing on the platform in num__20 sec . if the speed of the train is num__54 km / hr . what is the length of the platform ? <o> a ) num__250 <o> b ) num__400 <o> c ) num__150 <o> d ) num__300 <o> e ) num__450 |
speed = num__54 * num__0.277777777778 = num__15 m / sec . length of the train = num__15 * num__20 = num__300 m . let the length of the platform be x m . then ( x + num__300 ) / num__40 = num__15 = > x = num__300 m . answer : d <eor> d <eos> |
d |
multiply__20.0__15.0__ round__300.0__ |
multiply__20.0__15.0__ multiply__20.0__15.0__ |
| six women can do a work in num__10 days . ten men can complete the same work in num__4 days . what is the ratio between the capacity of a man and a woman ? <o> a ) num__1 : num__2 <o> b ) num__2 : num__1 <o> c ) num__2 : num__3 <o> d ) num__3 : num__2 <o> e ) none of these |
explanation : ( num__6 Ã — num__10 ) women can complete the work in num__1 day . â ˆ ´ num__1 woman ' s num__1 day ' s work = num__0.0166666666667 ( num__10 Ã — num__4 ) men can complete the work in num__1 day . â ˆ ´ num__1 man ' s num__1 day ' s work = num__0.025 so required ratio = num__0.0166666666667 : num__0.025 = num__3 : num__2 answer : d <eor> d <eos> |
d |
subtract__10.0__4.0__ subtract__4.0__1.0__ subtract__3.0__1.0__ round__3.0__ |
subtract__10.0__4.0__ subtract__4.0__1.0__ subtract__3.0__1.0__ round__3.0__ |
| how many pounds of salt at num__70 cents / lb must be mixed with num__45 lbs of salt that costs num__40 cents / lb so that a merchant will get num__20.0 profit by selling the mixture at num__48 cents / lb ? <o> a ) num__5 <o> b ) num__9 <o> c ) num__40 <o> d ) num__50 <o> e ) num__25 |
selling price is num__48 cents / lb for a num__20.0 profit cost price should be num__40 cents / lb ( cp * num__1.2 = num__48 ) basically you need to mix num__40 cents / lb ( salt num__1 ) with num__70 cents / lb ( salt num__2 ) to get a mixture costing num__45 cents / lb ( salt avg ) weight of salt num__1 / weight of salt num__2 = ( salt num__2 - saltavg ) / ( saltavg - salt num__1 ) = ( num__70 - num__45 ) / ( num__45 - num__40 ) = num__5.0 we know that weight of salt num__1 is num__45 lbs . weight of salt num__2 must be num__9 lbs . answer ( b ) <eor> b <eos> |
b |
percent__45.0__20.0__ percent__45.0__20.0__ |
percent__45.0__20.0__ percent__45.0__20.0__ |
| a car covers a certain distance at a speed of num__40 kmph in num__5 hours . to cover the same distance in num__2 hr it must travel at a speed of ? <o> a ) num__100 km / hr <o> b ) num__678 km / hr <o> c ) num__782 km / hr <o> d ) num__789 km / hr <o> e ) num__720 km / hr |
distance = ( num__40 x num__5 ) = num__200 km speed = distance / time speed = num__100.0 = num__100 km / hr answer : a <eor> a <eos> |
a |
multiply__40.0__5.0__ divide__200.0__2.0__ round__100.0__ |
multiply__40.0__5.0__ divide__200.0__2.0__ divide__200.0__2.0__ |
| the product of two numbers is num__266 and their difference is num__5 . what is the bigger number ? <o> a ) num__67 <o> b ) num__26 <o> c ) num__26 <o> d ) num__19 <o> e ) num__72 |
explanation : let the two numbers be a and b here a > b ab = num__266 b = num__266 / a - - - - - - - - - - - - - - - - - ( i ) given a – b = num__5 - - - - - - - - - - - ( ii ) substitute from ( i ) in ( ii ) we get a – num__266 / a = num__5 a num__2 – num__5 a + num__266 = num__0 ( a – num__19 ) ( a – num__14 ) = num__0 therefore a = num__19 or a = num__14 hence bigger number = a = num__19 answer : d <eor> d <eos> |
d |
divide__266.0__19.0__ divide__266.0__14.0__ |
divide__266.0__19.0__ divide__266.0__14.0__ |
| if a trader sold two cars each at rs . num__325475 and gains num__14.0 on the first and loses num__14.0 on the second then his profit or loss percent on the whole is ? <o> a ) num__1.44 <o> b ) num__1.74 <o> c ) num__1.84 <o> d ) num__1.96 <o> e ) num__1.24 % |
sp of each car is rs . num__325475 he gains num__14.0 on first car and losses num__14.0 on second car . in this case there will be loss and percentage of loss is given by = [ ( profit % ) ( loss % ) ] / num__100 = ( num__14 ) ( num__14 ) / num__100.0 = num__1.96 answer : d <eor> d <eos> |
d |
percent__100.0__1.96__ |
percent__100.0__1.96__ |
| the average ( arithmetic mean ) weight of num__5 friends is num__150 pounds . if joanne leaves the group and the new average of the group is num__162 pounds . what is joanne ' s weight ? <o> a ) num__60 <o> b ) num__78 <o> c ) num__102 <o> d ) num__105 <o> e ) num__110 |
the average ( arithmetic mean ) weight of num__5 friends is num__150 pounds . let w be sum of weights of the four others . and j be joanne ' s weight ( w + j ) / num__5 = num__150 when joanne exits the average is num__162 w / num__4 = num__162 w = num__648 w + j = num__750 j = num__102 answer : c <eor> c <eos> |
c |
multiply__162.0__4.0__ multiply__5.0__150.0__ subtract__750.0__648.0__ subtract__750.0__648.0__ |
multiply__162.0__4.0__ multiply__5.0__150.0__ subtract__750.0__648.0__ subtract__750.0__648.0__ |
| rob also compared the empire state building and the petronas towers . what is the height difference between the two if the empire state building is num__443 m tall and the petronas towers is num__452 m tall ? <o> a ) num__9 <o> b ) num__17 <o> c ) num__23 <o> d ) num__45 <o> e ) num__12 |
num__452 - num__443 = num__9 . answer is a . <eor> a <eos> |
a |
subtract__452.0__443.0__ subtract__452.0__443.0__ |
subtract__452.0__443.0__ subtract__452.0__443.0__ |
| num__40 people are sitting around a circular table . starting from num__1 every num__2 nd person is killed . this process continues till only one person remains . how is the survivor ? ( josephus riddle ) <o> a ) num__22 <o> b ) num__17 <o> c ) num__88 <o> d ) num__237 <o> e ) num__121 |
write the given number in binary form : num__101000 shift the left digit to the right hand side : num__010001 convert this number into decimal and this could be the number of the survivor : num__17 sol num__2 : write the given number as a sum of num__2 k + m where k should be maximum power less than the given number . survivor number is given by num__2 m + num__1 this this case num__40 < num__25 + num__8 so m = num__8 survivor number = num__2 ( num__8 ) + num__1 = num__17 answer : b <eor> b <eos> |
b |
square_perimeter__2.0__ multiply__1.0__17.0__ |
square_perimeter__2.0__ multiply__1.0__17.0__ |
| ab + cd = jjj where ab and cd are two - digit numbers and jjj is a three digit number ; a b c and d are distinct positive integers . in the addition problem above what is the value of c ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__7 <o> d ) num__9 <o> e ) can not be determined |
ab and cd are two digit integers their sum can give us only one three digit integer of a kind of jjj it ' s num__111 . so a = num__1 . num__1 b + cd = num__111 now c can not be less than num__9 because no to digit integer with first digit num__1 ( mean that it ' s < num__20 ) can be added to two digit integer less than num__90 to have the sum num__111 ( if cd < num__90 meaning c < num__9 cd + num__1 b < num__111 ) - - > c = num__9 answer : d . <eor> d <eos> |
d |
multiply__1.0__9.0__ |
multiply__1.0__9.0__ |
| a single discount equivalent to the discount series of num__20.0 num__15.0 and num__5.0 is ? <o> a ) num__31.6 <o> b ) num__35.4 <o> c ) num__31.6 <o> d ) num__31.1 <o> e ) num__31.5 |
num__100 * ( num__0.8 ) * ( num__0.85 ) * ( num__0.95 ) = num__68.4 num__100 - num__64.6 = num__35.4 answer : b <eor> b <eos> |
b |
percent__100.0__35.4__ |
percent__100.0__35.4__ |
| num__325.124 x num__12.98 Ã · num__3.001 + num__21.21 = ? <o> a ) num__1430.08 <o> b ) num__1420.06 <o> c ) num__781.189 <o> d ) num__656.112 <o> e ) num__456.512 |
explanation : ? = num__325.124 x num__12.98 Ã · num__3.001 + num__21.21 = ? â ‰ ˆ ( num__325.124 x num__4.33333333333 ) + num__21.21 â ‰ ˆ num__1408.87 + num__21.21 â ‰ ˆ num__1430.080 answer : option a <eor> a <eos> |
a |
add__21.21__1408.87__ add__21.21__1408.87__ |
add__21.21__1408.87__ add__21.21__1408.87__ |
| there is food for num__760 men for num__22 days . how many more men should join after two days so that the same food may last for num__19 days more ? <o> a ) num__16 <o> b ) num__40 <o> c ) num__18 <o> d ) num__19 <o> e ) num__11 |
num__760 - - - - num__22 num__760 - - - - num__20 x - - - - - num__19 x * num__19 = num__760 * num__20 x = num__800 num__760 - - - - - - - num__40 answer : b <eor> b <eos> |
b |
divide__760.0__19.0__ round__40.0__ |
subtract__800.0__760.0__ subtract__800.0__760.0__ |
| the difference between simple interest and c . i . at the same rate for rs . num__5000 for num__2 years in rs . num__72 . the rate of interest is ? <o> a ) num__11 <o> b ) num__12 <o> c ) num__77 <o> d ) num__85 <o> e ) num__65 |
num__5000 = num__72 ( num__100 / r ) num__2 num__5 r num__2 = num__720 = > r = num__12 . answer : b <eor> b <eos> |
b |
percent__2.0__5000.0__ percent__100.0__12.0__ |
percent__2.0__5000.0__ percent__100.0__12.0__ |
| num__10 x num__0.5 = ? <o> a ) num__0.0001 <o> b ) num__0.001 <o> c ) num__0.01 <o> d ) num__0.1 <o> e ) num__5 |
explanation : num__10 x num__5 = num__50 . sum of decimal places = num__1 num__10 x num__0.5 = num__5 answer - e <eor> e <eos> |
e |
multiply__10.0__0.5__ multiply__10.0__5.0__ multiply__10.0__0.5__ |
multiply__10.0__0.5__ multiply__10.0__5.0__ subtract__10.0__5.0__ |
| if ( t - num__8 ) is a factor of t ^ num__2 - kt - num__43 then k = <o> a ) num__16 <o> b ) num__12 <o> c ) num__2 <o> d ) num__3 <o> e ) num__14 |
t ^ num__2 - kt - num__48 = ( t - num__8 ) ( t + m ) where m is any positive integer . if num__6.0 = num__6 then we know as a matter of fact that : m = + num__6 and thus k = num__8 - num__6 = num__3 t ^ num__2 - kt - m = ( t - a ) ( t + m ) where a > m t ^ num__2 + kt - m = ( t - a ) ( t + m ) where a < m t ^ num__2 - kt + m = ( t - a ) ( t - m ) t ^ num__2 + kt + m = ( t + a ) ( t + m ) d <eor> d <eos> |
d |
subtract__8.0__2.0__ divide__6.0__2.0__ divide__6.0__2.0__ |
subtract__8.0__2.0__ divide__6.0__2.0__ subtract__6.0__3.0__ |
| find the compound interest on rs . num__10000 at num__12.0 rate of interest for num__1 year compounded half - yearly <o> a ) rs . num__1036 <o> b ) rs . num__1236 <o> c ) rs . num__1186 <o> d ) rs . num__1206 <o> e ) rs . num__1226 |
amount with ci = num__10000 [ num__1 + ( num__6.0 * num__100 ) ] num__2 = rs . num__11236 therefore ci = num__11236 – num__10000 = rs . num__1236 answer : b <eor> b <eos> |
b |
percent__1.0__10000.0__ percent__100.0__1236.0__ |
percent__1.0__10000.0__ percent__100.0__1236.0__ |
| in how many no . between num__100 and num__1000 exactly one of the digits is num__3 ? <o> a ) num__648 <o> b ) num__512 <o> c ) num__252 <o> d ) num__225 <o> e ) num__26 |
another approach can be total nos . between num__100 and num__1000 are num__899 nos . in which all digits are num__3 = num__1 nos . in which digit num__3 is in two places = num__8 + num__9 + num__9 = num__26 ( for thousand tens and once as different digit ) nos . in which no digit is num__3 = num__8 * num__9 * num__9 - num__1 = num__648 - num__1 ( for num__100 not to be included ) required nos = num__899 - num__1 - num__26 - num__647 = num__225 answer is d <eor> d <eos> |
d |
add__1.0__8.0__ subtract__648.0__1.0__ multiply__1.0__225.0__ |
add__1.0__8.0__ subtract__648.0__1.0__ multiply__1.0__225.0__ |
| a manufacturing company has num__15.0 cobalt num__25.0 led and num__60.0 of copper . if num__5 kg of led is used in a mixture how much copper we need to use <o> a ) num__12 kg <o> b ) num__13 kg <o> c ) num__22 kg <o> d ) num__14 kg <o> e ) num__15 kg |
num__25.0 = num__5 kg num__100.0 = num__20 kg num__60.0 of num__20 kg = num__12 kg num__12 kg answer : a <eor> a <eos> |
a |
add__15.0__5.0__ divide__60.0__5.0__ divide__60.0__5.0__ |
add__15.0__5.0__ divide__60.0__5.0__ divide__60.0__5.0__ |
| a person crosses a num__600 m long street in num__5 minutes . what is his speed in km per hour ? <o> a ) num__5.2 km / hr . <o> b ) num__8.2 km / hr . <o> c ) num__7.2 km / hr . <o> d ) num__3.2 km / hr . <o> e ) num__4.2 km / hr . |
c num__7.2 km / hr . speed = ( num__120.0 * num__60 ) m / sec = num__2 m / sec . converting m / sec to km / hr = ( num__2 * num__3.6 ) km / hr = num__7.2 km / hr . <eor> c <eos> |
c |
divide__600.0__5.0__ hour_to_min_conversion__ divide__120.0__60.0__ divide__7.2__2.0__ round__7.2__ |
divide__600.0__5.0__ hour_to_min_conversion__ divide__120.0__60.0__ divide__7.2__2.0__ multiply__3.6__2.0__ |
| the expression ( num__3 x - num__2 ) / ( x + num__3 ) is equivalent to which of the following ? <o> a ) ( num__5 - num__2 ) / num__3 <o> b ) num__5 – ( num__0.666666666667 ) <o> c ) num__3 – ( num__11 ) / ( x + num__3 ) <o> d ) num__5 – ( num__17 ) / ( x + num__3 ) <o> e ) num__5 + ( num__17 ) / ( x + num__3 ) |
easiest way is to pick any number and put in ( num__3 x - num__2 ) / ( x + num__3 ) . then check the options with the same number . i took x = num__1 and got c as the answer . <eor> c <eos> |
c |
subtract__3.0__2.0__ multiply__3.0__1.0__ |
subtract__3.0__2.0__ divide__3.0__1.0__ |
| in a num__100 m race sam beats john by num__4 seconds . on the contrary if sam allowed john to start num__24 m ahead of sam then sam and john reach the finishing point at the same time . how long does sam take to run the num__100 m race ? <o> a ) num__4 seconds <o> b ) num__25 seconds <o> c ) num__19 seconds <o> d ) num__21 seconds <o> e ) num__6.25 seconds |
their difference is num__4 second but this difference is num__0 if john allows sam to start the race from num__24 m ahead . that means jhon was num__24 m away from finishing line when they started together . so he will cover num__24 m in num__4 seconds . so his speed = num__6.0 = num__6 metre / second . so time taken = num__25.0 = num__25 seconds . so sam took = num__19 seconds . correct answer = c <eor> c <eos> |
c |
divide__24.0__4.0__ divide__100.0__4.0__ subtract__25.0__6.0__ round__19.0__ |
divide__24.0__4.0__ divide__100.0__4.0__ subtract__25.0__6.0__ round__19.0__ |
| a sum of num__13400 amounts to num__14400 in num__2 years at the rate of simple interest . what is the rate of interest ? <o> a ) num__2.0 <o> b ) num__1.0 <o> c ) num__6.0 <o> d ) num__4.0 <o> e ) num__8 % |
d num__4.0 s . i . = ( num__14400 - num__13400 ) = num__1000 . rate = ( num__100 x num__1000 ) / ( num__13400 x num__2 ) % = num__4.0 <eor> d <eos> |
d |
percent__4.0__100.0__ |
percent__4.0__100.0__ |
| when the smallest of num__3 consecutive odd integers is added to four times the largest it produces a result num__732 more than num__4 times the middle integer . find the numbers ? <o> a ) num__732 <o> b ) num__678 <o> c ) num__698 <o> d ) num__710 <o> e ) num__729 |
x + num__4 ( x + num__4 ) = num__732 + num__4 ( x + num__2 ) solve for x and find all three numbers x + num__4 x + num__16 = num__732 + num__4 x + num__8 x = num__724 x + num__2 = num__726 x + num__4 = num__728 check : the smallest is added to four times the largest num__724 + num__4 * num__728 = num__3636 four times the middle num__4 * num__726 = num__2904 num__3636 is more than num__2904 by num__3636 - num__2904 = num__732 a <eor> a <eos> |
a |
multiply__4.0__2.0__ subtract__732.0__8.0__ add__2.0__724.0__ subtract__732.0__4.0__ multiply__4.0__726.0__ add__4.0__728.0__ |
multiply__4.0__2.0__ subtract__732.0__8.0__ add__2.0__724.0__ subtract__732.0__4.0__ multiply__4.0__726.0__ add__4.0__728.0__ |
| a certain company retirement plan has arule of num__70 provision that allows an employee to retire when the employee ' s age plus years of employment with the company total at least num__70 . in what year could a female employee hired in num__1989 on her num__32 nd birthday first be eligible to retire under this provision ? <o> a ) num__2008 <o> b ) num__2004 <o> c ) num__2005 <o> d ) num__2006 <o> e ) num__2007 |
she must gain at least num__70 points now she has num__32 and every year gives her two more points : one for age and one for additional year of employment so num__32 + num__2 * ( # of years ) = num__70 - - > ( # of years ) = num__19 - - > num__1989 + num__19 = num__2008 . answer : a . <eor> a <eos> |
a |
add__1989.0__19.0__ add__1989.0__19.0__ |
add__1989.0__19.0__ add__1989.0__19.0__ |
| a and b complete a work in num__10 days . a alone can do it in num__14 days . if both together can do the work in how many days ? <o> a ) num__3.75 days <o> b ) num__5.8 days <o> c ) num__4.5 days <o> d ) num__4.7 days <o> e ) num__5.75 days |
num__0.1 + num__0.0714285714286 = num__0.171428571429 num__5.83333333333 = num__5.8 days answer : b <eor> b <eos> |
b |
add__0.0714__0.1__ round__5.8__ |
add__0.0714__0.1__ round__5.8__ |
| num__10 men working num__8 hours a day can complete a work in num__18 days . how many hours a day must num__15 men work to complete the work in num__12 days ? <o> a ) num__4 hours a day <o> b ) num__5 hours a day <o> c ) num__6 hours a day <o> d ) num__7 hours a day <o> e ) num__8 hours a day |
explanation : more men less hours { indirect proportion } less days more hours { indirect proportion } [ men num__15 num__10 days num__12 num__18 ] : : num__8 : x = > x â ˆ — num__15 â ˆ — num__12 = num__10 â ˆ — num__18 â ˆ — num__8 = > x = num__10 â ˆ — num__18 â ˆ — num__0.533333333333 â ˆ — num__12 = > x = num__8 option e <eor> e <eos> |
e |
divide__8.0__15.0__ round__8.0__ |
divide__8.0__15.0__ round__8.0__ |
| in one hour a boat goes num__14 km / hr along the stream and num__8 km / hr against the stream . the speed of the boat in still water ( in km / hr ) is : <o> a ) num__12 km / hr <o> b ) num__11 km / hr <o> c ) num__10 km / hr <o> d ) num__8 km / hr <o> e ) num__15 km / hr |
explanation let speed of the boat in still water = a and speed of the stream = b then a + b = num__14 a - b = num__8 adding these two equations we get num__2 a = num__22 = > a = num__11 ie speed of boat in still water = num__11 km / hr answer is b <eor> b <eos> |
b |
add__14.0__8.0__ divide__22.0__2.0__ round__11.0__ |
add__14.0__8.0__ divide__22.0__2.0__ divide__22.0__2.0__ |
| num__50 square stone slabs of equal size were needed to cover a floor area of num__72 sq . m . find the length of each stone slab ? <o> a ) num__120 cm <o> b ) num__767 cm <o> c ) num__88 cm <o> d ) num__666 cm <o> e ) num__776 cm |
area of each slab = num__1.44 m num__2 = num__1.44 m num__2 length of each slab √ num__1.44 = num__1.2 m = num__120 cm <eor> a <eos> |
a |
volume_rectangular_prism__50.0__1.2__2.0__ triangle_area__120.0__2.0__ |
volume_rectangular_prism__50.0__1.2__2.0__ triangle_area__120.0__2.0__ |
| if a : b = num__6 : num__4 b : c = num__7 : num__8 find a : b : c ? <o> a ) num__24 : num__28 : num__32 <o> b ) num__42 : num__28 : num__36 <o> c ) num__42 : num__28 : num__38 <o> d ) num__42 : num__28 : num__32 <o> e ) num__22 : num__28 : num__32 |
a : b = num__6 : num__4 b : c = num__7 : num__8 a : b : c = num__42 : num__28 : num__32 answer : d <eor> d <eos> |
d |
multiply__6.0__7.0__ multiply__4.0__7.0__ multiply__4.0__8.0__ multiply__6.0__7.0__ |
multiply__6.0__7.0__ multiply__4.0__7.0__ multiply__4.0__8.0__ multiply__6.0__7.0__ |
| an escalator is descending at constant speed . a walks down and takes num__50 steps to reach the bottom . b runs down and takes num__90 steps in the same time as a takes num__10 steps . how many steps are visible when the escalator is not operating ? <o> a ) num__80 <o> b ) num__90 <o> c ) num__100 <o> d ) num__110 <o> e ) num__120 |
lets suppose that a walks down num__1 step / min and escalator moves n steps / min it is given that a takes num__50 steps to reach the bottom in the same time escalator would have covered num__50 n steps so total steps on escalator is num__50 + num__50 n . again it is given that b takes num__90 steps to reach the bottom and time taken by him for this is equal to time taken by a to cover num__10 steps i . e num__10 minutes . so in this num__10 min escalator would have covered num__10 n steps . so total steps on escalatro is num__90 + num__10 n again equating num__50 + num__50 n = num__90 + num__10 n we get n = num__1 hence total no . of steps on escalator is num__100 . answer : c <eor> c <eos> |
c |
add__90.0__10.0__ round__100.0__ |
add__90.0__10.0__ add__90.0__10.0__ |
| the area of a triangle is with base num__4 m and height num__5 m ? <o> a ) num__20 sq m <o> b ) num__10 sq m <o> c ) num__5 sq m <o> d ) num__3 sq m <o> e ) num__12 sq m |
explanation : num__0.5 * num__4 * num__5 = num__10 m num__2 answer is b <eor> b <eos> |
b |
triangle_area__4.0__5.0__ square_perimeter__0.5__ triangle_area__4.0__5.0__ |
volume_rectangular_prism__4.0__5.0__0.5__ multiply__4.0__0.5__ multiply__5.0__2.0__ |
| a train num__700 m long is running at a speed of num__78 km / hr . if it crosses a tunnel in num__1 min then the length of the tunnel is ? <o> a ) num__298 m <o> b ) num__468 m <o> c ) num__600 m <o> d ) num__667 m <o> e ) num__781 m |
speed = num__78 * num__0.277777777778 = num__21.6666666667 m / sec . time = num__1 min = num__60 sec . let the length of the train be x meters . then ( num__700 + x ) / num__60 = num__21.6666666667 x = num__600 m . answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ round__600.0__ |
hour_to_min_conversion__ multiply__1.0__600.0__ |
| a grocer has a sale of rs . num__4000 rs . num__6524 rs . num__5689 rs . num__7230 and rs . num__6000 for num__5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . num__7000 ? <o> a ) s . num__12557 <o> b ) s . num__14993 <o> c ) s . num__15560 <o> d ) s . num__12589 <o> e ) s . num__12500 |
total sale for num__5 months = rs . ( num__4000 + num__6524 + num__5689 + num__7230 + num__6000 ) = rs . num__29443 required sale = rs . [ ( num__7000 x num__6 ) - num__29443 ] = rs . ( num__42000 - num__29443 ) = rs . num__12557 . option a <eor> a <eos> |
a |
multiply__7000.0__6.0__ subtract__42000.0__29443.0__ subtract__42000.0__29443.0__ |
multiply__7000.0__6.0__ subtract__42000.0__29443.0__ subtract__42000.0__29443.0__ |
| a person travels equal distances with speeds of num__3 km / hr num__4 km / hr and num__5 km / hr and takes a total time of num__47 minutes . the total distance is ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__4 <o> d ) num__45 <o> e ) num__4 |
let the total distance be num__3 x km . then x / num__3 + x / num__4 + x / num__5 = num__0.783333333333 num__47 x / num__60 = num__0.783333333333 = > x = num__1 . total distance = num__3 * num__1 = num__3 km . answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ subtract__4.0__3.0__ round__3.0__ |
hour_to_min_conversion__ subtract__4.0__3.0__ divide__3.0__1.0__ |
| x starts a business with rs . num__45000 . y joins in the business after num__3 months with rs . num__30000 . what will be the ratio in which they should share the profit at the end of the year ? <o> a ) num__1 : num__2 <o> b ) num__2 : num__1 <o> c ) num__1 : num__3 <o> d ) num__3 : num__1 <o> e ) none of these |
explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = num__45000 * num__12 : num__30000 * num__9 = num__45 * num__12 : num__30 * num__9 = num__3 * num__12 : num__2 * num__9 = num__2 : num__1 . answer : option b <eor> b <eos> |
b |
subtract__12.0__3.0__ subtract__3.0__2.0__ subtract__3.0__1.0__ |
subtract__12.0__3.0__ subtract__3.0__2.0__ multiply__1.0__2.0__ |
| an engineer undertakes a project to build a road num__10 km long in num__300 days and employs num__30 men for the purpose . after num__100 days he finds only num__2 km of the road has been completed . find the ( approximate ) number of extra men he must employ to finish the work in time . <o> a ) num__20 <o> b ) num__25 <o> c ) num__30 <o> d ) num__35 <o> e ) num__40 |
num__30 workers working already let x be the total men required to finish the task in next num__200 days num__2 km done hence remaining is num__8 km also work has to be completed in next num__200 days ( num__300 - num__100 = num__200 ) we know that proportion of men to distance is direct proportion and proportion of men to days is inverse proportion hence x = ( num__30 * num__8 * num__100 ) / ( num__2 * num__200 ) thus x = num__60 thus more men needed to finish the task = num__60 - num__30 = num__30 answer : c <eor> c <eos> |
c |
subtract__300.0__100.0__ subtract__10.0__2.0__ hour_to_min_conversion__ round__30.0__ |
subtract__300.0__100.0__ subtract__10.0__2.0__ multiply__30.0__2.0__ divide__300.0__10.0__ |
| what is the least number of square tiles required to pave the floor of a room num__11 m num__50 cm long and num__1 m num__50 cm broad ? <o> a ) num__724 <o> b ) num__804 <o> c ) num__69 <o> d ) num__844 <o> e ) none |
solution length of largest tile = h . c . f . of num__1150 cm & num__150 cm = num__50 cm . area of each tile = ( num__50 x num__50 ) cm num__2 ∴ required number of tiles = [ num__1150 x num__3.0 x num__50 ] = num__69 . answer c <eor> c <eos> |
c |
add__1.0__2.0__ round__69.0__ |
add__1.0__2.0__ round__69.0__ |
| how many positive even integers less than num__100 contain digits num__3 or num__7 ? <o> a ) num__16 <o> b ) num__17 <o> c ) num__18 <o> d ) num__10 <o> e ) num__20 |
two digit numbers : num__3 at tens place : num__30 num__3234 num__3638 num__7 at tens place : num__70 num__7274 num__7678 if num__3 and num__7 is at units place the number cant be even total : num__5 + num__5 = num__10 answer d <eor> d <eos> |
d |
subtract__100.0__30.0__ add__3.0__7.0__ divide__100.0__10.0__ |
subtract__100.0__30.0__ add__3.0__7.0__ add__3.0__7.0__ |
| a school has only four classes that contain num__10 num__20 num__30 and num__40 students respectively . the pass percentage of these classes are num__20.0 num__30.0 num__60.0 and num__100.0 respectively . find the pass % of the entire school . <o> a ) num__56.0 <o> b ) num__76.0 <o> c ) num__34.0 <o> d ) num__66.0 <o> e ) none of these |
solution : the number of pass candidates are num__2 + num__6 + num__18 + num__40 = num__66 out of total num__100 . hence pass pecentage = num__66.0 . answer : option d <eor> d <eos> |
d |
percent__10.0__20.0__ percent__10.0__60.0__ percent__30.0__60.0__ percent__100.0__66.0__ |
percent__10.0__20.0__ percent__10.0__60.0__ percent__30.0__60.0__ percent__100.0__66.0__ |
| today is jack ' s birthday . one year from today he will be twice as old as he was num__13 years ago . how old is jack today ? <o> a ) num__23 years <o> b ) num__24 years <o> c ) num__25 years <o> d ) num__26 years <o> e ) num__27 years |
let jack ' s age = x x + num__1 = num__2 ( x - num__13 ) x = num__27 answer : e <eor> e <eos> |
e |
round__27.0__ |
round__27.0__ |
| peter brought a scooter for a certain sum of money . he spent num__10.0 of the cost on repairs and sold the scooter for a profit of $ num__1100 . how much did he spend on repairs if he made a profit of num__20.0 ? <o> a ) num__100 <o> b ) num__200 <o> c ) num__300 <o> d ) num__400 <o> e ) num__500 |
e num__500 let the c . p . be $ x . then num__20.0 of x = num__1100 num__0.2 * x = num__1100 = > x = num__5500 c . p . = $ num__5500 expenditure on repairs = num__10.0 actual price = $ ( num__100 * num__5500 ) / num__110 = $ num__5000 expenditures on repairs = ( num__5500 - num__5000 ) = $ num__500 . <eor> e <eos> |
e |
percent__20.0__500.0__ percent__10.0__1100.0__ percent__10.0__5000.0__ |
percent__20.0__500.0__ percent__10.0__1100.0__ percent__10.0__5000.0__ |
| find the odd man out num__4 num__16 num__25 num__36 num__42 num__64 <o> a ) num__36 <o> b ) num__25 <o> c ) num__42 <o> d ) num__95 <o> e ) num__4 |
num__2 ^ num__2 = num__4 num__4 ^ num__2 = num__16 num__5 ^ num__2 = num__25 num__6 ^ num__2 = num__36 num__7 ^ num__2 = num__49 num__8 ^ num__2 = num__64 answer : c <eor> c <eos> |
c |
add__4.0__2.0__ divide__42.0__6.0__ add__42.0__7.0__ multiply__4.0__2.0__ add__36.0__6.0__ |
add__4.0__2.0__ divide__42.0__6.0__ power__7.0__2.0__ multiply__4.0__2.0__ add__36.0__6.0__ |
| the average weight of a group of boys is num__30 kg . after a boy of weight num__33 kg joins the group the average weight of the group goes up by num__1 kg . find the number of boys in the group originally ? <o> a ) num__4 <o> b ) num__8 <o> c ) num__5 <o> d ) num__2 <o> e ) num__9 |
let the number off boys in the group originally be x . total weight of the boys = num__30 x after the boy weighing num__33 kg joins the group total weight of boys = num__30 x + num__33 so num__30 x + num__33 + num__31 ( x + num__1 ) = > x = num__2 . answer : d <eor> d <eos> |
d |
add__30.0__1.0__ subtract__33.0__31.0__ subtract__33.0__31.0__ |
add__30.0__1.0__ subtract__33.0__31.0__ subtract__33.0__31.0__ |
| a man can swim in still water at num__4.5 km / h but takes twice as long to swim upstream than downstream . the speed of the stream is ? <o> a ) num__1.3 <o> b ) num__1.2 <o> c ) num__1.9 <o> d ) num__1.5 <o> e ) num__1.1 |
m = num__4.5 s = x ds = num__4.5 + x us = num__4.5 + x num__4.5 + x = ( num__4.5 - x ) num__2 num__4.5 + x = num__9 - num__2 x num__3 x = num__4.5 x = num__1.5 answer : d <eor> d <eos> |
d |
multiply__4.5__2.0__ divide__4.5__3.0__ round__1.5__ |
multiply__4.5__2.0__ subtract__4.5__3.0__ subtract__4.5__3.0__ |
| which of the following lines has x - intercept and y - intercept that are integers q ? <o> a ) y = num__3 x + num__1 <o> b ) y = √ x + num__1 <o> c ) y = - num__2 / x <o> d ) y = x ^ num__2 - num__1 <o> e ) xy = num__1 |
values that satisfy y = x ^ num__2 - num__1 are ( x y ) = ( - num__10 ) ( num__10 ) ( num__0 - num__1 ) . hence they are all integers q and the correct answer is d . <eor> d <eos> |
d |
multiply__1.0__2.0__ |
multiply__1.0__2.0__ |
| if a ^ num__2 + b ^ num__2 = num__177 and ab = num__54 then find the value of a + b / a - b ? <o> a ) num__5 <o> b ) num__15 <o> c ) num__6 <o> d ) num__3 <o> e ) num__7 |
( a + b ) ^ num__2 = a ^ num__2 + b ^ num__2 + num__2 ab = num__117 + num__2 * num__24 = num__225 a + b = num__15 ( a - b ) ^ num__2 = a ^ num__2 + b ^ num__2 - num__2 ab = num__117 - num__2 * num__54 a - b = num__3 a + b / a - b = num__5.0 = num__5 option a <eor> a <eos> |
a |
add__2.0__3.0__ add__2.0__3.0__ |
add__2.0__3.0__ add__2.0__3.0__ |
| the average of num__10 numbers is num__23 . if each number is increased by num__4 what will the new average be ? <o> a ) num__17 <o> b ) num__27 <o> c ) num__19 <o> d ) num__16 <o> e ) num__87 |
sum of the num__10 numbers = num__230 if each number is increased by num__4 the total increase = num__4 * num__10 = num__40 the new sum = num__230 + num__40 = num__270 the new average = num__27.0 = num__27 . answer : b <eor> b <eos> |
b |
multiply__10.0__23.0__ multiply__10.0__4.0__ add__230.0__40.0__ add__23.0__4.0__ add__23.0__4.0__ |
multiply__10.0__23.0__ multiply__10.0__4.0__ add__230.0__40.0__ add__23.0__4.0__ add__23.0__4.0__ |
| foodmart customers regularly buy at least one of the following products : milk chicken or apples . num__60.0 of shoppers buy milk num__50.0 buy chicken and num__35.0 buy apples . if num__15.0 of the customers buy all num__3 products what percentage of foodmart customers purchase exactly num__2 of the products listed above ? <o> a ) num__5.0 <o> b ) num__10.0 <o> c ) num__15.0 <o> d ) num__25.0 <o> e ) num__30 % |
num__60 - ( x + num__15 + z ) + num__50 - ( x + num__15 + y ) + num__35 - ( z + num__15 + y ) + x + + y + z + num__15 = num__100 where x = people who bought milkchicken y = people who bought chickenapples z = people who bought milk and apples x + y + z = the number of people who bought just exactly two products . hence solving the above equation we get num__115 - ( x + y + z ) = num__100 thus x + y + z = num__15 answer : c <eor> c <eos> |
c |
multiply__50.0__2.0__ add__15.0__100.0__ subtract__50.0__35.0__ |
multiply__50.0__2.0__ add__15.0__100.0__ subtract__50.0__35.0__ |
| ayesha ’ s father was num__38 years of age when she was born while her mother was num__34 years old when her brother four years younger to her was born . what is the difference between the ages of her parents ? <o> a ) num__2 years <o> b ) num__4 years <o> c ) num__6 years <o> d ) num__8 years <o> e ) none |
explanation mother ’ s age when ayesha ’ s brother was born = num__34 years . father ’ s age when ayesha ’ s brother was born = ( num__38 + num__4 ) years = num__42 years . required difference = ( num__42 – num__34 ) years = num__8 years . answer d <eor> d <eos> |
d |
subtract__38.0__34.0__ add__38.0__4.0__ subtract__42.0__34.0__ subtract__42.0__34.0__ |
subtract__38.0__34.0__ add__38.0__4.0__ subtract__42.0__34.0__ subtract__42.0__34.0__ |
| the ratio of boarders to day students at a school was originally num__2 to num__5 . however after a number of new boarders join the initial num__60 boarders the ratio changed to num__1 to num__2 . if no boarders became day students and vice versa and no students left the school how many new boarders joined the school ? <o> a ) num__12 <o> b ) num__15 <o> c ) num__20 <o> d ) num__24 <o> e ) num__30 |
let x be the number of new boarders . the ratio changed from num__2 : num__5 = num__4 : num__10 up to num__1 : num__2 = num__5 : num__10 . num__60 / ( num__60 + x ) = num__0.8 x = num__15 the answer is b . <eor> b <eos> |
b |
subtract__5.0__1.0__ multiply__2.0__5.0__ divide__4.0__5.0__ add__5.0__10.0__ round__15.0__ |
subtract__5.0__1.0__ multiply__2.0__5.0__ divide__4.0__5.0__ add__5.0__10.0__ add__5.0__10.0__ |
| julie put half of her savings in a savings account that pays an annual simple interest and half in a savings account that pays an annual compound interest . after two years she earned $ num__100 and $ num__105 from the simple interest account and the compound interest account respectively . if the interest rates for both accounts were the same what was the amount of julie ' s initial savings ? <o> a ) num__600 <o> b ) num__720 <o> c ) num__1000 <o> d ) num__1200 <o> e ) num__1440 |
$ num__100 for num__2 years = $ num__50 per year . extra $ num__5 yearned with the compound interest is the percent yearned on percent . so $ num__5 is yearned on $ num__50 which means that the interest = num__10.0 . this on the other hand means that half of the savings = num__50 * num__10 = $ num__500 . twice of that = $ num__1000 . answer : c . <eor> c <eos> |
c |
percent__100.0__1000.0__ |
percent__100.0__1000.0__ |
| the cost price of a radio is rs . num__1500 and it was sold for rs . num__1305 find the loss % ? <o> a ) num__13.0 <o> b ) num__16.0 <o> c ) num__17.0 <o> d ) num__78.0 <o> e ) num__28 % |
num__1500 - - - - num__195 num__100 - - - - ? = > num__13.0 answer : a <eor> a <eos> |
a |
percent__100.0__13.0__ |
percent__100.0__13.0__ |
| solve below question num__2 x + num__1 = - num__19 <o> a ) - num__8 <o> b ) - num__9 <o> c ) num__9 <o> d ) num__8 <o> e ) - num__10 |
num__2 x + num__1 = - num__19 x = - num__10 e <eor> e <eos> |
e |
multiply__1.0__10.0__ |
multiply__1.0__10.0__ |
| to prepare cake of type a it requires num__4 eggs . while to prepare cake of type b it requires num__6 eggs . if there are num__30 eggs available and at least one cake of each type needs to be prepared maximum how many cakes can be prepared of type b ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
if a number of type a cakes are prepared and b number of type b cakes are prepared . based on total number of eggs used num__4 a + num__6 b = num__30 . = > num__6 b = num__30 - num__4 a = > b = num__5 - ( num__0.666666666667 ) a = num__5 - ( num__0.666666666667 ) a to maximize b we have to minimize a and a has to be a multiple of num__3 . also since at least one cake of each type is to be prepared a > = num__1 . = > a = num__3 . b = num__5 - ( num__0.666666666667 ) num__3 = num__3 . answer : c <eor> c <eos> |
c |
divide__30.0__6.0__ divide__4.0__6.0__ subtract__4.0__3.0__ subtract__4.0__1.0__ |
divide__30.0__6.0__ divide__4.0__6.0__ subtract__4.0__3.0__ subtract__4.0__1.0__ |
| a jogger running at num__9 km / hr along side a railway track is num__240 m ahead of the engine of a num__120 m long train running at num__45 km / hr in the same direction . in how much time will the train pass the jogger ? <o> a ) num__16 <o> b ) num__11 <o> c ) num__36 <o> d ) num__17 <o> e ) num__12 |
speed of train relative to jogger = num__45 - num__9 = num__36 km / hr . = num__36 * num__0.277777777778 = num__10 m / sec . distance to be covered = num__240 + num__120 = num__360 m . time taken = num__36.0 = num__36 sec . answer : c <eor> c <eos> |
c |
subtract__45.0__9.0__ add__240.0__120.0__ subtract__45.0__9.0__ |
subtract__45.0__9.0__ add__240.0__120.0__ subtract__45.0__9.0__ |
| if x = num__2 + num__1 / ( num__1 + num__1 / ( num__1 - num__0.5 ) ) then the value of num__4 x + num__2.33333333333 <o> a ) num__12 num__0.666666666667 <o> b ) num__15 num__0.666666666667 <o> c ) num__18 num__0.666666666667 <o> d ) num__11 num__0.666666666667 <o> e ) num__17 num__0.666666666667 |
x = num__2 + num__1 / ( num__1 + num__1 / ( num__1 - num__0.5 ) ) = num__2 + num__1 / ( num__1 + num__2 ) = num__2 + num__0.333333333333 x = num__2.33333333333 = = > num__4 x + num__2.33333333333 = > num__4 ( num__2.33333333333 ) + num__2.33333333333 = > num__9.33333333333 + num__2.33333333333 = > num__11.6666666667 = > num__11 num__0.666666666667 . answer is d . <eor> d <eos> |
d |
subtract__2.3333__2.0__ round_down__11.6667__ subtract__1.0__0.3333__ round_down__11.6667__ |
subtract__2.3333__2.0__ round_down__11.6667__ subtract__1.0__0.3333__ divide__11.0__1.0__ |
| a thief goes away with a santro car at a speed of num__30 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at num__60 kmph when will the owner over take the thief from the start ? <o> a ) num__1 hours <o> b ) num__2 hours <o> c ) num__0.666666666667 hours <o> d ) num__0.333333333333 hours <o> e ) num__0.4 hours |
- - - - - - - - - - - num__30 - - - - - - - - - - - - - - - - - - - - | num__60 num__30 d = num__30 rs = num__60 â € “ num__30 = num__30 t = num__1.0 = num__1 hours answer : a <eor> a <eos> |
a |
round__1.0__ |
round__1.0__ |
| if x ^ num__2 + num__9 / x ^ num__2 = num__22 what is the value of x - num__3 / x <o> a ) num__36 <o> b ) num__25 <o> c ) num__9 <o> d ) num__5 <o> e ) num__4 |
to find : x - num__3 / x . let it be t . = > x - num__3 / x = t = > ( x ^ num__2 + num__9 / x ^ num__2 ) - num__2 * x * num__3 / x = t ^ num__2 ( squaring both sides ) . = > ( num__22 ) - num__2 * num__3 = num__16 = > t ^ num__2 = num__16 . thus t = num__4 or t = - num__4 . answer e <eor> e <eos> |
e |
divide__16.0__4.0__ |
divide__16.0__4.0__ |
| a certain class of students is being divided into teams . the class can either be divided into num__18 teams with an equal number of players on each team or num__24 teams with an equal number of players on each team . what is the lowest possible number of students in the class ? <o> a ) num__6 <o> b ) num__36 <o> c ) num__48 <o> d ) num__60 <o> e ) num__72 |
prime factorization of num__18 = num__2 * num__3 ^ num__2 prime factorization of num__24 = num__2 ^ num__3 * num__3 lcm of the given numbers = num__2 ^ num__3 * num__3 ^ num__2 = num__72 answer e <eor> e <eos> |
e |
multiply__24.0__3.0__ multiply__24.0__3.0__ |
multiply__24.0__3.0__ multiply__24.0__3.0__ |
| a man invests some money partly in num__9.0 stock at num__96 and partly in num__12.0 stock at num__120 . to obtain equal dividends from both he must invest the money in the ratio : <o> a ) num__3 : num__4 <o> b ) num__3 : num__5 <o> c ) num__4 : num__5 <o> d ) num__16 : num__15 <o> e ) num__12 : num__13 |
for an income of $ num__1 in num__9.0 stock at num__96 investment = $ ( num__10.6666666667 ) = num__10.6666666667 for an income of $ num__1 in num__12.0 stock at num__120 investment = $ ( num__10.0 ) = $ num__10 ratio of investment = num__10.6666666667 : num__10 = num__16 : num__15 option d <eor> d <eos> |
d |
divide__96.0__9.0__ round_down__10.6667__ subtract__16.0__1.0__ multiply__1.0__16.0__ |
divide__96.0__9.0__ round_down__10.6667__ subtract__16.0__1.0__ multiply__1.0__16.0__ |
| fresh grapes contain num__90.0 by weight while dried grapes contain num__20.0 water by weight . what is the weight of dry grapes available from num__5 kg of fresh grapes ? <o> a ) num__2 kg <o> b ) num__2.4 kg <o> c ) num__2.5 kg <o> d ) num__10 kg <o> e ) num__0.625 |
the weight of non - water in num__5 kg of fresh grapes ( which is num__100 - num__90 = num__10.0 of whole weight ) will be the same as the weight of non - water in x kg of dried grapes ( which is num__100 - num__20 = num__80.0 of whole weight ) so num__5 â ˆ — num__0.1 = x â ˆ — num__0.8 - - > x = num__0.625 answer : e . <eor> e <eos> |
e |
percent__0.625__100.0__ |
percent__0.625__100.0__ |
| a hat company ships its hats individually wrapped in num__8 - inch by num__10 - inch by num__12 - inch boxes . each hat is valued at $ num__7.50 . if the company ’ s latest order required a truck with at least num__384000 cubic inches of storage space in which to ship the hats in their boxes what was the minimum value of the order ? <o> a ) $ num__3000 <o> b ) $ num__1350 <o> c ) $ num__1725 <o> d ) $ num__2050 <o> e ) $ num__2 |
250 |
number of boxes = total volume / volume of one box = num__384000 / ( num__8 * num__10 * num__12 ) = num__400 one box costs num__7.50 so num__400 box will cost = num__400 * num__7.5 = num__3000 a is the answer <eor> a <eos> |
a |
a |
| a dealer purchases num__15 articles for rs . num__25 and sells num__12 articles for rs . num__30 . find the profit percentage ? <o> a ) num__60.0 <o> b ) num__50.0 <o> c ) num__80.0 <o> d ) num__70.0 <o> e ) num__53 % |
l . c . m of num__15 and num__12 = num__60 cp of num__60 articles = rs . num__100 ( num__25 * num__4 ) sp of num__60 articles = rs . num__150 ( num__30 * num__5 ) profit percentage = ( num__150 - num__100 ) / num__100 * num__100 = num__50.0 answer : b <eor> b <eos> |
b |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| two pipes can separately fill a tank in num__20 and num__30 hours respectively . both the pipes are opened to fill the tank but when the tank is full a leak develops in the tank through which one - third of water supplied by both the pipes goes out . what is the total time taken to fill the tank ? <o> a ) num__18 hrs <o> b ) num__16 hrs <o> c ) num__15 hrs <o> d ) num__12 hrs <o> e ) num__10 hrs |
explanation : num__0.05 + num__0.0333333333333 = num__0.0833333333333 num__1 + num__0.333333333333 = num__1.33333333333 num__1 - - - num__12 num__1.33333333333 - - - ? num__1.33333333333 * num__12 = num__16 hrs answer is b <eor> b <eos> |
b |
add__0.05__0.0333__ multiply__20.0__0.05__ add__1.0__0.3333__ round__16.0__ |
add__0.05__0.0333__ multiply__20.0__0.05__ add__1.0__0.3333__ round__16.0__ |
| a discount electronics store normally sells all merchandise at a discount of num__10 percent to num__30 percent off the suggested retail price . if during a special sale an additional num__20 percent were to be deducted from the discount price what would be the lowest possible price of an item costing $ num__400 before any discount ? <o> a ) $ num__230.00 <o> b ) $ num__224.00 <o> c ) $ num__263.80 <o> d ) $ num__282.00 <o> e ) $ num__210.00 |
since the question is essentially just about multiplication you can do the various mathstepsin a variety of ways ( depending on whichever method you find easiest ) . we ' re told that the first discount is num__10.0 to num__30.0 inclusive . we ' re told that the next discount is num__20.0 off of the discounted price . . . . we ' re told to maximize the discount ( thus num__30.0 off the original price and then num__20.0 off of the discounted price ) . thatmathcan be written in a number of different ways ( fractions decimals etc . ) : num__30.0 off = ( num__1 - . num__3 ) = ( num__1 - num__0.3 ) = ( . num__7 ) and the same can be done with the num__20.0 additional discount . . . the final price of an item that originally cost $ num__400 would be . . . . . ( $ num__400 ) ( . num__7 ) ( . num__8 ) = ( $ num__400 ) ( . num__56 ) = num__224 final answer : b <eor> b <eos> |
b |
percent__10.0__30.0__ percent__10.0__3.0__ percent__56.0__400.0__ percent__56.0__400.0__ |
percent__10.0__30.0__ percent__10.0__3.0__ percent__56.0__400.0__ percent__56.0__400.0__ |
| a company has two types of machines type r and type s . operating at a constant rate a machine of type r does a certain job in num__20 hrs and a machine of type s does the same job in num__30 hours . if the company used the same number of each type of machine to do the job in num__2 hours how many machines of type r were used ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
type r completes num__0.05 of the job each hour . type s completes num__0.0333333333333 of the job each hour . together r and s complete num__0.05 + num__0.0333333333333 = num__0.0833333333333 of the job each hour . let n be the number of each type of machine . num__2 * n * ( num__0.0833333333333 ) = num__1 job completed n = num__6 the answer is c . <eor> c <eos> |
c |
percent__20.0__30.0__ percent__20.0__30.0__ |
percent__20.0__30.0__ percent__20.0__30.0__ |
| the true discount on rs . num__1760 due after a certain time at num__12.0 per annum is rs . num__160 . the time after which it is due is : <o> a ) num__6 months <o> b ) num__8 months <o> c ) num__9 months <o> d ) num__10 months <o> e ) num__15 months |
td = amt * r * t / ( num__100 + rt ) given td = num__160 amt = num__1760 r = num__12 t = ? by substituting the values we get num__160 = ( num__1760 * num__12 t ) / ( num__100 + num__12 t ) num__160 ( num__100 + num__12 t ) = num__1760 * num__12 t num__100 + num__12 t = ( num__1760 * num__12 t ) / num__160 num__100 + num__12 t = num__132 t num__120 t = num__100 t = num__0.833333333333 = num__0.833333333333 yrs i . e ( num__0.833333333333 ) * num__12 = num__10 months answer : d <eor> d <eos> |
d |
percent__10.0__100.0__ |
percent__10.0__100.0__ |
| two trains of equal lengths are running at speeds of num__30 kmph and num__60 kmph . the two trains crossed each other in num__30 seconds when travelling in opposite direction . in what time will they cross each other when travelling in the same direction ? <o> a ) num__90 sec <o> b ) num__75 sec <o> c ) num__85 sec <o> d ) num__80 sec <o> e ) none of these |
let the length of each train be x m . ( x + x ) / ( num__60 + num__30 ) num__0.277777777778 = ( num__750 * num__18 ) / ( num__30 * num__5 ) = num__90 sec . answer : a <eor> a <eos> |
a |
add__30.0__60.0__ round__90.0__ |
add__30.0__60.0__ add__30.0__60.0__ |
| two cars cover the same distance at the speed of num__32 and num__64 kmps respectively . find the distance traveled by them if the slower car takes num__1 hour more than the faster car . <o> a ) num__73 <o> b ) num__38 <o> c ) num__64 <o> d ) num__83 <o> e ) num__93 |
explanation : num__32 ( x + num__1 ) = num__64 x x = num__1 num__32 * num__2 = num__64 km answer : option c <eor> c <eos> |
c |
divide__64.0__32.0__ round__64.0__ |
divide__64.0__32.0__ multiply__32.0__2.0__ |
| the average weight of num__8 people increases by num__2.5 kg when a new person comes in place of one of them weighing num__85 kg . what is the weight of the new person ? <o> a ) num__75 kg <o> b ) num__85 kg <o> c ) num__95 kg <o> d ) num__65 kg <o> e ) num__105 kg |
the total weight increase = ( num__8 x num__2.5 ) kg = num__20 kg weight of new person = ( num__85 + num__20 ) kg = num__105 kg the answer is e . <eor> e <eos> |
e |
multiply__8.0__2.5__ add__85.0__20.0__ add__85.0__20.0__ |
multiply__8.0__2.5__ add__85.0__20.0__ add__85.0__20.0__ |
| of the integers between num__200 and num__799 inclusive how many do not have digit num__2 and num__7 ? <o> a ) num__256 <o> b ) num__326 <o> c ) num__410 <o> d ) num__426 <o> e ) num__520 |
options for the first digit num__6 - num__2 ( num__2 and num__5 ) = num__4 ; options for the second digit num__10 - num__2 = num__8 ; options for the third digit num__10 - num__2 = num__8 ; total numbers possible num__4 * num__8 * num__8 = num__256 . answer : a . <eor> a <eos> |
a |
subtract__7.0__2.0__ subtract__6.0__2.0__ multiply__2.0__5.0__ multiply__2.0__4.0__ power__2.0__8.0__ power__2.0__8.0__ |
subtract__7.0__2.0__ subtract__6.0__2.0__ multiply__2.0__5.0__ multiply__2.0__4.0__ power__2.0__8.0__ power__2.0__8.0__ |
| num__3 x ^ num__2 - num__6 x + num__3 = num__0 find the above equation find the value of x <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
a = num__3 b = - num__6 c = num__3 x num__12 = ( num__6 ± √ ( ( - num__6 ) num__2 - num__4 × num__3 × num__3 ) ) / ( num__2 × num__3 ) = ( num__6 ± √ ( num__36 - num__36 ) ) / num__6 = ( num__6 ± num__0 ) / num__6 x num__1 = x num__2 = num__1 b <eor> b <eos> |
b |
multiply__2.0__6.0__ subtract__6.0__2.0__ multiply__3.0__12.0__ subtract__3.0__2.0__ reverse__1.0__ |
multiply__2.0__6.0__ subtract__6.0__2.0__ multiply__3.0__12.0__ subtract__3.0__2.0__ reverse__1.0__ |
| in a parallelogram the length of one diagonal and the perpendicular dropped on that diagonal are num__30 and num__20 metres respectively . find its area <o> a ) num__600 m num__2 <o> b ) num__540 m num__2 <o> c ) num__680 m num__2 <o> d ) num__574 m num__2 <o> e ) none of these |
in a parallelogram . area = diagonal × length of perpendicular on it . = num__30 × num__20 = num__600 m num__2 answer a <eor> a <eos> |
a |
multiply__30.0__20.0__ multiply__30.0__20.0__ |
multiply__30.0__20.0__ multiply__30.0__20.0__ |
| the average of num__7 numbers is num__24 . if each number be multiplied by num__5 . find the average of new set of numbers ? <o> a ) num__110 <o> b ) num__122 <o> c ) num__120 <o> d ) num__125 <o> e ) num__145 |
explanation : average of new numbers = num__24 * num__5 = num__120 answer : option c <eor> c <eos> |
c |
multiply__24.0__5.0__ multiply__24.0__5.0__ |
multiply__24.0__5.0__ multiply__24.0__5.0__ |
| in a college election between two rivals a candidate who got num__40.0 of the total votes polled was defeated by his rival by num__160 votes . the total number of votes polled was <o> a ) num__900 <o> b ) num__800 <o> c ) num__700 <o> d ) num__600 <o> e ) none of these |
let total number of votes polled be x . then votes polled by other candidate = ( num__100 – num__40 ) % of x = num__60.0 of x now num__60.0 of x – num__40.0 of x = num__160 ⇒ num__20 x ⁄ num__100 = num__160 ⇒ x = num__800 votes answer b <eor> b <eos> |
b |
percent__100.0__800.0__ |
percent__100.0__800.0__ |
| a rectangular - shaped carpet that measures x feet by y feet is priced at $ num__44 . what is the cost of the carpet in dollars per square yard ? ( num__1 square yard = num__9 square feet ) <o> a ) xy / num__360 <o> b ) num__9 xy / num__40 <o> c ) num__40 xy / num__9 <o> d ) num__396 xy <o> e ) num__396 / ( xy ) |
the area of the carpet in feet is xy . the area in square yards is xy / num__9 . the price per square yard is num__44 / ( xy / num__9 ) = num__396 / ( xy ) . the answer is e . <eor> e <eos> |
e |
multiply__44.0__9.0__ multiply__44.0__9.0__ |
multiply__44.0__9.0__ divide__396.0__1.0__ |
| the number obtained by interchanging the two digits of a two - digit number is less than the original number by num__45 . if the sum of the two digits of the number so obtained is num__13 then what is the original number ? <o> a ) num__94 <o> b ) num__66 <o> c ) num__39 <o> d ) num__87 <o> e ) num__58 |
a num__94 let the number be in the form of num__10 a + b number formed by interchanging a and b = num__10 b + a . a + b = num__13 - - - ( num__1 ) num__10 b + a = num__10 a + b - num__45 num__45 = num__9 a - num__9 b = > a - b = num__5 - - - ( num__2 ) adding ( num__1 ) and ( num__2 ) we get num__2 a = num__18 = > a = num__9 and b = num__4 the number is : num__94 . <eor> a <eos> |
a |
subtract__10.0__1.0__ divide__45.0__9.0__ divide__10.0__5.0__ add__13.0__5.0__ subtract__13.0__9.0__ multiply__1.0__94.0__ |
subtract__10.0__1.0__ divide__45.0__9.0__ divide__10.0__5.0__ add__13.0__5.0__ subtract__13.0__9.0__ multiply__1.0__94.0__ |
| a man can row upstream at num__20 kmph and downstream at num__80 kmph and then find the speed of the man in still water ? <o> a ) num__32 kmph <o> b ) num__50 kmph <o> c ) num__30 kmph <o> d ) num__45 kmph <o> e ) num__65 kmph |
us = num__20 ds = num__80 m = ( num__20 + num__80 ) / num__2 = num__50 answer : b <eor> b <eos> |
b |
round__50.0__ |
round__50.0__ |
| in an office num__30 percent of the workers have at least num__5 years of service and a total of num__16 workers have at least num__10 years of service . if num__90 percent of the workers have fewer than num__10 years of service how many of the workers have at least num__5 but fewer than num__10 years of service ? <o> a ) num__480 <o> b ) num__48 <o> c ) num__50 <o> d ) num__144 <o> e ) num__160 |
( num__0.1 ) workers = num__16 = > number of workers = num__160 ( num__0.3 ) * workers = x + num__16 = > x = num__48 answer b <eor> b <eos> |
b |
percent__30.0__160.0__ percent__30.0__160.0__ |
percent__30.0__160.0__ percent__30.0__160.0__ |
| a man can swim in still water at num__4.5 km / h but takes twice as long to swim upstream than downstream . the speed of the stream is ? <o> a ) num__5 <o> b ) num__8 <o> c ) num__1 <o> d ) num__6 <o> e ) num__1.5 |
explanation : m = num__4.5 s = x ds = num__4.5 + x us = num__4.5 + x num__4.5 + x = ( num__4.5 - x ) num__2 num__4.5 + x = num__9 - num__2 x num__3 x = num__4.5 x = num__1.5 answer : e <eor> e <eos> |
e |
multiply__4.5__2.0__ divide__4.5__3.0__ round__1.5__ |
multiply__4.5__2.0__ subtract__4.5__3.0__ subtract__4.5__3.0__ |
| a b c can complete a piece of work in num__24 num__812 days . working together they complete the same work in how many days ? <o> a ) num__4 <o> b ) num__0.8 <o> c ) num__0.777777777778 <o> d ) num__10 <o> e ) num__3.42857142857 |
a + b + c num__1 day work = num__0.0416666666667 + num__0.125 + num__0.0833333333333 = num__0.25 = num__0.25 a b c together will complete the job in num__4 days answer is a <eor> a <eos> |
a |
divide__1.0__24.0__ subtract__0.125__0.0417__ divide__1.0__0.25__ round__4.0__ |
divide__1.0__24.0__ subtract__0.125__0.0417__ divide__1.0__0.25__ round__4.0__ |
| the sum and the product of two numbers are num__16 and num__15 respectively the difference of the number is ? <o> a ) num__14 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__2 |
explanation : x + y = num__16 xy = num__15 ( x - y ) num__2 = ( x + y ) num__2 - num__4 xy ( x - y ) num__2 = num__256 - num__60 = > ( x - y ) = num__14 answer : a <eor> a <eos> |
a |
multiply__15.0__4.0__ subtract__16.0__2.0__ subtract__16.0__2.0__ |
multiply__15.0__4.0__ subtract__16.0__2.0__ subtract__16.0__2.0__ |
| in a class there are num__20 boys whose average age is decreased by num__2 months when one boy aged num__17 years replaced by a new boy . the age of the new boy is ? <o> a ) num__14 years num__8 months <o> b ) num__15 years <o> c ) num__13 years num__8 months <o> d ) num__17 years num__10 months <o> e ) num__17 years |
total decrease = ( num__20 x num__2 ) months = num__3 years num__4 months age of the new boy = num__17 years - num__3 years num__4 months . = num__13 years num__8 months . answer : c <eor> c <eos> |
c |
subtract__20.0__17.0__ subtract__17.0__4.0__ multiply__2.0__4.0__ subtract__17.0__4.0__ |
subtract__20.0__17.0__ subtract__17.0__4.0__ multiply__2.0__4.0__ subtract__17.0__4.0__ |
| the probability that a man will be alive for num__10 more yrs is num__0.25 & the probability that his wife will alive for num__10 more yrs is num__0.333333333333 . the probability that none of them will be alive for num__10 more yrs is <o> a ) num__0.5 <o> b ) num__0.666666666667 <o> c ) num__0.6 <o> d ) num__0.428571428571 <o> e ) num__0.375 |
sol . required probability = pg . ) x p ( b ) = ( num__1 — d x ( num__1 — i ) = : x num__1 = num__0.5 ans . ( a ) <eor> a <eos> |
a |
divide__0.25__0.5__ |
divide__0.25__0.5__ |
| during a pizza buffet where a eats more times num__2.4 than b and b eats num__6 times less than c . find the leat number of times all the three has to eat <o> a ) num__50 <o> b ) num__60 <o> c ) num__70 <o> d ) num__80 <o> e ) num__85 |
a eats more than b if b eats num__1 times than the ratio of a and b is a : b is num__2.4 : num__1 or num__12 : num__5 and as b eat num__6 times less the c the the ratio of b : c is num__5 : num__30 the the least number of times all three has eat is the lcm of a b c that is num__60 . . answer : b <eor> b <eos> |
b |
subtract__6.0__1.0__ multiply__6.0__5.0__ multiply__5.0__12.0__ multiply__1.0__60.0__ |
subtract__6.0__1.0__ multiply__6.0__5.0__ multiply__5.0__12.0__ multiply__1.0__60.0__ |
| x and y together can completes a work in num__12 days and y complete the same work in num__24 days . then the number of days x required to complete the work will be ? <o> a ) num__15 days . <o> b ) num__12 days . <o> c ) num__24 days . <o> d ) num__13 days . <o> e ) num__16 days . |
x and y complete in one day = num__0.0833333333333 part y completes in one day = num__0.0416666666667 parts . therefore x alone can do in one day = num__0.0416666666667 part . therefore the work can be finished by x = num__24 days . ans : c <eor> c <eos> |
c |
round__24.0__ |
round__24.0__ |
| if the average ( arithmetic mean ) of a and b is num__215 and the average of b and c is num__160 what is the value of a − c ? <o> a ) − num__220 <o> b ) − num__100 <o> c ) num__100 <o> d ) num__110 <o> e ) it can not be determined from the information given |
question : a - c = ? ( a + b ) / num__2 = num__215 = = = > a + b = num__430 ( b + c ) / num__2 = num__160 = = = > b + c = num__320 ( a + b ) - ( b + c ) = num__430 - num__320 = = = > a + b - b - c = num__110 = = = > a - c = num__110 answer : d <eor> d <eos> |
d |
multiply__215.0__2.0__ multiply__160.0__2.0__ subtract__430.0__320.0__ subtract__430.0__320.0__ |
multiply__215.0__2.0__ multiply__160.0__2.0__ subtract__430.0__320.0__ subtract__430.0__320.0__ |
| a brick measures num__20 cm * num__10 cm * num__7.5 cm how many bricks will be required for a wall num__27 m * num__2 m * num__0.75 m ? <o> a ) num__22377 <o> b ) num__27000 <o> c ) num__27891 <o> d ) num__25000 <o> e ) num__18771 |
num__27 * num__2 * num__0.75 = num__0.2 * num__0.1 * num__7.5 / num__100 * x num__27 = num__0.001 * x = > x = num__27000 answer : b <eor> b <eos> |
b |
divide__2.0__10.0__ divide__2.0__20.0__ divide__20.0__0.2__ divide__0.1__100.0__ round__27000.0__ |
divide__2.0__10.0__ divide__2.0__20.0__ divide__20.0__0.2__ divide__0.1__100.0__ round__27000.0__ |
| incomes of two companies a and b are in the ratio of num__5 : num__8 . had the income of company a been more by num__20 lakh the ratio of their incomes would have been num__5 : num__4 . what is the income of company b ? <o> a ) num__32 lakh <o> b ) num__50 lakh <o> c ) num__40 lakh <o> d ) num__60 lakh <o> e ) none of these |
let the incomes of two companies a and b be num__5 x and num__8 x respectively . from the question num__5 x + num__2.5 x = num__1.25 ⇒ num__20 x + num__80 = num__40 x ∴ x = num__4 ∴ income of company b = num__8 x = num__32 lakh answer a <eor> a <eos> |
a |
divide__20.0__8.0__ divide__5.0__4.0__ multiply__20.0__4.0__ multiply__5.0__8.0__ multiply__8.0__4.0__ multiply__8.0__4.0__ |
divide__20.0__8.0__ divide__5.0__4.0__ multiply__20.0__4.0__ multiply__5.0__8.0__ multiply__8.0__4.0__ multiply__8.0__4.0__ |
| resident of town x participated in a survey to determine the number of hours per week each resident spent watching television . the distribution of the results of the survey had a mean of num__21 hours and std deviation of num__8 hrs . the number of hours that pat a resident of town x watched television last week was between num__1 and num__2 standard deviation below the mean . which of the following could be the number of hours that pat watched television last week ? <o> a ) num__30 <o> b ) num__20 <o> c ) num__18 <o> d ) num__12 <o> e ) num__6 |
num__21 - num__12 < no of hrs < num__21 - num__8 num__9 < no of hrs < num__13 . d <eor> d <eos> |
d |
subtract__21.0__12.0__ subtract__21.0__8.0__ subtract__21.0__9.0__ |
subtract__21.0__12.0__ subtract__21.0__8.0__ subtract__21.0__9.0__ |
| meera is making telescopes each consisting of num__2 lenses num__1 tube and num__1 eyepiece . lenses can be purchased only in packs of num__50 tubes only in packs of num__10 and eyepieces only in packs of num__30 . however half of the lenses in each pack are not usable for telescopes . if all parts are used only for the telescopes what is the minimum number of lenses meera must purchase to make a set of telescopes with no leftover components other than the unusable lenses ? <o> a ) num__75 <o> b ) num__150 <o> c ) num__300 <o> d ) num__3600 <o> e ) num__7 |
500 |
i ' d use backsolving for this question . check d : num__3600 lenses = num__1800 usable lenses = num__900 tubes = num__900 eyepieces . we can buy num__900 tubes and num__900 eyepieces so that not to have leftovers . answer : d . <eor> d <eos> |
d |
d |
| two trains one from howrah to patna and the other from patna to howrah start simultaneously . after they meet the trains reach their destinations after num__4 hours and num__16 hours respectively . the ratio of their speeds is ? <o> a ) num__4 : num__6 <o> b ) num__4 : num__3 <o> c ) num__4 : num__9 <o> d ) num__4 : num__4 <o> e ) num__4 : num__2 |
let us name the trains a and b . then ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š num__16 : â ˆ š num__4 = num__4 : num__2 . answer : e <eor> e <eos> |
e |
round__4.0__ |
round__4.0__ |
| what least number must be subtracted from num__9671 so that the remaining number is divisible by num__5 ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__3 <o> d ) num__2 <o> e ) num__1 |
on dividing num__9671 by num__5 we get remainder = num__1 . required number be subtracted = num__1 answer : e <eor> e <eos> |
e |
reverse__1.0__ |
reverse__1.0__ |
| crazy eddie has a key chain factory . eddie managed to decrease the cost of manufacturing his key chains while keeping the same selling price and thus increased the profit from the sale of each key chain from num__40.0 of the selling price to num__50.0 of the selling price . if the manufacturing cost is now $ num__50 what was it before the decrease ? <o> a ) $ num__20 <o> b ) $ num__40 <o> c ) $ num__50 <o> d ) $ num__80 <o> e ) $ num__60 |
deargoodyear num__2013 i ' m happy to help . this is a relatively straightforward problem not very challenging . btw crazy eddiewas the actually name of an electronics chain on the east coast of the usa back in the num__1970 s . manufacturing now is $ num__50 . they now are making a num__50.0 profit so the selling price must be $ num__100 . they had this same selling price $ num__100 before they made the change and had a profit of num__40.0 so the manufacturing must have been $ num__60 . answer = ( e ) . <eor> e <eos> |
e |
subtract__100.0__40.0__ subtract__100.0__40.0__ |
subtract__100.0__40.0__ subtract__100.0__40.0__ |
| find out the square of a number which when doubled exceeds its one sixth by num__11 ? <o> a ) num__36 <o> b ) num__25 <o> c ) num__19 <o> d ) num__26 <o> e ) num__17 |
a let the number be p then the square will be p ^ num__2 according to question : num__2 p = ( p / num__6 ) + num__11 = > num__12 p = p + num__66 = > p = num__6 p ^ num__2 = num__6 ^ num__2 = num__36 . answer : a <eor> a <eos> |
a |
multiply__2.0__6.0__ multiply__11.0__6.0__ power__6.0__2.0__ triangle_area__2.0__36.0__ |
multiply__2.0__6.0__ multiply__11.0__6.0__ power__6.0__2.0__ power__6.0__2.0__ |
| a dinner has num__2 coffee pitchers ; each can hold num__8 cups of coffee . during monday ' s morning rush a customer orders num__10 cups of coffee . at the time this customer orders the coffee one coffee pitcher is num__0.25 full while the other is num__0.375 full . how much of the customer ' s order can be filled without making more coffee . <o> a ) num__0.333333333333 <o> b ) num__0.625 <o> c ) num__0.75 <o> d ) num__0.666666666667 <o> e ) num__0.5 |
one pitcher has num__2 cups of coffee and the other has num__3 cups of coffee . this means there is a total of num__5 cups of coffee . that means num__5 of the num__10 cups of coffee can be given to the customer before the dinner makes more coffee . num__0.5 simplifies to num__0.5 . ( answer : e ) <eor> e <eos> |
e |
multiply__8.0__0.375__ add__2.0__3.0__ reverse__2.0__ reverse__2.0__ |
multiply__8.0__0.375__ add__2.0__3.0__ reverse__2.0__ reverse__2.0__ |
| a picnic attracts num__240 persons . there are num__30 more men than women and num__30 more adults than children . how many men are at this picnic ? <o> a ) num__240 <o> b ) num__75 <o> c ) num__85 <o> d ) num__130 <o> e ) num__200 |
adult + children = num__240 let children = y then adult = y + num__30 i . e . y + ( y + num__30 ) = num__210 i . e . y = num__115 i . e . adult = num__115 + num__30 = num__145 adults include only men and women i . e . men + women = num__145 let women w = x then men m = x + num__30 i . e . x + ( x + num__30 ) = num__2 x + num__30 = num__145 i . e . x = num__55 i . e . men m = num__55 + num__30 = num__85 answer : option c <eor> c <eos> |
c |
subtract__240.0__30.0__ add__30.0__115.0__ add__30.0__55.0__ add__30.0__55.0__ |
subtract__240.0__30.0__ add__30.0__115.0__ add__30.0__55.0__ add__30.0__55.0__ |
| if a boat goes num__8 km upstream in num__80 minutes and the speed of the stream is num__5 kmph then the speed of the boat in still water is ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
rate upsteram = ( num__0.1 * num__60 ) kmph = num__6 kmph . speed of the stream = num__5 kmph let speed in still water be xkm / hr . then speed upstream = ( x - num__5 ) km / hr . x - num__5 = num__6 = = > x = num__11 km / hr answer ( d ) <eor> d <eos> |
d |
divide__8.0__80.0__ hour_to_min_conversion__ multiply__60.0__0.1__ add__5.0__6.0__ round__11.0__ |
divide__8.0__80.0__ hour_to_min_conversion__ multiply__60.0__0.1__ add__5.0__6.0__ round__11.0__ |
| a car runs num__375 km in num__3 hours . what ' s the car ' s speed ? <o> a ) num__124 <o> b ) num__125 <o> c ) num__126 <o> d ) num__127 <o> e ) none |
solution : num__375 ÷ num__3 = num__125 answer b <eor> b <eos> |
b |
divide__375.0__3.0__ round__125.0__ |
divide__375.0__3.0__ round__125.0__ |
| the ratio of spinsters to cats is num__2 to num__7 . if there are num__35 more cats than spinsters how many spinsters are there ? <o> a ) num__14 <o> b ) num__21 <o> c ) num__28 <o> d ) num__35 <o> e ) num__42 |
let num__2 x be the number of spinsters . then num__7 x is the number of cats . num__7 x - num__2 x = num__35 x = num__7 and the number of spinsters is num__2 ( num__7 ) = num__14 . the answer is a . <eor> a <eos> |
a |
multiply__2.0__7.0__ multiply__2.0__7.0__ |
multiply__2.0__7.0__ multiply__2.0__7.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__6 seconds . find the length of the train . <o> a ) num__110 meter <o> b ) num__120 meter <o> c ) num__100 meter <o> d ) num__90 meter <o> e ) num__105 meter |
speed = num__60 * ( num__0.277777777778 ) m / sec = num__16.6666666667 m / sec length of train ( distance ) = speed * time ( num__16.6666666667 ) * num__6 = num__100 meter answer : c <eor> c <eos> |
c |
round__100.0__ |
round__100.0__ |
| find the missing figures : num__0.1 of ? = num__0.24 <o> a ) num__12 <o> b ) num__120 <o> c ) num__24 <o> d ) num__240 <o> e ) num__36 |
let num__0.1 of x = num__0.24 . then num__0.1 * x / num__100 = num__0.24 x = [ ( num__0.24 * num__100 ) / num__0.1 ] = num__240 . answer is d . <eor> d <eos> |
d |
percent__100.0__240.0__ |
percent__100.0__240.0__ |
| a person took some amount with some interest for num__4 years but increase the interest for num__1.0 he paid rs . num__160 / - extra then how much amount he took ? <o> a ) rs . num__5500 / - <o> b ) rs . num__6000 / - <o> c ) rs . num__4000 / - <o> d ) rs . num__7000 / - <o> e ) none of these |
explanation : num__4 years = rs . num__160 / - year = num__40.0 rate of interest = num__1.0 num__100.0 % × num__40.0 = rs . num__4000 / - p = rs . num__4000 / - answer : option c <eor> c <eos> |
c |
divide__160.0__4.0__ multiply__40.0__100.0__ multiply__1.0__4000.0__ |
divide__160.0__4.0__ multiply__40.0__100.0__ divide__4000.0__1.0__ |
| a shopkeeper sold an article at $ num__100 with num__10.0 profit . then find its cost price ? <o> a ) $ num__120 <o> b ) $ num__100 <o> c ) $ num__91 <o> d ) $ num__72 <o> e ) $ num__69 |
cost price = selling price * num__100 / ( num__100 + profit ) c . p . = num__100 * num__0.909090909091 = $ num__91 ( approximately ) answer is c <eor> c <eos> |
c |
percent__100.0__91.0__ |
percent__100.0__91.0__ |
| pipe a can fill a tank in num__5 hours pipe b in num__10 hours and pipe c in num__30 hours . if all the pipes are open in how many hours will the tank be filled ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__3.5 <o> d ) num__4 <o> e ) num__4.5 |
part filled by a + b + c in num__1 hour = num__0.2 + num__0.1 + num__0.0333333333333 = num__0.333333333333 all the three pipes together will fill the tank in num__3 hours . answer b <eor> b <eos> |
b |
divide__1.0__5.0__ divide__1.0__10.0__ divide__1.0__30.0__ divide__10.0__30.0__ divide__30.0__10.0__ round__3.0__ |
divide__1.0__5.0__ divide__1.0__10.0__ divide__1.0__30.0__ divide__10.0__30.0__ divide__30.0__10.0__ round__3.0__ |
| the difference between the place values of num__8 and num__3 in the number num__587435 is <o> a ) num__75200 <o> b ) num__79970 <o> c ) num__75460 <o> d ) num__24580 <o> e ) none |
sol . = ( place value of num__8 ) – ( place value of num__3 ) = ( num__80000 - num__30 ) = num__79970 answer b <eor> b <eos> |
b |
subtract__80000.0__30.0__ subtract__80000.0__30.0__ |
subtract__80000.0__30.0__ subtract__80000.0__30.0__ |
| car dealership x the total profit from sales increased by num__10 percent over the previous years while the number of cars sold decreased by num__10 percent over the previous years . approximately what was the average percent increase in profit per car over the previous year . <o> a ) num__18.0 <o> b ) num__20.0 <o> c ) num__22.0 <o> d ) num__23.0 <o> e ) num__25 % |
let profit be x and cars sold be y originally profit / car = x / y now it is num__1.1 x / num__0.9 y = num__1.22222222222 ( x / y ) increase in profit per car = ( ( num__0.222222222222 ) ( x / y ) ) / ( x / y ) * num__100 = num__22.22 answer = c <eor> c <eos> |
c |
divide__1.1__0.9__ multiply__100.0__0.2222__ round_down__22.22__ |
divide__1.1__0.9__ multiply__100.0__0.2222__ round_down__22.22__ |
| at parkins gym class can be divided into num__8 teams with an equal number of players on each team or into num__12 teams with an equal number of players on each team . what is the lowest possible number of students in the class ? <o> a ) num__20 <o> b ) num__24 <o> c ) num__36 <o> d ) num__48 <o> e ) num__96 |
we are given that parkins gym class can be divided into num__8 teams or num__12 teams with an equal number of players on each team . translating this into two mathematical expressions we can say where g is the total number of students in the gym class that : g / num__8 = integer and g / num__12 = integer this means that g is a multiple of both num__8 and num__12 . we are asked to determine the lowest number of students in the class or the lowest value for variable “ g ” . because we know that g is a multiple of num__8 and of num__12 we need to find the least common multiple of num__8 and num__12 . although there are technical ways for determining the least common multiple the easiest method is to analyze the multiples of num__8 and num__12 until we find one in common . starting with num__8 we have : num__8 num__16 num__24 num__32 for num__12 we have : num__12 num__24 for the multiples of num__12 we stopped at num__24 because we see that num__24 is also a multiple of num__8 . thus num__24 is the least common multiple of num__8 and num__12 and therefore we know that the lowest possible number of students in the gym class is num__24 . answer b . <eor> b <eos> |
b |
add__8.0__16.0__ add__8.0__24.0__ add__8.0__16.0__ |
add__8.0__16.0__ add__8.0__24.0__ add__8.0__16.0__ |
| in a num__1300 m race usha beats shiny by num__50 m . in the same race by what time margin shiny beat mercy who runs at num__4 m / s ? <o> a ) num__65 sec . <o> b ) num__50 sec <o> c ) num__25 sec <o> d ) data not sufficient <o> e ) none of these |
speed of shiny = num__5.0 = num__5 m / s time taken by shiny to complete the race is b = num__260.0 = num__260 sec . time taken by baley to complete the race is d = num__325.0 = num__325 sec . hence d - b = num__65 sec answer : a <eor> a <eos> |
a |
divide__1300.0__5.0__ divide__1300.0__4.0__ divide__260.0__4.0__ round__65.0__ |
divide__1300.0__5.0__ divide__1300.0__4.0__ divide__260.0__4.0__ divide__260.0__4.0__ |
| a rectangular field is num__72 yards long and num__33 yards wide . a fence is needed for the perimeter of the field . fencing is also needed to divide the field into three square sections . how many feet of fencing are needed ? ( it is a good idea to make a drawing for this one . ) <o> a ) num__630 ft <o> b ) num__830 ft <o> c ) num__430 ft <o> d ) num__828 ft <o> e ) num__830 ft |
num__72 + num__33 + num__72 + num__33 = num__210 yds . for outside of field . num__33 + num__33 = num__66 yds . for dividing sections num__210 + num__66 = num__276 yds . of fencing num__276 x num__3 = num__828 ft . of fencing correct answer d <eor> d <eos> |
d |
rectangle_perimeter__72.0__33.0__ rectangle_perimeter__72.0__66.0__ multiply__3.0__276.0__ multiply__3.0__276.0__ |
rectangle_perimeter__72.0__33.0__ rectangle_perimeter__72.0__66.0__ multiply__3.0__276.0__ multiply__3.0__276.0__ |
| in a factory there are num__40.0 technicians and num__60.0 non - technicians . if the num__60.0 of the technicians and num__40.0 of non - technicians are permanent employees then the percentage of workers who are temporary is ? <o> a ) num__82.0 <o> b ) num__58.0 <o> c ) num__52.0 <o> d ) num__22.0 <o> e ) num__42 % |
total = num__100 t = num__40 nt = num__60 num__40 * ( num__0.6 ) = num__24 num__60 * ( num__0.4 ) = num__24 num__24 + num__24 = num__48 = > num__100 - num__48 = num__52.0 answer : c <eor> c <eos> |
c |
percent__40.0__60.0__ percent__100.0__52.0__ |
percent__40.0__60.0__ percent__100.0__52.0__ |
| the average height of num__35 boys in a class was calculated as num__180 cm . it has later found that the height of one of the boys in the class was wrongly written as num__166 cm whereas his actual height was num__106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ? <o> a ) num__187.89 cm <o> b ) num__178.29 cm <o> c ) num__123.98 cm <o> d ) num__149.98 cm <o> e ) num__146.89 cm |
calculated average height of num__35 boys = num__180 cm . wrong total height of num__35 boys = num__180 * num__35 cm . this was as a result of an actual height of num__106 cm being wrongly written as num__166 cm . correct total height of num__35 boys = num__180 * num__35 cm - num__166 cm + num__106 cm = num__180 * num__35 cm - num__166 cm + num__106 cm / num__35 = num__180 cm - num__1.71428571429 cm = num__180 cm - num__1.71 cm = num__178.29 cm . answer : b <eor> b <eos> |
b |
subtract__180.0__1.71__ subtract__180.0__1.71__ |
subtract__180.0__1.71__ subtract__180.0__1.71__ |
| if the average ( arithmetic mean ) of a and b is num__35 and the average of b and c is num__80 what is the value of c − a ? <o> a ) num__25 <o> b ) num__50 <o> c ) num__90 <o> d ) num__140 <o> e ) it can not be determined from the information given . |
- ( a + b = num__70 ) b + c = num__160 c - a = num__90 c . num__90 <eor> c <eos> |
c |
subtract__160.0__70.0__ subtract__160.0__70.0__ |
subtract__160.0__70.0__ subtract__160.0__70.0__ |
| two pipes can fill the cistern in num__10 hr and num__12 hr respectively while the third empty it in num__20 hr . if all pipes are opened simultaneously then the cistern will be filled in <o> a ) num__7.5 hr <o> b ) num__8 hr <o> c ) num__8.5 hr <o> d ) num__10 hr <o> e ) none of these |
solution : work done by all the tanks working together in num__1 hour . num__0.1 + num__0.0833333333333 − num__0.05 = num__0.133333333333 hence tank will be filled in num__7.5 = num__7.5 hour option ( a ) <eor> a <eos> |
a |
divide__1.0__10.0__ divide__1.0__12.0__ divide__1.0__20.0__ add__0.05__0.0833__ round__7.5__ |
divide__1.0__10.0__ divide__1.0__12.0__ divide__1.0__20.0__ add__0.05__0.0833__ round__7.5__ |
| if two numbers x a perfect square and y a perfect cube are added results a two digit number whose digits if reversed difference is num__27 find x and y ? <o> a ) x = num__4 y = num__8 <o> b ) x = num__4 y = num__9 <o> c ) x = num__4 y = num__7 <o> d ) x = num__4 y = num__9 <o> e ) x = num__9 y = num__27 |
num__9 + num__27 = num__36 when reversed num__63 num__63 - num__36 = num__27 x = num__9 y = num__27 answer : e <eor> e <eos> |
e |
add__27.0__9.0__ add__27.0__36.0__ subtract__36.0__27.0__ |
add__27.0__9.0__ add__27.0__36.0__ subtract__36.0__27.0__ |
| a thief goes away with a santro car at a speed of num__40 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at num__50 kmph when will the owner over take the thief from the start ? <o> a ) num__2 hours <o> b ) num__7 hours <o> c ) num__9 hours <o> d ) num__5 hours <o> e ) num__3 hours |
| - - - - - - - - - - - num__20 - - - - - - - - - - - - - - - - - - - - | num__50 num__40 d = num__20 rs = num__50 – num__40 = num__10 t = num__2.0 = num__2 hours answer : a <eor> a <eos> |
a |
subtract__50.0__40.0__ divide__40.0__20.0__ round__2.0__ |
subtract__50.0__40.0__ divide__40.0__20.0__ round__2.0__ |
| on a partly cloudy day john decides to walk back from work . when it is sunny he walks at a speed of s miles / hr ( s is an integer ) and when it gets cloudy he increases his speed to ( s + num__1 ) miles / hr . if his average speed for the entire distance is num__2.8 miles / hr what fraction of the total distance did he cover while the sun was shining on him ? <o> a ) num__0.125 <o> b ) num__0.142857142857 <o> c ) num__0.166666666667 <o> d ) num__0.2 <o> e ) num__0.25 |
if s is an integer and we know that the average speed is num__2.8 s must be = num__2 . that meanss + num__1 = num__3 . this implies that the ratio of time for s = num__2 is num__0.25 of the total time . the formula for distance / rate is d = rt . . . so the distance travelled when s = num__2 is num__2 t . the distance travelled for s + num__1 = num__3 is num__3 * num__4 t or num__12 t . therefore total distance covered while the sun was shining over him is num__0.142857142857 = num__0.142857142857 . answer : b <eor> b <eos> |
b |
round_down__2.8__ add__1.0__2.0__ reverse__0.25__ divide__3.0__0.25__ multiply__1.0__0.1429__ |
round_down__2.8__ add__1.0__2.0__ reverse__0.25__ divide__3.0__0.25__ multiply__1.0__0.1429__ |
| in a certain state gasoline stations compute the price per gallon p in dollars charged at the pump by adding a num__5 percent sales tax to the dealer ' s price per gallon d in dollars and then adding a gasoline tax of $ num__0.18 per gallon . which of the following gives the dealer ' s price per gallon d in terms of the price per gallon p charged at the pump ? <o> a ) d = ( p - num__0.18 ) / num__1.05 <o> b ) d = ( p - num__0.05 ) / num__1.18 <o> c ) d = p / num__1.23 <o> d ) d = p - num__0.23 <o> e ) d = p / num__1.05 - num__0.18 |
let dealers price ( d ) be num__1 . so adding num__5.0 to dealers price is d + num__5.0 of d . i . e . num__1 + num__5.0 of num__1 which is num__1 + num__0.05 . then add num__0.18 to the value . now num__1.05 + num__0.18 . this is now num__1.23 . you have the gasoline stations price ( p ) as num__1.23 dollars . now sub num__1.23 in the options to know which option gave you d = num__1 . d must equal num__1 because you earlier picked num__1 as the value of d in the question . ps : always remember to start from e upwards . answer : a <eor> a <eos> |
a |
add__1.0__0.05__ add__0.18__1.05__ multiply__0.18__1.0__ |
add__1.0__0.05__ add__0.18__1.05__ multiply__0.18__1.0__ |
| seven people are planning to share equally the cost of a rental car . if one person withdraws from the arrangement and the others share equally the entire cost of the car then the share of each of the remaining persons increased by : <o> a ) num__0.166666666667 <o> b ) num__0.285714285714 <o> c ) num__0.428571428571 <o> d ) num__0.666666666667 <o> e ) none of them |
original share of num__1 person = num__0.142857142857 new share of num__1 person = num__0.166666666667 increase = ( num__0.166666666667 - num__0.142857142857 = num__0.0238095238095 ) therefore required fraction = ( num__0.0238095238095 ) / ( num__0.142857142857 ) = ( num__0.0238095238095 ) x ( num__7.0 ) = num__0.166666666667 answer is a . <eor> a <eos> |
a |
subtract__0.1667__0.1429__ multiply__0.1667__1.0__ |
subtract__0.1667__0.1429__ divide__0.1667__1.0__ |
| at a small company num__58 percent of the employees are women and num__60 percent of the employees are married . if num__0.666666666667 of the men are single what fraction of the women are married ? <o> a ) num__0.3125 <o> b ) num__0.333333333333 <o> c ) num__0.45 <o> d ) num__0.7 <o> e ) num__0.706896551724 |
lets take total employees are num__100 . given that total women = num__58 and total married = num__60 . total men = num__100 - num__58 = num__42 and single men = num__0.666666666667 * num__42 = num__28 . married men = total men - single men = num__47 - num__28 = num__19 . married women = total married - married men = num__60 - num__19 = num__41 . fraction of women are married = married women / total women = num__0.706896551724 = num__0.714285714286 . ans e <eor> e <eos> |
e |
percent__100.0__0.7069__ |
percent__100.0__0.7069__ |
| here are num__6 periods in each working day of a school . in how many ways can one organize num__5 subjects such that each subject is allowed at least one period ? <o> a ) num__1235 <o> b ) num__4510 <o> c ) num__1203 <o> d ) num__1800 <o> e ) num__4512 |
num__5 subjects can be arranged in num__6 periods in num__6 p num__5 ways . any of the num__5 subjects can be organized in the remaining period ( num__5 c num__1 ways ) . two subjects are alike in each of the arrangement . so we need to divide by num__2 ! to avoid overcounting . total number of arrangements = num__6 p num__5 × num__5 c num__0.5 ! = num__1800 ans : d <eor> d <eos> |
d |
subtract__6.0__5.0__ divide__1.0__2.0__ round__1800.0__ |
subtract__6.0__5.0__ divide__1.0__2.0__ round__1800.0__ |
| during a certain season a team won num__60 percent of its first num__100 games and num__50 percent of its remaining games . if the team won num__70 percent of its games for the entire season what was the total number of games that the team played ? <o> a ) num__80 <o> b ) num__70 <o> c ) num__56 <o> d ) num__50 <o> e ) num__105 |
we are first given that a team won num__60 percent of its first num__100 games . this means the team won num__0.6 x num__100 = num__60 games out of its first num__100 games . we are next given that the team won num__50 percent of its remaining games . if we use variable t to represent the total number of games in the season then we can say t – num__100 equals the number of remaining games in the season . thus we can say : num__0.5 ( t – num__100 ) = number of wins for remaining games num__0.5 t – num__50 = number of wins for remaining games lastly we are given that team won num__70 percent of all games played in the season . that is they won num__0.7 t games in the entire season . with this we can set up the equation : number of first num__100 games won + number of games won for remaining games = total number of games won in the entire season num__60 + num__0.5 t – num__50 = num__0.7 t num__10 = num__0.2 t num__100 = num__2 t num__50 = t answer is d . <eor> d <eos> |
d |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| of the num__600 residents of clermontville num__35.0 watch the television show island survival num__40.0 watch lovelost lawyers and num__50.0 watch medical emergency . if all residents watch at least one of these three shows and num__18.0 watch exactly num__2 of these shows then how many clermontville residents v watch all of the shows ? <o> a ) num__150 <o> b ) num__108 <o> c ) num__42 <o> d ) num__21 <o> e ) - num__21 |
oa is d . num__100 = a + b + c - ab - ac - bc + abc which is the same as the following formula num__100 = a + b + c + ( - ab - ac - bc + abc + abc + abc ) - num__2 abc . the term between parantheses value num__18.0 so the equation to resolve is num__100 = num__35 + num__40 + num__50 - num__18 - num__2 abc therefore the value of abc is num__3.5 of num__600 v is num__21 . d is the correct answer <eor> d <eos> |
d |
percent__3.5__600.0__ percent__3.5__600.0__ |
percent__3.5__600.0__ percent__3.5__600.0__ |
| thabo owns exactly num__220 books and each book is either paperback fiction paperback nonfiction or hardcover nonfiction . if he owns num__20 more paperback nonfiction books than hardcover nonfiction books and twice as many paperback fiction books as paperback nonfiction books how many hardcover books nonfiction books does thabo own ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__30 <o> d ) num__40 <o> e ) num__50 |
i think we can use double - matrix method and solve using only one variable . our goal is to find the number of hardcover nonfiction books . let that number be x . we are given that all num__140 books are either paperback fiction paperback nonfiction or hardcover nonfiction . this implies that number of hardcover fiction books is num__0 . double - matrix : p = paperback ; h = hardcover ; f = fiction ; nf = nonfiction p h total f num__2 x + num__40 num__0 nf x + num__20 x total num__3 x + num__60 x num__220 num__3 x + num__60 + x = num__220 x = num__40 answer ( d . ) <eor> d <eos> |
d |
multiply__20.0__2.0__ multiply__20.0__3.0__ multiply__20.0__2.0__ |
multiply__20.0__2.0__ add__20.0__40.0__ subtract__60.0__20.0__ |
| in an examination a student scores num__4 marks for every correct answer and loses num__1 mark for every wrong answer . if he attempts all num__60 questions and secures num__130 marks the no of questions he attempts correctly is : <o> a ) num__35 <o> b ) num__38 <o> c ) num__40 <o> d ) num__42 <o> e ) num__44 |
explanation : let the number of correct answers be x . number of incorrect answers = ( num__60 – x ) . num__4 x – ( num__60 – x ) = num__130 = > num__5 x = num__190 = > x = num__38 answer : b <eor> b <eos> |
b |
add__4.0__1.0__ add__60.0__130.0__ divide__190.0__5.0__ round__38.0__ |
add__4.0__1.0__ add__60.0__130.0__ divide__190.0__5.0__ round__38.0__ |
| the average ( arithmetic mean ) of eight numbers is num__43.2 . if the sum of half of these numbers is num__154.4 what is the average of the other half ? <o> a ) num__44.2 <o> b ) num__47.8 <o> c ) num__50.5 <o> d ) num__53.6 <o> e ) num__56.9 |
the average of this half is num__154.4 / num__4 = num__38.6 this is num__4.6 below the overall average thus the average of the other half of the numbers must be num__4.6 above the overall age that is num__43.2 + num__4.6 = num__47.8 the answer is b . <eor> b <eos> |
b |
divide__154.4__4.0__ subtract__43.2__38.6__ add__43.2__4.6__ add__43.2__4.6__ |
divide__154.4__4.0__ subtract__43.2__38.6__ add__43.2__4.6__ add__43.2__4.6__ |
| express a speed of num__108 kmph in meters per second ? <o> a ) num__10 mps <o> b ) num__05 mps <o> c ) num__09 mps <o> d ) num__30 mps <o> e ) num__11 mps |
d num__30 mps num__108 * num__0.277777777778 = num__30 mps <eor> d <eos> |
d |
divide__30.0__108.0__ round__30.0__ |
divide__30.0__108.0__ round__30.0__ |
| a brick measures num__20 cm * num__10 cm * num__7.5 cm how many bricks will be required for a wall num__24 m * num__2 m * num__0.75 m ? <o> a ) num__22377 <o> b ) num__27782 <o> c ) num__27891 <o> d ) num__24000 <o> e ) num__18771 |
num__24 * num__2 * num__0.75 = num__0.2 * num__0.1 * num__7.5 / num__100 * x num__24 = num__0.01 * x = > x = num__24000 answer : d <eor> d <eos> |
d |
divide__2.0__10.0__ divide__2.0__20.0__ divide__20.0__0.2__ divide__0.2__20.0__ round__24000.0__ |
divide__2.0__10.0__ divide__2.0__20.0__ divide__20.0__0.2__ divide__0.2__20.0__ round__24000.0__ |
| if num__1 is added to the denominator of fraction the fraction becomes num__0.5 . if num__1 is added to the numerator the fraction becomes num__1 . the fraction is ? <o> a ) num__0.571428571429 <o> b ) num__0.555555555556 <o> c ) num__0.666666666667 <o> d ) num__0.909090909091 <o> e ) none |
answer let the required fraction be x / y then x / ( y + num__1 ) = num__0.5 ⇒ num__2 x - y = num__1 . . . ( num__1 ) ( x + num__1 ) / y = num__1 ⇒ x - y = - num__1 . . . ( num__2 ) solving eq . ( num__1 ) and ( num__2 ) we get x = num__2 y = num__3 ∴ the fraction is num__0.666666666667 correct option : c <eor> c <eos> |
c |
reverse__0.5__ add__1.0__2.0__ divide__2.0__3.0__ multiply__1.0__0.6667__ |
reverse__0.5__ add__1.0__2.0__ divide__2.0__3.0__ divide__2.0__3.0__ |
| the ratio of length to width of a rectangular showroom window is num__3.3 to num__2 . if the width of the window is num__8 feet what is the approximate length of the display in feet ? <o> a ) num__7 <o> b ) num__11 <o> c ) num__13 <o> d ) num__16 <o> e ) num__26 |
explanation : letting l be the length of the window the proportion for the ratio of the length to the width can be expressed in the following equation : num__3.3 / num__2 = l / num__8 num__26.4 = num__2 l num__13.2 = l answer : option c <eor> c <eos> |
c |
multiply__3.3__8.0__ divide__26.4__2.0__ round_down__13.2__ |
multiply__3.3__8.0__ divide__26.4__2.0__ round_down__13.2__ |
| what is the greatest prime factor of num__4 ^ num__15 - num__2 ^ num__28 ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__11 |
i ' m happy to help with this . we know num__4 = num__2 ^ num__2 so num__4 ^ num__15 = ( num__2 ^ num__2 ) ^ num__15 = num__2 ^ ( num__2 * num__15 ) = num__2 ^ num__30 that takes advantage of a law of exponents that says ( a ^ n ) ^ m = a ^ ( n * m ) so num__4 ^ num__15 - num__2 ^ num__28 = num__2 ^ num__30 - num__2 ^ num__28 = num__2 ^ ( num__28 + num__2 ) - num__2 ^ num__28 = ( num__2 ^ num__28 ) * ( num__2 * num__2 ) - num__2 ^ num__28 = ( num__2 ^ num__2 - num__1 ) * ( num__2 ^ num__28 ) = ( num__4 - num__1 ) * ( num__2 ^ num__28 ) = num__3 * ( num__2 ^ num__28 ) the prime factors of num__63 are num__3 so the largest prime factor is num__7 answer choice d . here ' s a blog you may find helpful . http : / / magoosh . com / gmat / num__2012 / gmat - math - factors / does all that make sense ? please let me know if you have any further questions . mike wow . i am floored by how great of an explanation you provided . posts like that make me really think that doing thousands of practice problems with good explanations beats out reading books on math every day of the week . b <eor> b <eos> |
b |
multiply__15.0__2.0__ subtract__4.0__1.0__ add__4.0__3.0__ subtract__4.0__1.0__ |
multiply__15.0__2.0__ subtract__4.0__1.0__ add__4.0__3.0__ subtract__4.0__1.0__ |
| a number of num__44 marbles is to be divided and contain with boxes . if each box is to contain num__3 num__4 or num__5 marbles what is the largest possible number of boxes ? <o> a ) num__10 <o> b ) num__14 <o> c ) num__15 <o> d ) num__16 <o> e ) num__17 |
to maximize # of boxes we should minimize marbles per box : num__13 * num__3 + num__1 * num__5 = num__44 - - > num__13 + num__1 = num__14 . answer b <eor> b <eos> |
b |
subtract__4.0__3.0__ add__1.0__13.0__ add__1.0__13.0__ |
subtract__4.0__3.0__ add__1.0__13.0__ add__1.0__13.0__ |
| a batsman makes a score of num__82 runs in the num__17 th inning and thus increases his averages by num__3 . what is his average after num__17 th inning ? <o> a ) num__25 <o> b ) num__31 <o> c ) num__27 <o> d ) num__29 <o> e ) num__34 |
let the average after num__17 innings = x total runs scored in num__17 innings = num__17 x average after num__16 innings = ( x - num__3 ) total runs scored in num__16 innings = num__16 ( x - num__3 ) total runs scored in num__16 innings + num__82 = total runs scored in num__17 innings = > num__16 ( x - num__3 ) + num__82 = num__17 x = > num__16 x - num__48 + num__82 = num__17 x = > x = num__34 answer is e . <eor> e <eos> |
e |
multiply__3.0__16.0__ subtract__82.0__48.0__ subtract__82.0__48.0__ |
multiply__3.0__16.0__ subtract__82.0__48.0__ subtract__82.0__48.0__ |
| which of the following describes all values of x for which num__1 – x ^ num__2 > num__0 ? <o> a ) x > = num__1 <o> b ) x < = – num__1 <o> c ) num__0 < = x < = num__1 <o> d ) x < = – num__1 or x > = num__1 <o> e ) – num__1 < x < num__1 |
which of the following describes all values of x for which num__1 – x ^ num__2 > = num__0 ? ( a ) x > = num__1 plugged in num__2 . num__1 – ( num__2 ) ^ num__2 > = num__0 - num__3 > = num__0 ? no . wrong ( b ) x < = – num__1 plugged in - num__2 . num__1 – ( - num__2 ) ^ num__2 > = num__0 - num__3 > = num__0 ? no . wrong ( c ) num__0 < = x < = num__1 plugged in num__0 num__1 and num__0.5 . all of them work . but e is better because it describes all the values of x ( d ) x < = – num__1 or x > = num__1 a and b answer this . wrong . ( e ) – num__1 < x < num__1 x is a positive or negative fraction . x = - num__0.5 x = num__0.5 x = num__0 plug all of them . they work . answer is e . <eor> e <eos> |
e |
add__1.0__2.0__ reverse__2.0__ reverse__1.0__ |
add__1.0__2.0__ reverse__2.0__ subtract__2.0__1.0__ |
| the sum of ages of num__5 children born num__1.5 years different each is num__50 years . what is the age of the elder child ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__16 <o> e ) num__13 |
let the ages of children be x ( x + num__1.5 ) ( x + num__3 ) ( x + num__4.5 ) and ( x + num__6 ) years . then x + ( x + num__1.5 ) + ( x + num__3 ) + ( x + num__4.5 ) + ( x + num__6 ) = num__50 num__5 x = num__35 x = num__7 . x + num__6 = num__7 + num__6 = num__13 answer : e <eor> e <eos> |
e |
add__1.5__3.0__ add__1.5__4.5__ divide__35.0__5.0__ add__6.0__7.0__ add__6.0__7.0__ |
add__1.5__3.0__ add__1.5__4.5__ divide__35.0__5.0__ add__6.0__7.0__ add__6.0__7.0__ |
| two wheels are connected via a conveyor belt . the larger wheel has a num__48 cm diameter and the smaller wheel has a num__32 cm diameter . in order for the conveyor belt to work smoothly each wheel must rotate the exact same number of centimetres per minute . if the larger wheel makes r revolution per minute how many revolutions does the smaller wheel make per hour in terms of r ? <o> a ) num__90 r <o> b ) num__75 r <o> c ) num__48 r <o> d ) num__24 r <o> e ) ( num__64 π ) / num__3 |
interesting to note that the larger wheel has a diameter of num__48 ( num__8 * num__6 ) while the smaller one has a diameter of num__32 ( num__8 * num__4 ) . . . if the large wheel has a diameter of num__40 and the small wheel num__32 then their circumferences are num__48 pi and num__32 pi respectively . in order for them to move the conveyor belt at the same rate the smaller wheel would need to rotate num__1.5 times as fast as the larger wheel . lets say the large wheel makes num__10 revolutions per minute the smaller wheel would need to make num__10 * num__1.5 = num__15 revolutions per minute . if the large wheel makes num__10 revolutions per minute it makes num__600 per hour . therefore the smaller wheel would need to make num__600 * num__1.5 = num__900 revolutions per hour . if r = num__10 then the answer choice must be b . a . num__90 r <eor> a <eos> |
a |
divide__48.0__8.0__ divide__32.0__8.0__ subtract__48.0__8.0__ divide__48.0__32.0__ add__4.0__6.0__ multiply__1.5__10.0__ multiply__15.0__40.0__ multiply__1.5__600.0__ multiply__6.0__15.0__ round__90.0__ |
divide__48.0__8.0__ divide__32.0__8.0__ subtract__48.0__8.0__ divide__48.0__32.0__ add__4.0__6.0__ multiply__1.5__10.0__ multiply__15.0__40.0__ multiply__1.5__600.0__ multiply__6.0__15.0__ multiply__6.0__15.0__ |
| a and b can do a piece of work in num__18 and num__9 days respectively . a does the work for a few days and left then b continued for num__6 days . how many days did a and b work together ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
a and b work together for num__2 days . a num__1 day work num__0.0555555555556 b num__1 day work num__0.111111111111 let a work x days . then equation form is ( x / num__18 ) + ( ( x + num__6 ) / num__9 ) = num__1 from this equation x value is num__2 . answer : b <eor> b <eos> |
b |
divide__18.0__9.0__ divide__1.0__18.0__ divide__1.0__9.0__ round__2.0__ |
divide__18.0__9.0__ divide__1.0__18.0__ divide__1.0__9.0__ divide__18.0__9.0__ |
| from the answers select the option that is near to √ num__5 ? <o> a ) num__2.2 <o> b ) num__2.3 <o> c ) num__3.0 <o> d ) num__3.8 <o> e ) num__1.9 |
num__2.2 is nearly equal to the square root of num__5 . num__2.2 * num__2.2 = num__4.4 num__2.3 * num__2.3 = num__5.29 difference num__5 - num__4.4 = num__0.6 num__5 - num__5.3 = num__0.7 since num__0.6 is lesser among both num__2.2 is closer . answer is a <eor> a <eos> |
a |
subtract__5.0__4.4__ subtract__4.4__2.2__ |
subtract__5.0__4.4__ subtract__4.4__2.2__ |
| the average weight of num__8 person ' s increases by num__2.5 kg when a new person comes in place of one of them weighing num__55 kg . what might be the weight of the new person ? <o> a ) num__56 kg <o> b ) num__90 kg <o> c ) num__75 kg <o> d ) data inadequate <o> e ) none of these |
c num__75 kg total weight increased = ( num__8 x num__2.5 ) kg = num__20 kg . weight of new person = ( num__55 + num__20 ) kg = num__75 kg . <eor> c <eos> |
c |
multiply__8.0__2.5__ add__55.0__20.0__ |
multiply__8.0__2.5__ add__55.0__20.0__ |
| if x is divisible by num__5 num__20 and num__125 which of the following must be the least multiple of x ? <o> a ) num__12500 <o> b ) num__1250 <o> c ) num__500 <o> d ) num__100 <o> e ) num__125 |
if x is divisible by num__5 num__20 and num__125 then x must be a multiple of num__54 and num__25 . hence the least multiple of x will be num__5 * num__4 * num__25 = num__500 . answer : c <eor> c <eos> |
c |
add__5.0__20.0__ divide__20.0__5.0__ multiply__20.0__25.0__ multiply__20.0__25.0__ |
add__5.0__20.0__ divide__20.0__5.0__ multiply__20.0__25.0__ multiply__20.0__25.0__ |
| a can knit a pair of socks in num__3 days . b can knit the same thing in num__6 days . if they are knitting together in how many days will they knit two pairs of socks ? <o> a ) num__4 days <o> b ) num__2 days <o> c ) num__4 num__1 ⁄ num__2 days <o> d ) num__3 days <o> e ) none of these |
explanationa ’ s one day ’ s work = num__1 ⁄ num__3 rd work . b ’ s one day ’ s work = num__1 ⁄ num__6 rd work . ( a + b ) ’ s one day ’ s work = num__1 ⁄ num__3 + num__1 ⁄ num__6 = num__1 ⁄ num__2 nd work . ∴ a and b together can complete the work ( knit a pair of socks ) in num__2 days . ∴ they together knit two pair of socks in num__4 days . answer a <eor> a <eos> |
a |
subtract__3.0__1.0__ add__3.0__1.0__ round__4.0__ |
subtract__3.0__1.0__ add__3.0__1.0__ add__3.0__1.0__ |
| if num__12 : num__8 : : x : num__16 then find the value of x <o> a ) num__18 <o> b ) num__24 <o> c ) num__28 <o> d ) num__16 <o> e ) num__20 |
explanation : treat num__12 : num__8 as num__1.5 and x : num__16 as x / num__16 treat : : as = so we get num__1.5 = x / num__16 = > num__8 x = num__192 = > x = num__24 option b <eor> b <eos> |
b |
divide__12.0__8.0__ multiply__12.0__16.0__ add__8.0__16.0__ add__8.0__16.0__ |
divide__12.0__8.0__ multiply__12.0__16.0__ divide__192.0__8.0__ divide__192.0__8.0__ |
| reema can complete a piece of work in num__12 days while seema can the same work in num__18 days . if they both work together then how many days will be required to finish the work ? <o> a ) num__6 days <o> b ) num__7.2 days <o> c ) num__9.5 days <o> d ) num__12 days <o> e ) num__13 days |
explanation : hint : a ' s one day work = num__0.0833333333333 b ' s one day work = num__0.0555555555556 ( a + b ) ' s one day work = num__0.0833333333333 + num__0.0555555555556 = ( num__18 + num__12 ) / ( num__12 x num__18 ) = num__0.138888888889 = num__1 / num__7.2 together a & b will finish the work in num__7.2 days . answer is b <eor> b <eos> |
b |
add__0.0556__0.0833__ round__7.2__ |
add__0.0556__0.0833__ divide__7.2__1.0__ |
| calculate the sum of first num__39 natural numbers . <o> a ) num__780 <o> b ) num__891 <o> c ) num__812 <o> d ) num__847 <o> e ) num__890 |
solution we know that ( num__1 + num__2 + num__3 + . . . . . + num__39 ) = n ( n + num__1 ) / num__2 therefore ( num__1 + num__2 + num__3 + . . . . + num__39 ) = ( num__39 × num__20.0 ) = num__780 . answer a <eor> a <eos> |
a |
add__1.0__2.0__ multiply__39.0__20.0__ multiply__39.0__20.0__ |
add__1.0__2.0__ multiply__39.0__20.0__ divide__780.0__1.0__ |
| excluding stoppages the speed of a train is num__45 kmph and including stoppages it is num__42 kmph . of how many minutes does the train stop per hour ? <o> a ) num__16 <o> b ) num__4 <o> c ) num__15 <o> d ) num__18 <o> e ) num__12 |
t = num__0.0666666666667 * num__60 = num__4 answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ round__4.0__ |
hour_to_min_conversion__ round__4.0__ |
| a sum of money is to be distributed among a b c d in the proportion of num__6 : num__2 : num__4 : num__3 . if c gets $ num__500 more than d what is d ' s share ? <o> a ) $ num__2000 <o> b ) $ num__2500 <o> c ) $ num__3000 <o> d ) $ num__3600 <o> e ) $ num__4200 |
let the shares of a b c and d be num__6 x num__2 x num__4 x and num__3 x respectively . then num__4 x - num__3 x = num__500 x = $ num__500 a ' s share = num__6 x = num__6 * $ num__500 = $ num__3000 the answer is c . <eor> c <eos> |
c |
multiply__6.0__500.0__ multiply__6.0__500.0__ |
multiply__6.0__500.0__ multiply__6.0__500.0__ |
| if a lends rs . num__2000 to b at num__10.0 per annum and b lends the same sum to c at num__11.5 per annum then the gain of b in a period of num__3 years is ? <o> a ) num__90 <o> b ) num__122.5 <o> c ) num__132.5 <o> d ) num__114.5 <o> e ) num__212.5 |
( num__2000 * num__1.5 * num__3 ) / num__100 = > num__90 answer : a <eor> a <eos> |
a |
percent__90.0__100.0__ |
percent__90.0__100.0__ |
| kim purchased n items from a catalog for $ num__8 each . postage and handling charges consisted of $ num__6 for the first item and $ num__1 for each additional item . which of the following gives the total dollar amount for kim ’ s purchase including postage and handling in terms of n ? <o> a ) num__8 n + num__2 <o> b ) num__8 n + num__4 <o> c ) num__9 n + num__2 <o> d ) num__9 n + num__5 <o> e ) num__9 n + num__4 |
its c n items for $ num__8 each total price $ num__8 n postage and handling of $ num__3 for num__1 st item and $ num__1 for the rest total postage and handling = $ [ num__6 + ( n - num__1 ) ] = $ n + num__5 total cost num__8 n + n + num__5 = num__9 n + num__5 d <eor> d <eos> |
d |
subtract__8.0__3.0__ add__8.0__1.0__ add__8.0__1.0__ |
subtract__8.0__3.0__ add__8.0__1.0__ add__8.0__1.0__ |
| claire has a total of num__92 pets consisting of gerbils and hamsters only . one - quarter of the gerbils are male and one - third of the hamsters are male . if there are num__25 males altogether how many gerbils does claire have ? <o> a ) num__39 <o> b ) num__68 <o> c ) num__54 <o> d ) num__57 <o> e ) num__60 |
g + h = num__92 . . . num__1 ; g / num__4 + h / num__3 = num__25 . . . . num__2 or num__3 g + num__4 h = num__25 * num__12 = num__300 g = num__92 - h or num__3 ( num__92 - h ) + num__4 h = num__300 h = num__300 - num__276 = num__24 then g = num__92 - num__24 = num__68 b <eor> b <eos> |
b |
subtract__4.0__1.0__ subtract__3.0__1.0__ multiply__3.0__4.0__ multiply__25.0__12.0__ multiply__92.0__3.0__ subtract__25.0__1.0__ subtract__92.0__24.0__ subtract__92.0__24.0__ |
subtract__4.0__1.0__ subtract__3.0__1.0__ multiply__3.0__4.0__ multiply__25.0__12.0__ multiply__92.0__3.0__ subtract__25.0__1.0__ subtract__92.0__24.0__ subtract__92.0__24.0__ |
| the mass of num__1 cubic meter of a substance is num__400 kg under certain conditions . what is the volume in cubic centimeters of num__1 gram of this substance under these conditions ? ( num__1 kg = num__1000 grams and num__1 cubic meter = num__1 num__000000 cubic centimeters ) <o> a ) num__1.5 <o> b ) num__2.5 <o> c ) num__3.5 <o> d ) num__4.5 <o> e ) num__5.5 |
num__400 kg - num__1 cubic meter ; num__400000 g - num__1 cubic meter ; num__400000 g - num__1 num__000000 cubic centimeters ; num__1 g - num__1 num__000000 / num__400000 = num__2.5 = num__2.5 cubic centimeters . answer : b . <eor> b <eos> |
b |
multiply__400.0__1000.0__ multiply__1.0__2.5__ |
multiply__400.0__1000.0__ multiply__1.0__2.5__ |
| a cistern is normally filled in num__8 hours but takes two hours longer to fill because of a leak in its bottom . if the cistern is full the leak will empty it in ? <o> a ) num__87 <o> b ) num__09 <o> c ) num__40 <o> d ) num__42 <o> e ) num__76 |
num__0.125 - num__1 / x = num__0.1 x = num__40 answer : c <eor> c <eos> |
c |
multiply__8.0__0.125__ round__40.0__ |
multiply__8.0__0.125__ divide__40.0__1.0__ |
| two trains num__500 m and num__750 m long run at the speed of num__60 km / hr and num__40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? <o> a ) num__10.6 <o> b ) num__45 <o> c ) num__10.4 <o> d ) num__10.8 <o> e ) num__10.1 |
relative speed = num__60 + num__40 = num__100 km / hr . = num__100 * num__0.277777777778 = num__27.7777777778 m / sec . distance covered in crossing each other = num__500 + num__750 = num__1250 m . required time = num__1250 * num__0.036 = num__45 sec . answer : b <eor> b <eos> |
b |
add__60.0__40.0__ add__500.0__750.0__ multiply__1250.0__0.036__ round__45.0__ |
add__60.0__40.0__ add__500.0__750.0__ multiply__1250.0__0.036__ multiply__1250.0__0.036__ |
| machine m n o working simultaneously machine m can produce x units in num__0.75 of the time it takes machine n to produce the same amount of units . machine n can produce x units in num__0.4 the time it takes machine o to produce that amount of units . if all three machines are working simultaneously what fraction of the total output is produced by machine n ? <o> a ) num__0.5 <o> b ) num__0.454545454545 <o> c ) num__0.307692307692 <o> d ) num__0.275862068966 <o> e ) num__0.181818181818 |
now ultimately the speed of every machine is given with respect to mach o . so lets assume the speed of o say num__12 hrs to make x units ( assuming num__6 because we can see we will need to divide by num__3 and num__4 mach o makes x units in num__12 hrs so mach n = num__0.4 of o = num__0.4 * num__12 = num__4.8 hrs to make x units and mach m = num__0.75 of n = num__0.75 * num__4.8 = num__6 hrs to make x units no they are running simultaneously . lets see how much each mach makes in num__1 hr mach o = x / num__12 units mach n = num__5 x / num__24 units mach m = x / num__6 units in num__1 hr together they make - x / num__12 + num__5 x / num__24 + x / num__6 = num__0.458333333333 so what ratio of this has mach n made ? ( num__5 x / num__24 ) / ( num__0.458333333333 ) = num__0.454545454545 ans : b = num__0.454545454545 <eor> b <eos> |
b |
divide__3.0__0.75__ multiply__0.4__12.0__ subtract__4.0__3.0__ add__1.0__4.0__ multiply__4.0__6.0__ multiply__1.0__0.4545__ |
divide__3.0__0.75__ multiply__0.4__12.0__ subtract__4.0__3.0__ add__1.0__4.0__ multiply__4.0__6.0__ multiply__1.0__0.4545__ |
| ramu bought an old car for rs . num__42000 . he spent rs . num__13000 on repairs and sold it for rs . num__60900 . what is his profit percent ? <o> a ) num__10.7 <o> b ) num__19.0 <o> c ) num__18.0 <o> d ) num__14.0 <o> e ) num__16 % |
total cp = rs . num__42000 + rs . num__13000 = rs . num__55000 and sp = rs . num__60900 profit ( % ) = ( num__60900 - num__55000 ) / num__55000 * num__100 = num__10.7 answer : a <eor> a <eos> |
a |
percent__10.7__100.0__ |
percent__10.7__100.0__ |
| the length of the bridge which a train num__110 metres long and travelling at num__45 km / hr can cross in num__30 seconds is ? <o> a ) num__255 <o> b ) num__265 <o> c ) num__245 <o> d ) num__277 <o> e ) num__211 |
speed = [ num__45 x num__0.277777777778 ] m / sec = [ num__12.5 ] m / sec time = num__30 sec let the length of bridge be x metres . then ( num__110 + x ) / num__30 = num__12.5 = > num__2 ( num__110 + x ) = num__750 = > x = num__265 m . answer : b <eor> b <eos> |
b |
round__265.0__ |
round__265.0__ |
| a shopkeeper loses num__15.0 if an article is sold for rs . num__102 . what should be the selling price of the article to gain num__20.0 ? <o> a ) rs . num__148 <o> b ) rs . num__144 <o> c ) rs . num__120 <o> d ) rs . num__129 <o> e ) rs . num__126 |
given that sp = rs . num__102 and loss = num__15.0 cp = [ num__100 ( sp ) ] / ( num__100 - l % ) = ( num__100 * num__102 ) / num__85 = num__20 * num__6 = rs . num__120 . to get num__20.0 profit new sp = [ ( num__100 + p % ) cp ] / num__100 = ( num__120 * num__120 ) / num__100 = rs . num__144 answer : b <eor> b <eos> |
b |
percent__100.0__144.0__ |
percent__100.0__144.0__ |
| given f ( x ) = num__3 x – num__5 for what value of x does num__2 * [ f ( x ) ] – num__19 = f ( x – num__4 ) ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
num__2 ( num__3 x - num__5 ) - num__19 = num__3 ( x - num__4 ) - num__5 num__3 x = num__12 x = num__4 the answer is b . <eor> b <eos> |
b |
multiply__3.0__4.0__ divide__12.0__3.0__ |
multiply__3.0__4.0__ divide__12.0__3.0__ |
| num__3 person can fill tank in num__25 min a can fill in num__30 min b can fill in num__35 min and c can empty the num__5 gallon per min then what is the capacity of tank ? <o> a ) num__208.26 gallons <o> b ) num__218.26 gallons <o> c ) num__228.26 gallons <o> d ) num__238.26 gallons <o> e ) num__248.26 gallons |
a can fill the tank in num__30 minutes b can fill the same tank in num__35 minutes a b c together fill the tank in num__25 minutes equating we get num__45.652173913 * num__5 = num__228.26 gallons answer : c <eor> c <eos> |
c |
round__228.26__ |
round__228.26__ |
| the edge of three cubes of metal is num__3 dm num__4 dm and num__5 dm . they are melted and formed into a single cube . find the edge of the new cube ? <o> a ) num__1 <o> b ) num__9 <o> c ) num__4 <o> d ) num__6 <o> e ) num__3 |
num__33 + num__43 + num__53 = a num__3 = > a = num__6 answer : d <eor> d <eos> |
d |
triangle_area__3.0__4.0__ triangle_area__3.0__4.0__ |
triangle_area__3.0__4.0__ triangle_area__3.0__4.0__ |
| a man gains num__30.0 by selling an article for a certain price . if he sells it at double the price the percentage of profit will be . <o> a ) num__130.0 <o> b ) num__140.0 <o> c ) num__150.0 <o> d ) num__260.0 <o> e ) num__170 % |
explanation : let the c . p . = x then s . p . = ( num__1.3 ) x = num__13 x / num__10 new s . p . = num__2 ( num__13 x / num__10 ) = num__26 x / num__10 profit = num__26 x / num__10 - x = num__26 x / num__10 profit % = ( profit / c . p . ) * num__100 = > ( num__26 x / num__10 ) * ( num__1 / x ) * num__100 = num__260.0 option d <eor> d <eos> |
d |
percent__100.0__260.0__ |
percent__100.0__260.0__ |
| a doctor prescribed num__20 cubic centimeters of a certain drug to a patient whose body weight was num__120 pounds . if the typical dosage is num__2 cubic centimeters per num__15 pounds of the body weight by what percent was the prescribed dosage greater than the typical dosage ? <o> a ) num__25.0 <o> b ) num__9.0 <o> c ) num__11.0 <o> d ) num__12.5 <o> e ) num__14.8 % |
typical dosage per num__15 pound of the body weight = num__2 c . c typical dosage per num__120 pound of the body weight = num__2 * ( num__8.0 ) = num__2 * num__8 = num__16 c . c dosage prescribed by doctor for num__120 pound patient = num__20 c . c % prescribed dosage greater than the typical dosage = ( num__20 - num__1.0 ) * num__100.0 = ( num__0.125 ) * num__100.0 = num__25.0 answer a <eor> a <eos> |
a |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| the price of lunch for num__15 people was $ num__208.00 including a num__15 percent gratuity for service . what was the average price per person excluding the gratuity ? <o> a ) $ num__11.73 <o> b ) $ num__12.60 <o> c ) $ num__13.80 <o> d ) $ num__14.00 <o> e ) $ num__15.87 |
take the initial price before the gratuity is num__100 the gratuity is calculated on the final price so as we assumed the final bill before adding gratuity is num__100 so gratuity is num__15.0 of num__100 is num__15 so the total price of meals is num__115 so the given amount i . e num__208 is for num__115 then we have to calculate for num__100 for num__115 num__208 for num__100 x so by cross multiplication we get num__115 x = num__100 * num__208 = > x = num__100 * num__1.89090909091 by simplifying we get x as num__189 which is the price of lunch before gratuity so the gratuity is num__19 so as the question ask the average price person excluding gratuity is num__12.6 = num__12.6 so our answer is b ) <eor> b <eos> |
b |
add__15.0__100.0__ subtract__208.0__189.0__ divide__189.0__15.0__ divide__189.0__15.0__ |
add__15.0__100.0__ subtract__208.0__189.0__ divide__189.0__15.0__ divide__189.0__15.0__ |
| an adventuring party consists of num__4 white mages num__6 warriors and num__4 archers . in how many ways can num__3 white mages num__2 warriors and num__2 archers be selected from the given group ? <o> a ) num__987 ways <o> b ) num__446 ways <o> c ) num__360 ways <o> d ) num__157 ways <o> e ) num__134 ways |
the number of ways of selecting three white mages two warriors and two archers is : = ⁴ c ₃ * ⁶ c ₂ * ⁴ c ₂ = ( num__4 * num__3 * num__2 ) / ( num__3 * num__2 * num__1 ) * ( num__6 * num__5 ) / ( num__2 * num__1 ) * ( num__4 * num__3 ) / ( num__2 * num__1 ) = num__4 * num__15 * num__6 = num__360 ways . answer : c <eor> c <eos> |
c |
vowel_space__ choose__5.0__2.0__ choose__5.0__2.0__ |
vowel_space__ choose__5.0__2.0__ choose__5.0__2.0__ |
| if x is equal to the sum of the integers from num__60 to num__80 inclusive and y is the number of even integers from num__60 to num__80 inclusive what is the value of x + y ? <o> a ) num__1361 <o> b ) num__1481 <o> c ) num__1601 <o> d ) num__1721 <o> e ) num__1841 |
x = num__60 + num__61 + . . . + num__80 = num__21 ( num__70 ) = num__1470 y = num__11 x + y = num__1481 the answer is b . <eor> b <eos> |
b |
multiply__70.0__21.0__ add__11.0__1470.0__ add__11.0__1470.0__ |
multiply__70.0__21.0__ add__11.0__1470.0__ add__11.0__1470.0__ |
| if num__2 a – b = num__3 c where a b and c are non - zero integers which of the following could be the average ( arithmetic mean ) of a and b if the average must itself be an integer ? <o> a ) - num__2 <o> b ) - num__1 <o> c ) num__1 <o> d ) num__10 <o> e ) num__12 |
let ' s say average of a & b is c a + b / num__2 = c a + b = num__2 c num__2 a - b = num__3 c a + b = num__2 c - - - - - - - - num__3 a = num__5 c a = num__5 c / num__3 we can plug in all answer choices for average - ` ` c ` ` and see which one gives us an integer as ` ` a ' ' is also an integer . num__5 ( - num__2 ) / num__3 = - num__3.33333333333 num__5 ( - num__1 ) / num__3 = - num__1.66666666667 num__5 ( num__1 ) / num__3 = num__1.66666666667 num__5 ( num__10 ) / num__3 = num__16.6666666667 num__5 ( num__12 ) / num__3 = num__20.0 = num__20 ( hence num__12 is the correct answer ) answer : e <eor> e <eos> |
e |
add__2.0__3.0__ subtract__3.0__2.0__ divide__5.0__3.0__ multiply__2.0__5.0__ add__2.0__10.0__ multiply__2.0__10.0__ add__2.0__10.0__ |
add__2.0__3.0__ subtract__3.0__2.0__ divide__5.0__3.0__ multiply__2.0__5.0__ add__2.0__10.0__ add__3.3333__16.6667__ add__2.0__10.0__ |
| a group of n students can be divided into equal groups of num__4 with num__2 student left over or equal groups of num__5 with num__2 students left over . what is the sum of the two smallest possible values of n ? <o> a ) num__33 <o> b ) num__46 <o> c ) num__49 <o> d ) num__64 <o> e ) num__86 |
num__4 x + num__2 = num__5 y + num__2 . . . . . . . . . . . ie : num__4 x - num__5 y = num__0 x y must be > num__1 and y is even ie ( num__2 num__46 . . etc ) if y = num__2 thus x is fraction if y = num__4 thus x = num__5 n = num__22 if y = num__6 thus x = not possible fraction if y = num__8 thus x = num__10 n = num__42 n = num__22 + num__42 = num__64 d <eor> d <eos> |
d |
subtract__5.0__4.0__ add__4.0__2.0__ multiply__4.0__2.0__ add__4.0__6.0__ subtract__46.0__4.0__ add__22.0__42.0__ multiply__1.0__64.0__ |
subtract__5.0__4.0__ add__4.0__2.0__ add__2.0__6.0__ add__4.0__6.0__ subtract__46.0__4.0__ add__22.0__42.0__ add__22.0__42.0__ |
| what is the number which when multiplied by num__13 is increased by num__180 ? <o> a ) num__13 <o> b ) num__15 <o> c ) num__23 <o> d ) num__35 <o> e ) num__45 |
num__13 × num__15 = num__195 answer : b <eor> b <eos> |
b |
multiply__13.0__15.0__ divide__195.0__13.0__ |
multiply__13.0__15.0__ divide__195.0__13.0__ |
| a man can do a piece of work in num__5 days but with the help of his son he can do it in num__3 days . in what time can the son do it alone ? <o> a ) num__712 days <o> b ) num__612 days <o> c ) num__512 days <o> d ) num__412 days <o> e ) none of these |
explanation : in this type of question where we have one person work and together work done . then we can easily get the other person work just by subtracting them . as son ' s one day work = ( num__13 − num__15 ) = ( num__5 − num__315 ) = num__215 so son will do whole work in num__7.5 days which is = num__712 days answer : a <eor> a <eos> |
a |
multiply__5.0__3.0__ round__712.0__ |
multiply__5.0__3.0__ round__712.0__ |
| present ages of sameer and anand are in the ratio of num__5 : num__4 respectively . eight years hence the ratio of their ages will become num__11 : num__9 respectively . what is anand ' s present age in years ? <o> a ) ca n ' t be determined <o> b ) num__40 <o> c ) num__27 <o> d ) num__64 <o> e ) none of these |
explanation : let the present ages of sameer and anand be num__5 x years and num__4 x years respectively . then ( num__5 x + num__8 ) / ( num__4 x + num__8 ) = num__1.22222222222 ⇒ num__45 x + num__72 = num__44 x + num__88 ⇒ num__9 ( num__5 x + num__8 ) = num__11 ( num__4 x + num__8 ) ⇒ num__45 x - num__44 x = num__88 - num__72 ⇒ x = num__16 . anand ' s present age = num__4 x = num__64 years . answer : d <eor> d <eos> |
d |
divide__11.0__9.0__ multiply__5.0__9.0__ multiply__9.0__8.0__ multiply__4.0__11.0__ multiply__11.0__8.0__ add__5.0__11.0__ multiply__4.0__16.0__ multiply__4.0__16.0__ |
divide__11.0__9.0__ multiply__5.0__9.0__ multiply__9.0__8.0__ multiply__4.0__11.0__ multiply__11.0__8.0__ add__5.0__11.0__ subtract__72.0__8.0__ subtract__72.0__8.0__ |
| an error num__2.0 in excess is made while measuring the side of a square . the percentage of error in the calculated area of the square is : <o> a ) num__4.0 <o> b ) num__4.14 <o> c ) num__4.04 <o> d ) num__4.26 <o> e ) num__4.27 % |
num__100 cm is read as num__102 cm . a num__1 = ( num__100 x num__100 ) cm num__2 and a num__2 ( num__102 x num__102 ) cm num__2 . ( a num__2 - a num__1 ) = [ ( num__102 ) num__2 - ( num__100 ) num__2 ] = ( num__102 + num__100 ) x ( num__102 - num__100 ) = num__404 cm num__2 . percentage error = num__404 x num__100.0 = num__4.04 num__100 x num__100 c <eor> c <eos> |
c |
percent__1.0__404.0__ percent__1.0__404.0__ |
percent__1.0__404.0__ percent__1.0__404.0__ |
| a man can row upstream at num__25 kmph and downstream at num__35 kmph and then find the speed of the man in still water ? <o> a ) num__18 kmph <o> b ) num__20 kmph <o> c ) num__30 kmph <o> d ) num__19 kmph <o> e ) num__10 kmph |
us = num__25 ds = num__35 m = ( num__35 + num__25 ) / num__2 = num__30 answer : c <eor> c <eos> |
c |
round__30.0__ |
round__30.0__ |
| the original price of a suit is $ num__200 . the price increased num__30.0 and after this increase the store published a num__30.0 off coupon for a one - day sale . given that the consumers who used the coupon on sale day were getting num__30.0 off the increased price how much did these consumers pay for the suit ? <o> a ) $ num__182 <o> b ) $ num__191 <o> c ) $ num__200 <o> d ) $ num__209 <o> e ) $ num__219 |
given the foregoing discussion it may be obvious now the trap - mistake answer is ( c ) . even if you can ’ t remember the correct thing to do at the very least learn to spot the trap ! the multiplier for a num__30.0 increases is num__1 + num__0.30 = num__1.3 and the multiplier for a num__30.0 decrease is num__1 – num__0.30 = num__0.70 so the combined change is num__1.3 * num__0.7 = num__0.91 num__91.0 percent of the original or a num__9.0 decreases . now multiply $ num__200 * num__0.91 = $ num__182 . answer = ( a ) . <eor> a <eos> |
a |
add__1.0__0.3__ subtract__1.0__0.3__ multiply__1.3__0.7__ multiply__30.0__0.3__ multiply__200.0__0.91__ multiply__200.0__0.91__ |
add__1.0__0.3__ subtract__1.0__0.3__ multiply__1.3__0.7__ multiply__30.0__0.3__ multiply__200.0__0.91__ multiply__200.0__0.91__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__240 m and they cross each other in num__12 sec then the speed of each train is ? <o> a ) num__72 <o> b ) num__89 <o> c ) num__36 <o> d ) num__34 <o> e ) num__23 |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__240 + num__240 ) / num__12 = > x = num__20 speed of each train = num__20 m / sec . = num__20 * num__3.6 = num__72 km / hr . answer : option a <eor> a <eos> |
a |
divide__240.0__12.0__ multiply__3.6__20.0__ round__72.0__ |
divide__240.0__12.0__ multiply__3.6__20.0__ multiply__3.6__20.0__ |
| the average weight of num__10 men is increased by num__1 ½ kg when one of the men who weighs num__48 kg is replaced by a new man . what is the weight of the new man ? <o> a ) num__80 kg <o> b ) num__63 kg <o> c ) num__70 kg <o> d ) num__75 kg <o> e ) num__85 kg |
since the average has increased by num__1.5 kg the weight of the man who stepped in must be equal to num__48 + num__10 x num__1.5 num__48 + num__15 = num__63 kg ans : ' b ' <eor> b <eos> |
b |
multiply__10.0__1.5__ add__48.0__15.0__ multiply__1.0__63.0__ |
multiply__10.0__1.5__ add__48.0__15.0__ add__48.0__15.0__ |
| if an article is sold at num__19.0 profit instead of num__12.0 profit then the profit would be rs . num__105 more . what is the cost price ? <o> a ) rs . num__1505 <o> b ) rs . num__1510 <o> c ) rs . num__1500 <o> d ) rs . num__1490 <o> e ) none of these |
let the cost price of an article be rs . x . ( num__19.0 of x ) - ( num__12.0 of x ) = num__105 num__19 x / num__100 - num__12 x / num__100 = num__105 = > num__7 x = num__105 * num__100 = > x = num__1500 cost price = rs . num__1500 answer : c <eor> c <eos> |
c |
percent__100.0__1500.0__ |
percent__100.0__1500.0__ |
| . num__005 / ? = . num__01 <o> a ) . num__5 <o> b ) . num__05 <o> c ) . num__005 <o> d ) . num__0005 <o> e ) none of them |
let . num__005 / x = . num__01 ; then x = . num__005 / . num__01 = . num__5.0 = . num__5 answer is a <eor> a <eos> |
a |
multiply__5.0__1.0__ |
divide__5.0__1.0__ |
| two trains are running at num__40 kmph and num__20 kmph respectively in the same direction . fast train completely passes a man sitting in the slower train in num__9 seconds . what is the length of the fast train ? <o> a ) num__23 m <o> b ) num__23 num__0.222222222222 m <o> c ) num__27 m <o> d ) num__46 m <o> e ) num__50 m |
relative speed = num__20 kmph = num__5.55555555556 m / sec length of the train = num__5.55555555556 * num__9 = num__50 m answer : e <eor> e <eos> |
e |
round__50.0__ |
round__50.0__ |
| the average weight of num__4 persons increases by num__1.5 kg . if a person weighing num__65 kg is replaced by a new person what could be the weight of the new person ? <o> a ) num__71 kg <o> b ) num__77 kg <o> c ) num__76.5 kg <o> d ) data inadequate <o> e ) none of these |
total weight increases = num__4 × num__1.5 = num__6 kg so the weight of new person = num__65 + num__6 = num__71 kg answer a <eor> a <eos> |
a |
multiply__4.0__1.5__ add__65.0__6.0__ add__65.0__6.0__ |
multiply__4.0__1.5__ add__65.0__6.0__ add__65.0__6.0__ |
| the numerator of a certain fraction is num__8 less than the denominator . if num__3 is added to the numerator and num__3 is subtracted from the denominator the fraction becomes num__0.75 . find the original fraction ? <o> a ) num__0.272727272727 <o> b ) num__0.363636363636 <o> c ) num__0.454545454545 <o> d ) num__0.461538461538 <o> e ) num__0.538461538462 |
explanation : the denominator be p the numerator will be ( p - num__8 ) . the fraction will be ( p - num__8 ) / p . adding num__3 to the numerator and subtracting num__3 from the denominator ( p - num__8 + num__3 ) / ( p - num__3 ) = num__0.75 . ( p - num__5 ) / ( p - num__3 ) = num__0.75 p = num__20 - num__9 = > p = num__11 . the fraction is : num__0.272727272727 . a ) <eor> a <eos> |
a |
subtract__8.0__3.0__ add__8.0__3.0__ divide__3.0__11.0__ divide__3.0__11.0__ |
subtract__8.0__3.0__ add__8.0__3.0__ divide__3.0__11.0__ divide__3.0__11.0__ |
| which of the following leads to the correct mathematical solution for the number of ways that the letters of the word radar could be arranged to create a five - letter code ? <o> a ) num__5 ! / ( num__2 ! x num__2 ! ) <o> b ) num__6 ! − ( num__3 ! + num__2 ! ) <o> c ) num__6 ! − ( num__3 ! × num__2 ! ) <o> d ) num__6 ! / ( num__3 ! + num__2 ! ) <o> e ) num__6 ! / ( num__3 ! × num__2 ! ) |
number of letters in word ' radar ' = num__5 . the letters ' a ' and ' r ' appear num__2 times and num__2 times respectively in the word ' radar ' . therefore the mathematical solution for number of ways that the letters of the word radar can be arranged to create five - letter code = num__5 ! / ( num__2 ! * num__2 ! ) answer : a <eor> a <eos> |
a |
vowel_space__ coin_space__ vowel_space__ |
vowel_space__ coin_space__ vowel_space__ |
| the captain of a cricket team of num__11 members is num__26 years old and the wicket keeper is num__2 years older . if the ages of these two are excluded the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? <o> a ) num__23 years <o> b ) num__22 years <o> c ) num__25 years <o> d ) num__26 years <o> e ) num__27 years |
explanation let the average age of the whole team by x years . num__11 x â € “ ( num__26 + num__28 ) = num__9 ( x - num__1 ) num__11 x â € “ num__9 x = num__45 num__2 x = num__45 x = num__22 . so average age of the team is num__22 years . answer b <eor> b <eos> |
b |
add__26.0__2.0__ subtract__11.0__2.0__ multiply__11.0__2.0__ multiply__11.0__2.0__ |
add__26.0__2.0__ subtract__11.0__2.0__ multiply__11.0__2.0__ multiply__11.0__2.0__ |
| in covering a distance of num__30 km a takes num__2 hours more than b . if a doubles his speed then he would take num__1 hour less than b . a ' s speed is : <o> a ) num__5 km / hr . <o> b ) num__3 km / hr . <o> c ) num__1 km / hr . <o> d ) num__7 km / hr . <o> e ) num__9 km / hr . |
c num__5 km / hr . let a ' s speed be x km / hr . then num__30 / x - num__15.0 x = num__3 num__6 x = num__30 x = num__5 km / hr . <eor> a <eos> |
a |
divide__30.0__2.0__ add__2.0__1.0__ divide__30.0__5.0__ round__5.0__ |
divide__30.0__2.0__ subtract__5.0__2.0__ divide__30.0__5.0__ divide__30.0__6.0__ |
| one half of a two digit number exceeds its one third by num__8 . what is the sum of the digits of the number ? <o> a ) a ) num__3 <o> b ) b ) num__5 <o> c ) c ) num__12 <o> d ) d ) num__9 <o> e ) e ) num__11 |
x / num__2 – x / num__3 = num__8 = > x = num__48 num__4 + num__8 = num__12 answer : c <eor> c <eos> |
c |
divide__8.0__2.0__ add__8.0__4.0__ add__8.0__4.0__ |
divide__8.0__2.0__ add__8.0__4.0__ add__8.0__4.0__ |
| patrick purchased num__90 pencils and sold them at a loss equal to the selling price of num__40 pencils . the cost of num__90 pencils is how many times the selling price of num__90 pencils ? <o> a ) num__0.75 <o> b ) num__0.8 <o> c ) num__1 <o> d ) num__1.2 <o> e ) num__1.44 |
say the cost price of num__90 pencils was $ num__90 ( $ num__1 per pencil ) and the selling price of num__1 pencil was p . selling at a loss : num__90 - num__90 p = num__40 p - - > p = num__0.692307692308 . ( cost price ) / ( selling price ) = num__1 / ( num__0.692307692308 ) = num__1.44444444444 = num__1.44 . answer : e . <eor> e <eos> |
e |
multiply__1.0__1.44__ |
divide__1.44__1.0__ |
| mandy drove from point a to point b at num__60 km / h . on her way back she drove at num__80 km / h and therefore her way back lasted one hour less . what is the distance ( in km ) between a and b ? <o> a ) num__140 . <o> b ) num__200 . <o> c ) num__240 . <o> d ) num__300 . <o> e ) num__600 . |
distance = speed * time d num__1 = s num__1 t num__1 d num__2 = s num__2 t num__2 the distance from point a to point b is the same for each trip so d num__1 = d num__2 and t num__2 = t num__1 - num__1 thus s num__1 t num__1 = s num__2 ( t num__1 - num__1 ) num__60 t num__1 = num__80 ( t num__1 - num__1 ) t num__1 = num__4 num__60 * num__4 = num__240 answer : c <eor> c <eos> |
c |
multiply__60.0__4.0__ round__240.0__ |
multiply__60.0__4.0__ multiply__60.0__4.0__ |
| in a certain apartment building there are one - bedroom and two - bedroom apartments . the rental prices of the apartment depend on a number of factors but on average two - bedroom apartments have higher rental prices than do one - bedroom apartments . let m be the average rental price for all apartments in the building . if m is $ num__1000 higher than the average rental price for all one - bedroom apartments and if the average rental price for all two - bedroom apartments is $ num__9000 higher that m then what percentage of apartments in the building are two - bedroom apartments ? <o> a ) num__50.0 <o> b ) num__40.0 <o> c ) num__20.0 <o> d ) num__30.0 <o> e ) num__10 % |
ratio of num__2 bedroom apartment : num__1 bedroom apartment = num__1000 : num__9000 - - - - - > num__1 : num__9 let total number of apartments be x no . of num__2 bedroom apartment = ( num__0.1 ) * x percentage of apartments in the building are two - bedroom apartments - - - - > ( num__0.1 ) * num__100 - - - > num__10.0 answer : e <eor> e <eos> |
e |
divide__9000.0__1000.0__ multiply__1000.0__0.1__ reverse__0.1__ reverse__0.1__ |
divide__9000.0__1000.0__ multiply__1000.0__0.1__ multiply__100.0__0.1__ multiply__100.0__0.1__ |
| find the number of square tiles to cover the floor of a room measuring num__10 m * num__12 m leaving num__0.25 m space around the room . a side of square tile is given to be num__50 cms ? <o> a ) num__277 <o> b ) num__476 <o> c ) num__437 <o> d ) num__257 <o> e ) num__212 |
num__9 num__0.5 * num__11 num__0.5 = num__0.5 * num__0.5 * x = > x = num__437 answer : c <eor> c <eos> |
c |
round__437.0__ |
round__437.0__ |
| find the value for x from below equation : x / num__3 = - num__2 ? <o> a ) - num__6 <o> b ) - num__4 <o> c ) num__4 <o> d ) num__5 <o> e ) - num__5 |
num__1 . multiply both sides by num__3 : x * num__1.0 = - num__0.666666666667 num__2 . simplify both sides : x = - num__6 a <eor> a <eos> |
a |
subtract__3.0__2.0__ divide__2.0__3.0__ multiply__3.0__2.0__ multiply__3.0__2.0__ |
subtract__3.0__2.0__ divide__2.0__3.0__ multiply__3.0__2.0__ multiply__3.0__2.0__ |
| if m = num__9 ^ ( x − num__1 ) then in terms of m num__3 ^ ( num__4 x − num__5 ) must be which of the following ? <o> a ) m / num__3 <o> b ) num__9 m <o> c ) num__9 m ^ num__2 <o> d ) m ^ num__0.666666666667 <o> e ) m ^ num__0.222222222222 |
m = num__9 ^ ( x - num__1 ) m = num__3 ^ ( num__2 x - num__2 ) m ^ num__2 = num__3 ^ ( num__4 x - num__4 ) m ^ num__0.666666666667 = num__3 ^ ( num__4 x - num__5 ) answer d <eor> d <eos> |
d |
subtract__3.0__1.0__ divide__2.0__3.0__ multiply__1.0__0.6667__ |
subtract__3.0__1.0__ divide__2.0__3.0__ multiply__1.0__0.6667__ |
| in the first half of the num__20 th century the population of a particular country increased by num__200 percent . in the second half of the century the population increased by num__300 percent . what was the percent increase for the num__20 th century as a whole ? <o> a ) num__500.0 <o> b ) num__600.0 <o> c ) num__800.0 <o> d ) num__1100.0 <o> e ) num__1200 % |
say initially population was num__100 . what is num__200.0 of num__100 ? it is num__2.0 * num__100 = num__200 . an increase of num__200.0 means the new population became num__100 + num__200 = num__300 what is num__300.0 of num__300 ? it is num__3.0 * num__300 = num__900 an increase of num__300.0 means the new population now is num__300 + num__900 = num__1200 so from num__100 the population increased to num__1200 i . e . an increase of num__1100 . num__1100 is what percent of num__100 ? num__1100 = x / num__100 * num__100 i . e . it is num__1100.0 d <eor> d <eos> |
d |
percent__100.0__1100.0__ |
percent__100.0__1100.0__ |
| find the average of all the number between num__6 and num__34 which are divisible by num__5 . <o> a ) num__18 <o> b ) num__20 <o> c ) num__24 <o> d ) num__30 <o> e ) num__32 |
solution average = ( num__10 + num__15 + num__20 + num__25 + num__6.0 ) = num__50.0 = num__20 answer b <eor> b <eos> |
b |
add__5.0__10.0__ add__5.0__15.0__ add__5.0__20.0__ multiply__5.0__10.0__ add__5.0__15.0__ |
add__5.0__10.0__ add__5.0__15.0__ add__5.0__20.0__ multiply__5.0__10.0__ add__5.0__15.0__ |
| num__1 num__6 num__13 num__22 num__33 ? <o> a ) num__46 <o> b ) num__44 <o> c ) num__35 <o> d ) num__65 <o> e ) num__31 |
a num__46 the pattern is + num__5 + num__7 + num__9 + num__11 . . . . <eor> a <eos> |
a |
add__13.0__33.0__ subtract__6.0__1.0__ add__1.0__6.0__ subtract__22.0__13.0__ add__6.0__5.0__ multiply__1.0__46.0__ |
add__13.0__33.0__ subtract__6.0__1.0__ add__1.0__6.0__ subtract__22.0__13.0__ add__6.0__5.0__ add__13.0__33.0__ |
| if scott has earned num__15 dollars by working num__4 days a week at a constant daily rate for num__1 weeks which of the following represents his daily wage ? <o> a ) num__0.266666666667 <o> b ) num__0.0166666666667 <o> c ) num__15.0 <o> d ) num__5.0 <o> e ) num__3.75 |
scott total earning = num__15 no days he worked = num__4 rate = num__1 week daily wage = total / ( rate * days ) = num__15 / ( num__4 * num__1 ) e is correct answer . . . <eor> e <eos> |
e |
divide__15.0__4.0__ |
divide__15.0__4.0__ |
| find the area of the quadrilateral of one of its diagonals is num__24 cm and its off sets num__9 cm and num__6 cm ? <o> a ) num__180 cm num__2 <o> b ) num__150 cm num__2 <o> c ) num__168 cm num__2 <o> d ) num__198 cm num__2 <o> e ) num__987 cm num__2 |
num__0.5 * num__24 ( num__9 + num__6 ) = num__180 cm num__2 answer : a <eor> a <eos> |
a |
square_perimeter__0.5__ triangle_area__2.0__180.0__ |
square_perimeter__0.5__ volume_rectangular_prism__0.5__2.0__180.0__ |
| on a wedding catering service an experienced chef can prepare a service for a wedding in num__4 hours while an novice chef would finish the preparations in num__16 hours . if the catering service employs the same number of novice and experienced chefs then how many chefs would it take to prepare a wedding service in num__1 hour and num__36 minutes ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
experienced chefs work = num__1 wedding / num__4 hours novice chefs work = num__1 wedding / num__16 hours since we do n ' t know the number of experienced or novice chefs but know that there is an equal number each let the number of chefs for each group equal x num__1 hr and num__36 mins = num__1.6 an hour x / num__4 + x / num__16 = num__1 wedding / ( num__1.6 ) x / num__4 + x / num__16 = num__0.625 x = num__2 so there are num__2 novice chefs and num__2 experienced chefs . in total there are num__4 . the answer is c . <eor> c <eos> |
c |
mile_to_km_conversion__ divide__1.0__1.6__ round__4.0__ |
mile_to_km_conversion__ divide__1.0__1.6__ divide__4.0__1.0__ |
| if num__8 parallel lines in a plane is intersected by a family of another num__8 parallel lines how many parallelograms are there in the network thus formed ? <o> a ) num__784 <o> b ) num__763 <o> c ) num__120 <o> d ) num__160 <o> e ) num__1260 |
parallelogram can formed by num__2 horizontal and num__2 vertical lines for horizontal num__8 c num__2 for vertical num__8 c num__2 total parallelogram is num__10 c num__2 * num__8 c num__2 = num__28 * num__28 = num__784 answer : a <eor> a <eos> |
a |
add__8.0__2.0__ round__784.0__ |
add__8.0__2.0__ round__784.0__ |
| a coin is tossed three times . what is the probability that there is at the least one tail ? <o> a ) num__0.875 <o> b ) num__0.837837837838 <o> c ) num__1.72222222222 <o> d ) num__3.1 <o> e ) num__1.63157894737 |
let p ( t ) be the probability of getting least one tail when the coin is tossed three times . = there is not even a single tail . i . e . all the outcomes are heads . = num__0.125 ; p ( t ) = num__1 - num__0.125 = num__0.875 answer : a <eor> a <eos> |
a |
negate_prob__0.125__ negate_prob__0.125__ |
negate_prob__0.125__ negate_prob__0.125__ |
| if x + y = num__2 x + num__2 z x - num__2 y = num__2 z and x + y + z = num__21 what is the value of y / z ? <o> a ) - num__4 . <o> b ) - num__2 . <o> c ) - num__1.7 . <o> d ) num__3 . <o> e ) num__2.5 . |
x + y = num__2 x + num__2 z y = x + num__2 z - - - - - - - - - - num__1 x - num__2 y = num__2 z x - num__2 z = num__2 y - - - - - - - - - num__2 subtracting equation num__1 from equation num__2 - num__4 z = y y / z = - num__4 a is the answer <eor> a <eos> |
a |
multiply__1.0__4.0__ |
divide__4.0__1.0__ |
| three photographers lisa mike and norm take photos of a wedding . the total of lisa and mikes photos is num__60 less than the sum of mike ' s and norms . if norms photos number num__10 more than twice the number of lisa ' s photos then how many photos did norm take ? <o> a ) num__40 <o> b ) num__50 <o> c ) num__60 <o> d ) num__110 <o> e ) num__80 |
l + m = m + n - num__60 / n = num__2 l + num__10 num__60 = m + n - l - m num__60 = n - l num__60 = num__2 l + num__10 - l num__50 = l num__2 ( num__50 ) + num__10 = num__110 d <eor> d <eos> |
d |
subtract__60.0__10.0__ add__60.0__50.0__ add__60.0__50.0__ |
subtract__60.0__10.0__ add__60.0__50.0__ add__60.0__50.0__ |
| in a camp there is a meal for num__120 men or num__200 children . if num__110 children have taken the meal how many men will be catered to with the remaining meal ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__40 <o> d ) num__50 <o> e ) num__54 |
solution there is a metal for num__200 children . num__110 children have taken the meal . remaining meal is to be catered to num__90 children . now num__200 children = num__120 men num__90 children = ( num__0.6 x num__90 ) men = num__54 men . answer e <eor> e <eos> |
e |
subtract__200.0__110.0__ divide__120.0__200.0__ multiply__0.6__90.0__ multiply__0.6__90.0__ |
subtract__200.0__110.0__ divide__120.0__200.0__ multiply__0.6__90.0__ multiply__0.6__90.0__ |
| a man invests in a num__16.0 stock at num__128 . the interest obtained by him is <o> a ) num__22.5 <o> b ) num__42.5 <o> c ) num__12.5 <o> d ) num__62.5 <o> e ) num__82.5 % |
by investing rs num__128 income derived = rs . num__16 by investing rs . num__100 income derived = = rs . num__12.5 interest obtained = num__12.5 answer : c <eor> c <eos> |
c |
percent__100.0__12.5__ |
percent__100.0__12.5__ |
| the product of two natural numbers is num__17 . then the sum of the reciprocals of their squares is : <o> a ) num__0.00346020761246 <o> b ) num__0.996551724138 <o> c ) num__1.00346020761 <o> d ) num__289 <o> e ) none of these |
let the numbers be a and b . then ab = num__17 a = num__1 and b = num__17 so num__1 / a num__2 + num__1 / b num__2 = ( a num__2 + b num__2 ) / a num__2 b num__2 = ( num__12 + num__172 ) / ( num__1 * num__17 ) num__2 = num__1.00346020761 answer : c <eor> c <eos> |
c |
multiply__1.0__1.0035__ |
multiply__1.0__1.0035__ |
| paul ' s income is num__40.0 less than rex ' s income quentin ' s income is num__20.0 less than paul ' s income and sam ' s income is num__40.0 less than paul ' s income . if rex gave num__60.0 of his income to sam and num__40.0 of his income to quentin quentin ' s new income would be what fraction of sam ' s new income ? <o> a ) num__0.916666666667 <o> b ) num__0.764705882353 <o> c ) num__0.684210526316 <o> d ) num__0.631578947368 <o> e ) num__0.578947368421 |
let rex ' s income = num__100 then : rex = num__100 paul = num__60 quentin = num__48 sam = num__36 rex gives num__60 dollars to sam so now sam has num__96 dollars rex gives num__40 dollars to quentin so now quentin has num__88 dollars the fraction q / s = num__0.916666666667 answer : a <eor> a <eos> |
a |
add__40.0__60.0__ add__60.0__36.0__ add__40.0__48.0__ divide__88.0__96.0__ divide__88.0__96.0__ |
add__40.0__60.0__ add__60.0__36.0__ add__40.0__48.0__ divide__88.0__96.0__ divide__88.0__96.0__ |
| ayesha ' s father was num__38 years of age when she was born while her mother was num__36 years old when her brother four years younger to her was born . what is the difference between the ages of her parents ? <o> a ) num__2 years <o> b ) num__4 years <o> c ) num__6 years <o> d ) num__8 years <o> e ) num__10 years |
mother ' s age when ayesha ' s brother was born = num__36 years . father ' s age when ayesha ' s brother was born = ( num__38 + num__4 ) = num__42 years . required difference = ( num__42 - num__36 ) = num__6 years . answer : c <eor> c <eos> |
c |
add__38.0__4.0__ subtract__42.0__36.0__ divide__36.0__6.0__ |
add__38.0__4.0__ subtract__42.0__36.0__ subtract__42.0__36.0__ |
| a num__300 m long train crosses a platform in num__39 sec while it crosses a signal pole in num__36 sec . what is the length of the platform ? <o> a ) num__389 m <o> b ) num__350 m <o> c ) num__289 m <o> d ) num__299 m <o> e ) num__25 m |
speed = num__8.33333333333 = num__8.33333333333 m / sec . let the length of the platform be x meters . then ( x + num__300 ) / num__39 = num__8.33333333333 = > x = num__25 m . answer : e <eor> e <eos> |
e |
divide__300.0__36.0__ round__25.0__ |
divide__300.0__36.0__ round__25.0__ |
| the numbers in which of the following pairs do not have a pair of distinct prime divisors in common ? <o> a ) num__10 and num__20 <o> b ) num__12 and num__18 <o> c ) num__20 and num__16 <o> d ) num__21 and num__63 <o> e ) num__22 and num__88 |
num__16 has only the prime num__2 ( num__16 = num__2 ^ num__4 ) in its prime factorization so it can not have a pair of distinct prime divisors in common with any of the numbers . the answer is c . <eor> c <eos> |
c |
add__16.0__4.0__ |
add__16.0__4.0__ |
| find the value of num__15 + num__2 â € ¢ ( num__8 â € “ num__3 ) <o> a ) num__25 <o> b ) num__13 <o> c ) num__17 <o> d ) num__24 <o> e ) num__15 |
num__15 + num__2 â € ¢ ( num__8 â € “ num__3 ) = num__15 + num__2 ( num__5 ) = num__15 + num__2 * num__5 = num__15 + num__10 = num__25 correct answer a <eor> a <eos> |
a |
divide__15.0__3.0__ subtract__15.0__5.0__ add__15.0__10.0__ add__15.0__10.0__ |
add__2.0__3.0__ add__2.0__8.0__ add__15.0__10.0__ add__15.0__10.0__ |
| num__5 blue marbles num__3 red marbles and num__4 purple marbles are placed in a bag . if num__4 marbles are drawn without replacement what is the probability that the result will not be num__2 blue and num__2 purple marbles ? <o> a ) num__0.121212121212 <o> b ) ( num__0.138888888889 ) ^ num__2 <o> c ) num__0.5 <o> d ) ( num__0.861111111111 ) ^ num__2 <o> e ) num__0.878787878788 |
answer is num__0.878787878788 . the probability of num__2 blue and num__2 purple marbles selected is num__5 c num__2.4 c num__0.166666666667 c num__4 = num__0.121212121212 . subtracting the above from num__1 we get num__0.878787878788 e <eor> e <eos> |
e |
negate_prob__0.8788__ negate_prob__0.1212__ |
negate_prob__0.8788__ negate_prob__0.1212__ |
| a train passes a station platform in num__36 sec and a man standing on the platform in num__20 sec . if the speed of the train is num__108 km / hr . what is the length of the platform ? <o> a ) num__229 <o> b ) num__240 <o> c ) num__288 <o> d ) num__480 <o> e ) num__221 |
speed = num__108 * num__0.277777777778 = num__30 m / sec . length of the train = num__30 * num__20 = num__600 m . let the length of the platform be x m . then ( x + num__600 ) / num__36 = num__30 = > x = num__480 m . answer : d <eor> d <eos> |
d |
multiply__20.0__30.0__ round__480.0__ |
multiply__20.0__30.0__ round__480.0__ |
| a man took some money for borrowed for num__3 years the total will be rs . num__5000 and num__5 years it will be rs . num__6000 / - . then how much amount he borrowed ? <o> a ) s . num__2000 / - <o> b ) s . num__2500 / - <o> c ) s . num__2770 / - <o> d ) s . num__2800 / - <o> e ) s . num__3500 / - |
num__3 years - - - - - - - - > rs . num__5000 / - num__5 years - - - - - - - - > rs . num__6000 / - ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - num__2 years - - - - - - - - - > rs . num__1000 / - num__1 year - - - - - - - - - - > rs . num__500 / - num__3 years * rs . num__500 / - = rs . num__1500 / - p = rs . num__5000 / - - rs . num__1500 / - = rs . num__3500 / - p = rs . num__3500 / - e <eor> e <eos> |
e |
subtract__5.0__3.0__ divide__5000.0__5.0__ subtract__3.0__2.0__ divide__1000.0__2.0__ multiply__3.0__500.0__ subtract__5000.0__1500.0__ subtract__5000.0__1500.0__ |
subtract__5.0__3.0__ divide__5000.0__5.0__ subtract__3.0__2.0__ divide__1000.0__2.0__ multiply__3.0__500.0__ subtract__5000.0__1500.0__ subtract__5000.0__1500.0__ |
| in june a baseball team that played num__60 games had won num__30.0 of its game played . after a phenomenal winning streak this team raised its average to num__50.0 . how many games must the team have won in a row to attain this average ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__24 <o> d ) num__12 <o> e ) num__26 |
explanation : number of matches won by the team in june : num__30.0 of num__60 = num__18 after a phenomenal winning streak the average is num__50.0 let the team played x matches continuously and won all of them to achieve an average winning percentage of num__50.0 . so total matches played by the team = ( num__60 + x ) total matches won by the team = ( num__18 + x ) total matches won = num__50.0 of total matches played = > ( num__18 + x ) = num__50.0 of ( num__60 + x ) = > num__18 + x = ( num__0.5 ) * ( num__60 + x ) = > num__18 + x = ( num__0.5 ) * ( num__60 + x ) = > num__36 + num__2 x = num__60 + x = > x = num__24 answer : c <eor> c <eos> |
c |
divide__30.0__60.0__ divide__18.0__0.5__ reverse__0.5__ subtract__60.0__36.0__ subtract__60.0__36.0__ |
divide__30.0__60.0__ divide__18.0__0.5__ reverse__0.5__ subtract__60.0__36.0__ subtract__60.0__36.0__ |
| a train is num__100 meter long and is running at the speed of num__30 km per hour . find the time it will take to pass a man standing at a crossing . <o> a ) num__10 seconds <o> b ) num__12 seconds <o> c ) num__14 seconds <o> d ) num__16 seconds <o> e ) none of these |
explanation : as we need to get answer in seconds so never forget to convert speed into meter per second . speed = num__30 km / hr = num__30 * num__0.277777777778 m / sec = num__8.33333333333 m / sec distance = length of train = num__100 meter required time = distancespeed = num__100253 = num__100 ∗ num__325 = num__12 sec answer : b <eor> b <eos> |
b |
divide__100.0__8.3333__ round__12.0__ |
divide__100.0__8.3333__ divide__100.0__8.3333__ |
| the difference between two numbers is num__2395 . when the larger number is divided by the smaller one the quotient is num__6 and the remainder is num__15 . the smaller number is <o> a ) num__120 <o> b ) num__239 <o> c ) num__476 <o> d ) num__523 <o> e ) num__526 |
let the smaller number be x . then larger number = ( num__2395 + x ) therefore num__2395 + x = ( num__6 x + num__15 ) ‹ = › num__5 x = num__2380 ‹ = › x = num__476 . answer : c <eor> c <eos> |
c |
subtract__2395.0__15.0__ divide__2380.0__5.0__ divide__2380.0__5.0__ |
subtract__2395.0__15.0__ divide__2380.0__5.0__ divide__2380.0__5.0__ |
| if num__8 ^ num__25 divided by num__7 then what is reminder <o> a ) num__25 <o> b ) num__1 <o> c ) num__0 <o> d ) num__2 <o> e ) num__3 |
num__8 ^ num__3.57142857143 num__8 can be written as num__7 * num__1 + num__1 ( num__7 * num__1 + num__1 ) ^ num__3.57142857143 ( ax + num__1 ) ^ n / a always gives the remainder num__1 here a = num__7 x = num__1 n = num__25 hence remainder is num__1 answer : b <eor> b <eos> |
b |
divide__25.0__7.0__ subtract__8.0__7.0__ reverse__1.0__ |
divide__25.0__7.0__ subtract__8.0__7.0__ reverse__1.0__ |
| a can do a work in num__15 days b in num__14 days and c in num__16 days . if they work on it together then in how many days required to complete the work ? <o> a ) num__16.0 <o> b ) num__160.0 <o> c ) num__4.98516320475 <o> d ) num__14.0 <o> e ) num__15.0 |
person ( a ) ( b ) ( c ) ( a + b + c ) time - ( num__15 ) ( num__14 ) ( num__16 ) rate - ( num__224 ) ( num__240 ) ( num__210 ) ( num__674 ) work - ( num__3360 ) ( num__3360 ) ( num__3360 ) ( num__3360 ) therefore a + b + c requires ( num__4.98516320475 ) days to complete entire work = num__4.98516320475 answer is c <eor> c <eos> |
c |
multiply__14.0__16.0__ multiply__15.0__16.0__ multiply__15.0__14.0__ multiply__15.0__224.0__ divide__3360.0__674.0__ divide__3360.0__674.0__ |
multiply__14.0__16.0__ add__16.0__224.0__ subtract__224.0__14.0__ multiply__15.0__224.0__ divide__3360.0__674.0__ divide__3360.0__674.0__ |
| num__8 litres are drawn from a cask full of wine and is then filled with water . this operation is performed three more times . the ratio of the quantity of wine now left in cask to that of the water is num__16 : num__65 . how much wine the cask hold originally ? <o> a ) num__18 litres <o> b ) num__24 litres of wine <o> c ) num__32 litres <o> d ) num__42 litres <o> e ) none |
solution let the quantity of the wine in the cask originally be x litres then quantity of wine left in cask after num__4 operations = [ x ( num__1 - num__8 / x ) num__4 ] litres . therefore x ( num__1 - num__8 / x ) num__4 / x = num__0.197530864198 = › ( num__1 - num__8 / x ) num__4 = ( num__0.666666666667 ) num__2 = › ( x - num__8 / x ) = num__0.666666666667 = › num__3 x - num__24 = num__2 x = › x = num__24 . answer b <eor> b <eos> |
b |
divide__8.0__4.0__ add__1.0__2.0__ add__8.0__16.0__ add__8.0__16.0__ |
divide__8.0__4.0__ subtract__4.0__1.0__ add__8.0__16.0__ divide__24.0__1.0__ |
| if num__148 is multiplied first by num__163 and then by num__236 which of the following numbers would come at the unit ' s place ? <o> a ) num__8 <o> b ) num__6 <o> c ) num__4 <o> d ) num__3 <o> e ) none of these |
explanation : num__148 * num__163 * num__236 = num__8 * num__3 * num__6 = num__4 at the unit ' s place . when the numbers are multiplied the unit digit of the final product is obtained as the product of the unit digits of individual numbers . answer : c <eor> c <eos> |
c |
subtract__8.0__4.0__ |
subtract__8.0__4.0__ |
| a pupil ' s marks were wrongly entered as num__85 instead of num__33 . due to that the average marks for the class got increased by half . the number of pupils in the class is : <o> a ) num__30 <o> b ) num__104 <o> c ) num__20 <o> d ) num__25 <o> e ) num__26 |
let there be x pupils in the class . total increase in marks = ( x * num__0.5 ) = x / num__2 . x / num__2 = ( num__85 - num__33 ) = > x / num__2 = num__52 = > x = num__104 . answer : b <eor> b <eos> |
b |
reverse__0.5__ subtract__85.0__33.0__ multiply__2.0__52.0__ multiply__2.0__52.0__ |
reverse__0.5__ subtract__85.0__33.0__ multiply__2.0__52.0__ multiply__2.0__52.0__ |
| amit and ananthu can do a work in num__10 days and num__20 days respectively . amit started the work and left after num__2 days . ananthu took over and completed the work . in how many days was the total work completed ? <o> a ) num__18 days <o> b ) num__20 days <o> c ) num__23 days <o> d ) num__25 days <o> e ) num__27 days |
amit ’ s one day ’ s work = num__0.1 amit ’ s num__2 day ’ s work = num__0.1 * num__2 = num__0.2 work left = num__1 - num__0.2 = num__0.8 ananthu ’ s one day ’ s work = num__0.05 ananthu can do work in = num__0.8 * num__20 = num__16 days so total days = num__16 + num__2 = num__18 days answer : a <eor> a <eos> |
a |
divide__2.0__20.0__ divide__2.0__10.0__ multiply__10.0__0.1__ subtract__1.0__0.2__ divide__1.0__20.0__ multiply__20.0__0.8__ subtract__20.0__2.0__ round__18.0__ |
divide__2.0__20.0__ multiply__2.0__0.1__ multiply__10.0__0.1__ subtract__1.0__0.2__ divide__1.0__20.0__ multiply__20.0__0.8__ subtract__20.0__2.0__ subtract__20.0__2.0__ |
| a and b go around a circular track of length num__600 m on a cycle at speeds of num__36 kmph and num__54 kmph . after how much time will they meet for the first time at the starting point ? <o> a ) num__120 sec <o> b ) num__198 sec <o> c ) num__178 sec <o> d ) num__665 sec <o> e ) num__276 sec |
time taken to meet for the first time at the starting point = lcm { length of the track / speed of a length of the track / speed of b } = lcm { num__600 / ( num__36 * num__0.277777777778 ) num__600 / ( num__54 * num__0.277777777778 ) } = lcm ( num__60 num__40 ) = num__120 sec . answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ round__120.0__ |
hour_to_min_conversion__ round__120.0__ |
| ramu bought an old car for rs . num__42000 . he spent rs . num__13000 on repairs and sold it for rs . num__64900 . what is his profit percent ? <o> a ) num__17.0 <o> b ) num__19.0 <o> c ) num__18.0 <o> d ) num__14.0 <o> e ) num__16 % |
total cp = rs . num__42000 + rs . num__13000 = rs . num__55000 and sp = rs . num__64900 profit ( % ) = ( num__64900 - num__55000 ) / num__55000 * num__100 = num__18.0 answer : c <eor> c <eos> |
c |
percent__18.0__100.0__ |
percent__18.0__100.0__ |
| hree unbiased coins are tossed . what is the probability of getting at least num__2 tails ? <o> a ) num__0.5 <o> b ) num__0.8 <o> c ) num__0.2 <o> d ) num__0.9 <o> e ) num__0.1 |
explanation : s = { hhh hht hth htt thh tht tth ttt } e = { htt tht tth ttt } n ( s ) = num__8 n ( e ) = num__4 p ( e ) = n ( e ) / n ( s ) = num__0.5 = num__0.5 answer : a <eor> a <eos> |
a |
negate_prob__0.5__ |
negate_prob__0.5__ |
| in a weight - lifting competition the total weight of joe ' s two lifts was num__1800 pounds . if twice the weight of his first lift was num__300 pounds more than the weight of his second lift what was the weight in pounds of his first lift ? <o> a ) num__225 <o> b ) num__275 <o> c ) num__325 <o> d ) num__350 <o> e ) num__700 |
this problem is a general word translation . we first define variables and then set up equations . we can define the following variables : f = the weight of the first lift s = the weight of the second lift we are given that the total weight of joe ' s two lifts was num__1800 pounds . we sum the two variables to obtain : f + s = num__1800 we are also given that twice the weight of his first lift was num__300 pounds more than the weight of his second lift . we express this as : num__2 f = num__300 + s num__2 f – num__300 = s we can now plug in ( num__2 f – num__300 ) for s into the first equation so we have : f + num__2 f – num__300 = num__1800 num__3 f = num__2100 f = num__700 answer is e . <eor> e <eos> |
e |
add__1800.0__300.0__ divide__2100.0__3.0__ divide__2100.0__3.0__ |
add__1800.0__300.0__ divide__2100.0__3.0__ divide__2100.0__3.0__ |
| ashwin rented a power tool from a rental shop . the rent for the tool was $ num__25 for the first hour and $ num__10 for each additional hour . if ashwin paid a total of $ num__125 excluding sales tax to rent the tool for how many hours did she rent it ? <o> a ) num__11 <o> b ) num__12 <o> c ) num__15 <o> d ) num__18 <o> e ) num__16 |
num__25 + num__10 n = num__125 n = num__10 total time = n + num__1 hrs = num__10 + num__1 hrs = num__11 hrs answer : a <eor> a <eos> |
a |
add__10.0__1.0__ add__10.0__1.0__ |
add__10.0__1.0__ add__10.0__1.0__ |
| a thief is noticed by a policeman from a distance of num__220 m . the thief starts running and the policeman chases him . the thief and the policeman run at the rate of num__10 km and num__11 km per hour respectively . what is the distance between them after num__6 minutes ? <o> a ) num__120 <o> b ) num__277 <o> c ) num__2987 <o> d ) num__278 <o> e ) num__271 |
relative speed of the thief and policeman = num__11 - num__10 = num__1 km / hr . distance covered in num__6 minutes = num__0.0166666666667 * num__6 = num__0.1 km = num__100 m . distance between the thief and policeman = num__220 - num__100 = num__120 m . answer : a <eor> a <eos> |
a |
subtract__11.0__10.0__ divide__1.0__10.0__ divide__10.0__0.1__ subtract__220.0__100.0__ round__120.0__ |
subtract__11.0__10.0__ divide__1.0__10.0__ divide__10.0__0.1__ subtract__220.0__100.0__ subtract__220.0__100.0__ |
| the average of num__1 st num__3 of num__4 numbers is num__16 and of the last num__3 are num__15 . if the sum of the first and the last number is num__13 . what is the last numbers ? <o> a ) num__7 <o> b ) num__6 <o> c ) num__5 <o> d ) num__2 <o> e ) num__4 |
a + b + c = num__48 b + c + d = num__45 a + d = num__13 a – d = num__3 a + d = num__13 num__2 d = num__10 d = num__5 answer : c <eor> c <eos> |
c |
multiply__3.0__16.0__ multiply__3.0__15.0__ subtract__3.0__1.0__ subtract__13.0__3.0__ add__1.0__4.0__ add__1.0__4.0__ |
multiply__3.0__16.0__ multiply__3.0__15.0__ subtract__3.0__1.0__ subtract__13.0__3.0__ add__1.0__4.0__ add__1.0__4.0__ |
| a bag contains num__3 red num__3 green and num__3 white balls . three balls are picked up one by one without replacement . what is the probability that there will be at least one red ball ? <o> a ) num__0.714285714286 <o> b ) num__0.785714285714 <o> c ) num__0.761904761905 <o> d ) num__0.678571428571 <o> e ) num__0.657142857143 |
p ( no red balls ) = num__0.666666666667 * num__0.625 * num__0.571428571429 = num__0.238095238095 p ( at least one red ball ) = num__1 - num__0.238095238095 = num__0.761904761905 the answer is c . <eor> c <eos> |
c |
union_prob__0.6667__0.5714__0.2381__ negate_prob__0.2381__ negate_prob__0.2381__ |
union_prob__0.6667__0.5714__0.2381__ negate_prob__0.2381__ negate_prob__0.2381__ |
| a candidate got num__35.0 of the votes polled and he lost to his rival by num__2250 votes . how many votes were cast ? <o> a ) num__7500 <o> b ) num__3600 <o> c ) num__8900 <o> d ) num__9000 <o> e ) num__5000 |
num__35.0 - - - - - - - - - - - l num__65.0 - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - num__30.0 - - - - - - - - - - num__2250 num__100.0 - - - - - - - - - ? = > num__7500 answer a <eor> a <eos> |
a |
percent__100.0__7500.0__ |
percent__100.0__7500.0__ |
| ten years ago q was half of r ' s age . if the ratio of their present ages is num__3 : num__4 what will be the total of their present ages ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__15 <o> d ) num__20 <o> e ) num__25 |
let present age of p and q be x and num__2 x respectively . ten years ago q was half of r ' s age ⇒ ( x − num__10 ) = num__2 ( num__2 x − num__10 ) ⇒ x − num__10 = num__4 x − num__20 ⇒ num__3 x = num__10 ⇒ x = num__3.33333333333 total of their present ages = x + num__2 x = > num__3.33333333333 + num__6.66666666667 = > num__10.0 = > num__10 answer : b <eor> b <eos> |
b |
multiply__10.0__2.0__ divide__10.0__3.0__ subtract__10.0__3.3333__ add__3.3333__6.6667__ |
multiply__10.0__2.0__ divide__10.0__3.0__ subtract__10.0__3.3333__ add__3.3333__6.6667__ |
| a train passes a station platform in num__36 sec and a man standing on the platform in num__20 sec . if the speed of the train is num__54 km / hr . what is the length of the platform ? <o> a ) num__227 <o> b ) num__240 <o> c ) num__987 <o> d ) num__178 <o> e ) num__171 |
speed = num__54 * num__0.277777777778 = num__15 m / sec . length of the train = num__15 * num__20 = num__300 m . let the length of the platform be x m . then ( x + num__300 ) / num__36 = num__15 = > x = num__240 m . answer : b <eor> b <eos> |
b |
multiply__20.0__15.0__ round__240.0__ |
multiply__20.0__15.0__ round__240.0__ |
| rs . num__630 / - distributed among a b and c such that on decreasing their shares by rs . num__10 rs . num__5 and rs . num__15 respectively the balance money would be divided among them in the ratio num__3 : num__4 : num__5 . then a ’ s share is : ? <o> a ) rs . num__160 / - <o> b ) rs . num__170 / - <o> c ) rs . num__180 / - <o> d ) rs . num__190 / - <o> e ) rs . num__200 / - |
a : b : c = num__3 : num__4 : num__5 total parts = num__12 a ' s share is = num__3 parts num__12 - - - - - > rs . num__600 / - num__3 - - - - - > rs . num__150 / - a ' s total = num__150 + num__10 = rs . num__160 / - a ) <eor> a <eos> |
a |
subtract__15.0__3.0__ multiply__10.0__15.0__ add__10.0__150.0__ add__10.0__150.0__ |
subtract__15.0__3.0__ divide__600.0__4.0__ add__10.0__150.0__ add__10.0__150.0__ |
| after working for num__6 days ashok finds that only num__0.333333333333 rd of the work has been done . he employs ravi who is num__60.0 as efficient as ashok . how many days more would ravi take to complete the work ? <o> a ) num__20 days <o> b ) num__19 days <o> c ) num__10 days <o> d ) num__12 days <o> e ) num__15 days |
num__0.333333333333 - - - - num__6 num__1 - - - - - - - ? a = num__18 r = num__0.0555555555556 * num__0.6 = num__0.0333333333333 num__1 - - - - - num__0.0333333333333 - - - num__0.666666666667 - - - - num__20 days answer a <eor> a <eos> |
a |
divide__1.0__18.0__ km_to_mile_conversion__ divide__0.6__18.0__ subtract__1.0__0.3333__ round__20.0__ |
divide__1.0__18.0__ km_to_mile_conversion__ divide__0.6__18.0__ subtract__1.0__0.3333__ round__20.0__ |
| after deducting a commission of num__5.0 a t . v . set costs rs . num__9595 . its gross value is ? <o> a ) rs . num__10000 <o> b ) rs . num__10074.75 <o> c ) rs . num__10100 <o> d ) none of these <o> e ) can not be determined |
answer ∵ num__95.0 of gross value = num__9595 ∴ gross value = ( num__9595 x num__100 ) / num__95 = num__10100 correct option : c <eor> c <eos> |
c |
percent__100.0__10100.0__ |
percent__100.0__10100.0__ |
| during a car trip maria stopped to rest after she traveled num__0.5 of the total distance to her destination . she stopped again after she traveled num__0.25 of the distance remaining between her first stop and her destination and then she drove the remaining num__105 miles to her detination . what was the total distance in miles from maria ' s starting point to her destination ? <o> a ) num__280 <o> b ) num__320 <o> c ) num__360 <o> d ) num__420 <o> e ) num__480 |
easy num__280 is the answer . num__0.75 ( x / num__2 ) = num__105 x = num__105 * num__2.66666666667 = num__280 . answer a <eor> a <eos> |
a |
add__0.5__0.25__ divide__0.5__0.25__ divide__280.0__105.0__ round__280.0__ |
add__0.5__0.25__ divide__0.5__0.25__ divide__280.0__105.0__ round__280.0__ |
| a train traveling at num__144 kmph crosses a platform in num__30 seconds and a man standing on the platform in num__12 seconds . what is the length of the platform in meters ? <o> a ) num__240 meters <o> b ) num__360 meters <o> c ) num__480 meters <o> d ) num__600 meters <o> e ) can not be determined |
answer distance covered by the train when crossing a man and when crossing a platform when a train crosses a man standing on a platform the distance covered by the train is equal to the length of the train . however when the same train crosses a platform the distance covered by the train is equal to the length of the train plus the length of the platform . the extra time that the train takes when crossing the platform is on account of the extra distance that it has to cover . i . e . length of the platform . compute length of platform length of the platform = speed of train * extra time taken to cross the platform . length of platform = num__144 kmph * num__12 seconds convert num__144 kmph into m / sec num__1 kmph = num__0.277777777778 m / s ( this can be easily derived . but if you can remember this conversion it saves a good num__30 seconds ) . ∴ num__144 kmph = num__0.277777777778 ∗ num__144 = num__40 m / sec therefore length of the platform = num__40 m / s * num__12 sec = num__480 meters . choice c <eor> c <eos> |
c |
multiply__12.0__40.0__ round__480.0__ |
multiply__12.0__40.0__ multiply__12.0__40.0__ |
| difference between the length & breadth of a rectangle is num__10 m . if its perimeter is num__80 m then its area is ? ? <o> a ) num__2000 m ^ num__2 <o> b ) num__2340 m ^ num__2 <o> c ) num__375 m ^ num__2 <o> d ) num__2556 m ^ num__2 <o> e ) num__2534 m ^ num__2 |
we have : ( l - b ) = num__10 and num__2 ( l + b ) = num__80 or ( l + b ) = num__40 num__2 l = num__50 = > l = num__25 & b = num__15 area = num__25 * num__15 = num__375 m ^ num__2 answer : c <eor> c <eos> |
c |
square_perimeter__10.0__ triangle_area__15.0__50.0__ triangle_area__2.0__375.0__ |
square_perimeter__10.0__ multiply__15.0__25.0__ multiply__15.0__25.0__ |
| what values of ' x ' will be the solution to the inequality num__15 x - num__2 / x > num__1 ? <o> a ) x > num__0.4 <o> b ) x < num__0.333333333333 <o> c ) - num__0.333333333333 < x < num__0.4 x > num__7.5 <o> d ) - num__0.333333333333 < x < num__0 x > num__0.4 <o> e ) x < - - num__0.333333333333 or x > num__0.4 |
explanatory answer we can rewrite the above inequality as num__15 x - num__2 / x - num__1 > num__0 i . e . num__15 x num__2 − num__2 − x / x > num__0 i . e . num__15 x num__2 − num__6 x + num__5 x − num__2 / x > num__0 factorize the quadratic expression : ( num__5 x − num__2 ) ( num__3 x + num__1 ) / x > num__0 the above inequality will hold good if the numerator and denominator are both positive or are both negative . possibility num__1 : when ( num__5 x - num__2 ) ( num__3 x + num__1 ) > num__0 and x > num__0 this will hold true for values of ' x ' that do not lie between - num__0.333333333333 and num__0.4 and for x > num__0 . combining all these conditions we get x > num__0.4 possibility num__2 : when ( num__5 x - num__2 ) ( num__3 x + num__1 ) < num__0 and x < num__0 . this will hold true for the following values of ' x ' : - num__0.333333333333 < x < num__0.4 and x < num__0 . combining we get - num__0.333333333333 < x < num__0 . therefore the range of values of x for which the above inequality will hold true is : - num__0.333333333333 < x < num__0 ∪ ∪ x > num__0.4 choice d <eor> d <eos> |
d |
subtract__6.0__1.0__ divide__15.0__5.0__ reverse__3.0__ divide__2.0__5.0__ reverse__3.0__ |
subtract__6.0__1.0__ divide__15.0__5.0__ reverse__3.0__ divide__2.0__5.0__ reverse__3.0__ |
| what is the sum of digits of ( num__5 ! * num__4 ! * num__3 ! * num__2 ! - num__144 * num__120 ) / num__5 ! <o> a ) num__22 <o> b ) num__4 <o> c ) num__12 <o> d ) num__3 <o> e ) num__7 |
( num__5 ! * num__4 ! * num__3 ! * num__2 ! - num__144 * num__120 ) / num__5 ! = ( num__5 ! * num__4 ! * num__3 ! * num__2 ! - num__24 * num__6 * num__5 ! ) / num__5 ! = ( num__5 ! * num__4 ! * num__3 ! * num__2 ! - num__4 ! * num__3 ! * num__5 ! ) / num__5 ! = num__5 ! * num__4 ! * num__3 ! ( num__2 ! - num__1 ) / num__5 ! = num__120 sum of digits = num__3 answer d <eor> d <eos> |
d |
subtract__144.0__120.0__ add__4.0__2.0__ subtract__5.0__4.0__ subtract__5.0__2.0__ |
subtract__144.0__120.0__ multiply__3.0__2.0__ subtract__5.0__4.0__ subtract__5.0__2.0__ |
| find the distance covered by a man walking for num__42 min at a speed of num__10 km / hr ? <o> a ) num__7 km <o> b ) num__3 km <o> c ) num__4 km <o> d ) num__5 km <o> e ) num__6 km |
distance = num__10 * num__0.7 = num__7 km answer is a <eor> a <eos> |
a |
multiply__10.0__0.7__ round__7.0__ |
multiply__10.0__0.7__ multiply__10.0__0.7__ |
| which expression has the greatest value ? <o> a ) num__0.999998482898 <o> b ) num__0.999998486735 <o> c ) num__0.999998486752 <o> d ) num__0.999998486728 <o> e ) num__0.999998486735 |
the difference between each numerator / denominator pair is three . the largest expression ( which in this case will also be the closest to one ) is the expression where ' num__3 ' is the smallest fraction of the numerator ( or the denominator ) . the largest numerator is num__1982487 or c and this will give an expression with a value closest to one . we can also demonstrate that this approach works with some simple logic . taking numerator / denominator pairs that differ by three . . . num__0.25 = num__0.25 num__0.4 = num__0.4 num__0.5 = num__0.5 . . . num__0.9 = num__0.9 . . . num__0.999998486752 = num__0.999998486 . . . answer : c <eor> c <eos> |
c |
add__0.5__0.4__ reverse__1.0__ |
add__0.5__0.4__ reverse__1.0__ |
| an engineer designed a ball so that when it was dropped it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a num__16 - meter platform and caught it after it had traveled num__44.5 meters . how many times did the ball bounce ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
division of total diatance travelled will be num__16 + num__16 + num__8 + num__4 + num__0.5 ans : num__4 <eor> a <eos> |
a |
divide__8.0__16.0__ divide__16.0__4.0__ |
divide__8.0__16.0__ divide__16.0__4.0__ |
| a can do a work in num__24 days and b can do it in num__8 days . in how many days a and b can do the work ? <o> a ) num__20 days <o> b ) num__10 days <o> c ) num__6 days <o> d ) num__8 days <o> e ) num__7 days |
explanation : a ' s num__1 day ' s work = num__0.0416666666667 b ' s num__1 day ' s work = num__0.125 they work together = num__0.0416666666667 + num__0.125 = num__6 days answer : option c <eor> c <eos> |
c |
divide__1.0__24.0__ divide__1.0__8.0__ round__6.0__ |
divide__1.0__24.0__ divide__1.0__8.0__ round__6.0__ |
| a gambler bought $ num__5000 worth of chips at a casino in denominations of $ num__20 and $ num__100 . that evening the gambler lost num__16 chips and then cashed in the remainder . if the number of $ num__20 chips lost was num__2 more or num__2 less than the number of $ num__100 chips lost what is the largest amount of money that the gambler could have received back ? <o> a ) $ num__4040 <o> b ) $ num__4120 <o> c ) $ num__1960 <o> d ) $ num__4920 <o> e ) $ num__1 |
400 |
in order to maximize the amount of money that the gambler kept we should maximize # of $ num__20 chips lost and minimize # of $ num__100 chips lost which means that # of $ num__20 chips lost must be num__2 more than # of $ num__100 chips lost . so if # of $ num__20 chips lost is x then # of $ num__100 chips lost should be x - num__2 . now given that total # of chips lost is num__16 : x + x - num__2 = num__16 - - > x = num__9 : num__9 $ num__20 chips were lost and num__9 - num__2 = num__7 $ num__100 chips were lost . total worth of chips lost is num__9 * num__20 + num__7 * num__100 = $ num__880 so the gambler kept $ num__5000 - $ num__880 = $ num__4120 . answer : b . <eor> b <eos> |
b |
b |
| what is the ratio whose term differ by num__50 and the measure of which is num__0.285714285714 ? <o> a ) a ) num__32 : num__23 <o> b ) b ) num__16 : num__56 <o> c ) c ) num__71 : num__85 <o> d ) d ) num__20 : num__70 <o> e ) e ) num__41 : num__52 |
let the ratio be x : ( x + num__50 ) then x / ( x + num__50 ) = num__0.285714285714 x = num__20 required ratio = num__20 : num__70 answer is d <eor> d <eos> |
d |
add__50.0__20.0__ subtract__70.0__50.0__ |
add__50.0__20.0__ subtract__70.0__50.0__ |
| the average ( arithmetic mean ) of num__16 students first quiz scores in a difficult english class is num__62.5 . when one student dropped the class the average of the remaining scores increased to num__65.0 . what is the quiz score of the student who dropped the class ? <o> a ) a ) num__10 <o> b ) b ) num__25 <o> c ) c ) num__40 <o> d ) d ) num__55 <o> e ) e ) num__70 |
total score of num__16 students is num__16 * num__62.50 = num__1000 total score of num__15 students is num__15 * num__65 = num__975 so the score of the person who left is num__25 ( num__1000 - num__975 ) answer will be ( b ) <eor> b <eos> |
b |
multiply__16.0__62.5__ multiply__65.0__15.0__ subtract__1000.0__975.0__ subtract__1000.0__975.0__ |
multiply__16.0__62.5__ multiply__65.0__15.0__ subtract__1000.0__975.0__ subtract__1000.0__975.0__ |
| the public library has exactly num__2 floors . it has num__0.666666666667 as many shelves on its num__2 nd floor as it does on its num__1 st . if the second floor has num__0.166666666667 as many books per shelf as the first floor what fraction of the library ' s books are on the first floor ? <o> a ) num__0.25 <o> b ) num__0.9 <o> c ) num__0.555555555556 <o> d ) num__0.642857142857 <o> e ) num__0.833333333333 |
let x be the no of shelves and y be the no of books per shelf on first floor . now no of shelves on num__2 nd floor = ( num__0.666666666667 ) * x no of books per shelf on num__2 nd floor = ( num__0.166666666667 ) * y so total no books on first floor = xy and total no of books on num__2 nd floor = ( num__0.666666666667 ) * x * ( num__0.166666666667 ) * y = ( num__0.111111111111 ) * xy fraction of library books on first floor = ( xy ) / ( xy + ( num__0.111111111111 ) * xy ) = num__1 / ( num__1 + ( num__0.111111111111 ) ) = num__0.9 so b <eor> b <eos> |
b |
multiply__0.6667__0.1667__ multiply__1.0__0.9__ |
multiply__0.6667__0.1667__ multiply__1.0__0.9__ |
| the two lines y = x and x = - num__4 intersect on the coordinate plane . if z represents the area of the figure formed by the intersecting lines and the x - axis what is the side length e of a cube whose surface area is equal to num__6 z ? <o> a ) e = num__16 <o> b ) e = num__8 √ num__2 <o> c ) e = num__8 <o> d ) e = num__2 √ num__2 <o> e ) ( √ num__2 ) / num__3 |
num__800 score official solution : the first step to solving this problem is to actually graph the two lines . the lines intersect at the point ( - num__4 - num__4 ) and form a right triangle whose base length and height are both equal to num__4 . as you know the area of a triangle is equal to one half the product of its base length and height : a = ( num__0.5 ) bh = ( num__0.5 ) ( num__4 × num__4 ) = num__8 ; so z = num__8 . the next step requires us to find the length of a side of a cube that has a face area equal to num__8 . as you know the num__6 faces of a cube are squares . so we can reduce the problem to finding the length of the side of a square that has an area of num__8 . since the area of a square is equal to s ² where s is the length of one of its side we can write and solve the equation s ² = num__8 . clearly s = √ num__8 = num__2 √ num__2 oranswer choice ( d ) . <eor> d <eos> |
d |
divide__4.0__0.5__ reverse__0.5__ reverse__0.5__ |
divide__4.0__0.5__ subtract__6.0__4.0__ subtract__4.0__2.0__ |
| the area of sector of a circle whose radius is num__12 metro and whose angle at the center is num__42 ° is ? <o> a ) num__52.6 <o> b ) num__52.9 <o> c ) num__52.8 <o> d ) num__52.1 <o> e ) num__52.2 |
num__0.116666666667 * num__3.14285714286 * num__12 * num__12 = num__52.8 m num__2 answer : c <eor> c <eos> |
c |
triangle_area__52.8__2.0__ |
triangle_area__52.8__2.0__ |
| find the value of * in the following ( num__1 num__1.66666666667 ) ÷ num__0.272727272727 × * / num__11 = ( num__2 num__0.666666666667 × num__1.4 × num__0.857142857143 ) <o> a ) num__3 <o> b ) num__3.2 <o> c ) num__3.6 <o> d ) num__3.8 <o> e ) num__3.5 |
num__3.6 option ' c ' <eor> c <eos> |
c |
multiply__1.0__3.6__ |
multiply__1.0__3.6__ |
| a car covers a distance of num__624 km in num__4 hours . find its speed ? <o> a ) num__104 kmph <o> b ) num__126 kmph <o> c ) num__156 kmph <o> d ) num__174 kmph <o> e ) num__101 kmph |
num__156.0 = num__156 kmph answer : c <eor> c <eos> |
c |
divide__624.0__4.0__ round__156.0__ |
divide__624.0__4.0__ round__156.0__ |
| num__7500 + ( num__25.0 ) <o> a ) num__7500 <o> b ) num__7525 <o> c ) num__7550 <o> d ) num__8000 <o> e ) none of these |
explanation : as per bodmas rule first we will solve the terms in the bracket then other . = num__7500 + ( num__25 ) = num__7525 option b <eor> b <eos> |
b |
add__7500.0__25.0__ add__7500.0__25.0__ |
add__7500.0__25.0__ add__7500.0__25.0__ |
| a driver paid n dollars for auto insurance for the year num__1997 . this annual premium was raised by p percent for the year num__1998 ; for each of the years num__1999 and num__2000 the premium was decreased by num__0.166666666667 from the previous year ’ s figure . if the driver ’ s insurance premium for the year num__2000 was again n dollars what is the value of p ? <o> a ) num__12 <o> b ) num__33 num__0.333333333333 <o> c ) num__36 <o> d ) num__44 <o> e ) num__50 |
premium in num__1997 = n ; premium in num__1998 = n ( num__1 + p / num__100 ) ; premium in num__1999 = num__0.833333333333 * n ( num__1 + p / num__100 ) ; premium in num__2000 = num__0.833333333333 * num__0.833333333333 * n ( num__1 + p / num__100 ) . given that premium in num__2000 = num__0.833333333333 * num__0.833333333333 * n ( num__1 + p / num__100 ) = n - - > num__0.833333333333 * num__0.833333333333 * ( num__1 + p / num__100 ) = num__1 - - > num__1 + p / num__100 = num__1.44 - - > p / num__100 = num__0.44 = num__0.44 - - > p = num__44 . answer : d . <eor> d <eos> |
d |
percent__100.0__44.0__ |
percent__100.0__44.0__ |
| which of the following is equal to the cube of a non - integer ? <o> a ) – num__64 <o> b ) – num__1 <o> c ) num__8 <o> d ) num__9 <o> e ) num__27 |
all except num__9 is cube of non integer as it is square of num__3 . a . – num__64 = - num__4 * - num__4 * - num__4 - - out b . – num__1 = - num__1 * - num__1 * - num__1 - - out c . num__8 = num__2 * num__2 * num__2 - - out d . num__9 - - left one . hence the answer e . num__27 = num__3 * num__3 * num__3 - - out answer : d <eor> d <eos> |
d |
subtract__4.0__3.0__ subtract__9.0__1.0__ subtract__3.0__1.0__ multiply__3.0__9.0__ add__1.0__8.0__ |
subtract__4.0__3.0__ subtract__9.0__1.0__ subtract__3.0__1.0__ multiply__3.0__9.0__ add__1.0__8.0__ |
| two sides of a triangle have lengths x and e and meet at a right angle . if the perimeter of the triangle is num__4 x what is the ratio of x to e ? <o> a ) a ) num__2 : num__3 <o> b ) b ) num__3 : num__4 <o> c ) c ) num__4 : num__3 <o> d ) d ) num__3 : num__2 <o> e ) e ) num__2 : num__1 |
ahhhh . . . . my bad . thank you you ' re correct . num__8 x = num__6 e - - > x / e = num__0.75 . actually b is the right answer . <eor> b <eos> |
b |
square_perimeter__0.75__ |
square_perimeter__0.75__ |
| a amount doubles itself in num__15 years . what is the rate of interest ? <o> a ) num__6.66666666667 % <o> b ) num__3.33333333333 % <o> c ) num__13.3333333333 % <o> d ) num__16.6666666667 % <o> e ) num__23.3333333333 % |
let amount be x & rate of interest be r % annual then ( num__2 x - x ) = ( x * num__15 * r ) / num__100 = > num__15 r = num__100 = > r = num__6.66666666667 % answer : a <eor> a <eos> |
a |
percent__100.0__6.6667__ |
percent__100.0__6.6667__ |
| the number of stamps that p and q had were in the ratio of num__7 : num__3 respectively . after p gave q num__13 stamps the ratio of the number of p ' s stamps to the number of q ' s stamps was num__5 : num__4 . as a result of the gift p had how many more stamps than q ? <o> a ) num__10 <o> b ) num__25 <o> c ) num__40 <o> d ) num__65 <o> e ) num__90 |
p started with num__7 k stamps and q started with num__3 k stamps . ( num__7 k - num__13 ) / ( num__3 k + num__13 ) = num__1.25 num__28 k - num__15 k = num__117 k = num__9 p has num__7 ( num__9 ) - num__13 = num__50 stamps and q has num__3 ( num__9 ) + num__13 = num__40 stamps . the answer is a . <eor> a <eos> |
a |
divide__5.0__4.0__ multiply__7.0__4.0__ multiply__3.0__5.0__ subtract__13.0__4.0__ divide__50.0__1.25__ add__7.0__3.0__ |
divide__5.0__4.0__ multiply__7.0__4.0__ subtract__28.0__13.0__ subtract__13.0__4.0__ divide__50.0__1.25__ add__7.0__3.0__ |
| solution y is num__30 percent liquid x and num__70 percent water . if num__3 kilograms of water evaporate from num__8 kilograms of solution y and num__3 kilograms of solution y are added to the remaining num__6 kilograms of liquid what percent of this new solution is liquid x ? <o> a ) num__35.0 <o> b ) num__37.0 <o> c ) num__41.3 <o> d ) num__42.0 <o> e ) num__45 % |
in num__8 kilograms of solution y there are num__0.3 * num__8 = num__2.4 kilograms of solution x ; after num__3 kilograms of water are replaced by num__3 kilograms of solution y to the existing num__2.4 kilograms of solution x num__0.3 * num__3 = num__0.9 kilograms of solution x are added so in the new solution of num__8 kilograms there are num__2.4 + num__0.9 = num__3.3 kilograms of solution x which is num__3.3 / num__8 * num__100 = num__41.3 of this new solution . answer : c . <eor> c <eos> |
c |
percent__30.0__8.0__ percent__30.0__3.0__ percent__100.0__41.3__ |
percent__30.0__8.0__ percent__30.0__3.0__ percent__100.0__41.3__ |
| the average of first num__22 natural numbers is ? <o> a ) num__11.6 <o> b ) num__11.5 <o> c ) num__12.9 <o> d ) num__15.4 <o> e ) num__15.1 |
sum of num__22 natural no . = num__253.0 = num__253 average = num__11.5 = num__11.5 answer : b <eor> b <eos> |
b |
divide__253.0__22.0__ divide__253.0__22.0__ |
divide__253.0__22.0__ divide__253.0__22.0__ |
| listed below are the mean temperatures measured in the tokharian desert in a certain month . what is the median temperature e in that particular month ? mean temp ( c ° ) no . of days num__29 - - - - - - - - - - - - - - - - - - - - > num__5 num__30 - - - - - - - - - - - - - - - - - - - - > num__11 num__31 - - - - - - - - - - - - - - - - - - - - > num__8 num__32 - - - - - - - - - - - - - - - - - - - - > num__6 <o> a ) num__29 <o> b ) num__30 <o> c ) num__30.5 <o> d ) num__31 <o> e ) num__32 |
to find out median = = = > they must be arranged in ascending order . num__29 num__29 num__29 . . . ( num__5 times ) num__30 num__30 num__30 num__30 . . . . . ( num__11 times ) num__31 num__31 num__31 . . . . . ( num__8 times ) num__32 num__32 num__32 . . . . . ( num__6 times ) total days = num__5 + num__11 + num__8 + num__6 = num__30 therefore num__30 numbers . . . median e will be = = > mean of num__15 and num__16 th place temperature = = = > ( num__30 + num__30 ) / num__2 = num__30 hence b <eor> b <eos> |
b |
add__5.0__11.0__ divide__30.0__15.0__ multiply__5.0__6.0__ |
add__5.0__11.0__ divide__30.0__15.0__ multiply__5.0__6.0__ |
| you collect baseball cards . suppose you start out with num__11 . maria takes half of one more than the number of baseball cards you have . since you ' re nice you give peter num__1 baseball card . since his father makes baseball cards paul decides to triple your baseball cards . how many baseball cards do you have at the end ? <o> a ) num__12 <o> b ) num__18 <o> c ) num__19 <o> d ) num__20 <o> e ) num__21 |
solution start with num__11 baseball cards . maria takes half of one more than the number of baseball cards you have . so maria takes half of num__11 + num__1 which is num__6 so you ' re left with num__11 - num__6 = num__5 . peter takes num__1 baseball card from you : num__5 - num__1 = num__4 baseball cards . paul triples the number of baseball cards you have : num__4 Ã — num__3 = num__12 baseball cards . so you have num__12 at the end . correct answer : a <eor> a <eos> |
a |
subtract__11.0__6.0__ subtract__5.0__1.0__ triple__1.0__ triple__4.0__ triple__4.0__ |
subtract__11.0__6.0__ subtract__5.0__1.0__ subtract__4.0__1.0__ add__11.0__1.0__ add__11.0__1.0__ |
| find the smallest number which when divided by num__11 and num__12 leaves respective remainders of num__2 and num__3 . <o> a ) num__131 <o> b ) num__197 <o> c ) num__207 <o> d ) num__219 <o> e ) num__227 |
let ' n ' is the smallest number which divided by num__11 and num__12 leaves respective remainders of num__2 and num__3 . required number = ( lcm of num__11 and num__12 ) - ( common difference of divisors and remainders ) = ( num__132 ) - ( num__1 ) = num__131 . answer : a <eor> a <eos> |
a |
multiply__11.0__12.0__ subtract__12.0__11.0__ subtract__132.0__1.0__ multiply__1.0__131.0__ |
multiply__11.0__12.0__ subtract__12.0__11.0__ subtract__132.0__1.0__ subtract__132.0__1.0__ |
| a two - digit number is seven times the sum of its digits . if each digit is increased by num__2 the number thus obtained is num__4 more than six times the sum of its digits . find the number . <o> a ) num__42 <o> b ) num__24 <o> c ) num__48 <o> d ) data inadequate <o> e ) none of these |
let the two - digit number be l num__0 x + y . num__10 x + y = num__7 ( x + y ) Þ x = num__2 y . . . ( i ) num__10 ( x + num__2 ) + y + num__2 = num__6 ( x + y + num__4 ) + num__4 or num__10 x + y + num__22 = num__6 x + num__6 y + num__28 Þ num__4 x – num__5 y = num__6 . . . ( ii ) solving equations ( i ) and ( ii ) we get x = num__4 and y = num__2 answer a <eor> a <eos> |
a |
add__2.0__4.0__ multiply__4.0__7.0__ subtract__7.0__2.0__ multiply__6.0__7.0__ |
add__2.0__4.0__ add__6.0__22.0__ subtract__7.0__2.0__ multiply__6.0__7.0__ |
| ross has num__20 shirts num__0.75 of the shirts are green and num__0.1 is without buttons . therefore ross has between ___ and ___ shirts with buttons that are not green . <o> a ) num__3 ; num__5 . <o> b ) num__3 ; num__6 . <o> c ) num__3 ; num__7 . <o> d ) num__3 ; num__8 . <o> e ) num__3 ; num__9 . |
total shirts = num__20 green shirts = num__0.75 * num__20 = num__15 non green shirts = num__5 shirts without button = num__0.1 * num__20 = num__2 shirts with button = num__18 required : range of shirts with buttons that are not green . maximum non green shirts with buttons = number of non green shirts = num__5 minimum non green shirts with buttons ( all without button shirts are non green ) = non green shirts - shirts without button = num__5 - num__2 = num__3 hence the range would be ( num__3 num__5 ) correct option : a <eor> a <eos> |
a |
multiply__20.0__0.75__ subtract__20.0__15.0__ multiply__20.0__0.1__ subtract__20.0__2.0__ subtract__5.0__2.0__ subtract__5.0__2.0__ |
multiply__20.0__0.75__ subtract__20.0__15.0__ multiply__20.0__0.1__ subtract__20.0__2.0__ subtract__5.0__2.0__ subtract__5.0__2.0__ |
| consider an equilateral triangle num__4 abc with unit side length . let m be the midpoint of side ab . a ball is released in a straight line from m and bounces o the side bc at a point d . then it bounces o the side ca at a point e and lands exactly at b . by the law of re ection we have \ bdm = \ cde and \ ced = \ aeb . calculate md + de + eb the distance that the ball travels before reaching b . <o> a ) √ num__6.0 <o> b ) √ num__5.5 <o> c ) √ num__6.5 <o> d ) √ num__7.0 <o> e ) √ num__8.5 |
the solution can be obtained by re ecting the triangle across edge cb to obtain another equilateral triangle num__4 a num__0 cb and then across a num__0 c to get num__4 a num__0 b num__0 c . the length of the line mb num__0 is the minimum distance from m to the vertex b that hits both sides bc and ca num__0 . the length of the line segment bb num__0 is √ num__1.5 thus the length of mb = √ num__1 + num__0.75 = √ num__6.5 correct answer c <eor> c <eos> |
c |
round_down__1.5__ multiply__6.5__1.0__ |
round_down__1.5__ multiply__6.5__1.0__ |
| ( num__786 × num__74 ) ÷ ? = num__1211.75 <o> a ) num__4.8 <o> b ) num__48 <o> c ) num__58 <o> d ) num__68 <o> e ) num__48 |
explanation : num__58164 / x = num__1211.75 = > x = num__58164 / num__1211.75 = num__48 answer : option b <eor> b <eos> |
b |
multiply__786.0__74.0__ divide__58164.0__1211.75__ divide__58164.0__1211.75__ |
multiply__786.0__74.0__ divide__58164.0__1211.75__ divide__58164.0__1211.75__ |
| two printing presses begin printing currency at the same time and at constant speeds . press f produces num__5 - dollar bills at the rate of num__1000 bills per minute . press t produces num__20 - dollar bills at the rate of num__200 bills per minute . once the machines have begun printing how many seconds does it take for press f to produce num__50 dollars more currency than press t ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
press f produces $ num__5 bills @ num__1000 per minute ( num__60 sec ) so $ num__500 in num__6 secs press t produces $ num__20 bills num__200 per minute ( num__60 sec ) so $ num__400 in num__6 secs . . so we can see f produces $ num__100 in num__6 secs or will print $ num__50 in num__3 secs . . ans num__3 secs . . b <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__5.0__20.0__ divide__60.0__20.0__ round__3.0__ |
hour_to_min_conversion__ multiply__5.0__20.0__ divide__60.0__20.0__ round__3.0__ |
| if the numerator of a fraction q is tripled and the denominator of a fraction is doubled the resulting fraction will reflect an increase of what percent ? <o> a ) num__16 num__1 ⁄ num__6.0 <o> b ) num__25.0 <o> c ) num__33 num__1 ⁄ num__3.0 <o> d ) num__50.0 <o> e ) num__66 num__2 ⁄ num__3 % |
this question can be dealt with conceptually or by testing values . we ' re told that the numerator of a fraction q is tripled and the denominator of that same fraction is doubled . we ' re asked for the resulting increase in percentage terms of that fraction . let ' s test num__2.0 if we triple the numerator and double the denominator we end up with . . . . ( num__2 x num__3 ) / ( num__1 x num__2 ) = num__3.0 = num__3 since we ' re increasing num__2 to num__3 we ' re increasing that fraction by ( num__3 - num__2 ) / num__2 = num__0.5 of itself = num__50.0 d <eor> d <eos> |
d |
subtract__3.0__2.0__ half__ multiply__1.0__50.0__ |
subtract__3.0__2.0__ divide__1.0__2.0__ divide__50.0__1.0__ |
| the price of pulses has fallen by num__22.0 . how many quintals can be bought for the same amount which was sufficient to buy num__20 quintals at the higher price ? <o> a ) num__20 <o> b ) num__22.5 <o> c ) num__25.6 <o> d ) num__27.8 <o> e ) num__32.5 |
num__80.0 of original price can buy = num__18 quintals . therefore he can buy ( num__20 x num__100 ) / num__78 = num__25.6 quintals at the lower price . answer : c <eor> c <eos> |
c |
percent__100.0__25.6__ |
percent__100.0__25.6__ |
| a man purchased num__3 blankets @ rs . num__100 each num__5 blankets @ rs . num__150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . num__150 . find the unknown rate of two blankets ? <o> a ) num__450 <o> b ) num__200 <o> c ) num__250 <o> d ) num__300 <o> e ) num__350 |
num__10 * num__150 = num__1500 num__3 * num__100 + num__5 * num__150 = num__1050 num__1500 â € “ num__1050 = num__450 answer a <eor> a <eos> |
a |
multiply__150.0__10.0__ multiply__3.0__150.0__ multiply__3.0__150.0__ |
multiply__150.0__10.0__ multiply__3.0__150.0__ multiply__3.0__150.0__ |
| a certain series is defined by the following recursive rule : sp = k ( sp - num__1 ) where k is a constant . if the num__1 st term of this series is num__64 and the num__25 th term is num__192 wha is the num__9 th term ? <o> a ) root num__2 <o> b ) root num__3 <o> c ) num__64 * root num__3 <o> d ) num__64 * num__3 ^ num__0.333333333333 <o> e ) num__64 * num__3 ^ num__24 |
ans . . . d no need for any gp formula here the rule is that pth term is k times the ( p - num__1 ) th term . num__1 st = num__64 num__2 nd = k . num__64 num__3 rd = k ^ num__2.64 . . . num__9 th term = k ^ num__8 * num__64 . . . so num__25 th = k ^ num__24 * num__64 using this solve for k and substitute k in the equation for the num__9 th term <eor> d <eos> |
d |
add__1.0__2.0__ subtract__9.0__1.0__ subtract__25.0__1.0__ multiply__1.0__64.0__ |
add__1.0__2.0__ subtract__9.0__1.0__ subtract__25.0__1.0__ multiply__1.0__64.0__ |
| the ratio between the length and the breadth of a rectangular park is num__3 : num__2 . if a man cycling alongthe oundary of the park at the speed of num__12 km / hr completes one round in num__8 min then the area of the park ( in sq . m ) is ? <o> a ) num__123600 m <o> b ) num__112600 m <o> c ) num__153600 m <o> d ) num__153500 m <o> e ) num__154300 m |
perimeter = distance covered in num__8 min . = num__12000 x num__8 m = num__1600 m . num__60 let length = num__3 x metres and breadth = num__2 x metres . then num__2 ( num__3 x + num__2 x ) = num__1600 or x = num__160 . length = num__480 m and breadth = num__320 m . area = ( num__480 x num__320 ) m num__2 = num__153600 m c <eor> c <eos> |
c |
multiply__3.0__160.0__ multiply__2.0__160.0__ surface_cube__160.0__ surface_cube__160.0__ |
multiply__3.0__160.0__ multiply__2.0__160.0__ surface_cube__160.0__ surface_cube__160.0__ |
| the difference between the place value and the face value of num__6 in the numeral num__856973 is <o> a ) num__5994 <o> b ) num__6973 <o> c ) num__973 <o> d ) num__1000 <o> e ) num__7000 |
( place value of num__6 ) - ( face value of num__6 ) = ( num__6000 - num__6 ) = num__5994 answer a <eor> a <eos> |
a |
subtract__6000.0__6.0__ subtract__6000.0__6.0__ |
subtract__6000.0__6.0__ subtract__6000.0__6.0__ |
| water is leaking out from a cylinder container at the rate of num__0.31 m ^ num__3 per minute . after num__10 minutes the water level decreases num__0.0625 meters . what is value of the radius in meters ? <o> a ) num__0.25 <o> b ) num__2 <o> c ) num__4 <o> d ) num__8 <o> e ) num__12 |
num__10 * num__0.31 = num__3.1 = pi * r ^ num__2 * h r ^ num__2 = num__3.1 / ( pi * num__0.0625 ) which is about num__16 r = num__4 the answer is c . <eor> c <eos> |
c |
multiply__0.31__10.0__ reverse__0.0625__ divide__16.0__4.0__ |
multiply__0.31__10.0__ reverse__0.0625__ divide__16.0__4.0__ |
| find the smallest number that should be multiplied with num__54000 to make it a perfect cube . <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__3 <o> e ) num__7 |
num__54000 = num__2 ^ num__1 * num__3 ^ num__3 * num__10 ^ num__3 to make it ps * num__2 ^ num__2 or num__4 answer : a <eor> a <eos> |
a |
rectangle_perimeter__2.0__3.0__ square_perimeter__1.0__ square_perimeter__1.0__ |
rectangle_perimeter__2.0__3.0__ square_perimeter__1.0__ power__4.0__1.0__ |
| a factory produces num__4340 toys per week . if the workers at this factory work num__2 days a week and if these workers make the same number of toys everyday how many toys are produced each day ? <o> a ) num__1375 toys <o> b ) num__2170 toys <o> c ) num__2375 toys <o> d ) num__2175 toys <o> e ) num__5375 toys |
to find the number of toys produced every day we divide the total number of toys produced in one week ( of num__2 days ) by num__2 . num__2170.0 = num__2170 toys correct answer b <eor> b <eos> |
b |
divide__4340.0__2.0__ round__2170.0__ |
divide__4340.0__2.0__ round__2170.0__ |
| if the terms of a series are either num__2 or num__24 and the sum of all the terms s of the series is num__124 then which of the following could be the number of num__2 sin the series ? <o> a ) num__26 <o> b ) num__29 <o> c ) num__35 <o> d ) num__40 <o> e ) num__48 |
ans : a solution : we are not certain how many num__2 or num__24 are there . but as given sum of all the terms is num__124 means num__24 * num__5 = num__120 so number of num__24 ca n ' t be more than num__5 so s = num__24 x + num__2 y = num__124 num__24 * num__5 + num__2 y = num__124 then y = num__2 num__24 * num__4 + num__2 y = num__124 then y = num__14 num__24 * num__3 + num__2 y = num__124 then y = num__26 num__24 * num__2 + num__2 y = num__124 then y = num__38 num__24 + num__2 y = num__124 then y = num__50 num__26 is the ans . <eor> a <eos> |
a |
multiply__24.0__5.0__ subtract__124.0__120.0__ subtract__5.0__2.0__ add__2.0__24.0__ add__24.0__14.0__ add__24.0__26.0__ add__2.0__24.0__ |
multiply__24.0__5.0__ subtract__124.0__120.0__ subtract__5.0__2.0__ add__2.0__24.0__ add__24.0__14.0__ add__24.0__26.0__ add__2.0__24.0__ |
| if x ≠ num__7 and ( x ^ num__2 - num__49 ) / ( num__2 y ) = ( x - num__7 ) / num__4 then in terms of y x = ? <o> a ) ( y - num__6 ) / num__2 <o> b ) ( y - num__3 ) / num__2 <o> c ) y - num__3 <o> d ) y - num__6 <o> e ) ( y - num__14 ) / num__2 |
since ( x ^ num__2 – num__49 ) = ( x - num__7 ) ( x + num__7 ) the original equation can be changed into num__4 * ( x - num__7 ) ( x + num__7 ) = num__2 y * ( x - num__7 ) . by cancelling num__2 ( x - num__7 ) ( we can do it because x ≠ num__7 ) we get : num__2 * ( x + num__7 ) = y . so x = ( y - num__14 ) / num__2 . so the answer is e ) . <eor> e <eos> |
e |
multiply__7.0__2.0__ multiply__7.0__2.0__ |
multiply__7.0__2.0__ multiply__7.0__2.0__ |
| the speed of a car is num__50 km in the first hour and num__60 km in the second hour . what is the average speed of the car ? <o> a ) num__79 kmph <o> b ) num__85 kmph <o> c ) num__55 kmph <o> d ) num__23 kmph <o> e ) num__14 kmph |
s = ( num__55 + num__60 ) / num__2 = num__55 kmph answer : c <eor> c <eos> |
c |
round__55.0__ |
round__55.0__ |
| ( √ num__27 + √ num__192 ) / √ num__54 = ? <o> a ) num__2 √ num__2 <o> b ) num__2 √ num__3 <o> c ) num__3 √ num__2 <o> d ) num__11 / ( num__3 * √ num__2 ) <o> e ) √ num__2 |
( √ num__27 + √ num__192 ) / √ num__54 = ( num__3 √ num__3 + num__8 √ num__3 ) / num__3 √ num__3 * num__2 = num__11 √ num__1.0 √ num__3 * num__2 = num__11 / ( num__3 * √ num__2 ) = num__11 / ( num__3 * √ num__2 ) . hence the correct answer is d . <eor> d <eos> |
d |
divide__54.0__27.0__ add__3.0__8.0__ subtract__3.0__2.0__ multiply__1.0__11.0__ |
divide__54.0__27.0__ add__3.0__8.0__ subtract__3.0__2.0__ multiply__1.0__11.0__ |
| num__300 metres long yard num__26 trees are palnted at equal distances one tree being at each end of the yard . what is the distance between num__2 consecutive trees <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__16 <o> e ) num__17 |
num__26 trees have num__25 gaps between them required distance ( num__12.0 ) = num__12 b <eor> b <eos> |
b |
divide__300.0__25.0__ round__12.0__ |
divide__300.0__25.0__ round__12.0__ |
| line l passes through the points ( - num__20 ) and ( num__0 a ) . line ll passes through the points ( num__40 ) and ( num__62 ) . what value of a makes the two lines parallel ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
find slopes first slope of l = ( a - num__0 ) / ( num__0 - ( - num__2 ) ) = a / num__2 slope of ll = ( num__2 - num__0 ) / ( num__6 - num__4 ) = num__1 for l and ll to be parallel their slopes must be equal a / num__2 = num__1 a = num__2 correct answer b <eor> b <eos> |
b |
rectangle_perimeter__0.0__2.0__ rectangle_perimeter__0.0__1.0__ |
rectangle_perimeter__0.0__2.0__ rectangle_perimeter__0.0__1.0__ |
| a and b can do a piece of work in num__6 days . with the help of c they finish the work in num__4 days . c alone can do that piece of work in ? <o> a ) num__40 days <o> b ) num__12 days <o> c ) num__15 days <o> d ) num__60 days <o> e ) num__40 days |
b num__12 days c = num__0.25 – num__0.166666666667 = num__0.0833333333333 = > num__12 days b <eor> b <eos> |
b |
subtract__0.25__0.1667__ round__12.0__ |
subtract__0.25__0.1667__ round__12.0__ |
| the average age of a husband and a wife is num__23 years when they were married five years ago but now the average age of the husband wife and child is num__20 years ( the child was born during the interval ) . what is the present age of the child ? <o> a ) num__2 years <o> b ) num__5 years <o> c ) num__9 years <o> d ) num__4 years <o> e ) num__1 years |
num__28 * num__2 = num__56 num__20 * num__3 = num__60 - - - - - - - - - - - num__4 years answer : d <eor> d <eos> |
d |
multiply__2.0__28.0__ subtract__23.0__20.0__ multiply__20.0__3.0__ subtract__60.0__56.0__ subtract__60.0__56.0__ |
multiply__2.0__28.0__ subtract__23.0__20.0__ multiply__20.0__3.0__ subtract__60.0__56.0__ subtract__60.0__56.0__ |
| the owner of a local jewellery store hired three watchmen to guard his diamonds but a thief still got in and stole some diamonds . on the way out the thief met each watchman one at a time . to each he gave num__0.333333333333 of the diamonds he had then and num__3 more besides . he escaped with num__3 diamonds . how many did he steal originally ? <o> a ) num__198 <o> b ) num__199 <o> c ) num__200 <o> d ) none of these <o> e ) can not be determined |
explanation : since thief escaped with num__3 diamond before num__3 rd watchman he had ( num__3 + num__3 ) x num__3 = num__18 diamonds . before num__2 nd watchman he had ( num__18 + num__3 ) x num__3 = num__63 diamonds . before num__1 st watchman he had ( num__63 + num__3 ) x num__3 = num__198 diamonds . answer : a <eor> a <eos> |
a |
subtract__3.0__2.0__ multiply__1.0__198.0__ |
subtract__3.0__2.0__ multiply__1.0__198.0__ |
| a ' and ' b ' are positive integers such that their lcm is num__20 and their hcf is num__1 . what is the addition of the maximum and minimum possible values of ' a + b ' ? <o> a ) num__21 <o> b ) num__34 <o> c ) num__30 <o> d ) num__25 <o> e ) num__12 |
possible values of a and b can be num__54 ; num__45 ( which are same for a + b ) and num__120 ; num__201 ( same result for a + b ) so num__21 + num__9 = num__30 ans c <eor> c <eos> |
c |
add__20.0__1.0__ subtract__54.0__45.0__ add__21.0__9.0__ multiply__1.0__30.0__ |
add__20.0__1.0__ subtract__54.0__45.0__ add__21.0__9.0__ add__21.0__9.0__ |
| on a map num__1 inch represents num__28 miles . how many b inches would be necessary to represent a distance of num__383.6 miles ? <o> a ) num__5.2 <o> b ) num__7.4 <o> c ) num__13.7 <o> d ) num__21.2 <o> e ) num__28.7 |
b inches necessary to represent a distance of num__383.6 miles = num__383.6 / num__28 = num__13.7 answer c <eor> c <eos> |
c |
divide__383.6__28.0__ round__13.7__ |
divide__383.6__28.0__ divide__383.6__28.0__ |
| on dividing a number by num__68 we get num__269 as quotient and num__0 as remainder . on dividing the same number by num__67 what will be the remainder ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
number = num__269 * num__68 + num__0 = num__18292 num__67 ) num__18292 ( num__273 num__18291 - - - - - - - - num__1 required number = num__1 . answer : b <eor> b <eos> |
b |
multiply__68.0__269.0__ multiply__67.0__273.0__ subtract__68.0__67.0__ reverse__1.0__ |
multiply__68.0__269.0__ multiply__67.0__273.0__ subtract__68.0__67.0__ subtract__68.0__67.0__ |
| a milkman sells his milk at cp but he mixes it with water and thereby gains num__25.0 . what is the % of water in the mixture ? <o> a ) num__10.0 <o> b ) num__15.0 <o> c ) num__20.0 <o> d ) num__25.0 <o> e ) num__30 % |
here gain = num__25.0 error = quantity of water he mixes in the milk = x true value = true quantity of milk = t so the formula becomes num__25 = x ( t − x ) × num__100 ⇒ num__1 = x ( t − x ) × num__4 ⇒ t − x = num__4 x ⇒ t = num__5 x percentage of water in the mixture = xt × num__100 = x num__5 x × num__100 = num__15 × num__100 = num__20.0 c <eor> c <eos> |
c |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| a broker invested her own money in the stock market . during the first year she increased her stock market wealth by num__90 percent . in the second year largely as a result of a slump in the stock market she suffered a num__50 percent decrease in the value of her stock investments . what was the net increase or decrease on her overall stock investment wealth by the end of the second year ? <o> a ) − num__5.0 <o> b ) num__5.0 <o> c ) num__15.0 <o> d ) num__20.0 <o> e ) num__80 % |
the actual answer is obtained by multiplying num__140.0 by num__70.0 and subtracting num__100.0 from this total . that is : num__190.0 × num__50.0 = num__95.0 ; num__95.0 − num__100.0 = - num__5.0 . answer : a <eor> a <eos> |
a |
percent__50.0__140.0__ percent__50.0__190.0__ percent__100.0__5.0__ |
percent__50.0__140.0__ percent__50.0__190.0__ percent__100.0__5.0__ |
| the average weight of num__8 person ' s increases by num__3.5 kg when a new person comes in place of one of them weighing num__65 kg . what is the weight of the new person ? <o> a ) num__93 kg <o> b ) num__50 kg <o> c ) num__85 kg <o> d ) num__80 kg <o> e ) num__60 kg |
explanation : total increase in weight = num__8 Ã — num__3.5 = num__28 if x is the weight of the new person total increase in weight = x â ˆ ’ num__65 = > num__28 = x - num__65 = > x = num__28 + num__65 = num__93 answer : option a <eor> a <eos> |
a |
multiply__8.0__3.5__ add__65.0__28.0__ add__65.0__28.0__ |
multiply__8.0__3.5__ add__65.0__28.0__ add__65.0__28.0__ |
| present ages of sameer and anand are in the ratio of num__5 : num__4 respectively . three years hence the ratio of their ages will become num__11 : num__9 respectively . what is anand ' s present age in years ? <o> a ) num__14 <o> b ) num__16 <o> c ) num__15 <o> d ) num__23 <o> e ) num__24 |
given the preset ratio of the ages of sameer and anand is num__5 : num__4 let age of sameer / num__5 = age of anand / num__4 = x present age of sameer = num__5 x present age of anand = num__4 x after num__3 years age of sameer woud be num__5 x + num__3 age of anand woud be num__4 x + num__3 so ( num__5 x + num__3 ) / ( num__4 x + num__3 ) = num__119 = x = num__6 hence age of anand will be num__6 x num__4 = num__24 years . answer : e <eor> e <eos> |
e |
subtract__11.0__5.0__ multiply__4.0__6.0__ multiply__4.0__6.0__ |
subtract__11.0__5.0__ multiply__4.0__6.0__ multiply__4.0__6.0__ |
| john bought blue and pink recorder for pupils in his music class in the ratio of num__6 : num__4 . if num__40.0 of the blue recorder and num__60.0 of the pink recorder were bought at half price how many recorders did john buy at half price ? <o> a ) num__48.0 <o> b ) num__52.0 <o> c ) num__44.0 <o> d ) num__46.0 <o> e ) num__47 % |
let total no of recorder bought be num__100 blue recorder num__60 and pink recorder num__40 so recorder bought at half price for both blue and pink recorder respectively num__60 * num__0.4 = num__24 and num__40 * num__0.6 = num__24 so total num__48 recorders were bought at half price out of num__100 blue and pink recorders . . so required % is num__48.0 answer : a <eor> a <eos> |
a |
percent__40.0__60.0__ percent__100.0__48.0__ |
percent__40.0__60.0__ percent__100.0__48.0__ |
| if i walk at num__3 kmph i miss the train by num__2 min if however i walk at num__4 kmph . i reach the station num__2 min before the arrival of the train . how far do i walk to reach the station ? <o> a ) num__0.8 km <o> b ) num__0.444444444444 <o> c ) num__0.666666666667 <o> d ) num__2.0 <o> e ) num__4.0 |
x / num__3 – x / num__4 = num__0.0666666666667 x = num__0.8 km answer : a <eor> a <eos> |
a |
round__0.8__ |
round__0.8__ |
| of the land owned by a farmer num__90 percent was cleared for planting . of the cleared land num__20 percent was planted with potato and num__70 percent of the cleared land was planted with tomato . if the remaining num__630 acres of cleared land was planted with corn how many acres did the farmer own ? <o> a ) num__7000 <o> b ) num__6800 <o> c ) num__8000 <o> d ) num__5400 <o> e ) num__7100 |
num__20.0 od num__90.0 = num__18.0 num__70.0 num__0 f num__90.0 = num__63.0 so the remaining num__90 - num__18 - num__63 = num__9.0 = num__630 acres or num__10.0 of num__90.0 = num__9.0 - - > num__70.0 * num__100 = num__7000 acres answer ( a ) <eor> a <eos> |
a |
percent__90.0__20.0__ percent__90.0__70.0__ percent__100.0__7000.0__ |
percent__90.0__20.0__ percent__90.0__70.0__ percent__100.0__7000.0__ |
| a customer asks the clerk for a paintbrush and a bucket of whitewash for a total price of b dollars . if the paintbrush costs num__700 cents less than twice the price of a bucket of whitewash what is the price of half a bucket of whitewash in dollars ? <o> a ) ( b + num__200 ) / num__2 . <o> b ) ( b + num__2 ) / num__6 . <o> c ) ( num__2 b + num__200 ) / num__3 . <o> d ) ( b + num__2 ) / num__3 . <o> e ) ( b - num__2 ) / num__6 . |
p = num__2 w - num__200 thus total price in cents as given should be num__2 w - num__200 + w = num__100 b num__3 w - num__200 = num__100 b num__3 w = num__100 b + num__200 num__3 w = b + num__2 ( in dollars ) w / num__2 = ( b + num__2 ) / num__6 ans : c <eor> c <eos> |
c |
divide__200.0__2.0__ multiply__2.0__3.0__ divide__200.0__100.0__ |
divide__200.0__2.0__ multiply__2.0__3.0__ divide__200.0__100.0__ |
| tom traveled the entire num__60 miles trip . if he did the first num__12 miles of at a constant rate num__24 miles per hour and the remaining trip of at a constant rate num__48 miles per hour what is the his average speed in miles per hour ? <o> a ) num__20 mph <o> b ) num__24 mph <o> c ) num__30 mph <o> d ) num__40 mph <o> e ) num__42 mph |
avg speed = total distance / total time = ( d num__1 + d num__2 ) / ( t num__1 + t num__2 ) = ( num__12 + num__48 ) / ( ( num__0.5 ) + ( num__1.0 ) ) = num__60 * num__0.666666666667 = num__40 mph d <eor> d <eos> |
d |
divide__24.0__12.0__ divide__12.0__24.0__ round__40.0__ |
divide__24.0__12.0__ divide__12.0__24.0__ divide__40.0__1.0__ |
| find the number of square tiles to cover the floor of a room measuring num__4 m * num__9 m leaving num__0.25 m space around the room . a side of square tile is given to be num__25 cms ? <o> a ) num__299 <o> b ) num__476 <o> c ) num__277 <o> d ) num__268 <o> e ) num__982 |
num__3 num__0.5 * num__8 num__0.5 = num__0.25 * num__0.25 * x = > x = num__476 answer : b <eor> b <eos> |
b |
divide__4.0__0.5__ round__476.0__ |
divide__4.0__0.5__ round__476.0__ |
| if a train travelling at a speed of num__80 kmph crosses a pole in num__5 sec then the length of train is ? <o> a ) num__110.111 m <o> b ) num__121.111 m <o> c ) num__111.11 m <o> d ) num__141.111 m <o> e ) num__181.111 m |
d = num__80 * num__0.277777777778 * num__5 = num__111.11 m answer : c <eor> c <eos> |
c |
round__111.11__ |
round__111.11__ |
| for each hour of production a certain factory requires num__1 assembly line worker for every num__25 units to be produced in that hour . the factory also requires num__8 managers regardless of the number of units to be produced . which of the following expressions represents the total number of assembly line workers and managers that this factory requires to produce num__50 n in one hour where n is a positive integer ? <o> a ) num__12 + num__50 n <o> b ) num__12 + num__2 n <o> c ) num__8 + num__2 n <o> d ) num__37 n <o> e ) num__25 |
num__25 units = num__1 worker ; num__50 n units = num__50 n / num__25 = num__2 n workers . so the answer is num__2 n workers plus num__8 managers . answer : c . <eor> c <eos> |
c |
divide__50.0__25.0__ multiply__1.0__8.0__ |
divide__50.0__25.0__ divide__8.0__1.0__ |
| a company organized a recruiting process for num__3 vacant positions of assistant manager for its product launches . the company ' s efforts yielded num__15 eligible candidates . how many sets of num__3 candidates can the company choose to fill the num__3 assistant manager positions ? <o> a ) num__2060 <o> b ) num__1320 <o> c ) num__455 <o> d ) num__315 <o> e ) num__220 |
num__15 * num__14 * num__4.33333333333 * num__2 * num__1 = num__455 c <eor> c <eos> |
c |
subtract__3.0__2.0__ multiply__1.0__455.0__ |
subtract__3.0__2.0__ multiply__1.0__455.0__ |
| a single discount equivalent to the discount series of num__30.0 num__10.0 and num__8.0 is ? <o> a ) num__31.8 <o> b ) num__42.04 <o> c ) num__31.6 <o> d ) num__31.2 <o> e ) num__31.9 |
explanation : num__100 * ( num__0.7 ) * ( num__0.9 ) * ( num__0.92 ) = num__57.96 num__100 - num__57.96 = num__42.04 answer : b <eor> b <eos> |
b |
percent__42.04__100.0__ |
percent__42.04__100.0__ |
| the ages of two persons differ by num__16 years . num__6 years ago the elder one was num__3 times as old as the younger one . what are their present ages of the elder person <o> a ) num__15 <o> b ) num__20 <o> c ) num__25 <o> d ) num__30 <o> e ) num__28 % |
explanation : let ' s take the present age of the elder person = x and the present age of the younger person = x – num__16 ( x – num__6 ) = num__3 ( x - num__16 - num__6 ) = > x – num__6 = num__3 x – num__66 = > num__2 x = num__60 = > x = num__30.0 = num__30 option d <eor> d <eos> |
d |
divide__6.0__3.0__ subtract__66.0__6.0__ divide__60.0__2.0__ divide__60.0__2.0__ |
divide__6.0__3.0__ subtract__66.0__6.0__ divide__60.0__2.0__ subtract__60.0__30.0__ |
| out of a total of num__1000 employees at a certain corporation num__52 percent are female and num__40 percent of these females work in research . if num__65 percent of the total number of employees work in research how many male employees do not work in research ? <o> a ) num__520 <o> b ) num__480 <o> c ) num__392 <o> d ) num__208 <o> e ) num__38 |
total number of female employees = num__52.0 = num__520 female employees working in research = ( num__0.4 ) * num__520 = num__208 total no of employees working in research = num__65.0 = num__650 total male employees = num__48.0 = num__480 male employees working in research = num__650 - num__208 = num__442 male employees not working in research = num__480 - num__442 = num__38 answer e <eor> e <eos> |
e |
multiply__0.4__520.0__ subtract__1000.0__520.0__ subtract__650.0__208.0__ subtract__480.0__442.0__ subtract__480.0__442.0__ |
multiply__0.4__520.0__ subtract__1000.0__520.0__ subtract__650.0__208.0__ subtract__480.0__442.0__ subtract__480.0__442.0__ |
| a shopkeeper sold an article offering a discount of num__5.0 and earned a profit of num__19.7 . what would have been the percentage of profit earned if no discount was offered ? <o> a ) num__60.0 <o> b ) num__26.0 <o> c ) num__30.0 <o> d ) num__56.0 <o> e ) num__73 % |
let c . p . be rs . num__100 . then s . p . = rs . num__19.70 let marked price be rs . x . then num__0.95 x = num__119.70 x = num__126.0 = rs . num__126 now s . p . = rs . num__126 c . p . = rs . num__100 profit % = num__26.0 . answer : b <eor> b <eos> |
b |
percent__100.0__26.0__ |
percent__100.0__26.0__ |
| if the speed of x meters per minute is equivalent to the speed of y kilometers per hour what is y in terms of x ? ( num__1 kilometer = num__1000 meters ) <o> a ) num__15 x / num__18 <o> b ) num__6 x / num__5 <o> c ) num__18 x / num__5 <o> d ) num__0.06 x <o> e ) num__3600000 x |
x = num__1000 y / num__60 y = num__0.06 x answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ divide__60.0__1000.0__ round__0.06__ |
hour_to_min_conversion__ divide__60.0__1000.0__ divide__60.0__1000.0__ |
| if a number p is prime and num__2 p + num__1 = q where q is also prime then the decimal expansion of num__1 / q will produce a decimal with q - num__1 digits . if this method produces a decimal with num__166 digits what is the units digit of the product of p and q <o> a ) num__1 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__9 |
num__0.428571428571 = num__0.428571 . . . ( a repeating pattern one digit long ) b <eor> b <eos> |
b |
add__2.0__1.0__ |
add__2.0__1.0__ |
| two trains travel in opposite directions at num__36 kmph and num__45 kmph and a man sitting in slower train passes the faster train in num__10 seconds . the length of the faster train is <o> a ) num__225 m <o> b ) num__100 m <o> c ) num__120 m <o> d ) num__180 m <o> e ) none |
solution relative speed = ( num__36 + num__45 ) km / hr = ( num__81 x num__0.277777777778 ) m / sec = ( num__22.5 ) m / sec length of the train = ( num__22.5 x num__10 ) m = num__225 m . answer a <eor> a <eos> |
a |
add__36.0__45.0__ divide__10.0__36.0__ multiply__10.0__22.5__ round__225.0__ |
add__36.0__45.0__ divide__10.0__36.0__ multiply__10.0__22.5__ round__225.0__ |
| the simple interest on rs . num__11000 at a certain rate of interest in five years is rs . num__7200 . find the compound interest on the same amount for two years at the same rate of interest . <o> a ) num__3068.82 <o> b ) num__3052.89 <o> c ) num__3052.85 <o> d ) num__3068.51 <o> e ) num__3068.81 |
r = num__100 i / pt = > r = ( num__100 * num__7200 ) / ( num__11000 * num__5 ) = num__13.09 ci = p { [ num__1 + r / num__100 ] n - num__1 } = num__11000 { [ num__1 + num__13.09 / num__100 ] ^ num__2 - num__1 } = rs . num__3068.51 answer : d <eor> d <eos> |
d |
percent__100.0__3068.51__ |
percent__100.0__3068.51__ |
| in a factory there are num__90.0 technicians and num__10.0 non - technicians . if the num__90.0 of the technicians and num__10.0 of non - technicians are permanent employees then the percentage of workers who are temporary is ? <o> a ) num__62.0 <o> b ) num__57.0 <o> c ) num__52.0 <o> d ) num__82.0 <o> e ) num__42 % |
total = num__100 t = num__90 nt = num__10 num__90 * ( num__0.1 ) = num__910 * ( num__0.9 ) = num__9 num__9 + num__9 = num__18 = > num__100 - num__18 = num__82.0 answer : d <eor> d <eos> |
d |
percent__90.0__10.0__ percent__100.0__82.0__ |
percent__90.0__10.0__ percent__100.0__82.0__ |
| a mixture contains milk and water in the ratio num__5 : num__2 . on adding num__10 liters of water the ratio of milk to water becomes num__5 : num__3 . the quantity of milk in the original mixture is ? <o> a ) num__40 <o> b ) num__48 <o> c ) num__52 <o> d ) num__56 <o> e ) num__70 |
milk : water = num__5 : num__2 num__5 x : num__2 x + num__10 = num__5 : num__3 num__3 [ num__5 x ] = num__5 [ num__2 x + num__10 ] num__15 x = num__10 x + num__50 num__15 x - num__10 x = num__50 x = num__10 the quantity of milk in the original mixture is = num__5 : num__2 = num__5 + num__2 = num__7 num__7 x = num__70 e <eor> e <eos> |
e |
add__5.0__10.0__ multiply__5.0__10.0__ add__5.0__2.0__ multiply__10.0__7.0__ multiply__10.0__7.0__ |
add__5.0__10.0__ multiply__5.0__10.0__ add__5.0__2.0__ multiply__10.0__7.0__ multiply__10.0__7.0__ |
| a man can row downstream at num__18 kmph and upstream at num__10 kmph . find the speed of the man in still water and the speed of stream respectively ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__8 <o> d ) num__4 <o> e ) num__7 |
let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = num__18 - - - ( num__1 ) and x - y = num__10 - - - ( num__2 ) from ( num__1 ) & ( num__2 ) num__2 x = num__28 = > x = num__14 y = num__4 . answer : d <eor> d <eos> |
d |
add__18.0__10.0__ divide__28.0__2.0__ subtract__18.0__14.0__ round__4.0__ |
add__18.0__10.0__ divide__28.0__2.0__ subtract__18.0__14.0__ subtract__18.0__14.0__ |
| the fraction num__101 num__0.00027 in decimal for is : <o> a ) num__101.27 <o> b ) num__101.00027 <o> c ) num__101.0027 <o> d ) num__101.027 <o> e ) none of them |
num__101 num__0.00027 = num__101 + num__0.00027 = num__101 + . num__00027 = num__101.00027 answer is b . <eor> b <eos> |
b |
add__101.0__0.0003__ add__101.0__0.0003__ |
add__101.0__0.0003__ add__101.0__0.0003__ |
| if exactly num__7 guests drank only one of the two types of drinks how many guests drank both types of drinks ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__4 <o> e ) num__2 |
say x guests drank both drinks . ( num__7 - x ) + ( num__4 - x ) = num__7 - - > x = num__2 . answer : e . <eor> e <eos> |
e |
coin_space__ coin_space__ |
coin_space__ coin_space__ |
| sara is younger than robert by num__8 years . if their ages are in the respective ratio of num__3 : num__5 how old is sara ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__15 <o> e ) num__20 |
let robert ' s age be x years then sara ' s age = x - num__8 years ( x - num__8 ) / x = num__0.6 x = num__20 sara ' s age = x - num__8 = num__12 years answer is c <eor> c <eos> |
c |
divide__3.0__5.0__ subtract__20.0__8.0__ subtract__20.0__8.0__ |
divide__3.0__5.0__ subtract__20.0__8.0__ subtract__20.0__8.0__ |
| a car covers a distance of num__520 km in num__4 hours . find its speed ? <o> a ) num__104 <o> b ) num__130 <o> c ) num__298 <o> d ) num__269 <o> e ) num__213 |
num__130.0 = num__130 kmph answer : b <eor> b <eos> |
b |
divide__520.0__4.0__ round__130.0__ |
divide__520.0__4.0__ round__130.0__ |
| a student travels from his house to school at num__10 km / hr and reaches school num__2 hours late . the next day he travels num__16 km / hr and reaches school num__1 hour early . what is the distance between his house and the school ? <o> a ) num__70 <o> b ) num__75 <o> c ) num__80 <o> d ) num__85 <o> e ) num__90 |
let x be the distance from his house to the school . x / num__10 = x / num__16 + num__3 num__8 x = num__5 x + num__240 num__3 x = num__240 x = num__80 km the answer is c . <eor> c <eos> |
c |
add__2.0__1.0__ subtract__10.0__2.0__ divide__10.0__2.0__ multiply__10.0__8.0__ round__80.0__ |
add__2.0__1.0__ divide__16.0__2.0__ divide__10.0__2.0__ divide__240.0__3.0__ divide__240.0__3.0__ |
| a sum fetched a total simple interest of rs . num__4016.25 at the rate of num__5.0 p . a . in num__5 years . what is the sum ? <o> a ) num__12021 <o> b ) num__14520 <o> c ) num__16065 <o> d ) num__18925 <o> e ) num__12887 |
principal = ( num__100 * num__4016.25 ) / ( num__5 * num__5 ) = rs . num__16065 . answer : c <eor> c <eos> |
c |
percent__100.0__16065.0__ |
percent__100.0__16065.0__ |
| a sequence consists of num__16 consecutive even integers written in increasing order . the sum of the first num__8 of these even integers is num__504 . what is the sum of the last num__8 of the even integers ? <o> a ) num__608 <o> b ) num__614 <o> c ) num__620 <o> d ) num__626 <o> e ) num__632 |
the sum of the first num__8 numbers is : x + ( x + num__2 ) + . . . + ( x + num__14 ) = num__504 the sum of the next num__8 numbers is : ( x + num__16 ) + ( x + num__2 + num__16 ) + . . . + ( x + num__14 + num__16 ) = num__504 + num__8 ( num__16 ) = num__504 + num__128 = num__632 the answer is e . <eor> e <eos> |
e |
divide__16.0__8.0__ subtract__16.0__2.0__ multiply__16.0__8.0__ add__504.0__128.0__ add__504.0__128.0__ |
divide__16.0__8.0__ subtract__16.0__2.0__ multiply__16.0__8.0__ add__504.0__128.0__ add__504.0__128.0__ |
| at an international conference “ red ” world countries and “ blue ” world countries are the only participants . the ratio of “ red ” world participants to “ blue ” world participants is num__2 : num__1 . if one - third of “ red ” world participants are left - handed and two - thirds of “ blue ” world participants are left - handed then what is the fraction of the participants who are left - handed ? <o> a ) num__0.666666666667 <o> b ) num__0.8 <o> c ) num__3.0 <o> d ) num__2.5 <o> e ) num__0.444444444444 |
red : blue = num__2 : num__1 let red = num__2 x and blue = num__1 x num__0.333333333333 of red are left handed = > num__0.333333333333 * num__2 x = num__2 x / num__3 red left handed num__0.666666666667 of blue are left handed = > num__0.666666666667 * num__1 x = num__2 x / num__3 blue left handed fraction of participants who are left handed = total left handed / total participants = ( red left handed + blue left handed ) / total participants = ( num__2 x / num__3 + num__2 x / num__3 ) / ( num__2 x + num__1 x ) = num__0.444444444444 answer : e <eor> e <eos> |
e |
add__2.0__1.0__ divide__2.0__3.0__ multiply__1.0__0.4444__ |
add__2.0__1.0__ divide__2.0__3.0__ multiply__1.0__0.4444__ |
| if a man crosses a num__600 m long street in num__5 minutes . what is his speed ? <o> a ) num__7 km / hr <o> b ) num__7.1 km / hr <o> c ) num__7.2 km / hr <o> d ) num__8 km / hr <o> e ) num__8.5 km / hr |
distance = num__600 meter time = num__5 minutes = num__5 x num__60 seconds = num__300 seconds speed = distancetime = num__600300 = num__2 m / s = num__2 × num__185 km / hr = num__365 km / hr = num__7.2 km / hr c <eor> c <eos> |
c |
hour_to_min_conversion__ multiply__5.0__60.0__ divide__600.0__300.0__ round__7.2__ |
hour_to_min_conversion__ multiply__5.0__60.0__ divide__600.0__300.0__ round__7.2__ |
| nr books bought nr of people num__5 num__7 num__6 num__3 num__7 num__3 num__8 num__2 what is the median of books bought per person ? <o> a ) a ) num__2 <o> b ) b ) num__4 <o> c ) c ) num__6 <o> d ) d ) num__8 <o> e ) e ) num__18 |
num__5 num__55 num__55 num__55 num__66 num__67 num__77 num__88 so you will observer that the median of the list is num__6 . ans c <eor> c <eos> |
c |
multiply__3.0__2.0__ |
multiply__3.0__2.0__ |
| a customer bought a product at the shop . however the shopkeeper increased the price of the product by num__20.0 so that the customer could not buy the required amount of the product . the customer managed to buy only num__80.0 of the required amount . what is the difference in the amount of money that the customer paid for the second purchase compared to the first purchase ? <o> a ) num__10.0 <o> b ) num__8.0 <o> c ) num__6.0 <o> d ) num__4.0 <o> e ) num__2 % |
let x be the amount of money paid for the first purchase . the second time the customer paid num__0.8 ( num__1.2 x ) = num__0.96 x . the difference is num__4.0 . the answer is d . <eor> d <eos> |
d |
multiply__1.2__0.8__ divide__80.0__20.0__ divide__80.0__20.0__ |
multiply__1.2__0.8__ divide__80.0__20.0__ divide__80.0__20.0__ |
| the g . c . d of num__1.08 num__0.36 and num__1.5 is <o> a ) num__0.06 <o> b ) num__0.1 <o> c ) num__0.18 <o> d ) num__0.11 <o> e ) num__0.12 |
explanation : given numbers are num__1.08 num__0.36 and num__1.5 h . c . f of num__108 num__36 and num__150 is num__6 [ because g . c . d is nothing but h . c . f ] therefore h . c . f of given numbers = num__0.06 answer : a <eor> a <eos> |
a |
gcd__36.0__150.0__ divide__0.36__6.0__ divide__0.36__6.0__ |
gcd__36.0__150.0__ divide__0.36__6.0__ divide__0.36__6.0__ |
| company x sells a selection of products at various price points . listed below are unit sales made for one particular day . how many unit sales on that day were greater than the mean sale price but less than the median sale price ? $ num__50 $ num__50 $ num__97 $ num__97 $ num__97 $ num__120 $ num__155 $ num__155 $ num__199 $ num__199 $ num__219 <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
taking the prices of products in ascending order ( already arranged ) $ num__50 $ num__50 $ num__97 $ num__97 $ num__97 $ num__120 $ num__125 $ num__155 $ num__199 $ num__199 $ num__219 we see that median value is the num__6 th value as there in total num__11 values given arithmetic mean = total / number of entries = num__128.0 = num__128 we are asked to find how many unit sales on that day were greater than the mean sale price but less than the median sale price as we can clearly see that there is one value between $ num__120 and $ num__128 answer is num__1 unit correct answer - b <eor> b <eos> |
b |
reverse__1.0__ |
reverse__1.0__ |
| a bookseller has two display windows . she plans to display num__4 new fiction books in the left window and num__5 new non - fiction books in the right window . assuming she can put the five fiction books in any order and separately the three non - fiction books in any order how many total configurations will there be for the two display windows ? <o> a ) num__2400 <o> b ) num__720 <o> c ) num__2880 <o> d ) num__336 <o> e ) num__420 |
two displays books could be arranged n ! # of total arrangements are combinations multiplied together num__4 ! * num__5 ! = num__24 * num__120 = num__2880 answer is c <eor> c <eos> |
c |
multiply__5.0__24.0__ multiply__24.0__120.0__ multiply__24.0__120.0__ |
multiply__5.0__24.0__ multiply__24.0__120.0__ multiply__24.0__120.0__ |
| the larger of two negative consecutive even integers num__2 t and num__2 ( t - num__1 ) is multiplied by num__6 and then added to the smaller of the two original integers . which of the following represents this operation ? <o> a ) num__6 t - num__2 <o> b ) num__8 t - num__2 <o> c ) num__14 t - num__2 <o> d ) num__3 <o> e ) - num__2 - num__4 t ^ num__2 |
larger of num__2 t - num__2 num__2 t ( when they are negative ) is num__2 t . the smallest of num__2 t and num__2 t - num__2 when they are positive is num__2 t - num__2 . therefore num__6 * num__2 t + num__2 t - num__2 = num__14 t - num__2 . option c . <eor> c <eos> |
c |
multiply__1.0__14.0__ |
multiply__1.0__14.0__ |
| the price of pulses has fallen by num__20.0 . how many quintals can be bought for the same amount which was sufficient to buy num__18 quintals at the higher price ? <o> a ) num__20 <o> b ) num__22.5 <o> c ) num__25 <o> d ) num__30 <o> e ) num__32.5 |
num__80.0 of original price can buy = num__18 quintals . therefore he can buy ( num__18 x num__100 ) / num__80 = num__22.5 quintals at the lower price . answer : b <eor> b <eos> |
b |
percent__22.5__100.0__ |
percent__22.5__100.0__ |
| there are num__6 boxes numbered num__1 num__23 num__45 and num__6 . each box is to be filled up either with a red or a green ball in such a way that at least num__1 box contains a green ball & the boxes containing green balls are consecutively numbered . the total no . of ways in which this can be done is ? <o> a ) num__18 <o> b ) num__19 <o> c ) num__21 <o> d ) num__23 <o> e ) num__25 |
if only one of the boxes has a green ball it can be any of the num__6 boxes . so this can be achieved in num__6 ways . if two of the boxes have green balls and then there are num__5 consecutive sets of num__2 boxes . num__12 num__23 num__34 num__45 num__56 . similarly if num__3 of the boxes have green balls there will be num__4 options . if num__4 boxes have green balls there will be num__3 options . if num__5 boxes have green balls then there will be num__2 options . if all num__6 boxes have green balls then there will be just num__1 options . total number of options = num__6 + num__5 + num__4 + num__3 + num__2 + num__1 = num__21 c <eor> c <eos> |
c |
subtract__6.0__1.0__ multiply__6.0__2.0__ divide__6.0__2.0__ subtract__6.0__2.0__ subtract__23.0__2.0__ multiply__1.0__21.0__ |
subtract__6.0__1.0__ multiply__6.0__2.0__ add__1.0__2.0__ add__1.0__3.0__ subtract__23.0__2.0__ multiply__1.0__21.0__ |
| a cubical tank is filled with water to a level of num__2 feet . if the water in the tank occupies num__50 cubic feet to what fraction of its capacity is the tank filled with water ? <o> a ) num__0.333333333333 <o> b ) num__0.666666666667 <o> c ) num__0.75 <o> d ) num__0.4 <o> e ) num__0.833333333333 |
the volume of water in the tank is h * l * b = num__50 cubic feet . since h = num__2 then l * b = num__25 and l = b = num__5 . since the tank is cubical the capacity of the tank is num__5 * num__5 * num__5 = num__125 . the ratio of the water in the tank to the capacity is num__0.4 = num__0.4 the answer is d . <eor> d <eos> |
d |
divide__50.0__2.0__ multiply__5.0__25.0__ divide__2.0__5.0__ round__0.4__ |
divide__50.0__2.0__ multiply__5.0__25.0__ divide__2.0__5.0__ round__0.4__ |
| num__5 is added to a certain number the sum is multiplied by num__7 the product is divided by num__5 and num__5 is subtracted from the quotient . the remainder left is half of num__66 . what is the number ? <o> a ) num__21 <o> b ) num__20 <o> c ) num__22 <o> d ) num__30 <o> e ) num__45 |
let number is x . when num__5 added to it = ( x + num__5 ) num__7 multiplied to sum = num__7 * ( x + num__5 ) now = [ { num__7 * ( x + num__5 ) } / num__5 ] and = [ { num__7 * ( x + num__5 ) } / num__5 ] - num__5 according to question [ { num__7 * ( x + num__10 ) } / num__5 ] - num__5 = half of num__66 [ ( num__7 x + num__35 ) / num__5 ) = num__33 + num__5 num__7 x + num__35 = num__38 * num__5 x = num__22.1428571429 x = num__22.14 = num__22 so required number is : num__22 . answer : c <eor> c <eos> |
c |
multiply__5.0__7.0__ add__5.0__33.0__ round_down__22.1429__ round_down__22.1429__ |
multiply__5.0__7.0__ add__5.0__33.0__ round_down__22.1429__ round_down__22.1429__ |
| the greatest number of four digits which is divisible by num__15 num__25 num__40 and num__75 is : <o> a ) num__9000 <o> b ) num__9400 <o> c ) num__9600 <o> d ) num__9800 <o> e ) num__7500 |
explanation : greatest number of num__4 - digits is num__9999 . l . c . m . of num__15 num__25 num__40 and num__75 is num__600 . on dividing num__9999 by num__600 the remainder is num__399 . required number ( num__9999 - num__399 ) = num__9600 . answer is c <eor> c <eos> |
c |
multiply__15.0__40.0__ subtract__9999.0__399.0__ subtract__9999.0__399.0__ |
multiply__15.0__40.0__ subtract__9999.0__399.0__ subtract__9999.0__399.0__ |
| in an xy - coordinate plane a line is defined by y = kx + num__1 . if ( num__4 b ) ( a num__5 ) and ( a b + num__1 ) are three points on the line where a and b are unknown then k = ? <o> a ) num__0.5 <o> b ) num__1 <o> c ) num__0.75 <o> d ) num__2 <o> e ) num__2.5 |
b = num__4 k + num__1 . . . ( num__1 ) b + num__1 = ak + num__1 . . . ( num__2 ) num__5 = ak + num__1 . . . ( num__3 ) taking ( num__2 ) and ( num__3 ) num__5 = b + num__1 b = num__4 taking ( num__1 ) num__4 = num__4 k + num__1 k = num__0.75 answer : c <eor> c <eos> |
c |
add__1.0__2.0__ divide__3.0__4.0__ multiply__1.0__0.75__ |
add__1.0__2.0__ divide__3.0__4.0__ multiply__1.0__0.75__ |
| a searchlight on top of the watchtower makes num__2 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least num__6 seconds ? <o> a ) num__0.833333333333 <o> b ) num__0.8 <o> c ) num__0.75 <o> d ) num__0.666666666667 <o> e ) num__0.5 |
the searchlight completes one revolution every num__30 seconds . the probability that the man ' s area will be lit up is num__0.2 = num__0.2 . the probability that he will stay in the dark is num__1 - num__0.2 = num__0.8 the answer is b . <eor> b <eos> |
b |
divide__6.0__30.0__ subtract__1.0__0.2__ round__0.8__ |
divide__6.0__30.0__ subtract__1.0__0.2__ subtract__1.0__0.2__ |
| what is the probability of always having x + y > num__0 where y < num__0 <o> a ) num__0.5 <o> b ) num__0.125 <o> c ) num__0.75 <o> d ) num__0.25 <o> e ) none |
total value of x can be : num__1 . x < num__0 ; num__2 . x = num__0 ; num__3 . x > num__0 but less than y in magnitude num__4 . x > num__0 as well as | y | so probability of having x + y > num__0 would be num__0.25 answer : d <eor> d <eos> |
d |
add__1.0__2.0__ add__1.0__3.0__ reverse__4.0__ reverse__4.0__ |
add__1.0__2.0__ add__1.0__3.0__ reverse__4.0__ reverse__4.0__ |
| in a game of billiards a can give b num__20 points in num__60 and he can give c num__30 points in num__60 . how many points can b give c in a game of num__100 ? <o> a ) num__43 <o> b ) num__54 <o> c ) num__25 <o> d ) num__86 <o> e ) num__23 |
a scores num__60 while b score num__40 and c scores num__30 . the number of points that c scores when b scores num__100 = ( num__100 * num__30 ) / num__40 = num__25 * num__3 = num__75 . in a game of num__100 points b gives ( num__100 - num__75 ) = num__25 points to c . answer : c <eor> c <eos> |
c |
subtract__60.0__20.0__ divide__60.0__20.0__ subtract__100.0__25.0__ subtract__100.0__75.0__ |
subtract__60.0__20.0__ divide__60.0__20.0__ subtract__100.0__25.0__ subtract__100.0__75.0__ |
| how many positive integers will divide evenly into num__310 ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__8 <o> d ) num__12 <o> e ) num__16 |
the question is asking how many factors num__310 has . num__310 = num__2 * num__5 * num__31 the number of factors is num__2 ^ num__3 = num__8 the answer is c . <eor> c <eos> |
c |
subtract__5.0__2.0__ add__3.0__5.0__ add__3.0__5.0__ |
subtract__5.0__2.0__ add__3.0__5.0__ add__3.0__5.0__ |
| num__3 num__712 . . . . . . . . . . . . . . num__4 th terms <o> a ) num__43 <o> b ) num__53 <o> c ) num__18 <o> d ) num__65 <o> e ) num__78 |
num__3 + num__4 = num__7 num__7 + num__5 = num__12 num__12 + num__6 = num__18 answer : c <eor> c <eos> |
c |
add__3.0__4.0__ multiply__3.0__4.0__ multiply__3.0__6.0__ multiply__3.0__6.0__ |
add__3.0__4.0__ add__5.0__7.0__ add__6.0__12.0__ add__6.0__12.0__ |
| divide rs . num__1700 among a b and c so that a receives num__0.333333333333 as much as b and c together and b receives num__0.666666666667 as a and c together . a ' s share is ? <o> a ) num__346 <o> b ) num__425 <o> c ) num__375 <o> d ) num__337 <o> e ) num__639 |
a + b + c = num__1700 a = num__0.333333333333 ( b + c ) ; b = num__0.666666666667 ( a + c ) a / ( b + c ) = num__0.333333333333 a = num__0.25 * num__1700 = > num__425 answer : b <eor> b <eos> |
b |
multiply__1700.0__0.25__ multiply__1700.0__0.25__ |
multiply__1700.0__0.25__ multiply__1700.0__0.25__ |
| jim bought edging to go around a circular garden with a radius of num__10 feet . later he decided to double the diameter of the garden . how many more feet of edging must he buy ? <o> a ) num__28.84 ' additional edging <o> b ) num__48.84 ' additional edging <o> c ) num__38.84 ' additional edging <o> d ) num__62.80 ' additional edging <o> e ) num__58.84 ' additional edging |
circumference of small garden = num__2 x num__3.14 x num__10 = num__62.80 ' double of the circumference of small garden = num__2 x num__62.80 ' = num__125.60 ' more feet to be buy = num__125.60 - num__62.80 = num__62.80 answer : d <eor> d <eos> |
d |
volume_rectangular_prism__10.0__2.0__3.14__ multiply__62.8__2.0__ triangle_area__62.8__2.0__ |
volume_rectangular_prism__10.0__2.0__3.14__ multiply__62.8__2.0__ triangle_area__62.8__2.0__ |
| the sum of the present ages of two persons a and b is num__90 . if the age of a is twice that of b find the sum of their ages num__5 years hence ? <o> a ) num__50 <o> b ) num__60 <o> c ) num__70 <o> d ) num__80 <o> e ) num__100 |
a + b = num__90 a = num__2 b num__2 b + b = num__90 = > b = num__30 then a = num__60 . num__5 years their ages will be num__65 and num__35 . sum of their ages = num__65 + num__35 = num__100 . answer : e <eor> e <eos> |
e |
subtract__90.0__30.0__ add__5.0__60.0__ add__5.0__30.0__ add__65.0__35.0__ add__65.0__35.0__ |
subtract__90.0__30.0__ add__5.0__60.0__ add__5.0__30.0__ add__65.0__35.0__ add__65.0__35.0__ |
| the area of circle o is added to its diameter . if the circumference of circle o is then subtracted from this total the result is num__10 . what is the radius of circle o ? <o> a ) – num__2 / pi <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
pi * r ^ num__2 + num__2 r - num__2 * pi * r = num__10 simplifying the equation : pi * r ( r - num__2 ) + num__2 r = num__10 without much algebraic : we can test the answers quickly then num__5 is the only possible answer that will eliminate pi from equation . answer is e <eor> e <eos> |
e |
triangle_area__2.0__5.0__ |
triangle_area__2.0__5.0__ |
| how many pieces of num__85 cm length can be cut from a rod of num__29.75 meters long ? <o> a ) num__55 <o> b ) num__45 <o> c ) num__35 <o> d ) num__25 <o> e ) num__15 |
number of pieces = num__35.0 = num__35 the answer is c . <eor> c <eos> |
c |
round__35.0__ |
round__35.0__ |
| num__26 an uneducated retailar marks all its goods at num__50.0 above the cost price and thinking that he will still make num__25.0 profit offers a discount of num__25.0 on the marked price . what is the actual profit on the sales ? <o> a ) num__5.0 <o> b ) num__10.5 <o> c ) num__9.5 <o> d ) num__12.5 <o> e ) num__10 % |
let c . p = rs num__100 . then marked price = rs num__100 s . p = num__75.0 of rs num__150 = rs num__112.50 hence gain % = num__12.5 answer is d . <eor> d <eos> |
d |
percent__75.0__150.0__ percent__50.0__25.0__ percent__50.0__25.0__ |
percent__75.0__150.0__ percent__50.0__25.0__ percent__50.0__25.0__ |
| the length of a room is num__5.5 m and width is num__3.75 m . what is the cost of paying the floor by slabs at the rate of rs . num__800 per sq . meter . <o> a ) rs . num__12000 <o> b ) rs . num__19500 <o> c ) rs . num__18000 <o> d ) rs . num__16500 <o> e ) rs . num__17500 |
explanation : area = num__5.5 × num__3.75 sq . metre . cost for num__11 sq . metre . = rs . num__800 hence total cost = num__5.5 × num__3.75 × num__800 = num__5.5 × num__3000 = rs . num__16500 answer : option d <eor> d <eos> |
d |
multiply__3.75__800.0__ multiply__5.5__3000.0__ round__16500.0__ |
multiply__3.75__800.0__ multiply__5.5__3000.0__ round__16500.0__ |
| in six successive games a baseball team scored num__1 runs once num__4 runs twice and num__5 runs three times . what was the average ( arithmetic mean ) number of runs the team scored per game over the six - game period ? <o> a ) num__8 <o> b ) num__7 <o> c ) num__6 <o> d ) num__5 <o> e ) num__4 |
num__1 runs once ( one game ) num__4 runs twice ( two game ) num__5 runs three times ( three games ) so we have ( num__1 * num__1 + num__4 * num__2 + num__5 * num__3 ) / num__6 = num__4.0 = num__4 . ans : e . <eor> e <eos> |
e |
add__1.0__2.0__ add__1.0__5.0__ multiply__1.0__4.0__ |
add__1.0__2.0__ add__1.0__5.0__ multiply__1.0__4.0__ |
| your town has a population of num__1260 . every three people have a car . how many cars are in your town ? <o> a ) num__400 <o> b ) num__410 <o> c ) num__420 <o> d ) num__430 <o> e ) num__450 |
every three person has a car then the number of cars in the town could be calculated by the division of number of people over the person - to - car ratio . the number of cars in the town = num__1260 / ( num__3.0 ) = num__420.0 = num__420 cars so the correct choice is c <eor> c <eos> |
c |
divide__1260.0__3.0__ divide__1260.0__3.0__ |
divide__1260.0__3.0__ divide__1260.0__3.0__ |
| tough and tricky questions : work / rate problems . a group of num__4 junior lawyers require num__7 hours to complete a legal research assignment . how many hours would it take a group of three legal assistants to complete the same research assignment assuming that a legal assistant works at two - thirds the rate of a junior lawyer ? source : chili hot gmat <o> a ) num__14 <o> b ) num__10 <o> c ) num__9 <o> d ) num__6 <o> e ) num__5 |
# of people times the # of hours : num__4 * num__7 = num__28 - - > num__4 lawyers do num__28 worksin num__7 hours . num__3 * num__4.66666666667 = num__14 - - > num__3 assistants do num__14 worksin num__4 hours so since the amount of work the assistants do is half the work the lawyers do the time will be double soans a <eor> a <eos> |
a |
multiply__4.0__7.0__ subtract__7.0__4.0__ subtract__28.0__14.0__ |
multiply__4.0__7.0__ subtract__7.0__4.0__ subtract__28.0__14.0__ |
| the mass of num__1 cubic meter of a substance is num__500 kg under certain conditions . what is the volume in cubic centimeters of num__1 gram of this substance under these conditions ? ( num__1 kg = num__1000 grams and num__1 cubic meter = num__1 num__000000 cubic centimeters ) <o> a ) num__0.1 <o> b ) num__0 <o> c ) num__1 <o> d ) num__2 <o> e ) num__3 |
num__500 kg - num__1 cubic meter ; num__500000 g - num__1 cubic meter ; num__500000 g - num__1 num__000000 cubic centimeters ; num__1 g - num__1 num__000000 / num__500000 = num__2.0 = num__2 cubic centimeters . answer : d . <eor> d <eos> |
d |
multiply__500.0__1000.0__ rectangle_perimeter__1.0__0.0__ rectangle_perimeter__1.0__0.0__ |
multiply__500.0__1000.0__ rectangle_perimeter__1.0__0.0__ rectangle_perimeter__1.0__0.0__ |
| for a road work a contractor was given a deadline to complete the work in num__25 days . hence he employed a certain number of workers to accomplish the job . but from the very first day num__9 workers were absent and the remaining workers could finish the work in num__40 days . what number of workers did the contractor hired in the beginning ? <o> a ) num__24 <o> b ) num__37 <o> c ) num__26 <o> d ) num__19 <o> e ) num__17 |
explanation : let the number of workers initially hired be x . hence x workers can complete the work in num__25 days and ( x - num__9 ) workers can complete the work in num__40 days . now the number of men and the amount of work are in direct proportion and number of men and number of days in indirect proportion . x * num__25 = ( x – num__9 ) * num__40 num__25 x = num__40 x – num__360 num__15 x = num__360 x = num__24 answer : a <eor> a <eos> |
a |
multiply__9.0__40.0__ subtract__40.0__25.0__ add__9.0__15.0__ round__24.0__ |
multiply__9.0__40.0__ subtract__40.0__25.0__ add__9.0__15.0__ round__24.0__ |
| two taps can separately fill a cistern num__20 minutes and num__20 minutes respectively and when the waste pipe is open they can together fill it in num__18 minutes . the waste pipe can empty the full cistern in ? <o> a ) num__22.8 <o> b ) num__22.7 <o> c ) num__22.6 <o> d ) num__22.9 <o> e ) num__22.5 |
num__0.05 + num__0.05 - num__1 / x = num__0.0555555555556 x = num__22.5 answer : e <eor> e <eos> |
e |
multiply__20.0__0.05__ divide__1.0__18.0__ round__22.5__ |
multiply__20.0__0.05__ divide__1.0__18.0__ divide__22.5__1.0__ |
| anna and carol buy cds and tapes at a music store that sells each of its cds for a certain price and each of its tapes for a certain price . anna spends twice as much as carol spends buying three times as many cds and the same number of tapes . if carol spends $ num__50.00 on four cds and five tapes how much does one tape cost ? <o> a ) $ num__5.00 <o> b ) $ num__6.25 <o> c ) $ num__12.00 <o> d ) $ num__25.00 <o> e ) $ num__100.00 |
cd = > c type = > t carol : num__4 c + num__5 t = num__50 anna : num__12 c + num__5 t = num__100 num__12 c - num__4 c = num__50 num__8 c = num__50 c = num__6.25 num__4 c + num__5 t = num__50 num__25 + num__5 t = num__50 num__5 t = num__25 t = num__5 answer is a <eor> a <eos> |
a |
subtract__12.0__4.0__ divide__50.0__8.0__ multiply__6.25__4.0__ divide__25.0__5.0__ |
subtract__12.0__4.0__ divide__50.0__8.0__ multiply__6.25__4.0__ divide__25.0__5.0__ |
| if a / x = num__0.333333333333 and a / y = num__0.166666666667 then ( x + y ) = <o> a ) a ) num__2 a <o> b ) b ) a / num__2 <o> c ) c ) num__6 a <o> d ) d ) num__7 a <o> e ) e ) num__9 a |
ratio num__1 : num__3 a = x ratio num__2 : num__6 a = y x + y = num__9 a answer is e <eor> e <eos> |
e |
subtract__3.0__1.0__ multiply__2.0__3.0__ add__3.0__6.0__ multiply__1.0__9.0__ |
subtract__3.0__1.0__ multiply__2.0__3.0__ add__3.0__6.0__ add__3.0__6.0__ |
| jack has two dice one has six equally probable sides labeled num__1 num__2 num__3 num__4 num__5 num__6 and the other has seven equally probable sides labeled num__1 num__2 num__3 num__4 num__5 num__6 num__7 . if jack rolls both dice what is the probability that both of the numbers will be num__7 ? <o> a ) num__0.214285714286 <o> b ) num__0.142857142857 <o> c ) num__0 <o> d ) num__0.5 <o> e ) num__0.571428571429 |
method - num__1 probability that the number on first die is num__7 = num__0 probability that the number on second die is num__7 = num__0.142857142857 [ because num__1 out of num__7 faces is num__7 ] probability that both dice result in odd numbers = ( num__0 ) * ( num__0.142857142857 ) = num__0 answer : option c <eor> c <eos> |
c |
reverse__7.0__ round_down__0.1429__ |
reverse__7.0__ multiply__1.0__0.0__ |
| rajan travelled for num__12 hours . he covered the first half of the distance at num__60 kmph and remaining half of the distance at num__40 kmph . find the distance travelled by rajan ? <o> a ) num__376 <o> b ) num__576 <o> c ) num__476 <o> d ) num__566 <o> e ) num__676 |
let the distance travelled be x km . total time = ( x / num__2 ) / num__60 + ( x / num__2 ) / num__40 = num__12 = > x / num__120 + x / num__80 = num__12 = > ( num__2 x + num__3 x ) / num__240 = num__12 = > x = num__576 km answer : b <eor> b <eos> |
b |
multiply__60.0__2.0__ multiply__40.0__2.0__ divide__120.0__40.0__ multiply__2.0__120.0__ round__576.0__ |
multiply__60.0__2.0__ multiply__40.0__2.0__ divide__120.0__40.0__ multiply__2.0__120.0__ round__576.0__ |
| num__2500 - ( num__1002 / num__20.04 ) = ? <o> a ) num__2984 <o> b ) num__2983 <o> c ) num__2982 <o> d ) num__2450 <o> e ) none of these |
num__2500 - num__50 = num__2450 answer : d <eor> d <eos> |
d |
divide__1002.0__20.04__ subtract__2500.0__50.0__ subtract__2500.0__50.0__ |
divide__1002.0__20.04__ subtract__2500.0__50.0__ subtract__2500.0__50.0__ |
| a num__270 metres long train running at the speed of num__120 kmph crosses another train running in opposite direction at the speed of num__80 kmph in num__9 seconds . what is the length of the other train ? <o> a ) num__230 m <o> b ) num__240 m <o> c ) num__260 m <o> d ) num__320 m <o> e ) none of these |
explanation : relative speed = ( num__120 + num__80 ) km / hr = ( num__200 x num__0.277777777778 ) m / sec = ( num__55.5555555556 ) m / sec let the length of the other train be x metres . then x + num__30.0 = num__55.5555555556 x + num__270 = num__500 x = num__230 . answer is a <eor> a <eos> |
a |
add__120.0__80.0__ divide__270.0__9.0__ add__200.0__30.0__ round__230.0__ |
add__120.0__80.0__ divide__270.0__9.0__ add__200.0__30.0__ add__200.0__30.0__ |
| zachary is helping his younger brother sterling learn his multiplication tables . for every question that sterling answers correctly zachary gives him num__3 pieces of candy . for every question that sterling answers incorrectly zachary takes away two pieces of candy . after num__7 questions if sterling had answered num__2 more questions correctly he would have earned num__31 pieces of candy . how many of the num__7 questions did zachary answer correctly ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
i got two equations : num__3 x - num__2 y = num__25 x + y = num__7 num__3 x - num__2 ( num__7 - x ) = num__25 num__3 x - num__14 + num__2 x = num__25 num__5 x = num__39 x = num__7.8 or between num__7 and num__8 . ( ans b ) <eor> b <eos> |
b |
multiply__7.0__2.0__ add__3.0__2.0__ add__14.0__25.0__ divide__39.0__5.0__ add__3.0__5.0__ round_down__7.8__ |
multiply__7.0__2.0__ add__3.0__2.0__ add__14.0__25.0__ divide__39.0__5.0__ add__3.0__5.0__ add__2.0__5.0__ |
| for each color copy print shop x charges $ num__1.20 and print shop y charges $ num__1.70 . how much greater is the charge for num__70 color copies at print shop y than at print shop x ? <o> a ) $ num__27 <o> b ) $ num__29 <o> c ) $ num__31 <o> d ) $ num__33 <o> e ) $ num__35 |
the difference in the two prices is $ num__1.70 - $ num__1.20 = $ num__0.50 for each color copy . each color copy will cost an extra $ num__0.50 at print shop y . num__70 * $ num__0.50 = $ num__35 the answer is e . <eor> e <eos> |
e |
subtract__1.7__1.2__ multiply__70.0__0.5__ multiply__70.0__0.5__ |
subtract__1.7__1.2__ multiply__70.0__0.5__ multiply__70.0__0.5__ |
| if rs . num__10 be allowed as true discount on bill of rs . num__110 due at the end of a certain time then the discount allowedon the same sum due at the end of double the time is : <o> a ) num__16 <o> b ) num__18.33 <o> c ) num__20.21 <o> d ) num__26 <o> e ) none |
sol . s . i . on rs . ( num__110 - num__10 ) for a certian time = rs . num__10 . s . i . on rs . num__100 for double the time = rs . num__20 . t . d . on rs . num__120 = rs . ( num__120 - num__100 ) = rs . num__20 t . d . on rs . num__110 = rs . [ num__0.166666666667 * num__110 ] = rs . num__18.33 answer b <eor> b <eos> |
b |
percent__100.0__18.33__ |
percent__100.0__18.33__ |
| find the least number when successively divided by num__2 num__3 and num__7 leaves remainders num__1 num__2 and num__3 respectively ? <o> a ) num__22 <o> b ) num__65 <o> c ) num__28 <o> d ) num__27 <o> e ) num__19 |
explanation : num__2 ) num__65 ( num__32 num__3 ) num__32 ( num__10 num__7 ) num__10 ( num__1 num__64 num__30 num__7 - - - - - - - - - - - - - - - - - - - - - - num__1 num__2 num__3 = > num__65 answer : b <eor> b <eos> |
b |
add__3.0__7.0__ multiply__2.0__32.0__ multiply__3.0__10.0__ multiply__1.0__65.0__ |
add__3.0__7.0__ subtract__65.0__1.0__ subtract__32.0__2.0__ multiply__1.0__65.0__ |
| carl can wash all the windows of his house in num__8 hours . his wife maggie can wash all the windows in num__4 hours . how many hours will it take for both of them working together to wash all the windows ? <o> a ) num__2 <o> b ) num__2 num__0.25 <o> c ) num__3 num__1.0 <o> d ) num__4 num__0.5 <o> e ) num__5 |
work hrs = ab / ( a + b ) = num__2.66666666667 = num__3 num__1.0 answer is c <eor> c <eos> |
c |
divide__8.0__2.6667__ subtract__4.0__3.0__ round__3.0__ |
divide__8.0__2.6667__ subtract__4.0__3.0__ divide__8.0__2.6667__ |
| a train num__110 metres long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__5 sec <o> b ) num__6 sec <o> c ) num__7 sec <o> d ) num__10 sec <o> e ) none |
solution speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr = ( num__66 x num__0.277777777778 ) m / sec = ( num__18.3333333333 ) m / sec ∴ time taken to pass the man = ( num__110 x num__0.0545454545455 ) sec = num__6 sec answer b <eor> b <eos> |
b |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ round__6.0__ |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ divide__110.0__18.3333__ |
| jeevan bought an article with num__30 per cent discount on the labelled price . he sold the article with num__12 per cent profit on the labelled price . what was his per cent profit on the price he bought ? <o> a ) num__40 <o> b ) num__50 <o> c ) num__60 <o> d ) data inadequate <o> e ) none of these |
let the labelled price of the article = ₹ num__100 then cp = ₹ num__70 and sp = ₹ num__112 . ∴ read profit percent = num__112 − num__1.0 × num__100 = num__6.0 × num__10 = num__60 answer c <eor> c <eos> |
c |
percent__100.0__60.0__ |
percent__100.0__60.0__ |
| find the area of trapezium whose parallel sides are num__20 cm and num__18 cm long and the distance between them is num__15 cm <o> a ) num__178 cm num__2 <o> b ) num__179 cm num__2 <o> c ) num__285 cm num__2 <o> d ) num__167 cm num__2 <o> e ) num__197 cm num__2 |
area of a trapezium = num__0.5 ( sum of parallel sides ) * ( perpendicular distance between them ) = num__0.5 ( num__20 + num__18 ) * ( num__15 ) = num__285 cm num__2 answer : c <eor> c <eos> |
c |
subtract__20.0__18.0__ round__285.0__ |
subtract__20.0__18.0__ round__285.0__ |
| what least number must be added to num__1056 so that the sum is completely divisible by num__23 ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
num__23 ) num__1056 ( num__45 num__92 - - - num__136 num__115 - - - num__21 - - - required number = ( num__23 - num__21 ) = num__2 . a ) <eor> a <eos> |
a |
add__23.0__92.0__ subtract__136.0__115.0__ subtract__23.0__21.0__ subtract__23.0__21.0__ |
add__23.0__92.0__ subtract__136.0__115.0__ subtract__23.0__21.0__ subtract__23.0__21.0__ |
| each week a restaurant serving mexican food uses the same volume of chili paste which comes in either num__25 - ounce cans or num__15 - ounce cans of chili paste . if the restaurant must order num__40 more of the smaller cans than the larger cans to fulfill its weekly needs then how many larger cans are required to fulfill its weekly needs ? <o> a ) num__40 <o> b ) num__30 <o> c ) num__60 <o> d ) num__45 <o> e ) num__55 |
let x be the number of num__25 ounce cans . therefore ( x + num__40 ) is the number of num__15 ounce cans . total volume is same therefore num__25 x = num__15 ( x + num__40 ) num__10 x = num__600 x = num__60 ans is c <eor> c <eos> |
c |
subtract__25.0__15.0__ multiply__15.0__40.0__ divide__600.0__10.0__ divide__600.0__10.0__ |
subtract__25.0__15.0__ multiply__15.0__40.0__ divide__600.0__10.0__ divide__600.0__10.0__ |
| pascal has num__96 miles remaining to complete his cycling trip . if he reduced his current speed by num__4 miles per hour the remainder of the trip would take him num__16 hours longer than it would if he increased his speed by num__50.0 . what is his current speed x ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__12 <o> e ) num__16 |
let the current speed be x miles per hour . time taken if speed is num__50.0 faster ( i . e . num__3 x / num__2 = num__1.5 x ) = num__96 / num__1.5 x time taken if speed is reduced by num__4 miles / hr ( i . e . ( x - num__4 ) ) = num__96 / ( x - num__4 ) as per question num__96 / ( x - num__4 ) - num__96 / num__1.5 x = num__16 solving this x we get x = num__8 . b . <eor> b <eos> |
b |
divide__3.0__2.0__ multiply__4.0__2.0__ round__8.0__ |
divide__3.0__2.0__ divide__16.0__2.0__ divide__16.0__2.0__ |
| an amount of money lent against simple interest at num__4 num__0.5 % per annum amounts to rs . num__381 after num__6 years . find the sum . <o> a ) num__100 <o> b ) num__200 <o> c ) num__300 <o> d ) num__400 <o> e ) num__500 |
given si + p = num__381 si = num__381 - p si = ptr / num__100 num__381 - p = p * num__6 * num__4.5 / num__100 p = num__300 answer : c <eor> c <eos> |
c |
percent__100.0__300.0__ |
percent__100.0__300.0__ |
| the perimeter of a square is equal to the perimeter of a rectangle of length num__19 cm and breadth num__14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) <o> a ) num__22.78 <o> b ) num__23.54 <o> c ) num__23.5 <o> d ) num__24.55 <o> e ) num__25.93 |
let the side of the square be a cm . perimeter of the rectangle = num__2 ( num__19 + num__14 ) = num__66 cm perimeter of the square = num__66 cm i . e . num__4 a = num__66 a = num__16.5 diameter of the semicircle = num__16.5 cm circumference of the semicircle = num__0.5 ( ∏ ) ( num__16.5 ) = num__0.5 ( num__3.14285714286 ) ( num__16.5 ) = num__25.93 cm to two decimal places answer : e <eor> e <eos> |
e |
rectangle_perimeter__19.0__14.0__ triangle_area__2.0__25.93__ |
rectangle_perimeter__19.0__14.0__ triangle_area__2.0__25.93__ |
| amar takes as much time in running num__24 meters as a car takes in covering num__60 meters . what will be the distance covered by amar during the time the car covers num__2.2 km ? <o> a ) num__700 m <o> b ) num__500 m <o> c ) num__870 m <o> d ) num__880 m <o> e ) num__840 m |
distance covered by amar = num__0.4 ( num__2.2 km ) = num__0.4 ( num__2200 ) = num__880 m answer : d <eor> d <eos> |
d |
divide__24.0__60.0__ multiply__2200.0__0.4__ round__880.0__ |
divide__24.0__60.0__ multiply__2200.0__0.4__ round__880.0__ |
| in a bag there are num__2400 ball and their colors are red green blue . . the ratio of the balls are num__15 : num__13 : num__17 . then how many red color balls are available in the bag ? <o> a ) num__900 <o> b ) num__1600 <o> c ) num__750 <o> d ) num__890 <o> e ) num__1010 |
red : green : blue = num__15 + num__13 + num__17 = num__45 ; ratio of the red balls = num__0.375 simplify = num__0.375 * num__2400 = num__900 . answer = a <eor> a <eos> |
a |
multiply__2400.0__0.375__ multiply__2400.0__0.375__ |
multiply__2400.0__0.375__ multiply__2400.0__0.375__ |
| there is a total of num__120 marbles in a box each of which is red green blue or white . if one marble is drawn from the box at random the probability that it will be white is num__0.25 and the probability that it will be green is num__0.333333333333 . what is the probability that the marble will be either red or blue ? <o> a ) num__0.166666666667 <o> b ) num__0.25 <o> c ) num__0.285714285714 <o> d ) num__0.333333333333 <o> e ) num__0.416666666667 |
total marbles in the box = num__120 white marbles = num__30.0 = num__30 green marbles = num__40.0 = num__40 w + g = num__70 red + blue = num__50 p ( red or blue ) = num__0.416666666667 = num__0.416666666667 answer : e <eor> e <eos> |
e |
multiply__120.0__0.25__ add__40.0__30.0__ subtract__120.0__70.0__ divide__50.0__120.0__ divide__50.0__120.0__ |
multiply__120.0__0.25__ add__40.0__30.0__ subtract__120.0__70.0__ divide__50.0__120.0__ divide__50.0__120.0__ |
| ramu bought an old car for rs . num__42000 . he spent rs . num__15000 on repairs and sold it for rs . num__64900 . what is his profit percent ? <o> a ) num__12.0 <o> b ) num__13.85 <o> c ) num__18.0 <o> d ) num__82.0 <o> e ) num__23 % |
total cp = rs . num__42000 + rs . num__15000 = rs . num__57000 and sp = rs . num__64900 profit ( % ) = ( num__64900 - num__57000 ) / num__57000 * num__100 = num__13.85 answer : b <eor> b <eos> |
b |
percent__13.85__100.0__ |
percent__13.85__100.0__ |
| how many litres of pure acid are there in num__10 litres of a num__25.0 solution <o> a ) num__1.5 <o> b ) num__1.6 <o> c ) num__2.5 <o> d ) num__1.8 <o> e ) num__1.9 |
explanation : question of this type looks a bit typical but it is too simple as below . . . it will be num__10 * num__0.25 = num__2.5 answer : option c <eor> c <eos> |
c |
percent__10.0__25.0__ percent__10.0__25.0__ |
percent__10.0__25.0__ percent__10.0__25.0__ |
| last year bill ' s farm produced num__768000 oranges . if bill distributed his apples evenly across num__9 different retail shops how many apples remained with him . <o> a ) num__1 orange <o> b ) num__2 oranges <o> c ) num__3 oranges <o> d ) num__4 oranges <o> e ) num__5 oranges |
this is equivalent to finding the first number that is divisible by num__9 that occurs before num__768000 oranges . in order to divide the sum in num__9 parts the amount must be divisible by num__9 divisibility rule of num__9 : the sum of the digits must be divisible by num__9 sum of digits of num__768000 = num__21 and num__18 is divisible by num__9 . hence we need to remove num__3 to this number for it to be divisible by num__9 correct option : c <eor> c <eos> |
c |
subtract__21.0__18.0__ divide__9.0__3.0__ |
subtract__21.0__18.0__ divide__9.0__3.0__ |
| sara ' s mother is num__6 times older than sara . next year however she will be only num__5 times older than her daughter . how old is sara ' s mother . <o> a ) num__24 years <o> b ) num__25 years <o> c ) num__26 years <o> d ) num__27 years <o> e ) num__28 years |
at the present sara ' s age is x . sara ' s mother ' s age is y = num__6 x . next year sara ' s age will be x + num__1 . sara ' s mother ' s age will be y + num__1 = num__5 ( x + num__1 ) . from the two equations we can calculate sara ' s age : y = num__6 x y + num__1 = num__5 ( x + num__1 ) num__6 x + num__1 = num__5 x + num__5 num__6 x - num__5 x = num__5 - num__1 x = num__4 if sara is currently num__4 years old and her mother is num__6 times older then her age is num__6 x = num__6 × num__4 = num__24 . answer : a <eor> a <eos> |
a |
subtract__6.0__5.0__ subtract__5.0__1.0__ multiply__6.0__4.0__ multiply__6.0__4.0__ |
subtract__6.0__5.0__ subtract__5.0__1.0__ multiply__6.0__4.0__ multiply__6.0__4.0__ |
| num__48 persons can repair a road in num__12 days working num__5 hours a day . in how many days will num__30 persons working num__6 hours a day complete the work ? <o> a ) num__10 <o> b ) num__16 <o> c ) num__13 <o> d ) num__18 <o> e ) num__19 |
let the required number of days be x . less persons more days ( indirect proportion ) more working hours per day less days ( indirect proportion ) persons num__30 : num__48 : : num__12 : x working hours / day num__6 : num__5 num__30 x num__6 x x = num__48 x num__5 x num__12 x = ( num__48 x num__5 x num__12 ) / ( num__30 x num__6 ) x = num__16 answer b <eor> b <eos> |
b |
round__16.0__ |
round__16.0__ |
| find the area of the quadrilateral of one of its diagonals is num__20 cm and its off sets num__9 cm and num__6 cm ? <o> a ) num__328 <o> b ) num__150 <o> c ) num__279 <o> d ) num__277 <o> e ) num__267 |
num__0.5 * num__20 ( num__9 + num__6 ) = num__150 cm num__2 answer : b <eor> b <eos> |
b |
square_perimeter__0.5__ triangle_area__2.0__150.0__ |
square_perimeter__0.5__ volume_rectangular_prism__0.5__2.0__150.0__ |
| a certain musical scale has has num__13 notes each having a different frequency measured in cycles per second . in the scale the notes are ordered by increasing frequency and the highest frequency is twice the lowest . for each of the num__12 lower frequencies the ratio of a frequency to the next higher frequency is a fixed constant . if the lowest frequency is num__330 cycles per second then the frequency of the num__7 th note in the scale is how many cycles per second ? <o> a ) num__330 * sqrt num__2 <o> b ) num__330 * sqrt ( num__2 ^ num__7 ) <o> c ) num__330 * sqrt ( num__2 ^ num__12 ) <o> d ) num__330 * the twelfth root of ( num__2 ^ num__7 ) <o> e ) num__330 * the seventh root of ( num__2 ^ num__12 ) |
let the constant be k . f num__1 = num__330 f num__2 = num__330 k f num__3 = num__330 k * k = num__330 * k ^ num__2 f num__13 = num__330 * k ^ num__12 we know f num__13 = num__2 * f num__1 = num__2 * num__330 = num__660 num__2.0 = k ^ num__12 k = twelfth root of num__2 for f num__7 . . . f num__7 = num__330 * k ^ num__6 ( as we wrote for f num__2 and f num__3 ) f num__7 = num__330 * ( twelfth root of num__2 ) ^ num__6 f num__7 = num__330 * sqrt ( num__2 ) the answer is a . <eor> a <eos> |
a |
subtract__13.0__12.0__ add__1.0__2.0__ multiply__330.0__2.0__ subtract__13.0__7.0__ multiply__330.0__1.0__ |
subtract__13.0__12.0__ add__1.0__2.0__ multiply__330.0__2.0__ multiply__2.0__3.0__ multiply__330.0__1.0__ |
| find the average of first num__50 natural numbers . <o> a ) a ) num__25.5 <o> b ) b ) num__20.5 <o> c ) c ) num__24.5 <o> d ) d ) num__19.9 <o> e ) e ) num__25 |
sum of first n natural numbers = n ( n + num__1 ) / num__2 sum of first num__50 natural numbers = num__50 * num__25.5 = num__1275 average = num__25.5 = num__25.5 answer is a <eor> a <eos> |
a |
multiply__50.0__25.5__ multiply__25.5__1.0__ |
multiply__50.0__25.5__ multiply__25.5__1.0__ |
| if x dollars is invested at num__9 percent for one year and y dollars is invested at num__8 percent for one year the annual income from the num__9 percent investment will exceed the annual income from the num__8 percent investment by $ num__48 . if $ num__2000 is the total amount invested how much is invested at num__9 percent ? <o> a ) num__300 <o> b ) num__250 <o> c ) num__1223.53 <o> d ) num__1133.24 <o> e ) num__776.47 |
num__2 equations with num__2 unknowns num__9 x / num__100 - num__8 y / num__100 = num__48 and x + y = num__2000 solving these num__2 equations x = num__1223.53 and y = num__776.47 answer - c <eor> c <eos> |
c |
subtract__2000.0__1223.53__ subtract__2000.0__776.47__ |
subtract__2000.0__1223.53__ subtract__2000.0__776.47__ |
| the parameter of a square is double the perimeter of a rectangle . the area of the rectangle is num__480 sq cm . find the area of the square ? <o> a ) num__288 <o> b ) num__270 <o> c ) num__289 <o> d ) num__480 <o> e ) num__211 |
let the side of the square be a cm . let the length and the breadth of the rectangle be l cm and b cm respectively . num__4 a = num__2 ( l + b ) num__2 a = l + b l . b = num__480 we can not find ( l + b ) only with the help of l . b . therefore a can not be found . area of the square can not be found . answer : d <eor> d <eos> |
d |
triangle_area__480.0__2.0__ |
triangle_area__480.0__2.0__ |
| a dishonest shopkeeper professes to sell pulses at the cost price but he uses a false weight of num__970 gm . for a kg . his gain is … % . <o> a ) num__3.09 <o> b ) num__5.36 <o> c ) num__4.26 <o> d ) num__6.26 <o> e ) num__7.26 % |
his percentage gain is num__100 * num__0.0309278350515 as he is gaining num__30 units for his purchase of num__970 units . so num__3.09 . answer : a <eor> a <eos> |
a |
percent__100.0__3.09__ |
percent__100.0__3.09__ |
| arun and tarun can do a work in num__10 days . after num__4 days tarun went to his village . how many days are required to complete the remaining work by arun alone . arun can do the work alone in num__30 days . <o> a ) num__16 days . <o> b ) num__17 days . <o> c ) num__18 days . <o> d ) num__19 days . <o> e ) num__20 days . |
they together completed num__0.4 work in num__4 days . balance num__0.6 work will be completed by arun alone in num__30 * num__0.6 = num__18 days . answer : c <eor> c <eos> |
c |
divide__4.0__10.0__ km_to_mile_conversion__ multiply__30.0__0.6__ round__18.0__ |
divide__4.0__10.0__ km_to_mile_conversion__ multiply__30.0__0.6__ multiply__30.0__0.6__ |
| two pipes p and q can fill a cistern in num__12 and num__15 minutes respectively . both are opened together but at the end of num__4 minutes the first is turned off . how many more minutes will it take for the cistern to fill after the first pipe is turned off ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
let x be the total time it takes for the cistern to fill . num__0.333333333333 + x / num__15 = num__1 x / num__15 = num__0.666666666667 x = num__10 after the first pipe is turned off it takes num__6 more minutes to fill the cistern . the answer is a . <eor> a <eos> |
a |
divide__4.0__12.0__ subtract__1.0__0.3333__ subtract__10.0__4.0__ round__6.0__ |
divide__4.0__12.0__ subtract__1.0__0.3333__ subtract__10.0__4.0__ divide__6.0__1.0__ |
| gaurav spends num__30.0 of his monthly income on food articles num__40.0 of the remaining on conveyance and clothes and saves num__50.0 of the remaining . if his monthly salary is rs . num__18400 how much money does he save every month ? <o> a ) num__3864 <o> b ) num__2338 <o> c ) num__3799 <o> d ) num__1267 <o> e ) num__2991 |
explanation : saving = num__50.0 of ( num__100 - num__40 ) % of ( num__100 - num__30 ) % of rs . num__18400 = rs . ( num__0.5 * num__0.6 * num__0.7 * num__18400 ) = rs . num__3864 . answer : a ) num__3864 <eor> a <eos> |
a |
percent__100.0__3864.0__ |
percent__100.0__3864.0__ |
| in a company of num__15 employees num__6 employees earn $ num__37000 num__5 employees earn $ num__42000 and the num__4 highest - paid employees earn the same amount . if the average annual salary for the num__15 employees is $ num__45000 what is the annual salary for each of the highest - paid employees ? <o> a ) $ num__56250 <o> b ) $ num__57750 <o> c ) $ num__59250 <o> d ) $ num__60750 <o> e ) $ num__62 |
250 |
num__6 * num__37000 + num__5 * num__42000 + num__4 x = num__15 * num__45000 num__4 x = num__675000 - num__222000 - num__210000 num__4 x = num__243000 x = num__60750 the answer is d . <eor> d <eos> |
d |
d |
| $ num__500 is divided amongst a b and c so that a may get num__0.666666666667 as much as b and c together b may get num__0.666666666667 as much as a and c together then the share of a is <o> a ) $ num__100 <o> b ) $ num__150 <o> c ) $ num__125 <o> d ) $ num__200 <o> e ) $ num__250 |
a : ( b + c ) = num__2 : num__3 a ' s share = num__500 * num__0.4 = $ num__200 answer is d <eor> d <eos> |
d |
multiply__500.0__0.4__ multiply__500.0__0.4__ |
multiply__500.0__0.4__ multiply__500.0__0.4__ |
| eight years ago p was half of q in age . if the ratio of their present ages is num__3 : num__4 what will be the total of their present ages <o> a ) a ) num__35 <o> b ) b ) num__28 <o> c ) c ) num__45 <o> d ) d ) num__25 <o> e ) e ) num__26 |
explanation : let the present age of p and q be num__3 x and num__4 x respectively . eight years ago p was half of q in age = > num__2 ( num__3 x – num__8 ) = ( num__4 x – num__8 ) = > num__6 x – num__16 = num__4 x – num__8 = > num__2 x = num__8 = > x = num__4 num__7 x = num__7 * num__4 = num__28 answer : option b <eor> b <eos> |
b |
multiply__4.0__2.0__ multiply__3.0__2.0__ multiply__2.0__8.0__ add__3.0__4.0__ multiply__4.0__7.0__ multiply__4.0__7.0__ |
multiply__4.0__2.0__ multiply__3.0__2.0__ multiply__2.0__8.0__ add__3.0__4.0__ multiply__4.0__7.0__ multiply__4.0__7.0__ |
| num__28 is divided into two parts in such a way that seventh part of first and ninth part of second are equal . find the smallest part ? <o> a ) num__13.25 <o> b ) num__38.25 <o> c ) num__33.25 <o> d ) num__12.25 <o> e ) num__31.25 |
x / num__7 = y / num__9 = > x : y = num__7 : num__9 num__0.4375 * num__28 = num__12.25 answer : d <eor> d <eos> |
d |
multiply__28.0__0.4375__ multiply__28.0__0.4375__ |
multiply__28.0__0.4375__ multiply__28.0__0.4375__ |
| a grocer has a sale of rs . num__3435 rs . num__3927 rs . num__3855 rs . num__4230 and rs . num__3562 for num__5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . num__3500 ? <o> a ) s . num__1991 <o> b ) s . num__2991 <o> c ) s . num__3991 <o> d ) s . num__4991 <o> e ) s . num__5991 |
explanation : total sale for num__5 months = rs . ( num__3435 + num__3927 + num__3855 + num__4230 + num__3562 ) = rs . num__19009 . required sale = rs . [ ( num__3500 x num__6 ) â € “ num__19009 ] = rs . ( num__21000 â € “ num__19009 ) = rs . num__1991 . answer a <eor> a <eos> |
a |
multiply__3500.0__6.0__ subtract__21000.0__19009.0__ subtract__21000.0__19009.0__ |
multiply__3500.0__6.0__ subtract__21000.0__19009.0__ subtract__21000.0__19009.0__ |
| a man buys an article and sells it at a profit of num__20.0 . if he had bought it at num__20.0 less and sold it for rs . num__74 less he could have gained num__25.0 . what is the cost price ? <o> a ) s . num__370 <o> b ) s . num__375 <o> c ) s . num__375 <o> d ) s . num__350 <o> e ) s . num__300 |
cp num__1 = num__100 sp num__1 = num__120 cp num__2 = num__80 sp num__2 = num__80 * ( num__1.25 ) = num__100 num__20 - - - - - num__100 num__74 - - - - - ? = > num__370 answer : a <eor> a <eos> |
a |
percent__100.0__370.0__ |
percent__100.0__370.0__ |
| ian is paid an hourly wage totaling $ num__400 for h hours of work in a week where h > num__0 . if ian ' s hourly wage increases by num__25.0 and ian decides to work num__25.0 fewer hours each week how much will ian be paid in a week ? <o> a ) $ num__200 <o> b ) $ num__375 <o> c ) $ num__400 <o> d ) $ num__425 <o> e ) $ num__600 |
let no of hours ian worked = h and wage per hour = w therefore w * h = num__400 if he decides to work num__25.0 fewer hours = num__0.75 * h and num__25.0 more age is paid = num__1.25 * w therefore num__1.25 * w * num__0.75 * h = num__1.25 * num__0.75 * num__400 = num__375 answer : b <eor> b <eos> |
b |
subtract__400.0__25.0__ subtract__400.0__25.0__ |
subtract__400.0__25.0__ subtract__400.0__25.0__ |
| a and b go around a circular track of length num__400 m on a cycle at speeds of num__36 kmph and num__36 kmph . after how much time will they meet for the first time at the starting point ? <o> a ) num__40 sec <o> b ) num__198 sec <o> c ) num__178 sec <o> d ) num__665 sec <o> e ) num__276 sec |
time taken to meet for the first time at the starting point = lcm { length of the track / speed of a length of the track / speed of b } = lcm { num__400 / ( num__36 * num__0.277777777778 ) num__400 / ( num__36 * num__0.277777777778 ) } = lcm ( num__40 num__40 ) = num__40 sec . answer : a <eor> a <eos> |
a |
round__40.0__ |
round__40.0__ |
| if num__821562 is to be divisible by num__5 what is the least whole number that should be added to it ? <o> a ) num__6 <o> b ) num__4 <o> c ) num__5 <o> d ) num__2 <o> e ) num__3 |
a number is divisible by num__8 if the number formed by the last three digits is divisible by num__8 . here num__821562 = num__562 the next multiple of num__9 is num__568 . num__6 must be added to num__821562 to make it divisible by num__8 a <eor> a <eos> |
a |
subtract__568.0__562.0__ subtract__568.0__562.0__ |
subtract__568.0__562.0__ subtract__568.0__562.0__ |
| if a : b = num__2 : num__3 and b : c = num__2 : num__4 find a : b : c ? <o> a ) num__6 : num__2 : num__3 <o> b ) num__2 : num__6 : num__3 <o> c ) num__2 : num__3 : num__6 <o> d ) num__3 : num__2 : num__6 <o> e ) num__6 : num__3 : num__2 |
a : b = num__2 : num__3 b : c = num__2 : num__4 num__2 : num__3 num__2 : num__4 ( a = a × b b = b × b and c = b × c ) a : b : c = num__2 : num__3 : num__6 c <eor> c <eos> |
c |
multiply__2.0__3.0__ subtract__4.0__2.0__ |
multiply__2.0__3.0__ subtract__4.0__2.0__ |
| if x = - num__2 and y = - num__4 what is the value of num__5 ( x - y ) ^ num__2 - xy ? <o> a ) num__12 <o> b ) num__13 <o> c ) num__15 <o> d ) num__18 <o> e ) num__22 |
x = - num__2 and y = - num__4 x - y = - num__2 - ( - num__4 ) = - num__2 + num__4 = num__2 x * y = - num__2 * - num__4 = num__8 now we apply it in the equation num__5 ( x - y ) ^ num__2 - xy = num__5 ( num__2 ) ^ num__2 - num__8 = = > num__5 * num__4 - num__8 = num__20 - num__8 = num__12 answer : a <eor> a <eos> |
a |
multiply__2.0__4.0__ multiply__4.0__5.0__ add__4.0__8.0__ add__4.0__8.0__ |
multiply__2.0__4.0__ multiply__4.0__5.0__ add__4.0__8.0__ add__4.0__8.0__ |
| country x taxes each of its citizens an amount equal to num__11 percent of the first $ num__40000 of income plus num__20 percent of all income in excess of $ num__40000 . if a citizen of country x is taxed a total of $ num__8000 what is her income ? <o> a ) $ num__40000 <o> b ) $ num__56000 <o> c ) $ num__64000 <o> d ) $ num__58000 <o> e ) $ num__80 |
000 |
equation is correct so math must be a problem . num__0.11 * num__40000 + num__0.2 * ( x - num__40000 ) = num__8000 - - > num__4400 + num__0.2 x - num__8000 = num__8000 - - > num__0.2 x = num__11600 - - > x = num__58000 . answer : d . <eor> d <eos> |
d |
d |
| if any amount is distributed among a b and c so that the part of a is doubled that of the part of b and part of b is num__3 time the part of c . find the ratio among their parts . <o> a ) num__1 : num__2 : num__4 <o> b ) num__1 : num__4 : num__1 <o> c ) num__8 : num__4 : num__1 <o> d ) num__2 : num__4 : num__1 <o> e ) num__6 : num__3 : num__1 |
by the assumptions as mentioned we have a = num__2 b and b = num__3 c . so a = num__2 ( num__3 c ) and we get a = num__6 c . we have therefore a : b : c = num__6 c : num__3 c : c . by cancelling the ratio with c we have num__6 : num__3 : num__1 . the answer is therefore e . <eor> e <eos> |
e |
multiply__3.0__2.0__ subtract__3.0__2.0__ multiply__3.0__2.0__ |
multiply__3.0__2.0__ subtract__3.0__2.0__ multiply__3.0__2.0__ |
| a shopkeeper buys two articles for rs . num__1000 each and then sells them making num__20.0 profit on the first article and num__20.0 loss on second article . find the net profit or loss percent ? <o> a ) num__200 <o> b ) num__962 <o> c ) num__267 <o> d ) num__258 <o> e ) num__268 |
profit on first article = num__20.0 of num__1000 = num__200 . this is equal to the loss he makes on the second article . that is he makes neither profit nor loss . answer : a <eor> a <eos> |
a |
percent__20.0__1000.0__ percent__20.0__1000.0__ |
percent__20.0__1000.0__ percent__20.0__1000.0__ |
| at a loading dock each worker on the night crew loaded num__0.75 as many boxes as each worker on the day crew . if the night crew has num__0.833333333333 as many workers as the day crew what fraction of all the boxes loaded by the two crews did the day crew load ? <o> a ) num__0.571428571429 <o> b ) num__0.555555555556 <o> c ) num__0.545454545455 <o> d ) num__0.583333333333 <o> e ) num__0.615384615385 |
let x be the number of workers on the day crew . let y be the number of boxes loaded by each member of the day crew . then the number of boxes loaded by the day crew is xy . the number of boxes loaded by the night crew is ( num__5 x / num__6 ) ( num__3 y / num__4 ) = num__5 xy / num__8 the total number of boxes is xy + num__5 xy / num__8 = num__13 xy / num__8 the fraction loaded by the day crew is xy / ( num__13 xy / num__8 ) = num__0.615384615385 the answer is e . <eor> e <eos> |
e |
divide__3.0__0.75__ add__3.0__5.0__ add__5.0__8.0__ divide__8.0__13.0__ divide__8.0__13.0__ |
divide__3.0__0.75__ add__3.0__5.0__ add__5.0__8.0__ divide__8.0__13.0__ divide__8.0__13.0__ |
| a man swims downstream num__72 km and upstream num__45 km taking num__9 hours each time ; what is the speed of the current ? <o> a ) num__1.8 <o> b ) num__1.5 <o> c ) num__1.3 <o> d ) num__1.2 <o> e ) num__1.6 |
num__72 - - - num__9 ds = num__8 ? - - - - num__1 num__45 - - - - num__9 us = num__5 ? - - - - num__1 s = ? s = ( num__8 - num__5 ) / num__2 = num__1.5 answer : b <eor> b <eos> |
b |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ round__1.5__ |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ divide__1.5__1.0__ |
| if c + d = num__12 and c and d are positive integers which of the following is a possible value for num__5 c + num__8 d ? <o> a ) num__111 <o> b ) num__121 <o> c ) num__150 <o> d ) num__171 <o> e ) num__180 |
c + d = num__11 num__5 c + num__8 d = num__5 c + num__5 d + num__3 d = num__5 ( c + d ) + num__3 d = num__60 + num__3 d = d = num__30 d = num__30 c + d = num__12 = c + num__30 = num__9 = c = num__12 - num__30 = - num__18 c = - num__18 num__5 c + num__8 d = num__5 ( - num__18 ) + num__8 ( num__30 ) = - num__90 + num__240 = num__150 hence c is the right answer . <eor> c <eos> |
c |
subtract__8.0__5.0__ multiply__12.0__5.0__ subtract__12.0__3.0__ subtract__30.0__12.0__ multiply__5.0__18.0__ multiply__8.0__30.0__ multiply__5.0__30.0__ multiply__5.0__30.0__ |
subtract__8.0__5.0__ multiply__12.0__5.0__ subtract__12.0__3.0__ subtract__30.0__12.0__ add__60.0__30.0__ multiply__8.0__30.0__ subtract__240.0__90.0__ subtract__240.0__90.0__ |
| a tank is filled in eight hours by three pipes a b and c . pipe a is twice as fast as pipe b and b is twice as fast as c . how much time will pipe b alone take to fill the tank ? <o> a ) num__16 hours <o> b ) num__28 hours <o> c ) num__18 hours <o> d ) num__19 hours <o> e ) num__10 hours |
num__1 / a + num__1 / b + num__1 / c = num__0.125 ( given ) also given that a = num__2 b and b = num__2 c = > num__0.5 b + num__1 / b + num__2 / b = num__0.125 = > ( num__1 + num__2 + num__4 ) / num__2 b = num__0.125 = > num__2 b / num__7 = num__8 = > b = num__28 hours . answer : b <eor> b <eos> |
b |
divide__1.0__2.0__ divide__2.0__0.5__ divide__1.0__0.125__ multiply__4.0__7.0__ round__28.0__ |
divide__1.0__2.0__ divide__2.0__0.5__ divide__1.0__0.125__ multiply__4.0__7.0__ divide__28.0__1.0__ |
| how many positive integers p less than num__20 are either a multiple of num__2 an odd multiple of num__9 or the sum of a positive multiple of num__2 and a positive multiple of num__9 ? <o> a ) num__19 <o> b ) num__18 <o> c ) num__17 <o> d ) num__16 <o> e ) num__15 |
we ' re asked to deal with the positive integers less than num__20 . there are only num__19 numbers in that group ( num__1 to num__19 inclusive ) . we ' re asked to find all of the numbers that fit one ( or more ) of the given descriptions . looking at the answer choices we have every value form num__15 to num__19 inclusive so most ( if not all ) of the numbers from num__1 to num__19 fit one ( or more ) of the descriptions . how long would it take you to find the ones that do n ' t fit . . . . . ? the first several should be pretty easy to find - the prompt wants us to focus on multiples of num__2 and num__9 ( and sums of those multiples ) . so what odd numbers are less than num__9 ? num__1 num__3 num__5 num__7 none of these values fit the given descriptions . that ' s num__4 out of num__19 that we know for sure do not fit . num__19 - num__4 = num__15 = e <eor> e <eos> |
e |
subtract__20.0__19.0__ add__2.0__1.0__ subtract__20.0__15.0__ add__2.0__5.0__ divide__20.0__5.0__ subtract__20.0__5.0__ |
subtract__20.0__19.0__ add__2.0__1.0__ subtract__20.0__15.0__ subtract__9.0__2.0__ subtract__9.0__5.0__ subtract__20.0__5.0__ |
| the speed of a boat in upstream is num__70 kmph and the speed of the boat downstream is num__80 kmph . find the speed of the boat in still water and the speed of the stream ? <o> a ) num__10 kmph <o> b ) num__67 kmph <o> c ) num__5 kmph <o> d ) num__88 kmph <o> e ) num__12 kmph |
speed of the boat in still water = ( num__70 + num__80 ) / num__2 = num__75 kmph . speed of the stream = ( num__80 - num__70 ) / num__2 = num__5 kmph . answer : c <eor> c <eos> |
c |
subtract__80.0__75.0__ round__5.0__ |
subtract__80.0__75.0__ subtract__80.0__75.0__ |
| while driving from a - ville to b - town harriet drove at a constant speed of num__100 kilometers per hour . upon arriving in b - town harriet immediately turned and drove back to a - ville at a constant speed of num__150 kilometers per hour . if the entire trip took num__5 hours how many minutes did it take harriet to drive from a - ville to b - town ? <o> a ) num__138 <o> b ) num__148 <o> c ) num__150 <o> d ) num__162 <o> e ) num__180 |
num__5 hr = num__300 min . if harriet spend equal hrs on each leg she will spend num__150 min on each . since speed a - b is less than speed b - a and distance on each leg is the same time spent on a - b is more than num__150 min which mean we can eliminate ans . a b and c . now let plug in ans . d or e and verify which one give same distance on each leg . e . t = num__180 min * leg a - b - - - > d = num__100.180 / num__60 = num__300.0 * leg b - a - - - - > d = num__150 * num__2.0 = num__300.0 so the correct ans . ise <eor> e <eos> |
e |
hour_to_min_conversion__ divide__300.0__150.0__ round__180.0__ |
divide__300.0__5.0__ divide__300.0__150.0__ round__180.0__ |
| two trains of length num__120 m and num__280 m are running towards each other on parallel lines at num__42 kmph and num__36 kmph respectively . in what time will they be clear of each other from the moment they meet ? <o> a ) num__18.4 sec <o> b ) num__77 sec <o> c ) num__76 sec <o> d ) num__20 sec <o> e ) num__66 sec |
relative speed = ( num__42 + num__36 ) * num__0.277777777778 = num__21.7 mps . distance covered in passing each other = num__120 + num__280 = num__400 m . the time required = d / s = num__400 / num__21.7 = num__18.4 sec . answer : a <eor> a <eos> |
a |
add__120.0__280.0__ round__18.4__ |
add__120.0__280.0__ round__18.4__ |
| what will be the remainder when num__8 - num__3 + num__3 ^ num__5 + num__2 ^ num__10 is divided by num__4 ? <o> a ) num__12 <o> b ) num__3 <o> c ) num__16 <o> d ) num__0 <o> e ) num__9 |
the multiplication ( e . g . num__2 ^ num__10 ) is done first and then the sum ( e . g . num__15 + num__20 ) and subtraction ( e . g . num__8 - num__3 ) and after all this expression it should be divided by num__4 and the answer is num__0 option d . <eor> d <eos> |
d |
multiply__3.0__5.0__ multiply__5.0__4.0__ multiply__8.0__0.0__ |
add__5.0__10.0__ add__5.0__15.0__ multiply__8.0__0.0__ |
| find the numbers which are in the ratio num__3 : num__2 : num__4 such that the sum of the first and the second added to the difference of the third and the second is num__28 ? <o> a ) num__4 num__322 <o> b ) num__4 num__422 <o> c ) num__12 num__616 <o> d ) num__9 num__612 <o> e ) num__9 num__2 |
23 |
let the numbers be a b and c . a : b : c = num__3 : num__2 : num__4 given ( a + b ) + ( c - b ) = num__21 = > a + c = num__28 = > num__3 x + num__4 x = num__28 = > x = num__4 a b c are num__3 x num__2 x num__4 x a b c are num__12 num__8 num__16 . answer : c <eor> c <eos> |
c |
c |
| the length of the bridge which a train num__130 metres long and travelling at num__45 km / hr can cross in num__30 seconds is : <o> a ) num__200 m <o> b ) num__225 m <o> c ) num__245 m <o> d ) num__250 m <o> e ) num__300 m |
explanation : speed = num__45 x num__0.277777777778 m / sec = num__12.5 m / sec time = num__30 sec . let the length of bridge be x metres . then num__130 + x / num__30 = num__12.5 num__2 ( num__130 + x ) = num__750 x = num__245 m . answer is c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| find the number of different prime factors of num__25650 <o> a ) num__4 <o> b ) num__2 <o> c ) num__3 <o> d ) num__5 <o> e ) num__6 |
explanation : l . c . m of num__25650 = num__2 x num__3 x num__3 x num__3 x num__5 x num__5 x num__19 num__3 num__2 num__519 number of different prime factors is num__4 . answer : option a <eor> a <eos> |
a |
add__2.0__3.0__ lcm__2.0__4.0__ |
add__2.0__3.0__ lcm__2.0__4.0__ |
| if y = x ^ num__2 + ux + v y is minimum when x is : <o> a ) u / u <o> b ) - u / u <o> c ) - u / num__2 <o> d ) - v / num__2 <o> e ) v / u |
we ' re given the equation y = x ^ num__2 + ux + v . if . . we use a simple classic quadratic . . . . u = num__2 v = num__1 y = x ^ num__2 + num__2 x + num__1 we can then go about finding the answer that yields the minimum result when x = . . . answer a : u / v = num__2.0 = num__2 - - > num__4 + num__4 + num__1 = + num__9 answer b : - u / v = - num__2.0 = - num__2 - - > num__4 - num__4 + num__1 = + num__1 answer c : - u / num__2 = - num__1.0 = - num__1 - - > num__1 - num__2 + num__1 = num__0 answer d : - v / num__2 = - num__0.5 - - > ( num__0.25 ) - num__1 + num__1 = + num__0.25 answer e : v / u = num__0.5 - - > ( num__0.25 ) + num__1 + num__1 = + num__2 num__0.25 from these results we can see the minimum result : c <eor> c <eos> |
c |
reverse__2.0__ reverse__4.0__ reverse__0.5__ |
reverse__2.0__ reverse__4.0__ reverse__0.5__ |
| num__164 pencils are distributed to children with the same number . what can ’ t be the range including the number of children ? <o> a ) num__1 ~ num__10 <o> b ) num__10 ~ num__20 <o> c ) num__50 ~ num__60 <o> d ) num__20 ~ num__30 <o> e ) num__70 ~ num__80 |
in num__164 = num__1 * num__164 = num__2 * num__82 = num__4 * num__41 = num__6 * num__24 you can come up with num__4 cases . amongst these cases what is not included is c . num__50 ~ num__60 . therefore the answer is c . <eor> c <eos> |
c |
divide__164.0__2.0__ divide__164.0__4.0__ add__2.0__4.0__ multiply__6.0__4.0__ multiply__1.0__50.0__ |
divide__164.0__2.0__ divide__164.0__4.0__ add__2.0__4.0__ multiply__6.0__4.0__ multiply__1.0__50.0__ |
| there are seven pairs of black shoes and num__5 pairs of white shoes . tey all are put into a box and shoes are drawn at a time . to ensure that at leat one pair of black shoes are taken out what is the minimum number of shoes required to be drawn out so as to get at leat num__1 pair of correct shoes . <o> a ) num__13 <o> b ) num__14 <o> c ) num__15 <o> d ) num__16 <o> e ) num__17 |
all the one sided ( assume only left ) white shoes total draws required : num__5 all the one sided ( assume only left ) black shoes total draws required : num__7 so total draws are : num__5 + num__7 = num__12 next draw or num__13 th draw will contain any color right footed shoe i . e . making at least one correct pair . so minimum number of shoes required to be drawn out is : num__13 answer : a <eor> a <eos> |
a |
add__5.0__7.0__ add__1.0__12.0__ multiply__1.0__13.0__ |
add__5.0__7.0__ add__1.0__12.0__ add__1.0__12.0__ |
| the dimensions of a room are num__25 feet * num__15 feet * num__12 feet . what is the cost of white washing the four walls of the room at rs . num__5 per square feet if there is one door of dimensions num__6 feet * num__3 feet and three windows of dimensions num__4 feet * num__3 feet each ? <o> a ) rs . num__4529 <o> b ) rs . num__4586 <o> c ) rs . num__4597 <o> d ) rs . num__4530 <o> e ) rs . num__4528 |
area of the four walls = num__2 h ( l + b ) since there are doors and windows area of the walls = num__2 * num__12 ( num__15 + num__25 ) - ( num__6 * num__3 ) - num__3 ( num__4 * num__3 ) = num__906 sq . ft . total cost = num__906 * num__5 = rs . num__4530 answer : d <eor> d <eos> |
d |
multiply__5.0__906.0__ multiply__5.0__906.0__ |
multiply__5.0__906.0__ multiply__5.0__906.0__ |
| if a num__1 mm thick paper is folded so that the area is halved at every fold then what would be the thickness of the pile after num__50 folds ? <o> a ) num__100 km <o> b ) num__1000 km <o> c ) num__1 million km <o> d ) num__1 billion km <o> e ) none |
after num__1 hold thickness will be num__2 mm after num__2 hold thich ness will be num__4 mm after num__50 th hold thick ness will be num__2 ^ num__50 mm . . converting to km . . . . gives num__1 billion km answer : d <eor> d <eos> |
d |
square_perimeter__1.0__ volume_cube__1.0__ |
square_perimeter__1.0__ power__1.0__2.0__ |
| | - num__4 | ( | - num__25 | - | num__15 | ) = ? ? source : preparation material mba center <o> a ) num__40 <o> b ) – num__60 <o> c ) num__60 <o> d ) num__75 <o> e ) num__100 |
absolute value will turn negatives into their positive ' equivalents ' and will leave positives unchanged so | - num__4 | = num__4 | - num__25 | = num__25 and | num__15 | = num__15 . getting rid of our absolute values we have : | - num__4 | ( | - num__25 | - | num__15 | ) = ( num__4 ) ( num__25 - num__15 ) = num__4 * num__10 = num__40 <eor> a <eos> |
a |
subtract__25.0__15.0__ multiply__4.0__10.0__ multiply__4.0__10.0__ |
subtract__25.0__15.0__ multiply__4.0__10.0__ multiply__4.0__10.0__ |
| what distance ( in meters ) will be covered by a bus moving at num__54 km / hr in num__20 seconds ? <o> a ) num__200 <o> b ) num__250 <o> c ) num__300 <o> d ) num__350 <o> e ) num__400 |
num__54 km / hr = num__54 * num__0.277777777778 = num__15 m / s distance = num__15 * num__20 = num__300 meters the answer is c . <eor> c <eos> |
c |
multiply__20.0__15.0__ round__300.0__ |
multiply__20.0__15.0__ multiply__20.0__15.0__ |
| a watch was sold at a loss of num__10.0 . if it was sold for rs . num__140 more there would have been a gain of num__4.0 . what is the cost price ? <o> a ) rs . num__1000 <o> b ) rs . num__1028 <o> c ) rs . num__1029 <o> d ) rs . num__1030 <o> e ) rs . num__1028 |
num__90.0 num__104.0 - - - - - - - - num__14.0 - - - - num__140 num__100.0 - - - - ? = > rs . num__1000 answer : a <eor> a <eos> |
a |
percent__10.0__140.0__ percent__100.0__1000.0__ |
percent__10.0__140.0__ percent__100.0__1000.0__ |
| a bicycle wheel has a diameter of num__0.75 m . how many complete revolutions does it make in num__1 km ? <o> a ) num__424 <o> b ) num__448 <o> c ) num__1408 <o> d ) num__710 <o> e ) num__223 |
num__1 revolution = num__3.14 * diameter . number of revolutions in num__1 km = num__1000 m / ( num__3.14 * num__0.75 m ) = num__424.6 . hence num__424 complete revolutions . answer a <eor> a <eos> |
a |
round__424.0__ |
multiply__1.0__424.0__ |
| a cycle is bought for rs . num__1500 and sold for rs . num__1620 find the gain percent ? <o> a ) num__11 <o> b ) num__8 <o> c ) num__10 <o> d ) num__20 <o> e ) num__12 |
num__1500 - - - - num__120 num__100 - - - - ? = > num__8.0 answer : b <eor> b <eos> |
b |
percent__100.0__8.0__ |
percent__100.0__8.0__ |
| a basketball is dropped from a height of num__20 feet . if it bounces back up to a height that is exactly half of its previous height and it stops bouncing after hitting the ground for the fourth time then how many total feet will the ball have traveled after num__2 full bounces . <o> a ) num__50 <o> b ) num__55 <o> c ) num__60 <o> d ) num__75 <o> e ) num__80 |
initial distance = num__20 feet first bounce = num__10 feet up + num__10 feet down = num__20 feet second bouche = num__5 feet up + num__5 feet down = num__10 feet total distance covered = num__20 + num__20 + num__10 + = num__50 answer is a <eor> a <eos> |
a |
divide__20.0__2.0__ divide__10.0__2.0__ multiply__10.0__5.0__ multiply__10.0__5.0__ |
divide__20.0__2.0__ divide__10.0__2.0__ multiply__10.0__5.0__ multiply__10.0__5.0__ |
| the hcf of two numbers is num__42 and the other two factors of their lcm are num__10 and num__20 . what is the largest number . <o> a ) num__462 <o> b ) num__450 <o> c ) num__840 <o> d ) num__504 <o> e ) num__555 |
explanation : hcf of the two numbers = num__42 hcf will be always a factor of lcm num__42 is factor of lcm other two factors are num__10 & num__20 then the numbers are ( num__42 * num__10 ) and ( num__42 x num__20 ) = num__420 and num__840 answer : option c <eor> c <eos> |
c |
multiply__42.0__10.0__ multiply__42.0__20.0__ multiply__42.0__20.0__ |
multiply__42.0__10.0__ multiply__42.0__20.0__ multiply__42.0__20.0__ |
| solution x is num__10 percent alcohol by volume and solution y is num__30 percent alcohol by volume . how many milliliters of solution y must be added to num__200 milliliters of solution x to create a solution that is num__14 percent alcohol by volume ? <o> a ) num__50 <o> b ) num__100 <o> c ) num__150 <o> d ) num__200 <o> e ) num__250 |
num__14.0 is num__4.0 - points higher than num__10.0 but num__16.0 - points lower than num__30.0 . thus there should be num__4 parts of solution x for num__1 part of solution y . we should add num__50 ml of solution y . the answer is a . <eor> a <eos> |
a |
subtract__14.0__10.0__ subtract__30.0__14.0__ divide__200.0__4.0__ divide__200.0__4.0__ |
subtract__14.0__10.0__ subtract__30.0__14.0__ divide__200.0__4.0__ divide__200.0__4.0__ |
| nhai employs num__100 men to build a highway of num__2 km in num__50 days working num__8 hours a day . if in num__25 days they completed num__0.333333333333 part of work . than how many more employees should nhai hire to finish it in time working num__10 hours a day ? <o> a ) num__55 <o> b ) num__60 <o> c ) num__70 <o> d ) num__50 <o> e ) num__65 |
explanation : the given problem can be written in a tabular form like below : - men days hour work planned num__100 num__50 num__8 num__1 actual num__100 num__25 num__8 num__0.333333333333 remaining ? num__25 num__10 num__0.666666666667 we can apply chain rule now . total men required to complete the remaining work = num__100 × ( num__1.0 ) × ( num__0.8 ) × ( num__0.666666666667 ) / ( num__0.333333333333 ) = num__160 . so additional men required = num__160 - num__100 = num__60 . answer is b <eor> b <eos> |
b |
subtract__1.0__0.3333__ divide__8.0__10.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
subtract__1.0__0.3333__ divide__8.0__10.0__ subtract__160.0__100.0__ subtract__160.0__100.0__ |
| the cost of num__100 articles is equal to selling price of num__90 . find the gain or loss percent ? <o> a ) num__11.0 <o> b ) num__20.0 <o> c ) num__27.0 <o> d ) num__32.0 <o> e ) num__49 % |
c . p . of each article be $ num__1 c . p . of num__90 articles = num__100 gain % = num__0.111111111111 * num__100 = num__11.0 approximately answer is a <eor> a <eos> |
a |
percent__100.0__11.0__ |
percent__100.0__11.0__ |
| the probability that a speaks truth is num__0.6 and that of b speaking truth is num__0.571428571429 . what is the probability that they agree in stating the same fact ? <o> a ) num__0.514285714286 <o> b ) num__0.461538461538 <o> c ) num__0.5 <o> d ) num__0.545454545455 <o> e ) num__0.5625 |
if both agree stating the same fact either both of them speak truth of both speak false . probability = num__0.6 * num__0.571428571429 + num__0.4 * num__0.428571428571 = num__0.342857142857 + num__0.171428571429 = num__0.514285714286 answer : a <eor> a <eos> |
a |
negate_prob__0.6__ negate_prob__0.5714__ union_prob__0.6__0.3429__0.4286__ union_prob__0.6__0.4286__0.5143__ |
negate_prob__0.6__ negate_prob__0.5714__ union_prob__0.6__0.3429__0.4286__ union_prob__0.6__0.4286__0.5143__ |
| a canteen requires num__112 kgs of wheat for one week . how many kgs of wheat will it require for num__69 days ? <o> a ) num__1204 kgs <o> b ) num__1401 kgs <o> c ) num__1104 kgs <o> d ) num__1014 kgs <o> e ) none |
quantity of wheat for num__7 days = num__112 kg quantity of wheat for one day = num__16.0 kg quantity of wheat for num__69 days = num__16.0 * num__69 = num__1104 kg answer c <eor> c <eos> |
c |
divide__112.0__7.0__ multiply__69.0__16.0__ round__1104.0__ |
divide__112.0__7.0__ multiply__69.0__16.0__ multiply__69.0__16.0__ |
| k - numbers are positive integers with only num__2 ' s as their digits . for example num__2 num__22 and num__222 are k - numbers . the k - weight of a number n is the minimum number of k - numbers that must be added together to equal n . for example the k - weight of num__50 is num__5 because num__50 = num__22 + num__22 + num__2 + num__2 + num__2 . what is the k - weight of num__750 ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__12 <o> d ) num__13 <o> e ) num__15 |
num__750 = num__3 * num__222 + ( num__22 * num__3 ) + num__9 * num__2 k weight of num__600 = num__3 + num__3 + num__9 = num__15 answer e <eor> e <eos> |
e |
subtract__5.0__2.0__ multiply__5.0__3.0__ multiply__5.0__3.0__ |
subtract__5.0__2.0__ multiply__5.0__3.0__ multiply__5.0__3.0__ |
| three candidates contested an election and received num__1136 num__7636 and num__11628 votes respectively . what percentage of the total votes did the winning candidate get ? <o> a ) num__57.0 <o> b ) num__60.0 <o> c ) num__90.0 <o> d ) num__65.0 <o> e ) num__70 % |
req percentage = num__0.57 * num__100 = num__57.0 answer a <eor> a <eos> |
a |
multiply__100.0__0.57__ multiply__100.0__0.57__ |
multiply__100.0__0.57__ multiply__100.0__0.57__ |
| in a group of cows and hens the number of legs are num__10 more than twice the number of heads . the number of cows is : <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__10 <o> e ) num__12 |
let no of cows be x no of hens be y . so heads = x + y legs = num__4 x + num__2 y now num__4 x + num__2 y = num__2 ( x + y ) + num__10 num__2 x = num__10 x = num__5 . answer : a <eor> a <eos> |
a |
coin_space__ vowel_space__ vowel_space__ |
coin_space__ vowel_space__ vowel_space__ |
| a company d has num__45 percent of the employees are secretaries and num__45 percent are salespeople . if there are num__50 other employees of company d how many employees does company d have ? <o> a ) num__500 <o> b ) num__162 <o> c ) num__180 <o> d ) num__152 <o> e ) num__250 |
let the total number of employees in the company be x % of secretaries = num__45.0 % of salespeople = num__45.0 % of of employees other than secretaries and salespeople = num__100 - num__90 = num__10.0 but this number is given as num__50 so num__10.0 of x = num__50 x = num__500 therefore there a total of num__500 employees in the company d correct answer - a <eor> a <eos> |
a |
percent__100.0__500.0__ |
percent__100.0__500.0__ |
| to a sugar solution of num__10 liters containing num__50.0 sugar num__10 liter of water is added . the percentage of sugar in the new solution is ? <o> a ) num__25.0 <o> b ) num__30.0 <o> c ) num__15.0 <o> d ) num__20.0 <o> e ) num__18 % |
quantity of sugar = num__50 * num__0.1 = num__5 kg new percentage = num__0.25 * num__100 = num__25.0 answer is a <eor> a <eos> |
a |
percent__10.0__50.0__ percent__100.0__25.0__ |
percent__10.0__50.0__ percent__100.0__25.0__ |
| three runners w b and c run a race with runner w finishing num__12 m ahead of runner b and num__18 m ahead of runner c while runner b finishes num__8 m ahead of runner c . each runner travels entire distance at a constant speed . what was the length of the race ? <o> a ) num__36 m <o> b ) num__48 m <o> c ) num__60 m <o> d ) num__72 m <o> e ) num__84 m |
let distance of race be x mtrs . then when w finishes x m b has run ( x - num__12 ) mtrs and c has run x - num__18 mtrs . so at this point b is num__6 m ahead of c . now to finish race b needs to run another num__12 m so he runs another num__12 m . when b finishes race he is num__8 m ahead of c . so last num__12 m b has run c has run num__10 m . as speeds are constant we have equation x - num__12 / x - num__18 = num__1.2 > x = num__48 . answer b <eor> b <eos> |
b |
subtract__18.0__12.0__ subtract__18.0__8.0__ divide__12.0__10.0__ multiply__8.0__6.0__ round__48.0__ |
subtract__18.0__12.0__ subtract__18.0__8.0__ divide__12.0__10.0__ multiply__8.0__6.0__ round__48.0__ |
| the value of ( num__1 + . num__1 + . num__01 + . num__001 ) is : <o> a ) num__1.001 <o> b ) num__1.011 <o> c ) num__1.003 <o> d ) num__1.111 <o> e ) none of them |
num__1.0 num__0.1 num__0.01 num__0.001 - - - - - - - num__1.111 answer is d <eor> d <eos> |
d |
multiply__0.01__0.1__ multiply__1.0__1.111__ |
multiply__0.01__0.1__ multiply__1.0__1.111__ |
| how much is num__70.0 of num__40 is greater than num__0.8 of num__25 ? <o> a ) num__12 <o> b ) num__27 <o> c ) num__18 <o> d ) num__12 <o> e ) num__8 |
( num__0.7 ) * num__40 â € “ ( num__0.8 ) * num__25 num__28 - num__20 = num__8 answer : e <eor> e <eos> |
e |
multiply__40.0__0.7__ multiply__0.8__25.0__ subtract__28.0__20.0__ subtract__28.0__20.0__ |
multiply__40.0__0.7__ multiply__0.8__25.0__ subtract__28.0__20.0__ subtract__28.0__20.0__ |
| two numbers are less than a third number by num__40.0 and num__47.0 respectively . how much per cent is the second number less than the first ? <o> a ) num__95.0 <o> b ) num__88.0 <o> c ) num__85.0 <o> d ) num__90.0 <o> e ) none of these |
here x = num__40 and y = num__47 therefore second number = [ [ ( num__100 - y ) / ( num__100 - x ) ] x num__100 ] % of first number = [ [ ( num__100 - num__47 ) / ( num__100 - num__40 ) ] x num__100 ] % of first number i . e num__88.3 of the first . answer : b <eor> b <eos> |
b |
round_down__88.3__ |
round_down__88.3__ |
| two goods trains each num__500 m long are running in opposite directions on parallel tracks . their speeds are num__60 km / hr and num__90 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? <o> a ) num__48 <o> b ) num__93 <o> c ) num__24 <o> d ) num__23 <o> e ) num__12 |
relative speed = num__60 + num__90 = num__150 km / hr . num__150 * num__0.277777777778 = num__41.6666666667 m / sec . distance covered = num__500 + num__500 = num__1000 m . required time = num__1000 * num__0.024 = num__24 sec . answer : c <eor> c <eos> |
c |
add__60.0__90.0__ multiply__1000.0__0.024__ round__24.0__ |
add__60.0__90.0__ multiply__1000.0__0.024__ multiply__1000.0__0.024__ |
| one man traveled a distance of num__61 km in num__9 hrs . he traveled partly on footat num__4 km / hr and partly on bicycle ta num__9 km / hr . the distance traveled on foot is ? <o> a ) num__8 km <o> b ) num__12 km <o> c ) num__16 km <o> d ) num__18 km <o> e ) num__20 km |
let the distance travelled on foot be x km . then distance travelled on bicycle = ( num__61 - x ) km . so x + ( num__61 - x ) = num__9 num__4 num__9 num__9 x + num__4 ( num__61 - x ) = num__9 x num__36 num__5 x = num__80 x = num__16 km . c <eor> c <eos> |
c |
multiply__9.0__4.0__ subtract__9.0__4.0__ divide__80.0__5.0__ round__16.0__ |
multiply__9.0__4.0__ subtract__9.0__4.0__ divide__80.0__5.0__ round__16.0__ |
| an express traveled at an average speed of num__100 km / hr stopping for num__4 min after every num__75 km . how long did it take to reach its destination num__150 km from the starting point ? <o> a ) num__8 hrs num__29 min <o> b ) num__6 hrs num__28 min <o> c ) num__2 hrs num__28 min <o> d ) num__6 hrs num__28 min <o> e ) num__1 hr num__34 min |
explanation : time taken to cover num__150 km = num__1.5 = num__1 hr num__30 mins number of stoppages = num__2.0 - num__1 = num__1 total time of stoppages = num__4 x num__1 = num__4 min hence total time taken = num__1 hr num__34 min . answer : e <eor> e <eos> |
e |
divide__150.0__100.0__ round_down__1.5__ divide__150.0__75.0__ add__4.0__30.0__ round_down__1.5__ |
divide__150.0__100.0__ round_down__1.5__ divide__150.0__75.0__ add__4.0__30.0__ subtract__2.0__1.0__ |
| how many quarters are equal to num__7 dollars ? <o> a ) num__1 <o> b ) num__28 <o> c ) num__12 <o> d ) num__9 <o> e ) num__7 |
num__7 * num__4 = num__28 quarters answer : b <eor> b <eos> |
b |
multiply__7.0__4.0__ multiply__7.0__4.0__ |
multiply__7.0__4.0__ multiply__7.0__4.0__ |
| what is the next number in this sequence : num__39 num__52 num__65 num__78 num__91 ? <o> a ) num__107 <o> b ) num__112 <o> c ) num__104 <o> d ) num__99 <o> e ) num__109 |
by subtracting each number from the one after it we can determine that the numbers are all multiples of num__13 . by adding num__13 to the last number you get the answer . num__91 + num__13 = num__104 answer : c <eor> c <eos> |
c |
subtract__52.0__39.0__ add__39.0__65.0__ add__39.0__65.0__ |
subtract__52.0__39.0__ add__39.0__65.0__ add__39.0__65.0__ |
| the length of a rectangle is increased by num__25.0 and its breadth is decreased by num__20.0 . what is the effect on its area ? <o> a ) num__10000 <o> b ) num__10020 <o> c ) num__10208 <o> d ) num__10029 <o> e ) num__10182 |
num__100 * num__100 = num__10000 num__125 * num__80 = num__10000 answer : a <eor> a <eos> |
a |
square_perimeter__25.0__ surface_rectangular_prism__25.0__20.0__100.0__ square_perimeter__20.0__ multiply__80.0__125.0__ |
square_perimeter__25.0__ surface_rectangular_prism__25.0__20.0__100.0__ square_perimeter__20.0__ multiply__80.0__125.0__ |
| solve this num__6 + num__7 = num__12 num__8 + num__9 = num__16 num__5 + num__6 = num__10 num__7 + num__8 = num__14 then num__9 + num__10 = ? ? <o> a ) num__11 <o> b ) num__12 <o> c ) num__13 <o> d ) num__14 <o> e ) num__18 |
num__18 answer : e <eor> e <eos> |
e |
add__6.0__12.0__ add__6.0__12.0__ |
add__6.0__12.0__ add__6.0__12.0__ |
| num__9.009 / num__1.001 <o> a ) num__0.009 <o> b ) num__0.09 <o> c ) num__0.9 <o> d ) num__9 <o> e ) num__90 |
answer is num__9 move the decimal forward three places for both numerator and denominator or just multiply both by a thousand . the result is num__9.0 = num__9 answer d <eor> d <eos> |
d |
round_down__9.009__ round_down__9.009__ |
round_down__9.009__ round_down__9.009__ |
| cheese bologna and peanut butter sandwiches were made for a picnic in a ratio of num__5 to num__7 to num__8 . if a total of num__160 sandwiches were made how many bologna sandwiches were made ? <o> a ) num__15 <o> b ) num__30 <o> c ) num__38 <o> d ) num__42 <o> e ) num__56 |
for deciding such task we should calculate all parts num__5 + num__7 + num__8 = num__20 parts and we should calculate how many sandwiches holds num__1 part : num__8.0 = num__8 sandwiches in one part for bologna we have num__8 parts so : num__7 * num__8 = num__56 answer is e <eor> e <eos> |
e |
divide__160.0__8.0__ subtract__8.0__7.0__ multiply__7.0__8.0__ multiply__7.0__8.0__ |
divide__160.0__8.0__ subtract__8.0__7.0__ multiply__7.0__8.0__ multiply__7.0__8.0__ |
| the length of the bridge which a train num__140 meters long and travelling at num__45 km / hr can cross in num__30 seconds is ? <o> a ) num__388 <o> b ) num__267 <o> c ) num__235 <o> d ) num__288 <o> e ) num__261 |
speed = ( num__45 * num__0.277777777778 ) m / sec = ( num__12.5 ) m / sec . time = num__30 sec . let the length of bridge be x meters . then ( num__140 + x ) / num__30 = num__12.5 = = > x = num__235 m . answer : c <eor> c <eos> |
c |
round__235.0__ |
round__235.0__ |
| for any positive integer n the sum of the first n positive integers equals [ n ( n + num__1 ) ] / num__2 . what is the sum of all the even integers between num__99 and num__201 ? <o> a ) num__6540 <o> b ) num__7650 <o> c ) num__8760 <o> d ) num__9870 <o> e ) num__10 |
980 |
num__100 + num__102 + . . . + num__200 = num__51 * num__100 + ( num__2 + num__4 . . . + num__100 ) = num__51 * num__100 + num__2 ( num__1 + num__2 + . . . + num__50 ) = num__51 * num__100 + num__2 ( num__50 ) ( num__51 ) / num__2 = num__150 * num__51 = num__7650 the answer is b . <eor> b <eos> |
b |
b |
| what is the solution of the equations x - y = num__0.9 and num__11 ( x + y ) - num__1 = num__2 ? <o> a ) x = num__3.2 y = num__2.3 <o> b ) x = num__1 y = num__0.1 <o> c ) x = num__2 y = num__1.1 <o> d ) x = num__1.2 y = num__0.3 <o> e ) none |
answer x - y = num__0.9 . . . ( i ) and num__11 ( x + y ) - num__1 = num__2 ⇒ num__11 / ( x + y ) = num__2 ⇒ num__2 ( x + y ) = num__11 ⇒ x + y = num__5.5 . . . ( ii ) on solving eqs . ( i ) and ( ii ) we get x = num__3.2 and y = num__2.3 correct option : a <eor> a <eos> |
a |
divide__11.0__2.0__ subtract__5.5__3.2__ add__0.9__2.3__ |
divide__11.0__2.0__ subtract__5.5__3.2__ add__0.9__2.3__ |
| a certain college ' s enrollment at the beginning of num__1992 was num__20 percent greater than it was at the beginning of num__1991 and its enrollment at the beginning of num__1993 was num__10 percent greater than it was at the beginning of num__1992 . the college ' s enrollment at the beginning of num__1993 was what percent greater than its enrollment at the beginning of num__1991 ? <o> a ) num__17.5 <o> b ) num__24.0 <o> c ) num__30.0 <o> d ) num__32.0 <o> e ) num__38 % |
suppose enrollment in num__1991 was num__100 then enrollment in num__1992 will be num__120 and enrollment in num__1993 will be num__120 * num__1.10 = num__132 increase in num__1993 from num__1991 = num__132 - num__100 = num__32 answer : d <eor> d <eos> |
d |
percent__32.0__100.0__ |
percent__32.0__100.0__ |
| at what rate of annual simple interest a certain sum will amount to four times in num__15 year ? <o> a ) num__15.0 <o> b ) num__17.5 <o> c ) num__20.0 <o> d ) num__25.0 <o> e ) num__35 % |
r = ? t = num__15 yrs si = p * r * t / num__100 amount = si + p = num__4 p ( p * r * num__0.15 ) + p = num__4 p p ( num__15 r / num__100 + num__1 ) = num__4 p num__15 r + num__100 = num__400 num__15 r = num__300 r = num__20.0 answer : c <eor> c <eos> |
c |
percent__20.0__100.0__ |
percent__20.0__100.0__ |
| the area of a parallelogram is num__147 sq m and its base is thrice the corresponding height . then the length of the base is ? <o> a ) num__14 m <o> b ) num__29 m <o> c ) num__49 m <o> d ) num__7 <o> e ) none |
b * h = num__147 b * h = num__147 = > b ^ num__2 = num__49 m . b = num__7 answer : ( d ) <eor> d <eos> |
d |
triangle_area__2.0__7.0__ |
triangle_area__2.0__7.0__ |
| the sum of three numbers is num__98 . if the ratio of the first to the second is num__2 : num__3 . and that of the second to the third is num__5 : num__8 then the second number is : <o> a ) num__17 <o> b ) num__30 <o> c ) num__87 <o> d ) num__27 <o> e ) num__82 |
let the three parts a b c . then a : b = num__2 : num__3 and b : c = num__5 : num__8 = ( num__5 * num__0.6 ) : ( num__8 * num__0.6 ) = num__3 : num__4.8 a : b : c = num__2 : num__3 : num__4.8 = num__10 : num__15 : num__24 = > b = num__98 * num__0.30612244898 = num__30 . answer : b <eor> b <eos> |
b |
divide__3.0__5.0__ multiply__8.0__0.6__ multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__3.0__8.0__ multiply__2.0__15.0__ multiply__2.0__15.0__ |
divide__3.0__5.0__ multiply__8.0__0.6__ multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__3.0__8.0__ multiply__2.0__15.0__ multiply__2.0__15.0__ |
| a train is num__360 meter long is running at a speed of num__60 km / hour . in what time will it pass a bridge of num__140 meter length ? <o> a ) num__87 <o> b ) num__69 <o> c ) num__30 <o> d ) num__72 <o> e ) num__21 |
speed = num__60 km / hr = num__60 * ( num__0.277777777778 ) m / sec = num__16.6666666667 m / sec total distance = num__360 + num__140 = num__500 meter time = distance / speed = num__500 * ( num__0.06 ) = num__30 seconds answer : c <eor> c <eos> |
c |
add__360.0__140.0__ multiply__0.06__500.0__ round__30.0__ |
add__360.0__140.0__ multiply__0.06__500.0__ multiply__0.06__500.0__ |
| the current of a stream runs at the rate of num__4 kmph . a boat goes num__6 km and back to the starting point in num__2 hours then find the speed of the boat in still water ? <o> a ) num__9 <o> b ) num__4 <o> c ) num__8 <o> d ) num__5 <o> e ) num__3 |
s = num__4 m = x ds = x + num__4 us = x - num__4 num__6 / ( x + num__4 ) + num__6 / ( x - num__4 ) = num__2 x = num__8 answer : c <eor> c <eos> |
c |
multiply__4.0__2.0__ round__8.0__ |
add__6.0__2.0__ add__6.0__2.0__ |
| if a is thrice as fast as b and together can do a work in num__15 days . in how many days a alone can do the work ? <o> a ) num__36 <o> b ) num__20 <o> c ) num__28 <o> d ) num__54 <o> e ) num__45 |
a ’ s one day ’ s work = num__1 / x b ’ s one day ’ s work = num__0.333333333333 x a + b ’ s one day ’ s work = num__1 / x + num__0.333333333333 x = num__0.0666666666667 = num__3 + num__0.333333333333 x = num__1.33333333333 x = num__0.0666666666667 x = num__15 * num__1.33333333333 = num__20 answer : b <eor> b <eos> |
b |
divide__1.0__15.0__ add__1.0__0.3333__ round__20.0__ |
divide__1.0__15.0__ add__1.0__0.3333__ divide__20.0__1.0__ |
| num__32.0 of num__1500 is <o> a ) num__480 <o> b ) num__440 <o> c ) num__370 <o> d ) num__415 <o> e ) num__435 |
num__1.0 of num__1500 is = num__15 num__32.0 of num__1500 is = num__32 * num__15 = num__480 answer : a <eor> a <eos> |
a |
percent__1.0__1500.0__ percent__32.0__1500.0__ percent__32.0__1500.0__ |
percent__1.0__1500.0__ percent__32.0__1500.0__ percent__32.0__1500.0__ |
| if num__4 men can colour num__48 m long cloth in num__2 days then num__6 men can colour num__36 m long cloth in <o> a ) num__1 day <o> b ) num__2 days <o> c ) num__4 days <o> d ) num__3 days <o> e ) none |
the length of cloth painted by one man in one day = num__12.0 × num__2 = num__6 m no . of days required to paint num__36 m cloth by num__6 men = num__6.0 × num__6 = num__1 day . option ' a ' <eor> a <eos> |
a |
divide__48.0__4.0__ round__1.0__ |
divide__48.0__4.0__ round__1.0__ |
| a cyclist covers a distnce of num__750 m in num__2 min num__30 sec . what is the speed in km / hr of the cyclist ? <o> a ) num__28 km / hr <o> b ) num__18 km / hr <o> c ) num__20 km / hr <o> d ) num__22 km / hr <o> e ) num__24 km / hr |
speed = { num__750 } m / sec = num__5 m / sec = { num__5 * num__18 } km / hr = num__18 km / hr answer b <eor> b <eos> |
b |
round__18.0__ |
round__18.0__ |
| pipes p and q would fill a cistern num__18 and num__24 minutes respectively . both pipes being opened find when the first pipe must be turned off so that the cistern may be just filled in num__12 minutes ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__2 <o> d ) num__7 <o> e ) num__5 |
x / num__18 + num__0.5 = num__1 x = num__9 answer : b <eor> b <eos> |
b |
divide__12.0__24.0__ multiply__18.0__0.5__ round__9.0__ |
divide__12.0__24.0__ multiply__18.0__0.5__ divide__9.0__1.0__ |
| a ship going at the speed of num__18 km / hr crosses a light house in num__20 seconds . what is the length of the train ? <o> a ) num__120 <o> b ) num__150 <o> c ) num__100 <o> d ) num__240 <o> e ) num__200 |
speed = ( num__18 x num__0.277777777778 ) = num__5 m / sec . length of the train = ( speed x time ) . length of the train = num__5 x num__20 m = num__100 m . answer : c <eor> c <eos> |
c |
multiply__20.0__5.0__ round__100.0__ |
multiply__20.0__5.0__ round__100.0__ |
| the ratio of the incomes of x and y is num__6 : num__4 and the ratio of their expenditure is num__5 : num__3 . if at the end of the year each saves $ num__2000 then the income of x is ? <o> a ) $ num__11000 <o> b ) $ num__10000 <o> c ) $ num__12000 <o> d ) $ num__10500 <o> e ) $ num__10600 |
let the income of x and y be $ num__6 x and $ num__4 x let their expenditures be $ num__5 y and $ num__3 y num__6 x - num__5 y = num__2000 - - - - - - - num__1 ) num__4 x - num__3 y = num__2000 - - - - - - - num__2 ) from num__1 ) and num__2 ) num__2 x - num__2 y = num__0 = > x = y x = num__2000 income of x num__5 * num__2000 = $ num__10000 answer : b <eor> b <eos> |
b |
subtract__6.0__5.0__ subtract__6.0__4.0__ multiply__5.0__2000.0__ multiply__5.0__2000.0__ |
subtract__6.0__5.0__ subtract__6.0__4.0__ multiply__5.0__2000.0__ multiply__5.0__2000.0__ |
| there is a square of side num__6 cm . a circle is inscribed inside the square . find the ratio of the area of circle to square . <o> a ) num__0.785714285714 <o> b ) num__0.0714285714286 <o> c ) num__0.928571428571 <o> d ) num__0.857142857143 <o> e ) num__0.357142857143 |
let the side of square = a unit so area of square = a ^ num__2 diameter of inscribed circle = side of square = a radius of circle = a / num__2 area of circle = ( num__3.14285714286 ) ( a / num__2 ) ^ num__2 ratio of area of circle to square = { ( num__3.14285714286 ) ( a / num__2 ) ^ num__2 } / a ^ num__2 = num__0.785714285714 answer : a <eor> a <eos> |
a |
triangle_area__2.0__0.7857__ |
triangle_area__2.0__0.7857__ |
| the length of a room is num__8 m and width is num__4.75 m . what is the cost of paying the floor by slabs at the rate of rs . num__900 per sq . metre . <o> a ) num__25650 <o> b ) num__25750 <o> c ) num__26550 <o> d ) num__26750 <o> e ) num__34200 |
area = num__8 × num__4.75 sq . metre . cost for num__1 sq . metre . = rs . num__900 hence total cost = num__8 × num__4.75 × num__900 = num__8 × num__4275 = rs . num__34200 answer is e . <eor> e <eos> |
e |
multiply__4.75__900.0__ multiply__8.0__4275.0__ round__34200.0__ |
multiply__4.75__900.0__ multiply__8.0__4275.0__ round__34200.0__ |
| a certain pilot flew num__400 miles to city k at an average speed of num__350 miles per hour with the wind and made the trip back at an average speed of num__450 miles per hour against the wind . which of the following is closest to the pilot ’ s average speed in miles per hour for the round - trip ? <o> a ) num__390 <o> b ) num__290 <o> c ) num__300 <o> d ) num__310 <o> e ) num__320 |
avg speed = total distance / total time total distance = num__800 total time = num__1.14285714286 + num__0.888888888889 = num__2.03174603175 = > avg speed = ( num__800 * num__63 ) / num__128 = num__390 ( approx ) ans is a <eor> a <eos> |
a |
add__350.0__450.0__ divide__400.0__350.0__ divide__400.0__450.0__ round__390.0__ |
add__350.0__450.0__ divide__400.0__350.0__ divide__400.0__450.0__ round__390.0__ |
| a boat can move upstream at num__15 kmph and downstream at num__23 kmph then the speed of the current is ? <o> a ) num__5 kmph <o> b ) num__7 kmph <o> c ) num__8 kmph <o> d ) num__9 kmph <o> e ) num__4 kmph |
us = num__15 ds = num__23 m = ( num__23 - num__15 ) / num__2 = num__4 answer : e <eor> e <eos> |
e |
round__4.0__ |
round__4.0__ |
| train x crosses a stationary train y in num__60 seconds and a pole in num__25 seconds with the same speed . the length of the train x is num__300 m . what is the length of the stationary train y ? <o> a ) num__277 m <o> b ) num__420 m <o> c ) num__617 m <o> d ) num__868 m <o> e ) num__651 m |
let the length of the stationary train y be ly given that length of train x lx = num__300 m let the speed of train x be v . since the train x crosses train y and a pole in num__60 seconds and num__25 seconds respectively . = > num__300 / v = num__25 - - - > ( num__1 ) ( num__300 + ly ) / v = num__60 - - - > ( num__2 ) from ( num__1 ) v = num__12.0 = num__12 m / sec . from ( num__2 ) ( num__300 + ly ) / num__12 = num__60 = > num__300 + ly = num__60 ( num__12 ) = num__720 = > ly = num__720 - num__300 = num__420 m length of the stationary train = num__420 m answer : b <eor> b <eos> |
b |
divide__300.0__25.0__ multiply__60.0__12.0__ subtract__720.0__300.0__ round__420.0__ |
divide__300.0__25.0__ multiply__60.0__12.0__ subtract__720.0__300.0__ divide__420.0__1.0__ |
| num__16.02 × num__0.001 = ? <o> a ) num__0.1602 <o> b ) num__0.001602 <o> c ) num__1.6021 <o> d ) num__0.01602 <o> e ) none of these |
num__16.02 × num__0.001 = ? or ? = num__0.01602 answer d <eor> d <eos> |
d |
multiply__16.02__0.001__ multiply__16.02__0.001__ |
multiply__16.02__0.001__ multiply__16.02__0.001__ |
| in the coordinate plane points ( x num__8 ) and ( num__20 y ) are on line k . if line k passes through the origin and has slope num__0.25 then x * y = <o> a ) num__120 <o> b ) num__100 <o> c ) num__135 <o> d ) num__140 <o> e ) num__160 |
line k passes through the origin and has slope num__0.25 means that its equation is y = num__0.25 * x . thus : ( x num__8 ) = ( num__32 num__8 ) and ( num__20 y ) = ( num__205 ) - - > x * y = num__32 * num__5 = num__160 . answer : e <eor> e <eos> |
e |
divide__8.0__0.25__ multiply__20.0__0.25__ multiply__8.0__20.0__ multiply__8.0__20.0__ |
divide__8.0__0.25__ multiply__20.0__0.25__ multiply__8.0__20.0__ multiply__8.0__20.0__ |
| four people are planning to share equally the cost of a rental car . if one person withdraws from the arrangement and the others share equally the entire cost of the car then the share of each of the remaining persons increased by : <o> a ) num__0.333333333333 <o> b ) num__0.285714285714 <o> c ) num__0.428571428571 <o> d ) num__0.571428571429 <o> e ) none of them |
original share of num__1 person = num__0.25 new share of num__1 person = num__0.333333333333 increase = ( num__0.333333333333 - num__0.25 = num__0.0833333333333 ) therefore required fraction = ( num__0.0833333333333 ) / ( num__0.25 ) = ( num__0.0833333333333 ) x ( num__4.0 ) = num__0.333333333333 answer is a . <eor> a <eos> |
a |
multiply__0.25__0.3333__ reverse__0.25__ add__0.25__0.0833__ |
subtract__0.3333__0.25__ reverse__0.25__ divide__0.3333__1.0__ |
| if the sum of a number and its square is num__306 what is the number ? <o> a ) num__16 <o> b ) num__77 <o> c ) num__25 <o> d ) num__17 <o> e ) num__171 |
explanation : let the integer be x . then x + x num__2 = num__306 x num__2 + x - num__306 = num__0 ( x + num__18 ) ( x – num__17 ) = num__0 x = num__17 answer : d <eor> d <eos> |
d |
side_by_diagonal__17.0__0.0__ |
side_by_diagonal__17.0__0.0__ |
| at exactly what time past num__4 : num__00 will the minute and hour hands of an accurate working clock be precisely perpendicular to each other for the first time ? <o> a ) num__20 num__0.619047619048 minutes past num__7 : num__00 <o> b ) num__20 num__0.764705882353 minutes past num__7 : num__00 <o> c ) num__21 num__0.130434782609 minutes past num__7 : num__00 <o> d ) num__21 num__0.818181818182 minutes past num__4 : num__00 <o> e ) num__22 num__0.444444444444 minutes past num__7 : num__00 |
num__5.5 is the angle between minute n hour this is what i was taught . . . so should n ' t it be solve by dividing num__90 with num__5.5 ? that would have been the case if your initial difference between the hour and the minute hand was = num__0 degrees or in other words both minute and hour hands were at the same location . but as per the question you are asked for time after num__4 : num__00 . at num__4 : num__00 the angle between the hour and the minute hand is num__210 degrees . you need to take this into account as well . so in order for the difference to decrease to num__90 degrees the minute hand must eat away this difference of num__210 - num__90 = num__120 degree at the rate of num__5.5 degrees per minute - - - > num__120 / num__5.5 = num__21 num__0.818181818182 minutes . thus d is the correct answer . <eor> d <eos> |
d |
subtract__210.0__90.0__ round__21.0__ |
subtract__210.0__90.0__ round__21.0__ |
| right triangle rst can be constructed in the xy - plane such that rs is perpendicular to the y - axis and the right angle is at r . the x and y - coordinates of r s and t are to be nonzero integers that satisfy the inequalities − num__3 ≤ x ≤ num__4 and − num__7 ≤ y ≤ num__3 . given these restrictions how many different triangles can be constructed ? <o> a ) num__3780 <o> b ) num__4200 <o> c ) num__4900 <o> d ) num__6160 <o> e ) num__7744 |
− num__3 ≤ x ≤ num__4 and − num__7 ≤ y ≤ num__3 gives a rectangle with num__8 * num__11 dimensions ( num__8 horizontal and num__11 vertical dots ) . we are given that rs is parallel to x - axis rt is parallel to y - axis and the right angle is at r . choose the ( x y ) coordinates for vertex r : num__7 c num__1 * num__10 c num__1 ( we are told that coordinates of r s and t must be nonzero integers so we are choosing from num__7 and num__10 instead of num__8 and num__11 because we should exclude num__0 ) ; choose the x coordinate for vertex s ( as y coordinate is fixed by r ) : num__6 c num__1 ( num__7 - num__1 = num__6 as num__1 horizontal dot is already occupied by r ) ; choose the y coordinate for vertex t ( as x coordinate is fixed by r ) : num__9 c num__1 ( num__10 - num__1 = num__9 as num__1 vertical dot is already occupied by b ) . num__7 c num__1 * num__10 c num__1 * num__6 c num__1 * num__9 c num__1 = num__3780 . answer : a . <eor> a <eos> |
a |
add__3.0__8.0__ subtract__4.0__3.0__ add__3.0__7.0__ subtract__7.0__1.0__ add__3.0__6.0__ multiply__1.0__3780.0__ |
add__3.0__8.0__ subtract__4.0__3.0__ subtract__11.0__1.0__ subtract__7.0__1.0__ subtract__10.0__1.0__ multiply__1.0__3780.0__ |
| working alone at its constant rate machine k took num__3 hours to produce num__0.125 of the units produced last friday . then machine m started working and the two machines working simultaneously at their respective constant rates took num__9 hours to produce the rest of the units produced last friday . how many hours would it have taken machine m working alone at its constant rate to produce all of the units produced last friday ? <o> a ) num__8 <o> b ) num__12 <o> c ) num__18 <o> d ) num__24 <o> e ) num__30 |
machine k works at a rate of num__0.0416666666667 of the units per hour . the rate of k + m together is num__0.875 * num__0.111111111111 = num__0.0972222222222 of the units per hour . the rate of machine m is num__0.0972222222222 - num__0.0416666666667 = num__0.0555555555556 . it would have taken machine m a total time of num__18 hours . the answer is c . <eor> c <eos> |
c |
divide__0.125__3.0__ divide__0.875__9.0__ round__18.0__ |
divide__0.125__3.0__ multiply__0.875__0.1111__ round__18.0__ |
| if num__5 women or num__8 girls can do a work in num__84 days . in how many days can num__10 women and num__5 girls can do the same work ? <o> a ) num__32 days <o> b ) num__48 days <o> c ) num__52 days <o> d ) num__38 days <o> e ) num__28 days |
given that num__5 women is equal to num__8 girls to complete a work . so num__10 women = num__16 girls . therefore num__10 women + num__5 girls = num__16 girls + num__5 girls = num__21 girls . num__8 girls can do a work in num__84 days then num__21 girls can do a work in ( num__8 * num__4.0 ) = num__32 days . therefore num__10 women and num__5 girls can a work in num__32 days answer : a . <eor> a <eos> |
a |
add__5.0__16.0__ divide__84.0__21.0__ multiply__8.0__4.0__ round__32.0__ |
add__5.0__16.0__ divide__84.0__21.0__ multiply__8.0__4.0__ multiply__8.0__4.0__ |
| p q and r can do a work in num__20 num__30 and num__60 days respectively . how many days does it need to complete the work if p does the work and he is assisted by q and r on every third day ? <o> a ) num__15 days <o> b ) num__17 days <o> c ) num__25 days <o> d ) num__27 days <o> e ) num__29 days |
amount of work p can do in num__1 day = num__0.05 amount of work q can do in num__1 day = num__0.0333333333333 amount of work r can do in num__1 day = num__0.0166666666667 p is working alone and every third day q and r is helping him work completed in every three days = num__2 × ( num__0.05 ) + ( num__0.05 + num__0.0333333333333 + num__0.0166666666667 ) = num__0.2 so work completed in num__15 days = num__5 × num__0.2 = num__1 ie the work will be done in num__15 days a ) <eor> a <eos> |
a |
divide__1.0__20.0__ divide__1.0__30.0__ divide__1.0__60.0__ divide__60.0__30.0__ divide__30.0__2.0__ subtract__20.0__15.0__ round__15.0__ |
divide__1.0__20.0__ divide__1.0__30.0__ divide__1.0__60.0__ divide__60.0__30.0__ divide__30.0__2.0__ subtract__20.0__15.0__ round__15.0__ |
| the sum of three integers a b and c is num__60 . a is one third of the sum of b and c and b is one fifth of the sum of a and c . what is c ? <o> a ) num__40 <o> b ) num__45 <o> c ) num__35 <o> d ) num__50 <o> e ) num__55 |
a is one third of the sum of b and c . . or a = num__0.333333333333 ( b + c ) . . . but a + b + c = num__60 so num__0.333333333333 ( b + c ) + b + c = num__60 num__4 ( b + c ) / num__3 = num__60 b + c = num__60 * num__0.75 = num__45 a = num__60 - num__45 = num__15 num__2 ) b is one fifth of the sum of a and c or b = num__0.2 ( a + c ) . . . but a + b + c = num__60 so num__0.2 ( a + c ) + a + c = num__60 a + c = num__60 * num__0.833333333333 = num__50 but a = num__15 so c = num__50 - num__15 = num__35 answer : c <eor> c <eos> |
c |
divide__3.0__4.0__ multiply__60.0__0.75__ divide__60.0__4.0__ divide__3.0__15.0__ subtract__50.0__15.0__ subtract__50.0__15.0__ |
divide__3.0__4.0__ multiply__60.0__0.75__ divide__60.0__4.0__ divide__3.0__15.0__ subtract__50.0__15.0__ subtract__50.0__15.0__ |
| find the number difference between number and its num__0.6 is num__64 . <o> a ) num__160 <o> b ) num__163 <o> c ) num__164 <o> d ) num__165 <o> e ) num__166 |
explanation : let the number = x then x - ( num__0.6 ) x = num__64 = > ( num__0.4 ) x = num__64 = > num__2 x = num__64 * num__5 = > x = num__160 answer : option a <eor> a <eos> |
a |
divide__2.0__0.4__ divide__64.0__0.4__ divide__64.0__0.4__ |
divide__2.0__0.4__ divide__64.0__0.4__ divide__64.0__0.4__ |
| in a mixture num__60 litres the ratio of milk and water num__2 : num__1 . if this ratio is to be num__1 : num__2 then the quanity of water to be further added is : <o> a ) num__40 <o> b ) num__20 <o> c ) num__30 <o> d ) num__80 <o> e ) num__60 |
quantity of milk = ( num__60 x num__0.666666666667 ) litres = num__40 litres . quantity of water in it = ( num__60 - num__40 ) litres = num__20 litres . new ratio = num__1 : num__2 let quantity of water to be added further be x litres . then milk : water = ( num__40 / ( num__20 + x ) ) now ( num__40 / ( num__20 + x ) ) = num__0.5 = num__20 + x = num__80 = x = num__60 . therefore quantity of water to be added = num__60 litres . answer is e . <eor> e <eos> |
e |
subtract__60.0__40.0__ reverse__2.0__ add__60.0__20.0__ multiply__60.0__1.0__ |
subtract__60.0__40.0__ reverse__2.0__ add__60.0__20.0__ divide__60.0__1.0__ |
| a ( n + num__1 ) = num__1 + ( num__1 / a ( n ) ) and a ( num__1 ) = num__1 . what is the value of a ( num__7 ) ? <o> a ) num__2.14285714286 <o> b ) num__1.88888888889 <o> c ) num__1.72727272727 <o> d ) num__1.61538461538 <o> e ) num__1.53333333333 |
a ( n + num__1 ) = num__1 + ( num__1 / a ( n ) ) a ( num__1 ) = num__1 a ( num__2 ) = num__2 a ( num__3 ) = num__1.5 a ( num__4 ) = num__1.66666666667 a ( num__5 ) = num__1.6 a ( num__6 ) = num__1.625 a ( num__7 ) = num__1.61538461538 the answer is d . <eor> d <eos> |
d |
add__1.0__2.0__ divide__3.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ add__1.0__5.0__ multiply__1.0__1.6154__ |
add__1.0__2.0__ divide__3.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ add__1.0__5.0__ divide__1.6154__1.0__ |
| two fill pipes a and b can fill a cistern in num__12 and num__16 minutes respectively . both fill pipes are opened together but num__4 minutes before the cistern is full one pipe a is closed . how much time will the cistern take to fill ? <o> a ) num__9 num__1 ⁄ num__7 min . <o> b ) num__3 num__1 ⁄ num__3 min . <o> c ) num__5 min . <o> d ) num__3 min . <o> e ) none of these |
let cistern will be full in x min . then part filled by b in x min + part filled by a in ( x – num__4 ) min = num__1 ⇒ x / num__16 + x − num__0.333333333333 = num__1 ⇒ x = num__64 ⁄ num__7 = num__91 ⁄ num__7 hours . answer a <eor> a <eos> |
a |
divide__4.0__12.0__ multiply__16.0__4.0__ subtract__16.0__7.0__ |
divide__4.0__12.0__ multiply__16.0__4.0__ subtract__16.0__7.0__ |
| what is the sum of the different positive prime factors of num__540 ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__15 <o> d ) num__16 <o> e ) num__18 |
num__540 num__54 * num__10 - - > num__3 * num__3 * num__3 * num__2 * num__2 * num__5 - - > num__3 + num__5 + num__2 = num__10 . remember its asking for the different positive prime factors of num__540 . a <eor> a <eos> |
a |
divide__540.0__54.0__ gcd__10.0__54.0__ add__2.0__3.0__ gcd__540.0__10.0__ |
divide__540.0__54.0__ gcd__10.0__54.0__ add__2.0__3.0__ multiply__2.0__5.0__ |
| mysoon collects glass ornaments . ten more than num__0.166666666667 of the ornaments in her collection are handmade and num__0.5 of the handmade ornaments are antiques . if num__0.25 of the ornaments in her collection are handmade antiques how many ornaments are in her collection ? <o> a ) num__30 <o> b ) num__60 <o> c ) num__108 <o> d ) num__144 <o> e ) num__180 |
the number of ornaments = a ten more than num__0.166666666667 of the ornaments in her collection are handmade = > handmade = num__10 + a / num__6 num__0.5 of the handmade ornaments are antiques = > handmade ornaments = num__0.5 * ( num__10 + a / num__6 ) = num__5 + a / num__12 num__0.25 of the ornaments in her collection are handmade antiques = > handmade ornaments = a / num__4 = > num__5 + a / num__12 = a / num__4 = > a = num__30 ans : a <eor> a <eos> |
a |
multiply__0.5__10.0__ divide__6.0__0.5__ reverse__0.25__ multiply__5.0__6.0__ multiply__5.0__6.0__ |
multiply__0.5__10.0__ divide__6.0__0.5__ reverse__0.25__ multiply__5.0__6.0__ multiply__5.0__6.0__ |
| a ferry can transport num__85 tons of vehicles . automobiles range in weight from num__1800 to num__3200 pounds . what is the greatest number of automobiles that can be loaded onto the ferry ? <o> a ) num__23 <o> b ) num__41 <o> c ) num__48 <o> d ) num__90 <o> e ) num__86 |
to get maximum vehicles we must take into consideration the minimum weight i . e num__1800 pounds here since num__1 ton = num__2000 pounds num__85 tons will be num__170000 pounds from the answer choices : let max number of vehicles be num__86 total weight will be = num__90 * num__1800 = num__162000 pounds which is lesser than the maximum weight allowed . ans : d <eor> d <eos> |
d |
multiply__85.0__2000.0__ add__85.0__1.0__ multiply__1800.0__90.0__ multiply__1.0__90.0__ |
multiply__85.0__2000.0__ add__85.0__1.0__ multiply__1800.0__90.0__ multiply__1.0__90.0__ |
| insert the missing number num__4 - num__8 num__16 - num__32 num__64 ( . . . . ) <o> a ) num__128 <o> b ) - num__128 <o> c ) num__192 <o> d ) - num__132 <o> e ) num__0 |
num__4 * - num__2 = - num__8 - num__8 * - num__2 = num__16 num__16 * - num__2 = - num__32 - num__32 * - num__2 = num__64 num__64 * - num__2 = - num__128 answer : b <eor> b <eos> |
b |
divide__8.0__4.0__ multiply__4.0__32.0__ multiply__4.0__32.0__ |
divide__8.0__4.0__ multiply__4.0__32.0__ multiply__4.0__32.0__ |
| in a question on division with zero remainder a candidate took num__12 as divisor instead of num__21 . the quotient obtained by him was num__35 . the correct quotient is ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__24 <o> d ) num__26 <o> e ) num__28 |
number = ( num__35 * num__12 ) = num__420 correct quotient = num__20.0 = num__20 b ) <eor> b <eos> |
b |
multiply__12.0__35.0__ divide__420.0__21.0__ divide__420.0__21.0__ |
multiply__12.0__35.0__ divide__420.0__21.0__ divide__420.0__21.0__ |
| a train num__110 m long is running with a speed of num__60 km / hr . in what time will it pass a man who is running at num__6 km / hr in the direction opposite to that in which the train is going ? <o> a ) num__9 <o> b ) num__7 <o> c ) num__5 <o> d ) num__4 <o> e ) num__6 |
speed of train relative to man = num__60 + num__6 = num__66 km / hr . = num__66 * num__0.277777777778 = num__18.3333333333 m / sec . time taken to pass the men = num__110 * num__0.0545454545455 = num__6 sec . answer : e <eor> e <eos> |
e |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ round__6.0__ |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ divide__110.0__18.3333__ |
| a tourist purchased a total of $ num__1450 worth of traveler ’ s checks in $ num__10 and $ num__50 denominations during the trip the tourist cashed num__7 checks and then lost all of the rest . if the number of $ num__10 checks cashed was one more or one less than the number of $ num__50 checks cashed what is the minimum possible value of the checks that were lost ? <o> a ) $ num__1430 <o> b ) $ num__1310 <o> c ) $ num__1290 <o> d ) $ num__1220 <o> e ) $ num__1 |
150 |
since total cashed checks = num__7 ans $ num__10 check cashed = one less or more than $ num__50 checks therefore either $ num__10 checks = num__3 and $ num__50 checks = num__4 or $ num__10 checks = num__4 and $ num__50 checks = num__3 using this information num__1 st option gives us the maximum cashed value therefore if we negate this value from total value we will get minimum value of which checks have lost therfore num__1450 - num__230 = num__1220 hence answer is d <eor> d <eos> |
d |
d |
| the sum of number of boys and girls in a school is num__150 . if the number of boys is x then the number of girls becomes x % of the total number of students . the number of boys is ? <o> a ) num__50 <o> b ) num__40 <o> c ) num__60 <o> d ) num__100 <o> e ) num__70 |
we have x + x % of num__150 = num__150 x + x / num__100 * num__150 = num__150 num__2.5 * x = num__150 x = num__150 * num__0.4 = num__60 answer is c <eor> c <eos> |
c |
reverse__2.5__ divide__150.0__2.5__ divide__150.0__2.5__ |
reverse__2.5__ divide__150.0__2.5__ divide__150.0__2.5__ |
| the average of num__7 numbers is num__15 . if each number be multiplied by num__5 . find the average of new set of numbers ? <o> a ) a ) num__110 <o> b ) b ) num__122 <o> c ) c ) num__90 <o> d ) d ) num__85 <o> e ) e ) num__75 |
explanation : average of new numbers = num__15 * num__5 = num__75 answer : option e <eor> e <eos> |
e |
multiply__15.0__5.0__ multiply__15.0__5.0__ |
multiply__15.0__5.0__ multiply__15.0__5.0__ |
| a farmer with num__1350 acres of land had planted his fields with corn sugar cane and tobacco in the ratio of num__5 : num__2 : num__2 respectively but he wanted to make more money so he shifted the ratio to num__2 : num__3 : num__4 respectively . how many more acres of land were planted with tobacco under the new system ? <o> a ) num__75 <o> b ) num__150 <o> c ) num__300 <o> d ) num__450 <o> e ) num__600 |
originally ( num__0.222222222222 ) * num__1350 = num__300 acres were planted with tobacco . in the new system ( num__0.444444444444 ) * num__1350 = num__600 acres were planted with tobacco . thus num__600 - num__300 = num__300 more acres were planted with tobacco . the answer is c . <eor> c <eos> |
c |
multiply__2.0__0.2222__ multiply__2.0__300.0__ divide__600.0__2.0__ |
multiply__2.0__0.2222__ multiply__2.0__300.0__ subtract__600.0__300.0__ |
| a positive integer is divisible by num__3 if and only if the sum of its digits is divisible by num__3 . if the six - digit integer is divisible by num__3 and n is of the form num__1 k num__1 k num__34 where k represents a digit that occurs twice how many values could n have ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__10 |
num__1 k num__1 k num__34 taking sum of the numericals = num__1 + num__1 + num__3 + num__4 = num__9 we require the values of k such that they are also divisible by num__3 num__101034 num__131334 num__161634 num__191934 answer = num__4 = d <eor> d <eos> |
d |
add__3.0__1.0__ add__3.0__1.0__ |
add__3.0__1.0__ add__3.0__1.0__ |
| the average price of three items of furniture is rs . num__15000 . if their prices are in the ratio num__2 : num__4 : num__8 the price of the cheapest item is ? <o> a ) num__2379 <o> b ) num__2889 <o> c ) num__5625 <o> d ) num__9000 <o> e ) num__28311 |
let their prices be num__3 x num__5 x and num__7 x . then num__2 x + num__6 x + num__8 x = ( num__15000 * num__3 ) or x = num__2812.5 . cost of cheapest item = num__2 x = rs . num__5625 . answer : c <eor> c <eos> |
c |
add__2.0__3.0__ add__2.0__5.0__ add__2.0__4.0__ multiply__2.0__2812.5__ multiply__2.0__2812.5__ |
add__2.0__3.0__ add__2.0__5.0__ add__2.0__4.0__ multiply__2.0__2812.5__ multiply__2.0__2812.5__ |
| working at their respective constant rates paul abdul and adam alone can finish a certain work in num__3 num__4 and num__5 hours respectively . if all three work together to finish the work what fraction w of the work will be done by adam ? <o> a ) num__0.25 <o> b ) num__0.255319148936 <o> c ) num__0.333333333333 <o> d ) num__0.416666666667 <o> e ) num__0.425531914894 |
let the total work be num__60 units . pual does num__20.0 = num__20 units of work per hr . abdul does num__15 units per hr and adam does num__12 units per hr . if all work together they do ( num__20 + num__15 + num__12 ) units per hr = num__47 units per hr . so the time taken to finish the work = num__1.27659574468 hrs . adam will do num__1.27659574468 * num__12 units of work in num__1.27659574468 hr . fraction of work adam does = work done by adam / total work w > ( num__1.27659574468 * num__12 ) / num__60 = num__0.255319148936 answer b <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__4.0__5.0__ multiply__3.0__5.0__ multiply__3.0__4.0__ divide__60.0__47.0__ divide__1.2766__5.0__ divide__1.2766__5.0__ |
hour_to_min_conversion__ multiply__4.0__5.0__ multiply__3.0__5.0__ multiply__3.0__4.0__ divide__60.0__47.0__ divide__1.2766__5.0__ divide__1.2766__5.0__ |
| if price of t . v set is reduced by num__10.0 then its sale increases by num__85.0 find net effect on sale value <o> a ) num__44 <o> b ) num__45 <o> c ) num__46 <o> d ) num__67 <o> e ) num__48 |
- a + b + ( ( - a ) ( b ) / num__100 ) = - num__10 + num__85 + ( - num__10 * num__85 ) / num__100 = - num__10 + num__85 - num__8.5 = num__67 answer : d <eor> d <eos> |
d |
percent__10.0__85.0__ percent__67.0__100.0__ |
percent__10.0__85.0__ percent__67.0__100.0__ |
| two trains one from howrah to patna and the other from patna to howrah start simultaneously . after they meet the trains reach their destinations after num__64 hours and num__25 hours respectively . the ratio of their speeds is ? <o> a ) num__4 : num__9 <o> b ) num__4 : num__3 <o> c ) num__4 : num__5 <o> d ) num__5 : num__8 <o> e ) num__4 : num__2 |
let us name the trains a and b . then ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ num__25 : √ num__64 = num__5 : num__8 answer : d <eor> d <eos> |
d |
round__5.0__ |
round__5.0__ |
| the percentage profit earned by selling an article for rs . num__1920 is equal to the percentage loss incurred by selling the same article for rs . num__1280 . at what price should the article be sold to make num__40.0 profit ? <o> a ) num__2000 <o> b ) num__2778 <o> c ) num__2299 <o> d ) num__2778 <o> e ) num__2240 |
let c . p . be rs . x . then ( num__1920 - x ) / x * num__100 = ( x - num__1280 ) / x * num__100 num__1920 - x = x - num__1280 num__2 x = num__3200 = > x = num__1600 required s . p . = num__140.0 of rs . num__1600 = num__1.4 * num__1600 = rs . num__2240 . answer : e <eor> e <eos> |
e |
percent__100.0__2240.0__ |
percent__100.0__2240.0__ |
| the ‘ moving walkway ’ is a num__300 - foot long walkway consisting of a conveyor belt that moves continuously at num__3 feet per second . when bill steps on the walkway a group of people that are also on the walkway stands num__120 feet in front of him . he walks toward the group at a rate of num__3 feet per second . once bill reaches the group of people he stops walking and stands with them until the walkway ends . what is bill ’ s average rate q of movement for his trip along the moving walkway ? <o> a ) num__2 feet per second <o> b ) num__2.5 feet per second <o> c ) num__3 feet per second <o> d ) num__4 feet per second <o> e ) num__5 feet per second |
a b and c do n ' t make any sense ! we know for some time . . . ( a majority of the num__300 ft ) bill will walk at a pace of num__6 ft per second . . . and then at some later time he will stop and move at num__3 ft per second . . . the average must bebetween num__3 - num__6 ft per second or the earth is flat ! ! ! . so we are down to de d does n ' t make sense on a weighted averages level because we know when bill gets on the walkway the patrons are already num__40.0 of the way down the num__300 ft walkway and they are still moving at half the rate of bill ! ! so for bill to average num__4 ft per second he would have had to spend a majority of the num__300 ft at num__3 ft per second because num__4 is much closer ( when you factor in the size of the values we ' re dealing with num__2 is double num__1 ) to num__3 than to num__6 . we know from the information that is n ' t possible . billmust havespent the majority of his time at num__6 ft q per second before he stopped walking . that leaves only answer e as plausible . <eor> e <eos> |
e |
divide__120.0__3.0__ divide__6.0__3.0__ subtract__3.0__2.0__ add__3.0__2.0__ |
divide__120.0__3.0__ subtract__6.0__4.0__ subtract__3.0__2.0__ subtract__6.0__1.0__ |
| if a tire rotates at num__400 revolutions per minute when the car is traveling num__72 km / h what is the circumference of the tire ? <o> a ) num__2 <o> b ) num__1 <o> c ) num__4 <o> d ) num__3 <o> e ) num__5 |
num__400 rev / minute = num__400 * num__60 rev / num__60 minutes = num__24000 rev / hour num__24000 * c = num__72000 m : c is the circumference c = num__3 meters correct answer d <eor> d <eos> |
d |
hour_to_min_conversion__ multiply__400.0__60.0__ divide__72000.0__24000.0__ round__3.0__ |
hour_to_min_conversion__ multiply__400.0__60.0__ divide__72000.0__24000.0__ divide__72000.0__24000.0__ |
| the difference between the place values of two sevens in the numerical num__69758472 is <o> a ) num__0 <o> b ) num__6993 <o> c ) num__699930 <o> d ) num__996330 <o> e ) none of them |
required difference = ( num__700000 - num__70 ) = num__699930 . answer is c <eor> c <eos> |
c |
subtract__700000.0__70.0__ subtract__700000.0__70.0__ |
subtract__700000.0__70.0__ subtract__700000.0__70.0__ |
| what is the middle number in a sequence of num__11 integers with a total sum num__99 ? <o> a ) num__12 <o> b ) num__9 <o> c ) num__5.5 <o> d ) num__7 <o> e ) num__8 |
the average or median is : num__9.0 = num__9 which is the middle in the sequence . answer : b <eor> b <eos> |
b |
divide__99.0__11.0__ divide__99.0__11.0__ |
divide__99.0__11.0__ divide__99.0__11.0__ |
| num__8597 - ? = num__7429 - num__4358 <o> a ) num__5526 <o> b ) num__5536 <o> c ) num__5556 <o> d ) num__5576 <o> e ) num__5586 |
let num__8597 - x = num__7429 - num__4358 . then x = ( num__8597 + num__4358 ) - num__7429 = num__12955 - num__7429 = num__5526 . answer is a <eor> a <eos> |
a |
add__8597.0__4358.0__ subtract__12955.0__7429.0__ subtract__12955.0__7429.0__ |
add__8597.0__4358.0__ subtract__12955.0__7429.0__ subtract__12955.0__7429.0__ |
| a five - digit number divisible by num__3 is to be formed using numerical num__0 num__1 num__2 num__3 num__4 and num__5 without repetition . the total number t of ways this can be done is : <o> a ) num__122 <o> b ) num__210 <o> c ) num__216 <o> d ) num__217 <o> e ) num__225 |
we should determine which num__5 digits from given num__6 would form the num__5 digit number divisible by num__3 . we have six digits : num__0 num__1 num__2 num__3 num__4 num__5 . their sum = num__15 . for a number to be divisible by num__3 the sum of the digits must be divisible by num__3 . as the sum of the six given numbers is num__15 ( divisible by num__3 ) only num__5 digits good to form our num__5 digit number would be num__15 - num__0 = { num__1 num__2 num__3 num__4 num__5 } and num__15 - num__3 = { num__0 num__1 num__2 num__4 num__5 } . meaning that no other num__5 from given six will total the number divisible by num__3 . second step : we have two set of numbers : num__1 num__2 num__3 num__4 num__5 and num__0 num__1 num__2 num__4 num__5 . how many num__5 digit numbers can be formed using these two sets : num__1 num__2 num__3 num__4 num__5 - - > num__5 ! as any combination of these digits would give us num__5 digit number divisible by num__3 . num__5 ! = num__120 . num__0 num__1 num__2 num__4 num__5 - - > here we can not use num__0 as the first digit otherwise number wo n ' t be any more num__5 digit and become num__4 digit . so desired # would be total combinations num__5 ! minus combinations with num__0 as the first digit ( combination of num__4 ) num__4 ! - - > num__5 ! - num__4 ! = num__4 ! ( num__5 - num__1 ) = num__4 ! * num__4 = num__96 num__120 + num__96 = num__216 = y answer : c . <eor> c <eos> |
c |
multiply__3.0__2.0__ multiply__3.0__5.0__ add__96.0__120.0__ multiply__1.0__216.0__ |
multiply__3.0__2.0__ multiply__3.0__5.0__ add__96.0__120.0__ multiply__1.0__216.0__ |
| a factory producing tennis balls stores them in either big boxes num__25 balls per box or small boxes num__20 balls per box . if num__104 freshly manufactured balls are to be stored what is the least number of balls that can be left unboxed ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
we have to work with multiples of num__20 and num__25 . first we must know the limits of this multiples so : num__4.2 = num__4 . . . . so the max is num__4 num__5.25 = num__5 . . . so the max is num__5 num__105 - num__100 = num__5 answer : e <eor> e <eos> |
e |
round_down__4.2__ round_down__5.25__ multiply__25.0__4.2__ multiply__25.0__4.0__ round_down__5.25__ |
round_down__4.2__ subtract__25.0__20.0__ multiply__25.0__4.2__ subtract__104.0__4.0__ subtract__25.0__20.0__ |
| a train num__100 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is ? <o> a ) num__28 <o> b ) num__50 <o> c ) num__88 <o> d ) num__41 <o> e ) num__12 |
speed of the train relative to man = ( num__10.0 ) m / sec = ( num__10 ) m / sec . [ ( num__10 ) * ( num__3.6 ) ] km / hr = num__36 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__36 = = > x = num__41 km / hr . answer : d <eor> d <eos> |
d |
multiply__10.0__3.6__ add__5.0__36.0__ round__41.0__ |
multiply__10.0__3.6__ add__5.0__36.0__ round__41.0__ |
| a boat covers a certain distance downstream in num__4 hours but takes num__6 hours to return upstream to the starting point . if the speed of the stream be num__3 km / hr find the speed of the boat in still water <o> a ) num__15 km / hr <o> b ) num__12 km / hr <o> c ) num__13 km / hr <o> d ) num__14 km / hr <o> e ) none of these |
explanation : let the speed of the water in still water = x given that speed of the stream = num__3 kmph speed downstream = ( x + num__3 ) kmph speed upstream = ( x - num__3 ) kmph he travels a certain distance downstream in num__4 hour and come back in num__6 hour . ie distance travelled downstream in num__4 hour = distance travelled upstream in num__6 hour since distance = speed × time we have ( x + num__3 ) num__4 = ( x − num__3 ) num__6 = > ( x + num__3 ) num__2 = ( x - num__3 ) num__3 = > num__2 x + num__6 = num__3 x - num__9 = > x = num__6 + num__9 = num__15 kmph . answer : option a <eor> a <eos> |
a |
subtract__6.0__4.0__ add__6.0__3.0__ add__6.0__9.0__ round__15.0__ |
subtract__6.0__4.0__ add__6.0__3.0__ add__6.0__9.0__ add__6.0__9.0__ |
| carmen made a sculpture from small pieces of wood . the sculpture is num__2 feet num__10 inches tall . carmen places her sculpture on a base that is num__4 inches tall . how tall are the sculpture andbase together ? <o> a ) num__3.17 feet <o> b ) num__3.2 feet <o> c ) num__3.3 feet <o> d ) num__3.4 feet <o> e ) num__3.5 feet |
we know num__1 feet = num__12 inch then num__2 feet = num__24 inch num__24 + num__10 = num__34 then num__34 + num__4 = num__38 num__3.16666666667 = num__3.17 feet answer : a <eor> a <eos> |
a |
rectangle_perimeter__2.0__4.0__ surface_cube__2.0__ triangle_perimeter__2.0__12.0__24.0__ triangle_area__2.0__3.17__ |
rectangle_perimeter__2.0__4.0__ surface_cube__2.0__ triangle_perimeter__2.0__12.0__24.0__ triangle_area__2.0__3.17__ |
| by investing in num__1623.0 stock at num__64 one earns rs . num__1800 . the investment made is <o> a ) s . num__6500 <o> b ) s . num__7500 <o> c ) s . num__5640 <o> d ) s . num__5760 <o> e ) s . num__6912 |
explanation : market value = rs . num__64 face value is not given and hence take it as rs . num__100 num__16 num__0.666666666667 % of the face value = num__16.6666666667 ie to earn num__16.6666666667 investment = rs . num__64 hence to earn rs . num__1800 investment needed = num__64 × num__3 × num__36.0 = num__6912 answer : option e <eor> e <eos> |
e |
percent__100.0__6912.0__ |
percent__100.0__6912.0__ |
| two stations a and b are num__200 km apart on a straight track . one train starts from a at num__7 a . m . and travels towards b at num__20 kmph . another train starts from b at num__8 a . m . and travels towards a at a speed of num__25 kmph . at what time will they meet ? <o> a ) num__02 p . m . <o> b ) num__06 p . m . <o> c ) num__10 p . m . <o> d ) num__12 p . m . <o> e ) num__11 p . m . |
explanation : assume both trains meet after ' p ' hours after num__7 a . m . distance covered by train starting from a in ' p ' hours = num__20 p km distance covered by train starting from b in ( p - num__1 ) hours = num__25 ( p - num__1 ) total distance = num__200 = > num__20 x + num__25 ( x - num__1 ) = num__200 = > num__45 x = num__225 = > p = num__5 means they meet after num__5 hours after num__7 am ie they meet at num__12 p . m . answer : d <eor> d <eos> |
d |
subtract__8.0__7.0__ add__20.0__25.0__ add__200.0__25.0__ subtract__25.0__20.0__ add__7.0__5.0__ round__12.0__ |
subtract__8.0__7.0__ add__20.0__25.0__ add__200.0__25.0__ subtract__25.0__20.0__ add__7.0__5.0__ add__7.0__5.0__ |
| num__0 num__1 num__4 num__9 num__16 num__25 num__36 num__49 ? what number should replace the question mark ? <o> a ) num__81 <o> b ) num__74 <o> c ) num__64 <o> d ) num__24 <o> e ) num__48 |
c num__64 add num__0 ^ num__21 ^ num__22 ^ num__2 . . . . . . <eor> c <eos> |
c |
multiply__4.0__16.0__ subtract__25.0__4.0__ add__1.0__21.0__ multiply__1.0__64.0__ |
multiply__4.0__16.0__ subtract__25.0__4.0__ add__1.0__21.0__ power__64.0__1.0__ |
| a can contains a mixture of liquids a and b is the ratio num__7 : num__5 . when num__3 litres of mixture are drawn off and the can is filled with b the ratio of a and b becomes num__7 : num__9 . how many liter of liquid a was contained by the can initially ? <o> a ) num__7 <o> b ) num__10 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
ci * vi = cf * vf ( num__0.583333333333 ) * ( v num__1 - num__3 ) = ( num__0.4375 ) * v num__1 ( v num__1 - num__3 ) / v num__1 = num__0.75 num__3 accounts for the difference of num__1 on ratio scale so initial volume = v num__1 = num__4 * num__3 = num__12 litres . num__0.583333333333 of the initial mixture was liquid a so liquid a was ( num__0.583333333333 ) * num__12 = num__7 litres . answer : a <eor> a <eos> |
a |
divide__0.4375__0.5833__ subtract__7.0__3.0__ add__7.0__5.0__ multiply__7.0__1.0__ |
divide__0.4375__0.5833__ subtract__7.0__3.0__ multiply__3.0__4.0__ multiply__7.0__1.0__ |
| if the area of a square with sides of length num__11 centimeters is equal to the area of a rectangle with a width of num__11 centimeters what is the length of the rectangle in centimeters ? <o> a ) num__4 <o> b ) num__8 <o> c ) num__11 <o> d ) num__16 <o> e ) num__18 |
let length of rectangle = l num__11 ^ num__2 = l * num__11 = > l = num__11.0 = num__11 answer c <eor> c <eos> |
c |
triangle_area__11.0__2.0__ |
triangle_area__11.0__2.0__ |
| ( num__3080 + num__6160 ) / num__28 <o> a ) num__380 <o> b ) num__350 <o> c ) num__330 <o> d ) num__310 <o> e ) none of these |
explanation : as per bodmas rule first we will solve the equation in bracket then we will go for division = ( num__9240 ) / num__28 = num__330 option c <eor> c <eos> |
c |
add__3080.0__6160.0__ divide__9240.0__28.0__ divide__9240.0__28.0__ |
add__3080.0__6160.0__ divide__9240.0__28.0__ divide__9240.0__28.0__ |
| how many num__3 digit numbers are divisible by num__6 in all ? <o> a ) num__150 <o> b ) num__111 <o> c ) num__120 <o> d ) num__243 <o> e ) num__345 |
a num__150 required numbers are num__102 num__108 num__114 . . . num__996 this is an a . p . in which a = num__102 d = num__6 and l = num__996 let the number of terms be n . then a + ( n - num__1 ) d = num__996 num__102 + ( n - num__1 ) x num__6 = num__996 num__6 x ( n - num__1 ) = num__894 ( n - num__1 ) = num__149 n = num__150 . <eor> a <eos> |
a |
add__6.0__102.0__ add__6.0__108.0__ subtract__996.0__102.0__ subtract__150.0__1.0__ add__1.0__149.0__ |
add__6.0__102.0__ add__6.0__108.0__ subtract__996.0__102.0__ subtract__150.0__1.0__ add__1.0__149.0__ |
| during a pizza buffet where a eats more times num__4 than b and b eats num__8 times less than c . find the least number of times all the three has to eat <o> a ) num__7 <o> b ) num__5 <o> c ) num__9 <o> d ) num__2 <o> e ) num__4 |
a eats more than b if b eats num__1 times than the ratio of a and b is a : b is num__4 : num__1 and as b eat num__8 times less the c the the ratio of b : c is num__1 : num__8 the the least number of times all three has eat is the lcm of a b c that is num__8 . . answer : e <eor> e <eos> |
e |
multiply__4.0__1.0__ |
multiply__4.0__1.0__ |
| the ratio between the present ages of p and q is num__5 : num__7 respectively . if the difference between q ' s present age and p ' s age after num__6 years is num__2 . what is the total of p ' s and q ' s present ages ? <o> a ) num__48 years <o> b ) num__65 years <o> c ) num__87 years <o> d ) num__20 years <o> e ) num__26 years |
let the present ages of p and q be num__5 x and num__7 x years respectively . then num__7 x - ( num__5 x + num__6 ) = num__2 num__2 x = num__8 = > x = num__4 required sum = num__5 x + num__7 x = num__12 x = num__48 years . answer : a <eor> a <eos> |
a |
add__6.0__2.0__ subtract__6.0__2.0__ add__5.0__7.0__ multiply__6.0__8.0__ multiply__6.0__8.0__ |
add__6.0__2.0__ subtract__6.0__2.0__ add__5.0__7.0__ multiply__6.0__8.0__ multiply__6.0__8.0__ |
| a pipe can fill a cistern in num__12 minutes whereas the cistern when fill can be emptied by a leak in num__18 minutes . when both pipes are opened find when the cistern will be full ? <o> a ) num__17 <o> b ) num__16 <o> c ) num__70 <o> d ) num__36 <o> e ) num__12 |
num__0.0833333333333 - num__0.0555555555556 = num__0.0277777777778 num__36 minutes answer : d <eor> d <eos> |
d |
round__36.0__ |
round__36.0__ |
| tommo spent six lucky days in las vegas . on his first day he won a net amount of only $ num__20 but on each of the following days the daily net amount he won grew by d dollars . if tommo won a total net amount of $ num__1620 during his stay in las vegas how much did he win on the last day ? <o> a ) num__330 <o> b ) num__500 <o> c ) num__520 <o> d ) num__540 <o> e ) num__620 |
i drew a diagram : num__1 - $ num__20 num__2 - num__3 - num__4 - $ num__320 num__5 - num__6 - total : $ num__1620 between num__1 and num__6 are num__5 days where he won $ num__1600 . this means he averaged $ num__320 per day ( num__320.0 ) . you can put $ num__320 by num__4 because it ' s the middle number . now you just find the two points betwenn $ num__20 and $ num__320 ( num__320 - num__20 = num__100.0 = num__100 ) . so each day he earned $ num__100 more . this means on day num__6 tommo earned $ num__520 . answer choice c . <eor> c <eos> |
c |
add__1.0__2.0__ add__1.0__3.0__ divide__20.0__4.0__ add__1.0__5.0__ subtract__1620.0__20.0__ multiply__20.0__5.0__ multiply__1.0__520.0__ |
add__1.0__2.0__ add__1.0__3.0__ divide__20.0__4.0__ add__1.0__5.0__ subtract__1620.0__20.0__ multiply__20.0__5.0__ multiply__1.0__520.0__ |
| a bus moving at a speed of num__45 km / hr . overtakes a truck num__150 metres ahead going in the same directions in num__30 seconds . the speed of the truck is <o> a ) num__27 km / hr <o> b ) num__24 km / hr <o> c ) num__25 km / hr <o> d ) num__28 km / hr <o> e ) none of these |
explanation : the lag of num__150 m has been covered by the bus in num__30 seconds . therefore the relative speed of the bus and truck is = ( num__5 x num__18 ) / num__5 = num__18 km / h the speed of truck is num__45 - num__18 = num__27 km / hr answer a <eor> a <eos> |
a |
divide__150.0__30.0__ subtract__45.0__18.0__ round__27.0__ |
divide__150.0__30.0__ subtract__45.0__18.0__ subtract__45.0__18.0__ |
| in triangle pqr the angle q = num__90 degree pq = num__3 cm qr = num__8 cm . x is a variable point on pq . the line through x parallel to qr intersects pr at y and the line through y parallel to pq intersects qr at z . find the least possible length of xz <o> a ) num__3.6 cm <o> b ) num__2.4 cm <o> c ) num__4.8 cm <o> d ) num__2.16 cm <o> e ) num__3.2 cm |
look at the diagram below : now in case when qy is perpendicular to pr two right triangles pqr and pqy are similar : qy : qp = qr : pr - - > qy : num__3 = num__8 : num__10 - - > qy = num__2.4 . answer : b . <eor> b <eos> |
b |
round__2.4__ |
round__2.4__ |
| if the integer n = num__139 k num__50 where k is the hundreds digit n can never be divisible by which of the following ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__6 <o> e ) num__9 |
it is divisible by num__2 irrespective of k value as unit digit ' s value is even ( num__0 ) . it can be divisible by num__3 for k values num__2 num__58 as sum will be divisible by num__3 it can be divisible by num__6 if divisible by num__2 and num__3 . it can be divisible by num__9 for k value num__2 as sum will be divisible by num__9 . therefore answer is num__4 as number is divisible if last two digits are divisible . in this case num__50 ( no ) hence answer c <eor> c <eos> |
c |
multiply__2.0__3.0__ add__3.0__6.0__ subtract__6.0__2.0__ subtract__6.0__2.0__ |
multiply__2.0__3.0__ add__3.0__6.0__ subtract__6.0__2.0__ subtract__6.0__2.0__ |
| find the value of ( num__55 + num__1.44871794872 ) × num__78 <o> a ) num__4423 <o> b ) num__4403 <o> c ) num__4413 <o> d ) num__2403 <o> e ) num__4375 |
= ( num__55 + num__1.44871794872 ) × num__78 = ( num__4290 + num__113 ) / num__78 × num__78 = num__56.4487179487 × num__78 = num__4403 answer is b . <eor> b <eos> |
b |
multiply__55.0__78.0__ add__55.0__1.4487__ add__4290.0__113.0__ add__4290.0__113.0__ |
multiply__55.0__78.0__ add__55.0__1.4487__ add__4290.0__113.0__ add__4290.0__113.0__ |
| num__3 men can complete a piece of work in num__6 days . two days after they started the work num__3 more men joined them . how many days will they take to complete the remaining work ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
explanation : num__3 man num__1 day work = num__1616 num__3 man num__2 days work = num__2626 remaining work = ( num__1 – num__2626 ) = num__0.666666666667 parts . num__6 man together perform the work in num__1 day is = num__1616 + num__1616 = num__2626 parts num__2626 parts completed in num__1 day num__2323 parts will b completed in num__2 days answer : a <eor> a <eos> |
a |
subtract__3.0__1.0__ divide__2.0__3.0__ round__2.0__ |
subtract__3.0__1.0__ divide__2.0__3.0__ round__2.0__ |
| if n is a prime number which of the following could be true <o> a ) n ^ n = n <o> b ) n ^ n / num__4 = num__1 ^ ( n - num__1 ) <o> c ) ( n ) ( n ^ n ) = is negative <o> d ) n ^ num__2 + n ^ num__3 = n ^ num__5 <o> e ) n ^ num__0.5 = is even |
b should be the answer . num__2 ^ num__0.5 = num__1 ^ ( num__2 - num__1 ) <eor> b <eos> |
b |
reverse__2.0__ multiply__0.5__2.0__ divide__2.0__0.5__ |
reverse__2.0__ multiply__0.5__2.0__ divide__2.0__0.5__ |
| the vertex of a parallelogram are ( num__1 num__0 ) ( num__3 num__0 ) ( num__1 num__1 ) and ( num__3 num__1 ) respectively . if line l passes through the origin and divided the parallelogram into two identical quadrilaterals what is the slope of line l ? <o> a ) . num__0.5 <o> b ) num__2 <o> c ) num__0.25 <o> d ) num__3 <o> e ) num__0.75 |
soln : - added the line be ( the required line that divides the given parallelogram into num__2 identical quadrilaterals ) . careful observation reveals that slope of the diagonal ad = ( num__1 - num__0 ) / ( num__3 - num__1 ) = num__0.5 . thus the slope of line be should be < slope of ad only option c is < num__0.5 and is thus the correct answer . <eor> c <eos> |
c |
subtract__3.0__1.0__ reverse__2.0__ divide__0.5__2.0__ |
subtract__3.0__1.0__ reverse__2.0__ divide__0.5__2.0__ |
| if a man can cover num__14 metres in one second how many kilometres can he cover in num__3 hours num__45 minutes ? <o> a ) num__288 <o> b ) num__162 <o> c ) num__189 <o> d ) num__168 <o> e ) num__122 |
num__14 m / s = num__14 * num__3.6 kmph num__3 hours num__45 minutes = num__3 num__0.75 hours = num__3.75 hours distance = speed * time = num__14 * num__3.6 * num__3.75 km = num__189 km . answer : c <eor> c <eos> |
c |
add__3.0__0.75__ round__189.0__ |
add__3.0__0.75__ round__189.0__ |
| an aeroplane covers a certain distance of num__420 kmph in num__6 hours . to cover the same distance in num__4 num__0.666666666667 hours it must travel at a speed of <o> a ) num__580 kmph <o> b ) num__550 kmph <o> c ) num__540 kmph <o> d ) num__530 kmph <o> e ) num__520 kmph |
explanation : speed of aeroplane = num__420 kmph distance travelled in num__6 hours = num__420 * num__6 = num__2520 km speed of aeroplane to acver num__2520 km in num__4.66666666667 = num__2520 * num__0.214285714286 = num__540 km answer : option c <eor> c <eos> |
c |
multiply__420.0__6.0__ add__4.0__0.6667__ round__540.0__ |
multiply__420.0__6.0__ add__4.0__0.6667__ round__540.0__ |
| the area of a triangle is with base num__4 m and height num__5 m ? <o> a ) num__76 m num__2 <o> b ) num__10 m num__2 <o> c ) num__27 m num__2 <o> d ) num__18 m num__2 <o> e ) num__17 m num__2 |
num__0.5 * num__4 * num__5 = num__10 m num__2 answer : b <eor> b <eos> |
b |
triangle_area__4.0__5.0__ square_perimeter__0.5__ triangle_area__4.0__5.0__ |
volume_rectangular_prism__4.0__5.0__0.5__ multiply__4.0__0.5__ multiply__5.0__2.0__ |
| the ratio between the speeds of two bullet trains is num__7 : num__8 . if the second bullet train runs num__520 kms in num__4 hours then the speed of the first train is : <o> a ) num__89.25 km / hr . <o> b ) num__37.25 km / hr . <o> c ) num__90.25 km / hr . <o> d ) num__113.75 km / hr . <o> e ) num__96.25 km / hr . |
let the speed of two bullet trains be num__7 x and num__8 x km / hr . then num__8 x = ( num__130.0 ) = num__130 x = ( num__16.25 ) = num__16.25 hence speed of first bullet train = ( num__7 x num__16.25 ) km / hr = num__113.75 km / hr . d <eor> d <eos> |
d |
divide__520.0__4.0__ divide__130.0__8.0__ multiply__7.0__16.25__ round__113.75__ |
divide__520.0__4.0__ divide__130.0__8.0__ multiply__7.0__16.25__ round__113.75__ |
| of the num__600 residents of clermontville num__35.0 watch the television show island survival num__40.0 watch lovelost lawyers and num__50.0 watch medical emergency . if all residents watch at least one of these three shows and num__18.0 watch exactly num__2 of these shows then how many clermontville residents z watch all of the shows ? <o> a ) num__150 <o> b ) num__108 <o> c ) num__42 <o> d ) num__21 <o> e ) - num__21 |
oa is d . num__100 = a + b + c - ab - ac - bc + abc which is the same as the following formula num__100 = a + b + c + ( - ab - ac - bc + abc + abc + abc ) - num__2 abc . the term between parantheses value num__18.0 so the equation to resolve is num__100 = num__35 + num__40 + num__50 - num__18 - num__2 abc therefore the value of abc is z = num__3.5 of num__600 is num__21 . d is the correct answer <eor> d <eos> |
d |
percent__3.5__600.0__ percent__3.5__600.0__ |
percent__3.5__600.0__ percent__3.5__600.0__ |
| if a and b are both negative and ab < b ^ num__2 which of the following must be true ? <o> a ) a < b < a ^ num__2 < b ^ num__2 <o> b ) a < b < b ^ num__2 < a ^ num__2 <o> c ) b < a < a ^ num__2 < b ^ num__2 <o> d ) a ^ num__2 < b ^ num__2 < b < a <o> e ) b ^ num__2 < a ^ num__2 < b < a |
because ab < b ^ num__2 and both are negative i thought a < b . so i crossed off answers c ) d ) and e ) . and because a < b a ^ num__2 < b ^ num__2 ans c <eor> c <eos> |
c |
coin_space__ |
coin_space__ |
| the sale price of an article including the sales tax is rs . num__616 . the rate of sales tax is num__10.0 . if the shopkeeper has made a profit of num__17.0 then the cost price of the article is : <o> a ) num__500 <o> b ) num__334 <o> c ) num__555 <o> d ) num__664 <o> e ) num__526 |
num__110.0 of s . p . = num__616 s . p . = ( num__616 * num__100 ) / num__110 = rs . num__560 c . p = ( num__110 * num__560 ) / num__117 = rs . num__526 answer : option e <eor> e <eos> |
e |
percent__100.0__526.0__ |
percent__100.0__526.0__ |
| num__1000 men have provisions for num__15 days . if num__200 more men join them for how many days will the provisions last now ? <o> a ) num__12.9 <o> b ) num__12.0 <o> c ) num__12.5 <o> d ) num__12.2 <o> e ) num__12.1 |
num__1000 * num__15 = num__1200 * x x = num__12.5 answer : c <eor> c <eos> |
c |
add__1000.0__200.0__ round__12.5__ |
add__1000.0__200.0__ round__12.5__ |
| a fort had provision of food for num__150 men for num__45 days . after num__10 days num__25 men left the fort . the number of days for which the remaining food will last is : <o> a ) num__29 <o> b ) num__37 <o> c ) num__54 <o> d ) num__42 <o> e ) num__60 |
after num__10 days : num__150 men had food for num__35 days . suppose num__125 men had food for x days . now less men more days ( indirect proportion ) num__125 : num__150 : : num__35 : x num__125 * x = num__150 x num__35 x = ( num__150 x num__35 ) / num__125 x = num__42 . answer is d . <eor> d <eos> |
d |
subtract__45.0__10.0__ subtract__150.0__25.0__ round__42.0__ |
subtract__45.0__10.0__ subtract__150.0__25.0__ round__42.0__ |
| the average age of students of a class is num__15.7 years . the average age of boys in the class is num__16.4 years and that of the girls is num__15.4 years . the ration of the number of boys to the number of girls in the class is ? <o> a ) num__3 : num__7 <o> b ) num__7 : num__3 <o> c ) num__2 : num__5 <o> d ) num__2 : num__1 <o> e ) num__2 : num__4 |
let the ratio be k : num__1 . then k * num__16.4 + num__1 * num__15.4 = ( k + num__1 ) * num__15.7 = ( num__16.4 - num__15.7 ) k = ( num__15.7 - num__15.4 ) = k = num__0.3 / num__0.7 = num__0.428571428571 required ratio = num__0.428571428571 : num__1 = num__3 : num__7 . answer : a <eor> a <eos> |
a |
subtract__16.4__15.4__ subtract__15.7__15.4__ subtract__16.4__15.7__ divide__0.3__0.7__ multiply__1.0__3.0__ |
subtract__16.4__15.4__ subtract__15.7__15.4__ subtract__16.4__15.7__ divide__0.3__0.7__ multiply__1.0__3.0__ |
| when the smallest of num__3 consecutive odd integers is added to four times the largest it produces a result num__731 more than num__4 times the middle integer . find the numbers ? <o> a ) num__650 <o> b ) num__731 <o> c ) num__698 <o> d ) num__710 <o> e ) num__729 |
x + num__4 ( x + num__4 ) = num__731 + num__4 ( x + num__2 ) solve for x and find all three numbers x + num__4 x + num__16 = num__731 + num__4 x + num__8 x = num__723 x + num__2 = num__725 x + num__4 = num__727 check : the smallest is added to four times the largest num__723 + num__4 * num__727 = num__3631 four times the middle num__4 * num__725 = num__2900 num__3631 is more than num__2900 by num__3631 - num__2900 = num__731 b <eor> b <eos> |
b |
multiply__4.0__2.0__ subtract__731.0__8.0__ add__2.0__723.0__ subtract__731.0__4.0__ multiply__4.0__725.0__ add__4.0__727.0__ |
multiply__4.0__2.0__ subtract__731.0__8.0__ add__2.0__723.0__ subtract__731.0__4.0__ multiply__4.0__725.0__ add__4.0__727.0__ |
| three positive integers a b and c are such that their average is num__22 and a ≤ b ≤ c . if the median is ( a + num__13 ) what is the least possible value of c ? <o> a ) num__27 <o> b ) num__26 <o> c ) num__25 <o> d ) num__24 <o> e ) num__23 |
solution : we know that the average value is num__22 therefore ( a + b + c ) / num__3 = num__22 so a + b + c = num__66 and b = a + num__13 therefore a + ( a + num__13 ) + c = num__66 the least value of c is when c = b so take c = a + num__13 hence a + ( a + num__13 ) + ( a + num__13 ) = num__66 i . e . a = num__13.3333333333 a non integer therefore c is not equal to b so take c = b + num__1 therefore c = a + num__14 a + ( a + num__13 ) + ( a + num__14 ) = num__66 i . e . a = num__13.0 = num__13 hence a = num__13 b = num__26 and c = num__27 answer : a <eor> a <eos> |
a |
multiply__22.0__3.0__ add__13.0__1.0__ add__13.0__14.0__ add__13.0__14.0__ |
multiply__22.0__3.0__ add__13.0__1.0__ add__13.0__14.0__ add__13.0__14.0__ |
| all of the stocks on the over - the - counter market are designated by either a num__5 - letter or a num__5 - letter code that is created by using the num__26 letters of the alphabet . which of the following gives the maximum number of different stocks that can be designated with these codes ? <o> a ) num__2 ( num__26 ^ num__5 ) <o> b ) num__26 ( num__26 ^ num__4 ) <o> c ) num__27 ( num__26 ^ num__4 ) <o> d ) num__26 ( num__26 ^ num__5 ) <o> e ) num__27 ( num__26 ^ num__5 ) |
number of num__4 - letter codes : num__26 * num__26 * num__26 * num__26 * num__26 = num__26 ^ num__5 number of num__5 - letter codes : num__26 * num__26 * num__26 * num__26 * num__26 = num__26 ^ num__5 total number of codes : num__26 ^ num__5 + num__26 ^ num__5 = num__2 * ( num__26 ^ num__5 ) therefore the answer isa : num__2 * ( num__26 ^ num__5 ) . <eor> a <eos> |
a |
subtract__4.0__2.0__ |
subtract__4.0__2.0__ |
| find the odd man out num__3 num__7 num__15 num__27 num__63 num__127 num__255 <o> a ) num__7 <o> b ) num__15 <o> c ) num__27 <o> d ) num__63 <o> e ) num__54 |
num__1 * num__2 + num__1 = num__3 num__3 * num__2 + num__1 = num__7 num__7 * num__2 + num__1 = num__15 num__15 * num__2 + num__1 = num__31 num__31 * num__2 + num__1 = num__63 num__63 * num__2 + num__1 = num__127 num__127 * num__2 + num__1 = num__255 answer : c <eor> c <eos> |
c |
subtract__3.0__1.0__ multiply__27.0__1.0__ |
subtract__3.0__1.0__ multiply__27.0__1.0__ |
| how many possible ways can num__4 girls ( rebecca kate ashley ) go on a date with num__4 boys ( peter kyle sam ) ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__24 |
any one girl can go with num__4 boys . num__2 nd girl can go with remaining num__3 boys num__3 nd girl can go with remaining num__2 boys and num__3 rd girl can go with remaining num__1 boy num__4 ! i . e num__24 ( e ) is the answer <eor> e <eos> |
e |
coin_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
coin_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
| the least number which when divided by num__12 num__31 num__20 and num__54 leaves in each case a remainder of num__8 is : <o> a ) num__448 <o> b ) num__488 <o> c ) num__16748 <o> d ) num__548 <o> e ) num__560 |
required number = ( l . c . m . of num__12 num__31 num__20 num__54 ) + num__8 = num__16740 + num__8 = num__16748 . answer : c <eor> c <eos> |
c |
add__8.0__16740.0__ add__8.0__16740.0__ |
add__8.0__16740.0__ add__8.0__16740.0__ |
| if a job is complete by a and b in z days b and c in x days and a and c in y days . in how many days that a b and c can be work simultaneously ? <o> a ) num__2 ( xyz / xy + yz + xz ) <o> b ) xyz / xy + yz + xz <o> c ) ( xy + yz + xz ) / xyz <o> d ) can not be determined <o> e ) num__2 ( xy + yz + xz ) / xyz |
option a is answer . one day work of a + b b + c and c + a is num__1 / z num__1 / x and num__1 / y . that is num__2 ( a + b + c ) = ( yz + xz + xy ) / xyz . a + b + c = ( yz + xz + xy ) / num__2 xyz . ( one day work ) . total work done by a b and c is num__2 xyz / ( xy + yz + xz ) . <eor> a <eos> |
a |
round__2.0__ |
divide__2.0__1.0__ |
| a plane flies num__660 km with the wind and num__540 km against the wind in the same length of time . if the speed of the wind is num__20 km / h what is the speed of the plane in still air ? <o> a ) num__160 km / h <o> b ) num__180 km / h <o> c ) num__200 km / h <o> d ) num__220 km / h <o> e ) num__240 km / h |
the speed of the plane in still air = x km / h the speed of the wind is num__20 km / h speed with the wind = ( x + num__20 ) km / h speed against the wind = ( x â € “ num__20 ) km / h time = distance / speed num__660 / ( x + num__20 ) = num__540 / ( x - num__20 ) num__660 ( x - num__20 ) = num__540 ( x + num__20 ) num__66 x - num__1320 = num__54 x + num__1080 num__12 x = num__2400 x = num__200 therefore the speed of the plane in still air is num__200 km / h . the answer is c . <eor> c <eos> |
c |
multiply__20.0__66.0__ multiply__20.0__54.0__ subtract__66.0__54.0__ add__1320.0__1080.0__ divide__2400.0__12.0__ round__200.0__ |
multiply__20.0__66.0__ multiply__20.0__54.0__ subtract__66.0__54.0__ add__1320.0__1080.0__ divide__2400.0__12.0__ divide__2400.0__12.0__ |
| a cistern has a leak which would empty the cistern in num__20 minutes . a tap is turned on which admits num__2 liters a minute into the cistern and it is emptied in num__24 minutes . how many liters does the cistern hold ? <o> a ) num__480 <o> b ) num__240 <o> c ) num__289 <o> d ) num__270 <o> e ) num__927 |
num__1 / x - num__0.05 = - num__0.0416666666667 x = num__120 num__120 * num__2 = num__240 answer : b <eor> b <eos> |
b |
divide__1.0__20.0__ divide__1.0__24.0__ multiply__2.0__120.0__ round__240.0__ |
divide__1.0__20.0__ divide__1.0__24.0__ multiply__2.0__120.0__ multiply__2.0__120.0__ |
| what sum of money will produce rs . num__70 as simple interest in num__4 years at num__3 num__0.5 percent ? <o> a ) num__176 <o> b ) num__500 <o> c ) num__278 <o> d ) num__270 <o> e ) num__279 |
num__70 = ( p * num__4 * num__3.5 ) / num__100 p = num__500 answer : b <eor> b <eos> |
b |
percent__100.0__500.0__ |
percent__100.0__500.0__ |
| a train running at the speed of num__90 km / hr crosses a pole in num__9 sec . what is the length of the train ? <o> a ) num__298 m <o> b ) num__225 m <o> c ) num__208 m <o> d ) num__988 m <o> e ) num__299 m |
speed = num__90 * num__0.277777777778 = num__25.0 m / sec length of the train = speed * time = num__25 * num__9 = num__225 m answer : b <eor> b <eos> |
b |
multiply__9.0__25.0__ round__225.0__ |
multiply__9.0__25.0__ multiply__9.0__25.0__ |
| two girls started running simultaneously around a circular track of length num__1200 m from the same point at speeds of num__30 km / hr and num__50 km / hr . when will they meet for the first time any where on the track if they are moving in opposite directions ? <o> a ) num__50 <o> b ) num__51 <o> c ) num__52 <o> d ) num__53 <o> e ) num__54 |
time taken to meet for the first time anywhere on the track = length of the track / relative speed = num__1200 / ( num__30 + num__50 ) num__0.277777777778 = num__1200 * num__0.225 * num__5 = num__54 seconds . answer : e <eor> e <eos> |
e |
round__54.0__ |
round__54.0__ |
| num__0.375 of num__168 * num__3.0 + x = num__61.0 + num__275 <o> a ) num__147 <o> b ) num__149 <o> c ) num__268 <o> d ) num__2696 <o> e ) num__2976 |
explanation : let num__0.375 of num__168 * num__3.0 + x = num__61.0 + num__275 then num__63 * num__3.0 + x = num__61 + num__275 num__63 * num__3 + x = num__336 num__189 + x = num__336 x = num__147 answer : a <eor> a <eos> |
a |
multiply__0.375__168.0__ add__61.0__275.0__ multiply__3.0__63.0__ subtract__336.0__189.0__ subtract__336.0__189.0__ |
multiply__0.375__168.0__ add__61.0__275.0__ multiply__3.0__63.0__ subtract__336.0__189.0__ subtract__336.0__189.0__ |
| the rental charge for a car is num__34 cents for the first num__0.25 mile driven and num__6 cents for every num__0.2 mile driven over the initial num__0.25 mile . if a man paid $ num__1.24 in rental charges how many miles did he drive ? <o> a ) num__2.5 <o> b ) num__3.00 <o> c ) num__3.25 <o> d ) num__3.75 <o> e ) num__4.0 |
for first num__0.25 miles num__34 cents . so the person is left with num__1.24 - num__0.34 = num__0.9 so the person can travel next ( num__15.0 ) times num__0.2 miles i . e num__15 * num__0.2 miles = num__3 miles . so total distance is num__3 + num__0.25 = num__3.25 miles ( c ) <eor> c <eos> |
c |
subtract__1.24__0.34__ multiply__0.2__15.0__ add__0.25__3.0__ round__3.25__ |
subtract__1.24__0.34__ multiply__0.2__15.0__ add__0.25__3.0__ add__0.25__3.0__ |
| in how many ways can the letters of the word education be rearranged so that the relative position of the vowels and consonants remain the same as in the word education ? <o> a ) num__4 ! x num__9 ! <o> b ) num__4 ! x num__2 ! <o> c ) num__4 ! x num__5 ! <o> d ) num__4 ! x num__6 ! <o> e ) num__7 ! x num__9 ! |
explanation : the word education is a num__9 letter word with none of the letters repeating . the vowels occupy num__3 rd num__5 th num__7 th and num__8 th position in the word and the remaining num__5 positions are occupied by consonants as the relative position of the vowels and consonants in any arrangement should remain the same as in the word education the vowels can occupy only the afore mentioned num__4 places and the consonants can occupy num__1 st num__2 nd num__4 th num__6 th and num__9 th positions . the num__4 vowels can be arranged in the num__3 rd num__5 th num__7 th and num__8 th position in num__4 ! ways . similarly the num__5 consonants can be arranged in num__1 st num__2 nd num__4 th num__6 th and num__9 th position in num__5 ! ways . hence the total number of ways = num__4 ! × num__5 ! answer : c ) num__4 ! x num__5 ! <eor> c <eos> |
c |
vowel_space__ coin_space__ die_space__ choose__4.0__3.0__ |
vowel_space__ coin_space__ die_space__ choose__4.0__3.0__ |
| a train passes a station platform in num__50 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__177 m <o> b ) num__176 m <o> c ) num__450 m <o> d ) num__187 m <o> e ) num__186 m |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__20 ) m = num__300 m . let the length of the platform be x meters . then x + num__6.0 = num__15 x + num__300 = num__750 x = num__450 m . answer : c <eor> c <eos> |
c |
multiply__20.0__15.0__ divide__300.0__50.0__ multiply__50.0__15.0__ subtract__750.0__300.0__ round__450.0__ |
multiply__20.0__15.0__ divide__300.0__50.0__ multiply__50.0__15.0__ subtract__750.0__300.0__ round__450.0__ |
| x y and z can do a work in num__6 num__9 and num__10 days respectively . they completed the work and got rs . num__775 . what is the share of y ? <o> a ) num__250 <o> b ) num__240 <o> c ) num__260 <o> d ) num__230 <o> e ) num__245 |
the ratio of their working rates = num__0.166666666667 : num__0.111111111111 : num__0.1 = num__15 : num__10 : num__6 . since they work together the share of y = num__0.322580645161 * num__775 = rs . num__250 answer : a <eor> a <eos> |
a |
add__6.0__9.0__ round__250.0__ |
add__6.0__9.0__ round__250.0__ |
| a certain bus driver is paid a regular rate of $ num__18 per hour for any number of hours that does not exceed num__40 hours per week . for any overtime hours worked in excess of num__40 hours per week the bus driver is paid a rate that is num__75.0 higher than his regular rate . if last week the bus driver earned $ num__976 in total compensation how many total hours did he work that week ? <o> a ) num__36 <o> b ) num__40 <o> c ) num__44 <o> d ) num__48 <o> e ) num__52 |
for num__40 hrs = num__40 * num__18 = num__720 excess = num__976 - num__720 = num__252 for extra hours = . num__75 ( num__18 ) = num__13.5 + num__18 = num__31.5 number of extra hrs = num__252 / num__31.5 = num__8 total hrs = num__40 + num__8 = num__48 answer d num__48 <eor> d <eos> |
d |
multiply__18.0__40.0__ add__18.0__13.5__ divide__252.0__31.5__ add__40.0__8.0__ add__40.0__8.0__ |
multiply__18.0__40.0__ add__18.0__13.5__ divide__252.0__31.5__ add__40.0__8.0__ add__40.0__8.0__ |
| in a group of cows and chickens the number of legs was num__16 more than twice the number of heads . the number of cows was : <o> a ) num__8 <o> b ) num__7 <o> c ) num__10 <o> d ) num__12 <o> e ) num__14 |
let the number of cows be x and their legs be num__4 x . let the number of chicken be y and their legs be num__2 x . total number of legs = num__4 x + num__2 y . total number of heads = x + y . the number of legs was num__16 more than twice the number of heads . therefore num__2 × ( x + y ) + num__16 = num__4 x + num__2 y . or num__2 x + num__2 y + num__16 = num__4 x + num__2 y . or num__2 x + num__16 = num__4 x [ subtracting num__2 y from both sides ] . or num__16 = num__4 x – num__2 x [ subtracting num__2 x from both sides ] . or num__16 = num__2 x . or x = num__8 [ dividing by num__2 on both sides ] . therefore the number of cows = num__7 . correct answer : a ) num__8 <eor> a <eos> |
a |
coin_space__ choose__8.0__7.0__ |
coin_space__ choose__8.0__7.0__ |
| calculate the area of a triangle if the sides are num__13 cm num__12 cm and num__5 cm what is its area ? <o> a ) num__36 cm num__2 <o> b ) num__35 cm num__2 <o> c ) num__30 cm num__2 <o> d ) num__32 cm num__2 <o> e ) num__31 cm num__2 |
the triangle with sides num__13 cm num__12 cm and num__5 cm is right angled where the hypotenuse is num__13 cm . area of the triangle = num__0.5 * num__12 * num__5 = num__30 cm num__2 answer : c <eor> c <eos> |
c |
triangle_area__12.0__5.0__ square_perimeter__0.5__ rectangle_perimeter__13.0__2.0__ |
volume_rectangular_prism__12.0__5.0__0.5__ square_perimeter__0.5__ volume_rectangular_prism__12.0__5.0__0.5__ |
| there are num__100 students in a class . if num__14.0 are absent on a particular day find the number of students present in the class . <o> a ) num__43 <o> b ) num__36 <o> c ) num__86 <o> d ) num__129 <o> e ) num__11 |
number of students absent on a particular day = num__14.0 of num__100 i . e . num__0.14 × num__100 = num__14 therefore the number of students present = num__100 - num__14 = num__86 students . answer : c <eor> c <eos> |
c |
percent__100.0__86.0__ |
percent__100.0__86.0__ |
| if x / num__6 and x / num__15 are integers then x must be divisible by - <o> a ) num__12 <o> b ) num__18 <o> c ) num__30 <o> d ) num__45 <o> e ) num__60 |
factors of num__6 num__2 and num__3 factors of num__15 are num__3 and num__5 so number is num__2 * num__3 * num__5 = num__30 answer : c <eor> c <eos> |
c |
divide__6.0__2.0__ divide__15.0__3.0__ multiply__6.0__5.0__ multiply__6.0__5.0__ |
divide__6.0__2.0__ divide__15.0__3.0__ multiply__6.0__5.0__ multiply__6.0__5.0__ |
| if num__5 machines can produce num__20 units in num__10 hours how long would it take num__20 machines to produce num__140 units ? <o> a ) num__50 hours <o> b ) num__40 hours <o> c ) num__17.5 hours <o> d ) num__12 hours <o> e ) num__8 hours |
here we ' re told that num__5 machines can produce num__20 units in num__10 hours . . . . that means that each machine works for num__10 hours apiece . since there are num__5 machines ( and we ' re meant to assume that each machine does the same amount of work ) then the num__5 machines equally created the num__20 units . num__20 units / num__5 machines = num__4 units are made by each machine every num__10 hours now that we know how long it takes each machine to make num__4 units we can break this down further if we choose to . . . num__10 hours / num__4 units = num__2.5 hours per unit when num__1 machine is working . the prompt asks us how long would it take num__20 machines to produce num__140 units . if num__20 machines each work for num__2.5 hours then we ' ll have num__20 units . since num__140 units is ' num__7 times ' num__20 we need ' num__7 times ' more time . ( num__2.5 hours ) ( num__7 times ) = num__17.5 hours final answer : [ reveal ] spoiler : c <eor> c <eos> |
c |
divide__20.0__5.0__ divide__10.0__4.0__ subtract__5.0__4.0__ divide__140.0__20.0__ subtract__20.0__2.5__ round__17.5__ |
divide__20.0__5.0__ divide__10.0__4.0__ subtract__5.0__4.0__ divide__140.0__20.0__ subtract__20.0__2.5__ divide__17.5__1.0__ |
| a sum of rs . num__2665 is lent into two parts so that the interest on the first part for num__8 years at num__3.0 per annum may be equal to the interest on the second part for num__3 years at num__5.0 per annum . find the second sum ? <o> a ) num__1642 <o> b ) num__1640 <o> c ) num__1632 <o> d ) num__2789 <o> e ) num__6386 |
( x * num__8 * num__3 ) / num__100 = ( ( num__2665 - x ) * num__3 * num__5 ) / num__100 num__24 x / num__100 = num__399.75 - num__15 x / num__100 num__39 x = num__39975 = > x = num__1025 second sum = num__2665 – num__1025 = num__1640 answer : b <eor> b <eos> |
b |
percent__100.0__1640.0__ |
percent__100.0__1640.0__ |
| the average of six number is num__3.95 . the average of two of them is num__3.4 while the average of the other two is num__3.62 . what is the average of the remaining two number ? <o> a ) num__4.5 <o> b ) num__4.6 <o> c ) num__4.7 <o> d ) num__4.83 <o> e ) none of these |
solution : sum of the remaining two numbers = ( num__3.95 × num__6 ) - [ ( num__3.4 × num__2 ) + ( num__3.62 × num__2 ) ] = num__23.70 - ( num__6.8 + num__7.24 ) = num__23.70 - num__14.04 = num__9.66 . ∴ required average = num__9.66 / num__2 = num__4.83 answer d <eor> d <eos> |
d |
multiply__3.95__6.0__ multiply__3.4__2.0__ multiply__3.62__2.0__ add__6.8__7.24__ subtract__23.7__14.04__ divide__9.66__2.0__ divide__9.66__2.0__ |
multiply__3.95__6.0__ multiply__3.4__2.0__ multiply__3.62__2.0__ add__6.8__7.24__ subtract__23.7__14.04__ divide__9.66__2.0__ divide__9.66__2.0__ |
| a began business with rs . num__27000 and was joined afterwards by b with rs . num__54000 . when did b join if the profits at the end of the year were divided in the ratio of num__2 : num__1 ? <o> a ) num__9 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__2 |
num__27 * num__12 : num__54 * x = num__2 : num__1 x = num__3 num__12 - num__3 = num__9 answer : a <eor> a <eos> |
a |
multiply__2.0__27.0__ add__2.0__1.0__ subtract__12.0__3.0__ multiply__1.0__9.0__ |
multiply__2.0__27.0__ add__2.0__1.0__ subtract__12.0__3.0__ multiply__1.0__9.0__ |
| running at their respective constant rate machine x takes num__2 days longer to produce w widgets than machines y . at these rates if the two machines together produce num__5 w / num__4 widgets in num__3 days how many days would it take machine x alone to produce num__6 w widgets . <o> a ) num__4 <o> b ) num__36 <o> c ) num__8 <o> d ) num__10 <o> e ) num__12 |
i am getting num__12 . e . hope havent done any calculation errors . . approach . . let y = no . of days taken by y to do w widgets . then x will take y + num__2 days . num__1 / ( y + num__2 ) + num__1 / y = num__0.416666666667 ( num__0.416666666667 is because ( num__1.25 ) w widgets are done in num__3 days . so x widgets will be done in num__2.4 days or num__0.416666666667 th of a widget in a day ) solving we have y = num__4 = > x takes num__6 days to doing x widgets . so he will take num__36 days to doing num__6 w widgets . answer : b <eor> b <eos> |
b |
multiply__2.0__6.0__ subtract__5.0__4.0__ divide__5.0__12.0__ divide__5.0__4.0__ divide__3.0__1.25__ multiply__3.0__12.0__ multiply__3.0__12.0__ |
multiply__2.0__6.0__ subtract__5.0__4.0__ divide__5.0__12.0__ divide__5.0__4.0__ divide__3.0__1.25__ multiply__3.0__12.0__ divide__36.0__1.0__ |
| james works in a bakery . he made cookies that cost $ num__2 and made $ num__200 . how many customer did he have ? <o> a ) num__200 customers <o> b ) num__85 customers <o> c ) num__250 customers <o> d ) num__100 customers <o> e ) num__170 customers |
a cookie costs $ num__2 adding another one is $ num__4 . num__200 divided by num__3 is num__50 x num__2 is num__100 . he had num__100 customers . the correct answer is d . <eor> d <eos> |
d |
divide__200.0__4.0__ multiply__2.0__50.0__ round__100.0__ |
divide__200.0__4.0__ multiply__2.0__50.0__ round__100.0__ |
| the product of a and b is equal to num__11 more than twice the sum of a and b . if b = num__7 what is the value of b - a ? <o> a ) num__2 <o> b ) num__5 <o> c ) num__7 <o> d ) num__24 <o> e ) num__35 |
given : ab = num__2 ( a + b ) + num__11 and b = num__7 hence num__7 a = num__2 a + num__14 + num__11 num__5 a = num__25 a = num__5 ( a potential trap . do not get excited here . we need to find b - a not a ) b - a = num__7 - num__5 = num__2 correct option : a <eor> a <eos> |
a |
multiply__7.0__2.0__ subtract__7.0__2.0__ add__11.0__14.0__ subtract__7.0__5.0__ |
multiply__7.0__2.0__ subtract__7.0__2.0__ add__11.0__14.0__ subtract__7.0__5.0__ |
| in year y imported machine tools accounted for num__20 percent of total machine - tools sales in the united states and japanese imports accounted for num__40 percent of the sales of imported machine tools . if the total sales of machine tools imported from japan that year was x billion dollars then the total sales of all machine tools in the united states was how many billion dollars ? <o> a ) num__15 x <o> b ) num__13 x <o> c ) num__12.5 x <o> d ) num__14 x <o> e ) num__14.5 x |
sales of imported tools = num__20.0 of the total sales = num__0.20 * { total } = { total } / num__5 ; sales of japanese tools = num__40.0 of the sales of imported tools = num__0.4 * { total } / num__5 = num__2 { total } / num__25 since the sales of of japanese tools was x then num__2 { total } / num__25 = x - - > num__2 { total } = num__25 x { total } = num__25 x / num__2 = num__12.5 x answer : c <eor> c <eos> |
c |
reverse__0.2__ divide__40.0__20.0__ add__20.0__5.0__ divide__5.0__0.4__ divide__5.0__0.4__ |
reverse__0.2__ divide__40.0__20.0__ divide__5.0__0.2__ divide__5.0__0.4__ divide__5.0__0.4__ |
| in how many ways can the integer num__84 be expressed as a product of two different positive integers ? <o> a ) num__10 <o> b ) num__8 <o> c ) num__5 <o> d ) num__4 <o> e ) num__2 |
num__84 = ( num__2 ^ num__2 ) * num__3 * num__7 since num__84 is not a perfect square no of ways = num__4 answer d <eor> d <eos> |
d |
subtract__7.0__3.0__ subtract__7.0__3.0__ |
subtract__7.0__3.0__ subtract__7.0__3.0__ |
| in measuring the sides of a rectangle one side is taken num__8.0 in excess and the other num__7.0 in deficit . find the error percent in the area calculated from these measurements . <o> a ) num__0.11 <o> b ) num__0.7 <o> c ) num__0.44 <o> d ) num__0.6 <o> e ) num__0.8 % |
let x and y be the sides of the rectangle . then correct area = xy . calculated area = ( num__1.08 ) x ( num__0.93023255814 ) y = ( num__1.00440044004 ) ( xy ) error in measurement = ( num__1.00440044004 ) xy - xy = ( num__0.004400440044 ) xy error percentage = [ ( num__0.004400440044 ) xy ( num__1 / xy ) num__100 ] % = ( num__0.44 ) % = num__0.44 . answer is c . <eor> c <eos> |
c |
percent__100.0__0.44__ |
percent__100.0__0.44__ |
| __ num__2 a x __ b ____ cc in the multiplication problem above a b and c represent distinct digits . if the sum of a and b is equal to num__4.6 what is the value of c ? <o> a ) num__6 <o> b ) num__5.2 <o> c ) num__4 <o> d ) num__3 <o> e ) num__2 |
if a + b = num__4.6 assuming a and b are positive then a * b < num__10 ( they could be either num__12 num__34 ) therefore a * b = c num__2 * b = c a + b = num__4.6 three simple equations - divide the num__1 st / num__2 nd - - > a = num__2 plug it the num__3 rd - - > b = num__2.6 - - > c = num__5.2 ( answer b ) <eor> b <eos> |
b |
add__2.0__10.0__ add__2.0__1.0__ subtract__4.6__2.0__ multiply__2.0__2.6__ multiply__2.0__2.6__ |
add__2.0__10.0__ add__2.0__1.0__ subtract__4.6__2.0__ multiply__2.0__2.6__ multiply__2.0__2.6__ |
| a metallic sphere of radius num__12 cm is melted and drawn into a wire whose radius of cross section is num__16 cm . what is the length of the wire ? <o> a ) num__6 cm <o> b ) num__4 cm <o> c ) num__8 cm <o> d ) num__3 cm <o> e ) num__9 cm |
volume of the wire ( in cylindrical shape ) is equal to the volume of the sphere . π ( num__16 ) num__2 * h = ( num__1.33333333333 ) π ( num__12 ) num__3 = > h = num__9 cm answer : e <eor> e <eos> |
e |
divide__16.0__12.0__ subtract__12.0__3.0__ round__9.0__ |
divide__16.0__12.0__ subtract__12.0__3.0__ round__9.0__ |
| a train num__420 m long running with a speed of num__63 km / hr will pass a tree in ? <o> a ) num__36 sec <o> b ) num__16 sec <o> c ) num__17 sec <o> d ) num__88 sec <o> e ) num__12 sec |
speed = num__63 * num__0.277777777778 = num__17.5 m / sec time taken = num__420 * num__0.0571428571429 = num__36 sec answer : a <eor> a <eos> |
a |
round__36.0__ |
round__36.0__ |
| if a square mirror has a num__20 - inch diagonal what is the approximate perimeter t of the mirror in inches ? <o> a ) num__40 <o> b ) num__60 <o> c ) num__80 <o> d ) num__100 <o> e ) num__120 |
if you draw the square and diagonal inside the square . u can see square becomes part of two triangles opposite to each other . and we know the property of the triangle addition of two sides of triangle must be greater than its diagonal in order to complete the triangle . and each side must be less than num__20 and perimeter t must be less than num__80 so we can eliminate answer choice c d and e . so side num__1 + side num__2 > num__20 that means side num__1 or side num__2 must be > num__10 . so we can eliminate the answer choice a . now we are left with is b <eor> b <eos> |
b |
square_perimeter__20.0__ triangle_area__20.0__1.0__ rectangle_perimeter__20.0__10.0__ |
square_perimeter__20.0__ triangle_area__20.0__1.0__ rectangle_perimeter__20.0__10.0__ |
| in a t . v . factory an average of num__60 tvs are produced per day for the fist num__25 days of the months . a few workers fellill for the next five daysreducing the daily avg for the month to num__58 sets / day . the average production per day for day last num__5 days is ? <o> a ) num__35 <o> b ) num__39 <o> c ) num__48 <o> d ) num__50 <o> e ) num__52 |
production during these num__5 days = total production in a month - production in first num__25 days . = num__30 x num__58 - num__25 x num__60 = num__240 ∴ average for last num__5 days = num__48.0 = num__48 c <eor> c <eos> |
c |
add__25.0__5.0__ divide__240.0__5.0__ divide__240.0__5.0__ |
add__25.0__5.0__ divide__240.0__5.0__ divide__240.0__5.0__ |
| positive integer a gives the remainder of num__32 when divided by another positive integer b . if a / b = num__147.64 what is the value of b ? <o> a ) num__96 <o> b ) num__75 <o> c ) num__48 <o> d ) num__50 <o> e ) num__12 |
. num__64 of b = remainder . num__64 of b = num__32 b = ( num__32 * num__100 ) / num__64 = num__50 . d <eor> d <eos> |
d |
subtract__100.0__50.0__ |
subtract__100.0__50.0__ |
| there is a total of num__90 marbles in a box each of which is red green blue or white . if one marble is drawn from the box at random the probability that it will be white is num__0.333333333333 and the probability that it will be green is num__0.2 . what is the probability that the marble will be either red or blue ? <o> a ) num__0.333333333333 <o> b ) num__0.6 <o> c ) num__0.466666666667 <o> d ) num__0.366666666667 <o> e ) num__0.566666666667 |
p ( red or blue ) = num__1 - p ( white ) - p ( green ) = num__1.0 - num__0.333333333333 - num__0.2 = num__0.466666666667 the answer is c . <eor> c <eos> |
c |
multiply__1.0__0.4667__ |
multiply__1.0__0.4667__ |
| a can finish a work in num__18 days and b can do the same work in half the time taken by a . then working together what part of the same work they can finish in a day ? <o> a ) num__0.5 <o> b ) num__0.333333333333 <o> c ) num__0.166666666667 <o> d ) num__0.4 <o> e ) num__0.222222222222 |
num__1 days work of a + b = ( num__18 + num__9 ) / num__18 * num__9 = num__0.166666666667 answer is c <eor> c <eos> |
c |
divide__0.1667__1.0__ |
divide__0.1667__1.0__ |
| the perimeter of one face of a cube is num__24 cm . its volume will be : <o> a ) num__216 cm num__3 <o> b ) num__400 cm num__3 <o> c ) num__250 cm num__3 <o> d ) num__625 cm num__3 <o> e ) none of these |
explanation : edge of cude = num__6.0 = num__6 cm volume = a * a * a = num__6 * num__6 * num__6 = num__216 cm cube option a <eor> a <eos> |
a |
volume_cube__6.0__ volume_cube__6.0__ |
volume_cube__6.0__ volume_cube__6.0__ |
| ram professes to sell his goods at the cost price but he made use of num__850 grms instead of a kg what is the gain percent ? a . num__11.0 <o> a ) num__11 num__0.111111111111 % <o> b ) num__11 num__0.125 % <o> c ) num__17.65 <o> d ) num__11 num__0.555555555556 % <o> e ) num__11 num__0.2 % |
num__850 - - - num__150 num__100 - - - ? = > num__17.65 answer : c <eor> c <eos> |
c |
percent__17.65__100.0__ |
percent__17.65__100.0__ |
| the list price of an article is rs . num__65 . a customer pays rs . num__56.16 for it . he was given two successive discounts one of them being num__10.0 . the other discount is ? <o> a ) num__8.0 <o> b ) num__7.0 <o> c ) num__10.0 <o> d ) num__12.0 <o> e ) num__4 % |
option e explanation : num__65 * ( num__0.9 ) * ( ( num__100 - x ) / num__100 ) = num__56.16 x = num__4.0 <eor> e <eos> |
e |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| num__62467 × num__9998 = ? <o> a ) num__624545037 <o> b ) num__627745452 <o> c ) num__624545077 <o> d ) num__624545066 <o> e ) num__625454211 |
d num__624545066 num__62467 × num__9998 = num__62467 × ( num__10000 - num__2 ) = num__62467 × num__10000 - num__62467 × num__2 = num__624670000 - num__124934 = num__624545066 <eor> d <eos> |
d |
multiply__62467.0__9998.0__ subtract__10000.0__9998.0__ multiply__62467.0__10000.0__ multiply__62467.0__2.0__ multiply__62467.0__9998.0__ |
multiply__62467.0__9998.0__ subtract__10000.0__9998.0__ multiply__62467.0__10000.0__ subtract__624670000.0__624545066.0__ subtract__624670000.0__124934.0__ |
| a car travels num__25 km an hour faster than a bus for a journey of num__500 km . if the bus takes num__10 hours more than the car then the speeds of the bus and the car are <o> a ) num__25 km / h and num__40 km / h respectively <o> b ) num__25 km / h and num__60 km / h respectively <o> c ) num__25 km / h and num__50 km / h respectively <o> d ) num__25 km / h and num__70 km / h respectively <o> e ) none of these |
let the speed of the bus be x km / h . then speed of the car = ( x + num__25 ) km / h ∴ num__500 / x = num__500 / x + num__25 + num__10 ⇒ x num__2 + num__25 x – num__1250 = num__0 ⇒ x = num__25 thus speed of the bus = num__25 km / h speed of the car = num__50 km / h answer c <eor> c <eos> |
c |
multiply__25.0__2.0__ round__25.0__ |
divide__500.0__10.0__ divide__1250.0__50.0__ |
| machine m n o working simultaneously machine m can produce x units in num__0.75 of the time it takes machine n to produce the same amount of units . machine n can produce x units in num__0.5 the time it takes machine o to produce that amount of units . if all three machines are working simultaneously what fraction of the total output is produced by machine n ? <o> a ) num__0.4 <o> b ) num__0.333333333333 <o> c ) num__0.307692307692 <o> d ) num__0.275862068966 <o> e ) num__0.181818181818 |
now ultimately the speed of every machine is given with respect to mach o . so lets assume the speed of o say num__12 hrs to make x units ( assuming num__6 because we can see we will need to divide by num__3 and num__4 mach o makes x units in num__12 hrs so mach n = num__0.5 of o = num__0.5 * num__12 = num__6 hrs to make x units and mach m = num__0.75 of n = num__0.75 * num__6 = num__6 hrs to make x units no they are running simultaneously . lets see how much each mach makes in num__1 hr mach o = x / num__12 units mach n = x / num__6 units mach m = x / num__6 units in num__1 hr together they make - x / num__12 + x / num__6 + x / num__6 = num__5 x / num__12 so what ratio of this has mach n made ? ( x / num__6 ) / ( num__5 x / num__12 ) = num__0.4 ans : a = num__0.4 <eor> a <eos> |
a |
multiply__0.5__12.0__ multiply__0.5__6.0__ divide__3.0__0.75__ subtract__4.0__3.0__ add__4.0__1.0__ multiply__1.0__0.4__ |
multiply__0.5__12.0__ multiply__0.5__6.0__ divide__3.0__0.75__ subtract__4.0__3.0__ add__4.0__1.0__ multiply__1.0__0.4__ |
| a polling company reports that there is a num__80.0 chance that a certain candidate will win the next election . if the candidate wins there is a num__60.0 chance that she will sign bill x and no other bills . if she decides not to sign bill x she will sign either bill y or bill z chosen randomly . what is the chance that the candidate will sign bill z ? <o> a ) num__10 <o> b ) num__16 <o> c ) num__6 <o> d ) num__4 <o> e ) num__5 |
num__80.0 - candidate elected num__100.0 - num__60.0 = num__40.0 - candidate doesnotsigh bill x num__50.0 - candidate randomly chooses between two bills . these are multiplicative : num__80.0 x num__40.0 x num__50.0 num__0.8 x num__0.4 x num__0.5 = num__0.16 = num__16.0 answer ( b ) <eor> b <eos> |
b |
percent__0.8__50.0__ percent__0.4__40.0__ percent__16.0__100.0__ |
percent__0.8__50.0__ percent__0.4__40.0__ percent__16.0__100.0__ |
| last year for every num__100 million vehicles that travelled on a certain highway num__92 vehicles were involved in accidents . if num__3 billion vehicles travelled on the highway last year how many of those vehicles were involved in accidents ? ( num__1 billion = num__1000 num__000000 ) <o> a ) num__288 <o> b ) num__320 <o> c ) num__2760 <o> d ) num__3200 <o> e ) num__28 |
800 |
to solve we will set up a proportion . we know that “ num__100 million vehicles is to num__92 accidents as num__3 billion vehicles is to x accidents ” . to express everything in terms of “ millions ” we can use num__3000 million rather than num__3 billion . creating a proportion we have : num__1.08695652174 = num__3000 / x cross multiplying gives us : num__100 x = num__3000 * num__92 x = num__30 * num__92 = num__2760 correct answer is c . <eor> c <eos> |
c |
c |
| in a school of num__900 students num__44.0 wear blue shirts num__28.0 wear red shirts num__10.0 wear green shirts and the remaining students wear other colors . how many students wear other colors ( not blue not red not green ) ? <o> a ) num__144 <o> b ) num__153 <o> c ) num__162 <o> d ) num__171 <o> e ) num__180 |
num__44 + num__28 + num__10 = num__82.0 num__100 – num__82 = num__18.0 num__900 * num__0.18 = num__162 the answer is c . <eor> c <eos> |
c |
percent__18.0__900.0__ percent__100.0__162.0__ |
percent__18.0__900.0__ percent__100.0__162.0__ |
| num__5301 x num__13 = ? <o> a ) num__68381 <o> b ) num__69831 <o> c ) num__68931 <o> d ) num__68319 <o> e ) num__68913 |
append num__0 before and after : num__053010 calculation : num__1 x num__3 + num__0 = num__3 ( take num__3 as ones digit of the product ) num__0 x num__3 + num__1 = num__1 ( take num__1 tens digit of the product ) num__3 x num__3 + num__0 = num__9 ( take num__9 hundreds digit of the product ) num__5 x num__3 + num__3 = num__18 ( take num__8 as thousands digit of the product carry over num__1 ) num__0 x num__3 + num__5 = num__5 ; num__5 + num__1 = num__6 ( take num__6 as ten thousands digits of the product ) so num__5301 x num__13 = num__68913 answer is e . <eor> e <eos> |
e |
add__13.0__5.0__ subtract__13.0__5.0__ add__1.0__5.0__ multiply__5301.0__13.0__ multiply__5301.0__13.0__ |
add__13.0__5.0__ add__3.0__5.0__ add__1.0__5.0__ multiply__5301.0__13.0__ multiply__5301.0__13.0__ |
| a can do a piece of work in num__5 days of num__9 hours each and b alone can do it in num__5 days of num__3 hours each . how long will they take it to do working together num__2 num__0.25 hours a day ? <o> a ) num__2 days <o> b ) num__3 days <o> c ) num__4 days <o> d ) num__5 days <o> e ) num__6 days |
a ' s work per hour = num__0.0222222222222 b ' s work per hour = num__0.0666666666667 a & b ' s work per hour together = ( num__0.0222222222222 ) + ( num__0.0666666666667 ) = num__0.0888888888889 so a & b together complete the work in num__11.25 hours . . . if they work num__2 num__0.25 = num__2.25 hours a day it will take ( num__11.25 ) / ( num__2.25 ) days = num__5 days . . . answer : d <eor> d <eos> |
d |
add__0.0667__0.0222__ multiply__9.0__0.25__ round__5.0__ |
add__0.0667__0.0222__ add__2.0__0.25__ add__3.0__2.0__ |
| one bottle is half - full of oil and another bottle with twice the capacity is one quarter full of oil . if water is added so that both the bottles are full and the contents of both are then poured into a third bottle that is empty and large enough to hold the contents of both what fractions of the contents in the third bottle is oil ? <o> a ) num__0.25 <o> b ) num__0.333333333333 <o> c ) num__0.375 <o> d ) num__0.666666666667 <o> e ) num__0.5 |
let assume that num__1 st bottle capacity = num__1 litre ; therefore it contain num__0.5 litre oil and num__0.5 litre water . ( i . e ) water = num__0.5 litre oil = num__0.5 litre num__2 nd bottle is twice the capacity i . e . num__2 litre capacity it means num__0.5 litre filled with oil and num__1 ( num__0.5 ) litre filled with water . i . e . oil = num__0.5 liter water = num__1 ( num__0.5 ) litre in total the third bottle will contain num__1 litre oil and num__2 litre water . i . e . oil = num__1 litre water = num__2 liter therefore oil proportion is num__0.333333333333 answer : b <eor> b <eos> |
b |
half__ twice__1.0__ multiply__1.0__0.3333__ |
half__ twice__1.0__ multiply__1.0__0.3333__ |
| the perimeter of a semi circle is num__144 cm then the radius is ? <o> a ) num__22 <o> b ) num__28 <o> c ) num__99 <o> d ) num__77 <o> e ) num__21 |
num__5.14285714286 r = num__144 = > r = num__28 answer : b <eor> b <eos> |
b |
round__28.0__ |
round__28.0__ |
| a furniture manufacturer has two machines but only one can be used at a time . machine e is utilized during the first shift and machine b during the second shift while both work half of the third shift . if machine e can do the job in num__12 days working two shifts and machine b can do the job in num__15 days working two shifts how many days will it take to do the job with the current work schedule ? <o> a ) num__14 <o> b ) num__13 <o> c ) num__11 <o> d ) num__9 <o> e ) num__7 |
machine e finish the job in num__2 * num__12 shifts = num__24 shifts machine b finish the job in num__2 * num__15 shifts = num__30 shifts lets assume total work require num__120 shifts therefore rate of e = num__5 shifts / day rate of b = num__4 shifts / day rate of ( e + b ) = num__9 shifts / day according to current schedule work complete in a day = num__5 + num__4 + ( num__4.5 ) = num__13.5 shifts / day therefore time required to finish num__120 shifts = ( num__120 / num__13.5 ) = num__8.88 . . days ~ num__9 days = d <eor> d <eos> |
d |
multiply__12.0__2.0__ multiply__15.0__2.0__ divide__120.0__24.0__ divide__120.0__30.0__ add__4.0__5.0__ divide__9.0__2.0__ add__4.5__9.0__ round__9.0__ |
multiply__12.0__2.0__ multiply__15.0__2.0__ divide__120.0__24.0__ divide__120.0__30.0__ add__4.0__5.0__ divide__9.0__2.0__ add__4.5__9.0__ multiply__4.5__2.0__ |
| num__9 + num__2 + num__2 ^ num__2 + num__2 ^ num__3 + num__2 ^ num__4 + num__2 ^ num__5 = ? <o> a ) ( num__2 ^ num__3 - num__1 ) ( num__2 ^ num__3 + num__1 ) <o> b ) num__2 ^ num__6 + num__1 <o> c ) num__2 ^ num__5 - num__1 <o> d ) num__2 ^ num__6 + num__7 <o> e ) num__2 ^ num__5 - num__2 |
from num__9 + num__2 + num__2 ^ num__2 + num__2 ^ num__3 + num__2 ^ num__4 + num__2 ^ num__5 = num__71 num__2 ^ num__6 + num__7 = num__71 the correct answer is d . <eor> d <eos> |
d |
subtract__9.0__3.0__ subtract__9.0__2.0__ subtract__9.0__7.0__ |
add__2.0__4.0__ add__2.0__5.0__ subtract__9.0__7.0__ |
| a salesman â € ™ s terms were changed from a flat commission of num__5.0 on all his sales to a fixed salary of rs . num__1000 plus num__2.5 commission on all sales exceeding rs . num__4000 . if his remuneration as per new scheme was rs . num__800 more than that by the previous schema his sales were worth ? <o> a ) s . num__4000 <o> b ) s . num__12000 <o> c ) s . num__30000 <o> d ) s . num__40000 <o> e ) s . num__50 |
000 |
[ num__1000 + ( x - num__4000 ) * ( num__2.5 / num__100 ) ] - x * ( num__0.05 ) = num__800 x = num__4000 answer : a <eor> a <eos> |
a |
a |
| if integer k is equal to the sum of all even multiples of num__35 between num__280 and num__630 what is the greatest prime factor of k ? <o> a ) num__5 <o> b ) num__7 <o> c ) num__11 <o> d ) num__13 <o> e ) num__11 * num__455 |
if we break down what the stem is asking what is the sum of all mult of num__30 between num__300 and num__600 . using arithmetic progression to find n : num__630 = num__280 + ( n - num__1 ) num__35 num__350 + num__35 = num__35 n num__385 = num__35 n = > n = num__11 the sum would be : num__11 * mean mean = [ num__630 + num__280 ] / num__2 = num__455 num__11 * num__455 = e <eor> e <eos> |
e |
subtract__630.0__30.0__ subtract__630.0__280.0__ add__35.0__350.0__ divide__385.0__35.0__ divide__600.0__300.0__ multiply__1.0__11.0__ |
subtract__630.0__30.0__ subtract__630.0__280.0__ add__35.0__350.0__ divide__385.0__35.0__ divide__600.0__300.0__ multiply__1.0__11.0__ |
| elena ’ s bread recipe calls for num__3 ounces of butter for each num__4 cups of flour used . she needs to make num__6 times the original recipe . if num__12 ounces of butter is used then how many cups of flour are needed ? <o> a ) num__1 <o> b ) num__4 <o> c ) num__9 <o> d ) num__13 <o> e ) num__24 |
solving through algebra route : num__3 b + num__4 f = x amount if we multiply this equation with num__6 we get : num__18 b + num__24 f = num__6 x therefore we got num__18 ounces of butter and num__6 x amount of quantity when we use num__24 ounces of floor . ans : e <eor> e <eos> |
e |
multiply__3.0__6.0__ multiply__4.0__6.0__ multiply__4.0__6.0__ |
add__6.0__12.0__ add__6.0__18.0__ add__6.0__18.0__ |
| in the coordinate plane the points f ( - num__21 ) g ( num__16 ) and h ( num__61 ) lie on a circle with center p . what are the coordinates of point p ? <o> a ) ( num__00 ) <o> b ) ( num__11 ) <o> c ) ( num__22 ) <o> d ) ( num__1 - num__2 ) <o> e ) ( num__2.5 - num__2.5 ) |
i thought i need to find the num__4 th point that would lie on the circle hence my answer was d ( num__1 - num__2 ) but the the center of the circle will lie on ( num__22 ) c <eor> c <eos> |
c |
add__21.0__1.0__ round__22.0__ |
add__21.0__1.0__ round__22.0__ |
| if num__2 x + y = num__6 and x + num__2 y = num__5 then ( x + y ) / num__3 = <o> a ) num__1 <o> b ) num__1.33333333333 <o> c ) num__3.4 <o> d ) num__1.22222222222 <o> e ) num__4 |
we have two equations : num__2 x + y = num__6 x + num__2 y = num__5 notice that something nice happens when we add them . we get : num__3 x + num__3 y = num__11 divide both sides by num__3 to get : x + y = num__3.66666666667 so ( x + y ) / num__3 = num__1.22222222222 answer : d <eor> d <eos> |
d |
add__6.0__5.0__ divide__11.0__3.0__ divide__3.6667__3.0__ divide__3.6667__3.0__ |
add__6.0__5.0__ divide__11.0__3.0__ divide__3.6667__3.0__ divide__3.6667__3.0__ |
| a is twice as good as workman as b and together they finish a piece of work in num__18 days . in how many days will b alone finish the work . <o> a ) num__27 days <o> b ) num__54 days <o> c ) num__56 days <o> d ) num__68 days <o> e ) num__78 days |
explanation : as per question a do twice the work as done by b . so a : b = num__2 : num__1 also ( a + b ) one day work = num__0.0555555555556 to get days in which b will finish the work lets calculate work done by b in num__1 day = = ( num__0.0555555555556 ∗ num__0.333333333333 ) = num__0.0185185185185 [ please note we multiplied by num__0.333333333333 as per b share and total of ratio is num__0.333333333333 ] so b will finish the work in num__54 days option b <eor> b <eos> |
b |
divide__1.0__18.0__ multiply__0.0556__0.3333__ round__54.0__ |
divide__1.0__18.0__ multiply__0.0556__0.3333__ round__54.0__ |
| num__30.0 loss on cost price in what percent loss on selling price ? <o> a ) num__28.44 <o> b ) num__38.0 <o> c ) num__40.35 <o> d ) num__42.85 <o> e ) none |
sol . loss = num__30.0 on cp i . e . num__0.3 cp ⇒ sp = num__0.7 cp loss % on sp = loss / sp * num__100 = num__0.3 cp / num__0.7 cp * num__100 = num__42.85 . answer d <eor> d <eos> |
d |
percent__42.85__100.0__ |
percent__42.85__100.0__ |
| five years ago the average age of a b c and d was num__45 yr . with e joining them now the average of all the five is num__42 yr . how old is e ? <o> a ) num__10 yr <o> b ) num__40 yr <o> c ) num__45 yr <o> d ) num__64 yr <o> e ) none |
solution : total present age of a b c and d = ( num__45 * num__4 ) + ( num__4 * num__5 ) = num__200 yr ; total age present age of a b c d and e = num__42 * num__5 = num__210 yr . so age of e = num__10 yr . answer : option a <eor> a <eos> |
a |
multiply__42.0__5.0__ subtract__210.0__200.0__ subtract__210.0__200.0__ |
multiply__42.0__5.0__ subtract__210.0__200.0__ subtract__210.0__200.0__ |
| how much interest can a person get on rs . num__8400 at num__17.5 p . a . simple interest for a period of two years and six months ? <o> a ) num__3587.58 <o> b ) num__3587.59 <o> c ) num__3675 <o> d ) num__3587.52 <o> e ) num__3587.56 |
i = ( num__8400 * num__2.5 * num__17.5 ) / num__100 = ( num__8400 * num__5 * num__35 ) / ( num__100 * num__2 * num__2 ) = rs . num__3675 . answer : c <eor> c <eos> |
c |
percent__100.0__3675.0__ |
percent__100.0__3675.0__ |
| a family consists of two grandparents two parents and three grandchildren . the average age of the grandparents is num__67 years that of the parents is num__35 years and that of the grandchildren is num__6 years . what is the average age of the family ? <o> a ) num__28 num__0.571428571429 years <o> b ) num__31 num__0.714285714286 years <o> c ) num__32 num__0.142857142857 years <o> d ) none of these <o> e ) num__31 |
required average = ( ( num__67 x num__2 + num__35 x num__2 + num__6 x num__3 ) / ( num__2 + num__2 + num__3 ) ) = ( num__134 + num__70 + num__18 ) / num__7 = num__31.7142857143 = num__31 num__0.714285714286 years . answer : b <eor> b <eos> |
b |
divide__6.0__2.0__ multiply__67.0__2.0__ add__67.0__3.0__ multiply__6.0__3.0__ round_down__31.7143__ subtract__31.7143__31.0__ round_down__31.7143__ |
divide__6.0__2.0__ multiply__67.0__2.0__ add__67.0__3.0__ multiply__6.0__3.0__ round_down__31.7143__ subtract__31.7143__31.0__ round_down__31.7143__ |
| a train leaves delhi at num__7 a . m . at a speed of num__30 kmph . another train leaves at num__2 p . m . at a speed of num__40 kmph on the same day and in the same direction . how far from delhi will the two trains meet ? <o> a ) num__229 <o> b ) num__840 <o> c ) num__600 <o> d ) num__888 <o> e ) num__121 |
d = num__30 * num__7 = num__210 rs = num__40 – num__30 = num__10 t = num__21.0 = num__21 d = num__40 * num__21 = num__840 km answer : b <eor> b <eos> |
b |
multiply__7.0__30.0__ subtract__40.0__30.0__ divide__210.0__10.0__ multiply__40.0__21.0__ round__840.0__ |
multiply__7.0__30.0__ subtract__40.0__30.0__ divide__210.0__10.0__ multiply__40.0__21.0__ multiply__40.0__21.0__ |
| if num__102 y = num__25 then num__10 - y equals : <o> a ) a ) num__3 <o> b ) b ) num__2 <o> c ) c ) num__6 <o> d ) d ) num__9 <o> e ) e ) num__0.2 |
num__102 y = num__25 ( num__10 y ) num__2 = num__52 num__10 y = num__5 num__0.1 y = num__0.2 num__10 - y = num__0.2 answer : e <eor> e <eos> |
e |
divide__10.0__2.0__ reverse__10.0__ reverse__5.0__ reverse__5.0__ |
divide__10.0__2.0__ reverse__10.0__ reverse__5.0__ reverse__5.0__ |
| sreenivas sells a table to shiva at num__10.0 profit and shiva sells it to mahesh at num__10.0 loss . at what price did sreenivas purchase the table if mahesh paid rs . num__2178 ? <o> a ) rs . num__2220 <o> b ) rs . num__2229 <o> c ) rs . num__2200 <o> d ) rs . num__2297 <o> e ) rs . num__2218 |
let the cost price of table for sreenivas be rs . x and given that cost price of table for mahesh = rs . num__2178 . = > ( num__90.0 ) of ( num__110.0 ) of x = rs . num__2178 . = > ( num__0.9 ) ( num__1.1 ) x = num__2178 = > x = ( num__2178 * num__100 ) / ( num__9 * num__11 ) = > x = rs . num__2200 answer : c <eor> c <eos> |
c |
percent__10.0__90.0__ percent__10.0__110.0__ percent__100.0__2200.0__ |
percent__10.0__90.0__ percent__10.0__110.0__ percent__100.0__2200.0__ |
| a fruit seller had some oranges . he sells num__40.0 oranges and still has num__420 oranges . how many oranges he had originally ? <o> a ) num__500 <o> b ) num__700 <o> c ) num__750 <o> d ) num__900 <o> e ) num__950 |
he sells num__40.0 of oranges and still there are num__420 oranges remaining . = > num__60.0 of oranges = num__420 = > total oranges × num__60 num__100 = num__420 × num__60100 = num__420 = > total oranges = num__420 × num__100 num__60 = num__700 option b <eor> b <eos> |
b |
percent__100.0__700.0__ |
percent__100.0__700.0__ |
| the h . c . f of two numbers is num__11 and their l . c . m is num__7700 . if one of the numbers is num__275 then the other is <o> a ) num__279 <o> b ) num__283 <o> c ) num__308 <o> d ) num__318 <o> e ) none |
solution other number = ( num__11 × num__28.0 ) = num__308 . answer c <eor> c <eos> |
c |
divide__7700.0__275.0__ multiply__11.0__28.0__ lcm__11.0__308.0__ |
divide__7700.0__275.0__ multiply__11.0__28.0__ lcm__11.0__308.0__ |
| fred and sam are standing num__55 miles apart and they start walking in a straight line toward each other at the same time . if fred walks at a constant speed of num__6 miles per hour and sam walks at a constant speed of num__5 miles per hour how many miles has sam walked when they meet ? <o> a ) num__5 <o> b ) num__9 <o> c ) num__25 <o> d ) num__30 <o> e ) num__45 |
relative distance = num__55 miles relative speed = num__6 + num__5 = num__11 miles per hour time taken = num__5.0 = num__5 hours distance travelled by sam = num__5 * num__5 = num__25 miles = c <eor> c <eos> |
c |
divide__55.0__5.0__ round__25.0__ |
add__6.0__5.0__ round__25.0__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__360 m and they cross each other in num__12 sec then the speed of each train is ? <o> a ) num__11 <o> b ) num__27 <o> c ) num__36 <o> d ) num__99 <o> e ) num__108 |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__360 + num__360 ) / num__12 = > x = num__30 speed of each train = num__10 m / sec . = num__30 * num__3.6 = num__108 km / hr . answer : e <eor> e <eos> |
e |
divide__360.0__12.0__ subtract__12.0__2.0__ multiply__3.6__30.0__ round__108.0__ |
divide__360.0__12.0__ subtract__12.0__2.0__ multiply__3.6__30.0__ multiply__3.6__30.0__ |
| an bus covers a certain distance at a speed of num__240 kmph in num__5 hours . to cover the samedistance in num__1 hr it must travel at a speed of ? <o> a ) num__700 km / hr <o> b ) num__720 km / hr <o> c ) num__870 km / hr <o> d ) num__869 km / hr <o> e ) num__890 km / hr |
distance = ( num__240 x num__5 ) = num__1200 km . speed = distance / time speed = num__1200 / ( num__1.66666666667 ) km / hr . [ we can write num__1 hours as num__1.66666666667 hours ] required speed = num__1200 x num__3 km / hr = num__720 km / hr . b <eor> b <eos> |
b |
multiply__240.0__5.0__ multiply__240.0__3.0__ round__720.0__ |
multiply__240.0__5.0__ multiply__240.0__3.0__ divide__720.0__1.0__ |
| two bullet train s are moving in opposite directions at num__60 km / hr and num__90 km / hr . their lengths are num__1.10 km and num__0.9 km respectively . the time taken by the slower bullet train to cross the faster bullet train in seconds is : <o> a ) num__58 sec . <o> b ) num__78 sec . <o> c ) num__68 sec . <o> d ) num__48 sec . <o> e ) num__88 sec . |
d num__48 sec . relative speed = ( num__60 + num__90 ) km / hr = num__150 x num__0.277777777778 = num__40.0 m / sec distance covered = ( num__1.10 + num__0.9 ) km = num__2 km = num__2000 m . required time = num__2000 x num__0.024 = num__48 sec . <eor> d <eos> |
d |
add__60.0__90.0__ add__1.1__0.9__ divide__48.0__2000.0__ round__48.0__ |
add__60.0__90.0__ add__1.1__0.9__ divide__48.0__2000.0__ round__48.0__ |
| a bus starts from city x . the number of women in the bus is half of the number of men . in city y num__16 men leave the bus and eight women enter . now number of men and women is equal . in the beginning how many passengers entered the bus ? <o> a ) num__15 <o> b ) num__30 <o> c ) num__36 <o> d ) num__72 <o> e ) num__46 |
explanation : originally let number of women = x . then number of men = num__2 x . so in city y we have : ( num__2 x - num__16 ) = ( x + num__8 ) or x = num__24 . therefore total number of passengers in the beginning = ( x + num__2 x ) = num__3 x = num__72 . answer : d <eor> d <eos> |
d |
divide__16.0__2.0__ add__16.0__8.0__ divide__24.0__8.0__ multiply__3.0__24.0__ multiply__3.0__24.0__ |
divide__16.0__2.0__ add__16.0__8.0__ divide__24.0__8.0__ multiply__3.0__24.0__ multiply__3.0__24.0__ |
| the length of the bridge which a train num__130 metres long and travelling at num__45 km / hr can cross in num__30 seconds is : <o> a ) num__234 <o> b ) num__230 <o> c ) num__245 <o> d ) num__250 <o> e ) num__260 |
speed = [ num__45 x num__0.277777777778 ] m / sec = [ num__12.5 ] m / sec time = num__30 sec let the length of bridge be x metres . then ( num__130 + x ) / num__30 = num__12.5 = > num__2 ( num__130 + x ) = num__750 = > x = num__245 m . answer : option c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| how many positive integers less than num__50 are there such that they are multiples of num__9 ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__4 <o> d ) num__8 <o> e ) num__3 |
number of multiples of num__9 = > num__45 - num__1.0 + num__1 = num__5 answer is a <eor> a <eos> |
a |
subtract__50.0__45.0__ subtract__50.0__45.0__ |
subtract__50.0__45.0__ subtract__50.0__45.0__ |
| the average of first num__10 odd numbers is ? <o> a ) num__11 <o> b ) num__10 <o> c ) num__88 <o> d ) num__176 <o> e ) num__12 |
sum of num__10 odd no . = num__100 average = num__10.0 = num__10 answer : b <eor> b <eos> |
b |
divide__100.0__10.0__ |
divide__100.0__10.0__ |
| of the num__11 temporary employees in a certain company num__4 will be hired as permanent employees . if num__5 of the num__11 temporary employees are women how many of the possible groups of num__4 employees consist of num__3 women and one man ? <o> a ) num__22 <o> b ) num__35 <o> c ) num__60 <o> d ) num__70 <o> e ) num__105 |
another approach : one scenario : [ wwwm ] now we have num__5 ways for the first women num__4 ways for the num__2 nd women and num__3 ways for the third women . and num__6 ways to select one man so num__5 * num__4 * num__3 * num__6 = num__360 now the three women are same so combinations with win them are irrelevant the num__3 women can we arranged in num__3 ! ways hence num__120.0 ! = num__60.0 = num__60 c <eor> c <eos> |
c |
coin_space__ die_space__ choose__5.0__2.0__ choose__6.0__4.0__ choose__6.0__4.0__ |
coin_space__ die_space__ choose__5.0__2.0__ choose__6.0__4.0__ choose__6.0__4.0__ |
| a certain industrial loom weaves num__0.126 meters of cloth every second . approximately how many seconds will it take for the loom to weave num__15 meters of cloth ? <o> a ) num__114 <o> b ) num__115 <o> c ) num__116 <o> d ) num__117 <o> e ) num__119 |
let the required number of seconds be x more cloth more time ( direct proportion ) hence we can write as ( cloth ) num__0.126 : num__15 : : num__1 : x = > num__0.126 * x = num__15 = > x = num__15 / num__0.126 = > x = num__119 answer : e <eor> e <eos> |
e |
round__119.0__ |
divide__119.0__1.0__ |
| the difference between a two - digit number and the number obtained by interchanging the digits is num__36 . what is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is num__1 : num__2 ? <o> a ) num__4 <o> b ) num__8 <o> c ) num__16 <o> d ) num__20 <o> e ) none |
since the number is greater than the number obtained on reversing the digits so the ten ' s digit is greater than the unit ' s digit . let ten ' s and unit ' s digits be num__2 x and x respectively . then ( num__10 x num__2 x + x ) - ( num__10 x + num__2 x ) = num__36 num__9 x = num__36 x = num__4 . required difference = ( num__2 x + x ) - ( num__2 x - x ) = num__2 x = num__8 . answer = b <eor> b <eos> |
b |
subtract__10.0__1.0__ divide__36.0__9.0__ multiply__2.0__4.0__ multiply__1.0__8.0__ |
subtract__10.0__1.0__ divide__36.0__9.0__ subtract__9.0__1.0__ subtract__9.0__1.0__ |
| the area of a rectangular field is equal to num__300 square meters . its perimeter is equal to num__70 meters . find the length and width of this rectangle . <o> a ) w = num__35 and l = num__31 <o> b ) w = num__15 and l = num__20 <o> c ) w = num__25 and l = num__42 <o> d ) w = num__27 and l = num__37 <o> e ) w = num__42 and l = num__13 |
l * w = num__300 : area l is the length and w is the width . num__2 l + num__2 w = num__70 : perimeter l = num__35 - w : solve for l ( num__35 - w ) * w = num__300 : substitute in the area equation w = num__15 and l = num__20 : solve for w and find l using l = num__35 - w . correct answer b <eor> b <eos> |
b |
triangle_area__2.0__15.0__ |
triangle_area__2.0__15.0__ |
| calculate num__24 x num__99 <o> a ) num__2673 <o> b ) num__2763 <o> c ) num__2637 <o> d ) num__2736 <o> e ) num__2376 |
= num__24 - num__1 = num__23 = ( decrement each digit of the number obtained from num__9 ) here we got num__23 . now = num__9 - num__2 = num__7 and num__9 - num__3 = num__6 . so we have num__76 just write these numbers together . that is we have num__23 and num__76 . hence answer is num__2376 answer is e . <eor> e <eos> |
e |
subtract__24.0__1.0__ subtract__9.0__2.0__ add__1.0__2.0__ multiply__2.0__3.0__ subtract__99.0__23.0__ multiply__24.0__99.0__ multiply__24.0__99.0__ |
subtract__24.0__1.0__ subtract__9.0__2.0__ add__1.0__2.0__ subtract__7.0__1.0__ subtract__99.0__23.0__ multiply__24.0__99.0__ multiply__24.0__99.0__ |
| a man buys num__50 pens at marked price of num__46 pens from a whole seller . if he sells these pens giving a discount of num__1.0 what is the profit percent ? <o> a ) num__5.3 <o> b ) num__7.6 <o> c ) num__10.6 <o> d ) num__12.0 <o> e ) none of these |
explanation : let marked price be re . num__1 each c . p . of num__50 pens = rs . num__46 s . p . of num__50 pens = num__99.0 of rs . num__50 = rs . num__49.50 profit % = ( profit / c . p . ) x num__100 profit % = ( num__3.50 / num__46 ) x num__100 = num__7.6 answer b <eor> b <eos> |
b |
percent__50.0__99.0__ percent__100.0__7.6__ |
percent__50.0__99.0__ percent__100.0__7.6__ |
| the sum of all consecutive odd integers from − num__23 to num__31 inclusive is <o> a ) num__96 <o> b ) num__112 <o> c ) num__158 <o> d ) num__192 <o> e ) num__235 |
the sum of the odd numbers from - num__23 to + num__23 is num__0 . let ' s add the remaining numbers . num__25 + num__27 + num__29 + num__31 = num__112 the answer is b . <eor> b <eos> |
b |
round__112.0__ |
round__112.0__ |
| jack ' s toy box contains num__13 toy soldiers num__7 model airplanes and num__5 dolls . what is the probability that a randomly chosen toy from the box will be either a model airplane or a doll ? <o> a ) num__0.138888888889 <o> b ) num__0.48 <o> c ) ( num__0.28 ) + ( num__0.208333333333 ) <o> d ) num__0.8 <o> e ) num__12 |
probability of selecting either a model airplanes or a doll = num__1 - ( probability of selecting a toy ) probability of selecting either a model airplanes or a doll = num__1 - num__13 c num__0.04 c num__1 probability of selecting either a model airplanes or a doll = num__1 - num__0.52 probability of selecting either a model airplanes or a doll = num__0.48 hence answer will be ( b ) <eor> b <eos> |
b |
negate_prob__0.52__ negate_prob__0.52__ |
negate_prob__0.52__ negate_prob__0.52__ |
| if - num__2 < = x < = num__2 and num__3 < = y < = num__5 which of the following represents the range of all possible values of y - x ? <o> a ) num__5 < = y - x < = num__6 <o> b ) num__1 < = y - x < = num__7 <o> c ) num__1 < = y - x < = num__6 <o> d ) num__5 < = y - x < = num__10 <o> e ) num__1 < = y - x < = num__10 |
maximum value of y - x = num__5 - ( - num__2 ) = num__7 minimum value of y - x = num__3 - num__2 = num__1 b mentions the correct range <eor> b <eos> |
b |
add__2.0__5.0__ subtract__3.0__2.0__ reverse__1.0__ |
add__2.0__5.0__ subtract__3.0__2.0__ subtract__2.0__1.0__ |
| for what value of x between − num__8 and num__8 inclusive is the value of x ^ num__2 − num__10 x + num__16 the greatest ? <o> a ) − num__8 <o> b ) − num__2 <o> c ) num__0 <o> d ) num__2 <o> e ) num__8 |
we can see from the statement that two terms containing x x ^ num__2 will always be positive and - num__10 x will be positive if x is - ive . . so the equation will have greatest value if x is - ive and lower the value of x greater is the equation . so - num__8 will give the greatest value . . ans a <eor> a <eos> |
a |
subtract__10.0__2.0__ |
subtract__10.0__2.0__ |
| donovan and michael are racing around a circular num__500 - meter track . if donovan runs each lap in num__45 seconds and michael runs each lap in num__40 seconds how many laps will michael have to complete in order to pass donovan assuming they start at the same time ? <o> a ) num__9 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__13 |
one way of approaching this question is by relative speed method num__1 . speed / rate of donovan = distance / time = > num__11.1111111111 = > num__11.1111111111 num__2 . speed / rate of michael = distance / time = > num__12.5 = > num__12.5 relative speed between them = num__12.5 - num__11.1111111111 = > num__0.694444444444 ( we subtract the rates if moving in the same direction and add the rates if moving in the opposite direction ) in order to pass donovan - distance to be covered = num__500 relative rate = num__1.38888888889 total time taken by micheal to surpass donovan = distance / rate = > num__500 * num__0.72 = > num__360.0 = > num__360 no . of laps taken by michael = total time / michael ' s rate = > num__9.0 = > num__9 hence correct answer is num__9 laps . a <eor> a <eos> |
a |
divide__500.0__45.0__ divide__500.0__40.0__ subtract__12.5__11.1111__ divide__1.0__1.3889__ multiply__500.0__0.72__ multiply__12.5__0.72__ round__9.0__ |
divide__500.0__45.0__ divide__500.0__40.0__ subtract__12.5__11.1111__ divide__1.0__1.3889__ multiply__500.0__0.72__ multiply__12.5__0.72__ multiply__12.5__0.72__ |
| a crate measures num__5 feet by num__8 feet by num__12 feet on the inside . a stone pillar in the shape of a right circular cylinder must fit into the crate for shipping so that it rests upright when the crate sits on at least one of its six sides . what is the radius in feet of the pillar with the largest volume that could still fit in the crate ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__5 <o> d ) num__8 <o> e ) num__12 |
we can find the radius of all the three cases of cylinders . the only crux to find the answer faster is that : voulme is pi * r ^ num__2 * h . the volume is a function of r ^ num__2 . so r has to be the highest to find the largest volume . so r = num__5 for the surface num__8 * num__12 face . volume = num__125 pi answer c <eor> c <eos> |
c |
volume_cube__5.0__ triangle_area__5.0__2.0__ |
volume_cube__5.0__ triangle_area__5.0__2.0__ |
| two persons start running simultaneously around a circular track of length num__300 m from the same point at speeds of num__10 km / hr and num__20 km / hr . when will they meet for the first time any where on the track if they are moving in opposite directions ? <o> a ) num__27 sec <o> b ) num__87 sec <o> c ) num__67 sec <o> d ) num__36 sec <o> e ) num__46 sec |
explanation : time taken to meet for the first time anywhere on the track = length of the track / relative speed = num__300 / ( num__10 + num__20 ) num__0.277777777778 = num__300 x num__0.6 x num__5 = num__36 seconds . answer : d <eor> d <eos> |
d |
km_to_mile_conversion__ round__36.0__ |
km_to_mile_conversion__ round__36.0__ |
| if num__50.0 of a number is equal to one - third of another number what is the ratio of first number to the second number ? <o> a ) num__3 : num__2 <o> b ) num__2 : num__3 <o> c ) num__1 : num__2 <o> d ) num__4 : num__5 <o> e ) num__5 : num__3 |
let num__50.0 of a = num__0.333333333333 b . then num__50 a / num__100 = num__1 b / num__3 = > num__1 a / num__2 = num__1 b / num__3 a / b = ( num__0.333333333333 * num__2.0 ) = num__0.666666666667 a : b = num__2 : num__3 . answer : b <eor> b <eos> |
b |
subtract__3.0__1.0__ subtract__1.0__0.3333__ multiply__1.0__2.0__ |
divide__100.0__50.0__ divide__2.0__3.0__ multiply__1.0__2.0__ |
| the average of num__5 numbers is num__7.4 . if one of the numbers is multiplied by a factor of num__3 the average of the numbers increases to num__9.2 . what number is multiplied by num__3 ? <o> a ) num__1.5 <o> b ) num__3.0 <o> c ) num__3.9 <o> d ) num__4.5 <o> e ) num__6.0 |
the average of num__5 numbers is num__7.4 the sum of num__5 numbers will be num__7.4 x num__5 = num__37 the average of num__5 number after one of the number is multiplied by num__3 is num__9.2 the sum of the numbers will now be num__9.2 x num__5 = num__46 so the sum has increased by num__46 - num__37 = num__9 let the number multiplied by num__3 be n then num__3 n = n + num__9 or num__2 n = num__9 or n = num__4.5 answer : - d <eor> d <eos> |
d |
multiply__5.0__7.4__ multiply__5.0__9.2__ round_down__9.2__ subtract__5.0__3.0__ divide__9.0__2.0__ subtract__9.0__4.5__ |
multiply__5.0__7.4__ multiply__5.0__9.2__ subtract__46.0__37.0__ subtract__5.0__3.0__ divide__9.0__2.0__ subtract__9.0__4.5__ |
| carl drove from his home to the beach at an average speed of num__80 kilometers per hour and returned home by the same route at an average speed of num__70 kilometers per hour . if the trip home took num__0.5 hour longer than the trip to the beach how many kilometers e did carl drive each way ? <o> a ) num__350 <o> b ) num__345 <o> c ) num__320 <o> d ) num__280 <o> e ) num__240 |
let us backsolve here . the answer option has to be divisible by num__7 to give us num__0.5 . let us try e = num__280 km . time taken will be num__3.5 hours and num__4 hours . hence d is the answer . <eor> d <eos> |
d |
multiply__0.5__7.0__ add__0.5__3.5__ round__280.0__ |
multiply__0.5__7.0__ add__0.5__3.5__ round__280.0__ |
| if the average ( arithmetic mean ) of the four numbers k num__2 k + num__3 num__3 k – num__5 and num__5 k + num__1 is num__41 what is the value of k ? <o> a ) num__41 <o> b ) num__15 num__0.75 <o> c ) num__22 <o> d ) num__23 <o> e ) num__25 num__0.3 |
k + num__2 k + num__3 + num__3 k - num__5 + num__5 k + num__1 = num__11 k - num__1 ( num__11 k - num__1 ) / num__4 = num__41 num__11 k = num__41 * num__4 + num__1 = num__164 + num__1 = num__165 k = num__15.0 = num__15 . answer a . <eor> a <eos> |
a |
add__3.0__1.0__ multiply__41.0__4.0__ add__1.0__164.0__ multiply__3.0__5.0__ multiply__1.0__41.0__ |
add__3.0__1.0__ multiply__41.0__4.0__ add__1.0__164.0__ multiply__3.0__5.0__ multiply__1.0__41.0__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__120 m <o> b ) num__240 m <o> c ) num__300 m <o> d ) num__360 m <o> e ) num__280 m |
speed = num__54 x num__0.277777777778 m / sec = num__15 m / sec . length of the train = ( num__15 x num__20 ) m = num__300 m . let the length of the platform be x metres . then ( x + num__300 ) / num__36 = num__15 x + num__300 = num__540 x = num__240 m . answer : option b <eor> b <eos> |
b |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
| rectangle dimensions were given num__2 doors dimensions given num__2 windows dimensions given . num__70 cm wide wallpaper to be covered on the walls of this room . cost of wallpaper is num__60 per meter . . calculate total cost <o> a ) num__20 <o> b ) num__25 <o> c ) num__30 <o> d ) num__60 <o> e ) num__70 |
area of room = ( l * b ) m ^ num__2 area of door + window = ( d + w ) m ^ num__2 remaining area = ( ( l * b ) - ( d + w ) ) m ^ num__2 so length = remaining area / num__0.70 m total cost = length * num__60 - - - ans and total cost = num__60 answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ |
hour_to_min_conversion__ |
| a waiter ' s salary consists of his salary and tips . during one week his tips were num__1.66666666667 of his salary . what fraction of his income came from tips ? <o> a ) num__0.444444444444 <o> b ) num__1.25 <o> c ) num__0.625 <o> d ) num__0.555555555556 <o> e ) num__0.666666666667 |
income = salary ( s ) + tips = s + s * num__1.66666666667 = s * num__2.66666666667 tips = s * num__2.66666666667 fraction of his income came from tips = ( s * num__1.66666666667 ) / ( s * num__2.66666666667 ) = num__0.625 answer : c <eor> c <eos> |
c |
divide__1.6667__2.6667__ divide__1.6667__2.6667__ |
divide__1.6667__2.6667__ divide__1.6667__2.6667__ |
| what is the tens ' digit e of the sum of the first num__40 terms of num__1 num__11 num__111 num__1111 num__11111 num__111111 . . . ? <o> a ) e = num__2 <o> b ) e = num__3 <o> c ) num__4 <o> d ) num__8 <o> e ) num__9 |
all of the first num__40 terms have num__1 at the units place . except the first term the tens digit of all the remaining num__39 terms is num__1 . so now if you do the addition num__1 num__1 num__1 num__1 . . . . . . . . . num__1 num__1 num__0 num__1 num__3 num__0 answer is b <eor> b <eos> |
b |
subtract__40.0__1.0__ reverse__111111.0__ multiply__1.0__3.0__ |
subtract__40.0__1.0__ reverse__111111.0__ multiply__1.0__3.0__ |
| can c and can В are both right circular cylinders . the radius of can c is twice the radius of can b while the height of can c is half the height of can b . if it costs $ num__4.00 to fill half of can b with a certain brand of gasoline how much would it cost to completely fill can c with the same brand of gasoline ? <o> a ) $ num__1 <o> b ) $ num__2 <o> c ) $ num__4 <o> d ) $ num__8 <o> e ) $ num__16 |
let x be the radius of b and num__2 h be the height of b . therefore radius of c = num__2 x and height = h vol of b = num__3.14 * x ^ num__2 * num__2 h vol of a = num__3.14 * num__4 x ^ num__2 * h cost to fill half of b = $ num__4 - - > cost to fill full b = $ num__8 - - > num__3.14 * x ^ num__2 * num__2 h = num__8 - - > num__3.14 * x ^ num__2 * h = num__4 - - > num__4 * ( num__3.14 * x ^ num__2 * h ) = $ num__16 ans e <eor> e <eos> |
e |
square_perimeter__2.0__ square_perimeter__4.0__ square_perimeter__4.0__ |
multiply__4.0__2.0__ power__4.0__2.0__ power__4.0__2.0__ |
| the average height of num__35 boys in a class was calculated as num__184 cm . it has later found that the height of one of the boys in the class was wrongly written as num__166 cm whereas his actual height was num__106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ? <o> a ) num__187.89 cm <o> b ) num__182.29 cm <o> c ) num__123.98 cm <o> d ) num__149.98 cm <o> e ) num__146.89 cm |
calculated average height of num__35 boys = num__184 cm . wrong total height of num__35 boys = num__184 * num__35 cm . this was as a result of an actual height of num__106 cm being wrongly written as num__166 cm . correct total height of num__35 boys = num__183 cm - ( num__166 cm + num__106 cm ) / num__35 = num__184 cm - num__1.71428571429 cm = num__184 cm - num__1.71 cm = num__182.29 cm . answer : b <eor> b <eos> |
b |
subtract__184.0__1.71__ subtract__184.0__1.71__ |
subtract__184.0__1.71__ subtract__184.0__1.71__ |
| a watch was sold at a loss of num__10.0 . if it was sold for rs . num__140 more there would have been a gain of num__4.0 . what is the cost price ? <o> a ) num__1000 <o> b ) num__2881 <o> c ) num__2771 <o> d ) num__1111 <o> e ) num__7121 |
num__90.0 num__104.0 - - - - - - - - num__14.0 - - - - num__140 num__100.0 - - - - ? = > rs . num__1000 answer : a <eor> a <eos> |
a |
percent__10.0__140.0__ percent__100.0__1000.0__ |
percent__10.0__140.0__ percent__100.0__1000.0__ |
| if num__5 machines can produce num__20 units in num__10 hours how long would it take num__20 machines to produce num__180 units ? <o> a ) num__50 hours <o> b ) num__40 hours <o> c ) num__22.5 hours <o> d ) num__12 hours <o> e ) num__8 hours |
here we ' re told that num__5 machines can produce num__20 units in num__10 hours . . . . that means that each machine works for num__10 hours apiece . since there are num__5 machines ( and we ' re meant to assume that each machine does the same amount of work ) then the num__5 machines equally created the num__20 units . num__20 units / num__5 machines = num__4 units are made by each machine every num__10 hours now that we know how long it takes each machine to make num__4 units we can break this down further if we choose to . . . num__10 hours / num__4 units = num__2.5 hours per unit when num__1 machine is working . the prompt asks us how long would it take num__20 machines to produce num__180 units . if num__20 machines each work for num__2.5 hours then we ' ll have num__20 units . since num__180 units is ' num__9 times ' num__20 we need ' num__9 times ' more time . ( num__2.5 hours ) ( num__9 times ) = num__22.5 hours final answer : [ reveal ] spoiler : c <eor> c <eos> |
c |
divide__20.0__5.0__ divide__10.0__4.0__ subtract__5.0__4.0__ add__5.0__4.0__ add__20.0__2.5__ round__22.5__ |
divide__20.0__5.0__ divide__10.0__4.0__ subtract__5.0__4.0__ divide__180.0__20.0__ add__20.0__2.5__ divide__22.5__1.0__ |
| the ratio of males to females in a class is num__2 : num__3 . the career preferences of the students in the class are represented in a circle graph . if the area of the graph allocated to each career preference is proportional to the number of students who have that career preference how many degrees of the circle should be used to represent a career that is preferred by one - fourth of the males and three - fourths of the females in the class ? <o> a ) num__186 <o> b ) num__192 <o> c ) num__198 <o> d ) num__204 <o> e ) num__210 |
num__0.25 * num__0.4 + num__0.75 * num__0.6 = num__0.1 + num__0.45 = num__0.55 the number of degrees is num__0.55 * num__360 = num__198 degrees the answer is c . <eor> c <eos> |
c |
multiply__3.0__0.25__ triangle_area__3.0__0.4__ multiply__0.25__0.4__ multiply__0.75__0.6__ multiply__360.0__0.55__ triangle_area__2.0__198.0__ |
multiply__3.0__0.25__ volume_rectangular_prism__2.0__0.75__0.4__ multiply__0.25__0.4__ multiply__0.75__0.6__ multiply__360.0__0.55__ multiply__360.0__0.55__ |
| when a number is divided by num__6 & then multiply by num__12 the answer is num__18 what is the no . ? <o> a ) num__4.5 <o> b ) num__5 <o> c ) num__5.6 <o> d ) num__5.7 <o> e ) num__9 |
if $ x $ is the number x / num__6 * num__12 = num__18 = > num__2 x = num__18 = > x = num__9 e <eor> e <eos> |
e |
divide__12.0__6.0__ divide__18.0__2.0__ subtract__18.0__9.0__ |
divide__12.0__6.0__ divide__18.0__2.0__ divide__18.0__2.0__ |
| a cube has four of its faces painted half red and half white . the other faces are completely painted white . what is the ratio between the red painted areas and the white painted areas of the cube ? <o> a ) num__1 : num__4 <o> b ) num__1 : num__3 <o> c ) num__1 : num__2 <o> d ) num__2 : num__5 <o> e ) num__2 : num__11 |
let x be the area of each face of the cube . the area painted red is num__4 ( x / num__2 ) = num__2 x the area painted white is num__4 ( x / num__2 ) + num__2 x = num__4 x the ratio of red to white is num__2 x : num__4 x which is num__1 : num__2 . the answer is c . <eor> c <eos> |
c |
volume_cube__1.0__ |
volume_cube__1.0__ |
| if the range w of the six numbers num__4 num__314 num__710 and x is num__12 what is the difference between the greatest possible value of x and least possible value of x ? <o> a ) num__0 <o> b ) num__2 <o> c ) num__12 <o> d ) num__13 <o> e ) num__15 |
the range w of a set is the difference between the largest and smallest elements of a set . without x the difference between the largest and smallest elements of a set is num__14 - num__3 = num__11 < num__12 which means that in order num__12 to be the range of the set x must be either the smallest element so that num__14 - x = num__12 - - - > x = num__2 or x must the largest element so that x - num__3 = num__12 - - > x = num__15 . the the difference between the greatest possible value of x and least possible value of x is num__15 - num__2 = num__13 . answer : d . <eor> d <eos> |
d |
divide__12.0__4.0__ subtract__14.0__3.0__ subtract__14.0__12.0__ add__4.0__11.0__ add__2.0__11.0__ add__2.0__11.0__ |
divide__12.0__4.0__ subtract__14.0__3.0__ subtract__14.0__12.0__ add__4.0__11.0__ subtract__15.0__2.0__ subtract__15.0__2.0__ |
| a man can row num__4 kmph is still water . if the river is running at num__2 kmph it takes num__90 min to row to a place and back . how far is the place <o> a ) num__2 km <o> b ) num__4 km <o> c ) num__5 km <o> d ) num__2.25 km <o> e ) none of these |
explanation : speed in still water = num__4 kmph speed of the stream = num__2 kmph speed upstream = ( num__4 - num__2 ) = num__2 kmph speed downstream = ( num__4 + num__2 ) = num__6 kmph total time = num__90 minutes = num__90 ⁄ num__60 hour = num__3 ⁄ num__2 hour let l be the distance . then ( l / num__6 ) + ( l / num__2 ) = num__32 = > l + num__3 l = num__9 = > num__4 l = num__9 = > l = num__9 ⁄ num__4 = num__2.25 km . answer : option d <eor> d <eos> |
d |
add__4.0__2.0__ hour_to_min_conversion__ divide__6.0__2.0__ add__3.0__6.0__ divide__9.0__4.0__ round__2.25__ |
add__4.0__2.0__ hour_to_min_conversion__ divide__6.0__2.0__ add__3.0__6.0__ divide__9.0__4.0__ divide__9.0__4.0__ |
| the distance light travels in one year is approximately num__5 num__870000 num__000000 miles . the distance light travels in num__100 years is : <o> a ) × num__108 miles <o> b ) × num__1010 miles <o> c ) × num__10 - num__10 miles <o> d ) × num__1012 miles <o> e ) × num__10 - num__12 miles |
solution : the distance of the light travels in num__100 years is : num__5 num__870000 num__000000 × num__100 miles . = num__587 num__000000 num__000000 miles . = num__587 × num__1012 miles . answer : d <eor> d <eos> |
d |
round__1012.0__ |
round__1012.0__ |
| a b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = num__2 : num__3 and b : c = num__2 : num__5 . if the total runs scored by all of them are num__100 the runs scored by a are ? <o> a ) num__11 <o> b ) num__18 <o> c ) num__13 <o> d ) num__16 <o> e ) num__12 |
a : b = num__2 : num__3 b : c = num__2 : num__5 a : b : c = num__4 : num__6 : num__15 num__0.16 * num__100 = num__16 answer : d <eor> d <eos> |
d |
multiply__2.0__3.0__ multiply__3.0__5.0__ multiply__100.0__0.16__ multiply__100.0__0.16__ |
multiply__2.0__3.0__ multiply__3.0__5.0__ multiply__100.0__0.16__ multiply__100.0__0.16__ |
| a single discount equivalent to the discount series of num__20.0 num__10.0 and num__5.0 is ? <o> a ) num__31.7 <o> b ) num__31.21 <o> c ) num__31.6 <o> d ) num__31.37 <o> e ) num__31.25 |
num__100 * ( num__0.8 ) * ( num__0.9 ) * ( num__0.95 ) = num__68.4 num__100 - num__68.4 = num__31.6 answer : c <eor> c <eos> |
c |
percent__100.0__31.6__ |
percent__100.0__31.6__ |
| a and b can do a piece of work in num__3 days b and c in num__4 days c and a in num__6 days . how long will c take to do it ? <o> a ) num__24 <o> b ) num__18 <o> c ) num__20 <o> d ) num__30 <o> e ) num__15 |
num__2 c = Â ¼ + num__0.166666666667 â € “ num__0.333333333333 = num__0.0833333333333 c = num__0.0416666666667 = > num__24 days answer a <eor> a <eos> |
a |
divide__6.0__3.0__ divide__2.0__6.0__ divide__0.1667__2.0__ divide__0.1667__4.0__ multiply__4.0__6.0__ round__24.0__ |
divide__6.0__3.0__ divide__2.0__6.0__ divide__0.1667__2.0__ divide__0.1667__4.0__ multiply__4.0__6.0__ round__24.0__ |
| of num__15 applicants for a job num__7 had at least num__4 years ' experience num__9 had degrees and num__1 had less than num__4 years ' experience and did not have a degree . how many of the applicants had at least num__4 years ' experience and a degree ? <o> a ) num__14 <o> b ) num__13 <o> c ) num__9 <o> d ) num__7 <o> e ) num__2 |
num__15 - num__1 = num__14 num__14 - num__7 - num__9 = - num__2 then num__2 are in the intersection between num__4 years experience and degree . answer : e <eor> e <eos> |
e |
subtract__15.0__1.0__ subtract__9.0__7.0__ subtract__4.0__2.0__ |
subtract__15.0__1.0__ subtract__9.0__7.0__ subtract__4.0__2.0__ |
| num__0.002 x num__0.5 = ? <o> a ) num__0.0001 <o> b ) num__1.0 <o> c ) num__0.1 <o> d ) num__0.01 <o> e ) num__0.001 |
num__2 x num__5 = num__10 . sum of decimal places = num__4 num__0.002 x num__0.5 = num__0.001 answer is e . <eor> e <eos> |
e |
reverse__0.5__ multiply__2.0__5.0__ divide__2.0__0.5__ multiply__0.002__0.5__ multiply__0.002__0.5__ |
reverse__0.5__ multiply__2.0__5.0__ divide__2.0__0.5__ multiply__0.002__0.5__ multiply__0.002__0.5__ |
| a number is four less than two times the other number . if their difference is num__21 what is the greater number ? <o> a ) num__50 <o> b ) num__46 <o> c ) num__31 <o> d ) num__21 <o> e ) num__15 |
a = one number b = other number a = - num__4 + num__2 b a - b = num__21 solve for a by adding b to each side a = b + num__21 substitute ( b + num__21 ) for a in a = - num__4 + num__2 b ( b + num__21 ) = - num__4 + num__2 b add - b to each side num__21 = - num__4 + b add num__4 to each side num__25 = b since a = b + num__21 a = num__25 + num__21 a = num__46 answer : b <eor> b <eos> |
b |
add__21.0__4.0__ add__21.0__25.0__ add__21.0__25.0__ |
add__21.0__4.0__ add__21.0__25.0__ add__21.0__25.0__ |
| a man travelled a distance of num__48 km in num__7 hours . he travelled partly on foot at num__4 km / hr and partly on bicycle at num__9 km / hr . what is the distance travelled on foot ? <o> a ) num__16 km <o> b ) num__20 km <o> c ) num__12 km <o> d ) num__10 km <o> e ) none of these |
explanation : let the time in which he travelled on foot = x hour time for travelling on bicycle = ( num__9 - x ) hr distance = speed * time and total distance = num__48 km so num__4 x + num__7 ( num__9 - x ) = num__48 = > num__3 x = num__15 = > x = num__5 so distance traveled on foot = num__4 ( num__5 ) = num__20 km answer : option b <eor> b <eos> |
b |
subtract__7.0__4.0__ subtract__9.0__4.0__ multiply__4.0__5.0__ round__20.0__ |
subtract__7.0__4.0__ subtract__9.0__4.0__ multiply__4.0__5.0__ multiply__4.0__5.0__ |
| company c produces toy trucks at a cost of $ num__5.00 each for the first num__100 trucks and $ num__3.50 for each additional truck . if num__500 toy trucks were produced by company c and sold for $ num__8.00 each what was company c ’ s gross profit ? <o> a ) $ num__2250 <o> b ) $ num__2500 <o> c ) $ num__2100 <o> d ) $ num__2150 <o> e ) $ num__2 |
700 |
cost of num__500 trucks : ( num__100 * num__5 ) + ( num__400 * num__3.5 ) = num__500 + num__1400 = $ num__1900 revenue : num__500 * num__8 = $ num__4000 profit : num__4000 - num__1900 = $ num__2100 option c is correct <eor> c <eos> |
c |
c |
| a boat covers a certain distance downstream in num__1 hour while it comes back in num__1.30 hours . if the speed of the stream be num__3 kmph what is the speed of the boat in still water ? <o> a ) num__10 <o> b ) num__13 <o> c ) num__11 <o> d ) num__08 <o> e ) num__15 |
let the speed of the water in still water = x given that speed of the stream = num__3 kmph speed downstream = ( x + num__3 ) kmph speed upstream = ( x − num__3 ) kmph he travels a certain distance downstream in num__1 hour and come back in num__1 num__1 ⁄ num__2 hour . i . e . distance travelled downstream in num__1 hour = distance travelled upstream in num__1 num__1 ⁄ num__2 hour since distance = speed × time we have ( x + num__3 ) × num__1 = ( x − num__3 ) num__1.5 ⇒ num__2 ( x + num__3 ) = num__3 ( x − num__3 ) ⇒ num__2 x + num__6 = num__3 x − num__9 ⇒ x = num__6 + num__9 = num__15 kmph answer is e . <eor> e <eos> |
e |
subtract__3.0__1.0__ divide__3.0__2.0__ multiply__3.0__2.0__ add__3.0__6.0__ add__6.0__9.0__ round__15.0__ |
subtract__3.0__1.0__ divide__3.0__2.0__ multiply__3.0__2.0__ add__3.0__6.0__ add__6.0__9.0__ add__6.0__9.0__ |
| steve traveled the first num__2 hours of his journey at num__30 mph and the remaining num__3 hours of his journey at num__80 mph . what is his average speed for the entire journey ? <o> a ) num__60 mph <o> b ) num__56.67 mph <o> c ) num__53.33 mph <o> d ) num__64 mph <o> e ) num__66.67 mph |
distance traveled in num__2 hours = num__2 * num__30 = num__60 m distance traveled in num__3 hours = num__3 * num__80 = num__240 m total distance covered = num__240 + num__60 = num__300 m total time = num__2 + num__3 = num__5 h hence avg speed = total distance covered / total time taken = num__60.0 = num__60 mph answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ multiply__3.0__80.0__ add__240.0__60.0__ add__2.0__3.0__ hour_to_min_conversion__ |
multiply__2.0__30.0__ multiply__3.0__80.0__ add__240.0__60.0__ add__2.0__3.0__ multiply__2.0__30.0__ |
| find the no . of ways of arranging the boy and num__5 guests at a circular table so that the boy always sits in a particular seat ? <o> a ) num__7 ! <o> b ) num__5 ! <o> c ) num__9 ! <o> d ) num__10 ! <o> e ) num__11 ! |
ans . ( b ) sol . total number of persons = num__6 host can sit in a particular seat in one way . now remaining positions are defined relative to the host . hence the remaining can sit in num__5 places in num__5 p num__5 = num__5 ! ways . . . the number of required arrangements = num__5 ! x num__1 = num__5 ! = num__5 ! ways <eor> b <eos> |
b |
die_space__ vowel_space__ |
die_space__ vowel_space__ |
| use distributive property to solve the problem below : maria bought num__10 notebooks and num__5 pens costing num__2 dollars each . how much did maria pay ? <o> a ) num__30 dollars <o> b ) num__40 dollars <o> c ) num__50 dollars <o> d ) num__60 dollars <o> e ) num__70 dollars |
solution num__2 × ( num__10 + num__5 ) = num__2 × num__10 + num__2 × num__5 = num__20 + num__10 = num__30 dollars answer a <eor> a <eos> |
a |
multiply__10.0__2.0__ add__10.0__20.0__ add__10.0__20.0__ |
multiply__10.0__2.0__ add__10.0__20.0__ add__10.0__20.0__ |
| a person x working alone can complete a work in num__5 days . a person y completes the same amount of work in num__20 days and a person z when working alone can complete the same amount of work in num__30 days . all three people work together for num__2 days but then x and y leave . how many more days does z need to complete the work after x and y leave ? <o> a ) num__13 <o> b ) num__14 <o> c ) num__15 <o> d ) num__16 <o> e ) num__17 |
the fraction of work completed in two days is num__0.4 + num__0.1 + num__0.0666666666667 = num__0.566666666667 the fraction of work remaining is num__1 - num__0.566666666667 = num__0.433333333333 the number of days required for z is num__13 days . the answer is a . <eor> a <eos> |
a |
divide__2.0__5.0__ divide__2.0__20.0__ divide__2.0__30.0__ subtract__1.0__0.5667__ round__13.0__ |
divide__2.0__5.0__ divide__2.0__20.0__ divide__2.0__30.0__ subtract__1.0__0.5667__ round__13.0__ |
| in traveling from a dormitory to a certain city a student went num__0.2 of the way by foot num__0.666666666667 of the way by bus and the remaining num__10 kilometers by car . what is the distance in kilometers from the dormitory to the city ? <o> a ) num__30 <o> b ) num__45 <o> c ) num__60 <o> d ) num__75 <o> e ) num__120 |
whole trip = distance by foot + distance by bus + distance by car x = num__0.2 x + num__0.666666666667 x + num__10 x - num__0.866666666667 x = num__10 num__0.133333333333 x = num__10 = > so x = ( num__7.5 ) * num__10 = num__75 km answer d <eor> d <eos> |
d |
add__0.2__0.6667__ multiply__0.2__0.6667__ multiply__10.0__7.5__ round__75.0__ |
add__0.2__0.6667__ multiply__0.2__0.6667__ multiply__10.0__7.5__ multiply__10.0__7.5__ |
| if num__1 tic equals num__5 tacs and num__2 tacs equal num__8 tocs what is the ratio of one tic to one toc ? <o> a ) num__7.5 <o> b ) num__20.0 <o> c ) num__0.833333333333 <o> d ) num__0.3 <o> e ) num__0.0666666666667 |
tic = num__5 * tac and num__2 * tac = num__8 * toc ; num__2 * tic = num__10 * tac and num__10 * tac = num__40 * toc - - > num__2 * tic = num__40 * toc - - > tic / toc = num__20.0 = num__20.0 . answer : b . <eor> b <eos> |
b |
multiply__5.0__2.0__ multiply__5.0__8.0__ multiply__2.0__10.0__ multiply__1.0__20.0__ |
multiply__5.0__2.0__ multiply__5.0__8.0__ multiply__2.0__10.0__ multiply__1.0__20.0__ |
| a motor cyclist participant of a race says ` ` we drove with the speed of num__10 miles an hour one way but while returning because of less traffic we drove on the same route with num__18 miles per hour . ' ' what was their average speed in the whole journey ? <o> a ) num__10 miles / hour <o> b ) num__12 miles / hour <o> c ) num__11 miles / hour <o> d ) num__13 miles / hour <o> e ) num__12.9 miles / hour |
num__2 xy / x + y = num__12.9 miles / hour answer : e <eor> e <eos> |
e |
round__12.9__ |
round__12.9__ |
| two trains are moving at num__50 kmph and num__70 kmph in opposite directions . their lengths are num__150 m and num__100 m respectively . the time they will take to pass each other completely is ? <o> a ) num__6 ½ sec <o> b ) num__9 ½ sec <o> c ) num__3 ½ sec <o> d ) num__7 ½ sec <o> e ) num__2 ½ sec |
\ num__70 + num__50 = num__120 * num__0.277777777778 = num__33.3333333333 mps d = num__150 + num__100 = num__250 m t = num__250 * num__0.03 = num__7.5 = num__7 ½ sec answer : d <eor> d <eos> |
d |
add__50.0__70.0__ add__150.0__100.0__ multiply__0.03__250.0__ round__7.0__ |
add__50.0__70.0__ add__150.0__100.0__ multiply__0.03__250.0__ round__7.0__ |
| a train num__360 m long is running at a speed of num__36 km / hr . in what time will it pass a bridge num__140 m long ? <o> a ) num__40 sec <o> b ) num__50 sec <o> c ) num__88 sec <o> d ) num__19 sec <o> e ) num__10 sec |
speed = num__36 * num__0.277777777778 = num__10 m / sec total distance covered = num__360 + num__140 = num__500 m required time = num__500 * num__0.1 = num__50 sec answer : b <eor> b <eos> |
b |
divide__360.0__36.0__ add__360.0__140.0__ divide__36.0__360.0__ multiply__0.1__500.0__ round__50.0__ |
divide__360.0__36.0__ add__360.0__140.0__ divide__36.0__360.0__ multiply__0.1__500.0__ multiply__0.1__500.0__ |
| a man took a loan at rate of num__12.0 per annum simple interest . after num__3 years he had to pay num__7200 interest . the principal amount borrowed by him was . <o> a ) rs num__14000 <o> b ) rs num__15000 <o> c ) rs num__16000 <o> d ) rs num__20000 <o> e ) none of these |
explanation : s . i . = p â ˆ — r â ˆ — t / num__100 = > p = s . i . â ˆ — num__100 / r â ˆ — t = > p = num__7200 â ˆ — num__8.33333333333 â ˆ — num__3 = rs num__20000 option d <eor> d <eos> |
d |
percent__100.0__20000.0__ |
percent__100.0__20000.0__ |
| what approximate value should come in place of the question mark ( ? ) in each of the following equations ? ( num__9.009 ) num__0.5 + ( num__16.0001 ) num__0.25 = ( ? ) % of num__50.5 <o> a ) num__7 <o> b ) num__8 <o> c ) num__5 <o> d ) num__10 <o> e ) num__11 |
explanation : num__92 + num__162 = ( ? ) % of ( num__50.5 ) num__3 + num__2 = ( num__2 ) % of ( num__50.5 ) num__5 = num__10.0 of ( num__50.5 ) answer : option d <eor> d <eos> |
d |
reverse__0.5__ add__3.0__2.0__ divide__5.0__0.5__ divide__5.0__0.5__ |
reverse__0.5__ add__3.0__2.0__ divide__5.0__0.5__ divide__5.0__0.5__ |
| the average monthly income of p and q is rs . num__5050 . the average monthly income of q and r is num__6250 and the average monthly income of p and r is rs . num__5200 . the monthly income of p is ? <o> a ) rs . num__4078 <o> b ) rs . num__4000 <o> c ) rs . num__4029 <o> d ) rs . num__4027 <o> e ) rs . num__4020 |
let p q and r represent their respective monthly incomes . then we have : p + q = ( num__5050 * num__2 ) = num__10100 - - - ( i ) q + r = ( num__6250 * num__2 ) = num__12500 - - - ( ii ) p + r = ( num__5200 * num__2 ) = num__10400 - - - ( iii ) adding ( i ) ( ii ) and ( iii ) we get : num__2 ( p + q + r ) = num__33000 = p + q + r = num__16500 - - - ( iv ) subtracting ( ii ) from ( iv ) we get p = num__4000 . p ' s monthly income = rs . num__4000 . answer : b <eor> b <eos> |
b |
multiply__5050.0__2.0__ multiply__6250.0__2.0__ multiply__5200.0__2.0__ divide__33000.0__2.0__ subtract__16500.0__12500.0__ subtract__16500.0__12500.0__ |
multiply__5050.0__2.0__ multiply__6250.0__2.0__ multiply__5200.0__2.0__ divide__33000.0__2.0__ subtract__16500.0__12500.0__ subtract__16500.0__12500.0__ |
| at a dinner party num__6 people are to be seated around a circular table . two seating arrangements are considered different only when the positions of the people are different relative to each other . what is the total number of different possible seating arrangements for the group ? <o> a ) num__1 ) num__5 <o> b ) num__2 ) num__10 <o> c ) num__3 ) num__24 <o> d ) num__4 ) num__32 <o> e ) num__5 ) num__120 |
for number of distinct arrangements of ' n ' items around a circular table we get the arrangements by way of ( n - num__1 ) ! in this case there are num__6 guests so number of distinct arrangements = num__5 ! = num__120 answer : e <eor> e <eos> |
e |
vowel_space__ vowel_space__ |
vowel_space__ vowel_space__ |
| how many num__7 ' s are there in the following series which are preceded by num__6 which is not preceded by num__8 ? num__8 num__7 num__6 num__7 num__8 num__6 num__7 num__5 num__6 num__7 num__9 num__7 num__6 num__1 num__5 num__7 num__7 num__6 num__8 num__8 num__6 num__9 num__7 num__6 num__8 num__7 <o> a ) num__2 <o> b ) num__1 <o> c ) num__5 <o> d ) num__3 <o> e ) none |
total num__2 are there answer : a <eor> a <eos> |
a |
subtract__7.0__5.0__ subtract__7.0__5.0__ |
subtract__7.0__5.0__ subtract__7.0__5.0__ |
| rahul ' s mathematics test had num__75 problems num__10 arithmetic num__30 algebra num__35 geometry problems . although he answered num__70.0 of arithmetic num__40.0 of arithmetic and num__60.0 of geometry problems correctly still he got less than num__60.0 problems right . how many more questions he would have to answer more to get passed <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
explanation : number of questions attempted correctly = ( num__70.0 of num__10 + num__40.0 of num__30 + num__60.0 of num__35 ) = num__7 + num__12 + num__21 = num__40 . questions to be answered correctly for num__60.0 = num__60.0 of total quations = num__60.0 of num__75 = num__45 . he would have to answer num__45 - num__40 = num__5 answer : option a <eor> a <eos> |
a |
divide__70.0__10.0__ subtract__75.0__30.0__ subtract__75.0__70.0__ subtract__75.0__70.0__ |
divide__70.0__10.0__ subtract__75.0__30.0__ subtract__75.0__70.0__ subtract__75.0__70.0__ |
| ram professes to sell his goods at the cost price but he made use of num__950 grms instead of a kg what is the gain percent ? a . num__11.0 <o> a ) num__11 num__0.111111111111 % <o> b ) num__5.26 <o> c ) num__11 num__1.0 % <o> d ) num__11 num__0.555555555556 % <o> e ) num__11 num__0.2 % |
num__950 - - - num__50 num__100 - - - ? = > num__5.26 answer : b <eor> b <eos> |
b |
percent__100.0__5.26__ |
percent__100.0__5.26__ |
| in a group of hens and cows the total number of legs are num__24 more than twice the number of heads . how many cows are there in the group ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
explanation : let the number of cows be x and the number of hens be y . then = > num__4 x + num__2 y = num__2 ( x + y ) + num__24 = > num__2 x = num__24 = > x = num__12 . answer : c <eor> c <eos> |
c |
divide__24.0__2.0__ subtract__24.0__12.0__ |
divide__24.0__2.0__ subtract__24.0__12.0__ |
| lines y = num__4 √ num__7 · x − num__3 and y = num__3 √ num__7 · x − num__1 intersect at what height above the x axis ? <o> a ) num__7 √ num__5 <o> b ) num__12 <o> c ) num__5 √ num__7 <o> d ) num__0 <o> e ) num__5 |
solve two equations for y y = num__4 √ num__7 · x − num__3 y = num__3 √ num__7 · x − num__1 set the two equations equal to each other and solve for x num__4 √ num__7 · x − num__3 = num__3 √ num__7 · x − num__1 √ num__7 · x = num__2 x = num__2 / √ num__7 plug x = num__2 / √ num__7 into any one of the original equations to find the y value ( height above x axis ) where the two lines intersect y = num__4 √ num__7 · x − num__3 y = ( num__4 √ num__7 ) · ( num__2 / √ num__7 ) − num__3 y = num__4 · num__2 - num__3 y = num__8 - num__3 y = num__5 final answer : e ) num__5 <eor> e <eos> |
e |
subtract__3.0__1.0__ multiply__4.0__2.0__ add__4.0__1.0__ add__4.0__1.0__ |
subtract__3.0__1.0__ multiply__4.0__2.0__ subtract__7.0__2.0__ subtract__7.0__2.0__ |
| for all even numbers n h ( n ) is defined to be the sum of the even numbers between num__2 and n inclusive . what is the value of h ( num__16 ) / h ( num__10 ) ? <o> a ) num__1.8 <o> b ) num__3 <o> c ) num__2.4 <o> d ) num__18 <o> e ) num__60 |
concept : when terms are in arithmetic progression ( a . p . ) i . e . terms are equally spaced then mean = median = ( first + last ) / num__2 and sum = mean * number of terms h ( num__16 ) = [ ( num__2 + num__16 ) / num__2 ] * num__8 = num__72 h ( num__10 ) = ( num__2 + num__10 ) / num__2 ] * num__5 = num__30 h ( num__16 ) / h ( num__10 ) = ( num__72 ) / ( num__30 ) = num__2.4 answer : c <eor> c <eos> |
c |
divide__16.0__2.0__ divide__10.0__2.0__ divide__72.0__30.0__ divide__72.0__30.0__ |
divide__16.0__2.0__ divide__10.0__2.0__ divide__72.0__30.0__ divide__72.0__30.0__ |
| what is the least number of square tiles required to pave the floor of a room num__13 m num__44 cm long and num__4 m num__44 cm broad ? <o> a ) num__7724 <o> b ) num__7804 <o> c ) num__4144 <o> d ) num__7844 <o> e ) none |
solution length of largest tile = h . c . f . of num__1344 cm & num__444 cm = num__12 cm . area of each tile = ( num__12 x num__12 ) cm num__2 ∴ required number of tiles = [ num__1344 x num__37.0 x num__12 ] = num__4144 . answer c <eor> c <eos> |
c |
divide__444.0__12.0__ round__4144.0__ |
divide__444.0__12.0__ round__4144.0__ |
| find the value of num__80641 x num__9999 = m ? <o> a ) num__807518799 <o> b ) num__806436469 <o> c ) num__807538799 <o> d ) num__806329359 <o> e ) num__817431046 |
num__80641 x num__9999 = num__80641 x ( num__10000 - num__1 ) = num__80641 x num__10000 - num__80641 x num__1 = num__806410000 - num__80641 = num__806329359 d <eor> d <eos> |
d |
subtract__10000.0__9999.0__ multiply__80641.0__10000.0__ multiply__80641.0__9999.0__ multiply__80641.0__9999.0__ |
subtract__10000.0__9999.0__ multiply__80641.0__10000.0__ subtract__806410000.0__80641.0__ subtract__806410000.0__80641.0__ |
| num__45 persons can repair a road in num__12 days working num__5 hours a day . in how many days will num__30 persons working num__6 hours a day complete the work ? <o> a ) num__10 <o> b ) num__13 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
according to the chain rule m num__1 x t num__1 = m num__2 x t num__2 therefore num__45 x num__12 x num__5 = num__30 x num__6 x x x = num__15 hence the number of days = num__15 . answer : d <eor> d <eos> |
d |
subtract__6.0__5.0__ divide__12.0__6.0__ subtract__45.0__30.0__ round__15.0__ |
subtract__6.0__5.0__ divide__12.0__6.0__ subtract__45.0__30.0__ round__15.0__ |
| a tradesman by means of his false balance defrauds to the extent of num__20.0 ? in buying goods as well as by selling the goods . what percent does he gain on his outlay ? <o> a ) num__49.0 <o> b ) num__47.0 <o> c ) num__44.0 <o> d ) num__46.0 <o> e ) num__43 % |
g % = num__20 + num__20 + ( num__20 * num__20 ) / num__100 = num__44.0 answer : c <eor> c <eos> |
c |
percent__100.0__44.0__ |
percent__100.0__44.0__ |
| in the following sequence : [ x y z num__5 num__7 num__16 num__28 ] each number is equal to the sum of the three numbers succeding it . what is x + y ? <o> a ) num__113 <o> b ) num__213 <o> c ) num__163 <o> d ) num__213 <o> e ) num__131 |
lets start from z as per the question stem z = num__5 + num__7 + num__16 = num__28 - - > > z = num__28 similarly y = num__28 + num__5 + num__7 - - > y = num__40 similarly x = num__40 + num__28 + num__5 - - > x = num__73 hence x + y = num__73 + num__40 = num__113 - - > x + y = num__113 answer : a <eor> a <eos> |
a |
add__40.0__73.0__ add__40.0__73.0__ |
add__40.0__73.0__ add__40.0__73.0__ |
| find the odd man out . num__1 num__27 num__216 num__512 num__1024 num__1331 <o> a ) num__1024 <o> b ) num__512 <o> c ) num__27 <o> d ) num__1 <o> e ) num__2 |
explanation : all given numbers except num__1024 are perfect cubes . answer : option a <eor> a <eos> |
a |
multiply__1.0__1024.0__ |
multiply__1.0__1024.0__ |
| num__2 ^ ( num__10 ) + num__2 ^ ( num__11 ) + num__2 ^ ( num__12 ) + . . . + num__2 ^ ( num__24 ) + num__2 ^ ( num__25 ) = <o> a ) num__2 ^ num__9 ( num__2 ^ ( num__10 ) - num__1 ) <o> b ) num__2 ^ num__9 ( num__2 ^ ( num__12 ) - num__1 ) <o> c ) num__2 ^ num__10 ( num__2 ^ ( num__14 ) - num__1 ) <o> d ) num__2 ^ num__10 ( num__2 ^ ( num__16 ) - num__1 ) <o> e ) num__2 ^ num__11 ( num__2 ^ ( num__17 ) - num__1 ) |
num__2 ^ ( num__10 ) + num__2 ^ ( num__11 ) + num__2 ^ ( num__12 ) + . . . + num__2 ^ ( num__24 ) + num__2 ^ ( num__25 ) = num__2 ^ ( num__10 ) * ( num__1 + num__2 + num__4 + . . . + num__2 ^ ( num__15 ) ) = num__2 ^ ( num__10 ) * ( num__2 ^ ( num__16 ) - num__1 ) the answer is d . <eor> d <eos> |
d |
subtract__11.0__10.0__ add__11.0__4.0__ add__12.0__4.0__ multiply__2.0__1.0__ |
subtract__11.0__10.0__ add__11.0__4.0__ add__12.0__4.0__ multiply__2.0__1.0__ |
| the distance between two cities a and b is num__600 km . a train starts from a at num__8 a . m . and travel towards b at num__60 km / hr . another train starts from b at num__9 a . m and travels towards a at num__75 km / hr . at what time do they meet ? <o> a ) num__09 am <o> b ) num__07 am <o> c ) num__01 pm <o> d ) num__05 pm <o> e ) num__03 pm |
explanation : suppose they meet x hrs after num__8 a . m then [ distance moved by first in x hrs ] + [ distance moved by second in ( x - num__1 ) hrs ] = num__600 . therefore num__60 x + num__75 ( x - num__1 ) = num__600 . = > x = num__5 . so they meet at ( num__8 + num__5 ) i . e num__1 p . m . answer : c ) <eor> c <eos> |
c |
subtract__9.0__8.0__ round__1.0__ |
subtract__9.0__8.0__ subtract__9.0__8.0__ |
| three numbers are in the ratio num__3 : num__4 : num__5 . the largest number value is num__40 . find difference between smallest & largest number is ? <o> a ) num__20 <o> b ) num__24 <o> c ) num__16 <o> d ) num__28 <o> e ) num__30 |
= = num__3 : num__4 : num__5 total parts = num__12 = the largest number value is num__40 = the largest number is = num__5 = then num__5 parts - - - - - > num__40 ( num__5 * num__8 = num__40 ) = smallest number = num__3 & largest number = num__5 = difference between smallest number & largest number is = num__5 - num__3 = num__2 = then num__2 parts - - - - - > num__16 ( num__2 * num__8 = num__16 ) c <eor> c <eos> |
c |
multiply__3.0__4.0__ add__3.0__5.0__ subtract__5.0__3.0__ add__4.0__12.0__ add__4.0__12.0__ |
multiply__3.0__4.0__ subtract__12.0__4.0__ subtract__5.0__3.0__ multiply__2.0__8.0__ multiply__2.0__8.0__ |
| two tains of equal lengths take num__10 seconds and num__15 seconds respectively to cross a telegraph post . if the length of each train be num__100 metres in what time ( in seconds ) will they cross each other travelling in opposite direction ? <o> a ) num__12 <o> b ) num__11.9 <o> c ) num__16 <o> d ) num__20 <o> e ) num__18 |
sol . speed of the first train = [ num__10.0 ] m / sec = num__10 m / sec . speed of the second train = [ num__6.66666666667 ] m / sec = num__6.7 m / sec . relative speed = ( num__10 + num__6.7 ) = m / sec = num__16.7 m / sec . ∴ required time = ( num__100 + num__100 ) / num__16.7 secc = num__11.9 sec . answer b <eor> b <eos> |
b |
divide__100.0__15.0__ add__10.0__6.7__ round__11.9__ |
divide__100.0__15.0__ add__10.0__6.7__ round__11.9__ |
| the sum of all consecutive odd integers from − num__31 to num__41 inclusive is <o> a ) num__110 <o> b ) num__135 <o> c ) num__150 <o> d ) num__165 <o> e ) num__185 |
the sum of the odd numbers from - num__31 to + num__31 is num__0 . let ' s add the remaining numbers . num__33 + num__35 + num__37 + num__39 + num__41 = num__5 ( num__37 ) = num__185 the answer is e . <eor> e <eos> |
e |
multiply__37.0__5.0__ round__185.0__ |
multiply__37.0__5.0__ round__185.0__ |
| a retailer buys num__40 pens at the market price of num__36 pens from a wholesaler if he sells these pens giving a discount of num__1.0 what is the profit % ? <o> a ) num__10 <o> b ) num__15 <o> c ) num__12 <o> d ) num__18 <o> e ) num__20 |
let the market price of each pen be $ num__1 then cost price of num__40 pens = $ num__36 selling price of num__40 pens = num__99.0 of $ num__40 = $ num__39.60 profit % = ( ( num__3.60 * num__100 ) / num__36 ) % = num__10.0 answer a ) num__10 <eor> a <eos> |
a |
percent__40.0__99.0__ percent__10.0__100.0__ |
percent__40.0__99.0__ percent__10.0__100.0__ |
| the diagonal of a rhombus are num__25 m and num__50 m . its area is : <o> a ) num__900 <o> b ) num__800 <o> c ) num__700 <o> d ) num__625 <o> e ) num__650 |
area of the rhombus = num__0.5 d num__1 d num__2 = ( num__0.5 Ã — num__25 Ã — num__50 ) cm ( power ) num__2 = num__25 Ã — num__25 = num__625 cm ( power ) num__2 answer is d . <eor> d <eos> |
d |
square_perimeter__0.5__ triangle_area__25.0__50.0__ triangle_area__25.0__50.0__ |
square_perimeter__0.5__ triangle_area__25.0__50.0__ triangle_area__25.0__50.0__ |
| in the first num__10 overs of a cricket game the run rate was only num__3.2 . what should be the run rate in the remaining num__50 overs to reach the target of num__282 runs ? <o> a ) num__6.25 <o> b ) num__5.5 <o> c ) num__7.4 <o> d ) num__5 <o> e ) num__6 |
explanation : runs scored in the first num__10 overs = num__10 × num__3.2 = num__32 total runs = num__282 remaining runs to be scored = num__282 - num__32 = num__250 remaining overs = num__50 run rate needed = num__5.0 = num__5 answer : option d <eor> d <eos> |
d |
percent__10.0__50.0__ percent__10.0__50.0__ |
percent__10.0__50.0__ percent__10.0__50.0__ |
| a certain bus driver is paid a regular rate of $ num__16 per hour for any number of hours that does not exceed num__40 hours per week . for any overtime hours worked in excess of num__40 hours per week the bus driver is paid a rate that is num__75.0 higher than his regular rate . if last week the bus driver earned $ num__1200 in total compensation how many total hours did he work that week ? <o> a ) num__60 <o> b ) num__40 <o> c ) num__44 <o> d ) num__48 <o> e ) num__52 |
for num__40 hrs = num__40 * num__16 = num__640 excess = num__1200 - num__640 = num__560 for extra hours = . num__75 ( num__16 ) = num__12 + num__16 = num__28 number of extra hrs = num__20.0 = num__20 total hrs = num__40 + num__20 = num__60 answer a num__60 <eor> a <eos> |
a |
multiply__16.0__40.0__ subtract__1200.0__640.0__ add__16.0__12.0__ divide__560.0__28.0__ add__40.0__20.0__ add__40.0__20.0__ |
multiply__16.0__40.0__ subtract__1200.0__640.0__ add__16.0__12.0__ divide__560.0__28.0__ add__40.0__20.0__ add__40.0__20.0__ |
| a lent rs . num__5000 to b for num__2 years and rs . num__3000 to c for num__4 years on simple interest at the same rate of interest and received rs . num__2200 in all from both of them as interest . the rate of interest per annum is ? <o> a ) num__33487 <o> b ) num__2778 <o> c ) num__2788 <o> d ) num__2200 <o> e ) num__2344 |
let the rate be r % p . a . then ( num__5000 * r * num__2 ) / num__100 + ( num__3000 * r * num__4 ) / num__100 = num__2200 num__100 r + num__120 r = num__2200 r = num__10.0 answer : d <eor> d <eos> |
d |
percent__2.0__5000.0__ percent__4.0__3000.0__ percent__100.0__2200.0__ |
percent__2.0__5000.0__ percent__4.0__3000.0__ percent__100.0__2200.0__ |
| what will come in place of the x in the following number series ? num__1 num__2 num__12 num__7 num__23 num__12 num__34 x <o> a ) num__14 <o> b ) num__15 <o> c ) num__16 <o> d ) num__17 <o> e ) num__18 |
there are two series here num__1 num__12 num__23 num__34 . . . ( increase by num__11 ) num__2 num__7 num__12 . . . ( increase by num__5 ) hence next term is num__12 + num__5 = num__17 correct answer choice is d <eor> d <eos> |
d |
subtract__12.0__1.0__ subtract__12.0__7.0__ add__12.0__5.0__ multiply__1.0__17.0__ |
subtract__12.0__1.0__ subtract__12.0__7.0__ add__12.0__5.0__ add__12.0__5.0__ |
| a man swims downstream num__24 km and upstream num__12 km taking num__6 hours each time what is the speed of the man in still water ? <o> a ) num__5 <o> b ) num__8 <o> c ) num__10 <o> d ) num__4 <o> e ) num__3 |
num__24 - - - num__6 ds = num__4 ? - - - - num__1 num__12 - - - - num__6 us = num__2 ? - - - - num__1 m = ? m = ( num__4 + num__2 ) / num__2 = num__3 answer : e <eor> e <eos> |
e |
divide__24.0__6.0__ divide__24.0__12.0__ divide__12.0__4.0__ round__3.0__ |
divide__24.0__6.0__ divide__24.0__12.0__ divide__12.0__4.0__ divide__12.0__4.0__ |
| a bowl of nuts is prepared for a party . brand p mixed nuts are num__20.0 almonds and brand q ' s deluxe nuts are num__25.0 almonds . if a bowl contains a total of num__66 ounces of nuts representing a mixture of both brands and num__15 ounces of the mixture are almonds how many ounces of brand q ' s deluxe mixed nuts are used ? <o> a ) num__16 <o> b ) num__20 <o> c ) num__36 <o> d ) num__44 <o> e ) num__48 |
lets say x ounces of p is mixed with q . = > num__66 - x ounces of q is present in the mixture ( as the total = num__66 ounces ) given total almond weight = num__15 ounces ( num__20 x / num__100 ) + ( num__0.25 ) ( num__66 - x ) = num__15 = > x = num__30 = > num__66 - num__30 = num__36 ounces of q is present in the mixture . answer is c . <eor> c <eos> |
c |
divide__25.0__100.0__ subtract__66.0__30.0__ subtract__66.0__30.0__ |
divide__25.0__100.0__ subtract__66.0__30.0__ subtract__66.0__30.0__ |
| the product of three consecutive numbers is num__210 . then the sum of the smallest two numbers is ? <o> a ) num__11 <o> b ) num__15 <o> c ) num__20 <o> d ) num__38 <o> e ) num__56 |
product of three numbers = num__210 num__210 = num__2 * num__3 * num__5 * num__7 = num__5 * num__6 * num__7 . so the three numbers are num__5 num__6 and num__7 . and sum of smallest of these two = num__5 + num__6 = num__11 . answer : option a <eor> a <eos> |
a |
add__2.0__3.0__ add__2.0__5.0__ multiply__2.0__3.0__ add__5.0__6.0__ add__5.0__6.0__ |
add__2.0__3.0__ add__2.0__5.0__ multiply__2.0__3.0__ add__5.0__6.0__ add__5.0__6.0__ |
| a vessel of capacity num__90 litres is fully filled with pure milk . nine litres of milk is removed from the vessel and replaced with water . nine litres of the solution thus formed is removed and replaced with water . find the quantity of pure milk in the final milk solution ? <o> a ) num__23 : num__89 <o> b ) num__72.9 <o> c ) num__38 : num__3 <o> d ) num__78 : num__3 <o> e ) num__79 : num__3 |
explanation : let the initial quantity of milk in vessel be t litres . let us say y litres of the mixture is taken out and replaced by water for n times alternatively . quantity of milk finally in the vessel is then given by [ ( t - y ) / t ] n * t for the given problem t = num__90 y = num__9 and n = num__2 . hence quantity of milk finally in the vessel = [ ( num__90 - num__9 ) / num__90 ] num__2 ( num__90 ) = num__72.9 litres . answer : option b <eor> b <eos> |
b |
round__72.9__ |
round__72.9__ |
| if x dollars is invested at num__9 percent for one year and y dollars is invested at num__8 percent for one year the annual income from the num__9 percent investment will exceed the annual income from the num__8 percent investment by $ num__48 . if $ num__2000 is the total amount invested how much is invested at num__8 percent ? <o> a ) a . $ num__280 <o> b ) b . $ num__776.47 <o> c ) c . $ num__892 <o> d ) d . $ num__1108 <o> e ) e . $ num__1200 |
num__2 equations with num__2 unknowns num__9 x / num__100 - num__8 y / num__100 = num__48 and x + y = num__2000 solving these num__2 equations x = num__1223.53 and y = num__776.47 answer b . <eor> b <eos> |
b |
subtract__2000.0__1223.53__ subtract__2000.0__1223.53__ |
subtract__2000.0__1223.53__ subtract__2000.0__1223.53__ |
| if x / num__5 + num__9 / x = num__2.8 what are the values of num__3 x - num__7 ? <o> a ) num__8 and num__9 <o> b ) num__8 and num__20 <o> c ) num__17 and num__21 <o> d ) num__12 and num__29 <o> e ) num__17 and num__29 |
i got the same thing b is the answer num__8 or num__20 <eor> b <eos> |
b |
add__5.0__3.0__ add__5.0__3.0__ |
add__5.0__3.0__ add__5.0__3.0__ |
| in a certain game a large bag is filled with blue green purple and red chips worth num__1 num__5 x and num__11 points each respectively . the purple chips are worth more than the green chips but less than the red chips . a certain number of chips are then selected from the bag . if the product of the point values of the selected chips is num__2200 how many purple chips were selected ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
num__2200 = num__1 * num__5 ^ num__2 * num__8 * num__11 the factor of num__8 must come from the purple point value so there is num__1 purple chip . the answer is a . <eor> a <eos> |
a |
reverse__1.0__ |
reverse__1.0__ |
| increasing of length and breadth is proposional . . length = num__6 which changes into num__21 and breadth changes to num__14 . . what was the previous value of breadth ? ? <o> a ) num__4 <o> b ) num__3 <o> c ) num__2 <o> d ) num__5 <o> e ) num__6 |
let previous breadth be ` ` b ' ' previous ratio of length : breadth = = > num__6 : b let increased proportion be ` ` x ' ' new ratio is = = > num__6 x : bx = = > num__21 : num__14 length num__6 x = num__21 x = num__3.5 = > x = num__3.5 breadth bx = num__14 b * num__3.5 = num__14 b = num__4.0 = > num__4 previous breadth = num__4 answer : a <eor> a <eos> |
a |
divide__21.0__6.0__ divide__14.0__3.5__ divide__14.0__3.5__ |
divide__21.0__6.0__ divide__14.0__3.5__ divide__14.0__3.5__ |
| arun and varun appeared at an examination . arun secured num__9 marks more than varun and his marks was num__56.0 of the sum of their marks . what are the marks obtained by them <o> a ) num__42 num__33 <o> b ) num__42 num__36 <o> c ) num__44 num__33 <o> d ) num__44 num__36 <o> e ) num__44 num__37 |
explanation : let the marks secured by them be x and ( x + num__9 ) then sum of their marks = x + ( x + num__9 ) = num__2 x + num__9 given that ( x + num__9 ) was num__56.0 of the sum of their marks = > ( x + num__9 ) = num__0.56 ( num__2 x + num__9 ) = > ( x + num__9 ) = num__0.56 ( num__2 x + num__9 ) = > num__25 x + num__225 = num__28 x + num__126 = > num__3 x = num__99 = > x = num__33 then ( x + num__9 ) = num__33 + num__9 = num__42 hence their marks are num__42 and num__33 answer : a <eor> a <eos> |
a |
multiply__9.0__25.0__ divide__56.0__2.0__ multiply__225.0__0.56__ subtract__28.0__25.0__ subtract__225.0__126.0__ divide__99.0__3.0__ add__9.0__33.0__ add__9.0__33.0__ |
multiply__9.0__25.0__ divide__56.0__2.0__ multiply__225.0__0.56__ subtract__28.0__25.0__ subtract__225.0__126.0__ divide__99.0__3.0__ add__9.0__33.0__ add__9.0__33.0__ |
| tap ' a ' can fill the tank completely in num__6 hrs while tap ' b ' can empty it by num__12 hrs . by mistake the person forgot to close the tap ' b ' as a result both the taps remained open . after num__4 hrs the person realized the mistake and immediately closed the tap ' b ' . in how much time now onwards would the tank be full ? <o> a ) num__2 hours <o> b ) num__4 hours <o> c ) num__5 hours <o> d ) num__1 hour <o> e ) none of these |
explanation : tap a can fill the tank completely in num__6 hours = > in num__1 hour tap a can fill num__1 ⁄ num__6 of the tank tap b can empty the tank completely in num__12 hours = > in num__1 hour tap b can empty num__1 ⁄ num__12 of the tank i . e . in one hour tank a and b together can effectively fill num__1 ⁄ num__6 - num__1 ⁄ num__12 = num__1 ⁄ num__12 of the tank = > in num__4 hours tank a and b can effectively fill num__1 ⁄ num__12 × num__4 = num__1 ⁄ num__3 of the tank . time taken to fill the remaining num__1 − ( num__0.333333333333 ) = num__0.666666666667 of the tank = ( num__0.666666666667 ) / ( num__0.166666666667 ) = num__4 hours . answer : option b <eor> b <eos> |
b |
negate_prob__0.3333__ choose__4.0__3.0__ |
negate_prob__0.3333__ choose__4.0__3.0__ |
| mr . ortiz has to successfully interview num__90.0 of his assigned households . he was assigned num__500 households . he has interviewed num__430 households so far . has he met his goal ? <o> a ) num__450 <o> b ) num__550 <o> c ) num__650 <o> d ) num__750 <o> e ) num__850 |
no he has not met his goal . he has only interviewed num__530 households . he needs num__90.0 of num__500 . multiply : num__500 x . num__90 answer : num__450 correct answer a <eor> a <eos> |
a |
percent__90.0__500.0__ percent__90.0__500.0__ |
percent__90.0__500.0__ percent__90.0__500.0__ |
| find the volume and surface area of a cuboid num__16 m long num__14 m broad and num__7 m high . <o> a ) num__878 cm ^ num__2 <o> b ) num__858 cm ^ num__2 <o> c ) num__838 cm ^ num__2 <o> d ) num__868 cm ^ num__2 <o> e ) none of them |
volume = ( num__16 x num__14 x num__7 ) m ^ num__3 = num__1568 m ^ num__3 . surface area = [ num__2 ( num__16 x num__14 + num__14 x num__7 + num__16 x num__7 ) ] cm ^ num__2 = ( num__2 x num__434 ) cm ^ num__2 = num__868 cm ^ num__2 . answer is d <eor> d <eos> |
d |
volume_rectangular_prism__16.0__14.0__7.0__ multiply__2.0__434.0__ triangle_area__2.0__868.0__ |
volume_rectangular_prism__16.0__14.0__7.0__ multiply__2.0__434.0__ triangle_area__2.0__868.0__ |
| the least number which should be added to num__1920 so that the sum is exactly divisible by num__5 num__64 and num__3 is : <o> a ) num__3 <o> b ) num__13 <o> c ) num__23 <o> d ) num__33 <o> e ) num__28 |
solution l . c . m . of num__5 num__64 and num__3 = num__60 . on dividing num__1920 by num__60 the remainder is num__32 . ∴ number to be added = ( num__60 - num__32 ) = num__28 . answer e <eor> e <eos> |
e |
divide__1920.0__60.0__ subtract__60.0__32.0__ subtract__60.0__32.0__ |
divide__1920.0__60.0__ subtract__60.0__32.0__ subtract__60.0__32.0__ |
| marco rode his dirt bike at num__40 miles per hour ( mph ) for two hours . if he then continued to ride at a different constant rate for another three hours and at the end of the three hours his average speed for the entire five hour ride was num__40 mph what was his average speed over the three hour portion of his ride ? <o> a ) num__14 mph <o> b ) num__40 mph <o> c ) num__23.3333333333 mph <o> d ) num__26.6666666667 mph <o> e ) num__34 mph |
average speed for first two hours s num__1 = num__40 mph distance travelled in these two hours d num__1 = num__80 miles average speed for the entire num__5 hour ride s = num__40 mph total distance traveller in the entire num__5 hour ride d = num__40 x num__5 = num__200 miles . hence distance traveller in the latter num__3 hour period d num__2 = d - d num__1 = num__200 - num__80 = num__120 average speed for the latter num__3 hour period s num__2 = d num__0.666666666667 = num__40.0 = num__40 hence the correct answer is b <eor> b <eos> |
b |
multiply__40.0__5.0__ subtract__3.0__1.0__ multiply__40.0__3.0__ divide__2.0__3.0__ round__40.0__ |
multiply__40.0__5.0__ subtract__3.0__1.0__ subtract__200.0__80.0__ divide__2.0__3.0__ subtract__80.0__40.0__ |
| carmelo and lebron participate in a six - person footrace on the basketball court during all - star weekend . if all six contestants finish ( including charles barkley ) and there are no ties how many different arrangements of finishes are there in which carmelo defeats lebron ? <o> a ) num__5040 <o> b ) num__2520 <o> c ) num__360 <o> d ) num__120 <o> e ) num__42 |
num__6 ! = num__720 num__360.0 = num__360 half the time carmelo is in front vice versa answer : ( c ) num__360 <eor> c <eos> |
c |
subtract__720.0__360.0__ |
subtract__720.0__360.0__ |
| num__60.0 of num__2 is equal to <o> a ) num__1.2 <o> b ) num__0.4 <o> c ) num__0.6 <o> d ) num__0.7 <o> e ) num__0.9 |
num__60.0 of num__2 = ( num__0.6 ) * num__2 = num__1.2 answer : option a <eor> a <eos> |
a |
percent__60.0__2.0__ percent__60.0__2.0__ |
percent__60.0__2.0__ percent__60.0__2.0__ |
| find the cost price of an item if by selling at rs . num__320 a profit of num__6.0 is made ? <o> a ) rs . num__300 <o> b ) rs . num__200 <o> c ) rs . num__500 <o> d ) rs . num__100 <o> e ) rs . num__310 |
sp = num__320 cp = ( sp ) * [ num__100 / ( num__100 + p ) ] = num__320 * [ num__100 / ( num__100 + num__6 ) ] = rs . num__300 answer : a <eor> a <eos> |
a |
percent__100.0__300.0__ |
percent__100.0__300.0__ |
| the speed at which a man can row a boat in still water is num__6 km / hr . if he rows downstream where the speed of current is num__3 km / hr how many seconds will he take to cover num__80 meters ? <o> a ) num__28 <o> b ) num__32 <o> c ) num__36 <o> d ) num__40 <o> e ) num__44 |
the speed of the boat downstream = num__6 + num__3 = num__9 km / hr num__9 km / hr * num__0.277777777778 = num__2.5 m / s the time taken to cover num__80 meters = num__80 / num__2.5 = num__32 seconds . the answer is b . <eor> b <eos> |
b |
add__6.0__3.0__ divide__80.0__2.5__ round__32.0__ |
add__6.0__3.0__ divide__80.0__2.5__ divide__80.0__2.5__ |
| a shopkeeper sold an article at $ num__100 with num__15.0 profit . then find its cost price ? <o> a ) $ num__120 <o> b ) $ num__100 <o> c ) $ num__91 <o> d ) $ num__87 <o> e ) $ num__69 |
cost price = selling price * num__100 / ( num__100 + profit ) c . p . = num__100 * num__0.869565217391 = $ num__87 ( approximately ) answer is d <eor> d <eos> |
d |
percent__100.0__87.0__ |
percent__100.0__87.0__ |
| bill is golfing with three friends and can either buy generic golf tees that are packaged by the dozen or the higher quality aero flight tees that come by the pair . what is the minimum number of packages of aero flight tees bill must purchase to ensure that he has at least num__20 golf tees for each member of his foursome if he will buy no more than num__3 packages of the generic golf tees ? <o> a ) num__16 <o> b ) num__10 <o> c ) num__8 <o> d ) num__22 <o> e ) num__2 |
at least num__20 golf tees for each member of his foursome = total of at least num__4 * num__20 = num__80 tees . num__3 packages of the generic golf tees that are packaged by the dozen = num__3 * num__12 = num__36 tees . so bill must by at least num__44 aero tees . they come by the pair hence he must by at least num__22.0 = num__22 packages of aero flight tees . answer : d <eor> d <eos> |
d |
multiply__20.0__4.0__ multiply__3.0__4.0__ multiply__3.0__12.0__ subtract__80.0__36.0__ subtract__44.0__22.0__ |
multiply__20.0__4.0__ multiply__3.0__4.0__ multiply__3.0__12.0__ subtract__80.0__36.0__ subtract__44.0__22.0__ |
| the average age of a husband and a wife is num__23 years when they were married five years ago but now the average age of the husband wife and child is num__20 years ( the child was born during the interval ) . what is the present age of the child ? <o> a ) num__2 years <o> b ) num__8 years <o> c ) num__3 years <o> d ) num__4 years <o> e ) num__1 years |
num__28 * num__2 = num__56 num__20 * num__3 = num__60 - - - - - - - - - - - num__4 years answer : d <eor> d <eos> |
d |
multiply__2.0__28.0__ subtract__23.0__20.0__ multiply__20.0__3.0__ subtract__60.0__56.0__ subtract__60.0__56.0__ |
multiply__2.0__28.0__ subtract__23.0__20.0__ multiply__20.0__3.0__ subtract__60.0__56.0__ subtract__60.0__56.0__ |
| which of the following can not be a value of num__5 / ( x – num__5 ) ? <o> a ) - num__1 <o> b ) - num__0.5 <o> c ) num__0 <o> d ) num__2 <o> e ) num__5 |
the question does n ' t ask for the value of x here but for the possible results of the equation . if num__5 / ( x - num__5 ) = num__0 then the numerator must be num__0 . but since the numerator is num__5 the fraction can not be equal to num__0 . the answer is c . <eor> c <eos> |
c |
multiply__5.0__0.0__ |
divide__0.0__5.0__ |
| the average weight of num__4 person ' s increases by num__3 kg when a new person comes in place of one of them weighing num__70 kg . what might be the weight of the new person ? <o> a ) num__75 kg <o> b ) num__82 kg <o> c ) num__86 kg <o> d ) data inadequate <o> e ) none of these |
total weight increased = ( num__4 x num__3 ) kg = num__12 kg . weight of new person = ( num__70 + num__12 ) kg = num__82 kg . answer : option b <eor> b <eos> |
b |
multiply__4.0__3.0__ add__70.0__12.0__ add__70.0__12.0__ |
multiply__4.0__3.0__ add__70.0__12.0__ add__70.0__12.0__ |
| when a number is divided by num__6 & then multiply by num__12 the answer is num__9 what is the no . ? <o> a ) num__4.5 <o> b ) num__5 <o> c ) num__5.6 <o> d ) num__7 <o> e ) num__7.2 |
if $ x $ is the number x / num__6 * num__12 = num__9 = > num__2 x = num__9 = > x = num__4.5 a <eor> a <eos> |
a |
divide__12.0__6.0__ divide__9.0__2.0__ subtract__9.0__4.5__ |
divide__12.0__6.0__ divide__9.0__2.0__ divide__9.0__2.0__ |
| trapezoid abcd has ab parallel to cd with \ adc = num__90 o . given that ad = num__5 bc = num__13 and dc = num__18 compute the area of the trapezoid . <o> a ) num__30 <o> b ) num__40 <o> c ) num__50 <o> d ) num__60 <o> e ) num__70 |
the trapezoid can be divided into a right triangle with legs of length num__5 and num__12 and a rectangle with side lengths num__5 and num__6 . this gives an area of num__60 . correct answer d <eor> d <eos> |
d |
side_by_diagonal__13.0__5.0__ multiply__5.0__12.0__ multiply__5.0__12.0__ |
side_by_diagonal__13.0__5.0__ multiply__5.0__12.0__ multiply__5.0__12.0__ |
| if a man can cover num__14 metres in one second how many kilometres can he cover in num__3 hours num__30 minutes ? <o> a ) num__165 km <o> b ) num__170 km <o> c ) num__176.4 km <o> d ) num__180 km <o> e ) num__190 km |
num__14 m / s = num__14 * num__3.6 kmph num__3 hours num__30 minutes = num__3 num__0.5 hours = num__3.5 hours distance = speed * time = num__14 * num__3.6 * num__3.5 km = num__176.4 km . answer : c <eor> c <eos> |
c |
add__3.0__0.5__ round__176.4__ |
add__3.0__0.5__ round__176.4__ |
| if num__20 m − n = p then which of the following represents the average ( arithmetic mean ) of m n and p in terms of m ? <o> a ) num__2 m − num__1 <o> b ) num__4 m <o> c ) num__6 m <o> d ) num__7 m <o> e ) num__6 m / num__5 |
num__20 m - n = p add n to both sides : num__20 m = n + p add m to both sides : num__21 m = n + p + m now divide both sides by num__3 num__7 m = ( n + p + m ) / num__3 = the average of n p and m d <eor> d <eos> |
d |
divide__21.0__3.0__ divide__21.0__3.0__ |
divide__21.0__3.0__ divide__21.0__3.0__ |
| the difference between the place values of num__9 and num__4 in the number num__529435 is <o> a ) num__8500 <o> b ) num__8900 <o> c ) num__8600 <o> d ) num__6970 <o> e ) none |
sol . = ( place value of num__9 ) – ( place value of num__4 ) = ( num__9000 - num__400 ) = num__8600 answer c <eor> c <eos> |
c |
subtract__9000.0__400.0__ subtract__9000.0__400.0__ |
subtract__9000.0__400.0__ subtract__9000.0__400.0__ |
| car x began traveling at an average speed of num__35 miles per hour . after num__72 minutes car y began traveling at an average speed of num__38 miles per hour . when both cars had traveled the same distance both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ? <o> a ) num__405 <o> b ) num__420 <o> c ) num__440 <o> d ) num__447 <o> e ) num__490 |
car y began travelling after num__72 minutes or num__1.2 hours . let t be the time for which car y travelled before it stopped . both cars stop when they have travelled the same distance . so num__35 ( t + num__1.2 ) = num__38 t t = num__14 distance traveled by car x from the time car y began traveling until both cars stopped is num__35 x num__14 = num__490 miles answer : - e <eor> e <eos> |
e |
multiply__35.0__14.0__ round__490.0__ |
multiply__35.0__14.0__ round__490.0__ |
| an article when sold for num__200 fetches num__25 per cent profit . what would be the percentage profit / loss if num__6 such articles are sold for num__1056 ? <o> a ) num__10 per cent loss <o> b ) num__10 per cent profit <o> c ) num__5 per cent loss <o> d ) num__5 per cent profit <o> e ) none of these |
cp = num__200 ⁄ num__125 × num__100 = num__160 ∴ cp of num__6 articles = num__6 × num__160 = num__960 ∴ profit = num__1056 – num__960 = num__96 percentage profit = num__96 ⁄ num__960 × num__100 = num__10.0 answer b <eor> b <eos> |
b |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| there are num__4 girls and num__4 boys . if they are to be seated in a row what is the probability that no two girls sit together ? <o> a ) a ) num__0.125 ! <o> b ) b ) num__6 ! num__5 ! / num__8 ! <o> c ) c ) num__2 * num__4 ! num__4 ! / num__8 ! <o> d ) d ) num__2 ∗ num__6 ! num__6 ! / num__12 ! <o> e ) e ) num__6 ! num__7 ! / num__12 ! |
if girls are alternate then only no num__2 girls can be together . so among num__8 places girls can be seated in num__4 alternate places in num__4 ! ways . and boys can be seated in the remaining num__4 places in num__4 ! ways . therefore total number of ways = num__4 ! * num__4 ! but there could be num__2 such arrangements - - > num__1 . when first place is filled by a girl num__2 . when first place is filled by a boy therefore total number of ways = num__2 * num__4 ! * num__4 ! also total number of ways to fill num__8 places = num__8 ! hence probability = num__2 ∗ num__4 ! num__4 ! / num__8 ! answer : c <eor> c <eos> |
c |
coin_space__ coin_space__ |
coin_space__ coin_space__ |
| in a village they organize num__4 programs for an festival . the festival occurs in num__2 days . in how many ways that the num__4 programs can be divided for num__2 days festival ? <o> a ) num__12 <o> b ) num__10 <o> c ) num__6 <o> d ) num__4 <o> e ) num__2 |
here num__4 programs organize for num__2 day festival . so num__4 c num__2 ways to solve this problem . num__4 c num__2 = num__4 ! / num__2 ! * num__2 ! num__4 c num__2 = num__4 * num__3 * num__2 ! / num__2 ! * num__2 * num__1 = num__6 there are num__6 ways to divided the programs . answer : option c <eor> c <eos> |
c |
subtract__4.0__3.0__ add__4.0__2.0__ add__4.0__2.0__ |
subtract__4.0__3.0__ multiply__2.0__3.0__ multiply__2.0__3.0__ |
| which of the following inequalities is equivalent to – num__2 < x < num__4 ? <o> a ) | x – num__2 | < num__4 <o> b ) | x – num__1 | < num__3 <o> c ) | x + num__1 | < num__3 <o> d ) | x + num__2 | < num__4 <o> e ) none of the above |
first remove the mod then set the expression to num__0 then get a value of x then add and subtract the value from the given number on the rhs lets try with an example ( b ) - - - - - - - - - - - - - - > | x – num__1 | < num__3 first remove the mod - - - - - - - - - - - - - - - - - - - > x - num__1 then set the expression to num__0 - - - - - - - - - - - - - - > x - num__1 = num__0 then get a value of x - - - - - - - - - - - - - - - - - - - - - > x = num__1 then add and subtract the value from the given number on the rhs num__1 + num__3 = num__4 and num__1 - num__3 = - num__2 the range become - num__2 to num__4 correct answer now try its with an incorrect option ( c ) | - - - - - - - - - - > | x + num__1 | < num__3 first remove the mod - - - - - - - - - - - - - - - - - - - - - - - - - - > x + num__1 then set the expression to num__0 - - - - - - - - - - - - - - - - - - - - - - > x + num__1 = num__0 then get a value of x - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - > x = - num__1 then add and subtract the value from the given number on the rhs - num__1 + num__3 = num__2 and - num__1 - num__3 = - num__4 the range become - num__4 to num__2 incorrect the correct answer is b <eor> b <eos> |
b |
add__2.0__1.0__ reverse__1.0__ |
add__2.0__1.0__ subtract__2.0__1.0__ |
| a sum was put at simple interest at certain rate for num__3 years . had it been put at num__1.0 higher rate it would have fetched rs . num__69 more . the sum is : a . rs . num__2400 b . rs . num__2100 c . rs . num__2200 d . rs . num__2480 <o> a ) num__2000 <o> b ) num__2100 <o> c ) num__2200 <o> d ) num__2300 <o> e ) num__2400 |
num__1 percent for num__3 years = num__69 num__1 percent for num__1 year = num__23 = > num__100 percent = num__2300 answer : d <eor> d <eos> |
d |
percent__100.0__2300.0__ |
percent__100.0__2300.0__ |
| e = √ [ num__2 √ num__63 + num__2 / ( num__8 + num__3 √ num__7 ) ] = <o> a ) num__8 + num__3 √ num__7 <o> b ) num__4 + num__3 √ num__7 <o> c ) num__8 <o> d ) num__4 <o> e ) √ num__7 |
in these type of question u multiply the nominator and denominator with conjugate . . . conjugate of num__8 + num__3 sqrt ( num__7 ) is num__8 - num__3 sqrt ( num__7 ) sqrt [ num__2 sqrt ( num__63 ) + num__2 { num__8 - num__3 sqrt ( num__7 ) } / { num__64 - num__63 } ] e = sqrt [ num__2 sqrt ( num__63 ) + num__16 - num__2 sqrt ( num__63 ) ] = num__4 answer is num__4 . d <eor> d <eos> |
d |
multiply__2.0__8.0__ divide__8.0__2.0__ divide__8.0__2.0__ |
multiply__2.0__8.0__ divide__8.0__2.0__ divide__8.0__2.0__ |
| t is an even integer greater than num__300000 and smaller than num__1 num__000000 . how many numbers can t be ? <o> a ) num__300000 <o> b ) num__349999 <o> c ) num__350000 <o> d ) num__399999 <o> e ) num__400 |
000 |
t num__1 num__000000 - num__300000 = num__700000 integers num__350000.0 = num__350000 even integers . num__350000 + num__1 inclusive . but since num__1 num__000000 and num__300000 are not included . num__350001 - num__2 = num__349999 b <eor> b <eos> |
b |
b |
| a box of paper clips contains num__100 clips . if an office uses paper clips at the rate of num__15 per num__100 pieces of outgoing mail how many boxes will be required for the num__4000 pieces of outgoing mail ? <o> a ) num__4 boxes <o> b ) num__5 boxes <o> c ) num__6 boxes <o> d ) num__7 boxes <o> e ) num__8 boxes |
num__15 clips - - - num__100 pieces x clips - - - num__4000 pieces x = num__40.0 * num__15 = num__600 clips num__1 box - - num__100 clips y box - - num__600 clips y = num__6.0 = num__6 boxes answer : c <eor> c <eos> |
c |
percent__15.0__4000.0__ percent__15.0__40.0__ percent__100.0__6.0__ |
percent__15.0__4000.0__ percent__15.0__40.0__ percent__100.0__6.0__ |
| a “ sophie germain ” prime is any positive prime number p for which num__2 p + num__1 is also prime . the product of all the possible units digits of sophie germain primes greater than num__3 is <o> a ) num__3 <o> b ) num__189 <o> c ) num__221 <o> d ) num__227 <o> e ) num__230 |
in that case the sophie prime numbers greater than num__5 are num__7 num__1123 num__4759 . . which yields units digit as num__1 num__37 and num__9 product would be num__1 x num__3 x num__7 x num__9 = num__189 answer should be b . <eor> b <eos> |
b |
add__2.0__3.0__ add__2.0__5.0__ add__2.0__7.0__ multiply__1.0__189.0__ |
add__2.0__3.0__ add__2.0__5.0__ add__2.0__7.0__ multiply__1.0__189.0__ |
| a camera lens filter kit containing num__5 filters sells for $ num__52.50 . if the filters are purchased individually num__2 of them are priced at $ num__6.45 each num__2 at $ num__11.05 each num__1 at $ num__15.50 . the amount saved by purchasing the kit is what percent of the total price of the num__5 filters purchased individually ? <o> a ) num__3.76 <o> b ) num__4.76 <o> c ) num__5.76 <o> d ) num__6.76 <o> e ) num__7.76 % |
cost of kit = $ num__52.50 if filters are purchased individually - $ num__6.45 * num__2 + $ num__11.05 * num__2 + $ num__15.50 = $ num__50.50 amount saved = $ num__52.50 - $ num__50.50 = $ num__2.50 required % age = ( $ num__2.50 / $ num__52.50 ) * num__100 = num__4.76 so the correct answer is b . <eor> b <eos> |
b |
subtract__52.5__2.0__ divide__5.0__2.0__ multiply__1.0__4.76__ |
subtract__52.5__2.0__ divide__5.0__2.0__ multiply__1.0__4.76__ |
| in an objective competitive exam a correct answer score num__4 marks and on a wrong answer num__2 marks are negatively added . a student scores num__420 marks from num__150 question . how many answers were correct ? <o> a ) num__150 <o> b ) num__155 <o> c ) num__140 <o> d ) num__120 <o> e ) num__165 |
let x be the correct answer and y be the wrong answer so the total number of questions is ( x + y ) = num__150 . . . . . ( num__1 ) = > num__4 x - num__2 y = num__420 . . . . . ( num__2 ) by solving ( num__1 ) & ( num__2 ) we get = > num__6 x = num__720 hence x = num__120 therefore number of correct answers are num__120 . answer d <eor> d <eos> |
d |
add__4.0__2.0__ divide__720.0__6.0__ multiply__1.0__120.0__ |
add__4.0__2.0__ divide__720.0__6.0__ multiply__1.0__120.0__ |
| two pipes function simultaneously the reservoir will be filled in num__12 hours . one pipe fills reservoir num__10 hours faster than the other . how many hours does the faster pipe take to fill the reservoir ? <o> a ) num__33 <o> b ) num__88 <o> c ) num__20 <o> d ) num__72 <o> e ) num__27 |
num__1 / x + num__1 / ( x + num__10 ) = num__0.0833333333333 x = num__20 answer : c <eor> c <eos> |
c |
divide__1.0__12.0__ round__20.0__ |
divide__1.0__12.0__ divide__20.0__1.0__ |
| income and expenditure of a person are in the ratio num__5 : num__4 . if the income of the person is rs . num__16000 then find his savings ? <o> a ) num__6769 <o> b ) num__3200 <o> c ) num__3201 <o> d ) num__2319 <o> e ) num__7591 |
let the income and the expenditure of the person be rs . num__5 x and rs . num__4 x respectively . income num__5 x = num__16000 = > x = num__3200 savings = income - expenditure = num__5 x - num__4 x = x so savings = rs . num__3200 . answer : b <eor> b <eos> |
b |
divide__16000.0__5.0__ divide__16000.0__5.0__ |
divide__16000.0__5.0__ divide__16000.0__5.0__ |
| the average age of num__15 students of a class is num__16 years . out of these the average age of num__5 students is num__14 years and that of the other num__9 students is num__16 years . the age of the num__15 th student is ? <o> a ) num__11 years <o> b ) num__17 years <o> c ) num__26 years <o> d ) num__14 years <o> e ) num__12 years |
age of the num__15 th student = [ num__15 * num__16 - ( num__14 * num__5 + num__16 * num__9 ) ] = ( num__240 - num__214 ) = num__26 years . answer : c <eor> c <eos> |
c |
multiply__15.0__16.0__ subtract__240.0__214.0__ subtract__240.0__214.0__ |
multiply__15.0__16.0__ subtract__240.0__214.0__ subtract__240.0__214.0__ |
| a num__6 - liter solution is num__25.0 alcohol . how many liters of pure alcohol must be added to produce a solution that is num__50.0 alcohol ? <o> a ) num__2.1 <o> b ) num__2.4 <o> c ) num__2.7 <o> d ) num__3.0 <o> e ) num__3.3 |
let x be the amount of pure alcohol required . num__0.25 ( num__6 ) + x = num__0.5 ( x + num__6 ) num__0.5 x = num__3 - num__1.5 x = num__3 liters the answer is d . <eor> d <eos> |
d |
divide__25.0__50.0__ multiply__6.0__0.5__ multiply__6.0__0.25__ multiply__6.0__0.5__ |
divide__25.0__50.0__ multiply__6.0__0.5__ multiply__6.0__0.25__ subtract__6.0__3.0__ |
| how many integers are divisible by num__3 between num__10 ! and num__10 ! + num__20 inclusive ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
num__0 ! is divisible by num__3 - the way i look factorials is that any number included will also be divisible by the product . num__109 num__87 num__65 num__43 num__21 are all divisors of num__10 ! there are num__6 numbers between num__10 ! and num__10 ! + num__20 that are divisible by num__3 . hence b : num__7 <eor> b <eos> |
b |
subtract__10.0__3.0__ subtract__10.0__3.0__ |
subtract__10.0__3.0__ subtract__10.0__3.0__ |
| a gets num__3 times as much money as b gets b gets only rs . num__25 more then what c gets . the three gets rs . num__675 in all . find the share of b ? <o> a ) num__297 <o> b ) num__278 <o> c ) num__287 <o> d ) num__140 <o> e ) num__272 |
a + b + c = num__675 a = num__3 b num__3 b + b + b - num__25 = num__675 num__5 b = num__700 b = num__140 answer : d <eor> d <eos> |
d |
add__25.0__675.0__ divide__700.0__5.0__ divide__700.0__5.0__ |
add__25.0__675.0__ divide__700.0__5.0__ divide__700.0__5.0__ |
| a can do double work of b . so a take num__15 days less than b in doing a work . then both together in how much days will finish that work ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__15 <o> d ) num__8 <o> e ) num__16 |
let alone finish a work in x days . then b will take num__2 x days num__2 x - x = num__15 x = num__15 . so a alone finish this work in num__15 days and b alone finish this work in num__30 days work of a in num__1 days = num__0.0666666666667 work of b in num__1 days = num__0.0333333333333 ( a + b ) work in num__1 days = num__0.1 so together they finish the work in num__10 days answer a <eor> a <eos> |
a |
multiply__15.0__2.0__ divide__1.0__15.0__ divide__1.0__30.0__ add__0.0333__0.0667__ divide__1.0__0.1__ round__10.0__ |
multiply__15.0__2.0__ divide__1.0__15.0__ divide__1.0__30.0__ add__0.0333__0.0667__ divide__1.0__0.1__ round__10.0__ |
| what is the units digit of ( num__7 ! * num__6 ! + num__6 ! * num__4 ! ) / num__2 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
( num__7 ! * num__6 ! + num__6 ! * num__4 ! ) / num__2 = num__6 ! ( num__7 ! + num__4 ! ) / num__2 = num__240 ( num__1680 + num__24 ) / num__2 = num__204480 units digit of the above product will be equal to num__0 answer a <eor> a <eos> |
a |
multiply__7.0__240.0__ multiply__6.0__4.0__ reverse__204480.0__ reverse__204480.0__ |
multiply__7.0__240.0__ multiply__6.0__4.0__ reverse__204480.0__ reverse__204480.0__ |
| insert the missing number num__3 num__7 num__6 num__5 num__9 num__3 num__12 num__1 num__15 . . . <o> a ) num__18 <o> b ) num__33 <o> c ) - num__1 <o> d ) num__3 <o> e ) - num__3 |
num__3 num__7 num__6 num__5 num__9 num__3 num__12 num__1 num__15 . . . has two different series num__1 ) num__3 num__6 num__9 num__12 num__15 . . . . . num__2 ) num__7 num__5 num__3 num__1 . . . . . - num__1 is the next number answer : c <eor> c <eos> |
c |
subtract__3.0__1.0__ reverse__1.0__ |
subtract__3.0__1.0__ subtract__3.0__2.0__ |
| a number increased by num__15.0 gives num__1150 . the number is <o> a ) num__250 <o> b ) num__1000 <o> c ) num__450 <o> d ) num__500 <o> e ) num__520 |
formula = total = num__100.0 increse = ` ` + ' ' decrease = ` ` - ' ' a number means = num__100.0 that same number increased by num__15.0 = num__115.0 num__115.0 - - - - - - - > num__1150 ( num__115 × num__100 = num__1150 ) num__100.0 - - - - - - - > num__1000 ( num__100 × num__100 = num__1000 ) b ) <eor> b <eos> |
b |
percent__100.0__1000.0__ |
percent__100.0__1000.0__ |
| the radius of a circle is increased by num__1.0 . find how much % does its area increases ? <o> a ) num__2.07 <o> b ) num__2.02 <o> c ) num__2.21 <o> d ) num__2.01 <o> e ) num__2.08 % |
r = num__100 r = num__101 r num__2 = num__10000 r num__2 = num__10201 num__10000 - - - - num__201 num__100 - - - - ? = > num__2.01 answer : d <eor> d <eos> |
d |
power__100.0__2.0__ power__101.0__2.0__ multiply__1.0__2.01__ |
power__100.0__2.0__ power__101.0__2.0__ multiply__1.0__2.01__ |
| a train is running at a speed of num__40 km / hr and it crosses a post in num__18 seconds . what is the length of the train ? <o> a ) num__140 m <o> b ) num__200 m <o> c ) num__400 m <o> d ) num__600 m <o> e ) num__120 m |
speed = num__40 km / hr = num__40 Ã — num__0.277777777778 = num__11.1111111111 m / s time = num__18 seconds distance covered = num__11.1111111111 Ã — num__18 = num__200 m length of the train = num__200 m answer : b <eor> b <eos> |
b |
round__200.0__ |
round__200.0__ |
| the average of num__2 num__8 num__9 num__10 num__1 and x is num__6 . what is the value of x ? <o> a ) num__2 <o> b ) num__8 <o> c ) num__7 <o> d ) num__6 <o> e ) num__3 |
x = num__6 because : ( num__2 + num__8 + num__9 + num__10 + num__1 + x ) / num__6 = num__6 ( num__2 + num__8 + num__9 + num__10 + num__1 + x ) / num__6 * num__6 = num__6 * num__6 ( num__2 + num__8 + num__9 + num__10 + num__1 + x ) = num__36 ( num__2 + num__8 + num__9 + num__10 + num__1 + x ) - num__30 = num__36 - num__30 x = num__6 therefore the correct answer is d num__6 . <eor> d <eos> |
d |
subtract__36.0__6.0__ subtract__8.0__2.0__ |
subtract__36.0__6.0__ subtract__8.0__2.0__ |
| if x gets num__25.0 more than y and y gets num__20.0 more than z the share of z out of rs . num__555 will be : <o> a ) rs . num__300 <o> b ) rs . num__200 <o> c ) rs . num__240 <o> d ) rs . num__150 <o> e ) none of these |
z share = z y = num__1.2 z x = num__1.25 Ã — num__1.2 z x + y + z = num__555 ( num__1.25 Ã — num__1.2 + num__1.2 + num__1 ) z = num__55 num__3.7 z = num__555 z = num__150 answer : . d <eor> d <eos> |
d |
divide__25.0__20.0__ round_down__1.25__ divide__555.0__3.7__ divide__555.0__3.7__ |
divide__25.0__20.0__ round_down__1.25__ divide__555.0__3.7__ divide__555.0__3.7__ |
| a b and c started a business by investing rs . num__45000 rs . num__55000 and rs . num__60000 respectively . at the end of a year they got a total profit of rs . num__11200 . find how much b gets more than a in the profit ? <o> a ) rs . num__700 <o> b ) rs . num__750 <o> c ) rs . num__710 <o> d ) rs . num__780 <o> e ) none of these |
explanation : ratio of shares of a b and c = num__45000 : num__55000 : num__60000 = num__9 : num__11 : num__12 a ' s share = rs . [ ( num__0.28125 ) × num__11200 ] = rs . num__3150 b ' s share = rs . [ ( num__0.34375 ) × num__11200 ] = rs . num__3850 . : b ' s share is more than a by rs . ( num__3850 - num__3150 ) = rs . num__700 answer : option a <eor> a <eos> |
a |
subtract__3850.0__3150.0__ subtract__3850.0__3150.0__ |
subtract__3850.0__3150.0__ subtract__3850.0__3150.0__ |
| the length of a rectangle is two - seventh of the radius of a circle . the radius of the circle is equal to the side of the square whose area is num__5929 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is num__25 units ? <o> a ) num__660 sq . units <o> b ) num__440 sq . units <o> c ) num__770 sq . units <o> d ) num__550 sq . units <o> e ) num__220 sq . units |
given that the area of the square = num__5929 sq . units = > side of square = â ˆ š num__5929 = num__77 units the radius of the circle = side of the square = num__77 units length of the rectangle = num__0.285714285714 * num__77 = num__22 units given that breadth = num__25 units area of the rectangle = lb = num__22 * num__25 = num__550 sq . units answer : d <eor> d <eos> |
d |
multiply__25.0__22.0__ multiply__25.0__22.0__ |
multiply__25.0__22.0__ multiply__25.0__22.0__ |
| a boat goes num__100 km downstream in num__10 hours and num__60 km upstream in num__30 hours . the speed of the stream is ? <o> a ) num__5 km / h <o> b ) num__2 km / h <o> c ) num__3 km / h <o> d ) ( num__0.5 ) km / h <o> e ) e ) num__4 |
num__100 - - - num__10 ds = num__10 ? - - - - num__1 num__60 - - - - num__30 us = num__2 ? - - - - - num__1 s = ( num__10 - num__2 ) / num__2 = num__4 kmph answer : e <eor> e <eos> |
e |
divide__60.0__30.0__ round__4.0__ |
divide__60.0__30.0__ divide__4.0__1.0__ |
| a certain research group plans to create computer models of x % of a list of num__10000 bacterial species known to inhabit the human body . after a budget cut the group finds it must reduce this selection by ( x − num__9 ) % . in terms of x how many species of bacteria will the group be able to model ? <o> a ) x * x – num__5 x <o> b ) ( x ) ( num__105 – x ) <o> c ) ( num__100 ) ( num__105 – x ) <o> d ) ( x ) * ( num__109 - x ) <o> e ) ( x - num__5 ) / num__100 |
initial : ( x / num__100 ) * num__10000 = num__100 x ( bacterial species ) after reduce by ( x - num__9 ) % the percentage of bacterial species = num__1 - ( x - num__9 ) % = num__1 - ( x - num__9 ) / num__100 = ( num__109 - x ) / num__100 note : difference between reduce to [ means : the remain ] and reduce by [ means : the remain = num__1 - reduce by ] so the number of bacterial species after reducing : num__100 x * ( num__109 - x ) / num__100 = ( x ) * ( num__109 - x ) ans is d . <eor> d <eos> |
d |
percent__100.0__109.0__ |
percent__100.0__109.0__ |
| express num__15 mps in kmph ? <o> a ) num__54 <o> b ) num__88 <o> c ) num__90 <o> d ) num__27 <o> e ) num__21 |
num__15 * num__3.6 = num__54 kmph answer : a <eor> a <eos> |
a |
multiply__15.0__3.6__ round__54.0__ |
multiply__15.0__3.6__ multiply__15.0__3.6__ |
| a group of n students can be divided into equal groups of num__4 with num__1 student left over or equal groups of num__5 with num__2 students left over . what is the sum of the two smallest possible values of n ? <o> a ) num__33 <o> b ) num__46 <o> c ) num__49 <o> d ) num__54 <o> e ) num__86 |
n = num__4 k + num__1 = num__5 j + num__2 let ' s start at num__1 = num__4 ( num__0 ) + num__1 and keep adding num__4 until we find a number in the form num__5 j + num__2 . num__1 num__5 num__9 num__13 num__17 = num__5 ( num__3 ) + num__2 the next such number is num__17 + num__4 * num__5 = num__37 . num__17 + num__37 = num__54 the answer is d . <eor> d <eos> |
d |
add__4.0__5.0__ add__4.0__9.0__ add__4.0__13.0__ subtract__4.0__1.0__ add__17.0__37.0__ multiply__1.0__54.0__ |
add__4.0__5.0__ add__4.0__9.0__ add__4.0__13.0__ add__1.0__2.0__ add__17.0__37.0__ multiply__1.0__54.0__ |
| due to construction the speed limit along an num__7 - mile section of highway is reduced from num__55 miles per hour to num__35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? <o> a ) num__4.37 <o> b ) num__8 <o> c ) num__10 <o> d ) num__15 <o> e ) num__24 |
old time in minutes to cross num__7 miles stretch = num__7 * num__1.09090909091 = num__7 * num__1.09090909091 = num__7.63 new time in minutes to cross num__7 miles stretch = num__7 * num__1.71428571429 = num__7 * num__1.71428571429 = num__12 time difference = num__4.37 ans : a <eor> a <eos> |
a |
subtract__12.0__7.63__ round__4.37__ |
subtract__12.0__7.63__ round__4.37__ |
| jane covered a distance of num__340 miles between city a and city b taking a total of num__5 hours . if part of the distance was covered at num__60 miles per hour speed and the balance at num__80 miles per hour speed how many hours did she travel at num__60 miles per hour ? <o> a ) num__2 hours num__30 minutes <o> b ) num__3 hours <o> c ) num__2 hours <o> d ) num__1 hour num__45 minutes <o> e ) none of these |
explanatory answer let jane travel ' x ' at num__60 miles per hour . as the total time taken to cover num__340 miles is num__5 hours jane would have traveled ( num__5 - x ) hours at num__80 miles per hour . distance covered at num__60 miles per hour = speed * time = num__60 * x = num__60 x miles . distance covered at num__80 miles per hour = speed * time = num__80 ( num__5 - x ) = num__400 - num__80 x miles . total distance covered = distance covered at num__60 miles per hour + distance covered at num__80 miles per hour . therefore total distance = num__60 x + num__400 - num__80 x . total distance travelled = num__340 miles . therefore num__340 = num__60 x + num__400 - num__80 x num__20 x = num__60 or x = num__3 hours . choice b is the correct answer . <eor> b <eos> |
b |
add__340.0__60.0__ subtract__80.0__60.0__ divide__60.0__20.0__ round__3.0__ |
add__340.0__60.0__ subtract__80.0__60.0__ divide__60.0__20.0__ round__3.0__ |
| if num__100 < x < num__199 and num__20 < y < num__100 then the product xy can not be equal to : <o> a ) num__1900 <o> b ) num__19303 <o> c ) num__19 num__356.732 <o> d ) num__19502 <o> e ) num__19 |
909 |
correct answer : ( a ) determine the range of xy by multiplying the two extremes of each individual range together . the smallest value of xy must be greater than num__100 * num__20 . the largest value must be less than num__199 * num__100 . this means that num__2000 < xy < num__19900 . ( a ) is outside of this range so it is not a possible product of xy . <eor> a <eos> |
a |
a |
| num__26 buckets of water fill a tank when the capacity of each bucket is num__13.5 litres . how many buckets will be required to fill the same tank if the capacity of each bucket is num__9 litres ? <o> a ) num__30 <o> b ) num__39 <o> c ) num__60 <o> d ) data inadequate <o> e ) none of these |
capacity of the tank = num__26 Ã — num__13.5 = num__351 litres when the capacity of each bucket = num__9 litres then the required no . of buckets = num__351 â „ num__9 = num__39 answer b <eor> b <eos> |
b |
multiply__26.0__13.5__ divide__351.0__9.0__ round__39.0__ |
multiply__26.0__13.5__ divide__351.0__9.0__ round__39.0__ |
| a train of length num__250 m crosses a bridge of length num__120 m in num__20 seconds . what is the speed of train ? <o> a ) num__33 <o> b ) num__27 <o> c ) num__66.6 <o> d ) num__22 <o> e ) num__72 |
sol : ( length of train + length of bridge ) = speed of train x time ( num__250 + num__120 ) = num__20 x speed speed = num__18.5 = num__18.5 m / s = num__66.6 km / h answer = c <eor> c <eos> |
c |
round__66.6__ |
round__66.6__ |
| two parallel sides of a trapezium are num__4 cm and num__5 cm respectively . the perpendicular distance between the parallel sides is num__6 cm . find the area of the trapezium . <o> a ) num__24 cm num__2 <o> b ) num__25 cm num__2 <o> c ) num__26 cm num__2 <o> d ) num__27 cm num__2 <o> e ) num__28 cm num__2 |
area of trapezium when height and two parallel sides are given = num__0.5 × h × ( a + b ) = num__0.5 x num__6 x ( num__4 + num__5 ) = num__27 cm num__2 answer d <eor> d <eos> |
d |
multiply__4.0__0.5__ round__27.0__ |
multiply__4.0__0.5__ round__27.0__ |
| in a certain experiment the data collected is the number of organisms per sample and this data follows a normal distribution . if the sample of data has a mean of num__60 and a standard deviation of num__12 which of the following is exactly num__1.95 standard deviations more than the mean ? <o> a ) a ) num__48 <o> b ) b ) num__60 <o> c ) c ) num__72 <o> d ) d ) num__77.5 <o> e ) e ) num__83.4 |
standard deviation is a relatively rare category in the quant section although you ' re like to be tested on it num__1 time on test day . you ' ll never be asked to calculate sd though so you really just need to learn the basic ' concepts ' behind it . here we ' re told two things about a group of numbers : num__1 ) the average of the group is num__60 num__2 ) the standard deviation of the group is num__12 if you go num__1 sdupfrom the average you hit . . . . . num__60 + num__12 = num__72 if you go num__1 sddownfrom the average you hit . . . . . num__60 - num__12 = num__48 if you go num__2 sdsupfrom the average you hit . . . . . num__60 + num__2 ( num__12 ) = num__84 if you go num__2 sdsdownfrom the average you hit . . . . . num__60 - num__2 ( num__12 ) = num__36 etc . here we ' re asked for the number that is exactly num__1.95 sds above the mean . . . . num__1.95 sdsupwould be . . . . . num__60 + num__1.95 ( num__12 ) = num__83.4 e <eor> e <eos> |
e |
round_down__1.95__ add__60.0__12.0__ subtract__60.0__12.0__ add__12.0__72.0__ divide__72.0__2.0__ multiply__1.0__83.4__ |
round_down__1.95__ add__60.0__12.0__ subtract__60.0__12.0__ add__12.0__72.0__ subtract__48.0__12.0__ multiply__1.0__83.4__ |
| what number times ( num__1 ⁄ num__6 ) ^ num__2 will give the value of num__6 ^ num__3 ? <o> a ) num__6 <o> b ) num__36 <o> c ) num__216 <o> d ) num__1296 <o> e ) num__7776 |
x * ( num__0.166666666667 ) ^ num__2 = num__6 ^ num__3 x = num__6 ^ num__2 * num__6 ^ num__3 = num__6 ^ num__5 = num__7776 the answer is e . <eor> e <eos> |
e |
reverse__6.0__ subtract__6.0__1.0__ multiply__1.0__7776.0__ |
reverse__6.0__ subtract__6.0__1.0__ multiply__1.0__7776.0__ |
| find the area of a parallelogram with base num__12 cm and height num__10 cm ? <o> a ) num__297 cm num__2 <o> b ) num__384 cm num__2 <o> c ) num__120 cm num__2 <o> d ) num__267 cm num__2 <o> e ) num__186 cm num__2 |
area of a parallelogram = base * height = num__12 * num__10 = num__120 cm num__2 answer : c <eor> c <eos> |
c |
multiply__12.0__10.0__ multiply__12.0__10.0__ |
multiply__12.0__10.0__ multiply__12.0__10.0__ |
| a license plate in the country kerrania consists of four digits followed by two letters . the letters a b and c are used only by government vehicles while the letters d through z are used by non - government vehicles . kerrania ' s intelligence agency has recently captured a message from the country gonzalia indicating that an electronic transmitter has been installed in a kerrania government vehicle with a license plate starting with num__79 . if it takes the police num__5 minutes to inspect each vehicle what is the probability that the police will find the transmitter within three hours ? <o> a ) num__0.227848101266 <o> b ) num__0.166666666667 <o> c ) num__0.04 <o> d ) num__0.02 <o> e ) num__0.00111111111111 |
if it takes num__5 minutes to inspect one vehicle the # of vehicles that can be inspected in num__3 hours ( num__180 minutes ) = num__36.0 = num__36 . hence for calculating the probability that the police will find the transmitter within three hours the favorable cases = num__36 . now we need to figure out the total # of cases . the total # of cases = total # of such cars possible . the details given about the car is that it starts with num__79 which leaves num__2 more digits both of which can be filled by all num__10 numbers ( num__0 - num__9 ) . in addition we have num__3 letters each of which can be filled by any from the set { a b c } . hence the total # of such cars possible = num__10 * num__10 * num__3 * num__3 = num__900 so the probability that the police will find the transmitter within three hours = num__0.04 = num__0.04 . option c <eor> c <eos> |
c |
divide__180.0__5.0__ subtract__5.0__3.0__ multiply__5.0__2.0__ multiply__5.0__180.0__ divide__36.0__900.0__ round__0.04__ |
divide__180.0__5.0__ subtract__5.0__3.0__ multiply__5.0__2.0__ multiply__5.0__180.0__ divide__36.0__900.0__ round__0.04__ |
| pipe p can drain the liquid from a tank in num__0.75 the time that it takes pipe q to drain it and in num__0.333333333333 the time that it takes pipe r to do it . if all num__3 pipes operating simultaneously but independently are used to drain liquid from the tank then pipe q drains what portion of the liquid from the tank ? <o> a ) num__0.310344827586 <o> b ) num__0.347826086957 <o> c ) num__0.36 <o> d ) num__0.586206896552 <o> e ) num__0.75 |
suppose q can drain in num__1 hr . so rq = num__1.0 = num__1 so rp = num__1 / [ ( num__0.75 ) rq ] = num__1.33333333333 also rp = rr / ( num__0.333333333333 ) = > num__1.33333333333 = rr / ( num__0.333333333333 ) = > rr = num__0.444444444444 let h is the time it takes to drain by running all num__3 pipes simultaneously so combined rate = rc = num__1 / h = num__1 + num__1.33333333333 + num__0.444444444444 = num__2.77777777778 = num__1 / ( num__0.36 ) thus running simultaneously pipe q will drain num__0.36 of the liquid . thus answer = c . <eor> c <eos> |
c |
add__0.3333__1.0__ divide__0.3333__0.75__ divide__1.0__2.7778__ round__0.36__ |
add__0.3333__1.0__ divide__0.3333__0.75__ divide__1.0__2.7778__ divide__1.0__2.7778__ |
| in the kitchen of a busy restaurant it takes m minutes to wash p pots . at this rate how many hours does it take to wash num__8 p pots ? <o> a ) num__60 / m <o> b ) num__8 m <o> c ) num__2 m / num__15 <o> d ) num__480 m <o> e ) m / num__60 |
the time it takes to wash num__8 p pots is num__8 m minutes which is num__8 m / num__60 = num__2 m / num__15 hours . the answer is c . <eor> c <eos> |
c |
hour_to_min_conversion__ round__2.0__ |
hour_to_min_conversion__ round__2.0__ |
| the perimeter of a triangle is num__28 cm and the inradius of the triangle is num__2.5 cm . what is the area of the triangle ? <o> a ) num__76 cm num__2 <o> b ) num__15 cm num__2 <o> c ) num__76 cm num__2 <o> d ) num__17 cm num__2 <o> e ) num__35 cm num__2 |
area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = num__2.5 * num__14.0 = num__35 cm num__2 answer : e <eor> e <eos> |
e |
triangle_area__28.0__2.5__ triangle_area__28.0__2.5__ |
multiply__2.5__14.0__ multiply__2.5__14.0__ |
| sunil start from home at num__4 pm and reaches school everyday at num__5 pm to pick up his children . on saturday the school got over ai num__4 pm and children started walking towards home . sunil met them on the way and returned home num__15 min early . if speed of sunil is num__35 km / hr then for how many min did children walk ? what is speed of children ? ? ? <o> a ) num__4 km / hr <o> b ) num__5 km / hr <o> c ) num__6 km / hr <o> d ) num__7 km / hr <o> e ) num__8 km / hr |
sunil reach num__15 min early . it is time save at both for going to school and coming to home . so sunil meet children befor num__7.5 = num__7.5 min . speed of sunil is num__35 km / hr i . e . num__583.33 m / min . remaining distance to cover by sunil or else we can say total distance cover by children is num__7.5 * num__583.33 = num__4375 m sunil meet children after num__52.5 min i . e . ( num__1 hr - num__7.5 min ) . so num__4375 is distance cover by children in num__52.5 min . so speed of children is num__4375 / num__52.5 = num__83.33 m / min = num__5 km / hr . answer : b <eor> b <eos> |
b |
subtract__5.0__4.0__ round__5.0__ |
subtract__5.0__4.0__ divide__5.0__1.0__ |
| a train num__75 m long is running with a speed of num__54 km per hour . in what time will it pass a telegraph post ? <o> a ) num__11 s <o> b ) num__5 s <o> c ) num__7 s <o> d ) num__12 s <o> e ) none of these |
we know from the formula time = distance / speed thus time = num__1.38888888889 x num__0.277777777778 or time = num__5 sec . answer : b <eor> b <eos> |
b |
divide__75.0__54.0__ round__5.0__ |
divide__75.0__54.0__ round__5.0__ |
| a car travels num__25 km an hour faster than a bus for a journey of num__500 km . if the bus takes num__10 hours more than the car then the speeds of the bus and the car are <o> a ) num__25 km / hr and num__40 km / hr respectively <o> b ) num__25 km / hr and num__60 km / hr respectively <o> c ) num__25 km / hr and num__50 km / hr respectively <o> d ) none of these <o> e ) can not be determined |
let the speeds of bus and car be x and y . here y = x + num__25 as per the question num__500 / x - num__500 / x + num__25 = num__10 ⇒ num__500 ( x + num__25 ) − num__500 ( x ) = num__10 x ( x + num__25 ) ⇒ x = num__25 y = num__50 . answer : c . <eor> c <eos> |
c |
divide__500.0__10.0__ round__25.0__ |
divide__500.0__10.0__ subtract__50.0__25.0__ |
| a recycling facility is staffed by num__7 floor workers and one manager . all of the floor workers are paid equal wages but the manager is paid n times as much as a floor worker . if the manager ’ s wages account for num__0.142857142857 of all wages paid at the facility what is the value of n ? <o> a ) num__0.428571428571 <o> b ) num__0.8 <o> c ) num__1.4 <o> d ) num__1.16666666667 <o> e ) num__1.75 |
say each floor worker is paid $ x then the manager is paid $ xn . total salary would be num__7 x + xn and we are told that it equals to num__7 xn : num__7 x + xn = num__7 xn - - > reduce by x : num__7 + n = num__7 n - - > num__6 n = num__7 n = num__1.16666666667 answer : d <eor> d <eos> |
d |
divide__7.0__6.0__ divide__7.0__6.0__ |
divide__7.0__6.0__ divide__7.0__6.0__ |
| the difference between the squares of two consecutive numbers is num__37 . find the numbers . <o> a ) num__14 - num__15 <o> b ) num__20 - num__23 <o> c ) num__11 - num__12 <o> d ) num__18 - num__19 <o> e ) num__13 - num__14 |
if the difference between the squares of two consecutive numbers is x then the numbers are ( x - num__1 ) / num__2 and ( x + num__1 ) / num__2 ( num__37 - num__1 ) / num__2 and ( num__37 + num__1 ) / num__2 num__18.0 and num__19.0 therefore the required answer = num__18 and num__19 . <eor> d <eos> |
d |
subtract__37.0__18.0__ round__18.0__ |
subtract__37.0__18.0__ subtract__37.0__19.0__ |
| a sells a bicycle to b at a profit of num__75.0 and b sells it to c at a loss of num__40.0 . find the resultant profit or loss . <o> a ) - num__4.0 <o> b ) num__5.0 <o> c ) - num__5.0 <o> d ) num__6.0 <o> e ) - num__7 % |
the resultant profit or loss = num__75 - num__40 - ( num__75 * num__40 ) / num__100 = num__5.0 profit = num__5.0 answer is b <eor> b <eos> |
b |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| out of num__17 applicants num__8 boys and num__9 girls . two persons are to be selected for the job . find the probability that at least one of the selected persons will be a girl . <o> a ) num__0.794117647059 <o> b ) num__0.735294117647 <o> c ) num__0.558823529412 <o> d ) num__0.617647058824 <o> e ) num__0.396226415094 |
explanation : the probability that no girl is selected is : - = > num__8 c num__0.117647058824 c num__2 . = > num__0.205882352941 . = > num__0.205882352941 . hence the probability that atleast one girl is selected is : - = > num__1 - num__0.205882352941 . = > num__0.794117647059 . answer : a <eor> a <eos> |
a |
coin_space__ negate_prob__0.2059__ negate_prob__0.2059__ |
coin_space__ negate_prob__0.2059__ negate_prob__0.2059__ |
| a num__270 metres long train running at the speed of num__120 kmph crosses another train running in opposite direction at the speed of num__80 kmph in num__9 seconds . what is the length of the other train ? <o> a ) num__220 meter <o> b ) num__225 meter <o> c ) num__230 meter <o> d ) num__235 meter <o> e ) num__240 meter |
explanation : as trains are running in opposite directions so their relative speed will get added so relative speed = num__120 + num__80 = num__200 kmph = num__200 * ( num__0.277777777778 ) = num__55.5555555556 m / sec let the length of other train is x meter then x + num__30.0 = num__55.5555555556 = > x + num__270 = num__500 = > x = num__230 so the length of the train is num__230 meters answer is c <eor> c <eos> |
c |
add__120.0__80.0__ divide__270.0__9.0__ add__200.0__30.0__ round__230.0__ |
add__120.0__80.0__ divide__270.0__9.0__ add__200.0__30.0__ add__200.0__30.0__ |
| ayesha ' s father was num__38 years of age when she was born while her mother was num__36 years old when her brother four years younger to her was born . what is the difference between the ages of her parents ? <o> a ) num__2 years <o> b ) num__4 years <o> c ) num__6 years <o> d ) num__8 years <o> e ) num__9 years |
mother ' s age when ayesha ' s brother was born = num__36 years . father ' s age when ayesha ' s brother was born = ( num__38 + num__4 ) years = num__42 years . required difference = ( num__42 - num__36 ) years = num__6 years answer : option c <eor> c <eos> |
c |
add__38.0__4.0__ subtract__42.0__36.0__ divide__36.0__6.0__ |
add__38.0__4.0__ subtract__42.0__36.0__ subtract__42.0__36.0__ |
| a salesman sold twice as much pears in the afternoon than in the morning . if he sold $ num__360 kilograms of pears that day how many kilograms did he sell in the morning and how many in the afternoon ? <o> a ) num__120 <o> b ) num__180 <o> c ) num__240 <o> d ) num__280 <o> e ) num__320 |
let x be the number of kilograms he sold in the morning . then in the afternoon he sold num__2 x kilograms . so the total is x + num__2 x = num__3 x . this must be equal to num__360 . num__3 x = num__360 x = num__3603 x = num__120 therefore the salesman sold num__120 kg in the morning and num__2 ⋅ num__120 = num__240 kg in the afternoon . so answer is c . <eor> c <eos> |
c |
divide__360.0__3.0__ subtract__360.0__120.0__ round__240.0__ |
divide__360.0__3.0__ subtract__360.0__120.0__ round__240.0__ |
| find the simple interest on $ num__10000 at num__8.0 per annum for num__12 months ? <o> a ) $ num__410 <o> b ) $ num__500 <o> c ) $ num__650 <o> d ) $ num__800 <o> e ) $ num__1000 |
p = $ num__10000 r = num__8.0 t = num__1.0 years = num__1 year s . i . = p * r * t / num__100 = num__10000 * num__8 * num__0.01 = $ num__800 answer is d <eor> d <eos> |
d |
percent__1.0__10000.0__ percent__8.0__10000.0__ percent__8.0__10000.0__ |
percent__1.0__10000.0__ percent__8.0__10000.0__ percent__8.0__10000.0__ |
| johnny borrows $ num__50000 from tony at num__3.0 daily interest compounded daily . how much does johnny owe tony after num__4 weeks ? <o> a ) num__50000 * num__0.02 ^ num__28 <o> b ) num__50000 * num__1.12 <o> c ) num__50000 * num__2.12 <o> d ) num__50000 * num__1.03 ^ num__28 <o> e ) ( num__50000 * num__1.02 ) ^ num__28 |
a = a = p ( num__1 + r num__100 ) ^ n so a = num__50000 ( num__1 + num__0.03 ) ^ num__7 â ˆ — num__4 so a = num__50000 ( num__1.03 ) ^ num__28 hence answer will be ( d ) <eor> d <eos> |
d |
percent__3.0__1.0__ percent__100.0__50000.0__ |
percent__3.0__1.0__ percent__100.0__50000.0__ |
| num__893.7 – num__573.07 – num__95.007 = ? <o> a ) num__225.623 <o> b ) num__224.777 <o> c ) num__233.523 <o> d ) num__414.637 <o> e ) none of these |
solution given expression = num__893.7 - ( num__573.07 + num__95.007 ) = num__893.7 - num__668.077 = num__225.623 . answer a <eor> a <eos> |
a |
add__573.07__95.007__ subtract__893.7__668.077__ subtract__893.7__668.077__ |
add__573.07__95.007__ subtract__893.7__668.077__ subtract__893.7__668.077__ |
| a snail climbing a num__36 feet high wall climbs up num__8 feet on the first day but slides down num__4 feet on the second . it climbs num__8 feet on the third day and slides down again num__4 feet on the fourth day . if this pattern continues how many days will it take the snail to reach the top of the wall ? <o> a ) num__12 <o> b ) num__14 <o> c ) num__15 <o> d ) num__20 <o> e ) num__21 |
total transaction in two days = num__8 - num__4 = num__4 feet in num__15 days it will climb num__36 feet thus reaching the top therefore total no of days required = num__15 c <eor> c <eos> |
c |
round__15.0__ |
round__15.0__ |
| a sum of rs . num__125000 amounts to rs . num__15500 in num__4 years at the rate of simple interest . what is the rate of interest ? <o> a ) num__3.0 <o> b ) num__2.0 <o> c ) num__1.0 <o> d ) num__6.0 <o> e ) num__7 % |
s . i . = ( num__15500 - num__12500 ) = rs . num__3000 \ rate = ( num__100 * num__3000 ) / ( num__12500 * num__4 ) = num__6.0 answer : d <eor> d <eos> |
d |
percent__100.0__6.0__ |
percent__100.0__6.0__ |
| the difference between a two - digit number and the number obtained by interchanging the two digits is num__63 . which is the smaller of the two numbers ? <o> a ) num__3 <o> b ) num__6 <o> c ) num__5 <o> d ) num__7 <o> e ) num__9 |
explanation : let the ten ' s digit be x and unit ' s digit be y . then ( num__10 x + y ) - ( num__10 y + x ) = num__63 = > num__9 ( x - y ) = num__63 = > x - y = num__7 . thus none of the numbers can be deermined . . answer : d <eor> d <eos> |
d |
divide__63.0__9.0__ divide__63.0__9.0__ |
divide__63.0__9.0__ divide__63.0__9.0__ |
| a goods train runs at the speed of num__72 km / hr and crosses a num__250 m long platform in num__26 sec . what is the length of the goods train ? <o> a ) num__287 m <o> b ) num__266 m <o> c ) num__987 m <o> d ) num__270 m <o> e ) num__256 m |
speed = num__72 * num__0.277777777778 = num__20 m / sec . time = num__26 sec . let the length of the train be x meters . then ( x + num__250 ) / num__26 = num__20 x = num__270 m . answer : d <eor> d <eos> |
d |
add__250.0__20.0__ round__270.0__ |
add__250.0__20.0__ add__250.0__20.0__ |
| num__4 shooters are hitting num__1 out of num__4 different targets . what is the probability that exactly one shooter will hit each target . <o> a ) num__1.0 ^ num__4 <o> b ) num__4 ^ num__1.0 <o> c ) num__4 ! / num__4 <o> d ) num__4 ! / num__4 ^ num__4 <o> e ) num__0.25 ! |
each shooter out of num__4 has num__4 options hence total # of outcomes is num__4 ^ num__4 ; favorable outcomes will be num__4 ! which is # of ways to assign num__4 different targets to num__4 shooters : num__1 - num__2 - num__3 - num__4 ( targets ) a - b - c - d ( shooters ) b - a - b - c ( shooters ) c - b - a - b ( shooters ) . . . so basically # of arrangements of num__4 distinct objects : num__4 ! . p = favorable / total = num__4 ! / num__4 ^ num__4 answer : d . <eor> d <eos> |
d |
coin_space__ choose__4.0__3.0__ |
coin_space__ choose__4.0__3.0__ |
| working alone printers x y and z can do a certain printing job consisting of a large number of pages in num__15 num__12 and num__18 hours respectively . what is the ratio of the time it takes printer x to do the job working alone at its rate to the time it takes printers y and z to do the job working together at their individual rates ? <o> a ) num__0.266666666667 <o> b ) num__0.5 <o> c ) num__0.681818181818 <o> d ) num__2.08333333333 <o> e ) num__2.75 |
the time it takes printer x is num__15 hours . the combined rate of y and z is num__0.0833333333333 + num__0.0555555555556 = num__0.138888888889 the time it takes y and z is num__7.2 the ratio of times is num__15 / ( num__7.2 ) = num__5 * num__0.416666666667 = num__2.08333333333 the answer is d . <eor> d <eos> |
d |
reverse__12.0__ reverse__18.0__ add__0.0556__0.0833__ divide__5.0__12.0__ divide__15.0__7.2__ divide__15.0__7.2__ |
reverse__12.0__ reverse__18.0__ add__0.0556__0.0833__ divide__5.0__12.0__ divide__15.0__7.2__ divide__15.0__7.2__ |
| the average age num__15 members of a committee are the same as it was num__2 years ago because an old number has been replaced by a younger number . find how much younger is the new member than the old number ? <o> a ) num__18 <o> b ) num__99 <o> c ) num__30 <o> d ) num__26 <o> e ) num__12 |
num__15 * num__2 = num__30 years answer : c <eor> c <eos> |
c |
multiply__15.0__2.0__ multiply__15.0__2.0__ |
multiply__15.0__2.0__ multiply__15.0__2.0__ |
| the circulation for magazine y in num__1971 was num__5 times the average ( arithmetic mean ) yearly circulation for magazine y for the years num__1972 - num__1980 . what is the ratio of the circulation in num__1971 to the total circulation during num__1971 - num__1980 for magazine y ? <o> a ) num__0.214285714286 <o> b ) num__0.357142857143 <o> c ) num__0.642857142857 <o> d ) num__0.785714285714 <o> e ) num__0.5 |
there are num__9 years from num__1972 - num__1980 inclusive . let ' s say the average circulation every year between num__1972 - num__1980 inclusive is x . so the total circulation is num__9 x from num__1972 - num__1980 inclusive . in num__1971 the circulation is num__5 x . so total circulation for num__1971 - num__1980 is num__5 x + num__9 x = num__14 x . ratio of circulation in num__1971 to total circulation during num__1971 - num__1980 is num__5 x to num__14 x or num__0.357142857143 . answer : b <eor> b <eos> |
b |
subtract__1980.0__1971.0__ add__5.0__9.0__ divide__5.0__14.0__ divide__5.0__14.0__ |
subtract__1980.0__1971.0__ add__5.0__9.0__ divide__5.0__14.0__ divide__5.0__14.0__ |
| john has only pennies dimes and nickels in a jar . the jar has at least num__1 but no more than num__4 pennies . if the jar has at least num__1 nickel and num__1 dime which of the following could not be the total amount of money in the jar ? <o> a ) num__54 <o> b ) num__55 <o> c ) num__56 <o> d ) num__57 <o> e ) num__58 |
let there be a pennies b nickels and c dimes so total amount can be num__1 + num__5 b + num__10 c cents to num__4 + num__5 b + num__10 c as you can see the equation of total whenever divided by num__5 would leave a remainder from num__1 to num__4 ( as pennies can only be from num__1 to num__4 and hence a is limited to values from num__1 to num__4 ) so the total can never be divisible by num__5 and hence only num__55 that is c is the option which is divisible by num__5 . so answer is b <eor> b <eos> |
b |
add__1.0__4.0__ multiply__1.0__55.0__ |
add__1.0__4.0__ multiply__1.0__55.0__ |
| the average age of three boys is num__28 years and their ages are in the proportion num__4 : num__6 : num__8 . the age of the youngest boy is : <o> a ) num__21 years <o> b ) num__18 years <o> c ) num__14 years <o> d ) num__9 years <o> e ) none of these |
total age of num__3 boys = ( num__28 Ã — num__3 ) years = num__84 years . ratio of their ages = num__4 : num__6 : num__8 . age of the youngest = ( num__84 Ã — num__3 â „ num__18 ) years = num__14 years . answer c <eor> c <eos> |
c |
multiply__28.0__3.0__ multiply__6.0__3.0__ add__6.0__8.0__ subtract__28.0__14.0__ |
multiply__28.0__3.0__ multiply__6.0__3.0__ add__6.0__8.0__ subtract__28.0__14.0__ |
| the number num__152 is equal to the sum of the cubes of two integers . what is the product of those integers ? <o> a ) num__8 <o> b ) num__15 <o> c ) num__21 <o> d ) num__27 <o> e ) num__39 |
actually decomposition into factors can easily give you the answer here . you should just do the decomposition of the right thing i . e . the options since they represent the product of those integers . since the sum of cubes is num__152 the numbers can not be larger than num__5 since num__6 ^ num__3 itself is num__216 . num__21 num__27 num__39 - the factors are too large so ignore num__8 - ( num__2 num__4 ) does not satisfy num__15 - ( num__3 num__5 ) yes . num__3 ^ num__3 + num__5 ^ num__3 = num__152 - answer answer : b <eor> b <eos> |
b |
add__6.0__21.0__ add__3.0__5.0__ subtract__5.0__3.0__ subtract__6.0__2.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
add__6.0__21.0__ add__3.0__5.0__ subtract__5.0__3.0__ subtract__6.0__2.0__ subtract__21.0__6.0__ subtract__21.0__6.0__ |
| the current ages of num__3 individuals are in the ratio num__4 : num__7 : num__9 . num__0.8 decades ago the sum of their ages was num__56 . find their current ages in years <o> a ) num__16 num__28 num__36 <o> b ) num__17 num__28 num__39 <o> c ) num__18 num__30 num__38 <o> d ) num__20 num__32 num__46 <o> e ) none of these |
a num__16 num__28 num__36 let their current ages are num__4 a num__7 a and num__9 a years correspondingly . then ( num__4 a – num__8 ) + ( num__7 a – num__8 ) + ( num__9 a – num__8 ) = num__56 num__20 a = num__80 a = num__4 their current ages are num__16 years num__28 years and num__36 years correspondingly . <eor> a <eos> |
a |
add__7.0__9.0__ multiply__4.0__7.0__ multiply__4.0__9.0__ divide__56.0__7.0__ add__4.0__16.0__ multiply__4.0__20.0__ add__7.0__9.0__ |
add__7.0__9.0__ multiply__4.0__7.0__ multiply__4.0__9.0__ divide__56.0__7.0__ add__4.0__16.0__ multiply__4.0__20.0__ add__7.0__9.0__ |
| the population of a bacteria culture doubles every num__2 minutes . approximately how many minutes will it take for the population to grow from num__1000 to num__1 num__000000 bacteria <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__16 <o> e ) num__20 |
the question basically asks how many minutes it takes for a population to increase by factor num__1000 ( num__1 num__000000 / num__1000 = num__1000 ) . now you know that every two minutes the population doubles i . e . is multiplied by num__2 . so the equation becomes : num__2 ^ x > = num__1000 where x represents the number of times the population doubles . a lot of people remember that num__2 ^ num__10 = num__1024 . i . e . the population has to double num__10 times . since it takes the population num__2 minutes to double once it takes num__10 * num__2 minutes = num__20 minutes to double nine times . thus solution e = num__20 is correct . <eor> e <eos> |
e |
multiply__2.0__10.0__ round__20.0__ |
multiply__2.0__10.0__ multiply__2.0__10.0__ |
| a single discount equivalent to the discount series of num__20.0 num__10.0 and num__5.0 is ? <o> a ) num__31.2 <o> b ) num__31.2 <o> c ) num__31.6 <o> d ) num__31.3 <o> e ) num__31.1 |
num__100 * ( num__0.8 ) * ( num__0.9 ) * ( num__0.95 ) = num__68.4 num__100 - num__68.4 = num__31.6 answer : c <eor> c <eos> |
c |
percent__100.0__31.6__ |
percent__100.0__31.6__ |
| a certain clock rings two notes at quarter past the hour four notes at half past and six notes at three - quarters past . on the hour it rings eight notes plus an additional number of notes equal to whatever hour it is . how many notes will the clock ring from num__1 : num__00 p . m . through num__5 : num__00 p . m . including the rings at num__1 : num__00 and num__5 : num__00 ? <o> a ) num__87 <o> b ) num__95 <o> c ) num__102 <o> d ) num__103 <o> e ) num__115 |
form num__1 pm to num__5 pm . excluding the actual hour chime we have num__20 ( num__1 pm ) + num__20 ( num__2 pm ) + num__20 ( num__3 pm ) + num__20 ( num__4 pm ) + num__8 ( num__5 pm ) = num__88 now the hour chimes are num__1 + num__2 + num__3 + num__4 + num__5 = num__15 total = num__88 + num__15 = num__103 answer d . <eor> d <eos> |
d |
add__1.0__2.0__ add__1.0__3.0__ add__5.0__3.0__ multiply__5.0__3.0__ add__15.0__88.0__ multiply__1.0__103.0__ |
add__1.0__2.0__ add__1.0__3.0__ add__5.0__3.0__ multiply__5.0__3.0__ add__15.0__88.0__ add__15.0__88.0__ |
| if ( a + b ) = num__4 ( b + c ) = num__7 and ( c + d ) = num__5 what is the value of ( a + d ) ? <o> a ) num__16 . <o> b ) num__8 . <o> c ) num__7 . <o> d ) num__2 . <o> e ) - num__2 . |
given a + b = num__4 b + c = num__7 c + d = num__5 ( a + b ) - ( b + c ) + ( c + d ) = ( a + d ) = > num__4 - num__7 + num__5 = num__2 . option d . . . <eor> d <eos> |
d |
subtract__7.0__5.0__ subtract__4.0__2.0__ |
subtract__7.0__5.0__ subtract__4.0__2.0__ |
| a man buys an item at rs . num__750 and sells it at the loss of num__20 percent . then what is the selling price of that item <o> a ) rs . num__660 <o> b ) rs . num__760 <o> c ) rs . num__600 <o> d ) rs . num__960 <o> e ) none of these |
explanation : here always remember when ever x % loss it means s . p . = ( num__100 - x ) % of c . p when ever x % profit it means s . p . = ( num__100 + x ) % of c . p so here will be ( num__100 - x ) % of c . p . = num__80.0 of num__750 = num__0.8 * num__750 = num__600 option c <eor> c <eos> |
c |
percent__80.0__750.0__ percent__80.0__750.0__ |
percent__80.0__750.0__ percent__80.0__750.0__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__130 m and they cross each other in num__12 sec then the speed of each train is ? <o> a ) num__30 <o> b ) num__35 <o> c ) num__36 <o> d ) num__37 <o> e ) num__39 |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__130 + num__130 ) / num__12 = > x = num__10.83 speed of each train = num__10 m / sec . = num__10.83 * num__3.6 = - num__39 km / hr . answer : option e <eor> e <eos> |
e |
subtract__12.0__2.0__ round__39.0__ |
subtract__12.0__2.0__ round__39.0__ |
| two trains one from howrah to patna and the other from patna to howrah start simultaneously . after they meet the trains reach their destinations after num__32 hours and num__16 hours respectively . the ratio of their speeds is <o> a ) num__4 : num__9 <o> b ) num__4 : num__3 <o> c ) num__4 : num__12 <o> d ) num__1 : num__2 <o> e ) num__4 : num__5 |
let us name the trains as a and b . then ( a ' s speed ) : ( b ' s speed ) = b : a = num__16 : num__32 = num__1 : num__2 . answer : d <eor> d <eos> |
d |
divide__32.0__16.0__ round__1.0__ |
divide__32.0__16.0__ round__1.0__ |
| num__50 percent of andrea ' s living room floor is covered by a carpet that is num__4 feet by num__9 feet . what is the area of her living room floor ? <o> a ) num__14.4 <o> b ) num__25.7142857143 <o> c ) num__50.4 <o> d ) num__64.8 <o> e ) num__90 |
num__50.0 of area of the floor = num__4 * num__9 square feet = num__36 square feet i . e . num__100.0 area of floor = ( num__0.72 ) * num__100 = num__64.8 square feet answer : option d <eor> d <eos> |
d |
percent__64.8__100.0__ |
percent__64.8__100.0__ |
| a person purchased a tv set for rs . num__16000 and a dvd player for rs . num__6250 . he sold both the items together for rs . num__31150 . what percentage of profit did he make ? <o> a ) num__10.0 <o> b ) num__30.0 <o> c ) num__50.0 <o> d ) num__40.0 <o> e ) none of these |
explanation : he total cp = rs . num__16000 + rs . num__6250 = rs . num__22250 and sp = rs . num__31150 profit ( % ) = ( num__31150 - num__22250 ) / num__22250 * num__100 = num__40.0 answer : d <eor> d <eos> |
d |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| in a photography exhibition some photographs were taken by octavia the photographer and some photographs were framed by jack the framer . jack had framed num__24 photographs taken by octavia and num__12 photographs taken by other photographers . if num__36 of the photographs in the exhibition were taken by octavia how many photographs were either framed by jack or taken by octavia ? <o> a ) num__36 <o> b ) num__48 <o> c ) num__72 <o> d ) num__96 <o> e ) num__108 |
the number of photographs either framed by jack or taken by octavia = { jack } + { octavia } - { overlap } = num__36 + num__36 - num__24 = num__48 ( { overlap } is green box ) . answer : b <eor> b <eos> |
b |
square_perimeter__12.0__ square_perimeter__12.0__ |
square_perimeter__12.0__ square_perimeter__12.0__ |
| at a local appliance manufacturing facility the workers received a num__40.0 hourly pay raise due to extraordinary performance . if one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged by approximately what percent would he reduce the number of hours that he worked ? <o> a ) num__83.0 <o> b ) num__28.0 <o> c ) num__20.0 <o> d ) num__17.0 <o> e ) num__12 % |
let ' s say he works usually num__40 hours and earns num__100 per hour . num__10 * num__100 = num__1000 num__10 * num__140 = num__1400 ( this are the new earnings after the raise ) to figure out how much he needs to work with the new salary in order to earn the original num__1000 : num__7.14285714286 = num__7.14 so he can reduce his work by num__2.86 hours . which is > num__28.0 . answer b <eor> b <eos> |
b |
percent__100.0__28.0__ |
percent__100.0__28.0__ |
| a circular path of num__16 m radius has marginal walk num__3 m wide all round it . find the cost of leveling the walk at rs . num__2 per m num__2 ? <o> a ) rs . num__660 <o> b ) rs . num__40 <o> c ) rs . num__44 <o> d ) rs . num__42 <o> e ) rs . num__43 |
explanation : π ( num__19 ^ num__2 - num__16 ^ num__2 ) = num__3.14285714286 * ( num__361 - num__256 ) = num__330 num__330 * num__2 = rs . num__660 answer : option a <eor> a <eos> |
a |
power__19.0__2.0__ power__16.0__2.0__ multiply__2.0__330.0__ multiply__2.0__330.0__ |
power__19.0__2.0__ power__16.0__2.0__ multiply__2.0__330.0__ multiply__2.0__330.0__ |
| by giving rs . num__50 to m a would have the amount equal to what m had earlier . if the sum of the amounts with a and m is rs . num__650 . what is the ratio of the amount with a to that with m earlier ? <o> a ) num__7 : num__4 <o> b ) num__5 : num__3 <o> c ) num__2 : num__1 <o> d ) num__7 : num__6 <o> e ) none of these |
explanation : let the amounts with a and m be rs . “ x ” and rs . “ y ” respectively . thus we have x + y = num__650 and x – num__50 = y x – y = num__50 . hence x = num__350 & y = num__300 thus the required ratio is num__350 : num__300 = num__7 : num__6 answer d <eor> d <eos> |
d |
subtract__650.0__350.0__ divide__350.0__50.0__ divide__300.0__50.0__ divide__350.0__50.0__ |
subtract__650.0__350.0__ divide__350.0__50.0__ divide__300.0__50.0__ divide__350.0__50.0__ |
| at amiee ’ s pet shop num__10 cups of bird seed are used every num__4 days to feed num__6 parakeets . how many cups of bird seed would be required to feed num__10 parakeets for num__5 days ? <o> a ) num__28.7 <o> b ) num__35.4 <o> c ) num__2 num__0.5 <o> d ) num__50 <o> e ) num__25 |
let ' s go step - by - step . num__10 cups are used over a num__4 day period which means num__2 num__0.5 cups a day . this feeds num__6 parakeets which means each parakeet needs . num__41 of a cup every day . for num__10 parakeets we need num__10 * . num__41 cups = num__4.10 cups a day . over num__7 days we need num__28.70 cups . choice a . <eor> a <eos> |
a |
divide__10.0__5.0__ divide__5.0__10.0__ divide__41.0__10.0__ add__5.0__2.0__ multiply__7.0__4.1__ round__28.7__ |
subtract__6.0__4.0__ divide__5.0__10.0__ divide__41.0__10.0__ add__5.0__2.0__ multiply__7.0__4.1__ multiply__7.0__4.1__ |
| the diagonals of two squares are in the ratio of num__2 : num__5 . find the ratio of their areas . <o> a ) num__4 : num__25 <o> b ) num__4 : num__28 <o> c ) num__4 : num__29 <o> d ) num__4 : num__25 <o> e ) num__4 : num__21 |
let the diagonals of the squares be num__2 x and num__5 x . then ratio of their areas will be area of square = num__0.5 ∗ diagonal num__2 num__0.5 ∗ num__2 x num__2 : num__0.5 ∗ num__5 x num__24 x num__2 : num__25 x num__2 = num__4 : num__25 answer : a <eor> a <eos> |
a |
surface_cube__2.0__ power__5.0__2.0__ triangle_area__2.0__4.0__ |
surface_cube__2.0__ power__5.0__2.0__ triangle_area__2.0__4.0__ |
| the length of a train and that of a platform are equal . if with a speed of num__90 k / hr the train crosses the platform in one minute then the length of the train ( in meters ) is ? <o> a ) num__278 <o> b ) num__272 <o> c ) num__892 <o> d ) num__750 <o> e ) num__298 |
speed = [ num__90 * num__0.277777777778 ] m / sec = num__25 m / sec ; time = num__1 min . = num__60 sec . let the length of the train and that of the platform be x meters . then num__2 x / num__60 = num__25 è x = num__25 * num__30.0 = num__750 answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ subtract__90.0__60.0__ multiply__25.0__30.0__ round__750.0__ |
hour_to_min_conversion__ divide__60.0__2.0__ multiply__25.0__30.0__ divide__750.0__1.0__ |
| in a certain lottery drawing two balls are selected at random from a container with num__50 balls numbered from num__1 to num__50 inclusive . if the winner of the lottery is awarded a cash prize in the amount of $ num__1000 times the product of the numbers on the two selected balls which of the following is a possible amount of the cash prize given to the winner ? <o> a ) $ num__9.85 x num__10 ^ num__6 <o> b ) $ num__2.45 x num__10 ^ num__6 <o> c ) $ num__1.00 x num__10 ^ num__7 <o> d ) $ num__1.05 x num__10 ^ num__7 <o> e ) $ num__9.90 x num__10 ^ num__7 |
the max product of any two balls can be num__49 * num__50 = num__2450 = num__2.45 * num__10 ^ num__3 if you multiply it by $ num__1000 the max amount will be num__2.45 * num__10 ^ num__6 . b <eor> b <eos> |
b |
subtract__50.0__1.0__ multiply__50.0__49.0__ divide__2450.0__1000.0__ multiply__1.0__2.45__ |
subtract__50.0__1.0__ multiply__50.0__49.0__ divide__2450.0__1000.0__ multiply__1.0__2.45__ |
| a man walking at a constant rate of num__6 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of num__12 miles per hour . the woman stops to wait for the man num__10 minutes after passing him while the man continues to walk at his constant rate . how many minutes must the woman wait until the man catches up ? <o> a ) num__5 <o> b ) num__15 <o> c ) num__10 <o> d ) num__20 <o> e ) num__25 |
when the woman passes the man they are aligned ( m and w ) . they are moving in the same direction . after num__5 minutes the woman ( w ) will be ahead the man ( m ) : m - - - - - - m - - - - - - - - - - - - - - - w w in the num__5 minutes after passing the man the woman walks the distance mw = ww which is num__10 * num__0.2 = num__2 miles and the man walks the distance mm which is num__10 * num__0.1 = num__1 mile . the difference of num__2 - num__1 = num__1 miles ( mw ) will be covered by the man in ( num__1 ) / num__6 = num__0.166666666667 of an hour which is num__10 minutes . answer c . <eor> c <eos> |
c |
divide__12.0__6.0__ divide__0.2__2.0__ subtract__6.0__5.0__ divide__1.0__6.0__ round__10.0__ |
divide__12.0__6.0__ divide__0.2__2.0__ subtract__6.0__5.0__ divide__1.0__6.0__ subtract__12.0__2.0__ |
| in how many different ways can the letters of the word ‘ red ’ be arranged ? <o> a ) num__0 <o> b ) num__6 <o> c ) num__24 <o> d ) num__120 <o> e ) num__720 |
b num__6 required number of arrangements = num__3 ! = num__6 <eor> b <eos> |
b |
die_space__ die_space__ |
die_space__ die_space__ |
| what is the area d of the square with the following coordinates : ( x y ) ( num__20 num__20 ) ( num__20 num__5 ) ( x num__5 ) ? <o> a ) num__60 . <o> b ) num__85 . <o> c ) num__125 . <o> d ) num__225 . <o> e ) it can not be determined from the information given |
length of one side = num__15 ( num__20 - num__5 ) since its a square the area will be d = num__15 ^ num__2 = num__225 d is the answer <eor> d <eos> |
d |
power__15.0__2.0__ triangle_area__225.0__2.0__ |
power__15.0__2.0__ power__15.0__2.0__ |
| an ice cream cone is three inches tall and its top is a circle with diameter two inches . the cone is lled with ice cream such that the interior of the cone is entirely full . the cone is topped with a hemisphere of ice cream with diameter two inches . what is the total volume in cubic inches of the ice cream in and atop the cone ? <o> a ) π <o> b ) num__1.33333333333 π <o> c ) num__1.5 π <o> d ) num__1.66666666667 π <o> e ) num__2 π |
the volume consists of that of the cone plus that of the hemisphere on top . the volume of a cone of radius r and height h is num__0.333333333333 π r num__2 h so with h = num__3 and r = num__2 num__2 = num__1 this volume is num__1 num__312 num__3 = . the volume of a sphere of radius r is num__4 num__3 r num__3 and we have half this or num__2 num__3 : the total volume is num__1.66666666667 π correct answer d <eor> d <eos> |
d |
square_perimeter__1.0__ multiply__1.0__1.6667__ |
square_perimeter__1.0__ multiply__1.0__1.6667__ |
| pipe a can fill a tank in num__20 hours pipe b in num__20 hours and pipe c in num__30 hours . if all the pipes are open in how many hours will the tank be filled ? <o> a ) num__2 <o> b ) num__2.5 <o> c ) num__7 <o> d ) num__7.5 <o> e ) none |
sol . part filled by ( a + b + c ) in num__1 hour = ( num__0.05 + num__0.05 + num__0.0333333333333 ) = num__0.133333333333 . â ˆ ´ all the three pipes together will fill the tank in num__7.5 hours . answer d <eor> d <eos> |
d |
divide__1.0__20.0__ divide__1.0__30.0__ round__7.5__ |
divide__1.0__20.0__ divide__1.0__30.0__ round__7.5__ |
| given f ( x ) = num__3 x – num__5 for what value of x does num__2 * [ f ( x ) ] – num__16 = f ( x – num__6 ) ? <o> a ) num__1 <o> b ) num__4 <o> c ) num__6 <o> d ) num__7 <o> e ) num__13 |
num__2 ( num__3 x - num__5 ) - num__16 = num__3 ( x - num__6 ) - num__5 num__3 x = num__3 x = num__1 the answer is a . <eor> a <eos> |
a |
subtract__3.0__2.0__ reverse__1.0__ |
subtract__3.0__2.0__ subtract__3.0__2.0__ |
| if rs . num__450 amount to rs . num__540 in num__4 years what will it amount to in num__6 years at the same rate % per annum ? <o> a ) s . num__575 <o> b ) s . num__595 <o> c ) s . num__590 <o> d ) s . num__580 <o> e ) s . num__585 |
num__80 = ( num__450 * num__4 * r ) / num__100 r = num__5.0 i = ( num__450 * num__6 * num__5 ) / num__100 = num__135 num__450 + num__135 = num__585 answer : e <eor> e <eos> |
e |
percent__100.0__585.0__ |
percent__100.0__585.0__ |
| for which of the following functions does f ( − q ) = − f ( q ) for all real number values of x ? <o> a ) q ^ num__8 − q ^ num__4 <o> b ) q ^ num__2 − q ^ num__6 <o> c ) q ^ num__5 / q ^ num__7 <o> d ) q ^ num__9 / ( q ^ num__5 + num__1 ) <o> e ) q ^ num__5 / ( q ^ num__2 + num__1 ) |
q ^ num__5 / ( q ^ num__2 + num__1 ) e <eor> e <eos> |
e |
multiply__1.0__5.0__ |
divide__5.0__1.0__ |
| if the product of the integers j k l and m is num__770 and if num__1 < j < k < l < m what is the value of j + m ? <o> a ) num__10 <o> b ) num__13 <o> c ) num__16 <o> d ) num__18 <o> e ) num__21 |
num__770 = num__2 * num__5 * num__7 * num__11 so j = num__2 k = num__5 l = num__7 m = num__11 j + m = num__2 + num__11 = num__13 answer - b <eor> b <eos> |
b |
add__5.0__2.0__ add__11.0__2.0__ multiply__1.0__13.0__ |
add__5.0__2.0__ add__11.0__2.0__ multiply__1.0__13.0__ |
| for prime number y y > num__3 . which of the following could be the remainder when y ^ num__3 is divided by num__12 ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
let ’ s test the first prime number greater than num__3 which is num__5 . num__5 ^ num__3 = num__125 num__10.4166666667 = num__10 remainder num__5 answer : d <eor> d <eos> |
d |
divide__125.0__12.0__ round_down__10.4167__ subtract__10.0__5.0__ |
divide__125.0__12.0__ round_down__10.4167__ subtract__10.0__5.0__ |
| it costs $ num__2 for the first num__0.25 hour to use the bathroom . after the first ¼ hour it costs $ num__8 per hour . if a certain customer uses the bathroom for num__1 hour how much will it cost him ? <o> a ) $ num__8 <o> b ) $ num__3 <o> c ) $ num__26 <o> d ) $ num__7 <o> e ) $ num__2.99 |
num__1 hour = num__60 minutes first num__15 min - - - - - - > $ num__2 time left is num__45 min . . . now num__60 min costs $ num__8 num__1 min costs $ num__0.133333333333 num__45 min costs $ num__0.133333333333 * num__45 = > $ num__6 so total cost will be $ num__6 + $ num__2 = > $ num__8 the answer will be ( a ) $ num__8 <eor> a <eos> |
a |
multiply__0.25__60.0__ subtract__60.0__15.0__ divide__2.0__15.0__ subtract__8.0__2.0__ divide__2.0__0.25__ |
multiply__0.25__60.0__ subtract__60.0__15.0__ divide__2.0__15.0__ subtract__8.0__2.0__ add__2.0__6.0__ |
| a single discount equivalent to the discount series of num__14.0 num__10.0 and num__5.0 is ? <o> a ) num__26.47 <o> b ) num__36.97 <o> c ) num__31.67 <o> d ) num__31.47 <o> e ) num__26.17 |
num__100 * ( num__0.86 ) * ( num__0.9 ) * ( num__0.95 ) = num__73.53 num__100 - num__73.53 = num__26.47 answer : a <eor> a <eos> |
a |
percent__100.0__26.47__ |
percent__100.0__26.47__ |
| what will be the remainder if num__2 ^ num__300 is divided by num__4 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
d num__4 now if you go on calculating num__2 ^ num__300 you will lose so much time and it might not even be feasible to carry out so long of calculations . thus we will make use of a trick here . we will calculate the remainder of each power of num__2 till we come across a pattern . num__2 ^ num__1 divided by num__4 leaves the remainder num__2 . num__2 ^ num__2 divided by num__4 leaves the remainder num__0 . num__2 ^ num__3 divided by num__4 leaves the remainder num__0 . num__2 ^ num__4 divided by num__4 leaves the remainder num__0 . you can see that all the following powers of two will be divisible by num__4 . therefore the remainder when num__2 ^ num__300 is divided by num__4 will be num__0 only . <eor> d <eos> |
d |
add__2.0__1.0__ multiply__4.0__1.0__ |
add__2.0__1.0__ multiply__4.0__1.0__ |
| if re . num__1 amounts to rs . num__9 over a period of num__20 years . what is the rate of simple interest ? <o> a ) num__26 num__0.666666666667 % <o> b ) num__30.0 <o> c ) num__27 num__0.5 % <o> d ) num__40.0 <o> e ) num__50 % |
explanation : num__8 = ( num__1 * num__20 * r ) / num__100 r = num__40.0 answer is d <eor> d <eos> |
d |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| two trains one from howrah to patna and the other from patna to howrah start simultaneously . after they meet the trains reach their destinations after num__9 hours and num__49 hours respectively . the ratio of their speeds is ? <o> a ) num__4 : num__7 <o> b ) num__7 : num__3 <o> c ) num__4 : num__9 <o> d ) num__4 : num__4 <o> e ) num__4 : num__2 |
let us name the trains a and b . then ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š num__49 : â ˆ š num__9 = num__7 : num__3 . answer : b <eor> b <eos> |
b |
round__7.0__ |
round__7.0__ |
| the radius of a wheel is num__24.2 cm . what is the distance covered by the wheel in making num__500 resolutions ? <o> a ) num__708 m <o> b ) num__704 m <o> c ) num__774 m <o> d ) num__760.57 m <o> e ) num__744 m |
in one resolution the distance covered by the wheel is its own circumference . distance covered in num__500 resolutions . = num__500 * num__2 * num__3.14285714286 * num__24.2 = num__76057 cm = num__760.57 m answer : d <eor> d <eos> |
d |
round__760.57__ |
round__760.57__ |
| in a house a hall is num__20 m long num__15 m wide and num__5 m high . its interior has to be covered with mat . what will be the total expenditure if it costs rs . num__40 per square m ? <o> a ) s . num__43000 <o> b ) s . num__50000 <o> c ) s . num__57000 <o> d ) s . num__38000 <o> e ) s . num__62000 |
length ( l ) = num__20 m breadth ( b ) = num__15 m and height ( h ) = num__5 m total area of the hall to be covered with mat = num__2 ( lb + bh + hl ) = num__2 ( num__20 * num__15 + num__15 * num__5 + num__5 * num__20 ) = num__2 ( num__300 + num__75 + num__100 ) = num__2 * num__475 = num__950 sq . m total expenditure = num__40 * num__950 = rs . num__38000 d <eor> d <eos> |
d |
divide__40.0__20.0__ multiply__20.0__15.0__ multiply__15.0__5.0__ multiply__20.0__5.0__ multiply__2.0__475.0__ multiply__40.0__950.0__ multiply__40.0__950.0__ |
divide__40.0__20.0__ multiply__20.0__15.0__ multiply__15.0__5.0__ multiply__20.0__5.0__ multiply__2.0__475.0__ multiply__40.0__950.0__ multiply__40.0__950.0__ |
| a train is num__360 meter long is running at a speed of num__45 km / hour . in what time will it pass a bridge of num__140 meter length <o> a ) num__20 seconds <o> b ) num__30 seconds <o> c ) num__40 seconds <o> d ) num__50 seconds <o> e ) none of these |
explanation : speed = num__45 km / hr = num__45 * ( num__0.277777777778 ) m / sec = num__12.5 m / sec total distance = num__360 + num__140 = num__500 meter time = distance / speed = num__500 ∗ num__0.08 = num__40 seconds option c <eor> c <eos> |
c |
add__360.0__140.0__ divide__500.0__12.5__ round__40.0__ |
add__360.0__140.0__ divide__500.0__12.5__ divide__500.0__12.5__ |
| in an office in singapore there are num__60.0 female employees . num__50.0 of all the male employees are computer literate . if there are total num__62.0 employees computer literate out of total num__1500 employees then the no . of female employees who are computer literate ? <o> a ) num__630 <o> b ) num__674 <o> c ) num__672 <o> d ) num__960 <o> e ) none |
solution : total employees = num__1500 female employees num__60.0 of num__1500 . = ( num__60 * num__1500 ) / num__100 = num__900 . then male employees = num__600 num__50.0 of male are computer literate = num__300 male computer literate . num__62.0 of total employees are computer literate = ( num__62 * num__1500 ) / num__100 = num__930 computer literate . thus female computer literate = num__930 - num__300 = num__630 answer : option a <eor> a <eos> |
a |
percent__60.0__1500.0__ percent__50.0__600.0__ percent__62.0__1500.0__ percent__100.0__630.0__ |
percent__60.0__1500.0__ percent__50.0__600.0__ percent__62.0__1500.0__ percent__100.0__630.0__ |
| num__5 n - num__3 > num__12 and num__7 n - num__5 < num__44 ; n must be between which numbers ? <o> a ) num__1 and num__8 <o> b ) num__2 and num__6 <o> c ) num__0 and num__9 <o> d ) num__3 and num__7 <o> e ) num__2 and num__9 |
num__5 n - num__3 > num__12 num__5 n > num__15 n > num__3 num__7 n - num__5 < num__44 num__7 n < num__49 n < num__7 so n must be between num__3 and num__7 num__3 < n < num__7 correct answer d <eor> d <eos> |
d |
multiply__5.0__3.0__ add__5.0__44.0__ divide__15.0__5.0__ |
multiply__5.0__3.0__ add__5.0__44.0__ subtract__15.0__12.0__ |
| num__4.036 divided by num__0.04 gives : <o> a ) num__100.8 <o> b ) num__100.2 <o> c ) num__100.5 <o> d ) num__100.9 <o> e ) num__100.7 |
= num__4.036 / num__0.04 = num__403.6 / num__4 = num__100.9 answer is d . <eor> d <eos> |
d |
round_down__4.036__ divide__4.036__0.04__ divide__4.036__0.04__ |
round_down__4.036__ divide__4.036__0.04__ divide__4.036__0.04__ |
| the positive difference between sam and lucy ’ s ages is a and the sum of their ages is z . if lucy is older than sam then which of the following represents lucy ’ s age ? <o> a ) ( z - a ) / num__2 <o> b ) a - z / num__2 <o> c ) num__2 a + z <o> d ) ( z + a ) / num__2 <o> e ) ( a - z ) / num__2 |
let lucy ' s age be l and sam ' s age be s as given l - s = a - - num__1 l + s = z - - num__2 adding both the equations num__2 l = a + z l = ( a + z ) / num__2 answer is d <eor> d <eos> |
d |
multiply__1.0__2.0__ |
divide__2.0__1.0__ |
| num__12 + num__13 + num__14 + . . . num__61 + num__62 + num__63 = ? <o> a ) num__1361 <o> b ) num__1362 <o> c ) num__1950 <o> d ) num__1364 <o> e ) num__1365 |
sum = num__12 + num__13 + num__14 + . . . num__61 + num__62 + num__63 sum of n consecutive positive integers starting from num__1 is given as n ( n + num__1 ) / num__2 sum of first num__63 positive integers = num__63 * num__32.0 sum of first num__11 positive integers = num__11 * num__6.0 sum = num__12 + num__13 + num__14 + . . . num__61 + num__62 + num__63 = num__63 * num__32.0 - num__11 * num__6.0 = num__1950 answer : c <eor> c <eos> |
c |
subtract__13.0__12.0__ subtract__14.0__12.0__ subtract__12.0__1.0__ divide__12.0__2.0__ multiply__1.0__1950.0__ |
subtract__13.0__12.0__ subtract__14.0__12.0__ subtract__12.0__1.0__ divide__12.0__2.0__ multiply__1.0__1950.0__ |
| a sum of num__725 is lent in the beginning of a year at a certain rate of interest . after num__8 months a sum of num__362.50 more is lent but at the rate twice the former . at the end of the year num__33.50 is earned as interest from both the loans . what was the original rate of interest ? <o> a ) num__3.6 <o> b ) num__4.5 <o> c ) num__5.0 <o> d ) num__3.46 <o> e ) none of these |
let the original rate be r % . then new rate = ( num__2 r ) % ∴ ( num__725 × r × num__0.01 ) + ( num__362.50 × num__2 r × num__0.01 × num__3 ) = num__33.50 ⇒ ( num__2175 + num__725 ) r = num__33.50 × num__100 × num__3 = num__10050 ⇒ r = num__10050 ⁄ num__2900 = num__3.46 answer d <eor> d <eos> |
d |
percent__100.0__3.46__ |
percent__100.0__3.46__ |
| num__2 no ' s are respectively num__40.0 & num__60.0 more than num__3 rdnumber . find the ration of two numbers ? <o> a ) num__3 : num__4 <o> b ) num__4 : num__5 <o> c ) num__7 : num__8 <o> d ) num__9 : num__7 <o> e ) num__11 : num__9 |
step num__1 : let the third number is a then first number is num__140.0 of a = num__140 x a / num__100 = num__7 a / num__5 and second number is num__160.0 of b = num__160 x b / num__100 = num__8 b / num__5 . step num__2 : now ratio of first and second number is num__7 a / num__5 : num__8 b / num__5 = num__35 a : num__40 b = num__7 : num__8 . c <eor> c <eos> |
c |
subtract__3.0__2.0__ add__40.0__60.0__ add__2.0__3.0__ add__60.0__100.0__ divide__40.0__5.0__ subtract__40.0__5.0__ add__2.0__5.0__ |
subtract__3.0__2.0__ add__40.0__60.0__ add__2.0__3.0__ add__60.0__100.0__ divide__40.0__5.0__ subtract__40.0__5.0__ divide__35.0__5.0__ |
| the two trains of lengths num__400 m num__600 m respectively running at same directions . the faster train can cross the slower train in num__180 sec the speed of the slower train is num__48 km . then find the speed of the faster train ? <o> a ) num__29 kmph <o> b ) num__27 kmph <o> c ) num__68 kmph <o> d ) num__22 kmph <o> e ) num__12 kmph |
length of the two trains = num__600 m + num__400 m speed of the first train = x speed of the second train = num__48 kmph num__1000 / x - num__48 = num__180 num__1000 / x - num__48 * num__0.277777777778 = num__180 num__50 = num__9 x - num__120 x = num__68 kmph answer : c <eor> c <eos> |
c |
add__400.0__600.0__ round__68.0__ |
add__400.0__600.0__ round__68.0__ |
| the diameter of a ball is num__20 inches . if the surface area of the diameter is the same as the surface area of a cube what does num__1 side of the cube measure ? round to the nearest whole number . <o> a ) num__10 <o> b ) num__5 <o> c ) num__8 <o> d ) num__4 <o> e ) num__3 |
the surface area of a sphere is determined by squaring the radius multiply by pi and multiply by num__4 . the radius is half of the diameter of a circle . so the radius of the ball is num__10 inches . multiply num__10 by pi then multiply the answer by num__4 will calculate to num__125.663 inches as the surface area of the ball . since a cube has num__6 evenly measured sides we can divide the surface are by num__6 . this comes to num__20.943 inches as the surface area for each square side of the cube . to find out the measurement for each side of the cube you will need to find out the square root of num__20.943 inches . this comes to num__4.576 inches . rounding to the nearest whole number each side of the cube measures num__5 inches . the correct answer is ( b ) . <eor> b <eos> |
b |
square_perimeter__1.0__ triangle_area__20.0__1.0__ surface_cube__1.0__ triangle_area__1.0__10.0__ triangle_area__1.0__10.0__ |
square_perimeter__1.0__ triangle_area__20.0__1.0__ surface_cube__1.0__ triangle_area__1.0__10.0__ triangle_area__1.0__10.0__ |
| the sum of the ages of mother and her daughter is num__50 years . also num__5 years ago the mother ' s age was num__7 times the age of the daughter . what is the age of the mother ? <o> a ) num__50 years <o> b ) num__60 years <o> c ) num__30 years <o> d ) num__25 years <o> e ) num__40 years |
let the age of the daughter be x years the age of the mother is num__50 - x years num__5 years ago num__7 ( x - num__5 ) = num__50 - x - num__5 x = num__10 mother ' s age = num__50 - num__10 = num__40 years answer is e <eor> e <eos> |
e |
divide__50.0__5.0__ subtract__50.0__10.0__ subtract__50.0__10.0__ |
divide__50.0__5.0__ subtract__50.0__10.0__ subtract__50.0__10.0__ |
| a hiker walked for num__3 days . she walked num__18 miles on the first day walking num__3 miles per hour . on the second day she walked for one less hour but she walked one mile per hour faster than on the first day . on the third day she walked at num__4 miles per hour for num__5 hours . how many miles in total did she walk ? <o> a ) num__24 <o> b ) num__44 <o> c ) num__58 <o> d ) num__60 <o> e ) num__62 |
first day - num__18 miles with num__3 miles per hours then total - num__6 hours for that day second day - num__4 miles per hour and num__5 hours - num__20 miles third day - num__4 miles per hour and num__5 hours - num__20 miles total num__18 + num__20 + num__20 = num__58 answer : option c <eor> c <eos> |
c |
divide__18.0__3.0__ multiply__4.0__5.0__ round__58.0__ |
divide__18.0__3.0__ multiply__4.0__5.0__ round__58.0__ |
| jerry an electrician worked num__8 months out of the year . what percent of the year did he work ? ( round answer to the nearest hundredth ) what percent num__12 is num__8 ? num__12 months = num__1 year <o> a ) num__58.33 <o> b ) num__66.67 <o> c ) num__78.33 <o> d ) num__88.33 <o> e ) num__98.33 % |
num__1 . multiply the opposites num__8 x num__100 = num__800 num__100 = num__0.666666666667 divide by the remaining number num__66.67 ( rounded to hundredth ) correct answer b <eor> b <eos> |
b |
percent__100.0__66.67__ |
percent__100.0__66.67__ |
| two trains are running at num__40 kmph and num__20 kmph respectively in the same direction . fast train completely passes a man sitting in the slower train in num__8 seconds . what is the length of the fast train ? <o> a ) num__23 m <o> b ) num__23 num__0.222222222222 m <o> c ) num__47 m <o> d ) num__37 num__0.777777777778 m <o> e ) num__44 num__0.444444444444 m |
relative speed = num__20 kmph = num__5.55555555556 m / sec length of the train = num__5.55555555556 * num__8 = num__44 num__0.444444444444 m answer : e <eor> e <eos> |
e |
round__44.0__ |
round__44.0__ |
| murali travelled from city a to city b at a speed of num__40 kmph and from city b to city c at num__60 kmph . what is the average speed of murali from a to c given that the ratio of distances between a to b and b to c is num__8 : num__4 ? <o> a ) num__16 <o> b ) num__45 <o> c ) num__277 <o> d ) num__92 <o> e ) num__11 |
let the distances between city a to b and b to c be num__8 x km and num__4 x km respectively . total time taken to cover from a to c = ( num__8 x ) / num__40 + ( num__4 x ) / num__60 = ( num__4 x + num__8 x ) / num__120 = num__32 x / num__120 = num__4 x / num__15 average speed = ( num__8 * x + num__4 x ) / ( num__4 x / num__15 ) = num__45 kmph . answer : b <eor> b <eos> |
b |
subtract__40.0__8.0__ divide__60.0__4.0__ subtract__60.0__15.0__ round__45.0__ |
multiply__8.0__4.0__ divide__60.0__4.0__ subtract__60.0__15.0__ round__45.0__ |
| if the quantity ( p − num__4 ) is num__5 times the quantity ( q − num__5 ) then what is the relationship between p and q ? <o> a ) p = num__5 q + num__30 <o> b ) p = num__5 q − num__21 <o> c ) num__5 p = q + num__20 <o> d ) p = num__5 q <o> e ) p = num__5 q − num__20 |
p - num__4 = num__5 ( q - num__5 ) p - num__4 = num__5 q - num__25 p = num__5 q - num__21 option b <eor> b <eos> |
b |
subtract__25.0__4.0__ divide__25.0__5.0__ |
subtract__25.0__4.0__ divide__25.0__5.0__ |
| kyle david and catherine each try independently to solve a problem . if their individual probabilities for success are num__0.125 num__0.25 and num__0.6 respectively what is the probability no one will solve the problem ? <o> a ) num__0.0375 <o> b ) num__0.2625 <o> c ) num__0.3125 <o> d ) num__0.328947368421 <o> e ) num__0.1125 |
p ( kyle will not solve ) = num__1 - num__0.125 = num__0.875 p ( david will not solve ) = num__1 - num__0.25 = num__0.75 p ( catherine will not solve ) = num__1 - num__0.6 = num__0.4 p = ( num__0.875 ) * ( num__0.75 ) * ( num__0.4 ) = num__0.2625 = num__0.2625 answer : b <eor> b <eos> |
b |
subtract__1.0__0.125__ subtract__1.0__0.25__ subtract__1.0__0.6__ multiply__1.0__0.2625__ |
subtract__1.0__0.125__ subtract__1.0__0.25__ subtract__1.0__0.6__ multiply__1.0__0.2625__ |
| what percent of a day is nine hours ? <o> a ) num__20.83 <o> b ) num__16 x num__0.5 % <o> c ) num__18 x num__0.666666666667 % <o> d ) num__37.5 <o> e ) none of these |
explanation : required percentage = ( num__0.000373443983402 ) % = num__37.5 . answer : d <eor> d <eos> |
d |
round__37.5__ |
round__37.5__ |
| audrey num__4 hours to complete a certain job . ferris can do the same job in num__3 hours . audrey and ferris decided to collaborate on the job working at their respective rates . while audrey worked continuously ferris took num__4 breaks of equal length . if the two completed the job together in num__2 hours how many minutes long was each of ferris ’ breaks ? <o> a ) num__5 <o> b ) num__7 <o> c ) num__15 <o> d ) num__20 <o> e ) num__25 |
audery and ferris collective work rate : num__0.25 + num__0.333333333333 = num__0.583333333333 collective work time = num__1.71428571429 = num__1.7 hrs job was actually done in = num__2 ( includes breaks ) breaks = actual time taken - collective work time = num__2 - num__1.7 = . num__3 hrs = num__0.5 so ferrais took num__4 breaks = . num__0.75 = . num__07 hrs = num__7 m so answer is b ) num__10 mins <eor> b <eos> |
b |
add__0.25__0.3333__ divide__2.0__4.0__ divide__3.0__4.0__ add__4.0__3.0__ add__3.0__7.0__ round__7.0__ |
add__0.25__0.3333__ divide__2.0__4.0__ add__0.25__0.5__ add__4.0__3.0__ add__3.0__7.0__ add__4.0__3.0__ |
| num__1370 num__1320 x - num__180 - num__6430 <o> a ) num__1070 <o> b ) num__6530 <o> c ) num__6630 <o> d ) num__6730 <o> e ) num__6830 |
num__1370 - num__50 * ( num__5 ^ num__0 ) = num__1320 num__1320 - num__50 * ( num__5 ^ num__1 ) = num__1070 num__1070 - num__50 * ( num__5 ^ num__2 ) = - num__180 - num__180 - num__50 * ( num__5 ^ num__3 ) = - num__6430 answer : a . <eor> a <eos> |
a |
subtract__1370.0__1320.0__ add__1.0__2.0__ multiply__1.0__1070.0__ |
subtract__1370.0__1320.0__ subtract__5.0__2.0__ multiply__1.0__1070.0__ |
| if x is a positive integer which of the following must be odd ? <o> a ) x + num__1 <o> b ) x ^ num__2 + x <o> c ) x ^ num__2 + x + num__11 <o> d ) x ^ num__2 − num__1 <o> e ) num__3 x ^ num__2 − num__3 |
a . x + num__1 = can be odd or even . since o + o = e or e + o = o b . x ^ num__2 + x = x ( x + num__1 ) . since from the above derivation we already know the term x + num__1 can be odd or even directly substitute here . x ( odd ) = even ( when x is even ) or x ( even ) = even [ when x is odd ] c . here ' s the answer . since we know the term x ^ num__2 + x can always take a even number even + num__11 = odd hence c . <eor> c <eos> |
c |
multiply__1.0__2.0__ |
power__2.0__1.0__ |
| a garrison of num__2000 men has provisions for num__54 days . at the end of num__18 days a reinforcement arrives and it is now found that the provisions will last only for num__20 days more . what is the reinforcement ? <o> a ) num__1977 <o> b ) num__1600 <o> c ) num__1979 <o> d ) num__1900 <o> e ) num__1278 |
num__2000 - - - - num__54 num__2000 - - - - num__36 x - - - - - num__20 x * num__20 = num__2000 * num__36 x = num__3600 num__2000 - - - - - - - num__1600 answer : b <eor> b <eos> |
b |
subtract__54.0__18.0__ subtract__3600.0__2000.0__ round__1600.0__ |
subtract__54.0__18.0__ subtract__3600.0__2000.0__ subtract__3600.0__2000.0__ |
| in a group of cows and hens the number of legs are num__12 more than twice the number of heads . the number of cows is : <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__10 <o> e ) num__12 |
let no of cows be x no of hens be y . so heads = x + y legs = num__4 x + num__2 y now num__4 x + num__2 y = num__2 ( x + y ) + num__12 num__2 x = num__12 x = num__6 . answer : b <eor> b <eos> |
b |
coin_space__ die_space__ die_space__ |
coin_space__ die_space__ die_space__ |
| the average of num__11 numbers is num__10.9 . if the average of first six is num__10.5 and that of the last six is num__11.7 the sixth number is ? <o> a ) num__11.4 <o> b ) num__11.3 <o> c ) num__13.3 <o> d ) num__11.5 <o> e ) num__11.1 |
num__1 to num__11 = num__11 * num__10.9 = num__119.9 num__1 to num__6 = num__6 * num__10.5 = num__63 num__6 to num__11 = num__6 * num__11.7 = num__70.2 num__63 + num__70.2 = num__133.2 – num__119.9 = num__13.3 num__6 th number = num__13.3 answer : c <eor> c <eos> |
c |
multiply__11.0__10.9__ multiply__10.5__6.0__ multiply__11.7__6.0__ add__70.2__63.0__ subtract__133.2__119.9__ multiply__1.0__13.3__ |
multiply__11.0__10.9__ multiply__10.5__6.0__ multiply__11.7__6.0__ add__70.2__63.0__ subtract__133.2__119.9__ multiply__1.0__13.3__ |
| a reduction of num__23.0 in the price of salt enables a lady to obtain num__10 kgs more for rs . num__100 find the original price per kg ? <o> a ) s . num__2.99 <o> b ) s . num__2.4 <o> c ) s . num__2.5 <o> d ) s . num__2.2 <o> e ) s . num__2.1 |
num__100 * ( num__0.23 ) = num__23 - - - num__10 ? - - - num__1 = > rs . num__2.3 num__100 - - - num__77 ? - - - num__2.3 = > rs . num__2.99 answer : a <eor> a <eos> |
a |
percent__23.0__10.0__ percent__100.0__2.99__ |
percent__23.0__10.0__ percent__100.0__2.99__ |
| if a lends rs . num__3200 to b at num__12.0 per annum and b lends the same sum to c at num__14.5 per annum then the gain of b in a period of num__5 years is ? <o> a ) num__157.78 <o> b ) num__157.98 <o> c ) num__400 <o> d ) num__420 <o> e ) num__430 |
( num__3200 * num__2.5 * num__5 ) / num__100 = > num__400 answer : c <eor> c <eos> |
c |
percent__100.0__400.0__ |
percent__100.0__400.0__ |
| a train covers a distance in num__50 min if it runs at a speed of num__48 kmph on an average . the speed at which the train must run to reduce the time of journey to num__45 min will be . <o> a ) num__60 km / h <o> b ) num__53 km / h <o> c ) num__40 km / h <o> d ) num__70 km / h <o> e ) num__65 km / h |
time = num__0.833333333333 hr = num__0.833333333333 hr speed = num__48 mph distance = s * t = num__48 * num__0.833333333333 = num__40 km time = num__0.75 hr = num__0.75 hr new speed = num__40 * num__1.33333333333 kmph = num__53 kmph answer : b <eor> b <eos> |
b |
round__53.0__ |
round__53.0__ |
| a specialized type of sand consists of num__40.0 mineral x by volume and num__60.0 mineral y by volume . if mineral x weighs num__2.5 grams per cubic centimeter and mineral y weighs num__3 grams per cubic centimeter how many grams does a cubic meter of specialized sand combination weigh ? ( num__1 meter = num__100 centimeters ) <o> a ) num__5 num__500000 <o> b ) num__2 num__800000 <o> c ) num__55000 <o> d ) num__28000 <o> e ) num__280 |
let the volume be num__1 m ^ num__3 = num__1 m * num__1 m * num__1 m = num__100 cm * num__100 cm * num__100 cm = num__1 num__000000 cm ^ num__3 by volume num__40.0 is x = num__400000 cm ^ num__3 num__60.0 is y = num__600000 cm ^ num__3 by weight in num__1 cm ^ num__3 x is num__2.5 gms in num__400000 cm ^ num__3 x = num__2.5 * num__400000 = num__1 num__000000 grams in num__1 cm ^ num__3 y is num__3 gms in num__600000 cm ^ num__3 y = num__3 * num__600000 = num__1 num__800000 gms total gms in num__1 m ^ num__3 = num__1 num__000000 + num__1 num__800000 = num__2 num__800000 answer : b <eor> b <eos> |
b |
triangle_area__3.0__400000.0__ rectangle_perimeter__0.0__400000.0__ rectangle_perimeter__1.0__0.0__ rectangle_perimeter__1.0__0.0__ |
triangle_area__3.0__400000.0__ rectangle_perimeter__0.0__400000.0__ rectangle_perimeter__1.0__0.0__ power__2.0__1.0__ |
| a b and c have rs . num__900 between them a and c together have rs . num__400 and b and c rs . num__750 . how much does c have ? <o> a ) num__100 <o> b ) num__150 <o> c ) num__200 <o> d ) num__250 <o> e ) num__350 |
a + b + c = num__900 a + c = num__400 b + c = num__750 - - - - - - - - - - - - - - a + b + num__2 c = num__1150 a + b + c = num__900 - - - - - - - - - - - - - - - - c = num__250 answer : d <eor> d <eos> |
d |
add__400.0__750.0__ subtract__1150.0__900.0__ subtract__1150.0__900.0__ |
add__400.0__750.0__ subtract__1150.0__900.0__ subtract__1150.0__900.0__ |
| if | num__20 x - num__10 | = num__170 then find the product of the values of x ? <o> a ) - num__45 <o> b ) num__50 <o> c ) - num__72 <o> d ) num__35 <o> e ) - num__30 |
| num__20 x - num__10 | = num__170 num__20 x - num__10 = num__170 or num__20 x - num__10 = - num__170 num__20 x = num__180 or num__20 x = - num__160 x = num__9 or x = - num__8 product = - num__8 * num__9 = - num__72 answer is c <eor> c <eos> |
c |
add__10.0__170.0__ subtract__170.0__10.0__ divide__180.0__20.0__ divide__160.0__20.0__ multiply__8.0__9.0__ multiply__8.0__9.0__ |
add__10.0__170.0__ subtract__170.0__10.0__ divide__180.0__20.0__ divide__160.0__20.0__ multiply__8.0__9.0__ multiply__8.0__9.0__ |
| num__0.35 represents what percent of num__0.7 ? <o> a ) num__0.05 <o> b ) num__0.5 <o> c ) num__50.0 <o> d ) num__500.0 <o> e ) num__5000 % |
one more method num__0.35 represents what percent of num__0.7 ? adjusting the decimal num__3.5 represents what percent of num__7 ? divide by num__7 num__0.5 represents what percent of num__1 ? answer = num__0.5 * num__100 = num__50.0 = c <eor> c <eos> |
c |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| a pipe can fill a tank in num__15 minutes and another one in num__10 minutes . a third pipe can empty the tank in num__5 minutes . the first two pipes are kept open for num__4 minutes in the beginning and then the third pipe is also opened . in what time will the tank be empited ? <o> a ) num__35 min <o> b ) num__15 min <o> c ) num__20 min <o> d ) can not be emptied <o> e ) none of these |
proportion of the volume of the tank filled by both the pipes in num__4 min = num__4 ( num__1 ⁄ num__15 + num__1 ⁄ num__10 ) = num__2 ⁄ num__3 rd of the tank volume of the tank filled by all the pipes working together = num__1 ⁄ num__15 + num__1 ⁄ num__10 - num__1 ⁄ num__5 = - num__1 ⁄ num__30 i . e . num__1 ⁄ num__30 tank is emptied in num__1 min . ∴ num__2 ⁄ num__3 rd of the tank can be emptied in num__2 × num__10.0 = num__20 min answer c <eor> c <eos> |
c |
subtract__5.0__4.0__ divide__10.0__5.0__ divide__15.0__5.0__ multiply__15.0__2.0__ add__15.0__5.0__ round__20.0__ |
subtract__5.0__4.0__ divide__10.0__5.0__ subtract__5.0__2.0__ multiply__15.0__2.0__ add__15.0__5.0__ add__15.0__5.0__ |
| find the value of y from given equation : ( num__12 ) ^ num__3 x num__6 ^ num__4 ÷ num__432 = y ? <o> a ) num__2345 <o> b ) num__2790 <o> c ) num__3490 <o> d ) num__7389 <o> e ) num__5184 |
given exp . = ( num__12 ) num__3 x num__64 = ( num__12 ) num__3 x num__64 = ( num__12 ) num__2 x num__62 = ( num__72 ) num__2 = num__5184 num__432 num__12 x num__62 e <eor> e <eos> |
e |
divide__12.0__6.0__ subtract__64.0__2.0__ multiply__12.0__6.0__ multiply__12.0__432.0__ multiply__12.0__432.0__ |
divide__12.0__6.0__ subtract__64.0__2.0__ multiply__12.0__6.0__ multiply__12.0__432.0__ multiply__12.0__432.0__ |
| a volume of num__11780 l water is in a container of sphere . how many hemisphere of volume num__4 l each will be required to transfer all the water into the small hemispheres ? <o> a ) num__2812 <o> b ) num__8231 <o> c ) num__2734 <o> d ) num__4222 <o> e ) num__2945 |
a volume of num__4 l can be kept in num__1 hemisphere therefore a volume of num__11780 l can be kept in ( num__2945.0 ) hemispheres ans . num__2945 answer : e <eor> e <eos> |
e |
multiply__1.0__2945.0__ |
multiply__1.0__2945.0__ |
| two heavily loaded sixteen - wheeler transport trucks are num__855 kilometers apart sitting at two rest stops on opposite sides of the same highway . driver a begins heading down the highway driving at an average speed of num__90 kilometers per hour . exactly one hour later driver b starts down the highway toward driver a maintaining an average speed of num__80 kilometers per hour . how many kilometers farther than driver b will driver a have driven when they meet and pass each other on the highway ? <o> a ) num__90 <o> b ) num__130 <o> c ) num__135 <o> d ) num__320 <o> e ) num__450 |
i ' ve been reading the website for a while and i ' m always keen to see different approaches so i would like to share one that works for me : short version : truck a travels for an hour . distance remaining = num__855 - num__90 = num__765 k ratio of speeds num__9 : num__8 - > num__45.0 = num__45 truck a = num__90 + num__45 * num__9 = num__495 truck b = num__45 * num__8 = num__360 delta = num__135 km answer c <eor> c <eos> |
c |
subtract__855.0__90.0__ subtract__855.0__495.0__ add__90.0__45.0__ add__90.0__45.0__ |
subtract__855.0__90.0__ subtract__855.0__495.0__ add__90.0__45.0__ add__90.0__45.0__ |
| for a certain set the value range of its members is num__96.8 . a new set is created from the members of the old set as follows : num__12 is subtracted from a member of the old set and the result is divided by num__4 . the resulting value is a member of the new set . if this operation is done for each member of the old set what is the range of values of the members of the new set ? <o> a ) num__21.2 <o> b ) num__24.2 <o> c ) num__36.3 <o> d ) num__48.4 <o> e ) num__96.8 |
let x and z be the smallest and largest of the original set respectively . z - x = num__96.8 the smallest and largest members of the new set will be ( x - num__12 ) / num__4 and ( z - num__12 ) / num__4 . then the range is ( z - num__12 ) / num__4 - ( x - num__12 ) / num__4 = ( z - x ) / num__4 = num__96.8 / num__4 = num__24.2 the answer is b . <eor> b <eos> |
b |
divide__96.8__4.0__ divide__96.8__4.0__ |
divide__96.8__4.0__ divide__96.8__4.0__ |
| a regular hexagon is there . the mid points of the sides were joined and formed another hexagon . then what is the percentage reduction in area . <o> a ) num__25.0 <o> b ) num__35.0 <o> c ) num__45.0 <o> d ) num__55.0 <o> e ) num__65 % |
let abcdef be the regular hexagon with sides ab and ed parallel . p and q are mid points of af and ef resp . ae = root ( num__3 ) * s ( s is side of hexagon ) then pq = ( root ( num__3 ) * s ) / num__2 mid point theorem now u know side of inner hexagon use formula area of hexagon = ( num__3 * root ( num__3 ) ) num__2 * s ^ num__2 and find diff in area of two hexagon multiply by num__100 and divide the product by area of bigger hexagon ans num__25.0 answer : a <eor> a <eos> |
a |
triangle_area__25.0__2.0__ |
triangle_area__25.0__2.0__ |
| in a garden num__26 trees are planted at equal distances along a yard num__400 metres long one tree being at each end of the yard . what is the distance between two consecutive trees ? <o> a ) num__10 <o> b ) num__8 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
num__26 trees have num__25 gaps between them . length of each gap = num__16.0 = num__16 i . e . distance between two consecutive trees = num__16 answer is e . <eor> e <eos> |
e |
divide__400.0__25.0__ round__16.0__ |
divide__400.0__25.0__ round__16.0__ |
| the average weight of a class of num__20 boys was calculated to be num__58.4 kgs and it was later found that one weight was misread as num__56 kg instead of num__68 kg . what is the correct weight ? <o> a ) num__58 kgs <o> b ) num__58.85 kgs <o> c ) num__58.95 kgs <o> d ) num__59 kgs <o> e ) num__59.85 kgs |
actual total weight is ( num__20 x num__58.4 - num__56 + num__68 ) = num__1180 kgs actual average weight is num__59.0 = num__59 kgs d <eor> d <eos> |
d |
divide__1180.0__20.0__ divide__1180.0__20.0__ |
divide__1180.0__20.0__ divide__1180.0__20.0__ |
| a train covers a distance in num__50 minutes if it runs at a speed of num__48 kmph on an average . find the speed at which the train must run to reduce the time of journey to num__40 minutes . <o> a ) num__50 km / hr <o> b ) num__60 km / hr <o> c ) num__65 km / hr <o> d ) num__70 km / hr <o> e ) none of these |
explanation : we are having time and speed given so first we will calculate the distance . then we can get new speed for given time and distance . lets solve it . time = num__0.833333333333 hr = num__0.833333333333 hr speed = num__48 mph distance = s × t = num__48 × num__0.833333333333 = num__40 km new time will be num__40 minutes so time = num__0.666666666667 hr = num__0.666666666667 hr now we know speed = distance / time new speed = num__40 × num__1.5 kmph = num__60 kmph answer : b <eor> b <eos> |
b |
divide__40.0__48.0__ add__0.8333__0.6667__ hour_to_min_conversion__ hour_to_min_conversion__ |
divide__40.0__48.0__ add__0.8333__0.6667__ hour_to_min_conversion__ hour_to_min_conversion__ |
| if all of the telephone extensions in a certain company must be even numbers and if each of the extensions uses all four of the digits num__1 num__2 num__6 and num__7 what is the greatest number of four - digit extensions that the company can have ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__11 <o> d ) num__12 <o> e ) num__24 |
since the phone number must be even the unit ' s digit can be either num__2 or num__6 . when the unit ' s digit is num__2 - - > number of possibilities is num__3 ! = num__6 when the unit ' s digit is num__6 - - > number of possibilities is num__3 ! = num__6 largest number of extensions = num__6 + num__6 = num__12 answer : d <eor> d <eos> |
d |
add__1.0__2.0__ multiply__2.0__6.0__ multiply__1.0__12.0__ |
add__1.0__2.0__ multiply__2.0__6.0__ multiply__1.0__12.0__ |
| alice is now num__10 years older than bob . if in num__6 years alice will be twice as old as bob how old will alice be in num__5 years ? <o> a ) num__19 <o> b ) num__21 <o> c ) num__23 <o> d ) num__25 <o> e ) num__27 |
a = b + num__10 so b = a - num__10 . a + num__6 = num__2 ( b + num__6 ) . a + num__6 = num__2 ( a - num__10 + num__6 ) . a + num__6 = num__2 a - num__8 . a = num__14 . in num__5 years alice will be num__19 years old . the answer is a . <eor> a <eos> |
a |
divide__10.0__5.0__ subtract__10.0__2.0__ add__6.0__8.0__ add__5.0__14.0__ add__5.0__14.0__ |
divide__10.0__5.0__ subtract__10.0__2.0__ add__6.0__8.0__ add__5.0__14.0__ add__5.0__14.0__ |
| the sides of a square region measured to the nearest centimeter are num__4 centimeters long . the least possible value of the actual area of the square region is <o> a ) num__15.25 sq cm <o> b ) num__18.25 sq cm <o> c ) num__12.25 sq cm <o> d ) num__21.25 sq cm <o> e ) num__25.25 sq cm |
though there might be some technicalities concerning the termnearest ( as num__3.5 is equidistant from both num__3 and num__4 ) the answer still should be : num__3.5 ^ num__2 = num__12.25 . answer : c <eor> c <eos> |
c |
power__3.5__2.0__ power__3.5__2.0__ |
power__3.5__2.0__ power__3.5__2.0__ |
| a b c d e f g and h are all integers listed in order of increasing size . when these numbers are arranged on a number line the distance between any two consecutive numbers is constant . if g and h are equal to num__5 ^ num__12 and num__5 ^ num__16 respectively what is the value of a ? <o> a ) - num__24 ( num__5 ^ num__12 ) <o> b ) - num__27 ( num__5 ^ num__12 ) <o> c ) - num__24 ( num__5 ^ num__6 ) <o> d ) num__23 ( num__5 ^ num__12 ) <o> e ) num__24 ( num__5 ^ num__12 ) |
ans : b assume that the numbers appear as shown below on the number line a - - - - - b - - - - - c - - - - - d - - - - - e - - - - - f - - - - - g - - - - - h ( num__5 ^ num__12 ) ( num__5 ^ num__13 ) as the values for g and h are given we can calculate the difference between any two terms of the series . common difference d = ( num__5 ^ num__13 ) - ( num__5 ^ num__12 ) = ( num__5 ^ num__12 ) * [ num__5 - num__1 ] = ( num__5 ^ num__12 ) * ( num__4 ) also f + d = g as the terms are in equidistant and in increasing order . so f + ( num__5 ^ num__12 ) * ( num__4 ) = ( num__5 ^ num__12 ) . that is f = ( num__5 ^ num__12 ) - ( num__5 ^ num__12 ) * ( num__4 ) = ( num__5 ^ num__12 ) [ num__1 - num__4 ] = ( num__5 ^ num__12 ) ( - num__3 ) similarly e = f - d = ( num__5 ^ num__12 ) [ - num__3 - num__4 ] = ( num__5 ^ num__12 ) * ( - num__7 ) you can see a - num__4 getting added to the non - exponent part of the values . that is according to the pattern d should be ( num__5 ^ num__12 ) * ( - num__7 - num__4 ) = ( num__5 ^ num__12 ) * ( - num__11 ) following this pattern a = ( num__5 ^ num__12 ) * ( - num__27 ) b <eor> b <eos> |
b |
subtract__13.0__12.0__ subtract__5.0__1.0__ divide__12.0__4.0__ subtract__12.0__5.0__ subtract__12.0__1.0__ add__16.0__11.0__ add__16.0__11.0__ |
subtract__13.0__12.0__ subtract__5.0__1.0__ subtract__16.0__13.0__ subtract__12.0__5.0__ subtract__12.0__1.0__ add__16.0__11.0__ add__16.0__11.0__ |
| the smallest number which when diminished by num__7 is divisible by num__12 num__16 num__18 num__21 and num__28 is <o> a ) num__2333 <o> b ) num__1015 <o> c ) num__2683 <o> d ) num__2693 <o> e ) num__1113 |
explanation : required number = ( l . c . m of num__12 num__16 num__18 num__2128 ) + num__7 = num__1008 + num__7 = num__1015 answer : b ) num__1015 <eor> b <eos> |
b |
add__7.0__1008.0__ add__7.0__1008.0__ |
add__7.0__1008.0__ add__7.0__1008.0__ |
| what are the least number of square tiles required to pave the floor of a room num__15 m num__17 cm long and num__9 m num__2 cm broad ? <o> a ) num__714 <o> b ) num__814 <o> c ) num__850 <o> d ) num__866 <o> e ) none of these |
explanation : in this type of questions first we need to calculate the area of tiles . with we can get by obtaining the length of largest tile . length of largest tile can be obtained from hcf of length and breadth . so lets solve this length of largest tile = hcf of ( num__1517 cm and num__902 cm ) = num__41 cm required number of tiles = area of floor / area of tile = ( num__1517 × num__22.0 × num__41 ) = num__814 option b <eor> b <eos> |
b |
divide__902.0__41.0__ round__814.0__ |
divide__902.0__41.0__ round__814.0__ |
| a father said to his son ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is num__38 years now the son ' s age five years back was : <o> a ) num__14 years <o> b ) num__10 years <o> c ) num__8 years <o> d ) num__17 years <o> e ) none of these |
explanation : let the ages of children be x ( x + num__3 ) ( x + num__6 ) ( x + num__9 ) and ( x + num__12 ) years . then x + ( x + num__3 ) + ( x + num__6 ) + ( x + num__9 ) + ( x + num__12 ) = num__50 ⇒ num__5 x = num__20 ⇒ x = num__4 . ⇒ age of the youngest child = x = num__4 years . answer : a <eor> a <eos> |
a |
add__3.0__6.0__ add__3.0__9.0__ add__38.0__12.0__ subtract__9.0__5.0__ add__5.0__9.0__ |
add__3.0__6.0__ add__3.0__9.0__ add__38.0__12.0__ subtract__9.0__5.0__ add__5.0__9.0__ |
| if you spend num__0.25 of your salary during the first week of the month and num__20.0 of it during each of the coming num__3 weeks what part ( fraction ) of your salary will leave unspent at the end of the month ? <o> a ) num__0.1 <o> b ) num__0.15 <o> c ) num__0.2 <o> d ) num__0.25 <o> e ) none |
solution : num__0.25 + num__3 ( num__0.2 ) = num__0.25 + num__0.6 = num__5 + num__0.6 = num__0.85 the sallary that will be spent = num__0.85 unspend will be = num__1 − num__0.85 = num__20 − num__0.85 = num__0.15 answer b <eor> b <eos> |
b |
percent__20.0__3.0__ percent__20.0__5.0__ percent__3.0__5.0__ percent__3.0__5.0__ |
percent__20.0__3.0__ percent__20.0__5.0__ percent__3.0__5.0__ percent__3.0__5.0__ |
| given the sum num__195 of num__13 consecutive integers find the sum of num__4 th number and num__7 th number of the sequence <o> a ) num__65 <o> b ) num__54 <o> c ) num__26 <o> d ) num__30 <o> e ) num__27 |
the median is : num__15.0 = num__15 . the middle position is : num__6.5 = num__6.5 rounded up : num__7 th position . so num__15 is the value of the num__7 th position difference between num__7 th and num__4 th position = num__3 positions less num__15 - num__3 = num__12 = is the value of the num__4 th position num__15 + num__12 = num__27 = sum of num__4 th and num__7 th number . answer : e <eor> e <eos> |
e |
divide__195.0__13.0__ subtract__7.0__4.0__ multiply__4.0__3.0__ add__12.0__15.0__ add__12.0__15.0__ |
divide__195.0__13.0__ subtract__7.0__4.0__ subtract__15.0__3.0__ add__12.0__15.0__ add__12.0__15.0__ |
| a set of football matches is to be organized in a ` ` round - robin ' ' fashion i . e . every participating team plays a match against every other team once and only once . if num__2 matches are totally played how many teams participated ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__2 <o> e ) num__3 |
num__2 ways to solve this problem . . summation ( x ) = n ( n - num__1 ) / num__2 n ( n - num__1 ) / num__2 = num__2 ; n ^ num__2 - n - num__4 = num__0 ans : num__2 answer : d <eor> d <eos> |
d |
multiply__2.0__1.0__ |
divide__2.0__1.0__ |
| the length of a rectangular plot is thrice its width . if the area of the rectangular plot is num__432 sq meters then what is the width ( in meters ) of the rectangular plot ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__12 <o> d ) num__13 <o> e ) num__14 |
area = l * w = num__3 w ^ num__2 = num__432 w ^ num__2 = num__144 w = num__12 the answer is c . <eor> c <eos> |
c |
square_perimeter__3.0__ square_perimeter__3.0__ |
square_perimeter__3.0__ square_perimeter__3.0__ |
| the total number of digits used in numbering the pages of a book having num__360 pages is <o> a ) num__972 <o> b ) num__990 <o> c ) num__1098 <o> d ) num__1305 <o> e ) num__1405 |
total number of digits = ( no . of digits in num__1 - digit page nos . + no . of digits in num__2 - digit page nos . + no . of digits in num__3 - digit page nos . ) = ( num__1 x num__9 + num__2 x num__90 + num__3 x num__261 ) = ( num__9 + num__180 + num__783 ) = num__972 . answer : a <eor> a <eos> |
a |
add__1.0__2.0__ divide__360.0__2.0__ multiply__3.0__261.0__ multiply__1.0__972.0__ |
add__1.0__2.0__ divide__360.0__2.0__ multiply__3.0__261.0__ multiply__1.0__972.0__ |
| gold is num__19 times as heavy as water and copper is num__9 times as heavy as water . in what ratio should these be mixed to get an alloy num__12 times as heavy as water ? <o> a ) num__1 : num__2 <o> b ) num__3 : num__7 <o> c ) num__4 : num__1 <o> d ) num__5 : num__2 <o> e ) num__6 : num__5 |
g = num__19 w c = num__9 w let num__1 gm of gold mixed with x gm of copper to get num__1 + x gm of the alloy num__1 gm gold + x gm copper = x + num__1 gm of alloy num__19 w + num__9 wx = x + num__1 * num__12 w num__19 + num__9 x = num__12 ( x + num__1 ) x = num__2.33333333333 ratio of gold with copper = num__1 : num__2.33333333333 = num__3 : num__7 answer is b <eor> b <eos> |
b |
subtract__12.0__9.0__ subtract__19.0__12.0__ divide__9.0__3.0__ |
subtract__12.0__9.0__ subtract__19.0__12.0__ multiply__1.0__3.0__ |
| the sum of two numbers is num__20 and the sum of their squares is num__140 . find the product of the two numbers . <o> a ) num__130 <o> b ) num__140 <o> c ) num__120 <o> d ) num__145 <o> e ) num__150 |
let a and b be the two numbers ( a + b ) ^ num__2 = a ^ num__2 + num__2 ab + b ^ num__2 given ( a + b ) = num__20 a ^ num__2 + b ^ num__2 = num__140 so num__20 ^ num__2 = num__140 + num__2 ab num__2 ab = num__400 - num__140 num__2 ab = num__260 ab = num__130 ans a <eor> a <eos> |
a |
subtract__400.0__140.0__ divide__260.0__2.0__ divide__260.0__2.0__ |
subtract__400.0__140.0__ divide__260.0__2.0__ subtract__260.0__130.0__ |
| a river num__4 m deep and num__22 m wide is flowing at the rate of num__2 kmph the amount of water that runs into the sea per minute is ? <o> a ) num__4500 <o> b ) num__2678 <o> c ) num__2933 <o> d ) num__2761 <o> e ) num__2882 |
( num__2000 * num__4 * num__22 ) / num__60 = num__2933 m num__3 answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ round__2933.0__ |
hour_to_min_conversion__ round__2933.0__ |
| sum of products of three numbers taken two at a time is num__131 . sum of squares of the three numbers is num__138 . what is the sum of the three numbers ? <o> a ) num__20 <o> b ) num__88 <o> c ) num__26 <o> d ) num__18 <o> e ) num__17 |
explanation : all the data given in the question points to the equation mentioned below : considering the three numbers to be a b and c : ( a + b + c ) num__2 = a num__2 + b num__2 + c num__2 + num__2 ab + num__2 bc + num__2 ca accordingly substitute for a b and c and solve : = > a num__2 + b num__2 + c num__2 = num__138 ab + bc + ac = num__131 . thus num__2 ( ab + bc + ac ) = num__2 * num__131 = num__262 ( a + b + c ) num__2 = num__138 + num__262 = num__400 . therefore a + b + c = num__20 answer : a <eor> a <eos> |
a |
multiply__131.0__2.0__ add__138.0__262.0__ divide__400.0__20.0__ |
multiply__131.0__2.0__ add__138.0__262.0__ divide__400.0__20.0__ |
| it takes greg num__6 minutes to run a mile and it takes pete num__8 minutes . if greg and pete both begin running at exactly num__8 : num__30 a . m . without pausing when is the first time at which they will finish running a mile simultaneously ? <o> a ) num__8 : num__36 am <o> b ) num__8 : num__38 am <o> c ) num__8 : num__48 am <o> d ) num__8 : num__54 am <o> e ) num__9 : num__00 am |
since greg will run s miles after s × num__6 minutes and pete will run c miles after c × num__8 minutes they both will finish running a mile at the same time when s × num__6 = c × num__8 . since s and c must be integers ( they represent the number of miles finished ) this question is asking you to find a common multiple of num__6 and num__8 . the question asks for the first time they will finish running a mile simultaneously so you must find the least common multiple . the least common multiple of num__6 and num__8 is num__24 so in the context of the question this would be num__24 minutes . therefore the first time they will finish running a mile at the same time is num__8 : num__30 + num__24 minutes or num__8 : num__54 a . m . the answer is ( d ) . <eor> d <eos> |
d |
subtract__30.0__6.0__ add__30.0__24.0__ round__8.0__ |
subtract__30.0__6.0__ add__30.0__24.0__ round__8.0__ |
| the greatest number of four digits which is divisible by num__15 num__25 num__40 and num__75 is : <o> a ) num__9000 <o> b ) num__9400 <o> c ) num__9600 <o> d ) num__9800 <o> e ) num__9900 |
greatest number of num__4 - digits is num__9999 . l . c . m . of num__15 num__25 num__40 and num__75 is num__600 . on dividing num__9999 by num__600 the remainder is num__399 . required number ( num__9999 - num__399 ) = num__9600 . answer : option c <eor> c <eos> |
c |
multiply__15.0__40.0__ subtract__9999.0__399.0__ subtract__9999.0__399.0__ |
multiply__15.0__40.0__ subtract__9999.0__399.0__ subtract__9999.0__399.0__ |
| if the man walks at the rate of num__5 kmph he misses a train by num__7 minutes . however if he walks at the rate of num__6 kmph he reaches the station num__5 minutes before the arrival of the train . find the distance covered by him to reach the station . <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
let the distance be x difference between time = num__12 min which is num__0.2 in hour which is num__0.2 now x / num__5 - x / num__6 = num__0.2 num__6 x - num__5 x / num__30 = num__0.2 x / num__30 = num__0.2 x = num__6.0 x = num__6 so the required distance = num__6 answer : c <eor> c <eos> |
c |
add__5.0__7.0__ multiply__5.0__6.0__ round__6.0__ |
add__5.0__7.0__ divide__6.0__0.2__ divide__30.0__5.0__ |
| a person has to cover a distance of num__6 km in num__45 minutes . if he covers one - half of the distance in two - thirds of the total time ; to cover the remaining distance in the remaining time what should be his speed in km / hr ? <o> a ) num__12 km / hr <o> b ) num__14 km / hr <o> c ) num__10 km / hr <o> d ) num__8 km / hr <o> e ) num__9 km / hr |
explanation : the person needs to cover num__6 km in num__45 minutes given that he covers one - half of the distance in two - thirds of the total time = > he covers half of num__6 km in two - thirds of num__45 minutes = > he covers num__3 km in num__30 minutes hence now heneed to cover the remaining num__3 km in the remaining num__15 minutes distance = num__3 km time = num__15 minutes = num__0.25 hour required speed = distance / time = num__3 / ( num__0.25 ) = num__12 km / hr answer : option a <eor> a <eos> |
a |
divide__45.0__3.0__ divide__3.0__0.25__ round__12.0__ |
divide__45.0__3.0__ divide__3.0__0.25__ divide__3.0__0.25__ |
| the units digit of ( num__3 ) ^ ( num__44 ) + ( num__10 ) ^ ( num__46 ) is : <o> a ) num__2 <o> b ) num__4 <o> c ) num__1 <o> d ) num__8 <o> e ) num__0 |
any power of anything ending in zero always has a units digit of num__0 . so the first term has a units digit of num__0 . the period is num__4 . this means num__3 to the power of any multiple of num__4 will have a units digit of num__1 . num__3 ^ num__44 has a units digit of num__1 . of course num__0 + num__1 = num__1 c <eor> c <eos> |
c |
subtract__4.0__3.0__ reverse__1.0__ |
subtract__4.0__3.0__ reverse__1.0__ |
| what is the greatest prime factor of num__1 + num__2 + num__3 + 。 。 。 + num__22 ? <o> a ) num__17 <o> b ) num__29 <o> c ) num__23 <o> d ) num__37 <o> e ) num__11 |
find the sum from direct formula = n ( n + num__1 ) / num__2 = num__22 ( num__22 + num__1 ) / num__2 = num__11 * num__23 . . . num__23 is a prime number so ans is num__23 . . c is the answer <eor> c <eos> |
c |
divide__22.0__2.0__ add__1.0__22.0__ add__1.0__22.0__ |
divide__22.0__2.0__ add__1.0__22.0__ add__1.0__22.0__ |
| a mixture of num__150 liters of wine and water contains num__20.0 water . how much more water should be added so that water becomes num__25.0 of the new mixture ? <o> a ) num__5 <o> b ) num__18 <o> c ) num__12 <o> d ) num__24 <o> e ) num__10 |
explanation : number of liters of water in num__150 liters of the mixture = num__20.0 of num__150 = num__0.2 * num__150 = num__30 liters . p liters of water added to the mixture to make water num__25.0 of the new mixture . total amount of water becomes ( num__30 + p ) and total volume of mixture is ( num__150 + p ) . ( num__30 + p ) = num__0.25 * ( num__150 + p ) num__120 + num__4 p = num__150 + p = > p = num__10 liters . correct option : e <eor> e <eos> |
e |
multiply__150.0__0.2__ subtract__150.0__30.0__ reverse__0.25__ subtract__30.0__20.0__ subtract__20.0__10.0__ |
multiply__150.0__0.2__ subtract__150.0__30.0__ multiply__20.0__0.2__ subtract__30.0__20.0__ subtract__20.0__10.0__ |
| what is num__82.0 of num__0.6 ? <o> a ) num__6.9 <o> b ) num__0.492 <o> c ) num__0.6845 <o> d ) num__0.6859 <o> e ) num__0.69 |
num__82.0 * ( num__0.6 ) = num__0.82 * num__0.6 = num__0.492 answer : b <eor> b <eos> |
b |
percent__82.0__0.6__ percent__82.0__0.6__ |
percent__82.0__0.6__ percent__82.0__0.6__ |
| a train num__100 m long is running with a speed of num__68 kmph . in what time will it pass a man who is running at num__8 kmph in the same direction in which the train is going ? <o> a ) num__6 sec . <o> b ) num__7 sec . <o> c ) num__9 sec . <o> d ) num__11 sec . <o> e ) none |
solution speed of the train relative to man = ( num__68 - num__8 ) = num__60 kmph = num__60 x num__0.277777777778 = num__16.6666666667 m / sec . time taken by it to cover num__100 m at ( num__16.6666666667 ) m / sec = ( num__100 x num__0.06 ) sec = num__6 sec . answer a <eor> a <eos> |
a |
hour_to_min_conversion__ divide__100.0__16.6667__ round__6.0__ |
subtract__68.0__8.0__ divide__100.0__16.6667__ divide__100.0__16.6667__ |
| when num__242 is divided by a certain divisor the remainder obtained is num__15 . when num__698 is divided by the same divisor the remainder obtained is num__27 . however when the sum of the two numbers num__242 and num__698 is divided by the divisor the remainder obtained is num__5 . what is the value of the divisor ? <o> a ) num__11 <o> b ) num__17 <o> c ) num__13 <o> d ) num__23 <o> e ) num__37 |
let that divisor be x since remainder is num__15 or num__27 it means divisor is greater than num__27 . now num__242 - num__15 = num__227 = kx ( k is an integer and num__234 is divisble by x ) similarly num__698 - num__27 = num__671 = lx ( l is an integer and num__689 is divisible by x ) adding both num__698 and num__242 = ( num__227 + num__671 ) + num__15 + num__27 = x ( k + l ) + num__42 when we divide this number by x then remainder will be equal to remainder of ( num__42 divided by x ) = num__5 hence x = num__42 - num__5 = num__37 hence e <eor> e <eos> |
e |
subtract__242.0__15.0__ subtract__698.0__27.0__ add__15.0__27.0__ subtract__42.0__5.0__ subtract__42.0__5.0__ |
subtract__242.0__15.0__ subtract__698.0__27.0__ add__15.0__27.0__ subtract__42.0__5.0__ subtract__42.0__5.0__ |
| the difference of two numbers is num__1365 . on dividing the larger number by the smaller we get num__6 as quotient and the num__15 as remainder . what is the smaller number ? <o> a ) num__240 <o> b ) num__270 <o> c ) num__295 <o> d ) num__360 <o> e ) num__300 |
explanation : let the smaller number be x . then larger number = ( x + num__1365 ) . x + num__1365 = num__6 x + num__15 num__5 x = num__1350 x = num__270 smaller number = num__270 . answer is b <eor> b <eos> |
b |
subtract__1365.0__15.0__ divide__1350.0__5.0__ divide__1350.0__5.0__ |
subtract__1365.0__15.0__ divide__1350.0__5.0__ divide__1350.0__5.0__ |
| valerie buys num__3 new dresses on average every year for the last num__5 years . in order to increase her average to num__4 dresses per year how many dresses should valerie buy next year ? <o> a ) num__1 <o> b ) num__5 <o> c ) num__6 <o> d ) num__9 <o> e ) num__12 |
solution average of num__6 years = num__4 . â ˆ ´ required number of dresses = ( num__6 x num__4 ) - ( num__5 x num__3 ) = num__24 - num__15 = num__9 . answer d <eor> d <eos> |
d |
multiply__4.0__6.0__ multiply__3.0__5.0__ add__3.0__6.0__ add__3.0__6.0__ |
multiply__4.0__6.0__ multiply__3.0__5.0__ subtract__15.0__6.0__ subtract__15.0__6.0__ |
| sangakara and ponting select batting by using a dice but dice is biased . so to resolve ponting takes out a coin . what is the probability that coin shows correct option ? <o> a ) num__0.5 <o> b ) num__0.166666666667 <o> c ) num__0.0833333333333 <o> d ) num__0.6 <o> e ) num__0.3 |
as coin has only two faces then probability must be num__0.5 answer : a <eor> a <eos> |
a |
negate_prob__0.5__ |
negate_prob__0.5__ |
| the average temperature for monday tuesday wednesday and thursday was num__48 degrees and for tuesday wednesday thursday and friday was num__46 degrees . if the temperature on monday was num__42 degrees . find the temperature on friday ? <o> a ) num__34 <o> b ) num__66 <o> c ) num__25 <o> d ) num__18 <o> e ) num__12 |
explanation : m + tu + w + th = num__4 x num__48 = num__192 tu + w + th + f = num__4 x num__46 = num__184 m = num__42 tu + w + th = num__192 - num__42 = num__150 f = num__184 â € “ num__150 = num__34 answer : a <eor> a <eos> |
a |
subtract__46.0__42.0__ multiply__48.0__4.0__ multiply__46.0__4.0__ subtract__192.0__42.0__ subtract__184.0__150.0__ subtract__184.0__150.0__ |
subtract__46.0__42.0__ multiply__48.0__4.0__ multiply__46.0__4.0__ subtract__192.0__42.0__ subtract__184.0__150.0__ subtract__184.0__150.0__ |
| arnold and danny are two twin brothers that are celebrating their birthday . the product of their ages today is smaller by num__11 from the product of their ages a year from today . what is their age today ? <o> a ) num__2 . <o> b ) num__4 . <o> c ) num__5 . <o> d ) num__7 . <o> e ) num__9 . |
ad = ( a + num__1 ) ( d + num__1 ) - num__11 num__0 = a + d - num__10 a + d = num__10 a = d ( as they are twin brothers ) a = d = num__5 c is the answer <eor> c <eos> |
c |
subtract__11.0__1.0__ multiply__1.0__5.0__ |
subtract__11.0__1.0__ subtract__10.0__5.0__ |
| what is the remainder when num__1201 × num__1202 × num__1203 × num__1207 is divided by num__6 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
the remainders when dividing each number by six are : num__1 num__2 num__3 and num__1 . the product is num__1 * num__2 * num__3 * num__1 = num__6 the remainder when dividing num__6 by num__6 is num__0 . the answer is a . <eor> a <eos> |
a |
subtract__1202.0__1201.0__ subtract__1203.0__1201.0__ divide__6.0__2.0__ multiply__1201.0__0.0__ |
subtract__1202.0__1201.0__ subtract__1203.0__1201.0__ divide__6.0__2.0__ multiply__1201.0__0.0__ |
| if f ( x ) = num__7 x + num__12 what is f - num__1 ( x ) ( the inverse function ) ? <o> a ) ( x - num__12 ) / num__7 <o> b ) ( x - num__12 ) / num__6 <o> c ) ( x - num__12 ) / num__5 <o> d ) ( x - num__12 ) / num__4 <o> e ) ( x - num__12 ) / num__3 |
f ( x ) = num__7 x + num__12 take both side f - num__1 ( x ) f ( f - num__1 ( x ) ) = num__7 f - num__1 ( x ) + num__12 x = num__7 ( f - num__1 ( x ) ) + num__12 f - num__1 ( x ) = ( x - num__12 ) / num__7 answer : a <eor> a <eos> |
a |
multiply__12.0__1.0__ |
divide__12.0__1.0__ |
| the balance of a trader weighs num__10.0 less than it should . still the trader marks up his goods to get an overall profit of num__40.0 . what is the mark up on the cost price ? <o> a ) num__40.0 <o> b ) num__8.0 <o> c ) num__25.0 <o> d ) num__16.66 <o> e ) num__26 % |
the most natural way to deal with ' weights ' questions is by assuming values . say the trader ' s balance shows num__100 gms . it is actually num__90 gms because it weighs num__10.0 less . say the cost price is $ num__90 ( $ num__1 / gm ) . since he gets a profit of num__40.0 the selling price must be num__90 + ( num__0.4 ) * num__90 = $ num__126 since the cost price is actually supposed to be $ num__100 ( for num__100 gms ) and the selling price is $ num__126 the mark up is simply num__26.0 . ans : e <eor> e <eos> |
e |
percent__40.0__1.0__ percent__100.0__26.0__ |
percent__40.0__1.0__ percent__100.0__26.0__ |
| if m > num__1 and n = num__2 ^ ( m − num__3 ) then num__4 ^ m = <o> a ) num__16 n ^ num__2 <o> b ) num__4 n ^ num__2 <o> c ) num__2 n ^ num__2 <o> d ) num__8 n ^ num__2 <o> e ) num__64 n ^ num__2 |
n = num__2 ^ ( m - num__3 ) = num__2 ^ m / num__8 num__2 ^ m = num__8 n num__4 ^ m = ( num__2 ^ m ) ^ num__2 = ( num__8 n ) ^ num__2 = num__64 n ^ num__2 the answer is e . <eor> e <eos> |
e |
multiply__2.0__4.0__ multiply__1.0__64.0__ |
multiply__2.0__4.0__ divide__64.0__1.0__ |
| a furniture store owner decided to drop the price of her recliners by num__20.0 to spur business . by the end of the week she had sold num__60.0 more recliners . what is the percentage increase of the gross ? <o> a ) num__10.0 <o> b ) num__15.0 <o> c ) num__20.0 <o> d ) num__28.0 <o> e ) num__50 % |
say a recliner is actually worth $ num__100 if she sells num__100 recliners then she earns $ num__10000 after the discount of num__20.0 she will earn $ num__80 per recliner and she sells num__60.0 more ie . num__160 recliners hence her sales tields num__160 * num__80 = $ num__12800 increase in sales = num__12800 - num__10000 = $ num__2800 so % increase = num__2800 * num__0.01 = num__28.0 d is the answer <eor> d <eos> |
d |
add__20.0__60.0__ add__60.0__100.0__ multiply__160.0__80.0__ subtract__12800.0__10000.0__ reverse__100.0__ divide__2800.0__100.0__ divide__2800.0__100.0__ |
subtract__100.0__20.0__ add__60.0__100.0__ multiply__160.0__80.0__ subtract__12800.0__10000.0__ reverse__100.0__ multiply__2800.0__0.01__ multiply__2800.0__0.01__ |
| in an election only two candidates contested . a candidate secured num__70.0 of the valid votes and won by a majority of num__178 votes . find the total number of valid votes ? <o> a ) num__445 <o> b ) num__570 <o> c ) num__480 <o> d ) num__520 <o> e ) num__550 |
let the total number of valid votes be x . num__70.0 of x = num__0.7 * x = num__7 x / num__10 number of votes secured by the other candidate = x - num__7 x / num__100 = num__3 x / num__10 given num__7 x / num__10 - num__3 x / num__10 = num__178 = > num__4 x / num__10 = num__178 = > num__4 x = num__1780 = > x = num__445 . answer : a <eor> a <eos> |
a |
percent__100.0__445.0__ |
percent__100.0__445.0__ |
| if jack walked num__6 miles in num__1 hour and num__15 minutes what was his rate of walking in miles per hour ? <o> a ) num__4 <o> b ) num__4.8 <o> c ) num__6 <o> d ) num__6.25 <o> e ) num__15 |
distance walked in num__1 hour and num__15 mins = num__6 miles speed per hour = distance / time = num__6 / ( num__1.25 ) = num__4.8 miles per hour answer b <eor> b <eos> |
b |
divide__6.0__1.25__ round__4.8__ |
divide__6.0__1.25__ divide__6.0__1.25__ |
| a man has some hens and cows . if the number of heads be num__100 and the number of feet equals num__180 then the number of hens will be : <o> a ) num__22 <o> b ) num__23 <o> c ) num__80 <o> d ) num__90 <o> e ) num__28 |
explanation : let the number of hens be x and the number of cows be y . then x + y = num__100 . . . . ( i ) and num__2 x + num__4 y = num__220 x + num__2 y = num__110 . . . . ( ii ) solving ( i ) and ( ii ) we get : x = num__90 y = num__10 . the required answer = num__90 . answer : d <eor> d <eos> |
d |
divide__220.0__2.0__ divide__180.0__2.0__ subtract__100.0__90.0__ subtract__100.0__10.0__ |
divide__220.0__2.0__ divide__180.0__2.0__ subtract__100.0__90.0__ subtract__100.0__10.0__ |
| a and b together can complete a piece of work in num__4 days . if a alone can complete the same work in num__12 days in how many days can b alone complete that work ? <o> a ) num__22 days <o> b ) num__4 days <o> c ) num__13 days <o> d ) num__6 days <o> e ) num__9 days |
( a + b ) ' s num__1 day ' s work = ( num__0.25 ) . a ' s num__1 day ' s work = ( num__0.0833333333333 ) . b ' s num__1 day ' s work = ( ( num__0.25 ) - ( num__0.0833333333333 ) ) = ( num__0.166666666667 ) hence b alone can complete the work in num__6 days . answer is d . <eor> d <eos> |
d |
divide__1.0__4.0__ divide__1.0__12.0__ subtract__0.25__0.0833__ round__6.0__ |
divide__1.0__4.0__ divide__1.0__12.0__ subtract__0.25__0.0833__ subtract__12.0__6.0__ |
| bill downloads the movierevenge of the avengersto his computer in num__2.5 hours using a download manager that downloads from num__3 sources marked a b and c . each source provides download at a constant rate but the rates of different sources are not necessarily identical . if the movie was downloaded from sources a and c alone it would take num__4 hours to complete the download . the next day source b is available but the other sources are inactive . how long will it take to download the trailer of the movie a file that is num__16 times smaller from source b alone ? <o> a ) num__6 hours and num__40 minutes <o> b ) num__15 minutes <o> c ) num__12 minutes <o> d ) num__10 minutes <o> e ) num__25 minutes |
let the movie size be num__400 u . given a + c = num__4 hrs . a + c = num__100 u / hr and a + b + c = num__2.5 hrs or num__400 / num__2.5 = num__160 u / hr b alone = num__160 - num__100 = num__60 u / hr trailer = num__40 times smaller or num__25.0 = num__25 u b will take num__0.416666666667 hrs or num__25 minutes . ans e <eor> e <eos> |
e |
divide__400.0__4.0__ divide__400.0__2.5__ subtract__160.0__100.0__ multiply__2.5__16.0__ divide__100.0__4.0__ divide__25.0__60.0__ divide__100.0__4.0__ |
divide__400.0__4.0__ divide__400.0__2.5__ subtract__160.0__100.0__ divide__160.0__4.0__ divide__100.0__4.0__ divide__25.0__60.0__ divide__100.0__4.0__ |
| one pipe can fill a pool three times faster than a second pipe . when both pipes are opened they fill the pool in seven hours . how long would it take to fill the pool if only the faster pipe is used ? <o> a ) num__3.6 hr <o> b ) num__4.7 hr <o> c ) num__5.1 hr <o> d ) num__9.3 hr <o> e ) num__9.7 hr |
fast pipe + slow pipe = together one pipe fills the pool three times faster than the other pipe so hourly fill rates : num__1 / x + num__0.333333333333 x = num__0.142857142857 let x = number of hours it would take the fast pipe to fill the pool thus x = num__9.3 hours answer : d <eor> d <eos> |
d |
round__9.3__ |
divide__9.3__1.0__ |
| num__1 num__13 num__21 num__53 num__410 num__42 ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__12 <o> d ) num__13 <o> e ) num__14 |
num__1 + num__1 + num__1 = num__3 num__2 + num__1 + num__2 = num__5 num__3 + num__4 + num__3 = num__10 num__4 + num__2 + num__4 = num__10 answer : a <eor> a <eos> |
a |
divide__42.0__21.0__ add__2.0__3.0__ add__1.0__3.0__ subtract__13.0__3.0__ multiply__1.0__10.0__ |
divide__42.0__21.0__ add__2.0__3.0__ add__1.0__3.0__ subtract__13.0__3.0__ multiply__1.0__10.0__ |
| what is the least common multiple of num__12 num__18 and num__24 ? <o> a ) num__72 <o> b ) num__120 <o> c ) num__240 <o> d ) num__360 <o> e ) num__720 |
let us first write the numbers in the form of prime factors : num__12 = num__2 ^ num__2 * num__3 ^ num__1 num__18 = num__2 ^ num__1 * num__3 ^ num__2 num__24 = num__2 * num__17 ^ num__1 the lcm would be the largest powers of the prime numbers from all these three numbers . hence lcm = num__72 option a <eor> a <eos> |
a |
divide__24.0__12.0__ subtract__3.0__2.0__ subtract__18.0__1.0__ multiply__24.0__3.0__ multiply__24.0__3.0__ |
divide__24.0__12.0__ subtract__3.0__2.0__ subtract__18.0__1.0__ multiply__24.0__3.0__ multiply__24.0__3.0__ |
| p alone can complete a piece of work in num__6 days . work done by q alone in one day is equal to one - third of the work done by p alone in one day . in how many days can the work be completed if p and q work together ? <o> a ) num__4 ( num__0.75 ) <o> b ) num__7 <o> c ) num__4 <o> d ) num__5 <o> e ) num__4 ( num__0.5 ) |
option e explanation : work done by p alone in one day = num__0.166666666667 th of the total work done by q alone in one day = num__0.333333333333 ( of that done by p in one day ) = num__0.333333333333 ( num__0.166666666667 of the total ) = num__0.0555555555556 of the total . work done by p and q working together in one day = num__0.166666666667 + num__0.0555555555556 = num__0.222222222222 = num__0.222222222222 of the total they would take num__4.5 days = num__4 ( num__0.5 ) days to complete the work working together . <eor> e <eos> |
e |
multiply__0.1667__0.3333__ subtract__4.5__4.0__ round__4.0__ |
multiply__0.1667__0.3333__ add__0.1667__0.3333__ round__4.0__ |
| if integer k is equal to the sum of all even multiples of num__15 between num__390 and num__615 what is the greatest prime factor of k ? <o> a ) num__5 <o> b ) num__7 <o> c ) num__8 <o> d ) num__13 <o> e ) num__17 |
if we break down what the stem is asking what is the sum of all mult of num__30 between num__390 and num__600 . using arithmetic progression to find n : num__600 = num__390 + ( n - num__1 ) num__30 num__210 + num__30 = num__30 n num__240 = num__3 n = > n = num__8 the sum would be : num__8 * mean mean = [ num__600 + num__390 ] / num__2 = num__495 num__8 * num__495 c <eor> c <eos> |
c |
subtract__615.0__15.0__ subtract__600.0__390.0__ add__210.0__30.0__ divide__240.0__30.0__ subtract__3.0__1.0__ multiply__1.0__8.0__ |
subtract__615.0__15.0__ subtract__600.0__390.0__ add__210.0__30.0__ divide__240.0__30.0__ subtract__3.0__1.0__ multiply__1.0__8.0__ |
| if num__6 men and num__2 boys working together can do four times as much work per hour as a man and a boy together . find the ratio of the work done by a man and that of a boy for a given time ? <o> a ) num__3 : num__2 <o> b ) num__2 : num__2 <o> c ) num__3 : num__4 <o> d ) num__1 : num__2 <o> e ) num__6 : num__1 |
b num__2 : num__2 num__6 m + num__2 b = num__4 ( num__1 m + num__1 b ) num__6 m + num__2 b = num__4 m + num__4 b num__2 m = num__2 b the required ratio of work done by a man and a boy = num__2 : num__2 <eor> b <eos> |
b |
subtract__6.0__2.0__ round__2.0__ |
subtract__6.0__2.0__ round__2.0__ |
| how long does a train num__150 m long traveling at num__50 kmph takes to cross a bridge of num__250 m in length ? <o> a ) num__28.8 sec <o> b ) num__16.8 sec <o> c ) num__15.2 sec <o> d ) num__25.4 sec <o> e ) num__16.2 sec |
d = num__150 + num__250 = num__400 m s = num__50 * num__0.277777777778 = num__13.8888888889 t = num__400 * num__13.8888888889 = num__28.8 sec answer : a <eor> a <eos> |
a |
add__150.0__250.0__ divide__400.0__13.8889__ round__28.8__ |
add__150.0__250.0__ divide__400.0__13.8889__ round__28.8__ |
| how many integers are divisible by num__5 between num__10 ! and num__10 ! + num__20 inclusive ? <o> a ) num__5 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
a - num__7 num__10 ! is divisible by num__5 there are num__4 numbers between num__10 ! and num__10 ! + num__20 that are divisible by num__5 . hence num__5 <eor> a <eos> |
a |
divide__20.0__5.0__ subtract__10.0__5.0__ |
divide__20.0__5.0__ subtract__10.0__5.0__ |
| the average of first num__17 even numbers is ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__18 <o> d ) num__20 <o> e ) num__24 |
sum of num__17 even numbers = num__17 * num__18 = num__306 average = num__18.0 = num__18 answer : c <eor> c <eos> |
c |
multiply__17.0__18.0__ divide__306.0__17.0__ |
multiply__17.0__18.0__ divide__306.0__17.0__ |
| on the coordinate plane points p and u are defined by the coordinates ( - num__10 ) and ( num__33 ) respectively and are connected to form a chord of a circle which also lies on the plane . if the area of the circle is ( num__6.25 ) π what are the coordinates of the center of the circle ? <o> a ) ( num__1.5 num__1 ) <o> b ) ( num__2 - num__5 ) <o> c ) ( num__00 ) <o> d ) ( num__1 num__1.5 ) <o> e ) ( num__22 ) |
although it took me num__3 mins to solve this question using all those equations later i thought this question can be solved easily using options . one property to keep in mind - a line passing through the centre of the circle bisects the chord ( or passes from the mid point of the chord ) . now mid point of chord here is ( - num__1 + num__3 ) / num__2 ( num__3 + num__0 ) / num__2 i . e . ( num__1 num__1.5 ) now luckily we have this in our ans . choice . so definitely this is the ans . it also indictaes that pu is the diameter of the circle . there can be a case when pu is not a diameter but in that case also the y - coordinate will remain same as it is the midpoint of the chord and we are moving up in the st . line to locate the centre of the circle . if ans choices are all distinct ( y cordinates ) only check for y cordinate and mark the ans = d <eor> d <eos> |
d |
triangle_area__1.0__3.0__ volume_cube__1.0__ |
triangle_area__1.0__3.0__ volume_cube__1.0__ |
| cheese bologna and peanut butter sandwiches were made for a picnic in a ratio of num__5 to num__7 to num__8 . if a total of num__180 sandwiches were made how many bologna sandwiches were made ? <o> a ) num__63 <o> b ) num__30 <o> c ) num__38 <o> d ) num__42 <o> e ) num__48 |
for deciding such task we should calculate all parts num__5 + num__7 + num__8 = num__20 parts and we should calculate how many sandwiches holds num__1 part : num__9.0 = num__9 sandwiches in one part for bologna we have num__7 parts so : num__7 * num__9 = num__63 answer is a <eor> a <eos> |
a |
subtract__8.0__7.0__ add__8.0__1.0__ multiply__7.0__9.0__ multiply__7.0__9.0__ |
subtract__8.0__7.0__ add__8.0__1.0__ multiply__7.0__9.0__ multiply__7.0__9.0__ |
| at num__6 ' o clock clock ticks num__6 times . the time between first and last ticks was num__30 sec . how much time it takes at num__10 ' o clock . <o> a ) num__60 sec <o> b ) num__62 sec <o> c ) num__66 sec <o> d ) num__64 sec <o> e ) num__54 sec |
at num__6 ' num__0 clock clock ticks num__6 times . so there must be num__5 intervals between clock ticks . time between first and last ticks = num__30 sec so num__1 interval = num__6.0 = num__6 sec so num__6 ' o clock num__5 * num__6 = num__30 sec num__7 ' o clock num__6 * num__6 = num__36 sec . . num__9 ' o clock num__8 * num__6 = num__48 sec num__10 ' o clock num__9 * num__6 = num__54 sec so num__54 sec at num__10 ' o clock . answer : e <eor> e <eos> |
e |
divide__30.0__6.0__ subtract__6.0__5.0__ add__6.0__1.0__ add__6.0__30.0__ subtract__10.0__1.0__ add__1.0__7.0__ multiply__6.0__8.0__ multiply__6.0__9.0__ multiply__6.0__9.0__ |
divide__30.0__6.0__ subtract__6.0__5.0__ add__6.0__1.0__ add__6.0__30.0__ subtract__10.0__1.0__ add__1.0__7.0__ multiply__6.0__8.0__ multiply__6.0__9.0__ multiply__6.0__9.0__ |
| if one length of a square is tripled and the original parameter is num__4 y . what is the new length of one of the sides ? <o> a ) num__12 y / num__3 <o> b ) num__2 y / num__2 <o> c ) num__6 y / num__3 <o> d ) num__12 y / num__4 <o> e ) num__4 y / num__3 |
if the original parameter is num__4 y each length of the square is num__4 y / num__4 . if the parameter is tripled then each length becomes num__3 ( num__4 y ) / num__4 . this is num__12 y / num__4 . answer option d . <eor> d <eos> |
d |
multiply__4.0__3.0__ round__12.0__ |
multiply__4.0__3.0__ round__12.0__ |
| if log num__27 = num__1.434 then the value of log num__9 is : <o> a ) num__0.934 <o> b ) num__0.945 <o> c ) num__0.956 <o> d ) num__0.958 <o> e ) num__0.987 |
log num__27 = num__1.434 log ( num__33 ) = num__1.434 num__3 log num__3 = num__1.434 log num__3 = num__0.478 log num__9 = log ( num__32 ) = num__2 log num__3 = ( num__2 x num__0.478 ) = num__0.956 . answer : option c <eor> c <eos> |
c |
divide__27.0__9.0__ divide__1.434__3.0__ subtract__1.434__0.478__ subtract__1.434__0.478__ |
divide__27.0__9.0__ divide__1.434__3.0__ subtract__1.434__0.478__ subtract__1.434__0.478__ |
| a boat travels num__72 km downstream in num__8 hours and num__84 km upstream in num__12 hours . find the speed of the boat in still water and the speed of the water current ? <o> a ) num__6 and num__1 kmph <o> b ) num__8 and num__2 kmph <o> c ) num__8 and num__1 kmph <o> d ) num__9 and num__1 kmph <o> e ) num__2 and num__1 kmph |
explanation : downstream speed = num__72 km / num__8 hrs = num__9 kmph upstream speed = num__84 km / num__12 hrs = num__7 kmph speed of boat = avg of downstream and upstream speeds speed of boat = ( num__9 + num__7 ) / num__2 kmph = num__8 kmph . current speed = half of the difference of downstream and upstream speeds currend speed = ( num__9 - num__7 ) / num__2 kmph = num__1 kmph answer : c <eor> c <eos> |
c |
divide__72.0__8.0__ divide__84.0__12.0__ subtract__9.0__7.0__ subtract__8.0__7.0__ round__8.0__ |
divide__72.0__8.0__ divide__84.0__12.0__ subtract__9.0__7.0__ subtract__8.0__7.0__ divide__72.0__9.0__ |
| if rupee one produces rupees nine over a period of num__40 years find the rate of simple interest ? <o> a ) num__22 num__0.125 % <o> b ) num__22 num__4.5 % <o> c ) num__22 num__0.142857142857 % <o> d ) num__22 num__0.5 % <o> e ) num__21 num__0.5 % |
num__9 = ( num__1 * num__40 * r ) / num__100 r = num__22 num__0.5 % answer : d <eor> d <eos> |
d |
percent__100.0__22.0__ |
percent__100.0__22.0__ |
| solve the equation for x : num__6 x - num__27 + num__3 x = num__4 + num__9 - x <o> a ) num__4 <o> b ) num__5 <o> c ) num__7 <o> d ) num__9 <o> e ) num__3 |
a num__4 num__9 x + x = num__13 + num__27 num__10 x = num__40 = > x = num__4 <eor> a <eos> |
a |
add__4.0__9.0__ add__6.0__4.0__ add__27.0__13.0__ divide__40.0__10.0__ |
add__4.0__9.0__ add__6.0__4.0__ add__27.0__13.0__ divide__40.0__10.0__ |
| in a fraction if num__1 is added to both the numerator at the denominator the fraction becomes num__0.5 . if numerator is subtracted from the denominator the fraction becomes num__0.75 . find the fraction . <o> a ) num__0.428571428571 <o> b ) num__0.571428571429 <o> c ) num__0.714285714286 <o> d ) num__0.857142857143 <o> e ) num__1.14285714286 |
let fraction be a / b where a and b are integers . ` ` if num__1 is added to both the numerator and denominator the fraction becomes num__0.5 ' ' ( a + num__1 ) / ( b + num__1 ) = num__0.5 num__2 ( a + num__1 ) = b + num__1 num__2 a + num__2 = b + num__1 num__2 a - b = - num__1 ` ` if numerator is subtracted from the denominator the fraction becomes num__0.75 ' ' a / ( b - a ) = num__0.75 num__4 a = num__3 ( b - a ) num__4 a = num__3 b - num__3 a num__4 a + num__3 a - num__3 b = num__0 num__7 a - num__3 b = num__0 now we just need to solve system of linear equations : num__2 a - b = - num__1 . . . [ num__1 ] num__7 a - num__3 b = num__0 . . . [ num__2 ] from [ num__1 ] we get : b = num__2 a + num__1 substituting in this value of b into [ num__2 ] we get num__7 a - num__3 ( num__2 a + num__1 ) = num__0 num__7 a - num__6 a - num__3 = num__0 a - num__3 = num__0 a = num__3 b = num__2 ( num__3 ) + num__1 b = num__7 so original fraction is a / b = num__0.428571428571 answer : a <eor> a <eos> |
a |
reverse__0.5__ divide__2.0__0.5__ add__1.0__2.0__ round_down__0.5__ add__3.0__4.0__ multiply__2.0__3.0__ divide__3.0__7.0__ multiply__1.0__0.4286__ |
reverse__0.5__ divide__2.0__0.5__ add__1.0__2.0__ round_down__0.5__ add__3.0__4.0__ add__2.0__4.0__ divide__3.0__7.0__ divide__3.0__7.0__ |
| given f ( x ) = num__3 x – num__5 for what value of x does num__2 * [ f ( x ) ] – num__10 = f ( x – num__2 ) ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
num__2 ( num__3 x - num__5 ) - num__10 = num__3 ( x - num__2 ) - num__5 num__3 x = num__9 x = num__3 the answer is c . <eor> c <eos> |
c |
subtract__5.0__2.0__ |
subtract__5.0__2.0__ |
| can you get total of num__720 by using six zeros num__00 num__00 num__00 and any maths operators . <o> a ) num__650 <o> b ) num__550 <o> c ) num__700 <o> d ) num__720 <o> e ) num__500 |
d num__720 ( num__0 ! + num__0 ! + num__0 ! + num__0 ! + num__0 ! + num__0 ! ) ! = ( num__1 + num__1 + num__1 + num__1 + num__1 + num__1 ) ! = ( num__5 ) ! = num__6 ! is num__6 * num__5 * num__4 * num__3 * num__2 * num__1 = num__720 <eor> d <eos> |
d |
add__1.0__5.0__ subtract__5.0__1.0__ subtract__4.0__1.0__ subtract__3.0__1.0__ multiply__720.0__1.0__ |
add__1.0__5.0__ subtract__5.0__1.0__ subtract__4.0__1.0__ subtract__3.0__1.0__ multiply__720.0__1.0__ |
| a four - digit code can consist of the digits num__0 ~ num__9 except that it does not contain the digits num__1 num__2 and num__4 at all . if repeated digits are allowed what is the probability that it has at least one even digit ? <o> a ) num__0.361596009975 <o> b ) num__0.937846836848 <o> c ) num__0.81727337616 <o> d ) num__0.893377759267 <o> e ) num__0.92472802117 |
p ( no even digits ) = num__0.571428571429 * num__0.571428571429 * num__0.571428571429 * num__0.571428571429 = num__0.106622240733 p ( at least one even digit ) = num__1 - num__0.106622240733 = num__0.893377759267 the answer is d . <eor> d <eos> |
d |
subtract__1.0__0.1066__ multiply__1.0__0.8934__ |
subtract__1.0__0.1066__ multiply__1.0__0.8934__ |
| an average age of a b and c is num__45 . if the averageage of a and b be num__40 and that of b and c be num__43 then the weight of b is <o> a ) num__25 <o> b ) num__29 <o> c ) num__31 <o> d ) num__35 <o> e ) num__36 |
let a b c represent their respective age ' s . then we have : a + b + c = ( num__45 x num__3 ) = num__135 . . . . ( i ) a + b = ( num__40 x num__2 ) = num__80 . . . . ( ii ) b + c = ( num__43 x num__2 ) = num__86 . . . . ( iii ) adding ( ii ) and ( iii ) we get : a + num__2 b + c = num__166 . . . . ( iv ) subtracting ( i ) from ( iv ) we get : b = num__31 . b ' s age = num__31 c <eor> c <eos> |
c |
subtract__43.0__40.0__ multiply__45.0__3.0__ subtract__45.0__43.0__ multiply__40.0__2.0__ multiply__43.0__2.0__ add__80.0__86.0__ subtract__166.0__135.0__ subtract__166.0__135.0__ |
subtract__43.0__40.0__ multiply__45.0__3.0__ subtract__45.0__43.0__ multiply__40.0__2.0__ multiply__43.0__2.0__ add__80.0__86.0__ subtract__166.0__135.0__ subtract__166.0__135.0__ |
| find the principle on a certain sum of money at num__5.0 per annum for num__2 num__0.4 years if the amount being rs . num__1456 ? <o> a ) num__1000 <o> b ) num__2217 <o> c ) num__2889 <o> d ) num__2777 <o> e ) num__1300 |
num__1456 = p [ num__1 + ( num__5 * num__2.4 ) / num__100 ] p = num__1300 answer : e <eor> e <eos> |
e |
add__2.0__0.4__ multiply__1.0__1300.0__ |
add__2.0__0.4__ multiply__1.0__1300.0__ |
| a bar over a sequence of digits in a decimal indicates that the sequence repeats indefinitely . what is the value of ( num__10 ^ num__4 - num__10 ^ num__2 ) ( num__0.0018 ) ? <o> a ) num__16.82 <o> b ) num__19.82 <o> c ) num__17.82 <o> d ) num__20 <o> e ) num__22 |
you get num__10 ^ num__2 ( num__100 - num__1 ) ( . num__0018 ) we know num__0.0018 = . num__0018 num__10 ^ num__2 ( num__99 ) * ( num__0.0018 ) num__10 ^ num__2 = num__100 and num__100 ^ num__2 = num__10000 cancel out the num__100 with the num__10 ^ num__2 left with num__99 ( num__0.18 ) . num__0.99 * num__18 = num__17.82 answer : c <eor> c <eos> |
c |
power__10.0__2.0__ subtract__100.0__1.0__ power__10.0__4.0__ multiply__0.0018__100.0__ divide__99.0__100.0__ multiply__99.0__0.18__ multiply__1.0__17.82__ |
power__10.0__2.0__ subtract__100.0__1.0__ power__10.0__4.0__ multiply__0.0018__100.0__ divide__99.0__100.0__ multiply__99.0__0.18__ multiply__1.0__17.82__ |
| find the option to replace the question mark in the series below num__5 ? num__15 num__75 num__525 num__4725 <o> a ) num__2 <o> b ) num__6 <o> c ) num__7 <o> d ) num__5 <o> e ) num__8 |
ol : num__5 x num__1 = num__5 num__5 x num__3 = num__15 num__15 x num__5 = num__75 num__75 x num__7 = num__525 num__525 x num__9 = num__4725 so ? = num__5 answer : d <eor> d <eos> |
d |
divide__15.0__5.0__ divide__525.0__75.0__ divide__4725.0__525.0__ multiply__5.0__1.0__ |
divide__15.0__5.0__ divide__525.0__75.0__ divide__4725.0__525.0__ multiply__5.0__1.0__ |
| if one and a half women drink one and a half tea in one and a half minutes . how many tea can num__9 women drink in num__3 minutes ? <o> a ) num__11 <o> b ) num__8 <o> c ) num__37 <o> d ) num__15 <o> e ) num__18 |
e num__18 explanation : more minutes implies more tea . more women implies more tea . the time become twice ( num__3 minutes / num__1.5 minutes ) . women become six times ( num__9 / num__1.5 ) . number of tea = num__2 * num__6 * original ( num__1.5 ) <eor> e <eos> |
e |
divide__3.0__1.5__ subtract__9.0__3.0__ round__18.0__ |
divide__3.0__1.5__ divide__9.0__1.5__ multiply__9.0__2.0__ |
| a shop sells bicycles and tricycles . in total there are num__10 cycles ( cycles include both bicycles and tricycles ) and num__21 wheels . determine how many of each there are if a bicycle has two wheels and a tricycle has three wheels . <o> a ) b = num__9 t = num__1 <o> b ) b = num__7 t = num__3 <o> c ) b = num__6 t = num__2 <o> d ) b = num__2 t = num__5 <o> e ) b = num__6 t = num__4 |
let b be the number of bicycles and let t be the number of tricycles . set up the equations b + t = num__10 . . . . . . . . . . . . ( num__1 ) num__2 b + num__3 t = num__21 . . . . . . . . . . . . ( num__2 ) rearrange equation ( num__1 ) and substitute into equation ( num__2 ) num__2 b + num__30 - num__3 b = num__21 - b = - num__9 b = num__9 calculate the number of tricycles t t = num__10 − b = num__10 − num__9 = num__1 there are num__1 tricycles and num__9 bicycles . answer is a . <eor> a <eos> |
a |
add__1.0__2.0__ multiply__10.0__3.0__ subtract__10.0__1.0__ subtract__10.0__1.0__ |
add__1.0__2.0__ multiply__10.0__3.0__ subtract__10.0__1.0__ subtract__10.0__1.0__ |
| which of the following are roots of an equation ( x ^ - num__2 ) + ( num__3 x ^ - num__1 ) - num__10 = num__0 <o> a ) num__0.2 and - num__0.5 <o> b ) - num__0.2 and num__0.5 <o> c ) num__0.2 and num__0.5 <o> d ) - num__0.2 and - num__0.5 <o> e ) - num__2.5 and - num__0.5 |
given : ( x ^ - num__2 ) + ( num__3 x ^ - num__1 ) - num__10 = num__0 rewrite as : num__1 / ( x ² ) + num__3 / x - num__10 = num__0 remove fractions by multiplying both sides by x ² to get : num__1 + num__3 x - num__10 x ² = num__0 rearrange to get : num__10 x ² - num__3 x - num__1 = num__0 factor to get : ( num__5 x + num__1 ) ( num__2 x - num__1 ) = num__0 so either num__5 x + num__1 or num__2 x - num__1 = num__0 if num__5 x + num__1 = num__0 then x = - num__0.2 if num__2 x - num__1 = num__0 then x = num__0.5 so the roots ( solutions ) are - num__0.2 and num__0.5 the answer is b . <eor> b <eos> |
b |
add__2.0__3.0__ reverse__5.0__ reverse__2.0__ reverse__5.0__ |
add__2.0__3.0__ reverse__5.0__ reverse__2.0__ reverse__5.0__ |
| rose is two years older than bruce who is twice as old as chris . if the total of the ages of rose b and chris be num__27 the how old is bruce ? <o> a ) num__8 years <o> b ) num__10 years <o> c ) num__12 years <o> d ) num__13 years <o> e ) num__14 years |
let chris ' s age be x years . then bruce ' s age = num__2 x years . rose ' s age = ( num__2 x + num__2 ) years . ( num__2 x + num__2 ) + num__2 x + x = num__27 num__5 x = num__25 x = num__5 . hence bruce ' s age = num__2 x = num__10 years . b <eor> b <eos> |
b |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
| how many shots of num__1 cm radius can be prepared from a sphere of num__7 cm radius ? <o> a ) num__343 <o> b ) num__388 <o> c ) num__327 <o> d ) num__88 <o> e ) num__99 |
num__1.33333333333 Ï € * num__7 * num__7 * num__7 = num__1.33333333333 Ï € * num__1 * num__1 * num__1 * x x = num__343 answer : a <eor> a <eos> |
a |
round__343.0__ |
multiply__1.0__343.0__ |
| seed mixture x is num__40 percent ryegrass and num__60 percent bluegrass by weight ; seed mixture y is num__25 percent ryegrass and num__75.0 fescue . if a mixture of x and y contains num__30.0 ryegrass what percent of the weight of the mixture is x ? <o> a ) num__10.0 <o> b ) num__33 num__0.333333333333 % <o> c ) num__40.0 <o> d ) num__50.0 <o> e ) num__66 num__0.666666666667 % |
assuming the weight of the mixture to be num__100 g * * then the weight of ryegrass in the mixture would be num__30 g . also assume the weight mixture x used in the mixture is xg then the weight of mixture y used in the mixture would be ( num__100 - x ) g . so we can now equate the parts of the ryegrass in the mixture as : num__0.4 x + num__0.25 ( num__100 - x ) = num__30 num__0.4 x + num__25 - num__0.25 x = num__30 num__0.15 x = num__5 x = num__5 / num__0.15 = num__33.3333333333 = num__33.3333333333 so the weight of mixture x as a percentage of the weight of the mixture = ( weight of x / weight of mixture ) * num__100.0 = ( num__33.3333333333 ) / num__100 * num__100.0 = num__33.0 answer : b <eor> b <eos> |
b |
percent__60.0__0.25__ percent__33.0__100.0__ |
percent__60.0__0.25__ percent__33.0__100.0__ |
| the area of a square is equal to five times the area of a rectangle of dimensions num__125 cm * num__64 cm . what is the perimeter of the square ? <o> a ) num__700 <o> b ) num__800 <o> c ) num__900 <o> d ) num__600 <o> e ) num__500 |
area of the square = s * s = num__5 ( num__125 * num__64 ) = > s = num__25 * num__8 = num__200 cm perimeter of the square = num__4 * num__200 = num__800 cm . answer : option b <eor> b <eos> |
b |
multiply__8.0__25.0__ square_perimeter__200.0__ square_perimeter__200.0__ |
multiply__8.0__25.0__ multiply__4.0__200.0__ multiply__4.0__200.0__ |
| the expression ( num__11.98 x num__11.98 + num__11.98 x x + num__0.02 x num__0.02 ) will be a perfect square for x equal to : <o> a ) num__0.02 <o> b ) num__0.2 <o> c ) num__0.04 <o> d ) num__0.4 <o> e ) none |
explanation given expression = ( num__11.98 ) num__2 + ( num__0.02 ) num__2 + num__11.98 x x . for the given expression to be a perfect square we must have num__11.98 x x = num__2 x num__11.98 x num__0.02 or x = num__0.04 answer c <eor> c <eos> |
c |
multiply__0.02__2.0__ multiply__0.02__2.0__ |
multiply__0.02__2.0__ multiply__0.02__2.0__ |
| sreenivas sells a table to shiva at num__10.0 profit and shiva sells it to mahesh at num__10.0 loss . at what price did sreenivas purchase the table if mahesh paid rs . num__2178 ? <o> a ) num__2887 <o> b ) num__2665 <o> c ) num__2200 <o> d ) num__2665 <o> e ) num__2771 |
let the cost price of table for sreenivas be rs . x and given that cost price of table for mahesh = rs . num__2178 . = > ( num__90.0 ) of ( num__110.0 ) of x = rs . num__2178 . = > ( num__0.9 ) ( num__1.1 ) x = num__2178 = > x = ( num__2178 * num__100 ) / ( num__9 * num__11 ) = > x = rs . num__2200 answer : c <eor> c <eos> |
c |
percent__10.0__90.0__ percent__10.0__110.0__ percent__100.0__2200.0__ |
percent__10.0__90.0__ percent__10.0__110.0__ percent__100.0__2200.0__ |
| the ages of x and y are in the proportion of num__6 : num__5 and total of their ages is num__44 years . the proportion of their ages after num__8 years will be <o> a ) num__3 : num__6 <o> b ) num__6 : num__3 <o> c ) num__8 : num__7 <o> d ) num__9 : num__5 <o> e ) none of these |
explanation : let current ages of x and y correspondingly is num__6 a & num__5 a given : num__6 a + num__5 a = num__44 = > a = num__4 proportion of ages after num__0.8 decades will be num__6 a + num__8 : num__5 a + num__8 num__32 : num__28 ( or ) num__8 : num__7 . answer : c <eor> c <eos> |
c |
divide__4.0__5.0__ multiply__8.0__4.0__ subtract__32.0__4.0__ divide__28.0__4.0__ divide__32.0__4.0__ |
divide__4.0__5.0__ multiply__8.0__4.0__ subtract__32.0__4.0__ divide__28.0__4.0__ divide__32.0__4.0__ |
| the sum of the present ages of a b c is num__33 years . three years ago their ages were in the ratio num__1 : num__2 : num__3 . what is the present age of a ? <o> a ) num__5 <o> b ) num__7 <o> c ) num__9 <o> d ) num__11 <o> e ) num__13 |
three years ago : a : b : c = num__1 : num__2 : num__3 let a = num__1 x b = num__2 x and c = num__3 x . today : ( x + num__3 ) + ( num__2 x + num__3 ) + ( num__3 x + num__3 ) = num__33 x = num__4 so the present age of a is x + num__3 = num__7 the answer is b . <eor> b <eos> |
b |
add__1.0__3.0__ add__3.0__4.0__ multiply__1.0__7.0__ |
add__1.0__3.0__ add__3.0__4.0__ add__3.0__4.0__ |
| a shop owner has a sale of rs . num__6435 rs . num__6927 rs . num__6855 rs . num__7230 and rs . num__6562 for num__5 years . how much sale must she have in the num__6 th year so that she gets an average sale of rs . num__6500 ? <o> a ) num__5000 <o> b ) num__5500 <o> c ) num__4991 <o> d ) num__5600 <o> e ) num__5700 |
total sale for num__5 months = rs . ( num__6435 + num__6927 + num__6855 + num__7230 + num__6562 ) = rs . num__34009 . required sale = rs . [ ( num__6500 x num__6 ) - num__34009 ] = rs . ( num__39000 - num__34009 ) = rs . num__4991 c <eor> c <eos> |
c |
multiply__6.0__6500.0__ subtract__39000.0__34009.0__ subtract__39000.0__34009.0__ |
multiply__6.0__6500.0__ subtract__39000.0__34009.0__ subtract__39000.0__34009.0__ |
| the average age of husband wife and their child num__3 years ago was num__27 years and that of wife and the child num__7 years ago was num__20 years . the present age of the husband is : <o> a ) num__20 years <o> b ) num__30 years <o> c ) num__60 years <o> d ) num__36 years <o> e ) num__35 years |
d num__36 years sum of the present ages of husband wife and child = ( num__27 x num__3 + num__3 x num__3 ) years = num__90 years . sum of the present ages of wife and child = ( num__20 x num__2 + num__7 x num__2 ) years = num__54 years . husband ' s present age = ( num__90 - num__54 ) years = num__36 years . <eor> d <eos> |
d |
multiply__27.0__2.0__ subtract__90.0__54.0__ |
subtract__90.0__36.0__ subtract__90.0__54.0__ |
| in one hour a boat goes num__15 km / hr along the stream and num__9 km / hr against the stream . the speed of the boat in still water ( in km / hr ) is : <o> a ) num__12 kmph <o> b ) num__13 kmph <o> c ) num__14 kmph <o> d ) num__15 kmph <o> e ) num__16 kmph |
explanation : let the speed downstream be a km / hr and the speed upstream be b km / hr then speed in still water = num__0.5 ( a + b ) km / hr rate of stream = num__0.5 ( a − b ) km / hr speed in still water = num__0.5 ( num__15 + num__9 ) kmph = num__12 kmph . answer : option a <eor> a <eos> |
a |
round__12.0__ |
round__12.0__ |
| nitrous oxide ( laughing gas ) worth rs num__126 per kg and rs num__135 per kg are mixed with a third variety in the ratio num__1 : num__1 : num__2 . if the mixture is worth rs num__153 per kg then the price of the third variety per kg will be <o> a ) rs num__169.50 <o> b ) rs num__170 <o> c ) rs num__175.50 <o> d ) rs num__180 <o> e ) rs num__190 |
suppose the quantities of nitrous oxide worth rs num__126 per kg rs num__135 per kg and rs . x per kg purchased are y y and num__2 y kg respectively : ( num__126 y + num__135 y + num__2 xy ) / num__4 y = num__153 = > num__261 + num__2 x = num__612 = > x = num__175.5 = num__175.50 nitrous oxide of the third varity is purchased at the rate of rs num__175.50 per kg answer : c <eor> c <eos> |
c |
add__126.0__135.0__ multiply__153.0__4.0__ multiply__1.0__175.5__ |
add__126.0__135.0__ multiply__153.0__4.0__ divide__175.5__1.0__ |
| walking at num__25.0 of his usual speed a man takes num__24 minutes more to cover a distance . what is his usual time to cover this distance ? <o> a ) num__8 <o> b ) num__16 <o> c ) num__42 <o> d ) num__48 <o> e ) num__54 |
speed is inversly proprtional to time walking at num__25.0 of speed meand num__0.25 s takes num__4 t . it takes num__24 minutes extra to cover the distance . then num__4 t = t + num__24 num__3 t = num__24 t = num__8 . option a is correct <eor> a <eos> |
a |
divide__24.0__3.0__ round__8.0__ |
divide__24.0__3.0__ round__8.0__ |
| what is the least integer greater than – num__4 + num__0.5 ? <o> a ) – num__3 <o> b ) – num__1 <o> c ) num__0 <o> d ) num__1 <o> e ) num__2 |
this question is just about doing careful arithmetic and remembering what makes a numberbiggerorsmallercompared to another number . first let ' s take care of the arithmetic : ( - num__4 ) + ( num__0.5 ) = - num__3.5 on a number line since we ' re adding + . num__5 to a number the total moves to the right ( so we ' re moving from - num__4 to - num__3.5 ) . next the question asks for the least integer that is greater than - num__3.5 again we can use a number line . numbers become greater as you move to the right . the first integer to the right of - num__3.5 is - num__3 . final answer : a <eor> a <eos> |
a |
subtract__4.0__0.5__ round_down__3.5__ round_down__3.5__ |
subtract__4.0__0.5__ subtract__3.5__0.5__ subtract__3.5__0.5__ |
| a train passes a station platform in num__28 seconds and a man standing on the platform in num__10 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__270 <o> b ) num__277 <o> c ) num__240 <o> d ) num__88 <o> e ) num__112 |
speed = [ num__54 * num__0.277777777778 ] m / sec = num__15 m / sec . length of the train = ( num__15 * num__10 ) m = num__150 m . let the length of the platform be x meters . then x + num__5.35714285714 = num__15 x + num__150 = num__420 x = num__270 m . answer : a <eor> a <eos> |
a |
multiply__10.0__15.0__ divide__150.0__28.0__ multiply__28.0__15.0__ subtract__420.0__150.0__ round__270.0__ |
multiply__10.0__15.0__ divide__150.0__28.0__ multiply__28.0__15.0__ subtract__420.0__150.0__ round__270.0__ |
| the average of first num__10 odd numbers is ? <o> a ) num__11 <o> b ) num__10 <o> c ) num__18 <o> d ) num__12 <o> e ) num__19 |
explanation : sum of num__10 odd no . = num__100 average = num__10.0 = num__10 answer : b <eor> b <eos> |
b |
divide__100.0__10.0__ |
divide__100.0__10.0__ |
| if re . num__1 amounts to rs . num__9 over a period of num__20 years . what is the rate of simple interest ? <o> a ) num__27 <o> b ) num__29 <o> c ) num__26 <o> d ) num__40 <o> e ) num__12 |
num__8 = ( num__1 * num__20 * r ) / num__100 r = num__40.0 . answer : d <eor> d <eos> |
d |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| david completes a piece of work in num__5 days john completes the same work in num__9 days . if both of them work together then the number of days required to complete the work is ? <o> a ) num__3.2 days <o> b ) num__8 days <o> c ) num__10 days <o> d ) num__12 days <o> e ) num__14 days |
ifa can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days . that is the required no . of days = num__5 × num__0.642857142857 = num__3.2 days answer : a <eor> a <eos> |
a |
round__3.2__ |
round__3.2__ |
| car x began traveling at an average speed of num__35 miles per hour . after num__72 minutes car y began traveling at an average speed of num__49 miles per hour . when both cars had traveled the same distance both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ? <o> a ) num__105 <o> b ) num__120 <o> c ) num__140 <o> d ) num__147 <o> e ) num__168 |
car y began travelling after num__72 minutes or num__1.2 hours . let t be the time for which car y travelled before it stopped . both cars stop when they have travelled the same distance . so num__35 ( t + num__1.2 ) = num__49 t or num__5 t + num__6 = num__7 t or t = num__3 distance travelled by car x from the time car y began traveling until both cars stopped is num__35 x num__3 = num__105 miles answer : - a <eor> a <eos> |
a |
multiply__5.0__1.2__ divide__35.0__5.0__ multiply__35.0__3.0__ round__105.0__ |
multiply__5.0__1.2__ divide__35.0__5.0__ multiply__35.0__3.0__ round__105.0__ |
| the cost of fencing a square field @ rs . num__20 per metre is rs . num__10.080 . how much will it cost to lay a three meter wide pavement along the fencing inside the field @ rs . num__50 per sq m <o> a ) num__73889 <o> b ) num__27788 <o> c ) num__73879 <o> d ) num__73800 <o> e ) num__2799 |
explanation : perimeter = total cost / cost per m = num__504.0 = num__504 m side of the square = num__126.0 = num__126 m breadth of the pavement = num__3 m side of inner square = num__126 - num__6 = num__120 m area of the pavement = ( num__126 x num__126 ) - ( num__120 x num__120 ) = num__246 x num__6 sq m cost of pavement = num__246 * num__6 * num__50 = rs . num__73800 answer : d ) <eor> d <eos> |
d |
multiply__10.08__50.0__ multiply__20.0__6.0__ add__120.0__126.0__ round__73800.0__ |
multiply__10.08__50.0__ multiply__20.0__6.0__ add__120.0__126.0__ round__73800.0__ |
| an oratorical society consists of six members and at an upcoming meeting the members will present a total of four speeches . if no member presents more than two of the four speeches in how many different orders could the members give speeches ? <o> a ) num__720 <o> b ) num__1080 <o> c ) num__1170 <o> d ) num__1470 <o> e ) num__1560 |
you need to remember that of the num__3 people num__1 is going twice and two are going once . since one of the num__3 is going once there are num__3 c num__1 possibilities for the double speaker . so to convert your work to the correct answer you need to multiply : num__240 * num__3 c num__1 = num__240 * num__3 = num__720 . ans : a <eor> a <eos> |
a |
multiply__240.0__3.0__ multiply__240.0__3.0__ |
multiply__240.0__3.0__ multiply__240.0__3.0__ |
| andrew investe num__12 percent of his march earnings . in april he earned num__20 percent more than he did in march and he invested num__9 percent of his april earnings . the amount he invested in april was what percent of the amount he invested in march ? <o> a ) num__80.0 <o> b ) num__90.0 <o> c ) num__100.0 <o> d ) num__110.0 <o> e ) num__120 % |
let andrew ' s march earnings be num__100 - - > amount invested = num__12.0 of num__100 = num__12 andrews april earning = num__1.2 * num__100 = num__120 - - > amount invested = num__9.0 of num__120 = num__10.8 amount invested in april = ( num__10.8 / num__12 ) * num__100 = ( num__0.9 ) * num__100 = num__90.0 of the amount invested in march answer : b <eor> b <eos> |
b |
percent__9.0__120.0__ percent__100.0__90.0__ |
percent__9.0__120.0__ percent__100.0__90.0__ |
| kelly has had num__3 pay cuts in her salary in the past num__6 months . if the first pay cut was num__8.0 the second pay cut was num__14.0 and the third was num__18.0 . what will be the percentage decrease if the salary is decreased in a single shot ? <o> a ) num__35.9 <o> b ) num__34.12 <o> c ) num__32.12 <o> d ) num__35.12 <o> e ) num__31.12 % |
let rs . num__100 be initial salary . salary after num__1 st decrease num__8.0 = num__92 salary after num__2 nd decrease num__14.0 = num__79.12 i . e . reduced by num__14 percent of num__92 salary after num__3 rd decrease num__18.0 = num__64.88 i . e . reduced by num__18 percent of num__79.12 so if its decreased in single shot = i = ( ( b - a ) / b ) * num__100 = num__35.12 answer : d <eor> d <eos> |
d |
percent__100.0__35.12__ |
percent__100.0__35.12__ |
| find the num__12 th term of an arithmetic progression whose first term is num__2 and the common difference is num__8 . <o> a ) num__45 <o> b ) num__38 <o> c ) num__44 <o> d ) num__90 <o> e ) num__96 |
n th term of a . p = a + ( n - num__1 ) * d = num__2 + ( num__12 - num__1 ) * num__8 = num__2 + num__88 = num__90 . answer : d <eor> d <eos> |
d |
add__2.0__88.0__ add__2.0__88.0__ |
add__2.0__88.0__ add__2.0__88.0__ |
| a boat having a length num__3 m and breadth num__2 m is floating on a lake . the boat sinks by num__2 cm when a man gets on it . the mass of man is <o> a ) num__12 kg <o> b ) num__60 kg <o> c ) num__72 kg <o> d ) num__120 kg <o> e ) none |
solution volume of water displaced = ( num__3 x num__2 x num__0.02 ) m num__3 = num__0.12 m num__3 . mass of man = volume of water displaced × density of water = ( num__0.12 × num__1000 ) kg = num__120 kg . answer d <eor> d <eos> |
d |
multiply__1000.0__0.12__ round__120.0__ |
multiply__1000.0__0.12__ round__120.0__ |
| the numerator of a certain fraction is num__8 less than the denominator . if num__3 is added to the numerator and num__3 is subtracted from the denominator the fraction becomes num__0.75 . find the original fraction ? <o> a ) a ) num__24.5 <o> b ) b ) num__28.5 <o> c ) c ) num__30 <o> d ) inadequate <o> e ) of these |
explanation : let c . p . be rs . num__100 . then s . p . = rs . num__123.50 let marked price be rs . x . then num__0.95 x = num__123.50 x = num__130.0 = rs . num__130 now s . p . = rs . num__130 c . p . = rs . num__100 profit % = num__30.0 . answer : option c <eor> c <eos> |
c |
divide__123.5__0.95__ subtract__130.0__100.0__ subtract__130.0__100.0__ |
divide__123.5__0.95__ subtract__130.0__100.0__ subtract__130.0__100.0__ |
| a cistern can be filled by a tap in num__4 hours while it can be emptied by another tap in num__9 hours . if both the taps are opened simultaneously then after how much time will the cistern get filled ? <o> a ) num__7.5 <o> b ) num__7.4 <o> c ) num__7.9 <o> d ) num__7.2 <o> e ) num__7.1 |
net part filled in num__1 hour = ( num__0.25 - num__0.111111111111 ) = num__0.138888888889 the cistern will be filled in num__7.2 hrs i . e . num__7.2 hrs . answer : d <eor> d <eos> |
d |
divide__1.0__4.0__ divide__1.0__9.0__ subtract__0.25__0.1111__ round__7.2__ |
divide__1.0__4.0__ divide__1.0__9.0__ subtract__0.25__0.1111__ round__7.2__ |
| if z is a multiple of num__9702 what is the remainder when z ^ num__2 is divided by num__4 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) it depends on z . |
num__2 is a factor of num__9702 so num__2 is a factor of z . then num__2 ^ num__2 = num__4 is a factor of z ^ num__2 . then the remainder when z ^ num__2 is divided by num__4 is num__0 . the answer is a . <eor> a <eos> |
a |
multiply__9702.0__0.0__ |
multiply__9702.0__0.0__ |
| in a division when a student mistook the divisor as num__12 instead of num__21 and obtained num__35 as quotient the remainder is num__0 . what is the correct quotient ? <o> a ) num__14 <o> b ) num__18 <o> c ) num__20 <o> d ) num__24 <o> e ) num__29 |
let x be the number x ÷ num__12 = num__35 remainder = num__0 = > x = num__35 × num__12 correct quotient = x num__21 = num__35 × num__1221 = num__5 × num__123 = num__5 × num__4 = num__20 c <eor> c <eos> |
c |
multiply__4.0__5.0__ multiply__4.0__5.0__ |
multiply__4.0__5.0__ multiply__4.0__5.0__ |
| what are the next two numbers in this sequence ? num__7 num__14 num__17 num__21 num__27 num__28 num__35 num__37 ? <o> a ) num__40 <o> b ) num__41 <o> c ) num__42 <o> d ) num__43 <o> e ) num__44 |
these are digits which are either multiples of num__7 or containing the digit num__7 . s num__0 next numbers are num__42 answer : c <eor> c <eos> |
c |
add__7.0__35.0__ add__7.0__35.0__ |
add__7.0__35.0__ add__7.0__35.0__ |
| how many num__3 - digit numbers can be formed from the digits num__2 num__3 num__5 num__6 num__7 and num__9 which are divisible by num__5 and none of the digits is repeated ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__15 <o> d ) num__20 <o> e ) num__25 |
explanation : since each desired number is divisible by num__5 so we must have num__5 at the unit place . so there is num__1 way of doing it . the tens place can now be filled by any of the remaining num__5 digits ( num__2 num__3 num__6 num__7 num__9 ) . so there are num__5 ways of filling the tens place . the hundreds place can now be filled by any of the remaining num__4 digits . so there are num__4 ways of filling it . required number of numbers = ( num__1 x num__5 x num__4 ) = num__20 . answer : d <eor> d <eos> |
d |
subtract__3.0__2.0__ add__3.0__1.0__ multiply__5.0__4.0__ multiply__5.0__4.0__ |
subtract__3.0__2.0__ add__3.0__1.0__ multiply__5.0__4.0__ multiply__5.0__4.0__ |
| in a party every person shakes hands with every other person . if there are num__105 hands shakes find the number of person in the party . <o> a ) num__15 <o> b ) num__14 <o> c ) num__21 <o> d ) num__25 <o> e ) num__18 |
solution : let n be the number of persons in the party . number of hands shake = num__105 ; total number of hands shake is given by nc num__2 . now according to the question nc num__2 = num__105 ; or n ! / [ num__2 ! * ( n - num__2 ) ! ] = num__105 ; or n * ( n - num__1 ) / num__2 = num__105 ; or n num__2 - n = num__210 ; or n num__2 - n - num__210 = num__0 ; or n = num__15 - num__14 ; but we can not take negative value of n . so n = num__15 i . e . number of persons in the party = num__15 . answer : option a <eor> a <eos> |
a |
multiply__105.0__2.0__ subtract__15.0__1.0__ add__1.0__14.0__ |
multiply__105.0__2.0__ subtract__15.0__1.0__ divide__15.0__1.0__ |
| the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is num__42 kmph find the speed of the stream ? <o> a ) num__22 kmph <o> b ) num__77 kmph <o> c ) num__14 kmph <o> d ) num__88 kmph <o> e ) num__12 kmph |
the ratio of the times taken is num__2 : num__1 . the ratio of the speed of the boat in still water to the speed of the stream = ( num__2 + num__1 ) / ( num__2 - num__1 ) = num__3.0 = num__3 : num__1 speed of the stream = num__14.0 = num__14 kmph . answer : c <eor> c <eos> |
c |
add__1.0__2.0__ divide__42.0__3.0__ round__14.0__ |
add__1.0__2.0__ divide__42.0__3.0__ divide__42.0__3.0__ |
| which of these must the factor m of the product of four consecutive even integers : - num__1 ) num__48 num__2 ) num__64 num__3 ) num__96 num__4 ) num__192 num__5 ) num__80 <o> a ) m = num__12 only <o> b ) m = num__23 only <o> c ) m = num__1 num__23 only <o> d ) m = num__12 num__34 only <o> e ) all of them |
let the four variables be a < b < c < d . assume a worst case scenario where the a equal to a prime number ( hence odd ) . therefore a = divisible by num__1 b = a + num__1 ( divisible by num__2 ) c = a + num__3 ( divisible by num__3 ) d = a + num__4 ( divisible by num__4 ) therefore each answer choice must be divisible by num__2 x num__3 x num__4 = num__24 only num__80 is not divisible . therefore answer = d <eor> d <eos> |
d |
divide__48.0__2.0__ divide__48.0__4.0__ |
divide__48.0__2.0__ divide__48.0__4.0__ |
| of the num__800 employees in a certain company num__70.0 have serviced more than num__10 years . a number of y of those who have serviced more than num__10 years will retire and no fresh employees join in . what is y if the num__10 years employees become num__60.0 of the total employees ? <o> a ) num__150 <o> b ) num__200 <o> c ) num__250 <o> d ) num__450 <o> e ) num__550 |
total num__10 yrs original : num__800 num__560 new : num__800 - y num__560 - y num__560 - y = . num__6 ( num__800 - y ) num__560 - y = num__480 - . num__6 y num__80 = . num__4 y num__200.0 = y y = num__200 answer : num__200 answer : b <eor> b <eos> |
b |
divide__60.0__10.0__ divide__800.0__10.0__ subtract__10.0__6.0__ divide__800.0__4.0__ divide__800.0__4.0__ |
divide__60.0__10.0__ subtract__560.0__480.0__ subtract__10.0__6.0__ divide__800.0__4.0__ divide__800.0__4.0__ |
| a is two years older than b who is twice as old as c . if the total of the ages of a b and c be num__27 the how old is b ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__8 <o> d ) num__10 <o> e ) num__12 |
let c ' s age be x years . then b ' s age = num__2 x years . a ' s age = ( num__2 x + num__2 ) years . ( num__2 x + num__2 ) + num__2 x + x = num__27 num__5 x = num__25 x = num__5 . hence b ' s age = num__2 x = num__10 years . answer : d <eor> d <eos> |
d |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
| if q is an odd number and the median of q consecutive integers is num__130 what is the largest of these integers ? <o> a ) ( q + num__130 ) / num__2 <o> b ) q / num__2 + num__129 <o> c ) q / num__2 + num__130 <o> d ) ( q + num__129 ) / num__2 <o> e ) ( q - num__1 ) / num__2 + num__130 |
consider the easiest case say q = num__3 then ; set = { num__129 num__130 num__131 } ; the largest integer = num__131 . now plug q = num__3 into the answers to see which yields num__131 . answer : e . <eor> e <eos> |
e |
subtract__130.0__129.0__ |
subtract__130.0__129.0__ |
| an article is bought for rs . num__675 and sold for rs . num__1215 find the gain percent ? <o> a ) num__33 num__0.111111111111 % <o> b ) num__33 num__2.66666666667 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__33 num__0.666666666667 % <o> e ) num__80 % |
num__675 - - - - num__540 num__100 - - - - ? = > num__80.0 answer : e <eor> e <eos> |
e |
percent__80.0__100.0__ |
percent__80.0__100.0__ |
| a car traveled from san diego to san francisco at an average speed of num__54 miles per hour . if the journey back took twice as long what was the average speed of the trip ? <o> a ) num__24 . <o> b ) num__32 . <o> c ) num__36 . <o> d ) num__42 . <o> e ) num__44 . |
let the time taken be = x one way distance = num__54 x total distance traveled = num__2 * num__54 x = num__108 x total time taken = x + num__2 x = num__3 x average speed = num__108 x / num__3 x = num__36 answer : c <eor> c <eos> |
c |
multiply__54.0__2.0__ divide__108.0__3.0__ round__36.0__ |
multiply__54.0__2.0__ divide__108.0__3.0__ divide__108.0__3.0__ |
| if ( num__400 ) ( num__7000 ) = ( num__28000 ) ( num__100 ^ x ) what is the value of x ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__3 <o> d ) num__2 <o> e ) num__1 |
( num__400 ) ( num__7000 ) = ( num__2800 ) ( num__100 ^ x ) = > ( num__400 ) ( num__7000 ) / num__2800 = num__100 ^ x = > num__100 = num__100 ^ x = > num__100 ^ num__1 = num__100 ^ x since base is same so powers will be same too . so x = num__1 answer will be e <eor> e <eos> |
e |
reverse__1.0__ |
reverse__1.0__ |
| in g is a positive integer smaller than num__200 and ( num__14 g ) / num__60 is an integer then g has how many different positive prime factors ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__5 <o> d ) num__6 <o> e ) num__8 |
num__7 g / num__30 tells us that g is a factor of num__30 which has num__2 num__3 and num__5 as its prime factors . each of them is distinct . moreover all multiples of num__30 less than num__200 can be derived by multiplying these prime factors alone . thus number of different prime factors is num__3 . answer : b <eor> b <eos> |
b |
gcd__200.0__14.0__ add__2.0__3.0__ gcd__60.0__3.0__ |
divide__14.0__7.0__ add__2.0__3.0__ gcd__60.0__3.0__ |
| a monkey start climbing up a tree num__21 ft tall . each hour it hops num__3 ft and slips back num__2 ft . how much time would it take the monkey to reach the top . <o> a ) num__15 hrs <o> b ) num__18 hrs <o> c ) num__19 hrs <o> d ) num__17 hrs <o> e ) num__16 hrs |
if monkey hops num__3 ft and slips back num__2 ft in a hour it means the monkey hops ( num__3 ft - num__2 ft ) = num__1 ft / hr . similarly in num__18 hrs it wil be num__18 ft . but since the height of the tree is num__21 ft so if the monkey hops up the tree in the next hr i . e num__19 th hr then it reaches at the top of the tree . hence it takes num__19 hrs for monkey to reach at the top answer : c <eor> c <eos> |
c |
subtract__3.0__2.0__ subtract__21.0__3.0__ subtract__21.0__2.0__ round__19.0__ |
subtract__3.0__2.0__ subtract__21.0__3.0__ subtract__21.0__2.0__ subtract__21.0__2.0__ |
| a grocery shop has a sale of rs . num__6735 rs . num__6927 rs . num__6855 rs . num__7230 and rs . num__6562 for num__5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . num__6500 ? <o> a ) num__4857 <o> b ) num__4184 <o> c ) num__4012 <o> d ) num__4691 <o> e ) num__5291 |
let the sale in the sixth month = x then ( num__6735 + num__6927 + num__6855 + num__7230 + num__6562 + x ) / num__6 = num__6500 = > num__6735 + num__6927 + num__6855 + num__7230 + num__6562 + x = num__6 × num__6500 = > num__34309 + x = num__39000 = > x = num__39000 − num__34309 = num__4691 answer : d <eor> d <eos> |
d |
multiply__6500.0__6.0__ subtract__39000.0__34309.0__ subtract__39000.0__34309.0__ |
multiply__6500.0__6.0__ subtract__39000.0__34309.0__ subtract__39000.0__34309.0__ |
| mayank bothra purchased num__20 dozens of toys at the rate of num__375 rs per dozen . he sold each one of them at the rate of rs num__33 . what was his percentage profit ? <o> a ) num__5.6 <o> b ) num__6.5 <o> c ) num__7.5 <o> d ) num__7.6 <o> e ) none |
sol . cost = num__375 * num__20 = num__7500 sale price num__33 * num__20 * num__12 = num__7920 profit = num__420 profit in % = num__0.056 * num__100 = num__5.6 ans : a . <eor> a <eos> |
a |
percent__5.6__100.0__ |
percent__5.6__100.0__ |
| nine persons went to a hotel for taking their meals . eight of them spent rs . num__12 each on their meals and the ninth spent rs . num__8 more than the average expenditure of all the nine . what was the total money spent by them . <o> a ) rs . num__115 <o> b ) rs . num__116 <o> c ) rs . num__117 <o> d ) rs . num__118 <o> e ) none |
solution : let the average expenditure of all the nine be x . then num__12 × num__8 + ( x + num__8 ) = num__9 x . therefore x = num__13 . total money spent = num__9 x = rs . ( num__9 × num__13 ) = rs . num__117 answer : option c <eor> c <eos> |
c |
multiply__9.0__13.0__ multiply__9.0__13.0__ |
multiply__9.0__13.0__ multiply__9.0__13.0__ |
| n and m are each num__3 - digit integers . each of the numbers num__34 num__5 num__6 num__7 and num__8 is a digit of either n or m . what is the smallest possible positive difference between n and m ? <o> a ) num__59 <o> b ) num__49 <o> c ) num__58 <o> d ) num__113 <o> e ) num__131 |
you have num__6 digits : num__3 num__4 num__5 num__6 num__7 num__8 each digit needs to be used to make two num__3 digit numbers . this means that we will use each of the digits only once and in only one of the numbers . the numbers need to be as close to each other as possible . the numbers can not be equal so the greater number needs to be as small as possible and the smaller number needs to be as large as possible to be close to each other . the first digit ( hundreds digit ) of both numbers should be consecutive integers now let ' s think about the next digit ( the tens digit ) . to minimize the difference between the numbers the tens digit of the greater number should be as small as possible and the tens digit of the smaller number should be as large as possible . so let ' s not use num__3 and num__8 in the hundreds places and reserve them for the tens places . now what are the options ? try and make a pair with ( num__4 * * and num__5 * * ) . make the num__4 * * number as large as possible and make the num__5 * * number as small as possible . num__487 and num__536 ( difference is num__49 ) or try and make a pair with ( num__6 * * and num__7 * * ) . make the num__6 * * number as large as possible and make the num__7 * * number as small as possible . we get num__685 and num__734 ( difference is num__49 ) b <eor> b <eos> |
b |
subtract__7.0__3.0__ subtract__536.0__487.0__ add__685.0__49.0__ subtract__536.0__487.0__ |
subtract__7.0__3.0__ subtract__536.0__487.0__ add__685.0__49.0__ subtract__536.0__487.0__ |
| a sum of money deposited at c . i . amounts to rs . num__1820 in num__4 years and to rs . num__2402 in num__5 years . find the rate percent ? <o> a ) num__30.0 <o> b ) num__31.0 <o> c ) num__34.0 <o> d ) num__19.0 <o> e ) num__50 % |
num__1820 - - - num__582 num__100 - - - ? = > num__31.0 answer : b <eor> b <eos> |
b |
percent__100.0__31.0__ |
percent__100.0__31.0__ |
| the marks obtained by vijay and amith are in the ratio num__4 : num__5 and those obtained by amith and abhishek in the ratio of num__3 : num__2 . the marks obtained by vijay and abhishek are in the ratio of ? <o> a ) num__6 : num__3 <o> b ) num__6 : num__9 <o> c ) num__6 : num__5 <o> d ) num__6 : num__1 <o> e ) num__6 : num__4 |
num__4 : num__5 num__3 : num__2 - - - - - - - num__12 : num__15 : num__10 num__12 : num__10 num__6 : num__5 answer : c <eor> c <eos> |
c |
multiply__4.0__3.0__ multiply__5.0__3.0__ multiply__5.0__2.0__ add__4.0__2.0__ add__4.0__2.0__ |
multiply__4.0__3.0__ multiply__5.0__3.0__ subtract__12.0__2.0__ subtract__10.0__4.0__ subtract__10.0__4.0__ |
| a certain characteristic in a large population has a distribution that is symmetric about the mean m . num__80 percent of the distribution lies within one standard deviation d of the mean . if the shelf ’ s average life is num__8.6 years and the standard deviation is num__3.4 years what percent of the distribution has more than num__12.0 years as a shelf ’ s average life ? <o> a ) num__23.0 <o> b ) num__22.0 <o> c ) num__21.0 <o> d ) num__20.0 <o> e ) num__19 % |
average = num__8.6 sd = num__3.4 num__8.6 - num__3.4 < num__80.0 of distribution < num__8.6 + num__3.4 num__5.2 < num__80.0 of distribution < num__12.0 num__40.0 is outside this range . given : distribution is symmetric . so num__20.0 of distribution is less than num__5.2 and the other num__20.0 of distribution is greater than num__12.0 . answer : d <eor> d <eos> |
d |
subtract__8.6__3.4__ subtract__40.0__20.0__ |
subtract__8.6__3.4__ subtract__40.0__20.0__ |
| the simple interest on a sum of money will be rs . num__600 after num__10 years . if the principal is trebled after num__5 years what will be the total interest at the end of the tenth year ? <o> a ) rs . num__600 <o> b ) rs . num__900 <o> c ) rs . num__1200 <o> d ) rs . num__1500 <o> e ) none |
solution num__22.22 let the sum be rs . x . now s . i . = rs . num__600 t = num__10 years . rate = % = ( num__100 x num__600 / x x num__10 ) % = ( num__6000 / x ) % s . i . for first num__5 years = rs . ( x x num__5 x num__6000 / x x num__100 ) = rs . num__300 . s . i . for first num__5 years = rs . ( num__3 x x num__5 x num__6000 / x x num__100 ) = rs . num__900 . ∴ total interest = rs . num__1200 . answer c <eor> c <eos> |
c |
multiply__600.0__10.0__ divide__300.0__100.0__ add__600.0__300.0__ add__300.0__900.0__ add__300.0__900.0__ |
multiply__600.0__10.0__ divide__300.0__100.0__ add__600.0__300.0__ divide__6000.0__5.0__ divide__6000.0__5.0__ |
| if the lcm of a and b is num__72 and their hcf is num__6 what is the product of a & b ? <o> a ) num__432 <o> b ) num__433 <o> c ) num__434 <o> d ) num__435 <o> e ) num__436 |
the most important property of lcm ( the least common multiple ) and gcd ( the greatest common divisors ) is : for any positive integers x and y x ∗ y = gcd ( x y ) ∗ lcm ( x y ) so ab = num__72 * num__6 = num__432 . answer : a <eor> a <eos> |
a |
multiply__72.0__6.0__ round__432.0__ |
multiply__72.0__6.0__ multiply__72.0__6.0__ |
| sonika bought a v . c . r . at the list price of num__18500 . if the rate of sales tax was num__8.0 find the amount she had to pay for purchasing the v . c . r . <o> a ) num__19980 <o> b ) num__19780 <o> c ) num__19680 <o> d ) num__19380 <o> e ) none of these |
sol . list price of v . c . r . = num__18500 rate of sales tax = num__8.0 ∴ sales tax = num__8.0 of num__18500 = num__8 ⁄ num__100 × num__18500 = num__1480 so total amount which sonika had to pay for purchasing the v . c . r . = num__18500 + num__1480 = num__19980 . answer a <eor> a <eos> |
a |
percent__8.0__18500.0__ percent__100.0__19980.0__ |
percent__8.0__18500.0__ percent__100.0__19980.0__ |
| find the simple interest on $ num__10000 at num__4.0 per annum for num__12 months ? <o> a ) $ num__400 <o> b ) $ num__500 <o> c ) $ num__650 <o> d ) $ num__710 <o> e ) $ num__1000 |
p = $ num__10000 r = num__4.0 t = num__1.0 years = num__1 year s . i . = p * r * t / num__100 = num__10000 * num__4 * num__0.01 = $ num__400 answer is a <eor> a <eos> |
a |
percent__1.0__10000.0__ percent__4.0__10000.0__ percent__4.0__10000.0__ |
percent__1.0__10000.0__ percent__4.0__10000.0__ percent__4.0__10000.0__ |
| a train num__360 m long is running at a speed of num__45 km / hr . in what time will it pass a bridge num__140 m long ? <o> a ) num__40 sec <o> b ) num__29 sec <o> c ) num__26 sec <o> d ) num__27 sec <o> e ) num__34 sec |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec total distance covered = num__360 + num__140 = num__500 m required time = num__500 * num__0.08 = num__40 sec answer : a <eor> a <eos> |
a |
add__360.0__140.0__ divide__500.0__12.5__ round__40.0__ |
add__360.0__140.0__ divide__500.0__12.5__ divide__500.0__12.5__ |
| consider a set s given as { num__11 num__2233 num__4455 num__6677 num__8899 } . if each element of the data set is multiplied by - num__9 then by what factors will the mean variance and standard deviation respectively change ? <o> a ) - num__9 - num__9 - num__9 <o> b ) - num__9 num__99 <o> c ) num__1 num__11 <o> d ) - num__9 num__819 <o> e ) num__9 num__9 |
81 |
mean : if all values are multiplied by - num__9 then their sum will also be multiplied by - num__9 so mean will also get - num__9 variance : variance is equal to ( sum of ( square of ( the difference between mean and individual values ) ) / total number of values when we multiply all the values by - num__9 then the difference between the mean and the individual values also gets multiplied by - num__9 and the square of that difference becomes num__81 . so the variance gets multiplied by a factor of num__81 sd : standard deviation is the square root of variance so sqrt ( num__81 ) = num__9 . so sd gets multiplied by a factor of num__9 . so answer will be d . <eor> d <eos> |
d |
d |
| rs . num__700 is divided among a b c so that a receives half as much as b and b half as much as c . then c ' s share is <o> a ) rs num__200 <o> b ) rs num__300 <o> c ) rs num__400 <o> d ) rs num__500 <o> e ) rs num__600 |
explanation : let c = x . then b = x / num__2 and a = x / num__4 a : b : c = num__1 : num__2 : num__4 . c ' s share rs . [ ( num__0.571428571429 ) * num__700 ) = num__400 option c <eor> c <eos> |
c |
multiply__1.0__400.0__ |
multiply__1.0__400.0__ |
| a number is increased by num__30.0 and then decreased by num__30.0 . find the net increase or decrease per cent . <o> a ) num__9.0 <o> b ) num__8.0 <o> c ) num__7.0 <o> d ) num__3.0 <o> e ) num__6 % |
let the number be num__100 . increase in the number = num__30.0 = num__30.0 of num__100 = ( num__0.3 Ã — num__100 ) = num__30 therefore increased number = num__100 + num__30 = num__130 this number is decreased by num__30.0 therefore decrease in number = num__30.0 of num__130 = ( num__0.3 Ã — num__130 ) = num__39.0 = num__39 therefore new number = num__130 - num__39 = num__91 thus net decreases = num__100 - num__91 = num__9 hence net percentage decrease = ( num__0.09 Ã — num__100 ) % = ( num__9.0 ) % = num__9.0 answer : a <eor> a <eos> |
a |
divide__30.0__100.0__ add__30.0__100.0__ multiply__130.0__0.3__ subtract__130.0__39.0__ multiply__30.0__0.3__ divide__9.0__100.0__ multiply__30.0__0.3__ |
divide__30.0__100.0__ add__30.0__100.0__ multiply__130.0__0.3__ subtract__130.0__39.0__ subtract__100.0__91.0__ divide__9.0__100.0__ subtract__100.0__91.0__ |
| two pipes can fill a tank in num__18 minutes and num__15 minutes . an outlet pipe can empty the tank in num__45 minutes . if all the pipes are opened when the tank is empty then how many minutes will it take to fill the tank ? <o> a ) num__10 minutes <o> b ) num__20 minutes <o> c ) num__15 minutes <o> d ) num__12 minutes <o> e ) num__18 minutes |
part of the filled by all the three pipes in one minute = num__0.0555555555556 + num__0.0666666666667 - num__0.0222222222222 = ( num__5 + num__6 - num__2 ) / num__90 = num__0.1 = num__0.1 so the tank becomes full in num__10 minutes . answer : a <eor> a <eos> |
a |
multiply__18.0__5.0__ subtract__15.0__5.0__ round__10.0__ |
multiply__18.0__5.0__ subtract__15.0__5.0__ subtract__15.0__5.0__ |
| if a rectangular room measures num__8 meters by num__5 meters by num__4 meters what is the volume of the room in cubic centimeters ? ( num__1 meter = num__100 centimeters ) <o> a ) num__24000 <o> b ) num__16 num__000000 <o> c ) num__2 num__400000 <o> d ) num__24 num__000000 <o> e ) num__240 num__000 |
000 |
b . num__160 num__000000 num__8 * num__100 * num__5 * num__100 * num__4 * num__100 = num__160 num__000000 <eor> b <eos> |
b |
b |
| in num__26 minutes the minute hand gains over the hour hand by <o> a ) num__116 ° <o> b ) num__125 ° <o> c ) num__143 ° <o> d ) num__158 ° <o> e ) num__172 ° |
in one hour the hour hand moves num__30 ° which is num__0.5 ° each minute . in one hour the minute hand moves num__360 ° which is num__6 ° each minutes . the minute hand gains num__5.5 ° each minute . in num__26 minutes the minute hand gains num__26 * num__5.5 ° = num__143 ° . the answer is c . <eor> c <eos> |
c |
subtract__6.0__0.5__ multiply__26.0__5.5__ round__143.0__ |
subtract__6.0__0.5__ multiply__26.0__5.5__ multiply__26.0__5.5__ |
| ben earns $ num__12800 a year . about num__15.0 is taken out for taxes . how much is taken out for taxes ? <o> a ) $ num__19200 <o> b ) $ num__11920 <o> c ) $ num__1920 <o> d ) $ num__19820 <o> e ) $ num__10920 |
num__0.15 = x / num__12.800 divide by the remaining number : num__0.01 num__100 $ num__1920 correct answer c <eor> c <eos> |
c |
divide__0.15__15.0__ reverse__0.01__ multiply__12800.0__0.15__ multiply__12800.0__0.15__ |
divide__0.15__15.0__ reverse__0.01__ multiply__12800.0__0.15__ multiply__12800.0__0.15__ |
| ram professes to sell his goods at the cost price but he made use of num__900 grms instead of a kg what is the gain percent ? <o> a ) num__11 num__0.125 % <o> b ) num__11 num__0.2 % <o> c ) num__11 num__0.111111111111 % <o> d ) num__11 num__12.0 <o> e ) num__11 num__0.555555555556 % |
num__900 - - - num__100 num__100 - - - ? = > num__11 num__0.111111111111 % answer : c <eor> c <eos> |
c |
percent__100.0__11.0__ |
percent__100.0__11.0__ |
| by selling an shirt for $ num__300 a shop keeper gains num__20.0 . during a clearance sale the shopkeeper allows a discount of num__10.0 on the marked price . his gain percent during the sale is : <o> a ) num__9.0 <o> b ) num__2.0 <o> c ) num__8.0 <o> d ) num__5.0 <o> e ) num__6 % |
c num__8.0 marked price = $ num__300 c . p . = num__0.833333333333 * num__300 = $ num__250 sale price = num__90.0 of $ num__300 = $ num__270 required gain % = num__0.08 * num__100 = num__8.0 . <eor> c <eos> |
c |
percent__90.0__300.0__ percent__100.0__8.0__ |
percent__90.0__300.0__ percent__100.0__8.0__ |
| a number whose fifth part increased by num__3 is equal to its fourth part diminished by num__3 is ? <o> a ) num__120 <o> b ) num__180 <o> c ) num__200 <o> d ) num__220 <o> e ) none |
answer let the number be n . then ( n / num__5 ) + num__3 = ( n / num__4 ) - num__3 â ‡ ’ ( n / num__4 ) - ( n / num__5 ) = num__6 â ‡ ’ ( num__5 n - num__4 n ) / num__20 = num__6 â ˆ ´ n = num__120 option : a <eor> a <eos> |
a |
multiply__4.0__5.0__ multiply__6.0__20.0__ multiply__6.0__20.0__ |
multiply__4.0__5.0__ multiply__6.0__20.0__ multiply__6.0__20.0__ |
| tickets numbered num__1 to num__20 are mixed up and then a ticked is drawn at random . what is the probability that the ticket drawn bears a number which is a multiple of num__3 ? <o> a ) num__0.3 <o> b ) num__0.15 <o> c ) num__0.4 <o> d ) num__0.5 <o> e ) num__0.333333333333 |
s = { num__1 num__23 … . num__20 } e = { num__3 num__6 num__912 num__1518 } p ( e ) = num__0.3 = num__0.3 option a <eor> a <eos> |
a |
add__20.0__3.0__ divide__6.0__20.0__ multiply__1.0__0.3__ |
add__20.0__3.0__ divide__6.0__20.0__ multiply__1.0__0.3__ |
| a cicketer bas a certain average for num__10 innings in the eleventh inning he scorod num__128 runs thereby increasing his average by num__6 runs . his new average is <o> a ) num__18 runs <o> b ) num__28 runs <o> c ) num__48 runs <o> d ) num__42 runs <o> e ) num__68 runs |
explanation : let average for num__10 innings be x . then ( num__10 x + num__108 ) / num__11 = x + num__6 = > num__11 x + num__66 = num__10 x + num__128 = > x = num__62 . new average = ( x + num__6 ) = num__68 runs . answer : e <eor> e <eos> |
e |
multiply__6.0__11.0__ subtract__128.0__66.0__ add__6.0__62.0__ add__6.0__62.0__ |
multiply__6.0__11.0__ subtract__128.0__66.0__ add__6.0__62.0__ add__6.0__62.0__ |
| twelve people are planning to share equally the cost of a rental van . if one person withdraws from the arrangement and the others share equally the cost of the rental van then the share of each of the remaining people will increase by . . . ? <o> a ) num__0.0833333333333 <o> b ) num__0.125 <o> c ) num__0.0909090909091 <o> d ) num__1.09090909091 <o> e ) num__0.909090909091 |
let p = total cost of the rental van . the original share per person is p / num__12 . the new share per person is p / num__11 . p / num__11 = p / num__12 * num__1.09090909091 = ( num__1 + num__0.0909090909091 ) * original share the answer is c . <eor> c <eos> |
c |
divide__12.0__11.0__ round_down__1.0909__ reverse__11.0__ reverse__11.0__ |
divide__12.0__11.0__ round_down__1.0909__ reverse__11.0__ reverse__11.0__ |
| in a trapezium abcd ab is parallel to dc ab = num__3 * dc and the diagonals of the trapezium intersect at o . the ratio of the area of ∆ ocd to the area of ∆ oab is : <o> a ) num__1 : num__9 <o> b ) num__1 : num__3 <o> c ) num__3 : num__1 <o> d ) num__1 : num__10 <o> e ) can not be determined |
triangles are similar so ratio will be num__1 : num__9 a <eor> a <eos> |
a |
volume_cube__1.0__ |
volume_cube__1.0__ |
| the average ( arithmetic mean ) monthly income of four workers is $ num__3000 . after one worker ’ s income increases by num__20 percent the new average income is $ num__3300 . what was the original income of the worker whose monthly income increased ? <o> a ) $ num__6500 <o> b ) $ num__6400 <o> c ) $ num__6000 <o> d ) $ num__6200 <o> e ) $ num__6 |
600 |
increase in total income was num__300 * num__4 = $ num__1200 we know that this increase was num__20.0 ( num__0.2 th ) of the workers original income thus his / her original income was num__1200 * num__5 = $ num__6000 . answer : c <eor> c <eos> |
c |
c |
| a can give b num__100 meters start and c num__150 meters start in a kilometer race . how much start can b give c in a kilometer race ? <o> a ) num__10.55 meters <o> b ) num__11.55 meters <o> c ) num__33.55 meters <o> d ) num__55.55 meters <o> e ) none of these |
explanation : a runs num__1000 meters while b runs num__900 meters and c runs num__850 meters . therefore b runs num__900 meters while c runs num__850 meters . so the number of meters that c runs when b runs num__1000 meters = ( num__1000 x num__850 ) / num__900 = num__944.44 meters thus b can give c ( num__1000 - num__944.44 ) = num__55.55 meters start answer : d <eor> d <eos> |
d |
subtract__1000.0__100.0__ subtract__1000.0__150.0__ round__55.55__ |
subtract__1000.0__100.0__ subtract__1000.0__150.0__ round__55.55__ |
| the length of the bridge which a train num__150 m long and traveling at num__45 km / hr can cross in num__30 sec is ? <o> a ) num__377 <o> b ) num__225 <o> c ) num__237 <o> d ) num__245 <o> e ) num__267 |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec . time = num__30 sec let the length of bridge be x meters . then ( num__150 + x ) / num__30 = num__12.5 x = num__225 m . answer : b <eor> b <eos> |
b |
round__225.0__ |
round__225.0__ |
| if a = num__1 what is the value of – ( a ^ num__2 + a ^ num__3 + a ^ num__4 + a ^ num__5 + a ^ num__6 ) ? <o> a ) - num__14 <o> b ) - num__4 <o> c ) num__0 <o> d ) num__4 <o> e ) - num__5 |
if a = - num__1 then putting values in equation = - [ ( num__1 ) ^ num__2 + ( num__1 ) ^ num__3 + ( num__1 ^ num__4 ) + ( num__1 ^ num__5 ) + ( num__1 ^ num__6 ) ] = - [ num__1 + num__1 + num__1 + num__1 + num__1 ] = - num__5 answer = e = - num__5 <eor> e <eos> |
e |
add__1.0__4.0__ |
add__1.0__4.0__ |
| a town ' s oldest inhabitant is x years older than the sum of the ages of the lee triplets . if the oldest inhabitants is now j years old how old will one of the triplets w be in num__20 years ? j - x - num__13.3333333333 is my answers . j = x + l + l + l is the initial situation after num__20 years j + num__20 = x + l + l + l + num__60 . . . num__20 years for each triplet so num__60 years totally . j - x - num__13.3333333333 = l is my answer . what wrong am i doing ? since the age asked is after num__20 years i also consider adding num__20 years to j . <o> a ) ( j - num__50 ) / num__3 <o> b ) num__3 ( j + num__20 ) / x <o> c ) ( j + x - num__50 ) / num__3 <o> d ) ( j - x + num__60 ) / num__3 <o> e ) ( j + x - num__20 ) / num__3 |
here it goes : oldest inhabitant = sum of age of triplets + x j = num__3 l + x so l = ( j - x ) / num__3 after num__20 years = l + num__20 = ( j - x ) / num__3 + num__20 w = ( j - x + num__60 ) / num__3 = d <eor> d <eos> |
d |
divide__60.0__20.0__ multiply__20.0__3.0__ |
divide__60.0__20.0__ multiply__20.0__3.0__ |
| if john takes num__7 minutes to eat x raisins how many seconds will it take him to eat z raisins assuming he continues to eat at the same rate ? <o> a ) num__420 xz <o> b ) num__7 z / x <o> c ) num__420 z / x <o> d ) num__7 xz <o> e ) num__420 x / z |
it will take ( z / x ) ( num__7 ) minutes which is ( z / x ) ( num__7 ) ( num__60 ) seconds = num__420 z / x seconds . the answer is c . <eor> c <eos> |
c |
hour_to_min_conversion__ multiply__7.0__60.0__ round__420.0__ |
hour_to_min_conversion__ multiply__7.0__60.0__ round__420.0__ |
| the average of first num__25 natural numbers is ? <o> a ) num__14 <o> b ) num__13 <o> c ) num__15 <o> d ) num__18 <o> e ) num__12 |
sum of num__25 natural no . = num__325.0 = num__325 average = num__13.0 = num__13 answer : b <eor> b <eos> |
b |
divide__325.0__25.0__ divide__325.0__25.0__ |
divide__325.0__25.0__ divide__325.0__25.0__ |
| the distance from steve ' s house to work is num__30 km . on the way back steve drives twice as fast as he did on the way to work . altogether steve is spending num__6 hours a day on the roads . what is steve ' s speed on the way back from work ? <o> a ) num__5 . <o> b ) num__10 . <o> c ) num__14 . <o> d ) num__15 <o> e ) num__20 . |
time is in the ratio num__2 : num__1 : : to : fro office therefore num__2 x + num__1 x = num__6 hrs time take to come back - num__2 hrs distance travelled - num__30 km = > speed = num__15 kmph answer : d <eor> d <eos> |
d |
divide__30.0__2.0__ round__15.0__ |
divide__30.0__2.0__ subtract__30.0__15.0__ |
| the length of a rectangular garden is num__2 feet longer than num__3 times its width . if the perimeter of the garden is num__100 feet find the width and the length of the garden . <o> a ) num__80 <o> b ) num__90 <o> c ) num__100 <o> d ) num__70 <o> e ) num__110 |
let l and w be the length and width of the garden . the statement ` ` the length of a rectangular garden is num__2 feet longer than num__3 times its width ' ' may be formulated by l = num__2 + num__3 w the formula for the perimeter is given by p = num__2 l + num__2 w substitute p and l in the above equation by num__100 and num__2 + num__3 w respectively to obtain num__100 = num__2 ( num__2 + num__3 w ) + num__2 w solve for w and l w = num__12 and l = num__2 + num__3 w = num__38 . check that the perimeter of the rectangular garden is num__100 p = num__2 l + num__2 w = num__76 + num__24 = num__100 answer c <eor> c <eos> |
c |
square_perimeter__3.0__ multiply__2.0__38.0__ surface_cube__2.0__ triangle_area__2.0__100.0__ |
square_perimeter__3.0__ multiply__2.0__38.0__ surface_cube__2.0__ triangle_area__2.0__100.0__ |
| a train num__125 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is : <o> a ) num__45 km / hr <o> b ) num__55 km / hr <o> c ) num__35 km / hr <o> d ) num__50 km / hr <o> e ) num__56 km / hr |
speed of the train relative to man = num__12.5 m / sec num__12.5 m / sec . num__12.5 x num__1.2 km / hr = num__45 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__45 x = num__50 km / hr . answer : d <eor> d <eos> |
d |
divide__125.0__10.0__ multiply__5.0__10.0__ round__50.0__ |
divide__125.0__10.0__ multiply__5.0__10.0__ round__50.0__ |
| what is the probability of getting a number less than num__4 when a die is rolled ? <o> a ) num__2 <o> b ) num__1 <o> c ) num__1.5 <o> d ) num__0.666666666667 <o> e ) num__0.5 |
total number of outcomes possible when a die is rolled = num__6 that is any one face out of the num__6 faces is n ( s ) = num__6 e = getting a number less than num__4 = { num__1 num__23 } . hence n ( e ) = num__3 p ( e ) = n ( e ) / n ( s ) = num__0.5 = num__0.5 answer : option e <eor> e <eos> |
e |
die_space__ negate_prob__0.5__ |
die_space__ negate_prob__0.5__ |
| two trains start from stations a and b spaced num__50 kms apart at the same time and speed . as the trains start a bird flies from one train towards the other and on reaching the second train it flies back to the first train . this is repeated till the trains collide . if the speed of the trains is num__25 km / h and that of the bird is num__100 km / h . how much did the bird travel till the collision . <o> a ) num__100 <o> b ) num__288 <o> c ) num__127 <o> d ) num__27 <o> e ) num__2691 |
ince the trains is travelling at num__25 kmph at each other the relative speed is num__50 kmph . speed = num__50 kmph distance = num__50 km time to collision = distance / speed = num__1 hr speed of bird = num__100 kmph time flying = num__1 hr ( the bird is flying till the trains collide ) distance travelled = speed × time = num__100 km answer : a <eor> a <eos> |
a |
round__100.0__ |
divide__100.0__1.0__ |
| s num__1 = num__6 s num__2 = num__9 . . . sn = num__3 n + num__4 for the sequence above in which any term n is defined as num__3 n + num__4 what is the value of n for the first term in the sequence to exceed num__50 ? <o> a ) num__14 <o> b ) num__12 <o> c ) num__13 <o> d ) num__16 <o> e ) num__15 |
term num__16 = num__3 * num__16 + num__4 = num__52 answer : d <eor> d <eos> |
d |
power__2.0__4.0__ add__2.0__50.0__ multiply__1.0__16.0__ |
power__2.0__4.0__ add__2.0__50.0__ multiply__1.0__16.0__ |
| the distance between delhi and mathura is num__215 kms . a starts from delhi with a speed of num__20 kmph at num__7 a . m . for mathura and b starts from mathura with a speed of num__25 kmph at num__8 p . m . from delhi . when will they meet ? <o> a ) num__10.50 a . m . <o> b ) num__00.20 p . m . <o> c ) num__10.30 a . m . <o> d ) num__11.40 a . m . <o> e ) num__1.40 p . m . |
d = num__215 â € “ num__20 = num__195 rs = num__20 + num__25 = num__45 t = num__4.33333333333 = num__4 num__0.333333333333 hours num__8 a . m . + num__4 hrs num__20 min = num__00.20 p . m . answer : b <eor> b <eos> |
b |
subtract__215.0__20.0__ add__20.0__25.0__ divide__195.0__45.0__ subtract__4.3333__4.0__ divide__4.0__20.0__ round__0.2__ |
subtract__215.0__20.0__ add__20.0__25.0__ divide__195.0__45.0__ subtract__4.3333__4.0__ divide__4.0__20.0__ round__0.2__ |
| a b and c can do a work in num__9 days num__8 days and num__12 days respectively . in how many days can all three of them working together complete the work ? <o> a ) num__3 num__0.130434782609 <o> b ) num__3 num__0.107142857143 <o> c ) num__2 num__2 / num__0 <o> d ) num__2 num__1.0 <o> e ) num__2 num__2.0 |
work done by all three of them in one day = num__0.111111111111 + num__0.125 + num__0.0833333333333 = num__0.319444444444 . the number of days required = num__3.13043478261 = num__3 num__0.130434782609 days . answer : a <eor> a <eos> |
a |
subtract__12.0__9.0__ subtract__3.1304__3.0__ round__3.0__ |
subtract__12.0__9.0__ subtract__3.1304__3.0__ round__3.0__ |
| if the number of boys in a class are num__8 times the number of girls . if the number of people in the class are num__45 how many girls are in the class ? <o> a ) a - num__5 <o> b ) b - num__6 <o> c ) c - num__7 <o> d ) d - num__8 <o> e ) e - num__9 |
let the number of girls = x and the number of boys = num__8 x then total number of students = x + num__8 x = num__9 x = num__45 hence x = num__5.0 = num__5 i . e . the total number of girls in the class are num__5 . answer is a . <eor> a <eos> |
a |
divide__45.0__9.0__ divide__45.0__9.0__ |
divide__45.0__9.0__ divide__45.0__9.0__ |
| two spies agreed to meet at a gas station between noon and num__1 pm but they have both forgotten the arranged time . each arrives at a random time between noon and num__1 pm and stays for num__6 minutes unless the other is there before the num__6 minutes are up . assuming all random times are equally likely what is the probability that they will meet within the hour ( noon to num__1 pm ) ? <o> a ) num__0.12 <o> b ) num__0.15 <o> c ) num__0.17 <o> d ) num__0.19 <o> e ) num__0.25 |
the probability that they do not meet is represented by the total of the areas of the two outer triangles in the figure below which is num__0.81 . so the probability of a meeting is num__1 - num__0.81 = num__0 . num__19 . correct answer d <eor> d <eos> |
d |
round_down__0.81__ subtract__1.0__0.81__ |
round_down__0.81__ subtract__1.0__0.81__ |
| the sum of the present ages of a father and his son is num__50 years . five years ago father ' s age was five times the age of the son . after num__5 years son ' s age will be : <o> a ) num__16 <o> b ) num__16.2 <o> c ) num__16.1 <o> d ) num__16.66 <o> e ) num__17 |
let the present ages of son and father be x and ( num__50 - x ) years respectively . then ( num__50 - x ) - num__5 = num__5 ( x - num__5 ) num__45 - x = num__5 x - num__25 num__6 x = num__70 x = num__11.6666666667 = num__11.66 years son ' s age after num__6 years = ( x + num__5 ) = num__16.66 years . answer : d <eor> d <eos> |
d |
subtract__50.0__5.0__ add__45.0__25.0__ divide__70.0__6.0__ add__5.0__11.66__ add__5.0__11.66__ |
subtract__50.0__5.0__ add__45.0__25.0__ divide__70.0__6.0__ add__5.0__11.66__ add__5.0__11.66__ |
| a committee of two people is to be chosen from num__4 married couples . what is the number of different committees that can be chosen if two people who are married to each other can not both serve on the committee ? <o> a ) num__16 <o> b ) num__24 <o> c ) num__26 <o> d ) num__30 <o> e ) num__32 |
one of the approaches : each couple can send only onerepresentativeto the committee . let ' s see in how many ways we can choose num__2 couples ( as there should be num__2 members ) out of num__4 to send only onerepresentativesto the committee : num__4 c num__2 = num__6 . but each of these num__2 couples can send two persons ( husband or wife ) : num__2 * num__2 = num__2 ^ num__2 = num__4 . total # of ways : num__4 c num__2 * num__2 ^ num__2 = num__24 . answer : b . <eor> b <eos> |
b |
coin_space__ die_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
coin_space__ die_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
| of num__25 applicants for a job num__14 had at least num__4 years ' experience num__18 had degrees and num__3 had less than num__4 years ' experience and did not have a degree . how many of the applicants had at least num__4 years ' experience and a degree ? <o> a ) num__14 <o> b ) num__10 <o> c ) num__9 <o> d ) num__7 <o> e ) num__5 |
set a : people with more than num__4 years exp set b : people with degree aub = total - ( less than num__4 exp and no degree ) aub = num__25 - num__3 = num__22 aub = a + b - aib aib = num__18 + num__14 - num__22 = num__10 answer b <eor> b <eos> |
b |
subtract__25.0__3.0__ subtract__14.0__4.0__ subtract__14.0__4.0__ |
subtract__25.0__3.0__ subtract__14.0__4.0__ subtract__14.0__4.0__ |
| working alone at its constant rate machine a produces x boxes in num__10 minutes and working alone at its constant rate machine b produces num__2 x boxes in num__5 minutes . how many minutes does it take machines a and b working simultaneously at their respective constant rates to produce num__5 x boxes ? <o> a ) num__3 minutes <o> b ) num__4 minutes <o> c ) num__5 minutes <o> d ) num__10 minutes <o> e ) num__12 minutes |
rate = work / time given rate of machine a = x / num__10 min machine b produces num__2 x boxes in num__5 min hence machine b produces num__4 x boxes in num__10 min . rate of machine b = num__4 x / num__10 we need tofind the combined time that machines a and b working simultaneouslytakeat their respective constant rates let ' s first find the combined rate of machine a and b rate of machine a = x / num__10 min + rate of machine b = num__4 x / num__10 = num__5 x / num__10 now combine time = combine work needs to be done / combine rate = num__5 x / num__5 x * num__10 = num__10 min ans : d <eor> d <eos> |
d |
round__10.0__ |
multiply__2.0__5.0__ |
| working alone mary can pave a driveway in num__4 hours and hillary can pave the same driveway in num__3 hours . when they work together mary thrives on teamwork so her rate increases by num__33.33 but hillary becomes distracted and her rate decreases by num__50.0 . if they both work together how many hours will it take to pave the driveway ? <o> a ) num__2 hours <o> b ) num__4 hours <o> c ) num__5 hours <o> d ) num__6 hours <o> e ) num__7 hours |
initial working rates : mary = num__0.25 per hour hillary = num__0.333333333333 per hour rate when working together : mary = num__0.25 + ( num__0.333333333333 * num__0.25 ) = num__0.333333333333 per hour hillary = num__0.333333333333 - ( num__0.5 * num__0.333333333333 ) = num__0.166666666667 per hour together they work num__0.333333333333 + num__0.166666666667 = num__0.5 per hour so they will need num__2 hours to complete the driveway . the correct answer is a . <eor> a <eos> |
a |
percent__4.0__50.0__ percent__4.0__50.0__ |
percent__4.0__50.0__ percent__4.0__50.0__ |
| a and b can do a piece of work in num__12 days and num__16 days respectively . both work for num__3 days and then a goes away . find how long will b take to complete the remaining work ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__5 <o> d ) num__9 <o> e ) num__3 |
num__0.25 + ( num__3 + x ) / num__16 = num__1 x = num__9 days answer : d <eor> d <eos> |
d |
divide__3.0__12.0__ subtract__12.0__3.0__ round__9.0__ |
divide__3.0__12.0__ subtract__12.0__3.0__ divide__9.0__1.0__ |
| the edge of a cube is num__9 a cm . find its surface ? <o> a ) num__153 a num__2 cm num__2 <o> b ) num__143 a num__2 cm num__2 <o> c ) num__233 a num__2 cm num__2 <o> d ) num__243 a num__2 cm num__2 <o> e ) num__253 a num__2 cm num__2 |
num__6 a num__2 = num__6 * num__9 a * num__9 a = num__243 a num__2 answer : d <eor> d <eos> |
d |
triangle_area__2.0__243.0__ |
triangle_area__2.0__243.0__ |
| if a lends rs . num__2500 to b at num__10.0 per annum and b lends the same sum to c at num__11.5 per annum then the gain of b in a period of num__3 years is ? a . rs . num__107.50 <o> a ) num__112.5 <o> b ) num__122.5 <o> c ) num__132.5 <o> d ) num__114.5 <o> e ) num__212.5 |
( num__2500 * num__1.5 * num__3 ) / num__100 = > num__112.50 answer : a <eor> a <eos> |
a |
percent__100.0__112.5__ |
percent__100.0__112.5__ |
| the daily hotel rates for five hotels john was considering were $ num__60 $ num__125 $ num__80 $ num__70 and $ x . if the median price was $ num__5 lower than the average ( arithmetic mean ) price which of the following could be $ x ? <o> a ) $ num__80 <o> b ) $ num__85 <o> c ) $ num__90 <o> d ) $ num__95 <o> e ) $ num__100 |
look at the options available value of x seems to $ num__80 < = $ x < = $ num__100 a . $ num__80 b . $ num__85 c . $ num__90 d . $ num__95 e . $ num__100 so position of $ x is fourth $ num__60 $ num__70 $ num__80 $ x $ num__125 median = $ num__80 median price was $ num__5 lower than the average so average = $ num__85 $ num__60 + $ num__70 + $ num__80 + x + $ num__125 = $ num__425 $ x = $ num__90 = answer = c <eor> c <eos> |
c |
add__80.0__5.0__ add__5.0__85.0__ add__5.0__90.0__ multiply__5.0__85.0__ add__5.0__85.0__ |
add__80.0__5.0__ add__5.0__85.0__ add__5.0__90.0__ multiply__5.0__85.0__ add__5.0__85.0__ |
| in one hour a boat goes num__11 km along the stream and num__5 km against the stream . the speed of the boat in still water is : <o> a ) num__3 <o> b ) num__5 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
solution speed in still water = num__0.5 ( num__11 + num__5 ) kmph = num__8 kmph . answer c <eor> c <eos> |
c |
round__8.0__ |
round__8.0__ |
| the cost price of a radio is rs . num__1680 and it was sold for rs . num__1450 find the loss % ? <o> a ) num__18 <o> b ) num__13 <o> c ) num__15 <o> d ) num__18 <o> e ) num__41 |
num__1680 - - - - num__230 num__100 - - - - ? = > num__13.0 answer : b <eor> b <eos> |
b |
percent__100.0__13.0__ |
percent__100.0__13.0__ |
| out of num__15 students in a class num__10 are wearing blue shirts num__2 are wearing green shirts and num__3 are wearing red shirts . four students are to be selected at random . what is the probability that at least one is wearing a green shirt ? <o> a ) num__0.47619047619 <o> b ) num__0.645161290323 <o> c ) num__0.731707317073 <o> d ) num__0.78431372549 <o> e ) num__0.819672131148 |
total possible ways to choose num__4 students out of num__15 = num__15 c num__4 = num__1365 the number of ways to choose num__4 students with no green shirts = num__13 c num__4 = num__715 p ( no green shirts ) = num__0.52380952381 = num__0.52380952381 p ( at least num__1 green shirt ) = num__1 - num__0.52380952381 = num__0.47619047619 the answer is a . <eor> a <eos> |
a |
negate_prob__0.5238__ negate_prob__0.5238__ |
negate_prob__0.5238__ negate_prob__0.5238__ |
| cole drove from home to work at an average speed of num__80 kmh . he then returned home at an average speed of num__120 kmh . if the round trip took a total of num__3 hours how many minutes did it take cole to drive to work ? <o> a ) num__66 <o> b ) num__70 <o> c ) num__72 <o> d ) num__85 <o> e ) num__108 |
first round distance travelled ( say ) = d speed = num__80 k / h time taken t num__2 = d / num__80 hr second round distance traveled = d ( same distance ) speed = num__120 k / h time taken t num__2 = d / num__120 hr total time taken = num__3 hrs therefore num__3 = d / num__80 + d / num__120 lcm of num__80 and num__120 = num__240 num__3 = d / num__80 + d / num__120 = > num__3 = num__3 d / num__240 + num__2 d / num__240 = > d = num__240 * num__0.4 km therefore t num__1 = d / num__80 = > t num__1 = ( num__240 * num__3 ) / ( num__5 x num__80 ) = > t num__1 = ( num__9 x num__60 ) / num__5 - - in minutes = > t num__1 = num__108 minutes . e <eor> e <eos> |
e |
multiply__80.0__3.0__ subtract__3.0__2.0__ add__3.0__2.0__ hour_to_min_conversion__ round__108.0__ |
multiply__80.0__3.0__ subtract__3.0__2.0__ add__3.0__2.0__ divide__120.0__2.0__ divide__108.0__1.0__ |
| on a sum of money the s . i . for num__2 years is $ num__600 while the c . i . is $ num__612 the rate of interest being the same in both the cases . the rate of interest is ? <o> a ) num__3.0 <o> b ) num__4.0 <o> c ) num__5.0 <o> d ) num__6.0 <o> e ) num__7 % |
difference in c . i . and s . i for num__2 years = $ num__612 - $ num__600 = $ num__12 s . i for one year = $ num__300 s . i . on $ num__300 for num__1 year = $ num__12 rate = ( num__100 * num__12 ) / ( num__300 ) = num__4.0 the answer is b . <eor> b <eos> |
b |
percent__2.0__600.0__ percent__100.0__4.0__ |
percent__2.0__600.0__ percent__100.0__4.0__ |
| the sum of number of boys and girls in a school is num__300 . if the number of boys is x then the number of girls becomes x % of the total number of students . the number of boys is ? <o> a ) num__50 <o> b ) num__40 <o> c ) num__60 <o> d ) num__100 <o> e ) num__75 |
we have x + x % of num__300 = num__300 x + x / num__100 * num__300 = num__300 num__4 * x = num__300 x = num__75 answer is e <eor> e <eos> |
e |
divide__300.0__4.0__ divide__300.0__4.0__ |
divide__300.0__4.0__ divide__300.0__4.0__ |
| john and steve are speed walkers in a race . john is num__15 meters behind steve when he begins his final push . john blazes to the finish at a pace of num__4.2 m / s while steve maintains a blistering num__3.8 m / s speed . if john finishes the race num__2 meters ahead of steve how long was john ’ s final push ? <o> a ) num__13 seconds <o> b ) num__17 seconds <o> c ) num__26 seconds <o> d ) num__34 seconds <o> e ) num__42.5 seconds |
let t be the time that john spent for his final push . thus per the question num__4.2 t = num__3.8 t + num__15 + num__2 - - - > num__0.4 t = num__17 - - - > t = num__42.5 seconds . e is the correct answer . <eor> e <eos> |
e |
subtract__4.2__3.8__ add__15.0__2.0__ divide__17.0__0.4__ round__42.5__ |
subtract__4.2__3.8__ add__15.0__2.0__ divide__17.0__0.4__ round__42.5__ |
| in a box of num__8 pens a total of num__2 are defective . if a customer buys num__2 pens selected at random from the box what is the probability that neither pen will be defective ? <o> a ) num__0.545454545455 <o> b ) num__0.533333333333 <o> c ) num__0.45 <o> d ) num__0.44 <o> e ) num__0.535714285714 |
p ( neither pen is defective ) = num__0.75 * num__0.714285714286 = num__0.535714285714 the answer is e . <eor> e <eos> |
e |
multiply__0.75__0.7143__ multiply__0.75__0.7143__ |
multiply__0.75__0.7143__ multiply__0.75__0.7143__ |
| â ˆ š num__1.5625 = ? <o> a ) num__1.05 <o> b ) num__1.25 <o> c ) num__1.45 <o> d ) num__1.55 <o> e ) num__1.65 |
explanation : num__1 | num__1.5625 ( num__1.25 | num__1 | - - - - - - - num__22 | num__56 | num__44 | - - - - - - - num__245 | num__1225 | num__1225 | - - - - - - - | x | - - - - - - - num__1.5625 = num__1.25 . answer is b <eor> b <eos> |
b |
round_down__1.5625__ divide__1.5625__1.25__ |
round_down__1.5625__ divide__1.5625__1.25__ |
| the sum of three consecutive integers is num__312 . what is the sum of the next three consecutive integers ? <o> a ) num__315 <o> b ) num__321 <o> c ) num__330 <o> d ) num__415 <o> e ) num__424 |
a + ( a + num__1 ) + ( a + num__2 ) = num__3 a + num__3 = num__312 ( a + num__3 ) + ( a + num__4 ) + ( a + num__5 ) = ( num__3 a + num__3 ) + num__9 = num__312 + num__9 = num__321 answer : b . <eor> b <eos> |
b |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ add__4.0__5.0__ add__312.0__9.0__ add__312.0__9.0__ |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ add__4.0__5.0__ add__312.0__9.0__ add__312.0__9.0__ |
| an epidemic is reported to have broken out in florida . the number of detected instances of a certain disease is reported to have increased by num__85.0 in the last year . what is the lowest number of newly detected instances possible ? <o> a ) num__1 <o> b ) num__5 <o> c ) num__11 <o> d ) num__15 <o> e ) num__17 |
can ( num__0.85 * n ) be num__1 ? no because in that case n will not be an integer . can ( num__0.85 * n ) be num__5 ? no because in that case n will not be an integer . notice that num__85 = num__5 * num__17 . so num__0.85 = num__0.85 ( num__0.85 * n ) = num__17 n = num__20 ( an integer ) we will not get n as integer in any of the other options . answer : e <eor> e <eos> |
e |
percent__85.0__20.0__ |
percent__85.0__20.0__ |
| the value of ( ( x – y ) ³ + ( y - z ) ³ + ( z – x ) ³ ) / ( num__9 ( x – y ) ( y – z ) ( z – x ) ) is equal to : <o> a ) num__0 <o> b ) num__0.0833333333333 <o> c ) num__1 <o> d ) num__0.25 <o> e ) num__0.333333333333 |
since ( x – y ) + ( y – z ) + ( z – x ) = num__0 so ( x – y ) ³ + ( y – z ) ³ + ( z – x ) ³ = num__3 ( x – y ) ( y – z ) ( z – x ) . ( num__3 ( x – y ) ( y – z ) ( z – x ) ) / ( num__9 ( x – y ) ( y – z ) ( z – x ) ) = num__0.333333333333 . answer : e <eor> e <eos> |
e |
reverse__3.0__ reverse__3.0__ |
reverse__3.0__ reverse__3.0__ |
| the least common multiple of positive integer d and num__3 - digit integer n is num__690 . if n is not divisible by num__3 and d is not divisible by num__2 what is the value of n ? <o> a ) num__115 <o> b ) num__230 <o> c ) num__460 <o> d ) num__575 <o> e ) num__690 |
the lcm of n and d is num__690 = num__2 * num__3 * num__5 * num__23 . d is not divisible by num__2 thus num__2 goes to n n is not divisible by num__3 thus num__3 goes to d . from above : n must be divisible by num__2 and not divisible by num__3 : n = num__2 * . . . in order n to be a num__3 - digit number it must take all other primes too : n = num__2 * num__5 * num__23 = num__230 . answer : b . <eor> b <eos> |
b |
add__3.0__2.0__ divide__690.0__3.0__ divide__690.0__3.0__ |
add__3.0__2.0__ divide__690.0__3.0__ divide__690.0__3.0__ |
| two stations a and b are num__110 km apart on a straight line . one train starts from a at num__7 a . m . and travels towards b at num__20 kmph . another train starts from b at num__8 a . m . and travels towards a at a speed of num__25 kmph . at what time will they meet ? <o> a ) num__18 a . m <o> b ) num__10 a . m <o> c ) num__167 a . m <o> d ) num__18 a . m <o> e ) num__11 a . m |
suppose they meet x hours after num__7 a . m . distance covered by a in x hours = num__20 x km . distance covered by b in ( x - num__1 ) hours = num__25 ( x - num__1 ) km . therefore num__20 x + num__25 ( x - num__1 ) = num__110 num__45 x = num__135 x = num__3 . so they meet at num__10 a . m . answer : b <eor> b <eos> |
b |
subtract__8.0__7.0__ add__20.0__25.0__ add__110.0__25.0__ divide__135.0__45.0__ add__7.0__3.0__ round__10.0__ |
subtract__8.0__7.0__ add__20.0__25.0__ add__110.0__25.0__ divide__135.0__45.0__ add__7.0__3.0__ add__7.0__3.0__ |
| a man purchased num__3 blankets @ rs . num__100 each num__5 blankets @ rs . num__150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . num__157 . find the unknown rate of two blankets ? <o> a ) num__420 <o> b ) num__550 <o> c ) num__490 <o> d ) num__450 <o> e ) num__520 |
num__10 * num__157 = num__1570 num__3 * num__100 + num__5 * num__150 = num__1050 num__1570 – num__1050 = num__520 answer : e <eor> e <eos> |
e |
multiply__157.0__10.0__ subtract__1570.0__1050.0__ subtract__1570.0__1050.0__ |
multiply__157.0__10.0__ subtract__1570.0__1050.0__ subtract__1570.0__1050.0__ |
| a train moves fast a telegraph post and a bridge num__264 m long in num__8 sec and num__20 sec respectively . what is the speed of the train ? <o> a ) num__79.7 <o> b ) num__79.8 <o> c ) num__76.2 <o> d ) num__79.2 <o> e ) num__29.2 |
let the length of the train be x m and its speed be y m / sec . then x / y = num__8 = > x = num__8 y ( x + num__264 ) / num__20 = y y = num__22 speed = num__22 m / sec = num__22 * num__3.6 = num__79.2 km / hr . answer : d <eor> d <eos> |
d |
multiply__3.6__22.0__ round__79.2__ |
multiply__3.6__22.0__ multiply__3.6__22.0__ |
| if the tens digit x and the units digit y of a positive integer n are reversed the resulting integer is num__81 more than n . what is y in terms of x ? <o> a ) num__10 - x <o> b ) num__9 - x <o> c ) x + num__9 <o> d ) x - num__1 <o> e ) x + num__1 |
original digits = xy i . e . number = num__10 x + y after reversing the digits : digits = yx i . e . number = num__10 y + x num__10 y + x is num__9 more than num__10 x + y num__10 x + y + num__81 = num__10 y + x num__10 x - x + num__81 = num__10 y - y num__9 x + num__81 = num__9 y x + num__9 = y or y = x + num__9 ans : c <eor> c <eos> |
c |
divide__81.0__9.0__ |
divide__81.0__9.0__ |
| a trader mixes num__26 kg of rice at rs . num__20 per kg with num__30 kg of rice of other variety at rs . num__36 per kg and sells the mixture at rs . num__30 per kg . his profit percent is : <o> a ) no profit no loss <o> b ) num__5.0 <o> c ) num__8.0 <o> d ) num__10.0 <o> e ) none of the above |
c . p . of num__56 kg rice = rs . ( num__26 x num__20 + num__30 x num__36 ) = rs . ( num__520 + num__1080 ) = rs . num__1600 . s . p . of num__56 kg rice = rs . ( num__56 x num__30 ) = rs . num__1680 gain = ( num__0.05 * num__100 ) % = num__5.0 answer = b <eor> b <eos> |
b |
percent__5.0__100.0__ |
percent__5.0__100.0__ |
| how many such letters pairs r in word frontier having same no . of letters left between them in the word as they have in the series <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
the no of letters between the specific two letters when we write in the alphabetical order should be same as the number of letters between those two letters in the given word two are possible num__1 ) letters ' o ' and ' n ' num__2 ) ' n ' and ' r ' . . . . . . ( middle letters are o p q i . e num__3 letters in alphabetical order & in the given word t i e i . e num__3 letters ) so num__2 such pairs are possible answer : b <eor> b <eos> |
b |
add__1.0__2.0__ multiply__1.0__2.0__ |
add__1.0__2.0__ multiply__1.0__2.0__ |
| convert num__100 miles into inches ? <o> a ) num__6336000 <o> b ) num__6542000 <o> c ) num__5462300 <o> d ) num__6213000 <o> e ) num__6120330 |
num__1 feet = num__12 inches num__1 mile = num__5280 feet num__100 mile = num__5280 * num__12 * num__100 = num__6336000 ans : a <eor> a <eos> |
a |
round__6336000.0__ |
round__6336000.0__ |
| curie bakes num__5 cookies for num__5 friends . num__2 of the friend arrive early and eat num__3 cookies each . how many cookies are left over if the other friends all have the same number of cookies ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
curie made num__5 * num__5 cookies = num__25 cookies . num__2 friends eat num__3 cookies leaving num__9 remaining for num__2 friends . num__9 is not divisible by num__2 . the num__2 friends ate a cumulative number of cookies less than num__9 that must be divisible by num__2 because the number is the same . the nearest multiple of num__2 is num__8 . num__9 - num__8 = num__1 a is the correct answer <eor> a <eos> |
a |
power__5.0__2.0__ power__3.0__2.0__ add__5.0__3.0__ subtract__3.0__2.0__ reverse__1.0__ |
power__5.0__2.0__ power__3.0__2.0__ add__5.0__3.0__ subtract__3.0__2.0__ subtract__2.0__1.0__ |
| a person want to give his money of $ num__12000 to his num__5 children a b c d e in the ratio num__2 : num__4 : num__3 : num__1 : num__5 . what is the d + e share ? <o> a ) $ num__1800 <o> b ) $ num__2800 <o> c ) $ num__3800 <o> d ) $ num__5800 <o> e ) $ num__4800 |
d ' s share = num__12000 * num__0.0666666666667 = $ num__800 d ' s share = num__12000 * num__0.333333333333 = $ num__4000 a + d = $ num__4800 answer is e <eor> e <eos> |
e |
reverse__3.0__ divide__12000.0__3.0__ add__800.0__4000.0__ multiply__1.0__4800.0__ |
reverse__3.0__ multiply__5.0__800.0__ add__800.0__4000.0__ multiply__1.0__4800.0__ |
| the h . c . f of two numbers is num__23 and the other two factors of their l . c . m are num__11 and num__12 . the larger of the two numbers is : <o> a ) num__338 <o> b ) num__276 <o> c ) num__322 <o> d ) num__231 <o> e ) num__121 |
clearly the numbers are ( num__23 * num__11 ) and ( num__23 * num__12 ) . larger number = ( num__23 * num__12 ) = num__276 . answer : b <eor> b <eos> |
b |
multiply__23.0__12.0__ multiply__23.0__12.0__ |
multiply__23.0__12.0__ multiply__23.0__12.0__ |
| the ratio of buses to cars on river road is num__1 to num__3 . if there are num__20 fewer buses than cars on river road how many cars are on river road ? <o> a ) num__100 <o> b ) num__120 <o> c ) num__140 <o> d ) num__30 <o> e ) num__150 |
b / c = num__0.333333333333 c - b = num__20 . . . . . . . . . > b = c - num__20 ( c - num__20 ) / c = num__0.333333333333 testing answers . clearly eliminate abce put c = num__30 . . . . . . . . . > ( num__30 - num__20 ) / num__30 = num__0.333333333333 = num__0.333333333333 answer : d <eor> d <eos> |
d |
reverse__3.0__ multiply__1.0__30.0__ |
reverse__3.0__ divide__30.0__1.0__ |
| a man can row downstream at num__16 kmph and upstream at num__10 kmph . find the speed of the man in still water and the speed of stream respectively ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__4 <o> e ) num__9 |
let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = num__16 - - - ( num__1 ) and x - y = num__10 - - - ( num__2 ) from ( num__1 ) & ( num__2 ) num__2 x = num__26 = > x = num__13 y = num__3 . answer : a <eor> a <eos> |
a |
add__16.0__10.0__ divide__26.0__2.0__ subtract__16.0__13.0__ round__3.0__ |
add__16.0__10.0__ divide__26.0__2.0__ subtract__16.0__13.0__ subtract__16.0__13.0__ |
| num__85 num__80 num__25 num__70 num__55 num__25 num__35 ? ? <o> a ) num__10 num__25 <o> b ) num__25 num__85 <o> c ) num__35 num__25 <o> d ) num__85 num__35 <o> e ) num__25 num__75 |
num__85 - num__5 = num__80 again num__25 num__80 - num__10 = num__70 num__70 - num__15 = num__55 again num__25 num__55 - num__20 = num__35 num__35 - num__25 = num__10 again num__25 then answer is num__10 num__25 answer : a <eor> a <eos> |
a |
subtract__85.0__80.0__ subtract__80.0__70.0__ subtract__85.0__70.0__ subtract__25.0__5.0__ subtract__80.0__70.0__ |
subtract__85.0__80.0__ subtract__80.0__70.0__ subtract__85.0__70.0__ subtract__25.0__5.0__ subtract__80.0__70.0__ |
| when num__2 is added to half of one - third of one - fifth of a number the result is one - fifteenth of the number . find the number ? <o> a ) num__40 <o> b ) num__50 <o> c ) num__60 <o> d ) num__70 <o> e ) num__80 |
let the number be num__2 + num__0.5 [ num__0.333333333333 ( a / num__5 ) ] = a / num__15 = > num__2 = a / num__30 = > a = num__60 answer : c <eor> c <eos> |
c |
reverse__2.0__ multiply__2.0__15.0__ multiply__2.0__30.0__ multiply__2.0__30.0__ |
reverse__2.0__ divide__15.0__0.5__ divide__30.0__0.5__ divide__30.0__0.5__ |
| x # is defined for every positive even integer x as the product of all even integers from num__2 to x . what is the smallest possible prime factor of ( x # + num__7 ) ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__11 |
since smallest possible prime factor is required . let x = num__2 then - - > num__2 # = num__2 x # + num__7 = = > num__2 # + num__7 = num__2 + num__7 = num__9 = num__3 * num__3 . thus num__3 is the smallest possible prime factor num__2 can not be the possible prime factor because - - > x # is always even since it is product of even numbers even + num__7 = odd . hence num__2 can not be prime factor of this result answer : b <eor> b <eos> |
b |
add__2.0__7.0__ divide__9.0__3.0__ |
add__2.0__7.0__ divide__9.0__3.0__ |
| rs . num__875 becomes rs . num__956 in num__3 years at a certain rate of simple interest . if the rate of interest is increased by num__4.0 what amount will rs . num__875 become in num__3 years ? <o> a ) rs . num__1020.80 <o> b ) rs . num__1025 <o> c ) rs . num__1061 <o> d ) data inadequate <o> e ) none of these |
solution s . i . = rs . ( num__956 - num__875 ) = rs . num__81 rate = ( num__100 x num__0.0925714285714 x num__3 ) = num__3.08571428571 % new rate = ( num__3.08571428571 + num__4 ) % = num__7.08571428571 % new s . i . = rs . ( num__875 x num__7.08571428571 x num__0.03 ) rs . num__186 . ∴ new amount = rs . ( num__875 + num__186 ) = rs . num__1061 . answer c <eor> c <eos> |
c |
percent__100.0__1061.0__ |
percent__100.0__1061.0__ |
| sakshi can do a piece of work in num__20 days . tanya is num__25.0 more efficient than sakshi . the number of days taken by tanya to do the same piece of work is : <o> a ) num__14 <o> b ) num__15 <o> c ) num__16 <o> d ) num__18 <o> e ) num__19 |
ratio of times taken by sakshi and tanya = num__125 : num__100 = num__5 : num__4 . suppose tanya takes x days to do the work . num__5 : num__4 : : num__20 : x x = ( num__4 x num__20 ) / num__5 x = num__16 days . hence tanya takes num__16 days to complete the work . answer : c <eor> c <eos> |
c |
subtract__125.0__25.0__ subtract__25.0__20.0__ divide__20.0__5.0__ subtract__20.0__4.0__ round__16.0__ |
subtract__125.0__25.0__ divide__100.0__20.0__ divide__20.0__5.0__ subtract__20.0__4.0__ round__16.0__ |
| the least common multiple of num__2 ^ num__6 − num__1 and num__2 ^ num__9 − num__1 is <o> a ) num__2 ^ num__12 + num__27 * num__2 ^ num__9 - num__217 <o> b ) num__2 ^ num__12 + num__63 * num__2 ^ num__3 - num__1 <o> c ) num__2 ^ num__12 + num__5 ^ num__29 - num__1 <o> d ) num__2 ^ num__12 + num__9 * num__2 ^ num__8 - num__1 <o> e ) none of these . |
the best approach in my opinion is what zarroli has suggested above . lcm * gcf = product of the numbers = ( num__2 ^ num__6 − num__1 ) ∗ ( num__2 ^ num__9 − num__1 ) = ( num__2 ^ num__3 − num__1 ) ( num__2 ^ num__3 + num__1 ) ∗ ( num__2 ^ num__3 − num__1 ) ( num__2 ^ num__6 + num__1 + num__2 ^ num__3 ) notice that the only common factor between them is ( num__23 − num__1 ) ( num__23 − num__1 ) so this must be the gcf . hence lcm will be the rest of the product . lcm = ( num__2 ^ num__3 + num__1 ) ∗ ( num__2 ^ num__3 − num__1 ) ( num__2 ^ num__6 + num__1 + num__2 ^ num__3 ) = ( num__2 ^ num__6 − num__1 ) ( num__2 ^ num__6 + num__1 + num__2 ^ num__3 ) now how do you get it in the format in the options ? almost all options have num__2 ^ num__12 and num__1 so retain those two and club everything else together . lcm = num__2 ^ num__12 − num__1 + ( num__2 ^ num__6 + num__2 ^ num__9 − num__2 ^ num__6 − num__2 ^ num__3 ) = num__2 ^ num__12 − num__1 + num__2 ^ num__3 ( num__2 ^ num__6 − num__1 ) = num__2 ^ num__12 − num__1 + num__2 ^ num__3 ∗ num__63 answer ( b ) <eor> b <eos> |
b |
add__2.0__1.0__ multiply__2.0__6.0__ multiply__2.0__1.0__ |
add__2.0__1.0__ multiply__2.0__6.0__ multiply__2.0__1.0__ |
| a fair coin is tossed repeatedly . if head appears on the first four tosses then the probability of appearance of tail on the fifth toss is . <o> a ) num__0.142857142857 <o> b ) num__0.5 <o> c ) num__0.428571428571 <o> d ) num__0.666666666667 <o> e ) num__0.333333333333 |
since the coin is a fair coin every toss is independent of any other toss . hence the probability of appearance of tail on the fifth toss is = num__0.5 answer : b <eor> b <eos> |
b |
negate_prob__0.5__ |
negate_prob__0.5__ |
| a train num__220 m long passed a pole in num__12 sec . how long will it take to pass a platform num__600 m long ? <o> a ) num__80 <o> b ) num__45 <o> c ) num__49 <o> d ) num__50 <o> e ) num__56 |
speed = num__18.3333333333 = num__18 m / sec . required time = ( num__220 + num__600 ) / num__18 = num__45 sec . answer : option b <eor> b <eos> |
b |
divide__220.0__12.0__ round__45.0__ |
divide__220.0__12.0__ round__45.0__ |
| john want to buy a $ num__100 trouser at the store but he think it â € ™ s too expensive . finally it goes on sale for $ num__80 . what is the percent decrease ? <o> a ) num__20.0 <o> b ) num__30.0 <o> c ) num__40.0 <o> d ) num__50.0 <o> e ) num__60 % |
the is always the difference between our starting and ending points . in this case it â € ™ s num__100 â € “ num__80 = num__20 . the â € œ original â € is our starting point ; in this case it â € ™ s num__100 . ( num__0.2 ) * num__100 = ( num__0.2 ) * num__100 = num__20.0 . a <eor> a <eos> |
a |
subtract__100.0__80.0__ divide__20.0__100.0__ subtract__100.0__80.0__ |
subtract__100.0__80.0__ divide__20.0__100.0__ multiply__100.0__0.2__ |
| suppose you have three identical prisms with congruent equilateral triangles as the end - polygons . suppose you attach them by the rectangular faces so they are perfectly aligned . there will be some large faces created by two or more co - planar faces of the individual prisms : count each such large face as one . given that how many faces does the resultant solid have <o> a ) num__4 <o> b ) num__9 <o> c ) num__6 <o> d ) num__10 <o> e ) num__12 |
to arrange them as described i did as follows on my scratch paper ( see the file attached ) where i counted num__4 sides and added top and bottom hence num__4 + num__2 = num__6 . c <eor> c <eos> |
c |
add__2.0__4.0__ add__2.0__4.0__ |
add__2.0__4.0__ add__2.0__4.0__ |
| two identical machines have the ability to produce both nuts and bolts . however it takes num__1 second to produce a bolt but num__2 seconds to produce a nut . what is the fastest the two machines working together can produce num__1200 nuts and num__1200 bolts ? <o> a ) num__1250 seconds <o> b ) num__1500 seconds <o> c ) num__1800 seconds <o> d ) num__2000 seconds <o> e ) num__3000 seconds |
i used the rate formula ( r = \ frac { num__1 } { t } ) rate of num__1 st machine = rate of making num__1 bolt + rate of making num__1 nut = num__1 + \ frac { num__1 } { num__2 } rate of num__1 st machine = rate of num__2 nd machine after this i got lost . please can you help how to approach using algebra ? i would n ' t complicate : num__1 machine needs num__1200 * num__1 seconds to produce num__1200 bolts so num__2 need half of that so num__600.0 seconds . = num__600 sec num__1 machine needs num__1200 * num__2 seconds to produce num__1200 nuts so num__2 need half of that so num__1200 * num__1.0 seconds . = num__1200 sec total = num__1800 c <eor> c <eos> |
c |
divide__1200.0__2.0__ add__1200.0__600.0__ round__1800.0__ |
divide__1200.0__2.0__ add__1200.0__600.0__ multiply__1.0__1800.0__ |
| a goods train leaves a station at a certain time and at a fixed speed . after ^ hours an express train leaves the same station and moves in the same direction at a uniform speed of num__90 kmph . this train catches up the goods train in num__4 hours . find the speed of the goods train . <o> a ) num__26 <o> b ) num__16 <o> c ) num__36 <o> d ) num__46 <o> e ) num__13 |
let the speed of the goods train be x kmph . distance covered by goods train in num__10 hours = distance covered by express train in num__4 hours num__10 x = num__4 x num__90 or x = num__36 . so speed of goods train = num__36 kmph . answer c <eor> c <eos> |
c |
round__36.0__ |
round__36.0__ |
| ten people are planning to share equally the cost of a rental van . if one person withdraws from the arrangement and the others share equally the cost of the rental van then the share of each of the remaining people will increase by . . . ? <o> a ) num__0.111111111111 <o> b ) num__0.125 <o> c ) num__0.142857142857 <o> d ) num__0.875 <o> e ) num__1.14285714286 |
let p = total cost of the rental van . the original share per person is p / num__10 . the new share per person is p / num__9 . p / num__9 = p / num__10 * num__1.11111111111 = ( num__1 + num__0.111111111111 ) * original share the answer is a . <eor> a <eos> |
a |
divide__10.0__9.0__ round_down__1.1111__ reverse__9.0__ reverse__9.0__ |
divide__10.0__9.0__ round_down__1.1111__ reverse__9.0__ reverse__9.0__ |
| the length of a rectangular landscape is num__4 times its breadth . there is a playground in it whose area is num__1200 square mtr & which is num__0.333333333333 rd of the total landscape . what is the length of the landscape ? <o> a ) num__100 <o> b ) num__110 <o> c ) num__120 <o> d ) num__140 <o> e ) num__150 |
sol . x * num__4 x = num__3 * num__1200 x = num__30 length = num__4 * num__30 = num__120 c <eor> c <eos> |
c |
square_perimeter__30.0__ square_perimeter__30.0__ |
multiply__4.0__30.0__ multiply__4.0__30.0__ |
| how many digits are in ( num__8 × num__10 ^ num__14 ) ( num__10 × num__10 ^ num__10 ) ? <o> a ) num__24 <o> b ) num__25 <o> c ) num__26 <o> d ) num__27 <o> e ) num__28 |
he question simplifies to ( num__8 × num__10 ^ num__14 ) ( num__10 ^ num__11 ) = > num__8 * num__10 ^ num__25 = > will contain num__25 zeros + num__1 digit num__8 = > num__26 ans c <eor> c <eos> |
c |
add__14.0__11.0__ subtract__11.0__10.0__ add__1.0__25.0__ add__1.0__25.0__ |
add__14.0__11.0__ subtract__11.0__10.0__ add__1.0__25.0__ add__1.0__25.0__ |
| a man can row num__6 kmph in still water . when the river is running at num__1.2 kmph it takes him num__1 hour to row to a place and black . how far is the place ? <o> a ) num__2.89 <o> b ) num__2.88 <o> c ) num__2.89 <o> d ) num__2.82 <o> e ) num__2.12 |
m = num__6 s = num__1.2 ds = num__6 + num__1.2 = num__7.2 us = num__6 - num__1.2 = num__4.8 x / num__7.2 + x / num__4.8 = num__1 x = num__2.88 answer : b <eor> b <eos> |
b |
add__6.0__1.2__ subtract__6.0__1.2__ round__2.88__ |
add__6.0__1.2__ subtract__6.0__1.2__ divide__2.88__1.0__ |
| last year department store x had a sales total for december that was num__2 times the average ( arithmetic mean ) of the monthly sales totals for january through november . the sales total for december was what fraction of the sales total for the year ? <o> a ) num__0.25 <o> b ) num__0.266666666667 <o> c ) num__0.153846153846 <o> d ) num__0.363636363636 <o> e ) num__0.8 |
let avg for num__11 mos . = num__10 therefore dec = num__20 year total = num__11 * num__10 + num__20 = num__130 answer = num__0.153846153846 = num__0.153846153846 = c <eor> c <eos> |
c |
multiply__2.0__10.0__ divide__20.0__130.0__ divide__20.0__130.0__ |
multiply__2.0__10.0__ divide__20.0__130.0__ divide__20.0__130.0__ |
| the area of a rectangular field is equal to num__750 square meters . its perimeter is equal to num__110 meters . find the width of this rectangle . <o> a ) num__5 <o> b ) num__10 <o> c ) num__15 <o> d ) num__20 <o> e ) num__25 |
l * w = num__750 : area l is the length and w is the width . num__2 l + num__2 w = num__110 : perimeter l = num__55 - w : solve for l ( num__55 - w ) * w = num__750 : substitute in the area equation w = num__25 and l = num__30 correct answer e <eor> e <eos> |
e |
triangle_area__2.0__25.0__ |
triangle_area__2.0__25.0__ |
| a number is said to be a “ digifac ” if each of its digits is a factor of the number itself . what is the sum m of the missing digits of the following five - digit digifac : num__9 num__5 num__3 _ _ ? <o> a ) num__5 <o> b ) num__7 <o> c ) num__9 <o> d ) num__10 <o> e ) num__14 |
here the term “ digifac ” should look intimidating . you probably haven ’ t studied digifacs before so how should you approach this problem ? well keep in mind that digifacs aren ’ t being tested ; in fact the author of this question just made that term up and then defined it for you . what makes this question hard is that the non - challenge - seeker ( i think i just made that term up too … ) will see the unfamiliar term “ digifac ” and lose faith immediately . “ i don ’ t know what that is ! ” she who finds the challenge in the gmat fun however will read the definition and think “ got it – i need to find the two digits that ensure that num__9 num__5 and num__3 are both factors of the overall number and that the remaining two digits are also factors ” . and work from there . the number must be divisible by num__5 so the only units digits that work are num__0 or num__5 . and the number must be divisible by num__9 ( and also num__3 ) so we need the sum m of all digits to be a multiple of num__9 . num__9 + num__5 + num__3 = num__17 so our only options are to get the sum to num__18 ( by adding num__1 ) or to num__27 ( by adding num__10 ) . a quick glance at the answer choices shows that num__0 num__1 isn ’ t an option . why not ? that would require num__0 to be one of the digits … and num__0 isn ’ t a factor of anything . so the units digit must be num__5 making the tens digit num__5 and we have num__95355 . that number is a multiple of num__5 num__3 and num__9 so it works : the correct answer is d and more importantly this fun challenge required no “ trivial ” information about digifacs … that term only existed to obscure the link between the given information and the path to the answer . d <eor> d <eos> |
d |
subtract__18.0__17.0__ multiply__9.0__3.0__ add__9.0__1.0__ add__9.0__1.0__ |
subtract__18.0__17.0__ add__9.0__18.0__ add__9.0__1.0__ add__9.0__1.0__ |
| the food in a camp lasts for num__35 men for num__40 days . if ten more men join how many days will the food last ? <o> a ) num__40 days <o> b ) num__20 days <o> c ) num__31 days <o> d ) num__50 days <o> e ) num__45 days |
one man can consume the same food in num__35 * num__40 = num__1400 days . num__10 more men join the total number of men = num__45 the number of days the food will last = num__31.1111111111 = num__31 days . answer : c <eor> c <eos> |
c |
multiply__35.0__40.0__ add__35.0__10.0__ divide__1400.0__45.0__ round__31.0__ |
multiply__35.0__40.0__ add__35.0__10.0__ divide__1400.0__45.0__ round__31.0__ |
| if an amount of rs num__42240 is distributed equally amongst num__22 persons how much amount would each person get ? <o> a ) rs num__1905 <o> b ) rs num__1920 <o> c ) rs num__745 <o> d ) rs num__765 <o> e ) none |
required amount = num__1920.0 = rs num__1920 answer b <eor> b <eos> |
b |
divide__42240.0__22.0__ divide__42240.0__22.0__ |
divide__42240.0__22.0__ divide__42240.0__22.0__ |
| a num__160 meter long train crosses a man standing on the platform in num__18 sec . what is the speed of the train ? <o> a ) num__96 kmph <o> b ) num__94 kmph <o> c ) num__32 kmph <o> d ) num__56 kmph <o> e ) num__76 kmph |
s = num__8.88888888889 * num__3.6 = num__32 kmph answer : c <eor> c <eos> |
c |
divide__160.0__18.0__ multiply__8.8889__3.6__ round__32.0__ |
divide__160.0__18.0__ multiply__8.8889__3.6__ multiply__8.8889__3.6__ |
| if negative integers k and p are not both even which of the following must be odd ? <o> a ) kp <o> b ) num__4 ( k + p ) <o> c ) k – p <o> d ) k + num__1 – p <o> e ) num__2 ( k + p ) – num__1 |
if negative integers k and p are not both even which of the following must be odd ? ( a ) kp - - > can be even if either k or p is even . ( b ) num__4 ( k + p ) - - > always even . ( c ) k – p - - > can be even as well as odd . for example consider k - p = odd - odd = even and k - p = odd - even = odd . ( d ) k + num__1 – p - - > can be even as well as odd . for example consider k + num__1 - p = odd + odd - odd = odd and k + num__1 - p = odd + odd - even = even . ( e ) num__2 ( k + p ) – num__1 - - > always odd because : num__2 ( k + p ) - num__1 = num__2 * integer - odd = even - odd = odd . answer : e . <eor> e <eos> |
e |
coin_space__ coin_space__ |
coin_space__ coin_space__ |
| difference between the length & breadth of a rectangle is num__30 m . if its perimeter is num__300 m then its area is ? ? we have : ( l - b ) = num__30 and num__2 ( l + b ) = num__300 or ( l + b ) = num__150 ? <o> a ) num__5425 m ^ num__2 <o> b ) num__5250 m ^ num__2 <o> c ) num__5300 m ^ num__2 <o> d ) num__5200 m ^ num__2 <o> e ) num__5400 m ^ num__2 |
solving the two equations we get : l = num__90 and b = num__60 . area = ( l x b ) = ( num__90 x num__60 ) m num__2 = num__5400 m ^ num__2 e <eor> e <eos> |
e |
multiply__30.0__2.0__ multiply__90.0__60.0__ multiply__90.0__60.0__ |
multiply__30.0__2.0__ multiply__90.0__60.0__ multiply__90.0__60.0__ |
| an athlete runs num__200 metres race in num__24 seconds . what is his speed ? <o> a ) num__20 km / hr <o> b ) num__30 km / hr <o> c ) num__25 km / hr <o> d ) num__27.5 km / hr <o> e ) none of these |
explanation : speed = distance / time = num__8.33333333333 m / s = num__8.33333333333 × num__3.6 km / hr = num__40 × num__34 km / hr = num__10 × num__3 km / hr = num__30 km / hr answer : option b <eor> b <eos> |
b |
divide__200.0__24.0__ subtract__34.0__24.0__ multiply__3.0__10.0__ round__30.0__ |
divide__200.0__24.0__ subtract__34.0__24.0__ multiply__3.0__10.0__ round__30.0__ |
| a reduction of num__25.0 in the price of oil enables a house wife to obtain num__5 kgs more for rs . num__800 what is the reduced price for kg ? <o> a ) num__72 <o> b ) num__27 <o> c ) num__40 <o> d ) num__28 <o> e ) num__20 |
num__800 * ( num__0.25 ) = num__200 - - - - num__5 ? - - - - num__1 = > rs . num__40 answer : c <eor> c <eos> |
c |
percent__25.0__800.0__ percent__5.0__800.0__ percent__5.0__800.0__ |
percent__25.0__800.0__ percent__5.0__800.0__ percent__5.0__800.0__ |
| a train crosses a bridge of length num__2500 m in num__120 seconds and a lamp post on the bridge in num__30 seconds . what is the length of the train in metres ? <o> a ) num__375 m <o> b ) num__250.0 m <o> c ) num__833.333333333 m <o> d ) num__800 m <o> e ) num__300 m |
let length of train = l case - num__1 : distance = num__2500 + l ( while crossing the bridge ) time = num__120 seconds i . e . speed = distance / time = ( num__2500 + l ) / num__120 case - num__2 : distance = l ( while passing the lamp post ) time = num__30 seconds i . e . speed = distance / time = ( l ) / num__30 but since speed has to be same in both cases so ( num__2500 + l ) / num__120 = ( l ) / num__30 i . e . num__4 l = num__2500 + l i . e . num__3 l = num__2500 i . e . l = num__833.333333333 answer : option c <eor> c <eos> |
c |
divide__120.0__30.0__ add__1.0__2.0__ divide__2500.0__3.0__ divide__2500.0__3.0__ |
divide__120.0__30.0__ add__1.0__2.0__ divide__2500.0__3.0__ divide__2500.0__3.0__ |
| in what time will a railway train num__60 m long moving at the rate of num__54 kmph pass a telegraph post on its way ? <o> a ) num__4 sec <o> b ) num__7 sec <o> c ) num__2 sec <o> d ) num__6 sec <o> e ) num__9 sec |
t = num__1.11111111111 * num__3.6 = num__4 sec answer : a <eor> a <eos> |
a |
divide__60.0__54.0__ multiply__3.6__1.1111__ round__4.0__ |
divide__60.0__54.0__ multiply__3.6__1.1111__ multiply__3.6__1.1111__ |
| a boatman goes num__2 km against the current of the stream in num__1 hour and goes num__1 km along the current in num__10 minutes . how long will it take to go num__5 km in stationary water ? <o> a ) num__1 hr num__15 min <o> b ) num__1 hr num__30 min <o> c ) num__1 hr <o> d ) num__2 hr <o> e ) num__2 hr num__30 min |
upstream speed = u - v ; downstream speed = u + v ; . according to formula s = d / t . u - v = num__2.0 and u + v = num__6.0 . on solving these equations we get . u = num__4 and v = num__2 . therefore speed of boatman = num__4 km / hr . hence t = d / s . t = num__1.25 . we get num__1 hr num__15 min is answer . answer : a <eor> a <eos> |
a |
add__1.0__5.0__ subtract__10.0__6.0__ divide__5.0__4.0__ add__10.0__5.0__ round__1.0__ |
add__1.0__5.0__ subtract__10.0__6.0__ divide__5.0__4.0__ add__10.0__5.0__ subtract__2.0__1.0__ |
| a train num__90 meters long completely crosses a num__270 - meter long bridge in num__36 seconds . what is the speed of the train in km / h ? <o> a ) num__28 km / h <o> b ) num__32 km / h <o> c ) num__36 km / h <o> d ) num__40 km / h <o> e ) num__44 km / h |
speed = distance / time = ( num__90 + num__270 ) / num__36 = num__10.0 ( m / s ) * num__3.6 = num__36 km / h the answer is c . <eor> c <eos> |
c |
divide__36.0__10.0__ round__36.0__ |
divide__36.0__10.0__ multiply__3.6__10.0__ |
| num__36 men can complete a piece of work in num__18 days . in how many days will num__27 men complete the same work ? <o> a ) num__26 <o> b ) num__22 <o> c ) num__12 <o> d ) num__24 <o> e ) num__20 |
explanation : let the required number of days be x more men less days ( indirect proportion ) hence we can write as ( men ) num__36 : num__27 : : x : num__18 ⇒ num__36 × num__18 = num__27 × x ⇒ num__12 × num__18 = num__9 × x ⇒ num__12 × num__2 = x ⇒ x = num__24 answer : option d <eor> d <eos> |
d |
subtract__36.0__27.0__ divide__36.0__18.0__ subtract__36.0__12.0__ round__24.0__ |
subtract__36.0__27.0__ divide__36.0__18.0__ subtract__36.0__12.0__ round__24.0__ |
| in your classes you counted num__122 hands . how many students were at the class ? <o> a ) num__60 <o> b ) num__61 <o> c ) num__85 <o> d ) num__86 <o> e ) num__88 |
total number of hands = num__122 we have num__2 hands . to find how many students were at the class divide the total number of hands by the number of hands we have . we get divide num__122 by num__2 num__122 ÷ num__2 = num__61 therefore there were num__61 students at the class . answer is b <eor> b <eos> |
b |
divide__122.0__2.0__ divide__122.0__2.0__ |
divide__122.0__2.0__ divide__122.0__2.0__ |
| the distance between west - town to east - town is num__20 kilometers . two birds start flying simultaneously towards one another the first leaving from west - town at a speed of num__4 kilometers per minute and the second bird leaving from east - town at a speed of num__1 kilometers per minute . what will be the distance in kilometers between the meeting point and west - town ? <o> a ) num__3 . <o> b ) num__7 . <o> c ) num__10 . <o> d ) num__16 . <o> e ) num__15 . |
time taken by the birds to meet = num__20 / ( num__4 + num__1 ) = num__4 mins distance traveled by the bird traveling from west - town = num__4 * num__4 = num__16 answer : d <eor> d <eos> |
d |
subtract__20.0__4.0__ round__16.0__ |
subtract__20.0__4.0__ subtract__20.0__4.0__ |
| in how many different number of ways num__4 boys and num__2 girls can sit on a bench ? <o> a ) num__700 <o> b ) num__710 <o> c ) num__720 <o> d ) num__730 <o> e ) num__740 |
npn = n ! num__6 p num__6 = num__6 × num__5 × num__4 × num__3 × num__2 × num__1 = num__720 c <eor> c <eos> |
c |
die_space__ vowel_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
die_space__ vowel_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| two sides of a triangle have lengths x and w and meet at a right angle . if the perimeter of the triangle is num__4 x what is the ratio of x to w ? <o> a ) a ) num__2 : num__3 <o> b ) b ) num__3 : num__4 <o> c ) c ) num__4 : num__3 <o> d ) d ) num__3 : num__2 <o> e ) e ) num__2 : num__1 |
ahhhh . . . . my bad . thank you you ' re correct . num__8 x = num__6 w - - > x / w = num__0.75 . actually b is the right answer . <eor> b <eos> |
b |
square_perimeter__0.75__ |
square_perimeter__0.75__ |
| two trains are traveling on parallel tracks in the same direction . the faster train travels at num__130 miles per hour while the slower train travels at num__100 miles per hour . at num__3 o ’ clock the faster train is num__15 miles behind the slower one . how far apart are the two trains at num__5 o ' clock ? <o> a ) num__45 miles <o> b ) num__80 miles <o> c ) num__90 miles <o> d ) num__120 miles <o> e ) num__400 miles |
answer = a . num__45 miles relational speed = num__130 - num__100 = num__30 miles per hour in num__2 hours difference = num__30 * num__2 = num__60 miles fast train trailing num__15 miles so effective difference = num__60 - num__15 = num__45 miles <eor> a <eos> |
a |
multiply__3.0__15.0__ subtract__130.0__100.0__ subtract__5.0__3.0__ hour_to_min_conversion__ round__45.0__ |
multiply__3.0__15.0__ subtract__130.0__100.0__ subtract__5.0__3.0__ multiply__2.0__30.0__ multiply__3.0__15.0__ |
| if m and n are whole numbers such that mn = num__121 the value of ( m – num__1 ) n + num__1 is : <o> a ) num__1 <o> b ) num__10 <o> c ) num__121 <o> d ) num__1000 <o> e ) num__131 |
explanation we know that num__112 = num__121 . putting m = num__11 and n = num__2 we get : ( m – num__1 ) n + num__1 = ( num__11 – num__1 ) ( num__2 + num__1 ) = num__10 ( num__3 ) = num__1000 . answer d <eor> d <eos> |
d |
subtract__11.0__1.0__ add__1.0__2.0__ multiply__1.0__1000.0__ |
subtract__11.0__1.0__ add__1.0__2.0__ multiply__1.0__1000.0__ |
| the diameter of a wheel of cycle is num__21 cm . it moves slowly along a road . how far will it go in num__500 revolutions ? <o> a ) num__260 m <o> b ) num__290 m <o> c ) num__320 m <o> d ) num__330 m <o> e ) num__380 m |
in revolution distance that wheel covers = circumference of wheel diameter of wheel = num__21 cm therefore circumference of wheel = π d = num__3.14285714286 × num__21 = num__66 cm so in num__1 revolution distance covered = num__66 cm in num__500 revolution distance covered = num__66 × num__500 cm = num__33000 cm = num__330.0 m = num__330 m answer : option d <eor> d <eos> |
d |
multiply__500.0__66.0__ round__330.0__ |
multiply__500.0__66.0__ round__330.0__ |
| the value of num__1 + [ num__1 / ( num__8 x num__2 ) ] + [ num__1 / ( num__8 x num__2 ² ) ] + [ num__1 / ( num__8 x num__2 ³ ) ] is : <o> a ) num__1.109375 <o> b ) num__0.0625 <o> c ) num__0.25 <o> d ) num__0.333333333333 <o> e ) none of these |
( num__8 x num__2 ³ + num__2 ² + num__2 + num__1 ) / ( num__8 x num__2 ³ ) = ( num__64 + num__4 + num__2 + num__1 ) / num__64 = num__1.109375 . answer : a <eor> a <eos> |
a |
divide__8.0__2.0__ multiply__1.0__1.1094__ |
divide__8.0__2.0__ divide__1.1094__1.0__ |
| how many num__3 - digit numbers can be formed from the digits num__2 num__3 num__5 num__6 num__7 and num__9 which are divisible by num__5 and none of the digits is repeated ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__51 <o> d ) num__15 <o> e ) num__20 |
since each desired number is divisible by num__5 so we must have num__5 at the unit place . so there is num__1 way of doing it . the tens place can now be filled by any of the remaining num__5 digits ( num__2 num__3 num__6 num__7 num__9 ) . so there are num__5 ways of filling the tens place . the hundreds place can now be filled by any of the remaining num__4 digits . so there are num__4 ways of filling it . required number of numbers = ( num__1 x num__5 x num__4 ) = num__20 . option e <eor> e <eos> |
e |
subtract__3.0__2.0__ add__3.0__1.0__ multiply__5.0__4.0__ multiply__5.0__4.0__ |
subtract__3.0__2.0__ add__3.0__1.0__ multiply__5.0__4.0__ multiply__5.0__4.0__ |
| if num__40.0 of a number is num__28 then the number is <o> a ) number is num__60 <o> b ) number is num__70 <o> c ) number is num__90 <o> d ) number is num__80 <o> e ) none of these |
solution : let ` ` a ' ' be the number we seek . we get num__40.0 × a = num__28 num__2 × num__40.0 × a = num__2 × num__28 = num__56 num__80.0 × a = num__56 num__10.0 × a = num__56 : num__8 = num__7 num__100.0 × a = num__7 × num__10 = num__70 a = num__70 answer b <eor> b <eos> |
b |
percent__10.0__80.0__ percent__100.0__70.0__ |
percent__10.0__80.0__ percent__100.0__70.0__ |
| a man purchases num__8 pens for rs . num__9 and sells num__9 pens for rs . num__8 how much profit or loss does he make ? <o> a ) num__20.97 <o> b ) num__20.98 <o> c ) num__90.97 <o> d ) num__20.95 <o> e ) num__20.91 |
num__81 - - - - num__17 num__100 - - - - - ? è num__20.98 loss answer : option b explanatib <eor> b <eos> |
b |
percent__100.0__20.98__ |
percent__100.0__20.98__ |
| in the city of san durango num__90 people own cats dogs or rabbits . if num__30 people owned cats num__30 owned dogs num__50 owned rabbits and num__14 owned exactly two of the three types of pet how many people owned all three ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
total = c + d + r - ( cd + dr + cr ) - num__2 cdr num__90 = num__30 + num__30 + num__50 - ( num__14 ) - num__2 x x = num__3 so answer will be c <eor> c <eos> |
c |
divide__90.0__30.0__ divide__90.0__30.0__ |
divide__90.0__30.0__ divide__90.0__30.0__ |
| if the speed of x inch per second is equivalent to the speed of y foot per hour what is y in terms of x ? ( num__1 foot = num__12 inch ) <o> a ) num__120 x <o> b ) num__240 x <o> c ) num__360 x <o> d ) num__300 x <o> e ) num__100 x |
x inch per second - - > - - > num__3600 x inch per hour ( as there are num__3600 seconds in one hour ) ; - - > num__3600 x / num__12 = num__300 x foot per hour ( as there are num__12 inches in one foot ) . answer : d . <eor> d <eos> |
d |
divide__3600.0__12.0__ round__300.0__ |
divide__3600.0__12.0__ divide__3600.0__12.0__ |
| which of the following is between num__0.757142857143 and num__0.833333333333 ? <o> a ) num__0.333333333333 <o> b ) num__0.375 <o> c ) num__0.5 <o> d ) num__0.625 <o> e ) num__0.777777777778 |
i see this as a poe ( process of elimination ) and ballparking ( estimation ) question . not sure if this is the most efficient but it worked : num__1 ) i estimate num__0.757142857143 to be ~ num__0.75 + ( approximately slightly greater than num__0.75 ) num__2 ) i estimate num__0.833333333333 to be ~ num__0.8 + ( approximately slightly greater than num__0.8 ) num__3 ) so now i ' m looking for an answer choice that is . num__75 < x < . num__80 it should be pretty easy to identify which answer choices are less than . num__75 ( or num__0.75 ) if you have had practice with fractions . a ) clearly wrong - it ' s less than num__0.75 b ) also clearly wrong - it ' s less than num__0.75 c ) also clearly wrong - it ' s less than num__0.75 d ) also clearly wrong - it ' s less than num__0.75 i would just stop here and select e . but to verify : e ) num__0.777777777778 ~ . num__77 which is greater than . num__75 and less than . num__80 <eor> e <eos> |
e |
add__1.0__2.0__ add__2.0__75.0__ multiply__1.0__0.7778__ |
add__1.0__2.0__ add__2.0__75.0__ multiply__1.0__0.7778__ |
| what is the least number should be added to num__1056 so the sum of the number is completely divisible by num__28 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__8 |
( num__37.7142857143 ) gives remainder num__20 num__20 + num__8 = num__28 so we need to add num__8 e <eor> e <eos> |
e |
divide__1056.0__28.0__ subtract__28.0__20.0__ subtract__28.0__20.0__ |
divide__1056.0__28.0__ subtract__28.0__20.0__ subtract__28.0__20.0__ |
| ajith and rana walk around a circular course num__115 km in circumference starting together from the same point . if they walk at speed of num__4 and num__5 kmph respectively in the same direction when will they meet ? <o> a ) after num__20 hours <o> b ) after num__115 hours <o> c ) after num__115 minutes <o> d ) after num__20 minutes <o> e ) after num__30 minutes |
rana is the faster person . he gains num__1 km in num__1 hour . so rana will gain one complete round over ajith in num__115 hours . i . e . they will meet after num__115 hours . answer : b <eor> b <eos> |
b |
subtract__5.0__4.0__ round__115.0__ |
subtract__5.0__4.0__ round__115.0__ |
| how much time will take for an amount of rs . num__360 to yield rs . num__81 as interest at num__4.5 per annum of simple interest ? <o> a ) num__5 years <o> b ) num__6 years <o> c ) num__7 years <o> d ) num__12 years <o> e ) num__15 years |
time = ( num__100 * num__81 ) / ( num__360 * num__4.5 ) = num__5 years answer : a <eor> a <eos> |
a |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| the average age of a class of num__32 students is num__16 yrs . if the teacher ' s age is also included the average increases by one year . find the age of the teacher <o> a ) num__34 years <o> b ) num__37 years <o> c ) num__49 years <o> d ) num__51 years <o> e ) num__57 years |
total age of students is num__32 x num__16 = num__512 years total age inclusive of teacher = num__33 x ( num__16 + num__1 ) = num__561 so teacher ' s age is num__561 - num__512 = num__49 yrs there is a shortcut for these type of problems teacher ' s age is num__16 + ( num__33 x num__1 ) = num__49 years c <eor> c <eos> |
c |
multiply__32.0__16.0__ subtract__33.0__32.0__ add__16.0__33.0__ add__16.0__33.0__ |
multiply__32.0__16.0__ subtract__33.0__32.0__ add__16.0__33.0__ add__16.0__33.0__ |
| in a family num__7 children dont eat spinach num__6 dont eat carrot num__5 dont eat beans num__4 dont eat spinach & carrots num__3 dont eat carrot & beans num__2 dont eat beans & spinach . one doesnt eat all num__3 . find the no . of children . <o> a ) num__9 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__13 |
do not spinach only = num__2 do not carrot only = num__0 do not beans only = num__1 do not spinach & carrots only = num__3 do not beans & carrots only = num__2 do not spinach & beans only = num__1 do not all three = num__1 total num__10 children ( num__2 + num__0 + num__1 + num__3 + num__2 + num__1 + num__1 ) answer : b <eor> b <eos> |
b |
subtract__7.0__6.0__ add__7.0__3.0__ add__7.0__3.0__ |
subtract__7.0__6.0__ add__7.0__3.0__ add__7.0__3.0__ |
| to deliver an order on time a company has to make num__25 parts a day . after making num__25 parts per day for num__3 days the company started to produce num__5 more parts per day and by the last day of work num__100 more parts than planned were produced . find how many parts the company made and how many days this took . <o> a ) num__530 <o> b ) num__675 <o> c ) num__330 <o> d ) num__987 <o> e ) num__854 |
let x e the number of days the company worked . then num__25 x is the number of parts they planned to make . at the new production rate they made : num__3 ⋅ num__25 + ( x − num__3 ) ⋅ num__30 = num__75 + num__30 ( x − num__3 ) therefore : num__25 x = num__75 + num__30 ( x − num__3 ) − num__100 num__25 x = num__75 + num__30 x − num__90 − num__100 num__190 − num__75 = num__30 x − num__25 num__115 = num__5 x x = num__23 so the company worked num__23 days and they made num__23 ⋅ num__25 + num__100 = num__675 pieces . <eor> b <eos> |
b |
add__25.0__5.0__ multiply__25.0__3.0__ multiply__3.0__30.0__ add__100.0__90.0__ add__25.0__90.0__ divide__115.0__5.0__ round__675.0__ |
add__25.0__5.0__ multiply__25.0__3.0__ multiply__3.0__30.0__ add__100.0__90.0__ add__25.0__90.0__ divide__115.0__5.0__ round__675.0__ |
| a man has rs . num__480 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__45 <o> b ) num__60 <o> c ) num__75 <o> d ) num__90 <o> e ) none |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__480 num__16 x = num__480 x = num__30 . hence total number of notes = num__3 x = num__90 . answer d <eor> d <eos> |
d |
divide__480.0__16.0__ divide__30.0__10.0__ multiply__3.0__30.0__ multiply__3.0__30.0__ |
divide__480.0__16.0__ divide__30.0__10.0__ multiply__3.0__30.0__ multiply__3.0__30.0__ |
| from a group of boys and girls num__15 girls leave . there are then left num__2 boys for each girl . after this num__45 boys leave . there are then num__5 girls for each boy . find the number of girls in the beginning ? <o> a ) num__40 <o> b ) num__10 <o> c ) num__50 <o> d ) num__30 <o> e ) num__20 |
let at present there be x boys . then no of girls at present = num__5 x before the boys had left : no of boys = x + num__45 and no of girls = num__5 x x + num__45 = num__2 * num__5 x num__9 x = num__45 x = num__5 no of girls in the beginning = num__25 + num__15 = num__40 answer is a <eor> a <eos> |
a |
divide__45.0__5.0__ power__5.0__2.0__ add__15.0__25.0__ add__15.0__25.0__ |
divide__45.0__5.0__ power__5.0__2.0__ add__15.0__25.0__ add__15.0__25.0__ |
| the average temperature for monday tuesday wednesday and thursday was num__48 degrees and for tuesday wednesday thursday and friday was num__46 degrees . if the temperature on monday was num__42 degrees . find the temperature on friday ? <o> a ) num__65 degrees <o> b ) num__73 degrees <o> c ) num__37 degrees <o> d ) num__34 degrees <o> e ) num__74 degrees |
m + tu + w + th = num__4 * num__48 = num__192 tu + w + th + f = num__4 * num__46 = num__184 m = num__42 tu + w + th = num__192 - num__42 = num__150 f = num__184 – num__150 = num__34 answer : d <eor> d <eos> |
d |
subtract__46.0__42.0__ multiply__48.0__4.0__ multiply__46.0__4.0__ subtract__192.0__42.0__ subtract__184.0__150.0__ subtract__184.0__150.0__ |
subtract__46.0__42.0__ multiply__48.0__4.0__ multiply__46.0__4.0__ subtract__192.0__42.0__ subtract__184.0__150.0__ subtract__184.0__150.0__ |
| the perimeter of a triangle is num__32 cm and the inradius of the triangle is num__2.5 cm . what is the area of the triangle ? <o> a ) num__46 cm num__2 <o> b ) num__42 cm num__2 <o> c ) num__29 cm num__2 <o> d ) num__25 cm num__2 <o> e ) num__40 cm num__2 |
area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = num__2.5 * num__16.0 = num__40 cm num__2 answer : e <eor> e <eos> |
e |
triangle_area__32.0__2.5__ triangle_area__32.0__2.5__ |
multiply__2.5__16.0__ multiply__2.5__16.0__ |
| two digits number is divided by the product of their digit and the result is num__3 . now num__18 is added to the number the digit will be inter changed . what is the number . <o> a ) num__16 <o> b ) num__18 <o> c ) num__20 <o> d ) num__22 <o> e ) num__24 |
let tenth digit of the number is x and unit digit of that number is y then the number will be ( num__10 x + y ) . now according to question ( num__10 x + y ) / xy = num__3 ( num__10 x + y ) = num__3 xy - - - - - - - - - - - - - - - - - - - - - - - - - - - ( num__1 ) and num__10 x + y + num__18 = num__10 y + x y - x = num__2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ( num__2 ) y = num__2 + x - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ( num__3 ) on solving all the equations we get number is num__24 . answer : e <eor> e <eos> |
e |
subtract__3.0__1.0__ multiply__1.0__24.0__ |
subtract__3.0__1.0__ divide__24.0__1.0__ |
| two persons start running simultaneously around a circular track of length num__300 m from the same point at speeds of num__15 km / hr and num__25 km / hr . when will they meet for the first time any where on the track if they are moving in opposite directions ? <o> a ) num__11 <o> b ) num__10 <o> c ) num__28 <o> d ) num__27 <o> e ) num__12 |
time taken to meet for the first time anywhere on the track = length of the track / relative speed = num__300 / ( num__15 + num__25 ) num__0.277777777778 = num__300 * num__0.45 * num__5 = num__27 seconds . answer : d <eor> d <eos> |
d |
round__27.0__ |
round__27.0__ |
| two employees x and y are paid a total of rs . num__800 per week by their employer . if x is paid num__120 percent of the sum paid to y how much is y paid per week ? <o> a ) s . num__200.63 <o> b ) s . num__250.63 <o> c ) s . num__290.63 <o> d ) s . num__300.63 <o> e ) s . num__363.63 |
let the amount paid to x per week = x and the amount paid to y per week = y then x + y = num__800 but x = num__120.0 of y = num__120 y / num__100 = num__12 y / num__10 ∴ num__12 y / num__10 + y = num__800 ⇒ y [ num__1.2 + num__1 ] = num__800 ⇒ num__22 y / num__10 = num__800 ⇒ num__22 y = num__8000 ⇒ y = num__363.636363636 = rs . num__363.63 e ) <eor> e <eos> |
e |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__12.0__10.0__ multiply__800.0__10.0__ divide__8000.0__22.0__ multiply__1.0__363.63__ |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__12.0__10.0__ multiply__800.0__10.0__ divide__8000.0__22.0__ divide__363.63__1.0__ |
| a num__300 meter long train crosses a platform in num__39 seconds while it crosses a signal pole in num__18 seconds . what is the length of the platform ? <o> a ) num__399 <o> b ) num__267 <o> c ) num__350 <o> d ) num__788 <o> e ) num__277 |
speed = [ num__16.6666666667 ] m / sec = num__16.6666666667 m / sec . let the length of the platform be x meters . then x + num__7.69230769231 = num__16.6666666667 num__3 ( x + num__300 ) = num__1950 è x = num__350 m . answer : c <eor> c <eos> |
c |
divide__300.0__18.0__ divide__300.0__39.0__ round__350.0__ |
divide__300.0__18.0__ divide__300.0__39.0__ round__350.0__ |
| how many ltrs of water must beadded to num__16 ltrs of milk and water containing num__10.0 water to make it num__20.0 water in it ? <o> a ) num__1 liter <o> b ) num__2 liter <o> c ) num__3 liter <o> d ) num__5 liter <o> e ) num__7 liter |
by rule of alligation % concentration of water in pure water ( num__100 ) % concentration of water in the given mixture ( num__10 ) mean % concentration ( num__20 ) num__20 - num__10 = num__10 num__100 - num__20 = num__80 = > quantity of water : quantity of the mixture = num__10 : num__80 = num__1 : num__8 here quantity of the mixture = num__16 litres = > quantity of water : num__16 = num__1 : num__8 = > quantity of water = num__16 × num__18 = num__2 liter b <eor> b <eos> |
b |
percent__10.0__80.0__ percent__10.0__20.0__ percent__10.0__20.0__ |
percent__10.0__80.0__ percent__10.0__20.0__ percent__10.0__20.0__ |
| the average age of a class of num__12 students is num__19 years . the average increased by num__2 when the teacher ' s age also included . what is the age of the teacher ? <o> a ) num__40 years <o> b ) num__34 years <o> c ) num__42 years <o> d ) num__43 years <o> e ) num__44 years |
if age of the teacher was num__12 average would not have changed . since average increased by num__2 age of the teacher = num__12 + num__22 × num__1 = num__34 answer : b <eor> b <eos> |
b |
add__12.0__22.0__ add__12.0__22.0__ |
add__12.0__22.0__ add__12.0__22.0__ |
| a car is traveling num__75 kilometers per hour . how many meters does the car travel in one minute ? <o> a ) num__1250 m / min <o> b ) num__1360 m / min <o> c ) num__2560 m / min <o> d ) num__1670 m / min <o> e ) num__1890 m / min |
convert hour into minutes ( num__1 hour = num__60 minutes ) and kilometers into meters ( num__1 km = num__1000 m ) and simplify num__75 kilometers per hour = num__75 km / hr = ( num__75 × num__1000 meters ) / ( num__60 minutes ) = num__1250 meters / minute correct answer is a ) num__1250 m / min <eor> a <eos> |
a |
hour_to_min_conversion__ round__1250.0__ |
hour_to_min_conversion__ divide__1250.0__1.0__ |
| for how many prime pairs ( p q ) does there exist an integer n such that ( p num__2 + num__1 ) * ( q num__2 + num__1 ) = n num__2 + num__1 ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) infinitely many |
first we show that one of p and q is num__2 . if both p and q are odd then n num__2 + num__1 = ( p num__2 + num__1 ) ( q num__2 + num__1 ) is even so that n is also odd . hence there is an integer t such that n = num__2 t + num__1 so that n num__2 + num__1 = num__4 t num__2 + num__4 t + num__2 . but this means n num__2 + num__1 is not divisible by num__4 whereas ( p num__2 + num__1 ) ( q num__2 + num__1 ) clearly is . thus we must have one of p or q is num__2 . we now have n num__2 + num__1 = num__5 ( x num__2 + num__1 ) where either x = p or x = q . note that x num__3 . we deduce that num__5 x num__2 = n num__2 − num__4 = ( n − num__2 ) ( n + num__2 ) . since x is prime and n + num__2 > n − num__2 we obtain that n − num__2 num__2 { num__1 num__5 x } and n + num__2 = num__5 x num__2 / ( n − num__2 ) . since also ( n + num__2 ) − ( n − num__2 ) = num__4 we get that one of num__5 x num__2 − num__1 x num__2 − num__5 and num__4 x equals num__4 . as x is prime we obtain x = num__3 . thus ( p q ) = ( num__2 num__3 ) or ( num__3 num__2 ) . each pair gives a solution to the equation n num__2 + num__1 = ( p num__2 + num__1 ) ( q num__2 + num__1 ) implying the answer . correct answer a <eor> a <eos> |
a |
add__1.0__4.0__ add__2.0__1.0__ multiply__2.0__1.0__ |
add__1.0__4.0__ add__2.0__1.0__ divide__2.0__1.0__ |
| there are num__6 boxes numbered num__1 num__2 . . . num__6 . each box is to be filled up either with a red or a green ball in such a way that at least num__1 box contains a green ball and the boxes containing green balls are consecutively numbered . the total number of ways in which this can be done is <o> a ) num__5 <o> b ) num__21 <o> c ) num__33 <o> d ) num__60 <o> e ) num__40 |
given the constraints of this question we have to consider num__6 possible outcomes . . . num__1 green ball : num__6 options grrrrr rgrrrr rrgrrr rrrgrr rrrrgr rrrrrg num__2 green balls : num__5 options ggrrrr rggrrr rrggrr rrrggr rrrrgg num__3 green balls : num__4 options gggrrr rgggrr rrgggr rrrggg num__4 green balls : num__3 options ggggrr rggggr rrgggg num__5 green balls : num__2 options gggggr rggggg num__6 green balls : num__1 option gggggg num__6 + num__5 + num__4 + num__3 + num__2 + num__1 = num__21 options . b <eor> b <eos> |
b |
subtract__6.0__1.0__ divide__6.0__2.0__ subtract__6.0__2.0__ multiply__1.0__21.0__ |
subtract__6.0__1.0__ add__1.0__2.0__ add__1.0__3.0__ multiply__1.0__21.0__ |
| if ( num__18 ^ a ) * num__9 ^ ( num__3 a – num__1 ) = ( num__2 ^ num__3 ) ( num__3 ^ b ) and a and b are positive integers what is the value of a ? <o> a ) num__22 <o> b ) num__11 <o> c ) num__9 <o> d ) num__6 <o> e ) num__3 |
num__18 ^ a ) * num__9 ^ ( num__3 a – num__1 ) = ( num__2 ^ num__3 ) ( num__3 ^ b ) = num__2 ^ a . num__9 ^ a . num__9 ^ ( num__3 a – num__1 ) = ( num__2 ^ num__3 ) ( num__3 ^ b ) just compare powers of num__2 from both sides ( no need to calculate powers of num__3 num__9 as value of b is not asked ) answer = num__3 answer : e <eor> e <eos> |
e |
divide__9.0__3.0__ |
divide__9.0__3.0__ |
| if n is the square of a positive integer which of the following must be equal to the square of the next positive integer ? . <o> a ) √ n + num__1 <o> b ) n + num__1 <o> c ) n ^ num__2 + num__1 <o> d ) n + num__2 √ n + num__1 <o> e ) n ^ num__2 + num__2 n + num__1 |
let a = + ve integer such that n = a ^ num__2 ( a + num__1 ) = next + ve integer ( a + num__1 ) ^ num__2 = a ^ num__2 + num__1 + num__2 a and a = n ^ ( num__0.5 ) therefore ( a + num__1 ) ^ num__2 = n + num__1 + num__2 * n ^ ( num__0.5 ) answer d <eor> d <eos> |
d |
reverse__2.0__ reverse__0.5__ |
reverse__2.0__ reverse__0.5__ |
| excluding stoppages the speed of a bus is num__64 kmph and including stoppages it is num__50 kmph . for how many minutes does the bus stop per hour ? <o> a ) num__15.5 <o> b ) num__12.3 <o> c ) num__10.5 <o> d ) num__10.12 <o> e ) num__13.12 |
due to stoppages it covers num__14 km less . time taken to cover num__14 km = ( ( num__0.21875 ) Ã — num__60 ) = num__13.12 min . option ( e ) is correct <eor> e <eos> |
e |
subtract__64.0__50.0__ divide__14.0__64.0__ hour_to_min_conversion__ round__13.12__ |
subtract__64.0__50.0__ divide__14.0__64.0__ hour_to_min_conversion__ round__13.12__ |
| a rectangular field num__30 m long and num__20 m broad . how much deep it should be dug so that from the earth taken out a platform can be formed which is num__8 m long num__5.5 m broad and num__1.5 m high where as the earth taken out is increase by num__2.0 ? <o> a ) num__37 cm <o> b ) num__10 cm <o> c ) num__17 cm <o> d ) num__88 cm <o> e ) num__87 cm |
num__30 * num__20 * x = ( num__8 * num__5.5 * num__1.5 ) / num__2 \ answer : b <eor> b <eos> |
b |
subtract__30.0__20.0__ |
divide__20.0__2.0__ |
| if the price of a certain computer increased num__30 percent from d dollars to num__351 dollars then num__2 d = <o> a ) num__540 <o> b ) num__570 <o> c ) num__619 <o> d ) num__649 <o> e ) num__700 |
before price increase price = d after num__30.0 price increase price = d + ( num__0.3 ) * d = num__1.3 d = num__351 ( given ) i . e . d = num__351 / num__1.3 = $ num__270 i . e . num__2 d = num__2 * num__270 = num__540 answer : option a <eor> a <eos> |
a |
divide__351.0__1.3__ multiply__2.0__270.0__ multiply__2.0__270.0__ |
divide__351.0__1.3__ multiply__2.0__270.0__ multiply__2.0__270.0__ |
| a certain drink of type a is prepared by mixing num__4 parts milk with num__3 parts fruit juice . another drink of type b is prepared by mixing num__4 parts of fruit juice and num__3 parts of milk . how many liters of fruit juice must be added to num__21 liters of drink a to convert it to drink b ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
in num__21 liters of drink a there are num__12 liters of milk and num__9 liters of juice . with num__12 liters of milk we need a total of num__16 liters of juice to make drink b . we need to add num__7 liters of juice . the answer is e . <eor> e <eos> |
e |
multiply__4.0__3.0__ subtract__21.0__12.0__ add__4.0__12.0__ add__4.0__3.0__ add__4.0__3.0__ |
multiply__4.0__3.0__ subtract__21.0__12.0__ add__4.0__12.0__ add__4.0__3.0__ add__4.0__3.0__ |
| a is half good a work man as b and together they finish a job in num__10 days . in how many days working alone b finish the job ? <o> a ) num__98 days <o> b ) num__21 days <o> c ) num__17 days <o> d ) num__18 days <o> e ) num__15 days |
wc = num__1 : num__2 num__2 x + x = num__0.1 = > x = num__0.0333333333333 num__2 x = num__0.0333333333333 = > num__15 days answer : e <eor> e <eos> |
e |
divide__1.0__10.0__ round__15.0__ |
divide__1.0__10.0__ round__15.0__ |
| we have a rectangular metallic piece of paper that covers exactly the area of a cube . the length of the piece of paper is num__120 inches and the width is num__108 inches . what is the volume of the cube in cubic feet is num__1 feet is num__12 inches ? <o> a ) a num__125 <o> b ) b num__120 <o> c ) c num__70 <o> d ) d num__40 <o> e ) e num__10 |
l = num__10.0 = num__10 ft w = num__9.0 = num__9 ft area of paper = num__90 area of cube = num__10 * side ^ num__2 side of cube = num__5 v of cube = num__125 <eor> a <eos> |
a |
multiply__9.0__10.0__ triangle_area__1.0__10.0__ volume_cube__5.0__ volume_cube__5.0__ |
multiply__9.0__10.0__ triangle_area__1.0__10.0__ volume_cube__5.0__ multiply__1.0__125.0__ |
| a b c can complete a piece of work in num__39 and num__15 days respectively . working together they will complete the same work in how many days ? <o> a ) num__1.95652173913 <o> b ) num__2.39130434783 <o> c ) num__1.0701754386 <o> d ) num__0.923076923077 <o> e ) num__0.95652173913 |
( a + b + c ) ' s num__1 day work = ( num__0.333333333333 ) + ( num__0.111111111111 ) + ( num__0.00869565217391 ) = num__0.511111111111 a b c together will complete the work in num__1.95652173913 days answer is a <eor> a <eos> |
a |
multiply__1.0__1.9565__ |
multiply__1.0__1.9565__ |
| a person got rs . num__48 more when he invested a certain sum at compound interest instead of simple interest for two years at num__8.0 p . a . find the sum ? <o> a ) rs . num__7500 <o> b ) rs . num__7503 <o> c ) rs . num__7227 <o> d ) rs . num__7528 <o> e ) rs . num__1281 |
explanation : p = ( d * num__1002 ) / r num__2 = > ( num__48 * num__100 * num__100 ) / num__8 * num__8 = rs . num__7500 answer : a <eor> a <eos> |
a |
percent__100.0__7500.0__ |
percent__100.0__7500.0__ |
| what is the remainder when the number t = num__14 ^ num__2 * num__15 ^ num__8 is divided by num__5 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__4 <o> e ) num__5 |
num__14 ^ num__2 has units digit num__6 num__15 ^ num__8 has units digit num__5 thus t = num__14 ^ num__2 * num__15 ^ num__8 has units digit num__0 and will be divisible by num__5 . the remainder will be zero answer : ( a ) <eor> a <eos> |
a |
subtract__14.0__8.0__ multiply__14.0__0.0__ |
subtract__14.0__8.0__ multiply__14.0__0.0__ |
| how many num__2 digit number contain number num__3 ? <o> a ) num__18 <o> b ) num__28 <o> c ) num__38 <o> d ) num__68 <o> e ) num__58 |
total num__2 digit no . = num__9 * num__10 * = num__90 not containing num__3 = num__8 * num__9 = num__72 total num__2 digit number contain num__3 = num__90 - num__72 = num__18 answer : a <eor> a <eos> |
a |
power__3.0__2.0__ multiply__9.0__10.0__ power__2.0__3.0__ multiply__8.0__9.0__ multiply__2.0__9.0__ multiply__2.0__9.0__ |
power__3.0__2.0__ multiply__9.0__10.0__ subtract__10.0__2.0__ multiply__8.0__9.0__ multiply__2.0__9.0__ multiply__2.0__9.0__ |
| half a number plus num__6 is num__11 . what is the number ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
let x be the number . always replace ` ` is ' ' with an equal sign ( num__0.5 ) x + num__6 = num__11 ( num__0.5 ) x = num__11 - num__6 ( num__0.5 ) x = num__5 x = num__10 correct answer is c <eor> c <eos> |
c |
subtract__11.0__6.0__ divide__5.0__0.5__ divide__5.0__0.5__ |
subtract__11.0__6.0__ divide__5.0__0.5__ divide__5.0__0.5__ |
| what is the present worth of rs . num__264 due in num__2 years at num__5.0 simple interest per annum ? <o> a ) num__128 <o> b ) num__277 <o> c ) num__120 <o> d ) num__240 <o> e ) num__171 |
let the present worth be rs . x . then s . i . = ( num__264 - x ) ( x * num__5 * num__2 ) / num__100 = ( num__264 - x ) num__10 x = num__26400 - num__100 x num__110 x = num__26400 = > x = num__240 . answer : d <eor> d <eos> |
d |
percent__100.0__240.0__ |
percent__100.0__240.0__ |
| the current of a stream at num__1 kmph . a motor boat goes num__35 km upstream and back to the starting point in num__12 hours . the speed of the motor boat in still water is ? <o> a ) num__6 <o> b ) num__9 <o> c ) num__5 <o> d ) num__1 <o> e ) num__4 |
s = num__1 m = x ds = x + num__1 us = x - num__1 num__35 / ( x + num__1 ) + num__35 / ( x - num__1 ) = num__12 x = num__6 answer : a <eor> a <eos> |
a |
round__6.0__ |
subtract__12.0__6.0__ |
| in the seaside summer camp there are num__50 children . num__90.0 of the children are boys and the rest are girls . the camp administrator decided to make the number of girls only num__5.0 of the total number of children in the camp . how many more boys must she bring to make that happen ? <o> a ) num__50 . <o> b ) num__45 . <o> c ) num__40 . <o> d ) num__30 . <o> e ) num__25 . |
given there are num__50 students in the seaside summer camp num__90.0 of num__50 = num__45 boys and remaining num__5 girls . now here num__90.0 are boys and num__10.0 are girls . now question is asking about how many boys do we need to add to make the girls percentage to num__5 or num__5.0 . . if we add num__50 to existing num__45 then the count will be num__95 and the girls number will be num__5 as it . now boys are num__95.0 and girls are num__5.0 . ( out of num__100 students = num__95 boys + num__5 girls ) . imo option a is correct . <eor> a <eos> |
a |
subtract__50.0__5.0__ divide__50.0__5.0__ add__50.0__45.0__ add__90.0__10.0__ multiply__5.0__10.0__ |
subtract__50.0__5.0__ divide__50.0__5.0__ add__50.0__45.0__ add__90.0__10.0__ add__5.0__45.0__ |
| if in a game of num__80 p can give num__16 points to q and r can give num__20 points to p then in a game of num__150 how many points can r give to q ? <o> a ) num__60 <o> b ) num__28 <o> c ) num__26 <o> d ) num__17 <o> e ) num__81 |
explanation : when p scores num__80 q scores num__64 . when r scores num__80 p scores num__60 hence when r scores num__150 q scores ( num__60 * num__64 * num__150 ) / ( num__80 * num__80 ) = num__90 therefore in a game of num__150 r can give num__60 points to q . answer : a <eor> a <eos> |
a |
subtract__80.0__16.0__ subtract__80.0__20.0__ subtract__150.0__60.0__ subtract__80.0__20.0__ |
subtract__80.0__16.0__ subtract__80.0__20.0__ subtract__150.0__60.0__ subtract__80.0__20.0__ |
| in township k each property is taxed at num__10 percent of its assessed value . if the assessed value of a property in township k is increased from $ num__20000 to $ num__24000 by how much will the property tax increase ? <o> a ) $ num__32 <o> b ) $ num__50 <o> c ) $ num__320 <o> d ) $ num__400 <o> e ) $ num__500 |
increase in house value = $ num__24000 - $ num__20000 = $ num__4000 so tax increase = num__10.0 of $ num__4000 = $ num__400 answer : d <eor> d <eos> |
d |
subtract__24000.0__20000.0__ divide__4000.0__10.0__ divide__4000.0__10.0__ |
subtract__24000.0__20000.0__ divide__4000.0__10.0__ divide__4000.0__10.0__ |
| a circular rim num__35 inches in diameter rotates the same number of inches per second as a circular rim num__42 inches in diameter . if the smaller rim makes x revolutions per second how many revolutions per minute does the larger rim makes in terms of x ? <o> a ) num__48 pi / x <o> b ) num__50 x <o> c ) num__48 x <o> d ) num__24 x <o> e ) x / num__75 |
c = ( pi ) d c ( small ) : ( pi ) * num__35 c ( large ) : ( pi ) * num__42 lets say the time horizon is num__60 seconds so during that time the smaller rim covers a distance of ( pi ) * num__35 * num__60 = ( pi ) * ( num__2100 ) inches ( pi ) * ( num__2100 ) = ( pi ) * ( num__42 ) ( x ) pi * ( num__50 ) = pi * ( x ) num__50 = x answer : b . <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__35.0__60.0__ divide__2100.0__42.0__ round__50.0__ |
hour_to_min_conversion__ multiply__35.0__60.0__ divide__2100.0__42.0__ round__50.0__ |
| if the integer p is rounded to the nearest hundred the result represents a num__66 num__0.666666666667 % increase over the value obtained when p is rounded to the nearest ten . which of the following is a possible value of p ? <o> a ) num__64 <o> b ) num__67 <o> c ) num__99 <o> d ) num__133 <o> e ) num__147 |
we have to check options num__1 by num__1 . num__64 rounded to nearest hundred = num__100 num__64 rounded to nearest ten = num__60 now difference = num__40 the result represents a num__66 num__0.666666666667 % increase over the value obtained when x is rounded to the nearest ten . therefore num__0.666666666667 ) * num__100 = num__66 num__0.666666666667 = = > this is the answer . num__99 rounded to nearest ten = num__100 num__133 rounded to nearest ten = num__130 num__133 rounded to nearest hundred p = num__100 = a <eor> a <eos> |
a |
subtract__100.0__60.0__ subtract__100.0__1.0__ add__66.0__64.0__ multiply__64.0__1.0__ |
subtract__100.0__60.0__ subtract__100.0__1.0__ add__66.0__64.0__ multiply__64.0__1.0__ |
| a parking garage rents parking spaces for $ num__10 per week or $ num__25 per month . how much does a person save in a year by renting by the month rather than by the week ? <o> a ) $ num__120 <o> b ) $ num__160 <o> c ) $ num__220 <o> d ) $ num__240 <o> e ) $ num__260 |
num__10 $ per week ! an year has num__52 weeks . annual charges per year = num__52 * num__10 = num__520 $ num__30 $ per month ! an year has num__12 months . annual charges per year = num__12 * num__25 = num__300 $ num__520 - num__300 = num__120 ans a <eor> a <eos> |
a |
multiply__10.0__52.0__ multiply__10.0__30.0__ multiply__10.0__12.0__ multiply__10.0__12.0__ |
multiply__10.0__52.0__ multiply__10.0__30.0__ multiply__10.0__12.0__ multiply__10.0__12.0__ |
| a two years older than b who is twice as old as c . if the total of the ages of a b and c be num__27 the how old b ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__12 |
c ' s age be x years . the b ' s age = num__2 x years a ' s age = ( num__2 x + num__2 ) years therefore ( num__2 x + num__2 ) + num__2 x + x = num__27 num__5 x = num__25 x = num__5 b ' s age num__2 x = num__10 years . correct answer ( d ) <eor> d <eos> |
d |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
| the symphony sells two kinds of tickets : orchestra for $ num__30 and upper tiers for $ num__12 . on a certain night the symphony sells num__90 tickets and gets $ num__1836 in revenue from the sales . how many orchestra tickets did they sell ? <o> a ) num__38 <o> b ) num__42 <o> c ) num__46 <o> d ) num__50 <o> e ) num__54 |
let x be the number of orchestra tickets . then num__90 - x is the number of upper tier tickets . num__30 x + num__12 ( num__90 - x ) = num__1836 num__18 x = num__1836 - num__1080 x = num__42 the answer is b . <eor> b <eos> |
b |
subtract__30.0__12.0__ multiply__12.0__90.0__ add__30.0__12.0__ add__30.0__12.0__ |
subtract__30.0__12.0__ multiply__12.0__90.0__ add__30.0__12.0__ add__30.0__12.0__ |
| pipe a can fill a tank in num__8 hours pipe b in num__16 hours and pipe c in num__16 hours . if all the pipes are open in how many hours will the tank be filled ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
part filled by a + b + c in num__1 hour = num__0.125 + num__0.0625 + num__0.0625 = num__0.25 all the three pipes together will fill the tank in num__4 hours . answer : c <eor> c <eos> |
c |
divide__1.0__8.0__ divide__1.0__16.0__ multiply__16.0__0.25__ round__4.0__ |
divide__1.0__8.0__ divide__1.0__16.0__ multiply__16.0__0.25__ round__4.0__ |
| if x ^ num__2 + num__9 / x ^ num__2 = num__42 what is the value of x - num__3 / x <o> a ) num__6 <o> b ) num__25 <o> c ) num__9 <o> d ) num__5 <o> e ) num__3 |
to find : x - num__3 / x . let it be t . = > x - num__3 / x = t = > ( x ^ num__2 + num__9 / x ^ num__2 ) - num__2 * x * num__3 / x = t ^ num__2 ( squaring both sides ) . = > ( num__42 ) - num__2 * num__3 = num__36 = > t ^ num__2 = num__36 . thus t = num__6 or t = - num__6 . answer a <eor> a <eos> |
a |
multiply__2.0__3.0__ multiply__2.0__3.0__ |
multiply__2.0__3.0__ multiply__2.0__3.0__ |
| express num__25 mps in kmph ? <o> a ) num__299 <o> b ) num__29 <o> c ) num__90 <o> d ) num__77 <o> e ) num__11 |
num__25 * num__3.6 = num__90 kmph answer : c <eor> c <eos> |
c |
multiply__25.0__3.6__ round__90.0__ |
multiply__25.0__3.6__ multiply__25.0__3.6__ |
| x y and z are different prime numbers . the product x ^ num__3 * y ^ num__2 * z ^ num__2 is divisible by how many different positive numbers ? <o> a ) num__24 <o> b ) num__36 <o> c ) num__48 <o> d ) num__60 <o> e ) num__72 |
the exponents of x ^ num__3 * y ^ num__2 * z ^ num__2 are num__3 num__2 and num__2 . the number of factors is ( num__3 + num__1 ) ( num__2 + num__1 ) ( num__2 + num__1 ) = num__36 the answer is b . <eor> b <eos> |
b |
subtract__3.0__2.0__ multiply__1.0__36.0__ |
subtract__3.0__2.0__ multiply__1.0__36.0__ |
| if x is an integer then x ( x - num__1 ) ( x - b ) must be evenly divisible by three when b is any of the following values except <o> a ) - num__4 <o> b ) - num__2 <o> c ) - num__1 <o> d ) num__2 <o> e ) num__5 |
here ' s another approach x ( x - num__1 ) ( x - b ) all three are consecutive so the product must be a multiple of num__3 we do n ' t know the value of b just yet . . . so let ' s extend the series . . . the extension itself reveals the answers . . ( x - num__5 ) . . ( x - num__2 ) ( x - num__1 ) x ( x + num__1 ) . . ( x + num__4 ) . . we can see the possible values of b too from the series b = num__2 or num__2 + num__3 n [ num__25 ] b = - num__1 or - num__1 + num__3 n [ - num__1 - num__4 ] b i . e . - num__2 does not fit in any value of b so b it is <eor> b <eos> |
b |
subtract__3.0__1.0__ add__1.0__3.0__ multiply__1.0__2.0__ |
subtract__3.0__1.0__ add__1.0__3.0__ subtract__3.0__1.0__ |
| in num__1982 and num__1983 company b ’ s operating expenses were $ num__12.0 million and $ num__14.0 million respectively and its revenues were $ num__15.6 million and $ num__18.8 million respectively . what was the percent increase in company b ’ s profit ( revenues minus operating expenses ) from num__1982 to num__1983 ? <o> a ) num__3.0 <o> b ) num__16 num__0.666666666667 % <o> c ) num__25.0 <o> d ) num__33 num__0.333333333333 % <o> e ) num__60 % |
profit in num__1982 = num__15.6 - num__12.0 = num__3.6 profit in num__1983 = num__18.8 - num__14.0 = num__4.8 increase in profit = num__4.8 - num__3.6 = num__1.2 percentage increase num__12 * num__2.77777777778 = num__33.33 ; answer : d <eor> d <eos> |
d |
subtract__15.6__12.0__ subtract__18.8__14.0__ subtract__4.8__3.6__ round_down__33.33__ |
subtract__15.6__12.0__ subtract__18.8__14.0__ subtract__4.8__3.6__ round_down__33.33__ |
| two trains of equal lengths are running at speeds of num__30 kmph and num__60 kmph . the two trains crossed each other in num__30 seconds when travelling in opposite direction . in what time will they cross each other when travelling in the same direction ? <o> a ) num__90 sec <o> b ) num__18 sec <o> c ) num__19 sec <o> d ) num__15 sec <o> e ) num__11 sec |
let the length of each train be x m . ( x + x ) / ( num__60 + num__30 ) num__0.277777777778 = ( num__750 * num__18 ) / ( num__30 * num__5 ) = num__90 sec . answer : a <eor> a <eos> |
a |
add__30.0__60.0__ round__90.0__ |
add__30.0__60.0__ add__30.0__60.0__ |
| an accurate clock shows num__8 o ’ clock in the morning . through how many degrees will the hour hand rotate when the clock shows num__2 o ’ clock in the afternoon ? <o> a ) num__144 ° <o> b ) num__150 ° <o> c ) num__168 ° <o> d ) num__180 ° <o> e ) none of these |
angle traced by the hour hand in num__6 hours = ( num__360 ⁄ num__12 × num__6 ) ° = num__180 ° answer b <eor> b <eos> |
b |
radians_to_degree__2.0__ straight_angle__ clock_small_arm_angle__2.0__180.0__ |
radians_to_degree__2.0__ straight_angle__ clock_small_arm_angle__2.0__180.0__ |
| num__5355 x num__51 = ? <o> a ) num__273105 <o> b ) num__273243 <o> c ) num__273247 <o> d ) num__273250 <o> e ) num__273258 |
num__5355 x num__51 = num__5355 x ( num__50 + num__1 ) = num__5355 x num__50 + num__5355 x num__1 = num__267750 + num__5355 = num__273105 a <eor> a <eos> |
a |
subtract__51.0__50.0__ multiply__5355.0__50.0__ multiply__5355.0__51.0__ multiply__5355.0__51.0__ |
subtract__51.0__50.0__ multiply__5355.0__50.0__ add__5355.0__267750.0__ add__5355.0__267750.0__ |
| ( num__786 × num__74 ) ÷ ? = num__3635.25 <o> a ) a ) num__4.8 <o> b ) b ) num__48 <o> c ) c ) num__16 <o> d ) d ) num__68 <o> e ) e ) num__48 |
explanation : num__58164 / x = num__3635.25 = > x = num__58164 / num__3635.25 = num__16 answer : option c <eor> c <eos> |
c |
multiply__786.0__74.0__ divide__58164.0__3635.25__ divide__58164.0__3635.25__ |
multiply__786.0__74.0__ divide__58164.0__3635.25__ divide__58164.0__3635.25__ |
| sides of a rectangular park are in the ratio num__3 : num__2 and its area is num__3750 sq m the cost of fencing it at num__50 ps per meter is ? <o> a ) rs . num__150 <o> b ) rs . num__100 <o> c ) rs . num__125 <o> d ) rs . num__175 <o> e ) rs . num__275 |
num__3 x * num__2 x = num__3750 = > x = num__25 num__2 ( num__75 + num__50 ) = num__250 m num__250 * num__0.5 = rs . num__125 answer : c <eor> c <eos> |
c |
triangle_area__3.0__50.0__ rectangle_perimeter__50.0__75.0__ multiply__0.5__250.0__ triangle_area__2.0__125.0__ |
multiply__3.0__25.0__ rectangle_perimeter__50.0__75.0__ multiply__0.5__250.0__ multiply__0.5__250.0__ |
| there are two cars . one is num__300 miles north of the other . simultaneously the car to the north is driven westward at num__20 miles per hour and the other car is driven eastward at num__60 miles per hour . how many miles apart are the cars after num__5 hours ? <o> a ) num__300 <o> b ) num__360 <o> c ) num__400 <o> d ) num__450 <o> e ) num__500 |
here drawing a quick sketch of the ' actions ' described will end in a diagonal line that you canbuilda right triangle around : the right triangle will have a base of num__400 and a height of num__300 . the hidden pattern here is a num__0.75 / num__5 right triangle ( the num__300 lines up with the ' num__3 ' and the num__400 lines up with the ' num__4 ' ) . in this way you can deduce that each side is ' num__100 times ' bigger than it ' s corresponding side : num__0.75 / num__5 becomes num__0.75 / num__500 thus the distance between the two cars is the length of the hypotenuse of this larger right triangle . . . final answer : e <eor> e <eos> |
e |
divide__300.0__400.0__ divide__60.0__20.0__ divide__20.0__5.0__ divide__300.0__3.0__ multiply__5.0__100.0__ round__500.0__ |
divide__300.0__400.0__ divide__60.0__20.0__ divide__20.0__5.0__ divide__300.0__3.0__ multiply__5.0__100.0__ round__500.0__ |
| a and b go around a circular track of length num__600 m on a cycle at speeds of num__36 kmph and num__60 kmph . after how much time will they meet for the first time at the starting point ? <o> a ) num__120 sec <o> b ) num__176 sec <o> c ) num__178 sec <o> d ) num__187 sec <o> e ) num__180 sec |
time taken to meet for the first time at the starting point = lcm { length of the track / speed of a length of the track / speed of b } = lcm { num__600 / ( num__36 * num__0.277777777778 ) num__600 / ( num__60 * num__0.277777777778 ) } = num__180 sec . answer : e <eor> e <eos> |
e |
round__180.0__ |
round__180.0__ |
| a train num__240 m long passes a pole in num__24 seconds . how long will it take to pass a platform num__650 m long ? <o> a ) num__80 sec <o> b ) num__89 sec <o> c ) num__99 sec <o> d ) num__86 sec <o> e ) num__85 sec |
explanation : speed = ( num__10.0 ) m / sec = num__10 m / sec . required time = [ ( num__240 + num__650 ) / num__10 ) ] sec = num__89 sec . answer : b <eor> b <eos> |
b |
divide__240.0__24.0__ round__89.0__ |
divide__240.0__24.0__ round__89.0__ |
| a circular rim num__20 inches in diameter rotates the same number of inches per second as a circular rim num__30 inches in diameter . if the smaller rim makes x revolutions per second how many revolutions per minute does the larger rim makes in terms of x ? <o> a ) num__40 x <o> b ) num__75 x <o> c ) num__48 x <o> d ) num__24 x <o> e ) x / num__75 |
revolutions - circumference of the rim . = > num__20 pi * x * num__60 = num__30 pi * revolutions = > num__0.666666666667 * x * num__60 = revolutions = > num__40 x - option a <eor> a <eos> |
a |
hour_to_min_conversion__ divide__20.0__30.0__ subtract__60.0__20.0__ round__40.0__ |
hour_to_min_conversion__ divide__20.0__30.0__ subtract__60.0__20.0__ subtract__60.0__20.0__ |
| a man sitting in a bullet train which is travelling at num__50 kmph observes that a goods bullet train travelling in opposite direction takes num__9 seconds to pass him . if the goods bullet train is num__280 m long find its speed . ? <o> a ) num__44 kmph . <o> b ) num__52 kmph . <o> c ) num__75 kmph . <o> d ) num__83 kmph . <o> e ) num__62 kmph . |
e num__62 kmph . relative speed = num__31.1111111111 m / sec = ( ( num__31.1111111111 ) * ( num__3.6 ) ) kmph = num__112 kmph . speed of goods train = ( num__112 - num__50 ) kmph = num__62 kmph . <eor> e <eos> |
e |
divide__280.0__9.0__ add__50.0__62.0__ round__62.0__ |
divide__280.0__9.0__ multiply__31.1111__3.6__ subtract__112.0__50.0__ |
| what is the smallest integer n for which num__25 ^ n > num__5 ^ num__12 ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
to solve we want to get the bases the same . thus we need to break num__25 ^ n into prime factors . num__25 ^ n = ( num__5 ^ num__2 ) ^ n = num__5 ^ ( num__2 n ) ( remember that when we have a power to a power we multiply the exponents . ) we can use the new value in the given inequality : num__5 ^ ( num__2 n ) > num__5 ^ num__12 since we have the same bases on either side of the inequality we can drop the bases and set up an equation involving just the exponents . num__2 n > num__12 n > num__6 because n is greater than num__6 the smallest integer that satisfies the inequality num__25 ^ n > num__5 ^ num__12 is num__7 . the answer is b . <eor> b <eos> |
b |
divide__12.0__2.0__ add__5.0__2.0__ add__5.0__2.0__ |
divide__12.0__2.0__ add__5.0__2.0__ add__5.0__2.0__ |
| on a sum of money the simple interest for num__2 years is rs . num__324 while the compound interest is rs . num__340 the rate of interest being the same in both the cases . the rate of interest is <o> a ) num__15.0 <o> b ) num__14.25 <o> c ) num__9.87 <o> d ) num__10.5 <o> e ) num__11.5 % |
the difference between compound interest and simple interest on rs . p for num__2 years at r % per annum = ( r à — si ) / ( num__2 à — num__100 ) difference between the compound interest and simple interest = num__340 - num__324 = num__16 ( r à — si ) / ( num__2 à — num__100 ) = num__16 ( r à — num__324 ) / ( num__2 à — num__100 ) = num__16 r = num__9.87 answer : option c <eor> c <eos> |
c |
percent__9.87__100.0__ |
percent__9.87__100.0__ |
| the two trains of lengths num__400 m num__600 m respectively running at same directions . the faster train can cross the slower train in num__180 sec the speed of the slower train is num__48 km . then find the speed of the faster train ? <o> a ) num__60 <o> b ) num__68 <o> c ) num__67 <o> d ) num__69 <o> e ) num__70 |
length of the two trains = num__600 m + num__400 m speed of the first train = x speed of the second train = num__48 kmph num__1000 / x - num__48 = num__180 num__1000 / x - num__48 * num__0.277777777778 = num__180 num__50 = num__9 x - num__120 x = num__68 kmph answer : option b <eor> b <eos> |
b |
add__400.0__600.0__ round__68.0__ |
add__400.0__600.0__ round__68.0__ |
| find the principle on a certain sum of money at num__5.0 per annum for num__2 num__0.4 years if the amount being rs . num__1120 ? <o> a ) num__1000 <o> b ) num__2217 <o> c ) num__2889 <o> d ) num__2777 <o> e ) num__2991 |
num__1120 = p [ num__1 + ( num__5 * num__2.4 ) / num__100 ] p = num__1000 answer : a <eor> a <eos> |
a |
add__2.0__0.4__ multiply__1.0__1000.0__ |
add__2.0__0.4__ multiply__1.0__1000.0__ |
| today my car meter reads as num__72927 kms . i notes that this is a palindrome . how many minimum kms i need to travel so my car meter find another palindrom . <o> a ) num__100 kms <o> b ) num__115 kms <o> c ) num__110 kms <o> d ) num__210 kms <o> e ) num__120 kms |
c num__110 kms num__72927 + num__110 = num__73037 a palindrome <eor> c <eos> |
c |
add__72927.0__110.0__ round__110.0__ |
add__72927.0__110.0__ round__110.0__ |
| the difference between the place value and the face value of num__9 in the numeral num__856973 is <o> a ) num__973 <o> b ) num__891 <o> c ) num__5994 <o> d ) num__6084 <o> e ) none of these |
( place value of num__9 ) - ( face value of num__9 ) = ( num__900 - num__9 ) = num__891 answer : option b <eor> b <eos> |
b |
subtract__900.0__9.0__ subtract__900.0__9.0__ |
subtract__900.0__9.0__ subtract__900.0__9.0__ |
| a certain sum of money at simple interest amounted rs . num__1200 in num__12 years at num__6.0 per annum find the sum ? <o> a ) num__337 <o> b ) num__268 <o> c ) num__198 <o> d ) num__697 <o> e ) num__168 |
num__1200 = p [ num__1 + ( num__12 * num__6 ) / num__100 ] p = num__697 ' answer : d <eor> d <eos> |
d |
percent__100.0__697.0__ |
percent__100.0__697.0__ |
| look at this series : num__2 num__3 num__4 num__9 num__8 . . . what number should come next ? <o> a ) num__128 <o> b ) num__64 <o> c ) num__16 <o> d ) num__32 <o> e ) num__27 |
explanation : the given series is obtained by increasing powers of num__2 and num__3 . num__2 * num__1 = num__2 num__3 * num__1 = num__3 num__2 * num__2 = num__4 num__3 * num__3 = num__9 num__2 * num__2 * num__2 = num__8 num__3 * num__3 * num__3 = num__27 and so on . answer : option e <eor> e <eos> |
e |
subtract__3.0__2.0__ multiply__3.0__9.0__ multiply__3.0__9.0__ |
subtract__3.0__2.0__ multiply__3.0__9.0__ multiply__3.0__9.0__ |
| tough and tricky questions : decimals . the value of x is derived by summing a b and c and then rounding the result to the tenths place . the value of y is derived by first rounding a b and c to the tenths place and then summing the resulting values . if a = num__5.45 b = num__2.95 and c = num__3.74 what is y – x ? <o> a ) - num__0.1 <o> b ) num__0 <o> c ) num__0.05 <o> d ) num__0.2 <o> e ) num__0.3 |
to find x we first sum a b and c then round to the tenths place . num__5.45 + num__2.95 + num__3.74 = num__12.14 which rounds to num__12.2 . to find y we first round a b and c to the tenths place and them sum them . so num__5.4 + num__2.9 + num__3.7 = num__12.0 . we are looking for y - x which gives us num__12.2 - num__12.0 = num__0.2 or answer choice d . <eor> d <eos> |
d |
round_down__12.14__ subtract__12.2__12.0__ subtract__12.2__12.0__ |
round_down__12.14__ subtract__12.2__12.0__ subtract__12.2__12.0__ |
| two trains num__111 meters and num__165 meters in length respectively are running in opposite directions one at the rate of num__100 km and the other at the rate of num__120 kmph . in what time will they be completely clear of each other from the moment they meet ? <o> a ) num__4.85 <o> b ) num__7.85 <o> c ) num__4.51 <o> d ) num__5.85 <o> e ) num__6.15 |
t = ( num__111 + num__165 ) / ( num__100 + num__120 ) * num__3.6 t = num__4.51 answer : c <eor> c <eos> |
c |
round__4.51__ |
round__4.51__ |
| a can give b num__100 meters start and c num__200 meters start in a kilometer race . how much start can b give c in a kilometer race ? <o> a ) num__111.12 m <o> b ) num__161.12 m <o> c ) num__111.82 m <o> d ) num__111.19 m <o> e ) num__131.12 m |
a runs num__1000 m while b runs num__900 m and c runs num__800 m . the number of meters that c runs when b runs num__1000 m = ( num__1000 * num__800 ) / num__900 = num__888.888888889 = num__888.88 m . b can give c = num__1000 - num__888.88 = num__111.12 m . answer : a <eor> a <eos> |
a |
subtract__1000.0__100.0__ subtract__900.0__100.0__ subtract__1000.0__888.88__ round__111.12__ |
subtract__1000.0__100.0__ subtract__900.0__100.0__ subtract__1000.0__888.88__ subtract__1000.0__888.88__ |
| complete the following series num__2 num__4 num__6 num__10 . . . . . num__26 <o> a ) num__16 <o> b ) num__20 <o> c ) num__12 <o> d ) num__14 <o> e ) num__18 |
in the series we can get the next term by adding the previous num__2 terms . so to get the answer we have to add num__6 + num__10 = num__16 so the answer is option a ) num__16 . <eor> a <eos> |
a |
power__2.0__4.0__ power__2.0__4.0__ |
add__6.0__10.0__ add__6.0__10.0__ |
| num__106 x num__106 - num__94 x num__94 = ? <o> a ) num__2400 <o> b ) num__2000 <o> c ) num__1904 <o> d ) num__1906 <o> e ) none of them |
= ( num__106 ) ^ num__2 - ( num__94 ) ^ num__2 = ( num__106 + num__94 ) ( num__106 - num__94 ) = ( num__200 x num__12 ) = num__2400 answer is a <eor> a <eos> |
a |
add__106.0__94.0__ subtract__106.0__94.0__ multiply__200.0__12.0__ multiply__200.0__12.0__ |
add__106.0__94.0__ subtract__106.0__94.0__ multiply__200.0__12.0__ multiply__200.0__12.0__ |
| what is the value of n if the sum of the consecutive odd integers from num__1 to n equals num__169 ? <o> a ) num__47 <o> b ) num__25 <o> c ) num__37 <o> d ) num__33 <o> e ) num__29 |
# of terms = ( n - num__0.5 ) + num__1 { ( last term - first term ) / num__2 + num__1 | sum = ( num__1 + n ) / num__2 * # of terms = ( n + num__1 ) ^ num__0.5 = num__169 n + num__1 = num__13 * num__2 n + num__1 = num__26 n = num__25 . answer : b <eor> b <eos> |
b |
reverse__0.5__ multiply__2.0__13.0__ subtract__26.0__1.0__ multiply__1.0__25.0__ |
reverse__0.5__ multiply__2.0__13.0__ subtract__26.0__1.0__ multiply__1.0__25.0__ |
| which of the following describes all values of x for which num__4 - x ^ num__2 ≥ num__0 ? <o> a ) x ≥ num__1 <o> b ) x ≤ - num__1 <o> c ) num__0 ≤ x ≤ num__1 <o> d ) x ≤ - num__1 or x ≥ num__1 <o> e ) - num__2 ≤ x ≤ num__2 |
num__4 - x ^ num__2 > = num__0 - - - > x ^ num__2 - num__4 < = num__0 - - > ( x + num__2 ) ( x - num__2 ) < = num__0 above equation true for i ) x + num__2 < = num__0 and x - num__2 > = num__0 - - - > x < = - num__2 and x > = num__2 - - - > this is not possible - - - strike out this solution ii ) x + num__2 > = num__0 and x - num__2 < = num__0 - - - > x > = - num__2 and x < = num__2 - - > - num__2 < = x < = num__2 answer e <eor> e <eos> |
e |
subtract__4.0__2.0__ |
subtract__4.0__2.0__ |
| the length of the bridge which a train num__130 m long and traveling at num__45 km / hr can cross in num__30 sec is ? <o> a ) num__276 m <o> b ) num__287 m <o> c ) num__245 m <o> d ) num__276 m <o> e ) num__237 m |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec . time = num__30 sec let the length of bridge be x meters . then ( num__130 + x ) / num__30 = num__12.5 x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| if n is a positive integer and the greatest common divisor of n and num__16 is num__4 and the greatest common divisor of n and num__15 is num__3 which of the following is the greatest common divisor of n and num__72 ? <o> a ) num__24 <o> b ) num__36 <o> c ) num__12 <o> d ) num__18 <o> e ) num__16 |
the greatest common divisor of n and num__16 is num__4 - - > n is a multiple of num__4 but not a multiple of num__8 . the greatest common divisor of n and num__15 is num__3 - - > n is a multiple of num__3 but not a multiple of num__5 . num__72 = num__2 * num__3 * num__3 * num__2 * num__2 is divisible by num__12 . therefore the greatest common divisor is num__12 c <eor> c <eos> |
c |
divide__15.0__3.0__ divide__16.0__8.0__ subtract__16.0__4.0__ subtract__16.0__4.0__ |
subtract__8.0__3.0__ subtract__5.0__3.0__ subtract__16.0__4.0__ subtract__16.0__4.0__ |
| if tier is written as num__7183 and brain is written as num__23415 how is rent coded ? <o> a ) num__3653 <o> b ) num__3657 <o> c ) num__3857 <o> d ) num__2790 <o> e ) num__2711 |
explanation : given : letter : t i e r b a n code : num__7 num__1 num__8 num__3 num__2 num__4 num__5 thus the code for rent is num__3857 . answer : c <eor> c <eos> |
c |
add__1.0__7.0__ subtract__3.0__1.0__ add__1.0__3.0__ add__1.0__4.0__ multiply__1.0__3857.0__ |
add__1.0__7.0__ subtract__3.0__1.0__ add__1.0__3.0__ add__1.0__4.0__ multiply__1.0__3857.0__ |
| p q and r can complete a work in num__24 num__6 and num__12 days respectively . the work will be completed in - - - days if all of them are working together . <o> a ) num__2 <o> b ) num__3 num__3 ⁄ num__7 <o> c ) num__4 ¼ <o> d ) num__5 <o> e ) num__5 ¼ |
explanation : work done by p in num__1 day = num__0.0416666666667 work done by q in num__1 day = num__0.166666666667 work done by r in num__1 day = num__0.0833333333333 work done by p q and r in num__1 day = num__0.0416666666667 + num__0.166666666667 + num__0.0833333333333 = num__0.291666666667 = > working together they will complete the work in num__3.42857142857 days = num__3 num__3 ⁄ num__7 days answer : option b <eor> b <eos> |
b |
divide__1.0__24.0__ divide__1.0__6.0__ divide__1.0__12.0__ add__6.0__1.0__ round__3.0__ |
divide__1.0__24.0__ divide__1.0__6.0__ divide__1.0__12.0__ add__6.0__1.0__ round__3.0__ |
| in a kilometer race a beats b by num__50 meters or num__10 seconds . what time does a take to complete the race ? <o> a ) num__900 <o> b ) num__190 <o> c ) num__277 <o> d ) num__262 <o> e ) num__223 |
time taken by b run num__1000 meters = ( num__1000 * num__10 ) / num__50 = num__200 sec . time taken by a = num__200 - num__10 = num__190 sec . answer : b <eor> b <eos> |
b |
subtract__200.0__10.0__ round__190.0__ |
subtract__200.0__10.0__ subtract__200.0__10.0__ |
| a train num__290 m long passed a pole in num__29 sec . how long will it take to pass a platform num__650 m long ? <o> a ) num__94 sec <o> b ) num__89 sec <o> c ) num__54 sec <o> d ) num__27 sec <o> e ) num__22 sec |
speed = num__10.0 = num__10 m / sec . required time = ( num__290 + num__650 ) / num__10 = num__94 sec . answer : a <eor> a <eos> |
a |
divide__290.0__29.0__ round__94.0__ |
divide__290.0__29.0__ round__94.0__ |
| if num__100000 microns = num__1 decimeter and num__10000 num__000000 angstroms = num__1 decimeter how many angstroms equal num__1 micron ? <o> a ) num__1.0 e - num__05 <o> b ) num__0.0001 <o> c ) num__0.001 <o> d ) num__10000 <o> e ) num__100 |
000 |
given that num__100000 microns = num__1 decimeter = num__10000 num__000000 angstroms so num__1 micron = num__10000 num__000000 / num__100000 = num__100000 answer : e <eor> e <eos> |
e |
e |
| a specialized type of sand consists of num__40.0 mineral x by volume and num__60.0 mineral y by volume . if mineral x weighs num__1.5 grams per cubic centimeter and mineral y weighs num__3 grams per cubic centimeter how many grams does a cubic meter of specialized sand combination weigh ? ( num__1 meter = num__100 centimeters ) <o> a ) num__2 num__400000 <o> b ) num__2 num__800000 <o> c ) num__55000 <o> d ) num__28000 <o> e ) num__280 |
let the volume be num__1 m ^ num__3 = num__1 m * num__1 m * num__1 m = num__100 cm * num__100 cm * num__100 cm = num__1 num__000000 cm ^ num__3 by volume num__40.0 is x = num__400000 cm ^ num__3 num__60.0 is y = num__600000 cm ^ num__3 by weight in num__1 cm ^ num__3 x is num__1.5 gms in num__400000 cm ^ num__3 x = num__1.5 * num__400000 = num__600000 grams in num__1 cm ^ num__3 y is num__3 gms in num__600000 cm ^ num__3 y = num__3 * num__600000 = num__1 num__800000 gms total gms in num__1 m ^ num__3 = num__600000 + num__1 num__800000 = num__2 num__400000 answer : a <eor> a <eos> |
a |
multiply__1.5__400000.0__ rectangle_perimeter__0.0__400000.0__ rectangle_perimeter__1.0__0.0__ rectangle_perimeter__1.0__0.0__ |
multiply__1.5__400000.0__ rectangle_perimeter__0.0__400000.0__ rectangle_perimeter__1.0__0.0__ power__2.0__1.0__ |
| i chose a number and divide it by num__5 . then i subtracted num__154 from the result and got num__6 . what was the number i chose ? <o> a ) num__200 <o> b ) num__600 <o> c ) num__300 <o> d ) num__800 <o> e ) num__1000 |
let x be the number i chose then x / num__5 − num__154 = num__6 x / num__5 = num__160 x = num__800 correct answer d <eor> d <eos> |
d |
add__154.0__6.0__ multiply__5.0__160.0__ multiply__5.0__160.0__ |
add__154.0__6.0__ multiply__5.0__160.0__ multiply__5.0__160.0__ |
| the marks obtained by vijay and amith are in the ratio num__4 : num__5 and those obtained by amith and abhishek in the ratio of num__3 : num__2 . the marks obtained by vijay and abhishek are in the ratio of ? <o> a ) num__6 : num__7 <o> b ) num__6 : num__9 <o> c ) num__6 : num__5 <o> d ) num__6 : num__1 <o> e ) num__6 : num__2 |
num__4 : num__5 num__3 : num__2 - - - - - - - num__12 : num__15 : num__10 num__12 : num__10 num__6 : num__5 answer : c <eor> c <eos> |
c |
multiply__4.0__3.0__ multiply__5.0__3.0__ multiply__5.0__2.0__ add__4.0__2.0__ add__4.0__2.0__ |
multiply__4.0__3.0__ multiply__5.0__3.0__ subtract__12.0__2.0__ subtract__10.0__4.0__ subtract__10.0__4.0__ |
| find the principal if at num__9.0 per annum simple interest after num__3 years the interest amounts to rs . num__2336 less than the principal ? <o> a ) num__1200 <o> b ) num__3450 <o> c ) num__3200 <o> d ) num__3500 <o> e ) num__3300 |
p - num__2336 = ( p * num__9 * num__3 ) / num__100 p = num__3200 answer : c <eor> c <eos> |
c |
percent__100.0__3200.0__ |
percent__100.0__3200.0__ |
| a train num__110 m long is running with a speed of num__60 km / hr . in what time will it pass a man who is running at num__6 km / hr in the direction opposite to that in which the train is going ? <o> a ) num__7 sec <o> b ) num__6 sec <o> c ) num__8 sec <o> d ) num__4 sec <o> e ) num__2 sec |
speed of train relative to man = num__60 + num__6 = num__66 km / hr . = num__66 * num__0.277777777778 = num__18.3333333333 m / sec . time taken to pass the men = num__110 * num__0.0545454545455 = num__6 sec . answer : b <eor> b <eos> |
b |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ round__6.0__ |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ divide__110.0__18.3333__ |
| mary ' s income is num__60.0 more than tim ' s income and tim ' s income is num__30.0 less than juan ' s income . what % of juan ' s income is mary ' s income . <o> a ) num__112.0 <o> b ) b . num__100.0 <o> c ) num__96.0 <o> d ) num__80.0 <o> e ) num__64 % |
even i got num__96.0 j = num__100 t = num__100 * num__0.7 = num__70 m = num__70 * num__1.6 = num__112 if mary ' s income is x percent of j m = j * x / num__100 x = m * num__100 / j = num__112 * num__1.0 = num__112 ans : a <eor> a <eos> |
a |
subtract__100.0__30.0__ divide__96.0__60.0__ multiply__70.0__1.6__ round_down__1.6__ multiply__1.0__112.0__ |
multiply__100.0__0.7__ divide__96.0__60.0__ multiply__70.0__1.6__ round_down__1.6__ multiply__1.0__112.0__ |
| of num__45 applicants for a job num__21 had at least num__4 years ' experience num__27 had degrees and num__5 had less than num__4 years ' experience and did not have a degree . how many of the applicants had at least num__4 years ' experience and a degree ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__4 <o> d ) num__3 <o> e ) num__2 |
num__45 - num__5 = num__40 num__40 - num__21 - num__27 = - num__8 then num__8 are in the intersection between num__4 years experience and degree . answer : b <eor> b <eos> |
b |
subtract__45.0__5.0__ divide__40.0__5.0__ divide__40.0__5.0__ |
subtract__45.0__5.0__ divide__40.0__5.0__ divide__40.0__5.0__ |
| if z is a positive integer which of the following must be odd ? <o> a ) x + num__1 <o> b ) z ^ num__2 + z <o> c ) z ^ num__2 + z + num__1 <o> d ) x ^ num__2 − num__1 <o> e ) num__3 x ^ num__2 − num__3 |
a . z + num__1 = can be odd or even . since o + o = e or e + o = o b . z ^ num__2 + z = z ( z + num__1 ) . since from the above derivation we already know the term z + num__1 can be odd or even directly substitute here . z ( odd ) = even ( when z is even ) or z ( even ) = even [ when z is odd ] c . here ' s the answer . since we know the term z ^ num__2 + z can always take a even number even + num__1 = odd hence c . <eor> c <eos> |
c |
multiply__1.0__2.0__ |
power__2.0__1.0__ |
| ( √ num__97 + √ num__486 ) / √ num__54 = ? <o> a ) num__2 √ num__2 <o> b ) num__2 √ num__3 <o> c ) num__3 √ num__2 <o> d ) num__3 √ num__3 <o> e ) num__4.33333333333 |
( √ num__96 + √ num__486 ) / √ num__54 = ( num__4 √ num__6 + num__9 √ num__6 ) / num__3 √ num__6 = num__13 √ num__2.0 √ num__6 = num__4.33333333333 hence the correct answer is e . <eor> e <eos> |
e |
divide__486.0__54.0__ subtract__9.0__6.0__ add__4.0__9.0__ divide__6.0__3.0__ divide__13.0__3.0__ divide__13.0__3.0__ |
divide__486.0__54.0__ subtract__9.0__6.0__ add__4.0__9.0__ divide__6.0__3.0__ divide__13.0__3.0__ divide__13.0__3.0__ |
| if x * y = xy + num__2 ( x + y ) for all integers x and y then num__2 * ( – num__3 ) = <o> a ) – num__16 <o> b ) – num__11 <o> c ) – num__8 <o> d ) num__4 <o> e ) num__16 |
num__2 * ( - num__3 ) = num__2 * ( - num__3 ) + num__2 ( num__2 + ( - num__3 ) ) = - num__6 - num__2 = - num__8 option ( c ) <eor> c <eos> |
c |
multiply__2.0__3.0__ add__2.0__6.0__ add__2.0__6.0__ |
multiply__2.0__3.0__ add__2.0__6.0__ add__2.0__6.0__ |
| a thief is spotted by a policeman from a distance of num__225 meters . when the policeman starts the chase the thief also starts running . if the speed of the thief be num__8 km / hr and that of the policeman num__10 km / hr how far the thief will have run before he is overtaken ? <o> a ) num__350 m <o> b ) num__200 m <o> c ) num__400 m <o> d ) num__900 m <o> e ) none of them |
relative speed of the policeman = ( num__10 - num__8 ) km / hr = num__2 km / hr . time taken by police man to cover ( num__225 m / num__1000 ) x num__0.5 hr = num__0.1125 hr . in num__0.1125 hrs the thief covers a distance of num__8 x num__0.1125 km = num__0.9 km = num__900 m answer is d . <eor> d <eos> |
d |
subtract__10.0__8.0__ multiply__8.0__0.1125__ multiply__0.9__1000.0__ round__900.0__ |
subtract__10.0__8.0__ multiply__8.0__0.1125__ multiply__0.9__1000.0__ round__900.0__ |
| if x # y is defined to equal x ^ num__2 / y for all x and y then ( - num__1 # num__3 ) # num__3 = <o> a ) num__1.33333333333 <o> b ) num__0.333333333333 <o> c ) num__0.037037037037 <o> d ) - num__0.0833333333333 <o> e ) - num__1.33333333333 |
( - num__1 ) ^ num__0.666666666667 = num__0.333333333333 ( num__0.333333333333 ) ^ num__0.666666666667 = num__0.037037037037 so c is my answer <eor> c <eos> |
c |
divide__2.0__3.0__ reverse__3.0__ multiply__1.0__0.037__ |
divide__2.0__3.0__ subtract__1.0__0.6667__ multiply__1.0__0.037__ |
| if the radius of a cylinder is doubled and so is the height what is the new volume of the cylinder divided by the old one ? <o> a ) num__8 . <o> b ) num__2 <o> c ) num__6 <o> d ) num__4 <o> e ) num__10 |
let the radius be r and the the height be h . new radius = num__2 r and height = num__2 h . area ( new ) : area ( old ) = pi ∗ ( num__2 r ) ^ num__2 ∗ num__2 h / pi ∗ r ^ num__2 ∗ h = num__8 : num__1 . hence the answer is a . <eor> a <eos> |
a |
square_perimeter__2.0__ square_perimeter__2.0__ |
square_perimeter__2.0__ power__8.0__1.0__ |
| the length of a rectangle is halved while its breadth is tripled . watis the % change in area ? <o> a ) num__30.0 <o> b ) num__40.0 <o> c ) num__50 percent <o> d ) num__60.0 <o> e ) num__70 % |
let original length = x and original breadth = y . original area = xy . new length = x . num__2 new breadth = num__3 y . new area = x x num__3 y = num__3 xy . num__2 num__2 increase % = num__1 xy x num__1 x num__100.0 = num__50.0 . num__2 xy c <eor> c <eos> |
c |
triangle_area__1.0__100.0__ triangle_area__1.0__100.0__ |
triangle_area__1.0__100.0__ triangle_area__1.0__100.0__ |
| q : robert purchased $ num__3500 worth of us saving bonds . if bonds are sold in $ num__50 or $ num__100 denominations only which of the following can not be the number of us saving bonds that robert purchased ? <o> a ) num__20 <o> b ) num__27 <o> c ) num__50 <o> d ) num__70 <o> e ) num__80 |
i started off with looking at the answer choices . . . num__20 - num__27 - num__50 are in a tight range so my guess was to look and rule out the bigger numbers . the total amount is $ num__3500 . so with num__80 bonds each $ num__50 = $ num__4000 we already exceed the $ num__3500 limit . hence e is the answer . <eor> e <eos> |
e |
subtract__100.0__20.0__ multiply__50.0__80.0__ subtract__100.0__20.0__ |
subtract__100.0__20.0__ multiply__50.0__80.0__ subtract__100.0__20.0__ |
| num__24 men working num__8 hours a day can finish a work in num__10 days . working at the rate of num__10 hours a day the number of men required to finish the same work in num__6 days is <o> a ) num__30 <o> b ) num__36 <o> c ) num__34 <o> d ) num__32 <o> e ) none of these |
m num__1 × d num__1 × t num__1 × w num__2 = m num__2 × d num__2 × t num__2 × w num__1 num__24 × num__10 × num__8 × num__1 = m num__2 × num__6 × num__10 × num__1 ⇒ m num__2 = num__24 × num__10 × num__1.33333333333 × num__10 = num__32 men answer d <eor> d <eos> |
d |
subtract__8.0__6.0__ divide__8.0__6.0__ add__24.0__8.0__ round__32.0__ |
subtract__8.0__6.0__ divide__8.0__6.0__ add__24.0__8.0__ round__32.0__ |
| a polygon has num__104 diagonals . how many sides does it have ? self made <o> a ) num__12 <o> b ) num__13 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
the best way to solve this problem is to use the formula : an n sided polygon can have n * ( n - num__3 ) / num__2 diagonals n * ( n - num__3 ) / num__2 = num__104 n * ( n - num__3 ) = num__208 substitute n from the answer choices . n = num__16 answer : e <eor> e <eos> |
e |
multiply__104.0__2.0__ triangle_area__2.0__16.0__ |
multiply__104.0__2.0__ triangle_area__2.0__16.0__ |
| num__2525 * num__9 <o> a ) num__22725 <o> b ) num__25675 <o> c ) num__22655 <o> d ) num__27575 <o> e ) none of these |
explanation : num__2525 * ( num__10 - num__1 ) = num__25250 - num__2525 = num__22725 option a <eor> a <eos> |
a |
subtract__10.0__9.0__ multiply__2525.0__10.0__ multiply__2525.0__9.0__ multiply__2525.0__9.0__ |
subtract__10.0__9.0__ multiply__2525.0__10.0__ multiply__2525.0__9.0__ multiply__2525.0__9.0__ |
| what will come after num__1 num__39 num__31129 <o> a ) num__451 <o> b ) num__551 <o> c ) num__651 <o> d ) num__751 <o> e ) num__851 |
num__1 * num__1 + num__2 = num__3 num__3 * num__2 + num__3 = num__9 num__9 * num__3 + num__4 = num__31 num__31 * num__4 + num__5 = num__129 in same logic next number will be - num__129 * num__5 + num__6 = num__651 answer : c <eor> c <eos> |
c |
add__1.0__2.0__ power__3.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ add__1.0__5.0__ multiply__1.0__651.0__ |
add__1.0__2.0__ power__3.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ add__1.0__5.0__ multiply__1.0__651.0__ |
| num__7 ^ num__6 n - num__1 ^ num__6 n when n is an integer > num__0 is divisible by <o> a ) num__112 <o> b ) num__127 <o> c ) num__134 <o> d ) num__154 <o> e ) num__156 |
num__127 b <eor> b <eos> |
b |
multiply__1.0__127.0__ |
multiply__1.0__127.0__ |
| the “ length of integer x ” refers to the number of prime factors not necessarily distinct that x has . ( if x = num__60 the length of x would be num__4 because num__60 = num__2 × num__2 × num__3 × num__5 . ) what is the greatest possible length of integer z if z < num__1000 ? <o> a ) num__13 <o> b ) num__10 <o> c ) num__9 <o> d ) num__7 <o> e ) num__6 |
to maximize the length of z we should minimize its prime base . the smallest prime is num__2 and since num__2 ^ num__9 = num__512 < num__1000 then the greatest possible length of integer z is num__9 . the answer is c . <eor> c <eos> |
c |
add__4.0__5.0__ add__4.0__5.0__ |
add__4.0__5.0__ add__4.0__5.0__ |
| mona and sona go around a circular track of length num__400 m on a bike at speeds of num__18 kmph and num__36 kmph . after how much time will they meet for the first time at the starting point ? <o> a ) num__120 <o> b ) num__80 <o> c ) num__60 <o> d ) num__136 <o> e ) num__150 |
time taken to meet for the first time at the starting point = lcm { length of the track / speed of mona length of the track / speed of sona } = lcm { num__400 / ( num__18 * num__0.277777777778 ) num__400 / ( num__36 * num__0.277777777778 ) } = lcm ( num__80 num__40 ) = num__80 sec . answer : b <eor> b <eos> |
b |
round__80.0__ |
round__80.0__ |
| a walks around a circular field at the rate of one round per hour while b runs around it at the rate of five rounds per hour . they start in the same direction from the same point at num__7.30 a . m . they shall first cross each other after ___ minutes ? <o> a ) num__11 min <o> b ) num__12 min <o> c ) num__10 min <o> d ) num__15 min <o> e ) num__16 min |
since a and b move in the same direction along the circle so they will first meet each other when there is a difference of one round between the two . relative speed of a and b = num__5 - num__1 = num__4 rounds per hour . time taken to complete one round at this speed = num__0.25 hr = num__15 min . d <eor> d <eos> |
d |
subtract__5.0__1.0__ divide__1.0__4.0__ round__15.0__ |
subtract__5.0__1.0__ divide__1.0__4.0__ round__15.0__ |
| if n is a natural number then ( num__6 n ^ num__2 + num__6 n ) is always divisible by ? <o> a ) num__6 only <o> b ) both num__6 and num__12 <o> c ) num__12 only <o> d ) num__6 or num__12 <o> e ) none |
num__6 n ^ num__2 + num__6 n = num__6 n ( n + num__1 ) n ( n + num__1 ) is always even when n is a natural number hence num__6 n ^ num__2 + num__6 n is always divisible by num__6 and num__12 b ) <eor> b <eos> |
b |
multiply__6.0__2.0__ multiply__6.0__1.0__ |
multiply__6.0__2.0__ multiply__6.0__1.0__ |
| sheila works num__8 hours per day on monday wednesday and friday and num__6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ num__396 per week . how much does she earn in dollars per hour ? <o> a ) num__2 <o> b ) num__8 <o> c ) num__9 <o> d ) num__11 <o> e ) num__2 |
explanation : total hours worked = num__8 x num__3 + num__6 x num__2 = num__36 total earned = num__396 . hourly wage = num__11.0 = num__11 answer : d <eor> d <eos> |
d |
subtract__8.0__6.0__ add__8.0__3.0__ round__11.0__ |
subtract__8.0__6.0__ add__8.0__3.0__ add__8.0__3.0__ |
| a num__25 cm wide path is to be made around a circular garden having a diameter of num__4 meters . approximate area of the path is square meters is ? <o> a ) num__3.34 sq m <o> b ) num__2.98 sq m <o> c ) num__9.98 sq m <o> d ) num__1.98 sq m <o> e ) num__3.78 sq m |
area of the path = area of the outer circle - area of the inner circle = ∏ { num__2.0 + num__0.25 } num__2 - ∏ [ num__2.0 ] num__2 = ∏ [ num__2.252 - num__22 ] = ∏ ( num__0.25 ) ( num__4.25 ) { ( a num__2 - b num__2 = ( a - b ) ( a + b ) } = ( num__3.14 ) ( num__0.25 ) ( num__4.25 ) = num__53.38 / num__16 = num__3.34 sq m answer : a <eor> a <eos> |
a |
volume_rectangular_prism__4.0__4.25__3.14__ square_perimeter__4.0__ triangle_area__2.0__3.34__ |
volume_rectangular_prism__4.0__4.25__3.14__ square_perimeter__4.0__ triangle_area__2.0__3.34__ |
| find the value of ( x ) in the given equation ? num__35.0 of num__1500 + x = num__45.0 of num__4200 – num__320 <o> a ) num__910 <o> b ) num__980 <o> c ) num__1012 <o> d ) num__1045 <o> e ) none of these |
explanation : num__35.0 of num__1500 + x = num__45.0 of num__4200 – num__320 ( num__0.35 * num__1500 ) + x = ( num__0.45 * num__4200 ) – num__320 num__525 + x = num__1890 – num__320 x = num__1890 – num__320 – num__525 x = num__1890 – num__845 x = num__1045 answer : d <eor> d <eos> |
d |
multiply__1500.0__0.35__ multiply__4200.0__0.45__ add__320.0__525.0__ subtract__1890.0__845.0__ subtract__1890.0__845.0__ |
multiply__1500.0__0.35__ multiply__4200.0__0.45__ add__320.0__525.0__ subtract__1890.0__845.0__ subtract__1890.0__845.0__ |
| the average of five consecutive odd numbers is num__61 . what is the difference between the highest and the lowest number ? <o> a ) num__8 <o> b ) num__2 <o> c ) num__5 <o> d ) can not be determined <o> e ) none of these |
suppose the consecutive odd numbers are : x x + num__2 x + num__4 x + num__6 and x + num__8 therefore the required difference = x + num__8 – x = num__8 note that answering the above question does not require the average of the five consecutive odd numbers . answer a <eor> a <eos> |
a |
add__2.0__4.0__ multiply__2.0__4.0__ multiply__2.0__4.0__ |
add__2.0__4.0__ add__2.0__6.0__ add__2.0__6.0__ |
| robert can row a boat at num__15 kmph in still water . if the speed of the stream is num__4 kmph what is the time taken to row a distance of num__80 km downstream ? <o> a ) num__5.33333333333 <o> b ) num__4.70588235294 <o> c ) num__4.21052631579 <o> d ) num__3.80952380952 <o> e ) num__3.47826086957 |
speed downstream = num__15 + num__4 = num__19 kmph . time required to cover num__60 km downstream = d / s = num__4.21052631579 hours . answer : c <eor> c <eos> |
c |
add__15.0__4.0__ hour_to_min_conversion__ divide__80.0__19.0__ divide__80.0__19.0__ |
add__15.0__4.0__ hour_to_min_conversion__ divide__80.0__19.0__ divide__80.0__19.0__ |
| at a certain supplier a machine of type a costs $ num__20000 and a machine of type b costs $ num__65000 . each machine can be purchased by making a num__20 percent down payment and repaying the remainder of the cost and the finance charges over a period of time . if the finance charges are equal to num__40 percent of the remainder of the cost how much less would num__2 machines of type a cost than num__1 machine of type b under this arrangement ? <o> a ) $ num__10000 <o> b ) $ num__11200 <o> c ) $ num__12000 <o> d ) $ num__12800 <o> e ) $ num__33 |
000 |
total cost of num__2 machines of type a = num__20.0 of ( cost of num__2 machine a ) + remainder + num__40.0 remainder = num__20.0 of num__40000 + ( num__40000 - num__20.0 of num__40000 ) + num__40.0 of ( num__40000 - num__20.0 of num__40000 ) = num__52800 total cost of num__1 machine of type b = num__20.0 of ( cost of num__1 machine b ) + remainder + num__40.0 remainder = num__20.0 of num__65000 + ( num__65000 - num__20.0 of num__65000 ) + num__40.0 of ( num__65000 - num__20.0 of num__65000 ) = num__85800 diff = num__66000 - num__85800 = num__33000 hence e . <eor> e <eos> |
e |
e |
| the s . i . on a certain sum of money for num__2 years at num__6.0 per annum is half the c . i . on rs . num__4000 for num__2 years at num__10.0 per annum . the sum placed on s . i . is ? <o> a ) num__2197 <o> b ) num__1267 <o> c ) num__3500 <o> d ) num__2267 <o> e ) num__1262 |
explanation : c . i . = [ num__4000 * ( num__1 + num__0.1 ) num__2 - num__4000 ] = ( num__4000 * num__1.1 * num__1.1 - num__4000 ) = rs . num__840 . sum = ( num__420 * num__100 ) / ( num__2 * num__6 ) = rs . num__3500 answer : c <eor> c <eos> |
c |
percent__10.0__1.0__ percent__100.0__3500.0__ |
percent__10.0__1.0__ percent__100.0__3500.0__ |
| if s and t are positive integers st + s + t can not be <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
let st + t + s = x add num__1 on both sides : st + t + s + num__1 = x + num__1 t ( s + num__1 ) + s + num__1 = x + num__1 ( s + num__1 ) ( t + num__1 ) = x + num__1 minimum value of ( s + num__1 ) = num__2 minimum value of ( t + num__1 ) = num__2 hence x + num__1 can not be prime substitute x from the given options : num__6 + num__1 = num__7 - - > prime - - > st + t + s can not be num__6 answer : b <eor> b <eos> |
b |
add__1.0__6.0__ multiply__1.0__6.0__ |
add__1.0__6.0__ subtract__7.0__1.0__ |
| the average of num__20 numbers is zero . of them at the most how many may be greater than zero ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__10 <o> d ) num__19 <o> e ) num__20 |
solution average of num__20 numbers = num__0 sum of num__20 numbers = ( num__0 x num__20 ) = num__0 . it is quite possible that num__19 of these numbers may be positive and if there sum id a then num__20 th number is ( - a ) . answer a <eor> a <eos> |
a |
multiply__20.0__0.0__ |
multiply__20.0__0.0__ |
| a number x is num__12 times another number y . the percentage that y is less than x is <o> a ) num__91.6 <o> b ) num__87.5 <o> c ) num__80.0 <o> d ) num__11.0 <o> e ) num__1 % |
say y = num__1 and x = num__12 . then y = num__1 is less than x = num__12 by ( num__12 - num__1 ) / num__12 * num__100 = num__0.916666666667 * num__100 = num__91.6 . answer : a . <eor> a <eos> |
a |
multiply__1.0__91.6__ |
multiply__1.0__91.6__ |
| at a local beach the ratio of little dogs to average dogs to enormous dogs is num__3 : num__8 : num__9 . late in the afternoon the ratio of little dogs to average dogs doubles and the ratio of little dogs to enormous dogs increases . if the new percentage of little dogs and the new percentage of average dogs are both integers and there are fewer than num__30 total dogs at the beach which of the following represents a possible new percentage of enormous dogs ? <o> a ) num__25.0 <o> b ) num__40.0 <o> c ) num__30.0 <o> d ) num__55.0 <o> e ) num__70 % |
little dogs ( l ) average dogs ( a ) and enormous dogs ( e ) the initial ratio for l : a : e : : num__3 : num__8 : num__9 initial total dogs = num__20 x ( x assumed ; num__3 + num__8 + num__9 = num__20 ) since the total dogs are less than num__30 therefore initial total value has to be num__20 l = num__2 a = num__5 e = num__8 l : a = num__2 : num__5 this ratio doubles hence new dog count is l = num__6 a = num__8 e = x : also num__6 + num__8 + x < num__30 we need to find x * num__100 / ( num__6 + num__8 + x ) now it says that new percentage of little dogs and average dogs is an integer % l = num__6 * num__100 / ( num__14 + x ) % a = num__8 * num__100 / ( num__14 + x ) ; only value for x is num__6 ; num__14 + x < num__30 and % integer therefore enormus dogs % is = num__6 * num__100 / ( num__20 ) = num__30.0 c <eor> c <eos> |
c |
add__3.0__2.0__ multiply__3.0__2.0__ multiply__5.0__20.0__ add__8.0__6.0__ multiply__5.0__6.0__ |
add__3.0__2.0__ multiply__3.0__2.0__ multiply__5.0__20.0__ add__8.0__6.0__ multiply__5.0__6.0__ |
| a man can reach certain place in num__10 hours . if he reduces his speed by num__0.2 th he goes num__10 km less in time . find his speed ? <o> a ) num__5 km / hr <o> b ) num__22.5 km / hr <o> c ) num__10.78 km / hr <o> d ) num__14 km / hr <o> e ) num__43.15 km / hr |
let the speed be x km / hr num__10 x - num__10 * num__0.8 * x = num__10 num__10 x - num__8 x = num__10 num__2 x = num__10 x = num__5 km / hr answer is a <eor> a <eos> |
a |
multiply__10.0__0.8__ multiply__10.0__0.2__ divide__10.0__2.0__ round__5.0__ |
multiply__10.0__0.8__ multiply__10.0__0.2__ divide__10.0__2.0__ divide__10.0__2.0__ |
| if the product num__4864 x num__9 p num__2 is divisible by num__12 then the value of p is : <o> a ) num__3 <o> b ) num__2 <o> c ) num__1 <o> d ) num__6 <o> e ) num__5 |
c num__1 clearly num__4864 is divisible by num__4 . so num__9 p num__2 must be divisible by num__3 . so ( num__9 + p + num__2 ) must be divisible by num__3 . p = num__1 . <eor> c <eos> |
c |
add__2.0__1.0__ reverse__1.0__ |
add__2.0__1.0__ reverse__1.0__ |
| ( num__2 ^ num__2 - num__1 ) ( num__2 ^ num__2 + num__1 ) ( num__2 ^ num__4 + num__1 ) ( num__2 ^ num__9 + num__1 ) = <o> a ) num__2 ^ num__17 - num__1 <o> b ) num__2 ^ num__16 + num__1 <o> c ) num__2 ^ num__32 - num__1 <o> d ) num__2 ^ num__128 - num__1 <o> e ) num__2 ^ num__16 ( num__2 ^ num__16 - num__1 ) |
a : is correct ( num__2 ^ num__2 - num__1 ) * ( num__2 ^ num__2 + num__1 ) = ( num__2 ^ num__4 - num__1 ) ( num__2 ^ num__4 - num__1 ) * ( num__2 ^ num__4 + num__1 ) = num__2 ^ num__8 - num__1 ( num__2 ^ num__8 - num__1 ) * ( num__2 ^ num__9 + num__1 ) = num__2 ^ num__17 - num__1 you can apply to this formula : a ^ num__2 - b ^ num__2 = ( a - b ) * ( a + b ) <eor> a <eos> |
a |
multiply__2.0__4.0__ add__9.0__8.0__ multiply__2.0__1.0__ |
multiply__2.0__4.0__ add__9.0__8.0__ multiply__2.0__1.0__ |
| if x ^ num__2 = num__3 x + num__1 then x ^ num__3 = ? <o> a ) num__8 x + num__2 <o> b ) num__10 x + num__3 <o> c ) num__6 x + num__1 <o> d ) num__2 x + num__3 <o> e ) num__6 x + num__4 |
x ^ num__2 = num__3 x + num__1 x ^ num__3 = x * x ^ num__2 = x * ( num__3 x + num__1 ) = num__3 x ^ num__2 + x = num__3 ( num__3 x + num__1 ) + x = num__10 x + num__3 the answer is b . <eor> b <eos> |
b |
multiply__1.0__10.0__ |
multiply__1.0__10.0__ |
| a reduction of num__20.0 in the price of oil enables a house wife to obtain num__10 kgs more for rs . num__1500 what is the reduced price for kg ? <o> a ) rs . num__20 <o> b ) rs . num__25 <o> c ) rs . num__30 <o> d ) rs . num__35 <o> e ) rs . num__40 |
explanation : num__1500 * ( num__0.2 ) = num__300 - - - - num__10 ? - - - - num__1 = > rs . num__30 answer : c <eor> c <eos> |
c |
percent__20.0__1500.0__ percent__10.0__300.0__ percent__10.0__300.0__ |
percent__20.0__1500.0__ percent__10.0__300.0__ percent__10.0__300.0__ |
| the average number of shirts with salman ambani and dalmiya is num__40 if all of them reached a shopping mall in delhi and purchased num__14 shirts each of them then average number of shirt each of them now has <o> a ) num__66 <o> b ) num__54 <o> c ) num__67 <o> d ) ca n ' t be determined <o> e ) none of these |
answer required average = old average + new average = num__40 + num__14 = num__54 correct option : b <eor> b <eos> |
b |
add__40.0__14.0__ add__40.0__14.0__ |
add__40.0__14.0__ add__40.0__14.0__ |
| the average weight of a b and c is num__60 kg . if the average weight of a and b be num__70 kg and that of b and c be num__50 kg then the weight of b is : <o> a ) num__50 kg <o> b ) num__60 kg <o> c ) num__55 kg <o> d ) num__57 kg <o> e ) num__62 kg |
explanation let a b c represent their respective weights . then we have : a + b + c = ( num__60 x num__3 ) = num__180 â € ¦ . ( i ) a + b = ( num__70 x num__2 ) = num__140 â € ¦ . ( ii ) b + c = ( num__50 x num__2 ) = num__100 â € ¦ . ( iii ) adding ( ii ) and ( iii ) we get : a + num__2 b + c = num__240 â € ¦ . ( iv ) subtracting ( i ) from ( iv ) we get : b = num__60 . b â € ™ s weight = num__60 kg . answer b <eor> b <eos> |
b |
multiply__60.0__3.0__ multiply__70.0__2.0__ multiply__50.0__2.0__ add__60.0__180.0__ subtract__240.0__180.0__ |
multiply__60.0__3.0__ multiply__70.0__2.0__ multiply__50.0__2.0__ add__60.0__180.0__ subtract__240.0__180.0__ |
| if the perimeter of a rectangular garden is num__600 m its length when its breadth is num__100 m is ? <o> a ) num__299 m <o> b ) num__777 m <o> c ) num__200 m <o> d ) num__167 m <o> e ) num__128 m |
num__2 ( l + num__100 ) = num__600 = > l = num__200 m answer : c <eor> c <eos> |
c |
multiply__100.0__2.0__ round__200.0__ |
multiply__100.0__2.0__ round__200.0__ |
| maxwell leaves his home and walks toward brad ' s house . one hour later brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is num__14 kilometers maxwell ' s walking speed is num__4 km / h and brad ' s running speed is num__6 km / h . what is the total time it takes maxwell before he meets up with brad ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__8 |
total distance = num__14 kms maxwell speed = num__4 kms / hr maxwell travelled for num__1 hour before brad started therefore maxwell traveled for num__4 kms in num__1 hour . time taken = total distance / relative speed total distance after brad started = num__10 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = num__6 + num__4 = num__10 kms / hr time taken to meet brad after brad started = num__1.0 = num__1 hrs distance traveled by maxwell = maxwell ' s speed * time taken = num__4 * num__1 = num__4 + num__4 = num__8 kms . . . therefore total time taken by maxwell to meet brad = distance travelled by maxwell / maxwell ' s speed = num__2.0 = num__2 hrs . . . answer a <eor> a <eos> |
a |
subtract__14.0__4.0__ subtract__14.0__6.0__ subtract__6.0__4.0__ round__2.0__ |
add__4.0__6.0__ subtract__14.0__6.0__ divide__8.0__4.0__ divide__4.0__2.0__ |
| two trains one from howrah to patna and the other from patna to howrah start simultaneously . after they meet the trains reach their destinations after num__9 hours and num__16 hours respectively . the ratio of their speeds is ? <o> a ) num__4 : num__6 <o> b ) num__4 : num__3 <o> c ) num__4 : num__4 <o> d ) num__4 : num__7 <o> e ) num__4 : num__2 |
let us name the trains a and b . then ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ num__16 : √ num__9 = num__4 : num__3 answer : b <eor> b <eos> |
b |
round__4.0__ |
round__4.0__ |
| num__40.0 of the employees in a factory are workers . all the remaining employees are executives . the annual income of each worker is rs . num__390 . the annual income of each executive is rs . num__420 . what is the average annual income of all the employees in the factory together ? <o> a ) num__390 <o> b ) num__405 <o> c ) num__408 <o> d ) num__415 <o> e ) none |
solution : let x be the number of employees . we are given that num__40.0 of the employees are workers . now num__40.0 of x is ( num__0.4 ) * x = num__0.4 x . hence the number of workers is num__2 x / num__5 . all the remaining employees are executives so the number of executives equals ( the number of employees ) – ( the number of workers ) = x – ( num__2 x / num__5 ) = ( num__3 x / num__5 ) the annual income of each worker is rs . num__390 . hence the total annual income of all the workers together is = ( num__2 x / num__5 ) * num__390 = num__156 x . also the annual income of each executive is rs . num__420 . hence the total income of all the executives together is ( num__3 x / num__5 ) × num__420 = num__252 x hence the total income of the employees is = num__156 x + num__252 x = num__408 x . the average income of all the employees together equals answer : option c <eor> c <eos> |
c |
divide__2.0__0.4__ subtract__5.0__2.0__ multiply__390.0__0.4__ add__252.0__156.0__ add__252.0__156.0__ |
divide__2.0__0.4__ subtract__5.0__2.0__ multiply__390.0__0.4__ add__252.0__156.0__ add__252.0__156.0__ |
| a highway is to be divided into num__3 lanes . for this purpose two yellow stripes are painted so that stripes divide the highway into num__3 lanes . if num__3 gallons of paint cover an area of num__4 p square feet of highway how many gallons of paint are needed to paint two stripes x inches wide on a stretch of highway m miles long ? ( num__1 mile = num__5280 feet and num__1 feet = num__12 inches ) <o> a ) num__220 p / mx <o> b ) num__660 mx / p <o> c ) num__880 mx / p <o> d ) num__440 p / mx <o> e ) num__330 mx / p |
num__1 square foot needs num__0.75 p gallons of paint . the width of each stripe is x / num__12 feet . the length of each stripe is num__5280 m feet . the area of each stripe is ( num__5280 m ) ( x / num__12 ) = num__440 mx square feet . for two stripes the total area is num__880 mx square feet . the number of gallons needed is ( num__880 mx ) * ( num__0.75 p ) = num__660 mx / p gallons . the answer is b . <eor> b <eos> |
b |
divide__3.0__4.0__ divide__5280.0__12.0__ multiply__0.75__880.0__ multiply__1.0__660.0__ |
divide__3.0__4.0__ divide__5280.0__12.0__ multiply__0.75__880.0__ multiply__1.0__660.0__ |
| when a person sells two items $ num__800 each one at a gain of num__20.0 and other at a loss of num__20.0 . then the seller incures a loss or gain of : <o> a ) no loss no gain <o> b ) num__4.0 loss <o> c ) num__5.0 gain <o> d ) num__2.0 gain <o> e ) num__1.0 loss |
when two items are sold at the same price one at a profit of a % and other at a loss of a % ( which means the cost price of one item was lower than the selling price and the cost price of the other item was higher than the selling price ) there will always be a loss of ( a ^ num__2 ) / num__100.0 . e . g . here a = num__20 so loss % = ( num__20 ) ^ num__0.02 % = num__4.0 . answer : b <eor> b <eos> |
b |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| num__1500 men have provisions for num__17 days . if num__320 more men join them for how many days will the provisions last now ? <o> a ) num__12.9 <o> b ) num__12.0 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
num__1500 * num__17 = num__1820 * x x = num__14 answer : c <eor> c <eos> |
c |
add__1500.0__320.0__ round__14.0__ |
add__1500.0__320.0__ round__14.0__ |
| pavan travelled for num__11 hours . he covered the first half of the distance at num__30 kmph and remaining half of the distance at num__25 kmph . find the distance travelled by pavan ? <o> a ) num__288 km <o> b ) num__886 km <o> c ) num__866 km <o> d ) num__300 km <o> e ) num__261 km |
let the distance travelled be x km . total time = ( x / num__2 ) / num__30 + ( x / num__2 ) / num__25 = num__11 = > x / num__60 + x / num__50 = num__11 = > ( num__5 x + num__6 x ) / num__300 = num__11 = > x = num__300 km answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ multiply__25.0__2.0__ subtract__30.0__25.0__ subtract__11.0__5.0__ multiply__5.0__60.0__ round__300.0__ |
hour_to_min_conversion__ multiply__25.0__2.0__ subtract__30.0__25.0__ divide__30.0__5.0__ multiply__5.0__60.0__ round__300.0__ |
| in num__1992 a total of num__550 earthquakes occurred worldwide some but not all of which occurred in asia . if num__150 of these earthquakes occurred in asia which of the following represents the ratio of the number of earthquakes that occurred in asia to the number that did not occur in asia ? <o> a ) num__0.25 <o> b ) num__0.125 <o> c ) num__0.375 <o> d ) num__0.625 <o> e ) num__0.875 |
we ' re given a couple of facts to work with : num__1 ) there were a total of num__550 earthquakes num__2 ) of those earthquakes num__150 of them occurred in asia we ' re asked for the ratio of the number of earthquakes that occurred in asia to the number of earthquakes that did not occur in asia . thus there were num__400 earthquakes that did not occur in asia . . . . the answer to the question is num__0.375 = num__0.375 answer : c <eor> c <eos> |
c |
subtract__550.0__150.0__ divide__150.0__400.0__ divide__150.0__400.0__ |
subtract__550.0__150.0__ divide__150.0__400.0__ divide__150.0__400.0__ |
| the radii of two cones are in the ratio num__2 : num__1 their volumes are equal . find the ratio of their heights . <o> a ) num__0.25 <o> b ) num__0.125 <o> c ) num__0.5 <o> d ) num__4.0 <o> e ) none |
sol . let their radii be num__2 x x and their heights be h and h respectively . then num__0.333333333333 * ∏ * ( num__2 x ) ² * h = num__0.333333333333 * ∏ * x ² * h or h / h = num__0.25 . answer a <eor> a <eos> |
a |
triangle_area__2.0__0.25__ |
triangle_area__2.0__0.25__ |
| company c sells a line of num__25 products with an average retail price of $ num__1000 . if none of these products sells for less than $ num__400 and exactly num__10 of the products sell for less than $ num__1000 what is the greatest possible selling price of the most expensive product ? <o> a ) num__9000 <o> b ) num__8000 <o> c ) num__7000 <o> d ) num__6000 <o> e ) num__5000 |
the average price of num__25 products is $ num__1000 means that the total price of num__25 products is num__25 * num__1000 = $ num__25000 . next since exactly num__10 of the products sell for less than $ num__1000 then let ' s make these num__10 items to be at $ num__400 each ( min possible ) . now the remaining num__14 items can not be priced less than $ num__1000 thus the minimum possible price of each of these num__14 items is $ num__1000 . thus the minimum possible value of num__24 products is num__10 * num__400 + num__14 * num__1000 = $ num__18000 . therefore the greatest possible selling price of the most expensive product is $ num__25000 - $ num__18200 = $ num__7000 . answer : c . <eor> c <eos> |
c |
multiply__25.0__1000.0__ add__10.0__14.0__ subtract__25000.0__18000.0__ subtract__25000.0__18000.0__ |
multiply__25.0__1000.0__ add__10.0__14.0__ subtract__25000.0__18000.0__ subtract__25000.0__18000.0__ |
| how many integers between num__324700 and num__448600 have tens digit num__1 and units digit num__3 ? <o> a ) num__10300 <o> b ) num__8030 <o> c ) num__1253 <o> d ) num__1252 <o> e ) num__1 |
239 |
the integers are : num__324713 num__324813 etc . . . num__448513 the number of integers is num__4486 - num__3247 = num__1239 the answer is e . <eor> e <eos> |
e |
e |
| the total number of digits used in numbering the pages of a book having num__266 pages is <o> a ) num__732 <o> b ) num__990 <o> c ) num__109 <o> d ) num__130 <o> e ) num__690 |
total number of digits = ( no . of digits in num__1 - digit page nos . + no . of digits in num__2 - digit page nos . + no . of digits in num__3 - digit page nos . ) = ( num__1 x num__9 + num__2 x num__90 + num__3 x num__167 ) = ( num__9 + num__180 + num__501 ) = num__690 . answer : e <eor> e <eos> |
e |
add__1.0__2.0__ multiply__2.0__90.0__ multiply__3.0__167.0__ multiply__1.0__690.0__ |
add__1.0__2.0__ multiply__2.0__90.0__ multiply__3.0__167.0__ multiply__1.0__690.0__ |
| a rectangular - shaped carpet that measures x feet by y feet is priced at $ num__12 . what is the cost of the carpet in dollars per square yard ? ( num__1 square yard = num__9 square feet ) <o> a ) num__108 / ( xy ) <o> b ) num__90 xy <o> c ) xy / num__90 <o> d ) xy / num__10 <o> e ) num__10 / ( xy ) |
the area of the carpet in feet is xy . the area in square yards is xy / num__9 . the price per square yard is num__12 / ( xy / num__9 ) = num__108 / ( xy ) . the answer is a . <eor> a <eos> |
a |
multiply__12.0__9.0__ multiply__12.0__9.0__ |
multiply__12.0__9.0__ divide__108.0__1.0__ |
| if it takes num__4 identical printing presses exactly num__4 hours to print num__8000 newspapers how long would it take num__2 of these presses to print num__6000 newspapers ? <o> a ) num__6 hours <o> b ) num__8 hours <o> c ) num__7 hours <o> d ) num__5 hours <o> e ) num__9 hours |
num__4 presses - num__8000 newspapers - num__4 hours ; num__2 presses - num__4000 newspapers - num__4 hours ; ( num__240 mins ) num__2 presses - num__6000 newspapers - num__0.06 * num__6000 = num__360 mins = num__6 hrs answer : a <eor> a <eos> |
a |
divide__8000.0__2.0__ divide__240.0__4000.0__ multiply__6000.0__0.06__ add__4.0__2.0__ round__6.0__ |
divide__8000.0__2.0__ divide__240.0__4000.0__ multiply__6000.0__0.06__ add__4.0__2.0__ round__6.0__ |
| three numbers are in the ratio num__2 : num__3 : num__5 and their h . c . f is num__30 . the numbers are : <o> a ) num__2 num__3 num__5 <o> b ) num__5 num__10 num__15 <o> c ) num__10 num__20 num__30 <o> d ) num__60 num__90 num__150 <o> e ) num__12 num__24 num__39 |
let the required numbers be num__2 x num__3 x and num__5 x . then their h . c . f = x . so x = num__30 . the numbers are num__60 num__90 num__150 . answer : d <eor> d <eos> |
d |
multiply__2.0__30.0__ multiply__3.0__30.0__ multiply__5.0__30.0__ multiply__2.0__30.0__ |
multiply__2.0__30.0__ multiply__3.0__30.0__ multiply__5.0__30.0__ multiply__2.0__30.0__ |
| tom traveled the entire num__100 miles trip . if he did the first num__50 miles of at a constant rate num__20 miles per hour and the remaining trip of at a constant rate num__50 miles per hour what is the his average speed in miles per hour ? <o> a ) num__28.36 mph <o> b ) num__26.55 mph <o> c ) num__28.57 mph <o> d ) num__25.56 mph <o> e ) num__28.45 mph |
avg speed = total distance / total time = ( d num__1 + d num__2 ) / ( t num__1 + t num__2 ) = ( num__50 + num__50 ) / ( ( num__2.5 ) + ( num__1.0 ) ) = num__60 * num__0.666666666667 = num__28.57 mph c <eor> c <eos> |
c |
divide__100.0__50.0__ divide__50.0__20.0__ hour_to_min_conversion__ round__28.57__ |
divide__100.0__50.0__ divide__50.0__20.0__ hour_to_min_conversion__ divide__28.57__1.0__ |
| two trains each num__100 m long moving in opposite directions cross other in num__10 sec . if one is moving twice as fast the other then the speed of the faster train is ? <o> a ) num__76 km / hr <o> b ) num__66 km / hr <o> c ) num__48 km / hr <o> d ) num__67 km / hr <o> e ) num__22 km / hr |
let the speed of the slower train be x m / sec . then speed of the train = num__2 x m / sec . relative speed = ( x + num__2 x ) = num__3 x m / sec . ( num__100 + num__100 ) / num__10 = num__3 x = > x = num__6.66666666667 . so speed of the faster train = num__13.3333333333 = num__13.3333333333 * num__3.6 = num__48 km / hr . answer : c <eor> c <eos> |
c |
round__48.0__ |
round__48.0__ |
| a large research project was scheduled over the course of a num__100 - month period and a budget of $ num__250000 was allotted for the course of the entire project . after num__10 months the project had spent $ num__26000 . was the project at this point over budget or under budget and by how much ? <o> a ) under $ num__4200 <o> b ) under $ num__1200 <o> c ) under $ num__700 <o> d ) over $ num__1800 <o> e ) over $ num__1000 |
cost per month = num__2500.0 = num__2500 . now multiply by seven months : num__10 * ( $ num__2500 ) = $ num__25000 . that would be the budgeted amount the amount theoretically supplied to the project . the real amount spent is more than this so it is over budget . $ num__25000 â € “ # num__26000 = $ num__1000 answer = ( e ) <eor> e <eos> |
e |
divide__250000.0__100.0__ divide__250000.0__10.0__ multiply__100.0__10.0__ multiply__100.0__10.0__ |
divide__250000.0__100.0__ multiply__10.0__2500.0__ multiply__100.0__10.0__ multiply__100.0__10.0__ |
| the sum of all consecutive odd integers from − num__25 to num__35 inclusive is <o> a ) num__130 <o> b ) num__135 <o> c ) num__155 <o> d ) num__195 <o> e ) num__235 |
the sum of the odd numbers from - num__25 to + num__25 is num__0 . let ' s add the remaining numbers . num__27 + num__29 + num__31 + num__33 + num__35 = num__5 ( num__31 ) = num__155 the answer is c . <eor> c <eos> |
c |
multiply__5.0__31.0__ round__155.0__ |
multiply__5.0__31.0__ round__155.0__ |
| the present ages of three persons are in proportions num__4 : num__7 : num__9 . eight years ago the sum of their ages was num__136 . find their present ages . <o> a ) num__20 num__28 <o> b ) num__28 num__36 <o> c ) num__35 num__45 <o> d ) num__56 num__72 <o> e ) of these |
let their present ages be num__4 x num__7 x and num__9 x years respectively . then ( num__4 x - num__8 ) + ( num__7 x - num__8 ) + ( num__9 x - num__8 ) = num__136 num__20 x = num__160 = > x = num__8 their present ages are num__32 num__56 and num__72 years respectively . answer : d <eor> d <eos> |
d |
multiply__8.0__20.0__ multiply__4.0__8.0__ multiply__7.0__8.0__ multiply__9.0__8.0__ multiply__7.0__8.0__ |
multiply__8.0__20.0__ multiply__4.0__8.0__ multiply__7.0__8.0__ multiply__9.0__8.0__ multiply__7.0__8.0__ |
| the distance between delhi and mathura is num__110 kms . a starts from delhi with a speed of num__20 kmph at num__7 a . m . for mathura and b starts from mathura with a speed of num__25 kmph at num__8 p . m . from delhi . when will they meet ? <o> a ) num__10 a . m <o> b ) num__10 a . m <o> c ) num__90 a . m <o> d ) num__18 a . m <o> e ) num__20 a . m |
d = num__110 – num__20 = num__90 rs = num__20 + num__25 = num__45 t = num__2.0 = num__2 hours num__8 a . m . + num__2 = num__10 a . m . answer : b <eor> b <eos> |
b |
subtract__110.0__20.0__ add__20.0__25.0__ divide__90.0__45.0__ divide__20.0__2.0__ round__10.0__ |
subtract__110.0__20.0__ add__20.0__25.0__ divide__90.0__45.0__ add__8.0__2.0__ add__8.0__2.0__ |
| a man can row downstream at num__28 kmph and upstream at num__16 kmph . find the speed of stream ? <o> a ) num__8 <o> b ) num__5 <o> c ) num__4 <o> d ) num__6 <o> e ) num__3 |
let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = num__28 - - - ( num__1 ) and x - y = num__16 - - - ( num__2 ) from ( num__1 ) & ( num__2 ) num__2 x = num__44 = > x = num__22 y = num__6 . answer : d <eor> d <eos> |
d |
add__28.0__16.0__ divide__44.0__2.0__ subtract__28.0__22.0__ round__6.0__ |
add__28.0__16.0__ divide__44.0__2.0__ subtract__28.0__22.0__ subtract__28.0__22.0__ |
| a farmer travelled a distance of num__61 km in num__9 hours . he travelled partly on foot at num__4 km / hr and partly on bicycle at num__9 km / hr . the distance travelled on foot is : <o> a ) num__1 hr num__16 min <o> b ) num__1 hr num__17 min <o> c ) num__1 hr num__18 min <o> d ) num__1 hr num__19 min <o> e ) none |
sol . let the distance travelled on foot be x km . then distance travelled on bicycle = ( num__61 - x ) km . so x / num__4 + ( num__61 - x ) / num__9 = num__9 ⇔ num__9 x + num__4 ( num__61 - x ) = num__9 * num__36 ⇔ num__5 x = num__80 ⇔ x = num__16 km . answer a <eor> a <eos> |
a |
multiply__9.0__4.0__ subtract__9.0__4.0__ divide__80.0__5.0__ subtract__5.0__4.0__ |
multiply__9.0__4.0__ subtract__9.0__4.0__ divide__80.0__5.0__ subtract__5.0__4.0__ |
| a store has announced a num__10.0 rebate for the entire day . if amy only has rs . num__500 in her pocket and needs rs . num__20 to get back home how many bottles of juice can she buy at the most from the store with each juice costing rs . num__50 before the discount ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
explanation : amy has num__500 - num__20 = num__480 to spend . each juice costs num__50 * ( num__1 - num__10.0 ) = num__50 * num__0.90 = num__45 . num__10.6666666667 = num__10.67 since you can not buy juice in halves amy can buy num__10 bottles at the most . correct option : c <eor> c <eos> |
c |
percent__20.0__50.0__ |
percent__20.0__50.0__ |
| a grocer has a sale of rs . num__5700 rs . num__8550 rs . num__6855 rs . num__3850 for num__4 consecutive months . how much sale must he have in the fifth month so that he gets an average sale of rs . num__7800 ? <o> a ) s . num__14991 <o> b ) s . num__49930 <o> c ) s . num__14045 <o> d ) s . num__14999 <o> e ) s . num__14578 |
total sale for num__4 months = rs . ( num__5700 + num__8550 + num__6855 + num__3850 ) = rs . num__24955 required sale = rs . [ ( num__7800 x num__5 ) - num__24955 ] = rs . ( num__39000 - num__24955 ) = rs . num__14045 option c <eor> c <eos> |
c |
multiply__7800.0__5.0__ subtract__39000.0__24955.0__ subtract__39000.0__24955.0__ |
multiply__7800.0__5.0__ subtract__39000.0__24955.0__ subtract__39000.0__24955.0__ |
| for every $ num__20 that a billionaire spends a millionaire spends the equivalent of num__20 cents . for every $ num__2 that a millionaire spends a yuppie spends the equivalent of $ num__1 . the ratio of money spent by a yuppie millionaire and billionaire can be expressed as <o> a ) num__1 : num__4 : num__400 <o> b ) num__1 : num__2 : num__200 <o> c ) num__20 : num__4 : num__1 <o> d ) num__100 : num__4 : num__1 <o> e ) num__400 : num__4 : num__1 |
b . . . . . . . . . m . . . . . . . . y num__20 . . . . . num__020 . . . . . . y b . . . . . . . . . num__2 . . . . . . . . num__1 what i did first was to turn num__0.20 to num__2 ( by multiplying by num__10 ) so that it is easy to find the lcm . this led me to this : b . . . . . . . . . m . . . . . . . . y num__200 . . . . . num__2 . . . . . . . . . y b . . . . . . . . num__20 . . . . . . num__10 then i multiplied every row by num__20 ( the lcm of num__2 and num__20 ) which led me to this : b . . . . . . . . . m . . . . . . . . y num__2000 . . . num__20 . . . . . . . num__10 then i got rid of the extra zero and in the correct order this is y : m : b = num__1 : num__2 : num__200 ans b <eor> b <eos> |
b |
divide__20.0__2.0__ multiply__20.0__10.0__ multiply__200.0__10.0__ reverse__1.0__ |
divide__20.0__2.0__ multiply__20.0__10.0__ multiply__200.0__10.0__ reverse__1.0__ |
| a train num__100 m long crosses a platform num__200 m long in num__15 sec ; find the speed of the train ? <o> a ) num__94 kmph <o> b ) num__58 kmph <o> c ) num__72 kmph <o> d ) num__94 kmph <o> e ) num__59 kmph |
d = num__100 + num__200 = num__300 t = num__15 s = num__20.0 * num__3.6 = num__72 kmph answer : c <eor> c <eos> |
c |
add__100.0__200.0__ divide__300.0__15.0__ multiply__3.6__20.0__ round__72.0__ |
add__100.0__200.0__ divide__300.0__15.0__ multiply__3.6__20.0__ multiply__3.6__20.0__ |
| find the average of all prime numbers between num__1 and num__10 <o> a ) num__2.5 <o> b ) num__1.35 <o> c ) num__4.25 <o> d ) num__6.12 <o> e ) num__7.13 |
prime numbers between num__1 and num__10 are num__23 num__57 required average = ( num__2 + num__3 + num__5 + num__7 ) / num__4 = num__4.25 = num__4.25 answer is c <eor> c <eos> |
c |
add__1.0__2.0__ divide__10.0__2.0__ subtract__10.0__3.0__ add__1.0__3.0__ multiply__1.0__4.25__ |
add__1.0__2.0__ divide__10.0__2.0__ add__2.0__5.0__ add__1.0__3.0__ divide__4.25__1.0__ |
| the ratio between the length and the breadth of a rectangular park is num__3 : num__2 . if a man cycling alongthe oundary of the park at the speed of num__12 km / hr completes one round in num__8 minutes then the area of the park ( in sq . m ) is <o> a ) num__124345 m <o> b ) num__233234 m <o> c ) num__153600 m <o> d ) num__123567 m <o> e ) num__168999 m |
perimeter = distance covered in num__8 min . = num__12000 x num__8 m = num__1600 m . num__60 let length = num__3 x metres and breadth = num__2 x metres . then num__2 ( num__3 x + num__2 x ) = num__1600 or x = num__160 . length = num__480 m and breadth = num__320 m . area = ( num__480 x num__320 ) m num__2 = num__153600 m c <eor> c <eos> |
c |
hour_to_min_conversion__ multiply__3.0__160.0__ multiply__2.0__160.0__ multiply__320.0__480.0__ round__153600.0__ |
hour_to_min_conversion__ multiply__3.0__160.0__ multiply__2.0__160.0__ multiply__320.0__480.0__ round__153600.0__ |
| a person travels equal distances with speeds of num__2 km / hr num__4 km / hr num__6 km / hr . and takes a total time of num__11 minutes . find the total distance ? <o> a ) num__1 km <o> b ) num__500 mts <o> c ) num__600 mts <o> d ) num__2 km <o> e ) num__250 mts |
let the each distance be x km total distance = num__3 x then total time ( x / num__2 ) + ( x / num__4 ) + ( x / num__6 ) = num__0.183333333333 x = num__0.2 total distance = num__3 * num__0.2 = num__0.6 km = num__600 meters correct option is c <eor> c <eos> |
c |
divide__6.0__2.0__ km_to_mile_conversion__ round__600.0__ |
divide__6.0__2.0__ multiply__3.0__0.2__ round__600.0__ |
| the length of a room is num__5.5 m and width is num__3.75 m . find the cost of paying the floor by slabs at the rate of rs . num__1400 per sq . metre . <o> a ) rs . num__15000 <o> b ) rs . num__15500 <o> c ) rs . num__15600 <o> d ) rs . num__28875 <o> e ) none |
solution area of the floor = ( num__5.5 x num__3.75 ) m ² = num__20.635 m ² cost of paying = rs . ( num__1400 x num__20.625 ) = rs . num__28875 . answer d <eor> d <eos> |
d |
multiply__5.5__3.75__ multiply__1400.0__20.625__ round__28875.0__ |
multiply__5.5__3.75__ multiply__1400.0__20.625__ round__28875.0__ |
| the difference between a number and its two - fifth is num__510 . what is num__10.0 of that number ? <o> a ) num__19 <o> b ) num__85 <o> c ) num__77 <o> d ) num__76 <o> e ) num__78 |
let the number be x . then x - num__0.4 x = num__510 x = ( num__510 * num__5 ) / num__3 = num__850 num__10.0 of num__850 = num__85 . answer : b <eor> b <eos> |
b |
divide__850.0__10.0__ divide__850.0__10.0__ |
divide__850.0__10.0__ divide__850.0__10.0__ |
| in a garment industry num__12 men working num__8 hours per day complete a piece of work in num__10 days . to complete the same work in num__8 days working num__20 hours a day the number of men required is : <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
explanation : let the required number of men be x . less days more men ( indirect proportion ) more working hrs per day less men ( indirect proportion ) days num__8 : num__10 working hrs num__20 : num__8 : : num__12 : x = > num__8 x num__20 x x = num__10 x num__8 x num__12 = > x = num__10 x num__8 x num__12 / ( num__8 x num__20 ) = > x = num__6 answer : c <eor> c <eos> |
c |
round__6.0__ |
round__6.0__ |
| ( num__786 × num__74 ) ÷ ? = num__2423.5 <o> a ) a ) num__24 <o> b ) b ) num__48 <o> c ) c ) num__58 <o> d ) d ) num__68 <o> e ) e ) num__48 |
explanation : num__58164 / x = num__2423.5 = > x = num__58164 / num__2423.5 = num__24 answer : option a <eor> a <eos> |
a |
multiply__786.0__74.0__ divide__58164.0__2423.5__ divide__58164.0__2423.5__ |
multiply__786.0__74.0__ divide__58164.0__2423.5__ divide__58164.0__2423.5__ |
| what will come in place of the x in the following number series ? num__65536 num__32768 num__16384 num__8192 num__4096 x <o> a ) num__2048 <o> b ) num__2046 <o> c ) num__2045 <o> d ) num__2044 <o> e ) num__2043 |
go on dividing by num__2 to the next number answer : a <eor> a <eos> |
a |
divide__65536.0__32768.0__ divide__4096.0__2.0__ |
divide__65536.0__32768.0__ divide__4096.0__2.0__ |
| a ball dropped from h height and moves num__80.0 of height each time . total distance covered is <o> a ) num__4 h <o> b ) num__5 h <o> c ) num__7 h <o> d ) num__9 h <o> e ) num__8 h |
first time distance is h second time num__80 h / num__100 = num__4 h / num__5 similarly third time num__80.0 of num__4 h / num__5 = h ( num__4 ^ num__2 ) / ( num__5 ^ num__2 ) and so on . . . this will lead to infinite terms of geometric progression i . e h + num__2 * num__4 h / num__5 + num__2 * num__16 h / num__25 . . . . . . . . . . . . sum = h + num__2 * num__4 h / ( num__5 ( num__1 - num__0.8 ) ) = num__9 h answer : d <eor> d <eos> |
d |
percent__4.0__25.0__ percent__80.0__1.0__ percent__9.0__100.0__ |
percent__4.0__25.0__ percent__80.0__1.0__ percent__9.0__100.0__ |
| a boat having a length num__3 m and breadth num__2 m is floating on a lake . the boat sinks by num__1 cm when a man gets on it . the mass of the man is : <o> a ) num__12 kg <o> b ) num__60 kg <o> c ) num__72 kg <o> d ) num__85 kg <o> e ) num__96 kg |
volume of water displaced = ( num__3 x num__2 x num__0.01 ) m num__3 = num__0.06 m num__3 . mass of man = volume of water displaced x density of water = ( num__0.06 x num__1000 ) kg = num__60 kg . answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ hour_to_min_conversion__ |
hour_to_min_conversion__ hour_to_min_conversion__ |
| chris mixed num__4 pounds of raisins with num__4 pounds of nuts . if a pound of nuts costs num__3 times as much as a pound of raisins then the total cost of the raisins was what fraction of the total cost of the mixture ? <o> a ) num__0.142857142857 <o> b ) num__0.2 <o> c ) num__0.25 <o> d ) num__0.333333333333 <o> e ) num__0.428571428571 |
num__1 lbs of raisin = $ num__1 num__4 lbs of raisin = $ num__4 num__1 lbs of nuts = $ num__3 num__4 lbs of nuts = $ num__12 total value of mixture = num__12 + num__4 = num__16 fraction of the value of raisin = num__0.25 = num__0.25 ans : c <eor> c <eos> |
c |
subtract__4.0__3.0__ multiply__4.0__3.0__ add__4.0__12.0__ reverse__4.0__ reverse__4.0__ |
subtract__4.0__3.0__ multiply__4.0__3.0__ add__4.0__12.0__ reverse__4.0__ reverse__4.0__ |
| steve traveled the first num__2 hours of his journey at num__40 mph and the last num__3 hours of his journey at num__80 mph . what is his average speed of travel for the entire journey ? <o> a ) num__60 mph <o> b ) num__56.67 mph <o> c ) num__53.33 mph <o> d ) num__64 mph <o> e ) num__66.67 mph |
answer average speed of travel = total distance travelled / total time taken total distance traveled by steve = distance covered in the first num__2 hours + distance covered in the next num__3 hours . distance covered in the first num__2 hours = speed * time = num__40 * num__2 = num__80 miles . distance covered in the next num__3 hours = speed * time = num__80 * num__3 = num__240 miles . therefore total distance covered = num__80 + num__240 = num__320 miles . total time taken = num__2 + num__3 = num__5 hours . hence average speed = total distance travelled / total time taken = num__64.0 = num__64 miles per hour . choice d <eor> d <eos> |
d |
multiply__3.0__80.0__ add__80.0__240.0__ add__2.0__3.0__ divide__320.0__5.0__ round__64.0__ |
multiply__3.0__80.0__ add__80.0__240.0__ add__2.0__3.0__ divide__320.0__5.0__ divide__320.0__5.0__ |
| what is the average of odd numbers from num__1 to num__79 ? <o> a ) num__30 <o> b ) num__25 <o> c ) num__35 <o> d ) num__40 <o> e ) num__45 |
sum of n odd natural number is given by sum of n odd natural numbers = n * n average of n odd natural numbers is given by = ( n * n ) / n = n . here n = number of terms = ( num__79 + num__1 ) / num__2 = num__40 . so required average = num__40 . answer : d <eor> d <eos> |
d |
multiply__1.0__40.0__ |
multiply__1.0__40.0__ |
| . two pipes a and b can fill a cistern in num__20 and num__30 minutes respectively and a third pipe c can empty it in num__40 minutes . how long will it take to fill the cistern if all the three are opened at the same time ? <o> a ) num__17 num__0.125 <o> b ) num__17 num__1.0 <o> c ) num__17 num__0.142857142857 <o> d ) num__17 num__0.111111111111 <o> e ) num__17 num__0.5 |
explanation : num__0.05 + num__0.0333333333333 - num__0.025 = num__0.0583333333333 num__17.1428571429 = num__17 num__0.142857142857 answer : c <eor> c <eos> |
c |
add__0.025__0.0333__ subtract__17.1429__17.0__ round__17.0__ |
add__0.025__0.0333__ subtract__17.1429__17.0__ subtract__17.1429__0.1429__ |
| an electric pump can fill a tank in num__3 hours . because of a leak in the tank it took num__3 hours num__30 min to fill the tank . in what time the leak can drain out all the water of the tank and will make tank empty ? <o> a ) num__10 hours <o> b ) num__13 hours <o> c ) num__17 hours <o> d ) num__21 hours <o> e ) num__23 hours |
explanation : we can get the answer by subtrating work done by leak in one hour by subtraction of filling for num__1 hour without leak and with leak as work done for num__1 hour without leak = num__0.333333333333 work done with leak = num__3 num__0.5 = num__3.5 work done with leak in num__1 hr = num__0.285714285714 work done by leak in num__1 hr = num__0.333333333333 − num__0.285714285714 = num__0.047619047619 so tank will be empty by the leak in num__21 hours . option d <eor> d <eos> |
d |
divide__1.0__3.0__ add__3.0__0.5__ divide__1.0__3.5__ subtract__0.3333__0.2857__ round__21.0__ |
divide__1.0__3.0__ add__3.0__0.5__ divide__1.0__3.5__ subtract__0.3333__0.2857__ round__21.0__ |
| two numbers n and num__16 have lcm = num__52 and gcf = num__8 . find n . <o> a ) num__35 <o> b ) num__26 <o> c ) num__76 <o> d ) num__87 <o> e ) num__24 |
the product of two integers is equal to the product of their lcm and gcf . hence . num__16 * n = num__52 * num__8 n = num__52 * num__0.5 = num__26 correct answer b <eor> b <eos> |
b |
divide__8.0__16.0__ multiply__52.0__0.5__ round__26.0__ |
divide__8.0__16.0__ multiply__52.0__0.5__ multiply__52.0__0.5__ |
| a man took a loan at rate of num__12.0 per annum simple interest . after num__3 years he had to pay num__5400 interest . the principal amount borrowed by him was . <o> a ) rs num__14000 <o> b ) rs num__15000 <o> c ) rs num__16000 <o> d ) rs num__17000 <o> e ) none of these |
explanation : s . i . = p ∗ r ∗ t / num__100 = > p = s . i . ∗ num__100 / r ∗ t = > p = num__5400 ∗ num__8.33333333333 ∗ num__3 = rs num__15000 option b <eor> b <eos> |
b |
percent__100.0__15000.0__ |
percent__100.0__15000.0__ |
| if the area of a circle is num__100 pi square feet find its circumference . <o> a ) num__65 pi feet <o> b ) num__20 pi feet <o> c ) num__42 pi feet <o> d ) num__18 pi feet <o> e ) num__64 pi feet |
the area is given by pi * r * r . hence pi * r * r = num__100 pi r * r = num__100 ; hence r = num__10 feet the circumference is given by num__2 * pi * r = num__2 * pi * num__10 = num__20 pi feet correct answer b <eor> b <eos> |
b |
multiply__10.0__2.0__ triangle_area__20.0__2.0__ |
multiply__10.0__2.0__ multiply__10.0__2.0__ |
| zachary is helping his younger brother sterling learn his multiplication tables . for every question that sterling answers correctly zachary gives him num__3 pieces of candy . for every question that sterling answers incorrectly zachary takes away two pieces of candy . after num__9 questions if sterling had answered num__2 more questions correctly he would have earned num__31 pieces of candy . how many of the num__9 questions did zachary answer correctly ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
i got two equations : num__3 x - num__2 y = num__25 x + y = num__9 num__3 x - num__2 ( num__9 - x ) = num__25 num__3 x - num__18 + num__2 x = num__25 num__5 x = num__43 x = num__8.6 or between num__8 and num__9 . ( ans c ) <eor> c <eos> |
c |
multiply__9.0__2.0__ add__3.0__2.0__ add__18.0__25.0__ divide__43.0__5.0__ round_down__8.6__ round_down__8.6__ |
multiply__9.0__2.0__ add__3.0__2.0__ add__18.0__25.0__ divide__43.0__5.0__ add__3.0__5.0__ add__3.0__5.0__ |
| let q represent a set of three distinct prime numbers . if the sum of the numbers in q is even and x is a member of q then what is the least possible value that x can be ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__5 <o> e ) num__7 |
q = p num__1 + p num__2 + p num__3 = even ( and all primes are distinct ) if the least prime is num__2 then we have sum of q = even . ans . b . num__2 <eor> b <eos> |
b |
add__1.0__2.0__ multiply__1.0__2.0__ |
add__1.0__2.0__ multiply__1.0__2.0__ |
| alex and brenda both stand at point x . alex begins to walk away from brenda in a straight line at a rate of num__2 miles per hour . one hour later brenda begins to ride a bicycle in a straight line in the opposite direction at a rate of r miles per hour . if r > num__8 which of the following represents the amount of time in terms of r that alex will have been walking when brenda has covered twice as much distance as alex ? <o> a ) r - num__4 <o> b ) r / ( r + num__4 ) <o> c ) r / ( r - num__4 ) <o> d ) num__8 / ( r - num__8 ) <o> e ) num__2 r - num__4 |
let t be the time thatalexwill have been walking when brenda has covered twice as much distance as alex . in t hours alex will cover num__4 t miles ; since brenda begins her journey num__1 hour later than alex then total time for her will be t - num__1 hours and the distance covered in that time will be r ( t - num__1 ) ; we want the distance covered by brenda to be twice as much as that of alex : num__2 * num__2 t = r ( t - num__1 ) - - > num__4 t = rt - r - - > t = r / ( r - num__4 ) . answer : c . <eor> c <eos> |
c |
divide__8.0__2.0__ round__4.0__ |
divide__8.0__2.0__ divide__8.0__2.0__ |
| num__24 persons can manufacture num__1000 soaps in num__15 days working num__7 hours a day . in how many days will num__18 persons working num__5 hours a day manufacture the same num__1000 soaps ? <o> a ) num__30 <o> b ) num__28 <o> c ) num__35 <o> d ) num__14 <o> e ) num__21 |
explanation : solution : let the required number of days be x . less persons more days ( indirect proportion ) more working hrs per day less days ( indirect proportion ) persons num__18 : num__24 } : : num__15 : x . working hrs per day less days num__5 : num__7 . ' . num__18 * num__5 * x = num__24 * num__7 * num__15 < = > x = num__28 . answer : b <eor> b <eos> |
b |
round__28.0__ |
round__28.0__ |
| lucy deposited $ num__62500 in an investment fund that provided num__16 percent annual return compounded quarterly . if she made no other transactions with the fund in how much time in years did her investment earn a total interest of $ num__5100 ? <o> a ) num__0.5 <o> b ) num__2 <o> c ) num__3 <o> d ) num__6 <o> e ) num__6.1 |
a = p + i = num__62500 + num__5100 = num__67600 num__67600 = num__62500 ( num__1 + num__4.0 * num__100 ) ^ ( num__4 t ) ( num__1.0816 ) = ( num__1.04 ) ^ ( num__4 t ) ( num__1.04 ) ^ num__2 = ( num__1.04 ) ^ num__4 t t = num__0.5 yrs answer : a <eor> a <eos> |
a |
percent__0.5__100.0__ |
percent__0.5__100.0__ |
| complete the numerical series with the correct number num__5 + num__3 + num__1 = num__15517 num__9 + num__2 + num__2 = num__181834 num__5 + num__6 + num__4 = num__302044 num__5 + num__4 + num__2 = num__201026 num__9 + num__3 + num__5 = ? <o> a ) num__468645 <o> b ) num__454586 <o> c ) num__274569 <o> d ) num__454386 <o> e ) num__444586 |
num__9 + num__3 + num__5 = num__274569 answer : c <eor> c <eos> |
c |
multiply__1.0__274569.0__ |
multiply__1.0__274569.0__ |
| if num__4 ^ k = num__5 then num__4 ^ ( num__2 k + num__2 ) = <o> a ) num__400 <o> b ) num__540 <o> c ) num__100 <o> d ) num__830 <o> e ) num__420 |
num__4 ^ k = num__5 num__4 ^ num__2 k = num__5 ^ num__2 num__4 ^ num__2 k = num__25 num__4 ^ ( num__2 k + num__2 ) = num__4 ^ num__2 k * num__4 ^ num__2 = num__25 * num__16 = num__400 answer : a <eor> a <eos> |
a |
multiply__16.0__25.0__ multiply__16.0__25.0__ |
multiply__16.0__25.0__ multiply__16.0__25.0__ |
| a certain elevator has a safe weight limit of num__2500 pounds . what is the greatest possible number of people who can safely ride on the elevator at one time with the average ( arithmetic mean ) weight of half the riders being num__180 pounds and the average weight of the others being num__230 pounds ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__12 |
lets assume there are num__2 x people . half of them have average weight of num__180 and other half has num__230 . maximum weight is = num__2500 so num__180 * x + num__230 * x = num__2500 = > num__410 x = num__2500 = > x is approximately equal to num__6 . so total people is num__2 * num__6 = num__12 answer e . <eor> e <eos> |
e |
add__180.0__230.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
add__180.0__230.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
| sum of two numbers is num__15 . two times of the first exceeds by num__5 from the three times of the other . then the numbers will be ? <o> a ) num__6 num__9 <o> b ) num__10 num__5 <o> c ) num__7 num__8 <o> d ) num__9 num__6 <o> e ) num__8 num__5 |
explanation : x + y = num__15 num__2 x – num__3 y = num__5 x = num__10 y = num__5 answer is b <eor> b <eos> |
b |
divide__15.0__5.0__ subtract__15.0__5.0__ subtract__15.0__5.0__ |
divide__15.0__5.0__ subtract__15.0__5.0__ subtract__15.0__5.0__ |
| two trains travel in opposite directions at num__36 kmph and num__45 kmph and a man sitting in slower train passes the faster train in num__6 seconds . the length of the faster train is <o> a ) num__80 m <o> b ) num__100 m <o> c ) num__120 m <o> d ) num__135 m <o> e ) none |
solution relative speed = ( num__36 + num__45 ) km / hr = ( num__81 x num__0.277777777778 ) m / sec = ( num__22.5 ) m / sec length of the train = ( num__22.5 x num__6 ) m = num__135 m . answer d <eor> d <eos> |
d |
add__36.0__45.0__ multiply__6.0__22.5__ round__135.0__ |
add__36.0__45.0__ multiply__6.0__22.5__ round__135.0__ |
| a number whose fifth part increased by num__6 is equal to its fourth part diminished by num__6 is ? <o> a ) num__160 <o> b ) num__180 <o> c ) num__200 <o> d ) num__240 <o> e ) none |
answer let the number be n . then ( n / num__5 ) + num__6 = ( n / num__4 ) - num__6 â ‡ ’ ( n / num__4 ) - ( n / num__5 ) = num__12 â ‡ ’ ( num__5 n - num__4 n ) / num__20 = num__12 â ˆ ´ n = num__240 option : d <eor> d <eos> |
d |
multiply__4.0__5.0__ multiply__12.0__20.0__ multiply__12.0__20.0__ |
multiply__4.0__5.0__ multiply__12.0__20.0__ multiply__12.0__20.0__ |
| the speed of a boat in upstream is num__35 kmph and the speed of the boat downstream is num__50 kmph . find the speed of the boat in still water and the speed of the stream ? <o> a ) num__5.5 kmph <o> b ) num__8.5 kmph <o> c ) num__7.15 kmph <o> d ) num__7.5 kmph <o> e ) num__17.5 kmph |
speed of the boat in still water = ( num__35 + num__50 ) / num__2 = num__42.5 kmph . speed of the stream = ( num__50 - num__35 ) / num__2 = num__7.5 kmph . answer : d <eor> d <eos> |
d |
subtract__50.0__42.5__ round__7.5__ |
subtract__50.0__42.5__ subtract__50.0__42.5__ |
| a city with a population of num__160080 is to be divided into num__8 voting districts and no district is to have a population that is more than num__10 percent greater than the population of any other district . what is the minimum possible population that the least populated district could have ? <o> a ) num__18200 <o> b ) num__18300 <o> c ) num__18400 <o> d ) num__18500 <o> e ) num__18 |
600 |
the minimum possible population occurs when all the other districts have a population that is num__10.0 greater than the least populated district . let p be the population of the least populated district . then num__160080 = p + num__7 ( num__1.1 ) p num__8.7 p = num__160080 p = num__18400 the answer is c . <eor> c <eos> |
c |
c |
| in a bucket there are num__3 blue balls and num__2 red balls . what is the probability of drawing at least one blue ball when drawing two consecutive balls randomly ? <o> a ) num__0.9 <o> b ) num__0.8 <o> c ) num__0.4 <o> d ) num__0.6 <o> e ) ½ |
p ( at least one blue ) = num__1 - p ( no blue so num__2 red ) = num__1 - num__0.4 * num__0.25 = num__0.9 . answer : a . <eor> a <eos> |
a |
subtract__3.0__2.0__ round__0.9__ |
subtract__3.0__2.0__ round__0.9__ |
| what will be the cost of house to paint which area equal to num__484 sq ft if the price per foot of building is rs . num__20 <o> a ) num__1800 <o> b ) num__1760 <o> c ) num__1400 <o> d ) num__2600 <o> e ) num__3600 |
let the side of the square plot be a ft . a num__2 = num__484 = > a = num__22 length of the fence = perimeter of the plot = num__4 a = num__88 ft . cost of building the fence = num__88 * num__20 = rs . num__1760 . answer : b <eor> b <eos> |
b |
square_perimeter__22.0__ multiply__20.0__88.0__ multiply__20.0__88.0__ |
multiply__4.0__22.0__ multiply__20.0__88.0__ multiply__20.0__88.0__ |
| walking at the rate of num__5 kmph a man cover certain distance in num__5 hr . running at a speed of num__15 kmph the man will cover the same distance in . <o> a ) num__12 min <o> b ) num__36 min <o> c ) num__40 min <o> d ) num__48 min <o> e ) num__60 min |
distance = speed * time num__5 * num__5 = num__25 km new speed = num__15 kmph therefore time = num__1.66666666667 = num__1.66666666667 = num__36 min answer : b <eor> b <eos> |
b |
divide__25.0__15.0__ round__36.0__ |
divide__25.0__15.0__ round__36.0__ |
| there are num__4 people of different heights standing in order of increasing height . the difference is num__2 inches between the first person and the second person and also between the second person and the third person . the difference between the third person and the fourth person is num__6 inches and the average height is num__78 . how tall is the fourth person ? <o> a ) num__82 <o> b ) num__84 <o> c ) num__86 <o> d ) num__88 <o> e ) num__90 |
let x be the height of the first person . then the heights are x x + num__2 x + num__4 and x + num__10 . num__4 x + num__16 = num__4 ( num__78 ) = num__312 x = num__74 and the fourth person has a height of num__74 + num__10 = num__84 inches the answer is b . <eor> b <eos> |
b |
add__4.0__6.0__ add__6.0__10.0__ multiply__4.0__78.0__ subtract__78.0__4.0__ add__6.0__78.0__ add__6.0__78.0__ |
add__4.0__6.0__ add__6.0__10.0__ multiply__4.0__78.0__ subtract__78.0__4.0__ add__6.0__78.0__ add__6.0__78.0__ |
| how long does a train num__60 m long travelling at num__60 kmph takes to cross a bridge of num__80 m in length ? <o> a ) num__5.8 sec <o> b ) num__8.4 sec <o> c ) num__12.4 sec <o> d ) num__6.8 sec <o> e ) num__1.8 sec |
b num__16.8 sec d = num__60 + num__80 = num__140 m s = num__60 * num__0.277777777778 = num__16.6666666667 t = num__140 * num__0.06 = num__8.4 sec answer is b <eor> b <eos> |
b |
add__60.0__80.0__ divide__140.0__16.6667__ round__8.4__ |
add__60.0__80.0__ multiply__140.0__0.06__ multiply__140.0__0.06__ |
| five pieces of wood have an average length of num__132 cm and a median length of num__140 cm . what is the maximum possible length in cm of the shortest piece of wood ? <o> a ) a ) num__90 <o> b ) b ) num__100 <o> c ) c ) num__110 <o> d ) d ) num__120 <o> e ) e ) num__140 |
d . num__120 sum of all lengths of all num__5 pieces of wood = num__132 * num__5 = num__660 num__3 rd piece ( sorted in increasing length ) length = num__140 ( median ) for sum of first num__2 wood length to become max last two should be least . let num__4 th num__5 th wood also have length num__140 each . total of last num__3 = num__140 * num__3 = num__420 sum of first num__2 = num__660 - num__420 = num__240 each of these num__2 will have length num__120.0 = num__120 answer d <eor> d <eos> |
d |
multiply__132.0__5.0__ subtract__5.0__3.0__ multiply__140.0__3.0__ multiply__2.0__120.0__ divide__240.0__2.0__ |
multiply__132.0__5.0__ subtract__5.0__3.0__ multiply__140.0__3.0__ multiply__2.0__120.0__ subtract__240.0__120.0__ |
| the measurement of a rectangular box with lid is num__25 cmx num__12 cmx num__18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of π cm num__3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere ) <o> a ) num__288 <o> b ) num__48 <o> c ) num__72 <o> d ) num__864 <o> e ) num__964 |
d = num__12 r = num__6 ; volume of the largest sphere = num__1.33333333333 π r num__3 = num__1.33333333333 * π * num__6 * num__6 * num__6 = num__288 π cm num__3 answer : a <eor> a <eos> |
a |
subtract__18.0__12.0__ round__288.0__ |
subtract__18.0__12.0__ round__288.0__ |
| two trains one from howrah to patna and the other from patna to howrah start simultaneously . after they meet the trains reach their destinations after num__4 hours and num__49 hours respectively . the ratio of their speeds is : <o> a ) num__2 : num__3 <o> b ) num__4 : num__3 <o> c ) num__6 : num__7 <o> d ) num__7 : num__2 <o> e ) none of these |
let us name the trains as a and b . then ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š num__49 : â ˆ š num__4 = num__7 : num__2 . answer d <eor> d <eos> |
d |
round__7.0__ |
round__7.0__ |
| what is the remainder when num__9 ^ num__19 is divided by num__5 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
num__9 ^ num__0 = num__1 num__9 ^ num__1 = num__9 num__9 ^ num__2 = num__81 num__9 ^ num__3 = num__729 num__9 ^ num__4 = . . . num__1 num__9 ^ num__5 = . . . . . num__9 and so on in a distinct last - digit pattern of num__91 num__91 . . . after ignoring the first last - digit of num__1 num__9.5 = num__9 ( with a remainder of num__1 ) so the last number is a num__9 . alternatively num__19 is an odd number and hence the last digit is a num__9 when num__9 ^ num__19 is divided by num__5 the remainder is num__4 ans = e <eor> e <eos> |
e |
subtract__5.0__2.0__ multiply__9.0__81.0__ subtract__9.0__5.0__ divide__19.0__2.0__ subtract__9.0__5.0__ |
subtract__5.0__2.0__ multiply__9.0__81.0__ subtract__9.0__5.0__ divide__19.0__2.0__ subtract__9.0__5.0__ |
| the length of the bridge which a train num__140 meters long and travelling at num__45 km / hr can cross in num__30 seconds is : <o> a ) num__239 <o> b ) num__277 <o> c ) num__235 <o> d ) num__88 <o> e ) num__232 |
speed = ( num__45 * num__0.277777777778 ) m / sec = ( num__12.5 ) m / sec . time = num__30 sec . let the length of bridge be x meters . then ( num__140 + x ) / num__30 = num__12.5 = = > num__2 ( num__140 + x ) = num__750 = = > x = num__235 m . answer : c <eor> c <eos> |
c |
round__235.0__ |
round__235.0__ |
| the h . c . f . of two numbers is num__23 and the other two factors of their l . c . m . are num__13 and num__16 . the larger of the two numbers is : <o> a ) num__276 <o> b ) num__299 <o> c ) num__322 <o> d ) num__345 <o> e ) num__368 |
clearly the numbers are ( num__23 x num__13 ) and ( num__23 x num__16 ) . larger number = ( num__23 x num__16 ) = num__368 . answer : option e <eor> e <eos> |
e |
multiply__23.0__16.0__ multiply__23.0__16.0__ |
multiply__23.0__16.0__ multiply__23.0__16.0__ |
| num__7 + num__3 x num__14 + num__6 x num__7 ^ num__2 + num__6 x num__7 ^ num__3 + num__6 x num__7 ^ num__4 + num__6 x num__7 ^ num__5 + num__6 x num__7 ^ num__6 + num__6 x num__7 ^ num__7 = <o> a ) num__7 ^ num__21 <o> b ) num__7 ^ num__14 <o> c ) num__7 ^ num__10 <o> d ) num__7 ^ num__8 <o> e ) none of the above |
let ' s see that num__7 + num__3 x num__14 is num__49 or num__7 ^ num__2 then : num__7 ^ num__2 + num__6 x num__7 ^ num__2 becomes num__7 x num__7 ^ num__2 that is num__7 ^ num__3 . so we follow with the next element : num__7 ^ num__3 + num__6 x num__7 ^ num__3 is equal to num__7 x num__7 ^ num__3 that is num__7 ^ num__4 . then is assumed right before num__6 x num__7 ^ num__7 : num__7 ^ num__7 + num__6 x num__7 ^ num__7 becomes num__7 x num__7 ^ num__7 that is num__7 ^ num__8 . thus the correct answer is the option d num__7 ^ num__8 . <eor> d <eos> |
d |
add__3.0__5.0__ add__3.0__4.0__ |
add__3.0__5.0__ add__3.0__4.0__ |
| find the area between two concentric circles defined by x num__2 + y num__2 - num__2 x + num__4 y + num__1 = num__0 x num__2 + y num__2 - num__2 x + num__4 y - num__11 = num__0 <o> a ) num__40 pi <o> b ) num__19 pi <o> c ) num__12 pi <o> d ) num__46 pi <o> e ) num__22 pi |
rewrite equations of circles in standard form . hence equation x num__2 + y num__2 - num__2 x + num__4 y + num__1 = num__0 may be written as ( x - num__1 ) num__2 + ( y + num__2 ) num__2 = num__4 = num__22 and equation x num__2 + y num__2 - num__2 x + num__4 y - num__11 = num__0 as ( x - num__1 ) num__2 + ( y + num__2 ) num__2 = num__16 = num__42 knowing the radii the area of the ring is pi ( num__4 ) num__2 - pi ( num__2 ) num__2 = num__12 pi correct answer c <eor> c <eos> |
c |
multiply__2.0__11.0__ add__1.0__11.0__ add__1.0__11.0__ |
multiply__2.0__11.0__ add__1.0__11.0__ add__1.0__11.0__ |
| if num__5 men and num__2 boys working together can do three times as much work per hour as a man and a boy together . find the ratio of the work done by a man and that of a boy for a given time ? <o> a ) num__3 : num__2 <o> b ) num__2 : num__1 <o> c ) num__3 : num__3 <o> d ) num__1 : num__2 <o> e ) num__6 : num__1 |
num__5 m + num__2 b = num__3 ( num__1 m + num__1 b ) num__5 m + num__2 b = num__3 m + num__3 b num__2 m = num__1 b the required ratio of work done by a man and a boy = num__1 : num__2 d <eor> d <eos> |
d |
subtract__5.0__2.0__ subtract__3.0__2.0__ round__1.0__ |
subtract__5.0__2.0__ subtract__3.0__2.0__ round__1.0__ |
| a shopkeeper buys mangoes at the rate of num__4 a rupee and sells them at num__3 a rupee . find his net profit or loss percent ? <o> a ) num__33 num__0.125 % <o> b ) num__33 num__2.33333333333 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__33 num__0.625 % <o> e ) num__34 num__0.333333333333 % |
the total number of mangoes bought by the shopkeeper be num__12 . if he buys num__4 a rupee his cp = num__3 he selling at num__3 a rupee his sp = num__4 profit = sp - cp = num__4 - num__3 = num__1 profit percent = num__0.333333333333 * num__100 = num__33 num__0.333333333333 % answer : c <eor> c <eos> |
c |
percent__33.0__100.0__ |
percent__33.0__100.0__ |
| num__3 * num__13 + num__3 * num__14 + num__3 * num__17 + num__11 = ? <o> a ) num__125 <o> b ) num__126 <o> c ) num__130 <o> d ) num__143 <o> e ) num__151 |
num__3 * num__13 + num__3 * num__14 + num__3 * num__17 + num__11 = num__39 + num__42 + num__51 + num__11 = num__143 the answer is d . <eor> d <eos> |
d |
multiply__3.0__13.0__ multiply__3.0__14.0__ multiply__3.0__17.0__ multiply__13.0__11.0__ multiply__13.0__11.0__ |
multiply__3.0__13.0__ multiply__3.0__14.0__ multiply__3.0__17.0__ multiply__13.0__11.0__ multiply__13.0__11.0__ |
| pipe a that can fill a tank in two hour and pipe b that can fill the tank in an hour are opened simultaneously when the tank is empty . pipe b is shut num__30 minutes before the tank overflows . when will the tank overflow ? <o> a ) num__36 mins <o> b ) num__35 mins <o> c ) num__40 mins <o> d ) num__32 mins <o> e ) num__30 mins |
the last num__30 minutes only pipe a was open . since it needs num__2 hour to fill the tank then in num__30 minutes it fills num__0.25 th of the tank thus num__0.75 of the tank is filled with both pipes open . the combined rate of two pipes is num__1 + num__2 = num__3 tanks / hour therefore to fill num__0.75 th of the tank they need ( time ) = ( work ) / ( rate ) = ( num__0.75 ) / num__3 = num__0.25 hours = num__15 minutes . total time = num__15 + num__15 = num__30 minutes . answer : e <eor> e <eos> |
e |
add__0.25__0.75__ divide__0.75__0.25__ divide__30.0__2.0__ round__30.0__ |
add__0.25__0.75__ divide__0.75__0.25__ divide__30.0__2.0__ divide__30.0__1.0__ |
| if x is num__12 percent greater than num__80 then x = <o> a ) num__89.6 <o> b ) num__91.0 <o> c ) num__88.0 <o> d ) num__70.9 <o> e ) num__71.2 |
num__12.0 of num__80 = ( num__80 * num__0.12 ) = num__9.6 num__12.0 greater than num__80 = num__80 + num__9.6 = num__89.6 answer is clearly a . <eor> a <eos> |
a |
multiply__80.0__0.12__ add__80.0__9.6__ add__80.0__9.6__ |
multiply__80.0__0.12__ add__80.0__9.6__ add__80.0__9.6__ |
| what is the measure of the radius of the circle that circumscribes a triangle whose sides measure num__3 num__4 and num__5 ? <o> a ) num__3.5 <o> b ) num__3 <o> c ) num__5 <o> d ) num__4 <o> e ) num__2.5 |
some of pyhtagron triplets we need to keep it in mind . like { ( num__2 num__35 ) ( num__5 num__1213 ) ( num__7 num__2425 ) ( num__11 num__6061 ) . so now we know the triangle is an right angle triangle . the circle circumscribes the triangle . the circum raduis of the circle that circumscribes the right angle triangle = hypotanse / num__2 = num__2.5 = num__2.5 ans . e <eor> e <eos> |
e |
triangle_perimeter__4.0__5.0__2.0__ triangle_area__2.5__2.0__ |
triangle_perimeter__4.0__5.0__2.0__ triangle_area__2.5__2.0__ |
| num__10 men do a work in num__10 days . how many men are needed to finish the work in num__4 days ? <o> a ) num__25 <o> b ) num__20 <o> c ) num__30 <o> d ) num__10 <o> e ) num__15 |
men required to finish the work in num__4 days = num__10 * num__2.5 = num__25 answer is a <eor> a <eos> |
a |
divide__10.0__4.0__ multiply__10.0__2.5__ round__25.0__ |
divide__10.0__4.0__ multiply__10.0__2.5__ multiply__10.0__2.5__ |
| a river num__2 m deep and num__45 m wide is flowing at the rate of num__3 kmph the amount of water that runs into the sea per minute is ? <o> a ) num__1390 <o> b ) num__2887 <o> c ) num__2778 <o> d ) num__2779 <o> e ) num__2121 |
num__2 ( num__25 * num__15 + num__15 * num__8 + num__25 * num__8 ) num__2 ( num__375 + num__120 + num__200 ) = > num__1390 answer : a <eor> a <eos> |
a |
divide__45.0__3.0__ multiply__15.0__25.0__ multiply__8.0__15.0__ multiply__8.0__25.0__ round__1390.0__ |
divide__45.0__3.0__ multiply__15.0__25.0__ multiply__8.0__15.0__ multiply__8.0__25.0__ round__1390.0__ |
| the basic one - way air fare for a child aged between num__3 and num__10 years costs half the regular fare for an adult plus a reservation charge that is the same on the child ' s ticket as on the adult ' s ticket . one reserved ticket for an adult costs $ num__216 and the cost of a reserved ticket for an adult and a child ( aged between num__3 and num__10 ) costs $ num__327 . what is the basic fare for the journey for an adult ? <o> a ) $ num__111 <o> b ) $ num__52.5 <o> c ) $ num__210 <o> d ) $ num__58.5 <o> e ) $ num__6 |
explanatory answer step num__1 : assign variables and frame equations let the basic fare for the child be $ x . basic one - way air fare of a child costs half the regular fare for an adult therefore the basic fare for an adult = num__2 ( basic one - way airfare for a child ) = $ num__2 x . reservation charge is the same on the child ' s ticket as on the adult ' s ticket . let the reservation charge per ticket be $ y hence an adult ticket will cost ( basic fare + reservation charges ) = num__2 x + y = $ num__216 . . . . . ( num__1 ) and ticket for an adult and a child will cost ( num__2 x + y ) + ( x + y ) = num__3 x + num__2 y = $ num__327 . . . . ( num__2 ) step num__2 : solve the equations and determine basic fare for an adult multiply equation ( num__1 ) by num__2 : num__4 x + num__2 y = num__432 . . . . ( num__3 ) subtract equation ( num__2 ) from equation ( num__3 ) : ( num__4 x + num__2 y = num__432 ) - ( num__3 x + num__2 y = num__327 ) ∴ x = $ num__105 . the basic fare of an adult ticket = num__2 x = num__2 * num__105 = $ num__210 answer c <eor> c <eos> |
c |
subtract__3.0__1.0__ add__3.0__1.0__ multiply__216.0__2.0__ subtract__432.0__327.0__ multiply__2.0__105.0__ multiply__1.0__210.0__ |
subtract__3.0__1.0__ add__3.0__1.0__ multiply__216.0__2.0__ subtract__432.0__327.0__ multiply__2.0__105.0__ multiply__1.0__210.0__ |
| three years ago the average age of a family of num__5 members was num__17 . a baby having been born the average age of the family is the same today . what is the age of the child ? <o> a ) num__3 <o> b ) num__2 <o> c ) num__4 <o> d ) num__1 <o> e ) num__5 |
present age of num__5 members = num__5 x num__17 + num__3 x num__5 = num__100 years also present ages of num__5 members + age of the baby = num__6 x num__17 = num__102 years age of the baby = num__102 – num__100 = num__2 years . answer : b <eor> b <eos> |
b |
multiply__17.0__6.0__ subtract__5.0__3.0__ subtract__5.0__3.0__ |
multiply__17.0__6.0__ subtract__5.0__3.0__ subtract__5.0__3.0__ |
| p alone can complete a job in num__6 days . the work done by q alone in one day is equal to one - fifth of the work done by p alone in one day . in how many days can the work be completed if p and q work together ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
p ' s rate is num__0.166666666667 q ' s rate is num__0.0333333333333 the combined rate is num__0.166666666667 + num__0.0333333333333 = num__0.2 if they work together the job will take num__5 days . the answer is e . <eor> e <eos> |
e |
add__0.1667__0.0333__ round__5.0__ |
add__0.1667__0.0333__ round__5.0__ |
| if a truck is traveling at a constant rate of num__72 kilometers per hour how many seconds will it take the truck to travel a distance of num__600 meters ? ( num__1 kilometer = num__1000 meters ) <o> a ) num__18 <o> b ) num__24 <o> c ) num__30 <o> d ) num__36 <o> e ) num__48 |
speed = num__72 km / hr = > num__72000 m / hr in one minute = > num__1200.0 = num__1200 meters in one sec = > num__20.0 = num__20 meters time = total distance need to be covered / avg . speed = > num__30.0 = num__30 and hence the answer : c <eor> c <eos> |
c |
multiply__72.0__1000.0__ divide__600.0__20.0__ round__30.0__ |
multiply__72.0__1000.0__ divide__600.0__20.0__ divide__600.0__20.0__ |
| the g . c . d . of num__1.08 num__0.35 and num__0.9 is : <o> a ) num__0.01 <o> b ) num__0.9 <o> c ) num__0.18 <o> d ) num__0.108 <o> e ) none |
explanation given numbers are num__1.08 num__0.35 and num__0.90 . h . c . f . of num__108 num__35 and num__90 is num__1 h . c . f . of given numbers = num__0.01 . answer a <eor> a <eos> |
a |
divide__1.08__108.0__ divide__1.08__108.0__ |
divide__1.08__108.0__ divide__1.08__108.0__ |
| a completes num__80.0 of a work in num__20 days . then b also joins and a and b together finish the remaining work in num__3 days . how long does it need for b if he alone completes the work ? <o> a ) num__35 days <o> b ) num__36.5 days <o> c ) num__37 days <o> d ) num__37.5 days <o> e ) num__38 days |
work done by a in num__20 days = num__0.8 = num__0.8 = num__0.8 work done by a in num__1 day = ( num__0.8 ) / num__20 = num__0.04 = num__0.04 - - - ( num__1 ) work done by a and b in num__3 days = num__0.2 = num__0.2 ( because remaining num__20.0 is done in num__3 days by a and b ) work done by a and b in num__1 day = num__0.0666666666667 - - - ( num__2 ) work done by b in num__1 day = num__0.0666666666667 – num__0.04 = num__0.0266666666667 = > b can complete the work in num__37.5 days = num__37 ½ days answer is d . <eor> d <eos> |
d |
divide__0.8__20.0__ subtract__1.0__0.8__ divide__0.2__3.0__ subtract__3.0__1.0__ subtract__0.0667__0.04__ round__37.5__ |
divide__0.8__20.0__ subtract__1.0__0.8__ divide__0.2__3.0__ subtract__3.0__1.0__ subtract__0.0667__0.04__ divide__37.5__1.0__ |
| eleven bags are bought for rs . num__1000 and sold at num__10 for rs . num__1600 . what is the gain or loss in percentage ? <o> a ) num__10.0 <o> b ) num__26.0 <o> c ) num__25.0 <o> d ) num__20.0 <o> e ) none of these |
explanation : as selling price is rs . num__1600 for num__10 bags each bag is sold at rs . num__160 . hence the profit is rs . num__100 after selling num__10 bags plus the selling price of num__1 bag = num__100 + num__160 = rs . num__260 . % profit = num__0.26 * num__100 = num__26.0 answer b <eor> b <eos> |
b |
percent__10.0__1600.0__ percent__10.0__1000.0__ percent__10.0__260.0__ percent__10.0__260.0__ |
percent__10.0__1600.0__ percent__10.0__1000.0__ percent__10.0__260.0__ percent__10.0__260.0__ |
| rs . num__2500 is divided into two parts such that if one part be put out at num__5.0 simple interest and the other at num__6.0 the yearly annual income may be rs . num__125 . how much was lent at num__5.0 ? <o> a ) num__2333 <o> b ) num__2777 <o> c ) num__2688 <o> d ) num__1000 <o> e ) num__2500 |
( x * num__5 * num__1 ) / num__100 + [ ( num__2500 - x ) * num__6 * num__1 ] / num__100 = num__125 x = num__2500 answer : e <eor> e <eos> |
e |
percent__100.0__2500.0__ |
percent__100.0__2500.0__ |
| a train num__110 m long is running with a speed of num__60 km / hr . in what time will it pass a man who is running at num__6 km / hr in the direction opposite to that in which the train is going ? <o> a ) num__4 sec <o> b ) num__6 sec <o> c ) num__7 sec <o> d ) num__3 sec <o> e ) num__2 sec |
speed of train relative to man = num__60 + num__6 = num__66 km / hr . = num__66 * num__0.277777777778 = num__18.3333333333 m / sec . time taken to pass the men = num__110 * num__0.0545454545455 = num__6 sec . answer : b <eor> b <eos> |
b |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ round__6.0__ |
add__60.0__6.0__ divide__110.0__6.0__ divide__6.0__110.0__ divide__110.0__18.3333__ |
| two cars car num__1 and car num__2 move towards each other from e and y respectively with respective speeds of num__20 m / s and num__15 m / s . after meeting each other car num__1 reaches y in num__10 seconds . in how many seconds does car num__2 reach e starting from y ? <o> a ) num__15.5 sec <o> b ) num__8.4 sec <o> c ) num__33.6 sec <o> d ) num__31.11 sec <o> e ) num__16.8 sec |
e - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - | - - - - - - - - - - - - - - - - - - - - - - - - - - - - y car a ( num__20 mps ) - - - - - - - - - - - - - - - - - - - - - - - - - > p < - - - - - - - - - - - - - - - car b ( num__15 mps ) let num__2 cars meet each other at point p in t seconds . car num__1 covers distance = num__20 t . car num__2 covers distance = num__15 t . so total distance ey = num__35 t . from p car num__1 reaches onto y in num__10 secs . so it covers num__15 t further . so num__15 t / num__20 = num__10 so t = num__13.3333333333 sec and total distance = ( num__35 * num__40 ) / num__3 hence car num__2 will cover total distance in ( num__35 * num__40 ) / ( num__3 * num__15 ) = num__31.11 sec approx . answer d <eor> d <eos> |
d |
add__20.0__15.0__ multiply__2.0__20.0__ add__1.0__2.0__ round__31.11__ |
add__20.0__15.0__ multiply__2.0__20.0__ divide__40.0__13.3333__ multiply__1.0__31.11__ |
| num__5 people can write num__25 book in num__20 days working num__8 hour a day . then in how many day num__125 can be written by num__75 people ? <o> a ) num__8.33333333333 <o> b ) num__6.33333333333 <o> c ) num__6.66666666667 <o> d ) num__8.33333333333 <o> e ) num__7.33333333333 |
work per day epr hour per person = num__25 / ( num__20 * num__8 * num__5 ) / / eq - num__1 people = num__75 ; let suppose day = p ; per day work for num__8 hours acc . to condition work per day epr hour per person = num__125 / ( p * num__8 * num__75 ) / / eq - num__2 eq - num__1 = = eq - num__2 ; p = num__6.66666666667 answer : c <eor> c <eos> |
c |
multiply__1.0__6.6667__ |
multiply__1.0__6.6667__ |
| john and roger can finish the work num__24 days if they work together . they worked together for num__16 days and then roger left . john finished the remaining work in another num__16 days . in how many days john alone can finish the work ? <o> a ) num__30 days <o> b ) num__48 days <o> c ) num__70 days <o> d ) num__80 days <o> e ) num__90 days |
amount of work done by john and roger in num__1 day = num__0.0416666666667 amount of work done by john and roger in num__16 days = num__16 Ã — ( num__0.0416666666667 ) = num__0.666666666667 remaining work â € “ num__1 â € “ num__0.666666666667 = num__0.333333333333 john completes num__0.333333333333 work in num__16 days amount of work john can do in num__1 day = ( num__0.333333333333 ) / num__16 = num__0.0208333333333 = > john can complete the work in num__48 days answer : b <eor> b <eos> |
b |
divide__1.0__24.0__ divide__16.0__24.0__ subtract__1.0__0.6667__ divide__0.3333__16.0__ round__48.0__ |
divide__1.0__24.0__ divide__16.0__24.0__ subtract__1.0__0.6667__ divide__0.3333__16.0__ divide__48.0__1.0__ |
| a group of num__3 investors and num__3 clients recently frequented the chinese luyang restaurant . the total bill for the meal including num__20.0 gratuity came to $ num__720 . on average how much did the meal of each individual cost before gratuity ? <o> a ) $ num__160 <o> b ) $ num__96 <o> c ) $ num__90 <o> d ) $ num__80 <o> e ) $ num__100 |
num__3 investors and num__3 clients - so total num__6 people the bill $ num__720 includes num__20.0 gratuity . . . so the actual cost of dinner was $ num__600 now the cost per person will be $ num__100.0 which is $ num__100 option e <eor> e <eos> |
e |
divide__600.0__6.0__ divide__600.0__6.0__ |
divide__600.0__6.0__ divide__600.0__6.0__ |
| working at a constant rate p can finish a job in num__3 hours . q also working at a constant rate can finish the same job in num__9 hours . if they work together for num__2 hours how many more minutes will it take p to finish the job working alone at his constant rate ? <o> a ) num__20 <o> b ) num__30 <o> c ) num__40 <o> d ) num__50 <o> e ) num__60 |
each hour they complete num__0.333333333333 + num__0.111111111111 = num__0.444444444444 of the job . in num__2 hours they complete num__2 ( num__0.444444444444 ) = num__0.888888888889 of the job . the time for p to finish is ( num__0.111111111111 ) / ( num__0.333333333333 ) = ( num__0.333333333333 ) hour = num__20 minutes the answer is a . <eor> a <eos> |
a |
divide__3.0__9.0__ divide__0.3333__3.0__ add__0.1111__0.3333__ round__20.0__ |
divide__3.0__9.0__ divide__0.3333__3.0__ add__0.1111__0.3333__ round__20.0__ |
| in measuring the sides of a rectangle one side is taken num__9.0 in excess and the other num__8.0 in deficit . find the error percent in the area calculated from these measurements . <o> a ) num__0.11 <o> b ) num__0.7 <o> c ) num__0.4 <o> d ) num__0.6 <o> e ) num__0.28 % |
let x and y be the sides of the rectangle . then correct area = xy . calculated area = ( num__1.09090909091 ) x ( num__0.92 ) y = ( num__1.00280112045 ) ( xy ) error in measurement = ( num__1.00280112045 ) xy - xy = ( num__0.00280112044818 ) xy error percentage = [ ( num__0.00280112044818 ) xy ( num__1 / xy ) num__100 ] % = ( num__0.28 ) % = num__0.28 . answer is e . <eor> e <eos> |
e |
percent__100.0__0.28__ |
percent__100.0__0.28__ |
| the speed of a boat in upstream is num__85 kmph and the speed of the boat downstream is num__155 kmph . find the speed of the boat in still water and the speed of the stream ? <o> a ) num__10 kmph <o> b ) num__11 kmph <o> c ) num__16 kmph <o> d ) num__18 kmph <o> e ) num__35 kmph |
speed of the boat in still water = ( num__85 + num__155 ) / num__2 = num__120 kmph . speed of the stream = ( num__155 - num__85 ) / num__2 = num__35 kmph . answer : e <eor> e <eos> |
e |
subtract__155.0__120.0__ round__35.0__ |
subtract__155.0__120.0__ subtract__155.0__120.0__ |
| from a group of boys and girls num__15 girls leave . there are then left num__2 boys for each girl . after this num__45 boys leave . there are then num__5 girls for each boy . find the number of girls in the beginning . <o> a ) num__36 <o> b ) num__40 <o> c ) num__56 <o> d ) num__65 <o> e ) num__28 |
let the present boys be x boys . then the number of girls at present = num__5 x . before the boys had left : number of boys = x + num__45 and number of girls = num__5 x . x + num__45 = num__2 * num__5 x = = = > x + num__45 = num__10 x = = > num__9 x = num__45 x = num__5 . hence the number of girls in the beginning = num__5 x + num__15 = num__25 + num__15 = num__40 so the correct answer is option b ) num__40 . <eor> b <eos> |
b |
subtract__15.0__5.0__ divide__45.0__5.0__ add__15.0__10.0__ add__15.0__25.0__ add__15.0__25.0__ |
multiply__2.0__5.0__ divide__45.0__5.0__ add__15.0__10.0__ add__15.0__25.0__ add__15.0__25.0__ |
| a completes a work in num__10 days and b complete the same work in num__20 days . if both of them work together then the number of days required to complete the work will be <o> a ) num__8 days <o> b ) num__6.67 days <o> c ) num__10 days <o> d ) num__12 days <o> e ) num__13 days |
if a can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days therefore here the required number of days = num__10 × num__0.666666666667 = num__6.67 days . answer : b <eor> b <eos> |
b |
round__6.67__ |
round__6.67__ |
| the ratio of the volumes of two cubes is num__125 : num__1000 . what is the ratio of their total surface areas ? <o> a ) num__81 : num__121 <o> b ) num__9 : num__11 <o> c ) num__729 : num__1331 <o> d ) num__1 : num__4 <o> e ) none of these |
ratio of the sides = Â ³ â ˆ š num__125 : Â ³ â ˆ š num__1000 = num__5 : num__10 ratio of surface areas = num__5 ^ num__2 : num__10 ^ num__2 = num__25 : num__100 = num__1 : num__4 answer : d <eor> d <eos> |
d |
power__5.0__2.0__ square_perimeter__25.0__ square_perimeter__1.0__ volume_cube__1.0__ |
power__5.0__2.0__ power__10.0__2.0__ square_perimeter__1.0__ power__1.0__2.0__ |
| if p and q are positive integers how many integers are larger than pq and smaller than p ( q + num__4 ) ? <o> a ) num__4 p - num__1 <o> b ) p + num__2 <o> c ) p – num__2 <o> d ) num__2 p – num__1 <o> e ) num__2 p + num__1 |
the number of integers between x and y where x > y is ( x - y ) - num__1 . for example the number of integers between num__1 and num__5 is ( num__5 - num__1 ) - num__1 = num__3 : num__2 num__3 and num__4 . thus the number of integers between pq and p ( q + num__4 ) = pq + num__4 p is ( pq + num__4 p - pq ) - num__1 = num__4 p - num__1 . answer : a . <eor> a <eos> |
a |
add__4.0__1.0__ subtract__4.0__1.0__ subtract__3.0__1.0__ multiply__4.0__1.0__ |
add__4.0__1.0__ subtract__4.0__1.0__ subtract__3.0__1.0__ add__1.0__3.0__ |
| if the compound interest on a certain sum of money for num__2 years at num__10.0 per annum be rs . num__993 what would be the simple interest ? <o> a ) rs . num__600 <o> b ) rs . num__890 <o> c ) rs . num__895 <o> d ) rs . num__900 <o> e ) none |
let p = principal a - amount we have a = p ( num__1 + r / num__100 ) num__3 and ci = a - p atq num__993 = p ( num__1 + r / num__100 ) num__3 - p ? p = num__3000 / - now si @ num__10.0 on num__3000 / - for num__2 yrs = ( num__3000 x num__10 x num__2 ) / num__100 = num__600 / - answer : a . <eor> a <eos> |
a |
percent__100.0__600.0__ |
percent__100.0__600.0__ |
| a can do a piece of work in num__20 days . b in num__15 days a and c in num__12 days . in how many days can a finish the work if he is assisted by b on one day and c on the next alternately ? <o> a ) num__9 days <o> b ) num__8 days <o> c ) num__5 days <o> d ) num__54 days <o> e ) num__2 days |
a + b = num__0.05 + num__0.0666666666667 = num__0.116666666667 a + c = num__0.05 + num__0.0833333333333 = num__0.133333333333 num__0.116666666667 + num__0.133333333333 = num__0.25 = num__0.25 num__4 days * num__2 = num__8 days answer : b <eor> b <eos> |
b |
add__0.05__0.0667__ add__0.05__0.0833__ add__0.1333__0.1167__ subtract__20.0__12.0__ round__8.0__ |
add__0.05__0.0667__ add__0.05__0.0833__ add__0.1333__0.1167__ multiply__2.0__4.0__ multiply__2.0__4.0__ |
| the ratio of spinsters to cats is num__2 to num__9 . if there are num__42 more cats than spinsters how many spinsters are there ? <o> a ) num__9 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__13 |
let num__2 x be the number of spinsters . then num__9 x is the number of cats . num__9 x - num__2 x = num__42 x = num__6 and the number of spinsters is num__2 ( num__6 ) = num__12 . the answer is d . <eor> d <eos> |
d |
multiply__2.0__6.0__ multiply__2.0__6.0__ |
multiply__2.0__6.0__ multiply__2.0__6.0__ |
| if x and y are positive integers and x ^ num__3 * y ^ num__4 = num__27783 which of the following is the value of xy ? <o> a ) num__6 <o> b ) num__10 <o> c ) num__14 <o> d ) num__15 <o> e ) num__21 |
num__27783 = num__3 * num__9261 = num__3 ^ num__2 * num__3087 = num__3 ^ num__3 * num__1029 = num__3 ^ num__4 * num__343 = num__3 ^ num__4 * num__7 ^ num__3 xy = num__3 * num__7 = num__21 the answer is e . <eor> e <eos> |
e |
divide__27783.0__3.0__ divide__9261.0__3.0__ divide__3087.0__3.0__ divide__1029.0__3.0__ add__3.0__4.0__ multiply__3.0__7.0__ multiply__3.0__7.0__ |
divide__27783.0__3.0__ divide__9261.0__3.0__ divide__3087.0__3.0__ divide__1029.0__3.0__ add__3.0__4.0__ multiply__3.0__7.0__ multiply__3.0__7.0__ |
| find out the c . i on rs . num__5000 at num__4.0 p . a . compound half - yearly for num__1 num__0.5 years . <o> a ) num__306.03 <o> b ) num__306.01 <o> c ) num__306.04 <o> d ) num__306.0 <o> e ) num__306.06 |
a = num__5000 ( num__1.02 ) num__3 = num__5306.04 num__5000 - - - - - - - - - - - num__306.04 answer : c <eor> c <eos> |
c |
subtract__4.0__1.0__ subtract__5306.04__5000.0__ multiply__1.0__306.04__ |
subtract__4.0__1.0__ subtract__5306.04__5000.0__ subtract__5306.04__5000.0__ |
| a train leaves mumabai at num__9 am at a speed of num__40 kmph . after one hour another train leaves mumbai in the same direction as that of the first train at a speed of num__50 kmph . when and at what distance from mumbai do the two trains meet ? <o> a ) num__2 : num__00 pm num__400 km <o> b ) num__8 : num__00 pm num__200 km <o> c ) num__2 : num__00 pm num__200 km <o> d ) num__2 : num__00 pm num__900 km <o> e ) num__6 : num__00 pm num__200 km |
when the second train leaves mumbai the first train covers num__40 * num__1 = num__40 km so the distance between first train and second train is num__40 km at num__10.00 am time taken by the trains to meet = distance / relative speed = num__40 / ( num__50 - num__40 ) = num__4 hours so the two trains meet at num__2 p . m . the two trains meet num__4 * num__50 = num__200 km away from mumbai . answer : c <eor> c <eos> |
c |
add__9.0__1.0__ divide__40.0__10.0__ multiply__50.0__4.0__ round__2.0__ |
subtract__50.0__40.0__ divide__40.0__10.0__ multiply__50.0__4.0__ divide__2.0__1.0__ |
| in a regular week there are num__5 working days and for each day the working hours are num__8 . a man gets rs . num__2.40 per hour for regular work and rs . num__3.20 per hours for overtime . if he earns rs . num__432 in num__4 weeks then how many hours does he work for ? <o> a ) num__160 <o> b ) num__175 <o> c ) num__180 <o> d ) num__195 <o> e ) num__185 |
suppose the man works overtime for x hours . now working hours in num__4 weeks = ( num__5 x num__8 x num__4 ) = num__160 . therefore num__160 x num__2.40 + x x num__3.20 = num__432 < = > num__3.20 x = num__432 - num__384 = num__48 x = num__15 . hence total hours of work = ( num__160 + num__15 ) = num__175 . answer : b <eor> b <eos> |
b |
multiply__2.4__160.0__ subtract__432.0__384.0__ divide__48.0__3.2__ add__160.0__15.0__ round__175.0__ |
multiply__2.4__160.0__ subtract__432.0__384.0__ divide__48.0__3.2__ add__160.0__15.0__ add__160.0__15.0__ |
| a polling company surveyed a certain country and it found that num__35.0 of that country ’ s registered voters had an unfavorable impression of both of that state ’ s major political parties and that num__20.0 had a favorable impression only of party e . if one registered voter has a favorable impression of both parties for every two registered voters who have a favorable impression only of party b then what percentage of the country ’ s registered voters have a favorable impression of both parties ( assuming that respondents to the poll were given a choice between favorable and unfavorable impressions only ) ? <o> a ) num__15 <o> b ) num__20 <o> c ) num__30 <o> d ) num__35 <o> e ) num__45 |
s = num__100 not ( e and b ) = num__35 only e = num__20 ( e and b ) / b = num__0.5 let ( e and b ) = x only b = num__2 x so now num__20 + num__35 + x + num__2 x = num__100 x = num__15 a ans <eor> a <eos> |
a |
percent__100.0__15.0__ |
percent__100.0__15.0__ |
| bird is flying num__120 km / hr between b to r . two trains at b to r at num__60 kmph . the distance traveled by the bird before it is killed ? <o> a ) num__90 kms . <o> b ) num__100 kms . <o> c ) num__110 kms . <o> d ) num__120 kms . <o> e ) num__130 kms . |
it will depend on the distance between b and r . if distance between b and r is num__120 kms time taken by trains before collision = num__120 / ( num__60 + num__60 ) = num__1 hr distance traveled by bird in num__1 hr = num__120 kms . so the distance traveled by the bird before it is killed = num__120 kms . answer : d <eor> d <eos> |
d |
round__120.0__ |
divide__120.0__1.0__ |
| if the price of a tv is first decreased by num__10.0 and then increased by num__30.0 then the net change in the price will be : <o> a ) num__4.0 increase <o> b ) num__17.0 increase <o> c ) num__10.0 decrease <o> d ) num__6.0 increase <o> e ) none of these |
explanation : solution : let the original price be rs . num__100 . new final price = num__130.0 of ( num__10.0 of num__100 ) = rs . num__1.3 * num__0.9 * num__100 = rs . num__117 . . ' . increase = num__17.0 answer : b <eor> b <eos> |
b |
add__30.0__100.0__ divide__130.0__100.0__ multiply__130.0__0.9__ subtract__117.0__100.0__ subtract__117.0__100.0__ |
add__30.0__100.0__ divide__130.0__100.0__ multiply__130.0__0.9__ subtract__117.0__100.0__ subtract__117.0__100.0__ |
| at a local supermarket a box of cereal usually costs num__19 dollars . this week the supermarket sells the box of cereal for num__5 dollars . how much money will you save if you buy this cereal at this supermarket ? <o> a ) num__15 dollars <o> b ) num__13 dollars <o> c ) num__14 dollars <o> d ) num__12 dollars <o> e ) num__11 dollar |
solution : in this situation there is a decrease in the price . saving = num__19 - num__5 = num__14 dollars option c <eor> c <eos> |
c |
subtract__19.0__5.0__ subtract__19.0__5.0__ |
subtract__19.0__5.0__ subtract__19.0__5.0__ |
| find the odd man out . num__1 num__3 num__9 num__12 num__19 num__29 <o> a ) num__12 <o> b ) num__9 <o> c ) num__1 <o> d ) num__3 <o> e ) num__6 |
explanation : num__12 is an even number . all other given numbers are odd answer : option a <eor> a <eos> |
a |
multiply__1.0__12.0__ |
multiply__1.0__12.0__ |
| the positive numbers w x y and z are such that x is num__10 percent greater than y y is num__20 percent greater than z and w is num__20 percent less than x . what percent greater than z is w ? <o> a ) num__15.2 <o> b ) num__16.0 <o> c ) num__20.0 <o> d ) num__23.2 <o> e ) num__24.8 % |
my strategy is same as thedobermanbut instead take z = num__100 which makes life a bit easy . as : z = num__100 y = num__120 ( num__20.0 greater than z ) z = num__144 ( num__20.0 greater than y ) now calculate w num__20.0 less than z = num__144 * num__0.8 = num__115.2 now by just looking relation between w and z : w - z / z * num__100 = num__16 - answer b <eor> b <eos> |
b |
add__20.0__100.0__ multiply__0.8__144.0__ multiply__20.0__0.8__ multiply__20.0__0.8__ |
add__20.0__100.0__ multiply__0.8__144.0__ multiply__20.0__0.8__ multiply__20.0__0.8__ |
| the sides of a triangle are in the ratio num__5 : num__12 : num__13 and its perimeter is num__300 m its area is ? <o> a ) num__150 <o> b ) num__288 <o> c ) num__278 <o> d ) num__111 <o> e ) num__112 |
num__5 x + num__12 x + num__13 x = num__300 = > x = num__10 a = num__50 b = num__120 c = num__130 s = ( num__50 + num__120 + num__130 ) / num__2 = num__150 answer : a <eor> a <eos> |
a |
multiply__5.0__10.0__ multiply__12.0__10.0__ multiply__13.0__10.0__ surface_cube__5.0__ surface_cube__5.0__ |
multiply__5.0__10.0__ multiply__12.0__10.0__ multiply__13.0__10.0__ surface_cube__5.0__ surface_cube__5.0__ |
| in right triangle abc ac is the hypotenuse . if ac is num__50 and ab + bc = num__70 what is the area of the triangle abc ? <o> a ) num__225 <o> b ) num__450 <o> c ) num__25 √ num__2 <o> d ) num__200 <o> e ) num__600 |
square ab + bc = num__70 : ( ab ) ^ num__2 + num__2 * ab * bc + ( bc ) ^ num__2 = num__4900 . since ( ac ) ^ num__2 = ( ab ) ^ num__2 + ( bc ) ^ num__2 = num__50 ^ num__2 = num__2500 then ( ab ) ^ num__2 + num__2 * ab * bc + ( bc ) ^ num__2 = num__2500 + num__2 * ab * bc = num__4900 . num__2500 + num__2 * ab * bc = num__4900 . ab * bc = num__1200 . the area = num__0.5 * ab * bc = num__600 . answer : e . <eor> e <eos> |
e |
power__70.0__2.0__ power__50.0__2.0__ multiply__0.5__1200.0__ multiply__0.5__1200.0__ |
power__70.0__2.0__ power__50.0__2.0__ multiply__0.5__1200.0__ multiply__0.5__1200.0__ |
| find the probability that a leap year selected at random will not have num__53 sundays or mondays . <o> a ) num__0.857142857143 <o> b ) num__0.714285714286 <o> c ) num__0.571428571429 <o> d ) num__0.428571428571 <o> e ) num__0.285714285714 |
there are num__366 days in a leap year : num__52 weeks and num__2 more days . so num__52 sundays num__52 mondays and num__2 days . these num__2 days can be : { mon tue } { tue wed } { wed thu } { thu fri } { fri sat } { sat sun } and { sun mon } ( num__7 cases ) . in order to have num__53 sundays or num__53 mondays we should have either { mon tue } { sat sun } or { sun mon } case . the probability of that is num__0.428571428571 . so probability of not having num__53 sundays or mondays probability is num__1 - num__0.428571428571 = num__0.571428571429 answer : c . <eor> c <eos> |
c |
card_space__ coin_space__ negate_prob__0.4286__ negate_prob__0.4286__ |
card_space__ coin_space__ negate_prob__0.4286__ negate_prob__0.4286__ |
| how many bricks each measuring num__25 cm x num__11.25 cm x num__6 cm will be needed to build a wall of num__8.5 m x num__6 m x num__22.5 cm ? <o> a ) num__6400 <o> b ) num__6410 <o> c ) num__6440 <o> d ) num__6500 <o> e ) num__6800 |
number of bricks = volume of wall / volume of bricks = num__850 x num__600 x num__22.5 / num__25 x num__11.25 x num__6 = = num__6800 answer : e <eor> e <eos> |
e |
round__6800.0__ |
round__6800.0__ |
| positive integer x is num__50 percent of num__50 percent of positive integer y and x percent of y equals num__100 . what is the value of y ? <o> a ) num__50 <o> b ) num__100 <o> c ) num__400 <o> d ) num__200 <o> e ) num__2 |
000 |
x = num__50.0 of num__50.0 num__0 f y = y / num__4 and x / num__100 of y = num__100 x / num__100 * num__4 x = num__100 x = num__50 and y = num__200 answer - d <eor> d <eos> |
d |
d |
| a train passes a station platform in num__35 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__248 m <o> b ) num__225 m <o> c ) num__277 m <o> d ) num__765 m <o> e ) num__176 m |
speed = ( num__54 * num__0.277777777778 ) m / sec = num__15 m / sec . length of the train = ( num__15 x num__20 ) m = num__300 m . let the length of the platform be x meters . then ( x + num__300 ) / num__35 = num__15 = = > x + num__300 = num__525 = = > x = num__225 m . answer : b <eor> b <eos> |
b |
subtract__35.0__20.0__ multiply__20.0__15.0__ multiply__35.0__15.0__ subtract__525.0__300.0__ round__225.0__ |
subtract__35.0__20.0__ multiply__20.0__15.0__ multiply__35.0__15.0__ subtract__525.0__300.0__ round__225.0__ |
| the average of first four multiples of num__4 is : <o> a ) num__3 <o> b ) num__9 <o> c ) num__10 <o> d ) num__15 <o> e ) num__17 |
solution average = num__4 ( num__1 + num__2 + num__3 + num__4 ) / num__4 = num__10.0 = num__10 answer c <eor> c <eos> |
c |
subtract__4.0__1.0__ multiply__1.0__10.0__ |
add__1.0__2.0__ divide__10.0__1.0__ |
| how many integers from num__101 to num__900 inclusive remains the value unchanged when the digits were reversed ? <o> a ) num__50 <o> b ) num__60 <o> c ) num__70 <o> d ) num__80 <o> e ) num__90 |
question is asking for palindrome first digit possibilities - num__1 through num__8 = num__8 num__9 is not possible here because it would result in a number greater than num__9 ( i . e num__909 num__919 . . ) second digit possibilities - num__0 though num__9 = num__10 third digit is same as first digit = > total possible number meeting the given conditions = num__8 * num__10 = num__80 answer is d . <eor> d <eos> |
d |
add__1.0__8.0__ multiply__101.0__9.0__ add__1.0__9.0__ multiply__8.0__10.0__ multiply__1.0__80.0__ |
add__1.0__8.0__ multiply__101.0__9.0__ subtract__919.0__909.0__ multiply__8.0__10.0__ multiply__1.0__80.0__ |
| two numbers a and b are such that the sum of num__8.0 of a and num__4.0 of b is two - third of the sum of num__6.0 of a and num__8.0 of b . find the ratio of a : b . <o> a ) num__2 : num__1 <o> b ) num__1 : num__2 <o> c ) num__4 : num__3 <o> d ) num__1 : num__3 <o> e ) num__3 : num__2 |
explanation : num__8.0 of a + num__4.0 of b = num__0.666666666667 ( num__6.0 of a + num__8.0 of b ) num__8 a / num__100 + num__4 b / num__100 = num__0.666666666667 ( num__6 a / num__100 + num__8 b / num__100 ) ⇒ num__8 a + num__4 b = num__0.666666666667 ( num__6 a + num__8 b ) ⇒ num__24 a + num__12 b = num__12 a + num__16 b ⇒ num__12 a = num__4 b ⇒ ab = num__0.333333333333 ⇒ a : b = num__1 : num__3 answer : option d <eor> d <eos> |
d |
divide__4.0__6.0__ multiply__4.0__6.0__ add__8.0__4.0__ add__4.0__12.0__ divide__8.0__24.0__ add__0.6667__0.3333__ subtract__4.0__1.0__ reverse__1.0__ |
divide__4.0__6.0__ multiply__4.0__6.0__ add__8.0__4.0__ add__4.0__12.0__ divide__8.0__24.0__ add__0.6667__0.3333__ divide__12.0__4.0__ reverse__1.0__ |
| three models ( m n and o ) of cars are distributed among three showrooms . the number of cars in each showrooms must be equal and each model must be represented by at least one car in every showroom . there are num__19 cars of model m num__17 cars of model n and num__15 cars of model o . what is the maximum number of cars of model m in any showroom ? <o> a ) num__17 <o> b ) num__16 <o> c ) num__15 <o> d ) num__14 <o> e ) num__13 |
the total number of cars is num__51 . so each showroom has num__17 cars ( since the number of cars in each showrooms should be equal num__17.0 = num__17 ) . moreover that the number of model m is maximum means that the numbers of model n and o should be minimum . since each model must be represented by at least one car in every showroom that minimum number should be num__1 . so maximum number of model m is num__17 - num__2 = num__15 . the answer is ( c ) <eor> c <eos> |
c |
subtract__19.0__17.0__ subtract__17.0__2.0__ |
subtract__19.0__17.0__ subtract__17.0__2.0__ |
| if x is an integer and ( x ) ( x ^ num__2 ) ( x ^ num__3 ) is positive which of the following could be negative ? <o> a ) x ^ num__4 <o> b ) ( x ) ( x ^ num__3 ) <o> c ) ( x ) ( x ) <o> d ) x + x ^ num__3 <o> e ) num__3 x ^ num__2 |
assuming ) x is negative ( makes it easier ) ( - num__2 ) ( - num__2 ^ num__2 ) ( - num__2 ^ num__3 ) ( - num__2 ) ^ num__3 . ( - num__2 ) ^ num__3 the two negative numbers will remain negative so they will both cancel out . so we need an equation where x ^ n with n being odd . a ) not sufficient - > will always be positive b ) not sufficient - > equals x ^ num__4 which will always be positive c ) not sufficient - > equals x ^ num__2 which will always be positive d ) - num__2 + - num__8 = - num__10 ! ! ! e ) num__3 * num__4 = num__12 is positive ( pemdas ! ) answer is d <eor> d <eos> |
d |
multiply__2.0__4.0__ add__2.0__8.0__ add__2.0__10.0__ divide__12.0__4.0__ |
multiply__2.0__4.0__ add__2.0__8.0__ add__2.0__10.0__ divide__12.0__4.0__ |
| a bat is bought for rs . num__420 and sold at a gain of num__20.0 find its selling price <o> a ) s . num__430 / - <o> b ) s . num__480 / - <o> c ) s . num__400 / - <o> d ) s . num__504 / - <o> e ) s . num__600 / - |
num__100.0 - - - - - - > num__420 num__120.0 - - - - - - > num__420 * num__1.2 selling price = rs . num__504 d <eor> d <eos> |
d |
percent__100.0__504.0__ |
percent__100.0__504.0__ |
| the diagonals of a rhombus are num__14 cm and num__20 cm . find its area ? <o> a ) num__176 <o> b ) num__140 <o> c ) num__150 <o> d ) num__179 <o> e ) num__172 |
num__0.5 * num__14 * num__20 = num__140 answer : b <eor> b <eos> |
b |
triangle_area__14.0__20.0__ triangle_area__14.0__20.0__ |
volume_rectangular_prism__14.0__20.0__0.5__ volume_rectangular_prism__14.0__20.0__0.5__ |
| the length of the bridge which a train num__130 metres long and travelling at num__45 km / hr can cross in num__30 seconds is : <o> a ) num__230 <o> b ) num__240 <o> c ) num__245 <o> d ) num__250 <o> e ) num__260 |
speed = [ num__45 x num__0.277777777778 ] m / sec = [ num__12.5 ] m / sec time = num__30 sec let the length of bridge be x metres . then ( num__130 + x ) / num__30 = num__12.5 = > num__2 ( num__130 + x ) = num__750 = > x = num__245 m . answer : option c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| the difference between a two - digit number and the number obtained by interchanging the two digits is num__63 . which is the smaller of the two numbers ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__9 <o> d ) num__7 <o> e ) num__1 |
explanation : let the ten ' s digit be x and units digit by y . then ( num__10 x + y ) - ( num__10 y + x ) = num__63 num__9 ( x - y ) = num__63 x - y = num__7 thus none of the numbers can be determined . answer : d <eor> d <eos> |
d |
divide__63.0__9.0__ divide__63.0__9.0__ |
divide__63.0__9.0__ divide__63.0__9.0__ |
| by walking at num__0.75 th of his usual speed a man reaches office num__5 minutes later than usual . what is his usual time ? <o> a ) num__15 min <o> b ) num__50 min <o> c ) num__65 min <o> d ) num__60 min <o> e ) num__70 min |
let t be the usual time . time spent = num__4 t / num__3 therefore num__4 t / num__3 = t + num__5 num__4 t = num__3 t + num__15 . . . therefor t = num__15 min answer : a <eor> a <eos> |
a |
multiply__0.75__4.0__ multiply__5.0__3.0__ round__15.0__ |
multiply__0.75__4.0__ multiply__5.0__3.0__ round__15.0__ |
| i live x floors above the ground floor of a high - rise building . it takes me num__30 s per floor to walk down the steps and num__2 s per floor to ride the lift . what is x if the time taken to walk down the steps to the ground floor is the same as to wait for the lift for num__7 min and then ride down ? <o> a ) num__4 <o> b ) num__7 <o> c ) num__14 <o> d ) num__15 <o> e ) num__17 |
explanation : since i live x floors above the ground floor and it takes me num__30 s per floor to walk and num__2 s per floor to ride it takes num__30 x s to walk down and num__2 x s to ride down after waiting num__420 s = > num__30 x = num__2 x + num__420 = > x = num__15 . answer : d <eor> d <eos> |
d |
divide__30.0__2.0__ divide__30.0__2.0__ |
divide__30.0__2.0__ divide__30.0__2.0__ |
| tough and tricky questions : remainders . num__1 ^ num__1 + num__2 ^ num__2 + num__3 ^ num__3 + . . . + num__7 ^ num__7 is divided by num__7 . what is the remainder ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
a number is divisible by num__5 if its last digit is divisible by num__5 let ' s look into the sum of last digits of each term of the given expression num__1 ^ num__1 = num__1 num__2 ^ num__2 = num__4 num__3 ^ num__3 = num__7 num__4 ^ num__4 = num__6 num__5 ^ num__5 = num__5 num__6 ^ num__6 = num__6 num__7 ^ num__7 = num__3 adding all these numbers we get num__32 which gives a remainder of num__4 when divided by num__7 . so answer must be num__4 . bunuel can you please confirm the answer of this question . yes the oa is c . clicked the wrong button when posting . edited . thank you . e <eor> e <eos> |
e |
add__2.0__3.0__ add__1.0__3.0__ add__1.0__5.0__ add__1.0__3.0__ |
add__2.0__3.0__ add__1.0__3.0__ add__1.0__5.0__ add__1.0__3.0__ |
| what is the next number : num__2 num__17 num__257 __ <o> a ) num__2977 <o> b ) num__3837 <o> c ) num__4097 <o> d ) num__4847 <o> e ) num__5387 |
num__4 ^ num__0 + num__1 = num__2 num__4 ^ num__2 + num__1 = num__17 num__4 ^ num__4 + num__1 = num__257 num__4 ^ num__6 + num__1 = num__4097 the answer is c . <eor> c <eos> |
c |
add__2.0__4.0__ multiply__1.0__4097.0__ |
add__2.0__4.0__ multiply__1.0__4097.0__ |
| square x is inscribed in circle y . if the perimeter of x is num__64 what is the circumference of y ? <o> a ) num__16 √ num__2 π <o> b ) num__8 √ num__2 π <o> c ) num__22 √ num__2 π <o> d ) num__12 √ num__2 π <o> e ) num__9 |
square forms two right angled triangles . any time we have a right angle triangle inside a circle the hypotenuse is the diameter . hypotenuse here = diagonal of the square = num__16 sqrt ( num__2 ) = diameter = > radius = num__8 sqrt ( num__2 ) circumference of the circle = num__2 pi r = num__16 pi sqrt ( num__2 ) answer is a . <eor> a <eos> |
a |
square_perimeter__2.0__ triangle_area__16.0__2.0__ |
square_perimeter__2.0__ triangle_area__16.0__2.0__ |
| the sum of three integers a b and c is num__300 . a is one third of the sum of b and c and b is one fifth of the sum of a and c . what is c ? <o> a ) num__175 <o> b ) num__180 <o> c ) num__185 <o> d ) num__190 <o> e ) num__195 |
a is one third of the sum of b and c . . or a = num__0.333333333333 ( b + c ) . . . but a + b + c = num__300 so num__0.333333333333 ( b + c ) + b + c = num__300 num__4 ( b + c ) / num__3 = num__300 b + c = num__300 * num__0.75 = num__225 a = num__300 - num__225 = num__75 num__2 ) b is one fifth of the sum of a and c or b = num__0.2 ( a + c ) . . . but a + b + c = num__300 so num__0.2 ( a + c ) + a + c = num__300 a + c = num__300 * num__0.833333333333 = num__250 but a = num__75 so c = num__250 - num__75 = num__175 answer : a <eor> a <eos> |
a |
divide__3.0__4.0__ multiply__300.0__0.75__ subtract__300.0__225.0__ subtract__250.0__75.0__ subtract__250.0__75.0__ |
divide__3.0__4.0__ multiply__300.0__0.75__ subtract__300.0__225.0__ subtract__250.0__75.0__ subtract__250.0__75.0__ |
| for every $ num__20 that a billionaire spends a millionaire spends the equivalent of num__20 cents . for every $ num__8 that a millionaire spends a yuppie spends the equivalent of $ num__1 . the ratio of money spent by a yuppie millionaire and billionaire can be expressed as <o> a ) num__1 : num__8 : num__800 <o> b ) num__1 : num__4 : num__100 <o> c ) num__20 : num__4 : num__1 <o> d ) num__100 : num__4 : num__1 <o> e ) num__400 : num__4 : num__1 |
b . . . . . . . . . m . . . . . . . . y num__20 . . . . . num__020 . . . . . . y b . . . . . . . . . num__8 . . . . . . . . num__1 what i did first was to turn num__0.20 to num__2 ( by multiplying by num__10 ) so that it is easy to find the lcm . this led me to this : b . . . . . . . . . m . . . . . . . . y num__200 . . . . . num__2 . . . . . . . . . y b . . . . . . . . num__80 . . . . . . num__10 then i multiplied every row by num__80 ( the lcm of num__2 and num__80 ) which led me to this : b . . . . . . . . . m . . . . . . . . y num__8000 . . . num__80 . . . . . . . num__10 then i got rid of the extra zero and in the correct order this is y : m : b = num__1 : num__8 : num__800 ans a <eor> a <eos> |
a |
divide__20.0__2.0__ multiply__20.0__10.0__ multiply__8.0__10.0__ divide__8000.0__10.0__ reverse__1.0__ |
divide__20.0__2.0__ multiply__20.0__10.0__ multiply__8.0__10.0__ divide__8000.0__10.0__ reverse__1.0__ |
| in assembling a bluetooth device a factory uses one of two kinds of modules . one module costs $ num__10 and the other one that is cheaper costs $ num__2.5 . the factory holds a $ num__62.50 worth stock of num__22 modules . how many of the modules in the stock are of the cheaper kind ? <o> a ) num__31 <o> b ) num__35 <o> c ) num__21 <o> d ) num__40 <o> e ) num__45 |
so the number of $ num__2.50 modules must be num__21 so that the leftover num__1 modules are of $ num__10 which will give a total value $ num__62.50 . num__21 * num__2.50 + num__1 * num__10 = num__52.50 + num__10 = num__62.50 answer : c <eor> c <eos> |
c |
subtract__22.0__21.0__ multiply__2.5__21.0__ subtract__22.0__1.0__ |
subtract__22.0__21.0__ multiply__2.5__21.0__ multiply__1.0__21.0__ |
| a train travelled from station p to q in num__8 hours and came back from station q to p is num__6 hours . what would be the ratio of the speed of the train while traveling from station p to q to that from station q to p ? <o> a ) num__3 : num__5 <o> b ) num__3 : num__3 <o> c ) num__3 : num__1 <o> d ) num__3 : num__4 <o> e ) num__3 : num__2 |
: since s # num__1 / t s num__1 : s num__2 = num__1 / t num__1 : num__1 / t num__2 = num__0.125 : num__0.166666666667 = num__3 : num__4 answer : d <eor> d <eos> |
d |
subtract__8.0__6.0__ divide__1.0__8.0__ divide__1.0__6.0__ divide__6.0__2.0__ divide__8.0__2.0__ round__3.0__ |
subtract__8.0__6.0__ divide__1.0__8.0__ divide__1.0__6.0__ divide__6.0__2.0__ divide__8.0__2.0__ divide__6.0__2.0__ |
| a train having a length of num__240 m passes a post in num__24 seconds . how long will it take to pass a platform having a length of num__670 m ? <o> a ) num__120 sec <o> b ) num__91 sec <o> c ) num__89 sec <o> d ) num__80 sec <o> e ) num__85 sec |
explanation : v = num__10.0 ( where v is the speed of the train ) = num__10 m / s t = ( num__240 + num__670 ) / num__10 = num__91 seconds answer : option b <eor> b <eos> |
b |
divide__240.0__24.0__ round__91.0__ |
divide__240.0__24.0__ round__91.0__ |
| a train passes a man standing on the platform . if the train is num__245 meters long and its speed is num__126 kmph how much time it took in doing so ? <o> a ) num__2 sec <o> b ) num__67 sec <o> c ) num__5 sec <o> d ) num__9 sec <o> e ) num__7 sec |
d = num__245 s = num__126 * num__0.277777777778 = num__35 mps t = num__7.0 = num__7 sec answer : e <eor> e <eos> |
e |
divide__245.0__35.0__ round__7.0__ |
divide__245.0__35.0__ round__7.0__ |
| the age of man is three times the sum of the ages of his two sons . nine years hence his age will be double of the sum of the ages of his sons . the father â € ™ s present age is : <o> a ) num__40 years <o> b ) num__45 years <o> c ) num__50 years <o> d ) num__55 years <o> e ) num__81 years |
solution let the sum of present ages of the two sons be x years . then father ' s present age = num__3 x years . â ˆ ´ ( num__3 x + num__9 ) = num__2 ( x + num__18 ) â ‡ ” num__3 x + num__9 = num__2 x + num__36 â ‡ ” x = num__27 hence father ' s present age = num__81 years . answer e <eor> e <eos> |
e |
multiply__2.0__9.0__ multiply__2.0__18.0__ multiply__3.0__9.0__ multiply__3.0__27.0__ multiply__3.0__27.0__ |
multiply__2.0__9.0__ multiply__2.0__18.0__ add__9.0__18.0__ multiply__3.0__27.0__ multiply__3.0__27.0__ |
| if a radio is purchased for rs num__490 and sold for rs num__465.50 find the loss % ? <o> a ) num__3.0 <o> b ) num__4.0 <o> c ) num__5.0 <o> d ) num__6.0 <o> e ) none of these |
explanation : loss = num__490 – num__465.5 = num__24.5 loss in % = num__24.5 / num__490 x num__100 = num__5.0 . answer : c <eor> c <eos> |
c |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| a batsman in his num__17 th innings makes a score of num__85 and their by increasing his average by num__3 . what is his average after the num__17 th innings ? <o> a ) num__12 <o> b ) num__13 <o> c ) num__46 <o> d ) num__37 <o> e ) num__26 |
num__16 x + num__85 = num__17 ( x + num__3 ) x = num__34 + num__3 = num__37 answer : d <eor> d <eos> |
d |
add__3.0__34.0__ add__3.0__34.0__ |
add__3.0__34.0__ add__3.0__34.0__ |
| if page of chemistry book is num__30 more than page of physics book . if ratio of pages is num__5 : num__3 . a child torn num__15 pages of chemistry book . calculate new ratio . <o> a ) num__1 : num__1 <o> b ) num__2 : num__1 <o> c ) num__5 : num__4 <o> d ) num__4 : num__3 <o> e ) num__2 : num__3 |
chemistry page = num__5 x physics page = num__3 x num__5 x - num__3 x = num__302 x = num__30 x = num__15 chemistry page = num__75 physics page = num__45 after torn num__15 page of chemistry = num__60 ratio become num__1.33333333333 = num__4 : num__3 answer d <eor> d <eos> |
d |
multiply__5.0__15.0__ add__30.0__15.0__ add__15.0__45.0__ divide__60.0__45.0__ divide__60.0__15.0__ divide__60.0__15.0__ |
multiply__5.0__15.0__ subtract__75.0__30.0__ subtract__75.0__15.0__ divide__60.0__45.0__ divide__60.0__15.0__ divide__60.0__15.0__ |
| two trains are moving in opposite directions at num__60 km / hr and num__90 km / hr . their lengths are num__1.10 km and num__0.9 km respectively . the time taken by the slower train to cross the faster train in seconds is ? <o> a ) num__12 <o> b ) num__77 <o> c ) num__48 <o> d ) num__99 <o> e ) num__11 |
relative speed = num__60 + num__90 = num__150 km / hr . = num__150 * num__0.277777777778 = num__41.6666666667 m / sec . distance covered = num__1.10 + num__0.9 = num__2 km = num__2000 m . required time = num__2000 * num__0.024 = num__48 sec . answer : c <eor> c <eos> |
c |
add__60.0__90.0__ add__1.1__0.9__ multiply__2000.0__0.024__ round__48.0__ |
add__60.0__90.0__ add__1.1__0.9__ multiply__2000.0__0.024__ multiply__2000.0__0.024__ |
| in the equation num__2 x â € “ cy = num__18 c is a constant . if the value of y is num__2 when x is num__6 what is the value of x when y is num__4 ? <o> a ) â ˆ ’ num__4.5 <o> b ) - num__4 <o> c ) num__3 <o> d ) num__4 <o> e ) num__4.5 |
num__2 x - cy = num__18 x = num__6 y = num__2 ; num__12 - num__2 c = num__18 c = - num__3 num__2 x - cy = num__18 when y = num__4 c = - num__3 x = x = num__3 answer : c <eor> c <eos> |
c |
multiply__2.0__6.0__ divide__18.0__6.0__ divide__18.0__6.0__ |
subtract__18.0__6.0__ divide__18.0__6.0__ subtract__6.0__3.0__ |
| a is the average ( arithmetic mean ) of the first num__7 positive multiples of num__5 and b is the median of the first num__3 positive multiples of positive integer n . if the value of a ^ num__2 – b ^ num__2 is zero what is the value of n ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__10 <o> d ) num__15 <o> e ) num__20 |
if a ^ num__2 - b ^ num__2 = num__0 then let ' s assume that a = b . a must equal the num__4 th positive multiple of num__4 thus a = num__20 which also equals b . b is the second positive multiple of n thus n = num__10.0 = num__10 . the answer is c . <eor> c <eos> |
c |
subtract__7.0__3.0__ multiply__5.0__4.0__ add__7.0__3.0__ add__7.0__3.0__ |
subtract__7.0__3.0__ multiply__5.0__4.0__ add__7.0__3.0__ subtract__20.0__10.0__ |
| how much pure alcohol should be added to num__400 ml of a num__15.0 solution to make the strength of solution num__25.0 ? <o> a ) num__100 ml <o> b ) num__60 ml <o> c ) num__125 ml <o> d ) num__130 ml <o> e ) num__150 ml |
we can also go by answer choices tke num__100 ml for eg num__400 ( old ) + num__100 ( new concentr ) ml num__500 * num__0.25 = num__125 ml ( num__25 ml is de old concentration + num__100 ml ( newly added ) answer c <eor> c <eos> |
c |
add__400.0__100.0__ divide__25.0__100.0__ add__25.0__100.0__ add__25.0__100.0__ |
add__400.0__100.0__ divide__25.0__100.0__ add__25.0__100.0__ add__25.0__100.0__ |
| find the one which does not belong to that group ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__9 <o> e ) num__7 |
explanation : num__3 num__5 num__9 and num__7 are odd numbers but not num__4 . answer is b <eor> b <eos> |
b |
subtract__9.0__5.0__ subtract__9.0__5.0__ |
subtract__9.0__5.0__ subtract__9.0__5.0__ |
| if log num__32 x = num__0.8 then x is equal to <o> a ) num__25.6 <o> b ) num__16 <o> c ) num__10 <o> d ) num__12.8 <o> e ) none |
solution log num__32 x = num__0.8 . x = ( num__32 ) num__0.8 ‹ = › ( num__25 ) num__0.8 ‹ = › num__24 ‹ = › num__16 . answer b <eor> b <eos> |
b |
subtract__32.0__16.0__ |
subtract__32.0__16.0__ |
| when running a mile during a recent track meet nuria was initially credited with a final time of num__5 minutes num__42 seconds . shortly after her run officials realized that the timing mechanism malfunctioned . the stopwatch did not begin timing her until num__0.44 of a minute after she began to run . if the time was otherwise correct how long did it actually take nuria to run the mile ? <o> a ) num__5 minutes num__17.6 seconds <o> b ) num__5 minutes num__21.8 seconds <o> c ) num__5 minutes num__43.56 seconds <o> d ) num__5 minutes num__44.44 seconds <o> e ) num__6 minutes num__10.4 seconds |
one approach : the watch starts to work after nuria began his running . it means the time should be greater than credited num__5 minutes num__42 seconds . the only number is num__6 minutes num__10.4 seconds . another approach : num__0.44 close to num__30 second when added to the num__5 minutes num__44 seconds it means it passes num__6 minute . answer : e <eor> e <eos> |
e |
multiply__5.0__6.0__ round__6.0__ |
multiply__5.0__6.0__ round__6.0__ |
| when a plot is sold for rs . num__18700 the owner loses num__15.0 . at what price must that plot be sold in order to gain num__15.0 ? <o> a ) num__25300 <o> b ) num__22000 <o> c ) num__21000 <o> d ) num__23560 <o> e ) num__24540 |
sell at num__18700 he got a loss of num__15.0 means this num__18700 is num__85.0 not num__100.0 . to get num__1.0 : num__220.0 = num__220 ; then num__220 is num__1.0 . to get num__100.0 : num__220 * num__100 = num__22000 . to get num__115.0 : num__220 * num__115 = num__25300 . answer : a <eor> a <eos> |
a |
percent__100.0__25300.0__ |
percent__100.0__25300.0__ |
| wo trains of equal length running in opposite directions pass a pole in num__18 and num__12 seconds . the trains will cross each other in <o> a ) num__14.4 sec <o> b ) num__14.9 sec <o> c ) num__12.4 sec <o> d ) num__14.8 sec <o> e ) num__15.4 sec |
explanation : let the length of the train be l metres speed of first train = \ inline \ frac { num__3600 } { num__18 } \ times l m / hour speed of secxond train = \ inline \ frac { num__3600 } { num__12 } \ times l m / hour when running in opposite directions relative speed = num__200 l + num__300 l m / hour distance to be covered = l + l = num__2 l metre time taken = \ inline \ frac { num__2 l } { num__500 l } \ times num__3600 sec = num__14.4 sec answer : a ) num__14.4 sec <eor> a <eos> |
a |
divide__3600.0__18.0__ divide__3600.0__12.0__ add__200.0__300.0__ round__14.4__ |
divide__3600.0__18.0__ divide__3600.0__12.0__ add__200.0__300.0__ round__14.4__ |
| a train num__165 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__9 <o> d ) num__8 <o> e ) num__5 |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__165 * num__0.0545454545455 ] sec = num__9 sec answer : c <eor> c <eos> |
c |
add__60.0__6.0__ divide__165.0__18.3333__ round__9.0__ |
add__60.0__6.0__ divide__165.0__18.3333__ divide__165.0__18.3333__ |
| two trains each num__250 m in length are running on the same parallel lines in opposite directions with the speed of num__80 kmph and num__20 kmph respectively . in what time will they cross each other completely ? <o> a ) num__15 sec <o> b ) num__18 sec <o> c ) num__12 sec <o> d ) num__10 sec <o> e ) num__11 sec |
explanation : d = num__250 m + num__250 m = num__500 m rs = num__80 + num__20 = num__100 * num__0.277777777778 = num__27.7777777778 t = num__500 * num__0.036 = num__18 sec answer : option b <eor> b <eos> |
b |
add__80.0__20.0__ multiply__0.036__500.0__ round__18.0__ |
add__80.0__20.0__ multiply__0.036__500.0__ multiply__0.036__500.0__ |
| what is num__0.01 percent of num__12356 ? <o> a ) num__0.12356 <o> b ) num__1.2356 <o> c ) num__12.356 <o> d ) num__0.012356 <o> e ) num__0.0012356 |
soln : - num__0.01 of num__12356 = num__0 . num__011000.01100 x num__12356 = num__1100 ∗ num__1001100 ∗ num__100 x num__12356 = num__12356100 ∗ num__10012356100 ∗ num__100 = num__1.2356 answer : b <eor> b <eos> |
b |
percent__0.01__12356.0__ percent__0.01__12356.0__ |
percent__0.01__12356.0__ percent__0.01__12356.0__ |
| the age of a person is thrice the total ages of his num__2 daughters . num__0.5 decades hence his age will be twice of the total ages of his daughters . then what is the father ’ s current age ? [ num__0.5 decades = num__5 years ] <o> a ) num__35 years <o> b ) num__40 years <o> c ) num__45 years <o> d ) num__47 years <o> e ) none of these |
explanation : let total of current ages of the num__2 daughters is a years . then father ’ s current age = num__3 a years . ( num__3 a + num__5 ) = num__2 ( a + num__10 ) num__3 a + num__5 = num__2 a + num__20 a = num__15 therefore father ’ s current age = num__45 years . answer : c <eor> c <eos> |
c |
subtract__5.0__2.0__ multiply__2.0__5.0__ multiply__2.0__10.0__ multiply__5.0__3.0__ multiply__3.0__15.0__ multiply__3.0__15.0__ |
subtract__5.0__2.0__ multiply__2.0__5.0__ multiply__2.0__10.0__ add__5.0__10.0__ multiply__3.0__15.0__ multiply__3.0__15.0__ |
| in orange county one tenth of the people are gathering mushrooms and one seventh of the people are collecting apples . what can be the number of people in orange county ? <o> a ) num__60 . <o> b ) num__42 . <o> c ) num__85 . <o> d ) num__70 . <o> e ) num__252 . |
the number must be an integer and multiple of both num__10 and num__7 . only option d fulfill this condition and hence is the answer . <eor> d <eos> |
d |
multiply__10.0__7.0__ |
multiply__10.0__7.0__ |
| what is the radius of a circle that has a circumference of num__3.14 meters ? <o> a ) num__1.0 <o> b ) num__2.5 <o> c ) num__1.3 <o> d ) num__0.5 <o> e ) num__0.7 |
circumference of a circle = num__2 π r . given circumference = num__3.14 meters . therefore num__2 π r = circumference of a circle or num__2 π r = num__3.14 . or num__2 * num__3.14 r = num__3.14 [ putting the value of pi ( π ) = num__3.14 ] . or num__6.28 r = num__3.14 . or r = num__3.14 / num__6.28 . or r = num__0.5 . answer : num__0.5 meter . correct answer d <eor> d <eos> |
d |
multiply__3.14__2.0__ triangle_area__0.5__2.0__ |
multiply__3.14__2.0__ triangle_area__0.5__2.0__ |
| a train is running at a speed of num__40 km / hr and it crosses a post in num__17.1 seconds . what is the length of the train ? <o> a ) num__190 metres <o> b ) num__160 metres <o> c ) num__200 metres <o> d ) num__120 metres <o> e ) num__250 metres |
speed of the train v = num__40 km / hr = num__11.1111111111 m / s = num__11.1111111111 m / s time taken to cross t = num__17.1 s distance covered d = vt = ( num__11.1111111111 ) Ã — num__17.1 = num__190 m distance covered is equal to the length of the train = num__190 m correct answer is num__190 metres a <eor> a <eos> |
a |
round__190.0__ |
round__190.0__ |
| buses a and b start from a common bus stop x . bus a begins to travel in a straight line away from bus b at a constant rate of num__30 miles per hour . one hour later bus b begins to travel in a straight line in the exact opposite direction at a constant rate of num__50 miles per hour . if both buses travel indefinitely what is the positive difference in minutes between the amount of time it takes bus b to cover the exact distance that bus a has covered and the amount of time it takes bus b to cover twice the distance that bus a has covered ? <o> a ) num__36 <o> b ) num__84 <o> c ) num__132 <o> d ) num__144 <o> e ) num__180 |
num__1 st part : - in num__1 hr bus a covers num__30 miles . relative speed of bus abus b is ( num__80 - num__30 ) = num__50 mph . so time required for bus b to cover the exact distance as a is num__50 * t = num__30 t = num__0.6 = num__36 min num__2 nd part num__80 * t = num__2 d - b has to cover twice the distance num__30 * ( t + num__1 ) = d - a traveled num__1 hr more and has to travel only only d so d / num__30 - num__2 d / num__80 = num__1 d = num__120 t = num__3 hrs = num__180 min question asks for + ve difference between part num__1 and part num__2 in minutes = num__180 - num__36 = num__84 min b <eor> b <eos> |
b |
add__30.0__50.0__ km_to_mile_conversion__ add__1.0__2.0__ subtract__120.0__36.0__ round__84.0__ |
add__30.0__50.0__ divide__30.0__50.0__ add__1.0__2.0__ subtract__120.0__36.0__ subtract__120.0__36.0__ |
| there are num__10 students named alphabetically from a to j . what is the probability that a and d do not sit together if all num__10 sit around a circular table ? <o> a ) num__0.222222222222 <o> b ) num__0.4 <o> c ) num__0.777777777778 <o> d ) num__0.8 <o> e ) num__0.888888888889 |
number of students = num__10 number of ways num__10 students can sit around a circular table = ( num__10 - num__1 ) ! = num__9 ! number of ways a and d sit together ( consider a and d as one entity ) = ( num__9 - num__1 ) ! = num__8 ! * num__2 number of ways a and d do not sit together = num__9 ! - ( num__8 ! * num__2 ) probability = ( num__9 ! - ( num__8 ! * num__2 ) ) / num__9 ! = num__1 - num__0.222222222222 = num__0.777777777778 answer : c <eor> c <eos> |
c |
coin_space__ negate_prob__0.2222__ negate_prob__0.2222__ |
coin_space__ negate_prob__0.2222__ negate_prob__0.2222__ |
| a car covers a distance of num__660 km in num__6 hours . find its speed ? <o> a ) num__104 <o> b ) num__7778 <o> c ) num__266 <o> d ) num__110 <o> e ) num__121 |
num__110.0 = num__110 kmph answer : d <eor> d <eos> |
d |
divide__660.0__6.0__ round__110.0__ |
divide__660.0__6.0__ round__110.0__ |
| if jos é reads at a constant rate of num__5 pages every num__5 minutes how many seconds will it take him to read n pages ? <o> a ) num__60 <o> b ) num__2 n <o> c ) num__2.5 * n <o> d ) num__24 n <o> e ) num__150 |
jose would read num__1 page in num__1.0 min jose would read n page in ( num__1.0 ) * n min i . e . ( num__1.0 ) * n * num__60 seconds = num__60 n seconds . option a is the correct answer . <eor> a <eos> |
a |
hour_to_min_conversion__ hour_to_min_conversion__ |
hour_to_min_conversion__ hour_to_min_conversion__ |
| calculate the percentage gain if a trader bought a bicycle for rs num__950 and sold it for rs . num__1500 ? <o> a ) num__56.89 <o> b ) num__52.89 <o> c ) num__59.89 <o> d ) num__56.89 <o> e ) num__57.89 % |
c . p . = num__950 s . p . = num__1500 gain = num__1500 - num__950 - - - - num__550.0 gain = > num__0.578947368421 * num__100 = > num__57.89 answer : e <eor> e <eos> |
e |
percent__100.0__57.89__ |
percent__100.0__57.89__ |
| num__10 men can complete a piece of work in num__15 days and num__15 women can complete the same work in num__12 days . if all the num__10 men and num__15 women work together in how many days will the work get completed ? <o> a ) num__6 <o> b ) num__7 num__2 ⁄ num__3 <o> c ) num__6 num__2 ⁄ num__3 <o> d ) num__6 num__1 ⁄ num__3 <o> e ) none of these |
num__10 men + num__15 women in num__1 day do num__1 ⁄ num__15 + num__1 ⁄ num__12 = num__9 ⁄ num__60 work \ time taken = num__60 ⁄ num__9 days = num__62 ⁄ num__3 days answer c <eor> c <eos> |
c |
subtract__10.0__1.0__ hour_to_min_conversion__ subtract__15.0__12.0__ subtract__15.0__9.0__ |
subtract__10.0__1.0__ hour_to_min_conversion__ subtract__15.0__12.0__ subtract__15.0__9.0__ |
| there are two batches a and b of a class . batch a consists of num__36 students and batch b consists of num__44 students . find the average weight of whole class if average weight of batch a is num__40 kg and that of batch b is num__35 kg . <o> a ) num__29.23 kg <o> b ) num__32.56 kg <o> c ) num__35.66 kg <o> d ) num__37.25 kg <o> e ) none of the above |
given : average weight of batch a = num__40 kg average weight of batch b = num__35 kg num__1 ) first find the total weight of all students - weight of batch a = ( num__36 x num__40 ) = num__1440 - weight of batch b = ( num__44 x num__35 ) = num__1540 total weight of all students = ( num__1440 + num__1540 ) = num__2980 kg num__2 ) find average weight of whole class ( batch a + batch b ) students = ( num__36 + num__44 ) = num__80 students average weight = total weight of all the students / no . of students = num__37.25 = num__37.25 kg answer is d <eor> d <eos> |
d |
subtract__36.0__35.0__ multiply__36.0__40.0__ multiply__44.0__35.0__ add__1440.0__1540.0__ add__36.0__44.0__ divide__2980.0__80.0__ multiply__1.0__37.25__ |
subtract__36.0__35.0__ multiply__36.0__40.0__ multiply__44.0__35.0__ add__1440.0__1540.0__ add__36.0__44.0__ divide__2980.0__80.0__ divide__37.25__1.0__ |
| num__30 men can do a work in num__40 days . when should num__20 men leave the work so that the entire work is completed in num__40 days after they leave the work ? <o> a ) num__87 days <o> b ) num__10 days <o> c ) num__55 days <o> d ) num__44 days <o> e ) num__22 days |
total work to be done = num__30 * num__40 = num__1200 let num__20 men leave the work after ' p ' days so that the remaining work is completed in num__40 days after they leave the work . num__40 p + ( num__20 * num__40 ) = num__1200 num__40 p = num__400 = > p = num__10 days answer : b <eor> b <eos> |
b |
multiply__30.0__40.0__ subtract__30.0__20.0__ round__10.0__ |
multiply__30.0__40.0__ subtract__30.0__20.0__ round__10.0__ |
| the length of the bridge which a train num__130 metres long and travelling at num__45 km / hr can cross in num__30 seconds is ? <o> a ) num__38 <o> b ) num__2767 <o> c ) num__245 <o> d ) num__277 <o> e ) num__232 |
speed = [ num__45 x num__0.277777777778 ] m / sec = [ num__12.5 ] m / sec time = num__30 sec let the length of bridge be x metres . then ( num__130 + x ) / num__30 = num__12.5 = > num__2 ( num__130 + x ) = num__750 = > x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| suppose two numbers are in the ratio num__2 : num__3 . if num__5 is subtracted from both of the numbers then the ratio becomes num__1 : num__2 then find the greatest number ? <o> a ) num__5 <o> b ) num__15 <o> c ) num__10 <o> d ) num__20 <o> e ) num__25 |
given ratio - > num__2 : num__3 num__2 x - num__5 : num__3 x - num__10 = num__1 : num__2 num__2 [ num__2 x - num__5 ] = num__1 [ num__3 x - num__5 ] num__4 x - num__10 = num__3 x - num__5 num__4 x - num__3 x = num__10 - num__5 x = num__5 then greatest number is = num__3 x num__3 x = num__15 ans - b <eor> b <eos> |
b |
multiply__2.0__5.0__ add__3.0__1.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
multiply__2.0__5.0__ subtract__5.0__1.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
| a restaurant buys fruit in cans containing num__3 num__0.5 cups of fruit each . if the restaurant uses num__0.5 cup of the fruit in each serving of its fruit compote what is the least number of cans needed to prepare num__60 servings of the compote ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__12 |
num__3.5 cups = num__1 can num__0.5 cup of each serving . so num__7 serving per num__1 can . least number is num__7 * num__9 = num__63 = num__60 approx answer - num__9 answer : c <eor> c <eos> |
c |
add__3.0__0.5__ divide__3.5__0.5__ add__3.0__60.0__ multiply__1.0__9.0__ |
add__3.0__0.5__ divide__3.5__0.5__ multiply__7.0__9.0__ multiply__1.0__9.0__ |
| two trains running in opposite directions cross a man standing on the platform in num__27 seconds and num__17 seconds respectively and they cross each other in num__23 seconds . the ratio of their speeds is ? <o> a ) num__3.0 <o> b ) num__1.5 <o> c ) num__0.333333333333 <o> d ) num__0.6 <o> e ) num__3.5 |
let the speeds of the two trains be x m / sec and y m / sec respectively . then length of the first train = num__27 x meters and length of the second train = num__17 y meters . ( num__27 x + num__17 y ) / ( x + y ) = num__23 = = > num__27 x + num__17 y = num__23 x + num__23 y = = > num__4 x = num__6 y = = > x / y = num__1.5 . answer : b <eor> b <eos> |
b |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ round__1.5__ |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ divide__6.0__4.0__ |
| sum of num__19 odd numbers is ? <o> a ) num__341 <o> b ) num__351 <o> c ) num__361 <o> d ) num__371 <o> e ) num__381 |
sum of num__1 st n odd no . s = num__1 + num__3 + num__5 + num__7 + . . . = n ^ num__2 so sum of num__1 st num__19 odd numbers = num__19 ^ num__2 = num__361 answer : c <eor> c <eos> |
c |
subtract__3.0__1.0__ multiply__1.0__361.0__ |
subtract__3.0__1.0__ multiply__1.0__361.0__ |
| the difference between the place values of num__6 and num__5 in the number num__826533 is <o> a ) num__5500 <o> b ) num__4500 <o> c ) num__2500 <o> d ) num__6970 <o> e ) none |
sol . = ( place value of num__6 ) – ( place value of num__5 ) = ( num__6000 - num__500 ) = num__5500 answer a <eor> a <eos> |
a |
subtract__6000.0__500.0__ subtract__6000.0__500.0__ |
subtract__6000.0__500.0__ subtract__6000.0__500.0__ |
| in a flight of num__600 km an aircraft was slowed down due to bad weather . its average speed for the trip was reduced by num__200 km / hr and the time of flight increased by num__30 minutes . the duration of the flight is : <o> a ) num__2 hours <o> b ) num__3 hours <o> c ) num__4 hours <o> d ) num__1 hour <o> e ) num__5 hours |
let the duration of the flight be x hours . then num__600 / x - num__600 / ( x + num__0.5 ) = num__200 = = > num__600 / x - num__1200 / ( num__2 x + num__1 ) = num__200 = = > x ( num__2 x + num__1 ) = num__3 = = > num__2 x sqr . + x - num__3 = num__0 = = > ( num__2 x + num__3 ) ( x - num__1 ) = num__0 . by neglecting negative value of x we got x = num__1 hr . answer d ) num__1 hour . <eor> d <eos> |
d |
divide__600.0__0.5__ reverse__0.5__ multiply__0.5__2.0__ divide__600.0__200.0__ round_down__0.5__ reverse__1.0__ |
divide__600.0__0.5__ reverse__0.5__ multiply__0.5__2.0__ divide__600.0__200.0__ round_down__0.5__ reverse__1.0__ |
| a tank is filled in four hours by three pipes a b and c . pipe a is twice as fast as pipe b and b is twice as fast as c . how much time will pipe b alone take to fill the tank ? <o> a ) num__56 hours <o> b ) num__28 hours <o> c ) num__55 hours <o> d ) num__14 hours <o> e ) num__47 hours |
num__1 / a + num__1 / b + num__1 / c = num__0.25 ( given ) also given that a = num__2 b and b = num__2 c = > num__0.5 b + num__1 / b + num__2 / b = num__0.25 = > ( num__1 + num__2 + num__4 ) / num__2 b = num__0.25 = > num__2 b / num__7 = num__4 = > b = num__14 hours . answer : d <eor> d <eos> |
d |
multiply__0.25__2.0__ divide__1.0__0.25__ multiply__2.0__7.0__ round__14.0__ |
divide__1.0__2.0__ divide__1.0__0.25__ divide__7.0__0.5__ divide__7.0__0.5__ |
| what is the ratio whose term differ by num__35 and the measure of which is num__0.285714285714 ? <o> a ) a ) num__32 : num__23 <o> b ) b ) num__14 : num__49 <o> c ) c ) num__71 : num__85 <o> d ) d ) num__32 : num__39 <o> e ) e ) num__41 : num__52 |
let the ratio be x : ( x + num__35 ) then x / ( x + num__35 ) = num__0.285714285714 x = num__14 required ratio = num__14 : num__49 answer is b <eor> b <eos> |
b |
add__35.0__14.0__ subtract__49.0__35.0__ |
add__35.0__14.0__ subtract__49.0__35.0__ |
| if x gets num__25.0 more than y and y gets num__20.0 more than z the share of z out of rs . num__1110 will be : <o> a ) rs . num__300 <o> b ) rs . num__200 <o> c ) rs . num__240 <o> d ) rs . num__350 <o> e ) none of these |
z share = z y = num__1.2 z x = num__1.25 Ã — num__1.2 z x + y + z = num__111 ( num__1.25 Ã — num__1.2 + num__1.2 + num__1 ) z = num__1110 num__3.7 z = num__1110 z = num__300 answer : . a <eor> a <eos> |
a |
divide__25.0__20.0__ round_down__1.25__ divide__1110.0__3.7__ divide__1110.0__3.7__ |
divide__25.0__20.0__ round_down__1.25__ divide__1110.0__3.7__ divide__1110.0__3.7__ |
| of the num__180 people at a party num__70 were women and num__30 women tried the appetizer . if num__50 people did not try the appetizer what is the total number of men who tried the appetizer ? <o> a ) num__40 <o> b ) num__50 <o> c ) num__60 <o> d ) num__70 <o> e ) num__100 |
total people at party = num__180 women = num__70 so men num__180 - num__70 = num__110 no . of pple who tried appetizer = num__180 - num__50 ( given info ) = num__130 no of women who tried appetizer = num__30 so remaining ppl ( men ) who tried the appetizer = num__130 - num__30 = num__100 correct option e <eor> e <eos> |
e |
subtract__180.0__70.0__ subtract__180.0__50.0__ add__70.0__30.0__ add__70.0__30.0__ |
subtract__180.0__70.0__ subtract__180.0__50.0__ subtract__130.0__30.0__ subtract__130.0__30.0__ |
| what is the value of ( p + q ) / ( p - q ) if p / q is num__4 ? <o> a ) num__1.66666666667 <o> b ) num__0.666666666667 <o> c ) num__0.333333333333 <o> d ) num__0.875 <o> e ) num__1.14285714286 |
( p + q ) / ( p - q ) = [ ( p / q ) + num__1 ] / [ ( p / q ) - num__1 ] = ( num__4 + num__1 ) / ( num__4 - num__1 ) = num__1.66666666667 = num__1.66666666667 answer : a <eor> a <eos> |
a |
multiply__1.0__1.6667__ |
divide__1.6667__1.0__ |
| in a coconut grove ( x + num__2 ) trees yield num__40 nuts per year x trees yield num__120 nuts per year and ( x – num__2 ) trees yield num__180 nuts per year . if the average yield per year per tree be num__100 find x . <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
( x + num__2 ) × num__40 + x × num__120 + ( x − num__2 ) × num__180 / ( x + num__2 ) + x + ( x − num__2 ) = num__100 ⇒ num__340 x − num__93.3333333333 x = num__100 ⇒ num__40 x = num__280 ⇒ x = num__7 answer e <eor> e <eos> |
e |
add__180.0__100.0__ divide__280.0__40.0__ divide__280.0__40.0__ |
add__180.0__100.0__ divide__280.0__40.0__ divide__280.0__40.0__ |
| given that e and f are events such that p ( e ) = num__0.16 p ( f ) = num__0.4 and p ( e n f ) = num__0.4 find p ( e | f ) and p ( f | e ) ? <o> a ) num__1 <o> b ) num__0.25 <o> c ) num__2 <o> d ) num__0.4 <o> e ) none |
here e and f are events p ( e | f ) = p ( enf ) / p ( f ) = num__0.4 / num__0.4 = num__1 p ( f | e ) = p ( enf ) / p ( e ) = num__0.4 / num__0.16 = num__0.25 . option is b <eor> b <eos> |
b |
multiply__0.25__1.0__ |
divide__0.25__1.0__ |
| what amount does an investor receive if the investor invests $ num__4000 at num__10.0 p . a . compound interest for two years compounding done annually ? <o> a ) $ num__4720 <o> b ) $ num__4840 <o> c ) $ num__4920 <o> d ) $ num__5080 <o> e ) $ num__5160 |
a = ( num__1 + r / num__100 ) ^ n * p ( num__1.1 ) ^ num__2 * num__5000 = num__1.21 * num__5000 = num__4840 the answer is b . <eor> b <eos> |
b |
percent__100.0__4840.0__ |
percent__100.0__4840.0__ |
| antonio works in a bakery . he made cookies that cost $ num__2 and made $ num__420 . how many customer did he have ? <o> a ) num__200 customers <o> b ) num__210 customers <o> c ) num__250 customers <o> d ) num__230 customers <o> e ) num__170 customers |
a cookie costs $ num__2 adding another one is $ num__4 . num__420 divided by num__4 is num__105 x num__2 is num__210 . he had num__210 customers . the correct answer is b . <eor> b <eos> |
b |
divide__420.0__4.0__ multiply__2.0__105.0__ round__210.0__ |
divide__420.0__4.0__ multiply__2.0__105.0__ round__210.0__ |
| gold is num__11 times as heavy as water and copper is num__5 times as heavy as water . in what ratio should these be mixed to get an alloy num__8 times as heavy as water ? <o> a ) num__3 : num__2 <o> b ) num__1 : num__1 <o> c ) num__3 : num__1 <o> d ) num__5 : num__2 <o> e ) num__4 : num__3 |
g = num__11 w c = num__5 w let num__1 gm of gold mixed with x gm of copper to get num__1 + x gm of the alloy num__1 gm gold + x gm copper = x + num__1 gm of alloy num__11 w + num__5 wx = x + num__1 * num__8 w num__11 + num__5 x = num__8 ( x + num__1 ) x = num__1.0 ratio of gold with copper = num__1 : num__1 = num__1 : num__1 answer is b <eor> b <eos> |
b |
reverse__1.0__ |
reverse__1.0__ |
| a man can reach certain place in num__50 hours . if he reduces his speed by num__0.1 th he goes num__450 km less in time . find his speed ? <o> a ) num__20 km / hr <o> b ) num__30 km / hr <o> c ) num__40 km / hr <o> d ) num__50 km / hr <o> e ) num__90 km / hr |
let the speed be x km / hr num__50 x - num__50 * num__0.9 * x = num__450 num__50 x - num__45 x = num__450 x = num__90 km / hr answer is e <eor> e <eos> |
e |
multiply__50.0__0.9__ round__90.0__ |
multiply__50.0__0.9__ round__90.0__ |
| five pieces of wood have an average length of num__128 cm and a median length of num__140 cm . what is the maximum possible length in cm of the shortest piece of wood ? <o> a ) a ) num__90 <o> b ) b ) num__100 <o> c ) c ) num__110 <o> d ) d ) num__130 <o> e ) e ) num__140 |
c . num__110 sum of all lengths of all num__5 pieces of wood = num__128 * num__5 = num__640 num__3 rd piece ( sorted in increasing length ) length = num__140 ( median ) for sum of first num__2 wood length to become max last two should be least . let num__4 th num__5 th wood also have length num__140 each . total of last num__3 = num__140 * num__3 = num__420 sum of first num__2 = num__640 - num__420 = num__220 each of these num__2 will have length num__110.0 = num__110 answer c <eor> c <eos> |
c |
multiply__128.0__5.0__ subtract__5.0__3.0__ multiply__140.0__3.0__ subtract__640.0__420.0__ divide__220.0__2.0__ |
multiply__128.0__5.0__ subtract__5.0__3.0__ multiply__140.0__3.0__ subtract__640.0__420.0__ subtract__220.0__110.0__ |
| a num__600 meter long train crosses a signal post in num__40 seconds . how long will it take to cross a num__18 kilometer long bridge at the same speed ? <o> a ) num__14 min <o> b ) num__20 min <o> c ) num__18 min <o> d ) num__19 min <o> e ) num__13 min |
s = num__15.0 = num__15 mps s = num__1200.0 = num__1200 sec = num__20 min . answer : b <eor> b <eos> |
b |
divide__600.0__40.0__ round__20.0__ |
divide__600.0__40.0__ round__20.0__ |
| if x and y are integers such that x ^ num__2 = y and xy = num__343 then x – y = ? <o> a ) - num__42 <o> b ) - num__20 <o> c ) - num__5 <o> d ) num__5 <o> e ) num__20 |
here x and y are integers . x ^ num__2 = y xy = num__343 . substitute x ^ num__2 = y in xy = > x ^ num__3 = num__343 . here x ^ num__3 is positive x is also positive . x = num__7 then y = num__49 . x - y = - num__42 so option a is correct <eor> a <eos> |
a |
divide__343.0__7.0__ subtract__49.0__7.0__ subtract__49.0__7.0__ |
divide__343.0__7.0__ subtract__49.0__7.0__ subtract__49.0__7.0__ |
| what is the sum of all the composite numbers up to num__18 ? <o> a ) num__228 <o> b ) num__112 <o> c ) num__80 <o> d ) num__96 <o> e ) num__100 |
explanation : num__4 + num__6 + num__8 + num__9 + num__10 + num__12 + num__14 + num__15 + num__16 + num__18 = num__112 answer : b <eor> b <eos> |
b |
subtract__18.0__8.0__ subtract__18.0__6.0__ subtract__18.0__4.0__ add__6.0__9.0__ add__4.0__12.0__ multiply__8.0__14.0__ multiply__8.0__14.0__ |
add__4.0__6.0__ add__4.0__8.0__ add__4.0__10.0__ add__6.0__9.0__ add__4.0__12.0__ multiply__8.0__14.0__ multiply__8.0__14.0__ |
| the average successfull search time for sequential search on n items is <o> a ) n / num__2 <o> b ) ( n - num__1 ) / num__2 <o> c ) ( n + num__1 ) / num__2 <o> d ) log ( n ) + num__1 <o> e ) ( n ) + num__1 |
to understand this think as there is an array and now you are searching for an element which is a [ num__0 ] element in that arrary . . . so for this you need to do num__1 comparison . now similarly for sequential searching suppose your are looking for an element which is present at num__2 nd position i . e a [ num__1 ] position of that array then you will get answer within num__2 comparisons . now similarly if you have n elements and your desired element is the nth element then you will have to do n comparisons . so we can write total search time will be : ( num__1 + num__2 + num__3 + . . . . . . . . + n ) average search time will be ( num__1 + num__2 + num__3 + . . . . . . . . + n ) / n now we know sum of n natural numbers i . e num__1 + num__2 + num__3 + . . . . . n = { n ( n + num__1 ) } / num__2 answer : c <eor> c <eos> |
c |
add__1.0__2.0__ reverse__1.0__ |
add__1.0__2.0__ reverse__1.0__ |
| a and b invest in a business in the ratio num__3 : num__2 . if num__5.0 of the total profit goes to charity and a ' s share is rs . num__855 the total profit is : <o> a ) rs . num__1425 <o> b ) rs . num__1500 <o> c ) rs . num__1537.50 <o> d ) rs . num__1576 <o> e ) rs . num__1643 |
let the total profit be rs . num__100 . after paying to charity a ' s share = rs . ( num__95 x num__0.6 ) = rs . num__57 . if a ' s share is rs . num__57 total profit = rs . num__100 . if a ' s share rs . num__855 total profit = ( num__100 x num__15.0 ) = num__1500 . the correct answer is b <eor> b <eos> |
b |
percent__100.0__1500.0__ |
percent__100.0__1500.0__ |
| in the parking lot there are num__50 vehicles num__30 of them are buses and the rest are cars . the color of num__30 vehicles is red of which num__18 are buses . how many cars can be found in the parking lot which are not colored red ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
the number of cars is num__50 - num__30 = num__20 . the number of red cars is num__30 - num__18 = num__12 . the number of cars which are not red is num__20 - num__12 = num__8 . the answer is e . <eor> e <eos> |
e |
subtract__50.0__30.0__ subtract__30.0__18.0__ subtract__20.0__12.0__ subtract__20.0__12.0__ |
subtract__50.0__30.0__ subtract__30.0__18.0__ subtract__20.0__12.0__ subtract__20.0__12.0__ |
| find the simple interest on rs . num__71200 at num__16 num__0.666666666667 % per annum for num__9 months . <o> a ) s . num__8500 <o> b ) s . num__8900 <o> c ) s . num__7500 <o> d ) s . num__7000 <o> e ) s . num__6500 |
p = rs . num__71200 r = num__16.6666666667 % p . a and t = num__0.75 years = num__0.75 years . s . i . = ( p * r * t ) / num__100 = rs . ( num__71200 * ( num__16.6666666667 ) * ( num__0.75 ) * ( num__0.01 ) ) = rs . num__8900 answer is b . <eor> b <eos> |
b |
percent__100.0__8900.0__ |
percent__100.0__8900.0__ |
| find the unit ' s digit in the product ( num__2467 ) ^ num__153 * ( num__341 ) ^ num__72 <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__7 <o> e ) num__9 |
unit digit of num__341 ^ num__72 will be num__1 . unit digit of ( num__2467 ) ^ num__153 will be same as that of num__7 ^ num__1 ( num__153 mod num__4 = num__1 ) answer num__7 answer : d <eor> d <eos> |
d |
multiply__1.0__7.0__ |
multiply__1.0__7.0__ |
| the ratio of two numbers is num__3 : num__4 and their h . c . f . is num__4 . their l . c . m . is : <o> a ) num__12 <o> b ) num__16 <o> c ) num__20 <o> d ) num__24 <o> e ) num__48 |
the ratio of two numbers is num__3 : num__4 and their h . c . f . is num__4 . their l . c . m . is : let the numbers be num__3 x and num__4 x . then their h . c . f . = x . so x = num__4 . so the numbers num__12 and num__16 . l . c . m . of num__12 and num__16 = num__48 . answer : e <eor> e <eos> |
e |
multiply__3.0__4.0__ add__4.0__12.0__ multiply__3.0__16.0__ multiply__3.0__16.0__ |
multiply__3.0__4.0__ add__4.0__12.0__ multiply__3.0__16.0__ multiply__3.0__16.0__ |
| the average age of num__39 students in a group is num__10 years . when teacher ’ s age is included to it the average increases by one . what is the teacher ’ s age in years ? <o> a ) num__50 <o> b ) num__55 <o> c ) num__60 <o> d ) can not be determined <o> e ) none of these |
age of the teacher = ( num__40 × num__11 – num__39 × num__10 ) years = num__50 years . answer a <eor> a <eos> |
a |
add__39.0__11.0__ add__39.0__11.0__ |
add__39.0__11.0__ add__39.0__11.0__ |
| rs . num__2500 is divided into two parts such that if one part be put out at num__5.0 simple interest and the other at num__6.0 the yearly annual income may be rs . num__135 . how much was lent at num__5.0 ? <o> a ) num__2333 <o> b ) num__2777 <o> c ) num__2688 <o> d ) num__1500 <o> e ) num__2871 |
( x * num__5 * num__1 ) / num__100 + [ ( num__2500 - x ) * num__6 * num__1 ] / num__100 = num__135 x = num__1500 answer : d <eor> d <eos> |
d |
percent__100.0__1500.0__ |
percent__100.0__1500.0__ |
| a man can row num__30 km downstream and num__20 km upstream in num__4 hours . he can row num__45 km downstream and num__40 km upstream in num__7 hours . find the speed of man in still water ? <o> a ) num__12.8 kmph <o> b ) num__12.7 kmph <o> c ) num__12.6 kmph <o> d ) num__12.5 kilometre per hour <o> e ) num__12.3 kmph |
let the speed of the man in still water be a kmph and let the speed of the stream be b kmph . now num__30 / ( a + b ) + num__20 / ( a - b ) = num__4 and num__45 / ( a + b ) + num__40 / ( a - b ) = num__7 solving the equation the speed of man in still water is num__12.5 kmph . answer : d <eor> d <eos> |
d |
round__12.5__ |
round__12.5__ |
| in a party every person shakes hands with every other person . if there were a total of num__136 handshakes in the party then what is the number of persons present in the party ? <o> a ) num__15 <o> b ) num__16 <o> c ) num__17 <o> d ) num__18 <o> e ) num__19 |
explanation : let the number of persons be n â ˆ ´ total handshakes = nc num__2 = num__136 n ( n - num__1 ) / num__2 = num__136 â ˆ ´ n = num__17 answer : c <eor> c <eos> |
c |
multiply__1.0__17.0__ |
divide__17.0__1.0__ |
| a certain list consists of num__11 different numbers . if n is in the list and n is num__5 times the average ( arithmetic mean ) of the other num__10 numbers in the list then n is what fraction of the sum of the num__11 numbers in the list ? <o> a ) num__0.333333333333 <o> b ) num__0.4 <o> c ) num__0.25 <o> d ) num__0.6 <o> e ) num__0.166666666667 |
series : a num__1 a num__2 . . . . a num__10 n sum of a num__1 + a num__2 + . . . + a num__10 = num__10 * x ( x = average ) so n = num__5 * x hence a num__1 + a num__2 + . . + a num__10 + n = num__15 x so the fraction asked = num__5 x / num__15 x = num__0.333333333333 answer is a <eor> a <eos> |
a |
subtract__11.0__10.0__ divide__10.0__5.0__ add__5.0__10.0__ divide__5.0__15.0__ divide__5.0__15.0__ |
subtract__11.0__10.0__ divide__10.0__5.0__ add__5.0__10.0__ divide__5.0__15.0__ divide__5.0__15.0__ |
| a clock gains num__5 minutes . in num__24 hours . it was set right at num__10 a . m . on monday . what will be the true time when the clock indicates num__10 : num__30 a . m . on the next sunday ? <o> a ) num__10 a . m <o> b ) num__11 a . m <o> c ) num__25 minutes past num__10 a . m . <o> d ) num__5 minutes to num__11 a . m . <o> e ) none of these |
time between num__10 a . m . on monday to num__10 : num__30 a . m . on sunday = num__144 num__1 ⁄ num__2 num__24 num__1 ⁄ num__2 hours of incorrect clock = num__24 hours of correct time . ∴ num__1441 ⁄ num__2 hours of incorrect clock = x hours of correct time . ∴ num__144 num__0.5 × num__1.0 num__0.5 = num__144 hoursi . e the true time is num__10 a . m . on sunday . answer a <eor> a <eos> |
a |
divide__10.0__5.0__ divide__5.0__10.0__ round__10.0__ |
divide__10.0__5.0__ divide__5.0__10.0__ round__10.0__ |
| if num__893 × num__78 = p which of the following is equal to num__893 × num__79 ? <o> a ) p + num__1 <o> b ) p + num__78 <o> c ) p + num__79 <o> d ) p + num__893 <o> e ) p + num__894 |
since num__893 × num__78 = p and num__79 = num__78 + num__1 we have num__893 × num__79 = num__893 × ( num__78 + num__1 ) = num__893 × num__78 + num__893 × num__1 = p + num__893 . the answer is ( d ) . <eor> d <eos> |
d |
subtract__79.0__78.0__ multiply__893.0__1.0__ |
subtract__79.0__78.0__ multiply__893.0__1.0__ |
| a train can travel num__50.0 faster than a car . both start from point a at the same time and reach point b num__75 kms away from a at the same time . on the way however the train lost about num__12.5 minutes while stopping at the stations . the speed of the car is : <o> a ) num__130 kmph <o> b ) num__110 kmph <o> c ) num__150 kmph <o> d ) num__120 kmph <o> e ) num__170 kmph |
d num__120 kmph let speed of the car be x kmph . then speed of the train = ( num__1.5 ) * x = ( num__1.5 ) * x kmph ( num__75 / x ) - ( num__75 / ( num__1.5 x ) ) = num__125 / ( num__10 * num__60 ) ( num__75 / x ) - ( num__50 / x ) = num__0.208333333333 x = num__25 x num__4.8 = num__120 kmph . <eor> d <eos> |
d |
divide__75.0__50.0__ add__50.0__75.0__ divide__125.0__12.5__ hour_to_min_conversion__ divide__12.5__60.0__ subtract__75.0__50.0__ divide__120.0__25.0__ round__120.0__ |
divide__75.0__50.0__ add__50.0__75.0__ divide__125.0__12.5__ hour_to_min_conversion__ divide__12.5__60.0__ subtract__75.0__50.0__ divide__120.0__25.0__ multiply__4.8__25.0__ |
| a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . num__51.84 the number of the member is the group is : <o> a ) num__57 <o> b ) num__67 <o> c ) num__72 <o> d ) num__87 <o> e ) num__97 |
explanation : money collected = ( num__51.84 x num__100 ) paise = num__5184 paise . ∴ number of members = √ ( num__5184 ) = num__72 . answer : c <eor> c <eos> |
c |
multiply__51.84__100.0__ divide__5184.0__72.0__ |
multiply__51.84__100.0__ divide__5184.0__72.0__ |
| convert the following unit : num__5 hectares in m  ² <o> a ) num__50000 m  ² <o> b ) num__40000 m  ² <o> c ) num__60000 m  ² <o> d ) num__70000 m  ² <o> e ) num__55000 m  ² |
num__5 hectares in m  ² num__1 hectare = num__10000 m  ² therefore num__5 hectares = num__5 à — num__10000 m  ² = num__50000 m  ² answer : option a <eor> a <eos> |
a |
multiply__5.0__10000.0__ multiply__5.0__10000.0__ |
multiply__5.0__10000.0__ multiply__5.0__10000.0__ |
| a train of num__30 carriages each of num__60 meters length when an engine also of num__60 meters length is running at a speed of num__60 kmph . in what time will the train cross a bridge num__1.5 km long ? <o> a ) num__4 <o> b ) num__3.3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__9 |
d = num__30 * num__60 + num__1500 = num__3300 m t = num__55.0 * num__3.6 = num__198 sec = num__3.3 mins answer : b <eor> b <eos> |
b |
divide__3300.0__60.0__ multiply__3.6__55.0__ divide__198.0__60.0__ round__3.3__ |
divide__3300.0__60.0__ multiply__3.6__55.0__ divide__198.0__60.0__ round__3.3__ |
| the population of a bacteria culture doubles every num__6 minutes . approximately how many minutes will it take for the population to grow from num__1000 to num__500000 bacteria <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__54 <o> e ) num__18 |
this one ' s easy . num__1000 * num__2 ^ t = num__500000 num__2 ^ t = num__500 now gauging since num__2 ^ num__8 = num__256 then num__2 ^ num__9 = num__512 so t = num__9 but be careful ' t ' is in time intervals of num__6 minutes so answer is num__9 * num__6 = num__54 minutes answer ( d ) <eor> d <eos> |
d |
divide__1000.0__2.0__ add__6.0__2.0__ multiply__256.0__2.0__ multiply__6.0__9.0__ round__54.0__ |
divide__1000.0__2.0__ add__6.0__2.0__ multiply__256.0__2.0__ multiply__6.0__9.0__ multiply__6.0__9.0__ |
| working alone mary can pave a driveway in num__3 hours and hillary can pave the same driveway in num__6 hours . when they work together mary thrives on teamwork so her rate increases by num__10.0 but hillary becomes distracted and her rate decreases by num__20.0 . if they both work together how many hours will it take to pave the driveway ? <o> a ) num__2 hours <o> b ) num__4 hours <o> c ) num__5 hours <o> d ) num__6 hours <o> e ) num__7 hours |
initial working rates : mary = num__0.333333333333 per hour hillary = num__0.166666666667 per hour rate when working together : mary = num__0.333333333333 + ( num__0.1 * num__0.333333333333 ) = num__0.375 per hour hillary = num__0.166666666667 - ( num__0.2 * num__0.166666666667 ) = num__0.133333333333 per hour together they work num__0.375 + num__0.133333333333 = num__0.5 per hour so they will need num__2 hours to complete the driveway . the correct answer is a . <eor> a <eos> |
a |
percent__10.0__20.0__ percent__10.0__20.0__ |
percent__10.0__20.0__ percent__10.0__20.0__ |
| two pipes a and b can fill a cistern in num__12 and num__15 minutes respectively . both are opened together but after num__3 minutes a is turned off . after how much more time will the cistern be filled ? <o> a ) num__8 num__0.125 <o> b ) num__8 num__0.111111111111 <o> c ) num__8 num__0.25 <o> d ) num__8 num__1.0 <o> e ) num__8 num__0.2 |
num__0.25 + ( num__3 + x ) / num__15 = num__1 x = num__8 num__0.25 answer : c <eor> c <eos> |
c |
divide__3.0__12.0__ round__8.0__ |
divide__3.0__12.0__ divide__8.0__1.0__ |
| four horses are tethered at four corners of a square plot of side num__63 metres so that they just can not reach one another . the area left ungrazed is : <o> a ) num__675.5 m num__2 <o> b ) num__780.6 m num__2 <o> c ) num__785.8 m num__2 <o> d ) num__850.5 m num__2 <o> e ) num__950.5 m num__2 |
required area = ( num__63 * num__63 – num__4 * num__0.25 * num__3.14285714286 * num__31.5 * num__31.5 ) = num__850.5 m num__2 answer : d <eor> d <eos> |
d |
triangle_area__2.0__850.5__ |
volume_rectangular_prism__0.25__4.0__850.5__ |
| out of four numbers the average of first three is num__16 and that of the last three is num__15 . if the last number is num__23 the first number is : <o> a ) num__22 <o> b ) num__21 <o> c ) num__77 <o> d ) num__99 <o> e ) num__26 |
explanation : let the numbers be a b c d given a + b + c = num__48 b + c + d = num__45 now d = num__23 thus b + c + num__23 = num__45 ⇒ b + c = num__22 putting the value of b + c in a + b + c = num__48 a + num__22 = num__48 ⇒ a = num__26 answer : e <eor> e <eos> |
e |
subtract__45.0__23.0__ subtract__48.0__22.0__ subtract__48.0__22.0__ |
subtract__45.0__23.0__ subtract__48.0__22.0__ subtract__48.0__22.0__ |
| a man can row upstream at num__8 kmph and downstream at num__13 kmph . the speed of the stream is ? <o> a ) num__3 kmph <o> b ) num__2.5 kmph <o> c ) num__3.5 kmph <o> d ) num__4.3 kmph <o> e ) num__5 kmph |
speed of stream = num__0.5 ( num__13 - num__8 ) = num__2.5 kmph answer is b <eor> b <eos> |
b |
round__2.5__ |
round__2.5__ |
| fresh fruit contains num__68.0 water and dry fruit contains num__20.0 water . how much dry fruit can be obtained from num__100 kg of fresh fruits ? <o> a ) num__10 kg <o> b ) num__20 kg <o> c ) num__30 kg <o> d ) num__40 kg <o> e ) num__50 kg |
the fruit content in both the fresh fruit and dry fruit is the same . given fresh fruit has num__68.0 water . so remaining num__32.0 is fruit content . weight of fresh fruits is num__100 kg dry fruit has num__20.0 water . so remaining num__80.0 is fruit content . let weight if dry fruit be y kg . fruit % in fresh fruit = fruit % in dry fruit ( num__0.32 ) * num__100 = ( num__0.8 ) * y we get y = num__40 kg answer : d <eor> d <eos> |
d |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| find the fourth proportion to num__3 num__124 <o> a ) num__18 <o> b ) num__16 <o> c ) num__10 <o> d ) num__12 <o> e ) num__14 |
explanation : num__3 : num__12 : : num__4 : x = > num__0.25 = num__4 / x = > x = num__16.0 = > x = num__16 option b <eor> b <eos> |
b |
divide__12.0__3.0__ reverse__4.0__ divide__4.0__0.25__ divide__4.0__0.25__ |
divide__12.0__3.0__ reverse__4.0__ divide__4.0__0.25__ divide__4.0__0.25__ |
| charles walks over a railway - bridge . at the moment that he is just ten meters away from the middle of the bridge he hears a train coming from behind . at that moment the train which travels at a speed of num__90 km / h is exactly as far away from the bridge as the bridge measures in length . without hesitation charles rushes straight towards the train to get off the bridge . in this way he misses the train by just four meters ! if charles would however have rushed exactly as fast in the other direction the train would have hit him eight meters before the end of the bridge . what is the length of the railway - bridge ? <o> a ) num__40 meters . <o> b ) num__41 meters . <o> c ) num__42 meters . <o> d ) num__43 meters . <o> e ) num__44 meters . |
solution : let the length of the bridge be x meters . running towards the train charles covers num__0.5 x - num__10 meters in the time that the train travels x - num__4 meters . running away from the train charles covers num__0.5 x + num__2 meters in the time that the train travels num__2 x - num__8 meters . because their speeds are constant the following holds : ( num__0.5 x - num__10 ) / ( x - num__4 ) = ( num__0.5 x + num__2 ) / ( num__2 x - num__8 ) which can be rewritten to num__0.5 x num__2 - num__24 x + num__88 = num__0 using the abc formula we find that x = num__44 so the railway - bridge has a length of num__44 meters . answer e <eor> e <eos> |
e |
reverse__0.5__ multiply__2.0__4.0__ subtract__90.0__2.0__ round_down__0.5__ multiply__0.5__88.0__ multiply__0.5__88.0__ |
reverse__0.5__ divide__4.0__0.5__ subtract__90.0__2.0__ round_down__0.5__ divide__88.0__2.0__ divide__88.0__2.0__ |
| c and d represents num__2 train stations which are num__320 km apart . if num__2 trains leave each station and travel towards each other at what time will they meet each other given that the first train left the station at num__11 a . m . and travels at a speed of num__85 kmph while the second train left the station at num__12 p . m . and travels at a speed of num__75 kmph . <o> a ) num__1.00 p . m <o> b ) num__2.00 p . m <o> c ) num__12.00 p . m <o> d ) num__12.30 p . m <o> e ) num__11.00 p . m |
suppose they meet x hours after num__11 a . m . distance moved by first train in x hours + distance moved by num__2 nd train in x - num__1 hrs = num__320 num__85 x + num__75 ( x - num__1 ) = num__320 num__85 x + num__75 x - num__75 = num__320 num__160 x = num__320 x = num__2 so they meet num__2 hr after num__10 a . m . which is at num__12.00 p . m . answer is c <eor> c <eos> |
c |
subtract__12.0__11.0__ divide__320.0__2.0__ subtract__11.0__1.0__ round__12.0__ |
subtract__12.0__11.0__ add__85.0__75.0__ subtract__11.0__1.0__ add__2.0__10.0__ |
| if x is the median of the set { num__4.5 num__3.66666666667 x num__3.11111111111 num__4.2 } x could be <o> a ) num__3.2 <o> b ) num__3.4 <o> c ) num__0.571428571429 <o> d ) num__4.28571428571 <o> e ) num__4 |
the median is the middle number once all the numbers are arranged in increasing / decreasing order . we see that num__3.66666666667 = num__3 . something num__3.11111111111 = num__3 . something num__4.2 = num__4 . something num__4.5 = num__4 . something so x should greater than the smallest two numbers and smaller than the greatest two numbers . we can see that x = num__4 is possible . ( first look at the simplest option or the middle option since options are usually arranged in increasing / decreasing order ) answer ( e ) <eor> e <eos> |
e |
round_down__3.6667__ round_down__4.5__ round_down__4.5__ |
round_down__3.6667__ round_down__4.5__ round_down__4.5__ |
| a library has an average of num__175 visitors on sundays and num__75 on other days . the average number of visitors per day in a month of num__30 days beginning with a sunday is : <o> a ) num__80 <o> b ) num__90 <o> c ) num__100 <o> d ) num__110 <o> e ) num__120 |
explanation : since the month begins with a sunday so there will be five sundays in the month required average = ( num__165 * num__5 + num__75 * num__25 ) / num__30 = num__90.0 = num__90 answer : b ) num__90 <eor> b <eos> |
b |
subtract__30.0__5.0__ subtract__165.0__75.0__ subtract__165.0__75.0__ |
subtract__30.0__5.0__ subtract__165.0__75.0__ subtract__165.0__75.0__ |
| tickets for all but num__100 seats in a num__10000 - seat stadium were sold . of the tickets sold num__15.0 were sold at half price and the remaining tickets were sold at the full price of $ num__2 . what was the total revenue from ticket sales ? <o> a ) $ num__15840 <o> b ) $ num__17820 <o> c ) $ num__18000 <o> d ) $ num__18315 <o> e ) $ num__21 |
780 |
num__10000 seats - - > full price : half price = num__8500 : num__1500 price when all seats are filled = num__17000 + num__1500 = num__18500 num__100 seats are unsold - - > loss due to unfilled seats = num__15 + num__2 * num__85 = num__185 revenue = num__18500 - num__185 = num__18315 answer : d <eor> d <eos> |
d |
d |
| suppose you have access to a large vat of distilled water several gallons large . you have two precise measuring pipettes one to measure exactly num__0.25 of an ounce and one to measure exactly num__0.2 of an ounce . you can pour precisely measured amounts into a beaker which initially is empty . you can use either pipette to remove distilled water from the vat or from the beaker and use either pipette to dispense water into either of those receptacles but you can not use either pipette to take any quantity of distilled water other than the amount for which it is designed . which of the following represents in ounces a precise amount of distilled water you can transfer from the vat to the beaker ? i . num__0.166666666667 ii . num__0.142857142857 iii . num__0.0833333333333 <o> a ) i only <o> b ) num__0.05 and num__0.45 <o> c ) i and iii only <o> d ) ii and iii only <o> e ) i ii and iii |
num__0.25 - num__0.2 = num__0.05 num__0.25 + num__0.2 = num__0.45 answer : b <eor> b <eos> |
b |
subtract__0.25__0.2__ add__0.25__0.2__ subtract__0.25__0.2__ |
subtract__0.25__0.2__ add__0.25__0.2__ subtract__0.25__0.2__ |
| a train is running at a speed of num__40 km / hr and it crosses a post in num__16.2 seconds . what is the length of the train ? <o> a ) num__190 metres <o> b ) num__160 metres <o> c ) num__200 metres <o> d ) num__120 metres <o> e ) num__250 metres |
speed of the train v = num__40 km / hr = num__11.1111111111 m / s = num__11.1111111111 m / s time taken to cross t = num__16.2 s distance covered d = vt = ( num__11.1111111111 ) Ã — num__16.2 = num__160 m distance covered is equal to the length of the train = num__160 m correct answer is num__160 metres b <eor> b <eos> |
b |
round__160.0__ |
round__160.0__ |
| the two trains of lengths num__400 m num__600 m respectively running at same directions . the faster train can cross the slower train in num__180 sec the speed of the slower train is num__48 km . then find the speed of the faster train ? <o> a ) num__22 <o> b ) num__68 <o> c ) num__26 <o> d ) num__28 <o> e ) num__12 |
length of the two trains = num__600 m + num__400 m speed of the first train = x speed of the second train = num__48 kmph num__1000 / x - num__48 = num__180 num__1000 / x - num__48 * num__0.277777777778 = num__180 num__50 = num__9 x - num__120 x = num__68 kmph answer : b <eor> b <eos> |
b |
add__400.0__600.0__ round__68.0__ |
add__400.0__600.0__ round__68.0__ |
| what sum of money put at c . i amounts in num__2 years to rs . num__7000 and in num__3 years to rs . num__9261 ? <o> a ) num__4000 <o> b ) num__8877 <o> c ) num__2877 <o> d ) num__2678 <o> e ) num__1011 |
num__7000 - - - - num__2261 num__100 - - - - ? = > num__32.3 x * num__13.23 * num__13.23 = num__7000 x * num__1.75 = num__7000 x = num__7000 / num__1.75 = > num__3999.25 answer : a <eor> a <eos> |
a |
subtract__9261.0__7000.0__ divide__7000.0__1.75__ |
subtract__9261.0__7000.0__ divide__7000.0__1.75__ |
| three dice are thrown together find the probability of getting a total of at least num__5 ? <o> a ) num__0.285714285714 <o> b ) num__0.6 <o> c ) num__0.980769230769 <o> d ) num__0.5 <o> e ) num__0.981481481481 |
three different dice are thrown at the same time . therefore total number of possible outcomes will be num__63 = ( num__6 × num__6 × num__6 ) = num__216 . number of events of getting a total of less than num__5 = num__4 i . e . ( num__1 num__1 num__1 ) ( num__1 num__1 num__2 ) ( num__1 num__2 num__1 ) and ( num__2 num__1 num__1 ) . therefore probability of getting a total of less than num__5 p ( e ) = number of favorable outcomes / total number of possible outcome = num__0.0185185185185 = num__0.0185185185185 therefore probability of getting a total of at least num__5 = num__1 - p ( getting a total of less than num__5 ) = num__1 - num__0.0185185185185 = ( num__54 - num__1 ) / num__54 = num__0.981481481481 <eor> e <eos> |
e |
die_space__ coin_space__ negate_prob__0.0185__ negate_prob__0.0185__ |
die_space__ coin_space__ negate_prob__0.0185__ negate_prob__0.0185__ |
| the average ( arithmetic mean ) of eight numbers is num__44.1 . if the sum of half of these numbers is num__158.4 what is the average of the other half ? <o> a ) num__12.8 <o> b ) num__24.2 <o> c ) num__48.6 <o> d ) num__72.1 <o> e ) num__96.8 |
arithmetic mean = sum / total numbers sum = num__44.1 * num__8 = num__352.8 sum of half of these numbers is num__158.4 . so num__4 numbers sum is num__158.4 . rest num__4 numbers sum = num__352.8 - num__158.4 = num__194.4 arithmetic mean of the num__4 nos = num__194.4 / num__4 = num__48.6 hence c is the answer . <eor> c <eos> |
c |
multiply__44.1__8.0__ subtract__352.8__158.4__ divide__194.4__4.0__ divide__194.4__4.0__ |
multiply__44.1__8.0__ subtract__352.8__158.4__ divide__194.4__4.0__ divide__194.4__4.0__ |
| the price of the jewel passing through three hands rises on the whole by num__80.0 . if the first and the second sellers num__20.0 and num__25.0 profit respectively find the percentage profit earned by the third seller . <o> a ) num__20 <o> b ) num__10 <o> c ) num__15 <o> d ) num__25 <o> e ) num__30 |
let the original price of the jewel be $ p and let the profit earned by the third seller be x % then ( num__100 + x ) % of num__125.0 of num__120.0 of p = num__180.0 of p ( ( num__100 + x ) / num__100 * num__1.25 * num__1.2 * p ) = ( num__1.8 * p ) = = > ( num__100 + x ) = ( num__180 * num__100 * num__100 ) / ( num__125 * num__120 ) = num__120 = > x = num__20.0 answer a <eor> a <eos> |
a |
percent__80.0__25.0__ |
percent__80.0__25.0__ |
| a man buys rs . num__44 shares paying num__9.0 dividend . the man wants to have an interest of num__12.0 on his money . the market value of each share is : <o> a ) s . num__12 <o> b ) s . num__15 <o> c ) s . num__18 <o> d ) s . num__21 <o> e ) s . num__33 |
dividend on rs . num__44 = rs . num__0.09 x num__44 = rs . num__3.96 . rs . num__12 is an income on rs . num__100 . rs . num__3.96 is an income on rs . num__8.33333333333 x num__3.96 = rs . num__33 . answer : option e <eor> e <eos> |
e |
percent__44.0__9.0__ percent__33.0__100.0__ |
percent__44.0__9.0__ percent__33.0__100.0__ |
| ron walks to a viewpoint and returns to the starting point by his car and thus takes a total time of num__6 hours num__45 minutes . he would have gained num__2 hours by driving both ways . how long r would it have taken for him to walk both ways . <o> a ) num__8 h num__45 min <o> b ) num__7 h num__45 min <o> c ) num__6 h num__45 min <o> d ) num__5 h num__30 min <o> e ) none of these |
num__1 . walking to to a viewpoint + driving back = num__6 hours num__45 minutes num__2 . driving to a viewpoint + driving back = num__6 hours num__45 minutes - num__2 hours = num__4 hours num__45 minutes thereforeone way driving = num__4 hours num__45 minutes / num__2 = num__2 hours num__22.5 minutes . num__3 . from num__1 . one way driving = num__6 hours num__45 minutes - num__2 hours num__22.5 minutes = num__4 hours num__22.5 minutes . num__4 . walking to to a viewpoint + walking back r = num__4 hours num__22.5 minutes + num__4 hours num__22.5 minutes = num__8 hours num__45 minutes . answer : a . <eor> a <eos> |
a |
subtract__6.0__2.0__ divide__45.0__2.0__ divide__6.0__2.0__ add__6.0__2.0__ round__8.0__ |
subtract__6.0__2.0__ divide__45.0__2.0__ divide__6.0__2.0__ add__6.0__2.0__ add__6.0__2.0__ |
| a work can be finished in num__14 days by thirty women . the same work can be finished in fifteen days by forteen men . the ratio between the capacity of a man and a woman is <o> a ) num__3 : num__4 <o> b ) num__4 : num__3 <o> c ) num__2 : num__1 <o> d ) num__3 : num__2 <o> e ) num__4 : num__5 |
work done by num__20 women in num__1 day = num__0.0714285714286 work done by num__1 woman in num__1 day = num__1 / ( num__14 Ã — num__30 ) work done by num__14 men in num__1 day = num__0.0666666666667 work done by num__1 man in num__1 day = num__1 / ( num__15 Ã — num__14 ) ratio of the capacity of a man and woman = num__1 / ( num__15 Ã — num__14 ) : num__1 / ( num__14 Ã — num__30 ) = num__0.0666666666667 : num__0.0333333333333 = num__0.333333333333 : num__0.166666666667 = num__1.0 = num__0.5 = num__2 : num__1 answer is c <eor> c <eos> |
c |
divide__1.0__14.0__ add__14.0__1.0__ divide__1.0__30.0__ add__0.3333__0.1667__ divide__1.0__0.5__ round__2.0__ |
divide__1.0__14.0__ add__14.0__1.0__ divide__1.0__30.0__ divide__15.0__30.0__ divide__1.0__0.5__ divide__1.0__0.5__ |
| a train moves fast a telegraph post and a bridge num__264 m long in num__8 sec and num__20 sec respectively . what is the speed of the train ? <o> a ) num__0.833333333333 km / hr <o> b ) num__25 km / hr <o> c ) num__79.2 km / hr <o> d ) num__15 km / hr <o> e ) num__38 km / hr |
explanation : let the length of the train be x m and its speed be y m / sec . then x / y = num__8 = > x = num__8 y ( x + num__264 ) / num__20 = y y = num__22 speed = num__22 m / sec = num__22 * num__3.6 = num__79.2 km / hr answer : c <eor> c <eos> |
c |
multiply__3.6__22.0__ round__79.2__ |
multiply__3.6__22.0__ multiply__3.6__22.0__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__228 m <o> b ) num__240 m <o> c ) num__227 m <o> d ) num__167 m <o> e ) num__186 m |
speed = ( num__54 * num__0.277777777778 ) m / sec = num__15 m / sec . length of the train = ( num__15 x num__20 ) m = num__300 m . let the length of the platform be x meters . then ( x + num__300 ) / num__36 = num__15 = = > x + num__300 = num__540 = = > x = num__240 m . answer : b <eor> b <eos> |
b |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
| if raj shares all of his amount equally among his five friends then each of his friend ' s amount increased by $ num__10 then raj had how much ? <o> a ) $ num__10 <o> b ) $ num__20 <o> c ) $ num__30 <o> d ) $ num__40 <o> e ) $ num__50 |
five friends amount increased by $ num__10 and raj shared equally so amount = num__5 * num__10 = num__50 answer : e <eor> e <eos> |
e |
multiply__10.0__5.0__ multiply__10.0__5.0__ |
multiply__10.0__5.0__ multiply__10.0__5.0__ |
| the public library has exactly num__2 floors . it has num__0.75 as many shelves on its num__2 nd floor as it does on its num__1 st . if the second floor has num__0.833333333333 as many books per shelf as the first floor what fraction of the library ' s books are on the first floor ? <o> a ) num__0.615384615385 <o> b ) num__0.357142857143 <o> c ) num__0.555555555556 <o> d ) num__0.642857142857 <o> e ) num__0.833333333333 |
let x be the no of shelves and y be the no of books per shelf on first floor . now no of shelves on num__2 nd floor = ( num__0.75 ) * x no of books per shelf on num__2 nd floor = ( num__0.833333333333 ) * y so total no books on first floor = xy and total no of books on num__2 nd floor = ( num__0.75 ) * x * ( num__0.833333333333 ) * y = ( num__0.625 ) * xy fraction of library books on first floor = ( xy ) / ( xy + ( num__0.625 ) * xy ) = num__1 / ( num__1 + ( num__0.625 ) ) = num__0.615384615385 so a <eor> a <eos> |
a |
multiply__0.75__0.8333__ multiply__1.0__0.6154__ |
multiply__0.75__0.8333__ multiply__1.0__0.6154__ |
| the population of a bacteria colony doubles every day . if it was started num__6 days ago with num__2 bacteria and each bacteria lives for num__12 days how large is the colony today ? <o> a ) num__64 <o> b ) num__128 <o> c ) num__256 <o> d ) num__512 <o> e ) num__1024 |
num__2 ^ num__6 ( num__2 ) = num__2 ^ num__7 = num__128 the answer is b . <eor> b <eos> |
b |
round__128.0__ |
round__128.0__ |
| square a has an area of num__25 square centimeters . square b has a perimeter of num__12 centimeters . if square b is placed within square a and a random point is chosen within square a what is the probability the point is not within square b ? <o> a ) num__0.36 <o> b ) num__0.2 <o> c ) num__0.64 <o> d ) num__0.6 <o> e ) num__0.24 |
i guess it ' s mean that square b is placed within square aentirely . since the perimeter of b is num__12 then its side is num__3.0 = num__3 and the area is num__3 ^ num__2 = num__9 empty space between the squares is num__25 - num__9 = num__16 square centimeters so if a random point is in this area then it wo n ' t be within square b : p = favorable / total = num__0.64 . answer : c <eor> c <eos> |
c |
power__3.0__2.0__ triangle_area__2.0__0.64__ |
power__3.0__2.0__ triangle_area__2.0__0.64__ |
| if num__2 x + y = num__7 and x + num__2 y = num__5 then num__4 xy / num__3 = ? <o> a ) a ) num__1 <o> b ) b ) num__2 <o> c ) c ) num__3.4 <o> d ) d ) num__3.6 <o> e ) e ) num__4 |
num__2 * ( x + num__2 y = num__5 ) equals num__2 x + num__4 y = num__10 num__2 x + num__4 y = num__10 - num__2 x + y = num__7 = num__3 y = num__3 therefore y = num__1 plug and solve . . . num__2 x + num__1 = num__7 num__2 x = num__6 x = num__3 ( num__4 * num__3 * num__1 ) / num__3 = num__4.0 = num__4 e <eor> e <eos> |
e |
multiply__2.0__5.0__ subtract__5.0__4.0__ add__2.0__4.0__ subtract__7.0__3.0__ |
multiply__2.0__5.0__ subtract__5.0__4.0__ add__2.0__4.0__ subtract__7.0__3.0__ |
| num__10 men can complete a piece of work in num__15 days and num__15 women can complete the same work in num__12 days . if all the num__10 men and num__15 women work together in how many days will the work get completed ? <o> a ) num__6 <o> b ) num__6 num__1 ⁄ num__3 <o> c ) num__6 num__2 ⁄ num__3 <o> d ) num__7 num__2 ⁄ num__3 <o> e ) none of these |
num__10 men ’ s num__1 day ’ s work = num__1 ⁄ num__15 num__15 women ’ s num__1 day ’ s work = num__1 ⁄ num__12 ( num__10 men + num__15 women ) ’ s num__1 day ’ s work = ( num__1 ⁄ num__15 + num__1 ⁄ num__12 ) = num__9 ⁄ num__60 = num__3 ⁄ num__20 ∴ num__10 men and num__15 women will complete the work in num__20 ⁄ num__3 = num__62 ⁄ num__3 days . answer c <eor> c <eos> |
c |
subtract__10.0__1.0__ hour_to_min_conversion__ subtract__15.0__12.0__ divide__60.0__3.0__ subtract__15.0__9.0__ |
subtract__10.0__1.0__ hour_to_min_conversion__ subtract__15.0__12.0__ divide__60.0__3.0__ subtract__15.0__9.0__ |
| rohan spends num__40.0 of his salary on food num__20.0 on house rent num__10.0 on entertainment and num__10.0 on conveyance . if his savings at the end of a month are rs . num__1500 . then his monthly salary is <o> a ) rs . num__6000 <o> b ) rs . num__7500 <o> c ) rs . num__8000 <o> d ) rs . num__10000 <o> e ) none |
sol . saving = [ num__100 - ( num__40 + num__20 + num__10 + num__10 ] % = num__20.0 . let the monthly salary be rs . x . then num__20.0 of x = num__1500 ⇔ num__0.2 x = num__1500 ⇔ x = num__1500 × num__5 = num__7500 . answer b <eor> b <eos> |
b |
percent__100.0__7500.0__ |
percent__100.0__7500.0__ |
| ( num__4300631 ) - ? = num__2535618 <o> a ) num__1865113 <o> b ) num__1775123 <o> c ) num__1765013 <o> d ) num__1675123 <o> e ) none of them |
let num__4300631 - x = num__2535618 then x = num__4300631 - num__2535618 = num__1765013 answer is c <eor> c <eos> |
c |
subtract__4300631.0__2535618.0__ subtract__4300631.0__2535618.0__ |
subtract__4300631.0__2535618.0__ subtract__4300631.0__2535618.0__ |
| krishna ' s monthly expenditure was decreased by num__50.0 and subsequently increased by num__50.0 . how much percent does he loss ? <o> a ) num__25.0 <o> b ) num__50.0 <o> c ) num__75.0 <o> d ) num__85.0 <o> e ) none of these |
explanation : let the original expenditure = rs . num__100 . new final expenditure = num__150.0 of ( num__50.0 of rs . num__100 ) = rs . ( num__0.0149253731343 / num__100100 ) = rs . num__75 . decrease = num__100 - num__75 = num__25.0 answer : a <eor> a <eos> |
a |
add__50.0__100.0__ subtract__75.0__50.0__ subtract__50.0__25.0__ |
add__50.0__100.0__ subtract__75.0__50.0__ subtract__50.0__25.0__ |
| in a school with num__5 classes each class has num__2 students less then the previous class . how many students are there in the largest class if the total number of students at school is num__95 ? <o> a ) num__17 <o> b ) num__19 <o> c ) num__21 <o> d ) num__23 <o> e ) num__25 |
total classes = num__5 total students = num__95 . average = num__19 if the classes had average number of students : num__19 num__19 num__19 num__19 num__19 given case = num__23 num__21 num__19 num__17 num__15 hence number of students in the largest class = num__23 correct option : d <eor> d <eos> |
d |
divide__95.0__5.0__ add__2.0__19.0__ subtract__19.0__2.0__ subtract__17.0__2.0__ add__2.0__21.0__ |
divide__95.0__5.0__ add__2.0__19.0__ subtract__19.0__2.0__ subtract__17.0__2.0__ add__2.0__21.0__ |
| a lent rs . num__5000 to b for num__2 years and rs . num__3000 to c for num__4 years on simple interest at the same rate of interest and received rs . num__2200 in all from both of them as interest . the rate of interest per annum is ? <o> a ) num__16.0 <o> b ) num__12.0 <o> c ) num__74.0 <o> d ) num__10.0 <o> e ) num__45 % |
let the rate be r % p . a . then ( num__5000 * r * num__2 ) / num__100 + ( num__3000 * r * num__4 ) / num__100 = num__2200 num__100 r + num__120 r = num__2200 r = num__10.0 answer : d <eor> d <eos> |
d |
percent__2.0__5000.0__ percent__4.0__3000.0__ percent__100.0__10.0__ |
percent__2.0__5000.0__ percent__4.0__3000.0__ percent__100.0__10.0__ |
| a merchant gains or loses in a bargain a certain sum . in a second bargain he gains num__280 dollars and in a third loses num__20 . in the end he finds he has gained num__120 dollars by the three together . how much did he gain or lose bv the first ? <o> a ) num__80 <o> b ) - num__140 <o> c ) num__140 <o> d ) num__120 <o> e ) none |
in this sum as the profit and loss are opposite in their nature they must be distinguished by contrary signs . if the profit is marked + the loss must be - . let x = the sum required . then according to the statement x + num__280 - num__20 = num__120 and x = - num__140 . answer b <eor> b <eos> |
b |
add__20.0__120.0__ subtract__280.0__140.0__ |
add__20.0__120.0__ subtract__280.0__140.0__ |
| set j consists of the terms { a b c d e } where e > d > c > b > a > num__3 . which of the following operations would decrease the standard deviation of set j ? <o> a ) multiply each term by e / d <o> b ) divide each term by b / c <o> c ) multiply each term by − num__0.5 <o> d ) divide each term by d / e <o> e ) multiply each term by c / e |
concept : standard deviation is defined as average deviation of terms in the set from the mean value of the set . i . e . num__1 ) it depends on the separation between the successive terms of the set num__2 ) if a constant value is added / subtracted in every terms of set then the separation between successive terms does not change hence s . d . remains constant num__3 ) if a constant value is multiplied in every terms then the separation between succesive terms gets multiplied by the constant hence s . d . remains gets multiplied by same number c <eor> c <eos> |
c |
subtract__3.0__1.0__ reverse__2.0__ |
subtract__3.0__1.0__ reverse__2.0__ |
| two trains are running at num__40 km / hr and num__20 km / hr respectively in the same direction . fast train completely passes a man sitting in the slower train in num__9 sec . what is the length of the fast train ? <o> a ) num__27 num__1.16666666667 <o> b ) num__27 <o> c ) num__27 num__3.5 <o> d ) num__50 <o> e ) num__27 num__7.0 |
relative speed = ( num__40 - num__20 ) = num__20 km / hr . = num__20 * num__0.277777777778 = num__5.55555555556 m / sec . length of faster train = num__5.55555555556 * num__9 = num__50 m answer : d <eor> d <eos> |
d |
round__50.0__ |
round__50.0__ |
| to apply for the position of photographer at a local magazine a photographer needs to include three or four photos in an envelope accompanying the application . if the photographer has pre - selected seven photos representative of her work how many choices does she have to provide the photos for the magazine ? <o> a ) num__54 <o> b ) num__60 <o> c ) num__64 <o> d ) num__70 <o> e ) num__75 |
num__7 c num__3 + num__7 c num__4 = num__35 + num__35 = num__70 the answer is d . <eor> d <eos> |
d |
subtract__7.0__3.0__ round__70.0__ |
subtract__7.0__3.0__ round__70.0__ |
| a can do a work in num__7 days and b in num__12 days . if they work on it together then in how many days required to complete the work ? <o> a ) num__4.36842105263 <o> b ) num__4.42105263158 <o> c ) num__4.26315789474 <o> d ) num__4.05263157895 <o> e ) num__4.57894736842 |
person ( a ) ( b ) ( a + b ) time - ( num__7 ) ( num__12 ) ( num__4.42105263158 ) rate - ( num__12 ) ( num__7 ) ( num__19 ) work - ( num__84 ) ( num__84 ) ( num__84 ) therefore a + b requires ( num__4.42105263158 ) days to complete entire work = num__4.42105263158 answer is b <eor> b <eos> |
b |
add__7.0__12.0__ multiply__7.0__12.0__ divide__84.0__19.0__ |
add__7.0__12.0__ multiply__7.0__12.0__ divide__84.0__19.0__ |
| in a certain game each player scores either num__2 or num__5 points . if n players score num__2 points and m players score num__5 points and the total number of points scored is num__55 what is the least possible difference between n and m . <o> a ) a . num__1 <o> b ) b . num__3 <o> c ) c . num__4 <o> d ) d . num__7 <o> e ) e . num__9 |
num__2 n + num__5 m = num__55 try with n = num__5 and m = num__8 the equation will be satisfied num__2 ( num__5 ) + num__5 ( num__9 ) = num__55 so difference between m and n is num__4 answer c <eor> c <eos> |
c |
divide__8.0__2.0__ divide__8.0__2.0__ |
divide__8.0__2.0__ divide__8.0__2.0__ |
| hendrix works num__20 days a month at d dollars per day for m months out of the year . which of the following represents his num__5 year pay ? <o> a ) m / ( num__20 d * num__5 ) <o> b ) num__100 md <o> c ) num__50 md / num__6 <o> d ) num__20 d / num__5 m <o> e ) num__20 d / num__5 |
total income of amber in a year = d * num__20 * m dollars . now num__5 year income = total * num__5 = d * num__20 * m * num__5 = num__100 md answer is b <eor> b <eos> |
b |
multiply__20.0__5.0__ round__100.0__ |
multiply__20.0__5.0__ multiply__20.0__5.0__ |
| a plane was originally flying at an altitude of x feet when it ascended num__2000 feet and then descended num__5000 feet . if the plane ' s altitude after these two changes was num__0.5 its original altitude then the solution of which of the following equations gives the plane ' s original altitude in feet ? <o> a ) x + num__2000 = num__0.333333333333 * ( x - num__3000 ) <o> b ) num__0.333333333333 * ( x - num__3000 ) = x <o> c ) x - num__3000 = num__0.5 * x <o> d ) x - num__7000 = num__0.333333333333 * x <o> e ) x - num__3000 = num__0.333333333333 * x |
plane ' s original altitude = x plane ' s new altitude after ascending num__2000 ft = x + num__2000 plane ' s new altitude after descending num__5000 ft from previous altitude = x + num__2000 - num__5000 = x - num__3000 so after two changes plane is at num__0.5 its original altitude = > x - num__3000 = x / num__2 answer ( c ) <eor> c <eos> |
c |
subtract__5000.0__2000.0__ reverse__0.5__ subtract__5000.0__2000.0__ |
subtract__5000.0__2000.0__ reverse__0.5__ subtract__5000.0__2000.0__ |
| if a light flashes every num__6 seconds how many times will it flash in num__0.4 of an hour ? <o> a ) num__181 <o> b ) num__211 <o> c ) num__241 <o> d ) num__271 <o> e ) num__301 |
in num__0.4 of an hour there are num__24 * num__60 = num__1440 seconds the number of num__6 - second intervals = num__240.0 = num__240 after the first flash there will be num__240 more flashes for a total of num__241 . the answer is c . <eor> c <eos> |
c |
hour_to_min_conversion__ multiply__24.0__60.0__ divide__1440.0__6.0__ round__241.0__ |
hour_to_min_conversion__ multiply__24.0__60.0__ divide__1440.0__6.0__ round__241.0__ |
| find a if ( ax num__3 + num__3 x num__2 - num__3 ) and ( num__2 x num__3 - num__5 x + a ) when divided by ( x - num__4 ) leave same rmainder ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
remainders are r num__1 = f ( num__4 ) = a ( num__4 ) num__3 + num__3 ( num__4 ) num__2 - num__3 = num__64 a + num__45 r num__2 = f ( num__4 ) = num__2 ( num__4 ) num__3 - num__5 ( num__4 ) + a = a + num__108 since r num__1 = r num__2 num__64 a + num__45 = a + num__108 = > num__63 a = num__63 = > a = num__1 a <eor> a <eos> |
a |
subtract__3.0__2.0__ subtract__64.0__1.0__ reverse__1.0__ |
subtract__3.0__2.0__ subtract__64.0__1.0__ subtract__3.0__2.0__ |
| find the percentage gain if a shop owner sells num__18 mtr of fabric and gains cost price of num__3 meters ? <o> a ) num__17.67 <o> b ) num__19.67 <o> c ) num__11.67 <o> d ) num__16.67 <o> e ) num__15.67 % |
let cost of each metre be rs . num__100 . therefore cost price of num__3 m cloth = num__3 * num__100 = num__300 cost price of num__18 m cloth = num__18 * num__100 = num__1800 since the trader made the cp of num__3 m from the sale of num__18 m therefore profit on sale of num__18 m = cost price of num__3 m cloth = num__300 selling price of num__18 m = num__1800 + num__300 = num__2100 profit % = num__16.67 d <eor> d <eos> |
d |
percent__100.0__16.67__ |
percent__100.0__16.67__ |
| find the third proportional to num__3 and num__9 ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__27 <o> d ) num__19 <o> e ) num__20 |
formula = third proportional = ( b × b ) / a a = num__3 and b = num__9 ( num__9 x num__9 ) / num__3 = num__27.0 = num__27 c <eor> c <eos> |
c |
multiply__3.0__9.0__ multiply__3.0__9.0__ |
multiply__3.0__9.0__ multiply__3.0__9.0__ |
| in the third quarter of num__2015 atlantis law firm had num__20 lawyers and num__400 cases . in the fourth quarter of num__2015 the firm had num__22 lawyers and num__500 cases . by approximately what percent did the ratio of lawyers to cases decrease from the third quarter to the fourth quarter . <o> a ) num__10.0 <o> b ) num__11.0 <o> c ) num__12.0 <o> d ) num__13.0 <o> e ) num__14.0 % |
num__3 rd quarter : lawyers / cases = num__0.05 = num__0.05 num__4 th quarter : lawyers / cases = num__0.044 = num__0.044 aside : it ' s useful to write both ratios with the same denominator . this allows us to ignore the denominator and focus solely on the numerators . so our ratio went from num__0.05 to num__0.044 ignoring the denominators we went from num__100 to num__88 the percent change = num__100 ( difference in values ) / ( original value ) = ( num__100 ) ( num__100 - num__88 ) / num__100 = ( num__100 ) ( num__12 ) / num__100 ) = num__12.0 answer : c <eor> c <eos> |
c |
reverse__20.0__ divide__22.0__500.0__ divide__400.0__4.0__ multiply__22.0__4.0__ multiply__3.0__4.0__ multiply__3.0__4.0__ |
reverse__20.0__ divide__22.0__500.0__ divide__400.0__4.0__ multiply__22.0__4.0__ subtract__100.0__88.0__ subtract__100.0__88.0__ |
| q b and c are positive integers . if q b and c are assembled into the six - digit number qbcqbc which one of the following must be a factor of qbcqbc ? <o> a ) num__16 <o> b ) num__13 <o> c ) num__5 <o> d ) num__3 <o> e ) none of the above |
plug in some values and check - qbcqbc = num__123123 not divisible by num__16 and num__5 let qbcqbc = num__125125 not divisible by num__3 only option ( b ) and ( e ) is left in both the cases . . . check once more to marke ( b ) as correct answer let qbcqbc = num__135135 again divisible by num__13 so mark answer as ( b ) num__13 <eor> b <eos> |
b |
subtract__16.0__3.0__ subtract__16.0__3.0__ |
subtract__16.0__3.0__ subtract__16.0__3.0__ |
| in the xy - plane the point ( num__1 num__3 ) is the center of a circle . the point ( - num__2 num__3 ) lies inside the circle and the point ( num__1 - num__2 ) lies outside the circle . if the radius r of the circle is an integer then r = <o> a ) num__6 <o> b ) num__5 <o> c ) num__4 <o> d ) num__3 <o> e ) num__2 |
an easy way to solve this question will be just to mark the points on the coordinate plane . you ' ll see that the distance between the center ( num__1 num__3 ) and the point inside the circle ( - num__2 num__3 ) is num__3 units ( both points are on y = num__3 line so the distance will simply be num__1 - ( - num__2 ) = num__3 ) so the radius must be more than num__3 units . the distance between the center ( num__13 ) and the point outside the circle ( num__1 - num__2 ) is num__5 units ( both points are on x = num__1 line so the distance will simply be num__3 - ( - num__2 ) = num__5 ) so the radius must be less than num__5 units which implies num__3 < r < num__5 thus as r is an integer then r = num__4 . answer : c . <eor> c <eos> |
c |
add__3.0__2.0__ add__1.0__3.0__ add__1.0__3.0__ |
add__3.0__2.0__ subtract__5.0__1.0__ subtract__5.0__1.0__ |
| a does half as much work as band c does half as much work as a and b together . if c alone can finish the work in num__60 days then together all will finish the work in : <o> a ) num__13 num__0.333333333333 days <o> b ) num__12 num__0.333333333333 days <o> c ) num__15 days <o> d ) num__20 days <o> e ) num__30 days |
c alone can finish the work in num__60 days . ( a + b ) can do it in num__20 days ( a + b ) s num__1 days wok = num__0.05 . as num__1 days work : bs num__1 days work = num__0.5 : num__1 = num__1 : num__2 . a â € ™ s num__1 day â € ™ s work = ( num__0.05 ) * ( num__0.333333333333 ) = ( num__0.0166666666667 ) . [ divide num__0.05 in the raio num__1 : num__2 ] bs num__1 days work = ( num__0.05 ) * ( num__0.666666666667 ) = num__0.0333333333333 ( a + b + c ) s num__1 day â € ™ s work = ( num__0.0166666666667 ) + ( num__0.0333333333333 ) + ( num__0.0166666666667 ) = num__0.0666666666667 all the three together will finish it in num__15 days . answer : c <eor> c <eos> |
c |
divide__1.0__20.0__ divide__1.0__0.5__ divide__20.0__60.0__ divide__1.0__60.0__ subtract__1.0__0.3333__ divide__2.0__60.0__ add__0.05__0.0167__ round__15.0__ |
divide__1.0__20.0__ divide__1.0__0.5__ divide__20.0__60.0__ multiply__0.05__0.3333__ subtract__1.0__0.3333__ multiply__0.05__0.6667__ add__0.05__0.0167__ round__15.0__ |
| what is the relationship between the fractions num__0.933333333333 and num__0.925 . <o> a ) num__0.933333333333 = num__0.925 <o> b ) num__0.933333333333 < num__0.925 <o> c ) num__0.933333333333 > num__0.925 <o> d ) can not bedetermined <o> e ) none |
take lcm of both num__15 num__40 we get num__120 ( num__14 * num__8 ) / ( num__15 * num__8 ) = num__0.933333333333 ( num__37 * num__3 ) / ( num__40 * num__3 ) = num__0.925 so num__0.933333333333 > num__0.925 answer : c <eor> c <eos> |
c |
divide__120.0__15.0__ multiply__0.925__40.0__ subtract__40.0__37.0__ divide__14.0__15.0__ |
divide__120.0__15.0__ multiply__0.925__40.0__ divide__120.0__40.0__ divide__14.0__15.0__ |
| a man can row downstream at num__18 kmph and upstream at num__10 kmph . find the speed of the man in still water and the speed of stream respectively ? <o> a ) num__14 num__5 <o> b ) num__14 num__9 <o> c ) num__14 num__6 <o> d ) num__14 num__4 <o> e ) num__14 num__2 |
let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = num__18 - - - ( num__1 ) and x - y = num__10 - - - ( num__2 ) from ( num__1 ) & ( num__2 ) num__2 x = num__28 = > x = num__14 y = num__4 . answer : d <eor> d <eos> |
d |
add__18.0__10.0__ divide__28.0__2.0__ subtract__18.0__14.0__ round__14.0__ |
add__18.0__10.0__ divide__28.0__2.0__ subtract__18.0__14.0__ subtract__18.0__4.0__ |
| the current in a river is num__10 mph . a boat can travel num__50 mph in still water . how far up the river can the boat travel if the round trip is to take num__5 hours ? <o> a ) num__120 <o> b ) num__100 <o> c ) num__96 <o> d ) num__85 <o> e ) num__150 |
upstream speed = num__50 - num__10 = num__40 mph downstream speed = num__50 + num__10 = num__60 mph d / num__40 + d / num__60 = num__5 hours solving for d we get d = num__120 answer : a <eor> a <eos> |
a |
subtract__50.0__10.0__ hour_to_min_conversion__ round__120.0__ |
subtract__50.0__10.0__ add__10.0__50.0__ round__120.0__ |
| at num__1 : num__00 annie starts to bicycle along a num__45 mile road at a constant speed of num__14 miles per hour . thirty minutes earlier scott started bicycling towards annie on the same road at a constant speed of num__12 miles per hour . at what time will they meet ? <o> a ) num__2 : num__00 <o> b ) num__2 : num__30 <o> c ) num__3 : num__00 <o> d ) num__3 : num__30 <o> e ) num__4 : num__00 |
in the first num__30 minutes scott can travel num__6 miles so there are num__39 miles left . together annie and scott can complete num__26 miles . num__1.5 = num__1.5 so they will meet num__1.5 hours after num__1 : num__00 . the answer is b . <eor> b <eos> |
b |
subtract__45.0__6.0__ add__14.0__12.0__ divide__45.0__30.0__ subtract__14.0__12.0__ |
subtract__45.0__6.0__ add__14.0__12.0__ divide__45.0__30.0__ subtract__14.0__12.0__ |
| in how many different ways can the letters of the word ' airport ' be arranged in such a way that the vowels always come together ? <o> a ) a ) num__360 <o> b ) b ) num__720 <o> c ) c ) num__480 <o> d ) d ) num__5040 <o> e ) of these |
the word ' airport ' has num__7 different letters . when the vowels aio are always together they can be supposed to form one letter . then we have to arrange the letters rprt ( aio ) . now num__5 ( num__4 + num__1 = num__5 ) letters can be arranged in num__5 ! = num__120 ways . the vowels ( aio ) can be arranged among themselves in num__3 ! = num__6 ways . required number of ways = ( num__120 x num__6 ) = num__720 . answer : b <eor> b <eos> |
b |
vowel_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
vowel_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| what is the least value of x so that num__2 x num__5452 is divisible by num__9 <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__3 <o> e ) num__0 |
explanation : the sum of the digits of the number is divisible by num__9 . then the number is divisible by num__9 . num__2 + x + num__5 + num__4 + num__5 + num__2 = num__18 + x least value of x may be ' num__0 ' so that the total num__18 + num__0 = num__18 is divisible by num__9 . answer : option e <eor> e <eos> |
e |
subtract__9.0__5.0__ multiply__2.0__9.0__ multiply__2.0__0.0__ |
subtract__9.0__5.0__ multiply__2.0__9.0__ multiply__2.0__0.0__ |
| if num__5 women or num__8 girls can do a work in num__84 days . in how many days can num__10 women and num__5 girls can do the same work ? <o> a ) num__32 <o> b ) num__33 <o> c ) num__34 <o> d ) num__35 <o> e ) can not be determined |
given that num__5 women is equal to num__8 girls to complete work / so num__10 women = num__16 girls . num__10 women + num__5 girls = num__16 girls + num__5 girls = num__21 girls . num__8 girls can do a work in num__84 days then num__21 girls can do a work in ( num__8 * num__84 ) / num__21 = num__32 days therefore num__10 women and num__5 girls can a work in num__32 days . answer is option a <eor> a <eos> |
a |
add__5.0__16.0__ round__32.0__ |
add__5.0__16.0__ round__32.0__ |
| what number is obtained by adding the units digits of num__734 ^ num__100 and num__347 ^ num__83 ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__11 |
the units digit of num__734 ^ num__100 is num__6 because num__4 raised to the power of an even integer ends in num__6 . the units digit of num__347 ^ num__83 is num__3 because powers of num__7 end in num__7 num__9 num__3 or num__1 cyclically . since num__83 is in the form num__4 n + num__3 the units digit is num__3 . then num__6 + num__3 = num__9 . the answer is c . <eor> c <eos> |
c |
add__3.0__4.0__ add__3.0__6.0__ subtract__4.0__3.0__ multiply__1.0__9.0__ |
add__3.0__4.0__ add__3.0__6.0__ subtract__4.0__3.0__ add__3.0__6.0__ |
| a shopkeeper give num__12.0 additional discount on the discounted price after giving an initial discount of num__20.0 on the labelled price of a radio . if the final sale price of the radio is rs . num__704 then what is its labelled price ? <o> a ) rs . num__844.80 <o> b ) rs . num__929.28 <o> c ) rs . num__1000 <o> d ) rs . num__1044.80 <o> e ) none |
solution let the labelled price be rs . x . num__88.0 of num__80.0 of x = num__704 x = ( num__704 x num__100 x num__1.13636363636 x num__80 ) = num__1000 . answer c <eor> c <eos> |
c |
percent__100.0__1000.0__ |
percent__100.0__1000.0__ |
| what is the smallest number which when increased by num__3 is divisible by num__9 num__35 num__25 and num__21 ? <o> a ) num__1572 <o> b ) num__1782 <o> c ) num__1992 <o> d ) num__2122 <o> e ) num__2342 |
when increased by num__3 the number must include at least num__3 ^ num__2 * num__5 ^ num__2 * num__7 = num__1575 the answer is a . <eor> a <eos> |
a |
add__3.0__2.0__ subtract__9.0__2.0__ subtract__1575.0__3.0__ |
add__3.0__2.0__ subtract__9.0__2.0__ subtract__1575.0__3.0__ |
| two trains num__140 m and num__160 m long run at the speed of num__60 km / hr and num__40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? <o> a ) num__17.9 sec <o> b ) num__11.8 sec <o> c ) num__27.7 sec <o> d ) num__10.8 sec <o> e ) num__17.8 sec |
relative speed = num__60 + num__40 = num__100 km / hr . = num__100 * num__0.277777777778 = num__27.7777777778 m / sec . distance covered in crossing each other = num__140 + num__160 = num__300 m . required time = num__300 * num__0.036 = num__10.8 = num__10.8 sec . answer : d <eor> d <eos> |
d |
subtract__140.0__40.0__ add__140.0__160.0__ multiply__300.0__0.036__ round__10.8__ |
add__60.0__40.0__ add__140.0__160.0__ multiply__300.0__0.036__ multiply__300.0__0.036__ |
| if x < num__0 and num__0 < y < num__1 which of the following has the greatest value h ? <o> a ) x ^ num__2 <o> b ) ( xy ) ^ num__2 <o> c ) ( x / y ) ^ num__2 <o> d ) x ^ num__2 / y <o> e ) x ^ num__2 * y |
given x < num__0 and num__0 < y < num__1 let x = - num__2 and y = num__0.5 a . x ^ num__2 = ( - num__2 ) ^ num__2 = num__4 b . ( xy ) ^ num__2 = ( - num__2 * num__0.5 ) ^ num__2 = num__1 c . ( x / y ) ^ num__2 = { - num__2 / ( num__0.5 ) } ^ num__2 = ( - num__4 ) ^ num__2 = num__16 d . x ^ num__2 / y = ( - num__2 ) ^ num__2 / ( num__0.5 ) = num__4 * num__2 = num__8 e . x ^ num__2 * y = ( - num__2 ) ^ num__2 * ( num__0.5 ) = num__2 answer : option c <eor> c <eos> |
c |
reverse__2.0__ divide__2.0__0.5__ power__2.0__4.0__ multiply__2.0__4.0__ reverse__0.5__ |
reverse__2.0__ divide__2.0__0.5__ power__2.0__4.0__ multiply__2.0__4.0__ reverse__0.5__ |
| the average mark of the students of a class in a particular exam is num__80 . if num__5 students whose average mark in that exam is num__40 are excluded the average mark of the remaining will be num__90 . find the number of students who wrote the exam . <o> a ) num__78 <o> b ) num__77 <o> c ) num__25 <o> d ) num__77 <o> e ) num__62 |
let the number of students who wrote the exam be x . total marks of students = num__80 x . total marks of ( x - num__5 ) students = num__90 ( x - num__5 ) num__80 x - ( num__5 * num__40 ) = num__90 ( x - num__5 ) num__250 = num__10 x = > x = num__25 answer : c <eor> c <eos> |
c |
subtract__90.0__80.0__ divide__250.0__10.0__ divide__250.0__10.0__ |
subtract__90.0__80.0__ divide__250.0__10.0__ divide__250.0__10.0__ |
| the age of father num__8 years ago was thrice the age of his son . five years hence father ' s age will be twice that of his son . the ratio of their present ages is : <o> a ) num__47 : num__11 <o> b ) num__37 : num__21 <o> c ) num__27 : num__21 <o> d ) num__47 : num__21 <o> e ) num__47 : num__31 |
let the ages of father and son num__8 years ago be num__3 x and x years respectively . then ( num__3 x + num__8 ) + num__5 = num__2 [ ( x + num__8 ) + num__5 ] num__3 x + num__13 = num__2 x + num__26 x = num__13 . required ratio = ( num__3 x + num__8 ) : ( x + num__8 ) = num__47 : num__21 . answer : option d <eor> d <eos> |
d |
subtract__8.0__3.0__ subtract__5.0__3.0__ add__8.0__5.0__ multiply__2.0__13.0__ add__8.0__13.0__ add__21.0__26.0__ |
subtract__8.0__3.0__ subtract__5.0__3.0__ add__8.0__5.0__ multiply__2.0__13.0__ add__8.0__13.0__ add__21.0__26.0__ |
| how many even integers n such that num__20 < = n < = num__200 are of the form num__3 k + num__4 where k is any natural number ? <o> a ) num__26 <o> b ) num__30 <o> c ) num__34 <o> d ) num__38 <o> e ) num__42 |
the first number is num__22 = num__16 + num__6 ( num__1 ) . we can continue adding num__6 to make a list : num__22 num__28 num__34 . . . the last number is num__196 = num__16 + num__6 ( num__30 ) there are num__30 numbers in the list . the answer is b . <eor> b <eos> |
b |
subtract__20.0__4.0__ subtract__22.0__16.0__ subtract__4.0__3.0__ add__6.0__22.0__ add__6.0__28.0__ subtract__200.0__4.0__ subtract__34.0__4.0__ multiply__1.0__30.0__ |
subtract__20.0__4.0__ subtract__22.0__16.0__ subtract__4.0__3.0__ add__6.0__22.0__ add__6.0__28.0__ subtract__200.0__4.0__ subtract__34.0__4.0__ multiply__1.0__30.0__ |
| a and b can finish num__5 piece of work in num__10 days each . c can do num__40.0 faster than a and b . approximately how many days would it take to finish num__51 piece of work with all three working simultaneously ? <o> a ) num__38 <o> b ) num__42 <o> c ) num__46 <o> d ) num__50 <o> e ) num__51 |
since c is num__40.0 faster thus it will create num__7 pieces of work in num__10 days in num__10 days total number of work by all three will be : num__5 + num__5 + num__7 = num__17 in num__20 days : num__17 + num__17 = num__34 in num__25 days : num__34 + num__17 = num__51 thus in all num__51 days . ans : e <eor> e <eos> |
e |
add__10.0__7.0__ subtract__51.0__17.0__ add__5.0__20.0__ round__51.0__ |
add__10.0__7.0__ subtract__51.0__17.0__ add__5.0__20.0__ add__34.0__17.0__ |
| on the independence day bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day num__390 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ? <o> a ) num__237 <o> b ) num__780 <o> c ) num__197 <o> d ) num__287 <o> e ) num__720 |
explanation : let the number of children in the school be x . since each child gets num__2 bananas total number of bananas = num__2 x . num__2 x / ( x - num__390 ) = num__2 + num__2 ( extra ) = > num__2 x - num__780 = x = > x = num__780 . answer : b <eor> b <eos> |
b |
multiply__390.0__2.0__ multiply__390.0__2.0__ |
multiply__390.0__2.0__ multiply__390.0__2.0__ |
| a certain social security recipient will receive an annual benefit of $ num__12000 provided he has annual earnings of $ num__9360 or less but the benefit will be reduced by $ num__1 for every $ num__3 of annual earnings over $ num__9360 . what amount of total annual earnings would result in a num__65 percent reduction in the recipient ' s annual social security benefit ? ( assume social security benefits are not counted as part of annual earnings . ) <o> a ) $ num__15360 <o> b ) $ num__17360 <o> c ) $ num__18000 <o> d ) $ num__21960 <o> e ) $ num__27 |
360 |
for every $ num__3 earn above $ num__9360 the recipient loses $ num__1 of benefit . or for every $ num__1 loss in the benefit the recipient earns $ num__3 above $ num__9360 if earning is ; num__9360 + num__3 x benefit = num__12000 - x or the vice versa if benefit is num__12000 - x the earning becomes num__9360 + num__3 x he lost num__50.0 of the benefit ; benefit received = num__12000 - num__0.65 * num__12000 = num__12000 - num__7800 x = num__4200 earning becomes num__9360 + num__3 x = num__9360 + num__3 * num__4200 = num__21960 ans : d <eor> d <eos> |
d |
d |
| x / ( y / z ) in the expression above x y and z are different numbers and each is one of the numbers num__1 num__5 or num__6 . what is the least possible value of the expression ? <o> a ) num__1.2 <o> b ) num__0.133333333333 <o> c ) num__0.833333333333 <o> d ) num__0.3 <o> e ) num__1.0 |
x / ( y / z ) = ( x * z ) / y the expression will have the least value when numerator ( x * z ) is the smallest . = ( num__1 * num__5 ) / num__6 = num__0.833333333333 answer c <eor> c <eos> |
c |
divide__5.0__6.0__ multiply__1.0__0.8333__ |
divide__5.0__6.0__ multiply__1.0__0.8333__ |
| each of the integers from num__1 to num__15 is written on the a seperate index card and placed in a box . if the cards are drawn from the box at random without replecement how many cards must be drawn to ensure that the product of all the integers drawn is even ? <o> a ) num__9 <o> b ) num__12 <o> c ) num__11 <o> d ) num__10 <o> e ) num__3 |
out of the num__15 integers : num__8 are odd and num__7 are even . if we need to make sure that the product of all the integers withdrawn is even then we need to make sure that we have at least one even number . in the worst case : num__1 . we will end up picking odd numbers one by one so we will pick all num__8 odd numbers first num__2 . num__9 th number will be the first even number so we need to withdraw at least num__9 numbers to make sure that we get one even number and the product of all the integers picked is even . so answer will be num__9 . ( a ) <eor> a <eos> |
a |
subtract__15.0__8.0__ add__1.0__8.0__ add__1.0__8.0__ |
subtract__15.0__8.0__ add__1.0__8.0__ add__1.0__8.0__ |
| the present ages of three persons are in proportions num__4 : num__7 : num__9 . eight years ago the sum of their ages was num__56 . find their present ages . <o> a ) num__16 num__18 num__36 <o> b ) num__16 num__28 num__36 <o> c ) num__8 num__20 num__28 <o> d ) num__20 num__35 num__45 <o> e ) none of these |
answer : option b let their present ages be num__4 x num__7 x and num__9 x years respectively . then ( num__4 x - num__8 ) + ( num__7 x - num__8 ) + ( num__9 x - num__8 ) = num__56 num__20 x = num__80 = > x = num__4 their present ages are num__16 num__28 and num__36 years respectively . <eor> b <eos> |
b |
divide__56.0__7.0__ multiply__4.0__20.0__ add__7.0__9.0__ multiply__4.0__7.0__ multiply__4.0__9.0__ add__7.0__9.0__ |
divide__56.0__7.0__ multiply__4.0__20.0__ add__7.0__9.0__ add__8.0__20.0__ subtract__56.0__20.0__ add__7.0__9.0__ |
| a van takes num__6 hours to cover a distance of num__540 km . how much should the speed in kmph be maintained to cover the same direction in num__1.5 th of the previous time ? <o> a ) num__50 kmph <o> b ) num__60 kmph <o> c ) num__70 kmph <o> d ) num__80 kmph <o> e ) num__90 kmph |
time = num__6 distence = num__540 num__1.5 of num__6 hours = num__6 * num__1.5 = num__9 hours required speed = num__60.0 = num__60 kmph b <eor> b <eos> |
b |
multiply__6.0__1.5__ hour_to_min_conversion__ hour_to_min_conversion__ |
multiply__6.0__1.5__ hour_to_min_conversion__ hour_to_min_conversion__ |
| in the quiet town ofnothintodothere are num__600 inhabitants num__200 are unemployed and num__300 are somnambulists . if half of the somnambulists are unemployed how many are employed and are not somnambulists ? <o> a ) num__50 . <o> b ) num__100 . <o> c ) num__150 . <o> d ) num__250 . <o> e ) num__300 . |
total = num__600 unemployed = num__200 employed = num__600 - num__200 = num__400 som = num__300 unemployed som = num__150.0 = num__150 employed som = num__150 employed that are not som = num__400 - num__150 = num__250 answer : d <eor> d <eos> |
d |
twice__200.0__ subtract__400.0__150.0__ subtract__400.0__150.0__ |
subtract__600.0__200.0__ subtract__400.0__150.0__ subtract__400.0__150.0__ |
| num__617 x num__617 + num__583 x num__583 = ? <o> a ) num__720578 <o> b ) num__80578 <o> c ) num__80698 <o> d ) num__81268 <o> e ) none of them |
= ( num__617 ) ^ num__2 + ( num__583 ) ^ num__2 = ( num__600 + num__17 ) ^ num__2 + ( num__600 - num__17 ) ^ num__2 = num__2 [ ( num__600 ) ^ num__2 + ( num__17 ) ^ num__2 ] = num__2 [ num__360000 + num__289 ] = num__2 x num__360289 = num__720578 answer is a <eor> a <eos> |
a |
subtract__617.0__600.0__ add__360000.0__289.0__ multiply__2.0__360289.0__ multiply__2.0__360289.0__ |
subtract__617.0__600.0__ add__360000.0__289.0__ multiply__2.0__360289.0__ multiply__2.0__360289.0__ |
| what amount does kiran get if he invests rs . num__8000 at num__10.0 p . a . compound interest for two years compounding done annually ? <o> a ) rs . num__9689 <o> b ) rs . num__9678 <o> c ) rs . num__9688 <o> d ) rs . num__9678 <o> e ) rs . num__9680 |
a = p { num__1 + r / num__100 } n = > num__8000 { num__1 + num__0.1 } num__2 = rs . num__9680 answer : e <eor> e <eos> |
e |
percent__10.0__1.0__ percent__100.0__9680.0__ |
percent__10.0__1.0__ percent__100.0__9680.0__ |
| num__5 men and num__12 boys finish a piece of work in num__4 days num__7 men and num__6 boys do it in num__5 days . the ratio between the efficiencies of a man and boy is ? <o> a ) num__1 : num__2 <o> b ) num__2 : num__1 <o> c ) num__2 : num__3 <o> d ) num__6 : num__5 <o> e ) num__5 : num__6 |
num__7 m + num__6 b - - - - - - - num__5 days num__20 m + num__48 b = num__35 m + num__30 b num__18 b = num__15 m = > num__5 m = num__6 b m : b = num__6 : num__5 answer d <eor> d <eos> |
d |
multiply__5.0__4.0__ multiply__12.0__4.0__ multiply__5.0__7.0__ multiply__5.0__6.0__ add__12.0__6.0__ subtract__35.0__20.0__ round__6.0__ |
multiply__5.0__4.0__ multiply__12.0__4.0__ multiply__5.0__7.0__ subtract__35.0__5.0__ add__12.0__6.0__ subtract__35.0__20.0__ subtract__12.0__6.0__ |
| both robert and alice leave from the same location at num__7 : num__00 a . m . driving in the same direction but in separate cars . robert drives num__50 miles per hour while alice drives num__80 miles per hour . after num__5 hours alice ’ s car stops . at what time will robert ’ s car reach alice ’ s car ? <o> a ) num__2 p . m . <o> b ) num__3 p . m . <o> c ) num__4 p . m . <o> d ) num__8 p . m . <o> e ) num__9 p . m . |
num__7 : num__00 am so num__4 hours later is num__11 : num__00 am in five hours robert will have driven num__5 * num__50 = num__250 miles in five hours alive will have driven num__5 * num__80 = num__400 miles so robert needs num__400 - num__250 = num__150 miles do catch alice up . so at num__50 mph he will need num__3 hours num__11 : num__00 am + num__3 hours = num__2 : num__00 pm ans : a <eor> a <eos> |
a |
add__7.0__4.0__ multiply__50.0__5.0__ multiply__80.0__5.0__ subtract__400.0__250.0__ subtract__7.0__4.0__ subtract__7.0__5.0__ round__2.0__ |
add__7.0__4.0__ multiply__50.0__5.0__ multiply__80.0__5.0__ subtract__400.0__250.0__ subtract__7.0__4.0__ subtract__7.0__5.0__ subtract__7.0__5.0__ |
| two trains each of which is num__100 m long moving in opposite direction to one another cross each other taking num__8 seconds . if speed of one train is twice the speed of other train find the speed of the faster train . <o> a ) num__60 km / hr <o> b ) num__76 km / hr <o> c ) num__98 km / hr <o> d ) num__9 km / hr <o> e ) num__77 km / hr |
explanation : no explanation is available for this question ! answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ |
hour_to_min_conversion__ |
| a is two years older than b who is twice as old as c . if the total of the ages of a b and c be num__27 the how old is b ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__11 |
let c ' s age be x years . then b ' s age = num__2 x years . a ' s age = ( num__2 x + num__2 ) years . ( num__2 x + num__2 ) + num__2 x + x = num__27 num__5 x = num__25 x = num__5 . hence b ' s age = num__2 x = num__10 years . answer : option d <eor> d <eos> |
d |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
| a and b can finish a work in num__16 days while a alone can do the same work in num__24 days . in how many days b alone will complete the work ? <o> a ) num__24 days <o> b ) num__48 days <o> c ) num__22 days <o> d ) num__77 days <o> e ) num__55 days |
b = num__0.0625 – num__0.0416666666667 = num__0.0208333333333 = > num__48 days answer : b <eor> b <eos> |
b |
subtract__0.0625__0.0417__ round__48.0__ |
subtract__0.0625__0.0417__ round__48.0__ |
| a wheel that has num__6 cogs is meshed with a larger wheel of num__14 cogs . when the smaller wheel has made num__21 revolutions then the number of revolutions made by the larger wheel is : <o> a ) num__4 <o> b ) num__9 <o> c ) num__12 <o> d ) num__49 <o> e ) num__59 |
solution let the required number of revolutions made by larger wheel be x . then . more cogs less revolutions ( indirect proporation ) ∴ num__14 : num__6 : : num__21 : x ⇔ num__14 × x = num__6 × num__21 ⇔ x = ( num__6 x num__1.5 ) = num__9 . answer b <eor> b <eos> |
b |
divide__21.0__14.0__ multiply__6.0__1.5__ multiply__6.0__1.5__ |
divide__21.0__14.0__ multiply__6.0__1.5__ multiply__6.0__1.5__ |
| a b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = num__2 : num__3 and b : c = num__2 : num__5 . if the total runs scored by all of them are num__75 the runs scored by b are ? <o> a ) num__10 <o> b ) num__18 <o> c ) num__20 <o> d ) num__25 <o> e ) num__30 |
a : b = num__2 : num__3 b : c = num__2 : num__5 a : b : c = num__4 : num__6 : num__15 num__0.24 * num__75 = num__18 answer b <eor> b <eos> |
b |
multiply__2.0__3.0__ multiply__3.0__5.0__ multiply__3.0__6.0__ multiply__3.0__6.0__ |
multiply__2.0__3.0__ multiply__3.0__5.0__ multiply__3.0__6.0__ multiply__3.0__6.0__ |
| if the average of num__10 consecutive integers is num__21.5 then the num__10 th integer is : - <o> a ) num__15 <o> b ) num__20 <o> c ) num__23 <o> d ) num__26 <o> e ) num__25 |
the average falls between the num__5 th and num__6 th integers integer num__5 = num__21 integer num__6 = num__22 . counting up to the tenth integer we get num__26 . answer : d <eor> d <eos> |
d |
round_down__21.5__ add__5.0__21.0__ add__5.0__21.0__ |
round_down__21.5__ add__5.0__21.0__ add__5.0__21.0__ |
| a car is bought for rs . num__60000 / - and sold at a loss of num__30.0 find its selling price ? <o> a ) s . num__45400 / - <o> b ) s . num__40000 / - <o> c ) s . num__48080 / - <o> d ) s . num__45590 / - <o> e ) s . num__42000 / - |
num__100.0 - - - - - - > num__60000 ( num__100 * num__600 = num__60000 ) num__70.0 - - - - - - > num__42000 ( num__70 * num__600 = num__42000 ) selling price = rs . num__42000 / - option ' e ' <eor> e <eos> |
e |
percent__70.0__60000.0__ percent__100.0__42000.0__ |
percent__70.0__60000.0__ percent__100.0__42000.0__ |
| the current in a river is num__3 mph . a boat can travel num__21 mph in still water . how far up the river can the boat travel if the round trip is to take num__10 hours ? <o> a ) num__69 miles <o> b ) num__88 miles <o> c ) num__96 miles <o> d ) num__102.8 miles <o> e ) num__112 miles |
upstream speed = num__21 - num__3 = num__18 mph downstream speed = num__21 + num__3 = num__24 mph d / num__18 + d / num__24 = num__10 hours solving for d we get d = num__102.8 answer : d <eor> d <eos> |
d |
subtract__21.0__3.0__ add__3.0__21.0__ round__102.8__ |
subtract__21.0__3.0__ add__3.0__21.0__ round__102.8__ |
| solve below question num__2 x + num__1 = - num__23 <o> a ) - num__8 <o> b ) - num__9 <o> c ) - num__12 <o> d ) - num__4 <o> e ) num__12 |
num__1 . subtract num__1 from both sides : num__2 x + num__1 - num__1 = - num__23 - num__1 num__2 . simplify both sides : num__2 x = - num__24 num__3 . divide both sides by num__2 : num__4 . simplify both sides : x = - num__12 c <eor> c <eos> |
c |
add__1.0__23.0__ add__2.0__1.0__ add__1.0__3.0__ multiply__3.0__4.0__ multiply__1.0__12.0__ |
add__1.0__23.0__ add__2.0__1.0__ add__1.0__3.0__ multiply__3.0__4.0__ subtract__24.0__12.0__ |
| the age of father num__10 years ago was thrice the age of his son . ten years hence father ' s age will be twice that of his son . the ratio of their present ages is : <o> a ) num__7 : num__3 <o> b ) num__7 : num__4 <o> c ) num__7 : num__5 <o> d ) num__7 : num__65 <o> e ) num__7 : num__9 |
let the ages of father and son num__10 years ago be num__3 x and x years respectively . then ( num__3 x + num__10 ) + num__10 = num__2 [ ( x + num__10 ) + num__10 ] num__3 x + num__20 = num__2 x + num__40 x = num__20 . required ratio = ( num__3 x + num__10 ) : ( x + num__10 ) = num__70 : num__30 = num__7 : num__3 . answer : a <eor> a <eos> |
a |
multiply__10.0__2.0__ multiply__2.0__20.0__ multiply__10.0__3.0__ subtract__10.0__3.0__ subtract__10.0__3.0__ |
multiply__10.0__2.0__ multiply__2.0__20.0__ add__10.0__20.0__ subtract__10.0__3.0__ subtract__10.0__3.0__ |
| the length of the bridge which a train num__130 meters long and travelling at num__45 km / hr can cross in num__30 seconds is ? <o> a ) num__766 m <o> b ) num__156 m <o> c ) num__245 m <o> d ) num__156 m <o> e ) num__156 m |
speed = ( num__45 * num__0.277777777778 ) m / sec = ( num__12.5 ) m / sec . time = num__30 sec . let the length of bridge be x meters . then ( num__130 + x ) / num__30 = num__12.5 = = > num__2 ( num__130 + x ) = num__750 = = > x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| how many integers are there between num__35 and num__108 inclusive ? <o> a ) num__51 <o> b ) num__55 <o> c ) num__56 <o> d ) num__57 <o> e ) num__74 |
i guess the easiest way to answer this is - say you have two positive integers x and y where y > x then the number of integers between x and y is given by - ( y - x ) + num__1 in this case it ' s ( num__108 - num__35 ) + num__1 = num__74 . option e <eor> e <eos> |
e |
multiply__1.0__74.0__ |
multiply__1.0__74.0__ |
| on one side of a coin there is the number num__0 and on the other side the number num__1 . what is the probability that the sum of three coin tosses will be num__0 if a num__1 is the value of the first coin flip ? <o> a ) num__0.0 . <o> b ) num__0.5 . <o> c ) num__0.2 . <o> d ) num__0.375 . <o> e ) num__0.333333333333 . |
assume the coin is unbiased . possible sums from num__3 tosses = num__01 num__23 o and num__3 are possible in only num__1 case each . ( num__0 num__00 or num__1 num__11 ) num__1 is possible in num__3 c num__1 = num__3 cases . ( num__1 num__00 ; num__0 num__10 or num__0 num__01 ) or similarly num__2 is possible in num__3 c num__2 = num__3 cases ( num__1 num__01 ; num__1 num__10 ; num__0 num__11 ) so answer will be num__0.0 . option a . its impossible to get a total value of num__0 if a num__1 has been showed <eor> a <eos> |
a |
coin_space__ negate_prob__1.0__ |
coin_space__ negate_prob__1.0__ |
| in a certain economy c represents the total amount of consumption in millions of dollars y represents the total national income in millions of dollars and the relationship between these two values is given by the equation c = num__90 + num__9 y / num__11 . if the total amount of consumption in the economy increases by num__99 million dollars what is the increase in the total national income in millions of dollars ? <o> a ) num__11 <o> b ) num__22 <o> c ) num__99 <o> d ) num__121 <o> e ) num__171 |
just assume y = num__11 and c = num__99 then c = num__198 and y = num__132 so y increases by num__121 answer : d <eor> d <eos> |
d |
subtract__132.0__11.0__ subtract__132.0__11.0__ |
subtract__132.0__11.0__ subtract__132.0__11.0__ |
| how much interest can a person get on rs . num__8200 at num__17.5 p . a . simple interest for a period of two years and six months ? <o> a ) num__3587.58 <o> b ) num__3587.52 <o> c ) num__3587.5 <o> d ) num__3587.59 <o> e ) num__3582.52 |
i = ( num__8200 * num__2.5 * num__17.5 ) / num__100 = ( num__8200 * num__5 * num__35 ) / ( num__100 * num__2 * num__2 ) = rs . num__3587.50 . answer : c <eor> c <eos> |
c |
percent__100.0__3587.5__ |
percent__100.0__3587.5__ |
| a company wants to spend equal amounts of money for the purchase of two types of computer printers costing $ num__350 and $ num__200 per unit respectively . what is the fewest number of computer printers that the company can purchase ? <o> a ) num__9 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__13 |
the smallest amount that the company can spend is the lcm of num__350 and num__200 which is num__1400 for each which is total num__2800 . the number of num__1 st type of computers which costing $ num__350 = num__4.0 = num__4 . the number of num__2 nd type of computers which costing $ num__200 = num__7.0 = num__7 . total = num__4 + num__7 = num__11 answer is c . <eor> c <eos> |
c |
divide__1400.0__350.0__ divide__2800.0__1400.0__ divide__1400.0__200.0__ add__4.0__7.0__ multiply__1.0__11.0__ |
divide__1400.0__350.0__ divide__2800.0__1400.0__ divide__1400.0__200.0__ add__4.0__7.0__ add__4.0__7.0__ |
| which of the following equations is not equivalent to num__4 x ^ num__2 = a ^ num__2 - num__9 ? <o> a ) num__4 x ^ num__2 + num__9 = a ^ num__2 <o> b ) num__4 x ^ num__2 - a ^ num__2 = - num__9 <o> c ) num__4 x ^ num__2 = ( a + num__3 ) ( a - num__3 ) <o> d ) num__2 x = a - num__3 <o> e ) x ^ num__2 = ( a ^ num__2 - num__9 ) / num__4 |
num__4 x ^ num__2 = a ^ num__2 - num__9 the basic rule when dealing with equations is that you can do anything to both sides of an equation as long as you do it equally to both sides . everything gets a bit more complex when you ' re dealing with variables in the denominator of a fraction and / or inequalities but neither of those subjects is a factor in this prompt . looking at answer d we have . . . . num__2 x = a - num__3 while you could take the square root of both sides of the original equation it ' s important to note that the square root of a ^ num__2 - num__9 is not ( a - num__3 ) . a ^ num__2 - num__9 can be factored into ( a - num__3 ) ( a + num__3 ) but neither of these parentheses is the square root of a ^ num__2 - num__9 . you can see the proof when you square either of the two parentheses : ( a - num__3 ) ^ num__2 = a ^ num__2 - num__6 x + num__9 ( a + num__3 ) ^ num__2 = a ^ num__2 + num__6 x + num__9 thus d is not equivalent to the prompt . <eor> d <eos> |
d |
add__4.0__2.0__ subtract__4.0__2.0__ |
add__4.0__2.0__ subtract__4.0__2.0__ |
| the table shows the number of calls received by each of five operators during each of num__5 one - hour periods . for which operator was the standard deviation of the numbers of calls received during these num__5 periods the least ? <o> a ) operator a : num__3 num__7 num__7 num__3 num__4 <o> b ) operator b : num__4 num__5 num__5 num__6 num__5 <o> c ) operator c : num__8 num__2 num__5 num__5 num__6 <o> d ) operator d : num__6 num__4 num__4 num__6 num__6 <o> e ) operator e : num__3 num__4 num__5 num__8 num__7 |
a it has num__2 numbers right at the average a has none . <eor> a <eos> |
a |
subtract__5.0__2.0__ |
subtract__5.0__2.0__ |
| a is two years older than b who is twice as old as c . if the total of the ages of a b and c be num__27 then how old is b ? <o> a ) num__10 <o> b ) num__5 <o> c ) num__8 <o> d ) num__6 <o> e ) num__11 |
c ' s age be x years . then b ' s age = num__2 x years . a ' s age = ( num__2 x + num__2 ) years . ( num__2 x + num__2 ) + num__2 x + x = num__27 num__5 x = num__25 = > x = num__5 hence b ' s age = num__2 x = num__10 years . answer a <eor> a <eos> |
a |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
| an aeroplane covers a certain distance at a speed of num__150 kmph in num__2 hours . to cover the same distance in num__1 num__0.666666666667 hours it must travel at a speed of : <o> a ) num__1300 kmph <o> b ) num__160 kmph <o> c ) num__180 kmph <o> d ) num__1080 kmph <o> e ) none |
explanation : distance = ( num__150 x num__2 ) = num__300 km . speed = distance / time speed = num__300 / ( num__1.66666666667 ) km / hr . [ we can write num__1 num__0.666666666667 hours as num__1.66666666667 hours ] required speed = num__300 x num__0.6 km / hr = num__180 km / hr . answer : option c <eor> c <eos> |
c |
multiply__150.0__2.0__ add__1.0__0.6667__ km_to_mile_conversion__ multiply__0.6__300.0__ round__180.0__ |
multiply__150.0__2.0__ add__1.0__0.6667__ divide__1.0__1.6667__ multiply__0.6__300.0__ divide__180.0__1.0__ |
| two pipes a and b can separately fill a cistern in num__10 and num__15 minutes respectively . a person opens both the pipes together when the cistern should have been was full he finds the waste pipe open . he then closes the waste pipe and in another num__4 minutes the cistern was full . in what time can the waste pipe empty the cistern when fill ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__6 <o> d ) num__5 <o> e ) num__1 |
explanation : num__0.1 + num__0.0666666666667 = num__0.166666666667 * num__4 = num__0.666666666667 num__1 - num__0.666666666667 = num__0.333333333333 num__0.1 + num__0.0666666666667 - num__1 / x = num__0.333333333333 x = num__8 answer : b <eor> b <eos> |
b |
add__0.1__0.0667__ divide__10.0__15.0__ multiply__10.0__0.1__ subtract__1.0__0.6667__ round__8.0__ |
add__0.1__0.0667__ divide__10.0__15.0__ multiply__10.0__0.1__ subtract__1.0__0.6667__ divide__8.0__1.0__ |
| find the cost of fencing around a circular field of diameter num__32 m at the rate of rs . num__2 a meter ? <o> a ) num__201 <o> b ) num__132 <o> c ) num__772 <o> d ) num__592 <o> e ) num__261 |
num__2 * num__3.14285714286 * num__16 = num__100.5 num__100.5 * num__2 = rs . num__201 answer : a <eor> a <eos> |
a |
divide__32.0__2.0__ multiply__2.0__100.5__ round__201.0__ |
divide__32.0__2.0__ multiply__2.0__100.5__ multiply__2.0__100.5__ |
| a computer wholesaler sells ten different computers and each is priced differently . if the wholesaler chooses three computers for display at a trade show what is the probability ( all things being equal ) that the two most expensive computers will be among the three chosen for display ? <o> a ) num__0.267857142857 <o> b ) num__0.107142857143 <o> c ) num__0.0357142857143 <o> d ) num__0.0666666666667 <o> e ) num__0.00595238095238 |
since two of the choices are prefixed we are free to choose num__1 from the rest of the num__8 avilable . so num__8 c num__1 is the numerator . total no of ways we can choose num__3 from num__10 is num__10 c num__3 which is the denominator . so the probability : num__8 c num__0.1 c num__3 = num__0.0666666666667 ans is d . <eor> d <eos> |
d |
reverse__10.0__ multiply__1.0__0.0667__ |
reverse__10.0__ multiply__1.0__0.0667__ |
| on a saturday night each of the rooms at a certain motel was rented for either $ num__50 or $ num__60 . if num__10 of the rooms that were rented for $ num__60 had instead been rented for $ num__50 then the total rent the motel charged for that night would have been reduced by num__25 percent . what was the total rent the motel actually charged for that night ? <o> a ) $ num__400 <o> b ) $ num__800 <o> c ) $ num__1000 <o> d ) $ num__1600 <o> e ) $ num__2 |
400 |
let total rent the motel charge for all rooms = x if num__10 rooms that were rented for num__60 $ had instead been rented for num__50 $ then total difference in prices = num__10 $ * num__10 = num__100 $ total rent the motel charged would have been reduced by num__25.0 . num__25 x = num__100 = > x = num__400 answer a <eor> a <eos> |
a |
a |
| for any positive number x the function [ x ] denotes the greatest integer less than or equal to x . for example [ num__1 ] = num__1 [ num__1.567 ] = num__1 and [ num__1.999 ] = num__1 . if k is a positive integer such that k ^ num__2 is divisible by num__45 and num__80 what is the units digit of k ^ num__0.00075 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__27 <o> d ) num__0 <o> e ) can not be determined |
k = [ lcm of num__80 and num__45 ] * ( any integer ) however minimum value of k is sq . rt of num__3 ^ num__2 * num__4 ^ num__2 * num__5 ^ num__2 = num__60 * any integer for value of k ( num__60 ) * any integer unit value will be always zero . d <eor> d <eos> |
d |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ round_down__0.0008__ |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ round_down__0.0008__ |
| two trains of length num__210 m and num__120 m are num__160 m apart . they start moving towards each other on parallel tracks at speeds num__74 kmph and num__92 kmph . after how much time will the trains meet ? <o> a ) num__3.5 <o> b ) num__3.1 <o> c ) num__3.1 <o> d ) num__3.6 <o> e ) num__3.7 |
they are moving in opposite directions relative speed is equal to the sum of their speeds . relative speed = ( num__74 + num__92 ) * num__0.277777777778 = num__46.1 mps . the time required = d / s = num__160 / num__46.1 = num__3.5 sec . answer : a <eor> a <eos> |
a |
round__3.5__ |
round__3.5__ |
| ayush was born two years after his father ' s marriage . his mother is five years younger than his father but num__20 years older than ayush who is num__15 years old . at what age did the father get married ? <o> a ) num__20 years <o> b ) num__21 years <o> c ) num__22 years <o> d ) num__23 years <o> e ) num__25 years |
explanation : ayush ' s present age = num__15 years . his mother ' s present age = ( num__15 + num__20 ) years = num__35 years . ayush ' s father ' s present age = ( num__35 + num__5 ) years = num__40 years . ayush ' s father ' s age at the time of ayush ' s birth = ( num__40 - num__15 ) years = num__25 years . therefore ayush ' s father ' s age at the time of marriage = ( num__25 - num__2 ) years = num__23 years . answer : d ) num__23 year <eor> d <eos> |
d |
add__20.0__15.0__ subtract__20.0__15.0__ add__35.0__5.0__ add__20.0__5.0__ divide__40.0__20.0__ subtract__25.0__2.0__ subtract__25.0__2.0__ |
add__20.0__15.0__ subtract__20.0__15.0__ add__35.0__5.0__ add__20.0__5.0__ divide__40.0__20.0__ subtract__25.0__2.0__ subtract__25.0__2.0__ |
| the present worth of rs . num__242 due in num__2 years at num__10.0 per annum compound interest is : <o> a ) rs . num__180 <o> b ) rs . num__240 <o> c ) rs . num__220 <o> d ) rs . num__200 <o> e ) rs . num__300 |
explanation : present worth of rs . x due t years hence is given by present worth ( pw ) = x / ( num__1 + r / num__100 ) ^ t present worth ( pw ) = num__242 / ( num__1 + num__0.1 ) ^ num__2 = num__242 / ( num__1.1 ) ^ num__2 = rs . num__200 answer : option d <eor> d <eos> |
d |
percent__10.0__1.0__ percent__100.0__200.0__ |
percent__10.0__1.0__ percent__100.0__200.0__ |
| a man can row a boat at num__10 kmph in still water . if the speed of the stream is num__6 kmph what is the time taken to row a distance of num__60 km downstream ? <o> a ) num__0.361445783133 hours <o> b ) num__1.15384615385 hours <o> c ) num__3.75 hours <o> d ) num__1.57894736842 hours <o> e ) num__2.46153846154 hours |
speed downstream = num__10 + num__6 = num__16 kmph . time required to cover num__60 km downstream = d / s = num__3.75 = num__3.75 hours . answer : c <eor> c <eos> |
c |
add__10.0__6.0__ divide__60.0__16.0__ round__3.75__ |
add__10.0__6.0__ divide__60.0__16.0__ divide__60.0__16.0__ |
| a man jumps into a river from a overbridge at the same time a hat drops in the river which flows with the stream of the water . that man travels num__10 minute upstream and returns back where he was asked to catch the hat . he catches hat at a distance of num__1000 yard down at a second bridge . find the speed of the river . <o> a ) num__50 <o> b ) num__60 <o> c ) num__55 <o> d ) num__70 <o> e ) num__75 |
explanation : suppose x is the distance travelled by the man in num__10 min upstream . then = > x / ( v - u ) = num__10 where v is man ' s speed and u is river speed ) total time taken to swim by the man upstream and down stream is equal to the time taken by the hat to travel num__1000 yards with the speed of river : = > num__1000 / u = num__10 + ( x + num__1000 ) / ( u + v ) = > now substitute x = num__10 ( v - u ) and simplify we will get = > num__100 v = num__2 uv = > num__2 u = num__100 = > u = num__50 yards per minute hence ( a ) is the correct answer . answer : a <eor> a <eos> |
a |
divide__1000.0__10.0__ divide__100.0__2.0__ round__50.0__ |
divide__1000.0__10.0__ divide__100.0__2.0__ subtract__100.0__50.0__ |
| the volumes of two cones are in the ratio num__1 : num__10 and the radii of the cones are in the ratio of num__1 : num__2 . what is the length of the wire ? <o> a ) num__2 : num__5 <o> b ) num__2 : num__7 <o> c ) num__2 : num__1 <o> d ) num__2 : num__3 <o> e ) num__2 : num__2 |
the volume of the cone = ( num__0.333333333333 ) π r num__2 h only radius ( r ) and height ( h ) are varying . hence ( num__0.333333333333 ) π may be ignored . v num__1 / v num__2 = r num__12 h num__1 / r num__22 h num__2 = > num__0.1 = ( num__1 ) num__2 h num__1 / ( num__2 ) num__2 h num__2 = > h num__1 / h num__2 = num__0.4 i . e . h num__1 : h num__2 = num__2 : num__5 answer : a <eor> a <eos> |
a |
rectangle_perimeter__1.0__10.0__ square_perimeter__0.1__ triangle_area__1.0__10.0__ multiply__1.0__2.0__ |
rectangle_perimeter__1.0__10.0__ square_perimeter__0.1__ triangle_area__1.0__10.0__ multiply__1.0__2.0__ |
| a man swims downstream num__72 km and upstream num__45 km taking num__9 hours each time ; what is the speed of the current ? <o> a ) num__1.9 <o> b ) num__1.5 <o> c ) num__2.8 <o> d ) num__3.0 <o> e ) num__3.9 |
num__72 - - - num__9 ds = num__8 ? - - - - num__1 num__45 - - - - num__9 us = num__5 ? - - - - num__1 s = ? s = ( num__8 - num__5 ) / num__2 = num__1.5 answer : b <eor> b <eos> |
b |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ round__1.5__ |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ divide__1.5__1.0__ |
| excluding stoppages the speed of the bus is num__54 kms / hr and with stoppages it is num__45 kmph . for how many minutes does the bus stop per hour ? <o> a ) num__9 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__20 |
by including stoppages bus will cover num__9 km less . . ( num__54 km - num__45 km = num__9 km ) excluding stoppages x = num__54 kmph including stoppages y = num__45 kmph difference z = num__9 km time taken to cover that num__9 km ( bus stops per hour ( num__60 minutes ) ) = z / x ( num__60 minutes ) = num__0.166666666667 ( num__60 ) = num__0.166666666667 ( num__60 ) = num__10 minutes . answer : b <eor> b <eos> |
b |
subtract__54.0__45.0__ hour_to_min_conversion__ divide__9.0__54.0__ round__10.0__ |
subtract__54.0__45.0__ hour_to_min_conversion__ divide__9.0__54.0__ round__10.0__ |
| of the five numbers in a sequence the first term is num__10000 and each of the following terms is num__70.0 of the previous term . what is the value range of the five numbers ? <o> a ) num__7599 <o> b ) num__9750 <o> c ) num__9975 <o> d ) num__9984 <o> e ) num__10 |
736 |
num__1 st number = num__10000 num__2 nd number = num__70.0 of num__10000 = num__7000 num__3 rd number = num__70.0 of num__7000 = num__4900 num__4 th number = num__70.0 of num__4900 = num__3430 num__5 th number = num__70.0 of num__3430 = num__2401 range = num__10000 - num__2401 = num__7599 a is the answer <eor> a <eos> |
a |
a |
| the average of seven results is num__156 and that of the first six is num__169 . the fifth result is ? <o> a ) num__120 <o> b ) num__160 <o> c ) num__89 <o> d ) num__78 <o> e ) num__148 |
num__7 * num__156 â € “ num__6 * num__169 = num__78 answer : d <eor> d <eos> |
d |
subtract__156.0__78.0__ |
subtract__156.0__78.0__ |
| a man complete a journey in num__1010 hours . he travels first half of the journey at the rate of num__2121 km / hr and second half at the rate of num__2424 km / hr . find the total journey in km . <o> a ) num__121 km <o> b ) num__242 km <o> c ) num__224 km <o> d ) num__112 km <o> e ) num__110 km |
solution num__1 average speed = num__2 Ã — num__21 Ã — num__2421 + num__24 = num__22.4 km / hr = num__2 Ã — num__21 Ã — num__2421 + num__24 = num__22.4 km / hr total distance = num__22.4 Ã — num__10 = num__224 km answer is c <eor> c <eos> |
c |
multiply__10.0__22.4__ round__224.0__ |
multiply__10.0__22.4__ divide__224.0__1.0__ |
| points a b and c lie in that order on a straight railroad track . the distance from point a to point b is twice the distance from point b to point c . a train traveled from point a to point c without stopping . the train ' s average speed when traveling from point a to point b was num__150 miles per hour and the train ' s average speed when traveling from point b to point c was num__100 miles per hour . what was the train ' s average speed in miles per hour when traveling from point a to point c ? <o> a ) num__22.5 <o> b ) num__21.43 <o> c ) num__25.5 <o> d ) num__26.5 <o> e ) num__27.5 |
average speed = distance / time because we are looking for average speed we can pick a distance for the variable d . speed a - b = num__150 speed b - c = num__100 average speed = total distance / total rate rate = distance / time a = = = = = = = = = = = = = = = = = = = = b = = = = = = = = = = c if a - b is twice the length of b - c then let a - b = num__2 d and let b - c = d average speed = num__3 d / ( num__2 d / num__150 ) + ( d / num__100 ) num__3 d / ( num__4 d / num__300 + ( num__3 d / num__300 ) num__3 d / ( num__7 d / num__50 ) num__150 d / num__7 d average speed = num__21.43 answer : b <eor> b <eos> |
b |
multiply__150.0__2.0__ add__3.0__4.0__ subtract__150.0__100.0__ round__21.43__ |
multiply__150.0__2.0__ add__3.0__4.0__ subtract__150.0__100.0__ round__21.43__ |
| a father is now three times as old as his son . six years back he was four times as old as his son . the age of the son ( in years ) is <o> a ) num__12 <o> b ) num__15 <o> c ) num__18 <o> d ) num__20 <o> e ) num__25 |
if father ' s age is f and son ' s age is s then f = num__3 s and f - num__6 = num__4 * ( s - num__6 ) num__3 s - num__6 = num__4 s - num__24 s = num__18 f = num__54 years answer : c <eor> c <eos> |
c |
multiply__4.0__6.0__ multiply__3.0__6.0__ multiply__3.0__18.0__ multiply__3.0__6.0__ |
multiply__4.0__6.0__ multiply__3.0__6.0__ multiply__3.0__18.0__ multiply__3.0__6.0__ |
| a biker covered half the distance between two towns in num__2 hr num__30 min . after that he increased his speed by num__2 km / hr . he covered the second half of the distance in num__2 hr num__20 min . find the distance between the two towns and the initial speed of the biker . <o> a ) num__140 <o> b ) num__260 <o> c ) num__320 <o> d ) num__480 <o> e ) num__500 |
let x km / hr be the initial speed of the biker then his speed during the second part of the trip is x + num__2 km / hr . half the distance between two cities equals num__2 * num__0.5 ⋅ x km and num__2 * num__0.333333333333 ⋅ ( x + num__2 ) km . from the equation : num__2 * num__0.5 ⋅ x = num__2 * num__0.333333333333 ⋅ ( x + num__2 ) we get x = num__28 km / hr . the intial speed of the biker is num__28 km / h . half the distance between the two towns is num__2 h num__30 min × num__28 = num__2.5 × num__28 = num__70 . so the distance is num__2 × num__70 = num__140 km <eor> a <eos> |
a |
subtract__30.0__2.0__ add__2.0__0.5__ multiply__2.5__28.0__ multiply__2.0__70.0__ round__140.0__ |
subtract__30.0__2.0__ add__2.0__0.5__ multiply__2.5__28.0__ multiply__2.0__70.0__ multiply__2.0__70.0__ |
| a train leaves mumabai at num__9 am at a speed of num__40 kmph . after one hour another train leaves mumbai in the same direction as that of the first train at a speed of num__50 kmph . when and at what distance from mumbai do the two trains meet ? <o> a ) num__287 <o> b ) num__279 <o> c ) num__200 <o> d ) num__278 <o> e ) num__379 |
when the second train leaves mumbai the first train covers num__40 * num__1 = num__40 km so the distance between first train and second train is num__40 km at num__10.00 am time taken by the trains to meet = distance / relative speed = num__40 / ( num__50 - num__40 ) = num__4 hours so the two trains meet at num__2 p . m . the two trains meet num__4 * num__50 = num__200 km away from mumbai . answer : c <eor> c <eos> |
c |
add__9.0__1.0__ divide__40.0__10.0__ multiply__50.0__4.0__ round__200.0__ |
subtract__50.0__40.0__ divide__40.0__10.0__ multiply__50.0__4.0__ multiply__50.0__4.0__ |
| the avg weight of a b & c is num__84 kg . if d joins the group the avg weight of the group becomes num__80 kg . if another man e who weights is num__7 kg more than d replaces a then the avgof b c d & e becomes num__79 kg . what is the weight of a ? <o> a ) num__45 <o> b ) num__65 <o> c ) num__75 <o> d ) num__79 <o> e ) num__90 |
a + b + c = num__3 * num__84 = num__252 a + b + c + d = num__4 * num__80 = num__320 - - - - ( i ) so d = num__68 & e = num__68 + num__7 = num__75 b + c + d + e = num__79 * num__4 = num__316 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = num__320 – num__316 = num__4 a = e + num__4 = num__75 + num__4 = num__79 d <eor> d <eos> |
d |
multiply__84.0__3.0__ subtract__84.0__80.0__ multiply__80.0__4.0__ subtract__320.0__252.0__ add__7.0__68.0__ multiply__79.0__4.0__ add__4.0__75.0__ |
multiply__84.0__3.0__ subtract__84.0__80.0__ multiply__80.0__4.0__ subtract__320.0__252.0__ add__7.0__68.0__ multiply__79.0__4.0__ add__4.0__75.0__ |
| cole drove from home to work at an average speed of num__75 kmh . he then returned home at an average speed of num__105 kmh . if the round trip took a total of num__1 hours how many minutes did it take cole to drive to work ? <o> a ) num__35 <o> b ) num__50 <o> c ) num__62 <o> d ) num__75 <o> e ) num__78 |
first round distance travelled ( say ) = d speed = num__75 k / h time taken t num__2 = d / num__75 hr second round distance traveled = d ( same distance ) speed = num__105 k / h time taken t num__2 = d / num__105 hr total time taken = num__1 hrs therefore num__1 = d / num__75 + d / num__105 lcm of num__75 and num__105 = num__525 num__1 = d / num__75 + d / num__105 = > num__1 = num__7 d / num__525 + num__5 d / num__525 = > d = num__43.75 km therefore t num__1 = d / num__75 = > t num__1 = num__525 / ( num__12 x num__75 ) = > t num__1 = ( num__7 x num__60 ) / num__12 - - in minutes = > t num__1 = num__35 minutes . a <eor> a <eos> |
a |
divide__525.0__75.0__ subtract__7.0__2.0__ add__5.0__7.0__ hour_to_min_conversion__ multiply__5.0__7.0__ round__35.0__ |
divide__525.0__75.0__ subtract__7.0__2.0__ add__5.0__7.0__ hour_to_min_conversion__ multiply__5.0__7.0__ divide__35.0__1.0__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__360 m and they cross each other in num__24 sec then the speed of each train is ? <o> a ) num__78 <o> b ) num__89 <o> c ) num__36 <o> d ) num__54 <o> e ) num__23 |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__360 + num__360 ) / num__24 = > x = num__15 speed of each train = num__15 m / sec . = num__15 * num__3.6 = num__54 km / hr . answer : option d <eor> d <eos> |
d |
divide__360.0__24.0__ multiply__15.0__3.6__ round__54.0__ |
divide__360.0__24.0__ multiply__15.0__3.6__ multiply__15.0__3.6__ |
| if deepesh had walked num__20 km / h faster he woulh have saved num__1 hour in the distance of num__60 km . what is the usual speed of deepesh ? <o> a ) num__100 <o> b ) num__299 <o> c ) num__208 <o> d ) num__273 <o> e ) num__222 |
explanation : let the original speed be x km / h then \ inline \ frac { num__600 } { x } - \ frac { num__600 } { ( x + num__20 ) } = num__1 \ inline \ rightarrow num__600 \ left ( \ frac { x + num__20 - x } { x ( x + num__20 ) } \ right ) = num__1 \ inline \ rightarrow \ : \ : x ^ { num__2 } + num__120 x - num__12000 = num__0 \ inline \ rightarrow \ : \ : ( x - num__120 ) ( x - num__100 ) = num__0 \ inline \ rightarrow \ : \ : x = num__100 \ : \ : and \ : \ : x = num__120 \ inline \ therefore original speed = num__100 km / h answer : a <eor> a <eos> |
a |
multiply__60.0__2.0__ multiply__20.0__600.0__ divide__12000.0__120.0__ round__100.0__ |
multiply__60.0__2.0__ multiply__20.0__600.0__ divide__12000.0__120.0__ divide__12000.0__120.0__ |
| in the xy - plane the points ( c d ) ( c - d ) and ( - c - d ) are three vertices of a certain square . if c < num__0 and d > num__0 which of the following points t is in the same quadrant as the fourth vertex of the square ? <o> a ) ( - num__5 - num__3 ) <o> b ) ( - num__5 num__3 ) <o> c ) ( num__5 - num__3 ) <o> d ) ( num__3 - num__5 ) <o> e ) ( num__3 num__5 ) |
the question : in the xy - plane the points ( c d ) ( c - d ) and ( - c - d ) are three vertices of a certain square . if c < num__0 and d > num__0 which of the following points t is in the same quadrant as the fourth vertex of the square ? i marked the tricky part in red . it seems c is anegativenumber and d is a positive number . this means vertex # num__1 = ( c d ) is in qii ( that is negative x and positive y ) vertex # num__2 = ( c - d ) is in qiii ( that is both xy negative ) vertex # num__3 = ( - c - d ) is in qiv ( that is y is negative but x is positive ) that means the last vertex should be in the first quadrant - - - the only first quadrant point is ( num__5 num__3 ) answer = e . <eor> e <eos> |
e |
add__1.0__2.0__ add__2.0__3.0__ add__1.0__2.0__ |
add__1.0__2.0__ add__2.0__3.0__ subtract__5.0__2.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__12 sec . what is the length of the train ? <o> a ) num__356 <o> b ) num__200 <o> c ) num__127 <o> d ) num__279 <o> e ) num__150 |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__12 = num__200 m answer : b <eor> b <eos> |
b |
round__200.0__ |
round__200.0__ |
| a square mirror has exactly half the area of the rectangular wall on which it is hung . if each side of the mirror is num__54 inches and the width of the wall is num__68 inches what is the length of the wall in inches ? <o> a ) num__25.7 <o> b ) num__72.7 <o> c ) num__70.7 <o> d ) num__85.7 <o> e ) num__56.7 |
since the mirror is num__42 inches in all sides it must be a square . area of a square is a = a ^ num__2 ; num__54 ^ num__2 = num__2916 . area of rectangle is double of that num__2 * num__2916 = num__5832 . now a = lw and we need find w so a / l = w ; num__85.7647058824 = num__85.7 answer ! answer is d <eor> d <eos> |
d |
power__54.0__2.0__ multiply__2.0__2916.0__ triangle_area__2.0__85.7__ |
power__54.0__2.0__ multiply__2.0__2916.0__ triangle_area__2.0__85.7__ |
| what is the area of a square with perimeter num__8 p ? <o> a ) num__16 p ^ num__2 <o> b ) num__4 p ^ num__2 <o> c ) p ^ num__0.5 <o> d ) p / num__16 <o> e ) p ^ num__0.125 |
each side is num__2 p a = ( num__2 p ) ^ num__2 = num__4 ( p ^ num__2 ) answer b <eor> b <eos> |
b |
triangle_area__2.0__4.0__ |
triangle_area__2.0__4.0__ |
| insert the missing number num__121 num__112 . . . num__97 num__91 num__86 <o> a ) num__102 <o> b ) num__108 <o> c ) num__99 <o> d ) num__104 <o> e ) num__106 |
first and second number difference is starting from num__9 and further decresed by num__1 by each iteration . num__121 num__112 . . . num__97 num__91 num__86 . . . . difference between two numbers will be as num__9 num__8 num__7 num__6 num__5 . . . answer : d <eor> d <eos> |
d |
subtract__121.0__112.0__ subtract__9.0__1.0__ subtract__8.0__1.0__ subtract__97.0__91.0__ subtract__91.0__86.0__ subtract__112.0__8.0__ |
subtract__121.0__112.0__ subtract__9.0__1.0__ subtract__8.0__1.0__ subtract__97.0__91.0__ subtract__91.0__86.0__ subtract__112.0__8.0__ |
| how many seconds will a num__700 m long train take to cross a man walking with a speed of num__3 km / hr in the direction of the moving train if the speed of the train is num__63 km / hr ? <o> a ) num__22 <o> b ) num__42 <o> c ) num__99 <o> d ) num__77 <o> e ) num__12 |
speed of train relative to man = num__63 - num__3 = num__60 km / hr . = num__60 * num__0.277777777778 = num__16.6666666667 m / sec . time taken to pass the man = num__700 * num__0.06 = num__42 sec . answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__700.0__0.06__ round__42.0__ |
subtract__63.0__3.0__ multiply__700.0__0.06__ multiply__700.0__0.06__ |
| if num__500 gm of salt solution has num__30.0 salt in it how much salt must be added to make the concentration of salt num__50.0 in the solution ? <o> a ) num__200 gm <o> b ) num__280 gm <o> c ) num__202 gm <o> d ) num__700 gm <o> e ) num__206 gm |
explanation : amount of water in solution = num__100.0 - num__30.0 = num__70.0 of total amount of water = num__70.0 x num__500 = num__350 gm amount of water remains same for both the solutions . therefore let amount of new solution be x gm . x = num__350 x ( num__2.0 ) x = num__700 salt added = ( num__700 – num__500 ) = num__200 gm alternate solution : quantity of sugar to be added : = [ solution ( required % value – present % value ) ] / ( num__100 – required % value ) = [ num__500 ( num__50 – num__30 ) ] / num__100 – num__50 = ( num__500 x num__20 ) / num__50 = num__200 gm answer : a <eor> a <eos> |
a |
subtract__100.0__30.0__ divide__100.0__50.0__ multiply__2.0__350.0__ multiply__2.0__100.0__ subtract__50.0__30.0__ multiply__2.0__100.0__ |
subtract__100.0__30.0__ divide__100.0__50.0__ multiply__2.0__350.0__ subtract__700.0__500.0__ subtract__50.0__30.0__ subtract__700.0__500.0__ |
| a person can swim in still water at num__4 km / h . if the speed of water num__2 km / h how many hours will the man take to swim back against the current for num__6 km ? <o> a ) num__5 <o> b ) num__7 <o> c ) num__3 <o> d ) num__0.5 <o> e ) num__4 |
explanation : m = num__4 s = num__2 us = num__4 - num__2 = num__2 d = num__6 t = num__3.0 = num__3 answer : c <eor> c <eos> |
c |
divide__6.0__2.0__ round__3.0__ |
divide__6.0__2.0__ subtract__6.0__3.0__ |
| a man can row num__6 kmph in still water . when the river is running at num__1.2 kmph it takes him num__1 hour to row to a place and black . what is the total distance traveled by the man ? <o> a ) num__5.29 <o> b ) num__5.2 <o> c ) num__5.76 <o> d ) num__5.12 <o> e ) num__5.93 |
m = num__6 s = num__1.2 ds = num__7.2 us = num__4.8 x / num__7.2 + x / num__4.8 = num__1 x = num__2.88 d = num__2.88 * num__2 = num__5.76 answer : c <eor> c <eos> |
c |
add__6.0__1.2__ subtract__6.0__1.2__ multiply__1.2__4.8__ round__5.76__ |
add__6.0__1.2__ subtract__6.0__1.2__ multiply__1.2__4.8__ multiply__1.2__4.8__ |
| equal weights of two alloys containing tin copper and lead in the ratio num__7 : num__4 : num__5 and num__5 : num__3 : num__8 are melted and mixed together . what is the ratio of tin copper and lead in the resultant alloy ? <o> a ) num__41 : num__81 : num__37 <o> b ) num__33 : num__91 : num__81 <o> c ) num__16 : num__7 : num__12 <o> d ) num__12 : num__70 : num__16 <o> e ) num__12 : num__7 : num__16 |
let the weight of the two alloys be w each required ratio = ( num__7 w / num__16 + num__5 w / num__16 ) : ( num__4 w / num__16 + num__3 w / num__16 ) : ( num__5 w / num__16 + num__6 w / num__16 ) = num__12 w / num__16 : num__7 w / num__16 : num__11 w / num__16 = num__12 : num__7 : num__16 answer : e <eor> e <eos> |
e |
add__7.0__5.0__ add__7.0__4.0__ add__7.0__5.0__ |
add__7.0__5.0__ add__7.0__4.0__ add__7.0__5.0__ |
| pipe a can fill a tank in num__2 hours . due to a leak at the bottom it takes num__3 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? <o> a ) num__10 <o> b ) num__9 <o> c ) num__7 <o> d ) num__6 <o> e ) num__5 |
let the leak can empty the full tank in x hours num__0.5 - num__1 / x = num__0.333333333333 = > num__1 / x = num__0.5 - num__0.333333333333 = ( num__3 - num__2 ) / num__6 = num__0.166666666667 = > x = num__6 . answer : d <eor> d <eos> |
d |
multiply__2.0__0.5__ divide__1.0__3.0__ multiply__2.0__3.0__ divide__0.5__3.0__ round__6.0__ |
subtract__3.0__2.0__ divide__1.0__3.0__ divide__3.0__0.5__ divide__0.5__3.0__ divide__3.0__0.5__ |
| each day a man meets his wife at the train station after work and then she drives him home . she always arrives exactly on time to pick him up . one day he catches an earlier train and arrives at the station an hour early . he immediately begins walking home along the same route the wife drives . eventually his wife sees him on her way to the station and drives him the rest of the way home . when they arrive home the man notices that they arrived num__16 minutes earlier than usual . how much time did the man spend walking ? <o> a ) num__45 minutes <o> b ) num__50 minutes <o> c ) num__40 minutes <o> d ) num__55 minutes <o> e ) num__52 minutes |
as they arrived num__16 minutes earlier than usual they saved num__16 minutes on round trip from home to station ( home - station - home ) - - > num__8 minutes in each direction ( home - station ) - - > wife meets husband num__8 minutes earlier the usual meeting time - - > husband arrived an hour earlier the usual meeting time so he must have spent waking the rest of the time before their meeting which is hour - num__8 minutes = num__52 minutes . answer : e <eor> e <eos> |
e |
round__52.0__ |
round__52.0__ |
| a person purchased a tv set for rs . num__16000 and a dvd player for rs . num__6250 . he sold both the items together for rs . num__31750 . what percentage of profit did he make ? <o> a ) num__42.7 <o> b ) num__27 <o> c ) num__40 <o> d ) num__26 <o> e ) num__11 |
the total cp = rs . num__16000 + rs . num__6250 = rs . num__22250 and sp = rs . num__31750 profit ( % ) = ( num__31750 - num__22250 ) / num__22250 * num__100 = num__42.7 . answer : a <eor> a <eos> |
a |
percent__100.0__42.7__ |
percent__100.0__42.7__ |
| a certain list consists of num__21 different numbers . if n is in the list and n is num__5 times the average ( arithmetic mean ) of the other num__20 numbers in the list then n is what fraction of the sum of the num__21 numbers in the list ? <o> a ) num__0.05 <o> b ) num__0.166666666667 <o> c ) num__0.2 <o> d ) num__0.190476190476 <o> e ) num__0.238095238095 |
series : a num__1 a num__2 . . . . a num__20 n sum of a num__1 + a num__2 + . . . + a num__20 = num__20 * x ( x = average ) so n = num__5 * x hence a num__1 + a num__2 + . . + a num__20 + n = num__25 x so the fraction asked = num__5 x / num__25 x = num__0.2 answer c <eor> c <eos> |
c |
subtract__21.0__20.0__ add__5.0__20.0__ reverse__5.0__ reverse__5.0__ |
subtract__21.0__20.0__ add__5.0__20.0__ reverse__5.0__ reverse__5.0__ |
| a sum of rs . num__96000 is divided into three parts such that the simple interests accrued on them for eight two and four years respectively may be equal . find the amount deposited for num__8 years . <o> a ) num__14000 <o> b ) num__16000 <o> c ) num__12000 <o> d ) num__20000 <o> e ) num__13000 |
let the amounts be x y z in ascending order of value . as the interest rate and interest accrued are same for num__2 years num__4 years and num__8 years i . e . num__2 x = num__4 y = num__8 z = k . l . c . m . of num__2 num__48 = num__8 so x : y : z : = num__48000 : num__24000 : num__12000 the amount deposited for num__11 years = num__12000 answer : c <eor> c <eos> |
c |
divide__8.0__2.0__ divide__96000.0__2.0__ divide__96000.0__4.0__ divide__96000.0__8.0__ divide__96000.0__8.0__ |
divide__8.0__2.0__ divide__96000.0__2.0__ divide__96000.0__4.0__ divide__96000.0__8.0__ divide__96000.0__8.0__ |
| in an examination num__30.0 of total students failed in hindi num__42.0 failed in english and num__28.0 in both . the percentage of these who passed in both the subjects is : <o> a ) num__23 <o> b ) num__37 <o> c ) num__56 <o> d ) num__40 <o> e ) num__81 |
explanation : formula n ( a â ˆ ª b ) = n ( a ) + n ( b ) â ˆ ’ n ( a â ˆ © b ) fail in hindi or english = num__30 + num__42 â € “ num__28 = num__44 therefore students who passed = num__100 â € “ num__44 = num__56 . answer : c <eor> c <eos> |
c |
subtract__100.0__44.0__ subtract__100.0__44.0__ |
subtract__100.0__44.0__ subtract__100.0__44.0__ |
| the average weight of num__8 person ' s increases by num__2.5 kg when a new person comes in place of one of them weighing num__65 kg . what might be the weight of the new person ? <o> a ) num__85 kg <o> b ) num__87 kg <o> c ) num__90 kg <o> d ) num__91 kg <o> e ) num__93 kg |
total weight increased = ( num__8 x num__2.5 ) kg = num__20 kg . weight of new person = ( num__65 + num__20 ) kg = num__85 kg . answer is a <eor> a <eos> |
a |
multiply__8.0__2.5__ add__65.0__20.0__ add__65.0__20.0__ |
multiply__8.0__2.5__ add__65.0__20.0__ add__65.0__20.0__ |
| num__25 liters of a mixture is created by mixing liquid p and liquid q in the ratio num__3 : num__2 . how many liters of liquid q must be added to make the ratio num__5 : num__4 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
let x be the amount of liquid q to be added . ( num__0.4 ) * num__25 + x = ( num__0.444444444444 ) * ( num__25 + x ) num__450 + num__45 x = num__500 + num__20 x num__25 x = num__50 x = num__2 the answer is b . <eor> b <eos> |
b |
divide__2.0__5.0__ subtract__25.0__5.0__ multiply__25.0__2.0__ subtract__5.0__3.0__ |
divide__2.0__5.0__ multiply__5.0__4.0__ multiply__25.0__2.0__ multiply__5.0__0.4__ |
| what is the place value of num__7 in the numeral num__2734 ? <o> a ) num__600 <o> b ) num__650 <o> c ) num__700 <o> d ) num__800 <o> e ) num__850 |
option ' c ' num__7 * num__100 = num__700 <eor> c <eos> |
c |
multiply__7.0__100.0__ multiply__7.0__100.0__ |
multiply__7.0__100.0__ multiply__7.0__100.0__ |
| lali builds a tower using only red green and blue toy bricks in a ratio of num__4 : num__3 : num__1 . she then removes num__0.5 of the green bricks and adds num__0.333333333333 more blue bricks reducing the size of the tower by num__14 bricks . how many red bricks will she need to add in order to double the total number of bricks used to build the original tower ? <o> a ) num__82 <o> b ) num__96 <o> c ) num__110 <o> d ) num__120 <o> e ) num__192 |
lali want to double the original amount of blocks not just the red blocks the original tower had num__96 bricks the new tower has num__82 bricks and the question is asking how many extra red blocks are needed to build a tower of num__192 bricks ( double the original : num__96 * num__2 ) . num__192 - num__82 = num__110 so num__110 bricks need to be added to the new num__82 brick tower to double the original . they worded the question ashow many red bricks need to be addedbut really it could be any colour just that num__110 more bricks need to be there to equal num__192 = e <eor> e <eos> |
e |
subtract__96.0__14.0__ divide__96.0__0.5__ reverse__0.5__ add__14.0__96.0__ multiply__1.0__192.0__ |
subtract__96.0__14.0__ divide__96.0__0.5__ multiply__4.0__0.5__ subtract__192.0__82.0__ multiply__1.0__192.0__ |
| the h . c . f . of two numbers is num__23 and the other two factors of their l . c . m . are num__13 and num__14 . the larger of the two numbers is : <o> a ) num__276 <o> b ) num__299 <o> c ) num__322 <o> d ) num__345 <o> e ) num__390 |
the numbers are ( num__23 x num__13 ) and ( num__23 x num__14 ) . larger number = ( num__23 x num__14 ) = num__322 . answer : option c <eor> c <eos> |
c |
multiply__23.0__14.0__ multiply__23.0__14.0__ |
multiply__23.0__14.0__ multiply__23.0__14.0__ |
| in a recent election ms . robbins received num__7000 votes cast by independent voters that is voters not registered with a specific political party . she also received num__20 percent of the votes cast by those voters registered with a political party . if n is the total number of votes cast in the election and num__50 percent of the votes cast were cast by independent voters which of the following represents the number of votes that ms . robbins received ? <o> a ) num__0.06 n + num__3200 <o> b ) num__0.1 n + num__7000 <o> c ) num__0.4 n + num__7200 <o> d ) num__0.1 n + num__8000 <o> e ) num__0.06 n + num__8 |
000 |
i will go with option b according to question num__50 percent of total votes ( n ) cas were cast by independent voters . this implies num__50 percent of n are voters that belong to political party . and she received num__7000 votes from independent voters and num__20 percent from registered voters ( i . e num__0.2 * num__0.5 * n = num__0.1 n ) . hence total votes = num__7000 + num__0.1 n . . . . option b <eor> b <eos> |
b |
b |
| a man can row with a speed of num__15 kmph in still water . if the stream flows at num__15 kmph then the speed in downstream is ? <o> a ) num__30 kmph <o> b ) num__5 kmph <o> c ) num__20 kmph <o> d ) num__22 kmph <o> e ) num__25 kmph |
m = num__15 s = num__15 ds = num__15 + num__15 = num__30 answer : a <eor> a <eos> |
a |
round__30.0__ |
round__30.0__ |
| num__562 Ã — num__99 = ? <o> a ) num__55684 <o> b ) num__55432 <o> c ) num__55638 <o> d ) num__44530 <o> e ) num__44538 |
num__562 Ã — ( num__100 - num__1 ) num__56200 - num__562 = num__55638 answer c <eor> c <eos> |
c |
subtract__100.0__99.0__ multiply__562.0__100.0__ multiply__562.0__99.0__ multiply__562.0__99.0__ |
subtract__100.0__99.0__ multiply__562.0__100.0__ subtract__56200.0__562.0__ subtract__56200.0__562.0__ |
| what will be the ci on a sum of rs . num__25000 after num__3 years at the rate of num__12 per year ? <o> a ) rs . num__10110.20 <o> b ) rs . num__10120 <o> c ) rs . num__10123.20 <o> d ) rs . num__10124 <o> e ) rs . num__10125.12 |
amount = rs . num__25000 x num__1 + num__12 num__3 num__100 = rs . num__25000 x num__28 x num__28 x num__28 num__25 num__25 num__25 = rs . num__35123.20 c . i . = rs . ( num__35123.20 - num__25000 ) = rs . num__10123.20 c <eor> c <eos> |
c |
percent__100.0__10123.2__ |
percent__100.0__10123.2__ |
| the average age of num__12 men is increased by years when two of them whose ages are num__21 years and num__23 years are replaced by two new men . the average age of the two new men is <o> a ) num__22 <o> b ) num__30 <o> c ) num__34 <o> d ) num__38 <o> e ) num__27 |
total age increased = ( num__12 * num__2 ) years = num__24 years . sum of ages of two new men = ( num__21 + num__23 + num__24 ) years = num__68 years average age of two new men = ( num__34.0 ) years = num__34 years . answer : c <eor> c <eos> |
c |
subtract__23.0__21.0__ multiply__12.0__2.0__ divide__68.0__2.0__ divide__68.0__2.0__ |
subtract__23.0__21.0__ multiply__12.0__2.0__ divide__68.0__2.0__ divide__68.0__2.0__ |
| what percent is num__2 gm of num__1 kg ? <o> a ) num__0.5 <o> b ) num__0.2 <o> c ) num__1.5 <o> d ) num__2.0 <o> e ) num__3 % |
num__1 kg = num__1000 gm num__0.002 × num__100 = num__0.2 = num__0.2 = num__0.2 b ) <eor> b <eos> |
b |
percent__100.0__0.2__ |
percent__100.0__0.2__ |
| what is the value of ( num__2 ) ^ - num__3 ? <o> a ) num__0.166666666667 <o> b ) num__0.2 <o> c ) num__0.125 <o> d ) num__0.111111111111 <o> e ) num__0.142857142857 |
num__2 ^ - num__3 = num__1 / ( num__2 ) ^ num__3 = num__0.125 answer : c <eor> c <eos> |
c |
subtract__3.0__2.0__ multiply__0.125__1.0__ |
subtract__3.0__2.0__ divide__0.125__1.0__ |
| if a card is drawn from a well shuffled deck of cards what is the probability of drawing a black card or a face card ? <o> a ) num__0.538461538462 <o> b ) num__0.576923076923 <o> c ) num__0.75 <o> d ) num__0.615384615385 <o> e ) num__0.557692307692 |
p ( b á ´ œ f ) = p ( b ) + p ( f ) - p ( b â ˆ © f ) where b denotes black cards and f denotes face cards . p ( b á ´ œ f ) = num__0.5 + num__0.230769230769 - num__0.115384615385 = num__0.615384615385 answer : d <eor> d <eos> |
d |
union_prob__0.5__0.2308__0.1154__ union_prob__0.5__0.2308__0.1154__ |
union_prob__0.5__0.2308__0.1154__ union_prob__0.5__0.2308__0.1154__ |
| the sum of num__3 consecutive numbers is definitely <o> a ) positive . <o> b ) divisible by num__3 . <o> c ) divisible by num__2 . <o> d ) divisible by num__4 . <o> e ) divisible by num__5 . |
if num__1 st term is x : x + ( x + num__1 ) + ( x + num__2 ) = num__3 x + num__3 - - - > always divisible by num__3 if num__2 nd term is x : ( x - num__1 ) + x + ( x + num__1 ) = num__3 x - - - > always divisible by num__3 if num__3 rd term is x : ( x - num__2 ) + ( x - num__1 ) + x = num__3 x - num__3 - - - > always divisible by num__3 answer : b <eor> b <eos> |
b |
subtract__3.0__1.0__ round__3.0__ |
subtract__3.0__1.0__ add__1.0__2.0__ |
| if w x and y are consecutive even positive integers and w < x < y which of the following could be equal to y - x - w ? <o> a ) - num__4 <o> b ) - num__2 <o> c ) - num__1 <o> d ) num__0 <o> e ) num__3 |
since w x and y are consecutive even positive integers and w < x < y we can start with the easiest set of values that comes to mind : w = num__2 x = num__4 y = num__6 the question asks for what could be the value of y - x - w . in this case the value would be . . . num__6 - num__4 - num__2 = num__0 num__0 is among the answer choices so we ' re done . d <eor> d <eos> |
d |
add__2.0__4.0__ multiply__0.0__2.0__ |
add__2.0__4.0__ multiply__0.0__2.0__ |
| num__75.0 of a number when added to num__75 is equal to the number . the number is ? <o> a ) num__150 <o> b ) num__200 <o> c ) num__225 <o> d ) num__300 <o> e ) none |
answer num__75 + num__75.0 of n = n ⇒ n - num__3 n / num__4 = num__75 ⇒ n / num__4 = num__75 ∴ n = ( num__75 x num__4 ) = num__300 correct option : d <eor> d <eos> |
d |
multiply__75.0__4.0__ multiply__75.0__4.0__ |
multiply__75.0__4.0__ multiply__75.0__4.0__ |
| given two sets a = { num__11 num__66 num__77 } and b = { num__01 } if one number is chosen from each set at random what is the probability that the sum of both numbers is an even number <o> a ) num__0.25 <o> b ) num__0.166666666667 <o> c ) num__0.5 <o> d ) num__0.333333333333 <o> e ) num__0.2 |
one way to look at it : the number from set a can be anything . the number selected from set b will determine whether the sum is odd or even . for example if a num__6 is selected from set a we need a num__0 from set b to get an even sum . if a num__7 is selected from set a we need a num__1 from set b to get an even sum . and so on . so p ( sum is even ) = p ( select any number from set aandselect the number from set b that makes the sum even ) = p ( select any number from set a ) xp ( select the number from set b that makes the sum even ) = num__1 x num__0.5 = num__0.5 = c <eor> c <eos> |
c |
divide__66.0__11.0__ divide__77.0__11.0__ subtract__1.0__0.5__ |
divide__66.0__11.0__ divide__77.0__11.0__ subtract__1.0__0.5__ |
| an outlet pipe empties a tank which is full in num__5 hours . if the inlet pipe is kept open which lets water in at the rate of num__4 litres / min then outlet pipe would take num__3 hours longer . find the capacity of the tank . <o> a ) num__8600 litres <o> b ) num__3200 litres <o> c ) num__12800 litres <o> d ) num__11200 litres <o> e ) num__13200 litres |
let the rate of outlet pipe be x liters / hour ; rate of inlet pipe is num__8 litres / min or num__4 * num__60 = num__240 liters / hour ; net outflow rate when both pipes operate would be x - num__240 liters / hour . capacity of the tank = x * num__5 hours = ( x - num__240 ) * ( num__5 + num__3 ) hours num__5 x = ( x - num__240 ) * num__8 - - > x = num__640 - - > capacity = num__5 x = num__3200 liters . answer : b . <eor> b <eos> |
b |
add__5.0__3.0__ hour_to_min_conversion__ multiply__4.0__60.0__ multiply__5.0__640.0__ round__3200.0__ |
add__5.0__3.0__ hour_to_min_conversion__ multiply__4.0__60.0__ multiply__5.0__640.0__ multiply__5.0__640.0__ |
| some persons can do a piece of work in num__20 days . two times the number of these people will do half of that work in ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__8 |
num__20 / ( num__2 * num__2 ) = num__5 days answer : c <eor> c <eos> |
c |
round__5.0__ |
round__5.0__ |
| if log ( a / b ) + log ( b / a ) = log ( a + b ) then : <o> a ) a = b <o> b ) a + b = num__0 <o> c ) a - b = num__1 <o> d ) a + b = num__1 <o> e ) a - b = num__0 |
log ( a / b ) + log ( b / a ) = log ( a + b ) log ( a + b ) = log { ( a / b ) x ( b / a ) } = log num__1 . so a + b = num__1 . answer is d <eor> d <eos> |
d |
reverse__1.0__ |
reverse__1.0__ |
| a man can row with a speed of num__15 kmph in still water . if the stream flows at num__5 kmph then the speed in downstream is ? <o> a ) num__27 <o> b ) num__27 <o> c ) num__20 <o> d ) num__99 <o> e ) num__2 |
m = num__15 s = num__5 ds = num__15 + num__5 = num__20 ' answer : c <eor> c <eos> |
c |
add__15.0__5.0__ round__20.0__ |
add__15.0__5.0__ add__15.0__5.0__ |
| aishwarya was first to board to her flight to delhi . she forgot her seat number and picks a random seat for herself . after this every single person who get to the flight sits on his seat if its available else chooses any available seat at random . abhishek is last to enter the flight and at that moment num__0.99 seats were occupied . whats the probability what abhishek gets to sit in his own seat ? <o> a ) num__0.5 <o> b ) num__0.2 <o> c ) num__0.222222222222 <o> d ) num__0.571428571429 <o> e ) num__0.5 |
a num__0.5 one of two is the possibility num__1 . if any of the first num__99 people sit in abhishek seat abhishek will not get to sit in his own seat . num__2 . if any of the first num__99 people sit in aishwarya ' s seat abhishek will get to sit in his seat . <eor> a <eos> |
a |
coin_space__ negate_prob__0.5__ |
coin_space__ negate_prob__0.5__ |
| a diagonal of a polygon is an segment between two non - adjacent vertices of the polygon . how many diagonals does a regular num__10 - sided polygon have ? <o> a ) num__875 <o> b ) num__35 <o> c ) num__1425 <o> d ) num__2025 <o> e ) num__2500 |
there ' s a direct formula for this . number of diagonals in a regular polygon = [ n * ( n - num__3 ) ] / num__2 n = number of sides of the regular polygon . here n = num__10 . plugging it in we get num__35 diagonals ! answer ( b ) . <eor> b <eos> |
b |
triangle_area__35.0__2.0__ |
triangle_area__35.0__2.0__ |
| the length of a rectangular plot is num__20 metres more than its breadth . if the cost of fencing the plot @ rs . num__26.50 per metre is rs . num__9540 what is the length of the plot in metres ? <o> a ) num__20 <o> b ) num__200 <o> c ) num__300 <o> d ) num__400 <o> e ) num__100 |
let length of plot = l meters then breadth = l - num__20 meters and perimeter = num__2 [ l + l - num__20 ] = [ num__4 l - num__40 ] meters [ num__4 l - num__40 ] * num__26.50 = num__9540 [ num__4 l - num__40 ] = num__9540 / num__26.50 = num__360 num__4 l = num__400 l = num__100.0 = num__100 meters . answer : e <eor> e <eos> |
e |
multiply__20.0__2.0__ divide__9540.0__26.5__ add__40.0__360.0__ divide__400.0__4.0__ round__100.0__ |
multiply__20.0__2.0__ divide__9540.0__26.5__ add__40.0__360.0__ divide__400.0__4.0__ divide__400.0__4.0__ |
| every day daniel drives num__64 miles back from work . on sunday daniel drove all the way back from work at a constant speed of x miles per hour . on monday daniel drove the first num__32 miles back from work at ( num__2 x ) miles per hour and the rest of the way at ( x / num__2 ) miles per hour . the time it took daniel to drive back from work on monday is longer than the time it took him to drive back from work on sunday by what percent ? <o> a ) num__15.0 <o> b ) num__25.0 <o> c ) num__35.0 <o> d ) num__45.0 <o> e ) num__55 % |
let ' s test x = num__4 . . . . on sunday daniel drove num__64 miles at num__4 miles / hour . d = ( r ) ( t ) num__64 = ( num__4 ) ( t ) num__16.0 = num__16 = t it takes num__16 hours to drive home on monday daniel drove the first num__32 miles at ( num__2 ) ( num__4 ) = num__8 miles / hour and the rest of the way ( num__32 miles ) at num__2.0 = num__2 miles / hour d = ( r ) ( t ) num__32 = ( num__8 ) ( t ) num__4.0 = num__4 = t it takes num__4 hours for the first part d = ( r ) ( t ) num__32 = ( num__2 ) ( t ) num__16.0 = num__16 = t it takes num__16 hours for the second part total time to drive home on monday = num__4 + num__16 = num__20 hours we ' re asked by what percent num__20 hours is greater than num__16 hours . num__1.25 = num__1.25 so it is num__25.0 greater . b <eor> b <eos> |
b |
divide__64.0__4.0__ divide__32.0__4.0__ add__4.0__16.0__ divide__20.0__16.0__ multiply__1.25__20.0__ round__25.0__ |
divide__64.0__4.0__ divide__32.0__4.0__ add__4.0__16.0__ divide__20.0__16.0__ multiply__1.25__20.0__ round__25.0__ |
| three fifth of the square of a certain number is num__126.15 . what is the number ? <o> a ) num__210.25 <o> b ) num__75.69 <o> c ) num__14.5 <o> d ) num__145 <o> e ) num__150 |
let the number be x given num__0.6 x ^ num__2 = num__126.5 x ^ num__2 = num__126.15 * num__1.66666666667 x ^ num__2 = num__42.05 * num__5 = num__210.25 x = num__14.5 answer c . <eor> c <eos> |
c |
multiply__5.0__42.05__ triangle_area__2.0__14.5__ |
multiply__5.0__42.05__ triangle_area__2.0__14.5__ |
| the greatest common factor of positive integers m and n is num__12 . what is the greatest common factor of ( num__2 m ^ num__2 num__2 n ^ num__2 ) ? <o> a ) num__2 <o> b ) num__12 <o> c ) num__24 <o> d ) num__144 <o> e ) num__288 |
given : gcd of ( m and n ) = num__12 = num__2 ^ num__2 * num__3 i . e . m and n are both multiples of num__2 ^ num__2 * num__3 i . e . m ^ num__2 and n ^ num__2 will both be multiples of ( num__2 ^ num__2 * num__3 ) ^ num__2 = num__2 ^ num__4 * num__3 ^ num__2 i . e . num__2 m ^ num__2 and num__2 n ^ num__2 will both be multiples of num__2 ( num__2 ^ num__2 * num__3 ) ^ num__2 = num__2 ^ num__5 * num__3 ^ num__2 = num__288 answer : option e <eor> e <eos> |
e |
divide__12.0__3.0__ add__2.0__3.0__ lcm__12.0__288.0__ |
divide__12.0__3.0__ add__2.0__3.0__ lcm__12.0__288.0__ |
| age of raghav who is num__30 years old is num__0.6 times of ram and ram is older than mohan by num__35 years . if vijay ' s age is just between the age of ramesh and and mohan and ramesh is num__21 years old . what is age of vijay ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__18 <o> d ) num__17 <o> e ) num__16 |
solution : raghav ' s age = num__0.6 ram ' s age = > num__30 = num__0.6 ram ' s age = > ram ' s age = num__10.0 * num__5 = num__50 ram ' s age = mohan ' s age + num__35 = > mohan ' s age = ram ' s age - num__35 = > mohan ' s age = num__50 - num__35 = > mohan ' s age = num__15 vijay ' s age = ( ramesh ' s age + mohan ' s age ) / num__2 = > vijay ' s age = ( num__21 + num__15 ) / num__2 = > vijay ' s age = num__18 years . answer c <eor> c <eos> |
c |
subtract__35.0__30.0__ divide__30.0__0.6__ add__5.0__10.0__ divide__30.0__15.0__ multiply__30.0__0.6__ multiply__30.0__0.6__ |
subtract__35.0__30.0__ divide__30.0__0.6__ add__5.0__10.0__ divide__30.0__15.0__ multiply__30.0__0.6__ multiply__30.0__0.6__ |
| a man walking at the rate of num__6 km per hour crosses a square field diagonally in num__9 seconds the area of the field is <o> a ) num__110 sq . m <o> b ) num__111.5 aq . m <o> c ) num__112.5 sq . m <o> d ) num__114 sq . m <o> e ) none of these |
explanation : distance covered in num__9 seconds = ( num__6 × num__0.277777777778 ) × num__9 = num__15 m diagonal of square field = num__15 m side of square = a then diagonal of that square = √ num__2 a hence area of the square = a num__2 = ( num__152 ) / num__2 = num__112.5 sq . m answer : option c <eor> c <eos> |
c |
triangle_area__112.5__2.0__ |
triangle_area__112.5__2.0__ |
| a is half good a work man as b and together they finish a job in num__26 days . in how many days working alone b finish the job ? <o> a ) num__98 days <o> b ) num__21 days <o> c ) num__17 days <o> d ) num__39 days <o> e ) num__19 days |
wc = num__1 : num__2 num__2 x + x = num__0.0384615384615 = > x = num__0.0128205128205 num__2 x = num__0.0128205128205 = > num__39 days answer : d <eor> d <eos> |
d |
divide__1.0__26.0__ round__39.0__ |
divide__1.0__26.0__ round__39.0__ |
| a and b can do a piece of work in num__6 days b and c in num__7 days c and a in num__8 days . how long will c take to do it ? <o> a ) num__12.6 days <o> b ) num__14.4 days <o> c ) num__15.2 days <o> d ) num__16.8 days <o> e ) num__18.2 days |
num__2 c = num__0.142857142857 + num__0.125 – num__0.166666666667 = num__0.119047619048 = num__0.119047619048 c = num__0.0595238095238 = > num__16.8 = num__16.8 days the answer is d . <eor> d <eos> |
d |
subtract__8.0__6.0__ divide__0.119__2.0__ round__16.8__ |
subtract__8.0__6.0__ divide__0.119__2.0__ round__16.8__ |
| two brothers ram and ravi appeared for an exam . the probability of selection of ram is num__0.714285714286 and that of ravi is num__0.2 . find the probability that both of them are selected . <o> a ) num__0.0571428571429 <o> b ) num__0.666666666667 <o> c ) num__0.2 <o> d ) num__0.142857142857 <o> e ) num__1.4 |
let a be the event that ram is selected and b is the event that ravi is selected . p ( a ) = num__0.714285714286 p ( b ) = num__0.2 let c be the event that both are selected . p ( c ) = p ( a ) x p ( b ) as a and b are independent events : = num__0.714285714286 x num__0.2 = num__0.142857142857 answer : d <eor> d <eos> |
d |
multiply__0.7143__0.2__ multiply__0.7143__0.2__ |
multiply__0.7143__0.2__ multiply__0.7143__0.2__ |
| three business people who wish to invest in a new company . each person is willing to pay one third of the total investment . . after careful calculation they realize that each of them would pay $ num__7000 less if they could find two more equal investors . how much is the total investment in the new business . <o> a ) a ) $ num__64000 <o> b ) b ) $ num__54000 <o> c ) c ) $ num__21000 <o> d ) d ) $ num__52500 <o> e ) e ) $ num__3 |
600 |
initially each invest in x . hence total investment is num__3 x . total investment is also num__5 ( x - num__7000 ) . num__3 x = num__5 ( x - num__7000 ) x = num__5 * num__3500.0 = num__17500 num__3 x = num__52500 and the answer is d . <eor> d <eos> |
d |
d |
| ron walks to a viewpoint and returns to the starting point by his car and thus takes a total time of num__6 hours num__45 minutes . he would have gained num__2 hours by driving both ways . how long w would it have taken for him to walk both ways . <o> a ) num__8 h num__45 min <o> b ) num__7 h num__45 min <o> c ) num__6 h num__45 min <o> d ) num__5 h num__30 min <o> e ) none of these |
num__1 . walking to to a viewpoint + driving back = num__6 hours num__45 minutes num__2 . driving to a viewpoint + driving back = num__6 hours num__45 minutes - num__2 hours = num__4 hours num__45 minutes thereforeone way driving = num__4 hours num__45 minutes / num__2 = num__2 hours num__22.5 minutes . num__3 . from num__1 . one way driving = num__6 hours num__45 minutes - num__2 hours num__22.5 minutes = num__4 hours num__22.5 minutes . num__4 . walking to to a viewpoint + walking back w = num__4 hours num__22.5 minutes + num__4 hours num__22.5 minutes = num__8 hours num__45 minutes . answer : a . <eor> a <eos> |
a |
subtract__6.0__2.0__ divide__45.0__2.0__ divide__6.0__2.0__ add__6.0__2.0__ round__8.0__ |
subtract__6.0__2.0__ divide__45.0__2.0__ divide__6.0__2.0__ add__6.0__2.0__ add__6.0__2.0__ |
| a and b can do a work in num__12 days and num__36 days respectively . if they work on alternate days beginning with b in how many days will the work be completed ? <o> a ) num__11 <o> b ) num__17 <o> c ) num__10 <o> d ) num__17 <o> e ) num__18 |
the work done in the first two days = num__0.0833333333333 + num__0.0277777777778 = num__0.111111111111 so num__9 such two days are required to finish the work . i . e . num__18 days are required to finish the work . answer : e <eor> e <eos> |
e |
add__0.0278__0.0833__ round__18.0__ |
add__0.0278__0.0833__ round__18.0__ |
| if the range e of the six numbers num__4 num__314 num__710 and x is num__12 what is the difference between the greatest possible value of x and least possible value of x ? <o> a ) num__0 <o> b ) num__2 <o> c ) num__12 <o> d ) num__13 <o> e ) num__15 |
the range e of a set is the difference between the largest and smallest elements of a set . without x the difference between the largest and smallest elements of a set is num__14 - num__3 = num__11 < num__12 which means that in order num__12 to be the range of the set x must be either the smallest element so that num__14 - x = num__12 - - - > x = num__2 or x must the largest element so that x - num__3 = num__12 - - > x = num__15 . the the difference between the greatest possible value of x and least possible value of x is num__15 - num__2 = num__13 . answer : d . <eor> d <eos> |
d |
divide__12.0__4.0__ subtract__14.0__3.0__ subtract__14.0__12.0__ add__4.0__11.0__ add__2.0__11.0__ add__2.0__11.0__ |
divide__12.0__4.0__ subtract__14.0__3.0__ subtract__14.0__12.0__ add__4.0__11.0__ subtract__15.0__2.0__ subtract__15.0__2.0__ |
| num__8 identical machines working alone and at their constant rates take num__6 hours to complete a job lot . how long would it take for num__5 such machines to perform the same job ? <o> a ) num__2.25 hours <o> b ) num__8.75 hours <o> c ) num__12 hours <o> d ) num__14.25 hours <o> e ) num__9.6 hours |
let each machine do num__1 unit of work for num__1 hour num__8 machines - - > num__8 units of work in num__1 hour for num__6 hours = num__8 * num__6 = num__48 units of total work is done . now this num__48 units of total work must be done by num__5 machines num__5 units of work ( num__5 machines ) - - - > num__1 hour for num__48 units of work num__5 * num__9.6 - - - > num__1 * num__9.6 hours e num__9.6 hours <eor> e <eos> |
e |
subtract__6.0__5.0__ multiply__8.0__6.0__ divide__48.0__5.0__ round__9.6__ |
subtract__6.0__5.0__ multiply__8.0__6.0__ divide__48.0__5.0__ round__9.6__ |
| the mean of num__50 observations is num__200 . but later he found that there is decrements of num__47 from each observations . what is the the updated mean is ? <o> a ) num__165 <o> b ) num__185 <o> c ) num__153 <o> d ) num__198 <o> e ) num__199 |
num__153 answer is c <eor> c <eos> |
c |
subtract__200.0__47.0__ subtract__200.0__47.0__ |
subtract__200.0__47.0__ subtract__200.0__47.0__ |
| car r and car y traveled the same num__80 - mile route . if car r took num__2 hours and car y traveled at an average speed that was num__50 percent faster than the average speed of car r how many hours did it take car y to travel the route ? <o> a ) num__0.666666666667 <o> b ) num__1 <o> c ) num__1.33333333333 <o> d ) num__1.6 <o> e ) num__3 |
the speed of car r is ( distance ) / ( time ) = num__40.0 = num__40 miles per hour . the speed of car y = num__1.5 * num__40 = num__60 miles per hour - - > ( time ) = ( distance ) / ( speed ) = num__1.33333333333 = num__1.33333333333 hours . answer : c . or : to cover the same distance at num__1.5 as fast rate num__0.666666666667 as much time is needed - - > ( time ) * num__0.666666666667 = num__2 * num__0.666666666667 = num__1.33333333333 hours . answer : c . <eor> c <eos> |
c |
divide__80.0__2.0__ hour_to_min_conversion__ divide__80.0__60.0__ subtract__2.0__1.3333__ divide__80.0__60.0__ |
divide__80.0__2.0__ multiply__1.5__40.0__ divide__80.0__60.0__ subtract__2.0__1.3333__ divide__80.0__60.0__ |
| a motorboat whose speed in num__15 km / hr in still water goes num__30 km downstream and comes back in a total of num__4 hours num__30 minutes . the speed of the stream ( in km / hr ) is : <o> a ) num__10 <o> b ) num__6 <o> c ) num__5 <o> d ) num__4 <o> e ) none of these |
explanation : speed of the motor boat = num__15 km / hr let speed of the stream = v speed downstream = num__15 + v km / hr speed upstream = num__15 - v km / hr time taken downstream = num__2.0 + v time taken upstream = num__2.0 - v total time = num__2.0 + v + num__2.0 - v given that total time is num__4 hours num__30 minutes = num__4 num__0.5 hour = num__4.5 hour answer is c <eor> c <eos> |
c |
divide__30.0__15.0__ divide__15.0__30.0__ add__4.0__0.5__ add__0.5__4.5__ |
divide__30.0__15.0__ divide__15.0__30.0__ add__4.0__0.5__ add__0.5__4.5__ |
| find the missing number in the series . num__2 num__3 num__6 num__0 num__10 - num__3 num__14 . . . <o> a ) - num__6 <o> b ) num__8 <o> c ) num__7 <o> d ) num__67 <o> e ) num__65 |
explanation : there are two series in the question num__1 . add num__4 : num__2 + num__3 = num__6 num__6 + num__4 = num__10 num__2 . subtract num__3 : num__3 - num__3 = num__0 num__0 - num__3 - - num__3 thus following the series : - num__3 - num__3 = - num__6 answer : a <eor> a <eos> |
a |
subtract__3.0__2.0__ add__3.0__1.0__ multiply__2.0__3.0__ |
subtract__3.0__2.0__ add__3.0__1.0__ add__2.0__4.0__ |
| a car travels at a speed of num__65 miles per hour . how far will it travel in num__7 hours ? <o> a ) num__125 miles <o> b ) num__225 miles <o> c ) num__455 miles <o> d ) num__425 miles <o> e ) num__525 miles |
during each hour the car travels num__65 miles . for num__7 hours it will travel num__65 + num__65 + num__65 + num__65 + num__65 + num__65 + num__65 = num__7 × num__65 = num__455 miles correct answer is c ) num__455 miles <eor> c <eos> |
c |
multiply__65.0__7.0__ round__455.0__ |
multiply__65.0__7.0__ round__455.0__ |
| triangle atriangle b are similar triangles with areas num__1536 units square and num__1734 units square respectively . the ratio of there corresponding height would be <o> a ) num__9 : num__10 <o> b ) num__17 : num__19 <o> c ) num__23 : num__27 <o> d ) num__16 : num__17 <o> e ) num__15 : num__23 |
let x be the height of triangle a and y be the height of triangle of b . since triangles are similar ratio of area of a and b is in the ratio of x ^ num__2 / y ^ num__2 therefore ( x ^ num__2 / y ^ num__2 ) = num__0.885813148789 ( x ^ num__2 / y ^ num__2 ) = ( num__16 * num__16 * num__6 ) / ( num__17 * num__17 * num__7 ) ( x ^ num__2 / y ^ num__2 ) = num__16 ^ num__0.117647058824 ^ num__2 x / y = num__0.941176470588 ans = d <eor> d <eos> |
d |
rectangle_perimeter__2.0__6.0__ |
rectangle_perimeter__2.0__6.0__ |
| kyle david and catherine each try independently to solve a problem . if their individual probabilities for success are num__0.333333333333 num__0.285714285714 and num__0.555555555556 respectively what is the probability that kyle and catherine but not david will solve the problem ? <o> a ) num__0.185185185185 <o> b ) num__0.255813953488 <o> c ) num__0.132275132275 <o> d ) num__0.212389380531 <o> e ) num__0.10582010582 |
p ( kyle will solve ) = num__0.333333333333 p ( david will not solve ) = num__1 - num__0.285714285714 = num__0.714285714286 p ( catherine will solve ) = num__0.555555555556 p = ( num__0.333333333333 ) * ( num__0.714285714286 ) * ( num__0.555555555556 ) = num__0.132275132275 answer : c <eor> c <eos> |
c |
subtract__1.0__0.2857__ multiply__1.0__0.1323__ |
subtract__1.0__0.2857__ multiply__1.0__0.1323__ |
| a car traveled from san diego to san francisco at an average speed of num__63 miles per hour . if the journey back took twice as long what was the average speed of the trip ? <o> a ) num__24 . <o> b ) num__32 . <o> c ) num__36 . <o> d ) num__42 . <o> e ) num__44 . |
let the time taken be = x one way distance = num__63 x total distance traveled = num__2 * num__63 x = num__126 x total time taken = x + num__2 x = num__3 x average speed = num__126 x / num__3 x = num__42 answer : d <eor> d <eos> |
d |
multiply__63.0__2.0__ divide__126.0__3.0__ round__42.0__ |
multiply__63.0__2.0__ divide__126.0__3.0__ divide__126.0__3.0__ |
| mike took a taxi to the airport and paid $ num__2.50 to start plus $ num__0.25 per mile . annie took a different route to the airport and paid $ num__2.50 plus $ num__5.00 in bridge toll fees plus $ num__0.25 per mile . if each was charged exactly the same amount and annie ' s ride was num__14 miles how many miles was mike ' s ride ? <o> a ) num__28 <o> b ) num__30 <o> c ) num__32 <o> d ) num__34 <o> e ) num__36 |
the cost of annie ' s ride was num__2.5 + num__5 + ( num__0.25 * num__14 ) = $ num__11 let x be the distance of mike ' s ride . the cost of mike ' s ride is num__2.5 + ( num__0.25 * x ) = num__11 num__0.25 * x = num__8.5 x = num__34 miles the answer is d . <eor> d <eos> |
d |
subtract__11.0__2.5__ divide__8.5__0.25__ divide__8.5__0.25__ |
subtract__11.0__2.5__ divide__8.5__0.25__ divide__8.5__0.25__ |
| the rowing athletes in a rowing conference voted for coach of the year . each rower who voted chose exactly num__3 coaches to vote for among the num__36 coaches in the conference . if each of the num__36 coaches received exactly num__5 votes ( a num__36 way tie ) how many rowers voted for coach of the year ? <o> a ) num__60 <o> b ) num__70 <o> c ) num__75 <o> d ) num__84 <o> e ) num__90 |
there were num__36 * num__5 = num__180 total votes made . if each voter picked num__3 coaches there were num__60.0 = num__60 voters . a <eor> a <eos> |
a |
multiply__36.0__5.0__ divide__180.0__3.0__ divide__180.0__3.0__ |
multiply__36.0__5.0__ divide__180.0__3.0__ divide__180.0__3.0__ |
| a seller uses num__900 gm in place of one kg to sell his goods . find his actual % profit or loss when he sells his articles at num__5.0 profit on cost price ? <o> a ) num__15.5 <o> b ) num__16.5 <o> c ) num__16.67 <o> d ) num__17.0 <o> e ) none of these |
explanation : = [ num__100 - x ] * z / y - num__100 = [ num__100 + num__5 ] num__1.11111111111 - num__100 = num__16.67 answer : c <eor> c <eos> |
c |
percent__16.67__100.0__ |
percent__16.67__100.0__ |
| due to num__10.0 decrease in the price of sugar and john can buy num__5 kg more sugar in rs num__100 then find the cp of sugar ? <o> a ) rs . num__1 ( num__0.285714285714 ) <o> b ) rs . num__2 ( num__0.111111111111 ) <o> c ) rs . num__3 ( num__0.181818181818 ) <o> d ) rs . num__2 ( num__0.153846153846 ) <o> e ) rs . num__2 ( num__0.285714285714 ) |
here r = num__10.0 x = num__100 and a = num__5 kg actual price of sugar = num__10 * num__100 / ( ( num__100 - num__10 ) * num__5 ) = rs . num__2 ( num__0.222222222222 ) b <eor> b <eos> |
b |
divide__10.0__5.0__ divide__10.0__5.0__ |
divide__10.0__5.0__ divide__10.0__5.0__ |
| by selling an shirt for $ num__300 a shop keeper gains num__20.0 . during a clearance sale the shopkeeper allows a discount of num__10.0 on the marked price . his gain percent during the sale is : <o> a ) num__2 <o> b ) num__14 <o> c ) num__8.0 <o> d ) num__12 <o> e ) num__10 |
c num__8.0 marked price = $ num__300 c . p . = num__0.833333333333 * num__300 = $ num__250 sale price = num__90.0 of $ num__300 = $ num__270 required gain % = num__0.08 * num__100 = num__8.0 . <eor> c <eos> |
c |
percent__90.0__300.0__ percent__100.0__8.0__ |
percent__90.0__300.0__ percent__100.0__8.0__ |
| the probability that a will not tell truth is num__75.0 and the probability that b will tell truth is num__60.0 . what is the probability that an incident described by them is true ? <o> a ) num__10.0 <o> b ) num__12.0 <o> c ) num__15.0 <o> d ) num__17.0 <o> e ) num__19 % |
a truth . . . . > num__25.0 b truth . . . . . > num__60.0 a & b truth . . . > num__0.25 * num__0.6 = num__0.15 = num__15.0 answer : c <eor> c <eos> |
c |
percent__60.0__0.25__ percent__60.0__25.0__ percent__60.0__25.0__ |
percent__60.0__0.25__ percent__60.0__25.0__ percent__60.0__25.0__ |
| how long does a train num__130 m long running at the speed of num__65 km / hr takes to cross a bridge num__150 m length ? <o> a ) num__15.7 sec <o> b ) num__15.1 sec <o> c ) num__15.5 sec <o> d ) num__17.1 sec <o> e ) num__16.7 sec |
speed = num__65 * num__0.277777777778 = num__18.1 m / sec total distance covered = num__130 + num__150 = num__280 m . required time = num__280 / num__18.1 ' = num__15.5 sec . answer : c <eor> c <eos> |
c |
add__130.0__150.0__ round__15.5__ |
add__130.0__150.0__ round__15.5__ |
| find the value of num__72519 x num__9999 = m ? <o> a ) num__723437481 <o> b ) num__725465481 <o> c ) num__625117481 <o> d ) num__725117481 <o> e ) num__725118081 |
num__72519 x num__9999 = num__72519 x ( num__10000 - num__1 ) = num__72519 x num__10000 - num__72519 x num__1 = num__725190000 - num__72519 = num__725117481 d <eor> d <eos> |
d |
subtract__10000.0__9999.0__ multiply__72519.0__10000.0__ multiply__72519.0__9999.0__ multiply__72519.0__9999.0__ |
subtract__10000.0__9999.0__ multiply__72519.0__10000.0__ subtract__725190000.0__72519.0__ subtract__725190000.0__72519.0__ |
| in how many ways can num__4 rings be arranged in num__5 finger ( no finger has more than num__1 ring ) ? <o> a ) num__5 c num__4 <o> b ) num__5 p num__4 <o> c ) ( num__5 c num__4 ) num__4 ! <o> d ) b and c <o> e ) none of the above |
no . of ways of selecting num__4 fingers from num__5 ( considering a non - mutant ) = num__5 c num__4 . no . of ways of arranging the num__4 rings = num__4 ! answer : ( num__5 c num__4 ) . num__4 ! answer : c <eor> c <eos> |
c |
vowel_space__ |
vowel_space__ |
| in num__20 years a will be twice as old as b was num__20 years ago . if a is now num__10 years older than b the present age of b is ? <o> a ) num__35 yrs <o> b ) num__40 yrs <o> c ) num__50 yrs <o> d ) num__60 yrs <o> e ) num__70 yrs |
let b ' s present age = x years then a ' s present age = x + num__10 years x + num__10 + num__20 = num__2 ( x - num__20 ) x + num__30 = num__2 x - num__40 x = num__70 years answer is e <eor> e <eos> |
e |
divide__20.0__10.0__ add__20.0__10.0__ multiply__20.0__2.0__ add__40.0__30.0__ add__40.0__30.0__ |
divide__20.0__10.0__ add__20.0__10.0__ add__10.0__30.0__ add__40.0__30.0__ add__40.0__30.0__ |
| in a new housing development trees are to be planted along the sidewalk of a certain street . each tree takes up one square foot of sidewalk space and there are to be num__9 feet between each tree . how many trees can be planted if the road is num__151 feet long ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__16 |
let t be the number of trees . then the length required for trees on the sidewalk will be num__1 * t = t to maximize the number of trees the number of num__14 feet spaces between trees should be num__1 less than total number of trees . for example if there are num__3 trees then there should be num__2 spaces between them . so the number of num__9 feet spaces will be t - num__1 . then the length of sidewalk required for num__9 feet spaces will be num__9 * ( t - num__1 ) it is given that total lenght of sidewalk is num__151 feet . or num__9 ( t - num__1 ) + t = num__151 or num__9 t - num__9 + t = num__151 or num__10 t = num__160 or t = num__16 answer : - e <eor> e <eos> |
e |
rectangle_perimeter__2.0__3.0__ multiply__1.0__16.0__ |
rectangle_perimeter__2.0__3.0__ multiply__1.0__16.0__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__120 metres and they cross each other in num__12 seconds then the speed of each train ( in km / hr ) is : <o> a ) num__31 km / hr . <o> b ) num__32 km / hr . <o> c ) num__34 km / hr . <o> d ) num__35 km / hr . <o> e ) num__36 km / hr . |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__120 + num__120 ) / num__12 num__2 x = num__20 x = num__10 . speed of each train = num__10 m / sec = num__10 x num__3.6 km / hr = num__36 km / hr . answer : e <eor> e <eos> |
e |
divide__120.0__12.0__ multiply__10.0__3.6__ round__36.0__ |
divide__120.0__12.0__ multiply__10.0__3.6__ round__36.0__ |
| by selling a bat for rs . num__450 / - a man loses num__10.0 what is that bat cost price <o> a ) s . num__600 / - <o> b ) s . num__500 / - <o> c ) s . num__700 / - <o> d ) s . num__780 / - <o> e ) s . num__800 / - |
num__90.0 - - - - - - > num__450 ( num__90 * num__5 = num__450 ) num__100.0 - - - - - - > num__500 ( num__100 * num__5 = num__500 ) cost price = rs . num__500 / - b ) <eor> b <eos> |
b |
divide__450.0__90.0__ add__10.0__90.0__ multiply__100.0__5.0__ multiply__100.0__5.0__ |
divide__450.0__90.0__ add__10.0__90.0__ multiply__100.0__5.0__ multiply__100.0__5.0__ |
| the sum of the first num__40 positive even integers is num__1640 . what is the sum of the first num__40 odd integers ? <o> a ) num__1500 <o> b ) num__1600 <o> c ) num__1400 <o> d ) num__1200 <o> e ) num__1000 |
sum of first n even numbers = n ( n + num__1 ) = num__1640 sum of first n odd numbers = n ^ num__2 = num__40 * num__40 = num__1600 ( here n = num__40 ) answer : b <eor> b <eos> |
b |
subtract__1640.0__40.0__ subtract__1640.0__40.0__ |
subtract__1640.0__40.0__ multiply__1.0__1600.0__ |
| if a = ( num__0.5 ) b and c = num__8 a then which of the following represents the average ( arithmetic mean ) of a b and c in terms of a ? <o> a ) a + num__4 <o> b ) ( num__3.66666666667 ) a <o> c ) num__4 a <o> d ) ( num__4 num__0.142857142857 ) a <o> e ) ( num__7 num__0.25 ) a |
official answer : b the average of the three variables isa + b + c / num__3 . however we need to solve in terms of a which means we must convert b and c into something in terms of a . were told that a = num__0.5 b which is equivalent to b = num__2 a . we can plug that in and simplify the average to : a + num__2 a + c / num__3 we also know that c = num__8 a which we can plug directly into the average expression : a + num__2 a + num__8 a / num__3 = num__11 a / num__3 = ( num__3.66666666667 ) a choiceb . <eor> b <eos> |
b |
reverse__0.5__ add__8.0__3.0__ divide__11.0__3.0__ divide__11.0__3.0__ |
reverse__0.5__ add__8.0__3.0__ divide__11.0__3.0__ divide__11.0__3.0__ |
| an article is bought for rs . num__675 and sold for rs . num__900 find the gain percent ? <o> a ) num__33 num__0.111111111111 % <o> b ) num__33 num__0.5 % <o> c ) num__33 num__0.333333333333 % <o> d ) num__33 num__1.66666666667 % <o> e ) num__35 num__0.333333333333 % |
num__675 - - - - num__225 num__100 - - - - ? = > num__33 num__0.333333333333 % answer : c <eor> c <eos> |
c |
percent__100.0__33.0__ |
percent__100.0__33.0__ |
| num__5358 x num__56 = ? <o> a ) num__272258 <o> b ) num__272358 <o> c ) num__300048 <o> d ) num__274258 <o> e ) num__274358 |
num__5358 x num__51 = num__5358 x ( num__50 + num__6 ) = num__5358 x num__50 + num__5358 x num__6 = num__267900 + num__32148 = num__300048 . c ) <eor> c <eos> |
c |
subtract__56.0__50.0__ multiply__5358.0__50.0__ multiply__5358.0__6.0__ multiply__5358.0__56.0__ multiply__5358.0__56.0__ |
subtract__56.0__50.0__ multiply__5358.0__50.0__ multiply__5358.0__6.0__ add__32148.0__267900.0__ add__32148.0__267900.0__ |
| if @ is a binary operation defined as the difference between an integer n and the product of n and num__5 then what is the largest positive integer n such that the outcome of the binary operation of n is less than num__10 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
the product of n and num__5 = num__5 n . . the difference between an integer n and the product of n and num__5 = num__5 n - n = num__4 n . . the outcome of the binary operation of n is less than num__10 ? so num__4 n < num__10 . . . max value that n can take as an integer is num__2 answer : b <eor> b <eos> |
b |
divide__10.0__5.0__ divide__10.0__5.0__ |
divide__10.0__5.0__ subtract__4.0__2.0__ |
| if num__5 x = num__3 y and xy ≠ num__0 what is the ratio of num__0.2 * x to num__0.166666666667 * y ? <o> a ) num__4.16666666667 <o> b ) num__1.44 <o> c ) num__0.72 <o> d ) num__0.833333333333 <o> e ) num__0.694444444444 |
num__5 x = num__3 y = > x / y = num__0.6 num__0.2 * x to num__0.166666666667 * y = x / y * num__1.2 = ( num__0.6 ) * ( num__1.2 ) = num__0.72 ans : c <eor> c <eos> |
c |
divide__3.0__5.0__ multiply__0.6__1.2__ multiply__0.6__1.2__ |
divide__3.0__5.0__ multiply__0.6__1.2__ multiply__0.6__1.2__ |
| how many num__6 digit number contain number num__4 ? <o> a ) num__317608 <o> b ) num__327608 <o> c ) num__337608 <o> d ) num__427608 <o> e ) num__357608 |
total num__6 digit no . = num__9 * num__10 * num__10 * num__10 * num__10 * num__10 = num__900000 not containing num__4 = num__8 * num__9 * num__9 * num__9 * num__9 * num__9 = num__472392 total num__6 digit number contain num__4 = num__900000 - num__472392 = num__427608 answer : d <eor> d <eos> |
d |
add__6.0__4.0__ subtract__900000.0__472392.0__ subtract__900000.0__472392.0__ |
add__6.0__4.0__ subtract__900000.0__472392.0__ subtract__900000.0__472392.0__ |
| a mixture of num__70 liters of wine and water contains num__10.0 water . how much water must be added to make water num__12 ½ % of the total mixture ? <o> a ) num__6 liters <o> b ) num__8 liters <o> c ) num__4 liters <o> d ) num__2 liters <o> e ) num__1 liters |
num__70 * ( num__0.1 ) = num__7 wine water num__87 num__0.5 num__12.0 num__0.5 num__87.0 num__0.5 % - - - - - - - num__63 num__12 num__0.5 % - - - - - - - ? = > num__9 - num__7 = num__2 answer : d <eor> d <eos> |
d |
reverse__10.0__ divide__70.0__10.0__ subtract__70.0__7.0__ divide__63.0__7.0__ reverse__0.5__ reverse__0.5__ |
reverse__10.0__ multiply__70.0__0.1__ subtract__70.0__7.0__ divide__63.0__7.0__ subtract__12.0__10.0__ subtract__12.0__10.0__ |
| if x ^ num__2 + ( num__1 / x ^ num__2 ) = num__7 x ^ num__4 + ( num__1 / x ^ num__4 ) = ? <o> a ) num__10 <o> b ) num__47 <o> c ) num__12 <o> d ) num__14 <o> e ) num__15 |
- > x ^ num__4 + ( num__1 / x ^ num__4 ) = ( x ^ num__2 ) ^ num__2 + ( num__1 / x ^ num__2 ) ^ num__2 = ( x ^ num__2 + num__1 / x ^ num__2 ) ^ num__2 - num__2 x ^ num__2 ( num__1 / x ^ num__2 ) = num__7 ^ num__2 - num__2 = num__47 . thus the answer is b . <eor> b <eos> |
b |
multiply__1.0__47.0__ |
divide__47.0__1.0__ |
| a man ordered num__4 pairs of black socks and some pair of brown socks . the price of a black socks is double that of a brown pair while preparing the bill the clerk interchanged the number of black and brown pairs by mistake which increased the bill by num__50.0 . the ratio of the number of black and brown pairs of socks in the original order was <o> a ) num__1 : num__4 <o> b ) num__1 : num__2 <o> c ) num__1 : num__3 <o> d ) num__1 : num__5 <o> e ) num__1 : num__7 |
let the price of a pair of brown socks = x then price of a pair of black socks = num__2 x if in original order there are ' a ' pairs of brown socks then total price = num__4 * num__2 x + a * x with interchanging number of black and brown pairs new price = a * num__2 x + num__4 * x given new price is num__50.0 more than the original so a * num__2 x + num__4 * x = ( num__1.5 ) ( num__4 * num__2 x + a * x ) = > num__2 ax + num__4 x = ( num__1.5 ) ( num__8 x + ax ) = > num__4 ax - num__3 ax = num__16 x = > a = num__16 = original number of pairs of brown socks so ratio of number of black and brown pairs of socks in original order = num__4 : num__16 = num__1 : num__4 answer : a <eor> a <eos> |
a |
multiply__4.0__2.0__ multiply__1.5__2.0__ multiply__2.0__8.0__ round_down__1.5__ round_down__1.5__ |
multiply__4.0__2.0__ multiply__1.5__2.0__ multiply__2.0__8.0__ subtract__4.0__3.0__ subtract__4.0__3.0__ |
| a can do a piece of work in num__4 days . b can do it in num__5 days . with the assistance of c they completed the work in num__2 days . find in how many days can c alone do it ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__30 <o> d ) num__40 <o> e ) num__15 |
c = num__0.5 - num__0.25 - num__0.2 = num__0.05 = num__20 days answer b <eor> b <eos> |
b |
divide__2.0__4.0__ divide__0.5__2.0__ divide__0.25__5.0__ multiply__4.0__5.0__ round__20.0__ |
divide__2.0__4.0__ divide__0.5__2.0__ subtract__0.25__0.2__ multiply__4.0__5.0__ round__20.0__ |
| of a group of people num__10 play piano num__11 play guitar num__14 play violin num__3 play all the instruments num__20 play only one instrument . how many play num__2 instruments ? <o> a ) num__3 <o> b ) num__6 <o> c ) num__9 <o> d ) num__12 <o> e ) num__15 |
let single piano players = p let single guitar players = g let single violin players = v given p + g + v + num__3 + num__20 = num__10 + num__11 + num__14 . hence p + g + v = num__35 - num__23 = num__12 answer : d <eor> d <eos> |
d |
add__3.0__20.0__ add__10.0__2.0__ add__10.0__2.0__ |
add__3.0__20.0__ add__10.0__2.0__ add__10.0__2.0__ |
| jaya invests in a new mutual fund . the fund averages num__10.0 growth annually for the first three years but it loses num__30.0 of its value in the fourth year . at the end of four years the value of the mutual fund is approximately what percent of the amount jaya originally paid ? <o> a ) num__93.0 <o> b ) num__90.0 <o> c ) num__88.0 <o> d ) num__85.0 <o> e ) num__80 % |
let the initial amount be num__100 . since it grew on constant basis as num__10.0 for three years = num__100 * num__1.1 ^ num__3 = num__11 * num__11 * num__1.1 = num__132 [ taking num__1.1 as num__1 ] it eroded by num__30.0 = num__132 * . num__7 = num__92 . something approximated to num__93.0 so the correct answer will be ( a ) num__93.0 <eor> a <eos> |
a |
percent__10.0__30.0__ percent__100.0__93.0__ |
percent__10.0__30.0__ percent__100.0__93.0__ |
| when num__0.1 percent of num__2000 is subtracted from num__0.1 of num__2000 the difference is <o> a ) num__0 <o> b ) num__198 <o> c ) num__450 <o> d ) num__495 <o> e ) num__500 |
we can break this problem into two parts : num__1 ) what is num__0.1 percent of num__2000 ? num__2 ) what is num__0.1 of num__2000 ? to calculate num__0.1 percent of num__2000 we must first remember to divide num__0.1 by num__100 . so we have : ( num__0.1 ) / ( num__100 ) to divide a number by num__100 means to multiply it by num__0.01 so we have : num__0.1 x num__0.01 = num__1 / num__1000 thus num__0.1 percent of num__2000 = num__1 / num__1000 x num__2000 = num__2 . now let ' s concentrate on part num__2 . we need to calculate num__0.1 of num__2000 . to do this we simply multiply num__0.1 by num__2000 . num__0.1 x num__2000 = num__200 the answer to part num__1 is num__2 and the answer to part num__2 is num__200 . their difference is num__200 – num__2 = num__198 . answer b . <eor> b <eos> |
b |
reverse__100.0__ divide__2000.0__2.0__ multiply__0.1__2000.0__ subtract__200.0__2.0__ multiply__1.0__198.0__ |
reverse__100.0__ divide__2000.0__2.0__ divide__2.0__0.01__ subtract__200.0__2.0__ divide__198.0__1.0__ |
| if num__4 - x < ( num__2 + num__5 x ) / num__3 which of the following is correct ? <o> a ) x < - num__5 . <o> b ) x > - num__5 . <o> c ) x > num__1.25 . <o> d ) - num__5 < x < num__0 . <o> e ) num__0 < x < num__5 . |
num__4 - x < ( num__2 + num__5 x ) / num__3 num__12 - num__3 x < num__2 + num__5 x num__10 < num__8 x num__1.25 < x num__1.25 < x c is the answer <eor> c <eos> |
c |
multiply__4.0__3.0__ multiply__2.0__5.0__ multiply__4.0__2.0__ divide__5.0__4.0__ divide__5.0__4.0__ |
multiply__4.0__3.0__ subtract__12.0__2.0__ add__5.0__3.0__ divide__5.0__4.0__ divide__5.0__4.0__ |
| the hiker walking at a constant rate of num__4 miles per hour is passed by a bike traveling in the same direction along the same path at num__20 miles per hour . the bike stops to wait for the hiker num__5 minutes after passing her while the hiker continues to walk at her constant rate how many minutes must the bike wait until the hiker catches up ? <o> a ) num__16 <o> b ) num__14 <o> c ) num__10 <o> d ) num__15 <o> e ) num__20 |
after passing the hiker the cyclist travels for num__5 minutes at a rate of num__20 miles / hour . in those num__5 mins the cyclist travels a distance of num__1.66666666667 miles . in those num__5 mins the hiker travels a distance of num__0.333333333333 miles . so the hiker still has to cover num__1.33333333333 miles to meet the waiting cyclist . the hiker will need num__0.333333333333 hours or num__20 mins to cover the remaining num__1.33333333333 miles . so the answer is e . <eor> e <eos> |
e |
divide__1.6667__5.0__ round__20.0__ |
divide__1.6667__5.0__ round__20.0__ |
| a box contains eight bulbs out of which num__4 are defective . if four bulbs are chosen at random find the probability that at least one bulb is good ? <o> a ) num__0.8125 <o> b ) num__0.5 <o> c ) num__0.9375 <o> d ) num__0.6875 <o> e ) num__0.5625 |
required probability = num__1 - num__0.0625 = num__0.9375 answer : c <eor> c <eos> |
c |
negate_prob__0.0625__ negate_prob__0.0625__ |
negate_prob__0.0625__ negate_prob__0.0625__ |
| a tap can fill a tank in num__14 hours . after half the tank is filled three more similar taps are opened . what is the total time taken to fill the tank completely ? <o> a ) num__11 hrs num__45 mins <o> b ) num__12 hrs num__45 mins <o> c ) num__13 hrs num__45 mins <o> d ) num__14 hrs num__45 mins <o> e ) num__15 hrs num__45 mins |
time taken by one tap to fill the tank = num__7 hrs . part filled by the taps in num__1 hour = num__4 * num__0.0714285714286 = num__0.285714285714 remaining part = num__1 - num__0.5 = num__0.5 num__0.285714285714 : num__0.5 : : num__1 : x x = num__0.5 * num__1 * num__3.5 = num__1.75 = num__105 mins = num__1 hr num__45 mins so total time taken = num__15 hrs num__45 mins answer : e <eor> e <eos> |
e |
divide__1.0__14.0__ divide__4.0__14.0__ divide__7.0__14.0__ divide__14.0__4.0__ multiply__0.5__3.5__ add__14.0__1.0__ round__15.0__ |
divide__1.0__14.0__ divide__4.0__14.0__ divide__7.0__14.0__ multiply__0.5__7.0__ multiply__0.5__3.5__ add__14.0__1.0__ round__15.0__ |
| a fruit juice is composed of pineapple strawberry juice and cherry juice . how many ml of cherry juice is required to make num__6 liters of fruit juice containing twice as much pineapple juice as strawberry juice and three times as much strawberry juice as cherry juice ? ( num__1 l = num__1000 ml ) <o> a ) num__1800 ml <o> b ) num__600 ml <o> c ) num__1000 ml <o> d ) num__200 ml <o> e ) num__50 ml |
num__1 l = num__1000 ml num__6 l = num__6000 ml pineapple = num__2 strawberry strawberry = num__3 cherry pineapple + strawberry + cherry = num__6000 num__2 strawberry + strawberry + num__0.333333333333 strawberry = num__6000 num__10 strawberry = num__18000 strawberry = num__1800 ml answer a <eor> a <eos> |
a |
multiply__6.0__1000.0__ divide__6.0__2.0__ reverse__3.0__ multiply__3.0__6000.0__ divide__18000.0__10.0__ multiply__1.0__1800.0__ |
multiply__6.0__1000.0__ add__1.0__2.0__ reverse__3.0__ multiply__3.0__6000.0__ divide__18000.0__10.0__ multiply__1.0__1800.0__ |
| find the number of square tiles to cover the floor of a room measuring num__6.5 m * num__9.5 m leaving num__0.25 m space around the room . a side of square tile is given to be num__25 cms ? <o> a ) num__422 <o> b ) num__476 <o> c ) num__429 <o> d ) num__864 <o> e ) num__413 |
floor area to be covered by tiles = num__6 * num__9 = num__54 tiles area = num__0.25 * num__0.25 = num__0.0625 no . of tiles = num__54 / num__0.0625 = num__864 answer : d <eor> d <eos> |
d |
multiply__6.0__9.0__ divide__54.0__0.0625__ round__864.0__ |
multiply__6.0__9.0__ divide__54.0__0.0625__ divide__54.0__0.0625__ |
| the average of nine numbers is num__14 . the average of first four numbers is num__11.5 and the average of last four numbers is num__16.5 . what is the middle number ? <o> a ) num__14 <o> b ) num__16 <o> c ) num__15 <o> d ) num__18 <o> e ) num__17 |
the total of nine numbers = num__9 x num__14 = num__126 the total of first num__4 and last num__4 numbers is = num__4 x num__11.5 + num__4 x num__16.5 = num__112 so the middle number is ( num__126 - num__112 ) = num__14 a <eor> a <eos> |
a |
multiply__14.0__9.0__ subtract__126.0__14.0__ divide__126.0__9.0__ |
multiply__14.0__9.0__ subtract__126.0__14.0__ subtract__126.0__112.0__ |
| a can do a piece of work in num__15 days and b in num__20 days . they began the work together but num__5 days before the completion of the work a leaves . the work was completed in ? <o> a ) num__11 num__0.666666666667 days <o> b ) num__11 num__0.428571428571 days <o> c ) num__8 days <o> d ) num__6 days <o> e ) num__5 days |
a ' s num__1 day ' s work = num__0.0666666666667 b ' s num__1 day ' s work = num__0.05 num__5 days before the completion of the work a leaves = x - num__0.333333333333 ( x â € “ num__5 ) / num__15 + x / num__20 = num__1 x = num__11 num__0.428571428571 days answer : option b <eor> b <eos> |
b |
divide__1.0__15.0__ divide__1.0__20.0__ divide__5.0__15.0__ round__11.0__ |
divide__1.0__15.0__ divide__1.0__20.0__ divide__5.0__15.0__ divide__11.0__1.0__ |
| a man performs num__0.6 of the total journey by rail num__0.85 by bus and the remaining num__6.5 km on foot . his total journey is ? <o> a ) num__130 <o> b ) num__152 <o> c ) num__225 <o> d ) num__260 <o> e ) num__180 |
let the total journey be x km then ( num__3 x / num__5 ) + ( num__7 x / num__20 ) + num__6.5 = x num__12 x + num__7 x + num__20 * num__6.5 = num__20 x x = num__130 km answer is a <eor> a <eos> |
a |
divide__3.0__0.6__ multiply__0.6__20.0__ multiply__6.5__20.0__ multiply__6.5__20.0__ |
divide__3.0__0.6__ multiply__0.6__20.0__ multiply__6.5__20.0__ multiply__6.5__20.0__ |
| what is the perimeter of a square field whose diagonal is num__4 √ num__2 ? <o> a ) num__16 m <o> b ) num__32 m <o> c ) num__27 m <o> d ) num__26 m <o> e ) num__76 m |
num__4 a = num__16 m answer : a <eor> a <eos> |
a |
square_perimeter__4.0__ square_perimeter__4.0__ |
square_perimeter__4.0__ square_perimeter__4.0__ |
| what percent of num__120 are num__90 ? <o> a ) num__12 <o> b ) num__33 <o> c ) num__75 <o> d ) num__88 <o> e ) num__11 |
( ? % / num__100 ) * num__120 = num__90 ? = num__75.0 answer : c <eor> c <eos> |
c |
percent__75.0__100.0__ |
percent__75.0__100.0__ |
| mr . zimmer has num__237 papers to hand out to his class . the class consists of num__30 students and each student will need num__8 papers . how many more papers will he need to be able to give num__8 papers to each num__30 student ? <o> a ) num__3 <o> b ) num__7 <o> c ) num__21 <o> d ) num__78 <o> e ) num__4 |
num__8 * num__30 = num__240 not num__237 . therefore num__3 papers will need to be added in order for num__30 students to get num__8 papers each . therefore a is correct <eor> a <eos> |
a |
multiply__30.0__8.0__ subtract__240.0__237.0__ subtract__240.0__237.0__ |
multiply__30.0__8.0__ subtract__240.0__237.0__ subtract__240.0__237.0__ |
| | - num__4 | ( | - num__25 | - | num__5 | ) = ? ? source : preparation material mba center <o> a ) num__80 <o> b ) num__100 <o> c ) num__160 <o> d ) num__175 <o> e ) num__200 |
absolute value will turn negatives into their positive ' equivalents ' and will leave positives unchanged so | - num__4 | = num__4 | - num__25 | = num__25 and | num__5 | = num__5 . getting rid of our absolute values we have : | - num__4 | ( | - num__25 | - | num__5 | ) = ( num__4 ) ( num__25 - num__5 ) = num__4 * num__20 = num__80 <eor> a <eos> |
a |
multiply__4.0__5.0__ multiply__4.0__20.0__ multiply__4.0__20.0__ |
multiply__4.0__5.0__ multiply__4.0__20.0__ multiply__4.0__20.0__ |
| x is what percent of y percent of z in terms of x y and z ? p be the value expressed in terms of x y and z . <o> a ) xyz <o> b ) num__1 / xyz <o> c ) num__10000 x / yz <o> d ) num__1000 * xyz <o> e ) num__100 x / yz |
algebraically y % of z is just ( y / num__100 ) * z or yz / num__100 . so we want to answer the question : x is what percent of yz / num__100 ? say x is equal to p % of yz / num__100 ; then we have : x = ( p / num__100 ) * ( yz / num__100 ) num__10000 x / yz = p c <eor> c <eos> |
c |
percent__100.0__10000.0__ |
percent__100.0__10000.0__ |
| a company wants to manufacture computer models of x % of a list of num__10000 models from the market . after a budget cut the company finds it must reduce this selection by ( x − num__6 ) % . in terms of x how many computer model will the group be able to manufacture ? <o> a ) x * x – num__6 x <o> b ) ( x ) ( num__106 – x ) <o> c ) ( num__100 ) ( num__106 – x ) <o> d ) ( num__100 ) ( num__94 – x ) <o> e ) ( x - num__6 ) / num__100 |
based on the answer choices and the question this question begs the use of x = num__6 as a sample number . initial = num__6.0 * num__10000 = num__600 reduction = num__6 - num__6 = num__0.0 so no math required here to calculate the reduction ; just make sure that you can calculate num__600 in your answer . a . x * x – num__6 x = num__0 ; no b . ( x ) ( num__106 – x ) = num__600 ; winner ! c . ( num__100 ) ( num__106 – x ) > num__600 ; no d . ( num__100 ) ( num__94 – x ) > num__600 ; no e . ( x - num__6 ) / num__100 = num__0 ; no b <eor> b <eos> |
b |
percent__6.0__10000.0__ percent__100.0__106.0__ |
percent__6.0__10000.0__ percent__100.0__106.0__ |
| if num__10 typists can type num__30 letters in num__30 minutes then how many letters will num__40 typists working at the same rate complete in num__1 hour ? <o> a ) num__200 <o> b ) num__210 <o> c ) num__205 <o> d ) num__225 <o> e ) num__240 |
no . of letters typing by num__10 typists in num__30 minutes = num__30 no . of letters typing by num__10 typists in num__60 minutes = num__30 * num__2 = num__60 no . of letters typing by num__40 typists in num__60 minutes = num__6.0 * num__40 = num__240 answer : e <eor> e <eos> |
e |
hour_to_min_conversion__ divide__60.0__30.0__ divide__60.0__10.0__ multiply__40.0__6.0__ round__240.0__ |
hour_to_min_conversion__ divide__60.0__30.0__ divide__60.0__10.0__ multiply__40.0__6.0__ multiply__40.0__6.0__ |
| workers decided to raise rs . num__3 lacs by equal contribution from each . had they contributed rs . num__50 eachextra the contribution would have been rs . num__3.50 lacs . how many workers were they ? <o> a ) num__220 <o> b ) num__230 <o> c ) num__500 <o> d ) num__560 <o> e ) num__1000 |
n * num__50 = ( num__350000 - num__300000 ) = num__50000 n = num__1000.0 = num__1000 e <eor> e <eos> |
e |
subtract__350000.0__300000.0__ divide__50000.0__50.0__ divide__50000.0__50.0__ |
subtract__350000.0__300000.0__ divide__50000.0__50.0__ divide__50000.0__50.0__ |
| two trains of equal length running with the speeds of num__60 and num__20 kmph take num__50 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? <o> a ) num__12.5 <o> b ) num__8.01 <o> c ) num__7.5 <o> d ) num__26 <o> e ) num__22 |
rs = num__60 - num__40 = num__20 * num__0.277777777778 = num__5.55555555556 t = num__50 d = num__50 * num__5.55555555556 = num__277.777777778 rs = num__60 + num__20 = num__80 * num__0.277777777778 t = num__277.777777778 * num__0.045 = num__12.5 sec . answer : a <eor> a <eos> |
a |
subtract__60.0__20.0__ add__60.0__20.0__ multiply__0.045__277.7778__ round__12.5__ |
subtract__60.0__20.0__ add__60.0__20.0__ multiply__0.045__277.7778__ multiply__0.045__277.7778__ |
| how many numbers from num__9 to num__79 are exactly divisible by num__11 ? <o> a ) num__5 <o> b ) num__7 <o> c ) num__9 <o> d ) num__11 <o> e ) num__12 |
num__0.818181818182 = num__0 and num__7.18181818182 = num__7 = = > num__7 - num__0 = num__7 numbers answer : b <eor> b <eos> |
b |
divide__9.0__11.0__ round_down__0.8182__ divide__79.0__11.0__ round_down__7.1818__ round_down__7.1818__ |
divide__9.0__11.0__ round_down__0.8182__ divide__79.0__11.0__ round_down__7.1818__ round_down__7.1818__ |
| two trains start from p and q respectively and travel towards each other at a speed of num__50 km / hr and num__40 km / hr respectively . by the time they meet the first train has travelled num__1 ookm more than the second . the distance between p and q is <o> a ) num__500 km <o> b ) num__600 km <o> c ) num__700 km <o> d ) num__900 km <o> e ) num__400 km |
explanation : at the time of meeting let the distance travelled by the second train be x km then distance covered by the first train is ( x + num__100 ) km = > x / num__40 = ( x + num__100 ) / num__50 = > num__50 x = num__40 x + num__4000 = > x = num__400 km so distance between p and q = ( x + x + num__100 ) km = num__900 km answer : option d <eor> d <eos> |
d |
multiply__40.0__100.0__ round__900.0__ |
multiply__40.0__100.0__ divide__900.0__1.0__ |
| if x ⁄ num__2 + num__5 ⁄ num__4 = num__5 ⁄ num__4 what is the value of x ? <o> a ) – num__2 <o> b ) – num__1 <o> c ) num__0 <o> d ) num__1 <o> e ) num__2 |
x ⁄ num__2 + num__5 ⁄ num__4 = num__5 ⁄ num__4 let ' s multiply both sides by num__4 . num__2 x + num__5 = num__5 num__2 x = num__0 x = num__0 the answer is c . <eor> c <eos> |
c |
multiply__2.0__0.0__ |
multiply__2.0__0.0__ |
| lucy deposited $ num__31250 in an investment fund that provided num__16 percent annual return compounded quarterly . if she made no other transactions with the fund in how much time in months did her investment earn a total interest of $ num__2550 ? <o> a ) num__0.5 <o> b ) num__2 <o> c ) num__3.0 <o> d ) num__6.0 <o> e ) num__6.1 |
a = p + i = num__31250 + num__2550 = num__33800 num__33800 = num__31250 ( num__1 + num__4.0 * num__100 ) ^ ( num__4 t ) ( num__1.0816 ) = ( num__1.04 ) ^ ( num__4 t ) ( num__1.04 ) ^ num__2 = ( num__1.04 ) ^ num__4 t t = num__0.5 yrs = num__6.0 months answer : d <eor> d <eos> |
d |
percent__6.0__100.0__ |
percent__6.0__100.0__ |
| what number comes next in this number series . num__5 num__5 num__3 num__4 num__4 num__6 num__9 ? * hint : check calender <o> a ) num__7 <o> b ) num__8 <o> c ) num__6 <o> d ) num__1 <o> e ) num__8 |
a num__7 ( the number of letters in the month october ) march : num__5 letters april : num__5 letters may : num__3 letters june : num__4 letters july : num__4 letters august : num__6 letters september : num__9 letters <eor> a <eos> |
a |
add__3.0__4.0__ add__3.0__4.0__ |
add__3.0__4.0__ add__3.0__4.0__ |
| aa is two digit number then m times its cube of aa has num__1 in its tens place then what is m . <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
when aa = num__11 num__7 * ( num__11 ) ^ num__3 = num__7 * num__1331 = num__9317 m = num__7 answer : e <eor> e <eos> |
e |
multiply__1331.0__7.0__ round__7.0__ |
multiply__1331.0__7.0__ multiply__1.0__7.0__ |
| in a certain sequence of num__7 numbers each number after the first is num__1 more than the previous number . if the first number is − num__5 how many of the numbers in the sequence are positive ? <o> a ) none <o> b ) one <o> c ) two <o> d ) three <o> e ) four |
non - negative ( zero to positive infinity ) solution to the question : - num__5 - num__4 - num__3 - num__2 - num__10 num__1 hence ans is num__1 . . option b <eor> b <eos> |
b |
subtract__5.0__1.0__ subtract__7.0__4.0__ subtract__7.0__5.0__ add__7.0__3.0__ reverse__1.0__ |
subtract__5.0__1.0__ subtract__7.0__4.0__ subtract__7.0__5.0__ add__7.0__3.0__ subtract__5.0__4.0__ |
| a dishonest dealer claims to sell a product at its cost price . he uses a counterfeit weight which is num__20.0 less than the real weight . further greed overtook him and he added num__30.0 impurities to the product . find the net profit percentage of the dealer ? <o> a ) num__44.0 <o> b ) num__40.0 <o> c ) num__50.0 <o> d ) num__56.25 <o> e ) num__37.5 % |
the dealer uses weight which is num__20.0 less than the real weight . or ( num__1 - num__0.2 ) or num__0.8 of real weight . it means that he is selling $ num__4 worth of product for $ num__5 . the dealer then further added num__30.0 impurities to the product . it means that he is selling $ num__5 worth of product for $ num__5.5 . so his profit is $ num__5.5 - $ num__4 = $ num__1.5 and his profit percent is ( num__1.5 / num__4 ) * num__100 = num__37.5 answer : - e <eor> e <eos> |
e |
subtract__1.0__0.2__ multiply__20.0__0.2__ reverse__0.2__ divide__30.0__20.0__ multiply__20.0__5.0__ divide__30.0__0.8__ divide__30.0__0.8__ |
subtract__1.0__0.2__ multiply__20.0__0.2__ reverse__0.2__ divide__30.0__20.0__ multiply__20.0__5.0__ divide__30.0__0.8__ divide__30.0__0.8__ |
| a batsman makes a score of num__64 runs in the num__16 th innings and thus increased his average by num__3 . find his average after the num__16 th inning ? <o> a ) num__15 <o> b ) num__14 <o> c ) num__18 <o> d ) num__19 <o> e ) num__20 |
let the average after the num__16 th inning be p . so the average after the num__15 th inning will be ( p - num__3 ) hence num__15 ( p - num__30 ) + num__64 = num__16 p = > p = num__19 . answer : d <eor> d <eos> |
d |
add__16.0__3.0__ add__16.0__3.0__ |
add__16.0__3.0__ add__16.0__3.0__ |
| every day fred takes the train to travel from his work back to alkmaar his place of residence . usually he arrives at the station of alkmaar at six o ' clock and exactly at that moment his wife picks him up by car . yesterday evening fred took an earlier train without informing his wife and therefore he already was at the station of alkmaar at five o ' clock . he decided to walk part of the way to meet his wife . when he met the car with his wife he drove home with her . in this way they were home ten minutes earlier than normal . fred ' s wife always drives the entire way between home and station at the same constant speed . how long did fred walk yesterday evening ? <o> a ) num__55 <o> b ) num__45 <o> c ) num__58 <o> d ) num__65 <o> e ) num__59 |
fred ' s wife needed to drive ten minutes less than usual which is five minutes less in the direction of the station and five minutes less back home . therefore fred met his wife at five minutes to six . he had walked already since five o ' clock so he walked num__55 minutes in total . answer a <eor> a <eos> |
a |
round__55.0__ |
round__55.0__ |
| an article is bought for rs . num__1200 and sold for rs . num__800 find the loss percent ? <o> a ) num__16.0 <o> b ) num__35.0 <o> c ) num__33.0 <o> d ) num__18.0 <o> e ) num__12 % |
num__1200 - - - - num__400 num__100 - - - - ? = > num__33.0 answer : c <eor> c <eos> |
c |
percent__100.0__33.0__ |
percent__100.0__33.0__ |
| what will be the cost of building a fence around a square plot with area equal to num__289 sq ft if the price per foot of building the fence is rs . num__48 ? <o> a ) a ) rs . num__3944 <o> b ) b ) rs . num__3948 <o> c ) c ) rs . num__3264 <o> d ) d ) rs . num__3949 <o> e ) e ) rs . num__3923 |
let the side of the square plot be a ft . a num__2 = num__289 = > a = num__17 length of the fence = perimeter of the plot = num__4 a = num__68 ft . cost of building the fence = num__68 * num__48 = rs . num__3264 . answer : c <eor> c <eos> |
c |
square_perimeter__17.0__ multiply__48.0__68.0__ multiply__48.0__68.0__ |
multiply__17.0__4.0__ multiply__48.0__68.0__ multiply__48.0__68.0__ |
| the city of boston decided to reconstruct its major tunnels . it estimated the job would require num__612 mini projects spread evenly over an num__18 month plan of completion . only num__108 mini projects had been successfully completed after num__8 months . at this time the construction was behind schedule by how many projects ? <o> a ) num__34 <o> b ) num__96 <o> c ) num__198 <o> d ) num__164 <o> e ) num__504 |
project / month : num__34.0 = num__34 project in num__6 month to be completed = num__34 * num__8 = num__272 lag : num__272 - num__108 = num__164 d is the answer <eor> d <eos> |
d |
divide__612.0__18.0__ divide__108.0__18.0__ multiply__8.0__34.0__ subtract__272.0__108.0__ subtract__272.0__108.0__ |
divide__612.0__18.0__ divide__108.0__18.0__ multiply__8.0__34.0__ subtract__272.0__108.0__ subtract__272.0__108.0__ |
| in an election candidate a got num__65.0 of the total valid votes . if num__15.0 of the total votes were declared invalid and the total numbers of votes is num__560000 find the number of valid vote polled in favor of candidate ? <o> a ) num__355600 <o> b ) num__355800 <o> c ) num__356500 <o> d ) num__309400 <o> e ) num__357000 |
total number of invalid votes = num__15.0 of num__560000 = num__0.15 × num__560000 = num__84000.0 = num__84000 total number of valid votes num__560000 – num__84000 = num__476000 percentage of votes polled in favour of candidate a = num__65.0 therefore the number of valid votes polled in favour of candidate a = num__65.0 of num__476000 = num__0.65 × num__476000 = num__309400.0 = num__309400 d ) <eor> d <eos> |
d |
percent__15.0__560000.0__ percent__65.0__476000.0__ percent__65.0__476000.0__ |
percent__15.0__560000.0__ percent__65.0__476000.0__ percent__65.0__476000.0__ |
| if xy = num__2 and x ^ num__2 + y ^ num__2 = num__12 then x / y + y / x = <o> a ) num__6 <o> b ) num__3 num__0.142857142857 <o> c ) num__5 num__0.333333333333 <o> d ) num__7 <o> e ) num__60 |
we can make simplifying of question and get it in view : ( x ^ num__2 + y ^ num__2 ) / xy and as we know the meaning of this parts : x ^ num__2 + y ^ num__2 = num__12 xy = num__2 we can calculate the answer num__6.0 - > num__6 so answer is a <eor> a <eos> |
a |
divide__12.0__2.0__ divide__12.0__2.0__ |
divide__12.0__2.0__ divide__12.0__2.0__ |
| the length of the bridge which a train num__130 meters long and travelling at num__45 km / hr can cross in num__30 seconds is : <o> a ) num__266 <o> b ) num__288 <o> c ) num__245 <o> d ) num__211 <o> e ) num__776 |
speed = ( num__45 * num__0.277777777778 ) m / sec = ( num__12.5 ) m / sec . time = num__30 sec . let the length of bridge be x meters . then ( num__130 + x ) / num__30 = num__12.5 = = > num__2 ( num__130 + x ) = num__750 = = > x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| one fourth of one third of two fifth of a number is num__10 . what will be num__40.0 of that number <o> a ) a ) num__140 <o> b ) b ) num__150 <o> c ) c ) num__120 <o> d ) d ) num__200 <o> e ) e ) num__220 |
explanation : ( num__0.25 ) * ( num__0.333333333333 ) * ( num__0.4 ) * x = num__10 then x = num__10 * num__30 = num__300 num__40.0 of num__300 = num__120 answer : option c <eor> c <eos> |
c |
percent__40.0__300.0__ percent__40.0__300.0__ |
percent__40.0__300.0__ percent__40.0__300.0__ |
| a fruit seller had some apples . he sells num__40.0 apples and still has num__420 apples . originally he had : <o> a ) num__588 apples <o> b ) num__600 apples <o> c ) num__672 apples <o> d ) num__700 apples <o> e ) num__720 apples |
explanation : suppose originally he had x apples . then ( num__100 - num__40 ) % of x = num__420 . num__60 x x = num__420 num__100 x = ( num__420 * num__1.66666666667 ) = num__700 . answer is d <eor> d <eos> |
d |
percent__100.0__700.0__ |
percent__100.0__700.0__ |
| a meal cost $ num__48.75 adn there was no tax . if the tip was more than num__5 pc but less than num__20 pc of the price then the total amount paid should be : <o> a ) num__60 - num__65 <o> b ) num__55 - num__58 <o> c ) num__51 - num__59 <o> d ) num__39 - num__41 <o> e ) num__38 - num__40 |
num__5.0 ( num__48.75 ) = num__2.4375 num__20.0 ( num__48.75 ) = num__9.75 total amount could have been num__48.75 + num__2.4375 and num__48.75 + num__9.75 = > could have been between num__51.1875 and num__58.5 = > approximately between num__51 and num__59 answer is c . <eor> c <eos> |
c |
divide__48.75__20.0__ divide__48.75__5.0__ add__48.75__2.4375__ add__48.75__9.75__ round_down__51.1875__ round_down__51.1875__ |
divide__48.75__20.0__ divide__48.75__5.0__ add__48.75__2.4375__ add__48.75__9.75__ round_down__51.1875__ round_down__51.1875__ |
| solve below question num__2 x + num__1 = - num__17 <o> a ) - num__8 <o> b ) - num__9 <o> c ) - num__5 <o> d ) - num__4 <o> e ) num__1 |
num__1 . subtract num__1 from both sides : num__2 x + num__1 - num__1 = - num__17 - num__1 num__2 . simplify both sides : num__2 x = - num__18 num__3 . divide both sides by num__2 : num__4 . simplify both sides : x = - num__9 b <eor> b <eos> |
b |
add__1.0__17.0__ add__2.0__1.0__ add__1.0__3.0__ divide__18.0__2.0__ multiply__1.0__9.0__ |
add__1.0__17.0__ add__2.0__1.0__ add__1.0__3.0__ divide__18.0__2.0__ subtract__18.0__9.0__ |
| maths physics and chemistry books are stored on a library shelf that can accommodate num__25 books . currently num__20.0 of the shelf spots remain empty . there are twice as many maths books as physics books and the number of physics books is num__4 greater than that of chemistry books . among all the books num__12 books are soft cover and the remaining are hard - cover . if there are a total of num__7 hard - cover books among the maths and physics books . what is the probability t that a book selected at random is either a hard cover book or a chemistry book ? <o> a ) num__0.1 <o> b ) num__0.15 <o> c ) num__0.2 <o> d ) num__0.25 <o> e ) num__0.45 |
first phase of this problem requires you to determine how many mathematics and chemistry books are even on the shelf . to do so you have the equations : m + p + c = num__20 ( since num__0.8 of the num__25 spots are full of books ) m = num__2 p p = num__4 + c from that you can use substitution to get everything down to one variable . c = p - num__4 m = num__2 p p = p then ( p - num__4 ) + num__2 p + p = num__20 so num__4 p = num__24 and p = num__6 . that means that there are num__12 math num__6 physics and num__2 chemistry books on the shelf . with those numbers you also know that there are num__8 total hardcovers num__1 of which is chemistry . so if your goal is to get either a hardcover or a chemistry there are num__9 ways towin - either one of the num__7 hardcovers that are n ' t chemistry or the two chemistry books . so out of the num__20 total t = num__9 provide the desired outcome making the answer e . <eor> e <eos> |
e |
divide__20.0__25.0__ add__20.0__4.0__ add__4.0__2.0__ subtract__20.0__12.0__ subtract__25.0__24.0__ add__7.0__2.0__ divide__9.0__20.0__ |
divide__20.0__25.0__ add__20.0__4.0__ add__4.0__2.0__ subtract__20.0__12.0__ subtract__25.0__24.0__ add__7.0__2.0__ divide__9.0__20.0__ |
| num__80.0 of the population of a village is num__32000 . the total population of the village is ? <o> a ) num__40000 <o> b ) num__24000 <o> c ) num__26682 <o> d ) num__29973 <o> e ) num__12312 |
x * ( num__0.8 ) = num__32000 x = num__400 * num__100 x = num__40000 answer : a <eor> a <eos> |
a |
divide__32000.0__80.0__ divide__80.0__0.8__ divide__32000.0__0.8__ divide__32000.0__0.8__ |
divide__32000.0__80.0__ divide__80.0__0.8__ multiply__400.0__100.0__ multiply__400.0__100.0__ |
| what is the % change in the area of a rectangle when its length increases by num__20.0 and its width decreases by num__20.0 ? <o> a ) num__0.0 <o> b ) num__20.0 increase <o> c ) num__20.0 decrease <o> d ) num__4.0 decrease <o> e ) insufficient data |
( num__1.2 ) * ( num__0.8 ) = num__0.96 of original area num__0.96 is a num__4.0 decrease from num__1.0 - > d <eor> d <eos> |
d |
multiply__1.2__0.8__ rectangle_perimeter__1.2__0.8__ square_perimeter__1.0__ |
multiply__1.2__0.8__ rectangle_perimeter__1.2__0.8__ square_perimeter__1.0__ |
| a num__160 meter long train crosses a man standing on the platform in num__12 sec . what is the speed of the train ? <o> a ) num__48 kmph <o> b ) num__54 kmph <o> c ) num__92 kmph <o> d ) num__86 kmph <o> e ) num__76 kmph |
s = num__13.3333333333 * num__3.6 = num__48 kmph answer : a <eor> a <eos> |
a |
divide__160.0__12.0__ round__48.0__ |
divide__160.0__12.0__ round__48.0__ |
| two persons start running simultaneously around a circular track of length num__700 m from the same point at speeds of num__15 km / hr and num__25 km / hr . when will they meet for the first time any where on the track if they are moving in opposite directions ? <o> a ) num__63 <o> b ) num__10 <o> c ) num__28 <o> d ) num__27 <o> e ) num__12 |
time taken to meet for the first time anywhere on the track = length of the track / relative speed = num__700 / ( num__15 + num__25 ) num__0.277777777778 = num__700 * num__0.45 * num__5 = num__63 seconds . answer : a <eor> a <eos> |
a |
round__63.0__ |
round__63.0__ |
| if a num__10 percent deposit that has been paid toward the purchase of a certain product is $ num__105 how much more remains to be paid ? <o> a ) $ num__945 <o> b ) $ num__965 <o> c ) $ num__1025 <o> d ) $ num__1050 <o> e ) $ num__1105 |
num__90.0 remains to be paid so the remaining amount is num__9 * num__105 = $ num__945 . the answer is a . <eor> a <eos> |
a |
divide__90.0__10.0__ multiply__105.0__9.0__ multiply__105.0__9.0__ |
divide__90.0__10.0__ multiply__105.0__9.0__ multiply__105.0__9.0__ |
| how much is num__60.0 of num__40 is greater than num__0.8 of num__25 ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__10 <o> d ) num__8 <o> e ) num__13 |
num__0.6 ) * num__40 â € “ ( num__0.8 ) * num__25 num__24 - num__20 = num__4 answer : a <eor> a <eos> |
a |
multiply__40.0__0.6__ subtract__60.0__40.0__ subtract__24.0__20.0__ subtract__24.0__20.0__ |
multiply__40.0__0.6__ subtract__60.0__40.0__ subtract__24.0__20.0__ subtract__24.0__20.0__ |
| a and b together can do a piece of work in num__8 days . if a alone can do the same work in num__12 days then b alone can do the same work in ? <o> a ) num__10 days <o> b ) num__5 days <o> c ) num__6 days <o> d ) num__24 days <o> e ) num__18 days |
explanation : b = num__0.125 – num__0.5 = num__0.0416666666667 = > num__24 days answer d <eor> d <eos> |
d |
divide__0.5__12.0__ divide__12.0__0.5__ round__24.0__ |
divide__0.5__12.0__ divide__12.0__0.5__ round__24.0__ |
| if p is a prime number greater than num__7 what is the remainder when p ^ num__2 is divided by num__8 . <o> a ) num__4 <o> b ) num__3 <o> c ) num__1 <o> d ) num__2 <o> e ) can not be determined |
take square of any prime number remainder will be num__1 ans c <eor> c <eos> |
c |
subtract__8.0__7.0__ reverse__1.0__ |
subtract__8.0__7.0__ reverse__1.0__ |
| for any number z z * is defined as the greatest positive even integer less than or equal to y . what is the value of num__6.35 – num__6.35 * ? <o> a ) num__0.35 <o> b ) num__0.5 <o> c ) num__6.25 <o> d ) num__0.25 <o> e ) num__6.0 |
since z * is defined as the greatest positive even integer less than or equal to z then num__6.35 * = num__6 ( the greatest positive even integer less than or equal to num__6.35 is num__6 ) . hence num__6.35 – num__6.35 * = num__6.35 - num__6 = num__0.35 answer : a . <eor> a <eos> |
a |
round_down__6.35__ subtract__6.35__6.0__ subtract__6.35__6.0__ |
round_down__6.35__ subtract__6.35__6.0__ subtract__6.35__6.0__ |
| margaret is num__9 years more than twice the age of his son . the age of son is num__12 . find the age of mother and find the difference between their ages <o> a ) num__21 yrs <o> b ) num__20 yrs <o> c ) num__30 yrs <o> d ) num__40 yrs <o> e ) num__50 yrs |
let age of son x = num__12 margaret is num__9 years more than twice the age of his son y = num__9 + num__2 ( x ) = num__9 + num__2 ( num__12 ) = num__9 + num__24 = num__33 yrs difference = num__33 - num__12 = num__21 yrs answer : a <eor> a <eos> |
a |
multiply__12.0__2.0__ add__9.0__24.0__ add__9.0__12.0__ add__9.0__12.0__ |
multiply__12.0__2.0__ add__9.0__24.0__ add__9.0__12.0__ add__9.0__12.0__ |
| two pipes a and b can fill a cistern in num__12 and num__15 minutes respectively . both are opened together but after num__3 minutes a is turned off . after how much more time will the cistern be filled ? <o> a ) num__8 num__0.142857142857 <o> b ) num__8 num__1.0 <o> c ) num__8 num__0.25 <o> d ) num__8 num__0.5 <o> e ) num__8 num__0.428571428571 |
num__0.25 + ( num__3 + x ) / num__15 = num__1 x = num__8 num__0.25 answer : c <eor> c <eos> |
c |
divide__3.0__12.0__ round__8.0__ |
divide__3.0__12.0__ divide__8.0__1.0__ |
| if num__3 persons can do num__3 times of a particular work in num__3 days then num__7 persons can do num__7 times of that work in ? <o> a ) num__3 days <o> b ) num__4 days <o> c ) num__6 days <o> d ) num__7 days <o> e ) num__8 days |
that is num__1 person can do one time of the work in num__3 days . therefore num__7 persons can do num__7 times work in the same num__3 days itself a ) <eor> a <eos> |
a |
round__3.0__ |
round__3.0__ |
| the new york classical group is designing the liner notes for an upcoming cd release . there are num__10 soloists featured on the album but the liner notes are only num__5 pages long and therefore only have room for num__5 of the soloists . the soloists are fighting over which of them will appear in the liner notes though not about which page they appear on . how many different combinations w of soloists can appear in the liner notes ? <o> a ) num__5 ! <o> b ) num__10 ! / ( num__5 ! num__5 ! ) <o> c ) num__10 ! / num__5 ! <o> d ) num__10 ! <o> e ) num__10 ! * num__5 ! |
total soloists to choose from = num__10 soloist required = num__5 question : how many different combinations w of soloists can appear in the liner notes ? the language of the question clearly mentions that only the combinations need to be calculated i . e . arrangement of the soloists does not matter ( refer thehighlightedpart of the question ) method - num__1 : total ways to choose num__5 out of num__10 soloists = num__10 c num__5 = num__10 ! / ( num__5 ! * num__5 ! ) method - num__2 : no . of ways to select and arrange num__5 soloist on num__5 page = num__10 * num__9 * num__8 * num__7 * num__6 since the arrangement of num__5 selected notes ( which can happen in num__5 ! ways ) does n ' t not matter therefore total ways to pick num__5 out of num__10 soloists = num__10 * num__9 * num__8 * num__7 * num__1.2 ! = num__10 * num__9 * num__8 * num__7 * num__6 * num__5 ! / ( num__5 ! * num__5 ! ) = num__10 ! / ( num__5 ! * num__5 ! ) answer : option b <eor> b <eos> |
b |
coin_space__ die_space__ choose__10.0__9.0__ |
coin_space__ die_space__ choose__10.0__9.0__ |
| two employees x and y are paid a total of rs . num__770 per week by their employer . if x is paid num__120 percent of the sum paid to y how much is y paid per week ? <o> a ) s . num__150 <o> b ) s . num__200 <o> c ) s . num__250 <o> d ) s . num__350 <o> e ) s . num__400 |
let the amount paid to x per week = x and the amount paid to y per week = y then x + y = num__770 but x = num__120.0 of y = num__120 y / num__100 = num__12 y / num__10 â ˆ ´ num__12 y / num__10 + y = num__770 â ‡ ’ y [ num__1.2 + num__1 ] = num__770 â ‡ ’ num__22 y / num__10 = num__770 â ‡ ’ num__22 y = num__7700 â ‡ ’ y = num__31.8181818182 = num__350.0 = rs . num__350 d ) <eor> d <eos> |
d |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__10.0__12.0__ multiply__770.0__10.0__ divide__7700.0__22.0__ multiply__1.0__350.0__ |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__10.0__12.0__ multiply__770.0__10.0__ divide__7700.0__22.0__ divide__7700.0__22.0__ |
| a single discount % equal to three successive discounts of num__30.0 num__20.0 and num__10.0 . <o> a ) num__49.8 <o> b ) num__49.5 <o> c ) num__49.3 <o> d ) num__49.6 <o> e ) num__49.1 |
explanation : let the initial price be num__100 . num__30.0 discount on num__100 is num__30 ( num__100 – num__30 ) = num__70 num__20.0 discount on the num__70 is num__14 ( num__70 – num__14 ) = num__56 num__10.0 discount on the num__56 is num__5.6 so the answer is num__30 + num__14 + num__5.6 = num__49.6 answer : d <eor> d <eos> |
d |
percent__20.0__70.0__ percent__10.0__56.0__ percent__100.0__49.6__ |
percent__20.0__70.0__ percent__10.0__56.0__ percent__100.0__49.6__ |
| a cube is divided into num__64 identical cubelets . each cut is made parallel to some surface of the cube . but before doing that the cube is painted with green on one set of opposite faces red on another set of opposite faces and blue on the third set of opposite faces . how many cubelets are painted with exactly one colour ? <o> a ) num__12 <o> b ) num__16 <o> c ) num__20 <o> d ) num__24 <o> e ) num__30 |
each side of the cube has num__4 x num__4 = num__16 cubelets . only the interior cubelets are painted one colour . on each side num__2 x num__2 = num__4 cubelets are painted one colour . since the cube has six sides the number of cubes with one colour is num__6 * num__4 = num__24 the answer is d . <eor> d <eos> |
d |
square_perimeter__4.0__ surface_cube__2.0__ surface_cube__2.0__ |
square_perimeter__4.0__ multiply__4.0__6.0__ multiply__4.0__6.0__ |
| in a college the ratio of the numbers of boys to the girls is num__5 : num__7 . if there are num__140 girls the total number of students in the college is ? <o> a ) num__200 <o> b ) num__240 <o> c ) num__160 <o> d ) num__250 <o> e ) num__310 |
let the number of boys and girls be num__5 x and num__7 x then num__7 x = num__140 x = num__20 total number of students = num__12 x = num__12 * num__20 = num__240 answer is b <eor> b <eos> |
b |
divide__140.0__7.0__ add__5.0__7.0__ multiply__12.0__20.0__ multiply__12.0__20.0__ |
divide__140.0__7.0__ add__5.0__7.0__ multiply__12.0__20.0__ multiply__12.0__20.0__ |
| num__32 meters of wire is available to fence off a flower bed in the form of a circular sector . what must the radius of the circle in meters be if we wish to have a flower bed with the greatest possible surface area ? <o> a ) num__2 √ num__2 <o> b ) num__2 √ num__5 <o> c ) num__5 <o> d ) num__4 √ num__2 <o> e ) num__8 |
area of sector a = x / num__360 * pi * r ^ num__2 circumference of the sector = num__32 = > x / num__360 * num__2 * pi * r + num__2 r = num__32 = > num__2 a / r + num__2 r = num__32 = > a = r num__16 - r ^ num__2 = r num__16 - r ^ num__2 we will now max using derivations max value of a will found at a = num__0 i . e num__16 - num__2 r = num__0 r = num__8 e <eor> e <eos> |
e |
square_perimeter__2.0__ square_perimeter__2.0__ |
square_perimeter__2.0__ square_perimeter__2.0__ |
| a train is moving at a speed of num__132 km / hr . if the length of the train is num__200 meters how long will it take to cross a railway platform num__240 meters long <o> a ) num__6 ½ sec <o> b ) num__7 ½ sec <o> c ) num__12 sec <o> d ) num__8 ½ sec <o> e ) num__9 sec |
explanation : speed of train = num__132 × ( num__0.277777777778 ) m / sec = num__36.6666666667 m / sec . distance covered in passing the platform = ( num__200 + num__240 ) m = num__440 m . time taken = num__440 × ( num__0.0272727272727 ) sec = num__12 sec answer : option c <eor> c <eos> |
c |
add__200.0__240.0__ divide__440.0__36.6667__ round__12.0__ |
add__200.0__240.0__ divide__440.0__36.6667__ divide__440.0__36.6667__ |
| the distance between west - town to east - town is num__10 kilometers . two birds start flying simultaneously towards one another the first leaving from west - town at a speed of num__4 kilometers per minute and the second bird leaving from east - town at a speed of num__1 kilometers per minute . what will be the distance in kilometers between the meeting point and west - town ? <o> a ) num__3 . <o> b ) num__8 . <o> c ) num__10 . <o> d ) num__12 . <o> e ) num__15 . |
time taken by the birds to meet = num__10 / ( num__4 + num__1 ) = num__2 mins distance traveled by the bird traveling from west - town = num__4 * num__2 = num__8 answer : b <eor> b <eos> |
b |
subtract__10.0__2.0__ round__8.0__ |
subtract__10.0__2.0__ subtract__10.0__2.0__ |
| a vessel of capacity num__2 litre has num__16.0 of alcohol and another vessel of capacity num__6 litre had num__40.0 alcohol . the total liquid of num__8 litre was poured out in a vessel of capacity num__10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? <o> a ) num__31.0 . <o> b ) num__71.0 . <o> c ) num__49.0 . <o> d ) num__27.2 . <o> e ) num__51.0 . |
num__16.0 of num__2 litres = num__0.32 litres num__40.0 of num__6 litres = num__2.4 litres therefore total quantity of alcohol is num__2.72 litres . this mixture is in a num__10 litre vessel . hence the concentration of alcohol in this num__10 litre vessel is num__27.2 answer : d <eor> d <eos> |
d |
add__2.4__0.32__ multiply__10.0__2.72__ multiply__10.0__2.72__ |
add__2.4__0.32__ multiply__10.0__2.72__ multiply__10.0__2.72__ |
| num__5 men and num__2 boys working together can do four times as much work as a man and a boy . working capacity of man and boy is in the ratio <o> a ) num__1 : num__2 <o> b ) num__1 : num__3 <o> c ) num__2 : num__1 <o> d ) num__2 : num__3 <o> e ) none of these |
explanation : let num__1 man num__1 day work = x num__1 boy num__1 day work = y then num__5 x + num__2 y = num__4 ( x + y ) = > x = num__2 y = > x / y = num__2.0 = > x : y = num__2 : num__1 option c <eor> c <eos> |
c |
subtract__5.0__1.0__ multiply__2.0__1.0__ |
subtract__5.0__1.0__ divide__2.0__1.0__ |
| a group of ants each num__0.333333333333 centimeters in length crosses a sidewalk in single file each right behind the next at a speed of num__30 cm / min . if the sidewalk is num__1.5 meters wide how long does it take for num__30 ants to all make it across ? <o> a ) num__2 min <o> b ) num__3 min <o> c ) num__4 min <o> d ) num__5 min <o> e ) num__6 min |
e num__6 min d = num__30 * num__0.333333333333 cm + num__150 cm = num__180 cm t = num__180 cm / num__30 cm / min = num__6 min <eor> e <eos> |
e |
add__30.0__150.0__ round__6.0__ |
add__30.0__150.0__ divide__180.0__30.0__ |
| how many possible integer values are there for x if | num__5 x - num__3 | < num__6 ? <o> a ) one <o> b ) two <o> c ) three <o> d ) four <o> e ) five |
solution : | num__5 x - num__3 | < num__6 let num__5 x = a therefore we have | a - num__3 | < num__6 = = > read this as origin is at + num__3 and we have to move + num__6 to the right and - num__6 to the left ( the less than sign represents that the a must be within boundaries ) ( num__3 - num__6 ) - - - - - - - - - - num__3 - - - - - - - - - - ( num__3 + num__6 ) now we have - num__3 < a < num__9 but a = num__5 x = = > - num__3 < num__5 x < num__9 dividing all values by + num__4 we have - num__0.6 < x < num__1.8 now question says integer values ( not rational ) therefore we have num__01 hence num__2 b <eor> b <eos> |
b |
coin_space__ coin_space__ |
coin_space__ coin_space__ |
| a b and c are partners . a receives num__0.666666666667 of profits b and c dividing the remainder equally . a ' s income is increased by rs . num__200 when the rate to profit rises from num__5 to num__7 percent . find the capital of b ? <o> a ) num__2655 <o> b ) num__3527 <o> c ) num__2500 <o> d ) num__2567 <o> e ) num__2571 |
a : b : c = num__0.666666666667 : num__0.166666666667 : num__0.166666666667 = num__4 : num__1 : num__1 x * num__0.02 * num__0.666666666667 = num__200 b capital = num__15000 * num__0.166666666667 = num__2500 answer : c <eor> c <eos> |
c |
subtract__5.0__4.0__ divide__4.0__200.0__ multiply__1.0__2500.0__ |
subtract__5.0__4.0__ divide__4.0__200.0__ multiply__1.0__2500.0__ |
| a company d has num__15 percent of the employees are secretaries and num__45 percent are salespeople . if there are num__50 other employees of company d how many employees does company d have ? <o> a ) num__200 <o> b ) num__125 <o> c ) num__180 <o> d ) num__152 <o> e ) num__250 |
let the total number of employees in the company be x % of secretaries = num__15.0 % of salespeople = num__45.0 % of of employees other than secretaries and salespeople = num__100 - num__60 = num__40.0 but this number is given as num__50 so num__40.0 of x = num__50 x = num__125 therefore there a total of num__125 employees in the company d correct answer - b <eor> b <eos> |
b |
percent__100.0__125.0__ |
percent__100.0__125.0__ |
| in a certain school num__20.0 of students are below num__8 years of age . the number of students above num__8 years of age is num__0.666666666667 of the number of students of num__8 years of age which is num__48 . what is the total number of students in the school ? <o> a ) num__72 <o> b ) num__80 <o> c ) num__120 <o> d ) num__150 <o> e ) num__100 |
let the number of students be x . then number of students above num__8 years of age = ( num__100 - num__20 ) % of x = num__80.0 of x . num__80.0 of x = num__48 + num__0.666666666667 of num__48 num__0.8 x = num__80 x = num__100 . answer : option e <eor> e <eos> |
e |
subtract__100.0__20.0__ divide__80.0__100.0__ add__20.0__80.0__ |
subtract__100.0__20.0__ divide__80.0__100.0__ add__20.0__80.0__ |
| in what ratio should a mixture of milk and water in the ratio of num__3 : num__4 be mixed with another mixture of milk and water in the ratio num__5 : num__2 so that the resultant mixture contains equal quantities of milk and water ? <o> a ) num__3 : num__1 <o> b ) num__3 : num__5 <o> c ) num__7 : num__8 <o> d ) num__7 : num__9 <o> e ) num__9 : num__11 |
first mixture : milk and water in the ratio of num__3 x : num__4 x . second mixture : milk and water in the ratio num__5 y : num__2 y . we want their mixture to have equal quantities of milk and water : num__3 x + num__5 y = num__4 x + num__2 y - - > x / y = num__3 . answer : a . <eor> a <eos> |
a |
subtract__5.0__2.0__ |
subtract__5.0__2.0__ |
| a man takes num__3 hours num__45 minutes to row a boat num__15 km downstream of a river and num__2 hours num__30 minutes to cover a distance of num__5 km upstream . find the speed of the river current in km / hr . <o> a ) num__1 km / hr <o> b ) num__2 km / hr <o> c ) num__3 km / hr <o> d ) num__4 km / hr <o> e ) num__5 km / hr |
rate downstream = ( num__5.0 Â ¾ ) km / hr = ( num__15 * num__0.266666666667 ) km / hr = num__4 km / hr . rate upstream = ( num__2.5 Â ½ ) km / hr = ( num__5 * num__0.4 ) km / hr = num__2 km / hr . speed of current = num__0.5 ( num__4 - num__2 ) km / hr = num__1 km / hr answer a <eor> a <eos> |
a |
divide__5.0__2.0__ divide__2.0__5.0__ subtract__3.0__2.5__ subtract__3.0__2.0__ round__1.0__ |
divide__5.0__2.0__ divide__2.0__5.0__ subtract__3.0__2.5__ subtract__3.0__2.0__ subtract__3.0__2.0__ |
| two numbers are respectively num__17.0 and num__30.0 more than a third number . the percentage that is first of the second is ? <o> a ) num__75.0 <o> b ) num__80.0 <o> c ) num__85.0 <o> d ) num__70.0 <o> e ) num__90 % |
i ii iii num__117 num__130 num__100 num__130 - - - - - - - - - - num__117 num__100 - - - - - - - - - - - ? = > num__90.0 answer : e <eor> e <eos> |
e |
percent__90.0__100.0__ |
percent__90.0__100.0__ |
| in one hour a boat goes num__8 km / hr along the stream and num__2 km / hr against the stream . the speed of the boat in still water ( in km / hr ) is : <o> a ) num__3 km / hr <o> b ) num__4 km / hr <o> c ) num__5 km / hr <o> d ) num__6 km / hr <o> e ) num__8 km / hr |
upstream relative speed is u + v = num__8 km / hr downstream speed is u - v = num__2 where u = speed of boat in still water and v is speed of stream then adding two equations u + v + u - v = num__8 + num__2 num__2 u = num__10 finally u = num__5 . answer : c <eor> c <eos> |
c |
add__8.0__2.0__ divide__10.0__2.0__ round__5.0__ |
add__8.0__2.0__ divide__10.0__2.0__ divide__10.0__2.0__ |
| the diagonals of a rhombus are num__15 cm and num__20 cm . find its area ? <o> a ) num__176 <o> b ) num__186 <o> c ) num__150 <o> d ) num__179 <o> e ) num__172 |
num__0.5 * num__15 * num__20 = num__150 answer : c <eor> c <eos> |
c |
triangle_area__15.0__20.0__ triangle_area__15.0__20.0__ |
volume_rectangular_prism__15.0__20.0__0.5__ volume_rectangular_prism__15.0__20.0__0.5__ |
| maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is num__70 kilometers maxwell ' s walking speed is num__4 km / h and brad ' s running speed is num__6 km / h what is the distance traveled by brad ? <o> a ) num__42 <o> b ) num__18 <o> c ) num__20 <o> d ) num__24 <o> e ) num__30 |
time taken = total distance / relative speed total distance = num__70 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = num__6 + num__4 = num__10 kms / hr time taken = num__7.0 = num__7 hrs distance traveled by brad = brad ' s speed * time taken = num__6 * num__7 = num__42 kms . . . answer - a <eor> a <eos> |
a |
add__4.0__6.0__ divide__70.0__10.0__ multiply__6.0__7.0__ round__42.0__ |
add__4.0__6.0__ divide__70.0__10.0__ multiply__6.0__7.0__ multiply__6.0__7.0__ |
| a train num__100 meters long completely crosses a num__300 meters long bridge in num__15 seconds . what is the speed of the train is ? <o> a ) num__32 kmph <o> b ) num__96 kmph <o> c ) num__34 kmph <o> d ) num__43 kmph <o> e ) num__40 kmph |
s = ( num__100 + num__300 ) / num__45 = num__26.6666666667 * num__3.6 = num__96 answer : b <eor> b <eos> |
b |
round__96.0__ |
round__96.0__ |
| dan ' s age after num__16 years will be num__6 times his age num__4 years ago . what is the present age of dan ? <o> a ) num__8 <o> b ) num__10 <o> c ) num__12 <o> d ) num__14 <o> e ) num__16 |
let dan ' s present age be x . x + num__16 = num__6 ( x - num__4 ) num__5 x = num__40 x = num__8 the answer is a . <eor> a <eos> |
a |
divide__40.0__5.0__ subtract__16.0__8.0__ |
divide__40.0__5.0__ subtract__16.0__8.0__ |
| what is the value of ( p + q ) / ( p - q ) if p / q is num__5 ? <o> a ) num__1.33333333333 <o> b ) num__1.5 <o> c ) num__0.333333333333 <o> d ) num__0.875 <o> e ) num__1.14285714286 |
( p + q ) / ( p - q ) = [ ( p / q ) + num__1 ] / [ ( p / q ) - num__1 ] = ( num__5 + num__1 ) / ( num__5 - num__1 ) = num__1.5 = num__1.5 answer : b <eor> b <eos> |
b |
multiply__1.5__1.0__ |
divide__1.5__1.0__ |
| ♠ n denotes the number obtained when n is rounded to the nearest tenth . for example ♠ num__4.31 = num__4.3 ♠ num__0.089 - ♠ num__1.235 = <o> a ) num__1.05 <o> b ) num__1.04 <o> c ) - num__1.05 <o> d ) - num__1.0 <o> e ) - num__1.1 |
♠ num__0.089 - ♠ num__1.135 = num__0.1 - num__1.2 = - num__1.1 answer : e <eor> e <eos> |
e |
subtract__1.235__1.135__ subtract__1.2__0.1__ subtract__1.2__0.1__ |
subtract__1.235__1.135__ subtract__1.2__0.1__ subtract__1.2__0.1__ |
| it takes joey the postman num__1 hours to run a num__3 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is num__6 mile / hour what is the speed with which joey returns ? <o> a ) num__11 <o> b ) num__12 <o> c ) num__13 <o> d ) num__14 <o> e ) num__18 |
let his speed for one half of the journey be num__3 miles an hour let the other half be x miles an hour now avg speed = num__6 mile an hour num__2 * num__3 * x / num__3 + x = num__6 num__6 x = num__6 x + num__18 = > x = num__18 e <eor> e <eos> |
e |
subtract__3.0__1.0__ multiply__3.0__6.0__ round__18.0__ |
divide__6.0__3.0__ multiply__3.0__6.0__ multiply__1.0__18.0__ |
| my grandson is about as many days as my son in weeks and my grandson is as many months as i am in years . my grandson my son and i together are num__140 years . can you tell me my age in years ? <o> a ) num__80 <o> b ) num__90 <o> c ) num__85 <o> d ) num__84 <o> e ) num__89 |
let the age of the man = x years so according to the given condition age of my grandson = x / num__12 years and age of my son = x num__0.576923076923 = num__15 x / num__26 and x + ( x / num__12 ) + ( num__15 x / num__26 ) = num__140 or ( num__156 x + num__13 x + num__90 x ) / num__156 = num__140 or num__259 x = num__156 * num__140 or x = num__156 * num__0.540540540541 = num__84 answer : d <eor> d <eos> |
d |
divide__156.0__12.0__ divide__140.0__259.0__ round__84.0__ |
divide__156.0__12.0__ divide__140.0__259.0__ round__84.0__ |
| a man buys an article for $ num__100 . and sells it for $ num__110 . find the gain percent ? <o> a ) num__10.0 <o> b ) num__15.0 <o> c ) num__25.0 <o> d ) num__20.0 <o> e ) num__30 % |
c . p . = $ num__100 s . p . = $ num__110 gain = $ num__10 gain % = num__0.1 * num__100 = num__10.0 answer is a <eor> a <eos> |
a |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| ram who is half as efficient as krish will take num__18 days to complete a task if he worked alone . if ram and krish worked together how long will they take to complete the task ? <o> a ) num__16 days <o> b ) num__12 days <o> c ) num__8 days <o> d ) num__6 days <o> e ) num__18 days |
number of days taken by ram to complete task = num__18 since ram is half as efficient as krish amount of work done by krish in num__1 day = amount of work done by ram in num__2 days if total work done by ram in num__18 days is num__18 w amount of work done by ram in num__1 day = w amount of work done by krish in num__1 day = num__2 w total amount of work done by krish and ram in a day = num__3 w total amount of time needed by krish and ram to complete task = num__18 w / num__3 w = num__6 days answer d <eor> d <eos> |
d |
add__1.0__2.0__ divide__18.0__3.0__ round__6.0__ |
add__1.0__2.0__ divide__18.0__3.0__ divide__18.0__3.0__ |
| if x is a positive odd number then each of the following is odd except <o> a ) ( x + num__3 ) ( x + num__5 ) <o> b ) x ^ num__2 + num__5 <o> c ) x ^ num__2 + num__6 x + num__9 <o> d ) num__3 x ^ num__2 + num__4 <o> e ) num__5 ( x + num__3 ) |
since it is given tht x is even number so any integer multiplied with x will also be even . . so we should concentrate only on other terms . . lets see the choices . . a . ( x + num__3 ) ( x + num__5 ) we have two terms with x and each is added with a odd number . . each bracket becomes odd and odd * odd = odd b . x ^ num__2 + num__5 here we are adding an odd number to even . . so e + o = o c . x ^ num__2 + num__6 x + num__9 here we are again adding an odd number to even . . so e + e + o = o d . num__3 x ^ num__2 + num__4 here we are adding an even number to even . . so e + e = e . . so tjis is our answer e . num__5 ( x + num__3 ) again o * o = o b <eor> b <eos> |
b |
subtract__5.0__3.0__ multiply__2.0__3.0__ power__3.0__2.0__ subtract__6.0__2.0__ subtract__4.0__2.0__ |
subtract__5.0__3.0__ multiply__2.0__3.0__ power__3.0__2.0__ subtract__6.0__2.0__ subtract__4.0__2.0__ |
| a man gave num__0.125 of his property to his wife and num__0.25 of the remaining to his daughter . what fraction of his property is he left with after giving to his wife and his daughter ? <o> a ) num__0.3125 <o> b ) num__0.25 <o> c ) num__0.65625 <o> d ) num__0.5 <o> e ) num__0.8 |
a man gave num__0.125 of his property to his wife so remaining is num__0.875 he gave num__0.25 of the remaining property to his daughter num__0.25 * num__0.875 = num__0.21875 total property given to his wife and daughter is num__0.125 + num__0.21875 to make the denominator common we use num__32 = num__0.125 num__0.125 + num__0.21875 = num__0.34375 is given to his wife and daughter remaining property left with him is num__0.65625 answer is c <eor> c <eos> |
c |
multiply__0.25__0.875__ add__0.125__0.2188__ subtract__0.875__0.2188__ subtract__0.875__0.2188__ |
multiply__0.25__0.875__ add__0.125__0.2188__ subtract__0.875__0.2188__ subtract__0.875__0.2188__ |
| the average of num__4 num__812 and x is num__10 . find the value of x ? <o> a ) num__10 <o> b ) num__16 <o> c ) num__15 <o> d ) num__24 <o> e ) num__12 |
average = ( num__4 + num__8 + num__12 + x ) / num__4 = num__10 x = num__16 answer is b <eor> b <eos> |
b |
add__4.0__8.0__ add__4.0__12.0__ add__4.0__12.0__ |
add__4.0__8.0__ add__4.0__12.0__ add__4.0__12.0__ |
| two pumps are connected to an empty tank . pump x fills the tank with water at a constant rate while pump y drains water out of the tank at a constant rate . the two pumps finish filling the tank in four times the duration it would take pump x alone to fill the tank . if pump y alone can empty a whole tank in num__44 minutes then how many minutes does it take pump x alone to fill the tank ? <o> a ) num__27 <o> b ) num__30 <o> c ) num__33 <o> d ) num__36 <o> e ) num__39 |
let v be the volume of the tank . let r be the rate per minute that pump x fills the tank . let t be the time it takes pump x to fill the tank . the rate at which pump y empties the tank is v / num__44 per minute . ( r - v / num__44 ) * num__4 t = v = rt . ( r - v / num__44 ) * num__4 = r num__3 r = v / num__11 . r = v / num__33 . it takes pump x num__33 minutes to fill the tank . the answer is c . <eor> c <eos> |
c |
divide__44.0__4.0__ subtract__44.0__11.0__ subtract__44.0__11.0__ |
divide__44.0__4.0__ subtract__44.0__11.0__ subtract__44.0__11.0__ |
| num__12.1212 + num__17.0005 - num__9.1105 = ? <o> a ) num__20.0015 <o> b ) num__20.0105 <o> c ) num__20.0115 <o> d ) num__20.1015 <o> e ) num__20.0112 |
solution given expression = ( num__12.1212 + num__17.0005 ) - num__9.1105 = ( num__29.1217 - num__9.1105 ) = num__20.0112 . answer e <eor> e <eos> |
e |
add__12.1212__17.0005__ subtract__29.1217__9.1105__ subtract__29.1217__9.1105__ |
add__12.1212__17.0005__ subtract__29.1217__9.1105__ subtract__29.1217__9.1105__ |
| how long does a train num__110 m long traveling at num__90 kmph takes to cross a bridge of num__170 m in length ? <o> a ) num__16.5 <o> b ) num__16.0 <o> c ) num__16.4 <o> d ) num__16.8 <o> e ) num__11.2 |
d = num__110 + num__170 = num__280 m s = num__90 * num__0.277777777778 = num__25 t = num__280 * num__0.04 = num__11.2 sec answer : e <eor> e <eos> |
e |
add__110.0__170.0__ divide__280.0__25.0__ round__11.2__ |
add__110.0__170.0__ multiply__280.0__0.04__ multiply__280.0__0.04__ |
| at a dinner party num__4 people are to be seated around a circular table . num__2 seating arrangements are considered different only when the positions of the people are different relative to each other . what is the total number of different possible seating arrangements for the group ? <o> a ) num__6 <o> b ) num__10 <o> c ) num__24 <o> d ) num__32 <o> e ) num__120 |
soln : since the arrangement is circular and num__2 seating arrangements are considered different only when the positions of the people are different relative to each other we can find the total number of possible seating arrangements by fixing one person ' s position and arranging the others . thus if one person ' s position is fixed the others can be arranged in num__3 ! = num__6 ways . ans is a . <eor> a <eos> |
a |
die_space__ die_space__ |
die_space__ die_space__ |
| a train moves fast a telegraph post and a bridge num__264 m long in num__8 sec and num__20 sec respectively . what is the speed of the train ? <o> a ) num__89 <o> b ) num__23 <o> c ) num__56 <o> d ) num__79.2 <o> e ) num__67 |
let the length of the train be x m and its speed be y m / sec . then x / y = num__8 = > x = num__8 y ( x + num__264 ) / num__20 = y y = num__22 speed = num__22 m / sec = num__22 * num__3.6 = num__79.2 km / hr . answer : option d <eor> d <eos> |
d |
multiply__3.6__22.0__ round__79.2__ |
multiply__3.6__22.0__ multiply__3.6__22.0__ |
| what is the units digit of the expression num__17 ^ num__7 - num__2 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__7 <o> e ) num__9 |
num__7 ^ num__1 = num__7 num__7 ^ num__2 = num__49 num__7 ^ num__3 = num__343 num__7 ^ num__4 = num__1 ( last digit ) num__7 ^ num__5 = num__7 ( last digit ) and the cycle repeats after every num__4 powers therefore last digit of num__17 ^ num__7 = num__3 num__3 - num__2 = num__1 answer a <eor> a <eos> |
a |
add__2.0__1.0__ multiply__7.0__49.0__ subtract__7.0__3.0__ subtract__7.0__2.0__ reverse__1.0__ |
add__2.0__1.0__ multiply__7.0__49.0__ subtract__7.0__3.0__ subtract__7.0__2.0__ subtract__2.0__1.0__ |
| num__5 years ago the age of anand was one - forth of the age of bala at that time . the present age of bala is num__13 years more than the present age of anand . find the present age of anand ? <o> a ) num__11 <o> b ) num__21 <o> c ) num__19 <o> d ) num__22 <o> e ) num__14 |
explanation : let the present ages of anand and bala be ' a ' and ' b ' respectively . a - num__5 = num__0.333333333333 ( b - num__5 ) - - - ( num__1 ) b = a + num__13 substituting b = a + num__8 in first equation a - num__10 = num__0.333333333333 ( a + num__8 ) = > num__3 a - num__30 = a + num__8 = > num__2 a = num__38 = > a = num__19 . answer : c <eor> c <eos> |
c |
subtract__13.0__5.0__ subtract__13.0__10.0__ multiply__3.0__10.0__ subtract__5.0__3.0__ add__8.0__30.0__ divide__38.0__2.0__ multiply__1.0__19.0__ |
subtract__13.0__5.0__ subtract__13.0__10.0__ multiply__3.0__10.0__ subtract__5.0__3.0__ add__8.0__30.0__ divide__38.0__2.0__ subtract__38.0__19.0__ |
| a man can row a boat at num__20 kmph in still water . if the speed of the stream is num__8 kmph what is the time taken to row a distance of num__60 km downstream ? <o> a ) num__1.66666666667 hours <o> b ) num__2.14285714286 hours <o> c ) num__6.15384615385 hours <o> d ) num__2.30769230769 hours <o> e ) num__0.697674418605 hours |
speed downstream = num__20 + num__8 = num__28 kmph . time required to cover num__60 km downstream = d / s = num__2.14285714286 = num__2.14285714286 hours . answer : b <eor> b <eos> |
b |
add__20.0__8.0__ divide__60.0__28.0__ divide__60.0__28.0__ |
add__20.0__8.0__ divide__60.0__28.0__ divide__60.0__28.0__ |
| simple interest on a sum at num__2.0 per annum for num__2 years is rs . num__100 . the c . i . on the same sum for the same period is ? <o> a ) num__101.8 <o> b ) num__101.4 <o> c ) num__101.2 <o> d ) num__100 <o> e ) num__110.2 |
si = num__50 + num__50 ci = num__50 + num__50 + num__1.8 = num__101.8 answer : a <eor> a <eos> |
a |
percent__100.0__101.8__ |
percent__100.0__101.8__ |
| which of the following expressions can be equal to num__0 when x ^ num__2 − num__16 = num__0 ? <o> a ) x ^ num__2 − num__6 x + num__9 <o> b ) x ^ num__2 − num__4 x + num__3 <o> c ) x ^ num__2 − x − num__2 <o> d ) x ^ num__2 − num__7 x + num__6 <o> e ) x ^ num__2 − num__5 x + num__4 |
x ^ num__2 − num__16 = num__0 x = - num__4 or x = num__4 onle one choice results in a possibility of num__0 when you plug in - num__4 or num__4 for x except rest of all x ^ num__2 − num__5 x + num__4 answer is e <eor> e <eos> |
e |
subtract__4.0__2.0__ |
subtract__4.0__2.0__ |
| if num__100.0 of j is equal to num__25.0 of k num__150.0 of k is equal to num__50.0 of l and num__125.0 of l is equal to num__75.0 of m then num__20.0 of m is equal to what percent of num__200.0 of j ? <o> a ) num__0.35 <o> b ) num__3.5 <o> c ) num__350 <o> d ) num__35 <o> e ) num__3500 |
imo answer should be num__350 . . . consider j = num__10 then k = num__50 l = num__150 and m = num__350 . . . . num__20.0 of num__350 comes out to be num__70 . . . . num__200.0 of num__10 is num__20 . . . . ( num__70 * num__100 ) / num__20 = num__350 . . . . ans : c <eor> c <eos> |
c |
percent__50.0__20.0__ percent__20.0__350.0__ percent__100.0__350.0__ |
percent__50.0__20.0__ percent__20.0__350.0__ percent__100.0__350.0__ |
| w and b are integers . the expression ( w + num__1 ) ( b + num__1 ) is even . what can be said about w and b ? <o> a ) they are both even numbers . <o> b ) at least one of them is even . <o> c ) at least one of them is odd . <o> d ) they are both odd . <o> e ) nothing can be said surly on w and b |
odd x odd = odd odd x even = even even x even = even to fulfill condition either ( w + num__1 ) or ( b + num__1 ) needs to be even so either w or b needs to be odd or at least one of them is odd . at least one of them is odd = c <eor> c <eos> |
c |
reverse__1.0__ |
reverse__1.0__ |
| the pilot of a small aircraft with a num__40 - gallon fuel tank wants to fly to cleveland which is num__480 miles away . the pilot recognizes that the current engine which can fly only num__7 miles per gallon will not get him there . by how many miles per gallon must the aircraft ’ s fuel efficiency be improved to make the flight to cleveland possible ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__5 <o> d ) num__40 <o> e ) num__160 |
actual miles / gallon is = num__12.0 = num__12 miles / gallon . current engine miles / gallon is num__7 miles / gallon . additional num__5 miles / gallon is required to match the actual mileage . imo option c . <eor> c <eos> |
c |
divide__480.0__40.0__ subtract__12.0__7.0__ round__5.0__ |
divide__480.0__40.0__ subtract__12.0__7.0__ round__5.0__ |
| john has taken four ( num__4 ) tests that have an average of num__85 . in order to bring his course grade up to a ‘ b ’ he will need to have a final average of num__89 . what will he need to average on his final two tests to achieve this grade ? <o> a ) num__87 <o> b ) num__90 <o> c ) num__92 <o> d ) num__97 <o> e ) num__98 |
traditional method : total scored till now num__85 * num__4 = num__340 total score to avg num__89 in num__6 tests = num__89 * num__6 = num__534 total to be scored on num__2 tests = num__534 - num__340 = num__194 avg on num__2 tests = num__97.0 = num__97 answer : d <eor> d <eos> |
d |
multiply__4.0__85.0__ multiply__89.0__6.0__ subtract__6.0__4.0__ subtract__534.0__340.0__ divide__194.0__2.0__ subtract__194.0__97.0__ |
multiply__4.0__85.0__ multiply__89.0__6.0__ subtract__6.0__4.0__ subtract__534.0__340.0__ divide__194.0__2.0__ subtract__194.0__97.0__ |
| if the price of an article went up by num__25.0 then by what percent should it be brought down to bring it back to its original price ? <o> a ) num__20.0 <o> b ) num__16 num__1.33333333333 % <o> c ) num__16 num__2.33333333333 % <o> d ) num__16 num__2.0 % <o> e ) num__16 num__0.666666666667 % |
let the price of the article be rs . num__100 . num__25.0 of num__100 = num__25 . new price = num__100 + num__25 = rs . num__125 required percentage = ( num__125 - num__100 ) / num__125 * num__100 = num__0.2 * num__100 = num__20.0 answer : a <eor> a <eos> |
a |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| a man has rs . num__480 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__33 <o> b ) num__38 <o> c ) num__37 <o> d ) num__90 <o> e ) num__28 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__480 ⇒ num__16 x = num__480 ∴ x = num__30 . hence total number of notes = num__3 x = num__90 answer : d <eor> d <eos> |
d |
divide__480.0__16.0__ divide__30.0__10.0__ multiply__3.0__30.0__ multiply__3.0__30.0__ |
divide__480.0__16.0__ divide__30.0__10.0__ multiply__3.0__30.0__ multiply__3.0__30.0__ |
| the length of a room is num__9 m and width is num__4.75 m . what is the cost of paying the floor by slabs at the rate of rs . num__900 per sq . metre . <o> a ) num__25650 <o> b ) num__25750 <o> c ) num__26550 <o> d ) num__30750 <o> e ) num__38475 |
area = num__9 × num__4.75 sq . metre . cost for num__1 sq . metre . = rs . num__900 hence total cost = num__9 × num__4.75 × num__900 = num__9 × num__4275 = rs . num__38475 answer is e . <eor> e <eos> |
e |
multiply__4.75__900.0__ multiply__9.0__4275.0__ round__38475.0__ |
multiply__4.75__900.0__ multiply__9.0__4275.0__ round__38475.0__ |
| praveen starts business with rs . num__3920 and after num__5 months hari joins with praveen as his partner . after a year the profit is divided in the ratio num__2 : num__3 . what is hari ’ s contribution in the capital ? <o> a ) s . num__10080 <o> b ) s . num__8000 <o> c ) s . num__8500 <o> d ) s . num__9000 <o> e ) s . num__6000 |
let hari ’ s capital be rs . x . then num__3920 * num__1.71428571429 x = num__0.666666666667 = > num__14 x = num__141120 = > x = num__10080 . answer : a <eor> a <eos> |
a |
divide__2.0__3.0__ divide__141120.0__14.0__ divide__141120.0__14.0__ |
divide__2.0__3.0__ divide__141120.0__14.0__ divide__141120.0__14.0__ |
| how many digits are required to number a book containing num__300 pages ? <o> a ) num__792 <o> b ) num__684 <o> c ) num__492 <o> d ) num__372 <o> e ) num__340 |
num__9 pages from num__1 to num__9 will require num__9 digits . num__90 pages from num__10 to num__99 will require num__90 * num__2 = num__180 digits . num__300 - ( num__90 + num__9 ) = num__201 pages will require num__201 * num__3 = num__603 digits . the total number of digits is num__9 + num__180 + num__603 = num__792 . the answer is a . <eor> a <eos> |
a |
add__1.0__9.0__ add__9.0__90.0__ multiply__2.0__90.0__ subtract__300.0__99.0__ add__1.0__2.0__ multiply__3.0__201.0__ multiply__1.0__792.0__ |
add__1.0__9.0__ add__9.0__90.0__ multiply__2.0__90.0__ subtract__300.0__99.0__ add__1.0__2.0__ multiply__3.0__201.0__ multiply__1.0__792.0__ |
| a train num__110 m long is running with a speed of num__60 km / hr . in what time will it pass a trolley that is running with a speed of num__12 km / hr in the direction opposite to that in which the train is going ? <o> a ) num__4.31 <o> b ) num__6.75 <o> c ) num__7.92 <o> d ) num__5.5 <o> e ) num__6.5 |
speed of train relative to trolley = num__60 + num__12 = num__72 km / hr . = num__72 * num__0.277777777778 = num__20 m / sec . time taken to pass the trolley = num__110 * num__0.05 = num__5.5 sec . answer : d <eor> d <eos> |
d |
add__60.0__12.0__ multiply__110.0__0.05__ round__5.5__ |
add__60.0__12.0__ multiply__110.0__0.05__ multiply__110.0__0.05__ |
| two trains of length num__100 m and num__220 m are running towards each other on parallel lines at num__42 kmph and num__30 kmph respectively . in what time will they be clear of each other from the moment they meet ? <o> a ) num__18 sec <o> b ) num__70 sec <o> c ) num__21 sec <o> d ) num__20 sec <o> e ) num__16 sec |
relative speed = ( num__42 + num__30 ) * num__0.277777777778 = num__4 * num__5 = num__20 mps . distance covered in passing each other = num__100 + num__220 = num__320 m . the time required = d / s = num__16.0 = num__16 sec . answer : e <eor> e <eos> |
e |
divide__100.0__5.0__ add__100.0__220.0__ divide__320.0__20.0__ round__16.0__ |
divide__100.0__5.0__ add__100.0__220.0__ divide__320.0__20.0__ divide__320.0__20.0__ |
| a football field is num__9600 square yards . if num__800 pounds of fertilizer are spread evenly across the entire field how many pounds of fertilizer were spread over an area of the field totaling num__3600 square yards ? <o> a ) num__450 <o> b ) num__600 <o> c ) num__750 <o> d ) num__2400 <o> e ) num__300 |
answer e ) num__9600 yards need num__1200 lbs num__1 yard will need num__0.0833333333333 = num__0.0833333333333 lbs num__3600 yards will need num__0.0833333333333 * num__3600 yards = num__300 lbs <eor> e <eos> |
e |
multiply__1.0__300.0__ |
multiply__1.0__300.0__ |
| what is the least number which when divided by num__12 num__2135 will leave in each case the same reminder num__6 ? <o> a ) num__210 <o> b ) num__420 <o> c ) num__414 <o> d ) num__426 <o> e ) num__454 |
here least number means lcm hence lcm of num__12 num__3521 = num__2 ^ num__2 x num__7 x num__3 x num__5 = num__420 therefore required number is num__420 + num__6 = num__426 ans - d <eor> d <eos> |
d |
divide__12.0__6.0__ divide__6.0__2.0__ subtract__12.0__7.0__ add__6.0__420.0__ add__6.0__420.0__ |
divide__12.0__6.0__ divide__6.0__2.0__ subtract__12.0__7.0__ add__6.0__420.0__ add__6.0__420.0__ |
| ( x + num__3 ) is a factor in x ^ num__2 - mx - num__15 . what is the value of m ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
i solved the second degree equation and found it like this : x ^ num__2 - mx - num__15 = num__0 ( x - num__5 ) ( x + num__3 ) = num__0 x = num__5 or x = - num__3 substituting both values for x in the equation we find : x ^ num__2 - mx - num__15 = > ( - num__3 ) ^ num__2 - m ( - num__3 ) = num__15 = > num__9 + num__3 m = num__15 = > num__3 m = num__15 - num__9 = num__6 = > m = num__2 and with num__5 using a similar process we end up with : ( num__5 ) ^ num__2 - m ( num__5 ) = num__15 - num__5 m = num__15 - num__25 = - num__10 m = num__2 ao ans b <eor> b <eos> |
b |
add__3.0__2.0__ multiply__3.0__2.0__ multiply__2.0__5.0__ subtract__5.0__3.0__ |
add__3.0__2.0__ subtract__15.0__9.0__ subtract__15.0__5.0__ subtract__5.0__3.0__ |
| if each digit in the set of a = ( num__23 num__45 ) is exactly used once in how many ways can the digits be arranged such that the num__3 and num__4 are not adjacent ? <o> a ) num__12 <o> b ) num__15 <o> c ) num__18 <o> d ) num__22 <o> e ) num__24 |
the answer should be = total arrangements - arrangements when num__2 and num__5 are together . = num__4 ! - num__3 ! * num__2 ! = num__24 - num__12 = num__12 take num__25 ( num__34 ) here there are num__3 entities whose number of arrangements are num__3 ! and num__2 ! is the arrangement between num__3 and num__4 answer is a <eor> a <eos> |
a |
add__3.0__2.0__ multiply__3.0__4.0__ add__23.0__2.0__ multiply__3.0__4.0__ |
add__3.0__2.0__ multiply__3.0__4.0__ add__23.0__2.0__ multiply__3.0__4.0__ |
| mary ’ s annual income is $ num__15000 and john ’ s annual income is $ num__18000 . by how much must mary ’ s annual income increase so that it constitutes num__60.0 of mary and john ’ s combined income ? <o> a ) $ num__3000 <o> b ) $ num__4000 <o> c ) $ num__7000 <o> d ) $ num__12000 <o> e ) $ num__25 |
000 |
let mary ' s income increase by x then the equation will be num__15000 + x = ( num__0.6 ) * ( num__15000 + x + num__18000 ) num__15000 + x = ( num__0.6 ) * ( num__33000 + x ) num__75000 + num__5 x = num__3 x + num__99000 num__2 x = num__24000 x = num__12000 so answer will be d <eor> d <eos> |
d |
d |
| the diagonals of two squares are in the ratio of num__2 : num__5 . find the ratio of their areas . <o> a ) num__5 : num__20 <o> b ) num__5 : num__25 <o> c ) num__4 : num__25 <o> d ) num__3 : num__15 <o> e ) num__3 : num__20 |
let the diagonals of the squares be num__2 x and num__5 x respectively . ratio of their areas = ( num__0.5 ) * ( num__2 x ) num__2 : ( num__0.5 ) * ( num__5 x ) num__2 = num__4 x num__2 : num__25 x num__2 = num__4 : num__25 . option c <eor> c <eos> |
c |
power__5.0__2.0__ triangle_area__2.0__4.0__ |
power__5.0__2.0__ volume_rectangular_prism__2.0__0.5__4.0__ |
| in covering a distance of num__30 km arun takes num__2 hours more than anil . if arun doubles his speed then he would take num__1 hour less than anil . what is arun ' s speed ? <o> a ) num__2 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
let the speed of arun = x kmph and the speed of anil = y kmph distance = num__30 km we know that distancespeed = timehence num__30 x − num__30 y = num__2 . . . . . . . . . . . ( equation num__1 ) num__30 y − num__302 x = num__1 . . . . . . . . . . . ( equation num__2 ) equation num__1 + equation num__2 ⇒ num__30 x − num__302 x = num__3 ⇒ num__302 x = num__3 ⇒ num__15 x = num__3 ⇒ num__5 x = num__1 ⇒ x = num__5 hence arun ' s speed = num__5 kmph answer : b <eor> b <eos> |
b |
add__2.0__1.0__ divide__30.0__2.0__ add__2.0__3.0__ round__5.0__ |
add__2.0__1.0__ divide__30.0__2.0__ add__2.0__3.0__ add__2.0__3.0__ |
| a and b walk around a circular track . they start at num__8 a . m . from the same point in the opposite directions . a and b walk at a speed of num__2 rounds per hour and num__3 rounds per hour respectively . how many times shall they cross each other before num__8.30 a . m . ? <o> a ) num__7 <o> b ) num__2 <o> c ) num__9 <o> d ) num__2 <o> e ) num__8 |
b num__2 relative speed = num__2 + num__3 = num__5 rounds per hour . so they cross each other num__5 times in an hour and num__2 times in half an hour . hence they cross each other num__2 times before num__8.30 a . m . <eor> b <eos> |
b |
subtract__8.0__3.0__ round__2.0__ |
add__2.0__3.0__ round__2.0__ |
| rahul can do a work in num__3 days while rajesh can do the same work in num__2 days . both of them finish the work together and get $ num__2250 . what is the share of rahul ? <o> a ) $ num__50 <o> b ) $ num__40 <o> c ) $ num__60 <o> d ) $ num__100 <o> e ) $ num__900 |
rahul ' s wages : rajesh ' s wages = num__0.333333333333 : num__0.5 = num__2 : num__3 rahul ' s share = num__2250 * num__0.4 = $ num__900 answer is e <eor> e <eos> |
e |
multiply__2250.0__0.4__ round__900.0__ |
multiply__2250.0__0.4__ multiply__2250.0__0.4__ |
| working alone pump a can empty a pool in num__6 hours . working alone pump b can empty the same pool in num__9 hours . working together how many minutes will it take pump a and pump b to empty the pool ? <o> a ) num__192 <o> b ) num__200 <o> c ) num__208 <o> d ) num__216 <o> e ) num__224 |
pump a can empty ( num__0.166666666667 ) of the pool per hour . pump b can empty ( num__0.111111111111 ) of the pool per hour . together the pumps can empty num__0.166666666667 + num__0.111111111111 = num__0.277777777778 of the pool per hour . num__1 pool / ( num__0.277777777778 ) pool per hour = num__3.6 hours = num__216 minutes . the answer is d . <eor> d <eos> |
d |
add__0.1111__0.1667__ round__216.0__ |
add__0.1111__0.1667__ divide__216.0__1.0__ |
| in a can there is a mixture of milk and water in the ratio num__4 : num__5 . if it is filled with an additional num__8 litres of milk the can would be full and ratio of milk and water would become num__6 : num__5 . find the capacity of the can ? <o> a ) num__40 <o> b ) num__44 <o> c ) num__48 <o> d ) num__47 <o> e ) num__50 |
let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = num__0.444444444444 ( t - num__8 ) after adding milk quantity of milk in the mixture = num__0.545454545455 t . num__6 t / num__11 - num__8 = num__0.444444444444 ( t - num__8 ) num__10 t = num__792 - num__352 = > t = num__44 . answer b <eor> b <eos> |
b |
add__5.0__6.0__ add__4.0__6.0__ multiply__4.0__11.0__ multiply__4.0__11.0__ |
add__5.0__6.0__ add__4.0__6.0__ divide__352.0__8.0__ divide__352.0__8.0__ |
| the h . c . f . of two numbers is num__40 and the other two factors of their l . c . m . are num__11 and num__15 . the larger of the two numbers is : <o> a ) num__276 <o> b ) num__600 <o> c ) num__699 <o> d ) num__722 <o> e ) num__745 |
the numbers are ( num__40 x num__11 ) and ( num__40 x num__15 ) . larger number = ( num__40 x num__15 ) = num__600 . answer : b <eor> b <eos> |
b |
multiply__40.0__15.0__ multiply__40.0__15.0__ |
multiply__40.0__15.0__ multiply__40.0__15.0__ |
| a bus trip of num__450 miles would have taken num__1 hour less if the average speed s for the trip had been greater by num__5 miles per hour . what was the average speed s in miles per hour for the trip ? <o> a ) num__10 <o> b ) num__40 <o> c ) num__45 <o> d ) num__50 <o> e ) num__55 |
( s + num__5 ) ( t - num__1 ) = num__450 s * t = num__450 solving both the equations we get : s = num__45 or - num__50 since speed should be positive s = num__45 hence option c <eor> c <eos> |
c |
add__5.0__45.0__ round__45.0__ |
add__5.0__45.0__ multiply__1.0__45.0__ |
| if num__7 / w + num__7 / x = num__7 / y and wx = y then the average ( arithmetic mean ) of w and x is <o> a ) num__0.5 <o> b ) num__1 <o> c ) num__2 <o> d ) num__4 <o> e ) num__8 |
given : num__7 / w + num__7 / x = num__7 / ywx = y find : ( w + x ) / num__2 = ? num__7 ( num__1 / w + num__1 / x ) = num__7 ( num__1 / y ) - divide both sides by num__7 ( num__1 / w + num__1 / x ) = num__1 / y ( x + w ) / wx = num__1 / wx - sub ' d in y = wx x + w - num__1 = num__0 x + w = num__1 therefore ( w + x ) / num__2 = num__0.5 ans : a <eor> a <eos> |
a |
reverse__2.0__ reverse__2.0__ |
reverse__2.0__ reverse__2.0__ |
| in an it company there are a total of num__90 employees including num__50 programmers . the number of male employees is num__80 including num__35 male programmers . how many employees must be selected to guaranty that we have num__3 programmers of the same sex ? <o> a ) num__10 <o> b ) num__45 <o> c ) num__55 <o> d ) num__35 <o> e ) num__65 |
you could pick num__40 non - programmers num__2 male programmers and num__2 female programmers and still not have num__3 programmers of the same sex . but if you pick one more person you must either pick a male or a female programmer so the answer is num__45 . b <eor> b <eos> |
b |
subtract__90.0__50.0__ divide__80.0__40.0__ divide__90.0__2.0__ divide__90.0__2.0__ |
subtract__90.0__50.0__ divide__80.0__40.0__ subtract__80.0__35.0__ subtract__90.0__45.0__ |
| the number of singles that a baseball player hit increased num__15 percent from her first season to her second season and the number of doubles that she hit in the same time period decreased by num__5 percent . if the total number of singles and doubles that she hit increased num__4 percent from her first season to her second season what is the ratio of the number of singles she hit in her first season to the number of doubles she hit in her first season ? <o> a ) num__3 : num__5 <o> b ) num__5 : num__7 <o> c ) num__7 : num__11 <o> d ) num__9 : num__11 <o> e ) num__11 : num__15 |
soln : - num__1.15 s + num__0.95 d = num__1.04 [ s + d ] num__0.11 s = num__0.09 d s / d = num__0.818181818182 answer : d <eor> d <eos> |
d |
subtract__1.15__1.04__ subtract__1.04__0.95__ divide__0.09__0.11__ add__5.0__4.0__ |
subtract__1.15__1.04__ subtract__1.04__0.95__ divide__0.09__0.11__ add__5.0__4.0__ |
| how many ways are there to split a group of num__4 boys into two groups of num__2 boys each ? ( the order of the groups does not matter ) <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__3 <o> e ) num__8 |
the combination is num__4 c num__1.0 = num__4 ! / num__2 ! * num__2 ! * num__2 = num__3 d <eor> d <eos> |
d |
choose__3.0__2.0__ |
choose__3.0__2.0__ |
| in a num__1400 m race usha beats shiny by num__50 m . in the same race by what time margin shiny beat mercy who runs at num__4 m / s ? <o> a ) num__100 sec . <o> b ) num__70 sec <o> c ) num__25 sec <o> d ) data not sufficient <o> e ) none of these |
speed of shiny = num__5.0 = num__5 m / s time taken by shiny to complete the race is b = num__280.0 = num__280 sec . time taken by baley to complete the race is d = num__350.0 = num__350 sec . hence d - b = num__70 sec answer : b <eor> b <eos> |
b |
divide__1400.0__5.0__ divide__1400.0__4.0__ divide__280.0__4.0__ round__70.0__ |
divide__1400.0__5.0__ divide__1400.0__4.0__ divide__280.0__4.0__ divide__280.0__4.0__ |
| lucy invested $ num__10000 in a new mutual fund account exactly three years ago . the value of the account increased by num__10 percent during the first year increased by num__5 percent during the second year and decreased by num__5 percent during the third year . what is the value of the account today ? <o> a ) $ num__10350 <o> b ) $ num__10 num__972.50 <o> c ) $ num__10500 <o> d ) $ num__11500 <o> e ) $ num__12 |
705 |
value after num__1 year : num__10000 * num__1.1 = num__11000 value after num__2 years : num__11000 * num__1.05 = num__11550 value today : num__11550 * num__0.95 = num__10395 answer b is correct . <eor> b <eos> |
b |
b |
| what is the greatest possible length which can be used to measure exactly the lengths num__8 m num__4 m num__20 cm and num__12 m num__20 cm ? <o> a ) num__10 cm <o> b ) num__30 cm <o> c ) num__25 cm <o> d ) num__20 cm <o> e ) num__35 cm |
required length = hcf of num__800 cm num__420 cm num__1220 cm = num__20 cm answer : option d <eor> d <eos> |
d |
add__800.0__420.0__ round__20.0__ |
add__800.0__420.0__ round__20.0__ |
| out of first num__20 natural numbers one number is selected at random . the probability that it is either an even number or a prime number is <o> a ) num__0.68 <o> b ) num__0.708333333333 <o> c ) num__0.739130434783 <o> d ) num__0.85 <o> e ) num__0.809523809524 |
n ( s ) = num__20 n ( even no ) = num__10 = n ( e ) n ( prime no ) = num__8 = n ( p ) p ( e ᴜ p ) = num__0.5 + num__0.4 - num__0.05 = num__0.85 answer : d <eor> d <eos> |
d |
union_prob__0.5__0.4__0.05__ union_prob__0.5__0.4__0.05__ |
union_prob__0.5__0.4__0.05__ union_prob__0.5__0.4__0.05__ |
| aishwarya ’ s mom was num__42 years of age at the time of her birth while her mom was num__32 years old when her num__4 years younger sister was born . the variance between the parents ages is : <o> a ) num__6 years <o> b ) num__14 years <o> c ) num__10 years <o> d ) num__9 years <o> e ) num__4 years |
b num__14 years mom ’ s age when aishwarya ’ s sister was born = num__32 years . dad ’ s age when aishwarya ’ s sister was born = ( num__42 + num__4 ) years = num__46 years . needed variance = ( num__46 – num__32 ) years = num__14 years . answer is b <eor> b <eos> |
b |
add__42.0__4.0__ subtract__46.0__32.0__ |
add__42.0__4.0__ subtract__46.0__32.0__ |
| if x = a ! which of these following values of a is the least possible value ( in this list ) for which the last num__7 digits of the integer x will all be zero ? <o> a ) num__28 <o> b ) num__32 <o> c ) num__36 <o> d ) num__40 <o> e ) num__44 |
num__2 * num__5 will give one num__0 at the end . we need num__2 ^ num__7 * num__5 ^ num__7 to get num__7 zeroes at the end . there are many num__2 ' s so we need to find the first num__7 appearances of num__5 . these multiples of num__5 are : num__5 num__10 num__15 num__20 num__5 * num__5 num__30 . . . which gives us num__7 . to have at least num__7 zeroes at the end a = > num__30 . the answer is b . <eor> b <eos> |
b |
subtract__7.0__2.0__ multiply__2.0__5.0__ add__5.0__10.0__ multiply__2.0__10.0__ multiply__2.0__15.0__ add__2.0__30.0__ |
subtract__7.0__2.0__ multiply__2.0__5.0__ add__5.0__10.0__ multiply__2.0__10.0__ multiply__2.0__15.0__ add__2.0__30.0__ |
| by selling num__80 bananas a fruit - seller gains the selling price of num__10 bananas . find the gain percent ? <o> a ) num__25.32 <o> b ) num__17.46 <o> c ) num__15.55 <o> d ) num__14.28 <o> e ) num__56.25 % |
sp = cp + g num__80 sp = num__80 cp + num__10 sp num__70 sp = num__80 cp num__70 - - - num__10 cp num__100 - - - ? = > num__14.28 answer : d <eor> d <eos> |
d |
percent__14.28__100.0__ |
percent__14.28__100.0__ |
| on a certain date pat invested $ num__9000 at x percent annual interest compounded annually . if the total value of the investment plus interest at the end of num__10 years will be $ num__36000 in how many years total will the total value of the investment plus interest increase to $ num__72000 ? <o> a ) num__15 <o> b ) num__16 <o> c ) num__18 <o> d ) num__20 <o> e ) num__24 |
num__36 |
000 = num__9000 ( num__1 + x ) ^ num__10 num__4 = ( num__1 + x ) ^ num__10 = num__2 ^ num__2 ( num__1 + x ) ^ num__10 = ( ( num__1 + x ) ^ num__5 ) ^ num__2 = num__2 ^ num__2 therefore ( num__1 + x ) ^ num__5 = num__2 num__72000 = num__9000 ( num__1 + x ) ^ n num__8 = ( num__1 + x ) ^ n num__2 ^ num__3 = ( num__1 + x ) ^ n ( num__1 + x ) ^ n = ( ( num__1 + x ) ^ num__5 ) ^ num__3 = ( num__1 + x ) ^ num__15 therefore n = num__15 . the answer is a . <eor> a <eos> |
a |
a |
| a box contains num__100 balls numbered from num__1 to num__100 . if three balls are selected at random and with replacement from the box what is the probability w that the sum of the three numbers on the balls selected from the box will be odd ? <o> a ) num__0.25 <o> b ) num__0.375 <o> c ) num__0.5 <o> d ) num__0.625 <o> e ) num__0.75 |
the sum of the three numbers on the balls selected from the box to be odd one should select either three odd numbered balls ( odd + odd + odd = odd ) or two even numbered balls and one odd numbered ball ( even + even + odd = odd ) ; p ( ooo ) = ( num__0.5 ) ^ num__3 ; p ( eeo ) = num__3 * ( num__0.5 ) ^ num__2 * num__0.5 = num__0.375 ( you should multiply by num__3 as the scenario of two even numbered balls and one odd numbered ball can occur in num__3 different ways : eeo eoe or oee ) ; so finally w = num__0.125 + num__0.375 = num__0.5 . answer : c . <eor> c <eos> |
c |
reverse__0.5__ subtract__0.5__0.375__ reverse__2.0__ |
reverse__0.5__ subtract__0.5__0.375__ add__0.375__0.125__ |
| a worker earns $ num__24 on the first day and spends $ num__18 on the second day . the worker earns $ num__24 on the third day and spends $ num__18 on the fourth day . if this pattern continues on which day will the worker first reach a net total of $ num__48 ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__9 <o> d ) num__12 <o> e ) num__16 |
every two days the net total is $ num__6 . after num__8 days the worker will have $ num__24 . on day num__9 the worker will receive $ num__24 for a net total of $ num__48 . the answer is c . <eor> c <eos> |
c |
subtract__24.0__18.0__ divide__48.0__6.0__ round__9.0__ |
subtract__24.0__18.0__ divide__48.0__6.0__ round__9.0__ |
| two trains leave the same train station at num__6 : num__00 am and num__6 : num__45 am and they travel at num__100 kph and num__120 kph respectively . how many kilometers from the train station will the two trains be together ? <o> a ) num__420 <o> b ) num__450 <o> c ) num__480 <o> d ) num__510 <o> e ) num__540 |
at num__6 : num__45 the first train is num__75 km ahead . the second train gains num__20 km each hour . the time it takes the second train to catch the first train is num__3.75 = num__3.75 hours . in this time the second train travels num__3.75 * num__120 = num__450 km . the answer is b . <eor> b <eos> |
b |
subtract__120.0__45.0__ divide__120.0__6.0__ divide__75.0__20.0__ multiply__6.0__75.0__ round__450.0__ |
subtract__120.0__45.0__ divide__120.0__6.0__ divide__75.0__20.0__ multiply__6.0__75.0__ multiply__6.0__75.0__ |
| a work can be finished in num__16 days by twenty women . the same work can be finished in fifteen days by sixteen men . the ratio between the capacity of a man and a woman is <o> a ) num__1 : num__3 <o> b ) num__4 : num__3 <o> c ) num__2 : num__3 <o> d ) num__2 : num__1 <o> e ) num__3 : num__1 |
explanation : work done by num__20 women in num__1 day = num__0.0625 work done by num__1 woman in num__1 day = num__1 / ( num__16 × num__20 ) work done by num__16 men in num__1 day = num__0.0666666666667 work done by num__1 man in num__1 day = num__1 / ( num__15 × num__16 ) ratio of the capacity of a man and woman = num__1 / ( num__15 × num__16 ) : num__1 / ( num__16 × num__20 ) = num__0.0666666666667 : num__0.05 = num__0.333333333333 : num__0.25 = num__4 : num__3 answer : option b <eor> b <eos> |
b |
divide__1.0__16.0__ subtract__16.0__1.0__ divide__1.0__20.0__ multiply__16.0__0.25__ subtract__4.0__1.0__ round__4.0__ |
divide__1.0__16.0__ subtract__16.0__1.0__ divide__1.0__20.0__ divide__1.0__0.25__ subtract__4.0__1.0__ divide__16.0__4.0__ |
| num__30 square stone slabs of equal size were needed to cover a floor area of num__58.8 sq . m . find the length of each stone slab ? <o> a ) num__140 cm <o> b ) num__767 cm <o> c ) num__88 cm <o> d ) num__666 cm <o> e ) num__776 cm |
area of each slab = num__58.8 / num__30 m num__2 = num__1.96 m num__2 length of each slab √ num__1.96 = num__1.4 m = num__140 cm <eor> a <eos> |
a |
triangle_area__2.0__140.0__ |
triangle_area__2.0__140.0__ |
| a water tank is two - fifth full . pipe a can fill a tank in num__10 minutes and pipe b can empty it in num__6 minutes . if both the pipes are open how long will it take to empty or fill the tank completely ? <o> a ) num__6 min <o> b ) num__8 min <o> c ) num__7 min <o> d ) num__9 min <o> e ) num__1 min |
clearly pipe b is faster than pipe a and so the tank will be emptied . part to be emptied = num__0.4 part emptied by ( a + b ) in num__1 minute = ( num__0.166666666667 - num__0.1 ) = num__0.0666666666667 num__0.0666666666667 : num__0.4 : : num__1 : x x = ( num__0.4 * num__1 * num__15 ) = num__6 min . so the tank will be emptied in num__6 min . answer : a <eor> a <eos> |
a |
divide__1.0__6.0__ divide__1.0__10.0__ divide__0.4__6.0__ divide__6.0__0.4__ round__6.0__ |
divide__1.0__6.0__ divide__1.0__10.0__ multiply__0.4__0.1667__ divide__6.0__0.4__ multiply__0.4__15.0__ |
| what is the least number of square tiles required to pave the floor of a room num__10 m num__17 cm long and num__9 m num__9 cm broad ? <o> a ) num__724 <o> b ) num__804 <o> c ) num__11413 <o> d ) num__844 <o> e ) none |
solution length of largest tile = h . c . f . of num__1017 cm & num__909 cm = num__9 cm . area of each tile = ( num__9 x num__9 ) cm num__2 ∴ required number of tiles = [ num__1017 x num__101.0 x num__9 ] = num__11413 . answer c <eor> c <eos> |
c |
divide__909.0__9.0__ round__11413.0__ |
divide__909.0__9.0__ round__11413.0__ |
| in a bag of small balls num__0.25 are green num__0.125 are blue num__0.0833333333333 are yellow and the remaining num__26 white . how many balls are blue ? <o> a ) num__5 balls . <o> b ) num__9 balls . <o> c ) num__10 balls . <o> d ) num__7 balls . <o> e ) num__6 balls . |
let us first find the fraction of green blue and yellow balls num__0.25 + num__0.125 + num__0.0833333333333 = num__0.25 + num__0.125 + num__0.0833333333333 common denominator = num__0.458333333333 add numerators the fraction of white balls is given by num__1.0 - num__0.458333333333 = num__0.541666666667 so the fraction num__0.541666666667 corresponds to num__26 balls . if x is the total number of balls then ( num__0.541666666667 ) of x = num__26 balls or ( num__0.541666666667 ) ? x = num__26 x = num__26 ? ( num__1.84615384615 ) = num__48 total number of balls the fraction of blue balls is num__0.125 of x . the number of blue balls is given by ( num__0.125 ) of num__48 = num__6 balls . correct answer e <eor> e <eos> |
e |
subtract__1.0__0.4583__ multiply__0.125__48.0__ multiply__0.125__48.0__ |
subtract__1.0__0.4583__ multiply__0.125__48.0__ multiply__0.125__48.0__ |
| the average ( arithmetic mean ) of four numbers is num__4 x + num__2 . if one of the numbers is x what is the average of the other three numbers ? <o> a ) x + num__1 <o> b ) num__3 x + num__3 <o> c ) num__5 x + num__1 <o> d ) num__5 x + num__2.66666666667 <o> e ) num__15 x + num__12 |
if the average of four numbers is num__4 x + num__3 then the sum of all four numbers if num__4 * ( num__4 x + num__2 ) = num__16 x + num__8 . if one of the numbers is x then the average of the remaining three numbers would be : ( num__16 x + num__8 - x ) / num__3 = ( num__15 x + num__8 ) / num__3 = num__5 x + num__2.66666666667 . answer is d . <eor> d <eos> |
d |
multiply__4.0__2.0__ add__2.0__3.0__ divide__8.0__3.0__ add__2.0__3.0__ |
multiply__4.0__2.0__ add__2.0__3.0__ divide__8.0__3.0__ add__2.0__3.0__ |
| by weight liquid x makes up num__0.8 percent of solution a and num__1.8 percent of solution b . if num__600 grams of solution a are mixed with num__700 grams of solution b then liquid x accounts for what percent of the weight of the resulting solution ? <o> a ) num__1.74 <o> b ) num__1.94 <o> c ) num__10.0 <o> d ) num__15.0 <o> e ) num__19 % |
i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = num__0.8 of weight of a + num__1.8 of weight of b when num__600 gms of a and num__700 gms of b is mixed : weight of liquid x = ( num__0.8 * num__600 ) / num__100 + ( num__1.8 * num__700 ) / num__100 = num__17.4 gms % of liquid x in resultant mixture = ( num__17.4 / num__1000 ) * num__100 = num__1.74 a <eor> a <eos> |
a |
percent__100.0__1.74__ |
percent__100.0__1.74__ |
| a circular path of num__13 m radius has marginal walk num__2 m wide all round it . find the cost of leveling the walk at num__25 p per m num__2 ? <o> a ) rs . num__18 <o> b ) rs . num__47 <o> c ) rs . num__44 <o> d ) rs . num__28 <o> e ) rs . num__28 |
π ( num__152 - num__132 ) = num__176 num__176 * num__0.25 = rs . num__44 answer : c <eor> c <eos> |
c |
multiply__176.0__0.25__ triangle_area__2.0__44.0__ |
multiply__176.0__0.25__ multiply__176.0__0.25__ |
| find the quadratic equations whose roots are the reciprocals of the roots of num__2 x num__2 + num__5 x + num__3 = num__0 ? <o> a ) + num__5 x - num__2 = num__0 <o> b ) + num__5 x + num__2 = num__0 <o> c ) - num__5 x + num__2 = num__0 <o> d ) - num__5 x - num__2 = num__0 <o> e ) of these |
the quadratic equation whose roots are reciprocal of num__2 x num__2 + num__5 x + num__3 = num__0 can be obtained by replacing x by num__1 / x . hence num__2 ( num__1 / x ) num__2 + num__5 ( num__1 / x ) + num__3 = num__0 = > num__3 x num__2 + num__5 x + num__2 = num__0 answer : b <eor> b <eos> |
b |
subtract__3.0__2.0__ add__2.0__3.0__ |
subtract__3.0__2.0__ add__2.0__3.0__ |
| a can do a piece of work in num__10 days ; b can do the same in num__20 days . how many days will it take the work to get finished if a and b work together ? <o> a ) num__4.21 <o> b ) num__5.32 <o> c ) num__6.66 <o> d ) num__7.52 <o> e ) num__8.02 |
a completes num__0.1 th of the work in one day and b completes num__0.05 th . ( num__0.1 ) + ( num__0.05 ) = ( num__1 / x ) num__0.15 = num__1 / x = > x = num__6.66666666667 = num__6.66 days answer : c <eor> c <eos> |
c |
multiply__10.0__0.1__ add__0.1__0.05__ divide__1.0__0.15__ round__6.66__ |
multiply__10.0__0.1__ add__0.1__0.05__ divide__1.0__0.15__ divide__6.66__1.0__ |
| a bullet train num__220 m long is running with a speed of num__59 kmph . in what time will it pass a man who is running at num__7 kmph in the direction opposite to that in which the bullet train is going ? <o> a ) num__23 sec <o> b ) num__15 sec <o> c ) num__12 sec <o> d ) num__11 sec <o> e ) num__16 sec |
c num__12 sec speed of the bullet train relative to man = ( num__59 + num__7 ) kmph = num__66 * num__0.277777777778 m / sec = num__18.3333333333 m / sec . time taken by the bullet train to cross the man = time taken by it to cover num__220 m at ( num__18.3333333333 ) m / sec = ( num__220 * num__0.0545454545455 ) sec = num__12 sec <eor> c <eos> |
c |
add__59.0__7.0__ divide__220.0__12.0__ divide__12.0__220.0__ round__12.0__ |
add__59.0__7.0__ divide__220.0__12.0__ divide__12.0__220.0__ divide__220.0__18.3333__ |
| in an entrance exam num__3 marks is awarded for every correct answer and ( - num__1 ) for every wrong answer . if a student gets num__38 marks after attempting all questions find the number of questions answered correctly if the total questions were num__70 . <o> a ) num__27 <o> b ) num__88 <o> c ) num__26 <o> d ) num__29 <o> e ) num__71 |
explanation : let x be the number of questions answered correctly . hence number of questions whose answer were wrong = ( num__70 – x ) num__3 * x + ( num__70 – x ) * ( - num__1 ) = num__38 num__4 x = num__38 + num__70 = num__108 x = num__27 answer : a <eor> a <eos> |
a |
add__3.0__1.0__ add__38.0__70.0__ divide__108.0__4.0__ multiply__1.0__27.0__ |
add__3.0__1.0__ add__38.0__70.0__ divide__108.0__4.0__ multiply__1.0__27.0__ |
| the mean of num__50 observations is num__200 . but later he found that there is decrements of num__15 from each observations . what is the the updated mean is ? <o> a ) num__165 <o> b ) num__185 <o> c ) num__190 <o> d ) num__198 <o> e ) num__199 |
num__185 answer is b <eor> b <eos> |
b |
subtract__200.0__15.0__ subtract__200.0__15.0__ |
subtract__200.0__15.0__ subtract__200.0__15.0__ |
| a bag contains an equal number of one rupee num__50 paise and num__25 paise coins respectively . if the total value is num__35 how many coins of each type are there ? <o> a ) num__20 coin <o> b ) num__30 coins <o> c ) num__28 coins <o> d ) num__25 coins <o> e ) none of these |
let number of each type of coin = x . then num__1 × x + . num__50 × x + . num__25 x = num__35 ⇒ num__1.75 x = num__35 ⇒ x = num__20 coin answer a <eor> a <eos> |
a |
divide__35.0__1.75__ divide__35.0__1.75__ |
divide__35.0__1.75__ divide__35.0__1.75__ |
| a basket of num__2 apple is to be chosen from four apples . what is the probability ways of apples that can be chosen ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__4 <o> e ) num__2 |
the question is out of num__4 apples num__2 apples should be chosen . required probability is num__4 c num__2 = num__4 ! / num__2 ! * num__2 ! = num__6 the option a is correct . <eor> a <eos> |
a |
die_space__ die_space__ |
die_space__ die_space__ |
| ab + ba - - - - - - - - - - - aac in the correctly worked addition problem shown here where the sum of the two - digit positive integers ab and ba is the three - digit integer aac and a b and c are different digits what is the units digit of the integer aac ? <o> a ) num__9 <o> b ) num__6 <o> c ) num__3 <o> d ) num__0 <o> e ) num__2 |
ab + ba = aac - > eq num__1 eq num__1 implies : b + a = num__10 + c - eq num__2 ( difficult to interpret . i agree . just sum up the units digit in eq num__1 ) now split up eq num__1 num__10 a + b + num__10 b + a = num__100 a + num__10 a + c - eq num__3 eq num__3 implies : num__11 ( a + b ) = num__110 a + c substitute eq num__2 ( b + a = num__10 + c ) in above num__11 ( num__10 + c ) = num__110 a + c = > num__110 + num__11 c = num__110 a + c implies a = num__1 c = num__0 d <eor> d <eos> |
d |
add__1.0__2.0__ add__1.0__10.0__ add__100.0__10.0__ multiply__0.0__1.0__ |
add__1.0__2.0__ add__1.0__10.0__ add__100.0__10.0__ multiply__0.0__1.0__ |
| triangle abc will be constructed in a xy - plane according to the following conditions : angle abc is num__90 degrees and ab is parallel to the y - axis ; for each of points a b and c both the x - coordinate and the y - coordinate must be integers ; the range of possible x - coordinates is num__0 = < x = < num__5 and the range of possible y - coordinates is - num__4 = < y = < num__6 . any two triangles with non - identical vertices are considered different . given these constraints how many different triangles could be constructed ? <o> a ) num__50 <o> b ) num__66 <o> c ) num__2500 <o> d ) num__3300 <o> e ) num__4356 |
since give that num__0 = < x = < num__5 and - num__4 = < y = < num__6 then we have a rectangle with dimensions num__6 * num__11 ( num__6 horizontal and num__11 vertical dots ) . ab is parallel to y - axis bc is parallel to x - axis and the right angle is at b . choose the ( x y ) coordinates for vertex b : num__6 c num__1 * num__11 c num__1 ; choose the x coordinate for vertex c ( as y coordinate is fixed by b ) : num__5 c num__1 ( num__6 - num__1 = num__5 as num__1 horizontal dot is already occupied by b ) ; choose the y coordinate for vertex a ( as x coordinate is fixed by b ) : num__10 c num__1 ( num__11 - num__1 = num__10 as num__1 vertical dot is already occupied by b ) . num__6 c num__1 * num__11 c num__1 * num__5 c num__1 * num__10 c num__1 = num__3300 . answer : d <eor> d <eos> |
d |
add__5.0__6.0__ subtract__5.0__4.0__ add__4.0__6.0__ multiply__1.0__3300.0__ |
add__5.0__6.0__ subtract__5.0__4.0__ subtract__11.0__1.0__ multiply__1.0__3300.0__ |
| two trains of equal lengths take num__10 sec and num__15 sec respectively to cross a telegraph post . if the length of each train be num__120 m in what time will they cross other travelling in opposite direction ? <o> a ) num__22 <o> b ) num__12 <o> c ) num__77 <o> d ) num__99 <o> e ) num__21 |
speed of the first train = num__12.0 = num__12 m / sec . speed of the second train = num__24.0 = num__8 m / sec . relative speed = num__12 + num__8 = num__20 m / sec . required time = ( num__120 + num__120 ) / num__20 = num__12 sec . answer : b <eor> b <eos> |
b |
divide__120.0__10.0__ divide__120.0__15.0__ add__8.0__12.0__ round__12.0__ |
divide__120.0__10.0__ divide__120.0__15.0__ add__8.0__12.0__ divide__120.0__10.0__ |
| the milk level in a rectangular box measuring num__62 feet by num__25 feet is to be lowered by num__6 inches . how many gallons of milk must be removed ? ( num__1 cu ft = num__7.5 gallons ) <o> a ) num__100 <o> b ) num__250 <o> c ) num__750 <o> d ) num__5812.5 <o> e ) num__5835.5 |
num__6 inches = num__0.5 feet ( there are num__12 inches in a foot . ) so num__62 * num__25 * num__0.5 = num__775 feet ^ num__3 of milk must be removed which equals to num__775 * num__7.5 = num__5812.5 gallons . answer : d . <eor> d <eos> |
d |
divide__6.0__0.5__ multiply__6.0__0.5__ multiply__7.5__775.0__ multiply__1.0__5812.5__ |
divide__6.0__0.5__ multiply__6.0__0.5__ multiply__7.5__775.0__ multiply__1.0__5812.5__ |
| the ratio of the money with ravi and giri is num__6 : num__7 and that with giri and kiran is num__6 : num__15 . if ravi has $ num__36 how much money does kiran have ? <o> a ) $ num__36 <o> b ) $ num__42 <o> c ) $ num__60 <o> d ) $ num__75 <o> e ) $ num__105 |
rita : sita : kavita num__6 : num__7 num__6 : num__15 num__36 : num__42 : num__105 the ratio of money with ravi giri and kiran is num__36 : num__42 : num__105 we see that num__36 = $ num__36 then num__105 = $ num__105 answer is e <eor> e <eos> |
e |
multiply__6.0__7.0__ multiply__7.0__15.0__ multiply__7.0__15.0__ |
multiply__6.0__7.0__ multiply__7.0__15.0__ multiply__7.0__15.0__ |
| two trains one from howrah to patna and the other from patna to howrah start simultaneously . after they meet the trains reach their destinations after num__16 hours and num__4 hours respectively . the ratio of their speeds is : <o> a ) num__2 : num__4 <o> b ) num__4 : num__3 <o> c ) num__6 : num__7 <o> d ) num__9 : num__16 <o> e ) none of these |
let us name the trains as a and b . then ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š num__4 : â ˆ š num__16 = num__2 : num__4 . answer a <eor> a <eos> |
a |
round__2.0__ |
round__2.0__ |
| one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill tank in num__37 min then the slower pipe alone will be able to fill the tank in ? <o> a ) num__229 <o> b ) num__787 <o> c ) num__144 <o> d ) num__148 <o> e ) num__121 |
let the slower pipe alone fill the tank in x min . then faster pipe will fill it in x / num__3 min . num__1 / x + num__3 / x = num__0.027027027027 num__4 / x = num__0.027027027027 = > x = num__148 min . answer : d <eor> d <eos> |
d |
divide__1.0__37.0__ add__1.0__3.0__ multiply__37.0__4.0__ round__148.0__ |
divide__1.0__37.0__ add__1.0__3.0__ multiply__37.0__4.0__ divide__148.0__1.0__ |
| if a person has two rectangular fields . the larger field has thrice the length and num__4 times the width of the smaller field . if the smaller field has a length num__50.0 more than the width . if a person takes num__20 minutes to complete one round of a smaller field then what is the time required to complete one round of a larger field ? <o> a ) num__69 minutes <o> b ) num__68 minutes <o> c ) num__58 minutes <o> d ) num__48 minutes <o> e ) num__67 minutes |
let the width of the smaller rectangle is num__4 units then the length of the smaller rectangle is num__6 units ( that is num__50.0 more than the width ) now the perimeter of the rectangle is num__2 ( num__6 + num__4 ) = num__20 units so num__20 units is covered in num__20 units implies covers one unit in one minute so now coming to the larger rectangle the width = num__16 units length = num__18 units perimeter = num__2 ( num__16 + num__18 ) num__68 units thus to cover the larger field num__68 minutes required to cover the rectangle . answer : b <eor> b <eos> |
b |
subtract__6.0__4.0__ subtract__20.0__4.0__ subtract__20.0__2.0__ add__50.0__18.0__ add__50.0__18.0__ |
subtract__6.0__4.0__ subtract__20.0__4.0__ add__2.0__16.0__ add__50.0__18.0__ add__50.0__18.0__ |
| three years ago the average age of a family of eight members was num__20 years . a boy have been born the average age of the family is the same today . what is the age of the boy ? <o> a ) a ) num__4 <o> b ) b ) num__3 <o> c ) c ) num__2 <o> d ) d ) num__1 <o> e ) e ) num__5 |
num__8 * num__23 = num__184 num__9 * num__20 = num__180 - - - - - - - - - - - - - - num__4 answer : a <eor> a <eos> |
a |
multiply__8.0__23.0__ multiply__20.0__9.0__ subtract__184.0__180.0__ subtract__8.0__4.0__ |
multiply__8.0__23.0__ multiply__20.0__9.0__ subtract__184.0__180.0__ subtract__8.0__4.0__ |
| find the highest common factor of num__36 and num__84 . <o> a ) num__4 <o> b ) num__6 <o> c ) num__12 <o> d ) num__16 <o> e ) num__18 |
num__36 = num__22 x num__32 num__84 = num__22 x num__3 x num__7 h . c . f . = num__22 x num__3 = num__12 . answer : option c <eor> c <eos> |
c |
gcd__36.0__84.0__ gcd__36.0__84.0__ |
gcd__36.0__84.0__ gcd__36.0__84.0__ |
| a and b complete a work in num__10 days . a alone can do it in num__40 days . if both together can do the work in how many days ? <o> a ) num__1.0875 days <o> b ) num__0.125 days <o> c ) num__0.0675 days <o> d ) num__0.0875 days <o> e ) num__0.0775 days |
num__0.1 + num__0.025 = num__0.125 days answer : b <eor> b <eos> |
b |
add__0.025__0.1__ add__0.025__0.1__ |
add__0.025__0.1__ add__0.025__0.1__ |
| what will be the ratio of simple interest earned by certain amount at the same rate of interest for num__4 years and that for num__9 years ? <o> a ) num__1 : num__3 <o> b ) num__2 : num__3 <o> c ) num__1 : num__2 <o> d ) num__4 : num__9 <o> e ) num__1 : num__4 |
let the principal be p and rate of interest be r % . required ratio = ( p x r x num__4 ) / num__100 divided by ( p x r x num__9 ) / num__100 = num__4 pr / num__9 pr = num__0.444444444444 = num__4 : num__9 . answer : d <eor> d <eos> |
d |
percent__4.0__100.0__ |
percent__4.0__100.0__ |
| peter mixed num__24 kg of butter at $ num__150 per kg with num__36 kg butter at the rate of $ num__125 per kg . at what price per kg should he sell the mixture to make a profit of num__40.0 in the transaction ? <o> a ) $ num__156 <o> b ) $ num__177 <o> c ) $ num__189 <o> d ) $ num__190 <o> e ) $ num__153 |
c $ num__189 cp per kg of mixture = [ num__24 ( num__150 ) + num__36 ( num__125 ) ] / ( num__24 + num__36 ) = $ num__135 sp = cp [ ( num__100 + profit % ) / num__100 ] = num__135 * [ ( num__100 + num__40 ) / num__100 ] = $ num__189 . <eor> c <eos> |
c |
percent__100.0__189.0__ |
percent__100.0__189.0__ |
| if a family spends num__40 percent of its household budget on rent and utilities num__30 percent to food and num__20 percent on other bills what fraction of their monthly budget is available for savings ? <o> a ) num__0.8 <o> b ) num__0.1 <o> c ) num__0.6 <o> d ) num__0.1 <o> e ) num__0.3 |
num__100.0 - ( num__40.0 + num__30.0 + num__20.0 ) = num__10.0 of the budget is available - - > num__10.0 = num__0.1 = num__0.1 . answer : d . <eor> d <eos> |
d |
percent__100.0__0.1__ |
percent__100.0__0.1__ |
| a boat can move upstream at num__25 kmph and downstream at num__35 kmph then the speed of the current is ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__7 <o> e ) num__1 |
us = num__25 ds = num__35 m = ( num__35 - num__25 ) / num__2 = num__5 answer : a <eor> a <eos> |
a |
round__5.0__ |
divide__25.0__5.0__ |
| a rectangular tiled patio is composed of num__40 square tiles . the rectangular patio will be rearranged so that there will be num__2 fewer columns of tiles and num__4 more rows of tiles . after the change in layout the patio will still have num__40 tiles and it will still be rectangular . how many rows are in the tile patio before the change in layout ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__10 <o> d ) num__13 <o> e ) num__28 |
r * c = num__70 and ( num__6 + num__4 ) ( c - num__2 ) = num__70 - - > r = num__6 and c = num__6 . answer : b . <eor> b <eos> |
b |
triangle_area__2.0__6.0__ |
triangle_area__2.0__6.0__ |
| krishna has num__20 apples . ram has num__15 apples . totally how many apples they have in all ? <o> a ) num__15 <o> b ) num__25 <o> c ) num__55 <o> d ) num__15 <o> e ) num__35 |
num__20 + num__15 = num__35 . answer is e <eor> e <eos> |
e |
add__20.0__15.0__ add__20.0__15.0__ |
add__20.0__15.0__ add__20.0__15.0__ |
| ram is able to sell a hand - carved statue for $ num__770 which was a num__10.0 profit over his cost . how much did the statue originally cost him ? <o> a ) $ num__496.30 <o> b ) $ num__512.40 <o> c ) $ num__555.40 <o> d ) $ num__700 <o> e ) $ num__588.20 |
cost price = num__7.0 * num__100 = num__700 answer : d <eor> d <eos> |
d |
percent__100.0__700.0__ |
percent__100.0__700.0__ |
| a certain number of workers can do a work in num__45 days . if there were num__10 workers more it could be finished in num__10 days less . how many workers are there ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__28 <o> d ) num__24 <o> e ) num__35 |
number of workers = num__10 * ( num__45 - num__10 ) / num__10 = num__35 answer is e <eor> e <eos> |
e |
subtract__45.0__10.0__ round__35.0__ |
subtract__45.0__10.0__ subtract__45.0__10.0__ |
| a train num__575 m long crosses a tunnel of length num__325 in num__90 sec . what is the speed of the train in kmph . <o> a ) num__28 <o> b ) num__32 <o> c ) num__36 <o> d ) num__24 <o> e ) num__30 |
total distance traveled = length of train + length of tunnel = num__575 + num__325 = num__900 time taken to cross the tunnel = num__90 sec . speed in kmph = distance / time * num__3.6 = num__10.0 * num__3.6 = num__36.0 = num__36 kmph answer : c <eor> c <eos> |
c |
add__575.0__325.0__ divide__900.0__90.0__ multiply__10.0__3.6__ round__36.0__ |
add__575.0__325.0__ divide__900.0__90.0__ multiply__10.0__3.6__ multiply__10.0__3.6__ |
| if x and y are both positive two - digit integers and share no digits and x > y what is the smallest possible value of x - y ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__9 |
answer = b = num__1 num__20 - num__19 = num__30 - num__29 = num__40 - num__39 = num__50 - num__49 = num__1 <eor> b <eos> |
b |
subtract__20.0__1.0__ subtract__30.0__1.0__ subtract__40.0__1.0__ add__20.0__30.0__ subtract__50.0__1.0__ reverse__1.0__ |
subtract__20.0__1.0__ subtract__30.0__1.0__ subtract__40.0__1.0__ add__20.0__30.0__ subtract__50.0__1.0__ subtract__40.0__39.0__ |
| a table is bought for rs . num__500 / - and sold at a loss of num__10.0 find its selling price <o> a ) s . num__450 / - <o> b ) s . num__530 / - <o> c ) s . num__540 / - <o> d ) s . num__600 / - <o> e ) s . num__700 / - |
num__100.0 - - - - - - > num__500 ( num__100 * num__5 = num__500 ) num__90.0 - - - - - - > num__450 ( num__90 * num__5 = num__450 ) selling price = rs . num__450 / - a <eor> a <eos> |
a |
percent__90.0__500.0__ percent__90.0__500.0__ |
percent__90.0__500.0__ percent__90.0__500.0__ |
| if the remainder is num__11 when the integer n is divided by num__20 what is the remainder when num__2 n is divided by num__10 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
n = num__20 k + num__11 num__2 n = num__2 ( num__20 k + num__11 ) = num__4 k * num__10 + num__22 = num__4 k * num__10 + num__2 * num__10 + num__2 = num__10 j + num__2 the answer is c . <eor> c <eos> |
c |
multiply__11.0__2.0__ divide__20.0__10.0__ |
multiply__11.0__2.0__ divide__20.0__10.0__ |
| the average age of num__15 students of a class is num__15 years . out of these the average age of num__5 students is num__14 years and that of the other num__9 students is num__16 years . tee age of the num__15 th student is ? <o> a ) num__11 years <o> b ) num__17 years <o> c ) num__67 years <o> d ) num__14 years <o> e ) num__12 years |
age of the num__15 th student = [ num__15 * num__15 - ( num__14 * num__5 + num__16 * num__9 ) ] = ( num__225 - num__214 ) = num__11 years . answer : a <eor> a <eos> |
a |
subtract__16.0__5.0__ subtract__16.0__5.0__ |
subtract__16.0__5.0__ subtract__16.0__5.0__ |
| on children ' s day sweets were to be equally distributed among num__175 children in a school . actually on the children ' s day num__35 children were absent and therefore each child got num__4 sweets extra . total how many sweets were available for distribution ? <o> a ) num__2400 <o> b ) num__2480 <o> c ) num__2680 <o> d ) num__2800 <o> e ) num__2900 |
let take x is the sweets received by each student . actually on the children ' s day num__35 children were absent and therefore each child got num__4 sweets extra . means num__35 * x = num__140 * num__4 = num__560 . x = num__16 . total sweets are num__175 * num__16 = num__2800 answer : d <eor> d <eos> |
d |
subtract__175.0__35.0__ multiply__4.0__140.0__ divide__560.0__35.0__ multiply__175.0__16.0__ multiply__175.0__16.0__ |
multiply__35.0__4.0__ multiply__4.0__140.0__ divide__560.0__35.0__ multiply__175.0__16.0__ multiply__175.0__16.0__ |
| a circular well with a diameter of num__6 metres is dug to a depth of num__24 metres . what is the volume of the earth dug out ? <o> a ) num__678.6 m num__3 <o> b ) num__36 m num__3 <o> c ) num__40 m num__3 <o> d ) num__44 m num__3 <o> e ) none |
solution volume = π r num__2 h ‹ = › ( num__3.14285714286 × num__3 × num__3 × num__24 ) m num__3 ‹ = › num__678.6 m num__3 . answer a <eor> a <eos> |
a |
triangle_area__678.6__2.0__ |
triangle_area__678.6__2.0__ |
| a total of num__3000 chocolates were distributed among num__120 boys and girls such that each boy received num__2 chocolates and each girl received num__3 chocolates . find the respective number of boys and girls ? <o> a ) num__97 <o> b ) num__60 <o> c ) num__91 <o> d ) num__26 <o> e ) num__21 |
let the number of boys be x . number of girls is num__120 - x . total number of chocolates received by boys and girls = num__2 x + num__3 ( num__120 - x ) = num__300 = > num__360 - x = num__300 = > x = num__60 . so the number of boys or girls is num__60 . answer : b <eor> b <eos> |
b |
multiply__120.0__3.0__ divide__120.0__2.0__ divide__120.0__2.0__ |
multiply__120.0__3.0__ subtract__360.0__300.0__ subtract__120.0__60.0__ |
| a boat having a length num__3 m and breadth num__2 m is floating on a lake . the boat sinks by num__1 cm when a man gets on it . the mass of the man is : <o> a ) num__12 kg <o> b ) num__60 kg <o> c ) num__72 kg <o> d ) num__80 kg <o> e ) num__96 kg |
volume of water displaced = ( num__3 x num__2 x num__0.01 ) m num__3 = num__0.06 m num__3 . mass of man = volume of water displaced x density of water = ( num__0.06 x num__1000 ) kg = num__60 kg . answer : option b <eor> b <eos> |
b |
hour_to_min_conversion__ hour_to_min_conversion__ |
hour_to_min_conversion__ hour_to_min_conversion__ |
| a train travels at an average of num__50 miles per hour for num__2 num__0.5 hours and then travels at a speed of num__70 miles per hour for num__1 num__0.5 hours . how far did the train travel in the entire num__4 hours ? <o> a ) num__200 miles <o> b ) num__230 miles <o> c ) num__250 miles <o> d ) num__150 miles <o> e ) num__300 miles |
total distance traveled = ( num__50 * num__2 num__0.5 ) + ( num__70 * num__1 num__0.5 ) = num__125 + num__105 = num__230 miles answer is b <eor> b <eos> |
b |
add__105.0__125.0__ round__230.0__ |
add__105.0__125.0__ multiply__1.0__230.0__ |
| num__12 men work num__8 hours per day to complete the work in num__10 days . to complete the same work in num__10 days working num__12 hours a day the number of men required ? <o> a ) num__10 days <o> b ) num__9 days <o> c ) num__6 days <o> d ) num__8 days <o> e ) num__7 days |
that is num__1 work done = num__12 × num__8 × num__10 then num__12 num__8 × num__10 = ? × num__12 × num__10 ? ( i . e . no . of men required ) = num__12 × num__8 × num__0.833333333333 × num__10 = num__8 days d ) <eor> d <eos> |
d |
divide__10.0__12.0__ round__8.0__ |
divide__10.0__12.0__ round__8.0__ |
| which of the following numbers is not prime ? <o> a ) num__5 + num__6 <o> b ) num__20 - num__1 <o> c ) num__3 + num__4 <o> d ) num__59 - num__6 <o> e ) num__9 + num__6 |
choice a is excluded because num__5 can not be factored . choice b is excluded because num__1 has no factors . in choice c the number num__3 is not able to be factored . choice d num__59 can not be factored . notice we can factor num__3 out of both num__9 + num__6 . - - > num__9 + num__6 = num__3 * ( num__3 + num__2 ) . the correct answer is e . <eor> e <eos> |
e |
add__1.0__5.0__ subtract__3.0__1.0__ multiply__1.0__9.0__ |
add__1.0__5.0__ subtract__3.0__1.0__ multiply__1.0__9.0__ |
| a is thrice as efficient as b and is therefore able to finish a piece of work num__10 days earlier than b . in how many days a and b will finish it together ? <o> a ) num__87 days <o> b ) num__77 days <o> c ) num__8 days <o> d ) num__5 days <o> e ) num__6 days |
wc = num__3 : num__1 wt = num__1 : num__3 x num__3 x num__1 / x – num__0.333333333333 x = num__0.1 x = num__6.66666666667 num__0.15 + num__0.05 = num__0.2 = > num__5 days answer : d <eor> d <eos> |
d |
divide__1.0__3.0__ divide__1.0__10.0__ divide__1.0__6.6667__ divide__0.15__3.0__ add__0.05__0.15__ divide__1.0__0.2__ round__5.0__ |
divide__1.0__3.0__ divide__1.0__10.0__ divide__1.0__6.6667__ divide__0.15__3.0__ add__0.05__0.15__ divide__1.0__0.2__ divide__1.0__0.2__ |
| num__5 n + num__2 > num__12 and num__7 n - num__12 < num__44 ; n must be between which numbers ? <o> a ) num__2 and num__8 <o> b ) num__2 and num__6 <o> c ) num__0 and num__9 <o> d ) num__2 and num__7 <o> e ) num__2 and num__9 |
num__5 n + num__2 > num__12 num__5 n > num__10 n > num__2 num__7 n - num__12 < num__44 num__7 n < num__56 n < num__8 so n must be between num__2 and num__8 num__2 < n < num__8 correct answer a <eor> a <eos> |
a |
multiply__5.0__2.0__ add__12.0__44.0__ subtract__10.0__2.0__ subtract__12.0__10.0__ |
subtract__12.0__2.0__ add__12.0__44.0__ subtract__10.0__2.0__ subtract__12.0__10.0__ |
| a cistern is filled by pipe a in num__8 hours and the full cistern can be leaked out by an exhaust pipe b in num__12 hours . if both the pipes are opened in what time the cistern is full ? <o> a ) num__24 hrs <o> b ) num__60 hrs <o> c ) num__70 hrs <o> d ) num__80 hrs <o> e ) num__90 hrs |
time taken to full the cistern = ( num__0.125 - num__0.0833333333333 ) hrs = num__0.0416666666667 = num__24 hrs answer : a <eor> a <eos> |
a |
subtract__0.125__0.0833__ round__24.0__ |
subtract__0.125__0.0833__ round__24.0__ |
| when sold at a num__60.0 discount a sweater nets the merchant a num__20.0 profit on the wholesale cost at which he initially purchased the item . by what % is the sweater marked up from wholesale at its normal retail price ? <o> a ) num__20.0 <o> b ) num__40.0 <o> c ) num__66.67 <o> d ) num__80.0 <o> e ) num__100 % |
we should be careful about what are we measuring % on / what is the base . . let the marked up price = num__100 . . selling price = num__100 - num__60.0 of num__100 = num__40 . . profit = num__20.0 . . therefore the wholesale purchase cost = x . . . . num__1.2 x = num__40 or x = num__33.33 . . . marked price was num__100 so . . . so answer is num__66.67 . . c <eor> c <eos> |
c |
percent__100.0__66.67__ |
percent__100.0__66.67__ |
| if | x - num__16 | = num__2 x then x = ? <o> a ) num__10 <o> b ) num__15 <o> c ) - num__16 <o> d ) num__16 <o> e ) num__5 |
| x - num__16 | = num__2 x . . . ( given ) x ^ num__2 - num__32 x + num__256 = num__4 x ^ num__2 num__3 * x ^ num__2 + num__32 * x - num__256 = num__0 . . . . ( by solving above eq . we get ) x = - num__16 or num__5.33 = = = > ans - c <eor> c <eos> |
c |
multiply__16.0__2.0__ subtract__32.0__16.0__ |
multiply__16.0__2.0__ subtract__32.0__16.0__ |
| insert the missing number num__121 num__112 . . . num__97 num__91 num__86 <o> a ) num__102 <o> b ) num__108 <o> c ) num__99 <o> d ) num__104 <o> e ) num__114 |
first and second number difference is starting from num__9 and further decresed by num__1 by each iteration . num__121 num__112 . . . num__97 num__91 num__86 . . . . dfference between two numbers will be as num__9 num__8 num__7 num__6 num__5 . . . answer : d <eor> d <eos> |
d |
subtract__121.0__112.0__ subtract__9.0__1.0__ subtract__8.0__1.0__ subtract__97.0__91.0__ subtract__91.0__86.0__ subtract__112.0__8.0__ |
subtract__121.0__112.0__ subtract__9.0__1.0__ subtract__8.0__1.0__ subtract__97.0__91.0__ subtract__91.0__86.0__ subtract__112.0__8.0__ |
| the distance between two stars is num__2.5 × num__10 ^ num__5 light years . what is the distance between the two stars in parsecs ? ( num__1 parsec = num__3.26 light years ) <o> a ) num__5.5 × num__10 ^ num__3 <o> b ) num__3.3 × num__10 ^ num__4 <o> c ) num__7.7 × num__10 ^ num__4 <o> d ) num__9.9 × num__10 ^ num__5 <o> e ) num__1.1 × num__10 ^ num__6 |
num__2.5 × num__10 ^ num__5 ly / ( num__3.26 ly / parsec ) = ( num__2.5 / num__3.26 ) x num__10 ^ num__5 = num__0.77 x num__10 ^ num__5 = num__7.7 x num__10 ^ num__4 parsec the answer is c . <eor> c <eos> |
c |
multiply__10.0__0.77__ divide__10.0__2.5__ multiply__10.0__0.77__ |
multiply__10.0__0.77__ divide__10.0__2.5__ divide__7.7__1.0__ |
| a can complete a work in num__12 days and b can do the same work in num__6 days . if a after doing num__3 days leaves the work find in how many days b will do the remaining work ? <o> a ) num__2 num__0.5 days <o> b ) num__4 num__0.5 days <o> c ) num__6 num__0.5 days <o> d ) num__7 num__0.5 days <o> e ) num__10 num__0.5 days |
the required answer = ( num__12 - num__3 ) * num__0.5 = num__4.5 = num__4 num__0.5 days answer is b <eor> b <eos> |
b |
divide__6.0__12.0__ divide__12.0__3.0__ round__4.0__ |
divide__6.0__12.0__ subtract__4.5__0.5__ subtract__4.5__0.5__ |
| two numbers are respectively num__20.0 and num__25.0 more than a third number . the percentage that is first of the second is ? <o> a ) num__76.0 <o> b ) num__86.0 <o> c ) num__96.0 <o> d ) num__56.0 <o> e ) num__92 % |
i ii iii num__120 num__125 num__100 num__125 - - - - - - - - - - num__120 num__100 - - - - - - - - - - - ? = > num__96.0 answer : c <eor> c <eos> |
c |
percent__96.0__100.0__ |
percent__96.0__100.0__ |
| after decreasing num__24.0 in the price of an article costs rs . num__912 . find the actual cost of an article ? <o> a ) num__1288 <o> b ) num__1238 <o> c ) num__1200 <o> d ) num__1233 <o> e ) num__1234 |
cp * ( num__0.76 ) = num__912 cp = num__12 * num__100 = > cp = num__1200 answer : c <eor> c <eos> |
c |
percent__100.0__1200.0__ |
percent__100.0__1200.0__ |
| paul earns $ num__12000 an year from his job . his income increased and now makes $ num__13500 an year . what is the percent increase ? <o> a ) num__25.0 <o> b ) num__20.0 <o> c ) num__15.5 <o> d ) num__12.5 <o> e ) num__10.5 % |
increase = ( num__0.125 ) * num__100 = ( num__0.125 ) * num__100 = num__12.5 . d <eor> d <eos> |
d |
percent__12.5__100.0__ |
percent__12.5__100.0__ |
| a tank is filled in eight hours by three pipes a b and c . pipe a is twice as fast as pipe b and b is twice as fast as c . how much time will pipe b alone take to fill the tank ? <o> a ) num__56 hours <o> b ) num__28 hours <o> c ) num__55 hours <o> d ) num__66 hours <o> e ) num__47 hours |
num__1 / a + num__1 / b + num__1 / c = num__0.125 ( given ) also given that a = num__2 b and b = num__2 c = > num__0.5 b + num__1 / b + num__2 / b = num__0.125 = > ( num__1 + num__2 + num__4 ) / num__2 b = num__0.125 = > num__2 b / num__7 = num__8 = > b = num__28 hours . answer : b <eor> b <eos> |
b |
divide__1.0__2.0__ divide__2.0__0.5__ divide__1.0__0.125__ multiply__4.0__7.0__ round__28.0__ |
divide__1.0__2.0__ divide__2.0__0.5__ divide__1.0__0.125__ multiply__4.0__7.0__ divide__28.0__1.0__ |
| ( num__9568422 x num__558 ) = ? <o> a ) num__5846381256 <o> b ) num__5339179476 <o> c ) num__5346381356 <o> d ) num__5846381406 <o> e ) num__5346381456 |
num__9568422 x num__558 = num__5339179476 ans b <eor> b <eos> |
b |
multiply__9568422.0__558.0__ multiply__9568422.0__558.0__ |
multiply__9568422.0__558.0__ multiply__9568422.0__558.0__ |
| num__5 men are equal to as many women as are equal to num__8 boys . all of them earn rs . num__75 only . men â € ™ s wages are ? <o> a ) num__6 rs <o> b ) num__7 rs <o> c ) num__8 rs <o> d ) num__5 rs <o> e ) num__1 rs |
num__5 m = xw = num__8 b num__5 m + xw + num__8 b - - - - - num__75 rs . num__5 m + num__5 m + num__5 m - - - - - num__75 rs . num__15 m - - - - - - num__75 rs . = > num__1 m = num__5 rs . answer : d <eor> d <eos> |
d |
divide__75.0__5.0__ multiply__5.0__1.0__ |
divide__75.0__5.0__ multiply__5.0__1.0__ |
| if in a certain sequence of consecutive multiples of num__100 the median is num__750 and the greatest term is num__1200 how many terms that are smaller than num__650 are there in the sequence ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__12 <o> e ) num__13 |
since the median is num__750 we know there must be a even number of integers so the list around num__750 must go . num__700 ( num__750 ) num__800 num__900 num__1000 num__1100 num__1200 since we know there are num__5 numbers greater than num__750 then there must be num__5 numbers less then num__750 c <eor> c <eos> |
c |
add__100.0__700.0__ add__100.0__800.0__ add__100.0__900.0__ add__100.0__1000.0__ divide__800.0__100.0__ |
add__100.0__700.0__ add__100.0__800.0__ add__100.0__900.0__ add__100.0__1000.0__ divide__800.0__100.0__ |
| pipe a fills a tank in num__56 minutes . pipe b can fill the same tank num__7 times as fast as pipe a . if both the pipes are kept open when the tank is empty how many minutes will it take to fill the tank ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
a ' s rate is num__0.0178571428571 and b ' s rate is num__0.125 . the combined rate is num__0.0178571428571 + num__0.125 = num__0.142857142857 the pipes will fill the tank in num__7 minutes . the answer is d . <eor> d <eos> |
d |
divide__7.0__56.0__ add__0.125__0.0179__ round__7.0__ |
divide__7.0__56.0__ add__0.125__0.0179__ round__7.0__ |
| one side of a rectangle is num__3 cm shorter than the other side . if we increase the length of each side by num__1 cm then the area of the rectangle will increase by num__24 cm num__2 . find the lengths of all sides . <o> a ) num__10 and num__3 <o> b ) num__7 and num__10 <o> c ) num__10 and num__7 <o> d ) num__3 and num__10 <o> e ) num__13 and num__10 |
let x be the length of the longer side x > num__3 then the other side ' s length is x − num__3 cm . then the area is s num__1 = x ( x - num__3 ) cm num__2 . after we increase the lengths of the sides they will become ( x + num__1 ) and ( x − num__3 + num__1 ) = ( x − num__2 ) cm long . hence the area of the new rectangle will be a num__2 = ( x + num__1 ) ⋅ ( x − num__2 ) cm num__2 which is num__24 cm num__2 more than the first area . therefore a num__1 + num__24 = a num__2 x ( x − num__3 ) + num__24 = ( x + num__1 ) ( x − num__2 ) x num__2 − num__3 x + num__24 = x num__2 + x − num__2 x − num__2 num__2 x = num__26 x = num__13 . so the sides of the rectangle are num__13 cm and ( num__13 − num__3 ) = num__10 cm long . so answer is e . <eor> e <eos> |
e |
triangle_area__1.0__26.0__ rectangle_perimeter__3.0__2.0__ multiply__1.0__13.0__ |
triangle_area__1.0__26.0__ rectangle_perimeter__3.0__2.0__ multiply__1.0__13.0__ |
| the angles of a triangle are in the ratio num__1 : num__3 : num__8 . find the measurement of the three angles of triangle . <o> a ) num__90 ° <o> b ) num__100 ° <o> c ) num__120 ° <o> d ) num__140 ° <o> e ) num__160 ° |
if the ratio of the three angles is num__1 : num__3 : num__8 then the measures of these angles can be written as x num__3 x and num__8 x . also the sum of the three interior angles of a triangle is equal to num__180 ° . hence x + num__3 x + num__8 x = num__180 solve for x num__12 x = num__180 x = num__15 the measures of the three angles are x = num__15 ° num__3 x = num__3 × num__15 = num__45 ° num__8 x = num__8 × num__15 = num__120 ° c <eor> c <eos> |
c |
straight_angle__ clock_small_arm_angle__1.0__180.0__ clock_small_arm_angle__1.0__180.0__ |
straight_angle__ clock_small_arm_angle__1.0__180.0__ clock_small_arm_angle__1.0__180.0__ |
| cost of num__16 mirror and num__8 comb is rs . num__360 and the cost of num__4 mirror and num__4 comb is rs . num__96 . find the cost of each mirror ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__20 <o> d ) num__21 <o> e ) num__22 |
let the cost of each pen and pencil be ' p ' and ' q ' respectively . num__16 m + num__8 c = num__360 - - - ( num__1 ) num__4 m + num__4 c = num__96 num__8 m + num__8 c = num__192 - - - ( num__2 ) ( num__1 ) - ( num__2 ) = > num__8 p = num__168 = > m = num__21 d <eor> d <eos> |
d |
divide__16.0__8.0__ subtract__360.0__192.0__ divide__168.0__8.0__ multiply__1.0__21.0__ |
divide__16.0__8.0__ subtract__360.0__192.0__ divide__168.0__8.0__ multiply__1.0__21.0__ |
| in a market a dozen eggs cost as much as a pound of rice and a half - liter of kerosene costs as much as num__6 eggs . if the cost of each pound of rice is $ num__0.24 then how many cents does a liter of kerosene cost ? [ one dollar has num__100 cents . ] <o> a ) num__0.20 <o> b ) num__0.24 <o> c ) num__20 <o> d ) num__24 <o> e ) num__55 |
a dozen eggs cost as much as a pound of rice - - > num__12 eggs = num__1 pound of rice = num__24 cents ; a half - liter of kerosene costs as much as num__6 eggs - - > num__6 eggs = num__0.5 liters of kerosene . how many cents does a liter of kerosene cost - - > num__1 liter of kerosene = num__12 eggs = num__1.0 * num__24 = num__24 cents . answer : d . <eor> d <eos> |
d |
multiply__0.24__100.0__ divide__6.0__12.0__ multiply__0.24__100.0__ |
multiply__0.24__100.0__ divide__6.0__12.0__ multiply__0.24__100.0__ |
| a jogger running at num__9 km / hr along side a railway track is num__240 m ahead of the engine of a num__120 m long train running at num__45 km / hr in the same direction . in how much time will the train pass the jogger ? <o> a ) num__22 <o> b ) num__88 <o> c ) num__36 <o> d ) num__88 <o> e ) num__27 |
speed of train relative to jogger = num__45 - num__9 = num__36 km / hr . = num__36 * num__0.277777777778 = num__10 m / sec . distance to be covered = num__240 + num__120 = num__360 m . time taken = num__36.0 = num__36 sec . answer : c <eor> c <eos> |
c |
subtract__45.0__9.0__ add__240.0__120.0__ subtract__45.0__9.0__ |
subtract__45.0__9.0__ add__240.0__120.0__ subtract__45.0__9.0__ |
| the number of arrangements that can be made with the letters of the word copy so that the vowels occupy the even places ? <o> a ) num__6 <o> b ) num__144 <o> c ) num__777 <o> d ) num__992 <o> e ) num__221 |
the word meadows has num__7 letters of which num__3 are vowels . - v - v - v - as the vowels have to occupy even places they can be arranged in the num__1 even places in num__1 ! i . e . num__1 ways . while the consonants can be arranged among themselves in the remaining num__3 places in num__3 ! i . e . num__6 ways . hence the total ways are num__1 * num__6 = num__6 . answer : a <eor> a <eos> |
a |
die_space__ die_space__ |
die_space__ die_space__ |
| a train num__250 m long running at num__72 kmph crosses a platform in num__30 sec . what is the length of the platform ? <o> a ) num__350 m <o> b ) num__200 m <o> c ) num__250 m <o> d ) num__270 m <o> e ) num__300 m |
d = num__72 * num__0.277777777778 = num__30 = num__600 â € “ num__250 = num__350 m answer : a <eor> a <eos> |
a |
subtract__600.0__250.0__ round__350.0__ |
subtract__600.0__250.0__ round__350.0__ |
| a train num__300 m long passed a pole in num__22 sec . how long will it take to pass a platform num__725 m long ? <o> a ) num__12 sec <o> b ) num__59 sec <o> c ) num__73 sec <o> d ) num__67 sec <o> e ) num__76 sec |
speed = num__13.6363636364 = num__14 m / sec . required time = ( num__300 + num__725 ) / num__14 = num__73 sec . answer : c <eor> c <eos> |
c |
divide__300.0__22.0__ round__73.0__ |
divide__300.0__22.0__ round__73.0__ |
| a man can row a boat at num__14 kmph in still water . if the speed of the stream is num__5 kmph what is the time taken to row a distance of num__80 km downstream ? <o> a ) num__4.47368421053 <o> b ) num__5.0 <o> c ) num__4.21052631579 <o> d ) num__3.68421052632 <o> e ) num__3.15789473684 |
speed downstream = num__14 + num__5 = num__19 kmph . time required to cover num__80 km downstream = d / s = num__4.21052631579 = num__4.21052631579 hours . answer : c <eor> c <eos> |
c |
add__14.0__5.0__ divide__80.0__19.0__ divide__80.0__19.0__ |
add__14.0__5.0__ divide__80.0__19.0__ divide__80.0__19.0__ |
| num__2 men and num__3 boys can do a piece of work in num__10 days while num__3 men and num__2 boys can do the same work in num__8 days . in how many days can num__2 men and num__1 boy do the work ? <o> a ) num__14.5 <o> b ) num__10.5 <o> c ) num__13.5 <o> d ) num__11.5 <o> e ) num__12.5 |
let num__1 man ’ s num__1 day ’ s work = x and num__1 boy ’ s num__1 day ’ s work = y . then num__2 x + num__3 y = num__0.1 and num__3 x + num__2 y = num__0.125 solving we get : x = num__0.035 and y = num__0.01 ( num__2 men + num__1 boy ) ’ s num__1 day ’ s work = ( num__2 x num__0.035 + num__1 x num__0.01 ) = num__0.08 = num__0.08 so num__2 men and num__1 boy together can finish the work in num__12.5 days answer is e . <eor> e <eos> |
e |
divide__1.0__10.0__ divide__1.0__8.0__ divide__0.1__10.0__ multiply__8.0__0.01__ divide__1.0__0.08__ round__12.5__ |
divide__1.0__10.0__ divide__1.0__8.0__ divide__0.1__10.0__ multiply__8.0__0.01__ divide__1.0__0.08__ round__12.5__ |
| the ratio of the incomes of a and b is num__5 : num__4 and the ratio of their expenditure is num__3 : num__2 . if at the end of the year each saves $ num__1600 then the income of a is ? <o> a ) $ num__1500 <o> b ) $ num__4000 <o> c ) $ num__2000 <o> d ) $ num__2500 <o> e ) $ num__3200 |
let the income of a and b be $ num__5 x and $ num__4 x let their expenditures be $ num__3 y and $ num__2 y num__5 x - num__3 y = num__1600 - - - - - - - num__1 ) num__4 x - num__2 y = num__1600 - - - - - - - num__2 ) from num__1 ) and num__2 ) x = num__800 a ' s income = num__5 x = num__5 * num__800 = $ num__4000 answer is b <eor> b <eos> |
b |
subtract__5.0__4.0__ divide__1600.0__2.0__ multiply__5.0__800.0__ multiply__5.0__800.0__ |
subtract__5.0__4.0__ divide__1600.0__2.0__ multiply__5.0__800.0__ multiply__5.0__800.0__ |
| a salesman ’ s terms were changed from a flat commission of num__5.0 on all his sales to a fixed salary of rs . num__1000 plus num__2.5 commission on all sales exceeding rs . num__4000 . if his remuneration as per new scheme was rs . num__600 more than that by the previous schema his sales were worth ? <o> a ) num__18997 <o> b ) num__12000 <o> c ) num__28992 <o> d ) num__13009 <o> e ) num__28811 |
[ num__1000 + ( x - num__4000 ) * ( num__2.5 / num__100 ) ] - x * ( num__0.05 ) = num__600 x = num__12000 answer : b <eor> b <eos> |
b |
divide__5.0__100.0__ divide__600.0__0.05__ divide__600.0__0.05__ |
divide__5.0__100.0__ divide__600.0__0.05__ divide__600.0__0.05__ |
| the cross - section of a cannel is a trapezium in shape . if the cannel is num__10 m wide at the top and num__6 m wide at the bottom and the area of cross - section is num__640 sq m the depth of cannel is ? <o> a ) num__18 <o> b ) num__27 <o> c ) num__20 <o> d ) num__80 <o> e ) num__81 |
num__0.5 * d ( num__10 + num__6 ) = num__640 d = num__80 answer : d <eor> d <eos> |
d |
round__80.0__ |
round__80.0__ |
| there are num__250 scraps . for each num__11 cutting one extra can be obtained . how much maximum is possible if one cut by one unit ? <o> a ) num__271 <o> b ) num__272 <o> c ) num__273 <o> d ) num__274 <o> e ) num__275 |
num__242 is divisible by num__11 . s num__022 extra will come . and another num__8 remained . so num__242 + num__8 + num__22 = num__272 answer : b <eor> b <eos> |
b |
divide__242.0__11.0__ subtract__250.0__242.0__ add__250.0__22.0__ add__250.0__22.0__ |
divide__242.0__11.0__ subtract__250.0__242.0__ add__250.0__22.0__ add__250.0__22.0__ |
| a man have num__800 kg mango . he found num__8.0 of its mango are rotten . then how much mango is rotten ? <o> a ) num__64 <o> b ) num__72 <o> c ) num__68 <o> d ) num__65 <o> e ) num__82 |
num__8.0 of num__800 kg = ( num__800 * num__8 ) / num__100 . = num__64 mangoes answer a <eor> a <eos> |
a |
percent__8.0__800.0__ percent__8.0__800.0__ |
percent__8.0__800.0__ percent__8.0__800.0__ |
| if num__2 x = y = num__6 z what is x - z in terms of y ? <o> a ) y <o> b ) num__2 y <o> c ) y / num__2 <o> d ) y / num__3 <o> e ) y / num__4 |
num__2 x = y = num__6 z x = y / num__2 and z = y / num__6 x - z = y / num__2 - y / num__6 = y / num__3 answer is d <eor> d <eos> |
d |
divide__6.0__2.0__ divide__6.0__2.0__ |
divide__6.0__2.0__ divide__6.0__2.0__ |
| cole drove from home to work at an average speed of num__80 kmh . he then returned home at an average speed of num__120 kmh . if the round trip took a total of num__2 hours how many minutes did it take cole to drive to work ? <o> a ) num__66 <o> b ) num__70 <o> c ) num__72 <o> d ) num__75 <o> e ) num__78 |
first round distance travelled ( say ) = d speed = num__80 k / h time taken t num__2 = d / num__80 hr second round distance traveled = d ( same distance ) speed = num__120 k / h time taken t num__2 = d / num__120 hr total time taken = num__2 hrs therefore num__2 = d / num__80 + d / num__120 lcm of num__80 and num__120 = num__240 num__2 = d / num__80 + d / num__120 = > num__2 = num__3 d / num__240 + num__2 d / num__240 = > d = num__240 * num__0.4 km therefore t num__1 = d / num__80 = > t num__1 = ( num__240 * num__2 ) / ( num__5 x num__80 ) = > t num__1 = ( num__6 x num__60 ) / num__5 - - in minutes = > t num__1 = num__72 minutes . c <eor> c <eos> |
c |
multiply__120.0__2.0__ divide__240.0__80.0__ subtract__3.0__2.0__ add__2.0__3.0__ multiply__2.0__3.0__ hour_to_min_conversion__ round__72.0__ |
multiply__120.0__2.0__ divide__240.0__80.0__ subtract__3.0__2.0__ add__2.0__3.0__ multiply__2.0__3.0__ divide__120.0__2.0__ divide__72.0__1.0__ |
| cereal a is num__10.0 sugar by weight whereas healthier but less delicious cereal b is num__2.0 sugar by weight . to make a delicious and healthy mixture that is num__5.0 sugar what should be the ratio of cereal a to cereal b by weight ? <o> a ) num__2 : num__5 <o> b ) num__3 : num__5 <o> c ) num__4 : num__7 <o> d ) num__2 : num__6 <o> e ) num__1 : num__4 |
num__2.0 is num__3.0 - points below num__5.0 and num__10.0 is num__5.0 - points above num__5.0 . the ratio of a : b should be num__3 : num__5 . the answer is b . <eor> b <eos> |
b |
subtract__5.0__2.0__ subtract__5.0__2.0__ |
subtract__5.0__2.0__ subtract__5.0__2.0__ |
| what will be the cost of building a fence around a square plot with area equal to num__36 sq ft if the price per foot of building the fence is rs . num__58 ? <o> a ) num__3944 <o> b ) num__2287 <o> c ) num__2977 <o> d ) num__2668 <o> e ) num__1392 |
let the side of the square plot be a ft . a num__2 = num__36 = > a = num__6 length of the fence = perimeter of the plot = num__4 a = num__24 ft . cost of building the fence = num__24 * num__58 = rs . num__1392 . answer : e <eor> e <eos> |
e |
surface_cube__2.0__ multiply__58.0__24.0__ multiply__58.0__24.0__ |
multiply__4.0__6.0__ multiply__58.0__24.0__ multiply__58.0__24.0__ |
| num__9 men are equal to as many women as are equal to num__7 boys . all of them earn rs . num__216 only . men ’ s wages are ? <o> a ) num__6 <o> b ) num__5 <o> c ) num__2 <o> d ) num__8 <o> e ) num__2 |
num__9 m = xw = num__7 b num__9 m + xw + num__7 b - - - - - num__216 rs . num__9 m + num__9 m + num__9 m - - - - - num__216 rs . num__27 m - - - - - - num__216 rs . = > num__1 m = num__8 rs . answer : d <eor> d <eos> |
d |
subtract__9.0__1.0__ subtract__9.0__1.0__ |
subtract__9.0__1.0__ subtract__9.0__1.0__ |
| the average age of num__3 girls is num__24 years and their ages are in the proportion num__1 : num__2 : num__3 . the age of the youngest girl is ? <o> a ) num__10 years <o> b ) num__11 years <o> c ) num__12 years <o> d ) num__13 years <o> e ) num__14 years |
total age of num__3 boys = num__24 * num__3 = num__72 ratio of their ages = num__1 : num__2 : num__3 age of the youngest = num__72 * num__0.166666666667 = num__12 years answer is c <eor> c <eos> |
c |
multiply__3.0__24.0__ divide__24.0__2.0__ divide__24.0__2.0__ |
multiply__3.0__24.0__ divide__24.0__2.0__ divide__24.0__2.0__ |
| num__12 buckets of water fill a tank when the capacity of each bucket is num__49 litres . how many buckets will be needed to fill the same tank if the capacity of each bucket is num__7 litres ? <o> a ) num__39 <o> b ) num__84 <o> c ) num__80 <o> d ) num__82 <o> e ) num__86 |
capacity of the tank = ( num__12 Ã — num__49 ) litre number of buckets required of capacity of each bucket is num__17 litre = num__12 Ã — num__7.0 = num__12 Ã — num__7 = num__84 answer is b <eor> b <eos> |
b |
multiply__12.0__7.0__ round__84.0__ |
multiply__12.0__7.0__ round__84.0__ |
| a and b can finish a work in num__16 days while a alone can do the same work in num__24 days . in how many days b alone will complete the work ? <o> a ) num__87 days <o> b ) num__48 days <o> c ) num__17 days <o> d ) num__87 days <o> e ) num__12 days |
b = num__0.0625 – num__0.0416666666667 = num__0.0208333333333 = > num__48 days answer : b <eor> b <eos> |
b |
subtract__0.0625__0.0417__ round__48.0__ |
subtract__0.0625__0.0417__ round__48.0__ |
| of num__64 players on a cricket team num__37 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed how many right - handed players are there total ? <o> a ) num__54 <o> b ) num__55 <o> c ) num__59 <o> d ) num__71 <o> e ) num__92 |
total = num__64 thrower = num__37 rest = num__64 - num__37 = num__27 left handed = num__9.0 = num__9 right handed = num__18 if all thrower are right handed then total right handed is num__37 + num__18 = num__55 so b . num__55 is the right answer <eor> b <eos> |
b |
subtract__64.0__37.0__ subtract__27.0__9.0__ subtract__64.0__9.0__ subtract__64.0__9.0__ |
subtract__64.0__37.0__ subtract__27.0__9.0__ subtract__64.0__9.0__ subtract__64.0__9.0__ |
| roger can read a book in k minutes . what part of the book can he read in num__9 minutes ? ( k > num__8 ) <o> a ) num__8 + k <o> b ) num__8 / k <o> c ) k / num__8 <o> d ) ( k + num__8 ) / k <o> e ) ( k - num__8 ) / k |
let ' s sayk = num__24 that is it takes num__24 minutes to read the entire book . so in num__8 minutes roger can read num__0.333333333333 of the book so we ' re looking for the answer choice that yields an output of num__0.333333333333 whenk = num__24 a ) num__8 + num__24 = num__32 eliminate b ) num__0.333333333333 = num__0.333333333333 keep c ) num__3.0 = num__3 eliminate d ) ( num__24 + num__8 ) / num__24 = num__1.33333333333 = num__1.33333333333 eliminate e ) ( num__24 - num__8 ) / num__24 = num__0.666666666667 = num__0.666666666667 eliminate answer : c <eor> c <eos> |
c |
divide__8.0__24.0__ add__8.0__24.0__ divide__24.0__8.0__ divide__32.0__24.0__ round__8.0__ |
divide__8.0__24.0__ add__8.0__24.0__ divide__24.0__8.0__ divide__32.0__24.0__ subtract__32.0__24.0__ |
| what is the dividend . divisor num__14 the quotient is num__9 and the remainder is num__5 ? <o> a ) a ) num__130 <o> b ) b ) num__131 <o> c ) c ) num__148 <o> d ) d ) num__158 <o> e ) e ) num__160 |
d = d * q + r d = num__14 * num__9 + num__5 d = num__126 + num__5 d = num__131 answer b <eor> b <eos> |
b |
multiply__14.0__9.0__ add__5.0__126.0__ add__5.0__126.0__ |
multiply__14.0__9.0__ add__5.0__126.0__ add__5.0__126.0__ |
| a person covered one - fourth of the total distance at num__15 kmph and remaining distance at num__24 kmph . what is the average speed for the total distance ? <o> a ) num__20 ( num__0.125 ) kmph <o> b ) num__20 ( num__0.869565217391 ) kmph <o> c ) num__20 ( num__0.739130434783 ) kmph <o> d ) num__27 ( num__0.0434782608696 ) kmph <o> e ) num__21 ( num__0.125 ) kmph |
let the total distance be x km total time taken = ( x / num__4 ) / num__15 + ( num__3 x / num__4 ) / num__24 = x / num__60 + x / num__32 = num__23 x / num__480 average speed = x / ( num__23 x / num__480 ) = num__20.8695652174 kmph = num__20 ( num__0.869565217391 ) kmph . answer : b <eor> b <eos> |
b |
multiply__15.0__4.0__ multiply__15.0__32.0__ divide__480.0__23.0__ round_down__20.8696__ divide__20.8696__24.0__ round_down__20.8696__ |
multiply__15.0__4.0__ multiply__15.0__32.0__ divide__480.0__23.0__ divide__480.0__24.0__ divide__20.8696__24.0__ divide__480.0__24.0__ |
| car a runs at the speed of num__65 km / hr & reaches its destination in num__8 hr . car b runs at the speed of num__70 km / h & reaches its destination in num__4 h . what is the respective ratio of distances covered by car a & car b ? <o> a ) num__11 : num__6 <o> b ) num__12 : num__7 <o> c ) num__13 : num__7 <o> d ) num__15 : num__6 <o> e ) num__13 : num__6 |
sol . distance travelled by car a = num__65 × num__8 = num__520 km distance travelled by car b = num__70 × num__4 = num__280 km ratio = num__1.85714285714 = num__13 : num__7 c <eor> c <eos> |
c |
multiply__65.0__8.0__ multiply__70.0__4.0__ divide__520.0__280.0__ round__13.0__ |
multiply__65.0__8.0__ multiply__70.0__4.0__ divide__520.0__280.0__ round__13.0__ |
| if i walk at num__3 kmph i miss the train by num__2 min if however i walk at num__4 kmph . i reach the station num__2 min before the arrival of the train . how far do i walk to reach the station ? <o> a ) num__0.8 <o> b ) num__0.666666666667 <o> c ) num__4 / num__0 <o> d ) num__2.0 <o> e ) num__4.0 |
x / num__3 – x / num__4 = num__0.0666666666667 x = num__0.8 km answer : a <eor> a <eos> |
a |
round__0.8__ |
round__0.8__ |
| in a bag containing only blue red and green marbles all but num__15 are blue all but num__13 are red and all but num__12 are green . how many are red ? <o> a ) num__13 <o> b ) num__7 <o> c ) num__25 <o> d ) num__20 <o> e ) num__10 |
all but blue means number of marbles except blue . . hence let x be the red marble y be the green marble z be the blue marble . . . x + y = num__15 y + z = num__13 x + z = num__12 solving this eqns we get x = num__7 y = num__8 z = num__5 hence number of red marbles are num__7 answer : b <eor> b <eos> |
b |
subtract__15.0__7.0__ subtract__13.0__8.0__ subtract__15.0__8.0__ |
subtract__15.0__7.0__ subtract__13.0__8.0__ subtract__15.0__8.0__ |
| in an xy - coordinate plane a line is defined by y = kx + num__1 . if ( num__1 b ) ( a num__4 ) and ( a b + num__1 ) are three points on the line where a and b are unknown then k = ? <o> a ) num__0.5 <o> b ) num__1 <o> c ) num__1.5 <o> d ) num__2 <o> e ) num__2.5 |
b = k + num__1 . . . ( num__1 ) b + num__1 = ak + num__1 . . . ( num__2 ) num__4 = ak + num__1 . . . ( num__3 ) taking ( num__2 ) and ( num__3 ) num__4 = b + num__1 b = num__3 taking ( num__1 ) num__3 = k + num__1 k = num__2 answer : d <eor> d <eos> |
d |
add__1.0__2.0__ multiply__1.0__2.0__ |
add__1.0__2.0__ multiply__1.0__2.0__ |
| if a : b = num__2 : num__3 and b : c = num__4 : num__3 then find a : b : c <o> a ) num__8 : num__12 : num__9 <o> b ) num__2 : num__3 : num__8 <o> c ) num__2 : num__3 : num__9 <o> d ) num__2 : num__3 : num__12 <o> e ) none of these |
explanation : a : b = num__2 : num__3 b : c = num__4 : num__3 = ( num__4 ∗ num__0.75 : num__3 ∗ num__0.75 ) = num__3 : num__2.25 a : b : c = num__2 : num__3 : num__2.25 = num__8 : num__12 : num__9 option a <eor> a <eos> |
a |
divide__3.0__4.0__ subtract__3.0__0.75__ multiply__2.0__4.0__ multiply__3.0__4.0__ multiply__4.0__2.25__ multiply__2.0__4.0__ |
divide__3.0__4.0__ subtract__3.0__0.75__ multiply__2.0__4.0__ multiply__3.0__4.0__ multiply__4.0__2.25__ multiply__2.0__4.0__ |
| the current of a stream runs at the rate of num__4 kmph . a boat goes num__6 km and back to the starting point in num__2 hours then find the speed of the boat in still water ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__8 <o> d ) num__9 <o> e ) num__6 |
s = num__4 m = x ds = x + num__4 us = x - num__4 num__6 / ( x + num__4 ) + num__6 / ( x - num__4 ) = num__2 x = num__8 answer : c <eor> c <eos> |
c |
multiply__4.0__2.0__ round__8.0__ |
add__6.0__2.0__ add__6.0__2.0__ |
| the average of num__6 no . ' s is num__5.40 . the average of num__2 of them is num__5.2 while the average of the other num__2 is num__5.80 . what is the average of the remaining num__2 no ' s ? <o> a ) num__5.2 <o> b ) num__2.6 <o> c ) num__3.6 <o> d ) num__4.5 <o> e ) num__4.6 |
sum of the remaining two numbers = ( num__5.40 * num__6 ) - [ ( num__5.2 * num__2 ) + ( num__5.8 * num__2 ) ] = num__32.40 - ( num__10.4 + num__11.6 ) = num__32.40 - num__22.0 = num__10.40 . required average = ( num__10.4 / num__2 ) = num__5.2 answer : a <eor> a <eos> |
a |
multiply__6.0__5.4__ multiply__2.0__5.2__ multiply__2.0__5.8__ add__10.4__11.6__ divide__10.4__2.0__ |
multiply__6.0__5.4__ multiply__2.0__5.2__ multiply__2.0__5.8__ add__10.4__11.6__ divide__10.4__2.0__ |
| the simple interest on a certain sum of money at the rate of num__4.0 p . a . for num__5 years is rs . num__1680 . at what rate of interest the same amount of interest can be received on the same sum after num__4 years ? <o> a ) num__5.0 <o> b ) num__6.0 <o> c ) num__8.0 <o> d ) num__4.0 <o> e ) num__7 % |
s . i . = num__1680 r = num__4.0 t = num__5 years principal = ( num__100 * num__1680 ) / ( num__5 * num__4 ) = num__8400 so p = num__8400 rate = ( num__100 * num__1680 ) / ( num__8400 * num__4 ) = num__5.0 answer : a <eor> a <eos> |
a |
percent__5.0__100.0__ |
percent__5.0__100.0__ |
| a work crew of num__7 men takes num__10 days to complete one - half of a job . if num__8 men are then added to the crew and the men continue to work at the same rate how many days will it take the enlarged crew to do the rest of the job ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__3 num__0.333333333333 <o> d ) num__4 <o> e ) num__4 num__0.666666666667 |
suppose num__1 man can do work in x days . . so num__7 men will do in . . num__7 / x = num__0.1 * num__0.5 as half job is done x = num__140 now num__8 more are added then num__0.107142857143 = num__0.5 * num__1 / d for remaining half job d = num__4 num__0.666666666667 number of days e <eor> e <eos> |
e |
subtract__8.0__7.0__ divide__1.0__10.0__ multiply__8.0__0.5__ round__4.0__ |
subtract__8.0__7.0__ divide__1.0__10.0__ multiply__8.0__0.5__ multiply__8.0__0.5__ |
| if the annual interest on a principal is num__25.0 how many years before the amount is double ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__7 <o> d ) num__4 <o> e ) num__5 |
p = ( p * num__25 * r ) / num__100 r = num__4.0 answer : d <eor> d <eos> |
d |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| how many positive integers less than num__300 are there such that they are multiples of num__17 or multiples of num__16 ? <o> a ) num__31 <o> b ) num__32 <o> c ) num__33 <o> d ) num__34 <o> e ) num__35 |
num__17.6470588235 = num__17 ( plus remainder ) so there are num__17 multiples of num__17 num__18.75 = num__18 ( plus remainder ) so there are num__18 multiples of num__16 we need to subtract num__1 because num__17 * num__16 is a multiple of both so it was counted twice . the total is num__17 + num__18 - num__1 = num__34 the answer is d . <eor> d <eos> |
d |
divide__300.0__17.0__ divide__300.0__16.0__ round_down__18.75__ subtract__17.0__16.0__ add__16.0__18.0__ add__16.0__18.0__ |
divide__300.0__17.0__ divide__300.0__16.0__ round_down__18.75__ subtract__17.0__16.0__ add__16.0__18.0__ add__16.0__18.0__ |
| what is the unit ’ s digit of num__7 ^ num__8 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
num__7 ^ num__1 = num__7 num__7 ^ num__2 = num__49 num__7 ^ num__3 = num__343 num__7 ^ num__4 = num__1 ( last digit ) num__7 ^ num__5 = num__7 ( last digit ) and the cycle repeats after every num__4 powers therefore last digit of num__7 ^ num__8 = num__1 answer a <eor> a <eos> |
a |
subtract__8.0__7.0__ add__1.0__2.0__ multiply__7.0__49.0__ subtract__7.0__3.0__ subtract__7.0__2.0__ reverse__1.0__ |
subtract__8.0__7.0__ add__1.0__2.0__ multiply__7.0__49.0__ subtract__7.0__3.0__ subtract__7.0__2.0__ reverse__1.0__ |
| a student traveled num__25 percent of the distance of the trip alone continued another num__40 miles with a friend and then finished the last half of the trip alone . how many miles long was the trip ? <o> a ) num__120 <o> b ) num__140 <o> c ) num__160 <o> d ) num__180 <o> e ) num__200 |
let x be the total length of the trip . num__0.25 x + num__40 miles + num__0.5 x = x num__40 miles = num__0.25 x x = num__160 miles the answer is c . <eor> c <eos> |
c |
divide__40.0__0.25__ round__160.0__ |
divide__40.0__0.25__ round__160.0__ |
| in the xy - plane a triangle has vertices ( num__00 ) ( num__40 ) and ( num__48 ) . if a point ( a b ) is selected at random from the triangular region what is the probability that a - b > num__0 ? <o> a ) num__0.2 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__0.8 |
the area of the right triangle is ( num__0.5 ) * num__4 * num__8 = num__16 . only the points ( a b ) below the line y = x satisfy a - b > num__0 . the part of the triangle which is below the line y = x has an area of ( num__0.5 ) ( num__4 ) ( num__4 ) = num__8 . p ( a - b > num__0 ) = num__0.5 = num__0.5 the answer is c . <eor> c <eos> |
c |
subtract__48.0__40.0__ divide__8.0__0.5__ divide__8.0__16.0__ |
subtract__48.0__40.0__ divide__8.0__0.5__ divide__8.0__16.0__ |
| if a trader sold two cars each at rs . num__325475 and gains num__12.0 on the first and loses num__12.0 on the second then his profit or loss percent on the whole is ? <o> a ) num__1.44 <o> b ) num__2.45 <o> c ) num__7.44 <o> d ) num__8.44 <o> e ) num__2.44 % |
explanation : sp of each car is rs . num__325475 he gains num__12.0 on first car and losses num__12.0 on second car . in this case there will be loss and percentage of loss is given by = [ ( profit % ) ( loss % ) ] / num__100 = ( num__12 ) ( num__12 ) / num__100.0 = num__1.44 answer : a <eor> a <eos> |
a |
percent__100.0__1.44__ |
percent__100.0__1.44__ |
| if the average ( arithmetic mean ) of a and b is num__40 and the average of b and c is num__60 what is the value of c − a ? <o> a ) num__25 <o> b ) num__40 <o> c ) num__90 <o> d ) num__140 <o> e ) it can not be determined from the information given . |
- ( a + b = num__80 ) b + c = num__120 c - a = num__40 b . num__40 <eor> b <eos> |
b |
add__40.0__80.0__ subtract__80.0__40.0__ |
add__40.0__80.0__ subtract__80.0__40.0__ |
| calculate the speed of a boat in still water ( in km / hr ) if in one hour the boat goes at num__13 km / hr downstream and num__4 km / hr upstream . <o> a ) num__4.5 kmph <o> b ) num__9.5 kmph <o> c ) num__7.5 kmph <o> d ) num__8.5 kmph <o> e ) num__5.5 kmph |
speed in still water = ( num__13 + num__4 ) num__0.5 kmph = num__8.5 kmph . answer : d <eor> d <eos> |
d |
round__8.5__ |
round__8.5__ |
| a contractor undertakes to built a walls in num__50 days . he employs num__60 peoples for the same . however after num__25 days he finds that only num__40.0 of the work is complete . how many more man need to be employed to complete the work in time ? <o> a ) num__25 <o> b ) num__90 <o> c ) num__35 <o> d ) num__20 <o> e ) none of these |
num__60 men complete num__0.4 work in num__25 days . applying the work rule m num__1 × d num__1 × w num__2 = m num__2 × d num__2 × w num__1 we have num__60 × num__25 × num__0.6 = m num__2 × num__25 × num__0.4 or m num__2 = num__60 × num__25 × num__0.6 / num__25 × num__0.4 = num__90 men answerb <eor> b <eos> |
b |
divide__50.0__25.0__ km_to_mile_conversion__ add__50.0__40.0__ round__90.0__ |
divide__50.0__25.0__ km_to_mile_conversion__ add__50.0__40.0__ divide__90.0__1.0__ |
| fox jeans regularly sell for $ num__15 a pair and pony jeans regularly sell for $ num__18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ num__3 is saved by purchasing num__5 pairs of jeans : num__3 pairs of fox jeans and num__2 pairs of pony jeans . if the sum of the two discounts rates is num__18 percent what is the discount rate on pony jeans ? <o> a ) num__9.0 <o> b ) num__56.6 <o> c ) num__11.0 <o> d ) num__12.0 <o> e ) num__15 % |
you know that fox jeans costs $ num__15 and pony jeans costs $ num__18 you also know that num__3 pairs of fox jeans and num__2 pairs of pony jeans were purchased . so num__3 ( num__15 ) = num__45 - fox num__2 ( num__18 ) = num__36 - pony the total discount discount is $ num__3 and you are asked to find the percent discount of pony jeans so num__45 ( num__18 - x ) / num__100 + num__36 ( x ) / num__100 = num__3 or num__45 * num__18 - num__45 * x + num__36 * x = num__3 * num__100 or num__9 x = - num__3 * num__100 + num__45 * num__18 x = num__56.6666666667 = num__56.6 b <eor> b <eos> |
b |
percent__56.6__100.0__ |
percent__56.6__100.0__ |
| find the compound ratio of ( num__2 : num__5 ) ( num__6 : num__11 ) and ( num__11 : num__2 ) is <o> a ) num__3 : num__2 <o> b ) num__2 : num__1 <o> c ) num__1 : num__2 <o> d ) num__6 : num__5 <o> e ) num__2 : num__3 |
required ratio = num__0.4 * num__0.545454545455 * num__5.5 = num__2.0 = num__6 : num__5 answer is d <eor> d <eos> |
d |
divide__2.0__5.0__ divide__6.0__11.0__ divide__11.0__2.0__ subtract__11.0__5.0__ |
divide__2.0__5.0__ divide__6.0__11.0__ divide__11.0__2.0__ subtract__11.0__5.0__ |
| if two - third of a bucket is filled in num__6 minute then the time taken to fill the bucket completely will be . <o> a ) num__90 seconds <o> b ) num__70 seconds <o> c ) num__60 seconds <o> d ) num__9 minutes <o> e ) num__120 seconds |
num__0.666666666667 filled in num__6 mint num__0.333333333333 filled in num__3 mint thn num__0.666666666667 + num__0.333333333333 = num__6 + num__3 = num__9 minutes answer : d <eor> d <eos> |
d |
add__6.0__3.0__ round__9.0__ |
add__6.0__3.0__ add__6.0__3.0__ |
| a jogger running at num__9 km / hr along side a railway track is num__240 m ahead of the engine of a num__120 m long train running at num__45 km / hr in the same direction . in how much time will the train pass the jogger ? <o> a ) num__88 <o> b ) num__27 <o> c ) num__36 <o> d ) num__88 <o> e ) num__12 |
speed of train relative to jogger = num__45 - num__9 = num__36 km / hr . = num__36 * num__0.277777777778 = num__10 m / sec . distance to be covered = num__240 + num__120 = num__360 m . time taken = num__36.0 = num__36 sec . answer : c <eor> c <eos> |
c |
subtract__45.0__9.0__ add__240.0__120.0__ subtract__45.0__9.0__ |
subtract__45.0__9.0__ add__240.0__120.0__ subtract__45.0__9.0__ |
| what will be the cost of building a fence around a square plot with area equal to num__81 sq ft if the price per foot of building the fence is rs . num__58 ? <o> a ) num__2088 <o> b ) num__2882 <o> c ) num__2999 <o> d ) num__2667 <o> e ) num__2121 |
let the side of the square plot be a ft . a num__2 = num__81 = > a = num__9 length of the fence = perimeter of the plot = num__4 a = num__36 ft . cost of building the fence = num__36 * num__58 = rs . num__2088 . answer : a <eor> a <eos> |
a |
square_perimeter__9.0__ multiply__58.0__36.0__ multiply__58.0__36.0__ |
multiply__4.0__9.0__ multiply__58.0__36.0__ multiply__58.0__36.0__ |
| six years ago the ratio of the ages of kunal and sagar was num__6 : num__5 four years hence the ratio of their ages will be num__11 : num__10 . what is sagar age at present <o> a ) num__22 <o> b ) num__17 <o> c ) num__15 <o> d ) num__16 <o> e ) num__12 |
let six years ago the age of kunal and sagar are num__6 x and num__5 x resp . then = > ( num__6 x + num__6 ) + num__4 ( num__5 x + num__6 ) + num__4 = num__1110 = > ( num__6 x + num__6 ) + num__4 ( num__5 x + num__6 ) + num__4 = num__1110 < = > num__10 ( num__6 x + num__10 ) = num__11 ( num__5 x + num__10 ) < = > num__10 ( num__6 x + num__10 ) = num__11 ( num__5 x + num__10 ) < = > num__5 x = num__10 = > x = num__2 < = > num__5 x = num__10 = > x = num__2 so sagar age is ( num__5 x + num__6 ) = num__16 answer : d <eor> d <eos> |
d |
subtract__10.0__6.0__ subtract__6.0__4.0__ add__6.0__10.0__ add__6.0__10.0__ |
subtract__10.0__6.0__ subtract__6.0__4.0__ add__6.0__10.0__ add__6.0__10.0__ |
| a can complete a project in num__20 days and b can complete the same project in num__20 days . if a and b start working on the project together and a quits num__10 days before the project is completed in how many days will the project be completed ? <o> a ) num__18 days <o> b ) num__27 days <o> c ) num__26.67 days <o> d ) num__16 days <o> e ) num__15 days |
let x = the number of days taken to complete the project . the amount of work done by a is ( x - num__10 ) * ( num__0.05 ) . the amount of work done by b is ( x ) * ( num__0.0333333333333 ) . ( num__0.05 ) * ( x - num__10 ) + ( num__0.05 ) * ( x ) = num__1 ( x / num__20 ) + ( x / num__20 ) - ( num__0.5 ) = num__1 x / num__10 = num__1.5 x = num__15 therefore the answer is e : num__15 . <eor> e <eos> |
e |
multiply__20.0__0.05__ divide__10.0__20.0__ add__0.5__1.0__ multiply__10.0__1.5__ round__15.0__ |
multiply__20.0__0.05__ divide__10.0__20.0__ add__0.5__1.0__ multiply__10.0__1.5__ multiply__10.0__1.5__ |
| ashok secured average of num__72 marks in num__6 subjects . if the average of marks in num__5 subjects is num__74 how many marks did he secure in the num__6 th subject ? <o> a ) num__62 <o> b ) num__74 <o> c ) num__78 <o> d ) num__80 <o> e ) none of these |
explanation : number of subjects = num__6 average of marks in num__6 subjects = num__72 therefore total marks in num__6 subjects = num__72 * num__6 = num__432 now no . of subjects = num__5 total marks in num__5 subjects = num__74 * num__5 = num__370 therefore marks in num__6 th subject = num__432 – num__370 = num__62 answer a <eor> a <eos> |
a |
multiply__72.0__6.0__ multiply__5.0__74.0__ subtract__432.0__370.0__ subtract__432.0__370.0__ |
multiply__72.0__6.0__ multiply__5.0__74.0__ subtract__432.0__370.0__ subtract__432.0__370.0__ |
| a cube has a volume of num__64 cubic feet . if a similar cube is twice as long twice as wide and twice as high then the volume in cubic feet of such cube is ? <o> a ) num__24 <o> b ) num__48 <o> c ) num__64 <o> d ) num__80 <o> e ) num__512 |
volume = num__64 = side ^ num__3 i . e . side of cube = num__4 new cube has dimensions num__8 num__8 and num__8 as all sides are twice of teh side of first cube volume = num__8 * num__8 * num__8 = num__512 square feet answer : option e <eor> e <eos> |
e |
volume_cube__8.0__ volume_cube__8.0__ |
multiply__64.0__8.0__ multiply__64.0__8.0__ |
| john traveled num__80.0 of the way from yellow - town to green - fields by train at an average speed of num__80 miles per hour . the rest of the way john traveled by car at an average speed of v miles per hour . if the average speed for the entire trip was num__50 miles per hour what is v in miles per hour ? <o> a ) num__20 <o> b ) num__40 <o> c ) num__50 <o> d ) num__55 <o> e ) num__70 |
hibunuel the question seems incorrect . it should not be num__80.0 at the speed of num__80 . however if it ' s num__20.0 at the speed of num__80 answer comes out num__55 . the question is correct . here ' s the explanation : let distance be d . we can find the total timeequate it which comes as : num__0.8 d / num__80 + num__0.2 d / v = d / num__50 = > v = num__20 ( option a ) . <eor> a <eos> |
a |
round__20.0__ |
round__20.0__ |
| if a person moves num__15 km straight and turns right and moves num__15 km straight then how much distance he needs to walk to reach starting point ? <o> a ) num__20.2 <o> b ) num__21.2 <o> c ) num__23.2 <o> d ) num__24.2 <o> e ) num__26.2 |
person will move along right angled traingle whose smaller sides are num__15 km & num__15 km reqd distance = hypotenuse of triangle = √ ( num__15 ^ num__2 + num__15 ^ num__2 ) = num__15 √ num__2 = num__21.2 km answer : b <eor> b <eos> |
b |
round__21.2__ |
round__21.2__ |
| tomeka is playing a dice game . if she rolls the same number on her second roll as she rolls on her first she wins . each roll is with two fair six - sided dice . if tomeka rolled a seven on her first roll what is the probability that she will win on her second roll ? <o> a ) num__0.25 <o> b ) num__0.2 <o> c ) num__0.166666666667 <o> d ) num__0.111111111111 <o> e ) num__0.0833333333333 |
there are num__6 ways to roll a seven : num__1 and num__6 num__6 and num__1 num__2 and num__5 num__5 and num__2 num__3 and num__4 num__4 and num__3 . there are num__6 * num__6 = num__36 ways to roll two six - sided dice . thus the probability of winning by rolling a seven on the second roll given a seven on the first roll is num__0.166666666667 = num__0.166666666667 c <eor> c <eos> |
c |
subtract__6.0__1.0__ add__1.0__2.0__ add__1.0__3.0__ reverse__6.0__ reverse__6.0__ |
subtract__6.0__1.0__ subtract__5.0__2.0__ subtract__5.0__1.0__ reverse__6.0__ reverse__6.0__ |
| num__0.666666666667 rd of the boys and num__0.75 th of the girls of a school participate in a function . if the no . of participating students is num__550 out of which num__150 are girls what is the total no . of students in the school ? <o> a ) num__300 <o> b ) num__400 <o> c ) num__800 <o> d ) num__900 <o> e ) num__1000 |
let total number of boys be x and total number of girls be y . y = num__400 = > x = num__600 and ^ = i num__50 = > y = num__200 now x + y = num__800 c <eor> c <eos> |
c |
subtract__550.0__150.0__ subtract__600.0__550.0__ divide__150.0__0.75__ divide__600.0__0.75__ divide__600.0__0.75__ |
subtract__550.0__150.0__ subtract__600.0__550.0__ add__150.0__50.0__ add__600.0__200.0__ add__600.0__200.0__ |
| if x and y are positive integers such that x / y = num__6.8 which of the following numbers could be y ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__2 <o> d ) num__6 <o> e ) num__10 |
with the given information we are trying to find the value of y . to achieve this we must convert the given decimal to a fraction and reduce it to it ' s lowest integer form . take the decimal x / y = num__6.8 and because it ' s in the tens decimal place put the given value ( x / y = num__6.8 / num__1 ) and multiply it by num__10 . take the resulting fraction ( x / y = num__6.8 ) and reduce it to it ' s lowest form . both of the numbers are divisible by num__2 . the result is x / y = num__6.8 . thus y = num__5 . the correct answer is c . <eor> c <eos> |
c |
divide__10.0__2.0__ multiply__1.0__2.0__ |
divide__10.0__2.0__ divide__10.0__5.0__ |
| if two dice are thrown together the probability of getting an even number on one die and an odd number on the other is ? <o> a ) num__0.125 <o> b ) num__0.5 <o> c ) num__1.0 <o> d ) num__0.166666666667 <o> e ) num__0.333333333333 |
the number of exhaustive outcomes is num__36 . let e be the event of getting an even number on one die and an odd number on the other . let the event of getting either both even or both odd then = num__0.5 = num__0.5 p ( e ) = num__1 - num__0.5 = num__0.5 . answer : b <eor> b <eos> |
b |
negate_prob__0.5__ |
negate_prob__0.5__ |
| the ratio by weight measured in pounds of books to clothes to electronics in a suitcase initially stands at num__7 : num__4 : num__3 . someone removes num__8 pounds of clothing from the suitcase thereby doubling the ratio of books to clothes . how many pounds do the electronics in the suitcase weigh ? <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__12 <o> e ) num__14 |
the weights of the items in the suitcase are num__7 k num__4 k and num__3 k . if removing num__8 pounds of clothes doubles the ratio of books to clothes then num__8 pounds represents half the weight of the clothes . num__2 k = num__8 pounds and then k = num__4 pounds . the electronics weigh num__3 ( num__4 ) = num__12 pounds . the answer is d . <eor> d <eos> |
d |
divide__8.0__4.0__ multiply__4.0__3.0__ multiply__4.0__3.0__ |
divide__8.0__4.0__ multiply__4.0__3.0__ multiply__4.0__3.0__ |
| num__1 num__0.5 num__0.25 num__0.125 num__0.0625 ? <o> a ) num__0.015625 <o> b ) num__0.03125 <o> c ) num__0.0078125 <o> d ) none of these <o> e ) num__0.025874 |
explanation : divide by num__2 to get the next term answer : option b <eor> b <eos> |
b |
reverse__0.5__ multiply__0.5__0.0625__ |
reverse__0.5__ multiply__0.5__0.0625__ |
| a and b can together complete a piece of work in num__4 days . if a alone can complete the same work in num__12 days in how many days can b alone complete that work ? <o> a ) num__4 days <o> b ) num__5 days <o> c ) num__6 days <o> d ) num__7 days <o> e ) none of these |
explanation : ( a + b ) ' s num__1 day work = num__0.25 a ' s num__1 day work = num__0.0833333333333 b ' s num__1 day work = ( num__14 − num__112 ) = num__3 − num__112 = num__16 so b alone can complete the work in num__6 days answer : c <eor> c <eos> |
c |
divide__1.0__4.0__ divide__1.0__12.0__ subtract__4.0__1.0__ add__4.0__12.0__ round__6.0__ |
divide__1.0__4.0__ divide__1.0__12.0__ subtract__4.0__1.0__ add__4.0__12.0__ round__6.0__ |
| what is the rate percent when the simple interest on rs . num__800 amount to rs . num__160 in num__5 years ? <o> a ) num__5.0 <o> b ) num__7.0 <o> c ) num__9.0 <o> d ) num__2.0 <o> e ) num__4 % |
num__160 = ( num__800 * num__5 * r ) / num__100 r = num__4.0 answer : e <eor> e <eos> |
e |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| the difference of two numbers is num__1415 . on dividing the larger number by the smaller we get num__6 as quotient and the num__15 as remainder . what is the smaller number ? <o> a ) a ) num__270 <o> b ) b ) num__280 <o> c ) c ) num__290 <o> d ) d ) num__300 <o> e ) e ) num__310 |
let the smaller number be x . then larger number = ( x + num__1415 ) . x + num__1415 = num__6 x + num__15 num__5 x = num__1400 x = num__280 smaller number = num__280 . b ) <eor> b <eos> |
b |
subtract__1415.0__15.0__ divide__1400.0__5.0__ divide__1400.0__5.0__ |
subtract__1415.0__15.0__ divide__1400.0__5.0__ divide__1400.0__5.0__ |
| a train traveled the first d miles of its journey it an average speed of num__60 miles per hour the next d miles of its journey at an average speed of y miles per hour and the final d miles of its journey at an average speed of num__160 miles per hour . if the train ’ s average speed over the total distance was num__94 miles per hour what is the value of y ? <o> a ) num__98 <o> b ) num__84 <o> c ) num__90 <o> d ) num__120 <o> e ) num__135 |
average speed = total distance traveled / total time taken num__3 d / d / num__60 + d / y + d / num__160 = num__94 solving for d and y num__15 y = num__11 y + num__480 num__5 y = num__480 y = num__98 answer a <eor> a <eos> |
a |
multiply__160.0__3.0__ divide__15.0__3.0__ round__98.0__ |
multiply__160.0__3.0__ divide__15.0__3.0__ round__98.0__ |
| country x imported approximately $ num__1.44 billion of goods in num__1996 . if country x imported $ num__488 million of goods in the first two months of num__1997 and continued to import goods at the same rate for the rest of the year by how much would country xs num__1997 imports exceed those of num__1996 ? <o> a ) $ num__1124 million <o> b ) $ num__1120 million <o> c ) $ num__1144 million <o> d ) $ num__1240 million <o> e ) $ num__1488 million |
convert units to millions as answer is in millions num__1996 imports = $ num__1.44 bill = $ num__1440 mill i . e . num__120.0 = $ num__120 mill / month num__1997 imports = $ num__488 mill / num__2 month i . e . $ num__244 mill / month difference / month = num__244 - num__120 = num__124 difference / year = $ num__124 mill * num__12 = $ num__1488 mill answer : e <eor> e <eos> |
e |
divide__488.0__2.0__ subtract__244.0__120.0__ divide__1440.0__120.0__ multiply__12.0__124.0__ multiply__12.0__124.0__ |
divide__488.0__2.0__ subtract__244.0__120.0__ divide__1440.0__120.0__ multiply__12.0__124.0__ multiply__12.0__124.0__ |
| a certain sum amounts to rs . num__1725 in num__3 years and rs . num__1875 in num__5 years . find the rate % per annum ? <o> a ) num__6.0 <o> b ) num__5.0 <o> c ) num__3.0 <o> d ) num__9.0 <o> e ) num__2 % |
num__3 - - - num__1725 num__5 - - - num__1875 - - - - - - - - - - - - - - num__2 - - - num__150 n = num__1 i = num__75 r = ? p = num__1725 - num__225 = num__1500 num__75 = ( num__1500 * num__1 * r ) / num__100 r = num__5.0 answer : b <eor> b <eos> |
b |
percent__5.0__100.0__ |
percent__5.0__100.0__ |
| one pump drains one - half of a pond in num__4 hours and then a second pump starts draining the pond . the two pumps working together finish emptying the pond in one - half hour . how long would it take the second pump to drain the pond if it had to do the job alone ? <o> a ) num__1 hour <o> b ) num__1.1 hour <o> c ) num__3 hours <o> d ) num__5 hours <o> e ) num__6 hours |
first pump drains num__0.5 of the tank in num__4 hours so num__8 hours it will take to drain the full tank . let num__2 nd pump drains the full tank in a hours so both together can drain ( num__0.125 + num__1 / a ) part in num__1 hour son in num__0.5 hour they drain num__0.5 * ( num__0.125 + num__1 / a ) part of the tank given that in num__0.5 hour they drain num__0.5 of the tank hence we can say num__0.5 * ( num__0.125 + num__1 / a ) = num__0.5 solving u get a = num__1.14285714286 = num__1.1 hence answer is b <eor> b <eos> |
b |
divide__4.0__0.5__ multiply__4.0__0.5__ divide__0.5__4.0__ multiply__0.5__2.0__ round__1.1__ |
divide__4.0__0.5__ multiply__4.0__0.5__ divide__0.5__4.0__ multiply__0.5__2.0__ divide__1.1__1.0__ |
| a solid yellow stripe is to be painted in the middle of a certain highway . if num__1 gallon of paint covers an area of t square feet of highway how many gallons of paint will be needed to paint a stripe of t inches wide on a stretch of highway m miles long ? ( num__1 mile = num__5280 feet and num__1 foot = num__12 inches ) <o> a ) ( num__5280 mt ) / num__12 t <o> b ) ( num__5280 pt ) / num__12 m <o> c ) ( num__5280 pmt ) / num__12 <o> d ) ( num__5280 ) ( num__12 m ) / pt <o> e ) ( num__5280 ) ( num__12 p ) / mt |
given that : num__1 gallon of paint covers an area oftsquare feet . question : how many gallonsof paint will be needed . . . in any case you will have : ( total area in square feet ) / ( gallons per feet ) = ( total area in square feet ) / t so t must be in the denominator : eliminate all but a and d . now lets see where should be t : ( area in square feet ) = ( width in feet ) * ( length in feet ) - - > width = tinchesas num__1 feet = num__12 inchesthent inches = t / num__12 feet so ( area in square feet ) = ( t / num__12 ) * ( length in feet ) so t must be in the nominator : only a is left . answer : a . <eor> a <eos> |
a |
round__5280.0__ |
multiply__1.0__5280.0__ |
| on an order of num__5 dozen boxes of a consumer product a retailer receives an extra dozen free . this is equivalent to allowing him a discount of : <o> a ) num__26 num__0.666666666667 % . <o> b ) num__18 num__0.25 % . <o> c ) num__16 num__0.666666666667 % . <o> d ) num__18 num__0.666666666667 % . <o> e ) num__19 num__0.666666666667 % . |
explanation : clearly the retailer gets num__1 dozen out of num__6 dozens free . equivalent discount = num__0.166666666667 * num__100 = num__16 num__0.666666666667 % . answer : c <eor> c <eos> |
c |
add__5.0__1.0__ reverse__6.0__ multiply__1.0__16.0__ |
add__5.0__1.0__ reverse__6.0__ multiply__1.0__16.0__ |
| at a certain college num__80 percent of the total number of students are freshmen . if num__60 percent of the fresh - men are enrolled in the school of liberal arts and of these num__50 percent are psychology majors what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal arts ? <o> a ) num__25.0 <o> b ) num__30.0 <o> c ) num__24.0 <o> d ) num__28.0 <o> e ) num__20 % |
let ' s say there is a total of num__100 students at this college . num__80 percent of the total number of students are freshmen . # of freshmen = num__80.0 of num__100 = num__80 num__60 percent of the fresh - men are enrolled in the school of liberal arts . . . number of liberal arts freshmen = num__60.0 of num__80 = num__48 . . . and of these num__50 percent are psychology majors . . . number of liberal arts freshmen who are psychology majors = num__50.0 of num__48 = num__24 what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal arts ? num__0.24 = num__24.0 answer : c <eor> c <eos> |
c |
divide__24.0__100.0__ subtract__48.0__24.0__ |
divide__24.0__100.0__ subtract__48.0__24.0__ |
| working at their respective constant rates machine a makes num__100 copies in num__12 minutes and machine b makes num__150 copies in num__15 minutes . if these machines work simultaneously at their respective rates for num__30 minutes what is the total number of copies that they will produce ? <o> a ) num__250 <o> b ) num__425 <o> c ) num__550 <o> d ) num__700 <o> e ) num__750 |
machine a can produce num__100 * num__2.5 = num__250 copies and machine b can produce num__150 * num__2.0 = num__300 copies total producing num__550 copies . c is the answer <eor> c <eos> |
c |
divide__30.0__12.0__ add__100.0__150.0__ divide__30.0__15.0__ multiply__150.0__2.0__ add__300.0__250.0__ round__550.0__ |
divide__30.0__12.0__ multiply__100.0__2.5__ divide__30.0__15.0__ multiply__150.0__2.0__ add__300.0__250.0__ round__550.0__ |
| how much time will it take for an amount of rs . num__400 to yield rs . num__60 as interest at num__12.0 per annum of simple interest ? <o> a ) num__2.35 years <o> b ) num__4.25 years <o> c ) num__7 years <o> d ) num__1.25 years <o> e ) num__2 years |
explanation : time = ( num__100 x num__60 ) / ( num__400 x num__12 ) years = num__1.25 years . answer : d <eor> d <eos> |
d |
percent__100.0__1.25__ |
percent__100.0__1.25__ |
| for every positive even integer n the function h ( n ) is defined to be the product of all the even integers from num__2 to n inclusive . if p is the smallest prime factor of h ( num__100 ) + num__2 then p is ? <o> a ) between num__2 and num__20 <o> b ) between num__10 and num__20 <o> c ) between num__20 and num__30 <o> d ) between num__30 and num__40 <o> e ) num__2 |
h ( num__100 ) + num__2 = num__2 ∗ num__4 ∗ num__6 ∗ . . . ∗ num__100 + num__2 h ( num__100 ) + num__2 = num__2 ∗ num__4 ∗ num__6 ∗ . . . ∗ num__100 + num__2 . notice that we can factor out num__2 from h ( num__100 ) + num__2 thus the smallest prime factor of h ( num__100 ) + num__2 is num__2 : h ( num__100 ) + num__2 = num__2 ∗ ( num__4 ∗ num__6 ∗ . . . ∗ num__100 + num__1 ) h ( num__100 ) + num__2 = num__2 ∗ ( num__4 ∗ num__6 ∗ . . . ∗ num__100 + num__1 ) . answer : e . <eor> e <eos> |
e |
add__2.0__4.0__ multiply__2.0__1.0__ |
add__2.0__4.0__ multiply__2.0__1.0__ |
| the operation is defined for all integers a and b by the equation ab = ( a - num__1 ) ( b - num__1 ) . if x num__9 = num__160 what is the value of x ? <o> a ) num__18 <o> b ) num__15 <o> c ) num__17 <o> d ) num__19 <o> e ) num__21 |
ab = ( a - num__1 ) ( b - num__1 ) x num__9 = ( x - num__1 ) ( num__9 - num__1 ) = num__160 - - > x - num__1 = num__20 - - > x = num__21 answer : e <eor> e <eos> |
e |
add__1.0__20.0__ add__1.0__20.0__ |
add__1.0__20.0__ add__1.0__20.0__ |
| a train has a length of num__150 meters . it is passing a man who is moving at num__7 km / hr in the same direction of the train in num__3 seconds . find out the speed of the train . <o> a ) num__182 km / hr <o> b ) num__180 km / hr <o> c ) num__152 km / hr <o> d ) num__169 km / hr <o> e ) num__187 km / hr |
explanation : length of the train l = num__150 m speed of the man vm = num__7 km / hr relative speed vr = total distance / time = ( num__50.0 ) m / s = ( num__50.0 ) × ( num__3.6 ) = num__180 km / hr relative speed = speed of train vt - speed of man ( as both are moving in the same direction ) = > num__180 = vt - num__7 = > vt = num__180 + num__7 = num__187 km / hr answer : option e <eor> e <eos> |
e |
divide__150.0__3.0__ multiply__50.0__3.6__ add__7.0__180.0__ round__187.0__ |
divide__150.0__3.0__ multiply__50.0__3.6__ add__7.0__180.0__ add__7.0__180.0__ |
| the inner circumference of a circular race track num__14 m wide is num__440 m . find radius of the outer circle . <o> a ) num__84 <o> b ) num__86 <o> c ) num__82 <o> d ) num__80 <o> e ) none of them |
let inner radius be r metres . then num__2 ( num__3.14285714286 ) r = num__440 = r = ( num__440 x ( num__0.159090909091 ) ) = num__70 m . radius of outer circle = ( num__70 + num__14 ) m = num__84 m . answer is a <eor> a <eos> |
a |
triangle_area__2.0__84.0__ |
triangle_area__2.0__84.0__ |
| in how many different ways can the letters of the word ' leading ' be arranged such that the vowels should always come together ? <o> a ) num__650 <o> b ) num__720 <o> c ) num__420 <o> d ) num__122 <o> e ) num__450 |
explanation : the word ' leading ' has num__7 letters . it has the vowels ' e ' ' a ' ' i ' in it and these num__3 vowels should always come together . hence these num__3 vowels can be grouped and considered as a single letter . that is ldng ( eai ) . hence we can assume total letters as num__5 and all these letters are different . number of ways to arrange these letters = num__5 ! = num__5 Ã — num__4 Ã — num__3 Ã — num__2 Ã — num__1 = num__120 = num__5 ! = num__5 Ã — num__4 Ã — num__3 Ã — num__2 Ã — num__1 = num__120 in the num__3 vowels ( eai ) all the vowels are different . number of ways to arrange these vowels among themselves = num__3 ! = num__3 Ã — num__2 Ã — num__1 = num__6 = num__3 ! = num__3 Ã — num__2 Ã — num__1 = num__6 hence required number of ways = num__120 Ã — num__6 = num__720 answer is b <eor> b <eos> |
b |
vowel_space__ coin_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
vowel_space__ coin_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| if a and b is a positive integer then num__2 ^ a + num__2 ^ ( b + num__1 ) = <o> a ) can not be determined <o> b ) num__2 ^ ( a + b ) <o> c ) num__2 a + b <o> d ) num__2 ^ ( a + num__1 + b ) <o> e ) none |
if a and b is a positive integer then num__2 ^ a + num__2 ^ ( b + num__1 ) = num__2 ^ a + num__2 ^ ( b + num__1 ) num__2 ^ ( num__1 + a + b ) answer : option d <eor> d <eos> |
d |
multiply__2.0__1.0__ |
multiply__2.0__1.0__ |
| by selling an article at rs . num__600 a profit of num__30.0 is made . find its cost price ? <o> a ) num__400 <o> b ) num__267 <o> c ) num__287 <o> d ) num__480 <o> e ) num__811 |
sp = num__600 cp = ( sp ) * [ num__100 / ( num__100 + p ) ] = num__600 * [ num__100 / ( num__100 + num__50 ) ] = num__600 * [ num__0.666666666667 ] = rs . num__400 answer : a <eor> a <eos> |
a |
percent__100.0__400.0__ |
percent__100.0__400.0__ |
| excluding stoppages the average speed of a bus is num__60 km / hr and including stoppages the average speed of the bus is num__15 km / hr . for how many minutes does the bus stop per hour ? <o> a ) num__22 <o> b ) num__88 <o> c ) num__77 <o> d ) num__20 <o> e ) num__45 |
in num__1 hr the bus covers num__60 km without stoppages and num__15 km with stoppages . stoppage time = time take to travel ( num__60 - num__15 ) km i . e num__45 km at num__60 km / hr . stoppage time = num__0.75 hrs = num__45 min . answer : e <eor> e <eos> |
e |
subtract__60.0__15.0__ divide__45.0__60.0__ subtract__60.0__15.0__ |
subtract__60.0__15.0__ divide__45.0__60.0__ subtract__60.0__15.0__ |
| the sum of the squares of two consecutive positive integers exceeds their product by num__91 . find the integers ? <o> a ) num__65 <o> b ) num__58 <o> c ) num__23 <o> d ) num__98 <o> e ) num__9 |
10 |
explanation : let the two consecutive positive integers be x and x + num__1 x num__2 + ( x + num__1 ) num__2 - x ( x + num__1 ) = num__91 x num__2 + x - num__90 = num__0 ( x + num__10 ) ( x - num__9 ) = num__0 = > x = - num__10 or num__9 . as x is positive x = num__9 hence the two consecutive positive integers are num__9 and num__10 . answer e <eor> e <eos> |
e |
e |
| a num__260 m long train running at the speed of num__120 km / hr crosses another train running in opposite direction at the speed of num__80 km / hr in num__9 sec . what is the length of the other train ? <o> a ) num__230 <o> b ) num__240 <o> c ) num__267 <o> d ) num__256 <o> e ) num__298 |
relative speed = num__120 + num__80 = num__200 km / hr . = num__200 * num__0.277777777778 = num__55.5555555556 m / sec . let the length of the other train be x m . then ( x + num__260 ) / num__9 = num__55.5555555556 = > x = num__240 . answer : b <eor> b <eos> |
b |
add__120.0__80.0__ round__240.0__ |
add__120.0__80.0__ round__240.0__ |
| a train consists of num__12 boggies each boggy num__15 metres long . the train crosses a telegraph post in num__9 seconds . due to some problem two boggies were detached . the train now crosses a telegraph post in <o> a ) num__18 sec <o> b ) num__7.5 sec <o> c ) num__15 sec <o> d ) num__20 sec <o> e ) none of these |
length of train = num__12 Ã — num__15 = num__180 m . then speed of train = num__180 â „ num__9 = num__20 m / s now length of train = num__10 Ã — num__15 = num__150 m â ˆ ´ required time = num__150 â „ num__20 = num__7.5 sec . answer b <eor> b <eos> |
b |
multiply__12.0__15.0__ divide__180.0__9.0__ multiply__15.0__10.0__ divide__150.0__20.0__ round__7.5__ |
multiply__12.0__15.0__ divide__180.0__9.0__ multiply__15.0__10.0__ divide__150.0__20.0__ divide__150.0__20.0__ |
| there are num__2 kinds of staff members in a certain company part time and full time . num__25 percent of the total members are part time members others are full time members . the work time of part time members is num__0.6 of the full time members . wage per hour is same . what is the ratio of total wage of part time members to total wage of all members . <o> a ) a . num__0.25 <o> b ) b . num__0.2 <o> c ) num__0.166666666667 <o> d ) num__0.142857142857 <o> e ) num__0.125 |
just plug in assuming there are num__100 employees num__25.0 are pt = num__25 num__75.0 ft = num__75 assume that the total number of hrs ft employees have to work = num__10 pt will work = num__0.6 * num__10 = num__6 since the wages are same there will no affect on ratios total number of hrs ft employees work = num__75 * num__10 = num__750 pt = num__25 * num__6 = num__150 now the ratio is num__0.166666666667 = num__0.166666666667 answer is c <eor> c <eos> |
c |
subtract__100.0__25.0__ multiply__0.6__10.0__ multiply__10.0__75.0__ multiply__2.0__75.0__ reverse__6.0__ reverse__6.0__ |
subtract__100.0__25.0__ multiply__0.6__10.0__ multiply__10.0__75.0__ multiply__2.0__75.0__ reverse__6.0__ reverse__6.0__ |
| a cistern can be filled by a tap in num__4 hours while it can be emptied by another tap in num__9 hours . if both the taps are opened simultaneously then after how much time will the cistern get filled ? <o> a ) num__7.3 hrs <o> b ) num__7.9 hrs <o> c ) num__7.1 hrs <o> d ) num__7.2 hrs <o> e ) num__7.8 hrs |
net part filled in num__1 hour = ( num__0.25 - num__0.111111111111 ) = num__0.138888888889 the cistern will be filled in num__7.2 hrs i . e . num__7.2 hrs . answer : d <eor> d <eos> |
d |
divide__1.0__4.0__ divide__1.0__9.0__ subtract__0.25__0.1111__ round__7.2__ |
divide__1.0__4.0__ divide__1.0__9.0__ subtract__0.25__0.1111__ round__7.2__ |
| how long does a train num__100 m long traveling at num__72 kmph takes to cross a tunnel of num__1400 m in length ? <o> a ) num__70 sec <o> b ) num__60 sec <o> c ) num__82 sec <o> d ) num__75 sec <o> e ) num__62 sec |
d = num__1400 + num__100 = num__1500 m s = num__72 * num__0.277777777778 = num__20 t = num__1500 * num__0.05 = num__75 sec answer : d <eor> d <eos> |
d |
add__100.0__1400.0__ multiply__0.05__1500.0__ round__75.0__ |
add__100.0__1400.0__ multiply__0.05__1500.0__ multiply__0.05__1500.0__ |
| in a triangle abc ab = num__6 bc = num__8 and ac = num__10 . a perpendicular dropped from b meets the side ac at d . a circle of radius bd ( with centre b ) is drawn . if the circle cuts ab and bc at p and q respectively then ap : qc is equal to <o> a ) num__1 : num__1 <o> b ) num__3 : num__2 <o> c ) num__4 : num__1 <o> d ) num__3 : num__8 <o> e ) num__2 : num__5 |
explanation : triangle abc and triangle adb are similar ac / ab = bc / bd num__1.66666666667 = num__8 / r r = num__4.8 bp = bq = bd = r = num__4.8 ap = ab - r = num__6 - num__4.8 = num__1.2 cq = bc - r = num__8 - num__4.8 = num__3.2 ap : cq = num__0.375 = num__3 : num__8 answer : d <eor> d <eos> |
d |
multiply__8.0__0.375__ multiply__8.0__0.375__ |
multiply__8.0__0.375__ multiply__8.0__0.375__ |
| three pipes a b & c are attached to a tank . a & b can fill it in num__20 & num__30 minutes respectively while c can empty it in num__20 minutes . if a b & c are kept open successively for num__1.5 minute each how soon will the tank be filled ? <o> a ) num__2 hours <o> b ) num__4 hours <o> c ) num__5 hours <o> d ) num__2 hours num__15 min <o> e ) num__3 hours |
in three minute num__0.05 + num__0.0333333333333 - num__0.05 = num__0.0333333333333 = num__0.0333333333333 part is filled num__4.5 min - - - - - - - - num__0.0333333333333 parts x min - - - - - - - - - num__1 part ( full ) x = num__135 min = num__2 hours num__15 min answer : d <eor> d <eos> |
d |
divide__1.5__30.0__ divide__0.05__1.5__ multiply__20.0__0.05__ multiply__30.0__4.5__ divide__30.0__2.0__ round__2.0__ |
divide__1.5__30.0__ divide__0.05__1.5__ multiply__20.0__0.05__ multiply__30.0__4.5__ divide__30.0__2.0__ round__2.0__ |
| if the time is currently num__1 : num__30 pm what time will it be in exactly num__644 hours ? <o> a ) num__7 : num__30 am <o> b ) num__8 : num__30 am <o> c ) num__9 : num__30 am <o> d ) num__10 : num__30 am <o> e ) num__11 : num__30 am |
num__644 = num__26 ( num__24 ) + num__0.833333333333 the time will be num__20 hours later than num__1 : num__30 pm which is num__9 : num__30 am . the answer is c . <eor> c <eos> |
c |
round__9.0__ |
round__9.0__ |
| a river num__5 m deep and num__35 m wide is flowing at the rate of num__2 kmph calculate the amount of water that runs into the sea per minute ? <o> a ) num__5832.75 <o> b ) num__5839.75 <o> c ) num__5837.75 <o> d ) num__5222.75 <o> e ) num__5835.75 |
rate of water flow - num__2 kmph - - num__33.3333333333 - - num__33.33 m / min depth of river - - num__5 m width of river - - num__35 m vol of water per min - - num__33.33 * num__5 * num__35 - - - num__5832.75 answer a <eor> a <eos> |
a |
round__33.3333__ round__5832.75__ |
round__33.3333__ round__5832.75__ |
| marketing executives for a certain chewing gum company projected a num__25 percent increase in revenue this year over that of last year but revenue this year actually decreased by num__25.0 . what percent of the projected revenue was the actual revenue ? <o> a ) num__53.0 <o> b ) num__58.0 <o> c ) num__60.0 <o> d ) num__64.0 <o> e ) num__75 % |
last year revenue = num__100 ( assume ) ; this year revenue = num__75 ; projected revenue = num__125 . actual / projected * num__100 = num__0.6 * num__100 = num__60.0 . answer : c . <eor> c <eos> |
c |
subtract__100.0__25.0__ add__25.0__100.0__ divide__75.0__125.0__ multiply__0.6__100.0__ multiply__0.6__100.0__ |
subtract__100.0__25.0__ add__25.0__100.0__ divide__75.0__125.0__ multiply__0.6__100.0__ multiply__0.6__100.0__ |
| a batsman makes a score of num__87 runs in the num__17 th inning and thus increases his average by num__3 . find his average after num__17 th inning ? <o> a ) num__38 <o> b ) num__178 <o> c ) num__39 <o> d ) num__27 <o> e ) num__11 |
let the average after num__7 th inning = x then average after num__16 th inning = x - num__3 num__16 ( x - num__3 ) + num__87 = num__17 x x = num__87 - num__48 = num__39 answer : c <eor> c <eos> |
c |
multiply__3.0__16.0__ subtract__87.0__48.0__ subtract__87.0__48.0__ |
multiply__3.0__16.0__ subtract__87.0__48.0__ subtract__87.0__48.0__ |
| the average of first five multiples of num__6 is ? <o> a ) num__6 <o> b ) num__18 <o> c ) num__9 <o> d ) num__5 <o> e ) num__7 |
average = num__6 ( num__1 + num__2 + num__3 + num__4 + num__5 ) / num__5 = num__18.0 = num__18 . answer : b <eor> b <eos> |
b |
divide__6.0__2.0__ subtract__6.0__2.0__ subtract__6.0__1.0__ multiply__6.0__3.0__ multiply__6.0__3.0__ |
divide__6.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ multiply__6.0__3.0__ divide__18.0__1.0__ |
| the average marks obtained by num__100 candidates in an examination is num__50 . find how many candidates have passed in the examination if the average marks of candidates who passed is num__70 and that of failed candidates is num__20 . <o> a ) num__90 <o> b ) num__75 <o> c ) num__60 <o> d ) num__58 <o> e ) num__65 |
explanation : let the number of students who passed the examination be x number of students failed = ( num__100 - x ) total marks of students who have passed = num__70 x total marks of num__100 students = num__100 * num__50 = num__5000 total marks of students who have failed = num__20 ( num__100 – x ) num__20 ( num__100 – x ) + num__70 x = num__5000 num__2000 – num__20 x + num__70 x = num__5000 num__50 x = num__3000 x = num__60 answer : c <eor> c <eos> |
c |
multiply__100.0__50.0__ multiply__100.0__20.0__ subtract__5000.0__2000.0__ divide__3000.0__50.0__ divide__3000.0__50.0__ |
multiply__100.0__50.0__ multiply__100.0__20.0__ subtract__5000.0__2000.0__ divide__3000.0__50.0__ divide__3000.0__50.0__ |
| what is the next number in this sequence : num__1 num__11 num__21 num__1211 num__111221 num__312211 num__13112221 _________ ? <o> a ) num__1113213211 <o> b ) num__1113213244 <o> c ) num__1113213233 <o> d ) num__1113213222 <o> e ) num__1113213255 |
a num__1113213211 the next number in sequence is num__1113213211 . explanation - the sequence goes like you need to spell the no of digits . e . g . take the no num__1 num__1121 take num__1 . it spells as one num__1 so we get the no num__11 take num__11 we spell it as two ones ‘ and we get num__21 . so take the last given no in the sequence num__13112221 so we spell it as one num__1 one num__3 two num__1 three num__2 ’ s one num__1 . so the next no in sequence is num__1113213211 . <eor> a <eos> |
a |
subtract__3.0__1.0__ multiply__1.0__1113213211.0__ |
subtract__3.0__1.0__ multiply__1.0__1113213211.0__ |
| divide rs . num__1500 among a b and c so that a receives num__0.333333333333 as much as b and c together and b receives num__0.666666666667 as a and c together . a ' s share is ? <o> a ) num__297 <o> b ) num__268 <o> c ) num__375 <o> d ) num__367 <o> e ) num__363 |
a + b + c = num__1500 a = num__0.333333333333 ( b + c ) ; b = num__0.666666666667 ( a + c ) a / ( b + c ) = num__0.333333333333 a = num__0.25 * num__1500 = > num__375 answer : c <eor> c <eos> |
c |
multiply__1500.0__0.25__ multiply__1500.0__0.25__ |
multiply__1500.0__0.25__ multiply__1500.0__0.25__ |
| two trains of equal length running with the speeds of num__60 and num__40 kmph take num__55 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? <o> a ) num__10 sec <o> b ) num__16 sec <o> c ) num__13 sec <o> d ) num__67 sec <o> e ) num__11 sec |
rs = num__60 - num__40 = num__20 * num__0.277777777778 = num__5.55555555556 t = num__55 d = num__55 * num__5.55555555556 = num__305.555555556 rs = num__60 + num__40 = num__100 * num__0.277777777778 t = num__305.555555556 * num__0.036 = num__11 sec answer : e <eor> e <eos> |
e |
subtract__60.0__40.0__ add__60.0__40.0__ multiply__0.036__305.5556__ round__11.0__ |
subtract__60.0__40.0__ add__60.0__40.0__ multiply__0.036__305.5556__ multiply__0.036__305.5556__ |
| mark bought a set of num__6 flower pots of different sizes at a total cost of $ num__8.25 . each pot cost num__0.15 more than the next one below it in size . what was the cost in dollars of the largest pot ? <o> a ) $ num__1.75 <o> b ) $ num__1.85 <o> c ) $ num__2.00 <o> d ) $ num__2.15 <o> e ) $ num__2.30 |
this question can be solved with a handful of different algebra approaches ( as has been shown in the various posts ) . since the question asks for the price of the largest pot and the answers are prices we can test the answers . we ' re told that there are num__6 pots and that each pot costs num__25 cents more than the next . the total price of the pots is $ num__8.25 . we ' re asked for the price of the largest ( most expensive ) pot . since the total price is $ num__8.25 ( a num__15 - cent increment ) and the the difference in sequential prices of the pots is num__15 cents the largest pot probably has a price that is a num__15 - cent increment . from the answer choices i would then test answer a first if . . . . the largest pot = $ num__1.75 num__1 num__1.15 num__1.3 num__1.45 num__1.6 num__1.75 total = $ num__8.25 so this must be the answer . a <eor> a <eos> |
a |
round_down__1.75__ add__0.15__1.0__ add__0.15__1.15__ add__0.15__1.3__ add__0.15__1.45__ add__0.15__1.6__ |
round_down__1.75__ add__0.15__1.0__ add__0.15__1.15__ add__0.15__1.3__ subtract__1.75__0.15__ add__0.15__1.6__ |
| s is a set of num__64 consecutive multiples of num__4 . if the smallest number in s is num__68 then the greatest number in s is <o> a ) num__198 <o> b ) num__297 <o> c ) num__320 <o> d ) num__402 <o> e ) num__405 |
last term = first term + ( total no . of terms - num__1 ) consecutive difference s is a set of num__64 consecutive multiples of num__4 . if the smallest number in s is num__68 then the greatest number in s is first term = num__68 ; total terms = num__64 ; difference = num__4 num__68 + ( num__63 ) num__4 = num__320 ans c <eor> c <eos> |
c |
subtract__64.0__1.0__ multiply__1.0__320.0__ |
subtract__64.0__1.0__ multiply__1.0__320.0__ |
| ajay buys num__6 dozen eggs for rs num__10.80 and num__12 eggs are found rotten and the rest are sold at num__5 eggs per rupee . find his percentage gain or loss . <o> a ) num__11 num__0.111111111111 % gain <o> b ) num__11 num__0.111111111111 % loss <o> c ) num__9 num__0.0909090909091 % gain <o> d ) num__9 num__0.0909090909091 % loss <o> e ) none of these |
num__6 dozen egg cost = rs num__10.80 since one dozen is rotten he sells num__5 dozen at num__5 eggs per rupee . hence s . p = rs num__12 thus gain % = [ ( num__12 - num__10.8 ) / num__10.8 ] x num__100 = num__11 num__0.111111111111 % answer : a <eor> a <eos> |
a |
percent__100.0__11.0__ |
percent__100.0__11.0__ |
| | - num__5 | ( | - num__20 | - | num__4 | ) = ? ? source : preparation material mba center <o> a ) num__64 <o> b ) – num__60 <o> c ) num__44 <o> d ) num__75 <o> e ) num__100 |
absolute value will turn negatives into their positive ' equivalents ' and will leave positives unchanged so | - num__5 | = num__5 | - num__20 | = num__20 and | num__4 | = num__4 . getting rid of our absolute values we have : | - num__5 | ( | - num__20 | - | num__4 | ) = ( num__5 ) ( num__20 - num__4 ) = num__4 * num__16 = num__64 <eor> a <eos> |
a |
subtract__20.0__4.0__ multiply__4.0__16.0__ multiply__4.0__16.0__ |
subtract__20.0__4.0__ multiply__4.0__16.0__ multiply__4.0__16.0__ |
| a train crosses a platform of num__150 m in num__15 sec same train crosses another platform of length num__250 m in num__20 sec . then find the length of the train ? <o> a ) num__150 metres <o> b ) num__157 m <o> c ) num__750 m <o> d ) num__850 m <o> e ) num__350 m |
length of the train be ‘ x ’ x + num__10.0 = x + num__12.5 num__4 x + num__600 = num__3 x + num__750 x = num__150 m answer : a : <eor> a <eos> |
a |
divide__150.0__15.0__ divide__250.0__20.0__ multiply__150.0__4.0__ add__150.0__600.0__ round__150.0__ |
divide__150.0__15.0__ divide__250.0__20.0__ multiply__150.0__4.0__ add__150.0__600.0__ round__150.0__ |
| the markup on a television set is num__20 percent of the cost . the markup is what percent of the selling price ? ( markup = selling price - cost ) <o> a ) num__8.0 <o> b ) num__10.0 <o> c ) num__12 num__0.5 % <o> d ) num__15.0 <o> e ) num__16 num__0.666666666667 % |
sp = num__1.2 cp markup / sp = num__0.2 cp / num__1.2 cp = num__0.166666666667 * num__100 = num__16.6666666667 = num__16 num__0.666666666667 % answer - e <eor> e <eos> |
e |
divide__0.2__1.2__ divide__20.0__0.2__ divide__20.0__1.2__ round_down__16.6667__ subtract__16.6667__16.0__ round_down__16.6667__ |
divide__0.2__1.2__ divide__20.0__0.2__ divide__20.0__1.2__ round_down__16.6667__ subtract__16.6667__16.0__ subtract__16.6667__0.6667__ |
| if ( log num__5 num__5 ) ( log num__4 num__9 ) ( log num__3 num__2 ) is equal to <o> a ) - num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__1 |
= > num__1 * log num__9 / log num__4 * log num__2 / log num__3 = > log num__3 ^ num__2 / log num__2 ^ num__2 * log num__2 / log num__3 = > num__2 log num__1.5 log num__2 * log num__2 / log num__3 = num__1 answer a <eor> a <eos> |
a |
subtract__5.0__4.0__ divide__3.0__2.0__ round_down__1.5__ |
subtract__5.0__4.0__ divide__3.0__2.0__ reverse__1.0__ |
| a company pays num__12.5 dividend to its investors . if an investor buys rs . num__40 shares and gets num__25.0 on investment at what price did the investor buy the shares ? <o> a ) num__20 <o> b ) num__66 <o> c ) num__18 <o> d ) num__19 <o> e ) num__01 |
explanation : dividend on num__1 share = ( num__12.5 * num__40 ) / num__100 = rs . num__5 rs . num__25 is income on an investment of rs . num__100 rs . num__5 is income on an investment of rs . ( num__5 * num__100 ) / num__25 = rs . num__20 answer : a <eor> a <eos> |
a |
percent__12.5__40.0__ percent__100.0__20.0__ |
percent__12.5__40.0__ percent__100.0__20.0__ |
| the simple interest on a sum of money is num__0.444444444444 of the principal . find the rate percentage and time if both are numerically equal . <o> a ) num__6 num__0.333333333333 % and num__6 num__0.666666666667 years <o> b ) num__6 num__0.666666666667 % and num__6 num__0.333333333333 years <o> c ) num__6 num__0.666666666667 % and num__6 num__0.666666666667 years <o> d ) num__6 num__0.333333333333 % and num__6 num__0.333333333333 years <o> e ) num__6.0 and num__6 num__0.666666666667 years |
let sum = rs . x . then s . l . = rs . num__4 x / num__9 let rate = r % and time = r years . then ( x * r * r ) / num__100 = num__4 x / num__9 or r ^ num__2 = num__44.4444444444 or r = num__6.66666666667 = num__6 num__0.666666666667 . rate = num__6 num__0.666666666667 % and time = num__6 num__0.666666666667 years = num__6 years num__8 months answer is c <eor> c <eos> |
c |
round_down__6.6667__ divide__4.0__6.0__ multiply__2.0__4.0__ round_down__6.6667__ |
round_down__6.6667__ divide__4.0__6.0__ multiply__2.0__4.0__ round_down__6.6667__ |
| if x ^ num__2 + ( num__1 / x ^ num__2 ) = num__7 x ^ num__4 + ( num__1 / x ^ num__4 ) = ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__12 <o> d ) num__14 <o> e ) num__47 |
- > x ^ num__4 + ( num__1 / x ^ num__4 ) = ( x ^ num__2 ) ^ num__2 + ( num__1 / x ^ num__2 ) ^ num__2 = ( x ^ num__2 + num__1 / x ^ num__2 ) ^ num__2 - num__2 x ^ num__2 ( num__1 / x ^ num__2 ) = num__7 ^ num__2 - num__2 = num__47 . thus the answer is e . <eor> e <eos> |
e |
multiply__1.0__47.0__ |
divide__47.0__1.0__ |
| a man can row at num__6 kmph in still water . if the velocity of current is num__2 kmph and it takes him num__2 hour to row to a place and come back how far is the place ? <o> a ) num__5.33 km <o> b ) num__6.33 km <o> c ) num__5.63 km <o> d ) num__7.33 km <o> e ) num__5.93 km |
speed in still water = num__6 kmph speed of the current = num__2 kmph speed downstream = ( num__6 + num__2 ) = num__8 kmph speed upstream = ( num__6 - num__2 ) = num__4 kmph let the required distance be x km total time taken = num__2 hour x / num__8 + x / num__4 = num__2 x + num__2 x / num__8 = num__2 num__3 x / num__8 = num__2 num__3 x = num__16 x = num__5.33 km answer : a <eor> a <eos> |
a |
add__6.0__2.0__ subtract__6.0__2.0__ divide__6.0__2.0__ multiply__2.0__8.0__ round__5.33__ |
add__6.0__2.0__ subtract__6.0__2.0__ divide__6.0__2.0__ multiply__2.0__8.0__ round__5.33__ |
| joe ’ s average ( arithmetic mean ) test score across num__4 equally weighted tests was num__70 . he was allowed to drop his lowest score . after doing so his average test score improved to num__75 . what is the lowest test score that was dropped ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__55 <o> d ) num__65 <o> e ) num__80 |
the arithmetic mean of num__4 equally weighted tests was num__70 . so what we can assume is that we have num__4 test scores each num__70 . he dropped his lowest score and the avg went to num__75 . this means that the lowest score was not num__70 and other three scores had given the lowest score num__5 each to make it up to num__70 too . when the lowest score was removed the other num__3 scores got their num__5 back . so the lowest score was num__3 * num__5 = num__15 less than num__70 . so the lowest score = num__70 - num__15 = num__55 answer ( c ) <eor> c <eos> |
c |
subtract__75.0__70.0__ divide__75.0__5.0__ subtract__70.0__15.0__ subtract__70.0__15.0__ |
subtract__75.0__70.0__ multiply__3.0__5.0__ subtract__70.0__15.0__ subtract__70.0__15.0__ |
| in a series of consecutive numbers num__27 is the eighth smallest number and num__20 is the tenth largest number . what is the range of the series ? <o> a ) num__32 <o> b ) num__30 <o> c ) num__28 <o> d ) num__16 <o> e ) num__25 |
imp : all are consecutive odd numbers . num__27 is the num__8 th smallest there are num__7 numbers smaller then num__27 smallest number : num__27 - num__14 = num__13 num__20 is the num__10 th largest so there are num__9 numbers larger then num__20 largest number : num__20 + num__18 = num__38 range : num__38 - num__13 = num__25 ans : e <eor> e <eos> |
e |
subtract__27.0__20.0__ subtract__27.0__14.0__ subtract__27.0__9.0__ add__20.0__18.0__ subtract__38.0__13.0__ subtract__38.0__13.0__ |
subtract__27.0__20.0__ subtract__27.0__14.0__ subtract__27.0__9.0__ add__20.0__18.0__ subtract__38.0__13.0__ subtract__38.0__13.0__ |
| the area of a triangle is with base num__4 m and height num__5 m ? <o> a ) num__88 m num__2 <o> b ) num__10 m num__2 <o> c ) num__66 m num__2 <o> d ) num__77 m num__2 <o> e ) num__31 m num__2 |
num__0.5 * num__4 * num__5 = num__10 m num__2 answer : b <eor> b <eos> |
b |
triangle_area__4.0__5.0__ square_perimeter__0.5__ triangle_area__4.0__5.0__ |
volume_rectangular_prism__4.0__5.0__0.5__ multiply__4.0__0.5__ multiply__5.0__2.0__ |
| a alone can do a piece of work in num__6 days and b alone in num__8 days . a and b undertook to do it for rs . num__3200 . with the help of c they completed the work in num__3 days . how much is to be paid to c ? <o> a ) rs . num__375 <o> b ) rs . num__400 <o> c ) rs . num__600 <o> d ) rs . num__800 <o> e ) rs . num__850 |
c ' s num__1 day ' s work = num__0.333333333333 - ( num__0.166666666667 + num__0.125 ) = num__0.333333333333 - num__0.291666666667 = num__0.0416666666667 a ' s wages : b ' s wages : c ' s wages = num__0.166666666667 : num__0.125 : num__0.0416666666667 = num__4 : num__3 : num__1 c ' s share ( for num__3 days ) = rs . ( num__3 * num__0.0416666666667 * num__3200 ) = rs . num__400 answer = b <eor> b <eos> |
b |
divide__1.0__3.0__ divide__1.0__6.0__ divide__1.0__8.0__ add__0.125__0.1667__ divide__0.125__3.0__ add__3.0__1.0__ divide__3200.0__8.0__ round__400.0__ |
divide__1.0__3.0__ divide__1.0__6.0__ divide__1.0__8.0__ add__0.125__0.1667__ multiply__0.125__0.3333__ add__3.0__1.0__ multiply__3200.0__0.125__ multiply__3200.0__0.125__ |
| an amount of money is to be distributed among faruk vasim and ranjith in the ratio num__3 : num__5 : num__7 . if vasims share is rs . num__1500 what is the difference between faruk ' s and ranjith ' s shares ? <o> a ) rs num__1200 <o> b ) rs num__1500 <o> c ) rs num__1600 <o> d ) rs num__1900 <o> e ) rs num__1700 |
explanation : let p = faruk q = vasim r = ranjith let p = num__3 x q = num__5 x and r = num__7 x . then num__5 x = num__1500 ? x = num__300 . p = num__900 q = num__1500 and r = num__2100 . hence ( r - p ) = ( num__2100 - num__900 ) = num__1200 answer : a <eor> a <eos> |
a |
divide__1500.0__5.0__ multiply__3.0__300.0__ multiply__7.0__300.0__ subtract__1500.0__300.0__ subtract__1500.0__300.0__ |
divide__1500.0__5.0__ multiply__3.0__300.0__ multiply__7.0__300.0__ subtract__1500.0__300.0__ subtract__1500.0__300.0__ |
| what is the remainder when num__12346 is divided by num__9 ? <o> a ) num__7 <o> b ) num__6 <o> c ) num__5 <o> d ) num__4 <o> e ) num__3 |
if the sum of the digits of a number is divisible by num__9 the number will be divisible by num__9 . num__1 + num__2 + num__3 + num__4 + num__6 = num__16 the nearest multiple of num__9 is num__9 . num__16 - num__9 = num__7 . answer : a <eor> a <eos> |
a |
add__1.0__2.0__ add__1.0__3.0__ subtract__9.0__3.0__ subtract__9.0__2.0__ subtract__9.0__2.0__ |
add__1.0__2.0__ add__1.0__3.0__ subtract__9.0__3.0__ subtract__9.0__2.0__ subtract__9.0__2.0__ |
| a ’ s speed is num__1.76470588235 times that of b . if a and b run a race what part of the length of the race should a give b as a head start so that the race ends in a dead heat ? <o> a ) num__0.0588235294118 <o> b ) num__0.433333333333 <o> c ) num__0.566666666667 <o> d ) num__0.366666666667 <o> e ) num__0.846153846154 |
we have the ratio of a ’ s speed and b ’ s speed . this means we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers num__30 meters b covers num__17 meters in that time . so if the race is num__30 meters long when a reaches the finish line b would be num__13 meters behind him . if we want the race to end in a dead heat we want b to be at the finish line too at the same time . this means b should get a head start of num__13 meters so that he doesn ’ t need to cover that . in that case the time required by a ( to cover num__30 meters ) would be the same as the time required by b ( to cover num__17 meters ) to reach the finish line . so b should get a head start of num__0.433333333333 th of the race . answer ( b ) <eor> b <eos> |
b |
subtract__30.0__17.0__ divide__13.0__30.0__ divide__13.0__30.0__ |
subtract__30.0__17.0__ divide__13.0__30.0__ divide__13.0__30.0__ |
| rs . num__700 is divided among a b c so that a receives half as much as b and b half as much as c . then c ' s share is : <o> a ) rs num__200 <o> b ) rs num__300 <o> c ) rs num__400 <o> d ) rs num__500 <o> e ) rs num__600 |
let c = x . then b = x / num__2 and a = x / num__4 a : b : c = num__1 : num__2 : num__4 . c ' s share rs . [ ( num__0.571428571429 ) x num__700 ) = num__400 answer : c <eor> c <eos> |
c |
multiply__1.0__400.0__ |
divide__400.0__1.0__ |
| the current of a stream runs at the rate of num__4 kmph . a boat goes num__6 km and back to the starting point in num__2 hours then find the speed of the boat in still water ? <o> a ) num__9 <o> b ) num__7 <o> c ) num__8 <o> d ) num__6 <o> e ) num__4 |
s = num__4 m = x ds = x + num__4 us = x - num__4 num__6 / ( x + num__4 ) + num__6 / ( x - num__4 ) = num__2 x = num__8 answer : c <eor> c <eos> |
c |
multiply__4.0__2.0__ round__8.0__ |
add__6.0__2.0__ add__6.0__2.0__ |
| how many different positive integers are factors of num__64 ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__14 <o> d ) num__16 <o> e ) num__18 |
num__8 × num__8 = num__4 ^ num__2 × num__2 ^ num__2 so total factors = ( num__3 + num__1 ) ( num__3 + num__1 ) = num__16 answer : d <eor> d <eos> |
d |
divide__8.0__4.0__ subtract__3.0__2.0__ divide__64.0__4.0__ divide__64.0__4.0__ |
divide__8.0__4.0__ subtract__3.0__2.0__ divide__64.0__4.0__ divide__64.0__4.0__ |
| at a school two - fifths of the male students and two - thirds of the female students speak a foreign language . if the number of males is three - fourths the number of females what fraction of the students speak a foreign language ? <o> a ) num__0.466666666667 <o> b ) num__0.533333333333 <o> c ) num__0.514285714286 <o> d ) num__0.628571428571 <o> e ) num__0.552380952381 |
let x be the number of students in the school . the number of males who speak a foreign language is ( num__0.4 ) ( num__0.428571428571 ) x = ( num__0.171428571429 ) x the number of females who speak a foreign language is ( num__0.666666666667 ) ( num__0.571428571429 ) x = ( num__0.380952380952 ) x the total number of students who speak a foreign language is ( num__0.171428571429 ) x + ( num__0.380952380952 ) x = ( num__0.552380952381 ) x the answer is e . <eor> e <eos> |
e |
multiply__0.4__0.4286__ add__0.1714__0.4__ multiply__0.6667__0.5714__ add__0.1714__0.381__ add__0.1714__0.381__ |
multiply__0.4__0.4286__ add__0.1714__0.4__ multiply__0.6667__0.5714__ add__0.1714__0.381__ add__0.1714__0.381__ |
| if num__3 y + num__7 = x ^ num__2 + p = num__7 x + num__5 what is the value of p ? <o> a ) num__8 ( num__0.25 ) <o> b ) num__8 ( num__0.5 ) <o> c ) num__8 ( num__0.333333333333 ) <o> d ) num__8 ( num__0.2 ) <o> e ) none of these |
explanation : num__3 x + num__7 = num__7 x + num__5 = > num__7 x - num__3 x = num__2 = > num__4 x = num__2 = > x = num__0.5 . now num__3 x + num__7 = x ^ num__2 + p = > num__1.5 + num__7 = num__0.25 + p = > p = num__8 * num__0.25 . answer : a <eor> a <eos> |
a |
subtract__7.0__3.0__ reverse__2.0__ divide__3.0__2.0__ reverse__4.0__ add__3.0__5.0__ add__3.0__5.0__ |
subtract__7.0__3.0__ reverse__2.0__ multiply__3.0__0.5__ reverse__4.0__ add__3.0__5.0__ add__3.0__5.0__ |
| if b is an integer what is the greatest possible value for b that would still make the following statement true : num__11 * num__10 ^ b < num__0.1 ? <o> a ) – num__4 <o> b ) – num__3 <o> c ) – num__2 <o> d ) – num__1 <o> e ) num__0 |
just putting values the lhs becomes b = num__0 - - > num__11 b = - num__1 - - > num__1.1 b = - num__2 - - > num__0.11 b = - num__3 - - > num__0.011 anything lower will be smaller than num__0.011 . b = - num__2 equality does not hold but it does for b = - num__3 . answer is b . <eor> b <eos> |
b |
round_down__0.1__ subtract__11.0__10.0__ divide__11.0__10.0__ multiply__0.1__1.1__ add__1.0__2.0__ multiply__0.1__0.11__ add__1.0__2.0__ |
round_down__0.1__ subtract__11.0__10.0__ divide__11.0__10.0__ multiply__0.1__1.1__ add__1.0__2.0__ multiply__0.1__0.11__ add__1.0__2.0__ |
| ram singh goes to his office in the city every day from his suburban house . his driver mangaram drops him at the railway station in the morning and picks him up in the evening . every evening ram singh reaches the station at num__5 o ' clock . mangaram also reaches at the same time . one day ramsingh started early from his office and came to the station at num__4 o ' clock . not wanting to wait for the car he starts walking home . mangaram starts at normal time picks him up on the way and takes him back house half an hour early . how much time did ram singh walk . <o> a ) num__0.25 hours <o> b ) num__0.5 hours <o> c ) num__0.75 hours <o> d ) num__1.25 hours <o> e ) num__1.75 hours |
let the time required for walk is t hours then ram singh will reach at the car in time ( num__4 + t ) due to this walking of ram singh time saving = [ num__5 - ( num__4 + t ) ] * num__2 here num__2 is due to forward motion upto railway station and backward from railway station to the point where the car met to ram singh . it is given that time saving = num__0.5 hour so num__2 [ num__1 - t ] = num__0.5 on solving we get t = num__0.75 hours answer : c <eor> c <eos> |
c |
divide__2.0__4.0__ subtract__5.0__4.0__ round__0.75__ |
divide__2.0__4.0__ subtract__5.0__4.0__ round__0.75__ |
| a and b start from house at num__10 am . they travel on the mg road at num__20 kmph and num__40 kmph . there is a junction t on their path . a turns left at t junction at num__12 : num__00 noon b reaches t earlier and turns right . both of them continue to travel till num__2 pm . what is the distance between a and b at num__2 pm <o> a ) num__160 km <o> b ) num__170 km <o> c ) num__180 km <o> d ) num__190 km <o> e ) num__150 km |
at num__12 a will travel = num__20 * num__2 = num__40 km b will travel this num__40 km in num__1.0 = num__1 hr i . e . by num__11 am . after t junction for a - distance travelled = num__2 * num__20 = num__40 km for b distance travelled = num__40 * num__3 = num__120 so distance between a & b is = num__120 + num__40 = num__160 km answer : a <eor> a <eos> |
a |
add__10.0__1.0__ add__2.0__1.0__ multiply__10.0__12.0__ add__40.0__120.0__ round__160.0__ |
add__10.0__1.0__ add__2.0__1.0__ multiply__10.0__12.0__ add__40.0__120.0__ add__40.0__120.0__ |
| the average ( arithmetic mean ) of five numbers is num__5 . if num__2 is subtracted from each of three of the numbers what is the new average ? <o> a ) num__3.8 <o> b ) num__8.9 <o> c ) num__4.5 <o> d ) num__7.5 <o> e ) num__2.9 |
sum of num__5 numbers = num__5 * num__5 = num__25 if num__2 is subtracted from each of three of the numbers we subtract num__2 * num__3 = num__6 from the total sum sum of num__5 number after subtracting num__2 from each of three of the numbers = num__25 - num__6 = num__19 new average = num__3.8 = num__3.8 answer a <eor> a <eos> |
a |
subtract__5.0__2.0__ multiply__2.0__3.0__ subtract__25.0__6.0__ divide__19.0__5.0__ divide__19.0__5.0__ |
subtract__5.0__2.0__ multiply__2.0__3.0__ subtract__25.0__6.0__ divide__19.0__5.0__ divide__19.0__5.0__ |
| light glows for every num__30 seconds . how many max . times did it glow between num__1 : num__57 : num__58 and num__3 : num__20 : num__47 am . <o> a ) num__380 times <o> b ) num__142 times <o> c ) num__382 times <o> d ) num__392 times <o> e ) num__165 times |
time difference is num__1 hr num__22 min num__49 sec = num__4969 sec . so light glows floor ( num__165.633333333 ) = num__165 times . answer : e <eor> e <eos> |
e |
divide__4969.0__30.0__ round_down__165.6333__ round_down__165.6333__ |
divide__4969.0__30.0__ round_down__165.6333__ round_down__165.6333__ |
| three runners r b and c run a race with runner r finishing num__12 m ahead of runner b and num__18 m ahead of runner c while runner b finishes num__8 m ahead of runner c . each runner travels entire distance at a constant speed . what was the length of the race ? <o> a ) num__36 m <o> b ) num__48 m <o> c ) num__60 m <o> d ) num__72 m <o> e ) num__84 m |
let distance of race be x mtrs . then when r finishes x m b has run ( x - num__12 ) mtrs and c has run x - num__18 mtrs . so at this point b is num__6 m ahead of c . now to finish race b needs to run another num__12 m so he runs another num__12 m . when b finishes race he is num__8 m ahead of c . so last num__12 m b has run c has run num__10 m . as speeds are constant we have equation x - num__12 / x - num__18 = num__1.2 > x = num__48 . answer b <eor> b <eos> |
b |
subtract__18.0__12.0__ subtract__18.0__8.0__ divide__12.0__10.0__ multiply__8.0__6.0__ round__48.0__ |
subtract__18.0__12.0__ subtract__18.0__8.0__ divide__12.0__10.0__ multiply__8.0__6.0__ round__48.0__ |
| jake remembers only the last four letters of his five - letter klingon name . if he is sure that the first letter is neither ` ` n ' ' nor ` ` z ' ' and assuming that there are only num__10 letters in the klingon alphabet what is the probability that he will give the correct name when asked for it by the space attendant ? <o> a ) a ) num__0.08 <o> b ) b ) num__0.125 <o> c ) c ) num__0.111111111111 <o> d ) d ) num__0.8 <o> e ) e ) num__0.9 |
total letters num__10 out of which the first letter is anything of num__10 but n or x . . . so there are num__8 possible letters for first place . . and only one of them is correct . . so num__0.125 . . answer : b <eor> b <eos> |
b |
reverse__8.0__ reverse__8.0__ |
reverse__8.0__ reverse__8.0__ |
| a certain musical scale has has num__13 notes each having a different frequency measured in cycles per second . in the scale the notes are ordered by increasing frequency and the highest frequency is twice the lowest . for each of the num__12 lower frequencies the ratio of a frequency to the next higher frequency is a fixed constant . if the lowest frequency is num__110 cycles per second then the frequency of the num__7 th note in the scale is how many cycles per second ? <o> a ) num__110 * sqrt num__2 <o> b ) num__110 * sqrt ( num__2 ^ num__7 ) <o> c ) num__110 * sqrt ( num__2 ^ num__12 ) <o> d ) num__110 * the twelfth root of ( num__2 ^ num__7 ) <o> e ) num__110 * the seventh root of ( num__2 ^ num__12 ) |
let the constant be k . f num__1 = num__110 f num__2 = num__110 k f num__3 = num__110 k * k = num__110 * k ^ num__2 f num__13 = num__110 * k ^ num__12 we know f num__13 = num__2 * f num__1 = num__2 * num__110 = num__220 num__2.0 = k ^ num__12 k = twelfth root of num__2 for f num__7 . . . f num__7 = num__110 * k ^ num__6 ( as we wrote for f num__2 and f num__3 ) f num__7 = num__110 * ( twelfth root of num__2 ) ^ num__6 f num__7 = num__110 * sqrt ( num__2 ) the answer is a . <eor> a <eos> |
a |
subtract__13.0__12.0__ add__1.0__2.0__ multiply__110.0__2.0__ subtract__13.0__7.0__ multiply__110.0__1.0__ |
subtract__13.0__12.0__ add__1.0__2.0__ multiply__110.0__2.0__ multiply__2.0__3.0__ multiply__110.0__1.0__ |
| danny is sitting on a rectangular box . the area of the front face of the box is half the area of the top face and the area of the top face is num__1.5 times the area of the side face . if the volume of the box is num__648 what is the area of the side face of the box ? <o> a ) num__32 <o> b ) num__78 <o> c ) num__84 <o> d ) num__90 <o> e ) num__72 |
lets suppose length = l breadth = b depth = d front face area = l * w = num__0.5 w * d ( l = num__0.5 d or d = num__2 l ) top face area = w * d side face area = w * d = num__1.5 d * l ( w = num__1.5 l ) volume = l * w * d = num__648 l * num__1.5 l * num__2 l = num__648 l = num__6 side face area = l * d = l * num__2 l = num__6 * num__2 * num__6 = num__72 e is the answer <eor> e <eos> |
e |
square_perimeter__0.5__ square_perimeter__1.5__ triangle_area__2.0__72.0__ |
square_perimeter__0.5__ square_perimeter__1.5__ volume_rectangular_prism__0.5__2.0__72.0__ |
| if num__3 x + num__4 y = num__88 and num__3 x - num__4 y = num__14 what is the value of x subs y ? <o> a ) num__4.25 <o> b ) num__6.25 <o> c ) num__6.75 <o> d ) num__6.25 <o> e ) num__4.75 |
given num__3 x + num__4 y = num__88 - - - eq num__1 num__3 x - num__4 y = num__14 - - eq num__2 sum both eqns we get num__6 x = num__102 = > x = num__17 sum num__3 x in eq num__2 = > num__51 - num__4 y = num__14 . = > y = num__9.25 now xy = num__14 - num__9.25 = num__4.75 . option e is correct answer . <eor> e <eos> |
e |
subtract__4.0__3.0__ subtract__3.0__1.0__ multiply__3.0__2.0__ add__88.0__14.0__ add__3.0__14.0__ multiply__3.0__17.0__ subtract__14.0__9.25__ subtract__14.0__9.25__ |
subtract__4.0__3.0__ subtract__3.0__1.0__ add__4.0__2.0__ add__88.0__14.0__ add__3.0__14.0__ multiply__3.0__17.0__ subtract__14.0__9.25__ subtract__14.0__9.25__ |
| a trader bought a car at num__20.0 discount on its original price . he sold it at a num__40.0 increase on the price he bought it . what percent of profit did he make on the original price ? <o> a ) num__19.0 <o> b ) num__72.0 <o> c ) num__12.0 <o> d ) num__16.0 <o> e ) num__32 % |
original price = num__100 cp = num__80 s = num__80 * ( num__1.4 ) = num__112 num__100 - num__112 = num__12.0 answer : c <eor> c <eos> |
c |
percent__100.0__12.0__ |
percent__100.0__12.0__ |
| the total of the ages of amar akbar and anthony is num__62 years . what was the total of their ages three years ago ? <o> a ) num__51 years <o> b ) num__55 years <o> c ) num__74 years <o> d ) num__53 years <o> e ) num__78 years |
explanation : required sum = ( num__62 - num__3 x num__3 ) years = ( num__62 - num__9 ) years = num__53 years . answer : option d <eor> d <eos> |
d |
subtract__62.0__9.0__ subtract__62.0__9.0__ |
subtract__62.0__9.0__ subtract__62.0__9.0__ |
| a and b can do a work in num__4 days b and c in num__5 days c and a in num__7 days . if a b and c work together they will complete the work in ? <o> a ) num__3 days <o> b ) num__5 days <o> c ) num__7 days <o> d ) num__1 days <o> e ) num__2 dasy |
a + b num__1 day work = num__0.25 b + c num__1 day work = num__0.2 c + a num__1 day work = num__0.142857142857 adding we get num__2 ( a + b + c ) = num__0.25 + num__0.2 + num__0.142857142857 = num__0.592857142857 a + b + c num__1 day work = num__0.296428571429 a b c can finish the work in num__3.3734939759 days = num__3 days approximately answer is a <eor> a <eos> |
a |
subtract__5.0__4.0__ divide__1.0__4.0__ divide__1.0__5.0__ divide__1.0__7.0__ subtract__7.0__5.0__ divide__0.5929__2.0__ subtract__4.0__1.0__ round__3.0__ |
subtract__5.0__4.0__ divide__1.0__4.0__ divide__1.0__5.0__ divide__1.0__7.0__ subtract__7.0__5.0__ divide__0.5929__2.0__ add__1.0__2.0__ add__1.0__2.0__ |
| what is the greatest possible straight line distance in inches between any two points on a rectangular box that is num__5 inches wide num__3 inches long and num__4 inches high ? <o> a ) num__2 √ num__5 <o> b ) num__5 <o> c ) num__5 √ num__2 <o> d ) num__10 <o> e ) num__50 |
the maximum possible distance in a cuboid is the diagonal . diagonal length = sqrt ( length ^ num__2 + breadth ^ num__2 + height ^ num__2 ) = sqrt ( num__50 ) = num__5 √ num__2 ( option c ) <eor> c <eos> |
c |
coin_space__ vowel_space__ |
coin_space__ vowel_space__ |
| suresh can complete a job in num__15 hours . ashutosh alone can complete the same job in num__30 hours . suresh works for num__9 hours and then the remaining job is completed by ashutosh . how many hours will it take ashutosh to complete the remaining job alone ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__12 <o> e ) none of these |
the part of job that suresh completes in num__9 hours = num__9 â „ num__15 = num__3 â „ num__5 remaining job = num__1 - num__3 â „ num__5 = num__2 â „ num__5 remaining job can be done by ashutosh in num__2 â „ num__5 Ã — num__30 = num__12 hours answer d <eor> d <eos> |
d |
divide__15.0__3.0__ divide__30.0__15.0__ subtract__15.0__3.0__ round__12.0__ |
divide__15.0__3.0__ subtract__3.0__1.0__ subtract__15.0__3.0__ subtract__15.0__3.0__ |
| an article is bought for rs . num__675 and sold for rs . num__800 find the gain percent ? <o> a ) num__38.0 <o> b ) num__28.0 <o> c ) num__18.0 <o> d ) num__22.0 <o> e ) num__13 % |
num__675 - - - - num__125 num__100 - - - - ? = > = num__18.0 answer : c <eor> c <eos> |
c |
percent__18.0__100.0__ |
percent__18.0__100.0__ |
| if x and y are perfect squares then which five of the following is not necessarily a perfect square ? <o> a ) x ^ num__2 <o> b ) xy <o> c ) num__4 x <o> d ) x + y <o> e ) x ^ num__5 |
if x = y = num__1 ^ num__2 = num__1 then each option but a is a perfect square therefore a is not necessarily a perfect square . answer : a . <eor> a <eos> |
a |
multiply__1.0__2.0__ |
power__2.0__1.0__ |
| a train is num__435 meter long is running at a speed of num__45 km / hour . in what time will it pass a bridge of num__140 meter length <o> a ) num__20 seconds <o> b ) num__30 seconds <o> c ) num__46 seconds <o> d ) num__50 seconds <o> e ) none of these |
explanation : speed = num__45 km / hr = num__45 * ( num__0.277777777778 ) m / sec = num__12.5 m / sec total distance = num__435 + num__140 = num__575 meter time = distance / speed = num__575 ∗ num__0.08 = num__46 seconds option c <eor> c <eos> |
c |
add__435.0__140.0__ multiply__0.08__575.0__ round__46.0__ |
add__435.0__140.0__ multiply__0.08__575.0__ multiply__0.08__575.0__ |
| the angle between the minute hand and the hour hand of a clock when the time is num__4.20 is : <o> a ) num__0 º <o> b ) num__10 º <o> c ) num__5 º <o> d ) num__20 º <o> e ) num__30 º |
angle traced by hour hand in num__4.33333333333 hrs = ( num__30.0 x num__4.33333333333 ) º = num__130 º . angle traced by min . hand in num__20 min . = ( num__6.0 x num__20 ) º = num__120 º . required angle = ( num__130 - num__120 ) º = num__10 º answer : b <eor> b <eos> |
b |
multiply__6.0__20.0__ subtract__130.0__120.0__ round__10.0__ |
multiply__6.0__20.0__ subtract__130.0__120.0__ subtract__130.0__120.0__ |
| speed of a boat in standing water is num__8 kmph and the speed of the stream is num__2 kmph . a man rows to place at a distance of num__210 km and comes back to the starting point . the total time taken by him is : <o> a ) num__48 hours <o> b ) num__51 hours <o> c ) num__36 hours <o> d ) num__56 hours <o> e ) none |
sol . speed upstream = num__6 kmph ; speed downstream = num__10 kmph . ∴ total time taken = [ num__35.0 + num__21.0 ] hours = num__56 hours . answer d <eor> d <eos> |
d |
subtract__8.0__2.0__ add__8.0__2.0__ divide__210.0__6.0__ divide__210.0__10.0__ add__35.0__21.0__ round__56.0__ |
subtract__8.0__2.0__ add__8.0__2.0__ divide__210.0__6.0__ divide__210.0__10.0__ add__35.0__21.0__ add__35.0__21.0__ |
| if n is an integer which of the following can not be a factor of num__3 n + num__4 ? <o> a ) num__4 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
num__3 n + num__4 = num__3 ( n + num__1 ) + num__1 can not be a multiple of num__3 it ' s num__1 more than multiple of num__3 : . . . num__4 num__7 num__10 num__13 num__16 . . . hence it can not be a multiple of num__2 * num__3 = num__6 as well . answer : c . <eor> c <eos> |
c |
subtract__4.0__3.0__ add__3.0__4.0__ add__3.0__7.0__ add__3.0__10.0__ add__3.0__13.0__ subtract__3.0__1.0__ multiply__3.0__2.0__ multiply__3.0__2.0__ |
subtract__4.0__3.0__ add__3.0__4.0__ add__3.0__7.0__ add__3.0__10.0__ add__3.0__13.0__ subtract__3.0__1.0__ multiply__3.0__2.0__ multiply__3.0__2.0__ |
| the parameter of a square is equal to the perimeter of a rectangle of length num__22 cm and breadth num__16 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) <o> a ) num__77.14 cm <o> b ) num__47.14 cm <o> c ) num__84.92 cm <o> d ) num__29.85 cm <o> e ) num__23.57 cm |
let the side of the square be a cm . parameter of the rectangle = num__2 ( num__22 + num__16 ) = num__76 cm parameter of the square = num__76 cm i . e . num__4 a = num__76 a = num__19 diameter of the semicircle = num__19 cm circimference of the semicircle = num__0.5 ( ∏ ) ( num__19 ) = num__0.5 ( num__3.14285714286 ) ( num__19 ) = num__29.8571428571 = num__29.85 cm to two decimal places answer : d <eor> d <eos> |
d |
rectangle_perimeter__22.0__16.0__ triangle_area__2.0__29.85__ |
rectangle_perimeter__22.0__16.0__ triangle_area__2.0__29.85__ |
| special codes are designated by either a num__3 - letter or a num__4 - letter code that is created by using the num__26 letters of the alphabet . which of the following gives the maximum number of different codes that can be designated with this system ? <o> a ) num__2 ( num__26 ) ^ num__4 <o> b ) num__26 ( num__26 ) ^ num__3 <o> c ) num__27 ( num__26 ) ^ num__3 <o> d ) num__26 ( num__26 ) ^ num__4 <o> e ) num__27 ( num__26 ) ^ num__4 |
num__26 ^ num__3 + num__26 ^ num__4 = num__26 ^ num__3 ( num__1 + num__26 ) = num__27 * num__26 ^ num__3 the answer is c . <eor> c <eos> |
c |
subtract__4.0__3.0__ add__26.0__1.0__ add__26.0__1.0__ |
subtract__4.0__3.0__ add__26.0__1.0__ add__26.0__1.0__ |
| the manager of a produce market purchased a quantity of tomatoes for $ num__0.80 per pound . due to improper handling num__10 percent of the tomatoes by weight were ruined and discarded . at what price per pound should the manager sell the remaining tomatoes if she wishes to make a profit on the sale of the tomatoes equal to num__12 percent of the cost of the tomatoes . <o> a ) $ num__0.94 <o> b ) $ num__0.96 <o> c ) $ num__0.99 <o> d ) $ num__1.00 <o> e ) $ num__1.20 |
assume the manager bought num__100 tomatoes . cost price = num__80 given : num__10.0 are damaged - - > available tomatoes to sell = num__90 num__90 * x - num__80 = num__0.12 * num__80 num__90 x - num__80 = num__9.6 num__90 x = num__89.6 x = num__89.6 / num__90 = num__0.995 x is slightly under num__0.9955 = num__0.99 answer : c <eor> c <eos> |
c |
percent__12.0__80.0__ percent__100.0__0.99__ |
percent__12.0__80.0__ percent__100.0__0.99__ |
| for all numbers a and b the operationis defined by ab = ( a + num__2 ) ( b – num__3 ) . if num__3 x = – num__75 then x = <o> a ) – num__15 <o> b ) – num__13 <o> c ) - num__14 <o> d ) - num__12 <o> e ) num__15 |
( num__3 + num__2 ) ( x - num__3 ) = - num__75 . . x - num__3 = - num__15 . . x = - num__12 d <eor> d <eos> |
d |
subtract__15.0__3.0__ subtract__15.0__3.0__ |
subtract__15.0__3.0__ subtract__15.0__3.0__ |
| find the value of num__1 / ( num__3 + num__1 / ( num__3 + num__1 / ( num__3 - num__0.333333333333 ) ) ) <o> a ) num__0.23595505618 <o> b ) num__3.2962962963 <o> c ) num__0.303370786517 <o> d ) num__0.0714285714286 <o> e ) num__0.0178571428571 |
num__1 / [ num__3 + ( num__1 / ( num__3 + num__1 / ( num__3 - num__0.333333333333 ) ) ) ] = > num__1 / [ num__3 + num__1 / ( num__3 + num__1 / ( num__2.66666666667 ) ) ] = > num__1 / [ num__3 + num__1 / ( num__3 + num__0.375 ) ] = > num__1 / [ num__3 + num__0.296296296296 ] = > num__1 / ( num__3.2962962963 ) = > num__0.303370786517 c <eor> c <eos> |
c |
subtract__3.0__0.3333__ reverse__2.6667__ add__3.0__0.2963__ reverse__3.2963__ reverse__3.2963__ |
subtract__3.0__0.3333__ reverse__2.6667__ add__3.0__0.2963__ reverse__3.2963__ reverse__3.2963__ |
| a man can swim in still water at num__3 km / h but takes twice as long to swim upstream than downstream . the speed of the stream is ? <o> a ) num__1.7 <o> b ) num__1.0 <o> c ) num__1.2 <o> d ) num__1.5 <o> e ) num__1.16 |
m = num__3 s = x ds = num__3 + x us = num__3 + x num__3 + x = ( num__3 - x ) num__2 num__3 + x = num__6 - num__2 x num__3 x = num__3 x = num__1.0 answer : b <eor> b <eos> |
b |
multiply__3.0__2.0__ subtract__3.0__2.0__ round__1.0__ |
multiply__3.0__2.0__ subtract__3.0__2.0__ subtract__3.0__2.0__ |
| john buys num__100 shares of par value rs . num__20 each of a company which pays an annual dividend of num__12.0 at such a price that he gets num__10.0 on his investment . find the market value of a share . <o> a ) num__2 <o> b ) num__24 <o> c ) num__5 <o> d ) num__6 <o> e ) num__8 |
face value of each share = rs . num__20 total dividend received by john = num__100 à — num__20 à — num__0.12 = rs . num__240 let market value of num__100 shares = rs . x x à — num__0.1 = num__240 x = num__2400 ie market value of num__100 shares = rs . num__2400 hence market value of each share = rs . num__24 answer is b <eor> b <eos> |
b |
percent__10.0__240.0__ percent__100.0__24.0__ |
percent__10.0__240.0__ percent__100.0__24.0__ |
| what is the sum of the different positive prime factors of num__660 ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__15 <o> d ) num__17 <o> e ) num__21 |
prime factorization of num__660 = num__66 * num__10 = num__6 * num__11 * num__10 = num__2 * num__3 * num__2 * num__5 * num__11 = num__2 ^ num__2 * num__3 * num__5 * num__11 sum of the different positive prime factors of num__660 = num__2 + num__3 + num__5 + num__11 = num__21 answer e <eor> e <eos> |
e |
divide__660.0__66.0__ divide__66.0__6.0__ gcd__6.0__10.0__ divide__6.0__2.0__ add__2.0__3.0__ add__10.0__11.0__ lcm__3.0__21.0__ |
divide__660.0__66.0__ divide__66.0__6.0__ gcd__6.0__10.0__ divide__6.0__2.0__ add__2.0__3.0__ add__10.0__11.0__ add__10.0__11.0__ |
| of the num__65 cars on a car lot num__45 have air - bag num__30 have power windows and num__12 have both air - bag and power windows . how many of the cars on the lot have neither air - bag nor power windows ? <o> a ) num__8 <o> b ) num__2 <o> c ) num__10 <o> d ) num__15 <o> e ) num__18 |
total - neither = all air bag + all power windows - both or num__65 - neither = num__45 + num__30 - num__12 = num__63 . = > neither = num__2 hence b . answer : b <eor> b <eos> |
b |
coin_space__ coin_space__ |
coin_space__ coin_space__ |
| a person has num__100 $ in num__10 $ and num__5 $ bill . if the num__5 $ bill quantity is twice that of num__10 $ bill . what is quantity of num__10 $ . <o> a ) num__2 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__5 |
let amount of num__10 $ be x . then amount of num__5 $ be num__2 x . now num__5 * num__2 x + num__10 * x = num__100 . thus x = num__5 . answer : e <eor> e <eos> |
e |
divide__10.0__5.0__ subtract__10.0__5.0__ |
divide__10.0__5.0__ subtract__10.0__5.0__ |
| if a boat goes num__7 km upstream in num__42 minutes and the speed of the stream is num__5 kmph then the speed of the boat in still water is : <o> a ) num__3.5 km / hr <o> b ) num__4.2 km / hr <o> c ) num__5 km / hr <o> d ) num__10.5 km / hr <o> e ) none of these |
solution speed of stream = num__0.5 ( num__15 - num__8 ) km / hr = num__3.5 kmph . answer a <eor> a <eos> |
a |
subtract__15.0__7.0__ multiply__7.0__0.5__ round__3.5__ |
subtract__15.0__7.0__ multiply__7.0__0.5__ subtract__7.0__3.5__ |
| in a certain game a large bag is filled with blue green purple and red chips worth num__1 num__5 x and num__11 points each respectively . the purple chips are worth more than the green chips but less than the red chips . a certain number of chips are then selected from the bag . if the product of the point values of the selected chips is num__140800 how many purple chips were selected ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
num__140800 = num__1 * num__5 ^ num__2 * num__8 ^ num__3 * num__11 the factors of num__8 must come from the purple point value so there are num__3 purple chips . the answer is c . <eor> c <eos> |
c |
add__1.0__2.0__ add__1.0__2.0__ |
add__1.0__2.0__ add__1.0__2.0__ |
| two numbers are less than a third number by num__30.0 and num__37.0 . how much percent is the second number is less than the first ? <o> a ) num__10.0 <o> b ) num__15.0 <o> c ) num__20.0 <o> d ) num__25.0 <o> e ) num__30 % |
let the third number be x first number = num__70.0 of x = num__7 x / num__10 ; second number = num__63.0 of x = num__63 x / num__100 ; difference = num__7 x / num__10 - num__63 x / num__100 = num__7 x / num__100 ; required percentage = num__7 x / num__100 * num__1.42857142857 x * num__100 = num__10.0 answer is a <eor> a <eos> |
a |
percent__100.0__10.0__ |
percent__100.0__10.0__ |
| a single discount equivalent to the discount series of num__20.0 num__10.0 and num__8.0 is ? <o> a ) num__31.5 <o> b ) num__33.8 <o> c ) num__31.6 <o> d ) num__31.3 <o> e ) num__31.1 |
num__100 * ( num__0.8 ) * ( num__0.9 ) * ( num__0.92 ) = num__66.2 num__100 - num__66.2 = num__33.8 answer : b <eor> b <eos> |
b |
percent__10.0__8.0__ percent__100.0__33.8__ |
percent__10.0__8.0__ percent__100.0__33.8__ |
| b completes a work in num__6 days . a alone can do it in num__10 days . if both work together the work can be completed in how many days ? <o> a ) num__3.33 days <o> b ) num__4.33 days <o> c ) num__5.33 days <o> d ) num__6.33 days <o> e ) num__7.33 days |
num__0.2 + num__0.1 = num__0.3 num__3.33333333333 = num__3.33 days answer : a <eor> a <eos> |
a |
add__0.1__0.2__ round__3.3333__ round__3.33__ |
add__0.1__0.2__ round__3.3333__ round__3.33__ |
| the h . c . f . of two numbers is num__23 and the other two factors of their l . c . m . are num__13 and num__14 . the larger of the two numbers is : <o> a ) num__276 <o> b ) num__299 <o> c ) num__322 <o> d ) num__345 <o> e ) num__365 |
clearly the numbers are ( num__23 x num__13 ) and ( num__23 x num__14 ) . larger number = ( num__23 x num__14 ) = num__322 . answer : option c <eor> c <eos> |
c |
multiply__23.0__14.0__ multiply__23.0__14.0__ |
multiply__23.0__14.0__ multiply__23.0__14.0__ |
| which of the following reaches the largest value over the range - num__1 < = r < = num__1 ? <o> a ) ( r + num__3 ) ^ num__100 <o> b ) ( r + num__3 ) / ( r ^ num__100 ) <o> c ) ( r + num__1000 ) <o> d ) num__97 ^ num__97 <o> e ) ( r / num__2 r ) |
( r + num__3 ) / ( r ^ num__100 ) can be equal any arbitrarily large number . some of the other options are large but they all have a definite maximum . ( r + num__3 ) / ( r ^ num__100 ) does not so the correct answer is b . <eor> b <eos> |
b |
multiply__1.0__3.0__ |
divide__3.0__1.0__ |
| pipes a and b can fill a tank in num__6 and num__4 hours . pipe c can empty it in num__12 hours . if all pipes are opened together then the tank will be filled in ? <o> a ) num__3 hr <o> b ) num__2 hr <o> c ) num__1 hr <o> d ) num__1.28571428571 hr <o> e ) num__1.66666666667 hr |
net part filled in num__1 hour = num__0.166666666667 + num__0.25 - num__0.0833333333333 = num__0.333333333333 the tank will be full in num__3 hr answer is a <eor> a <eos> |
a |
divide__1.0__6.0__ divide__1.0__4.0__ subtract__0.25__0.1667__ divide__4.0__12.0__ subtract__4.0__1.0__ round__3.0__ |
divide__1.0__6.0__ divide__1.0__4.0__ subtract__0.25__0.1667__ add__0.25__0.0833__ subtract__4.0__1.0__ subtract__6.0__3.0__ |
| in a certain warehouse num__90 percent of the packages weigh less than num__75 pounds and a total of num__48 packages weigh less than num__25 pounds . if num__80 percent of the packages weigh at least num__25 pounds how many of the packages weigh at least num__25 pounds but less than num__75 pounds ? <o> a ) num__8 <o> b ) num__64 <o> c ) num__168 <o> d ) num__102 <o> e ) num__144 |
if num__80.0 of the packages weigh at least num__25 pounds this means that num__20.0 of the packages weigh less than num__25 pounds let t = total number of packages so num__20.0 of t = # of packages that weigh less than num__25 pounds num__48 packages weigh less than num__25 pounds great . so num__20.0 of t = num__48 rewrite to get : num__0.2 t = num__48 solve : t = num__240 num__90.0 of the packages weigh less than num__75 pounds so num__90.0 oft = number of packages that weigh less than num__75 pounds num__90.0 of num__240 = num__216 so num__216 packages weigh less than num__75 pounds of those num__216 packages that weigh less than num__75 pounds num__48 packages weigh less than num__25 pounds . so the number of packages that weight between num__25 and num__75 pounds = num__216 - num__48 = num__168 = c <eor> c <eos> |
c |
divide__48.0__0.2__ subtract__216.0__48.0__ subtract__216.0__48.0__ |
divide__48.0__0.2__ subtract__216.0__48.0__ subtract__216.0__48.0__ |
| paul and jack are num__2 mechanics who work num__8 hours per day changing oil in cars at the oil spot . if paul can change oil in num__2 cars per hour and jack can change oil in num__3 cars per hour . what is the least number of cars per work day the num__2 men can finish ? <o> a ) num__50 <o> b ) num__40 <o> c ) num__35 <o> d ) num__10 <o> e ) num__90 |
the proportion equation can be used . let x = number cars and c = least number of cars per day the num__2 men can finish ( c = x / ( num__8 hours ) p = cars per hour paul can change and j = cars per hour jack can change . note : num__8 hours = a work day p + j = c ( num__2 cars / hour ) + ( num__3 cars / hour ) = x / ( num__8 hours ) proportion : num__2.0 + num__3.0 / = x / num__300 x = num__8 ( num__2 + num__3 ) = num__40 . at least num__40 cars per work day . answer is b <eor> b <eos> |
b |
round__40.0__ |
round__40.0__ |
| in the interior of a forest a certain number of apes equal to the square of one - eighth of the total number are playing and having great fun . the remaining sixteen apes are on a hill and the echo of their shrieks by the adjoining hills frightens them . they came and join the apes in the forest and play with enthusiasm . what is the total number of apes ? <o> a ) num__48 <o> b ) num__16 <o> c ) num__64 <o> d ) num__80 <o> e ) num__32 |
let total number be x no in the interior = ( x / num__8 ) ^ num__2 no outside = num__16 so : x - ( x / num__8 ) ^ num__2 = num__16 x ^ num__2 - num__64 x + num__32 ^ num__2 = num__0 ( x - num__32 ) so either x = num__32 e <eor> e <eos> |
e |
multiply__2.0__8.0__ divide__64.0__2.0__ divide__64.0__2.0__ |
multiply__2.0__8.0__ divide__64.0__2.0__ divide__64.0__2.0__ |
| find compound interest on rs . num__8000 at num__15.0 per annum for num__2 years num__4 months compounded annually . <o> a ) rs num__309 <o> b ) rs num__319 <o> c ) rs num__3109 <o> d ) rs num__109 <o> e ) rs num__209 |
time = num__2 years num__4 months = num__2 ( num__0.333333333333 ) years = num__2 ( num__0.333333333333 ) years . amount = rs ' . [ num__8000 x ( num__1 + ¬ ( num__0.15 ) ) num__2 x ( num__1 + ( ( num__0.333333333333 ) * num__15 ) / num__100 ) ] = rs . [ num__8000 * ( num__1.15 ) * ( num__1.15 ) * ( num__1.05 ) ] = rs . num__11109 . . : . c . i . = rs . ( num__11109 - num__8000 ) = rs . num__3109 . answer c num__3109 <eor> c <eos> |
c |
percent__15.0__1.0__ percent__100.0__3109.0__ |
percent__15.0__1.0__ percent__100.0__3109.0__ |
| a mixture of num__150 liters of wine and water contains num__20.0 water . how much more water should be added so that water becomes num__25.0 of the new mixture ? <o> a ) num__22 <o> b ) num__77 <o> c ) num__10 <o> d ) num__27 <o> e ) num__18 |
number of liters of water in num__150 liters of the mixture = num__20.0 of num__150 = num__0.2 * num__150 = num__30 liters . p liters of water added to the mixture to make water num__25.0 of the new mixture . total amount of water becomes ( num__30 + p ) and total volume of mixture is ( num__150 + p ) . ( num__30 + p ) = num__0.25 * ( num__150 + p ) num__120 + num__4 p = num__150 + p = > p = num__10 liters . answer : c <eor> c <eos> |
c |
multiply__150.0__0.2__ subtract__150.0__30.0__ reverse__0.25__ subtract__30.0__20.0__ subtract__20.0__10.0__ |
multiply__150.0__0.2__ subtract__150.0__30.0__ multiply__20.0__0.2__ subtract__30.0__20.0__ subtract__20.0__10.0__ |
| reena took a loan of rs . num__1200 with simple interest for as many years as the rate of interest . if she paid rs . num__300 as interest at the end of the loan period what was the rate of interest ? <o> a ) num__3.6 <o> b ) num__6 <o> c ) num__5 <o> d ) num__24 <o> e ) none of these |
let rate = r % and time = r years . then ( num__1200 x r x r ) / num__100 = num__300 num__12 r num__2 = num__300 r num__2 = num__25 r = num__5 . answer : option c <eor> c <eos> |
c |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| the integer x is divisible by both num__7 and num__21 . which of the following must be an integer ? <o> a ) x / num__25 <o> b ) x / num__34 <o> c ) x / num__52 <o> d ) x / num__45 <o> e ) x / num__21 |
prime factorization of num__7 = num__1 * num__7 prime factorization of num__21 = num__3 * num__7 lcm of num__7 and num__21 = num__7 * num__3 = num__21 therefore x / num__21 must be an integer answer e <eor> e <eos> |
e |
divide__21.0__7.0__ multiply__7.0__3.0__ |
divide__21.0__7.0__ multiply__7.0__3.0__ |
| tanks a and e are each in the shape of a right circular cylinder . the interior of tank a has a height of num__10 meters and a circumference of num__8 meters and the interior of tank e has a height of num__8 meters and a circumference of num__10 meters . the capacity of tank a is what percent of the capacity of tank e ? <o> a ) num__75.0 <o> b ) num__80.0 <o> c ) num__100.0 <o> d ) num__120.0 <o> e ) num__125 % |
b . for a r = num__4.0 pi . its capacity = ( num__4 pi ) ^ num__2 * num__10 = num__160 pi for e r = num__10 / pi . its capacity = ( num__5 pi ) ^ num__2 * num__8 = num__200 pi a / e = num__160 pi / num__200 pi = num__0.8 = b . <eor> b <eos> |
b |
subtract__10.0__8.0__ divide__10.0__2.0__ divide__8.0__10.0__ multiply__10.0__8.0__ |
divide__8.0__4.0__ divide__10.0__2.0__ divide__8.0__10.0__ multiply__10.0__8.0__ |
| the average age of a group of num__10 students was num__20 . the average age increased by num__2 years when two new students joined the group . what is the average age of the two new students who joined the group ? <o> a ) num__22 years <o> b ) num__30 years <o> c ) num__44 years <o> d ) num__32 years <o> e ) none of these |
answer the average age of a group of num__10 students is num__20 . therefore the sum of the ages of all num__10 of them = num__10 * num__20 = num__200 when two new students join the group the average age increases by num__2 . new average = num__22 . now there are num__12 students . therefore the sum of the ages of all num__12 of them = num__12 * num__22 = num__264 therefore the sum of the ages of the two new students who joined = num__264 - num__200 = num__64 and the average age of each of the two new students = num__32.0 = num__32 years . answer d <eor> d <eos> |
d |
multiply__10.0__20.0__ add__20.0__2.0__ add__10.0__2.0__ multiply__12.0__22.0__ subtract__264.0__200.0__ add__10.0__22.0__ add__10.0__22.0__ |
multiply__10.0__20.0__ add__20.0__2.0__ subtract__22.0__10.0__ multiply__12.0__22.0__ subtract__264.0__200.0__ add__10.0__22.0__ subtract__64.0__32.0__ |
| if a car had traveled num__30 kmh faster than it actually did the trip would have lasted num__30 minutes less . if the car went exactly num__90 km at what speed did it travel ? <o> a ) num__35 kmh <o> b ) num__45 kmh <o> c ) num__60 kmh <o> d ) num__75 kmh <o> e ) num__90 kmh |
time = distance / speed difference in time = num__0.5 hrs num__90 / x - num__90 / ( x + num__30 ) = num__0.5 substitute the value of x from the options . - - > x = num__60 - - > num__1.5 - num__1.0 = num__1.5 - num__1 = num__0.5 answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ divide__90.0__60.0__ subtract__1.5__0.5__ hour_to_min_conversion__ |
divide__30.0__0.5__ divide__90.0__60.0__ subtract__1.5__0.5__ divide__30.0__0.5__ |
| find the smallest number which when divided by num__11 and num__13 leaves respective remainders of num__4 and num__6 . <o> a ) num__187 <o> b ) num__197 <o> c ) num__207 <o> d ) num__136 <o> e ) num__227 |
let ' n ' is the smallest number which divided by num__11 and num__13 leaves respective remainders of num__4 and num__6 . required number = ( lcm of num__11 and num__13 ) - ( common difference of divisors and remainders ) = ( num__143 ) - ( num__7 ) = num__136 . answer : d <eor> d <eos> |
d |
multiply__11.0__13.0__ subtract__11.0__4.0__ subtract__143.0__7.0__ subtract__143.0__7.0__ |
multiply__11.0__13.0__ subtract__11.0__4.0__ subtract__143.0__7.0__ subtract__143.0__7.0__ |
| the area of a square is num__4096 sq cm . find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is num__24 cm less than the side of the square ? <o> a ) num__5 : num__16 <o> b ) num__5 : num__12 <o> c ) num__5 : num__19 <o> d ) num__5 : num__18 <o> e ) num__5 : num__14 |
let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a num__2 = num__4096 = num__212 a = ( num__212 ) num__0.5 = num__26 = num__64 l = num__2 a and b = a - num__24 b : l = a - num__24 : num__2 a = num__40 : num__128 = num__5 : num__16 answer : e <eor> e <eos> |
e |
power__4096.0__0.5__ rectangle_perimeter__24.0__40.0__ rectangle_perimeter__0.5__2.0__ triangle_area__0.5__64.0__ rectangle_perimeter__0.5__2.0__ |
power__4096.0__0.5__ rectangle_perimeter__24.0__40.0__ rectangle_perimeter__0.5__2.0__ triangle_area__0.5__64.0__ rectangle_perimeter__0.5__2.0__ |
| the number of degrees that the hour hand of a clock moves through between noon and num__2.30 in the afternoon of the same day is ? <o> a ) num__38 <o> b ) num__27 <o> c ) num__75 <o> d ) num__28 <o> e ) num__21 |
explanation : the hour hand moves from pointing to num__12 to pointing to half way between num__2 and num__3 . the angle covered between each hour marking on the clock is num__30.0 = num__30 . since the hand has covered num__2.5 of these divisions the angle moved through is num__75 . answer : c <eor> c <eos> |
c |
divide__30.0__12.0__ multiply__2.5__30.0__ round__75.0__ |
divide__30.0__12.0__ multiply__2.5__30.0__ round__75.0__ |
| num__3034 - ( num__1002 / num__200.4 ) = ? <o> a ) num__2984 <o> b ) num__3029 <o> c ) num__2982 <o> d ) num__2981 <o> e ) none of these |
num__3034 - num__5 = num__3029 answer : b <eor> b <eos> |
b |
divide__1002.0__200.4__ subtract__3034.0__5.0__ subtract__3034.0__5.0__ |
divide__1002.0__200.4__ subtract__3034.0__5.0__ subtract__3034.0__5.0__ |
| after driving to a riverfront parking lot bob plans to run south along the river turn around and return to the parking lot running north along the same path . after running num__3.25 miles south he decides to run for only num__50 minutes more . if bob runs at a constant rate of num__8 minutes per mile how many miles farther south can he run and still be able to return to the parking lot in num__50 minutes ? <o> a ) num__1.5 <o> b ) num__2.25 <o> c ) num__3.0 <o> d ) num__3.25 <o> e ) num__4.75 |
for the first num__3.25 miles bob takes num__3.25 x num__8 = num__26 minutes say he runs ' x ' miles further south . so the total distance he covers is ( num__3.25 + num__2 x + num__3.25 ) in ( num__50 + num__26 ) minutes so we have num__6.5 + num__2 x = num__9.5 = > x = num__1.5 answer a <eor> a <eos> |
a |
multiply__3.25__8.0__ multiply__3.25__2.0__ subtract__8.0__6.5__ round__1.5__ |
multiply__3.25__8.0__ multiply__3.25__2.0__ subtract__8.0__6.5__ round__1.5__ |
| there is num__60.0 increase in an amount in num__6 years at s . i . what will be the c . i . of rs . num__12000 after num__3 years at the same rate ? <o> a ) rs . num__3389 <o> b ) rs . num__3923 <o> c ) rs . num__3972 <o> d ) rs . num__3928 <o> e ) rs . num__3929 |
let p = rs . num__100 . then s . i . rs . num__60 and t = num__6 years . r = ( num__100 * num__60 ) / ( num__100 * num__6 ) = num__10.0 p . a . now p = rs . num__12000 t = num__3 years and r = num__10.0 p . a . c . i . = [ num__12000 * { ( num__1 + num__0.1 ) num__3 - num__1 } ] = num__12000 * num__0.331 = rs . num__3972 answer : c <eor> c <eos> |
c |
percent__1.0__10.0__ percent__100.0__3972.0__ |
percent__1.0__10.0__ percent__100.0__3972.0__ |
| a garrison of num__2000 men has provisions for num__54 days . at the end of num__15 days a reinforcement arrives and it is now found that the provisions will last only for num__20 days more . what is the reinforcement ? <o> a ) num__1898 <o> b ) num__9801 <o> c ) num__1987 <o> d ) num__1900 <o> e ) num__8373 |
num__2000 - - - - num__54 num__2000 - - - - num__39 x - - - - - num__20 x * num__20 = num__2000 * num__39 x = num__3900 num__2000 - - - - - - - num__1900 answer : d <eor> d <eos> |
d |
subtract__54.0__15.0__ subtract__3900.0__2000.0__ round__1900.0__ |
subtract__54.0__15.0__ subtract__3900.0__2000.0__ subtract__3900.0__2000.0__ |
| find the length of the wire required to go num__15 times round a square field containing num__69696 m num__2 . <o> a ) num__15840 m <o> b ) num__16840 m <o> c ) num__15820 m <o> d ) num__15640 m <o> e ) num__16640 m |
a num__2 = num__69696 = > a = num__264 num__4 a = num__1056 num__1056 * num__15 = num__15840 answer : a <eor> a <eos> |
a |
square_perimeter__264.0__ multiply__15.0__1056.0__ multiply__15.0__1056.0__ |
multiply__4.0__264.0__ multiply__15.0__1056.0__ multiply__15.0__1056.0__ |
| ram get num__450 marks in his exam which is num__90.0 of total marks . what is the total marks ? <o> a ) num__475 <o> b ) num__600 <o> c ) num__550 <o> d ) num__500 <o> e ) num__525 |
x * ( num__0.9 ) = num__450 x = num__5 * num__100 x = num__500 answer : d <eor> d <eos> |
d |
divide__450.0__90.0__ divide__90.0__0.9__ divide__450.0__0.9__ divide__450.0__0.9__ |
divide__450.0__90.0__ divide__90.0__0.9__ multiply__100.0__5.0__ multiply__100.0__5.0__ |
| if jos é reads at a constant rate of num__2 pages every num__6 minutes how many seconds will it take him to read n pages ? <o> a ) num__180 <o> b ) num__2 n <o> c ) num__2.5 * n <o> d ) num__24 n <o> e ) num__150 |
jose would read num__1 page in num__3.0 min jose would read n page in ( num__3.0 ) * n min i . e . ( num__3.0 ) * n * num__60 seconds = num__180 n seconds . option a is the correct answer . <eor> a <eos> |
a |
add__2.0__1.0__ hour_to_min_conversion__ multiply__3.0__60.0__ round__180.0__ |
add__2.0__1.0__ hour_to_min_conversion__ multiply__3.0__60.0__ multiply__3.0__60.0__ |
| how much time will take for an amount of rs . num__320 to yield rs . num__81 as interest at num__4.5 per annum of simple interest ? <o> a ) num__8 years <o> b ) num__4 years <o> c ) num__3 years <o> d ) num__5 years num__7 months <o> e ) num__5 years |
time = ( num__100 * num__81 ) / ( num__320 * num__4.5 ) = num__5 years num__7 months answer : d <eor> d <eos> |
d |
percent__100.0__5.0__ |
percent__100.0__5.0__ |
| a sum of money is to be distributed among a b c d in the proportion of num__5 : num__2 : num__4 : num__3 . if c gets $ num__500 more than d what is a ' s share ? <o> a ) $ num__1200 <o> b ) $ num__1600 <o> c ) $ num__2000 <o> d ) $ num__2500 <o> e ) $ num__3000 |
let the shares of a b c and d be num__5 x num__2 x num__4 x and num__3 x respectively . then num__4 x - num__3 x = num__500 x = $ num__500 a ' s share = num__5 x = num__5 * $ num__500 = $ num__2500 the answer is d . <eor> d <eos> |
d |
multiply__5.0__500.0__ multiply__5.0__500.0__ |
multiply__5.0__500.0__ multiply__5.0__500.0__ |
| for each num__6 - month period during a light bulb ' s life span the odds of it not burning out from over - use are half what they were in the previous num__6 - month period . if the odds of a light bulb burning out during the first num__6 - month period following its purchase are num__0.666666666667 what are the odds of it burning out during the period from num__6 months to num__1 year following its purchase ? <o> a ) num__0.185185185185 <o> b ) num__0.222222222222 <o> c ) num__0.333333333333 <o> d ) num__0.444444444444 <o> e ) num__0.285714285714 |
p ( of not burning out in a six mnth period ) = num__0.5 of p ( of not burning out in prev num__6 mnth period ) p ( of burning out in num__1 st num__6 mnth ) = num__0.666666666667 - - - > p ( of not burning out in num__1 st num__6 mnth ) = num__1 - num__0.666666666667 = num__0.333333333333 - - - - > p ( of not burning out in a six mnth period ) = num__0.5 * num__0.333333333333 = num__0.166666666667 - - - > p ( of burning out in a six mnth period ) = num__1 - num__0.333333333333 = num__0.666666666667 now p ( of burning out in num__2 nd six mnth period ) = p ( of not burning out in num__1 st six mnth ) * p ( of burning out in a six mnth ) = num__0.666666666667 * num__0.166666666667 = num__0.285714285714 ans e <eor> e <eos> |
e |
multiply__0.6667__0.5__ reverse__6.0__ reverse__0.5__ multiply__1.0__0.2857__ |
multiply__0.6667__0.5__ subtract__0.6667__0.5__ reverse__0.5__ multiply__1.0__0.2857__ |
| a train covers a distance of num__6 km in num__10 min . if it takes num__12 sec to pass a telegraph post then the length of the train is ? <o> a ) num__120 m <o> b ) num__180 m <o> c ) num__240 m <o> d ) num__220 m <o> e ) num__280 m |
speed = ( num__0.6 * num__60 ) km / hr = ( num__36 * num__0.277777777778 ) m / sec = num__10 m / sec . length of the train = num__10 * num__12 = num__120 m . answer : a <eor> a <eos> |
a |
km_to_mile_conversion__ hour_to_min_conversion__ multiply__0.6__60.0__ divide__10.0__36.0__ multiply__10.0__12.0__ round__120.0__ |
divide__6.0__10.0__ multiply__6.0__10.0__ multiply__0.6__60.0__ divide__10.0__36.0__ multiply__10.0__12.0__ multiply__10.0__12.0__ |
| during num__2005 a company produced an average of num__2500 products per month . how many products will the company need to produce from num__2006 through num__2008 in order to increase its monthly average for the period from num__2005 through num__2008 by num__300.0 over its num__2005 average ? <o> a ) num__450000 <o> b ) num__475000 <o> c ) num__500000 <o> d ) num__525000 <o> e ) num__550 |
000 |
company produced num__12 * num__2500 = num__30000 products in num__2005 . if company produces x products from num__2006 to num__2008 then total amount of product produced in num__4 years ( num__2005 through num__2008 ) is x + num__30000 . the gives the average of ( x + num__30000 ) / num__4 . this average needs to be num__200.0 higher than that in num__2005 . in math terms num__30000 + num__300.0 ( num__30000 ) = num__120000 . so : ( x + num__30000 ) / num__4 = num__120000 x + num__30000 = num__480000 x = num__450000 the answer is a . <eor> a <eos> |
a |
a |
| yearly subscription to professional magazines cost a company $ num__940.00 . to make a num__50.0 cut in the magazine budget how much less must be spent ? <o> a ) num__654 <o> b ) num__655 <o> c ) num__656 <o> d ) num__657 <o> e ) num__470 |
total cost num__940 num__940 * num__0.5 = num__470 so the cut in amount is num__470 the less amount to be spend is num__940 - num__470 = num__470 answer : e <eor> e <eos> |
e |
multiply__940.0__0.5__ multiply__940.0__0.5__ |
multiply__940.0__0.5__ multiply__940.0__0.5__ |
| the weight of four dogs is determined to be num__42 pounds num__38 pounds num__35 pounds and num__41 pounds respectively . the weight of a fifth dog is determined to be y pounds . if the average ( arithmetic mean ) weight of the first four dogs is the same as that of all five dogs what is the value of y ? <o> a ) num__31 <o> b ) num__33 <o> c ) num__35 <o> d ) num__37 <o> e ) num__39 |
total weight of the num__4 dogs = ( num__42 + num__38 + num__35 + num__41 ) = num__156 avg = num__39.0 = num__39 total weight of num__5 dogs = num__156 + y or num__4 ( num__39 ) + y average of num__5 dogs as per question = num__39 equation : num__4 ( num__39 ) + y = num__5 ( num__39 ) or y = num__39 . choose e <eor> e <eos> |
e |
subtract__42.0__38.0__ add__35.0__4.0__ add__35.0__4.0__ |
subtract__42.0__38.0__ add__35.0__4.0__ add__35.0__4.0__ |
| by selling num__150 mangoes a fruit - seller gains the selling price of num__30 mangoes . find the gain percent ? <o> a ) num__22 <o> b ) num__25 <o> c ) num__77 <o> d ) num__88 <o> e ) num__99 |
sp = cp + g num__150 sp = num__150 cp + num__30 sp num__120 sp = num__150 cp num__120 - - - num__30 cp num__100 - - - ? = > num__25.0 answer : b <eor> b <eos> |
b |
percent__25.0__100.0__ |
percent__25.0__100.0__ |
| a qualified worker digs a well in num__5 hours . he invites num__2 apprentices both capable of working num__0.75 as fast and num__3 trainees both working num__0.2 as fast as he . if the five - person team digs the same well how much time does the team need to finish the job ? <o> a ) num__1 : num__56 <o> b ) num__1 : num__34 <o> c ) num__1 : num__44 <o> d ) num__1 : num__54 <o> e ) num__2 : num__14 |
and answer is correct as you have calculated : num__2.08333333333 hrs that is num__1.88 hrs - num__1 hr and num__56 minutes so num__1 : num__56 hence a <eor> a <eos> |
a |
round_down__1.88__ reverse__1.0__ |
subtract__3.0__2.0__ subtract__2.0__1.0__ |
| due to construction the speed limit along an num__5 - mile section of highway is reduced from num__40 miles per hour to num__22 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? <o> a ) a ) num__3.12 <o> b ) b ) num__8 <o> c ) c ) num__10 <o> d ) d ) num__15 <o> e ) e ) num__6.13 |
old time in minutes to cross num__5 miles stretch = num__5 * num__1.5 = num__5 * num__1.5 = num__7.5 new time in minutes to cross num__5 miles stretch = num__5 * num__2.72727272727 = num__5 * num__2.72727272727 = num__13.63 time difference = num__6.13 ans : e <eor> e <eos> |
e |
multiply__5.0__1.5__ subtract__13.63__7.5__ round__6.13__ |
multiply__5.0__1.5__ subtract__13.63__7.5__ round__6.13__ |
| a train num__400 m long can cross an electric pole in num__20 sec and then find the speed of the train ? <o> a ) num__76 kmph <o> b ) num__88 kmph <o> c ) num__72 kmph <o> d ) num__55 kmph <o> e ) num__44 kmph |
length = speed * time speed = l / t s = num__20.0 s = num__20 m / sec speed = num__20 * num__3.6 ( to convert m / sec in to kmph multiply by num__3.6 ) speed = num__72 kmph answer : c <eor> c <eos> |
c |
multiply__20.0__3.6__ round__72.0__ |
multiply__20.0__3.6__ multiply__20.0__3.6__ |
| a fruit seller had some apples . he sells num__40.0 apples and still has num__420 apples . originally he had : <o> a ) num__701 apples <o> b ) num__708 apples <o> c ) num__720 apples <o> d ) num__730 apples <o> e ) num__700 apples |
e suppose originally he had x apples . then ( num__100 - num__40 ) % of x = num__420 . num__0.6 x x = num__420 x = ( num__420 x num__100 ) / num__60 = num__700 . <eor> e <eos> |
e |
percent__100.0__700.0__ |
percent__100.0__700.0__ |
| a person crosses a num__500 m long street in num__5 minnutes . what is his speed in km per hour ? <o> a ) num__6.01 km / hr <o> b ) num__7.2 km / hr <o> c ) num__9 km / hr <o> d ) num__2.5 km / hr <o> e ) num__3 km / hr |
speed = num__100.0 * num__60 = num__1.67 m / sec = num__1.67 * num__3.6 = num__6.01 km / hr answer is a <eor> a <eos> |
a |
divide__500.0__5.0__ hour_to_min_conversion__ round__6.01__ |
divide__500.0__5.0__ hour_to_min_conversion__ round__6.01__ |
| if num__8 workers can build num__8 cars in num__8 days then how many days would it take num__4 workers to build num__4 cars ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__8 <o> d ) num__12 <o> e ) num__16 |
num__8 workers can build num__1 car per day on average . num__1 worker can build num__0.125 of a car per day . num__4 workers can build num__0.5 car per day . the time required to build num__4 cars is num__4 / ( num__0.5 ) = num__8 days the answer is c . <eor> c <eos> |
c |
divide__1.0__8.0__ divide__4.0__8.0__ round__8.0__ |
divide__1.0__8.0__ divide__4.0__8.0__ divide__8.0__1.0__ |
| a watch was sold at a loss of num__10.0 . if it was sold for rs . num__140 more there would have been a gain of num__4.0 . what is the cost price ? <o> a ) num__1000 <o> b ) num__2996 <o> c ) num__2686 <o> d ) num__1028 <o> e ) num__1013 |
num__90.0 num__104.0 - - - - - - - - num__14.0 - - - - num__140 num__100.0 - - - - ? = > rs . num__1000 answer : a <eor> a <eos> |
a |
percent__10.0__140.0__ percent__100.0__1000.0__ |
percent__10.0__140.0__ percent__100.0__1000.0__ |
| the owner of a furniture shop charges his customer num__18.0 more than the cost price . if a customer paid rs . num__6000 for a computer table then what was the cost price of the computer table ? <o> a ) rs . num__6289 <o> b ) rs . num__6298 <o> c ) rs . num__6290 <o> d ) rs . num__5084 <o> e ) rs . num__6708 |
cp = sp * ( num__100 / ( num__100 + profit % ) ) = num__6000 ( num__0.847457627119 ) = rs . num__5084 . answer : d <eor> d <eos> |
d |
percent__100.0__5084.0__ |
percent__100.0__5084.0__ |
| if b is the center of the circle in the figure above and the area of the shaded region is num__16 π what is the length of arc adc ? <o> a ) num__4 <o> b ) num__4 π <o> c ) num__16 <o> d ) num__8 π <o> e ) num__16 |
area of num__0.25 th of the circle is num__16 pi so area of the circle = num__64 pi radius of circle = num__8 units circumference = num__2 pi * radius = num__16 pi num__0.25 th the circumference = required length of adc = num__4 pi ans : b <eor> b <eos> |
b |
square_perimeter__16.0__ triangle_area__0.25__64.0__ power__16.0__0.25__ multiply__16.0__0.25__ multiply__16.0__0.25__ |
square_perimeter__16.0__ triangle_area__0.25__64.0__ multiply__0.25__8.0__ multiply__16.0__0.25__ multiply__16.0__0.25__ |
| two trains num__137 meters and num__163 meters in length are running towards each other on parallel lines one at the rate of num__42 kmph and another at num__48 kmpb . in what time will they be clear of each other from the moment they meet <o> a ) num__8 sec <o> b ) num__10 sec <o> c ) num__12 sec <o> d ) num__14 sec <o> e ) num__16 sec |
explanation : relative speed of the trains = ( num__42 + num__48 ) kmph = num__90 kmph = ( num__90 × num__0.277777777778 ) m / sec = num__25 m / sec . time taken by the trains to ‘ pass each other = time taken to cover ( num__137 + num__163 ) m at num__25 m / sec = ( num__12.0 ) sec = num__12 sec answer : option c <eor> c <eos> |
c |
add__42.0__48.0__ round__12.0__ |
add__42.0__48.0__ round__12.0__ |
| rodrick mixes a martini that has a volume of ' n ' ounces having num__37.0 vermouth and num__60.0 gin by volume . he wants to change it so that the martini is num__25.0 vermouth by volume . how many ounces of gin must he add ? <o> a ) n / num__6 <o> b ) n / num__2 <o> c ) num__3 n / num__5 <o> d ) num__5 n / num__6 <o> e ) num__8 n / num__5 |
total v g num__1 ounce num__0.37 num__0.6 n ounce num__0.37 n num__0.6 n - - - - - - - - - - - - - initial expression lets say g ounces of gin is added to this mixture n + g num__0.37 n num__0.6 n + g - - - - - - - - - - - - - - final expression given that after adding g ounces of gin v should become num__25.0 of the total volume . = > volume of v / total volume = num__0.25 = > num__0.37 n / n + g = num__0.25 = > num__1.48 n = n + g = > g = n / num__2 answer is b . note that after we add pure gin the volume of vermouth will remain the same . based on this set the equation : num__0.37 n = num__0.25 ( n + g ) - - > g = n / num__2 answer : b . <eor> b <eos> |
b |
divide__37.0__25.0__ multiply__1.0__2.0__ |
divide__37.0__25.0__ divide__2.0__1.0__ |
| the ages of smirthi and samantha vary by num__16 years six years ago deepika ’ s age was three times as that of smirthi ’ s find their current ages . <o> a ) num__20 <o> b ) num__50 <o> c ) num__40 <o> d ) num__30 <o> e ) num__45 |
d num__30 let smirthi ’ s age = a years so deepika ’ s age = ( a + num__16 ) years also num__3 ( a – num__6 ) = a + num__16 – num__6 or a = num__14 smirthi ’ s age = num__14 years and deepika ’ s age = num__14 + num__16 = num__30 years <eor> d <eos> |
d |
subtract__30.0__16.0__ add__16.0__14.0__ |
subtract__30.0__16.0__ add__16.0__14.0__ |
| what is the sum of two consecutive even numbers the difference of whose squares is num__84 ? <o> a ) num__34 <o> b ) num__38 <o> c ) num__42 <o> d ) num__46 <o> e ) none |
explanation let the numbers be x and x + num__2 . then ( x + num__2 ) num__2 – x num__2 = num__84 num__4 x + num__4 = num__84 num__4 x = num__80 x = num__20 . the required sum = x + ( x + num__2 ) = num__2 x + num__2 = num__42 . answer c <eor> c <eos> |
c |
subtract__84.0__4.0__ divide__80.0__4.0__ divide__84.0__2.0__ divide__84.0__2.0__ |
subtract__84.0__4.0__ divide__80.0__4.0__ divide__84.0__2.0__ divide__84.0__2.0__ |
| four friends patricia melyssa tania and cassandra are pooling their money to buy a $ num__1100 item . patricia has three times as much money as melyssa . tania has $ num__20 more than patricia . cassandra has num__25.0 more than tania . if they put all their money together and spend the $ num__1100 they will have $ num__19 left . how much money does patricia have ? <o> a ) $ num__275 <o> b ) $ num__300 <o> c ) $ num__400 <o> d ) $ num__350 <o> e ) $ num__220 |
p = num__3 m ; t = p + num__20 ; c = num__1.25 ( t ) p + m + t + c - num__1100 = num__19 p + ( num__0.333333333333 ) p + p + num__20 + num__1.25 ( p + num__20 ) = num__1119 p + ( num__0.333333333333 ) p + p + num__20 + num__1.25 p + num__25 = num__1119 p + ( num__0.333333333333 ) p + p + num__1.25 p + num__45 = num__1119 p + ( num__0.333333333333 ) p + p + num__1.25 p = num__1074 num__3.58 p = num__1074 p = num__300 answer : b <eor> b <eos> |
b |
divide__25.0__20.0__ reverse__3.0__ add__1100.0__19.0__ add__20.0__25.0__ subtract__1119.0__45.0__ divide__1074.0__3.58__ divide__1074.0__3.58__ |
divide__25.0__20.0__ reverse__3.0__ add__1100.0__19.0__ add__20.0__25.0__ subtract__1119.0__45.0__ divide__1074.0__3.58__ divide__1074.0__3.58__ |
| num__24 lbs of coffee p and num__25 lbs of coffee v are mixed to make coffee x and y . the ratio of p to v in coffee x is num__4 to num__1 in y is num__1 to num__5 . how much of p is contained in the mixture x ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__30 <o> d ) num__40 <o> e ) num__50 |
let there be num__4 x lbs of p in x and x lbs of v in x let there be y lbs of p in y and num__5 y lbs of v in y now total p is num__4 x + y = num__24 and total q is x + num__5 y = num__25 solving the two equations we get x = num__5 and y = num__4 mixture x has num__4 x lbs of p = num__4 x = num__4 * num__5 = num__20 lbs answer : b <eor> b <eos> |
b |
subtract__24.0__4.0__ subtract__24.0__4.0__ |
multiply__4.0__5.0__ multiply__4.0__5.0__ |
| num__3550 - ( num__1002 / num__20.04 ) = ? <o> a ) num__2984 <o> b ) num__2983 <o> c ) num__3500 <o> d ) num__2981 <o> e ) none of these |
num__3550 - num__50 = num__3500 answer : c <eor> c <eos> |
c |
divide__1002.0__20.04__ subtract__3550.0__50.0__ subtract__3550.0__50.0__ |
divide__1002.0__20.04__ subtract__3550.0__50.0__ subtract__3550.0__50.0__ |
| if num__5 ^ num__11 * num__4 ^ num__6 = num__2 * num__10 ^ n what is the value of n ? <o> a ) num__11 <o> b ) num__21 <o> c ) num__22 <o> d ) num__23 <o> e ) num__32 |
my attempt : num__5 ^ ( num__11 ) * num__4 ^ ( num__6 ) = num__2 * num__10 ^ p num__5 ^ ( num__11 ) * num__2 ^ ( num__12 ) = num__2 * ( num__2 * num__5 ) ^ p num__5 ^ ( num__11 ) * num__2 ^ ( num__12 ) = num__2 * num__2 ^ ( p ) * num__5 ^ ( p ) num__5 ^ ( num__11 ) * num__2 ^ ( num__12 ) = num__2 ^ ( p + num__1 ) * num__5 ^ ( p ) = > ( p + num__1 ) = num__12 = > p = num__11 answer : num__11 ( a ) <eor> a <eos> |
a |
multiply__6.0__2.0__ subtract__5.0__4.0__ add__5.0__6.0__ |
multiply__6.0__2.0__ subtract__5.0__4.0__ add__5.0__6.0__ |
| which of the following equations represents a line that is perpendicular to the line described by the equation num__4 x + num__3 y = num__12 ? <o> a ) num__3 x + num__4 y = num__11 <o> b ) num__3 x – num__4 y = num__12 <o> c ) num__5 y – num__3 x = num__6 <o> d ) num__1.5 y + num__2 x = num__8 <o> e ) num__8 x – num__6 y = num__2 |
perpendicular lines have slopes which are negative reciprocals of each other . the line num__4 x + num__3 y = num__12 in slope - intercept form is y = - ( num__1.33333333333 ) x + num__4 the slope of this line is - num__1.33333333333 . the slope of a perpendicular line is num__0.75 . num__3 x - num__4 y = num__12 in slope - intercept form is y = ( num__0.75 ) x - num__3 the answer is b . <eor> b <eos> |
b |
divide__4.0__3.0__ reverse__1.3333__ multiply__4.0__0.75__ |
divide__4.0__3.0__ reverse__1.3333__ multiply__4.0__0.75__ |
| convert num__500 miles into meters ? <o> a ) num__784596 <o> b ) num__845796 <o> c ) num__804670 <o> d ) num__784596 <o> e ) num__864520 |
num__1 mile = num__1609.34 meters num__500 mile = num__500 * num__1609.34 = num__804670 meters answer is c <eor> c <eos> |
c |
multiply__500.0__1609.34__ round__804670.0__ |
multiply__500.0__1609.34__ multiply__500.0__1609.34__ |
| square a has an area of num__121 square centimeters . square b has a perimeter of num__16 centimeters . if square b is placed within square a and a random point is chosen within square a what is the probability the point is not within square b ? <o> a ) num__0.36 <o> b ) num__0.2 <o> c ) num__0.132231404959 <o> d ) num__0.867768595041 <o> e ) num__0.24 |
i guess it ' s mean that square b is placed within square aentirely . since the perimeter of b is num__16 then its side is num__4.0 = num__4 and the area is num__4 ^ num__2 = num__16 ; empty space between the squares is num__121 - num__16 = num__105 square centimeters so if a random point is in this area then it wo n ' t be within square b : p = favorable / total = num__0.867768595041 . answer : d . <eor> d <eos> |
d |
triangle_area__2.0__0.8678__ |
triangle_area__2.0__0.8678__ |
| speed of a boat in standing water is num__9 kmph and speed of the stream is num__1.5 kmph . a man can rows to a place at a distance of num__105 km and comes back to the starting point . the total time taken by him is ? <o> a ) num__12 hours <o> b ) num__24 hours <o> c ) num__36 hours <o> d ) num__10 hours <o> e ) num__15 hours |
speed upstream = num__7.5 kmph speed downstream = num__10.5 kmph total time taken = num__105 / num__7.5 + num__105 / num__10.5 = num__24 hours answer is b <eor> b <eos> |
b |
subtract__9.0__1.5__ add__9.0__1.5__ round__24.0__ |
subtract__9.0__1.5__ add__9.0__1.5__ round__24.0__ |
| look at this series : num__80 num__10 num__70 num__15 num__60 . . . what number should come next ? <o> a ) num__20 <o> b ) num__23 <o> c ) num__25 <o> d ) num__30 <o> e ) num__35 |
this is an alternating addition and subtraction series . in the first pattern num__10 is subtracted from each number to arrive at the next . in the second num__5 is added to each number to arrive at the next . answer a <eor> a <eos> |
a |
subtract__15.0__10.0__ subtract__80.0__60.0__ |
subtract__15.0__10.0__ subtract__80.0__60.0__ |
| a thief steals a car at num__2.30 pm and drives it at num__60 kmph . the theft is discovered at num__3 pm and the owner sets off in another car at num__75 kmph when will he overtake the thief <o> a ) num__3.3 pm <o> b ) num__5.5 pm <o> c ) num__4.9 pm <o> d ) num__5 pm <o> e ) num__5.2 pm |
let the thief is overtaken x hrs after num__2.30 pm distance covered by the thief in x hrs = distance covered by the owner in x - num__0.5 hr num__60 x = num__75 ( x - ½ ) x = num__2.5 hr thief is overtaken at num__2.30 pm + num__2 ½ hr = num__5 pm answer is d . <eor> d <eos> |
d |
subtract__3.0__0.5__ subtract__2.5__0.5__ add__3.0__2.0__ round__5.0__ |
subtract__3.0__0.5__ subtract__2.5__0.5__ add__3.0__2.0__ add__3.0__2.0__ |
| a searchlight on top of the watch - tower makes num__3 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least num__10 seconds ? <o> a ) num__0.25 <o> b ) num__0.333333333333 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__0.75 |
num__3 revolutions per minute = num__1 revolution every num__20 seconds so no matter what anybody appearing at the tower can not stay in the dark for more than num__20 seconds . this will be our total number of possibilities i . e the denominator . p ( man in dark for at least num__5 seconds ) = num__1 - p ( man in dark for max of num__10 seconds ) = num__1 - num__0.5 = num__1 - num__0.5 = num__0.5 or the other way would be : p ( man in dark for at least num__5 seconds ) is like saying he can be in dark for num__5 num__67 . . . all the way to num__20 seconds because that is the max . in this approach it would be num__0.5 seconds = num__0.5 . answer is c <eor> c <eos> |
c |
divide__10.0__20.0__ round__0.5__ |
divide__10.0__20.0__ subtract__1.0__0.5__ |
| a number when divided by a divisor leaves a remainder of num__24 . when twice the original number is divided by the same divisor the remainder is num__11 . what is the value of the divisor ? <o> a ) num__29 <o> b ) num__33 <o> c ) num__37 <o> d ) num__43 <o> e ) num__45 |
let the original number be ' a ' let the divisor be ' d ' let the quotient of the division of aa by dd be ' x ' therefore we can write the relation as a / d = x and the remainder is num__24 . i . e . a = dx + num__24 when twice the original number is divided by d num__2 a is divided by d . we know that a = dx + num__24 . therefore num__2 a = num__2 dx + num__48 the problem states that ( num__2 dx + num__48 ) / d leaves a remainder of num__11 . num__2 dx num__2 dx is perfectly divisible by d and will therefore not leave a remainder . the remainder of num__11 was obtained by dividing num__48 by d . when num__48 is divided by num__37 the remainder that one will obtain is num__11 . hence the divisor is num__37 . c ) <eor> c <eos> |
c |
multiply__24.0__2.0__ subtract__48.0__11.0__ subtract__48.0__11.0__ |
multiply__24.0__2.0__ subtract__48.0__11.0__ subtract__48.0__11.0__ |
| the average age of an adult class is num__40 years . num__12 new students with an avg age of num__32 years join the class . thereforedecreasing the average by num__4 year . find what was theoriginal strength of class ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__16 <o> d ) num__20 <o> e ) num__22 |
let original strength = y then num__40 y + num__12 x num__32 = ( y + num__12 ) x num__36 ⇒ num__40 y + num__384 = num__36 y + num__432 ⇒ num__4 y = num__48 ∴ y = num__12 b <eor> b <eos> |
b |
subtract__40.0__4.0__ multiply__12.0__32.0__ multiply__12.0__36.0__ multiply__12.0__4.0__ divide__384.0__32.0__ |
add__32.0__4.0__ multiply__12.0__32.0__ multiply__12.0__36.0__ add__12.0__36.0__ divide__384.0__32.0__ |
| how many integers from num__0 to num__59 inclusive have a remainder of num__1 when divided by num__3 ? <o> a ) num__15 <o> b ) num__16 <o> c ) num__17 <o> d ) num__18 <o> e ) num__20 |
my ans is also c . num__17 . explanation : num__1 also gives num__1 remainder when divided by num__3 another number is num__4 then num__7 and so on . hence we have an arithmetic progression : num__1 num__4 num__7 num__10 . . . . . num__58 which are in the form num__3 n + num__1 . now we have to find out number of terms . tn = a + ( n - num__1 ) d where tn is the nth term of an ap a is the first term and d is the common difference . so num__58 = num__1 + ( n - num__1 ) num__3 or ( n - num__1 ) num__3 = num__57 or n - num__1 = num__19 or n = num__20 e <eor> e <eos> |
e |
add__1.0__3.0__ add__3.0__4.0__ add__3.0__7.0__ subtract__59.0__1.0__ subtract__58.0__1.0__ divide__57.0__3.0__ add__1.0__19.0__ add__1.0__19.0__ |
add__1.0__3.0__ add__3.0__4.0__ add__3.0__7.0__ subtract__59.0__1.0__ subtract__58.0__1.0__ divide__57.0__3.0__ add__1.0__19.0__ add__1.0__19.0__ |
| ravi can do a piece of work in num__25 days while prakash can do it in num__40 days . in how many days will they finish it together ? <o> a ) num__15 num__0.384615384615 days <o> b ) num__16 num__0.0769230769231 days <o> c ) num__17 num__0.0769230769231 days <o> d ) num__17 num__0.466666666667 days <o> e ) num__18 num__0.0769230769231 days |
num__0.04 + num__0.025 = num__0.065 num__15.3846153846 = num__15 num__0.384615384615 days answer : a <eor> a <eos> |
a |
add__0.025__0.04__ subtract__40.0__25.0__ divide__0.025__0.065__ round__15.0__ |
add__0.025__0.04__ subtract__40.0__25.0__ divide__0.025__0.065__ round__15.0__ |
| once a week susan collects wildflowers around her neighborhood to display in the vase on her family ' s kitchen table . when her dog rocko tags along she gets distracted by his barking so she collects about num__0.75 as many flowers as she usually would when she comes alone . when her father comes along he collects twice as many as she does . last week susan brought rocko with her and collected num__39 flowers . if susan and her father go next week without rocco how many total flowers will they bring home ? <o> a ) num__156 <o> b ) num__104 <o> c ) num__52 <o> d ) num__78 <o> e ) num__117 |
x equals number of flowers susan collects while alone knowing num__39 is num__3 equal fourths of x i can divide num__39 by num__3 to find the value of each fourth . num__0.25 of x = num__39 divided by num__3 num__0.25 of x = num__13 x = num__13 x num__4 x = num__52 susan ' s father collects twice as many flowers as susan which would be num__52 x num__2 = num__104 final calculation is susan ' s flowers plus her father ' s flowers num__52 + num__104 = num__156 flowers the correction answer is option a ) num__156 <eor> a <eos> |
a |
divide__0.75__3.0__ divide__39.0__3.0__ reverse__0.25__ divide__39.0__0.75__ multiply__2.0__52.0__ divide__39.0__0.25__ divide__39.0__0.25__ |
divide__0.75__3.0__ divide__39.0__3.0__ reverse__0.25__ add__39.0__13.0__ multiply__2.0__52.0__ add__104.0__52.0__ add__104.0__52.0__ |
| the areas of the two spheres are in the ratio num__1 : num__4 . the ratio of their volume is ? <o> a ) num__1 : num__9 <o> b ) num__1 : num__4 <o> c ) num__1 : num__8 <o> d ) num__1 : num__1 <o> e ) num__1 : num__2 |
num__4 π r num__12 : num__4 π r num__22 = num__1 : num__4 r num__1 : r num__2 = num__1 : num__2 num__1.33333333333 π r num__13 : num__1.33333333333 π r num__23 r num__13 : r num__23 = num__1 : num__8 answer : c <eor> c <eos> |
c |
triangle_area__1.0__4.0__ square_perimeter__2.0__ volume_cube__1.0__ |
triangle_area__1.0__4.0__ square_perimeter__2.0__ volume_cube__1.0__ |
| a car gets num__40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel num__180 kilometers ? <o> a ) num__4.5 <o> b ) num__5.5 <o> c ) num__6.5 <o> d ) num__7.5 <o> e ) num__8.5 |
each num__40 kilometers num__1 gallon is needed . we need to know how many num__40 kilometers are there in num__180 kilometers ? num__180 ÷ num__40 = num__4.5 × num__1 gallon = num__4.5 gallons correct answer a <eor> a <eos> |
a |
divide__180.0__40.0__ round__4.5__ |
divide__180.0__40.0__ round__4.5__ |
| a car during its journey travels num__1 hour at a speed of num__40 kmph another num__30 minutes at a speed of num__60 kmph and num__2 hours at a speed of num__60 kmph . the average speed of the car is <o> a ) num__63.07 kmph <o> b ) num__54.28 kmph <o> c ) num__62.02 kmph <o> d ) num__64.02 kmph <o> e ) none of these |
first car travels num__1 hrs at speed of num__40 kmph distance = num__40 x num__1 = num__40 m then car travels num__30 min at a speed of num__60 kmph distance = num__30 min at speed of num__60 kmph distance = num__60 x num__0.5 = num__30 m at last it travels num__2 hours at speed of num__60 kmph distance = num__60 x num__2 = num__120 m total distance = num__40 + num__30 + num__120 = num__190 total time = num__1 + num__0.5 + num__2 = num__3.50 average speed of the car = num__190 / num__3.25 = num__54.28 answer : b <eor> b <eos> |
b |
divide__1.0__2.0__ multiply__60.0__2.0__ round__54.28__ |
divide__1.0__2.0__ divide__60.0__0.5__ divide__54.28__1.0__ |
| if num__12 men do a work in num__80 days in how many days will num__16 men do it ? <o> a ) num__33 days <o> b ) num__77 days <o> c ) num__99 days <o> d ) num__60 days <o> e ) num__19 days |
num__12 * num__80 = num__16 * x x = num__60 days answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ hour_to_min_conversion__ |
hour_to_min_conversion__ hour_to_min_conversion__ |
| a person purchased a computer set for rs . num__16000 and a dvd drive for rs . num__6250 . he sold both the items together for rs . num__31150 . what percentage of profit did he make ? <o> a ) num__38.0 <o> b ) num__74.0 <o> c ) num__40.0 <o> d ) num__33.0 <o> e ) num__60 % |
c num__40.0 the total cp = rs . num__16000 + rs . num__6250 = rs . num__22250 and sp = rs . num__31150 profit ( % ) = ( num__31150 - num__22250 ) / num__22250 * num__100 = num__40.0 <eor> c <eos> |
c |
percent__100.0__40.0__ |
percent__100.0__40.0__ |
| the expression ( num__12.86 × num__12.86 + num__12.86 × p + num__0.14 × num__0.14 ) will be a perfect square for p equal to <o> a ) num__0.28 <o> b ) num__0.26 <o> c ) num__1 <o> d ) num__0 <o> e ) num__2 |
explanation : num__12.86 × num__12.86 + num__12.86 × p + num__0.14 × num__0.14 = ( num__12.86 ) num__2 + num__12.86 × p + ( num__0.14 ) num__2 this can be written as ( num__12.86 + num__0.14 ) num__2 = num__132 if num__12.86 × p = num__2 × num__12.86 × num__0.14 i . e . if p = num__2 × num__0.14 = num__0.28 hence p = num__0.28 . answer : option a <eor> a <eos> |
a |
multiply__0.14__2.0__ multiply__0.14__2.0__ |
multiply__0.14__2.0__ multiply__0.14__2.0__ |
| find the largest num__4 digit number which isexactly divisible by num__88 ? <o> a ) num__8765 <o> b ) num__9543 <o> c ) num__9944 <o> d ) num__1012 <o> e ) num__2465 |
largest num__4 digit number is num__9999 after doing num__9999 ÷ num__88 we get remainder num__55 hence largest num__4 digit number exactly divisible by num__88 = num__9999 - num__55 = num__9944 c <eor> c <eos> |
c |
subtract__9999.0__55.0__ subtract__9999.0__55.0__ |
subtract__9999.0__55.0__ subtract__9999.0__55.0__ |
| it is being given that ( num__232 + num__1 ) is completely divisible by a whole number . which of the following numbers is completely divisible by this number ? <o> a ) ( num__216 + num__1 ) <o> b ) ( num__216 - num__1 ) <o> c ) ( num__7 x num__223 ) <o> d ) ( num__296 + num__1 ) <o> e ) none |
let num__232 = x . then ( num__232 + num__1 ) = ( x + num__1 ) . let ( x + num__1 ) be completely divisible by the natural number n . then ( num__296 + num__1 ) = [ ( num__232 ) num__3 ] = ( x num__3 + num__1 ) = ( x + num__1 ) ( x num__2 - x + num__1 ) which is completely divisible by n since ( x + num__1 ) is divisible by n answer : d <eor> d <eos> |
d |
subtract__3.0__1.0__ multiply__1.0__296.0__ |
subtract__3.0__1.0__ multiply__1.0__296.0__ |
| a school having four classes only have student strength of num__10 num__40 num__30 and num__20 respectively . the pass percentages of these classes are num__20.0 num__30.0 num__60.0 and num__100.0 respectively . what is the pass percentage for the entire school ? <o> a ) num__56.0 <o> b ) num__76.0 <o> c ) num__52.0 <o> d ) num__66.0 <o> e ) num__46 % |
num__20.0 of num__10 + num__30.0 of num__40 + num__60.0 of num__30 + num__100.0 of num__20 = num__2 + num__12 + num__18 + num__20 = now num__52 of num__100 = num__52.0 answer : c <eor> c <eos> |
c |
percent__10.0__20.0__ percent__40.0__30.0__ percent__30.0__60.0__ percent__100.0__52.0__ |
percent__10.0__20.0__ percent__40.0__30.0__ percent__30.0__60.0__ percent__100.0__52.0__ |
| the list price of an article is rs . num__65 . a customer pays rs . num__56.16 for it . he was given two successive discounts one of them being num__10.0 . the other discount is ? <o> a ) num__9.0 <o> b ) num__4.0 <o> c ) num__2.0 <o> d ) num__1.0 <o> e ) num__6 % |
num__65 * ( num__0.9 ) * ( ( num__100 - x ) / num__100 ) = num__56.16 x = num__4.0 answer : b <eor> b <eos> |
b |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| three friends are buying a gift for a friend . declan contributes num__4 dollars more than num__0.25 the cost of the gift ed contributes num__1 dollar less than num__0.333333333333 the cost of the gift and frank contributes the remaining num__17 dollars . what is the cost of the gift ? <o> a ) num__48 <o> b ) num__54 <o> c ) num__60 <o> d ) num__66 <o> e ) num__72 |
declan = d ed = e frank = f t = total d + e + f = t ( t / num__4 + num__4 ) + ( t / num__3 - num__1 ) + num__17 = t t = num__20 + ( num__7 t / num__12 ) num__12 t = num__20 ( num__12 ) + num__7 t num__5 t = num__5 * num__4 ( num__12 ) t = num__48 the correct answer is a . <eor> a <eos> |
a |
subtract__4.0__1.0__ add__17.0__3.0__ add__4.0__3.0__ multiply__4.0__3.0__ add__4.0__1.0__ multiply__4.0__12.0__ multiply__4.0__12.0__ |
subtract__4.0__1.0__ add__17.0__3.0__ add__4.0__3.0__ multiply__4.0__3.0__ add__4.0__1.0__ multiply__4.0__12.0__ multiply__4.0__12.0__ |
| a began business with rs . num__45000 and was joined afterwards by b with rs . num__5400 . when did b join if the profits at the end of the year were divided in the ratio of num__2 : num__1 ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
num__45 * num__12 : num__54 * x = num__2 : num__1 x = num__5 num__12 - num__5 = num__7 answer : c <eor> c <eos> |
c |
add__2.0__5.0__ add__2.0__5.0__ |
subtract__12.0__5.0__ subtract__12.0__5.0__ |
| a person got rs . num__48 more when he invested a certain sum at compound interest instead of simple interest for two years at num__8.0 p . a . find the sum ? <o> a ) num__7500 rupees <o> b ) rs . num__7000 <o> c ) rs . num__8000 <o> d ) rs . num__6500 <o> e ) none of these |
p = ( d * num__1002 ) / r num__2 = > ( num__48 * num__100 * num__100 ) / num__8 * num__8 = rs . num__7500 answer : a <eor> a <eos> |
a |
percent__100.0__7500.0__ |
percent__100.0__7500.0__ |
| x y a and b are positive integers of v . when x is divided by y the remainder is num__6 . when a is divided by b the remainder is num__9 . which of the following is not a possible value for y + b ? <o> a ) num__24 <o> b ) num__21 <o> c ) num__20 <o> d ) num__17 <o> e ) num__15 |
x y a and b are positive integers of v . when x is divided by y the remainder is num__6 . when a is divided by b the remainder is num__9 . answer : e . <eor> e <eos> |
e |
add__6.0__9.0__ |
add__6.0__9.0__ |
| a woman takes certain sum of money at num__10.0 p . a . simple interest . if she pays twice the amount takes how many years has she been paying the interest ? <o> a ) num__5 years num__7 months <o> b ) num__10 years <o> c ) num__15 years <o> d ) num__6 years <o> e ) num__20 years |
solution : simple interest question b = a [ num__1 + ( tr / num__100 ) ] ; where a is principal amount b final amount t time in years and r interest rate after putting the values num__10 num__3 = [ num__1 + ( num__10 t / num__100 ) ] t = num__20 which is num__20 years answer is e <eor> e <eos> |
e |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| find large number from below question the difference of two numbers is num__1365 . on dividing the larger number by the smaller we get num__7 as quotient and the num__15 as remainder <o> a ) num__1235 <o> b ) num__1346 <o> c ) num__1590 <o> d ) num__1635 <o> e ) num__1489 |
let the smaller number be x . then larger number = ( x + num__1365 ) . x + num__1365 = num__7 x + num__15 num__6 x = num__1350 x = num__225 large number = num__225 + num__1365 = num__1590 c <eor> c <eos> |
c |
subtract__1365.0__15.0__ divide__1350.0__6.0__ add__1365.0__225.0__ add__1365.0__225.0__ |
subtract__1365.0__15.0__ divide__1350.0__6.0__ add__1365.0__225.0__ add__1365.0__225.0__ |
| the value of num__3 z + z ( zz ) ( zz ) when z = num__5 is : <o> a ) num__3150 <o> b ) num__3200 <o> c ) num__3140 <o> d ) num__3650 <o> e ) num__3104 |
solution : num__3 z + z ( zz ) ( zz ) put the value of z = num__5 in the above expression we get num__15 + num__5 ( num__55 ) ( num__55 ) = num__15 + num__5 ( num__5 Ã — num__5 ) ( num__5 Ã — num__5 ) = num__15 + num__3125 = num__3140 = num__3140 answer : c <eor> c <eos> |
c |
multiply__3.0__5.0__ add__15.0__3125.0__ add__15.0__3125.0__ |
multiply__3.0__5.0__ add__15.0__3125.0__ add__15.0__3125.0__ |
| in the figure shown the outer square has an area of num__9 cm num__2 the inner square has an area of num__1 cm num__2 and the four rectangles are identical . what is the perimeter of one of the four identical rectangles ? <o> a ) num__6 cm <o> b ) num__8 cm <o> c ) num__10 cm <o> d ) num__9 cm <o> e ) num__7 cm |
the outer square has an area of num__9 cm num__2 so the sides of this outer square have length num__3 cm ( since num__3 num__3 = num__9 ) and thus pn = num__3 cm . the inner square has an area of num__1 cm num__2 so the sides of this inner square have length num__1 cm ( since num__1 num__1 = num__1 ) and thus mr = num__1 cm . since pn = num__3 cm then ps + sn = num__3 cm and so qr + sn = num__3 cm ( since qr = ps ) . but qr = qm + mr so then qm + mr + sn = num__3 cm or qm + num__1 + sn = num__3 cm ( since mr = num__1 cm ) . from this last equation we get qm + sn = num__2 cm . p q r s m n since each of qm and sn is the width of an identical rectangle then qm = sn = num__1 cm . using ps + sn = num__3 cm we get ps + num__1 = num__3 cm and so ps = num__2 cm . since the rectangles are identical then sn = pq = num__1 cm . the perimeter of rectangle pqrs is num__2 ( ps + pq ) = num__2 x ( num__2 + num__1 ) = num__2 x num__3 = num__6 cm . correct answer a <eor> a <eos> |
a |
surface_cube__1.0__ surface_cube__1.0__ |
surface_cube__1.0__ surface_cube__1.0__ |
| of the people who responded to a market survey num__240 preferred brand x and the rest preferred brand y . if the respondents indicated a preference for brand x over brand y by ratio of num__6 to num__1 how many people responded to the survey ? <o> a ) num__80 <o> b ) num__160 <o> c ) num__280 <o> d ) num__360 <o> e ) num__480 |
ratio = num__6 : num__1 = > num__6 x respondents preferred brand x and x preferred brand y since no . of respondents who preferred brand x = num__240 = > num__6 x = num__240 = > x = num__40 hence total no . of respondents = num__240 + num__40 = num__280 hence c is the answer . <eor> c <eos> |
c |
divide__240.0__6.0__ add__240.0__40.0__ add__240.0__40.0__ |
divide__240.0__6.0__ add__240.0__40.0__ add__240.0__40.0__ |
| in a group of ducks and goats the total number of legs are num__32 more than twice the number of heads . find the total number of goats . <o> a ) a ) num__14 <o> b ) b ) num__12 <o> c ) c ) num__6 <o> d ) d ) num__8 <o> e ) e ) num__16 |
explanation : let the number of ducks be d and number of goats be c then total number of legs = num__2 d + num__4 c = num__2 ( d + num__2 c ) total number of heads = c + d given that total number of legs are num__32 more than twice the number of heads = > num__2 ( d + num__2 c ) = num__32 + num__2 ( c + d ) = > d + num__2 c = num__16 + c + d = > num__2 c = num__16 + c = > c = num__16 i . e . total number of goats = num__16 answer : e <eor> e <eos> |
e |
divide__32.0__2.0__ subtract__32.0__16.0__ |
divide__32.0__2.0__ subtract__32.0__16.0__ |
| you collect baseball cards . suppose you start out with num__17 . maria takes half of one more than the number of baseball cards you have . since you ' re nice you give peter num__1 baseball card . since his father makes baseball cards paul decides to triple your baseball cards . how many baseball cards do you have at the end ? <o> a ) num__17 <o> b ) num__18 <o> c ) num__19 <o> d ) num__20 <o> e ) num__21 |
solution start with num__17 baseball cards . maria takes half of one more than the number of baseball cards you have . so maria takes half of num__17 + num__1 which is num__9 so you ' re left with num__17 - num__9 = num__8 . peter takes num__1 baseball card from you : num__8 - num__1 = num__7 baseball cards . paul triples the number of baseball cards you have : num__7 Ã — num__3 = num__21 baseball cards . so you have num__21 at the end . correct answer : e <eor> e <eos> |
e |
subtract__17.0__9.0__ subtract__8.0__1.0__ triple__1.0__ triple__7.0__ triple__7.0__ |
subtract__17.0__9.0__ subtract__8.0__1.0__ triple__1.0__ triple__7.0__ triple__7.0__ |
| two trains of equal length running with the speeds of num__60 and num__16 kmph take num__50 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? <o> a ) num__10.11 <o> b ) num__8.11 <o> c ) num__77.2 <o> d ) num__13.15 <o> e ) num__22.22 |
rs = num__60 - num__40 = num__20 * num__0.277777777778 = num__5.55555555556 t = num__50 d = num__50 * num__5.55555555556 = num__277.777777778 rs = num__60 + num__16 = num__76 * num__0.277777777778 t = num__277.777777778 * num__0.0473684210526 = num__13.15 sec . answer : d <eor> d <eos> |
d |
subtract__60.0__40.0__ add__60.0__16.0__ round__13.15__ |
subtract__60.0__40.0__ add__60.0__16.0__ round__13.15__ |
| a can do a piece of work in num__40 days ; b can do the same in num__30 days . a started alone but left the work after num__10 days then b worked at it for num__10 days . c finished the remaining work in num__10 days . c alone can do the whole work in ? <o> a ) num__21 days <o> b ) num__24 days <o> c ) num__23 days <o> d ) num__25 days <o> e ) num__26 days |
b num__24 days num__0.25 + num__0.333333333333 + num__10 / x = num__1 x = num__24 days <eor> b <eos> |
b |
divide__10.0__40.0__ divide__10.0__30.0__ round__24.0__ |
divide__10.0__40.0__ divide__10.0__30.0__ divide__24.0__1.0__ |
| in the manufacture of a certain product num__7 percent of the units produced are defective and num__5 percent of the defective units are shipped for sale . what percent of the units produced are defective units that are shipped for sale ? <o> a ) num__0.125 <o> b ) num__0.35 <o> c ) num__0.8 <o> d ) num__1.25 <o> e ) num__2.0 % |
num__0.07 * num__0.05 = num__0.0035 = num__0.35 the answer is b . <eor> b <eos> |
b |
percent__7.0__0.05__ percent__7.0__5.0__ percent__7.0__5.0__ |
percent__7.0__0.05__ percent__7.0__5.0__ percent__7.0__5.0__ |
| a sum of money amounts to num__9800 after num__5 years and num__12005 after num__8 years at the same rate of simple interest . the rate of interest per annum is : <o> a ) num__10 <o> b ) num__18 <o> c ) num__11 <o> d ) num__12 <o> e ) num__15 |
d num__12 s . i . for num__3 years = ( num__12005 - num__9800 ) = num__2205 . s . i . for num__5 years = ( num__735.0 ) x num__5 = num__3675 principal = ( num__9800 - num__3675 ) = num__6125 . hence rate = ( num__100 x num__3675 ) / ( num__6125 x num__5 ) % = num__12.0 <eor> d <eos> |
d |
percent__100.0__12.0__ |
percent__100.0__12.0__ |
| in covering a distance of num__30 km abhay takes num__2 hours more than sameer . if abhay doubles his speed then he would take num__1 hour less than sameer . abhay ' s speed is : <o> a ) num__5 kmph <o> b ) num__6 kmph <o> c ) num__6.25 kmph <o> d ) num__7.5 kmph <o> e ) num__7 kmph |
let abhay ' s speed be x km / hr . then num__30 / x - num__15.0 x = num__3 num__6 x = num__30 x = num__5 km / hr answer : a <eor> a <eos> |
a |
divide__30.0__2.0__ add__2.0__1.0__ multiply__2.0__3.0__ divide__30.0__6.0__ round__5.0__ |
divide__30.0__2.0__ add__2.0__1.0__ multiply__2.0__3.0__ divide__30.0__6.0__ divide__30.0__6.0__ |
| when a certain number x is divided by num__66 the remainder is num__14 . what is the remainder when x is divided by num__11 ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__7 <o> d ) num__9 <o> e ) num__10 |
let possible value of x is num__80 least possible value of x / num__11 is num__7.27272727273 = > num__7 quotient with remainder num__3 thus answer is ( b ) num__3 <eor> b <eos> |
b |
add__66.0__14.0__ divide__80.0__11.0__ round_down__7.2727__ subtract__14.0__11.0__ subtract__14.0__11.0__ |
add__66.0__14.0__ divide__80.0__11.0__ round_down__7.2727__ subtract__14.0__11.0__ subtract__14.0__11.0__ |
| can you replace the question mark with the number to complete the series provided the pair of numbers exhibits a similar relationship ? ? : num__4623 : : num__9 : num__647 <o> a ) num__16 <o> b ) num__17 <o> c ) num__18 <o> d ) num__19 <o> e ) num__20 |
solution : num__17 explanation : the relationship holds for below formula : pow ( x num__3 ) - pow ( x num__2 ) - num__1 = > num__9 * num__9 * num__9 - num__9 * num__9 - num__1 = > num__729 - num__81 - num__1 = > num__647 similarly num__17 * num__17 * num__17 - num__17 * num__17 - num__1 = > num__4913 - num__289 - num__1 = > num__4623 answer b <eor> b <eos> |
b |
subtract__3.0__2.0__ power__9.0__3.0__ power__9.0__2.0__ power__17.0__3.0__ power__17.0__2.0__ multiply__1.0__17.0__ |
subtract__3.0__2.0__ power__9.0__3.0__ power__9.0__2.0__ power__17.0__3.0__ power__17.0__2.0__ multiply__1.0__17.0__ |
| a hare and a jackal are running a race . three leaps of the hare are equal to four leaps of the jackal . for every two leaps of the hare the jackal takes three leaps . find the ratio of the speed of the hare to the speed of the jackal . <o> a ) num__64 : num__25 <o> b ) num__8 : num__9 <o> c ) num__5 : num__8 <o> d ) num__25 : num__64 <o> e ) num__4 : num__7 |
the hare takes num__2 leaps and the jackal takes num__3 leaps . num__1 hare leap = num__1.33333333333 jackal leaps thus the hare ' s num__2 leaps = num__2 * ( num__1.33333333333 ) = num__2.66666666667 jackal leaps . the ratio of their speeds is num__2.66666666667 : num__3 = num__8 : num__9 . the answer is b . <eor> b <eos> |
b |
subtract__3.0__2.0__ add__1.0__8.0__ multiply__1.0__8.0__ |
subtract__3.0__2.0__ add__1.0__8.0__ multiply__1.0__8.0__ |
| two machines y and z work at constant rates producing identical items . machine y produces num__23 items in the same time machine z produces num__21 items . if machine y takes num__21 minutes to produce a batch of items how many minutes does it take for machine z to produce the same number of items ? <o> a ) num__6 <o> b ) num__9 <o> c ) num__9 num__0.5 <o> d ) num__12 <o> e ) num__23 |
rate z / rate y = num__0.913043478261 time z / time y = num__1.09523809524 ( num__1.09523809524 ) * num__21 = num__23 minutes e <eor> e <eos> |
e |
divide__21.0__23.0__ divide__23.0__21.0__ round__23.0__ |
divide__21.0__23.0__ divide__23.0__21.0__ round__23.0__ |
| in cliff ’ s impressive rock collection there are half as many igneous rocks as sedimentary rocks . of the igneous rocks num__0.666666666667 are shiny and the rest are matte while num__0.2 of the sedimentary rocks are shiny . if there are num__40 shiny igneous rocks how many total rocks does cliff have ? <o> a ) num__180 <o> b ) num__45 <o> c ) num__60 <o> d ) num__90 <o> e ) num__135 |
we can start with the known quantity and then go on to find the others . shiny igneous ricks are num__40 . these are ( num__0.666666666667 ) of total igneous rocks . ( num__0.666666666667 ) * total igneous rocks = num__40 total igneous rocks = num__40 * ( num__1.5 ) = num__60 total sedimentary rocks = num__2 * total igneous rocks = num__2 * num__60 = num__120 total number of rocks = num__60 + num__120 = num__180 answer ( a ) <eor> a <eos> |
a |
multiply__40.0__1.5__ multiply__2.0__60.0__ multiply__1.5__120.0__ multiply__1.5__120.0__ |
multiply__40.0__1.5__ multiply__2.0__60.0__ multiply__1.5__120.0__ multiply__1.5__120.0__ |
| from an island it is possible to reach the mainland by either ferry p or ferry q . ferry p travels for num__2 hours at num__8 kilometers per hour while ferry q takes a route that is three times longer . if ferry p is slower than ferry q by num__4 kilometers per hour how many hours longer is the journey of ferry q compared with the journey of ferry p ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
the distance traveled by ferry p is num__16 km . then the distance traveled by ferry q is num__48 km . ferry q travels at a speed of num__12 kph . the time of the journey for ferry q is num__4.0 = num__4 hours which is num__2 hours more than ferry p . the answer is b . <eor> b <eos> |
b |
multiply__2.0__8.0__ add__8.0__4.0__ round__2.0__ |
multiply__2.0__8.0__ add__8.0__4.0__ round__2.0__ |
| a reduction of num__20.0 in the price of salt enables a lady to obtain num__10 kgs more for rs . num__200 find the original price per kg ? <o> a ) s . num__7 <o> b ) s . num__4 <o> c ) s . num__5 <o> d ) s . num__1 <o> e ) s . num__9 |
num__100 * ( num__0.2 ) = num__20 - - - num__10 ? - - - num__1 = > rs . num__2 num__200 - - - num__80 ? - - - num__2 = > rs . num__5 answer : c <eor> c <eos> |
c |
percent__20.0__10.0__ percent__100.0__5.0__ |
percent__20.0__10.0__ percent__100.0__5.0__ |
| what percent of the different arrangements of the letters of the word ajicux are those in which the vowels appear together ? <o> a ) num__25.0 <o> b ) num__30.0 <o> c ) num__40.0 <o> d ) num__20.0 <o> e ) num__50 % |
let ' s determine the number of total possibilities in arranging the letters . there are six spaces so the total number of arrangements is num__6 ! or num__360 . next we need to figure out how to determine the number of ways that we can arrange the num__3 vowels together - simply place them together ( as in aau ) and call that a single place . next we must determine the number of ways to arrange the now num__4 units ( i . e . aau b c s ) . like above there are num__4 units and num__4 places so the number of arrangements is num__4 ! or num__24 . finally we need to account for the number of ways we can arrange aau . we can either write out each unique iteration ( aau aua and uaa ) or calculate as num__3 ! / num__2 ! and get num__3 . putting this all together we get the number of ways to arrange the letters so that the vowels are together is num__4 ! x num__3 = = > num__72 the number of total arrangements of all the letters is num__6 ! = = > num__360 num__0.2 = num__0.2 or num__20.0 correct answer is d <eor> d <eos> |
d |
multiply__4.0__6.0__ divide__6.0__3.0__ multiply__3.0__24.0__ divide__72.0__360.0__ divide__4.0__0.2__ divide__4.0__0.2__ |
multiply__4.0__6.0__ divide__6.0__3.0__ multiply__3.0__24.0__ divide__72.0__360.0__ divide__4.0__0.2__ divide__4.0__0.2__ |
| difference between the length & breadth of a rectangle is num__23 m . if its perimeter is num__206 m then its area is ? <o> a ) num__2400 m ^ num__2 <o> b ) num__2500 m ^ num__2 <o> c ) num__2520 m ^ num__2 <o> d ) num__2560 m ^ num__2 <o> e ) num__2580 m ^ num__2 |
solving the two equations we get : l = num__63 and b = num__40 . area = ( l x b ) = ( num__63 x num__40 ) m num__2 = num__2520 m ^ num__2 c <eor> c <eos> |
c |
multiply__40.0__63.0__ multiply__40.0__63.0__ |
multiply__40.0__63.0__ multiply__40.0__63.0__ |
| num__575.104 x num__15.98 Ã · num__9.001 + num__21.25 = ? <o> a ) num__983.578 <o> b ) num__1043.65 <o> c ) num__1033.65 <o> d ) num__656.112 <o> e ) num__456.512 |
explanation : ? = num__575.104 x num__15.98 Ã · num__9.001 + num__21.25 â ‰ ˆ ( num__575.104 x num__1.77777777778 ) + num__21.25 â ‰ ˆ num__1022.40 + num__21.25 â ‰ ˆ num__1043.65 answer : option b <eor> b <eos> |
b |
add__21.25__1022.4__ add__21.25__1022.4__ |
add__21.25__1022.4__ add__21.25__1022.4__ |
| by selling num__45 books a shopkeeper gains the price of num__12 books . his gain per cent is <o> a ) num__35.35 <o> b ) num__37.37 <o> c ) num__38.38 <o> d ) num__36.36 <o> e ) num__39.39 % |
suppose s . p of num__45 books = rs num__45 therefore gain = rs num__12 = > c . p = rs num__33 gain % = num__0.363636363636 x num__100 = num__36.36 gain = num__36.36 answer : d <eor> d <eos> |
d |
percent__100.0__36.36__ |
percent__100.0__36.36__ |
| two cars are traveling in the same direction along the same route . the red car travels at a constant speed of num__40 miles per hour and the black car is traveling at a constant speed of num__50 miles per hour . if the red car is num__30 miles ahead of the black car how many hours will it take the black car to overtake the red car ? <o> a ) num__0.1 <o> b ) num__0.6 <o> c ) num__1 <o> d ) num__1.2 <o> e ) num__3 |
option e num__20 + num__40 t = num__50 t t = num__3 <eor> e <eos> |
e |
subtract__50.0__30.0__ round__3.0__ |
subtract__50.0__30.0__ round__3.0__ |
| the side of a square is increased by num__25.0 then how much % does its area increases ? <o> a ) num__56.28 <o> b ) num__56.25 <o> c ) num__56.75 <o> d ) num__52.25 <o> e ) num__56.25 % |
a = num__100 a num__2 = num__10000 a = num__125 a num__2 = num__15625 - - - - - - - - - - - - - - - - num__10000 - - - - - - - - - num__5625 num__100 - - - - - - - ? = > num__56.25 answer : b <eor> b <eos> |
b |
square_perimeter__25.0__ power__100.0__2.0__ volume_cube__25.0__ triangle_area__56.25__2.0__ |
square_perimeter__25.0__ power__100.0__2.0__ volume_cube__25.0__ triangle_area__56.25__2.0__ |
| the greatest number of four digits which is divisible by num__25 num__36 num__40 and num__75 is : <o> a ) num__9000 <o> b ) num__9400 <o> c ) num__9600 <o> d ) num__9800 <o> e ) num__9200 |
greatest number of num__4 - digits is num__9999 . l . c . m . of num__25 num__36 num__40 and num__75 is num__1800 . on dividing num__9999 by num__1800 the remainder is num__999 . required number ( num__9999 - num__999 ) = num__9000 . answer : option a <eor> a <eos> |
a |
subtract__40.0__36.0__ subtract__9999.0__999.0__ subtract__9999.0__999.0__ |
subtract__40.0__36.0__ subtract__9999.0__999.0__ subtract__9999.0__999.0__ |
| in a factory an average of num__70 tv ' s are produced per day for the fist num__25 days of the months . a few workers fell ill for the next num__5 days reducing the daily avg for the month to num__68 sets / day . the average production per day for day last num__5 days is ? <o> a ) num__20 <o> b ) num__36 <o> c ) num__48 <o> d ) num__50 <o> e ) num__58 |
production during these num__5 days = total production in a month - production in first num__25 days . = num__30 x num__68 - num__25 x num__70 = num__290 ∴ average for last num__5 days = num__58.0 = num__58 e <eor> e <eos> |
e |
add__25.0__5.0__ divide__290.0__5.0__ divide__290.0__5.0__ |
add__25.0__5.0__ divide__290.0__5.0__ divide__290.0__5.0__ |
| what number should replace the question mark ? ? num__87 num__70 num__52 num__35 <o> a ) num__49 <o> b ) num__37 <o> c ) num__105 <o> d ) num__55 <o> e ) num__83 |
c num__105 the sequence progresses - num__18 - num__17 - num__18 - num__17 . <eor> c <eos> |
c |
add__70.0__35.0__ subtract__70.0__52.0__ subtract__87.0__70.0__ add__87.0__18.0__ |
add__70.0__35.0__ subtract__70.0__52.0__ subtract__87.0__70.0__ add__87.0__18.0__ |
| kim has num__40 percent more money than sal and sal has num__20 percent less money than phil . if sal and phil have a combined total of $ num__1.80 how much money does kim have ? <o> a ) $ num__1.00 <o> b ) $ num__1.12 <o> c ) $ num__1.20 <o> d ) $ num__1.32 <o> e ) $ num__1.40 |
phil = x sal = num__0.8 x kim = ( num__1.4 ) * num__0.8 x = num__1.12 x x + num__0.8 x = num__1.8 - - > x = num__1 kim = num__1.12 answer : b <eor> b <eos> |
b |
multiply__1.4__0.8__ round_down__1.8__ multiply__1.4__0.8__ |
multiply__1.4__0.8__ subtract__1.8__0.8__ multiply__1.4__0.8__ |
| the remainder when dividing the expression ( x + y ) by num__5 is num__4 . the remainder of x divided by num__10 is num__2 . what is the remainder f of y divided by num__5 ? <o> a ) f = num__1 . <o> b ) f = num__2 . <o> c ) num__3 . <o> d ) num__4 . <o> e ) num__5 . |
x divided by num__10 gives reminder of num__2 . x can be num__0.166666666667 / num__22 . . . . x + y divided by num__5 is num__4 . x + y can be num__0.444444444444 / num__19 . . . . if x + y = num__4 and x is num__2 then y = num__2 and y / num__5 will give a reminder of num__2 similarly if x + y = num__9 and x = num__2 then y / num__5 will give a reminder of num__2 hence the answer must be num__2 ( b ) also . . . . x + y = num__5 m + num__4 and x = num__10 k + num__2 hence num__10 k + num__2 + y = num__5 m + num__4 or y = num__5 ( m - num__2 k ) + num__2 m - num__2 k being a constant remainder is num__2 hence answer is b <eor> b <eos> |
b |
add__5.0__4.0__ subtract__4.0__2.0__ |
add__5.0__4.0__ subtract__4.0__2.0__ |
| in one year the population of a village increased by num__10.0 and in the next year it decreased by num__10.0 . if at the end of num__2 nd year the population was num__7920 what was it in the beginning ? <o> a ) num__2889 <o> b ) num__8000 <o> c ) num__2777 <o> d ) num__2999 <o> e ) num__2212 |
x * num__1.1 * num__0.9 = num__7920 x * num__0.99 = num__7920 x = num__7920 / num__0.99 = > num__8000 . answer : b <eor> b <eos> |
b |
subtract__2.0__1.1__ multiply__0.9__1.1__ divide__7920.0__0.99__ divide__7920.0__0.99__ |
subtract__2.0__1.1__ multiply__0.9__1.1__ divide__7920.0__0.99__ divide__7920.0__0.99__ |
| a b and c can do a work in num__6 num__8 and num__12 days respectively doing the work together and get a payment of rs . num__1800 . what is b ’ s share ? <o> a ) num__600 <o> b ) num__882 <o> c ) num__266 <o> d ) num__288 <o> e ) num__279 |
wc = num__0.166666666667 : num__0.125 : num__0.0833333333333 = > num__4 : num__3 : num__2 num__0.333333333333 * num__1800 = num__600 answer : a <eor> a <eos> |
a |
subtract__12.0__8.0__ divide__12.0__4.0__ divide__6.0__3.0__ divide__2.0__6.0__ divide__1800.0__3.0__ round__600.0__ |
subtract__12.0__8.0__ divide__12.0__4.0__ divide__6.0__3.0__ divide__2.0__6.0__ divide__1800.0__3.0__ round__600.0__ |
| in each of the following questions a number series is given with one term missing . choose the correct alternative that will continue the same pattern and fill in the blank spaces . ? num__7 num__14 num__23 num__34 num__47 <o> a ) num__31 <o> b ) num__2 <o> c ) num__36 <o> d ) num__31 <o> e ) num__33 |
b num__2 the given sequence is + num__5 + num__7 + num__9 — — ie . num__2 + num__5 = num__7 num__7 + num__7 = num__14 num__14 + num__9 = num__23 <eor> b <eos> |
b |
divide__14.0__7.0__ subtract__7.0__2.0__ add__7.0__2.0__ subtract__7.0__5.0__ |
divide__14.0__7.0__ subtract__7.0__2.0__ add__7.0__2.0__ subtract__7.0__5.0__ |
| a is two years older than b who is twice as old as c . if the total of the ages of a b and c be num__52 then how old is b ? <o> a ) num__17 years <o> b ) num__20 years <o> c ) num__29 years <o> d ) num__10 years <o> e ) num__12 years |
let c ' s age be x years . then b ' s age = num__2 x years . a ' s age = ( num__2 x + num__2 ) years . ( num__2 x + num__2 ) + num__2 x + x = num__52 num__5 x = num__50 = > x = num__10 hence b ' s age = num__2 x = num__20 years . answer : b <eor> b <eos> |
b |
subtract__52.0__2.0__ multiply__2.0__5.0__ multiply__2.0__10.0__ multiply__2.0__10.0__ |
subtract__52.0__2.0__ multiply__2.0__5.0__ multiply__2.0__10.0__ multiply__2.0__10.0__ |
| if a and b are multiples of num__8 then which are all the multiples of num__8 num__1 ) a * b num__2 ) a - b num__3 ) a + b num__4 ) a / b num__5 ) a ^ b <o> a ) num__1 num__25 <o> b ) num__13 <o> c ) num__23 num__45 <o> d ) num__12 num__35 <o> e ) all of these |
a * b a ^ b a + b a - b are multiples of num__6 except a / b answer : d <eor> d <eos> |
d |
subtract__8.0__2.0__ add__8.0__4.0__ |
subtract__8.0__2.0__ add__8.0__4.0__ |
| two men start together to walk a certain distance one at num__4 kmph and another at num__3 kmph . the former arrives half an hour before the latter . find the distance . <o> a ) num__6 kilometre <o> b ) num__61 kilometre <o> c ) num__8 km <o> d ) num__9 km <o> e ) none of these |
let the distance be x km . then x / num__3 - x / num__4 = num__0.5 ( num__4 x - num__3 x ) / num__12 = num__0.5 x = num__6 km answer : a <eor> a <eos> |
a |
multiply__4.0__3.0__ divide__3.0__0.5__ round__6.0__ |
multiply__4.0__3.0__ divide__3.0__0.5__ divide__3.0__0.5__ |
| jane invested $ num__1500 in fund a and $ num__1000 in fund b . over the next two years the money in fund a earned a total interest of num__12 percent for the two years combined and the money in fund b earned num__30 percent annual interest compounded annually . two years after jane made these investments . jane ' s investment in fund b was worth how much more than her investment in fund a ? <o> a ) num__1 . $ num__200 <o> b ) num__2 . $ num__100 <o> c ) num__3 . $ num__10 <o> d ) num__4 . $ num__150 <o> e ) num__5 . $ num__400 |
interest on fund a will be num__180 . rate of interest will be num__6.0 per annum simple interest as num__12.0 is for num__2 year . this will make investment a num__1680 at the end of num__2 nd year . interest on fund b will be num__690 on num__30.0 interest compounded annually . this will make investment b num__1690 at the end of num__2 nd year . difference in investment b and investment a = num__1690 - num__1680 = num__10 answer is c . <eor> c <eos> |
c |
divide__180.0__30.0__ divide__12.0__6.0__ add__1500.0__180.0__ add__1000.0__690.0__ subtract__12.0__2.0__ divide__30.0__10.0__ |
divide__180.0__30.0__ divide__12.0__6.0__ add__1500.0__180.0__ add__1000.0__690.0__ subtract__12.0__2.0__ divide__30.0__10.0__ |
| a and b started a partnership business investing some amount in the ratio of num__4 : num__5 . c joined then after six months with an amount equal to that of b . in what proportion should the profit at the end of one year be distributed among a b and c ? <o> a ) num__4 : num__9 : num__8 <o> b ) num__5 : num__7 : num__4 <o> c ) num__6 : num__10 : num__5 <o> d ) num__7 : num__9 : num__4 <o> e ) num__8 : num__10 : num__5 |
let the initial investments of a and b be num__4 x and num__5 x . a : b : c = ( num__4 x x num__12 ) : ( num__5 x x num__12 ) : ( num__5 x x num__6 ) = num__48 : num__60 : num__30 = num__8 : num__10 : num__5 . answer : e <eor> e <eos> |
e |
multiply__4.0__12.0__ multiply__5.0__12.0__ multiply__5.0__6.0__ subtract__12.0__4.0__ add__4.0__6.0__ subtract__12.0__4.0__ |
multiply__4.0__12.0__ multiply__5.0__12.0__ multiply__5.0__6.0__ subtract__12.0__4.0__ add__4.0__6.0__ subtract__12.0__4.0__ |
| fifty applicants for a job were given scores from num__1 to num__5 on their interview performance . their scores are shown in the table above . what was the average score for the group ? <o> a ) num__2.79 <o> b ) num__2.86 <o> c ) num__2.91 <o> d ) num__2.99 <o> e ) num__3.03 |
multiply score * number of application = num__10 + num__48 + num__63 + num__14 + num__8 = num__143 . number of applications is num__50 . num__2.86 = num__2.86 . answer : option b is correct . . <eor> b <eos> |
b |
multiply__5.0__10.0__ divide__143.0__50.0__ multiply__1.0__2.86__ |
multiply__5.0__10.0__ divide__143.0__50.0__ multiply__1.0__2.86__ |
| a man traveled a total distance of num__900 km . he traveled one - third of the whole trip by plane and the distance traveled by train is one - half of the distance traveled by bus . if he traveled by train plane and bus how many kilometers did he travel by bus ? <o> a ) num__400 km <o> b ) num__450 km <o> c ) num__500 km <o> d ) num__550 km <o> e ) num__600 km |
total distance traveled = num__900 km . distance traveled by plane = num__300 km . distance traveled by bus = x distance traveled by train = x / num__2 x + x / num__2 + num__300 = num__900 num__3 x / num__2 = num__600 x = num__400 km the answer is a . <eor> a <eos> |
a |
divide__900.0__300.0__ subtract__900.0__300.0__ round__400.0__ |
divide__900.0__300.0__ subtract__900.0__300.0__ round__400.0__ |
| if both num__5 ^ num__3 and num__3 ^ num__3 are factors of n x ( num__2 ^ num__5 ) x ( num__12 ^ num__2 ) x ( num__7 ^ num__3 ) x ( num__10 ^ num__2 ) what is the smallest possible positive value of n ? <o> a ) num__15 <o> b ) num__45 <o> c ) num__75 <o> d ) num__125 <o> e ) num__150 |
( num__2 ^ num__5 ) x ( num__12 ^ num__2 ) x ( num__7 ^ num__3 ) x ( num__10 ) has two appearances of num__3 ( in num__12 ^ num__2 ) and two appearances of num__5 ( in num__10 ^ num__2 ) . thus n must include at least num__3 * num__5 = num__15 the answer is a . <eor> a <eos> |
a |
multiply__5.0__3.0__ multiply__5.0__3.0__ |
multiply__5.0__3.0__ multiply__5.0__3.0__ |
| a can do a work in num__12 days . when he had worked for num__3 days b joinedhim . if they complete the work in num__3 more days in how manydays can balone finish the work ? <o> a ) num__5 days <o> b ) num__4 days <o> c ) num__6 days <o> d ) num__10 days <o> e ) num__11 days |
sax work done by afar num__3 days : i j . . remzming war — num__1 . work done by ( a + b ) for num__1 day : . work done by a for num__1 day num__6 days c <eor> c <eos> |
c |
round__6.0__ |
round__6.0__ |
| i chose a number and divide it by num__12 . then i subtracted num__240 from the result and got num__8 . what was the number i chose ? <o> a ) num__2976 <o> b ) num__2989 <o> c ) num__2789 <o> d ) num__2888 <o> e ) num__2098 |
let x be the number i chose then x / num__12 − num__240 = num__8 x / num__12 = num__248 x = num__2976 answer is a . <eor> a <eos> |
a |
add__240.0__8.0__ multiply__12.0__248.0__ multiply__12.0__248.0__ |
add__240.0__8.0__ multiply__12.0__248.0__ multiply__12.0__248.0__ |
| a worker ' s take - home pay last year was the same each month and she saved the same fraction of her take - home pay each month . the total amount of money that she had saved at the end of the year was num__5 times the amount of that portion of her monthly take - home pay that she did not save . if all the money that she saved last year was from her take - home pay what fraction of her take - home pay did she save each month ? <o> a ) num__0.181818181818 <o> b ) num__0.214285714286 <o> c ) num__0.294117647059 <o> d ) num__0.35 <o> e ) num__0.366666666667 |
let x be the fraction of her take - home pay that the worker saved . let p be the monthly pay . num__12 xp = num__5 ( num__1 - x ) p num__12 xp = num__5 p - num__5 xp num__17 xp = num__5 p x = num__0.294117647059 the answer is c . <eor> c <eos> |
c |
add__5.0__12.0__ divide__5.0__17.0__ divide__5.0__17.0__ |
add__5.0__12.0__ divide__5.0__17.0__ divide__5.0__17.0__ |
| a dishonest dealer professes to sell goods at the cost price but uses a weight of num__800 grams per kg what is his percent ? <o> a ) num__11.0 <o> b ) num__25.0 <o> c ) num__87.0 <o> d ) num__19.0 <o> e ) num__12 % |
num__800 - - - num__200 num__100 - - - ? = > num__25.0 answer : b <eor> b <eos> |
b |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| num__95 white and black tiles will be used to form a num__10 x num__10 square pattern . if there must be at least one black tile in every row and at least one white tile in every column what is the maximum difference between the number of black and white tiles that can be used ? <o> a ) num__75 <o> b ) num__80 <o> c ) num__85 <o> d ) num__90 <o> e ) num__95 |
answer = c please refer diagram below num__95 - num__10 = num__85 <eor> c <eos> |
c |
subtract__95.0__10.0__ subtract__95.0__10.0__ |
subtract__95.0__10.0__ subtract__95.0__10.0__ |
| connie paid a sales tax of num__8 percent on her purchase . if the sales tax had been only num__5 percent she would have paid $ num__12 less in sales tax on her purchase . what was the total amount that connie paid for her purchase including sales tax ? <o> a ) $ num__368 <o> b ) $ num__380 <o> c ) $ num__400 <o> d ) $ num__420 <o> e ) $ num__432 |
connie paid x dollar before sales tax so num__1.08 x - num__1.05 x = num__12 num__0.03 x = num__12 x = num__400 sales tax num__400 * num__8.0 = num__32 connie paid = num__400 + num__32 = $ num__432 ans . e <eor> e <eos> |
e |
subtract__1.08__1.05__ divide__12.0__0.03__ add__32.0__400.0__ add__32.0__400.0__ |
subtract__1.08__1.05__ divide__12.0__0.03__ add__32.0__400.0__ add__32.0__400.0__ |
| the avg . age of a group of num__20 students is num__20 years . if num__4 more students join the group the avg age increases by num__1 year . the avg age of the new student is ? <o> a ) num__22 years <o> b ) num__23 years <o> c ) num__24 years <o> d ) num__25 years <o> e ) num__26 years |
total age of num__20 students = num__20 * num__20 = num__400 if total age of num__4 students = x then ( num__400 + x ) / ( num__20 + num__4 ) = ( num__20 + num__1 ) x = num__104 so average age of new students = num__26.0 = num__26 years answer : e <eor> e <eos> |
e |
divide__104.0__4.0__ multiply__1.0__26.0__ |
divide__104.0__4.0__ multiply__1.0__26.0__ |
| what should be the least number to be added to the num__1202 number to make it divisible by num__4 ? <o> a ) num__12 <o> b ) num__17 <o> c ) num__18 <o> d ) num__77 <o> e ) num__2 |
answer : num__2 option : e <eor> e <eos> |
e |
subtract__4.0__2.0__ |
subtract__4.0__2.0__ |
| mrs . rodger got a weekly raise of $ num__145 . if she gets paid every other week write an integer describing how the raise will affect her paycheck . <o> a ) $ num__145 . <o> b ) $ num__146 <o> c ) $ num__147 <o> d ) $ num__148 <o> e ) none |
solution : let the num__1 st paycheck be x ( integer ) . mrs . rodger got a weekly raise of $ num__145 . so after completing the num__1 st week she will get $ ( x + num__145 ) . similarly after completing the num__2 nd week she will get $ ( x + num__145 ) + $ num__145 . = $ ( x + num__145 + num__145 ) = $ ( x + num__290 ) so in this way end of every week her salary will increase by $ num__145 . answer a <eor> a <eos> |
a |
multiply__145.0__2.0__ multiply__145.0__1.0__ |
multiply__145.0__2.0__ multiply__145.0__1.0__ |
| how long does a train num__110 m long traveling at num__60 kmph takes to cross a bridge of num__190 m in length ? <o> a ) num__18.9 sec <o> b ) num__88.9 sec <o> c ) num__22.9 sec <o> d ) num__18.0 sec <o> e ) num__72.0 sec |
d = num__110 + num__190 = num__300 m s = num__60 * num__0.277777777778 = num__16.6666666667 t = num__300 * num__0.06 = num__18.0 sec answer : d <eor> d <eos> |
d |
add__110.0__190.0__ divide__300.0__16.6667__ round__18.0__ |
add__110.0__190.0__ multiply__300.0__0.06__ multiply__300.0__0.06__ |
| the amounts of time that three secretaries worked on a special project are in the ratio of num__1 to num__2 to num__5 . if they worked a combined total of num__120 hours how many hours did the secretary who worked the longest spend on the project ? <o> a ) num__80 <o> b ) num__70 <o> c ) num__75 <o> d ) num__16 <o> e ) num__14 |
num__8 x = num__120 = > x = num__15 therefore the secretary who worked the longest spent num__15 x num__5 = num__75 hours on the project option ( c ) <eor> c <eos> |
c |
divide__120.0__8.0__ multiply__5.0__15.0__ round__75.0__ |
divide__120.0__8.0__ multiply__5.0__15.0__ round__75.0__ |
| the ratio between x and y is num__1.4 ; x is multiplied by y and y is multiplied by x what is the ratio between the new values of x and y ? <o> a ) num__1.4 <o> b ) num__0.714285714286 <o> c ) num__1 <o> d ) num__1.96 <o> e ) it can not be determined |
ratio = num__7 k / num__5 k = num__1.4 num__1.4 etc . x is multiplied by y and y is multiplied by x - - > ( num__7 k * num__9 k ) / ( num__9 k * num__7 k ) = num__1 answer : c <eor> c <eos> |
c |
divide__7.0__1.4__ round_down__1.4__ round_down__1.4__ |
divide__7.0__1.4__ round_down__1.4__ reverse__1.0__ |
| along a yard num__225 metres long num__26 trees are planted at equal distances one tree bring at each end of the yard . what is the distance between two consecutive trees ? long num__26 trees <o> a ) num__8 metres <o> b ) num__9 metres <o> c ) num__10 metres <o> d ) num__15 metres <o> e ) none of these |
solution num__26 trees have num__25 gaps between them . hence required distance = ( num__9.0 ) m = num__9 m answer b <eor> b <eos> |
b |
divide__225.0__25.0__ round__9.0__ |
divide__225.0__25.0__ round__9.0__ |
| working together printer a and printer b would finish the task in num__15 minutes . printer a alone would finish the task in num__45 minutes . how many pages does the task contain if printer b prints num__3 pages a minute more than printer a ? <o> a ) num__125 <o> b ) num__135 <o> c ) num__145 <o> d ) num__155 <o> e ) num__165 |
num__15 * a + num__15 * b = x pages in num__15 mins printer a will print = num__0.333333333333 * x pages = num__0.333333333333 * x pages thus in num__15 mins printer printer b will print x - num__0.333333333333 * x = num__0.666666666667 * x pages also it is given that printer b prints num__3 more pages per min that printer a . in num__15 mins printer b will print num__45 more pages than printer a thus num__0.666666666667 * x - num__0.333333333333 * x = num__45 = > x = num__135 pages answer : b <eor> b <eos> |
b |
divide__15.0__45.0__ multiply__45.0__3.0__ round__135.0__ |
divide__15.0__45.0__ multiply__45.0__3.0__ multiply__45.0__3.0__ |
| a train num__240 meters long is running with a speed of num__60 kmph . in what time will it pass a man who is running at num__6 kmph in the direction opposite to that in which the train is going ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__12 <o> d ) num__9 <o> e ) num__13 |
speed of train relative to man = ( num__60 + num__6 ) km / hr = num__66 km / hr [ num__66 * num__0.277777777778 ] m / sec = [ num__18.3333333333 ] m / sec . time taken to pass the man = [ num__240 * num__0.0545454545455 ] sec = num__13 sec answer : e <eor> e <eos> |
e |
add__60.0__6.0__ round__13.0__ |
add__60.0__6.0__ round__13.0__ |
| on dividing a number by num__56 we get num__28 as remainder . on dividing the same number by num__8 what will be the remainder ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
number = num__56 x + num__28 ( ∵ since the number gives num__28 as remainder on dividing by num__56 ) = ( num__7 × num__8 × x ) + ( num__3 × num__8 ) + num__4 hence if the number is divided by num__8 we will get num__4 as remainder . answer : c <eor> c <eos> |
c |
divide__56.0__8.0__ divide__28.0__7.0__ divide__28.0__7.0__ |
divide__56.0__8.0__ divide__28.0__7.0__ divide__28.0__7.0__ |
| a train num__240 m long passes a pole in num__24 seconds . how long will it take to pass a platform num__650 m long ? <o> a ) num__65 sec <o> b ) num__89 sec <o> c ) num__100 sec <o> d ) num__150 sec <o> e ) num__160 sec |
explanation : speed = num__10.0 m / sec = num__10 m / sec required time = ( num__240 + num__65.0 ) sec = num__89 sec answer is b <eor> b <eos> |
b |
divide__240.0__24.0__ divide__650.0__10.0__ add__24.0__65.0__ round__89.0__ |
divide__240.0__24.0__ divide__650.0__10.0__ add__24.0__65.0__ add__24.0__65.0__ |
| a car travels from point a to point b . the average speed of the car is num__60 km / hr and it travels the first half of the trip at a speed of num__50 km / hr . what is the speed of the car in the second half of the trip ? <o> a ) num__55 <o> b ) num__60 <o> c ) num__75 <o> d ) num__80 <o> e ) num__90 |
let d be the distance and let v be the speed in the second half . the total time = t num__1 + t num__2 d / num__60 = d / num__100 + ( d / num__2 ) / v d / num__150 = d / num__2 v and so v = num__75 km / hr the answer is c . <eor> c <eos> |
c |
multiply__50.0__2.0__ add__50.0__100.0__ divide__150.0__2.0__ multiply__1.0__75.0__ |
multiply__50.0__2.0__ add__50.0__100.0__ divide__150.0__2.0__ divide__75.0__1.0__ |
| rs num__80000 is divided into two parts one part is given to a person with num__10.0 interest and another part is given to a person with num__20.0 interest . at the end of first year he gets profit num__9000 find money given by num__10.0 ? <o> a ) num__30000 <o> b ) num__40000 <o> c ) num__50000 <o> d ) num__60000 <o> e ) num__70000 |
let first parrt is x and second part is y then x + y = num__80000 - - - - - - - - - - eq num__1 total profit = profit on x + profit on y num__9000 = ( x * num__10 * num__1 ) / num__100 + ( y * num__20 * num__1 ) / num__100 num__90000 = x + num__2 y - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - eq num__2 num__90000 = num__80000 + y so y = num__10000 then x = num__80000 - num__10000 = num__70000 first part = num__70000 answer : e <eor> e <eos> |
e |
percent__10.0__20.0__ percent__100.0__70000.0__ |
percent__10.0__20.0__ percent__100.0__70000.0__ |
| the amount of time that three people worked on a special project was in the ratio of num__2 to num__4 to num__6 . if the project took num__144 hours how many more hours did the hardest working person work than the person who worked the least ? <o> a ) num__47 hours <o> b ) num__45 hours <o> c ) num__48 hours <o> d ) num__49 hours <o> e ) num__50 hours |
let the persons be a b c . hours worked : a = num__2 * num__12.0 = num__24 hours b = num__4 * num__12.0 = num__48 hours c = num__6 * num__12.0 = num__72 hours c is the hardest worker and a worked for the least number of hours . so the difference is num__72 - num__24 = num__48 hours . answer : c <eor> c <eos> |
c |
multiply__2.0__6.0__ multiply__2.0__12.0__ multiply__2.0__24.0__ multiply__6.0__12.0__ round__48.0__ |
multiply__2.0__6.0__ multiply__2.0__12.0__ multiply__2.0__24.0__ multiply__6.0__12.0__ multiply__2.0__24.0__ |
| if num__49 x = num__343 ^ y which of the following expresses x in terms of y ? <o> a ) ( num__7.0 ) ^ y <o> b ) y ^ ( num__7.0 ) <o> c ) y ^ ( num__343 - num__49 ) <o> d ) num__7 ^ ( num__3 y - num__2 ) <o> e ) num__7 ^ ( num__2 y - num__1 ) |
by exponential simplification . num__49 = num__7 ^ num__2 and num__343 = num__7 ^ num__3 therefore ( num__7 ^ num__2 ) x = ( num__7 ^ num__3 ) ^ y gives x = ( num__7 ^ num__3 y ) / ( num__7 ^ num__2 ) further simplified to x = ( num__7 ^ num__3 y ) ( num__7 ^ - num__2 ) which gives x = num__7 ^ ( num__3 y - num__2 ) ( because exponential is additive in multiplication . i . e . a ^ b * a ^ c = a ^ ( b + c ) . answer : d <eor> d <eos> |
d |
divide__343.0__49.0__ divide__49.0__7.0__ |
divide__343.0__49.0__ divide__49.0__7.0__ |
| find the smallest number which when divided by num__13 and num__15 leaves respective remainders of num__2 and num__4 <o> a ) num__187 <o> b ) num__197 <o> c ) num__184 <o> d ) num__219 <o> e ) num__227 |
let ' n ' is the smallest number which divided by num__13 and num__15 leaves respective remainders of num__2 and num__4 . required number = ( lcm of num__13 and num__15 ) - ( common difference of divisors and remainders ) = ( num__195 ) - ( num__11 ) = num__184 . answer : c <eor> c <eos> |
c |
multiply__13.0__15.0__ subtract__13.0__2.0__ subtract__195.0__11.0__ subtract__195.0__11.0__ |
multiply__13.0__15.0__ subtract__13.0__2.0__ subtract__195.0__11.0__ subtract__195.0__11.0__ |
| a train running at the speed of num__54 km / hr crosses a pole in num__7 seconds . find the length of the train . <o> a ) num__150 meter <o> b ) num__105 meter <o> c ) num__140 meter <o> d ) num__135 meter <o> e ) none of these |
explanation : speed = num__54 * ( num__0.277777777778 ) m / sec = num__15 m / sec length of train ( distance ) = speed * time = num__15 * num__7 = num__105 meter option b <eor> b <eos> |
b |
multiply__7.0__15.0__ round__105.0__ |
multiply__7.0__15.0__ multiply__7.0__15.0__ |
| at a party everyone shook hands with everybody else . there were num__66 handshakes . how many people were at the party ? <o> a ) num__12 <o> b ) num__11 <o> c ) num__10 <o> d ) num__9 <o> e ) num__5 |
a num__12 in general with n + num__1 people the number of handshakes is the sum of the first n consecutive numbers : num__1 + num__2 + num__3 + . . . + n . since this sum is n ( n + num__1 ) / num__2 we need to solve the equation n ( n + num__1 ) / num__2 = num__66 . this is the quadratic equation n num__2 + n - num__132 = num__0 . solving for n we obtain num__11 as the answer and deduce that there were num__12 people at the party . since num__66 is a relatively small number you can also solve this problem with a hand calculator . add num__1 + num__2 = + num__3 = + . . . etc . until the total is num__66 . the last number that you entered ( num__11 ) is n . <eor> a <eos> |
a |
add__1.0__2.0__ multiply__66.0__2.0__ divide__132.0__12.0__ add__1.0__11.0__ |
add__1.0__2.0__ multiply__66.0__2.0__ divide__132.0__12.0__ add__1.0__11.0__ |
| the num__15 homes in a new development are each to be sold for one of three different prices so that the developer receives an average ( arithmetic mean ) of $ num__200000 per home . if num__4 of the homes are to be sold for $ num__170000 each and num__5 are to be sold for $ num__200000 each what will be the selling price of each of the remaining num__10 homes ? <o> a ) $ num__200000 <o> b ) $ num__212000 <o> c ) $ num__215000 <o> d ) $ num__220000 <o> e ) $ num__230 |
000 |
imo the answer has to be d . num__5 houses are being sold for num__200000 . num__4 houses are being sold for $ num__30000 less resulting in a loss of $ num__120000 . to make the average selling price intact i . e . $ num__200000 the remaining num__10 houses must be sold at such a profit that it compensates for the loss of num__120000 . hence num__10 x = num__120000 . x = num__12000 . the num__10 houses are sold at $ num__12000 profit or at $ num__212000 . ( answer b ) <eor> b <eos> |
b |
b |
| sachin is younger than rahul by num__7 years . if their ages are in the respective ratio of num__7 : num__9 how old is sachin ? <o> a ) num__23 <o> b ) num__23.5 <o> c ) num__24 <o> d ) num__24.5 <o> e ) num__25 |
let the present ages of sachin and rahul be num__7 x and num__9 x sachin is younger than rahul in num__7 years so num__9 x - num__7 x = num__7 num__2 x = num__7 x = num__3.5 sachin ; s age = num__7 x = num__7 * num__3.5 = num__24.5 answer : d <eor> d <eos> |
d |
subtract__9.0__7.0__ divide__7.0__2.0__ multiply__7.0__3.5__ multiply__7.0__3.5__ |
subtract__9.0__7.0__ divide__7.0__2.0__ multiply__7.0__3.5__ multiply__7.0__3.5__ |
| a team of eight entered for a shooting competition . the best marks man scored num__82 points . if he had scored num__92 points the average scores for . the team would have been num__86 . how many points altogether did the team score ? <o> a ) num__662 <o> b ) num__678 <o> c ) num__652 <o> d ) num__642 <o> e ) num__721 |
num__8 * num__86 = num__688 – num__10 = num__678 answer : b <eor> b <eos> |
b |
multiply__86.0__8.0__ subtract__92.0__82.0__ subtract__688.0__10.0__ subtract__688.0__10.0__ |
multiply__86.0__8.0__ subtract__92.0__82.0__ subtract__688.0__10.0__ subtract__688.0__10.0__ |
| the mean of ( num__54824 ) ^ num__2 and ( num__54826 ) ^ num__2 = <o> a ) ( num__54821 ) ^ num__2 <o> b ) ( num__54 num__821.5 ) ^ num__2 <o> c ) ( num__54 num__820.5 ) ^ num__2 <o> d ) ( num__54825 ) ^ num__2 + num__1 <o> e ) ( num__54821 ) ^ num__2 – num__1 |
num__54824 ^ num__2 = ( num__54825 - num__1 ) ^ num__2 = num__54825 ^ num__2 + num__1 ^ num__2 - num__2 * num__54825 * num__1 num__54826 ^ num__2 = ( num__54825 + num__1 ) ^ num__2 = num__54825 ^ num__2 + num__1 ^ num__2 + num__2 * num__54825 * num__1 taking the average of above num__2 we get ( num__54825 ) ^ num__2 + num__1 hence the answer is d <eor> d <eos> |
d |
divide__54824.0__54826.0__ add__54824.0__1.0__ |
subtract__54826.0__54825.0__ add__54824.0__1.0__ |
| there are num__16 bees in the hive then num__10 more fly . how many bees are there in all ? <o> a ) num__7 <o> b ) num__33 <o> c ) num__12 <o> d ) num__17 <o> e ) num__26 |
num__16 + num__10 = num__26 . answer is e . <eor> e <eos> |
e |
add__16.0__10.0__ add__16.0__10.0__ |
add__16.0__10.0__ add__16.0__10.0__ |
| walking at the rate of num__3 kmph a man cover certain distance in num__3 hr num__20 min . running at a speed of num__9 kmph the man will cover the same distance in . <o> a ) num__45 min <o> b ) num__50 min <o> c ) num__40 min <o> d ) num__48 min <o> e ) num__54 min |
distance = speed * time num__3 * num__3.33333333333 = num__10 km new speed = num__9 kmph therefore time = d / s = num__1.11111111111 = num__54 min answer : e . <eor> e <eos> |
e |
divide__3.3333__3.0__ round__54.0__ |
divide__3.3333__3.0__ round__54.0__ |
| what is the average ( arithmetic mean ) of all the multiples of ten from num__10 to num__110 inclusive ? <o> a ) num__60 <o> b ) num__95 <o> c ) num__100 <o> d ) num__105 <o> e ) num__110 |
the multiples of ten from num__10 to num__110 inclusive would be an evenly spaced set with num__11 terms - num__10 num__20 num__30 . . . . . . . num__110 so average = ( first term + last term ) / num__2 = ( num__10 + num__110 ) / num__2 = num__60.0 = num__60 hence the correct answer is a . <eor> a <eos> |
a |
divide__110.0__10.0__ add__10.0__20.0__ divide__20.0__10.0__ multiply__2.0__30.0__ multiply__2.0__30.0__ |
divide__110.0__10.0__ add__10.0__20.0__ divide__20.0__10.0__ multiply__2.0__30.0__ multiply__2.0__30.0__ |
| a motorcycle importer is planning on increasing the price of a certain model by $ num__1000 . at this new price num__9 fewer motorcycles will be sold per month but the total revenues will increase by $ num__26000 to $ num__594000 . what is the number of motorcycles the manufacturer will sell at this new price <o> a ) num__70 <o> b ) num__68 <o> c ) num__66 <o> d ) num__64 <o> e ) num__62 |
responding to a pm : to solve this question i will make an equation in x and then make educated guesses . here ' s how : assuming x motorcycles were sold every month initially . ( num__568000 / x + num__1000 ) ( x - num__9 ) = num__594000 ( num__568 / x + num__1 ) ( x - num__9 ) = num__594 now num__568 = num__8 * num__71 assuming x = num__71 ( we have all integers so it is obvious that num__568 / x should be an integer . we get num__9 * num__66 = num__594 ( matches ) so he will sell num__71 - num__9 = num__62 bikes this month answer ( e ) <eor> e <eos> |
e |
subtract__594000.0__26000.0__ divide__568000.0__1000.0__ divide__594000.0__1000.0__ subtract__9.0__1.0__ divide__568.0__8.0__ divide__594.0__9.0__ subtract__71.0__9.0__ multiply__1.0__62.0__ |
subtract__594000.0__26000.0__ divide__568000.0__1000.0__ divide__594000.0__1000.0__ subtract__9.0__1.0__ divide__568.0__8.0__ divide__594.0__9.0__ subtract__71.0__9.0__ multiply__1.0__62.0__ |
| a bag contains equal number of rs . num__5 rs . num__2 and re . num__1 coins . if the total amount in the bag is rs . num__1152 find the number of coins of each kind ? <o> a ) num__432 <o> b ) num__288 <o> c ) num__144 <o> d ) num__72 <o> e ) num__152 |
let the number of coins of each kind be x . = > num__5 x + num__2 x + num__1 x = num__1152 = > num__8 x = num__1152 = > x = num__144 answer : c <eor> c <eos> |
c |
divide__1152.0__8.0__ multiply__1.0__144.0__ |
divide__1152.0__8.0__ multiply__1.0__144.0__ |
| if money is invested at r percent interest compounded annually the amount of the investment will double in approximately num__54 / r years . if pat ' s parents invested $ num__6000 in a long - term bond that pays num__6 percent interest compounded annually what will be the approximate total amount of the investment num__18 years later when pat is ready for college ? <o> a ) $ num__20000 <o> b ) $ num__15000 <o> c ) $ num__12000 <o> d ) $ num__10000 <o> e ) $ num__9000 |
since investment doubles in num__54 / r years then for r = num__6 it ' ll double in num__9.0 = ~ num__9 years ( we are not asked about the exact amount so such an approximation will do ) . thus after num__18 years investment will become $ num__6000 * num__2 = $ num__12000 . answer : c . <eor> c <eos> |
c |
divide__54.0__6.0__ divide__18.0__9.0__ multiply__6000.0__2.0__ multiply__6000.0__2.0__ |
divide__54.0__6.0__ divide__18.0__9.0__ multiply__6000.0__2.0__ multiply__6000.0__2.0__ |
| in a bag of small balls num__0.25 are green num__0.125 are blue num__0.0833333333333 are yellow and the remaining num__26 white . how many balls are blue ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
let us first find the fraction of green blue and yellow balls num__0.25 + num__0.125 + num__0.0833333333333 = num__0.25 + num__0.125 + num__0.0833333333333 common denominator = num__0.458333333333 add numerators the fraction of white balls is given by num__1.0 - num__0.458333333333 = num__0.541666666667 so the fraction num__0.541666666667 corresponds to num__26 balls . if x is the total number of balls then ( num__0.541666666667 ) of x = num__26 balls or ( num__0.541666666667 ) × x = num__26 x = num__26 × ( num__1.84615384615 ) = num__48 total number of balls the fraction of blue balls is num__0.125 of x . the number of blue balls is given by ( num__0.125 ) of num__48 = num__6 balls . correct answer is c ) num__6 <eor> c <eos> |
c |
subtract__1.0__0.4583__ multiply__0.125__48.0__ multiply__0.125__48.0__ |
subtract__1.0__0.4583__ multiply__0.125__48.0__ multiply__0.125__48.0__ |
| what least whole no . should be added to num__532869 to make it divisible by num__9 ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__9 <o> d ) num__10 <o> e ) num__12 |
if a number is divisible by num__9 the sum of its digits must be a multiple of num__9 . here num__5 + num__3 + num__2 + num__8 + num__6 + num__9 = num__33 the next multiple of num__9 is num__36 . num__3 must be added to num__532869 to make it divisible by num__9 c <eor> c <eos> |
c |
subtract__5.0__3.0__ add__3.0__5.0__ subtract__9.0__3.0__ add__33.0__3.0__ add__3.0__6.0__ |
subtract__5.0__3.0__ add__3.0__5.0__ subtract__9.0__3.0__ add__33.0__3.0__ add__3.0__6.0__ |
| if d is the sum of consecutive even integers w x y and z where w < x < y < z all of the following must be true except <o> a ) z - w = num__3 ( y - x ) <o> b ) d is divisible by num__8 <o> c ) the average of w x y and z is odd <o> d ) d is divisible by num__4 <o> e ) w + x + num__8 = y + z |
just assume that the numbers are num__2 a num__2 a + num__2 num__2 a + num__4 and num__2 a + num__6 . d = num__2 a + num__2 a + num__2 + num__2 a + num__4 + num__2 a + num__6 = num__8 a + num__12 focus on the easiest options first . you see that d is not divisible by num__8 . answer ( b ) <eor> b <eos> |
b |
add__2.0__4.0__ multiply__2.0__4.0__ add__8.0__4.0__ subtract__12.0__4.0__ |
add__2.0__4.0__ add__2.0__6.0__ add__8.0__4.0__ add__2.0__6.0__ |
| for a finite sequence of non zero numbers the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative . what is the number of variations in sign for the sequence num__1 - num__3 num__2 num__5 - num__4 - num__6 - num__8 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
given sequence : { num__1 - num__3 num__2 num__5 - num__4 - num__6 - num__8 } the questions basically asks : how manypairs of consecutive termsare there in the sequence such that the product of these consecutive terms is negative . num__1 * ( - num__3 ) = - num__3 = negative ; - num__3 * num__2 = - num__6 = negative ; num__2 * num__5 = num__10 = positive ; num__5 * ( - num__4 ) = - num__20 = negative ; ( - num__4 ) * ( - num__6 ) = num__24 = positive . so there are num__3 pairs of consecutive terms of the sequence for which the product is negative . answer : c . <eor> c <eos> |
c |
multiply__2.0__5.0__ multiply__2.0__10.0__ multiply__3.0__8.0__ multiply__1.0__3.0__ |
multiply__2.0__5.0__ multiply__2.0__10.0__ multiply__3.0__8.0__ multiply__1.0__3.0__ |
| line w has the equation num__3 x + y = num__7 . which of the following lines is perpendicular to line w ? <o> a ) y = num__3 x + num__4 <o> b ) y = – num__3 x – num__6 <o> c ) y = ( num__0.333333333333 ) x – num__1 <o> d ) y = ( – num__0.333333333333 ) x + num__2 <o> e ) y = ( – num__2.33333333333 ) x – num__5 |
i first rewrote the equation in the standard y = mx + b form . therefore line w as presented num__3 x + y = num__7 can be rewritten as follows : y = - num__3 x + num__7 . thought process next is what line would be perpendicular to line w ? any line with a reciprocal of the slope but in the opposite direction . the reciprocal of any fraction / integer is num__1 over that number / integer . therefore the reciprocal of - num__3 is - num__0.333333333333 - need to drop the negative sign because the line would kinda run parallel and we want perpendicular . scan the answers choices and notice c as the only one . <eor> c <eos> |
c |
reverse__3.0__ reverse__3.0__ |
reverse__3.0__ reverse__3.0__ |
| a man took loan from a bank at the rate of num__12.0 p . a . simple interest . after num__3 years he had to pay rs . num__5400 interest only for the period . the principal amount borrowed by him was . <o> a ) num__15000 <o> b ) num__2655 <o> c ) num__16888 <o> d ) num__6677 <o> e ) num__1871 |
principal = rs . ( num__100 x num__5400 ) / ( num__12 * num__3 ) = rs . num__15000 . answer : a <eor> a <eos> |
a |
percent__100.0__15000.0__ |
percent__100.0__15000.0__ |
| a can do a work in num__30 days b can do it in num__20 days . they together under took to do a piece of work for rs . num__1000 what is the share of b ? <o> a ) rs . num__500 <o> b ) rs . num__400 <o> c ) rs . num__800 <o> d ) rs . num__600 <o> e ) rs . num__200 |
explanation : share of their work days = num__30 : num__20 share of their work = num__20 : num__30 share of b ’ s money = num__0.6 * num__1000 = num__600 answer : option d <eor> d <eos> |
d |
km_to_mile_conversion__ multiply__30.0__20.0__ round__600.0__ |
km_to_mile_conversion__ multiply__30.0__20.0__ multiply__30.0__20.0__ |
| what least number should be added to num__1056 so that the sum is completely divisible by num__23 <o> a ) num__4 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__5 |
explanation : ( num__45.9130434783 ) gives remainder num__21 num__21 + num__2 = num__23 so we need to add num__2 answer : option c <eor> c <eos> |
c |
divide__1056.0__23.0__ subtract__23.0__21.0__ subtract__23.0__21.0__ |
divide__1056.0__23.0__ subtract__23.0__21.0__ subtract__23.0__21.0__ |
| cereal a is num__10.0 sugar by weight whereas healthier but less delicious cereal b is num__2.0 sugar by weight . to make a delicious and healthy mixture that is num__6.0 sugar what should be the ratio of cereal a to cereal b by weight ? <o> a ) num__2 : num__9 <o> b ) num__2 : num__7 <o> c ) num__1 : num__6 <o> d ) num__1 : num__1 <o> e ) num__1 : num__3 |
( num__0.1 ) a + ( num__0.02 ) b = ( num__0.06 ) ( a + b ) num__4 a = num__4 b = > a / b = num__1.0 answer is d . <eor> d <eos> |
d |
reverse__10.0__ subtract__10.0__6.0__ multiply__10.0__0.1__ reverse__1.0__ |
reverse__10.0__ subtract__10.0__6.0__ multiply__10.0__0.1__ reverse__1.0__ |
| in a division sum the remainder is num__0 . as student mistook the divisor by num__12 instead of num__21 and obtained num__56 as quotient . what is the correct quotient ? <o> a ) num__0 <o> b ) num__12 <o> c ) num__32 <o> d ) num__20 <o> e ) num__25 |
num__12 * num__56 = num__672 num__672.0 num__21 = num__32 answer : c <eor> c <eos> |
c |
multiply__12.0__56.0__ divide__672.0__21.0__ divide__672.0__21.0__ |
multiply__12.0__56.0__ divide__672.0__21.0__ divide__672.0__21.0__ |
| how many minutes is it before num__12 noon if num__16 minutes ago it was three times as many minutes after num__9 am ? <o> a ) num__41 minutes <o> b ) num__49 minutes <o> c ) num__40 minutes <o> d ) num__42 minutes <o> e ) num__44 minutes |
a num__41 minutes <eor> a <eos> |
a |
round__41.0__ |
round__41.0__ |
| num__1394 x num__1394 = ? <o> a ) a ) num__1951609 <o> b ) b ) num__1951601 <o> c ) c ) num__1951602 <o> d ) d ) num__1943236 <o> e ) e ) num__1951604 |
num__1394 x num__1394 = ( num__1394 ) num__2 = ( num__1400 - num__6 ) num__2 = ( num__1400 ) num__2 + ( num__6 ) num__2 - ( num__2 x num__1400 x num__6 ) = num__1960000 + num__36 - num__16800 = num__1960036 - num__16800 = num__1943236 . answer : d <eor> d <eos> |
d |
subtract__1400.0__1394.0__ add__1960000.0__36.0__ subtract__1960036.0__16800.0__ subtract__1960036.0__16800.0__ |
subtract__1400.0__1394.0__ add__1960000.0__36.0__ subtract__1960036.0__16800.0__ subtract__1960036.0__16800.0__ |
| which of the following is true about num__0 < | x | - x < num__5 ? <o> a ) - num__1 < x < num__0 <o> b ) num__0 < x < num__1 <o> c ) num__1 < x < num__2 <o> d ) num__2 < x < num__3 <o> e ) num__3 < x < num__4 |
when can be | x | - x be greater than num__0 . . never when x is positive . . only a is left . . you can also substitute any value within the given range to check for answer example - put x as - num__0.5 for a . - num__1 < x < num__0 . . so | x | - x = | - num__0.5 | - ( - num__0.5 ) = num__1 and it satisfies num__0 < | x | - num__4 x < num__5 a <eor> a <eos> |
a |
subtract__5.0__1.0__ reverse__1.0__ |
subtract__5.0__1.0__ subtract__5.0__4.0__ |
| if logx y = num__10 and log num__2 x = num__1000 what is the value of y ? <o> a ) num__210002 <o> b ) num__277889 <o> c ) num__210000 <o> d ) num__27788 <o> e ) num__788291 |
explanation : log num__2 x = num__1000 = > x = num__21000 logx y = num__10 = > y = x num__10 = ( num__21000 ) num__10 = num__2 ( num__1000 × num__10 ) = num__210000 answer : option c <eor> c <eos> |
c |
multiply__10.0__21000.0__ multiply__10.0__21000.0__ |
multiply__10.0__21000.0__ multiply__10.0__21000.0__ |
| a man swims downstream num__72 km and upstream num__45 km taking num__9 hours each time ; what is the speed of the current ? <o> a ) num__1.5 <o> b ) num__1.8 <o> c ) num__1.6 <o> d ) num__1.3 <o> e ) num__1.1 |
num__72 - - - num__9 ds = num__8 ? - - - - num__1 num__45 - - - - num__9 us = num__5 ? - - - - num__1 s = ? s = ( num__8 - num__5 ) / num__2 = num__1.5 answer : b <eor> b <eos> |
b |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ divide__9.0__5.0__ |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ divide__9.0__5.0__ |
| a shop owner sells num__30 mtr of cloth and gains sp of num__10 metres . find the gain % ? <o> a ) num__40.0 <o> b ) num__45.0 <o> c ) num__50.0 <o> d ) num__65.0 <o> e ) num__70 % |
here selling price of num__10 m cloth is obtained as profit . profit of num__10 m cloth = ( s . p . of num__30 m cloth ) – ( c . p . of num__30 m cloth ) selling price of num__20 m cloth = selling price of num__30 m of cloth let cost of each metre be rs . num__100 . therefore cost price of num__20 m cloth = rs . num__2000 and s . p . of num__20 m cloth = rs . rs . num__3000 profit % = num__10 × num__100 = num__50.0 num__20 profit of num__50.0 was made by the merchant . c <eor> c <eos> |
c |
percent__100.0__50.0__ |
percent__100.0__50.0__ |
| what least number must be added to num__1145 so that the sum is completely divisible by num__23 ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__8 |
num__50 * num__23 = num__1150 num__1150 - num__1145 = num__5 answer : d <eor> d <eos> |
d |
multiply__23.0__50.0__ subtract__1150.0__1145.0__ subtract__1150.0__1145.0__ |
multiply__23.0__50.0__ subtract__1150.0__1145.0__ subtract__1150.0__1145.0__ |
| a number consists of num__3 digits whose sum is num__10 . the middle digit is equal to the sum of the other two and the number will be increased by num__99 if its digits are reversed . the number is <o> a ) num__145 <o> b ) num__185 <o> c ) num__253 <o> d ) num__370 <o> e ) none |
solution let the number be x . then num__2 x = num__10 or x = num__5 . so the number is either num__253 or num__352 . since the number increases on reversing the digits so the hundred ' s digit is smaller than the units digit . hence required number = num__253 . answer c <eor> c <eos> |
c |
add__3.0__2.0__ add__99.0__253.0__ subtract__352.0__99.0__ |
add__3.0__2.0__ add__99.0__253.0__ subtract__352.0__99.0__ |
| a telephone company needs to create a set of num__3 - digit area codes . the company is entitled to use only digits num__2 num__4 and num__5 which can be repeated . if the product of the digits in the area code must be even how many different codes can be created ? <o> a ) num__20 <o> b ) num__22 <o> c ) num__24 <o> d ) num__26 <o> e ) num__30 |
we have num__3 places ( - - - ) to fill & we can use only digits num__2 num__4 num__5 first place can be filled in num__3 ways num__2 nd place can be filled in num__3 ways because we can repeat the digits num__3 rd place can be filled in num__3 ways because we can repeat the digits total no of codes = num__3 x num__3 x num__3 = num__27 but one combination ( num__555 ) will not be even so total no of even codes = num__27 - num__1 = num__26 answer d <eor> d <eos> |
d |
subtract__3.0__2.0__ subtract__27.0__1.0__ multiply__1.0__26.0__ |
subtract__3.0__2.0__ subtract__27.0__1.0__ subtract__27.0__1.0__ |
| a set of numbers has the property that for any number x in the set x + num__2 is also in the set . if - num__2 is in the set which of the following must also be in the set num__1 . - num__4 num__2 . num__4 num__3 . num__0 <o> a ) num__1 only <o> b ) num__2 only <o> c ) num__1 and num__2 only <o> d ) num__1 and num__3 only <o> e ) num__12 and num__3 |
if x is in the set than x + num__2 is also in the set if - num__2 in the set than - num__2 + num__2 = num__0 is also in the set since + num__2 is present num__2 + num__2 = num__4 is also present in the set . therefore ans d <eor> d <eos> |
d |
reverse__1.0__ |
subtract__2.0__1.0__ |
| a man sells two articles for rs . num__3600 each and he gains num__48.0 on the first and loses num__48.0 on the next . find his total gain or loss ? <o> a ) num__9.0 loss <o> b ) num__400 <o> c ) num__4000 <o> d ) num__324 <o> e ) num__23.04 loss |
( num__48 * num__48 ) / num__100 = num__23.04 loss answer : e <eor> e <eos> |
e |
percent__23.04__100.0__ |
percent__23.04__100.0__ |
| x can finish a work in num__18 days . y can finish the same work in num__15 days . yworked for num__10 days and left the job . how many days does x alone need to finish the remaining work ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
work done by x in num__1 day = num__0.0555555555556 work done by y in num__1 day = num__0.0666666666667 work done by y in num__10 days = num__0.666666666667 = num__0.666666666667 remaining work = num__1 – num__0.666666666667 = num__0.333333333333 number of days in which x can finish the remaining work = ( num__0.333333333333 ) / ( num__0.0555555555556 ) = num__6 c <eor> c <eos> |
c |
divide__1.0__18.0__ divide__1.0__15.0__ divide__10.0__15.0__ subtract__1.0__0.6667__ round__6.0__ |
divide__1.0__18.0__ divide__1.0__15.0__ divide__10.0__15.0__ subtract__1.0__0.6667__ divide__6.0__1.0__ |
| the sum of five numbers is num__655 . the average of the first two numbers is num__85 and the third number is num__125 . find the average of the two numbers ? <o> a ) num__180 <o> b ) num__787 <o> c ) num__178 <o> d ) num__179 <o> e ) num__172 |
let the five numbers be p q r s and t . = > p + q + r + s + t = num__655 . ( p + q ) / num__2 = num__85 and r = num__125 p + q = num__170 and r = num__125 p + q + r = num__295 s + t = num__655 - ( p + q + r ) = num__360 average of the last two numbers = ( s + t ) / num__2 = num__180 . answer : a <eor> a <eos> |
a |
multiply__85.0__2.0__ add__125.0__170.0__ subtract__655.0__295.0__ divide__360.0__2.0__ divide__360.0__2.0__ |
multiply__85.0__2.0__ add__125.0__170.0__ subtract__655.0__295.0__ divide__360.0__2.0__ divide__360.0__2.0__ |
| when a number is divided by num__6 & then multiply by num__12 the answer is num__13 what is the no . ? <o> a ) num__6.5 <o> b ) num__5 <o> c ) num__5.5 <o> d ) num__5.8 <o> e ) num__6 |
if $ x $ is the number x / num__6 * num__12 = num__13 = > num__2 x = num__13 = > x = num__6.5 a <eor> a <eos> |
a |
divide__12.0__6.0__ divide__13.0__2.0__ subtract__13.0__6.5__ |
divide__12.0__6.0__ divide__13.0__2.0__ divide__13.0__2.0__ |
| a grocer has a sale of rs . num__5420 rs . num__5660 rs . num__6200 rs . num__6350 and rs . num__6500 for num__5 consecutive months . find the sale he should have in the sixth month so that he gets an average sale of rs . num__6500 ? <o> a ) rs . num__5870 <o> b ) rs . num__5991 <o> c ) rs . num__8870 <o> d ) rs . num__6850 <o> e ) none of these |
explanation : total sale for num__5 months = rs . ( num__5420 + num__5660 + num__6200 + num__6350 + num__6500 ) = rs . num__30130 therefore required sale = rs . [ ( num__6500 * num__6 ) – num__30130 ] = rs . ( num__39000 – num__30130 ) = rs . num__8870 answer c <eor> c <eos> |
c |
multiply__6500.0__6.0__ subtract__39000.0__30130.0__ subtract__39000.0__30130.0__ |
multiply__6500.0__6.0__ subtract__39000.0__30130.0__ subtract__39000.0__30130.0__ |
| a b c can do a work in num__15 num__2045 days respectively . they get rs num__500 for their work . what is the share of a ? <o> a ) rs . num__240 <o> b ) rs . num__340 <o> c ) rs . num__260 <o> d ) rs . num__280 <o> e ) rs . num__440 |
lcm = num__180 share of a = ( lcm / a x total amount ) / lcm / a + lcm / b + lcm / c = ( num__12.0 ) / ( num__12.0 + num__9.0 + num__4.0 ) = ( num__0.48 ) * num__500 = rs . num__240 answer : a <eor> a <eos> |
a |
divide__180.0__15.0__ multiply__500.0__0.48__ round__240.0__ |
divide__180.0__15.0__ multiply__500.0__0.48__ multiply__500.0__0.48__ |
| a train num__300 m long is running with a speed of num__68 kmph . in what time will it pass a man who is running at num__8 kmph in the same direction in which the train is going ? <o> a ) num__6 sec . <o> b ) num__7 sec . <o> c ) num__9 sec . <o> d ) num__18 sec . <o> e ) none |
solution speed of the train relative to man = ( num__68 - num__8 ) = num__60 kmph = num__60 x num__0.277777777778 = num__16.6666666667 m / sec . time taken by it to cover num__300 m at ( num__16.6666666667 ) m / sec = ( num__300 x num__0.06 ) sec = num__18 sec . answer d <eor> d <eos> |
d |
hour_to_min_conversion__ divide__300.0__16.6667__ round__18.0__ |
subtract__68.0__8.0__ divide__300.0__16.6667__ divide__300.0__16.6667__ |
| if the sides of a triangle are num__26 cm num__24 cm and num__10 cm what is its area ? <o> a ) num__120 <o> b ) num__139 <o> c ) num__140 <o> d ) num__122 <o> e ) num__178 |
the triangle with sides num__26 cm num__24 cm and num__10 cm is right angled where the hypotenuse is num__26 cm . area of the triangle = num__0.5 * num__24 * num__10 = num__120 cm num__2 answer : option a <eor> a <eos> |
a |
triangle_area__24.0__10.0__ square_perimeter__0.5__ triangle_area__24.0__10.0__ |
volume_rectangular_prism__24.0__10.0__0.5__ square_perimeter__0.5__ volume_rectangular_prism__24.0__10.0__0.5__ |
| the grade point average of one third of the classroom is num__54 ; the grade point average of the rest is num__45 . what is the grade point average of the whole class ? <o> a ) num__52 <o> b ) num__54 <o> c ) num__48 <o> d ) num__58 <o> e ) num__60 |
let n = total students in class total points for num__0.333333333333 class = num__54 n / num__3 = num__18 n total points for num__0.666666666667 class = num__45 * num__2 n / num__3 = num__30 n total points for whole class = num__18 n + num__30 n = num__48 n num__48 n total class points / n total students = num__48 grade point average for total class answer : c <eor> c <eos> |
c |
divide__54.0__3.0__ add__45.0__3.0__ add__45.0__3.0__ |
divide__54.0__3.0__ add__45.0__3.0__ add__45.0__3.0__ |
| the speed of a car is num__98 km in the first hour and num__60 km in the second hour . what is the average speed of the car ? <o> a ) num__76 kmph <o> b ) num__75 kmph <o> c ) num__87 kmph <o> d ) num__79 kmph <o> e ) num__86 kmph |
s = ( num__98 + num__60 ) / num__2 = num__79 kmph answer : d <eor> d <eos> |
d |
round__79.0__ |
round__79.0__ |
| if x and y are integers and x + y = num__7 which of the following must be true ? <o> a ) x and y are consecutive integers . <o> b ) both x and y are less than num__5 . <o> c ) if x > num__0 then y < num__0 . <o> d ) both x and y are even . <o> e ) if x < num__0 then y > num__0 . |
in option c if x > num__0 then y < num__0 means x is + ve and y is - ve in option b if x < num__0 then y > num__0 means y is + ve and x is - ve both are same and satisfying x + y = num__5 . how option e is right could you explain <eor> e <eos> |
e |
multiply__7.0__0.0__ |
multiply__7.0__0.0__ |
| the visitors of a modern art museum who watched a certain picasso painting were asked to fill in a short questionnaire indicating whether they had enjoyed looking at the picture and whether they felt they had understood it . according to the results of the survey all num__140 visitors who did not enjoy the painting also did not feel they had understood the painting and the number of visitors who enjoyed the painting was equal to the number of visitors who felt they had understood the painting . if num__0.75 of the visitors who answered the questionnaire both enjoyed the painting and felt they had understood the painting then how many visitors answered the questionnaire ? <o> a ) num__90 <o> b ) num__120 <o> c ) num__160 <o> d ) num__360 <o> e ) num__560 |
if we exclude those cases and take the question at face value then it seems straightforward . group # num__1 = ( did n ' t like did n ' t understand ) = num__120 group # num__2 = ( likeunderstood ) = num__0.75 ( num__0.25 ) n = num__560 n = num__480 answer = ( e ) <eor> e <eos> |
e |
subtract__1.0__0.75__ divide__140.0__0.25__ divide__120.0__0.25__ divide__140.0__0.25__ |
subtract__1.0__0.75__ divide__140.0__0.25__ divide__120.0__0.25__ divide__140.0__0.25__ |
| an amount of rs . num__100000 is invested in two types of shares . the first yields an interest of num__9.0 p . a and the second num__11.0 p . a . if the total interest at the end of one year is num__9 num__0.5 % then the amount invested at num__11.0 was ? <o> a ) num__23777 <o> b ) num__25000 <o> c ) num__29977 <o> d ) num__26777 <o> e ) num__19871 |
let the sum invested at num__9.0 be rs . x and that invested at num__11.0 be rs . ( num__100000 - x ) . then ( x * num__9 * num__1 ) / num__100 + [ ( num__100000 - x ) * num__11 * num__1 ] / num__100 = ( num__100000 * num__9.5 * num__0.01 ) ( num__9 x + num__1100000 - num__11 x ) = num__950000 x = num__75000 sum invested at num__9.0 = rs . num__75000 sum invested at num__11.0 = rs . ( num__100000 - num__75000 ) = rs . num__25000 . answer : b <eor> b <eos> |
b |
percent__100.0__25000.0__ |
percent__100.0__25000.0__ |
| there are num__100 fish in a pond . a fisherman put num__20 more fish in the pond . how many fish are in the pond <o> a ) num__0 <o> b ) num__19 <o> c ) num__178 <o> d ) num__140 <o> e ) num__120 |
if there are num__100 fish and you put num__20 more fish in the pond you will have num__120 fish num__100 + num__20 = num__120 e <eor> e <eos> |
e |
add__100.0__20.0__ add__100.0__20.0__ |
add__100.0__20.0__ add__100.0__20.0__ |
| a and b are two stations num__390 km apart . a train starts from a at num__10 a . m . and travels towards b at num__65 kmph . another train starts from b at num__11 a . m . and travels towards a at num__35 kmph . at what time do they meet ? <o> a ) num__2.15 p . m . <o> b ) num__1.15 p . m . <o> c ) num__4.15 p . m . <o> d ) num__3.15 p . m . <o> e ) num__12.15 p . m . |
suppose they meet x hours after num__10 a . m . then ( distance moved by first in x hrs ) + [ distance moved by second in ( x - num__1 ) hrs ] = num__390 . num__65 x + num__35 ( x - num__1 ) = num__390 = > num__100 x = num__425 = > x = num__4.25 so they meet num__4 hrs . num__15 min . after num__10 a . m i . e . at num__2.15 p . m . option a <eor> a <eos> |
a |
subtract__11.0__10.0__ add__65.0__35.0__ add__390.0__35.0__ divide__425.0__100.0__ add__11.0__4.0__ round__2.15__ |
subtract__11.0__10.0__ add__65.0__35.0__ add__390.0__35.0__ divide__425.0__100.0__ add__11.0__4.0__ round__2.15__ |
| let a be a positive integer . if n is divisible by num__2 ^ a and n is also divisible by num__3 ^ ( num__2 a ) then it is possible that n is not divisible by <o> a ) num__6 <o> b ) num__3 × num__2 ^ a <o> c ) num__2 × num__3 ^ ( num__2 a ) <o> d ) num__6 ^ a <o> e ) num__6 ^ ( num__2 a ) |
easiest method is to assume a value of a = num__1 given n is divisible by num__2 ^ num__2 - - - > n = num__2 ^ num__1 * p - - - > n = num__2 p also n is also divisible by num__3 ^ ( num__2 a ) - - - > n = num__3 ^ ( num__2 a ) * q = num__3 ^ num__2 * q = num__9 q lets look at a few numbers that are multiples of both num__2 and num__9 are num__18 num__3672 . . . . thus looking at the options a - d divide the numbers num__18 num__3672 . . . . while e ( = num__6 ^ ( num__2 ) = num__36 ) does not divide num__18 . thus e is the correct answer . <eor> e <eos> |
e |
subtract__3.0__2.0__ multiply__2.0__9.0__ multiply__2.0__3.0__ multiply__2.0__18.0__ multiply__2.0__3.0__ |
subtract__3.0__2.0__ multiply__2.0__9.0__ multiply__2.0__3.0__ multiply__2.0__18.0__ multiply__2.0__3.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__27 seconds . what is the length of the train ? <o> a ) num__186 m <o> b ) num__176 m <o> c ) num__450 m <o> d ) num__150 m <o> e ) num__765 m |
speed = ( num__60 * num__0.277777777778 ) m / sec = ( num__16.6666666667 ) m / sec length of the train = ( speed x time ) = ( num__16.6666666667 * num__27 ) m = num__450 m . answer : c <eor> c <eos> |
c |
round__450.0__ |
round__450.0__ |
| an alloy weighing num__48 ounces is num__25.0 gold . how many ounces of pure gold must be added to create an alloy that is num__40.0 gold ? <o> a ) num__15 <o> b ) num__20 <o> c ) num__12 <o> d ) num__14 <o> e ) num__7 |
an alloy of num__48 oz which is num__25.0 gold means there is num__12 oz of gold . to get to an alloy that is num__40.0 gold let ' s use this expression : ( num__12 + x ) / ( num__48 + x ) = num__0.40 with x representing the amount of pure gold that must be added to get to num__40.0 . the expression we are using represents the new total weight of pure gold over the new total weight of the alloy and this fraction should represent num__40.0 or num__0.4 . you will see that num__12 is the correct answer as num__0.4 = num__0.4 choose c <eor> c <eos> |
c |
percent__48.0__25.0__ percent__48.0__25.0__ |
percent__48.0__25.0__ percent__48.0__25.0__ |
| a tank has num__5 inlet pipes . three pipes are narrow and two are wide . each of the three narrow pipes works at num__0.5 the rate of each of the wide pipes . all the pipes working together will take what fraction r of time taken by the two wide pipes working together to fill the tank ? <o> a ) num__0.5 <o> b ) num__0.666666666667 <o> c ) num__0.75 <o> d ) num__0.428571428571 <o> e ) num__0.571428571429 |
i always plug in for these questions . . i find them easier let num__2 large pipes each work fill num__10 lts / hr small ones = num__0.5 * num__10 = num__5 total num__5 = num__10 + num__10 + num__5 + num__5 + num__5 = num__35 lts / hrs now assume the total capacity = num__140 lts ( lcm of num__2035 ) qn : all the pipes working together will take what fraction of time taken by the two wide pipes working together to fill the tank all working together will take num__4.0 = num__4 hrs two large pipes working will take num__7.0 = num__7 hrs hence ratio r = num__0.571428571429 = e <eor> e <eos> |
e |
divide__5.0__0.5__ divide__2.0__0.5__ add__5.0__2.0__ divide__4.0__7.0__ divide__4.0__7.0__ |
divide__5.0__0.5__ divide__2.0__0.5__ add__5.0__2.0__ divide__4.0__7.0__ divide__4.0__7.0__ |
| can you find the missing number in the sequence given below ? num__10 num__16 num__22 num__11 num__17 ? num__12 num__18 num__24 num__13 num__19 num__25 <o> a ) num__25 <o> b ) num__23 <o> c ) num__20 <o> d ) num__18 <o> e ) num__19 |
let ' s break the given series as below : num__10 num__16 num__22 num__11 num__17 ? num__12 num__18 num__24 num__13 num__19 num__25 now read the number from left hand side from top to bottom as : so the number that will replace ' ? ' is num__23 answer : b <eor> b <eos> |
b |
add__10.0__13.0__ add__10.0__13.0__ |
add__10.0__13.0__ add__10.0__13.0__ |
| two numbers differ by num__5 . if their product is num__336 then sum of two number is <o> a ) num__33 <o> b ) num__34 <o> c ) num__36 <o> d ) num__37 <o> e ) num__38 |
explanation : friends you remember = > ( x + y ) num__2 = ( x − y ) num__2 + num__4 xy = > ( x + y ) num__2 = ( num__5 ) num__2 + num__4 ( num__336 ) = > ( x + y ) = num__1369 − − − − √ = num__37 option d <eor> d <eos> |
d |
divide__1369.0__37.0__ |
divide__1369.0__37.0__ |
| which of the following numbers is between num__1 ⁄ num__3 and num__1 ⁄ num__2 ? <o> a ) . num__45 <o> b ) . num__32 <o> c ) . num__29 <o> d ) . num__22 <o> e ) . num__20 |
is n ' t it a ? num__0.333333333333 = . num__33 num__0.5 = . num__50 the only answer between that is a . num__45 . <eor> a <eos> |
a |
reverse__3.0__ reverse__2.0__ multiply__1.0__45.0__ |
reverse__3.0__ reverse__2.0__ multiply__1.0__45.0__ |
| two cars p and q start at the same time from a and b which are num__120 km apart . if the two cars travel in opposite directions they meet after one hour and if they travel in same direction ( from a towards b ) then p meets q after num__6 hours . what is the speed of car p ? <o> a ) num__87 km / hr <o> b ) num__70 km / hr <o> c ) num__76 km / hr <o> d ) num__97 km / hr <o> e ) num__57 km / hr |
let their speed be x km / hr and y km / he respectively . then num__120 / ( x + y ) = num__1 = > x + y = num__120 - - - ( i ) now when they move in same direction : ( distance traveled by p in num__6 hrs ) - ( distance traveled by q in num__6 hrs ) = num__120 km num__6 x - num__6 y = num__120 = > x - y = num__20 - - - ( ii ) sloving ( i ) and ( ii ) we get x = num__70 y = num__50 p ' s speed = num__70 km / hr . answer : b <eor> b <eos> |
b |
divide__120.0__6.0__ subtract__120.0__70.0__ round__70.0__ |
divide__120.0__6.0__ subtract__120.0__70.0__ subtract__120.0__50.0__ |
| when the positive integer k is divided by the positive integer n the remainder is num__11 . if k / n = num__81.1 what is the value of n ? <o> a ) num__9 <o> b ) num__20 <o> c ) num__55 <o> d ) num__70 <o> e ) num__81 |
here ' s an approach that ' s based on number properties and a bit ofbrute forcemath : we ' re told that k and n are both integers . since k / n = num__81.2 we can say that k = num__81.2 ( n ) n has tomultiply outthe . num__2 so that k becomes an integer . with the answers that we have to work with n has to be a multiple of num__5 . eliminate a and e . with the remaining answers we can test the answers and find the one that fits the rest of the info ( k / n = num__81.2 and k / n has a remainder of num__11 ) answer b : if n = num__20 then k = num__1624 ; num__81.2 has a remainder of num__4 not a match answer c : if n = num__55 then k = num__4466 ; num__81.2 has a remainder of num__11 match . final answer : e <eor> e <eos> |
e |
multiply__81.2__20.0__ divide__20.0__5.0__ multiply__11.0__5.0__ multiply__81.2__55.0__ round_down__81.1__ |
multiply__81.2__20.0__ divide__20.0__5.0__ multiply__11.0__5.0__ multiply__81.2__55.0__ round_down__81.1__ |
| on day one a store sold num__90 books . on each of the next d days the company sold num__62 books . if the average daily sales of books over this time period ( including day one ) is num__66 books what is the value of d ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
num__90 + num__62 d = num__66 ( d + num__1 ) . num__4 d = num__24 . d = num__6 . the answer is b . <eor> b <eos> |
b |
subtract__66.0__62.0__ subtract__90.0__66.0__ divide__24.0__4.0__ multiply__1.0__6.0__ |
subtract__66.0__62.0__ subtract__90.0__66.0__ divide__24.0__4.0__ multiply__1.0__6.0__ |
| a invested rs num__76000 in a business . after few months b joined him with rs num__57000 . the total profit was divided between them in the ratio num__2 : num__1 at the end of the year . after how many months did b join ? <o> a ) num__2 months <o> b ) num__3 months <o> c ) num__5 months <o> d ) num__4 months <o> e ) num__1 months |
suppose b was there in the business for x months . then a : b = num__76000 × num__12 : num__57000 × x therefore num__76000 × num__12 : num__57000 × x = num__2 : num__1 num__76 × num__12 : num__57 x = num__2 : num__1 num__76 × num__12 × num__1 = num__57 x × num__2 num__76 × num__4 = num__19 x × num__2 num__4 × num__4 = ( x ) × num__2 x = num__8 hence b was there in the business for num__8 months or joined after num__12 - num__8 = num__4 months answer is d . <eor> d <eos> |
d |
divide__76.0__4.0__ multiply__2.0__4.0__ multiply__1.0__4.0__ |
subtract__76.0__57.0__ subtract__12.0__4.0__ subtract__12.0__8.0__ |
| an article is bought for rs . num__600 and sold for rs . num__500 find the loss percent ? <o> a ) num__16 num__0.666666666667 % <o> b ) num__15 num__0.666666666667 % <o> c ) num__18 num__0.666666666667 % <o> d ) num__11 num__0.666666666667 % <o> e ) num__22 num__0.666666666667 % |
a num__16 num__0.666666666667 num__600.0 - - - - num__100 num__100 - - - - ? = > num__16 num__0.666666666667 % <eor> a <eos> |
a |
percent__16.0__100.0__ |
percent__16.0__100.0__ |
| a watch was sold at a loss of num__10.0 . if it was sold for rs . num__140 more there would have been a gain of num__4.0 . what is the cost price ? <o> a ) num__1000 <o> b ) num__2777 <o> c ) num__289 <o> d ) num__2779 <o> e ) num__27871 |
num__90.0 num__104.0 - - - - - - - - num__14.0 - - - - num__140 num__100.0 - - - - ? = > rs . num__1000 . answer : a <eor> a <eos> |
a |
percent__10.0__140.0__ percent__100.0__1000.0__ |
percent__10.0__140.0__ percent__100.0__1000.0__ |
| how much time will take for an amount of rs . num__200 to yield rs . num__81 as interest at num__4.5 per annum of simple interest ? <o> a ) num__7 <o> b ) num__4 <o> c ) num__5 <o> d ) num__3 <o> e ) num__9 |
time = ( num__100 * num__81 ) / ( num__200 * num__4.5 ) = num__4 years ' answer : e <eor> e <eos> |
e |
percent__4.5__200.0__ |
percent__4.5__200.0__ |
| p can lay railway track between two stations in num__16 days . q can do the same job in num__12 days . with the help of r they completes the job in num__4 days . how much days does it take for r alone to complete the work ? <o> a ) num__9 ( num__0.6 ) days <o> b ) num__9 ( num__0.2 ) days <o> c ) num__9 ( num__0.4 ) days <o> d ) num__10 days <o> e ) num__15 days |
explanation : amount of work p can do in num__1 day = num__0.0625 amount of work q can do in num__1 day = num__0.0833333333333 amount of work p q and r can together do in num__1 day = num__0.25 amount of work r can do in num__1 day = num__0.25 - ( num__0.0625 + num__0.0833333333333 ) = num__0.1875 â € “ num__0.0833333333333 = num__0.104166666667 = > hence r can do the job on num__9.6 days = num__9 ( num__0.6 ) days answer is a <eor> a <eos> |
a |
divide__1.0__16.0__ divide__1.0__12.0__ divide__4.0__16.0__ subtract__0.25__0.0625__ subtract__0.1875__0.0833__ km_to_mile_conversion__ round__9.0__ |
divide__1.0__16.0__ divide__1.0__12.0__ divide__4.0__16.0__ subtract__0.25__0.0625__ subtract__0.1875__0.0833__ subtract__9.6__9.0__ subtract__9.6__0.6__ |
| a contractor undertook to do a piece of work in num__6 days . he employed certain number of laboures but num__7 of them were absent from the very first day and the rest could finish the work in only num__10 days . find the number of men originally employed ? <o> a ) num__17.2 <o> b ) num__16.5 <o> c ) num__17.5 <o> d ) num__17.9 <o> e ) num__17.3 |
let the number of men originally employed be x . num__6 x = num__10 ( x â € “ num__7 ) or x = num__17.5 answer c <eor> c <eos> |
c |
round__17.5__ |
round__17.5__ |
| in n is a positive integer less than num__200 and num__14 n / num__60 is an integer then n has how many different positive prime factors q ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__5 <o> d ) num__6 <o> e ) num__8 |
i like to put the numbers in prime factors so it is easier and faster to visualize . num__14 * n / num__60 if we write the factors of num__14 - - > num__2 num__7 and the factors of num__60 - - > num__2 num__2 num__3 num__5 we have ( num__2 * num__7 * n ) / ( num__2 ^ num__2 * num__3 * num__5 ) simplifying num__7 * n / ( num__2 * num__3 * num__5 ) the only way the equation above has an integer value is if n has at least the factors num__2 num__3 and num__5 so we can simplify again and we have the number num__7 . the number could be num__2 * num__3 * num__5 or num__2 * num__3 * num__5 * num__2 or num__2 * num__3 * num__5 * . . . . . however to be less than num__200 we can not add any prime number . num__2 * num__3 * num__5 = num__120 if we added the next prime factor num__7 we would have q = num__2 * num__3 * num__5 * num__7 = num__840 thus answer b . <eor> b <eos> |
b |
divide__14.0__2.0__ add__2.0__3.0__ multiply__60.0__2.0__ multiply__14.0__60.0__ subtract__5.0__2.0__ |
divide__14.0__2.0__ subtract__7.0__2.0__ multiply__60.0__2.0__ multiply__14.0__60.0__ subtract__5.0__2.0__ |
| num__9000 + num__16 num__0.666666666667 % of ? = num__10500 <o> a ) num__1500 <o> b ) num__1750 <o> c ) num__9000 <o> d ) num__7500 <o> e ) none of these |
explanation : num__9000 + num__16 num__0.666666666667 % of ? = num__10500 = > num__9000 + num__16.6666666667 % of ? = num__10500 num__50 / ( num__3 * num__100 ) of ? = num__1500 = > ? = num__1500 * num__6 ? = num__9000 answer is c <eor> c <eos> |
c |
add__16.0__0.6667__ divide__50.0__16.6667__ subtract__10500.0__9000.0__ divide__9000.0__1500.0__ subtract__10500.0__1500.0__ |
add__16.0__0.6667__ divide__50.0__16.6667__ subtract__10500.0__9000.0__ divide__9000.0__1500.0__ multiply__6.0__1500.0__ |
| a tank is filled by num__3 pipes with uniform flow . the first two pipes operating simultaneously fill the tan in the same time during which the tank is filled by the third pipe alone . the num__2 nd pipe fills the tank num__5 hours faster than first pipe and num__4 hours slower than third pipe . the time required by first pipe is <o> a ) num__13 hours <o> b ) num__45 hours <o> c ) num__15 hours <o> d ) num__12 hours <o> e ) num__11 hours |
if first pipe fills tank in x hrs then second pipe fills tank in x - num__5 hrs and num__3 rd pipe fills tank in x - num__9 hrs then as per given condition num__1 / x + num__1 / ( x - num__5 ) = num__1 / ( x - num__9 ) solving it we get x = num__15 hrs answer : c <eor> c <eos> |
c |
add__5.0__4.0__ subtract__3.0__2.0__ multiply__3.0__5.0__ round__15.0__ |
add__5.0__4.0__ subtract__3.0__2.0__ multiply__3.0__5.0__ divide__15.0__1.0__ |
| a and b can finish a work in num__16 days while a alone can do the same work in num__32 days . in how many days b alone will complete the work ? <o> a ) num__76 days <o> b ) num__48 days <o> c ) num__98 days <o> d ) num__32 days <o> e ) num__22 days |
b = num__0.0625 – num__0.03125 = num__0.03125 = > num__32 days answer : d <eor> d <eos> |
d |
round__32.0__ |
round__32.0__ |
| convert num__0.4 to a decimal <o> a ) num__0.1 <o> b ) num__0.2 <o> c ) num__0.3 <o> d ) num__0.4 <o> e ) num__0.5 |
divide num__2 by num__5 : num__0.4 = num__0.4 answer : num__0.4 = num__0.4 ( d ) <eor> d <eos> |
d |
divide__2.0__0.4__ round__0.4__ |
divide__2.0__0.4__ round__0.4__ |
| the batting average of a particular batsman is num__62 runs in num__46 innings . if the difference in his highest and lowest score is num__150 runs and his average excluding these two innings is num__58 runs find his highest score . <o> a ) num__179 <o> b ) num__208 <o> c ) num__210 <o> d ) num__225 <o> e ) num__229 |
explanation : total runs scored by the batsman = num__62 * num__46 = num__2852 runs now excluding the two innings the runs scored = num__58 * num__44 = num__2552 runs hence the runs scored in the two innings = num__2852 – num__2552 = num__300 runs . let the highest score be x hence the lowest score = x – num__150 x + ( x - num__150 ) = num__300 num__2 x = num__450 x = num__225 runs answer d <eor> d <eos> |
d |
multiply__62.0__46.0__ multiply__58.0__44.0__ subtract__2852.0__2552.0__ subtract__46.0__44.0__ add__150.0__300.0__ divide__450.0__2.0__ subtract__450.0__225.0__ |
multiply__62.0__46.0__ multiply__58.0__44.0__ subtract__2852.0__2552.0__ subtract__46.0__44.0__ add__150.0__300.0__ divide__450.0__2.0__ subtract__450.0__225.0__ |
| a person bought num__118 glass bowls at a rate of rs . num__12 per bowl . he sold num__102 of them at rs . num__15 and the remaining broke . what is the percentage gain for a ? <o> a ) num__40 <o> b ) num__27.2727272727 <o> c ) num__34.7142857143 <o> d ) num__8.05084745763 <o> e ) num__34.875 |
cp = num__118 * num__12 = num__1416 and sp = num__102 * num__15 = num__1530 gain % = num__100 * ( num__1530 - num__1416 ) / num__1100 = num__8.05084745763 = num__8.05084745763 answer : d <eor> d <eos> |
d |
percent__100.0__8.0508__ |
percent__100.0__8.0508__ |
| there are three secretaries who work for four departments . if each of the four departments have one report to be typed out and the reports are randomly assigned to a secretary what is the probability t that all three secretary are assigned at least one report ? <o> a ) num__0.888888888889 <o> b ) num__0.79012345679 <o> c ) num__0.444444444444 <o> d ) num__0.197530864198 <o> e ) num__0.555555555556 |
i got the same answer : here ' s my reasoning first report you have num__3 choices second report you have num__2 choices third report you have num__1 choice fourth report num__3 choices again then total number of ways is : num__3 * num__2 * num__1 * num__3 = num__3 ^ num__2 * num__2 this is not correct . you have assumed that the num__4 th report must go to someone who already has a report . there is no such constraint . you can easily give the num__1 st and num__2 nd reports to secretary num__1 num__3 rd report to secretary num__2 and num__4 th report to secretary num__3 . but you have ignored all such cases . the number of ways of ensuring at least one report goes to each secretary is num__4 c num__2 ( select num__2 reports out of num__4 which go to the same person ) * num__3 c num__1 ( select the person who must type num__2 reports ) * num__2 ! ( since you have num__2 reports left which you must distribute to the num__2 remaining people such that each person gets one ) = num__36 required probability t = num__0.444444444444 . c <eor> c <eos> |
c |
subtract__3.0__2.0__ add__1.0__3.0__ multiply__1.0__0.4444__ |
subtract__3.0__2.0__ add__1.0__3.0__ multiply__1.0__0.4444__ |
| a bike covers a certain distance at the speed of num__64 km / h in num__8 hrs . if the bike was to cover the same distance in approximately num__6 hrs at what approximate speed should the bike travel ? <o> a ) num__80.56 km / h <o> b ) num__85.34 km / h <o> c ) num__95.05 km / h <o> d ) num__98 km / h <o> e ) num__99.56 km / h |
ans . ( b ) sol . total distance = num__64 × num__8 = num__512 km now speed = num__85.3333333333 = num__85.34 km / h <eor> b <eos> |
b |
multiply__64.0__8.0__ divide__512.0__6.0__ round__85.34__ |
multiply__64.0__8.0__ divide__512.0__6.0__ round__85.34__ |
| a train num__400 m long can cross an electric pole in num__20 sec and then find the speed of the train ? <o> a ) num__34 <o> b ) num__56 <o> c ) num__72 <o> d ) num__35 <o> e ) num__32 |
length = speed * time speed = l / t s = num__20.0 s = num__20 m / sec speed = num__20 * num__3.6 ( to convert m / sec in to kmph multiply by num__3.6 ) speed = num__72 kmph answer : option c <eor> c <eos> |
c |
multiply__20.0__3.6__ round__72.0__ |
multiply__20.0__3.6__ multiply__20.0__3.6__ |
| peter invested a certain sum of money in a simple interest bond whose value grew to $ num__400 at the end of num__4 years and to $ num__500 at the end of another num__2 years . what was the rate of interest in which he invested his sum ? <o> a ) num__12.0 <o> b ) num__12.5 <o> c ) num__67.0 <o> d ) num__25.0 <o> e ) num__33 % |
lets assume the principal amount ( initial amount invested ) to be p rate of interest to berand time as t . we need to find r now after a time of num__4 years the principal p amounts to $ num__400 and after a time of num__6 years ( question says after another num__6 years so num__4 + num__2 ) p becomes $ num__500 . formulating the above data amount ( a num__1 ) at end of num__4 years a num__1 = p ( num__1 + num__4 r / num__100 ) = num__400 amount ( a num__2 ) at end of num__6 years a num__2 = p ( num__1 + num__2 r / num__100 ) = num__500 dividing a num__2 by a num__1 we get ( num__1 + num__4 r / num__100 ) / ( num__1 + num__2 r / num__100 ) = num__1.25 after cross multiplication we are left with num__4 r = num__100 which gives r = num__25.0 option : d <eor> d <eos> |
d |
percent__25.0__100.0__ |
percent__25.0__100.0__ |
| if a positive integer n is divided by num__5 the remainder is num__3 . which of the number below yields a remainder of num__0 when it is divided by num__5 <o> a ) n + num__3 <o> b ) n + num__2 <o> c ) n - num__2 <o> d ) n - num__1 <o> e ) n + num__1 |
correct answer : b n is divided by num__5 the remainder is num__3 n = num__5 k + num__3 ( k is integer ) add num__2 to both sides of the above equation n + num__2 = num__5 k + num__5 = num__5 ( k + num__1 ) n + num__2 divided by num__5 yields a remainder to zero . <eor> b <eos> |
b |
subtract__5.0__3.0__ subtract__3.0__2.0__ subtract__5.0__3.0__ |
subtract__5.0__3.0__ subtract__3.0__2.0__ subtract__5.0__3.0__ |
| the second of two numbers is two less than three times the first . find the numbers if there sum is num__54 . <o> a ) num__7 - num__19 <o> b ) num__8 - num__20 <o> c ) num__10 - num__16 <o> d ) num__14 - num__40 <o> e ) num__20 - num__6 |
we are looking for two numbers . # num__1 - x # num__2 - num__3 x â € “ num__2 the sum is num__54 . # num__1 + # num__2 = num__54 substituting x + num__3 x â € “ num__2 = num__54 num__4 x â € “ num__2 = num__54 num__4 x = num__56 x = num__14 the first number is num__14 the second number is two less than three times num__14 or num__40 . correct answer d <eor> d <eos> |
d |
add__1.0__2.0__ add__1.0__3.0__ add__54.0__2.0__ divide__56.0__4.0__ subtract__54.0__14.0__ subtract__54.0__40.0__ |
add__1.0__2.0__ add__1.0__3.0__ add__54.0__2.0__ divide__56.0__4.0__ subtract__54.0__14.0__ subtract__54.0__40.0__ |
| of the total amount that jill spent on a shopping trip excluding taxes she spent num__40 percent on clothing num__30 percent on food and num__30 percent on other items . if jill paid a num__4 percent tax on the clothing no tax on the food and an num__8 percent tax on all other items then the total tax that she paid was what percent of the total amount that she spent excluding taxes ? <o> a ) num__2.8 <o> b ) num__3.6 <o> c ) num__4.0 <o> d ) num__5.2 <o> e ) num__6.0 % |
let amount spent by jill = num__100 clothing = num__40 food = num__30 others = num__30 tax on clothing = num__1.6 tax on others = num__2.4 percentage = num__0.04 = num__4.0 answer : c <eor> c <eos> |
c |
subtract__4.0__1.6__ divide__4.0__100.0__ subtract__8.0__4.0__ |
subtract__4.0__1.6__ divide__4.0__100.0__ subtract__8.0__4.0__ |
| a certain farmer pays $ num__30 per acre per month to rent farmland . how much does the farmer pay per month to rent a rectangular plot of farmland that is num__360 feet by num__1210 feet ? ( num__43560 square feet = num__1 acre ) <o> a ) $ num__5330 <o> b ) $ num__3360 <o> c ) $ num__1350 <o> d ) $ num__300 <o> e ) $ num__150 |
basically the question an error . num__1 acre = num__43560 square feet and if it is then the answer is num__300 ( d ) <eor> d <eos> |
d |
multiply__1.0__300.0__ |
multiply__1.0__300.0__ |
| with a uniform speed a car covers the distance in num__8 hours . had the speed been increased by num__2 km / hr the same distance could have been covered in num__7 num__0.5 hours . what is the distance covered ? <o> a ) num__187 km <o> b ) num__480 km <o> c ) num__278 km <o> d ) num__240 km <o> e ) num__671 km |
let the distance be x km . then x / ( num__7 num__0.5 ) - x / num__8 = num__2 num__2 x / num__15 - x / num__8 = num__2 = > x = num__240 km . answer : d <eor> d <eos> |
d |
add__8.0__7.0__ round__240.0__ |
add__8.0__7.0__ round__240.0__ |
| a class contains five juniors and five seniors . if one member of the class is assigned at random to present a paper on a certain subject and another member of the class is randomly assigned to assist him what is the probability t that both will be juniors ? <o> a ) num__0.1 <o> b ) num__0.2 <o> c ) num__0.222222222222 <o> d ) num__0.4 <o> e ) num__0.5 |
i think you ' re right . the other way to do it ( but the long way ) is to figure out the probability that it is not two juniors . num__2 seniors = p ( senior ) * p ( senior ) = num__0.222222222222 num__1 senior and num__1 junior = ( num__0.5 ) * ( num__0.555555555556 ) * num__2 = num__0.555555555556 probability that it is not two juniors is num__0.555555555556 + num__0.222222222222 = num__0.777777777778 so the probability t that it is two juniors is num__1 - ( num__0.777777777778 ) = num__0.222222222222 . c <eor> c <eos> |
c |
coin_space__ negate_prob__0.2222__ negate_prob__0.7778__ |
coin_space__ negate_prob__0.2222__ negate_prob__0.7778__ |
| if a tire rotates at num__400 revolutions per minute when the car is traveling num__120 km / h what is the circumference of the tire ? <o> a ) num__2 meters <o> b ) num__6 meters <o> c ) num__5 meters <o> d ) num__3 meters <o> e ) num__7 meters |
num__400 rev / minute = num__400 * num__60 rev / num__60 minutes = num__24000 rev / hour num__24000 * c = num__120000 m : c is the circumference c = num__5 meters correct answer c <eor> c <eos> |
c |
hour_to_min_conversion__ multiply__400.0__60.0__ divide__120000.0__24000.0__ round__5.0__ |
hour_to_min_conversion__ multiply__400.0__60.0__ divide__120000.0__24000.0__ divide__120000.0__24000.0__ |
| a can do a piece of work in num__15 days . a does the work for num__5 days only and leaves the job . b does the remaining work in num__12 days . in how many days b alone can do the work ? <o> a ) num__5 days <o> b ) num__7 days <o> c ) num__12 days <o> d ) num__9 days <o> e ) num__18 days |
explanation : a ’ s num__5 day work = num__5 * num__0.0666666666667 = num__0.333333333333 remaining work = num__1 - num__0.333333333333 = num__0.666666666667 b completes num__0.666666666667 work in num__6 days b alone can do in x days num__0.666666666667 * x = num__12 x = num__18 days answer : option e <eor> e <eos> |
e |
divide__5.0__15.0__ subtract__1.0__0.3333__ add__5.0__1.0__ add__12.0__6.0__ round__18.0__ |
divide__5.0__15.0__ subtract__1.0__0.3333__ add__5.0__1.0__ add__12.0__6.0__ round__18.0__ |
| solve : - num__666 x num__666 x num__666 + num__555 x num__555 x num__555 = ? ( num__666 x num__666 - num__666 x num__555 + num__555 x num__555 ) <o> a ) num__888 <o> b ) num__333 <o> c ) num__555 <o> d ) num__1221 <o> e ) num__889 |
given exp . = ( a num__3 + b num__3 ) = ( a + b ) = ( num__666 + num__555 ) = num__1221 ( a num__2 - ab + b num__2 ) answer d <eor> d <eos> |
d |
add__666.0__555.0__ add__666.0__555.0__ |
add__666.0__555.0__ add__666.0__555.0__ |
| sreedhar and sravan together can do a work in num__25 days . with the help of pavan they completed the work in num__8 days and earned rs . num__225 . what is the share of sravan if sreedhar alone can do the work in num__75 days ? <o> a ) num__33 <o> b ) num__77 <o> c ) num__48 <o> d ) num__99 <o> e ) num__12 |
sravan ' s one day ' s work = num__0.04 - num__0.0133333333333 = num__0.0266666666667 sravan worked for num__8 days . so his num__8 days work = num__8 * num__0.0266666666667 = num__0.213333333333 sravan completed num__0.213333333333 th of total work . so his share is num__0.213333333333 * num__225 = rs . num__48 . answer : c <eor> c <eos> |
c |
subtract__0.04__0.0133__ round__48.0__ |
subtract__0.04__0.0133__ round__48.0__ |
| a certain number of men can do a work in num__20 days . if there were num__4 men less it could be finished in num__5 days more . how many men are there ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__25 <o> d ) num__30 <o> e ) num__35 |
the original number of men = num__4 ( num__20 + num__5 ) / num__5 = num__20 men answer is b <eor> b <eos> |
b |
round__20.0__ |
round__20.0__ |
| janet covered a distance of num__340 miles between city a and city b taking a total of num__5 hours . if part of the distance was covered at num__60 miles per hour speed and the balance at num__80 miles per hour speed how many hours did she travel at num__60 miles per hour ? <o> a ) num__1 hour num__45 minutes <o> b ) num__2 hours <o> c ) num__2 hours num__30 minutes <o> d ) num__3 hours <o> e ) none of these |
janet covered a distance of num__340 miles between city a and city b taking a total of num__5 hours . time taken at num__60 : time taken at num__80 : : ( greater speed - average speed ) : ( average speed - lower speed ) average speed is num__68.0 = num__68 that gives the answer num__3 : num__2 = d <eor> d <eos> |
d |
divide__340.0__5.0__ subtract__5.0__3.0__ round__3.0__ |
divide__340.0__5.0__ subtract__5.0__3.0__ subtract__5.0__2.0__ |
| . a car covers a distance of num__742 km in num__7 hours . find its speed ? <o> a ) num__104 <o> b ) num__55 <o> c ) num__148 <o> d ) num__150 <o> e ) num__159 |
num__148.4 = num__148 kmph answer : c <eor> c <eos> |
c |
round__148.0__ |
round__148.0__ |
| num__5216 x num__51 = ? <o> a ) num__212016 <o> b ) num__266016 <o> c ) num__266436 <o> d ) num__216314 <o> e ) num__216318 |
normal way of multiplication may take time . here are one alternative . num__5216 x num__51 = ( num__5216 x num__50 ) + num__5216 = ( num__5216 x num__100 ) / num__2 + num__5216 = num__260800.0 + num__5216 = num__260800 + num__5216 = num__266016 answer is b <eor> b <eos> |
b |
divide__100.0__50.0__ multiply__5216.0__50.0__ multiply__5216.0__51.0__ multiply__5216.0__51.0__ |
divide__100.0__50.0__ multiply__5216.0__50.0__ add__5216.0__260800.0__ add__5216.0__260800.0__ |
| a certain deep blue paint contains num__45 percent blue pigment and num__55 percent red pigment by weight . a certain green paint contains num__35 percent blue pigment and num__65 percent yellow pigment . when these paints are mixed to produce a brown paint the brown paint contains num__40 percent blue pigment . if the brown paint weighs num__10 grams then the red pigment contributes how many grams of that weight ? <o> a ) num__1.5 <o> b ) num__2.5 <o> c ) num__3.5 <o> d ) num__2.75 <o> e ) num__4.5 |
num__10 grams of combined mixture and num__40.0 blue pigment means that the mixtures were mixed num__50.0 each . thus num__5 grams a piece . out of the num__5 grams of the dark blue paint num__60.0 is red . therefore num__5 * . num__55 = num__2.75 grams of red pigment <eor> d <eos> |
d |
percent__10.0__50.0__ percent__55.0__5.0__ percent__55.0__5.0__ |
percent__10.0__50.0__ percent__55.0__5.0__ percent__55.0__5.0__ |
| the sum of num__7 th and num__23 rd term of a . p . is equal to the sum of num__8 th num__15 th and num__13 th term . find the term which is num__0 <o> a ) num__6 <o> b ) num__8 <o> c ) num__10 <o> d ) num__12 <o> e ) num__14 |
t num__7 + t num__23 = t num__8 + t num__15 + t num__13 = > a + num__6 d + a + num__22 d = a + num__7 d + a + num__14 d + a + num__12 d = > a + num__5 d = num__0 = > t num__6 = num__0 i . e . num__6 th term is zero . answer : a <eor> a <eos> |
a |
subtract__13.0__7.0__ add__7.0__15.0__ add__8.0__6.0__ subtract__13.0__8.0__ subtract__13.0__7.0__ |
subtract__13.0__7.0__ add__7.0__15.0__ add__8.0__6.0__ subtract__13.0__8.0__ subtract__13.0__7.0__ |
| two bullet trains of equal lengths take num__10 seconds and num__20 seconds respectively to cross a telegraph post . if the length of each bullet train be num__120 metres in what time ( in seconds ) will they cross each other travelling in opposite direction ? <o> a ) num__11 sec . <o> b ) num__15 sec . <o> c ) num__13.3 sec . <o> d ) num__17 sec . <o> e ) num__19 sec . |
speed of the first bullet train = num__12.0 m / sec = num__12 m / sec . speed of the second bullet train = num__6.0 m / sec = num__6 m / sec . relative speed = ( num__12 + num__6 ) = num__18 m / sec . required time = ( num__120 + num__120 ) / num__18 sec = num__13.3 sec . c <eor> c <eos> |
c |
divide__120.0__10.0__ divide__120.0__20.0__ add__12.0__6.0__ round__13.3__ |
divide__120.0__10.0__ divide__120.0__20.0__ add__12.0__6.0__ round__13.3__ |
| a train crosses a pole in num__4 sec while travelling at a speed of num__135 kmph what is the length of the train ? <o> a ) num__135 m <o> b ) num__140 m <o> c ) num__145 m <o> d ) num__150 m <o> e ) num__155 m |
explanation : d = num__135 * num__0.277777777778 * num__4 = num__150 m answer : option d <eor> d <eos> |
d |
round__150.0__ |
round__150.0__ |
| average of first five multiples of num__5 is <o> a ) num__9 <o> b ) num__11 <o> c ) num__13 <o> d ) num__15 <o> e ) num__16 |
explanation : average = num__5 ( num__1 + num__2 + num__3 + num__4 + num__5 ) / num__5 = num__15 answer : option d <eor> d <eos> |
d |
subtract__5.0__2.0__ subtract__5.0__1.0__ multiply__5.0__3.0__ multiply__5.0__3.0__ |
add__1.0__2.0__ add__1.0__3.0__ multiply__5.0__3.0__ divide__15.0__1.0__ |
| earl can stuff advertising circulars into envelopes at the rate of num__36 envelopes per minutes and ellen requires a minutes and half to stuff the same number of envelops . working together how long will it take earl and ellen to stuff num__360 envelopes <o> a ) num__6 minutes <o> b ) num__5 minutes <o> c ) num__7 minutes <o> d ) num__3 minutes <o> e ) num__4 minutes |
earl takes num__1 min . for num__36 envelopes . ellen takes num__1.5 mins for the same . so ellen can stuff ( ( num__36 ) / ( num__1.5 ) ) in num__1 min . i . e . num__24 envlpes a min . so both of them when work together can stuff num__36 + num__24 = num__60 envelopes in num__1 min . for num__360 envelopes they will take num__6.0 mins . i . e . num__6 mins . answer : a <eor> a <eos> |
a |
divide__36.0__1.5__ hour_to_min_conversion__ divide__360.0__60.0__ round__6.0__ |
divide__36.0__1.5__ add__36.0__24.0__ divide__360.0__60.0__ divide__36.0__6.0__ |
| how many integers from num__22 to num__160 inclusive are divisible by num__3 but not divisible by num__7 ? <o> a ) num__30 <o> b ) num__35 <o> c ) num__40 <o> d ) num__50 <o> e ) num__60 |
we should find # of integers divisible by num__3 but not by num__3 * num__7 = num__21 . # of multiples of num__21 in the range from num__22 to num__160 inclusive is ( num__147 - num__42 ) / num__21 + num__1 = num__6 ; num__46 - num__6 = num__40 . answer : c . <eor> c <eos> |
c |
multiply__3.0__7.0__ multiply__7.0__21.0__ subtract__22.0__21.0__ subtract__7.0__1.0__ subtract__46.0__6.0__ multiply__1.0__40.0__ |
multiply__3.0__7.0__ multiply__7.0__21.0__ subtract__22.0__21.0__ subtract__7.0__1.0__ subtract__46.0__6.0__ multiply__1.0__40.0__ |
| a is twice efficient as b and together they do the same work in as much time as c and d together . if c and d can complete the work in num__20 and num__30 daysrespectively working alone then in how many days a can complete the work individually : <o> a ) num__12 days <o> b ) num__18 days <o> c ) num__24 days <o> d ) num__30 days <o> e ) none of these |
explanation : a + b = c + d | | | | ratio of efficiency num__10 x + num__5 x num__9 x + num__6 x | ________| | _________| num__15 x num__15 x therefore ratio of efficiency of a : c = num__10 : num__9 therefore ratio of days taken by a : c = num__9 : num__10 therefore number of days taken by a = num__18 days answer : b <eor> b <eos> |
b |
subtract__30.0__20.0__ divide__30.0__5.0__ subtract__20.0__5.0__ round__18.0__ |
subtract__30.0__20.0__ divide__30.0__5.0__ add__5.0__10.0__ round__18.0__ |
| consider the equation k = ( mv ^ num__2 ) / num__2 . if v and m are both doubled by what factor is k increased ? <o> a ) num__2 <o> b ) num__6 <o> c ) num__8 <o> d ) num__10 <o> e ) num__4 |
we have k = ( mv ^ num__2 ) / num__2 on doubling both m and v k = [ ( num__2 m ) * ( num__2 v ) ^ num__2 ] / num__2 k = ( num__2 m * num__4 v ^ num__2 ) / num__2 k = ( num__8 mv ^ num__2 ) / num__2 k = num__8 * ( mv ^ num__2 ) / num__2 initially k = ( mv ^ num__2 ) / num__2 . after changing m and v k = num__8 * ( mv ^ num__2 ) / num__2 hence k increases by num__8 times . answer = c = num__8 <eor> c <eos> |
c |
multiply__2.0__4.0__ multiply__2.0__4.0__ |
multiply__2.0__4.0__ multiply__2.0__4.0__ |
| danny is sitting on a rectangular box . the area of the front face of the box is half the area of the top face and the area of the top face is num__1.5 times the area of the side face . if the volume of the box is num__1536 what is the area of the side face of the box ? <o> a ) num__34 <o> b ) num__65 <o> c ) num__88 <o> d ) num__90 <o> e ) num__128 |
lets suppose length = l breadth = b depth = d front face area = l * w = num__0.5 w * d ( l = num__0.5 d or d = num__2 l ) top face area = w * d side face area = w * d = num__1.5 d * l ( w = num__1.5 l ) volume = l * w * d = num__1536 l * num__1.5 l * num__2 l = num__1536 l = num__8 side face area = l * d = l * num__2 l = num__8 * num__2 * num__8 = num__128 e is the answer <eor> e <eos> |
e |
square_perimeter__0.5__ square_perimeter__2.0__ triangle_area__2.0__128.0__ |
square_perimeter__0.5__ square_perimeter__2.0__ volume_rectangular_prism__0.5__2.0__128.0__ |
| bay is going with num__8 friends on a trip to new york for spring break . airfare and hotel costs a total of $ num__7200.00 for the group of num__9 friends . how much does each person have to pay for their hotel and airfare ? <o> a ) $ num__720 <o> b ) $ num__800 <o> c ) $ num__810 <o> d ) $ num__700 <o> e ) $ num__750 |
answer = b the total cost of the trip ( $ num__7200.00 ) divided by num__9 equals $ num__800.00 . <eor> b <eos> |
b |
divide__7200.0__9.0__ divide__7200.0__9.0__ |
divide__7200.0__9.0__ divide__7200.0__9.0__ |
| the area of a triangle will be when a = num__2 m b = num__5 m c = num__7 m a b c being lengths of respective sides ? <o> a ) num__2 <o> b ) num__7 <o> c ) num__5 <o> d ) num__3 <o> e ) num__4 |
s = ( num__2 + num__5 + num__7 ) / num__2 = num__7 answer : b <eor> b <eos> |
b |
triangle_area__2.0__7.0__ |
triangle_area__2.0__7.0__ |
| a certain volleyball team played seven games and scored an average of num__80 points per game . if in the team ’ s first five games it scored num__72 num__65 num__71 num__70 and num__83 points what was the average ( arithmetic mean ) number of points scored over the last two games ? <o> a ) num__65 <o> b ) num__81 <o> c ) num__82 <o> d ) num__119.5 <o> e ) can not be determined from the information given . |
avg points for total of num__7 games = num__80 so a + b + c + d + e + f + g / num__7 = num__80 and a + b + c + d + e + f + g = num__560 given a + b + c + d + e = num__321 so f + g = num__239 so average ( arithmetic mean ) number of points scored over the last two games is num__119.5 = num__119.5 correct answer is d ) num__119.5 <eor> d <eos> |
d |
subtract__72.0__65.0__ multiply__80.0__7.0__ subtract__560.0__321.0__ subtract__239.0__119.5__ |
subtract__72.0__65.0__ multiply__80.0__7.0__ subtract__560.0__321.0__ subtract__239.0__119.5__ |
| a and b started a partnership business investing rs . num__20000 and rs . num__15000 respectively . c joined them with rs . num__20000 after six months . calculate b ' s share in total profit of rs . num__25000 earned at the end of num__2 years from the starting of the business ? <o> a ) num__7500 <o> b ) num__8500 <o> c ) num__9000 <o> d ) num__8000 <o> e ) num__9500 |
explanation : a : b : c = num__20000 Ã — num__24 : num__15000 Ã — num__24 : num__20000 Ã — num__18 = num__20 Ã — num__4 : num__15 Ã — num__4 : num__20 Ã — num__3 = num__4 Ã — num__4 : num__3 Ã — num__4 : num__4 Ã — num__3 = num__4 : num__3 : num__3 = num__20000 Ã — num__24 : num__15000 Ã — num__24 : num__20000 Ã — num__18 = num__20 Ã — num__4 : num__15 Ã — num__4 : num__20 Ã — num__3 = num__4 Ã — num__4 : num__3 Ã — num__4 : num__4 Ã — num__3 = num__4 : num__3 : num__3 b ' s share = num__25000 Ã — num__310 = num__7500 answer is a <eor> a <eos> |
a |
add__2.0__18.0__ subtract__24.0__20.0__ subtract__18.0__15.0__ divide__15000.0__2.0__ divide__15000.0__2.0__ |
add__2.0__18.0__ subtract__24.0__20.0__ subtract__18.0__15.0__ divide__15000.0__2.0__ divide__15000.0__2.0__ |
| robert walks from point a to point b at an average speed of num__10 kilometers per hour . at what speed in kilometers per hour must robert walk from point b to point a so that her average speed for the entire trip is num__12 kilometers per hour ? <o> a ) num__15 km / hr <o> b ) num__21 km / hr <o> c ) num__16 km / hr <o> d ) num__20 km / hr <o> e ) num__17 km / hr |
let ' s suppose that speed while returning was x km / h since the distance is same we can apply the formula of avg speed avg speed = num__2 s num__1 s num__2 / s num__1 + s num__2 num__12 = num__2 * num__10 * x / num__10 + x x = num__15 km / hr answer is a <eor> a <eos> |
a |
subtract__12.0__10.0__ round__15.0__ |
subtract__12.0__10.0__ divide__15.0__1.0__ |
| if p and q are both negative and pq < q ^ num__2 which of the following must be true ? <o> a ) p < q < p ^ num__2 < q ^ num__2 <o> b ) q < p < p ^ num__2 < q ^ num__2 <o> c ) p < q < q ^ num__2 < p ^ num__2 <o> d ) p ^ num__2 < q ^ num__2 < q < p <o> e ) q ^ num__2 < p ^ num__2 < q < p |
because pq < q ^ num__2 and both are negative i thought p < q . so i crossed off answers c ) d ) and e ) . and because p < q p ^ num__2 < q ^ num__2 ans b <eor> b <eos> |
b |
coin_space__ |
coin_space__ |
| the average ( arithmetic mean ) of four distinct positive integers is num__5 . if the average of the smaller two of these four integers is num__6 which of the following represents the maximum possible value of the largest integer ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__3 <o> d ) num__8 <o> e ) num__1 |
let the distinct number be a b c and d its given a > b > c > d also a + b + c + d = num__20 and a + b = num__6 means c + d = num__14 since the question ask for the largest possible number we should choose the least value for a and b c . so d should be num__8 answer : d <eor> d <eos> |
d |
subtract__20.0__6.0__ subtract__14.0__6.0__ subtract__14.0__6.0__ |
subtract__20.0__6.0__ subtract__14.0__6.0__ subtract__14.0__6.0__ |
| a train passes a station platform in num__36 sec and a man standing on the platform in num__20 sec . if the speed of the train is num__54 km / hr . what is the length of the platform ? <o> a ) num__288 <o> b ) num__240 <o> c ) num__881 <o> d ) num__1277 <o> e ) num__121 |
speed = num__54 * num__0.277777777778 = num__15 m / sec . length of the train = num__15 * num__20 = num__300 m . let the length of the platform be x m . then ( x + num__300 ) / num__36 = num__15 = > x = num__240 m . answer : b <eor> b <eos> |
b |
multiply__20.0__15.0__ round__240.0__ |
multiply__20.0__15.0__ round__240.0__ |
| a train crosses a platform of num__150 m in num__15 sec same train crosses another platform of length num__220 m in num__18 sec . then find the length of the train ? <o> a ) num__200 m <o> b ) num__180 m <o> c ) num__272 m <o> d ) num__210 m <o> e ) num__267 m |
length of the train be ‘ x ’ ( x + num__150 ) / num__15 = ( x + num__220 ) / num__18 num__5 x + num__1100 = num__6 x + num__900 x = num__200 m answer : a <eor> a <eos> |
a |
multiply__220.0__5.0__ multiply__150.0__6.0__ subtract__1100.0__900.0__ round__200.0__ |
multiply__220.0__5.0__ multiply__150.0__6.0__ subtract__1100.0__900.0__ round__200.0__ |
| brenda and sally run in opposite direction on a circular track starting at diametrically opposite points . they first meet after brenda has run num__150 meters . they next meet after sally has run num__200 meters past their first meeting point . each girl runs at a constant speed . what is the length of the track in meters ? <o> a ) num__250 <o> b ) num__300 <o> c ) num__350 <o> d ) num__400 <o> e ) num__500 |
nice problem . + num__1 . first timetogetherthey run half of the circumference . second timetogetherthey run full circumference . first time brenda runs num__100 meters thus second time she runs num__2 * num__150 = num__300 meters . since second time ( when they run full circumference ) brenda runs num__300 meters and sally runs num__200 meters thus the circumference is num__300 + num__200 = num__500 meters . answer : e . <eor> e <eos> |
e |
divide__200.0__100.0__ multiply__150.0__2.0__ add__200.0__300.0__ round__500.0__ |
divide__200.0__100.0__ multiply__150.0__2.0__ add__200.0__300.0__ add__200.0__300.0__ |
| if the average ( arithmetic mean ) of the four numbers num__3 num__16 num__33 and ( n + num__1 ) is num__20 then n = <o> a ) num__19 <o> b ) num__20 <o> c ) num__21 <o> d ) num__27 <o> e ) num__29 |
num__3 + num__16 + num__33 + n + num__1 = num__20 x num__4 = num__80 = > n + num__53 = num__80 = > n = num__27 answer d <eor> d <eos> |
d |
add__3.0__1.0__ multiply__20.0__4.0__ add__33.0__20.0__ subtract__80.0__53.0__ multiply__1.0__27.0__ |
add__3.0__1.0__ multiply__20.0__4.0__ add__33.0__20.0__ subtract__80.0__53.0__ multiply__1.0__27.0__ |
| a train num__120 m long passed a pole in num__6 sec . how long will it take to pass a platform num__800 m long ? <o> a ) num__42 <o> b ) num__44 <o> c ) num__46 <o> d ) num__48 <o> e ) num__50 |
speed = num__20.0 = num__20 m / sec . required time = ( num__120 + num__800 ) / num__20 = num__46 sec . answer : c <eor> c <eos> |
c |
divide__120.0__6.0__ round__46.0__ |
divide__120.0__6.0__ round__46.0__ |
| the cost price of num__14 articles is equal to the selling price of num__11 articles . find the profit percent ? <o> a ) num__18 num__0.133333333333 <o> b ) num__27 num__0.272727272727 <o> c ) num__36 num__0.142857142857 <o> d ) num__32 num__0.125 <o> e ) num__12 num__0.166666666667 |
num__14 cp = num__11 sp num__11 - - - num__2 cp num__100 - - - ? = > num__27 num__0.272727272727 % . answer : b <eor> b <eos> |
b |
percent__100.0__27.0__ |
percent__100.0__27.0__ |
| a and b go cycling in the same direction with speeds of num__6 km / hr and num__12 km / hr . a car from behind passes them in num__9 and num__10 seconds respectively . what is the speed of the car ? <o> a ) num__22 km / hr <o> b ) num__33 km / hr <o> c ) num__66 km / hr <o> d ) num__44 km / hr <o> e ) none of these |
explanation : the relative speed of a and b is num__6 km / hr = num__1.67 m / s as the car passes a after num__10 s the distance between a and b after num__10 s ( i . e . at num__11 th second ) is the distance covered by car in num__1 second . therefore at t = num__11 d = num__1.67 * num__11 d = num__18.33 m v = d / t = num__18.33 / num__1 = num__18.33 m / s v = num__66 km / hr answer c <eor> c <eos> |
c |
subtract__12.0__11.0__ multiply__6.0__11.0__ round__66.0__ |
subtract__12.0__11.0__ multiply__6.0__11.0__ multiply__6.0__11.0__ |
| if the ratio of x to y is num__3 to num__5 then ( y - x ) / ( x + y ) ? <o> a ) - num__2.33333333333 <o> b ) - num__0.25 <o> c ) num__0.25 <o> d ) num__1 <o> e ) num__2.33333333333 |
i believe this problem can be solved using algebra . the ratio of x to y is num__3 to num__5 means ( x / y ) = ( num__0.6 ) cross - multiply num__3 x = num__5 y x = ( num__0.6 ) * y then substitute x in the original equation in the problem with the value we just found . ( y - ( num__0.6 ) * y ) / ( ( num__0.6 ) * y + y ) = ( ( num__0.4 ) * y ) / ( ( num__1.6 ) * y ) simplify by canceling the y in the numerator with the y in the denominator . ( num__0.4 ) / ( num__1.6 ) = ( num__0.4 ) * ( num__1.6 ) the num__5 s cancel each other out and you are left with num__0.25 . <eor> c <eos> |
c |
divide__3.0__5.0__ divide__0.4__1.6__ divide__0.4__1.6__ |
divide__3.0__5.0__ divide__0.4__1.6__ divide__0.4__1.6__ |
| num__16 men can complete a piece of work in num__25 days . in how many days can num__20 men complete that piece of work ? <o> a ) num__12 days <o> b ) num__9 days <o> c ) num__20 days <o> d ) num__14 days <o> e ) num__17 days |
answer : option c num__16 * num__25 = num__20 * x = > x = num__20 days <eor> c <eos> |
c |
round__20.0__ |
round__20.0__ |
| the average weight of num__50 students in a class is num__52 kg . num__5 of them whose average weight is num__48 kg leave the class and other num__5 students whose average weight is num__54 kg join the class . what is the new average weight ( in kg ) of the class ? <o> a ) num__51 num__1 ⁄ num__3 <o> b ) num__52 num__0.6 <o> c ) num__52 num__5 ⁄ num__3 <o> d ) num__43.42 <o> e ) none of these |
total weight of num__50 students = num__50 × num__52 = num__2340 kg total weight of num__5 students who leave = num__5 × num__48 = num__240 kg total weight of num__5 students who join = num__5 × num__54 = num__270 kg therefore new total weight of num__50 students = num__2600 – num__240 + num__270 = num__2630 ⇒ new average weight = num__2630 ⁄ num__50 = num__52 num__0.6 kg answer b <eor> b <eos> |
b |
multiply__5.0__48.0__ multiply__5.0__54.0__ multiply__50.0__52.0__ divide__2600.0__50.0__ |
multiply__5.0__48.0__ multiply__5.0__54.0__ multiply__50.0__52.0__ divide__2600.0__50.0__ |
| num__2 num__5 num__10 num__17 . . <o> a ) num__5 <o> b ) num__8 <o> c ) num__26 <o> d ) num__7 <o> e ) num__12 |
explanation : numbers are ( num__1 * num__1 ) + num__1 = num__2 ( num__2 * num__2 ) + num__1 = num__5 ( num__3 * num__3 ) + num__1 = num__10 ( num__4 * num__4 ) + num__1 = num__17 ( num__5 * num__5 ) + num__1 = num__26 answer : c <eor> c <eos> |
c |
add__2.0__1.0__ subtract__5.0__1.0__ multiply__1.0__26.0__ |
add__2.0__1.0__ add__1.0__3.0__ multiply__1.0__26.0__ |
| if ( a – b ) is num__9 more than ( c + d ) and ( a + b ) is num__3 less than ( c – d ) then ( a – c ) is : <o> a ) num__6 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
( a – b ) – ( c + d ) = num__9 and ( c – d ) – ( a + b ) = num__3 = > ( a – c ) – ( b + d ) = num__9 and ( c – a ) – ( b + d ) = num__3 = > ( b + d ) = ( a – c ) – num__9 and ( b + d ) = ( c – a ) – num__3 = > ( a – c ) – num__9 = ( c – a ) – num__3 = > num__2 ( a – c ) = num__6 = > ( a – c ) = num__3 answer : c <eor> c <eos> |
c |
subtract__9.0__3.0__ divide__9.0__3.0__ |
subtract__9.0__3.0__ divide__9.0__3.0__ |
| if teena is driving at num__55 miles per hour and is currently num__7.5 miles behind loe who is driving at num__40 miles per hour in the same direction then in how many minutes will teena be num__15 miles ahead of loe ? <o> a ) num__15 <o> b ) num__60 <o> c ) num__75 <o> d ) num__90 <o> e ) num__105 |
this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra num__30 - num__40 seconds for a difficult one . teena covers num__55 miles in num__60 mins . loe covers num__40 miles in num__60 mins so teena gains num__15 miles every num__60 mins teena need to cover num__7.5 + num__15 miles . teena can cover num__7.5 miles in num__30 mins teena will cover num__15 miles in num__60 mins so answer num__30 + num__60 = num__90 mins . d <eor> d <eos> |
d |
hour_to_min_conversion__ add__60.0__30.0__ round__90.0__ |
hour_to_min_conversion__ add__60.0__30.0__ add__60.0__30.0__ |
| the ratio between the present ages of a and b is num__5 : num__3 respectively . the ratio between a ' s age num__4 years ago and b ' s age num__4 years hence is num__1 : num__1 . what is the ratio between a ' s age num__4 years hence and b ' s age num__4 years ago ? <o> a ) num__1 : num__3 <o> b ) num__2 : num__3 <o> c ) num__3 : num__1 <o> d ) num__4 : num__1 <o> e ) num__4 : num__3 |
let the present ages of a and b be num__5 x and num__3 x years respectively . then ( num__5 x - num__4 ) / ( num__3 x + num__4 ) = num__1.0 num__2 x = num__8 = > x = num__4 required ratio = ( num__5 x + num__4 ) : ( num__3 x - num__4 ) = num__24 : num__8 = num__3 : num__1 answer c <eor> c <eos> |
c |
subtract__5.0__3.0__ add__5.0__3.0__ multiply__3.0__8.0__ subtract__5.0__2.0__ |
subtract__5.0__3.0__ add__5.0__3.0__ multiply__3.0__8.0__ subtract__5.0__2.0__ |
| the area of a square is equal to five times the area of a rectangle of dimensions num__60 cm * num__12 cm . what is the perimeter of the square ? <o> a ) num__289 cm <o> b ) num__800 cm <o> c ) num__829 cm <o> d ) num__288 cm <o> e ) num__240 cm |
area of the square = s * s = num__5 ( num__60 * num__12 ) = > s = num__60 = num__60 cm perimeter of the square = num__4 * num__60 = num__240 cm . answer : e <eor> e <eos> |
e |
square_perimeter__60.0__ square_perimeter__60.0__ |
multiply__60.0__4.0__ multiply__60.0__4.0__ |
| num__57 persons can repair a road in num__12 days working num__5 hours a day . in how many days will num__30 persons working num__6 hours a day complete the work ? <o> a ) num__10 <o> b ) num__13 <o> c ) num__14 <o> d ) num__15 <o> e ) num__19 |
according to the chain rule m num__1 x t num__1 = m num__2 x t num__2 therefore num__57 x num__12 x num__5 = num__30 x num__6 x x x = num__19 hence the number of days = num__19 . answer : e <eor> e <eos> |
e |
subtract__6.0__5.0__ divide__12.0__6.0__ round__19.0__ |
subtract__6.0__5.0__ divide__12.0__6.0__ round__19.0__ |
| of the five numbers in a sequence the first term is num__7000 and each of the following terms is num__20.0 of the previous term . what is the value range of the five numbers ? <o> a ) num__6988.8 <o> b ) num__9 num__750.90 <o> c ) num__9 num__975.15 <o> d ) num__9 num__984.30 <o> e ) num__10 num__736.25 |
num__1 st number = num__7000 num__2 nd number = num__20.0 of num__7000 = num__1400 num__3 rd number = num__20.0 of num__1400 = num__280 num__4 th number = num__20.0 of num__280 = num__56 num__5 th number = num__20.0 of num__56 = num__11.2 range = num__7000 - num__11.2 = num__6988.8 a is the answer <eor> a <eos> |
a |
add__1.0__2.0__ add__1.0__3.0__ divide__7000.0__1400.0__ divide__56.0__5.0__ subtract__7000.0__11.2__ subtract__7000.0__11.2__ |
add__1.0__2.0__ add__1.0__3.0__ divide__7000.0__1400.0__ divide__56.0__5.0__ subtract__7000.0__11.2__ subtract__7000.0__11.2__ |
| in smithtown the ratio of right - handed people to left - handed people is num__3 to num__1 and the ratio of men to women is num__3 to num__2 . if the number of right - handed men is maximized then what z percent of all the people in smithtown are left - handed women ? <o> a ) num__50.0 <o> b ) num__40.0 <o> c ) num__25.0 <o> d ) num__20.0 <o> e ) num__10 % |
looking at the ratio we can take total number of people = num__20 . . ans num__0.25 or num__25.0 c <eor> c <eos> |
c |
multiply__1.0__25.0__ |
multiply__1.0__25.0__ |
| bill downloads the movierevenge of the avengersto his computer in num__2.5 hours using a download manager that downloads from num__3 sources marked a b and c . each source provides download at a constant rate but the rates of different sources are not necessarily identical . if the movie was downloaded from sources a and c alone it would take num__4 hours to complete the download . the next day source b is available but the other sources are inactive . how long will it take to download the trailer of the movie a file that is num__20 times smaller from source b alone ? <o> a ) num__6 hours and num__40 minutes <o> b ) num__15 minutes <o> c ) num__20 minutes <o> d ) num__10 minutes <o> e ) num__3 minutes |
let the movie size be num__400 u . given a + c = num__4 hrs . a + c = num__100 u / hr and a + b + c = num__2.5 hrs or num__400 / num__2.5 = num__160 u / hr b alone = num__160 - num__100 = num__60 u / hr trailer = num__20 times smaller or num__20.0 = num__20 u b will take num__0.333333333333 hrs or num__20 minutes . ans c <eor> c <eos> |
c |
divide__400.0__4.0__ divide__400.0__2.5__ multiply__3.0__20.0__ reverse__3.0__ divide__400.0__20.0__ |
divide__400.0__4.0__ divide__400.0__2.5__ subtract__160.0__100.0__ reverse__3.0__ divide__400.0__20.0__ |
| if a student loses num__5 kilograms he will weigh twice as much as his sister . together they now weigh num__104 kilograms . what is the student ' s present weight in kilograms ? <o> a ) num__70 <o> b ) num__71 <o> c ) num__72 <o> d ) num__73 <o> e ) num__74 |
let x be the weight of the sister . then the student ' s weight is num__2 x + num__5 . x + ( num__2 x + num__5 ) = num__104 num__3 x = num__99 x = num__33 kg then the student ' s weight is num__71 kg . the answer is b . <eor> b <eos> |
b |
subtract__5.0__2.0__ subtract__104.0__5.0__ divide__99.0__3.0__ subtract__104.0__33.0__ subtract__104.0__33.0__ |
subtract__5.0__2.0__ subtract__104.0__5.0__ divide__99.0__3.0__ subtract__104.0__33.0__ subtract__104.0__33.0__ |
| a num__12.0 stock yields num__8.0 . the market value of the stock is : <o> a ) rs . num__72 <o> b ) rs . num__150 <o> c ) rs . num__112.50 <o> d ) rs . num__116.50 <o> e ) none of these |
solution to obtain rs . num__8 investment = rs . num__100 . to obtain rs . num__12 investment = rs . ( num__12.5 x num__12 ) = rs . num__150 ∴ market value of rs . num__100 stock = rs . num__150 answer b <eor> b <eos> |
b |
percent__100.0__150.0__ |
percent__100.0__150.0__ |
| in a certain city each of num__25 girl scout troops is represented by a colored flag . each flag consists of either a single color or a pair of two different colors or a combination of num__3 colors . if each troop has a different flag what is the minimum number of colors needed for the flags . ( assume that the order of colors in a pair on a flag does not matter . ) <o> a ) num__7 <o> b ) num__3 <o> c ) num__6 <o> d ) num__4 <o> e ) num__5 |
e . num__5 : if we have num__5 colors we can choose pairs in num__5 c num__2 + num__5 c num__3 ways = num__10 + num__10 and single color flags are num__5 . therefore total number of flags = num__25 . ans e . <eor> e <eos> |
e |
vowel_space__ coin_space__ vowel_space__ |
vowel_space__ coin_space__ vowel_space__ |
| in what time will a train num__600 meters long cross an electric pole if its speed is num__144 km / hr <o> a ) num__5 seconds <o> b ) num__4.5 seconds <o> c ) num__3 seconds <o> d ) num__15 seconds <o> e ) none of these |
explanation : first convert speed into m / sec speed = num__144 * ( num__0.277777777778 ) = num__40 m / sec time = distance / speed = num__15.0 = num__15 seconds answer : d <eor> d <eos> |
d |
divide__600.0__40.0__ round__15.0__ |
divide__600.0__40.0__ divide__600.0__40.0__ |
| if num__28 less than five times a certain number is num__232 . what is the number ? <o> a ) num__32 <o> b ) num__42 <o> c ) num__52 <o> d ) num__62 <o> e ) num__72 |
num__5 x − num__28 subtraction is built backwards multiply the unknown by num__5 num__5 x − num__28 = num__232 is translates to equals + num__28 + num__28 add num__28 to both sides num__5 x = num__260 the variable ismultiplied by num__5 num__5 num__5 divide both sides by num__5 x = num__52 the number is num__52 . correct answer c <eor> c <eos> |
c |
add__28.0__232.0__ divide__260.0__5.0__ divide__260.0__5.0__ |
add__28.0__232.0__ divide__260.0__5.0__ divide__260.0__5.0__ |
| the sum of the mean the median and the range of the set { num__2 num__4 num__6 num__810 } equals which one of the following values ? <o> a ) num__21 <o> b ) num__42 <o> c ) num__61 <o> d ) num__84 <o> e ) num__10 |
set { num__2 num__4 num__6 num__8 num__10 } mean = ( num__2 + num__4 + num__6 + num__8 + num__10 ) / num__5 = num__6 meadian = middle term = num__6 range = highest - lowest = num__10 - num__1 = num__9 mean + median + range = num__6 + num__6 + num__9 = num__21 answer : option a <eor> a <eos> |
a |
multiply__2.0__4.0__ add__2.0__8.0__ divide__10.0__2.0__ subtract__6.0__5.0__ add__4.0__5.0__ multiply__1.0__21.0__ |
add__2.0__6.0__ add__2.0__8.0__ divide__10.0__2.0__ subtract__6.0__5.0__ add__4.0__5.0__ divide__21.0__1.0__ |
| in a regular week there are num__5 working days and for each day the working hours are num__8 . a man gets rs . num__2.40 per hour for regular work and rs . num__3.20 per hours for overtime . if he earns rs . num__432 in num__4 weeks then how many hours does he work for ? <o> a ) num__160 <o> b ) num__175 <o> c ) num__180 <o> d ) num__195 <o> e ) num__215 |
suppose the man works overtime for x hours . now working hours in num__4 weeks = ( num__5 x num__8 x num__4 ) = num__160 . num__160 x num__2.40 + x x num__3.20 = num__432 num__3.20 x = num__432 - num__384 = num__48 x = num__15 . hence total hours of work = ( num__160 + num__15 ) = num__175 answer : option b <eor> b <eos> |
b |
multiply__2.4__160.0__ subtract__432.0__384.0__ divide__48.0__3.2__ add__160.0__15.0__ round__175.0__ |
multiply__2.4__160.0__ subtract__432.0__384.0__ divide__48.0__3.2__ add__160.0__15.0__ add__160.0__15.0__ |
| how many diagonals does a num__60 - sided convex polygon have ? <o> a ) num__960 <o> b ) num__1280 <o> c ) num__1710 <o> d ) num__2560 <o> e ) num__3420 |
a num__60 - sided convex polygon has num__60 vertices . if we examine a single vertex we can see that we can connect it with num__57 other vertices to create a diagonal . note that we ca n ' t connect the vertex to itself and we ca n ' t connect it to its adjacent vertices since this would not create a diagonal . if each of the num__60 vertices can be connected with num__57 vertices to create a diagonal then the total number of diagonals would be ( num__60 ) ( num__57 ) = num__3420 however we must recognize that we have counted every diagonal twice . to account for counting each diagonal twice we must divide num__3420 by num__2 to get num__1710 . the answer is c . <eor> c <eos> |
c |
multiply__60.0__57.0__ divide__3420.0__2.0__ divide__3420.0__2.0__ |
multiply__60.0__57.0__ divide__3420.0__2.0__ subtract__3420.0__1710.0__ |
| find the number of days required by z number of workers working z hours a day to complete z units of work if x workers working x hours a day complete x units of work in x days . <o> a ) x num__3 / z <o> b ) x num__2 / z <o> c ) x num__3 / z num__2 <o> d ) x num__2 / z num__2 <o> e ) none of these |
explanation : m num__1 d num__1 t num__1 w num__2 = m num__2 d num__2 t num__2 w num__1 where m – number of men d – number of days t – time duration w – amount of work x * x * x * z = z * d num__2 * z * x d num__2 = x num__2 / z answer : b <eor> b <eos> |
b |
round__2.0__ |
divide__2.0__1.0__ |
| a number is increased by num__50.0 and then decreased by num__50.0 the final value of the number is ? <o> a ) decrease by num__20.0 <o> b ) decrease by num__22.0 <o> c ) decrease by num__23.0 <o> d ) decrease by num__21.0 <o> e ) decrease by num__25 % |
here x = num__50 and y = - num__50 therefore the net % change in value = ( x + y + xy / num__100 ) % = [ num__50 - num__50 + ( num__50 * - num__50 ) / num__100 ] % or - num__25.0 since the sign is negative there is a decrease in value by num__25.0 . answer : e <eor> e <eos> |
e |
subtract__50.0__25.0__ |
subtract__50.0__25.0__ |
| with out any halt a train travels a certain distance with an average speed of num__55 km ph and with halts it covers the same distance at an average speed of num__40 kmph . when it is traveling with halts how many minutes / per hour does the train halt on an average ? <o> a ) num__48 min . <o> b ) num__6 min . <o> c ) num__15 min . <o> d ) num__18 min . <o> e ) num__19 min . |
with halt in num__1 hour the train travels num__60 km with out halt for traveling same distance it take num__1.1 = num__1.1 = num__66 minutes : . num__6 minutes is the halting time per hour answer : b <eor> b <eos> |
b |
hour_to_min_conversion__ multiply__60.0__1.1__ subtract__66.0__60.0__ round__6.0__ |
hour_to_min_conversion__ multiply__60.0__1.1__ subtract__66.0__60.0__ round__6.0__ |
| what will be the difference between simple and compound interest at num__14.0 per annum on a sum of rs . num__1000 after num__4 years ? <o> a ) num__129 <o> b ) num__130 <o> c ) num__124 <o> d ) num__133 <o> e ) num__145 |
s . i . = ( num__1000 * num__14 * num__4 ) / num__100 = rs . num__560 c . i . = [ num__1000 * ( num__1 + num__0.14 ) num__4 - num__1000 ] = rs . num__689 difference = ( num__689 - num__560 ) = rs . num__129 answer : a <eor> a <eos> |
a |
percent__14.0__1.0__ percent__100.0__129.0__ |
percent__14.0__1.0__ percent__100.0__129.0__ |
| if num__3 people can do num__3 times of a particular work in num__3 days then how many days would it take num__8 people to do num__8 times of that particular work ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__8 <o> e ) num__11 |
num__3 people can do the work one time in one day . num__1 person can do num__0.333333333333 of the work in one day . num__8 people can do num__2.66666666667 of the work in one day . num__8 people can do num__8 times the work in num__3 days . the answer is c . <eor> c <eos> |
c |
divide__1.0__3.0__ subtract__3.0__0.3333__ round__3.0__ |
divide__1.0__3.0__ subtract__3.0__0.3333__ round__3.0__ |
| find the average of first num__30 natural numbers . <o> a ) num__12 <o> b ) num__15.5 <o> c ) num__14.5 <o> d ) num__16 <o> e ) num__17 |
explanation : sum of first n natural numbers = n ( n + num__1 ) / num__2 hence sum of first num__30 natural numbers = ( num__30 x num__31 ) / num__2 = num__465 therefore required average of = num__15.5 = num__15.5 answer b <eor> b <eos> |
b |
add__30.0__1.0__ divide__465.0__30.0__ multiply__15.5__1.0__ |
add__30.0__1.0__ divide__465.0__30.0__ divide__15.5__1.0__ |
| what is the area of a square field whose diagonal of length num__20 m ? <o> a ) num__10 sq m <o> b ) num__200 sq m <o> c ) num__400 sq m <o> d ) num__500 sq m <o> e ) num__156 sq m |
explanation : d ^ num__1.0 = ( num__20 * num__20 ) / num__2 = num__200 sq m answer : b <eor> b <eos> |
b |
multiply__200.0__1.0__ |
multiply__200.0__1.0__ |
| a single discount equivalent to the discount series of num__20.0 num__10.0 and num__5.0 is ? <o> a ) num__31.8 <o> b ) num__31.1 <o> c ) num__31.6 <o> d ) num__31.2 <o> e ) num__31.9 |
explanation : num__100 * ( num__0.8 ) * ( num__0.9 ) * ( num__0.95 ) = num__68.4 num__100 - num__68.4 = num__31.6 answer : c <eor> c <eos> |
c |
percent__100.0__31.6__ |
percent__100.0__31.6__ |
| a boat can travel with a speed of num__13 km / hr in still water . if the speed of the stream is num__4 km / hr . find the time taken by the boat to go num__72 km downstream ? <o> a ) num__2 hours <o> b ) num__4 hours num__12 minutes <o> c ) num__4 hours <o> d ) num__5 hours <o> e ) none |
solution speed downstream = ( num__13 + num__4 ) km / hr = num__17 km / hr . time taken to travel num__72 km downstream = ( num__4.23529411765 ) hrs = num__4 hrs num__12 minutes . answer b <eor> b <eos> |
b |
add__13.0__4.0__ divide__72.0__17.0__ round__4.0__ |
add__13.0__4.0__ divide__72.0__17.0__ round__4.0__ |
| cole drove from home to work at an average speed of num__50 kmh . he then returned home at an average speed of num__110 kmh . if the round trip took a total of num__2 hours how many minutes did it take cole to drive to work ? <o> a ) num__66.5 <o> b ) num__70.5 <o> c ) num__72.5 <o> d ) num__75.5 <o> e ) num__82.5 |
let the distance one way be x time from home to work = x / num__50 time from work to home = x / num__110 total time = num__2 hrs ( x / num__50 ) + ( x / num__110 ) = num__2 solving for x we get x = num__68.75 time from home to work in minutes = ( num__68.75 ) * num__1.2 = num__82.5 minutes ans = e <eor> e <eos> |
e |
multiply__68.75__1.2__ round__82.5__ |
multiply__68.75__1.2__ multiply__68.75__1.2__ |
| log num__3 n + log num__5 n what is num__3 digit number n that will be whole number <o> a ) num__629 <o> b ) num__729 <o> c ) num__829 <o> d ) num__125 <o> e ) num__727 |
no of values n can take is num__1 num__5 ^ num__3 = num__125 answer : d <eor> d <eos> |
d |
multiply__1.0__125.0__ |
multiply__1.0__125.0__ |
| there were num__35 students in a hostel . due to the admission of num__7 new students he expenses of the mess were increased by rs . num__42 per day while the average expenditure per head diminished by rs num__1 . what was the original expenditure of the mess ? <o> a ) num__125 <o> b ) num__321 <o> c ) num__561 <o> d ) num__852 <o> e ) num__420 |
let the original average expenditure be rs . x . then num__42 ( x - num__1 ) - num__35 x = num__42 ; num__7 x = num__84 ; x = num__12 . original expenditure = rs . ( num__35 x num__12 ) = rs . num__420 . . ans : e <eor> e <eos> |
e |
divide__84.0__7.0__ multiply__35.0__12.0__ multiply__35.0__12.0__ |
divide__84.0__7.0__ multiply__35.0__12.0__ multiply__35.0__12.0__ |
| a retail item is offered at a discount of a percent ( where a > num__10 ) with a num__5.0 state sales tax assessed on the discounted purchase price . if the state sales tax were not assessed what percent discount from the item ’ s original retail price in terms of a would result in the same final price ? <o> a ) p + num__5 / num__1.05 <o> b ) p / num__1.05 + num__5 <o> c ) num__1.05 a - num__5 <o> d ) a - num__5 / num__1.05 <o> e ) num__1.05 ( a – num__5 ) |
let x be the price of the item . final price after discount and sales tax = x * ( num__1 - a / num__100 ) * num__1.05 let b be the percent discount which would result in the same final price . then x * ( num__1 - a / num__100 ) * num__1.05 = x * ( num__1 - b / num__100 ) = > num__1.05 - num__1.05 a / num__100 = num__1 - b / num__100 = > b / num__100 = num__1.05 a / num__100 - . num__05 = > b = num__1.05 a - num__5 hence option c is correct . <eor> c <eos> |
c |
percent__1.05__100.0__ |
percent__1.05__100.0__ |
| a can do a job in num__18 days and b can do it in num__30 days . a and b working together will finish twice the amount of work in - - - - - - - days ? <o> a ) num__22 num__0.125 <o> b ) num__22 num__0.5 <o> c ) num__22 num__1.0 <o> d ) num__22 num__0.2 <o> e ) num__22 num__0.111111111111 |
num__0.0555555555556 + num__0.0333333333333 = num__0.0888888888889 = num__0.0888888888889 num__11.25 = num__11 ¼ * num__2 = num__22 num__0.5 days answer : b <eor> b <eos> |
b |
add__0.0556__0.0333__ multiply__2.0__11.0__ divide__11.0__22.0__ round__22.0__ |
add__0.0556__0.0333__ multiply__2.0__11.0__ divide__11.0__22.0__ multiply__2.0__11.0__ |
| the area of a triangle will be when a = num__1 m b = num__2 m c = num__3 m a b c being lengths of respective sides ? <o> a ) num__3 <o> b ) num__7 <o> c ) num__5 <o> d ) num__3 <o> e ) num__4 |
s = ( num__1 + num__2 + num__3 ) / num__2 = num__3 answer : a <eor> a <eos> |
a |
multiply__1.0__3.0__ |
multiply__1.0__3.0__ |
| if a and s are positive integers such that ( num__2 ^ a ) ( num__4 ^ s ) = num__16 then num__2 a + s = <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
lets get the eq into simplest orm . . ( num__2 ^ a ) ( num__4 ^ s ) = num__16 . . ( num__2 ^ a ) ( num__2 ^ num__2 s ) = num__2 ^ num__4 . . or a + num__2 s = num__4 . . since a and s are positive integers only a as num__2 and s as num__1 satisfy the equation . . so num__2 a + s = num__2 * num__2 + num__1 = num__5 . . d <eor> d <eos> |
d |
add__4.0__1.0__ add__4.0__1.0__ |
add__4.0__1.0__ add__4.0__1.0__ |
| the average age num__9 members of a committee are the same as it was num__2 years ago because an old number has been replaced by a younger number . find how much younger is the new member than the old number ? <o> a ) num__12 <o> b ) num__18 <o> c ) num__22 <o> d ) num__24 <o> e ) num__26 |
num__9 * num__2 = num__18 years b <eor> b <eos> |
b |
multiply__9.0__2.0__ multiply__9.0__2.0__ |
multiply__9.0__2.0__ multiply__9.0__2.0__ |
| a car traveling at a certain constant speed takes num__15 seconds longer to travel num__1 kilometer than it would take to travel num__1 kilometer at num__60 kilometers per hour . at what speed in kilometers per hour is the car traveling ? <o> a ) num__48 <o> b ) num__50 <o> c ) num__52 <o> d ) num__54 <o> e ) num__56 |
num__60 * t = num__1 km = > t = num__0.0166666666667 km / h v * ( t + num__0.00416666666667 ) = num__1 v ( num__0.0166666666667 + num__0.00416666666667 ) = num__1 v ( num__0.0208333333333 ) = num__1 v = num__48 km / h the answer is a . <eor> a <eos> |
a |
divide__1.0__60.0__ round__48.0__ |
divide__1.0__60.0__ multiply__1.0__48.0__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__9 sec . what is the length of the train ? <o> a ) num__350 m <o> b ) num__250 m <o> c ) num__300 m <o> d ) num__148 m <o> e ) num__150 m |
speed = num__60 * num__0.277777777778 = num__16.6666666667 m / sec length of the train = speed * time = num__16.6666666667 * num__9 = num__150 m answer : e <eor> e <eos> |
e |
round__150.0__ |
round__150.0__ |
| a is twice as good a work man as b and together they finish the work in num__12 days . in how many days a alone can finish the work ? <o> a ) num__23 <o> b ) num__18 <o> c ) num__77 <o> d ) num__92 <o> e ) num__61 |
wc = num__2 : num__1 num__2 x + x = num__0.0833333333333 = > x = num__0.0277777777778 num__2 x = num__0.0555555555556 a can do the work in num__18 days . answer : b <eor> b <eos> |
b |
divide__1.0__12.0__ multiply__2.0__0.0278__ round__18.0__ |
divide__1.0__12.0__ multiply__2.0__0.0278__ round__18.0__ |
| monica planned her birthday party . she prepared num__5 muffins for each of her guests and kept aside two additional muffins in case someone will want extra . after the party it turned out that one of the guests did n ' t come but every one of the guests that did come ate six muffins and num__2 muffins remained . how many guests did monica plan on ? <o> a ) num__3 . <o> b ) num__4 . <o> c ) num__6 . <o> d ) num__8 . <o> e ) num__7 . |
num__5 n + num__2 = num__2 + num__6 ( n - num__1 ) solving n = num__6 hence c <eor> c <eos> |
c |
subtract__6.0__5.0__ add__5.0__1.0__ |
subtract__6.0__5.0__ add__5.0__1.0__ |
| num__34 men can complete a piece of work in num__18 days . in how many days will num__17 men complete the same work ? <o> a ) num__36 <o> b ) num__26 <o> c ) num__97 <o> d ) num__26 <o> e ) num__19 |
explanation : let the required number of days be a . then less men more days ( indirect proportion ) num__17 : num__34 : : num__18 : a num__17 x a = num__34 x num__18 a = ( num__34 x num__18 ) / num__17 a = num__36 answer : a <eor> a <eos> |
a |
round__36.0__ |
round__36.0__ |
| a fruits marks his fruit baskets num__40.0 more than the real price and allows num__15.0 discount . his profit is : <o> a ) num__16.0 <o> b ) num__17.0 <o> c ) num__18.0 <o> d ) num__19.0 <o> e ) num__20 % |
let the cp = num__100 rs . mark price = num__140 discount = num__15.0 selling price num__0.85 Ã — num__140 hence profit = num__19.0 answer : d . <eor> d <eos> |
d |
percent__19.0__100.0__ |
percent__19.0__100.0__ |
| two trains move in the same direction at speeds num__50 kmph and num__32 kmph respectively . a man in the slower train observes that num__15 seconds elapse before the faster train completely passes by him . what is the length of faster train ? <o> a ) num__100 m <o> b ) num__75 m <o> c ) num__120 m <o> d ) num__50 m <o> e ) num__40 m |
as both trains move in same direction the relative velocity is ( num__50 - num__32 ) km / hr = > velocity = num__18 km / hr = num__18 * ( num__0.277777777778 ) m / s = num__5 m / s as num__15 seconds is the total time taken by the train = > length of train = num__5 * num__15 = num__75 m answer : b <eor> b <eos> |
b |
subtract__50.0__32.0__ multiply__15.0__5.0__ round__75.0__ |
subtract__50.0__32.0__ multiply__15.0__5.0__ multiply__15.0__5.0__ |
| the average salary of all the workers in a workshop is rs . num__8000 . the average salary of num__7 technicians is rs . num__12000 and the average salary of the rest is rs . num__6000 . the total number of workers in the workshop is : <o> a ) num__21 <o> b ) num__20 <o> c ) num__25 <o> d ) num__30 <o> e ) num__27 |
let the total number of workers be x . then num__8000 x = ( num__12000 * num__7 ) + num__6000 ( x - num__7 ) = num__2000 x = num__42000 = x = num__21 . answer a <eor> a <eos> |
a |
subtract__8000.0__6000.0__ multiply__7.0__6000.0__ divide__42000.0__2000.0__ divide__42000.0__2000.0__ |
subtract__8000.0__6000.0__ multiply__7.0__6000.0__ divide__42000.0__2000.0__ divide__42000.0__2000.0__ |
| enrollment in city college in num__1980 was num__210 percent of enrollment in num__1990 . what was the percent increase in the college ’ s enrollment from num__1980 to num__1990 ? <o> a ) - num__10.0 <o> b ) - num__16 num__0.666666666667 % <o> c ) - num__52.38 <o> d ) - num__25.0 <o> e ) - num__183 num__0.333333333333 % |
assume num__100 enrollments present in num__1990 num__210.0 of num__1990 = enrollments on num__1980 enrollments on num__1980 = num__210.0 change = ( enrollment in num__1990 - enrollment in num__1980 ) * num__100 / ( enrollment in num__1980 ) = ( num__100 - num__210 ) * num__0.47619047619 = - num__52.380952381 = - num__52.38 ans - c <eor> c <eos> |
c |
percent__100.0__52.38__ |
percent__100.0__52.38__ |
| two trains one from p to q and the other from q to p start simultaneously . after they meet the trains reach their destinations after num__9 hours and num__16 hours respectively . the ratio of their speeds is <o> a ) num__4 : num__1 <o> b ) num__4 : num__2 <o> c ) num__4 : num__5 <o> d ) num__4 : num__3 <o> e ) num__4 : num__6 |
ratio of their speeds = speed of first train : speed of second train = num__16 − − √ num__9 √ = num__4 : num__3 answer is d . <eor> d <eos> |
d |
round__4.0__ |
round__4.0__ |
| it is given that num__2 ^ num__32 + num__1 is exactly divisible by a certain number . which one of the following is also divisible by the same number b ? <o> a ) a . num__2 ^ num__96 + num__1 <o> b ) b . num__2 ^ num__16 - num__1 <o> c ) c . num__2 ^ num__16 + num__1 <o> d ) d . num__7 * num__2 ^ num__33 <o> e ) e . num__2 ^ num__64 + num__1 |
a ³ + b ³ = ( a + b ) ( a ² - ab + b ² ) now let ( num__2 ^ num__32 + num__1 ) be ( a + b ) a ³ + b ³ = ( num__2 ^ num__96 + num__1 ) now as mentioned in formula above a ³ + b ³ is always divisible by ( a + b ) so any factor of b = ( a + b ) is a factor of ( a ³ + b ³ ) henca a <eor> a <eos> |
a |
multiply__2.0__1.0__ |
multiply__2.0__1.0__ |
| if | x | = num__7 x - num__5 then x = ? <o> a ) num__1 <o> b ) num__0.5 <o> c ) num__1 and num__0.5 <o> d ) num__2 <o> e ) - num__1 |
answer : a approach : substituted option a i . e x = num__1 . inequality satisfied . d <eor> d <eos> |
d |
subtract__7.0__5.0__ |
subtract__7.0__5.0__ |
| the distance between delhi and mathura is num__140 kms . a starts from delhi with a speed of num__20 kmph at num__7 a . m . for mathura and b starts from mathura with a speed of num__25 kmph at num__8 p . m . from delhi . when will they meet ? <o> a ) num__10.50 a . m . <o> b ) num__10.40 a . m . <o> c ) num__10.30 a . m . <o> d ) num__11.40 a . m . <o> e ) num__1.40 a . m . |
d = num__140 – num__20 = num__120 rs = num__20 + num__25 = num__45 t = num__2.66666666667 = num__2 num__0.666666666667 hours num__8 a . m . + num__2 hrs num__40 min = num__10.40 a . m . answer : b <eor> b <eos> |
b |
subtract__140.0__20.0__ add__20.0__25.0__ divide__120.0__45.0__ subtract__2.6667__2.0__ multiply__20.0__2.0__ round__10.4__ |
subtract__140.0__20.0__ add__20.0__25.0__ divide__120.0__45.0__ subtract__2.6667__2.0__ multiply__20.0__2.0__ round__10.4__ |
| a batsman scored num__100 runs which included num__3 boundaries and num__8 sixes . what percent of his total score did he make by running between the wickets ? <o> a ) num__50.0 <o> b ) num__40.0 <o> c ) num__60.0 <o> d ) num__70.0 <o> e ) num__80 % |
number of runs made by running = num__100 - ( num__3 x num__4 + num__8 x num__6 ) = num__100 - ( num__60 ) = num__40 now we need to calculate num__40 is what percent of num__100 . = > num__0.4 x num__100 = num__40.0 answer : b <eor> b <eos> |
b |
subtract__100.0__60.0__ divide__40.0__100.0__ multiply__100.0__0.4__ |
subtract__100.0__60.0__ divide__40.0__100.0__ subtract__100.0__60.0__ |
| there are num__10000 families having num__4 children for each family . in how many families num__2 daughters are probable . <o> a ) num__625 <o> b ) num__529 <o> c ) num__425 <o> d ) num__354 <o> e ) num__725 |
probability to have num__2 daughter = num__0.5 ^ num__4 = num__0.0625 so answer = num__10000 * num__0.0625 = num__625 answer : a <eor> a <eos> |
a |
reverse__2.0__ power__0.5__4.0__ multiply__10000.0__0.0625__ multiply__10000.0__0.0625__ |
reverse__2.0__ power__0.5__4.0__ multiply__10000.0__0.0625__ multiply__10000.0__0.0625__ |
| how many odd three - digit integers greater than num__600 are there such that all their digits are different ? <o> a ) num__40 <o> b ) num__56 <o> c ) num__72 <o> d ) num__81 <o> e ) num__120 |
num__3 ( hundred ' s digit ) * num__9 ( ten ' s digit ) * num__8 ( unit ' s digit ) = num__216 now take numbers in the range num__600 - num__900 . total numbers where all digits are different = num__216 ( as before ) [ highlight ] number of odd numbers = num__3 * num__8 * num__5 = num__120 [ / highlight ] ( now there are num__5 possibilities for the unit ' s digit ) e <eor> e <eos> |
e |
subtract__8.0__3.0__ divide__600.0__5.0__ divide__600.0__5.0__ |
subtract__8.0__3.0__ divide__600.0__5.0__ divide__600.0__5.0__ |
| let y be the smallest positive integer such that num__12 and num__8 are factors of num__20 * y so the value of y is : <o> a ) num__7 <o> b ) num__9 <o> c ) num__11 <o> d ) num__13 <o> e ) num__15 |
you can calculate the lcm ( num__128 ) ( least commom multiple of num__12 and num__8 ) = num__36 and calculate the lcm ( num__3620 ) = num__180 and making num__9.0 we obtain num__9 . so the answer is b . <eor> b <eos> |
b |
lcm__20.0__36.0__ divide__180.0__20.0__ gcd__9.0__180.0__ |
lcm__20.0__36.0__ divide__180.0__20.0__ gcd__9.0__180.0__ |
| a number x is chosen at random from the set of positive integers less than num__20 . what is the probability that ( num__18 / x ) > x ? <o> a ) num__0.2 <o> b ) num__0.222222222222 <o> c ) num__0.157894736842 <o> d ) num__0.210526315789 <o> e ) num__0.777777777778 |
number x has to be chosen from numbers num__1 - num__19 ( num__18 / x ) > x = > num__18 > x ^ num__2 = > x ^ num__2 - num__18 < num__0 x can have num__2 values only num__1 num__2 num__3 num__4 therefore probability = num__0.210526315789 answer d <eor> d <eos> |
d |
subtract__20.0__1.0__ subtract__20.0__18.0__ add__1.0__2.0__ add__1.0__3.0__ divide__4.0__19.0__ multiply__1.0__0.2105__ |
subtract__20.0__1.0__ subtract__20.0__18.0__ add__1.0__2.0__ add__1.0__3.0__ divide__4.0__19.0__ divide__4.0__19.0__ |
| a train covers a distance of num__12 km in num__10 min . if it takes num__3 sec to pass a telegraph post then the length of the train is ? <o> a ) num__60 m <o> b ) num__188 m <o> c ) num__120 m <o> d ) num__80 m <o> e ) num__189 m |
speed = ( num__1.2 * num__60 ) km / hr = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . length of the train = num__20 * num__3 = num__60 m . answer : a <eor> a <eos> |
a |
divide__12.0__10.0__ hour_to_min_conversion__ add__12.0__60.0__ divide__60.0__3.0__ hour_to_min_conversion__ |
divide__12.0__10.0__ hour_to_min_conversion__ multiply__1.2__60.0__ divide__60.0__3.0__ multiply__3.0__20.0__ |
| in final exam of class ix there are num__50 students num__10.0 students failed . how many students passed to class x ? <o> a ) num__33 <o> b ) num__378 <o> c ) num__37 <o> d ) num__39 <o> e ) num__45 |
percentage of students passed to class x = num__100.0 - num__10.0 = num__90.0 num__90.0 of num__50 = num__0.9 × num__50 = num__45.0 = num__45 therefore num__45 students passed to class x . answer : e <eor> e <eos> |
e |
percent__50.0__90.0__ percent__50.0__90.0__ |
percent__50.0__90.0__ percent__50.0__90.0__ |
| num__16 men can complete a piece of work in num__25 days . in how many days can num__20 men complete that piece of work ? <o> a ) num__55 days <o> b ) num__77 days <o> c ) num__20 days <o> d ) num__88 days <o> e ) num__44 days |
num__16 * num__25 = num__20 * x = > x = num__20 days answer : c <eor> c <eos> |
c |
round__20.0__ |
round__20.0__ |
| if log num__72 = m then log num__4928 is equal to <o> a ) num__2 ( num__1 + num__2 m ) <o> b ) ( num__1 + num__2 m ) / num__2 <o> c ) num__2 / ( num__1 + num__2 m ) <o> d ) num__1 + m <o> e ) num__1 + num__2 m |
log num__4928 = num__0.5 log num__7 ( num__7 x num__4 ) = num__0.5 ( num__1 + log num__74 ) = num__0.5 + ( num__0.5 ) . num__2 log num__72 = num__0.5 + log num__72 = num__0.5 + m = ( num__1 + num__2 m ) / num__2 . answer : b <eor> b <eos> |
b |
reverse__0.5__ reverse__1.0__ |
reverse__0.5__ reverse__1.0__ |
| rs . num__2500 is divided into two parts such that if one part be put out at num__5.0 simple interest and the other at num__6.0 the yearly annual income may be rs . num__140 . how much was lent at num__5.0 ? <o> a ) num__2333 <o> b ) num__2777 <o> c ) num__2688 <o> d ) num__1000 <o> e ) num__2871 |
( x * num__5 * num__1 ) / num__100 + [ ( num__2500 - x ) * num__6 * num__1 ] / num__100 = num__140 x = num__1000 answer : d <eor> d <eos> |
d |
percent__100.0__1000.0__ |
percent__100.0__1000.0__ |
| a person purchased a tv set for rs . num__13000 and a dvd player for rs . num__5000 . he sold both the items together for rs . num__21600 . what percentage of profit did he make ? <o> a ) num__48.0 <o> b ) num__70.0 <o> c ) num__40.0 <o> d ) num__45.0 <o> e ) num__20 % |
the total cp = rs . num__13000 + rs . num__5000 = rs . num__18000 and sp = rs . num__21600 profit ( % ) = ( num__21600 - num__18000 ) / num__18000 * num__100 = num__20.0 answer : e <eor> e <eos> |
e |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| a man owns num__0.666666666667 of market reserch beauro buzness and sells num__0.75 of his shares for num__45000 rs what is the value of buzness ? <o> a ) num__150000 <o> b ) num__90000 <o> c ) num__85000 <o> d ) num__190000 <o> e ) num__250000 |
if value of business = x total sell ( num__2 x / num__3 ) ( num__0.75 ) = num__45000 - > x = num__150000 answer : b <eor> b <eos> |
b |
multiply__45000.0__2.0__ |
multiply__45000.0__2.0__ |
| here ' s an easy question of averages but let ' s try to see innovative ways of solving this . a class has num__12 boys and x girls . average score of boys and girls is num__84 and num__92 respectively . the average of the whole class is num__86 what is the value of x ? <o> a ) a ) num__6 <o> b ) b ) num__4 <o> c ) c ) num__8 <o> d ) d ) num__10 <o> e ) e ) num__12 |
num__12 ( num__84 ) + num__92 x / num__12 + x = num__86 num__1008 + num__92 x / num__12 + x = num__86 num__1008 + num__92 x = num__86 ( num__12 + x ) num__1008 + num__92 x = num__1032 + num__86 x x ' s one side numbers one side we get num__92 x - num__86 x = num__1032 - num__1008 num__6 x = num__24 hence x = num__4 answer b <eor> b <eos> |
b |
multiply__12.0__84.0__ multiply__12.0__86.0__ subtract__92.0__86.0__ subtract__1032.0__1008.0__ divide__24.0__6.0__ divide__24.0__6.0__ |
multiply__12.0__84.0__ multiply__12.0__86.0__ subtract__92.0__86.0__ subtract__1032.0__1008.0__ divide__24.0__6.0__ divide__24.0__6.0__ |
| in the quadratic equation ax num__2 - num__11 x + num__40 = num__0 if the sum of two roots is num__1.1 what is the product of the two roots ? <o> a ) num__7 <o> b ) num__6 <o> c ) num__5 <o> d ) num__4 <o> e ) num__3 |
explanation : the sum of the roots of the quadratic equation ax num__2 - bx + c = num__0 are ( - b / a ) and the product of the roots are ( c / a ) . thus in the equation ax num__2 - num__11 x + num__40 = num__0 where a = a b = - num__11 and c = num__40 . we get sum of the roots = - ( - num__11 ) / a = num__1.1 a = num__11 / num__1.1 = num__10 product of the roots = num__4.0 = num__4 answer : d <eor> d <eos> |
d |
divide__11.0__1.1__ divide__40.0__10.0__ divide__40.0__10.0__ |
divide__11.0__1.1__ divide__40.0__10.0__ divide__40.0__10.0__ |
| rashmi walks to her bus stop at num__5 kmph and reaches there late by num__10 minutes . on the next day she increases her speed to num__6 kmph and reaches the bus stop num__10 minutes early . how far is the bus stop ? <o> a ) num__13 <o> b ) num__12 <o> c ) num__11 <o> d ) num__10 <o> e ) num__9 |
let the distance to the bus stop is ` ` d ' ' km time taken by her ( at num__5 kmph ) = d / num__5 hour and time taken by her ( at num__6 kmph ) = d / num__6 hour as per given condition d / num__5 - d / num__6 = num__0.333333333333 d = num__10 kms answer : d <eor> d <eos> |
d |
round__10.0__ |
round__10.0__ |
| there are two sticks of which the longer one is num__10 cm longer than the short stick . find the length of the shorter stick if the total length of both the sticks is num__24 cm ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__12 <o> e ) num__14 |
since num__24 is the total length . shorter is any one from answers . try back solving from first answer i . e . num__6 . num__6 + ( num__6 + num__10 ) = num__22 < num__24 hence select next greater no . : num__7 + ( num__7 + num__10 ) = num__24 . so num__7 is the answer <eor> b <eos> |
b |
die_space__ choose__7.0__6.0__ |
die_space__ choose__7.0__6.0__ |
| at a florist shop on a certain day all corsages sold for either $ num__20 or $ num__30 . if num__10 of the corsages that sold for $ num__30 had instead sold for $ num__20 then the store ' s revenue from corsages that day would have been reduced by num__20 percent . what was the store ' s actual revenue from corsages that day ? <o> a ) $ num__200 <o> b ) $ num__400 <o> c ) $ num__500 <o> d ) $ num__800 <o> e ) $ num__1000 |
let no . of corsages @ $ num__20 = x no . of corsages @ $ num__30 = y and revenue = r so num__20 x + num__30 y = r . . . . . . . . . ( num__1 ) now given the situation num__20 ( x + num__10 ) + num__30 ( y - num__10 ) = r - . num__2 r = > num__20 x + num__200 + num__30 y - num__300 = . num__8 r = > num__20 x + num__30 y = . num__8 r + num__100 . . . . . . . . . . . . ( num__2 ) so r = . num__8 r + num__100 = > r = num__500 the answer is c . <eor> c <eos> |
c |
percent__20.0__10.0__ percent__100.0__500.0__ |
percent__20.0__10.0__ percent__100.0__500.0__ |
| if a rod is sold for rs . num__34.80 there is a loss of num__25.0 . find out the cp of the rod ? <o> a ) num__45 <o> b ) num__46 <o> c ) num__46.4 <o> d ) num__47.23 <o> e ) num__47.34 |
sp = num__34.80 loss = num__25.0 cp = num__100 ( num__100 − loss % ) × sp = num__100 ( num__100 − num__25 ) × num__34.80 = num__10075 × num__34.80 = num__4 × num__34.803 = num__4 × num__11.60 = num__46.40 c <eor> c <eos> |
c |
percent__100.0__46.4__ |
percent__100.0__46.4__ |
| a shopkeeper sold an article offering a discount of num__5.0 and earned a profit of num__23.5 . what would have been the percentage of profit earned if no discount was offered ? <o> a ) num__24 <o> b ) num__28 <o> c ) num__30 <o> d ) num__32 <o> e ) num__34 |
let c . p . be rs . num__100 . then s . p . = rs . num__123.50 let marked price be rs . x . then num__0.95 x = num__123.50 x = num__130.0 = rs . num__130 now s . p . = rs . num__130 c . p . = rs . num__100 profit % = num__30.0 . answer : c <eor> c <eos> |
c |
percent__100.0__30.0__ |
percent__100.0__30.0__ |
| a man complete a journey in num__10 hours . he travels first half of the journey at the rate of num__21 km / hr and second half at the rate of num__24 km / hr . find the total journey in km <o> a ) num__200 km <o> b ) num__222 km <o> c ) num__224 km <o> d ) num__248 km <o> e ) none of these |
explanation : let time taken to travel the first half = x hr then time taken to travel the second half = ( num__10 - x ) hr distance covered in the the first half = num__21 x [ because distance = time * speed ] distance covered in the the second half = num__24 ( num__10 - x ) distance covered in the the first half = distance covered in the the second half so num__21 x = num__24 ( num__10 - x ) = > num__45 x = num__240 = > x = num__5.33333333333 total distance = num__2 * num__21 ( num__5.33333333333 ) = num__224 km [ multiplied by num__2 as num__21 x was distance of half way ] option c <eor> c <eos> |
c |
add__21.0__24.0__ multiply__10.0__24.0__ divide__240.0__45.0__ round__224.0__ |
add__21.0__24.0__ multiply__10.0__24.0__ divide__240.0__45.0__ round__224.0__ |
| ravi bought num__5 pencil and num__10 eraser for rs num__4 . sheetal bought num__3 pencil and num__9 eraser for rs num__3 then what would be the cost of pencil and eraser . <o> a ) . num__15 p . num__30 p <o> b ) . num__50 p . num__27 p <o> c ) . num__40 p . num__20 p <o> d ) . num__20 p . num__40 p <o> e ) . num__50 p . num__40 p |
let pencil be x and eraser be y so num__5 x + num__10 y = num__4 num__3 x + num__9 y = num__3 by solving the above two equations we get x = . num__40 p and y = . num__20 p correct ans is c <eor> c <eos> |
c |
multiply__10.0__4.0__ multiply__5.0__4.0__ multiply__10.0__4.0__ |
multiply__10.0__4.0__ multiply__5.0__4.0__ multiply__10.0__4.0__ |
| a coin is tossed live times . what is the probability that there is at the least one tail ? <o> a ) num__0.96875 <o> b ) num__1.82352941176 <o> c ) num__1.63157894737 <o> d ) num__2.58333333333 <o> e ) num__2.38461538462 |
let p ( t ) be the probability of getting least one tail when the coin is tossed five times . = there is not even a single tail . i . e . all the outcomes are heads . = num__0.03125 ; p ( t ) = num__1 - num__0.03125 = num__0.96875 answer : a <eor> a <eos> |
a |
negate_prob__0.0312__ negate_prob__0.0312__ |
negate_prob__0.0312__ negate_prob__0.0312__ |
| the average of a batsman for num__40 innings is num__50 runs . his highest score exceeds his lowest score by num__172 runs . if these two innings are excluded his average drops by num__2 runs . find his highest score . <o> a ) num__172 <o> b ) num__173 <o> c ) num__174 <o> d ) num__175 <o> e ) num__176 |
total runs = num__40 × num__50 = num__2000 let his highest score be = x then his lowest score = x – num__172 now num__200 − x − ( x − num__172 ) / num__38 = num__48 ⇒ num__2 x = num__2172 – num__1824 ⇒ x = num__174 answer c <eor> c <eos> |
c |
multiply__40.0__50.0__ subtract__40.0__2.0__ subtract__50.0__2.0__ add__172.0__2000.0__ multiply__48.0__38.0__ add__172.0__2.0__ add__172.0__2.0__ |
multiply__40.0__50.0__ subtract__40.0__2.0__ subtract__50.0__2.0__ add__172.0__2000.0__ multiply__48.0__38.0__ add__172.0__2.0__ add__172.0__2.0__ |
| a man swims downstream num__72 km and upstream num__45 km taking num__9 hours each time ; what is the speed of the current ? <o> a ) num__1.6 <o> b ) num__1.5 <o> c ) num__1.9 <o> d ) num__1.2 <o> e ) num__1.7 |
num__72 - - - num__9 ds = num__8 ? - - - - num__1 num__45 - - - - num__9 us = num__5 ? - - - - num__1 s = ? s = ( num__8 - num__5 ) / num__2 = num__1.5 answer : b <eor> b <eos> |
b |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ round__1.5__ |
divide__72.0__9.0__ subtract__9.0__8.0__ divide__45.0__9.0__ divide__1.5__1.0__ |
| a boats man can row in still water at speed of num__7 km / hr . it takes num__6 hours more to travel the same distance in upstream than in downstream if the speed of the river is num__3 km / hr . what is the distance between the two destinations ? <o> a ) num__40 <o> b ) num__87 <o> c ) num__126 <o> d ) num__18 <o> e ) num__91 |
explanation : x = num__7 km / hr ; y = num__3 km / hr ds = num__10 km / hr ; us = num__4 km / hr distance ( d ) is same . therefore if time taken for downstream is t hours the time for upstream is ( t + num__6 ) hours . num__10 * t = num__4 * ( t + num__6 ) num__6 t = num__24 ; t = num__4 hours d = num__10 * num__4 = num__40 km answer : a <eor> a <eos> |
a |
add__7.0__3.0__ subtract__7.0__3.0__ multiply__6.0__4.0__ multiply__4.0__10.0__ round__40.0__ |
add__7.0__3.0__ subtract__7.0__3.0__ multiply__6.0__4.0__ multiply__4.0__10.0__ multiply__4.0__10.0__ |
| if num__7 ^ r is a factor of the product of the integers from num__1 to num__100 inclusive what is the largest value of r ? <o> a ) num__12 <o> b ) num__13 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
so the question just means that we have to find all the multiples r of num__7 between num__1 to num__100 so there are num__14 multiples of num__7 ( num__7 - num__98 ) but num__49 and num__98 contain two num__7 ' s as factors so we have to add num__14 + num__2 = num__16 e <eor> e <eos> |
e |
multiply__7.0__14.0__ subtract__100.0__98.0__ add__2.0__14.0__ multiply__1.0__16.0__ |
multiply__7.0__14.0__ subtract__100.0__98.0__ add__2.0__14.0__ add__2.0__14.0__ |
| each of num__435 bags contains at least one of the following three items : raisins almonds and peanuts . the number of bags that contain only raisins is num__10 times the number of bags that contain only peanuts . the number of bags that contain only almonds is num__20 times the number of bags that contain only raisins and peanuts . the number of bags that contain only peanuts is one - fifth the number of bags that contain only almonds . num__210 bags contain almonds . how many bags contain only one kind of item ? <o> a ) num__256 <o> b ) num__260 <o> c ) num__316 <o> d ) num__320 <o> e ) num__350 |
there are total of num__435 bags and num__210 bags contain almonds . from the diagram num__20 y = num__5 x - - > y = x / num__4 . now total = num__435 = { almonds } + num__10 x + y + x - - > num__435 = num__210 + num__10 x + x / num__4 + x - - > x = num__20 - - > # of bags that contain only one kind of item is the sum of yellow segments : num__10 x + x + num__5 x = num__16 x = num__320 . answer : d . <eor> d <eos> |
d |
divide__20.0__5.0__ subtract__20.0__4.0__ multiply__20.0__16.0__ multiply__20.0__16.0__ |
divide__20.0__5.0__ subtract__20.0__4.0__ multiply__20.0__16.0__ multiply__20.0__16.0__ |
| if the price of a certain bond on may num__1 st was num__0.666666666667 the price of the bond on june num__1 st and the price of the bond on july num__1 st was num__25.0 greater than the price of the bond on may num__1 st . then the price of the bond on june num__1 st st was what percent of the average ( arithmetic mean ) price of the bond on may num__1 st and july num__1 st ? <o> a ) num__50.0 <o> b ) num__75.0 <o> c ) num__120.0 <o> d ) num__133 num__0.333333333333 % <o> e ) num__150 % |
the price on june num__1 st = num__12 ( assume ) ; the price on may num__1 st = num__0.666666666667 * num__12 = num__8 ; the price on july num__1 st = num__8 * num__1.25 = num__10 . the average price of the bond on may num__1 st and july num__1 st = ( num__8 + num__10 ) / num__2 = num__9 . the price of the bond on june num__1 st ( num__12 ) is num__1.33333333333 times ( num__134.0 ) the average price of the bond on may num__1 st and july num__1 st . answer : d . <eor> d <eos> |
d |
multiply__1.25__8.0__ subtract__10.0__8.0__ add__1.0__8.0__ subtract__2.0__0.6667__ subtract__134.0__1.0__ |
multiply__1.25__8.0__ subtract__10.0__8.0__ add__1.0__8.0__ divide__12.0__9.0__ subtract__134.0__1.0__ |
| alice and heather are num__20 miles apart and walk towards each other along the same route . alice walks at constant rate that is num__1 mile per hour faster than heather ' s constant rate of num__5 miles / hour . if heather starts her journey num__24 minutes after alice how far from the original destination has heather walked when the two meet ? <o> a ) num__4 miles <o> b ) num__8 miles <o> c ) num__10 miles <o> d ) num__6 miles <o> e ) num__12 mile |
original distance between s and h = num__20 miles . speed of a = num__5 + num__1 = num__6 mph speed of h = num__5 mph . time traveled by h = t hours - - - > time traveled by a = t + num__0.4 = t + num__0.4 hours . now the total distances traveled by a and h = num__20 miles - - - > num__6 * ( t + num__0.4 ) + num__5 * t = num__20 - - - > t = num__1.6 hours . thus h has traveled for num__1.6 hours giving you a total distance for h = num__5 * num__1.6 = num__8 miles . b is thus the correct answer . p . s . : based on the wording of the question you should calculatehow far from theoriginal destination has heather walkedwhen the two meet . ' original destination ' for h does not make any sense . original destination for h was situated at a distance of num__20 miles . <eor> b <eos> |
b |
add__1.0__5.0__ mile_to_km_conversion__ multiply__20.0__0.4__ round__8.0__ |
add__1.0__5.0__ mile_to_km_conversion__ multiply__20.0__0.4__ multiply__20.0__0.4__ |
| if x and y are integers such that | y + num__3 | ≤ num__3 and num__2 y – num__3 x + num__6 = num__0 what is the least possible value w of the product xy ? <o> a ) - num__12 <o> b ) - num__3 <o> c ) num__0 <o> d ) num__2 <o> e ) none of the above |
how to deal with inequalities involving absolute values ? first example shows us the so callednumber case in this case we have | y + num__3 | ≤ num__3 which is generalized | something | ≤ some number . first we solve as if there were no absolute value brackets : y + num__3 ≤ num__3 y ≤ num__0 so y is num__0 or negative second scenario - remove the absolute value brackets . put a negative sign around the other side of the inequality andflip the sign : y + num__3 > = - num__3 y > = - num__6 therefore we have a possible range for y : - num__6 = < y < = num__0 ok so far so good we ' re half way through . what about x ? here ' s the formula : num__2 y – num__3 x + num__6 = num__0 rewrite it as num__2 y + num__6 = num__3 x . you can say that num__2 y + num__6 is a multiple of num__3 ( = num__3 x ) . so all values which must be integer must also satisfy this constraint . i ' m just saying that so it ' s easier to evaluate all the possible numbers w ( - num__6 - num__3 num__0 ) . if you plug in y = num__0 x will be num__2 and xy = num__0 as the lowest possible value . hence answer choice c is the one to go . <eor> c <eos> |
c |
multiply__3.0__0.0__ |
multiply__3.0__0.0__ |
| one - third of num__1206 is what percent of num__200 ? <o> a ) num__3 <o> b ) num__201 <o> c ) num__300 <o> d ) none of these <o> e ) can not be determined |
answer let one - third of num__1206 is n % of num__200 . ∵ num__402.0 = ( n x num__200 ) / num__100 ∴ n = ( num__402 x num__100 ) / num__200 = num__201 correct option : b <eor> b <eos> |
b |
percent__100.0__201.0__ |
percent__100.0__201.0__ |
| in how many different ways can the letters of the word ‘ prepty ’ be arranged ? <o> a ) num__120 <o> b ) num__230 <o> c ) num__330 <o> d ) num__340 <o> e ) num__360 |
number of ways = num__2.90476190476 num__6 x num__5 x num__4 x num__3 x num__2 x num__1 - num__360 e <eor> e <eos> |
e |
die_space__ vowel_space__ coin_space__ choose__5.0__2.0__ choose__5.0__2.0__ |
die_space__ vowel_space__ coin_space__ choose__5.0__2.0__ choose__5.0__2.0__ |
| double of quarter of num__1 percent written as a decimal is : <o> a ) num__0.003 <o> b ) num__0.0005 <o> c ) num__0.25 <o> d ) num__0.005 <o> e ) none of these |
explanation : solution : ( num__2 ) * ( num__0.25 ) * num__1.0 = num__2 * ( num__0.25 * num__0.01 ) = num__0.0025 = num__2 * num__0.0025 = num__0.005 . answer : d <eor> d <eos> |
d |
twice__1.0__ quarter__ multiply__0.25__0.01__ twice__0.0025__ twice__0.0025__ |
twice__1.0__ quarter__ multiply__0.25__0.01__ twice__0.0025__ twice__0.0025__ |
| a cubical block of metal weighs num__6 pounds . how much will another cube of the same metal weigh if its sides are twice as long ? <o> a ) num__48 <o> b ) num__12 <o> c ) num__36 <o> d ) num__60 <o> e ) none of these |
explanation : if you double the sides of a cube the ratio of the surface areas of the old and new cubes will be num__1 : num__4 . the ratio of the volumes of the old and new cubes will be num__1 : num__8 . weight is proportional to volume . so if the first weighs num__6 pounds the second weighs num__6 x num__8 pounds = num__48 . answer : a <eor> a <eos> |
a |
square_perimeter__1.0__ multiply__6.0__8.0__ multiply__6.0__8.0__ |
square_perimeter__1.0__ multiply__6.0__8.0__ multiply__6.0__8.0__ |
| num__105 candies are distributed to children with the same number of candies for each child . what can ’ t be the range which includes the number of children ? <o> a ) num__1 ~ num__10 . <o> b ) num__10 ~ num__20 . <o> c ) num__20 ~ num__30 . <o> d ) num__30 ~ num__40 . <o> e ) num__40 ~ num__50 . |
num__105 = num__3 * num__5 * num__7 there are factors in each range except for num__40 ~ num__50 the answer is e . <eor> e <eos> |
e |
vowel_space__ choose__5.0__3.0__ choose__5.0__3.0__ |
vowel_space__ choose__5.0__3.0__ choose__5.0__3.0__ |
| the product of all the prime numbers less than num__12 is closest to which of the following powers of num__10 ? <o> a ) num__10 ^ num__9 <o> b ) num__10 ^ num__3 <o> c ) num__10 ^ num__7 <o> d ) num__10 ^ num__6 <o> e ) num__10 ^ num__5 |
p = num__2 * num__3 * num__5 * num__7 * num__11 ~ num__10 ^ num__3 answer : b <eor> b <eos> |
b |
subtract__12.0__10.0__ divide__10.0__2.0__ subtract__12.0__5.0__ subtract__12.0__2.0__ |
subtract__12.0__10.0__ divide__10.0__2.0__ subtract__12.0__5.0__ multiply__2.0__5.0__ |
| in a class of num__20 students num__3 students did not borrow any books from the library num__9 students each borrowed num__1 book num__4 students each borrowed num__2 books and the rest borrowed at least num__3 books . if the average number of books per student was num__2 what is the maximum number of books any single student could have borrowed ? <o> a ) num__13 <o> b ) num__14 <o> c ) num__15 <o> d ) num__16 <o> e ) num__17 |
the total number of books the students borrowed is num__20 * num__2 = num__40 . the students who borrowed zero one or two books borrowed num__9 * num__1 + num__4 * num__2 = num__17 books . the num__4 students who borrowed at least three books borrowed num__40 - num__17 = num__23 books . if num__3 of these students borrowed exactly num__3 books then the maximum that one student could have borrowed is num__23 - num__9 = num__14 books . the answer is b . <eor> b <eos> |
b |
multiply__20.0__2.0__ subtract__20.0__3.0__ add__20.0__3.0__ subtract__17.0__3.0__ multiply__1.0__14.0__ |
multiply__20.0__2.0__ subtract__20.0__3.0__ add__20.0__3.0__ subtract__17.0__3.0__ multiply__1.0__14.0__ |
| num__9773 + x = num__13200 then x is ? <o> a ) num__3427 <o> b ) num__3237 <o> c ) num__3337 <o> d ) num__2337 <o> e ) none of these |
answer x = num__13200 - num__9773 = num__3427 option : a <eor> a <eos> |
a |
subtract__13200.0__9773.0__ subtract__13200.0__9773.0__ |
subtract__13200.0__9773.0__ subtract__13200.0__9773.0__ |
| the average age of a b and c is num__29 years . if the average age of a and c is num__32 years what is the age of b in years ? <o> a ) num__19 <o> b ) num__35 <o> c ) num__20 <o> d ) num__32 <o> e ) num__23 |
age of b = age of ( a + b + c ) â € “ age of ( a + c ) = num__29 Ã — num__3 â € “ num__32 Ã — num__2 = num__87 â € “ num__64 = num__23 years e <eor> e <eos> |
e |
subtract__32.0__29.0__ multiply__29.0__3.0__ multiply__32.0__2.0__ subtract__87.0__64.0__ subtract__87.0__64.0__ |
subtract__32.0__29.0__ multiply__29.0__3.0__ multiply__32.0__2.0__ subtract__87.0__64.0__ subtract__87.0__64.0__ |
| the s . i . on a certain sum of money for num__4 years at num__12.0 per annum is half the c . i . on rs . num__6000 for num__2 years at num__15.0 per annum . the sum placed on s . i . is ? <o> a ) num__2019.625 <o> b ) num__2017.625 <o> c ) num__2013.625 <o> d ) num__2015.625 <o> e ) num__2011.625 |
c . i . = [ num__6000 * ( num__1 + num__0.15 ) num__2 - num__6000 ] = ( num__6000 * num__1.15 * num__1.15 - num__6000 ) = rs . num__1935 . sum = ( num__967.5 * num__100 ) / ( num__4 * num__12 ) = rs . num__2015.625 answer : d <eor> d <eos> |
d |
percent__15.0__1.0__ percent__100.0__2015.625__ |
percent__15.0__1.0__ percent__100.0__2015.625__ |
| $ num__300 is divided amongst a b and c so that a may get num__0.666666666667 as much as b and c together b may get num__0.666666666667 as much as a and c together then the share of a is <o> a ) $ num__100 <o> b ) $ num__150 <o> c ) $ num__125 <o> d ) $ num__120 <o> e ) $ num__250 |
a : ( b + c ) = num__2 : num__3 a ' s share = num__300 * num__0.4 = $ num__120 answer is d <eor> d <eos> |
d |
multiply__300.0__0.4__ multiply__300.0__0.4__ |
multiply__300.0__0.4__ multiply__300.0__0.4__ |
| an equal number of desks and bookcases are to be placed along a library wall that is num__15 meters long . each desk is num__2 meters long and each bookshelf is num__1.5 meters long . if the maximum possible number of desks and bookcases are to be placed along the wall then the space along the wall that is left over will be how many meters w long ? <o> a ) num__0.5 <o> b ) num__1 <o> c ) num__1.5 <o> d ) num__2 <o> e ) num__3 |
let x be the number of desks and bookcases that are placed along the library wall . num__2 x + num__1.5 x < num__15 num__3.5 x < num__15 since x is a non negative integer the largest number x can be is num__4 . when x is num__4 the desks and bookcases take up num__3.5 * num__4 = num__14 m leaving num__1 m of empty space . thus i believe the answer is b ) num__1 <eor> b <eos> |
b |
add__2.0__1.5__ multiply__3.5__4.0__ round_down__1.5__ round_down__1.5__ |
add__2.0__1.5__ multiply__3.5__4.0__ round_down__1.5__ round_down__1.5__ |
| some persons can do a piece of work in num__16 days . two times the number of these people will do half of that work in ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__8 |
num__16 / ( num__2 * num__2 ) = num__4 days answer : b <eor> b <eos> |
b |
round__4.0__ |
divide__16.0__4.0__ |
| the arithmetic mean and standard deviation of a certain normal distribution are num__10.5 and num__1 respectively . what value is exactly num__2 standard deviations less than the mean ? <o> a ) num__6 <o> b ) num__8.5 <o> c ) num__11.5 <o> d ) num__12 <o> e ) num__12.5 |
mean = num__10.5 two standard deviations is num__1 + num__1 = num__2 there could be two calues for this . mean + two standard deviations = num__12.5 mean - two standard deviations = num__8.5 answer choice has num__8.5 and so b is the answer . <eor> b <eos> |
b |
add__10.5__2.0__ subtract__10.5__2.0__ subtract__10.5__2.0__ |
add__10.5__2.0__ subtract__10.5__2.0__ subtract__10.5__2.0__ |
| a box contains num__60 balls numbered from num__1 to num__60 . if three balls are selected at random and with replacement from the box what is the probability that the sum of the three numbers on the balls selected from the box will be odd ? <o> a ) num__0.25 <o> b ) num__0.375 <o> c ) num__0.5 <o> d ) num__0.625 <o> e ) num__0.75 |
i do n ' t think order matters in this case because num__2 + num__2 + num__1 = num__2 + num__1 + num__2 my answer is : num__0.25 a <eor> a <eos> |
a |
multiply__1.0__0.25__ |
multiply__1.0__0.25__ |
| a boy has rs . num__720 in the denominations of one - rupee notes five - rupee notes and ten - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? <o> a ) num__90 <o> b ) num__110 <o> c ) num__140 <o> d ) num__135 <o> e ) num__120 |
let number of notes of each denomination be x . then x + num__5 x + num__10 x = num__720 num__16 x = num__720 x = num__45 hence total number of notes = num__3 x = num__135 d <eor> d <eos> |
d |
divide__720.0__16.0__ multiply__3.0__45.0__ multiply__3.0__45.0__ |
divide__720.0__16.0__ multiply__3.0__45.0__ multiply__3.0__45.0__ |
| fox jeans regularly sell for $ num__15 a pair and pony jeans regularly sell for $ num__18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ num__4 is saved by purchasing num__5 pairs of jeans : num__3 pairs of fox jeans and num__2 pairs of pony jeans . if the sum of the two discounts rates is num__18 percent what is the discount rate on pony jeans ? <o> a ) num__45.0 <o> b ) num__13.0 <o> c ) num__11.0 <o> d ) num__12.0 <o> e ) num__15 % |
you know that fox jeans costs $ num__15 and pony jeans costs $ num__18 you also know that num__3 pairs of fox jeans and num__2 pairs of pony jeans were purchased . so num__3 ( num__15 ) = num__45 - fox num__2 ( num__18 ) = num__36 - pony the total discount discount is $ num__4 and you are asked to find the percent discount of pony jeans so num__45 ( num__18 - x ) / num__100 + num__36 ( x ) / num__100 = num__4 or num__45 * num__18 - num__45 * x + num__36 * x = num__400 or - num__9 x = num__400 - num__45 * num__18 or x = num__45.5555555556 = num__45.0 = num__45.0 a <eor> a <eos> |
a |
percent__100.0__45.0__ |
percent__100.0__45.0__ |
| a batsman scored num__1 runs in his num__18 th innings and that makes his average num__18 . find his average upto the num__17 th innings ? <o> a ) num__19 <o> b ) num__20 <o> c ) num__21 <o> d ) num__22 <o> e ) num__23 |
avg = sum of the value / no . of the value num__18 = ( sum of the first num__17 innings score + num__1 ) / num__18 ( num__18 * num__18 - num__1 ) = sum of the first num__17 innings score sum of the first num__17 innings score = num__323 ave upto num__17 innings = num__19.0 = num__19 answer : a <eor> a <eos> |
a |
add__1.0__18.0__ add__1.0__18.0__ |
add__1.0__18.0__ add__1.0__18.0__ |
| the perimeter of a semi circle is num__140 cm then the radius is ? <o> a ) num__27 <o> b ) num__28 <o> c ) num__19 <o> d ) num__11 <o> e ) num__12 |
num__5.14285714286 r = num__140 = > r = num__27 answer : a <eor> a <eos> |
a |
round__27.0__ |
round__27.0__ |
| according to the directions on the can of frozen orange juice concentrate num__1 can of concentrate is to be mixed with num__3 cans of water to make orange juice . how many num__12 ounces cans of the concentrate are required to prepare num__280 num__6 ounces servings of orange juice ? <o> a ) a ) num__35 <o> b ) b ) num__34 <o> c ) c ) num__50 <o> d ) d ) num__67 <o> e ) e ) num__100 |
its a . total juice rquired = num__280 * num__6 = num__1680 ounce num__12 ounce concentate makes = num__12 * num__4 = num__48 ounce juice total cans required = num__35.0 = num__35 . answer a <eor> a <eos> |
a |
multiply__280.0__6.0__ add__1.0__3.0__ multiply__12.0__4.0__ divide__1680.0__48.0__ multiply__1.0__35.0__ |
multiply__280.0__6.0__ add__1.0__3.0__ multiply__12.0__4.0__ divide__1680.0__48.0__ multiply__1.0__35.0__ |
| fill in the blank with the correct number in this number series : num__20 num__65 num__27 num__61 __ num__57 num__41 ? <o> a ) num__34 <o> b ) num__35 <o> c ) num__38 <o> d ) num__25 <o> e ) num__56 |
this is a simple alternating subtraction and addition series ; the first series begins with num__20 and adds num__7 ; the second begins with num__65 and subtracts num__4 and so on . answer : a <eor> a <eos> |
a |
subtract__27.0__20.0__ subtract__65.0__61.0__ add__27.0__7.0__ |
subtract__27.0__20.0__ subtract__65.0__61.0__ add__27.0__7.0__ |
| the racing magic takes num__50 seconds to circle the racing track once . the charging bull makes num__40 rounds of the track in an hour . if they left the starting point together how many minutes will it take for them to meet at the starting point for the second time ? <o> a ) num__3 <o> b ) num__6 <o> c ) num__9 <o> d ) num__12 <o> e ) num__15 |
time taken by racing magic to make one circle = num__50 seconds time taken bycharging bullto make one circle = num__60 mins / num__40 = num__1.5 mins = num__90 seconds lcm of num__90 and num__50 seconds = num__450 seconds time taken for them to meet at the starting point for the second time = num__450 * num__2 = num__900 seconds = num__15 mins answer e <eor> e <eos> |
e |
hour_to_min_conversion__ divide__60.0__40.0__ add__50.0__40.0__ multiply__450.0__2.0__ divide__900.0__60.0__ round__15.0__ |
hour_to_min_conversion__ divide__60.0__40.0__ multiply__1.5__60.0__ multiply__450.0__2.0__ divide__900.0__60.0__ divide__900.0__60.0__ |
| a sum of rs . num__12500 amounts to rs . num__15500 in num__4 years at the rate of simple interest . what is the rate of interest ? <o> a ) num__4.0 <o> b ) num__5.0 <o> c ) num__6.0 <o> d ) num__8.0 <o> e ) num__14 % |
s . i . = ( num__15500 - num__12500 ) = rs . num__3000 / - rate = ( num__100 * num__3000 ) / ( num__12500 * num__4 ) = num__6.0 answer : c <eor> c <eos> |
c |
percent__100.0__6.0__ |
percent__100.0__6.0__ |
| a tightrope approximately num__320 m long is suspended between two poles . during a performance a break occurs in the line . assume that the line has an equal chance of breaking anywhere along its length . what is the probability that the break happened in the first num__50 meters of the rope ? <o> a ) num__0.84375 <o> b ) num__0.5 <o> c ) num__0.15625 <o> d ) num__0.185185185185 <o> e ) . num__0.666666666667 |
simly num__0.15625 = num__0.15625 answer will be ( c ) <eor> c <eos> |
c |
divide__50.0__320.0__ divide__50.0__320.0__ |
divide__50.0__320.0__ divide__50.0__320.0__ |
| num__8.036 divided by num__0.04 gives : <o> a ) num__200.9 <o> b ) num__2.06 <o> c ) num__20.06 <o> d ) num__100.9 <o> e ) num__200.6 |
= num__8.036 / num__0.04 = num__803.6 / num__4 = num__200.9 answer is a . <eor> a <eos> |
a |
divide__8.036__0.04__ divide__8.036__0.04__ |
divide__8.036__0.04__ divide__8.036__0.04__ |
| the ratio between the length and the breadth of a rectangular park is num__3 : num__2 . if a man cycling alongthe oundary of the park at the speed of num__12 km / hr completes one round in num__8 min then the area of the park ( in sq . m ) is ? <o> a ) num__1345460 m <o> b ) num__134500 m <o> c ) num__153600 m <o> d ) num__167360 m <o> e ) num__173600 m |
perimeter = distance covered in num__8 min . = num__12000 x num__8 m = num__1600 m . num__60 let length = num__3 x metres and breadth = num__2 x metres . then num__2 ( num__3 x + num__2 x ) = num__1600 or x = num__160 . length = num__480 m and breadth = num__320 m . area = ( num__480 x num__320 ) m num__2 = num__153600 m c <eor> c <eos> |
c |
multiply__3.0__160.0__ multiply__2.0__160.0__ surface_cube__160.0__ surface_cube__160.0__ |
multiply__3.0__160.0__ multiply__2.0__160.0__ surface_cube__160.0__ surface_cube__160.0__ |
| if a survey shows that num__56 citizens out of a sample of num__200 support a particular senate bill what percent of the sample does not support the bill ? <o> a ) num__56.0 <o> b ) num__64.0 <o> c ) num__72.0 <o> d ) num__82.0 <o> e ) num__86 % |
num__56 out of num__200 supports the bill and we are asked to find the percent who the does not support the bill . num__56 out of num__200 supports the bill = num__28 out of num__100 supports the bill ( taking in % terms ) num__28 out of num__100 supports the bill = num__72 ( num__100 - num__28 ) does n ' t support the bill so answer is ( c ) num__72.0 <eor> c <eos> |
c |
percent__100.0__72.0__ |
percent__100.0__72.0__ |
| the cross - section of a cannel is a trapezium in shape . if the cannel is num__15 m wide at the top and num__4 m wide at the bottom and the area of cross - section is num__1900 sq m the depth of cannel is ? <o> a ) num__150 <o> b ) num__200 <o> c ) num__140 <o> d ) num__240 <o> e ) num__145 |
num__0.5 * d ( num__15 + num__4 ) = num__1900 d = num__200 answer : b <eor> b <eos> |
b |
round__200.0__ |
round__200.0__ |
| num__40.0 of ram ' s marks is equal to num__20.0 of rahim ' s marks which percent is equal to num__30.0 of robert ' s marks . if robert ' s marks is num__80 then find the average marks of ram and rahim ? <o> a ) num__12 <o> b ) num__61 <o> c ) num__17 <o> d ) num__90 <o> e ) num__15 |
given num__40.0 of ram ' s marks = num__20.0 of rahim ' s marks = num__30.0 of robert ' s marks . given marks of robert = num__80 num__30.0 of num__80 = num__0.3 * num__8 = num__24 given num__40.0 of ram ' s marks = num__24 . = > ram ' s marks = ( num__24 * num__100 ) / num__40 = num__60 also num__20.0 of rahim ' s marks = num__24 = > rahim ' s marks = ( num__24 * num__100 ) / num__20 = num__120 average marks of ram and rahim = ( num__60 + num__120 ) / num__2 = num__90 . answer : d <eor> d <eos> |
d |
multiply__80.0__0.3__ add__20.0__80.0__ add__40.0__20.0__ add__40.0__80.0__ divide__40.0__20.0__ add__30.0__60.0__ add__30.0__60.0__ |
multiply__80.0__0.3__ add__20.0__80.0__ add__40.0__20.0__ add__40.0__80.0__ divide__40.0__20.0__ add__30.0__60.0__ add__30.0__60.0__ |
| num__3 pumps working num__8 hours a day can empty a tank in num__2 days . how many hours a day must num__4 pumps work to empty the tank in num__1 day ? <o> a ) num__9 <o> b ) num__10 <o> c ) num__11 <o> d ) num__12 <o> e ) num__13 |
let the required number of working hours per day be x . more pumps less working hours per day ( indirect proportion ) less days more working hours per day ( indirect proportion ) pumps num__4 : num__3 : : num__8 : x days num__1 : num__2 therefore num__4 x num__1 x x = num__3 x num__2 x num__8 = > x = ( num__3 x num__2 x num__8 ) / ( num__4 ) x = num__12 . correct answer is d <eor> d <eos> |
d |
multiply__3.0__4.0__ round__12.0__ |
multiply__3.0__4.0__ divide__12.0__1.0__ |
| the sum of number of boys and girls in a school is num__1150 . if the number of boys is x then the number of girls becomes x % of the total number of students . the number of boys is ? <o> a ) num__50 <o> b ) num__40 <o> c ) num__60 <o> d ) num__92 <o> e ) num__70 |
we have x + x % of num__1150 = num__1150 x + x / num__100 * num__1150 = num__1150 num__12.5 * x = num__1150 x = num__1150 * num__0.08 = num__92 answer is d <eor> d <eos> |
d |
reverse__12.5__ divide__1150.0__12.5__ divide__1150.0__12.5__ |
reverse__12.5__ divide__1150.0__12.5__ divide__1150.0__12.5__ |
| if the sum of the num__4 th term and the num__12 th term of an arithmetic progression is num__10 what is the sum of the first num__15 terms of the progression ? <o> a ) num__65 <o> b ) num__75 <o> c ) num__60 <o> d ) num__55 <o> e ) num__50 |
num__4 th term + num__12 th term = num__10 i . e . ( a + num__3 d ) + ( a + num__11 d ) = num__10 now sum of first num__15 terms = ( num__7.5 ) * [ num__2 a + ( num__15 - num__1 ) d ] = ( num__7.5 ) * [ num__2 a + num__14 d ] = ( num__7.5 ) * num__10 - - - - - - - - - - - - - - - from ( num__1 ) = num__75 answer : b <eor> b <eos> |
b |
divide__12.0__4.0__ subtract__15.0__4.0__ subtract__12.0__10.0__ subtract__4.0__3.0__ add__4.0__10.0__ multiply__10.0__7.5__ multiply__10.0__7.5__ |
subtract__15.0__12.0__ subtract__15.0__4.0__ subtract__12.0__10.0__ subtract__4.0__3.0__ add__4.0__10.0__ multiply__10.0__7.5__ multiply__10.0__7.5__ |
| the dimensions of a room are num__25 feet * num__15 feet * num__12 feet . what is the cost of white washing the four walls of the room at rs . num__5 per square feet if there is one door of dimensions num__6 feet * num__3 feet and three windows of dimensions num__4 feet * num__3 feet each ? <o> a ) num__4000 <o> b ) num__345 <o> c ) num__5673 <o> d ) num__4530 <o> e ) num__4566 |
area of the four walls = num__2 h ( l + b ) since there are doors and windows area of the walls = num__2 * num__12 ( num__15 + num__25 ) - ( num__6 * num__3 ) - num__3 ( num__4 * num__3 ) = num__906 sq . ft . total cost = num__906 * num__5 = rs . num__4530 answer : option d <eor> d <eos> |
d |
multiply__5.0__906.0__ multiply__5.0__906.0__ |
multiply__5.0__906.0__ multiply__5.0__906.0__ |
| what amount does an investor receive if the investor invests $ num__5000 at num__10.0 p . a . compound interest for two years compounding done annually ? <o> a ) $ num__5850 <o> b ) $ num__5950 <o> c ) $ num__6050 <o> d ) $ num__6150 <o> e ) $ num__6250 |
a = ( num__1 + r / num__100 ) ^ n * p ( num__1.1 ) ^ num__2 * num__5000 = num__1.21 * num__5000 = num__6050 the answer is c . <eor> c <eos> |
c |
percent__100.0__6050.0__ |
percent__100.0__6050.0__ |
| a cycle is bought for rs . num__670 and sold for rs . num__880 find the gain percent ? <o> a ) num__31.0 <o> b ) num__35.0 <o> c ) num__54.0 <o> d ) num__38.0 <o> e ) num__80 % |
explanation : num__670 - - - - num__210 num__100 - - - - ? = > num__31.0 answer : a <eor> a <eos> |
a |
percent__100.0__31.0__ |
percent__100.0__31.0__ |
| the sale price of an article including the sales tax is rs . num__616 . the rate of sales tax is num__10.0 . if the shopkeeper has made a profit of num__15.0 then the cost price of the article is : <o> a ) num__487 <o> b ) num__277 <o> c ) num__222 <o> d ) num__297 <o> e ) num__111 |
explanation : num__120.0 of s . p . = num__616 s . p . = ( num__616 * num__100 ) / num__120 = rs . num__560 c . p = ( num__100 * num__560 ) / num__115 = rs . num__487 answer : a <eor> a <eos> |
a |
percent__100.0__487.0__ |
percent__100.0__487.0__ |
| what number is that which is to num__12 increased by three times the number as num__2 to num__9 ? <o> a ) num__6 <o> b ) num__7 <o> c ) num__8 <o> d ) num__9 <o> e ) num__10 |
let x = the number sought . then x / num__12 + x + num__12 = num__64 . and x - num__48.0 = num__48 . <eor> a <eos> |
a |
divide__12.0__2.0__ |
divide__12.0__2.0__ |
| one hour after yolanda started walking from x to y a distance of num__52 miles bob started walking along the same road from y to x . if yolanda ' s walking rate was num__3 miles per hour and bob т ' s was num__4 miles per hour how many miles had bob walked when they met ? <o> a ) num__28 <o> b ) num__23 <o> c ) num__22 <o> d ) num__21 <o> e ) num__19.5 |
when b started walking y already has covered num__3 miles out of num__52 hence the distance at that time between them was num__52 - num__3 = num__49 miles . combined rate of b and y was num__3 + num__4 = num__7 miles per hour hence they would meet each other in num__7.0 = num__7 hours . in num__6 hours b walked num__7 * num__4 = num__28 miles . answer : a . <eor> a <eos> |
a |
subtract__52.0__3.0__ add__3.0__4.0__ multiply__4.0__7.0__ round__28.0__ |
subtract__52.0__3.0__ add__3.0__4.0__ multiply__4.0__7.0__ multiply__4.0__7.0__ |
| amanda has had num__3 pay cuts in her salary in the past num__6 months . if the first pay cut was num__10.0 the second pay cut was num__10.0 and the third was num__10.0 . what will be the percentage decrease if the salary is decreased in a single shot ? <o> a ) num__27.1 <o> b ) num__29.1 <o> c ) num__21.1 <o> d ) num__23.1 <o> e ) num__25.1 % |
let rs . num__100 be initial salary . salary after num__1 st decrease num__10.0 = num__90 salary after num__2 nd decrease num__10.0 = num__81 i . e . reduced by num__19 percent of num__90 salary after num__3 rd decrease num__10.0 = num__72.9 i . e . reduced by num__10 percent of num__81 so if its decreased in single shot = i = ( ( b - a ) / b ) * num__100 = num__27.1 answer : a <eor> a <eos> |
a |
percent__81.0__90.0__ percent__100.0__27.1__ |
percent__81.0__90.0__ percent__100.0__27.1__ |
| the tax on a commodity is diminished by num__10.0 and its consumption increased by num__20.0 . the effect on revenue is ? <o> a ) num__2.0 decrease <o> b ) num__8.0 increase <o> c ) num__9.0 decrease <o> d ) num__3.0 decrease <o> e ) num__2.0 decrease |
num__100 * num__100 = num__10000 num__90 * num__120 = num__10800 - - - - - - - - - - - num__10000 - - - - - - - - - - - num__800 num__100 - - - - - - - - - - - ? = > num__8.0 increase answer : b <eor> b <eos> |
b |
subtract__100.0__10.0__ add__20.0__100.0__ multiply__120.0__90.0__ subtract__10800.0__10000.0__ divide__800.0__100.0__ divide__800.0__100.0__ |
subtract__100.0__10.0__ add__20.0__100.0__ multiply__120.0__90.0__ subtract__10800.0__10000.0__ divide__800.0__100.0__ divide__800.0__100.0__ |
| present ages of sameer and anand are in the ratio of num__5 : num__4 respectively . five years hence the ratio of their ages will become num__11 : num__9 respectively . what is anand ' s present age in years ? <o> a ) a ) num__24 <o> b ) b ) num__89 <o> c ) c ) num__40 <o> d ) d ) num__56 <o> e ) e ) num__45 |
let the present ages of sameer and anand be num__5 x years and num__4 x years respectively . then ( num__5 x + num__1.25 x + num__5 ) = num__1.22222222222 num__9 ( num__5 x + num__5 ) = num__11 ( num__4 x + num__5 ) num__45 x + num__45 = num__44 x + num__55 num__45 x - num__44 x = num__55 - num__45 x = num__10 . anand ' s present age = num__4 x = num__40 years . answer : c <eor> c <eos> |
c |
divide__5.0__4.0__ divide__11.0__9.0__ multiply__5.0__9.0__ multiply__4.0__11.0__ multiply__5.0__11.0__ subtract__55.0__45.0__ multiply__4.0__10.0__ multiply__4.0__10.0__ |
divide__5.0__4.0__ divide__11.0__9.0__ multiply__5.0__9.0__ multiply__4.0__11.0__ add__11.0__44.0__ subtract__55.0__45.0__ subtract__44.0__4.0__ subtract__44.0__4.0__ |
| tom and jerry are running on the same road towards each other . if tom is running at a speed of num__5 meters per second and jerry is running num__50.0 slower how much time will it take them to meet if the initial distance between the two is num__200 meters and tom started running num__40 seconds before jerry did ? <o> a ) num__40 sec <o> b ) num__35 sec <o> c ) num__1 mnt <o> d ) num__52 sec <o> e ) num__30 sec |
tom is running alone for num__40 seconds . so he will cover a distance of num__40 * num__5 = num__200 m in num__40 seconds . basically jerry has not started yet and tom has covered the distance alone and met jerry on the other side answer is a <eor> a <eos> |
a |
round__40.0__ |
round__40.0__ |
| if x dollars is invested at num__10 percent for one year and y dollars is invested at num__8 percent for one year the annual income from the num__10 percent investment will exceed the annual income from the num__8 percent investment by $ num__48 . if $ num__1500 is the total amount invested how much is invested at num__8 percent ? <o> a ) a . $ num__280 <o> b ) b . $ num__566.67 <o> c ) c . $ num__892 <o> d ) d . $ num__1108 <o> e ) e . $ num__1200 |
num__2 equations with num__2 unknowns num__10 x / num__100 - num__8 y / num__100 = num__48 and x + y = num__1500 solving these num__2 equations x = num__933.33 and y = num__566.67 answer b . <eor> b <eos> |
b |
subtract__10.0__8.0__ subtract__1500.0__933.33__ subtract__1500.0__933.33__ |
subtract__10.0__8.0__ subtract__1500.0__933.33__ subtract__1500.0__933.33__ |
| jack jill and sandy each have one try to make a basket from half court . if their individual probabilities of making the basket are num__0.166666666667 num__0.142857142857 and num__0.125 respectively what is the probability that jack and jill will make a basket but sandy will miss ? <o> a ) num__0.125 <o> b ) num__0.0625 <o> c ) num__0.0208333333333 <o> d ) num__0.0104166666667 <o> e ) num__0.00595238095238 |
the probability that jack and jill will make a basket but sandy will miss is num__0.166666666667 * num__0.142857142857 * num__0.875 = num__0.0208333333333 . the answer is c . <eor> c <eos> |
c |
multiply__0.1667__0.125__ multiply__0.1667__0.125__ |
multiply__0.1667__0.125__ multiply__0.1667__0.125__ |
| num__6000 − ( num__105 ÷ num__21.0 ) = ? <o> a ) num__6100 <o> b ) num__5100 <o> c ) num__550 <o> d ) num__598 <o> e ) num__5995 |
explanation : = num__6000 − ( num__5.0 ) = num__6000 − num__5 = num__5995 option e <eor> e <eos> |
e |
divide__105.0__21.0__ subtract__6000.0__5.0__ subtract__6000.0__5.0__ |
divide__105.0__21.0__ subtract__6000.0__5.0__ subtract__6000.0__5.0__ |
| if num__16 x = num__64 ^ y which of the following expresses x in terms of y ? <o> a ) num__4 ^ ( num__3 y - num__2 ) <o> b ) num__4 ^ ( y - num__1 ) <o> c ) num__4 ^ ( num__2 y - num__1 ) <o> d ) num__4 ^ ( num__2 y - num__3 ) <o> e ) num__4 ^ y |
by exponential simplification . num__16 = num__4 ^ num__2 and num__64 = num__4 ^ num__3 therefore ( num__4 ^ num__2 ) x = ( num__4 ^ num__3 ) ^ y gives x = ( num__4 ^ num__3 y ) / ( num__4 ^ num__2 ) further simplified to x = ( num__4 ^ num__3 y ) ( num__4 ^ - num__2 ) which gives x = num__4 ^ ( num__3 y - num__2 ) because exponential is additive in multiplication . i . e . a ^ b * a ^ c = a ^ ( b + c ) . answer : a <eor> a <eos> |
a |
divide__64.0__16.0__ divide__16.0__4.0__ |
divide__64.0__16.0__ divide__16.0__4.0__ |
| a num__270 m long train running at the speed of num__120 km / hr crosses another train running in opposite direction at the speed of num__80 km / hr in num__9 sec . what is the length of the other train ? <o> a ) num__230 <o> b ) num__240 <o> c ) num__250 <o> d ) num__260 <o> e ) num__279 |
relative speed = num__120 + num__80 = num__200 km / hr . = num__200 * num__0.277777777778 = num__55.5555555556 m / sec . let the length of the other train be x m . then ( x + num__270 ) / num__9 = num__55.5555555556 = > x = num__230 . answer : option a <eor> a <eos> |
a |
add__120.0__80.0__ round__230.0__ |
add__120.0__80.0__ round__230.0__ |
| integer x represents the product of all integers between num__1 to num__28 ( inclusive ) . the smallest prime factor of ( x + num__1 ) must be <o> a ) between num__1 to num__10 <o> b ) between num__11 to num__15 <o> c ) between num__15 to num__20 <o> d ) between num__20 to num__28 <o> e ) greater than num__28 |
answer = e = greater than num__28 this problem is asking smallest prime factor of ( num__28 ! + num__1 ) num__28 ! already have there prime factors num__23 num__57 num__1113 . . . . . . . . . . so on upto num__23 ( num__1 can not be considered prime factor ) just adding num__1 to num__28 ! will remove all the factors stated above ; so the smallest possible prime factor has to be greater than num__28 answer = e <eor> e <eos> |
e |
multiply__1.0__28.0__ |
multiply__1.0__28.0__ |
| num__45 men working num__8 hours per day dig num__30 m deep . how many extra men should be put to dig to a depth of num__50 m working num__6 hours per day ? <o> a ) num__74 <o> b ) num__55 <o> c ) num__44 <o> d ) num__75 <o> e ) num__72 |
b num__55 ( num__45 * num__8 ) / num__30 = ( x * num__6 ) / num__50 = > x = num__100 num__100 – num__45 = num__55 <eor> b <eos> |
b |
add__45.0__55.0__ round__55.0__ |
add__45.0__55.0__ round__55.0__ |
| john works at a resort from the beginning of march to the end of september . during the month of august this past year he made num__6 times the average ( arithmetic mean ) of his monthly totals in tips for the other months . his total tips for august were what fraction of his total tips for all of the months he worked ? <o> a ) num__0.125 <o> b ) num__0.142857142857 <o> c ) num__0.5 <o> d ) num__0.25 <o> e ) num__0.166666666667 |
first notice the number of months for which he worked - march to sept i . e . num__7 months avg of monthly totals in tips for months other than august = x tips in august = num__6 x total tips for all months = num__6 * x + num__6 x = num__12 x tips for august / total tips for all months = num__6 x / num__12 x = num__0.5 answer : c <eor> c <eos> |
c |
divide__6.0__12.0__ divide__6.0__12.0__ |
divide__6.0__12.0__ divide__6.0__12.0__ |
| when the airplane moves at an average speed of num__450 kmph it reaches its destination on time . when its average speed becomes num__400 kmph then it reaches its destination num__30 minutes late . find the length of journey . <o> a ) num__1250 km <o> b ) num__1500 km <o> c ) num__1100 km <o> d ) num__1800 km <o> e ) none |
sol . difference between timings = num__30 min = num__0.5 hr . let the length of journey be x km . then x / num__400 - x / num__450 = num__0.5 â ‡ ” num__9 x - num__8 x = num__1800 â ‡ ” x = num__1800 km . answer d <eor> d <eos> |
d |
round__1800.0__ |
round__1800.0__ |
| a num__250 m long train is running at a speed of num__55 km / hr . it crossed a platform of length num__520 m in ? <o> a ) num__30.4 sec <o> b ) num__40.4 sec <o> c ) num__60.4 sec <o> d ) num__50.4 sec <o> e ) num__80.4 sec |
d num__50.4 sec speed = num__55 km / hr ( to convert km / hr in to m / s ) = num__55 x num__0.277777777778 m / s distance = num__250 m + num__520 m ( if questions is about train crossing a post you need to consider only the length of train ) = num__770 m time = distance / speed = num__770 x num__18 / ( num__5 x num__55 ) = num__50.4 sec <eor> d <eos> |
d |
add__250.0__520.0__ round__50.4__ |
add__250.0__520.0__ round__50.4__ |
| annie and sam set out together on bicycles traveling at num__15 and num__12 km per hour respectively . after num__40 minutes annie stops to fix a flat tire . if it takes annie num__35 minutes to fix the flat tire and sam continues to ride during this time how many minutes will it take annie to catch up with sam assuming that annie resumes riding at num__15 km per hour ? <o> a ) num__20 <o> b ) num__40 <o> c ) num__60 <o> d ) num__80 <o> e ) num__100 |
annie gains num__3 km per hour ( or num__1 km every num__20 minutes ) on sam . after num__40 minutes annie is num__2 km ahead . sam rides num__1 km every num__5 minutes . in the next num__35 minutes sam rides num__7 km so sam will be num__5 km ahead . it will take annie num__100 minutes to catch sam . the answer is e . <eor> e <eos> |
e |
subtract__15.0__12.0__ subtract__35.0__15.0__ divide__40.0__20.0__ divide__15.0__3.0__ subtract__12.0__5.0__ multiply__5.0__20.0__ round__100.0__ |
subtract__15.0__12.0__ subtract__35.0__15.0__ divide__40.0__20.0__ divide__15.0__3.0__ subtract__12.0__5.0__ multiply__5.0__20.0__ round__100.0__ |
| a dishonest shopkeeper professes to sell pulses at the cost price but he uses a false weight of num__930 gm . for a kg . his gain is … % . <o> a ) num__5.26 <o> b ) num__5.36 <o> c ) num__4.26 <o> d ) num__6.26 <o> e ) num__7.53 % |
his percentage gain is num__100 * num__0.00752688172043 as he is gaining num__70 units for his purchase of num__930 units . so num__7.53 . answer : e <eor> e <eos> |
e |
percent__100.0__7.53__ |
percent__100.0__7.53__ |
| a watch was sold at a loss of num__36.0 . if it was sold for rs . num__140 more there would have been a gain of num__4.0 . what is the cost price ? <o> a ) num__350 <o> b ) num__288 <o> c ) num__799 <o> d ) num__778 <o> e ) num__901 |
num__64.0 num__104.0 - - - - - - - - num__40.0 - - - - num__140 num__100.0 - - - - ? = > rs . num__350 answer : a <eor> a <eos> |
a |
percent__100.0__350.0__ |
percent__100.0__350.0__ |
| three people took gmat practice tests in which the minimum and maximum possible scores are num__200 and num__800 respectively . they each took a test num__3 times and no one scored below num__510 or over num__780 . if the individual ranges of the three people ' s scores in those num__3 practice tests were num__20 num__90 and num__130 what is the difference between the maximum and minimum possible ranges of all their scores put together . <o> a ) num__70 <o> b ) num__80 <o> c ) num__90 <o> d ) num__140 <o> e ) num__170 |
according to the information in the question the maximum range would be between num__510 and num__780 ( num__270 ) . the minimum range would be the greatest range for any one individual which is listed in the problem as num__130 . so the difference between these would be num__270 - num__130 or num__140 . answer : d <eor> d <eos> |
d |
multiply__3.0__90.0__ subtract__270.0__130.0__ subtract__270.0__130.0__ |
subtract__780.0__510.0__ subtract__270.0__130.0__ subtract__270.0__130.0__ |
| the speed of the boat in still water in num__12 kmph . it can travel downstream through num__48 kms in num__3 hrs . in what time would it cover the same distance upstream ? <o> a ) num__8 hours <o> b ) num__6 hours <o> c ) num__4 hours <o> d ) num__5 hours <o> e ) num__6 hours |
still water = num__12 km / hr downstream = num__16.0 = num__16 km / hr upstream = > > still water = ( u + v / num__2 ) = > > num__12 = u + num__8.0 = num__8 km / hr so time taken in upstream = num__6.0 = num__6 hrs answer : e <eor> e <eos> |
e |
divide__48.0__3.0__ divide__16.0__2.0__ divide__12.0__2.0__ round__6.0__ |
divide__48.0__3.0__ divide__16.0__2.0__ divide__12.0__2.0__ divide__12.0__2.0__ |
| sonika deposited rs . num__8000 which amounted to rs . num__9200 after num__3 years at simple interest . had the interest been num__2.0 more . she would get how much ? <o> a ) num__9680 <o> b ) num__2288 <o> c ) num__7799 <o> d ) num__1777 <o> e ) num__2991 |
( num__8000 * num__3 * num__2 ) / num__100 = num__480 num__9200 - - - - - - - - num__9680 answer : a <eor> a <eos> |
a |
percent__100.0__9680.0__ |
percent__100.0__9680.0__ |
| fine the volume of cuboid num__15 m length num__9 m breadth and num__2 m height . <o> a ) num__1440 <o> b ) num__1540 <o> c ) num__1140 <o> d ) num__1220 <o> e ) num__1330 |
l × b × h = > num__15 × num__12 × num__8 = num__1440 m ( power num__3 ) answer is a <eor> a <eos> |
a |
subtract__15.0__12.0__ round__1440.0__ |
subtract__15.0__12.0__ round__1440.0__ |
| if all of the telephone extensions in a certain company must be even numbers and if each of the extensions uses all four of the digits num__1 num__2 num__4 and num__5 what is the greatest number of four - digit extensions that the company can have ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__12 <o> d ) num__16 <o> e ) num__24 |
since the phone number must be even the unit ' s digit can be either num__2 or num__4 . when the unit ' s digit is num__2 - - > number of possibilities is num__3 ! = num__6 when the unit ' s digit is num__4 - - > number of possibilities is num__3 ! = num__6 largest number of extensions = num__6 + num__6 = num__12 answer : c <eor> c <eos> |
c |
add__1.0__2.0__ add__1.0__5.0__ multiply__2.0__6.0__ multiply__1.0__12.0__ |
add__1.0__2.0__ add__1.0__5.0__ multiply__2.0__6.0__ multiply__1.0__12.0__ |
| if ( n + num__2 ) ! / n ! = num__156 n = ? <o> a ) num__0.0152671755725 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
( n + num__2 ) ! / n ! = num__156 rewrite as : [ ( n + num__2 ) ( n + num__1 ) ( n ) ( n - num__1 ) ( n - num__2 ) . . . . ( num__3 ) ( num__2 ) ( num__1 ) ] / [ ( n ) ( n - num__1 ) ( n - num__2 ) . . . . ( num__3 ) ( num__2 ) ( num__1 ) ] = num__132 cancel out terms : ( n + num__2 ) ( n + num__1 ) = num__156 from here we might just test the answer choices . since ( num__13 ) ( num__12 ) = num__156 we can see that n = num__11 d <eor> d <eos> |
d |
add__2.0__1.0__ divide__156.0__13.0__ divide__132.0__12.0__ multiply__1.0__11.0__ |
add__2.0__1.0__ divide__156.0__13.0__ divide__132.0__12.0__ divide__132.0__12.0__ |
| what is the length of a bridge ( in meters ) which a train num__156 meters long and travelling at num__45 km / h can cross in num__30 seconds ? <o> a ) num__203 <o> b ) num__211 <o> c ) num__219 <o> d ) num__227 <o> e ) num__235 |
speed = num__45 km / h = num__45000 m / num__3600 s = num__12.5 m / s in num__30 seconds the train can travel num__12.5 * num__30 = num__375 meters num__375 = length of train + length of bridge length of bridge = num__375 - num__156 = num__219 meters the answer is c . <eor> c <eos> |
c |
divide__45000.0__3600.0__ multiply__30.0__12.5__ subtract__375.0__156.0__ round__219.0__ |
divide__45000.0__3600.0__ multiply__30.0__12.5__ subtract__375.0__156.0__ subtract__375.0__156.0__ |
| there are num__18 balls in a jar . you take out num__3 blue balls without putting them back inside and now the probability of pulling out a blue ball is num__0.2 . how many blue balls were there in the beginning ? <o> a ) num__12 <o> b ) num__9 <o> c ) num__8 <o> d ) num__7 <o> e ) num__6 |
there are num__18 balls in a jar . you take out num__3 blue balls without putting num__3 blue balls + num__3.0 = num__6 answer : e <eor> e <eos> |
e |
die_space__ die_space__ |
die_space__ die_space__ |
| how long does a train num__110 metres long running at the speed of num__72 km / hr take to cross a bridge num__132 metres in length ? <o> a ) num__9.8 sec <o> b ) num__12.1 sec <o> c ) num__12.42 sec <o> d ) num__14.3 sec <o> e ) none |
solution speed = ( num__72 x num__0.277777777778 ) m / sec = num__20 m / sec total distance covered = ( num__110 + num__132 ) m = num__242 m required time = ( num__4.03333333333 ) sec = num__12.1 sec answer b <eor> b <eos> |
b |
add__110.0__132.0__ divide__242.0__20.0__ round__12.1__ |
add__110.0__132.0__ divide__242.0__20.0__ divide__242.0__20.0__ |
| solve : - num__222 x num__222 x num__222 + num__555 x num__555 x num__555 = ? ( num__222 x num__222 - num__222 x num__555 + num__555 x num__555 ) <o> a ) num__888 <o> b ) num__333 <o> c ) num__555 <o> d ) num__988 <o> e ) num__777 |
given exp . = ( a num__3 + b num__3 ) = ( a + b ) = ( num__222 + num__555 ) = num__777 ( a num__2 - ab + b num__2 ) answer e <eor> e <eos> |
e |
add__222.0__555.0__ add__222.0__555.0__ |
add__222.0__555.0__ add__222.0__555.0__ |
| rodrick mixes a martini that has a volume of ' n ' ounces having num__35.0 vermouth and num__60.0 gin by volume . he wants to change it so that the martini is num__25.0 vermouth by volume . how many ounces of gin must he add ? <o> a ) n / num__6 <o> b ) n / num__3 <o> c ) num__2 n / num__5 <o> d ) num__5 n / num__6 <o> e ) num__8 n / num__5 |
total v g num__1 ounce num__0.35 num__0.6 n ounce num__0.35 n num__0.6 n - - - - - - - - - - - - - initial expression lets say g ounces of gin is added to this mixture n + g num__0.35 n num__0.6 n + g - - - - - - - - - - - - - - final expression given that after adding g ounces of gin v should become num__25.0 of the total volume . = > volume of v / total volume = num__0.25 = > num__0.35 n / n + g = num__0.25 = > num__1.4 n = n + g = > g = num__2 n / num__5 answer is c . note that after we add pure gin the volume of vermouth will remain the same . based on this set the equation : num__0.35 n = num__0.25 ( n + g ) - - > g = num__2 n / num__5 answer : c . <eor> c <eos> |
c |
subtract__0.6__0.35__ divide__35.0__25.0__ add__0.6__1.4__ multiply__1.0__2.0__ |
subtract__0.6__0.35__ divide__35.0__25.0__ add__0.6__1.4__ divide__2.0__1.0__ |
| a boat moves upstream at the rate of num__1 km in num__20 minutes and down stream num__1 km in num__15 minutes . then the speed of the current is : <o> a ) num__1 kmph <o> b ) num__0.5 kmph <o> c ) num__3 kmph <o> d ) num__2.5 kmph <o> e ) num__3.5 kmph |
rate upstream = ( num__0.05 * num__60 ) = num__3 kmph rate down stream = num__0.0666666666667 * num__60 = num__4 kmph rate of the current = ½ ( num__4 - num__3 ) = num__0.5 kmph answer : b <eor> b <eos> |
b |
divide__1.0__20.0__ hour_to_min_conversion__ multiply__0.05__60.0__ divide__1.0__15.0__ add__1.0__3.0__ round__0.5__ |
divide__1.0__20.0__ hour_to_min_conversion__ multiply__0.05__60.0__ divide__1.0__15.0__ add__1.0__3.0__ subtract__1.0__0.5__ |
| a motorcyclist made num__3 roundtrips to the park until he ran out of gas back at home . te park is num__1500 meters away from his home . his speedometer was broke and he wanted to calculate his speed . it took him two and half hours to complete the journey . how fast was the motorcyclist traveling ? <o> a ) num__25 <o> b ) num__60 <o> c ) num__36 <o> d ) num__45 <o> e ) num__32 |
the distance from the park to the motorcyclist home was num__15000 meters . he made the trip num__3 times in total with each trip being num__30000 meters . multiply num__3 by num__30000 to get num__90000 meters or num__90 km in total . d = st num__90000 / num__2.5 = t which comes to num__36000 meters or num__36 km per hour correct answer = num__36 <eor> c <eos> |
c |
multiply__3.0__30000.0__ divide__90000.0__2.5__ divide__90.0__2.5__ round__36.0__ |
multiply__3.0__30000.0__ divide__90000.0__2.5__ divide__90.0__2.5__ divide__90.0__2.5__ |
| a table is bought for rs . num__900 / - and sold at rs . num__810 / - find the loss percentage <o> a ) num__5.0 <o> b ) num__10.0 <o> c ) num__15.0 <o> d ) num__20.0 <o> e ) num__22 % |
formula = ( selling price ~ cost price ) / cost price * num__100 = ( num__810 ~ num__900 ) / num__900 = num__10.0 loss b <eor> b <eos> |
b |
divide__100.0__10.0__ |
divide__100.0__10.0__ |
| the average of num__5 numbers is num__6.8 . if one of the numbers is multiplied by a factor of num__3 the average of the numbers increases to num__12.8 . what number is multiplied by num__3 ? <o> a ) num__15.0 <o> b ) num__13.0 <o> c ) num__13.9 <o> d ) num__10.0 <o> e ) num__6.0 |
the average of num__5 numbers is num__6.8 the sum of num__5 numbers will be num__6.8 x num__5 = num__34 the average of num__5 number after one of the number is multiplied by num__3 is num__12.8 the sum of the numbers will now be num__12.8 x num__5 = num__64 so the sum has increased by num__64 - num__34 = num__30 let the number multiplied by num__3 be n then num__3 n = n + num__30 or num__2 n = num__30 or n = num__15 answer : - a <eor> a <eos> |
a |
multiply__5.0__6.8__ multiply__5.0__12.8__ subtract__64.0__34.0__ subtract__5.0__3.0__ multiply__5.0__3.0__ multiply__5.0__3.0__ |
multiply__5.0__6.8__ multiply__5.0__12.8__ subtract__64.0__34.0__ subtract__5.0__3.0__ multiply__5.0__3.0__ subtract__30.0__15.0__ |
| two trains are moving in the same direction at num__108 kmph and num__36 kmph . the faster train crosses the slower train in num__17 seconds . find the length of the faster train in meters . <o> a ) num__270 m <o> b ) num__340 m <o> c ) num__310 m <o> d ) num__350 m <o> e ) num__327 m |
relative speed = ( num__108 - num__36 ) * num__0.277777777778 = num__4 * num__5 = num__20 mps . distance covered in num__17 sec = num__17 * num__20 = num__340 m . the length of the faster train = num__340 m . answer : b <eor> b <eos> |
b |
multiply__5.0__4.0__ multiply__17.0__20.0__ round__340.0__ |
multiply__5.0__4.0__ multiply__17.0__20.0__ multiply__17.0__20.0__ |
| a train of num__25 carriages each of num__60 meters length when an engine also of num__60 meters length is running at a speed of num__60 kmph . in what time will the train cross a bridge num__1.0 km long ? <o> a ) num__2 mins num__14 sec <o> b ) num__2 mins num__34 sec <o> c ) num__150 sec <o> d ) num__164 sec <o> e ) num__2 mins num__44 sec |
d = num__25 * num__60 + num__1000 = num__2500 m t = num__41.6666666667 * num__3.6 = num__150 sec answer : c <eor> c <eos> |
c |
divide__2500.0__60.0__ round__150.0__ |
divide__2500.0__60.0__ multiply__1.0__150.0__ |
| two cyclist start from the same places in opposite directions . one is going towards north at num__10 kmph and the other is going towards south num__15 kmph . what time will they take to be num__50 km apart ? <o> a ) num__1 hr <o> b ) num__2 hrs <o> c ) num__3 hrs <o> d ) num__5 hrs <o> e ) num__6 hrs |
to be ( num__10 + num__15 ) km apart they take num__1 hour to be num__50 km apart they take num__0.04 * num__50 = num__2 hrs answer is b <eor> b <eos> |
b |
multiply__50.0__0.04__ round__2.0__ |
multiply__50.0__0.04__ multiply__50.0__0.04__ |
| square abcd is the base of the cube while square efgh is the cube ' s top face such that point e is above point a point f is above point b etc . what is the distance between the midpoint of edge ab and the midpoint of edge eh if the area of square abcd is num__2 ? <o> a ) num__1 / sqrt num__2 <o> b ) num__1 <o> c ) sqrt num__2 <o> d ) sqrt num__3 <o> e ) num__2 sqrt num__3 |
distance from mid point of ab to ad = sqrt [ ( num__1 / sqrt num__2 ) ^ num__2 + ( num__1 / sqrt num__2 ) ^ num__2 ] = num__1 the distance between the midpoint of edge ab and the midpoint of edge eh = sqrt [ num__1 ^ num__2 + ( sqrt num__2 ) ^ num__2 ] = sqrt num__3 . answer : d <eor> d <eos> |
d |
triangle_area__2.0__3.0__ |
power__3.0__1.0__ |
| the surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are num__10 cm each . the radius of the sphere is <o> a ) num__3 cm <o> b ) num__5 cm <o> c ) num__6 cm <o> d ) num__8 cm <o> e ) none |
solution num__4 Î r num__2 = num__2 Î num__5 x num__10 â ‡ ’ r num__2 = ( num__5 x num__5.0 ) â ‡ ’ num__25 â ‡ ’ r = num__5 cm . answer b <eor> b <eos> |
b |
triangle_area__10.0__5.0__ triangle_area__2.0__5.0__ |
triangle_area__10.0__5.0__ triangle_area__2.0__5.0__ |
| a train passes a man standing on the platform . if the train is num__170 meters long and its speed is num__72 kmph how much time it took in doing so ? <o> a ) num__6 ½ sec <o> b ) num__4 ½ sec <o> c ) num__9 ½ sec <o> d ) num__8 ½ sec <o> e ) num__2 ½ sec |
d = num__170 s = num__72 * num__0.277777777778 = num__20 mps t = num__8.5 = num__8 ½ sec answer : d <eor> d <eos> |
d |
divide__170.0__20.0__ round__8.0__ |
divide__170.0__20.0__ round__8.0__ |
| how much time will a train of length num__300 m moving at a speed of num__72 kmph take to cross another train of length num__500 m moving at num__36 kmph in the same direction ? <o> a ) num__50 <o> b ) num__80 <o> c ) num__88 <o> d ) num__76 <o> e ) num__12 |
the distance to be covered = sum of their lengths = num__300 + num__500 = num__800 m . relative speed = num__72 - num__36 = num__36 kmph = num__36 * num__0.277777777778 = num__10 mps . time required = d / s = num__80.0 = num__80 sec . answer : b <eor> b <eos> |
b |
add__300.0__500.0__ divide__800.0__10.0__ round__80.0__ |
add__300.0__500.0__ divide__800.0__10.0__ divide__800.0__10.0__ |
| in a recent election james received num__0.5 percent of the num__2000 votes cast . to win the election a candidate needed to receive more than num__50 percent of the vote . how many additional votes would james have needed to win the election ? <o> a ) num__901 <o> b ) num__989 <o> c ) num__990 <o> d ) num__991 <o> e ) num__1 |
001 |
james = ( num__0.5 / num__100 ) * num__2000 = num__10 votes to win = ( num__0.5 ) * total votes + num__1 = ( num__0.5 ) * num__2000 + num__1 = num__1001 remaining voted needed to win election = num__1001 - num__10 = num__991 answer : option d <eor> d <eos> |
d |
d |
| if num__2 x - y = num__4 then num__6 x - num__3 y = ? <o> a ) num__15 <o> b ) num__12 <o> c ) num__18 <o> d ) num__10 <o> e ) num__14 |
num__6 x - num__3 y = num__3 * ( num__2 x - y ) = num__3 * num__4 = num__12 answer : b <eor> b <eos> |
b |
multiply__2.0__6.0__ multiply__2.0__6.0__ |
multiply__2.0__6.0__ multiply__2.0__6.0__ |
| an assembly line produces num__20 cogs per hour until an initial order of num__60 cogs is completed . the speed of the assembly line is then immediately increased so that it can produce num__60 cogs per hour until another num__60 cogs are produced . what is the overall average output in cogs per hour for the assembly line during this whole time ? <o> a ) num__25 <o> b ) num__30 <o> c ) num__35 <o> d ) num__40 <o> e ) num__45 |
the time to produce the first num__60 cogs is num__3.0 = num__3 hours . the time to produce the next num__60 cogs is num__1.0 = num__1 hour . the average output is num__120 cogs / num__4 hours = num__30 cogs per hour . the answer is b . <eor> b <eos> |
b |
divide__60.0__20.0__ add__1.0__3.0__ divide__120.0__4.0__ subtract__60.0__30.0__ |
divide__60.0__20.0__ add__1.0__3.0__ divide__120.0__4.0__ divide__120.0__4.0__ |
| a dual sink has a right side and left side . working alone the right side can fill up in num__5 hours . working alone the left side can empty in num__15 hours . if it is desired that the sink should be filled from empty exactly num__6 hours after the right side is turned on at num__9 : num__30 am then at what time should the left side also be turned on ? <o> a ) num__10 : num__00 am <o> b ) num__10 : num__45 am <o> c ) num__12 : num__00 pm <o> d ) num__12 : num__30 pm <o> e ) num__1 : num__30 pm |
in num__6 hours the right side will fill up with water equal to num__6 * num__0.2 = num__1.2 of the sink . so num__0.2 more than necessary which should be compensated by the left side of the sink . to pump out num__0.2 of the water the left side needs num__3.0 = num__3 hours . therefore the left side of the sink should be turned on at num__9 : num__30 am + num__6 hours - num__3 hours = num__12 : num__30 pm . answer : d . <eor> d <eos> |
d |
divide__6.0__30.0__ divide__6.0__5.0__ divide__15.0__5.0__ subtract__15.0__3.0__ round__12.0__ |
divide__6.0__30.0__ multiply__6.0__0.2__ multiply__15.0__0.2__ subtract__15.0__3.0__ subtract__15.0__3.0__ |
| some persons can do a piece of work in num__8 days . two times the number of these people will do half of that work in ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__8 |
num__8 / ( num__2 * num__2 ) = num__2 days answer : a <eor> a <eos> |
a |
round__2.0__ |
round__2.0__ |
| after working for num__6 days ashok finds that only num__0.333333333333 rd of the work has been done . he employs ravi who is num__60.0 as efficient as ashok . how many days more would ravi take to complete the work ? <o> a ) num__27 <o> b ) num__28 <o> c ) num__20 <o> d ) num__24 <o> e ) num__12 |
num__0.333333333333 - - - - num__6 num__1 - - - - - - - ? a = num__18 r = num__0.0555555555556 * num__0.6 = num__0.0333333333333 num__1 - - - - - num__0.0333333333333 num__0.666666666667 - - - - ? = > num__20 days answer : c <eor> c <eos> |
c |
divide__1.0__18.0__ km_to_mile_conversion__ divide__0.6__18.0__ subtract__1.0__0.3333__ round__20.0__ |
divide__1.0__18.0__ km_to_mile_conversion__ divide__0.6__18.0__ subtract__1.0__0.3333__ round__20.0__ |
| at what rate p . a a sum of rs . num__2400 will become rs . num__3600 / - in num__5 years . <o> a ) num__10.0 <o> b ) num__15.0 <o> c ) num__20.0 <o> d ) num__12.0 <o> e ) num__22 % |
explanation : principal p = num__2400 / - rate of interest r = ? time t = num__5 years accumulated amount a = rs . num__10400 / - a = p + i a = p + ptr / num__100 a = p ( num__1 + tr / num__100 ) num__2400 [ num__1 + ( num__5 x r ) / num__100 ] = num__3600 num__2 [ ( num__20 + r ) / num__20 ] = num__3 num__20 + r = num__30 r = num__10.0 answer : option a <eor> a <eos> |
a |
divide__100.0__5.0__ subtract__5.0__2.0__ multiply__5.0__2.0__ multiply__5.0__2.0__ |
divide__100.0__5.0__ subtract__5.0__2.0__ divide__20.0__2.0__ divide__100.0__10.0__ |
| two - third of a positive number and num__0.0740740740741 of its reciprocal are equal . find the positive number . <o> a ) num__0.333333333333 <o> b ) num__0.235294117647 <o> c ) num__0.266666666667 <o> d ) num__0.363636363636 <o> e ) num__1.33333333333 |
explanation : let the positive number be x . then num__0.666666666667 x = num__0.0740740740741 * num__1 / x x num__2 = num__0.0740740740741 * num__1.5 = num__0.111111111111 x = √ num__0.111111111111 = num__0.333333333333 . answer : a <eor> a <eos> |
a |
multiply__0.0741__1.5__ subtract__1.0__0.6667__ subtract__1.0__0.6667__ |
multiply__0.0741__1.5__ divide__0.6667__2.0__ divide__0.1111__0.3333__ |
| a b and c completed a piece of work a worked for num__16 days b for num__9 days and c for num__4 days . their daily wages were in the ratio of num__3 : num__4 : num__5 . find the daily wages of c if their total earning was rs . num__1480 ? <o> a ) rs . num__80 <o> b ) rs . num__120 <o> c ) rs . num__71.15 <o> d ) rs . num__51.15 <o> e ) rs . num__180 |
num__3 x num__4 x num__5 x num__16 num__9 num__4 num__48 x + num__36 x + num__20 x = num__1480 num__104 x = num__1480 = > x = num__14.23 num__5 x = num__71.15 rs . answer : c <eor> c <eos> |
c |
multiply__16.0__3.0__ multiply__9.0__4.0__ add__16.0__4.0__ multiply__5.0__14.23__ multiply__5.0__14.23__ |
multiply__16.0__3.0__ multiply__9.0__4.0__ add__16.0__4.0__ multiply__5.0__14.23__ multiply__5.0__14.23__ |
| in the rectangular coordinate system point o has coordinates ( num__00 ) and point b has coordinates ( num__55 ) and if point a is equidistant from points o and b and the area of the triangle oab is num__16 which of the following are the possible coordinates of point a . <o> a ) ( - num__26 ) <o> b ) ( num__04 ) <o> c ) ( num__2 - num__6 ) <o> d ) ( num__26 ) <o> e ) ( num__40 ) |
area . pngsince a is equidistant from points o and b then it must be somewhere on the green line ( perpendicular bisector of ob ) . ( num__2 - num__6 ) and ( num__26 ) are not on that line . if a is at ( num__05 ) or ( num__50 ) then the area is num__0.5 * num__5 * num__5 = num__12.5 . thus a must be at ( num__26 ) . answer : d . <eor> d <eos> |
d |
triangle_area__0.5__50.0__ rectangle_perimeter__0.5__12.5__ |
triangle_area__0.5__50.0__ volume_rectangular_prism__0.5__2.0__26.0__ |
| there are some pigeons and hares in a zoo . if heads are counted there are num__200 . if legs are counted there are num__580 . the number of hares in the zoo is ? <o> a ) num__87 <o> b ) num__27 <o> c ) num__90 <o> d ) num__26 <o> e ) num__29 |
num__200 * num__2 = num__400 num__580 - - - - - num__180 num__1 - - - - num__2 ? - - - - num__180 = num__90 answer : c <eor> c <eos> |
c |
multiply__200.0__2.0__ subtract__580.0__400.0__ divide__180.0__2.0__ multiply__1.0__90.0__ |
multiply__200.0__2.0__ subtract__580.0__400.0__ divide__180.0__2.0__ multiply__1.0__90.0__ |
| john has $ num__1600 at the beginning of his trip after spending money he still has exactly $ num__800 less than he spent on the trip . how much money does john still have ? <o> a ) $ num__200 <o> b ) $ num__400 <o> c ) $ num__600 <o> d ) $ num__800 <o> e ) $ num__1 |
200 |
let the money spent be x money he is left with after spending = x - num__800 total money - - > x + ( x - num__800 ) = num__1600 solving for x will give x = num__1200 therefore the money he is left with = x - num__800 = num__1200 - num__800 = num__400 answer : b <eor> b <eos> |
b |
b |
| the present ratio of students to teachers at a certain school is num__35 to num__1 . if the student enrollment were to increase by num__50 students and the number of teachers were to increase by num__5 the ratio of students to teachers would then be num__25 to num__1 . what is the present number of teachers ? <o> a ) num__5 <o> b ) num__8 <o> c ) num__10 <o> d ) num__12 <o> e ) num__15 |
we are given that the ratio of students to teacher is num__35 to num__1 . we can rewrite this using variable multipliers . students : teachers = num__35 x : x we are next given that student enrollment increases by num__50 and the number of teachers increases by num__5 . with this change the new ratio becomes num__25 to num__1 . we can put all this into an equation : students / teachers num__25.0 = ( num__35 x + num__50 ) / ( x + num__5 ) if we cross multiply we have : num__25 ( x + num__5 ) = num__35 x + num__50 num__25 x + num__125 = num__35 x + num__50 num__7.5 = x x ~ num__8 since x is the present number of teachers currently there are num__8 teachers . answer b . <eor> b <eos> |
b |
multiply__5.0__25.0__ multiply__1.0__8.0__ |
multiply__5.0__25.0__ divide__8.0__1.0__ |
| anil can do a work in num__15 days while sunil can do it in num__25 days . how long will they take if both work together ? <o> a ) num__12 days <o> b ) num__5 days <o> c ) num__9 num__0.375 days <o> d ) num__8 days <o> e ) num__6 days |
explanation : num__0.0666666666667 + num__0.04 = num__0.106666666667 num__9.375 = num__9 num__0.375 days answer : c <eor> c <eos> |
c |
add__0.0667__0.04__ divide__9.375__25.0__ round__9.0__ |
add__0.0667__0.04__ divide__9.375__25.0__ round__9.0__ |
| tom working alone can paint a room in num__6 hours . peter and john working independently can paint the same room in num__3 hours and num__6 hours respectively . tom starts painting the room and works on his own for one hour . he is then joined by peter and they work together for an hour . finally john joins them and the three of them work together to finish the room each one working at his respective rate . what fraction of the whole job was done by peter ? <o> a ) num__0.5 <o> b ) num__0.333333333333 <o> c ) num__0.25 <o> d ) num__0.666666666667 <o> e ) num__0.4 |
tom paints num__0.166666666667 of the room in the first hour . tom and peter paint num__0.166666666667 + num__0.333333333333 = num__0.5 of the room in the next hour for a total of num__0.666666666667 . the three people then paint the remaining num__0.333333333333 in a time of ( num__0.333333333333 ) / ( num__0.666666666667 ) = num__0.5 hours peter worked for num__1.5 hours so he painted num__1.5 * num__0.333333333333 = num__0.5 of the room . the answer is a . <eor> a <eos> |
a |
divide__3.0__6.0__ add__0.1667__0.5__ multiply__3.0__0.5__ round__0.5__ |
divide__3.0__6.0__ add__0.1667__0.5__ multiply__3.0__0.5__ divide__3.0__6.0__ |
| two trains of equal lengths take num__10 sec and num__15 sec respectively to cross a telegraph post . if the length of each train be num__120 m in what time will they cross other travelling in opposite direction ? <o> a ) num__76 sec <o> b ) num__12 sec <o> c ) num__15 sec <o> d ) num__14 sec <o> e ) num__13 sec |
speed of the first train = num__12.0 = num__12 m / sec . speed of the second train = num__24.0 = num__8 m / sec . relative speed = num__12 + num__8 = num__20 m / sec . required time = ( num__120 + num__120 ) / num__20 = num__12 sec . answer : b <eor> b <eos> |
b |
divide__120.0__10.0__ divide__120.0__15.0__ add__8.0__12.0__ round__12.0__ |
divide__120.0__10.0__ divide__120.0__15.0__ add__8.0__12.0__ divide__120.0__10.0__ |
| a train passes a platform in num__18 seconds . the same train passes a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr the length of the platform is <o> a ) num__120 <o> b ) num__250 <o> c ) num__260 <o> d ) num__230 <o> e ) num__220 |
speed of the train = num__54 km / hr = ( num__54 × num__10 ) / num__36 m / s = num__15 m / s length of the train = speed × time taken to cross the man = num__15 × num__20 = num__300 m let the length of the platform = l time taken to cross the platform = ( num__300 + l ) / num__15 = > ( num__300 + l ) / num__15 = num__18 = > num__300 + l = num__15 × num__18 = num__270 = > l = num__390 - num__270 = num__120 meter answer is a . <eor> a <eos> |
a |
subtract__54.0__18.0__ multiply__20.0__15.0__ multiply__18.0__15.0__ subtract__390.0__270.0__ round__120.0__ |
subtract__54.0__18.0__ multiply__20.0__15.0__ multiply__18.0__15.0__ subtract__390.0__270.0__ subtract__390.0__270.0__ |
| when m divided by num__288 the remainder is num__47 . find the remainder when the same m is divided by num__24 ? <o> a ) num__20 <o> b ) num__21 <o> c ) num__22 <o> d ) num__23 <o> e ) num__26 |
num__23 option d <eor> d <eos> |
d |
subtract__47.0__24.0__ round__23.0__ |
subtract__47.0__24.0__ round__23.0__ |
| two good train each num__650 m long are running in opposite directions on parallel tracks . their speeds are num__45 km / hr and num__30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one . <o> a ) num__12.4 sec <o> b ) num__24.3 sec <o> c ) num__62.4 sec <o> d ) num__60.1 sec <o> e ) none |
sol . relative speed = ( num__45 + num__30 ) km / hr = ( num__75 x num__0.277777777778 ) m / sec = ( num__20.8333333333 ) m / sec . distance covered = ( num__650 + num__650 ) m = num__1300 m . required time = ( num__1300 x num__0.048 ) sec = num__62.4 sec . answer c <eor> c <eos> |
c |
add__45.0__30.0__ multiply__0.048__1300.0__ round__62.4__ |
add__45.0__30.0__ multiply__0.048__1300.0__ round__62.4__ |
| num__19 times a number gives num__342 . the number is <o> a ) num__18 <o> b ) num__19 <o> c ) num__16 <o> d ) none <o> e ) can not be determined |
explanation : let the number be ' n ' num__19 × n = num__342 ⇒ n = num__18 correct option : a <eor> a <eos> |
a |
divide__342.0__19.0__ divide__342.0__19.0__ |
divide__342.0__19.0__ divide__342.0__19.0__ |
| a can do a piece of work in num__10 days and b can do the same work in num__12 days . a and b worked together for num__5 days . how many more days are required to complete the remaining work if they work together ? <o> a ) num__3 ( num__0.277777777778 ) <o> b ) num__3 ( num__0.3125 ) <o> c ) num__3 ( num__0.454545454545 ) <o> d ) num__3 ( num__0.384615384615 ) <o> e ) num__0.454545454545 |
a can do num__0.1 of the work in a day . b can do num__0.0833333333333 of the work in a num__1 day . both of them together can do ( num__0.1 + num__0.0833333333333 ) part of work in num__1 day = ( num__6 + num__5 ) / num__60 = num__0.183333333333 they take num__5.45454545455 days to complete the work together . given that they already worked for num__5 days . the number of days required to complete remaining work = > num__5.45454545455 - num__5 = num__0.454545454545 = num__0.454545454545 days . answer : e <eor> e <eos> |
e |
multiply__10.0__0.1__ add__5.0__1.0__ hour_to_min_conversion__ add__0.1__0.0833__ divide__5.4545__12.0__ multiply__1.0__0.4545__ |
multiply__10.0__0.1__ add__5.0__1.0__ divide__6.0__0.1__ add__0.1__0.0833__ divide__5.4545__12.0__ divide__5.4545__12.0__ |
| if - num__1 < x < num__0 what is the median n of these six numbers listed below num__1 / x ^ num__31 / x ^ num__2 num__1 / x x x ^ num__2 x ^ num__3 <o> a ) num__1 / x <o> b ) x ^ num__2 <o> c ) x ^ num__2 ( x + num__1 ) / num__2 <o> d ) n = x ( x ^ num__2 + num__1 ) / num__2 <o> e ) x ^ num__1.0 x |
my approach : put any value between - num__1 and num__0 . you will get the answer . lets say num__0.5 . so the list will be as follows : - num__8 - num__2 - num__0.5 - num__0.125 and last two numbers will be positive so need to see that . median will be between - num__0.5 and - num__0.125 . so ( x ^ num__3 + x ) / num__2 ( d ) . <eor> d <eos> |
d |
reverse__2.0__ reverse__8.0__ reverse__0.5__ |
reverse__2.0__ reverse__8.0__ reverse__0.5__ |
| the function f is defined for all positive integers n by the following rule : f ( n ) is the number of position integer each of which is less than n and has no position factor in common with n other than num__3 . if p is any prime number then f ( p ) = <o> a ) p - num__1 <o> b ) p - num__2 <o> c ) ( p + num__1 ) / num__2 <o> d ) ( p - num__1 ) / num__2 <o> e ) num__2 |
the confusing moment in this question is its wording . basically question is : how many positive integers are less than given prime number p which has no common factor with p except num__1 . well as p is a prime all positive numbers less than p have no common factors with p ( except common factor num__1 ) . so there would be p - num__1 such numbers ( as we are looking number of integers less than p ) . if we consider p = num__7 how many numbers are less than num__7 having no common factors with num__7 : num__1 num__2 num__3 num__4 num__5 num__6 - - > num__7 - num__1 = num__6 . answer : d . <eor> d <eos> |
d |
subtract__3.0__1.0__ add__3.0__1.0__ add__3.0__2.0__ multiply__3.0__2.0__ reverse__1.0__ |
subtract__3.0__1.0__ subtract__7.0__3.0__ subtract__7.0__2.0__ subtract__7.0__1.0__ subtract__3.0__2.0__ |
| a car gets num__40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel num__210 kilometers ? <o> a ) num__8.25 gallons <o> b ) num__7.5 gallons <o> c ) num__6.55 gallons <o> d ) num__5.25 gallons <o> e ) num__4.5 gallons |
each num__40 kilometers num__1 gallon is needed . we need to know how many num__40 kilometers are there in num__210 kilometers ? num__5.25 = num__5.25 * num__1 gallon = num__5.25 gallons correct answer d <eor> d <eos> |
d |
divide__210.0__40.0__ round__5.25__ |
divide__210.0__40.0__ round__5.25__ |
| carolyn bought num__16 gumballs lew bought num__12 gumballs and bob bought x gumballs . the average ( arithmetic mean ) number of gumballs the three bought is between num__19 and num__25 inclusive . what is the difference between the greatest number and the smallest number of gumballs carey could have bought ? <o> a ) num__20 <o> b ) num__22 <o> c ) num__24 <o> d ) num__26 <o> e ) num__18 |
smallest gumballs = ( num__19 - num__16 ) + ( num__19 - num__12 ) + num__19 = num__29 largest gumballs = ( num__25 - num__16 ) + ( num__25 - num__12 ) + num__25 = num__47 difference = num__47 - num__29 = num__18 e <eor> e <eos> |
e |
subtract__47.0__29.0__ subtract__47.0__29.0__ |
subtract__47.0__29.0__ subtract__47.0__29.0__ |
| exactly num__28.0 of the reporters for a certain wire service cover local politics in country x . if num__30.0 of the reporters who cover politics for the wire service do not cover local politics in country x what percent of the reporters for the wire service do not cover politics ? <o> a ) num__20.0 <o> b ) num__42.0 <o> c ) num__44.0 <o> d ) num__60.0 <o> e ) num__84 % |
let ' s assume there are num__100 reporters - - > num__28 reporters cover local politics . now as num__30.0 of the reporters who cover all politics do not cover local politics then the rest num__70.0 of the reporters who cover politics do cover local politics so if there are x reporters who cover politics then num__70.0 of them equal to num__28 ( # of reporters who cover local politics ) : num__0.7 x = num__28 - - > x = num__40 hence num__40 reporters cover politics and the rest num__100 - num__40 = num__60 reporters do not cover politics at all . answer : d . <eor> d <eos> |
d |
percent__100.0__60.0__ |
percent__100.0__60.0__ |
| the no . of gift pack bought by dexter is num__1 more than the price in rupees of each gift pack . the amount of rs . num__380 which dexter had fell short of the required amount . what is theamount by which he might have fallen short . <o> a ) num__20 <o> b ) num__40 <o> c ) num__50 <o> d ) num__60 <o> e ) num__90 |
let the price of gift pack be ' aa ' then number of packs bought = a + num__1 = a + num__1 hence total cost is a ( a + num__1 ) a ( a + num__1 ) it is given that num__380 < a ( a + num__1 ) num__380 < a ( a + num__1 ) if a = num__19 a = num__19 the total cost = num__19 × num__20 = num__380 = num__19 × num__20 = num__380 dexter would not have fallen short if : a = num__20 a = num__20 a ( a + num__1 ) = num__420 a ( a + num__1 ) = num__420 so he would have fallen short by rs num__40 . this is the minimum amount by which he may have fallen short . b <eor> b <eos> |
b |
add__1.0__19.0__ subtract__420.0__380.0__ multiply__1.0__40.0__ |
add__1.0__19.0__ subtract__420.0__380.0__ multiply__1.0__40.0__ |
| if x and y are integers such that ( x + num__1 ) ^ num__2 is less than or equal to num__9 and ( y - num__1 ) ^ num__2 is less than num__64 what is the sum of the maximum possible value of xy and the minimum possible value of xy ? <o> a ) - num__16 <o> b ) - num__8 <o> c ) num__0 <o> d ) num__14 <o> e ) num__16 |
( x + num__1 ) ^ num__2 < = num__9 x < = num__2 x > = - num__4 ( y - num__1 ) ^ num__2 < num__64 y < num__9 y > - num__7 max possible value of xy is - num__4 × - num__6 = num__24 minimum possible value of xy is - num__4 × num__8 = - num__32 - num__32 + num__24 = - num__8 answer : b <eor> b <eos> |
b |
subtract__9.0__2.0__ add__2.0__4.0__ multiply__4.0__6.0__ add__1.0__7.0__ divide__64.0__2.0__ add__1.0__7.0__ |
subtract__9.0__2.0__ add__2.0__4.0__ multiply__4.0__6.0__ add__1.0__7.0__ add__8.0__24.0__ add__1.0__7.0__ |
| suppose you have access to a large vat of distilled water several gallons large . you have two precise measuring pipettes one to measure exactly num__0.2 of an ounce and one to measure exactly num__0.166666666667 of an ounce . you can pour precisely measured amounts into a beaker which initially is empty . you can use either pipette to remove distilled water from the vat or from the beaker and use either pipette to dispense water into either of those receptacles but you can not use either pipette to take any quantity of distilled water other than the amount for which it is designed . which of the following represents in ounces a precise amount of distilled water you can transfer from the vat to the beaker ? i . num__0.166666666667 ii . num__0.142857142857 iii . num__0.0833333333333 <o> a ) i only <o> b ) iii only <o> c ) num__0.0333333333333 and num__0.366666666667 <o> d ) ii and iii only <o> e ) i ii and iii |
num__0.2 - num__0.166666666667 = num__0.0333333333333 num__0.2 + num__0.166666666667 = num__0.366666666667 answer : c <eor> c <eos> |
c |
subtract__0.2__0.1667__ add__0.2__0.1667__ subtract__0.2__0.1667__ |
subtract__0.2__0.1667__ add__0.2__0.1667__ subtract__0.2__0.1667__ |
| the total marks obtained by a student in mathematics and physics is num__60 and his score in chemistry is num__20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together . <o> a ) num__40 <o> b ) num__99 <o> c ) num__77 <o> d ) num__66 <o> e ) num__12 |
let the marks obtained by the student in mathematics physics and chemistry be m p and c respectively . given m + c = num__60 and c - p = num__20 m + c / num__2 = [ ( m + p ) + ( c - p ) ] / num__2 = ( num__60 + num__20 ) / num__2 = num__40 . answer : a <eor> a <eos> |
a |
subtract__60.0__20.0__ subtract__60.0__20.0__ |
subtract__60.0__20.0__ subtract__60.0__20.0__ |
| a van takes num__5 hours to cover a distance of num__435 km . what speed in kph should the van maintain to cover the same distance in num__1.5 of the previous time ? <o> a ) num__50 <o> b ) num__52 <o> c ) num__54 <o> d ) num__56 <o> e ) num__58 |
( num__1.5 ) * num__5 = num__7.5 hours num__435 / num__7.5 = num__58 kph the answer is e . <eor> e <eos> |
e |
multiply__5.0__1.5__ divide__435.0__7.5__ round__58.0__ |
multiply__5.0__1.5__ divide__435.0__7.5__ divide__435.0__7.5__ |
| what is the cp of rs num__100 stock at num__8 discount with num__0.2 % brokerage ? <o> a ) num__99.6 <o> b ) num__96.2 <o> c ) num__97.5 <o> d ) num__92.2 <o> e ) none of these |
explanation : use the formula cp = num__100 â € “ discount + brokerage % cp = num__100 - num__8 + num__0.2 num__92.2 thus the cp is rs num__92.2 . answer d <eor> d <eos> |
d |
percent__100.0__92.2__ |
percent__100.0__92.2__ |
| a tv is purchased at rs . num__5000 and sold at rs . num__4000 find the lost percent . <o> a ) num__10.0 <o> b ) num__20.0 <o> c ) num__25.0 <o> d ) num__28.0 <o> e ) num__30 % |
explanation : we know c . p . = num__5000 s . p . = num__4000 loss = num__5000 - num__4000 = num__1000 loss % = ( losscost ∗ num__100 ) % = ( num__10005000 ∗ num__100 ) % = num__20.0 answer is b <eor> b <eos> |
b |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| a glass was filled with num__10 ounces of water and num__0.03 ounce of the water evaporated each day during a num__20 - day period . what percent of the original amount of water evaporated during this period ? <o> a ) num__0.006 <o> b ) num__0.06 <o> c ) num__0.6 <o> d ) num__6.0 <o> e ) num__60 % |
we are given that num__0.03 ounces of water evaporated each day . furthermore we know that this process happened over a num__20 - day period . to calculate the total amount of water that evaporated during this time frame we need to multiply num__0.01 by num__20 . this gives us : num__0.03 x num__20 = num__0.6 ounces finally we are asked for “ what percent ” of the original amount of water evaporated during this period . to determine this percentage we have to make sure we translate the expression correctly . we can translate it to : ( amount evaporated / original amount ) x num__100.0 ( num__0.6 / num__10 ) x num__100.0 ( num__0.06 ) x num__100.0 = num__6.0 answer d <eor> d <eos> |
d |
percent__10.0__0.6__ percent__100.0__6.0__ |
percent__10.0__0.6__ percent__100.0__6.0__ |
| an investor purchased num__100 shares of stock x at num__6 num__0.125 dollars per share and sold them all a year later at num__12 dollars per share . if the investor paid a num__2 percent brokerage fee on both the total purchase price and the total selling price which of the following is closest to the investor ' s percent gain on this investment ? <o> a ) num__192.0 <o> b ) num__94.0 <o> c ) num__280.0 <o> d ) num__300.0 <o> e ) num__380 % |
if the purchase price was num__6 $ per share then the investor would have made a profit of num__100.0 . since the purchase price is slightly more than num__6 $ the profit would be slightly less than num__100.0 . also a num__2.0 brokerage is negligible and it brings down the profit percentage only by a small value . approximation is very useful to solve these kind of problems as the answer choices are far apart . answer : b <eor> b <eos> |
b |
subtract__100.0__6.0__ |
subtract__100.0__6.0__ |
| in a mixture of num__13 litres the ratio of milk and water is num__3 : num__2 . if num__3 liters of this mixture is replaced by num__3 liters of milk then what will be the ratio of milk and water in the newly formed mixture ? <o> a ) num__10 : num__3 <o> b ) num__8 : num__5 <o> c ) num__9 : num__4 <o> d ) num__1 : num__1 <o> e ) num__2 : num__3 |
explanation : given : total quantity of mixture = num__13 liters num__3 litres of mixture is removed from the container – so let ' s forget this altogether ! now you are left with only num__10 litres of mixture in num__3 : num__2 ratio . milk in num__10 litres mix = num__10 x num__3 / ( num__2 + num__3 ) = num__6 litres water in num__10 litres mix = num__10 x num__2 / ( num__2 + num__3 ) = num__4 litres we add num__3 litres milk to this . so milk in new mix is = num__6 liters + num__3 litres = num__9 litres water = num__4 litres ratio of milk : water = num__9 : num__4 answer is c <eor> c <eos> |
c |
subtract__13.0__3.0__ multiply__3.0__2.0__ subtract__6.0__2.0__ subtract__13.0__4.0__ subtract__13.0__4.0__ |
subtract__13.0__3.0__ multiply__3.0__2.0__ subtract__6.0__2.0__ add__3.0__6.0__ add__3.0__6.0__ |
| num__39 persons can repair a road in num__12 days working num__5 hours a day . in how many days will num__30 persons working num__3 hours a day complete the work ? <o> a ) num__10 <o> b ) num__26 <o> c ) num__14 <o> d ) num__15 <o> e ) num__16 |
let the required number of days be x . less persons more days ( indirect proportion ) more working hours per day less days ( indirect proportion ) persons num__30 : num__39 : : num__12 : x working hours / day num__3 : num__5 num__30 x num__3 x x = num__39 x num__5 x num__12 x = ( num__39 x num__5 x num__12 ) / ( num__30 x num__3 ) x = num__26 . answer : b <eor> b <eos> |
b |
round__26.0__ |
round__26.0__ |
| carol and jordan draw rectangles of equal area . if carol ' s rectangle measures num__5 inches by num__24 inches and jordan ' s rectangle is num__2 inches long how wide is jordan ' s rectangle in inches ? <o> a ) num__65 <o> b ) num__63 <o> c ) num__52 <o> d ) num__60 <o> e ) num__68 |
area of carol ' s rectangle = num__24 * num__5 = num__120 let width of jordan ' s rectangle = w since the areas are equal num__2 w = num__120 = > w = num__60 answer d <eor> d <eos> |
d |
multiply__5.0__24.0__ triangle_area__5.0__24.0__ triangle_area__5.0__24.0__ |
multiply__5.0__24.0__ triangle_area__5.0__24.0__ triangle_area__5.0__24.0__ |
| find the equation of a line which passes through a ( num__4 - num__1 ) and is parallel to x axis . <o> a ) y = - num__2 <o> b ) y = num__3 <o> c ) y = num__2 <o> d ) y = - num__1 <o> e ) y = num__0 |
the slope of a line which is parallel to x - axis is zero . so y + num__1 = num__0 ( x − num__4 ) → y + num__1 = num__0 → y = − num__1 answer : d y = - num__1 <eor> d <eos> |
d |
reverse__1.0__ |
reverse__1.0__ |
| the value of p when num__4864 x num__9 p num__2 is divisible by num__12 is : <o> a ) num__2 <o> b ) num__5 <o> c ) num__8 <o> d ) num__9 <o> e ) num__1 |
since num__4864 is divisible by num__4 so num__9 p num__2 must be divisible by num__3 . therefore ( num__11 + p ) must be divisible by num__3 . therefore least value of p is num__1 answer : e <eor> e <eos> |
e |
subtract__12.0__9.0__ add__9.0__2.0__ subtract__12.0__11.0__ reverse__1.0__ |
subtract__12.0__9.0__ add__9.0__2.0__ subtract__12.0__11.0__ reverse__1.0__ |
| the length of a rectangle is increased to num__2 times its original size and its width is increased to num__3 times its original size . if the area of the new rectangle is equal to num__1800 square meters what is the area of the original rectangle ? <o> a ) num__300 square meters <o> b ) num__400 square meters <o> c ) num__500 square meters <o> d ) num__600 square meters <o> e ) num__700 square meters |
if l and w be the original length and width of the rectangle and its area is given by l ? w after increase the length becomes num__2 l and the width becomes num__3 w . the area is then given by ( num__2 l ) ? ( num__3 w ) and is known . hence ( num__2 l ) ? ( num__3 w ) = num__1800 solve the above equation to find l ? w num__6 l ? w = num__1800 l ? w = num__300.0 = num__300 square meters area of original rectangle correct answer a <eor> a <eos> |
a |
multiply__2.0__3.0__ triangle_area__2.0__300.0__ |
multiply__2.0__3.0__ triangle_area__2.0__300.0__ |
| if a - b = num__3 and a ^ num__2 + b ^ num__2 = num__27 find the value of ab . <o> a ) num__12 <o> b ) num__15 <o> c ) num__10 <o> d ) num__18 <o> e ) num__9 |
num__2 ab = ( a ^ num__2 + b ^ num__2 ) - ( a - b ) ^ num__2 = num__27 - num__9 = num__18 ab = num__9 . answer is e . <eor> e <eos> |
e |
divide__27.0__3.0__ multiply__2.0__9.0__ divide__27.0__3.0__ |
divide__27.0__3.0__ subtract__27.0__9.0__ subtract__27.0__18.0__ |
| the value of num__0.1 x num__0.1 x num__0.1 + num__0.02 x num__0.02 x num__0.02 is : num__0.2 x num__0.2 x num__0.2 + num__0.04 x num__0.04 x num__0.04 <o> a ) num__0.0125 <o> b ) num__0.125 <o> c ) num__0.25 <o> d ) num__0.5 <o> e ) none of these |
explanation : given expression = ( num__0.1 ) num__3 + ( num__0.02 ) num__3 = num__1 = num__0.125 num__23 [ ( num__0.1 ) num__3 + ( num__0.02 ) num__3 ] = num__8 answer - b <eor> b <eos> |
b |
reverse__0.125__ reverse__8.0__ |
reverse__0.125__ reverse__8.0__ |
| what is the area of a square field whose diagonal of length num__18 m ? <o> a ) num__120 sq m <o> b ) num__250 sq m <o> c ) num__200 sq m <o> d ) num__180 sq m <o> e ) num__100 sq m |
d num__1.0 = ( num__18 * num__18 ) / num__2 = num__180 answer : d <eor> d <eos> |
d |
multiply__1.0__180.0__ |
multiply__1.0__180.0__ |
| the number x is num__200.0 of y . the number z increased by num__200.0 is num__50.0 of x . what percentage of z is y ? <o> a ) num__33.3 <o> b ) num__100.0 <o> c ) num__200.0 <o> d ) num__300.0 <o> e ) num__333 % |
picking numbers : start with z as question asks in terms of z . let z = num__100 thus x = num__600 and y = num__300 from question stems . hence y is num__300.0 of z . ans d . <eor> d <eos> |
d |
add__200.0__100.0__ add__200.0__100.0__ |
add__200.0__100.0__ add__200.0__100.0__ |
| find the area of trapezium whose parallel sides are num__22 cm and num__18 cm long and the distance between them is num__15 cm . <o> a ) num__227 <o> b ) num__299 <o> c ) num__300 <o> d ) num__161 <o> e ) num__212 |
area of a trapezium = num__0.5 ( sum of parallel sides ) * ( perpendicular distance between them ) = num__0.5 ( num__22 + num__18 ) * ( num__15 ) = num__300 cm num__2 answer : c <eor> c <eos> |
c |
round__300.0__ |
round__300.0__ |
| find the principle on a certain sum of money at num__5.0 per annum for num__2 num__0.4 years if the amount being rs . num__896 ? <o> a ) num__1000 <o> b ) num__800 <o> c ) num__2889 <o> d ) num__2777 <o> e ) num__2991 |
num__896 = p [ num__1 + ( num__5 * num__2.4 ) / num__100 ] p = num__800 answer : b <eor> b <eos> |
b |
add__2.0__0.4__ multiply__1.0__800.0__ |
add__2.0__0.4__ multiply__1.0__800.0__ |
| there are a red yellow blue and green tennis balls respectively in a set . how many subsets of these balls can i get without using the green ball ? <o> a ) num__9 <o> b ) num__7 <o> c ) num__8 <o> d ) num__6 <o> e ) num__10 |
this is a combination solution since we have num__3 different kinds of balls other than the green ball we can take any ball from the set of num__3 to make a subset as follows . num__3 c num__1 + num__3 c num__2 + num__3 c num__3 = num__7 plus the one set that is null that is the set having no element in it which is num__3 c num__0 = num__1 = num__7 + num__1 = num__8 c = num__8 <eor> c <eos> |
c |
subtract__3.0__1.0__ add__1.0__7.0__ add__1.0__7.0__ |
subtract__3.0__1.0__ add__1.0__7.0__ add__1.0__7.0__ |
| if x = a / num__3 + b / num__3 ^ num__2 + c / num__3 ^ num__3 where a b and c are each equal to num__0 or num__1 then x could be each of the following except : <o> a ) num__0.037037037037 <o> b ) num__0.111111111111 <o> c ) num__0.148148148148 <o> d ) num__0.222222222222 <o> e ) num__0.444444444444 |
x = a / num__3 + b / num__3 ^ num__2 + c / num__3 ^ num__3 = a / num__3 + b / num__9 + c / num__27 = ( num__9 a + num__3 b + c ) / num__27 depending on whether a b and c take num__0 or num__1 the numerator can take num__1 num__3 num__4 and num__12 num__0.111111111111 = num__0.111111111111 num__0.444444444444 = num__0.444444444444 answer d <eor> d <eos> |
d |
multiply__3.0__9.0__ add__3.0__1.0__ multiply__3.0__4.0__ reverse__9.0__ divide__4.0__9.0__ divide__2.0__9.0__ |
multiply__3.0__9.0__ add__3.0__1.0__ add__3.0__9.0__ reverse__9.0__ divide__4.0__9.0__ divide__2.0__9.0__ |
| a part - time employee ’ s hourly wage was increased by num__50.0 . she decided to decrease the number of hours worked per week so that her total income did not change . by approximately what percent should the number of hours worked be decreased ? <o> a ) num__9.0 <o> b ) num__15.0 <o> c ) num__25.0 <o> d ) num__33.0 <o> e ) num__50 % |
let ' s plug in somenicenumbers and see what ' s needed . let ' s say the employee used to make $ num__1 / hour and worked num__100 hours / week so the total weekly income was $ num__100 / week after the num__50.0 wage increase the employee makes $ num__1.50 / hour we want the employee ' s income to remain at $ num__100 / week . so we want ( $ num__1.50 / hour ) ( new # of hours ) = $ num__100 divide both sides by num__1.50 to get : new # of hours = num__100 / num__1.50 ≈ num__67 hours so the number of hours decreases from num__100 hours to ( approximately ) num__67 hours . this represents a num__33.0 decrease ( approximately ) . answer : d <eor> d <eos> |
d |
subtract__100.0__67.0__ multiply__1.0__33.0__ |
subtract__100.0__67.0__ divide__33.0__1.0__ |
| if ajay drives at num__0.8 th of his usual speed to his office he is num__6 minutes late . what is his usual time to reach his office ? <o> a ) num__38 <o> b ) num__27 <o> c ) num__24 <o> d ) num__19 <o> e ) num__17 |
explanation : let t be his usual time to reach his office and v be his usual speed . v = d / t … … … . ( d is the distance ajay travels while going to his office ) vt = d at v num__1 = num__4 v / num__5 ; t num__1 = t + num__6 num__4 v / num__5 = d / ( t + num__6 ) num__4 v / num__5 * ( t + num__6 ) = d num__4 v / num__5 * ( t + num__6 ) = vt on solving we get t = num__24 minutes answer : c <eor> c <eos> |
c |
subtract__6.0__1.0__ multiply__6.0__4.0__ round__24.0__ |
add__1.0__4.0__ multiply__6.0__4.0__ multiply__6.0__4.0__ |
| what is the value of ( num__3 ) ^ - num__4 ? <o> a ) num__0.25 <o> b ) num__0.0123456790123 <o> c ) num__0.0153846153846 <o> d ) num__0.111111111111 <o> e ) num__0.333333333333 |
num__3 ^ - num__4 = num__1 / ( num__3 ) ^ num__4 = num__0.0123456790123 answer : b <eor> b <eos> |
b |
subtract__4.0__3.0__ multiply__1.0__0.0123__ |
subtract__4.0__3.0__ divide__0.0123__1.0__ |
| rabert ’ s age after num__15 years will be num__5 times his age num__5 years back . what is the present age of rajeev ? <o> a ) num__10 years . <o> b ) num__11 years . <o> c ) num__12 years . <o> d ) num__13 years . <o> e ) num__14 years . |
sol . let rabert ’ s present age be x years . then rabert ’ s age after num__15 years = ( x + num__15 ) years . rabert ’ s age num__5 years back = ( x - num__5 ) years . ∴ x + num__15 = num__5 ( x - num__5 ) ⇔ x + num__15 = num__5 x - num__25 ⇔ num__4 x = num__40 ⇔ x = num__10 . hence rabert ’ present age = num__10 years . answer a <eor> a <eos> |
a |
add__15.0__25.0__ subtract__15.0__5.0__ subtract__15.0__5.0__ |
add__15.0__25.0__ subtract__15.0__5.0__ subtract__15.0__5.0__ |
| a machine fills num__150 bottles of water every num__8 minutes . how many minutes it takes this machine to fill num__675 bottles ? <o> a ) num__54 <o> b ) num__27 <o> c ) num__36 <o> d ) num__57 <o> e ) num__87 |
num__8 minutes are needed to fill num__150 bottles . how many groups of num__150 bottles are there in num__675 bottles ? num__675 ÷ num__150 = num__4.5 = num__4 and num__0.5 for each of these groups num__8 minutes are needed . for num__4 groups and num__0.5 num__8 × num__4 + num__4 = num__32 + num__4 = num__36 minutes . ( num__4 is for num__0.5 a group that needs half time ) correct answer c <eor> c <eos> |
c |
divide__675.0__150.0__ subtract__4.5__4.0__ multiply__8.0__4.0__ multiply__8.0__4.5__ round__36.0__ |
divide__675.0__150.0__ subtract__4.5__4.0__ multiply__8.0__4.0__ add__4.0__32.0__ add__4.0__32.0__ |
| for every order a certain mail - order company charges a shipping fee of $ num__3 plus an additional $ num__2 if the value of the order is over $ num__50 but not over $ num__100 or an additional $ num__3 if the value of the order is over $ num__100 . how much greater are the total shipping fees for num__2 orders of $ num__75 each than the total shipping fee for num__3 order of $ num__150 ? <o> a ) $ num__1 <o> b ) $ num__2 <o> c ) $ num__3 <o> d ) $ num__4 <o> e ) $ num__8 |
total cost involved for each num__75 $ shipment : num__3 + num__2 = num__5 $ . thus for two such orders = num__10 $ total cost involved for a num__150 $ shipment : num__3 + num__3 = num__6 $ . thus for two such orders = num__18 $ the difference = num__8 $ . e . <eor> e <eos> |
e |
add__3.0__2.0__ multiply__2.0__5.0__ multiply__3.0__2.0__ multiply__3.0__6.0__ add__3.0__5.0__ add__3.0__5.0__ |
add__3.0__2.0__ multiply__2.0__5.0__ multiply__3.0__2.0__ multiply__3.0__6.0__ add__3.0__5.0__ add__3.0__5.0__ |
| a sum of money triples itself in ten years at simple interest . find the rate of interest ? <o> a ) num__12 num__0.5 % p . a . <o> b ) num__14 num__0.5 % p . a . <o> c ) num__20.0 p . a . <o> d ) num__22.0 p . a . <o> e ) num__25.0 p . a . |
let the pricipal be rs . x then amount = num__3 x ( where r = rate of interest ) = > interest = num__3 x - x = rs . num__2 x r = ( num__100 * num__2 x ) / ( x * num__10 ) = num__20.0 answer : c <eor> c <eos> |
c |
percent__20.0__100.0__ |
percent__20.0__100.0__ |
| the banker ' s discount on rs . num__1800 at num__16.0 per annum is equal to the true discount on rs . num__1872 for the same time at the same rate . find the time ? <o> a ) num__3 months <o> b ) num__4 months <o> c ) num__5 months <o> d ) num__6 months <o> e ) num__7 months |
solution s . i on rs . num__1800 = t . d on rs . num__1872 . p . w on rs . num__1872 is rs . num__1800 . rs . num__72 is s . i on rs . num__1800 at num__16.0 . time = ( num__100 x num__4.5 x num__1800 ) = num__0.25 year = num__3 months . answer a <eor> a <eos> |
a |
percent__3.0__100.0__ |
percent__3.0__100.0__ |
| p cando a work in num__15 days and q cando the same work in num__20 days . if they can work together for num__4 days what is the fraction of work left ? <o> a ) num__0.285714285714 <o> b ) num__0.272727272727 <o> c ) num__0.533333333333 <o> d ) num__0.466666666667 <o> e ) none |
amount of work p can do in num__1 day = num__0.0666666666667 amount of work q can do in num__1 day = num__0.05 amount of work p and q can do in num__1 day = num__0.0666666666667 + num__0.05 = num__0.116666666667 amount of work p and q can together do in num__4 days = num__4 × ( num__0.116666666667 ) = num__0.466666666667 fraction of work left = num__1 – num__0.466666666667 = num__0.533333333333 c <eor> c <eos> |
c |
divide__1.0__15.0__ divide__1.0__20.0__ add__0.05__0.0667__ subtract__1.0__0.4667__ multiply__1.0__0.5333__ |
divide__1.0__15.0__ divide__1.0__20.0__ add__0.05__0.0667__ subtract__1.0__0.4667__ multiply__1.0__0.5333__ |
| the cross - section of a cannel is a trapezium in shape . if the cannel is num__10 m wide at the top and num__6 m wide at the bottom and the area of cross - section is num__640 sq m the depth of cannel is ? <o> a ) num__22 <o> b ) num__288 <o> c ) num__29 <o> d ) num__60 <o> e ) num__12 |
num__0.5 * d ( num__10 + num__6 ) = num__640 d = num__80 answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ |
multiply__10.0__6.0__ |
| num__12.1212 + num__17.0005 - num__9.1103 = ? <o> a ) num__20.0015 <o> b ) num__20.0105 <o> c ) num__20.0115 <o> d ) num__20.0114 <o> e ) none |
solution given expression = ( num__12.1212 + num__17.0005 ) - num__9.1103 = ( num__29.1217 - num__9.1103 ) = num__20.0114 . answer d <eor> d <eos> |
d |
add__12.1212__17.0005__ subtract__29.1217__9.1103__ subtract__29.1217__9.1103__ |
add__12.1212__17.0005__ subtract__29.1217__9.1103__ subtract__29.1217__9.1103__ |
| if you write down all the numbers from num__1 to num__25 then how many times do you write num__3 ? <o> a ) a ) num__11 <o> b ) b ) num__3 <o> c ) c ) num__2 <o> d ) d ) num__1 <o> e ) e ) num__6 |
explanation : explanation : clearly from num__1 to num__25 there are ten numbers with num__3 as the unit ' s digit - num__3 num__13 num__23 so required number = num__3 answer : b <eor> b <eos> |
b |
multiply__1.0__3.0__ |
multiply__1.0__3.0__ |
| two trains are moving in opposite directions at num__60 km / hr and num__90 km / hr . their lengths are num__1.10 km and num__0.9 km respectively . the time taken by the slower train to cross the faster train in seconds is ? <o> a ) num__42 sec <o> b ) num__44 sec <o> c ) num__46 sec <o> d ) num__48 sec <o> e ) num__50 sec |
explanation : relative speed = num__60 + num__90 = num__150 km / hr . = num__150 x num__0.277777777778 = num__41.6666666667 m / sec . distance covered = num__1.10 + num__0.9 = num__2 km = num__2000 m . required time = num__2000 x num__0.024 = num__48 sec . answer is d <eor> d <eos> |
d |
add__60.0__90.0__ add__1.1__0.9__ multiply__2000.0__0.024__ round__48.0__ |
add__60.0__90.0__ add__1.1__0.9__ divide__2000.0__41.6667__ divide__2000.0__41.6667__ |
| a train travels num__225 km in num__3.5 hours and num__370 km in num__5 hours . find the average speed of train . <o> a ) num__80 kmph <o> b ) num__60 kmph <o> c ) num__70 kmph <o> d ) num__90 kmph <o> e ) none of these |
as we know that speed = distance / time for average speed = total distance / total time taken thus total distance = num__225 + num__370 = num__595 km thus total speed = num__8.5 hrs or average speed = num__595 / num__8.5 or num__70 kmph . answer : c <eor> c <eos> |
c |
add__225.0__370.0__ add__3.5__5.0__ divide__595.0__8.5__ round__70.0__ |
add__225.0__370.0__ add__3.5__5.0__ divide__595.0__8.5__ divide__595.0__8.5__ |
| if - num__4 < x < num__5 and - num__6 < y < num__3 which of the following specifies all the possible values of xy ? <o> a ) - num__42 < xy < num__21 <o> b ) - num__42 < xy < num__24 <o> c ) - num__30 < xy < num__24 <o> d ) - num__24 < xy < num__21 <o> e ) - num__24 < xy < num__24 |
the least value of xy is a bit more than num__5 * ( - num__6 ) = - num__30 and the largest value of xy is a bit less than ( - num__4 ) * ( - num__6 ) = num__24 . therefore - num__30 < xy < num__24 . answer : c . <eor> c <eos> |
c |
multiply__5.0__6.0__ multiply__4.0__6.0__ multiply__5.0__6.0__ |
multiply__5.0__6.0__ multiply__4.0__6.0__ multiply__5.0__6.0__ |
| salaries of ravi and sumit are in the ration num__2 : num__3 . it the salary of each in increased by $ num__4000 the new ration becomes num__40 : num__57 what is sumit ’ s present salary ? <o> a ) $ num__17000 <o> b ) $ num__20000 <o> c ) $ num__25500 <o> d ) $ num__38000 <o> e ) $ num__28500 |
let the original salaries of ravi and sumit be $ num__2 x and $ num__3 x respectively . then ( num__2 x + num__4000 ) / ( num__3 x + num__4000 ) = num__0.701754385965 num__57 ( num__2 x + num__4000 ) = num__40 ) ( num__3 x + num__4000 ) num__6 x = num__68000 num__3 x = num__34000 sumit ’ s present salary = num__3 x + num__4000 = num__34000 + num__4000 = $ num__38000 answer : d $ num__38000 <eor> d <eos> |
d |
divide__40.0__57.0__ multiply__2.0__3.0__ divide__68000.0__2.0__ add__4000.0__34000.0__ add__4000.0__34000.0__ |
divide__40.0__57.0__ multiply__2.0__3.0__ divide__68000.0__2.0__ add__4000.0__34000.0__ add__4000.0__34000.0__ |
| an empty bucket being filled with paint at a constant rate takes num__6 minutes to be filled to num__0.7 of its capacity . how much more time s will it take to fill the bucket to full capacity ? <o> a ) num__0.388888888889 <o> b ) num__0.5 <o> c ) num__2 <o> d ) num__2.57142857143 <o> e ) num__3.6 |
solution - work and time are directly proportional . w num__1 / w num__2 = t num__1 / t num__2 num__0.7 work in num__6 mins num__1 work in t mins ( num__0.7 ) / num__1 = num__6 / t - > t = num__8.57142857143 mins . remaining minutes to fill the tank s = num__8.57142857143 - num__6 = num__2.57142857143 mins . ans d . <eor> d <eos> |
d |
divide__6.0__0.7__ subtract__8.5714__6.0__ multiply__1.0__2.5714__ |
divide__6.0__0.7__ subtract__8.5714__6.0__ divide__2.5714__1.0__ |
| a cycle is bought for rs . num__900 and sold for rs . num__1080 find the gain percent ? <o> a ) num__39.0 <o> b ) num__20.0 <o> c ) num__23.0 <o> d ) num__74.0 <o> e ) num__83 % |
num__900 - - - - num__180 num__100 - - - - ? = > num__20.0 answer : b <eor> b <eos> |
b |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| if a solid sphere of radius num__10 cms is moulded into num__8 spherical solid balls of equal radius then surface area of each ball ( in cm num__2 ) is ? <o> a ) num__100 pi <o> b ) num__300 pi <o> c ) num__70 pi <o> d ) num__90 pi <o> e ) num__200 pi |
num__1.33333333333 Ï € * num__10 * num__10 * num__10 = num__8 * num__1.33333333333 Ï € r num__3 r = num__5 num__4 Ï € * num__5 * num__5 = num__100 Ï € answer a <eor> a <eos> |
a |
side_by_diagonal__5.0__3.0__ power__10.0__2.0__ power__10.0__2.0__ |
side_by_diagonal__5.0__3.0__ volume_rectangular_prism__10.0__2.0__5.0__ volume_rectangular_prism__10.0__2.0__5.0__ |
| find the value of num__1 / ( num__3 + num__1 / ( num__3 + num__1 / ( num__3 - num__0.333333333333 ) ) ) <o> a ) num__0.280898876404 <o> b ) num__0.268041237113 <o> c ) num__0.303370786517 <o> d ) num__3.33333333333 <o> e ) num__3.2962962963 |
num__1 / [ num__3 + ( num__1 / ( num__3 + num__1 / ( num__3 - num__0.333333333333 ) ) ) ] = > num__1 / [ num__3 + num__1 / ( num__3 + num__1 / ( num__2.66666666667 ) ) ] = > num__1 / [ num__3 + num__1 / ( num__3 + num__0.375 ) ] = > num__1 / [ num__3 + num__0.296296296296 ] = > num__1 / ( num__3.2962962963 ) = > num__0.303370786517 c ) <eor> c <eos> |
c |
subtract__3.0__0.3333__ reverse__2.6667__ add__3.0__0.2963__ reverse__3.2963__ reverse__3.2963__ |
subtract__3.0__0.3333__ reverse__2.6667__ add__3.0__0.2963__ reverse__3.2963__ reverse__3.2963__ |
| the total of num__35 students in a class is num__560 years . the total age of num__21 students is num__294 . what is the total age of remaining num__14 students ? <o> a ) num__260 <o> b ) num__270 <o> c ) num__266 <o> d ) num__294 <o> e ) num__244 |
the total age of num__35 students = num__560 the total age of num__21 students = num__294 the total age of remaining num__14 students = ( num__560 - num__294 ) = num__266 answer : c <eor> c <eos> |
c |
subtract__560.0__294.0__ subtract__560.0__294.0__ |
subtract__560.0__294.0__ subtract__560.0__294.0__ |
| if the average of num__42 num__87 num__36 num__78 num__99 num__22 and x is num__59 what is x ? <o> a ) num__49 <o> b ) num__42 <o> c ) num__59 <o> d ) num__78 <o> e ) num__44 |
the answer is num__49 because : ( num__42 + num__87 + num__36 + num__78 + num__99 + num__22 + x ) / num__7 = num__59 ( num__42 + num__87 + num__36 + num__78 + num__99 + num__22 + x ) / num__7 * num__7 = num__59 * num__7 ( num__42 + num__87 + num__36 + num__78 + num__99 + num__22 + x ) = num__413 ( num__42 + num__87 + num__36 + num__78 + num__99 + num__22 + x ) - num__364 = num__413 - num__364 x = num__49 so the correct answer is a num__49 . <eor> a <eos> |
a |
subtract__49.0__42.0__ multiply__59.0__7.0__ subtract__413.0__49.0__ add__42.0__7.0__ |
subtract__49.0__42.0__ multiply__59.0__7.0__ subtract__413.0__49.0__ add__42.0__7.0__ |
| harold works at a resort from the beginning of march to the end of september . during the month of august this past year he made num__2 times the average ( arithmetic mean ) of his monthly totals in tips for the other months . his total tips for august were what fraction of his total tips for all of the months he worked ? <o> a ) num__0.333333333333 <o> b ) num__0.4 <o> c ) num__0.25 <o> d ) num__0.5 <o> e ) num__0.571428571429 |
the time from beginning of march to the end of september is num__7 months . if x is the average monthly tip for all months other than august then his august month tip will be num__2 * x his total tip for the num__7 months = num__6 * ( average tip for the months other than august ) + num__2 x = num__8 x august tips as a fraction of total tips = num__2 x / num__8 x = num__0.25 . so c <eor> c <eos> |
c |
add__2.0__6.0__ divide__2.0__8.0__ divide__2.0__8.0__ |
add__2.0__6.0__ divide__2.0__8.0__ divide__2.0__8.0__ |
| given an integer x let f ( x ) be the sum of the digits of x . compute the number of positive integers less than num__1000 where f ( x ) = num__2 . <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
we do casework on the number of digits : ( a ) the number has num__1 digit . the only valid number is num__2 . ( b ) the number has num__2 digits . the only valid numbers are num__11 and num__20 . ( c ) the number has num__3 digits . the only valid numbers are num__101 num__110 and num__200 . this gives us num__6 numbers . correct answer c <eor> c <eos> |
c |
add__2.0__1.0__ multiply__2.0__3.0__ multiply__2.0__3.0__ |
add__2.0__1.0__ multiply__2.0__3.0__ multiply__2.0__3.0__ |
| an amount of rs . num__3000 becomes rs . num__3500 in four years at simple interest . if the rate of interest was num__1.0 more then what was be the total amount ? <o> a ) rs . num__4500 . <o> b ) rs . num__3800 <o> c ) rs . num__3700 . <o> d ) rs . num__2500 . <o> e ) rs . num__3500 . |
a = p ( num__1 + tr / num__100 ) = > num__3500 = num__3000 [ num__1 + ( num__4 * r ) / num__100 ] = > r = num__4.16 now r = num__5.16 = > a = num__3000 [ num__1 + ( num__4 * num__5.16 ) / num__100 ] = rs . num__3500 . answer : e <eor> e <eos> |
e |
percent__100.0__3500.0__ |
percent__100.0__3500.0__ |
| for each color copy print shop x charges $ num__1.20 and print shop y charges $ num__1.70 . how much greater is the charge for num__40 color copies at print shop y than at print shop x ? <o> a ) $ num__14 <o> b ) $ num__16 <o> c ) $ num__18 <o> d ) $ num__20 <o> e ) $ num__22 |
the difference in the two prices is $ num__1.70 - $ num__1.20 = $ num__0.50 for each color copy . each color copy will cost an extra $ num__0.50 at print shop y . num__40 * $ num__0.50 = $ num__20 the answer is d . <eor> d <eos> |
d |
subtract__1.7__1.2__ multiply__40.0__0.5__ multiply__40.0__0.5__ |
subtract__1.7__1.2__ multiply__40.0__0.5__ multiply__40.0__0.5__ |
| a can finish a work in num__9 days and b can do the same work in num__15 days . b worked for num__10 days and left the job . in how many days a alone can finish the remaining work ? <o> a ) num__6 <o> b ) num__3 <o> c ) num__5.5 <o> d ) num__7 <o> e ) num__8 |
b ' s num__10 day ' s work = ( num__1 x num__10 ) = num__2 . num__15 num__3 remaining work = ( num__1 - num__2 ) = num__1 . num__3 num__3 now num__1 work is done by a in num__1 day . num__9 therefore num__1 work is done by a in ( num__9 x num__1 ) = num__3 days . b <eor> b <eos> |
b |
subtract__10.0__9.0__ add__1.0__2.0__ round__3.0__ |
subtract__10.0__9.0__ add__1.0__2.0__ round__3.0__ |
| a num__140 meter long train crosses a man standing on the platform in num__6 sec . what is the speed of the train ? <o> a ) num__228 <o> b ) num__108 <o> c ) num__84 <o> d ) num__188 <o> e ) num__211 |
s = num__23.3333333333 * num__3.6 = num__84 kmph answer : c <eor> c <eos> |
c |
divide__140.0__6.0__ round__84.0__ |
divide__140.0__6.0__ round__84.0__ |
| a car moves at num__80 km / hr . what is the speed of the car in meters per second ? <o> a ) num__20 num__0.222222222222 m sec <o> b ) num__22 num__0.222222222222 m sec <o> c ) num__24 num__0.222222222222 m sec <o> d ) num__26 num__0.222222222222 m sec <o> e ) num__28 num__0.222222222222 m sec |
explanation : speed = ( num__80 ∗ num__0.277777777778 ) m / sec = num__22.2222222222 m / sec = num__22 num__0.222222222222 msec option b <eor> b <eos> |
b |
subtract__22.2222__22.0__ round__22.0__ |
subtract__22.2222__22.0__ round__22.0__ |
| if n is the smallest integer such that num__432 times n is the square of an integer what is the value of n ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__6 <o> d ) num__12 <o> e ) num__24 |
the prime factorization of a square has to have even powers of all its prime factors . if the original number has a factor say of num__7 then when it ’ s squared the square will have a factor of num__7 ^ num__2 . another way to say that is : any positive integer all of whose prime factors have even powers must be a perfect square of some other integer . look at the prime factorization of num__432 num__432 = ( num__2 ^ num__4 ) * ( num__3 ^ num__3 ) the factor of num__2 already has an even power — - that ’ s all set . the factor of num__3 currently has an odd power . if n = num__3 then num__432 * n would have an even power of num__2 and an even power of num__3 ; therefore it would be a perfect square . thus n = num__3 is a choice that makes num__432 * n a perfect square . answer : b . <eor> b <eos> |
b |
triangle_area__2.0__3.0__ |
triangle_area__2.0__3.0__ |
| if num__15.0 of a is the same as num__20.0 of b then a : b is : <o> a ) num__3 : num__4 <o> b ) num__4 : num__3 <o> c ) num__17 : num__16 <o> d ) num__16 : num__17 <o> e ) num__1 : num__2 |
num__15.0 of a i = num__20.0 of b num__15 a / num__100 = num__20 b / num__100 = num__1.33333333333 = num__4 : num__3 answer : b <eor> b <eos> |
b |
percent__15.0__20.0__ percent__100.0__4.0__ |
percent__15.0__20.0__ percent__100.0__4.0__ |
| if the area of circle is num__616 sq cm then its circumference ? <o> a ) num__78 m <o> b ) num__88 m <o> c ) num__75 m <o> d ) num__70 m <o> e ) num__80 m |
area of circle = ∏ r num__2 = num__3.14285714286 ( r num__2 ) = num__616 = r num__2 = ( num__616 ) ( num__0.318181818182 ) = r = num__14 circumference = num__2 ∏ r = ( num__2 ) ( num__3.14285714286 ) ( num__14 ) = num__88 m answer : b <eor> b <eos> |
b |
triangle_area__2.0__88.0__ |
triangle_area__2.0__88.0__ |
| a certain computer manufacturing firm last year produced num__82 percent of its computers using parts made by its subsidiary company . if the remaining num__10890 computers were produced using parts purchased from another parts manufacturer how many computers were produced by the computer manufacturing firm altogether ? <o> a ) num__60000 <o> b ) num__60100 <o> c ) num__60500 <o> d ) num__72000 <o> e ) num__74000 |
num__82.0 parts used of subsidary company so num__18.0 parts used of other companies . now num__18.0 parts = num__10890 computers so num__1.0 parts = num__605.0 or num__100.0 parts = num__60500 computers . hence answer is ( c ) <eor> c <eos> |
c |
percent__100.0__60500.0__ |
percent__100.0__60500.0__ |
| in n is a positive integer less than num__200 and num__18 n / num__60 is an integer then n has how many different positive prime factors ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__5 <o> d ) num__6 <o> e ) num__8 |
( a ) . num__18 n / num__60 must be an integer . = > num__3 n / num__10 must be an integer . hence n must be a multiple of num__2 * num__5 . = > n has num__2 different prime integers . <eor> a <eos> |
a |
add__2.0__3.0__ subtract__5.0__3.0__ |
divide__10.0__2.0__ divide__10.0__5.0__ |
| a bus travels first half distance between two places with a speed of num__40 kmph and the rest half distance with a speed of num__60 kmph . the average speed of the bus is ? <o> a ) num__48 kmph <o> b ) num__78 kmph <o> c ) num__28 kmph <o> d ) num__18 kmph <o> e ) num__98 kmph |
explanation : we know that speed = distance traveled / time taken let the total distance traveled by the car is num__2 x km . then time taken by it to cover first half is hour . and for second half is hour . then average speed = total distance travelled / total time taken . i . e . = > = num__48 kmph . answer : a <eor> a <eos> |
a |
round__48.0__ |
round__48.0__ |
| the arithmetic mean of the num__9 consecutive integers starting with ' s ' is ' a ' . what is the arithmetic mean of num__9 consecutive integers that start with s + num__4 ? <o> a ) num__2 + s + a <o> b ) num__22 + a <o> c ) num__2 s <o> d ) num__2 a + num__2 <o> e ) num__2 + a |
( num__9 s + num__36 ) / num__9 = s + num__4 = a ( num__9 s + num__54 ) / num__9 = s + num__6 = num__2 + a e <eor> e <eos> |
e |
multiply__9.0__4.0__ divide__54.0__9.0__ subtract__6.0__4.0__ subtract__4.0__2.0__ |
multiply__9.0__4.0__ divide__54.0__9.0__ subtract__6.0__4.0__ divide__4.0__2.0__ |
| a type q machine can complete a job in num__5 hours and a type b machine can complete the job in num__7 hours . how many hours will it take num__2 type q machines and num__3 type b machines working together and independently to complete the job ? <o> a ) num__0.2 <o> b ) num__0.828571428571 <o> c ) num__0.833333333333 <o> d ) num__1.20689655172 <o> e ) num__2.91666666667 |
now d should be the answer . q need num__5 hours to complete and b needs num__7 hours to compete so num__2 q + num__3 b will complete num__0.4 + num__0.428571428571 or num__0.828571428571 portion of the job in num__1 hour so the whole job will take num__1.20689655172 hours . . . . = d <eor> d <eos> |
d |
divide__2.0__5.0__ divide__3.0__7.0__ add__0.4__0.4286__ subtract__3.0__2.0__ divide__1.0__0.8286__ divide__1.0__0.8286__ |
divide__2.0__5.0__ divide__3.0__7.0__ add__0.4__0.4286__ subtract__3.0__2.0__ divide__1.0__0.8286__ divide__1.0__0.8286__ |
| bullock likes to keep a spare tyre in his car every time . on a certain day he travels num__1 num__00000 km and just to make the most of all the tyres he changes the tyres between his journey such that each tyre runs the same distance . what is the distance traveled by each tyre ? <o> a ) num__70000 <o> b ) num__60000 <o> c ) num__80000 <o> d ) num__90000 <o> e ) num__10 |
000 |
c num__80000 . the distance traveled by each tyre : num__0.8 * num__1 num__00 num__000 km = num__80000 km . <eor> c <eos> |
c |
c |
| pam and cathy begin running at the same time on a straight path . pam runs at num__10 miles per hour and cathy runs at num__8 miles per hour . after num__45 minutes pam stops to stretch . if it takes pam num__30 minutes to stretch and cathy continues to run during this time how many minutes will it take pam to catch up to cathy assuming pam resumes running at num__10 miles per hour ? <o> a ) num__30 minutes <o> b ) num__40 minutes <o> c ) num__45 minutes <o> d ) num__60 minutes <o> e ) num__75 minutes |
after num__45 mins pam would have covered num__0.75 * num__10 mph = num__7.5 miles cathy would have covered num__0.75 * num__8 mph = num__6 miles pam does num__30 minutes of stretching . . so after num__75 mins pam would still be at num__7.5 miles cathy would have covered num__1.25 * num__8 = num__10 miles . so after num__75 mins cathy is num__2.5 miles ahead . answer : e <eor> e <eos> |
e |
multiply__10.0__0.75__ multiply__8.0__0.75__ multiply__10.0__7.5__ divide__10.0__8.0__ subtract__10.0__7.5__ round__75.0__ |
multiply__10.0__0.75__ multiply__8.0__0.75__ multiply__10.0__7.5__ divide__10.0__8.0__ subtract__10.0__7.5__ multiply__10.0__7.5__ |
| there are two examinations rooms a and b . if num__10 students are sent from a to b then the number of students in each room is the same . if num__20 candidates are sent from b to a then the number of students in a is double the number of students in b . the number of students in room a is : <o> a ) num__50 <o> b ) num__120 <o> c ) num__100 <o> d ) num__60 <o> e ) num__40 |
explanation : let the number of students in rooms a and b be x and y respectively . then x - num__10 = y + num__10 x - y = num__20 . . . . ( i ) and x + num__20 = num__2 ( y - num__20 ) x - num__2 y = - num__60 . . . . ( ii ) solving ( i ) and ( ii ) we get : x = num__100 y = num__80 . the required answer a = num__100 . answer : c <eor> c <eos> |
c |
divide__20.0__10.0__ add__20.0__60.0__ add__20.0__80.0__ |
divide__20.0__10.0__ add__20.0__60.0__ add__20.0__80.0__ |
| stuart bought a sweater on sale for num__30.0 off the original price and another num__25.0 off the discounted price . if the original price of the sweater was $ num__100 what was the final price of the sweater ? <o> a ) $ num__25.75 <o> b ) $ num__52.5 <o> c ) $ num__15.75 <o> d ) $ num__45.75 <o> e ) $ num__18.75 |
the price with num__30.0 off num__100 - num__30.0 of num__100 = num__100 - ( num__0.3 ) * num__100 = num__70 the price with another num__25.0 off num__70 - num__25.0 of num__70 = num__70 - ( num__0.25 ) * num__70 = $ num__52.5 correct answer b <eor> b <eos> |
b |
percent__100.0__52.5__ |
percent__100.0__52.5__ |
| three numbers are in the ratio num__4 : num__5 : num__6 and their average is num__30 . the largest number is : <o> a ) num__28 <o> b ) num__32 <o> c ) num__36 <o> d ) num__42 <o> e ) num__45 |
explanation : let the numbers be num__4 x num__5 x and num__6 x . therefore ( num__4 x + num__5 x + num__6 x ) / num__3 = num__30 num__15 x = num__90 x = num__6 largest number = num__6 x = num__36 . answer c <eor> c <eos> |
c |
multiply__5.0__3.0__ multiply__6.0__15.0__ add__6.0__30.0__ add__6.0__30.0__ |
multiply__5.0__3.0__ multiply__6.0__15.0__ add__6.0__30.0__ add__6.0__30.0__ |
| a local restaurant recently renovated its dining space purchasing new tables and chairs to use in addition to the original tables and chairs . the new tables each seat six customers while the original tables each seat four customers . altogether the restaurant now has num__40 tables and is capable of seating num__208 customers . how many more new tables than original tables does the restaurant have ? <o> a ) num__8 <o> b ) num__12 <o> c ) num__16 <o> d ) num__20 <o> e ) num__24 |
if all the tables seated num__4 the number of customers could be num__4 * num__40 = num__160 . num__208 - num__160 = num__48 so num__24.0 = num__24 tables must be new tables seating num__6 people . the number of tables seating num__4 people is num__40 - num__24 = num__16 . the number of new tables is num__24 - num__16 = num__8 more than the number of old tables . the answer is a . <eor> a <eos> |
a |
multiply__40.0__4.0__ subtract__208.0__160.0__ divide__24.0__4.0__ subtract__40.0__24.0__ subtract__48.0__40.0__ subtract__48.0__40.0__ |
multiply__40.0__4.0__ subtract__208.0__160.0__ divide__24.0__4.0__ subtract__40.0__24.0__ subtract__48.0__40.0__ subtract__48.0__40.0__ |
| what is the least number of squares tiles required to pave the floor of a room num__15 m num__17 cm long and num__9 m num__2 cm broad ? <o> a ) num__814 <o> b ) num__820 <o> c ) num__840 <o> d ) num__844 <o> e ) num__548 |
length of largest tile = h . c . f . of num__1517 cm and num__902 cm = num__41 cm . area of each tile = ( num__41 x num__41 ) cm num__2 . required number of tiles = num__1517 x num__902 / ( num__41 ^ num__2 ) = num__814 . answer : a <eor> a <eos> |
a |
round__814.0__ |
round__814.0__ |
| given num__3 lines in the plane such that the points of intersection form a triangle with sides of length num__20 num__20 and num__30 the number of points equidistant from all the num__3 lines is <o> a ) num__1 <o> b ) num__3 <o> c ) num__4 <o> d ) num__0 <o> e ) num__2 |
num__4 points are possible in which num__1 point is the incentre of the circle and other num__3 are excentres answer : c <eor> c <eos> |
c |
subtract__4.0__3.0__ round__4.0__ |
subtract__4.0__3.0__ round__4.0__ |
| if f = ( num__10 ^ num__10 ) - num__47 what is the sum of all the digit of f ? <o> a ) num__40 <o> b ) num__45 <o> c ) num__50 <o> d ) num__55 <o> e ) num__80 |
start small and check for any pattern : num__10 ^ num__2 - num__47 = num__53 - > number of digits = power of num__10 ; number of num__9 s = ( power of num__10 ) - num__2 ; num__53 is the last num__2 digits num__10 ^ num__3 - num__47 = num__953 - > number of digits = power of num__10 ; number of num__9 s = ( power of num__10 ) - num__2 ; num__53 is the last num__2 digits . . . extrapolate num__10 ^ num__10 - num__47 - > number of digits = num__10 ; number of num__9 s = num__8 ; num__53 will be the last num__2 digits f = num__9 * num__8 + num__5 + num__3 = num__80 . answer ( e ) . <eor> e <eos> |
e |
subtract__10.0__2.0__ divide__10.0__2.0__ multiply__10.0__8.0__ multiply__10.0__8.0__ |
subtract__10.0__2.0__ add__2.0__3.0__ multiply__10.0__8.0__ multiply__10.0__8.0__ |
| a man rows his boat num__85 km downstream and num__45 km upstream taking num__2 num__0.5 hours each time . find the speed of the stream ? <o> a ) num__6 kmph <o> b ) num__9 kmph <o> c ) num__3 kmph <o> d ) num__8 kmph <o> e ) num__1 kmph |
speed downstream = d / t = num__85 / ( num__2 num__0.5 ) = num__34 kmph speed upstream = d / t = num__45 / ( num__2 num__0.5 ) = num__18 kmph the speed of the stream = ( num__34 - num__18 ) / num__2 = num__8 kmph answer : d <eor> d <eos> |
d |
round__8.0__ |
round__8.0__ |
| how long does a train num__110 m long running at the speed of num__72 km / hr takes to cross a bridge num__138 m length ? <o> a ) num__82.1 sec . <o> b ) num__12.4 sec <o> c ) num__19.1 sec . <o> d ) num__17.1 sec . <o> e ) num__42.1 sec . |
speed = num__72 * num__0.277777777778 = num__20 m / sec total distance covered = num__110 + num__138 = num__248 m . required time = num__12.4 = num__12.4 sec . answer : b <eor> b <eos> |
b |
add__110.0__138.0__ divide__248.0__20.0__ round__12.4__ |
add__110.0__138.0__ divide__248.0__20.0__ divide__248.0__20.0__ |
| there are num__44 students in a hostel due to the administration num__15 new students has joined . the expense of the mess increase by rs . num__33 per day . while the average expenditure per head diminished by rs . num__3 what was the original expenditure of the mess ? <o> a ) rs . num__618 <o> b ) rs . num__612 <o> c ) rs . num__622 <o> d ) rs . num__616 <o> e ) rs . num__630 |
explanation : let the average expenditure per head be rs . p now the expenditure of the mess for old students is rs . num__44 p after joining of num__15 more students the average expenditure per head is decreased by rs . num__3 = > p - num__3 here given the expenditure of the mess for ( num__44 + num__15 = num__59 ) students is increased by rs . num__33 therefore num__59 ( p - num__3 ) = num__44 p + num__33 num__59 p - num__177 = num__44 p + num__33 num__15 p = num__210 = > p = num__14 thus the expenditure of the mess for old students is rs . num__44 p = num__44 x num__14 = rs . num__616 . answer : d <eor> d <eos> |
d |
add__44.0__15.0__ multiply__3.0__59.0__ add__33.0__177.0__ divide__210.0__15.0__ multiply__44.0__14.0__ multiply__44.0__14.0__ |
add__44.0__15.0__ multiply__3.0__59.0__ add__33.0__177.0__ divide__210.0__15.0__ multiply__44.0__14.0__ multiply__44.0__14.0__ |
| a truck carrying cans of paint contains p stacks . each stack contains p / num__4 cases and each case contains num__12 cans . how many cans of paint are contained in num__7 trucks ? <o> a ) p ^ num__0.0952380952381 <o> b ) p ^ num__0.00595238095238 <o> c ) num__336 <o> d ) num__21 / p ^ num__2 <o> e ) num__21 p ^ num__2 |
there are num__12 cans in each case . there are num__3 p cans in each stack . there are num__3 p ^ num__2 cans in each truck . there are num__21 p ^ num__2 cans in num__7 trucks . the answer is e . <eor> e <eos> |
e |
divide__12.0__4.0__ multiply__7.0__3.0__ multiply__7.0__3.0__ |
divide__12.0__4.0__ multiply__7.0__3.0__ multiply__7.0__3.0__ |
| in a clothing store there are six different colored neckties ( orange yellow green blue and indigo ) and five different colored shirts ( orange yellow green blue and indigo ) that must be packed into boxes for gifts . if each box can only fit one necktie and one shirt what is the probability that all of the boxes will contain a necktie and a shirt of the same color ? <o> a ) num__0.998611111111 <o> b ) num__0.00833333333333 <o> c ) num__0.00858369098712 <o> d ) num__0.00552486187845 <o> e ) num__0.00138888888889 |
num__5 ties and num__5 shirts . . . red tie can take any of num__5 shirts . . orange can take any of the remaining num__4 shirts yellow any of remaining num__3 . . and so on till last indigo chooses the num__1 remaining . . total ways = num__5 * num__4 * num__3 * num__2 * num__1 = num__120 out of this num__120 only num__1 way will have same colour tie and shirt . . prob = num__0.00833333333333 b <eor> b <eos> |
b |
subtract__4.0__3.0__ subtract__3.0__1.0__ reverse__120.0__ reverse__120.0__ |
subtract__4.0__3.0__ subtract__3.0__1.0__ reverse__120.0__ reverse__120.0__ |
| a man invests some money partly in num__9.0 stock at num__96 and partly in num__12.0 stock at num__120 . to obtain equal dividends from both he must invest the money in the ratio ? <o> a ) num__16 : num__18 <o> b ) num__16 : num__13 <o> c ) num__16 : num__15 <o> d ) num__16 : num__12 <o> e ) num__16 : num__11 |
for an income of re . num__1 in num__9.0 stock at num__96 investment = rs . num__10.6666666667 = rs . num__10.6666666667 for an income re . num__1 in num__12.0 stock at num__120 investment = rs . num__10.0 = rs . num__10 . ratio of investments = ( num__10.6666666667 ) : num__10 = num__32 : num__30 = num__16 : num__15 answer : c <eor> c <eos> |
c |
divide__96.0__9.0__ round_down__10.6667__ subtract__16.0__1.0__ multiply__1.0__16.0__ |
divide__96.0__9.0__ round_down__10.6667__ subtract__16.0__1.0__ multiply__1.0__16.0__ |
| the length of the bridge which a train num__140 m long and traveling at num__45 km / hr can cross in num__30 sec is ? <o> a ) num__377 <o> b ) num__367 <o> c ) num__235 <o> d ) num__245 <o> e ) num__267 |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec . time = num__30 sec let the length of bridge be x meters . then ( num__140 + x ) / num__30 = num__12.5 x = num__235 m . answer : c <eor> c <eos> |
c |
round__235.0__ |
round__235.0__ |
| in a bus left side are num__15 seats available num__3 few seats in right side because in rear exit door . each seat hold num__3 people . in addition there is a seat back can sit num__12 people all together . how many people can sit in a bus ? <o> a ) num__52 <o> b ) num__49 <o> c ) num__95 <o> d ) num__88 <o> e ) num__93 |
right side = num__15 seat left side = num__15 - num__3 ( num__3 few seat in right side ) = num__12 seat total = num__15 + num__12 = num__27 people can seat in num__27 seat = num__27 * num__3 = num__81 people can seat in last seat = num__12 total people can seat = num__81 + num__12 = num__93 answer : e <eor> e <eos> |
e |
add__15.0__12.0__ multiply__3.0__27.0__ add__12.0__81.0__ add__12.0__81.0__ |
add__15.0__12.0__ multiply__3.0__27.0__ add__12.0__81.0__ add__12.0__81.0__ |
| one half of a two digit number exceeds its one third by num__6 . what is the sum of the digits of the number ? <o> a ) num__8 <o> b ) num__7 <o> c ) num__1 <o> d ) num__9 <o> e ) num__6 |
explanation : x / num__2 â € “ x / num__3 = num__6 = > x = num__6 num__3 + num__6 = num__9 answer : d <eor> d <eos> |
d |
divide__6.0__2.0__ add__6.0__3.0__ add__6.0__3.0__ |
divide__6.0__2.0__ add__6.0__3.0__ add__6.0__3.0__ |
| a and b are working on an assignment . a takes num__6 hours to type num__32 pages on a computer while b takes num__5 hours to type num__40 pages . how much time will they take working together on two different computers to type an assignment of num__110 pages ? <o> a ) num__2 hours <o> b ) num__5 hours <o> c ) num__6 hours <o> d ) num__8 hours <o> e ) num__3 hours |
number of pages typed by a in one hour = num__5.33333333333 = num__5.33333333333 number of pages typed by b in one hour = num__8.0 = num__8 number of pages typed by both in one hour = ( ( num__5.33333333333 ) + num__8 ) = num__13.3333333333 time taken by both to type num__110 pages = num__110 * num__0.075 = num__8 hours . answer : d . <eor> d <eos> |
d |
divide__32.0__6.0__ subtract__40.0__32.0__ add__8.0__5.3333__ round__8.0__ |
divide__32.0__6.0__ subtract__40.0__32.0__ add__8.0__5.3333__ round__8.0__ |
| in an election between two candidates num__75.0 of the voters cast their votes out of which num__2.0 of the votes were declared invalid . a candidate got num__9261 votes which were num__75.0 of the total valid votes . find the total number of votes <o> a ) num__16800 <o> b ) num__15800 <o> c ) num__16700 <o> d ) num__15700 <o> e ) none of these |
explanation : let the total number of votes enrolled are x . number of votes cast = num__75.0 of x valid votes = num__98.0 of num__75.0 of x now as num__9261 is the num__75.0 of valid casted votes so num__75.0 of num__98.0 of num__75.0 of x = num__9261 [ imporant ] = > num__75 × num__98 × num__75 × x / num__100 × num__100 × num__100 = num__9261 = > x = num__16800 option a <eor> a <eos> |
a |
percent__100.0__16800.0__ |
percent__100.0__16800.0__ |
| a man can row with a speed of num__15 kmph in still water . if the stream flows at num__15 kmph then the speed in downstream is ? <o> a ) a ) num__25 <o> b ) b ) num__30 <o> c ) c ) num__20 <o> d ) d ) num__21 <o> e ) e ) num__18 |
m = num__15 s = num__15 ds = num__15 + num__15 = num__30 answer : b <eor> b <eos> |
b |
round__30.0__ |
round__30.0__ |
| a number is increased by num__40.0 and then decreased by num__40.0 . find the net increase or decrease per cent . <o> a ) num__19.0 <o> b ) num__18.0 <o> c ) num__17.0 <o> d ) num__13.0 <o> e ) num__16 % |
let the number be num__100 . increase in the number = num__40.0 = num__40.0 of num__100 = ( num__0.4 × num__100 ) = num__40 therefore increased number = num__100 + num__40 = num__140 this number is decreased by num__40.0 therefore decrease in number = num__40.0 of num__140 = ( num__0.4 × num__140 ) = num__56.0 = num__56 therefore new number = num__140 - num__56 = num__84 thus net decreases = num__100 - num__84 = num__16 hence net percentage decrease = ( num__0.16 × num__100 ) % = ( num__16.0 ) % = num__16.0 answer : e <eor> e <eos> |
e |
divide__40.0__100.0__ add__40.0__100.0__ multiply__0.4__140.0__ subtract__140.0__56.0__ multiply__40.0__0.4__ divide__16.0__100.0__ multiply__40.0__0.4__ |
divide__40.0__100.0__ add__40.0__100.0__ multiply__0.4__140.0__ subtract__140.0__56.0__ subtract__100.0__84.0__ divide__16.0__100.0__ subtract__100.0__84.0__ |
| if c is a whole number c + num__1 is a whole number after that . if a is a whole number what is a whole number before that ? <o> a ) num__7 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__0 |
we know that the number ‘ num__0 ’ together with the natural numbers gives us the numbers num__0 num__1 num__2 num__3 num__4 num__5 … … … … … which are called whole numbers . if a is a whole number and it is greater than num__0 then the whole number before that is a – num__1 which will always be a whole number . again when a is a whole number and it is equal to num__0 then the whole number before that is a – num__1 which means num__0 – num__1 = - num__1 which is not a whole number . therefore if a is a whole number then the number before that will not always be a whole number . answer : b <eor> b <eos> |
b |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ reverse__1.0__ |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ subtract__2.0__1.0__ |
| the current of a stream runs at the rate of num__4 kmph . a boat goes num__6 km and back to the starting point in num__2 hours then find the speed of the boat in still water ? <o> a ) num__6 <o> b ) num__5 <o> c ) num__8 <o> d ) num__9 <o> e ) num__1 |
s = num__4 m = x ds = x + num__4 us = x - num__4 num__6 / ( x + num__4 ) + num__6 / ( x - num__4 ) = num__2 x = num__8 answer : c <eor> c <eos> |
c |
multiply__4.0__2.0__ round__8.0__ |
add__6.0__2.0__ add__6.0__2.0__ |
| reena took a loan of rs . num__1200 with simple interest for as many years as the rate of interest . if she paid rs . num__432 as interest at the end of the loan period what was the rate of interest ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__1 <o> e ) num__89 |
let rate = r % and time = r years . then ( num__1200 * r * r ) / num__100 = num__432 num__12 r num__2 = num__432 r num__2 = num__36 = > r = num__6 . answer : b <eor> b <eos> |
b |
percent__100.0__6.0__ |
percent__100.0__6.0__ |
| if num__6 men or num__8 women take num__6 days to complete some work how many days will num__8 men and num__4 women working together take to complete the work ? can anyone explain please ? the oa is num__2.25 <o> a ) num__1 . num__3.27272727273 days <o> b ) num__2 . num__2 days <o> c ) num__3 . num__2.25 days <o> d ) num__4.5 / num__2 days <o> e ) num__5.3 days |
num__6 men or num__8 women takes num__6 days to complete the work so num__6 m = num__8 w w = num__0.75 m num__1 man in num__1 day does num__1 unit of work so num__6 m in num__6 m = num__36 units of work now num__8 m and num__4 w = num__8 m + num__4 ( num__0.75 ) m = num__8 m + num__3 m = num__11 m since both set do the same work num__11 m in n days = num__36 units of work n = num__3.27272727273 days . a <eor> a <eos> |
a |
divide__6.0__8.0__ multiply__4.0__0.75__ add__8.0__3.0__ divide__36.0__11.0__ round__1.0__ |
divide__6.0__8.0__ add__2.25__0.75__ add__8.0__3.0__ divide__36.0__11.0__ round__1.0__ |
| if num__6 / w + num__6 / x = num__6 / y and wx = y then the average ( arithmetic mean ) of w and x is <o> a ) num__0.5 <o> b ) num__1 <o> c ) num__2 <o> d ) num__4 <o> e ) num__8 |
given : num__6 / w + num__6 / x = num__6 / ywx = y find : ( w + x ) / num__2 = ? num__6 ( num__1 / w + num__1 / x ) = num__6 ( num__1 / y ) - divide both sides by num__6 ( num__1 / w + num__1 / x ) = num__1 / y ( x + w ) / wx = num__1 / wx - sub ' d in y = wx x + w - num__1 = num__0 x + w = num__1 therefore ( w + x ) / num__2 = num__0.5 ans : a <eor> a <eos> |
a |
reverse__2.0__ reverse__2.0__ |
reverse__2.0__ reverse__2.0__ |
| a can do a work in num__14 days and working together a and b can do the same work in num__10 days . in what time can b alone do the work ? <o> a ) num__25 days <o> b ) num__30 days <o> c ) num__23 days <o> d ) num__35 days <o> e ) num__45 days |
work done by b in num__1 day = num__0.1 - num__0.0714285714286 = ( num__7 - num__5 ) / num__70 = num__0.0285714285714 so b alone can do the work in num__35 days . answer : d <eor> d <eos> |
d |
divide__1.0__10.0__ divide__1.0__14.0__ multiply__14.0__5.0__ subtract__0.1__0.0714__ multiply__5.0__7.0__ round__35.0__ |
divide__1.0__10.0__ divide__1.0__14.0__ divide__7.0__0.1__ subtract__0.1__0.0714__ multiply__5.0__7.0__ subtract__70.0__35.0__ |
| machine a working alone can complete a job in num__12 hours . machine a + b working alone can do the same job in num__15 hours . how long will it take machine b working at their respective constant rates to complete the job ? <o> a ) num__80 <o> b ) num__70 <o> c ) num__60 <o> d ) num__50 <o> e ) num__90 |
machines ( a ) ( b ) ( a + b ) - - - time - num__12 - ( - ) - ( num__15 ) - - - rate num__5 - - - num__1 - - num__4 - - work num__60 - - num__60 - num__60 a + b = ( num__60.0 ) = num__60 c <eor> c <eos> |
c |
subtract__5.0__1.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
subtract__5.0__1.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
| kim finds a num__3 - meter tree branch and marks it off in thirds and fifths . she then breaks the branch along all the markings and removes one piece of every distinct length . what fraction of the original branch remains ? <o> a ) num__0.4 <o> b ) num__0.6 <o> c ) num__0.533333333333 <o> d ) num__0.5 <o> e ) num__1.4 |
num__3 pieces of num__0.2 length and two piece each of num__0.0666666666667 and num__0.133333333333 lengths . removing one piece each from pieces of each kind of lengths the all that will remain will be num__2 pieces of num__0.2 i . e num__0.4 num__1 piece of num__0.0666666666667 and num__1 piece of num__0.133333333333 which gives us num__0.4 + num__0.0666666666667 + num__0.133333333333 - - - - - > num__0.6 answer is b <eor> b <eos> |
b |
divide__0.2__3.0__ subtract__0.2__0.0667__ multiply__2.0__0.2__ subtract__3.0__2.0__ km_to_mile_conversion__ km_to_mile_conversion__ |
divide__0.2__3.0__ subtract__0.2__0.0667__ multiply__2.0__0.2__ subtract__3.0__2.0__ subtract__1.0__0.4__ subtract__1.0__0.4__ |
| if [ y ] denotes the least integer greater than or equal to y and [ y ] = num__0 which of the following statements must be true ? <o> a ) y = num__0 <o> b ) num__0 < = y < num__1 <o> c ) num__0 < y < = num__1 <o> d ) - num__1 < y < = num__0 <o> e ) - num__1 < = y < num__0 |
the answer for this is num__0 < = y < = num__1 however it is not available in the options . the match to the above is - num__1 < y < = num__0 answer = d <eor> d <eos> |
d |
reverse__1.0__ |
reverse__1.0__ |
| an industrial loom weaves num__0.128 metres of cloth every second . approximately how many seconds will it take for the loom to weave num__28 metre of cloth ? <o> a ) num__175 seconds <o> b ) num__219 seconds <o> c ) num__155 seconds <o> d ) num__115 seconds <o> e ) num__115 seconds |
explanation : let the time required by x seconds . then more cloth means more time ( direct proportion ) so num__0.128 : num__1 : : num__28 : x = > x = { \ color { blue } \ frac { num__28 \ times num__1 } { num__0.128 } } = > x = num__218.75 so time will be approx num__219 seconds answer : b <eor> b <eos> |
b |
divide__28.0__0.128__ round__219.0__ |
divide__28.0__0.128__ round__219.0__ |
| the value of x + x ( x ^ x ) when x = num__5 is <o> a ) num__15630 <o> b ) num__14000 <o> c ) num__15400 <o> d ) num__15560 <o> e ) num__15605 |
solution : x + x ( x ^ x ) put the value of x = num__5 in the above expression we get num__5 + num__5 ( num__5 ^ num__5 ) = num__5 + num__5 ( num__5 Ã — num__5 x num__5 x num__5 x num__5 ) = num__5 + num__2 ( num__3125 ) = num__5 + num__15625 = num__15630 answer : ( a ) <eor> a <eos> |
a |
multiply__5.0__3125.0__ add__5.0__15625.0__ add__5.0__15625.0__ |
multiply__5.0__3125.0__ add__5.0__15625.0__ add__5.0__15625.0__ |
| find the missing number in the given sequence : num__15 num__913 ? num__2125 ? <o> a ) num__14 & num__26 <o> b ) num__15 & num__27 <o> c ) num__16 & num__28 <o> d ) num__17 & num__29 <o> e ) num__15 & num__30 |
num__1 + num__4 = num__5 num__5 + num__4 = num__9 num__9 + num__4 = num__13 num__13 + num__4 = num__17 num__17 + num__4 = num__21 num__21 + num__4 = num__25 num__25 + num__4 = num__29 answer : d <eor> d <eos> |
d |
add__1.0__4.0__ add__4.0__5.0__ add__4.0__9.0__ add__4.0__13.0__ add__4.0__17.0__ add__4.0__21.0__ add__4.0__25.0__ multiply__1.0__17.0__ |
add__1.0__4.0__ add__4.0__5.0__ add__4.0__9.0__ add__4.0__13.0__ add__4.0__17.0__ add__4.0__21.0__ add__4.0__25.0__ add__4.0__13.0__ |
| a person crosses a num__600 m long street in num__5 minutes . what is his speed in km per hour ? <o> a ) num__3.6 <o> b ) num__7.2 <o> c ) num__8.4 <o> d ) num__10 <o> e ) num__12 |
explanation : speed = ( num__120.0 x num__60 ) m / sec = num__2 m / sec . converting m / sec to km / hr = ( num__2 x num__3.6 ) km / hr = num__7.2 km / hr . answer is b <eor> b <eos> |
b |
divide__600.0__5.0__ hour_to_min_conversion__ divide__120.0__60.0__ multiply__2.0__3.6__ round__7.2__ |
divide__600.0__5.0__ hour_to_min_conversion__ divide__120.0__60.0__ multiply__2.0__3.6__ round__7.2__ |
| how much pure alcohol should be added to num__400 ml of a num__15.0 solution to make the strength of solution num__26.0 ? <o> a ) num__100 ml <o> b ) num__60 ml <o> c ) num__120 ml <o> d ) num__130 ml <o> e ) num__150 ml |
we can also go by answer choices tke num__100 ml for eg num__400 ( old ) + num__100 ( new concentr ) ml num__500 * num__0.26 = num__130 ml ( num__30 ml is de old concentration + num__100 ml ( newly added ) answer d <eor> d <eos> |
d |
add__400.0__100.0__ divide__26.0__100.0__ multiply__0.26__500.0__ subtract__130.0__100.0__ add__100.0__30.0__ |
add__400.0__100.0__ divide__26.0__100.0__ multiply__0.26__500.0__ subtract__130.0__100.0__ add__100.0__30.0__ |
| a room of num__6 m num__00 cm long and num__8 m num__00 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor . <o> a ) num__12 <o> b ) num__200 <o> c ) num__600 <o> d ) num__800 <o> e ) num__20 |
explanation : area of the room = num__600 * num__800 sq cm size of largest square tile = h . c . f of num__600 cm and num__800 cm = num__200 cm area of num__1 tile = num__200 * num__200 sq cm no . of tiles required = ( num__600 * num__800 ) / ( num__200 * num__200 ) = num__12 answer : a ) num__12 <eor> a <eos> |
a |
subtract__800.0__600.0__ round__12.0__ |
subtract__800.0__600.0__ divide__12.0__1.0__ |
| a num__21.0 stock yielding num__10.0 is quoted at : <o> a ) num__83.33 <o> b ) num__210 <o> c ) num__112 <o> d ) num__120 <o> e ) num__160 |
solution to earn rs . num__10 money invested = rs . num__100 . to earn rs . num__21 money invested = rs . ( num__10.0 x num__21 ) = rs . num__210 . â ˆ ´ market value of rs . num__100 stock = rs . num__210 answer b <eor> b <eos> |
b |
percent__100.0__210.0__ |
percent__100.0__210.0__ |
| five years ago the ratio of ages of p and q was num__3 : num__2 . ten years from now the ratio of their ages will be num__6 : num__5 . what is p ' s age at present ? <o> a ) num__20 <o> b ) num__25 <o> c ) num__30 <o> d ) num__35 <o> e ) num__40 |
num__2 ( p - num__5 ) / num__3 = ( q - num__5 ) . then q = num__2 ( p - num__5 ) / num__3 + num__5 num__5 ( p + num__10 ) / num__6 = q + num__10 num__5 ( p + num__10 ) / num__6 = num__2 ( p - num__5 ) / num__3 + num__15 num__5 p + num__50 = num__4 p + num__70 p = num__20 the answer is a . <eor> a <eos> |
a |
multiply__2.0__5.0__ multiply__3.0__5.0__ multiply__5.0__10.0__ subtract__6.0__2.0__ multiply__2.0__10.0__ multiply__2.0__10.0__ |
multiply__2.0__5.0__ add__5.0__10.0__ multiply__5.0__10.0__ subtract__6.0__2.0__ add__5.0__15.0__ add__5.0__15.0__ |
| two persons start running simultaneously around a circular track of length num__500 m from the same point at speeds of num__16 kmph and num__20 kmph . when will they meet for the first time any where on the track if they are moving in the opposite direction ? <o> a ) num__144 <o> b ) num__36 <o> c ) num__50 <o> d ) num__32 <o> e ) num__38 |
time taken to meet the first time = length of track / relative speed = num__500 / ( num__16 + num__20 ) ( num__0.277777777778 ) = num__13.8888888889 * ( num__3.6 ) = num__50 sec . answer : c <eor> c <eos> |
c |
multiply__3.6__13.8889__ round__50.0__ |
multiply__3.6__13.8889__ multiply__3.6__13.8889__ |
| if tier is written as num__7168 and brain is written as num__28415 how is rent coded ? <o> a ) num__3653 <o> b ) num__8657 <o> c ) num__2977 <o> d ) num__2790 <o> e ) num__2711 |
explanation : given : letter : t i e r b a n code : num__7 num__1 num__6 num__8 num__2 num__4 num__5 thus the code for rent is num__8657 . answer : b <eor> b <eos> |
b |
subtract__7.0__1.0__ add__1.0__7.0__ subtract__8.0__6.0__ subtract__6.0__2.0__ add__1.0__4.0__ multiply__1.0__8657.0__ |
subtract__7.0__1.0__ add__1.0__7.0__ subtract__8.0__6.0__ subtract__6.0__2.0__ add__1.0__4.0__ multiply__1.0__8657.0__ |
| a person want to give his money of $ num__4500 to his num__4 children a b c d in the ratio num__2 : num__4 : num__5 : num__4 . what is the a + b share ? <o> a ) $ num__1000 <o> b ) $ num__2000 <o> c ) $ num__1800 <o> d ) $ num__1500 <o> e ) $ num__1600 |
a ' s share = num__4500 * num__0.133333333333 = $ num__600 b ' s share = num__4500 * num__0.266666666667 = $ num__1200 a + b = $ num__1800 answer is c <eor> c <eos> |
c |
multiply__2.0__600.0__ add__1200.0__600.0__ add__1200.0__600.0__ |
multiply__2.0__600.0__ add__1200.0__600.0__ add__1200.0__600.0__ |
| a man â € ™ s current age is ( num__0.4 ) of the age of his father . after num__14 years he will be ( num__0.5 ) of the age of his father . what is the age of father at now ? <o> a ) num__40 <o> b ) num__45 <o> c ) num__38 <o> d ) num__50 <o> e ) num__70 |
let father â € ™ s current age is a years . then man â € ™ s current age = [ ( num__0.4 ) a ] years . therefore [ ( num__0.4 ) a + num__14 ] = ( num__0.5 ) ( a + num__14 ) num__2 ( num__2 a + num__70 ) = num__5 ( a + num__14 ) a = num__70 e <eor> e <eos> |
e |
reverse__0.5__ divide__2.0__0.4__ multiply__14.0__5.0__ |
reverse__0.5__ divide__2.0__0.4__ multiply__14.0__5.0__ |
| two students appeared at an examination . one of them secured num__9 marks more than the other and his marks was num__56.0 of the sum of their marks . what are the marks obtained by them ? <o> a ) num__42 and num__33 <o> b ) num__42 num__36 <o> c ) num__44 num__33 <o> d ) num__44 num__36 <o> e ) none of these |
explanation : let the marks secured by them be x and ( x + num__9 ) then sum of their marks = x + ( x + num__9 ) = num__2 x + num__9 given that ( x + num__9 ) was num__56.0 of the sum of their marks = > ( x + num__9 ) = ( num__0.56 ) ( num__2 x + num__9 ) = > ( x + num__9 ) = ( num__0.56 ) ( num__2 x + num__9 ) = > num__25 x + num__225 = num__28 x + num__126 = > num__3 x = num__99 = > x = num__33 then ( x + num__9 ) = num__33 + num__9 = num__42 hence their marks are num__33 and num__42 answer : option a <eor> a <eos> |
a |
multiply__9.0__25.0__ divide__56.0__2.0__ multiply__225.0__0.56__ subtract__28.0__25.0__ subtract__225.0__126.0__ divide__99.0__3.0__ add__9.0__33.0__ add__9.0__33.0__ |
multiply__9.0__25.0__ divide__56.0__2.0__ multiply__225.0__0.56__ subtract__28.0__25.0__ subtract__225.0__126.0__ divide__99.0__3.0__ add__9.0__33.0__ add__9.0__33.0__ |
| the digit in unit ' s place of the product ( num__459 × num__46 × num__28 × num__484 ) is num__2 the digit in place of * is <o> a ) num__3 <o> b ) num__5 <o> c ) num__7 <o> d ) num__9 <o> e ) num__11 |
solution ( num__9 × num__6 × num__4 ) = num__216 . in order to obtain num__2 at the unit place we must multiply num__216 by num__2 or num__7 . answer c <eor> c <eos> |
c |
subtract__6.0__2.0__ divide__28.0__4.0__ divide__28.0__4.0__ |
subtract__6.0__2.0__ divide__28.0__4.0__ divide__28.0__4.0__ |
| two trains are running in opposite directions with the same speed . if the length of each train is num__300 m and they cross each other in num__12 sec then the speed of each train is ? <o> a ) num__78 <o> b ) num__90 <o> c ) num__36 <o> d ) num__34 <o> e ) num__23 |
let the speed of each train be x m / sec . then relative speed of the two trains = num__2 x m / sec . so num__2 x = ( num__300 + num__300 ) / num__12 = > x = num__25 speed of each train = num__25 m / sec . = num__25 * num__3.6 = num__90 km / hr . answer : option b <eor> b <eos> |
b |
divide__300.0__12.0__ multiply__3.6__25.0__ round__90.0__ |
divide__300.0__12.0__ multiply__3.6__25.0__ multiply__3.6__25.0__ |
| the sum an integer n and its reciprocal is equal to num__5.2 . what is the value of n ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__15 <o> d ) num__20 <o> e ) num__25 |
write equation in n as follows n + num__1 / n = num__5.2 multiply all terms by n obtain a quadratic equation and solve to obtain n = num__5 . correct answer a <eor> a <eos> |
a |
round_down__5.2__ round_down__5.2__ |
round_down__5.2__ divide__5.0__1.0__ |
| which of the following no must added to num__5678 to give a remainder of num__35 when divided by num__460 ? <o> a ) num__955 <o> b ) num__797 <o> c ) num__618 <o> d ) num__980 <o> e ) num__680 |
to get a remainder of num__35 while dividing by num__460 the number should have num__5 at the unit digit . num__5678 ends with num__8 . so amongst the options the number which when added with num__5678 gives num__5 at the unit place will be the ans . here num__797 + num__5678 = num__6475 so ( b ) is the correct option . answer : b <eor> b <eos> |
b |
add__5678.0__797.0__ subtract__6475.0__5678.0__ |
add__5678.0__797.0__ subtract__6475.0__5678.0__ |
| a diagonal of a polygon is an segment between two non - adjacent vertices of the polygon . how many diagonals does a regular num__45 - sided polygon have ? <o> a ) num__875 <o> b ) num__945 <o> c ) num__1425 <o> d ) num__2025 <o> e ) num__2500 |
there ' s a direct formula for this . number of diagonals in a regular polygon = [ n * ( n - num__3 ) ] / num__2 n = number of sides of the regular polygon . here n = num__45 . plugging it in we get num__945 diagonals ! answer ( b ) . <eor> b <eos> |
b |
triangle_area__945.0__2.0__ |
triangle_area__945.0__2.0__ |
| a shipment of num__8 tv sets contains num__3 black and white sets and num__5 color sets . if num__2 tv sets are to be chosen at random from this shipment what is the probability that at least num__1 of the num__2 sets chosen will be a black and white set ? <o> a ) num__0.142857142857 <o> b ) num__0.25 <o> c ) num__0.642857142857 <o> d ) num__0.392857142857 <o> e ) num__0.464285714286 |
num__0.625 * num__0.571428571429 = num__0.357142857143 num__1 - ( num__0.357142857143 ) = num__0.642857142857 = num__0.642857142857 answer c <eor> c <eos> |
c |
negate_prob__0.3571__ negate_prob__0.3571__ |
negate_prob__0.3571__ negate_prob__0.3571__ |
| tough and tricky questions : statistics . set x consists of prime numbers { num__3 num__11 num__7 f num__17 num__19 } . if integer y represents the product of all elements in set x and if num__11 y is an even number what is the range of set x ? <o> a ) num__14 <o> b ) num__16 <o> c ) num__17 <o> d ) num__20 <o> e ) num__26 |
since num__11 y = even therefore y has to beevensince num__11 is a odd integer ( even * odd = even ) similarly y is the product of all integers in set x but all integers in set x are odd except the unknown f and since x contains only prime numbers f has to equal to num__2 . . . ( num__2 is the only even prime number and the product of all prime numbers in set x has to be even even * odd = even ) since you know value of f you can calculate the range = largest integer in the set minus smallest integer in the set = num__19 - num__2 = num__17 answer is c <eor> c <eos> |
c |
subtract__19.0__17.0__ subtract__19.0__2.0__ |
subtract__19.0__17.0__ subtract__19.0__2.0__ |
| dylan is on a road trip . he planned on arriving at his destination num__2150 miles away in num__6 days . in the first day he traveled num__310 miles . in the next three days he travels num__1140 miles . how many miles on average does dylan need to travel in the next two days to reach his destination ? <o> a ) num__330 <o> b ) num__340 <o> c ) num__350 <o> d ) num__360 <o> e ) num__370 |
he still has to travel num__2150 - num__300 - num__1130 = num__720 miles num__720 miles / num__2 days = num__360 miles per day . correct answer d <eor> d <eos> |
d |
divide__720.0__2.0__ divide__720.0__2.0__ |
divide__720.0__2.0__ divide__720.0__2.0__ |
| man is num__36 years older than his son . in two years his age will be twice the age of his son . the present age of his son is : <o> a ) num__14 years <o> b ) num__26 years <o> c ) num__17 years <o> d ) num__34 years <o> e ) num__22 years |
let the son ' s present age be x years . then man ' s present age = ( x + num__20 ) years . ( x + num__36 ) + num__2 = num__2 ( x + num__2 ) x + num__38 = num__2 x + num__4 x = num__34 . answer : d <eor> d <eos> |
d |
add__36.0__2.0__ subtract__36.0__2.0__ subtract__36.0__2.0__ |
add__36.0__2.0__ subtract__36.0__2.0__ subtract__36.0__2.0__ |
| if n is a positive integer and n ^ num__2 is divisible by num__212 then the largest positive integer that must divide n is <o> a ) num__6 <o> b ) num__12 <o> c ) num__24 <o> d ) num__36 <o> e ) num__48 |
the question asks aboutthe largest positive integer that must divide n not could divide n . since the least value of n for which n ^ num__2 is a multiple of num__72 is num__12 then the largest positive integer that must divide n is num__12 . complete solution of this question is given above . please ask if anything remains unclear . i spent a few hours on this one alone and i ' m still not clear . i chose num__12 at first but then changed to num__48 . i ' m not a native speaker so here is how i interpreted this question : the largest positive integer that must divide n = the largest positive factor of n . since n is a variable ( i . e . n is moving ) so is its largest factor . please correct if i ' m wrong here . i know that if n = num__12 n ^ num__2 = num__144 = num__2 * num__72 ( satisfy the condition ) . when n = num__12 the largest factor of n is n itself which is num__12 . check : num__12 is the largest positive number that must divide num__12 - - > true however if n = num__48 n ^ num__2 = num__48 * num__48 = num__32 * num__72 ( satisfy the condition too ) . when n = num__48 the largest factor of n is n itself which is num__48 . check : num__48 is the largest positive number that must divide num__48 - - > true so i also notice that the keyword ismust notcould . the question is why is num__48 notmust divide num__48 but instead onlycould divide num__48 ? i ' m not clear right here . why is num__12 must divide num__12 ? what ' s the difference between them ? only restriction we have on positive integer n is that n ^ num__2 is divisible by num__72 . the least value of n for which n ^ num__2 is divisible by num__72 is num__12 thus nmustbe divisible by num__12 ( n is in any case divisible by num__12 ) . for all other values of n for which n ^ num__2 is divisible by num__72 n will still be divisible by num__12 . this means that n is always divisible by num__12 if n ^ num__2 is divisible by num__72 . now ask yourself : if n = num__36 is n divisible by num__48 ? no . so n is not always divisible by num__48 . d <eor> d <eos> |
d |
multiply__2.0__72.0__ divide__72.0__2.0__ divide__72.0__2.0__ |
multiply__2.0__72.0__ subtract__48.0__12.0__ subtract__72.0__36.0__ |
| in a num__280 meters race a beats b by num__56 m or num__7 seconds . a ' s time over the course is : <o> a ) num__22 seconds <o> b ) num__12 seconds <o> c ) num__10 seconds <o> d ) num__18 seconds <o> e ) num__28 seconds |
b runs num__56 m in num__7 sec . = > b runs num__280 m in num__0.125 * num__280 = num__35 seconds since a beats b by num__7 seconds a runs num__280 m in ( num__35 - num__7 ) = num__28 seconds hence a ' s time over the course = num__28 seconds answer : e <eor> e <eos> |
e |
divide__7.0__56.0__ multiply__280.0__0.125__ subtract__35.0__7.0__ round__28.0__ |
divide__7.0__56.0__ multiply__280.0__0.125__ subtract__35.0__7.0__ subtract__56.0__28.0__ |
| there are two numbers . if num__20.0 of the first number is added to the second number then the second number increases to its five - fourth . find the ratio of the first number to the second number ? <o> a ) num__1.25 <o> b ) num__0.714285714286 <o> c ) num__1.66666666667 <o> d ) num__0.625 <o> e ) num__5.0 |
let the two numbers be x and y . num__0.2 * x + y = num__1.25 y = > num__0.2 x = num__0.25 y = > x / y = num__1.25 answer : a <eor> a <eos> |
a |
multiply__1.25__0.2__ divide__0.25__0.2__ |
multiply__1.25__0.2__ divide__0.25__0.2__ |
| the speeds of num__3 cars in the ratio num__2 : num__3 : num__4 . the ratio between time taken by them to travel the same distance is ? <o> a ) num__2 : num__3 : num__4 <o> b ) num__6 : num__4 : num__3 <o> c ) num__8 : num__7 : num__5 <o> d ) num__11 : num__12 : num__15 <o> e ) num__9 : num__2 : num__5 |
ratio of time taken is = num__0.5 : num__0.333333333333 : num__0.25 = num__6 : num__4 : num__3 answer is b <eor> b <eos> |
b |
divide__2.0__4.0__ divide__0.5__2.0__ multiply__3.0__2.0__ round__6.0__ |
divide__2.0__4.0__ divide__0.5__2.0__ multiply__3.0__2.0__ round__6.0__ |
| what will come in place of the x in the following number series ? num__65536 num__32768 num__16384 num__8192 x <o> a ) num__4090 <o> b ) num__4098 <o> c ) num__4094 <o> d ) num__4092 <o> e ) num__4096 |
go on dividing by num__2 to the next number answer : e <eor> e <eos> |
e |
divide__65536.0__32768.0__ divide__8192.0__2.0__ |
divide__65536.0__32768.0__ divide__8192.0__2.0__ |
| a train num__100 meters long completely crosses a num__350 - meter long bridge in num__40 seconds . what is the speed of the train in km / h ? <o> a ) num__37.5 km / h <o> b ) num__40.5 km / h <o> c ) num__43.5 km / h <o> d ) num__46.5 km / h <o> e ) num__49.5 km / h |
speed = distance / time = ( num__100 + num__350 ) / num__40 = num__11.25 ( m / s ) * num__3.6 = num__40.5 km / h the answer is b . <eor> b <eos> |
b |
multiply__11.25__3.6__ round__40.5__ |
multiply__11.25__3.6__ multiply__11.25__3.6__ |
| the average age of num__3 boys is num__15 years and their ages are in the proportion num__1 : num__2 : num__3 . the age of the youngest boy is ? <o> a ) num__10 years <o> b ) num__12 years <o> c ) num__15 years <o> d ) num__18 years <o> e ) num__7.5 years |
total age of num__3 boys = num__15 * num__3 = num__45 ratio of their ages = num__1 : num__2 : num__3 age of the youngest = num__45 * num__0.166666666667 = num__7.5 years answer is e <eor> e <eos> |
e |
multiply__3.0__15.0__ divide__15.0__2.0__ divide__15.0__2.0__ |
multiply__3.0__15.0__ divide__15.0__2.0__ divide__15.0__2.0__ |
| rohan ranked eleventh from the top and twenty - seventh from the bottom among the students who passed the annual examination in a class . if the number of students who failed in the examination was num__12 how many students appeared for the examination ? <o> a ) num__48 <o> b ) num__49 <o> c ) num__50 <o> d ) can not be determined <o> e ) none of these |
here . . from top its num__11 th and from bottom its num__27 th that means at top num__10 members are there and bottom num__26 members are there so num__26 + num__10 + num__1 = num__37 . num__37 memebers are passed in above question num__12 members are failed in examination so therefore num__37 + num__12 = num__49 . answer : b <eor> b <eos> |
b |
subtract__12.0__11.0__ add__10.0__27.0__ add__12.0__37.0__ add__12.0__37.0__ |
subtract__12.0__11.0__ add__10.0__27.0__ add__12.0__37.0__ add__12.0__37.0__ |
| in a weight - lifting competition the total weight of joe ' s two lifts was num__600 pounds . if twice the weight of his first lift was num__300 pounds more than the weight of his second lift what was the weight in pounds of his first lift ? <o> a ) num__225 <o> b ) num__275 <o> c ) num__300 <o> d ) num__350 <o> e ) num__400 |
this problem is a general word translation . we first define variables and then set up equations . we can define the following variables : f = the weight of the first lift s = the weight of the second lift we are given that the total weight of joe ' s two lifts was num__600 pounds . we sum the two variables to obtain : f + s = num__600 we are also given that twice the weight of his first lift was num__300 pounds more than the weight of his second lift . we express this as : num__2 f = num__300 + s num__2 f – num__300 = s we can now plug in ( num__2 f – num__300 ) for s into the first equation so we have : f + num__2 f – num__300 = num__600 num__3 f = num__900 f = num__300 answer is c <eor> c <eos> |
c |
divide__600.0__300.0__ add__600.0__300.0__ subtract__600.0__300.0__ |
divide__600.0__300.0__ add__600.0__300.0__ subtract__600.0__300.0__ |
| if num__5 ^ num__10 x num__2 ^ num__10 = num__10 ^ n what is the value of n ? <o> a ) num__16 <o> b ) num__14 <o> c ) num__12 <o> d ) num__10 <o> e ) num__18 |
num__5 ^ num__10 * num__2 ^ num__10 = num__10 ^ n or num__10 ^ num__10 = num__10 ^ n n = num__10 d <eor> d <eos> |
d |
multiply__5.0__2.0__ |
multiply__5.0__2.0__ |
| in how many different ways can the letters of the word ' optical ' be arranged so that the vowels always come together ? <o> a ) num__620 <o> b ) num__640 <o> c ) num__690 <o> d ) num__710 <o> e ) num__720 |
the word ' optical ' has num__7 letters . it has the vowels ' o ' ' i ' ' a ' in it and these num__3 vowels should always come together . hence these three vowels can be grouped and considered as a single letter . that is ptcl ( oia ) . hence we can assume total letters as num__5 and all these letters are different . number of ways to arrange these letters = num__5 ! = num__5 × num__4 × num__3 × num__2 × num__1 = num__120 = num__5 ! = num__5 × num__4 × num__3 × num__2 × num__1 = num__120 all the num__3 vowels ( oia ) are different number of ways to arrange these vowels among themselves = num__3 ! = num__3 × num__2 × num__1 = num__6 = num__3 ! = num__3 × num__2 × num__1 = num__6 hence required number of ways = num__120 × num__6 = num__720 e ) <eor> e <eos> |
e |
vowel_space__ coin_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
vowel_space__ coin_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| pipe a can fill a tank in num__7 hours pipe b in num__14 hours and pipe c in num__42 hours . if all the pipes are open in how many hours will the tank be filled ? <o> a ) num__4 <o> b ) num__4.2 <o> c ) num__4.5 <o> d ) num__4.3 <o> e ) num__4.1 |
num__0.142857142857 + num__0.0714285714286 + num__0.0238095238095 = num__0.238095238095 = num__1 / num__4.2 . so num__4.2 hrs answer : b <eor> b <eos> |
b |
round__4.2__ |
divide__4.2__1.0__ |
| a num__1200 m long train crosses a tree in num__120 sec how much time will i take to pass a platform num__700 m long ? <o> a ) num__288 <o> b ) num__190 <o> c ) num__188 <o> d ) num__188 <o> e ) num__12 |
l = s * t s = num__10.0 s = num__10 m / sec . total length ( d ) = num__1900 m t = d / s t = num__190.0 t = num__190 sec answer : b <eor> b <eos> |
b |
divide__1200.0__120.0__ add__1200.0__700.0__ divide__1900.0__10.0__ round__190.0__ |
divide__1200.0__120.0__ add__1200.0__700.0__ divide__1900.0__10.0__ divide__1900.0__10.0__ |
| the wheels revolve round a common horizontal axis . they make num__15 num__20 and num__48 revolutions in a minute respectively . starting with a certain point on the circumference down wards . after what interval of time will they come together in the same position ? <o> a ) num__1 min <o> b ) num__2 min <o> c ) num__3 min <o> d ) num__4 min <o> e ) none |
time for one revolution = num__4.0 = num__4 num__3.0 = num__3 num__1.25 = num__1.25 lcm of num__4 num__3 num__1.25 lcm of numerators / hcf of denominators = num__60.0 = num__60 answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ subtract__4.0__3.0__ |
hour_to_min_conversion__ subtract__4.0__3.0__ |
| the hcf and lcm of two numbers m and n are respectively num__6 and num__210 . if m + n = num__60 then num__1 / m + num__1 / n is equal to <o> a ) num__0.0285714285714 <o> b ) num__0.0857142857143 <o> c ) num__0.047619047619 <o> d ) num__0.0571428571429 <o> e ) none |
answer we have m x n = num__6 x num__210 = num__1260 â ˆ ´ num__1 / m + num__1 / n = ( m + n ) / mn = num__0.047619047619 = num__0.047619047619 correct option : c <eor> c <eos> |
c |
multiply__6.0__210.0__ divide__60.0__1260.0__ divide__60.0__1260.0__ |
multiply__6.0__210.0__ divide__60.0__1260.0__ divide__60.0__1260.0__ |
| if x = – num__2 what is the value of num__4 x ^ num__3 ? <o> a ) – num__32 <o> b ) – num__24 <o> c ) – num__8 <o> d ) num__8 <o> e ) num__32 |
num__4 x ^ num__3 = num__4 ( − num__2 ) num__34 ( − num__2 ) num__3 = num__4 * - num__8 = - num__32 . answer is - num__32 answer : a <eor> a <eos> |
a |
multiply__2.0__4.0__ multiply__4.0__8.0__ multiply__4.0__8.0__ |
multiply__2.0__4.0__ multiply__4.0__8.0__ multiply__4.0__8.0__ |
| a table is bought for rs . num__800 / - and sold at rs . num__600 / - find gain or loss percentage <o> a ) num__25.0 gain <o> b ) num__20.0 loss <o> c ) num__20.0 gain <o> d ) num__25.0 loss <o> e ) none |
formula = ( selling price ~ cost price ) / cost price * num__100 = ( num__600 - num__800 ) / num__800 = num__25.0 loss d <eor> d <eos> |
d |
percent__100.0__25.0__ |
percent__100.0__25.0__ |
| two numbers are less than third number by num__34.0 and num__37.0 respectively . how much percent is the second number less than by the first <o> a ) num__8.0 <o> b ) num__10.0 <o> c ) num__9.0 <o> d ) num__11.0 <o> e ) num__22 % |
let the third number is x . then first number = ( num__100 - num__34 ) % of x = num__66.0 of x = num__66 x / num__100 second number is ( num__63 x / num__100 ) difference = num__66 x / num__100 - num__63 x / num__100 = num__3 x / num__100 so required percentage is difference is what percent of first number ( num__3 x / num__100 * num__1.5873015873 x * num__100 ) % = num__22.0 answer : e <eor> e <eos> |
e |
percent__100.0__22.0__ |
percent__100.0__22.0__ |
| the average of first num__10 prime numbers is ? <o> a ) num__12.9 <o> b ) num__12.6 <o> c ) num__12.3 <o> d ) num__512.4 <o> e ) num__12.1 |
sum of num__10 prime no . = num__129 average = num__12.9 = num__12.9 answer : a <eor> a <eos> |
a |
divide__129.0__10.0__ divide__129.0__10.0__ |
divide__129.0__10.0__ divide__129.0__10.0__ |
| the speed of a boat in still water is num__12 kmph . what is the speed of the stream if the boat can cover num__32 km downstream or num__16 km upstream in the same time ? <o> a ) num__5 kmph <o> b ) num__6 kmph <o> c ) num__7 kmph <o> d ) num__4 kmph <o> e ) num__10 kmph |
x = the speed of the stream ( num__12 + x ) / ( num__12 - x ) = num__2.0 num__12 + x = num__24 - num__2 x num__3 x = num__12 x = num__4 km / hour if the speed of the stream is num__4 km / hour then the ' downstream ' speed of the boat is num__12 + num__4 = num__16 km / hour and the ' upstream ' speed of the boat is num__12 - num__4 = num__8 km / hour . in that way if the boat traveled for num__2 hours it would travel num__2 x num__8 = num__16 km downstream and num__2 x num__16 = num__32 km / hour upstream . answer : d <eor> d <eos> |
d |
divide__32.0__16.0__ multiply__12.0__2.0__ divide__12.0__3.0__ subtract__12.0__4.0__ round__4.0__ |
divide__32.0__16.0__ multiply__12.0__2.0__ divide__12.0__3.0__ subtract__12.0__4.0__ divide__12.0__3.0__ |
| calculate the value of num__8.14 x num__10 ^ num__3 ? <o> a ) num__81400000 <o> b ) num__8140000 <o> c ) num__814000 <o> d ) num__8140 <o> e ) num__814 |
num__8.14 x num__10 ^ num__6 = num__8.14 x num__1000 = num__8140 . answer = d <eor> d <eos> |
d |
multiply__8.14__1000.0__ multiply__8.14__1000.0__ |
multiply__8.14__1000.0__ multiply__8.14__1000.0__ |
| the ratio of two numbers is num__4 : num__5 and their h . c . f . is num__4 . their l . c . m . is <o> a ) num__48 <o> b ) num__22 <o> c ) num__80 <o> d ) num__27 <o> e ) num__35 |
explanation : let the numbers be num__4 x and num__5 x . then their h . c . f . = x . so x = num__4 . so the numbers num__16 and num__20 . l . c . m . of num__16 and num__20 = num__80 . answer : option c <eor> c <eos> |
c |
multiply__4.0__5.0__ multiply__4.0__20.0__ multiply__4.0__20.0__ |
multiply__4.0__5.0__ multiply__4.0__20.0__ multiply__4.0__20.0__ |
| a certain debt will be paid in num__52 installments from january num__1 to december num__31 of a certain year . each of the first num__12 payments is to be $ num__410 ; each of the remaining payments is to be $ num__65 more than each of the first num__12 payments . what is the average ( arithmetic mean ) payment that will be made on the debt for the year ? <o> a ) num__443 <o> b ) num__450 <o> c ) num__460 <o> d ) num__468 <o> e ) num__475 |
total number of installments = num__52 payment per installment for the first num__12 installments = num__410 payment per installment for the remaining num__32 installments = num__410 + num__65 = num__475 average = ( num__12 * num__410 + num__40 * num__475 ) / num__52 = num__460 answer c <eor> c <eos> |
c |
add__1.0__31.0__ add__410.0__65.0__ subtract__52.0__12.0__ multiply__1.0__460.0__ |
add__1.0__31.0__ add__410.0__65.0__ subtract__52.0__12.0__ multiply__1.0__460.0__ |
| when a certain tree was first planted it was num__4 feet tall and the height of the tree increased by a constant amount each year for the next num__6 years . at the end of the num__6 th year the tree was num__0.25 taller than it was at the end of the num__4 th year . by how many feet did the height of the tree increase each year ? <o> a ) num__1 <o> b ) num__0.4 <o> c ) num__0.5 <o> d ) num__0.666666666667 <o> e ) num__1.2 |
say the tree grows by x feet every year . then num__4 + num__6 x = ( num__1 + num__0.25 ) ( num__4 + num__4 x ) or x = num__1 answer a <eor> a <eos> |
a |
multiply__4.0__0.25__ reverse__1.0__ |
multiply__4.0__0.25__ reverse__1.0__ |
| if y > num__0 ( num__1 y ) / num__20 + ( num__3 y ) / num__10 is what percent of y ? <o> a ) num__35.0 <o> b ) num__50.0 <o> c ) num__60.0 <o> d ) num__70.0 <o> e ) num__80 % |
can be reduced to y / num__20 + num__3 y / num__10 = num__7 y / num__20 = num__35.0 a <eor> a <eos> |
a |
subtract__10.0__3.0__ multiply__1.0__35.0__ |
subtract__10.0__3.0__ divide__35.0__1.0__ |
| this topic is locked . if you want to discuss this question please re - post it in the respective forum . matt and peter can do together a piece of work in num__20 days . after they have worked together for num__12 days matt stops and peter completes the remaining work in num__14 days . in how many days peter complete the work separately . <o> a ) num__26 days <o> b ) num__27 days <o> c ) num__23 days <o> d ) num__35 days <o> e ) num__24 days |
together they complete the job in num__20 days means they complete num__0.6 of the job after num__12 days . peter completes the remaining ( num__0.4 ) of the job in num__14 days which means that the whole job ( num__1 ) can be completed in x days . < = > num__0.4 - > num__14 < = > x = num__14 / ( num__0.4 ) = num__35 thus the answer is d . <eor> d <eos> |
d |
km_to_mile_conversion__ add__0.4__0.6__ divide__14.0__0.4__ round__35.0__ |
divide__12.0__20.0__ add__0.4__0.6__ divide__14.0__0.4__ divide__14.0__0.4__ |
| the average weight of num__6 person ' s increases by num__4.5 kg when a new person comes in place of one of them weighing num__75 kg . what might be the weight of the new person ? <o> a ) num__160 kg <o> b ) num__175 kg <o> c ) num__180 kg <o> d ) num__102 kg <o> e ) num__190 kg |
total weight increased = ( num__6 x num__4.5 ) kg = num__27 kg . weight of new person = ( num__75 + num__27 ) kg = num__102 kg option d <eor> d <eos> |
d |
multiply__6.0__4.5__ add__75.0__27.0__ add__75.0__27.0__ |
multiply__6.0__4.5__ add__75.0__27.0__ add__75.0__27.0__ |
| - num__88 × num__39 + num__312 = ? <o> a ) - num__3200 <o> b ) num__3120 <o> c ) - num__3120 <o> d ) num__3200 <o> e ) num__3208 |
- num__88 × num__39 + num__312 = - num__88 × ( num__40 - num__1 ) + num__312 = - num__88 × num__40 + num__88 + num__312 = - num__3520 + num__88 + num__312 = - num__3120 answer is c <eor> c <eos> |
c |
subtract__40.0__39.0__ multiply__88.0__40.0__ multiply__1.0__3120.0__ |
subtract__40.0__39.0__ multiply__88.0__40.0__ multiply__1.0__3120.0__ |
| rob traveled a distance of num__30 km covering the first num__10 km in x minutes the next num__10 km in y minutes and the last num__10 km in z minutes . if he totally took num__3 y minutes to cover the whole distance then which of the following can not be true ? assume x y and z are different . <o> a ) z = num__3 x <o> b ) x = num__3 z <o> c ) y = num__2 x <o> d ) x = num__2 y <o> e ) y = num__3 x |
rob travelled for x y and for z min . total time : x + y + z which is equal to : num__3 y equating both sides we get x + y + z = num__3 y = > x + z = num__2 y . . . . . . . . eqn num__1 looking out at options d says x = num__2 y using it in eqn num__1 num__2 y + z = num__2 y = > z = num__0 mins which i guess is not possible . the ans isd <eor> d <eos> |
d |
subtract__3.0__2.0__ subtract__3.0__1.0__ |
subtract__3.0__2.0__ subtract__3.0__1.0__ |
| an inspector rejects num__0.08 of the meters as defective . how many will be examine to project ? <o> a ) num__5500 <o> b ) num__7500 <o> c ) num__3500 <o> d ) num__2500 <o> e ) num__4500 |
let the number of meters to be examined be x . then num__0.08 of x = num__2 [ ( num__0.08 ) * ( num__0.01 ) * x ] = num__2 x = [ ( num__2 * num__100 * num__100 ) / num__8 ] = num__2500 . answer is d . <eor> d <eos> |
d |
percent__100.0__2500.0__ |
percent__100.0__2500.0__ |
| what is the difference between the compound interest on rs . num__12000 at num__20.0 p . a . for one year when compounded yearly and half yearly ? <o> a ) num__337 <o> b ) num__120 <o> c ) num__266 <o> d ) num__287 <o> e ) num__612 |
when compounded annually interest = num__12000 [ num__1 + num__0.2 ] num__1 - num__12000 = rs . num__2400 when compounded semi - annually interest = num__12000 [ num__1 + num__0.1 ] num__2 - num__12000 = rs . num__2520 required difference = num__2520 - num__2400 = rs . num__120 answer : b <eor> b <eos> |
b |
percent__20.0__1.0__ percent__20.0__12000.0__ percent__1.0__12000.0__ percent__1.0__12000.0__ |
percent__20.0__1.0__ percent__20.0__12000.0__ percent__1.0__12000.0__ percent__1.0__12000.0__ |
| compound interest earned on a sum for the second and the third years are rs . num__2500 and rs . num__2625 respectively . find the rate of interest ? <o> a ) num__10.0 p . a . <o> b ) num__5.0 p . a . <o> c ) num__15.0 p . a . <o> d ) num__12.5 p . a . <o> e ) none of these . |
rs . num__2625 - num__2500 = rs . num__125 is the interest on rs . num__2500 for one year . rate of interest = ( num__0.05 ) * num__100 = num__5.0 p . a answer : b <eor> b <eos> |
b |
percent__5.0__100.0__ |
percent__5.0__100.0__ |
| the two trains of lengths num__400 m num__600 m respectively running at same directions . the faster train can cross the slower train in num__180 sec the speed of the slower train is num__48 km . then find the speed of the faster train ? <o> a ) num__60 <o> b ) num__68 <o> c ) num__67 <o> d ) num__69 <o> e ) num__54 |
length of the two trains = num__600 m + num__400 m speed of the first train = x speed of the second train = num__48 kmph num__1000 / x - num__48 = num__180 num__1000 / x - num__48 * num__0.277777777778 = num__180 num__50 = num__9 x - num__120 x = num__68 kmph answer : option b <eor> b <eos> |
b |
add__400.0__600.0__ round__68.0__ |
add__400.0__600.0__ round__68.0__ |
| jimmy has num__18.0 more cookies than janice . what percentage of cookies should jimmy give to janice so that both jimmy and janice have the same number of cookies ? <o> a ) num__7.6 <o> b ) num__8.5 <o> c ) num__9.0 <o> d ) num__11.0 <o> e ) num__12.2 % |
an easy way to solve this question is by number plugging . assume janice has num__100 cookies then jimmy has num__118 cookies . now for both jimmy and janice to have equal amounts of cookies we should give janice num__9 of jimmy ' s cookies which is num__0.0762711864407 = ~ num__7.6 of jimmy ' s cookies . answer : a . <eor> a <eos> |
a |
percent__100.0__7.6__ |
percent__100.0__7.6__ |
| the greatest common factor of num__16 and the positive integer n is num__4 and the greatest common factor of n and num__45 is num__3 . which of the following could be the greatest common factor of n and num__60 ? <o> a ) num__3 <o> b ) num__30 <o> c ) num__40 <o> d ) num__42 <o> e ) num__70 |
greatest common factor n and num__16 is num__4 = num__2 * num__2 * num__2 * num__2 greatest common factor n and num__45 is num__3 = num__3 * num__3 * num__5 greatest common factor n and num__60 is = num__2 * num__3 * num__5 = num__30 answer : b . <eor> b <eos> |
b |
add__3.0__2.0__ divide__60.0__2.0__ divide__60.0__2.0__ |
add__3.0__2.0__ divide__60.0__2.0__ divide__60.0__2.0__ |
| a car is running at a speed of num__80 kmph . what distance will it cover in num__20 sec ? <o> a ) num__100 m <o> b ) num__440 m <o> c ) num__180 m <o> d ) num__200 m <o> e ) num__250 m |
speed = num__80 kmph = num__80 * num__0.277777777778 = num__22 m / s distance covered in num__20 sec = num__22 * num__10 = num__440 m answer is b <eor> b <eos> |
b |
multiply__20.0__22.0__ round__440.0__ |
multiply__20.0__22.0__ multiply__20.0__22.0__ |
| in what time will a train num__120 m long cross an electric pole it its speed be num__121 km / hr ? <o> a ) num__2.5 <o> b ) num__2.7 <o> c ) num__2.9 <o> d ) num__3.6 <o> e ) num__2.1 |
speed = num__121 * num__0.277777777778 = num__33.6 m / sec time taken = num__120 / num__33.6 = num__3.6 sec . answer : d <eor> d <eos> |
d |
round__3.6__ |
round__3.6__ |
| average of money that group of num__4 friends pay for rent each month is $ num__800 . after one persons rent is increased by num__20.0 the new mean is $ num__870 . what was original rent of friend whose rent is increased ? <o> a ) num__800 <o> b ) num__1400 <o> c ) num__1000 <o> d ) num__1100 <o> e ) num__1200 |
num__0.2 x = num__4 ( num__870 - num__800 ) num__0.2 x = num__280 x = num__1400 answer b <eor> b <eos> |
b |
divide__4.0__20.0__ divide__280.0__0.2__ divide__280.0__0.2__ |
divide__4.0__20.0__ divide__280.0__0.2__ divide__280.0__0.2__ |
| find the num__25.0 of rs . num__800 . <o> a ) s . num__50 <o> b ) s . num__200 <o> c ) s . num__100 <o> d ) s . num__80 <o> e ) s . num__60 |
explanation : num__25.0 of num__800 = > num__0.25 * num__800 = rs . num__200 answer : b <eor> b <eos> |
b |
percent__25.0__800.0__ percent__25.0__800.0__ |
percent__25.0__800.0__ percent__25.0__800.0__ |
| find out the square of a number which when doubled exceeds its quarter by num__7 ? <o> a ) num__16 <o> b ) num__25 <o> c ) num__19 <o> d ) num__26 <o> e ) num__17 |
a let the number be p then the square will be p ^ num__2 according to question : num__2 p = ( p / num__4 ) + num__7 = > num__8 p = p + num__28 = > p = num__4 p ^ num__2 = num__4 ^ num__2 = num__16 . <eor> a <eos> |
a |
twice__2.0__ twice__4.0__ multiply__7.0__4.0__ twice__8.0__ twice__8.0__ |
twice__2.0__ twice__4.0__ multiply__7.0__4.0__ twice__8.0__ twice__8.0__ |
| in order to fence a square manish fixed num__48 poles . if the distance between two poles is num__6 metres then what will be the area of the square so formed ? <o> a ) can not be determined <o> b ) num__2600 cm num__2 <o> c ) num__5184 cm num__2 <o> d ) num__3025 cm num__2 <o> e ) none of these |
let the side of the square be x m . ∴ perimeter of the square = num__48 × num__6 = num__4 x ∴ x = num__72 m ∴ area = ( num__72 ) num__2 = num__5184 m num__2 answer c <eor> c <eos> |
c |
subtract__6.0__4.0__ round__5184.0__ |
subtract__6.0__4.0__ round__5184.0__ |
| a certain car dealership sells economy cars luxury cars and sport utility vehicles . the ratio of economy to luxury cars is num__5 : num__4 . the ratio of economy cars to sport utility vehicles is num__3 : num__2 . what is the ratio of luxury cars to sport utility vehicles ? <o> a ) num__9 : num__8 <o> b ) num__8 : num__9 <o> c ) num__6 : num__5 <o> d ) num__2 : num__3 <o> e ) num__1 : num__2 |
the ratio of economy to luxury cars is num__5 : num__4 - - > e : l = num__5 : num__4 = num__15 : num__12 . the ratio of economy cars to sport utility vehicles is num__3 : num__2 - - > e : s = num__3 : num__2 = num__15 : num__10 . thus l : s = num__12 : num__10 = num__6 : num__5 . answer : c . <eor> c <eos> |
c |
multiply__5.0__3.0__ multiply__4.0__3.0__ multiply__5.0__2.0__ add__4.0__2.0__ add__4.0__2.0__ |
multiply__5.0__3.0__ subtract__15.0__3.0__ subtract__12.0__2.0__ subtract__10.0__4.0__ subtract__10.0__4.0__ |
| a machine can finish a work in num__15 days and num__2 nd machine can finish the work in num__30 days . num__2 nd machine worked for num__10 days and left the job . in how many days a alone can finish the remaining work ? <o> a ) num__5 days <o> b ) num__6 days <o> c ) num__10 days <o> d ) num__12 days <o> e ) num__15 days |
num__2 nd machine num__10 days work = num__0.0333333333333 * num__10 = num__0.333333333333 remaining work = num__1 - num__0.333333333333 = num__0.666666666667 now num__0.0666666666667 work is done by num__1 st machine in num__1 day num__0.666666666667 work is done by num__1 machine in num__15 * num__0.666666666667 = num__10 days . answer is c <eor> c <eos> |
c |
divide__10.0__30.0__ divide__10.0__15.0__ divide__2.0__30.0__ round__10.0__ |
divide__10.0__30.0__ subtract__1.0__0.3333__ divide__2.0__30.0__ round__10.0__ |
| the average age of a group of num__5 members is num__20 years . two years later a new member joins the group . the average age of the group becomes num__21 years . what is the age of the new member ? <o> a ) num__16 years <o> b ) num__21 years <o> c ) num__22 years <o> d ) num__23 years <o> e ) num__24 years |
the average age of the num__5 members is num__20 years = > the sum of the ages of the num__5 members is num__20 * num__5 = num__100 years once the new member joins the average age becomes num__21 years . let the age of the new member be x . then ( num__110 + x ) / num__6 = num__21 = > x = num__16 years answer a <eor> a <eos> |
a |
multiply__5.0__20.0__ subtract__21.0__5.0__ subtract__21.0__5.0__ |
multiply__5.0__20.0__ subtract__21.0__5.0__ subtract__21.0__5.0__ |
| if x is an integer then x ( x - num__1 ) ( x - k ) must be evenly divisible by three when k is any of the following values except <o> a ) - num__4 <o> b ) - num__8 <o> c ) - num__1 <o> d ) num__2 <o> e ) num__5 |
if we put k = - num__1 we get : x ( x - num__1 ) ( x + num__1 ) rearrange : ( x - num__1 ) x ( x + num__1 ) so it looks like a sequenc if we assume that x = num__2 and put number from the answer then we get : ( x – num__1 ) x ( x – k ) k = num__5 = num__1.2 . - num__3 k = num__2 = num__1 . num__2.0 k = - num__1 = num__1 . num__2.3 k = - num__4 = num__1 . num__2.6 but when we put k = - num__8 = num__1 . num__2.10 not satisfied so the answer is b <eor> b <eos> |
b |
add__1.0__2.0__ add__1.0__3.0__ multiply__2.0__4.0__ multiply__1.0__8.0__ |
add__1.0__2.0__ add__1.0__3.0__ add__3.0__5.0__ add__3.0__5.0__ |
| the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square whose area is num__1225 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is num__10 units ? <o> a ) num__140 <o> b ) num__99 <o> c ) num__88 <o> d ) num__72 <o> e ) num__12 |
given that the area of the square = num__1225 sq . units = > side of square = √ num__1225 = num__35 units the radius of the circle = side of the square = num__35 units length of the rectangle = num__0.4 * num__35 = num__14 units given that breadth = num__10 units area of the rectangle = lb = num__14 * num__10 = num__140 sq . units answer : a <eor> a <eos> |
a |
multiply__35.0__0.4__ square_perimeter__35.0__ square_perimeter__35.0__ |
multiply__35.0__0.4__ multiply__10.0__14.0__ multiply__10.0__14.0__ |
| the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is num__675 sq m then what is the breadth of the rectangular plot ? <o> a ) num__11 <o> b ) num__15 <o> c ) num__18 <o> d ) num__101 <o> e ) num__1322 |
let the breadth of the plot be b m . length of the plot = num__3 b m ( num__3 b ) ( b ) = num__675 num__3 b num__2 = num__675 b num__2 = num__225 = num__15 ( b > num__0 ) b = num__15 m . answer : b <eor> b <eos> |
b |
triangle_area__2.0__15.0__ |
triangle_area__2.0__15.0__ |
| the sum an integer n and its reciprocal is equal to num__5.2 . what is the value of n ? <o> a ) num__9 <o> b ) num__8 <o> c ) num__7 <o> d ) num__6 <o> e ) num__5 |
write equation in n as follows n + num__1 / n = num__5.2 multiply all terms by n obtain a quadratic equation and solve to obtain n = num__5 correct answer e <eor> e <eos> |
e |
round_down__5.2__ round_down__5.2__ |
round_down__5.2__ divide__5.0__1.0__ |
| in one hour a boat goes num__11 km / hr along the stream and num__5 km / hr against the stream . the speed of the boat in still water ( in km / hr ) is : <o> a ) num__3 km / hr <o> b ) num__4 km / hr <o> c ) num__5 km / hr <o> d ) num__6 km / hr <o> e ) num__8 km / hr |
upstream relative speed is u + v = num__11 km / hr downstream speed is u - v = num__5 where u = speed of boat in still water and v is speed of stream then adding two equations u + v + u - v = num__11 + num__5 num__2 u = num__16 finally u = num__8 . answer : e <eor> e <eos> |
e |
add__11.0__5.0__ divide__16.0__2.0__ round__8.0__ |
add__11.0__5.0__ divide__16.0__2.0__ subtract__16.0__8.0__ |
| two airplanes take off from one airfield at noon . one flies due east at num__201 miles per hour while the other flies directly northeast at num__283 miles per hour . approximately how many miles apart are the airplanes at num__2 p . m . ? <o> a ) num__166 <o> b ) num__332 <o> c ) num__400 <o> d ) num__402 <o> e ) num__566 |
d in two hours : the plane flying east will be num__402 miles away from airport . the other plane will be num__566 miles away from airport . num__1.407960199 = ~ num__1.4 = ~ sqrt ( num__2 ) this means that planes formed a right isocheles triangle = > sides of such triangles relate as num__1 : num__1 : sqrt ( num__2 ) = > the planes are num__402 miles apart . d <eor> d <eos> |
d |
multiply__201.0__2.0__ multiply__283.0__2.0__ divide__283.0__201.0__ round__402.0__ |
multiply__201.0__2.0__ multiply__283.0__2.0__ divide__283.0__201.0__ round__402.0__ |
| a dishonest shopkeeper professes to sell pulses at the cost price but he uses a false weight of num__950 gm . for a kg . his gain is … % . <o> a ) num__5.26 <o> b ) num__5.36 <o> c ) num__4.26 <o> d ) num__6.26 <o> e ) num__7.26 % |
his percentage gain is num__100 * num__0.0526315789474 as he is gaining num__50 units for his purchase of num__950 units . so num__5.26 . answer : a <eor> a <eos> |
a |
percent__5.26__100.0__ |
percent__5.26__100.0__ |
| two trains one from howrah to patna and the other from patna to howrah start simultaneously . after they meet the trains reach their destinations after num__16 hours and num__9 hours respectively . the ratio of their speeds is ? <o> a ) num__4 : num__5 <o> b ) num__4 : num__3 <o> c ) num__4 : num__4 <o> d ) num__4 : num__9 <o> e ) num__3 : num__4 |
let us name the trains a and b . then ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ num__9 : √ num__16 = num__3 : num__4 answer : e <eor> e <eos> |
e |
round__3.0__ |
round__3.0__ |
| a man can swim in still water at num__4 km / h but takes twice as long to swim upstream than downstream . the speed of the stream is ? <o> a ) num__1.3 <o> b ) num__7.5 <o> c ) num__2.25 <o> d ) num__1.5 <o> e ) num__4 |
m = num__4 s = x ds = num__4 + x us = num__4 - x num__4 + x = ( num__4 - x ) num__2 num__4 + x = num__8 - num__2 x num__3 x = num__4 x = num__1.3 answer : a <eor> a <eos> |
a |
multiply__4.0__2.0__ round__1.3__ |
multiply__4.0__2.0__ round__1.3__ |
| a train speeds past a pole in num__15 seconds and a pla ƞ orm num__100 meter long in num__25 seconds . what is length of the train ? <o> a ) num__140 meter <o> b ) num__145 meter <o> c ) num__150 meter <o> d ) num__155 meter <o> e ) none of these |
explanation : let the length of the train is x meter and speed of the train is y meter / second then x / y = num__15 [ because distance / speed = Ɵ me ] = > y = num__15 / x = > x + num__10025 = x num__15 x = num__150 meters so length of the train is num__150 meters answer : c <eor> c <eos> |
c |
round__150.0__ |
round__150.0__ |
| from a group of num__3 boys and num__3 men four are to be randomly selected . what is the probability that equal numbers of boys and men will be selected ? <o> a ) num__0.2 <o> b ) num__0.6 <o> c ) num__0.8 <o> d ) num__0.4 <o> e ) num__0.6 |
using the first example here is the probability of that exact sequence occurring : bbmm = ( num__0.5 ) ( num__0.4 ) ( num__0.75 ) ( num__0.666666666667 ) = num__0.1 = num__0.1 each of the other num__5 options will yield the exact same probability . . . . eg bmbm = ( num__0.5 ) ( num__0.6 ) ( num__0.5 ) ( num__0.666666666667 ) = num__0.1 = num__0.1 so we have num__6 different options that each produce a num__0.1 chance of occurring . num__6 ( num__0.1 ) = num__0.6 = num__0.6 final answer : b <eor> b <eos> |
b |
vowel_space__ negate_prob__0.4__ die_space__ negate_prob__0.4__ |
vowel_space__ negate_prob__0.4__ die_space__ negate_prob__0.4__ |
| a man borrowed some amount after num__3 years he paid rs . num__10400 / - with num__10.0 interest then how much amount he borrowed ? <o> a ) rs . num__6000 / - <o> b ) rs . num__6500 / - <o> c ) rs . num__7000 / - <o> d ) rs . num__7200 / - <o> e ) rs . num__8000 / - |
principal p = ? rate of interest r = num__10.0 time t = num__3 years accumulated amount a = rs . num__10400 / - a = p + i a = p + ptr / num__100 a = p ( num__1 + tr / num__100 ) rs . num__10400 / - = p [ num__1 + num__3 * num__0.1 ] rs . num__10400 / - = p [ num__1.0 + num__0.3 ] rs . num__10400 / - = p [ num__10 + num__0.3 ] = p [ num__1.3 ] rs . num__10400 / - = p * num__1.3 = = > p = rs . num__8000.0 = = > p = rs . num__8000 / - e <eor> e <eos> |
e |
reverse__10.0__ divide__3.0__10.0__ add__1.0__0.3__ divide__10400.0__1.3__ divide__10400.0__1.3__ |
reverse__10.0__ divide__3.0__10.0__ add__1.0__0.3__ divide__10400.0__1.3__ divide__10400.0__1.3__ |
| the boys ordered several pizzas for the weekend . when the first evening was over the following amounts of pizza were left over : num__0.25 of the pepperoni pizza num__0.5 of the cheese pizza num__0.75 of the mushroom pizza and num__0.25 of the supreme pizza . the next morning each boy ate the equivalent of num__0.25 of a pizza for breakfast . if that finished the pizza how many boys were there ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__9 |
num__0.25 + num__0.5 + num__0.75 + num__0.25 = num__0.25 + num__0.5 + num__0.75 + num__0.25 = num__1.75 or num__1 num__0.75 pizza left over num__1.75 ÷ num__0.25 = num__1.75 x num__4.0 = num__7 boys correct answer c <eor> c <eos> |
c |
round_down__1.75__ reverse__0.25__ multiply__4.0__1.75__ multiply__4.0__1.75__ |
add__0.25__0.75__ reverse__0.25__ multiply__4.0__1.75__ multiply__4.0__1.75__ |
| let f ( x ) = x ^ num__2 + bx + c . if f ( num__1 ) = num__0 and f ( - num__6 ) = num__0 then f ( x ) crosses the y - axis at what y - coordinate ? <o> a ) - num__6 <o> b ) - num__1 <o> c ) num__0 <o> d ) num__1 <o> e ) num__6 |
when x = num__1 and when x = - num__6 the expression f ( x ) = x ² + bx + c equals num__0 . then f ( x ) = ( x - num__1 ) ( x + num__6 ) f ( num__0 ) = - num__6 the answer is a . <eor> a <eos> |
a |
multiply__1.0__6.0__ |
multiply__1.0__6.0__ |
| selling an kite for rs . num__30 a shop keeper gains num__15.0 . during a clearance sale the shopkeeper allows a discount of num__10.0 on the marked price . his gain percent during the sale is ? <o> a ) num__3.5 <o> b ) num__10.0 <o> c ) num__11.0 <o> d ) num__15.0 <o> e ) num__20 % |
explanation : marked price = rs . num__30 c . p . = num__0.869565217391 * num__30 = rs . num__26.08 sale price = num__90.0 of rs . num__30 = rs . num__27 required gain % = num__0.92 / num__26.08 * num__100 = num__3.5 . answer : a <eor> a <eos> |
a |
percent__30.0__90.0__ percent__3.5__100.0__ |
percent__30.0__90.0__ percent__3.5__100.0__ |
| the sum of the ages of num__5 children born at the intervals of num__2 years each is num__50 years . what is the age of the youngest child ? <o> a ) num__6 <o> b ) num__18 <o> c ) num__10 <o> d ) num__99 <o> e ) num__38 |
let x = the youngest child . each of the other four children will then be x + num__2 x + num__4 x + num__6 x + num__8 . we know that the sum of their ages is num__50 . so x + ( x + num__2 ) + ( x + num__4 ) + ( x + num__6 ) + ( x + num__8 ) = num__50 therefore the youngest child is num__6 years old answer : a <eor> a <eos> |
a |
add__2.0__4.0__ multiply__2.0__4.0__ add__2.0__4.0__ |
add__2.0__4.0__ add__2.0__6.0__ add__2.0__4.0__ |
| sonika deposited rs . num__8000 which amounted to rs . num__9200 after num__3 years at simple interest . had the interest been num__2.0 more . she would get how much ? <o> a ) num__9680 <o> b ) num__2899 <o> c ) num__2893 <o> d ) num__2790 <o> e ) num__2702 |
( num__8000 * num__3 * num__2 ) / num__100 = num__480 num__9200 - - - - - - - - num__9680 answer : a <eor> a <eos> |
a |
percent__100.0__9680.0__ |
percent__100.0__9680.0__ |
| a train num__600 m long can cross an electric pole in num__20 sec and then find the speed of the train ? <o> a ) num__76 kmph <o> b ) num__54 kmph <o> c ) num__62 kmph <o> d ) num__72 kmph <o> e ) num__108 kmph |
length = speed * time speed = l / t s = num__30.0 s = num__30 m / sec speed = num__30 * num__3.6 ( to convert m / sec in to kmph multiply by num__3.6 ) speed = num__108 kmph answer : e <eor> e <eos> |
e |
divide__600.0__20.0__ multiply__3.6__30.0__ round__108.0__ |
divide__600.0__20.0__ multiply__3.6__30.0__ multiply__3.6__30.0__ |
| the flow of water through a drainage pipe was monitored for a num__3 - hour period . in the second hour the rate of flow was num__30 gallons per hour which was num__50 percent faster than the rate of flow for the first hour . if num__25 percent more water flowed through the pipe in the third hour than it did in the second how many gallons of water flowed through the pipe during the entire three hours ? <o> a ) num__81.25 <o> b ) num__82.5 <o> c ) num__83.75 <o> d ) num__85.0 <o> e ) num__87.5 |
rate of flow of water in second hour = num__30 gallons per hour rate of flow of water in first hour = num__30 / ( num__1.5 ) = num__20 gallons per hour rate of flow of water in third hour = ( num__1.25 ) * num__30 = ( num__1.25 ) * num__30 = num__37.5 gallons per hour number of gallons of water that flowed through the pipe during the entire num__3 hours = num__20 + num__30 + num__37.5 = num__87.5 gallons answer e <eor> e <eos> |
e |
divide__30.0__1.5__ divide__25.0__20.0__ multiply__30.0__1.25__ add__50.0__37.5__ add__50.0__37.5__ |
divide__30.0__1.5__ divide__25.0__20.0__ multiply__30.0__1.25__ add__50.0__37.5__ add__50.0__37.5__ |
| john and karen begin running at opposite ends of a trail until they meet somewhere in between their starting points . they each run at their respective constant rates until john gets a cramp and stops . if karen runs num__50.0 faster than john who is only able to cover num__25.0 of the distance r before he stops what percent longer would karen have run than she would have had john been able to maintain his constant rate until they met . <o> a ) num__25.0 <o> b ) num__50.0 <o> c ) num__75.0 <o> d ) num__100.0 <o> e ) num__200 % |
john and karen begin running at opposite ends of a trail until they meet somewhere in between their starting points . they each run at their respective constant rates until john gets a cramp and stops . if karen runs num__50.0 faster than john who is only able to cover num__25.0 of the distance r before he stops what percent longer would karen have run than she would have had john been able to maintain his constant rate until they met . lets say the distance of the trail is num__100 miles . lets also say that j rate = num__10 miles / hour and k rate = num__15 miles / hour . if john stops at the num__25.0 mark that means he travels num__25 miles in num__2.5 hours . it would take karen t = d / r t = num__5.0 = num__5 hours to reach john . if john had not stopped their combined rate would num__10 + num__15 = num__25 miles / hour meaning they would have met in num__4 hours . therefore she ran one hour longer ( num__25.0 ) longer than she would have needed to if john ran for the entire time . answer : a ) num__25.0 <eor> a <eos> |
a |
percent__25.0__10.0__ percent__50.0__10.0__ percent__25.0__100.0__ |
percent__25.0__10.0__ percent__50.0__10.0__ percent__25.0__100.0__ |
| a coin is tossed live times . what is the probability that there is at the least one tail ? <o> a ) num__0.96875 <o> b ) num__1.36363636364 <o> c ) num__0.933333333333 <o> d ) num__4.36363636364 <o> e ) num__5.5 |
let p ( t ) be the probability of getting least one tail when the coin is tossed five times . = there is not even a single tail . i . e . all the outcomes are heads . = num__0.03125 ; p ( t ) = num__1 - num__0.03125 = num__0.96875 answer : a <eor> a <eos> |
a |
negate_prob__0.0312__ negate_prob__0.0312__ |
negate_prob__0.0312__ negate_prob__0.0312__ |
| if a fair die is rolled four time what is the probability that a num__3 occurs on at least one roll ? <o> a ) num__0.694444444444 <o> b ) num__0.578703703704 <o> c ) num__0.421296296296 <o> d ) num__0.305555555556 <o> e ) num__0.517745302714 |
questions such as these that talk about at least or maximum or minimum in probability questions should make sure realize that probability of any event ( n ) to occur = num__1 - p ( not n ) thus the probability of at least num__1 roll = num__1 - probability of no num__3 s = num__1 - ( num__0.833333333333 ) ( num__0.833333333333 ) ( num__0.833333333333 ) ( num__0.833333333333 ) = num__1 - num__0.482254697286 = num__0.517745302714 . num__0.833333333333 is the probability of not getting a num__3 in any num__1 roll with num__5 allowed numbers ( = num__1 num__24 num__56 ) out of a total of num__6 possibilities . e is thus the correct answer . <eor> e <eos> |
e |
negate_prob__0.4823__ vowel_space__ die_space__ negate_prob__0.4823__ |
negate_prob__0.4823__ vowel_space__ die_space__ negate_prob__0.4823__ |
| predict the next number num__2 num__2010 num__10050 <o> a ) num__200 <o> b ) num__300 <o> c ) num__400 <o> d ) num__500 <o> e ) num__600 |
look upon the series carefully num__2 num__2010 num__10050 . . . num__2 * num__10 = num__20 num__10.0 = num__10 num__10 * num__10 = num__100 num__50.0 = num__50 similarly num__50 * num__10 = num__500 answer : d <eor> d <eos> |
d |
multiply__2.0__10.0__ divide__100.0__2.0__ multiply__10.0__50.0__ multiply__10.0__50.0__ |
multiply__2.0__10.0__ divide__100.0__2.0__ multiply__10.0__50.0__ multiply__10.0__50.0__ |
| what will be the percentage increase in the area of the cube ' s surface if each of the cube ' s edges grows by num__30.0 ? <o> a ) num__70.0 <o> b ) num__69.0 <o> c ) num__80.0 <o> d ) num__82.0 <o> e ) num__55 % |
the question is very easy . my logic is the following : a surface = num__6 * a ^ num__2 after num__30.0 increase a surface = num__6 * ( ( num__1.3 a ) ^ num__2 ) = num__6 * num__1.69 * a ^ num__2 the increase in the surface area = ( num__6 * num__1.69 * a ^ num__2 - num__6 * a ^ num__2 ) / num__6 * a ^ num__2 = ( num__6 * a ^ num__2 ( num__1.69 - num__1 ) ) / ( num__6 * a ^ num__2 ) = num__1.69 - num__1 = num__0.69 = num__69.0 answer : b <eor> b <eos> |
b |
power__1.3__2.0__ multiply__1.0__69.0__ |
power__1.3__2.0__ multiply__1.0__69.0__ |
| two numbers are in respectively num__20.0 and num__50.0 more than a third number . the ratio of the two numbers is ? <o> a ) num__2 : num__5 <o> b ) num__1 : num__2 <o> c ) num__4 : num__5 <o> d ) num__3 : num__7 <o> e ) num__5 : num__6 |
let the num__3 rd number be x then first number = num__120.0 of x = num__120 x / num__100 = num__6 x / num__5 second number = num__150.0 of x = num__150 x / num__100 = num__3 x / num__2 ratio of first two numbers = num__6 x / num__5 : num__3 x / num__2 = num__12 x : num__15 x = num__4 : num__5 answer is c <eor> c <eos> |
c |
subtract__120.0__20.0__ divide__120.0__20.0__ divide__100.0__20.0__ multiply__50.0__3.0__ divide__100.0__50.0__ multiply__2.0__6.0__ subtract__20.0__5.0__ divide__20.0__5.0__ divide__20.0__5.0__ |
subtract__120.0__20.0__ divide__120.0__20.0__ divide__100.0__20.0__ multiply__50.0__3.0__ divide__100.0__50.0__ multiply__2.0__6.0__ subtract__20.0__5.0__ divide__20.0__5.0__ divide__20.0__5.0__ |
| a is thrice as good a workman as b and takes num__10 days less to do a piece of work than b takes . b alone can do the whole work in <o> a ) num__15 days <o> b ) num__10 days <o> c ) num__9 days <o> d ) num__8 days <o> e ) num__7 days |
explanation : ratio of times taken by a and b = num__1 : num__3 means b will take num__3 times which a will do in num__1 time if difference of time is num__2 days b takes num__3 days if difference of time is num__10 days b takes ( num__1.5 ) * num__10 = num__15 days option a <eor> a <eos> |
a |
subtract__3.0__1.0__ divide__3.0__2.0__ multiply__10.0__1.5__ round__15.0__ |
subtract__3.0__1.0__ divide__3.0__2.0__ multiply__10.0__1.5__ multiply__10.0__1.5__ |
| the product of two successive numbers is num__462 . which is the smaller of the two numbers ? <o> a ) num__21 <o> b ) num__23 <o> c ) num__26 <o> d ) num__27 <o> e ) num__29 |
a num__21 from the given alternatives num__21 × num__22 = num__462 ∴ smaller number = num__21 <eor> a <eos> |
a |
divide__462.0__21.0__ divide__462.0__22.0__ |
divide__462.0__21.0__ divide__462.0__22.0__ |
| a boat can move upstream at num__30 kmph and downstream at num__15 kmph then the speed of the current is ? <o> a ) num__5 <o> b ) num__7.5 <o> c ) num__32.5 <o> d ) num__9.2 <o> e ) num__5.8 |
us = num__30 ds = num__15 m = ( num__30 - num__15 ) / num__2 = num__7.5 answer : b <eor> b <eos> |
b |
divide__30.0__15.0__ divide__15.0__2.0__ round__7.5__ |
divide__30.0__15.0__ divide__15.0__2.0__ subtract__15.0__7.5__ |
| a tank is num__25 m long num__12 m wide and num__6 m deep . the cost of plastering its walls and bottom at num__25 paise per sq m is <o> a ) rs . num__186 <o> b ) rs . num__258 <o> c ) rs . num__486 <o> d ) rs . num__586 <o> e ) none of these |
explanation : area to be plastered = [ num__2 ( l + b ) à — h ] + ( l à — b ) = [ num__2 ( num__25 + num__12 ) à — num__6 ] + ( num__25 à — num__12 ) = num__744 sq m cost of plastering = num__744 à — ( num__0.25 ) = rs . num__186 answer : a <eor> a <eos> |
a |
divide__12.0__6.0__ multiply__0.25__744.0__ round__186.0__ |
divide__12.0__6.0__ multiply__0.25__744.0__ round__186.0__ |
| if a = { num__17 num__27 num__31 num__53 num__61 } what is the sum of mean and median of the numbers in a ? <o> a ) num__69 <o> b ) num__75 <o> c ) num__82 <o> d ) num__91 <o> e ) num__56 |
mean = ( num__17 + num__27 + num__31 + num__53 + num__62 ) / num__5 = num__38 median = num__31 sum = num__38 + num__31 = num__69 option a <eor> a <eos> |
a |
add__31.0__38.0__ add__31.0__38.0__ |
add__31.0__38.0__ add__31.0__38.0__ |
| a train passes a station platform in num__39 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__285 m <o> b ) num__240 m <o> c ) num__277 m <o> d ) num__765 m <o> e ) num__176 m |
speed = ( num__54 * num__0.277777777778 ) m / sec = num__15 m / sec . length of the train = ( num__15 x num__20 ) m = num__300 m . let the length of the platform be x meters . then ( x + num__300 ) / num__39 = num__15 = = > x + num__300 = num__585 = = > x = num__285 m . answer : a <eor> a <eos> |
a |
subtract__54.0__39.0__ multiply__20.0__15.0__ multiply__39.0__15.0__ subtract__585.0__300.0__ round__285.0__ |
subtract__54.0__39.0__ multiply__20.0__15.0__ multiply__39.0__15.0__ subtract__585.0__300.0__ round__285.0__ |
| if â € œ * â € is called â € œ + â € â € œ / â € is called â € œ * â € â € œ - â € is called â € œ / â € â € œ + â € is called â € œ - â € . num__2.0 â € “ num__10 * num__20 + num__10 = ? <o> a ) num__185 <o> b ) num__160 <o> c ) num__170 <o> d ) num__145 <o> e ) num__225 |
explanation : given : num__2.0 â € “ num__10 * num__20 + num__10 = ? substituting the coded symbols for mathematical operations we get num__80 * num__2.0 + num__20 â € “ num__10 = ? num__80 * num__2 + num__20 â € “ num__10 = ? num__160 + num__20 â € “ num__10 = ? num__180 â € “ num__10 = num__170 answer : c <eor> c <eos> |
c |
multiply__2.0__80.0__ add__20.0__160.0__ add__10.0__160.0__ add__10.0__160.0__ |
multiply__2.0__80.0__ add__20.0__160.0__ add__10.0__160.0__ add__10.0__160.0__ |
| a focus group is currently made up of x men and y women . if num__3 men and num__7 women are added to the group and if one person is selected at random from the larger focus group then what is the probability that a woman is selected ? <o> a ) x / ( x + num__7 ) <o> b ) y / ( x + y ) <o> c ) ( y + num__3 ) / ( x + y + num__7 ) <o> d ) ( y + num__7 ) / ( x + y + num__10 ) <o> e ) ( x + num__3 ) / ( x + y + num__10 ) |
x men y women x + num__3 + y + num__7 is the total = x + y + num__10 y + num__7 is the number of men so ( y + num__7 ) / ( x + y + num__10 ) d . ( y + num__7 ) / ( x + y + num__10 ) <eor> d <eos> |
d |
add__3.0__7.0__ subtract__10.0__3.0__ |
add__3.0__7.0__ subtract__10.0__3.0__ |
| how many times in a day are the hands of a clock in straight line but opposite in direction ? <o> a ) num__20 <o> b ) num__22 <o> c ) num__24 <o> d ) num__48 <o> e ) num__52 |
the hands of a clock point in opposite directions ( in the same straight line ) num__11 times in every num__12 hours . ( because between num__5 and num__7 they point in opposite directions at num__6 o ' clcok only ) . so in a day the hands point in the opposite directions num__22 times . answer : option b <eor> b <eos> |
b |
subtract__12.0__5.0__ subtract__11.0__5.0__ round__22.0__ |
subtract__12.0__5.0__ subtract__11.0__5.0__ round__22.0__ |
| the dimensions of a room are num__10 m x num__7 m x num__5 m . there are num__2 doors and num__3 windows in the room . the dimensions of the doors are num__1 m x num__3 m . one window is of size num__2 m x num__1.5 m and the other num__2 windows are of size num__1 m x num__1.5 m . the cost of painting the walls at rs . num__3 per sq m is ? <o> a ) rs . num__229 <o> b ) rs . num__429 <o> c ) rs . num__129 <o> d ) rs . num__474 <o> e ) rs . num__111 |
area of num__4 walls = num__2 ( l + b ) h = num__2 ( num__10 + num__7 ) x num__5 = num__170 sq m area of num__2 doors and num__3 windows = num__2 ( num__1 x num__3 ) + ( num__2 x num__1.5 ) + num__2 ( num__1 x num__1.5 ) = num__12 sq m area to be planted = num__170 - num__12 = num__158 sq m cost of painting = rs . num__158 x num__3 = rs . num__474 answer : d <eor> d <eos> |
d |
subtract__7.0__3.0__ add__10.0__2.0__ subtract__170.0__12.0__ multiply__3.0__158.0__ round__474.0__ |
subtract__7.0__3.0__ add__10.0__2.0__ subtract__170.0__12.0__ multiply__3.0__158.0__ round__474.0__ |
| i travel num__20 miles towards north and then travel num__25 miles eastward . i then travel num__40 miles rightwards then travel num__30 miles towards left and then travels num__12 miles to the left and finally num__20 miles northwards . how far am i pproximately from my original destination and in what direction ? <o> a ) num__20 miles towards south <o> b ) num__13 miles towards south - west <o> c ) num__12 miles towards east <o> d ) num__13 miles towards north - east <o> e ) none of these |
explanation : using pythagorus rule x num__2 = ( num__12 ) num__2 + ( num__5 ) num__2 x = num__13 answer is b <eor> b <eos> |
b |
divide__40.0__20.0__ subtract__25.0__20.0__ subtract__25.0__12.0__ round__13.0__ |
divide__40.0__20.0__ subtract__25.0__20.0__ subtract__25.0__12.0__ round__13.0__ |
| a number is num__101102103104 . . . num__150 . as num__101 num__102 num__103 num__103 . . . . num__150 . what is reminder when divided by num__3 ? <o> a ) num__2 <o> b ) num__5 <o> c ) num__6 <o> d ) num__7 <o> e ) num__8 |
divisibility rule for num__3 also same as num__9 . so from the above discussion sum of the digits = num__380 and remainder = num__126.666666667 = num__2 . answer : a <eor> a <eos> |
a |
divide__380.0__3.0__ subtract__103.0__101.0__ subtract__103.0__101.0__ |
divide__380.0__3.0__ subtract__103.0__101.0__ subtract__103.0__101.0__ |
| a bowl was half full of water . num__4 cups of water were then added to the bowl filling the bowl to num__70.0 of its capacity . how many cups of water are now in the bowl ? <o> a ) num__14 <o> b ) num__15 <o> c ) num__16 <o> d ) num__17 <o> e ) num__18 |
lets say total volume of the container = v initial water content is half of total volume = v / num__2 then num__4 cups of water were added . current water content = ( v / num__2 ) + num__4 cups = ( num__0.7 ) v = > v = num__20 cups = > current water content is equivalent to = v / num__2 + num__4 cups = num__10.0 + num__4 = num__14 cups ; answer : a <eor> a <eos> |
a |
percent__70.0__20.0__ percent__70.0__20.0__ |
percent__70.0__20.0__ percent__70.0__20.0__ |
| there are eight boxes of chocolates each box containing distinct number of chocolates from num__1 to num__8 . in how many ways four of these boxes can be given to four persons ( one boxes to each ) such that the first person gets more chocolates than each of the three the second person gets more chocolates than the third as well as the fourth persons and the third person gets more chocolates than fourth person ? <o> a ) num__35 <o> b ) num__70 <o> c ) num__105 <o> d ) num__210 <o> e ) none |
solution : all the boxes contain distinct number of chocolates . for each combination of num__4 out of num__8 boxes the box with the greatest number has to be given to the first person the box with the second highest to the second person and so on . the number of ways of giving num__4 boxes to the num__4 person is num__8 c num__4 = num__70 . answer : option b <eor> b <eos> |
b |
multiply__1.0__70.0__ |
multiply__1.0__70.0__ |
| in right triangle abc ac is the hypotenuse . if ac is num__20 and ab + bc = num__50 what is the area of the triangle abc ? <o> a ) num__225 <o> b ) num__525 <o> c ) num__25 √ num__2 <o> d ) num__200 <o> e ) num__200 √ num__2 |
square ab + bc = num__50 : ( ab ) ^ num__2 + num__2 * ab * bc + ( bc ) ^ num__2 = num__2500 . since ( ac ) ^ num__2 = ( ab ) ^ num__2 + ( bc ) ^ num__2 = num__20 ^ num__2 = num__400 then ( ab ) ^ num__2 + num__2 * ab * bc + ( bc ) ^ num__2 = num__400 + num__2 * ab * bc = num__2500 . num__400 + num__2 * ab * bc = num__2500 . ab * bc = num__1050 . the area = num__0.5 * ab * bc = num__525 . answer : b . <eor> b <eos> |
b |
power__50.0__2.0__ power__20.0__2.0__ multiply__0.5__1050.0__ multiply__0.5__1050.0__ |
power__50.0__2.0__ power__20.0__2.0__ multiply__0.5__1050.0__ multiply__0.5__1050.0__ |
| michael completes a piece of work in num__6 days madan completes the same work in num__18 days . if both of them work together then the number of days required to complete the work is ? <o> a ) num__4.5 days <o> b ) num__8 days <o> c ) num__10 days <o> d ) num__12 days <o> e ) num__14 days |
if a can complete a work in x days and b can complete the same work in y days then both of them together can complete the work in x y / x + y days . that is the required no . of days = num__6 × num__0.75 = num__4.5 days answer : a <eor> a <eos> |
a |
multiply__6.0__0.75__ round__4.5__ |
multiply__6.0__0.75__ round__4.5__ |
| mysoon collects glass ornaments . ten more than num__0.166666666667 of the ornaments in her collection are handmade and num__0.5 of the handmade ornaments are antiques . if num__0.125 of the ornaments in her collection are handmade antiques how many ornaments are in her collection ? <o> a ) num__36 <o> b ) num__60 <o> c ) num__120 <o> d ) num__144 <o> e ) num__180 |
the number of ornaments = a ten more than num__0.166666666667 of the ornaments in her collection are handmade = > handmade = num__10 + a / num__6 num__0.5 of the handmade ornaments are antiques = > handmade ornaments = num__0.5 * ( num__10 + a / num__6 ) = num__5 + a / num__12 num__0.125 of the ornaments in her collection are handmade antiques = > handmade ornaments = a / num__8 = > num__5 + a / num__12 = a / num__8 = > a = num__120 ans : c <eor> c <eos> |
c |
multiply__0.5__10.0__ divide__6.0__0.5__ reverse__0.125__ multiply__10.0__12.0__ multiply__10.0__12.0__ |
multiply__0.5__10.0__ divide__6.0__0.5__ reverse__0.125__ multiply__10.0__12.0__ multiply__10.0__12.0__ |
| if n divided by num__5 has a remainder of num__4 what is the remainder when num__3 times n is divided by num__5 ? <o> a ) num__1 <o> b ) num__4 <o> c ) num__3 <o> d ) num__2 <o> e ) num__0 |
as per question = > n = num__5 p + num__4 for some integer p hence num__3 n = > num__15 q + num__12 but again num__12 can be divided by num__5 to get remainder num__2 for some integer q hence d <eor> d <eos> |
d |
multiply__5.0__3.0__ multiply__4.0__3.0__ subtract__5.0__3.0__ subtract__5.0__3.0__ |
multiply__5.0__3.0__ multiply__4.0__3.0__ subtract__5.0__3.0__ subtract__5.0__3.0__ |
| which of the following is odd if r is even and t is odd ? <o> a ) rt <o> b ) num__5 rt <o> c ) num__6 r + num__5 t <o> d ) num__5 r + num__6 t <o> e ) num__6 ( r ^ num__2 ) t |
just checking options ( in case you have difficulty then choose values of r = num__2 and t = num__1 ) r = even t = odd a . rt = even * odd = evenincorrect b . num__5 rt = odd * even * odd = evenincorrect e . num__6 ( r ^ num__2 ) t = num__6 * ( even ^ num__2 ) * odd = evenincorrect d . num__5 r + num__6 t = num__5 * even + num__6 * odd = even + even = evenincorrect c . num__6 r + num__5 t = num__6 * even + num__5 * odd = even + odd = oddcorrect answer : option c <eor> c <eos> |
c |
add__1.0__5.0__ add__1.0__5.0__ |
add__1.0__5.0__ add__1.0__5.0__ |
| in triangle pqr the angle q = num__90 degree pq = num__5 cm qr = num__8 cm . x is a variable point on pq . the line through x parallel to qr intersects pr at y and the line through y parallel to pq intersects qr at z . find the least possible length of xz <o> a ) num__3.6 cm <o> b ) num__2.4 cm <o> c ) num__4.0 cm <o> d ) num__2.16 cm <o> e ) num__3.2 cm |
look at the diagram below : now in case when qy is perpendicular to pr two right triangles pqr and pqy are similar : qy : qp = qr : pr - - > qy : num__5 = num__8 : num__10 - - > qy = num__4.0 . answer : c . <eor> c <eos> |
c |
round__4.0__ |
subtract__8.0__4.0__ |
| a year ago the ratio of betty ' s and chi ' s age was num__6 : num__7 respectively . four years hence this ratio would become num__7 : num__8 . how old is betty ? <o> a ) num__34 years <o> b ) num__29 yeras <o> c ) num__25 years <o> d ) num__30 years <o> e ) num__40 years |
( b - num__1 ) / ( c - num__1 ) = num__0.857142857143 ; ( b - num__1 + num__4 ) / ( c - num__1 + num__4 ) = num__0.875 ; by solving them we get num__8 b - num__7 c = - num__3 num__7 b - num__6 c = num__1 then solving these eq we get c = num__29 b = num__25 answer : c <eor> c <eos> |
c |
subtract__7.0__6.0__ divide__6.0__7.0__ divide__7.0__8.0__ subtract__7.0__4.0__ subtract__29.0__4.0__ multiply__1.0__25.0__ |
subtract__7.0__6.0__ divide__6.0__7.0__ divide__7.0__8.0__ subtract__7.0__4.0__ subtract__29.0__4.0__ divide__25.0__1.0__ |
| in num__10 year a willbe twice as old as b was num__10 year ago . if a is now num__9 year older then b the present age of b is <o> a ) num__19 years <o> b ) num__29 years <o> c ) num__39 years <o> d ) num__49 years <o> e ) none of these |
solution let b ' s present age = x years . then a ' s present age = ( x + num__9 ) years . ∴ ( x + num__9 ) + num__10 = num__2 ( x - num__10 ) ⇔ x + num__19 = num__2 x - num__20 ⇔ x = num__39 . answer c <eor> c <eos> |
c |
add__10.0__9.0__ multiply__10.0__2.0__ add__19.0__20.0__ add__19.0__20.0__ |
add__10.0__9.0__ multiply__10.0__2.0__ add__19.0__20.0__ add__19.0__20.0__ |
| the sum of the present ages of a b c is num__51 years . three years ago their ages were in the ratio num__1 : num__2 : num__3 . what is the present age of a ? <o> a ) num__7 <o> b ) num__8 <o> c ) num__9 <o> d ) num__10 <o> e ) num__11 |
three years ago : a : b : c = num__1 : num__2 : num__3 let a = num__1 x b = num__2 x and c = num__3 x . today : ( x + num__3 ) + ( num__2 x + num__3 ) + ( num__3 x + num__3 ) = num__51 x = num__7 so the present age of a is x + num__3 = num__10 the answer is d . <eor> d <eos> |
d |
add__3.0__7.0__ multiply__1.0__10.0__ |
add__3.0__7.0__ add__3.0__7.0__ |
| insert the missing number num__9 num__12 num__11 num__14 num__13 . . . num__15 <o> a ) num__12 <o> b ) num__16 <o> c ) num__10 <o> d ) num__17 <o> e ) num__18 |
the sequence is consider into two series . . first series ( odd ) starting with num__9 and increasing num__2 as follows . second series ( even ) starting with num__12 and increasing with num__2 as follows . i . e . num__9 = num__9 ( odd ) num__12 = num__12 ( even ) num__9 + num__2 = num__11 ( odd series ) num__12 + num__2 = num__14 ( even series ) num__11 + num__2 = num__13 ( odd series ) num__14 + num__2 = num__16 ( even series ) answer : b <eor> b <eos> |
b |
subtract__11.0__9.0__ add__14.0__2.0__ add__14.0__2.0__ |
subtract__11.0__9.0__ add__14.0__2.0__ add__14.0__2.0__ |
| what is the unit digit in ( num__4137 ) ^ num__754 ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__7 <o> d ) num__9 <o> e ) none of them |
unit digit in ( num__4137 ) ^ num__754 = unit digit in [ ( num__4137 ) ^ num__4 ] ^ num__188 x ( num__4137 ) ^ num__2 = ( num__1 x num__9 ) = num__9 answer is d <eor> d <eos> |
d |
multiply__1.0__9.0__ |
multiply__1.0__9.0__ |
| find the value of x : x ² − num__3 x + num__2 <o> a ) - num__1 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
( x − num__1 ) ( x − num__2 ) x = num__1 or num__2 . c <eor> c <eos> |
c |
subtract__3.0__2.0__ subtract__3.0__1.0__ |
subtract__3.0__2.0__ subtract__3.0__1.0__ |
| a train num__1200 m long is running at a speed of num__78 km / hr . if it crosses a tunnel in num__1 min then the length of the tunnel is ? <o> a ) num__2898 <o> b ) num__277 <o> c ) num__500 <o> d ) num__100 <o> e ) num__435 |
speed = num__78 * num__0.277777777778 = num__21.6666666667 m / sec . time = num__1 min = num__60 sec . let the length of the train be x meters . then ( num__1200 + x ) / num__60 = num__21.6666666667 x = num__100 m . answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ round__100.0__ |
hour_to_min_conversion__ multiply__1.0__100.0__ |
| if num__1 pound of dried apricots that cost x dollars per pound are mixed with num__2 pounds of prunes that cost y dollars per pound what is the cost in dollars per pound of the mixture ? <o> a ) ( x + num__2 y ) / num__5 <o> b ) ( x + num__2 y ) / ( x + y ) <o> c ) ( x + num__2 y ) / ( xy ) <o> d ) num__5 ( x + num__2 y ) <o> e ) x + num__2 y |
total cost = weight ( in pounds ) * price / pound ; to find total cost / pound divide by total pounds . cost of dried apricots = x ; cost of prunes = num__2 y ; cost per pound = ( x + num__2 y ) / num__5 ; ans is ( a ) . <eor> a <eos> |
a |
multiply__1.0__2.0__ |
multiply__1.0__2.0__ |
| two trains are moving in opposite directions at num__60 km / hr and num__90 km / hr . their lengths are num__1.75 km and num__1.25 km respectively . the time taken by the slower train to cross the faster train in seconds is ? <o> a ) num__12 <o> b ) num__72 <o> c ) num__48 <o> d ) num__99 <o> e ) num__11 |
relative speed = num__60 + num__90 = num__150 km / hr . = num__150 * num__0.277777777778 = num__41.6666666667 m / sec . distance covered = num__1.75 + num__1.25 = num__3 km = num__3000 m . required time = num__3000 * num__0.024 = num__72 sec . answer : b <eor> b <eos> |
b |
add__60.0__90.0__ add__1.75__1.25__ divide__90.0__1.25__ round__72.0__ |
add__60.0__90.0__ add__1.75__1.25__ divide__90.0__1.25__ divide__90.0__1.25__ |
| every digit of a number written in binary is either num__0 or num__1 . to translate a number from binary multiply the nth digit ( reading from right to left ) by num__2 ^ ( n - num__1 ) what is the largest prime number ( written in binary ) that is a factor of both num__1010000 and num__10100000 ? <o> a ) num__10 <o> b ) num__11 <o> c ) num__101 <o> d ) num__1011 <o> e ) num__10001 |
binary divison can provide a quick answer if you are comfortable with it . as option e is the biggest binary number we try with it first : num__990198.019802 = num__10000 num__9901980.19802 = num__100000 so answer is option is c <eor> c <eos> |
c |
divide__1010000.0__10000.0__ |
divide__1010000.0__10000.0__ |
| an optometrist charges $ num__150 per pair for soft contact lenses and $ num__85 per pair for hard contact lenses . last week she sold num__5 more pairs of soft lenses than hard lenses . if her total sales for pairs of contact lenses last week were $ num__2160 what was the total number of pairs of contact lenses that she sold ? <o> a ) num__11 <o> b ) num__13 <o> c ) num__15 <o> d ) num__17 <o> e ) num__19 |
( x + num__5 ) * num__150 + x * num__85 = num__2160 = > x = num__6 total lens = num__6 + ( num__6 + num__5 ) = num__17 answer d <eor> d <eos> |
d |
divide__85.0__5.0__ divide__85.0__5.0__ |
divide__85.0__5.0__ divide__85.0__5.0__ |
| one pump drains one - half of a pond in num__2 hours and then a second pump starts draining the pond . the two pumps working together finish emptying the pond in one - half hour . how long would it take the second pump to drain the pond if it had to do the job alone ? <o> a ) num__1.3 hour <o> b ) num__1.2 hour <o> c ) num__3 hours <o> d ) num__5 hours <o> e ) num__6 hours |
the tricky part here i believed is one half hour = num__0.5 . then everything would be easy . we have the num__1 st pump working rate / hour = num__0.5 : num__2 = num__0.25 working rate of num__2 pumps : num__0.5 : num__0.5 = num__1 . working rate of num__2 nd pump : num__1 - num__0.25 = num__0.75 - - > time taken for the num__2 nd pump to finish : num__1 : num__0.75 = num__1.33333333333 = num__1.3 hours . a <eor> a <eos> |
a |
multiply__2.0__0.5__ divide__0.5__2.0__ add__0.5__0.25__ divide__1.0__0.75__ round__1.3__ |
multiply__2.0__0.5__ divide__0.5__2.0__ subtract__1.0__0.25__ divide__1.0__0.75__ divide__1.3__1.0__ |
| huey ' s hip pizza sells two sizes of square pizzas : a small pizza that measures num__14 inches on a side and costs $ num__10 and a large pizza that measures num__21 inches on a side and costs $ num__20 . if two friends go to huey ' s with $ num__30 apiece how many more square inches of pizza can they buy if they pool their money than if they each purchase pizza alone ? <o> a ) num__5 square inches <o> b ) num__10 square inches <o> c ) num__49 square inches <o> d ) num__25 square inches <o> e ) num__350 square inches |
in the first case each can buy one pizza of $ num__10 and one pizza of $ num__20 . in square inches that would be ( num__14 * num__14 = num__196 ) for the small pizza and ( num__21 * num__21 = num__441 ) for the large pizza . in total sq inches that would be ( num__196 + num__441 ) * num__2 = num__1274 sq inches . in the second case if they pool their money together they can buy num__3 large pizzas . in terms of square inches that would be num__3 * num__441 = num__1323 sq inches . hence the difference is num__49 square inches more ( num__1323 - num__1274 ) . the correct answer is c <eor> c <eos> |
c |
rectangle_perimeter__196.0__441.0__ multiply__3.0__441.0__ triangle_area__2.0__49.0__ |
rectangle_perimeter__196.0__441.0__ multiply__3.0__441.0__ triangle_area__2.0__49.0__ |
| find a sum for num__1 st num__7 prime number ' s ? <o> a ) num__25 <o> b ) num__28 <o> c ) num__58 <o> d ) num__34 <o> e ) num__36 |
required sum = ( num__2 + num__3 + num__5 + num__7 + num__11 + num__13 + num__17 ) = num__58 note : num__1 is not a prime number option c <eor> c <eos> |
c |
add__1.0__2.0__ subtract__7.0__2.0__ add__2.0__11.0__ multiply__1.0__58.0__ |
add__1.0__2.0__ add__2.0__3.0__ add__2.0__11.0__ multiply__1.0__58.0__ |
| if x is a positive number and num__2 the square root of x is equal to x / num__2 then x = <o> a ) num__0.0625 <o> b ) num__4 <o> c ) num__1 <o> d ) num__2 <o> e ) num__8 |
num__2 of sqrt ( x ) = x / num__2 which means that sqrt ( x ) = x or x = x ^ num__2 - > divide by x num__1 = x x = num__1 c . <eor> c <eos> |
c |
reverse__1.0__ |
reverse__1.0__ |
| there are two circles of different radii . the are of a square is num__784 sq cm and its side is twice the radius of the larger circle . the radius of the larger circle is seven - third that of the smaller circle . find the circumference of the smaller circle ? <o> a ) num__18 <o> b ) num__16 <o> c ) num__12 <o> d ) num__10 <o> e ) num__14 |
: let the radii of the larger and the smaller circles be l cm and s cm respectively . let the side of the square be a cm . a num__2 = num__784 = ( num__4 ) ( num__196 ) = ( num__22 ) . ( num__142 ) a = ( num__2 ) ( num__14 ) = num__28 a = num__2 l l = a / num__2 = num__14 l = ( num__2.33333333333 ) s therefore s = ( num__0.428571428571 ) ( l ) = num__6 circumference of the smaller circle = num__2 ∏ s = num__12 ∏ cm . answer : c <eor> c <eos> |
c |
multiply__2.0__14.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
multiply__2.0__14.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
| if p is not equal to zero and p + num__1 / p = num__5 then what is the value of p ^ num__4 + ( num__1 / p ) ^ num__4 ? <o> a ) num__527 <o> b ) num__247 <o> c ) num__547 <o> d ) num__527 <o> e ) num__537 |
p + num__1 / p = num__5 we square both sides so we have p ^ num__2 + num__1 / p ^ num__2 + num__2 = num__25 or p ^ num__2 + num__1 / p ^ num__2 = num__23 squaring again we have p ^ num__4 + num__1 / p ^ num__4 + num__2 = num__529 or p ^ num__4 + num__1 / p ^ num__4 = num__527 answer = num__527 ( d ) <eor> d <eos> |
d |
subtract__25.0__2.0__ subtract__529.0__2.0__ multiply__1.0__527.0__ |
subtract__25.0__2.0__ subtract__529.0__2.0__ divide__527.0__1.0__ |
| sequence w consists of num__16 consecutive integers . if - num__6 is the least integer in sequence w what is the range of the positive integers in sequence w ? <o> a ) num__16 <o> b ) num__15 <o> c ) num__9 <o> d ) num__8 <o> e ) num__7 |
since set w consists of num__16 consecutive integers and - num__6 is the least integer then set w is consecutive integers from - num__6 to num__9 inclusive : num__9 - ( - num__6 ) + num__1 = num__16 . the range of positive integers in sequence w is num__9 - num__1 = num__8 . answer : d . <eor> d <eos> |
d |
subtract__9.0__1.0__ subtract__16.0__8.0__ |
subtract__9.0__1.0__ subtract__16.0__8.0__ |
| a person can swim in still water at num__4 km / h . if the speed of water num__2 km / h how many hours will the man take to swim back against the current for num__6 km ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__8 <o> e ) num__2 |
m = num__4 s = num__2 us = num__4 - num__2 = num__2 d = num__6 t = num__3.0 = num__3 answer : a <eor> a <eos> |
a |
divide__6.0__2.0__ round__3.0__ |
divide__6.0__2.0__ subtract__6.0__3.0__ |
| for any real number x the operatoris defined as : ( x ) = x ( num__4 − x ) if p + num__1 = ( p + num__1 ) then p = <o> a ) − num__4 <o> b ) num__0 <o> c ) num__1 <o> d ) num__2 <o> e ) num__3 |
( x ) = x ( num__4 − x ) ( p + num__1 ) = ( p + num__1 ) ( num__4 - p - num__1 ) = ( num__3 - p ) ( p + num__1 ) we are given that p + num__1 = ( p + num__1 ) therefore ( num__3 - p ) ( p + num__1 ) = ( p + num__1 ) or ( p + num__1 ) + ( num__3 - p ) ( p + num__1 ) = num__0 ( p + num__1 ) ( p - num__3 ) = num__0 p = - num__1 p = num__3 option e <eor> e <eos> |
e |
subtract__4.0__1.0__ subtract__4.0__1.0__ |
subtract__4.0__1.0__ subtract__4.0__1.0__ |
| a pineapple costs rs num__7 each and a watermelon costs rs . num__5 each . if i spend rs num__38 on total what is the number of pineapple i purchased ? <o> a ) num__7 <o> b ) num__6 <o> c ) num__5 <o> d ) num__2 pineapples <o> e ) num__3 |
explanation : the equation for this problem can be made as : num__5 x + num__7 y = num__38 where x is the number of watermelons and y is the number of pineapples . now test for num__2 num__3 and num__4 : for y = num__2 num__5 x + num__14 = num__38 x is not an integer for y = num__3 num__5 x = num__17 x not an integer for y = num__4 x = num__2 so num__4 pineapples and num__2 watermelons can be bought by num__38 rs . answer : d <eor> d <eos> |
d |
subtract__7.0__5.0__ subtract__5.0__2.0__ subtract__7.0__3.0__ multiply__7.0__2.0__ add__3.0__14.0__ subtract__7.0__5.0__ |
subtract__7.0__5.0__ subtract__5.0__2.0__ subtract__7.0__3.0__ multiply__7.0__2.0__ add__3.0__14.0__ subtract__7.0__5.0__ |
| if f ( x ) = num__2 x ^ num__2 - num__3 x + num__5 and g ( y ) = num__2 y - num__4 then g ( f ( x ) ) <o> a ) num__7 x ^ num__2 + num__45 x + num__9 <o> b ) num__9 x ^ num__2 - num__15 x + num__4 <o> c ) num__4 x ^ num__2 - num__6 x + num__6 <o> d ) num__8 x ^ num__2 - num__9 x + num__6 <o> e ) num__9 x ^ num__2 + num__79 x + num__78 |
g ( f ( x ) ) = num__2 ( f ( x ) ) - num__4 = num__2 ( num__2 x ^ num__2 - num__3 x + num__5 ) - num__4 = num__4 x ^ num__2 - num__6 x + num__10 - num__4 = num__4 x ^ num__2 - num__6 x + num__6 the answer is c <eor> c <eos> |
c |
multiply__2.0__3.0__ multiply__2.0__5.0__ subtract__6.0__2.0__ |
add__2.0__4.0__ add__4.0__6.0__ subtract__6.0__2.0__ |
| a lent rs . num__5000 to b for num__2 years and rs . num__3000 to c for num__4 years on simple interest at the same rate of interest and received rs . num__3300 in all from both of them as interest . the rate of interest per annum is ? <o> a ) num__16.0 <o> b ) num__12.0 <o> c ) num__74.0 <o> d ) num__10.0 <o> e ) num__15 % |
let the rate be r % p . a . then ( num__5000 * r * num__2 ) / num__100 + ( num__3000 * r * num__4 ) / num__100 = num__3300 num__100 r + num__120 r = num__3300 r = num__15.0 answer : e <eor> e <eos> |
e |
percent__2.0__5000.0__ percent__4.0__3000.0__ percent__100.0__15.0__ |
percent__2.0__5000.0__ percent__4.0__3000.0__ percent__100.0__15.0__ |
| sripad has scored average of num__60 marks in three objects . in no subjects has he secured less than num__58 marks . he has secured more marks in maths than other two subjects . what could be his maximum score in maths ? <o> a ) num__79 <o> b ) num__64 <o> c ) num__38 <o> d ) num__27 <o> e ) num__21 |
assuming sripad has scored the least marks in subject other than science then the marks he could secure in other two are num__58 each . since the average mark of all the num__3 subject is num__60 . i . e ( num__58 + num__58 + x ) / num__3 = num__60 num__116 + x = num__180 x = num__64 marks . therefore the maximum marks he can score in maths is num__64 . answer : b <eor> b <eos> |
b |
multiply__60.0__3.0__ subtract__180.0__116.0__ subtract__180.0__116.0__ |
multiply__60.0__3.0__ subtract__180.0__116.0__ subtract__180.0__116.0__ |
| ( a / num__2 ) / ( b / c ) in the expression above a b and c are different numbers and each is one of the numbers num__1 num__2 or num__4 . what is the largest possible value of the expression ? <o> a ) num__4.0 <o> b ) num__0.5 <o> c ) num__0.125 <o> d ) num__0.5 <o> e ) num__0.666666666667 |
( a / num__2 ) / ( b / c ) = ( a * c ) / num__2 b the expression will have the largest value when numerator ( a * c ) is the largest . = ( num__2 * num__4 ) / num__2 * num__1 = num__4.0 answer a <eor> a <eos> |
a |
multiply__1.0__4.0__ |
multiply__1.0__4.0__ |
| find the unit ' s digit in ( num__264 ) ^ num__102 + ( num__264 ) ^ num__103 <o> a ) num__0 <o> b ) num__2 <o> c ) num__5 <o> d ) num__3 <o> e ) num__4 |
required unit ' s digit = unit ' s digit in ( num__4 ) ^ num__102 + ( num__4 ) ^ num__103 . now num__4 ^ num__2 gives unit digit num__6 . ( num__4 ) num__102 gives unjt digit num__6 . ( num__4 ) num__103 gives unit digit of the product ( num__6 x num__4 ) i . e . num__4 . hence unit ' s digit in ( num__264 ) m + ( num__264 ) num__103 = unit ' s digit in ( num__6 + num__4 ) = num__0 . answer is a <eor> a <eos> |
a |
add__2.0__4.0__ multiply__264.0__0.0__ |
add__2.0__4.0__ multiply__264.0__0.0__ |
| if x is an integer and ( x ) ( x ^ num__2 ) ( x ^ num__3 ) ( x ^ num__4 ) is positive which of the following could be negative ? <o> a ) x ^ num__4 <o> b ) ( x ) ( x ^ num__3 ) <o> c ) ( x ) ( x ) <o> d ) num__3 x ^ num__2 <o> e ) x + x ^ num__5 |
a ) x ^ num__4 - always positive ( or num__0 ) . b ) x ( x ^ num__3 ) = x ^ num__4 - again always positive ( or num__0 ) . c ) ( x ) ( x ) = x ^ num__2 - again always positive ( or num__0 ) . d ) num__3 x ^ num__2 - again always positive ( or num__0 ) e ) x + x ^ num__5 - could be - ve if x is - ve . answer e . <eor> e <eos> |
e |
add__2.0__3.0__ add__2.0__3.0__ |
add__2.0__3.0__ add__2.0__3.0__ |
| a polling company reports that there is a num__20.0 chance that a certain candidate will win the next election . if the candidate wins there is a num__60.0 chance that she will sign bill x and no other bills . if she decides not to sign bill x she will sign either bill y or bill z chosen randomly . what is the chance that the candidate will sign bill z ? <o> a ) num__10 <o> b ) num__8 <o> c ) num__6 <o> d ) num__4 <o> e ) num__5 |
num__20.0 - candidate elected num__100.0 - num__60.0 = num__40.0 - candidate doesnotsigh bill x num__50.0 - candidate randomly chooses between two bills . these are multiplicative : num__20.0 x num__40.0 x num__50.0 num__0.2 x num__0.4 x num__0.5 = num__0.04 = num__4.0 answer ( d ) <eor> d <eos> |
d |
percent__20.0__0.2__ percent__100.0__4.0__ |
percent__20.0__0.2__ percent__100.0__4.0__ |
| two pipes a and b can fill a cistern in num__22 and num__44 minutes respectively and a third pipe c can empty it in num__48 minutes . how long will it take to fill the cistern if all the three are opened at the same time ? <o> a ) num__18 num__0.111111111111 min <o> b ) num__21 num__0.125 min <o> c ) num__21 num__0.12 min <o> d ) num__18 num__1.14285714286 min <o> e ) num__19 num__0.428571428571 min |
num__0.0454545454545 + num__0.0227272727273 - num__0.0208333333333 = num__0.0473484848485 num__21.12 = num__21 num__0.12 answer : c <eor> c <eos> |
c |
subtract__21.12__21.0__ round__21.0__ |
subtract__21.12__21.0__ subtract__21.12__0.12__ |
| how many positive integers will divide evenly into num__230 ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__8 <o> d ) num__12 <o> e ) num__16 |
the question is asking how many factors num__230 has . num__230 = num__2 * num__5 * num__23 the number of factors is num__2 ^ num__3 = num__8 the answer is c . <eor> c <eos> |
c |
subtract__5.0__2.0__ add__3.0__5.0__ add__3.0__5.0__ |
subtract__5.0__2.0__ add__3.0__5.0__ add__3.0__5.0__ |
| find the next number in the sequence num__8 num__18 num__48 num__98 ? <o> a ) num__163 <o> b ) num__368 <o> c ) num__169 <o> d ) num__162 <o> e ) num__168 |
+ num__10 = > num__18 + num__30 = > num__48 + num__50 = > num__98 + num__70 = > num__168 answer is e <eor> e <eos> |
e |
subtract__18.0__8.0__ subtract__48.0__18.0__ subtract__98.0__48.0__ add__98.0__70.0__ add__98.0__70.0__ |
subtract__18.0__8.0__ subtract__48.0__18.0__ subtract__98.0__48.0__ add__98.0__70.0__ add__98.0__70.0__ |
| if $ num__0.40 is the commission for sales of $ num__1000 what percent of the sales amount is the commission ? <o> a ) num__4.0 <o> b ) num__0.4 <o> c ) num__0.04 <o> d ) num__0.004 <o> e ) num__0.0004 % |
% of sales amount of commission = ( commission / total value ) * num__100 = ( num__0.4 / num__1000 ) * num__100 = num__0.04 the answer is c . <eor> c <eos> |
c |
percent__100.0__0.04__ |
percent__100.0__0.04__ |
| a reduction of num__20.0 in the price of salt enables a lady to obtain num__10 kgs more for rs . num__100 find the original price per kg ? <o> a ) num__2.6 <o> b ) num__2.0 <o> c ) num__2.5 <o> d ) num__2.1 <o> e ) num__2.3 |
num__100 * ( num__0.2 ) = num__20 - - - num__10 ? - - - num__1 = > rs . num__2 num__100 - - - num__80 ? - - - num__2 = > rs . num__2.5 . answer : c <eor> c <eos> |
c |
percent__20.0__10.0__ percent__100.0__2.5__ |
percent__20.0__10.0__ percent__100.0__2.5__ |
| a problem in mathematics is given to three students a b and c whose chances of solving it are num__0.5 num__0.75 and num__0.25 respectively . what is the probability that the problem will be solved if all of them try independently . <o> a ) num__0.90625 <o> b ) num__30 <o> c ) num__0.939393939394 <o> d ) num__0.885714285714 <o> e ) num__33 |
the probability that the problem is not solved by any of them : ( num__1 - p ( a ) ) * ( num__1 - p ( b ) ) * ( num__1 - p ( c ) ) = ( num__1 - num__0.5 ) ( num__1 - num__0.75 ) ( num__1 - num__0.25 ) = num__0.5 * num__0.25 * num__0.75 = num__0.09375 hence the probability that the problem will be solved is num__1 - num__0.09375 = num__0.90625 . a ) <eor> a <eos> |
a |
union_prob__0.5__0.75__0.25__ negate_prob__0.0938__ negate_prob__0.0938__ |
union_prob__0.5__0.75__0.25__ negate_prob__0.0938__ negate_prob__0.0938__ |
| the distance traveled by helicopter in one year approximately num__750 num__000000 miles . the distance traveled by helicopter in num__15 years ? <o> a ) num__10 < num__4 > <o> b ) num__10 < num__6 > <o> c ) num__10 < num__7 > <o> d ) num__10 < num__9 > <o> e ) num__10 < num__16 > |
the distance traveled by helicopter in one year = num__75 num__000000 miles the distance traveled by helicopter in num__15 years = num__75 num__000000 miles * num__15 years = num__1125000000 = num__1125 * num__10 < num__6 > ( exponent miles ) answer = b <eor> b <eos> |
b |
multiply__15.0__75.0__ divide__750.0__75.0__ round__10.0__ |
multiply__15.0__75.0__ divide__750.0__75.0__ round__10.0__ |
| a does a work in num__10 days and b does the same work in num__15 days . in how many days they together will do the same work ? <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
a + b can finish = ab / a + b = num__10 * num__15 / ( num__10 + num__15 ) = num__6 days answer is c <eor> c <eos> |
c |
round__6.0__ |
round__6.0__ |
| a set consist of num__2 k - num__1 element . what is the number of subsets of this set which contain at most k - num__1 elements ? <o> a ) num__2 ^ ( num__2 k - num__2 ) <o> b ) num__2 ^ ( num__2 k ) - num__2 <o> c ) num__2 ^ ( num__2 k ) - num__1 <o> d ) num__2 ^ ( num__2 k ) <o> e ) num__2 ^ ( num__2 k - num__1 ) |
i used k = num__3 so then we have num__5 ! / num__2 ! num__3 ! + num__5 ! / num__4 ! num__1 ! + num__5 ! / num__0 ! num__5 ! num__10 + num__5 + num__1 = num__16 so our target is num__16 now replace in answer choices a gives us num__2 ^ num__4 = num__16 hence a is the correct option read carefully it says at most so keep in mind that picking a small number such as num__3 will help you save time since you have to list fewer outcomes avoid num__2 since you will get num__1 arrangement ( k - num__1 ) and may be risky since num__1 is a number with certain unique properties <eor> a <eos> |
a |
add__2.0__1.0__ add__2.0__3.0__ add__1.0__3.0__ multiply__2.0__5.0__ multiply__2.0__1.0__ |
add__2.0__1.0__ add__2.0__3.0__ add__1.0__3.0__ multiply__2.0__5.0__ divide__2.0__1.0__ |
| a box contains num__10 tablets of medicine a and num__10 tablets of medicine b . what is the least number of tablets that should be taken from the box to ensure that at least two tablets of each kind are among the extracted . <o> a ) num__12 <o> b ) num__15 <o> c ) num__17 <o> d ) num__19 <o> e ) num__21 |
the worst case scenario will be if we remove all num__15 tablets of medicine b first . the next num__2 tablets we remove have to be of medicine a so to guarantee that at least two tablets of each kind will be taken we should remove minimum of num__10 + num__2 = num__12 tablets . answer : a . <eor> a <eos> |
a |
add__10.0__2.0__ add__10.0__2.0__ |
add__10.0__2.0__ add__10.0__2.0__ |
| two trains each num__170 m in length each are running on two parallel lines in opposite directions . if one goes at the speed of num__55 km / h while the other travels at num__50 km / h . how long will it take for them to pass each other completely . <o> a ) num__15 sec <o> b ) num__11.6 sec <o> c ) num__31.6 sec <o> d ) num__12.6 sec <o> e ) num__23 sec |
explanation : d = num__170 m + num__170 m = num__340 m rs = num__55 + num__50 = num__105 * num__0.277777777778 = num__29.2 t = num__340 * num__0.0342465753425 = num__11.6 sec answer : option b <eor> b <eos> |
b |
add__55.0__50.0__ round__11.6__ |
add__55.0__50.0__ round__11.6__ |
| of num__60 children num__30 are happy num__10 are sad and num__20 are neither happy nor sad . there are num__19 boys and num__41 girls . if there are num__6 happy boys and num__4 sad girls how many boys are neither happy nor sad ? <o> a ) num__2 <o> b ) num__7 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
venn diagrams are useful for multiple values of a single variable e . g . state of mind - happy / sad / neither . when you have two or more variables such as here where you have gender - boy / girl too it becomes unwieldy . in this case either use the table or logic . table method is shown above ; here is how you will use logic : there are num__6 happy boys . there are num__4 sad girls but total num__10 sad children . so rest num__6 sad children must be sad boys . we have num__6 happy boys and num__6 sad boys . total we have num__19 boys . so num__19 - num__6 - num__6 = num__7 boys must be neither happy nor sad . answer ( b ) <eor> b <eos> |
b |
choose__7.0__6.0__ |
choose__7.0__6.0__ |
| a began business with rs . num__36000 and was joined afterwards by b with rs . num__54000 . when did b join if the profits at the end of the year were divided in the ratio of num__2 : num__1 ? <o> a ) num__1 <o> b ) num__6 <o> c ) num__7 <o> d ) num__8 <o> e ) num__2 |
num__36 * num__12 : num__54 * x = num__2 : num__1 x = num__4 num__12 - num__4 = num__8 answer : d <eor> d <eos> |
d |
multiply__2.0__4.0__ multiply__2.0__4.0__ |
multiply__2.0__4.0__ multiply__2.0__4.0__ |
| a district hospital treated num__290 patients today and each of them were either man woman or child . if it treated num__10 more men than of children and in equal as many men as women how many children were treated ? <o> a ) num__90 <o> b ) num__80 <o> c ) num__100 <o> d ) num__110 <o> e ) num__70 |
x = the number of men y = the number of women z = the number of children from the first sentence we have equation # num__1 : x + y + z = num__290 . . . a district hospital treated num__70 more men than children . . . equation # num__2 : x = num__10 + z . . . in equal as many men as women . . . equation # num__3 : x = y now we can replace x with y in equation # num__2 equation # num__4 : x - z = num__10 now we can replace x with y in equation # num__1 y + y + z = num__290 equation # num__5 : num__2 y + z = num__290 by adding equation # num__4 and equation # num__5 we get x = num__100 there are num__100 men treated . this is num__10 more than the number of children treated so z = num__90 . that ' s the answer . just as a check y = num__100 and num__100 + num__100 + num__90 = num__290 . answer = num__90 ( a ) <eor> a <eos> |
a |
add__1.0__2.0__ add__1.0__3.0__ divide__10.0__2.0__ subtract__100.0__10.0__ multiply__1.0__90.0__ |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ subtract__100.0__10.0__ subtract__100.0__10.0__ |
| what annual installment will discharge a debt of rs . num__1008 due in num__3 years at num__12.0 simple interest ? <o> a ) num__300 <o> b ) num__345 <o> c ) num__325 <o> d ) num__335 <o> e ) none of them |
let each installment be rs . x then ( x + ( ( x * num__12 * num__1 ) / num__100 ) ) + ( x + ( ( x * num__12 * num__2 ) / num__100 ) ) + x = num__1008 = ( ( num__28 x / num__25 ) + ( num__31 x / num__25 ) + x ) = num__1008 ï ƒ › ( num__28 x + num__31 x + num__25 x ) = ( num__1008 * num__25 ) x = ( num__1008 * num__25 ) / num__84 = rs . num__300 . therefore each installment = rs . num__300 . answer is a . <eor> a <eos> |
a |
percent__100.0__300.0__ |
percent__100.0__300.0__ |
| the daily wage is increased by num__50.0 and a person now gets rs num__25 per day . what was his daily wage before the increase ? <o> a ) rs num__20 <o> b ) rs num__16.6 <o> c ) rs num__18 <o> d ) rs num__19 <o> e ) none of these |
let a be the daily wages before the increase a ( num__1 + x / num__100 ) = num__25 ( given ) here x = num__50 therefore a ( num__1 + num__0.5 ) = num__25 a = ( num__25 x num__100 ) / ( num__150 ) = rs num__16.6 . answer : b <eor> b <eos> |
b |
divide__50.0__100.0__ add__50.0__100.0__ multiply__1.0__16.6__ |
divide__50.0__100.0__ add__50.0__100.0__ divide__16.6__1.0__ |
| if a card is drawn from a well shuffled pack of cards the probability of drawing a spade or a king is ? <o> a ) num__0.266666666667 <o> b ) num__0.210526315789 <o> c ) num__0.363636363636 <o> d ) num__0.307692307692 <o> e ) num__0.235294117647 |
p ( s ᴜ k ) = p ( s ) + p ( k ) - p ( s ∩ k ) where s denotes spade and k denotes king . p ( s ᴜ k ) = num__0.25 + num__0.0769230769231 - num__0.0192307692308 = num__0.307692307692 answer : d <eor> d <eos> |
d |
union_prob__0.25__0.0769__0.0192__ union_prob__0.25__0.0769__0.0192__ |
union_prob__0.25__0.0769__0.0192__ union_prob__0.25__0.0769__0.0192__ |
| a b and c can do a piece of work in num__24 num__30 and num__40 days respectively . they start the work together but c leaves num__4 days before the completion of the work . in how many days is the work done ? <o> a ) num__16 days <o> b ) num__18 days <o> c ) num__19 days <o> d ) num__11 days <o> e ) num__14 days |
x / num__24 + x / num__30 + x / num__40 = num__1 x = num__11 days answer : d <eor> d <eos> |
d |
round__11.0__ |
divide__11.0__1.0__ |
| the average of num__20 numbers is zero . of them at the most how many may be greater than zero ? <o> a ) num__0 <o> b ) num__9 <o> c ) num__2 <o> d ) num__7 <o> e ) num__1 |
average of num__20 numbers = num__0 . sum of num__20 numbers = ( num__0 * num__20 ) = num__0 . it is quite possible that num__19 of these numbers may be positive and if their sum is a then num__20 th number is ( - a ) . answer : a <eor> a <eos> |
a |
multiply__20.0__0.0__ |
multiply__20.0__0.0__ |
| how many seconds will a num__650 meter long train moving with a speed of num__63 km / hr take to cross a man walking with a speed of num__3 km / hr in the direction of the train ? <o> a ) num__48 <o> b ) num__36 <o> c ) num__26 <o> d ) num__39 <o> e ) num__18 |
explanation : here distance d = num__650 mts speed s = num__63 - num__3 = num__60 kmph = num__60 x num__0.277777777778 m / s time t = = num__39 sec . answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ round__39.0__ |
subtract__63.0__3.0__ round__39.0__ |
| the average age of the district level hockey team of eleven is num__22 years . the average age gets increased by num__1 year when the coach age is also included . what is the age of the coach ? <o> a ) num__84 yrs <o> b ) num__34 yrs <o> c ) num__64 yrs <o> d ) num__24 yrs <o> e ) num__44 yrs |
explanation : total age of players in team = num__22 x num__11 = num__242 when coach is included total members = num__12 avg age increases by one becomes num__23 total age when coach is included = num__23 x num__12 = num__276 age of coach = num__276 - num__242 = num__34 yrs . answer : b <eor> b <eos> |
b |
multiply__22.0__11.0__ add__1.0__11.0__ add__22.0__1.0__ multiply__12.0__23.0__ add__22.0__12.0__ add__22.0__12.0__ |
multiply__22.0__11.0__ add__1.0__11.0__ add__22.0__1.0__ multiply__12.0__23.0__ subtract__276.0__242.0__ subtract__276.0__242.0__ |
| a man can swim in still water at num__12 km / h but takes twice as long to swim upstream than downstream . the speed of the stream is ? <o> a ) num__4 <o> b ) num__4.2 <o> c ) num__5.3 <o> d ) num__1.5 <o> e ) num__5.2 |
m = num__12 s = x ds = num__12 + x us = num__12 - x num__12 + x = ( num__12 - x ) num__2 num__12 + x = num__24 - num__2 x num__3 x = num__12 x = num__4 answer : a <eor> a <eos> |
a |
multiply__12.0__2.0__ divide__12.0__3.0__ round__4.0__ |
multiply__12.0__2.0__ divide__12.0__3.0__ round__4.0__ |
| two trains running in opposite directions cross a man standing on the platform in num__27 seconds and num__17 seconds respectively and they cross each other in num__23 seconds . the ratio of their speeds is ? <o> a ) num__0.375 <o> b ) num__1.5 <o> c ) num__1.0 <o> d ) num__0.6 <o> e ) num__0.428571428571 |
let the speeds of the two trains be x m / sec and y m / sec respectively . then length of the first train = num__27 x meters and length of the second train = num__17 y meters . ( num__27 x + num__17 y ) / ( x + y ) = num__23 = = > num__27 x + num__17 y = num__23 x + num__23 y = = > num__4 x = num__6 y = = > x / y = num__1.5 . answer : b <eor> b <eos> |
b |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ round__1.5__ |
subtract__27.0__23.0__ subtract__23.0__17.0__ divide__6.0__4.0__ divide__6.0__4.0__ |
| a plane flies num__500 miles at a speed of r mph . if the plane flies at a speed of r + num__250 miles per hour it would fly the same distance in num__1 hour less . what is the value of r ? <o> a ) num__250 mph <o> b ) num__200 mph <o> c ) num__205 mph <o> d ) num__300 mph <o> e ) num__425 mph |
equation : num__500 / r = num__500 / ( r + num__50 ) + num__1 plug options into the equation a fits : num__2.0 = num__500 / ( num__250 + num__250 ) + num__1 answer : a <eor> a <eos> |
a |
divide__500.0__250.0__ round__250.0__ |
divide__500.0__250.0__ divide__500.0__2.0__ |
| oliver travelled for num__15 hours . he covered the first half of the distance at num__40 kmph and remaining half of the distance at num__45 kmph . find the distance travelled by oliver ? <o> a ) num__280 km <o> b ) num__634 km <o> c ) num__760 km <o> d ) num__720 km <o> e ) num__612 km |
let the distance travelled be x km . total time = ( x / num__2 ) / num__40 + ( x / num__2 ) / num__45 = num__15 = > x / num__80 + x / num__90 = num__15 = > ( num__9 x + num__8 x ) / num__720 = num__15 = > x = num__720 km answer : d <eor> d <eos> |
d |
multiply__40.0__2.0__ multiply__45.0__2.0__ multiply__9.0__80.0__ round__720.0__ |
multiply__40.0__2.0__ multiply__45.0__2.0__ multiply__9.0__80.0__ round__720.0__ |
| the element being searched for is not found in an array of num__100 elements . what is the average number of comparisons needed in a sequential search to determine that the element is not there if the elements are completely unordered ? <o> a ) num__25 <o> b ) num__50 <o> c ) num__75 <o> d ) num__100 <o> e ) num__125 |
let us assume k be the element we need to search then let array be n num__1 n num__2 n num__3 . . . . . . . . . . . . . . n num__100 now to prove that the element searched is not there in the array we have to compare the element k with every element in the array . so totally we need num__100 comparisons . answer : d <eor> d <eos> |
d |
add__1.0__2.0__ multiply__100.0__1.0__ |
add__1.0__2.0__ multiply__100.0__1.0__ |
| if ( a + b ) = num__11 ( b + c ) = num__9 and ( c + d ) = num__3 what is the value of ( a + d ) ? <o> a ) num__16 . <o> b ) num__8 . <o> c ) num__5 . <o> d ) num__2 . <o> e ) - num__2 . |
given a + b = num__11 = > a = num__11 - b - - > eq num__1 b + c = num__9 c + d = num__3 = > d = num__3 - c - - > eq num__2 then eqs num__1 + num__2 = > a + d = num__11 - b + num__3 - c = > num__14 - ( b + c ) = > num__14 - num__9 = num__5 . option c . . . <eor> c <eos> |
c |
subtract__11.0__9.0__ add__11.0__3.0__ add__3.0__2.0__ add__3.0__2.0__ |
subtract__11.0__9.0__ add__11.0__3.0__ add__3.0__2.0__ add__3.0__2.0__ |
| a is two years older than b who is twice as old as c . if the total of the ages of a b and c be num__27 then how old is b ? <o> a ) num__76 years <o> b ) num__88 years <o> c ) num__55 years <o> d ) num__10 years <o> e ) num__15 years |
let c ' s age be x years . then b ' s age = num__2 x years . a ' s age = ( num__2 x + num__2 ) years . ( num__2 x + num__2 ) + num__2 x + x = num__27 num__5 x = num__25 = > x = num__5 hence b ' s age = num__2 x = num__10 years . answer : d <eor> d <eos> |
d |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
subtract__27.0__2.0__ multiply__2.0__5.0__ multiply__2.0__5.0__ |
| in how many years will a sum of money doubles itself at num__5.0 per annum on simple interest ? <o> a ) num__27.0 <o> b ) num__70.0 <o> c ) num__29.0 <o> d ) num__20.0 <o> e ) num__50 % |
p = ( p * num__5 * r ) / num__100 r = num__20.0 answer : d <eor> d <eos> |
d |
percent__20.0__100.0__ |
percent__20.0__100.0__ |
| a department of ten people - six men and four women - needs to send a team of five to a conference . if they want to make sure that there are no more than three members of the team from any one gender how many distinct groups are possible to send ? <o> a ) num__120 <o> b ) num__150 <o> c ) num__180 <o> d ) num__210 <o> e ) num__240 |
they can make a team of num__3 men and num__2 women . the number of ways to do this is num__6 c num__3 * num__4 c num__2 = num__20 * num__6 = num__120 they can make a team of num__2 men and num__3 women . the number of ways to do this is num__6 c num__2 * num__4 c num__3 = num__15 * num__4 = num__60 the total number of distinct groups is num__180 . the answer is c . <eor> c <eos> |
c |
multiply__2.0__3.0__ subtract__6.0__2.0__ multiply__6.0__20.0__ multiply__3.0__20.0__ multiply__3.0__60.0__ multiply__3.0__60.0__ |
multiply__2.0__3.0__ subtract__6.0__2.0__ multiply__6.0__20.0__ multiply__3.0__20.0__ multiply__3.0__60.0__ multiply__3.0__60.0__ |
| a batsman scored num__120 runs which included num__3 boundaries and num__8 sixes . what percent of his total score did he make by running between the wickets ? <o> a ) num__50.0 <o> b ) num__40.0 <o> c ) num__60.0 <o> d ) num__70.0 <o> e ) num__80 % |
number of runs made by running = num__110 - ( num__3 x num__4 + num__8 x num__6 ) = num__120 - ( num__60 ) = num__60 now we need to calculate num__60 is what percent of num__120 . = > num__0.5 x num__100 = num__50.0 answer : a <eor> a <eos> |
a |
divide__3.0__6.0__ multiply__0.5__100.0__ multiply__0.5__100.0__ |
divide__3.0__6.0__ subtract__110.0__60.0__ subtract__110.0__60.0__ |
| in a certain game of dice the player ’ s score is determined as a sum of two throws of a single die . the player with the highest score wins the round . if more than one player has the highest score the winnings of the round are divided equally among these players . if john plays this game against num__22 other players what is the probability of the minimum score that will guarantee john some monetary payoff ? <o> a ) num__0.82 <o> b ) num__0.00452488687783 <o> c ) num__0.00462962962963 <o> d ) num__0.0277777777778 <o> e ) num__0.0238095238095 |
to guarantee that john will get some monetary payoff he must score the maximum score of num__6 + num__6 = num__12 because if he gets even one less than that so num__11 someone can get num__12 and john will get nothing . p ( num__12 ) = num__0.166666666667 ^ num__2 = num__0.0277777777778 . answer : d . <eor> d <eos> |
d |
reverse__6.0__ divide__22.0__11.0__ divide__0.1667__6.0__ divide__0.1667__6.0__ |
reverse__6.0__ divide__22.0__11.0__ divide__0.1667__6.0__ divide__0.1667__6.0__ |
| in a two - digit if it is known that its unit ' s digit exceeds its ten ' s digit by num__2 and that the product of the given number and the sum of its digits is equal to num__144 then the number is : <o> a ) num__24 <o> b ) num__26 <o> c ) num__32 <o> d ) num__42 <o> e ) num__46 |
let the ten ' s digit be x . then unit ' s digit = x + num__2 . number = num__10 x + ( x + num__2 ) = num__11 x + num__2 . sum of digits = x + ( x + num__2 ) = num__2 x + num__2 . ( num__11 x + num__2 ) ( num__2 x + num__2 ) = num__144 num__22 x num__2 + num__26 x - num__140 = num__0 num__11 x num__2 + num__13 x - num__70 = num__0 ( x - num__2 ) ( num__11 x + num__35 ) = num__0 x = num__2 . hence required number = num__11 x + num__2 = num__24 . answer : a <eor> a <eos> |
a |
multiply__2.0__11.0__ add__2.0__11.0__ divide__140.0__2.0__ divide__70.0__2.0__ add__2.0__22.0__ add__2.0__22.0__ |
multiply__2.0__11.0__ add__2.0__11.0__ divide__140.0__2.0__ add__13.0__22.0__ add__2.0__22.0__ add__2.0__22.0__ |
| x starts a business with rs . num__45000 . y joins in the business after num__3 months with rs . num__45000 . what will be the ratio in which they should share the profit at the end of the year ? <o> a ) num__1 : num__2 <o> b ) num__2 : num__1 <o> c ) num__3 : num__2 <o> d ) num__2 : num__3 <o> e ) num__4 : num__3 |
ratio in which they should share the profit = ratio of the investments multiplied by the time period = num__45000 × num__12 : num__45000 × num__9 = num__45 × num__12 : num__45 × num__9 = num__3 × num__12 : num__2 × num__9 = num__4 : num__3 answer is e . <eor> e <eos> |
e |
subtract__12.0__3.0__ divide__12.0__3.0__ divide__12.0__3.0__ |
subtract__12.0__3.0__ divide__12.0__3.0__ divide__12.0__3.0__ |
| last year company x paid out a total of $ num__1 num__050000 in salaries to its num__43 employees . if no employee earned a salary that is more than num__20.0 greater than any other employee what is the lowest possible salary that any one employee earned ? <o> a ) $ num__20 num__428.01 <o> b ) $ num__41667 <o> c ) $ num__42000 <o> d ) $ num__50000 <o> e ) $ num__60 |
000 |
employee num__1 earned $ x ( say ) employee num__2 will not earn more than $ num__1.2 x therfore to minimize the salary of any one employee we need to maximize the salaries of the other num__42 employees ( num__1.2 x * num__42 ) + x = num__1 num__050000 solving for x = $ num__20 num__428.01 answer a <eor> a <eos> |
a |
a |
| when n liters of fuel was added to a tank that was already num__0.666666666667 full the tank was filled to num__0.888888888889 of its capacity . in terms of n what is the capacity of the tank in liters ? <o> a ) num__1.11111111111 n <o> b ) num__1.33333333333 n <o> c ) num__4.5 n <o> d ) num__2.25 n <o> e ) num__2.33333333333 n |
c is the answer . to solve this problem draw diagram or use algebra . i ' m more comfortable with algebra . given - tank was already num__0.666666666667 full . - when n ltr added it became num__0.888888888889 so num__0.666666666667 + n = num__0.888888888889 n = num__0.888888888889 - num__0.666666666667 n = num__0.222222222222 capacity of the tank is full fraction . i . e . num__1 i . e . num__1.0 so the question is num__1.0 is how much times n = num__1.0 = num__1 = xn so x = num__4.5 and total = num__4.5 n <eor> c <eos> |
c |
subtract__0.8889__0.6667__ multiply__4.5__1.0__ |
subtract__0.8889__0.6667__ multiply__4.5__1.0__ |
| bob invested one half of his savings in a bond that paid simple interest for num__2 years and received $ num__500 as interest . he invested the remaining in a bond that paid compound interest ( compounded annually ) for the same num__2 years at the same rate of interest and received $ num__600 as interest . what was the annual rate of interest ? <o> a ) num__5.0 <o> b ) num__10.0 <o> c ) num__12.0 <o> d ) num__4.0 <o> e ) num__20 % |
first divide $ num__250.0 years to get num__250 $ / yr non - compounding interest . next subtract from the total compound after num__2 years to see the interest gained in the second year . this can be done because the same amount has been invested in both accounts and the interest gained for the first year will be the same . so $ num__600 - $ num__250 = $ num__350 . to find the difference in interest gained $ num__350 - $ num__250 = $ num__100 increase in interest after the first year through compounding . from this information we can set up the equation $ num__250 * x % = num__4.0 giving us answer choice d . <eor> d <eos> |
d |
percent__4.0__100.0__ |
percent__4.0__100.0__ |
| a fill pipe can fill num__0.2 of tank in num__25 minutes in how many minutes it can fill num__0.8 of the cistern ? <o> a ) num__48 min <o> b ) num__80 min <o> c ) num__100 min <o> d ) num__50 min <o> e ) num__56 min |
num__0.2 of the cistern can fill in num__25 min num__0.8 of the tank can fill in = num__25 * num__5 * num__0.8 = num__100 min answer is c <eor> c <eos> |
c |
multiply__0.2__25.0__ round__100.0__ |
multiply__0.2__25.0__ round__100.0__ |
| at a dinner party num__5 people are to be seated around a circular table . num__2 seating arrangements are considered different only when the positions of the people are different relative to each other . what is the total number w of different possible seating arrangements for the group ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__24 <o> d ) num__32 <o> e ) num__120 |
circular arrangements = ( n - num__1 ) ! hence ( num__4 ! ) = num__24 = w answer is c <eor> c <eos> |
c |
choose__4.0__2.0__ choose__4.0__2.0__ |
choose__4.0__2.0__ choose__4.0__2.0__ |
| if both num__5 ^ num__2 and num__3 ^ num__3 are factors of n x ( num__2 ^ num__5 ) x ( num__12 ^ num__2 ) x ( num__7 ^ num__3 ) x ( num__10 ) what is the smallest possible positive value of n ? <o> a ) num__15 <o> b ) num__45 <o> c ) num__75 <o> d ) num__125 <o> e ) num__150 |
( num__2 ^ num__5 ) x ( num__12 ^ num__2 ) x ( num__7 ^ num__3 ) x ( num__10 ) has two appearances of num__3 ( in num__12 ^ num__2 ) and one appearance of num__5 ( in the num__10 ) . thus n must include at least num__3 * num__5 = num__15 the answer is a . <eor> a <eos> |
a |
multiply__5.0__3.0__ multiply__5.0__3.0__ |
multiply__5.0__3.0__ multiply__5.0__3.0__ |
| if num__2 ^ num__4 num__3 ^ num__3 and num__10 ^ num__3 are factors of the product of num__1452 and w where w is a positive integer what is the smallest possible value of w ? <o> a ) num__198 <o> b ) num__288 <o> c ) num__360 <o> d ) num__396 <o> e ) num__484 |
i will go with c ( pending elements to match is num__2 ^ num__2 * num__3 ^ num__2 * num__10 ^ num__1 = num__360 <eor> c <eos> |
c |
subtract__4.0__3.0__ multiply__360.0__1.0__ |
subtract__4.0__3.0__ multiply__360.0__1.0__ |
| a certain type of concrete mixture is to be made of cement sand and graved ( gravel ? ) in a ratio num__1 : num__3 : num__5 by weight . what is the greatest number of kilograms of this mixture that can be made with num__5 kilograms of cement ? <o> a ) num__13 num__0.5 <o> b ) num__15 <o> c ) num__25 <o> d ) num__40 <o> e ) num__45 |
we ' re given a ratio of cement sand and gravel as num__1 : num__3 : num__5 by weight . this means that for every num__1 kg of cement we ' ll mix in num__3 kg of sand and num__5 kg of gravel . . . . num__1 + num__3 + num__5 = num__9 kg of concrete we ' re asked for the greatest number of kilograms of this mixture that can be made with num__5 kilograms of cement . . . from the prior work we know that num__1 kg of cement will get us num__9 km of concrete . since we ' re dealing with num__5 kg of cement we ' ll end up with ( num__5 ) ( num__9 ) = num__45 kg of concrete . final answer : e <eor> e <eos> |
e |
multiply__5.0__9.0__ multiply__1.0__45.0__ |
multiply__5.0__9.0__ multiply__1.0__45.0__ |
| an engineer designed a ball so that when it was dropped it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a num__16 - meter platform and caught it after it had traveled num__44 meters . how many times did the ball bounce ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__7 <o> d ) num__3 <o> e ) num__9 |
going down = num__16 m going up = num__8 - - > total = num__24 going down = num__8 - - > total = num__32 going up = num__4 - - > total = num__36 going down = num__4 - - > total = num__40 going up = num__2 - - > total = num__42 going down = num__2 - - > total = num__44 ( caught ) no of bounce = num__3 . . answer : d <eor> d <eos> |
d |
add__16.0__8.0__ add__8.0__24.0__ divide__32.0__8.0__ subtract__44.0__8.0__ add__16.0__24.0__ divide__16.0__8.0__ subtract__44.0__2.0__ divide__24.0__8.0__ divide__24.0__8.0__ |
add__16.0__8.0__ add__8.0__24.0__ divide__32.0__8.0__ subtract__44.0__8.0__ subtract__44.0__4.0__ divide__16.0__8.0__ subtract__44.0__2.0__ divide__24.0__8.0__ divide__24.0__8.0__ |
| the proportion of copper and zinc in the brass is num__13 : num__7 . how much zinc will there be in num__100 kg of brass ? <o> a ) num__73 <o> b ) num__35 <o> c ) num__83 <o> d ) num__63 <o> e ) num__83 |
explanation : num__0.35 * num__100 = num__35 answer : option b <eor> b <eos> |
b |
multiply__100.0__0.35__ multiply__100.0__0.35__ |
multiply__100.0__0.35__ multiply__100.0__0.35__ |
| the ratio of cost price to the marked price of an article is num__4 : num__5 . if the ratio of the profit percentage on selling the article to the discount allow ed on it is num__5 : num__4 w hat is the profit percentage ? <o> a ) num__10.0 <o> b ) num__12.0 <o> c ) num__12.5 <o> d ) num__15.0 <o> e ) num__17 % |
cp : mp = num__4 : num__5 so assume cp = num__400 and mp = num__500 . profit % : discount = num__5 : num__4 . so let num__5 x and num__4 x . now from the problem after giving discount on marked price still he has a profit too . num__500 - ( num__4 x % of num__500 ) = num__400 + ( num__5 x % of num__400 ) num__500 - num__2 x = num__400 + num__2 x solving we get x = num__2.5 profit percentage = num__5 x = num__5 ( num__2.5 ) = num__12.5 answer : c <eor> c <eos> |
c |
percent__2.5__500.0__ percent__2.5__500.0__ |
percent__2.5__500.0__ percent__2.5__500.0__ |
| a sum of money amounts to rs . num__6690 after num__3 years and to rs . num__10035 after num__6 years on compound interest . find the sum . <o> a ) num__4360 <o> b ) num__4460 <o> c ) num__4580 <o> d ) num__4690 <o> e ) num__5480 |
let the sum be rs . p . then p ( num__1 + r / num__100 ) ^ num__3 = num__6690 … ( i ) and p ( num__1 + r / num__100 ) ^ num__6 = num__10035 … ( ii ) on dividing we get ( num__1 + r / num__100 ) ^ num__3 = num__1.49850523169 = num__1.5 . substituting this value in ( i ) we get : p * ( num__1.5 ) = num__6690 or p = ( num__6690 * num__0.666666666667 ) = num__4460 hence the sum is rs . num__4460 . answer : b <eor> b <eos> |
b |
percent__100.0__4460.0__ |
percent__100.0__4460.0__ |
| the lenght of a room is num__5.5 m and width is num__4 m . find the cost of paving the floor by slabs at the rate of rs . num__800 per sq . metre . <o> a ) rs . num__15550 <o> b ) rs . num__15600 <o> c ) rs . num__16500 <o> d ) rs . num__17600 <o> e ) rs . num__17 |
900 |
area of the floor = ( num__5.5 × num__4 ) m num__2 = num__22 m num__2 . cost of paving = rs . ( num__800 × num__22 ) = rs . num__17600 . answer : option d <eor> d <eos> |
d |
d |
| a is half good a work man as b and together they finish a job in num__18 days . in how many days working alone b finish the job ? <o> a ) num__27 days <o> b ) num__21 days <o> c ) num__17 days <o> d ) num__18 days <o> e ) num__19 days |
wc = num__1 : num__2 num__2 x + x = num__0.0555555555556 = > x = num__0.0185185185185 num__2 x = num__0.0185185185185 = > num__27 days answer : a <eor> a <eos> |
a |
divide__1.0__18.0__ round__27.0__ |
divide__1.0__18.0__ round__27.0__ |
| the area of a square field num__3136 sq m if the length of cost of drawing barbed wire num__3 m around the field at the rate of rs . num__2.0 per meter . two gates of num__1 m width each are to be left for entrance . what is the total cost ? <o> a ) s . num__1014 <o> b ) s . num__1332 <o> c ) s . num__999 <o> d ) s . num__1085 <o> e ) s . num__1020 |
a num__2 = num__3136 = > a = num__56 num__56 * num__4 * num__3 = num__672 – num__6 = num__666 * num__2.0 = num__1332 answer : b <eor> b <eos> |
b |
add__3.0__1.0__ multiply__3.0__2.0__ subtract__672.0__6.0__ multiply__2.0__666.0__ round__1332.0__ |
add__3.0__1.0__ multiply__3.0__2.0__ subtract__672.0__6.0__ multiply__2.0__666.0__ multiply__2.0__666.0__ |
| ross has num__80 shirts num__0.75 of the shirts are green and num__0.1 is without buttons . therefore ross has between ___ and ___ shirts with buttons that are not green . <o> a ) num__12 ; num__20 . <o> b ) num__12 ; num__19 . <o> c ) num__12 ; num__18 . <o> d ) num__12 ; num__17 . <o> e ) num__12 ; num__16 . |
total shirts = num__80 green shirts = num__0.75 * num__80 = num__60 non green shirts = num__20 shirts without button = num__0.1 * num__80 = num__8 shirts with button = num__72 required : range of shirts with buttons that are not green . maximum non green shirts with buttons = number of non green shirts = num__20 minimum non green shirts with buttons ( all without button shirts are non green ) = non green shirts - shirts without button = num__20 - num__8 = num__12 hence the range would be ( num__12 num__20 ) correct option : a <eor> a <eos> |
a |
multiply__80.0__0.75__ subtract__80.0__60.0__ multiply__80.0__0.1__ subtract__80.0__8.0__ subtract__72.0__60.0__ subtract__72.0__60.0__ |
multiply__80.0__0.75__ subtract__80.0__60.0__ multiply__80.0__0.1__ subtract__80.0__8.0__ subtract__72.0__60.0__ subtract__72.0__60.0__ |
| tea worth rs . num__126 per kg and rs . num__135 per kg are mixed with a third variety of tea in the ratio num__1 : num__1 : num__2 . if the mixture is worth rs . num__153 per kg what is the price of the third variety per kg ? <o> a ) num__175.5 <o> b ) num__182.5 <o> c ) num__170.0 <o> d ) num__180.0 <o> e ) num__190.0 |
tea worth rs . num__126 ratio num__1 : num__1 average price = ( num__126 + num__135 ) / num__2 = num__130.5 mean price = ( x - num__153 ) : num__22.50 = > x - num__153 = num__22.50 x = num__175.5 answer a <eor> a <eos> |
a |
subtract__153.0__130.5__ add__153.0__22.5__ multiply__1.0__175.5__ |
subtract__153.0__130.5__ add__153.0__22.5__ add__153.0__22.5__ |
| a sum of money deposited at c . i . amounts to rs . num__2420 in num__2 years and to rs . num__3267 in num__3 years . find the rate percent ? <o> a ) num__11 <o> b ) num__10 <o> c ) num__28 <o> d ) num__24 <o> e ) num__35 |
explanation : num__2420 - - - num__847 num__100 - - - ? = > num__35.0 answer : option e <eor> e <eos> |
e |
percent__100.0__35.0__ |
percent__100.0__35.0__ |
| on a scale that measures the intensity of a certain phenomenon a reading of d + num__1 corresponds to an intensity that is num__10 times the intensity corresponding to a reading of d . on that scale the intensity corresponding to a reading of num__8 is how many times as great as the intensity corresponding to a reading of num__3 ? <o> a ) num__5 <o> b ) num__50 <o> c ) num__10 ^ num__5 <o> d ) num__5 ^ num__10 <o> e ) num__8 ^ num__10 - num__3 ^ num__10 |
to solve this problem we need to examine the information in the first sentence . we are told that “ a reading of d + num__1 corresponds to an intensity that is num__10 times the intensity corresponding to a reading of d . ” let ’ s practice this idea with some real numbers . let ’ s say d is num__2 . this means that d + num__1 = num__3 . with the information we were given we can say that a reading of num__3 is ten times as great as the intensity of a reading of num__2 . furthermore we can say that a reading of num__4 is actually num__10 x num__10 = num__10 ^ num__2 times as great as the intensity of a reading of num__2 . increasing one more unit we can say that a reading of num__5 is num__10 x num__10 x num__10 = num__10 ^ num__3 times as great as the intensity of a reading of num__2 . we have found a pattern which can be applied to the problem presented in the stem : num__3 is “ one ” unit away from num__2 and thus a reading of num__3 is num__10 ^ num__1 times as great as the intensity of a reading of num__2 . num__4 is “ two ” units away from num__2 and thus a reading of num__4 is num__10 ^ num__2 times as great as the intensity of a reading of num__2 . num__5 is “ three ” units away from num__2 and thus a reading of num__5 is num__10 ^ num__3 times as great as the intensity of a measure of num__2 . we can use this pattern to easily answer the question . here we are being asked for the number of times the intensity corresponding to a reading of num__8 is as great as the intensity corresponding to a reading of num__3 . because num__8 is num__5 units greater than num__3 a reading of num__8 is num__10 ^ num__5 times as great as the intensity corresponding to a reading of num__3 . answer c . <eor> c <eos> |
c |
subtract__10.0__8.0__ add__1.0__3.0__ add__1.0__4.0__ multiply__1.0__10.0__ |
subtract__10.0__8.0__ add__1.0__3.0__ add__1.0__4.0__ add__8.0__2.0__ |
| which of the following points is closest to line y = x <o> a ) ( num__2 - num__1 ) <o> b ) ( num__2 num__4 ) <o> c ) ( - num__1 - num__2 ) <o> d ) ( - num__2 num__1 ) <o> e ) ( - num__2 num__0 ) |
attachment : m num__12 - num__20 . pngas you can see point ( - num__1 - num__2 ) is the closest to line y = x . answer : c . <eor> c <eos> |
c |
reverse__1.0__ |
subtract__2.0__1.0__ |
| a train num__150 m long passes a man running at num__5 km / hr in the same direction in which the train is going in num__10 seconds . the speed of the train is ? <o> a ) num__36 <o> b ) num__50 <o> c ) num__88 <o> d ) num__59 <o> e ) num__22 |
speed of the train relative to man = ( num__15.0 ) m / sec = num__15 m / sec . [ num__15 * ( num__3.6 ) ] km / hr = num__54 km / hr . let the speed of the train be x km / hr . then relative speed = ( x - num__5 ) km / hr . x - num__5 = num__54 = = > x = num__59 km / hr . answer : d <eor> d <eos> |
d |
divide__150.0__10.0__ multiply__3.6__15.0__ add__5.0__54.0__ round__59.0__ |
divide__150.0__10.0__ multiply__3.6__15.0__ add__5.0__54.0__ round__59.0__ |
| the radius of a cylindrical water tank is reduced by num__50.0 . however the speed by which water is filled into the tank is also decreased by num__50.0 . how much more or less time q will it take to fill the tank now ? <o> a ) num__50.0 less time <o> b ) num__50.0 more time <o> c ) num__75.0 less time <o> d ) num__75.0 more time <o> e ) num__100.0 more time |
( vc ) volume of the cylinderical vessal is directly proportional to r ^ num__2 . so if radius is num__50.0 less volume will be num__0.25 th of the original volume . ( vc / num__4 ) now if with velocity v tank can be filled in t num__1 time of volume vc so now velocity is num__50.0 less i . . e v / num__2 so time taken to fill the capacity vc / num__4 by v / num__2 velocity is t num__2 . vt num__1 = vc v / num__2 * t num__2 = vc / num__4 so t num__1 / t num__2 = num__0.5 so tank will be filled in less time . that is q = num__50.0 less . a <eor> a <eos> |
a |
reverse__0.25__ multiply__0.25__4.0__ reverse__2.0__ multiply__50.0__1.0__ |
reverse__0.25__ multiply__0.25__4.0__ reverse__2.0__ multiply__50.0__1.0__ |
| num__2 num__3 num__6 num__0 num__10 - num__3 num__14 ( . . . ) <o> a ) num__6 <o> b ) num__2 <o> c ) - num__2 <o> d ) num__0 <o> e ) - num__6 |
there are two series num__2 num__6 num__10 num__14 . . . ( adding num__4 ) num__3 num__0 - num__3 . . . ( subtracting num__3 ) hence next term is - num__3 - num__3 = - num__6 answer is e . <eor> e <eos> |
e |
subtract__6.0__2.0__ multiply__2.0__3.0__ |
subtract__6.0__2.0__ subtract__10.0__4.0__ |
| during a certain week a seal ate num__60.0 of the first num__80 smelt it came across and num__30.0 of the remaining smelt it came across . if the seal ate num__40.0 of the smelt it came across during the entire week how many smelt did it eat ? <o> a ) num__32 <o> b ) num__40 <o> c ) num__55 <o> d ) num__64 <o> e ) num__96 |
total smelt = x . then num__0.6 * num__80 + num__0.3 ( x - num__80 ) = num__0.4 * x - - > x = num__240 - - > num__0.4 * x = num__96 . answer : e . <eor> e <eos> |
e |
percent__40.0__240.0__ percent__40.0__240.0__ |
percent__40.0__240.0__ percent__40.0__240.0__ |
| a number when divided by a certain divisor left remainder num__241 when twice the number was divided by the same divisor the remainder was num__115 . find the divisor ? <o> a ) num__370 <o> b ) num__365 <o> c ) num__380 <o> d ) num__367 <o> e ) num__460 |
easy solution : n = dq num__1 + num__241 num__2 n = num__2 dq num__1 + num__482 - ( num__1 ) num__2 n = dq num__2 + num__115 - ( num__2 ) as ( num__1 ) = ( num__2 ) = num__2 n d * ( q num__2 - num__2 q num__1 ) = num__367 d * some integer = num__367 checking all options only ( d ) syncs with it . answer d <eor> d <eos> |
d |
multiply__241.0__2.0__ subtract__482.0__115.0__ multiply__1.0__367.0__ |
multiply__241.0__2.0__ subtract__482.0__115.0__ multiply__1.0__367.0__ |
| if a lends rs . num__3500 to b at num__10.0 per annum and b lends the same sum to c at num__13.0 per annum then the gain of b in a period of num__3 years is ? <o> a ) num__320 <o> b ) num__300 <o> c ) num__310 <o> d ) num__315 <o> e ) num__330 |
( num__3500 * num__3 * num__3 ) / num__100 = > num__315 answer : d <eor> d <eos> |
d |
percent__100.0__315.0__ |
percent__100.0__315.0__ |
| how many minutes does aditya take to cover a distance of num__400 m if he runs at a speed of num__20 km / hr ? <o> a ) num__1 num__0.2 min <o> b ) num__4 num__0.2 min <o> c ) num__3 num__0.2 min <o> d ) num__2 num__0.2 min <o> e ) none of these |
explanation : we know that time = distance / speedspeed = num__20 km / hr = num__20 x num__0.277777777778 m / sec = num__509 m / sectime = ( num__400 x num__0.18 ) = num__72 sec = num__1 num__0.2 min answer : a <eor> a <eos> |
a |
multiply__400.0__0.18__ round__1.0__ |
multiply__400.0__0.18__ round__1.0__ |
| how many num__8 ' s are there preceded by num__0 but not followed by num__1 ? num__5 num__0 num__8 num__1 num__2 num__7 num__4 num__2 num__6 num__9 num__7 num__4 num__6 num__1 num__3 num__0 num__8 num__7 num__4 num__1 num__0 num__8 num__3 num__2 num__5 num__6 num__7 num__4 num__3 num__9 num__5 num__8 num__2 num__0 num__8 num__2 num__7 num__4 num__6 num__3 <o> a ) num__3 <o> b ) num__5 <o> c ) num__6 <o> d ) num__8 <o> e ) num__9 |
num__0 num__8 num__2 num__0 num__8 num__7 num__0 num__8 num__3 only at these places num__3 is preceded by num__0 but not followed by num__1 answer : a <eor> a <eos> |
a |
subtract__8.0__5.0__ |
subtract__8.0__5.0__ |
| shawn is planning a bus trip across town that involves three buses . bus num__1 travels between shawn ’ s house and downtown and it leaves every half - hour starting at num__7 : num__20 am . shawn will need to be on bus num__1 for num__1.2 hours . bus num__2 travels between downtown and uptown every half - hour starting at num__7 : num__10 am . shawn will need to be on bus num__2 for num__0.666666666667 hour . lastly bus num__3 travels between uptown and shawn ’ s destination every hour starting at num__9 am . assuming all buses stay on schedule what is the least amount of time shawn must spend waiting for buses ? <o> a ) num__12 minutes <o> b ) num__18 minutes <o> c ) num__48 minutes <o> d ) num__1 hour num__12 minutes <o> e ) num__1 hour num__20 minutes |
here ' s my solution . my first approach was to brute force the objective function in this optimization problem minimize time spent waiting between buses givens : num__1.2 hour = num__1 hour num__12 minutes num__2 num__0.666666666667 hour = num__2 hour num__40 minutes once you start filling out a table you realize that there ' s only num__2 solutions to this problem leaving at the num__20 minute mark of any hour versus leaving at the num__50 minute mark of any hour . [ reveal ] spoiler : answer b <eor> b <eos> |
b |
multiply__1.2__10.0__ multiply__20.0__2.0__ add__10.0__40.0__ subtract__20.0__2.0__ |
multiply__1.2__10.0__ multiply__20.0__2.0__ add__10.0__40.0__ subtract__20.0__2.0__ |
| what will be the product of ( num__25 - num__1 ) * ( num__25 - num__2 ) * ( num__25 - num__3 ) * . . . . . . . . . . . . . . * ( num__25 - num__99 ) * ( num__25 - num__100 ) ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__625 <o> d ) less than - num__100000 <o> e ) more than num__100 |
000 |
one of the terms is ( num__25 - num__25 ) so the product is num__0 . the answer is a . <eor> a <eos> |
a |
a |
| a crate measures num__4 feet by num__8 feet by num__12 feet on the inside . a stone pillar in the shape of a right circular cylinder must fit into the crate for shipping so that it rests upright when the crate sits on at least one of its six sides . what is the radius in feet of the pillar with the largest volume that could still fit in the crate ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__12 |
to fit the cylinder with largest radius inside this cuboid we should make the base of the crate as wide as possible so we will take the base as num__12 feet by num__8 feet now since the limiting number in the base is num__8 feet ; therefore a cylinder { we can visualise that a cylinder ' s width is its diameter } can only fit inside the crate if it is num__8 feet or less . therefore the radius of the cylinder will become diameter num__2 = = = > num__4.0 = num__4 diameter answer : d <eor> d <eos> |
d |
square_perimeter__2.0__ |
square_perimeter__2.0__ |
| if a certain number of people can dig earth num__100 m deep num__25 m long and num__30 m broad in num__12 days then find the number of days the same number of people will require to dig earth num__75 m deep num__20 m long and num__50 m broad . <o> a ) num__18 <o> b ) num__16 <o> c ) num__12 <o> d ) num__99 <o> e ) num__13 |
explanation : more number of days means – more length more depth and more width . hence it ’ s a direct proportion . ( num__100 * num__25 * num__30 ) : ( num__75 * num__20 * num__50 ) : : num__12 : x num__75000 : num__75000 : : num__12 : x x = num__12 answer : c <eor> c <eos> |
c |
round__12.0__ |
round__12.0__ |
| it takes ten minutes to fry a steak ( five minutes for each side ) . you are frying the steaks in a pan that can accommodate only two steaks at one time . what is the least amount of time by which you can fry all the three steaks you have ? <o> a ) num__15 minutes . <o> b ) num__17 minutes . <o> c ) num__19 minutes . <o> d ) num__15 minutes . <o> e ) num__12 minutes . |
d num__15 minutes . after frying for five minutes you can take out one steak and put the third one inside while turning the other one inside . after five more minutes one steak is fried from both sides and you can take it out . now put in the steak that we removed and turn the other one inside . after five more minutes they both will also be fried from both sides . <eor> a <eos> |
a |
round__15.0__ |
round__15.0__ |
| a can finish a work in num__18 days and b can do the same work in half the time taken by a . then working together what part of the same work they can finish in a day ? <o> a ) num__0.25 <o> b ) num__0.166666666667 <o> c ) num__0.666666666667 <o> d ) num__0.5 <o> e ) num__0.125 |
given that b alone can complete the same work in half the time taken by a = num__9 days a ’ s one day work = num__0.0555555555556 b ’ s one day work = num__0.111111111111 ( a + b ) ’ s one day work = num__0.0555555555556 + num__0.111111111111 = num__0.166666666667 . answer is b . <eor> b <eos> |
b |
add__0.1111__0.0556__ add__0.1111__0.0556__ |
add__0.1111__0.0556__ add__0.1111__0.0556__ |
| simplify ( num__3 ^ y + num__3 ^ y + num__3 ^ y + num__3 ^ y + num__3 ^ y ) ( num__5 ^ y + num__5 ^ y + num__5 ^ y ) <o> a ) num__15 ^ y + num__1 <o> b ) num__12 ^ y + num__1 <o> c ) num__16 ^ y + num__9 ^ y <o> d ) num__12 ^ y <o> e ) num__4 ^ y * num__12 ^ y |
a ( num__3 ^ y + num__3 ^ y + num__3 ^ y + num__3 ^ y + num__3 ^ y ) ( num__5 ^ y + num__5 ^ y + num__5 ^ y ) ( num__5 * num__3 ^ y ) ( num__3 * num__5 ^ y ) num__15 * num__15 ^ y = num__15 ^ ( y + num__1 ) . . <eor> a <eos> |
a |
multiply__3.0__5.0__ multiply__3.0__5.0__ |
multiply__3.0__5.0__ multiply__3.0__5.0__ |
| if x and y are different integers and x ^ num__2 = xy which of the following must be true ? <o> a ) x = num__0 <o> b ) x = num__6 <o> c ) x = num__3 <o> d ) x = num__2 <o> e ) x = num__1 |
( num__1 ) when x is num__0 above eqn is satisfied regardless of y - true ( num__2 ) when x is not zero the eqn is not satified eg whn x = num__2 num__2 * num__2 is not equal to num__2 * num__0 - false ( num__3 ) when x = - y clearly product xy is negative . but x ^ num__2 is positive - doesnt satify eqn answer a <eor> a <eos> |
a |
add__2.0__1.0__ multiply__2.0__0.0__ |
add__2.0__1.0__ multiply__2.0__0.0__ |
| a is the average ( arithmetic mean ) of the first num__7 positive multiples of num__7 and b is the median of the first num__3 positive multiples of positive integer n . if the value of a ^ num__2 – b ^ num__2 is zero what is the value of n ? <o> a ) num__4 <o> b ) num__12 <o> c ) num__14 <o> d ) num__21 <o> e ) num__28 |
if a ^ num__2 - b ^ num__2 = num__0 then let ' s assume that a = b . a must equal the num__4 th positive multiple of num__4 thus a = num__28 which also equals b . b is the second positive multiple of n thus n = num__14.0 = num__14 . the answer is c . <eor> c <eos> |
c |
subtract__7.0__3.0__ multiply__7.0__4.0__ multiply__7.0__2.0__ multiply__7.0__2.0__ |
subtract__7.0__3.0__ multiply__7.0__4.0__ multiply__7.0__2.0__ subtract__28.0__14.0__ |
| two trains num__160 m and num__160 m long run at the speed of num__60 km / hr and num__40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? <o> a ) num__10.7 <o> b ) num__10.9 <o> c ) num__10.52 <o> d ) num__11.52 <o> e ) num__18.82 |
relative speed = num__60 + num__40 = num__100 km / hr . = num__100 * num__0.277777777778 = num__27.7777777778 m / sec . distance covered in crossing each other = num__160 + num__160 = num__320 m . required time = num__320 * num__0.036 = num__11.52 sec . answer : d <eor> d <eos> |
d |
subtract__160.0__60.0__ multiply__320.0__0.036__ round__11.52__ |
add__60.0__40.0__ multiply__320.0__0.036__ multiply__320.0__0.036__ |
| half of num__7 percent written as decimal is <o> a ) num__5 <o> b ) num__0.5 <o> c ) num__0.035 <o> d ) num__0.025 <o> e ) none of these |
explanation : it will be num__0.5 ( num__7.0 ) = num__0.5 ( num__0.07 ) = num__0.035 = num__0.035 answer : option c <eor> c <eos> |
c |
percent__7.0__0.5__ percent__7.0__0.5__ |
percent__7.0__0.5__ percent__7.0__0.5__ |
| fathers age is num__5 times his son ' s age . num__4 years back the father was num__9 times older than son . find the fathers ' present age . <o> a ) num__41 <o> b ) num__40 <o> c ) num__39 <o> d ) num__45 <o> e ) num__30 |
fathers age = x sons age = y x = num__5 y ; x - num__4 = num__9 ( y - num__4 ) num__5 y - num__4 = num__9 y - num__36 y = num__8 so x = num__40 answer : b <eor> b <eos> |
b |
multiply__4.0__9.0__ multiply__5.0__8.0__ multiply__5.0__8.0__ |
multiply__4.0__9.0__ multiply__5.0__8.0__ multiply__5.0__8.0__ |
| if $ x is invested at a constant annually compound interest rate of k percent what is the ratio of the total amount including interest after num__5 n years to that after num__4 n years ? <o> a ) ( num__1 + k / num__100 ) ^ n <o> b ) ( num__1 + k ) ^ n <o> c ) ( num__1 + kn / num__100 ) <o> d ) ( num__1 + n / num__100 ) ^ k <o> e ) ( num__1 + kn ) |
amount after num__5 n years will be = x ( num__1 + k / num__100 ) ^ num__5 n amount after num__4 n years will be = x ( num__1 + k / num__100 ) ^ num__4 n so required ratio will be - { x ( num__1 + k / num__100 ) ^ num__5 n } / { x ( num__1 + k / num__100 ) ^ num__4 n } = ( num__1 + k / num__100 ) ^ n hence answer will be a . ( num__1 + k / num__100 ) ^ n a <eor> a <eos> |
a |
percent__1.0__100.0__ |
percent__1.0__100.0__ |
| at a certain company each employee has a salary grade s that is at least num__1 and at most num__6 . each employee receives an hourly wage p in dollars determined by the formula p = num__9.50 + num__0.25 ( s – num__1 ) . an employee with a salary grade of num__6 receives how many more dollars per hour than an employee with a salary grade of num__1 ? <o> a ) $ num__0.50 <o> b ) $ num__1.00 <o> c ) $ num__1.25 <o> d ) $ num__1.50 <o> e ) $ num__1.75 |
salary grade of num__6 is p ( num__6 ) = num__9.50 + num__0.25 ( num__6 – num__1 ) = num__9.50 + num__0.25 * num__5 ; salary grade of num__1 is p ( num__1 ) = num__9.50 + num__0.25 ( num__1 – num__1 ) = num__9.50 ; p ( num__6 ) - p ( num__1 ) = num__9.50 + num__0.25 * num__5 - num__9.50 = num__1.25 . answer : c . <eor> c <eos> |
c |
subtract__6.0__1.0__ add__1.0__0.25__ add__1.0__0.25__ |
subtract__6.0__1.0__ add__1.0__0.25__ add__1.0__0.25__ |
| at a recent small town election for mayor a total of num__971 votes were cast for the four candidates the winner exceeding his opponents by num__53 num__79 and num__105 votes respectively . how many votes were cast for the candidate in fourth place ? <o> a ) num__134 <o> b ) num__178 <o> c ) num__197 <o> d ) num__166 <o> e ) num__194 |
the number of votes the winning candidate received was num__971 + num__53 + num__79 + num__26.25 = num__302 . the second received num__302 – num__53 = num__249 the fourth place received num__302 – num__105 = num__197 . c <eor> c <eos> |
c |
subtract__302.0__53.0__ subtract__302.0__105.0__ subtract__302.0__105.0__ |
subtract__302.0__53.0__ subtract__302.0__105.0__ subtract__302.0__105.0__ |
| - num__84 * num__29 + num__365 = ? <o> a ) num__2436 <o> b ) num__2801 <o> c ) - num__2801 <o> d ) - num__2071 <o> e ) none of these |
= > - num__84 * ( num__30 - num__1 ) + num__365 ; = > - ( num__84 * num__30 ) + num__84 + num__365 ; = > - num__2520 + num__449 = - num__2071 . correct option : d <eor> d <eos> |
d |
subtract__30.0__29.0__ multiply__84.0__30.0__ add__84.0__365.0__ subtract__2520.0__449.0__ multiply__1.0__2071.0__ |
subtract__30.0__29.0__ multiply__84.0__30.0__ add__84.0__365.0__ subtract__2520.0__449.0__ multiply__1.0__2071.0__ |
| what amount does kiran get if he invests rs . num__8000 at num__10.0 p . a . compound interest for two years compounding done annually ? <o> a ) num__2888 <o> b ) num__2669 <o> c ) num__2662 <o> d ) num__2992 <o> e ) num__9680 |
a = p { num__1 + r / num__100 } n = > num__8000 { num__1 + num__0.1 } num__2 = rs . num__9680 . answer : e <eor> e <eos> |
e |
percent__10.0__1.0__ percent__100.0__9680.0__ |
percent__10.0__1.0__ percent__100.0__9680.0__ |
| num__9 people decided to split the travel expenditures evenly . if the bill was $ num__546874 dollars how much amount in dollars should be added to the bill to make it split evenly across everyone ? <o> a ) $ num__546872 <o> b ) $ num__546874 <o> c ) $ num__546876 <o> d ) $ num__546878 <o> e ) $ num__546 |
870 |
this is equivalent to finding the first number that is divisible by num__9 that occurs after num__546874 . in order to divide the sum in num__9 parts the amount must be divisible by num__9 divisibility rule of num__9 : the sum of the digits must be divisible by num__9 sum of digits of num__546874 = num__34 and num__18 is divisible by num__9 . hence we need to add num__2 to this number for it to be divisible by num__9 correct option : c <eor> c <eos> |
c |
c |
| the remainder when dividing the expression ( x + y ) by num__5 is num__4 . the remainder of x divided by num__10 is num__2 . what is the remainder a of y divided by num__5 ? <o> a ) a = num__1 . <o> b ) a = num__2 . <o> c ) num__3 . <o> d ) num__4 . <o> e ) num__5 . |
x divided by num__10 gives reminder of num__2 . x can be num__0.166666666667 / num__22 . . . . x + y divided by num__5 is num__4 . x + y can be num__0.444444444444 / num__19 . . . . if x + y = num__4 and x is num__2 then y = num__2 and y / num__5 will give a reminder of num__2 similarly if x + y = num__9 and x = num__2 then y / num__5 will give a reminder of num__2 hence the answer must be num__2 ( b ) also . . . . x + y = num__5 m + num__4 and x = num__10 k + num__2 hence num__10 k + num__2 + y = num__5 m + num__4 or y = num__5 ( m - num__2 k ) + num__2 m - num__2 k being a constant remainder is num__2 hence answer is b <eor> b <eos> |
b |
add__5.0__4.0__ subtract__4.0__2.0__ |
add__5.0__4.0__ subtract__4.0__2.0__ |
| monika purchased a pressure cooker at num__0.9 th of its selling price and sold it at num__8.0 more than its s . p . find her gain percent . <o> a ) num__12.0 <o> b ) num__20.0 <o> c ) num__15.0 <o> d ) num__36.0 <o> e ) num__51 % |
let the s . p be rs . x . then c . p = rs . num__9 x / num__10 receipt = num__108.0 of rs . x = rs num__27 x / num__25 gain = rs ( num__27 x / num__25 * num__9 x / num__10 ) = rs ( num__108 x - num__90 x / num__100 ) = rs num__18 x / num__100 gain % = ( num__18 x / num__100 * num__1.11111111111 x * num__100 ) % = num__20.0 ans : b <eor> b <eos> |
b |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| the average temperature for monday tuesday wednesday and thursday was num__48 degrees and for tuesday wednesday thursday and friday was num__46 degrees . if the temperature on monday was num__42 degrees . find the temperature on friday ? <o> a ) num__15 <o> b ) num__26 <o> c ) num__34 <o> d ) num__25 <o> e ) num__26 |
m + tu + w + th = num__4 * num__48 = num__192 tu + w + th + f = num__4 * num__46 = num__184 m = num__42 tu + w + th = num__192 - num__42 = num__150 f = num__184 – num__150 = num__34 c <eor> c <eos> |
c |
subtract__46.0__42.0__ multiply__48.0__4.0__ multiply__46.0__4.0__ subtract__192.0__42.0__ subtract__184.0__150.0__ subtract__184.0__150.0__ |
subtract__46.0__42.0__ multiply__48.0__4.0__ multiply__46.0__4.0__ subtract__192.0__42.0__ subtract__184.0__150.0__ subtract__184.0__150.0__ |
| jacob marks his goods up by num__75.0 and then offers a discount on the marked price . the profit that he makes after offering the discount is num__40.0 . what % discount did jacob offer ? <o> a ) num__15.0 <o> b ) num__20.0 <o> c ) num__25.0 <o> d ) num__35.0 <o> e ) num__75 % |
solution : let x is the cost price . so profit num__40.0 means final selling price = num__1.40 x marked up price = num__1.75 x let discount be r % so num__1.40 x = num__1.75 x ( num__1 - r / num__100 ) after solving the equation r = num__20.0 answer : b <eor> b <eos> |
b |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| tim and É lan are num__150 miles away from one another . they are starting to move towards each other simultaneously tim at a speed of num__10 mph and É lan at a speed of num__5 mph . if every hour they double their speeds what is the distance that tim will pass until he meets É lan ? <o> a ) num__30 miles . <o> b ) num__35 miles . <o> c ) num__45 miles . <o> d ) num__90 miles . <o> e ) num__100 miles . |
tim and elan will meet at the same time while their ratio of speed is num__2 : num__1 respectively . so their individual distance traveled ratio will be same . plugging in the answer choice only answer choice e meet the num__2 : num__1 ( tim : elan = num__100 : num__50 ) ratio of maintaining total distance traveled num__150 miles socorrect answer e <eor> e <eos> |
e |
divide__10.0__5.0__ subtract__150.0__100.0__ round__100.0__ |
divide__10.0__5.0__ subtract__150.0__100.0__ round__100.0__ |
| the sum of the first n positive perfect squares where n is a positive integer is given by the formula n ^ num__1.0 + c * n ^ num__2 + n / num__6 where c is a constant . what is the sum of the first num__18 positive perfect squares ? <o> a ) num__1010 <o> b ) num__1164 <o> c ) num__1240 <o> d ) num__1316 <o> e ) num__2 |
109 |
first we need to find the constant ' c ' . the easiest way to find this is for the sum of the first two perfect squares for num__1 and num__2 = num__1 and num__4 respectively . hence lhs = num__1 + num__4 and plug n = num__2 for rhs and simplify to get c = num__0.5 . plug values of n = num__18 and c = num__0.5 into the equation and simplify to get the answer num__2109 . option e . <eor> e <eos> |
e |
e |
| abby and bobby type at constant rates of num__80 words per minute and num__60 words per minute respectively . bobby begins typing before abby and has typed num__400 words when abby begins typing at num__1 : num__30 pm . if they continue typing at their respective rates at what time will abby have typed exactly num__200 more words than bobby ? <o> a ) num__1 : num__40 pm <o> b ) num__1 : num__50 pm <o> c ) num__2 : num__00 pm <o> d ) num__2 : num__10 pm <o> e ) num__2 : num__20 pm |
say time needed for abby to type num__200 more words than bobby is t . in that time she would type num__80 t words and bobby would type num__60 t words . now total words typed by bobby would be num__600 + num__60 t and we want that number to be num__200 less than num__80 t : num__400 + num__60 t = num__80 t - num__200 - - > t = num__30 . num__1 : num__30 pm + num__30 minutes = num__2 : num__00 pm . answer : c . <eor> c <eos> |
c |
add__400.0__200.0__ divide__60.0__30.0__ divide__60.0__30.0__ |
add__400.0__200.0__ divide__60.0__30.0__ divide__60.0__30.0__ |
| { - num__10 - num__6 - num__5 - num__4 - num__2.5 - num__1 num__0 num__2.5 num__4 num__6 num__7 num__10 } a number is to be selected at random from the set above . what is the probability that the number will be a solution to the equation ( x num__3 ) ( x + num__14 ) ( num__2 x + num__5 ) = num__0 ? <o> a ) num__0.0833333333333 <o> b ) num__0.166666666667 <o> c ) num__0.25 <o> d ) num__0.333333333333 <o> e ) num__0.5 |
x = - num__4 prob = num__0.0833333333333 answer - a <eor> a <eos> |
a |
multiply__1.0__0.0833__ |
multiply__1.0__0.0833__ |
| sarah made $ num__45 more than kara yesterday . today they each made the same amount of money as yesterday and it totaled $ num__834 . how much would kara make in num__5 days if she continues to make the same amount of money every day . <o> a ) $ num__225 <o> b ) $ num__879 <o> c ) $ num__930 <o> d ) $ num__2085 <o> e ) none of the above |
s - k = num__45 num__2 ( s + k ) = num__834 solving the num__2 equations you get s = $ num__231 and k = $ num__186 so $ num__186 x num__5 = $ num__930 correct answer : c <eor> c <eos> |
c |
subtract__231.0__45.0__ multiply__5.0__186.0__ multiply__5.0__186.0__ |
subtract__231.0__45.0__ multiply__5.0__186.0__ multiply__5.0__186.0__ |
| mary bernie and rose can complete a job all working together in num__4 hours . mary and bernie working together at their respective rates can complete the same job in num__5 hours . how long would it take rose working alone to complete the entire job ? <o> a ) num__8 hours <o> b ) num__10 hours <o> c ) num__12 hours <o> d ) num__20 hours <o> e ) num__21 hours |
if mary bernie and rosecan complete a job in num__4 hours they can complete num__0.25 of the job in an hour . furthermore if mary and bernie can complete the same job in num__5 hours they can do num__0.2 of the entire job in an hour . if the three of them can do num__0.25 of a job in an hour and without colleen the other two can do num__0.2 of a job in an hour then the amount of the job colleen can do in an hour is the difference of these results : num__0.25 – num__0.2 = num__0.25 – num__0.2 = num__0.05 . since colleen can do num__0.05 of the job in an hour it will take her num__20 hours to do the entire job by herself . the correct answer is choice ( d ) . <eor> d <eos> |
d |
divide__0.25__5.0__ multiply__4.0__5.0__ round__20.0__ |
divide__0.25__5.0__ multiply__4.0__5.0__ round__20.0__ |
| can you deduce the pattern and find the next number in the series ? num__6 num__14 num__26 num__98 __ ? <o> a ) num__276 <o> b ) num__277 <o> c ) num__278 <o> d ) num__279 <o> e ) none of these |
solution : num__276 explanation : num__6 = num__1 ^ num__1 + num__2 ^ num__1 + num__3 ^ num__1 num__14 = num__1 ^ num__2 + num__2 ^ num__2 + num__3 ^ num__2 num__36 = num__1 ^ num__3 + num__2 ^ num__3 + num__3 ^ num__3 num__98 = num__1 ^ num__4 + num__2 ^ num__4 + num__3 ^ num__4 thus the next number will be num__1 ^ num__5 + num__2 ^ num__5 + num__3 ^ num__5 = num__276 answer a <eor> a <eos> |
a |
divide__6.0__2.0__ power__6.0__2.0__ subtract__6.0__2.0__ subtract__6.0__1.0__ multiply__1.0__276.0__ |
add__1.0__2.0__ power__6.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ power__276.0__1.0__ |
| at what price must kantilal sell a mixture of num__80 kg . sugar at rs . num__6.75 per kg . with num__120 kg . at rs . num__8 per kg . to gain num__20.0 ? <o> a ) rs . num__7.50 per kg <o> b ) rs . num__8.20 per kg <o> c ) rs . num__8.85 per kg <o> d ) rs . num__9 per kg . <o> e ) none |
explanation : total c . p of num__200 kg of sugar = rs . ( num__80 × num__6.75 + num__120 × num__8 ) = rs . num__1500 c . p of num__1 kg = rs . [ num__7.5 ] = rs . num__7.50 gain required = num__20.0 s . p of num__1 kg = ( num__120.0 of rs . num__7.50 ) = rs . [ num__1.2 × num__7.50 ] = rs . num__9 per kg . correct option : d <eor> d <eos> |
d |
percent__1.0__120.0__ percent__7.5__120.0__ percent__7.5__120.0__ |
percent__1.0__120.0__ percent__7.5__120.0__ percent__7.5__120.0__ |
| a pump can fill a tank with water in num__2 hours . because of a leak it took num__2 num__0.2 hours to fill the tank . the leak can drain all the water of the tank in ? <o> a ) num__17 hr <o> b ) num__19 hr <o> c ) num__22 hr <o> d ) num__24 hr <o> e ) num__26 hr |
work done by the tank in num__1 hour = ( num__0.5 - num__2 num__0.2 ) = num__0.0454545454545 leak will empty the tank in num__22 hrs . answer : c <eor> c <eos> |
c |
divide__1.0__2.0__ round__22.0__ |
divide__1.0__2.0__ round__22.0__ |
| what is the smallest five digit number that is divisible by num__15 num__24 num__36 and num__54 ? <o> a ) num__10320 <o> b ) num__10800 <o> c ) num__10030 <o> d ) num__10380 <o> e ) num__10460 |
num__15 = num__3 * num__5 num__24 = num__2 ^ num__3 * num__3 num__36 = num__2 ^ num__2 * num__3 ^ num__2 num__54 = num__2 * num__3 ^ num__3 lcm = num__2 ^ num__3 * num__3 ^ num__3 * num__5 = num__1080 the smallest five - digit number that is a multiple of num__1080 is num__10 * num__1080 = num__10800 the answer is b . <eor> b <eos> |
b |
divide__15.0__3.0__ subtract__5.0__3.0__ subtract__15.0__5.0__ multiply__10.0__1080.0__ multiply__10.0__1080.0__ |
divide__15.0__3.0__ subtract__5.0__3.0__ subtract__15.0__5.0__ multiply__10.0__1080.0__ multiply__10.0__1080.0__ |
| find the sum of all num__3 - digit nos that can be formed by num__1 num__2 and num__3 <o> a ) num__5994 <o> b ) num__5042 <o> c ) num__6900 <o> d ) num__7002 <o> e ) num__3546 |
no of such num__3 - digit nos = num__27 ( which is ok ) num__1 st no = num__111 last no = num__333 ( these r also ok ) hence their average = ( num__111 + num__333 ) / num__2 = num__222 ( could n ' t understand how this formula is applied . i thought this holds true for an ap series only ) so sum = number of nos x average of the nos = num__27 x num__222 = num__5994 ans : a <eor> a <eos> |
a |
multiply__3.0__111.0__ multiply__2.0__111.0__ multiply__27.0__222.0__ multiply__1.0__5994.0__ |
multiply__3.0__111.0__ subtract__333.0__111.0__ multiply__27.0__222.0__ divide__5994.0__1.0__ |
| a river num__4 m deep and num__65 m wide is flowing at the rate of num__6 kmph the amount of water that runs into the sea per minute is ? <o> a ) num__25000 <o> b ) num__26000 <o> c ) num__27000 <o> d ) num__28000 <o> e ) num__29000 |
rate of water flow - num__6 kmph - - num__100.0 - - num__100 m / min depth of river - - num__4 m width of river - - num__65 m vol of water per min - - num__100 * num__4 * num__65 - - - num__26000 answer b <eor> b <eos> |
b |
round__26000.0__ |
round__26000.0__ |
| a boy runs num__200 metres in num__24 seconds . what is his speed ? <o> a ) num__20 km / hr <o> b ) num__24 km / hr <o> c ) num__30 km / hr <o> d ) num__32 km / hr <o> e ) num__34 km / hr |
num__8.33333333333 * num__3.6 = num__30 km / hr answer : c <eor> c <eos> |
c |
divide__200.0__24.0__ round__30.0__ |
divide__200.0__24.0__ round__30.0__ |
| if the sides of a triangle are num__28 cm num__24 cm and num__15 cm what is its area ? <o> a ) num__180 cm num__2 <o> b ) num__112 cm num__2 <o> c ) num__776 cm num__2 <o> d ) num__666 cm num__2 <o> e ) num__886 cm num__2 |
the triangle with sides num__28 cm num__24 cm and num__15 cm is right angled where the hypotenuse is num__28 cm . area of the triangle = num__0.5 * num__24 * num__15 = num__180 cm num__2 answer : a <eor> a <eos> |
a |
triangle_area__24.0__15.0__ square_perimeter__0.5__ triangle_area__24.0__15.0__ |
volume_rectangular_prism__24.0__15.0__0.5__ square_perimeter__0.5__ volume_rectangular_prism__24.0__15.0__0.5__ |
| an error num__8.0 in excess is made while measuring the side of a square . now what is the percentage of error in the calculated area of the square ? <o> a ) num__6.64 <o> b ) num__16.64 <o> c ) num__15.64 <o> d ) num__26.64 <o> e ) num__10.64 % |
percentage error in calculated area = ( num__8 + num__8 + ( num__8 × num__8 ) / num__100 ) % = num__16.64 answer : b <eor> b <eos> |
b |
percent__16.64__100.0__ |
percent__16.64__100.0__ |
| today jim is twice as old as fred and sam is num__2 years younger than fred . six years ago jim was num__7 times as old as sam . how old is jim now ? <o> a ) num__8 <o> b ) num__12 <o> c ) num__16 <o> d ) num__20 <o> e ) num__24 |
we ' re asked how old jim is now . we ' re given three facts to work with : num__1 ) today jim is twice as old as fred num__2 ) today sam is num__2 years younger than fred num__3 ) six years ago jim was num__7 times as old as sam . let ' s test answer d : num__20 if . . . . jim is currently num__20 years old . . . . fred is num__10 years old sam is num__8 years old num__6 years ago jim was num__14 and sam was num__2 so jim was num__7 times sam ' s age . this is an exact match for what we were told so this must be the answer . d <eor> d <eos> |
d |
add__2.0__1.0__ add__7.0__3.0__ add__7.0__1.0__ multiply__2.0__3.0__ multiply__2.0__7.0__ multiply__2.0__10.0__ |
add__2.0__1.0__ add__7.0__3.0__ add__7.0__1.0__ multiply__2.0__3.0__ multiply__2.0__7.0__ multiply__2.0__10.0__ |
| the cost of num__2 chairs and num__3 tables is rs . num__1600 . the cost of num__3 chairs and num__2 tables is rs . num__1200 . the cost of each table is more than that of each chair by ? <o> a ) num__228 <o> b ) num__287 <o> c ) num__277 <o> d ) num__188 <o> e ) num__400 |
explanation : num__2 c + num__3 t = num__1600 - - - ( num__1 ) num__3 c + num__3 t = num__1200 - - - ( num__2 ) subtracting num__2 nd from num__1 st we get - c + t = num__400 = > t - c = num__400 answer : e <eor> e <eos> |
e |
subtract__3.0__2.0__ subtract__1600.0__1200.0__ subtract__1600.0__1200.0__ |
subtract__3.0__2.0__ subtract__1600.0__1200.0__ subtract__1600.0__1200.0__ |
| one fast typist type some matter in num__2 hrs and another slow typist type the same matter in num__3 hrs . if both do combine in how much time they will take to finish . <o> a ) num__10 <o> b ) num__11 <o> c ) num__12 <o> d ) num__13 <o> e ) num__14 |
fast ' s num__1 hr work = num__0.5 slow ' s num__1 hr work = num__0.333333333333 num__0.5 + num__0.333333333333 = num__0.833333333333 they ll finish in num__1.2 hrs = num__1 num__0.2 = = > num__1 hr num__12 mins answer : c <eor> c <eos> |
c |
union_prob__1.0__0.3333__0.5__ choose__3.0__1.0__ choose__3.0__1.0__ |
union_prob__1.0__0.3333__0.5__ choose__3.0__1.0__ choose__3.0__1.0__ |
| in a certain diving competition num__5 judges score each dive on a scale from num__1 to num__10 . the point value of the dive is obtained by dropping the highest score and the lowest score and multiplying the sum of the remaining scores by the degree of difficulty . if a dive with a degree of difficulty of num__3.2 received scores of num__7.5 num__8.0 num__9.0 num__6.0 and num__8.8 what was the point value of the dive ? <o> a ) num__68.8 <o> b ) num__73.6 <o> c ) num__75.2 <o> d ) num__76.8 <o> e ) num__77.76 |
degree of difficulty of dive = num__3.2 scores are num__6.0 num__7.5 num__8.0 num__8.8 and num__9.0 we can drop num__6.0 and num__9.0 sum of the remaining scores = ( num__7.5 + num__8 + num__8.8 ) = num__24.3 point of value of the dive = num__24 * num__3.2 = num__77.76 answer e <eor> e <eos> |
e |
round_down__24.3__ multiply__3.2__24.3__ multiply__1.0__77.76__ |
multiply__3.2__7.5__ multiply__3.2__24.3__ multiply__1.0__77.76__ |
| the distance between two cities a and b is num__330 km . a train starts from a at num__8 a . m . and travel towards b at num__60 km / hr . another train starts from b at num__9 a . m and travels towards a at num__75 km / hr . at what time do they meet ? <o> a ) num__09 <o> b ) num__07 <o> c ) num__11 <o> d ) num__05 <o> e ) num__03 |
explanation : suppose they meet x hrs after num__8 a . m then [ distance moved by first in x hrs ] + [ distance moved by second in ( x - num__1 ) hrs ] = num__330 . therefore num__60 x + num__75 ( x - num__1 ) = num__330 . = > x = num__3 . so they meet at ( num__8 + num__3 ) i . e num__11 a . m . answer : c ) num__11 a . m <eor> c <eos> |
c |
subtract__9.0__8.0__ add__8.0__3.0__ round__11.0__ |
subtract__9.0__8.0__ add__8.0__3.0__ add__8.0__3.0__ |
| in a school of num__600 students num__45.0 wear blue shirts num__23.0 wear red shirts num__15.0 wear green shirts and the remaining students wear other colors . how many students wear other colors ( not blue not red not green ) ? <o> a ) num__102 <o> b ) num__112 <o> c ) num__122 <o> d ) num__132 <o> e ) num__142 |
num__45 + num__23 + num__15 = num__83.0 num__100 – num__83 = num__17.0 num__600 * num__0.17 = num__102 the answer is a . <eor> a <eos> |
a |
percent__17.0__600.0__ percent__100.0__102.0__ |
percent__17.0__600.0__ percent__100.0__102.0__ |
| the probability that a man will be alive for num__10 more yrs is num__0.25 & the probability that his wife will alive for num__10 more yrs is num__0.333333333333 . the probability that none of them will be alive for num__10 more yrs is <o> a ) num__0.5 <o> b ) num__0.75 <o> c ) num__0.6 <o> d ) num__0.428571428571 <o> e ) num__0.714285714286 |
sol . required probability = pg . ) x p ( b ) = ( num__1 — d x ( num__1 — i ) = : x num__1 = num__0.5 ans . ( a ) <eor> a <eos> |
a |
divide__0.25__0.5__ |
divide__0.25__0.5__ |
| { num__9 num__43 num__57 num__1210 } what value should be inserted into the above set of num__7 values to get a median value of num__6 ? <o> a ) num__5 <o> b ) num__4 <o> c ) num__3 <o> d ) num__7 <o> e ) num__6 |
median is the middle number of a sorted distribution . in case of even number of items median will be the average of middle two values . the question asks fora new number that will cause the median of new set at num__12 . mentionable that given set is already sorted and number of items of new set will be num__7 + num__1 = num__8 . to get median num__6 the sum of num__4 th num__5 th term should be num__12 and luckilyit ' s exactly same in the existing series i . e . num__5 + num__7 . so to keep the median num__6 the new item should be > = num__7 answer is d <eor> d <eos> |
d |
subtract__7.0__6.0__ subtract__9.0__1.0__ subtract__12.0__8.0__ subtract__9.0__4.0__ multiply__7.0__1.0__ |
subtract__7.0__6.0__ add__7.0__1.0__ subtract__12.0__8.0__ add__1.0__4.0__ add__6.0__1.0__ |
| one - fourth of the students at a nursery school are num__4 years old or older . if num__15 students have not yet reached their third birthday and a total of num__40 students are not between num__3 years old and num__4 years old how many children are in the nursery school ? <o> a ) num__80 <o> b ) num__90 <o> c ) num__100 <o> d ) num__110 <o> e ) num__120 |
x / num__4 students are > num__4 yrs num__15 students are < num__3 yrs x / num__4 + num__15 = num__40 x / num__4 = num__25 x = num__100 answer : c <eor> c <eos> |
c |
subtract__40.0__15.0__ multiply__4.0__25.0__ multiply__4.0__25.0__ |
subtract__40.0__15.0__ multiply__4.0__25.0__ multiply__4.0__25.0__ |
| let f ( x ) = x ^ num__2 + bx + c . if f ( num__5 ) = num__0 and f ( - num__3 ) = num__0 then b + c = <o> a ) num__18 <o> b ) - num__17 <o> c ) - num__15 <o> d ) - num__21 <o> e ) - num__24 |
f ( x ) = x ^ num__2 + bx + c . if f ( num__5 ) = num__0 and f ( - num__3 ) = num__0 then b + c = f ( num__5 ) = num__0 = num__25 + num__5 b + c - - - taking num__25 to the other side - > num__5 b + c = - num__25 f ( - num__3 ) = num__0 = num__9 - num__3 b + c - - - taking - num__3 b + c to the other side - > num__3 b - c = num__9 when we add these num__2 equations we get num__8 b = - num__16 - - - > b = - num__2 and while substituting b = - num__2 we get c = - num__15 . b + c = - num__17 - - - answer b <eor> b <eos> |
b |
add__5.0__3.0__ multiply__2.0__8.0__ multiply__5.0__3.0__ add__2.0__15.0__ add__2.0__15.0__ |
add__5.0__3.0__ subtract__25.0__9.0__ multiply__5.0__3.0__ add__2.0__15.0__ add__2.0__15.0__ |
| three numbers are in the ratio num__1 : num__4 : num__6 and their h . c . f is num__7 . the numbers are ? <o> a ) num__7 num__28 num__40 <o> b ) num__7 num__28 num__42 <o> c ) num__7 num__28 num__45 <o> d ) num__7 num__21 num__42 <o> e ) num__7 num__28 num__35 |
let the required numbers be x num__4 x and num__6 x . then their h . c . f = x . so x = num__7 . the numbers are num__7 num__28 num__42 . answer : b <eor> b <eos> |
b |
multiply__4.0__7.0__ multiply__6.0__7.0__ add__1.0__6.0__ |
multiply__4.0__7.0__ multiply__6.0__7.0__ add__1.0__6.0__ |
| a circular mat with diameter num__20 inches is placed on a square tabletop each of whose sides is num__25 inches long . which of the following is closest to the fraction of the tabletop covered by the mat ? <o> a ) num__0.5024 <o> b ) num__0.4 <o> c ) num__0.5 <o> d ) num__0.75 <o> e ) num__0.833333333333 |
it is a circle inscribed in a square . square side = num__25 - - - > square ( table ) area = num__25 ^ num__2 circle diameter = num__20 - - - > circle area = pir ^ num__2 = num__100 pi ( where pi = ~ num__3.14 ) covered fraction = num__100 * num__3.14 / num__25 * num__25 = ~ num__12.56 * num__25 = num__0.5024 a <eor> a <eos> |
a |
square_perimeter__25.0__ square_perimeter__3.14__ triangle_area__2.0__0.5024__ |
square_perimeter__25.0__ square_perimeter__3.14__ triangle_area__2.0__0.5024__ |
| the sum of the first k positive integers is equal to k ( k + num__1 ) / num__2 . what is the sum of the integers from y to z inclusive where num__0 < y < z ? <o> a ) z ( z + num__1 ) / num__2 - ( y + num__1 ) ( y + num__2 ) / num__2 <o> b ) z ( z + num__1 ) / num__2 - y ( y + num__1 ) / num__2 <o> c ) z ( z + num__1 ) / num__2 - ( y - num__1 ) y / num__2 <o> d ) ( z - num__1 ) z / num__2 - ( y + num__1 ) ( y + num__2 ) / num__2 <o> e ) ( z - num__1 ) z / num__2 - y ( y + num__1 ) / num__2 |
or try plug - in method : let z = num__4 and y = num__3 - - > then z + y = num__7 . let see which option yields num__7 . a . z ( z + num__1 ) / num__2 - ( y + num__1 ) ( y + num__2 ) / num__2 = num__10 - num__10 = num__0 ; b . z ( z + num__1 ) / num__2 - y ( n + num__1 ) / num__2 = num__10 - num__6 = num__4 ; c . z ( z + num__1 ) / num__2 - ( y - num__1 ) y / num__2 = num__10 - num__3 = num__7 - - > ok ; d . ( z - num__1 ) z / num__2 - ( y + num__1 ) ( y + num__2 ) / num__2 = num__6 - num__10 = - num__4 ; e . ( z - num__1 ) z / num__2 - y ( y + num__1 ) / num__2 = num__6 - num__6 = num__0 . answer : c . <eor> c <eos> |
c |
add__1.0__2.0__ add__3.0__4.0__ add__3.0__7.0__ multiply__2.0__3.0__ reverse__1.0__ |
add__1.0__2.0__ add__3.0__4.0__ add__3.0__7.0__ add__2.0__4.0__ reverse__1.0__ |
| a coin is tossed live times . what is the probability that there is at the least one tail ? <o> a ) num__0.96875 <o> b ) num__0.348314606742 <o> c ) num__0.319587628866 <o> d ) num__0.344444444444 <o> e ) num__0.553571428571 |
explanation : let p ( t ) be the probability of getting least one tail when the coin is tossed five times . = there is not even a single tail . i . e . all the outcomes are heads . = num__0.03125 ; p ( t ) = num__1 - num__0.03125 = num__0.96875 answer : option a <eor> a <eos> |
a |
negate_prob__0.0312__ negate_prob__0.0312__ |
negate_prob__0.0312__ negate_prob__0.0312__ |
| a train covers a distance of num__12 km in num__10 min . if it takes num__6 sec to pass a telegraph post then the length of the train is ? <o> a ) num__298 m <o> b ) num__888 m <o> c ) num__120 m <o> d ) num__217 m <o> e ) num__166 m |
speed = ( num__1.2 * num__60 ) km / hr = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . length of the train = num__20 * num__6 = num__120 m . answer : c <eor> c <eos> |
c |
divide__12.0__10.0__ hour_to_min_conversion__ multiply__12.0__6.0__ multiply__12.0__10.0__ round__120.0__ |
divide__12.0__10.0__ multiply__10.0__6.0__ multiply__12.0__6.0__ multiply__12.0__10.0__ multiply__12.0__10.0__ |
| what is the difference between the place value of num__2 in the numeral num__7559 ? <o> a ) num__160 <o> b ) num__165 <o> c ) num__180 <o> d ) num__190 <o> e ) num__450 |
answer : option ' e ' num__500 - num__50 = num__450 <eor> e <eos> |
e |
subtract__500.0__50.0__ subtract__500.0__50.0__ |
subtract__500.0__50.0__ subtract__500.0__50.0__ |
| joe needs to paint all the airplane hangars at the airport so he buys num__360 gallons of paint to do the job . during the first week he uses num__0.25 of all the paint . during the second week he uses num__0.166666666667 of the remaining paint . how many gallons of paint has joe used ? <o> a ) num__18 <o> b ) num__135 <o> c ) num__175 <o> d ) num__216 <o> e ) num__250 |
total paint initially = num__360 gallons paint used in the first week = ( num__0.25 ) * num__360 = num__90 gallons . remaning paint = num__270 gallons paint used in the second week = ( num__0.166666666667 ) * num__270 = num__45 gallons total paint used = num__135 gallons . option b <eor> b <eos> |
b |
multiply__360.0__0.25__ subtract__360.0__90.0__ add__45.0__90.0__ round__135.0__ |
multiply__360.0__0.25__ subtract__360.0__90.0__ add__45.0__90.0__ round__135.0__ |
| what is the total % gain or loss if a trader sold two items for rs . num__1500 each and makes a profit of num__20.0 from the first item and a loss of num__12.0 from the second item . <o> a ) num__2.4 <o> b ) num__4.4 <o> c ) num__1.4 <o> d ) num__12.4 <o> e ) num__20.5 |
( num__20 * num__12 ) / num__100 = num__2.4 answer : a <eor> a <eos> |
a |
percent__20.0__12.0__ percent__20.0__12.0__ |
percent__20.0__12.0__ percent__20.0__12.0__ |
| there are num__5 chess amateurs playing in villa ' s chess club tournament . if each chess amateur plays with exactly num__8 other amateurs what is the total number of chess games possible to be played in the tournament ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__40 <o> d ) num__60 <o> e ) num__120 |
method num__2 : each person is one participant of num__4 games . so there are in all num__8 * num__5 = num__40 instances of one participant games . but each game has num__2 participants so total number of games = num__20.0 = num__20 b <eor> b <eos> |
b |
divide__8.0__2.0__ multiply__5.0__8.0__ multiply__5.0__4.0__ multiply__5.0__4.0__ |
divide__8.0__2.0__ multiply__5.0__8.0__ multiply__5.0__4.0__ multiply__5.0__4.0__ |
| the food in a camp lasts for num__32 men for num__48 days . if eight more men join how many days will the food last ? <o> a ) num__80 days <o> b ) num__30 days <o> c ) num__38 days <o> d ) num__16 days <o> e ) num__15 days |
one man can consume the same food in num__32 * num__48 = num__1536 days . num__8 more men join the total number of men = num__40 the number of days the food will last = num__38.4 = num__38 days . answer : c <eor> c <eos> |
c |
multiply__32.0__48.0__ add__32.0__8.0__ divide__1536.0__40.0__ round__38.0__ |
multiply__32.0__48.0__ add__32.0__8.0__ divide__1536.0__40.0__ round__38.0__ |
| three grades of milk are num__1 percent num__33 percent and num__3 percent fat by volume . if x gallons of the num__1 percent grade y gallons of the num__2 percent grade and z gallons of the num__3 percent grade are mixed to give x + y + z gallons of a num__1.5 percent grade what is x in terms of y and z ? <o> a ) y + num__3 z <o> b ) ( y + z ) / num__4 <o> c ) num__2 y + num__3 z <o> d ) num__3 y + num__3 z <o> e ) num__3 y + num__4.5 z |
soln : the resulting equation is = > ( . num__01 x + . num__03 y + . num__03 z ) / ( x + y + z ) = num__1.5 / num__100 = > x + num__3 y + num__3 z = num__1.5 x + num__1.5 y + num__1.5 z taking x to one side and y and z to other side we get = > x = num__3 y + num__3 z ans is d <eor> d <eos> |
d |
multiply__1.0__3.0__ |
add__1.0__2.0__ |
| a man is standing on a railway bridge which is num__180 m long . he finds that a train crosses the bridge in num__20 seconds but himself in num__8 seconds . find the length of the train and its speed <o> a ) num__50 km <o> b ) num__52 km <o> c ) num__54 km <o> d ) num__56 km <o> e ) num__58 km |
explanation : let the length of the train be x meters then the train covers x meters in num__8 seconds and ( x + num__180 ) meters in num__20 sec x / num__8 = ( x + num__180 ) / num__20 ⇔ num__20 x = num__8 ( x + num__180 ) < = > x = num__120 length of the train = num__120 m . speed of the train = ( num__15.0 ) m / sec = m / sec = ( num__15 × num__3.6 ) kmph = num__54 km answer : option c <eor> c <eos> |
c |
divide__120.0__8.0__ multiply__15.0__3.6__ round__54.0__ |
divide__120.0__8.0__ multiply__15.0__3.6__ round__54.0__ |
| points a b c and d in that order lie on a line . if ab = num__2 cm ac = num__5 cm and bd = num__6 cm what is cd in centimeters ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
putting a value to each point lets use the following : a - num__0 b - num__2 ( ab = num__2 ) c - num__5 ( ac = num__5 ) d - num__8 ( bd = num__6 ) cd is num__8 - num__5 = num__3 . ans c <eor> c <eos> |
c |
add__2.0__6.0__ subtract__5.0__2.0__ round__3.0__ |
add__2.0__6.0__ subtract__5.0__2.0__ subtract__5.0__2.0__ |
| a b and c are integers and a < b < c . s is the set of all integers from a to b inclusive . q is the set of all integers from b to c inclusive . the median of set s is ( num__0.75 ) * b . the median of set q is ( num__7.5 / num__8 ) * c . if r is the set of all integers from a to c inclusive what fraction of c is the median of set r ? <o> a ) num__0.375 <o> b ) num__0.5 <o> c ) num__0.6875 <o> d ) num__0.71875 <o> e ) num__0.75 |
the answer isc : num__0.6875 . the key to this problem is remembering that the median for a consecutive set of numbers is equivalent to its mean . for example the mean and median of a set consisting of x x + num__1 x + num__2 . . . y will always be ( x + y ) / num__2 . for set s consisting of numbers ( a a + num__1 . . . b ) the median is given to be num__0.75 * b : ( a + b ) / num__2 = ( num__0.75 ) * b a = b / num__2 for set q consisting of numbers ( b b + num__1 . . . c ) the median is given to be num__7.5 / num__8 * c : ( b + c ) / num__2 = ( num__7.5 / num__8 ) * c b = ( num__0.875 ) * c for set r consisting of numbers ( a a + num__1 . . . c ) the median needs to be found : a = b / num__2 = ( num__0.875 * c ) / num__2 = ( num__0.4375 ) * c median = ( a + c ) / num__2 = ( num__0.4375 * c + c ) / num__2 = ( num__1.4375 ) * c / num__2 = ( num__0.71875 ) * c ( answer d ) <eor> d <eos> |
d |
divide__0.875__2.0__ add__0.75__0.6875__ divide__1.4375__2.0__ multiply__1.0__0.7188__ |
divide__0.875__2.0__ add__0.75__0.6875__ divide__1.4375__2.0__ multiply__1.0__0.7188__ |
| from the starting point in a boat race one competitor started to sail north at a speed of num__1.4 km / h the other competitor started to sail west at a speed of num__1.2 km / h . what is the distance in km between the two competitors after num__5 hours ? <o> a ) num__9.21 <o> b ) num__12 . <o> c ) num__12.5 . <o> d ) num__14 . <o> e ) num__15.4 . |
both competitors are sailing making angle of num__90 degrees . after num__5 hrs one competitor will cover a distance of = num__1.4 * num__5 = num__7 km and other competitor will cover a distance of = num__1.2 * num__5 = num__6 km distance between them after num__5 hrs = ( num__7 ^ num__2 + num__6 ^ num__2 ) ^ num__0.5 = num__9.21 km a is the answer <eor> a <eos> |
a |
multiply__1.4__5.0__ multiply__1.2__5.0__ subtract__7.0__5.0__ round__9.21__ |
multiply__1.4__5.0__ multiply__1.2__5.0__ subtract__7.0__5.0__ round__9.21__ |
| a kloke ' s train rails across an open track at num__250 kilometers per hour . a regular passenger train travels at num__68.0 of the kloke ' s train speed . if the two trains start moving from the same station at the same time how much time longer will it take the passenger train than the kloke ' s to travel num__850 kilometers ? <o> a ) num__1 hour and num__24 minutes . <o> b ) num__1 hour and num__36 minutes . <o> c ) num__2 hours and num__24 minutes . <o> d ) num__2 hours and num__36 minutes . <o> e ) num__5 hours . |
difference in time = time taken by passenger train - time taken by kloke ' s train num__850 / ( num__250 * num__68 ) * num__100 - num__3.4 num__850 ( num__0.4 * num__68 - num__0.004 ) num__850 * num__32 / ( num__250 * num__68 ) num__1.6 hrs or num__1 hr and num__36 mins b is the answer <eor> b <eos> |
b |
divide__850.0__250.0__ divide__100.0__250.0__ divide__0.4__100.0__ subtract__100.0__68.0__ mile_to_km_conversion__ multiply__250.0__0.004__ subtract__68.0__32.0__ round__1.0__ |
divide__850.0__250.0__ divide__100.0__250.0__ divide__0.4__100.0__ subtract__100.0__68.0__ mile_to_km_conversion__ multiply__250.0__0.004__ subtract__68.0__32.0__ multiply__250.0__0.004__ |
| pipe a can fill a tank in num__4 hours . due to a leak at the bottom it takes num__6 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? <o> a ) num__12 <o> b ) num__67 <o> c ) num__95 <o> d ) num__36 <o> e ) num__66 |
let the leak can empty the full tank in x hours num__0.25 - num__1 / x = num__0.166666666667 = > num__1 / x = num__0.25 - num__0.166666666667 = ( num__3 - num__2 ) / num__12 = num__0.0833333333333 = > x = num__12 . answer : a <eor> a <eos> |
a |
multiply__4.0__0.25__ divide__1.0__6.0__ subtract__4.0__1.0__ subtract__6.0__4.0__ multiply__4.0__3.0__ divide__0.25__3.0__ round__12.0__ |
multiply__4.0__0.25__ divide__1.0__6.0__ subtract__4.0__1.0__ subtract__6.0__4.0__ divide__3.0__0.25__ divide__0.25__3.0__ divide__3.0__0.25__ |
| the speed at which a woman can row a boat in still water is num__120 kmph . if she rows downstream where the speed of current is num__60 kmph what time will he take to cover num__500 metres ? <o> a ) num__8 <o> b ) num__9 <o> c ) num__10 <o> d ) num__11 <o> e ) num__12 |
speed of the boat downstream = num__120 + num__60 = num__180 kmph = num__180 * num__0.277777777778 = num__50 m / s hence time taken to cover num__500 m = num__10.0 = num__10 seconds . answer : c <eor> c <eos> |
c |
add__120.0__60.0__ subtract__60.0__50.0__ round__10.0__ |
add__120.0__60.0__ divide__500.0__50.0__ divide__500.0__50.0__ |
| a highway is to be divided into four lanes . for this purpose three yellow stripes are painted so that stripes divide the highway into four lanes . if num__3 gallons of paint cover an area of num__2 p square feet of highway how many gallons of paint are needed to paint two stripes x inches wide on a stretch of highway m miles long ? ( num__1 mile = num__5280 feet and num__1 feet = num__12 inches ) <o> a ) num__1320 mx / p <o> b ) num__880 mx / p <o> c ) num__1320 p / mx <o> d ) num__440 p / mx <o> e ) num__1980 mx / p |
num__1 square foot needs num__1.5 p gallons of paint . the width of each stripe is x / num__12 feet . the length of each stripe is num__5280 m feet . the area of each stripe is ( num__5280 m ) ( x / num__12 ) = num__440 mx square feet . for three stripes the total area is num__1320 mx square feet . the number of gallons needed is ( num__1320 mx ) * ( num__1.5 p ) = num__1980 mx / p gallons . the answer is e . <eor> e <eos> |
e |
divide__3.0__2.0__ divide__5280.0__12.0__ multiply__3.0__440.0__ multiply__1.5__1320.0__ multiply__1.0__1980.0__ |
divide__3.0__2.0__ divide__5280.0__12.0__ multiply__3.0__440.0__ multiply__1.5__1320.0__ multiply__1.0__1980.0__ |
| a train covers a distance of num__12 km in num__10 min . if it takes num__12 sec to pass a telegraph post then the length of the train is ? <o> a ) num__178 m <o> b ) num__240 m <o> c ) num__120 m <o> d ) num__168 m <o> e ) num__178 m |
speed = ( num__1.2 * num__60 ) km / hr = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . length of the train = num__20 * num__12 = num__240 m . answer : b <eor> b <eos> |
b |
divide__12.0__10.0__ hour_to_min_conversion__ add__12.0__60.0__ multiply__12.0__20.0__ round__240.0__ |
divide__12.0__10.0__ hour_to_min_conversion__ multiply__1.2__60.0__ multiply__12.0__20.0__ multiply__12.0__20.0__ |
| a jar contains num__18 balls . num__3 blue balls are removed from the jar and not replaced . now the probability of getting a blue ball is num__0.2 then how many blue balls jar contains initially ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
x / num__15 = num__0.2 x = num__3 num__3 + num__3 ( removed num__3 blue balls ) = num__6 answer : e <eor> e <eos> |
e |
die_space__ die_space__ |
die_space__ die_space__ |
| in a certain deck of cards each card has a positive integer written on it in a multiplication game a child draws a card and multiplies the integer on the card with the next large integer . if the each possible product is between num__15 and num__150 then the least and greatest integer on the card would be <o> a ) num__3 and num__15 <o> b ) num__3 and num__20 <o> c ) num__4 and num__12 <o> d ) num__4 and num__14 <o> e ) num__5 and num__14 |
given : num__15 < x ( x + num__1 ) < num__170 . now it ' s better to test the answer choices here rather than to solve : if x = num__4 then x ( x + num__1 ) = num__20 > num__15 - - > so the least value is num__4 . test for the largest value : if x = num__12 then x ( x + num__1 ) = num__12 * num__13 = num__156 > num__150 answer : c . <eor> c <eos> |
c |
subtract__170.0__150.0__ add__1.0__12.0__ multiply__12.0__13.0__ multiply__1.0__4.0__ |
subtract__170.0__150.0__ add__1.0__12.0__ multiply__12.0__13.0__ multiply__1.0__4.0__ |
| if a motor bike ' s wheel is num__12 cm diameter and rotating at the rate of x cm / min and another motor bike ' s wheel is num__6 cm diameter rotating at the rate of y cm / min and if both the bikes reach the distance at same time then what is the value of y in terms of x ? <o> a ) num__4 x / num__2 <o> b ) num__3 x / num__6 <o> c ) num__4 x / num__2 <o> d ) num__1 x / num__2 <o> e ) num__2 x / num__3 |
t = s num__1 / v num__1 = s num__2 / v num__2 or num__12 / x = num__6 / y or y = num__6 x / num__12 = num__1 x / num__2 answer is d <eor> d <eos> |
d |
divide__12.0__6.0__ round__1.0__ |
divide__12.0__6.0__ round__1.0__ |
| the diagonal of a rectangle is √ num__41 cm and its area is num__20 sq . cm . the perimeter of the rectangle must be : <o> a ) num__18 <o> b ) num__28 <o> c ) num__38 <o> d ) num__48 <o> e ) num__58 |
explanation : l num__2 + b num__2 = num__41 also lb = num__20 ( l + b ) num__2 = l num__2 + b num__2 + num__2 lb = num__41 + num__40 = num__81 ( l + b ) = num__9 perimeter = num__2 ( l + b ) = num__18 cm . answer : a <eor> a <eos> |
a |
multiply__20.0__2.0__ side_by_diagonal__41.0__40.0__ multiply__2.0__9.0__ triangle_area__2.0__18.0__ |
multiply__20.0__2.0__ side_by_diagonal__41.0__40.0__ multiply__2.0__9.0__ triangle_area__2.0__18.0__ |
| jaclyn buys $ num__40 num__000 worth of debentures in a company . she earns num__9.5 p . a . simple interest paid to her quarterly ( that is every num__3 months ) . if the agreed period of the debenture was num__18 months : calculate the amount of interest jaclyn will earn for each quarter <o> a ) num__950 <o> b ) num__1234 <o> c ) num__1289 <o> d ) num__1345 <o> e ) none of these |
explanation : i = ( p x r x t ) / num__100 = num__40000 * num__9.5 / num__100 * ( num__1.5 ) ^ num__0.166666666667 = num__950 answer : a <eor> a <eos> |
a |
percent__100.0__950.0__ |
percent__100.0__950.0__ |
| if s = t / p - num__1 < p < num__0 and num__4 < t which of the following is correct ? <o> a ) s > num__4 . <o> b ) num__0 < s < num__4 . <o> c ) - num__4 < s < num__0 . <o> d ) s < - num__4 . <o> e ) s < - num__20 . |
t is + vep is - ve eliminate ab for min . value max . numerator and minimize dinominator take t = num__4.1 s = - num__0.9 s = num__4.1 / - num__0.9 s < - num__4 ans d <eor> d <eos> |
d |
round_down__4.1__ |
divide__4.0__1.0__ |
| a train passes a station platform in num__36 seconds and a man standing on the platform in num__20 seconds . if the speed of the train is num__54 km / hr what is the length of the platform ? <o> a ) num__877 m <o> b ) num__240 m <o> c ) num__167 m <o> d ) num__887 m <o> e ) num__265 m |
speed = ( num__54 * num__0.277777777778 ) m / sec = num__15 m / sec . length of the train = ( num__15 x num__20 ) m = num__300 m . let the length of the platform be x meters . then ( x + num__300 ) / num__36 = num__15 = = > x + num__300 = num__540 = = > x = num__240 m . answer : b <eor> b <eos> |
b |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
multiply__20.0__15.0__ multiply__36.0__15.0__ subtract__540.0__300.0__ round__240.0__ |
| walking at the rate of num__8 kmph a man cover certain distance in num__4 hr num__45 min . running at a speed of num__19 kmph the man will cover the same distance in . <o> a ) num__100 min <o> b ) num__110 min <o> c ) num__140 min <o> d ) num__120 min <o> e ) num__150 min |
distance = speed * time num__8 * num__4.75 = num__38 km new speed = num__19 kmph therefore time = d / s = num__2.0 = num__120 min answer : d . <eor> d <eos> |
d |
divide__19.0__4.0__ multiply__8.0__4.75__ divide__8.0__4.0__ round__120.0__ |
divide__19.0__4.0__ multiply__8.0__4.75__ divide__8.0__4.0__ round__120.0__ |
| a palindrome is a number that reads the same forward and backward such as num__464 . how many odd five - digit numbers are palindromes ? <o> a ) num__40 <o> b ) num__400 <o> c ) num__500 <o> d ) num__5000 <o> e ) num__100 |
000 |
num__5 th digit num__1 num__35 num__79 num__4 th digit num__0 through num__9 num__3 rd digit num__0 through num__9 i . e num__10 * num__10 * num__5 = num__500 numbers first and second digit is going to be same as num__4 th and num__5 th . so it would still be num__400 numbers . answer is c . <eor> c <eos> |
c |
c |
| there are two circles of different radii . the are of a square is num__784 sq cm and its side is twice the radius of the larger circle . the radius of the larger circle is seven - third that of the smaller circle . find the circumference of the smaller circle ? <o> a ) num__13 <o> b ) num__14 <o> c ) num__12 <o> d ) num__16 <o> e ) num__17 |
let the radii of the larger and the smaller circles be l cm and s cm respectively . let the side of the square be a cm . a num__2 = num__784 = ( num__4 ) ( num__196 ) = ( num__22 ) . ( num__142 ) a = ( num__2 ) ( num__14 ) = num__28 a = num__2 l l = a / num__2 = num__14 l = ( num__2.33333333333 ) s therefore s = ( num__0.428571428571 ) ( l ) = num__6 circumference of the smaller circle = num__2 ∏ s = num__12 ∏ cm . answer : c <eor> c <eos> |
c |
multiply__2.0__14.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
multiply__2.0__14.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
| it takes one machine num__3 hours to complete a production order and another machine num__3 hours to complete the same order . how many hours would it take both amhcines working simultaneously at their respective rates to complete the order ? <o> a ) a ) num__0.583333333333 <o> b ) b ) num__1 num__0.5 <o> c ) c ) num__1 num__0.714285714286 <o> d ) d ) num__3 num__0.5 <o> e ) e ) num__1 num__0.5 |
rt = w given : r = num__0.333333333333 + num__0.333333333333 = num__0.666666666667 t = ? w = num__1 t = num__1 / ( num__0.666666666667 ) = num__1.5 = num__1 num__0.5 answer : e <eor> e <eos> |
e |
add__0.6667__0.3333__ divide__1.5__3.0__ round__1.0__ |
add__0.6667__0.3333__ divide__1.5__3.0__ add__0.6667__0.3333__ |
| there are num__6 tasks and num__6 persons . task num__1 can not be assigned either to person num__1 or to person num__2 ; task num__2 must be assigned to either person num__3 or person num__4 . every person is to be assigned one task . in how many ways can the assignment be done ? <o> a ) num__144 <o> b ) num__180 <o> c ) num__192 <o> d ) num__360 <o> e ) none of these |
task num__1 can not be assigned to either person num__1 or num__2 i . e . there are num__4 options . task num__2 can be assigned to num__3 or num__4 so there are only num__2 options for task num__2 . so required no . of ways = num__2 options for task num__2 × num__3 options for task num__1 × num__4 options for task num__3 × num__3 options for task num__4 × num__2 options for task num__5 × num__1 option for task num__6 . = num__2 × num__3 × num__4 × num__3 × num__2 × num__1 = num__144 answer a <eor> a <eos> |
a |
vowel_space__ choose__4.0__1.0__ choose__4.0__1.0__ |
vowel_space__ choose__4.0__1.0__ choose__4.0__1.0__ |
| a man purchases num__8 pens for rs . num__9 and sells num__9 pens for rs . num__8 how much profit or loss does he make ? <o> a ) num__29.98 <o> b ) num__20.98 <o> c ) num__20.92 <o> d ) num__20.96 <o> e ) num__20.91 |
num__81 - - - - num__17 num__100 - - - - - ? è num__20.98 loss answer : b <eor> b <eos> |
b |
percent__100.0__20.98__ |
percent__100.0__20.98__ |
| how much interest can a person get on rs . num__8800 at num__17.5 p . a . simple interest for a period of two years and six months ? <o> a ) num__3587.58 <o> b ) num__3587.59 <o> c ) num__3587.5 <o> d ) num__3587.52 <o> e ) num__3850 |
i = ( num__8800 * num__2.5 * num__17.5 ) / num__100 = ( num__8800 * num__5 * num__35 ) / ( num__100 * num__2 * num__2 ) = rs . num__3850 . answer : e <eor> e <eos> |
e |
percent__100.0__3850.0__ |
percent__100.0__3850.0__ |
| a person has to cover a distance of num__6 km in num__45 minutes . if he covers one half of the distance in two - thirds of the total time ; to cover the remaining distance in the remaining time his speed must be ? <o> a ) num__12 km per hr <o> b ) num__10 km / hr <o> c ) num__15 km / hr <o> d ) num__20 km / hr <o> e ) num__18 km / hr |
remaining distance = num__3 km remaining time = num__0.333333333333 * num__45 = num__15 min = num__0.25 hour required speed = num__3 * num__4 = num__12 km / hr answer is a <eor> a <eos> |
a |
divide__45.0__3.0__ divide__3.0__0.25__ round__12.0__ |
divide__45.0__3.0__ divide__3.0__0.25__ divide__3.0__0.25__ |
| two men a and b start from place x walking at num__4 ½ kmph and num__5 ¾ kmph respectively . how many km apart they are at the end of num__3 ½ hours if they are walking in the same direction ? <o> a ) num__4 num__0.333333333333 km <o> b ) num__5 num__0.375 km <o> c ) num__4 num__0.375 km <o> d ) num__6 num__0.375 km <o> e ) num__7 num__0.375 km |
rs = num__5 ¾ - num__4 ½ = num__1 ¼ t = num__3 ½ h . d = num__1.25 * num__3.5 = num__4.375 = num__4 num__0.375 km answer : c <eor> c <eos> |
c |
subtract__4.0__3.0__ divide__5.0__4.0__ multiply__1.25__3.5__ subtract__4.375__4.0__ round__4.0__ |
subtract__4.0__3.0__ divide__5.0__4.0__ multiply__1.25__3.5__ subtract__4.375__4.0__ subtract__5.0__1.0__ |
| what amount does an investor receive if the investor invests $ num__7000 at num__10.0 p . a . compound interest for two years compounding done annually ? <o> a ) $ num__8080 <o> b ) $ num__8130 <o> c ) $ num__8260 <o> d ) $ num__8320 <o> e ) $ num__8470 |
a = ( num__1 + r / num__100 ) ^ n * p ( num__1.1 ) ^ num__2 * num__7000 = num__1.21 * num__7000 = num__8470 the answer is e . <eor> e <eos> |
e |
percent__100.0__8470.0__ |
percent__100.0__8470.0__ |
| if num__16 men can reap num__80 hectares in num__24 days then how many hectares can num__36 men reap in num__30 days ? <o> a ) num__127 <o> b ) num__225 <o> c ) num__287 <o> d ) num__450 <o> e ) num__281 |
explanation : let the required no of hectares be x . then men - - - hectares - - - days num__16 - - - - - - - - - num__80 - - - - - - - - - num__24 num__36 - - - - - - - - - x - - - - - - - - - num__30 more men more hectares ( direct proportion ) more days more hectares ( direct proportion ) x = num__2.25 * num__1.25 * num__80 x = num__225 answer : b <eor> b <eos> |
b |
divide__36.0__16.0__ divide__30.0__24.0__ round__225.0__ |
divide__36.0__16.0__ divide__30.0__24.0__ round__225.0__ |
| if p is a positive integer and r is the units digit of num__3 ^ num__4 p × num__5 ^ num__5 p × num__6 ^ num__6 p × num__7 ^ num__7 p then r + num__2 is divisible by which of the following ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__9 |
the number will be an even number as it contains num__6 as a term in its product so units digit will be even . . even + num__2 will be another even and will be div by num__2 . . answer : a <eor> a <eos> |
a |
subtract__4.0__2.0__ |
subtract__4.0__2.0__ |
| due to num__10.0 decrease in the price of sugar and john can buy num__5 kg more sugar in rs num__100 then find the cp of sugar ? <o> a ) rs . num__2 ( num__0.111111111111 ) <o> b ) rs . num__2 ( num__0.222222222222 ) <o> c ) rs . num__3 ( num__0.111111111111 ) <o> d ) rs . num__1 ( num__0.111111111111 ) <o> e ) rs . num__2 ( num__0.4 ) |
here r = num__10.0 x = num__100 and a = num__5 kg actual price of sugar = num__10 * num__100 / ( ( num__100 - num__10 ) * num__5 ) = rs . num__2 ( num__0.222222222222 ) b <eor> b <eos> |
b |
divide__10.0__5.0__ divide__10.0__5.0__ |
divide__10.0__5.0__ divide__10.0__5.0__ |
| a man went into a bank to cash a check . in handing over the money the cashier by mistake gave him dollars for cents and cents for dollars . he pocketed the money without examining it and spent a nickel on his way home . he then found that he possessed exactly twice the amount of the check . he had no money in his pocket before going to the bank . what was the exact amount of that check ? <o> a ) $ num__31.63 <o> b ) $ num__35.63 <o> c ) $ num__32.63 <o> d ) $ num__37.63 <o> e ) $ num__33.63 |
a $ num__31.63 the amount must have been $ num__31.63 . he received $ num__63.31 . after he had spent a nickel there would remain the sum of $ num__63.26 which is twice the amount of the check . <eor> a <eos> |
a |
subtract__63.26__31.63__ |
subtract__63.26__31.63__ |
| by travelling at num__70 kmph a person reaches his destination on time . he covered two - third the total distance in one - third of the total time . what speed should he maintain for the remaining distance to reach his destination on time ? <o> a ) num__25 kmph <o> b ) num__20 kmph <o> c ) num__28 kmph <o> d ) num__35 kmph <o> e ) num__32 kmph |
let the time taken to reach the destination be num__3 x hours . total distance = num__70 * num__3 x = num__210 x km he covered num__0.666666666667 * num__210 x = num__140 x km in num__0.333333333333 * num__3 x = x hours so the remaining num__70 x km he has to cover in num__2 x hours . required speed = num__70 x / num__2 x = num__35 kmph . answer : d <eor> d <eos> |
d |
multiply__70.0__3.0__ subtract__210.0__70.0__ divide__70.0__210.0__ divide__140.0__70.0__ divide__70.0__2.0__ round__35.0__ |
multiply__70.0__3.0__ subtract__210.0__70.0__ divide__70.0__210.0__ divide__140.0__70.0__ divide__70.0__2.0__ divide__70.0__2.0__ |
| a is two years older than b who is twice as old as c . if the total of the ages of a b and c be num__32 the how old is b ? <o> a ) num__7 <o> b ) num__9 <o> c ) num__8 <o> d ) num__11 <o> e ) num__12 |
explanation : let c ' s age be x years . then b ' s age = num__2 x years . a ' s age = ( num__2 x + num__2 ) years . ( num__2 x + num__2 ) + num__2 x + x = num__32 ⇒ num__5 x = num__30 ⇒ x = num__6 . hence b ' s age = num__2 x = num__12 years . answer : e <eor> e <eos> |
e |
subtract__32.0__2.0__ divide__30.0__5.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
subtract__32.0__2.0__ divide__30.0__5.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
| a man performs num__0.5 of his journey by rail num__0.4 by bus and the remaining num__10 km on foot . his total journey is ? <o> a ) num__40 km <o> b ) num__50 km <o> c ) num__60 km <o> d ) num__70 km <o> e ) num__100 km |
let the total journey be x km x / num__2 + num__2 x / num__5 + num__10 = x num__9 x + num__50 = num__10 x x = num__50 km answer is b <eor> b <eos> |
b |
reverse__0.5__ multiply__0.5__10.0__ multiply__10.0__5.0__ multiply__10.0__5.0__ |
reverse__0.5__ divide__10.0__2.0__ multiply__10.0__5.0__ multiply__10.0__5.0__ |
| if n is a positive integer which of the following expressions must be even ? <o> a ) ( n − num__3 ) ( n + num__1 ) <o> b ) ( n − num__8 ) ( n + num__1 ) <o> c ) ( n − num__4 ) ( n + num__4 ) <o> d ) ( n − num__5 ) ( n + num__1 ) <o> e ) ( n − num__5 ) ( n + num__5 ) |
whether n is even or odd ( n - num__8 ) ( n + num__1 ) will have one odd factor and one even factor . the product will be even . the answer is b . <eor> b <eos> |
b |
multiply__8.0__1.0__ |
multiply__8.0__1.0__ |
| the average age of three boys is num__25 years and their ages are in the proportion num__3 : num__5 : num__7 . the age of the youngest boy is : <o> a ) num__21 years <o> b ) num__18 years <o> c ) num__15 years <o> d ) num__9 years <o> e ) none of these |
total age of num__3 boys = ( num__25 × num__3 ) years = num__75 years . ratio of their ages = num__3 : num__5 : num__7 . age of the youngest = ( num__75 × num__3 ⁄ num__15 ) years = num__15 years . answer c <eor> c <eos> |
c |
multiply__25.0__3.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
multiply__25.0__3.0__ multiply__3.0__5.0__ multiply__3.0__5.0__ |
| six animals of a circus has to be placed inside in six cages one in each cage . if num__4 of the cage are too small for num__6 of the animal then find the number of ways of caging the animal . <o> a ) a . num__240 <o> b ) b . num__808250 <o> c ) c . num__502450 <o> d ) d . num__784200 <o> e ) e . num__302400 |
ways for cages that are too small = num__5 * num__4 * num__3 * num__2 = num__120 since we have num__2 cages and num__2 animal left therefore ways for num__2 cages = num__2 ! total ways = num__120 * num__2 ! = num__604800 answer a <eor> a <eos> |
a |
subtract__6.0__4.0__ multiply__2.0__120.0__ |
subtract__6.0__4.0__ multiply__2.0__120.0__ |
| in a class there are a total of num__8 rows of desks and each desk can seat one student . there are num__10 desks in the first row . in each subsequent row there are num__2 more desks than in the previous row . find the maximum number of students seated in the class ? <o> a ) num__112 <o> b ) num__144 <o> c ) num__136 <o> d ) num__132 <o> e ) num__118 |
the num__8 rows form an arithmetic progression with first term num__10 and last num__24 . avg is num__17 . so num__17 * num__8 = num__136 answer is c . <eor> c <eos> |
c |
multiply__8.0__17.0__ multiply__8.0__17.0__ |
multiply__8.0__17.0__ multiply__8.0__17.0__ |
| in how many ways can the letters of word cremation be arranged such that no two vowels appear together ? <o> a ) num__9 ! <o> b ) num__5 ! * num__4 ! <o> c ) num__4 ! * num__5 ! <o> d ) num__5 ! * num__4 ! * num__2 ! <o> e ) num__6 ! * num__4 ! |
no num__2 vowels together = the only arrangement possible will be c v c v c v c v ( with v = vowel c = consonant ) . this is true as we have num__4 vowels and num__5 consonants and any other combination will force us to pair num__2 vowels together . thus the number of arrangements possible : num__5 * num__4 * num__4 * num__3 * num__3 * num__2 * num__2 * num__1 = num__4 ! * num__5 ! - - - - > c is the correct answer . <eor> c <eos> |
c |
coin_space__ vowel_space__ choose__4.0__3.0__ |
coin_space__ vowel_space__ choose__4.0__3.0__ |
| a factory produces num__4560 toys per week . if the workers at this factory work num__4 days a week and if these workers make the same number of toys everyday how many toys are produced each day ? <o> a ) num__1140 toys <o> b ) num__2375 toys <o> c ) num__3375 toys <o> d ) num__4375 toys <o> e ) num__5375 toys |
to find the number of toys produced every day we divide the total number of toys produced in one week ( of num__4 days ) by num__4 . num__1140.0 = num__1140 toys correct answer a <eor> a <eos> |
a |
divide__4560.0__4.0__ round__1140.0__ |
divide__4560.0__4.0__ round__1140.0__ |
| a and b together have rs . num__1210 . if num__0.266666666667 of a ' s amount is equal to num__0.4 of b ' s amount how much amount does b have ? <o> a ) rs . num__460 <o> b ) rs . num__484 <o> c ) rs . num__550 <o> d ) rs . num__664 <o> e ) none |
( num__0.266666666667 ) a = ( num__0.4 ) b a = ( num__0.4 * num__3.75 ) b a = ( num__1.5 ) b ( a / b ) = ( num__1.5 ) a : b = num__3 : num__2 b ' s share = rs . ( num__1210 * num__0.4 ) = rs . num__484 answer = b <eor> b <eos> |
b |
multiply__0.4__3.75__ round_down__3.75__ divide__3.0__1.5__ multiply__1210.0__0.4__ multiply__1210.0__0.4__ |
multiply__0.4__3.75__ round_down__3.75__ divide__3.0__1.5__ multiply__1210.0__0.4__ multiply__1210.0__0.4__ |
| train starts from amritsar to bombay at num__9 am . it reaches destination after num__3 days at num__9 : num__30 am . every day a train starts . how many trains does it come across on the way ? <o> a ) num__5 th <o> b ) num__4 th <o> c ) num__6 th <o> d ) num__7 th <o> e ) num__8 th |
because one train taking num__3 days so num__1 train first day another at second day and num__3 rd one at third day but as given train reaches at num__9.30 am insted of num__9 . oo am so d same day train already left that train will be num__4 th train . answer : b <eor> b <eos> |
b |
add__3.0__1.0__ round__4.0__ |
add__3.0__1.0__ round__4.0__ |
| for the past n days the average ( arithmetic mean ) daily production at a company was num__50 units . if today ' s production of num__60 units raises the average to num__55 units per day what is the value of n ? <o> a ) num__30 <o> b ) num__18 <o> c ) num__10 <o> d ) num__9 <o> e ) num__1 |
( average production for n days ) * n = ( total production for n days ) - - > num__50 n = ( total production for n days ) ; ( total production for n days ) + num__60 = ( average production for n + num__1 days ) * ( n + num__1 ) - - > num__50 n + num__60 = num__55 * ( n + num__1 ) - - > n = num__1 . answer : e . <eor> e <eos> |
e |
reverse__1.0__ |
reverse__1.0__ |
| by selling num__150 mangoes a fruit - seller gains the selling price of num__30 mangoes . find the gain percent ? <o> a ) num__26.0 <o> b ) num__25.0 <o> c ) num__15.0 <o> d ) num__28.0 <o> e ) num__55 % |
sp = cp + g num__150 sp = num__150 cp + num__30 sp num__120 sp = num__150 cp num__120 - - - num__30 cp num__100 - - - ? = > num__25.0 answer : b <eor> b <eos> |
b |
percent__25.0__100.0__ |
percent__25.0__100.0__ |
| rahul went to purchase a nokia mobile handset the shopkeeper told him to pay num__20.0 tax if he asked the bill . rahul manages to get the discount of num__5.0 on the actual saleprice of the mobile and he paid the shopkeeper rs . num__3325 without tax . besides he manages to avoid to pay num__20.0 tax on the already discounted price what is the amount of discount that he has gotten ? <o> a ) num__750 <o> b ) num__375 <o> c ) num__875 <o> d ) num__525 <o> e ) num__625 |
explanation : cp = num__100 sp ( with tax ) = num__120 new sp = num__100 - num__5 = num__95 \ therefore effective discount = num__120 - num__95 = num__25 so at sp of num__95 - - - - > discount = num__25 and at sp of num__3325 - - - - - > discount = num__0.263157894737 = num__3325 = num__875 . answer : c <eor> c <eos> |
c |
percent__100.0__875.0__ |
percent__100.0__875.0__ |
| farm tax is levied on the num__50.0 of the cultivated land . the tax department collected total $ num__3840 through the farm tax from the village of mr . william . mr . william paid only $ num__480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is : <o> a ) num__15.0 <o> b ) num__25.0 <o> c ) num__0.125 <o> d ) num__0.2083 <o> e ) none |
this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( num__0.125 ) x num__100 = num__12.5 in percentage terms . but the question asks ratio between his total land to total cultivated land . hence the answer is num__12.5 x ( num__2.0 ) = num__25.0 the correct answer is ( b ) . <eor> b <eos> |
b |
divide__480.0__3840.0__ multiply__0.125__100.0__ divide__100.0__50.0__ divide__50.0__2.0__ divide__50.0__2.0__ |
divide__480.0__3840.0__ multiply__0.125__100.0__ divide__100.0__50.0__ divide__50.0__2.0__ divide__50.0__2.0__ |
| two trains are moving in opposite directions at num__60 km / hr and num__90 km / hr . their lengths are num__1.15 km and num__0.6 km respectively . the time taken by the slower train to cross the faster train in seconds is ? <o> a ) num__42 <o> b ) num__9 <o> c ) num__7 <o> d ) num__67 <o> e ) num__15 |
: relative speed = num__60 + num__90 = num__150 km / hr . = num__150 * num__0.277777777778 = num__41.6666666667 m / sec . distance covered = num__1.15 + num__0.6 = num__1.75 km = num__1750 m . required time = num__1750 * num__0.024 = num__42 sec . answer : a <eor> a <eos> |
a |
add__60.0__90.0__ add__1.15__0.6__ multiply__1750.0__0.024__ round__42.0__ |
add__60.0__90.0__ add__1.15__0.6__ multiply__1750.0__0.024__ multiply__1750.0__0.024__ |
| sheila works num__8 hours per day on monday wednesday and friday and num__6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ num__468 per week . how much does she earn in dollars per hour ? <o> a ) num__2 <o> b ) num__8 <o> c ) num__9 <o> d ) num__11 <o> e ) num__13 |
explanation : total hours worked = num__8 x num__3 + num__6 x num__2 = num__36 total earned = num__468 . hourly wage = num__13.0 = num__13 answer : e <eor> e <eos> |
e |
subtract__8.0__6.0__ divide__468.0__36.0__ round__13.0__ |
subtract__8.0__6.0__ divide__468.0__36.0__ round__13.0__ |
| the shopkeeper increased the price of a product by num__22.0 so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only num__70.0 of the required amount . what is the net difference in the expenditure on that product ? <o> a ) a ) num__12.5 <o> b ) b ) num__12.2 <o> c ) c ) num__13.15 <o> d ) d ) num__14.0 <o> e ) e ) num__15 % |
quantity x rate = price num__1 x num__1 = num__1 num__0.7 x num__1.22 = num__0.854 decrease in price = ( num__0.122 / num__1 ) × num__100 = num__12.2 b ) <eor> b <eos> |
b |
multiply__1.22__0.7__ divide__70.0__0.7__ multiply__100.0__0.122__ multiply__1.0__12.2__ |
multiply__1.22__0.7__ divide__70.0__0.7__ multiply__100.0__0.122__ divide__12.2__1.0__ |
| in how many different ways can the letters of the word ' leading ' be arranged in such a way that the vowels always come together ? <o> a ) num__720 <o> b ) num__127 <o> c ) num__137 <o> d ) num__129 <o> e ) num__112 |
the word ' leading ' has num__7 different letters . when the vowels eai are always together they can be supposed to form one letter . then we have to arrange the letters lndg ( eai ) . now num__5 ( num__4 + num__1 ) letters can be arranged in num__5 ! = num__120 ways . the vowels ( eai ) can be arranged among themselves in num__3 ! = num__6 ways . required number of ways = ( num__120 x num__6 ) = num__720 . answer : a <eor> a <eos> |
a |
vowel_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
vowel_space__ die_space__ choose__6.0__3.0__ choose__6.0__3.0__ |
| in a num__1000 m race a beats b by num__50 m and b beats c by num__100 m . in the same race by how many meters does a beat c ? <o> a ) num__145 m <o> b ) num__829 m <o> c ) num__822 m <o> d ) num__929 m <o> e ) num__132 mj |
by the time a covers num__1000 m b covers ( num__1000 - num__50 ) = num__950 m . by the time b covers num__1000 m c covers ( num__1000 - num__100 ) = num__900 m . so the ratio of speeds of a and c = num__1.05263157895 * num__1.11111111111 = num__1.16959064327 so by the time a covers num__1000 m c covers num__855 m . so in num__1000 m race a beats c by num__1000 - num__855 = num__145 m . answer : a <eor> a <eos> |
a |
subtract__1000.0__50.0__ subtract__1000.0__100.0__ divide__1000.0__950.0__ divide__1000.0__900.0__ subtract__1000.0__855.0__ round__145.0__ |
subtract__1000.0__50.0__ subtract__1000.0__100.0__ divide__1000.0__950.0__ divide__1000.0__900.0__ subtract__1000.0__855.0__ subtract__1000.0__855.0__ |
| num__0.009 / x = num__0.05 . find the value of x <o> a ) num__0.00018 <o> b ) num__0.09 <o> c ) num__0.018 <o> d ) num__0.18 <o> e ) num__90 |
x = num__0.009 / num__0.18 = num__0.05 answer : d <eor> d <eos> |
d |
divide__0.009__0.05__ divide__0.009__0.05__ |
divide__0.009__0.05__ divide__0.009__0.05__ |
| if a : b = num__7 : num__5 b : c = num__9 : num__11 find a : b : c ? <o> a ) num__63 : num__45 : num__58 <o> b ) num__63 : num__45 : num__55 <o> c ) num__63 : num__45 : num__53 <o> d ) num__63 : num__45 : num__59 <o> e ) num__63 : num__45 : num__58 |
a : b = num__7 : num__5 b : c = num__9 : num__11 a : b : c = num__63 : num__45 : num__55 answer : b <eor> b <eos> |
b |
multiply__7.0__9.0__ multiply__5.0__9.0__ multiply__5.0__11.0__ multiply__7.0__9.0__ |
multiply__7.0__9.0__ multiply__5.0__9.0__ multiply__5.0__11.0__ multiply__7.0__9.0__ |
| a man swims downstream num__32 km and upstream num__24 km taking num__4 hours each time what is the speed of the man in still water ? <o> a ) num__9 <o> b ) num__8 <o> c ) num__6 <o> d ) num__7 <o> e ) num__2 |
num__32 - - - num__4 ds = num__8 ? - - - - num__1 num__24 - - - - num__4 us = num__6 ? - - - - num__1 m = ? m = ( num__8 + num__6 ) / num__2 = num__7 answer : d <eor> d <eos> |
d |
subtract__32.0__24.0__ divide__24.0__4.0__ subtract__6.0__4.0__ add__1.0__6.0__ round__7.0__ |
subtract__32.0__24.0__ divide__24.0__4.0__ subtract__6.0__4.0__ add__1.0__6.0__ add__1.0__6.0__ |
| how many number of three digit numbers can be formed usingzs numbers num__23 num__45 in which no repeatation . . ? <o> a ) num__24 <o> b ) num__34 <o> c ) num__18 <o> d ) num__16 <o> e ) num__12 |
num__4 * num__3 * num__2 = num__24 num__1 place of digit can be filled with num__4 no num__2 place of digit can be filled with num__3 no ( as num__1 is already filled and not available ) num__3 place of digit can be filled with num__2 no so num__24 ways answer : a <eor> a <eos> |
a |
coin_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
coin_space__ choose__4.0__2.0__ choose__4.0__2.0__ |
| a cycle is bought for rs . num__900 and sold for rs . num__1080 find the gain percent ? <o> a ) num__11 <o> b ) num__20 <o> c ) num__98 <o> d ) num__77 <o> e ) num__12 |
num__900 - - - - num__180 num__100 - - - - ? = > num__20.0 answer : b <eor> b <eos> |
b |
percent__100.0__20.0__ |
percent__100.0__20.0__ |
| the length of the bridge which a train num__130 m long and traveling at num__45 km / hr can cross in num__30 sec is ? <o> a ) num__299 <o> b ) num__249 <o> c ) num__245 <o> d ) num__289 <o> e ) num__271 |
speed = num__45 * num__0.277777777778 = num__12.5 m / sec . time = num__30 sec let the length of bridge be x meters . then ( num__130 + x ) / num__30 = num__12.5 x = num__245 m . answer : c <eor> c <eos> |
c |
round__245.0__ |
round__245.0__ |
| two trains num__151 meters and num__165 meters in length respectively are running in opposite directions one at the rate of num__80 km and the other at the rate of num__65 kmph . in what time will they be completely clear of each other from the moment they meet ? <o> a ) num__7.19 <o> b ) num__7.18 <o> c ) num__7.16 <o> d ) num__7.84 <o> e ) num__7.12 |
t = ( num__151 + num__165 ) / ( num__80 + num__65 ) * num__3.6 t = num__7.84 answer : d <eor> d <eos> |
d |
round__7.84__ |
round__7.84__ |
| let q represent a set of six distinct prime numbers . if the sum of the numbers in q is even and x is a member of q then what is the least possible value that x can be ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__5 <o> d ) num__3 <o> e ) num__7 |
q = p num__1 + p num__2 + p num__3 + p num__4 + p num__5 + p num__6 = even ( and all primes are distinct ) if the least prime is num__2 then we have sum of q = odd . . . . hence num__2 is not possible num__1 is not prime . next least prime is num__3 this satisfies the given condition that sum of q = even . ans . d . num__3 <eor> d <eos> |
d |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ add__1.0__5.0__ add__1.0__2.0__ |
add__1.0__2.0__ add__1.0__3.0__ add__1.0__4.0__ add__1.0__5.0__ add__1.0__2.0__ |
| if num__20.0 of a = b then b % of num__20 is the same as : <o> a ) num__4.0 of a <o> b ) num__5.0 of a <o> c ) num__20.0 of a <o> d ) none of these <o> e ) num__8.0 of a |
answer : option a num__20.0 of a = b = num__0.2 a = b b % of num__20 = b / num__100 * num__20 = num__0.2 a * num__0.01 * num__20 = num__0.04 a = num__4.0 of a <eor> a <eos> |
a |
percent__20.0__0.2__ percent__100.0__4.0__ |
percent__20.0__0.2__ percent__100.0__4.0__ |
| total num__100 members are writing exam . in the num__48 members are writing first exam . num__45 members are writing second exam . num__38 members are writing third exam . num__5 members are writing all the three exams . how many members are writing num__2 exams ? <o> a ) num__66 <o> b ) num__21 <o> c ) num__88 <o> d ) num__719 <o> e ) num__011 |
otal number of exams written by num__100 students = num__48 + num__45 + num__38 = num__131 now let us say x members are writing only num__1 exam y members are writing only num__2 exams z members are writing only num__3 exams . therefore x + num__2 y + num__3 z = num__131 also x + y + z = num__100 . given that z = num__5 . so x + num__2 y = num__116 and x + y = num__95 . solving we get y = num__21 . so num__21 members are writing exactly num__2 exams . answer : b <eor> b <eos> |
b |
subtract__48.0__45.0__ subtract__100.0__5.0__ subtract__116.0__95.0__ multiply__1.0__21.0__ |
add__2.0__1.0__ subtract__100.0__5.0__ subtract__116.0__95.0__ multiply__1.0__21.0__ |
| evaluate : ( num__1 - num__0.1 ) ( num__1 - num__0.0909090909091 ) ( num__1 - num__0.0833333333333 ) . . . ( num__1 - num__0.010101010101 ) ( num__1 - num__0.01 ) <o> a ) num__0.07 <o> b ) num__0.08 <o> c ) num__0.09 <o> d ) num__0.06 <o> e ) num__0.03 |
( num__1 - num__0.1 ) ( num__1 - num__0.0909090909091 ) ( num__1 - num__0.0833333333333 ) . . . ( num__1 - num__0.010101010101 ) ( num__1 - num__0.01 ) = ( num__0.9 ) ( num__0.909090909091 ) ( num__0.916666666667 ) . . . ( num__0.989898989899 ) ( num__0.99 ) = num__0.09 : simplify correct answer c <eor> c <eos> |
c |
subtract__1.0__0.1__ subtract__1.0__0.0909__ subtract__1.0__0.0833__ subtract__1.0__0.0101__ subtract__1.0__0.01__ subtract__0.1__0.01__ multiply__1.0__0.09__ |
subtract__1.0__0.1__ subtract__1.0__0.0909__ subtract__1.0__0.0833__ subtract__1.0__0.0101__ subtract__1.0__0.01__ subtract__0.1__0.01__ subtract__0.1__0.01__ |
| a pipe can fill a cistern in num__20 minutes whereas the cistern when fill can be emptied by a leak in num__28 minutes . when both pipes are opened find when the cistern will be full ? <o> a ) num__22 minutes <o> b ) num__88 minutes <o> c ) num__70 minutes <o> d ) num__77 minutes <o> e ) num__48 minutes |
num__0.05 - num__0.0357142857143 = num__0.0142857142857 num__70 minutes answer : c <eor> c <eos> |
c |
subtract__0.05__0.0357__ round__70.0__ |
subtract__0.05__0.0357__ round__70.0__ |
| shopkeeper rise price by num__34.0 and gives successive discount of num__10.0 and num__15.0 . what is overall % gain or loss ? <o> a ) num__2.51 <o> b ) num__4.51 <o> c ) num__3.51 <o> d ) num__6.51 <o> e ) num__7.51 % |
let d initial price be num__100 num__34.0 rise now price = num__1.34 * num__100 = num__134 num__10.0 discount then price = num__134 * num__0.9 = num__120.6 num__15.0 discount then price = num__120.6 * num__0.85 = num__102.51 so gain = num__102.51 - num__100 = num__2.51 gain % = gain * num__100 / cp = = > num__2.51 * num__1.0 = num__2.51 answer : a <eor> a <eos> |
a |
percent__100.0__2.51__ |
percent__100.0__2.51__ |
| on a test students receive num__5 points for each correct answer and are penalized by losing one point for each incorrect answer . there are num__6 questions on the test and each question has num__4 answer options a b c and d . it is known that num__5 of the questions have option b as the correct answer and one question has option c as the correct answer . if a student marks b for the first num__3 questions and c for the last num__3 questions what is the minimum possible score that student can receive ? <o> a ) - num__6 <o> b ) - num__1 <o> c ) num__0 <o> d ) num__1 <o> e ) num__6 |
for the minimum possible score let us take the worst case scenario suppose he gets all of the last num__3 qs wrong whose correct answer options are b and one of the first num__3 qs wrong whose correct option is c in that case he will get only num__2 out of the first num__3 qs right . therefore minimum possible score = ( num__5 * num__2 ) - ( num__4 * num__1 ) = num__6 answer e <eor> e <eos> |
e |
coin_space__ die_space__ |
coin_space__ die_space__ |
| a shade of paint is made by evenly mixing m gallons of white paint costing $ num__12 a gallon with n gallons of blue paint costing $ num__30 . what is the cost in dollars per gallon of the resulting mixture ? <o> a ) num__12 m + num__30 n <o> b ) num__42 ( num__12 m + num__30 n ) <o> c ) ( num__12 m + num__30 n ) / num__42 <o> d ) ( num__12 m + num__30 n ) / ( m + n ) <o> e ) num__42 ( m + n ) / ( num__12 m + num__30 n ) |
m gallons of white paint and n gallons of blue paint are combined to form a new mixture . the total number of gallons in the new mixture will be m + n gallons m gallons of white paint cost $ num__12 so the cost of adding this mixture is num__12 m n gallons of blue paint cost $ num__30 so the cost of adding this mixture is num__30 n the cost per gallon of the new mixture is equal to the total cost of all the paint used divided by the total number of gallons used to produce it : cost of new mixture : num__12 m + num__30 n total gallons used : m + n cost per gallon = total cost / total gallons used = ( num__12 m + num__30 n ) / ( m + n ) answer : d <eor> d <eos> |
d |
round__12.0__ |
round__12.0__ |
| if the population of a certain country increases at the rate of one person every num__20 seconds by how many persons does the population increase in num__30 minutes ? <o> a ) num__55 <o> b ) num__20 <o> c ) num__96 <o> d ) num__90 <o> e ) num__40 |
answer = num__3 * num__30 = num__90 answer = d <eor> d <eos> |
d |
multiply__30.0__3.0__ round__90.0__ |
multiply__30.0__3.0__ multiply__30.0__3.0__ |
| by selling a watch for rs . num__560 / - a man loses num__20.0 what is that watch cost price ? <o> a ) s . num__600 / - <o> b ) s . num__700 / - <o> c ) s . num__800 / - <o> d ) s . num__850 / - <o> e ) s . num__900 / - |
num__80.0 - - - - - - > num__560 ( num__80 * num__7 = num__640 ) num__100.0 - - - - - - > num__700 ( num__100 * num__7 = num__700 ) cost price = rs . num__700 / - b ) <eor> b <eos> |
b |
divide__560.0__80.0__ add__560.0__80.0__ add__20.0__80.0__ multiply__100.0__7.0__ multiply__100.0__7.0__ |
divide__560.0__80.0__ add__560.0__80.0__ add__20.0__80.0__ multiply__100.0__7.0__ multiply__100.0__7.0__ |
| in a division sum the remainder is num__0 . as student mistook the divisor by num__12 instead of num__21 and obtained num__35 as quotient . what is the correct quotient ? <o> a ) num__0 <o> b ) num__12 <o> c ) num__13 <o> d ) num__20 <o> e ) num__22 |
number = ( num__12 * num__35 ) = num__420 correct quotient = num__20.0 = num__20 answer : d <eor> d <eos> |
d |
multiply__12.0__35.0__ divide__420.0__21.0__ divide__420.0__21.0__ |
multiply__12.0__35.0__ divide__420.0__21.0__ divide__420.0__21.0__ |
| a person invested in all rs . num__2600 at num__4.0 num__6.0 and num__8.0 per annum simple interest . at the end of the year he got the same interest in all the three cases . the money invested at num__4.0 is ? <o> a ) num__3888 <o> b ) num__2799 <o> c ) num__2799 <o> d ) num__1200 <o> e ) num__2771 |
let the parts be x y and [ num__2600 - ( x + y ) ] . then ( x * num__4 * num__1 ) / num__100 = ( y * num__6 * num__1 ) / num__100 = { [ num__2600 - ( x + y ) ] * num__8 * num__1 } / num__100 y / x = num__0.666666666667 = num__0.666666666667 or y = num__0.666666666667 x so ( x * num__4 * num__1 ) / num__100 = [ ( num__2600 - num__1.66666666667 x ) * num__0.8 num__52 x = ( num__7800 * num__8 ) = > x = num__1200 money invested at num__4.0 = rs . num__1200 . answer : d <eor> d <eos> |
d |
percent__100.0__1200.0__ |
percent__100.0__1200.0__ |
| if the area of circle is num__452 sq cm then its circumference ? <o> a ) num__21 <o> b ) num__88 <o> c ) num__75 <o> d ) num__78 <o> e ) num__12 |
num__3.14285714286 r num__2 = num__452 = > r = num__12 num__2 * num__3.14285714286 * num__12 = num__75 answer : c <eor> c <eos> |
c |
triangle_area__2.0__75.0__ |
triangle_area__2.0__75.0__ |
| the diameter of a wheel is num__0.7 m . how far will it travel in num__200 revolutions <o> a ) num__4.4 m <o> b ) num__220 m <o> c ) num__440 m <o> d ) num__560 <o> e ) num__650 |
diameter ( d ) = num__0.7 m circumference of the wheel = num__2 * pi * r = d * pi = num__0.7 * num__3.14 = num__2.198 m . in one revolution wheel will covered = num__2.198 m so in num__200 revolution it will cover = num__200 * num__2.198 = num__4.396 m = num__440 m ( approx ) . answer : option c <eor> c <eos> |
c |
multiply__0.7__3.14__ multiply__2.0__2.198__ round__440.0__ |
multiply__0.7__3.14__ multiply__2.0__2.198__ round__440.0__ |
| john left home and drove at the rate of num__35 mph for num__2 hours . he stopped for lunch then drove for another num__3 hours at the rate of num__55 mph to reach his destination . how many miles did john drive ? <o> a ) num__235 miles . <o> b ) num__245 miles . <o> c ) num__255 miles . <o> d ) num__265 miles . <o> e ) num__275 miles . |
the total distance d traveled by john is given by d = num__35 * num__2 + num__3 * num__55 = num__235 miles . answer a <eor> a <eos> |
a |
round__235.0__ |
round__235.0__ |
| machine a working alone can complete a job in num__9 hours . machine b working alone can do the same job in num__12 hours . how long will it take both machines working together at their respective constant rates to complete the job ? <o> a ) num__2.625 <o> b ) num__5.4 <o> c ) num__1.71428571429 <o> d ) num__5.14285714286 <o> e ) num__2.14285714286 |
machines ( a ) ( b ) ( a + b ) - - - time - num__9 - ( num__12 ) - - - - rate num__4 - - - num__3 - - - num__7 - - work num__36 - - - num__36 - - num__36 a + b = ( num__5.14285714286 ) d <eor> d <eos> |
d |
subtract__12.0__9.0__ add__3.0__4.0__ multiply__9.0__4.0__ divide__36.0__7.0__ divide__36.0__7.0__ |
subtract__12.0__9.0__ add__3.0__4.0__ multiply__9.0__4.0__ divide__36.0__7.0__ divide__36.0__7.0__ |
| two trains are running at num__40 km / hr and num__20 km / hr respectively in the same direction . fast train completely passes a man sitting in the slower train in num__7 seconds . what is the length of the fast train ? <o> a ) num__23 m <o> b ) num__23 ( num__0.222222222222 ) m <o> c ) num__38 ( num__0.888888888889 ) m <o> d ) num__29 m <o> e ) num__28 m |
relative speed = ( num__40 - num__20 ) km / hr = ( num__20 x ( num__0.277777777778 ) ) m / sec = ( num__5.55555555556 ) m / sec . therefore length of faster train = ( ( num__5.55555555556 ) x num__7 ) m = ( num__38.8888888889 ) m = num__38 ( num__0.888888888889 ) m . c <eor> c <eos> |
c |
subtract__38.8889__38.0__ round__38.0__ |
subtract__38.8889__38.0__ subtract__38.8889__0.8889__ |
| johnny travels a total of one hour to and from school . on the way there he jogs at num__5 miles per hour and on the return trip he gets picked up by the bus and returns home at num__21 miles per hour . how far is it to the school ? <o> a ) num__2 miles <o> b ) num__4 miles <o> c ) num__6.5 miles <o> d ) num__8 miles <o> e ) num__10 miles |
answer : c ) num__6.5 miles . average speed for round trip = num__2 * a * b / ( a + b ) where a b are speeds so average speed was = num__2 * num__5 * num__21 / ( num__5 + num__21 ) = num__6.5 m / hr the distance between schoolhome should be half of that . ie . num__6.5 miles answer c <eor> c <eos> |
c |
round__6.5__ |
round__6.5__ |
| in the year num__2011 shantanu gets a total of rs . num__3832.5 as his pocket allowance . find his pocket allowance per day . ? <o> a ) rs . num__9.5 <o> b ) rs . num__10.5 <o> c ) rs . num__12.5 <o> d ) rs . num__11.5 <o> e ) none of these |
answer allowance per day = num__3832.5 / num__365 = rs . num__10.50 correct option : b <eor> b <eos> |
b |
divide__3832.5__365.0__ round__10.5__ |
divide__3832.5__365.0__ divide__3832.5__365.0__ |
| three numbers are in the ratio num__3 : num__5 : num__7 . the largest number is num__56 . what is the difference between smallest and largest number ? <o> a ) num__20 <o> b ) num__24 <o> c ) num__28 <o> d ) num__32 <o> e ) num__36 |
the three numbers are num__3 x num__5 x and num__7 x . the largest number is num__56 = num__7 * num__8 so x = num__8 . the smallest number is num__3 * num__8 = num__24 . num__56 - num__24 = num__32 the answer is d . <eor> d <eos> |
d |
add__3.0__5.0__ multiply__3.0__8.0__ subtract__56.0__24.0__ subtract__56.0__24.0__ |
add__3.0__5.0__ multiply__3.0__8.0__ subtract__56.0__24.0__ subtract__56.0__24.0__ |
| a and b can do a piece of work in num__40 days and num__50 days respectively . they work together for num__10 days and b leaves . in how many days the whole work is completed ? <o> a ) num__28 days <o> b ) num__60 days <o> c ) num__20 days <o> d ) num__30 days <o> e ) num__40 days |
explanation : ( a + b ) ’ s num__10 days work = num__10 [ num__0.025 + num__0.02 ] = num__10 [ num__5 + num__0.02 ] = num__0.45 a complete remaining work in num__0.45 * num__40 = num__18 total work = num__10 + num__18 = num__28 days answer : option a <eor> a <eos> |
a |
divide__50.0__10.0__ multiply__40.0__0.45__ add__10.0__18.0__ round__28.0__ |
divide__50.0__10.0__ multiply__40.0__0.45__ add__10.0__18.0__ add__10.0__18.0__ |
| the average salary of num__20 people in the shipping department at a certain firm is $ num__1000 . the salary of num__10 of the employees is $ num__600 each and the salary of num__6 of the employees is $ num__200 each . what is the average salary of the remaining employees ? <o> a ) $ num__2500 <o> b ) $ num__3200 <o> c ) $ num__4500 <o> d ) $ num__5000 <o> e ) $ num__5200 |
total salary . . . num__20 * num__1000 = num__20000 num__10 emp @ num__600 = num__6000 num__6 emp @ num__200 = num__1200 remaing num__4 emp salary = num__20000 - num__6000 - num__1200 = num__12800 average = num__3200.0 = num__3200 answer is b <eor> b <eos> |
b |
multiply__20.0__1000.0__ multiply__1000.0__6.0__ add__1000.0__200.0__ subtract__10.0__6.0__ divide__12800.0__4.0__ divide__12800.0__4.0__ |
multiply__20.0__1000.0__ multiply__1000.0__6.0__ multiply__6.0__200.0__ subtract__10.0__6.0__ divide__12800.0__4.0__ divide__12800.0__4.0__ |
| the perimeter of a triangle is num__48 cm and the inradius of the triangle is num__2.5 cm . what is the area of the triangle ? <o> a ) num__76 cm num__2 <o> b ) num__56 cm num__2 <o> c ) num__18 cm num__2 <o> d ) num__98 cm num__2 <o> e ) num__60 cm num__2 |
area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = num__2.5 * num__24.0 = num__60 cm num__2 answer : e <eor> e <eos> |
e |
triangle_area__48.0__2.5__ triangle_area__48.0__2.5__ |
multiply__2.5__24.0__ multiply__2.5__24.0__ |
| for the positive numbers n n + num__3 n + num__5 n + num__8 and n + num__9 the mean is how much greater than the median ? <o> a ) num__0 <o> b ) num__1 <o> c ) n + l <o> d ) n + num__2 <o> e ) n + num__3 |
let ’ s first calculate the mean ( arithmetic average ) . mean = sum / quantity mean = ( n + n + num__3 + n + num__5 + n + num__8 + n + num__9 ) / num__5 mean = ( num__5 n + num__25 ) / num__5 mean = n + num__5 next we determine the median . the median is the middle value when the terms are ordered from least to greatest . the terms ordered from least to greatest are as follows : n n + num__3 n + num__5 n + num__8 n + num__9 the median is n + num__5 . finally we are asked how much greater the mean is than the median . to determine the difference we can subtract the smaller value ( the median ) from the larger value ( the mean ) and we get : n + num__5 – ( n + num__5 ) = n + num__5 – n – num__5 = num__0 the answer is a <eor> a <eos> |
a |
multiply__3.0__0.0__ |
divide__0.0__3.0__ |
| if num__522 x is a num__3 digit no . with as a digit x . if the no . is divisible by num__6 what is the value of the digit x is ? <o> a ) num__4 <o> b ) num__6 <o> c ) num__9 <o> d ) num__10 <o> e ) num__11 |
if a number is divisible by num__6 it must be divisible by both num__2 and num__3 in num__522 x to this number be divisible by num__2 the value of x must be even . so it can be num__24 or num__6 from given options num__552 x is divisible by num__3 if sum of its digits is a multiple of num__3 . num__5 + num__5 + num__2 + x = num__12 + x if put x = num__2 num__12 + num__2 = num__14 not a multiple of num__3 if put x = num__4 num__12 + num__6 = num__18 is a multiple of num__3 if put x = num__6 num__12 + num__2 = num__14 not a multiple of num__3 the value of x is num__6 . b <eor> b <eos> |
b |
divide__6.0__3.0__ add__3.0__2.0__ multiply__6.0__2.0__ add__2.0__12.0__ subtract__6.0__2.0__ multiply__3.0__6.0__ multiply__3.0__2.0__ |
divide__6.0__3.0__ add__3.0__2.0__ multiply__6.0__2.0__ add__2.0__12.0__ subtract__6.0__2.0__ add__6.0__12.0__ add__2.0__4.0__ |
| if an amount of rs num__42680 is distributed equally amongst num__22 persons how much amount would each person get ? <o> a ) rs num__1905 <o> b ) rs num__2000 <o> c ) rs num__745 <o> d ) rs num__1940 <o> e ) none |
required amount = num__1940.0 = rs num__1940 answer d <eor> d <eos> |
d |
divide__42680.0__22.0__ divide__42680.0__22.0__ |
divide__42680.0__22.0__ divide__42680.0__22.0__ |
| how long will it take a train travelling at num__68 kmph to pass an electric pole if the train is num__170 m long <o> a ) num__6.99 <o> b ) num__5.99 <o> c ) num__8.99 <o> d ) num__4.99 <o> e ) num__9.99 |
sol . speed = [ num__68 x num__0.277777777778 ] m / sec = num__18.9 m / sec . time taken = ( num__170 / num__18.9 ) sec = num__8.99 sec . answer c <eor> c <eos> |
c |
round__8.99__ |
round__8.99__ |
| a person travels equal distances with speeds of num__3 km / hr num__4 km / hr and num__5 km / hr and takes a total time of num__52 minutes . the total distance is ? <o> a ) num__1 km <o> b ) num__2 km <o> c ) num__3.3 km <o> d ) num__4 km <o> e ) num__5 km |
c num__3 km let the total distance be num__3 x km . then x / num__3 + x / num__4 + x / num__5 = num__0.866666666667 num__47 x / num__60 = num__0.866666666667 = > x = num__1.1 total distance = num__3 * num__1.1 = num__3.3 km . <eor> c <eos> |
c |
subtract__52.0__5.0__ hour_to_min_conversion__ multiply__3.0__1.1__ round__3.3__ |
subtract__52.0__5.0__ hour_to_min_conversion__ multiply__3.0__1.1__ multiply__3.0__1.1__ |
| if the sides of a triangle are num__26 cm num__24 cm and num__10 cm what is its area ? <o> a ) num__120 cm num__2 <o> b ) num__176 cm num__2 <o> c ) num__267 cm num__2 <o> d ) num__268 cm num__2 <o> e ) num__976 cm num__2 |
the triangle with sides num__26 cm num__24 cm and num__10 cm is right angled where the hypotenuse is num__26 cm . area of the triangle = num__0.5 * num__24 * num__10 = num__120 cm num__2 answer : a <eor> a <eos> |
a |
triangle_area__24.0__10.0__ square_perimeter__0.5__ triangle_area__24.0__10.0__ |
volume_rectangular_prism__24.0__10.0__0.5__ square_perimeter__0.5__ volume_rectangular_prism__24.0__10.0__0.5__ |
| the speed of a boat in upstream is num__80 kmph and the speed of the boat downstream is num__120 kmph . find the speed of the boat in still water and the speed of the stream ? <o> a ) num__21 kmph . <o> b ) num__22 kmph . <o> c ) num__20 kmph . <o> d ) num__25 kmph . <o> e ) num__28 kmph . |
speed of the boat in still water = ( num__80 + num__120 ) / num__2 = num__100 kmph . speed of the stream = ( num__120 - num__80 ) / num__2 = num__20 kmph . answer : c <eor> c <eos> |
c |
subtract__120.0__100.0__ round__20.0__ |
subtract__120.0__100.0__ subtract__120.0__100.0__ |
| of num__61 players on a cricket team num__37 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed how many right - handed players are there total ? <o> a ) num__53 <o> b ) num__55 <o> c ) num__59 <o> d ) num__71 <o> e ) num__92 |
total = num__61 thrower = num__37 rest = num__61 - num__37 = num__24 left handed = num__8.0 = num__8 right handed = num__16 if all thrower are right handed then total right handed is num__37 + num__16 = num__53 so a . num__53 is the right answer <eor> a <eos> |
a |
subtract__61.0__37.0__ subtract__24.0__8.0__ subtract__61.0__8.0__ subtract__61.0__8.0__ |
subtract__61.0__37.0__ subtract__24.0__8.0__ subtract__61.0__8.0__ subtract__61.0__8.0__ |
| a company plans to gradually replace its fleet of num__20 cars with newer models . at the end of every year starting with num__2000 it retires num__5 of its old cars and buys num__6 new models . how many years did it take before the number of older cars left in the company ' s fleet was less than num__50 percent of the fleet ? <o> a ) num__6 <o> b ) num__3 <o> c ) num__2 <o> d ) num__4 <o> e ) num__5 |
total fleet - - num__20 end of year num__2000 - - ( retires num__5 and buys num__6 ) total fleet - - num__21 ( num__15 old num__6 new ) . end of year num__2001 - - ( retires num__5 and buys num__6 ) total fleet - - num__22 ( num__10 old num__12 new ) . answer is c - - after num__2 years the company has a total of num__22 cars ( num__10 old num__12 new ) the old cars are already less than num__50.0 of its fleet . <eor> c <eos> |
c |
percent__20.0__50.0__ percent__20.0__10.0__ percent__20.0__10.0__ |
percent__20.0__50.0__ percent__20.0__10.0__ percent__20.0__10.0__ |
| find the value of m num__32519 x num__9999 = m ? <o> a ) num__724533811 <o> b ) num__353654655 <o> c ) num__545463251 <o> d ) num__725117481 <o> e ) num__325157481 |
num__32519 x num__9999 = num__32519 x ( num__10000 - num__1 ) = num__32519 x num__10000 - num__32519 x num__1 = num__325190000 - num__32519 = num__325157481 e <eor> e <eos> |
e |
subtract__10000.0__9999.0__ multiply__32519.0__10000.0__ multiply__32519.0__9999.0__ multiply__32519.0__9999.0__ |
subtract__10000.0__9999.0__ multiply__32519.0__10000.0__ subtract__325190000.0__32519.0__ subtract__325190000.0__32519.0__ |
| an ant walks an average of num__600 meters in num__12 minutes . a beetle walks num__15.0 less distance at the same time on the average . assuming the beetle walks at her regular rate what is its speed in km / h ? <o> a ) num__2.215 . <o> b ) num__2.55 <o> c ) num__2.775 . <o> d ) num__3.2 . <o> e ) num__3.5 . |
the ant walks an average of num__600 meters in num__12 minutes num__600 meters in num__0.2 hours the beetle walks num__15.0 less distance = num__600 - num__90 = num__510 meters in num__12 minutes num__0.510 km in num__0.2 = num__0.2 hours speed = num__0.510 * num__5 = num__2.55 km / h correct answer b = num__2.55 <eor> b <eos> |
b |
subtract__600.0__90.0__ reverse__0.2__ multiply__5.0__0.51__ multiply__5.0__0.51__ |
subtract__600.0__90.0__ reverse__0.2__ multiply__5.0__0.51__ multiply__5.0__0.51__ |
| tough and tricky questions : statistics . set x consists of prime numbers { num__3 num__11 num__7 a num__17 num__19 } . if integer y represents the product of all elements in set x and if num__11 y is an even number what is the range of set x ? <o> a ) num__14 <o> b ) num__16 <o> c ) num__17 <o> d ) num__20 <o> e ) num__26 |
since num__11 y = even therefore y has to beevensince num__11 is a odd integer ( even * odd = even ) similarly y is the product of all integers in set x but all integers in set x are odd except the unknown a and since x contains only prime numbers a has to equal to num__2 . . . ( num__2 is the only even prime number and the product of all prime numbers in set x has to be even even * odd = even ) since you know value of a you can calculate the range = largest integer in the set minus smallest integer in the set = num__19 - num__2 = num__17 answer is c <eor> c <eos> |
c |
subtract__19.0__17.0__ subtract__19.0__2.0__ |
subtract__19.0__17.0__ subtract__19.0__2.0__ |
| a paper is in a square form whose one side is num__20 cm . two semi circles are drawn on its opposites as diameters . if these semi circles are cut down what is the area of the remaining paper ? <o> a ) ( num__400 - num__100 π ) cm num__2 <o> b ) ( num__400 - num__2 π ) cm num__2 <o> c ) ( num__400 – num__200 π ) cm num__2 <o> d ) num__200 π cm num__2 <o> e ) none of these |
explanation : num__400 - π * num__10 * num__10 = num__400 - num__100 π answer is a <eor> a <eos> |
a |
triangle_area__20.0__10.0__ square_perimeter__100.0__ |
triangle_area__20.0__10.0__ square_perimeter__100.0__ |
| a restaurant meal cost $ num__34.50 and there was no tax . if the tip was more than num__10 percent but less than num__15 percent of the cost of the meal then total amount paid must have been between : <o> a ) $ num__40 and $ num__42 <o> b ) $ num__39 and $ num__41 <o> c ) $ num__38 and num__40 <o> d ) $ num__37 and $ num__39 <o> e ) $ num__36 and $ num__37 |
let tip = t meal cost = num__34.50 range of tip = from num__10.0 of num__34.5 to num__15.0 of num__34.5 = num__3.55 to num__5.325 hence range of amount paid = num__34.5 + t = num__38.05 to num__39.825 answer : c <eor> c <eos> |
c |
add__34.5__3.55__ add__34.5__5.325__ round_down__38.05__ |
add__34.5__3.55__ add__34.5__5.325__ round_down__38.05__ |
| the ratio of two numbers is num__3 : num__4 and their sum is num__28 . the greater of the two numbers is ? <o> a ) num__12 <o> b ) num__16 <o> c ) num__18 <o> d ) num__19 <o> e ) num__20 |
num__3 : num__4 total parts = num__7 = num__7 parts - - > num__28 ( num__7 × num__4 = num__28 ) = num__1 part - - - - > num__4 ( num__1 × num__4 = num__4 ) = the greater of the two number is = num__4 = num__4 parts - - - - > num__16 ( num__4 × num__4 = num__16 ) b <eor> b <eos> |
b |
add__3.0__4.0__ subtract__4.0__3.0__ multiply__1.0__16.0__ |
add__3.0__4.0__ subtract__4.0__3.0__ multiply__1.0__16.0__ |
| the sum of three consecutive even numbers is num__42 . find the middle number of the three ? <o> a ) num__14 <o> b ) num__16 <o> c ) num__18 <o> d ) num__24 <o> e ) num__34 |
three consecutive even numbers ( num__2 p - num__2 ) num__2 p ( num__2 p + num__2 ) . ( num__2 p - num__2 ) + num__2 p + ( num__2 p + num__2 ) = num__42 num__6 p = num__42 = > p = num__7 . the middle number is : num__2 p = num__14 . answer : a <eor> a <eos> |
a |
divide__42.0__6.0__ multiply__2.0__7.0__ round__14.0__ |
divide__42.0__6.0__ multiply__2.0__7.0__ round__14.0__ |
| an empty pool being filled with water at a constant rate takes num__3 hours to fill to num__0.6 of its capacity . how much more time will it take to finish filling the pool ? <o> a ) num__5 hr num__30 min <o> b ) num__5 hr num__20 min <o> c ) num__4 hr num__48 min <o> d ) num__3 hr num__12 min <o> e ) num__2 hr num__40 min |
as pool is filled to num__0.6 of its capacity then num__0.4 of its capacity is left to fill . to fill num__0.6 of the pool took num__8 hours - - > to fill num__0.4 of the pool will take num__8 / ( num__0.6 ) * num__0.4 = num__5.33333333333 hours = num__5 hours num__20 minutes ( because if t is the time needed to fill the pool then t * num__0.6 = num__8 - - > t = num__8 * num__1.66666666667 hours - - > to fill num__0.4 of the pool num__8 * num__1.66666666667 * num__0.4 = num__5.33333333333 hours will be needed ) . or plug values : take the capacity of the pool to be num__5 liters - - > num__0.6 of the pool or num__3 liters is filled in num__8 hours which gives the rate of num__0.375 liters per hour - - > remaining num__2 liters will require : time = job / rate = num__2 / ( num__0.375 ) = num__5.33333333333 hours = num__2 hours num__40 minutes . answer : e . <eor> e <eos> |
e |
round_down__5.3333__ divide__8.0__0.4__ reverse__0.6__ divide__3.0__8.0__ multiply__0.375__5.3333__ multiply__2.0__20.0__ multiply__0.375__5.3333__ |
divide__3.0__0.6__ divide__8.0__0.4__ reverse__0.6__ divide__3.0__8.0__ multiply__0.375__5.3333__ multiply__2.0__20.0__ multiply__0.375__5.3333__ |
| sachin is younger than rahul by num__6 years . if the ratio of their ages is num__7 : num__9 find the age of sachin <o> a ) num__24.58 <o> b ) num__14 <o> c ) num__15 <o> d ) num__24.9 <o> e ) num__24.1 |
if rahul age is x then sachin age is x - num__6 so ( x - num__6 ) / x = num__0.777777777778 = > num__9 x - num__42 = num__7 x = > num__2 x = num__42 = > x = num__21 so sachin age is num__21 - num__6 = num__15 answer : c <eor> c <eos> |
c |
divide__7.0__9.0__ multiply__6.0__7.0__ subtract__9.0__7.0__ divide__42.0__2.0__ add__6.0__9.0__ add__6.0__9.0__ |
divide__7.0__9.0__ multiply__6.0__7.0__ subtract__9.0__7.0__ divide__42.0__2.0__ subtract__21.0__6.0__ subtract__21.0__6.0__ |
| which of the following could be the sum of the reciprocals of two prime numbers ? <o> a ) num__0.538461538462 <o> b ) num__0.916666666667 <o> c ) num__0.366666666667 <o> d ) num__0.46 <o> e ) num__0.424242424242 |
num__0.833333333333 it is num__0.424242424242 = ( num__0.333333333333 ) + ( num__0.0909090909091 ) answer - e <eor> e <eos> |
e |
subtract__0.4242__0.3333__ add__0.0909__0.3333__ |
subtract__0.4242__0.3333__ add__0.0909__0.3333__ |
| if num__3 x + y = num__40 num__2 x - y = num__20 for integers of x and y y ^ num__2 = ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__0 <o> d ) num__16 <o> e ) num__3 |
num__3 x + y = num__40 num__2 x - y = num__20 num__5 x = num__60 x = num__12 y = num__4 y ^ num__2 = num__16 answer is d <eor> d <eos> |
d |
add__3.0__2.0__ multiply__3.0__20.0__ divide__60.0__5.0__ divide__20.0__5.0__ subtract__20.0__4.0__ subtract__20.0__4.0__ |
add__3.0__2.0__ add__40.0__20.0__ divide__60.0__5.0__ divide__20.0__5.0__ subtract__20.0__4.0__ subtract__20.0__4.0__ |
| the average of nine numbers is num__15 . the average of first four numbers is num__11 and the average of last four numbers is num__14 . what is the middle number ? <o> a ) num__19 <o> b ) num__20 <o> c ) num__35 <o> d ) num__31 <o> e ) num__27 |
the total of nine numbers = num__9 x num__15 = num__135 the total of first num__4 and last num__4 numbers is = num__4 x num__11 + num__4 x num__14 = num__100 so the middle number is ( num__135 - num__100 ) = num__35 c <eor> c <eos> |
c |
multiply__15.0__9.0__ subtract__15.0__11.0__ subtract__135.0__100.0__ subtract__135.0__100.0__ |
multiply__15.0__9.0__ subtract__15.0__11.0__ subtract__135.0__100.0__ subtract__135.0__100.0__ |
| mother is aged three times more than her daughter rose . after num__8 years she would be two and a half times of rose ' s age . after further num__8 years how many times would he be of rose ' s age ? <o> a ) a ) num__15 <o> b ) b ) num__2 <o> c ) c ) num__20 <o> d ) d ) num__21 <o> e ) e ) num__26 |
let ronit ' s present age be x years . then father ' s present age = ( x + num__3 x ) years = num__4 x years . ( num__4 x + num__8 ) = num__5 ( x + num__8 ) num__2 num__8 x + num__16 = num__5 x + num__40 num__3 x = num__24 x = num__8 . hence required ratio = ( num__4 x + num__16 ) / ( x + num__16 ) = num__2.0 = num__2 . b <eor> b <eos> |
b |
subtract__8.0__3.0__ divide__8.0__4.0__ multiply__8.0__2.0__ multiply__8.0__5.0__ multiply__8.0__3.0__ divide__8.0__4.0__ |
subtract__8.0__3.0__ divide__8.0__4.0__ multiply__8.0__2.0__ multiply__8.0__5.0__ add__8.0__16.0__ divide__8.0__4.0__ |
| if f ( x ) = num__12 - x ^ num__1.0 and f ( num__2 k ) = num__4 k what is one possible value for k ? <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__6 <o> e ) num__8 |
first of all see thisgmat blog postand check the related lesson linked below for some background on function notation . we can plug anything in for x and get a result . you can find f ( num__1 ) for example by plugging in num__1 where x is and you would get num__12 - num__0.5 = num__11.5 . or we could find f ( num__2 ) which would be num__12 - num__2.0 = num__10 . so the notation f ( num__2 k ) means that we are going to plug a num__2 k in for x everywhere in the formula for f ( x ) . that would be : f ( num__2 k ) = num__12 - ( num__2 k ) ^ num__1.0 = num__12 - num__2 k ^ num__2 . remember that we have to square both the num__2 and the k to get num__4 k num__2 . now this expression the output we will set equal to num__2 k . num__12 - num__2 k ^ num__2 = num__2 k - - > k = - num__5 or k = num__4 . all the answers are positive so we choose k = num__4 . answer = c <eor> c <eos> |
c |
reverse__2.0__ subtract__12.0__0.5__ subtract__12.0__2.0__ add__1.0__4.0__ multiply__1.0__4.0__ |
reverse__2.0__ subtract__12.0__0.5__ subtract__12.0__2.0__ add__1.0__4.0__ subtract__5.0__1.0__ |
| two pipes a and b can fill a cistern in num__12 and num__15 minutes respectively . both are opened together but after num__2 minutes a is turned off . after how much more time will the cistern be filled ? <o> a ) num__8 num__0.142857142857 <o> b ) num__8 num__1.0 <o> c ) num__8 num__0.25 <o> d ) num__10 num__0.5 <o> e ) num__8 num__0.428571428571 |
num__0.166666666667 + ( num__2 + x ) / num__15 = num__1 x = num__10 num__0.5 answer : d <eor> d <eos> |
d |
divide__2.0__12.0__ subtract__12.0__2.0__ divide__1.0__2.0__ round__10.0__ |
divide__2.0__12.0__ subtract__12.0__2.0__ divide__1.0__2.0__ divide__10.0__1.0__ |
| if an object travels at eight feet per second how many feet does it travel in one hour ? <o> a ) num__34880 <o> b ) num__3778 <o> c ) num__12788 <o> d ) num__18000 <o> e ) num__28800 |
explanation : if an object travels at num__5 feet per second it covers num__5 x num__60 feet in one minute and num__5 x num__60 x num__60 feet in one hour . answer = num__28800 answer : e ) num__28800 <eor> e <eos> |
e |
hour_to_min_conversion__ round__28800.0__ |
hour_to_min_conversion__ round__28800.0__ |
| two pipes a and b can separately fill a cistern in num__10 and num__15 minutes respectively . a person opens both the pipes together when the cistern should have been was full he finds the waste pipe open . he then closes the waste pipe and in another num__4 minutes the cistern was full . in what time can the waste pipe empty the cistern when fill ? <o> a ) num__7 <o> b ) num__9 <o> c ) num__5 <o> d ) num__4 <o> e ) num__8 |
num__0.1 + num__0.0666666666667 = num__0.166666666667 * num__4 = num__0.666666666667 num__1 - num__0.666666666667 = num__0.333333333333 num__0.1 + num__0.0666666666667 - num__1 / x = num__0.333333333333 x = num__8 answer e <eor> e <eos> |
e |
add__0.1__0.0667__ divide__10.0__15.0__ multiply__10.0__0.1__ subtract__1.0__0.6667__ round__8.0__ |
add__0.1__0.0667__ divide__10.0__15.0__ multiply__10.0__0.1__ subtract__1.0__0.6667__ divide__8.0__1.0__ |
| two trains one from howrah to patna and the other from patna to howrah start simultaneously . after they meet the trains reach their destinations after num__9 hours and num__64 hours respectively . the ratio of their speeds is ? <o> a ) num__4 : num__5 <o> b ) num__8 : num__3 <o> c ) num__8 : num__4 <o> d ) num__4 : num__8 <o> e ) num__4 : num__1 |
let us name the trains a and b . then ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ num__64 : √ num__9 = num__8 : num__3 answer : b <eor> b <eos> |
b |
round__8.0__ |
round__8.0__ |
| find the area of a parallelogram with base num__15 cm and height num__40 cm ? <o> a ) num__200 cm num__2 <o> b ) num__100 cm num__2 <o> c ) num__42 cm num__2 <o> d ) num__600 cm num__2 <o> e ) num__230 cm num__2 |
area of a parallelogram = base * height = num__15 * num__40 = num__600 cm num__2 answer : d <eor> d <eos> |
d |
multiply__15.0__40.0__ multiply__15.0__40.0__ |
multiply__15.0__40.0__ multiply__15.0__40.0__ |
| a school ’ s annual budget for the purchase of student computers increased by num__50.0 this year over last year . if the price of student computers increased by num__20.0 this year then the number of computers it can purchase this year is what percent greater than the number of computers it purchased last year ? <o> a ) num__33.33 <o> b ) num__40.0 <o> c ) num__25.0 <o> d ) num__28.0 <o> e ) num__60 % |
lats year : budget = $ num__100 price = $ num__1 - - > the number of computers purchased = num__100.0 = num__100 ; this year : budget = $ num__150 price = $ num__1.2 - - > the number of computers purchased = num__150 / num__1.2 = num__125 . increase in the number of computers purchased = num__25.0 . answer : c . <eor> c <eos> |
c |
percent__20.0__125.0__ percent__20.0__125.0__ |
percent__20.0__125.0__ percent__20.0__125.0__ |
| a sells an article to b at a profit of num__10.0 b sells the article back to a at a loss of num__10.0 . in this transaction : <o> a ) a neither losses nor gains <o> b ) a makes a profit of num__11.0 <o> c ) a makes a profit of num__20.0 <o> d ) b loses num__20.0 <o> e ) none of these |
solution : first method let cp was num__100 for a originally . a sells article to b at num__10.0 profit cp for b = num__100 + num__10.0 of num__100 = num__110 . now b sells it a again with loss num__10.0 . now cp for a this time = num__110 - num__10.0 of num__110 = num__99 . a makes profit = num__110 - num__99 = num__11 . % profit for a = ( num__11 * num__100 ) / num__100 = num__10.0 . second method it could be easily shown by net percentage change graphic . num__100 ( a ) = = num__10.0 ( profit ) = = > num__110 ( b ) = = num__10.0 ( loss ) = = > num__99 ( a ) in this transaction a makes a profit of ( num__110 - num__99 = num__11.0 ) num__11.0 . [ num__10.0 on selling to b and num__1.0 profit on buying back from b ] . answer : option b <eor> b <eos> |
b |
percent__10.0__110.0__ percent__10.0__110.0__ |
percent__10.0__110.0__ percent__10.0__110.0__ |
| if the sides of a triangle are num__65 cm num__60 cm and num__25 cm what is its area ? <o> a ) num__350 cm ^ num__2 <o> b ) num__850 cm ^ num__2 <o> c ) num__550 cm ^ num__2 <o> d ) num__750 cm ^ num__2 <o> e ) num__650 cm ^ num__2 |
the triangle with sides num__65 cm num__60 cm and num__25 cm is right angled where the hypotenuse is num__65 cm . area of the triangle = num__0.5 * num__60 * num__25 = num__750 cm ^ num__2 answer : d <eor> d <eos> |
d |
triangle_area__60.0__25.0__ square_perimeter__0.5__ triangle_area__60.0__25.0__ |
volume_rectangular_prism__60.0__25.0__0.5__ square_perimeter__0.5__ volume_rectangular_prism__60.0__25.0__0.5__ |
| if tier is written as num__8163 and brain is written as num__23415 how is rent coded ? <o> a ) num__3653 <o> b ) num__3658 <o> c ) num__2977 <o> d ) num__2790 <o> e ) num__2711 |
explanation : given : letter : t i e r b a n code : num__8 num__1 num__6 num__3 num__2 num__4 num__5 thus the code for rent is num__3658 . answer : b <eor> b <eos> |
b |
subtract__3.0__1.0__ add__1.0__3.0__ add__1.0__4.0__ multiply__1.0__3658.0__ |
subtract__3.0__1.0__ add__1.0__3.0__ add__1.0__4.0__ multiply__1.0__3658.0__ |
| how many prime numbers f exist such that num__90 < f < num__106 and f is a factor of num__99999919 ? <o> a ) num__1 . zero <o> b ) num__2 . one <o> c ) num__3 . two <o> d ) num__4 . three <o> e ) num__5 . more than three |
i found that considering the constraint . . f is a prime num in between num__90 to num__106 not inclusive only num__3 numbers ( num__9197 and num__101 ) are present but those numbers are not the factors of the number mentioned . . . . therefore i doubt the answer . c <eor> c <eos> |
c |
gcd__90.0__3.0__ |
gcd__90.0__3.0__ |
| if x and y are positive integers and x ^ num__3 * y ^ num__4 = num__648 which of the following is the value of xy ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__6 <o> d ) num__8 <o> e ) num__10 |
num__648 = num__2 ^ num__3 * num__3 ^ num__4 xy = num__2 * num__3 = num__6 the answer is c . <eor> c <eos> |
c |
multiply__3.0__2.0__ multiply__3.0__2.0__ |
multiply__3.0__2.0__ multiply__3.0__2.0__ |
| a certain bus driver is paid a regular rate of $ num__16 per hour for any number of hours that does not exceed num__40 hours per week . for any overtime hours worked in excess of num__40 hours per week the bus driver is paid a rate that is num__75.0 higher than his regular rate . if last week the bus driver earned $ num__1004 in total compensation how many total hours did he work that week ? <o> a ) num__36 <o> b ) num__40 <o> c ) num__44 <o> d ) num__48 <o> e ) num__53 |
for num__40 hrs = num__40 * num__16 = num__640 excess = num__1004 - num__640 = num__364 for extra hours = . num__75 ( num__16 ) = num__12 + num__16 = num__28 number of extra hrs = num__13.0 = num__13 total hrs = num__40 + num__13 = num__53 answer e num__53 <eor> e <eos> |
e |
multiply__16.0__40.0__ subtract__1004.0__640.0__ add__16.0__12.0__ divide__364.0__28.0__ add__40.0__13.0__ add__40.0__13.0__ |
multiply__16.0__40.0__ subtract__1004.0__640.0__ add__16.0__12.0__ divide__364.0__28.0__ add__40.0__13.0__ add__40.0__13.0__ |
| in a certain furniture store each week nancy earns a salary of $ num__240 plus num__5.0 of the amount of her total sales that exceeds $ num__800 for the week . if nancy earned a total of $ num__450 one week what were her total sales that week ? <o> a ) $ num__2200 <o> b ) $ num__3450 <o> c ) $ num__4200 <o> d ) $ num__4250 <o> e ) $ num__5 |
000 |
num__450 = num__240 + ( num__0.05 ) * ( total sales - num__800 ) i . e . ( total sales - num__800 ) = num__20 * ( num__450 - num__240 ) = num__20 * num__210 = num__4200 i . e . total sales = num__4200 + num__800 = num__5000 answer : option e <eor> e <eos> |
e |
e |
| if b + a < b - a < a - b which of the following is correct ? <o> a ) a < b < num__0 <o> b ) b < a < num__0 <o> c ) b < num__0 < a <o> d ) num__0 < b < a . <o> e ) b > a > num__0 . |
check first inequality b + a < b - a a < - a a is - ve check another equality b - a < a - b - num__2 a < - num__2 b a > b combining both num__0 > a > b option b fits only ans b <eor> b <eos> |
b |
multiply__0.0__2.0__ |
multiply__0.0__2.0__ |
| num__80 num__87372 num__7264 num__8156 ? <o> a ) num__96 <o> b ) num__98 <o> c ) num__89 <o> d ) num__90 <o> e ) num__88 |
odd place numbers num__80 num__72 num__64 num__56 are divisible by num__8 . even place numbers num__873 num__72 num__81 are divisible by num__9 so required number is in even place from options num__90 is only value visible by num__9 so num__90 is the answer . answer : d <eor> d <eos> |
d |
subtract__80.0__72.0__ subtract__81.0__72.0__ add__81.0__9.0__ add__81.0__9.0__ |
subtract__80.0__72.0__ subtract__81.0__72.0__ add__81.0__9.0__ add__81.0__9.0__ |
| a table is bought for rs . num__600 / - and sold at a loss of num__12.0 find its selling price <o> a ) s . num__500 / - <o> b ) s . num__530 / - <o> c ) s . num__528 / - <o> d ) s . num__600 / - <o> e ) s . num__700 / - |
num__100.0 - - - - - - > num__600 ( num__100 * num__6 = num__600 ) num__88.0 - - - - - - > num__528 ( num__88 * num__6 = num__528 ) selling price = rs . num__528 / - c <eor> c <eos> |
c |
percent__88.0__600.0__ percent__88.0__600.0__ |
percent__88.0__600.0__ percent__88.0__600.0__ |
| train a & b leave from same station and reaches same destination but train a leaves half an after train b & travels num__2 times faster than b . total distance is num__100 miles & train b travels num__50 miles / hour . which train reaches the destination first & what is the time difference between the train <o> a ) train a reaches num__30 minutes before train b <o> b ) train b reaches num__30 minutes before train a <o> c ) train a reaches num__60 minutes before train b <o> d ) train b reaches num__60 minutes before train a <o> e ) none of the above |
time taken for train a = num__30 minutes + d num__1 / s num__1 = num__30 minutes + num__1.0 = num__1 hours num__30 minutes since train b travels num__50 miles / hour ; train a travels twice the speed of b which gives s num__1 = num__100 miles / hour time taken for train b = d num__2 / s num__2 = num__2.0 = num__2 hours so train a reaches num__30 minutes before train b answer : a <eor> a <eos> |
a |
multiply__1.0__30.0__ |
divide__30.0__1.0__ |
| if num__222 = num__6 num__333 = num__12 num__444 = num__20 then num__555 = ? ? <o> a ) num__30 <o> b ) num__35 <o> c ) num__40 <o> d ) num__45 <o> e ) num__50 |
num__222 = num__2 * num__2 + num__2 = num__6 num__333 = num__3 * num__3 + num__3 = num__12 num__444 = num__4 * num__4 + num__4 = num__20 num__555 = num__5 * num__5 + num__5 = num__30 answer : a <eor> a <eos> |
a |
divide__12.0__6.0__ divide__6.0__2.0__ subtract__6.0__2.0__ divide__20.0__4.0__ multiply__6.0__5.0__ multiply__6.0__5.0__ |
divide__12.0__6.0__ divide__6.0__2.0__ subtract__6.0__2.0__ add__2.0__3.0__ multiply__6.0__5.0__ multiply__6.0__5.0__ |
| a room is num__15 m long num__4 m broad and num__3 m height . find the cost of white washing its four walls at num__50 p per m num__2 ? <o> a ) num__28 <o> b ) num__57 <o> c ) num__58 <o> d ) num__83 <o> e ) num__73 |
explanation : num__2 * num__3 ( num__15 + num__4 ) = num__114 num__114 * num__0.5 = rs . num__57 answer : option b <eor> b <eos> |
b |
reverse__2.0__ multiply__0.5__114.0__ multiply__0.5__114.0__ |
reverse__2.0__ multiply__0.5__114.0__ multiply__0.5__114.0__ |
| if a num__2 + b num__2 + c num__2 = num__281 and ab + bc + ca = num__4 then a + b + c is <o> a ) num__16 <o> b ) num__18 <o> c ) num__22 <o> d ) num__17 <o> e ) num__20 |
by formula ( a + b + c ) ^ num__2 = a ^ num__2 + b ^ num__2 + c ^ num__2 + num__2 ( ab + bc + ca ) since a ^ num__2 + b ^ num__2 + c ^ num__2 = num__281 and ab + bc + ca = num__4 ( a + b + c ) ^ num__2 = num__281 + num__2 ( num__4 ) = num__289 = num__17 ^ num__2 therefore : a + b + c = num__17 answer : d <eor> d <eos> |
d |
divide__289.0__17.0__ |
divide__289.0__17.0__ |
| of all the students in a certain dormitory num__0.5 are first - year students and the rest are second - year students . if num__0.8 of the first - year students have not declared a major and if the fraction of second - year students who have declared a major is num__2 times the fraction of first - year students who have declared a major what fraction of all the students in the dormitory are second - year students who have not declared a major ? <o> a ) num__0.0666666666667 <o> b ) num__0.2 <o> c ) num__0.266666666667 <o> d ) num__0.3 <o> e ) num__0.4 |
tot students = x num__1 st year student = x / num__2 - - - - > non majaor = num__0.8 ( x / num__2 ) - - - - - > maj = num__0.2 ( x / num__2 ) num__2 nd year student = x / num__2 - - - - > maj = num__2 ( num__0.2 ( x / num__2 ) ) = num__0.2 ( x ) - - - > non major = x / num__2 - num__0.2 ( x ) = num__0.3 ( x ) hence num__0.3 d <eor> d <eos> |
d |
multiply__0.5__2.0__ subtract__1.0__0.8__ subtract__0.5__0.2__ subtract__0.5__0.2__ |
multiply__0.5__2.0__ subtract__1.0__0.8__ subtract__0.5__0.2__ subtract__0.5__0.2__ |
| how many integers between num__400 and num__1000 are there such that their unit digit is odd ? <o> a ) num__250 <o> b ) num__150 <o> c ) num__300 <o> d ) num__400 <o> e ) num__450 |
there are num__600 numbers from num__401 to num__1000 ( inclusive ) . half of the numbers are odd so there are num__300 odd numbers . the answer is c . <eor> c <eos> |
c |
subtract__1000.0__400.0__ subtract__600.0__300.0__ |
subtract__1000.0__400.0__ subtract__600.0__300.0__ |
| divide $ num__450 among a b in the ratio num__1 : num__4 . how many $ that a get ? <o> a ) $ num__90 <o> b ) $ num__500 <o> c ) $ num__150 <o> d ) $ num__250 <o> e ) $ num__600 |
sum of ratio terms = num__1 + num__4 = num__5 a = num__450 * num__0.2 = $ num__90 answer is a <eor> a <eos> |
a |
add__1.0__4.0__ reverse__5.0__ divide__450.0__5.0__ divide__450.0__5.0__ |
add__1.0__4.0__ reverse__5.0__ multiply__450.0__0.2__ multiply__450.0__0.2__ |
| if num__16666 – n is divisible by num__11 and num__0 < n < num__11 what is n ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__9 |
num__16 |
666 / num__11 = num__1515 with a remainder of num__1 . we need to subtract the remainder to get a multiple of num__11 . the answer is a . <eor> a <eos> |
a |
a |
| two trains num__150 m and num__160 m long run at the speed of num__60 km / hr and num__40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? <o> a ) num__10.7 <o> b ) num__10.9 <o> c ) num__10.26 <o> d ) num__11.16 <o> e ) num__18.8 |
relative speed = num__60 + num__40 = num__100 km / hr . = num__100 * num__0.277777777778 = num__27.7777777778 m / sec . distance covered in crossing each other = num__150 + num__160 = num__310 m . required time = num__310 * num__0.036 = num__11.16 sec . answer : d <eor> d <eos> |
d |
subtract__160.0__60.0__ add__150.0__160.0__ multiply__310.0__0.036__ round__11.16__ |
add__60.0__40.0__ add__150.0__160.0__ multiply__310.0__0.036__ multiply__310.0__0.036__ |
| in what time will a train num__120 m long cross an electric pole it its speed be num__185 km / hr ? <o> a ) num__2.3 sec <o> b ) num__2.9 sec <o> c ) num__2.7 sec <o> d ) num__8.7 sec <o> e ) num__8.5 sec |
speed = num__185 * num__0.277777777778 = num__51 m / sec time taken = num__2.35294117647 = num__2.3 sec . answer : a <eor> a <eos> |
a |
divide__120.0__51.0__ round__2.3__ |
divide__120.0__51.0__ round__2.3__ |
| how many positive three - digit integers are divisible by both num__2 and num__7 ? <o> a ) num__40 <o> b ) num__46 <o> c ) num__52 <o> d ) num__64 <o> e ) num__70 |
a number to be divisible by both num__2 and num__7 should be divisible by the least common multiple of num__2 and num__7 so by num__14 . multiples of num__14 between num__100 and num__999 inclusive is ( last - first ) / multiple + num__1 = ( num__994 - num__112 ) / num__14 + num__1 = num__63 + num__1 = num__64 answer : d . <eor> d <eos> |
d |
multiply__2.0__7.0__ add__1.0__63.0__ multiply__64.0__1.0__ |
multiply__2.0__7.0__ add__1.0__63.0__ divide__64.0__1.0__ |
| a bank offers an interest of num__5.0 per annum compounded annually on all its deposits . if $ num__10000 is deposited what will be the ratio of the interest earned in the num__4 th year to the interest earned in the num__5 th year ? <o> a ) num__1 : num__5 <o> b ) num__625 : num__3125 <o> c ) num__100 : num__105 <o> d ) num__100 ^ num__4 : num__105 ^ num__4 <o> e ) num__725 : num__3225 |
the interest earned in the num__1 st year = $ num__500 the interest earned in the num__2 nd year = $ num__500 * num__1.05 the interest earned in the num__3 rd year = $ num__500 * num__1.05 ^ num__2 the interest earned in the num__4 th year = $ num__500 * num__1.05 ^ num__3 the interest earned in the num__5 th year = $ num__500 * num__1.05 ^ num__4 ( num__500 * num__1.05 ^ num__3 ) / ( num__500 * num__1.05 ^ num__4 ) = num__1 / num__1.05 = num__0.952380952381 . answer : c . <eor> c <eos> |
c |
percent__5.0__10000.0__ percent__1.0__10000.0__ |
percent__5.0__10000.0__ percent__1.0__10000.0__ |
| the area of a square garden is q square feet and the perimeter is p feet . if q = p + num__21 what is the perimeter of the garden in feet ? <o> a ) num__24 <o> b ) num__28 <o> c ) num__32 <o> d ) num__36 <o> e ) num__40 |
let x be the length of one side of the square garden . x ^ num__2 = num__4 x + num__21 x ^ num__2 - num__4 x - num__21 = num__0 ( x - num__7 ) ( x + num__3 ) = num__0 x = num__7 - num__3 p = num__4 ( num__7 ) = num__28 the answer is b . <eor> b <eos> |
b |
square_perimeter__7.0__ square_perimeter__7.0__ |
square_perimeter__7.0__ square_perimeter__7.0__ |
| in a forest num__10 rabbits were caught tagged with electronic markers then released . a month later num__10 rabbits were captured in the same forest . of these num__10 rabbits it was found that num__2 had been tagged with the electronic markers . if the percentage of tagged rabbits in the second sample approximates the percentage of tagged rabbits in the forest and if no rabbits had either left or entered the forest over the preceding month what is the approximate number of rabbits in the forest ? <o> a ) num__150 <o> b ) num__70 <o> c ) num__50 <o> d ) num__1500 <o> e ) num__500 |
given that num__10 rabbits were caught and tagged in the first sample and num__10 were caught in the second sample of which num__2 had tags then num__0.2 approximates the fraction of the population of tagged rabbits in the forest . proportion : a / x = c / d a = rabbits tagged first sample c / d = the fraction of the population of tagged rabbits in the forest x = the population of rabbits in the forest proportion : num__10 / x = num__0.2 x = ( num__1 ) ( num__10 ) ( num__5 ) = num__50 . there are approximately num__50 rabbits in the forest . answer is c <eor> c <eos> |
c |
divide__2.0__10.0__ reverse__0.2__ multiply__10.0__5.0__ multiply__10.0__5.0__ |
divide__2.0__10.0__ reverse__0.2__ divide__10.0__0.2__ divide__10.0__0.2__ |
| sum of num__3 consecutive even no . ' s is num__34 more than the num__1 st no . of the series . find the middle no . ? <o> a ) num__8 <o> b ) num__12 <o> c ) num__14 <o> d ) num__16 <o> e ) num__18 |
let the numbers be x x + num__2 and x + num__4 then x + x + num__2 + x + num__4 = x + num__34 num__3 x + num__6 = x + num__34 num__2 x = num__28 \ inline \ therefore x = num__14 \ inline \ therefore middle number = x + num__2 = num__14 + num__2 = num__16 d <eor> d <eos> |
d |
subtract__3.0__1.0__ add__3.0__1.0__ multiply__3.0__2.0__ subtract__34.0__6.0__ divide__28.0__2.0__ add__2.0__14.0__ multiply__1.0__16.0__ |
subtract__3.0__1.0__ add__3.0__1.0__ add__2.0__4.0__ subtract__34.0__6.0__ divide__28.0__2.0__ add__2.0__14.0__ add__2.0__14.0__ |
| how many integers between num__265205 and num__758805 have tens digit num__1 and units digit num__3 ? <o> a ) num__4629 <o> b ) num__4639 <o> c ) num__3946 <o> d ) num__4926 <o> e ) num__4 |
936 |
there is one number in hundred with num__1 in the tens digit and num__3 in the units digit : num__13 num__113 num__213 num__313 . . . the difference between num__265205 and num__758805 is num__758805 - num__265205 = num__493600 - one number per each hundred gives num__1339.0 = num__4936 numbers . answer : e . <eor> e <eos> |
e |
e |
| at a certain supplier a machine of type a costs $ num__20000 and a machine of type b costs $ num__55000 . each machine can be purchased by making a num__20 percent down payment and repaying the remainder of the cost and the finance charges over a period of time . if the finance charges are equal to num__40 percent of the remainder of the cost how much less would num__2 machines of type a cost than num__1 machine of type b under this arrangement ? <o> a ) $ num__10000 <o> b ) $ num__11200 <o> c ) $ num__12000 <o> d ) $ num__12800 <o> e ) $ num__19 |
800 |
total cost of num__2 machines of type a = num__20.0 of ( cost of num__2 machine a ) + remainder + num__40.0 remainder = num__20.0 of num__40000 + ( num__40000 - num__20.0 of num__40000 ) + num__40.0 of ( num__40000 - num__20.0 of num__40000 ) = num__52800 total cost of num__1 machine of type b = num__20.0 of ( cost of num__1 machine b ) + remainder + num__40.0 remainder = num__20.0 of num__55000 + ( num__55000 - num__20.0 of num__55000 ) + num__40.0 of ( num__50000 - num__20.0 of num__55000 ) = num__72600 diff = num__72600 - num__52800 = num__19800 hence e . <eor> e <eos> |
e |
e |
| if an amount of rs num__13640 is divided equally among num__62 students approximately how much amount will each student get ? <o> a ) rs num__206 <o> b ) rs num__210 <o> c ) rs num__220 <o> d ) rs num__218 <o> e ) none |
amount received by each student = num__220.0 â ‰ ˆ rs num__220 answer c <eor> c <eos> |
c |
divide__13640.0__62.0__ divide__13640.0__62.0__ |
divide__13640.0__62.0__ divide__13640.0__62.0__ |
| a pupil ' s marks were wrongly entered as num__83 instead of num__63 . due to that the average marks for the class got increased by half ( num__0.5 ) . the number of pupils in the class is : <o> a ) num__38 <o> b ) num__39 <o> c ) num__40 <o> d ) num__46 <o> e ) num__49 |
let there be x pupils in the class . total increase in marks = ( x x num__0.5 ) = x / num__2 x / num__2 = ( num__83 - num__63 ) x / num__2 = num__20 x = num__40 answer is c <eor> c <eos> |
c |
reverse__0.5__ subtract__83.0__63.0__ multiply__2.0__20.0__ multiply__2.0__20.0__ |
reverse__0.5__ subtract__83.0__63.0__ divide__20.0__0.5__ divide__20.0__0.5__ |
| there are two circles of different radii . the are of a square is num__784 sq cm and its side is twice the radius of the larger circle . the radius of the larger circle is seven - third that of the smaller circle . find the circumference of the smaller circle ? <o> a ) num__16 <o> b ) num__15 <o> c ) num__12 <o> d ) num__87 <o> e ) num__16 |
let the radii of the larger and the smaller circles be l cm and s cm respectively . let the side of the square be a cm . a num__2 = num__784 = ( num__4 ) ( num__196 ) = ( num__22 ) . ( num__142 ) a = ( num__2 ) ( num__14 ) = num__28 a = num__2 l l = a / num__2 = num__14 l = ( num__2.33333333333 ) s therefore s = ( num__0.428571428571 ) ( l ) = num__6 circumference of the smaller circle = num__2 ∏ s = num__12 ∏ cm . answer : c <eor> c <eos> |
c |
multiply__2.0__14.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
multiply__2.0__14.0__ multiply__2.0__6.0__ multiply__2.0__6.0__ |
| a train running at the speed of num__90 km / hr crosses a pole in num__6 seconds . what is the length of the train ? <o> a ) num__150 <o> b ) num__2766 <o> c ) num__187 <o> d ) num__267 <o> e ) num__191 |
speed = ( num__60 * num__0.277777777778 ) m / sec = ( num__16.6666666667 ) m / sec length of the train = ( speed x time ) = ( num__16.6666666667 * num__9 ) m = num__150 m . answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ add__90.0__60.0__ round__150.0__ |
hour_to_min_conversion__ add__90.0__60.0__ round__150.0__ |
| in a graduate physics course num__60 percent of the students are male and num__30 percent of the students are married . if two - sevenths of the male students are married what fraction of the female students is single ? <o> a ) num__0.285714285714 <o> b ) num__0.566666666667 <o> c ) num__0.5 <o> d ) num__0.675 <o> e ) num__0.714285714286 |
let assume there are num__100 students of which num__60 are male and num__40 are females if num__30 are married then num__70 will be single . now its given that two - sevenths of the male students are married that means num__0.285714285714 of num__60 = num__17 males are married if num__30 is the total number of students who are married and out of that num__17 are males then the remaining num__13 will be females who are married . total females = num__40 married females = num__13 then single females = num__40 - num__13 = num__27 we need to find the fraction of female students who are single i . e single female students / total female student = num__0.675 [ d ] <eor> d <eos> |
d |
percent__100.0__0.675__ |
percent__100.0__0.675__ |
| compound x contains elements a and b at an approximate ratio by weight of num__2 : num__10 . approximately how many grams of element b are there in num__300 grams of compound x ? <o> a ) num__54 <o> b ) num__162 <o> c ) num__250 <o> d ) num__270 <o> e ) num__322 |
total number of fractions = num__2 + num__10 = num__12 element b constitutes = num__10 out of num__12 parts of x so in num__300 gms of x have num__300 * num__0.833333333333 = num__250 gms of b and num__300 - num__250 = num__50 gms of a . cross check : - a / b = num__0.2 = num__0.2 ( as given ) ans c <eor> c <eos> |
c |
add__2.0__10.0__ divide__10.0__12.0__ subtract__300.0__250.0__ divide__2.0__10.0__ subtract__300.0__50.0__ |
add__2.0__10.0__ divide__10.0__12.0__ subtract__300.0__250.0__ divide__2.0__10.0__ subtract__300.0__50.0__ |
| the sum of the present ages of two persons a and b is num__60 . if the age of a is twice that of b find the sum of their ages num__7 years hence ? <o> a ) num__58 <o> b ) num__74 <o> c ) num__62 <o> d ) num__70 <o> e ) num__74 |
explanation : a + b = num__60 a = num__2 b num__2 b + b = num__60 = > b = num__20 then a = num__40 . num__7 years their ages will be num__47 and num__27 . sum of their ages = num__47 + num__27 = num__74 . b ) <eor> b <eos> |
b |
subtract__60.0__20.0__ add__7.0__40.0__ add__7.0__20.0__ add__47.0__27.0__ add__47.0__27.0__ |
subtract__60.0__20.0__ add__7.0__40.0__ add__7.0__20.0__ add__47.0__27.0__ add__47.0__27.0__ |
| a trader bought a car at num__20.0 discount on its original price . he sold it at a num__40.0 increase on the price he bought it . what percent of profit did he make on the original price ? <o> a ) num__10.0 <o> b ) num__11.0 <o> c ) num__12.0 <o> d ) num__15.0 <o> e ) num__13 % |
original price = num__100 cp = num__80 s = num__80 * ( num__1.4 ) = num__112 num__100 - num__112 = num__12.0 answer : c <eor> c <eos> |
c |
percent__100.0__12.0__ |
percent__100.0__12.0__ |
| two hoses are pouring water into an empty pool . hose num__1 alone would fill up the pool in num__6 hours . hose num__2 alone would fill up the pool in num__3 hours . how long would it take for both hoses to fill up two - thirds of the pool ? <o> a ) num__0.416666666667 hours <o> b ) num__0.625 hours <o> c ) num__1.33333333333 hours <o> d ) num__1.71428571429 hours <o> e ) num__2.4 hours |
hose num__1 would fill the pool alone in num__6 hours i . e . num__1 hour work of hose num__1 = num__0.166666666667 hose num__2 would fill the pool alone in num__3 hours i . e . num__1 hour work of hose num__2 = num__0.333333333333 i . e . num__1 hour work of hose num__1 and hose num__2 together = ( num__0.166666666667 ) + ( num__0.333333333333 ) = ( num__0.5 ) i . e . ( num__0.5 ) work can be finished by hose num__1 and hose num__2 together in num__1 hour i . e . ( num__0.666666666667 ) work can be finished by hose num__1 and hose num__2 together in num__1 * ( num__2 ) * ( num__0.666666666667 ) = num__1.33333333333 hour answer : option c <eor> c <eos> |
c |
negate_prob__0.3333__ union_prob__1.0__0.5__0.1667__ union_prob__1.0__0.5__0.1667__ |
negate_prob__0.3333__ union_prob__1.0__0.5__0.1667__ union_prob__1.0__0.5__0.1667__ |
| two trains num__111 meters and num__165 meters in length respectively are running in opposite directions one at the rate of num__80 km and the other at the rate of num__65 kmph . in what time will they be completely clear of each other from the moment they meet ? <o> a ) num__6.85 <o> b ) num__7.18 <o> c ) num__7.16 <o> d ) num__7.15 <o> e ) num__7.12 |
t = ( num__111 + num__165 ) / ( num__80 + num__65 ) * num__3.6 t = num__6.85 answer : a <eor> a <eos> |
a |
round__6.85__ |
round__6.85__ |
| the market value of a num__10.5 stock in which an income of rs . num__756 is derived by investing rs . num__6000 brokerage being num__0.25 % is : <o> a ) num__83.08 <o> b ) num__114 <o> c ) num__114.75 <o> d ) num__124 <o> e ) num__124.75 |
face value = rs . num__6000 . dividend = num__10.5 . annual income = rs . num__756 . brokerage per rs . num__100 = rs . num__0.25 . dividend is always paid on the face value of a share . face value * dividend / ( market value + brokerage per rs . num__100 ) = annual income . = num__6000 * num__10.5 / num__756 = market value of rs . num__100 stock + brokerage per rs . num__100 . = market value of rs . num__100 stock + brokerage per rs . num__100 = rs . num__83.33 . = market value of rs . num__100 stock = rs . num__83.33 - re . num__0.25 . = market value of rs . num__100 stock = rs . num__83.08 answer : a <eor> a <eos> |
a |
percent__100.0__83.08__ |
percent__100.0__83.08__ |
| a train of num__24 carriages each of num__60 meters length when an engine also of num__60 meters length is running at a speed of num__60 kmph . in what time will the train cross a bridge num__1.5 km long ? <o> a ) num__1 Â ½ mins <o> b ) num__3 mins <o> c ) num__3 Â ½ mins <o> d ) num__7 mins <o> e ) num__5 mins |
explanation : d = num__25 * num__60 + num__1500 = num__3000 m t = num__50.0 * num__3.6 = num__180 sec = num__3 mins answer is b <eor> b <eos> |
b |
multiply__60.0__25.0__ divide__3000.0__60.0__ multiply__50.0__3.6__ divide__180.0__60.0__ round__3.0__ |
multiply__60.0__25.0__ divide__3000.0__60.0__ multiply__50.0__3.6__ divide__180.0__60.0__ round__3.0__ |
| which of the following equation is not equivalent to num__25 x ^ num__2 = y ^ num__2 - num__49 ? which of the following equation is not equivalent to num__25 x ^ num__2 = y ^ num__2 - num__4 ? <o> a ) num__25 x ^ num__2 + num__4 = y ^ num__2 <o> b ) num__75 x ^ num__2 = num__3 y ^ num__2 - num__12 <o> c ) num__25 x ^ num__2 = ( y + num__2 ) ( y - num__2 ) <o> d ) num__5 x = y - num__7 <o> e ) x ^ num__2 = ( y ^ num__2 - num__4 ) / num__25 |
num__1 ) take the original equation and try to manipulate it so that it becomes num__4 of the num__5 answers num__2 ) take each answer choice and see if you can manipulate it to make it into the original equation . with answer d we have . . . num__5 x = y - num__7 if we square both sides we end up with . . . . ( num__5 x ) ^ num__2 = ( y - num__7 ) ^ num__2 ( num__5 x ) ( num__5 x ) = ( y - num__7 ) ( y - num__7 ) num__25 x ^ num__2 = y ^ num__2 - num__14 y + num__49 this is not a match for the original equation so it ' s the answer that is not equivalent . d <eor> d <eos> |
d |
add__4.0__1.0__ add__2.0__5.0__ multiply__2.0__7.0__ divide__25.0__5.0__ |
add__4.0__1.0__ add__2.0__5.0__ multiply__2.0__7.0__ add__4.0__1.0__ |
| a certain bus driver is paid a regular rate of $ num__14 per hour for any number of hours that does not exceed num__40 hours per week . for any overtime hours worked in excess of num__40 hours per week the bus driver is paid a rate that is num__75.0 higher than his regular rate . if last week the bus driver earned $ num__976 in total compensation how many total hours did he work that week ? <o> a ) num__54 <o> b ) num__51 <o> c ) num__59 <o> d ) num__55 <o> e ) num__52 |
for num__40 hrs = num__40 * num__14 = num__560 excess = num__976 - num__560 = num__416 for extra hours = . num__75 ( num__14 ) = num__10.5 + num__14 = num__24.5 number of extra hrs = num__416 / num__24.5 = num__18.82 = num__19 approx . total hrs = num__40 + num__19 = num__59 answer c num__59 <eor> c <eos> |
c |
multiply__14.0__40.0__ subtract__976.0__560.0__ add__14.0__10.5__ add__40.0__19.0__ add__40.0__19.0__ |
multiply__14.0__40.0__ subtract__976.0__560.0__ add__14.0__10.5__ add__40.0__19.0__ add__40.0__19.0__ |
| num__24 oz of juice p and num__25 oz of juice f are mixed to make smothies x and y . the ratio of p to f in smothie x is num__4 is to num__1 and that in y is num__1 is to num__5 . how many ounces of juice p are contained in the smothie x ? <o> a ) num__5 <o> b ) num__10 <o> c ) num__15 <o> d ) num__20 <o> e ) num__25 |
easy way to solve this question is start from the answer and then conform the information provided in the question . we can start from option d i . e num__20 . . . as a quantity of juice p in x because it is the only one option that gets divided by num__4 is num__20 . . . since in the x the juice p to f ratio is num__4 : num__1 this gives us that quantity of juice p in x = num__20 therefore quantity of juice f will be num__5 . . . hence ratio = num__4 : num__1 this will lead to quantity of juice p in x = num__4 and quantity of juice f = num__20 . . . hence ratio num__1 : num__5 if we calculate total juice p = num__24 and total of juice v = num__25 it fits because totals are same as what mentioned in the question . . . thus ans is d <eor> d <eos> |
d |
subtract__24.0__4.0__ subtract__24.0__4.0__ |
subtract__24.0__4.0__ subtract__24.0__4.0__ |
| there are num__76 lights which are functional and each is controlled by a separate on / off switch . two children a and b start playing with the switches . a starts by pressing every third switch till he reaches the end . b thereafter presses every fifth switch till he too reaches the end . if all switches were in off position at the beggining how many lights are switched on by the end of this operation ? <o> a ) num__28 <o> b ) num__30 <o> c ) num__32 <o> d ) num__34 <o> e ) num__36 |
editing my solution : number of switches = num__76 number of switches turned on by a : num__3 num__6 . . . num__75 = num__25 number of switches turned on by b : num__5 num__10 . . . . num__75 = num__15 few switches are turned on by a and later turned off by b : lcm ( num__35 ) = num__15 x = num__15 num__30 . . . . num__90 = num__6 . subtract the above num__6 switches from both a and b as they are turned off . number of switches that are turned on = ( num__25 - num__6 ) + ( num__15 - num__6 ) = num__28 answer : a <eor> a <eos> |
a |
divide__75.0__3.0__ multiply__3.0__5.0__ add__10.0__25.0__ multiply__3.0__10.0__ multiply__3.0__30.0__ add__3.0__25.0__ add__3.0__25.0__ |
divide__75.0__3.0__ add__5.0__10.0__ add__10.0__25.0__ add__5.0__25.0__ add__75.0__15.0__ add__3.0__25.0__ add__3.0__25.0__ |
| a jar is filled of liquid which is num__3 parts water and num__5 parts alcohol . how much of this mixture should be drawn out and replaced such that this mixture may contain half water and half alcohol ? <o> a ) num__28.0 <o> b ) num__10.0 <o> c ) num__20.0 <o> d ) num__29.0 <o> e ) num__26 % |
explanation : let the jar initially contain num__8 litres of mixture : num__3 litres water and num__5 litres alcohol let x litres of this mixture is drawn out and is replaced by x litres of water . amount of water now in the solution : num__3 - ( num__0.375 ) x + x . amount of alcohol now in the solution : num__5 - ( num__0.625 ) x desired ratio : num__1.0 = > num__3 - ( num__0.375 ) x + x = num__5 - ( num__0.625 ) x x = num__1.6 litres % x = ( ( num__1.6 ) / num__8 ) * num__100 = num__20.0 answer : c <eor> c <eos> |
c |
add__3.0__5.0__ divide__3.0__8.0__ divide__5.0__8.0__ add__0.375__0.625__ reverse__0.625__ divide__100.0__5.0__ divide__100.0__5.0__ |
add__3.0__5.0__ divide__3.0__8.0__ divide__5.0__8.0__ add__0.375__0.625__ reverse__0.625__ divide__100.0__5.0__ divide__100.0__5.0__ |
| what is the rate percent when the simple interest on rs . num__1200 amount to rs . num__400 in num__4 years ? <o> a ) num__8.33 <o> b ) num__6.0 <o> c ) num__2.0 <o> d ) num__95.0 <o> e ) num__1 % |
interest for num__4 yrs = num__400 interest for num__1 yr = num__100 interest rate = num__0.0833333333333 x num__100 = num__8.33 answer : a <eor> a <eos> |
a |
percent__8.33__100.0__ |
percent__8.33__100.0__ |
| a train num__250 m long is running at a speed of num__68 kmph . how long does it take to pass a man who is running at num__8 kmph in the same direction as the train ? <o> a ) num__5 sec <o> b ) num__9 sec <o> c ) num__12 sec <o> d ) num__15 sec <o> e ) num__18 sec |
speed of the train relative to man = ( num__68 - num__8 ) kmph = ( num__60 * num__0.277777777778 ) m / sec = ( num__16.6666666667 ) m / sec time taken by the train to cross the man = time taken by it to cover num__250 m at num__16.6666666667 m / sec = num__250 * num__0.06 sec = num__15 sec answer : d <eor> d <eos> |
d |
hour_to_min_conversion__ divide__250.0__16.6667__ round__15.0__ |
subtract__68.0__8.0__ divide__250.0__16.6667__ divide__250.0__16.6667__ |
| ravi can do a piece of work in num__30 days while prakash can do it in num__40 days . in how many days will they finish it together ? <o> a ) num__17 num__0.142857142857 days <o> b ) num__17 num__0.125 days <o> c ) num__17 num__0.714285714286 days <o> d ) num__17 num__0.428571428571 days <o> e ) num__13 num__0.142857142857 days |
num__0.0333333333333 + num__0.025 = num__0.0583333333333 num__17.1428571429 = num__17 num__0.142857142857 days answer : a <eor> a <eos> |
a |
add__0.025__0.0333__ subtract__17.1429__17.0__ round__17.0__ |
add__0.025__0.0333__ subtract__17.1429__17.0__ round__17.0__ |
| determine the value of num__1.14814814815 divided by num__1.14814814815 <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) none |
solution : each number divided by itself equals num__1 . answer b <eor> b <eos> |
b |
round_down__1.1481__ round_down__1.1481__ |
round_down__1.1481__ round_down__1.1481__ |
| a train passes a station platform in num__30 sec and a man standing on the platform in num__20 sec . if the speed of the train is num__54 km / hr . what is the length of the platform ? <o> a ) num__120 m <o> b ) num__240 m <o> c ) num__168 m <o> d ) num__75 m <o> e ) num__60 m |
speed = num__54 * num__0.277777777778 = num__15 m / sec . length of the train = num__15 * num__20 = num__300 m . let the length of the platform be x m . then ( x + num__300 ) / num__30 = num__15 = > x = num__60 m answer : e <eor> e <eos> |
e |
multiply__20.0__15.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
multiply__20.0__15.0__ hour_to_min_conversion__ hour_to_min_conversion__ |
| with both valves open the pool will be filled with water in num__48 minutes . the first valve alone would fill the pool in num__2 hours . if the second valve emits num__50 cubic meters of water more than the first every minute then what is the capacity w of the pool ? <o> a ) num__9000 cubic meters <o> b ) num__10500 cubic meters <o> c ) num__11750 cubic meters <o> d ) num__12000 cubic meters <o> e ) num__12500 cubic meters |
d . num__12000 cubic meters . if both hte valves fill the pool in num__48 minutes and valve num__1 only fills in num__120 minutes then valve num__2 alone will fill the pool in ( num__48 * num__120 ) / ( num__120 - num__48 ) = num__80 minutes . now if valve num__1 admits x cubic meter of water per minute then the capacity of pool will be num__120 x and also num__80 ( x + num__50 ) . or num__120 x = num__80 ( x + num__50 ) . or x = num__100 . hence the capacity of pool = num__120 x = num__12000 cubic meters . <eor> d <eos> |
d |
multiply__2.0__50.0__ round__12000.0__ |
multiply__2.0__50.0__ divide__12000.0__1.0__ |
| what is the smallest number which when diminished by num__20 is divisible num__15 num__30 num__45 and num__60 ? <o> a ) num__200 <o> b ) num__220 <o> c ) num__240 <o> d ) num__322 <o> e ) num__342 |
required number = ( lcm of num__15 num__30 num__45 and num__60 ) + num__20 = num__180 + num__20 = num__200 option a <eor> a <eos> |
a |
add__20.0__180.0__ add__20.0__180.0__ |
add__20.0__180.0__ add__20.0__180.0__ |
| a train running at the speed of num__54 km / hr crosses a pole in num__9 sec . what is the length of the train ? <o> a ) num__135 m <o> b ) num__786 m <o> c ) num__566 m <o> d ) num__546 m <o> e ) num__445 m |
speed = num__54 * num__0.277777777778 = num__15 m / sec length of the train = speed * time = num__15 * num__9 = num__135 m answer : a <eor> a <eos> |
a |
multiply__9.0__15.0__ round__135.0__ |
multiply__9.0__15.0__ multiply__9.0__15.0__ |
| the dimensions of a room are num__25 feet * num__15 feet * num__12 feet . what is the cost of white washing the four walls of the room at rs . num__4 per square feet if there is one door of dimensions num__6 feet * num__3 feet and three windows of dimensions num__4 feet * num__3 feet each ? <o> a ) s . num__3624 <o> b ) s . num__4518 <o> c ) s . num__4518 <o> d ) s . num__4530 <o> e ) s . num__4517 |
area of the four walls = num__2 h ( l + b ) since there are doors and windows area of the walls = num__2 * num__12 ( num__15 + num__25 ) - ( num__6 * num__3 ) - num__3 ( num__4 * num__3 ) = num__906 sq . ft . total cost = num__906 * num__4 = rs . num__3624 answer : a <eor> a <eos> |
a |
square_perimeter__906.0__ square_perimeter__906.0__ |
multiply__4.0__906.0__ multiply__4.0__906.0__ |
| a train covers a distance of num__12 km in num__10 min . if it takes num__10 sec to pass a telegraph post then the length of the train is ? <o> a ) num__100 <o> b ) num__110 <o> c ) num__200 <o> d ) num__130 <o> e ) num__140 |
speed = ( num__1.2 * num__60 ) km / hr = ( num__72 * num__0.277777777778 ) m / sec = num__20 m / sec . length of the train = num__20 * num__10 = num__200 m . answer : option c <eor> c <eos> |
c |
divide__12.0__10.0__ hour_to_min_conversion__ add__12.0__60.0__ multiply__10.0__20.0__ round__200.0__ |
divide__12.0__10.0__ hour_to_min_conversion__ multiply__1.2__60.0__ multiply__10.0__20.0__ multiply__10.0__20.0__ |
| one year ago mrs promila was four times as old as her daughter swati . six year hence mrs . promila ’ s age will exceed her daughter ’ s age by num__9 years . the ratio of the present ages of promila and her daughter is <o> a ) num__9 : num__2 <o> b ) num__11 : num__3 <o> c ) num__12 : num__5 <o> d ) num__13 : num__4 <o> e ) none of these |
let swati ’ s age num__1 year ago = x then promila ’ s age num__1 year ago = num__4 x then ( num__4 x + num__6 ) - ( x + num__6 ) = num__9 = > num__3 x = num__9 = > x = num__3 present age of promila = ( num__12 + num__1 ) years = num__13 years present age of swati = ( num__3 + num__1 ) years = num__4 years thus ratio of their ages = num__13 : num__4 answer : d <eor> d <eos> |
d |
subtract__9.0__6.0__ add__9.0__3.0__ add__9.0__4.0__ add__9.0__4.0__ |
subtract__9.0__6.0__ add__9.0__3.0__ add__9.0__4.0__ add__9.0__4.0__ |
| a metallic sphere of radius num__12 cm is melted and drawn into a wire whose radius of cross section is num__8 cm . what is the length of the wire ? <o> a ) num__16 cm <o> b ) num__24 cm <o> c ) num__28 cm <o> d ) num__36 cm <o> e ) num__39 cm |
volume of the wire ( in cylindrical shape ) is equal to the volume of the sphere . π ( num__8 ) ^ num__2 * h = ( num__1.33333333333 ) π ( num__12 ) ^ num__3 = > h = num__36 cm answer : d <eor> d <eos> |
d |
multiply__12.0__3.0__ round__36.0__ |
multiply__12.0__3.0__ multiply__12.0__3.0__ |
| a man rows his boat num__90 km downstream and num__70 km upstream taking num__3 hours each time . find the speed of the stream ? <o> a ) num__76 kmph <o> b ) num__6 kmph <o> c ) num__3.5 kmph <o> d ) num__4.5 kmph <o> e ) num__4 kmph |
speed downstream = d / t = num__90 / ( num__3 ) = num__30 kmph speed upstream = d / t = num__70 / ( num__3 ) = num__23 kmph the speed of the stream = ( num__30 - num__23 ) / num__2 = num__3.5 kmph answer : c <eor> c <eos> |
c |
divide__90.0__3.0__ round__3.5__ |
divide__90.0__3.0__ round__3.5__ |
| claire has a total of num__86 pets consisting of gerbils and hamsters only . one - quarter of the gerbils are male and one - third of the hamsters are male . if there are num__25 males altogether how many gerbils does claire have ? <o> a ) num__39 <o> b ) num__50 <o> c ) num__60 <o> d ) num__54 <o> e ) num__60 |
g + h = num__86 . . . num__1 ; g / num__4 + h / num__3 = num__25 . . . . num__2 or num__3 g + num__4 h = num__25 * num__12 = num__300 g = num__86 - h or num__3 ( num__86 - h ) + num__4 h = num__300 h = num__300 - num__258 = num__42 then g = num__96 - num__42 = num__54 d <eor> d <eos> |
d |
subtract__4.0__1.0__ subtract__3.0__1.0__ multiply__3.0__4.0__ multiply__25.0__12.0__ multiply__86.0__3.0__ subtract__300.0__258.0__ subtract__96.0__42.0__ subtract__96.0__42.0__ |
subtract__4.0__1.0__ subtract__3.0__1.0__ multiply__3.0__4.0__ multiply__25.0__12.0__ multiply__86.0__3.0__ subtract__300.0__258.0__ subtract__96.0__42.0__ subtract__96.0__42.0__ |
| when a positive integer a is divided by num__11 and num__7 the remainders obtained are num__1 and num__2 respectively . when the positive integer b is divided by num__11 and num__7 the remainders obtained are num__1 and num__2 respectively . which of the following is a factor of ( a - b ) ? <o> a ) num__18 <o> b ) num__36 <o> c ) num__50 <o> d ) num__65 <o> e ) num__77 |
a = num__11 k + num__1 = num__7 j + num__2 b = num__11 m + num__1 = num__7 n + num__2 a - b = num__11 ( k - m ) = num__7 ( j - n ) a - b is a multiple of both num__11 and num__7 so it is a multiple of num__77 . the answer is e . <eor> e <eos> |
e |
multiply__11.0__7.0__ multiply__11.0__7.0__ |
multiply__11.0__7.0__ multiply__11.0__7.0__ |
| how many e ways are there of placing num__6 marbles in num__4 bowls if any number of them can be placed in each bowl ? <o> a ) num__6 c num__4 <o> b ) num__6 p num__4 <o> c ) num__4 ^ num__6 <o> d ) num__6 ^ num__4 <o> e ) num__6 ! |
each marble has num__4 options so there are total of e = num__4 * num__4 * num__4 * num__4 * num__4 * num__4 = num__4 ^ num__6 ways . answer : c . the total number of ways of dividing n identical items among r persons each one of whom can receive num__0 num__12 or more items is ( n + r - num__1 ) c ( r - num__1 ) . <eor> c <eos> |
c |
multiply__4.0__1.0__ |
multiply__4.0__1.0__ |
| wendy begins sanding a kitchen floor by herself and works for num__5 hours . she is then joined by bruce and together the two of them finish sanding the floor in num__2 hours . if bruce can sand the floor by himself in num__20 hours how long would it take wendy to sand the floor by herself ? <o> a ) num__0.1125 hours <o> b ) num__0.15 hours <o> c ) num__6.66666666667 hours <o> d ) num__8.88888888889 hours <o> e ) num__7.77142857143 hours |
let wendy finishes sanding the floor alone in w hours while b be the hours for bruce . thus in num__1 hour wendy finishes num__1 / w of the work while bruce finishes num__1 / b of the work . if wendy works for num__5 hours and is then joined by bruce to finish the work in num__2 more hours num__5 / w + num__2 / w + num__2 / b = num__1 ( num__1 denotes the total amount of work ) num__7 / w + num__2 / b = num__1 and given b = num__20 hours . thus w = num__7.77142857143 hours e is the correct answer . <eor> e <eos> |
e |
add__5.0__2.0__ multiply__1.0__7.7714__ |
add__5.0__2.0__ divide__7.7714__1.0__ |
| the floor of a rectangular room is num__19 m long and num__12 m wide . the room is surrounded by a veranda of width num__2 m on all its sides . the area of the veranda is : <o> a ) num__144 m num__2 <o> b ) num__140 m num__2 <o> c ) num__108 m num__2 <o> d ) num__158 m num__2 <o> e ) none of these |
area of the outer rectangle = num__23 Ã — num__16 = num__368 m num__2 area of the inner rectangle = num__19 Ã — num__12 = num__228 m num__2 required area = ( num__368 â € “ num__228 ) = num__140 m num__2 answer b <eor> b <eos> |
b |
multiply__16.0__23.0__ multiply__19.0__12.0__ triangle_area__2.0__140.0__ |
multiply__16.0__23.0__ multiply__19.0__12.0__ triangle_area__2.0__140.0__ |
| when a certain number x is divided by num__143 the remainder is num__21 . which of the following numbers if added to x would result in a number divisible by num__13 ? <o> a ) num__7 <o> b ) num__21 <o> c ) num__5 <o> d ) num__47 <o> e ) num__55 |
now x = num__143 q + num__45 = num__11 * num__13 * q + num__21 . . so we have to make only num__21 div by num__13 . . num__21 + num__5 = num__26 divisible by num__13 add num__5 ans c <eor> c <eos> |
c |
divide__143.0__13.0__ add__21.0__5.0__ subtract__26.0__21.0__ |
divide__143.0__13.0__ add__21.0__5.0__ subtract__26.0__21.0__ |
| how many odd integers between num__100 and num__300 are there such that their tens digit is even ? <o> a ) num__100 <o> b ) num__200 <o> c ) num__300 <o> d ) num__50 <o> e ) num__75 |
i guess the easier way to do this would be . num__200 numbers between - num__100 and num__300 out of which half would be even half odd . number of odd = num__100 . of this . . the one with even tenth digit would be half again . hence num__50 . correct option is d <eor> d <eos> |
d |
subtract__300.0__100.0__ subtract__100.0__50.0__ |
subtract__300.0__100.0__ subtract__100.0__50.0__ |
| if each participant of a chess tournament plays exactly one game with each of the remaining participants then num__210 games will be played during the tournament . what is the number of participants ? <o> a ) num__18 <o> b ) num__19 <o> c ) num__20 <o> d ) num__21 <o> e ) num__22 |
let n be the number of participants . the number of games is nc num__2 = n * ( n - num__1 ) / num__2 = num__210 n * ( n - num__1 ) = num__420 = num__21 * num__20 ( trial and error ) the answer is d . <eor> d <eos> |
d |
multiply__210.0__2.0__ divide__420.0__21.0__ add__1.0__20.0__ |
multiply__210.0__2.0__ divide__420.0__21.0__ divide__420.0__20.0__ |
| the h . c . f . of two numbers is num__20 and the other two factors of their l . c . m . are num__11 and num__15 . the larger of the two numbers is : <o> a ) num__276 <o> b ) num__300 <o> c ) num__299 <o> d ) num__322 <o> e ) num__345 |
the numbers are ( num__20 x num__11 ) and ( num__20 x num__15 ) . larger number = ( num__20 x num__15 ) = num__300 . answer : b <eor> b <eos> |
b |
multiply__20.0__15.0__ multiply__20.0__15.0__ |
multiply__20.0__15.0__ multiply__20.0__15.0__ |
| a water tank is half full . pipe a can fill a tank in num__10 minutes and pipe b can empty it in num__6 minutes . if both the pipes are open how long will it take to empty or fill the tank completely ? <o> a ) num__6.5 min . to empty <o> b ) num__9 min . to empty <o> c ) num__5.5 min . to empty <o> d ) num__4 min . to empty <o> e ) num__7.5 min . to empty |
explanation : clearly pipe b is faster than pipe a and so the tank will be emptied . part to be emptied = num__0.5 part emptied by ( a + b ) in num__1 minute = ( num__0.166666666667 - num__0.1 ) = num__0.0666666666667 so the tank will be emptied in num__7.5 min answer : e <eor> e <eos> |
e |
divide__1.0__6.0__ divide__1.0__10.0__ subtract__0.1667__0.1__ round__7.5__ |
divide__1.0__6.0__ divide__1.0__10.0__ subtract__0.1667__0.1__ round__7.5__ |
| what is the remainder when num__1201 × num__1202 × num__1205 × num__1207 is divided by num__6 ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
the remainders when dividing each number by six are : num__1 num__2 num__5 and num__1 . the product is num__1 * num__2 * num__5 * num__1 = num__10 the remainder when dividing num__10 by num__6 is num__4 . the answer is d . <eor> d <eos> |
d |
subtract__1202.0__1201.0__ subtract__1207.0__1205.0__ subtract__1207.0__1202.0__ multiply__2.0__5.0__ subtract__1205.0__1201.0__ subtract__1205.0__1201.0__ |
subtract__1202.0__1201.0__ subtract__1207.0__1205.0__ subtract__1207.0__1202.0__ multiply__2.0__5.0__ subtract__1205.0__1201.0__ multiply__1.0__4.0__ |
| adding num__40.0 of x to x is equivalent to multiplying x by which of the following ? <o> a ) num__12.5 <o> b ) num__1.05 <o> c ) num__1.15 <o> d ) num__1.2 <o> e ) num__1.4 |
num__140 x / num__100 = num__1.4 * x answer : e <eor> e <eos> |
e |
subtract__140.0__40.0__ divide__140.0__100.0__ divide__140.0__100.0__ |
subtract__140.0__40.0__ divide__140.0__100.0__ divide__140.0__100.0__ |
| a shopkeeper expects a gain of num__22.5 on his cost price . if in a week his sale was of rs . num__343 what was his profit ? <o> a ) s . num__63 <o> b ) s . num__69 <o> c ) s . num__72 <o> d ) s . num__75 <o> e ) s . num__90 |
c . p . = rs . ( num__100 / num__122.5 ) x num__343 = rs . ( num__0.816326530612 ) x num__343 = rs . num__280 profit = rs . ( num__343 - num__280 ) = rs . num__63 . answer : a <eor> a <eos> |
a |
percent__22.5__280.0__ percent__22.5__280.0__ |
percent__22.5__280.0__ percent__22.5__280.0__ |
| num__3 candidates in an election and received num__1136 num__7636 and num__10628 votes respectively . what % of the total votes did the winning candidate gotin that election ? <o> a ) num__40.0 <o> b ) num__55.0 <o> c ) num__54.8 <o> d ) num__60.0 <o> e ) num__62 % |
total number of votes polled = ( num__1136 + num__7636 + num__10628 ) = num__19400 so required percentage = num__0.547835051546 * num__100 = num__54.8 c <eor> c <eos> |
c |
percent__54.8__100.0__ |
percent__54.8__100.0__ |
| a sum of rs . num__13500 amounts to rs . num__15500 in num__4 years at the rate of simple interest . what is the rate of interest ? <o> a ) num__4.0 <o> b ) num__5.0 <o> c ) num__6.0 <o> d ) num__8.0 <o> e ) num__14 % |
s . i . = ( num__15500 - num__13500 ) = rs . num__2000 / - rate = ( num__100 * num__2000 ) / ( num__12500 * num__4 ) = num__4.0 answer : a <eor> a <eos> |
a |
percent__4.0__100.0__ |
percent__4.0__100.0__ |
| according to the directions on a can of frozen orange juice concentrate num__1 can of concentrate is to be mixed with num__4 cans of water to make orange juice . how many num__12 ounce cans of concentrate are required to prepare num__180 six - ounce servings of orange juice ? <o> a ) num__15 <o> b ) num__18 <o> c ) num__21 <o> d ) num__24 <o> e ) num__27 |
num__1 x num__12 ounce can of concentrate + num__4 x num__12 ounce cans of water = num__60 ounces of the mixture . num__60 ounces of mixture gives ( num__10.0 ) = num__10 servings . thus num__1 x num__12 ounce can of concentrate is used to produce num__10 servings of the mixture . to make num__180 servings of the mixture we need num__18.0 = num__18 cans of the concentrate . the answer is b . <eor> b <eos> |
b |
divide__180.0__10.0__ multiply__1.0__18.0__ |
divide__180.0__10.0__ multiply__1.0__18.0__ |
| num__10 women can complete a work in num__7 days and num__10 children take num__14 days to complete the work . how many days will num__5 women and num__10 children take to complete the work ? <o> a ) num__10 <o> b ) num__5 <o> c ) num__7 <o> d ) num__14 <o> e ) num__8 |
work done by num__1 woman in num__1 day = num__0.0142857142857 work done by num__1 children in num__1 day = num__0.00714285714286 work done by num__5 women = num__5 * num__0.0142857142857 = num__0.0714285714286 work done by num__10 children = num__10 * num__0.00714285714286 = num__0.0714285714286 amount of work num__5 women and num__10 children take to complete = num__0.0714285714286 + num__0.0714285714286 = num__0.142857142857 num__5 women and num__10 children will complete the work in num__7 days . c <eor> c <eos> |
c |
divide__1.0__14.0__ divide__1.0__7.0__ round__7.0__ |
divide__1.0__14.0__ divide__1.0__7.0__ round__7.0__ |
| two carpenters working in the same pace can build num__2 desks in two hours and a half . how many desks can num__4 carpenters build in num__4 hours ? <o> a ) num__2.4 . <o> b ) num__3.6 . <o> c ) num__4.2 <o> d ) num__5.5 <o> e ) num__6.4 |
w = num__2 desks t = num__2.5 hrs rate of num__2 carpenters = num__2 × r rate = work done / time num__2 xr = num__2 / num__2.5 r = num__1 / num__2.5 = num__0.4 ( this is the rate of each carpenter ) work done by num__4 carpenters in num__4 hrs = num__4 × rate of each carpenter x time = num__4 × num__0.4 × num__4 = num__6.4 desks e is the correct answer . <eor> e <eos> |
e |
divide__1.0__2.5__ round__6.4__ |
divide__1.0__2.5__ divide__6.4__1.0__ |
| num__16 boys or num__24 girls can construct the wall in num__6 days . the number of days that num__8 boys and num__4 girls will take to construct ? <o> a ) num__7 days <o> b ) num__14 days <o> c ) num__6 days <o> d ) num__8 days <o> e ) num__9 days |
explanation : num__16 boys = num__24 girls num__1 boy = num__1.5 girls num__1 boy = num__1.5 girls num__8 boys + num__4 girls = num__8 Ã — num__1.5 + num__12 = num__12 + num__4 = num__16 girls num__9 days to complete the work answer : option e <eor> e <eos> |
e |
divide__24.0__16.0__ subtract__16.0__4.0__ multiply__6.0__1.5__ round__9.0__ |
divide__24.0__16.0__ add__8.0__4.0__ add__8.0__1.0__ add__8.0__1.0__ |
| pipe a can fill a tank in num__6 minutes and pipe b cam empty it in num__24 minutes . if both the pipes are opened together after how many minutes should pipe b be closed so that the tank is filled in num__30 minutes ? <o> a ) num__18 <o> b ) num__27 <o> c ) num__96 <o> d ) num__27 <o> e ) num__21 |
let the pipe b be closed after x minutes . num__5.0 - x / num__24 = num__1 = > x / num__24 = num__5.0 - num__1 = num__4 = > x = num__4 * num__24 = num__96 . answer : c <eor> c <eos> |
c |
divide__30.0__6.0__ subtract__6.0__5.0__ divide__24.0__6.0__ multiply__24.0__4.0__ round__96.0__ |
divide__30.0__6.0__ subtract__6.0__5.0__ divide__24.0__6.0__ multiply__24.0__4.0__ multiply__24.0__4.0__ |
| a radio station surveyed num__190 students to determine the types of music they liked . the survey revealed that num__114 liked rock music num__50 liked folk music and num__41 liked classical music num__14 liked rock music and folk music num__15 liked rock music and classical music num__11 liked classical music and folk music . num__5 liked all the three types of music . how many liked rock music folk music but not classical only ? <o> a ) num__14 <o> b ) num__5 <o> c ) num__114 <o> d ) num__9 <o> e ) num__15 |
let r f and c represent the sets of students who liked rock music folk music and classical music respectively . r = num__114 f = num__50 c = num__41 r and f but not c = num__14 - num__5 = num__9 answer d <eor> d <eos> |
d |
subtract__50.0__41.0__ subtract__50.0__41.0__ |
subtract__50.0__41.0__ subtract__50.0__41.0__ |
| num__30.0 of num__2 is equal to <o> a ) num__0.2 <o> b ) num__0.4 <o> c ) num__0.6 <o> d ) num__0.7 <o> e ) num__0.9 |
num__30.0 of num__2 = ( num__0.3 ) * num__2 = num__0.6 answer : option c <eor> c <eos> |
c |
percent__30.0__2.0__ percent__30.0__2.0__ |
percent__30.0__2.0__ percent__30.0__2.0__ |
| an air - conditioning unit costs $ num__470 . on december there was a discount for christmas of num__16.0 . six months later the holiday season was over so the company raised the price of the air - conditioning by num__18.0 . how much will an air - conditioning unit cost in november ? <o> a ) $ num__466 <o> b ) $ num__470 <o> c ) $ num__472 <o> d ) $ num__484 <o> e ) $ num__491 |
if its prevoius november ( before discount ) then price is $ num__470 . but if its november of next year then num__16.0 discount on $ num__470 = num__470 ( num__1 - num__0.16 ) = $ num__394.8 again a corrected raised price of num__18.0 over $ num__394.8 = num__394.8 ( num__1 + num__0.18 ) = num__465.86 ~ $ num__466 ans a <eor> a <eos> |
a |
multiply__1.0__466.0__ |
multiply__1.0__466.0__ |
| water consists of hydrogen and oxygen and the approximate ratio by mass of hydrogen to oxygen is num__2 : num__16 . approximately how many grams of oxygen are there in num__144 grams of water ? <o> a ) num__16 <o> b ) num__72 <o> c ) num__112 <o> d ) num__128 <o> e ) num__142 |
since the ratio by mass of hydrogen to oxygen is num__2 : num__16 then oxygen is num__16 / ( num__2 + num__16 ) = num__0.888888888889 of mass of water . therefore there are num__144 * num__0.888888888889 = num__128 grams of oxygen in num__144 grams of water . answer : d . <eor> d <eos> |
d |
subtract__144.0__16.0__ subtract__144.0__16.0__ |
subtract__144.0__16.0__ subtract__144.0__16.0__ |
| num__105 num__10296 ? num__75 num__6042 <o> a ) num__80 <o> b ) num__75 <o> c ) num__60 <o> d ) num__82 <o> e ) num__87 |
the pattern is - num__3 - num__6 - num__9 - num__12 . . . num__96 - num__9 = num__87 answer : e <eor> e <eos> |
e |
add__3.0__6.0__ add__3.0__9.0__ subtract__105.0__9.0__ add__75.0__12.0__ add__75.0__12.0__ |
add__3.0__6.0__ add__3.0__9.0__ subtract__105.0__9.0__ subtract__96.0__9.0__ subtract__96.0__9.0__ |
| when w is divided by num__5 the remainder is num__3 . when y is divided by num__5 the remainder is num__4 . what is the remainder when w + y is divided by num__5 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
in my view the answer should be c w / num__5 has remainder = num__3 - > w = num__5 x q + num__3 y / num__5 has a remainder = num__4 - > y = num__5 x q + num__3 combining both ( ( num__5 x q num__1 + num__3 ) + ( num__5 x q num__2 + num__3 ) ) / num__5 = num__5 ( q num__1 + q num__2 ) / num__5 + num__1.4 = q num__1 + q num__2 + num__1.4 num__1.4 = num__1 + num__0.4 = > remainder num__2 answer c <eor> c <eos> |
c |
subtract__5.0__4.0__ subtract__5.0__3.0__ divide__2.0__5.0__ subtract__5.0__3.0__ |
subtract__5.0__4.0__ subtract__5.0__3.0__ divide__2.0__5.0__ subtract__5.0__3.0__ |
| list r contains five numbers that have an average value of num__65 . if the median of the numbers in the list is equal to the mean and the largest number is equal to num__20 more than two times the smallest number what is the smallest possible value in the list ? <o> a ) num__25 <o> b ) num__31 <o> c ) num__35 <o> d ) num__39 <o> e ) num__43 |
the middle number is num__65 . let the smallest number be x . then the largest number is num__2 x + num__20 . to make x as small as possible let ' s make the remaining numbers as large as possible . so the second largest = the largest = num__2 x + num__20 . the second smallest should be equal to the median . the numbers are x num__65 num__65 num__2 x + num__20 num__2 x + num__20 . x + num__65 + num__65 + num__2 x + num__20 + num__2 x + num__20 = num__5 * num__65 = num__325 num__5 x = num__155 x = num__31 the answer is b . <eor> b <eos> |
b |
multiply__65.0__5.0__ divide__155.0__5.0__ divide__155.0__5.0__ |
multiply__65.0__5.0__ divide__155.0__5.0__ divide__155.0__5.0__ |
| bradley owns b video game cartridges . if bradley â € ™ s total is quarter the total owned by andrew and four times the total owned by charlie how many video game cartridges do the three of them own altogether in terms of b ? <o> a ) num__5.33333333333 ) b <o> b ) num__4.25 ) b <o> c ) num__3.25 ) b <o> d ) num__5.25 ) b <o> e ) num__0.583333333333 ) b |
step num__1 : categorize the problem this problem is testing basic algebra and equations with an extra layer of complexity by having variables in the answer choices . step num__2 : think like the test maker what is the key detail or key details in the problem ? the key detail to immediately recognize is that the problem gives you andrew and charlie â € ™ s totals in relation to bradley â € ™ s total rather than giving you bradley â € ™ s total in relation to the other two ; and then the problem asks you to solve the problem in terms of b . this means that you have to relate andrew and charlie â € ™ s values in relation to bradley â € ™ s total . the test maker is attempting to force you to structure the problem in the opposite way that most people are used to structuring information . by gaining this insight it makes it easier to solve the problem . step num__3 : solve the problem b = bradley â € ™ s total num__4 b = andrew â € ™ s total ( num__0.25 ) b = charlie â € ™ s total add each total b + num__4 b + ( num__0.25 ) b = num__5 b + ( num__0.25 ) b = ( num__5.0 ) b + ( num__0.25 ) b = ( num__5.25 ) b therefore choose d . <eor> d <eos> |
d |
add__1.0__2.0__ add__1.0__3.0__ reverse__4.0__ add__1.0__4.0__ add__0.25__5.0__ add__0.25__5.0__ |
add__1.0__2.0__ add__1.0__3.0__ reverse__4.0__ add__1.0__4.0__ add__0.25__5.0__ add__0.25__5.0__ |
| in a country named ` ` gpuzzles ' ' num__0.5 of num__5 = num__3 . if the same proportion holds what is the value of num__0.333333333333 of num__10 ? <o> a ) num__2 <o> b ) num__5 <o> c ) num__6 <o> d ) num__9 <o> e ) num__4 |
e num__4 num__2.5 = num__3 and num__3.33333333333 = x ( num__3.33333333333 ) / ( num__2.5 ) = num__1.33333333333 = num__1.33333333333 so x / num__3 = num__1.33333333333 x = num__4 <eor> e <eos> |
e |
multiply__0.5__5.0__ add__3.0__0.3333__ divide__3.3333__2.5__ divide__10.0__2.5__ |
divide__10.0__4.0__ divide__10.0__3.0__ divide__3.3333__2.5__ divide__10.0__2.5__ |
| what is the place value of num__0 in the numeral num__2074 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__2 <o> d ) num__3 <o> e ) num__4 |
note : the place value of zero ( num__0 ) is always num__0 . it may hold any place in a number its value is always num__0 . a <eor> a <eos> |
a |
multiply__0.0__2074.0__ |
multiply__0.0__2074.0__ |
| how many distinct prime numbers are factors of num__33150 ? <o> a ) four <o> b ) five <o> c ) six <o> d ) seven <o> e ) eight |
start with the prime factorization : num__33150 = num__50 * num__663 = ( num__2 * num__5 * num__5 ) * num__3 * num__221 = ( num__2 ) * ( num__3 ) * ( num__5 ^ num__2 ) * ( num__13 ) * ( num__17 ) there are five distinct prime factors { num__2 num__3 num__5 num__13 and num__17 } answer : b . <eor> b <eos> |
b |
divide__33150.0__50.0__ subtract__5.0__2.0__ divide__663.0__3.0__ divide__221.0__13.0__ gcd__33150.0__5.0__ |
divide__33150.0__50.0__ subtract__5.0__2.0__ divide__663.0__3.0__ divide__221.0__13.0__ gcd__33150.0__5.0__ |
| the average age of students of a class is num__15.8 years . the average age of boys in the class is num__16.7 years and that of the girls is num__15.4 years . the ration of the number of boys to the number of girls in the class is : <o> a ) num__2 : num__5 <o> b ) num__2 : num__3 <o> c ) num__2 : num__4 <o> d ) num__2 : num__1 <o> e ) num__4 : num__9 |
let the ratio be k : num__1 . then k * num__16.7 + num__1 * num__15.4 = ( k + num__1 ) * num__15.8 = ( num__16.7 - num__15.8 ) k = ( num__15.8 - num__15.4 ) = k = num__0.4 / num__0.9 = num__0.444444444444 required ratio = num__0.444444444444 : num__1 = num__4 : num__9 . answer : e <eor> e <eos> |
e |
subtract__15.8__15.4__ subtract__16.7__15.8__ divide__0.4__0.9__ multiply__1.0__4.0__ |
subtract__15.8__15.4__ subtract__16.7__15.8__ divide__0.4__0.9__ multiply__1.0__4.0__ |
| the perimeter of a triangle is num__36 cm and the in radius of the triangle is num__2.5 cm . what is the area of the triangle ? <o> a ) num__76 <o> b ) num__88 <o> c ) num__66 <o> d ) num__55 <o> e ) num__45 |
area of a triangle = r * s where r is the in radius and s is the semi perimeter of the triangle . area of triangle = num__2.5 * num__18.0 = num__45 cm num__2 answer : e <eor> e <eos> |
e |
triangle_area__36.0__2.5__ triangle_area__36.0__2.5__ |
multiply__2.5__18.0__ multiply__2.5__18.0__ |
| what is greatest positive integer n such that num__2 ^ n is a factor of num__12 ^ num__9 ? <o> a ) a ) num__10 <o> b ) b ) num__18 <o> c ) c ) num__16 <o> d ) d ) num__20 <o> e ) e ) num__60 |
the given number is num__12 ^ num__9 = ( num__2 * num__2 * num__3 ) ^ num__9 = ( num__2 ^ num__18 ) * ( num__3 ^ num__9 ) so the greatest possible value for n such that num__2 ^ n can be factor of given number is num__18 . answer b <eor> b <eos> |
b |
gcd__12.0__9.0__ multiply__2.0__9.0__ multiply__2.0__9.0__ |
gcd__12.0__9.0__ multiply__2.0__9.0__ multiply__2.0__9.0__ |
| at a speed of num__60 miles per hour a certain car uses num__1 gallon of gasoline every num__30 miles . if the car starts with a full num__12 gallon tank of gasoline and travels for num__5 hours at num__60 miles per hour the amount of gasoline used would be what fraction of a full tank ? <o> a ) num__0.12 <o> b ) num__0.305555555556 <o> c ) num__0.583333333333 <o> d ) num__0.666666666667 <o> e ) num__0.833333333333 |
gas used = ( num__5 hours ) * ( num__60 miles / hour ) * ( num__1 gallon / num__30 miles ) = num__10 gallons portion used = ( num__10 ) / num__12 = num__0.833333333333 ans e <eor> e <eos> |
e |
divide__10.0__12.0__ multiply__1.0__0.8333__ |
divide__10.0__12.0__ multiply__1.0__0.8333__ |
| if num__16 ^ y = num__2 ^ num__16 what is y ? <o> a ) num__2 <o> b ) num__4 <o> c ) num__8 <o> d ) num__10 <o> e ) num__12 |
num__16 ^ y = num__2 ^ num__4 y = num__2 ^ num__16 num__4 y = num__16 y = num__4 the answer is b . <eor> b <eos> |
b |
divide__16.0__4.0__ |
divide__16.0__4.0__ |
| at what rate percent on simple interest will a sum of money double itself in num__40 years ? <o> a ) num__3 num__0.333333333333 % <o> b ) num__2 num__0.5 % <o> c ) num__3 num__3.0 % <o> d ) num__3 num__0.666666666667 % <o> e ) num__3 num__2.0 % |
p = ( p * num__40 * r ) / num__100 r = num__2 num__0.5 % answer : b <eor> b <eos> |
b |
percent__2.0__100.0__ |
percent__2.0__100.0__ |
| two bike riders ride in opposite directions around a circular track starting at the same time from the same point . biker a rides at a speed of num__16 kmph and biker b rides at a speed of num__14 kmph . if the track has a diameter of num__40 km after how much time ( in hours ) will the two bikers meet ? <o> a ) num__8.18 hrs . <o> b ) num__6.18 hrs . <o> c ) num__4.18 hrs . <o> d ) num__1.18 hrs . <o> e ) num__7.18 hrs . |
explanation : distance to be covered = = num__40 km relative speed of bikers = num__16 + num__14 = num__30 kmph . now = = num__4.18 hrs . answer : c <eor> c <eos> |
c |
add__16.0__14.0__ round__4.18__ |
add__16.0__14.0__ round__4.18__ |
| the average age of husband wife and their child num__3 years ago was num__24 years and that of wife and the child num__5 years ago was num__20 years . the present age of the husband is <o> a ) num__22 <o> b ) num__40 <o> c ) num__38 <o> d ) num__21 <o> e ) num__31 |
explanation : sum of the present ages of husband wife and child = ( num__24 x num__3 + num__3 x num__3 ) years = num__81 years . sum of the present ages of wife and child ( num__20 x num__2 + num__5 x num__2 ) years = num__50 years . husband ' s present age = ( num__81 - num__50 ) years = num__31 years . answer : e <eor> e <eos> |
e |
subtract__5.0__3.0__ subtract__81.0__50.0__ subtract__81.0__50.0__ |
subtract__5.0__3.0__ subtract__81.0__50.0__ subtract__81.0__50.0__ |
| if num__1.5 x = num__0.04 y then the value of ( y - x ) / ( y + x ) is <o> a ) num__9.48051948052 <o> b ) num__0.948051948052 <o> c ) num__7.3 / num__77 <o> d ) num__7.3 / num__770 <o> e ) num__7.3 / num__77 |
x / y = num__0.04 / num__1.5 y / x = num__1.5 / num__0.04 by componendo dividendo rule ( y + x ) / ( y - x ) = num__1.54 / num__1.46 ( y - x ) / ( y + x ) = num__1.46 / num__1.54 = num__0.948051948052 answer : b <eor> b <eos> |
b |
add__1.5__0.04__ subtract__1.5__0.04__ divide__1.46__1.54__ divide__1.46__1.54__ |
add__1.5__0.04__ subtract__1.5__0.04__ divide__1.46__1.54__ divide__1.46__1.54__ |
| num__5 out of num__1500 parts of earth is sulphur . what is the percentage of sulphur in earth <o> a ) num__0.22 <o> b ) num__0.222222222222 <o> c ) num__0.0222222222222 <o> d ) num__0.333333333333 <o> e ) none of these |
required percentage = ( num__0.00333333333333 * num__100 ) % = num__0.333333333333 % correct options : d <eor> d <eos> |
d |
percent__100.0__0.3333__ |
percent__100.0__0.3333__ |
| the speed of a car is num__100 km in the first hour and num__30 km in the second hour . what is the average speed of the car ? <o> a ) num__50 kmph <o> b ) num__65 kmph <o> c ) num__75 kmph <o> d ) num__85 kmph <o> e ) num__65 kmph |
explanation : s = ( num__100 + num__30 ) / num__2 = num__65 kmph e ) <eor> e <eos> |
e |
round__65.0__ |
round__65.0__ |
| walking with num__0.6 of my usual speed i miss the bus by num__5 minutes . what is my usual time ? <o> a ) num__16 min <o> b ) num__26 min <o> c ) num__34 min <o> d ) num__20 min <o> e ) num__15 min |
speed ratio = num__1 : num__0.6 = num__5 : num__3 time ratio = num__3 : num__5 num__1 - - - - - - - - num__5 num__3 - - - - - - - - - ? è num__15 answer : e <eor> e <eos> |
e |
multiply__0.6__5.0__ multiply__5.0__3.0__ round__15.0__ |
multiply__0.6__5.0__ multiply__5.0__3.0__ round__15.0__ |
| a car covers a distance of num__780 km in num__6 ½ hours . find its speed ? <o> a ) num__104 kmph <o> b ) num__187 kmph <o> c ) num__164 kmph <o> d ) num__130 kmph <o> e ) num__106 kmph |
num__130.0 = num__130 kmph answer : d <eor> d <eos> |
d |
divide__780.0__6.0__ round__130.0__ |
divide__780.0__6.0__ round__130.0__ |
| i remember during the school days the teacher asked the class ` ` can you tell me the sum of the first num__50 odd numbers ? ` ` . i ran quickly to the teacher and told her ` ` the answer is num__2500 ' ' . the teacher replied ` ` lucky guess ' ' . she then asked me ` ` can you tell me the sum of first num__74 odd numbers ? ` ` . i wait for approx num__10 seconds and replied with the correct answer . how can i answer so quickly and whats the correct answer ? <o> a ) num__8715 <o> b ) num__0152 <o> c ) num__3581 <o> d ) num__5476 <o> e ) num__9126 |
d num__5476 n ^ num__1 num__74 * num__74 = num__5476 ( sum of first num__74 odd numbers ) . num__50 * num__50 = num__2500 ( sum of first num__50 odd numbers ) . <eor> d <eos> |
d |
round__5476.0__ |
round__5476.0__ |
| the average ( arithmetic mean ) of num__4 different integers is num__68 . if the largest integer is num__90 what is the least possible value of the smallest integer ? <o> a ) num__5 <o> b ) num__19 <o> c ) num__29 <o> d ) num__30 <o> e ) num__33 |
total of integers = num__68 * num__4 = num__272 lowest of the least possible integer is when the middle num__2 intergers are at the maximum or equal to the highest possible integer . but all integers are distinct . so if the largest integer is num__90 then the middle num__2 will be num__88 and num__89 lowest of least possible integer = num__272 - ( num__90 + num__89 + num__88 ) = num__272 - num__267 = num__5 answer : a <eor> a <eos> |
a |
multiply__4.0__68.0__ subtract__90.0__2.0__ subtract__272.0__267.0__ subtract__272.0__267.0__ |
multiply__4.0__68.0__ subtract__90.0__2.0__ subtract__272.0__267.0__ subtract__272.0__267.0__ |
| the monthly incomes of a and b are in the ratio num__5 : num__2 . b ' s monthly income is num__12.0 more than c ' s monthly income . if c ' s monthly income is rs . num__13000 then find the annual income of a ? <o> a ) rs . num__420000 <o> b ) rs . num__180000 <o> c ) rs . num__436800 <o> d ) rs . num__504000 <o> e ) none of these |
b ' s monthly income = num__13000 * num__1.12 = rs . num__14560 b ' s monthly income = num__2 parts - - - - > rs . num__14560 a ' s monthly income = num__5 parts = num__2.5 * num__14560 = rs . num__36400 a ' s annual income = rs . num__36400 * num__12 = rs . num__436800 answer : c <eor> c <eos> |
c |
multiply__13000.0__1.12__ divide__5.0__2.0__ multiply__14560.0__2.5__ multiply__12.0__36400.0__ multiply__12.0__36400.0__ |
multiply__13000.0__1.12__ divide__5.0__2.0__ multiply__14560.0__2.5__ multiply__12.0__36400.0__ multiply__12.0__36400.0__ |
| a can lay railway track between two given stations in num__16 days and b can do the same job in num__12 days . with help of c they did the job in num__4 days only . then c alone can do the job in ? <o> a ) num__9 <o> b ) num__9 num__0.6 <o> c ) num__10 <o> d ) num__11 <o> e ) none of these |
( a + b + c ) ' s num__1 day ' s work = num__1 num__4 a ' s num__1 day ' s work = num__1 num__16 b ' s num__1 day ' s work = num__1 . num__12 therefore c ' s num__1 day ' s work = num__1 - ( num__1 + num__1 ) = ( num__1 - num__7 ) = num__5 . num__4 num__16 num__12 num__4 num__48 num__48 so c alone can do the work in num__48 = num__9 num__3 days . num__5 num__5 answer is b <eor> b <eos> |
b |
subtract__12.0__7.0__ multiply__12.0__4.0__ subtract__16.0__7.0__ divide__12.0__4.0__ round__9.0__ |
subtract__12.0__7.0__ multiply__12.0__4.0__ subtract__16.0__7.0__ subtract__12.0__9.0__ subtract__16.0__7.0__ |
| num__5 |
7 num__1119 ? num__67131 <o> a ) num__32 <o> b ) num__40 <o> c ) num__30 <o> d ) num__35 <o> e ) num__25 |
the pattern is + num__2 + num__4 + num__8 + num__16 . . . num__19 + num__16 = num__35 answer : d <eor> d <eos> |
d |
d |
| a ladder was rested along a wall of height num__5 m . if ladder slides num__2 m away from the wall then ladder touches the foot of the wall . what is the height of the ladder . <o> a ) num__6.25 mtrs <o> b ) num__7.25 mtrs <o> c ) num__8.25 mtrs <o> d ) num__9.25 mtrs <o> e ) num__5.25 mtrs |
suppose earlier foot of ladder was x mtres away from wall then length of ladder = x + num__2 mtrs hence ( x + num__2 ) ^ num__2 = x ^ num__2 + num__5 ^ num__2 x = num__5.25 = num__5.25 mtrs length of ladder = num__5.25 + num__2 = num__7.25 mtrs answer : b <eor> b <eos> |
b |
add__2.0__5.25__ add__2.0__5.25__ |
add__2.0__5.25__ add__2.0__5.25__ |
| a person purchased a tv set for rs . num__16000 and a dvd player for rs . num__6250 . he sold both the items together for rs . num__32150 . what percentage of profit did he make ? <o> a ) num__22 <o> b ) num__44.5 <o> c ) num__40 <o> d ) num__26 <o> e ) num__11 |
the total cp = rs . num__16000 + rs . num__6250 = rs . num__22250 and sp = rs . num__32150 profit ( % ) = ( num__32150 - num__22250 ) / num__22250 * num__100 = num__44.5 . answer : b <eor> b <eos> |
b |
percent__44.5__100.0__ |
percent__44.5__100.0__ |
| two persons start running simultaneously around a circular track of length num__380 m from the same point at speeds of num__24 kmph and num__28 kmph . when will they meet for the first time any where on the track if they are moving in the opposite direction ? <o> a ) num__144 <o> b ) num__36 <o> c ) num__124 <o> d ) num__26 <o> e ) num__38 |
time taken to meet the first time = length of track / relative speed = num__380 / ( num__24 + num__28 ) ( num__0.277777777778 ) = num__7.30769230769 * ( num__3.6 ) = num__26 sec . answer : d <eor> d <eos> |
d |
round__26.0__ |
round__26.0__ |
| the number num__341 is equal to the sum of the cubes of two integers . what is the product of those integers ? <o> a ) num__8 <o> b ) num__15 <o> c ) num__21 <o> d ) num__30 <o> e ) num__39 |
num__5 ^ num__3 + num__6 ^ num__3 = num__341 number is num__5 * num__6 = num__30 d <eor> d <eos> |
d |
multiply__5.0__6.0__ multiply__5.0__6.0__ |
multiply__5.0__6.0__ multiply__5.0__6.0__ |
| num__50 is divided into two parts such that the sum of their reciprocals is num__0.0833333333333 . find the two parts . <o> a ) num__4010 <o> b ) num__2228 <o> c ) num__2525 <o> d ) num__3515 <o> e ) num__30 |
20 |
let the two parts be x and ( num__50 - x ) . then num__1 / x + num__1 / ( num__50 – x ) = num__0.0833333333333 = > ( num__50 – x + x ) / x ( num__50 – x ) = num__0.0833333333333 = > x ^ num__2 – num__50 x + num__600 = num__0 = > ( x – num__30 ) ( x – num__20 ) = num__0 = > x = num__30 or x = num__20 . so the parts are num__30 and num__20 . answer e ) num__3020 . <eor> e <eos> |
e |
e |
| angelo and isabella are both salespersons . in any given week angelo makes $ num__550 in base salary plus num__8 percent of the portion of his sales above $ num__4000 for that week . isabella makes num__10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of money ? <o> a ) num__11500 <o> b ) num__24500 <o> c ) num__25500 <o> d ) num__26500 <o> e ) num__27 |
500 |
official solution : the problem asks for the amount of weekly sales it takes for angelo and isabella to earn the same amount of money . you can write an equation that sets angelo ’ s and isabella ’ s weekly earnings equal to each other with x representing weekly sales . weekly earnings for each salesperson equal base salary plus commission . so angelo ’ s earnings are num__550 + ( num__0.08 ) ( x – num__4000 ) and isabella ’ s are num__0.10 x . set up the equation and solve : num__550 + ( num__0.08 ) ( x – num__4000 ) = num__0.10 x distribute the num__0.08 : num__550 + num__0.08 x – num__320 = num__0.10 x combine terms and subtract num__0.08 x from both sides : num__230 = num__0.02 x divide both sides by num__0.02 : num__11500 = x your answer is a . <eor> a <eos> |
a |
a |
| what is the value of ( a y ) in ( num__13 ^ - num__20 - a * num__13 ^ y ) = num__168 * num__13 ^ - num__22 ? <o> a ) ( num__12 ) <o> b ) ( num__01 ) <o> c ) ( - num__122 ) <o> d ) ( num__1 - num__22 ) <o> e ) ( num__1 - num__20 ) |
( num__13 ^ - num__20 - a * num__13 ^ y ) = num__168 * num__13 ^ - num__22 divide the lhs and rhs by num__13 ^ - num__22 num__13 ^ num__2 - a * num__13 ^ ( y + num__22 ) = num__168 num__168 can be expressed as num__13 ^ num__2 - num__1 num__13 ^ num__2 - a * num__13 ^ ( y + num__22 ) = num__13 ^ num__2 - num__1 a * num__13 ^ ( y + num__22 ) = num__1 now look at options and substitute . . its easy for y = - num__22 . num__13 ^ num__0 = num__1 a also = num__1 answer : d <eor> d <eos> |
d |
subtract__22.0__20.0__ reverse__1.0__ |
subtract__22.0__20.0__ subtract__2.0__1.0__ |
| in an isoscele right angled triangle the perimeter is num__20 metre . find its area . <o> a ) num__9320 m num__2 <o> b ) num__8750 m num__2 <o> c ) num__7980 m num__2 <o> d ) num__6890 m num__2 <o> e ) none of these |
in an isoscele right angled triangle area = num__23.3 × perimeter num__2 = num__23.3 × num__20 ( num__2 ) = num__9320 m num__2 answer a <eor> a <eos> |
a |
triangle_area__9320.0__2.0__ |
triangle_area__9320.0__2.0__ |
| hcf and lcm two numbers are num__12 and num__396 respectively . if one of the numbers is num__36 then the other number is ? <o> a ) num__36 <o> b ) num__66 <o> c ) num__132 <o> d ) num__264 <o> e ) num__364 |
num__12 * num__396 = num__36 * x x = num__132 answer : c <eor> c <eos> |
c |
round__132.0__ |
round__132.0__ |
| there are num__8 students . num__4 of them are men and num__4 of them are women . if num__4 students are selected from the num__8 students . what is the probability r that the number of men is equal to that of women ? <o> a ) r = num__0.514285714286 <o> b ) num__0.457142857143 <o> c ) num__0.4 <o> d ) num__0.371428571429 <o> e ) num__0.342857142857 |
method - num__1 : favorable outcomes ( i . e . no . of men = no . of women = num__2 ) = num__4 c num__2 * num__4 c num__2 = num__6 * num__6 = num__36 total ways of selecting num__4 out of num__8 students = num__8 c num__4 = num__8 ! / ( num__4 ! * num__4 ! ) = num__70 probability = num__0.514285714286 = num__0.514285714286 a method - num__2 : also see the mistake done by the person who posted this question probability of first selected person being man = num__0.5 probability of second selected person being man = num__0.428571428571 probability of first selected person being woman = num__0.666666666667 probability of second selected person being woman = num__0.6 i . e . probability = ( num__0.5 ) * ( num__0.428571428571 ) * ( num__0.666666666667 ) * ( num__0.6 ) * [ num__4 ! / ( num__2 ! * num__2 ! ) ] = num__0.514285714286 the important part for the readers is to understand the reason of multiplying [ num__4 ! / ( num__2 ! * num__2 ! ) ] here when we take the probability of each case like we have have taken in this method then it always include arrangements as well and so we have to take every arrangement of the events as well the arrangement of these num__4 events can be done in num__4 ! ways but since the second man ca n ' t be selected before num__1 st so we have to exclude their arrangement by dividing by num__2 ! and similarly since the second woman ca n ' t be selected before num__1 st so we have to exclude their arrangement by dividing by num__2 ! a <eor> a <eos> |
a |
coin_space__ die_space__ union_prob__0.4286__0.6__0.5143__ |
coin_space__ die_space__ union_prob__0.4286__0.6__0.5143__ |
| a train running at the speed of num__60 km / hr crosses a pole in num__8 seconds . find the length of the train . <o> a ) num__133.33 <o> b ) num__882 <o> c ) num__772 <o> d ) num__252 <o> e ) num__121 |
speed = num__60 * ( num__0.277777777778 ) m / sec = num__16.6666666667 m / sec length of train ( distance ) = speed * time ( num__16.6666666667 ) * num__8 = num__133.33 meter . answer : a <eor> a <eos> |
a |
round__133.33__ |
round__133.33__ |
| a sum of money is to be distributed among a b c d in the proportion of num__5 : num__2 : num__4 : num__3 . if c gets rs . num__1000 more than d what is b ' s share ? <o> a ) num__1000 <o> b ) num__3000 <o> c ) num__2000 <o> d ) num__4000 <o> e ) num__5000 |
let the shares of a b c and d be rs . num__5 x rs . num__2 x rs . num__4 x and rs . num__3 x respectively . then num__4 x - num__3 x = num__1000 x = num__1000 . b ' s share = rs . num__2 x = rs . ( num__2 x num__1000 ) = rs . num__2000 . answer is c . <eor> c <eos> |
c |
multiply__2.0__1000.0__ multiply__2.0__1000.0__ |
multiply__2.0__1000.0__ multiply__2.0__1000.0__ |
| line e has the equation num__3 x + y = num__7 . which of the following lines is perpendicular to line e ? <o> a ) y = num__3 x + num__4 <o> b ) y = – num__3 x – num__6 <o> c ) y = ( num__0.333333333333 ) x – num__1 <o> d ) y = ( – num__0.333333333333 ) x + num__2 <o> e ) y = ( – num__2.33333333333 ) x – num__5 |
i first rewrote the equation in the standard y = mx + b form . therefore line e as presented num__3 x + y = num__7 can be rewritten as follows : y = - num__3 x + num__7 . thought process next is what line would be perpendicular to line e ? any line with a reciprocal of the slope but in the opposite direction . the reciprocal of any fraction / integer is num__1 over that number / integer . therefore the reciprocal of - num__3 is - num__0.333333333333 - need to drop the negative sign because the line would kinda run parallel and we want perpendicular . scan the answers choices and notice c as the only one . <eor> c <eos> |
c |
reverse__3.0__ reverse__3.0__ |
reverse__3.0__ reverse__3.0__ |
| if the simple interest on a sum of money for num__2 years at num__5.0 per annum is rs . num__50 what is the compound interest on the same sum at the rate and for the same time ? <o> a ) rs . num__51.25 <o> b ) rs . num__51.26 <o> c ) rs . num__51.22 <o> d ) rs . num__51.98 <o> e ) rs . num__51.11 |
sum = ( num__50 * num__100 ) / ( num__2 * num__5 ) = rs . num__500 amount = [ num__500 * ( num__1 + num__0.05 ) num__2 ] = rs . num__551.25 c . i . = ( num__551.25 - num__500 ) = rs . num__51.25 . answer : a <eor> a <eos> |
a |
percent__2.0__50.0__ percent__5.0__1.0__ percent__100.0__51.25__ |
percent__2.0__50.0__ percent__5.0__1.0__ percent__100.0__51.25__ |
| a train travels num__125 km in num__2.5 hours and num__270 km in num__3 hours . find the average speed of train . <o> a ) num__80 kmph <o> b ) num__60 kmph <o> c ) num__72 kmph <o> d ) num__90 kmph <o> e ) none of these |
as we know that speed = distance / time for average speed = total distance / total time taken thus total distance = num__125 + num__270 = num__395 km thus total speed = num__5.5 hrs or average speed = num__5 = num__395 / num__5.5 or num__72 kmph . answer : c <eor> c <eos> |
c |
add__125.0__270.0__ add__2.5__3.0__ round__72.0__ |
add__125.0__270.0__ add__2.5__3.0__ round__72.0__ |
| a car travelling with num__0.666666666667 km of its actual speed covers num__12 km in num__2 hr num__14 min num__28 sec find the actual speed of the car ? <o> a ) num__8.9 kmph <o> b ) num__2.96 kmph <o> c ) num__1.09 kmph <o> d ) num__45.9 kmph <o> e ) num__4.8 kmph |
time taken = num__2 hr num__14 min num__28 sec = num__16.46 hrs let the actual speed be x kmph then num__0.666666666667 x * num__16.46 = num__12 x = = num__1.09 kmph answer ( c ) <eor> c <eos> |
c |
round__1.09__ |
round__1.09__ |
| two alloys a and b are composed of two basic elements . the ratios of the compositions of the two basic elements in the two alloys are num__5 : num__3 and num__1 : num__1 respectively . a new alloy x is formed by mixing the two alloys a and b in the ratio num__4 : num__3 . what is the ratio of the composition of the two basic elements in alloy x ? <o> a ) num__1 : num__1 <o> b ) num__2 : num__3 <o> c ) num__5 : num__2 <o> d ) num__4 : num__3 <o> e ) num__7 : num__9 |
mixture a has a total of num__5 + num__3 = num__8 parts . if in the final mixture this represents num__4 parts then the total number of parts in mixture b should be ( num__2.0 ) * num__3 = num__6 . so we should take of mixture b a quantity with num__3 and num__3 parts respectively . this will give us in the final mixture ( num__5 + num__3 ) : ( num__3 + num__3 ) which means num__4 : num__3 answer d . <eor> d <eos> |
d |
add__5.0__3.0__ subtract__5.0__3.0__ add__5.0__1.0__ subtract__5.0__1.0__ |
add__5.0__3.0__ subtract__5.0__3.0__ add__5.0__1.0__ add__3.0__1.0__ |
| a certain college party is attended by both male and female students . the ratio of male to female students is num__3 to num__5 . if num__6 of the male students were to leave the party the ratio would change to num__1 to num__2 . how many total students are at the party ? <o> a ) num__64 <o> b ) num__72 <o> c ) num__80 <o> d ) num__88 <o> e ) num__96 |
the ratio is num__3 : num__5 = num__6 : num__10 so there are num__6 k males and num__10 k females . if num__6 males left the ratio would be num__1 : num__2 = num__5 : num__10 so there would be num__5 k males and num__10 k females . num__6 k - num__5 k = num__6 k = num__6 num__6 k + num__10 k = num__36 + num__60 = num__96 the answer is e . <eor> e <eos> |
e |
multiply__5.0__2.0__ multiply__6.0__10.0__ add__36.0__60.0__ multiply__1.0__96.0__ |
multiply__5.0__2.0__ multiply__6.0__10.0__ add__36.0__60.0__ add__36.0__60.0__ |
| a man divides $ num__8600 among num__4 sons num__4 daughters and num__2 nephews . if each daughter receives four times as much as each nephews and each son receives five times as much as each nephews how much does each daughter receive ? <o> a ) a ) $ num__200 <o> b ) b ) $ num__1000 <o> c ) c ) $ num__800 <o> d ) d ) $ num__1011.76 <o> e ) e ) $ num__400 |
let the share of each nephews be $ x . then share of each daughter = $ num__4 x share of each son = $ num__5 x . so num__4 * num__4 x + num__4 * num__4 x + num__2 * x = num__8600 num__16 x + num__16 x + num__2 x = num__8600 num__34 x = num__8600 x = num__252.94 . daughter receives four times of nephew so num__4 * num__252.94 = num__1011.76 . so each daughter receives $ num__800 . answer is option d ) $ num__1011.76 . <eor> d <eos> |
d |
multiply__4.0__252.94__ multiply__4.0__252.94__ |
multiply__4.0__252.94__ multiply__4.0__252.94__ |
| the area of a rectangle is num__15 square centimeters and the perimeter is num__16 square centimeters . what are the dimensions of the rectangle ? <o> a ) num__2 & num__4 <o> b ) num__3 & num__5 <o> c ) num__4 & num__6 <o> d ) num__5 & num__7 <o> e ) num__6 & num__8 |
let x and y be the length and width of the rectangle . using the formulas for the area and the perimeter we can write two equations . num__15 = x y and num__16 = num__2 x + num__2 y solve the second equation for x x = num__8 - y substitute x in the equation num__15 = x y by num__8 - y to rewrite the equation as num__15 = ( num__8 - y ) y solve for y to find y = num__3 and y = num__5 use x = num__8 - y to find x when y = num__3 x = num__5 and when y = num__5 x = num__3 . the dimensions of the rectangle are num__3 and num__5 . correct answer b <eor> b <eos> |
b |
square_perimeter__2.0__ triangle_area__2.0__3.0__ |
square_perimeter__2.0__ triangle_area__2.0__3.0__ |
| how many ways are there to split a group of num__4 students into two groups of num__2 students each ? ( the order of the groups does not matter ) <o> a ) num__2 <o> b ) num__3 <o> c ) num__4 <o> d ) num__5 <o> e ) num__6 |
num__4 c num__2 = num__6 if we consider these groups each group will be counted twice . the number of ways to choose num__2 groups of num__2 is num__3.0 = num__3 the answer is b . <eor> b <eos> |
b |
die_space__ choose__3.0__2.0__ |
die_space__ choose__3.0__2.0__ |
| if the sum of the num__4 th term and the num__12 th term of an arithmetic progression is num__12 what is the sum of the first num__15 terms of the progression ? <o> a ) num__90 <o> b ) num__80 <o> c ) num__70 <o> d ) num__60 <o> e ) num__50 |
num__4 th term + num__12 th term = num__12 i . e . ( a + num__3 d ) + ( a + num__11 d ) = num__12 now sum of first num__15 terms = ( num__7.5 ) * [ num__2 a + ( num__15 - num__1 ) d ] = ( num__7.5 ) * [ num__2 a + num__14 d ] = ( num__7.5 ) * num__12 - - - - - - - - - - - - - - - from ( num__1 ) = num__90 answer : a <eor> a <eos> |
a |
divide__12.0__4.0__ subtract__15.0__4.0__ divide__15.0__7.5__ subtract__4.0__3.0__ add__12.0__2.0__ multiply__12.0__7.5__ multiply__12.0__7.5__ |
subtract__15.0__12.0__ subtract__15.0__4.0__ divide__15.0__7.5__ subtract__4.0__3.0__ add__12.0__2.0__ multiply__12.0__7.5__ multiply__12.0__7.5__ |
| if r and p are two different prime numbers which of the following is the smallest possible value of r + p ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__5 <o> d ) num__8 <o> e ) num__10 |
by definition only positive numbers can be primes . two smallest primes are num__2 and num__3 - - > num__2 + num__3 = num__5 . answer : c <eor> c <eos> |
c |
add__2.0__3.0__ add__2.0__3.0__ |
add__2.0__3.0__ add__2.0__3.0__ |
| a man rows his boat num__70 km downstream and num__45 km upstream taking num__2 num__0.5 hours each time . find the speed of the stream ? <o> a ) num__1 kmph <o> b ) num__6 kmph <o> c ) num__5 kmph <o> d ) num__8 kmph <o> e ) num__7 kmph |
speed downstream = d / t = num__75 / ( num__2 num__0.5 ) = num__28 kmph speed upstream = d / t = num__45 / ( num__2 num__0.5 ) = num__18 kmph the speed of the stream = ( num__28 - num__18 ) / num__2 = num__5 kmph answer : c <eor> c <eos> |
c |
subtract__75.0__70.0__ round__5.0__ |
subtract__75.0__70.0__ subtract__75.0__70.0__ |
| a sum of rs . num__2500 amounts to rs . num__3500 in num__4 years at the rate of simple interest . what is the rate of interest ? <o> a ) num__7.0 <o> b ) num__9.0 <o> c ) num__5.0 <o> d ) num__6.0 <o> e ) num__10 % |
s . i . = ( num__3500 - num__2500 ) = rs . num__1000 rate = ( num__100 * num__1000 ) / ( num__2500 * num__4 ) = num__10.0 answer : e <eor> e <eos> |
e |
percent__4.0__2500.0__ percent__10.0__100.0__ |
percent__4.0__2500.0__ percent__10.0__100.0__ |
| the h . c . f of two numbers is num__19 and their l . c . m is num__4263 . if one of the numbers is num__399 then the other is ? <o> a ) num__202 <o> b ) num__203 <o> c ) num__204 <o> d ) num__205 <o> e ) num__206 |
other number = ( num__19 * num__4263 ) / num__399 = num__203 . answer : b <eor> b <eos> |
b |
gcd__4263.0__203.0__ |
gcd__4263.0__203.0__ |
| three years ago the average age of a family of seven members was num__20 years . a boy have been born the average age of the family is the same today . what is the age of the boy ? <o> a ) a ) num__8 <o> b ) b ) num__7 <o> c ) c ) num__6 <o> d ) d ) num__5 <o> e ) e ) num__4 |
num__7 * num__23 = num__161 num__7 * num__19 = num__153 - - - - - - - - - - - - - - num__8 answer : a <eor> a <eos> |
a |
multiply__7.0__23.0__ subtract__161.0__153.0__ subtract__161.0__153.0__ |
multiply__7.0__23.0__ subtract__161.0__153.0__ subtract__161.0__153.0__ |
| the value of ( num__68.237 ) num__2 – ( num__31.763 ) num__2 is : <o> a ) num__3.6474 <o> b ) num__36.474 <o> c ) num__364.74 <o> d ) num__3647.4 <o> e ) none of these |
solution given expression = ( a num__2 - b num__2 ) = ( a + b ) ( a - b ) = ( num__68.237 + num__31.763 ) ( num__68.237 - num__31.763 ) = ( num__100 × num__36.474 ) = num__3647.4 . answer d <eor> d <eos> |
d |
add__68.237__31.763__ subtract__68.237__31.763__ multiply__36.474__100.0__ multiply__36.474__100.0__ |
add__68.237__31.763__ subtract__68.237__31.763__ multiply__36.474__100.0__ multiply__36.474__100.0__ |
| the price of a mobile was increased by num__40 percent . the new price was then decreased by num__15 percent . a single increase of what percent is equivalent to these two successive changes ? <o> a ) num__25.0 <o> b ) num__19.0 <o> c ) num__30.0 <o> d ) num__22.5 <o> e ) num__15 % |
consider base price - $ num__100 num__25.0 increase = num__1.40 * num__100 = $ num__140 then a num__15.0 decrease on new price = num__0.85 * num__140 = $ num__119 so final price of radio - $ num__119 therefore a num__19.0 increase correct option - b <eor> b <eos> |
b |
subtract__40.0__15.0__ add__40.0__100.0__ multiply__140.0__0.85__ subtract__119.0__100.0__ subtract__119.0__100.0__ |
subtract__40.0__15.0__ multiply__100.0__1.4__ multiply__140.0__0.85__ subtract__119.0__100.0__ subtract__119.0__100.0__ |
| two employees x and y are paid a total of rs . num__550 per week by their employer . if x is paid num__120 percent of the sum paid to y how much is y paid per week ? <o> a ) a ) rs . num__200 <o> b ) b ) rs . num__220 <o> c ) c ) rs . num__250 <o> d ) d ) rs . num__300 <o> e ) e ) rs . num__350 |
let the amount paid to x per week = x and the amount paid to y per week = y then x + y = num__550 but x = num__120.0 of y = num__120 y / num__100 = num__12 y / num__10 ∴ num__12 y / num__10 + y = num__550 ⇒ y [ num__1.2 + num__1 ] = num__550 ⇒ num__22 y / num__10 = num__550 ⇒ num__22 y = num__5500 ⇒ y = num__250.0 = num__250.0 = rs . num__250 c <eor> c <eos> |
c |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__10.0__12.0__ multiply__550.0__10.0__ divide__5500.0__22.0__ multiply__1.0__250.0__ |
divide__120.0__12.0__ divide__120.0__100.0__ round_down__1.2__ add__10.0__12.0__ multiply__550.0__10.0__ divide__5500.0__22.0__ divide__250.0__1.0__ |
| when the positive integer y is divided by num__11 the quotient is z and the remainder num__3 . when y is divided by num__19 the remainder is also num__3 . what is the remainder when z is divided by num__19 ? <o> a ) num__4 <o> b ) num__3 <o> c ) num__2 <o> d ) num__1 <o> e ) num__0 |
any number which when divided by divisor d num__1 d num__2 etc . leaving same remainderrtakes the form ofk + r where k = lcm ( d num__1 d num__2 ) in this case the divisors are num__1119 and remainder is num__3 . so lcm ( num__1119 ) = num__209 so n = num__209 + num__3 = num__212 also y = d num__1 q + num__3 ; which means d num__1 q = num__209 d num__1 = num__11 therefore q = num__19 and ( z divided by num__19 ) num__1.0 leaves remainder num__0 . answer is e <eor> e <eos> |
e |
subtract__3.0__1.0__ multiply__11.0__19.0__ add__3.0__209.0__ multiply__11.0__0.0__ |
subtract__3.0__1.0__ multiply__11.0__19.0__ add__3.0__209.0__ multiply__11.0__0.0__ |
| if a b is num__20.0 num__30.0 smaller than c then how much percentage is b greater than a ? <o> a ) num__10.0 <o> b ) num__11.0 <o> c ) num__12 num__0.75 % <o> d ) num__12 num__0.5 % <o> e ) num__13 % |
a = num__80.0 b = num__70.0 c = num__100.0 a - b / a × num__100 = num__0.125 × num__100 = num__12 num__0.5 % greater than a d ) <eor> d <eos> |
d |
percent__100.0__12.0__ |
percent__100.0__12.0__ |
| what is the num__149 th digit to the right of the decimal point in the decimal form of num__0.909090909091 ? <o> a ) num__5 <o> b ) num__6 <o> c ) num__9 <o> d ) num__0 <o> e ) num__1 |
to solve we first have to use some long division . long division allows us to get num__0.909090909091 in decimal form which is num__0.909090 … where “ num__90 ” is repeating . we see that the num__1 st num__3 rd num__5 th digit to the right of the decimal point is a num__9 and that the num__2 nd num__4 th num__6 th digit to the right of the decimal point is a num__0 . in other words each odd - positioned digit is a num__9 and each even - positioned digit is a num__0 . then the num__149 th digit to the right of the decimal point and we see that num__149 is odd then num__149 th digit is a num__9 . answer c . <eor> c <eos> |
c |
subtract__3.0__1.0__ add__1.0__3.0__ add__1.0__5.0__ round_down__0.9091__ multiply__1.0__9.0__ |
subtract__3.0__1.0__ subtract__5.0__1.0__ subtract__9.0__3.0__ round_down__0.9091__ multiply__1.0__9.0__ |
| a b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = num__2 : num__3 and b : c = num__2 : num__5 . if the total runs scored by all of them are num__75 the runs scored by b are ? <o> a ) num__33 <o> b ) num__18 <o> c ) num__99 <o> d ) num__787 <o> e ) num__65 |
a : b = num__2 : num__3 b : c = num__2 : num__5 a : b : c = num__4 : num__6 : num__15 num__0.24 * num__75 = num__18 answer : b <eor> b <eos> |
b |
multiply__2.0__3.0__ multiply__3.0__5.0__ multiply__3.0__6.0__ multiply__3.0__6.0__ |
multiply__2.0__3.0__ multiply__3.0__5.0__ multiply__3.0__6.0__ multiply__3.0__6.0__ |
| the weight of a hollow sphere is directly dependent on its surface area . the surface area of a sphere is num__4 π · r ^ num__2 where r is the radius of the sphere . if a hollow sphere of radius num__0.15 cm made of a certain metal weighs num__8 grams a hollow sphere of radius num__0.3 cm made of the same metal would weigh how many q grams ? <o> a ) q = num__16 <o> b ) q = num__32 <o> c ) num__64 <o> d ) num__128 <o> e ) num__512 |
weight directly proportional to num__4 pi r ^ num__2 now num__4 pi is constant so weight is directly proportional to r ^ num__2 . when radius = num__0.15 weight = num__8 so ( num__0.15 ) ^ num__2 proportional to num__8 ; ( num__0.15 ) ^ num__2 * num__4 proportional to num__8 * num__4 solving further ( num__0.15 ) ^ num__2 * num__2 ^ num__2 = ( num__0.15 * num__2 ) ^ num__2 = num__0.3 ^ num__2 ; so answer = num__32 ( b ) <eor> b <eos> |
b |
square_perimeter__8.0__ square_perimeter__8.0__ |
multiply__4.0__8.0__ multiply__4.0__8.0__ |
| two trains running in the same direction with a speed of num__60 kph and num__50 kph having lengths of num__200 m each . how long it will take the trains to completely pass each other ( in seconds ) ? <o> a ) num__13.09 <o> b ) num__32.17 <o> c ) num__54.9 <o> d ) num__112.98 <o> e ) num__143.88 |
relative speed of trains with respect to each other = num__60 - num__50 = num__10 kph = num__2.78 m / s distance need to be covered = num__200 + num__200 = num__400 time taken to pass each other = num__400 / num__2.78 = num__143.88 s answer : e <eor> e <eos> |
e |
subtract__60.0__50.0__ round__143.88__ |
subtract__60.0__50.0__ round__143.88__ |
| a sum is divided among b c and d in such a way that for each rupee b gets c gets num__150 paisa and d gets num__50 paisa . if the share of c is rs . num__40 what is the total amount ? <o> a ) num__70 <o> b ) num__75 <o> c ) num__80 <o> d ) num__85 <o> e ) num__90 |
b : c : d = num__100 : num__150 : num__100 num__20 : num__30 : num__10 num__30 - - - num__40 num__60 - - - ? = > num__80 answer : c <eor> c <eos> |
c |
subtract__150.0__50.0__ subtract__50.0__20.0__ subtract__50.0__40.0__ add__50.0__10.0__ add__50.0__30.0__ add__50.0__30.0__ |
subtract__150.0__50.0__ subtract__50.0__20.0__ subtract__50.0__40.0__ subtract__100.0__40.0__ subtract__100.0__20.0__ subtract__100.0__20.0__ |
| how many integers from num__100 to num__800 inclusive remains the value unchanged when the digits were reversed ? <o> a ) num__50 <o> b ) num__60 <o> c ) num__80 <o> d ) num__70 <o> e ) num__90 |
question is asking for palindrome first digit possibilities - num__1 through num__7 = num__7 num__8 is not possible here because it would result in a number greater than num__8 ( i . e num__808 num__818 . . ) second digit possibilities - num__0 though num__9 = num__10 third digit is same as first digit = > total possible number meeting the given conditions = num__7 * num__10 = num__70 answer is d . <eor> d <eos> |
d |
divide__800.0__100.0__ add__800.0__8.0__ add__1.0__8.0__ add__1.0__9.0__ multiply__7.0__10.0__ multiply__1.0__70.0__ |
divide__800.0__100.0__ add__800.0__8.0__ add__1.0__8.0__ subtract__818.0__808.0__ multiply__7.0__10.0__ multiply__1.0__70.0__ |
| the product of two numbers is num__9375 and the quotient when the larger one is divided by the smaller is num__15 . the sum of the numbers is : <o> a ) num__380 <o> b ) num__395 <o> c ) num__400 <o> d ) num__425 <o> e ) either ( a ) or ( b ) |
let the numbers be x and y . then xy = num__9375 and x = num__15 . y xy = num__9375 ( x / y ) num__15 y num__2 = num__625 . y = num__25 . x = num__15 y = ( num__15 x num__25 ) = num__375 . sum of the numbers = x + y = num__375 + num__25 = num__400 . answer : c <eor> c <eos> |
c |
divide__9375.0__15.0__ divide__9375.0__25.0__ add__375.0__25.0__ add__375.0__25.0__ |
divide__9375.0__15.0__ divide__9375.0__25.0__ add__375.0__25.0__ add__375.0__25.0__ |
| set a = { num__1 num__2 num__3 num__4 num__5 num__6 y } which of the following possible values for y would cause set a to have the smallest standard deviation ? <o> a ) num__1 <o> b ) num__2.5 <o> c ) num__3 <o> d ) num__3.5 <o> e ) num__7 |
sd is minimum or least when the number is closest to the mean of the existing numbers calculating mean of the rest of the numbers gives mean = ( approx ) num__3.6 so the new number should be near to the mean and only number closest to the mean is num__3.5 answer : option d <eor> d <eos> |
d |
multiply__1.0__3.5__ |
multiply__1.0__3.5__ |
| num__35.0 of the employees of a company are men . num__60.0 of the men in the company speak french and num__40.0 of the employees of the company speak french . what is % of the women in the company who do not speak french ? <o> a ) num__4.0 <o> b ) num__10.0 <o> c ) num__96.0 <o> d ) num__90.12 <o> e ) num__70.77 % |
no of employees = num__100 ( say ) men = num__35 women = num__65 men speaking french = num__0.6 * num__35 = num__21 employees speaking french = num__0.4 * num__100 = num__40 therefore women speaking french = num__40 - num__21 = num__19 and women not speaking french = num__65 - num__19 = num__46.0 of women speaking french = num__0.707692307692 * num__100 = num__70.77 answer e <eor> e <eos> |
e |
percent__35.0__60.0__ percent__100.0__70.77__ |
percent__35.0__60.0__ percent__100.0__70.77__ |
| a builder decided to build a farm house in num__40 days . he employed num__100 men in the beginning and num__100 more after num__35 days and completed the construction in stipulated time . if he had not employed the additional men how many days behind schedule would it have been finished ? <o> a ) num__5 <o> b ) num__7 <o> c ) num__10 <o> d ) num__15 <o> e ) num__12 |
num__200 men do the rest of the work in num__40 - num__35 = num__5 days num__100 men can do the rest of the work in num__5 * num__2.0 = num__10 days required number of days = num__10 - num__5 = num__5 days answer is a <eor> a <eos> |
a |
subtract__40.0__35.0__ divide__200.0__100.0__ multiply__2.0__5.0__ subtract__40.0__35.0__ |
subtract__40.0__35.0__ divide__200.0__100.0__ multiply__2.0__5.0__ subtract__40.0__35.0__ |
| how many diagonals does a polygon with num__15 sides have if one of its vertices does not connect to any diagonal ? <o> a ) num__80 <o> b ) num__90 <o> c ) num__77 <o> d ) num__88 <o> e ) num__99 |
if i calculate it using the formulae # diagonals = n ( n - num__3 ) / num__2 each vertex sends of n - num__3 diagonals n = num__15 - num__1 then num__14 * ( num__14 - num__3 ) / num__2 = num__77 correct option : c <eor> c <eos> |
c |
multiply__1.0__77.0__ |
multiply__1.0__77.0__ |
| what is the least value of x . so that num__1894 x is divisible by num__3 ? <o> a ) num__2 <o> b ) num__5 <o> c ) num__1 <o> d ) num__6 <o> e ) num__7 |
the sum of the digits of the number is divisible by num__3 then the number is divisible by num__3 . num__1 + num__8 + num__9 + num__4 + x = num__22 + x least value of x may be num__2 therefore num__22 + num__2 = num__24 is divisible by num__3 . a <eor> a <eos> |
a |
add__1.0__8.0__ add__3.0__1.0__ subtract__3.0__1.0__ multiply__3.0__8.0__ subtract__3.0__1.0__ |
add__1.0__8.0__ add__3.0__1.0__ subtract__3.0__1.0__ add__2.0__22.0__ subtract__3.0__1.0__ |
| num__9.4 num__9.9 num__9.9 num__9.9 num__10.0 num__10.2 num__10.2 num__10.5 the mean and the standard deviation of the num__8 numbers shown above is num__10 and num__0.3 respectively . what percent of the num__8 numbers are within num__1 standard deviation of the mean ? <o> a ) num__90.0 <o> b ) num__85.0 <o> c ) num__80.0 <o> d ) num__75.0 <o> e ) num__70 % |
d . num__75.0 <eor> d <eos> |
d |
multiply__1.0__75.0__ |
multiply__1.0__75.0__ |
| what is the least number by which num__15625 must be divided to get a number perfect square <o> a ) num__21 <o> b ) num__25 <o> c ) num__24 <o> d ) num__40 <o> e ) num__45 |
num__15 |
625 / num__25 = num__625 num__625 = ( num__25 ) ^ num__2 ans - num__25 answer : b <eor> b <eos> |
b |
b |
| calculate the area of a triangle if the sides are num__30 cm num__21 cm and num__10 cm what is its area ? <o> a ) num__145 cm num__2 <o> b ) num__105 cm num__2 <o> c ) num__125 cm num__2 <o> d ) num__115 cm num__2 <o> e ) num__135 cm num__2 |
the triangle with sides num__30 cm num__21 cm and num__10 cm is right angled where the hypotenuse is num__30 cm . area of the triangle = num__0.5 * num__21 * num__10 = num__105 cm num__2 answer : b <eor> b <eos> |
b |
triangle_area__21.0__10.0__ square_perimeter__0.5__ triangle_area__21.0__10.0__ |
volume_rectangular_prism__21.0__10.0__0.5__ square_perimeter__0.5__ volume_rectangular_prism__21.0__10.0__0.5__ |
| a train num__60 m long is running with a speed of num__60 km / hr . in what time will it pass a man who is running at num__6 km / hr in the direction opposite to that in which the train is going ? <o> a ) num__7 <o> b ) num__6 <o> c ) num__8 <o> d ) num__2 <o> e ) num__3 |
speed of train relative to man = num__60 + num__6 = num__66 km / hr . = num__66 * num__0.277777777778 = num__18.3333333333 m / sec . time taken to pass the men = num__60 * num__0.0545454545455 = num__3 sec . answer e <eor> e <eos> |
e |
add__60.0__6.0__ round__3.0__ |
add__60.0__6.0__ round__3.0__ |
| a pizza which is thin is cut into ' x ' pieces by num__4 straight lines then the lowest possible value of x is ? <o> a ) num__3 <o> b ) num__4 <o> c ) num__5 <o> d ) num__6 <o> e ) num__7 |
num__4 parallel lines will give minimum number of pieces . num__5 pieces . answer : c <eor> c <eos> |
c |
vowel_space__ vowel_space__ |
vowel_space__ vowel_space__ |
| a survey was sent to num__70 customers num__7 of whom responded . then the survey was redesigned and sent to another num__63 customers num__9 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? <o> a ) num__2.0 <o> b ) num__6.0 <o> c ) num__4.0 <o> d ) num__28.0 <o> e ) num__63 % |
case num__1 : ( num__0.1 ) = x / num__100 x = num__10.0 case num__2 : ( num__0.142857142857 ) = y / num__100 y = num__14.0 so percent increase is = ( y - x ) = ( num__14 - num__10 ) % = num__4.0 answer is c <eor> c <eos> |
c |
percent__100.0__4.0__ |
percent__100.0__4.0__ |
| a box contains nine bulbs out of which num__4 are defective . if four bulbs are chosen at random find the probability that atleast one bulb is good . <o> a ) num__0.748502994012 <o> b ) num__0.672043010753 <o> c ) num__0.992063492063 <o> d ) num__0.672043010753 <o> e ) num__0.9765625 |
required probability = num__1 - num__0.00793650793651 = num__0.992063492063 answer : c <eor> c <eos> |
c |
negate_prob__0.0079__ negate_prob__0.0079__ |
negate_prob__0.0079__ negate_prob__0.0079__ |
| if there are num__210 workers in a factory and on a certain day num__198 were present . calculate the percentage that showed up for work ? ( round to the nearest tenth ) . <o> a ) num__94.3 <o> b ) num__95.3 <o> c ) num__93.3 <o> d ) num__92.3 <o> e ) num__91.3 % |
num__0.942857142857 * num__100 = num__94.28 num__94.3 correct answer a <eor> a <eos> |
a |
divide__198.0__210.0__ round__94.3__ |
divide__198.0__210.0__ round__94.3__ |
| a pentagon marked p q r s and t on its five sides . five colors black brown blue violet and white are used to paint the five sides of the pentagon . if the adjacent faces are painted with the different colors in how many ways can the pentagon be painted ? <o> a ) num__5 <o> b ) num__9 <o> c ) num__12 <o> d ) num__6 <o> e ) num__3 |
if the base of the pentagon is black then in order the adjacent faces to be painted with the different colors . num__4 side faces can be painted in brown - white - violet - blue brown - white - violet - blue or white - blue - violet - brown white - blue - violet - brown . but we can have the base painted in either of the four colors thus the total number of ways to paint the pentagon is num__5 * num__1 = num__5 . answer : a . <eor> a <eos> |
a |
vowel_space__ vowel_space__ |
vowel_space__ vowel_space__ |
| convert the num__0.361111111111 m / s into kilometers per hour ? <o> a ) num__1.6 kmph <o> b ) num__2.3 kmph <o> c ) num__1.3 kmph <o> d ) num__1.7 kmph <o> e ) num__3.3 kmph |
num__0.361111111111 m / s = num__0.361111111111 * num__3.6 = num__1.3 = num__1.3 kmph . answer : c <eor> c <eos> |
c |
multiply__0.3611__3.6__ round__1.3__ |
multiply__0.3611__3.6__ multiply__0.3611__3.6__ |
| there is a sequence such that each term is positive integer and each digit of the terms in the sequence has num__3 to be ordered what is the value k of num__100 th term ? <o> a ) num__126 <o> b ) num__192 <o> c ) num__232 <o> d ) num__252 <o> e ) num__342 |
( num__1 ~ num__99 ) num__19 ( num__100 ~ num__199 ) num__19 ( num__200 ~ num__299 ) num__19 . the num__100 th term should be num__100 - ( num__19 + num__19 + num__19 ) = num__43 th number . hence the correct answer k is num__342 . the correct answer is e . <eor> e <eos> |
e |
subtract__100.0__1.0__ add__100.0__99.0__ add__1.0__199.0__ add__100.0__199.0__ add__299.0__43.0__ multiply__1.0__342.0__ |
subtract__100.0__1.0__ add__100.0__99.0__ add__1.0__199.0__ add__100.0__199.0__ add__299.0__43.0__ add__299.0__43.0__ |
| in the game of dubblefud red chips blue chips and green chips are each worth num__2 num__4 and num__5 points respectively . in a certain selection of chips the product of the point values of the chips is num__12000 . if the number of blue chips in this selection equals the number of green chips how many red chips are in the selection ? <o> a ) num__1 <o> b ) num__2 <o> c ) num__3 <o> d ) num__4 <o> e ) num__5 |
this is equivalent to : - num__2 x * num__4 y * num__5 z = num__12000 y = z ( given ) num__2 x * num__4 y * num__5 y = num__12000 num__2 x * y ^ num__2 = num__600.0 num__2 x * y ^ num__2 = num__600 now from options given we will figure out which number will divide num__800 and gives us a perfect square : - which gives us x = num__3 as num__2 * num__3 * y ^ num__2 = num__600 y ^ num__2 = num__100 y = num__10 number of red chips = num__3 hence c <eor> c <eos> |
c |
subtract__5.0__2.0__ multiply__2.0__5.0__ subtract__5.0__2.0__ |
subtract__5.0__2.0__ multiply__2.0__5.0__ subtract__5.0__2.0__ |
| a train passes a station platform in num__36 sec and a man standing on the platform in num__24 sec . if the speed of the train is num__54 km / hr . what is the length of the platform ? <o> a ) num__288 <o> b ) num__180 <o> c ) num__881 <o> d ) num__1277 <o> e ) num__121 |
speed = num__54 * num__0.277777777778 = num__15 m / sec . length of the train = num__15 * num__24 = num__360 m . let the length of the platform be x m . then ( x + num__360 ) / num__36 = num__15 = > x = num__180 m . answer : b <eor> b <eos> |
b |
multiply__24.0__15.0__ round__180.0__ |
multiply__24.0__15.0__ round__180.0__ |
| find the value of ( num__950 + num__0.244444444444 ) × num__900 <o> a ) num__854542 <o> b ) num__856945 <o> c ) num__758965 <o> d ) num__855220 <o> e ) num__826450 |
( num__855000 + num__220 ) / num__900 * num__900 = num__855000 + num__220 = num__855220 answer : d <eor> d <eos> |
d |
multiply__950.0__900.0__ add__855000.0__220.0__ add__855000.0__220.0__ |
multiply__950.0__900.0__ add__855000.0__220.0__ add__855000.0__220.0__ |
| the ratio q by volume of soap to alcohol to water in a num__76 litre solution is num__2 : num__50 : num__100 . the solution is then altered by adding more soap alcohol and water . after this alteration the ratio by volume of soap to water in the solution doubles whereas the ratio by volume of soap to water remains the same as before . what could be the resulting volume . <o> a ) num__79 litres <o> b ) num__78 litres <o> c ) num__77 litres <o> d ) num__152 liters <o> e ) num__304 litres |
i guess it should be the ratio q by volume ofsoaptowaterin the solutiondoubleswhereas the ratio by volume ofalocoholtowaterremains thesameas before num__2 : num__50 : num__100 = > num__1 : num__25 : num__50 . if we add all the parts we get num__76 liters so we have num__1 liters of soap num__25 liters of alcohol and num__50 liters of water . now as per the question soap : water doubles but alcohol to water remains the same . so soap becomes num__2 liters alcohol remains num__25 liters and water remains at num__50 liters . hence num__77 liters - option c ) <eor> c <eos> |
c |
divide__50.0__2.0__ add__76.0__1.0__ add__76.0__1.0__ |
divide__50.0__2.0__ add__76.0__1.0__ add__76.0__1.0__ |
| a child must place num__4 different toys in num__5 different bins . if any of the toys can go in any of the bins in how many ways can the child place the toys into the bins ? <o> a ) num__5 ^ num__4 <o> b ) num__6 ^ num__4 <o> c ) num__4 ^ num__6 <o> d ) num__4 ^ num__4 <o> e ) num__6 ^ num__4 - num__4 ^ num__6 |
there are num__4 different bins and each of the toy can go to any of these bins . . so answer num__5 â ˆ — num__5 â ˆ — num__5 â ˆ — num__5 = num__5 ^ num__4 answer : a <eor> a <eos> |
a |
vowel_space__ |
vowel_space__ |
| a tradesman by means of his false balance defrauds to the extent of num__20.0 ? in buying goods as well as by selling the goods . what percent does he gain on his outlay ? <o> a ) num__48.0 <o> b ) num__24.0 <o> c ) num__44.0 <o> d ) num__46.0 <o> e ) num__84 % |
g % = num__20 + num__20 + ( num__20 * num__20 ) / num__100 = num__44.0 answer : c <eor> c <eos> |
c |
percent__100.0__44.0__ |
percent__100.0__44.0__ |
| rohan spends num__40.0 of his salary on food num__20.0 on house rent num__10.0 on entertainment and num__10.0 on conveyance . if his savings at the end of a month are rs . num__2000 . then his monthly salary is <o> a ) rs . num__10000 <o> b ) rs . num__6000 <o> c ) rs . num__8000 <o> d ) rs . num__4000 <o> e ) rs . num__2000 |
sol . saving = [ num__100 - ( num__40 + num__20 + num__10 + num__10 ] % = num__20.0 . let the monthly salary be rs . x . then num__20.0 of x = num__2000 â ‡ ” num__0.2 x = num__2000 â ‡ ” x = num__2000 Ã — num__5 = num__10000 . answer a <eor> a <eos> |
a |
percent__100.0__10000.0__ |
percent__100.0__10000.0__ |
| the operation is defined for all integers a and b by the equation ab = ( a - num__1 ) ( b - num__1 ) . if x num__20 = num__190 what is the value of x ? <o> a ) num__10 <o> b ) num__12 <o> c ) num__11 <o> d ) num__13 <o> e ) num__14 |
ab = ( a - num__1 ) ( b - num__1 ) x num__20 = ( x - num__1 ) ( num__20 - num__1 ) = num__190 - - > x - num__1 = num__10 - - > x = num__11 answer : c <eor> c <eos> |
c |
add__1.0__10.0__ add__1.0__10.0__ |
add__1.0__10.0__ add__1.0__10.0__ |
| an uneducated retailer marks all his goods at num__50.0 above the cost price and thinking that he will still make num__15.0 profit offers a discount of num__15.0 on the marked price . what is his actual profit on the sales ? <o> a ) num__12.5 <o> b ) num__27.5 <o> c ) num__14.0 <o> d ) num__14.5 <o> e ) none |
sol . let c . p . = rs . num__100 . then marked price = rs . num__150 . s . p . = num__85.0 of rs . num__150 = rs . num__127.5 . ∴ gain % = num__27.5 . answer b <eor> b <eos> |
b |
percent__85.0__150.0__ percent__27.5__100.0__ |
percent__85.0__150.0__ percent__27.5__100.0__ |
| julie had num__90 currency notes in all some of which are of rs num__80 denomination and the remaining of rs num__50 denomination . the total amount of all these currency notes was rs . num__6000 . how much amount ( in rs ) did she have in the denomination of rs num__50 ? <o> a ) num__56 <o> b ) num__35 <o> c ) num__30 <o> d ) num__45 <o> e ) num__40 |
let the number of num__50 - rupee notes = x then the number of num__80 - rupee notes = ( num__90 – x ) num__50 x + num__80 ( num__90 – x ) = num__6000 : x = num__40 answer : e <eor> e <eos> |
e |
subtract__90.0__50.0__ subtract__90.0__50.0__ |
subtract__90.0__50.0__ subtract__90.0__50.0__ |
| john bought bicycles at num__6 for rs num__10000 . what is the profit or loss percentage after selling bicycles at num__5 for rs num__10000 . <o> a ) num__30.0 <o> b ) num__50.0 <o> c ) num__20.0 <o> d ) num__60.0 <o> e ) none |
c . p . of num__6 bicycles = rs num__10000 c . p . of num__1 bicycle = rs num__1666.66666667 = rs num__1666.66666667 s . p . of num__5 bicycles = rs num__10000 s . p of num__1 bicycle = rs num__2000 therefore profit percentage = { ( num__2000 - num__1666.66666667 ) / ( num__1666.66666667 ) } num__100 = num__20.0 answer : option c <eor> c <eos> |
c |
percent__6.0__1666.6667__ percent__1.0__2000.0__ percent__1.0__2000.0__ |
percent__6.0__1666.6667__ percent__1.0__2000.0__ percent__1.0__2000.0__ |
| a container contains num__40 lit of milk . from this container num__4 lit of milk was taken out and replaced by water . this process was repeated further two times . how much milk is now contained by the container ? <o> a ) num__19.16 lit <o> b ) num__29.16 lit <o> c ) num__9.16 lit <o> d ) num__26.16 lit <o> e ) num__16.6 lit |
amount of milk left after num__3 operations = num__40 ( num__1 - num__0.1 ) num__3 lit = ( num__40 * num__0.9 * num__0.9 * num__0.9 ) = num__29.16 lit answer : b num__29.16 lit <eor> b <eos> |
b |
subtract__4.0__3.0__ divide__4.0__40.0__ subtract__1.0__0.1__ multiply__1.0__29.16__ |
subtract__4.0__3.0__ divide__4.0__40.0__ subtract__1.0__0.1__ multiply__1.0__29.16__ |
| if x % of y is num__100 and y % of z is num__200 then find the relation between x and z . <o> a ) z = x <o> b ) num__2 z = x <o> c ) z = num__2 x <o> d ) none of above <o> e ) z = num__2 y |
explanation : it is y % of z = num__2 ( x % of y ) = > yz / num__100 = num__2 xy / num__100 = > z = num__2 x option c <eor> c <eos> |
c |
percent__100.0__2.0__ |
percent__100.0__2.0__ |
| a trader bought a car at num__10.0 discount on its original price . he sold it at a num__40.0 increase on the price he bought it . what percent of profit did he make on the original price ? <o> a ) num__118 <o> b ) num__110 <o> c ) num__112 <o> d ) num__126 <o> e ) num__115 |
original price = num__100 cp = num__80 s = num__80 * ( num__1.4 ) = num__126 num__100 - num__126 = num__26.0 answer : d <eor> d <eos> |
d |
percent__100.0__126.0__ |
percent__100.0__126.0__ |
| the ratio of the volumes of two cubes is num__729 : num__343 . what is the ratio of their total surface areas ? <o> a ) num__81 : num__49 <o> b ) num__81 : num__122 <o> c ) num__81 : num__124 <o> d ) num__81 : num__126 <o> e ) num__81 : num__129 |
ratio of the sides = num__3 √ num__729 : num__3 √ num__343 = num__9 : num__7 ratio of surface areas = num__9 ^ num__2 : num__7 ^ num__2 = num__81 : num__49 answer : option a <eor> a <eos> |
a |
power__9.0__2.0__ power__7.0__2.0__ triangle_area__2.0__81.0__ |
power__9.0__2.0__ power__7.0__2.0__ power__9.0__2.0__ |
| a jogger running at num__9 km / hr along side a railway track is num__240 m ahead of the engine of a num__150 m long train running at num__45 km / hr in the same direction . in how much time will the train pass the jogger ? <o> a ) num__39 sec <o> b ) num__87 sec <o> c ) num__36 sec <o> d ) num__16 sec <o> e ) num__11 sec |
speed of train relative to jogger = num__45 - num__9 = num__36 km / hr . = num__36 * num__0.277777777778 = num__10 m / sec . distance to be covered = num__240 + num__150 = num__390 m . time taken = num__39.0 = num__39 sec . answer : a <eor> a <eos> |
a |
subtract__45.0__9.0__ add__240.0__150.0__ divide__390.0__10.0__ divide__390.0__10.0__ |
subtract__45.0__9.0__ add__240.0__150.0__ divide__390.0__10.0__ divide__390.0__10.0__ |
| in a shower num__5 cm of rain falls . the volume of water that falls on num__1.5 hectares of ground is <o> a ) num__75 cu . m <o> b ) num__750 cu . m <o> c ) num__7500 cu . m <o> d ) num__75000 cu . m <o> e ) none of these |
solution area = ( num__1.5 × num__10000 ) m num__2 = num__45000 m num__2 . depth = m = m . ∴ volume = ( area × depth ) = m num__3 = num__750 m num__3 . answer b <eor> b <eos> |
b |
subtract__5.0__2.0__ round__750.0__ |
subtract__5.0__2.0__ round__750.0__ |
| if a point ( x y ) is randomly selected within the square shown in the figure above ( the vertices are on num__11 num__1 - num__1 - num__11 - num__1 - num__1 ) what are the q odds that x ^ num__2 + y ^ num__2 > num__1 ? <o> a ) num__1 . pi / num__4 <o> b ) num__2 . pi / num__16 <o> c ) num__3 . q = num__1 - pi / num__4 <o> d ) num__4 . q = num__1 - pi / num__16 <o> e ) num__5.4 - pi |
area of the square will be num__4 and the area of circle with center ( num__00 ) and radius num__1 will be pi . then succesful outcomes = num__4 - pi ( the area where x num__2 + y num__2 will be greater than num__1 ) total outcomes = num__4 therefore probability = ( num__4 - pi ) / num__4 c <eor> c <eos> |
c |
add__1.0__2.0__ |
add__1.0__2.0__ |
| the measures of the num__2 acute angles of a triangle are in the ratio of num__4 : num__6 . what arethe measures of the num__2 angles ? <o> a ) num__20 ° <o> b ) num__70 ° <o> c ) num__110 ° <o> d ) num__120 ° <o> e ) num__54 ° |
if the ratio of the two angles is num__4 : num__6 then the measures of two angles can be written as num__4 x and num__6 x . also the two acute angles of a triangle is equal to num__90 ° . hence num__4 x + num__6 x = num__90 num__10 x = num__90 x = num__9 measures of the two acute angles are num__4 x = num__4 × num__9 = num__36 ° num__6 x = num__6 × num__9 = num__54 ° e <eor> e <eos> |
e |
square_perimeter__9.0__ multiply__6.0__9.0__ triangle_area__2.0__54.0__ |
square_perimeter__9.0__ multiply__6.0__9.0__ triangle_area__2.0__54.0__ |
| find the value of y from ( num__12 ) ^ num__3 x num__6 ^ num__4 ÷ num__432 = y ? <o> a ) num__2356 <o> b ) num__3454 <o> c ) num__4656 <o> d ) num__3456 <o> e ) num__5184 |
given exp . = ( num__12 ) num__3 x num__64 = ( num__12 ) num__3 x num__64 = ( num__12 ) num__2 x num__62 = ( num__72 ) num__2 = num__5184 num__432 num__12 x num__62 e <eor> e <eos> |
e |
divide__12.0__6.0__ subtract__64.0__2.0__ multiply__12.0__6.0__ multiply__12.0__432.0__ multiply__12.0__432.0__ |
divide__12.0__6.0__ subtract__64.0__2.0__ multiply__12.0__6.0__ multiply__12.0__432.0__ multiply__12.0__432.0__ |
| the proportion of water to alcohol in solution a is num__4 : num__1 and the proportion of water to alcohol in solution b is num__2 : num__3 . if an equal amount of each solution is mixed together what is the concentration of alcohol in the new solution ? <o> a ) num__39.0 <o> b ) num__40.0 <o> c ) num__41.0 <o> d ) num__42.0 <o> e ) num__43 % |
let v be the total volume of the new solution . then a volume of v / num__2 was added from each solution a and b . the amount of alcohol added to the new solution was : ( num__0.2 ) ( v / num__2 ) + ( num__0.6 ) ( v / num__2 ) = v / num__10 + num__3 v / num__10 = num__4 v / num__10 = num__2 v / num__5 . the concentration of alcohol is num__0.4 = num__40.0 the answer is b . <eor> b <eos> |
b |
multiply__3.0__0.2__ divide__2.0__0.2__ reverse__0.2__ divide__4.0__10.0__ multiply__4.0__10.0__ multiply__4.0__10.0__ |
multiply__3.0__0.2__ divide__2.0__0.2__ reverse__0.2__ divide__4.0__10.0__ multiply__4.0__10.0__ divide__40.0__1.0__ |
| the average marks scored by ganesh in english science mathematics and history is less than num__17 from that scored by him in english history geography and mathematics . what is the difference of marks in science and geography scored by him ? <o> a ) num__40 <o> b ) num__50 <o> c ) num__60 <o> d ) num__68 <o> e ) none of these |
e + h + g + m / num__4 â ˆ ’ e + s + m + h / num__4 = num__17 â ‡ ’ g - s = num__68 answer d <eor> d <eos> |
d |
multiply__17.0__4.0__ multiply__17.0__4.0__ |
multiply__17.0__4.0__ multiply__17.0__4.0__ |
| tea worth rs . num__126 per kg are mixed with a third variety in the ratio num__1 : num__1 : num__2 . if the mixture is worth rs . num__133 per kg the price of the third variety per kg will be <o> a ) rs . num__169.50 <o> b ) rs . num__1700 <o> c ) rs . num__175.50 <o> d ) rs . num__155.50 <o> e ) none |
solution since first second varieties are mixed in equal proportions so their average price = rs . ( num__126 + num__67.5 ) = rs . num__130.50 so the mixture is formed by mixing two varieties one at rs . num__130.50 per kg and the other at say rs . x per kg in the ratio num__2 : num__2 i . e . num__1 : num__1 . we have to find x . x - num__133 / num__22.50 = num__1 = â € º x - num__133 = num__22.50 = â € º x = num__155.50 . hence price of the third variety = rs . num__155.50 per kg . answer d <eor> d <eos> |
d |
add__133.0__22.5__ multiply__1.0__155.5__ |
add__133.0__22.5__ add__133.0__22.5__ |
| in an election candidate a got num__80.0 of the total valid votes . if num__15.0 of the total votes were declared invalid and the total numbers of votes is num__560000 find the number of valid vote polled in favor of candidate . <o> a ) num__330000 <o> b ) num__340000 <o> c ) num__380800 <o> d ) num__356000 <o> e ) num__357000 |
total number of invalid votes = num__15.0 of num__560000 = num__0.15 × num__560000 = num__84000.0 = num__84000 total number of valid votes num__560000 – num__84000 = num__476000 percentage of votes polled in favour of candidate a = num__80.0 therefore the number of valid votes polled in favour of candidate a = num__80.0 of num__476000 = num__0.8 × num__476000 = num__380800.0 = num__380800 c ) <eor> c <eos> |
c |
percent__15.0__560000.0__ percent__80.0__476000.0__ percent__80.0__476000.0__ |
percent__15.0__560000.0__ percent__80.0__476000.0__ percent__80.0__476000.0__ |
| monica and miya ' s age are in num__2 : num__5 . find teh age of monica if the sum of their age is num__28 . <o> a ) num__10 years <o> b ) num__14 years <o> c ) num__20 years <o> d ) num__8 years <o> e ) num__12 years |
let the age of monica be num__2 x and miya be num__5 x . num__2 x + num__5 x = num__28 x = num__4 monica ' s age is num__8 years answer : d <eor> d <eos> |
d |
multiply__2.0__4.0__ multiply__2.0__4.0__ |
multiply__2.0__4.0__ multiply__2.0__4.0__ |
| joe needs to paint all the airplane hangars at the airport so he buys num__360 gallons of paint to do the job . during the first week he uses num__0.25 of all the paint . during the second week he uses num__0.25 of the remaining paint . how many gallons of paint has joe used ? <o> a ) num__18 <o> b ) num__157 <o> c ) num__175 <o> d ) num__216 <o> e ) num__250 |
total paint initially = num__360 gallons paint used in the first week = ( num__0.25 ) * num__360 = num__90 gallons . remaning paint = num__270 gallons paint used in the second week = ( num__0.25 ) * num__270 = num__67 gallons total paint used = num__157 gallons . option b <eor> b <eos> |
b |
multiply__360.0__0.25__ subtract__360.0__90.0__ add__67.0__90.0__ round__157.0__ |
multiply__360.0__0.25__ subtract__360.0__90.0__ add__67.0__90.0__ round__157.0__ |
| if peter reads at a constant rate of num__1 pages every num__5 minutes how many seconds will it take him to read n pages ? <o> a ) num__300 <o> b ) num__2 n <o> c ) num__2.5 * n <o> d ) num__24 n <o> e ) num__150 |
peter would read num__1 page in num__5.0 minutes peter would read n page in ( num__5.0 ) * n min i . e . ( num__2.5 ) * n * num__60 seconds = num__300 n seconds . option a is the correct answer . <eor> a <eos> |
a |
hour_to_min_conversion__ multiply__5.0__60.0__ round__300.0__ |
hour_to_min_conversion__ multiply__5.0__60.0__ multiply__1.0__300.0__ |
| if y is num__60 percent greater than x then x is what percent less than y ? <o> a ) num__30.0 <o> b ) num__37.5 <o> c ) num__45.5 <o> d ) num__50.0 <o> e ) num__55.5 % |
y = num__1.6 x x = y / num__1.6 = num__10 y / num__16 = y - ( num__0.375 ) * y num__0.375 is num__37.5 . the answer is b . <eor> b <eos> |
b |
multiply__10.0__1.6__ divide__60.0__1.6__ divide__60.0__1.6__ |
multiply__10.0__1.6__ divide__60.0__1.6__ divide__60.0__1.6__ |
| a merchant has num__100 lbs of sugar part of which he sells at num__6.0 profit and the rest at num__17.0 profit . he gains num__10.0 on the whole . find how much is sold at num__6.0 profit ? <o> a ) num__70 lbs <o> b ) num__40 lbs <o> c ) num__30 lbs <o> d ) num__50 lbs <o> e ) num__64 lbs |
these types ofweighted averagequestions can be solved in a variety of ways so you can choose whichever method you find easiest / fastest . here ' s another variation on the weighted average formula : a = # of pounds sold at num__6.0 profit b = # of pounds sold at num__17.0 profit a + b = num__100 pounds ( . num__06 a + . num__17 b ) / ( a + b ) = . num__10 . num__06 a + . num__17 b = . num__1 a + . num__1 b . num__07 b = . num__04 a num__7 b = num__4 a num__1.75 = a / b so for every num__7 pounds of a we have num__4 pounds of b . with num__100 pounds total we have num__64 pounds of a and num__36 pounds of b . option e <eor> e <eos> |
e |
percent__100.0__64.0__ |
percent__100.0__64.0__ |
| when the positive integer x is divided by num__9 the remainder is num__5 . what is the remainder when num__6 x is divided by num__9 ? <o> a ) num__0 <o> b ) num__1 <o> c ) num__3 <o> d ) num__4 <o> e ) num__6 |
i tried plugging in numbers x = num__9 q + num__5 x = num__14 num__6 x = num__84 num__6 x / num__9 = num__9 * num__9 + num__3 remainder is num__3 . answer is c . <eor> c <eos> |
c |
add__9.0__5.0__ multiply__6.0__14.0__ subtract__9.0__6.0__ subtract__9.0__6.0__ |
add__9.0__5.0__ multiply__6.0__14.0__ subtract__9.0__6.0__ divide__9.0__3.0__ |
| a woman has $ num__252125 in her savings account . what is the least amount of money ( in whole number of dollars ) that she must add to her account if she wants to split this money evenly among her nine children ? <o> a ) $ num__1 <o> b ) $ num__2 <o> c ) $ num__3 <o> d ) $ num__4 <o> e ) $ num__11 |
to find the least amount the woman should add to her saving account to split the money evenly among her num__9 children she needs to make the total divisible by num__9 simply add the individual digits of the total = num__2 + num__5 + num__2 + num__1 + num__2 + num__5 = num__17 if you add num__1 the number is divisible by num__9 ( num__17 + num__1 ) correct option : a <eor> a <eos> |
a |
reverse__1.0__ |
reverse__1.0__ |
| x + ( num__1 / x ) = num__3.5 find x ^ num__2 + ( num__1 / x ^ num__2 ) <o> a ) num__2.25 <o> b ) num__3.25 <o> c ) num__10.25 <o> d ) num__5.25 <o> e ) num__6.25 |
squaring on both sides ( x + num__1 / x ) ^ num__2 = num__3.5 ^ num__2 x ^ num__2 + num__1 / x ^ num__2 = num__12.25 - num__2 x ^ num__2 + num__1 / x ^ num__2 = num__10.25 answer : c <eor> c <eos> |
c |
subtract__12.25__2.0__ multiply__1.0__10.25__ |
subtract__12.25__2.0__ subtract__12.25__2.0__ |
| tom drives his car at a speed of num__40 kmph and reaches his destination in num__5 hrs and john covers the same distance in num__8 hours . if speed of tomis increased by num__10 kmph and speed of ram is increased by num__15 kmph what would be the difference between the times taken by them to cover the distance ? <o> a ) num__2 hour <o> b ) num__1 hour <o> c ) num__3 hour <o> d ) num__4 hour <o> e ) num__5 hour |
distance covered by ravi at a speed of num__40 kmph in num__5 hours ð distance = speed x time = num__40 x num__5 = num__200 km ram covers same distance in num__8 hours = > speed of rama = distance time = num__25.0 = num__25 kmph . ravi speed is increased by num__10 kmph = > new speed of ravi = num__40 + num__10 = num__50 kmph ram speed is increased by num__15 kmph = > new speed of ram = num__25 + num__15 = num__40 kmph . therefore time taken by ravi to cover num__200 km at num__50 kmph = num__4.0 = num__4 hours time taken by ram to cover num__200 km at num__40 kmph = num__5.0 = num__5 hours b ) required time difference = num__5 hours – num__4 hours = num__1 hour b <eor> b <eos> |
b |
multiply__40.0__5.0__ subtract__40.0__15.0__ add__40.0__10.0__ divide__40.0__10.0__ subtract__5.0__4.0__ round__1.0__ |
multiply__40.0__5.0__ add__10.0__15.0__ add__40.0__10.0__ divide__40.0__10.0__ subtract__5.0__4.0__ round__1.0__ |
| a train num__490 m long running with a speed of num__63 km / hr will pass a tree in ? <o> a ) num__18 sec <o> b ) num__16 sec <o> c ) num__76 sec <o> d ) num__15 sec <o> e ) num__28 sec |
speed = num__63 * num__0.277777777778 = num__17.5 m / sec time taken = num__490 * num__0.0571428571429 = num__28 sec answer : e <eor> e <eos> |
e |
divide__490.0__17.5__ round__28.0__ |
divide__490.0__17.5__ divide__490.0__17.5__ |
| car a takes num__10 min lesser than car b which is travelling at an average speed of num__40 mph to cover a distance of num__60 miles . what is the average speed of car a ? <o> a ) num__30 <o> b ) num__35 <o> c ) num__45 <o> d ) num__50 <o> e ) num__55 |
time taken by car b to cover num__60 miles = num__1.5 = num__1.5 hrs = num__90 min time taken by car a to cover num__60 miles = num__80 min average speed of car a = num__0.75 = num__0.75 miles per min = num__45 miles per hour . answer : c <eor> c <eos> |
c |
divide__60.0__40.0__ multiply__60.0__1.5__ subtract__90.0__10.0__ divide__60.0__80.0__ multiply__60.0__0.75__ round__45.0__ |
divide__60.0__40.0__ multiply__60.0__1.5__ subtract__90.0__10.0__ divide__60.0__80.0__ multiply__60.0__0.75__ round__45.0__ |
| three competing juice makers conducted a blind taste test with mall shoppers . the shoppers could choose to taste any number of the three brands of juices but had to select at least one juice that they liked . if num__100 shoppers liked brand j num__200 shoppers liked brand k num__700 shoppers liked brand l num__350 shoppers liked exactly num__2 juices and num__50 shoppers liked all three juices how many shoppers took the taste test ? <o> a ) num__1300 <o> b ) num__1000 <o> c ) num__900 <o> d ) num__700 <o> e ) num__550 |
given : atleast num__1 juice was liked by the participants . - - > neither = num__0 assume i - - > no overlap between the sets ii - - > overlap between num__2 sets iii - - > overlap between num__3 sets i + num__2 * ( ii ) + num__3 * ( iii ) = num__100 + num__200 + num__700 i + num__2 * ( num__350 ) + num__3 * ( num__50 ) = num__1000 i = num__150 total number of shoppers who took the taste test = i + ii + iii = num__150 + num__350 + num__50 = num__550 answer : e <eor> e <eos> |
e |
add__2.0__1.0__ add__100.0__50.0__ add__200.0__350.0__ add__200.0__350.0__ |
add__2.0__1.0__ add__100.0__50.0__ add__200.0__350.0__ add__200.0__350.0__ |
| in an examination num__40.0 failed in hindi num__45.0 failed in english . if num__25.0 students failed in both the subjects . then the percentage of failed students is ? <o> a ) num__59.0 <o> b ) num__50.0 <o> c ) num__90.0 <o> d ) num__60.0 <o> e ) num__20 % |
male = num__45000 * num__0.555555555556 = num__25000 female = num__45000 * num__0.444444444444 = num__20000 married male = num__25000 * num__0.4 = num__10000 married female = num__10000 num__20000 - - - - - - - - - - - - num__10000 num__100 - - - - - - - - - - - - - ? = > num__50.0 answer : b <eor> b <eos> |
b |
percent__40.0__25000.0__ percent__0.4__25000.0__ percent__100.0__50.0__ |
percent__40.0__25000.0__ percent__0.4__25000.0__ percent__100.0__50.0__ |
| the daytime telephone rate between two cities is num__60 cents for the first num__3 minutes and c cents for each additional minute . the total charge is reduced num__50 percent on calls made after num__10 : num__00 p . m . the cost in dollars of a num__35 - minute call made at num__10 : num__30 p . m . between these two cities is : <o> a ) num__0.4 ( num__0.60 ) + num__35 c <o> b ) num__0.5 ( num__0.60 + num__32 c ) <o> c ) num__0.4 ( num__0.60 + num__9 c ) <o> d ) num__0.5 ( num__0.60 + num__0.32 c ) <o> e ) num__0.6 ( num__0.60 + num__0.35 c ) |
for first three minutes = num__60 cents remaining minutes = num__35 - num__3 = num__32 charge for num__32 minutes = num__32 c total cost ( if the call was made in daytime ) = num__0.60 + num__32 c num__50.0 chargereducedon night calls = > num__50.0 charge = > num__0.5 ( num__0.60 + num__32 c ) hence answer is d <eor> d <eos> |
d |
subtract__35.0__3.0__ divide__30.0__50.0__ divide__30.0__60.0__ divide__30.0__60.0__ |
subtract__35.0__3.0__ divide__30.0__50.0__ divide__30.0__60.0__ divide__30.0__60.0__ |
| a is faster than b . a and b each walk num__24 km . the sum of their speeds is num__7 km / hr and the sum of times taken by them is num__14 hours . then a ' s speed is equal to ? <o> a ) num__5 km / hr <o> b ) num__3 km / hr <o> c ) num__6 km / hr <o> d ) num__9 km / hr <o> e ) num__2 km / hr |
let a ' s speed = x km / hr . then b ' s speed = ( num__7 - x ) km / hr . so num__24 / x + num__24 / ( num__7 - x ) = num__14 x num__2 - num__98 x + num__168 = num__0 ( x - num__3 ) ( x - num__4 ) = num__0 = > x = num__3 or num__4 . since a is faster than b so a ' s speed = num__4 km / hr and b ' s speed = num__3 km / hr . answer : b <eor> b <eos> |
b |
divide__14.0__7.0__ multiply__7.0__14.0__ multiply__24.0__7.0__ subtract__7.0__3.0__ subtract__7.0__4.0__ |
divide__14.0__7.0__ multiply__7.0__14.0__ multiply__24.0__7.0__ subtract__7.0__3.0__ subtract__7.0__4.0__ |
| a and b put in rs . num__300 and rs . num__400 respectively into a business . a reinvests into the business his share of the first year ' s profit of rs . num__140 where as b does not . in what ratio should they divide the second year ' s profit ? <o> a ) num__8 : num__7 <o> b ) num__7 : num__5 <o> c ) num__1 : num__3 <o> d ) num__8 : num__7 <o> e ) num__9 : num__10 |
explanation : num__3 : num__4 a = num__0.428571428571 * num__140 = num__60 num__360 : num__400 num__39 : num__40 answer : e <eor> e <eos> |
e |
percent__3.0__300.0__ |
percent__3.0__300.0__ |
| num__0.4 of the fish in an aquarium is red and num__0.25 of the fish is brown . num__0.25 of the red fish and num__0.2 of the brown fish is male . if the rest of the fish in the aquarium is female what is the probability that a fish randomly picked from the aquarium is male ? <o> a ) num__0.15 <o> b ) num__0.233333333333 <o> c ) num__0.3125 <o> d ) num__0.333333333333 <o> e ) num__0.4 |
straight method . . . . num__0.25 of num__0.4 + num__0.2 of num__0.25 = num__0.1 + num__0.05 = num__0.15 answer : a <eor> a <eos> |
a |
union_prob__0.25__0.2__0.4__ union_prob__0.25__0.1__0.2__ union_prob__0.25__0.05__0.15__ |
union_prob__0.25__0.2__0.4__ union_prob__0.25__0.1__0.2__ union_prob__0.25__0.05__0.15__ |
| if a ^ num__2 + b ^ num__2 = num__177 and ab = num__54 then find the value of a + b / a - b ? <o> a ) num__1 <o> b ) num__3 <o> c ) num__5 <o> d ) num__7 <o> e ) num__9 |
( a + b ) ^ num__2 = a ^ num__2 + b ^ num__2 + num__2 ab = num__117 + num__2 * num__24 = num__225 a + b = num__15 ( a - b ) ^ num__2 = a ^ num__2 + b ^ num__2 - num__2 ab = num__117 - num__2 * num__54 a - b = num__3 a + b / a - b = num__5.0 = num__5 answer is c . <eor> c <eos> |
c |
add__2.0__3.0__ add__2.0__3.0__ |
add__2.0__3.0__ add__2.0__3.0__ |
| among all sales staff at listco corporation college graduates and those without college degrees are equally represented . each sales staff member is either a level - num__1 or level - num__2 employee . level - num__1 college graduates account for num__15.0 of listco ' s sales staff . listco employs num__60 level - num__1 employees num__30 of whom are college graduates . how many sales staff members without college degrees are level - num__2 employees ? <o> a ) num__46 <o> b ) num__42 <o> c ) num__56 <o> d ) num__70 <o> e ) num__58 |
i ' m going in on this one . so let ' s say that we have the following so we know that l num__1 = num__60 and that c and l num__1 = num__0.15 x we should set up a double set matrix btw but anyways i ' m just explaining the point with this problem . now we are told that num__0.15 x = num__30 therefore the grand total is num__200 . now we know that l num__2 is num__200 - num__60 = num__140 . we also learn that c and no c are equally represented thus num__100 each . therefore no c and no l num__2 will be num__100 - num__30 = num__70 . thus d is the correct answer choice <eor> d <eos> |
d |
divide__30.0__0.15__ subtract__200.0__60.0__ divide__15.0__0.15__ subtract__100.0__30.0__ multiply__1.0__70.0__ |
divide__30.0__0.15__ subtract__200.0__60.0__ divide__15.0__0.15__ subtract__100.0__30.0__ subtract__100.0__30.0__ |
| aishwarya was first to board to her flight to delhi . she forgot her seat number and picks a random seat for herself . after this every single person who get to the flight sits on his seat if its available else chooses any available seat at random . abhishek is last to enter the flight and at that moment num__0.99 seats were occupied . whats the probability what abhishek gets to sit in his own seat ? <o> a ) num__2 <o> b ) num__0.5 <o> c ) num__1.0 <o> d ) num__3 <o> e ) none |
solution : num__0.5 one of two is the possibility num__1 . if any of the first num__99 people sit in abhishek seat abhishek will not get to sit in his own seat . num__2 . if any of the first num__99 people sit in aishwarya ' s seat abhishek will get to sit in his seat . answer b <eor> b <eos> |
b |
coin_space__ negate_prob__0.5__ |
coin_space__ negate_prob__0.5__ |
| ajay can walk num__6 km in num__1 hour . in how many hours he can walk num__45 km ? <o> a ) num__5 hrs <o> b ) num__10 hrs <o> c ) num__7.5 hrs <o> d ) num__20 hrs <o> e ) num__30 hrs |
num__1 hour he walk num__6 km he walk num__45 km in = num__7.5 * num__1 = num__7.5 hours answer is c <eor> c <eos> |
c |
divide__45.0__6.0__ round__7.5__ |
divide__45.0__6.0__ multiply__1.0__7.5__ |
| the ratio of two numbers is num__3 : num__4 and their sum is num__28 . the greater of the two numbers is ? <o> a ) num__15 <o> b ) num__16 <o> c ) num__18 <o> d ) num__21 <o> e ) num__22 |
num__3 : num__4 total parts = num__7 = num__7 parts - - > num__28 ( num__7 × num__4 = num__28 ) = num__1 part - - - - > num__4 ( num__1 × num__4 = num__4 ) = the greater of the two number is = num__4 = num__4 parts - - - - > num__16 ( num__4 × num__4 = num__16 ) b <eor> b <eos> |
b |
add__3.0__4.0__ subtract__4.0__3.0__ multiply__1.0__16.0__ |
add__3.0__4.0__ subtract__4.0__3.0__ multiply__1.0__16.0__ |
| the sides of a square region measured to the nearest centimeter are num__10 centimeters long . the least possible value of the actual area of the square region is <o> a ) num__96.25 sq cm <o> b ) num__98.25 sq cm <o> c ) num__92.25 sq cm <o> d ) num__100.25 sq cm <o> e ) num__90.25 sq cm |
though there might be some technicalities concerning the termnearest ( as num__9.5 is equidistant from both num__9 and num__10 ) the answer still should be : num__9.5 ^ num__2 = num__90.25 . answer : e <eor> e <eos> |
e |
power__9.5__2.0__ power__9.5__2.0__ |
power__9.5__2.0__ power__9.5__2.0__ |
| let a be a positive integer . if w is divisible by num__2 ^ a and w is also divisible by num__3 ^ ( num__2 a ) then it is possible that w is not divisible by <o> a ) num__6 <o> b ) num__3 × num__2 ^ a <o> c ) num__2 × num__3 ^ ( num__2 a ) <o> d ) num__6 ^ a <o> e ) num__6 ^ ( num__2 a ) |
since w is divisible by num__2 ^ a and num__3 ^ ( num__2 a ) it must be divisible by num__6 . as least value of a = num__1 only for e num__6 ^ ( num__2 a ) does n ' t satisfy if a = num__1 and w = num__18 it is not divisible by num__6 ^ num__2 ( i . e num__36 ) hence answer is e <eor> e <eos> |
e |
multiply__2.0__3.0__ subtract__3.0__2.0__ multiply__3.0__6.0__ multiply__2.0__18.0__ multiply__2.0__3.0__ |
multiply__2.0__3.0__ subtract__3.0__2.0__ multiply__3.0__6.0__ multiply__2.0__18.0__ multiply__2.0__3.0__ |
| simplify : num__839478 x num__625 <o> a ) num__524673750 <o> b ) num__546273750 <o> c ) num__562473750 <o> d ) num__564273750 <o> e ) none of them |
num__839478 x num__625 = num__839478 x num__5 ^ num__4 = num__524673750.0 = num__524673750 . answer is a <eor> a <eos> |
a |
multiply__839478.0__625.0__ multiply__839478.0__625.0__ |
multiply__839478.0__625.0__ multiply__839478.0__625.0__ |
| the average age of the mother and her six children is num__13 years which is reduced by num__5 years if the age of the mother is excluded . how old is the mother ? <o> a ) num__40 <o> b ) num__41 <o> c ) num__42 <o> d ) num__43 <o> e ) num__44 |
total age of mother and num__6 children = avg x n = num__13 x ( num__6 + num__1 ) = num__91 if mother is excluded new average is ( num__13 - num__5 ) = num__8 so for num__6 children = num__6 x num__8 = num__48 so mother ' s age is num__91 - num__48 = num__43 answer : option d <eor> d <eos> |
d |
subtract__6.0__5.0__ subtract__13.0__5.0__ multiply__6.0__8.0__ subtract__48.0__5.0__ multiply__1.0__43.0__ |
subtract__6.0__5.0__ subtract__13.0__5.0__ multiply__6.0__8.0__ subtract__48.0__5.0__ subtract__48.0__5.0__ |
| a rectangular circuit board is designed to have a width of w inches a length of l inches a perimeter of p inches and an area of x square inches . which of the following equations must be true ? <o> a ) num__2 w ^ num__2 − pw + num__2 x = num__0 <o> b ) num__2 w ^ num__2 + pw + num__2 x = num__0 <o> c ) num__2 w ^ num__2 − pw − num__2 x = num__0 <o> d ) w ^ num__2 + pw + x = num__0 <o> e ) w ^ num__2 − pw + num__2 x = num__0 |
p = num__2 ( l + w ) - - - - - - - - - - - - - - - - - num__1 ) x = lw - - - - - - - - - - - - - - - - - - - - - - - - num__2 ) option a is not possible why ? ? because all the terms are positive . lets try option b put value of p and a from num__1 and num__2 we have num__2 w ^ num__2 - num__2 ( l + w ) w + num__2 ( lw ) num__2 w ^ num__2 - num__2 lw - num__2 w ^ num__2 + num__2 lw = num__0 . hence answer is a . <eor> a <eos> |
a |
rectangle_perimeter__0.0__1.0__ |
power__2.0__1.0__ |
| if x + y = num__4 x - y = num__36 for integers of x and y x = ? <o> a ) num__20 <o> b ) num__15 <o> c ) num__25 <o> d ) num__13 <o> e ) num__42 |
x + y = num__4 x - y = num__36 num__2 x = num__40 x = num__20 answer is a <eor> a <eos> |
a |
add__4.0__36.0__ divide__40.0__2.0__ divide__40.0__2.0__ |
add__4.0__36.0__ divide__40.0__2.0__ subtract__40.0__20.0__ |
| the length of a side of a hexagon is num__4 inches . what is the perimeter ? <o> a ) num__22 <o> b ) num__23 <o> c ) num__24 <o> d ) num__25 <o> e ) num__26 |
hexagon . it means num__6 equal sides . p = num__6 ( num__4 ) = num__24 inches answer c <eor> c <eos> |
c |
multiply__4.0__6.0__ round__24.0__ |
multiply__4.0__6.0__ round__24.0__ |
| for a positive integer n let pnpn denote the product of the digits of nn and snsn denote the sum of the digits of nn . the number of integers between num__10 and num__100 for which pn + sn = n <o> a ) num__9 <o> b ) num__8 <o> c ) num__7 <o> d ) num__6 <o> e ) num__5 |
let the two digit number be ' ab ' given a × b + a + b = num__10 a + b a × b = num__9 a b = num__9 therefore for b = num__9 the above condition satisfies . so all the two digit numbers for which units digit is num__9 are our solutions . they range from num__19 to num__99 . total num__9 numbers . answer : a <eor> a <eos> |
a |
add__10.0__9.0__ subtract__19.0__10.0__ |
add__10.0__9.0__ subtract__19.0__10.0__ |
| the profits of qrs company rose num__20.0 from march to april then dropped num__20.0 from april to may then rose num__50.0 from may to june . what was the percent increase for the whole quarter from march to june ? <o> a ) num__15.0 <o> b ) num__32.0 <o> c ) num__44.0 <o> d ) num__62.0 <o> e ) num__80 % |
assume num__100 in march then num__120 in april as num__20.0 increase then num__96 in may as num__20.0 decrease from april and then num__144 in june which is num__150.0 of num__96 . so overall increase is from num__100 to num__144 is num__44.0 answer c <eor> c <eos> |
c |
percent__100.0__44.0__ |
percent__100.0__44.0__ |
| a num__100 - litre mixture of milk and water contains num__36 litres of milk . ' x ' litres of this mixture is removed and replaced with an equal quantum of water . if the process is repeated once then the concentration of the milk stands reduced at num__16.0 . what is the value of x ? <o> a ) num__33.33 litres <o> b ) num__36.67 litres <o> c ) num__37.67 litres <o> d ) num__36.5 litres <o> e ) num__36 litres |
working formula . . . initial concentration * initial volume = final concentration * final volume . let x is the part removed from num__100 lts . num__36.0 ( num__1 - x / num__100 ) ^ num__2 = num__16.0 * num__100.0 ( num__1 - x / num__100 ) ^ num__2 = num__0.444444444444 - - - - - - > ( num__1 - x / num__100 ) ^ num__2 = ( num__0.666666666667 ) ^ num__2 num__100 - x = num__66.6666666667 x = num__33.33 . . . ans a <eor> a <eos> |
a |
divide__16.0__36.0__ multiply__1.0__33.33__ |
divide__16.0__36.0__ multiply__1.0__33.33__ |
| in a certain sequence of numbers a num__1 a num__2 a num__3 . . . an the average ( arithmetic mean ) of the first m consecutive terms starting with a num__1 is m for any positive integer m . if a num__1 = num__1 what is a num__13 ? <o> a ) num__100 <o> b ) num__55 <o> c ) num__25 <o> d ) num__19 <o> e ) num__1 |
as stated above ( a num__1 + a num__2 + a num__3 . . . . . . + a num__13 ) / num__13 = num__13 therefore a num__1 + a num__2 + a num__3 . . . . . . . a num__13 = num__169 ( num__1 ) using the same logic we got a num__1 + a num__2 + a num__3 . . . . . . . . . . + a num__12 = num__144 ( num__2 ) ( num__2 ) - ( num__1 ) we got a num__13 = num__25 c <eor> c <eos> |
c |
subtract__13.0__1.0__ add__13.0__12.0__ multiply__1.0__25.0__ |
subtract__13.0__1.0__ add__13.0__12.0__ add__13.0__12.0__ |
| a sixth grade teacher asked her students to draw rectangles with positive integer length and a perimeter of num__42 . the difference between the largest and smallest possible ares of the rectangles that the students could have come up with is ? <o> a ) num__50 sq units <o> b ) num__60 sq units <o> c ) num__70 sq units <o> d ) num__80 sq units <o> e ) num__90 sq units |
sum of length and breadth will be num__21 units . num__21.0 = num__21 area will be max when lxb = num__11 x num__10 = num__110 sq units area will be min when lxb = num__20 x num__1 = num__20 sq units . . the difference between the largest and smallest possible ares of the rectangles that the students could have come up with = num__110 - num__20 = num__90 sq units answer : e <eor> e <eos> |
e |
multiply__10.0__11.0__ multiply__1.0__90.0__ |
multiply__10.0__11.0__ multiply__1.0__90.0__ |
| working at constant rate pump x pumped out half of the water in a flooded basement in num__2 hours . the pump y was started and the two pumps working independently at their respective constant rates pumped out rest of the water in num__3 hours . how many hours would it have taken pump y operating alone at its own constant rate to pump out all of the water that was pumped out of the basement ? <o> a ) a . num__10 <o> b ) b . num__12 <o> c ) c . num__14 <o> d ) d . num__18 <o> e ) e . num__24 |
rate of x = num__0.125 rate of x + y = num__0.166666666667 rate of y = num__0.166666666667 - num__0.125 = num__0.0416666666667 num__18 hours d <eor> d <eos> |
d |
divide__0.125__3.0__ round__18.0__ |
subtract__0.1667__0.125__ round__18.0__ |
| donald duck can can swim his pool downstream ( with the pool current helping time ) in exact num__40 seconds and upstream ( against the pool current ) in a pool in exact num__60 seconds . the length of pool is num__2 kilometers . how long donald duck can cover distance of one side at a still pool ( with no current ) . <o> a ) num__48 minutes . <o> b ) num__58 minutes . <o> c ) num__68 minutes . <o> d ) num__88 minutes . <o> e ) num__78 minutes . |
solution : num__48 minutes . donald duck ' s speed = x km / seconds pool current speed = y km / seconds num__2 / ( x + y ) = num__40 num__2 / ( x - y ) = num__60 solving the simultaneous equations gives x = num__0.0416666666667 therefore to cover num__2 km will take num__2 / x = num__48 seconds answer a <eor> a <eos> |
a |
divide__2.0__48.0__ round__48.0__ |
divide__2.0__48.0__ round__48.0__ |
| if the given two numbers are respectively num__6.0 and num__18.0 of a third number then what percentage is the first of the second ? <o> a ) num__20.0 <o> b ) num__25.0 <o> c ) num__18.0 <o> d ) num__33 num__0.333333333333 % <o> e ) none of these |
here l = num__6 and m = num__18 therefore first number = l / m x num__100.0 of second number = num__0.333333333333 x num__100.0 of second number = num__33 num__0.333333333333 % of second number answer : d <eor> d <eos> |
d |
percent__33.0__100.0__ |
percent__33.0__100.0__ |
| a invested rs . num__76000 in a business . after few months b joined him rs . num__57000 . at the end of the year the total profit was divided between them in ratio num__2 : num__1 . after bow many months did b join ? <o> a ) num__4 <o> b ) num__3 <o> c ) num__2 <o> d ) num__1 <o> e ) none of them |
suppose b joined after x months . then b ' s money was invested for ( num__12 - x ) = ( num__76000 x num__12 ) / ( num__57000 x ( num__12 - x ) = num__2.0 = num__912000 = num__114000 ( num__12 - x ) = num__114 ( num__12 - x ) = num__912 = num__12 - x = num__8 = x = num__4 hence b joined after num__4 months answer is a . <eor> a <eos> |
a |
multiply__76000.0__12.0__ multiply__57000.0__2.0__ divide__912000.0__114000.0__ subtract__12.0__8.0__ multiply__1.0__4.0__ |
multiply__76000.0__12.0__ multiply__57000.0__2.0__ divide__912000.0__114000.0__ subtract__12.0__8.0__ divide__4.0__1.0__ |
| if n is a positive integer which of the following expressions must be even ? <o> a ) ( n − num__1 ) ( n + num__3 ) <o> b ) ( n − num__6 ) ( n + num__1 ) <o> c ) ( n − num__2 ) ( n + num__6 ) <o> d ) ( n − num__3 ) ( n + num__3 ) <o> e ) ( n − num__3 ) ( n + num__7 ) |
whether n is even or odd ( n - num__6 ) ( n + num__1 ) will have one odd factor and one even factor . the product will be even . the answer is b . <eor> b <eos> |
b |
multiply__1.0__6.0__ |
multiply__1.0__6.0__ |
| a train traveling at num__100 kmph overtakes a motorbike traveling at num__64 kmph in num__40 seconds . what is the length of the train in meters ? <o> a ) num__1777 meters <o> b ) num__1822 meters <o> c ) num__400 meters <o> d ) num__1111 mete <o> e ) none of these |
answer concept : distance covered by train when it crosses a moving object when a train overtakes another object such as a motorbike whose length is insignificant compared to the length of the train the distance traveled by the train is equal to the length of the train . because the motorbike is also moving we have to take the relative speed between the train and the motorbike and not just the speed of the train . the length of the train = distance traveled by the train while overtaking the motorbike = relative speed between the train and the motorbike * time taken in this case as both the objects i . e . the train and the motorbike are moving in the same direction the relative speed between them = difference between their respective speeds = num__100 - num__64 = num__36 kmph . distance traveled by the train while overtaking the motorbike = num__36 kmph * num__40 seconds . the final answer is in meters and the speed is given in kmph and the time in seconds . so let us convert the given speed from kmph to m / sec . num__1 kmph = num__0.277777777778 m / sec therefore num__36 kmph = num__36 * num__0.277777777778 = num__10 m / sec . relative speed = num__10 m / sec . time taken = num__40 seconds . therefore distance traveled = num__10 * num__40 = num__400 meters . choice c <eor> c <eos> |
c |
subtract__100.0__64.0__ multiply__40.0__10.0__ round__400.0__ |
subtract__100.0__64.0__ multiply__40.0__10.0__ multiply__40.0__10.0__ |
| simplfy b - [ b - ( a + b ) - { b - ( b - a + b ) } + num__2 a ] <o> a ) a <o> b ) num__2 a <o> c ) num__4 a <o> d ) num__0 <o> e ) num__1 |
explanation : b - [ b - ( a + b ) - { b - ( b - a + b ) } + num__2 a ] = b - [ b - a - b - { b - ( num__2 b - a ) } + num__2 a ] = b - [ - a - { b - num__2 b + a } + num__2 a ] = b - [ - a - { - b + a } + num__2 a ] = b - [ - a + b - a + num__2 a ] = b - [ - num__2 a + b + num__2 a ] = b - b = num__0 option d <eor> d <eos> |
d |
multiply__2.0__0.0__ |
multiply__2.0__0.0__ |
| if a die has num__34 and num__16 and num__25 opposite each other then how many such dies can be made . <o> a ) num__13 <o> b ) num__3 <o> c ) num__5 <o> d ) num__12 <o> e ) num__11 |
there are num__3 pairs . consider it a b c a - ( num__16 ) b - ( num__34 ) c - ( num__25 ) now the possible combinations to place a b c is num__6 . because for a - num__3 possible place b - num__2 possible place c - num__1 possible place num__3 * num__2 * num__1 = num__6 now a has two pair so possible combo for num__1 and num__6 is num__2 . . same for b and c num__2 + num__2 + num__2 = num__6 so num__6 + num__6 = num__12 answer : d <eor> d <eos> |
d |
die_space__ coin_space__ choose__3.0__1.0__ choose__3.0__1.0__ |
die_space__ coin_space__ choose__3.0__1.0__ choose__3.0__1.0__ |
| the radius of a circular wheel is num__1.75 m how many revolutions will it make in traveling num__1 km ? <o> a ) num__11000 <o> b ) num__11008 <o> c ) num__11006 <o> d ) num__110054 <o> e ) num__11001 |
num__2 * num__3.14285714286 * num__1.75 * x = num__11000 x = num__1000 answer : a <eor> a <eos> |
a |
multiply__1.0__11000.0__ |
multiply__1.0__11000.0__ |
| a cistern can be filled by a tap in num__4 hours while it can be emptied by another tap in num__10 hours . if both the taps are opened simultaneously then after how much time will the cistern get filled ? <o> a ) num__2.9 hrs <o> b ) num__8.9 hrs <o> c ) num__2.9 hrs <o> d ) num__6.7 hrs <o> e ) num__8.6 hrs |
net part filled in num__1 hour = ( num__0.25 - num__0.1 ) = num__0.15 the cistern will be filled in num__6.66666666667 hrs i . e . num__6.7 hrs . answer : d <eor> d <eos> |
d |
divide__1.0__4.0__ divide__1.0__10.0__ subtract__0.25__0.1__ divide__1.0__0.15__ round__6.7__ |
divide__1.0__4.0__ divide__1.0__10.0__ subtract__0.25__0.1__ divide__1.0__0.15__ round__6.7__ |
| two pipes a and b can separately fill a tank in num__2 minutes and num__15 minutes respectively . both the pipes are opened together but num__4 minutes after the start the pipe a is turned off . how much time will it take to fill the tank ? <o> a ) num__18 <o> b ) num__10 <o> c ) num__12 <o> d ) num__11 <o> e ) num__17 |
num__0.333333333333 + x / num__15 = num__1 x = num__10 answer : b <eor> b <eos> |
b |
round__10.0__ |
divide__10.0__1.0__ |
| num__0.333333333333 rd of work is completed by kiran in num__6 days . in how many days he can finish the remaining work ? <o> a ) num__6 days <o> b ) num__10 days <o> c ) num__12 days <o> d ) num__15 days <o> e ) num__20 days |
num__0.333333333333 rd work finished in num__6 days num__0.666666666667 rd work = num__0.666666666667 * num__3 * num__6 = num__12 days answer is c <eor> c <eos> |
c |
round__12.0__ |
round__12.0__ |
| if n is a positive integer and n ^ num__2 is divisible by num__72 then the largest positive integer that must divide n is <o> a ) num__6 <o> b ) num__12 <o> c ) num__24 <o> d ) num__36 <o> e ) num__48 |
possible values of n = num__12 num__24 num__36 and num__48 . but it is num__12 that can divide all the possible values of n . if we consider num__48 it will not divide num__12 num__24 and num__36 . hence num__12 is the value that must divide n . answer : b <eor> b <eos> |
b |
multiply__2.0__12.0__ divide__72.0__2.0__ multiply__2.0__24.0__ divide__24.0__2.0__ |
multiply__2.0__12.0__ divide__72.0__2.0__ multiply__2.0__24.0__ divide__24.0__2.0__ |
| a train of length num__150 metres takes num__40.5 seconds to cross a tunnel of length num__300 metres . what is the speed of the train in km / hr ? <o> a ) num__10 km / hr . <o> b ) num__20 km / hr . <o> c ) num__30 km / hr . <o> d ) num__40 km / hr . <o> e ) none |
sol . speed = [ num__150 + num__300 / num__40.5 ] m / sec = [ num__450 / num__40.5 * num__3.6 ] km / hr = num__40 km / hr . answer d <eor> d <eos> |
d |
add__150.0__300.0__ round__40.0__ |
add__150.0__300.0__ round__40.0__ |
| the average age of a class of num__32 students is num__15 yrs . if the teacher ' s age is also included the average increases by one year . find the age of the teacher <o> a ) num__45 years <o> b ) num__46 years <o> c ) num__49 years <o> d ) num__52 years <o> e ) num__48 years |
total age of students is num__32 x num__15 = num__480 years total age inclusive of teacher = num__33 x ( num__15 + num__1 ) = num__528 so teacher ' s age is num__528 - num__480 = num__48 yrs there is a shortcut for these type of problems teacher ' s age is num__15 + ( num__33 x num__1 ) = num__48 years answer : e <eor> e <eos> |
e |
multiply__32.0__15.0__ subtract__33.0__32.0__ add__15.0__33.0__ add__15.0__33.0__ |
multiply__32.0__15.0__ subtract__33.0__32.0__ add__15.0__33.0__ add__15.0__33.0__ |
| a car traveling at a certain constant speed takes num__12 seconds longer to travel num__1 kilometer than it would take to travel num__1 kilometer at num__75 kilometers per hour . at what speed in kilometers per hour is the car traveling ? <o> a ) num__60 <o> b ) num__72 <o> c ) num__72.5 <o> d ) num__73 <o> e ) num__73.5 |
many approaches are possible one of them : let the distance be num__1 kilometer . time to cover this distance at num__75 kilometers per hour is num__0.0133333333333 hours = num__48.0 seconds = num__48 seconds ; time to cover this distance at regular speed is num__48 + num__12 = num__60 seconds = num__60 / num__3600 hours = num__0.0166666666667 hours ; so we get that to cover num__1 kilometer num__0.0166666666667 hours is needed - - > regular speed num__60 kilometers per hour ( rate is a reciprocal of time or rate = distance / time ) . answer : a <eor> a <eos> |
a |
divide__1.0__75.0__ hour_to_min_conversion__ multiply__75.0__48.0__ divide__1.0__60.0__ hour_to_min_conversion__ |
divide__1.0__75.0__ add__12.0__48.0__ multiply__75.0__48.0__ divide__1.0__60.0__ add__12.0__48.0__ |
| a certain sum of money at simple interest amounted rs . num__1448 in num__9 years at num__4.0 per annum find the sum ? <o> a ) num__1064 <o> b ) num__1080 <o> c ) num__1980 <o> d ) num__646 <o> e ) num__168 |
num__840 = p [ num__1 + ( num__9 * num__4 ) / num__100 ] p = num__1064 answer : a <eor> a <eos> |
a |
percent__100.0__1064.0__ |
percent__100.0__1064.0__ |
| a person crosses a num__500 m long street in num__4 minutes . what is his speed in km per hour ? <o> a ) num__7.5 <o> b ) num__2.6 <o> c ) num__3.9 <o> d ) num__8.2 <o> e ) num__2.7 |
distance = num__500 meter time = num__4 minutes = num__4 x num__60 seconds = num__240 seconds speed = distance / time = num__2.08333333333 = num__2.08 m / s = num__2.08 Ã — num__3.6 km / hr = num__7.5 km / hr answer : a <eor> a <eos> |
a |
hour_to_min_conversion__ multiply__4.0__60.0__ divide__500.0__240.0__ round__2.0833__ round__7.5__ |
hour_to_min_conversion__ multiply__4.0__60.0__ divide__500.0__240.0__ round__2.0833__ round__7.5__ |
| ( num__1 ^ num__2 + num__2 ^ num__2 + num__3 ^ num__2 + . . . . . + num__11 ^ num__2 ) = ? <o> a ) num__506 <o> b ) num__345 <o> c ) num__365 <o> d ) num__385 <o> e ) none of them |
formula is ( num__1 ^ num__2 + num__2 ^ num__2 + num__3 ^ num__2 + . . . . + n ^ num__2 = num__0.166666666667 n ( n + num__1 ) ( num__2 n + num__1 ) n = num__11 = ( num__0.166666666667 x num__11 x num__12 x num__23 ) = num__506 answer is a . <eor> a <eos> |
a |
add__1.0__11.0__ add__11.0__12.0__ multiply__1.0__506.0__ |
add__1.0__11.0__ add__11.0__12.0__ multiply__1.0__506.0__ |
| the car a has an average speed of num__40 miles / hour and car b has an average speed of num__60 . how much time ( in minutes ) does the car a take more than car b to cover a distance of num__60 miles ? <o> a ) num__10 <o> b ) num__20 <o> c ) num__25 <o> d ) num__30 <o> e ) num__40 |
time taken by car a = num__60 miles / num__40 mph = num__1.5 hours = num__90 min time taken by car b = num__60 miles / num__60 mph = num__1 hour = num__60 min car a takes num__30 min more than car b answer : d <eor> d <eos> |
d |
divide__60.0__40.0__ multiply__60.0__1.5__ subtract__90.0__60.0__ round__30.0__ |
divide__60.0__40.0__ multiply__60.0__1.5__ subtract__90.0__60.0__ divide__30.0__1.0__ |
| in an election only two candidates contested . a candidate secured num__70.0 of the valid votes and won by a majority of num__180 votes . find the total number of valid votes ? <o> a ) num__430 <o> b ) num__450 <o> c ) num__436 <o> d ) num__434 <o> e ) num__422 |
let the total number of valid votes be x . num__70.0 of x = num__0.7 * x = num__7 x / num__10 number of votes secured by the other candidate = x - num__7 x / num__100 = num__3 x / num__10 given num__7 x / num__10 - num__3 x / num__10 = num__180 = > num__4 x / num__10 = num__180 = > num__4 x = num__1800 = > x = num__450 . answer : b <eor> b <eos> |
b |
percent__100.0__450.0__ |
percent__100.0__450.0__ |
| if x is a negative integer which of the following must be a negative integer ? <o> a ) x + num__1 <o> b ) x / num__2 + num__3 <o> c ) x ^ num__4 <o> d ) x – y <o> e ) x ^ num__11 |
we need to check each option here : a . x + num__1 this will depend on the value of x . nothing can be said b . x / num__2 + num__3 this will depend on the value of x . nothing can be said c . x ^ num__4 this will always be positive . even powers of negative numbers are positive and odd powers are negative d . x – y this will depend on the value of x and y . nothing can be said e . x ^ num__11 this will always be negative . even powers of negative numbers are positive and odd powers are negative correct option : e <eor> e <eos> |
e |
add__1.0__2.0__ add__1.0__3.0__ multiply__11.0__1.0__ |
add__1.0__2.0__ add__1.0__3.0__ divide__11.0__1.0__ |
| the food in a camp lasts for num__40 men for num__30 days . if ten more men join how many days will the food last ? <o> a ) num__32 days <o> b ) num__30 days <o> c ) num__28 days <o> d ) num__26 days <o> e ) num__24 days |
one man can consume the same food in num__40 * num__30 = num__1200 days . num__10 more men join the total number of men = num__50 the number of days the food will last = num__24.0 = num__24 days . answer : e <eor> e <eos> |
e |
multiply__40.0__30.0__ subtract__40.0__30.0__ add__40.0__10.0__ divide__1200.0__50.0__ round__24.0__ |
multiply__40.0__30.0__ subtract__40.0__30.0__ add__40.0__10.0__ divide__1200.0__50.0__ round__24.0__ |
| num__2 corner most boxes of a chess board ( diagonally opposite ) haven been cut out there ' s a rectangular block = num__2 sqaures of chess board how many such blocks can be placed on the chess board ? ” <o> a ) num__27 <o> b ) num__28 <o> c ) num__29 <o> d ) num__30 <o> e ) num__31 |
chess board has num__64 squares . so we can place num__32 rectangular blocks but num__2 are cut off from the corner . hence num__32 - num__2 = num__30 blocks answer : d <eor> d <eos> |
d |
triangle_area__2.0__30.0__ |
triangle_area__2.0__30.0__ |
| a can finish a piece of work in num__5 days . b can do it in num__15 days . they work together for two days and then a goes away . in how many days will b finish the work ? <o> a ) num__3 days <o> b ) num__5 days <o> c ) num__4 days <o> d ) num__6 days <o> e ) num__7 days |
num__0.4 + ( num__2 + x ) / num__15 = num__1 = > x = num__7 days answer : e <eor> e <eos> |
e |
multiply__5.0__0.4__ add__5.0__2.0__ round__7.0__ |
multiply__5.0__0.4__ add__5.0__2.0__ add__5.0__2.0__ |
| in plutarch enterprises num__70.0 of the employees are marketers num__20.0 are engineers and the rest are managers . marketers make an average salary of $ num__60000 a year and engineers make an average of $ num__80000 . what is the average salary for managers if the average for all employees is also $ num__80000 ? <o> a ) $ num__80000 <o> b ) $ num__130000 <o> c ) $ num__240000 <o> d ) $ num__290000 <o> e ) $ num__220 |
000 |
for sake of ease let ' s say there are num__10 employees : num__7 marketers num__2 engineers and num__1 manager . average company salary * number of employees = total company salary > > > $ num__80000 * num__10 = $ num__800000 subtract the combined salaries for the marketers ( num__7 * $ num__60000 ) and the engineers ( num__2 * $ num__80000 ) > > > $ num__800000 - $ num__420000 - $ num__160000 = $ num__220000 . the correct answer is e . <eor> e <eos> |
e |
e |
| if | x - num__20 | = num__40 what is the sum of all the values of x . <o> a ) num__0 <o> b ) num__60 <o> c ) - num__80 <o> d ) num__40 <o> e ) num__80 |
there will be two cases x - num__20 = num__40 and x - num__20 = - num__40 solve for x = > x = num__40 + num__20 = > x = num__60 or x = - num__40 + num__20 = > x = - num__20 the sum of both values will be num__60 + - num__20 = num__40 answer is d <eor> d <eos> |
d |
add__20.0__40.0__ subtract__60.0__20.0__ |
add__20.0__40.0__ subtract__60.0__20.0__ |
| if your mother asked you to bring her num__7 water bottles and one bottle costs num__1.5 $ how much money should she give you ? <o> a ) num__9 $ <o> b ) num__9.5 $ <o> c ) num__10 $ <o> d ) num__10.5 $ <o> e ) num__11 $ |
simply multiply the cost of one bottle by its number you would buy . the cost = num__7 * num__1.5 = num__10.5 $ so the correct choice is d <eor> d <eos> |
d |
multiply__7.0__1.5__ multiply__7.0__1.5__ |
multiply__7.0__1.5__ multiply__7.0__1.5__ |
| at what rate of interest ( compounded yearly ) will rs . num__10000 amount to rs . num__12100 in num__2 years ? <o> a ) num__11.0 <o> b ) num__8.0 <o> c ) num__10.0 <o> d ) num__9.0 <o> e ) num__7 % |
c . i = p ( num__1 + r / num__100 ) ^ n num__12100 = num__10000 ( num__1 + r / num__100 ) ^ num__2 num__1.21 = ( num__1 + r / num__100 ) ^ num__2 ( num__1.1 ) ^ num__2 = ( num__1 + r / num__100 ) r = num__10.0 answer : c <eor> c <eos> |
c |
percent__1.0__10000.0__ percent__10.0__100.0__ |
percent__1.0__10000.0__ percent__10.0__100.0__ |
| num__15 and num__5 are the first two terms in a geometric sequence . what is the arithmetic difference between the num__7 th term and the num__9 th term ? <o> a ) ( num__13.3333333333 ) * ( num__0.333333333333 ) ^ num__7 <o> b ) ( num__6.66666666667 ) * ( num__0.333333333333 ) ^ num__7 <o> c ) ( num__4.44444444444 ) * ( num__0.333333333333 ) ^ num__7 <o> d ) ( num__0.333333333333 ) ^ num__7 <o> e ) ( num__33.3333333333 ) * ( num__0.333333333333 ) ^ num__7 |
common ratio = num__0.333333333333 = num__0.333333333333 num__7 th term = num__15 * ( num__0.333333333333 ) ^ num__7 num__9 th term = num__15 * ( num__0.333333333333 ) ^ num__9 difference = num__15 * ( ( num__0.333333333333 ) ^ num__7 - ( num__0.333333333333 ) ^ num__9 ) = num__15 * ( num__0.333333333333 ) ^ num__7 * ( num__1 - ( num__0.333333333333 ) ^ num__2 ) = num__15 * ( num__0.333333333333 ) ^ num__7 * ( num__0.888888888889 ) = ( num__13.3333333333 ) * ( num__0.333333333333 ) ^ num__7 . . . . = ( num__13.3333333333 ) * ( num__0.333333333333 ) ^ num__7 a <eor> a <eos> |
a |
divide__5.0__15.0__ subtract__7.0__5.0__ multiply__1.0__13.3333__ |
divide__5.0__15.0__ subtract__7.0__5.0__ multiply__1.0__13.3333__ |
| the mean of num__1 ^ num__32 ^ num__33 ^ num__34 ^ num__35 ^ num__3 is ? <o> a ) num__30 <o> b ) num__45 <o> c ) num__50 <o> d ) num__60 <o> e ) num__75 |
num__1 ^ num__3 + num__2 ^ num__3 + num__3 ^ num__3 + - - - - - + n ^ num__3 = [ n ( n + num__1 ) / num__2 ] ^ num__2 num__1 ^ num__3 + num__2 ^ num__3 + num__3 ^ num__3 + - - - - - + num__5 ^ num__3 = ( num__5 * num__3.0 ) ^ num__2 = num__225 required average is = num__45.0 = num__45 answer is b <eor> b <eos> |
b |
subtract__34.0__32.0__ add__3.0__2.0__ divide__225.0__5.0__ multiply__1.0__45.0__ |
subtract__34.0__32.0__ add__3.0__2.0__ divide__225.0__5.0__ multiply__1.0__45.0__ |
| convert the num__0.428571428571 m / s into kilometers per hour ? <o> a ) num__3.5 kmph . <o> b ) num__2.5 kmph . <o> c ) num__1.2 kmph . <o> d ) num__1.54 kmph . <o> e ) num__1.9 kmph . |
num__0.428571428571 m / s = num__0.428571428571 * num__3.6 = num__1 ( num__0.54 ) = num__1.54 kmph . answer : d <eor> d <eos> |
d |
add__1.0__0.54__ round__1.54__ |
add__1.0__0.54__ divide__1.54__1.0__ |
| the ratio of salary of a worker in july to that in june was num__21 ⁄ num__2 : num__21 ⁄ num__4 by what % the salary of july more than salary of june . also find by what % salary of june was less than that of july . <o> a ) num__11 num__1 ⁄ num__9.0 and num__10.0 <o> b ) num__10.0 and num__11 num__1 ⁄ num__9.0 <o> c ) both num__10.0 <o> d ) both num__11 num__1 ⁄ num__9.0 <o> e ) none of these |
let the salary of july be num__5 ⁄ num__2 x and the salary of june be num__9 ⁄ num__4 x required percentages = num__2.5 x − num__2.25 x / num__2.25 x × num__100 and num__2.5 x − num__2.25 x / num__2.5 x × num__100 = num__100 ⁄ num__9.0 and num__100 ⁄ num__10.0 = num__11 num__1 ⁄ num__9.0 and num__10.0 answer a <eor> a <eos> |
a |
add__4.0__5.0__ divide__5.0__2.0__ divide__9.0__4.0__ multiply__2.0__5.0__ subtract__21.0__10.0__ subtract__5.0__4.0__ subtract__21.0__10.0__ |
add__4.0__5.0__ divide__5.0__2.0__ divide__9.0__4.0__ multiply__2.0__5.0__ subtract__21.0__10.0__ subtract__5.0__4.0__ divide__11.0__1.0__ |
| the probability that a arrow fired from a point will hit the target is num__0.25 . three such arrows are fired simultaneously towards the target from that very point . what is the probability that the target will be hit ? <o> a ) num__0.296875 <o> b ) num__0.359375 <o> c ) num__0.34328358209 <o> d ) num__0.578125 <o> e ) num__0.74 |
explanation : probability of a arrow not hitting the target = num__0.75 . probability that none of the num__3 arrow will hit the target : = > ( num__0.75 ) num__3 . = > num__0.421875 . probability that the target will hit at least once = > num__1 − ( num__0.421875 ) . = > num__0.578125 . answer : d <eor> d <eos> |
d |
negate_prob__0.25__ negate_prob__0.4219__ negate_prob__0.4219__ |
negate_prob__0.25__ negate_prob__0.4219__ negate_prob__0.4219__ |
| a and b together can do a piece of work in num__3 days . if a alone can do the same work in num__20 days then b alone can do the same work in ? <o> a ) num__0.28 days <o> b ) num__0.45 days <o> c ) num__0.55 days <o> d ) num__0.25 days <o> e ) num__0.15 days |
b = num__0.5 – num__0.05 = num__0.28 days answer : a <eor> a <eos> |
a |
kmh_to_ms__ kmh_to_ms__ |
kmh_to_ms__ kmh_to_ms__ |
| by travelling at num__60 kmph a person reaches his destination on time . he covered two - third the total distance in one - third of the total time . what speed should he maintain for the remaining distance to reach his destination on time ? <o> a ) num__30 kmph <o> b ) num__28 kmph <o> c ) num__26 kmph <o> d ) num__24 kmph <o> e ) num__22 kmph |
let the time taken to reach the destination be num__3 x hours . total distance = num__60 * num__3 x = num__180 x km he covered num__0.666666666667 * num__180 x = num__120 x km in num__0.333333333333 * num__3 x = x hours so the remaining num__60 x km he has to cover in num__2 x hours . required speed = num__60 x / num__2 x = num__30 kmph . answer : a <eor> a <eos> |
a |
multiply__60.0__3.0__ subtract__180.0__60.0__ divide__60.0__180.0__ divide__120.0__60.0__ divide__60.0__2.0__ round__30.0__ |
multiply__60.0__3.0__ subtract__180.0__60.0__ divide__60.0__180.0__ divide__120.0__60.0__ divide__60.0__2.0__ divide__60.0__2.0__ |
| find the invalid no . from the following series num__10 num__17 num__22 num__29 num__34 num__36 <o> a ) num__36 <o> b ) num__18 <o> c ) num__40 <o> d ) num__37 <o> e ) num__25 |
the differences between two successive terms from the beginning are num__7 num__5 num__7 num__5 num__7 num__5 . so num__36 is wrong . answer : a <eor> a <eos> |
a |
subtract__17.0__10.0__ subtract__22.0__17.0__ add__29.0__7.0__ |
subtract__17.0__10.0__ subtract__22.0__17.0__ add__29.0__7.0__ |
| present ages of x and y are in the ratio num__4 : num__5 respectively . num__6 years hence this ratio will become num__5 : num__6 respectively . what is x ' s present age in years ? <o> a ) num__35 <o> b ) num__24 <o> c ) num__37 <o> d ) num__39 <o> e ) num__40 |
let the present ages of x and y be num__4 x and num__5 x years respectively . then ( num__4 x + num__6 ) / ( num__5 x + num__6 ) = num__0.833333333333 num__6 ( num__4 x + num__6 ) = num__5 ( num__5 x + num__6 ) = > num__24 x + num__36 = num__25 x + num__30 = > x = num__6 x ' s present age = num__4 x = num__4 * num__6 = num__24 answer : b <eor> b <eos> |
b |
divide__5.0__6.0__ multiply__4.0__6.0__ multiply__5.0__6.0__ multiply__4.0__6.0__ |
divide__5.0__6.0__ multiply__4.0__6.0__ multiply__5.0__6.0__ multiply__4.0__6.0__ |
| in a num__2000 m race a beats b by num__100 meters or num__50 seconds . find the speed of b ? <o> a ) num__10 m / s <o> b ) num__2 m / s <o> c ) num__5 m / s <o> d ) num__7 m / s <o> e ) num__3 m / s |
since a beats b by num__100 m or num__50 seconds it implies that b covers num__100 m in num__50 seconds . hence speed of b = num__2.0 = num__2 m / s . answer : b <eor> b <eos> |
b |
divide__100.0__50.0__ round__2.0__ |
divide__100.0__50.0__ divide__100.0__50.0__ |
| two isosceles triangles have equal vertical angles and their areas are in the ratio num__16 : num__36 . find the ratio of their corresponding heights . <o> a ) num__0.666666666667 <o> b ) num__1.25 <o> c ) num__1.5 <o> d ) num__0.714285714286 <o> e ) num__0.666666666667 |
we are basically given that the triangles are similar . in two similar triangles the ratio of their areas is the square of the ratio of their sides and also the square of the ratio of their corresponding heights . therefore area / area = height ^ num__2 / height ^ num__2 = num__0.444444444444 - - > height / height = num__0.666666666667 . answer : a . <eor> a <eos> |
a |
triangle_area__2.0__0.6667__ |
triangle_area__2.0__0.6667__ |
| num__65.0 of x = num__20.0 of num__422.50 . find the value of x ? <o> a ) num__100 <o> b ) num__130 <o> c ) num__150 <o> d ) num__180 <o> e ) num__199 |
num__65.0 of x = num__20.0 of num__422.50 then num__0.65 * x = num__0.2 * num__422.5 x = num__84.5 * num__1.53846153846 = num__130 answer is b <eor> b <eos> |
b |
multiply__422.5__0.2__ reverse__0.65__ divide__84.5__0.65__ divide__84.5__0.65__ |
multiply__422.5__0.2__ reverse__0.65__ divide__84.5__0.65__ divide__84.5__0.65__ |
| the area of a square garden is q square feet and the perimeter is p feet . if q = num__2 p + num__48 what is the perimeter of the garden in feet ? <o> a ) num__36 <o> b ) num__40 <o> c ) num__44 <o> d ) num__48 <o> e ) num__52 |
let x be the length of one side of the square garden . x ^ num__2 = num__8 x + num__48 x ^ num__2 - num__8 x - num__48 = num__0 ( x - num__12 ) ( x + num__4 ) = num__0 x = num__12 - num__4 p = num__4 ( num__12 ) = num__48 the answer is d . <eor> d <eos> |
d |
square_perimeter__2.0__ rectangle_perimeter__2.0__0.0__ square_perimeter__12.0__ |
square_perimeter__2.0__ rectangle_perimeter__2.0__0.0__ square_perimeter__12.0__ |
| the perimeter of a triangle is num__24 cm and the inradius of the triangle is num__2.5 cm . what is the area of the triangle ? <o> a ) num__30 cm num__2 <o> b ) num__85 cm num__2 <o> c ) num__65 cm num__2 <o> d ) num__45 cm num__2 <o> e ) num__35 cm num__2 |
area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = num__2.5 * num__12.0 = num__30 cm num__2 answer : a <eor> a <eos> |
a |
triangle_area__24.0__2.5__ triangle_area__24.0__2.5__ |
multiply__2.5__12.0__ multiply__2.5__12.0__ |
| what is the rate percent when the simple interest on rs . num__9100 amount to rs . num__455 in num__3 years ? <o> a ) num__3.7 <o> b ) num__5.7 <o> c ) num__6.7 <o> d ) num__8.7 <o> e ) num__9.7 % |
num__455 = ( num__4100 * num__3 * r ) / num__100 r = num__3.7 answer : a <eor> a <eos> |
a |
percent__100.0__3.7__ |
percent__100.0__3.7__ |
| a thief goes away with a santro car at a speed of num__40 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at num__50 kmph when will the owner over take the thief from the start ? <o> a ) num__2 hours <o> b ) num__9 hours <o> c ) num__6 hours <o> d ) num__3 hours <o> e ) num__12 hours |
explanation : | - - - - - - - - - - - num__20 - - - - - - - - - - - - - - - - - - - - | num__50 num__40 d = num__20 rs = num__50 – num__40 = num__10 t = num__2.0 = num__2 hours answer : option a <eor> a <eos> |
a |
subtract__50.0__40.0__ divide__40.0__20.0__ round__2.0__ |
subtract__50.0__40.0__ divide__40.0__20.0__ round__2.0__ |
| if the radius of a circle is increased by num__20.0 then the area is increased by : <o> a ) num__44 <o> b ) num__219 <o> c ) num__16 <o> d ) num__18 <o> e ) num__11 |
explanation : let say radius = π r num__2 π r num__2 = π num__102 π num__102 = num__100 π num__100 π new radius = num__10 × num__120.0 = num__12 new area = π num__122 π num__122 = num__144 π num__144 π increment = num__144 π − num__100 π num__100 π × num__100144 π − num__100 π num__100 π × num__100 = num__44.0 answer : a <eor> a <eos> |
a |
triangle_perimeter__20.0__2.0__100.0__ power__12.0__2.0__ rectangle_perimeter__20.0__2.0__ rectangle_perimeter__20.0__2.0__ |
triangle_perimeter__20.0__2.0__100.0__ power__12.0__2.0__ rectangle_perimeter__20.0__2.0__ rectangle_perimeter__20.0__2.0__ |
| cathy and jim begin running at the same time and they start at the same place on a straight path . cathy runs at num__10 miles per hour and jim runs at num__6 miles per hour . after num__18 minutes cathy stops to stretch . if it takes cathy num__27 minutes to stretch and jim continues to run during this time how many minutes will it take cathy to catch up to jim ? <o> a ) num__20.5 <o> b ) num__21.5 <o> c ) num__22.5 <o> d ) num__23.5 <o> e ) num__24.5 |
in num__18 minutes cathy runs num__3 miles . in num__45 minutes jim runs num__4.5 miles . cathy can catch jim at a rate of num__4 miles per hour . since jim is ahead by num__1.5 miles it will take cathy num__1.5 / num__4 hours = num__22.5 minutes to catch jim . the answer is c . <eor> c <eos> |
c |
divide__18.0__6.0__ add__18.0__27.0__ divide__27.0__6.0__ subtract__10.0__6.0__ subtract__6.0__4.5__ add__18.0__4.5__ round__22.5__ |
divide__18.0__6.0__ add__18.0__27.0__ divide__27.0__6.0__ divide__18.0__4.5__ divide__6.0__4.0__ add__18.0__4.5__ round__22.5__ |
| a windmill is taking advantage of strong air currents in order to produce electrical energy . on a typical day the wind speed is around num__20 mph and in that speed the windmill produces num__600 kw / h ( kilowatts per hour ) . on a stormy day a windmill produces num__20.0 more energy . how much kw / h can three windmills produce in two hours on a stormy day ? <o> a ) num__2880 . <o> b ) num__4320 . <o> c ) num__5780 . <o> d ) num__5760 . <o> e ) num__6380 . |
efficiency per machine = num__600 watt / hr efficiency due to increase in wind speed = num__600 * num__120.0 = > num__720 watt / hr production by each machine in num__2 hours is num__720 * num__2 = > num__1440 watt production by num__3 machines = > num__1440 watt * num__3 = num__4320 watt answer will be ( b ) <eor> b <eos> |
b |
add__600.0__120.0__ multiply__2.0__720.0__ multiply__1440.0__3.0__ multiply__1440.0__3.0__ |
add__600.0__120.0__ multiply__2.0__720.0__ multiply__1440.0__3.0__ multiply__1440.0__3.0__ |
| a train num__800 metres long is running at a speed of num__78 km / hr . if it crosses a tunnel in num__1 minute then the length of the tunnel ( in meters ) is : <o> a ) num__130 <o> b ) num__360 <o> c ) num__500 <o> d ) num__540 <o> e ) num__580 |
l = num__800 m s = num__78 kmph = num__21.6666666667 mps time to pass tunnel = num__1 min = num__60 sec t . f ( num__800 + x ) / ( num__60 ) = ( num__21.6666666667 ) = num__500 answer : c <eor> c <eos> |
c |
hour_to_min_conversion__ round__500.0__ |
hour_to_min_conversion__ divide__500.0__1.0__ |
| find the sum of all num__2 digit numbers divisible by num__3 . <o> a ) num__99 <o> b ) num__199 <o> c ) num__1665 <o> d ) num__2452 <o> e ) num__2547 |
all num__2 digit numbers divisible by num__3 are : num__12 num__51 num__18 num__21 . . . num__99 . this is an a . p . with a = num__12 and d = num__3 . let it contain n terms . then num__12 + ( n - num__1 ) x num__3 = num__99 or n = num__30 . required sum = num__30 x ( num__12 + num__99 ) = num__1665 . answer c num__1665 <eor> c <eos> |
c |
add__3.0__18.0__ subtract__3.0__2.0__ add__12.0__18.0__ multiply__1.0__1665.0__ |
add__3.0__18.0__ subtract__3.0__2.0__ add__12.0__18.0__ multiply__1.0__1665.0__ |
| the speed of a boat in still water is num__60 kmph and the speed of the current is num__40 kmph . find the speed downstream and upstream ? <o> a ) num__87 kmph <o> b ) num__40 kmph <o> c ) num__16 kmph <o> d ) num__20 kmph <o> e ) num__18 kmph kmph |
speed downstream = num__60 + num__40 = num__100 kmph speed upstream = num__60 - num__40 = num__20 kmph answer : d <eor> d <eos> |
d |
add__60.0__40.0__ subtract__60.0__40.0__ round__20.0__ |
add__60.0__40.0__ subtract__60.0__40.0__ subtract__60.0__40.0__ |
| if pipe a can fill the tank in num__45 minutes and pipe b in num__30 minutes find the time to fill the tank if both the pipes are opened together . <o> a ) num__12 minutes <o> b ) num__20 minutes <o> c ) num__18 minutes <o> d ) num__15 minutes <o> e ) none of these |
explanation : in num__1 minute pipe a can fill num__0.0222222222222 th part of the tank and pipe b can fill num__0.0333333333333 th part of the tank . if they are opened simultaneously then in num__1 minute they can fill ( num__0.0222222222222 + num__0.0333333333333 ) part of the tank = num__0.0555555555556 th part of the tank . hence in num__18 minutes the tank gets filled if pipes a & b are opened together . answer c <eor> c <eos> |
c |
divide__1.0__45.0__ divide__1.0__30.0__ round__18.0__ |
divide__1.0__45.0__ divide__1.0__30.0__ round__18.0__ |